25-06-2025

2321CMD303032250001 MD

PHYSICS

1) The value of x on solving the quadratic equation 2x2 – 7x + 5 = 0 is :-

(1)
and 1

(2)
and 1

(3)
and 1

(4)
and – 1

2)

(1)

(2)

(3) 1
(4) Can't be determined

3)

cos215° – sin215° = .......... ?

(1)

(2) 1

(3)

(4)

4)

Which of following can be correct for any value of θ :-

(1) sin θ = –2
(2) cos θ = 2
(3) tan θ = 1000
(4) All
5) The slope of straight line is :-

(1)

(2)

(3)
(4) 3

6) If –3x – 5y = 15 then y – x graph is :-

(1)

(2)

(3)

(4)

7) If y = 5x2 – 2x then value of x for which, y is minimum is :-

(1)

(2)

(3)

(4)

8) If y = sin2x + cos2x, then is :-

(1) 2sin x + 2cos x

(2) 1
(3) 0
(4) 2sin x – 2cos x
9) The value of shaded area is :-

(1) 1
(2) –1
(3) 2
(4) –2

10) is equal to :-

(1) 2
(2) –2
(3) 6
(4) 8

11) Evaluate
Where KQ1Q2 is a constant.

(1)
KQ1Q2

(2)
KQ1Q2

(3)
KQ1Q2

(4)
KQ1Q2

12) Maximum value of 5 sinθ – 12 cosθ is:-

(1) 13
(2) 5
(3) 12
(4) None

13) (1+x)–n, where x << 1, can be written as :-

(1) 1 + nx
(2) 1 – nx
(3) x + n
(4) x – n

14)

1– + – + – +....... =?

(1) 3/2
(2) 2
(3) 1
(4) None of these

15) (0.97)1/3 is approximately equals to :

(1) 0.99
(2) 0.01
(3) 0.985
(4) None

16)

The value of sin (300°) :-

(1)

(2)

(3)

(4)

17) Value of sin 1° is approximately equal to:-

(1) 1

(2)

(3)

(4) None of these

18) If y = a(1 – cosx) where a is constant then find =?

(1) a (1 + sinx)
(2) a (sinx)
(3) –a cosx
(4) a (cos2x)

19) Find rate of change of volume of cube w.r.t. its side, where side is 3 m

(1) 9 m2
(2) 27 m2
(3) 24 m2
(4) zero

20) log10(1000)1/3 = ?

(1) 3

(2)

(3) 1
(4) –1

21) For the given graph, slope will be :–

(1) tan(30°)

(2)

(3) (1) and (2) both

(4) tan (150°)

22) Calculate the distance between two points (0, –1, 1) and (3, 3, 13) : –

(1) 5
(2) 12
(3) 13
(4) 17

23) Which of the following equation is the best representation of the given graph ?
(1) y = x2
(2) x = y2
(3) y = ex
(4) y = x

24) Which of two following graph is the best representation for the given equation.
y = sinx

(1)

(2)

(3)

(4)

25) Unit vector along will be

(1)

(2)

(3)

(4)

26) The unit vector along a vector is

(1)

(2)

(3)
(4)

27) The magnitude of vector is :-

(1)
(2)
(3)
(4)

28) If a unit vector is represented by then the value of q is :

(1)
(2)
(3)
(4)

29) If then -

(1) is parallel to
(2)
(3)
(4) and are mutually perpendicular

30) If then is :-

(1)
(2)
(3)

(4)

31) if = 5, = 12 and = 13 then find angle between and is :

(1)

(2)

(3)

(4)

32) The resultant of two equal forces is double of either forces. Angle between them is :

(1) 120°
(2) 90°
(3) 60°
(4) 0°

33) If = 2 and = and angle between and is 30°, then resultant will be :

(1)
(2)
(3)
(4)

34) If the direction of cosines of the vector are :-

(1)
and

(2)
and

(3)
and

(4)
and

35) Projection of vector on y-axis is :-

(1)

(2)

(3) 7
(4) 4

36) The angle made by the vector with x-axis is :-

(1) 90°
(2) 45°
(3) 22.5°
(4) 30°

37) The vector and are perpendicular to each other. The positive value
of a is :-

(1) 3
(2) 4
(3) 9
(4) 13

38) In a triangle, shown in figure :-

(1)
(2)
(3)
(4)

39)

Resultant force is :-

(1) 20 N
(2) 10 N
(3) 5 N
(4) None of these

40)

If = 1; then angle between is:-

(1) 0°
(2) 45°
(3) 90°
(4) 60°

41) Let cos sinθ, be any vector. Another vector which is normal to is :-

(1) cosθ + sinθ

(2) sin θ + cos θ
(3) sin θ – cos θ
(4) cos θ – sin θ
42) Find angle between minute hand and hour hand for 2:00 PM.

(1) 30º
(2) 45º
(3) 90º
(4) 60º

43) Find the vector that must be added to the vector and so that the resultant
vector is a unit vector along the y-axis.

(1)
(2)
(3)
(4)

44)

The angle between the vectors and is :

(1) 90°
(2) 60°
(3) 30°
(4) 45°

45) A force of 6 N and another of 8 N can be applied together to produce the effect of a single force
of :-

(1) 1 N
(2) 9 N
(3) 15 N
(4) 22 N

CHEMISTRY

1) Rearrange the following (I to IV) in the order of increasing masses:

(I) 0.5 mole of O3
(II) 0.5 g atoms of oxygen
(III) 3.01 × 1023 molecules of CO2
(IV) 5.6 L of CO2 at STP

(1) II < IV < III < I

(2) II < I < IV < III
(3) IV < II < III < I
(4) I < II < III < IV
2) Match the items in column-I with those in column-II.

Column-I Column-II

36.138 × 1023
(a) 8.8 g CO2(g) (p)
atoms of oxygen

5.6 L of CO2(g) 12.046 × 1022

(b) (q)
at STP molecules

(c) 96 g of O2(g) (r) 0.25 mole molecules

(d) 49 g of H2SO4 (s) 0.5 mole molecules

(1) (a)→(q), (b)→(r), (c)→(s), (d)→(p)
(2) (a)→(p), (b)→(r), (c)→(s), (d)→(q)
(3) (a)→(p), (b)→(r), (c)→(q), (d)→(s)
(4) (a)→(q), (b)→(r), (c)→(p), (d)→(s)

3) The number of electron in 3.1 mg NO3¯ is (NA = 6 × 1023)

(1) 32
(2) 1.6 × 10–3
(3) 9.6 × 1020
(4) 9.6 × 1023

4) The moles of O2 required for reacting with 6.8g ammonia

is

(1) 5
(2) 2.5
(3) 1
(4) 0.5

5) Match the column I with column II :

Column I Column II
(Amount of (Number of
substance) oxygen atoms)

(A) 98 mg H2SO4 (P) 2 NA

18 mL water
(B) (Q) 2 × 10–3 NA
(d = 1 g/mL)

(C) 60 g NO gas (R) 4 × 10–3 NA

44.8 mL of CO2 gas

(D) (S) NA
at 0.5 atm & 273 K

Correct match is :
(1) A → P, B → Q, C → R, D → S
(2) A → P, B → S, C → R, D → Q
(3) A → R, B → S, C → P, D → Q
(4) A → R, B → S, C → Q, D → P

6) Volume occupied by one molecule of water (density = 1 g cm–3) is (approximately) equal to :

(1) 3 × 10–23 cm3

(2) 5.5 × 10–23 cm3
(3) 9 × 10–23 cm3
(4) 6.023 × 10–23 cm3

7) Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous
hydrochloric acid according to the reaction
4HCl(aq)+MnO2(s) → 2H2O(ℓ) + MnCl2(aq) + Cl2(g)
How many gram of HCl react with 5.0 g of manganese dioxide ? (At. wt. of Mn = 55)

(1) 2.12 g
(2) 44.24 g
(3) 8.4 g
(4) 3.65 g

8) 120 g impure CaCO3 sample produces 56 g of CaO on strong heating, then % purity of CaCO3
sample will be :-

(1) 53.3%
(2) 83.3%
(3) 33.3%
(4) 60%

9) When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g) each at S.T.P, the moles of HCl(g)
formed is equal to :-

(1) 1 mol of HCl(g)

(2) 2 mol of HCl(g)
(3) 0.5 mol of HCl(g)
(4) 1.5 mol of HCl(g)

10) The equivalent weight of phosphoric acid (H3PO4) in the reaction,

NaOH + H3PO4 → NaH2PO4 + H2O is :-

(1) 59
(2) 49
(3) 25
(4) 98

11) Determine the moles of product formed by the reaction of 7 mol of A with 8 mol of B ?
A + 2B → 3C ?

(1) 7 mol
(2) 12 mol
(3) 21 mol
(4) 6 mol

12) 10 L of Butane (C4H10) on complete combustion will produce ?

(1) 4L of CO2
(2) 20L of CO2
(3) 10L of CO2
(4) 40L of CO2

13) Equal masses of O2 and N2 are reacted according to the equation : O2 + N2 —→ 2NO. Which
statement is true ?

(1) O2 is the limiting reagent and N2 is present in excess

(2) N2 is the limiting reagent and O2 is present in excess.
(3) All of O2 and N2 react and neither is in excess.
(4) Nothing can be said about the limiting reagent

14) When 3g of C2H6 is completely burnt then find out produced volume of water vapour at STP :-

(1) 0.672 lit

(2) 6.72 lit
(3) 67.2 lit
(4) 30 lit

15) Nitric acid (HNO3) is 1.6% H, 22.2%N and 76.2% O by mass. All pure samples of HNO3 have this
composition, according to the

(1) Law of multiple proportions

(2) Law of reciprocal proportions
(3) Law of definite proportions
(4) Law of conservation of mass

16) Which of the following pairs of compounds does not follow law of multiple proportions ?

(1) CO, CO2

(2) H2O, H2O2
(3) CaO, MgO
(4) N2O, N2O3

17)
The largest number of molecules are in :

(1) 36 g H2O
(2) 28 g CO
(3) 46 g C2H5OH
(4) 54 g N2O5

18) Which has the highest mass ?

(1) 56 g of Fe
(2) 1.5 mol N2
(3) 0.1 g atom of H
(4) 6 × 1023 atom of C

19) How many moles of O-atoms are there in 5 mol HNO3 :-

(1)

(2)

(3) 3 mol
(4) 15 mol

20) A contains equal mass of H2, O2 and O3 (ozone), the ratio of their number of atoms will be :-

(1) 2 : 2 : 3
(2) 1 : 1 : 1
(3) 16 : 1 : 1
(4) 1 : 1 : 3

21) Which of the following is correct for 21 mg of N3– (azide ion) ?

–3
(1) It has 1.5×10 NA nitrogen atoms
(2) It has approximately 21 coloumb charge
(3) It is equal to 0.25 mol
(4) All of these

22) The incorrect statement for 14g of CO is :-

(1) It occupies 2.24 litre at NTP

(2) It corresponds to 0.5 mol of CO
(3) It corresponds to same mole as in 14 g of N2
(4) It corresponds to 3.011 × 1023 molecules of CO

23) Caffeine has a molecular weight of 194. If it contains 28.9% by mass of nitrogen, the number of
atoms of nitrogen in one molecule of caffeine is:
(1) 4
(2) 6
(3) 2
(4) 3

24) 40% X & 60% Y by mass combine together to form compound having vapour density 100. If
atomic mass of X and Y is 10 amu and 20 amu respectively then molecular formula of compound is :-

(1) X4Y3
(2) X8Y6
(3) X2Y3
(4) XY

25) The vapour density of a gas is 11.2. The volume occupied by 1.6 g of gas at NTP would be :-

(1) 2.4 litre

(2) 3.2 litre
(3) 4.8 litre
(4) 1.6 litre

26) The percentage composition of a compound is H = 3.6%, O = 38.6%, and molar

mass = 166. What is its molecular formula ?

(1) C4H3O2
(2) C8H6O4
(3) C3H4O3
(4) CH3O2

27) 74.5 g of a metallic chloride contains 35.5 g of chlorine. The equivalent mass of metal is :-

(1) 19.5
(2) 35.5
(3) 39
(4) 74.5

28) Which of the following have maximum mass

(1) 1 Mole H2
(2) 11.2 L CO2 at STP
(3) atoms of carbon
(4) 1 gram atom of Nitrogen

29) Equivalent weight of nitrogen in NH3 is :-

(1) 14
(2)

(3)

(4) 28

30) Which of the following have lowest vapour density?

(1) C2H6
(2) CO2
(3) N2O
(4) N2

31) 10 gm metal displace 5.6 L H2 gas at STP from acidic solution. The equivalent weight of metal is

(1) 10
(2) 20
(3) 24
(4) 12

32) How many mole of NaOH are required to neutralize 2 mole H3PO3 solution

(1) 2 mole
(2) 4 mole
(3) 6 mole
(4) 1.5 mole

33) In the reaction 4A + 2B + 3C → A4B2C3 what will be the number of moles of product formed?
Starting from 2 moles of A, 1.2 moles of B and 1.44 moles of C.

(1) 0.5
(2) 0.6
(3) 0.48
(4) 4.64

34) Two elements X (at. mass = 75) and Y (at. mass = 16) combine to give a compound having 75.8%
of X. The formula of the compound is :

(1) XY
(2) X2Y
(3) X2Y2
(4) X2Y3

35) The percentage composition of various elements in a compound is as follows

Element Mass Percentage

 C 62
H 10.4
O 27.6

The empirical formula of compound is

(1) CH2O
(2) C2H4O
(3) CH3O
(4) C3H6O

36)

A gas is found to have the formula (CO)x. Its vapour density is 70. The value of x must be:

(1) 7
(2) 4
(3) 5
(4) 6

37) How many statements are not correct?

(A) Molar volume of an ideal gas at STP is
(B) When we use atomic masses of elements in calculation we always use the mass number of most
abundant isotope.
(C) An element contains particles of only one type, which may be atoms or molecules.
(D) Equal volumes of all gases at same temperature and pressure should have equal number of
molecules.

(1) One
(2) Two
(3) Three
(4) Four

38) Assertion (A) : The mass of one mole of a substance in grams is called molar mass.
Reason (R) : Molar mass of substance in gram is numerically equal to its molecular mass in u.

(1) Only A is correct

(2) Only R is correct
(3) Both A & R are correct
(4) Both A & R are incorrect.

39) Which of the following options contain the compound having same percentage composition:

(1) C6H12O6 & CH3COOH

(2) H2O2 & H2O
(3) HCHO & CH3CHO
(4) Both 1 & 3
40) What mass of 80% pure lime stone will be required to react completly with 0.02 mole of HCl
according to following equation?

(1) 1.0 g
(2) 1.25 g
(3) 2.50 g
(4) 1.40 g

41) Mass of a metal was increased by 40% due to oxide formation when it was heated in air. What
will be the equivalent mass of metal?

(1) 20
(2) 30
(3) 40
(4) 10

42) Molecular weight of dibasic acid is W. Its equivalent weight will be :

(1)

(2)

(3)
(4)

43) 74.5 g of metallic chloride contain 35.5 g of chlorine. The equivalent mass of metal is :

(1) 19.5
(2) 35.5
(3) 39
(4) 78

44) Equivalent weight of H3PO2 when it reacts with NaOH may be : (M = Molar mass of H3PO2)

(1) M

(2)

(3)

(4) All of these

45)

Statement : (1) Limiting reagent is the reactent which is consumed completely in the reaction.
Statement : (2) Limiting reagent limits the amount of product formed.
(1) Only statement 1 is correct
(2) Only statement 2 is correct
(3) Both statements are correct
(4) Both statements are incorrect

BOTANY

1) Which of the following is not correct ?

(1) Only Schawann formulated the plant cell theory

(2) Robert brown discovered the nucleus
(3) George palade discovered the ribosome
(4) A.V. Leeuwenhoek first saw and described a live cell

2) Choose the incorrect statement :-

(1) Unicellular organisms are capable of independent existence.

(2) Anything less than a complete structure of cell does not ensure independent living.
(3) Anton von Leuwenhoek first saw and described a live cell.
(4) All organisms are composed of a single cell.

3) Identify the correct match from the columns I, II and III :-

Column-I Column-II Column-III

1. Red blood cell a. Longest cell i. Plant cells

2. Nerve cell b. Smallest cell ii. diameter about 7µm

3. Mycoplasma c. Round and biconcave iii. Branched and long

4. Mesophyll cells d. Round and oval iv. 0.3 µm in length/diameter

(1) 1-d-ii, 2-a-iii, 3-b-iv, 4-c-i
(2) 1-c-ii, 2-b-iii, 3-a-iv, 4-d-i
(3) 1-a-ii, 2-c-iii, 3-b-iv, 4-d-i
(4) 1-c-ii, 2-a-iii, 3-b-iv, 4-d-i

4) Match the following :-

(a) Simple diffusion (I) Water

(b) Osmosis (II) Polar molecules

(c) Active transport (III) Neutral-solutes

(d) Facilitated diffusion (IV) Na+ – K+ pump

(1) a–I, b–III, c–II, d–IV
(2) a–I, b–III, c–IV, d–II
(3) a–III, b–I, c–IV, d–II
(4) a–III, b–I, c–II, d–IV

5) Mark the statements true (T) or false (F) with respect to the cell membrane :
(A) Cell membrane is composed of lipids that are arranged as monolayer
(B) Lipid component mainly consist of phosphoglycerides
(C) The membrane of erythrocyte has approximately 40% protein and 52% lipid
(D) Quasi-fluid nature of lipid enables lateral movement of proteins within the overall bilayer

A B C D

(1) T T T F

(2) F T T T

(3) T F F T

(4) F T F T
(1) 1
(2) 2
(3) 3
(4) 4

6) How many of the following statements are correct ?

(a) ER divide the intracellular space into two distinct compartments
(b) Only one chloroplast is present in Chlamydomonas
(c) Lysosomes is the double membrane bound vesicular structures formed by the process of
packaging in the golgi body
(d) In plant cells, the vacuole can occupy up to 90 percent of the volume of the cell

(1) Two
(2) One
(3) Four
(4) Three

7) Identify structure A,B,C and D in given diagram :-

A B C D

Peripheral Central Radial Central

(1)
microtubule sheath spoke microtubule

Interdoublet Central Central Central

(2)
bridge microtubule sheath bridge
 Interdoublet Central Radial Central
(3)
bridge sheath spoke microtubule

Interdoublet Radial Central Central

(4)
bridge spoke sheath microtubule
(1) 1
(2) 2
(3) 3
(4) 4

8) Match the following column–I with column–II and choose the correct option :-

Column-I Column-II

A Aleuroplast i Photosynthesis

B Elaioplast ii Storage of starch

C Chloroplast iii Storage of protein

D Amyloplast iv Storage of fat

(1) A–iii, B–ii, C–i, D–iv
(2) A–iv, B–iii, C–i, D–ii
(3) A–iii, B–iv, C–i, D–ii
(4) A–iv, B–iii, C–ii, D–i

9) Identify structure A,B,C and D is given below in diagram :-

A B C D

Stroma Outer
(1) Granum Stroma
lamella membrane

Outer Stroma
(2) Granum Stroma
membrane lamella

Outer Stroma
(3) Granum Stroma
membrane lamella

Stroma Outer
(4) Stroma Thylakoid
lamella membrane
(1) 1
(2) 2
(3) 3
(4) 4
10) Which one is wrong for given diagram ?

(1) Perform the function of packaging materials

(2) Important site of formation of glycoprotein and glycolipid
(3) Highly polymorphic
(4) It recieves the materials from E.R. through its cis–face

11) Identify structure A,B,C and D in given below:-

A B C D

Red White Columnar

Mesophyll
(1) blood blood epithelial
cells
cells cells cells

White Red Columnar

Mesophyll
(2) blood blood epithelial
cells
cells ​cells ​cells

Red White Columnar

Mesophyll
(3) blood ​blood epithelial
​cells
​cells cells ​cells
 White Red Columnar
Mesophyll
(4) ​blood blood epithelial
​cells
cells ​cells ​cells
(1) 1
(2) 2
(3) 3
(4) 4

12)

Match the following contents and select the correct one :

Column-I Column-II

Golgi Hydrolytic
(A) (i)
apparatus enzymes

Contractile
(B) (ii) Protists
vacuole

Glycolipid &
(C) Food vacuole (iii)
Glycoproteins

(D) Lysosomes (iv) Excretion

(1) A-i, B-ii, C-iii, D-iv
(2) A-iii, B-iv, C-ii, D-i
(3) A-iv, B-iii, C-ii, D-i
(4) A-ii, B-iii, C-i, D-iv

13) Which of the following statement is incorrect ?

(1) Nucleus was first described by Robert Brown

Nuclear pores are the passages through which movement of DNA and protein molecules takes
(2)
place in both directions between nucleus and cytoplasm
(3) Chromatin net and nucleolus are component of nucleoplasm
(4) Nucleolus is involved in rRNA sythesis

14) Which one of the following statement is/are not correct for 80s ribosome?
(a) Present in prokaryote and eukaryote
(b) Not surrounded by any membrane
(c) "S" is indirectly is a measurement of size and density
(d) Composed of two subunit 50s and 30s

(1) Only a
(2) Only b & c
(3) Only a and d
(4) Only a & c

15) Which of the following is not correctly matched?

(1) Cytoskeleton – maintenance of the shape of the cell
(2) Mitochondria – Enzymes of Calvin cycle
(3) Vacuoles – Anthocyanin pigment
(4) Microfilaments – Actin protein

16) Identify the A, B and C in the figure given below:-

A B C

Secondary
(1) Centromere Satellite
Constriction

Secondary
(2) Satellite Centromere
Constriction

Secondary
(3) Centromere Short arm
Constriction

Secondary
(4) Short arm Satellite
Constriction
(1) 1
(2) 2
(3) 3
(4) 4

17) Match the columns :-

Column-I Column-II

Terminal
a. Metacentric i.
centromere

Centromere is
b. Submetacentric ii. situated close to
its end

Middle
c. Acronematic iii.
centromere

Centromere is
d. Telocentric iv. slightly away
from middle
(1) a–i, b–iv, c–ii, d–iii
(2) a–ii, b–iv, c–i, d–iii
(3) a–iii, b–iv, c–ii, d–i
(4) a–iv, b–iii, c–i, d–ii
18) Which of the following is not the function of cell wall ?
(i) Provides shape to the cell.
(ii) Protects the cell from mechanical damage and infection.
(iii) Helps in cell to cell connection.
(iv) Provides barrier to undesirable macromolecules.
(v) Helps in cell furrow formation

(1) only (iii)

(2) only (iv)
(3) only (ii), (iii) and (v)
(4) only (v)

19)

Which of the following is correctly matched ?

Most dramatic
(A) M - Phase
period of cell cycle

Packaging of
(B) Golgi appratus
materials

Transport of
(C) Carrier protein
polar molecule

Several Ribosomes
(D) attached to a Polyribosome
single rRNA

Plastid store
(E) Amyloplast
starch
(1) A, B, C, E
(2) B, C, D, E
(3) A, D, E
(4) A, B, D, E

20) (i) Cilium/Flagellum contains an outer ring of nine doublet microtubules surrounding two singlet
microtubules.
(ii) Cilia are smaller which works like oars, causing the movement of either the cells or surrounding
fluid
(iii) Flagella are comparatively longer and responsible for cell movement
(iv) Cilium and flagellum are covered with plasma membrane
Which of the above statement is correct ?

(1) i and ii
(2) i, ii, iii and iv
(3) i and iv
(4) ii and iii

21) Select the correct statement for nucleolus :-

(1) It is a site for mRNA synthesis
(2) Large and more numerous nucleoli are present in cells actively carrying out protein synthesis
(3) Nucleolus contain nucleoplasm
(4) Nucleolus is a single membrane bound structure

22) Read the following statement (A-D)

(A) All living organism are composed of cells and product of cells
(B) All cell arise from pre existing cells
(C) According to Schleiden, all plants are composed of same type of cells which form tissues of the
plants
(D) According to Schwann animal cells had a thin outer layer which is known as plasma membrane
How many statement are correct ?

(1) One
(2) Three
(3) Two
(4) Four

23) Physico-chemical approach to study and understand living organism is called

(1) Comparative biology

(2) Reductionist biology
(3) Biophysics
(4) Biochemistry

24) Find the incorrect statement :

(1) Pili and fimbriae do not play a role in motility

(2) Bacterial cells are always motile in nature
(3) Bacterial cells attach with rock in stream through fimbriae
(4) Flagella helps in motility

25) Given below are two statements :

Statement I : In prokaryotes, ribosomes are associated with the plasma membrane of the cell.
Statement II : Several ribosomes may attach to a single t-RNA and form a chain called
polyribosomes or polysome.
In the light of the above statements, choose the correct answer from the options given below :

(1) Statement I is correct but Statement II is incorrect

(2) Statement I is incorrect but Statement II is correct
(3) Both Statement I and Statement II are correct
(4) Both Statement I and Statement II are incorrect

26) Assertion : A special membranous structure mesosome help in respiration.

Reason : Mesosome increase the surface area of the plasma membrane and enzymatic content.

(1) Both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
(2) Both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.
(3) Assertion is True but the Reason is False.
(4) Both Assertion & Reason are False.

27) Which of the following is incorrect about cell cycle ?

(1) All events occur in coordinated manner.

(2) All events are under genetic control.
(3) DNA synthesis occurs only during one specific stage in the cell cycle.
(4) Interphase is quiescent stage of cell cycle.

28) Assertion : G1 phase is the interval between mitosis and initiation of DNA replication.
Reason : During G1 phase the cell is metabolically active and have 2C DNA content.

(1) Both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
(2) Both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.
(3) Assertion is True but the Reason is False.
(4) Both Assertion & Reason are False.

29) In the cell cycle mitosis occurs between

(1) G1 and S phase

(2) S and G1 phase
(3) S and G2 phase
(4) G1 and G2 phase

30) A cell cycle includes

(1) Interphase and M phase/ division phase

(2) Prophase, metaphase, Anaphase and Telophase
(3) G1, S and G2 phase
(4) Karyokinesis and cytokinesis

31) Yeast completes cell cycle in

(1) 30 Minutes
(2) 60 minutes
(3) 90 minutes
(4) 120 minutes

32) What happens in S phase ?

(1) DNA replication

(2) In animal cells, replication of centriole
(3) Both (1) and (2)
(4) Separation of replicated DNA

33) Regarding to cell cycle which of the following statements is wrong ?

(1) Cytoplasm increase is a continuous process

(2) DNA synthesis occurs only during one specific stage
(3) Replicated chromosomes distributed to daughter nuclei by complex series of events
(4) Events for replicated chromosomes distribution are not under genetic control

34) Cell cycle includes :-

(1) Cell division

(2) DNA replication
(3) Cell growth
(4) All of the above

35) A stage in which cell remain metabolically active but no longer Proliferate is known as -

(1) G1 - phase
(2) S - phase
(3) G2 - phase
(4) G0 - phase

36) Which of the following statement(s) is/are correct about S-phase (synthetic phase)?
(i) It occurs between G1 and G2 phase.
(ii) It marks the period during which DNA replicates.
(iii) At the end of this phase, Amount of DNA becomes double but the number of chromosomes
remain unchanged.
(iv) As the DNA is doubled in this phase number of chromosomes is also doubled.
(v) In plants, Centrioles duplicates in cytoplasm.
(vi) Amount of DNA changes from 2C to 4C.

(1) (i), (ii), (iii), (v), (vi)

(2) (i), (ii), (iii), (vi) only
(3) Only (i), (ii) & (iv)
(4) Only (iv)

37) Statement I : The plasmid DNA confers certain unique phenotypic characters to such bacteria.
Statement II : Plasmid DNA is used to monitor bacterial transformation with foreign DNA.

(1) Both statements I and II are incorrect

(2) Both statements I and II are correct
(3) Only statement I is correct
(4) Only statement I is incorrect

38) The correct sequence of cell cycle is :-

(1) G1 → G2 → S → M
(2) G1 → S → G2 → M
(3) M → G2 → S → G1
(4) S → G1 → M → G2

39) Interphase is called the resting phase because.

(1) It is the most active phase of the cell cycle

(2) There is no apparent activity related to cell division
(3) It does not prepare cell for cell division
(4) It is the phase where cell rests before entering into mitosis

40) Cell theory which was given by Schleiden and Schwann, did not explain as to how the new cells
are formed ? .... modified the hypothesis of cell theory to give it a final shape?

(1) R. Hooks
(2) R. Virchow
(3) Mendel
(4) C.P. Swanson

41) The cell wall and middle lamellae may be traversed by _______ which connect the ________ of
neighbouring cells.

(1) Plasmodesmata; cytoplasm

(2) Plasmamembrane; cytoplasm
(3) Primary wall junction; protoplasm
(4) Plasmamembrane; protoplasm

42) Microbodies are … ?

(1) Membrane bound minute vesicles that contain various enzymes, are present in animal cells only.
Membrane bound Large vesicles that contain various enzymes, are present in both plant and
(2)
animal cells.
Membrane bound minute vesicles that contain various enzymes, are present in both plant and
(3)
animal cells.
Membrane bound large vesicles that contain various Hydrolytic enzymes, are present in both
(4)
plant and animal cells.

43) Main arena of cellular activity occurs in :

(1) Cell membrane

(2) Cytoplasm
(3) Nucleoplasm
(4) Membranes bounded cell organells

44) There is an extensive comapartmentalization of cytoplasm through membrane bound organelles

in all except :

(1) Plant
(2) Fungi
(3) Bacteria
(4) Animals

45) Statement-1 : Typically mitochondria is sausage shaped or cylindrical.

Statement-2 : Inner membrane of mitochondria divides it's lumen in two aqueous compartments
i.e., the outer compartment and the inner compartment.

(1) Both Statement-1 and 2 are correct.

(2) Statement-1 is correct but Statement-2 is incorrect
(3) Statement-1 is incorrect but Statement-2 is correct
(4) Both Statement-1 and 2 are incorrect.

ZOOLOGY

1) Tendon is an example of :-

(1) Dense regular connective tissue connecting muscle to skin

(2) Dense regular connective tissue connecting bone to bone
(3) Dense regular connective tissue connecting muscle to bone
(4) Dense irregular connective tissue connecting muscle to bone

2) Following diagram represent which type of example ?

(1) Tendon
(2) Adipose tissue
(3) Areolar connective tissue
(4) True vocal cords

3) The diagrams are labelled with alphabets. Identify and choose correct answer ?

(1) A = Columnar epithelium, B = Multicellular gland, C = Compound epithelium

(2) A = Unicellular gland, B = Multicellular gland, C = Compound epithelium
(3) A = Multicellular gland, B = Unicellular gland, C = Compound epithelium
(4) A = Compound epithelium, B = Unicellular gland, C = Multicellular gland

4) Read the following statements and find out the incorrect statement.

(1) The squamous epithelium is involved in functions like forming a diffusion boundary.
(2) The cuboidal and columnar epithelium help in secretion and absorption.
(3) The function of ciliated epithelium is to move particle or mucus in a specific direction.
(4) Compound epithelium play main role in secretion & absorption.

5) Read the following (A-D) statements :-

(A) Connective tissue are most abundant and widely distributed in the body of complex organism.
(B) Connective tissue helps in linking and supporting other tissue or organs in the body.
(C) Adipose tissue is a type of dense connective tisue.
(D) Connective tissue does not secrete its matrix.
Choose the incorrect statements :

(1) A, C & D
(2) B and D
(3) C and D
(4) Only D

6) Pharynx have which type of epithelium ?

(1)

(2)

(3)

(4)
7) Identify A, B and C in the given diagram of adipose tissue.

(1) A-Cytoplasm, B-Nucleus, C-cellwall

(2) A-Fat storage area, B-mast cell, C-plasma membrane
(3) A-cell fluid, B-collagen fibres, C-plasmalemma
(4) A-Fat storage area, B-nucleus, C-plasma membrane

8) Select the correct option about given diagram:-

(1) B = columnar epithelium - for protection and cover the dry surface of the skin
(2) B = squamous epithelium - found in tubular part of nephron
(3) A = squamous epithelium - found in wall of blood vessels
(4) A = cuboidal epithelium - found in air sac of lungs

9) Identify the connective tissue shown below as well as property of it's intercellular material and

select the right option for the two together.

Connective tissue Property of intercellular material

(1) Bone Solid and nonpliable
(2) Cartilage Solid and pliable
(3) Bone Solid and pliable
(4) Cartilage Solid and non-pliable

(1) 1
(2) 2
(3) 3
(4) 4

10) Consider the following four statements (A–D) and select the option which includes all the correct
ones only.
(A) Compound epithelium is made up of one layer of cells and has a role in secretion and absorption.
(B) The cuboidal epithelium is commonly found in ducts of glands and tubular parts of nephrons in
kidney.
(C) The main function of squamous epithelium is forming a diffusion boundary.
(D) The epithelial cells are loosly packed with intercellular matrix.
Options :-

(1) Statements (B), (C) and (D)

(2) Statements (A), (B)
(3) Statements (B), (C)
(4) Statements (A), (C) and (D)

11)

Our heart consists of :-

(i) Epithelial tissue (ii) Connective tissue
(iii) Muscular tissue (iv) Neural tissue

(1) Only ii
(2) i & iii only
(3) ii, iii & iv only
(4) All of these

12) ___(A)___ junction faciliate the cells to communicate with each other by connecting the
___(B)___of adjoining cells. Fill the blanks (A and B) with suitable options :-

(1) A : Tight, B : Cell wall

(2) A : Gap, B : Cytoplasm
(3) A : Adhering, B : Plasmamembrane
(4) A : Desmosomes, B : Cell wall

13) Consider the following figure & choose the correct option :-

Regular dense
(1) Tendon Join muscle to skin
connective tissue

Regular dense
(2) Ligament Join bone to bone
connective tissue

Irregular dense
(3) Tendon Join muscle to bone
connective tissue

Irregular dense
(4) Ligament Join muscle to bone
connective tissue
(1) 1
(2) 2
(3) 3
(4) 4
14) Find out incorrectly matched option.

(1) Microvilli – increase surface area for better absorption.

(2) Cilia – Move particles or mucus in a specific direction over the epithelium.
Gap junctions – facilitate the cells to communicate with each other by connecting the cytoplasm
(3)
of adjoining cells.
(4) Microvilli – Move particles or mucus in a specific direction over the epithelium.

15) Recognise the figure and find out correct match :-

(1) A-Mast cells, B-Fibroblast, C-Collagen fibres, D- Macrophages

(2) A-Macrophases, B-Fibroblast, C-Collagen fibres, D- Mast cells
(3) A-Macrophages, B-Collagen fibres, C-Fibroblast, D- Mast cells
(4) A-Fibroblast, B-Mast cells, C-Collagen fibres, D- Macrophages

16) Read the following statements : a. In unicellular organisms, all functions like digestion,
respiration and reproduction are not performed by a single cell.
b. All complex animals consist of only four basic types of tissues organized in specific proportion and
pattern.
c. In multicellular animals, group of similar cells without intercellular substances perform a specific
function.
d. The structure of the cells vary according to their function which leads to different types of
tissues.
(e) In connective tissues, the cells are compactly packed with little intercellular matrix.
Choose the correct answer from following given options.

(1) a, b and c are correct

(2) b, c and d are correct
(3) b and d are correct
(4) c, d and e are correct

17) Dense connective tissue and loose connective tissue are differentiated on the basis of :
a. Presence of fibroblast
b. Presence of blood vessels
c. Amount of intercellular matrix
d. Number of fibres

(1) a, b, c, d
(2) c, d only
(3) b, c only
(4) d only

18) Choose the correct option w.r.t common feature of the given tissues A and B.
(1) Location of tissues
(2) Branching
(3) Striated (striations)
(4) Number of nuclei per cell

19) Identify the given below figure with its correct example

(1) Squamous epithelium – Air sacs of lungs

(2) Cuboidal epithelium – Tubular part of nephron
(3) Columnar epithelium – Fallopian tube
(4) Cuboidal epithelium – Wall of blood vessels

20) Assertion :- Gap junctions help in rapid transfer of ions, small molecules and sometimes big
molecules. Reason :- Gap junction facilitate the cells to communicate with each other by connecting
the cytoplasm of adjoining cells.

(1) Assertion and Reason both are correct and reason is the correct explaination of assertion
(2) Assertion and Reason both are correct but reason is not correct explaination of assertion
(3) Assertion is correct but reason is not correct
(4) Assertion is not correct but reason is correct

21) How many of the statements are correct from following :-

(A) Exoskeleton of Arthropods made up of chitin.
(B) N - acetyl - D - Galactosamine is modified sugar.
(C) Starch forms helical structure.
(D) Cellulose is micromolecule.

(1) 1
(2) 2
(3) 3
(4) 4

22) Given below is the chemical formula of : –

(1) Palmitic acid
(2) Stearic acid
(3) Glycerol
(4) Galactose

23) How many of the following compounds are mono-saccharides :-

Glycerol, Lecithin, Cholesterol, Starch, Ribose, Triglyceride, Glucose, Mannose.

(1) 4
(2) 3
(3) 2
(4) 1

24) Which statement is a false statement?

(1) Glycerol is trihydroxy propane

(2) Cellulose does not show positive iodine test
(3) Starch shows iodine test
(4) Lipids are macromolecules

25) In the above diagrammatic representation of

glycogen, glycosidic bonds represented by A, B and C are :-

A B C

(1) α – 1' – 4" α – 1' – 6" α – 1' – 4"

(2) β – 1' – 4" β – 1' – 6" β – 1' – 4"

(3) α – 1' – 6" α – 1' – 4" α – 1' – 4"

(4) β – 1' – 6" α – 1' – 4" β – 1' – 4"

(1) 1
(2) 2
(3) 3
(4) 4

26) Match the column I with column II

 Column I Column II

A Phospholipid i β - D Glucose polymer

B Chitin ii α - D Glucose polymer

N - acetyl
C Glycogen iii
glucosamine

D Cellulose iv Lecithin
(1) A - iv, B - iii, C - ii, D - i
(2) A - i, B - ii, C - iii, D - iv
(3) A - ii, B - iii, C - i, D - iv
(4) A - iv, B - i, C - ii, D - iii

27) Match the columns I and II, and choose the correct combination from the options given:-

Column-II(% of the total cellular

Column-I(Component)
mass)

(a) Ions (i) 1

(b) Lipids (ii) 2

(c) Carbohydrates (iii) 3

(d) Nucleic acids (iv) 5–7

(e) Proteins (v) 10–15

(1) a–i, b–ii, c–iii, d–iv, e–v
(2) a–ii, b–iii, c–i, d–v, e–iv
(3) a–iii, b–i, c–ii, d–iv, e–v
(4) a–iv, b–ii, c–iii, d–v, e–i

28) When a tissue is fully burnt, ‘Ash’ is obtained which contains all of the following substances
except

(1) Calcium
(2) Magnesium
(3) Carbon
(4) Potassium

29) Which one of the following is a non-reducing carbohydrate?

(1) Maltose
(2) Lactose
(3) Sucrose
(4) Ribose

30) Statement, which is incorrect w.r.t lipids is

(1) Phospholipids have phosphorus and a phosphorylated organic compound in them
(2) Gingelly oil has higher melting point than fats
(3) Lecithin is an example of a phospholipid
(4) Phospholipids are found in cell membrane

31) Assertion: Lipids are not strictly macromolecules.

Reason: Lipid molecular weight do not exceed 800 Da.

(1) (A) is correct but (R) is not correct

(2) (A) is not correct but (R) is correct
(3) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(4) Both (A) and (R) are correct but (R) is not the correct explanation of (A)

32) Which element is present in negligible amount in human body ?

(1) N
(2) C
(3) Si
(4) O

33) Assertion : Polysaccharides are non-reducing .

Reason : In a polysaccharide chain, the right end is called the reducing end and the left end is
called the non-reducing end.

(1) Both Assertion and Reason are True and the Reason is a correct explanation of the Assertion.
(2) Both Assertion and Reason are True but Reason is not a correct explanation of the Assertion.
(3) Assertion is True but the Reason is False.
(4) Both Assertion and Reason are False.

34) The organic compound ‘X’ that is 5-7% of the total cellular mass in living organisms, constitute
monomeric units that are joined together with the help of

(1) Phosphodiester bonds

(2) Peptide bonds
(3) Glycosidic bonds
(4) Hydrogen bonds

35) Number of chiral carbons in β-D-(+)-glucose is

(1) five
(2) six
(3) three
(4) four

36) Statement I : The left end of polysaccharides is called the reducing end.
Statement II : Glycogen is a polysaccharide with right end as reducing end.
(1) Both the statements are correct.
(2) Only statement II is correct.
(3) Only statement I is correct.
(4) Both the statements are incorrect.

37) Correct for given figure–

(1) A is part of DNA

(2) A is polymer
(3) B is pentose sugar
(4) B is polymer

38) Read the following statements :

I. Right end of polysaccharide is called reducing end while left end is called non reducing end.
II. Starch can hold I2 molecules in its helical secondary structure but cellulose being non helical,
cannot hold I2
III. Starch and glycogen are branched molecule
IV. Starch in plant and glycogen in animal are store houses of energy
How many statements are correct

(1) 2
(2) 3
(3) 4
(4) 1

39) Which is not a polymer?

(1) Cellulose
(2) DNA
(3) Sucrose
(4) Protein

40) Assertion : When iodine comes into contact with starch, it forms a blue colour.
Reason : Starch have structure without helix , iodine trap in its branches to give blue colour.

(1) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(2) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(3) Assertion is true but Reason is false.
(4) Both Assertion and Reason are false.
41)
The above given figures A and B are diagrammatic structures of ___________.

(1) Glucose and amino acid

(2) Amino acid and glucose
(3) Glucose and ribose sugar
(4) Ribose sugar and glucose

42) Which of the following is the structure of

(1)

(2)

(3)
(4)

43) Which of the following about polysaccharides is not correct ?

A. Acid soluble pellet has polysaccharides as another class of macromolecules.
B. Cellulose is a homopolymer
C. Starch is main stored food in plants, glycogen is stored food in animal.
D. In a polysaccharide chain, the right end is called non-reducing end and the left end is called the
reducing end.

(1) A, B
(2) B, C
(3) A, D
(4) B, C, D

44) According to weight percentage, the first three elements in human body are

(1) C > H > O

(2) C > O > N
(3) O > N > C
(4) O > C > H

45) Stratified squamous epithelium is found in :-

(1) Trachea
(2) Epidermis
(3) Heart
(4) Lining of blood vessels
 ANSWER KEYS

PHYSICS

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. 1 1 3 3 1 1 2 3 4 2 1 1 2 4 1 1 2 2 2 3
Q. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
A. 3 3 1 1 1 3 2 3 4 4 3 4 1 1 2 2 1 3 1 3
Q. 41 42 43 44 45
A. 3 4 1 2 2

CHEMISTRY

Q. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65
A. 1 4 3 4 3 1 3 2 1 4 2 4 1 2 3 3 1 1 4 3
Q. 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85
A. 1 1 1 2 4 2 3 2 3 4 2 2 3 4 4 3 1 3 1 2
Q. 86 87 88 89 90
A. 1 2 3 4 3

BOTANY

Q. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110
A. 1 4 4 3 4 4 3 3 3 3 1 2 2 3 2 2 3 4 1 2
Q. 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130
A. 2 2 2 2 1 1 4 2 2 1 3 3 4 4 4 2 2 2 2 2
Q. 131 132 133 134 135
A. 1 3 2 3 1

ZOOLOGY

Q. 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155
A. 3 3 2 4 3 4 4 3 2 3 4 2 2 4 2 3 2 3 2 1
Q. 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175
A. 3 1 2 4 3 1 1 3 3 2 3 3 2 1 1 2 3 3 3 3
Q. 176 177 178 179 180
A. 3 1 3 4 2
 SOLUTIONS

PHYSICS

1) 2x2 – 7x + 5 = 0 2x2 – 2x – 5x + 5 = 0
2x(x – 1) – 5 (x – 1) = 0
(x – 1) (2x – 5)

x = 1 and x =

2)

3) cos215° – sin215°
cos2θ = cos2θ – sin2θ
∴ cos215° – sin215°
= cos30°
=

4)

choose options according to Range.

5)

slope = m =

6)
– 3x – 5y = 15
5y = – 3x – 15

y=

Here :- c = –3
7)

x=

At x = , y is minimum

ymin =

8) Explain - find
Concept - Differentation
Formula

Calculation

9) =

= –1–1
= –2

10)

11)

= K Q1 Q2

= K Q1 Q2

= K Q1 Q2
12)

Let y = 5sinθ – 12 cosθ

Compare with asinθ + bcosθ,
Maximum value =

= =13

13) 1. Explanation (30 words): We are asked to find the binomial expansion approximation
for The binomial expansion helps approximate powers of binomials.
2. Concept (20 words):
For large values of, the binomial expansion of (1 + x)-n simplifies to an approximation involving
powers of x.
3. Formula:
The binomial expansion for (1 + x)- n :

4. Calculation (Minimum Steps):

For we use the binomial approximation:
(for large x)

14)

15)

(0.97)1/3 = ( 1 – 0.03)1/3 =
= 0.99

16)

Concept:
Quadrants and Signs: The sine function is positive in the first and second quadrants and
negative in the third and fourth quadrants.
Mathematical Calculation:
1. Locate 300°: An angle of 300° lies in the fourth quadrant.
2. Find the reference angle: The reference angle for 300° is 360° - 300° = 60°.
3. Determine the sign: Since 300° is in the fourth quadrant, where the sine function is
negative, sin(300°) will be negative.
4. Evaluate: We know that sin(60°) = √3/2. Therefore, sin(300°) = sin(360-60°) = -sin(60°)= -
√3/2
Final Answer:

The value of sin(300°) is

option 1
Question Level: Easy
17)

sin 1° = sin
for angle to be small

18)

y = a – acosx

19)

V = a3

20)

B. Concept:

A. Logarithms: Logarithms are the inverse of exponents.

B. Properties of Logarithms:

A. log_b(b^x) = x

B. (log_b(x))^a = a * log_b(x)

C. Formulae:

A. log_b(b^x) = x

D. Calculation:

A. Evaluate log10(1000): log10(1000) = 3 (since 10^3 = 1000)

B. Apply the exponent: 3^(1/3) = 1

E. Final Answer:

log10(1000)^(1/3) = 1 option 1

21) Slope = tan θ = tan 30°

(Angle is measured anti clockwise is positive)

22)

23)

B. Concept:

A. Visual Recognition of Basic Functions: Identifying the shape and characteristics of

common graphs like linear, quadratic, exponential, and square root functions.

C. Formulae:

A. Linear: y = mx + b

B. Quadratic: y = ax²

C. Exponential: y = a^x

D. Square Root: y = √x

D. Calculation/Analysis:

A. Shape: The graph resembles a curve that starts at the origin and increases at an
increasing rate. This shape is characteristic of a quadratic function.its behaviour is
quadratic as y = ax²

2. Behavior: The graph is always positive (above the x-axis) for both positive and negative
x-values.

E. Final Answer:

The best representation of the given graph is y = x².

24)

Conceptual.

25)

26)

By theory

27)

Explain:
Find the magnitude of the vector 2i + 4j + 5k.

Concept: Pythagorean theorem.

Formula: Magnitude of vector ai + bj + ck =

Calculation: Magnitude =
Magnitude =
Magnitude =
Magnitude =

Hence, option (2) is correct.

28)

⇒
⇒ 0.2 + q2 = 1
⇒ q2 = 0.8

29)
⇒ θ = 90°

30)

Explain : a relationship between the magnitudes of the cross product and dot product of two
vectors and . find the magnitude of the vector .
Concept : Cross product and dot product.

Formula : Dot Product: | . | = abcos θ, where θ is the angle between and .

Cross Product: | . | = abcos θ
Magnitude of resultant of

Calculation : Given: | × | = ( . )

ab sinθ = ab cosθ

tanθ = cos θ

θ = 30°

Final Answer : (4)

31)

A2 + B2 +

cosθ = 0 = cos π/2

32)

R = 2A Given
So, = 0°

33)

Resultant
 34)

35) Answer:- Option (2)

Problem Explanation :-

A. Given ⇒ A vector is given in the form of standard orthogonal unit vectors i.e. =
B. To find ⇒ The projection of the vector on the y-axis ⇒ Ay = ?

Solution :-

A. Concept ⇒ To determine the component of the vector that lies along the y-axis.
B. Visual-Aid ⇒

A. Calculation ⇒ As we can see from the diagram, that the projection of vector =

Ay = units

36)
x axis =

cosθ =
37)

a2 – 2a – 3 = 0
on solving (a = 3, –1)

38) If initial and final position is same result will be zero

Using triangle law

39) R = 2A cos

⇒ R = 2 × 20 × = 20 N

40) at = 90°

41) For normal vectors, This is the case with the vector in option (3).
( cosθ + sin θ)· ( sinθ – cos θ) = AB sin θ cos θ – AB sin θ cos θ = 0

42)
12 hr → 360°

1 hr →

2 hr →

43) ⇒ r=

44)

where θ is the angle between two vectors and

∴

or
 45) Rmax = 8 + 6 = 14 N
Rmin = 8 – 6 = 2 N

CHEMISTRY

46)

(1) Mass of O3 = 0.5 × 48 = 24 g

(2) Mass of O = 0.5 × 16 = 8 g

(3) Moles of CO2 = = 6.023 × 1023 g

Mass of CO2 = = 22 g

(4) Moles of CO2 =

Mass of CO2 = × 44 = 11 g

47) NCERT Pg. # 15

48) Question Explanation :

Question is asking about find the number of electrons in 3.1 mg of nitrate ion (NO3–)

Concept :
This question is based on Moles = Weight / molar mass, Total No. of electrons = moles × NA ×
No. of electron in one ion

Solution :
Mass = 3.1 mg = 3.1 × 10–3 g
Moles = 3.1 × 10–3 / 62 = 5 × 10–5 mol
Total electrons = 5 × 10–5 × 6.022 × 1023 × 32 = 9.6 × 1020

Final Answer :
Option (3)

49)
50) (A) =
No. of oxygen atoms = 10–3 NA × 4
(B) 18 mL water ⇒ 18 g water
⇒ 1 mol H2O
⇒ NA atoms of oxygen

(C) nNO = = 2 mol

⇒ 2 NA molecules of NO
⇒ 2 NA atoms of oxygen
(D) PV = nRT

0.5 × = n × 0.0821 × 273

n = 10–3 mol
⇒ 10–3 NA molecules of CO2
⇒ 2 × 10–3 NA atoms of oxygen

51)
= 3 × 10–23

52) MnO2(s) + 4HCl(aq) →

MnCl2(aq) + 2H2O(ℓ) + Cl2(g)
1 mol MnO2 requires = 4 mol HCl
87 g MnO2 requires = 4 × 36.5 g HCl

1 g MnO2 requires = g HCl

5 g MnO2 requires = g HCl

= 8.39 g HCl

53) Question Explanation:

An impure sample of calcium carbonate (CaCO3) is heated strongly and produces 56 g of
calcium oxide (CaO). We are asked to find the percentage purity of CaCO3 in the original 120 g
sample.

Concept:
Percentage purity of CaCO3

Explanation
When CaCO3 is heated, it decomposes as follows:
CaCO3 → CaO + CO2
From the balanced equation. 100 g of pure CaCO3 gives 56 g of CaO. Using this stoichiometric
relationship we can calculate the amount of pure CaCO3 required to produce 56 g of CaO and
then determine the purity percentage.
Calculation:
Let the mass of pure CaCO3 in the sample be x g
According to the reaction
 100 g CaCO3 → 56 g CaO
So,
(56/100) × x = 56
⇒ x = 100g
% purity = (100/120) × 100 = 83.3%
Final Answer:
Percentage purity of CaCO3, = 83.3%
Correct Option: (2) 83.3%

54) Question Explanation

We need to determine the number of moles of HCl gas formed when 22.4 L of H2 and 11.2 L of
Cl2 are mixed at STP.
Given
Volume of H2 = 22.4L
Volume of Cl2 = 11.2L
Condition = STP (Standard Temperature and Pressure)

Concept
Number of Moles Calculation & Concept

Explanation
At STP, 1 mole of any gas occupies 22.4 L
The balanced chemical reaction for the formation of HCl is
H2(g) + Cl2(g) → 2HCl(g)
From the balanced equation. 1 mole of H2 reacts with 1 mole of Cl2 to form 2 of HCI.

Moles of H2 =

Moles of Cl2 = =0.5 mol

Since Cl2 is the limiting reagent (we have less of Cl2) the moles of HCI formed will be based on
the amount of Cl2, available.
From the equation. 1 mole of Cl2 produces 2 moles of HCl. Therefore. 0.5 moles of Cl2 will
produce
0.5 mol Cl2 × 2 = 1 mol HCl
Final Answer
The number of males of HCl formed is 1 mole.
Correct Answer: Option 1,

55)

A. Question
The equivalent weight of phosphoric acid (H₃PO₄) in the reaction,
NaOH + H₃PO₄ → NaH₂PO₄ + H₂O, is:

B. Given Data

A. Reaction: NaOH + H₃PO₄ → NaH₂PO₄ + H₂O

B. Phosphoric acid (H₃PO₄) is a triprotic acid, but in this reaction, only one proton is neutralized
(as indicated by the formation of NaH₂PO₄).
 C. Concept

A. Equivalent Weight: It is the mass of the acid that supplies one mole of replaceable hydrogen
ions (H⁺) in a reaction.
B. For an acid, equivalent weight = (Molar mass) / (Number of H⁺ ions replaced in the reaction).
C. Although H₃PO₄ has three acidic hydrogen atoms, the reaction provided shows only one proton
reacting with NaOH.

D. Mathematical Calculation

A. Determine the molar mass of H₃PO₄:

A. Hydrogen (H): 1 g/mol, there are 3 atoms → 3 × 1 = 3 g/mol
B. Phosphorus (P): 31 g/mol
C. Oxygen (O): 16 g/mol, there are 4 atoms → 4 × 16 = 64 g/mol
D. Total molar mass = 3 + 31 + 64 = 98 g/mol
B. Calculate the equivalent weight:
A. Since only 1 hydrogen ion is replaced, the equivalent weight = 98 g/mol ÷ 1 = 98
g/equiv

E. Final Answer
The equivalent weight of H₃PO₄ in this reaction is 98.
(Option 4)

F. Question Level
Easy

56)

Explanation: Question asks to find the moles of product C formed given moles of reactants A
and B and a balanced chemical equation.

Given data :

A. 7 mol A

B. 8 mol B

C. Balanced equation: A + 2B → 3C

Concept: Limiting reactant, Stoichiometry

Calculation:

A. 1 mol A needs 2 mol B.

B. 7 mol A needs 14 mol B, but only 8 mol B is present. B is Limiting reactant.

 C. 2 mol B produces 3 mol C.

D. 8 mol B produces (3/2) x 8 = 12 mol C

Final Answer: 12 mol

Correct option - 2

57) Question Explanation

We are asked how much CO2 gas is produced from the complete combustion of 10 L of butane
(C4H10)
Given
Volume of butane = 10 L
Reaction is complete combustion
Concept
Volume of gas after combustion of organic compound.
Explanation
Balanced combustion reaction of butane
2C4H10 + 13O2 → 8CO2 + 10H2O
From the equation :
2 volumes of C4H10 give 8 volumes of CO2
i.e. 1 volume of C4H10 gives 4 volumes of CO2
Given: 10 L of C4H10
CO2 produced = 10 × 4 = 40 L
5. Final Answer
40 L of CO2 is produced
Correct Answer: Option 4

58) Given data O2 + N2 → 2NO

Wa2 = WN2
Concept
o

LR ER
Correct option (1)

59) NCERT Pg # 25

moles of
moles of H2O(g) = 0.1 × 3 = 0.3
vol of H2O(g) at STP = 0.3 × 22.4 = 6.72 L

60) Question
 which of the following is true.

given data
In HNO3 pure sample 1.6% H, 22.2% N and 76.2% obg mass.

Concept
law of definite proportion:-
According to law of definite proportion, a compound can be obtained from different sources.
But the ratio of each component remains same it does not depend on the method of its
preparation or the source from which it has been obtained.

Conclusion
Hence, option (3) is correct.

61)

A. Question
Which of the following pairs of compounds does not follow the law of multiple proportions?

B. Given Data

A. Law of multiple proportions: When two elements form more than one compound, the masses of
one element that combine with a fixed mass of the other are in a simple whole-number ratio.
B. The law applies only to compounds composed of the same two elements.

C. Concept

A. For pairs like CO and CO₂, H₂O and H₂O₂, and N₂O and N₂O₃, the compounds are composed of
the same elements, so their mass ratios can be compared to see if they obey the law of
multiple proportions.
B. For a pair where the compounds do not contain the same two elements, the law cannot be
applied.

D. Mathematical Calculation / Comparison

A. CO and CO₂:
A. Both contain carbon (C) and oxygen (O).
B. In CO, the mass ratio of O to C is 16:12 = 4:3, while in CO₂ it is 32:12 = 8:3.
C. The ratio (4:3 to 8:3) is a simple whole-number multiple.
B. H₂O and H₂O₂:
A. Both contain hydrogen (H) and oxygen (O).
B. In H₂O, the oxygen-to-hydrogen mass ratio is 16:2 = 8:1, whereas in H₂O₂ it is 32:2 =
16:1.
C. The ratio of oxygen masses is exactly double (8:1 to 16:1), satisfying the law.
C. CaO and MgO:
A. CaO is made of calcium (Ca) and oxygen (O), whereas MgO is made of magnesium (Mg)
and oxygen (O).
B. They do not have the same pair of elements (the metal component differs), so the law of
multiple proportions does not apply here.
D. N₂O and N₂O₃:
 A. Both are nitrogen oxides containing nitrogen (N) and oxygen (O).
B. For N₂O, the mass of oxygen for a fixed amount of nitrogen is 16 g for 28 g of N (from 2
× 14 g). In N₂O₃, it is 48 g for 28 g of N.
C. The ratio of oxygen masses (16 vs 48) is a whole-number ratio (1:3).

E. Final Answer (Option 3)

F. Question Level
Easy

62) H2O = 2 mole; CO = 1 mole,

C2H5OH = 1 mole, N2O5 = 1/2 mole

63)

(1)
(2) 1.5 × 28 = 42 g
(3) 0.1 × 1 = 0.1 g
(4) 1 mole of c = 12 g

64)

1 mole HNO3 ≡ 3 mol O ; So 5 mole HNO3 contains 15 mol 'O' atoms.

65)

66) 21 mg =
= 0.5×10–3 mol ∴ No. of N atoms = 3 × 0.5 × 10–3 mol = 1.5 × 10–3 × NA atoms.

67)

NCERT Pg # 15

68)
[Hint : Mas of nitrogen in 194 amu caffeine =

One molecule of caffeine will contain atoms of nitrogen.

69) X :Y
mass 40 : 60

mole :
4 :3
E.F = X4Y3

z=

= =2
M.F. = (X4Y3)2
M.F. = X8Y6

70)

Mw = 2 × V.D. = 2 × 11.2 = 22.4

71)

NCERT Pg. # 19
C O H

: :
= 4.81 : 2.4 : 3.6

2 : 1 :
4 : 2 : 3
Empirical formula = C4H3O2
E.F. weight = 4(12) + 2(16) + 3(1)
= 48 +32 +3 = 83

Molecular formula = n[E.F.]

= 2[C4O2H3] = [C8O4H6]

72)

Weight of metal = 74.5 – 35.5 = 39 g

Equivalent weight of metal =

= × 35.5
= 39

73) Solution
Mass = mole x Mw

=
= 22

74)

in NH3, 3g hydrogen combines with 14 g nitrogen so, EW of nitrogen = g

75) Solution

Vapour density =

76)

E(M) = 20

77)
Mole × 1 = 2 × 2

78) 4A + 2B + 3C → A4B2C3
2 mole 1.2 1.44 —

Simple Ratio —
(L.R.) 0.48

79)

X Y

1 3/2
2 3 ⇒ X2Y3

80) [Old NCERT Page-19]

81)
 Molecular weight = 2 × VD
= 2 × 70 = 140
(CO)x = 140
28 × x =140
X=5

82) Solution:-
(B) We use average atomic mass.

83) Solution-
NCERT Page No. : 18

84) Solution :
C6H12O6 & CH3COOH have same emperical formula.

85) Solution :

Total required
= 1.25 g

86) Solution:

87) Solution:

Equivalent weight =

88) Solution-

89) Solution :

90) Solution-
Fact

BOTANY
91) NCERT Pg.#125

92) NCERT-XI Pg. # 125 IInd and IIIrd para

93) NCERT (XIth), Pg. # 126, 127

94) NCERT Pg. # 132

95)

NCERT Page no 93

96)

c is not correct

97) NCERT (XI) Pg. # 99

98) NCERT (XI) Pg. # 135, Para–3

99) NCERT (XI) Pg. # 136, Fig–8.8

100) NCERT (XII) Pg. # 133, Fig. 8.6

101) NCERT (XI) Pg. # 127, Fig. 8.1

102)

(A) Golgi Apparatus—Glycolipid and Glycoproteins

(B) Contractile Vacuole—Excretion
(C) Food Vacuole—Protists
(D) Lysosomes—Hydrolytic Enzymes

103) NCERT, Pg#138, para-8.5.10

104)

NCERT, Pg. # 98

105)

NCERT-XI Pg#97
106) NCERT, Pg. # 101

107) Module No.6 Pg.# 189

108) NCERT (XI) Pg. # 132, para-4

109)

NCERT Pg # 129–132–133–136

110) NCERT Pg. # 137

111)

NCERT Pg # 138

112)

NCERT Pg : 126, 2021 - 22

113) NCERT Pg # 123E

114) NCERT Pg.#129

115) NCERT 11th Class (Chapter-8) Page No. 129

116) NCERT Page No. # 128

117)

NCERT Pg # 163

118) NCERT Page No. 163

119) NCERT XI Page No. 163

120) NCERT XI Page no. 163

121)

NCERT page no. 163

 122) NCERT XI Page No. 163

123)

SOLUTION—

Events for replicated chromosomes distribution are not under genetic control:
This statement is incorrect. The events of the cell cycle, including the distribution of replicated
chromosomes, are tightly regulated by genetic factors.

The correct answer is:OPTION(4)Events for replicated

124)

SOLUTION—

The cell cycle includes cell division, DNA replication, and cell growth.

The correct answer is:OPTION(4)All of the above

125)

NCERT, Pg. # 164

126) NCERT XI PAGE NO. 163

127) Correct Answer: Option 2 - Both statements I and II are

correct
Explanation:
Statement I:

A. "The plasmid DNA confers certain unique phenotypic characters to such bacteria."
A. Correct: Plasmids are extrachromosomal DNA molecules that often carry genes
conferring antibiotic resistance, virulence factors, or metabolic capabilities to the host
bacteria. These unique traits can be expressed, influencing the phenotype of the
bacteria.

Statement II:

A. "Plasmid DNA is used to monitor bacterial transformation with foreign DNA."

A. Correct: Plasmids are commonly used in genetic engineering to introduce foreign DNA
into bacteria. They can carry genes of interest, and their presence allows for selection or
monitoring of transformed bacteria. For example, plasmids can carry antibiotic
resistance genes that help select for transformed bacteria.

Final Answer:

Both statements I and II are correct.

 128) NCERT Pg. # 163

129) NCERT Pg # 164

130)

131) NCERT Pg. # 132

132) NCERT 11th Class (Chapter-8) Page No. 140

133)

NCERT XI, Topic # 8.3

134) NCERT Pg.#129

135) NCERT-XII, Pg. # 97

ZOOLOGY

136) NCERT XI Pg # 103

137) NCERT XII, Pg # 103

138)

NCERT - XI. Pg # 102 Fig 7.2

139) NCERT (XI), Pg. # 101

140) NCERT (XI) Pg. # 102, 103

141) NCERT XI (E)Pg.# 102, figure 7.1(c), (d), (b), 7.3

NCERT XI (H)Pg.# 102, para 2
figure 7.1(c), (d), (b), 7.3
figure 7.1(c), (d), (b), 7.3

142) NCERT Pg# 103, fig.-7.4(b)

143)
 NCERT Page # 101

144)

The correct answer is: Option 2 (Cartilage - Solid and pliable).

Explanation:

A. Cartilage is a type of connective tissue that is solid yet flexible (pliable) due to the presence of
chondroitin sulfate in its matrix.
B. Unlike bone, cartilage lacks calcium salts, making it more flexible.

145) NCERT XI Pg.# 101, 103

146) Explanation:
The heart contains all four types of tissues: epithelial tissue, connective tissue, muscular
tissue, neural tissue

147) NCERT Pg # 102, Para 9

148) NCERT XI, Page # 103

149) NCERT XI, Page # 101, 102

150) NCERT XI Pg. # 103, fig 7.4 (a)

151)

NCERT Reference: Class XI, Page No. 100

152)

NCERT Reference: Class XI, Page No. 103

153)

NCERT Reference: Class XI, Page No. 105

154) NCERT-XI, Pg # 101, Para.-7.1.1

155) NCERT-XI, Pg # 102, Para.-7.1.1

156) NCERT XI Page # 148 & 149

 157) NCERT Pg. # 144, 2nd last para

158)

Concept:

A. Monosaccharides: Simple sugars that cannot be hydrolyzed into smaller carbohydrates.

Examples include glucose, ribose, and mannose.

Explanation:

A. Ribose, Glucose and Mannose are monosaccharides.

Final Answer (2) : 3

159)

Explanation:

Lipids are not considered macromolecules in the strict sense, as they are not made up of
repeating monomeric units like proteins, nucleic acids, or polysaccharides. While lipids can
form large complexes, they typically do not qualify as macromolecules.

Answer: 4. Lipids are macromolecules

160)

Concept:

A. Glycogen is a branched polysaccharide of glucose.

B. α-1,4-glycosidic bonds link glucose molecules in the linear chains.
C. α-1,6-glycosidic bonds occur at the branch points.
D. β-glycosidic bonds are not found in glycogen.

Explanation:

A. A and C represent the linear chains of glucose. Linear chains are formed by α-1,4 glycosidic
bonds.
B. B represents the branching point. Branch points are formed by α-1,6 glycosidic bonds.
C. Therefore, A and C are α-1,4, and B is α-1,6.
D. β-glycosidic bonds are not found in glycogen.

Final Answer: (3) : α – 1' – 6"α – 1' – 4"α – 1' – 4"

161) NCERT Page 148,149

162)

163) NCERT Page No.143

164) NCERT Page No.148

165) Concept:
Oils (like gingelly oil) are usually unsaturated with lower melting points than fats (which are
more saturated).
Explanation:
Gingelly (sesame) oil is an unsaturated fat and has a lower melting point than saturated fats —
making Option 2 incorrect.

Final Answer:
Option 2: Gingelly oil has higher melting point than fats.

166)

Explaining the question :- The question is asking about Assertion and reason

Concept :- Analysis of chemical analysis

Solution :- Lipids are not strictly macromolecules because lipids are not polymers built from
repeating monomer units like proteins or nucleic acids. They are biologically important but
structurally distinct from true macromolecules.

Most common lipids (e.g., fatty acids, triglycerides, phospholipids) have molecular weights
well below 800 Da.

Final Answer :- 3

167)

Explaining the question :- The question is asking about element is present in negligible
amount in human body

Concept :- Analysis of chemical composition

Solution :- Si (Silicon) is present in negligible amounts in the human body compared to

elements like nitrogen , carbon and oxygen.

Final Answer3

168) Explanation :- The question is asking about assertion and reason

Concept :- This question is based on Polysaccharide
Solution :- Most polysaccharides (like starch, glycogen, cellulose) are non-reducing because
their anomeric carbon atoms are involved in glycosidic bonds, making them unavailable to act
 as reducing agents.
In a polysaccharide chain, the right end is called the reducing end and the left end is called
the non-reducing end. The reason doesn't explain why polysaccharides are non-reducing.
Answer :- 2

169)

Concept:

A. The organic compound is likely nucleic acids (e.g., DNA or RNA), which are composed of
nucleotides.
B. Nucleotides are joined by phosphodiester bonds, which link the sugar of one nucleotide to the
phosphate group of the next.

Explanation:
The bond that joins monomeric units (nucleotides) in nucleic acids is the phosphodiester bond.

170)

Question Explanation: A chiral carbon is a carbon atom attached to four different groups. In
β-D-(+)-glucose, we need to identify the number of such carbons.

Explanation: In β-D-(+)-glucose, the chiral carbons are at positions C2, C3, C4, C5, and C6
(since C1 is anomeric in the cyclic form). Hence, there are five chiral carbons.

Correct Answer: 1 (five)

171)

The correct answer option 2 , "Only statement II is correct

All polysaccharide are non-reducing but in a polysaccharide chain, the right end is called the
reducing end and the left end is called the non-reducing end.

172)

Explanation: Figure B is a five-membered ring sugar, which is a pentose.

Answer: Option (3). B is pentose sugar

173) Solution/Explanation/Calculation:

A. I. Right end of polysaccharide is called reducing end while left end is called non
reducing end.

This statement is correct.

A. II. Starch can hold 12 molecules in its helical secondary structure but cellulose being
 non helical, cannot hold 12

This statement is also correct.

A. III. Starch and glycogen are branched molecule.

This is also correct.

A. IV. Starch in plant and glycogen in animal are store houses of energy.

This is correct.
Final Answer: 4, 4.

174)

Concept

A. Sucrose → a disaccharide (only two sugar units, glucose + fructose) → not a polymer

Explanation
Sucrose is a small molecule made of only two monomers, so it does not qualify as a polymer.

Final Answer : Option 3: Sucrose

175)

Concept:

A. Iodine and Starch Reaction: Iodine forms a blue color when it binds to starch,
specifically to the helical structure of amylose (a component of starch).

B. The reason provided incorrectly describes the structure of starch, suggesting no helix
and stating that iodine traps in branches, which is not accurate.

Explanation:

A. The Assertion is true: Iodine reacts with starch to give a blue color.

B. The Reason is incorrect: Starch has a helical structure in amylose, not without it. Iodine
fits into this helix to form the blue color.

176)

Question: The above given figures A and B are diagrammatic structures of _____.

Explanation: Figure A is a six-membered ring with specific -OH positions, Figure B is a five-
membered ring.

Answer: (3). Glucose and ribose sugar

177) Which of the following is the structure of

178) NCERT Page No. 110

179) NCERT Page No. 105

180)

Generated by Allie

Problem Statement: The question asks where stratified squamous epithelium is found by
providing options such as Pneumocyte, Epidermis of skin, Podocyte, and Inner lining of blood
vessels.

Underlying Concept: Stratified squamous epithelium is a type of tissue made up of several

layers of flat cells. It serves primarily as a protective barrier against mechanical stress,
dehydration, and microbial invasion. This tissue type is commonly found in areas subjected to
wear and tear, such as the skin's outer layer (epidermis).

Tips and Tricks: Remember that 'stratified' means multiple layers and 'squamous' means flat
cells-such tissues are generally located where protection is needed, such as the skin's surface.

Common Mistakes: Confusing cell types like pneumocytes and podocytes as epithelial
tissues; assuming all linings in the body are made of stratified squamous epithelium.

Why Other Options Are Incorrect?: Pneumocytes are cells lining the alveoli in the lungs and
are simple squamous epithelium, not stratified. Podocytes are specialized cells in the kidney's
glomerulus and not epithelial tissue. Inner lining of blood vessels is made of simple squamous
epithelium (endothelium), not stratified squamous.