Class-X

MATHEMATICS
REAL NUMBERS

PHYSICS
GEOMETRICAL OPTICS

CHEMISTRY
CARBON AND ITS COMPOUNDS

BIOLOGY
NUTRITION
 COURSE FOR NTSE/OLYMPIAD/IIT JEE - CLASS X - SYLLABUS
BOOK SUBJECT CHAPTER NAME
Science - 1 Physics Electricity
Chemistry Chemical Equation
Acid, Base & Salt
Biology Life Process: Nutrition
Life Process: Respiration
Science - 2 Physics Magnetic Effect of Electric Current
Chemistry Metals & Non-Metals
Biology Circulatory System & Transportation in Plant
Science - 3 Physics Source of Energy
Chemistry Periodic Properties
Biology Excretion
Control and Coordination
Reproduction
Science - 4 Physics Motion
Optical Instruments
Chemistry Carbon Compounds
Chemical Equilibrium and Chemical Kinetics
Solubility and Electrolysis
Biology Heredity
Power Sharing
Mathematics - 1 Mathematics Pair of Linear Equations in Two Variables
Polynomials
Real Numbers
Mathematics - 2 Mathematics Arithmetic Progression
Quadratic Equations
Triangles
Mathematics - 3 Mathematics Circle
Co-ordinate Geometry
Constructions
Introduction to Trigonometry
Mathematics - 4 Mathematics Area Related to Circles
Surface Area and Volume
Statistics
Probability
Mental Ability Mental Ability Alphabet Test
Coding Decoding
Direction Sense
Logical Diagrams
Mathematical Aptitude
Non Verbal Reasoning
Synonyms
Social Science Social Science India and the Contemporary World - II
Contemporary India - II
Democratic Politics – II
Understanding Economic Development
Contents
Chapter 1
Mathematics
Real Numbers 1-18
Exercise-I 19-20
Exercise-II 21-27
Exercise-III 28-31
Exercise-IV 32-45

Chapter 2
Physics
Geometrical Optics 46-85
Exercise-I 86-89
Exercise-II 90-97
Exercise-III 98-103
Exercise-IV 104-115

Chapter 3
Chemistry
Carbon and Its Compounds 116-147
Exercise-I 148-149
Exercise-II 150-153
Exercise-III 154-159
Exercise-IV 160-163

Chapter 4
Biology
Nutrition 164-173
Exercise-I 174-174
Exercise-II 175-178
Exercise-III 179-181
Exercise-IV 182-187

Answer Key 188-192

•
 Real Number

Real Number 1
l INTRODUCTION 1729 = 93 + 103,
Þ 1729 = 13 + 123 = 93 +103

e l RECALL
2/15 In our day to life, we deal with different types of
2/11
W numbers which can be broadly classified as follows.
1/3 Z
N
1,2,3,4,... Numbers
–5/6 Q
99,100,...
R Real Numbers Imaginary Numbers
0 53
5, Rational Numbers Irrational Numbers
3, +
2, 5, Integers
3, Fractions

“God gave us the natural number, all else is the work Negatives Whole Numbers
of man”. It was exclaimed by L. Kronecker (1823-1891).
the reputed German Mathematician. This statement Natural Numbers
Zero
reveals in a nut shell the significant role of the universe
of numbers played in the evolution of human though. CLASSIFICATION OF NUMBERS
N : The set of natural number, (a) Natural numbers (N) : N = {1, 2, 3, 4… ¥ }
W : The set of whole numbers, Remark :
Z : The ser of Integers , (i) The set N is infinite i.e. it has unlimited members.
Q : The set of rationales, (ii) N has the smallest element namely ‘1’.
R : The set of Real Numbers. (iii) N has no largest element. i.e., give me any natural
number, we can find the bigger number from the
l HISTORICAL FACTS given number.
Dedekind was the first modern mathematician to (iv) N does not contain ‘0’ as a member. i.e., ‘0’ is not a
publish in 1872 the mathematically rigorous definition member of the set N.
of irrational numbers. He gave explanation of their place
in the real Numbers System. He was able to demonstrate (b) Whole numbers (W) W = {0, 1, 2, 4… ¥ }
the completeness of the real number line. He filled in Remark :
the “holes’ in the system of Rational numbers with (i) The set of whole number is infinite (unlimited
irrational Numbers. This innovation the made Richard elements)
Dedekind an immortal figure in the history of (ii) This set has the smallest members as ‘0’. i.e. ‘0’ the
Mathematics. smallest whole number. i.e., set W contain ‘0’ as a
member.
Srinivasa Ramanujan (1887-1920) was one of the (iii) The set of whole numbers has no largest member.
most outstanding mathematician that India produced. (iv) Emery natural number is a whole number.
He worked on history of Numbers and discovered (v) Non-zero smallest whole number is ‘1’.
wonderful properties of numbers. He stated intuitively
many complicated result in mathematics. Once a great (c) Integers ( I or Z) :
mathematician Prof. Hardy come of India to see I or Z = { - ¥ …-3, -2, -1, 0, +2, +3…+ ¥ }
Ramanujan. Prof. Hardy remarked that the he has Positive integers : {1, 2, 3…}, Negative integers :
traveled in a taxi with a rather dull number viz. 1729. {…. -4, -3, -2, -1}
Ramanujan jumped up and said, Oh! No. 1729 is very
interesting number. It is the smallest number which Remark :
can be expressed as the sum of two cubes in two (i) This set Z is infinite .
different ways. (ii) It has neither the greatest nor the lest element.
viz 1729 = 13 +123, (iii) Every natural number is an integer.

3-1
Real Number

(iv) Every whole number is an integer. l NATURE OF THE DECIMAL EXPANSION OF

(v) The set of non-negative integer = {0, 1, 2, 3, 4,….} RATIONALNUMBERS
(vi) The set of non-positive integer
= {..-4, -3, -2, -1,0} Theorem -1 : Let x be a rational number whose decimal
expansion terminates. Then we can express x in the
(d) Rational numbers :- These are real numbers which
p
p form , where p and q are co-primes, and the prime
can be expressed in the form of , where p and q a
q factorisation of q is of the form 2m × 5n, where m, n are
are integers and q ¹ 0. non-negative integers.

2 37 - 17 p
Ex. , , ,-3, 0, 10, 4.33, 7.123123123.......... Theorem-2 : Let x = be a rational number, such
3 15 19 q
that the prime factorisation of q is the 2m × 5n, where
Remark : m, n are non-negative integers. Then , x has a decimal
(i) Every integer is a rational number. expansion which terminates.
(ii) Every terminating decimal is a rational number .
(iii) Every recurring decimal is a rational number. p
(iv) A non-terminating repeating decimal is called a Theorem-3 : Let x = be a rational number, such
q
recurring decimal.
(v) Between any two rational numbers there are an that the prime factorisation of q is not of the form 2m ×
infinite number of rational numbers. This property 5m, where m, n are non-negative integers. Then , x has
is known as the density rational numbers. a decimal expansion which is non-terminating
1 repeating
(vi) If a and b are two rational numbers then ( a + b)
2 189 189 189
E.x (i) = 3 = 0 3
1 125 5 2 ´5
lies between a and b. a < (a + b) < b n rational
2 we observe that prime factorization of the
number between two different rational numbers a denominators of these rational numbers are of the form
and b are : 2m × 5n, where m, n are non-negative integers. Hence,
(b - a ) 2(b - a) 3(b - a ) 4(b - a) n(b - a )
a+ ;a + ;a + ;a + ;............a + ; 189
n +1 n +1 n +1 n +1 n +1 has terminating decimal expansion.
125
(vii) Every rational number can be represented either 17 17
as a terminating decimal or a non-termination (ii) =
repeating (recurring) decimals.
6 2´3
(viii) Types of rational numbers :- we observe that the prime factorisation of the
denominator of these rational numbers are not of the
· Terminating decimal numbers and form 2m×5n, where m, n are non-negative integers.
· Non-termination repeating (recurring) decimal 17
numbers Hence has non-terminating and repeating decimal
6
(e) Irrational numbers :- A number is called irrational expansion

p 17 17
(iii) =
number , if it can not be written in the form
q
, 8 2 3 ´ 50

where p & q are integers and q ¹ 0. All Non- 17

So, the denominator 8 of is of the form 2m×5n, where
terminating & Non-repeating decimal numbers are 8
Irrational numbers. 17
m, n are non-negative integers. Hence has
Ex. 2, 3, 3 2 , 2 + 3 , 2 + 3 , p , e, etc 8
terminating decimal expansion.
(f) Real numbers :- The totality of rational numbers 64 64
(iv) =
and irrational numbers is called the set of real 455 5 ´ 7 ´ 13
numbers i.e. rational numbers and irrational Clearly, 455 is not of the form 2m × 5n, So, the decimal
numbers taken together are called real numbers .
Every real number is either a rational number or an 64
expansion of is non-terminating repeating.
irrational number. 455

3-2
 Real Number

l PROOF OF IRRATIONALITY OF 2, 3, 5,.......... Hence, 2 is irrational.

Example-2
SOLVED EXAMPLE
Prove that 3
3 is irrational .
Example-1
p
Prove that 2 is not a rational number or there is no
Sol. Let 3
3 be rational = q , where p and q Î Z and p, q
rational whose square is 2.[CBSE(outside Delhi) 2008] have no common factor except 1 also q > 1.
1.414215 p
\ =3 3
1 2.000000000000 q
+1 1 Cubing both sides
24 100
p3
4 96 =3
281 400 q3
+1 281 Multiply both sides by q2
2824 11900
p3
+4 11296 = 3q 2 , Clearly L.H.S is rational since p, q have no
28282 q
60400
+2 56564 common factor.
282841 383600 \ p3, q also have no common factor while R.H.S. is
Sol. +1 282841 an integer .
2828423 10075900 \ L.H.S. ¹ R.H.S. which contradicts our assumption
3 8485269 that 3
3 is Irrational .
28284265 159063100
+5 141421325 Example-3
28284270 17641775
Prove that 2 + 3 is irrational .
2 = 1.414215
Let us find the square root of 2 by long division method [Sample paper (CBSE) 2008]
as shown below.
Sol. Let 2 + 3 be a rational number equals to r
Clearly, the decimal representation of 2 is neither \ 2+ 3 = r
terminating nor repeating.
We shall prove this by the method of contradiction. 3 = r-2
If possible, let us assume that Here L.H.S. is an irrational number while R.H.S. r – 2 is
2 is a rational number.. rational. \ S.H.S. ¹ R.H.S
a
Then 2= where a, b are integers having no Hence it contradicts our assumption that 2 + 3 is
b
rational .
common factor other than 1.
\ 2 + 3 is irrational.
( 2)
2
2 æaö
Þ = ç ÷ (squaring both sides)
èbø
Example-4
a2
2= 2
b Prove that 2 + 3 is irrational.
a = 2b
2 2
Sol. Let 2 + 3 be rational number say ‘x’
Þ 2 divides a2
Þ 2 divides a Þx= 2+ 3
Therefore let a = 2c for some integer c. x 2 = 2 + 3 + 2 3. 2 = 5 + 2 6
Þ a2 = 4c2.
Þ 2b2 = 4c2 x2 - 5
Þ x2 = 5 + 2 6 Þ 6 =
Þ b2 = 2c2 2
Þ 2 divides b2 x2 - 5
Þ 2 divides b As x, 5 and 2 are rationales Þ is a rational number
2
Thus, 2 is a common factor of a and b.
.
But, it contradicts our assumption that a and b have
no common factor other than 1. x2 - 5
Þ 6= is a rational number
So, our assumption that 2 is a rational, is wrong. 2

3-3
Real Number

Which is contradiction of the fact that 6 is a irrational a

Which means n - 1 + n + 1 = … (i)
number. b
Hence our supposition is wrong Þ 2 + 3 is an 1
=
b
or a
irrational number . n -1 + n +1
Rationalizing LHS, we get
Example-5 1 n -1 - n +1 b
Þ ´ =
Prove that 2 is not a rational number.. n -1 + n +1 n -1 - n +1 a
Sol. Let 2 is a rational number
n -1 - n +1 b
p Þ =
2= n -1- n -1 a
\ q
n +1 - n -1 b
[p and q are co-prime and q¹ 0] Þ =
2 a
Squaring both sides
2b
p2 or n +1 - n -1 = …(ii)
2 = 2 Þ 2q 2 = p 2 a
q Adding and subtracting (i) and (ii) we get
Þ p is even or p is even.
2
2 n +1 =
a 2b
+
a 2b
and 2 n - 1 = -
Let p = 2r b a b a

Þ 2q 2 = (2r ) 2 = 4r 2 a 2 + 2b 2 a 2 - 2b 2
Þ n +1 = and n - 1 =
2ab 2ab
Þ q 2 = 2r 2
Þ RHS of both are rational.
Þ q 2 is even so q is even. \ n + 1 and n - 1 are also rational.
Hence, p and q have 2 as a common factor or p and q
Þ (n + 1) and (n - 1) are perfect squares of positive
are not co-prime.
So, our supposition is wrong. integers.
This is impossible as any two perfect squares differ at
\ 2 is not a rational number..
least by 3.
Hence, there is no positive integer n for which
Example-6
( n - 1 + n + 1 ) is rational.
Prove that 3+ 5 is an irrational number..
p
Sol. Suppose 3 + 5 is a rational number and can be taken Theorem:Let x = q be a rational number such that

as
a
, b¹ 0 and a, b are co-prime. q ¹ 0 and prime factorization of q is of the form
b
2n ´ 5m where m, n are non-negative integers then x
a
Þ 3+ 5= has a decimal representation which terminates.
b
For example:
[Rational]
Squaring both sides 275 52 ´ 11 11 11
0.275 = = = =
103 23 ´ 53 23 ´ 5 40
a2
3+5+ 2 3 5 = p
b2 Theorem:Let x = q be a rational number such that
a2 a 2 - 8b 2
Þ 8 + 2 15 =
b2
Þ 15 =
2b2
q ¹ 0 and prime factorization of q is not of the form

LHS is 15 which is irrational while RHS is rational. 2m ´ 5n , where m, n are non-negative integers, then x
has a decimal expansion which is non-terminating
So, our supposition is wrong. repeating.
Hence, 3 + 5 is not a rational number.. 5
For example : = 1.66666...
3
Example-7
Decimal
Show that there is no positive integer n for which Form of prime
Rational expansion of
factorisation of the
number rational
n-1+ n+1 is rational and can be expressed in the denominator
number
q = 2m5n where n
a Note: p terminating
form , b¹ 0 and a&b are co-prime. x = , where p and m are non-
b q negative integers
Sol. Let there be a positive integer n for which and q are co- q ¹ 2m5n where n
prime and q¹ 0 and m are non- non-terminating
n - 1 + n + 1 is rational. negative integers

3-4
 Real Number

SOME IDENTICAL SURDS

Example-8
(i) 3 4 is a surd as radicand is a rational number..
Without actually calculating, state whether the
3
following rational numbers have a terminating or non- Similar examples 5 , 4 12 , 5 7 , 12 ,.........
terminating repeating decimal expansion.
(ii) 2 3 is a surd (as surd + rational number will give
27 19
(i) (ii) a surd)
343 1600
Similar examples 3 + 1, 3 3 + 1,....
129
(iii)
2 ×55 ×32
2
(iii) 7 - 4 3 is a surd as 7 – 4 3 is a perfect square of

27 27 (2 - 3 )
Sol. (i) = Similar
343 73
examples 7 + 4 3 , 9 - 4 5 , 9 + 4 5 ,.........
Since q = 7 which is not of the form 2 ´ 5 .
3 m n 1
æ 1 ö3 1

\ It has non terminating decimal representation. (vi) 3

3 is a surd as
3
3 = ç32 ÷ = 36 = 6 3
ç ÷
è ø
19 19
(ii) = 6 2 Similar examples 3 3
5 , 4 5 6 ,....... ...
1600 2 ´ 5
Since q = 26 ´ 52 which is of the form 2m ´ 5n . SOME EXPRESSION ARE NOT SURDS
\ It has a terminating decimal representation. (i) 3 8 because 3 8 3
= 23 = 2, which is a rational
129 number.
(iii) (ii) because 2 + 3 is not a perfect square.
2 ´ 55 ´ 32
2 2+ 3

(iii) 3
1+ 3 because radicand is an irrational number..
Since q = 22 ´ 55 ´ 32 is not of the form 2m ´ 5n . It
has a non-terminating decimal representation. LAWS OF SURDS

Example-9
(i) ( a)n n
= n an = a

What can you say about the prime factorization of the e.g. (A) 3
8 = 23 = 2
3

denominators of the following rationales:

(B) 4
81 = 4
34 = 3
(i) 36.12345
(ii) 36.5678 (ii) a ´ n b = n ab
n

Sol. (i)Since 36.12345 has terminating decimal expansion. [Here order should be same]
So, its denominator is of the form 2 m ´ 5 n where m, n e.g. (A) 3 2 ´ 3 6 = 3 2 ´ 6 = 3 12
are non-negative integers. but, 3 3 ´ 4 6 ¹ 3 ´ 6 [Because order is not same]
(ii) Since 36.5678 has non terminating repeating decimal 1st make their order same and then you can multiply.
expansion. So, its denominator has factors other than a
(iii) n a ¸ n b = n
2 or 5. b

(iv) nm
a = nm
a= mn
a
SURDS
Any irrational number of the form n a is given a e.g. = 2 =88

special name surd. Where ‘a’ is called radicand, it n´p

(v) a =n
ap
should always be a rational number. Also the symbol
[Important for changing order of surds]
n is called the radical sign and the index n is called n ´p
or, n
am = a m ´p
order of the surd.
e.g. 3
6 2 make its order 6,
n
a is read as ‘nthroot a’ and can also be written as
then 3
62 =
3 ´2
6 2´ 2 = 6 4
6 .
1
a . n
e.g. 3
6 make its order 15, then 3 6 = 3´ 5 6 1´ 5 = 15 6 5 .

3-5
Real Number

OPERATION OF SURDS
ADDITION AND SUBTRACTION OF SURDS Example-13
Addition and subtraction of surds are possible only
6 ( 24 )3
when order and radicand are same i.e. only for surds. Divide 24 216
24 ¸ 3 200 = 3
= =6
200 6 ( 200)2 625
SOLVED EXAMPLE
COMPARISON OF SURDS
Example-10 It is clear that if x > y > 0 and n > 1 is a positive integer
Simplify then n
x >ny.
(i) 6 - 216 + 96 = 15 6 - 6 2 ´ 6 + 16 ´ 6

[Bring surd in simples form] SOLVED EXAMPLE

= 15 6 - 6 6 + 4 6
Example-14
= (15 - 6 + 4 ) 6 3 3
16 > 12 , 5 35 > 5
25 and so on.
= 13 6 Ans.
(ii) 5 3 250 + 7 3 16 - 143 54 Example-15
Which is greater in each of the following :
= 53 125 ´ 2 + 7 3 8 ´ 2 - 14 3 27 ´ 2

= 5 ´ 53 2 + 7 ´ 2 3 2 - 14 ´ 3 ´ 3 2 1 3
1
3
(i) 16 and 5 8 (ii) and
3
= ( 25 + 14 - 42 )3 2
2

= -3 3 2 Ans. Sol. L.C.M. of 3 and 5 15.

L.C.M. of 2 and 3 is 6.
(iii) 5 3 250 + 7 3 16 - 143 54
3´ 5
= 53 125 ´ 2 + 7 3 8 ´ 2 - 14 3 27 ´ 2
3
6 = 65 = 15
7776

= 5 ´ 53 2 + 7 ´ 2 3 2 - 14 ´ 3 ´ 3 2
3 2
æ1ö æ1ö
= ( 25 + 14 - 42 )3 2 6 ç ÷ and 3 ç ÷
è2ø è3ø
= -3 3 2 Ans.
3´ 5
5 1 5 1´ 3
5
8= 85 = 15 512
(iv) 4 3 + 3 48 - = 4 3 + 3 16 ´ 3 -
2 3 2 3´3

1 1 é 1 1ù
3 + 3´ 4 3 -
5 1
´
6 and 6
êAs 8 < 9 \ > ú
=4 2 3
3
8 9 ë 8 9û

5 \ 75 15
= 4 3 + 12 3 - 3 7776 > 512
6

æ 5ö 1 6 1 1 3 1
= ç 4 + 12 - ÷ 3 so, 6 > Þ 3
6 >58 Þ >
è 6ø 8 9 2 3

91
= 3 Ans.
6 Example-16
MULTIPLICATION AND DIVISION OF SURDS Arrange 2 , 3 3 and 4
5 is ascending order..
Sol. L.C.M. of 2, 3, 4 is 12.
Example-11
2´ 6
\ 2 = 2 6 = 12 64
(i) 3 3
4 ´ 22 = 3 3 3
4 ´ 22 = 2 ´ 11 = 2 11 3
3´ 4
3
3= 3 4 = 12 81
(ii) 3 2 ´ 4 3 = 12 2 4 ´ 12 3 3 = 12 2 4 ´ 3 3 = 12 16 ´ 27 = 12 432
4´ 3
5= 4
5 3 = 12 125
Example-12
As, 64 < 81 < 125.
Simplify 8a 5 b ´ 3 4a 2 b 2 \ 12
64 < 12 81 < 12
125

Hint: 6 8 3 a 15b 3 ´ 6 4 2 a 4 b 4 = 6 2 13 a 19 b7 = 6 2 ab . Þ 2< 3

3< 4
5

3-6
 Real Number

Example-17 SOLVED EXAMPLE

Which is greater 7 - 3 or 5 - 1?
Example-18
( 7 - 3 )( 7 + 3 ) 7-3 4 Find the R.F. (rationalizing factor) of the following :
Sol. 7- 3= = =
( 7 + 3) 7+ 3 7+ 3 (i) 10 (ii) 12

( 5 - 1)( 5 + 1) 5-1 4 (iii) 162 (iv) 3

4
And, 5 -1= = =
( 5 + 1) 5 +1 5 +1 (v) 3
16 (vi) 4
162
Now, we know that 7 > 5 and 3 > 1 , add (vii) 2 + 3 (viii) 7 - 4 3
So, 7+ 3 > 5 +1 (ix) 3 3 + 2 2 (x) 3 3 + 3 2
1 1 (xi) 1 + 2 + 3
Þ <
7+ 3 5 +1 (i) 10
4 4
Þ < Sol. [\ 10 ´ 10 = 10 ´ 10 = 10]
7+ 3 5 +1
as 10 is rational number.
Þ 7 - 3 < 5 -1 \ R.F. of 10 is 10 Ans.
So, 5 -1> 7 - 3 (ii) 12
Sol. First write its simplest form i.e. 2 3 .
RATIONALIZATION OF SURDS
Rationalizing factor product of two surds is a rational Now find R.F. (i.e. R.F. of 3 is 3)
number then each of them is called the rationalizing \ R.F. of 12 is 3 Ans.
factor (R.F.) of the other. The process of converting a
(iii) 162
surd to a rational number by using an appropriate
multiplier is known as rationalization. Sol. Simplest form of 162 is 9 2 .
R.F. of 2 is 2.
SOME EXAMPLES
\ R.F. of 162 is 2 Ans.
(i) R.F. of a is (
a \ a´ a =a . ) (iv) 3
4

a is a2 æç\ 3 a ´ a 2 = a3 = a ö÷ . Sol.
3 3 3 3 3
(ii) R.F. of
3 3
4 ´ 42 = 43 = 4
è ø
\ R.F. of 3
4 is 3
42 Ans.
(iii) R.F. of a + b is a - b & vice versa
(v). 3 16
[\ ( a + b )( a - b =a-b . ) ] Sol. Simplest form of 3
16 is 23 2

(iv) R.F. of a + b is a - b & vice versa Now R.F. of 3

2 is 3
22

[\ (a + b )(a - b ) = a 2
-b ] \ R.F. of 3
16 is 3
22 Ans.

3 (vi) 4 162
(v) R.F. of a + 3 b is
Sol. Simplest form of 4
162 is 34 2
æ3
è
2
øë
é
ç a - 3 ab + b ö÷ ê\
3 2
(
3
è
)
a + 3 b ç a - 3 ab + b ö÷ùú
æ3 2
øû
3 2
Now R.F. of 4
2 is 4 2 3

R.F. of ( 4 162 ) is 4
2 3 Ans.
é\
êë
( a) + ( b)
3 3 3 3
= a + bù which is rational.
úû (vii) 2 + 3

(vi) R.F. of Sol. ( )(

As 2 + 3 2 - 3 = ( 2 )2 - ) ( 3) 2
= 4 - 3 = 1,
( a + b + c is ) ( ) (
a + b - c nd a + b - c + 2 ab . ) which is rational.

3-7
Real Number

\ R.F. of (2 + 3 ) is (2 - 3 ) Ans. (iv) 7 - 4 3 , 7 + 4 3

(viii) 7 - 4 3 (v) 8 + 3 7 , 8 - 3 7 ............... and so on.

Sol. ( )(
As 7 - 4 3 7 + 4 3 = ( 7 )2 - 4 - 3 ) ( )
2
= 49 - 48 =
1, which is rational Example-21
\ R.F. of (7 - 4 3 ) is 7 + 4 3 Ans. ( ) Express the following surd with a rational denominator.

(ix). 3 3 + 2 2 8 8 é 15 + 1 + ( ) ( 5+ 3 ù)
Sol. 15 + 1 - 5 - 3
=
[( ) ( ´ê
15 + 1 - 15 + 3 êë 15 + 1 + )] ( ) ( )ú
5 + 3 úû
Sol. As (3 3 + 2 2 )(3 ) ( ) - (2 2 )
3 -2 2 = 3 3
2 2
= 27 - 8 = 19 ,
which is rational.
=
(
8 15 + 1 + 5 + 3 )
\ R.F. of (3 3 + 2 2 ) is (3 3 - 2 2 ) Ans. ( 15 + 1 - ) (
2
5+ 3)
2

(x) 3
3 +3 2
=
(
8 15 + 1 + 5 + 3 )
Sol. As ( 3
)
3 + 3 2 æç 3 32 - 3 3 ´ 3 2 + 2 2 ö÷ = æç 3 33 + 2 3 ö÷
3 3 15 + 1 + 2 15 - 5 + 3 + 2 15 ( )
è ø è ø
= 3 + 2 = 5, which is rational.
=
(
8 15 + 1 + 5 + 3 )
æ3 2 3 3 3 2 ö 8
\ R.F. of (3 3 + 3 2 ) is çè 3 - 3 ´ 2 + 2 ÷ø
= ( 15 + 1 + 5 + 3 ) Ans.
Ans.
(xi) 1 + 2 + 3 Example-22

Sol. (1 + )(
2 + 3 1+ 2 - 3 = 1+ 2 ) ( ) - ( 3)
2 2

Rationalize the denominator of

a2
a2 + b2 + b
+ ( 2 ) + 2( 1)( 2 ) - 3
2 2
= 1)

= 1+2+2 2 -3 a2 a2 a 2 + b2 - b
Sol. = ´
= 3+2 2 -3 a 2 + b2 + b a 2 + b2 + b a 2 + b2 - b

=2 2
= 2 2 ´ 2 = 2´2 = 4 a 2 æç a 2 + b 2 - b ö÷
è ø
\ R.F. of 1 + 2 + 3 is 1 + 2 - 3 and ( ) 2 Ans.
= 2
æç a 2 + b 2 ö÷ - ( b )2
è ø
Note: R.F. of a + b or a - b type surds are also called
conjugate surds & vice versa. a 2 æç a 2 + b 2 - b ö÷
= è ø = æ a2 + b2 - b ö Ans.
ç ÷
a2 + b2 - b2 è ø
Example-19
(i) 2 - 3 is conjugate of 2 + 3 Example-23
(ii) 5 + 1 is conjugate of 5 -1 3+2 2
If = a + b 2 , where a and b are rational
3- 2
Note: Sometimes conjugate surds and reciprocals are same.
then find the values of a and b.

3 + 2 2 (3 + 2 2 )(3 + 2 )
Example-20 Sol. L.H.S. =
3- 2 ( 3 - 2 )(3 + 2 )
(i) 2 + 3 , it’s conjugate is 2 - 3 , its reciprocal is
9+3 2 +6 2 +4
2 - 3 & vice versa. =
9-2
(ii) 5 - 2 6 , it’s conjugate is 5 + 2 6 , its reciprocal
13 + 9 2
is 5 - 2 6 & vice versa. =
7
(iii) 6 - 35 ,6 + 35

3-8
 Real Number

13 9
= + 2 Example-26
7 7
1
13 9 If = find the value of x3– x2– 11x + 3.
\ + 2 = a+b 2 2+ 3
7 7
1
Sol. As, x = =2- 3
Equating the rational and irrational parts 2+ 3
13 9 Þ x-2=- 3
We get a = ,b = Ans.
7 7
Þ (x - 2)2 = - 3( )2
[By squaring both sides]
Example-24 Þ x2 + 4 - 4x = 3
1 Þ x2 - 4x + 1 = 0
If 3 = 1.732 , find the value of Now, x3– x2 – 11x + 3 = x2– 4x2 + x + 3x2– 12x + 3
3 -1
= x (x2 - 4x + 1) + 3 (x2 - 4x + 1)
= x(0) + 3 (0)
1 1 3 +1 = 0 + 0 = 0 Ans.
Sol. = ´
3 -1 3 -1 3 +1
Example-27
3 +1 3 +1 1
= = 3
3-1 2 If x = 3 - 8 , find the value of x + 3 .
x
1.732 + 1 2.732 Sol. x=3- 8
= =
2 2
1 1
= 1.366 Ans. \ x = 3- 8

Example-25 1
Þ = 3+ 8
x
If 5 = 2.236 and 2 = 1.414, then
1
3 4 Now, x + = 3- 8 +3+ 8 = 6
Evaluate: + x
5+ 2 5- 2
3
3 1 æ 1ö 1æ 1ö
Þ x + = ç x + ÷ - 3x ç x + ÷
3
+
4
=
3 5 -3 2 + 4 ( 5+ 2 ) x 3
è xø xè xø
Sol. 5+ 2 5- 2 ( 5- 2 )( 5+ 2 ) 3
Þ x +
1
= (6)3 - 3(6)
3
x
3 5 -3 2 +4 5 +4 2 3 1
= Þ x + = 216 - 18
5-2 x3

3 1
7 5+ 2 Þ x + = 198 Ans.
= x3
5-2

7 5+ 2 Example-28
= If x = 1 + 21/3 + 22/3, show that x3 – 3x2 – 3x – 1 = 0
3
Sol. x = 1 + 21/3 + 22/3
7 ´ 2.236 + 1.414 Þ x – 1 = (21/3 + 22/3)
=
3 Þ (x – 1)3 = (21/3 + 22/3)3
15.652 + 1.414 Þ (x – 1)3= (21/3) + (22/3)3 + 3.21/3.22/3(21/3 + 21/3)
= Þ (x – 1)3 = 2 + 22 + 3.21 (x – 1)
3
Þ (x – 1)3 = 6 + 6 (x - 1)
17.066 Þ x3 – 3x2 + 3x – 1 = 6x
=
3
Þ x3 – 3x2 – 3x – 1 = 0 Ans.
= 5.689 (approximate)

3-9
Real Number

e.g. (i) Consider number 23 and 5, then :

Example-29 23 = 5 × 4 + 3
Solve: Comparing with a = bq + r
x + 3 + x - 2 = 5.
we get, a = 23, b = 5, q = 4, r = 3 and 0 £ r < b
Sol. x+3 = 5- x-2 (as 0 < 3 < 5 )
(ii) Consider positive integers 18 and 4
Þ ( x+3 ) = (5 -
2
x-2 )2
18 = 4 × 4 + 2
[By squaring both sides] For 18 ( = a) and 4 ( = b) we have q
Þ x + 3 = 25 + (x - 2) - 10 x - 2 = 4, r = 2 and 0 £ r < b
In the relations a = bq + r, where 0 £ r < b is nothing
Þ x + 3 = 25 + x - 2 - 10 x - 2 but a statement of the long division of number a by b
Þ 3 - 23 = - 10 x - 2 in which q is the quotient obtained and r is the
remainder.
Þ -20 = -10 x - 2
Þ 2= SOLVED EXAMPLE
x-2
Þ x-2=4 [By squaring both sides] Example-31
Þ x= 6
Use Euclid’s division lemma to show that the square
Ans. of any positive integer is either of the form 3m or 3m +1
for some integer m.
Example-30 Sol. Let a and b are two positive integers such that a is
If x = 1 + 2 + 3 , prove that x4 - 4x3 - 4x2 + 16 - 8 = 0. greater than b, then :
a = bq + r ; where q and r are also positive integers and
Sol. x=1 + 2+ 3 0£r < b
Þ x-1= 2+ 3 Taking b = 3, we get :
a = 3q + r ; where 0 £ r < 3
Þ ( x - 1 )2 = ( 2+ 3 )
2
Þ The value of positive integer a will be
3q + 0, 3q + 1 or 3q + 2
[By squaring both sides]
i.e., 3q, 3q + 1or 3q + 2.
Þ x2 + 1 - 2x = 2 + 3 + 2 6 Now we have to show that square of positive integers
3q, 3q + 1 and 3q + 2 can be expressed as 3m or 3m +1
Þ x2 - 2x - 4 = 2 6
for some integer m.
Þ (x2 - 2x - 4)2 = (2 6 )2 \ Square of 3q = (3q)2
= 9q2 = 3(3q2) = 3m ;
Þ x4 + 4x2 + 16 - 4x3 + 16x - 8x2 = 24
where m is some integer and m = 3q2
Þ x4 - 4x3 - 4x2 + 16x + 16 - 24 = 0 Square of 3q + 1 = (3q + 1)2
Þ x4 - 4x3 - 4x2 + 16x - 8 = 0 Ans. = 9q2 = 6q +1
= 3(3q2 + 2q) +1 = 3m + 1 for some integer and m 3q2 +
l EUCLID’S DIVISION LEMMA OR EUCLID’S 2q.
DIVISIONALGORITHM Square of 3q + 2 = (3q + 2)2
For any two positive integers a and b there exist unique = 9q2 = 12q + 4
integers q and r such that = 9q2 = 12q + 3 + 1
A = bq + r, where 0 £ r < b. = 3(3q2 + 4q + 1) +1 = 3m +1 for some integer and m =
Let us consider a = 217, b = 5 and make the division of 3q2 + 4q + 1.
217 by 5 as under : \ The square of any positive integer is either of the
Divined form 3m or 3m +1 for some integer m.
¯ Example-32
Divisor ® 5) 217(43 ¬ Quotient
Show that one and only out of n; + 2 or n + 4 is
20 divisible by 3, where is any positive integer.
17 Sol. Consider any two positive integers a and b such that a
15 is grater than b, then according to Euclid’s division
2 ¬ Remainder algorithm –
a = bq + r ; where q and r positive integers and
i.e. Dividends = Divisor × Quotient + Remainder 0 £ r<b
(a) = (b) × (q) + (r)
Let a = n and b = 3, then
a = bq + r Þ n = 3q + r ;

3-10
 Real Number

where 0 £ r < 3.
r = 0 Þ n = 3q + 0 = 3q Example-34
r = 1 Þ n = 3q + 1 Find the HCF of 1848, 3058 and 1331.
and r = 2 Þ n = 3q + 2 Sol. Two numbers 1848 and 3058 , where 3058 > 1848
if n = 3q ; n is divisible by 3 3058 = 1848 × 1 + 1210
If n = 3q +1 ; then n + 2 = 3q + 1 + 2 1848 = 1210 × 1 638 [Using Euclid’s division algorithm
= 3q +3 ; which is divisible by 3
to the given number 1848 and 3058]
Þ n + 2 is divisible by 3
If n = 3q + 2 ; then n + 4 = 3q + 2 + 4 1210 = 638 × 1 572
= 3q + 6 ; which is divisible by 3 638 = 572 × 1 + 66
Þ n + 4 is divisible by 3 527 = 66 × 8 + 44
Hence , if n is any positive integer, then one and only 66 = 44 × 1 + 22
one out n, n + 2 or n + 4 is divisible by 3. 44 = 22 × 2 + 0
Therefore HCF of 1848 and 3058 is 22.
l APPLICATION OF EUCLID’S DIVISION LEMMA HCF (1848 and 3058) = 22
FOR FINDING H.C.F. OF POSITIVE INTEGERS Let us find the HCF of the numbers 1331 and 22.
Algorithm : 1331 = 22 × 60 + 11
Consider positive integers 418 and 33 22 =11 × 2 + 10
Step. (a) \ HCF of 1331and 22 is 11
Taking bitter number (418) as a and smaller number Þ HCF (22, 1331) = 11
(33) as b. Hence the HCF of the given numbers 1848, 3058 and
Express the numbers are a = bq + r 1331 is 11.
418 = 33 × 12 +22 HCF (1848, 3058, 1331) = 11
Step. (b)
Now taking the divisor 33 and remainder 22, apply the Example-35
Euclid’s division method to get. Using Euclid’s division, find the HCF of 56, 96 and
33 = 22 × 1 + 11 [Expressing as a = bq + r ] 404. [Sample paper (CBSE)- 2008]
Sol. Using Euclid’s division algorithm, to 56 and 96.
Step. (c) 96 = 56 × 1 + 40
Again with new divisor 22 and new remainder 11, apply 56 = 40 × 1 + 16
40 = 16 × 2 + 8
the Euclid’s division algorithm to get
16 = 8 × 2 + 0
22 = 11 × 2 + 0 Now to find HCF of 8 and 404
Step. (d) We apply Euclid’s division algorithm to 404 and 8
Since, the remainder = 0 so we can not proceed further. 404 = 8 × 50 + 4
Step. (e) 8=4×2+0
The last divisor is 11 and we say H.C.F. of 418 and Hence 4 is the HCF of the given numbers 56, 96 and
33 = 11 404.

SOLVED EXAMPLE Example-36

Find the HCF of 128, 240 using Euclid’s Division
Example-33 Algorithm.
Use Euclid’s algorithm to find the HCF of 4052 and Sol. 240 = 128 × 1 + 112
12576. 128 = 112 × 1 + 16
Sol. Using a = bq + r, where 0 £ r < b. 112 = 16 × 7 + 0® Remainder is zero.
Clearly, 12576 > 4052 HCF of 112 and 16 is 16.
[a = 12576, b = 4051] HCF of 240 and 128 is 16.
Þ 12576 = 4051 × 3 + 420
Þ 4052 = 420 × 9 + 272 Example-37
Þ 402 = 272 × 1 + 148 A number when divided by 61 gives 27 as quotient
Þ 272 = 148 × 1 + 124 and 32 as remainder. Find the number.
Þ 148 = 124 × 1 + 24 Sol. a= ?, b = 61, q = 27, r = 32
Þ 124 = 24 × 5 + 4 Using Euclid’s Division Algorithm
Þ 24 = 4 × 6 + 0
The remainder at this stage is 0. So, the divisor at this a = bq + r = 61 ´ 27 + 32 = 1647 + 32 = 1679
stage, i.e., 4 is the HCF of 12576 and 4052. The required number is 1679.

3-11
Real Number

Example-38 Example-41
For a pair of integers 151, 16, find the quotient q and Use Euclid’s division algorithm to show that the cube
the remainder r when the larger number a is divided by of any positive integer is either of the form 9q, 9q + 1
or 9q + 8.
the smaller number b and verify that a = bq + r and
Sol. Let x be any positive integer.
0£ r <b. Then x = 3m, 3m +1, 3m + 2.
Sol. Given pair of integers = 151, 16 Let a = x, b = 3, then by Euclid’s Algorithm
When x = 3m
16 151 9
144 Þ x 3 = (3m)3 = 9 ´ 3m3 = 9q
7 [q = 3m3]
Suppose a = 151, b = 16 When x = 3m + 1
After dividing 151 by 16, x 3 = (3m + 1)3 = 27 m3 + 27 m2 + 9m + 1 = 9q + 1
we have q = 9 and r = 7
Which satisfies 0 £ r < 16 . [Let q = 3m3 + 3m 2 + m ]
It verifies a = bq + r When x = 3m + 2
Þ 151 = 16 ´ 9 + 7 Þ x3 = (3m + 2)3 = 27 m3 + 54m2 + 36m + 8
151 = 151
= 9(3m3 + 6m 2 + 4m) + 8 = 9q + 8
Example-39
[Let q = 3m3 + 6m 2 + 4m ]
Find the number by which 546 should be divided to
get 7 as quotient and 7 as remainder. Thus, we see that x3 may be of the form 9q, 9q + 1 and
Sol. a = 546, b= ?, q = 7, r = 7. 9q + 8 for any integer x.
Using Euclid’s Division Algorithm Example-42
a =b ´ q + r Show that only one of the numbers (n + 2),n and (n + 4)
is divisible by 3.
546 = b ´ 7 + 7
Sol. Let n be any positive integer.
546 - 7 = b ´ 7 On dividing n by 3, let q be the quotient and r be the
remainder.
539 Then, by Euclid’s division lemma, we have
=b
7 n = 3q + r , where 0 £ r < 3
77 = b
Þ n = 3q + r where r = 0, 1 or 2
546 should be divided by 77.
Þ n = 3q (where r = 0 ), n = 3q + 1 (when r = 1)
Example-40
and n = 3q + 2 (when r = 2)
Show that any positive odd integer is of the form
The following table shows
6q + 1, 6q + 3 or 6q + 5, where q is some integer.
Sol. Let us assume that a is a positive odd integer and the divisibility of n, n + 2 and n + 4 by 3.
b = 6 i.e. a is divided by 6. Positiveinteger (n) n n+2 n+4
Whenn = 3q 3q (3q) + 2 (3q) + 4 = 3(q + 1) + 1
Then, by Euclid’s algorithm. Case 1:
Divisionby 3 divisible
Leavesremainder 2 Leaves remainder 1
\ not divisible \ not divisible
a = 6q + r for some q³ 0 and 0 £r< 6. Whenn = 3q + 1 3q + 1 (3q + 1) + 2 = 3(q + 1)
(3q + 1) + 4 = 3(q + 1)
+2
Case 2: Leaves
\ a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 Division by 3 remainder 1 divisible
Leavesremainder2
\ not divisible
\ not divisible
or 6q + 5 Whenn = 3q + 2 3q + 2
(3q + 2) + 2 = 3(q + 1)
(3q + 2) + 4 = 3(q + 2)
+1
Þ If a is of the form 6q, 6q + 2, 6q + 4, then it is an Case 3: Leaves
Leaves remainder 1
Division by 3 remainder 2 divisible
\ not divisible
even integer. \ not divisible

Odd integer is formed by adding 1, 3, 5 to an even In case 1, n is divisible by 3 but n + 2 and n + 4 are not
integer. divisible by 3.
\ For a to be odd positive integer, it should be of the In case 2, n + 2 is divisible by 3 but n and n + 4 are not
form 6q + 1, 6q + 3 or 6q + 5. divisible by 3.

3-12
 Real Number

In case 3, n + 4 is divisible by 3 but n and n + 2 are not

divisible by 3. Example-45
Hence, one and only one out of n, n + 2 or n + 4 is If a = xq + ar , b = yq + br , c = zq + cr , then show
divisible by 3, where n is any positive integer.
a´b´c
Example-43 that remainder of is same as the remainder of
q
Show that any positive even integer is of the form 4q,
ar ´ br ´ cr 560
4q + 2, where q is some integer. . Also, find out the remainder of 3 .
Sol. Let a be any positive even integer and b = 4. q 8
By Euclid’s Division lemma, we have Sol. We have
a = 4q + r where 0 £r< 4
a ´ b ´ c = ( xq + ar ) ( yq + br ) ( zq + cr )
(i) a = 4q, when r = 0
or a = 2(2q) = ( xyq 2 + xbr q + yar q + ar br ) ( zq + cr )
\ It is a multiple of 2.
(ii) a = 4q + 1 when r = 1 = xyzq 3 + xzbr q 2 + yzar q 2 + ar br zq
or a = 2(2q) + 1
+ xycr q 2 + xbr cr q + yqar cr + ar br cr
\ It is not a multiple of 2.
(iii) a = 4q + 2 when r = 2
= q( xyzq 2 + xzbr q + yzar q + ar br z
or a = 2(2q + 1)
+ xycr q + xbr cr + yar cr ) + ar br cr
\ It is a multiple of 2.
(iv) a = 4q + 3 when r = 3 [Taking q common]

or a = 2(2q + 1) + 1 = qn + ar br cr
\ It is not a multiple of 2. where n =
Thus, a = 4q and a = 4q + 2 are positive even integers xyzq 2 + xzbr q + yzar q + ar br z + xycr q + xbr cr + yar cr
only. Now,

Example-44 a ´ b ´ c qn + ar br cr qn ar br cr abc
= = + =n+ r r r
q q q q q
Show that n2 - n is divisible by 2 for every +ve
integer n. abc
\ The remainder of
Sol. Any positive integer is of the form 2q or 2q + 1 for q is same as remainder of
some integer q.
ar br cr
Case I: When n = 2q .
q
Þ n2 – n = 4q 2 - 2q = 2q(2q - 1) = 2m
( )
280
3560 32 9 280
where m = q(2q - 1) Now = =
8 8 8
Hence n2 - n is divisible by 2.
[ (a m )n = a mn ]
Case II: When n = 2q + 1,
[9 ´ 9 ´ 9... ´ 9 (280 times)]
Þn2 – n = (2q + 1) 2 - (2q + 1) =
8
= 4q 2 + 1 + 4q - 2q - 1 Now using the above result, we get remainder for
3560 9 ´ 9 ´ 9... ´ 9 (280 times)
= 4q 2 + 2q =
8 8
= 2q(2q + 1)
1 ´ 1 ´ 1 ´ ... ´ 1 (280 times) 1
= =
= 2m¢ where m¢ = q (2q + 1) 8 8

Þ n2 - n is divisible by 2 also. 3560

\ The remainder of is 1.
8
Hence, n2 - n is divisible by 2 for every +ve integer
[i.e. numerator is the remainder]
m and m¢.

3-13
Real Number

l THE FUNDAMENTALTHEOREM OFARITHMETIC

Statement - “Every composite number can be Example-50
factorised as a product of prime numbers in a unique Given that HCF (306, 657) = 9. Find the LCM (306, 657)
way, except for the order in which the prime numbers Sol. HCF (306, 657) × LCM (306, 657) = 306 × 657
occur. Þ 9 × LCM (306, 657) = 306 × 657
e.g. (i) 90 = 2 × 3 × 3 × 5 = 2×32 × 5
(ii) 432 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 24 × 33 306 ´ 657
Þ LCM (306, 657) = = 34 ´ 657
(iii) 12600 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 7 9
= 2 3 × 32 × 52 × 7 Þ LCM (306, 657) = 22338
In general, a composite number is expressed as the
product of its prime factors written in ascending order Example-51
of their values. Check whether 6n can end with the digit 0, where n is a
natural number.
SOLVED EXAMPLE
Sol. Let the number 6n ends with the digit 0, where n is a
Example-46 natural number.
Is 7 × 11 × 13 + 11 a composite number? Þ 6n is divisible by 5.
Sol. 11 ´ (7 ´ 13 + 1) = 11 ´ (91 + 1) = 11 × 92 = 1012 Þ 5 is a prime factor of 6n.
It is a composite number which can be factorized into Þ But 2 and 3 are the only prime factors of 6n.
primes. Hence, the number 6n is divisible by 5 is not possible.
As the prime factorization of a number is always
Example-47 unique. Our supposition is wrong.
Find the missing numbers. Hence, 6n does not end with the digit 0, where n is a
Sol. Going upwards 165 × 2 = 330
natural number.

Example-52
Find the largest positive integer that will divide 398,
2 436 and 542 leaving remainder 7, 11 and 15 respectively.
Sol. Given condition is that on dividing 398 by the required
number, there is a remainder of 7 so that 398 – 7 = 391
2 165
is exactly divisible by the required number or the
330 × 2 = 660 required number is a factor of 391.
In the same way, required positive integer is a factor of
Example-48 436 – 11 = 425 and 542 – 15 = 527 also
Write the prime factorization of Clearly, required number is the HCF of 391, 425 and
(i) 72 527.
(ii) 5005 Using the factor tree for the prime factorisation of 391,
Sol. 425 and 527 are as under:
391 = 17 ´ 23, 425 = 5 2 ´ 17 and 527 = 17 ´ 31
72 5005
\ HCF of 391, 425 and 527 is 17
36 1001
1001
2 5 Hence, required number = 17
18 143
2 7
9 13
2
3
11 Example-53
3
Find the HCF and LCM of 26 and 91 and verify that
72 = 2 × 2 × 2 × 3 × 3 5005 = 5 × 7 × 11 × 13
L.C.M. × H.C.F.
= product of two numbers.
Sol. By prime factorization
Example-49 26 = 2 × 13
Find HCF and LCM of 45, 75 and 125. 91 = 7 × 13
Sol. 45 = 3 × 3 × 5 = 32 × 5 HCF (26, 91) = 13
75 = 3 × 5 × 5 = 3 × 52 LCM (26, 91) = 13 × 2 × 7 = 26 × 7 = 182
125 = 5 × 5 × 5 = 53 LCM × HCF = 13 × 182 = 2366
HCF = 51 = 5 Product of two numbers = 26 × 91 = 2366.
LCM = 32 × 53 = 9 × 125 = 1125 Hence, L.C.M. × H.C.F. = Product of two numbers.

3-14
 Real Number

Now, 8 = 56 ´ 4 + (–3) ´ 72
Example-54 8 = 56 ´ 4 + (–3) ´ 72 – 56 ´ 72 + 56 ´ 72
An army contingent of 616 members is to march behind Þ 8 = 56 ´ 4 – 56 ´ 72 + (–3) ´ 72 + 56 ´ 72
an army band of 32 members in a parade. The two Þ 8 = 56 ´ (4 – 72) {(–3) + 56} ´ 72
groups are to march in the same number of columns. Þ 8 = 56 ´ (–68) + (53) ´ 72
What is the maximum number of columns in which Þ x = –68 and y = 53.
they can march? Hence, x and y are not unique.
Sol. In order to find the maximum number of column in which
army contingent can march. We have to find the largest COMPETITION WINDOW
NUMBER OF FACTORS OFA NUMBER
number that divides 616 and 32.
Clearly, such a number is the HCF.
To get number of factors (or divisors) of a number N,
616 = 2 × 2 × 2 × 7 × 11 = 23 × 7 × 11
express N as
32 = 2 × 2 × 2 × 2 × 2 = 25 N = ap. bq. cr. ds………(a, b, c, d are prime numbers and
H.C.F. = 23 = 8 p, q, r, s are indices)
Hence, 8 is the maximum number of columns in which Then the number of total divisors or factors of
they can march. N = (p + 1) (q + 1) (r + 1) (s + 1) ……
Eg. 540 = 22 × 33 × 51
Example-55
\ total number of factors of
Find the greatest number of 6 digits exactly divisible 540 = (2 + 1) (3 + 1) (1 + 1) = 24
by 24, 15 and 36.
Sol. Greatest number of 6 digits is 999999. In order to find SUM OF FACTORS OFA NUMBER
the greatest 6 digit number divisible by 24, 15, 36, we The sum of all factors of
find their LCM.
( a p +1 - 1)(b q +1 - 1)(c r +1 - 1)(d s +1 - 1)
L.C.M. of 24, 15 and 36 = 360 N=
(a - 1)(b - 1)(c - 1)()d - 1
Eg. 270 = 2 × 33 × 5
360 999999 2777
\ Sum of factors of
720
2799 ( 21+1 - 1)(33+1 - 1)(51+1 - 1) 3 ´ 80 ´ 24
270 = = = 720
2520 (2 - 1)(3 - 1)()5 - 1 1´ 2 ´ 4
2799
2520 PRODUCT OF FACTORS
2799 The product of factors of composite number N = N n/2,
2520 where n is the total number of factors of N.
279 Eg. 360 = 23 × 32 × 51
Subtracting remainder i.e., 279 from 999999, we get \ No. of factors of 360 = (3 + 1) (2 + 1) (1 + 1) = 24
999999 – 279 = 999720 Thus, the product of factors = (360)24/2 = (360)12
Hence, 999720 is the largest 6 digit number divisible
by 24, 15 and 36. NUMBER OF ODD FACTORS OFA NUMBER
To get the number of odd factors of a number N, express
Example-56 N as
If d is the HCF of 56 and 72, find x, y satisfying N = ( p1a ´ p 2b ´ p3c ´ .........) ´ (e x )
d = 56x + 72y. Also, show that x and yare not unique. (where p1, p2, p3………are the odd prime factors and e
Sol. Applying Euclid’s division lemma to 56 and 72, we get is the even prime factor)
72 = 56 ´ 1+16 …(i) Then the total number of odd factors = (a + 1) (b + 1) (c
56 = 16 ´ 3 + 8 …(ii) + 1)…….
Eg. 90 = 21 × 32 × 51
16 = 8 ´ 2 + 0 …(iii)
\ Last divisor 8 is the HCF of 56 and 72.
\ Total number of odd factors of
90 = (2 + 1)(1 + 1) = 6
From (ii), we get
8 = 56 – 16 ´ 3 NUMBER OF EVEN FACTORS OFA NUMBER
Þ 8 = 56 – (72 – 56 ´ 1) ´ 3 Number of even factors of a number = Total number of
[From equation (i)] factors – Total number of odd factors.
Þ 8 = 56 – 3 ´ 72 + 56 ´ 3 NUMBER OF WAYS TO EXPRESS A NUMBER AS A
Þ 8 = 56 ´ 4 + (–3) ´ 72 PRODUCT OF TWO FACTORS
Comparing it with d = 56x + 72y, we get x = 4 and Let n be the number of total factors of a composite
y = –3. number .

3-15
Real Number

Case – 1 : If the composite number is not a perfect

square then number of ways of expressing the SOLVED EXAMPLE
n
composite number as a product of two factors = Example-57
2
Case – 2 : If the composite number is a perfect square Find the L.C.M. and H.C.F. of 1296 and 2520 by
then applying the fundamental theorem of arithmetic method
(a) Number of ways of expressing the composite i.e. using the prime factorisation method.
Sol. 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × = 24 × 34
n +1
number as a product of two factors = 2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 23 × 32 × 5 × 7
2
L.C.M. = 24 × 34 × 5 × 7 = 45360
(b) Number of ways of expressing the composite H.C.F. = 23 × 32 × = 72
number as a product of two distinct factors
( n - 1) Example-58
=
2 Given that H.C.F. (306, 657) = 9.
Find L.C.M. (306, 657)
l USING THE FUNDAMENTAL THEOREM OF Sol. H.C.F. (306, 657) = 9 means H.C.F. of 306
ARITHMETIC TO FIND H.C.F. AND L.C.M. and 657 = 9
For any two number a and b. Required L.C.M. (306, 657) means required L.C.M. of
(a) L.C.M. (Least common multiple) = Product of each 306 and 657.
prime factor with highest powers H.C.F. (a, b) For any two positive integers ;
Product of the numbers or (a×b) Product to the number
L.C.M. (a,b) = their L.C.M. =
H.C.F.(a, b) Their H.C.F.

(b) H.C.F. (Highest common factor) = Product of 306 ´ 657

common prime factor with lowest power. i.e., L.C.M. (306, 657) = = 22, 338
9
Product of the numbers or (a×b)
H.C.F. (a, b) =
L.C.M.(a, b) Example-59
Given that L.C.M. (150, 100) = 300,
Remark : The above relations hold only for two find H.C.F. (150, 100)
numbers. Sol. L.C.M. (150, 100) = 300
Þ L.C.M. of 150 and 100 = 300
COMPETITION WINDOW Since, the product of number 150 and 100
For any three positive integers p, q, r – = 150 × 100
HCF (P, q, r) × LCM (p, q, r) ¹ p × q × r And , we know : H.C.F. (150, 100) =
However , the following results hold good for the three Product of 150 and 100 150 ´ 100
positive integers p, q and r : = =50
L.C.M.(150. 100) 300

p.q.r.HCF ( p, q, r )
LCM (p, q, r) = HCF ( p, q ) . HCF ( q, r ) . HCF ( p, r ) Example-60
Explain why 7 × 13 +13 are composite numbers :
p.q.r.LCM ( p, q, r ) Sol. (i) Let 7 × 11 × 13 + 13 = (7 × 11 + 1) × 13
H.C.F. (p, q, r) = LCM ( p, q ) . LCM ( q, r ) . LCM ( p, r ) = (77 + 1) × 13 = 78 × 13
Þ 7 × 11 × 13 + 13 = 2 × 3 × 13 × 13
2 1296 = 2 × 3 × 132 is a composite number as powers of prime
2 2520
occur.
2 648
2 1260
2 324 COMPETITION WINDOW
2 630 HCFAND LCM OF FRACTIONS
2 162
3 315
3 81 HCF of Fractions : The greatest common fraction is
3 105 called the HCF of the given fractions.
3 27
5 35 HCF of Numerator
HCf of fractions =
3 9 LCM of Denominator
7
3
For example : The HCF of

3-16
 Real Number

4 4 2 36 HCF of 4, 4, 2, 36 2 TESTS OF DIVISIBILITY

, , , = = Test of Divisibility by 2: A number is divisible by 2, if
3 9 15 21 LCM of 3, 9,15, 21 315
its units digit is any of the digits 0, 2, 4, 6 and 8.
Called the LCM of the fractions.
Example.: Each of the numbers 24, 36, 78, 192, 310,
LCM of Numerator 214166 is divisible by 2.
LCM of fractions =
HCF of Denominator
Test of Divisibility by 3: A number is divisible by 3, if
For example : The LCM of the sum of its digits is divisible by 3.
4 4 2 36 LCM of 4, 4, 2, 36 36 Example: (i) Consider the number 349524.
, , , = = = 12
3 9 15 21 HCF of 3, 9,15, 21 3 Sum of its digits = (3 + 4 + 9 + 5 + 2 + 4) = 27, which is
divisible by 3.
HCFAND LCM OF DECIMALS \ 349524 is divisible by 3.
HCF (ii) Consider the number 871423
Step-1 : First of all equate the number of places in all Sum of its digits = (8 + 7 + 1 + 4 + 2 + 3) = 25, which is
the numbers by using zeros, wherever required . not divisible by 3.
Step-2 : Then considering these number as integers \ 871423 is not divisible by 3.
find the HCF of these numbers.
Step-3 : Put the decimal point in the resultant value as Test of Divisibility by 4: A number is divisible by 4, if
many places before the right most digit as that of in the number formed by its last two digits is divisible by
the every equated number. 4.
Examples: (i) Consider the number 15632.
SOLVED EXAMPLE The number formed by its last two digits is 32, which
is divisible by 4.
Example-61
Find the HCF of 0.0005, 0.005, 0.15, 0.175, 0.5 and 3.5
\ 15632 is divisible by 4.
(ii) Consider the number 19374.
Sol. 0.0005 Þ 5
The number formed by its last two digits is 74, which
0.0050 Þ 50 is not divisible by 4.
0.1500 Þ 15 \ 19374 is not divisible by 4.
0.1750 Þ 1750
0.5000 Þ 5000 Test of Divisibility by 5: A number is divisible by 5, if
3.5000 Þ 35000 its units digit is either 0 or 5.
Then the HCF of 5, 50, 1500, 1750, 5000, and 35000 is 5. Examples:
So, the HCF of the given numbers is 0.0005. The numbers 245, 16260, 27915, 411115 are all divisible
by 5.
LCM
Step-1: First of all equate the number of places in all Test of Divisibility by 6: A number is divisible by 6, if
the given numbers by putting the minimum possible it is divisible by 2 as well as 3.
number of zeros at the end of the decimal numbers. Examples: (i) Consider the number 753216. Since its
wherever even reired.
units digit is 6, so it is divisible by 2.
Step-2 : Now consider the equated numbers as integers
Sum of its digits = (7 + 5 + 3 + 2 + 1 + 6) = 24, which is
and then find the LCM of these numbers.
Step-3 : Put the decimal point in the LCM of the number divisible by 3.
as many places as that of in the equated numbers. \ 753216 is divisible by both 2 and 3 and is therefore
divisible by 6.
SOLVED EXAMPLE (ii) Consider the number 453212. Since, its unitsdigit is
2, so it is divisible by 2.
Example-62 Sum of its digits = (4 + 5 + 3 + 2 + 1 + 2) = 17, which is
Find LCM of 1.8, 0.54 and 7.2. not divisible by 3.
\ 453212 is not divisible by 6.
1.8 ü 1.80 ü 18 0
ï ï Test of Divisibility by 7: A number is divisible by 7, if
0.54ý 0.54ý 54
7.2 ïþ 7.20ïþ 720
the difference between twice the digit at unit place
and the number formed by other digits is either 0 or
Sol. Now the LCM of 180, 54 and 720 is 2160. Therefore the multiple of 7.
required LCM is 21.60.

3-17
Real Number

Example: Consider the number 658. = (1 + 2 + 7) – (1 + 2 + 0 + 5) = (10 – 8) = 2, which is not

65 – 2 × 8 = 65 – 16 = 49. As 49 is divisible by 7. Hence divisible by 11.
the number 658 is divisible by 7. \ 5702211 is not divisible by 11.
l SYNOPSIS
Note 1. Euclid Division Algorithm : Given any two positive
integers a and b, b ¹ 1. a > b and a is not divisible by
If a number is divisible by two co-primes, then it is
b, there exists two (unique) integers q and r such that
also divisible by their product.
a = q + r , where r < b
2. Prime Factorization Theorem : Every composite
number can be expressed as a product of prime factors,
Test of Divisibility by 8: A number is divisible by 8, if and the decomposition is unique, apart from the order
the number formed by its last three digits is divisible of factors.
by 8. (The fundamental Theorem of Arithmetic)
Examples: (i) Consider the number 29512. i.e. given any composite number x, we can find unique
The number formed by its last three digits is 512, which prime factors p1, p2, p3………pn such that
is divisible by 8. x = p1 × p2 × p3 ×……× pn
So, 29512 is divisible by 8.
(ii) Consider the number 16942. 3. HCF and LCM of two numbers : Let a, b be given
The number formed by its last three digits is 942, which numbers, Let each of these is expressed at a product
is not divisible by 8. of prime factors.
\ 16942 is not divisible by 8. (i) The product of the smaller powers of the common
Test of Divisibility by 9: A number is divisible by 9, if prime numbers is the HCF.
the sum of its digits is divisible by 9. (ii) The product of the prime numbers is either or both of
Examples: (i) Consider the number 517248. these expression taken with greater power is the
Sum of its digits = (5 + 1 + 7 + 2 + 4 + 8) = 27, which is required LCM.
divisible by 9. (iii) HCF (a, b) × LCM (a, b) = a × b
\ 517248 is divisible by 9. p
(ii) Consider the number 641857. 4. Rational Numbers , q ¹ 0 has a terminating decimal
Sum of its digits = (6 + 4 + 1 + 8 + 5 + 7) = 31, which is
q
not divisible by 9. expansion if the prime factors of q are only 2’s and 5’s
\ 641857 is not divisible by 9. or both .
p
Test of Divisibility by 10: A number is divisible by 10, 5. Let x = be a rational number such the prime
if its units digit is 0. q
Examples: The numbers 1110, 301020, 15670, 19250 are factorization of q is of the form 2n. 5m where n, m are
all divisible by 10. non-negative integers, then x ahs a decimal expansion
which terminates.
Test of Divisibility by 11:A number is divisible by 11, p
if the difference between the sum of its digits at odd 6. A rational number , q ¹ 0 has terminating repeating
places and sum of the digits at even places is either 0 q
or a number divisible by 11. decimal expansion if the prime factors of q are other
Examples: (i) Consider the number 749859. than 2 and 5 or both .
(Sum of digits at odd places) – (Sum of digits at even
p
places) 7. Let x = be a rational number such that the prime
= (9 + 8 + 4) – (5 + 9 + 7) = 0. q
\ 749859 is divisible by 11. factorisation of q is not of the form 2n . 5m, where n and
(ii) Consider the number 8192657. m are non negative integers, then x has a decimal
(Sum of digits at odd places) – (Sum of digits at even expansion which is non-terminating repeating .
places) 8. Irrational Numbers : 2 , 3, 5 ,3 3, 2 + 3, p , e are
= (7 + 6 + 9 + 8) – (5 + 2 + 1) = 30 – 8 = 22, which is all irrational numbers. numbers which are expressed as
divisible by 11. non-terminating and non-repeated decimals are called
\ 8192657 is divisible by 11. the irrational numbers.
(iii) Consider the number 5702211 9. Real Numbers are a combination of the rational numbers
(Sum of digits at even places) and the irrational numbers .
– (Sum of digits at odd places)

3-18
 Geometrical Optics

EXERCISE-I

CBSE Q.13 Find the largest number which divides 70 and 125
PREVIOUS YEAR'S leaving remainder 5 and 8 respectively.
[CBSE (Outside Delhi) 2015] [2]

43
Q.1 The decimal expansion of the rational number Q.14 Is 7 × 5 × 3 × 2 + 3 a composite number ? Justify your
2 4 53
answer.
will terminate after how many places of decimals. [CBSE (Outside Delhi) 2016] [2]
[CBSE (Delhi) 2013] [1]
Q.15 Can two numbers have 15 as their HCF and 175 as their
Q.2 Express 2717 as a product of its prime factors? LCM ? Given reasons.
[CBSE (Delhi) 2011][2] [CBSE (Outside Delhi) 2017] [2]

Q.3 Using Euclid’s division algorithm, check whether the Q.16 Find the sum of the first 25 terms of an A.P. whose n th
pair of numbers 50 and 20 are coprime or not. term is given by tn = 2 – 3n. [CBSE (Delhi) 2012] [2]
[CBSE (outside Delhi) 2012] [2]
Q.17 Find the sum of all three digit natural numbers, which
Q.4 How many two-digit numbers are divisible by 3? are multiples of 11. [CBSE (Delhi) 2012] [2]
[CBSE (outside Delhi) 2012] [2]
Q.18 Prove that 3 + 2 3 is an irrational number..
Q.5 Solve the following quadratic equation for x.
[CBSE (Delhi) 2014] [3]
4 3x 2 + 5x - 2 3 = 0 [CBSE (Delhi) 2013] [2]
Q.19 Show that one and only one out of n, (n + 1) and (n + 2)
Q.6 How many three-digit natural nubmers are divisible by is divisible by 3, where n is any positive integer.
7? [CBSE (Delhi) 2013] [2] [CBSE (Outside Delhi) 2015] [3]

Q.7 Find the HCF (865, 255( using Euclid’s division lemma. Q.20 Prove that 3 is an irrational number. Hence prove
[CBSE (Outside Delhi) 2013] [2]
that 3 - 5 is also an irrational number..
Q.8 Find the prime factorisation of the denominator of [CBSE (Delhi) 2016] [3]
rational number expressed as 6.12 in simplest form.
Q.21 Find the HCF and LCM of 306 and 657 and verify that
[CBSE (Delhi) 2014] [2]
LCM × HCF = Product of the two numbers.
[CBSE (Delhi) 2016] [3]
Q.9 Prove that 2 + 3 5 is an irrational number..
[CBSE (Outside Delhi) 2014] [2] Q.22 Show that any positive odd integer is of the form 4q +
1 or 4q + 3 where q is a positive integer.
Q.10 Find the number of natural numbers between 101 and [CBSE (Outside Delhi) 2016] [3]
999 which are divisible by both 2 and 5.
[CBSE (Outside Delhi) 2014] [2] Q.23 Three bells toll at intervals of 9, 12, 15 minutes
respectively. If they start tolling together, after what
Q.11 Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11 ) is a composite time will they next toll together?
number? [CBSE (Delhi) 2015] [2] [CBSE (Delhi) 2013] [3]

Q.12 Check whether 4n can end with the digit 0 for any Q.24 Show that 5 + 3 2 is an irrational number..
natural number n. [CBSE (Outside Delhi) 2015] [2]
[CBSE (outside Delhi) 2012] [3]

1-19
Geometrical Optics

Q.25 Check whether 6n can end with the digit 0 for any
natural number n ? [CBSE (Delhi) 2011][3]

Q.26 Find the largest number that divides 2053 and 967 and
leaves a remainder of 5 and 7 respectively.
[CBSE (Outside Delhi) 2011] [3]

Q.27 Prove that 5 is irrational and hence show that

3+ 5 is also irrational. [CBSE (Delhi) 2012] [3]

Q.28 In a school there are two sections A and B of class X.

There are 48 students in section A and 60 students in
section B.
Determine the least number of book required for the
library of the school so that the books can be
distributed equally among all students of each section.
[CBSE (Outside Delhi) 2017] [3]

Q.29 By using Euclid’s algorithm, find the largest number

which divides 650 and 1170.
[CBSE (Outside Delhi) 2017] [3]
Q.30 Prove that 3 + 2 5 is irrational.
[CBSE (Delhi) 2017] [4]

1-20
 Real Number

EXERCISE-II
Introduction of Numbers and their properties. Q.10 Number 6n can not end with
Q.1 The value of (A) 6 (B) 5
(C) Can not say (D) None of these.
1 3
+
1+ 1 3– 4
3– 4 3+ 1 61
Q.11 The decimal expansion of the rational number
1 2– 1 2 ´ 54
4
2+ 2
1
3– will terminate after:
2
(A) 1 place (B) 2 places (C) 3 places (D) 4 places

13 15 11 17
(A) (B) (C) (D) 97
7 7 21 28 Q.12 The decimal expansion of the rational number
2 ´ 54
Q.2 Find the value of will terminate after;
(A) One decimal place (B) Two decimal place
é 1ù é 1ù é 1ù é 1 ù
ê1 + 2 ú ê1 + 3 ú ê1 + 4 ú . . . . . ê1 + 2007 ú (C) Three decimal place (D) Four decimal place
ë û ë û ë û ë û

4 3 2008 2009 5
(A) (B) (C) (D) Q.13 The number when expressed as a decimal fraction
2007 2005 2 6 210
is
Q.3 The number of perfect squares below 2007 is (A) terminating
(A) 25 (B) 35 (C) 45 (D) 44 (B) recurring
(C) non terminating and non recurring
Q.4 Expressing 3185 as sum of two squares is - (D) none of these
(A) 542 + 52 (B) 562 + 72 (C) 482 + 62 (D) none
Q.14 The decimal fraction 0.1 2 is equal to
Q.5 Evaluate
2[(16 – 15)–1 + 25 (13 – 8)–2]–1 + (1024)° 1 11 14 5
(A) 2 (B) 3 (C) 1 (D) 5 (A) (B) (C) (D)
6 90 99 33
Q.6 If a is any positive integer and p is a prime number such
that p divides a2 then p will also divide Q.15 A rational number, in its lowest term can be expressed
as a terminating decimal iff the denominator has no
1
(A) a (B) a 3 prime factors other than
(A) 2 (B) 5 (C) 2 and 3 (D) 2 and 5
(C) a (D) None of these.

Q.16 A rational number in its lowest term, can be expressed

Q.7 For some integer ‘m’ every even integer is of the form
(A) 2m – 1 (B) 2m + 1 (C) 2(m + 1) (D) m as a non terminating recurring decimal iff the
denominator has prime factors other than
Q.8 For some integer ‘m’, every odd integer is of the form: (A) 2 (B) 5 (C) 2 and 5 (D) 2 and 4
(A) m (B) 2m + 1 (C) m + 1 (D) 3m +1
Q.17 The rational number 2.357 when expressed as the
131 quotient of two co-prime integers is
Q.9 The decimal form of is:
2 .52.71
3
2335 2379
(A) terminating (A) (B)
1001 997
(B) non-terminating non-repeating
(C) non-terminating repeating 2355
(D) none of three (C) (D) none of these
999

3-21
Real Number

Q.18 If x is a non-zero rational number and xy is irrational, Q.29 If a is a positive integer and p be a prime number and p
then y must be divides a2, then
(A) rational number (B) irrational number (A) a divides p (B) p divides a
(C) non zero (D) integer (C) p2 divides a (D) None of these

Q.19 If x is a positive real number, then the least value of

Q.30 If x = 0.16 , then 3x is –
1
x+ is (A) 0.48 (B) 0.49 (C) 0.5 (D) 0.5
x
(A) 2 (B) 4 (C) 11 (D) 1
Q.31 1.3 is equal to –
Q.20 The least positive prime number is (A) 3/4 (B) 2/3 (C) 4/3 (D) 2/5
(A) 1 (B) 3 (C) 2 (D) none of these
Q.32 Between any two distinct rational numbers –
Q.21 If a, b, c are rational numbers and ac = bc Þ a = b, then (A) There lie infinitely many rational numbers.
(A) c ³ 0 (B) c £ 0 (C) c = 0 (D) c ¹ 0 (B) There lies only one rational number.
(C) There lie only finitely many numbers.
Q.22 1.2348 is (D) There lie only rational numbers.
(A) an integer (B) an irrational number
(C) a rational number (D) none of these No. of Factors, Sum of factors, Product of factors
Q.33 Find the number of factors of the number 46200.
(A) 64 (B) 78 (C) 96 (D) 106
3 6 16
Q.23 How many times the HCF, of the fractions , and ,
14 35 21
Q.34 The number of odd factors (or divisors) of 24 is –
is their LCM? (A) 2 (B) 3
(A) 1440 (B) 144 (C) 210 (D) 48
(C) 1 (D) None of these

Circumference of the circle Q.35 The number of even factors (or divisors) of 24 is –
Q.24 p= Diameter of the circle (A) 6 (B) 4
(C) 8 (D) None of these
(A) A rational number (B) A whole number
(C) A positive integer (D) None of these
Q.36 In how many ways can 576 be expressed as a product
of two distinct factors ?
Q.25 HCF(p, q) × LCM (p, q) =
(A) 10 (B) 11
p (C) 21 (D) None of these
(A) p + q (B) (C) p × q (D) pq
q
Divisibility of Numbers
Q.37 The least number which is a perfect square and is
Q.26 HCF(p, q, r) . LCM (p, q, r) = divisible by each of 16, 20 and 24 is -
qr (A) 240 (B) 1600 (C) 2400 (D) 3600
pq
(A) (B)
r p Q.38 25102 – 4102 is divisible by .....
(C) p, q, r (D) None of these (A) 21 (B) 25 (C) 29 (D) 30

Q.27 0.737373… = Q.39 461 + 462 + 463 + 464 is divisible by -

(A) 3 (B) 10 (C) 11 (D) 13
73
(A) (0.73)3 (B)
100 Q.40 The least value of p such that the number 456p 8 is
divisible by 9
73 (A) 2 (B) 3 (C) 4 (D) 5
(C) (D) None of these
99
Q.41 If the number 243 × 51 is divisible by 9, then the value
Q.28 LCM of three numbers 28, 44, 132 is – of digit marked as × would be
(A) 528 (B) 231 (C) 462 (D) 924 (A) 3 (B) 1 (C) 2 (D) 4

3-22
 Real Number

Q.42 The nearest whole number to one million which is Q.53 Which of the following is not an irrational number:
divisible by 537 is -
(A) 3 + 3 (B) 4 + 27
(A) 1000106 (B) 999894
(C) 1000437 (D) 999563
(C) 9 + 27 (D) 26 - 46
Q.43 n2 – 1 is divisible by 8 if n is:
(A) an integer (B) an even integer Q.54 The product of a non-zero rational number and irrational
(C) a natural number (D) an odd integer number is
(A) always irrational (B) always rational
Q.44 The least number that is divisible by all the numbers (C) one (D) rational or irrational
from 1 to 10 (both inclusive) is
(A) 90 (B) 100 (C) 2520 (D) 360
Q.55 ( 6 + 3 3 ) - ( 4 - 5 3 ) is a
Q.45 For any positive integer n, n3 – n is always divisible by
(A) 7 (B) 6 (A) rational number (B) irrational number
(C) 5 (D) None of these (C) natural number (D) an integer

Q.46 If x and y are odd positive integers then x2 + y2 is not Q.56 Irrational numbers are
divisible by (A) terminating decimals
(A) 1 (B) 2 (B) non terminating, recurring decimals
(C) 4 (D) None of these (C) non terminating non recurring decimals
(D) none of these
Euclid’s division lemma and Algorithm
Q.47 The HCF of two numbers, obtained in three steps of
Q.57 3.24636363… is
division, is 7 and the first 3 quotients are 2, 4 and 6
(A) an integer (B) a rational number
respectively. Find the numbers -
(A) 175, 392 (B) 189, 392 (C) an irrational number (D) none of these
(C) 168, 385 (D) None of these
Q.58 If xy is rational, y is irrational and x is rational then
Q.48 Euclid’s division lemma states that for two positive (A) x> 0 (B) x < 0
integers a and b, there exist unique integers q and r (C) x = 0 (D) x¹ 0
such that a = bq + r. where r must satisfy:
(A) 0 £ r < b(B) 1 < r < b (C) 0 < r £ b (D) 0 < r < b Q.59 If x and y are positive real numbers, then

Q.49 If HCF of 210 and 55 is expressed in the form of 210 × 5 (A) x + y > x+y
+ 55y, then value of y2 is
(A) 381 (B) 368 (C) 361 (D) 19 (B) x + y < x+y

Q.50 If the HCF of 85 and 153 is expressible in the form 85 m (C) x + y = x+y
– 153, then value of m is:
(D) none of these
(A) 1 (B) 2 (C) 3 (D) 4

Q.60 2.13113111311113… is
Irrational Number
(A) a rational number (B) an irrational number

( 4)
1
2x+ 1 (C) an integer (D) none of these
Q.51 If 3 2 = , then x =
32
(A) – 2 (B) 4 (C) – 6 (D) – 4 Q.61 3 + 3 is
(A) a rational number (B) non-real number
(C) an integer (D) an irrational number
x 3 ´ 3 x5
Q.52 Evaluate ´ 30 x 77 =
5 3
x Q.62 Which of the following is not an irrational number?
(A) x 76/15
(B) x 78/15
(C) x 79/15
(D) x
77/15
(A) (B) 3 (C) 4 (D) 5
2

3-23
Real Number

Q.70 Find the square root of

6 1 4
Q.63 Simplify : - + 3
2 3+ 6 3- 2 6- 2 ( x + 4 ) + 2 x 2 + 19 x + 35
2
(A) 1 (B) 3

(C) 2 6 (D) 2 - 3 + 6 (A)

1
2
( x + 7 + 2x + 5 )
Q.64 If 2 = 1.414 and 3 = 1.732 find the value of (B)
1
2
( x - 7 + 2x + 5 )
(1 - 2 ) + 3 -1
(C)
1
( x + 7 + 2x + 5 )
3+ 2 2 3 +1 2

( )
(A) 1.076 (B) 0.673 1
(C) 0.198 (D) None of these (D) x + 7 + 2x - 5
2

1 + 4x 1 + 4x
Q.65 Find the value of + , when 1 1
1 + 1 + 4x 1 - 1 + 4 x Q.71 If x + = 4 , find the value of x 4 - 4 ( x > 1)
x x
x = ¼.
(A) 2 (B) – 2 (C) – 4 (D) 4 (A) 112 (B)112 3 (C) 224 (D) 224 3

Q.66 If x = 6 + 7 , then find the value of 1

x3 – 18x2 + 101x – 132 Q.72 is –
3
(A) 42 (B) 68 (C) 77 (D) 84
(A) A rational number (B) An irrational number
(C) a whole number (D) None of these
2+ 3 2- 3
Q.67 Simplify : +
2 + 2+ 3 2 + 2- 3 Q.73 5 - 3 is –
(A) An integer
1 (B) A rational number
(A) 1 (B) (C) An irrational number
2
(D) None of these

(
2 3 3 -1 ) Q.74 If 3
32 = 2 x then x is equal to
3( 3 + 1)
(C) 2 (D)
3 5
(A) 5 (B) 3 (C) (D)
5 3

10 2
Q.68 Simplify : + 30 - 10 5 Q.75 If p is a positive prime integer, then p is –
18 - 3 + 5
(A) A rational number (B) An irrational number
(A) 8 (B) 10 (C) 15 (D) 25 (C) a positive integer (D) None of these

Q.69 One of the values of

-2
æ1ö
x + 2y x - y + x - 2y x - y
2 2 2 2 2 2
is Q.76 Evaluate ç ÷
3

è6ø
(A) y2 + x2 (B) x2 – y2 (C) 2y (D) 2x
(A) 4 (B) 16 (C) 32 (D) 64

3-24
 Real Number

Q.77 The product of 4 6 and 3 24 is – Q.85 If ( 2 + 3 ) 2 = a + b 6 , where a, b Î Q, then –

(A) 124 (B) 134 (C) 144 (D) 154 (A) a = 5, b = 6 (B) a = 5, b = 2
(C) a = 6, b = 5 (D) None of these

1 Surds
If x = (7 + 4 3 ), then the value of x +
2
Q.78 is –
x2
(4 + ) (4 - )
2 3
(A) 193 (B) 194 (C) 195 (D) 196 Q.86 If 15 + 15 = a 5 , then find
a.
a+ 3 2- 3 (A) 5 (B) 7 2 (C) 7 (D)10 2
Q.79 If a = , b= then the value of a +
2- 3 2+ 3
b is – 3 4 6
Q.87 Simplify : 2572
(A) 14 (B) – 14 (C) 8 3 (D) - 3
(A) 4 (B) 5 (C) 1/5 (D) 2

Q.80 p and e are –

(A) Natural numbers (B) Integers 2+ 2 + 6
(C) Rational numbers (D) Irrational numbers. Q.88 Simplify
3+ 2 + 3 + 6
Q.81 If x is a non-zero rational number and xy is irrational ,
1
then y must by – (A) 1 (B) 2 (C) (D) 2
(A) a rational number (B) an irrational number 2
(C) non-zero (D) an integer

Q.82 The arithmetical fraction that exceeds it’s square by 1

the greatest quantity is – Q.89 Simplify : 4+ 2 3 + 3 3+ 2 3 ( )
1 1
(A) (B) (A) 2 - 3 (B) 2 + 3 (C) 3 - 2 (D) 3 +
4 2 2

3
(C) (D) None of these Q.90 Find the square root of 7 – 4 3 .
4
(A) 2 – 3 (B) 5 – 3
15 (C) 5 – 3 (D) None of these
Q.83 is equal to –
10 + 20 + 40 - 125
Q.91 The value of
(A) 5 (5 + 2 ) (B) 5 (2 + 2 )
5+2+ 5–2
– 3 – 2 2 is–
(C) 5 ( 2 + 1) (D) 5 (3 + 2 ) 5 +1

(A) 1 (B) 2 2 – 1
3 -1
Q.84 The expression is equal to –
2 2 - 3 -1 5
(C) (D) None of these
2
(A) 2+ 3+ 4+ 6
(B) 6- 4+ 3- 2 é 2 1ù
2

Q.92 The value of ê(23 + 2 2 ) 3 + (140 - 29) 2 ú is–

(C) 6- 4- 3+ 2 êë úû
(D) None of these
(A) 196 (B) 289 (C) 324 (D) 400

3-25
Real Number

Q.93 The value of æç 6 15 - 2 56 ö÷ . æç 3 7 - 2 2 ö÷ is – ( 3 x +1)

5 x - 37
è ø è ø æ 2ö æ 49 ö 2
Q.100 Solve for x if çç 4 ÷÷ = çç 3 ÷÷
(A) 0 (B) 1 (C) – 1 (D) 2
è 7ø è 4 ø

1
(A) 3 (B) 4
æ æ ö 3 4
(C) 5 (D) None of these
+ 27 3 ö÷ ÷÷ is–
1 1
The value of çç 5ç 8
3
Q.94
è è ø ø
(A) 3 (B) 6 (C) 5 (D) 4 Q.101 2 + 3 + 2 - 3 is equal to –

Q.95 Write the following surds in ascending order 3 2

3
(A) 3 (B) (C) (D) 6
11, 4 7, 5, 2 3
(A) 5, 4 7, 3 11 (B) 4 7, 5, 3 11
Miscellenous
(C) 4 3
7, 11, 5, (D) 5, 11, 73 4 Q.102 The number of numbers are there which are equal to
the sum of the digits of their cubes is -
(A) 5 (B) 6 (C) 7 (D) 8
1
Q.96 is equal to Q.103 Find the greatest four digit number which when divided
2 1/3
-1
by 18 and 12 leaves a remainder of 4 in each case -
(A) 3 2 + 1 (B)1 + 21/3 - 41/3 (A) 9976 (B) 9940 (C) 9904 (D) 9868
(C)1 - 21/ 3 - 41/3 (D)1 + 21/3 + 41/ 3
Q.104 The number of possible pairs of number, whose product
is 5400 and the HCF is 30 is -
Q.97 Find the positive square root of 45 - 40 (A) 1 (B) 2 (C) 3 (D) 4

(A) 5 ( 2 -1 ) (B) 5
1/ 4
( 2 -1 ) 1 1
Q.105 The H.C.F. of 10, 20 , is -
(C) 10 - 2 (D) 5 ( 2 -1 ) (A) 1 (B) 10
10 10 40
(C) 1020 (D) 10–40

Q.106 Find the largest four digit number which when reduced
( x - 3 )( 12 + 8 ) = 1 by 54, is perfectly divisible by all even natural numbers
Q.98 If
( 12 - 8 ) , then find the value of less than 20 -
(A) 5081 (B) 5094 (C) 5001 (D) 5196
x.
Q.107 Three bells, toll at intervals of 36 sec, 40 sec and 48 sec
(A) 5 - 2 3 + 6 (B) 5 + 2 6 + 3 respectively. They start ringing toll at particular time.
They next toll together after -
(C) 5 + 3 - 2 6 (D) 5 - 3 + 2 6
(A) 18 minutes (B) 12 minutes
(C) 6 minutes (D) 24 minutes
Q.99 Arrange the given surds is ascending order;
7 + 17, 8 + 14, 10 + 14 and 3 + 22 Q.108 The sum of the remainders of

(A) 8 + 14, 7 + 17, 10 + 14, 3 + 22 1555 ´1557 ´1559 1022 - 1012

and is -
13 17
(B) 8 + 14, 3 + 22, 10 + 14, 7 + 17
(A) 17 (B) 27 (C) 25 (D) 15
(C) 8 + 14, 3 + 22, 7 + 17, 10 + 14
Q.109 The unit digit of 6100000 is.
(D) 3 + 22, 8 + 14, 7 + 17, 10 + 14
(A) 2 (B) 3 (C) 6 (D) 8

3-26
 Real Number

Q.110 The largest number which divides 70 and 125 leaving

Q.120 If 16 ´ 8 n + 2 = 2 m , then m is equal to –
remainder 5 and 8 respectively is
(A) 11 (B) 13 (A) n + 8 (B) 2n +10 (C) 3n + 2 (D) 3n + 10
(C) 17 (D) None of these.
Q.121 The greatest possible number with which when we
divide 37 and 58, leaves the respective remainder of 2
( )
1/3
Simplify : ( 64 ) ´ ( 81) ´ ( 64 )
-1/ 6 3/ 4 2/3
Q.111 and 3, is -
(A) 2 (B) 5
(A) 2 (B) 6 (C) 16 (D) 22
(C) 10 (D) None of these

11 7
æ x 4 y -5 z 2 ö æ x 2 yz 3 ö Q.122 The largest possible number with which when 60 and
Q.112 Simplify ç -3 4 ÷ ¸ ç -8 2 5 ÷ 98 are divided, leaves the remainder 3 in each case, is –
è x yz ø èx y z ø (A) 38 (B) 18
(C) 19 (D) None of these
x7
(A) 59 8 (B) x 7 y 59 z 8
y z Q.123 The largest possible number with which when 38, 66
and 80 are divided the remainders remain the same is –
y 59 x 7 y 59 (A) 14 (B) 7
(C) (D)
x7 z 8 z8 (C) 28 (D) None of these

Q.113 Solve for x and y :xy = yx and y = x2 Q.124 If x divides y (written as x | y) and y | z, (x, y, z Î z) then–
(A) x = 0, y = 0 (B) x = 0, y = 2 (A) x | z (B) z | y
(C) x = 1, y = 2 (D) x = 2, y = 4 (C) z | x (D) None of these

Q.114 Simplify : Q.125 If x | y, where x > 0, y > 0 (x, y Î z) then –

(2 ) (A) x < y (B) x = y (C) x £ y (D) x ³ y

5/16
1/2 3/ 4 5/6 7 /8 9/10 11/12 13/14 15/16
.4 .8 .16 .32 .64 .128 .256
(A) 64 (B) 256 (C) 512 (D) 1024
Q.126 If a | b, then gcd of a and b is –
(A) a (B) b
Q.115 A number when divided by 928 leaves a remainder 244.
(C) ab (D) Can’t be determined
what would be the remainder when the number is
divided by 58?
(A) 8 (B) 12 (C) 17 (D) 23 Q.127 If gcd of b and c is g and d | b & d | c, then –
(A) d = g (B) g | d
Q.116 If a + b = 11, and ab = 7 then find the value of a 3 + b3 (C) d | g (D) None of these
(A) 1090 (B) 1100
(C) 1120 (D) None of these Q.128 If x Î R, then | x | =
(A) x (B) –x
Q.117 A number when successively divided by 7, 11 and 5 (C) max {x, -x} (D) min {x, -x}
leaves respective remainders of 5, 1 and 1.
Find the smallest such number. Q.129 If a, b, c Î R and a > b Þ ac < bc, then –
(A) 64 (B) 67 (C) 78 (D) 89
(A) c ³ 0 (B) c £ 0

Q.118 Find the largest four – digits numbers which when (C) c > 0 (D) c < 0
successively divided by 3, 4 and 5 leaves respective
remainders 2, 3 and 4. Q.130 If a, b, c Î R and ac = bc Þ a = b, then –
(A) 9985 (B) 9995 (C) 9996 (D) 9959 (A) c ³ 0 (B) c £ 0
(C) c = 0 (D) c ¹ 0
Q.119 Find the remainder when 350 is divided by 11.
(A) 1 (B) 6
(C) 7 (D) Cannot be determined

3-27
Real Number

EXERCISE-III

Q.1 If p, q and r are prime numbers such that r = q + 2 and Q.9 33 ! is divisible by
q = p + 2, then the number of triplets of the form (p, q, r) (A) 210 (B) 215 (C) 220 (D) None
is -
(A) 0 (B) 1 (C) 2 (D) 3 Q.10 If n is an odd natural number, 32n + 22n is always divisible
by
Q.2 The sum of all the factors of 72007 is (A) 13 (B) 5 (C) 17 (D) 19

1 2008 1 2008 Q.11 If 146! is divisible by 5n , then the maximum value of n

(A) (7 - 1) (B) (7 - 1)
2 6 is-
(A) 33 (B) 34 (C) 35 (D) 36
1 2007 1 2007
(C) (7 - 1) (D) (7 - 1)
5 4 Q.12 What will be the remainder when 5382 + 53 is divided by
54 ?
(A) 0 (B) 2 (C) 50 (D) 51
Q.3 The product of all the factors of 52005 is
(A) 51002 × 2004 (B) 52005 × 2006
Q.13 The remainder on dividing
(C) 52006 × 2007 (D) 51003 × 2005
121n – 25n + 1900n – (–4)n by 2000 is-
(A) 1000 (B) 1 (C) 0 (D) 8
Q.4 If n = 10800, then number of divisors of the form
4m + 2 is
Q.14 The remainder, when (1523– 2323) is divided by 19, is:
(A) 6 (B) 12 (C) 18 (D) 22
(A) 4 (B) 15 (C) 0 (D) 18

Q.5 If N = 75 38 54, the sum of divisors of N will be -

Q.15 If 2100 + 3100 + 4100 + 5100 is divided by 7, then the
remainder is
7 4 - 1 38 - 1 54 - 1
(A) ´ ´ (A) 5 (B) 3 (C) 2 (D) 1
7 -1 3 - 1 5 -1
Q.16 If 3243 + 2243 is divided by 5, then the remainder is -
7 - 1 3 -1 5 - 1
6 9 5
(A) 1 (B) 2 (C) 4 (D) 0
(B) ´ ´
7 +1 3 -1 5 -1

(1!) + ( 2!) + ( 3!) + ...... + (100!)

1! 2! 3! 100!
Q.17 If is
7 - 1 3 - 1 5 -1
4 7 5
(C) ´ ´
7 +1 3 +1 5 +1 divided by 5, then the remainder is
(A) 2 (B) 0 (C) 4 (C) None
(D) none of these
Q.18 The remainder when
Q.6 The total number of divisors of 10500 except 1 and
1! + 2! + 3! + 4! + ..... + 100! is divided by 120 is
itself is – (A) 1
(A) 48 (B) 50 (C) 46 (D) 56 (B) 7
(C) 33
Q.7 The sum is the factors of 19600 is – (D) cannot be determined
(A) 54777 (B) 33667
(C) 5428 (D) None of these Q.19 A hundred digit number is formed by writing first 54
natural numbers in front of each other as
Q.8 The product of divisors of 7056 is – 1234578910111213............ Find the remainder when this
(A) (84)48 (B) (84)44 number is divided by 8.
(C) (84)45 (D) None of these (A) 1 (B) 7 (C) 2 (D) 0

3-28
 Real Number

Q.20 What is the unit’s digit of the remainder when (1923 + Q.28 If k
1723) is divided by 36?
æ 16 1 öæ 16 1 öæ 8 1ö
(A) 5 (B) 8 (C) 7 (D) 0 çç a + a ÷ç÷ç a - a ÷ç÷ç a + a ÷÷
16 16 8

è øè øè ø

Q.21 Which of the following numbers p2 , 3

0.8 , æ4 1 öæ 1ö
çç a + 4 ÷÷ çç a + ÷ = 1 , then find k.
è a øè a ÷ø
4
0.00016 , 3
–1 , 0.001 is/are rational?

(A) 3 – 1 (B) p2 a a
(A) (B)
(C) 0.001 (D) All of these a +1
2
a -1
2

a2 a2
a (C) (D)
Q.22 If 0.4 ´ .4 ´ a = 0.04 × .4 × b , then b is- a2 + 1 a2 -1

(A) 16 × 10–3 (B) 16 × 10–4 Q.29 If x, y, z are real numbers such that
(C) 16 × 10–5 (D) None of these
x - 1 + y - 2 + z - 3 = 0 then the values of x, y, z
are respectively
Q.23 If x and y are rational number such that xy is irrational,
(A) 1, 2, 3 (B) 0, 0, 0
then x + y is (C) 2, 3, 1 (D) None of these
(A) rational (B) irrational
Q.30 Let a, b, c, p be rational numbers. Such that p is not a
(C) non real (D) none of these
perfect cube.
If a + bp1/3 + cp2/3 = 0 then.
1 1+ x + 1 - x (A) a = b = c (B) a = b ¹ c
Q.24 If x = , then find the value of (C) a ¹ b = c (D) a ¹ b ¹ c
2 1+ x - 1- x

(A)
1
2
(2 + 6 ) (B)
1
2
( 2+ 6 ) Q.31 11 11 11.......4 terms =
(C) 2 + 1 (D) 3 + 1
(A) 16
115 (B) 16 11

(C) 16 1114 (D) 16 1113

5 +1
Q.25 If x = , then find the value of 7x2 – 7x
5 -1
Q.32 If p + q + r = 0, then find the value of
(A) 4 (B) 7
qr 2 rp 2 pq 2
(C) 6 (D) None of these a p . aq . ar
(A) 1 (B) a3
(C)apqr (D) None of these
3+ 2 2 - 3- 2 2 x
Q.26 If =
3+ 2 2 + 3- 2 2 32 , then find the value of
12
x. Q.33 If = x + a 2 + b 5 + c 10 and
(A) 2 (B) 4 (C) 8 (D) 16 3+ 5 - 2 2
x, a, b and c are rational, then find the value of
Q.27 Find the value of x, if 3 x
x
= 81 x + a + b + c.
(A) 2 (B) 4
(A) 2 (B) 4 (C) 8 (D) 16 (C) 7 (D) None of these

3-29
Real Number

Q.44 The unit (ones) digit in the product

1æ 1 ö x2 - 1
Q.34 If x = ç 5 + ÷ , then find the value of (5 + 1) (52 + 1) (53 + 1) ... (52005 + 1) is -
2è 5ø x - x2 -1
(A) 1 (B) 2 (C) 5 (D) 6
(A) – 1 (B) 1 (C) – 2 (D) 2
129
Q.45 The last digit of 7 23 is -
Q.35 If 11 11 11.............¥ = 14641x , then find x.
(A) 7 (B) 9
(A) 11 (B) 4 (C) 1 (D) 3
(C) 1/4 (D) None of these
Q.46 The number of zeros at end of 2005 ! is
(A) 250 (B) 300
x- y x-y
Q.36 If + = 4 , then find the value (C) 410 (D) 500
x+ y x- y

of x. Q.47 Total number of zeros at the end of (5n – 1)! is

(A) 1
(B) 4 5n + 4n + 1 5n - 4n - 1
(A) (B)
(C) 6 4 4
(D) Cannot be determined uniquely
5n + 4n -1
Q.37 If 2200 – 2192. 31 + 2n is a perfect square, then n = (C) (D) None
4
(A) 190 (B) 198
(C) 200 (D) 206
Q.48 If 2x = 7y = 14z, then find the value of z in terms of x
Q.38 The natural number n for which and y.
39 + 312 + 315 + 3n is a perfect cube is
x+ y xy
(A) 11 (B) 12 (C) 13 (D) 14 (A) (B)
x-y x+ y
Q.39 The exponent of 6 in 33 ! is (C) x + y + xy (D)xy – ( x + y)
(A) 12 (B) 15 (C) 20 (D) 24
Q.49 If x = 4 + 4 1/3 + 4 2/3 then, find the value of
Q.40 If p = 3 and q = 2, the x3 – 12x2 + 36x + 8.
(3p – 4q)q– p ¸ (4p – 3q)2q – p =
(A) 24 (B) 32 (C) 36 (D) 44
1 2
(A) 1 (B) 6 (C) (D)
6 3 Q.50 If x and y are integers and 2x. 32y = 144, then what is the
value of ( x + y)?

Q.41 The unit’s place of the number represented by (A) 4 (B) 5 (C) 6 (D) 7
(795 – 358) is -
(A) 0 (B) 4 (C) 6 (D) 7
5x 2x
Q.51 If = 125, and = 2, then find the value of
Q.42 The unit digit of 5y 4y
1781 + 2781 + 3781................+ 9781 ?
(A) 1 (B) 3 (C) 5 (D) 7 æ 32 x ö
ç y + 4÷
è 27 ø
Q.43 The unit digit of 1 ! + 2 ! + 3 ! + 4 ! + .... + 1997 is
(A) 0 (B) 1 (C) 2 (D) 3 (A) 17 (B) 64 (C) 85 (D) 121

3-30
 Real Number

Q.52 If (1.761)x = (0.1761)y = 10z, then find the relationship Q.61 What is the least possible number which when divided
between x, y and z. by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5
respectively ?
1 1 1 1 1 1
(A) = - (B) = + (A) 39 (B) 48
z x y z x y (C) 59 (D) None of these

1 1 1 1 1 1
(C) = - (D) = + Q.62 How many numbers lie between 11 and 1111 which
y z x y x z when divided by 9 leave a remainder of 6 and when
divided by 21 leave a remainder of 12 ?
Q.53 If x is a positive integer, then the greatest number with (A) 18 (B) 28
which 5x + 5x+1 + 5x+ 2 would be always divisible is (C) 8 (D) None of these
(A) 31 (B) 155
(C) 225 (D) None of these Q.63 If x, y Î R and | x | + | y | = 0, then –
(A) x > 0, y < 0
Q.54 From a certain city, buses start for four different places (B) x < 0, y > 0
every 15, 20, 25 and 30 minutes starting from 8 am. At (C) x = 0, y = 0
what time, for the first time after 8 a.m ., would all the (D) None of these
buses start together again?
(A) 10 a .m (B) 12 noon Q.64 If a, b, c Î R and a2 + b2 +c2 = ab + bc + ca, then –
(C) 1 p.m (D) 2 p.m (A) a = b = c
(B) a = b = c = 0
Q.55 Find the unit’s digit in 57867192567 – 1452876. (C) a, b, c are distinct
(A) 3 (B) 6 (C) 5 (D) 7 (D) None of these

Q.56 What is the last digit of 424782 + 179137? Q.65 If x, y Î R and x < y Þ x2 > y2 then –
(A) 7 (B) 6 (C) 5 (D) 4
(A) x > 0 (B) y > 0
(C) x < 0 (D) y < 0
Q.57 What is the least possible number which when divided
by 24, 32 or 42 in each case it leaves the remainder 5 ?
(A) 557 (B) 677 Q.66 If x, y Î R and x > y Þ | x | > | y | , then –(A)(B)(C)(D)
(C) 777 (D) None of these (A) x > 0 (B) y > 0
(C) x < 0 (D) y < 0
Q.58 In Q.N. 4, how many numbers are possible between 666
and 8888 ? Q.67 If x, y Î R and x > y Þ | x | < | y | , then –
(A) 10 (B) 11 (C) 12 (D) 13 (A) x < 0 (B) x > 0
(C) y > 0 (D) y < 0
Q.59 What is the least number which when divided by 8, 12
and 16 leaves 3 as the remainder in each case, but when Q.68 If a, bÎ R and a < b, then –
divided by 7 leaves no remainder ?
(A) 147 (B) 145 1 1
(A) <
(C) 197 (D) None of these a b

Q.60 What is the least possible number which when divided 1 1

by 18, 35 or 42 leaves 2, 19, 26 as the remainders (B) >
a b
respectively?
(A) 514 (B) 614 (C) a2 > b2
(C) 314 (D) None of these (D) Nothing can be said

3-31
Real Number

EXERCISE-IV

NTSE STAGE-1 a b 1 1
PREVIOUS YEAR'S Q.9 If = , the value of + is :
b c b-c b-a
Q.1 The solution set of the equation x2/3 + x1/3 – 2 = 0 : [NTSE WB 2014-15]
[NTSE Chandigarh 2014-15]
1 1
(A) {8, 1} (B) {8, –1} (A) (B)
b a
(C) {–8, –1} (D) {–8, 1}
1 1
Q.2 The HCF of two expressions p and q is 1. Their LCM is (C) (D)
ab c
: [NTSE Chandigarh 2014-15]
(A) p + q (B) p – q
1 1 1
1 æ x b ö bc æ x c ö ca æ x a ö ab
(C) pq (D)
pq Q.10 The value of ç c ÷ . ç a ÷ . ç b ÷ is equal to :
èx ø èx ø èx ø
6
Q.3 Equivalent of is : [NTSE Chhatisgarh 2014-15] [NTSE Raj. 2014-15]
20
(A) 1 (B) –1
(A) 6% (B) 20%
(C) 0 (D) abc
(C) 26% (D) 30%
Q.11 The HCF of any two prime numbers a and b, is :
Q.4 3
64 = ? [NTSE Chhatisgarh 2014-15] [NTSE Raj. 2014-15]
(A) 2 (B) 4 (A) a (B) ab
(C) 6 (D) 8 (C) b (D) 1

Q.12 The total two-digit numbers which are divisible by 5,

Q.5 If a = 23, b = 27 and c = 50, the value of a3 + b3 – c3 + 3abc
are : [NTSE Raj. 2014-15]
is : [NTSE WB 2014-15]
(A) 17 (B) 18
(A) 100 (B) 73
(C) 19 (D) 20
(C) 77 (D) 0

Q.13 Let N be the set of natural numbers and P be the set of

Q.6 The identity a. b = ab is not true only when : prime integers in N. If A = {n/nÎN, n is a multiple of
[NTSE WB 2014-15] some prime pÎP}, then N – A = {n Î N/n Ï A} is :
(A) a > 0; b > 0 (B) a < 0; b < 0 [NTSE Tamilnadu 2014-15]
(C) a > 0; b < 0 (D) a < 0; b > 0 (A) Empty set
(B) Of cardinality 2
(C) A finite set of cardinality greater than 2

Q.7 The simplified value of

( 12 - 8 )( 3+ 2 ) is : (D) A singleton set

5 + 24
Q.14 The sum of the first k natural numbers is A, for a certain
[NTSE WB 2014-15]
k > 1; the sum of their cubes is B, then log A
B is :
(A) 6 -2 (B) 2 - 6
[NTSE Tamilnadu 2014-15]
(C) 0 (D) 1 (A) 4 (B) 3
(C) 2 (D) 1
Q.8 If x = 3 3, y = 4 4, z = 6 6 , then :
Q.15 The unit digit in the decimal expansion of 725 is :
[NTSE WB 2014-15]
[NTSE Tamilnadu 2014-15]
(A) x < y (B) y < x
(A) 1 (B) 3
(C) z < x (D) x = y = z
(C) 5 (D) 7

3-32
 Real Number

Q.16 If the sum S of three consecutive even numbers is a

perfect square between 200 and 400, then the square 1
Q.23 Simple form of is ....: [NTSE MP 2014-15]
root of S is : [NTSE Tamilnadu 2014-15] 1
3-
(A) 15 (B) 16 1
2-
7
(C) 18 (D) 19

13 32
1 (A) (B)
32 13
Q.17 If x = 3 + 8 , then x4 + 4 is :
x
7 13
[NTSE Tamilnadu 2014-15] (C) (D)
13 7
(A) 1056 (B) 1158
(C) 1156 (D) 1154
Q.24 The value of the following is :

(0.44) 2 + (0.06) 2 + (0.024) 2

Q.18 The value of 8 + 2 8 + 2 8 + 2 8 + 2 8 is : [NTSE MP 2014-15]
(0.044) 2 + (0.006) 2 + (0.0024) 2
[NTSE Odisha 2014-15] (A) 0.100 (B) 0.01
(A) 4 (B) 6 (C) 100 (D) 1
(C) 8 (D) 10
Q.25 The product of two numbers is 110 and the sum of their
squares is 264, then the sum of these numbers will be :
Q.19 The ascending order of 2, 3 4, 4 6 is :
[NTSE MP 2014-15]
[NTSE Odisha 2014-15] (A) 22 (B) 24
(A) 2, 3 4, 4 6 (B) 2, 4 6, 3 4 (C) 20 (D) 28

(C) 3 4, 2, 4 6 (D) 4 6, 3 4, 2 Q.26 The sum of the squares of two consecutive natural
numbers is 313. Then the numbers will be :
Q.20 The value of [NTSE MP 2014-15]
(A) 12, 13 (B) 13, –12
1 1 1 1
S= + + + ....... + is : (C) 12, –13 (D) –12, –13
2 ´ 7 7 ´ 12 12 ´17 252 ´ 257
[NTSE Odisha 2014-15] Q.27 The sum of two numbers is 100 and one number is two
less than twice the other number. Then the numbers
5 25
(A) (B) are : [NTSE MP 2014-15]
257 2 ´ 257
(A) 34, 66 (B) 24, 76
35 51 (C) 44, 56 (D) 46, 54
(C) (D)
2 ´ 257 2 ´ 257
Q.28 If the L.C.M. of two numbers is 2520 and H.C.F. is 12.
Its one number is 504, then the other number will be:
Q.21 999 ´1000 ´1001´ 1002 + 1 is : [NTSE MP 2014-15]
[NTSE Odisha 2014-15] (A) 50 (B) 65
(A) 1001001 (B) 1001999 (C) 40 (D) 60
(C) 1000999 (D) 1000989
Q.29 The sum of first n natural numbers is :
[NTSE MP 2014-15]
5 –2 5+2
Q.22 Simple form of + is :
5+2 5 -2 n n(n + 1)
(A) (B)
[NTSE MP 2014-15] 2 2

(A) 9 + 5 (B) 18 n +1
(C) (D) n + 1
2
(C) 18 + 5 (D) 9

3-33
Real Number

Q.38 The valuye of

1 1
If x = then the value of x - 2 is :
2
Q.30
2- 3 x (10 4 + 324) (22 4 + 324) (34 4 + 324) (46 4 + 324) (584 + 324)
[NTSE Karnataka 2014-15] (4 4 + 324) (16 4 + 324) (284 + 324) (40 4 + 324) (52 4 + 324)

(A) 12 3 (B) 8 3 is : [NTSE AP 2014-15]

(A) 324 (B) 400
(C) 14 (D) 12
(C) 373 (D) 1024

Q.31 The smallest number which when increased by 17 is

Q.39 Let x = 0.123456789101112................ 998999 where the
exactly divisible by both 520 and 468 is :
digits are obtained by writing the integers 1 through
[NTSE Karnataka 2014-15] 999 in order. Then the 2014th digit to right of the decimal
(A) 4697 (B) 4656 point is : [NTSE AP 2014-15]
(C) 4663 (D) 4680 (A) 7 (B) 6
(C) 5 (D) 9
1 1 1 1
Q.32 If x + = 5 , then x 3 - 5x 2 + x + 3 - 2 + = ...... Q.40 The value of
x x x x
[NTSE Bihar 2014-15] (20142 - 2020) (2014 2 + 4028 - 3) (2015)
(A) –5 (B) 0 is :
(2011) (2013) (2016) (2017)
(C) 5 (D) 10
[NTSE AP 2014-15]
(A) 2014 (B) 2015
Q.33 Given 5A9 + 3B7 + 2C8 = 1114, then the maximum value
(C) 2016 (D) 2017
of C is : [NTSE Bihar 2014-15]
(A) 5 (B) 7
3
(C) 8 (D) None of these æ 3 1 ö
Q.41 If x + x + 1 = 0, then what is the value of ç x + 3 ÷ ?
2
è x ø
Q.34 Find the smallest positive number from the numbers [NTSE AP 2014-15]
below : [NTSE AP 2014-15] (A) –8 (B) –1
(A) 10 - 3 11 (B) 3 11 - 10 (C) 0 (D) 1
Ans. (Bonus)
(C) 51 - 10 26 (D) 18 - 5 13
Q.42 If x + y = 1 then x3 + y3 + 3xy = ............... :
[NTSE Jharkhand 2014-15]
Q.35 If x = 9ab where a is an integer consists of a sequence (A) 0 (B) 1
of 2014 eights and the integer b consists of a sequence
(C) 2 (D) None of these
of 2014 fives. What is the sum of the digits of x ?
[NTSE AP 2014-15]
Q.43 The difference between a two digit given number and
(A) 9000 (B) 18135
the number obtained by interchanging the digits is 27.
(C) 18000 (D) 8585 The sum of the two digits is ............:
Ans. (Bonus) [NTSE Jharkhand 2014-15]
Q.36 If a2 + b2 + 2c2 – 4a + 2c – 2bc + 5 = 0 then the possible (A) 3 (B) 5
value of a + b – c : [NTSE AP 2014-15] (C) 7 (D) Cannot be found
(A) 1 (B) 2
(C) –1 (D) –2 Q.44 If 13 + 23 + ........... + 93 = 2025 then (0.11)3 + (0.22)3 +
.......... (0.99)3 will be : [NTSE UP 2014-15]
Q.37 a and b are both 4 digit numbers a > b and one is (A) 0.2695 (B) 2.695
obtained from the other by reversing the digits. Then (C) 3.695 (D) 0.3695
a + b b -1
the value of b if = is:
5 2 Q.45 If 2x–1 + 2x+1 = 320 then the value of x is :
[NTSE AP 2014-15] [NTSE UP 2014-15]
(A) 6 (B) 8
(A) 2003 (B) 1002
(C) 5 (D) 7
(C) 2005 (D) 2015

3-34
 Real Number

Q.53 The H.C.F. of expression (x + 1) (x –1)2 and (x +1)2

1 1 will be :
Q.46 If x + = 2 then x+ (x –1) is : [NTSE UP 2015-16]
x x (A) (x + 1) (x – 1) (B) (x + 1)2
[NTSE UP 2014-15] (C) (x – 1)2 (D) (x + 1)2 (x – 1)2
(A) 2 (B) 2
Q.54 If a, b and c are any positive real number then the value
(C) 2 + 1 (D) 1
of a -1b. b-1c. c -1a : [NTSE UP 2015-16]

Q.47 If x + y = 8, xy = 15, then the value of x 2 + y2 will be: (A) 1/2 (B) 0
[NTSE UP 2014-15] (C) 1 (D) –1
(A) 32 (B) 34
(C) 36 (D) 38 Q.55 If a nubmers is divided by 6, remainder is 3 then what
will be the remainder when t he square of the same
numbers is divided by 6 again : [NTSE UP 2015-16]
1 (A) 0 (B) 1
Q.48 If x = , then the value of x2 + 2x + 3 :
1+ 2 (C) 2 (D) 3
[NTSE Delhi 2014-15]
(A) 3 (B) 0 æ 2 1 ö
(C) 4 (D) 1 Q.56 ( )
If x = 3 + 8 , then ç x + 2 ÷ will be :
è x ø
[NTSE UP 2015-16]
Q.49 The value of
(A) 38 (B) 36
éæ 1 ö æ 2 ö æ n öù (C) 34 (D) 30
êç1 - n + 1 ÷ + ç1 - n + 1 ÷ + .......... + ç1 - n + 1 ÷ ú is :
ëè ø è ø è øû
[NTSE Delhi 2014-15] æaö
x -1
æbö
x -3
Q.57 If ç ÷ =ç ÷ then the value of x will be :
n èbø èaø
(A) n (B)
2 [NTSE UP 2015-16]
(C) n + 1 (D) 2n (A) –1 (B) 1
(C) 2 (D) 3
Q.50 If (12 + 22 + 32 + ..... + 122) = 650, then the value of
(22 + 42 + 62 + ...... + 242) is : Q.58 For how many values of n (where n is an integer), the
[NTSE Rajasthan 2015-16]
8(n 2 - 3n 2 + 5)
(A) 13000 (B) 2600 expression is an integer :
2n - 1
(C) 2500 (D) 42250
[NTSE Punjab 2015-16]
(A) 8 (B) 4
b2 b 2+ 2 ab a 2 - b2
Q.51 The square root of x x x is : (C) 11 (D) 28
[NTSE Rajasthan 2015-16]
a+b Q.59 The value of 4 3 + 128 - 72 3 is :
2(a + b)
(A) x (B) x 2
[NTSE Punjab 2015-16]
(a + b)2
(C) x 2 (D) xa+b (A) 9 3 (B) 9

(C) 9 + 3 (D) 9 - 3
Q.52 If the difference of two numbers is 5 and difference of
their squares is 300, then sum of the nubmers is:
[NTSE Rajasthan 2015-16] Q.60 If x– 2 x = 3 , then the value of x is :
(A) 1500 (B) 6 [NTSE West Bengal 2015-16]
(C) 12 (D) 60 (A) 1 (B) 3
(C) 9 (D) –1

3-35
Real Number

Q.61 The value of 5 - 2 6 is : Q.69 (1´ 2 ´ 3 ´ 4) + 1 = 5, (2 ´ 3 ´ 4 ´ 5) + 1 = 11,

[NTSE West Bengal 2015-16] (3 ´ 4 ´ 5 ´ 6) + 1 = 19 etc, t he vale of

(A) ± ( 3- 2 ) (B) 3 - 2 (50 ´ 51´ 52 ´ 53) + 1 is:[NTSE Telangana 2015-16]

(A) 2415 (B) 2661
(C) 2 - 3 (D) All of the above
(C) 2165 (D) 2225

Q.62 Among the numbers 2250, 2300, 2150 and 5100 , the greatest Q.70 If 23x = 64 –1 and 10 y = 0.01, then the value of
is : [NTSE Karanataka 2015-16] (50x)–1 × (10y)–1 is : [NTSE Telangana 2015-16]
(A) 2 250
(B) 3200 (A) – 1 (B) 1
(C) 4150 (D) 5100 (C) 1/2 (D) 2

Q.71 Given that a2 + 6b = – 14, b2 + 8c + 23 = 0 and

Q.63 8 + 2 15 - 8 - 2 15 is : c2 + 4a – 8 = 0, then the value of ab + bc + ca is :
[NTSE Karanataka 2015-16] [NTSE Telangana 2015-16]
(A) 26 (B) 32
(A) 2 15 (B) 8 (C) 16 (D) 25
(C) 12 (D) 5
3x 1 27 x
Q.72 If = , then the value of is :
1 + 3x 3 1 - 27 x
Q.64 Sum of the squares of two consecutive odd numbers [NTSE Telangana 2015-16]
added by 6 is always divisible by : (A) 2 (B) 3/4
[NTSE Karanataka 2015-16] (C) 1/7 (D) 1
(A) 5 (B) 6
(C) 8 (D) 9
2
æ6 3ö
Q.73 Simplify çç 27 - 6 ÷÷ :
Q.65 If x = (123456789) (76543211) + (23456789)2, then the è 4ø
number of zeros in 4
x is : [NTSE Telangana 2015-16]
[NTSE Telangana 2015-16]
3
(A) 5 (B) 4 (A) 3/4 (B)
2
(C) 6 (D) 2
3 3 3
(C) (D)
Q.66 30, 72 and x are three integers. Such that the product of 4 2
any two of them is divisible by the third, then the least
value of x is : [NTSE Telangana 2015-16] Q.74 For positive integers a, b, c and d have a product of 8!
(A) 60 (B) 40 and satisfy: [NTSE Telangana 2015-16]
(C) 96 (D) 84 ab + a + b = 524
bc + b + c = 146
Q.67 418 – 1 = 687194a6735, then the value of digit ‘a’ is: cd + c + d = 104
[NTSE Telangana 2015-16] then find the value of a – d (8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8)
(A) 7 (B) 8 (A) 4 (B) 6
(C) 5 (D) 3 (C) 8 (D) 10

Q.68 If xyz + xy + yz + x + y + z = 384 where x, y, z are positive n

integers, then the value of x + y + z is : Q.75 For how many integers n is the square of an
20 - n
[NTSE Telangana 2015-16] integer : [NTSE Telangana 2015-16]
(A) 23 (B) 18 (A) 1 (B) 2
(C) 20 (D) 35 (C) 3 (D) 4

3-36
 Real Number

Q.76 If a, b and c are positive real nubmers, such that a Q.84 If 4x1 = 5 , 5x 2 = 6 , 6x 3 = 7 , ......, 127x124 = 128 , what
( b + c = 152, (c + a) = 162 and c(a + b) = 110, then abc is is the value of x1 · x2· x3......x124 :
[NTSE Telangana 2015-16] [NTSE AP 2015-16]
(A) 672 (B) 688 (A) 2 (B) 5/2
(C) 704 (D) 720 (C) 3 (D) 7/2

Q.77 If 418 – 1 = 687194a6735, then the value of a is _____: 2

+8
Q.85 The sum of th digits in (10 4n + 1) 2 , where n is a
[NTSE AP 2015-16]
positive integer, is _____: [NTSE AP 2015-16]
(A) 6 (B) 3
(A) 4 (B) 4n
(C) 7 (D) 5
(C) 2 + 2n (D) 4n2

Q.78 If xyz + xy + xz + yz + x + y + z = 384, where x, y, z are

positive integers, the find the value of x + y + z : 32 23
Q.86 If = m , then the value of m is ____ :
[NTSE AP 2015-16] 500 5
(A) 20 (B) 17 [NTSE Tamilnadu 2015-16]
(C) 25 (D) 15 (A) 2 (B) 3
(C) 4 (D) 0
Q.79 If x = (123456789) (76543211) + (23456789)2 , the find the
number of zeros in 4
x. [NTSE AP 2015-16] 1 1
Q.87 If x2 + 2 = 23, x > 0 then x + is ____ :
(A) 3 (B) 4 x x
(C) 5 (D) 1 [NTSE Tamilnadu 2015-16]
(A) 2 (B) 3
(C) 4 (D) 5
1 1
Q.80 Express 1– + in the form a + b 3 , where
1 + 3 1- 3 Q.88 The rational form of 0.24 is ____:
a and b are rational numbers, then t he values of a and [NTSE Tamilnadu 2015-16]
b are _____: [NTSE AP 2015-16]
24 8
(A) 1,2 (B) 1, –1 (A) (B)
100 33
(C) 3, 1 (D) 2, 1
24 0.24
(C) (D)
Q.81 The sum of eleven consecutive Integers is 2002. What 1000 100
is the smallest of these integers?
[NTSE AP 2015-16] x y 1
(A) 163 (B) 177 Q.89 If P = ,Q= then the value of =
x+y x+y P-Q
(C) 145 (D) 156
2Q
is ______ : [NTSE Talmilnadu 2015-16]
Q.82 The G.C.D of the numbers 2100 – 1 and 2120 – 1 is ____ P - Q2
2

[NTSE AP 2015-16]
x+y
(A) 2 – 1
60
(B) 2 – 1
20
(A) x - y (B) 0
(C) 220
(D) 210
x-y
(C) 1 (D) x + y
Q.83 The First term of a sequence is 2005. Each succeeding
term is the sum of the cubes of the digits of the previous
term. What is the 2005th term of the sequence?: Q.90 The GCD of (x3 – 1) and (x4 – 1) is :
[NTSE AP 2015-16] [NTSE Talmilnadu 2015-16]
(A) 250 (B) 125 (A) x3 – 1 (B) x2 + 1
(C) 175 (D) 100 (C) x3 – 1 (D) x – 1

3-37
Real Number

Q.91 The LCM of a3b2 , abc is : Q.99 The sum of seven consecutive natural numbers is 1617.
[NTSE Talmilnadu 2015-16] How many of these are prime?
(A) c (B) a4b3c [NTSE Delhi 2015-16]
(A) 4 (B) 5
(C) ab (D) a3b2c
(C) 2 (D) 7

Q.92 Which the smallest number that devises 1512 is the

Q.100 If 2m– 2m–1– 4 = 0/ Then value of mm is :
complete the cube : [NTSE Gujarat 2015-16]
[NTSE Delhi 2015-16]
(A) 5 (B) 6 (A) 4 (B) 27
(C) 7 (D) 9 (C) 6 (D) 29

Q.101 Given that a2 + b2 = 1, c2 + d2 = 1, p2 + q2 = 1, where

5
Q.93 Which number is the inverse of the opposite of – ? a·b·c·d·p·q are all real numbers, then :
8
[NTSE Delhi 2015-16]
[NTSE Gujarat 2015-16] (A) ab + cd + pq £ 1 (B) ab + cd + pq ³ 3
5 3 1
(A) 8 (B) 1 (C) ab + cd + pq < 3 (D) ab + cd + pq £
8 5 2

2 8
(C) 2 (D) - xy xz yz
5 5
Q.102 If = a, = b and = c where a, b, c are
x+y x +z y+z

Q.94 What is value of (a10)3 + (a6)5? nonzero nubmers, t hen t he value of x ? :

[NTSE Jharkhand 2015-16]
[NTSE Gujarat 2015-16]
(A) a (B) a2 2abc 2abc
(A) (B)
(C) 1 (D) 1/a ab + ac - bc ac + bc - ab

abc 2abc
(C) (D)
Q.95 If x = 4 16 + 4 625 than what is x = ? : ab + bc + ac ab + bc - ac
[NTSE Gujarat 2015-16]
(A) 7 (B) 29 Q.103 If = 6 + 5 ; b = 6 - 5 , then find the value of
(C) 12 (D) 5 2a2 – 5ab + 2b2 ? [NTSE Maharastra 2015-16]
(A) 36 (B) 37
(C) 39 (D) 41
x x
Q.96 If + 7 = 15 - then find the solution?
3 5
( 3 + 1)
[NTSE Gujarat 2015-16] Q.104 If x = then the value of 4x3 + 2x2 – 8x + 7 is:
2
(A) x = 20 (B) x = 15
[NTSE Chandigarh 2015-16]
(C) x = 21 (D) x = 18
(A) 8 (B) 10
(C) 15 (D) 14
Q.97 The HCF is 6x2y2 and LCM 36x3y2 of two terms. If one
term is 12x3y2 what is the second term? : Q.105 The sum of ‘n’ terms of series
[NTSE Gujarat 2015-16]
æ 1ö æ 2ö æ 3ö
(A) 18 x y (B) 18 x3y2 ç 1 - ÷ + ç1 - ÷ + ç1 - ÷ + ____ will be -
2 2

è nø è nø è nø
(C) 36 x2y3 (D) 36 x3y2
[NTSE Chandigarh 2015-16]
Q.98 If x = (5)1/3, then value of x3 – 6x2 + 12x – 10 is : 1 1
(A) (n + 1) (B) (n - 1)
[NTSE Delhi 2015-16] n n
(A) 1 (B) – 2
1 1
(C) – 1 (D) 3 (C) (n + 1) (D) (n - 1)
2 2

3-38
 Real Number

Q.106 If xa = y, yb = z, zc = x then the correct statement will be:

4 2 1
[NTSE Chandigarh 2015-16] Q.113 The value of - + is :
9 9 9
(A) a.b.c = 1 (B) a.b.c = 0
[NTSE Haryana 2015-16]
(C) a + b + c =1 (D) a + b + c = 0
1
(A) 3 (B) 3 3
3
Q.107 The value of 10 + 25 + 108 + 154 + 225
3
3 3
will be : [NTSE Chandigarh 2015-16] (C) (D) 3
3
2 +1 2 +1
(A) 4 (B) 6
(C) 8 (D) 10
Q.114 If x = 5 + 21 and y = 5 - 21 then value of
Q.108 If x = 2 + 21/3 + 22/3 then x3 – 6x2 + 6x : (x – y) is : [NTSE Haryana 2015-16]
[NTSE Punjab 2015-16] (A) 2 21 (B) 10
(A) 2 (B) 1
(C) 6 (D) ± 6
(C) 4 (D) 3

Q.115 Let x be the greatest number by which if we divide 366,

Q.109 In a school the ratio of boys and girls in class VIII,
513 and 324, then in each case the remainder is the
class IX and class X are respectively 3 : 1, 5 : 3 and 7 : 5.
same. The sum of digits of x is :
If the number of students in each class is same, then
[NTSE Haryana 2015-16]
find the ratio of boys and girls in the school.
(A) 3 (B) 4
[NTSE Odisha 2015-16]
(C) 5 (D) 7
(A) 15 : 9 (B) 5 : 3
(C) 27 : 20 (D) 47 : 25
Q.116 If a number m is divided by 5 leaves a remainder 2,
while another number n is divided by 5 leaves a
Q.110 If pqr = 1, then the value of remainder 4, then the remainder, when (m + n) is divided
by 5 is : [NTSE Haryana 2015-16]
æ 1 1 1 ö
çç + + ÷ is : (A) 1 (B) 2
è 1+ p + q 1+ q + r
-1
1 + r + p ÷ø
-1
(C) 3 (D) 4
[NTSE Odisha 2015-16]
(A) 0 (B) pq 4xy a + 2x a + 2y
Q.117 If a = , the value of - + in most
1 x+y a 2x a - 2y
(C) 1 (D)
pq simplifid form is : [NTSE Delhi 2016-17]
(A) 0 (B) 1
(C) –1 (D) 2
Q.111 In a two digit number, the number of ten’s place is double
of the number of unit’s place. If we exchange the
Q.118 The value of the following expression is :
numbers mutually then the number decreases by 18,
then the number is - [NTSE MP 2015-16] é 1 ù é 1 ù é 1 ù é 1 ù
(A) 24 (B) 36 ê 2 ú+ê 2 ú+ê 2 ú + .... + ê 2 ú
ë (2 - 1) û ë (4 - 1) û ë (6 - 1) û ë (20 - 1) û
(C) 39 (D) 42
is : [NTSE Delhi 2016-17]

Q.112 The sum of sequares of the two consecutive natural 10 13

(A) (B)
numbers is 421, the numbers are : 21 27
[NTSE MP 2015-16]
15 8
(A) 14, 15 (B) 21, 22 (C) (D)
22 33
(C) 9, 10 (D) 17, 18

3-39
Real Number

Q.119 If we divide a two digit number by teh sum of its digits

Q.126 If y = 5, then what is the value of 10y y 3 - y 2 ?
we get 4 as quotient and 3 as remainder. Now if we
divide that two digit number by the product of its digits, [NTSE Punjab 2016-17]
we get 3 as quotient and 5 as remainder the two digit (A) 50 2 (B) 100
number is : [NTSE Delhi 2016-17]
(A) even (B) odd prime (C) 200 5 (D) 500
(C) odd composite (D) odd
Q.127 If the H.C.F. of the expression (a2 – 1) and pa2 – q
Q.120 Four positive integers sum to 125. If the first of these (a + 1) is (a – 1) then relation between p and q will be:
numbers is increased by 4, the second is decreased by [NTSE UP 2016-17]
4, the third is multiplied by 4 and the fourth is divided (A) p = q
by 4 we find four equal numbers then four original (B) p = 2q
integers are : [NTSE Delhi 2016-17] (C) p = 2q + 1
(A) 16, 24, 5, 80 (B) 8, 22, 38, 57 (D) p = q + 1
(C) 7, 12, 46, 53 (D) 12, 28, 40, 45

Q.121 If x% of y is equal to 1% of z, y% of z is equal to 1% of (0.7) 0 - (0.1) -1

-1 3 -1
x and z% of x is equal to 1% of y then the value of xy + Q.128 The value of æ 3 ö æ 3 ö + æ - 1 ö is:
yz + zx is : : [NTSE Delhi 2016-17] ç ÷ ç ÷ ç ÷
è8ø è 2ø è 3ø
(A) 1 (B) 2
[NTSE UP 2016-17]
(C) 3 (D) 4
3 2
(A) - (B)
Q.122 If x, y, z are real numbers such that 2 3
x - 1 + y - 2 + z - 3 = 0 then the values of x, y, z (C) 3 (D) 2
are respectively : [NTSE Delhi 2016-17]
(A) 1, 2, 3 (B) 0, 0, 0 Q.129 Sum of odd numbers between 0 and 50 is :
(C) 2. 3, 1 (D) 2, 4, 1 [NTSE UP 2016-17]
(A) 625 (B) 600
(C) 900 (D) 1200
37 1
Q.123 If = 2+ , Where x, y, z are integers, then
13 1
x+ 1 + a4
1
y+ Q.130 If a = 5 + 2 6 then =
z a2

the value of x + y + z is : [NTSE Punjab 2016-17] [NTSE Karnataka 2016-17]

(A) 6 (B) 8 (A) 4 6 (B) 10
(C) 7 (D) –2
(C) 5 (D) 2 6

Q.124 If 75% of a number is added to 75, then the result is

number itself. The number is : Q.131 Statement A : The rationalising factor of
[NTSE Punjab 2016-17] 3
a + 3 b is 3 a + 3 b .
(A) 50 (B) 60
Statement B : The product of ( 3 a - 3 b) and
(C) 300 (D) 400

( 3 a 2 + 3 b 2 + 3 ab ) is (a – b)
Q.125 Find the product of 3
3, 4 3, 12 243 : [NTSE Karnataka 2016-17]
[NTSE Punjab 2016-17] (A) both A and B statements are true
(A) (B) 3 (B) both A and B statements are false
3
(C) A is true and B is false
(C) 12 3 (D) 4 3 (D) A is false and B is true

3-40
 Real Number

Q.132 How many numbers lie between 10 to 300, which when Q.139 The missing term in the sequence 0, 3, 8, 15, 24, ...., 48
divided by 4 leave a remainder 3: is: [NTSE Bihar 2016-17]
[NTSE Jharkhand 2016-17] (A) 35 (B) 30
(A) 71 (B) 72 (C) 36 (D) 39
(C) 73 (D) 74
Q.140 If 13 + 23 + ....+ 103 = 3025 then 4 + 32 + 108 + ...+ 4000 is
5 equal to : [NTSE Bihar 2016-17]
Q.133 An equivalent expression of after rationlizing (A) 1200 (B) 12100
7+4 5
the donominator is ____: [NTSE Gujarat 2016-17] (C) 12200 (D) 12400

20 5 - 35 20 5 - 35
(A) (B) Q.141 What is the square root of 9 + 2 14 :
31 129
[NTSE Bihar 2016-17]
35 - 20 5 35 - 20 5 (A) 1 + 2 2 (B) 3 + 6
(C) (D)
31 121
(C) 2+ 7 (D) 2+ 5
Q.134 If a and b be the zeroes of the polynomial ax2 + bx + c,

a b 127
then the value of + is : Q.142 1- is equal to : [NTSE Bihar 2016-17]
b a 343
[NTSE Chhattisgarh 2016-17]
5 1
-b (A) (B) 1 -
9 7
(A) b (B)
ac
4 2
-b 1 (C) (D) 1 -
7 7
(C) (D)
ac ac

Q.143 If 30,72 and x are three integers, such that the product
Q.135 The value of of any two of them is divisible by the third, then the
6 + 6 + 6 + ......... is :
least value of x is : [NTSE AP 2016-17]
[NTSE Chhattisgarh 2016-17] (A) 45 (B) 60
(A) 4 (B) 3 (C) 48 (D) 24
(C) – 4 (D) 3.5
Q.144 Let a,b and c be real numbers, such that a – 7b + 8c = 4
(0.03) - (0.01)
2 2 and 8a + 4b – c = 7 then the value of a2 – b2 + c2 =
Q.136 The value of is : _____: [NTSE AP 2016-17]
0.03 - 0.01
(A) – 1 (B) 4
[NTSE Bihar 2016-17]
(A) 0.02 (B) 0.004 (C) –2 (D) 1
(C) 0.4 (D) 0.04
Q.145 If a < b < c < d < e are consecutive positive integers,
Q.137 If the sum of two numbers is 22 and sum of their squares such that b + c + d is a perfect square and a + b + c + d
is 404 then the product of the number is : + e is a perfect cube. What is the smallest possible
[NTSE Bihar 2016-17] value of c?: [NTSE AP 2016-17]
(A) 40 (B) 44 (A) 675 (B) 576
(C) 80 (D) 88 (C) 475 (D) 384

Q.138 What is the least number which divided by 42, 72 and Q.146 39 + 312 + 315 + 3n is a perfect cube, n Î N, then the value
84 leaves the remainder 25, 55 and 67 respectively: of n = _____ : [NTSE AP 2016-17]
[NTSE Bihar 2016-17] (A) 18 (B) 17
(A) 521 (B) 512 (C) 14 (D) 16
(C) 504 (D) 487

3-41
Real Number

Q.147 A four digit number has the following properties : Q.154 If the LCM of 12 and 42 is (10 m + 4) then the value of
[NTSE AP 2016-17] ‘m’ is : [NTSE Tamilnadu 2016-17]
(i) It is a perfect square. (A) 50 (B) 8
(ii) Its first two digits are equal to each other. 1
(C) (D) 1
(iii) Its last two digits are equal to each other. 5
Then the four digit number is ____
(A) 5566 (B) 7744 Q.155 Suppose that
(C) 2288 (D) 3399 4x1 = 5,5x 2 = 6

6 x3 = 7,.....127 x124 = 128

Q.148 Using the digits 1,2,3,4,5 without repetition, 120 five 0
What is x1, x2, x3 .... x124? [NTSE Telangana 2016-17]
digit numbers can be made. How many five - digit
numbers can be made using the digits 0, 1, 2, 3, 4 without 5
(A) 2 (B)
repletion : [NTSE Kerala 2016-17] 2
(A) 120 (B) 100
7
(C) 96 (D) 24 (C) 3 (D)
2

Q.149 The difference of the squares of two natural numbers Q.156 A positive integer n has 60 divisors and 7n has 80
is 101. What is the sum of their squares : divisors. hat is the greatest integer k such that 7k divides
[NTSE Kerala 2016-17] n? [NTSE Telangana 2016-17]
(A) 5000 (B) 5100 (A) 0 (B) 1
(C) 5101 (D) 5102 (C) 2 (D) 3

Q.150 The sum of a number of its reciprocal is 4. What is their Q.157 The such of 49 consecutive integers is 75 , what their
difference?: [NTSE Kerala 2016-17] median ? [NTSE Telangana 2016-17]
(A) 7 (B) 73
(A) 2 (B) 3 (C) 72
(D) 74
(C) 2 2 (D) 2 3
Q.158 What is the largest integer that is a divisor of (n+1)
(n+3) (n+5) (n+7) (n+9) for all positive even integers
Q.151 When 10x2 + x – 23 is divided by (2x + 3), the reminder ‘n’? [NTSE Telangana 2016-17]
is : [NTSE Tamilnadu 2016-17] (A) 3 (B) 5
(A) 1 (B) – 2 (C) 11 (D) 15
(C) 2 (D) 0
Q.159 The sum of 18 consecutive positive integers is perfect
square. The smallest possible vale of this sum is ____:
a3 b3 [NTSE Telangana 2016-17]
Q.152 The sum of and is :
b-a a -b (A) 169 (B) 225
(C) 289 (D) 361
[NTSE Tamilnadu 2016-17]
(A) a + ab + b
2 2
(B) – a2 – ab – b2
Q.160 The cube root of :
(C) a2 – ab + b2 (D) a3 – b3
x + y + 3x1/3 y1/3 (x1/3 + y1/3) is
[NTSE Rajasthan 2016-17]
Q.153 Which of the following are irrational numbers: (A) x + y (B) x1/3 + y1/3
[NTSE Tamilnadu 2016-17] (C) (x + y)1/3
(D) (x + y)3
(i) 2+ 3 (ii) 4 + 25
Q.161 Expressing 0.23 + 0.23 as a single decimal, we get:
(iii) 5 + 7
3 (iv) 6+ 8 3
[NTSE Rajasthan 2016-17]
(A) (i), (ii) (B) (iii), (iv) (A) 0.465 (B) 0.465
(C) (i), (iii) (D) (iv), (iv) (C) 0.465 (D) 0.4654

3-42
 Real Number

Q.162 If a + b = 3, ab = 2 and a > b, then what is the value of Q.170 If 3 3 × 33 + 3–1/2 = 3a+2, then a = ? :
2 :
a 3 - b1 [NTSE Odisha 2016-17]
[NTSE Chhatisgarh 2016-17]
(A) 32 (B) 64 (A) 2 (B) 1/2
(C) 128 (D) 256
(C) 4 (D) 0

a b c NTSE STAGE-2
Q.163 If x + y = y + z = z - x , then which of the following PREVIOUS YEAR'S
equations is true : [NTSE Maharashtra 2016-17]
(A) a = b + c (B) c = a + b Q.1 The sum of all the possible remainders, which can be
(C) b = a × c (D) b = a + c obtained when the cube of a natural number is divided
by 9, is : [NTSE 2017]
(A) 5 (B) 6
Q.164 m 4 n 4 ´ 6 m 2 n 2 ´ 3 m 2 n 2 = (m, n) k then find the (C) 8 (D) 9
value of k : [NTSE Maharashtra 2016-17]
(A) 6 (B) 3 Q.2 Positive integers from 1 to 21 are arranged in 3 groups
(C) 2 (D) 1 of 7 integers each, in some particular order. Then the
highest possible mean of the medians of these 3 groups
Q.165 If x3 + y = 2249, where x and y are natural nubmers and is ...... [NTSE 2017]
HCP of x and y is not 1, then the value of (A) 16 (B) 12.5
(x + y) is : [NTSE Haryana 2016-17]
(C) 11 (D) 14
(A) 26 (B) 39
(C) 52 (D) 65
Q.3 On dividing 2272 as well as 875 by a 3-digit number N,
we get the same remainder in each case. The sum of the
Q.166 If a = 6 - 11 and b = 6 + 11 then the value of (a digits of N is : [NTSE 2017]
+ b) is : [NTSE Haryana 2016-17] (A) 10 (B) 11
(A) 22 (B) 2 11 (C) 12 (D) 13

(C) 6 (D) 12
Q.4 On dividing a natural number by 13, the remainder is 3
and on dividing the same number by 21, the remainder
Q.167 The sum of squares of two successive natural nubmers is 11. If the number lies between 500 and 600, then the
is 145. The numbers will be : remainder on dividing the number by 19 is :
[NTSE Chhatisgarh 2016-17] [NTSE 2016]
(A) 6, 7 (B) 7, 8 (A) 4 (B) 6
(C) 8, 9 (D) 9, 10
(C) 9 (D) 13

1 1
Q.168 22 has how many ? Q.5 Expressing 0.34 + 0.34 as a single decimal, we get :
2 4
[NTSE 2016]
[NTSE Chhatisgarh 2016-17]
(A) 22 (B) 44 (A) 0.6788 (B) 0.689
(C) 45 (D) 90 (C) 0.6878 (D) 0.687

Q.169 H.C.F. and L.C.M. of two polynomials are x and

(x3 – 9x) respectively. If only polynomial is (x2 + 3x), Q.6 LCM of two numbers x and y is 720 and the LCM of
then second will be : numbers 12x and 5y is also 720. The number y is :
[NTSE Chhatisgarh 2016-17] [NTSE 2015]
(A) (x2 + 3x) (B) (x2 – 9x) (A) 180 (B) 144
(C) (x + 9x)
2
(D) (x2 – 3x) (C) 120 (D) 90

3-43
Real Number

Q.7 When a natural number x is divided by 5, the remainder

Q.15 (a - b) 2 + (b - a) 2 is : [NTSE 2013]
is 2. When a natural number y is divided by 5, the
remainder is 4. The remainder is z when x + y is divided (A) Always zero
(B) Never zero
2z - 5
by 5. The value of is : [NTSE 2015] (C) Positive if and only if a > b
3
(C) Positive only if a ¹ b
(A) –1 (B) 1
(C) –2 (D) 2 Q.16 Which of the following is an irrational number ?
(A) 41616 [NTSE 2013]
Q.8 x and y are two non-negative numbers such that
(B) 23.232323......
2x + y = 10. The sum of the maximum and minimum
values of (x + y) is : [NTSE 2015] (1 + 3)3 - (1 - 3) 3
(A) 6 (B) 9 (C)
3
(C) 10 (D) 15 (D) 23.10100100010000.......

Q.9 Value of the expression : [NTSE 2014] Q.17 Two positive numbers x and y are inversely
proportional. If x increases by 20%, then percentage
1 3 4
- - decrease in y is : [NTSE 2012]
11 - 2 30 7 - 2 10 8-4 3
2
(A) 20 (B) 16
(A) 30 (B) 2 10 3
(C) 1 (D) 0 9
(C) 5 (D) 1
11
Q.10 In how many ways can you position 6 into ordered
summars ? [For e.x. 3 can be positioned into 3 ways as Q.18 Given a number x = 248–1. Then between 5 and 10, x
1 + 2, 2 + 1, 1 + 1 + 1] : has/have : [NTSE 2012]
[NTSE 2014] (A) No factor (B) Only one factor
(A) 27 (B) 29 (C) Two factors (D) Three factors
(C) 31 (D) 33
Q.19 Given two 4-digit numbers abcd and dcba. If
Q.11 The no. of integers n (< 20) for which n – 3n + 3 is a
2 a + d = b + c = 7, then their sum is not divisible by:
perfect square is : [NTSE 2014] [NTSE 2012]
(A) 0 (B) 1 (A) 7 (B) 11
(C) 2 (D) 3 (C) 101 (D) 111

Q.12 Which of the following can be expressed as the sum of Q.20 Which is the greatest number amongst 21/2, 31/3, 81/8 and
square of two positive integers, as well as three positive 91/9 ? [NTSE 2011]
integers ? [NTSE 2014] (A) 91/9 (B) 81/8
(A) 75 (B) 192 (C) 31/2 (D) 21/2
(C) 250 (D) 100
Q.21 If the product of two numbers is 21 and their difference
is 4, then the ratio of the sum of their cubes to the
Q.13 Which of the following numbers is the fourth power of
difference of their cubes is : [NTSE 2011]
a natural number ? [NTSE 2013]
(A) 185 : 165 (B) 165 : 158
(A) 6765201 (B) 6765206
(C) 185 : 158 (D) 158 : 145
(C) 6765207 (D) 6765209
Q.22 The sum of any three distinct natural numbers arranged
Q.14 The square of an odd integer must be of the form: in ascending order is 200 such that the second number
(A) 6n + 1 [NTSE 2013] is a perfect cube. How many possible values are there
(B) 6n + 3 for this number ? [NTSE 2011]
(C) 8n + 1 (A) 4 (B) 3
(D) 4n + 1 but may not be 8n + 1 (C) 2 (D) 1

3-44
 Real Number

Q.23 If the digits of a three digit number are reserved, then

Q.29 3 + 2 2 - 3 - 2 2 is equal to : [NTSE 2010]
the number so obtained is less than the original number
by 297. If the sum of the digits of the number is 8 and its (A) 2 (B) 1
hundred’s digit has the largest possible value, then the (C) 2 2 (D) 6
ten’s digit of the number is : [NTSE 2011]
(A) 3 (B) 2
(C) 1 (D) 0 Q.30 If ax = b ,b =
y 3
c and c =
z
a then the value of xyz
is : [NTSE 2010]

Q.24 The value of the expression 34 - 24 2 ´ (4 + 3 2) 1 1

(A) (B)
2 3
is : [NTSE 2011]
(A) –2 (B) 2 1 1
(C) 3 (D) 4 (C) (D)
6 12

Q.25 If aabb is a four digit number and also a perfect square

then the value of a + b is : [NTSE 2011] 7 1
Q.31 If 5 + y = 12 , where fractions are in their lowest
(A) 12 (B) 11 x 13
(C) 10 (D) 9 terms, then x – y is equal to :
[NTSE 2010]
Q.26 If the sum of three consecutive odd numbers is a perfect (A) 2 (B) 4
square between 200 and 400, then the root of this sum (C) 7 (D) 9
is : [NTSE 2011]
(A) 15 (B) 16 Q.32 1x3y6 is a five digit number where x, y are digits and y
(C) 18 (D) 19 exceeds x by 6. If this number is divisible by 18, then
y
Q.27 If (x – 1)2 + (y – 3)2 + (z – 5)2 + (t – 7)2 = 0, then xyzt + 16 the value of is : [NTSE 2010]
x
is equal to : [NTSE 2010]
(A) 7 (B) 3
(A) 52 (B) 92
(C) 112 (D) 122 1 1
(C) (D)
3 7

810 + 410
Q.28 is equal to : [NTSE 2010]
642 + 49 ´ 16
(A) 8 (B) 16
(C) 36 (D) 256

3-45
Geometrical Optics

Geometrical Optics 2
DEFINITION
Light is form of energy which enables us to see objects
which emit or reflect light. O
Light is a type of (form of) energy which can produce
sensation in our eyes. So we can experience the Convergent beam of light: A beam of light that is going
sensation of vision. to meet (or converge) at a point is known as a convergent
It travel in straight line in form of particles and waves. beam. A parallel beam of light after passing through a
With the help of light we see all colours of nature. Our convex lens becomes a convergent beam.
eyes are mostly sensitive for yellow colour and least
sensitive for violet and red colour. Due to this reason
commercial vehicles are painted with yellow colour,
sodium lamps are used in road lights.
HOW WE SEE?
PROPERTIES OF LIGHT When a light ray is falling (strike) on the surface of
Light energy propagates (travels) via two processes. any object which reflect and reached to our eyes. Due
(i) The particles of the medium carry energy from one to this our eyes feel a sensation then we see the object.
point of the medium to another.
(ii) The particles transmit energy to the neighbouring REFLECTION OF LIGHT
particles and in this way energy propagates in the When rays of light falls on any object it return back in
form of a wave. the same medium from the surface this phenomenon is
(iii) It propagates in straight line. called reflection of light. Due to reflection of light we
(iv) It’s velocity in vacuum is maximum whose value is can see all the nature.
3 × 108 m/sec. (297489978 m/s)
(v) Light does not need a material medium to travel INCIDENT RAY
that is it can travel through a vacuum. The ray of light which falls on a polished surface (or a
(vi) It exhibits the phenomena of reflection, mirror) is called the incident ray of light.
interference, diffraction, polarization and double
reflection. REFLECTED RAY
The ray of light which gets reflected from a polished
RAY OF LIGHT surface (or a mirror) is called the reflected ray of light.
A straight line show the direction of light is called ray
of light.
NORMAL
BEAM OF LIGHT
A beam of light is a collection of these rays. There are plane normal to mirror
mainly three types of beams.
y
in

ra

Parallel beam of light: A search light and the headlight

cid

ed

i r
en

ct
tr

of a vehicle emit a parallel beam of light. The source of

fle
ay

re

light at a very large distance like sun effectively gives O

reflecting surface
a parallel beam. mirror

Divergent beam of light: The rays going out from a The normal is a line at right angle to the reflecting
point source generally form a divergent beam. surface.

1-46
 Geometrical Optics

LAWS OF REFLECTION IRREGULAR REFLECTION (DIFFUSED REFLECTION)

(i) The incident ray, the reflected ray and the normal When parallel light rays fall on a rough surface all the
to the surface at the point of incidence all lie in the rays of light are reflected in all possible (Different)
same plane. direction this is called diffused or irregular reflection.

i r
i r
O Diffused reflection
O
(a) (b)
OBJECT AND IMAGE
Object (O): Object is defined as point of intersection
of incident rays.

i r
O

O virtual point obje ct

(c) real point object

Let us call the side in which incident rays are present

(ii) The angle of incidence ( Ð i ) is always equal to
as incident side and the side in which reflected
the angle to reflection ( Ð r ) i.e. Ð i = Ð r (refracted) rays are present, as reflected (refracted) side.

Note
l When a ray of light falls on a mirror normally or at An object is called real if it lies on incident side
right angle it gets reflected back along the same otherwise it is called virtual.
path.
Image (I): Image is defined as point of intersection of
i = 0, r = 0
reflected rays (in case of reflection) or refracted rays
i = 0, r = 0
(in case of refraction).
C
C
i = 0, r = 0
Note
plane mirror
concave mirror convex mirror An image is called real if it lies on reflected or
refracted side otherwise it is called virtual.
Depending on the nature of the reflecting surface
there are two types of reflection:-
(i) Regular (specular) reflection MIRROR
(ii) Irregular (diffused) reflection A smooth, highly polished reflecting surface is called
a mirror. When a glass plate is polished on one sided
REGULAR REFLECTION with reflecting material such as silver or nickel then it
When parallel light rays fall on smooth plane surface becomes a mirror. From the reflecting surface of mirror
like mirror, if all rays of light are reflected paraller along there are two types of mirror.
a definite direction. Then this kind of reflection is called (i) Plane mirror
regular reflection. (ii) Spherical or curved mirror

(i) Plane mirror: A highly polished plane surface is

called a plane mirror or if a flat (totally plane) surface
of a glass plate is polished one side of reflecting
material is called plane mirror.

1-47
Geometrical Optics

FORMATION OF IMAGE BYA PLANE MIRROR When a watch placed in front of a plane mirror
P rays from N
object

Its
Reflected ray 01:45
Its
from mirror 10:15
A B
Mirror

(Virtual)
image P' N'
then watch is object and its time is object time and
Properties of image from flat (Plane) Mirror image of watch observed by a person standing in front
(i) Virtual and erect. of mirror then time seen by person.
(ii) Same in size of object. (i) Object Time = AH
(iii) The image is formed behind of the mirror (as far as Image Time = 12 – AH.
the object from the mirror).
(ii) Object Time = AH BM
(iv) The image formed is laterally inverted.
Image Time = 11 - 60' – AH BM
LATERAL INVERSION AND INVERSION (iii) Object Time = AH BMCS
The phenomenon due to which the image of an object Image Time = 11 - 59' - 60" – AHBMCS
turns through an angle of 1800 through vertical axis
rather than horizontal axis, such that the right side of
l Deviation : d is defined as the angle between directions
the image appears as left or vice versa is called lateral
of incident ray and emergent ray. So if light is incident
inversion.
at an angle of incidence i,
d = 180º – (Ði + Ðr) = (180º – 2i) [as Ði = Ðr]
INVERSION
PLANE MIRROR
¥ The image formed by a plane mirror suffers lateral-
r
inversion, i.e., in the image formed by a plane mirror left i
is turned into right and vice-versa with respect to object.

Plane mirror
So if light is incident at angle of 30º,
d = (180º – 2 × 30º) = 120º and for normal incidence
Ði = 0º, d = 180º [fig. A]

l Rotation of image : If keeping the incident ray

fixed, the mirror is rotated by an angle q, about an
axis in the plane of mirror, the reflected ray is
rotated through an angle 2q. This is illustrated in
fig. (b)
front
M q
Incident ray
q
q
ay
dr

mirror
cte

back
fle
Re

(B)

1-48
 Geometrical Optics

This fact is used in many optical instruments such as

periscope [fig. C)], sextant, optical–lever, etc.

l As very part of a mirror forms a complete image of q

an extended object and due to super–position of 90°– q
M1
image brightnes will depend on its light reflecting
area, a larger mirror gives more bright image than a
small one. This in turn also implies that if a portion q
of a mirror is obstructed, complete image will be
formed but of reduced brightness. M2

l Though every part of a mirror forms a complete l Number of Images : If there are two plane mirrors
image of an object, we usually see only that part inclined to each other at an angle q, the number of
of it from which light after reflection from the mirror images of a point object formed are determined as
reaches our eye. This is why : follows:
(i) If (360º/q) is even integer (say m) number of images
(i) To see his full image in a plane mirror a person formed
requires a mirror of at least half of his height n = (m – 1), for all positions of object
(ii) If (360º/q) is odd integer (say m) number of images
H formed
M n = m, if the object is not on the bisetor of mirrors
n = (m – 1), if the object is on the bisector of mirrors.
E (iii) If (360º/q) is a fraction, the number of images
H formed will be equal to its integral part.
M
SOLVED EXAMPLE
L
(A) Example-1
An object is placed in front of a plance mirror. If the
(ii) To see a complete wall behind himself a person mirror is moved away from the object through a
requires a mirror of at least (1/3) the height of wall distance x, by how much distance will the image move?
and he must be in the middle of wall and mirror Sol. Suppose the object O was initially at a distance d from
the plane mirror M as shown in fig. The image formed
at O’ is at a distance d behind the mirror. Now, the
A mirror is shifted by a distance x to M’ such that the
distance of the object from M’ becomes d + x. The
M image now formed at O” which is also at a distance
H E H/3 d + x from M’.
M
B
d d
(B) M M'
O O' O"
l If two plane mirrors are inclined to each other at
90º, the emergent ray is always antiparallel to
incident ray if it surffers one reflection from each
(as shown in fig.) whatever be the angle of
incidence. The same is found to hold good for
So, OM = MO’ = d
three–plane mirrors forming the corner of a cube if
OM’ = M’O” = d + x
the incident light suffers one reflection from each
Thus, OO” = M’O” + M’O” = 2(d + x) ...(1)
of them.

1-49
Geometrical Optics

when OO’ = OM + MO’ = 2d ...(2)

\ O’O” = OO” – OO’ Example-4
= 2(d + x) – 2d An object O is kept 80 cm in front of a plane mirror M.
= 2x If the mirror is moved 20 cm away from the object.
Thus, the image is shifted from O’ to O” by a distance Calculate the image shift from its previous position.
2x. Sol. The distance of object from the mirror = 80 cm
\ Distance between object and its image
Example-2 = 80 + 80 = 160 cm ...(1)
A ray of light is incident on a plane mirror. The mirror The new distance of object from the mirror
is tilted through an angle q. By how much angle will = 80 + 20 = 100 cm
the reflected ray be tilted? \ Now distance between object and image
Sol. Let AN be the normal to the mirror M1M2. A light ray = 100 + 100 = 200 cm
OA is incident on the mirror A and is reflected along \ Shift of the object = 200 – 160 = 40 cm
AB, therefore,
Example-5
Find graphically the position of the observer’s eye,
N which will allow him to see in a plane mirror of finite
N'
dimensions, the image of the straight line arranged as
O B
shown.

M'1 q B' B
i
x

A
M1 M2
q

A
B
M'2 A2
A
A1

ÐOAN = ÐNAB = Ði Sol.

Now, if the mirror is rotated through an angle q. the
C F E D
normal also rotates through the same angle q. AN¢ is
the new position of the normal, then
ÐNAN¢ = q
or ÐN¢AB = Ð NAB – Ð NAN¢ = Ði – q ...(1) A¢

B¢
Let AB¢ is the final reflected ray. Then
A¢ B ¢ is the reflected image of AB by CD. The reflected
ÐOAN¢= Ð N¢AB¢ = Ði + q ...(2)
rays EA2 and FA1 must reach the eye so that the whole
Thus, ÐBAB¢= ÐN¢AB¢ – ÐN¢AB
image can be seen.
= (Ði + q) – (Ði – q)
= 2q
Example-6
Thus, if the mirror is tilted through an angle q, the
What is the minimum length of a plane mirror required
reflected ray is rotated through angle 2q.
for a person of height 2m to see his full image? Is there
Example-3 any restriction on the position of the top edge of the
An insect is at a distance of 1.5m from a plane mirror. mirror?
Calculate the following? Sol. The man can view his entire image if the light rays from
(i) Distance at which the image of the insect is formed. the top of his head and from his feet reach his eye. Let
(ii) distance betweeen the insect and its image. AB be the mirror. PQ represents the man of height h
Sol. (i) The distance of insect from the mirror = 1.5 m and R is the position of his eyes. Light rays from P gets
\ The distance of insect from the mirror is also equal reflected at A and reach his eyes.
to 1.5 m. Light from Q gets reflected at B and reaches his eyes.
The image is formed at 1.5 m behind the mirror.
AM and BN are normals to the mirror AB.
(ii) The distance between the insect and image
Now, AB = MN = MR + RN
= 1.5 + 1.5 = 3m

1-50
 Geometrical Optics

(A) Convex mirror (B) Concave mirror

1
= (PR + RQ) (Q D APM @ D ARM ; DBQN @ D /// /
2 / / ////
/

/ //

/// /
//////
hollow hollow

//// / /// // ////

BRN) reflecting reflecting

// // // ////
sphere sphere
surface surface

// /
PQ h
/ // /

/
// / /
/ //

= = =1 m Concave mirror Convex mirror

2 2
(A) Concave (Converging) mirror: A spherical mirror
It is clear from the ray diagram that the top edge of the
whose inner hollow surface is the reflecting surface.
plane mirror (a) must be at a horizontal level half-way
(B) Convex (diverging) mirror: A spherical mirror
between the eyes (R) and the top of his head (P). whose outer bulging out surface is the reflecting
Head
P surface.
M A

Eye R
REFLECTIONAT SPHERICALSURFACE
SOME IMPORTANT DEFINITIONS
N
B
(i) Spherical Mirrors: A spherical mirror is a part of a
hollow sphere or a spherical surface. They are
Q
classified as concave or convex according to the
Feet
reflecting surface being concave or convex
respectively.
Example-7
Find the region on Y axis in which reflected rays are
present. Object is at A (2, 0) and MN is a plane mirror, as P
R
shown. C R P
C
y
y N'

N(4,3)
N(4,3)
(ii) Pole or Vertex: The geometrical centre of the
M'
spherical mirror is called the pole or vertex of the
M(4,2)
x M(4,2) mirror.
A (2,0) x
A (2,0) A' (6,0)

Sol. The image of point A, in the mirror is at A’ (6, 0).

P P
Join A’ M and extend to cut Y axis at M’ ( Ray originating
from A which strikes the mirror at M gets reflected as
the ray MM’ which appears to come from A’). Join A’N In the above figures, the point P is the pole.
and extend to cut Y axis at N’ (Ray originating from A (iii) Centre of curvature: The centre C of the sphere of
which strikes the mirror at N gets reflected as the ray which the spherical mirror is a part, is the centre of
NN’ which appears to come from A’). curvature of the mirror.
From Geometry.
M’ º (0, 6) (iv) Radius of curvature (R): Radius of curvature is the
N’º (0, 9). M’N’ is the region on Y axis in which reflected radius R of the sphere of which the mirror forms a
rays are present. part.

(ii) Spherical mirror: A mirror whose polished, P C

C P
reflection surface is a part of hollow sphere of glass R R
is called a spherical mirror. For a spherical mirror,
one of the two curved surfaces is coated with a
thin layer of silver followed by a coating of red (v) Principal axis: The line CP joining the pole and the
lead oxide paint. Thus one side of the spherical centre of curvature of the spherical mirror is called
mirror is made opaque and the other side acts as a the principal axis.
reflecting surface.
C P P C
For the polishing side there are two type of spherical Principal Principal
axis axis
mirror.

1-51
Geometrical Optics

(vi) Aperture: The aperture is the segment or area of RULES OF IMAGE FORMATION FROM THE SPHERICAL
the mirror which is available for reflecting light. MIRROR
The rules of reflection from the spherical mirror are
(vii) Paraxial ray: A paraxial ray is a ray which makes a based on incident and reflection angle.
small angle (q) to the optical axis of the system, and (i) A ray parallel to principal axis after reflection from
the mirror passes or appears to pass through its
lies close to the axis throughout the system.
focus by definition of focus.
M
(viii) Marginal ray: Ray which passes through an
optical system away from the optical axis, towards
C
the edge of the aperture. · P parallel ·
F f to axis
P focus F
centre of
curvature
(ix) Focus (F): If a parallel beam of rays, parallel to the M’

principal axis and close to it, is incident on a Concave mirror M’

Convex mirror
spherical mirror; the reflected rays converge to a
point F (in case of a concave mirror) or appear to (ii) A ray passing through or directed towards focus
diverge from a point F (in case of a convex mirror) after reflection from the mirror it will become parallel
to the principal axis
on the principal axis. The point F is called the focus
of the spherical mirror. M
parallel

C F P
focus

· P ·
F P F M’
(iii) A ray passing through or directed towards centre
of curvature after reflection from the mirror, retraces
Concave mirror
Convex mirror its path, as for it Ði = 0 and so Ðr = 0
M
(x) Focal Length (f): Focal length is the distance PF
between the pole P and focus F along the principle C F
axis. focus P
(i) Focal plane:- An imaginary plane passing through directed towards
the focus and at right angles to the principal axis. centre of curvature
M’

(j) Real image:- When the rays of light after getting (iv) Incident and reflected rays at the pole of a mirror
reflected from a mirror (or after getting refracted are symmetrical about the principal axis.
from a lens) – actually meet at a point, a real image M
is formed. A real image can be obtained on a screen.
i
(k) Virtual image: When the rays of light after getting r P
F
reflected from a mirror (or after getting refracted d
from a lens) appear to meet at a point, a virtual M’
image is formed. Such an image can only be seen Difference between Real and Imaginary image
through a mirror (or a lens) but cannot be obtained
S.No Real image Virtual image
on a screen. (1) When reflected or When reflected or
refracted light rays refracted light rays d o
/// ////////
///
//
// / ///
///
///
// //

A
////

/ ////

actually intersect at a not actually intersect at a

/////

////

B
///

/// ///

point. point but appear to meet

//////////////////// //

///////

at a point
//////////////

(2) It can be obtained on a It cannot be obtained o n

A'
////

screen. a screen.
///// /

// /

B'
///

///
///
/ ///

////
////

(3) It is always inverted. It is always erect.

//

/ /// //// // //
/// / / /

(a) Real image (b) Virtual image (4) It is always formed in It is always formed
front of mirror. behind the mirror.

1-52
 Geometrical Optics

Image formation from Concave mirror

S. No. Position of the Position of the Nature & size of Ray diagram
object image the image

Real, inverted and

1. At infinity At focus F highly diminished,
(point size)

Real, inverted and

Between infinity
2. Between C&F smaller than the
and C
object

Real, inverted and

3. At C At C
same size.

Between C and Real, inverted and

4. Between C&F
infinity. enlarged.

Real, inverted and

5. At F At infinity.
infinitely large.

Between focus and Virtual, erect and

6. Behind the mirror.
pole enlarged.

Use of Concave mirror (ii) Any ray of light which travellers towards the
(i) It is used as a shaving mirror. direction of principal focus of a convex mirror, after
(ii) It is used as a reflector in the head light of vehicles. reflection, it will travel for parallel to the principal
(iii) It is used by doctor to focus a parallel beam of axis of the mirror.
light on a small area.
r M
/// /

i
/////

parallel
///

FORMATION OF IMAGE FROM A CONVEX MIRROR

///// //////

to axis
There are rules of drawing images in convex mirror F C
///////// //// //

(i) Any ray of light travelling parallel to the principal P focus centre of
curvature
axis of a convex mirror of the appears to diverge
// ///

from the principal focus of the convex mirror.

/////

M'
///

r M
////

i
// //// //

parallel (iii) If ray of light which is incident along to the centre

///// //////

to axis
F C of curvature of a convex mirror after reflection it
///////// /// ////

P focus centre of returns back on the same path. It is because the

curvature
light ray strikes the convex mirror at right angle.
//////
///

M'
///

1-53
Geometrical Optics

Note
90° M

////
///// /
There are only two position of the object for

///////
showing the image formed by a convex mirror that

/ ////////////// /// /////

F C
P focus centre of is –
curvature
/// / /
//// (i) When the object is at a infinity.
M'
//

(ii) When the object is at a finite distance from the

pole of the convex mirror. Beside this positions
(iv) When the ray of light incident on the pole which is
reflects or returns back on same angle from principal are not possible because the focus and the
axis than it will reflect on the same angle of incident centre of curvature is behind the reflecting
Ð i = Ð r.. surface of the convex mirror.
M
/////
/ /////

Now we can study the image formation by following table

/////// //

S.No. Position of Position of Size of Nature of

F i
//////////// ///////

P the object the image image of the image

r
the image
(1) At infinity At F, behind Highly Virtual
//////

mirror diminished and erect.

/////

M' (2) Between Between P Diminished Virtual

/

infinity and & F behind and erect.

MAKING OF IMAGE FROM A CONVEX MIRROR pole of the mirror
(i) When the object is at infinity mirror
When the rays of light coming (diverging) from an
object, situated at infinity are always parallel these
parallel rays, strike the convex mirror, and reflected to Uses of Convex mirror
diverge outward from convex mirror. These rays seems (i) It is used as a rear view mirror in automobile.
(appear) to come from focus. (ii) It is used as a reflector for street light.

The characteristics of the image is virtual, erect,

Note: A plane mirror is not useful as a rear view mirror,
diminished to a point and formed at principal focus
behind the convex mirror. because its field of view is very small.

COMPETITION WINDOW
////
/////

virtual ray
SIGN CONVENTION OF SPHERICAL MIRROR
//////

l Whenever and wherever possible the ray of light

// /////////////// /// //////

virtual image
F
is taken to travel from left to right.
l The distances above principal axis are taken to be
incident ray positive while below it negative.
/////
////
/

Reflected ray

M
(ii) When the object is at a finite distance from the Object on the left
//////

pole then the image is formed between pole and A

Direction of
/////// /

incident light
principal focus behind the convex mirror and image Height
// ///////////////////////////////

is virtual, erect and diminished. upwards Distance towards Distance towards

(+ve) the left (–ve) P the right (+ve)
M X' X'
B B'
////

Object placed
// ///

in front of mirror Height

///// ///

downwards (–ve)
/////////////// // //////

A'
//////

O P I F C
/

Mirror N
/////

Virtual, erect, diminished

////

M' (m < +1), between P and F l Along principal axis, distances are +ve
//

1-54
 Geometrical Optics

RELATION FROM SPHERICALMIRROR

- h2 - v h v
= Þ 2 =-
h1 -u h1 u

/ / //
A Q

/////
Magnification

/
////// //
v v f f - v h2

/////////////// /////
C F 2 m=- Þ m =- = = =
P u u f -u f h1

f
POWER OFA MIRROR

///// //
The power of a mirror is defined as

/////
R

/
Relation between f and R for the spherical mirror 1 100
P=- =-
f ( m) f (cm)
QP
If Q is near to line P then from D QCP tan q » q =
R

QP

/// /
and from QFP tan 2q » 2q =

/////
field of view

////////
f C

/ ///////////////// /// /
/// //
/ ///
// //
//// /
/
//////// / /
/ ///
/

///
2QP QP R
so = Þf =

//////
R f 2

/////
Relation between u, v and f for curved mirror
If an object is placed at a distance u from the pole of a
l Convex mirrors give erect, virtual and diminished
mirror and its image is formed at a distance v (from the
image.
pole)
In convex mirror the field of view is increased as
compared to plane mirror.
A M It is used as rear-view mirror in vehicles.
/ / //
/////

Concave mirrors give enlarged, erect and virtual image,

/
object

////// //

so these are used by dentists for examining teeth. Due

h1
to their converging property concave mirror are also
////////////////////

C B' F
P used as reflectors in automobile head lights and search
B
image h2 light
A real image can be taken on a screen, but a virtual
//////

A' v image cannot be taken on a screen.

/////

u
//

M' R
As focal length of a spherical mirror f = depends
2
only on the radius of mirror and is independent of
MP MP MP
If angle is very small: a = ,b = ,g = wavelength of light and refractive index of medium so
u R v
the focal length of a spherical mirror in air or water and
from CMO, for red or blue light is same.
b = a+q Þ q = b-a

from D CMI, l = b + q Þ q = g - b l MIRROR FORMULA

so we can write b - a = g - b Þ 2b = g + a The relation between the distances of the object from
the pole of the spherical mirror (u), the distance of the
2 1 1 1 1 1 image from the pole of the spherical mirror (v) and its
\ = + Þ = + .
R v u f u v focal length (f) is given by the mathematical formula:
1 1 1
MAGNIFICATION + =
u v f
size of image I It must be remembered that focal length (f) of a spherical
Linear magnification m = size of object = O
mirror is half the radius of curvature (R).

D ABP and D A’B’P are similar R

Thus, (i) R = 2f, (ii) f =
2

1-55
Geometrical Optics

Important points in using the mirror formula

(i) Put the correct signs of known variables according SOLVED EXAMPLE
to the sign convention.
(ii) Do not put the sign of an unknown variable. The Example-8
sign will be automatically come up during A concave mirror is made up by cutting a portion of a
calculations. hollow glass sphere of radius 30 cm. Calculate the
(iii) If the calculated sign turns out to be positive, then focal length of the mirror.
the variable calculated is behind the mirror. Sol. The radius of corvature of the mirror = 30 cm
However,. if calculated sign turns out to be 30 cm
negative, then variable is to be in front of the mirror. Thus, the focal length of the mirror = = 15 cm
2

Linear magnification produced by spherical mirrors: Example-9

The ratio between the heights of the image produced
An object is placed at a distance of 15 cm from a
by the spherical mirror to the height of the object is concave mirror of focal length 10 cm. Find the position
called the linear magnification. of the image.
Height of the image 15cm
Thus, linear magnification =
Height of the object
A
hi B´
or m = h B F
0
A´ 10cm
Linear magnification when the image is real:
30cm
As we normally take object above the principal axis,
therefore, ho is always positive. The real image is always Sol. We have u = –15 cm and f = –10 cm
inverted and is formed below the principal axis. 1 1 1
Using the relation, + = ,
Therefore, hi is always negative. Thus, linear n u f
hi
magnification for real images = - h is always we get
1
+
1
=
1
0
n - 15 - 10
negative.
1 1 1 1
or = - =–
Linear magnification when the image is virtual: n 15 10 30
In case of virtual image. It is erect and formed above or v = –30 cm
the principal axis. Thus, ho and hi are both positive. So the image will be formed 30 cm from the mirror.
l The linear magnification produced by a spherical Since n has a negative sign, the image is formed to the
mirror is equal to the ratio of the distance of the left of the mirror, i.e. in front of the mirror as shown in
image from the pole of the mirror (v) to the distance fig.
of the object from the pole of the mirror (u) with a
minus sign. Example-10
A 3 cm long object is placed perpendicular to the
v
Liner magnification, m = - , principal axis of a concave mirror. The distance of the
u
object from the mirror is 15 cm, and its image is formed
hi v 30 cm from the mirror on the same side of the mirror as
Thus Linear magnification, m = h = - u . the object . Calculate the height of the image formed.
o
15cm
Important points in using magnification formula:
(i) Put the correct signs of known variables according A
to the sign convention. B´
(ii) If ‘m’ is known, take the sign for virtual image B F

positive and for real image negative. A´

(iii) Do not put the sign of unknown variables. The
30cm
sign will automatically come up during calculations. Sol. Here u = –15 cm and n = –30 cm

1-56
 Geometrical Optics

Size of the object, h = 2 cm - 30

Magnification, 3= Þ u = -20cm
- 30 - u
h' n Object must be at a distance of 20 cm in front of the
m=m= =–
h u mirror (in between F and P).

h' ( - 30 ) Example-12
or =– ( - 15 ) = 2
h A beam of light converges towards a point O, behind a
convex mirror of focal length 20 cm. Find the nature
or h’ = – 2 × h = –2 × 3
and position of image if the point O is (a) 10 cm behind
= –6 cm
the mirror (b) 30 cm behind the mirror.
So the height of the image is 6 cm. The minus sign
shows that it is on the lower side of the principal axis, M

////
/////
i.e. the image is inverted.

////////
///////////////////////
Example-11
The focal length of a concave mirror is 30cm. Find the I P O
position of the object in front of the mirror, so that the

/////
image is three times the size of the object.

/// /
M'

//
Sol. As the object is in front of the mirror it is real and for
Sol. (a) For the situation object will be virtual as shown in
real object the magnified image formed by concave
figure.
mirror can be inverted (i.e., real) or erect (i.e., virtual),
Here u = + 10cm and f = + 20cm.
so there are two possibilities.
(a) If the image is inverted (i.e., real) 1 1 1
\ + = i.e., v = -20 cm
v 10 + 20
i.e., the image will be at a distance of 20 cm in front of
////

M the
/////
///////

O mirror and will be real, erect and enlarged with

///////// //////////////

I C' F é 20 ù
P m = - ê - ú = +2
ë 10 û
40cm
(b) For this situation also object will be virtual as
//////

shown in Figure.
/////

120cm M'
/

Here, u = + 30 cm and f = + 20cm

1 1 1
\ + = i.e., v = + 60 cm
f - 30 v + 30 + 20
m= Þ -3 = Þ u = -40 cm
f -u - 30 - u
i.e., the image will be a distance of 60 cm behind the
Object must be at a distance of 40 cm in front of the mirror and will be virtual, inverted and enlarged with
mirror (in between C and F).
é 60 ù
(b) If the image is erect (i.e., virtual) m = - ê+ ú = -2
ë 30 û

M Example-13
/ / //

I
// ///

An object of size 7.0 cm is placed at 27cm in front of a

///////

O concave mirror of focal length 18 cm. At what distance

/// ////// ///// /// ///////

C P from the mirror should a screen be placed, so that a

F sharp focused image can be obtained? Find the size
20cm 60cm
and the nature of the image.
//////

30cm
Sol. Mirror: concave u=-27cm, f = -18 cm, v =?
/// //

M'
1 1 1
Using mirror formula + =
v u f
f
m=
f -u 1 1 1 1 1 -3+ 2 1
= - = + = =- cm
v f u - 18 27 54 54

1-57
Geometrical Optics

v = - 54 cm
The screen should be placed at a distance of 54 cm Example-16
from the mirror in front of it. A 1.4 cm long object is placed perpendicular to the
principal axis of a convex mirror of focal length 15 cm
v - 54
m=- =- = -2 at a distance of 10 cm from it. Calculate the following:
u - 27
(i) location of the image (ii) height of the image
Image is real and magnified, two times the object –
(iii) nature of the image
h'
2=
h
h ' = -2 ´ 7 cm = -14cm high A
The image is real, inverted, enlarged and 14 cm high. A´

B 10cm 6cm B´ F C
Example-14
An object 5.0 cm in length is placed at a distance of 20
cm in front of a convex morror of radius of curvature 15cm
30 cm. Find the position of the image, its nature and
size. Sol. (i) For a convex mirror, focal length is positive.
Sol. Radius of curvature of convex mirror (R) = 30 cm Therefore, f = +15 cm and u = –10 cm
\ Focal length of convex mirror 1 1 1
R 30cm Using the relation, + = , we get
(f)= = = 15 cm n u f
2 2
1 1 1 1 1 1
= + + =
Using mirror formula, n - 10 15
f u v
Object distance, u = - 20 cm 1 1 1 5 1
or = + = =
1 1 1 1 1 4+3 7 n 15 10 30 6
= - = + = = cm
v f u 15 20 60 60 or n = 6 cm
Since n is positive, the image is formed to the right of
60
v= cm = 8.57cm the mirror at a distance 6 cm from it.
7
(ii) Magnification,
The image is virtual, behind the mirror and erect.
h' 1
h' v h' - 60 / 7cm m= =-
m= =- = = h n
h u 5cm - 20cm

- 60 / 7cm ´ 5cm 3 15 h' 6

Þ h' = = ´ cm = cm or =- = + 0.6
- 20cm 7 7 h - 10
h’ = + 2.14 cm [It is erect and diminished]. or h’ = + 0.6 × h 0 = 0.6 × 1.4
= 0.84 cm.
Example-15 Thus, the height of the image is 0.84 cm.
An object is placed at a distance of 10 cm from a convex (iii) Since h e is positive, the image will be on the same
mirror of focal length 15 cm. Find the position and side of the principal axis as the object.
nature of the image. [NCERT] Hence, the image is virtual, erect and diminished.
Sol. Here, object distance, u = - 10 cm focal length, f =
15cm, image distance, v =? Example-17
1 1 1 Sunlight is incident on a concave mirror parallel to its
As + = , principal axis. The image of the sun is formed 15 cm
v u f
from the pole. Calculate the radius of curvature of the
1 1 1 1 1 5 1 mirror.
\ = - = + = = , f = 6 cm
v f u 15 10 30 6 Sol. The rays parallel to the principal axis pass through
Here, + sign of v indicates image is at the back of the the focus after reflection.
mirror. It must be virtual, erect and smaller in size than Therefore, image of the mirror, f = 15 cm
the object. Radius of curvature, r = 2f = 2 × 15 = 30 cm

1-58
 Geometrical Optics

\ Focal length of the mirror = 20 cm

Example-18
h' n
An object is placed at a distance of 40 cm from a Magnification. m= =-
h u
convex mirror of focal length 30 cm. Find the position
of image and its nature.
h' (-60)
or =-
3 (-30)

A or h’ = 3 × (–2) = –6 cm
A´ The height of the image is 6 cm. The negative sign
shows that the image is inverted.
B 40cm B´ F C
Example-20
30cm A 1 cm high object is placed at 20 cm in front of a
concave mirror of focal length 15 cm. Find the position
Sol. Here, object distance, u = –40 cm and nature of the image.
Focal length of convex mirror, f = +30 cm Sol. u = –20 cm, f = –15 cm, h 0 = 1 cm

1 1 1 Using mirror formula, 1 + 1 = 1 , we get

Now, using mirror formula, + = , n u f
n u f
1 1 1
+ =
1 1 1 n - 20 - 15
we get + =
n - 40 30
1 1 1 1
or =- + =–
1 1 1 7 n 15 2n 60
or = + = \ n = – 60 cm
n 40 30 120
The image is formed 60 cm from the mirror. Since, the
120 signs of u and n are the same, the object and image
or n = = 17.1 cm are formed on the same side of the mirror. Therefore,
7
the image is real.
The positive sign shows that the image is formed on
the right, i.e. behind the mirror. h' u -60 cm
Now, magnification, Now magnification, m = =- = = –3
h n - 20 cm
n 120 3 \ h’ = –3h = –3 × 1 cm = – 3cm
m= = - 7 ´ (-40) = + = + 0.42
u 7 The negative sign shows that the image is inverted.
Thus, the image is real, inverted and of size 3 cm and
Since, the magnification is positive, the image is erect.
formd 60 cm in front of the mirror.
Thus, the image is formed 17.1 cm behind the mirror.
The image is virtual, erect and diminished.
Example-21
An object 4 cm high is placed 25 cm in front of a concave
Example-19
mirror of focal length 15 cm. At what distance from the
A 3 cm high object is placed at a distance of 30 cm mirror should a screen be placed in order to obtain a
from a concave mirror. A real image is formed 60 cm sharp image? Find the nature and size of image.
from the mirror. Calculate the focal length of the mirror Sol. Here, u = –25 cm, f = –15 cm, h = +4 cm
and the size of the image.
Sol. Object distance, u = –30 1 1 1
Using the mirror formula, + = , we get
Image distance, n = –60 n u f
(real image is formed on the same side)
1 1 1
+ =
Now, using the mirror formula, 1 + 1 = 1 , we get n - 25 - 15
n u f
1 1 1 1 1 1 1 1 2
+ = or = – =– + =–
- 60 - 30 f n - 15 - 25 15 25 75
1 3 1 -75
or = =- = -37.5 cm
f 60 20 or n =
2
or f = –20 cm

1-59
Geometrical Optics

Thus, the screen must be placed 37.5 cm from the

mirror on the same side as the object. Example-23
A convex mirror used on a automobile has 3 m radius of
h' n curvature. If a bus is located 5 m from this mirror, find the
Now, magnification, m = =-
h u position, nature and size of image.
Sol. Here, u = –5 m, r = +3m
h' (-37.5)
or =– = -1.5
4.0 cm (-25) r 3
\ f= = + = 1.5 m
2 7
or h’ = – 1.5 × 4 = –6 cm
Negative sign shows that the image is inverted. Hence, 1 1 1
Using the relation, + = , we get
the image is real, inverted and of size 6 cm. n u f

Example-22 1 1 1
+ =
An object 5 cm high is placed at a distance of 20 cm n - 5 1.5
from a convex mirror of radius of curvature 30 cm.
1 1 1
Find the position, nature and size of image. or = + = + 1.15 m
n 1. 5 5
Sol. Here, u = –20 cm, h = 5 cm
The image is 1.15 m behind the mirror.
Radius of curvature, r = +30 cm
Magnification,
r 30
\ Focal length, f = = + = +15 cm h' n 1.15
2 2 m= =– =– = +0.23
h u (–5)
1 1 1 Thus, the image is virtual, erect and smaller in size
Using the mirror formula, + = , we get
n u f than the object.

1 1 1 Example-24
+ =
n - 20 + 15 A convex mirror used for rear-view on an automobile
has a radius of curvature of 3.00m. If a bus is located at
1 1 1 7 5.00 m from this mirror, find the position, nature and
or = + =
n 15 20 60 size of the image.
Sol. Radius of curvature, R = +3.00 m;
60 Object-distance, u = –5.00 m;
or n = cm = 8.5 cm. Image-distance, v = ?
7
Height of the image, h’ = ?
The image is formed 8.5 cm from the mirror. The
positive sign shows that the image is formed on the 3.00 m
Focal length, f = R/2 = + = + 1.50 m
other side or behind the mirror. So the image is virtual. 2
Magnification, 1 1 1
Since, + =
h' n v u f
m= =-
h u
1 1 1 1 1 1 1
or v = f - u = + 1.50 - ( -5.00) = 1.50 + 5.00
h' 60 / 7 60 3
or =– =+ =
5 (–20) 7 ´ 20) 7 =
5.00 + 1.50
7.50
3 15 +7.50
or h’ = 5 × = v = = +1.15 m
7 7 6.50
The image is 1.15 m at the back of the mirror.
= 2.1 cm
The height of the image is 2.1 cm. Positive sign shows h' v 1.15 m
Magnification, m = h = - u = - -5.00 m
that the image is erect.
The image is virtual, erect and smaller in size by a factor
of 0.23.

1-60
 Geometrical Optics

REFRACTION OF LIGHT
The bending of a ray of light as it passes from one inc normal i>r
ide
medium to another is called refraction. It is due to nt r
ay
change in velocity of light, While travelling from one i
air µR
medium to another.
(i) The maximum velocity of light is 3 × 108 m/sec in µD

ref
r

rac
vacuum or air.

ted
(ii) The velocity is less in denser medium.

ray
Water

N m 21 or 1 m 2
incident ray
Medium (I)
m2 (c / v2 ) V1
i or 1 m2 = = =
m1 (c / v1 ) V2
A B
l Bending of light ray
Medium (II) According to Snell’s law, m1 sin i = m2 sin r
(i) If light passes from rarer to denser medium
reflected ray m1= mR and m2= mD

inc
ide normal i>r i<r normal ray
nt ted
ray
e frac
air i r r
mR mR
Optically Denser mD mD

ref
Medium water

ay
r i

rac

r
nt
ted
A B

ide
O

ra

inc
water denser medium
r

y
R Optically Rarer
Medium
Normal
sin i mD
so that = >1Þ Ð i > Ðr
sin r mR
THE LAWS OF REFRACTION in passing from rarer to denser medium, the ray bends
(i) The ratio of sine of the angle of incidence to the towards the normal.
sine of the angle of refraction for a particular pair (ii) If light passes from denser to rarer medium m1 = mD
of media is constant. Thus if the angle of incidence and m2 = mR

is i, and that of refraction is r, then

sin i
= constant sin i mD
sin r = <1Þ Ð i < Ðr
sin r mR
= µ it is known as Snell’s law.
(ii) The incident ray, the refracted ray and the normal In passing from denser to rarer medium, the ray bends
at the point of incidence points, all lie in the same away from the normal
Refractive index depends on nature and density of
plane.
medium and colour of light refractive index is maximum
for violet and minimum of red light.
µ1
v1 COMPETITION WINDOW
v2 µ2 APPARENT DEPTH AND NORMAL SHIFT
If a point object in denser medium is observed from
rarer medium and boundary is plane, then from Snell’s
RELATIVE REFRACTIVE INDEX law we have
When light passes from one medium to the other, the m D sin i = m R sin r .....(i)
refractive index of medium 2 relative to 1 is written as If the rays OA and OB are close enough to reach the
1
m2 and is defined as eye.

1-61
Geometrical Optics

P P
sin i » tan i = sin r » tan r =
d ac and d ap
mR I image
So that eqn.(i) becomes x shift
r
dac = actual depth dap = apparent depth apparent dap O B
height dAC i
i dac actual
height
A B
mR A B
r r
p

r
mD
dap denser
I i medium
mD
dac i
x
O object
Object in a rarer medium is seen from a denser
medium
P P d ac m R m1
mD mR i.e., d = m = m d ac m1 m R 1
d ap d ac = = = (< 1)
ap D 2
d ap m 2 m D m
(If m R = 1, m D = m )
d ap = m d ac
d ac d ap > d ac
d ap = so dap < dac ……(ii)
m
A high flying object appears to be higher than in really
The distance between object and its image, called x = dap – dac
normal shift (x) Þ x = [ m -1 ]dac

Lateral Shift
The perpendicular distance between incident and
dap emergent ray is known as lateral shift.

dac image N
dac
dap= A
i
object
O D

r i–r t
C
D d
B

Lateral Shift d = BC and t = thickness of slab

m image dap In D BOC
dac
BC d
sin(i - r) = =
OB OB
object
Þ d = OBsin(i - r) ...(i)
d ac
x = dac – dap [Q d ap =
OD t
] In DOBD cos r = =
m OB OB

d ac é 1ù t
x = d ac - = d ac ê1 - ú …(iii) Þ OB = ...(ii)
m ë mû cos r

t
é 1ù From (i) and (ii) d= sin(i - r)
if d ac = d , x = d ê1 - ú cos r
ë mû

1-62
 Geometrical Optics

SOME ILLUSTRATIONS OF REFRACTION m R VD l D

l Bending of an object sin qC = = =
m D VR l R
l When light ray travel from a medium of refractive
index m to air then m R = 1, m D = m

1
sin qC =
m

1
Þ sin qC = µl Þ q(red) > qC (violet)
m

When a point object in a denser medium is seen from a 1

l For TIR, i > qC Þ sin i > sin qC Þ m >
sin i
d
Rarer medium it appears to be at a depth
m
æ 1ö
l Twinkling of stars çQ sin i = ÷
è m ø
Due to fluctuations in refractive index of atmosphere
the refraction become irregular and the light sometimes l When ray travel from glass to air
reaches the eye and sometimes it does not. This gives mR 1
rise to twinkling of stars. sin qC = = » 420
mD 3
2
COMPETITION WINDOW
Total Internal Reflection (TIR) Þ i > 420(TIR), i < 420 (refraction)
When light ray travel from denser to rarer medium it l When ray travel from water to air
bend away from the normal if the angle of incidenc is mR 1
increased angle of refraction will also increased. At a sin qC = = » 490
mD 4
particular value of angle of incidence the refracted ray 3
subtend 900 angle with the normal, this angle of incident
Þ i > 490(TIR), i < 490 (refraction)
is known as critical angle ( qC ). If angle of incidence l When ray travel from glass to water
further increase the ray come back in the same medium.
This phenomenon is known as total internal reflection. 4
mR 3
sin qC = = » 630
CONDITIONS mD 3
2
l Angle of incidence > critical angle [i > qC ]
l Light should travel from denser to rare medium Þ i > 630(TIR), i < 630 (refraction)
Þ Glass to air, water to air, Glass to water
l A point object is situated at the bottom of tank
µR rarer filled with a liquid of refractive index m upto height
r medium
r=90° h. It is found light from the source come out of
mR
x y liquid surface through a circular portion above the
qC

qC denser object then radius and area of circle

i2 TI R
i1 <

i1 medium
1 qC
i= i2> qC mD
r
O sin qC =
r + h2
2

Snell Law at boundary x – y,

1 1 r 1 r2
sin qC = Þ = Þ 2 =
m m m r +h
2 2 m r + h2
2
m D sin qC = mR sin 900 Þ sin qC = R
mD
m2 r 2 = r2 + h2
1 1
As mµ µ
l v Þ (m2 - 1)r 2 = h 2

1-63
Geometrical Optics

Similar to ‘mirage’ in deserts, in polar regions ‘looming’

h
r= takes place due to TIR. Here m decreases with height
Þ and area pr 2
m2 - 1 and so the image of an object is formed in air if (i>qc) as
shown in Fig.
r
Þ tan qC = Þ r = h tan qC
h

Angle which the eye of fish make = 2qC = 2 ´ 49 0 = 980 .

This angle does not depend on depth of liquid rare
sky
SOME ILLUSTRATIONS OF TOTAL INTERNAL denser
REFLECTION

l Sparkling of diamond
The sparkling of diamond is due to total internal
GOLDEN KEY POINTS
reflection inside it. As refractive index for diamond is
l A diver in water at a depth d sees the world outside
2.5 so qC = 240. Now the cutting of diamond are such
through a horizontal circle of radius. r = d tan q C .
that I > qc . So TIR take place again and again inside it.
l For total internal reflection to take place light must
The light which beams out from a few places in some
be propagating from denser to rarer medium.
specific directions makes it sparkle.
l In case of total internal reflection, as all (i.e. 100%)
incident light is reflected back into the same medium
l Optical Fiber
there is no loss of intensity while in case of
reflection from mirror or refraction from lenses there
is some loss of intensity as all light can never be
1 reflected or refracted. This is why images formed
1 > 2
by TIR are much brighter. than formed by mirrors
or lenses.

light pipe
SOLVED EXAMPLE
In it light through multiple total internal reflections is Example-25
propagated along the axis of a glass fiber of radius of
Speed of light in water is 2.25 × 108 m/s. Calculate the
few microns in which index of refraction of core is
refractive index of water.
greater than that of surroundings.
Sol. Refractive index is given by

l Mirage and looming speed of light in vaccum (c)

n=
Mirage is caused by total internal reflection in deserts speed of light in water (n )
where due to heating of the earth, refractive index of
air near the surface earth becomes lesser than above 3 ´ 10 8 m / s
= = 1 .33 .
it. Light from distant objects reaches the surface of 2.25 ´ 10 8 m / s
earth with i > qc so that TIR will take place and we see
the image of an object along with the object as shown
Example-26
in figure.
Refractive index of diamond is 2.42. Calculate the speed
cold air
of light in diamond.
Sol. We know that refractive index,

hot surface
hot air c speed of light in vaccum
n= =
n speed of light in diamond

3 ´ 108 3 ´ 108
or 2.42 = or n= = 1.24 × 108 m/s.
n 2.42

1-64
 Geometrical Optics

Sol. According to given problem, r + 900 + r’ = 1800

Example-27 i.e. r’ = 900 – r
The refractive index of diamond is 2.47 and that of Þ r’ = (900 – i )
glass is 1.51. How much faster does light travel in glass Þ sin i = m cos i[Q sin(90 - i ) = cos i]
than in diamond.?
Þ tan i = m
Sol. Let n1 and n2 be the refractive indices and v1 and v2 be
the velocity of light in diamond and glass respectively, Þ i = tan -1 m = tan -1 (1.62 = 58.30 ) .
then
Example-30
c c 3 ´ 10 8 m / s
n1 = or v1 = = = 1.215 ´ 10 8 m / s Light enters from air to glass having refractive index
v1 n1 2.47
1.50. What is the speed of light in the glass? The speed
of light in vacuum is 3 × 108m/s.
c c 3 ´ 10 8 m / s
n2 = or v 2 = = = 1.987 ´ 10 8 m / s
v2 n2 2.51 Sol. Refractive index m g = 1.5
Speed of light in vacuum c = 3 × 108 m/s
v 2 - v1 = (1.987 - 1.215) ´ 108 = 0.772 ´ 108 m / s
Speed of light in glass, v =?
= 7.72 ´ 10 7 m / s c
Thus light travels 7.72 ×107 m/s faster in glass than mg =
v
diamond.
c 3 ´108
v= = = 2 ´108 m / s
Example-28 mg 1.5
A tank is filled with water to a height of 12.5 cm. The
apparent depth of a needle lying at the bottom of the Example-31
tank is measured by a microscope to be 9.4 cm. What
A ray of light travelling in air falls on the surface of
is the refractive index of water? If water is replaced by
water. The angle of incidence is 60° with the normal to
a liquid of refractive index 1.63 upto the same height.
the surface. The refractive index of water = 4/3.
What will be apparent depth?
Calculate the angle of refraction.
Sol. Here, real depth = 12.5 cm: apparent depth = 9.4 cm:
sin i
m =? Sol. We know that =n
sin r
realdepth 125
Q m \ m= = 1.33 Here, i = 60°,
apparent depth 9 .4
n = 4/3
Now, in the second case, m = 1.63, real depth = 12.5 sin 60° 4
\ =
cm: apparent depth dap =? sin r 3

12.5 12.5 3/2 4

\ 1.63 = Þ d ap = = 7.67 cm or =
d ap 1.63 sin r 3

Example-29 3 3
A ray of light is incident on a transparent glass slab of or sin r = = 0.649
8
refractive index 1.62. If the reflected and refracted rays
\ r = 40.5°
are mutually perpendicular, what is the angle of
incidence? [tan-1(1.62) = 58.30] Example-32
If the refractive index of water is 4/3 and that of glass
is 3/2. Calculate the refractive index of glass with
i r respect to water.
µ=1 Sol. We known that
µ O mg
w
mg =
r' mw

w
where mg = refractive index of glass

1-65
Geometrical Optics

with respect to water m g = refractive index of glass

Example-36
= 3/2 A coin is placed in a tumbler, water is then filled in the
m w = refractive index of water = 4/3 tumbler to a height of 20 cm. If the refractive index of
water is 4/3, calculate the apparent depth of the coin.
g 3/ 2 9
\ mw = = = 1.1 Sol. Here,
4/3 8
Real depth, h = 20 cm
Example-33 Refractive index, n = 4/3
A ray of light is incident on the plane surface of a real depth
transparent medium at an angle 60° with the normal. Now, n =
apparent depth
The angle of refraction is 30°. Calculate the refractive
index of the transparent material.
4 20
Sol. Here, or =
3 apparent depth
Angle of incidence, i = 60°
Angle of refraction, r = 30°
20´ 3
or Apparent depth = = 15 cm
sin i sin 60° 3/2 4
Refractive index, n = = = = 3
sin r sin 30° 1 / 2
Example-37
Example-34 There is a black spot on a table. A glass slab of
A ray of light travelling in air falls on the surface of a thickness 6 cm is placed on the table over the spot.
glass slab at an angle 45° with the normal. The Refractive index of glass is 3/2. At what depth from
refractive index of glass is 1.5. Calculate the angle of the upper surface will the spot appear when viewed
refraction. from above?
Sol. Angle of incidence = 45° Sol. Real depth of the spot = 6 cm
Refractive index of glass, n = 1.5
3
Refractive index of glass, n =
sin i 2
Since n =
sin r
real depth
sin 45° Now, n =
apparent depth
or 1.5 =
sin r
3 6
sin 45° 1 / 2 1 or =
or sin r = = = 2 apparent depth
1.5 1.5 2 ´1.5

6´ 2
1 1 \ Apparent depth = = 4 cm
= = = 0.4728 3
1.41 ´ 1.5 2.115
\ r = 28°
Example-38
Example-35 Refractive index of diamond is 2.42 and that of glass
The refractive index of diamond is 2.42 and that of is 1.5. Calculate the critical angle for diamond-glass
carbon disulphide is 1.63. Calculate the refractive surface.
index of diamond with respect to carbon disulphide. Sol. Refractive index of diamond, n 1 = 2.42
Sol. Refractive index of carbon disulphide, n 1 = 1.63 Refractive index of glass, n2 = 1.5
Refractive index of diamond, n2 = 2.42
\ Refractive index of diamond with respect to carbon n2 1.5
Now, sin ic = = = 0.6198
disulphide, n1 2.42
n 2 2.42
1
n2 = = =1.48 \ ic = 38.3°
n1 1.63

1-66
 Geometrical Optics

Example-39
A
Refractive index of glass is 3/2. A ray of light travelling
N
in glass is incident on glass-water surface at an angle Angle of
incidence d N
30° with normal. Will it be able to come out into the Q R Angle of
i e emergence
water Refractive index of water = 4/3. P
Incident ray N
90° N 90° S
Sol. Refractive index of glass, n1 = 3/2 B C Emergent ray
Refraction trough a triangular glass prism
Refractive index of water, n 2 = 4/3

n2 4 / 3 8
Now, sin ic = = = = 0.88
n1 3 / 2 9 A prism is a transparent refracting medium bound by
\ ic = 62° two plane surfaces inclined to each other at certain
Since, the angle of incidence (30°) is less than the angle (commonly 600 or 450),
critical angle, the ray will be refracted into the water. l The faces ABED and ACFD are refracting surfaces
of the prism.
Example-40 l The face BEFC is the base of the prism.
The refractive index of dense flint glass is 1.65 and
The angle BAC is the prism angle.
that of alcohol is 1.36, both with respect to air. What
l The line of intersection of the two refracting
is the refractive index of flint glass with respect to
alcohol? surfaces is called refracting angle (the line AD in
Sol. Refractive index of flint glass, n2 = 1.65 the diagram) or the prism.
Refractive index of alcohol, n 1 = 1.36 l The face ABC is the principal section of the prism.
\ Refractive index of flint glass with respect to The principal section of a prism is perpendicular to
alcohol is given by its refracting edge.
n 2 1.65
1
n2 = = = 1.21 HOW DOES LIGHT GET REFRACTED BY A PRISM
n1 1.36
The incident ray suffers a deviation (or bending)
through an angle d due to refraction through the prism.
REFRACTION OF LIGHT THROUGH A PRISM
The angle d is called the angle of deviation.
What is a prism

Refracting edge Note

D l The prism is in the position of minimum
A deviation when angle of emergence = angle of
Pr ism Pr ism
angle angle incidence.
or Ð e = Ð i
l The refractive index of the material of prism is
given as

E Base F æ A +dm ö
B
C sin ç ÷
a. Three - dim ensionalview è 2 ø
of a glass prism m=
A
sin
Refracting angle
2
A

DISPERSION OF WHITE LIGHT BY A GLASS PRISM

What is meant by dispersion of light
The process of splitting white light into its seven
constituent colours is called dispersion of white light.
Base
The band of seven colours formed on a screen due to
B the dispersion of white light is called spectrum of
b. The principal sec tion
C
of a glass prism
visible light or spectrum of white light.

1-67
Geometrical Optics

Least bending
Original
direction of the
S lit Spectrum incident light
Slit
Red
Orange White
White Y ellow Re d ( le ast bent ray )
Lig ht G reen Light
Blue
Indigo
violet
Viole t ( most bent ra y )
Air
Most bend ing G lass

G las s prism Different colours bend

through different angle during
refraction through angle a glass prism
Other colours lie in between
What causes dispersion of white light? the two

Dispersion of white light into seven colours occurs

because the light of different colours has different
wavelength. In this band of seven colours, red light Recombination
is caused by prism P2
has the longest wavelength and violet has the shortest.
Lights of all colours travel at the same speed in R
White light
vacuum. But, in any transparent medium, such as glass
White light
or water, the lights of different colours travel with
different speeds. (Sunlight )
V
Due to difference in their speeds, the lights of different
colours bend through different angles. In any Dispersion is
Recomposing the dispersed white light
caused prism P1
transparent medium, the red light travels the fastest,
and the violet light the slowest of all the seven colours.
Therefore, the red light bends the least, and violet light How does a rainbow form
Rainbow is an example of the dispersion of white light.
bends the most.
Just after the rain, a large number of small droplets of
Thus, dispersion of white light into seven colours
water remain suspended in the air. Each drop acts like
occurs because the lights of different colour bend a small prism. When sunlight falls on these drops, the
through different angle while passing through a glass white light splits into seven colours. The dispersed
prism. light form a large number of drops forms a continuous
band of seven colours. This coloured band is called
What is meant by monochromatic and polychromatic rainbow. Thus, rainbow is produced due to dispersion
light? of white light by small raindrops hanging in the air
The light of one single colour, or one single after the rain.
The rainbow is seen when the sun is behind the
wavelength is called monochromatic light (chrome
observer.
means colour). Sodium light is golden yellow in colour.
So, sodium light is monochromatic light.
The light made up of many colours, or light consisting
Refraction (rarer to
of radiations of many wavelengths, is called Sunlight denser medium)
polychromatic light, white light is made up of seven
40°
colours. So, white light (or sunlight) is a polychromatic Red
42° Violet
light.
Total internal
Violet reflection
How the dispersed white light is recomposed Red
Refraction (denser
Recombination of the colours of the dispersed white to rare medium)

light to get white light is called recomposing of the Refraction of sunlight by a spherical raindrop
leading to the formation of the rainbow
dispersed white light.

1-68
 Geometrical Optics

LENSES
A lens is a piece of any transparent material bound by
two curved surfaces or by one curved and one plane v=
surface. Lens are of two types:
(i) Convex or convergent lens.
F1 O O F1
(ii) Concave or divergent lens.
convex concave
lens lens

While second focal point is an image point on the

R1 R2 R R ¥ R R1 R2
principal axis for which object lies at infinity

u=
Bi-convex equi-convex plano-convex cancavo-convex u=
F2 F2
O O
convex
lens concave
f f lens

R1 R2 R R R1 R2=¥ R1 R2

Focal Length f is defined as the distance between

optical centre of a lens and the point where the parallel
Bi-concave equi plano convexo beam of light converges or appears to converge.
concave concave concave

Aperture: In reference to a lens, aperture means the

l Optical Centre: O is a point for a given lens through effective diameter. Intensity of image formed by a lens
which any ray passes undeviated which depends on the light passing through the lens
will depend on th e square of aperture, i.e.,
I µ (Aperture)2
optical
centre O RULES FOR IMAGE FORMATION
C1 C2 A ray passing through optical center proceeds
to
ed ntre undeviated through the lens
t e
ec l c
dir tica A ray passing through first focus or directed to
op convex lens
towards it, after refraction from the lens, becomes
parallel to the principal axis.
A ray passing parallel to the principal axis after
refraction through it passes or appears to pass through
C1 O F2
C2
For Convergent or Convex Lens

concave lens Object Image Magnification

¥ F m << - 1
l Principal Axis: C1C2 is line passing through ¥ - 2F F – 2F m< -1
optical centre and perpendicular to the lens. 2F 2F m= -1
2F-F 2F - ¥ m> -1
l Principal Focus: A lens has two surfaces and F ¥ m >> -1
hence two focal points. First focal point is an object F–O In front of lens m>+1
point on the principal axis for which image is formed
at infinity.

1-69
Geometrical Optics

IMAGE FORMATION FOR CONVEX LENS (CONVERGENT

LENS)
(i) Object is placed at infinity h2
Image: at F real inverted very small in size m << - 1
2F F h1 O F 2F
I O

u
F O F v

(ii) Object is placed in between Given Real Virtual

u - -
h1 v + +
2F F O F I 2F
O
h2
h1 + +
h2 - +
u m - +
¥ - 2F
Image: real (F – 2F) inverted small in size
IMAGE FORMATION FOR CONCAVE LENS (DIVERGENT
(diminished)
LENS)
M
(iii) Object is placed at 2F
Image: real (at 2F) inverted equal (of same size)
(m = - 1) F
2F O
h1
2F F O F 2F
O
h2 N
I
(i)
u v

A M

A'
(iv) Object is placed in between 2F – F
Image: real (2F - ¥ ) inverted enlarged m > 1 B F B' O

h1 N
F O F 2F I
2F Image is virtual, diminished, erect, towards the object,
O
m = +ve
h2
(i) Object is placed at infinity
Image: At F virtual erected diminished (m << + 1)
u v
(ii) Object is placed infront of lens
(v) Object is placed at F Image: between F and optical centre virtual erected
M diminished (m < + 1)
A

B O F 2F Formula of lens: Focal length of a lens can be finding

2F F C2 out by the following formula.
F
1 1 1
N = -
f v u

(vi) Object is placed in between F – O Where – f = Focal length of lens.

Image: virtual (in front of lens) erected enlarge v = Distance of image from pole
(m > + 1) u = Distance of object from pole.

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 Geometrical Optics

Uses of this formula Height of the image (h1)

(i) Put the correct signs of known variables according Height of the object (h0)
to the sign conventions. Size of the image (I)
(ii) Do not put the sign of unknown variable. The sign Size of the object (O)
will automatically show up during calculations. Height of the image ( h1 )
M = Height of the object h
f1 f2 ( 0)
Size of the image ( I ) v
= Size of the object O =
( ) u

Uses of Lenses in Daily Life

Lenses are using is microscope, telescope. Prism,
(iii) If the calculated sign of a variable turns out
binoculars and slide projector etc.
positive, then the variable calculate is on the other
side of the lens, i.e., on the opposite side to the
RELATION BETWEEN FOCAL LENGTH (F) AND RADIUS
object. However if calculated variable is of negative OF CURVATURE (R)
sign, then it is on the same side as the object. Focal length of a lens depends on following factors:-
(i) Refractive index of the material of the lens.
COMBINATIONS OF LENSES (ii) Radius of curvature R1 and R2 of both in curved
Two thin lens are placed in contact to each other surfaces of the lens.
power of combination. (iii) Colour (wave length) of the light.
1 1 1
P = P1 + P2 = + Formula of making a spherical lens of specific focal
F f1 f 2
length.
f1 f2
1 æ 1 1 ö
= ( m - 1)çç - ÷÷
f R
è 1 R 2 ø

Where f is the average value of focal length for all the

colour. m is the refractive index of the material of the
lens with respect to air. R1 and R2 are the radius of
curvature of the curved surfaces respectively. R1 is
d
positive and R2 is negative for convex lens. R1 is
Use sign convention when solve numerical negative and R2 positive for concave lens.

Two thin lens are placed in a small distance d

SOLVED EXAMPLE
1 1 1 d
= + - Example-41
F f1 f 2 f1f 2
A lens has power of - 2.5 D. What is the focal length
P = P1 + P2 – d P1 P2 and nature of the lens?
Use sign convention when solve numerical Sol. P = – 2.5 D, f =?
Power of lens: The power of a lens is defined as
reciprocal of focal length of the lens. Focal length 1
From relation, P =
should always be measured in meters. f

1 u-v 1 1
P= f = = -0.4m = -40cm
f (m)
Þ P= P - 2 .5
uv
Negative sign indicates that it is a concave lens.
Unit of power of lens is 1/meter which is called
Example-42
DIOPTER . A doctor has prescribed a corrective lens of power
Magnification of lens: The magnification is defined as +1.5 D. Find the focal length of the lens Is the
the ratio of the height of the image and the height of prescribed lens diverging or converging? [NCERT]
the object it is represented by M. Sol. P = + 1.5D

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Geometrical Optics

1 10 2 1 1 1
f (in meters) = = = = 66.6cm - =
P 15 3 + 24 - 36 f
As the focal length and power of the lens is positive
therefore, lens is a convex (converging) lens). 1 1 1 5
or = + =
f 24 36 72
Example-43
An object 5 cm in length is held 25 cm away from a 72
\ f= = 14.4 cm
converging lens of length 10 cm. Draw the ray diagram 5
and find the position, size and the nature of the image
formed. [NCERT] Example-45
Sol. f = + 10 m A 2 cm long pin is placed perpendicular to the principal
axis of a lens of focal length 15 cm at distance of 25 cm
A from the lens. Find the position of image and its size.
Sol. Here, u = –25 cm, f = +15

B 10cm B' 1 1 1
Using the lens formula, - = we get
25cm 20cm 10cm O 2F2 n u f

1 1 1
A' or - =
n - 25 + 15

u = -25cm 1 1 1 2
or = - =
1 1 1 n 15 25 75
Using lens formula, - =
v u f
75
or n = = 37.5 cm
1 1 æ 1 ö 2
= -ç ÷
10 v è - 25 ø The positive sign shows that the image is formed on
the right-hand side of the lens.
1 1 1
= - Magnification is given by
v 10 25
h' n
1 3 50 m= =
= cm Þ v = = 16.7 m h u
v 50 3
The image is real at a distance of 16.7 cm behind the h ' 37.5
or = = –1.5
lens h - 25
h' v \ h = – 1.5 × h = –1.5 × 2 cm
m= =
h u = –3 cm
The image of the pin is 3 cm long. The negative sign
h' 50 / 3 50 / 3 ´ 5 10
= Þ h' = = - cm shows that it is formed below the principal axis, i.e.
5 - 25 - 25 3
the image is inverted.
Height of the image is 3.3 cm in height.
Example-46
Example-44
A point object is placed at a distance of 18 cm from a
An object is placed 36 cm from a convex lens. A real
convex lens on its principal axis. Its image is formed
image is formed 24 cm from the lens. Calculate the
on the other side of the lens at 27 cm. Calculate the
focal length of the lens.
focal length of the lens.
Sol. According to the sign convention the object is placed
Sol. According to the sign convention, the object is placed
on the left-hand side of the lens. So object distance
on the left-hand side of the lens, therefore object-
(u) is negative. Real image is formed on the other side
distance is negative, i.e. u = –18 cm. Since the image is
of the lens. So the image distance (n) is positive.
formed on the other side, the image-distance is
Thus, u = –36 cm, n = +24 cm, f = ?
positive, i.e., n = +27 cm. Using lens formula,
1 1 1 1 1 1
Using lens formula, - = , we get - = ,
n u f n u f

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 Geometrical Optics

we have
Example-49
1 1 1
- = An object is placed on the principal axis of a concave
+ 27 - 18 f
lens at a distance of 40 cm from it. If the focal length of
1 1 5 1 the lens is also 40 cm, find the location of the image and
or + = =
27 18 54 f the magnification.
Sol. For a concave lens focal length f is negative,
54 i.e. f = –40 cm. Since by convention, object is placed
or f = = 10.8 cm
5 on the left of the lens, so u = – 40 cm.

Example-47 1 1 1
Using the lens formula, – = ,
A convex lens forms an image of the same size as the n u f
object at a distance of 30 cm from the lens. Find the we get
focal length of the lens. Also find power of the lens. 1 1 1
What is the distance of the object from the lens? - =
n - 40 - 40
Sol. A convex lens forms the image of the same size as the
object only when the object is placed at a distance 2f 1 1 1 1
or =– – =–
from the lens. In this case the image is also equal to 2f n 40 40 20
from the lens. or n = – 20 cm
Hence, 2f = 30 cm The image is formed 20 cm from the lens. Minus sign
or f = 15 cm = 0.15 m shows that the image is formed on the same side of
Power of the lens, the lens as the object.

1 1 h' n -20 1
P= = D = 6.6D Now, magnification, m = = = =
f 0.15 h u - 40 2
The distance of the object from the lens is also Positive sign shows that the imag is erect.
2f = 30 cm.
Example-50
Example-48 A beam of light travelling parallel to the principal axis
A 3 cm high object is placed at a distance of 80 cm of a concave lens appears to diverge from a point 25
from a concave lens of focal length 20 cm. Find the cm behind the lens after refraction. Calculate the power
position and size of the image. of the lens.
Sol. Here, u = –80 cm, f = – 20 cm Sol. When a parallel beam after refraction through the lens
is incident on a concave lens, if appears to diverge
1 1 1 from the focus of the lens. Hence, the focal length of
Using the lens formula, - = ,
n u f the lens is 25 cm. According to sign convention, focal
we get length of a concave lens is negative.
\ f = –25 cm = –0.25 m
1 1 1
- =
n - 80 - 20 1 1
\ Power, P = = = – 4D
f - 0.25
1 1 1 -5 1
or =- – = =–
n 20 80 80 16 Example-51
or n = –16 cm A convex lens of power 5D is placed at a distance of
30 cm from a screen. At what distance from the lens
h' n -16 1 should the screen be placed so that its image is formed
Magnification, m = = = =
h u - 80 5 on the screen?
Sol. Power of the lens, P = +5D
h 3.0
or h’ = = = 0.6 cm
5 5 1 1
\ Focal length, f= = = 0.20 m = 20 cm
5D 5
Length of image is 0.6 cm. Positive sign shows that
the image is erect. Here, the screen is placed 30 cm from the lens.
\ n = +30 cm, f = +20 cm, u = ?

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Geometrical Optics

Power of the combination,

Using the lens formula, 1 - 1 = 1 , P = P1 + P2 = 2.5 – 4D = – 1.5D
n u f
Example-54
1 1 1
we get - = A concave lens has a focal length of 15 cm. At what
30 u 20
distance should the object be from the lens placed so
1 1 1 1 that if forms an image 10 cm from the lens? Also find
or = – =– the magnification.
u 30 20 60
Sol. A concave lens always forms a virtual, erect image on
or u = – 60 cm
the same side as the object.
Therefore, the screen should be placed at 60 cm from
Image distance, n = –10 cm
the lens.
Focal length f = –15 cm
Example-52 Object distance, u=?
A pin 3 cm long is placed at a distance of 24 cm from a
Using, the lens formula, 1 - 1 = 1 , we get
convex lens of focal length 18 cm. The pin is placed n u f
perpendicular to the principal axis. Find the position,
size and nature of the image. 1 1 1
or - =
Sol. Here, u = –24 cm, f = +18 cm, n = ? - 10 u - 15

1 1 1 1 2-3 1
Using the lens formula, - = , or – = =-
n u f 10 30 30
we get or u = –30 cm
1 1 1 Thus, the object should be placed 30 cm on the lens.
- =
n - 24 + 18 n -10 1
Magnification, m = = = = 0.33
u - 30 3
1 1 1 1
or = - = The positive sign shows that the image is erect and
n 18 24 72
virtual. The size of the image is one-third of that of the
or n = 72 cm
object.
The imag is formed 72 cm from the lens on the other
side. So the image is real. Example-55
h' n 72 A 2 cm tall object is placed perpendicular to the
Magnification, m = = = = –3
h u - 24 principal axis of a convex lens of focal length
10 cm. The distance of object from the lens is 15 cm.
or h’ = –3 × h = –3 × 3.0 = – 9 cm
The image is 9 cm in size. Negative sign shows that Find the position, nature and size of the image.
the image is inverted. Calculate the magnification of the lens.
Sol. Object distance, u = –15 cm
Example-53 Focal length, f = +10 cm
A convex lens of focal length 40 cm and a concave Object height, h = +2 cm
lens of focal length 25 cm are placed in contact in Image distance, n=?
such a way that they have the common principal axis. Image height, h’ = ?
Find the power of the combination.
1 1 1
Sol. Focal length of the convex lens, f1 = 40 cm = +0.4 m Using the lens formula, – =
n u f
\ Power of the convex lens,
we get
1
P1= = +2.5D
+ 0.40 1 1 1
= =
Focal length of the concave lens, n - 15 + 10
f2 = –25 cm = –0.25 m
\ Power of the concave lens, 1 1 1 +1
or = – =
n 10 15 30
1
P2 = = – 4D or n = +30 cm
- 0.25

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 Geometrical Optics

Positive sign of n shows that the image is formed at a

distance of 30 cm on the right side of the lens. Example-57
Therefore the image is real and inverted An object 25 cm high is placed in front of a convex
lens of focal length 30 cm. If the height of image formed
h' n is 50 cm find the distance between the object and the
Magnification, m = =
h u image?
Sol. As object is in front of the lens, is real and as
h' +30 h1 = 25cm, f = 30cm, h2 = -50cm
= = –2
2.0 - 15
h2 - 50
or h’ = –2 × 2 = –4 cm m= = = -2
h1 25
Magnification,
f 30
n 30 m= Þ -2=
m= = = –2 f +u 30 + u
u - 15
v v
Negative sign with the magnification and height of u = -45cm Þ m= Þ -2=
u - 45
the image shows that the image is inverted and real. Þ v = 90 cm
Thus, a real image of height 4 cm is formed at a distance As in this situation object and image are on opposite
of 30 cm on the right side of the lens. Image is inverted sides of lens, the distance between object and image
and twice the size of the object. d1 = u + v = 45 + 90 = 135 cm

Example-56 If the image is erect (i.e., virtual)

A magnifying lens has a focal length of 10cm. (a) f 30
Where should the object if the image is to be 30 cm m= Þ 2= Þ u = -15cm
f +u 30 + u
from the lens? (b) What will be the magnification?
Sol. (a) In case of magnifying lens, the lens is convergent v -v
Þ m=- Þ 2= Þ v = 30 cm.
and the image is erect, enlarged, and virtual, between u - 15
infinity As in the situation both image and object are in front
of the lens, the distance between object and image
d 2 = v - u = 3015 = 15 cm.
h2
F F Example-58
I O
A needle placed 45 cm from a lens forms an image on a
u screen placed 90 cm on other side of the lens. Identify
30cm the type of lens and determine its focal length. What
is the size of the image, it the size of the needle is 5 cm?
and object and on the same side of lens.
Sol. Here, u = - 45 cm, v = 90 cm, f =?, h 1 = 5 cm,
f = 10 cm and v = - 30 cm
1 1 1 1 1 1
1 1 1 Q - = \ + =
v u f 90 45 f
and hence from lens-formula, - =
v u f
1+ 2 1
Þ = or f = 30 cm
1 1 1 90 f
we have - = i.e., u = - 7.5 cm
- 30 u 10 As f is positive, the lens is converging
So the object must be placed in front of lens at a h2 v h2 90
distance of 7.5 cm (which is < f ) from it. Q = \ = = -2
h1 u 5 - 45
éh ù v -3 Þ h2 = – 10 cm.
(b) m = ê h ú = u = - 7.5 = 4 i.e., image is erect, virtual
2

ë 1û
Minus sign indicates that image is real and inverted

and four times the size of object.

P
2F F F 2F

45cm 90cm

12cm

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Geometrical Optics

v 1 .5 - x
Example-59 using m = , we get - 4 = Þ x = 0.3 meter
u -x
A beam of light converges to a point. P. A lens is placed
The lens is placed at a distance of 0.3 m from the object
in the path of the convergent beam 12 cm from P. At
(or 1.20m from the screen)
what point does the beam converge if the lens is (a) a
For focal length, we may use
convex lens of focal length 20 cm. (b) a concave lens
of focal length 16 cm. f
m=
Sol. Here, the point P on the right of the lens acts as a f +u
virtual object,
\ u = 12 cm, v =? f 1.2
or - 4 = Þ f = = 0.24 cm
f + (-0.3) 5
1 1 1
(a) f = 20 cm \ - =
v u f
Example-61
1 1 1 A concave lens of length 15 cm forms an image 10 cm
\ - =
v 12 20 from the lens. How far is the object placed from the
lens? Draw the ray diagram – [NCERT]

A
P
I A'
B F
25cm B' O
12cm 10cm
15cm
1 1 1 3+5 8
Þ = + = =
v 20 12 60 60 1 1 1
Sol. Using lens formula. - =
60 v u f
Þ v= = 7.5 cm
8
1 1 1
(b) f = -16 cm, u = 12 cm - =
- 10cm u - 15cm
1 1 1 1 1 3+4 1
\ = + = + = = 1 1 1 2-3 1
v f u - 16 12 48 48 = - = =-
u 15 10 30 30
Þ v = 48cm
Hence image is at 48 cm to the right of the lens, where u = -30cm
the beam would converge. Thus, the object is placed at a distance of 30 cm from
concave lens.
Example-60
An object is placed at a distance of 1.50 m from a screen Example-62
and a convex lens placed in between produces an image One-half of a convex lens is covered with a black paper.
magnified 4 times on the screen. What is the focal Will this lens produce a complete image of the object?
length and the position of the lens. Verify your answer experimentally. Explain your
observations. [NCERT]

B A

1.5 - x A'
A x C B'
B

B'
Black Paper A'

h2 Sol. Yes, it will produce a complete image of the object, as

Sol. m= = -4 shown in fig. This can be verified experimentally by
h1 observing the image of a distance object like tree on a
Let lens is placed at a distance of x from the object. screen, when half of the is covered with a black paper.
Then u = - x, However, the intensity of brightness of image will
and v = (1.5 - x) reduce.

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 Geometrical Optics

(ii) When the light is dim (as in a dark room), the iris
Example-63 makes the pupil to expand. As a result, more
An object is placed at a distance of 10 cm from a convex light can enter the eye.
mirror of focal length 15 cm. Find the position and
nature of the image. [NCERT] (c) EYE LENS: The eye lens is a convex lens
Sol. Here, object distance, u = - 10 cm focal length, f = 15 (converging lens) made of a transparent jelly-like
cm, image distance, v =? pertinacious material. The eye lens is hard at the
1 1 1 middle and gradually becomes soft towards the
As + = ,
v u f outer edges. The eye lens is held in position by
ciliary muscles. The curvature and therefore the
1 1 1 1 1 5 1
\ = - = + = = , f = 6 cm eye lens may be changed by the action of the ciliary
v f u 15 10 30 6
muscles.
Here, + sign of v indicates image is at the back of the (d) RETINA: The inside surface of the rear part of the
mirror. It must be virtual, erect and smaller in size than eye ball where the light entering the eye is focused
the object. is called retina. The surface of retina consist: of
about 125 million light sensitive receptors. These
Example-64
receptors are of two types - rods and cones. When
Find the focal length of a lens of power - 2.0 D. What
the light falls on these receptors, they send
type of lens is this?
electrical signals to the brain through the optic
Sol. Here, focal length f =?, power P = - 2.0 D
nerve.
100 The space between the eye lens and retina is filled with
As f = another liquid called vitreous humor. The image
P
formed on the retina is retained by it for about 1/
100 16th of a second.
\ f = = -50 cm
- 2. 0 (e) BUND SPOT: There are no rods and cones at the
point where optic nerves leave the eye ball to go to
Lens is concave lens
the brain. So, if any image is formed in this part of
the retina, then no signal is sent to the brain. As a
THE HUMAN EYE
result the object is not seen. This part is therefore
Our eye is the most important natural optical instrument.
called the blind spot of the eye.
The eye resembles a - camera in many ways. It has
nearly a spherical shape.
What are the essential parts of the eye and their
functions?
The essential parts of the eye are shown in Fig. and are
described below:
(a) CORNEA: The front of the eye is covered by a
transparent spherical membrane called the cornea.
Light enters the eye through cornea. The space
behind the cornea is filled with a clear liquid called
aqueous humor.

(b) IRIS AND PUPIL: The iris and pupil form the How does the eye work?
variable aperture system of the eye. Iris is a dark The light coming from an object enters the eye through
coloured muscular diaphragm which has a small cornea and the pupil. The eye lens converges these
circular opening in its middle. The central circular light rays to form a real, inverted and high diminished
aperture of iris is called pupil. image on the retina. The surface of retina consists of a
The iris regulates the amount of light entering the large number of light sensitive cells. When light falls
eye by adjusting the size of the pupil. This is on them, they get activated and generate electrical
explained below: signals. These signals are then sent to the brain by the
(i) When the light is very bright (as on a sunny day), optic nerves, and the observer sees the actual-sized,
the iris makes the pupil to contract. As a result, the erect image of the object.
amount of light entering the eye decreases.

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The property due to which eye lens is able to change

SOLVED EXAMPLE its focal length is called accommodation of the eye.

Example-65 When the eye is focused on distant objects (objects at

infinity), the ciliary muscle is fully relaxed. When the
Why does it take some time to see objects in a dim
ciliary muscle is in the relaxed state, the thickness of
room when you enter the room from bright sunlight the lens is minimum and the focal length maximum:
outside? equal to the distance of the retina from the eye lens.
Sol. In bright sunlight, iris makes the pupil to become small, Therefore, the parallel rays coming from any distant
so that only a small amount of light is able to enter & object are focused on the retina, and the object is seen
the eye. When someone enters a dim room, very little clearly.
light is available to the eye. The iris then makes the
pupil to gradually expand (open up) to allow more light When the eye is focused on a nearby object, the ciliary
muscle gets strained (tense). Tension in the ciliary
to enter the eye. This process takes some time. That is
muscle decreases the focal length of the eye lens by
why the pupil takes a little time to adjust itself to dim
slightly increasing its thickness in such a way that the
light. image is formed on the retina. Thus, the eye focuses on
the nearby objects by tensing the ciliary muscle.
What is meant by the far point, near point, and the These adjustments in the focal length of the eye lens
least distance of distinct vision? take place so fast that we do not realize such changes.
(a) FAR POINT: The farthest point up to which an eye
can see clearly is called the far point of the eye. For What is meant by the limit of accommodation?
a normal eye, the far point is at infinity. The eye can accommodate only up to a limit. A normal
eye can accommodate up to the least distance of
distinct vision: about 25 cm for a healthy adult. The
(b) NEAR POINT: The closest (nearest) point up to objects which are very close to the eye produce blurred
which an eye can see clearly is called the near point image. Thus, a normal eye can accommodate objects
of the eye. The near point for a healthy normal eye lying between infinity and the least distance of distinct
of an adult lies about 25 cm from the eye. vision.

(c) LEAST DISTANCE OF DISTINCT VISION: The What is the range of vision?
minimum distance up to which an eye can see The range of distance over which the eye can see clearly
clearly without any strain is called the least distance is called its range of vision. A normal human eye can
see objects clearly which lie between infinity and the
of distinct vision (denoted by d or D). The least
least distance of distinct vision. So, the range of vision
distance of distinct vision is thus equal to the of a normal healthy human eye is from infinity to the
distance between the eye and its near point. For a least distance of distinct vision, i.e. from infinity to
normal eye of an adult, the least distance of distinct about 25 cm from the eye.
vision is about 25 cm. This distance usually
increases with age. What is meant by the persistence of vision?
The image formed on the retina of the eyes is not
POWER OFACCOMMODATION OFTHE EYE permanent. It also does not fade away instantaneously
after the object is removed or we have stopped seeing
What is meant by accommodation of the eye?
the object. The image formed on the retina persists
A normal eye can see both the distant and the nearby
1/16th of a second even after we have stopped looking
objects clearly. For a clear vision, the image of any at the object. The ability of the eye to retain the image
object must fall on the retina. For a person, the distance (or the sensation caused by the light coming from the
between the retina and the eye lens is fixed. So, the object) for .about 1/16th of a second even after we have
distance of the image (v) from the eye lens is fixed. For stopped seeing the object is called persistence of vision.
objects at different distances, the values of u are The phenomenon of persistence of vision is made use
different. So, to get the images at the same v, the focal of in motion-picture projection (or cinematography). A
length of the eye lens should be different. sequence of still pictures is recorded on a film by a
movie camera. This recorded film is projected on a screen
The eye can focus the images of all the objects, distant
at the speed of about 24 pictures per second. Due to
or nearby, at the same place (on the retina) by changing
the persistent of vision, the successive image on the
the focal length of its lens. The eye lens can change its screen merges smoothly into one another giving an
focal length by changing its thickness with the help of impression of continuity. In this way, we are able to see
its ciliary muscles. the pictures in motion.

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COLOUR VISION may not have a particular kind of cones on its retina.
How do we see the colours? Then, such a person will not be able to distinguish
We see an object when its image is formed at the retina. between certain colours. For example, a person not
Our retina has a large number of light-sensitive cells. having cones sensitive to the blue light in his/her eye
These cells are of two shapes-rods and cones will not be sensitive to blue colour.
The rod-shaped cells respond to the intensity of light,
i.e. the degree of brightness or darkness is sensed by DEFECTS OFVISION AND THEIR CORRECTION
the rod-shaped cells on the retina of the eye. Abnormalities in the normal vision of the eye are called
The cone-shaped cells respond to the colours. These defects of vision or defects of eyes.
are sensitive to red, blue and green colours to different The most commonly observed defects of vision (or
extents. It is due to these cone-shaped cells that we are defects of eyes) are :
able to distinguish between different colours. (i) Myopia or shortsightedness or nearsightedness
The cone-shaped cells become active only in bright (ii) Hypermetropia or long sightedness or hyperopia or
light. That is why; we can’t differentiate between
farsightedness
colours in dim light. There are in fact three different
(iii) Astigmatism
kinds of the cone-shaped cells; one sensitive to red,
second to green and third to blue colour. So, depending
WHAT IS SHORT SIGHTEDNESS (OR MYOPIA)
upon the colour of the light entering the eye, one or
Short sightedness (or myopia) is the defect due to
more kinds of the cone-shaped cells get activated.
which the eye is not able to see the distant objects
clearly though it can see the nearby objects clearly.
How do certain animals including birds differ in their
So, a short sighted or myopic eye has its far point nearer
colours perception?
The structure and the number of rod-shaped and cone- than infinity.
shaped cells are different for different animals/birds.
As a result, therefore, their colours perceptions are also WHAT CAUSES SHORT SIGHTEDNESS (OR
different. For example MYOPIA)
Bees have cones that are sensitive to ultraviolet light Myopia or short sightedness is caused by the following
(light beyond violet colours). So, bees can see the reasons.
ultraviolet light present in sunlight. (a) Decrease of focal length of the eye lens, i.e. the eye
Human being cannot see the ultraviolet light because lens becomes more convergent.
the cones in their retina are not sensitive to the (b) Elongation of the eyeball, i.e. the increased length
ultraviolet light. of the eyeball.
The retina of chicken has mostly cone-shaped cells
and only a few rodshaped cells. It is because of this
reason that a chicken can see only in bright light and
wake up only the sun rise and goes to sleep at sunset

What is colour blindness?

Colour blindness is a defect of vision or defect of eye.
When a person is unable to distinguish between certain
colours, he/she is said to be colour blind. Thus colour
blindness is that defect of the eye due to which a person
is not able to distinguish between certain colours. How is short sightedness (or myopia) corrected?
Colour blindness is a genetic disorder, which occurs The short sightedness (myopia) can be corrected by
due to inheritance (from parents to their children) making the eye lens less convergent. This can be done
by placing a concave lens (divergent lens) of suitable
Cause of colour blindness focal length before the eye lens. The rays of light coming
The retina of human eye consists of three different from a distant object after passing through the concave
types of cone-shaped cells. Some are sensitive to blue (diverging) lens of the spectacles diverge slightly. As a
light, some to green light, and some others to red light. result, the rays entering the eye appear to come from
Thus, each colour of light is detected only by the cones the far point of the myopic eye, and therefore get
which are sensitive to that light. Sometimes, a person focused at the retina to form a clear image.

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Geometrical Optics

HOW IS LONG SIGHTEDNESS

T he ra ys g e t
(OR HYPERMETROPIA) CORRECTED
di ver g ed sl igh tly
du e t o the
Long sightedness (hypermetropia) can be corrected
c o nc a ve le ns by making the eye lens more convergent. This is
F a r po in t ( F ) or
t he m yop ic e ye generally done by placing a convex lens (converging
lens) of suitable focal length before the eye lens. This
R a ys f ro m a is shown in Fig.
f a r -o ff obje c t
The rays from a nearby object (about 25 cm) after
C o nc a ve
( di ve rg en t) l en s passing through the convex lens of the spectacles
C or r e ct ing myop ic
( sh or tig hte d ) e ye by usi n g a converge slightly. As a result, the rays entering the eye
c onc av e (d ive rge nt ) l en s
appear to come from the near point of the longsighted
eye, and therefore get focussed at the retina to form a
How to calculate the focal length and power of the lens clear image.
used for correcting a myopic eye
The corrective lens (concave lens) needed to correct a
myopic eye should form the image of the far-off object N
(e.g. at infinity) at the far point (d) of the myopic person. N

Thus, u = -¥, v = -d, f = ?

How to calculate the focal length and power of the lens
1 1 1 1 used for correcting a myopic eye
\ = - =- Þ f = -d
f v u d The corrective lens (a convex lens) needed to correct a
lens is concave. hypermetropic (or longsighted) eye should form the
image of the object placed at the normal near point (the
least distance of distinct vision is 25 cm) at the near
WHAT IS LONG SIGHTEDNESS
point of the hypermetropic person. Thus
(OR HYPERMETROPIA OR HYPEROPIA)
v = Near point distance of the hypermetropic eye
The long sightedness (or hypermetropia) is the defect
= –d
due to which the eye is not able to see clearly the nearby
u = Near point distance for th e normal eye
objects though it can see the distant objects clearly.
So, a longsighted eye has its near point farther away = - D = -25 cm . Using the lens formula
from the normal near point (about 25 cm for an adult).
1 1 1 1 1
= - = -
f v u -d 25cm
WHAT IS ASTIGMATISM
N A normal eye can focus all the light from any object at
the same point. Sometimes, the light coming in the
(a)
horizontal and vertical planes do not come to focus at
the same point. As a result, the horizontal and vertical
views of an object are not seen with the same clarity.
N N'
Such a defect of eye is called astigmatism. A person
(a)
suffering from astigmatism is not able to see in all the
directions equally well.
WHAT CAUSES LONG SIGHTEDNESS · Cause of astigmatism. Astigmatism occurs when
(OR HYPERMETROPIA) the cornea or the eye lens or both are not perfectly
Hypermetropia or long sightedness is caused due to spherical, i.e. the cornea or the eye lens or both are
the following reasons: more curved in one plane than in the other.
(i) Increase of the focal length of the eye lens, i.e. the
eye lens becoming less convergent. Correcting astigmatism. Astigmatism can be corrected
(ii) Shortening of the eye ball, i.e. the length of the eye by using cylindrical lenses. Cylindrical lenses have
ball has decreased. different curvature in the horizontal and vertical
directions.

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 Geometrical Optics

Simple microscope (simple magnifier)

n n
· Principle : A normal eye cannot see clearly the or =–1– ,
u f
object lying nearer than 25 cm (least distance of
distinct vision) from the eye. A single convex lens
æ Dö
helps the eye in seeing the objects from much i.e., M = – ç1 + ÷
è f ø
closer distances. It produces linear magnification
by increasing the visual angle. But here, n = D
· Diagram.
æ Dö
Hence, M = – ç1 + ÷
è f ø
Negative M means that image is virtual.

Compound Microscope
· Diagram :

· Construction : It is a single convex lens, held in a

rim with a handle. It is also called a magnifying
lens or a reading lens.
· Working : An object AB, very close to eye, is
seen by putting convex lens between eye and the
object. The lens makes its virtual image A1B1 at
the least distance of distinct vision (D). Eye sees
the object through its virtual image.
· Calculation : Object AB or image A1B1 subtend
angle b at the eye. For seeing the object directly it
must be in position A1D subtending angle a.

By definition magnifing power of visual optical

instrument.

size of image seen through the instrument

M=
size of object seen directly
When both are at same distance from the eye.
· Construction : It consists of two convex lenses
A1B1 A1B 2 AC -n n (each an achromatic doublet). One convex lens of
Hence, M = = = 1 = =
A1D AB AC -u u small focal length f0 and small aperture is fitted at
one end of a brass tube. This lens is near the
(For image on same side as object, n is –ve)
object to be seen, hence it is called objective.
1 1 1 The second convex lens of more focal length fe
From lens formula, – =
n u f and large aperture is fitted in a sliding tube. This
As n is –ve, tube can be moved in and out at other end of the
brass tube by a screw S. This lens is near the eye,
1 1 1
– = hence it is called eyepiece.
-n u f · Calculation : Magnifying power of compound
microscope = magnifying power of objective ×
n n
or – 1 – = magnifying power of eyepiece
u f
i.e., M = m0 × me

1-81
Geometrical Optics

Both lenses have a common axis, called axis of

n CA CC L
But m0 = 0 = 1 1 = 1 2 = telescope.
u0 C1A f0 f0 · Working : The object at infinity sends parallel
rays inclined to telescope axis. The objective forms
æ Dö D its real inverted image AB at its focus F.
and m0 = – çç1 + f ÷÷ = – f The eyepiece is moved inward to bring cross wires
è eø e
at image AB. The image comes at focus of the
Hence, M = m0 × me eyepiece. Final image is made at infinity. It is
virtual, inverted and much enlarged. Since eye
L D
M=- . seeing the final image at infinity is normal, this
f0 fe
adjustment of telescope is called normal
· Merits : adjustment.
(i) It has more magnifying power (about 200). The distance C1 C2 between the objective and the
(ii) It is free from effect of lens aberrations. eyepiece is called length of the telescope and is
represented by L.
Since final image is inverted, the telescope cannot
Astonomical telescope
be used for seeing terrestrial objects (objects on
· Diagram :
earth) which have shape. It is used for seeing
Objective Brass tube celestial heavenly objects which have no shape.
lens Sliding tube
Hence, it is called Astronomical Telescope.
Cross wires
Eyepiece · Calculation : Let the object subtend an angle a
on the objective. Same can be taken as the angle
Parallel rays A F C2 subtended at the eye, only a few cm behind.
a
C1 B b Let the image subtends an angle b on the eyepiece.
Eye
Same is taken as the angle subtended at the eye
just behind.
+S
To infinity in normal By definition,
adjustment Magnifying power
Astronomical telescope (Normal adustment)
size of image at inf inity seen by telescope
=
size of object at infinity seen by eye

tan gent of visual angle of image

M=
tan gent of visual angle of object

tan b AB / AC 2 in D AC2 B
= = [Q A is at F]
tan a AB / AC1 in D AC1B

AC1 FC1
= =
AC2 FC 2

· Construction : It consists of two convex lenses f0

or M=
(each an achromatic doublet). One convex lens of fe
large focal length f0 and large aperture is fitted at
Also as C1C2 = C1F + FC2
one end of a brass tube. This lens is towards the
object to be seen, hence it is called objective. Length of telescope, L = f 0 + f e

The second convex lens of less focal length fe and · Merits: It has a large field of view, because it uses
small aperture is fitted in a sliding tube. This tube can a converging (convex) lens as eyepiece.
be moved in and out at the other end of the brass tube · Demerits :
by a screw S. This lens is near the eye, hence it is (i) It makes final image inverted. hence not suitable
called eyepiece. Two perpendicular thin wires (cross for seeing terrestrial object which have shapes.
wires) are fixed in sliding tube in front of eyepiece at (ii) It has more length.
distance fe.

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 Geometrical Optics

SOLVED EXAMPLE Example-69

Find the focal length of a lens of power - 2.0 D. What
Example-66 type of lens is this?
A person cannot see objects lying beyond 2.5 m from Sol. Here, focal length f =?, power P = - 2.0 D
the eye. Calculate the power of the corrective lens he 100
As f =
should use. P
Sol. Since he cannot see the objects lying beyond 2.5 m,
100
he is suffering from myopia. The corretive lens he \ f = = -50 cm
- 2 .0
should use is a concave lens of focal length,
f = –2.5 m. Example-70
The far point of a myopic person is 80 cm in front of
1 1
\ Power of the lens =– =- = – 0.4D the eye. What is the nature and power of the lens
f 2.5 required to correct the problem?
Sol. Distance of far point, x = 80 cm, P =?
Example-67 For viewing distant objects, focal length of corrective
A person cannot see objects closer than 75 cm from lens,
f = – x = 80 cm
the eye. Calculate the power of the corrected lens
100 100
should he use. P= - = -1.25 D.
f - 80
Sol. Since the person cannot objects lying closer than
The lens is concave.
75cm, he suffers from hypermetropia. His near point
has shifted from 25 cm to 75 cm. The focal length of Example-71
the corrective lens can be calculated by considering Make a diagram to show how hypermetropia is
u = –25 cm, n = – 75 cm, f = ? corrected. The near point of a hypermetropic eye is
1m. What is the power of the lens required to correct
1 1 1 1 1 this defect? Assume that the near point of the normal
Now, = - = -
f n u - 75 - 25 eye is 25 cm.
Sol. Make diagram yourself
1 2 Here , x’ = 1m = 100 cm, d = 25, f =?
or =
f 75 x' d
From f =
x'-d
75 0.75
or f = cm = m 100 ´ 25
2 2 f = = 33.3 cm
100 - 25
1 2 8 100 100
\ Power = = D = + D = 2.66D P= = = 3D
f 0.75 3 f 33.3

Example-68 Example-72
The near point of a certain eye is 100cm in front of the Why is a normal eye not able to see clearly the objects
eye. What lens should be used to see clearly an object placed closer than 25 cm? [NCERT]
Sol. This is because the focal length of eye lens cannot be
25 cm in front of the eye?
decreased below a certain minimum limit.
Sol. u = - 25 cm v = - 100 cm
Using lens formula. Example-73
1 1 1 1 1 -1 + 4 3 The far point of a myopic person is 80 cm in front of
= - = - = = cm the eye. What is the nature and power of the lens
f v u - 100cm - 25cm 100cm 100
required to enable him to see very disant objects
distinctly?
100
f = cm = 33.3cm Sol. Since the person suffers from myopia, concave lens
3
of focal length 80 cm = –0.80 m should be used.
Hence a converging lens of focal length 33.3 cm is
1
required. \ P= = –1.25D
- 0.80

1-83
Geometrical Optics

Sol. If f0 is the focal length of the objective and fe is the

Example-74 focal length of the eyepiece, then magnification of a
A compound microscope has an objective of foal telescope for normal adjustment is given by
length 1 cm and an eyepice of focal length 4 cm. If
tube length is 20 cm, calculate the magnification of f 0 140
M= = = 28
the compound microscope. fe 5
Sol. Magnifying power of a compound microscope is given
Length of the telescope, L = f0 + fe = 140 + 5
by
= 145 cm
L D
M= ´ Example-78
f0 fe
The objective of a telescope has a focal length of 200
Here, L = 20 cm, f0 = 1, fe = 4 cm, D = 25 cm cm and the eyepiece has a focal length of 2 cm.
20 25 Calculate the magnification and tube length of this
\ M= ´ = 125.
1 4 telescope for normal adjustment.
Sol. For the normal adjustment the magnification of
Example-75 telescope is
A stamp collector uses a convex lens with a focal
f 0 200
length of 6.2 cm to examine a stamp.What is the M= = = 100
fe 2
magnification produced by the convex lens? Take the
least distance of distinct vision as 25 cm. Length of the telescope,
Sol. The magnification produced by a simple magnifier is L = f0 + fe = 200 + 2 = 202 cm
D
given by M = 1 + Example-79
f
The far point of a person suffering from myopia is 2 m
Here, D = 25 cm, f = 6.2 from the eye. Calculate the focal length and the power
25 of the corrective lens.
\ M=1 + = 1 + 4.03 = 5.03 Sol. The far point lies at 2 m. Therefore, a concave lens of
6.2
focal length 2 m should be used so that the objects
Example-76 lying at infinity can be focussed at the far point.
A compound microscope has an objective of focal \ For corrective lens, focal length,
length 0.5 cm and a tube length of 20 cm. If it produces f = –2 m
a magnification of 250, calculate the focal length of
1
the eyepiece. \ Power, P = = –0.5D
-2
L D
Sol. The magnification, M = × Example-80
f0 fe
The near point of an elderly person lies at 50 cm from
Here, L = 20 cm, f0 = 0.5 cm, D = 25 cm, M = 250.
the eye. Calculate the focal length and power of the
20 25 corrective lens.
\ 250 = 0.5 ´ f Sol. The person suffers from hypermetropia. His near point
e
lies at 50 cm. Therefore, a convex lens should be used
or fe = 4cm for the correction of his vision. The focal length of
the corrective lens is calculated by
Example-77
A telescope has an objective of focal length 140 cm 1 1 1
– =
and an eyepiece of focal length 5 cm. Calculate the - 25 - 50 f
magnification of the telescope for viewing distant
1 1
objects for normal adjustment and the separation or =
f 50
between the objective lens and the eyepiece.

1-84
 Geometrical Optics

or f = 50 cm = 0.5 m
\ Power of the corrective lens, Example-83
A 12 cm long microscope has an objective of focal
1 1
P= = = +2D length 1 cm and eyepiece of focal length 4 cm. What
f 0.5 is the magnification for normal adjustment?
Sol. For the normal adjustment, the magnification of the
Example-81 compound microscope is
The lens of a simple magnifier has a focal length
L D
of 2.5 cm. Calculate the angular magnification M= ´
f0 fe
produced when the image is at D and at infinity.
Sol. When the image s formed at D, the least distance of Here, L = 12 cm, f0 = 1 cm, fe = 4 cm, D = 25 cm
distinct vision, the angular magnification is 12 25
\ M= ´ = 75
D 25 1 4
M= 1+ =1+ = 11
f 2.5
Example-84
When the final image is at infinity, the angular
An astronomical telescope has an objective of length
magnification is
40 cm and eyepiece of focal length 2 cm. Calculate the
D 25 magnifying power and length of the telescope.
M= = = 10
f 2. 5 Sol. Magnifying power of telescope,
f 0 40
Example-82 M= = = 20
fe 2
The convex lens used in a simple microscope
Length of telescope, L = f0 + fe = 40 + 2 = 42 cm
produces a magnification of 5. The image is formed at
the least distance of distinct vision. Calculate the focal Example-85
length of the lens.
A telescope is set for normal adjustment. It has
Sol. When the image is formed at least distance of vision,
magnifying power 16 and length 85 cm. Calculate the
the angular magnification is
focal length of the objective and eyepiece.
D Sol. Magnifying power,
M=1+
f
f0
M= = 16 ...(1)
25 fe
or 5 = 1 +
1 Length, L = f0 + fe = 85 ...(2)
From equation (1),
25
or 4 = f0 = 1 + fe
f
Equation (2) gives
25 16 fe + fe = 85 or fe = 5 cm
or f = = 6.25 cm \ f0 = 5 × 10 = 80 cm.
4

1-85
Geometrical Optics

EXERCISE-I
CBSE Q.10 A student wants to project the image of a candle flame
PREVIOUS YEAR'S on a screen 60 cm in front of a mirror by keeping the
SECTION - A flame at a distance of 15 cm from its pole.
Q.1 List four characteristics of the images formed by plane (i) Write the type of mirror he should use.
mirrors. [CBSE (Delhi) 2014-15] [2] (ii) Find the linear magnification of the image
produced.
Q.2 “A concave mirror of focal length ‘f’ can form a (iii) What is the distance between the object and its
magnified erect as well as an inverted image of an image ?
object placed in front of it.” Justify this statement (iv) Draw a ray diagram to show the image formation in
stating the position of the object with respect to the this case. [CBSE (Outside Delhi) 2013-14] [3]
mirror in each case for obtaining these images.
[CBSE (Outside Delhi) 2012-13] [2] Q.11 A narrow beam PQ of white light is passing through a
glass prism ABC as shows in the diagram.
Q.3 An object is placed at a distance of 30 cm in front of a (i) Write the name and cause of the phenomenon
convex mirror of focal length 15 cm. Write four observed.
characteristics of the image formed by the mirror.
(ii) Where else in nature is this phenomenon
[CBSE (Delhi) 2017] [2]
observed?
(iii) Based on this observation, state the conclusion
Q.4 An object of height 6 cm is placed perpendicular to the
which can be drawn about the constituents of
principal axis of a concave lens of focal length 5 cm.
white light. [CBSE (Outside Delhi) 2013-14] [3]
Use lens formula to determine the position, size and
nature of the image if the distance of the object from
the lens is 10 cm. [CBSE (Delhi) 2012-13] [3] Q.12 If the image formed by a mirror for all positions of the
object placed in front of it is always erect and
Q.5 State the difference in colours of the Sun observed diminished, what type of mirror is it ? Draw a ray
during sunrise/sunset and noon. Give explanation for diagram to justify your answer. Where and why do we
each. [CBSE (Delhi) 2012-13] [3] generally use this type of mirror ?
[CBSE (Outside Delhi) 2014-15] [3]
Q.6 Name the type of mirror used (i) by dentists and (ii) in
solar furnaces. Give two reasons why such mirrors are Q.13 What is meant by scattering of light ? Use this
used in each case. phenomenon to explain why the clear sky appears blue
[CBSE (Outside Delhi) 2012-13] [3] or the sun appears reddish at sunrise.
[CBSE (Outside Delhi) 2014-15] [3]
Q.7 When and where do we see a rainbow ? How is a
rainbow formed ? Draw a labelled diagram to illustrate Q.14 Draw the following diagram, in which a ray of light is
the formation of a rainbow. incident on a concave/convex mirror, on your answer
[CBSE (Outside Delhi) 2012-13] [3] sheet. Show the path of this ray, after reflection, in
each case. [CBSE (Delhi) 2015-16] [3]
Q.8 A spherical mirror produces an image of magnification
–1 on a screen placed at a distance of 50 cm from the
////

///
////

////

mirror. [CBSE (Delhi) 2013-14] [3]

/////

/////
///

(a) Write the type of mirror.

////////////////
/////////////////

(b) Find the distance of the image from the object. (i) F (ii) F
(c) What is the focal length of the mirror ?
////

////
////

(d) Draw the ray diagram to show the image formation

///
/

in this case.
///
////

Q.9 Explain giving reason why the sky appears blue to an

/////

observer from the surface of the earth ? What will the

////////////////

colour of the sky be for an astronaut staying in the (iii) F

international space station orbiting the earth ? Justify
////

your answer giving reason.

///
/

[CBSE (Delhi) 2013-14] [3]

1-86
 Geometrical Optics

Q.15 Why does the sun appear reddish early in the Q.21 A student wants to project the image of a candle flame
morning? Will this phenomenon be observed by an on the walls of school laboratory by using a lens :
observer on the moon ? Justify your answer with a (a) Which type of lens should he use and why ?
reason. [CBSE (Delhi) 2015-16] [3] (b) At what distance in terms of focal length ‘F’ of the
lens should he place the candle flame so as to get
Q.16 The image of an object formed by a mirror is real, (i) a magnified, and
inverted and is of magnification –1. If the image is at a (ii) a diminished image respectively on the wall ?
distance of 40 cm from the mirror, where is the object
(c) Draw ray diagram to show the formation of the
placed ? Where would the image be if the object is
image in each case. [CBSE (Delhi) 2013-14] [5]
moved 20 cm towards the mirror ? State reason and
also draw ray diagram for the new position of the
Q.22 (a) Explain the following terms related to spherical
object to justify your answer.
[CBSE (Outside Delhi) 2015-16] [3] lenses:
(i) Optical center
Q.17 If the image formed by a lens for all positions of an (ii) Centres of curvature
object placed in front of it is always erect and (iii) Principal axis
diminished, what is the nature of this lens ? Draw a ray (iv) Aperture
diagram to justify your answer. If the numerical value (v) Principal focus
of the power of this lens is 10 D, what is its focal length (vi) focal length
in the Cartesian system ? (b) A converging lens has focal length of 12 cm.
[CBSE (Outside Delhi) 2017] [3] Calculate at what distance should the object be
placed from the lens so that it forms an image at 48
Q.18 (a) State the laws of refraction of light. Give an cm on the other side of the lens.
expression to relate the absolute refractive index of a [CBSE (Outside Delhi) 2013-14] [5]
medium with speed of light in vacuum.
(b) The refractive indices of water and glass with Q.23 “A convex lens can form a magnified erect as well as
respect to air are 4/3 and 3/2 respectively. If the magnified inverted image of an object placed in front of
speed of light in glass is 2 × 108 ms–1, find the speed
it. “ Draw ray diagram to justify this statement stating
of light in (i) air, (ii) water.
the position of the object with respect to the lens in
[CBSE (Delhi) 2012-13] [5]
each case.
Q.19 (a) To construct a ray diagram we use two light rays
which are so chosen that it is easy to know their An object of height 4 cm is placed at a distance of 20
directions after reflection from the mirror. List these cm from a concave lens of focal length 10 cm. Use lens
two rays and state the path of these rays after formula to determine the position of the image formed.
reflection. Use these two rays to locate the image of an [CBSE (Delhi) 2014-15] [5]
object placed between infinity and the centre of
curvature of a concave mirror. Q.24 What is meant by power of a lens ? Define its S.I. unit.
(b) Draw a ray diagram to show the formation of image You have two lenses A and B of focal lengths +10 cm
of an object placed between the pole and principal and –10 cm respectively. State the nature and power of
focus of a concave mirror. How will the nature and size each lens. Which of the two lenses will form a virtual
of the image formed change, if the mirror is replaced by and magnified image of an object placed 8 cm from the
a converging lens of same focal length? lens ? Draw a ray diagram to justify your answer.
[CBSE (Outside Delhi) 2012-13] [5] [CBSE (Outside Delhi) 2014-15] [5]

Q.20 (a) A student cannot see clearly a chart hanging on a Q.25 One half of a convex lens of focal length 10 cm is
wall placed at a distance of 3m from his eyes. Name covered with a black paper. Can such a lens produce an
the defect of vision he is suffering from. Draw a ray
image of a complete object placed at a distance of 30
diagram to illustrate this defect. List its two
cm from the lens ? Draw a ray diagram to justify your
possible causes.
answer. A 4 cm tall object is placed perpendicular to the
(b) Draw a ray diagram to show how this defect may be
principal axis of a convex lens of focal length 20 cm.
corrected using a lens of appropriate focal length.
(c) An eye donation camp is being organised by The distance of the object from the lens is 15 cm. Find
social workers in you locality. How and why would nature, position and size of the image.
you help in this cause ? [CBSE (Outside Delhi) 2014-15] [5]
[CBSE (Outside Delhi) 2012-13] [5]

1-87
Geometrical Optics

Q.26 It is desired to obtain an erect image of an object, using SECTION - B

concave mirror of focal length of 12 cm.
(i) What should be the range of distance of an object Q.1 Why does a ray of light while passing through a prism,
placed in front of the mirror ? bend towards its base ? [CBSE (Delhi) 2012-13] [2]
(ii) Will the image be smaller or larger than the object ?
Draw a ray diagram to show the formation of image Q.2 List the factors on which the angle of deviation through
in this case. a prism depend ?
(iii) Where will the image of this object be, if it is placed
24 cm in front of the mirror ? Draw a ray diagram for [CBSE (Outside Delhi) 2012-13] [2]
this situation also to justify your answer.
Show the positions of the pole, the principal focus and Q.3 For finding the focal length of a concave mirror, where
the centre of curvature in the above ray diagrams. do we keep the object / What is the position of image
[CBSE (Outside Delhi) 2015-16] [5]
formed ? On which structure we get the image ? What
Q.27 (a) To construct a ray diagram we use two rays which is the nature of the image formed ?
are so chosen that it is easy to know their [CBSE (Delhi) 2013-14] [2]
directions after reflection from the mirror. List two
such rays and state the path of these rays after Q.4 When a ray of light passes through a glass slab how
reflection in case of concave mirrors. Use these
many times does it change its path and why ?
two rays and draw ray diagram to locate the image
of an object placed between pole and focus of a [CBSE (Outside Delhi) 2013-14] [2]
concave mirror.
(b) A concave mirror produces three times magnified Q.5 To find the image-distance for varying object-distances
image on a screen. If the object is placed 20 cm in in case of a convex lens, a student obtains on a screen
front of the mirror, how far is the screen from the
a sharp image of a bright object placed very far from
object ? [CBSE (Delhi) 2017] [5]
the lens. After that he gradually moves the object
Q.28 (a) Draw a ray diagram to explain the term angle of towards the lens and each time focuses its image on
deviation. the screen.
(b) Why do the component colours of incident white (a) In which direction-towards or away from the lens,
light split into a spectrum while passing through a
glass prism, explain. does he move the screen to focus the object ?
(c) Draw a labelled ray diagram to show the formation (b) What happens to the size of image-does it increase
of a rainbow. [CBSE (Delhi) 2017] [5] or decrease ?
(c) What happen when he moves the object very close
Q.29 (a) If the image formed by a mirror for all positions of to the lens ? [CBSE (Delhi) 2014-15] [2]
the object placed in front of it is always diminished,
erect and virtual, state the type of the mirror and
also draw a ray diagram to justify your answer. Q.6 A 4 cm tall object is placed on the principal axis of a
Write one use such mirrors are put to and why ? convex lens. The distance of the object from the optical
(b) Define the radius of curvature of spherical mirrors. centre of the lens is 12 cm and its sharp image is formed
Find the nature and focal length of a spherical
at a distance of 24 cm from it on a screen on the other
mirror whose radius of curvature is +24 cm.
[CBSE (Outside Delhi) 2017] [5] side of the lens. If the object is now moved a little away
from the lens, in which way (towards the lens or away
Q.30 (a) A student suffering from myopia is not able to see from the lens) will he have to move the screen to get a
distinctly the objects placed beyond 5m. List two sharp image of the object on it again ? How will the
possible reasons due to which this defect of vision
magnification of the image be affect ?
may have arisen. With the help of ray diagrams,
explain. [CBSE (Outside Delhi) 2014-15] [2]
(i) Why the student is unable to see distinctly the
objects placed beyond 5m from his eyes. Q.7 An object of height 2.5 cm is placed at a distance of 15
(ii) The type of the corrective lens used to restore
proper vision and how this defect is corrected by cm from the optical centre ‘O’ of a convex lens of focal
the use of this lens. length 10 cm. Draw a ray diagram to find the position
(b) If, in this case, the numerical value of the focal and size of the image formed. Mark optical centre ‘O’,
length of the corrective lens is 5m, find the power
of the lens as per the n ew Cartesian sign principal focus F and height of the image on the diagram.
convention. [CBSE (Outside Delhi) 2017] [5] [CBSE (Outside Delhi) 2015-16] [2]

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Q.8 A student places a candle flame at a distance of about Q.9 A student focuses the image of a canle flame, placed at
60 cm from a convex lens of focal length 10 cm and about 2m from a convex lens of focal length 10 cm, on a
focuses the image of the flame on a screen. After that screen. After that he moves gradually the flame towards
he gradually moves the flame towards the lens and the lens and each time focuses its image on the screen.
each time focuses the image on the screen. (a) In which direction does he move the lens to focus
(a) In which direction-toward or away from the lens, the flame on the screen ?
does he move the screen to focus the image ? (b) What happens to the size of the image of the flame
(b) How does the size of the image changes ? formed on the screen ?
(c) How does the intensity of the image change as the (c) What is seen on the screen when the flame is very
flame moves towards the lens ? close (at about 5 cm) to the lens ?
(d) Approximately for what distance between the flame [CBSE (Outside Delhi) 2017] [2]
and the lens, the image formed on the screen is
inverted and of the same size ?
[CBSE (Delhi) 2017] [2]

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EXERCISE-II

Plane Mirror Q.7 A man of height 1.6m wishes to see his full image in a
Q.1 An object is placed between two parallel plane mirror. plane mirror. then minimum height needed for this
The number of image formed is purpose is :
(A) four (B) one (A) 3.2 m (B) 2.4 m (C) 1.6 m (D) 0.8 m
(C) two (D) infinite

Q.8 Two mirrors are inclined at an angle 60°, an object is

Q.2 An object is placed between two plane mirrors inclined
placed symmetrically between them. Then number of
at some angle to each other. If the number of images
images formed will be:
formed is 7 then angle of inclination is
(A) 150 (B) 300 (C) 450 (D) 600 (A) 6 (B) 5 (C) 7 (D) 9

Q.3 If an object is placed unsymmetrically between two Q.9 Find number of images formed according to given case
plane mirrors, inclined at an angle of 720, then the total
number of image formed is- ///
/
///
// M1 ////
///
/
///
/ M1
////
(A) 5 (B) 4 (C) 2 (D) Infinite /// / ///
/ ////
//// ///
O
/// /// /
/// ///
O
/// / ///
/// / ///
/ 10°
40°
/// / ///
Q.4 During regular reflection from a plane mirror, a parallel
/// ///
/// / ////
20°
///
40°
//// / ///
beam of light. ///////////////////////////////////////////////// M2 ///////////////////////////////////////////////// M2
(A) remains as parallel beam of light
(B) converges at a point (A) 8, 9 (B) 9, 8 (C) 9, 9 (D) 8, 8
(C) diverge from a point
(D) none of these
Q.10 To get 9 images of an object the angle between two
plane mirrors should be:
Q.5 A ray of light makes an angle of incidence of 25° with
(A) 60° (B) 36° (C) 50° (D) 90°
the plane mirror. If the plane mirror is turned through an
angle of 5º, the angle of reflection can be:
(A) 35° Q.11 An object A is placed at a distance d in front of a plane
(B) 15° mirror. If one stands directly behind the object at
(C) both 35° and 15° and depends on the direction of distance S from the mirror, then the distance of the
turn of mirror image of A from the individual is:
(D) none of these (A) 2S (B) 2d (C) S + d (D) S + 2d

Q.6 Two parallel plane mirrors A and B are placed at a

Q.12 A plane mirror is moved toward a stationary observers
separation 10 cm as shown in figure. A ray is incident
with a uniform speed of 2ms. The speed with which
on the corner of mirror B at an angle of incidence 45°.
the image will move toward observer will be:
Find the number of times this ray is reflected from mirror
(A) 2 ms (B) 1 ms (C) 6 ms (D) 4 ms
A:

A Q.13 An object 0.5 m tall is in front of a plane mirror at a

distance of 0.2 m. The size of the image formed is-
Incident ray
(A) 0.2 m (B) 0.5 m (C) 0.1 m (D) 1 m
45° 10cm

B Q.14 A ray of light is incident on a plane mirror at an angle of

100cm incidence of 30°. The deviation produced by the mirror
is-
(A) 4 (B) 5 (C) 6 (D) 7 (A) 30° (B) 60° (C) 90° (D) 120°

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Q.15 If two mirrors are kept at 60° to each other, then the Q.24 The mid-point of spherical mirror is called:
number of images formed by them is- (A) pole
(A) 5 (B) 6 (C) 7 (D) 8 (B) principal focus
(C) centre of curvature
Q.16 How many images of himself does an observer see if (D) none of the above.
two adjacent walls of rectangular room are mirror
surfaced? Q.25 If R the radius of curvature of spherical mirror and F the
(A) 3 (B) 5 (C) 7 (D) 9 focal length, then:
(A) R = f/ 2 (B) R = 2 / f (C) f = R/ 2 (D) f = 2R
Q.17 A plane mirror and an object approach each other with
speeds of 5m/s and 10 m/s respectively. The speed of Q.26 The point on the principal axis of a spherical mirror
the image will be where the rays parallel to principal axis actually meet or
(A) 5 m/s (B) 15 m/s (C) 20 m/s (D) 25 m/s appear to meet is called:
(A) centre (B) principal focus
Q.18 The angle of incidence is the angle between (C) pole (D) none of the above
(A) the incident ray and the surface of the mirror
(B) the reflected ray and the surface of the mirror Q.27 An object is placed in front of a concave mirror. It is
(C) the normal to the surface and the incident ray observed that its inverted images is formed on the object
(D) the normal to the surface and the reflected ray itself, which is of the same size as the object. The object
is at:
Q.19 The angle of reflection is the angle between (A) principal axis (B) centre of curvature
(A) the incident ray and the surface of the mirror (C) principal focus (D) none of the above
(B) the reflected ray and the surface of the mirror
(C) the normal to the surface and the incident ray Q.28 A real inverted and diminished than object, image is
(D) the normal to the surface and the reflected ray formed by concave mirror. The object is:
(A) at infinity (B) between infinity and C
Q.20 If you want to see your full image then the, minimum (C) between C and F (D) at F
size of the plane mirror
(A) should be of your height Q.29 A real inverted and magnified image is formed by a
(B) should be half of your height. concave mirror. The object is:
(C) should be twice of height. (A) between infinity and C
(D) depends upon your distance from the mirror (B) between P and F
(C) at F
Q.21 A light bulb is placed between two plane mirrors inclined (D) between C and F
at an angle of 600. the number of image formed are
(A) 6 (B) 5 (C) 4 (D) 2 Q.30 An object is placed in front of convex mirror. The
image formed is virtual, erect and smaller than object.
Q.22 Two plane mirrors inclined at an angle to one another The image is formed:
have an object placed between them. If five images of (A) between P and F (B) between F and C
the object are observed, the maximum possible angle (C) at F (D) at C
between the mirrors is
(A) 45º (B) 60º (C) 72º (D) 90º Q.31 Which of the following can produce virtual images
(A) convex mirror (B) concave mirror
Curved Mirror (C) plane mirror (D) all of the above
Q.23 The image of an object in a spherical mirror appears
diminished erect and behind it. The spherical mirror is: Q.32 The minimum distance between an object and its real
(A) concave (B) plane image in case of a concave mirror is ;
(C) convex (D) none of these (A) 0 (B) 1 (C) 4 (D) 2

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Geometrical Optics

Q.33 A concave mirror of focal length 10 cm produces an Q.40 Two plane mirrors are kept at an angle a. A light ray
image five times as large as the object. If the image is in striking the two mirrors successively suffers a deviation
front of the mirror, the distance of the object from the of 5p/6. The value of a is –
mirror will be
p 7p
(A) 10 cm (B) 12 cm (C) 16 cm (D) 20 cm (A) (B)
9 12
Q.34 A plane mirror and a concave mirror face each other 3p 9p
and are separated by a distance of 20 cm. An object is (C) (D)
5 11
placed at the mid-point of two mirrors. The virtual image
formed by the plane mirror acts as an object for concave
mirror. The final image formed by the concave mirror is Q.41 The distance between the image and an object kept on
at the mid-point of two mirror and coincides with the the principal axis of a spherical mirror of curvature
radius 80 cm is found to be 150 cm. How far is the image
original object. The focal length of concave mirror is:
from the centre of curvature–
(A) 10 cm (B) 5 cm (C) 20 cm (D) 25 cm
(A) 30 cm (B) 50 cm
(C) 120 cm (D) 160 cm
Q.35 A concave mirror produces three times magnified real
image of an object placed at 10 cm in front of it. The
Q.42 A person of height 1.8 standing at the centre of a room
image is located.
having equal dimensions of 10 m wishes to see the full
(A) 30 cm behind the mirror
image of the back wall in the mirror fixed on the front
(B) 30 cm in front of mirror
wall. The minimum height of the plane mirror needed
(C) 15 cm in front of mirror
for this purpose is
(D) 15 cm behind the mirror.
(A) 0.9 m (B) 1.8 m
(C) 10/3 m (D) 10 m
Q.36 An object 1 cm high produce a real image 1.5 cm high
when placed at a distance of 15 cm from concave mirror.
Q.43 A mirror forms a virtual image of a real object.
The position of image is:
(A) It must be a convex mirror.
(A) 22.5 cm in front of the mirror
(B) It must be a concave mirror.
(B) 22.5 cm behind the mirror
(C) It must be a plane mirror.
(C) 45 cm in front of the mirror
(D) It may be any of the mirror mentioned above.
(D) none of these
Q.44 An object is placed at the centre of curvature of a
Q.37 A convex mirror has focal length 20 cm. An object placed concave mirror. The distance between its image and
in front of it at a distance of 10 cm. Its image is formed: the pole is
(A) at a distance of 20 cm from pole (A) equal to f (B) between f and 2f
(B) at a distance of 10 cm from pole (C) equal to 2f (D) greater then 2f
(C) at a distance of less than 20 cm from pole
(D) at a distance more than 20 cm from pole Q.45 An object of size 2.0 cm is placed perpendicular to the
principal axis of a concave mirror. The distance of the
Q.38 An object is placed at a distance x from a convex mirror object from the mirror equals the radius of curvature.
of focal length 15 cm when an image is formed at The size of the image will be
distance of 6 cm behind the mirror. The value of x is: (A) 0.5 cm (B) 1.5 cm
(A) –15 cm (B) –10 cm (C) 1.0 cm (D) 2.0 cm
(C) –25 cm (D) none of these
Q.46 The magnification m of an image formed by a spherical
Q.39 A point object forms a real, enlarged image, in front of mirror is negative. It means, the image is
a concave mirror of curvature radius 40 cm. If the (A) smaller than the object
distance between the object and the image be 30 cm, (B) larger than the object
then the object distance will be– (C) erect
(A) 10 cm (B) 30 cm (C) 40 cm (D) 60 cm (D) inverted

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 Geometrical Optics

Q.47 A point object is placed on the principal axis of a Q.53 The focal length of concave mirror is f and the distance
spherical mirror. The object-distance u is from the object to the principal focus is x. The
(A) definitely negative magnification obtained will be
(A) (f + x)/f (B) f / x
(B) definitely positive
(C) positive if the object is to the left of the centre of f
curvature (C) (D) f2 / x2
x
(D) positive if the object is to the right of the centre of
curvature
Refraction at plane surface
Q.54 v1 is velocity of light in first medium, v2 is velocity of
R light in second medium, then refractive index of second
Q.48 f = is valid medium with respect to first medium is
2
(A) v1/v2 (B) v2/v1
(A) for convex mirrors but not for concave mirrors
(B) for concave mirror but not for convex mirrors (C) v1 / v 2 (D) v 2 / v1
(C) for both convex and concave mirrors
(D) neither for convex mirrors nor for concave mirror Q.55 Electromagnetic radiation of frequency n,
wavelength l , travelling with velocity v in air, enters a
Q.49 A ray of light is incident on a concave mirror. If it is glass slab of refractive index m . The frequency,,
parallel to the principal axis, the reflected ray will wavelength and velocity of light in the glass slab will
(A) pass through the focus be respectively:-
(B) pass through the centre of curvature
n l v l v
(C) pass through the pole (A) , , (B) n, ,
(D) retrace its path m m m m m

v n l
Q.50 If an incident ray passes through the centre of curvature (C) n, l, (D) , v
of a spherical mirror, the reflected ray will
m m m
(A) pass through the pole
(B) pass through the focus Q56 A ray of light while travelling obliquely from one
(C) retrace its path optically denser medium to an optically rarer medium
always:
(D) be parallel to the principal axis
(A) bends towards normal
(B) bends away from normal
Q.51 To get an image larger than the object, one can use
(C) travel along normal
(A) a convex mirror but not a concave mirror (D) none of these
(B) a concave mirror but not a convex mirror
(C) either a convex mirror or a concave mirror Q.57 A ray of light while travelling from medium P to Q
(D) a plane mirror deviates away from the normal. The speed of light
in :
Q.52 Indicate the only correct statement. (A) P is more than Q (B) Q is more than P
(C) P is same as Q (D) none of the above
(A) The image formed by a convex mirror can be taken
on the screen.
Q.58 Refractive index of material is:
(B) A convex mirror can produce a parallel beam of light
(A) velocity of light in the material ÷ velocity of light in
from a point source. vacuum
(C) The image of an object placed at the focus of a (B) velocity of light in vacuum × velocity of light in
convex mirror will be formed at infinity. material
(D) A concave mirror can never form a diminished virtual (C) velocity of light in vacuum ÷ velocity of light in
image. material
(D) none of these

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Geometrical Optics

Q.59 The velocity of light in vacuum is 3 × 108 and in glass is Q.67 In a museum a child walks towards a large concave
2 × 108 ms–1. The refractive index of glass is mirror. He will see that
(A) 2 (B) 1.5 (A) his real, erect image goes on decreasing in size.
(C) 1.33 (D) 1.8 (B) his virtual, erect image goes on increasing in size.
(C) his real, inverted image goes on diminishing in size
Q.60 A coin placed in at bottom of beaker appears and suddenly it becomes virtual, erect and
raised on account of magnified.
(A) reflection of light (B) diffraction of light
(D) his real, erect image goes on diminishing in size and
(C) dispersion of light (D) refraction of light
suddenly it becomes virtual, erect and magnified.

Q.61 A pencil held obliquely in water appears:

Q.68 The magnification of an object placed 10 cm from a
(A) bent and short
convex mirror of radius of curvature 20 cm will be
(B) bent and long
(C) straight and magnified (A) 0.2 (B) 0.5 (C) 1 (D) 20 cm
(D) none of the above
Q.69 A concave mirror of focal length 10 cm produces an
Q.62 A ray of light striking obliquely on the face of a glass image five times as large as large as the object. If the
slab emerge out of it. The incident ray is image is in front of the mirror, the distance of the object
(A) parallel to refracted ray from the mirror will be
(B) parallel to the emergent ray (A) 10 cm (B) 12 cm (C) 16 cm (D) 20 cm
(C) parallel to the face of glass slab
(D) none of the above Q.70 Refractive index of glass w.r.t. air is 3/2. What is the
refractive index of air w.r.t. glass?
Q.63 The wavelengths corresponding to violet, yellow and (A) 2/3 (B) 1 (C) Zero (D) (3/2)2
red lights are lv, ly and lr respectively.
(A) lv > ly > lr (B) lv < ly < lr Q.71 When monochromatic light passes from vacuum to a
(C) ly < lv < lr (D) ly < lr < lv material medium and vice versa: which of the following
characteristics of light beam does not changes?
Q.64 A ray of light travelling in air falls obliquely on the
(A) velocity (B) intensity
surface of a calm pond. It will
(C) wavelength (D) frequency
(A) go into the water without deviating from its path
(B) deviate away from the normal
Q.72 A piece of glass when immersed in a transparent
(C) deviate towards the normal
solution of refractive index 1.48 becomes almost
(D) turn back on its original path
invisible. The refractive index of glass used is
Q.65 A ray of light goes from a medium of refractive index (A) zero (B) 1
(C) 1.48 (D) infinite
m1 to a medium of refractive index m 2 . The angle of
incidence is i and the angle of refraction is r. Then, sin Q.73 Which of the following conditions are necessary for
i/sin r is equal to total internal reflection to take place at the boundary of
(A) m1 (B) m2 two optical media?
I. Light is passing from optically denser medium to
m1 m2 optically rarer medium.
(C) (D)
m2 m1 II. Light is passing from optically rarer medium to
optically denser medium.
III. Angle of incidence is greater than the critical angle.
Q.66 In case of a thick plane mirror multiple images arc formed. IV. Angle of incidence is less than the critical angle.
The brightest of all the images will be
(A) I and III only (B) II and IV only
(A) first (B) second
(C) III and IV only (D) I and IV only
(C) third (D) fourth

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 Geometrical Optics

Q.74 The speed of light in vacuum is 3.0 × 108 m/s. If the Prism
refractive index of a transparent liquid is 4/3, then the Q.82 When white light passes through a prism, it splits into
speed of light in the liquid is its component colours. This phenomenon is called
(A) 2.25 × 108 m/s (B) 3 × 108 m/s (A) spectrum (B) reflection
(C) 4 × 10 m/s
8
(D) 4.33 × 108 m/s (C) refraction (D) dispersion

Q.83 The number of surfaces bounding a prism is

Q.75 The refractive indices of water and glass are 4/3 and 3/
(A) 3 (B) 4 (C) 5 (D) 6
2 respectively. The refractive index of water with respect
to glass is Q.84 A deviation in the path of a ray of light can be produced
8 (A) by a glass prism but not by a rectangular glass slab
(A) (B) 2 (B) by a rectangular glass slab but not by a glass prism
9
(C) by a glass prism as well as a rectangular glass slab
2 1 (D) neither by a glass prism nor by a rectangular glass
(C) (D)
3 6 slab

Total Internal Reflection Q.85 Find the value of Ðr’ for the case shown in figure:
Q.76 What is the basic reason for the shining of a diamond?
(A) Reflection (B) Refraction 0 750
45 rr’ q
(C) Dispersion of light (D) Total internal reflection
m=1 m=Ö2
Q.77 Which of the following is used in optical fires?
(A) Total internal reflection (A) sin-1(0.5) (B) 75° - sin-1(0.5)
(B) Scattering (C) 90° (D) 60°
(C) Diffraction
(D) Refraction Lens
Q.86 When a news paper is seen through a lens. Its
Q.78 The angle beyond which when a ray is incident, it is print appears smaller. The Nature of the lens is.
completely reflected back is called: (A) convex (B) concave
(C) double convex (D) concavo convex
(A) angle of deviation (B) critical angle
(C) incident angle (D) angle of inversion
Q.87 A ray of light incident obliquely on a lens does not
suffer any refraction. The ray is passing through the:
Q.79 The mirage is formed due to (A) principal focus of lens
(A) reflection (B) centre of curvature of lens
(B) refraction (C) optical centre of lens
(C) total internal reflection (D) none of the above
(D) dispersion
Q.88 An object between infinity and 2F1 of a convex
Q.80 For total internal reflection, rays of light must go from: lens. The nature of image formed by the lens is:
(A) Rarer to denser medium (A) real, inverted and magnified
(B) Rarer to rarer medium (B) real, inverted and diminished
(C) Denser to rarer medium (C) real, erect and diminished
(D) virtual, erect and diminished
(D) Denser to denser medium

Q.89 A convex lens forms an image of an object in

Q.81 A glass prism has refractive index 1.5 and the refracting
between its 2F2 and F2. The position of the object with
angle 900 If a ray falls on it at an angle of incidence respect to the convex lens is:
(A) 60º (A) between F1 and 2F1
(B) 45º (B) between infinity and 2F
(C) 30º (C) at 2F1
(D) the ray will not emerge at all (D) at F1

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Geometrical Optics

Q.90 An object is located at 2F1 of a convex lens. The Q.97 A parallel beam of light is incident on a concave
nature of the image formed by the lens is: lens. The nature of image formed is:
(A) real, inverted and magnified (A) real, inverted and diminished at a point
(B) real, inverted and diminished (B) virtual erect and enlarged
(C) real, inverted and of same size as object (C) virtual, erect diminished to a point
(D) real, erect and magnified. (D) virtual inverted and enlarged

Q.91 A convex lens forms an image, which is located at Q.98 When an image is formed at F of a concave lens, t h e
the 2F2 of the lens. The position of object with respect object is at:
to convex lens is: (A) at 2F (B) F
(A) between infinity and 2F1 (C) at infinity (D) between F and 2F
(B) at 2F1
(C) optical centre and F1 Q.99 An object is situated anywhere between infinity and
(D) at F1 optical centre of a concave lens. The nature of image
formed by it is:
Q.92 An object is placed in between F1 and 2F1 of a (A) real, inverted and diminished
convex lens. The nature of image formed by the lens is: (B) virtual, erect and diminished
(A) real, inverted and magnified (C) virtual erect and enlarged
(B) real, erect and magnified (D) real, inverted and enlarged
(C) virtual, erect and magnified
(D) real, inverted and diminished Q.100 A lens has a power of +0.5 D. It is
(A) a concave lens of focal length 5 m
Q.93 A convex lens forms an image. Which located beyond (B) a convex lens of focal length 5 cm
2F2 but not infinity? The position of object with respect (C) a convex lens of focal length 2 m
to convex lens is: (D) a concave lens of focal length 2 m
(A) between infinity and 2F1
(B) between F1 and 2F1 Q.101 Two thin lenses of focal lengths 20 cm and 25 cm are
(C) at 2F1 placed in contact. The effective power of the
(D) between optical centre and F1 combination is
(A) 9 diopters (B) 45 diopters
Q.94 An object is placed at F1 of a convex lens. The (C) 6 diopters (D) 9 diopters
nature of the image found by it is:
(A) real, inverted and diminished Q.102 A thin lens and a spherical mirror have a focal length of
(B) real, inverted and highly magnified + 15 cm each.
(C) real, erect and highly magnified (A) both are convex.
(D) real, inverted and same size as object (B) the lens is convex and the mirror is concave.
(C) the lens is concave and the mirror is convex.
Q.95 A point source of light is placed in front of convex lens, (D) both are concave.
when a parallel beam of light emerges out of it. The
object with respect to lens is: Q.103 A convex lens forms a virtual image when an object is
(A) at F1 placed at a distance of 18 cm from it. The focal length
(B) at 2F1 must be
(C) at infinity (A) greater than 36 cm
(D)between F1 and optical centre (B) greater than 18 cm
(C) less than 36 cm
Q.96 An object is placed between the optical centre of (D) less than 18 cm
convex lens and F1. The nature of image formed by it is:
(A) real, magnified and erect Q.104 An object is placed before a convex lens. The image
(B) real, inverted and diminished formed
(C) virtual, magnified and erect (A) is always real (B) may be real or virtual
(D) virtual, diminished and erect (C) is always virtual (D) is always erect

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 Geometrical Optics

Q.105 An object is placed before a concave lens. The image Q.112 Myopia can be removed by using a lenses of
formed (A) concave lens
(A) is always erect (B) convex lens
(B) may be erect or inverted (C) cylindrical lens
(C) is always inverted (D) by surgical removal
(D) is always real

Q.113 A parallel beam of light falling on the eye gets focused

Q.106 A lens has a power of +0.5 D. It is
on the retina because of refractions at
(A) a concave lens of focal length 5 m
(A) the cornea
(B) a convex lens of focal length 5 cm
(C) a convex lens of focal length 2 m (B) the crystalline lens
(D) a concave lens of focal length 2 m (C) the vitreous humor
(D) various surfaces in the eye
Q.107 To form an image twice the size of the object, using a
convex lens of focal length 20 cm, the object distance Q.114 The combination responsible for admitting different
must be – amounts of light into the eye is
(A) < 20 cm (A) ciliary muscles and crystalline lens
(B) > 20 cm (B) ciliary muscles and pupil
(C) < 20 cm and between 20 cm and 40 cm (C) iris and pupil
(D) Cannot say (D) rods and cones

Q.108 Which of the following can be used to form a virtual

Q.115 The muscles of the iris control the
image of an object?
(A) focal length of the eye-lens
I. convex lens
(B) opening of the pupil
II. concave lens
III. concave mirror (C) shape of the crystalline lens
(A) II only (B) II and III only (D) optic nerve
(C) I and III only (D) I, II and III
Q.116 When the eye is focused on an object very far away,
Eye and defects the focal length of the eye-lens is
Q.109 Myopia is the defect of vision due to which a person (A) maximum
finds difficulty in seeing (B) minimum
(A) distance object (C) equal to that of the crystalline lens
(B) near object (D) half its maximum focal length
(C) object at all distances
(D) colours
Q.117 Other names for myopia are
(A) hyperopia and hypermetropia
Q.110 Loss of the ability of eye to focus on near and far object
(B) long-sightedness and hyperopia
with advancing age is called
(C) near-sightedness and presbyopia
(A) Presbyopia (B) Astigmatism
(C) Hypermetropia (D) Myopia (D) near-sightedness and short-sightedness

Q.111 A normal eye is not able to see objects closer than 25 Q.118 The inability among the elderly to see nearby objects
because clearly because of weakening of ciliary muscles is called
(A) The focal length of the eye is 25 cm (A) far-sightedness
(B) The distance of the retina form the lens is 25 cm (B) near-sightedness
(C) The eye is not able to decrease the distance between (C) presbyopia
the eye lens and the retina beyond a limit (D) astigmatism
(D) The eye is not able to decrease the focal length
beyond a limit

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Geometrical Optics

EXERCISE-III

Q.1 Which of the following letters do not surface lateral Q.5 A point source of light S is placed at a distance L in
inversion. front of the centre of a plane mirror PQ. of width d hung
vertically on a wall as shown in fig. A man walks in
(A) HGA (B) HOX
front of the mirror along a line parallel to the mirror at a
(C) VET (D) YUL
distance 2L from it as shown. The greatest distance
over which he can see the image of the light source in
Q.2 Two plane mirrors are arranged at right angles to each the mirror is –
other as shown in figure. A ray of light is incident on
the horizontal mirror at an angle q. For what value of q
the ray emerges parallel to the incoming ray after P

/ //////////////////////////
reflection from the vertical mirror
L
S
Q Man
q 2L

d
(A) (B) d (C) 2d (D) 3d
2
(A) 60° (B) 30°
(C) 45° (D) all of the above
Q.6 Two plane mirrors, each 1.6 m long, are help parallel

Q.3 Two plane mirrors are inclined at an angle 70º. A ray and facing each other at a separation of 20 3 cm. A
incident on one mirror at angle q after reflection falls on
ray of light is incident at a the end of one mirror at an
the second mirror and is reflected from there parallel to
angle of incidence of 300. The total number of reflections
the first mirror. q is:
the ray suffers before emerging from the system of
(A) 50º (B) 45º
mirrors is –
(C) 30º (D) 55º
(A) 9 (B) 12 (C) 14 (D) 16

Q.4 Two plane mirror AB and AC are inclined at an angle Q.7 Two plane mirrors A and B are aligned to each other, as
20°. A ray of light starting from point P is incident at shown in Fig. A light ray is incident at an angle of 300 at
point Q on the mirror AB, then at R on mirror AC and a point just inside one end of A. The plane of incidence
then at S on AB and finally the ray ST goes parallel to coincides with the plane of the figure. The maximum
mirror AC. The angle of incidence at point Q on mirror number of times the ray undergoes reflections
AB is (including the first one) before it emerges out is –

B B
S
T
Q
0
0.2m 30°
20
C
A P R

A
(A) 20° (B) 30°
(C) 40° (D) 60° (A) 28 (B) 30 (C) 32 (D) 34

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 Geometrical Optics

Q.8 In the figure shown, the maximum number of reflections Q.14 Two plane mirrors M1 and M2 each have length 1m and
will be: separated by 1 cm, A ray of light is incident on one end
of mirror M1 at angle 45º. How many reflections the ray
will have before going at from the other end

M2

(A) 2 (B) 3 (C) 4 (D) 1 45°

Q.9 A person’s eye is at a height of 1.5 m. He stands in front M1

of a 0.3m long plane mirror which is 0.8 m above the
ground. The length of the image he sees of himself is: (A) 50 (B) 51 (C) 100 (D) 101
(A) 1.5m (B) 1.0m (C) 0.8m (D) 0.6m
Q.15 A ray of light is incident at an angle of 30º on a plane
Q.10 Two plane mirrors are joined together as shown in
mirror M. Another plane mirror M2 is inclined at angle q
figure. Two point objects O 1 and O 2 are placed
symmetrically such that AO1 = AO2. The image of the to M1. What is the value of angle q so that light reflected
two objects is common if : from M2 is parallel to M1.

M2
T
0
30
q
M1
(A) q = 60° (B) q = 90°
(C) q = 30° (D) q = 45° (A) 60º (B) 75º
(C) 67.5º (D) none of these
Q.11 A person of height 1.8 standing at the centre of a room
having equal dimensions of 10 m wishes to see the full
Q.16 An unnumbered wall clock shows time 04: 25: 37, where
image of the back wall in the mirror fixed on the front
1st term represents hours, 2nd represents minutes &
wall. The minimum height of the plane mirror needed
for this purpose is the last term represents seconds. What time will its
image in a plane mirror show.
10
(A) 0.9 m (B) 1.8 m (C) m (D) 10 m (A) 08: 35: 23 (B) 07: 35: 23
3
(C) 07: 34: 23 (D) none of these
Q.12 At what angle must two plane mirrors be placed so that
incident and resulting reflected rays are always parallel Q.17 A concave mirror forms a real image three times larger
to each other than the object on a screen. Object and screen are moved
(A) 00 (B) 300 (C) 600 (D) 900 until the image becomes twice the size of object. If the
shift of object is 6 cm. The shift of the screen & focal
Q.13 Figure shows two plane mirrors parallel to each other length of mirror are
and an object O placed between them. Then the (A) 36 cm, 36cm (B) 36cm, 16cm
distance of the first three images from the mirror M2 will (C) 72cm, 36cm (D) none of these
be: (in cm)
Q.18 An infinitely long rod lies along the axis of a concave
mirror of focal length f. The near end of the rod is at
5cm a distance u > f from the mirror. Its image will have a
length
O
f2 uf
(A) (B)
15cm u- f u- f
M1 M2
f2 uf
(A) 5, 10, 15 (B) 5, 15, 30 (C) (D)
u+ f u+ f
(C) 5, 25, 35 (D) 5, 15, 25

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Geometrical Optics

Q.19 A concave mirror has a focal length 20 cm. The distance Q.25 The distance of a real object from the focus of a convex
between the two positions of the object for which the mirror of radius of curvature ‘a’ is ‘b’. Then the
image size is double of the object size is distance of the image from the focus is
(A) 20 cm (B) 40 cm
b2 a
(C) 30 cm (D) 60 cm (A) (B)
4a b2

Q.20 If an object is placed 10 cm in front of a concave mirror

a2
a focal length 20 cm, the image will be:- (C) (D) none of these
4b
(A) diminished, upright, virtual
(B) enlarged, upright, virtual
Q.26 Immiscible transparent liquids A, B, C, D and E are placed
(C) diminished, inverted, real
in a rectangular container of glass with the liquids
(D) enlarged, upright, real
making laxers according to their densities. The
refractive index of the liquids are shown in the adjoining
Q.21 The magnification m, the image position v and focal
diagram. The container is illuminated from the side and
length f are related to one another by the relation –
a small piece of glass having refractive index 1.61 is
f -v f gently dropped into the liquid also. The glass piece as
(A) m = (B) m = descends downwards will not be visible in:-
f f -v

f +v f µ = 1.64 A
(C) m = (D) m = µ = 1.63 B
f f -v
µ = 1.61 C
µ = 1.64 D
Q.22 The relation between magnification m, the object µ = 1.6 E
position u and focal length f of the mirror is
(A) Liquid A and B only
f -v f
(A) m = (B) m = (B) Liquid C only
f f -u
(C) Liquid D and E only
(D) Liquid A, B D and E
f +u f
(C) m = (D) m =
f f +u Q.27 A bird in air looks at a fish vertically below it and inside
water. x is the height of the bird above the surface of
water and y is the depth of the fish below the surface of
Q.23 A person is looking at the image of his face in a mirror
water. The distance of the fish as observed by the bird
by holding it close to his face. The image is virtual.
is: (Given m = refractive index of water w.r.t. air):-
When he moves the mirror away from his face, the image
is inverted. What type of mirror is he using? y
(A) Plane mirror (B) Convex mirror (A) x + y (B) x +
m
(C) Concave mirror (D) None of these
(C) m x+y (D) m x+m y
Q.24 Two objects A and B when placed in turn in front of a
concave mirror of focal length 7.5 cm, give images of Q.28 In the previous question, the distance of the bird as
equal size. If A is three times the size of B and is placed observed by the fish is:-
30 cm from mirror , what is the distance of B for the
y
(B) x +
mirror –
(A) x + y
(A) 10 cm (B) 12.5 cm m
(C) 15cm (D) 17.5 cm m x+y m x+m y
(C) (D)

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 Geometrical Optics

Q.29 If i m j represents refractive index when a light ray goes Q.34 An air bubble in a glass slab ( m = 1.5 ) is 6 cm deep
grom medium i to medium j, then the product when viewed through one face and 4 cm deep when
2 m1 ´3 m 2 ´ 4 m 3 is equal to
viewed through the opposite face. What is the
thickness of the slab?
1
(A) 7.0 cm (B) 7.5 cm
(A) 3 m1 (B) 3 m2 (C) (C) m2
1 m4
4
(C) 15 cm (D) 10.5 cm

Q.30 Light travels through a glass plate of thickness t and Q.35 A ray of light travels from an optical denser medium to
having refractive index n. If c is the velocity of light in
vacuum. The time taken by the light to travel this rarer medium. The critical angle for the two media is C.
thickness of glass is:- The maximum possible deviation of the refracted light
ray can be:
t nt tc
(A) (B) tnc (C) (D) (A) p – C (B) 2C
nc c n
p
Q.31 A ray of light passes through four transparent media (C) p – 2C (D) –C
2
with refractive indices m1 , m 2 , m 3 and m 4 as show
in the figure. The surfaces of all media are parallel. If Q.36 Total internal reflection of a ray of light is possible
the emergent ray CD is parallel to the incident ray AB,
we must have: when the (ic = critical angle, i = angle of incidence)
(A) Ray goes from denser medium to rarer medium and
µ1 µ2 µ3 µ4 i < ic
D (B) Ray goes from denser medium to rarer medium and
i > ic
B C (C) Ray goes from rarer medium to denser medium and
A
i > ic
(D) Ray goes from rarer medium to denser medium and
i < ic.
(A) m1 = m 2 (B) m 2 = m3
(C) m3 = m4 (D) m 4 = m1 Q.37 A ray of light is incident at the glass-water interface at
an angle i. It emerges finally parallel to the surface of
Q.32 A plane glass slab is kept over various coloured letters;
the letter which appears lest raised is water as shown in fig. the value of m g would be –
(A) blue (B) violet (C) green (D) red

Q.33 An observer can see through a pin-hole the top end of

a thin rod of height h, placed as shown in the figure, Air (µa=1)
the beaker height is 3h and its radius h. when the beaker
is filled with a liquid up to a height 2h, he can see the r
Water (µw=4/3)
lower end of the rod. Then the refractive index of the r
liquid is –
Glass (µs)
// / //// / // // //
//
/////
/ i
// /
////

3h æ4ö 1
(A) ç ÷ sin i (B)
h è3ø sin i

2h
2
(C) (D) 1.5
5 5 3 3 3 sin i
(A) (B) (C) (D)
2 2 2 2

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Geometrical Optics

Q.38 A ray of light in medium of refractive index m1 is partly Q.43 A ray of light falls on a transparent sphere with center
at C as shown in figure. The ray emerges from the
reflected and refracted at the boundary of a medium of sphere parallel to line AB. The refractive index of the
refractive index m 2 as shown fig. If Ð BOC = 900. The sphere is
value of angle i is given by – C
A
60° ·
B

A B (A) (B) (C) 3/2 (D) 1/2

2 3
i i
90° µ1 Q.44 An object is kept at a distance of 16 cm from a thin lens
O µ2
and the image formed is real. If the object is kept at a
r distance of 6 cm from the same lens the image formed is
C virtual. If the size of the images formed are equal, the
focal length of the lens will be
(A) 15 cm (B) 17 cm
(A) tan -1 ( m1 / m 2 ) (B) tan -1 ( m 2 / m1 ) (C) 21 cm (D) 11 cm

(C) sin -1 ( m 2 / m1 ) (D) cos -1 ( m1 / m 2 ) Q.45 Two point objects are placed on principal axis of a thin
converging lens. One is 20 cm from the lens and other
Q.39 Two transparent media A and B separated by a plane is on the other side of lens at a distance of 40 cm from
boundary. The speed of light in medium A is 2.0 × 108 the lens. The images of both objects coincide. The
and in medium B 2.5 × 108 ms-1. The critical angle for magnitude of focal length of lens is -
which a ray of light going from A to B it totally internally 80 40
reflected is – (A) cm (B) cm
3 3
-1 æ1ö -1 æ2ö 20
(A) sin ç ÷ (B) sin ç ÷ (C) 40 cm (D) cm
è2ø è5ø 3

-1 æ4ö Q.46 A plano convex glass lens (mg = 3/2) of radius of

(C) sin ç ÷ (D) None of these
è5ø curvature R = 10 cm is placed at a distance of ‘b’ from a
concave lens of focal length 20 cm. What should be
the distance ‘a’ of a point object O from the plano
Q.40 Angle of minimum deviation is equal to the angle of
convex lens so that the position of final image is
prism A of an equilateral glass prism. The angle
independent of ‘b’
Of incidence at which minimum deviation will be
obtained is
(A) 60° (B) 30° (C) 45° (D) sin–1 (2/3)
·
Q.41 An equilateral prism deviates a ray through 40° for two O
angles of incidence differing by 20°. The possible angles a b
of incidences are:
(A) 40°, 60° (B) 50°, 30°
(A) 40 cm (B) 60 cm (C) 30 cm (D) 20 cm
(C) 45°, 55° (D) 30°, 60°

Q.47 A convex lens of focal length A and a concave lens of

7 focal length B are placed in contact. The focal length of
Q.42 Refractive index of a prism is and the angle of the combination is
3
(A) A + B (B) (A – B)
prism is 60º. The minimum angle of incidence of a ray
that will be transmitted through the prism is: AB AB
(C) (D)
(A) 30º (B) 45º (C) 15º (D) 50º ( A + B) ( B - A)

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 Geometrical Optics

Q.48 A convex lens is making full image of an object. if half Q.53 The distance between the object and the real image
of lens is covered by an opaque object, then formed by a convex lens is d. if the magnification is m,
(A) half image is not seen the focal length of the lens is -
(B) full image of same intensity is seen md md
(C) full image of decreased intensity is seen (A) (B)
(m + 1) 2 (m + d )
(D) half image of same intensity is seen
md md
Q.49 When a thin convex lens is put in contact with a thin (C) (D)
(m - 1) 2 m -1
concave lens of the same focal length , the resultant
combination has a focal length equal to
Q.54 A parallel beam of light falls on a convex lens. The path
(A) f / 2 (B) 2f (C) 0 (D) ¥
of the rays is shown in fig. It follows that –

Q.50 Focal length of a convex lens will be maximum for

(A) blue light (B) yellow light
(C) green light (D) red light

µ1 µ µ2
Q.51 A convex lens has a focal length f. It is cut into two
parts along the dotted line as show in the figure. The
focal length of each part will be
(A) m1 > m > m 2
(B) m1 < m < m 2
(C) m1 = m < m 2
(D) m1 = m > m 2

Q.55 What is the relation between refractive indies m , m1

f 3
(A) (B) f (C) f (D) 2f
2 2 and m 2 if the behaviour of light rays is as shown in fig.

Q.52 A convex lens is made up of three different materials as

shown is the figure. For a point object placed on its µ1 µ µ2
axis, the number of image formed are

(A) m > m 2 > m1

(B) m < m 2 < m1
(C) m < m 2 : m = m1

(A) 1 (B) 3 (C) 4 (D) 5 (D) m 2 < m1 : m : m 2

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Geometrical Optics

EXERCISE-IV

NTSE STAGE-1 Q.7 A convex lens of focal length f1 is held in contact with
PREVIOUS YEAR'S a concave lens of focal length f2. We can find rough
focal length of the combination only when :
Q.1 Amount of light entering into the camera depends upon: [NTSE CG 2014-15]
[NTSE Delhi 2014-15]
(A) f1 = f2 (B) f1 < f2
(A) focal length of objective lens.
(C) f1 > f2 (D) None of these
(B) product of focal length & diameter of objective lens.
(C) distance of objective from camera.
Q.8 As shown in figure, a liquid of refractive index ‘n 2’ is
(D) aperture setting of the camera.
poured onto the concave surface of concav convex
Q.2 The resultant focal length of the lenses as shown in lens. R1 and R2 are the radii of curvature of convex and
the figure is : [NTSE Delhi 2014-15] concave surfaces of the lens respectively and
R1 = 2R2. The refractive index of material of lens is n 1.
f
(A) 2f (B) For which combination of n 1 and n2, the whole system
2
behaves as a diverging lens : [NTSE AP 2014-15]
f
(C) (D) f
4 R2
R1
Q.3 White colour of the cloud is due to :
[NTSE Delhi 2014-15] (A) n 1 = 1.2 and n2 = 1.8
(A) Reflection of seven colours of light (B) n1 = 1.63 and n2 = 1.35
(B) Refraction of seven colors of light (C) n1 = 1.56 and n2 = 1.33
(C) Scattering of seven colors of light
(D) n1 = 1.7 and n2 = 1.33
(D) Absorption of seven colours of light

Q.9 A convex lens forms a real image of a point object placed

Q.4 What is the angle of incidence when the incident ray is
on its principal axis. If the upper half of the lens is cut :
normal to the interface or boundary separating two
media ? [NTSE CG 2014-15] [NTSE Bihar 2014-15]
(A) 0º (B) 90º (A) The image will be shifted downward
(C) 180º (D) 45º (B) The image will be shifted upward
(C) The intensity of the image will decrease
Q.5 In an experiment with a rectangular glass slab, for an (D) None of the above
angle of incidence of 60º in air, angle of refraction is
measured to be r1. When the glass slab is replaced by Q.10 A screen is placed at a distance 40 cm away from an
a hollow slab filled with water, angle of refraction is
illuminated object. A converging lens if placed between
measured to be r 2. Then : [NTSE CG 2014-15]
the source and screen and attempt is made to form an
(A) r2 = r1 (B) r2 > r1
image on screen. If no position could be found. The
(C) r2 < r1 (D) Cannot say
focal length of the lens : [NTSE Bihar 2014-15]

Q.6 If angle of minimum deviation through an equilateral (A) Must be less than 10 cm
prism is 40º, angle of incidence (being equal to angle of (B) Must be greater than 20 cm
emergence) would be : [NTSE CG 2014-15] (C) Must not be greater than 20 cm
(A) 50º (B) 60º (D) Must not be less than 10 cm
(C) 40º (D) None of these

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 Geometrical Optics

Q.11 Which of the following correctly represents graphical

B
relation between the angle of incidence i and the angle
of reflection r ? [NTSE CG 2014-15] A F 2F
(c) 2F F A'
B'
i i
(A) (B) B A'
r r F 2F
(d) A
2F F
B'
i i
(C) (D) B
r r B' F 2F
(e) A
2F F A'
Q.12 The far point of a myopic person is 100 cm in front of
him. What is the power of the lens to correct the problem
B
of his eye : [NTSE Jhar. 2014-15]
(A) –1.0 D (B) +1.0 D A F 2F
(C) –1.25 D (D) +1.25 D (f) 2F F

Q.13 A vessel is 2 meter deep. How deep will it appear if it is

(A) a, b and e (B) b, c and e
filled with water and viewed from above ?
[NTSE Jhar. 2014-15] (C) d, c and f (D) a, e and f
(A) 2m (B) 3m
Q.17 Power of a lens is measured in ............... :
3 4 [NTSE MP 2014-15]
(C) m (D) m
2 3 (A) Meter (B) Centimeter
(C) Kilometer (D) Diopter
Q.14 A convex lens is made of a material (m = 1.2), both the
surfaces are convex. If it is dipped in water (m = 1.33) it Q.18 A convex spherical mirror is considered as a suitable
will behave like : [NTSE Jhar. 2014-15] rear view mirror for automobiles, because :
(A) Convergent lens (B) Divergent lens [NTSE Odisha 2014-15]
(C) A rectangular slab (D) A prism (A) It always produces virtual, erect and diminished
images.
Q.15 A ray of light passes from denser medium to rarer (B) It always produces real, erect and magnified images
medium, if thickness of the denser medium is doubled, (C) It always produces real, inverted and diminished
then the emerging angle is : [NTSE Kar. 2014-15]
images
(A) Also doubled (B) Reduced by its half
(D) It always produces virtual, inverted and magnified
(C) Not affected (D) Critical angle
images.
Q.16 Which of the following group of the ray diagram is
Q.19 The image of a candle formed by a convex lens is found
correct : [NTSE Kar. 2014-15]
to be real and inverted. What change would be found
B in the nature of this image if the lower half of the lens is
A F 2F convered by black paper ? [NTSE Odisha 2014-15]
(a) 2F A' (A) Only the upper half of the candle will be found in
B' the image
(B) There would be no change in the image of any kind.
(C) The image would now be real and erect at the same
B location.
A F 2F (D) The image would now be real and inverted at the
(b) 2F F B'
same location but its intensity would be reduced to
A'
half.

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Geometrical Optics

Q.20 The refractive index of glass with respect to air is 1.53 Q.26 A student was asked to draw a ray diagram for formation
and that of water with respect to air is 1.33. Then a of image by a convex lens for the following positions
convex lens whose focal length in air is 20 cm, when of the object : [NTSE Raj. 2014-15]
(A) Between F and 2F
fully immersed in water would have the focal length :
(B) At F
[NTSE Odisha 2014-15] (C) At 2F
(A) 20 cm (B) 3.9 cm (D) Between F and optical centre
(C) 70..49 cm (D) 7.8 cm The position for which virtual image can be formed
among these is :
Q.21 A myopic eye can be corrected by using a : (A) B (B) A
(C) C (D) D
[NTSE UP 2014-15]
(A) Convex lens (B) Concave lens
Q.27 The far point of a myopic person is 75 cm in front of the
(C) Plane mirror (D) Cylindrical lens eye. The nature and power of the lens required to correct
the problem, will be : [NTSE Raj. 2014-15]
Q.22 A concave mirror of focal length is 10 cm, produces an (A) Convex lens, –1.33 D
image five times large and real. The distance of object (B) Concave lens, –1.33 D
from the mirror will be : [NTSE UP 2014-15] (C) Concave lens, +1.33 D
(D) Convex lens, +1.33 D
(A) 10 cm (B) 12 cm
(C) 16 cm (D) 20 cm
Q.28 A student has been given a project to prepare an
astronomical telescope, she has to select two lenses of
Q.23 If a part of a convex lens is covered, its focal length type. [NTSE Karanataka 2016-17]
will: [NTSE WB 2014-15] (A) Concave lenses of different focal length.
(A) Remain unchanged (B) Concave lenses of same focal length.
(B) Become twice (C) Concave lenses of different focal length.
(C) Become half (D) Concave lenses of same focal length.

(D) Depend on the covered area

Q.29 Mirages are formed due to the natural phenomena.
[NTSE Karanataka 2016-17]
Q.24 The minimum distance between an object and its real (a) Earth’s terrostrical heating.
image in a convex lens is : [NTSE WB 2014-15] (b) Reflection of light.
(A) 2.5 times its focal length (c) Refraction of light.
(B) 2 times its focal length (d) Diffraction of light
(e) Total internal reflection of light.
(C) 4 times its focal length
(A) a, c and e (B) a, b and d
(D) Equal to its focal length
(C) b, c and d (D) c, d and e.

Q.25 The focal length of a concave mirror in air is f. If it is Q.30 If an object moves towards or away a plane mirror with
a velocity v then the image will approach or recede
æ 4ö
immersed in water ç n = ÷ , then the focal length will with velocity. [NTSE Jharkhand 2016-17]
è 3ø
(A) v (B) 2v
be : [NTSE Raj. 2014-15] (C) 3v (D) 4v

4 Q.31 The focal length of a concave mirror depends upon

(A) f (B) f
3 [NTSE Jharkhand 2016-17]
(A) The distance of the object from the mirror
3 (B) The distance of the image from the mirror
(C) f (D) 4f (C) The radius of curvature of the mirror
4
(D) None of the above

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Q.32 The rainbow is formed due to? Q.39 A human being has a horizontal field of view of
[NTSE Jharkhand 2016-17] about_______ with one eye and of about_______ with
(A) Refraction (B) Internal Reflection both eyes [NTSE MP 2016-17]
(C) Dispersion (D) all the above (A) 120°, 150° (B) 150°, 180°
(C) 180°, 210° (D) 210°,240°

Q.33 A person cannot see distincly any object placed

Q.40 1 dioptre is the power of a lens whose focal length is
beyond 40cm from his eye. What is the power of the
[NTSE Panjab 2016-17]
lens which will enable him to see distant stars clearly?
(A) 2 m (B) 1 m
[NTSE Jharkhand 2016-17] (C) 3 m (D) 1.5 m
(A) +2.5D (B) –2.5D
(C) +3.5D (D) –3.5D Q.41 Where should an object be placed in front of convex
lens to get a real image of the same size of the object
Q.34 The compound microscope consists of two convex [NTSE Panjab 2016-17]
lenses of 5cm and 20cm focal length, then which of (D) At the principal focus of the lens
them will be object lens and eye piece? (B) At twice the focal length
[NTSE Gujrat 2016-17] (C) At infinity
(A) Object lens with 20cm focal length and eye piece (D) Between optical centre of the lens and its principal
with 5cm focal length focus

(B) Object lens with 5cm focal length and eye piece
Q.42 The refractive index of water and glass with respect to
with 20cm focal length.
(C) Both should have 20cm focal length. 4 3
air are and respectively. The refractive index of
(D) Both should have 5cm focal length 3 2
glass with respect to water will be[NTSE UP 2016-17]
Q.35 The focal length of an eye lens is changed due to the
17 1
action of________ [NTSE Gujrat 2016-17] (A) (B)
6 6
(A) Pupil (B) Retina
(C) Ciliary muscles (D) Cornea 9
(C) 2 (D)
8
Q.36 At what distance should an object be placed to obtain
its real, inverted and of some height as the object by a Q.43 The total internal reflection of light is not possible,
convex lens? [NTSE Gujrat 2016-17] When light travels from [NTSE UP 2016-17]
(A) At focus (A) Glass to water (B) Water to glass
(B) Between focus and centre of curvature (C) Water to air (D) Glass to air
(C) At centre of curvature
(D) Between optical centre and focus Q.44 A radii of curvature of two faces of a biconvex lens of
refractive index are in the ratio of 2:3. The focal length
of the lens is 12 cm. The radius of curvature of the
Q.37 What is the focal length of convex lens having power
surface with low value of radius of curvature is…
+ 5.0 D? [NTSE Gujrat 2016-17]
[NTSE Telangana 2016-17]
(A) –10 cm (B) – 20 cm
(A) 5 cm (B) 10 cm
(C) +10 cm (D) + 20 cm (C) 15 cm (D) 20 cm

Q.38 Dentists use to see large image of the teeth of patients Q.45 The change in focal length of an eye lense is caused by
using [NTSE MP 2016-17] the action of… [NTSE Telangana 2016-17]
(A) Convex mirror (D) Convex lens (A) pupil (B) retina
(C) Concave lens (D) Concave mirror (C) ciliary muscles (D) iris

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Geometrical Optics

Q.46 If a ray of light incident on a plane mirror is such that it Q.49 Light rays from a very small object immersed in water
makes an angle fo 30° with the mirror, then the angle of falls on the bubble of radius R. Assume that the object
reflection will be: [NTSE Chattisgarh 2016-17] is very close to the surface of the bubble. Refractive
index of water is 4/3. Take the approximation in
(A) 30° (B) 45°
(C) 55° (D) 60° sin q @ q and cos q @ 1 where 'q' is angle and consider
the rays close to a diameter of the bubble. Use the
following formula to solve the problem
Q.47 System is shown in the figure. Light rays from a point
object are first deviated by a prism and then focused sin a sin b sin g
= = (See figure) [NTSE AP 2016-17]
by a thin lens of focal length f. The prism is made of a b c
material with refractive index 3/2 and has small apex
angle ie, small angle approximations are valid c b
sina @ a and sin (a + b) = a + b , whered a and b are
angles. The final image is ….. [NTSE AP 2016-17]
a

Point
Prism Lens
Object Air bubble
R
object
f 4/2 water

The image is …..(from the center of the bubble)

(A) virtual and formed at a distance 2f from the lens
(A) Virtual and formed at a distance 3R/2
(B) real and formed at a distance 2f from the lens
(B) Virtual and formed at a distance 2R/3
(C) real and formed at a distance 3f/2 (C) Virtual and formed at a distance 3R/5
(D) real and formed at a distance 3f from the lens (D) Real and formed at a distance 3R/2

Q.48 A narrow beam of light is incident on a 30°–60°– 90° Q.50 A light ray enters from medium A to medium B as shown
prism perpendicular to the surface AB. Assume that in figure below. The refractive index of medium B relative
light beam is close to A. The index of refraction of prism to A will be [NTSE Chandigarh 2016-17]

-1 æ 10 ö
is 2.1. See figure and take : sin ç ÷ = 28°26’. The
è 21 ø

beam emerges from the face……. [NTSE AP 2016-17] medium B

medium A

60° (A) Greater than unity (B) Less than unity

A B
(C) Equal to unity (D) Zero

Q.51 Which of the following defects can be rectified by using

(A) CB cylindrical lenses? [NTSE Chandigarh 2016-17]
(B) AB (A) Myopia (B) Presbyopia
(C) AC (C) Astigmatism (D) Hypermetropia
(D) Some light through AC and remaining light through
AB

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 Geometrical Optics

Q.52 Splitting of white light into its component is called Q.59 The image formed by a cancave mirror is observed to
[NTSE Chandigarh 2016-17] be real, inverted and large than the object. Where should
(A) Dispersion the position of the object? [NTSE Raj. 2016-17]
(B) Scattering (A) At the centre of curvature.
(C) Total internal Reflection (B) Between the principal focus and centre of curvature.
(D) Spectrum (C) Beyond the centre of curvature
(D) Between the pole of the mirror and its principal
Q.53 Formation of Rainbow is due to focus.
[NTSE Chandigarh 2016-17]
(A) Scattering Q.60 The path of ray of light in different media of refractive
(B) Dispersion
indics n1, n2, n3 and n4 is shown in figure. The velocity
(C) Atmospheric Refraction
of light will be maximum in the medium whose refractive
(D) Total internal Reflection
index is : [NTSE Raj. 2016-17]
Q.54 Speed of light is maximum in a medium whose refraction
index with respect to air is n1
[NTSE Chandigarh 2016-17] n2
(A) 1.33 (B) 1.5 n3
(C) 1.2 (D) 1.67
n4
Q.55 A beam of light traveling in air enters into a liquid. Its
speed reduces by 30%. The refractive index of liquid
with respect to air is [NTSE Bihar 2016-17] (A) n1 (B) n2
(A) 10/7 (B) 10/3 (C) n3 (D) n4
(C) 7/5 (D) 4/3
Q.61 Which one of the following phenomena is an example
Q.56 The refractive index of a liquid is 5/3. A ray of light of scattering of light ? [NTSE Raj. 2016-17]
traveling in this liquid falls at interface of liquid & air. (A) Bending of rod at interface of air and water
At what angle of incidence should it fall on liquid air (B) Twinkling of stars
interface so that it suffers total internal reflection? (C) Tyndal effect
[NTSE Bihar 2016-17] (D) Mirage in desert during summer.
(A) 53° (B) 24°
(C) 15° (D) 17°
Q.62 A convex lens of focal length f produces a real image of
size m-times the size of the object. Then the object
Q.57 A beam of light in air is incident upon the smooth plane
distance is: [NTSE Odisha 2016-17]
surface of a piece of flint glass making an angle of 30°
with its plane. If the reflected beam and refracted beam æ m +1 ö
are perpendicular to each other, what is the index of (A) ç ÷ (B) (m + 1)f
è m ø
refraction of flint glass with respect to air ?
[NTSE Bihar 2016-17] (m + 1) fm
(C) (D) (m+ 1)
3 f
(A) 1/2 (B)
2
Q.63 The absolute refractive index of a medium is 1.5. Then
2 what would be the velocity of light in this medium?
(C) (D) 3
3 [NTSE Odisha 2016-17]
(A) 2 × 10 m / s2 (B) 1.5 × 10 m / s2
Q.58 The image created by a converging lens is projected (C) 3.5 × 10 m /s (d) 2.5 × 10 m / s2
on a screen that is 60 cm away from the lens. If the
height of the image is one fourth the height of the object, Q.64 Two lenses of power +4 and –6 dioptres are placed in
what is focal length of the lens? contact with each other. The focal length of the
[NTSE Bihar 2016-17] combination will be: [NTSE Odisha 2016-17]
(A) 36 cm (B) 45 cm (A) 0.5 m (B) – 0.1 m
(C) 80 cm (D) 48 cm (C) – 0.5 m (D) 0.1 m

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Geometrical Optics

Q.65 Object placed _______ of lens or mirror give infinite Q.70 Two points P and Q lie on either side of an axis xy as
magnification. [NTSE Maharashtra 2016-17] shown. It is desired to produce an image of P at Q
(A) at focus (B) at infinite distance
using a spherical mirror with xy as the optic axis. The
(C) between (D) at 2F1
spherical mirror must be : [NTSE Haryana 2016-17]
Q.66 If a 3 cm tall object placed perpendicular to principal P
axis of a convex lens of focal length 15 cm produces a x y
Q
real inverted image of height 15 cm, then its object
distance (u) is ………. and image distance (v) is ……. (A) Converging and positioned with left of P.
[NTSE Maharashtra 2016-17] (B) Diverging and positioned with right of Q.
(A) u = – 18 m, v = + 90 m
(C) Diverging
(B) u = + 18 cm, v = – 90 cm
(D) Converging and positioned with right of Q.
(C) u = – 18 cm, v = + 90 cm
(D) u = + 18 cm, v = + 90 cm
Q.71 The refractive index of diamond with respect to glass is
Q.67 If the path of parallel light through a concave lens is as 1.6 and absolute refractive index of glass is 1.5. Then
shown in the figure, where h, h1 & h2 are refractive the absolute refractive index of diamond will be :
indices, then ……… [NTSE Maharashtra 2016-17] [NTSE Odisa 2015-16]
(A) 2.5 (B) 2.4
(C) 3 (D) 3.5

Q.72 An object is placed at a distance x1 from the focus on

the principal axis of a concave mirror. The image is
formed at a distance x2 from the focus. Then the focal
(A) h > h1 = h2 (B) h = h1 < h2 length of the mirror is : [NTSE Odisa 2015-16]
(C) h = h1 > h2 (D) h < h1 = h2
x1
(A) (B) x1x2
x2
Q.68 Distance covered by an object thrown upwards in the
last second ………… [NTSE Maharashtra 2016-17]
(A) depends on initial velocity x2
(C) (D) x1 x 2
(B) depends on mass x1
(C) depends on air velocity
(D) is always same
Q.73 Two thin lenses of focal lengths f1 and f2 are placed in
Q.69 A thin symmetric double convex lens of power P is cut contact with each other such that the combination
into three parts A, B and C as shwon. The power of behaves as a glass slab. Then how are f1 and f2 related
[NTSE Haryana 2016-17] to each other ? [NTSE Odisa 2015-16]

1
(A) f1 = (B) f2 = –f1
f2
A
(C) f1 = f2 (D) f1 = f2
B C

Q.74 A mirror forms a virtual, erect and small image then kind
P P of mirror is : [NTSE CG 2015-16]
(A) A is P, B is (B) A is 2P, B is
4 3 (A) Plane (B) Concave
(C) Convex (D) 1 and 2 both
P P P
(C) A is , B is (D) A is P, B is
2 4 2

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 Geometrical Optics

Q.75 If the upper half of convex lens is wrapped with a black Q.82 When light travels from medium X to medium Y as
paper, then how the image formed by it will be effected: shown: [NTSE Raj. 2015-16]
[NTSE CG 2015-16]

(A) The size of the image is reduced to one half (A) Both the speed and the frequency decrease
(B) The upper part of the image is reduced (B) Both the speed and the frequency increase
(C) The intensity of the image is reduced (C) Both the speed and the wavelength decrease
(D) There will be no effect on image (D) Both the wavelength and the frequency are
unchanged
Q.76 Identify the device used as spherical mirror or lens in
the following cases when the image formed is virtual & Q.83 A candle C is kept between two parallel mirrors, at a
erect in each case : [NTSE Haryana 2015-16] distance 0.2 d from the mirror 1. Here d is the distance
A. Object is placed between device & its focus, image between mirrors. Multiple images of the candle appear
is enlarged & behind it. in both mirrors. How far behind mirror 1 are the nearest
B. Object placed between focus & device image formed two images of the candle in that mirror ? :
is enlarged & on the same side as that of object.
[NTSE Raj. 2015-16]
A B
(A) Convex lens Concave mirror
(B) Convex mirror Convex lens

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
(C) Convex mirror Convex lens 0.2 d
(D) Concave mirror Concave lens
C
Q.77 Most of the refraction takes place at ................ when
light enter the eye : [NTSE Haryana 2015-16]
(A) Pupil
(B) Iris
(C) Outer surface of cornea d
(D) Crystalline lens
(A) 0.2 d, 1.8 d (B) 0.2 d, 2.2 d
Q.78 The focal length of a spherical mirror is 20 cm. The (C) 0.2 d, 0.8 d (D) 0.2 d, 1.2 d
radius of curvature of this mirror will be :
[NTSE MP 2015-16] Q.84 The sky appears blue due to : [NTSE UP 2015-16]
(A) 10 cm (B) 20 cm (A) Reflection of light
(C) 40 cm (D) 80 cm (B) Refraction of light
(C) Total internal reflection of light
Q.79 To correct the defect Myopia of near sightedness, we (D) Scattering of light
have to use : [NTSE MP 2015-16]
(A) Convex lens (B) Concave lens Q.85 The image produced by a concave lens is always :
(C) Plane glass (D) None of the above [NTSE UP 2015-16]
(A) Real (B) Virtual
Q.80 Which of the following colours is not a primary colour? (C) Inverted (D) Enlarged
[NTSE MP 2015-16]
(A) White (B) Green Q.86 A lens have power +5D. This lens will be :
(C) Red (D) Blue [NTSE UP 2015-16]
(A) A convex lens of focal length 0.20 m
Q.81 The image of an object in human eye is formed at : (B) A concave lens of focal length 0.20 m
[NTSE MP 2016-17] (C) A convex lens of focal length 0.20 m
(A) Cornea (B) Iris (D) A concave lens of focal length 0.05 m
(C) Pupil (D) Retina

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Geometrical Optics

Q.87 The nature of image formed when an object is placed Q.95 If speed of light travelling from air to a medium decreases
between Principal focus and optical centre of convex by 40%, find the refractive index of the medium with
lens : [NTSE Punjab 2015-16] respect to air. [NTSE MH 2015-16]
(A) Virtual, Diminished (B) Real, Magnified (A) 2.5 (B) 1.67
(C) Virtual, Magnified (D) Real, Diminished (C) 1.3 (D) 1.25

Q.88 The part of eye which changes in focal length of eye Q.96 Choose the wrong statement related to virtual image:
lens : [NTSE Punjab 2015-16] [NTSE MH 2015-16]
(A) Iris (B) Cornea (A) images are always produced by plane mirrors only
(C) Pupil (D) Ciliary muscles (B) images are always erect
(C) image cannot be obtained on the screen
Q.89 Refractive index of a medium with respect to air is
(D) imge is formed at point where extended and retraced
m= 2 , find the critical angle between the two medium rays appear to meet
[NTSE WB 2015-16]
(A) 30º (B) 90º Q.97 An object is placed at a distance of 20 cm infront of a
(C) 45º (D) 60º concave mirror of focal length 20 cm. The image formed
is, [NTSE Karnataka 2015-16]
Q.90 What will be the colour of the sky as seen from the (A) Virtual and inverted
earth if there is no atmosphere? [NTSE WB 2015-16] (B) Real, inverted and diminished
(A) Black (B) Blue (C) Real and erect
(C) Orange (D) Red (D) Real, inverted and same size as the object

Q.91 A convex lens of glass has power P in air. If it is immersed Q.98 A real and enlarged image can be formed by using a
in water, its power will be [NTSE WB 2015-16] [NTSE Jharkhand 2015-16]
(A) more than P
(A) convex mirror
(B) less than P
(B) plane mirror
(C) P
(C) concave mirror
(D) More the P for some colours and less than P for
(D) either convex or a plane mirror
others

Q.99 For an incident ray directed towards centre of curvature

Q.92 You are provided with a concave lens having focal
of a spherical mirror the reflected ray
length 15 cm. If it diminishes the image by one-third,
calculate the distance of image from lens? [NTSE Jharkhand 2015-16]
[NTSE CG 2015-16] (A) retraces its path
(A) 5 cm (B) 10 cm (B) passes through focus
(C) 5 m (D) 10 m (C) passes through the pole
(D) becomes parallel to the principal axis
Q.93 At the traffic signals, red light is used for stop due to
the reason that it can be seen from a distance. The Q.100 A mirror which can produce a magnification of +1 is
phenomenon involved is known as [NTSE Delhi 2015-16]
[NTSE CG 2015-16] (A) Convex mirror
(A) Reflection (B) Refraction (B)Concave mirror
(C) Dispersion (D) Scattering (C) Plane mirror
(D) Both concave mirror & plane mirror.
Q.94 Colour of scattered light depends ……..
[NTSE MH 2015-16] Q.101 The refractive index of the material of a double convex
(A) only on size of scattering particle lens is 1.5 and its focal length is 5 cm. If the radius of
(B) only on length of travelling light. curvatures of faces are equal then the value of its radius.
(C) both size of scattering particle and length of of curvature is [NTSE Telangana 2015-16]
travelling light (A) 5 cm (B) 6.5 cm
(D) on colour of incident light (C) 8 cm (D) 9.5 cm

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 Geometrical Optics

Q.102 The phenomenon used in optical fibers for transmission Q.4 You are standing on the shore of a lake. You spot a fish
of light energy is [NTSE Telangana 2015-16] swimming below the lake surface. You want to kill the
(A) refraction (B) scattering fish first by throwing a spear and next, by pointing a
(C) reflection (D) total internal reflection high-power laser torch. How should you aim the spear
and torch, respectively, from the options given below?
Q.103 A Converging bundle of light rays in the shape of cone [NTSE-2016]
with a vertex angle of 45° falls on a circular diaphragm I. above the apparent position of the fish
of 20 cm diameter. A lens with power 5D is fixed in the II. below the apparent position of the fish
diaphragm. Diameter of face of lens is equal to that of III. directly at the apparent position of the fish
diaphragm. If the vertex angle of new cone is q, then (A) SPEAR: II; LASER: III
tan q = …....…… [NTSE AP 2015-16]
(B) SPEAR: I; LASER: II
(A) 1.5 (B) 2
(C) SPEAR: II; LASER: II
(C) 3 (D) 3
(D) SPEAR: III; LASER: III
Q.104 A point source is located at a distance of 20 cm from
Q.5 A beam of light coming from a rarer medium is partially
the front surface of a glass bi-convex lens. The lens
reflected from the surface to a denser medium and
has a thickness of 5 cm and radius of curvature of it’s
surfaces is 5 cm. The refractive index of glass is 1.5. partially refracted into the denser medium. If the
The distance of image formed by it from the rear surface reflected and the refracted rays are perpendicular to
of this lens is ………. (approximately) each other and the ratio of the refractive indices of
[NTSE AP 2015-16] denser and rarer medium is 3 , the angle of refraction
(A) 9.2 cm (B) 6.3 cm will be- [NTSE-2016]
(C) 5.7 cm (D) 4 cm (A) 60º (B) 30º
(C) 45º (D) 41.5º
NTSE STAGE-2
PREVIOUS YEAR'S
Q.6 A person can see clearly only the objects situated in
the range 50 cm to 300 cm. He went to an Optometist
Q.1 To read a poster on a wall, a person with defective
who prescribed him a lens of certain power to increase
vision needs to stand at a distance of 0.4 m from the
the maximum distance of his vision to infinity, i.e., it
poster. A person with normal vision can read the poster
from a distance of 2.0 m. Which one of the following corrected the near-sightedness. However, upon using
lens may be used to correct the defective vision ? the prescribed lens the person discovered that the near
[NTSE-2017] point of his vision has shifted from 50 cm to a distance
(A) A concave lens of 0.5 D ‘d’. What is the value of d? [NTSE-2016]
(B) A concave lens of 1.0 D (A) 60cm (B) 100 cm
(C) A concave lens of 2.0 D (C) 40cm (D) 500 cm
(D) A convex lens of 2.0 D
Q.7 Match the following: [NTSE-2015]
Q.2 Foam of soap always appears white as [NTSE-2016]
(A) it contains large hydrocarbon chains.
Phenomenon Reason
(B) it absorbs red portion of the visible light
i Rainbow A Scattering of light
(C) it reflects light of all wavelengths.
ii Twinkling of stars B Dispersion of light
(D) it has one hydrophobic end, which is insoluble in
iii Blue colour of sky C Fluctuation of the
water. refraction index in
atmo sphere layers
Q.3 A ray of light in air is incident on an equilateral glass iv Advancement of D Refraction of light
prtsm at an angle qi to the normal. After refraction, the sunrise and delay
light travelled parallel to the base of prism and emerged of sunset
in air at an angle qe to the normal. If the angle between
the incident and the emergent rays is 60º, then the
(A) (i)-B,(ii)-D,(iii)-A,(iv)-C
refractive index of glass with respect to air is
[NTSE-2016] (B) (i)-B,(ii)-C,(iii)-A,(iv)-D
(A) 1.33 (B) 1.5 (C) (i)-B,(ii)-A,(iii)-C,(iv)-D
(C) 1.73 (D) 1.66 (D) (i)-D,(ii)-B,(iii)-A,(iv)-C

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Geometrical Optics

Q.8 A person is suffering from both near sightedness and Q.12 A ship sends a sonar wave to the sea bed which is flat
far sightedness. His spectacles would be made of and measured several times over a large area. One day
[NTSE-2015] the reflected sound wave takes longer time than in
previous measurements. The possible reason is:
(A) two convex lenses with the upper lens having a
[NTSE-2013]
larger focal length than the lower lens.
(A) the frequency of the sonar wave, generated by the
(B) two concave lenses with the upper lens having a
equipment is lower than previous measurements.
smaller focal length than the lower lens. (B) there is a solid object of large size in the path of
(C) a concave lens as the upper lens and a convex lens sonar wave.
as the lower lens (C) there is a huge air bubble in the path of sonar wave
(D) a convex lens as the upper lens and a concave lens (D) the loudness of the sonar wave, generated by the
as the lower lens equipment is lower than previous measurement.

Q.13 Which of the following ray diagram is correct?

Q.9 A ray of light is incident in medium 1 on a surface that
[NTSE-2013]
separates medium 1 from medium 2. Let v1 and v2
represent the velocity of light in medium 1 and medium
2 respectively. Also let n12 and n21 represent the refractive
index of medium 1 with respect to medium 2 and (A)
refractive index of medium 2 with respect to medium 1,
respectively. If i and r denote the angle of incidence
and angle of refraction, then- [NTSE-2014]

sin i v sin i v2
(A) = n 21 = 1 (B) sin r = n 21 = v (B)
sin r v2 1

sin i v1 sin i v
(C) sin r = n12 = v (D) = n12 = 2
2 sin r v1

(C)
Q.10 A convex lens has a focal length of 0.5 m. It has to be
combined with a second lens, so that the combination
has a power of 1.5 diopter. Which of the following could
be the second lens? [NTSE-2014]
(A) A concave lens of focal length 2 m.
(B) A convex lens of focal length 0.5 m.
(D)
(C) A concave lens of focal length 0.5 m.
(D) A convex lens of focal length 2 m.

Q.11 Which of the following statement is correct ‘?

Q.14 A concave lens always gives a virtual image. In optical
[NTSE-2014]
lenses worn by humans which of the following
(A) A person with myopia can see nearby objects clearly statements is true? [NTSE-2013]
(B) A person with hypermetropia can see nearby objects (l) The lens can never be concave.
clearly (B) In some cases the lens can be concave if the focal
(C) A person with myopia can see distant objects clearly length is much larger than 2.5 cm.
(D) A person with hypermetropia can not see distant (C) All focal’length concave lenses are possible.
objects clearly (D) All focal length’convex lenses are possible.

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 Geometrical Optics

Q.15 In the table below, column I lists various mirrors and Q.18 A printed page is seen through a glass slab place on it.
lenses and column II their uses, qualities or properties. The printed words appear raised. This is due to
[NTSE-2012] [NTSE-2011]
Column I Column II (l) refraction at the upper surface of the slab
I. Plane mirror A. Feels thinner (B) refraction at the lower surface of the slab
in the middle. (C) partial reflection at the upper surface of the slab
II. Concave mirror B. Always forms virtual (D) partial reflection at the lower surface of the slab
image of the same size.
III. Convex mirror C. Always forms virtual Q.19 Two mirrors Ml and M2 are placed at right angle to
image of smaller size. each others as shown. The total number of images of
IV. Concave lens D. Is used by dentists to
an object ‘O’ that can be seen are [NTSE-2010 & 11]
examine teeth.
V. Convex lens E. Can be used as reading
glass.
The correct matching is
( A) I-B, II-D, I I I-C, IV-A, V-E
(B) I-C, II-A, I II-E, IV-B, V-D
(C) I-A, I I-E, III-B, IV-D V-C
(D) I-A, I I-C, III-E, I V-B, V-D

(A) Two (B) Three

Q.16 A ray of light falls on a plane mirror A kept at an angle
(C) Four (D) Infinite
90' to mirror B as shown in the figure. The angle N is
[NTSE-2012]
Q.20 On passing through a prism, a parallel beam of sunlight
splits into lights of several colours. Take a combination
of two identical prisms as shown below. A parallel beam
of sunlight is incident on the face AD. The emergent
light from the face BC, consists of [NTSE-2010]

(A) 30º (B) 45º

(A) a parallel beam of light of several colours
(C) 60º (D) 90º
(B) a divergent beam of lights of several colours.
(C) a parallel beam of white light
Q.17 Atmospheric pressure in the center of a tropical cyclone
(D) a divergent beam of white light.
is [NTSE-2012]
(A) very low because of dense sinking air.
(B) very low because of rising warm air.
(C) very high because of sinking warm air.
(D) very high because of converging winds.

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Carbon and Its Compounds

CARBON & ITS COMPOUND

3
INTRODUCTION: — The combining atoms need 1, 2, 3, or 4 electrons to
The compound obtained from ‘Carbon’ are widely used complete their octet (hydrogen completes its duplet)
as clothes, medicines, books fertilizer, fuel etc. all living — The combining atoms should contribute equal number
structures are carbon based. of electrons to form pair of electrons to be shared.
The amount of carbon present in the earth’s crust and — After sharing the pair of electrons each combining
in the atmosphere is quite merging. The earth’s crust atoms should attain stable electronic configuration like
has only 0.02% carbon in the form of mineral (like its nearest noble gas.
carbonates, hydrogen-carbonates, coal and petroleum)
and the importance has 0.03% of carbon dioxide. In CLASSIFICATION OF COVALENT BOND:
spite of this small amount of carbon available in nature. On the basis of the number of electrons shared by two
the importance of carbon seems to be immense. combining atoms, the covalent bonds are of three
Carbon forms a large number of compounds with types.
hydrogen which are known as hydrocarbons. In
addition to hydrogen. Carbon compound may also — Single Covalent Bond: A single covalent bond is
contain some other element such as oxygen, halogen, formed by the sharing of one pair of electrons between
nitrogen, phosphorus, sulphur etc. the two atoms. It is represented by one short line (–)
The number of compounds of carbon is more than between the two atoms.
three million which is much larger than the compounds
formed by all other element put together. Example: H–H, Cl–Cl, H–Cl, CH3–CH3.
— Double Covalent Bond: A double covalent bond is
BONDING IN CARBON COMPOUNDS: formed by the sharing of two pairs of electron between
Carbon forms covalent bonds in its compounds with the two combining atoms. It is represented by putting
other atoms. In each compound the valency of carbon (=) two short lines between the two bonded atoms.
is four. That is, carbon has tetravalent character. But
what is covalent bond and what is meaning of Examples: O=O(O2), CO2(O=C=O), H2C = CH2
tetravalent? — Triple covalent bond: A triple bond is formed by the
sharing of three pair of electrons between the two
Why does a carbon atom form only covalent bond? combining atoms. It is represented by putting three
— The atomic number of carbon is 6 and first shell short line () between two bonded atom.
contains just two electrons and second shell Example: N2(N º N), CH º CH.
(Outermost shell).
Contains four electrons. FORMATION OF SINGLE CONVALENT COMPOUNDS:
— Carbon atom can attain the noble gas configuration
by sharing its valence electrons with other atoms of — Formation of hydrogen molecule (H2):
carbon or with atoms of other elements and form A molecule of hydrogen is composed to two H-atoms.
covalent bond. The electronic configuration of H-atom is.
Shell - K incomplete duplet (unstable)
CONVALENT BOND: Electron-1
A chemical bond formed between two atoms of the Electronic configuration of He atom
same element or two atoms of different elements by Shell - K Complete duplet (stable)
sharing of electron is called a covalent bond. Electrons-2
Necessary conditions of the formation of covalent H– + –H ¾® H : H ¾® H - H ¾® H2
bond: K(1) K(1) Shared Covalent Hydrogen
— The combining atoms should have nonmetallic electron
character. Atom Atom Pair Bond molecule
— The combining atoms should contain 4 to 7 electron H – H Bond in terms of energy shells (orbits)
as in their respective valence shell. — Formation of chlorine molecule (Cl2). The atomic
— In hydrogen there is only 1 valence electron, but it number of chlorine is 17, thus there 17 electrons in an
also forms covalent bond. atom of chlorine.

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 Carbon and Its Compounds

Electronic configuration of Cl atom – — Formation of ammonia molecule (NH3):

Shells K L M’ Incomplete octet The atomic number of N is 7. It’s electronic
Electrons 2 8 7 configuration is 2,5 there are 5 electrons in its valence
shell. It needs 3 electrons more to complete its octet
Electronic configuration of Ar atom – like noble gas neon (2,8).
Shells K L M Complete octet
Electrons 2 8 8 Unshared
Chlorine atom needs one electron more to complete its electron pair

octet –
Cl + Cl Cl Cl Cl–Cl Cl2 3H + N H N H H–N–H
Atom Atom Shared Covalent Chlorine Three Nitrogen H H
electrons Bond molecule hydrogen atom Shared electron Ammonia
Cl – Cl bond in terms of energy shell orbits atom pair in ammonia molecule

Cl Cl Cl Cl — Formation of H2O molecule:

The electronic configuration of hydrogen is K (1) and
Cl Atom Cl Atom Cl2 molecule
that of oxygen is K(2) L(6) thus each hydrogen require
one and oxygen required two electrons to achieve the
Formation of hydrochloric acid (HCl): stable electronic configuration.
— H atom has one valence electron. It needs 1 electron
Two unshared
more to complete its duplet and chlorine has 7 valence Pair of electron
electrons. It need 1 electron more to complete its octet
and acquire stable electronic configuration (2, 8, 8) H + O + H H H O H 2O
(Water
like noble gas argon. Hydrogen Oxygen Hydrogen H H molecule)
atom atom atom Shared pair
of electron
H + Cl H Cl H–Cl HCl
Shared pair Covalent Hydrogen
electrons bond chlorine molecule — Formation of CO2 molecule:
The atomic number of C is 6 and the electronic
— Formation of oxygen (O2): configuration of C is K(2). L(4) and that of oxygen is
The atomic number of O atom is 8. There are 6 electron K(2), L(6) thus each carbon require 4 and oxygen require
in the valence shell of oxygen atom it needs 2 more two electrons to achieve th e stable electronic
electrons to attain the nearest stable inert gas Neon configuration.
(2,8) configuration:
O + Cl + O O C O
Two pairs of K L K L K L
Shared electrons Double bond 2 6 2 6 2 6 Shared pair
of electron
CO2 O =C = O
O+ O O O O=O molecule Double covalent bond
K L K L
2 6 2 6
Oxygen Oxygen O2 molecule
atom atom — Formation of CH4 molecule:
Methane is a covalent compound containing 4 covalent
— Formation of nitrogen molecule (N2): bond. It contains one carbon atom and four hydrogen
The atomic number of nitrogen is 7 and its electronic atom covalently bonded to central carbon atom.
configuration is K(2). L(5). It needs 3 electrons more
to complete its octet like noble gas neon (2,8).
H H
Two pairs of C + 4H H C H H C H
Shared electrons Double bond
Carbon Hydrogen H H
atom atom
N+ N N N N=N Covalent bond
K L K L Shared electron in methane
2 5 2 5 pair in methane
Nitrogen Nitrogen N2 molecule
CH4 molecule
atom atom

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Carbon and Its Compounds

— Formation of carbon tetrachloride molecule (CCl4):

+
The electronic configuration of carbon and chlorine +

atoms are (2,4) and (2,8,7) respectively. Carbon atom H + Cl H Cl H Cl

needs four electrons and chlorine atom needs one Shared pair of electrons
electron to attain the stable electronic configuration. lie nearer to Cl because
Cl is more electronegative
Cl Cl Note : d means partial

C + 4 Cl Cl C Cl Cl C Cl CHARACTERISTICS OF COVALENT BOND AND

Carbon Chlorine Cl Cl COVALENT COMPOUNDS:
atom atom Characteristics of covalent bond:
Covalent bond
Shared in Carbon tetra chloride — Covalent bond are formed by mutual sharing of
electron pair
electrons
CCl4 molecule Note: Shared pair of electrons is also called bonding
pair of electrons.
— Formation of ethylene or ethane molecule (C2H4): — Covalent bond is directional in nature because shared
The electronic configuration of carbon atom is 2,4. pair of electrons remains localized in a definite space
There are 4 valence electrons in one C atom. Each H between the two atoms.
atom contains 1 valence electron. Thus, there are 12
valence electrons present in ethane molecule. Characteristics of covalent compounds:
Physical Store: The covalent compounds are generally
H H gases or liquids. but compounds with high molecular
H H masses are solids.
+ + H H
Example:
H + C + C +H H C C H C= C
Solid: Urea, Glucose, Naphthalene.
Carbon and hydrogen
with their valence electrons
Shared electron pair H H Liquids: Water, ethanol, benzene.
Double Covalent bond Gases: Methane, chlorine, hydrogen, oxygen
in ethane molecule

H2C = CH2 — Melting and boiling points: Covalent compounds have

Ethylene/Ethene molecule low melting and low boiling points because
intermolecular forces (cohesive forces) in covalent
compounds are weaker than th ose in ionic
— Formation of Acetylene or ethyne molecule (C2H2) compounds.
H + C + C +H H C C H
Note
H–C C– H
two carbon and two hydrogen triple covalent bond Some exceptions like diamond and graphite which
atoms with their valence electrons Shared pairs In ethane molecule
of electron are covalent solids have very high M.P. & B.P.

NON POLAR AND POLAR COVALENT COMPOUNDS: — Solubility:

Non polar covalent bond: Covalent compounds generally dissolve readily in
A covalent bond formed between two atoms of the
organic solvents but they are less soluble in water.
same element or same electronegativity is called a non-
For example: Naphthalene which is an organic
polar covalent bond. Example:
compound dissolves readily in organic solvents like
ether but is insoluble in water. However some covalent
H: H ; O O compounds like urea, glucose, sugar etc. are soluble
the shared pair electron lies exactly midway between in water. Some polar covalent compounds like ammonia
the two atoms
and hydrochloric acid are soluble in water.
Polar covalent bond:
The covalent bond between the atoms of two elements — Conductivity:
having different electronegativities is called a polar Covalent compounds do not conduct electricity
covalent bond. Molecules in which the atom is bonded because they contain neither the ions nor free electrons
by a polar covalent bond are called polar molecules. necessary for conduction. So they do not conduct
Note: In a polar covalent bond, the shared pair of electricity
electrons lies more toward the atom which is more For example: Covalent compounds like glucose,
electronegative. alcohol. Carbon tetrachlorides do not conduct
Example: HCl, H2O & NH3
electricity.

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 Carbon and Its Compounds

Structure: In diamond, each carbon atom is covalently

Differences between ionic and covalent compounds: bonded to four other carbon atoms in a tetrahedral
arrangement. This tetrahedral arrangement of carbon
S.N. Electrovalent Covalent Compounds
atoms gives a rigid; three dimensional structure to
(Ionic) Compounds
1 Formed by transfer Formed by sharing of diamond It is due to this rigid structure that diamond.
of electrons, (only electrons, (single, — Is very hard crystalline structure?
single bond network double & triple are — Has high melting point.
exist) formed). — Is non conductor of heat and electricity?
2 Usually crystalline Usually gasses or
solid liquid only a few of Properties: Pure diamond is a transparent and
them are solid
colourless solid.
3 Generally have high Generally have low
melting and boiling M.P. and B.P. — Polished diamond sparkles brightly because it reflects
points most of the light (refractive index of diamond is 245)
4 Soluble in water but Soluble in organic — Diamond is not attacked by acids. Alkalis and solvents
insoluble in organic solvent but insoluble like water, ether, benzene or carbon tetrachloride but
solvents or soluble in water diamond is attacked by fluorine at 7500C.
5 Conduct electricity Usually non conductor 0
in solution or molten of electricity C (Diamond) + 2 F2 ¾750
¾¾ C
® CF4
state Carbon Fluorine Carbon tetrafluoride
6 Highly polar and Usually Non-polar and
— The density of diamond is 3.51 g per cm3 at 200C.
ionize in water do not ionize in water
eg. NaCl®Na+ +Cl- but few compounds
are polar in nature and Uses:
ionise in water — A saw fitted with diamond is used for sawing marbles.
eg. HCl ® H+ + Cl- — A chip diamond is used for glass cutting.
— Black diamonds are used in making drill.
ORGANIC COMPOUNDS:
— Diamonds are used for making dice for drawing very
The chemical compounds which are present in living
thin wires of harder metals.
organisms (plant and animal) are called organic
— Diamonds are also used for making high precision tools
compounds. The belief that formation of organic
for use in surgery such as, for the removal of cataract.
compounds was possible only in plants and animals
— Diamond are used for making precision thermometers
led the scientists of early days to propose that Vital
and protective windows for space craft’s.
Force was necessary for the formation of such
compounds. But the experimental work of Friedrich
GRAPHITE:
Wohler (German chemist) denied the idea of vital force
Graphite is also known as black lead it marks paper
when he prepared urea in his laboratory. (Urea is an
work black. The name graphite has been taken from
organic compound and waste product of urine).
the greek word “graphein” (which means to write) in
reference to its uses as ’lead’ in lead pencils.
ALLOTROPY/ALLOTROPES OF CARBON:
The phenomenon of existence of allotropic forms of
Structure:
an element is called allotropy. Allotropes are the
Graphite is an opaque and dark grey solid. In a crystal
different forms of the same element having different
of graphite the carbon atoms are arranged in hexagonal
physical properties but almost similar chemical
patterns in parallel planes. In a layer of graphite each
properties. There are three allotropes of carbon these
carbon atom is strongly bonded to three carbon atoms
are diamond, graphite and fullerene.
by covalent bonds. Thus, one valence electron of each
DIAMOND
carbon atoms is free in every layer of graphite crystal.
Diamond is a crystalline allotrope of carbon. Its atomic
Thus free electron makes graphite a good conductor
symbol & empirical formula is ‘C’.
of electricity.

Structure of Diamond

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Carbon and Its Compounds

Each layer is bonded to the adjacent layers by weak — It consists of 12 pentagonal faces and 20 hexagonal
forces. As a result, each layer can easily slide over the faces giving it 60 corners. Thus, Buckminster fullerene
other. has a hollow, cage-like structure.
— In figure, ball like molecules containing C atoms.
Preparation:
Properties:
— By electrically heating a graphite rod in atmosphere of
— Graphite is grayish-black, opaque material having helium.
metallic (shiny) luster. — By vaporising graphite by using laser.
— It is soft and has a soapy (slippery) touch.
— Graphite is lighter than diamond. The density of graphite
is 2.26 g per cm3 at 20ºC. Properties:
— Graphite is a good conductor of heat and electricity. — Fullerene is soluble in benzene and forms deep violet
— Graphite has a very high melting point. colour solution.
— Graphite is insoluble in all common solvent. — Crystalline fullerene has semiconductor properties.
— Compounds of fullerene with alkali metals are called
fullerides and they are superconductors.
Uses:
— For making electrodes in dry cells and electric arc Uses:
furnaces. — As a superconductor.
— Graphite is a good dry lubricant for those parts of — As a semiconductor.
machines where grease and oil cannot be used. — As a lubricants and catalyst.
— For making crucibles for melting metals. — As fibres to reinforce plastics.
— For manufacturing lead pencils.
— Graphite is used as neutron moderator in nuclear VERSATILE NATURE OF CARBON:
reactors. About three million (or thirty lakh) compounds of
carbon are known. The existence of such a large number
— For the manufacture of gramophone records and in
of organic compounds is due to the following
electrotyping. characteristic features of carbon.
— For the manufacture of artificial diamond.
I. Catenation : Tendency to form Carbon-Carbon bond:
FULLERENE: “The property of forming bonds with atoms of the
— Fullerene was discovered in 1985 by Robert F. Curl Jr, same element is called catenation”.
Harold Kroto and Richard E.Smalley.
— This molecule containing sixty atoms of carbon has Carbon has the maximum tendency for catenation in
been named Buckminster fullerene. Fullerene has been the periodic table. This is because of strong
carbon-carbon bonds as compared to other atoms.
named after American architect and engineer R.
Buckminster-fuller whose geodesic domes follow
— When two or more carbon atoms combine with one
similar building principles. another, they form different types of chain such as
Type of fullerene: (i) Straight chains
(C)60
,C70,C74 and C78 are the members of the fullerene (ii) Branched chains
family. But C60 is the most stable and most studied (iii) Closed chain or ring chains
from of fullerenes.
| | |
Structure of fullerene:
-C -C -C -
| | |
Straight chain

|
-C -
| | | | |
-C -C -C -C -C -
| | | | |
-C -
Structure of fullerene |
Branched chain
— Buckminster fullerene molecule (C60) is nearly spherical.

1-120
 Carbon and Its Compounds

C C C2 H6O(Same Molecular Formulae)

Two Isomers
C

(Minimum 3C required in closed chain structure) C2H 5OH CH3-O-CH3

Ethyl alcohol Dimethyl ether

C
C C C5H12 (Pentane)
Three Isomers
C C
C
CH3-CH2-CH 2-CH 2-CH 3 CH3-CH-CH2-CH3 CH 3
Closed chain n-pentane
CH3 H3C-C-CH 3
Iso-pentane
II. Tetravalency of Carbon: CH 3
— The atomic number of carbon is 6. Neo- pentane
— The electronic configuration of carbon atom is
1s2,2s2,2p2. SOLVED EXAMPLE
— It has four electrons in the outermost shell, therefore
its valency is four. Thus carbon forms four covalent Example-1
bonds in its compounds. Allotropes differ in the :
(A) Number of neutrons
H
(B) Number of protons
C + 4H H C H
H (C) Way their atoms are arranged
Forms 4 Forms 1 (D) Number of electrons
covalent covalent Methane Ans. (C)
bond bond Molecule Sol. Way their atoms are averaged.

III Tendency to form multiple bonds: Example-2

Due to small size, carbon can easily form double or The number of free electrons in the outer most shell of
triple bonds (called multiple bonds) with itself and carbon atoms in diamond is :
with the atoms of other elements as nitrogen, oxygen, (A) Four (B) Zero (C) Two (D) Three
sulphur etc. Ans. (B)
H Sol. Free e–1s of C in the structure of diamond = 0;
| H H
H -C -H C =C Example-3
|
H H Steam reacts with red hot charcoal to form :
H
(A) H2 (B) H2 + O2 (C) H2O2 (D) CO + H2
Single bond Double bond Ans. (D)

C + H 2 O ¾¾® CO + H 2
C =O Sol. (Charcoal) (steam)
1424 3
Water gas

Example-4
H - C º C - H ,-C º N
Number of carbon atoms in one molecule of
Triple Bond
Buckminster fullerene is :
(A) 60 (B)1 7 (C) 76 (D) 78
IV. Isomerism:
Ans. (A)
Compounds having same molecular formula but
different structural formulae are known as Isomers and Sol. Buckminster fullerene contains 60 carbon in one
the phenomenon of existence of isomers is termed as molecular.
isomerism.

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Carbon and Its Compounds

Example-5 Example-12
The solid which is commonly called ‘dry ice’ is chemi- Graphite is used as lubricant because it is / has :
cally : (A) Greyish black (B) Insoluble in water
(C) High melting point (D) Soft and slippery
(A) CO (B) CO2 (C) CaCO3 (D) CaCl2
Ans. (D)
Ans. (B) Sol. Due to soft and slippery
Sol. Dry ice ¾¾® CO2
Example-13
Example-6 Coke is obtain from coal by :
In diamond all carbon atoms arranged as : (A) Fractional distillation (B) Destructive distillation
(A) Octahedral (B) Tetrahedral (C) Crystallization (D) Can’t be obtained
(C) Square planar (D) Linear Ans. (B)
Ans. (B) Sol. Destructive distillation
Sol. Tetrahedral structure is found in diamond.
Example-14
Example-7 Property of carbon due to which it forms large number
Two adjacent layers in graphite are bonded by com- of compounds is :
paratively ….................… forces : (A) Allotropy (B) Catenation
(A) Strong (B) Weak (C) Isomerism (D) None of these
(C) Both A and B (D) None of these Ans. (B)
Ans. (B) Sol. Properties to form long chain known as Catenation.
Sol. Adjacent layers in graphite are bonded by weak forces.
Example-15
Example-8 A chip diamond is used for :
In diamond the bonding between carbon atoms is : (A) Marble cutting (B) Metal cutting
(A) co-ordinate (B) ionic (C) Glass cutting (D) All are correct
(C) covalent (D) none of these Ans. (D)
Ans. (C) Sol. Chip diamond is used for marble cutting, metal cutting,
Sol. Convalent bond present in carbon. Glass cutting etc.

Example-9 Example-16
Which of the following is a crystalline form of carbon ? The covalency of carbon is :
(A) Charcoal (B) Graphite
(A) One (B) Two (C) Three (D) Four
(C) Coke (D) Coal
Ans. (D)
Ans. (B)
Sol. Graphite is a crystalline form of carbon. Sol. Covalency of carbon is four.

Example-10 Example-17
The number of free electrons in the outer most shell of Which of the following is not the combined form of
carbon atoms in diamond is : carbon :
(A) Calamine (B) Marble
(A) Four (B) Zero
(C) Diamond (D) Dolomite
(C) Two (D) Three
Ans. (C)
Ans. (B) Sol. Diamond is an element of carbon
Sol. Free electron in diamond structure = 0.
Example-18
Example-11 Carbon occurs in :
Graphite is used as lubricant because it is / has : (A) Free state (B) Combined state
(A) Greyish black (B) Insoluble in water (C) Both A and B (D) None of these
(C) High melting point (D) Soft and slippery Ans. (C)
Ans. (D) Sol. Carbon occurs in both free state and combined state.
Sol. Due to soft and slippery Graphite is used as lubricant.

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 Carbon and Its Compounds

HYDROCARBON:
Compounds formed from combination of carbon and hydrogen is known as hydrocarbon. Hydrocarbon the basis of
chain is mainly classified into two parts.
Hydrocarbon

Acyclic or Cyclic
open chainhydrocarbon Hydrocarbon
- C - C - C- C -
C
Saturated Unsaturated C C
Alkane
Single Bond C C
Alkene Alkyne C
(double bond) (triple bond)
Heterocyclic Homocyclic

Aromatic Alicyclic
(Benzene) (Cycloalkane)

SATURATEDAND UNSATURATED CARBON COMPOUNDS Step I: In this step we shall link two carbon atoms with
Hydrocarbons are of two types: a single covalent bond, leaving the rest of valencies of
SATURATED HYDROCARBONS carbon atoms free.
Alkanes: A hydrocarbon in which the carbon atoms
are connected by only single bonds is called a saturated
C C
hydrocarbon.
These are known as saturated compounds because of
Carbon atoms linked together
little chemical reactivity.
with a single covalent bond
Methane (CH4)] Ethane (C2H6), Propane (C3H8) are all
saturated hydrocarbons.
The general formula of saturated hydrocarbons is Step II : In this step we shall combine hydrogen atoms,
CnH2n+2 where n is no. of carbon stomas in one molecule so as to satisfy the unsatisfied valencies of both carbon
of the alkane.
H H
If n = 1
Then, C1H2 × 1 + 2 = CH4 (Methane) atoms as shown in Fig. H C C H
If n = 2
Then, C2 H2 × 2 + 2 = C2 H6 (Ethane) H H

Each carbon atom is bounded with 3 hydrogen atoms

H H H
with single covalent bonds
H C H H C C H
H H
H H H . .
(Methane) (Ethane) x x

x x .H
H .x C x C
H H H H H H H
x x
H . .
H C C C H H C C C C H H

H H H H H H H
(Propane) (Butane) Electron dot structure of ethane

Formation of Molecule of Ethane [C2H6] UNSATURATED HYDROCARBONS

Another compound of carbon and hydrogen is ethane. A hydrocarbon in which the two carbon atoms are
Its molecular formula is C2H6. Let us try to arrive at its connected by a ‘double bond’ or a ‘triple bond’ is called
structural dot diagram by taking following steps: an unsaturated hydrocarbons.

1-123
Carbon and Its Compounds

These compounds are more chemically reactive than General formula of alkynes: CnH2n-2
saturated compounds of carbon. Ethane, Propane, General formula of alkynes: CnH2n-2
Butane similarly Ethyne, propyne are all unsaturated No. of Name Formula Structure
hydrocarbons. ‘C’ atom
2 Ethyne or C2H2 or H–CºC–H
Unsaturated hydrocarbons are of two types: Acetylene HCºCH
3 Propyne or C3H4 or H
1. Those containing carbon carbon double bond Methl H – C – C –C– H
H3C–CºC–H
(Alkenes) acetylene H
2. Those containing carbon carbon triple bond
4 Butylene or C4H6 or H H
(Alkynes).
Dimethyle –C– C–H
H– C – C –
H3C–CºC–CH3
acetylene H H
The general formula of an alkene is CnH2n where n is
number of carbon atoms in its one molecule. Alkyl Groups: The group formed by removal of one
If n = 2 hydrogen atom from an alkane molecule is called an
Then, C2H2 × 2 = C2 H4 (Ethene) alkyl group, examples are methyl group ( CH3 ) , Ethyl
If n = 3
(C2H5) etc.
Then, C3H2×3 = C3H6 (Propane)
Alkane – H ® Alkyl group º CnH2n + 2 – 1
The general formula of alkynes is CnH2n–2 where n is the
number of carbon atoms in one molecule of the alkyne. e.g. CH 4 – H ® CH3 º Cn H 2n + 1
( Methane ) ( Methyl)
If n = 2
Then, C2 H2 × 2 – 2 = C2H2 (Ethyne)
STRAIGHT CHAINS, BRANCHED CHAINS AND RINGS IN
If n = 3
HYDROCARBONS
Then, C3 H2×3–2 = C3H4 (Propyne)
The skeleton of carbon atoms to which hydrogen atoms
are attached may be in the form of a straight chain, a
Saturated compounds are quite stable in nature while
branched chain or a ring.
the unsaturated hydrocarbons are unstable. They have
The chain of carbon atoms may have only single
always a tendency to change to saturated compounds
covalent bonds between them or may have a double
by combining with one or more molecules of other
covalent or triple covalent bond between the skeletons
reactants and become stable. This is why these are
of carbon atoms.
known unsaturated compounds.
For example, a molecule of ethene takes up a molecule
Straight chain saturated hydrocarbons: In this type of
of hydrogen as H2 and changes to ethane. hydrocarbons all the carbon atoms are arranged in the
form of a straight chain such that they are bound to
General formula of alkenes : CnH2n one another by single covalent bonds.
The straight chain of carbon atoms may consist of 2 to
No. of C Name Formula Structure 40 carbon atoms.
atoms E.g.
2. Ethene or C2H4 H H
C–C H H H
H H
Ethylene CH2=CH2
3. H H H H C H H C C H
H – C – C – C–H
Propene C3H6 H
H H H
or CH3-CH=CH2 (Methane) (Ethane)
Propylene H H H H
H – C – C = C– C– H
4. C4H8 H H H H H H H
Butene CH3-CH=CH-CH3 H H H H
or or H C C C H H C C C C H
H H H H
Butylene or
H – C = C – C– C– H
H H H H H H H
CH2=CH-CH2-CH3 H H
(Propane) (Butane)

Branched chain saturated hydrocarbons: In this type

of hydrocarbons, one or more carbon atoms are
(II) Alkyne ( -C º C -) attached to the main straight chain of carbon atoms,
— The hydrocarbon in which two carbon atoms are such that all carbon atoms are bound to one another
bonded by a triple bond are called alkyne. by a single covalent bond.
— Their general formula is CnH2n-2 where ‘n’ is the number E.g. The branched chain structure of Butane (C4H10) is
of carbon atoms. written as following.

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 Carbon and Its Compounds

Carbon atoms in the straight chain and 4th carbon in Branching can also be shown in the unsaturated
the side chain: hydrocarbons containing a minimum of four carbon
The straight chain of Butane is atoms.
H H H H

H C C C C H H H
H H H H
H H H H H H C
3
C2 C1 H
H C4 C3 C2 C1
Fig. (a) H
H H HH C
CH3
But-1-ene
Whereas the branched chain of Butane is H
2-Methylprop-1-ene or CH3 C = CH2
or CH3CH2CH = CH2
H H H
1 2 3 Open chain compounds a also known as aliphatic
H C C C H CH3 – CH – CH3
compounds because some of the originally known
H H H CH3 compounds were obtained from animal fats (In Greek ;
H C
4
H alei : animal and phato : fats)
H
Ring chain saturated hydrocarbons
Fig. (b) In this kind of hydrocarbons, the carbon atoms are
The formula with straight chain of carbon atoms is arranged in the form of a ring and are bounded by single
prefixed with “normal” i.e. Butane in structural formula covalent bonds.
is Fig. (a) is named as n-butane, where n stands for An example of the kind of hydrocarbons is cyclohexane
normal. of molecular formula C6H14. Fig (a) and Fig. (b) show
the structural and electron dot diagram of the molecule
The formula with branched chain of carbon atoms is of cyclohexane.
prefixed with “iso” i.e. the butane in structural formula
in Fig (b) is named as Iso-butane. H H H H
Branching of carbon atom can take place if a minimum H C C H
of Four carbon atoms are present in the organic C C
compound. H C C H
H H H H
Straight chain unsaturated hydrocarbon: In this type
of hydrocarbons all carbon atoms are arranged in the Structure of cyclohexane
form of a straight chain, such that there is either a double
covalent or triple covalent bond between a pair of H H H H Electron
carbon atoms in the chain. H
H C C
C C
C C H
1. Ethane (C2H4) is written as H
H H H H H
H

H C = C H or H2C = CH2
Electron dot diagram of cyclohexane
Aromatic hydrocarbon:
2. Propane (C3H6) is written as — These have at least one benzene ring in their molecules.
H H H — It is a special type of ring of six carbon atoms with
three double bonds in alternate positions.
H C = C C H or CH2 = CH – CH3

H H H
1 1
3. Ethyne (C2H2) is written as H C
2
H H C
2
H
6 6
H – C º C – H or HC º CH C C C C

4. Propyne (C3H4) is written as

C C3 C C3
H 5 or 5
H C H H C H
C H 4 4
HC C
H H
H

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Carbon and Its Compounds

FUNCTIONAL GROUP :
—. Carbon forms many compounds with hydrogen. But Example-20
carbon also forms bonds with other atoms such as Which of the following is not an electrolyte:
halogen, oxygen, nitrogen and sulphur. Therefore, (A) sugar (B) sodium chloride
carbon is said to be very friendly element. (C) copper sulphate (D) sulphuric acid
These compounds are obtained by replacing one or Ans. (A)
more hydrogen atoms by other atoms such that the Sol. Sugar is not an electrolyte because it does not ionize.
valency of carbon remains satisfied. The atom replacing
the hydrogen atom is called heteroatom or Functional Example-21
group. Arrange the following compounds in the increasing
— Different organic compounds having same functional
order of Carbon Carbon bond length : Ethene, Ethyne,
group have almost same properties these are called
Ethane.
families.
(A) Ethane, Ethyne, Ethene
Example: (B) Ethyne, Ethene, Ethane
H H H
| (C) Ethene, Ethyne, Ethane
| |
H - C - OH H - C - C - OH (D) None of these
| Functional
| | Ans. (B)
Functional
H group H H group
Methanol Ethanol 1
Sol. Bond length µ
no. of bond
— Properties of CH3–OH and CH3– CH–CH2OH are similar
and it is due to the presence of –OH (hydroxyl) group.
— This group is known as alcoholic group. Example-22
— Family of compounds having –OH group is called The solid which is commonly called ‘dry ice’ is chemi-
alcohols. cally :
(A) CO (B) CO2 (C) CaCO3 (D) CaCl2
Some Functional Groups in carbon Compounds Ans. (B)
Hetero atom Functional Formula of Functional Sol. Dry ice ¾¾® CO2
Group Group
Halogen atom Halo -X Example-23
(F, Cl, Br, I) (Fluoro,Chloro, (-F,-Cl,-Br,-I) Hydrocarbons are compounds of :
Bromo, Iodo) (A) C and O (B) C and Cl
Oxygen -OH (C) C and H (D) C and S
1. Alcohol H Ans. (C)
- C or - CHO Sol. Hydrocarbons are compounds of carbon and
2. Aldehydes hydrgoen.
O
| Example-24
C = O or - CO
3. Ketones Name of hydrocarbon C2H4 is :
O (A) Ethene (B) Acetylene
|| (C) Both (A) and (B) (D) Ethane
- C - OH or - COOH
4.Carboxylic acid Ans. (A)
-NO2
Nitrogen Sol. C2H4 ¾¾® ethene
1. Nitro
-NH2
2. Amines Example-25
General formula for Alkynes is :
(A) CnH2n (B) CnH2n+2 (C) CnH2n–2 (D) CnH2n+4
SOLVED EXAMPLE Ans. (C)
Sol. Alkynes ¾¾® CnH2n –2
Example-19
Methane is insoluble in which of the following
Example-26
solvents?
(A) Ether (B) Alcohol Chemical formula of fourth member of Alkenes is :
(C) Carbon tetrachloride (D) Water (A) C5H12 (B) C4H10 (C) C4H8 (D) C5H10
Ans. (D) Ans. (D)
Sol. Methane is non-polar compound that’s why not soluble
Sol. Alkenes ¾¾® CnH2n , n = 5 C5H10
in water (polar solvent).

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 Carbon and Its Compounds

Example-27 Example-33
The percentage by weight of carbon in ethyne is : Which of the following represents the correct order of
unsaturation :
(A) 33.3% (B) 85.71%
(A) Alkanes, alkenes, alkynes
(C) 50% (D) 27.35% (B) Alkanes, alkynes, alkenes
Ans. (B) (C) Alkenes, alkynes, alkanes
(D) Alkynes, alkanes, alkenes
ethyne(C2H2 ) 24 Ans. (A)
Sol. = ´ 100% = 92.3%
%of Carbon 26 alkanes,alkenes,alkynes
Sol. ¾¾¾¾¾¾¾¾¾¾¾¾
®
unsaturation increase
Example-28
Example-34
The formula C6H13 represents an :
Simplest hydrocarbon is :
(A) Alkane (B) Alkene (A) CH4 (B) C2H6 (C) C2H2 (D) C2H4
(C) Alkyne (D) Alkyl group Ans. (A)
Ans. (D) Sol. Simplest hydrocarbon is CH4
Sol. C6H13 ¾¾® alkyl group
Example-35
Example-29 Which of the following can have triple bond :
(A) C3H8 (B) C3H4 (C) C5H8 (D) C4H8
Formalin an aqueous solution of :
Ans. (C)
(A) Furfural (B) Fluroscein Sol. Alkyne have triple bond (C5H8)
(C) Formaldehyde (D) Formic acid
Ans. (C) Example-36
Sol. Formalin is the solution of formaldehyde Which of the following molecule is having triple bond:
(A) Ethyne (B) Nitrogen
Example-30 (C) Both A and B (D) None of these
Ans. (C)
Number of lone pair present in each molecules of H2O,
Sol. Ethyne and nitrogen both have triple bond.
NH3 and C2H4 respectively :
(A) 1, 2, 1 (B) 2, 1, 0 Example-37
(C) 1, 1, 1 (D) 2, 1, 2 Covalent bonds are of :
(A) Three types (B) Two types
Ans. (B)
(C) Four types (D) Five types
Sol. Compound Lone pair Ans. (A)
H2O 2 Sol. Covalent bond — Single, double, triple bond
NH3 1
Example-38
C2H4 0
Oxygen molecule involves formation of :
(A) Single covalent bond (B) Double covalent bond
Example-31
(C) Triple covalent bond (D) All are correct
Which one is not an organic compound : Ans. (B)
(A) Ammonium cyanate (B) Marsh gas Sol. O = O (oxygen have double bond)
(C) Urea (D) Cane sugar
Ans. (A) HOMOLOGOUS SERIES:
Sol. NH4CNO (ammonium cynate inorganic compound). “A series of organic compounds having similar
structures and similar chemical properties in which the
Example-32 successive members differ in their molecular formula
by -CH2 group”.
Which of the following is/are close chain compound :
The different members of the series are called
(A) Benzene (B) pyrrole homologous.
(C) Both (A) and (B) (D) Butane
Ans. (A) Characteristics of Homologous Series:
Sol. Benzene and pyrrole both are close chain compound. — All the member of a homologous series can be described
by a common general formula.

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Carbon and Its Compounds

Example: All alkane can be described by the general Homolgous Series Containing functional Groups.
formula CnH2n+2. — Aldehydes:
— Each member of a homologous series differs from its HCHO, CH3CHO, CH3CH2CHO, CH3CH2CH2CHO
higher and lowers neighbouring members by a common —. Carboxylic acids: HCOOH, CH3COOH, CH3CH2COOH,
difference of –CH2 group. CH3CH2CH2COOH
— Molecular masses of the two adjacent homologues — Amines: CH3NH2, CH3CH2NH2,CH3CH2CH2NH2.
differ by 14 mass units, because molecular mass of – —
CH2 group is 12 + 2 = 14. Ketones:CH3COCH3,CH3COCH2CH3,CH3COCH2CH2CH3
—
— All the members of a homologous series show similar
Haloalkanes:CH3X,CH3CH2X,CH3CH2CH2X,CH3CH2CH2-
chemical properties.
CH2X
— All the members of the series can be prepared by
similar methods known as the general method of
How do physical properties change in a homologous
preparation.
series of hydrocarbons.
Table: Some members of alkane, alkene and alkyne The physical properties of the various members of a
homologous series. homologous series change regularly with an increases
in the molecular mass.
Alkane Alkene Alkyne
(i) Melting and boiling points: Melting point and boiling
CnH2n+2 CnH2n CnH2n+2 of hydrocarbon in a homologous series increases with
an increase in molecular mass.
Homologous series Homologous series Homologous series
Name Formula Name Formula Name Formula
(ii) Physical State:
Methane CH4 - - - - — Hydrocarbons containing lesser number of carbon
atoms are gases.
Ethane C2H6 Ethene C2H4 Ethyne C2H2 — Hydrocarbons containing large number of carbon are
Propane C3H8 Propene C3H6 Propyne C3H4 solids.
— Hydrocarbon containing intermediate number of
Butane C4H10 Butene C4H8 Butyne C4H6 carbon atoms are liquid.
Pentane C5H12 Pentene C5H10 Pentyne C5H8 Example: Hydrocarbon containing 1-4 carbon atoms
are gases, these containing 5-13 carbon atoms are
Hexane C6H14 Hexene C6H12 Hexyne C6H10 liquid and those containing more than 14 carbon atoms
Activity: Calculate the difference in the formulae and are solids,
molecular masses for (a) CH3OH and C2H5OH (b) Nomenclature of carbon compounds:
C2H5OH and C3H7OH and (c) C3H7OH and C4H9OH — Carbon compounds can be called by their common
names, but, then remembering millions of compounds
by their individual names may be very difficult. Due to
Formula Molecular Mass Difference In this reason, the International Union of Pure and
(Calculated) Applied chemistry (IUPAC) has devised a very
Formula systematic method of naming these compounds.
Molecular mass Naming a carbon compound can be done by the
(a) CH3OH 12 + 3 + 16 + 1 = 32 -CH2 14 following methods.
C2H5OH 24 + 5 + 16 + 1 = 46
— The number of carbon atoms in the molecule of a
(b) C2H5OH 24 + 5 + 16 + 1 = 46 -CH2 14 hydrocarbon is indicated by the following stems.
No. of carbon
C3H7OH 36 + 7 + 16 + 1 = 60 Stem
atom:
(c) C3H7OH 36 + 7 + 16 + 1 = 60 -CH2 14 1 Meth
2 Eth
C4H9OH 49 + 9 + 16 + 1 = 74 3 Prop
4 But
Conclusion: 5 Pent
(i) Yes, all these compounds are the members of a 6 Hex
homologous series for alcohols. 7 Hept
(ii) CH3OH, C2H5OH, C3H7OH and C4H9OH –increasing 8 Oct
carbon atoms. These four compounds from a 9 Non
homologous series. 10 Dec.

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 Carbon and Its Compounds

Example; Saturated hydrocarbon. Functional Group:

Alkane ® Meth +ane = Methane “Functional group may be defined as an atom or a
group of atoms which is responsible for most of the
Unsaturated hydrocarbon
characteristic chemical properties of an organic
Alkene ® Eth + ene = Ethene compound”.
Alkyne ® Eth + yne = Ethyne
HALO Group
X (X can be Cl, Br or I) : The halo group can be Chloro
— In case of functional group is present, it is indicated in
-Cl, Bromo -Br or Iodo, -I depending upon whether a
the name of compound with either a prefix or a suffix. chlorine, bromine or iodine atom is linked to a carbon
chain of the organic compound.
— Identify the longest continuous chain of carbon atoms. The general formula of halogen compounds is R-X
This gives the name of parent hydrocarbon. where R stands for alkyl radical having general formula
CnH2n+1 and X stands for a halogen atom.
— In the case of any substituent appropriate prefix is A halogen compound is named in IUPAC system by
added before the name of parent hydrocarbon. placing a prefix; Chloro, bromo, fluro or iodo etc. in the
name of alkane.
Alkane + Chlorine ® chloro alkane
— In the case of a functional group. the ending ‘e’ in the The names and formulae of first five members of
name of the parent hydrocarbon is replaced by the halogen compounds of saturated hydrocarbons are
appropriate suffix. given in table.

Halogen Compounds of saturated hydrocarbons

Halogen IUPAC Common Molecular Condensed
Alkane Structural formula
group name name formula formula
H
Bromo- |
Methane Bromine Methylbromide CH3Br CH3–Br H - C - Br
methane |
H
H H
Chloro- | |
Ethane Chlorine Ethylchloride C2H5Cl CH3–CH2–Cl H - C - C - Cl
ethane | |
H H
H H H
Fluro- CH3–CH2– | | |
Propane Fluorine Propyl fluoride C3H7F H-C-C-C-F
propane CH2F | | |
H H H
H H H H
Iodo- CH3–CH2– | | | |
Butane Iodine Butyl iodide C4H9I H -C-C-C-C-I
butane CH2–CH2I | | | |
H H H H
H H H H H
CH3–CH2– | | | | |
Chloro- H - C - C - C - C - C - Cl
Pentane Chlorine Pentyl chloride C5H11Cl CH2–CH2–
pentane | | | | |
CH2Cl
H H H H H

ALCOHOL Group –OH In IUPAC system, the alcoholic homologous series,

When a hydrogen atom of an alkane is replaced by the name of a given member of alcohol can be derived
functional group (-OH), then the organic compound by replacing (-e) of the given alkane by (-ol).
so formed belongs to alcoholic group or hydroxyl e.g. Methane –e + ol ® Methanol
group.
The general formula of homologous series of alcoholic
group is R-OH, where R stands for alkyl radical, having
a general formula CnH2n+1.

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Carbon and Its Compounds

First Four Members of Alkanes and Alkanols (Alcohols)

Alkanes Alkanols (Alcohols)
IUPAC Formula IUPAC Formula
Name Name
Methane CH4 [CH3–H] Methanol CH3OH [CH3–OH]
Ethane C2H6 [CH3–CH2–H] Ethanol C2H5OH [CH3–CH2–OH]
Propane C3H8 [CH3–CH2–CH2–H] Propanol C3H7OH [CH3–CH2–CH2–OH]
Butane C4H10[CH3–CH2–CH2–CH2–H] Butanol C4H9OH [CH3–CH2–CH2–CH2–OH]

ALDEHYDE Group

When the hydrogen atom of an alkane is replaced by functional group (-CHO), then the organic compound so formed
belongs to aldehyde group.
The general formula of aldehyde homologous series is R-CHO; where R stands for alkyl radical having general formula
CnH2n+1.
In IUPAC system the aldehyde homologous series are called alkanal homologous series. The name of the given member
of an alkanal homologous series can be derived by replaced – e of a given alkane by –al. Eg. Alkane by –al e.g. Alkane +
al ® Alkanal.

First Four Members of Alkanes and Alkanals (Aldehydes)

Alkanes Alkanals (Aldehydes)
IUPAC IUPAC
Formula Common Name Formula
Name Name
O
Methane CH4 [H3C–H] Methanal Formaldehyde [H–CHO] or ||
H-C-H
O
Ethane C2H6 [CH3–CH3] Ethanal Accetaldehyde [CH3–CHO] or ||
CH 3 - C - H
[CH3–CH2–CHO] or
Propane C3H8 [CH3–CH2–CH3] Propanal – O
||
CH 3 - CH 2 - C - H
[CH3–CH2–CH2–CHO] or
Butane C4H10 [CH3–CH2–CH2–CH3] Butanal – O
||
CH 3 - CH 2 - CH2 - C - H

KETONE
Group >C = O Ketones are the >C = O group in which the carbon atom of >C=O is attached to two carbon atoms in the
carbon chain of alkyl rdical.
Then general formula of ketone homologous series is where R and R1 stand for same or different alkyl radicals,

having general formula CnH2n+1.

In IUPAC system the ketone homologous series is called alkanone homologous series. The name of a given number of
alkanone series can be derived by replacing –e of the given alkane by –one.
Examples.
Alkane – e + one ® Alkanone
Propane – e + one ® Propanone
Butane – e + one ® Butanone
The names and formulae of first six members of alkanes and corresponding alkanones are given in table.

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 Carbon and Its Compounds

First Six Members of Alkanes and Alkanones

IUPAC Name
Alkane Common Name Formula
of alkanone
NIL (Not possible as carbon
Methane Methanone
atoms are less than three.)
NIL (Not possible as carbon
Ethane Ethanone
atoms are less than three.)
O
Propane Propanone Acetone or dimethyl ketone ||
CH3 - C - CH3
O
2-Butanone/ ||
Butane Methyl ethyl ketone
Butanone 1
CH 3 – C– 3 CH 2 – 4 CH 3
2

O
2-Pentanone/ ||
Pentane Methyl propyl ketone
Pentanone 1
CH 3 – C– 3 CH 2 – 4 CH 2 – 5CH 3
2

O
2-Hexanone/ ||
Hexane Methyl butyl ketone
Hexanone 1
CH 3 – C– 3CH 2 – 4 CH 2 – 5CH 2 – 6 CH 3
2

CARBOXYLICACID Group
Carboxylic acids are the compounds containing -COOH ALKYNE Group – C º C–
group attached to the carbon atom of alkyl chain. When The alkyne group is a carbon-carbon triple bond. The
the hydrogen atom of an alkane is replaced by alkyne group is present in ethyne HC º CH . The
compounds containing alkyne group are known as
functional group , then the organic alkynes.

compound belongs to alkanoic acid group or carboxylic. SOLVED EXAMPLE

Example-39
First Four Members of Alkanes and Alkanoic Acid Nitrogen molecule involves formation of :
IUPAC
(A) Single covalent bond
Name of Common
Alkane Formula (B) Double covalent bond
Alkanoic Name
acid (C) Triple covalent bond
[HCOOH]
Methane
Methanonic Formic O (D) All are correct
acid acid or ||
H - C - OH Ans. (C)
[CH3COOH] Sol. Triple covalent bond present in N ºN.
Ethanonic O
Ethane Acetic acid or ||
acid
CH3 - C - OH Example-40
[C2H5COOH]
How many covalent bonds are there in methane :
Propanonic Propionic or
Propane O (A) Four (B) Two (C) Three (D) One
acid acid || Ans. (A)
CH3 - CH2 - C - OH
[C3H7COOH] H
or
Butanonic Butyric O Sol. H—C—H
Butane
acid acid ||
CH 3 - CH 2 - CH 2 - C - OH H

Example-41
Which of the following does not contain double
ALKENE Group C = C bond(s) :
The alkene group is a carbon-carbon double bond. The (A) H2O (B) CO2
alkene group is present in ethane (CH2 = CH2)., and (C) HCl (D) A and C both
propane (CH3 – CH = CH2). The compounds containing Ans. (D)
alkene group are known as alkenes. Sol. H — O — H, O = C = O, H — Cl

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Carbon and Its Compounds

Example-42 Example-47
Number of electrons shared between carbon-carbon The general formula of carboxylic acids is :
atoms in ethene is : (A) CnH2n+1 OH (B) CnH2nO
(A) 2 (B) 4 (C) 5 (D) 6 (C) CnH2nO2 (D) CnH2n+2
Ans. (B) Ans. (C)
H H Sol. The general formula of Carboxylic acid / ester – CnH2nO2
Sol. C C
H H Example-48
[4 electrons are shared between carbon–carbon atom] IUPAC name of the compounds
CH3 – CH – CH2 – CH(OH) – CH3 is :
Example-43
Which of the following molecule is not having double CH2
covalent bond : CH3
(A) CO2 (B) O2
(A) 4 –Ethyl – 2 – pentanol
(C) C2H4 (D) None of the above
(B) 4–Methyl – 2 – hexanol
Ans. (D)
(C) 2–Ethyl – 2 – pentanol
Sol. all compound have double bond in their composition
(D) 3 – Methyl – 2 – hexanol
H H Ans. (B)
O = C = O, O = O, C C .
H H Sol. CH3 – CH – CH2 – CH(OH) – CH3
CH2
Example-44
Double bond is ________ than a single bond CH3
(A) Weaker (B) Stronger 4 – Methyl – 2 – hexanol
(C) Longer (D) None of these
Ans. (B) Example-49
Sol. Bond strength : Triple bond > double bond > single The IUPAC name of CH3CH = CHCH(CH3)2 is :
bond. (A) 4–Methylpent–2-ene
(B) 4, 4–Dimethylbut–2–ene
Example-45 (C) Methyl isopropyne
IUPAC name of the compound is (D) 2–Methylpent–4–yne
CH3 – CH = C – CH3 Ans. (B)
CH3
CH2 –CH3 Sol. CH3 – CH = CH – CH 4–Methyl pent–2–ene
CH3
(A) 2 – Ethyl – 2 – butene
(B) 3 – Ethyl – 2 – butene
(C) 3 – Methyl – 3 – pentene Example-50
(D) 3 – Methyl – 2 – pentene Name of some compounds are given, which one is not
Ans. (D) true according to IUPAC system :
Sol. CH3 — CH = C — CH3 3 – Methyl – 2 – pentene (A) CH3 C º C – CH – (CH3 )2
4 – methyl – 2 – pentyne
CH2 — CH3
(B) CH3–CH2–C — CH–CH3
Example-46 CH2 CH3
The IUPAC name of the compound given below is : 2–Ethyl–3–methyl but–1–ene
CH3 – CH2 – C – CH2 – CH3 (C) CH3–CH — CH — CH3
O OH CH3
(A) 1 – Pentanone 3–methyl–2–butanol
(B) 2 – Pentanone
CH3
(C) 2 - Carboxy butane
(D) 3 – Pentanone (D) CH3–CH2–CH2–CH–CH–CH3
Ans. (D) CH2—CH3
Sol. CH3 – CH2 – C – CH2 – CH3 3–pentanone 5–Methyl–4–ethyl heptane
Ans. (D)
O

1-132
 Carbon and Its Compounds

CH3
Example-55
Sol. CH3 – CH2 – CH2 – CH – CH – CH3 The molecular formula of an ester is C3H7COOC2H5.
CH2 – CH3 The alcohol and the acid from which it might be prepared
4–ethyl –2–methyl hexane are:
(A) Propanol and Propanoic acid
Example-51 (B) Ethanol and Butanoic acid
(C) Ethyl alcohol and Propanoic acid
Isobutyl alcohol is :
(D) None of these
(A) Primary alcohol (B) Secondary alcohol
Ans. (B)
(C) Tertiary alcohol (D) Quaternary alcohol
+H2O
Ans. (A) Sol. C3H7 – C – OC2H5 C3H7 – C – OH + C2H2OH
Sol. CH3 — CH — CH2 —OH Isobutyl alcohol O O

CH3
Example-56
The IUPAC name of the compound CH2 = C(CH3)2 is :
Example-52 (A) 1, 1 Dimethylprop–2–ene
The IUPAC name of CH3 — CH2 — CH — CH3 is : (B) 2–Methyl prop–1–ene
(C) 2–Vinylpropane
CH3 (D) 1–Isopropylethane
Ans. (B)
(A) 3-methyl pentane (B) 3-methyl butane
CH3
Sol. CH 2 = C
(C) 2-methyl pentane (D*) 2-methyl butane CH3 2– methyle prop–1–ene
Ans. (D)
Example-57
Sol. CH3 — CH2 — CH — CH3 2-methyle butane. The compound having formula C9H20O is a/an :
CH3 (A) Ester (B) Aldehyde (C) Ketone (D) Alcohol
Ans. (D)
Sol. General formula of alcohol is CnH2n+2O (alkanols)
Example-53
CH3 Example-58
The IUPAC name of CH3 — CH2 — CH — C — CH3 is : Alcohols are isomeric with :
(A) Ketones (B) Ethers
CH3 CH3
(C) Esters (D) Aldehydes
(A) 2,2,3-trimethyl pentane Ans. (B)
(B) 3,4,4-trimethyl pentane Sol. Alcohol and ether have same general formula
(C) 2-ethyl-3,3-dimethyl butane
(D) 2,3-dimethyl hexane Example-59
Ans. (A) The functional group which always occur in the middle
CH3 of a carbon chain is :
Sol. CH3 — CH2 — CH — C — CH3 (A) Alcohol (B) Aldehyde
(C) Ketone (D) Carboxylic acids
CH3 CH3
Ans. (C)
2, 2, 3–trimethyl pentane Sol. Carbon of ketone group never come at the end of
Carbon chain
Example-54
The produce formed when etheyne reacts with excess Example-60
of chlorine is :
(A) 1, 1, 2, 2 – tetrachloroethane The molecular formula C 2H4 O 2 may contain the func-
(B) 1, 2 – dichloroethane tional group :
(C) 1, 1, 2 – trichloroethane (A) Carboxylic acid (B) Ester
(D) None of these
Ans. (A) (C) Both (A) and (B) (D) Neither A nor B
Cl
Ans. (C)
Cl
CH — CH Sol. The molecular formula C 2H4 O 2 represent acid as well
Sol. 1, 1, 2, 2–tetrachloroethane
Cl Cl as ester

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Carbon and Its Compounds

Example-61 Example-66
The functional group in aldehydes is : Which of the following is isobutyric acid :

—
(A) CH3 – CH2 – CH2 – COOH
(A) —CHO (B) —C —
—O
(C) —COOH (D) —COOR (B*) (CH3)2CH — COOH
Ans. (A) (C) CH3 – CH2 – CH2 – CH2 – COOH
Sol. Aldehyde = —CHO
CH3
Example-62
(D) CH3 C – COOH
The functional group which always occur in the middle CH3
of a carbon chain is :
(A) Alcohol (B) Aldehyde Ans. (B)
(C) Acid (D) Ketone Sol. (CH3)2CH — COOH
Ans. (D)
Sol. Ketonic group always occur in the middle of a carbon Example-67
chain. The correct IUPAC name of the following structure is-

Example-63 H3C – CH2

A person is suffering from Diabetes, which get con- CH – CH2 – CH3
firmed by testing his urine with Benedict’s reagent. The
organic compound and functional group present in the H3C – CH – CH3
(A) 3-lsopropylpentane
urine of suffering person is :
(B) 2-Methyl-3-ethylpentane
O (C*) 3-Ethyl-2-methylpentane
(A) Sucrose – COOH (B) Glucose, – C – (D) 2-Ethyl-4-methylpentane
Ans. (C)

O O H3C – CH2

(C) Glucose – C – H (D) Sucrose, – C – Sol. CH – CH2 – CH3 3– ethyl–2–methyl pentane

Ans. (C) H3C – CH – CH3
O
Sol. Glucose – C – H Example-68
A plastic bakelite is a compound of HCHO with :
Example-64 (A) Benzene (B) Phenol
The IUPAC nomenclature of CH 3 – CH – CH – CH 3 (C) Ammonia (D) Hydrocarbon
Ans. (B)
OH CH3
Sol. Phenol + formaldehyde ¾® Bakelite
is :
(A) 3-methyl butan-2-ol (B) 2-methyl butan-3-ol
Example-69
(C) Both A and B (D) None of these
Ans. (A) n-pentane, isopentane and neopentane show :
(A) Position isomerism (B) Functional isomerism
Sol. CH 3 – CH – CH – CH 3 3– methyl butan–2-ol
(C) Chain isomerism (D) None of these
OH CH3 Ans. (C)

Example-65 Sol. CH3 — CH2 — CH2 — CH2 — CH3

IUPAC name of tertiary butyl alcohol :
(A) 2-Butanol (B) 2-methyl propanol CH3 — CH — CH2 — CH3
(C) 2-methyl - 2- propanol (D) 2,2 - dimethyl propanol CH3 chain isomers
Ans. (C) CH3
CH3 CH3 — C — CH3

Sol. H3C — C — CH3 2–methyl –2–propanol CH3

OH

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 Carbon and Its Compounds

Sol. ketone and aldonyde are functional isomers

Example-70
Tautomerism is a type of : Example-77
(A) Stereoisomerism (B) Geometrical isomerism 1–Butanol and 2–Butanol are :
(C) Structural isomerism (D) Functional isomerism (A) Chain isomers (B) Position isomers
Ans. (C)
Sol. Structural isomerism (C) Functional isomers (D) None of these
Ans. (B)
Example-71 Sol. Position isomers
But-1,3-diene and But-2-yne are :
(A) Position isomers (B) Metamers Example-78
(C) Functional isomers (D) They are not isomers Which of the following is not a sub-division of
Ans. (C) structural isomerism :
H2C = CH — CH = CH2 (A) Tautomerism (B) Stereoisomerism
functional isomer (C) Metamerism (D) Position isomerism
Sol.
Ans. (B)
Sol. Stereoisomerism
Example-72
Ethoxyethane and 1-methoxy propane are : Example-79
(A) Position isomers (B) Metamers (i) CH3—CH—CH2—CH 3
(C) Functional isomers (D) Tautomers
Ans. (B)
CH3
CH3 — CH2 — O — CH2 — CH3
Sol. Metamers
CH3 — O — CH2 — CH2 — CH3 What is the IUPAC name of this compound?
(a) 3-methyl butane (b) 1-methyl butane
Example-73 (c) 1-pentane (d) 2-methyl butane
(ii) The functional group is methanol and methanal
The number of isomeric ketones with formula C5H10O
respectively are
are :
(A) 6 (B) 3 (C) 5 (D) 4 (a) -OH, - CHO (b) - CHO, - OH
Ans. (B)
(c) -OH, - COOH (d) -CHO, - COOH
Sol. HC3 — C — CH2 — CH2 — CH3
O
Sol. (i) (d) (ii) (a)
CH3 — CH2 — C — CH2 — CH3
O CHEMICAL PROPERTIES OF CARBON COMPOUND:
CH3 — C — CH — CH3
All carbon compounds sh ow more common
characteristic properties. As most of the fuels we use
O CH3
are either carbons or its compounds. Some such
properties are described here:
Example-74
Alkanols are represented by : COMBUSTION:
(A) CnH2n+1O (B) CnH2nO Combustion is a chemical process in which heat and
(C) CnH2nO2 (D) CnH2n+2O light (in the form of flame) are given out
Ans. (D) The process of combustion is a rapid oxidation reaction
Sol. General formula of alcohol is CnH2n+2O (alkanols) of any substance in which heat and light are produced.

Example-75 Combustion of some common substance:

Alcohols are isomeric with : — Combustion of Carbon: Carbon (or charcoal) burn in
(A) Ketones (B) Ethers air or oxygen to give CO2 production heat and light.
(C) Esters (D) Aldehydes C (s ) + O2 ( g ) ¾Combusion
¾ ¾¾® CO2 ( g ) + Heat + light
Ans. (B) Carbon Oxygen Carbon dioxide
Sol. Alcohol and either have same general formula
— Combustion of Hydro Carbon: Hydrocarbons burn to
Example-76 produce carbon dioxide (CO2), water (H2O) and heat
Propanone and propanal are : and light.
(A) Chain isomers (B) Position isomers CH 4 ( g ) + O2 ( g ) ¾Combusion
¾ ¾¾® CO 2 ( g ) + H 2O( g ) + Heat + Light
(C) Both A and B (D) Functional isomers Methane
Ans. (D)

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Carbon and Its Compounds

Note 2. Close the air hole and observe the colour of the
flame.
Natural gas and biogas contain methane. So, 3. Put a metal plate over it and observe the nature of
burning of natural gas and biogas are also deposit.
combustion reactions. 4. Open the air regulator to allow flow of air.
Burning of LPG (Butane) produces CO2, H2O heat 5. Observe the colour of flame.
and light. 6. Put a metal plate and observe the nature of deposit.
13
C 4 H 10 ( g ) + O2 ( g ) ¾Combustion
¾ ¾¾® 4CO2 ( g ) + 5 H 2 O( g ) + Heat + Light
2 Observation:
Butane/LPG
Air Closed Open
— Combustion of Cellulose : Regulator
Combustion of cellulose (like wood, cotton cloth and
paper) gives CO2, H2O heat and light. Cellulose is a Colour of Yellow Bluish flame
carbohydrate and can be described by the formula flame sooty flame
(C6H10O5)n.
(C6 H 10 O5 ) n (s ) + 6nO2 ( g ) ¾burn
¾®
¾ 6nCO2 ( g ) + 5nH 2 O ( g ) + Heat + light Nature of Black No black
Cellulose deposit carbon carbon
deposited deposited
— Combustion of alcohol: Nature of Reducing Oxidising
C2 H 5 OH (l ) + 3O2 ( g ) ¾burn
¾®
¾ 2CO2 ( g ) + 3H 2 O( g ) + Heat + light flame flame flame
Ethanol oxygen (in air) Temperature low High
Activity:To observe the combustion of given organic
compounds. Conclusion: Keep the air regulator open to get
Materials: Benzene, naphthalene, Camphor, alcohol oxidizing, non-sooty flame which has high temperature
(ethanol). Spirit, acetone. and does not lead to black deposits.
Procedure:
1. Take each compound on iron spatula and burn them Combustion and the Nature of Flame :
in Bunsen burner.
2. Record the type of flame produced. (i) Saturated hydrocarbon such as, methane, ethane,
3. Put a metal plate above the flame and observe propane, butane and natural gas and LPG burn
whether or not there in black carbon deposition. with a blue flame in the presence of sufficient/excess
of air/oxygen.
Observation:
(ii) In the presence of limited amount/of air/oxygen,
Compound Flame Produced Deposit
used saturated hydrocarbon, such as, methane, butane,
Benzene Smoky flame Carbon deposited etc give smoky flame.
Naphthalene Smoky flame Carbon deposited
Camphor Smoky flame Carbon deposited
Alcohol Non-Luminous flame No Carbon deposited (iii) Unsaturated hydrocarbon such as ethane, ethyne
Spirit Non-Luminous flame No Carbon deposited etc. burn with a luminous/yellow smoky flame.
Acetone Non-Luminous flame No Carbon deposited

Conclusion: (iv) The gas/kerosene stove used at home has inlets

Benzene, naphthalene, camphor burn with smoky flame for air so that a sufficiently oxygen rich mixture is
and carbon particles get deposited they undergo burnt to give a clean blue flame. If you carefully
incomplete combustion due to excess of carbon observe the bottoms of vessels getting blackened,
content. it is clear indication that the air holes are blocked
and the fuel is getting wasted.
— Alcohol, spirit and acetone burn with non-Luminous
flame and no carbon gets deposited. They under go (v) Fuels, such as coal and petroleum, have some
complete combustion, therefore produce more heat.
amount of nitrogen an d sulphur in them.
Activity: To study the different types of flames/
Combustion of coal and petroleum results in
presence of smoke.
Material required: Bunsen burner. formation of oxides of sulphur and nitrogen (such
Procedure: as sulphur dioxide, nitric oxide, nitrogen peroxide)
1. Light the Bunsen burner. which are major pollutants in the environment.

1-136
 Carbon and Its Compounds

Formation of Coal and Petroleum : Example:

Coal and petroleum have been formed from biomass Alcohols on oxidation with certain oxidising agents
which has been subjected to various biological can such as chromic anhydride in acetic acid, yield
geological processes. corresponding aldehydes, whereas on oxidation with
alkaline potassium permanganate (or acidified
Coal is a naturally occurring black mineral and is a potassium dichromate) corresponding carboxylic acid
mixture of free carbon and compounds of carbon is formed, as given below:
containing hydrogen, oxygen, nitrogen and sulphur.
It is not only a good fuel but is also a source of many CH 3 CH 2 OH (l ) + [O ] ¾CrO
¾¾ 3in
® CH 3 CHO + H 2 O
organic compounds. CH 3COOH

It is found in coal mines deep under the surface of Ethanol Nascent oxygen Ethanal (an aldehyde)
earth.
CH 3CH 2 OH (l ) + 2[O] ¾Alkaline
¾ ¾® CH 3COOH + H 2 O
KMnO 4 , Heat
Coal is believed to be formed from fossils which got
buried inside the earth during earthquakes and Ethanoic acid
volcanoes which occurred about 300 million years ago.
Due to huge pressure and temperature inside the earth Activity: To study the reaction of ethanol with alkaline
and in the absence of air the fossils fuels (vegetable potassium permanganate:
matter or wood, etc.) were converted into coal. The Material required: Ethanol, alkaline KMnO4, test tube.
slow chemical processes of the conversion of wood
into coal is called carbonization. Since coal is formed Procedure:
by slow carbonisation. of plants and fossils, it — Take about 3 ml of ethanol in a test tube.
produces many important carbonisation products like — Add 5% solution of alkaline KMnO4 drop by drop into
peat, lignite, bituminous and anthracite etc. and is itself this solution.
known as fossil fuel. Coal is also a non-renewable — Observe the colour of alkaline KMnO4 after adding
source of energy. initially as well as finally.
Petroleum is complex mixture containing various Observation: The colour of KMnO4 gets discharged
hydrocarbons (compounds of carbon and hydrogen) in the beginning. When excess of KMnO4 is added,
in addition to small amounts of other organic the colour of KMnO4 does not disappear because
compounds containing oxygen, nitrogen, and sulphur. whole of ethanol gets oxidised to ethanoic acid.
It is a dark coloured. viscous and foul smelling crude
oil. The name petroleum is derived from Latin words: CH 3 CH 2 OH ¾Alkaline
¾ ¾,¾ ¾¾ ¾® CH 3COOH + H 2 O
KMnO4 , Heat
or K 2Cr2O7 / H 2 SO4 , Heat )
“Petra” meaning rock and “oleum” meaning oil. Since
petroleum is found trapped between various rocks, it
ADDITION REACTION:
is also known as rock oil.
All unsaturated hydrocarbons (unsaturated carbon
compounds) react with a molecule like H2. X2. H2O etc.
OXIDATION:
to form another saturated compounds are called
Carbon and its compounds can be easily oxidised on
addition reactions.
combustion (or burning). During combustion/bruning,
the compounds gets oxidised completely to different
Unsaturated hydrocarbons add hydrogen, in the
products, depending upon the nature of the oxidizing
presence of catalysts, such as nickel or palladium to
agents.
give saturated hydrocarbons.
— Carbon gives carbon monoxide or carbon dioxide
depending upon the oxygen available.
Note: Catalysts are substance that cause a reaction to
2C (s ) + O2 ( g ) ® 2CO ( g )
occur or proceed at a different rate without the reaction
Carbon Oxygen(limited) Carbon monoxide
it say being affected.
— C ( s ) + O2 ( g ) ® CO2 ( g )
— Addition of hydrogen to ethene:
(excess) Carbon dioxide
H H H H
— Hydrocarbon when oxidised give different product as | | 0
follows: C=C + H 2 ( g ) ¾NI
¾,250
¾¾ C
® H -C -C - H
H H | |
CH 4 ( g ) + 2O 2 ( g ) ¾Complete
¾¾¾ ¾¾® CO 2 ( g ) + 2 H 2 O ( g )
oxidation

H H
Methane Oxygen(excess)
— 2CH 4 ( g ) + 3O2 ( g ) ¾Incomplete
¾ ¾ ¾oxidation
¾¾® 2CO ( g ) + 4 H 2 O( g ) Ethene (Double bond Ethane
Methane Oxygen (Limited) Containing unsaturated (Saturated hydrocarbon)
— Alcohols also give different products on oxidation Carbon compound)
depending upon the reaction conditions.

1-137
Carbon and Its Compounds

— Addition of hydrogen to a unsaturated carbon In the above reaction between methane and chlorine,
compound is called hydrogenation reaction. only one hydrogen atom of methane has been replaced
Certain vegetable oils such as ground nut oil, cotton by chlorine atom and we get chloromethane, CH3Cl. By
seed oil and mustard oil, contain double bonds (C = C) supplying more chlorine, it is possible to replace all the
are liquids at room temperature. Because of the hydrogen atoms of methane by chlorine, one by one.
unsaturation, the vegetable oils undergo In this way we can obtain three more compounds :
hydrogenation. like alkenes, to from saturated Dichloromethane or Methylene dichloride, CH2Cl2;
products called vanaspati ghee. Which is semi-solid Trichloromethane or Chloroform, CHCl 3 and
at room temperature. Tetrachloromethane or Carbon tetrachloride, CCl 4.
Vegetable oils (Unsaturated oil) + Hydrogen Methane (CH4), Ethane (C2H6), Propane (C3H8), Butane
(C4H10), Pentane (C5H12), and Hexane (C6H14), etc., are
¾Ni,
¾Heat
¾® Vanaspati Ghee (Saturated ghee)
all saturated hydrocarbons (or alkanes). So, all these
compounds will give substitution reactions (with
SUBSTITUTION REACTIONS
chlorine).
Reactions which involve the direct replacement
(displacement or substitution) of an atom or a ground
Substitution by bromine (Br) and iodine (I) atoms can
of atoms in an organic molecule by another atom or
be carried in the same manner. The chemical reactions
group of atoms without any change in the rest of the
carried in the presence of sunlight are known as
molecule are called substitution reactions.
photochemical reactions.
Saturated hydrocarbons (alkanes) are quite unreactive
From the above discussion, we conclude that
(because they contain only carbon-carbon single
1. Organic compounds, particularly the hydrocarbons
bonds) Being unreactive, saturated hydrocarbons do
readily undergo combustion reactions in air or
not react with many substances.
oxygen.
2. Saturated hydrocarbons take part in the substitution
Satur ated hydrocarbons, however, undergo
reactions in which one or more hydrogen atoms get
substitution reactions with chlorine in the presence of
replaced by suitable atoms/groups.
sunlight. Substitution reactions (like chlorination) are
3. Unsaturated hydrocarbons participate in the addition
a characteristic property of saturated hydrocarbons or
reactions in which the double bond and triple bond
alkanes (Unsaturated hydrocarbons do not give
change to single bonds.
substitution reactions with halogens, they give addition
reactions). The substitution reactions of saturated
hydrocarbons (alkanes) with chlorine take place in the Important Points
presence of sunlight. 1. Organic equations are normally not balanced to fulfill
the requirement of the Law or Mass Action because
Substitution Reaction of Methane with Chlorine. the products are formed in different proportions.
Methane reacts with chlorine in the presence of sunlight For example, in the chlorination of propane, both 1-
to form chloromethane and hydrogen chloride: chloropropane and 2-chloropropane are
simultaneously formed but in different percentages.
However, the total mass of the products species is
the same as that of the reacting species. Therefore,
there is no need to balance such equations.
2. Freon gas which is used as refrigerant to cause
cooling is chemically difluoro dichloromethane
(CF2Cl2). It has been formed by replacing two H
atoms in methane (CH4) with F atoms and two H
atoms with Cl atoms.

ELIMINATION REACTION
An elimination reaction is a type of organic reaction in
which two substituent’s are removed from a molecule
in either (a) one or two step mechanism.
For example, Dehydrogenation of Alkyl halide in
In this reaction, one H atom of methane has been presence of alcoholic KOH in formation of ethene.
substituted (replaced) by a Cl atom, converting CH4
b a
into CH3Cl. H - CH2 - CH2 - Br + KOH ( alc.) ¾¾
D
® CH 2 = CH2 + KBr + H2 O

1-138
 Carbon and Its Compounds

Ethanol is prepared on commercial scale by

SOLVED EXAMPLE fermentation of sugar Fermentation is allowed to take
place at 298-303 K in the absence of air. This is ethanol
Example-80 (ethyl alcohol) gets oxidised to ethanoic acid (acetic
acid) in the presence of air.
(i) An organic compound ‘A’ of molecular formula
C2H6O on oxidation with dilute alkaline KMnO4 C12 H 22 O11 + H 2 O ¾Invertase
¾¾ ¾® C 6 H 12 O6 + C 6 H 12 O6
solution gives an acid ‘B’ with the same number of Sucrose Glucose Fructose
carbon atoms. Compound ‘A’ is often used for C6 H 12 O6 ¾Zymase
¾¾® 2C 2 H 5 OH + 2CO2 ( g )
sterilization of skin by doctors. Name the Glucose/Fructose Ethanol (ethyl alcohol) Carbon dioxide
compounds. Write the chemical equation involved
in the formation of ‘B’ from ‘A’. Physical Properties :
(ii) What is addition reactions. Explain it by taking — Physical state/colour and odour: Pure ethanol is a
colourless liquid having a pleasant smell and a burning
example of reaction between ethane and hydrogen.
taste.
(iii) Ethene is produced when — Boiling and Freezing points: It is a volatile liquid with
(a) Ethanol reacts with ethanoic acid in presence of a a boiling point of 78.10C, and freezing point is -1180C.
few drops of conc. H2SO4 — Density: Ethanol is lighter than water as its density is
(b) Ethanol is oxidized with acidified potassium 0.79 g ml-1 at 293 K.
dichromate — Solubility: Ethanol is miscible with water in all
(c) Ethanol is heated with excess of conc. H2SO4 at proportion, due to the formation of hydrogen bonds
443 K with water molecules.
— Conductivity: Ethanol is a covalent compound and does
(d) Ethanol reacts with Na Metal
not ionise easily in water. Hence it is a neutral
Sol. (i) The compound ‘B’ should contain a —COOH group compound.
as it is an acid. Since it has only two carbon atoms, the — Action on Litmus: Ethanol is a neutral compound So,
other carbon atom must represent CH3 group. Thus, it has no effect on the colour of litmus.
compound ‘B’ is ethanoic acid (CH3COOH). The
compound ‘A’ used for the sterilization of skin by Chemical Properties of Ethanol :
doctors is C2H5OH (C2H6O). The chemical reaction — Combustion (or burning): Ethanol is highly
involved in the oxidation is : inflammable liquid and readily burn in air with blue
flame to form water vapour, carbon dioxide and
C 2 H5OH ( l ) + O2 ( g ) ¾¾¾¾
Baeyer's
reagent
® CH3COOH ( l ) + H 2 O ( l ) evolving heat. Thus, combustion of ethanol is an
Ethanol ( A ) Ethanoic acid ( B ) exothermic reaction.
(ii) Addition Reaction of Ethane with Hydrogen : Ethane Combustion
C6H5OH(l) + 3 O2(g) ¾¾¾¾¾
® 2CO2 + 3H2O (g) + Heat
reacts with hydrogen when heated in the presence
of nickel catalyst to form ethane : in this reaction,
— Reaction with sodium metal: Ethanol reacts with
one H atom adds to each C atom of ethane due to sodium metal to produce sodium ethoxide and
which the double bond opens up to form a single hydrogen gas is evolved.
bond in ethane. One molecule of hydrogen is added 2C 2 H 5 OH (l ) + 2 Na( s ) ® 2C 2 H 5 ONa + H 2 ( g )
to an unsaturated hydrocarbon ‘ethane’ having a
Ethanol Sodium metal Sodium ethoxide Hydrogen
double bond to form a saturated hydrocarbon Activity: To study the reaction of ethanol with sodium
‘ethane’ having a single bond. metal.
Materials: Ethanol, dry piece of sodium metal test tube.
Ni catalyst
Procedure:
CH 2 = CH 2 + H 2 ¾¾¾¾¾ Heat
® CH 3 — CH3 — Take ethanol in a test tube.
Ethene Hydrogen Ethane
( Unsaturated ) (Saturated ) — Add a dry piece of sodium metal.
— Bring a burning matchstick near the gas evolved to
(iii) (c)
test it and record observation.
Observation: The gas burns in air with a pop sound
SOME IMPORTANT CARBON COMPOUNDS:
which is the characteristics of hydrogen gas.
ETHANOL (Ethyl Alcohol, C2H5OH):
Conclusion: Alcohol reacts with sodium metal to
Ethanol is the second member of the homologous series
liberate hydrogen gas.
of alcohols.

Preparation: By the fermentation of carbohydrates — Reaction with ethanoic acid (Esterification reaction):
(sugar or starch). Conc. H2SO4, Heat

1-139
Carbon and Its Compounds

The reaction in which an alcohol reacts with acetic — When large quantities of ethanol are consumed, it
tends to slow metabolic processes and to depress the
acid in the presence of conc. H2SO4 to form an ester is
central nervous system. This results in lack of
called esterification. coordination, mental confusion, drowsiness, lowering
CH 3COOH (l ) + C 2 H 5 OH (l ) of normal inhibitions and finally stupor (unconscious
state of wild)
ethanoic acid ethanol
(acetic acid) — Drinking of alcohol over a long period of time damages
liver.
Heat Conc. H2SO 4
Denatured Alcohol:
Ethanol to which certain poisonous and nauseating
CH 3COOC 2 H 5 (l ) + H 2 O(l ) substances like methyl alcohol, pyridine etc. have been
ethyl ethanoate added is termed denatured alcohol.
(ethyl acetate) Note : To prevent the misuse of ethanol (Alcohol),
Sweet smelling compound industrial alcohol is coloured blue to that it can be
Note: Ester is sweet-smelling compounds and is used recognized easily.
for making perfumes.
Harmful effects of denatured alcohol:
— Reaction with conc. sulphuric acid (Dehydration): — Methanol is highly poisonous compounds for human
Ethanol when heated with excess of concentrated beings. Methanol when taken, even in small amount,
sulphuric acid at 443 K, gets dehydrated to give ethene. can cause death.
C 2 H 5 OH (l ) + H 2 SO4 (Conc.) ¾443
¾ ¾ K
® H 2 C = CH 2 ( g ) + H 2 O(l )
— Methanol gets oxidised to methanal in the liver, which
causes coagulation of protoplasm.
ethanol excess ethene
— Methanol also effects the optic nerve and cause
Note: The concentrated sulphuric acid can be regarded blindness.
as a dehydrating agent who removes water from
ethanol.
TESTS FOR AN ALCOHOL
Use of ethanol: An alcohol can be tested by any one of the following
— Ethanol is present in alcoholic beverages such as beer, tests:
wine, and whisky. Sodium Metal Test
— As a solvent for paints, varnishes-dyes, cosmetics, Add a small piece of sodium metal to the organic liquid
perfumes, soaps and synthetic rubber etc. (to be tested), taken in a dry test-tube. If bubbles (or
— Ethanol is used in cough syrups, digestive syrups and effervescence) of hydrogen gas are produced, it
tonics. indicates that the given organic liquid is an alcohol.
— A mixture of 80% rectified spirit and 20% petrol is called
power alcohol. It is used as fuel in cars and aeroplanes. Ester Test for ALcohols
— A mixture of ethanol and water has lower freezing point The organic compound (to be tested) is warmed with
some glacial ethanoic acid and a few drops of
than water this mixture is known as antifreezing and
concentrated sulphuric acid. A sweet smell (due to the
is used in radiators of vehicles in cold countries and at
formation of ester) indicates that the organic compound
hill stations.
is an alcohol.
— As an antiseptic to sterilize wounds and syringes in
hospitals. ETHANOIC ACID (ACETIC ACID) CH3COOH:
— For the manufacture of terylene and polythene. — Ethanoic acid is commonly called acetic acid and
— As a preservative for biological specimens. belongs to the homologous series of carboxylic acid
— Ethyl alcohol is used as hypnotic (induces-sleep). and is represented as CH3COOH.
— 5-8% solution of acetic acid in water is called vinegar
Harmful effects of Alcohols: and is used for preservation foods like sausage,
— Consumption of small quantities of dilute ethanol pickles.
causes drunkenness. Even though this practice is Physical properties:
condemned, it is a socially widespread practice. — At ordinary temperature, ethanoic acid is a colourless
However, intake of even a small quantity of pure liquid with a strong pungent smell and sour taste.
ethanol (called absolute alcohol) can be lethal. Also — Its boiling point is 391 K and its density at 273 K is
long-term consumption of alcohol leads to many health 1.08 (heavier than water).
problems. — It is miscible with water due to the formation of
hydrogen bonds with water molecules.

1-140
 Carbon and Its Compounds

— On cooling at 289.6K, it turns in ice-like crystals, hence The reaction of carboxylic acid with an alcohol to form
named as glacial acetic acid. an ester is called “esterification”.
— It dissolves sulphur, iodine and many other organic
compounds. Note: Ester can be hydrolysed in the presence of an
— It dimerise when dissolved in benzene. acid or a base to give back the parent carboxylic acid
— 2CH 3COOH ˆˆˆ
†
‡ˆˆ
ˆ (CH 3 COOH ) 2 and the alcohol.
Ethanoic acid Dimer
Example:
Activity: To determine pH of acetic acid and (i) Ethyl ethanoate on acid hydrolysis gives ethanoic
hydrochloric acid. acid and ethanol.
Material: Acetic acid (1M).HCl (1M), blue litmus paper, CH 3COOC 2 H 5 (l ) + H 2 O (l ) ® CH 3 COOH (aq.) + C 2 H 5OH
universal indicator. (ii) Hydrolysis of ester in the presence of base (alkali) is
Procedure: Take two strips of blue litmus paper.
called “Saponification reactions”.
CH 3COOC 2 H 5 (l ) + NaOH (aq ) ® CH 3COONa + C 2 H 5 OH
— Put a few drops of HCl on one of them and few drops
of acetic acid on the other. Ethyl ethanoate Sodium Sodium Ethanol
— Observe the change in colour. Hydroxide ethanoate
— Take 1 ml of acetic acid in a test tube and add a few
drops of universal indicator. Note : Alkaline hydrolysis of higher esters is used in
— Take 1ml of HCl in a test tube and add few drops of the manufacture of soaps.
universal indicator.

Observation: Both acetic acid and HCl turn blue litmus

red showing that they are acidic in nature. pH of acetic
acid and HCl are not equal.

Conclusion: HCl is strong acid than CH3COOH,

therefore, pH of HCl will be lower than that of acetic
acid.
Chemical Properties :
The chemical characteristics of the acid are mainly of
the carboxyl group.
Acidic nature: The acidic nature of ethanoic acid is
supported by litmus test. On adding a few drops of
blue litmus to the aqueous solution of the acid, the
solution acquires a red colour.

But ethanoic acid is a weaker acid than minerals like Activity: To study the esterification process using
H2SO4, HCl and HNO3. In fact, organic acids are weaker ethanol and acetic acid.
acids than inorganic acids since they are ionised to
smaller extent in solution.
Materials: Beaker, water, test tube, ethanol, acetic acid.
Conc. H2SO4 etc.
Reaction with metals: Ethanoic acid evolves hydrogen
on reacting with active metals present above hydrogen
in the reactivity series. As a result, metal salts are formed. Procedure: Take 2ml of ethanol in a test tube.
For example: — Add 2ml of ethanoic acid (acetic acid) in to it.
— Add few drops of conc. H2SO4.
2CH3COOH + 2Na ¾¾® 2CH3COONa + H 2
Sodium ethanoate — Warm it in a beaker containing water.
2CH3 COOH + Ca ¾¾® ( CH3 COOH )2 Ca + H 2 Observation: Pleasant fruity smelling compound
Calcium ethanoate
(called ester) is formed.
— Reaction with alcohols (Esterification reaction ):
Conclusion: Acetic acid reacts with alcohol in presence
Ethanoic acid reacts with ethanol in the presence of
of conc. H2SO4 which act as a dehydrating agent to
cons. H2SO4 to form ethyl ethanoate which is an ester.
form ester.
CH 3COOH (l ) + C 2 H 5 OH (l ) ¾Conc
¾¾ ¾¾ ¾® CH 3COOC2 H 5
, H 2 SO4 ,heat
+ H 2O
— Reaction with sodium carbonate and sodium hydrogen
Ethanoic acid Ethanol Ethyl ethanoate (ester)
carbonate:

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Carbon and Its Compounds

Ethanoic acid decomposes sodium hydrogen carbonate At the anode: The ethanoate ion (CH3COO–) releases
and sodium carbonate with a rapid evaluation of carbon one electron to form the radical CH 3COO . This×
dioxide gas.
NaHCO3 ( aq) + CH 3COOH (aq) ® CH 3COONa ( aq) + H 2 O(l ) + CO2 ( g )
radical forms a methyl radical (CH×3 ) and CO2.
Sodium Sodium ethanoate Hydrogen carbonate Two methyl radicals combine to form C2H6.
ethnic acid CH 3COO - ® CH 3COO× + e
Na 2 CO3 ( aq ) + 2CH 3 COOH ( aq ) ® 2CH 3 COONa( aq ) + H 2 O + CO2 ( g )
CH 3COO× ® CH3× +CO 2
Sodium carbonate Ethanoic acid Sodium ethanoate 2CH×3 ® C 2 H 6
Note: Reactions of ethanoic acid with NaOH, NaHCO3, At the cathode: Here Na+ and H+ ions (from water)
Na2CO3 and active metals show that the hydrogen compete for discharge. Ultimately, it is the H+ ion,
present in the carboxy (–COOH) group is acidic in being more electropositive, attracts the free electron
nature. and gets discharged to produce the hydrogen gas.
Activity: To study the reaction of carboxylic acid with
sodium carbonate and sodium hydrogen carbonate. H+ + e ® H
Material: Ethanoic acid, Sodium carbonate, Sodium 2H ® H 2
hydrogen carbonate.
Uses of Ethanoic Acid :
Procedure:
— Take 1g of Na2CO3 and 2ml of ethanoic acid into it. — Ethanoic acid is used in the manufacture of various
— Pass the gas formed through lime water and note down dyes, perfumes and rayon.
the observation. — It is used for making vinegar.
— Repeat the same procedure with sodium hydrogen — It is used for making white lead [2PbCO 3. Pb
carbonate and record observation. (OH)2]which is used in white paints.
Observation: Brisk effervescence due to carbon
— Its 5% solution is bactericidal (destroys bacteria).
dioxide formed which turns lime water milky.
Conclusion: Acetic acid reacts with Na 2CO3 and — It is used in preparation of cellulose acetate which is
NaHCO3 to liberate CO2 gas. used for making photographic film.
— It is used for coagulation of the latex.
Decarboxylation: — It is used for preparation of 2,4-dichloro phenoxy
(a) Decarboxylation of sodium ethanoate: ethanoic acid which is used as herbicide.
When sodium ethanoate is heated with soda lime, — Aluminium acetate and chromium acetate are used as
methane is formed. mordants in dyeing and water proofing of fabrics.
CH 3COONa + NaOH ® CH 4 + Na 2 CO3
sodiumethanoate methane SOLVED EXAMPLE
(The term ‘decarboxylation’ is used when the
elements of carbon dioxide are removed from a Example-81
molecule.) Which of the following is ethyl acetate :
(b) Decarboxylation of calcium ethanoate: (A*) CH3COOC2H5 (B) CH3COOCH3
When calcium ethanoate is heated (dry distilled), (C) CH3CH2COOCH3 (D) None of these
propanone is produced.
Ans. (A)
Sol. ethyl acetate ¾® CH3COOC2H5

Example-82
When ethyl alcohol undergoes oxidation in presence
of acidified K2Cr2O7 or alkalline KMnO4, it produces :
(A*) Acetic acid (B) Acetaldehyde
(C) Ethyl ethanoate (D) Propanal
(c) Electrolytic decarboxylation (Kolbe method): Ans. (A)
When a strong aqueous solution of sodium ethanoate acidified
K Cr O
is electrolyzed, ethane is formed. Sol. C2H5 OH ¾¾¾¾¾
2 2 7
® CH3COOH
alkalline
2CH3COO–Na + + 2H2O ® C 2H6 + 2CO2 + 2NaOH + H2 KMnO4

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 Carbon and Its Compounds

Example-83 Example-88
When ethanoic acid is treated with ethanol in presence Number of moles required for the complete combustion
of H2SO4, then the products is/are : of one mole of butane :
(A) 13 (B) 7 (C) 7.5 (D*) 6.5
(A) HCOOC2H5 and water
Ans. (D)
(B) C2H5COOCH3 and water
(C) C2H5COOC2H5 and water 13
Sol. C4H10 + O2 ¾¾® 4CO2 + 5H2O
(D*) CH3COOC2H5 and water 2
Ans. (D)
Example-89
Sol. CH3COOH + C2H5OH ¾® CH3COOC2H5
How many moles of oxygen are required for the complete
combustion of 4.4 gm of propane:
Example-84 (A) 10 moles (B) 5 moles
Which is more acidic : (C*) 0.5 moles (D) 0.75 moles
(A*) HCOOH Ans. (C)
(B) CH3COOH Sol. Protane = C3H8; at wt. = 36 + 8 = 44
4.4
(C) CH3 — CH2 — COOH Mole = =1
44
C3H8 + 5O2 ¾¾® 3CO2 + 4H2O
CH3
1 mole requires 5 moles of oxygen
(D) CH3 – CH2 – COOH
Ans. (A) Example-90
When compound ‘x’ is oxidised by acidified potassium
Sol. Formic acid is more acidic than its remaining
dichromate compound ‘y’ is formed. Compound ‘y’ on
homologues members reduction with LiAIH4 gives ‘x’. ‘x’ and ‘y’ respectively
are :
Example-85 (A*) C2H5OH, CH3COOH
Vinegar contains : (B) CH3COCH3, CH3COOH
(A) 100% acetic acid (B) 10–20% acetic acid (C) C2H5OH, CH3COCH3
(C) 1% acetic acid (D*) 7–8% acetic acid (D) CH3CHO, CH3COCH3
Ans. (D) Ans. (A)
Sol. Vinegar : 7–8% acetic acid acidified
k Cr O
2 2 7
Sol. alcohol ¾¾¾¾¾ ® acetic acid
Example-86 LiAlH4

Butanoic acid and methyl ethanoate are :

(A) Chain isomers of each other Example-91
(B) Position isomers of each other The gas evolved when formic acid react with sodium :
(C) Functional isomers of each other (A) CO2 (B*) H2
(D*) None of these (C) O2 (D) None of these
Ans. (D) Ans. (B)
Sol. Butanoic acid and methyl ethanoate having different Sol. 2HCOOH + 2Na ¾® 2HCOONa + H2­
no. of carbon so they are not isomer.
Example-92
Example-87 The gas evolved when acetic acid react with Na 2CO3 :
When 2- Butyne react with one molecule of chlorine (A*) CO2 (B) H2 (C) O2 (D) None
then product obtained is : Ans. (A)
(A) 1, 2 – Dichloro butane Sol. acetic acid + Na2CO3 ¾® CO2 ­
(B) 2, 3 – Dichloro butane
(C) 2, 2 – Dichloro – 2 – butene Example-93
(D*) 2, 3 – Dichloro – 2 – butene In esterification, concentrated H2SO4 acts as :
Ans. (D) (A) Catalyst
(B) Dehydrating agent
Cl (C) Hydrolysis agent
Cl2
H3C C C CH3 CH 3 C C CH3 (D*) Both as catalyst and dehydrating agent
Sol.
Ans. (D)
Cl Sol. H2SO4 in esterification

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Carbon and Its Compounds

Ans. (B)
Example-94 2Na 1
Sol. 2CH3 – OH ¾¾¾® 2CH3ONa + H2 ­
Which of the following has maximum boiling point : 2
(A) CH3COOH (B) HCOOH
(C) C2H5COOH (D*) C3H7COOH Example-100
Ans. (D) An organic compound A of molecular formula C2H6O
Sol. Mol. wt. µ boiling point on oxidation gives an acid B with the same no of carbon
atom as in the molecule A. Compound A is often used
as tincture of iodine :
Example-95 (A) CH3–O–CH3, CH3–COOH
Which of the following acid is present in vinegar : (B) CH3–O–CH3, COOH
(A) Hydrochloric acid (B*) Acetic acid (C*) CH3–CH2–OH, CH3COOH
(C) Tartaric acid (D) Citric acid (D) CH3–CH2–OH, HCOOH
Ans. (B) Ans. (C)
Sol. Acid present in vinegar is acetic acid. Sol. oxidation
C2H5 OH ¾¾¾¾¾
® CH3COOH

Example-96
SOAPAND DETERGENTS:
Vegetable oils (which are liquid) at room temperature
Soap and detergents are substances which are used
can be converted into vanaspati ghee (solid) by the
for cleaning. There are two types of detergents:
process of :
1. Soap 2. Synthetic detergents
(A) Dehydrogenation (B) Decolourisation
(C*) Hydrogenation (D) None of the above
Ans. (C) SOAP: A soap is the sodium or potassium salt of a
Sol. Vegetable oils (which are liquid) at room temperature long-chain fatty acids (carboxylic acid or glycerol).
can be converted into vanaspati ghee (solid) by the Activity:
process of Hydrogenation

Example-97
The following reaction is an example of :
CH4 (g) + 2O 2 (g) ¾¾® CO 2 (g) + 2H2O + Heat and light
(A) Addition reaction (B) Displacement reaction
(C*) Combustion reaction (D) Substitution reaction
Ans. (C)
Sol. Combustion reaction
CH4 (g) + 2O 2 (g) ¾¾® CO 2 (g) + 2H2O + Heat and light

Formation of micelles
Example-98
An organic compound gives hydrogen gas on reacting Formation of micelles
with Sodium metal. It also gives iodoform test and forms Take about 10mL of water each in two test tubes.
an aldehyde of molecular formula C2H4O on oxidation Add a drop of oil (cooking oil) to both the test tubes
with CrO3 in glacial acetic acid. The organic compound is: and label them as A and B.
(A) CH3OH (B) CH3COOH To test tube B, add a few drops of soap solution. Now
(C) CH3CHO (D) C2H5OH shake both the test tubes vigorously for the same
Ans. (D) period of time.
Sol. 2C2H5OH + 2Na ¾® 2C2H5ONa + H2­ Can you see the oil and water layers separately in both
C2H5OH ¾¾¾¾¾ 3 CrO in
® CH3CHO the test tubes immediately after you stop shaking
acetic acid
them.
Leave the test tubes undisturbed for some time add
Example-99 observe. Does the oil layer separate out? In which test
When a compound (A) reacts with Sodium metal, it tube does this happen first?
forms sodium salt along with a gas which has highest
This activity demonstrate effect of soap in cleansing
calorific value. This gas is also used along with Pd to
as we know that most of the dirt is oily in nature and
convert another compound (B) into the lower
homologue of compound (A). Compound (B) is the oil does it dissolve in water.
first member of its homologous series. Compound (B) But know the question arise what are soap? What is
is: the detergent which one is more effect? How the work.
(A) Methanol (B*) Methanal Soap is sodium or potassium salt a long change fatty
(C) Ethanol (D) Ethanal acid (Carboxylic acid or Glycerol)

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 Carbon and Its Compounds

Soap has large non ionic hydrocarbon group and an Cleansing action of soap:
ionic group. COONa. The molecules of soap are sodium or potassium salts
of long chain carboxylic acids, the ionic end of soap
Ex. of soap is: dissolves in water while the carbon chain dissolves in
(1) Sodium stearate (C17H35COONa) oil. The soap molecules thus form structures called
(2) Sodium palmitate (C15H31COONa) micelles where one end of the molecules is towards
the oil droplet while the ionic end faces outside. This
Soap is basic in nature so soap solution turns red form and emulsion in water. The soap micelle thus helps
litmus to blue. in dissolving the dirt in water and we can wash out
clothes clean.
Preparation of Soap:
The soap is prepared by heating animal fats or
vegetable oils (olive oils, castor oil or palm oil) with
sodium hydroxide or potassium hydroxide.
The process of formation of soap by the hydrolysis of
fat or oil with alkali is called saponification.
Oil or Fat + Sodium hydroxide ’! Soap + glycerol

Structure:
Sodium stearate(soap)
Sodium hydroxide
C17 H 35 COO - CH 2
|
C17 H 35 COO - CH + 3 NaOH ® 3C17 H 35COONa
| Sodium Sodium
hydroxide stearate (soap) Effect of soap in cleaning
C17 H 35 COO - CH 2
Glycerol stearate
(fatty acid)
Activity: Take two clean test tubes and label them as
‘A’ and ‘B’. Now put 10ml hard water in each of the
Structure: two test tubes. Add five drops of soap solution in test
— The hydrocarbon chain is non-polar and water-hating tube ‘A’ and five drops of detergent solution in test
(hydrophobic), while the other part is polar or water tube ‘B’ Shake the two test tubes for the same period
loving (hydrophilic). and observe if both observe if both the test tubes
have the same amount of foam. Find out in which of
— Hydrophilic part makes the soap soluble in water and the two test tubes a curdy white mass is formed.
hydrophobic part makes the soap insoluble.
O In which test due do you get more form?
=

+
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2 –C–O–Na
We get more foams in test tube……………..
Hydrophobic “Tail” Hydrophilic “head”
(Non-polar part) (Polar part) A white curdy precipitate is formed in test
tube……………….
— When soap is added to water, the soap molecules
assume a configuration which increases the interaction Result (Conclusion): Soaps are not effective in acidic
of the water loving heads with the water molecules, medium.
and decreases the interaction between the water hating
tails with the water molecules. When soaps is used for washing clothes with hard
water, a large amount of soap is wasted in reacting
— The hydrophobic part of the soap molecules traps the with the calcium and magnesium ions of hard water to
dirt and the hydrophilic part makes the entire molecules form an insoluble precipitate called scum, before it can
soluble in water. Thus, the dirt gets washed away with be used for the real purpose of washing soap. A large
the soap. amount of soap is needed for washing clothes. When
the water is hard.
— The water-hating, non polar tails clump together in a
radial fashion with the water-loving. Polar heads Activity:
remaining at the periphery of the clump, these clumps Take two test tubes with a about 10 mL of hard water
or droplets of soap molecules are called micelles. in each.

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Carbon and Its Compounds

Add five drops of soap solution to one and five drops You will observe that the amount of foam in the two
of detergent solution to the other. test tubes in different. The foam is formed to a greater
Shake both test tubes for the same period extent in test tube ‘B’ (containing detergent solution),
Do both test tubes have the same amount of foam? while formation of a curdy white mass will be observed
In which test tube is a curdy solid formed? in test tube ‘A’. This activity clearly indicates that
detergents can be used for cleansing purpose, even
Observation: with hard water.
Test tube in which …………………………..is present
contain more amount of foam.
SYNTHETIC DETERGENTS:
Curdy solids is form in the test tube containing
— Synthetic detergents are called soapless soap because
…………….
they are not prepared from fatty acid and alkali.
Conclusion: Detergents have better cleansing action
than soap. Detergents are generally ammonium or — Synthetic detergents are sodium salts of sulphonic
sulphonate salts of long chain carboxylic acids. The acids, i.e. detergents contain a sulphonic acid group (-
charged ends of these compounds do not form SO3H). instead of a carboxylic acid group (-COOH), on
insoluble precipitates with the calcium and magnesium one end of the hydrocarbon chain.
ions in hard water. Thus, they remain effective in hard
water. Detergents are usually used to make shampoos O

- -
and products for cleaning clothes. CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH4 - S - O - - Na+
O
Disadvantage of soap:

— Soaps are not effective in hard water: Hard water Hydrophobic “Tail” Hydrophilic “head”
contains calcium ions (Ca2+ ) and magnesium ions (Non-polar part) (Polar part)
(Mg2+) These ions react with the carboxylate ions
(RCCO-) of the soap forming an insoluble precipitate Properties of synthetic detergents:
called scum, For example, soap like sodium
stearate (C17H35COONa) reacts with calcium and — Synthetic detergents do not react with the ions present
magnesium ions as per the in hard water. Hence synthetic detergents have no
‘ following chemical equation. problem in forming lather with hard water, i.e. their
2C17 H 35 COONa + Ca 2+ ( aq) ® (C17 H 35 COO) 2 Ca ¯ + 2 Na + ( aq) efficiency is not affected by hard water.
Sodium stearate (In hard water) Calcium stearate Sodium ion
— Synthetic detergents can be used even in acidic
2C17 H 35 COONa + Mg 2 + (aq ) ® (C17 H 35 COO ) 2 Mg ¯ + 2 Na + (aq ) solution and sea water, whereas soap cannot be used
Sodium stearate (In hard water) Magnesium Sodium ion in the acidic solution (due to precipitation of free acids)
stearate (scum)
— Synthetic detergents do not form insoluble salts of
The scum gets attached to the clothes, utensils and calcium or magnesium with hard water. Hence, lesser
even skin and thus, interferes with the cleansing ability amounts of synthetic detergents are required for
of the additional soap and makes the cleansing of washing.
clothes difficult. Moreover, large amount of soap is
wasted in reacting with calcium and magnesium ions
present in hard water. Washing powder:
— Washing powders used for washing clothes contain
— Soaps are not effective in acidic medium: In presence only about 15 to 30 percent detergents by mass. The
of hydrogen ions (H ions), i.e. in acidic medium, the remaining part is made of the following.
carboxylate ions of soap (RCOO -ion)interact with (i) Sodium sulphate and sodium silicate which keep
hydrogen ions (H+) to form undissociated (free) fatty the power dry,
acid as represented below: (ii) Sodium tripolyphosphate or sodium carbonate
C17 H 35 COO - (aq ) + H + ˆˆˆ
†
‡ˆˆ
ˆ C17 H 35 COOH which maintains alkalinity for removing dirt.
(iii) carboxymethyl cellulose (CM- cellulose) which
Carboxylate ion Carboxylic acid(Unionised)
keep the dirt particle suspended in water.
As the fatty are weak acids, so they do not get ionised (iv) Sodium perborate (a mild bleaching agent) which
and hence, micelle formation is hindered, thus, impart whiteness to the materials (clothes, etc.)
adversely affecting the cleansing property of soaps. being washed.

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 Carbon and Its Compounds

Harmful effects of Detergents

A comparative study of the soaps and detergents has SO3Na
revealed that detergents are better than soaps in many
respects. However, there is one major disadvantage
also. Detergents as pointed above are non-
Branched
biodegradable whereas soaps are biodegradable. The Chain
widespread use of detergents over the years has These chains break with difficulty by the micro-
resulted in the pollution of water obtained from rivers, organisms that are present in sewage discharge, septic
lakes and other sources. The main reason for the non- tanks or other water bodies. They make the detergent
biodegradable nature of detergents is the presence of non-biodegradable and are responsible for the pollution
branching in the hydrocarbon chains that are attached problem. Chemists have been able to synthesise
to the sulphonic acid group or its sodium salt. detergents with minimum branching in order to solve
this pollution problem.

SOLVED EXAMPLE

Example-101
(i) Which of the following salts when dissolved in water produce hard water?
(a) Calcium sulphate
(b) Magnesium bicarbonate
(c) Calcium chloride
(d) any of the above
(ii) Soaps are
(a) Phosphorous salt of long chain carboxylic acid
(b) Sodium salts of long chain carboxylic acids
(c) Potassium salts of long chain carboxylic acids
(d) Both (b) and (c)
Sol. (i) (d) (ii) (d)

Differences between soaps and synthetic detergents:

Soaps Synthetic detergents
1. Soaps are sodium salts of long chain fatty acid 1. Synthetic detergents are the sodium salts of long-chain
(carboxylic acids) benzene sulphonic acids or the sodium salt of a long-chain
alkyl hydrogen sulphate

2. The ionic part of soap is –COO–Na+ 2. The ionic part in a synthetic detergents is –SO3–Na+

3. They are prepared from animal fats 3. They are prepared from hydrocarbons or plant based oils.
extracted from coal or petroleum.

4. Their efficiency decreases in hard water 4. Their efficiency is unaffected in hard water.

5. Soaps take more time to dissolve in water. 5. Synthetic detergents dissolve faster than soaps in water

6. They are biodegradable. 6. Some synthetic detergents are not biodegradable.

7. Examples: Sodium sterate, sodium palmitate 7. Example: Sodium lauryl sulphate, sodium dodecyl benzene
sulphonate.

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Carbon and Its Compounds

EXERCISE-I
CBSE BOARD
Q.10 State the meaning of functional group in a carbon
PREVIOUS YEAR'S
compound. Write the functional group present in (i)
Q.1 Write the number of covalent bonds in the molecule of ethanol and (ii) ethanoic aicd and also draw their
ethane. structures.
[CBSE (Outside Delhi) 2014-15] [1] [CBSE (Delhi) 2013-14] [3]
Q.2 Write the name and structure of an alcohol with three
carbon atoms in its molecule. Q.11 Why homologous series of carbon compounds are so
[CBSE (Outside Delhi) 2015-16] [1] called ? Write chemical formula of two consecutive
members of a homologous series and state the part of
Q.3 A student adds a spoon full of powdered sodium these compounds that determines their (i) physical
hydrogen carbonate to a flask containing ethanoic properties, and (ii) chemical properties.
acid. List two main observations, he must note in his [CBSE (Delhi) 2012-13] [3]
note book, about the reaction that takes place. Also
write chemical equation for the reaction. Q.12 A carboxylic acid C2H4O2 reacts with an alcohol in the
[CBSE (Outside Delhi) 2015-16] [2] presence of H2SO4 to form a compound ‘X’. The alcohol
on oxidation with alkaline KMnO4 gives the same
Q.4 List two observations which you make when you add carboxylic acid, C2H4O2. Write the name and structure
a pinch of sodium hydrogen carbonate to acetic acid of (i) Carboxylic acid, (ii) alcohol and (iii) the compound
in a test tube. Write the chemical equation for the ‘X’. [CBSE (Outside Delhi) 2012-13] [3]
reaction that occurs. [CBSE (Delhi) 2014-15] [2]
Q.13 An aldehyde as well as a ketone can be represented
Q.5 What happens when acetic acid is added in a solution by the same molecular formula, say C3H6O. Write their
of Na2CO3 in a test tube ? Write the equation for
structures and name them. State the relation between
detecting the gas evolved.
the two in the language of science.
[CBSE (Outside Delhi) 2013-14] [2]
[CBSE (Outside Delhi) 2015-16] [3]

Q.6 When you add sodium hydrogen carbonate to acetic

Q.14 Complete the following chemical equations :
acid in a test tube, a gas liberates immediately with a
brisk effevescence. Name this gas. Describe the (i) CH3COOC2H5 + NaOH ¾¾
®
method of testing this gas. (ii) CH3COOH + NaOH ¾¾
®
[CBSE (Outside Delhi) 2014-15] [2]
(iii) C2H5OH + CH3COOH ¾¾¾¾¾
Conc. H 2SO4
®
Q.7 Write any three characteristics of a good fuel. [CBSE (Delhi) 2017] [3]
[CBSE (Outside Delhi) 2014-15] [3]
Q.15 Write the structural formula of ethanol. What happens
Q.8 What are homologous series of carbon compounds ? when it is heated with excess of conc. H2SO4 acid at
Write the molecular formula of two consecutive 443 K ? Write the chemical equation for the reaction,
members of homologous series of aldehydes. State stating the role of conc. H2SO4 acid in this reaction.
which part of these compounds determines their (i) [CBSE (Outside Delhi) 2017] [3]
physical and (ii) chemical properties.
[CBSE (Outside Delhi) 2013-14] [3]
Q.16 List three techniques that have been developed to
prevent pregnancy. Which one of these techniques is
Q.9 With the help of an example, explain the process of
not meant for males ? How does the use of these
hydrogenation. Mention the essential conditions for
techniques have a direct impact on the health and
the reaction and state the change in physical property
prosperity of a family ?
with the formation of the product.
[CBSE (Outside Delhi) 2017] [3]
[CBSE (Delhi) 2014-15] [3]

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 Carbon and Its Compounds

Q.17 An organic compound ‘X’ on heating with conc. H2SO4 Q.21 (a) Give a chemical test to distinguish between
forms a compound ‘Y’ which on addition of one saturated and unsaturated hydrocarbon.
molecule of hydrogen in the presence of nickel forms a (b) Name the products formed when ethane burns in
air. Write the balanced chemical equation for the
compound ‘Z’. One molecule of compound ‘Z’ on
reaction showing the types of energies liberated.
combustion forms two molecules of CO2 and three
(c) Why is reaction between methane and chlorine in
molecules of H2O. Identify giving reasons the the presence of sunlight considered a substitution
compounds ‘X’, ‘Y’ and ‘Z’. Write the chemical reaction ?
equations for all the chemical reactions involved. [CBSE (Delhi) 2015-16] [5]
[CBSE (Outside Delhi) 2012-13] [5]
Q.22 A carbon compound ‘P’ on heating with excess conc.
Q.18 State the reason why carbon can neither form C 4+ H2SO4 forms another carbon compound ‘Q’ which on
cations nor C4– anions, but forms covalent compounds. addition of hydrogen in the presence of nickel catalyst
forms a saturated carbon compound ‘R’. One molecule
Also state reasons to explain why covalent
of ‘R’ on combustion, forms two molecules of carbon
compounds:
dioxide and three molecules of water. Identify P, Q and
(1) Are bad conductors of electricity ? R and write chemical equations for the reactions
(2) Have low melting and boiling points ? involved.
[CBSE (Delhi) 2013-14] [5] [CBSE (Outside Delhi) 2015-16] [5]

Q.23 Soaps and detergents are both types of salts. State

Q.19 Explain why carbon forms compounds mainly by the difference between the two. Write the mechanism
covalent bond. Explain in brief two main reasons for of cleansing action of soaps. Why do soaps not form
lather (foam) with hard water ? Mention any two
carbon forming a large number of compounds. Why
problems that arise due to the use of detergents instead
does carbon form strong bonds with most other
of soaps.
elements ? [CBSE (Delhi) 2017] [5]
[CBSE (Delhi) 2014-15] [5]
Q.24 Why are certain compounds called hydrocarbons ?
Q.20 Both soap and detergent are some type of salts. What Write the general formula for homologous series of
is the difference between them ? Describe in brief the alkanes, alkenes and alkynes and also draw the
cleasing action of soap. Why do soaps not form lather structure of the first member of each series. Write the
in hard water ? List two problems that arise due to the name of the reaction that converts alkenes into alkanes
and also write a chemical equation to show the
use of detergents instead of soaps.
necessary conditions for the reaction to occur.
[CBSE (Outside Delhi) 2014-15] [5]
[CBSE (Outside Delhi) 2017] [5]

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Carbon and Its Compounds

EXERCISE-II
Introduction
Q.1 Carbon forms a large number of compounds because it Q.11 Coke is obtained from coal by
has (A) Cracking
(A) fixed valency (B) Fractional distillation
(B) non-metallic nature
(C) Destructive distillation
(C) high ionization potential
(D) None of these
(D) property of catenation

Q.2 Which one among them is carbon compound? Q.12 The first organic substance ever prepared in the
(A) Kerosene (B) LPG laboratory from inorganic compound is
(C) Plastics (D) All of these (A) Urea (B) Ethyl alcohol
(C) Methane (D) Ethane
Q.3 Following elements are present in both CNG and LPG
(A) Carbon and hydrogen Q.13 Which of the following halide of carbon is used as
(B) Carbon and nitrogen refrigerant
(C) Carbon and helium
(A) CCl4 (B) CF4
(D) Carbon and oxygen
(C) CH2Cl2 (D) CH2F2
Q.4 Highest electronegativity among the following is for
(A) C (B) Si Q.14 Which of the following has both s and p bond
(C) Sn (D) Pb (A) C2H2 (B) C2H4
(C) C6H6 (D) C2H6
Q.5 A covalent bond is formed by
(A) complete transfer of electrons Allotropes of carbon
(B) one sided sharing of electron Q.15 Good conductor of heat and current is
(C) mutual sharing of electron (A) Anthracite (B) Diamond
(D) all of the three above.
(C) Charcoal (D) Graphite
Q.6 Carbon tetrachloride has zero dipole moment because
of Q.16 Diamonds can be synthesized by subjecting pure
(A) Planar structure carbon to very
(B) Smaller size of C and Cl atoms (A) high pressure and temperature
(C) Regular tetrahedral structure (B) low pressure and temperature
(D) None of these (C) moderate pressure & temperature
(D) both (B) & (C)
Q.7 The nature of chemical bonding in diamond is
(A) Metallic (B) Co-ordinate Q.17 Graphite is used in nuclear reactors
(C) Covalent (D) Ionic
(A) As a lubricant (B) As a fuel
(C) As moderator (D) None of these
Q.8 Carbon reacts with strong electropositive metal oxides
to form
(A) Carbide (B) Carbonate Q.18 The forces of attraction between the various layers of
(C) Hydroxide (D) Oxide carbon atoms in graphite is
(A) Covalent bond (B) Van der walls forces
Q.9 The coal form containing maximum percentage of (C) Ionic bond (D) None of these
carbon is
(A) Lignite (B) Anthracite Q.19 Which of the following is not an allotropic form of
(C) Bituminous (D) Peat carbon?
(A) Coal (B) Fullerene
Q.10 Which of the following represents Lewis structure of
(C) Diamond (D)Graphite
N2 molecule?
´´ ´´
(A) ´
´ N º N ´´ (B) ´´ N º N ´´ Q.20 The inert form of carbon is
´´ ´´ ´´ (A) Diamond (B) Graphite
´´
(C) ´´ N - N ´´ (D) ´´ N = N ´´ (C) Coal (D) Charcoal

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 Carbon and Its Compounds

Q.21 The most reactive form of carbon is Q.32 Which of the following represent the correct order of
(A) Diamond (B) Graphite unsaturation?
(C) Coal (D) Charcoal (A) Alkanes, alkenes, alkynes
(B) Alkanes, alkynes, alkenes
Q.22 Which is the least pure form of carbon (C) Alkenes, alkynes, alkenes
(A) Graphite (B) Lamp black (D) Alkynes, alkanes, alkenes
(C) Wood charcoal (D) Animal charcoal
Q.33 The molecular formula of a methyl group is
Q.23 Which form of carbon has a two-dimensional sheet- (A) CH4 (B) CH3 (C) CH (D) CH2
like structure
(A) Coal (B) Coke Q.34 The hydrocarbon with the general formula C n H 2 n + n is
(C) Diamond (D) Graphite
an-
Q.24 (C)rystalline form of carbon having 60 carbon atoms (A) Alkane
joined together (B) Alkene
(A) polymer (C) Alkyne
(B) Buckminster Fullerene (D) unsaturated compounds
(C) coal
(D) diamond Q.35 Which of the following is an alkyne?
(A) C 6 H 6 (B) C 6 H 12
Q.25 Graphite is a soft lubricant extremely difficult to melt.
The reason for this anomalous behaviour is that (C) C 6 H 10 (D) C 6 H 14
graphite
(A) has carbon atoms arranged in large plates of rings
Q.36 (C)ompounds made up of carbon and hydrogen only
of strongly bound carbon atoms with weak interplate
are called
bonds
(A) Alkanes (B) Alkenes
(B) is a non-crystalline substance
(C) Alkynes (D) Hydrocarbons
(C) is an allotropic form of carbon
(D) has only single bonds between carbon atoms
Q.37 Total no. of C-H bond in propane will be
(A) 2 (B) 4 (C) 6 (D) 8
Q.26 Which of the following are different forms of carbon
(A) Coke (B) Charcoal
Q.38 Which of the following compounds does not contain a
(C) Carbon black (D) All of these
multiple bond?
(A) Ethane (B) Ethene (C) Ethyne (D) Benzene
Hydrocarbon
Q.27 Which of the following can have double bond
(A) C2H4 (B) C4H6 (C) C4H8 (D) C3H8 Q.39 Open-chain saturated hydrocarbons are called
(A) Paraffins (B) Alkenes
Q.28 Ethane, with the molecular formula C2H6 has (NCERT) (C) Alkynes (D) Alkyl groups
(A) 6 covalent bonds (B) 7 covalent bonds
(C) 8 covalent bonds (D) 9 covalent bonds Q.40 Compounds made up of carbon and hydrogen only are
called
Q.29 Which of the following is not hydrocarbon (A) Alkanes (B) Alkenes
(A) methane (B)Benzene (C) Alkynes (D) Hydrocarbons
(C) Naphthalene (D) Alcohol
Q.41 Which of the following formula represent alkenes?
Q.30 Which of the following compounds does not contain (A) C n H 2 n (B) C n H 2 n + 2
a multiple bond?
(C) C n H 2 n -2 (D) C n H 2 n +1
(A) Ethane (B) Ethene
(C) Ethyne (D) Benzene
Q.42 CH3 – CH – CH2 – CH2 – CH3
Q.31 Benzene with molecular formula, C6H6, has |
(A) 6 single bonds and 6 double bonds CH3
(B) 12 single bonds and 3 double bonds What is the IUPAC name of this compound?
(C) 18 single bonds only (A) 3-methyl butane (B) 1-methyl butane
(D) 12 double bonds only (C) 1-pentane (D) 2-Methyl pentane

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Carbon and Its Compounds

Q.43 The general formula of cyclic alkanes is Q.54 Which of the following set of compounds have the
(A) C n H 2 n + 2 (B) C n H 2 n -2 same molecular formula?
(A) Butane and isobutane
(C) C n H 2 n -1 (D) C n H 2 n (B) Cyclohexane and hexene
(C) Propanal and propanone
Q.44 Which of the following compounds can have a triple (D) All the three
bond?
(A) C2H6 (B) C3H4 (C) C3H8 (D) C3H6 Q.55 The functional group present in carboxylic acids is
(A) –OH (B) –CHO
Q.45 What is the total no. of H-atom in following bond line (C) –COOH (D) –CO
notation
Q.56 A carboxylic group is present in
(A) 4 (B) 8 (C) 10 (D) 12 (A) Ethylene (B) Methanoic acid
(C) Formaldehyde (D) Ethanol
Q.46 Benzene with molecular formula, C6H6, has
(A) 6 single bonds and 6 double bonds Q.57 The functional group in aldehydes is
(B) 12 single bonds and 3 double bonds
(C) 18 single bonds only (A) - CHO (B) – C = O
(D) 12 double bonds only
(C) - COOH (D) - COOR

Homologues series and Functional group

Q.58 Which one of the following compounds is an isomer of
Q.47 The functional group in aldehydes is
CH3CH2CH2CH2OH?
(A) –CHO (B) –C = O (C) –COOH (D) –COOR
O OH
|| |
Q.48 Butanone is a four-carbon compound with the (A) CH 3CH 2CH 2 - C- H (B) CH3CH 2 - CH CH3
functional group (NCERT) (C) CH3OH (D) CH3CH2CH2OH
(A) Carboxylic acid (B) aldehyde
(C) ketone (D) alcohol Q.59 Carboxylic acids are obtained from alcohols by -
(A) Oxidation (B) Reduction
Q.49 A carboxylic group is present in (C) Hydrolysis (D) Pyrolysis
(A) Ethylene (B) Methanoic acid
(C) Formaldehyde (D) Ethanol Physical and chemical properties of organic Compound
Q.60 CH2 = CH2 + Br2 (aq) ¾¾® product is
Q.50 The functional group in an alcohol is
(A) CH2BrCH2Br (B) CH3CH3
O O (C) CH3Br (D) All of these
(A) –C–O– (B) –C–OH
H Q.61 Which of the following will react with sodium metal?
(A) Ethanol (B) Ethanal
(C) –OH (D) –C=O
(C) Ethene (D) Ethane

Q.51 The functional group in methanol and methanal

Q.62 CH4 + Cl2 ¾¾¾light
® gives
respectively are:
(A) CH3Cl (B) HCl
(A) - OH ,-CHO (B) - CHO ,-OH
(C) Both (A) & (B) (D) None of these
(C) - OH ,-COOH (D) - CHO,-COOH
Q.63 Which of the following will give a pleasant smell of
ester when heated with ethanol and a small quantity of
Q.52 The next higher homologue of ethane is
sulphuric acid?
(A) Butane (B) Pentane (A) CH3COOH (B) CH2CH2OH
(C) Propane (D) Methane (C) CH3OH (D) CH3CHO

Q.53 The functional group present in ethanol is Q.64 Acetic anhydride is used in laboratory as
(A) Alcohol (B) Aldehyde (A) Solvent (B) Dehydrating agent
(C) Carboxylic acid (D) Ketone (C) Acetylating agent (D) Both (B) and (C)

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 Carbon and Its Compounds

Q.65 Which of the following will not decolourise bromine Q.76 Which of the following salts when dissolved in water
water? produce hard water?
(A) C 4 H 8 (B) C 3 H 4 (A) Calcium sulphate
(B) Magnesium bicarbonate
(C) C 3 H 8 (D) C 4 H 6 (C) Calcium chloride
(D) Any of the above
Q.66 Alcohols can be produced by the hydration of:
(A) Alkenes (B) Alkynes Miscellenous
(C) Alkanes (D) Acids Q.77 The major constituent of biogas is
(A) Propane (B) Acetylene
Q.67 Ethanol on oxidation gives (C) Methane (D) Benzene
(A) Ethane (B) Formalin
(C) Ethanoic acid (D) Methane Q.78 The major constituent of natural gas is
(A) Butane (B) Methane
Q.68 Which among the following compounds has maximum (C) Propane (D) Ethane
boiling point
(A) Butane (B) Butanol Q.79 Ethyl Ethanoate is an
(C) Butanone (D) Butanol (A) Acid (B) Alcohol
(C) Ester (D) Aldehyde
Q.69 The formula and name of the ester formed when
ethanoic acid reacts with propanol is Q.80 CO2 is called dry ice or drikold because
(A) CH3CH2COOH2CN3 (ethyl propanoate) (A) It wets the surface
(B) CH3COOCH2CH2CH3 (propyl ethanoate) (B) It does not melt
(C) CH3CH2COOCH3 (methyl propanoate) (C) At atmospheric pressure solid CO2 changes
(D) none of these directly into the gas and the liquid phase is not
formed and does not wet the surface
Q.70 CH3COOH + Na ¾¾ ® CH3COONa + Y, Y is ______ (D) It is gaseous in nature
(A) H2O (B) H2 (C) CO2 (D) NaOH
Q.81 ‘Lead pencil’ contains
Q.71 Acid + Alcohol ¾¾ (A) Pb (B) FeS
® gives
(C) Graphite (D) PbS
(A) Ester (B) Water
(C) Both (A) & (B) (D) Alkane
Q.82 The main constituent of LPG is
(A) methane (B) ethane
Q.72 Esters are
(C) propane (D) butane
(A) Sour smelling substances
(B) sweet smelling substances
(C) used in making perfumes and flavoring agents Q.83 Carborundum is
(D) both (B) and (C) (A) Al2O3 (B) SiC
(C) BF3 (D) B4C
Soap and detergent
Q.73 Soaps are prepared by alkaline hydrolysis of - Q.84 Secondary forces present in the compound are
(A) Carboxylic acids (B) Lower esters (A) van der Waals attractions
(C) Higher esters (D) None of these (B) Dipole-Dipole attractions
(C) Hydrogen bonding
Q.74 Detergents are sodium or potassium salts of long chain (D) All of these
(A) aldehydes (B) ketones
(C) carboxylic acids (D) sulphonic acids Q.85 Which of the following is used as an antiknocking
material in gasoline-
Q.75 An example of soap is: (A) TEL (B) C2H5OH
(A) C15H31COONa (B) CH3COONa (C) Glyoxal (D) Freon
(C) C6H5COONa (D) C17H35OSO3Na

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Carbon and Its Compounds

EXERCISE-III

Q.1 Carbon forms a large number of organic compounds Q.12 n-butane and isobutane are
due to (A) Alkenes (B) Alkynes
(A) catenation (C) Isomers (D) None of these
(B) tendency to form multiple bonds
(C) phenomenon of isomerism Q.13 Which of the following is aromatic
(D) all the above (A) cyclohexane (B) benzene
(C) Naphthalene (D) cyclobutane
Q.2 What is the reason for forming a strong bond by carbon
atoms? Q.14 Which one of the following compounds has linear
(A) Small size of carbon atoms planar geometry
(A) Ethylene (B) Carbon dioxide
(B) Large size of carbon atoms
(C) Borontrifluoride (D) Sulphur dioxide
(C) Can’t say
(D) None of these
Q.15 Which of the following is not a saturated hydrocarbon?
(A) Cyclohexane (B) Benzene
Q.3 The hybridisation of carbon in carbon monoxide is (C) Butane (D) Isobutane
(A) sp3 (B) sp2 (C) sp (D) dsp2
Q.16 The IUPAC name of the compound
Q.4 Carbon in CO2 is
(A) sp-hybridised (B) sp2-hybridised CH 2 = C - CH 2 - C º CH is
|
(C) sp3-hybridised (D) dsp3-hybridised CH
3
Q.5 Which is a true acid anhydride (A) 4-Methyl-2-penten-1-yne
(A) Al2O3 (B) CO (C) CaO (D) CO2 (B) 4-Methyl-4-penten-1-yne
(C) 2-Methyl-2-penten-4-yne
(D) 2-Methyl-1-penten-4-yne
Q.6 The main constituents of coal gas are
(A) CH4 + CO + H2 (B) CO2 + CO + H2
Q.17 Which of the following has the shortest carbon-carbon
(C) CO + CO2 (D) CO + N2
bond length?

Q.7 The refractive index of diamond is highest among (A) C 2 H 2

solids. Its value is (B) C 2 H 4
(A) 2.225 (B) 3.235
(C) 2.15 (D) 2.417 (C) C 2 H 6
(D) All have the same bond length
Q.8 The various physical forms of elements can exist are
called? Q.18 Which of the following has the weakest carbon-carbon
(A) Allotropes (B) Isotopes bond strength?
(C) Isobar (D) Isomers
(A) C 2 H 2
Q.9 The arrangement of carbon atoms in diamond crystals (B) C 2 H 4
(A) Octahedral (B) Trigonal
(C) Hexagonal (D) Tetrahedral (C) C 2 H 6
(D) All have the same bond length
Q.10 Which of following is not a crystalline allotropic form
of carbon? Q.19 In which compound carbon-carbon double bond
(A) Diamond (B) Graphite length is maximum
(C) Fullerene (D) Coke (A) Ethene
(B) Propene
Q.11 Diamond is not a good conductor of electricity because: (C) 2-butene
(A) It is very hard (D) 2,3-dimethyl-2-butene
(B) Its structure is very compact
(C) It is not water soluble Q.20 The general formula of an alkene is
(D) It has no free electrons to conduct electric current (A) CnH2n-2 (B) CnH2n+2 (C) CnH2n (D) CnH2n+1

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 Carbon and Its Compounds

Q.21 Which of the following is not a saturated hydrocarbon? Q.29 The IUPAC name of CH3CHO is:
(A) Cyclohexane (B) Benzene (A) Acetaldehyde
(C) Butane (D) Isobutane (B) Formaldehyde
(C) Methyl formaldehyde
Q.22 The characteristic reaction of alkanes is (D) Ethanal
(A) Addition (B) Substitution
(C) Pollymerization (D) Isomerization Q.30 The general formula of alcohols is
(A) CnH2n+2 (B) CnH2n+1OH
(C) CnH2n (D) CnH2n+1COOH
Q.23 Which of the following is not chain isomer of C5 H12
Q.31 Methane is prepared in the Laboratory by heating a
(A) (B) mixture of sodium acetate and
(A) water (B) soda lime
(C) lime stone (D) sodium carbonate

(C) (D) Q.32 The functional group of methanol and ethanoic acid
respectively are
(A) -OH, - CHO (B) - CHO, - OH
Q.24 Which of the following is position isomer of (C) -OH, - COOH (D) -CHO, - COOH

Q.33 The IUPAC name of the given compound

Cl
CH3 — CH — C = CH — CH3 is -
| |
Cl OH CH 3
(A) (B) (A) 1,2-Dimethyl-2-butenol
Cl (B) 2,3-Dimethyl-3-pentenol
(C) 3,4-Dimethyl-2-buten-4-ol
Cl (D) 3-Methyl-3-penten-2-ol

(C) (D) Q.34 IUPAC name of the given compound

Cl
OH OH
| |
CH 3 — CH 2 — C = C — CH 3 is
Q.25 IUPAC name of the given compound
(A) 2-Butene-2,3-diol
CH3 (B) 2-Pentene-2,3-diol
| (C) 2-Hexene-2,3-diol
CH3 — (CH 2 )4 — CH — C — CH 2 —CH 2 — CH 3 is
| | (D) 2-Methyl-2-butene-2,3-diol
CH3 CH 2-CH3
Q.35 Which of the following will react with sodium metal?
(A) 4-Ethyl-4,5-dimethyldecane (A) Ethanol (B)Ethanal
(B) 3,4-Dimethyl-3-n-propylnonane (C) Ethene (D) Ethane
(C) 6,7-Dimethyl-7-n-propylnonane
(D) 6,7-Dimethyl-7-ethyldecane Q.36 Which one of the following is an acyl group

Q.26 The IUPAC name of propyl chloride is O

||
(A) Chlorobutane (B) Chloropropane (A) R —O— (B) R — C— O —
(C) Chlorohexane (D) Chloroethane
O O
|| ||

Q.27 IUPAC name of this molecule is _______ (C) R — C— (D) R — C— NH —

(A) 3-methyl 1-pentene (B) 3-ethyl butane
Q.37 Which of the following will give a pleasant smell of
(C) 3-ethyl butene (D) 2-ethyl butane
ester when heated with ethanol and a small quantity of
sulphuric acid?
Q.28 In order to form branching an organic compound must
have a minimum of _____ Carbon atoms (A) CH 3 COOH (B) CH 2 CH 2 OH
(A) 3 (B) 4 (C) 5 (D) 6
(C) CH 3 OH (D) CH 3 CHO

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Carbon and Its Compounds

Q.38 The IUPAC name of the given compound Q.48 Breaking of higher hydrocarbon into lower one by
heat is called
(A) Neutralisation (B) Polymerisation
is - (C) Cracking (D) Acetylation

Q.49 Which class of organic compounds give effervescence

(A) 3-Ethyl-1-cyclobuten-2-ol with NaHCO3 solution ?
(B) 4-Ethyl-2-cyclobuten-1-ol (A) Esters (B) Alcohols
(C) 2-Ethyl-3-cyclobuten-1-ol (C) Carboxylic acids (D) Aldehydes
(D) 4-Ethyl-1-cyclobuten-3-ol
Q.50 Ethene is produced when
Q.39 The functional group is methanol and ethanal (A) Ethanol reacts with ethanoic acid in presence of a
respectively are few drops of conc. H2SO4
(A) -OH, - CHO (B) - CHO, - OH (B) Ethanol is oxidized with acidified potassium
dichromate
(C) - OH, - COOH (D) - CHO, - COOH
(C) Ethanol is heated with excess of conc. H2SO4 at
443 K
Q.40 Which of the following does not give a positive test (D) Ethanol reacts with Na Metal
with alkaline solution of potassium permanganate
(A) C10H22 (B) C6H12 Q.51 Which of the following represents an esterification
(C) C10H18 (D) C10H20 reaction?
(A) CH3COOH + C2 H5 OH ¾¾¾¾®
conc.H SO
CH3COOC2 H5 + H2 O
2 4

Q.41 Which of the following substance is added to denature

ethanol? (B) CH 3 COOC 2 H 5 + NaOH ¾¾¾Heat
® CH 3 COONa + C 2 H 5 OH
(A) Methanol (B) Benzene (C) CH 3 COONa + NaOH ¾¾¾¾® CH 4 + Na 2 CO 3
CaO, 633K

(C) Copper nitrate (D) Poison

(D) CH 4 + Cl 2 ® CH 3 Cl + HCl
Q.42 Acetic acid can be used for the preparation of
(A) Ethane (B) Methane Q.52 Wh ich one of the following acids under goes
(C) Acetone (D) All of these decarboxylation on strong heating
(A) Benzoic acid (B) Acetic acid
(C) Formic acid (D) All of these
Q.43 Heating ethanol at 443 K with excess conc. H2SO4
results in the
(A) dehydration of ethanol to give ethyne Q.53 Which of the following cannot distinguish ethanol from
ethanoic acid:
(B) dehydration of ethanol to give ethane
(A) Blue litmus (B) Sodium carbonate
(C) dehydration of ethanol to give ethene
(C) Sodium bicarbonate (D) Sodium metal
(D) both (A) and (B)
Q.54 Saponification means:
Q.44 Which of the following substances cannot be used to
(A) Acid hydrolysis of ester
distinguish ethanol from ethanoic acid?
(B) Alkaline hydrolysis of ester
(A) Na metal
(C) Esterification
(B) NaHCO3
(D) Decarboxylation
(C) hot alkaline KMnO4
(D) hot acidified K2Cr2O7 solution
Q.55 Colour of bromine is discharged by
(A) Phenol (B) Formic acid
Q.45 Physical properties of a compound depend on
(C) Ethylene (D) All of these
(A) Primary forces present in the compound
(B) Secondary forces present in the compound
Q.56 Carbon monoxide is toxic as it combines with RBC to
(C) Bond angle of the molecule
form
(D) Shape of the molecule
(A) Oxyhaemoglobin
(B) Carboxyhaemoglobin
Q.46 Ethanol on complete oxidation gives (C) Deoxyhaemoglobin
(A) CO2 and water (B) Acetaldehyde (D) None of these
(C) Acetic acid (D) Acetone
Q.57 Which one of the following is most acidic
Q.47 Benzene reacts with Cl2 to form BHC in the presence (A) 2-propanol
of (B) 2-methyl -2-propanol
(A) Ni (B) AlCl3 (C) Ethanol
(C) Sunlight (D) Zn (D) Methanol

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 Carbon and Its Compounds

Q.58 Which of the following will undergo addition reactions? Q.69 The concentration of ethanoic acid in vinegar is:
(A) C2H4 (B) C2H6 (C) CH4 (D) C3H8 (A) 10 to 15 percent (B) 5 to 8 percent
(C) 20 to 30 percent (D) 40 to 50 percent
Q.59 p-nitrophenol has higher mp than o-nitrophenol, due
to Q.70 Chlorophyll is soluble in
(A) Intermolecular hydrogen bonding (A) Benzene (B) Water
(B) Intramolecular hydrogen bonding (C) Phenol (D) Aniline
(C) Hyperconjugation
(D) Resonance Q.71 A kettle which becomes furred-up in use has inside it,
a deposit of
Q.60 The reaction, (A) Calcium carbonate
(B) Magnesium bicarbonate
2C 2 H 5 OH + 2Na ¾¾
® 2C 2 H 5 ONa + H 2
(C) Magnesium sulphate
suggests that ethanol is (D) Sodium sulphate
(A) Acidic in nature (B) Basic in nature
(C) Amphoteric (D) Neutral Q.72 If carbon dioxide is passed in excess into limewater
the milkiness first formed disappears due to
Q.61 When formic acid is heated with conc.H2SO4, the gas (A) Reversal of the original reaction
coming out is (B) Formation of volatile calcium derivatives
(A) H2 (B) CO (C) Formation of water soluble calcium bicarbonate
(C) CO2 (D) O2 (D) The solution getting heated by exothermic reaction

Q.62 In the given reaction NaH + H 2 O ¾¾

® NaOH + H 2 Q.73 Wine contains
Hydrogen undergo (A) CH3OH (B) C6H5OH
(A) Oxidation only (B) Both oxidation and reduction (C) C2H5OH (D) CH3COOH
(C) Reduction only (D) none of these
Q.74 In laboratory burners we use contain
Q.63 Methane reacts with chlorine in the presence of (A) Producer gas (B) Oil gas
sunlight. If this reaction is allowed to continue, the (C) Coal gas (D) Gobar gas
end product formed will be-
(A) Carbon tetrachloride Q.75 Oil gas is obtained in laboratory by
(B) Chloroform (A) Heating of alcohol
(C) Ethylene Chloride (B) Cracking of kerosene
(D) Chalcogens (C) Heating castor oil
(D) Heating coconut oil
Q.64 An example of soap is
(A) CH3COONa (B) CH3ONa Q.76 When wine is put in air, it becomes sour due to :
(C) C17H35COONa (D) C17H35COOC2H5 (A) bacteria
(B) oxidation of C2H5OH to CH3COOH
Q.65 Which of the following incorrect (C) virus
(A) Soaps are sodium salts of fatty acid (D) formic acid formation
(B) Detergents are salts of sulphonic acid
(C) Detergents work well with soft water Q.77 The anhydride of carbonic acid H2CO3 is
(D) Soap work well with hard water (A) C2O2 (B) CO2
(C) CO (D) Na2CO3
Q.66 Detergents are sodium or potassium salts of long chain
(A) aldehydes (B) ketones Q.78 Ethyl alcohol is ______ to litmus:
(C) carboxylic acids (D) sulphonic acids (A) acidic (B) basic
(C) neutral (D) none of these
Q.67 Soaps are prepared by alkaline hydrolysis of-
(A) Carboxylic acids (B) Lower esters Q.79 Widespread deaths due to liquor poisoning is because
(C) Higher esters (D) None of these of
(A) Presence of bad compound in liquor
Q.68 The acid present in vinegar is (B) Presence of Methanol
(A) CH3COOH (B) HCOOH (C) presence of Ethanol
(C) CH3CH2COOH (D) CH3CH2CH2COOH (D) Eermentation

1-157
Carbon and Its Compounds

Q.80 The greatest percentage of CO is in Q.90 A Gaseous fuel which does not contain carbon
(A) Coal gas (B) Producer gas monoxide is
(C) Water gas (D) Oil gas (A) coal gas (B) natural gas
(C) water gas (D) producer gas
Q.81 Most favourable conditions for the alcoholic
fermentation of sugar are Q.91 Acid chloride can be reduced into
(A) High concentration of solution, low temperature, (A) Aldehyde (B) Primary alcohol
plenty of air supply (C) Ketone (D) Any one of these
(B) Low concentration of sugar solution, moderate
temperature, absence of air Q.92 Acetylation reaction with acetyl chloride is carried
(C) Low concentration of sugar solution, low out in the presence of which solvent
temperature, plenty of air (A) HOH (B) CH3OH
(D) None (C) Pyridine (D) Any one of these

Q.82 CO reacts with chlorine in presence of sunlight to Q.93 The metal which does not form a polynuclear carbonyl
give is
(A) COCl2 (B) CO2 (A) Sodium (B) Manganese
(C) CCl4 (D) CHCl3 (C) Iron (D) Cobalt

Q.83 While cooking. If the bottom of the vessel is getting Q.94 Animal charcoal is used for decolourisation of sugar
blackened on the outside, it means that (NCERT) because
(A) the food is not cooked completely (A) It oxidises coloured material
(B) the fuel is not burning completely (B) It reduces coloured material
(C) the fuel is wet (C) It converts coloured material into colourless
(D) the fuel is burning completely (D) It absorbs coloured material

Q.84 The chemical added to LPG to help detect it’s leakage Q.95 Wood charcoal is used in gas masks because it
is (A) Is poisonous (B) Liquifies gas
(A) isobutene (B) ethylamine (C) Is porous (D) Adsorbs gases
(C) methyl mercaptan (D) ethyl mercaptan
Q.96 Which gas is essential constituent of almost all fuel
Q.85 The fraction by volume of carbon monoxide in gases:
producer gas is about (A) CO2 (B) N2
(A) 1/2 (B) 1/3 (C) CO (D) H2O
(C) 1/4 (D) 2/3
Q.97 Hofmann rearrangement is given by
Q.86 Carbon monoxide on heating with sulphur gives (A) Amide (B) Acid chloride
(A) COS (B) SO2 (C) Acid anhydride (D) All of these
(C) SO3 (D) None of these
Q.98 Which one of the followin g compounds gives
Q.87 Synthesis gas is a mixture of carboxylic acid with HNO2
(A) Steam and carbon monoxide
(B) Carbon monoxide and nitrogen
O
||
(C) Hydrogen and carbon monoxide (A) C6 H 5 — C— Cl
(D) Hydrogen and methane
(B) C6H5CONH2
Q.88 Carbogen is O O
|| ||
(A) Mixture of O2 + 5-10% CO2
(B) Used by pneumonia patients for respiration
(C) C6 H 5 — C— O — C— CH 3
(C) Used by victims of CO for respiration
(D) All of these (D) CH3COOC2H5

Q.89 Sodium carbonate is Q.99 The major constituent of biogas is

(A) Efflorescent (B) Deliquescent (A) Propane (B) Acetylen
(C) Hygroscopic (D) Oxidant (C) Methane (D) Benzene

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 Carbon and Its Compounds

Q.100 Which one of the following dibasic acids on strong Q.111 What happens when steam is passed over red hot
heating gives CO2, CO and HOH carbon
(A) Formic acid (B) Oxalic acid
(A) C + 2H2O ¾¾
® CO2 + 2H2
(C) Malonic acid (D) Succinic acid
(B) C + H2O ¾¾ ® CO + H2
Q.101 mp of which compound is maximum (C) Water vapour dissociates into H 2 and O 2
(A) chlorobenzene (D) None of these
(B) o-dichlorobenzene
(C) m-dichlorobenzene Q.112 H2SO4 is not used for the preparation of CO2 from
(D) p-dichlorobenzene marble chips because
(A) it does not react
Q.102 Which of the following gives silver mirror test (B) huge amount of heat is evolved
(A) HCOOH (C) the reaction is vigorous
(B) COOH–COOH (D) calcium sulphate is sparingly soluble and get
(C) Citric acid deposited on marble chips and stops the reaction
(D) Formic and tartaric acid
Q.113 Benzoic acid on treatment with hydrazoic acid in the
Q.103 The hydrocarbon that is used for ripening fruits presence of concentrated sulphuric acid gives
(A) ethane (B) ethene (A) Benzamide (B) Sodium benzoate
(C) acetylene (D) none of these (C) Aniline (D) C6H5CON3

Q.104 Which of the following is an example of fossil fuel Q.114 Hunsdiecker reaction is given by
(A) Coal gas (B) Coke (A) Aldehydes (B) Ketones
(C) Natural gas (D) Producer gas (C) Carboxylic acids (D) All of these

Q.105 Carbon dioxide is isostructural with Q.115 Rochelle salt is salt of

(A) HgCl2 (B) SnCl2 (A) Tartaric acid (B) Citric acid
(C) C2H2 (D) NO2 (C) Oxalic acid (D) Salicylic acid

Q.106 Producer gas is a mixture of Q.116 The ability of a substance to assume two or more
(A) carbon monoxide and nitrogen gas crystalline structures is called
(B) carbon monoxide and hydrogen gas (A) Isomerism
(C) carbon monoxide and water vapour (B) Amorphism
(D) carbon monoxide and nitrous oxide (C) Polymorphism
(D) Isomorphism
Q.107 Methane is a major constituent of
(A) Coal gas (B) Water gas Q.117 Which of the following is incombustible
(C) Petroleum (D) Biogas (A) Carbon monoxide
(B) Hydrogen
Q.108 The chemical formula of phosgene or carbonyl (C) Carbon monoxide
chloride is (D) Wax
(A) PH3 (B) COCl2
(C) POCl3 (D) PCl3 Q.118 Carbon dioxide is isostructural with:
(A) HgCl2 (B) SnCl2 (C) C2H2 (D) NO2
Q.109 The gas used in the welding and cutting of metals is
(A) ethane (B) ethene Q.119 While cooking, if the bottom of the vessel is getting
(C) ethyne (D) propene blackened on the outside, it means that (NCERT)
(A) the food is not cooked completely
Q.110 The major constituent of natural gas is (B) the fuel is not burning completely
(A) Butane (B) Methane (C) the fuel is wet
(C) Propane (D) Ethane (D) the fuel is burning completely

1-159
Carbon and Its Compounds

EXERCISE-IV

NTSE STAGE-1 Q.10 The IUPAC name of

PREVIOUS YEAR'S CH 3CH 2 - CH 2 - CH - CH 2 CH 2 CH 3 is :
Q.1 An alkyne has 4 numbers of hydrogen atoms. What |
will be the number of carbon atoms in it ? CH = CH 2
[NTSE Rajasthan 2014-15] [NTSE AP 2014-15]
(A) Two (B) Three (A) 4-ethelene-1-heptane
(C) Four (D) Five (B) 3-propyl-hex-1-ene
(C) 4-propyl-hex-6-ene
Q.2 Identify the final product Z in the series of chemical (D) 3-propyl-1-heptane
reactions : [NTSE Odisha 2014-15]
CH3CN ¾¾¾¾® X ¾¾¾ Q.11 _____is the pollutant released from air-conditioner :
NaIC 2 H5 OH HNO2
® Y ¾¾¾¾
Cu /573K
®Z
[NTSE Tamilnadu 2014-15]
(A) CH3COOH (B) CH3CH2NHOH (A) Chlorofluoro carbons
(C) CH3CONH2 (D) CH3CHO (B) Carbodioxide
(C) Methane
Q.3 The number and type of bonds between carbon atoms (D) Carbon-monoxide
in CaC2 are : [NTSE Odisha 2014-15]
(A) One sigma (s) and one pi (p) bond Q.12 Which one of the following compound will give
(B) One sigma (s) and two pi (p) bonds addition reaction : [NTSE Chhattisgarh 2014-15]
(C) One sigma (s) and one half pi (s) bond (A) CH4 (B) C2H6
(D) Two sigma (s) bonds and one pi (s) bond (C) C2H4 (D) C3H8

Q.4 On heating an aliphatic primary amine with chloroform Q.13 The functional group of carboxylic acid is :
and ethanolic potassium hydroxide the organic [NTSE Chhattisgarh 2014-15]
compound formed is : [NTSE Bihar 2014-15] (A) –OH (B) –CHO
(A) An alkyl cyanide (B) An alkyl isocyanide (C) > C (D) –C – OH
(C) An alkanol (D) An alkanediol || ||
O O
Q.5 The IUPAC name of C6H5CN is :
[NTSE Bihar 2014-15] Q.14 Which statement is correct regarding ethane, ethene
(A) Pheny Cyanide (B) Phenylacetonitrile and ethyne ? [NTSE WB 2014-15]
(C) Benzene Cyanide (D) Benzonirile (A) Ethyne is weakest acid and C–H bond length is
longest
(B) Ethyne is strongest acid and C–H bond length is
Q.6 Which of the following aldehyde undergo cannizzaro
shortest
reaction ? [NTSE Jharkhand 2014-15]
(C) Ethane is strongest acid and C–H bond length is
(A) C3H7CHO (B) C6H5CHO longest
(C) CH3CHO (D) CH3CH2CHO (D) Ethene is the strongest acid and C–H bond length
is shortest
Q.7 Which of the following has shortest carbon-carbon
bond length ? [NTSE Delhi 2014-15] Q.15 1 mole of a hydrocarbon ‘X’ undergoes complete
(A) C2H2 (B) C2H4 saturation with 1 mole of hydrogen in presence of a
(C) C2H6 (D) C6H6 heated catalyst. What would be the formula of ‘X’?
[NTSE WB 2014-15]
Q.8 Which of the following may be isomer of aldehyde (A) C2H6 (B) C2H2
having general formula CnH2nO ? (C) C5H10 (D) C7H16
[NTSE Delhi 2014-15]
(A) Alcohol (B) Ether Q.16 Which of the following represent saponification
(C) Ester (D) Ketone reaction ? [NTSE Chandigarh 2014-15]
(A) CH3COONa + NaOH ¾¾¾ CaO
® Na2CO3
Q.9 How many number of ‘sigma’ bonds are present in (B) CH3COOH+C2H5OH ¾¾¾H 2SO4
® CH3COOC2H5+H2O
CH3 – C º N ? [NTSE AP 2014-15] (C) 2CH3COOH + 2Na ® CH3COONa + H2
(A) 4 (B) 3 (D) CH3COOC2H5 + NaOH ® CH3COONa + C2H5OH
(C) 2 (D) 5

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 Carbon and Its Compounds

Q.17 What is the amount of water produced by the complete

combustion of 16 gm of methane ? CH3
|
[NTSE Odisha 2015-16] Q.27 Write the IUPAC name of CH3 - C - CH 3 :
(A) 16 gm (B) 18 gm |
(C) 32 gm (D) 36 gm CH3

Q.18 The correct order of acid strength is : [NTSE MP 2015-16]

[NTSE Odisha 2015-16] (A) Neo-pentane (B) 2, 2-dimethyl propane
(A) HCOOH > C6H5COOH > CH3COOH (C) 2-methyl butane (D) 2, 3-dimethyl propane
(B) C6HCOOH > HCOOH > CH3COOH
(C) CH3COOH > HCOOH > C6H5COOH Q.28 What is the structure of aldehyde function group ?
(D) C6H5COOH > CH3COOH > HCOOH [NTSE MP 2015-16]
|
Q.19 Reaction of water with aluminium carbide gives a (A) –C = O (B) ® O = C - H
colourless gas. The gas is : [NTSE Odisha 2015-16]
(A) Methane (B) Acetylene O O
|| ||
(C) Ethane (D) Propane (C) -C - H (D) -C - O - H
Q.20 Which of the following is a natural polymer ?
[NTSE Odisha 2015-16] Q.29 Which of the following organic molecules does not
(A) Cellulose (B) Teflon have carbon-carbon double bond :
(C) Nylon (D) Terylene [NTSE Odisha 2016-17]
(A) C3H4O (B) C3H4O2
Q.21 Artificial soap is : [NTSE UP 2015-16] (C) C3H8O (D) C6H6O
(A) Sodium stearate (B) Iloril sulphuric acid
(C) Iloril alcohol (D) Sodium Iloril sulphate Q.30 Ethane with the molecular formula C2H6 has :
[NTSE Jharkhand 2016-17]
Q.22 Deuterium is an isotope of......... : (A) 6 Covalent bonds (B) 7 Covalent bonds
[NTSE Telangana 2015-16] (C) 8 Covalent bonds (D) 9 Covalent bonds
(A) Carbon (B) Hydrogen
(C) Cobalt (D) Cesium Q.31 The molecular formula of Sucrose is :
[NTSE MP 2016-17]
Q.23 A mixture of CH4, C2H4 and C2H2 is passed through a (A) CH2O (B) C6H12O6
basic copper (I) chloride solution. Whgich gas/gases (C) C12H22O11 (D) CH3COOH
will come out ? [NTSE WB 2015-16]
(A) Whole mixture (B) CH4 and C2H2 Q.32 Chemical formula of Teflon is : [NTSE MP 2016-17]
(C) CH4 and C2H4 (D) C2H4
æ CH ö
Q.24 Consider the chemical formulae CH3COOH and (A) (–CF2–CF2–)n (B) çç ........CH 2 - | ÷÷
è Cl øn
HCOOCH3 and choose the incorrect statement :
[NTSE Delhi 2015-16] (C) (–CH2–CH2–)n (D) None of these
(A) Both have the equal boiling point
(B) Both have the equal molecular weight Q.33 In the following structural formulae one IUPAC name
(C) Both have the equal number of covalent bonds is incorrect. Identify it :
(D) Both are not the same compound [NTSE Maharashtra 2016-17]
(A) CH3 - CH 2 - C - CH3 - Bu tan one
Q.25 What is the IUPAC name of the following compound?
||
[NTSE Maharashtra 2015-16] O
(A) Pent-2-ene (B) Pent-1-ene
(C) Pent-3-ene (D) 1-methyl-but-2-ene H
|
Q.26 Formula of Teflon is : [NTSE MP 2015-16] (B) CH3 - CH 2 - C = O. Pr opanal

é........CH 2 - CH ù O
(A) (–CH2–CH2–)n (B) ê | ún ||
êë Cl úû (C) CH3 - CH 2 - C - OH.Ethanoic acid
(C) (–CF2–CF2–)n (D) None of these (D) CH3–CH2–CH2–CH2–OH–Butanol

1-161
Carbon and Its Compounds

Q.34 What is the IUPAC name of the following compounds? Q.41 Teflon is a polymer of which of the following monomer
[NTSE Chhattisgarh 2016-17]
CH 3 - CH 2 - CH - C 3 H 7 (A) Mono fluoro ethene (B) Tetra fluoro ethene
|
Cl - C - Cl (C) Difluro ethene (D) Tri fluoro ethene
|
C2 H 5 Q.42 Identify X in the following reaction :

[NTSE Maharashtra 2016-17] CH3–CH2–OH ¾¾¾¾

Hot, conc.
® (X) + H2O
H2 SO 4
(A) 4-Ethyl-3, 3-dichloro heptanes [NTSE Rajasthan 2016-17]
(B) 4-Ethyl-3, 3-dichloro hexane (A) Ethane (B) Methane
(C) 4-Ethyl-3-chlorohexaneq (C) Ethene (D) Ethanol
(D) 3, 3-dichloro-4-butyl heptanes

Q.35 Match the following : [NTSE AP 2016-17] O O

Compound formula Class of the compound
Q.43 IUPAC name of H – C – C – H
(a) C2H4 (i) Alkane
(b) C7H12 (ii) Alkene [NTSE Delhi 2016-17]
(c) C13H28 (iii) Alkyne (A) Oxoethanal (B) Glyoxal
(C) Ethanedial (D) Ethanedione
(d) C5H10 (iv) Possible cyclo alkane
(A) a-iii, b-ii, c-i, d-iv (B) a-ii, b-iii, c-i, d-iv
Q.44 Out of the following pair of compounds the unsaturated
(C) a-iv, b-iii, c-ii, d-i (D) a-iii, b-ii, c-iv, d-i
compounds are- [NTSE Panjab 2016-17]
(A) C2H6 and C4H6 (B) C2H12 and C5H12
Q.36 Graphite is very soft as compared other substances
(C) C4H6 and C6H12 (D) C2H6 and C4H10
because...... : [NTSE AP 2016-17]
(A) Carbon atoms are arranged in such a way that they
Q.45 Which of the following is having maximum double
form flint layers bonds- [NTSE Panjab 2016-17]
(B) Carbon atoms are arranged in such a way that they (A) Propanone (B) Benzene
form flint layers (C) Propene (D) Propanol
(C) Linkages between atoms within layer of graphite
are weak Q.46 In the presence of concentrated sulphuric acid, acetic
(D) Linkages between atoms of the layers are weak acid react with ethyl alcohol to produce
[NTSE UP 2016-17]
Q.37 Property of self-combination of the atoms of the same (A) aldehyde (B) alcohol
element to form long chains is known as : (C) ester (D) carboxylic acid
[NTSE Chandigarh 2016-17]
(A) Protonation (B) Carbonation Q.47 The allotrope of Carbon in amorphous form among the
(C) Oronation (D) Catenation following is____ [NTSE Telangana 2016-17]
(A) Diamond (B) Graphite
Q.38 Hydrocarbon 2-methylbutane is an isomer of : (C) buckminsterfullerene (D) Coal
[NTSE Chandigarh 2016-17]
(A) n-pentane (B) n-butane Q.48 Which of the following hydrocarbons have acidic
(C) propane (D) iso-butane hydrogen? [NTSE Haryana 2016-17]
(A) CH2 = CH2 (B) HC = CH
Q.39 Which of the following hydrocarbon have a triple bond: (C) CH4 (D) C6H6
[NTSE Chhattisgarh 2016-17]
(A) C2H6 (B) C3H4 Q.49 An organs compound butanal has functional group:
(C) C3H6 (D) C3H6 [NTSE Haryana 2016-17]
(A) Aldehyde (B) Ketone
Q.40 The compound obtained by reaction of C2H5OH with (C) Alcohol (D) Amine
concentrated H2SO4 at 443 K is :
Q.50 NaOH reacts with CO at 200°C and 5 atmospheric
[NTSE Chhattisgarh 2016-17]
pressure to give [NTSE Bihar 2016-17]
(A) C2H4 (B) CH3CHO
(A) CH3CH2COONa (B) C6H5COONa
(C) CH3COOH (D) CH3COOH3
(C) HCOONa (D) CH3COONa

1-162
 Carbon and Its Compounds

NTSE STAGE-2 Q.5 The reaction between carbon and oxygen can be
PREVIOUS YEAR'S represented as :
C(s) + O2(g) ® CO2(g) + heat
Q.1 Match the following : In which of the following type(s), the above reaction
can be classified ? [NTSE 2013]
I. Combustion reaction
Contains repeating
i. Acrylic a. II. Displacement reaction
ester units
III. Endothermic reaction
Used for making IV. Combination reaction
ii. Cellulose b.
sweaters (A) I and III (B) I, III and IV
Made up of large (C) I and IV (D) I only
iii. Polythene c. number of glucose
units Q.6 An organic compound is a clear liquid having a molecular
Used for making formula C4H8O. It has an open chain structure. Without
iv. Terylene d.
electrical switches any carbon-carbon double bond. The compound can
be : [NTSE 2014]
Used for
e. (a) An alcohol (b) An ester
manufacturing toys
(c) An aldehyde (d) A ketone
(A) a and b (B) c and d
Which of the following is the correct matching ?
(C) b and d (D) d and a
[NTSE 2010]
(A) i-b, ii-a, iii-d, iv-c
(B) i-b, ii-c, iii-e, iv-a Q.7 A hydrocarbon has a molecular formula as C6H12. It
(C) i-d, ii-b, iii-e, iv-a does not react with hydrogen to give C6H14 nor does it
(D) i-d, ii-c, iii-b, iv-e react with chlorine to give C6H12Cl2. The hydrocarbon
C6H12 is : [NTSE 2014]
Q.2 Which of the following will be suitable for coating dress (a) A saturated hydrocarbon
materials of fire-men ? [NTSE 2011] (b) An unsaturated hydrocarbon
(A) Nylon (c) An open chain hydrocarbon
(B) Polyester (d) A cyclo-alkane
(A) a and b (B) c and d
(C) Melamine
(C) d and b (D) a and d
(D) Acrylic
Q.8 The total number of electrons and the number of
Q.3 Match each item in column I with appropriate item of electrons involved in the formation of various bonds
column II : [NTSE 2012] present in one molecule of propanal (C2H5CHO) are
Column I Column II respectively : [NTSE 2015]
(A) 32 and 20 (B) 24 and 20
I. Metamine A. Non-sticking cookwares
(C) 24 and 18 (D) 32 and 18
II. Nylon B. Easily biodegradable
III. Teflon C. Appears like silk Q.9 The number of structural isomers of the compound
IV. Cotton D. Flame resistant having molecular formula C4H9Br is : [NTSE 2015]
Select the correct alternative : (A) 3 (B) 5
(A) I-D, II-C, III-A, IV-B (C) 4 (D) 2
(B) I-A, II-B, III-C, IV-D
Q.10 What mass of CO2 will be formed when 6g of carbon is
(B) I-B, II-A, III-D, IV-C burnt in 32 g of oxygen ? [NTSE 2015]
(D) I-C, II-D, III-B, IV-A (A) 38 g (B) 12 g
(C) 26 g (D) 22 g
Q.4 Which of the following pairs of compounds of carbon
will undergo combustion as well as addition reaction? Q.11 The molecular formula of carboxylic acid that differs
from the rest is : [NTSE 2016]
(A) CH4 and C2H6 [NTSE 2013]
(A) C13H26O2 (B) C2H4O2
(B) C2H6O and C3H8O
(C) C9H18O2 (D) C7H12O2
(C) C2H4O2 and C3H6O
(D) C2H2 and C3H6

1-163
Nutrition

Nutrition 4
Living things are complex organizations of molecules, (A) Autotrophic :
which perform certain life processes such as growth, (Auto = self, trophic = food) It is a mode of nutrition in
metabolism, reproduction,etc., that distinguish them which organisms prepare their own food. Inorganic
from non-living matter. molecules like CO2 and H2O are converted into organic
Life processes are those basic functions of living molecules like carbohydrates in the presence of
beings which are essential for their survival. They sunlight and chlorophyll. e.g. Green plants. Autotrophs
are the same in all types of living forms whether are further categorized. as :
unicellular or multicellular, plants or animals. Energy (i) Photoautotroph : Those which utilize sunlight for
is required by all of them. preparing their food
Living organisms exhibit many activities like:- (ii) Chemoautotroph : Those which utilize chemical
(i) Nutrition energy for preparing their food.
(ii) Respiration
(iii) Transport of Material (i) Photoautotrophs:-
(iv) Blood Circulation Green plants synthesize food through the process of
(v) Excretion photosynthesis, using simple raw material like water,
(vi) Reproduction CO2 in the presence of sunlight,chlorophyll and
(vii) Control and Coordination chloroplasts. Chlorophyll present in the chloroplast
of green plants are the sites of photosynthesis.
NUTRITION : Chlorophyll
6CO 2 + 12H 2 O+ Light energy ¾¾¾¾¾ ®
“Nutrition” is a process of intake as well as utilization ( Carbon dioxide ) ( Water )
of nutrients by an organism. It is the process of
C6H12O6 +6O2 + 6H2O
breakdown of nutrients into smaller molecules and
Since autotrophic plants are able to produce food for
their absorption. Food provides us nutrition and
others they are also known as producers. Because
energy. It contains different types of nutrients in
the autotrophic nutrition is the characteristic of plants
varying amounts according to the need of our body.
therefore it is also called holophytic nutrition.
Nutrients :
(b) Chemoautotrophs:-
These are the substances required by our body for its
Some non green bacteria like sulphur bacteria use
growth, repair, work and maintenance of the body.
chemical energy to manufacture their food. This
Different types of nutrients are carbohydrates, fats,
energy is derived from chemical reactions occuring in
proteins, vitamins, mineral etc. Our daily energy need
the bacteria. This process is called chemosynthesis.
may vary according to our occupation, age, sex and
Chemosynthetic bacteria do not require light as the
under some specific conditions.
source of energy.

Modes of nutrition:- Mainly two modes of nutrition 2H 2S + O 2 ® 2H 2 O + S + 126 K.Cal

are there on the basis of the method of food 2S + 2H 2 + O 2 ® 2H 2SO 4 + 141.8K.Cal
procurement. (B) Heterotrophic :
(Hetero = different ; trophic = food) It is a mode of
nutrition in which organisms derive their food from
some other animals or plants. They cannot prepare
their own food e.g. human being.
Heterotrophs are further categorized depending on the
nature of food they consume :
(i) Herbivores : Animals which eat only plants, e.g.
Cow, goat etc.

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(ii) Carnivores : They feed on flesh of other animals, light energhy

Carbon dioxide + water¾¾¾¾¾ ® Glucose + Oxygen
e.g. Lion, vulture etc. chlorophyll

(iii) Omnivores : T hey feed on plants and animals The overall equation of photosynthesis is :
both e.g. Dog, human etc. Chlorophyll
(iv) Detritivores : Feed on detritus or dead organic 6CO 2 + 12H 2 O ¾¾¾¾¾
Solar energy
® C6 H12 O6 + 6H 2 O + 6O 2 ­
remains, e.g. Earthworm etc. The steps involved in photosynthesis are:
(v) Sanguivorous : Feed on blood e.g. Leech, female 1. Absorption of solar (light) energy by chlorophyll.
mosquito etc. 2. Conversion of light energy into chemical energy
(vi) Frugivorous : Feed on fruits, e.g. Parrot etc. and also splitting of water into oxygen and hydrogen
(vii) Insectivores : Feed on insects, e.g. Bats etc. by light energy.
3. Reduction of CO2 to carbohydrates by hydrogen by
(C) On the Basis of Mode of Feeding, Organisms are utilising chemical energy.
Categorised As : Sunlight :
(i) Holozoic : They ingest mostly solid but sometimes • For plants sun is the basic source of radiant energy.
liquid food. e.g., Amoeba, human etc.
(ii) Saprotrophic : they absorb organic matter from • Plants utilize the light in the visible region of solar
dead and decaying organisms with the help of their spectra (electromagnetic spectrum) which comes
enzymes. e.g., Bacteria, fungi etc. under the range of 390 nm - 780 nm.
(iii) Parasitic : They derive their nutrition from other • Visible region consists of white light which is a
living plants or animals e.g. Plasmodium. round mixture of 7 lights of different wavelengths.
worms etc.
(iv) Mutualistic nutrition:- The mutualistic nutrition
can be defined as the interdependent nutrition in
which each organism is dependent mutually on
the other.
Ex. The lichens share mutualistic nutrition between a
fungus and a cynobacterium. 380 nm 760 nm
Nutrition can be divided into two
categories on the basis of occurrence • Maximum photosynthesis occurs in red region.
¯
• There is minimum photosynthesis in green region
because green parts of plants reflect whole of the
Nutrition in plants Nutrition in animals green light.
Chlorophyll :
Autotrophic Nutrition in Plants :- These are the green pigments present in chloroplast.
Plants As Autotroph:- They are found in green leaves in a maximum amount
Most plants can prepare their food from inorganic as well as in other green aerial parts of plant. There are
substance. Plants are therfore, called autotrophs six different types of chlorophyll, they are chlorophyll
(auto Þ self; troph Þ food). They make their food by a,b,c,d,e and bacteriochlorophyll, amongst them
the process of photosynthesis. chlorophyll a and chlorophyll b are the most commonly
NUTRITION IN PLANTS occurring chlorophylls.
The process of absorption and conversion of light • Besides chlorophyll certain other pigments are also
energy into chemical energy by green plants is called present in plants like.
as photosynthesis. (i) Carotenes : Orange in colour e.g. Carrot.
The synthesis of organic compound like glucose from (ii) Xanthophylls : Orange yellow in colour e.g. Maize.
simple inorganic compound like CO2 & H2O by the (iii) Phycobilins : Different colour like red, violet e.g.
cell of green plants having chlorophyll in the presence Blue-green algae, brown algae etc.
of sunlight is called photosynthesis. Raw Materials of Photosynthesis :
Photosynthesis : (i) Carbondioxide : Terrestrial plants obtain carbon
Photosynthesis can also be defined as the process by dioxide from the atmosphere through the small
which green plants manufacture carbohydrates from openings present on leaves called as stomata.
atmospheric carbon dioxide and water from the soil, in ‘Stomata’ are the small pores present on the surface of
the presence of sunlight. leaves. They help in exchange of gases and water.

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Stomata opening is guarded by the presence of guard • Chloroplast also have variable shapes, for example
cells (kidney shaped). Aquatic plants obtain CO2 cup shaped, ribbon shaped etc. in algae while it is
dissolved in water through their general body surface discoidal in higher plants.
so they perform more photosynthesis than terrestrial
plants.

(ii) Water : Plants absorb water from the soil by the

process of osmosis. This water is transported to leaves
by a special type of tissue called as xylem.
• Plants utilize carbon dioxide during
photosynthesis, the intensity of light at which
amount of CO 2 used during photosynthesis
becomes equal to the amount of CO2 released
during respiration by plants in called as
Compensation point.
• Compensation point occurs at low light intensity
that is during morning and during evening hours.
Site of Photosynthesis :
Site of photosynthesis is different in prokaryotes and
eukaryotes.
• In prokaryotes : Photosynthesis occurs in lamellar
chromatophores.
• In eukaryotes : Photosynthesis occurs in
chloroplast.
• Exception : Fungi ( It lacks chlorophyll so no
photosynthesis occurs here). • A typical structure of chloroplast is a double
• In higher plants chloroplast is the main site of membranous structure having two parts.
photosynthesis. (i) Grana : It is a lamellar system consisting of stacks of
• Chloroplast is also called as green plastid. granum lamella each bounded by a membranous box
• Plastid was first observed by Haeckel. called as thylakoid. They are 40 - 60 per cell. Number
• Plastids are of 3 different types on the basis of of thylakoids per grana is 50 or more Chlorophyll
pigments present in them. molecules are found inside the thylakoid membrane
(i) Leucoplast : White in colour, found in underground where they trap solar energy in the form of small energy
parts, lacks and coloured pigment. Helps in storage packets called ‘photon’ or ‘quanta’. Grana are
of protein (Aleuroplst), oil (Elaioplast), starch interconnected to each other by a channel called as
(Amyloplst) stroma lamellae or Fret’s channel.
(ii) Chromoplast : Colour other than green found in (ii) Stroma : It is a non pigmented proteinaceous matrix in
aerial parts on the plants which grana remain embedded. It contain enzymes for
(iii) Chloroplast : Contain green pigment, called as dark reaction.
chlorophyll. Mechanism of Photosynthesis :
• Chloroplast was discovered by Schimper. (A) Light reaction :
• Number of chloroplasts is variable in different • It is also called as photochemical process.
species of plants. • It was discovered by ® ‘Robert Hill’ therefore it
• In lower plants like algae they are 1 or 2 number. is also called as Hill’s reaction/
• In higher plants their number varies from 40 -100 •Site : Grana of chloroplast.
per palisade cell or more. • Raw materials : Light and water.

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• Regulation : This process is regulated by some other factor such as carbon dioxide becomes
chlorophyll molecules. limiting. An increase in carbon dioxide concentration
• It consist of 3 steps : at this point increases the rate of photosynthesis which
(i) Photo excitation of chlorophyll molecule : During reaches an optium, till another factor becomes limiting.
this process chlorophyll molecule receives sunlight Extremely intense light may have an inhibitory effect
in the form of small energy bundles called as on photosynthesis.
photons and become excited to higher energy level. (ii) Quality of light:- The rate of photosynthesis is
higher in white light than in light of any particular
colour. (Light of one particular wavelength is called
monochromatic light.)
(ii) Photolysis : It is also called as photoxidation of However, maximum photosynthesis takes place in the
water, this takes place in presence of Mn +2 and Cl- red part of the spectrum followed by blue light and
ions. green light in decreasing order.
Temperature: The rate of photosynthesis increases
¾® 4H + + O 2 + 4e -
2H2O ¾ with rising temperature. Above 40°C, there is
O2 is liberated as by product and H+ ions are decrease in photosynthesis.
used for reduction of NADP Optimum range = 25°C to 30°C.
The temperature range over which plants can
2NADP + 4H+ ¾ ¾® 2NADPH2
photosynthesize is very wide. Certain bacteria can
(iii) Photophosphorylation : During this process photosynthesize even at 70°C while some coniferous
ATP are produced. It takes place in trees can photosynthesize at –6°C or below.
quantasomes. Carbon dioxide: The rate of photosynthesis
Mg+2 ions and inorganic phosphate is required increases with increasing CO2 concentration of the
to convert ADP ¾® ATP, ADP + iP ¾® ATP. environment. At very high concentration,
(B) Dark reaction : photosynthesis decreases as it causes closing of
• It is also called as thermo chemical reaction. stomata resulting in inhibition of gaseous exchange.
• It was discovered by Melvin calving and benson Its concentration affects the rate of photosynthesis
therefore it is also called as Calving cycle to a great extent. Since its concentration in the
Site = Stroma of chloroplast. atmosphere is very low (0.03%), it acts as a limiting
• Raw materials : They require CO2, NADPH2, ATP factor in natural photosynthesis. At optium
and Enzymes. temperature and light intensity, if the carbon dioxide
• Regulated by : Light reaction and enzymes. supply is increased, the rate of photosynthesis may
• It involves three basic steps : increase ten to thirty times the normal rate.
(i) Carboxylation : In this step CO2 is captured by CO2 Water: The rate of photosynthesis is lowered if the
acceptors like RUBP (C3 Plants) PET (C4Plants) with plants are inadequately supplied with water. Decrease
the help of carboxylase enzyme i.e. RUBISCO & PEPCO in water causes closure of stomata.
respectively.
(ii) Sugar Synthesis : This phase cap true CO 2 is Internal Factors :-
assimilated into glucose in the presence of = Chlorophyll content: Chlorophyll is essential for
phosphatase and isomerease enzymes and RUBP is photosynthesis. Photosynthesis does not take place
regenerated back. in non-green parts of variegated leaves and in etiolated
(iii) Regeneration of RUBP plants. (A plant is said to be etiolated when its leaves
Factors affecting Photosynthesis:- become yellow and the stem long and pale due to
Mechanism of phtosynthesis is affected by several absence of light over a prolonged period).
external and internal factors. These are: = Accumulation of end products of photosynthesis:
Ø External factors :- Accumulation of carbohydrates in the
Light: The rate of photosynthesis increases with photosynthesizing cells slows down photosynthesis.
increase in light intensity. = Anatomy of leaf: The rate of photosynthesis is greatly
Highest rate of photosynthesis : Red light influenced by the internal structure of the leaf.
Minimum photosynthesis : Green light Thickness of the cuticle and epidermis, number,
(i) Intensity of light:- With the increase in light structure and distribution of stomata and distribution
intensity, the rate of photosynthesis increases initially of various tissues in the leaf influence the amount of
and then reaches plateau because at this point, light and carbon dioxide entering the stomata.

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Adaptations of photosynthesis in leaves : = Egestion : The process by which undigested matter is

= It is thin, so that gases can diffuse in and out quickly. expelled out.
= It has a large surface area to maximize light absorption. = Digestive system is regulated by various hormones
= It has a thin layer of wax-the cuticle to reduce secreted by some endocrine glands.
evaporation from the upper surface. = Alimentary canal was first of all developed in the
= It has a rich network of xylem fibres to deliver water phylum Platyelminthes but only mouth was present in
and minerals, and pholem fibres to remove the product them.
of photosynthesis (mainly as sucrose). This vascular = Coiled and well developed alimentary canal was
tissue also gives the leaf more rigidity. developed in annelida till mammals.
= The cells of the upper epidermis have no chlolroplasts,
so that more light gets through to the pallisade cells. (A) Nutrition in Amoeba :
= The pallisade cells are deep and tightly packed. The It is a unicellular organism living in water.
chloroplasts within them can move so that all have a • Mode of nutrition of holozoic.
chance of obtaining all the light they need. • The process of obtaining food is the phagocytosis
= Guard cells surround pores called stomata (singular = (cell eating)
stoma; ‘hole’) opening in the day to allow carbon • Steps involved in digestion of amoeba are :
(i) Ingestion : Since it is unicellular so a single cell is
dioxide in, then closing at night to reduce water loss.
responsible for carrying out all the vital activities. Food
= Loosely packed spongy mesophyll cells create air
is ingested with the help of pseudopodia. Animal
spaces, making gas exchange more efficient between
engulfs the food particle lying near it by forming
the atmosphere and the pallisade cells. The air spaces
pseudopodia around it and forming a food vacuole while
also trap a little carbon dioxide, allowing some
is considered at its temporary stomach.
photosynthesis even when the stomata are closed.
(ii) Digestion : The enzymes from surrounding cytoplasm
Significance of Photosynthesis :
enter the food vacuole and break down the food into
= The whole of the metabolism of the green plant is
smaller & soluble forms.
sustained by the products of photosynthesis.
(iii) Absorption : The digested food is now absorbed by
= Sugar and other metabolites are transported from the
cytoplasm by simple diffusion and then the food
chloroplasts to all parts of the plant.
vacuole disappear.
= These products are used in the living cells as the (iv) Assimilation : The food absorbed in amoeba is used
building blocks. to obtain energy from respiration, for its growth and
= Conversion of sugars to amino acids, building blocks reproduction.
of proteins. (v) Egestion : Undigested food is thrown out of the cell.
= The products of photosynthesis also supply the energy
needed to carry out chemical change.
= The survival of life depends upon photosynthesis.
= This is because all organisms get their nutrients from
green plants, either directly or indirectly.

NUTRITION IN ANIMALS
= Animals have highly evolved digestive mechanism that
includes two basic components :
= Alimentary canal : Long, hollow, tubular structure
consisting of various organs for digestion. (B) Nutrition is Grasshopper :
= Digestive glands : They secrete enzymes/hormones • It has a well developed digestive system having
which help in digestion. an alimentary canal and digestive glands.
= Digestion in animals consist of following steps : • The various organs of digestive system of
= Ingestion : The process of intake of food. grasshopper are
= Digestion : It is the breakdown of large and complex Mouth ® Oesophagus ® Crop ® Gizzard
molecules into simpler, smaller and soluble forms. ® Stomach ® Ileum ® Colon ® Rectum.
= Absorption : Taking up of the digested food through • Glands associated with it are :
intestinal wall to blood. (i) Salivary glands (ii) Hepatic caeca
= Assimilation : In this process absorbed food in taken • Digestive system of a grasshopper can be divided
by body cells. into three parts.
(i) Foregut : mouth to gizzard

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(ii) Midgut : gizzard to ileum (actual stomach) (A) Alimentary canal:

• The alimentary canal is tubular structure which
extends from mouth to anus. It develops from ectoderm
& endoderm.
Ectoderm - upto hard palate
Endoderm - from soft palate to rectum
Ectoderm - from anal canal to Anus
• The alimentary canal is divided into following parts –
(i) Mouth (ii) Buccal cavity
(iii) Pharynx (iv) Oesophagus
(iii) Hindgut : stomach to anus. (v) Stomach (vi) Small intestine
The process involves: (vii) Large intestine
(i) Ingestion : If feeds on green leaves so it takes food (viii) Anus.
through its mouth with the help of it’s forelegs and
mouth parts.
(ii) Digestion:
(a) It starts from the mouth.
(b) A pair of salivary glands secretes saliva and release
it into the mouth through the salivary duct.
(c) Saliva mixed with food and lubricates and soften
the food.
(d) Digestion of starch begins here.
(e) This slightly digested food enters into the crop
through a food pipe i.e. esophagus.
(f) Crop stores the food temporarily.
(g) Now the food moves to gizzard. Here it is finally
crushed and masticated an then moves to stomach.
(h) In stomach hepatic caeca release its secretions in
the form of digestive enzymes, thus the food is then
completely digested at this site.
(iii) Absorption : The digested food moves to small
intestine (ileum) and absorbed through its walls.
(iv) Assimilation : Nutrients are assimilated whenever
required by the cells for the fulfillment of the growth,
energy and repair of the body.
(v) Egestion : Undigested food is then passed through
hindgut (where H2 O absorbed) and expelled out
through anus in the form of elongated dry faecal pallets.
• The excretory organ of the grasshopper is malphigian
tubules present at the junction of hindgut and midgut.

(C) Nutrition in Humans: (i) Mouth :

• To perform various functions of the body, energy is It is small slit through which food is ingested.
required, which is obtained from food. The process of (ii) Buccal Cavity :
conversion of complex indiffusible food material into Mouth opens into a chamber called as buccal cavity.
simple and diffusible food by mechanical and Roof of buccal cavity is called hard palate. At the
biochemical (hydrolysis) process, is termed as floor of this cavity thick muscular structure is present
Digestion. called tongue. it helps in chewing swallowing, testing
• Humans have highly evolved and complicated and speaking. Tongue has various types of papilla
digestive system consisting of having taste buds.
(A) An alimentary canal and • Jaws present in buccal cavity are provided with
(B) Different types of digestive glands. four different types of teeth :

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Nutrition

Ø Table: Number, structure and function of different SALIVARY GLANDS-

types of teeth in man. In mammals , 4 pairs of salivary glands are present.
Type of No. Appeara-nce Function Infra-orbital-glands
teeth in Gland is located just below the eye-orbit. The duct of
each these glands open in the upper-jaw near the 2nd molar
jaw teeth. These glands are absent in Human.
Incisors 4 Sharp and Biting and Parotid-glands (largest salivary glands) -
(i) chisel-shaped cutting of
Thess glands are located just below the auditory -
vegetable food
Canines 2 Sharp and Tearing of capsule (ear). Their duct is called Parotid duct/
(c) dagger- shaped animal food Stenson's duct which open in the vestibule of the
Premolars 4 Broad and Chewing and upper jaw i e the Buccal-vestibule. Whenever in
(pm) cusped grinding human, these glands are infected by viruses this
Molars 6 Broad and Chewing and disease is called as Mumps. Due to this, the gland
(m) cusped. grinding swells up.
Submaxillary or Submandibular glands-
• Dental formula of humans :
These are located at the junction of the upper and the
(a) Milk teeth ® These are temporary, arise at 6 - 111
month age, 20 in number lower jaw. Their duct is called Wharton's duct (largest
salivary duct). These ducts open in the lower jaw just
Half upper jaw 2102
= behind the Incisor teeth.
Hlaf lower jaw 2102
Sublingual glands These are the smallest salivary
(b) Permanent teeth ® arise at 6 – 12 years, 32 in glands. These glands are found in the lower jaw. Many
number
ducts arise from these glands called as the Ducts of
Half upper jaw 2123 Rivinus or Bartholin's ducts. These ducts open in the
=
Half lower jaw 2123 bucco-pharyngeal cavity on the ventral side of the
tongue.
Maximum saliva is secreted by the Sub-lingual glands.
(Smallest salivary duct) Salivary glands are Exocrine
glands. The secretion of salivary glands is termed as
the saliva.

(iii) Pharynx:-
It is a funnel-shaped common passage of respiratory
and digestive tracts which is also connected with
middle ear. At the end of mastication, nasal chambers
Incisor
are closed by the raising of uvula while glottis or
Canine
opening of trachea is closed by the raising of larynx
Premolars to come in contact with epiglottis. Masticated food
Molars
enters pharynx in the form of bolus. Soon the muscles
Socket
of jaw of pharynx contract to push the food into oesophagus.
Arrangement of different types of Th e act of pushin g into oesophagus is called
teeth in the jaws on one side and
the sockets on the other side
swallowing.

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(iv) Oesophagus (food pipe):- = Caecum:- The caecum is a small pouch that lies at the
It is long, narrow and tubular. It passes through junction of the small and large intestine. Caecum
neck and thorax and opens in stomach. It conducts posses a narrow tubular lymphoid outgrowth, 8-9 cm
the food by peristalsis. in length, called vermiform appendix. The appendix is
Waves of muscle contraction, called peristalsis, push a vestigial organ in man. Caecum leads into the colon.
the food ball down the oesophagus, the tube that = Colon:- The colon is a U-shaped tube, about 1 meter
links the pharynx to the stomach. long. It is the largest part of the large intestine. It is
(v) Stomach:- divisible into four parts- ascending colon, transverse
It is a thick, muscular and J-shaped sac present on colon, descen ding colon, an d sigmoid colon.
the left side of upper part of abdomen. It helps in Throughout its length, the colon bears alternate sac-
mechanical churning and chemical digestion of food.
like swellings called haustra. The colon is meant for
It also stores the food.
absorption of water from undigested food. The colon
When food arrives in the stomach, millions of tiny
harbours a large number of bacteria which feeds upon
glands in the stomach’s wall produce a liquid called
the undigested food.
gastric (stomach) juice. Gastric juice contains powerful
= Rectum:- Sigmoid colon leads into the rectum. Rectum
hydrochloric acid, which kills most harmful bacteria
is the terminal part of alimentary canal and opens to
in food and the enzyme pepsin, which digests proteins.
the exterior through the anus. It is about 20 cm long.
Gastric juice also contains mucus that covers the
It absorbs excess water from the undigested food,
stomach’s lining and stops it from being digested by
its own pepsin. secretes mucus and helps in egestion of faecal matter.
(vi) Small intestine:- (viii) Anus:- It lies at the base of trunk and is for egestion.
It is a long (about 6 meter) narrow, tubular and coiled
part. It is differentiated into anterior duodenum, middle (B) Digestive Glands :
jejunum and posterior ileum. It is mainly concerned (i) Salivary glands : 3 pairs of salivary glands are found
with completion of digestion and absorption of food. in mouth cavity. It helps in chemical digestion. They
The small intestine is the long tube that runs from the secret at enzyme called salivary amylase or ptyalin. It
stomach to the large intestine. As food passes along helps in digestion of starch.
the small intestine, it is broken down into very simple (ii) Gastric glands : Present in stomach. They secret
nutrients. These nutrients then pass into the blood hydrochloric acid, protein digesting enzymes and
stream and are carried to all the body’s cells. mucus.
= Duodenum:- It is the proximal part of the small (iii) Liver : It is the largest gland, secrets bile into the small
intestine and is about 25 cm in length and U-shaped intestine. Bile contains bile juice and bile pigments.
in appearance. It receives the opening of common Bile is alkaline in nature and it is temporarily stored in
bile and pancreatic duct. gall bladder and helps in digestion of fats, it also helps
= Jejunum:- It is the middle and highly coiled part of in absorption fats.
the intestine and is about 2.5 m in length. Jejunum is
thick-walled and narrower than the duodenum.
= Ileum:- It is the distal, thin-walled, highly coiled part
of the small intestine with a length of about 3.5 metres.
(vii) Large intestine:-
It is shorter and wider than small intestine. It is
differentiated into caecum, colon and rectum. It helps
in formation and temporary storages of faeces.
The large intestine receives watery waste from the
small intestine. This waste contains food, such as
fibre, that could not be digested. As waste passes
along the large intestine, water is absorbed back into
the blood. This process turns the liquid waste into
more solid faeces. Faeces are pushed out through the
anus.

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= Synthesis of Vitamin-A- The liver changes b -

carotene into vitamin -Allow part of fruits. It is
abundantly found in carrot.
= Liver stores vitamins A,D,E,K,B12
= torage of minerals- Liver stores iron, copper, zinc,
cobalt, molybdenum etc. Liver is a good source of
iron.
= Detoxification- The conversion of toxic substances
into non-toxic substance is done by liver. The toxic
substances are formed by metabolic activities of
the body. e.g. Prussic acid is converted into
Potassium sulfocynide (it is a non-toxic salt) by the
liver.
= Haemopoiesis- The formation of blood cells is called
haemopoesis. In embryonic stage R.B.C and WBC
are formed by liver
FUNCTIONS OF LIVER :- (Liver is known as = Yolk synthesis- Most of the yolk is synthesised in
biological and chemical factory of the body). liver.
Most of the biochemical functions of the body are done = Secretion of enzymes- Some enzymes are secreted
by the liver. by liver, participate in metabolism of proteins, fats
= Secretion & synthesis of bile - This is the main and carbohydrates e.g. Dehydrogenase,
cytochrome oxidase etc.
function of liver. Bile is yellowish-green, alkaline
= Prothrombin and fibrinogen proteins are also formed
fluid. In bile juice, bile salts, sodium bicarbonate,
in hepatic cells. These help in blood clotting.
glycocholate, taurocholate, bile pigments,
= Factors I, II, V, VII , IX and X are formed in liver,
cholesterol, Lecithin etc. are present.
which are responsible for blood clotting.
Bile salts help in emulsification of fats. Bile prevents (iv) Pancreas: It lies parallel to and below the stomach. It
the food from decomposition. It kills the harmful secrets pancreatic juice into small intestine. Pancre-
bacteria. atic juice contains tyrosine and pancreatic amylase.
= Carbohydrate Metabolism- The main centre of Besides these 2 enzymes pancreas secretes 2 hor-
carbohydrate metabolism is liver. mones also i.e. :- insulin and glucagons so it has both
exocrine as well as endocrine functions. Both bile and
= Storage of fats- Liver stores fats in a small amount.
pancreatic juice are released into the duodenum by a
Hepatic cells play an important role in fat
common duct.
metabolism.
= Deamination and Urea formation- Deamination of
amino acids is mainly done by liver (Amino acid
® NH3)
Liver converts ammonia (more toxic) into urea (less
toxic) through ornithine cycle .
= Purification of blood- Kupffer cells of liver &
splenocytes of spleen are the phagocytic cells,
helps in phagocytosis of dead blood cells and
Section of pancreas
bacteria from the blood.
(v) Intestine gland:- These lie in the wall of small intestine
= Synthesis of plasma proteins- Many types of and secrete intestinal juice.
proteins are present in blood plasma. All the Physiology of nutrition:-
proteins except Gama-globulins are synthesized in (i) Ingestion:-
the liver. Chemically antibodies are gama globulins = Man is omnivorous in feeding and is holozoic.
formed by lymphocytes. = Ingestion involves carrying the food to the mouth
= Synthesis of heparin- Heparin is a natural with the help of hands and cutting of food with
incisors or canines depending upon the nature of
anticoagulant (mucopolysaccharide).
food.

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 Nutrition

(ii) Digestion:- In man, digestion is started in buccal (c) In small intestine:-

cavity and completed in intestine. = The small intestine is the site of the complete digestion
(a) In buccal cavity:- Here, food is chewed with the help of carbohydrates, proteins and fats.
of premolars and molars which increases the rate of = Food is mixed with three digestive juices, bile juice,
action of salivary amylase. Food is moistened, pancreatic juice and intestinal juice.
crushed and paratially digested in buccal cavity due = Bile juice provide alkaline medium and emulsifies fat
to action of salivary amylase on starch. At the end it globules into smaller fat droplets but is a non
is rolled into a small ball or bolus. enzymatic digestive juice so has no chemical action
Salivary amylase on food.
Starch ¾¾¾¾¾¾ - ® Maltose + Isomaltose = Pancreatic juice contains trypsin, pancreatic amylase
6.8PH,Cl

(b) In stomach:- and pancreatic lipase enzymes which digest the

When food comes in stomach it get completely peptones, starch and fats into peptides, maltose,
churned and get changed into paste like form called isomaltose and fatty acids.
chyme. = Intestinal juice contains aminopeptidase, intestinal
= Food is mixed with gastric juice which contains mucus, amylase, maltase, isomaltase and lipase enzymes
hydrochloric acid, pepsin, renin and a weak lipase which hydrolyse following:-
enzyme. Amylase
Strach ¾¾¾¾¾
® Maltose
= Mucus, lubricates the food and protects the inner
lining of the stomach from the action of acids. Maltase
Maltose ¾¾¾¾® Two glucose
= Hydrochloric acid stops the action of saliva in
Lipase
stomach, kills the bacteria present in the food, Fats ¾¾¾¾
® Fatty acid + Glycerol
activates pepsin and provides acidic medium.
Lactase
= Pepsin hydrolyses proteins into proteoses and Lactose ¾¾¾¾ ® Glucose + Galactose
peptones. Sucrase
= Lipase enzymes hydrolyses small amounts of fats Sucrose ¾¾¾¾ ® Glucose + Fructose
into fatty acids and glycerol. aminopeptid ase
Peptides ¾¾¾¾¾¾ ® Amino acid
= Curdling of milk is done by the enzyme renin. (Renin
is not found in human beings, it found only in cattles). Enzymes Substrate End products
= In addition to chemical digestion, food also undergoes Maltase Maltose Glucose
mechanical churning inside the stomach.
Pepsinogen + HCl ® Pepsin Sucrase Sucrose Glucose + Fructose
Proteins ¾¾¾
Pep sin
® Proteoses and Peptones (Peptides )
Lactase Lactose Glucose + Galactose
Prorennin + HCl ® Rennin
Erepsin Peptones Amino acids
Casein ¾¾¾
Re nnin
® Paracasein + Whey Proteins
and
Paracasein + Calcium ® Calcium-Paracaseinate proteases
Peptidase Peptides Amino acids
Calcium Paracaseinate ¾¾¾
Pep sin
® Peptones
Lipase Fat Fatty acid + Glycerol

Nuclease DNA and Nucleotides and

RNA nitrogenous bases

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EXERCISES-I

CBSE BOARD Q.13 State the kind of chemical reactions in the

PREVIOUS YEAR'S following examples :
Q.1 Define photosynthesis. (i) Digestion of food in stomach
[CBSE (Delhi) 2015-16] [1] (ii) Combustion of coal in air
Q.2 Which type of digestion occurs in Amoeba and (iii) Heating of limestone.
Paramecium ? [1] [CBSE (Delhi) 2014-15] [3]

Q.3 Name a vestigeal structure in the alimentary canal

Q.14 Explain the nutrition process in Amoeba.
of man. [1]
[CBSE (Outside Delhi) 2017] [3]
Q.4 State two functions of stomach in man. [1]
Q.15 What is nutrition ? How do plants get their food?
Q.5 What is source of bile juice ? [1] [5]

Q.6 Name the duct of parotid salivary gland. [1] Q.16 Discuss the mechanism, how food (carbohydrate)
is synthesized during photosynthesis ? [5]
Q.7 What is photosynthesis ? Why do plants look
green? Write the chemical equation of Q.17 Define various steps of the process of nutrition in
photosynthesis. [2] the animals. [5]

Q.8 Describe the events which occur during the Q.18 Draw a labelled diagram of alimentary canal of
process of photosynthesis. [2]
man. [5]

Q.9 Describe ingestion and digestion of food in

Q.19 (a) Draw a diagram of human alimentary canal
Amoeba. [2]
and label the following parts :
Q.10 Bile juice has no enzyme but the digestion is not (i) Largest gland.
proper if it is absent. Why ? [2] (ii) Gland that secretes digestive enzymes and
hormone.
Q.11 What is main site of absorption of digested food (iii) Part where HCl is produced.
in man ? How is it adapted for complete (iv) Part where digested food is absorbed.
absorption? [2] (b) What are villi ? Explain their function in the
digestive system.
Q.12 Name three different glands associated with the [CBSE (Outside Delhi) 2012-13] [5]
digestive system in humans. Also name their
secretions.
[CBSE (Outside Delhi) 2012-13] [3]

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EXERCISES-II

Introduction, Nutrition, Modes of nutrition Q.10 This process is stopped at night because -
Q.1 Holophytic nutrition means- (A) CO2 increases
(A) autotrophism (B) heterotrophism (B) O2 decreases
(C) symbiotism (D) parasitism (C) Water is not transported
(D) Sunlight is not available
Q.2 Autotrophic nutrition occurs in
(A) Fungi Q.11 In which substance the chemical energy is stored by
(B) Plants the above process ?
(C) Some protists and prokaryotees (A) O2 (B) CO2
(D) Both B and C (C) C6H12O6 (D) H2O

Q.3 Mushroom, Rhizopus and Yeast are Q.12 Chlorophyll cannot absorb one of the following
(A) Chemosynthetic (B) Parasitic (A) red light (B) blue light
(C) Holozoic (D) Saprophytic (C) blue and red light (D) green light

Q.4 In bacterial photosynthesis, the hydrogen donor is- Q.13 The process in which water is split during
(A) H2O (B) H2SO4 photosynthesis is -
(C) NH3 (D) H2S (A) Photolysis (B) Hydrolysis
(C) Plasmolysis (D) Hemolysis
Nutrition in Plants
Q.5 Chlorophyll pigment are present in - Q.14 The oxygen in photosynthesis is released from-
(A) Plastids (B) Root hairs (A) CO2 (B) H2O
(C) Stomata (D) Lenticels (C) Carbohydrate (D) Chlorophyll

Q.6 Dark reaction is also called - Q.15 Dark reaction of photosynthesis occurs in
(A) Hill reaction (A) Grana (B) Stroma
(B) Calvin-benson cycle (C) Matrix (D) Cytoplasm
(C) Both
(D) None of these Q.16 Light waves where photosynthesis is maximum are
(A) Orange (B) Green
Q.7 The process represented by above equation is - (C) Violet (D) Violet-Blue and Red
(A) Photosynthesis
(B) Reduction of Carbohydrate Q.17 The carbohydrate reserve of plants is
(C) Respiration (A) Starch (B) Glycogen
(D) Protein Synthesis (C) Fat (D) Cellulose

Q.8 The gas produced in above process is - Q.18 In which part of chloroplast light reaction of
(A) Oxygen (B) Carbon di oxide photosynthesis occurs?
(C) Water Vapour (D) All above (A) Grana (B) Stroma
(C) Matrix (D) All the above
Q.9 The essential factors for above process are
(A) Temperature and Cytoplasm Nutrition in Animals
(B) Sunlight and Chlorophyll Q.19 Digestion of food in human starts from
(C) Chlorophyll and Humidity (A) duodenum (B) small intestine
(D) Sunlight and Air (C) mouth (D) large intestine

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Q.20 Bile is produced and secreted by - Q.32 The main organ for digestion and absorption of food
(A) Gall bladder (B) Pancreas is -
(C) Spleen (D) Liver (A) large intestine (B) small intestine
(C) stomach (D) liver
Q.21 Pepsin digests -
(A) Protein (B) Fat Q.33 Liver helps in -
(C) Carbohydrate (D) Cellulose (A) Digestion of food (B) Detoxification
(C) Secretion (D) All of these
Q.22 The end product of fat digestion is -
(A) Glucose (B) Fatty acids Q.34 Food pipe is the another name of -
(C) Amino acids (D) Alkaloids (A) Oesophagus (B) Bile duct
(C) Salivary gland (D) Pancreatic duct
Q.23 Alimentary canal is usually longer in
(A) carnivores (B) herbivores Q.35 The wave of contractions that pushes the food
(C) omnivores (D) insectivores through the alimentary canal is called
(A) peritoneum (B) peristalsis
Q.24 A good source of lipase is - (C) cyclosis (D) polarisation
(A) Saliva (B) Gastric juice Q.36 Starch is digested by -
(C) Bile (D) Pancreatic juice (A) Peptidase (B) Amylase
(C) Lipase (D) Proteinase
Q.25 Enzymes required for digestion of fat is -
(A) Amylase (B) Trypsin Q.37 Bile is produced by -
(C) Pepsin (D) Lipase (A) Stomach (B) Liver
(C) Gall bladder (D) Pancreas
Q.26 Ptyalin is an enzyme present in -
(A) Gastric juice (B) Pancreatic juice Q.38 The liver stores food in the form of
(C) Intestinal juice (D) Saliva (A) glucose (B) glycogen
(C) albumen (D) ATP
Q.27 Aminopeptidase, a digestive enzyme produces-
(A) Dipeptides (B) Smaller peptides Q.39 Vermiform appendix is a part of
(C) Peptones (D) Amino acids (A) alimentary canal (B) nervous system
(C) vascular system (D) reproductive system
Q.28 Which one does not produce any digestive enzyme?
(A) Pancreas (B) Liver Q.40 The hardest constituent of human tooth is
(C) Stomach (D) Duodenum (A) root (B) pulp cavity
(C) dentine (D) enamel
Q.29 The number of salivary glands present in human
beings is - Q.41 Wisdom teeth in man are
(A) 5 pairs (B) 4 pairs (A) incisor (B) canine
(C) 3 pairs (D) 2 pairs (C) last molars (D) all of these

Q.30 Largest gland in the body is - Q.42 Which reserve does a starving man first consume?
(A) Liver (B) Pancreas (A) fat (B) protein
(C) Gastric gland (D) Adrenal (C) glycogen (D) vitamin

Q.31 Which of the following has no digestive enzyme? Q.43 Digestive process in human beings is
(A) Saliva (B) Bile (A) intracellular (B) extracellular
(C) Gastric juice (D) Intestinal juice (C) both of these (D) none of these

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Q.44 In man, a significant role in digestion of milk is played Q.55 Total number of cannines in permanent dental set of
by human is -
(A) rennin (B) intestinal bacteria (A) 4 (B) 6 (C) 2 (D) 12
(C) invertase (D) pancreatic amylase
Q.56 In amoeba absorption of the digested nutrients
Q.45 The action of bile can be called - occurs in
(A) Oxidation (B) Emulsification (A) contractile vacuole (B) plasma membrane
(C) Esterification (D) Dehydrogenation (C) cytoplasm (D) pseudopodia

Q.46 In humans, digestion of protein starts in Q.57 Coiled and well developed alimentary canal first
(A) Stomach (B) Mouth developed in
(C) Duodenum (D) Ileum (A) Protozoans (B) Mammals
(C) Arthropods (D) Poriferans
Q.47 Emulsification of fat is carried out by
(A) Lipase (B) Bile Q.58 Digestion of starch starts from
(C) Gastric juice (D) Intestine juice (A) stomach (B) intestine
(C) esophagus (D) mouth
Q.48 Incisors take part in
(A) Grinding (B) Tearing Q.59 Teeth involved in cutting of food material are called
(C) Cutting (D) Crushing (A) canines (B) incisors
(B) molars (D) premolars
Q.49 Nutrition of Amoeba is
(A) Holozoic (B) Holophytic Q.60 Ptyalin enzyme is secreted by
(C) Saprophytic (D) Parasitic (A) salivary glands (B) mouth
(C) esophagus (D) stomach
Q.50 Proteolytic enzyme of pancreatic juice is
(A) Pepsin (B) Trypsin Q.61 Villi are present on
(C) Amylase (D) Lipase (A) stomach (B) large intestine
(C) small intestine (D) mouth
Q.51 Wavelength of visible light is
(A) 200 - 400 nm (B) 400 - 700 nm Q.62 What is vitamin C -
(C) 700 - 900 nm (D) 100 - 200 nm (A) Follic acid (B) Nicotinic acid
(C) Ascorbic acid (D) Citric acid
Q.52 Chemical reaction takes place during dark reaction of
photosynthesis is Q.63 Which vitamins are water soluble -
(A) photolysis (A) Vitamin B and C (B) Vitamin A and B
(B) hydrolysis (C) Vitamin A and C (D) Vitamin C and D
(C) carbon dioxide is bonded with RUBP
(D) nitrogen fixation Q.64 Rickets is caused by the deficiency of -
(A) Vitamin A (B) Vitamin B
Q.53 CO2 acceptor during dark reaction of photosynthesis (C) Vitamin C (D) Vitamin D
is
(A) RUBP (B) PEP Q.65 Beri-beri is caused by -
(C) NADPH (D) ATP (A) Vitamin B2 (B) Vitamin B5
(C) Vitamin B3 (D) Vitamin B1
Q.54 Gastric juice is -
(A) Acidic (B) Alkaline Q.66 Kwashiorkor is caused due to the deficiency of-
(C) Neutral (D) Slightly alkaline (A) Protein (B) Fats
(C) Carbohydrates (D) Minerals
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Q.67 Human saliva contains enzyme - Q.78 The structure which prevents entry of food into wind
(A) Pepsin (B) Trypsin pipe during swallowing in mammals is-
(C) Ptyalin (D) Amylase. (A) Larynx (B) Glottis
(C) Epiglottis (D) Pharynx
Q.68 Enzyme trypsin acts on -
(A) Sucrose (B) Fats Q.79 The hardest constituent of the tooth is-
(C) starch (D) Proteins (A) Enamel (B) Dentine
(C) Bone (D) Pulp
Q.69 Duodenum is the part of -
(A) Small intestine (B) Large intestine Q.80 In which part of stomach mainly digestion occurs-
(C) Oesophagous (D) Stomach (A) Cardiac region (B) Fundic region
(C) Pyloric region (D) All of the above
Q.70 Maximum absorption takes place in -
(A) Ileum (B) Colon Q.81 Parietal cells of mucosa in stomach is secrets :
(C) Stomach (D) Rectum (A) Mucin (B) Pepsin
(C) HCl (D) All of the above
Q.71 Prorennin occurs in -
(A) Saliva (B) Gastric juice Q.82 Parotid salivary gland are present :
(C) Pancreatic juice (D) Intestinal juice (A) Below the tongue
(B) Below the ear
Q.72 Trypsin changes - (C) Below the eye orbit
(A) Starch to sugar (D) In the angle between two jaws
(B) Fats to fatty acids and glycerol
(C) Proteins to peptides Q.83 Cells of liver which act as phagocytes are :
(D) None of the above (A) Dieter's cells (B) Kupffer's cells
(C) Hensen cells (D) Aciner cells
Q.73 The digestion of food starts in the mouth where starch
is converted into sugar by the action of- Q.84 Proteins are broken down into amino acids in-
(A) Pepsin (B) Trypsin (A) Buccal cavity (B) Stomach
(C) Ptyalin (D) Sucrose (C) Intestine (D) Rectum

Q.74 Liver secretes - Q.85 Some proteolytic enzymes are-

(A) Acids only (B) Bile (A) Trypsin, Erepsin, Pepsin
(C) Hormones (D) Digestive enzymes only (B) Amylase, Lipase, Zymase
(C) Ampylopsin, Steapsin, Ptyalin
Q.75 Trypsin act in a medium which is - (D) Urease, Dehydrogenase, Zymase
(A) Acidic (B) Slightly alkaline
(C) Neutral (D) Highly alkaline Q.86 Emulsification of fats by bile takes place in-
(A) Duodenum (B) Liver
Q.76 Milk proteins is - (C) Stomach (D) Intestine
(A) Galactose (B) Glycine
(C) Casein (D) Rennin Q.87 Pancreatic juice takes part in digestion of-
(A) Proteins carbohydrate and fats
Q.77 Dental formula of adult man is - (B) Proteins and fats
(C) Protein, carbohydrate
2,1,2, 3 2,1,2, 3
(A) (B) (D) Proteins only
2,1,2, 3 2,1,2, 2
Q.88 Enzyme trypsinogen is changed to trypsin by
2,1,2, 3 2,1,3,2
(C) (D) (A) Gastrin (B) Enterogastrone
2,1,2, 4 2,1,3,2
(C) Enterokinase (D) Secretin

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 Nutrition

EXERCISES-III

Q.1 Which of the follwing statements about the Q.7 Choose the event that does not occur in
autotrophs is incorrect ? photosynthesis
(A) They synthesize carbohydrates from carbon (A) Absorption of light energy by chlorophyll
dioxide and water in the presence of sunlight and (B) Reduction of carbon dioxide to carbohydrates
chlorophyll (C) Oxidation of carbon to carbon dioxide
(B) They store carbohydrates in the form of starch (D) Conversion of light energy to chemical energy
(C) They convert carbon dioxide and water into
carbohydrates in the absence of sunlight Q.8 Which of the following is the best equation
(D) They constitute the first trophic level in food representing photosynthesis ?
chains.
Chlorophyll
(A) energy + 6O2 + 6 H2O¾¾¾¾¾ ® C5HO12 + 6O2
Q.2 Select the correct statement
(A) Heterotrophs do not synthesise their own food (B) energy + CO2 + H2O ¾¾
® CH2O + O2
(B) Heterotrophs utilise solar energy for Chlorophyll / Light
(C) energy + 6 CO2 + 12 H2O ¾¾¾¾¾¾ ¾ ®
photosynthesis
(C) Heterotrophs synthesise their own food C6H12O6 + 6H2O + 6 O2
(D) Heterotrophs are capable of converting carbon Light / chlorophyll
(D) energy + 12 CO2 + 6 H2O ¾¾¾¾¾¾ ¾ ®
dioxide and water into carbohydrates.
C6H12O6 + 6H2O + 6 O2
Q.3 Among the following which is a parasitic plant ?
(A) Plasmodium (B) Cuscuta Q.9 The raw materials for photosynthesis are -
(C) Amoeba (D) Rhizobium (A) CO2 & O2 (B) sunlight and CO2
(C) water & chlorophyll (D) CO2 and water
Q.4 Heterotrophic nutrition means -
(A) Various types of nutrition taken by an animal Q.10 Plants are green in colour because -
(B) Preperation of nutrients by a plant with the help (A) they absorb green light only
of chlorophyll (B) they reflect green light
(C) Utilization of food by animals prepared by plants (C) they abosorb green light but reflect all other lights
(D) All of the above (D) none of the above are correct

Q.5 Chlorophyll is present Q.11 Most of the photosynthesis (80%) which takes place
(A) in the grana of chloroplast on this earth is carried out by
(B) on the surface of chloroplast (A) green plants on land
(C) stacks of thylakoid (B) algae present in fresh water
(D) none of these (C) algae found in ocean
(D) algae present in ocean and fresh water sources.
Q.6 Photosynthesis proceeds in sequence of -
(A) Dark phase and light phase
Q.12 Full name of N ADP is
(B) Light phase alone
(A) Nicotinamide dinucleotide phosphate
(C) Light phase and dark phase
(B) Nicotine adenine dinuceotide phosphate
(D) Dark phase alone
(C) Nicotinamide adenine dinucleotide phosphate
(D) None of the above

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Q.13 The presence of sugar in onion leaves can be tested Q.21 In amoeba the digestion is intracellular because
with (A) amoeba is unicellular
(A) iodine (B) copper sulphate solution (B) amoeba is multicellular
(C) lime water (D) benedict’s solution (C) amoeba is found is pond
(D) amoeba is microscopic animal
Q.14 Dark reaction and light reaction of photosynthesis
takes place in Q.22 The most important function of villi in the small
(A) stroma and grana of chloroplast respectively intestine is
(B) grana and stroma of chloroplast respectively (A) to provide strength to the intestine
(B) to provide space for capillaries and lacteals
(C) grana only
(C) to provide increased surface area for absorption
(D) stroma only
of digested food
(D) to provide habitat for bacteria
Q.15 Compensation point refers to the intensity of light at
which
Q.23 The final product of digestion of carbohydrates and
(A) Rate of respiration = rate of photosynthesis
proteins are
(B) Rate of respiration > rate of photosynthesis (A) glycerol and amino acid respectively
(C) Rate of respiration < Rate of photosynthesis (B) glucose and amino acids respectively
(D) None of the above is correct (C) amino acids and glycerol respectively
(D) amino acids and glucose respectively
Q.16 The inner lining of stomach is protected by one of the
following from hydrochloric acid. Choose the correct Q.24 Through mastication of food is essential because
one. (A) mastication of food makes the teeth stronger
(A) Pepsin (B) Mucus (B) it makes the process of swallowing the food easier
(C) Salivary amylase (D) Bile (C) by this process bigger pieces of food are broken
down into smaller pieces.
Q.17 Digested fat is absorbed in the intestine by (D) bigger pieces of food are broken down into smaller
(A) Blood capillaries (B) Blood arteriole pieces and saliva is properly mixed with it
(C) Blood venule (D) Lymph capillary
Q.25 The path taken by food material after ingestion is
Q.18 Which set is mixed with the food in small intestine? represented by
(A) Mouth ® Pharynx ® Oesophagus ® Stomach
(A) Saliva, gastric juice, bile
(B) Mouth ® Ph aryn x ® Oesopha gus ® Sm al l
(B) Gastric juice, bile, pancreatic juice
Intestine
(C) Bile, pancreatic juice, succus entericus
(C) Mouth ® Oesophagus ® Stomach ® Pharynx
(D) Bile, pancreatic juice and saliva
(D) Oesophagus ® Mouth ® Pharynx ® Stomach

Q.19 The main function of intestinal villi is -

Q.26 The richest sources of vitamin B12 are -
(A) Stimulate peristalsis (A) Carrot and chicken's breast
(B) Prevent antiperistalsis (B) Goat's liver and spirulina
(C) Provide large surface area of absorption (C) Chocolate and green gram
(D) Distribute digestive enzymes uniformly. (D) Rice and hen's egg
Q.27 Conjugated proteins occur in the protoplasm in the
Q.20 Major function of HCl of gastric juice is - form -
(A) Providing acidic medium for pepsin (A) Lipoproteins
(B) Kill microoganisms (B) Nucleoproteins
(C) Dissolve food (C) Glycoproteins
(D) Facilitate absorption of food (D) All of the above

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 Nutrition

Q.28 Which of the following organs are not directly Q.34 Match the column :
connected to each other - Column - I Column - II
(Vitamin) (Deficiency problems)
(A) Oesophagous – stomach
1. Vitamin A a. Scurvy
(B) Buccal cavity – stomach
2. Vitamin D b. Night blindness
(C) Colon – rectum 3. Vitamin C c. Rickets
(D) Stomach – duodenum 4. Vitamin B1 d. Beri-beri
5. Vitamin B4 e. Pernicious anaemia
Q.29 The process of digestion proceeds in this order- 6. Vitamin B12 f. Pellagra
(A) Ingestion ® digestion ® absorption and solution g. Megaloblastic anaemia
n. Haemorrhage
® egestion
(A) 1-b, 2-c, 3-a, 4-d, 5-f, 6-e
(B) Digestion ® Ingestion ® Solution ® absorption
(B) 1-e, 2-b, 3-c, 4-a, 5-d, 6-f
® egestion
(C) 1-f, 2-e, 3-a, 4-d, 5-b, 6-c
(C) Ingestion ® digestion ® absorption ® egestion (D) 1-b, 2-c, 3-a, 4-e, 5-d, 6-f
(D) Ingestion ® solution ® absorption ®
accumulation ® egestion Q.35 Match the column :
Column - I Column - II
Q.30 What is the important function of bile- 1. Pepsin a. Roundworm
2. Parasite b. Breaking down of protein
(A) For digestion by emulsification of fats
3. Grana c. light reaction
(B) Elimination of excretory products
4. Hill reaction d. Part of chloroplast
(C) For digestion by enzymes e. Dark reaction
(D) Coordination of digestive activities (A) 1-b, 2-a, 3-d, 4-c (B) 1-b, 2-c, 3-a, 4-d
(C) 1-c, 2-b, 3-a, 4-d (D) None of these
Q.31 Enzyme pepsin acts upon food at a pH of about-
(A) 3 to split proteins Q.36 Assertion : Sucrose is formed by condensation of
one molecule of glucose and one molecule of
(B) 2 to split carbohydrate
fructose.
(C) 7 to change protein into peptones
Reason : Because it is a disaccharides.
(D) 2 to change protein in amino acids (A) If both Assertion and Reason are true and
Reason is the correct explanation of Assertion.
Q.32 A carbohydrate splitting enzyme is secreted by - (B) If both Assertion and Reason are true but Reason
(A) Liver is not correct explanation of Assertion.
(B) Zymogen cells of gastic glands (C) If Assertion is false but Reason is true.
(D) If both Assertion and Reason are false.
(C) Spleen
(D) Crypts of Lieberkuhn
Q.37 Assertion : Deficiency of vitamin B1 causes sterlity.
Reason : Citrus fruits, tomato, cobbage are the richest
Q.33 Rennin acts on- source of vitamin B1 .
(A) Milk changing casein into calcium paracaseinate (A) If both Assertion and Reason are true and
at 7.2 - 8.2 PH Reason is the correct explanation of Assertion.
(B) Proteins in stomach (B) If both Assertion and Reason are true but Reason
is not correct explanation of Assertion.
(C) Fat in intestine
(C) If Assertion is false but Reason is true.
(D) Milk changing casein into calcium paracascinate
(D) If both Assertion and Reason are false.
at 1-3 pH

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EXERCISES-IV

NTSE STAGE-1 Q.9 Example of omnivorous animals is-

PREVIOUS YEAR'S [NTSE MP 2014-15]
(1) Sheep (2) Deer
Q.1 Which of the following products of light dependent
(3) Lion (4) Cockroach
phase are used during light independent phase of
photosynthesis? [NTSE WB 2014-15]
(1) RuBP &ATP (2) H2O & O2 Q.10 Which of the following is an example of Insectivorous
(3) NADPH2 & ATP (4) ATP & O2 plant - [NTSE MP 2014-15]
(1) Cuscuta (2) Rafflesia
Q.2 Limiting factors for the rate of photosynthesis are- (3) Drosera (4) Tulsi
(1) Oxygen and water [NTSE Chhattisgarh 2014-15]
(2) Mineral and salt Q.11 Sphincter muscles helps to release the food from
(3) Light intensity and temperature
(a) Large intestine to outside
(4) Structure of leaf and stem
(b) Small intestine to large intestine
Q.3 Rickets is caused by the deficiency of which of the (c) Oesophagus to stomach
following - [NTSE Chhattisgarh 2014-15] (d) Stomach to small intestine
(1) Carbohydrate (2) Vitamin A [NTSE Karnataka 2014-15]
(3) Vitamin D (4) Protein (1) (a) and (b) are correct (2) (b) and (c) are correct
(3) (c) and (d) are correct (4) (a) and (d) are correct
Q.4 The substance essential for photosynthesis is :
[NTSE Rajasthan 2014-15]
Q.12 The element present in chlorophyll :
(1) glucose (2) oxygen
[NTSE AP 2014-15]
(3) nitrogen (4) water
(1) iron (2) magnesium
Q.5 If salivary amylase is lacking in the saliva, which of (3) manganese (4) copper
the following functions in mouth cavity will be affect?
[NTSE Chandigarh 2014-15] Q.13 The longest part in human alimentary canal is____ :
(1) Proteins breaking down into amino acids [NTSE AP 2014-15]
(2) Starch breaking down into sugars (1) Oesophagus (2) Small intestine
(3) Fats breaking down into fatty and gyycerol
(3) Large intestine (4) Stomach
(4) None of these

Q.6 Movement of water and mineral salts in plant is : Q.14 In this disease, caused due to protein deficiency face
[NTSE Tamilnadu 2014-15] and limbs are swollen: [NTSE AP 2014-15]
(1) Oxmosis (2) Absorption (1) Kwashiorkor (2) Marasmus
(3) Ascent of sap (4) Active absroption (3) Rickets (4) Pellagra

Q.7 Pick up the incorrect pairing : Q.15 The site of photosynthesis in plant is
[NTSE Odisha 2014-15]
[NTSE Jharkhand 2014-15]
(1) Mouth cavity - Carbohydrate
(1) Mitochondria (2) Chloroplast
(2) Small intestine - Fat digestion
(3) pancreas - FAt digestion (3) Leucoplast (4) Dictyosomes
(4) Liver - Protein digestion
Q.16 Oxygen released during photosynthesis comes from
Q.8 Chlorophyll contains ____ [NTSE MP 2014-15] [NTSE Jharkhand 2014-15]
(1) Potassium (2) Iron (1) Water (2) Carbon dioxide
(3) Manganese (4) Magnesium (3) Glucose (4) Dictyosomes

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Q.17 During lack of oxygen in tissues of our body, the Q.26 Match the items in Column – I with Column – II
pyruvate is converted into lactic acid in: [NTSE AP 2015-16]
[NTSE Delhi 2014-15] Column-I Column-II
(1) Mitochondria (2) Nucleus (A) Riboflavin (i) Anemia
(3) Cytoplasm (4) Ribosome (B) Folic acid (ii) Scurvy
(C) Retinol (iii) Glossitis
Q.18 Which of the following is a micronutrienl element : (D) Ascorbic acid (iv) Night blindness
[NTSE UP 2014-15] Select the correct matching
(1) Mg (2) K (1) A-(iii), B-(i), C-(iv), D-(ii)
(3) Ca (4) Zn (2) A-(iv), B-(i), C-(ii), D-(iii)
(3) A-(iii), B-(iv), C-(i), D-(ii)
Q.19 A substance produced in liver which prevents the (4) A-(ii), B-(i), C-(iv), D-(iii)
freezing of blood is called - [NTSE UP 2014-15]
Q.27 Symptoms of B12 deficiency:
(1) Ptylin (2) Heparin
[NTSE Tamilnadu 2015-16]
(3) Trypsin (4) Insulin
(1) Nervous disorder (2) Dementia, dermatitis
(3) Destruction of RBC (4) Bleeding gums
Q.20 In plants transport of soluble products in teh process
of photosynthesis occurs in :
Q.28 Which of the following factors does a plant use for
[NTSE Rajasthan 2015-16]
the process of photosynthesis?
(1) xylem (2) phloem
[NTSE Gujarat 2015-16]
(3) both of these (4) non of these
(1) Sunlight (2) Chlorophyll
(3) CO2 and H2O (4) All of them
Q.21 The simplest amino acid is [NTSE WB 2015-16]
(1) Lysine (2) Leucine Q.29 Which is the longest organ of the digestive system?
(3) Glycine (4) Methionine [NTSE Gujarat 2015-16]
(1) Oesophagus (2) Stomach
Q.22 Gastric juice contains : [NTSE WB 2015-16] (3) Small Intestine (4) Large Intestine
(1) Pepsin and trypsin (2) Pepsin and HCl
(3) Trypsin ahd HCl (4) Amylase and Pepsin. Q.30 For which of the following diseases vaccine has not
been invented yet? [NTSE Gujarat 2015-16]
Q.23 In a food industry, food containing oil is prepared (1) Jaundice (2) cholera
and wants to avoid development of foul smell. The (3) Typhoid (4) Chikungunya
substance that need to be added
[NTSE Karnatka 2015-16] Q.31 If pepsin is lacking in gastric juice, then the event in
(1) Antioxidants (2) Preservatives the stomach will be affected: [NTSE Delhi 2015-16]
(3) Colourants (4) Flavourants (1) Digestion of starch into sugars
(2) Proteins break down into peptides
Q.24 Delay in blood clotting is due to the deficiency of the (3) Breaking of fats into glycerol and fatty acids
following vitamin [NTSE Telangana 2015-16] (4) Digestion of nucleic acids
(1) Ascorbic Acid (2) Pantothenic acid
(3) Biotin (4) Phylloquinone Q.32 The liver stores food in the form of
[NTSE Jharkhand 2015-16]
Q.25 Decrease of vitamin C and Carotene content in (1) Glucose (2) Glycogen
vegetables and fruits is due to_____ (3) Albumen (4) ATP
[NTSE AP 2015-16]
(1) Excess Potassium content in soil. Q.33 During photosynthesis, the oxygen gas comes from
(2) Excess Phosphate content in soil. [NTSE Jharkhand 2015-16]
(1) CO2 (2) Water
(3) Less Nitrogen content in soil.
(3) Both CO2 and water (4) Oxygen via air
(4) Less Potassium content in soil
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Nutrition

Q.34 Antioxidant vitamin is [NTSE Jharkhand 2015-16] Q.42 Select the correct sequence of the steps of human
(1) ‘A’ (2) ‘E’ nutrition : [NTSE Maharashtra 2016-17]
(3) ‘C’ (4) All of these (1) Ingestion ® Digestion ® Absorption ®
Assimilation ® Egestion
Q.35 Liver gland secretes_______ (2) Ingestion ® Digestion ® Assimilation ®
[NTSE Maharashtra 2015-16] Absorption ® Egestion
(1) Pancreatic juice (2) Billl juice (3) Ingestion ® Assimilation ® Digestion ®
(3) Gastric juice (4) Various digestive juice Absorption ® Egestion
(4) Ingestion ® Absorption ® Digestion ®
Q.36 Deficiency of Iodine in the body can cause Assimilation ® Egestion
[NTSE Panjab 2015-16]
(1) Kwashiorkar (2) Marasmus Q.43 Which one of the following does not contain any
(3) Goitre (4) Rickets enzyme? [NTSE Odisha 2016-17]
(1) Bile (2) Gastric Juice
Q.37 In plants, translocation of food and other substance (3) Saliva (4) Pancreatic Juice
take place through sieve tubes in
[NTSE Panjab 2015-16] Q.44 The lining of oesophagus and mouth is covered with
which type of tissues : [NTSE Rajasthan 2016-17]
(1) Upward direction
(1) Cuboidal epithelium
(2) Downward direction
(2) Squamous epithelium
(3) Lateral direction
(3) Columnar epithelium
(4) Both upward and downward direction:
(4) Strated squamous epithelium

Q.38 If a photosynthesizing plant releases oxygen gas

containing O18 . It is concluded that the plant has Q.45 6CO2 + ¾¾
?
® ¾¾¾®
Sunlight
? C6H12O6 + 6O2 + 6H2O
been supplied with : [NTSE Haryana 2015-16] Which two raw materials required for photosynthesis
(1) C6H12O6 containing O18 are missing in the above equations?
(2) CO2 containing O18 [NTSE Tamilnadu 2016-17]
(3) H2O containing O18 (1) Oxygen and Water
(4) Oxygen in the form of ozone (2) Oxygen and Calcium
(3) Water and Chlorophyll
Q.39 Deficiency of vitamin A causes:[NTSE MP 2015-16] (4) Chlorophyll and Oxygen
(1) Beri-Beri (2) Anaemia
(3) Night blindness (4) Scurvy Q.46 Select the process that occurs in dark reaction
(1) Light energy is converted into chemical energy
Q.40 Which one of the following statement is true for (2) Water splits into hydrogen and oxygen
photosynthesis : [NTSE Odisha 2015-16] (3) Hydrogen is added to carbon dioxide
(1) ATP is consumed in light reaction. (4) Oxygen is evolved [NTSE Kerala 2016-17]
(2) NADP is reduced in dark reaction.
(3) CO2 is required in the light reaction Q.47 Which Vitamin is present in Golden nee?
(4) O2 is produced in the light reaction. [NTSE Bihar 2016-17]
(1) Vitamin A (2) Vitamin B12
Q.41 If the saliva is lacking in salivery amylase then which (3) Vitamin C (4) Vitamin D
of the following processes taking place in the buccal
cavity will be affected : [NTSE Haryana 2016-17] Q.48 The process of photosynthesis is?
(1) Proteins breaking down into amino acids. [NTSE Jharkhand 2016-17]
(2) Starch breaking down into sugars. (1) Anabolic
(3) Fats breaking down into fatty acids and glyceral. (2) Catabolic
(4) Intestinal layers breaking down leading to ulcers. (3) Anabolic and Catabolic
(4) None of these

1-184
 Nutrition

Q.49 Which enzyme converts protein into peptone found Q.55 The number of salivary glands in human is
in our food? [NTSE Jharkhand 2016-17] [NTSE UP 2016-17]
(1) Ptyalin (2) Insulin (1) two pairs (2) three pairs
(3) Pepsin (4) None of these (3) four pairs (4) five pairs

Q.50 At the end of the experiment to prove that light is Q.56 Enzyme responsible for digestion of protein is
necessary for photosynthesis, when the leaf was [NTSE UP 2016-17]
tested with iodine, the ‘S’ shaped figure on the leaf (1) ptylin (2) pepsin
was found to be [NTSE AP 2016-17] (3) amylopsin (4) steapsin

Q.57 Calciferol is [NTSE UP 2016-17]

(1) vitamin A (2) vitamin B
(3) vitamin C (4) vitamin D:

(1) green – presence of starch

Q.58 The Pancreas secretes Pancreatic Juice which
(2) blue – black presence of starch
contains ____________ enzyme for digesting
(3) green – absence of starch
proteins. [NTSE Punjab 2016-17]
(4) blue black absence of starch
(1) Lipase (2) Amylase
(3) Zymase (4) Trypsin
Q.51 One boy as not able to see in the night. Which kind
of vitamins you suggest him to take ?
Q.59 Choose the incorrect statement:[NTSE Delhi 2016-17]
[NTSE AP 2016-17]
(1) lodine – value is a parameter to denote the degree
(1) Calciferol (2) Tocoferol
of unsaturation of fatty acids.
(3) Retinol (4) Riboflavin
(2) Cholesterol is not present in plant fats
(3) Rancidity is a reduction process of oily food
Q.52 The largest gland in human body is -
materials.
[NTSE MP 2016-17]
(4) Tocopherol is an antioxidant.
(1) Pituitarygland (2) Liver
(3) Adrenal gland (4) Thyroid gland
Q.60 Iodine present in iodised salt in our diet is essential
for [NTSE Delhi 2016-17]
Q.53 Which of the following is an example of Insectivorous
(1) Synthesis of insulin
plant - [NTSE MP 2016-17]
(2) Synthesis of thyroxine
(1) Amla (2) Baheda
(3) Synthesis of adrenalin
(3) Utricularia (4) Isoetes
(4) Synthesis of growth hormone

Q.54 Which of the following is made by hydrolysis of starch

Q.61 Out of the following, which enzyme is active in acidic
[NTSE UP 2016-17]
medium [NTSE Delhi 2016-17]
(1) glucose (2) fructose
(1) Pepsin (2) Trypsin
(3) sucrose (4) maltose :
(3) Lipase (4) Amylase

1-185
Nutrition

NTSE STAGE-2 Q.6 Photosynthesis in an aquatic plant was measured by

PREVIOUS YEAR'S counting the number of O2 bubbles coming out of the
cut end of the plant. What will happen to O 2
Q.1 Match the items in Column I with Column II : production if you use a pipe to blow air from your
mouth into water in the beaker : [NTSE 2013]
Column-I Column-II
A. Mouth a. Protein
B. Stomach b. Water
C. Small Intestine c. Starch
D. Large Intestine d. Fat

Select the alternative which shows the correct

matching : [NTSE 2010]
(1) A-d, B-c, C-a, D-b
(2) A-c, B-d, C-b, D-a
(3) A-c, B-a, C-d, D-b
(4) A-b, B-a, C-d, D-c

Q.2 Which of the following rotation of crops will reduce

dependence on the use of chemical fertilizers ? (1) Air from mouth contains O2 which is being added
[NTSE 2011] to the plant. Hence increase in O2 production
(1) Rice and Chilli (2) Wheat and Potato (2) Air from mouth contains CO2 which is utilized in
(3) Potato and Rice (4) Gram and Rice photosynthesis. Hence increase in O2 production
(3) Bacteria from mouth will infect plant. Hence
Q.3 Which of the following is a source of instant energy? reduction in O2 production
[NTSE 2011] (4) Water is already in contact with air. Hence air from
(1) Salt (2) Glucose mouth will have no effect.
(3) Water (4) Starch
Q.7 In nitrogen cycle, atmospheric nitrogen is fixed by
bacteria and converted into ammonia. Ammonia is
Q.4 Which of the following situations is likely to cause
further converted into other forms of nitrogen. At the
muscle cramps ? [NTSE 2012] end of the cycle it returns to the atmosphere by the
(1) Glucose ¾¾¾¾
No Oxygen
® Lactic acid process of : [NTSE 2014]
(1) Ammonification (2) Nitrification
(2) Glucose ¾¾¾
Oxygen
® Lactic acid
(3) Dentrification (4) Assimilation
(3) Glucose ¾¾¾¾
No Oxygen
® Alcohol + CO2
Q.8 Movement of food in digestive tract is due to :
(4) Glucose ¾¾¾
Oxygen
® CO2 + H2O [NTSE 2014]
(1) Concentration gradient
Q.5 Baker’s yeast with sugar is added to aquarium for the (2) Secretions
following reason : [NTSE 2012] (3) Peristalsis
(A) It provides minerals and metals. (4) Villi
(B) It absorbs heavy metals present in water and
purifies it. Q.9 Which one of the following statements is true with
(C) It helps the plants for photosynthesis through respect to photosynthesis : [NTSE 2015]
(1) Oxygen evolved during photosynthesis comes
aerobic respiration.
from CO2
(D) It helps the plants for photosynthesis through
(2) Chlorophyll a is the only photosynthetic pigment
anaerobic respiration.
in plants
Select the correct option :
(3) Photosynthesis occurs in stem of some plants
(1) A only (2) D only (4) Photosynthesis does not occur in red light.
(3) B and C (4) A and B

1-186
 Nutrition

Q.10 Pancreas is composed of : [NTSE 2016] Q.13 What is the main reason for increase in temperature
(1) Only exocrine cells in a glass house : [NTSE 2016]
(2) Only endocrine cell (1) Sunlight is completely absorbed by plants in the
(3) Both endocrine and exocrine glass house
(4) Nephrons (2) Radiation fails to escape from the glass house
completely
Q.11 In the synbiotic relationship between a bacterium and (3) Plant do not utilize sunlight in a glass house
a root of legume the : [NTSE 2016] (4) Plants produce heat inside the glass house
(1) Bacteria provide N 2 and the plant roots provide
Carbon Q.14 Pancreatic juice contains more than one enzyme.
(2) Roots provide NH4 and bacteria provide Carbon Which among the following combination is correct:
(3) Bacteria provide NH4 and the roots provide Carbon [NTSE 2017]
(4) Bacteria provide N2 and the roots provide NH4 (1) Pepsin and Lipase (2) Amylase and Pepsin
(3) Pepsin and Trypsin (4) Trypsin and Lipase

1-187
Real Number

MATHEMATICS
ANSWER KEY

EXERCISES-I

-2 3
Q.1 4 decimal places. Q.2 11 × 13 × 19 Q.3 Number are not co-prime. Q.4 30 Q.5 or
3 4
Q.6 128 Q.7 HCF = 5 Q.8 3 × 11 Q.10 89
Q.11 is divisible by 2, 11 and 256 and it has more than 2 prime factors Q.13 13 Q.14 3 × 71
Q.16 – 925 Q.17 44,550 Q.21 3 -5 Q.23 180 min. or 3 hrs.
Q.26 64 Q.28 240 Q.29 130

EXERCISES-II
Q.1 (B) Q.2 (C) Q.3 (D) Q.4 (B) Q.5 (A) Q.6 (C) Q.7 (C) Q.8 (B) Q.9 (C) Q.10 (B)
Q.11 (D) Q.12 (D) Q.13 (B) Q.14 (B) Q.15 (D) Q.16 (C) Q.17 (C) Q.18 (B) Q.19 (A) Q.20 (C)
Q.21 (D) Q.22 (C) Q.23 (A) Q.24 (D) Q.25 (C) Q.26 (D) Q.27 (C) Q.28 (D) Q.29 (B) Q.30 (A)
Q.31 (C) Q.32 (A) Q.33 (C) Q.34 (A) Q.35 (A) Q.36 (A) Q.37 (D) Q.38 (C) Q.39 (B) Q.40 (C)
Q.41 (A) Q.42 (B) Q.43 (D) Q.44 (C) Q.45 (B) Q.46 (C) Q.47 (A) Q.48 (A) Q.49 (C) Q.50 (B)
Q.51 (D) Q.52 (D) Q.53 (C) Q.54 (A) Q.55 (B) Q.56 (C) Q.57 (B) Q.58 (C) Q.59 (A) Q.60 (B)
Q.61 (D) Q.62 (C) Q.63 (B) Q.64 (C) Q.65 (C) Q.66 (A) Q.67 (D) Q.68 (B) Q.69 (C) Q.70 (C)
Q.71 (B) Q.72 (B) Q.73 (C) Q.74 (D) Q.75 (B) Q.76 (B) Q.77 (C) Q.78 (B) Q.79 (B) Q.80 (D)
Q.81 (B) Q.82 (B) Q.83 (D) Q.84 (A) Q.85 (B) Q.86 (B) Q.87 (B) Q.88 (B) Q.89 (A) Q.90 (A)
Q.91 (A) Q.92 (D) Q.93 (B) Q.94 (C) Q.95 (A) Q.96 (D) Q.97 (B) Q.98 (C) Q.99 (D) Q.100 (C)
Q.101 (D) Q.102 (D) Q.103 (A) Q.104 (D) Q.105 (D) Q.106 (B) Q.107 (C) Q.108 (B) Q.109 (C) Q.110 (B)
Q.111 (B) Q.112 (A) Q.113 (D) Q.114 (D) Q.115 (B) Q.116 (B) Q.117 (D) Q.118 (D) Q.119 (A) Q.120 (D)
Q.121 (B) Q.122 (C) Q.123 (A) Q.124 (A) Q.125 (C) Q.126 (A) Q.127 (C) Q.128 (C) Q.129 (D) Q.130 (D)

EXERCISE-III
Q.1 (B) Q.2 (B) Q.3 (D) Q.4 (B) Q.5 (D) Q.6 (C) Q.7 (A) Q.8 (C) Q.9 (B) Q.10 (A)
Q.11 (C) Q.12 (A) Q.13 (C) Q.14 (C) Q.15 (A) Q.16 (D) Q.17 (A) Q.18 (C) Q.19 (A) Q.20 (D)
Q.21 (A) Q.22 (C) Q.23 (B) Q.24 (B) Q.25 (B) Q.26 (D) Q.27 (B) Q.28 (B) Q.29 (A) Q.30 (A)
Q.31 (D) Q.32 (B) Q.33 (A) Q.34 (D) Q.35 (C) Q.36 (B) Q.37 (B) Q.38 (D) Q.39 (B) Q.40 (C)
Q.41 (B) Q.42 (C) Q.43 (D) Q.44 (D) Q.45 (A) Q.46 (D) Q.47 (D) Q.48 (B) Q.49 (D) Q.50 (B)
Q.51 (C) Q.52 (A) Q.53 (B) Q.54 (C) Q.55 (D) Q.56 (C) Q.57 (B) Q.58 (D) Q.59 (A) Q.60 (B)
Q.61 (C) Q.62 (A) Q.63 (C) Q.64 (A) Q.65 (D) Q.66 (B) Q.67 (A) Q.68 (D)

EXERCISE-IV
NTSE STAGE-1
PREVIOUS YEAR'S
Q.1 (D) Q.2 (C) Q.3 (D) Q.4 (B) Q.5 (D) Q.6 (B) Q.7 (A) Q.8 (C) Q.9 (A) Q.10 (A)
Q.11 (D) Q.12 (B) Q.13 (D) Q.14 (A) Q.15 (D) Q.16 (C) Q.17 (D) Q.18 (A) Q.19 (B) Q.20 (D)
Q.21 (C) Q.22 (B) Q.23 (A) Q.24 (C) Q.25 (A) Q.26 (A) Q.27 (A) Q.28 (D) Q.29 (B) Q.30 (B)
Q.31 (C) Q.32 (B) Q.33 (D) Q.34 (C) Q.35 (Bonus) Q.36 (B) Q.37 (A) Q.38 (C) Q.39 (A) Q.40 (B)

3-88
 Real Number

Q.41 (Bonus) Q.42 (B) Q.43 (D) Q.44 (B) Q.45 (D) Q.46 (B) Q.47 (B) Q.48 (C) Q.49 (B) Q.50 (B)
Q.51 (C) Q.52 (D) Q.53 (A) Q.54 (C) Q.55 (D) Q.56 (C) Q.57 (C) Q.58 (A) Q.59 (B) Q.60 (C)
Q.61 (B) Q.62 (B) Q.63 (C) Q.64 (C) Q.65 (B) Q.66 (A) Q.67 (A) Q.68 (C) Q.69 (B) Q.70 (A)
Q.71 (A) Q.72 (C) Q.73 (A) Q.74 (D) Q.75 (D) Q.76 (Bonous) Q.77 (C) Q.78 (A) Q.79 (B) Q.80 (B)
Q.81 (B) Q.82 (B) Q.83 (A) Q.84 (D) Q.85 (A) Q.86 (B) Q.87 (D) Q.88 (B) Q.89 (C) Q.90 (D)
Q.91 (D) Q.92 (C) Q.93 (B) Q.94 (C) Q.95 (D) Q.96 (B) Q.97 (D) Q.98 (D) Q.99 (C) Q.100 (B)
Q.101 (D) Q.102 (B) Q.103 (C) Q.104 (C) Q.105 (D) Q.106 (A) Q.107 (A) Q.108 (C) Q.109 (D) Q.110 (C)
Q.111 (D) Q.112 (A) Q.113 (C) Q.114 (C) Q.115 (A) Q.116 (A) Q.117 (D) Q.118 (A) Q.119 (B) Q.120 (A)
Q.121 (C) Q.122 (A) Q.123 (B) Q.124 (C) Q.125 (B) Q.126 (D) Q.127 (B) Q.128 (A) Q.129 (A) Q.130 (B)
Q.131 (D) Q.132 (C) Q.133 (A) Q.134 (B) Q.135 (B) Q.136 (D) Q.137 (A) Q.138 (D) Q.139 (A) Q.140 (B)
Q.141 (C) Q.142 (B) Q.143 (B) Q.144 (D) Q.145 (A) Q.146 (C) Q.147 (B) Q.148 (C) Q.149 (C) Q.150 (D)
Q.151 (B) Q.152 (B) Q.153(Bonous) Q.154 (B) Q.155 (D) Q.156 (C) Q.157 (B) Q.158 (D) Q.159 (B) Q.160 (D)
Q.161 (B) Q.162 (C) Q.163 (D) Q.164 (B) Q.165 (D) Q.166 (A) Q.167 (C) Q.168 (D) Q.169 (D) Q.170 (C)

NTSE STAGE-2
PREVIOUS YEAR'S
Q.1 (D) Q.2 (D) Q.3 (A) Q.4 (A) Q.5 (D) Q.6 (B) Q.7 (A) Q.8 (D) Q.9 (D) Q.10 (C)
Q.11 (C) Q.12 (C) Q.13 (A) Q.14 (D) Q.15 (D) Q.16 (D) Q.17 (B) Q.18 (C) Q.19 (C) Q.20 (C)
Q.21 (C) Q.22 (B) Q.23 (C) Q.24 (B) Q.25 (B) Q.26 (A) Q.27 (C) Q.28 (B) Q.29 (A) Q.30 (D)
Q.31 (C) Q.32 (A)

3-89
Real Number

PHYSICS
ANSWER KEY
EXERCISES-II

Q.1 (D) Q.2 (C) Q.3 (A) Q.4 (A) Q.5 (C) Q.6 (B) Q.7 (D) Q.8 (B) Q.9 (A) Q.10 (B)
Q.11 (C) Q.12 (D) Q.13 (B) Q.14 (D) Q.15 (A) Q.16 (A) Q.17 (C) Q.18 (C) Q.19 (D) Q.20 (B)
Q.21 (B) Q.22 (C) Q.23 (C) Q.24 (A) Q.25 (C) Q.26 (B) Q.27 (B) Q.28 (B) Q.29 (D) Q.30 (A)
Q.31 (D) Q.32 (A) Q.33 (B) Q.34 (B) Q.35 (B) Q.36 (A) Q.37 (C) Q.38 (B) Q.39 (B) Q.40 (B)
Q.41 (A) Q.42 (C) Q.43 (D) Q.44 (C) Q.45 (D) Q.46 (D) Q.47 (A) Q.48 (C) Q.49 (A) Q.50 (C)
Q.51 (B) Q.52 (D) Q.53 (B) Q.54 (A) Q.55 (B) Q56 (B) Q.57 (B) Q.58 (C) Q.59 (B) Q.60 (D)
Q.61 (A) Q.62 (B) Q.63 (B) Q.64 (C) Q.65 (D) Q.66 (B) Q.67 (C) Q.68 (B) Q.69 (B) Q.70 (A)
Q.71 (D) Q.72 (C) Q.73 (A) Q.74 (A) Q.75 (A) Q.76 (D) Q.77 (A) Q.78 (B) Q.79 (C) Q.80 (C)
Q.81 (D) Q.82 (D) Q.83 (C) Q.84 (A) Q.85 (A) Q.86 (B) Q.87 (C) Q.88 (B) Q.89 (B) Q.90 (C)
Q.91 (B) Q.92 (A) Q.93 (B) Q.94 (B) Q.95 (A) Q.96 (C) Q.97 (C) Q.98 (C) Q.99 (B) Q.100 (C)
Q.101 (D) Q.102 (A) Q.103 (B) Q.104 (B) Q.105 (A) Q.106 (C) Q.107 (C) Q.108 (D) Q.109 (A) Q.110 (A)
Q.111 (D) Q.112 (A) Q.113 (D) Q.114 (C) Q.115 (B) Q.116 (D) Q.118 (C)

EXERCISES-III
Q.1 (B) Q.2 (D) Q.3 (A) Q.4 (B) Q.5 (D) Q.6 (A) Q.7 (B) Q.8 (B) Q.9 (D) Q.10 (B)
Q.11 (C) Q.12 (D) Q.13 (C) Q.14 (D) Q.15 (A) Q.16 (C) Q.17 (A) Q.18 (A) Q.19 (A) Q.20 (B)
Q.21 (A) Q.22 (B) Q.23 (C) Q.24 (C) Q.25 (C) Q.26 (B) Q.27 (B) Q.28 (C) Q.29 (C) Q.30 (C)
Q.31 (D) Q.32 (D) Q.33 (B) Q.34 (C) Q.35 (D) Q.36 (B) Q.37 (B) Q.38 (B) Q.39 (C) Q.40 (A)
Q.41 (A) Q.42 (A) Q.43 (B) Q.44 (D) Q.45 (A) Q.46 (D) Q.47 (C) Q.48 (C) Q.49 (D) Q.50 (D)
Q.51 (D) Q.52 (B) Q.53 (A) Q.54 (C) Q.55 (C)

EXERCISES-IV
NTSE STAGE-1
PREVIOUS YEAR'S

Q.1 (D) Q.2 (B) Q.3 (C) Q.4 (A) Q.5 (B) Q.6 (A) Q.7 (A) Q.8 (D) Q.9 (C) Q.10 (A)
Q.11 (D) Q.12 (A) Q.13 (C) Q.14 (B) Q.15 (C) Q.16 (D) Q.17 (D) Q.18 (A) Q.19 (D) Q.20 (C)
Q.21 (B) Q.22 (B) Q.23 (A) Q.24 (C) Q.25 (A) Q.26 (D) Q.27 (B) Q.28 (C) Q.29 (A) Q.30 (A)
Q.31 (C) Q.32 (D) Q.33 (B) Q.34 (B) Q.35 (C) Q.36 (C) Q.37 (D) Q.38 (D) Q.39 (B) Q.40 (B)
Q.41 (B) Q.42 (D) Q.43 (B) Q.44 (Bbonous) Q.45 (C) Q.46 (B) Q.47 (D) Q.48 (A) Q.49 (B) Q.50 (A)
Q.51 (C) Q.52 (A) Q.53 (D) Q.54 (C) Q.55 (A) Q.56 (*) Q.57 (D) Q.58 (D) Q.59 (B) Q.60 (C)
Q.61 (C) Q.62 (A) Q.63 (A) Q.64 (C) Q.65 (A) Q.66 (C) Q.67 (B) Q.68 (A) Q.69 (D) Q.70 (A)
Q.71 (B) Q.72 (D) Q.73 (B) Q.74 (C) Q.75 (C) Q.76 (C) Q.77 (D) Q.78 (C) Q.79 (B) Q.80 (A)
Q.81 (D) Q.82 (C) Q.83 (A) Q.84 (D) Q.85 (B) Q.86 (C) Q.87 (C) Q.88 (D) Q.89 (C) Q.90 (A)
Q.91 (B) Q.92 (A) Q.93 (D) Q.94 (C) Q.95 (B) Q.96 (A) Q.97 (A) Q.98 (C) Q.99 (A) Q.100 (C)
Q.101 (A) Q.102 (D) Q.103 (1) Q.104 (D)

NTSE STAGE-2
PREVIOUS YEAR'S
Q.1 (C) Q.2 (C) Q.3 (C) Q.4 (B) Q.5 (C) Q.6 (A) Q.7 (B) Q.8 (C) Q.9 (A) Q.10 (D)
Q.11 (A) Q.12 (C) Q.13 (A) Q.14 (B) Q.15 (A) Q.16 (C) Q.17 (B) Q.18 (A) Q.19 (B) Q.20 (C)

3-90
 Real Number

CHEMISTRY
ANSWER KEY
EXERCISE-II

Q.1 (D) Q.2 (D) Q.3 (A) Q.4 (A) Q.5 (C) Q.6 (C) Q.7 (C) Q.8 (A) Q.9 (B) Q.10 (A)
Q.11 (C) Q.12 (A) Q.13 (D) Q.14 (ABC) Q.15 (D) Q.16 (A) Q.17 (C) Q.18 (B) Q.19 (A) Q.20 (A)
Q.21 (D) Q.22 (D) Q.23 (D) Q.24 (B) Q.25 (A) Q.26 (D) Q.27 (D) Q.28 (B) Q.29 (D) Q.30 (A)
Q.31 (B) Q.32 (A) Q.33 (B) Q.34 (A) Q.35 (C) Q.36 (D) Q.37 (D) Q.38 (A) Q.39 (A) Q.40 (D)
Q.41 (A) Q.42 (D) Q.43 (D) Q.44 (B) Q.45 (C) Q.46 (B) Q.47 (A) Q.48 (C) Q.49 (B) Q.50 (C)
Q.51 (A) Q.52 (C) Q.53 (A) Q.54 (D) Q.55 (C) Q.56 (B) Q.57 (A) Q.58 (B) Q.59 (A) Q.60 (A)
Q.61 (A) Q.62 (C) Q.63 (A) Q.64 (D) Q.65 (C) Q.66 (A) Q.67 (C) Q.68 (D) Q.69 (B) Q.70 (B)
Q.71 (C) Q.72 (D) Q.73 (C) Q.74 (D) Q.75 (A) Q.76 (D) Q.77 (C) Q.78 (B) Q.79 (C) Q.80 (C)
Q.81 (C) Q.82 (D) Q.83 (B) Q.84 (D) Q.85 (A)

EXERCISE-III

Q.1 (D) Q.2 (A) Q.3 (C) Q.4 (A) Q.5 (D) Q.6 (A) Q.7 (D) Q.8 (A) Q.9 (D) Q.10 (D)
Q.11 (D) Q.12 (C) Q.13 (B,C) Q.14 (B) Q.15 (B) Q.16 (D) Q.17 (A) Q.18 (C) Q.19 (D) Q.20 (C)
Q.21 (B) Q.22 (B) Q.23 (D) Q.24 (A) Q.25 (A) Q.26 (B) Q.27 (A) Q.28 (B) Q.29 (D) Q.30 (B)
Q.31 (B) Q.32 (C) Q.33 (D) Q.34 (B) Q.35 (A) Q.36 (C) Q.37 (A) Q.38 (D) Q.39 (A) Q.40 (A)
Q.41 (A) Q.42 (D) Q.43 (C) Q.44 (A) Q.45 (B) Q.46 (A) Q.47 (C) Q.48 (C) Q.49 (C) Q.50 (C)
Q.51 (A) Q.52 (C) Q.53 (D) Q.54 (B) Q.55 (D) Q.56 (B) Q.57 (B) Q.58 (A) Q.59 (A) Q.60 (A)
Q.61 (C) Q.62 (B) Q.63 (A) Q.64 (C) Q.65 (D) Q.66 (D) Q.67 (C) Q.68 (A) Q.69 (B) Q.70 (A)
Q.71 (A) Q.72 (C) Q.73 (C) Q.74 (B) Q.75 (B) Q.76 (B) Q.77 (B) Q.78 (C) Q.79 (B) Q.80 (C)
Q.81 (B) Q.82 (A) Q.83 (B) Q.84 (D) Q.85 (B) Q.86 (A) Q.87 (C) Q.88 (D) Q.89 (A) Q.90 (B)
Q.91 (D) Q.92 (C) Q.93 (A) Q.94 (D) Q.95 (D) Q.96 (C) Q.97 (A) Q.98 (B) Q.99 (C) Q.100 (B)
Q.101 (D) Q.102 (D) Q.103 (C) Q.104 (C) Q.105 (A,C) Q.106 (A) Q.107 (D) Q.108 (B) Q.109 (C) Q.110 (B)
Q.111 (B) Q.112 (D) Q.113 (C) Q.114 (C) Q.115 (A) Q.116 (C) Q.117 (C) Q.118 (A,C) Q.119 (B)

EXERCISE-IV
NTSE STAGE-1
PREVIOUS YEAR'S
Q.1 (B) Q.2 (D) Q.3 (B) Q.4 (2) Q.5 (4) Q.6 (B) Q.7 (A) Q.8 (D) Q.9 (D) Q.10 (B)
Q.11 (A) Q.12 (C) Q.13 (D) Q.14 (B) Q.15 (C) Q.16 (D) Q.17 (D) Q.18 (A) Q.19 (A) Q.20 (A)
Q.21 (D) Q.22 (B) Q.23 (C) Q.24 (A) Q.25 (A) Q.26 (C) Q.27 (B) Q.28 (C) Q.29 (C) Q.30 (B)
Q.31 (C) Q.32 (A) Q.33 (3) Q.34 ( ) Q.35 (B) Q.36 (D) Q.37 (D) Q.38 (A) Q.39 (B) Q.40 (A)
Q.41 (B) Q.42 (C) Q.43 (C) Q.44 (C) Q.45 (B) Q.46 (C) Q.47 (D) Q.48 (B) Q.49 (A) Q.50 (C)

NTSE STAGE-2
PREVIOUS YEAR'S
Q.1 (B) Q.2 (C) Q.3 (A) Q.4 (B) Q.5 (A) Q.6 (B) Q.7 (D) Q.8 (A) Q.9 (C) Q.10 (D)
Q.11 (D)

3-91
Real Number

BIOLOGY
ANSWER KEY

EXERCISES-II
Q.1 (A) Q.2 (D) Q.3 (D) Q.4 (D) Q.5 (A) Q.6 (B) Q.7 (A) Q.8 (A) Q.9 (B) Q.10 (D)
Q.11 (C) Q.12 (D) Q.13 (A) Q.14 (B) Q.15 (B) Q.16 (D) Q.17 (A) Q.18 (A) Q.19 (C) Q.20 (D)
Q.21 (A) Q.22 (B) Q.23 (B) Q.24 (D) Q.25 (D) Q.26 (D) Q.27 (D) Q.28 (B) Q.29 (C) Q.30 (A)
Q.31 (B) Q.32 (B) Q.33 (D) Q.34 (A) Q.35 (B) Q.36 (B) Q.37 (B) Q.38 (B) Q.39 (A) Q.40 (D)
Q.41 (C) Q.42 (C) Q.43 (B) Q.44 (A) Q.45 (B) Q.46 (A) Q.47 (B) Q.48 (C) Q.49 (A) Q.50 (B)
Q.51 (B) Q.52 (C) Q.53 (A) Q.54 (A) Q.55 (A) Q.56 (A) Q.57 (C) Q.58 (D) Q.59 (B) Q.60 (A)
Q.61 (C) Q.62 (C) Q.63 (A) Q.64 (D) Q.65 (D) Q.66 (A) Q.67 (C) Q.68 (D) Q.69 (A) Q.70 (A)
Q.71 (B) Q.72 (C) Q.73 (C) Q.74 (B) Q.75 (B) Q.76 (C) Q.77 (A) Q.78 (C) Q.79 (A) Q.80 (A)
Q.81 (C) Q.82 (B) Q.83 (B) Q.84 (C) Q.85 (A) Q.86 (A) Q.87 (A) Q.88 (C)

EXERCISES-III

Q.1 (C) Q.2 (A) Q.3 (B) Q.4 (C) Q.5 (A) Q.6 (C) Q.7 (C) Q.8 (C) Q.9 (D) Q.10 (D)
Q.11 (D) Q.12 (C) Q.13 (D) Q.14 (A) Q.15 (A) Q.16 (B) Q.17 (D) Q.18 (C) Q.19 (C) Q.20 (A)
Q.21 (A) Q.22 (C) Q.23 (B) Q.24 (D) Q.25 (A) Q.26 (B) Q.27 (D) Q.28 (B) Q.29 (C) Q.30 (A)
Q.31 (A) Q.32 (D) Q.33 (D) Q.34 (A) Q.35 (A) Q.36 (A) Q.37 (D)

EXERCISES-IV
NTSE STAGE-1
PREVIOUS YEAR'S
Q.1 (C) Q.2 (C) Q.3 (C) Q.4 (D) Q.5 (B) Q.6 (C) Q.7 (D) Q.8 (D) Q.9 (C) Q.10 (C)
Q.11 (C) Q.12 (B) Q.13 (B) Q.14 (A) Q.15 (B) Q.16 (A) Q.17 (C) Q.18 (D) Q.19 (B) Q.20 (A)
Q.21 (C) Q.22 (B) Q.23 (A) Q.24 (D) Q.25 (A) Q.26 (A) Q.27 (C) Q.28 (D) Q.29 (A) Q.30 (A)
Q.31 (B) Q.32 (B) Q.33 (B) Q.34 (A) Q.35 (B) Q.36 (C) Q.37 (D) Q.38 (C) Q.39 (C) Q.40 (D)
Q.41 (B) Q.42 (A) Q.43 (A) Q.44 (D) Q.45 (C) Q.46 (C) Q.47 (A) Q.48 (A) Q.49 (C) Q.50 (C)
Q.51 (C) Q.52 (B) Q.53 (C) Q.54 (A) Q.55 (B) Q.56 (B) Q.57 (D) Q.58 (D) Q.59 (C) Q.60 (B)
Q.61 (A)

NTSE STAGE-2
PREVIOUS YEAR'S
Q.1 (C) Q.2 (D) Q.3 (B) Q.4 (A) Q.5 (B) Q.6 (B) Q.7 (C) Q.8 (C) Q.9 (C) Q.10 (C)
Q.11 (C) Q.13 (B) Q.14 (D)

3-92
Real Number

MATHEMATICS
ANSWER KEY

EXERCISES-I

-2 3
Q.1 4 decimal places. Q.2 11 × 13 × 19 Q.3 Number are not co-prime. Q.4 30 Q.5 or
3 4
Q.6 128 Q.7 HCF = 5 Q.8 3 × 11 Q.10 89
Q.11 is divisible by 2, 11 and 256 and it has more than 2 prime factors Q.13 13 Q.14 3 × 71
Q.16 – 925 Q.17 44,550 Q.21 3 -5 Q.23 180 min. or 3 hrs.
Q.26 64 Q.28 240 Q.29 130

EXERCISES-II
Q.1 (B) Q.2 (C) Q.3 (D) Q.4 (B) Q.5 (A) Q.6 (C) Q.7 (C) Q.8 (B) Q.9 (C) Q.10 (B)
Q.11 (D) Q.12 (D) Q.13 (B) Q.14 (B) Q.15 (D) Q.16 (C) Q.17 (C) Q.18 (B) Q.19 (A) Q.20 (C)
Q.21 (D) Q.22 (C) Q.23 (A) Q.24 (D) Q.25 (C) Q.26 (D) Q.27 (C) Q.28 (D) Q.29 (B) Q.30 (A)
Q.31 (C) Q.32 (A) Q.33 (C) Q.34 (A) Q.35 (A) Q.36 (A) Q.37 (D) Q.38 (C) Q.39 (B) Q.40 (C)
Q.41 (A) Q.42 (B) Q.43 (D) Q.44 (C) Q.45 (B) Q.46 (C) Q.47 (A) Q.48 (A) Q.49 (C) Q.50 (B)
Q.51 (D) Q.52 (D) Q.53 (C) Q.54 (A) Q.55 (B) Q.56 (C) Q.57 (B) Q.58 (C) Q.59 (A) Q.60 (B)
Q.61 (D) Q.62 (C) Q.63 (B) Q.64 (C) Q.65 (C) Q.66 (A) Q.67 (D) Q.68 (B) Q.69 (C) Q.70 (C)
Q.71 (B) Q.72 (B) Q.73 (C) Q.74 (D) Q.75 (B) Q.76 (B) Q.77 (C) Q.78 (B) Q.79 (B) Q.80 (D)
Q.81 (B) Q.82 (B) Q.83 (D) Q.84 (A) Q.85 (B) Q.86 (B) Q.87 (B) Q.88 (B) Q.89 (A) Q.90 (A)
Q.91 (A) Q.92 (D) Q.93 (B) Q.94 (C) Q.95 (A) Q.96 (D) Q.97 (B) Q.98 (C) Q.99 (D) Q.100 (C)
Q.101 (D) Q.102 (D) Q.103 (A) Q.104 (D) Q.105 (D) Q.106 (B) Q.107 (C) Q.108 (B) Q.109 (C) Q.110 (B)
Q.111 (B) Q.112 (A) Q.113 (D) Q.114 (D) Q.115 (B) Q.116 (B) Q.117 (D) Q.118 (D) Q.119 (A) Q.120 (D)
Q.121 (B) Q.122 (C) Q.123 (A) Q.124 (A) Q.125 (C) Q.126 (A) Q.127 (C) Q.128 (C) Q.129 (D) Q.130 (D)

EXERCISE-III
Q.1 (B) Q.2 (B) Q.3 (D) Q.4 (B) Q.5 (D) Q.6 (C) Q.7 (A) Q.8 (C) Q.9 (B) Q.10 (A)
Q.11 (C) Q.12 (A) Q.13 (C) Q.14 (C) Q.15 (A) Q.16 (D) Q.17 (A) Q.18 (C) Q.19 (A) Q.20 (D)
Q.21 (A) Q.22 (C) Q.23 (B) Q.24 (B) Q.25 (B) Q.26 (D) Q.27 (B) Q.28 (B) Q.29 (A) Q.30 (A)
Q.31 (D) Q.32 (B) Q.33 (A) Q.34 (D) Q.35 (C) Q.36 (B) Q.37 (B) Q.38 (D) Q.39 (B) Q.40 (C)
Q.41 (B) Q.42 (C) Q.43 (D) Q.44 (D) Q.45 (A) Q.46 (D) Q.47 (D) Q.48 (B) Q.49 (D) Q.50 (B)
Q.51 (C) Q.52 (A) Q.53 (B) Q.54 (C) Q.55 (D) Q.56 (C) Q.57 (B) Q.58 (D) Q.59 (A) Q.60 (B)
Q.61 (C) Q.62 (A) Q.63 (C) Q.64 (A) Q.65 (D) Q.66 (B) Q.67 (A) Q.68 (D)

EXERCISE-IV
NTSE STAGE-1
PREVIOUS YEAR'S
Q.1 (D) Q.2 (C) Q.3 (D) Q.4 (B) Q.5 (D) Q.6 (B) Q.7 (A) Q.8 (C) Q.9 (A) Q.10 (A)
Q.11 (D) Q.12 (B) Q.13 (D) Q.14 (A) Q.15 (D) Q.16 (C) Q.17 (D) Q.18 (A) Q.19 (B) Q.20 (D)
Q.21 (C) Q.22 (B) Q.23 (A) Q.24 (C) Q.25 (A) Q.26 (A) Q.27 (A) Q.28 (D) Q.29 (B) Q.30 (B)
Q.31 (C) Q.32 (B) Q.33 (D) Q.34 (C) Q.35 (Bonus) Q.36 (B) Q.37 (A) Q.38 (C) Q.39 (A) Q.40 (B)

3-188
 Real Number

Q.41 (Bonus) Q.42 (B) Q.43 (D) Q.44 (B) Q.45 (D) Q.46 (B) Q.47 (B) Q.48 (C) Q.49 (B) Q.50 (B)
Q.51 (C) Q.52 (D) Q.53 (A) Q.54 (C) Q.55 (D) Q.56 (C) Q.57 (C) Q.58 (A) Q.59 (B) Q.60 (C)
Q.61 (B) Q.62 (B) Q.63 (C) Q.64 (C) Q.65 (B) Q.66 (A) Q.67 (A) Q.68 (C) Q.69 (B) Q.70 (A)
Q.71 (A) Q.72 (C) Q.73 (A) Q.74 (D) Q.75 (D) Q.76 (Bonous) Q.77 (C) Q.78 (A) Q.79 (B) Q.80 (B)
Q.81 (B) Q.82 (B) Q.83 (A) Q.84 (D) Q.85 (A) Q.86 (B) Q.87 (D) Q.88 (B) Q.89 (C) Q.90 (D)
Q.91 (D) Q.92 (C) Q.93 (B) Q.94 (C) Q.95 (D) Q.96 (B) Q.97 (D) Q.98 (D) Q.99 (C) Q.100 (B)
Q.101 (D) Q.102 (B) Q.103 (C) Q.104 (C) Q.105 (D) Q.106 (A) Q.107 (A) Q.108 (C) Q.109 (D) Q.110 (C)
Q.111 (D) Q.112 (A) Q.113 (C) Q.114 (C) Q.115 (A) Q.116 (A) Q.117 (D) Q.118 (A) Q.119 (B) Q.120 (A)
Q.121 (C) Q.122 (A) Q.123 (B) Q.124 (C) Q.125 (B) Q.126 (D) Q.127 (B) Q.128 (A) Q.129 (A) Q.130 (B)
Q.131 (D) Q.132 (C) Q.133 (A) Q.134 (B) Q.135 (B) Q.136 (D) Q.137 (A) Q.138 (D) Q.139 (A) Q.140 (B)
Q.141 (C) Q.142 (B) Q.143 (B) Q.144 (D) Q.145 (A) Q.146 (C) Q.147 (B) Q.148 (C) Q.149 (C) Q.150 (D)
Q.151 (B) Q.152 (B) Q.153(Bonous) Q.154 (B) Q.155 (D) Q.156 (C) Q.157 (B) Q.158 (D) Q.159 (B) Q.160 (D)
Q.161 (B) Q.162 (C) Q.163 (D) Q.164 (B) Q.165 (D) Q.166 (A) Q.167 (C) Q.168 (D) Q.169 (D) Q.170 (C)

NTSE STAGE-2
PREVIOUS YEAR'S
Q.1 (D) Q.2 (D) Q.3 (A) Q.4 (A) Q.5 (D) Q.6 (B) Q.7 (A) Q.8 (D) Q.9 (D) Q.10 (C)
Q.11 (C) Q.12 (C) Q.13 (A) Q.14 (D) Q.15 (D) Q.16 (D) Q.17 (B) Q.18 (C) Q.19 (C) Q.20 (C)
Q.21 (C) Q.22 (B) Q.23 (C) Q.24 (B) Q.25 (B) Q.26 (A) Q.27 (C) Q.28 (B) Q.29 (A) Q.30 (D)
Q.31 (C) Q.32 (A)

3-189
Real Number

PHYSICS
ANSWER KEY
EXERCISES-II

Q.1 (D) Q.2 (C) Q.3 (A) Q.4 (A) Q.5 (C) Q.6 (B) Q.7 (D) Q.8 (B) Q.9 (A) Q.10 (B)
Q.11 (C) Q.12 (D) Q.13 (B) Q.14 (D) Q.15 (A) Q.16 (A) Q.17 (C) Q.18 (C) Q.19 (D) Q.20 (B)
Q.21 (B) Q.22 (C) Q.23 (C) Q.24 (A) Q.25 (C) Q.26 (B) Q.27 (B) Q.28 (B) Q.29 (D) Q.30 (A)
Q.31 (D) Q.32 (A) Q.33 (B) Q.34 (B) Q.35 (B) Q.36 (A) Q.37 (C) Q.38 (B) Q.39 (B) Q.40 (B)
Q.41 (A) Q.42 (C) Q.43 (D) Q.44 (C) Q.45 (D) Q.46 (D) Q.47 (A) Q.48 (C) Q.49 (A) Q.50 (C)
Q.51 (B) Q.52 (D) Q.53 (B) Q.54 (A) Q.55 (B) Q56 (B) Q.57 (B) Q.58 (C) Q.59 (B) Q.60 (D)
Q.61 (A) Q.62 (B) Q.63 (B) Q.64 (C) Q.65 (D) Q.66 (B) Q.67 (C) Q.68 (B) Q.69 (B) Q.70 (A)
Q.71 (D) Q.72 (C) Q.73 (A) Q.74 (A) Q.75 (A) Q.76 (D) Q.77 (A) Q.78 (B) Q.79 (C) Q.80 (C)
Q.81 (D) Q.82 (D) Q.83 (C) Q.84 (A) Q.85 (A) Q.86 (B) Q.87 (C) Q.88 (B) Q.89 (B) Q.90 (C)
Q.91 (B) Q.92 (A) Q.93 (B) Q.94 (B) Q.95 (A) Q.96 (C) Q.97 (C) Q.98 (C) Q.99 (B) Q.100 (C)
Q.101 (D) Q.102 (A) Q.103 (B) Q.104 (B) Q.105 (A) Q.106 (C) Q.107 (C) Q.108 (D) Q.109 (A) Q.110 (A)
Q.111 (D) Q.112 (A) Q.113 (D) Q.114 (C) Q.115 (B) Q.116 (D) Q.118 (C)

EXERCISES-III
Q.1 (B) Q.2 (D) Q.3 (A) Q.4 (B) Q.5 (D) Q.6 (A) Q.7 (B) Q.8 (B) Q.9 (D) Q.10 (B)
Q.11 (C) Q.12 (D) Q.13 (C) Q.14 (D) Q.15 (A) Q.16 (C) Q.17 (A) Q.18 (A) Q.19 (A) Q.20 (B)
Q.21 (A) Q.22 (B) Q.23 (C) Q.24 (C) Q.25 (C) Q.26 (B) Q.27 (B) Q.28 (C) Q.29 (C) Q.30 (C)
Q.31 (D) Q.32 (D) Q.33 (B) Q.34 (C) Q.35 (D) Q.36 (B) Q.37 (B) Q.38 (B) Q.39 (C) Q.40 (A)
Q.41 (A) Q.42 (A) Q.43 (B) Q.44 (D) Q.45 (A) Q.46 (D) Q.47 (C) Q.48 (C) Q.49 (D) Q.50 (D)
Q.51 (D) Q.52 (B) Q.53 (A) Q.54 (C) Q.55 (C)

EXERCISES-IV
NTSE STAGE-1
PREVIOUS YEAR'S

Q.1 (D) Q.2 (B) Q.3 (C) Q.4 (A) Q.5 (B) Q.6 (A) Q.7 (A) Q.8 (D) Q.9 (C) Q.10 (A)
Q.11 (D) Q.12 (A) Q.13 (C) Q.14 (B) Q.15 (C) Q.16 (D) Q.17 (D) Q.18 (A) Q.19 (D) Q.20 (C)
Q.21 (B) Q.22 (B) Q.23 (A) Q.24 (C) Q.25 (A) Q.26 (D) Q.27 (B) Q.28 (C) Q.29 (A) Q.30 (A)
Q.31 (C) Q.32 (D) Q.33 (B) Q.34 (B) Q.35 (C) Q.36 (C) Q.37 (D) Q.38 (D) Q.39 (B) Q.40 (B)
Q.41 (B) Q.42 (D) Q.43 (B) Q.44 (Bbonous) Q.45 (C) Q.46 (B) Q.47 (D) Q.48 (A) Q.49 (B) Q.50 (A)
Q.51 (C) Q.52 (A) Q.53 (D) Q.54 (C) Q.55 (A) Q.56 (*) Q.57 (D) Q.58 (D) Q.59 (B) Q.60 (C)
Q.61 (C) Q.62 (A) Q.63 (A) Q.64 (C) Q.65 (A) Q.66 (C) Q.67 (B) Q.68 (A) Q.69 (D) Q.70 (A)
Q.71 (B) Q.72 (D) Q.73 (B) Q.74 (C) Q.75 (C) Q.76 (C) Q.77 (D) Q.78 (C) Q.79 (B) Q.80 (A)
Q.81 (D) Q.82 (C) Q.83 (A) Q.84 (D) Q.85 (B) Q.86 (C) Q.87 (C) Q.88 (D) Q.89 (C) Q.90 (A)
Q.91 (B) Q.92 (A) Q.93 (D) Q.94 (C) Q.95 (B) Q.96 (A) Q.97 (A) Q.98 (C) Q.99 (A) Q.100 (C)
Q.101 (A) Q.102 (D) Q.103 (1) Q.104 (D)

NTSE STAGE-2
PREVIOUS YEAR'S
Q.1 (C) Q.2 (C) Q.3 (C) Q.4 (B) Q.5 (C) Q.6 (A) Q.7 (B) Q.8 (C) Q.9 (A) Q.10 (D)
Q.11 (A) Q.12 (C) Q.13 (A) Q.14 (B) Q.15 (A) Q.16 (C) Q.17 (B) Q.18 (A) Q.19 (B) Q.20 (C)

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 Real Number

CHEMISTRY
ANSWER KEY
EXERCISE-II

Q.1 (D) Q.2 (D) Q.3 (A) Q.4 (A) Q.5 (C) Q.6 (C) Q.7 (C) Q.8 (A) Q.9 (B) Q.10 (A)
Q.11 (C) Q.12 (A) Q.13 (D) Q.14 (ABC) Q.15 (D) Q.16 (A) Q.17 (C) Q.18 (B) Q.19 (A) Q.20 (A)
Q.21 (D) Q.22 (D) Q.23 (D) Q.24 (B) Q.25 (A) Q.26 (D) Q.27 (D) Q.28 (B) Q.29 (D) Q.30 (A)
Q.31 (B) Q.32 (A) Q.33 (B) Q.34 (A) Q.35 (C) Q.36 (D) Q.37 (D) Q.38 (A) Q.39 (A) Q.40 (D)
Q.41 (A) Q.42 (D) Q.43 (D) Q.44 (B) Q.45 (C) Q.46 (B) Q.47 (A) Q.48 (C) Q.49 (B) Q.50 (C)
Q.51 (A) Q.52 (C) Q.53 (A) Q.54 (D) Q.55 (C) Q.56 (B) Q.57 (A) Q.58 (B) Q.59 (A) Q.60 (A)
Q.61 (A) Q.62 (C) Q.63 (A) Q.64 (D) Q.65 (C) Q.66 (A) Q.67 (C) Q.68 (D) Q.69 (B) Q.70 (B)
Q.71 (C) Q.72 (D) Q.73 (C) Q.74 (D) Q.75 (A) Q.76 (D) Q.77 (C) Q.78 (B) Q.79 (C) Q.80 (C)
Q.81 (C) Q.82 (D) Q.83 (B) Q.84 (D) Q.85 (A)

EXERCISE-III

Q.1 (D) Q.2 (A) Q.3 (C) Q.4 (A) Q.5 (D) Q.6 (A) Q.7 (D) Q.8 (A) Q.9 (D) Q.10 (D)
Q.11 (D) Q.12 (C) Q.13 (B,C) Q.14 (B) Q.15 (B) Q.16 (D) Q.17 (A) Q.18 (C) Q.19 (D) Q.20 (C)
Q.21 (B) Q.22 (B) Q.23 (D) Q.24 (A) Q.25 (A) Q.26 (B) Q.27 (A) Q.28 (B) Q.29 (D) Q.30 (B)
Q.31 (B) Q.32 (C) Q.33 (D) Q.34 (B) Q.35 (A) Q.36 (C) Q.37 (A) Q.38 (D) Q.39 (A) Q.40 (A)
Q.41 (A) Q.42 (D) Q.43 (C) Q.44 (A) Q.45 (B) Q.46 (A) Q.47 (C) Q.48 (C) Q.49 (C) Q.50 (C)
Q.51 (A) Q.52 (C) Q.53 (D) Q.54 (B) Q.55 (D) Q.56 (B) Q.57 (B) Q.58 (A) Q.59 (A) Q.60 (A)
Q.61 (C) Q.62 (B) Q.63 (A) Q.64 (C) Q.65 (D) Q.66 (D) Q.67 (C) Q.68 (A) Q.69 (B) Q.70 (A)
Q.71 (A) Q.72 (C) Q.73 (C) Q.74 (B) Q.75 (B) Q.76 (B) Q.77 (B) Q.78 (C) Q.79 (B) Q.80 (C)
Q.81 (B) Q.82 (A) Q.83 (B) Q.84 (D) Q.85 (B) Q.86 (A) Q.87 (C) Q.88 (D) Q.89 (A) Q.90 (B)
Q.91 (D) Q.92 (C) Q.93 (A) Q.94 (D) Q.95 (D) Q.96 (C) Q.97 (A) Q.98 (B) Q.99 (C) Q.100 (B)
Q.101 (D) Q.102 (D) Q.103 (C) Q.104 (C) Q.105 (A,C) Q.106 (A) Q.107 (D) Q.108 (B) Q.109 (C) Q.110 (B)
Q.111 (B) Q.112 (D) Q.113 (C) Q.114 (C) Q.115 (A) Q.116 (C) Q.117 (C) Q.118 (A,C) Q.119 (B)

EXERCISE-IV
NTSE STAGE-1
PREVIOUS YEAR'S
Q.1 (B) Q.2 (D) Q.3 (B) Q.4 (2) Q.5 (4) Q.6 (B) Q.7 (A) Q.8 (D) Q.9 (D) Q.10 (B)
Q.11 (A) Q.12 (C) Q.13 (D) Q.14 (B) Q.15 (C) Q.16 (D) Q.17 (D) Q.18 (A) Q.19 (A) Q.20 (A)
Q.21 (D) Q.22 (B) Q.23 (C) Q.24 (A) Q.25 (A) Q.26 (C) Q.27 (B) Q.28 (C) Q.29 (C) Q.30 (B)
Q.31 (C) Q.32 (A) Q.33 (3) Q.34 ( ) Q.35 (B) Q.36 (D) Q.37 (D) Q.38 (A) Q.39 (B) Q.40 (A)
Q.41 (B) Q.42 (C) Q.43 (C) Q.44 (C) Q.45 (B) Q.46 (C) Q.47 (D) Q.48 (B) Q.49 (A) Q.50 (C)

NTSE STAGE-2
PREVIOUS YEAR'S
Q.1 (B) Q.2 (C) Q.3 (A) Q.4 (B) Q.5 (A) Q.6 (B) Q.7 (D) Q.8 (A) Q.9 (C) Q.10 (D)
Q.11 (D)

3-191
Real Number

BIOLOGY
ANSWER KEY

EXERCISES-II
Q.1 (A) Q.2 (D) Q.3 (D) Q.4 (D) Q.5 (A) Q.6 (B) Q.7 (A) Q.8 (A) Q.9 (B) Q.10 (D)
Q.11 (C) Q.12 (D) Q.13 (A) Q.14 (B) Q.15 (B) Q.16 (D) Q.17 (A) Q.18 (A) Q.19 (C) Q.20 (D)
Q.21 (A) Q.22 (B) Q.23 (B) Q.24 (D) Q.25 (D) Q.26 (D) Q.27 (D) Q.28 (B) Q.29 (C) Q.30 (A)
Q.31 (B) Q.32 (B) Q.33 (D) Q.34 (A) Q.35 (B) Q.36 (B) Q.37 (B) Q.38 (B) Q.39 (A) Q.40 (D)
Q.41 (C) Q.42 (C) Q.43 (B) Q.44 (A) Q.45 (B) Q.46 (A) Q.47 (B) Q.48 (C) Q.49 (A) Q.50 (B)
Q.51 (B) Q.52 (C) Q.53 (A) Q.54 (A) Q.55 (A) Q.56 (A) Q.57 (C) Q.58 (D) Q.59 (B) Q.60 (A)
Q.61 (C) Q.62 (C) Q.63 (A) Q.64 (D) Q.65 (D) Q.66 (A) Q.67 (C) Q.68 (D) Q.69 (A) Q.70 (A)
Q.71 (B) Q.72 (C) Q.73 (C) Q.74 (B) Q.75 (B) Q.76 (C) Q.77 (A) Q.78 (C) Q.79 (A) Q.80 (A)
Q.81 (C) Q.82 (B) Q.83 (B) Q.84 (C) Q.85 (A) Q.86 (A) Q.87 (A) Q.88 (C)

EXERCISES-III

Q.1 (C) Q.2 (A) Q.3 (B) Q.4 (C) Q.5 (A) Q.6 (C) Q.7 (C) Q.8 (C) Q.9 (D) Q.10 (D)
Q.11 (D) Q.12 (C) Q.13 (D) Q.14 (A) Q.15 (A) Q.16 (B) Q.17 (D) Q.18 (C) Q.19 (C) Q.20 (A)
Q.21 (A) Q.22 (C) Q.23 (B) Q.24 (D) Q.25 (A) Q.26 (B) Q.27 (D) Q.28 (B) Q.29 (C) Q.30 (A)
Q.31 (A) Q.32 (D) Q.33 (D) Q.34 (A) Q.35 (A) Q.36 (A) Q.37 (D)

EXERCISES-IV
NTSE STAGE-1
PREVIOUS YEAR'S
Q.1 (C) Q.2 (C) Q.3 (C) Q.4 (D) Q.5 (B) Q.6 (C) Q.7 (D) Q.8 (D) Q.9 (C) Q.10 (C)
Q.11 (C) Q.12 (B) Q.13 (B) Q.14 (A) Q.15 (B) Q.16 (A) Q.17 (C) Q.18 (D) Q.19 (B) Q.20 (A)
Q.21 (C) Q.22 (B) Q.23 (A) Q.24 (D) Q.25 (A) Q.26 (A) Q.27 (C) Q.28 (D) Q.29 (A) Q.30 (A)
Q.31 (B) Q.32 (B) Q.33 (B) Q.34 (A) Q.35 (B) Q.36 (C) Q.37 (D) Q.38 (C) Q.39 (C) Q.40 (D)
Q.41 (B) Q.42 (A) Q.43 (A) Q.44 (D) Q.45 (C) Q.46 (C) Q.47 (A) Q.48 (A) Q.49 (C) Q.50 (C)
Q.51 (C) Q.52 (B) Q.53 (C) Q.54 (A) Q.55 (B) Q.56 (B) Q.57 (D) Q.58 (D) Q.59 (C) Q.60 (B)
Q.61 (A)

NTSE STAGE-2
PREVIOUS YEAR'S
Q.1 (C) Q.2 (D) Q.3 (B) Q.4 (A) Q.5 (B) Q.6 (B) Q.7 (C) Q.8 (C) Q.9 (C) Q.10 (C)
Q.11 (C) Q.13 (B) Q.14 (D)

3-192