SITEEE Mock Test 3

Total Time: 1 Hour Total Marks: 120

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 Physics
1. A particle starts from origin O from rest and moves with a uniform acceleration (+2)
along the positive x -axis. Identify all figures that correctly represent the motion
qualitatively. (a = acceleration, v = velocity, x = displacement, t = time)

a. (A), (B), (C)

b. (A)

c. (A), (B), (D)

d. (B), (C)

2. In a line of sight radio communication, a distance of about 50 km is kept (+2)

between the transmitting and receiving antennas. If the height of the receiving
antenna is 70 m , then the minimum height of the transmitting antenna should be
: (Radius of the Earth = 6.4 × 106 m ).

a. 40 m

b. 51 m

c. 32 m

d. 20 m

3. The magnetic field of an electromagnetic wave is given by : B = 1.6 × (+2)

10−6 cos (2 × 107 z + 6 × 1015 t) (2^i + ^j ) Wb2 The associated electric field will be :-
​

m
​

a. E = 4.8 × 102 cos (2 × 107 z + 6 × 1015 t) (i^ − 2^j ) m

​
V
​

b. E = 4.8 × 102 cos (2 × 107 z − 6 × 1015 t) (2^i + ^j ) m

​
V
​

c. E = 4.8 × 102 cos (2 × 107 z − 6 × 1015 t) (−2^i + ^j ) m

V
​ ​

d. E = 4.8 × 102 cos (2 × 107 z + 6 × 1015 t) (−1^i + 2^j ) m

V
​ ​
4. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room (+2)
temperature. It will emit :

a. 2 lines in the Lyman series and 1 line in the Balmar series

b. 3 lines in the Lyman series

c. 1 line in the Lyman series and 2 lines in the Balmar series

d. 3 lines in the Balmer series

5. A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27∘ C . (+2)

Amount of heat transferred to the gas, so that rms velocity of molecules is
doubled, is about : [Take R = 8.3 J/ K mole]

a. 10 kJ

b. 0.9 kJ

c. 6 kJ

d. 14 kJ

6. A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown (+2)

in figure. The time at which rate of dissipation of energy (joule's heat) across
resistance is equal to the rate at which magnetic energy is stored in the
inductor is :

a. 2
ℓn2
​

b. ℓn2

c. 2ℓn2

d. 1
2
​ℓn2

7. A 25 cm long solenoid has radius 2 cm and 500 total number of turns. It carries a (+2)
current of 15 A . If it is equivalent to a magnet of the same size and
magnetization ∣M ˉ ∣ (magnetic moment/volume), then ∣M ˉ ∣ is :
 a. 3πAm−1

b. 30000 Am−1

c. 300 Am−1

d. 30000 πAm−1

8. An LCR circuit contains resistance of 110Ω and a supply of 220V at 300rad/s (+2)
angular frequency. If only capacitance is removed from the circuit, current lags
behind the voltage by 45∘ . If on the other hand, only inductor is removed the
current leads by 45∘ with the applied voltage. The rms current flowing in the
circuit will be :

a. 1A

b. 2.5A

c. 1.5A

d. 2A

9. Determine the electric dipole moment of the system of three charges, placed on (+2)
the vertices of an equilateral triangle, as shown in the figure:

^ ^
a. (ql) i+2j ​
​

^ ^
b. 3ql j −2I
​
​

​
​

c. − 3ql^j ​ ​

d. 2ql^j ​

10. Determine the charge on the capacitor in the following circuit : (+2)

a. 2?C

b. 60?C
 c. 200?C

d. 10?C

11. If surface tension (S), Moment of inertia (I) and Planck's constant (h), were to be (+2)
taken as the fundamental units, the dimensional formula for linear momentum
would be : -

3 1
a. S 2 I 2 h0 ​ ​

1 1
b. S 2 I 2 h0 ​ ​

1 1
c. S 2 I 2 h−1
​ ​

1 3
d. S 2 I 2 h−1​ ​

12. In the circuit shown, the switch S1 is closed at time t = 0 and the switch S2 is
​ ​ (+2)
kept open. At some later time (t0 ) , the switch S1 is opened and S2 is closed. The
​ ​ ​

behavious of the current I as a function of time ′ t′ is given by :

a.

b.
 c.

d.

13. Proton with kinetic energy of 1 M eV moves from south to north It gets an (+2)
acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value
of magnetic field : (Rest mass of proton is 1.6 × 10−27 kg )

a. 7.1 mT

b. 71 mT

c. 0.071 mT

d. 0.71 mT

14. The actual value of resistance R , shown in the figure is 30Ω . This is measured in (+2)
an experiment as shown using the standard formula R = V
I
​
, where V and I are
the readings of the voltmeter and ammeter, respectively. If the measured value
of R is 5% less, then the internal resistance of the voltmeter is :

a. 350 Ω
 b. 570 Ω

c. 35 Ω

d. 600 Ω

15. The angular width of the central maximum in a single slit diffraction pattern is (+2)
60∘ . The width of the slit is 1μm . The slit is illuminated by monochromatic plane
waves. If another slit of same width is made near it, Young's fringes can be
observed on a screen placed at a distance 50 cm from the slits. If the observed
fringe width is 1 cm , what is slit separation distance ? (i.e. distance between the
centres of each slit.)

a. 25 μm

b. 50 μm

c. 75 μm

d. 100 μm
 Chemistry
16. An organic compound contains C, H and S. The minimum molecular weight of (+2)
the compound containing 8% sulphur is : (atomic weight of S=32 amu)

a. 200 g mol−1

b. 400 g mol−1

c. 600 g mol−1

d. 300 g mol−1

17. An organic compound 'X' showing the following solubility profile is - (+2)

a. m-Cresol

b. Oleic acid

c. o-Toluidine

d. Benzamide

18. Bouveault-Blanc reduction reaction involves : (+2)

a. Reduction of an acyl halide with H2/P d

b. Reduction of an ester with N a/C2H5OH

c. Reduction of a carbonyl compound with Na/Hg and HCl

d. Reduction of an anhydride with LiAlH4 ​

19. Bromination of cyclohexene under conditions given below yields : (+2)

 a.

b.

c.

d.

20. Calgon is used for water treatment. Which of the following statement is NOT (+2)
true about Calgon?

a. Calgon contains the 2nd most abundant element by weight in the Earth's
crust.
 b. It is polymeric compound and is water soluble.

c. It is also known as Graham's salt

d. It does not remove Ca2+ ion by precipitation.

21. For which of the following compounds Kjeldahl method can be used to (+2)
determine the percentage of Nitrogen ?

a. Nitrobenzene

b. Pyridine

c. Alanine

d. Diazomethane

22. If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium (+2)
constant (K) for the reaction Zn(s) + cu2+ (aq) <=> Zn2+ (aq) + Cu(s) at 300 K is
approximately. (R = 8 JK −1 mol−1 , F = 96000 C mol−1 )

a. e160

b. e320

c. e−160

d. e−80

23. In Freundlich adsorption isotherm, slope of AB line is : (+2)

a. log n with (n > 1)

b. n with (n, 0.1 to 0.5 )

c. log n1 with (n < 1)

​

d. 1
n
​ with ( n1 = 0 to 1)
​
24. In the cell Pt(s)∣H2 (g, 1bar∣HCl(aq)∣Ag(s)∣Pt(s) the cell potential is 0.92 when a
​ (+2)
10 −6
molal HCl solution is used. THe standard electrode potential of
(AgCl/Ag, Cl− ) electrode is : {given, 2.303RT
F
= 0.06V at 298K }
​

a. 0.20 V

b. 0.76 V

c. 0.40 V

d. 0.94 V

25. Solid Ba(N O3 )2 is gradually dissolved in a 1.0 × 10−4 M N a2 C O3 solution. At

​ ​ ​ ​ (+2)
which concentration of Ba2+ , precipitate of BaCO3 begins to form ? ( Ksp for
​ ​

BaCO3 = 5.1 × 10−9 )

​

a. 5.1 × 10−5 M

b. 7.1 × 10−8 M

c. 4.1 × 10−5 M

d. 8.1 × 10−7 M

26. sp3 d2 hybridization is not displayed by : (+2)

a. BrF 5

b. SF 6

c. [CrF 6]3−

d. P F5 ​

27. Stability of the species Li2 , Li−

2 , and Li2 increases in the order of
+
​ ​ ​ (+2)

a. Li < Li+ −
2 < Li2 ​ ​

b. Li− +
2 < Li2 < Li2
​ ​ ​
 c. Li2 < Li−
​
+
2 < Li2 ​ ​

d. Li− ​
+
2 < Li2 < Li2 ​ ​

28. The aerosol is a kind of colloid in which : (+2)

a. gas is dispersed in solid

b. solid is dispersed in gas

c. liquid is dispersed in water

d. gas is dispersed in liquid

29. The catenation tendency of C, Si and Ge is in the order Ge $ (+2)

a. 398180167

b. 348167180

c. 180167348

d. 167180348

30. The complex that can show fac- and mer- isomers is : (+2)

a. [Co(N H3 )3 (N O2 )3 ] ​ ​ ​ ​

b. [Pt(N H3 )2 C l2 ]
​ ​ ​

c. [Co(N H3 )4a2 ]+ ​ ​

d. [CoCl2 (en)2 ]
​ ​
 Mathematics
31. The length of the projection of the line segment joining the points (5, −1, 4) and (+2)
(4, −1, 3) on the plane, x + y + z = 7 is:

a. 2
3
​

2
b. 3
​

1
c. 3
​

d. 2
3
​ ​

32. The maximum area of a right angled triangle with hypotenuse h is : (+2)

h2
a. 2 2
​

h2
b. 2
​

h2
c. 2
​

h2
d. 4
​

33. The number of all possible positive integral values of α for which the roots of (+2)
the quadratic equation, 6x2 − 11x + a = 0 are rational numbers is :

a. 2

b. 5

c. 3

d. 4

34. The number of common tangents to the circles x2 + y 2 − 4x − 6y − 12 = 0 and (+2)

2 2
x + y + 6x + 18y + 26 = 0

a. 1
 b. 2

c. 3

d. 4

35. The plane through the intersection of the planes x + y + z = 1 and 2x + 3y − z + (+2)
4 = 0 and parallel to y-axis also passes through the point :

a. (-3, 0, -1)

b. (3, 3, -1)

c. (3, 2, 1)

d. (-3, 1, 1)

13 +23 13 +23 +33 13 +23 +33 +....+153

36. The sum 1 + 1+2
​ + 1+2+3
​ + .... + 1+2+3+...+15
​ − 12 (1 + 2 + 3 + ... + 15)
​ (+2)

a. 1240

b. 1860

c. 660

d. 620

37. The system of linear equations λx + 2y + 2z = 5 2λx + 3y + 5z = 8 4x + λy + 6z = (+2)

10 has :

a. no soiution when λ = 2

b. infinitely many solutions when λ = 2

c. no solution when λ = 8

d. a unique solution when λ = −8

38. Two integers are selected at random from the set {1, 2,...., 11}. Given that the sum (+2)
of selected numbers is even, the conditional probability that both the numbers
are even is :

a. 2
5
​

b. 1
2
​

c. 3
5
​

d. 7
10
​

39. Two sets A and B are as under: A = {(a, b) ∈ R × R : ∣a − 5∣ < 1 and ∣b − 5∣ < 1} ; (+2)
B = {(a, b) ∈ R × R : 4(a − 6)2 + 9(b − 5)2 ≤ 36} . Then

a. B ⊂ A

b. A ⊂ B

c. A ∩ B = ϕ (an empty set)

d. neither A ⊂ B nor B ⊂ A

40. The angle θ between the vectors a = 5^i − ^j + k

^ and b = ^i + ^j − k
​ ​
^ equals to (+2)

-1
a. cos θ ( 13 )

-1
b. cos θ ( 23 )

-1
c. cos θ ( 47 )

d. None of these

41. A tower stands at the center of a circular park. 𝐴 and 𝐵 are two points on the (+2)
∘
boundary of the park such that 𝐴𝐵 = 𝑎 subtends an angle 60 of at the foot of
∘
the tower and the angle of elevation of the top of the tower from 𝐴 or 𝐵 is 30 .
The height of the tower is
 a. 2a
3
​

b. 2a 3 ​

c. a
3
​

d. a 3 ​

𝑥+1 𝑦−1 𝑧−2

42. If the angle 𝜃 between the line 1 = 2 = 2 and the plane 2𝑥 − 𝑦 + √𝜆𝑧 + 4 = 0 (+2)
is such that sin⁡𝜃 = 13 , then value of 𝜆 is

a. (A) −3
5
​

b. (B) 5
3
​

c. (C) −4
3
​

d. (D ) 3
4
​

43. The rate of change of volume of a sphere with respect to its surface area when (+2)
the radius is 4 cm is:

a. (A) 4 cm3 / cm2

3 2
b. (B) 6 cm / cm

c. (C) 2 cm3 / cm2

d. (D) 8 cm3 / cm2

44. For any two statements 𝑝 and 𝑞, the negation of the expression 𝑝 ∨ (∼ 𝑝 ∧ 𝑞) is: (+2)

a. 𝑝 ∧ 𝑞
 b. 𝑝 ↔ 𝑞

c. ∼𝑝 ∨ ∼ 𝑞

d. ∼𝑝 ∧ ∼ 𝑞

0 2𝑞 𝑟
45. Let 𝐴 = ( 𝑝 𝑞 −𝑟 ). If 𝐴𝐴𝑇 = 𝐼3 , then |𝑝| is: (+2)
𝑝 −𝑞 𝑟

a. 1
√2

b. 1
√5

c. 1
√6

d. 1
√3

46. A function y = f(x) has a second order derivative f''(x) = 6(x - 1). If its graph (+2)
passes through the point (2, 1) and at that point the tangent to the graph is y =
3x -5. then the function is :

a. (A) (x - 1)2

b. (B) (x - 1)3

c. (C) (x + 1)3

d. (D) (x + 1)2

47. If sin 5𝑥 + sin 3𝑥 + sin 𝑥 = 0, then the value of x other than zero, lying (+2)
π
between π0 ≤ × ≤ 2
is;

a. (A) π π6
 π
b. (B) π 12

c. (C) π π3

d. (D) π π4

th 100
48. If an be the 𝑛 term of the GP of positive numbers. Let ∑𝑛 = 100 𝑎2𝑛 = 𝛼 and (+2)
∑100 𝑎2𝑛 − 1 = 𝛽 such that 𝛼 ≠ 𝛽 then the common ratio is:
𝑛=1

𝛼
a. (A) 𝛽

𝛽
b. (B) 𝛼

c. (C) √ 𝛼𝛽

d. (D) √ 𝛽𝛼

49. Let R be a relation on R , given by R = {(a, b) : 3a − 3b + 7 is an irrational

​ (+2)
number } Then R is

a. reflexive and symmetric but not transitive

b. reflexive and transitive but not symmetric

c. reflexive but neither symmetric nor transitive

d. an equivalence relation

50. The shortest distance between the lines x−5

1
​ = y−2
2
​ = z−4
−3
​ and x+3
1
​ = y+5
4
​ = z−1
−5
is (+2)

a. 7 3

b. 6 3

c. 4 3 ​

d. 5 3
51. In a binomial distribution B(n, p) , the sum and the product of the mean and the (+2)
variance are 5 and 6 respectively, then 6(n + p − q) is equal to

a. 50

b. 53

c. 52

d. 51

52. The negation of the expression q ∨ ((∼ q) ∧ p) is equivalent to (+2)

a. p ∧ (∼ q)

b. (∼ p) ∧ (∼ q)

c. (∼ p) ∨ (∼ q)

d. (∼ p) ∨ q

53. Let the plane P pass through the intersection of the planes 2x + 3y − z = 2 and (+2)
x + 2y + 3z = 6, and be perpendicular to the plane 2x + y − z + 1 = 0 If d is the
distance of P from the point (-7,1,1), then d2 is equal to :

a. 25
83
​

b. 250
83
​

c. 15
53
​

d. 250
82
​

54. A bag contains 6 balls Two balls are drawn from it at random and both are (+2)
found to be black The probability that the bag contains at least 5 black balls is

a. 2
7
​
 b. 3
7
​

c. 5
7
​

d. 5
6
​

⎛1 0 0⎞
55. Let A = 0 4 −1 Then the sum of the diagonal elements of the matrix (A + (+2)
⎝0 −3⎠
​ ​ ​ ​ ​

12
I)11 is equal to :

a. 2050

b. 4094

c. 6144

d. 4097

56. Let a1 , a2 , a3 , … be an AP If a7 = 3 , the product a1 a4 is minimum and the sum of

​ ​ ​ ​ ​ ​
(+2)
its first n terms is zero, then n! − 4an(n+2) is equal to : ​

381
a. 4
​

b. 9

33
c. 4
​

d. 24

57. The coefficient of x301 in (1 + x)500 + x(1 + x)499 + x2 (1 + x)498 + …… + x500 is : (+2)

a. 500
C301 ​

b. 501
C200 ​

c. 3
500
​
C300 ​

d. 501
C302 ​
 x3 ax2 x3
58. If the functions f (x) = 3
​ + 2bx + 2
​ and g(x) = 3
​ = 2b have a
+ ax + bx2 , a  (+2)
common extreme point, then a + 2b + 7 is equal to:

a. 3
2
​

b. 3

c. 6

d. 4

59. Let N be the sum of the numbers appeared when two fair dice are rolled and (+2)
let the probability that N − 2, 3N , N + 2 are in geometric progression be
​
k
48
​,
Then the value of k is

a. 16

b. 2

c. 8

d. 4

2
60. The minimum value of the function f (x) = ∫ e∣x−t∣ dt is : ​ (+2)
0

a. 2

b. 2(e − 1)

c. 2e − 1

d. e(e − 1)
 Answers

1. Answer: c

Explanation:

Given initial velocity u = 0 and acceleration is constant

At time t
v = 0 + at ⇒ v = at
also x = 0(t) + 12 at2 ⇒ x = 12 at2
​ ​

Graph (A) ; (B) and (D) are correct.

Concepts:

1. Motion in a straight line:

The motion in a straight line is an object changes its position with respect to its
surroundings with time, then it is called in motion. It is a change in the position of an
object over time. It is nothing but linear motion.

Types of Linear Motion:

Linear motion is also known as the Rectilinear Motion which are of two types:

1. Uniform linear motion with constant velocity or zero acceleration: If a body

travels in a straight line by covering an equal amount of distance in an equal
interval of time then it is said to have uniform motion.
2. Non-Uniform linear motion with variable velocity or non-zero acceleration: Not
like the uniform acceleration, the body is said to have a non-uniform motion
when the velocity of a body changes by unequal amounts in equal intervals of
time. The rate of change of its velocity changes at different points of time during
its movement.

2. Answer: c

Explanation:
 Range = 2RhT + ​ 2RhR ​

3
50 × 10 = 2 × 6400 × 103 × hT +
​ 2 × 6400 × 103 × 70 ​

by solving hT = 32 m
​

Concepts:

1. Communication Systems:

A system that describes the information exchange between two points is called the
communication system. The transmission and reception process of information is
called communication. The major elements of communication are such as:

The Transmitter of information

Channel or medium of communication
The Receiver of information

Examples Of Communication Systems:

The following are the examples of communication systems:

1. Internet
2. Public Switched Telephone network
3. Intranet and Extranet
4. Television

Types Of Communication Systems:

Turning on Signal specification or technology, the communication system is
categorized as follows:

1. Analog technology communicates data as electronic signals of differing

frequencies or amplitude. For instance Broadcast and telephone transmission.
2. In Digital technology, the data are created and processed in two states: High
(represented as 1) and Low (represented as 0). Also this technology stores and
transmits data in the form of 1's and 0's.

Elements of a Communication System:

3. Answer: a

Explanation:

If we use that direction of light propagation will be along E × B ? Then (4) option is
correct.
Detailed solution is as following.
magnitude of E = CB
E = 3 × 108 × 1.6 × 10−6 × 5 ​

2
E = 4.8 × 10 5 ​

E and B are perpendicular to each other

⇒ E .B = 0
⇒ either direction of E is ^i − 2^j or −^i + 2^j from given option
​ ​

Also wave propagation direction is parallel to

E × B which is −k^
⇒ E is along (−^i + 2^j ) ​

Concepts:

1. Electromagnetic waves:

The waves that are produced when an electric field comes into contact with a
magnetic field are known as Electromagnetic Waves or EM waves. The constitution of
an oscillating magnetic field and electric fields gives rise to electromagnetic waves.

Types of Electromagnetic Waves:

Electromagnetic waves can be grouped according to the direction of disturbance in
them and according to the range of their frequency. Recall that a wave transfers
energy from one point to another point in space. That means there are two things
 going on: the disturbance that defines a wave, and the propagation of wave. In this
context the waves are grouped into the following two categories:

Longitudinal waves: A wave is called a longitudinal wave when the disturbances

in the wave are parallel to the direction of propagation of the wave. For example,
sound waves are longitudinal waves because the change of pressure occurs
parallel to the direction of wave propagation.
Transverse waves: A wave is called a transverse wave when the disturbances in
the wave are perpendicular (at right angles) to the direction of propagation of
the wave.

4. Answer: a

Explanation:

6.62×10−34 ×3×108
E= hc
λ
​ ⇒λ= hc
E
​ = 12.5×1.6×10−19
​

= 993A?
1
λ
​
= R ( n12 − ​
1
n22
​
)
1 ​ ​

(where Rydberg constant ,R= 1.097 x 107 )

or, 1
993×10−10
​ = 1.097 × 107 ( 112 − ​
1
n22 ​
​ )
Solving we get n2 = 3 ​

Spectral lines Total number of spectral lines = 3

Two lines in Lyman series for n1 = 1, n2 = 2 ​ ​

and n1 = 1, n2 = 3 and one in Balmer series

​ ​

for n1 = 2, n2 = 3
​ ​

Concepts:

1. Atoms:

The smallest unit of matter indivisible by chemical means is known as an atom.

The fundamental building block of a chemical element.
The smallest possible unit of an element that still has all the chemical properties
of that element.
An atom is consisting of a nucleus surrounded by one or more shells of electron
s.
Word origin: from the Greek word atomos, which means uncuttable, something
that cannot be divided further.
 All matter we encounter in everyday life consists of smallest units called atoms – the
air we breath consists of a wildly careening crowd of little groups of atoms, my
computer’s keyboard of a tangle of atom chains, the metal surface it rests on is a
crystal lattice of atoms. All the variety of matter consists of less than hundred species
of atoms (in other words: less than a hundred different chemical elements).

Atom

Every atom consists of an nucleus surrounded by a cloud of electrons. Nearly all of

the atom’s mass is concentrated in its nucleus, while the structure of the electron
cloud determines how the atom can bind to other atoms (in other words: its
chemical properties). Every chemical element can be defined via a characteristic
number of protons in its nucleus. Atoms that have lost some of their usual number of
electrons are called ions. Atoms are extremely small (typical diameters are in the
region of tenths of a billionth of a metre = 10-10 metres), and to describe their
properties and behaviour, one has to resort to quantum theory.

5. Answer: a

Explanation:

Q = nCv ΔT as gas in closed vessel

​

15 5×R
Q= 28
​ × 2
​ × (4T − T )
Q = 10000 J = 10 kJ

Concepts:
 1. Kinetic Molecular Theory of Gases:

Postulates of Kinetic Theory of Gases:

Gases consist of particles in constant, random motion. They continue in a

straight line until they collide with each other or the walls of their container.
Particles are point masses with no volume. The particles are so small compared
to the space between them, that we do not consider their size in ideal gases.
Gas pressure is due to the molecules colliding with the walls of the container. All
of these collisions are perfectly elastic, meaning that there is no change in
energy of either the particles or the wall upon collision. No energy is lost or
gained from collisions. The time it takes to collide is negligible compared with
the time between collisions.
The kinetic energy of a gas is a measure of its Kelvin temperature. Individual gas
molecules have different speeds, but the temperature and
kinetic energy of the gas refer to the average of these speeds.
The average kinetic energy of a gas particle is directly proportional to the
temperature. An increase in temperature increases the speed in which the gas
molecules move.
All gases at a given temperature have the same average kinetic energy.
Lighter gas molecules move faster than heavier molecules.

6. Answer: c

Explanation:
 LIdI = I 2 R
L× E
10
​ (−e−t/2 ) × −1
2
​ = E
10
​ (1 − e−t/2 ) × 10
e−t/2 = 1 − e−t/2
t = 2ℓn2

Concepts:

1. Alternating Current:

An alternating current can be defined as a current that changes its magnitude and
polarity at regular intervals of time. It can also be defined as an electrical current that
repeatedly changes or reverses its direction opposite to that of Direct Current or DC
which always flows in a single direction as shown below.

Alternating Current Production

Alternating current can be produced or generated by using devices that are known
as alternators. However, alternating current can also be produced by different
methods where many circuits are used. One of the most common or simple ways of
generating AC is by using a basic single coil AC generator which consists of two-pole
magnets and a single loop of wire having a rectangular shape.

Application of Alternating Current

AC is the form of current that are mostly used in different appliances. Some of the
examples of alternating current include audio signal, radio signal, etc. An alternating
current has a wide advantage over DC as AC is able to transmit power over large
distances without great loss of energy.

7. Answer: b

Explanation:

ˉ∣ =
∣M Total magnetic moment
volume
​

NiA
= Al
= l = 500×15
​
Ni
0.25
​ = 30, 000 Am−1 ​

Concepts:
 1. Magnetism & Matter:

Magnets are used in many devices like electric bells, telephones, radio, loudspeakers,
motors, fans, screwdrivers, lifting heavy iron loads, super-fast trains, especially in
foreign countries, refrigerators, etc.

Magnetite is the world’s first magnet. This is also called a natural magnet. Though
magnets occur naturally, we can also impart magnetic properties to a substance. It
would be an artificial magnet in that case.

Read More: Magnetism and Matter

Some of the properties of the magnetic field lines are:

The lines and continuous and outside the magnet, the field lines originate from
the North pole and terminate at the South pole
They form closed loops traversing inside the magnet.
But here the lines seem to originate from the South pole and terminate at the
North pole to form closed loops.
More number of close lines indicate a stronger magnetic field
The lines do not intersect each other
The tangent drawn at the field line gives the direction of the field at that point.

8. Answer: d

Explanation:

1
tan 45∘ = ωCR
​ = ωL
R
​

⇒ XL = XC ​ ​

⇒ resonance
i= V
R
​

220
= 110
​ = 2A

Concepts:

1. LCR Circuit:

An LCR circuit, also known as a resonant circuit, or an RLC circuit, is an electrical

circuit consist of an inductor (L), capacitor (C) and resistor (R) connected in series or
parallel.

Series LCR circuit

When a constant voltage source is connected across a resistor a current is induced
in it. This current has a unique direction and flows from the negative to positive
terminal. Magnitude of current remains constant.

Alternating current is the current if the direction of current through this resistor
changes periodically. An AC generator or AC dynamo can be used as AC voltage
source.
9. Answer: c

Explanation:

∣P 1 ∣ = q ( d )
​

∣P2 ∣ = qd
​

∣Resultant∣ = 2P cos 30∘

2qd ( 2 )
3 ​

​ = 3qd
​

Concepts:

1. Electric Dipole:

An electric dipole is a pair of equal and opposite point charges -q and q, separated
by a distance of 2a. The direction from q to -q is said to be the direction in space.
 p=q×2a

where,

p denotes the electric dipole moment, pointing from the negative charge to the
positive charge.

Force Applied on Electric Dipole

10. Answer: c

Explanation:

Applying point potential method

q = cV
q = 10?F ? 20 = 200?C
Option (3)

Concepts:

1. Combination of Capacitors:

The total capacitance of this equivalent single capacitor depends both on the
individual capacitors and how they are connected. There are two simple and
 common types of connections, called series and parallel, for which we can easily
calculate the total capacitance.

Read Also: Combination of Capacitors

Series capacitors

When one terminal of a capacitor is connected to the terminal of another capacitors

, called series combination of capacitors.

Capacitors in Parallel

Capacitors can be connected in two types which are in series and in parallel. If
capacitors are connected one after the other in the form of a chain then it is in
series. In series, the capacitance is less.

When the capacitors are connected between two common points they are called to
be connected in parallel.

When the plates are connected in parallel the size of the plates gets doubled,
because of that the capacitance is doubled. So in a parallel combination of
capacitors, we get more capacitance.

Read More: Types of Capacitors

11. Answer: b

Explanation:

p = ksa I b hc

where k is dimensionless constant

a b c
M LT −1 = (M T −2 ) (M L2 ) (M L2 T −1 )

a+b+c=1

2b + 2c = 1

−2a − c = −1

1 1
a= 2
​ b= 2
​ c=0
 1 1
∴ S 2 I 2 h0
​ ​

1 1
Hence, Correct answer is option (B) : S 2 I 2 h0 .
​ ​

Concepts:

1. Dimensional Analysis:

Dimensional Analysis is a process which helps verify any formula by the using the pri
nciple of homogeneity. Basically dimensions of each term of a dimensional equation
on both sides should be the same.

Limitation of Dimensional Analysis: Dimensional analysis does not check for the
correctness of value of constants in an equation.

Using Dimensional Analysis to check the correctness of the equation

Let us understand this with an example:

Suppose we don’t know the correct formula relation between speed, distance and
time,

We don’t know whether

(i) Speed = Distance/Time is correct or

(ii) Speed =Time/Distance.

Now, we can use dimensional analysis to check whether this equation is correct or
not.

By reducing both sides of the equation in its fundamental units form, we get

(i) [L][T]-¹ = [L] / [T] (Right)

(ii) [L][T]-¹ = [T] / [L] (Wrong)

From the above example it is evident that the dimensional formula establishes the
correctness of an equation.
12. Answer: b

Explanation:

From time t = 0 to t = t0 , growth of current takes place and after that decay of
​

current takes place.

Concepts:

1. Capacitor:

Capacitors commonly known as Condensers are passive components, similar to a

resistor. In capacitors, charges are usually stored in the form of an "electrical field".
Electrical and electronic circuits depend on the same which is made up of two
parallel metal plates that are not connected to one another. The two plates are
separated by a non-conducting insulating medium called dielectric.

Uses of Capacitors:
DC blocking capacitors block the DC and allows only AC to certain parts of the
circuit.
These are main elements of filters.
They possess the ability to couple a section of the circuit to another.

Types of Capacitors:
Ceramic capacitors are created by covering two sides of their tiny ceramic disc
with silver and stacking them together.
Film Capacitors are commonly used capacitors that are made up of different
sets of capacitors.
In an electrolytic capacitor metallic anode coated with an oxidized layer used
as a dielectric.
A Paper capacitor is also known as a fixed capacitor in which paper is used as
the dielectric material.

Read More: Types of Capacitors

13. Answer: d

Explanation:

qvB
a= m
​

ma m
B = qv
ma
= ​

2k
​ ​

m3/2 a (1.6×10−27 )3/2 ×1012

= e 2k ​
​ = 1.6×10−19 2×1×106 ×1.6×10−19
​

= 0.71 m T

Concepts:

1. Magnetic Field:

The magnetic field is a field created by moving electric charges. It is a force field that
exerts a force on materials such as iron when they are placed in its vicinity. Magnetic
fields do not require a medium to propagate; they can even propagate in a vacuum.
Magnetic field also referred to as a vector field, describes the magnetic influence on
moving electric charges, magnetic materials, and electric currents.

A magnetic field can be presented in two ways.

Magnetic Field Vector: The magnetic field is described mathematically as a
vector field. This vector field can be plotted directly as a set of many vectors
drawn on a grid. Each vector points in the direction that a compass would point
and has length dependent on the strength of the magnetic force.
Magnetic Field Lines: An alternative way to represent the information contained
within a vector field is with the use of field lines. Here we dispense with the grid
pattern and connect the vectors with smooth lines.

Properties of Magnetic Field Lines

Magnetic field lines never cross each other
The density of the field lines indicates the strength of the field
Magnetic field lines always make closed-loops
Magnetic field lines always emerge or start from the north pole and terminate
at the south pole.
14. Answer: b

Explanation:

0.95R = RRv
R+Rv
​

​
​

0.95 × 30 = 0.05Rv ​

Rv = 19 × 30 = 570 Ω
​

Concepts:

1. Resistance:

Resistance is the measure of opposition applied by any object to the flow of electric
current. A resistor is an electronic constituent that is used in the circuit with the
purpose of offering that specific amount of resistance.

R=V/I

In this case,

v = Voltage across its ends

I = Current flowing through it

All materials resist current flow to some degree. They fall into one of two broad
categories:

Conductors: Materials that offer very little resistance where electrons can move
easily. Examples: silver, copper, gold and aluminum.
Insulators: Materials that present high resistance and restrict the flow of
electrons. Examples: Rubber, paper, glass, wood and plastic.

Resistance measurements are normally taken to indicate the condition of a

component or a circuit.

The higher the resistance, the lower the current flow. If abnormally high, one
possible cause (among many) could be damaged conductors due to burning
or corrosion. All conductors give off some degree of heat, so overheating is an
issue often associated with resistance.
The lower the resistance, the higher the current flow. Possible causes: insulators
damaged by moisture or overheating.
15. Answer: a

Explanation:

d sin θ = λ

λ= d
2 ​ [d = 1 × 10−6 m]
⇒ λ = 5000 ?
Fringe width, B = λD
d′ ​ (d′ is slit separation)
−10
10−2 = 5000×10
d′
×0.5
​

⇒ d′ = 25 × 10−6 m = 25 μ m

Concepts:

1. Young’s Double Slit Experiment:

Considering two waves interfering at point P, having different distances.

Consider a monochromatic light source ‘S’ kept at a relevant distance from two
slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume
that S1 and S2 are two coherent sources derived from S.
 The light passes through these slits and falls on the screen that is kept at the
distance D from both the slits S1 and S2. It is considered that d is the separation
between both the slits. The S1 is opened, S2 is closed and the screen opposite to
the S1 is closed, but the screen opposite to S2 is illuminating.
Thus, an interference pattern takes place when both the slits S1 and S2 are open.
When the slit separation ‘d ‘and the screen distance D are kept unchanged, to
reach point P the light waves from slits S1 and S2 must travel at different
distances. It implies that there is a path difference in the Young double-slit
experiment between the two slits S1 and S2.

Read More: Young’s Double Slit Experiment

16. Answer: b

Explanation:
wt of sulphur
%of sulphur = wt of compound
​ × 100 = 8
32 8
∴ wt. of compound
​ = 100
​ ∴ wt. of compound = 400 g mol−1

Concepts:

1. Stoichiometry:

Stoichiometry is founded on the law of conservation of mass where the total mass of
the reactants equals the total mass of the products, leading to the insight that the
relations among quantities of reactants and products typically form a ratio of
positive integers. This means that if the amounts of the separate reactants are
known, then the amount of the product can be calculated. Conversely, if one
reactant has a known quantity and the quantity of the products can be empirically
determined, then the amount of the other reactants can also be calculated.

Stoichiometry helps us determine how much substance is needed or is present.

Things that can be measured are;

1. Reactants and Products mass

2. Molecular weight
3. Chemical equations
4. Formulas
 ​Stoichiometric Coefficient

The Stoichiometric coefficient of any given component is the number of molecules

and/or formula units that participate in the reaction as written.

Mole Ratios

The mass of one mole of a substance in grams is called molar mass. The molar
mass of one mole of a substance is numerically equal to the atomic/molecular
formula mass.

17. Answer: a

Explanation:

∗ Oleic acid is also soluble in N aHCO3 ​

∗ o-toluidine is not soluble in N aOH as well as N aHCO3 ​

∗ Benzamide is also not soluble in N aOH N aHCO3 ​

Concepts:

1. Haloalkanes and Haloarenes - Chemical Reactions:

Chemical Reactions go with the breaking and bonding of covalent bonds which
involve of exchange of electrons. The functional groups of Organic compounds play
a consequential role in the process. Based on the above theory, reactions can be
classified into five main groups:

Rearrangement Reactions are the type of reactions in which products get formed
simply by the rearrangement of atoms and electrons in the reactant molecules.

||

NH4CNO → NH2 –C – NH2

Substitution Reactions are the reactions in which an atom or group of atoms is

replaced by some other atom or group of atoms without any change in the structure
of the remaining part of the molecule.
 CH3Br + KOH (aqueous) → CH3OH + KBr

Addition Reactions are the reactions in which products get formed by the addition of
some reagent to an unsaturated compound.

CH2 = CH2 + HCl → CH5Cl

Electrophilic Addition Reactions

Nucleophilic Addition Reactions
Free Radical Addition Reactions

Elimination Reactions are the reactions in which the products get formed by the loss
of simple molecules like HX from the reactant molecules.

C2H5OH → C2H4

EN1 (Nucleophilic Elimination Unimolecular)

EN2 (Nucleophilic Elimination Bimolecular)

A polymerization Reaction is the union of two or more molecules of a substance that

form a single molecule with higher molecular weight.

n (CH = CH2) → (-CH2 – CH2 -) n

18. Answer: b

Explanation:

Bouveault-Blanc reduction reaction involves reduction of an ester with N a/C2 H5 OH.

​ ​

Concepts:

1. Aldehydes, Ketones, and Carboxylic Acids:

Aldehydes, Ketones, and Carboxylic Acids are carbonyl compounds that contain a
carbon-oxygen double bond. These organic compounds are very important in the
field of organic chemistry and also have many industrial applications.

Aldehydes:
 Aldehydes are organic compounds that have the functional group -CHO.

Preparation of Aldehydes

Acid chlorides are reduced to aldehydes with hydrogen in the presence of palladium
catalyst spread on barium sulfate.

Ketones:
Ketones are organic compounds that have the functional group C=O and the
structure R-(C=O)-R’.

Preparation of Ketones

Acid chlorides on reaction with dialkyl cadmium produce ketones. Dialkyl cadmium
themselves are prepared from Grignard reagents.

Carboxylic Acid:
Carboxylic acids are organic compounds that contain a (C=O)OH group attached
to an R group (where R refers to the remaining part of the molecule).

Preparation of Carboxylic Acids

Primary alcohols are readily oxidized to carboxylic acids with common oxidizing
agents such as potassium permanganate in neutral acidic or alkaline media or by
potassium dichromate and chromium trioxide in acidic media.

19. Answer: b

Explanation:

In presence of u.v. light allylic C − H bond undergoes bromination.

Concepts:

1. Hydrocarbons:

Hydrocarbons can be described as organic compounds that consists only hydrogen

and carbon atoms. These compounds are of different types and thereby have
 distinct natures. Hydrocarbons are colorless gases and are known for discharging
faint odours. These have been categorized under four major classes named as
alkynes, alkanes, alkenes, and aromatic hydrocarbons.

Types of Hydrocarbons
1. Saturated hydrocarbons - Saturated hydrocarbons are those compounds
where there is a single bond exists between carbon atoms and are saturated
with atoms of hydrogen.
2. Unsaturated hydrocarbons - Hydrocarbons comprises of at least one double or
triple bond between carbon atoms are known as unsaturated hydrocarbons.
3. Aliphatic hydrocarbons - The term denotes the hydrocarbons formed as an
outcome of the chemical degradation of fats. Aliphatic hydrocarbons are
basically chemical compounds.
4. Aromatic hydrocarbons - They are distinguished because of the presence of
benzene rings in them. They give away distinct types of aroma. These
hydrocarbons comprises of only hydrogen and carbon atoms.

20. Answer: a

Explanation:

→ 2nd most abundant element is "Si" and it is not present in calgon N a6 P6 O18 = (
​ ​ ​

Graham's salt ) (Sodium hexametaphosphate)

→ It exist in polymeric form as (N aP O3 )6 and water soluble compound

​ ​

→ It removes Ca2+ in soluble ion but not by precipitation

Concepts:

1. Hydrogen Bonding:

Hydrogen bonding implies the formation of hydrogen bonds which are an attractive
intermolecular force. An example of hydrogen bonding is the bond between the H
atom and the O atom in water.
 A special type of intermolecular attractive force arises only in the compounds
having Hydrogen atoms bonded to an electronegative atom. This force is known as
the Hydrogen bond. For instance, in water molecules, the hydrogen atom is bonded
to a highly electronegative Oxygen.

The conditions for hydrogen bonding are:

1. The molecule must contain a strongly electronegative atom that is bound to
the hydrogen atom. The higher the electronegativity, the more polarized is the
molecule.
2. The electronegative atom must be small. The smaller the size, the greater the
electrostatic magnetism.

Effects of Hydrogen Bonding on Elements:

Association: The molecules of carboxylic acids exist as dimer because of the
hydrogen bonding. The molecular masses of such compounds are found to be
double than those calculated from their simple formula.

Dissociation: In aqueous solution, HF dissociates and gives the difluoride ion instead
of fluoride ion. This is due to hydrogen bonding in HF. The molecules of HCl, HBr, HI do
not form a hydrogen bond. This explains the non-existence of compounds like
KHCl2, KHBr2, KHI2.

Types of Hydrogen bonding

1. Intramolecular Hydrogen bonding: When hydrogen bonding takes place
between different molecules of the same or different compounds, it is called
intermolecular hydrogen bonding.
2. Intermolecular hydrogen bonding: The hydrogen bonding which takes place
within a molecule itself is called intramolecular hydrogen bonding.
3. Symmetrical Hydrogen bonding: The symmetric hydrogen bond is a type of a
three-centre four-electron bond.

21. Answer: c

Explanation:
 Kjeldahl?s method is not applicable for compounds containing nitrogen in nitro and
azo groups and nitrogen present in the ring. Because nitrogen of these compounds
does not change to ammonium sulphate under these conditions.
Hence only Alanine can be used to determine percentage of nitrogen.

Concepts:

1. Purification of Organic Compounds:

We commonly use these methods for the purification of substances:

Simple crystallization - The most common method that we use to purify organic
solids.
Fractional crystallization - It is the process of separation of different
components of a mixture by repeated crystallisations.
Sublimation - Certain organic solids on heating directly change from solid to
vapor state without passing through a liquid state. These substances are
sublimable. This process is sublimation.
Simple distillation - It is the joint process of vaporization and condensation.
Fractional distillation - It is the process to separate a mixture of two or more
miscible liquids which have boiling points close to each other
Distillation under reduced pressure
Steam distillation
Azeotropic distillation
Chromatography

22. Answer: a

Explanation:

ΔG∘ = −RT lnk = − nF Ecell

∘ ​

∘
n×F ×E 2×96000×2
Ink = R×T
​ = 8×300
​

Ink = 160
k = e160

Concepts:

1. Equilibrium Constant:
 The equilibrium constant may be defined as the ratio between the product of the
molar concentrations of the products to that of the product of the molar
concentrations of the reactants with each concentration term raised to a power
equal to the stoichiometric coefficient in the balanced chemical reaction.

The equilibrium constant at a given temperature is the ratio of the rate constant of
forwarding and backward reactions.

Equilibrium Constant Formula:

Kequ = kf/kb = [C]c [D]d/[A]a [B]b = Kc

where Kc, indicates the equilibrium constant measured in moles per litre.

For reactions involving gases: The equilibrium constant formula, in terms of partial
pressure will be:

Kequ = kf/kb = [[pC]c [pD]d]/[[pA]a [pB]b] = Kp

Where Kp indicates the equilibrium constant formula in terms of partial pressures.

Larger Kc/Kp values indicate higher product formation and higher percentage
conversion.
Lower Kc/Kp values indicate lower product formation and lower percentage
conversion.

Medium Kc/Kp values indicate optimum product formation.

Units of Equilibrium Constant:

The equilibrium constant is the ratio of the concentrations raised to the stoichiometr
ic coefficients. Therefore, the unit of the equilibrium constant = [Mole L-1]△n.

where, ∆n = sum of stoichiometric coefficients of products – a sum of stoichiometric

coefficients of reactants.

23. Answer: d

Explanation:
 x
m
​ = K(P )1/n
1
log ( m
x
) = log K +
​

n
​ log P
y = c + mx
m = 1/n so slope will be equal to 1/n

Concepts:

1. Adsorption:

Heinrich Kayser, the German physicist was the first to coin the term adsorption.
Adsorption can be explained as a surface phenomenon where particles remain
attached on the top of a material. Generally, it comprises the molecules, atoms,
liquid, solid in a dissolved stage, even the ions of a gas that are attached to the
surface. Much to our surprise, the consequence of surface energy i.e. adsorption is
present in biological, physical, chemical, and natural systems and are used in many
industrial applications.
24. Answer: a

Explanation:

Pt(s)H2 (g, 1bar)HCl(aq)AgCl(s)Ag(s)∣Pt(s)

​

Anode: H2 − > [10−6 m]2H + + 2e × 1

​

Cathode : e− + AgCl(s)− > Ag(s) + Cl− (aq)

H2 (g)l + AgCl(s)− > 2H + + 2Ag(s) + 2Cl− (aq)
​

0 0.06
Ecell = Ecell
​ − ​

2
​ log10 ((H + )2 .(Cl− )2 )
​

${.925 = (E^{0}_{H_2 /H^{+}}} + E^{0}_{AgCl / Ag, Cl^{-}}) - \frac{0.06}{2}

\log_{10} ((10^{-6})^{2} (10^{-6})^{2})}$
0 −6 4
.92 = 0 + EAgCl/Ag,Cl − − 0.03 log10 (10 ​) ​

0
EAgCl /Ag, Cl− = .9.2 + .03 × −24 = 0.2 V
​
Concepts:

1. Galvanic Cells:

Galvanic cells, also known as voltaic cells, are electrochemical cells in which
spontaneous oxidation-reduction reactions produce electrical energy. It converts
chemical energy to electrical energy.

It consists of two half cells and in each half cell, a suitable electrode is immersed.
The two half cells are connected through a salt bridge. The need for the salt bridge
is to keep the oxidation and reduction processes running simultaneously. Without it,
the electrons liberated at the anode would get attracted to the cathode thereby
stopping the reaction on the whole.

Working Principle Of Galvanic Cell:

1. Take two beakers containing electrolytic solutions of copper sulphate and zinc
sulphate are taken. It is connected via a salt bridge containing an aqueous
solution of potassium chloride.
2. Zinc and copper electrodes are immersed in the respective electrodes and
connected through a voltmeter to measure the electrical potential.
3. Zinc which acts as the anode readily undergoes an oxidation process and
acquires a negative charge.
4. The electrons travel through the salt bridge and undergo a reduction process
at the copper cathode.
5. Thus the cathode would acquire a positive charge.
6. This flow of electrons from the anode to the cathode induces a flow of electric
current in the opposite direction which shall be measured by the voltmeter.

Types of Voltaic Cell:

Primary Cell
Dry Cell
Mercury Cell
Alkaline Cell
Secondary Cell
Nickel-Cadmium Cell
Lead-Acid Cell
Lithium-Ion Cell
25. Answer: a

Explanation:

Given N a2 C O3 = 1.0 × 10−4 M

​ ​

∴ [CO3−− ] ​ = 1.0 × 10−4 M

i.e. s = 1.0 × 10−4 M
At equilibrium
[Ba++ ] [CO3−− ] = Ksp of BaCO3
​ ​ ​

++ Ksp 5.1×10−9
[Ba ]= ​

=
[CO3−− ] 1.0×10−4
​ ​

= 5.1 × 10−5 M

Concepts:

1. Acids and Bases:

Acid is any hydrogen-containing substance that is capable of donating a proton

(hydrogen ion) to another substance. Base is an ion or molecule capable of
accepting a hydrogen ion from acid.

Physical Properties of Acids and Bases

Physical Properties ACIDS BASES

Taste Sour Bitter

Colour on Litmus paper Turns blue litmus red Turns red litmus blue

Ions produced on dissociation H+ OH-

pH <7 (less than 7) >7 (more than 7)

Strong acids HCl, HNO3, H2SO4 NaOH, KOH

Weak Acids CH3COOH, H3PO4, H2CO3 NH4OH

Chemical Properties of Acids and Bases

Type of Reaction Acid Bases

Base + Metal → Salt +

Acid + Metal → Salt +
Hydrogen gas (H2)
Hydrogen gas (H2)
E.g.,
Reaction with Metals E.g.,
2NaOH +Zn →
Zn(s)+ dil. H2SO4 →
Na2ZnO2 (Sodium
ZnSO4 (Zinc Sulphate) + H2
zincate) + H2

Metal carbonate/Metal
hydrogen carbonate + Acid →
Salt + Carbon dioxide + Water

E.g., HCl+NaOH → NaCl+ H2O

Reaction with hydrogen 2. Na2CO3+ 2 HCl(aq)

Base+ Carbonate/
carbonates →2NaCl(aq)+ H2O(l) + CO2(g) bicarbonate → No
(bicarbonate) and
reaction
carbonates 3. Na2CO3+ 2H2SO4(aq)
→2Na2SO4(aq)+ H2O(l) +
CO2(g)

4. NaHCO3+ HCl → NaCl+ H2O+

CO2

Base + Acid → Salt +

Base + Acid → Salt + Water
Water
Neutralisation Reaction E.g., NaOH(aq) + HCl(aq) →
E.g., CaO+ HCl (l) →
NaCl(aq) + H2O(l)
CaCl2 (aq)+ H2O (l)

Reaction with Oxides

Metal oxide + Acid → Salt + Non- Metallic oxide +
Water Base → Salt + Water
 E.g., CaO+ HCl (l) → E.g., Ca(OH)2+ CO2 →
CaCl2 (aq)+ H2O (l) CaCO3+ H2O

Acid gives H+ ions in water.

Base gives OH- ions in
Dissolution in Water E.g., HCl → H+ + Cl-
water.
HCl + H2O → H3O+ + Cl–

Read more on Acids, Bases and Salts

26. Answer: d

Explanation:

sp3 d2 hybridization is not displayed by P F5 . ​

Therefore, the correct option is (D):P F5 ​

Concepts:

1. Hybridisation:

Hybridization refers to the concept of combining atomic orbitals in order to form

new hybrid orbitals that are appropriate to represent their bonding properties.
Hybridization influences the bond length and bond strength in organic compounds.

Types of Hybridization:

sp Hybridization

sp hybridization is observed while one s and one p orbital inside the identical
principal shell of an atom mix to shape two new equal orbitals. The new orbitals
formed are referred to as sp hybridized orbitals.

sp2 Hybridization
 sp2 hybridization is observed whilst ones and p orbitals of the same shell of an atom
blend to shape three equivalent orbitals. The new orbitals formed are referred to as
sp2 hybrid orbitals.

sp3 Hybridization

When one ‘s’ orbital and 3 ‘p’ orbitals belonging to the identical shell of an atom
blend together to shape 4 new equal orbitals, the sort of hybridization is referred to
as a tetrahedral hybridization or sp3.

sp3d Hybridization

sp3d hybridization involves the joining of 3p orbitals and 1d orbital to form 5 sp3d
hybridized orbitals of identical energy. They possess trigonal bipyramidal geometry.

sp3d2 Hybridization

With 1 s three p’s and two d’s, there is a formation of 6 new and identical sp3d2
orbitals.

27. Answer: b

Explanation:

The correct answer is B: Li− +

2 < Li2 < Li2
​ ​ ​

Electrons in Li2 = 6
​

Bond order = 4−2

2 ​ =1
Electrons for Li+
2 ​ is 5
Bond order = 23−2
​ = 0.5
Electrons for Li−
2 ​ is 7
Bond order = 24−3
​ = 0.5
More the electrons in the bonding orbitals the more is the stability.
Therefore, the order of stability is Li− ​
+
2 < Li2 < Li2​ ​

Concepts:

1. Molecular Orbital Theory:

 The Molecular Orbital Theory is a more sophisticated model of chemical bonding
where new molecular orbitals are generated using a mathematical process called Li
near Combination of Atomic Orbitals (LCAO).

Molecular Orbital theory is a chemical bonding theory that states that individual
atoms combine together to form molecular orbitals. Due to this arrangement in MOT
Theory, electrons associated with different nuclei can be found in different atomic or
bitals. In molecular orbital theory, the electrons present in a molecule are not
assigned to individual chemical bonds between the atoms. Rather, they are treated
as moving under the influence of the atomic nuclei in the entire molecule.

Molecular Orbital Theory

28. Answer: b

Explanation:

Aerosol is suspension of fine solid or liquid particles in air or other gas.

Ex. Fog, dust, smoke etc
 Concepts:

1. Colloids:

A colloid is a mixture in which one substance consisting of microscopically

dispersed insoluble particles is suspended throughout another substance. Some
definitions specify that the particles must be dispersed in a liquid, while others
extend the definition to include substances like aerosols and gels.

Example of Colloids:

1. Blood: A respiration pigment which has albumin protein in water. Pigment part
contains albumin that acts as the dispersed phase and the dispersion medium
is water. It is a hydrosol.
2. Cloud: It contains air which is the dispersion medium and droplets of water as
a dispersed phase. These are aerosol.
3. Gold sol: It is a metallic sol in which gold particles are dispersed in the water.

Read More: Classification of Colloids

Property of Colloid Particles:

Colloids are insoluble particle mixes (sometimes known as colloidal solutions
or colloidal systems). They depict the microscopic dispersion of one substance
as well as a suspension in another substance. In a colloid, the size of these
suspended particles can range from 1 to 1000 nanometres (10-9 meters). They
are heterogeneous in nature.
Colloidal dispersions are made up of particles that are substantially larger
than conventional solution solutes. Colloidal particles are large molecules or
clusters of smaller species that scatter light.

Purification of Colloids:
Colloids contain ionic impurities and other categories of impure substances that
decrease the quality of colloids used in various applications. Following are the
methods to purify the colloids.

29. Answer: a
Explanation:

The linking of identical atoms with each other to form long chains is called
catenation. However, this property decreases from carbon to lead. Decrease of this
property is associated with M-M bond energy which decreases from carbon to lead.

Concepts:

1. Group 14 Elements:

The group 14 elements are also known as the carbon group. Sometimes it is also
called crystallogens. The group 14 elements can be found on the right side of the
periodic table. The elements of the carbon group are:

Carbon - C
Silicon - Si
Germanium - Ge
Tin - Sn
Lead - Pb
Flerovium - FI

The group 14 elements or the carbon family belongs to the p-block of the periodic ta
ble. The carbon group is also the second group in the p-block.

Electronic Configuration:
The arrangement of electrons in the orbitals of a molecule or atom is known as elect
ronic configuration. The general electronic configuration of the group 14 elements is
ns2np2.
 All group 14 elements have 4 electrons in the outer shell. Hence, the valency of the
carbon family or group 14 elements is 4.

Properties of Group 14 Elements:

Covalent Radii
Ionization Enthalpy
Electronegativity
Metallic Character
Density
Four Covalent Compounds
The Melting and Boiling Points

30. Answer: a

Explanation:

[M a3 b3 ] type complex shows fac and mer isomerism.

​ ​

[Co(N H3 )3 (N O2 )3 ]
​ ​ ​ ​

Concepts:

1. Nomenclature of Coordination Compounds:

Nomenclature of Coordination Compounds is important in Coordination Chemistry

because of the need to have an unambiguous method of describing formulas and
writing systematic names, particularly when dealing with isomers.

We can apply the following formulas:

On the very first the central atom is listed.
Ligands are then listed in alphabetical order and their placement in the list
does not depend on their charge.
Polydentate ligands are also listed alphabetically. In such a case of an
abbreviated ligand, the first letter of the abbreviation is used to determine the
position of the ligand in alphabetical order.
The formula for the entire coordination entity is enclosed in square brackets
whether charged or not. The formulas are enclosed in parentheses when
 ligands are polyatomic. Ligand abbreviations are also enclosed in
parentheses.
Within a coordination sphere, there should be no space between the ligands
and the metal.
When the formula of a charged coordination entity is to be written without that
of the counter-ions, the charge is indicated outside the square brackets as a
right superscript with the number before the sign. For example, [Co(CN)6]3-,
[Cr(H2O)6]3+, etc.
The charge of the anion(s) balances the charge of the cation(s).

31. Answer: d

Explanation:

Normal to the plane x + y + z = 7 is n = ^i + ^j + k^ AB = −^i − k^ ⇒ ∣AB ∣ = AB = 2 BC =

​ ​

^ ∣ = ∣(−^i − k^) ⋅ (i+j +

^ ^ k^) ∣
Length of projection of AB on n = ∣AB ⋅ n ​

= 23 Length of projection of
∣ 3 ∣
​ ​ ​ ​

​ ​

the line segment on the plane is AC AC 2 = AB 2 − BC 2 = 2 − 43 = 23 AC 2 = 23​ ​ ​ ​

Concepts:

1. Plane:

A surface comprising all the straight lines that join any two points lying on it is called
a plane in geometry. A plane is defined through any of the following uniquely:
 Using three non-collinear points
Using a point and a line not on that line
Using two distinct intersecting lines
Using two separate parallel lines

Properties of a Plane:
In a three-dimensional space, if there are two different planes than they are
either parallel to each other or intersecting in a line.
A line could be parallel to a plane, intersects the plane at a single point or is
existing in the plane.
If there are two different lines that are perpendicular to the same plane then
they must be parallel to each other.
If there are two separate planes which are perpendicular to the same line then
they must be parallel to each other.

32. Answer: d

Explanation:

Let base = b Altitude (or perpendicular) p → (q → p) = h2 − b2 Area, A = ​

1
2
× base×
​

[ h2 − b2 + b. 2 ] [ h h−2b ] Put
2 2
altitude = 1
2
​ ×b× h2 − b 2 ⇒ ​
dA
db
​ = 1
2
​ ​
−2b
h2 −b2
​

​
= 1
2
​

2 −b2
​

​
dA
db
​ = 0, ⇒ b =
h
2
​

​
Maximum area = 1
2
​ × h
2
​

​
× h2 − h2
2
​
= h2
4
​ ​

Concepts:

1. Maxima and Minima:

What are Maxima and Minima of a Function?

The extrema of a function are very well known as Maxima and minima. Maxima is
the maximum and minima is the minimum value of a function within the given set
of ranges.
 There are two types of maxima and minima that exist in a function, such as:

Local Maxima and Minima

Absolute or Global Maxima and Minima

33. Answer: c

Explanation:

6x2 − 11x + α = 0 given roots are rational ⇒ D must be perfect square ⇒ 121 −
24α = λ2 ⇒ maximum value of ⇒ is 5 α = 1 ⇒ λ ∈
/I α=2⇒ λ∈
/I α=3⇒ λ∈
/
I ⇒ 3 integral values α = 4 ⇒ λ ∈
/I α=5⇒ λ∈
/I

Concepts:

1. Quadratic Equations:

A polynomial that has two roots or is of degree 2 is called a quadratic equation. The
general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real
numbers.

Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real
coefficients.

The solution of a quadratic equation can be found using the formula, x=((-
b±√(b²-4ac))/2a)

Two important points to keep in mind are:

A polynomial equation has at least one root.

 A polynomial equation of degree ‘n’ has ‘n’ roots.

Read More: Nature of Roots of Quadratic Equation

There are basically four methods of solving quadratic equations. They

are:

1. Factoring
2. Completing the square
3. Using Quadratic Formula
4. Taking the square root

34. Answer: c

Explanation:

x2 + y 2 − 4x − 6y − 12 = 0 C1 (2, 3) r1 =
​ ​ 22 + 32 + 12 = 5 x2 + y 2 + 6x + 18y + 26 = 0
​

C2 (−3, −9), r2 =
​ ​ 32 + 92 − 26 = 8 C1 C2 =
​ ​ ​ (52 + 122) = 13
​

Concepts:

1. Circle:

A circle can be geometrically defined as a combination of all the points which lie at
an equal distance from a fixed point called the centre. The concepts of the circle are
very important in building a strong foundation in units likes mensuration and
coordinate geometry. We use circle formulas in order to calculate the area,
diameter, and circumference of a circle. The length between any point on the circle
and its centre is its radius.
 Any line that passes through the centre of the circle and connects two points of the
circle is the diameter of the circle. The radius is half the length of the diameter of the
circle. The area of the circle describes the amount of space that is covered by the
circle and the circumference is the length of the boundary of the circle.

Also Check:

Areas Related to Circles Perimeter and Area of Circle Circles Revision Notes

35. Answer: c

Explanation:

Equation of plane (x + y + z − 1) + λ(2x + 3y − z + 4) = 0 ⇒ (1 + 2λ)x + (1 + 3λ)y + (1 −

λ)z − 1 + 4λ = 0 dr's of normal of the plane are 1 + 2λ, 1 + 3λ, 1 − λ Since plane is
parallel to y - axis, 1 + 3λ = 0 ⇒ λ = −1/3 So the equation of plane is x + 4z − 7 = 0
Point (3, 2, 1) satisfies this equation

Concepts:

1. Distance of a Point from a Plane:

The shortest perpendicular distance from the point to the given plane is the
distance between point and plane. In simple terms, the shortest distance from a
point to a plane is the length of the perpendicular parallel to the normal vector
dropped from the particular point to the particular plane. Let's see the formula for
the distance between point and plane.

Read More: Distance Between Two Points

36. Answer: d
 Explanation:

13 +23 +...n3
Sum = ∑15
n=1 ​

1+2+...+n ​
− 12 . 15.16
​

2
​ = ∑15
n=1 ​
n(n+1)
2 ​
15
− 60 = ∑n=1 ​
n(n+1)(n+2−(n−1))
6 ​ − 60 = 15.16.17
6 ​ −
60 = 620

Concepts:

1. Sum to n Terms of Special Series:

A sequence is a list of numbers in a certain or particular order. Each number in a

sequence is called a term. A series is the sum of all the terms of a given sequence is
called a series. A finite series with a countable number of terms is commonly known
as a finite series, and that with an infinite number of terms is called an infinite series.
The sum to n terms of a series is reflected by Sn.

In mathematics, we may come across distinct types of series such as geometric

series, arithmetic series, harmonic series, etc. Apart from these, we can notice some
special series for which we can find the sum of the terms using distinct techniques.

Some special series are given below:

Sum of first n natural numbers = 1 + 2 + 3 +…+ n

Sum of squares of the first n natural numbers = 12 + 22 + 32 +…+ n2
Sum of cubes of the first n natural numbers = 13 + 23 + 33 +…+ n3

37. Answer: a

Explanation:
∣λ 3 2∣
D = 2λ
​ ​ 3 ​ 5 = (λ + 8) (2 − λ) for λ = 2; D1 
​ ​ = 0 Hence, no solution for λ = 2
​

∣4 λ 6∣

Concepts:

1. Solution of System of Linear Inequalities in Two Variables:

A System of Linear Inequalities is a set of 2 or more linear inequalities which have the
same variables.
 Example

x+y ≥5

x–y≤3

Here are two inequalities having two same variables that are, x and y.

Solution of System of Linear Inequalities in Two Variables

The solution of a system of a linear inequality is the ordered pair which is the
solution of all inequalities in the studied system and the graph of the system of a
linear inequality is the graph of the common solution of the system.

Therefore, the Solution of the System of Linear Inequalities could be:

Graphical Method:

For the Solution of the System of Linear Inequalities, the Graphical Method is the
easiest method. In this method, the process of making a graph is entirely similar to
the graph of linear inequalities in two variables.

Non-Graphical Method:

In the Non-Graphical Method, there is no need to make a graph but we can find the
solution to the system of inequalities by finding the interval at which the system
persuades all the inequalities.

In this method, we have to find the point of intersection of the two inequalities by
resolving them. It could be feasible that there is no intersection point between them.

38. Answer: a

Explanation:

Since sum of two numbers is even so either both are odd or both are even. Hence
number of elements in reduced samples space = 5 C2 + 6 C2 so required probability =
​ ​

5C
2 ​

5 C +6 C ​

2
​
2 ​

Concepts:
1. Conditional Probability:

Conditional Probability is defined as the occurrence of any event which determines

the probability of happening of the other events. Let us imagine a situation, a
company allows two days’ holidays in a week apart from Sunday. If Saturday is
considered as a holiday, then what would be the probability of Tuesday being
considered a holiday as well? To find this out, we use the term Conditional
Probability.

Let’s discuss certain theorems of Conditional Probability:

1. Let us consider a random experiment where the sample space S is considered

as space and two events namely A and B happen there. Then, the formula
would be:

P(S | B) = P(B | B) = 1.

Proof of the same: P(S | B) = P(S ∩ B) ⁄ P(B) = P(B) ⁄ P(B) = 1.

[S ∩ B indicates the outcomes common in S and B equals the outcomes in B].

1. Now let us consider any two events namely A and B happening in a sample
space ‘s’, then, P(A ∩ B) = P(A).

P(B | A), P(A) >0 or, P(A ∩ B) = P(B).P(A | B), P(B) > 0.

This theorem is named as the Multiplication Theorem of Probability.

Proof of the same: As we all know that P(B | A) = P(B ∩ A) / P(A), P(A) ≠ 0.

We can also say that P(B|A) = P(A ∩ B) ⁄ P(A) (as A ∩ B = B ∩ A).

So, P(A ∩ B) = P(A). P(B | A).

Similarly, P(A ∩ B) = P(B). P(A | B).

The interesting information regarding the Multiplication Theorem is that it can

further be extended to more than two events and not just limited to the two events.
So, one can also use this theorem to find out the conditional probability in terms of
A, B, or C.

Read More: Types of Sets

 Sometimes students get confused between Conditional Probability and Joint
Probability. It is essential to know the differences between the two.

39. Answer: b

Explanation:
(a−6)2 (b−5)2
As, ∣a − 5∣ < 1 and ∣b − 5∣ < 1 ⇒ 4 < a, b < 6 and 9
​ + 4
​ ≤ 1 Taking axes as a -axis
and b -axis

The set A represents square PQRS inside set B representing ellipse and hence A ⊂
B.

Concepts:

1. Types of Sets:

Sets are of various types depending on their features. They are as follows:

Empty Set - It is a set that has no element in it. It is also called a null or void set
and is denoted by Φ or {}.
 Singleton Set - It is a set that contains only one element.
Finite Set - A set that has a finite number of elements in it.
Infinite Set - A set that has an infinite number of elements in it.
Equal Set - Sets in which elements of one set are similar to elements of another
set. The sequence of elements can be any but the same elements exist in both
sets.
Sub Set - Set X will be a subset of Y if all the elements of set X are the same as
the element of set Y.
Power Set - It is the collection of all subsets of a set X.
Universal Set - A basic set that has all the elements of other sets and forms the
base for all other sets.
Disjoint Set - If there is no common element between two sets, i.e if there is no
element of Set A present in Set B and vice versa, then they are called disjoint
sets.
Overlapping Set - It is the set of two sets that have at least one common
element, called overlapping sets.

40. Answer: a

Explanation:

The correct answer is (A): cos-1 θ ( 13 )

41. Answer: c

Explanation:

Explanation:
Let 𝑂𝐶 be the tower at the centre 𝑂 of the circular park.
𝐴 and 𝐵 are two points on the boundary of the park ( = 𝑎) subtends an angle of 60∘
at the foot of the tower and the angle of elevation of the top of the tower from 𝐴 or 𝐵
is
Let 'ℎ' be the height of the tower.
We have to find the value of 'ℎ'.
 Since, ∠𝐴𝑂𝐵 = 60∘ and ∠𝑂𝐴𝐵 = ∠𝑂𝐵𝐴
∴Δ𝑂𝐴𝐵 is an equilateral triangle.
∴𝑂𝐴 = 𝑂𝐵 = 𝐴𝐵 = 𝑎
Now in △OAC,
tan⁡30∘ = ℎ
𝑎
⇒ 1
√3
= ℎ
𝑎 [Using trigonometric ratios ]
𝑎
⇒ℎ= √3
𝑎
∴ The height of the tower is √3
Hence, the correct option is (C): a
3
​

​
.

Chapter Trigonometric Functions One Shot Video | Importa…

Importa…

Concepts:

1. Trigonometric Functions:
The relationship between the sides and angles of a right-angle triangle is described
by trigonometry functions, sometimes known as circular functions. These
trigonometric functions derive the relationship between the angles and sides of a
triangle. In trigonometry, there are three primary functions of sine (sin), cosine (cos),
tangent (tan). The other three main functions can be derived from the primary
functions as cotangent (cot), secant (sec), and cosecant (cosec).

Six Basic Trigonometric Functions:

Sine Function: The ratio between the length of the opposite side of the triangle
to the length of the hypotenuse of the triangle.

sin x = a/h

Cosine Function: The ratio between the length of the adjacent side of the
triangle to the length of the hypotenuse of the triangle.

cos x = b/h

Tangent Function: The ratio between the length of the opposite side of the
triangle to the adjacent side length.

tan x = a/b

Tan x can also be represented as sin x/cos x

Secant Function: The reciprocal of the cosine function.

sec x = 1/cosx = h/b

Cosecant Function: The reciprocal of the sine function.

cosec x = 1/sinx = h/a

Cotangent Function: The reciprocal of the tangent function.

cot x = 1/tan x = b/a

Formulas of Trigonometric Functions:

42. Answer: b

Explanation:

Explanation:
𝑥+1 𝑦−1 𝑧−2
Given:Equation of line, 1 = 2 = 2 … (𝑖)and equation of the plane,
2𝑥 − 𝑦 + √𝜆𝑧 + 4 = 0 … (𝑖𝑖)We have to find value of 𝜆 such that the angle 𝜃 between
line and plane is given by sin⁡𝜃 = 13 The angle 𝜃 between the line (i) and plane (ii) is
given by sin⁡𝜃 = | 1.2 + 2 ⋅ ( − 1) + 2 ⋅ (√𝜆)
√1 + 4 + 4 ⋅ √4 + 1 + 𝜆
|[Using formula of Angle Between a Line and a
1 2√𝜆
Plane] ⇒ 3 = | 3 ⋅ √4 + 1 + 𝜆
| [sin⁡𝜃 = 13 ( Given )] ⇒ √5 + 𝜆 = 2√𝜆 ⇒ 5 + 𝜆 = 4𝜆 [Squaring
5
both sides] ⇒ 3𝜆 = 5 ⇒ 𝜆 = 3
Hence, the correct option is (B).

43. Answer: c

Explanation:

Explanation:

3 2
We know that, volume of sphere V = 43 𝜋r Surface area of sphere S = 4𝜋r , where 𝑟 is
the radius of the sphere.So, rate of change of volume of sphere,
⇒ dVd 4 𝜋𝑟3 ) = 4 × 3 × 𝜋𝑟2 dr ⇒ dV = 4𝜋r2 dr
= ----(1)Rate of change of surface area of
dtdt ( 3 3 dt dt dt
2
sphere, ⇒ dS dt
d (4𝜋𝑟 ) = 4𝜋 × 2 × 𝑟 dr ⇒ dS = 8𝜋r dr
= dt dt d𝑡 dt
----(2)From equation (1) and
dV 2 dr
4𝜋r dt
equation (2), ⇒ dV 4r
dS = dS = 8𝜋r dr ⇒ 8 = 2 (∵𝑟 = 4 cm)
dt
Therefore, the rate of change of
dt dt
3 2
volume of a sphere with respect to its surface area is 2 cm / cm .Hence, the correct
option is (C).
 Second Order Derivatives | Class 12 Maths NCERT | IITJEE …

44. Answer: d

Explanation:

Explanation:
The given expression is 𝑝 ∨ ( ∼ 𝑝 ∧ 𝑞).
Thus, the negation of this expression can be written as:∼(𝑝 ∨ ( ∼ 𝑝 ∧ 𝑞))
=∼ 𝑝 ∧ ∼ ( ∼ 𝑝 ∧ 𝑞)[ ∵∼ (𝑎 ∨ 𝑏) ≡∼ 𝑎 ∧ ∼ 𝑏]
=∼ 𝑝 ∧ (𝑝 ∨ ∼ 𝑞)[ ∵∼ (𝑎 ∧ 𝑏) ≡∼ 𝑎 ∨ ∼ 𝑏]
= ( ∼ 𝑝 ∧ 𝑝) ∨ ( ∼ 𝑝 ∧ ∼ 𝑞)[∵𝑐 ∧ (𝑎 ∨ 𝑏) ≡ (𝑐 ∧ 𝑎) ∨ (𝑐 ∧ 𝑏)]
= 𝐹 ∨ ( ∼ 𝑝 ∧ ∼ 𝑞)[ ∵∼ 𝑎 ∧ 𝑎 ≡ 𝐹]
= ( ∼ 𝑝 ∧ ∼ 𝑞)[∵𝐹 ∨ 𝑎 ≡ 𝑎]
Hence, the correct option is (D).

Concepts:

1. Sets:

In mathematics, a set is a well-defined collection of objects. Sets are named and

demonstrated using capital letter. In the set theory, the elements that a set
comprises can be any sort of thing: people, numbers, letters of the alphabet,
shapes, variables, etc.

Read More: Set Theory

 Elements of a Set:

The items existing in a set are commonly known to be either elements or members
of a set. The elements of a set are bounded in curly brackets separated by
commas.

Read Also: Set Operation

Cardinal Number of a Set:

The cardinal number, cardinality, or order of a set indicates the total number of
elements in the set.

Read More: Types of Sets

45. Answer: a

Explanation:

Explanation:
0 2𝑞 𝑟
𝑇 𝑇
Given:𝐴 = ( 𝑝 𝑞 −𝑟 )We have to find the value of |𝑝| if 𝐴𝐴 = 𝐼3 Consider:𝐴𝐴 = 𝐼3
𝑝 −𝑞 𝑟
0 2𝑞 𝑟 0 𝑝 𝑝 1 0 0
( 𝑝 𝑞 −𝑟 )( 2𝑞 𝑞 −𝑞 ) = ( 0 1 0 )
𝑝 −𝑞 𝑟 𝑟 −𝑟 𝑟 0 0 1
2 2 2 2 2 2
4𝑞 + 𝑟 2𝑞 − 𝑟 −2𝑞 + 𝑟 1 0 0
2 2 2 2 2 2 2 2
( 2𝑞 − 𝑟 𝑝 +𝑞 +𝑟 𝑝 − 𝑞 − 𝑟 ) = ( 0 1 0 )After comparing, we get
2 2 2 2 2 2 2 2 0 0 1
−2𝑞 + 𝑟 𝑝 −𝑞 −𝑟 𝑝 +𝑞 +𝑟
𝑝2 + 𝑞2 + 𝑟2 = 1....(i)𝑝2 − 𝑞2 − 𝑟2 = 0....(ii)Adding equation (i) and (ii)2𝑝2 = 1𝑝2 = 12 𝑝 = 1
√2
Hence, the correct option is (A).

Concepts:

1. Transpose of a Matrix:

The matrix acquired by interchanging the rows and columns of the parent matrix is
called the Transpose matrix. The transpose matrix is also defined as - “A Matrix
 which is formed by transposing all the rows of a given matrix into columns and
vice-versa.”

The transpose matrix of A is represented by A’. It can be better understood by the

given example:

The transpose matrix of A

is denoted by A’

Now, in Matrix A, the number of rows was 4 and the number of columns was 3 but,
on taking the transpose of A we acquired A’ having 3 rows and 4 columns.
Consequently, the vertical Matrix gets converted into Horizontal Matrix.

Hence, we can say if the matrix before transposing was a vertical matrix, it will be
transposed to a horizontal matrix and vice-versa.

Read More: Transpose of a Matrix

46. Answer: b

Explanation:

Explanation:
Given :A function y = f(x) such that f''(x) = 6(x - 1)....... (A)The graph of f (x) passes
through the point (2, 1) at which the tangent to the graph is y = 3x - 5.We have to
find the value of f (x). We shall use the concept of tangents and normals in this
question.From eqn (A), we havef'(x) = 3(x -1)2 +C ........ (i)At the point (2, 1) the
𝑑𝑦
tangent to the graph is given byy = 3x - 5Slope of tangent i.e. 𝑑𝑥
= 3⇒ f'(2) =
3Therefore, f'(2) = 3(2 - 1)2 +C = 3 + C = 0.From equation (i), we getf'(x) = 3(x - 1)2⇒
f'(x) = 3(x - 1)2⇒ f(x) = (x - 1)3 + k ..........(ii)Since, graph of f(x) passes through (2, 1),
therefore1 = (2 - 1)3 + k⇒ k = 0Equation of funciton isf(x) = (x - 1)3Hence, the correct
option is (B).
47. Answer: c

Explanation:

Explanation:
Given:A trigonometric equation, sin⁡5𝑥 + sin3𝑥 + sin⁡𝑥 = 0,We have to find the value of 𝑥
other than zero lying between 0 ≤ 𝑥 ≤ 𝜋2 Consider,sin⁡5𝑥 + sin⁡3𝑥 + sin⁡𝑥 = 0
⇒ (sin⁡𝑥 + sin⁡5𝑥) + sin⁡3𝑥 = 0 ⇒ 2sin⁡3𝑥cos⁡2𝑥 + sin⁡3𝑥 = 0 [∵sin⁡𝐶 + sin⁡𝐷 = 2sin⁡( 𝐶 +2 𝐷 )cos⁡( 𝐶 −2 𝐷 )]
⇒ sin⁡3𝑥(2cos⁡2𝑥 + 1) = 0 ⇒ sin⁡3𝑥 = 0 or 2cos⁡2𝑥 + 1 = 0But cos⁡2𝑥 ≠ −1
2 for 𝑥 ∈ [0, 𝜋2 ]Thus
sin⁡3𝑥 = 0 ⇒ 3𝑥 = 𝜋 ⇒ 𝑥 = 𝜋3 Hence, the correct option is (C).

48. Answer: a

Explanation:

Explanation:
Given:A geometric progression such that 𝑛th term is 𝑎𝑛 and
∑100 𝑎2𝑛 = 𝛼, ∑100 𝑎2𝑛 − 1 = 𝛽We have to find the common ratio of given GP.Let a be
𝑛=1 𝑛=1
the first term, r be the common ratio of the given GP (𝑟 > 0).Then
2 3 199
a, ar, ar , ar , … . , ar are in GP.We shall use the formula of general term of GP.
100
𝛼= ∑𝑛 = 1 𝑎2𝑛 ⇒ 𝑎 = 𝑎2 + 𝑎4 + … … . + 𝑎200
3 199 2 198
⇒ 𝑎 = 𝑎𝑟 + 𝑎𝑟 + … … . . + 𝑎𝑟 = ar(1 + 𝑟 + … … + 𝑟 )and
100 2 198
𝛽= ∑𝑛 = 1 𝑎2𝑛 − 1 ⇒ 𝛽 = 𝑎1 + 𝑎3 + … … … + 𝑎199 ⇒ 𝛽 = 𝑎 + 𝑎𝑟 + … … . . + ar
⇒ 𝛽 = 𝑎(1 + 𝑟2 + … … . + 𝑟198 )Clearly, a / 𝛽 = 𝑟 = common ratioHence, the correct
option is (A).

49. Answer: c

Explanation:

Check for reflexivity:

As 3(a − a) + 7=
​ 7 ​

which belongs to relation so relation is reflexive

Check for symmetric:
Take a = 7
3 ,b
​

​
=0
Now (a, b) ∈ R but (b, a) ∈
/R
 As 3(b − a) + 7=0
​

which is rational so relation is not symmetric.

Check for Transitivity:
Take (a, b) as ( 3
7​

​, 1)
&(b, c) as (1, 2 3 7 ) ​

So now (a, b) ∈ R&(b, c) ∈ R but (a, c) ∈

/ R which means relation is not transitive

Concepts:

1. Operations on Sets:

Some important operations on sets include union, intersection, difference, and the
complement of a set, a brief explanation of operations on sets is as follows:

1. Union of Sets:

The union of sets lists the elements in set A and set B or the elements in both
set A and set B.
For example, {3,4} ∪ {1, 4} = {1, 3, 4}
It is denoted as “A U B”

2. Intersection of Sets:

Intersection of sets lists the common elements in set A and B.

For example, {3,4} ∪ {1, 4} = {4}
It is denoted as “A ∩ B”

3.Set Difference:

Set difference is the list of elements in set A which is not present in set B
For example, {3,4} - {1, 4} = {3}
It is denoted as “A - B”

4.Set Complement:

The set complement is the list of all elements present in the Universal set
except the elements present in set A
It is denoted as “U-A”
50. Answer: b

Explanation:

The correct answer is option (B) : 6 3 ​

Shortest distance between two lines

x−x1
a1
​

​ = y−y
a2
1
= z−z
a3
1
and x−x b1
2
= ​

​
​

​
​

​
y−y2
b2
​

​ = z−z2
b3
​

​ is given as
∣x 1 − x 2 y 1 − y 2 z 1 − z 2 ∣
​ ​ ​ ​ ​ ​

​ ​ ​ ​ ​ ​

a1 ​
a2 a3 ​
​
​
​
​
​ ​

= ∣ b1 b2 b3 ∣ ​ ​ ​

(a1 b3 −a3 b2 )2 +(a1 b3 −a3 b1 )2 +(a1 b2 −a2 b1 )2

​

​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

∣5 − (3) 2 − (−5) 4 − 1∣
1 ​
2 −3 ​ ​ ​ ​

1 4 −5 ∣
=∣
(−10+12)2 +(−5+3)2 +(4−2)2
​

∣8 7 3 ∣
1 2 −3 ​ ​ ​ ​ ​

1 4 −5∣
= ∣
(2)2 +(−2)2 +(2)2
​

∣8(−10 + 12) − 7(−5 + 3) + 3(4 − 2)∣

=
​

4+4+4
​

16 + 14 + 6∣
=∣
​

12
​

36
= 12
= 2363 ​

​
​

18
= 3
=6 3 ​

​
​

Concepts:

1. Three Dimensional Geometry:

Mathematically, Geometry is one of the most important topics. The concepts of

Geometry are derived w.r.t. the planes. So, Geometry is divided into three major
categories based on its dimensions which are one-dimensional geometry, two-
dimensional geometry, and three-dimensional geometry.

Direction Cosines and Direction Ratios of Line:

Consider a line L that is passing through the three-dimensional plane. Now, x,y and z
are the axes of the plane and α,β, and γ are the three angles the line makes with
these axes. These are commonly known as the direction angles of the plane. So,
appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the
given line L.
51. Answer: c

Explanation:

np + npq = 5, np ⋅ npq = 6
np(1 + q) = 5, n2 p2 q = 6
n2 p2 (1 + q)2 = 25, n2 p2 q = 6
6
q (1
​ + q)2 = 25
2
6q + 12q + 6 = 25q
6q 2 − 13q + 6 = 0
6q 2 − 9q − 4q + 6 = 0
(3q − 2)(2q − 3) = 0
q = 23 , 32 , q =
​ ​
2
3
​
is accepted
p = 13 ​
⇒n⋅ 1
3
​
+n⋅ 1
3
​
⋅ 2
3 =5
3n+2n
9
​
=5
n=9
1
So 6(n + p − q) = 6 (9 + 3
​
− 23 ) = 52
​

Concepts:

1. Binomial Distribution:

A common probability distribution that models the probability of obtaining one of

two outcomes under a given number of parameters is called the binomial
distribution. It summarizes the number of trials when each trial has the same
probability of attaining one specific outcome. The value of a binomial is acquired by
multiplying the number of independent trials by the successes.

Criteria of Binomial Distribution:

 Binomial distribution models the probability of happening an event when specific
criteria are met. In order to use the binomial probability formula, the binomial
distribution involves the following rules that must be present in the process:

1. Fixed trials
2. Independent trials
3. Fixed probability of success
4. Two mutually exclusive outcomes

52. Answer: b

Explanation:

∼ (q ∨ ((∼ q) ∧ p))
=∼ q∧ ∼ ((∼ q) ∧ p)
=∼ q ∧ (q∨ ∼ p)
= (∼ q ∧ q) ∨ (∼ q∧ ∼ p)
= (∼ q∧ ∼ p)

Concepts:

1. Binomial Distribution:

A common probability distribution that models the probability of obtaining one of

two outcomes under a given number of parameters is called the binomial
distribution. It summarizes the number of trials when each trial has the same
probability of attaining one specific outcome. The value of a binomial is acquired by
multiplying the number of independent trials by the successes.

Criteria of Binomial Distribution:

Binomial distribution models the probability of happening an event when specific

criteria are met. In order to use the binomial probability formula, the binomial
distribution involves the following rules that must be present in the process:

1. Fixed trials
2. Independent trials
3. Fixed probability of success
4. Two mutually exclusive outcomes
53. Answer: b

Explanation:

The plane P passes through the intersection of the planes:

P1 : 2x + 3y − z − 2 = 0
​ and P2 : x + 2y + 3z − 6 = 0.
​

The equation of P is:

P ≡ P1 + λP2 = 0. ​

Substituting the equations of P1 and P2 :

​ ​

(2 + λ)x + (3 + 2λ)y + (−1 + 3λ)z − (2 + 6λ) = 0. (1)

Step 2: Determine λ Since P is perpendicular to the plane P3 : 2x + y - z + 1 = 0, the ​

normal vector of P is perpendicular to the normal vector of P_3. This gives:

nP ⋅ n3 = 0,
​ ​

where nP = (2 + λ, 3 + 2λ, −1 + 3λ) and n3 = (2, 1, −1). Taking the dot product:
​ ​

(2 + λ)(2) + (3 + 2λ)(1) + (−1 + 3λ)(−1) = 0.

Simplify:

4 + 2λ + 3 + 2λ + 1 − 3λ = 0, 8 + λ = 0.

Solve for \lambda:

λ = −8. (2)

Step 3: Equation of Plane P Substitute \lambda = -8 into equation (1):

P = (−6)x − (13)y + (−25)z + 46 = 0. (3)

Step 4: Distance of Point (-7, 1, 1) from Plane P The distance d of a point (x_1, y_1, z_1)
from a plane ax + by + cz + d = 0 is:

∣ax1 + by1 + cz1 + d∣

d= .
​ ​ ​

a2 + b 2 + c 2
​

Here, (x_1, y_1, z_1) = (-7, 1, 1), and the equation of P is -6x - 13y - 25z + 46 = 0.
 Substituting:

∣(−6)(−7) + (−13)(1) + (−25)(1) + 46∣

d= .
(−6)2 + (−13)2 + (−25)2
​

Simplify:

∣42 − 13 − 25 + 46∣
d= ,
36 + 169 + 625
​

50
d= .
830
​

Step 5: Calculate d^2

2
50 2500 250
d =(
2
) = = .
830 830 83
​ ​ ​

Conclusion:

250
d2 = .
83
​

Concepts:

1. Distance between Two Points:

The distance between any two points is the length or distance of the line segment
joining the points. There is only one line that is passing through two points. So, the
distance between two points can be obtained by detecting the length of this line
segment joining these two points. The distance between two points using the given
coordinates can be obtained by applying the distance formula.

54. Answer: c

Explanation:
 The correct answer is (C) : 5
7
​

5 C +6 C
2 2
​ ​

2 C +3 C +4 C +5 C +6 C ​

2​
2 ​
2 2 ​
2 ​ ​

10+15 25
= 1+3+6+10+15
​ = 35
​

5
= 7
​

Concepts:

1. Probability:

Probability is defined as the extent to which an event is likely to happen. It is

measured by the ratio of the favorable outcome to the total number of possible
outcomes.

The definitions of some important terms related to probability are given

below:

Sample space

The set of possible results or outcomes in a trial is referred to as the sample space.
For instance, when we flip a coin, the possible outcomes are heads or tails. On the
other hand, when we roll a single die, the possible outcomes are 1, 2, 3, 4, 5, 6.

Sample point

In a sample space, a sample point is one of the possible results. For instance, when
using a deck of cards, as an outcome, a sample point would be the ace of spades
or the queen of hearts.

Experiment

When the results of a series of actions are always uncertain, this is referred to as a
trial or an experiment. For Instance, choosing a card from a deck, tossing a coin, or
rolling a die, the results are uncertain.

Event

An event is a single outcome that happens as a result of a trial or experiment. For

instance, getting a three on a die or an eight of clubs when selecting a card from a
deck are happenings of certain events.
 Outcome

A possible outcome of a trial or experiment is referred to as a result of an outcome.

For instance, tossing a coin could result in heads or tails. Here the possible
outcomes are heads or tails. While the possible outcomes of dice thrown are 1, 2, 3,
4, 5, or 6.

55. Answer: d

Explanation:

2
⎡1 0 0 ⎤ ⎡1 0 0⎤
A = 0 4 −1 0 4 −1
⎣0 12 −3⎦ ⎣0 12 −3⎦
​ ​ ​ ​ ​ ​ ​ ​ ​ ​

⎡1 0 0⎤
= 0 4 −1 = A
⎣0 12 −3⎦
​ ​ ​ ​ ​

⇒ A3 = A4 = …… = A
(A + I)11 = 11 C0 A11 + 11 C1 A10 + … . ⋅ 11 C10 A + 11 C11 I
​ ​ ​ ​

= (11 C0 + 11 C1 + … .11 C10 ) A + I

​ ​ ​

= (211 − 1) A + I = 2047A + I
∴ Sum of diagonal elements = 2047(1 + 4 − 3) + 3
= 4094 + 3 = 4097

Concepts:

1. Matrix Transformation:

The numbers or functions that are kept in a matrix are termed the elements or the
entries of the matrix.

Transpose Matrix:

The matrix acquired by interchanging the rows and columns of the parent matrix is
termed the Transpose matrix. The definition of a transpose matrix goes as follows -
“A Matrix which is devised by turning all the rows of a given matrix into columns and
vice-versa.”
56. Answer: d

Explanation:

The correct answer is (D) : 24

a + 6d = 3 .......................(1)
Z = a(a + 3d)
= (3 − 6d)(3 − 3d)
= 18d2 − 27d + 9
Differentiating with respect to d
⇒ 36d − 27 = 0
⇒ d = 34 , from (1)a =
​
−3
2
, (Z = minimum
​ )
Now, Sa = ​
n
2
​ (−3 + (n − 1) 34 ) = 0
​

⇒n=5
Now,
n! − 4an(n+2) = 120 − 4 (a35 )
​ ​

= 120 − 4(a + (35 − 1)d)

= 120 − 4 ( −3
2
+ 34 ⋅ ( 34 ))
​ ​

= 120 − 4 ( −6+102
4
) ​

= 120 − 96 = 24

Concepts:

1. Arithmetic Progression:

Arithmetic Progression (AP) is a mathematical series in which the difference

between any two subsequent numbers is a fixed value.

For example, the natural number sequence 1, 2, 3, 4, 5, 6,... is an AP because the

difference between two consecutive terms (say 1 and 2) is equal to one (2 -1). Even
when dealing with odd and even numbers, the common difference between two
consecutive words will be equal to 2.

In simpler words, an arithmetic progression is a collection of integers where each

term is resulted by adding a fixed number to the preceding term apart from the first
term.

For eg:- 4,6,8,10,12,14,16

 We can notice Arithmetic Progression in our day-to-day lives too, for eg:- the
number of days in a week, stacking chairs, etc.

Read More: Sum of First N Terms of an AP

57. Answer: b

Explanation:

We are asked to find the coefficient of x301 in the expansion of the series:

(1 + x)500 + x(1 + x)499 + x2 (1 + x)498 + ⋯ + x500 .

Step 1: Understanding the Series

The given series can be written as:

S = (1 + x)500 + x(1 + x)499 + x2 (1 + x)498 + ⋯ + x500 .

This is a sum of terms where each term involves (1 + x)500−n multiplied by xn , where
n ranges from 0 to 500.

The general term in the expansion is:

xn (1 + x)500−n .

Step 2: Finding the Coefficient of x301

We need to find the coefficient of x301 in the entire series. For each term xn (1 + x)500−n
, the exponent of x in the expanded form of (1 + x)500−n will be 500 − n . We are
interested in terms where the total exponent of x equals 301.

The exponent of x in each term is:

n + k = 301,

where k is the exponent of x in the expansion of (1 + x)500−n . The coefficient of x301 in

(1 + x)500−n is:
500−n
C301−n .
​

Step 3: Coefficient of x301 in the Series

 Thus, the required coefficient is the sum of the coefficients of x301 in all terms. After
simplifying, we get the coefficient of x301 as:
501
C200 .
​

Thus, the correct answer is 501

C200 .
​

Concepts:

1. Binomial Theorem:

The binomial theorem formula is used in the expansion of any power of a binomial
in the form of a series. The binomial theorem formula is

Properties of Binomial Theorem

The number of coefficients in the binomial expansion of (x + y)n is equal to (n +

1).
There are (n+1) terms in the expansion of (x+y)n.
The first and the last terms are xn and yn respectively.
From the beginning of the expansion, the powers of x, decrease from n up to 0,
and the powers of a, increase from 0 up to n.
The binomial coefficients in the expansion are arranged in an array, which is
called Pascal's triangle. This pattern developed is summed up by the binomial
theorem formula.

58. Answer: c

Explanation:

f ′ (x) = x2 + 2b + ax
g ′ (x) = x2 + a + 2bx
(2b − a) − x(2b − a) = 0
∴ x = 1 is the common root
 Put x = 1 in f ′ (x) = 0 or g ′ (x) = 0
1 + 2b + a = 0
7 + 2b + a = 6
So, the correct option is (C) : 6

Concepts:

1. Relations and functions:

A relation R from a non-empty set B is a subset of the cartesian product A × B. The

subset is derived by describing a relationship between the first element and the
second element of the ordered pairs in A × B.

A relation f from a set A to a set B is said to be a function if every element of set A

has one and only one image in set B. In other words, no two distinct elements of B
have the same pre-image.

Representation of Relation and Function

Relations and functions can be represented in different forms such as arrow
representation, algebraic form, set-builder form, graphically, roster form, and
tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4,
f(3) = 9. Now, represent this function in different forms.

1. Set-builder form - {(x, y): f(x) = y2, x ∈ A, y ∈ B}

2. Roster form - {(1, 1), (2, 4), (3, 9)}
3. Arrow Representation
59. Answer: d

Explanation:

For N − 2, 3N , N + 2 to be in geometric progression, the condition:

​

( 3N )2 = (N − 2)(N + 2)
​

3N 2 = N 2 − 4 ⇒ N 2 − 3N − 4 = 0

Solving this quadratic equation, we find:

N =4 (since N must be a possible sum of two dice)

The favorable outcomes for N = 4 when two dice are rolled are (1,3), (2,2), (3,1),
totaling 3 outcomes. Since there are 6 × 6 = 36 total outcomes:

3 1
P (A) = 36
​ = 12
​

Thus, k = 4 as 1
12
​ = k
48
​.

Concepts:
1. Probability:

Probability is defined as the extent to which an event is likely to happen. It is

measured by the ratio of the favorable outcome to the total number of possible
outcomes.

The definitions of some important terms related to probability are given

below:

Sample space

The set of possible results or outcomes in a trial is referred to as the sample space.
For instance, when we flip a coin, the possible outcomes are heads or tails. On the
other hand, when we roll a single die, the possible outcomes are 1, 2, 3, 4, 5, 6.

Sample point

In a sample space, a sample point is one of the possible results. For instance, when
using a deck of cards, as an outcome, a sample point would be the ace of spades
or the queen of hearts.

Experiment

When the results of a series of actions are always uncertain, this is referred to as a
trial or an experiment. For Instance, choosing a card from a deck, tossing a coin, or
rolling a die, the results are uncertain.

Event

An event is a single outcome that happens as a result of a trial or experiment. For

instance, getting a three on a die or an eight of clubs when selecting a card from a
deck are happenings of certain events.

Outcome

A possible outcome of a trial or experiment is referred to as a result of an outcome.

For instance, tossing a coin could result in heads or tails. Here the possible
outcomes are heads or tails. While the possible outcomes of dice thrown are 1, 2, 3,
4, 5, or 6.
60. Answer: b

Explanation:

For x = 0
2
f (x) = ? et-x dt = e-x (e2 - 1)
?

0
For 0 < x < 2
x
2
f (x) = ? ex-t dt + ?x et-x dt = ex + e2-x - 2
? ?

0
For x = 2
2
f (x) = ? ex-t dt = ex-2 (e2 - 1)
?

0
For x = 0, f (x) is ? and x = 2, f (x) is ?
? Minimum value of f (x) lies in x ? (0, 2)
Applying A.M = G.M ,
minimum value of f (x) is 2(e - 1)