KCET 2021

QUESTIONS

PHYSICS

1. The physical quantity which is measure in the unit of 𝑊𝑏𝐴−1 is

a. Self-inductance b. Mutual inductance c. Magnetic flux d. Both (a) and (b)
2. What will be the reading in the voltmeter and ammeter of the circuit shown?

a. 90 V, 2A b. 0 V, 2 A c. 90 V, 1 A d. 0 V, 1 A
3. 𝐿𝐶 oscillations are similar and analogous to the mechanical oscillations of a block attached to a
spring. The electrical equivalent of the force constant of the spring is
a. Reciprocal of capacitive reactance
b. capacitive reactance
c. Reciprocal of capacitance
d. Capacitance
4. In an oscillating 𝐿𝐶 circuit, 𝐿 = 3.00mH and 𝐶 = 2.70𝜇𝐹. At 𝑡 = 0 the charge on the capacitor is
zero and the current is 2.00 A. The maximum charge that will appear on the capacitor will be
a. 1.8 × 10−5 𝐶 b. 18 × 10−5 𝐶 c. 9 × 10−5 𝐶 d. 90 × 10−5 𝐶
5. Suppose that the electric field amplitude of electromagnetic wave is 𝐸0 = 120𝑁𝐶 −1 and its
frequency if 𝑓 = 50 𝑀𝐻𝑧. Then which of the following value incorrectly computed?
a. Magnetic field amplitude is 400 nT.
b. Angular frequency of EM wave is 𝜋 × 108 rad 𝑠 −1
c. Propagation constant (angular wave number) is 2.1 rad 𝑚−1
d. Wave length of EM wave is 6 m.
6. The source of electromagnetic wave can be a charge
a. Moving with a constant velocity
b. Moving in a circular orbit
c. At rest
d. Moving parallel to the magnetic field
7. In refraction, light waves are bend on passing from one medium to second medium becomes, in
the second medium
a. Frequency is different
b. Speed is different
c. Coefficient of elasticity is difference
d. Amplitude is smaller
8. If the refractive index from air to glass is 3/2 and that from air to water is 4/3, then the radio of
focal lengths of a glass lens in water and in air
a. 1 : 2 b. 2 : 1 c. 1 : 4 d. 4 : 1
9. Two thin biconvex lenses have focal lengths 𝑓1 and 𝑓2. A third thin biconcave lens has focal
length of 𝑓3. If the two biconvex lenses are in contact, the total power of the lenses is 𝑃1 . If the
first convex lenses is in contact with the third lens, the total power is 𝑃2 . If the second lens is in
contact with the third lens, the total power is 𝑃3 then
𝑓𝑓 𝑓𝑓 𝑓𝑓
a. 𝑃1 = 𝑓 1−𝑓2 , 𝑃2 = 𝑓 1−𝑓3 and 𝑃3 = 𝑓 2−𝑓3
1 2 3 1 3 2
𝑓1 −𝑓2 𝑓3 −𝑓1 𝑓3 −𝑓2
b. 𝑃1 = 𝑓1 𝑓2
, 𝑃2 = 𝑓3 +𝑓1
and 𝑃3 = 𝑓2 𝑓3
𝑓1 −𝑓2 𝑓3 −𝑓1 𝑓3 −𝑓2
c. 𝑃1 = 𝑓1 𝑓2
, 𝑃2 = 𝑓3 𝑓1
and 𝑃3 = 𝑓2 𝑓3
𝑓1 +𝑓2 𝑓3 −𝑓1 𝑓3 −𝑓2
d. 𝑃1 = 𝑓𝑓
, 𝑃2 = 𝑓3 𝑓1
and 𝑃3 = 𝑓2 𝑓3
1 2
10. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length
30 cm is 2 cm. if a concave lens of focal length 20 cm is placed between the convex lens and the
image at a distance of 26 cm from the lens, the new size of the image is
a. 1.25 cm b. 2.5 cm c. 1.05 cm d. 2 cm
11. A slit of width 𝑎 is illuminated by red light of wavelength 6500 Å. If the first diffraction minimum
fall at 30 Å, then the value of 𝑎 is
a. 6.5 × 10−4 mm b. 1.3 micron c. 3250 Å d. 2.6 × 10−4 cm
12. Which of the statements are correct with reference to single slit diffraction patern?
(i) Fringes are of unequal width
(ii) Fringes are of equal width
(iii) Light energy is conserved
(iv) Intensities of all bright fringes are equal
a. (i) and (iii) b. (i) and (iv) c. (ii) and (iv) d. (ii) and (iii)
13. In the young’s double slit experiment a monochromatic source of wavelength 𝜆 is used. The
intensity of light passing through each slit is 𝐼0 . The intensity of light the screen 𝑆𝐶 at a point P, a
distance 𝑥 from O is given by (Take d< < D)

𝜋𝐷 𝜋𝑑 𝜋𝑑 𝜋𝑑
a. 𝐼0 cos2 (𝜆𝑑 𝑥) b. 4𝐼0 cos2 (𝜆𝐷 𝑥) c. 𝐼0 sin2 (2𝜆𝐷 𝑥) d. 4𝐼0 cos (2𝜆𝐷 𝑥)
14. The work function of a metal is 1 eV. Light of wavelength 3000 Å is incident on this metal
surface. The velocity of emitted photoelectrons will be
 a. 10 𝑚𝑠 −1 b. 1 × 103 𝑚𝑠 −1 c. 1 × 104 𝑚𝑠 −1 d. 1 × 106 𝑚𝑠 −1
15. A proton moving with a momentum 𝑝1 has a kinetic energy 1/8th of its rest mass energy.
Another light photon having energy to the kinetic energy of the possesses a momentum 𝑝2 .
Then the ratio (𝑝1 − 𝑝2 )/𝑝1 is equal to
1 1 3
a. 1 b. 4 c. 2 d. 4
16. According to Einstein’s photoelectric equation to the graph between kinetic energy of
photoelectrons ejected and the frequency of incident radiation is
a. b.

c. d.

17. Energy of an electron in the second orbit of hydrogen atom is 𝐸2 . The energy of electron in the
third orbit of 𝐻𝑒 + will be
9 16 3 16
a. 𝐸
16 2
b. 𝐸
9 2
c. 16 𝐸2 d. 3 2
𝐸
18. The figure shows standing de Broglie waves due to the revolution of electron in a certain orbit of
hydrogen atom. Then the expression for the orbit radius is (all notations have their usual
meanings)

ℎ 2 𝜀0 4ℎ 2 𝜀 9ℎ 2𝜀 16ℎ2 𝜀0
a. 𝜋𝑚𝑒 2
b. 𝜋𝑚𝑒 20 c. 𝜋𝑚𝑒 20 d. 𝜋𝑚𝑒 2
19. An electron is an excited state of 𝐿𝑖 2+ion has angular momentum 3ℎ/2𝜋. The de Broglie
wavelength in this state is 𝑃𝜋𝑎0 (where 𝑎0 = Bohr radius). The value of 𝑃 is
a. 3 b. 2 c. 1 d. 4
20. Which graph in the following diagram correctly represents the potential energy of a pair of
nucleons as a function of their separation?
a. b.

c. d.

21. In a nuclear reactor heavy nuclei is not used as moderators because

a. They will break up
b. Elastic collision of neutrons with heavy nuclei will not slow them down
c. The net weight of the reactor would be unbearably high
d. Substance with heavy nuclei do not occur in liquid or gaseous state at room temperature
22. The circuit given represents which of the logic operation?

a. OR b. AND c. NOT d. NOR

23. Identify the incorrect statement:
a. When a p-n junction diode is forward biased, the width of the depletion region decreases.
b. When a p-n junction diode is reverse biased, the barrier potential increases
c. A photon diode is operated in the reverse bias
d. An LED is a lightly doped p-n junction diode which emits spontaneous radiation of forward
biasing.
24. Three photodiodes 𝐷1 , 𝐷2 and 𝐷3 are made of semiconductors having band gaps of 2.5 eV. 2.5
eV, 2 eV and 3 eV respectively. Which one will be able to detect light of wavelength 600 nm?
a. 𝐷1 only b. both 𝐷1 and 𝐷3 c. 𝐷2 only d. All the three diodes
25. For a body moving along a straight line, the following 𝑣 − 𝑡 graph is obtained. According to the
graph, the displacement during

a. Uniform acceleration is greater than that during uniform motion

b. Uniform acceleration is less than that during uniform motion
c. Uniform acceleration is equal to that during uniform motion
d. Uniform motion zero
26. A patticle starts from rest. Its acceleration 𝑎 versus time 𝑡 is shown in the figure. The maximum
speed of the particle will be:

a. 80 𝑚𝑠 −1 b. 40 𝑚𝑠 −1 c. 18 𝑚𝑠 −1 d. 2 𝑚𝑠 −1
27. The maximum range of a gun on horizontal plane is 16 km. if g=10 𝑚𝑠 −2, then muzzle velocity of
a shell is
a. 160 𝑚𝑠 −1 b. 200√2𝑚𝑠 −1 c. 400 𝑚𝑠 −1 d. 800 𝑚𝑠 −1
28. The trajectory of projectile is
a. Semicircle
b. An ellipse
c. A parabola always
d. A parabola in the absence of air resistance
29. For a projectile motion, the angle between the velocity and acceleration is minimum and acute
at
a. Only one point b. two points c. three points d. four points
−1
30. A particle starts from the origin at 𝑡 = 0 𝑠 with a velocity of 10𝑗̂𝑚𝑠 and move in the 𝑥 − 𝑦
plane with a constant acceleration of (8𝑖̂ + 2𝑗̂)𝑚𝑠 −2 . At an instant when the 𝑥 − coordinate of
the particle is 16 m, 𝑦 − coordinate of the particle is:
a. 16 m b. 28 m c. 36 m d. 24 m
31. A coin placed on a rotation turn table just slips if it is placed at a distance of 4 cm from the
center. If the angular velocity of the turn table is doubled it will just slip at a distance of
a. 1 cm b. 2 cm c. 4 cm d. 8 cm
32. A 1 kg ball moving at 12 𝑚𝑠 collides with a 2 kg ball moving in opposite direction at 24 𝑚𝑠 −1 .
−1

If the coefficient of restitution is 2/3, then their velocities after the collision are
a. -4 𝑚𝑠 −1, -28 𝑚𝑠 −1 b. -28 𝑚𝑠 −1 , -4 𝑚𝑠 −1
−1 −1
c. 4 𝑚𝑠 , 28 𝑚𝑠 d. 28 𝑚𝑠 −1 , 4 𝑚𝑠 −1
33. A ball hits the floor and rebounds after an inelastic. In this case
a. The momentum of the ball is conserved
b. The mechanical energy of the ball is conserved
c. The total momentum of the ball and the earth is conserved
d. The total mechanical energy of the ball and the earth is conserved
34. In a figure 𝐸 and 𝑣𝑐𝑚 represent the total energy and speed of center of mass of an object of
mass 1 kg in pure rolling. The object is:

a. Sphere b. Ring c. Disc d. Hollow Cylinder

35. Two bodies of masses 8 kg are placed at the vertices A and B of an equilateral triangle ABC. A
third body of mass 2 kg is placed at the centroid G of the triangle. If AG = BG = CG = 1 m, where
should a fourth body of mass 4 kg be placed so that the resultant force on the 2 kg body is zero?
a. At C
1
b. At a point P on the line CG such that PG = m
√2
c. At a point P on the line CG such that PG = 0.5 m
d. At a point P on the line CG such that PG = 2 m
36. Two capillary tubes P and Q are dipped vertically in water. The height of water level in capillary
tube P is 2/3rd of the height in capillary tube Q. the ratio of their diameter is
a. 2 : 3 b. 3 : 2 c. 3 : 4 d. 4 : 3
37. Which of the following curves represent the variation of coefficient of volume expansion of an
ideal gas at constant pressure?
 a. b.

c. d.

38. A number of Carnot engines are operated at identical cold reservoir temperature (𝑇𝐿 ).
However, their hot reservoir temperatures are kept different. A graph of the efficiency of the
engines versus hot reservoir temperature (𝑇𝐻 ) is plotted. The correct graphical representation is
a. b.
 c. d.

39. A gas mixture contains monoatomic and diatomic molecules of 2 moles each. The mixture has a
total internal energy of (symbols have usual meanings)
a. 3 RT b. 5 RT c. 8 RT d. 9 RT
40. A pendulum oscillates simple harmonically and only if
(i) The sizer of the bob of pendulum is negligible in comparison with the length of the
pendulum
(ii) The angular amplitude is less than 10°
a. Both i and ii are correct
b. Both I and ii are incorrect
c. Only I is correct
d. Only ii is correct
41. To propagate both longitudinal and transverse waves, a material must have
a. Bulk and shear moduli b. Only bulk modulus
c. Only shear modulus d. Young’s and Bulk modulus
42. A copper rod AB of length 𝑙 is rotated about end A with a constant angular velocity 𝜔. The
electric field at a distance 𝑥 from the axis of rotation is
𝑚𝜔2 𝑥 𝑚𝜔𝑥 𝑚𝑥 𝑚𝑒
a. 𝑒
b. 𝑒𝑙
c. 𝜔3 𝑙 d. 𝜔2 𝜋
43. Electric field due to infinite, straight uniformly charge wire given with distance 𝑟 as
1 1
a. 𝑟 b. 𝑟 c. 𝑟2 d. 𝑟 2
44. A 2 – gram object, located in a region of uniform electric field 𝐸⃗ = (300𝑁𝐶 −1 )𝑖̂ carries charge
𝑄. The object released from rest at 𝑥 = 0, has a kinetic energy of 0.12 J at 𝑥 = 0.5 m. Then 𝑄 is
a. 400𝜇𝐶 b. −400𝜇𝐶 c. 800𝜇𝐶 d. −800𝜇𝐶
−3
45. If a slab of insulating material (conceptual). 4 × 10 m thick is introduced between the plates
of a parallel plate capacitor, the separation between the plates has to be increased by
3.5 × 10−3 m to restore the capacity to original value. The dielectric constant of the material
will be
a. 6 b. 8 c. 10 d. 12
46. Eight drops of mercury of equal radii combine to form a big drop. The capacitance of a bigger
drop as compared to each smaller drop is
a. 2 times b. 8 times c. 4 times d. 16 times
47. Which of the statement is false in the case of polar molecules?
a. Centers of positive and negative charges are separated in the absence of external electric
field.
b. Centers of positive and negative charges are separated in the presence of external electric
field.
c. Do not possess permanent dipole moments.
d. Ionic molecule HCl is the example of polar molecule.
48. An electrician requires a capacitance of 6 𝜇𝐹 in a circuit across a potential difference of 1.5kV. A
large number of 2 𝜇𝐹 capacitors which can withstand a potential difference of not more than
500 V are available. The minimum number of capacitors required for the purpose is
a. 3 b. 9 c. 6 d. 27
49. In figure, charge on the capacitor is plotted against potential difference across the capacitor.
The capacitance and energy stored in the capacitor are respectively.

a. 12𝜇𝐹, 1200𝜇𝐽 b. 12𝜇𝐹, 600𝜇𝐽 c. 24𝜇𝐹, 600𝜇𝐽 d. 24𝜇𝐹, 1200𝜇𝐽

50. A wire of resistance 3Ω is stretched to twice its original length. The resistance of the new wire
will be
a. 1.5Ω b. 3Ω c. 6Ω d. 12Ω
51. In the given arrangement of experiment on metre bridge, if AD corresponding to null deflection
of the galvanometer is 𝑥, what would be its value if the radius of the wire AB is doubled?

𝑥
a. 𝑥 b. 4 c. 4𝑥 d. 2𝑥
52. A copper wire of length 1 m and uniform cross sectional area 5 × 10−7 𝑚2 carries a current of 1
A. Assuming that there are 8 × 108 free electrons per 𝑚2 in copper, how long will an electron
take drift from one end of the wire to the other?
a. 0.8 × 103s b. 1.6 × 103s c. 3.2 × 103 s d. 6.4 × 103s
53. Consider an electrical conductor connected across a potential difference 𝑉. Let ∆𝑞 be a small
charge moving through it in time ∆𝑡. If 𝐼 is the electric current through it,
(i) The kinetic energy of the charge increases by 𝐼𝑉∆𝑡.
(ii) The electric potential energy of the charge decreases by 𝐼𝑉∆𝑡.
(iii) The thermal energy of the conductor increases by 𝐼𝑉∆𝑡.

Then the correct statement is /are

a. (i) b. (i), (ii) c. (i) and (iii) d. (I), (ii)

54. A strong magnetic field on a stationary electron. Then the electron
a. Moves in the direction of the field
b. Moves in an opposite direction of the field
c. Remains stationary
d. Starts spinning
55. Two parallel wires in free space are 10 cm apart and each carries a current of 10 A in the same
direction. The force exerted by one wire on the other [per unit length] is
a. 2 × 10−4 𝑁𝑚−1 [attractive] b. 2 × 10−7 𝑁𝑚−1 [attractive]
c. 2 × 10−4 𝑁𝑚−1 [repulsive] d. 2 × 10−7 𝑁𝑚−1 [repulsive]
56. A toroid with thick winding of 𝑁 turns has inner and outer radii 𝑅1 and 𝑅2 respectively. If it
certain steady current 𝐼, the variation of the magnetic field due to the toroid with radial distance
is correctly graphed in
a. b.

c. D.
57. A tightly wound long solenoid has 𝑛 turns per unit length, a radius 𝑟 and carries a current 𝐼. A
particle having charge 𝑞 and mass 𝑚 is projected from a point on the axis in a direction
perpendicular to the axis. The maximum speed of the particle for which the particle does strike
the solenoid is
𝜇0 𝑛𝐼𝑞𝑟 𝜇0 𝑛𝐼𝑞𝑟 𝜇0 𝑛𝐼𝑞𝑟 𝜇0 𝑛𝐼𝑞𝑟
a. 𝑚
b. 2𝑚
c. 4𝑚
d. 8𝑚
58. Earth’s magnetic field always has a horizontal component except at
a. Equator b. magnetic poles c. a latitude of 60° d. an altitude of 60°
59. Which of the field pattern given below is valid for electric field as well as for magnetic field?
a. b. c. d.

60. The current following through an inductance coil self-inductance 6 mH at different time instant
is as shown. The emf induced between 𝑡 = 20𝑠 and 𝑡 = 40 𝑠 is nearly

a. 2 × 10−2 𝑉 b. 3 × 10−2 𝑉 c. 4 × 10−3 𝑉 d. 30 × 102 𝑉

CHEMISTRY

1. Zeta potential is
a. Potential required to bring about coagulation of a colloidal sol
b. Potential required to give the particle a speed of 1 cm s-1
 c. Potential difference between fixed charged layer and the diffused layer having opposite
charges.
d. Potential energy of the colloidal particles
2. Which of the following compound on heating given N2O?
a. Pb(NO3)2 b. NH4NO3 c. NH4NO2 d. NaNO3
3. Which of the following property is true for the given sequence
NH3 > PH3 > AsH3 > SbH3 > BiH3
a. Reducing property b. Thermal stability c. Bond angle d. Acidic character
4. The correct order of boiling point in the following compound is
a. HF > H2 O > NH3 b. H2 O > HF > NH3
c. NH3 > H2 O > HF d. NH3 > HF > H2 O
5. XeF6 on partial hydrolysis give a compound X, which has square pyramidal geometry. The ‘X’ is
a. XeO3 b. XeO4 c. XeOF4 d. XeO2 F2
6. A colourless, neutral paramagnetic oxide of nitrogen ‘P’ on oxidation gives reddish brown gas Q.
Q on cooling gives colourless gas R, R on reaction with P gives blue solid S. Identify P, Q, R, S,
respectively.
a. N2 O NO NO2 N2 O3 b. N2 O NO2 N2 O4 N2 O3
b. NO NO2 N2 O4 N2 O3 d. NO NO2 N2 O4 N2 O5
7. Which of the following does not represent property stated against it?
a. CO2+ < Fe2+ < Mn2+ − Ionic size
b. Ti < V < Mn − Number of oxidation states
2+ 2+ 2+
c. Cr < Mn < Fe − Paramagnetic behaviour
d. Sc > Cr > Fe − Density
8. Which one of the following is correct for all elements form Sc to Cu?
a. The lowest oxidation state shown by them is +2.
b. 4s orbital is completely filled in the ground state
c. 3d orbital is not completely filled in the ground state.
d. The ions in +2 oxidation states are paramagnetic
9. When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of
gas
a. Will be half of original volume b. will be 4 times the original volume
c. Will be 2 times the original volume d. will be ¼ times the original volume
10. Which of the following pairs has both the ions coloured in aqueous solution? [Atomic numbers
of Sc = 21, Ti = 22, Ni = 28, Cu = 29, Mn = 25]
a. Sc 3+ Mn2+ b. Ni2+ , Ti4 c. Ti3+ , Cu+ d. Mn2+ , Ti3+
11. For the crystal field splitting in octahedral complexes,
3
a. The energy of the eg orbitals will decrease by (5) ∆0 and that of the t 2g will increase by (2/5)
∆o .
3
b. The energy of the eg orbitals will increase by (5) ∆0 and that of the t 2g will decrease by (2/5)
∆o
 3
c. The energy of the eg orbitals will increase by ( ) ∆0 and that of the t 2g will increase by (2/5)
5
∆o
3
d. The energy of the eg orbitals will decrease by ( ) ∆0 and that of the t 2g will decrease by
5
(2/5) ∆o
12. Peroxide effect is observed with the addition of HBr but not with the addition of HI to
unsymmetrical alkene because
a. H − I bond is strong that H − Br and is not cleaved by the free radical.
b. H − I bond is weaker than H − Br bond so that iodine free radicals combine to form iodine
molecules.
c. Bond strength of HI and HBr are same but free radicals are formed in HBr
d. All of these.
13. The IUPAC name of [Co(NH3 )5 (CO3 )]Cl is
a. Pentaamminecarbonatocobalt (III) Chloride
b. Carbonatopentamminecobalt (III) Chloride
c. Pentaamminecarbonatocobaltate (III) Chloride
d. Pentaammine cobalt (III) Carbonate Chloride
14. Homoleptic complexes among the following are
i. K 3 [Al(C2 O4 )]3 ii. [CoCl2 (en)2 ]+ iii. K 2 [Zn(OH)4 ]
a. (i) only b. (i) and (ii) only c. (i) and (iii) only d. (iii) only

15. The correct order for wavelength of light absorbed in the complex ions
[CoCl(NH3 )5 ]2+ , [Co(NH3 )6 ]3+ and [Co(CN)6 ]3− is
a. [CoCl(NH3 )5 ]2+ > [Co(NH3 )6 ]3+ > [Co(CN)6 ]3−
b. [Co(NH3 )6 ]3+ > [Co(CN)6 ]3− > [CoCl(NH3 )5 ]2+
c. [Co(CN)6 ]3− > [CoCl(NH3 )5 ]2+ > [Co(CN)6 ]3−
d. [Co(NH3 )6 ]3+ > [CoCl(NH3 )5 ] > [Co(CN)6 ]3−
16.

The compound A (major product) is

a.

b.

c.
 d.

17. Bond enthalpies of A2 , B2 and AB are in the ratio 2: 1: 2. If bond enthalpy of formation of AB is
−100 kJ mol− . The bond enthalpy of B2 is
a. 100 kJ mol−1 b. 50 kJ mol−1 c. 200 kJ mol−1 d. 150 kJ mol−1
18. The order of reactivity of the compounds
C6 H5 CH2 Br, C6 H5 CH(C6 H5 )Br, C6 H5 CH(CH3 )Br and C6 H5 C(CH3 )(C6 H5 )Br in SN 2 reaction is
a. CH3 H H H
| | | |
C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br
| | | |
C6 H5 C6 H5 CH3 H
b. H H H CH3
| | | |
C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br
| | | |
H CH3 C6 H5 C 6 H5
c. H H H CH3
| | | |
C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br
| | | |
CH3 H C6 H5 C 6 H5
d. H H H CH3
| | | |
C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br
| | | |
C6 H5 H CH3 C 6 H5
HBr
19. The major product of the following reaction is CH2 = CH − CH2 − OH →
product
Excess
a. CH3 − CHBr − CH2 Br b. CH2 = CH − CH2 Br
c. CH3 − CHBr − CH2 − OH d. CH3 − CHOH − CH2 OH
20.
 The product ‘A’ gives white precipitate when treated with bromine water. The product ‘B’ is
treated with Barium hydroxide to give the product C. The compound C is heated strongly to
form product D. The product D is
a. 4 − methylpent − 3 − en − 2 − one b. but − 2 − enal
c. 3 − methylpent − 3 − en − 2 − one d. 2 − methylbut − 2 − enal
21. For the reaction A(g) + B(g) ⇌ C(g) + D(g); ∆H = −Q kJ
The equilibrium constant cannot be disturbed by
a. Addition of A b. Addition of D
c. Increasing of pressure d. Increasing of temperature
22. An organic compound ‘X’ on treatment with PCC in dichloromethane gives the compound Y.
Compound ‘Y’ reacts with I2 and alkali to form yellow precipitate of triiodomethane. The
compound X is
a. CH3 CHO b. CH3 COCH3 c. CH3 CH2 OH d. CH3 COOH
23. A compound ‘A’ (C7 H8 O) is insoluble in NaHCO3 solution but dissolve in NaOH and gives a
characteristics colour with neutral Fecl3 solution. When treated with Bromine water compound
‘A’ forms the compound B with the formula C7 H5 OBr3 . ‘A’ is
a. b.

c. d.

24. In set of reactions, identify D

SOCl2 Benzene HCN H2 O
CH3 COOH → A B → C → D
→
Anh AlCl3
a. b.
 c. d.

25. K a values for acids H2 SO3 , HNO2 , CH3 COOH and HCN are respectively 1.3 × 10−2 , 4 ×
10−4 , 1.8 × 10−5 and 4 × 10−10 , which of the above acids produces stronger conjugate base in
aqueous solution?
a. H2 SO3 b. HNO2 c. CH3 COOH d. HCN
26.

A, B and C respectively are

a. Ethanol, ethane nitrile and ethyne b. ethane nitrile, ethanol and ethyne
c. Ethyne, ethanol and ethane nitrile d. ethyne, ethane nitrile and ethanol
27. The reagent which can do the conversion CH3 COOH → CH3 − CH2 − OH is
a. LiAlH4 /ether b. H2 , Pt c. NaBH4 d. Na and C2 H5 OH
(i) CH3 MgBr Conc.H2 SO4 (i)B2 H6
28. CH3 CHO ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ A ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ B ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
(ii) H2 O,OH−
C
(ii) H3 O+ ∆
A and C are
a. Identical b. Position isomers
c. Functional isomers d. Optical isomers
29. Which of the following is not true for oxidation?
a. Addition of oxygen b. addition of electronegative element
c. Removal of hydrogen d. Removal of electronegative element
30. Which is the most suitable reagent for the following conversion?
O O
∥ ∥
CH3 − CH = CH − CH2 − C − CH3 → CH3 − CH = CH − CH2 − C − OH
 a. Tollen’s reagent b. Benzoyl peroxide
c. I2 and NaOH solution with subsequent acidification
d. Sn and NaOH solution
alc.NH3 2CH3CL
31. C6 H5 CH2 Cl → A→ B
The product B is
a. N, N – dimethyl phenyl methenamine
b. N, N – dimethyl benzenamine
c. N – Benyl – N – methyl methanamine
d. Phenyl -N, N- dimethyl methanamine
32. The method by which aniline cannot be prepared is
a. Nitration of benzene followed by reduction with Sn and con HCl
b. Degradation of benzamide with bromine in alkaline solution
c. Reduction of nitrobenzene with H2 /pd is ethanol.
d. Potassium salt of phthalimide treated with chlorobenzene followed by the hydrolysis with
aqueous NaOH solution.

33. Permanent hardness cannot be removed by

a. Using washing soda b. Calgon’s method
c. Clark’s method d. Ion exchange method
34. A hydrocarbon A (C4 H8 ) on reaction with HCl gives a compound B (C4 H9 Cl) which on reaction
with 1 mol of NH3 gives compound C(C4 H10 N). On reacting with NaNO2 and HCl followed by
treatment with water, compound C yields an optically active compound D. The D is
CH2 − CH3 CH2 − CH3
| |
a. CH3 − C − H b. CH3 − C − H
| |
Cl OH
CH2 − CH3 CH2 − CH3
| |
c. CH3 − C − H d. CH3 − C − H
| |
NH2 H
35. RNA and DNA are chiral molecules, their chirality is due to the presence of
a. D – sugar component b. L -Sugar component
c. Chiral bases d. Chiral phosphate ester unit
36. The property of the alkaline earth metals that increases with their atomic number is
a. Ionisation enthalpy b. electronegativity
c. Solubility of their hydroxide in water
d. Solubility of their sulphate in water
37. Primary structure in a nuclei acid chain contains bases as GATGC ________. The chain which is
complementary to this chain is
 a. G G T G A b. T G A A G c. C T A C G d. T T T A G
38. In the detection of II group acid radical, the salt containing chloride is treated with concentrated
sulphuric acid, the colour-less gas is liberated. The name of the gas is
a. Hydrogen chloride gas b. Chlorine gas
c. Sulphur dioxide gas d. hydrogen gas
39. The number of six membered and five membered rings in Buckminster Fullerence respectively is
a. 20, 12 b. 12, 20 c. 14, 18 d. 14,11
40. In Chrysoberyl, a compound containing Beryllium, Aluminum and oxygen, oxide ions form cubic
close packed structure. Aluminum ions occupy 1/4th of tetrahedral voids and Beryllium ions
occupy 1/4th of octahedral voids. The formula of the compound is
a. BeAlO4 b. BeAl2 O4 c. Be2 AlO2 d. BeAlO2
41. The correct statement regarding defects in solids is
a. Frenkel defect is a vacancy defect.
b. Schottky defect is a dislocation defect.
c. Trapping of an electron in the lattices leads to the formation of F – centre.
d. Schottky defect has no effect on density.
42. A metal crystallises in BCC lattice with unit cell edge length of 300 pm and density 6.15 gcm−3 .
The molar mass of the metal is
a. 50 g mol−1 b. 60 g mol−1 c. 40 g mol−1 d. 70 g mol−1
43. Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole
fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at
298 K and 5 atm pressure is
a. 4.0 × 10−4 b. 4.0 × 10−5 c. 5.0 × 10−4 d. 4.0 × 10−6
44. A pure compound contains 2.4 g of C, 1.2 × 1023 atoms of H, 0.2 moles of oxygen atoms. Its
empirical formula is
a. C2 HO b. C2 H2 O2 c. CH2 O d. CHO
45. Choose the correct statement.
a. k H value is same for a gas in any solvent.
b. Higher the k H value more the solubility of gas.
c. k H value increases on increasing the temperature of the solution
d. Easily liquefiable gases usually has lesser k H values.
46. The k H value (K bar) of Argon (I), carbon dioxide (II) form- aldehyde (III) and methane (IV) are
respectively 40.3, 1.67, 1.83 × 10−5 and 0.413 at 298 K. The increasing order of solubility of gas
in liquid is
a. I < II < IV < III b. III < IV < II < I
c. I < III < II < IV d. I < IV < II < III
47. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg at 350 K respectively. If the
total vapour pressure of the mixture is 600 mm Hg, the composition of the mixture in the
solution is
a. xA = 0.4, xB = 0.6 b. xA = 0.6, xB = 0.4
c. xA = 0.3, xB = 0.7 d. xA = 0.7, xB = 0.3
48. Consider the following electrodes
P = Zn2+ (0.0001 M)/Zn
Q = Zn2+ (0.1 M)/Zn
R = Zn2+ (0.01 M)/Zn
S = Zn2+ (0.001 M)/Zn
0
Ezn/zn2+ = −0.76 V Electrode potentials of the above electrodes in volts are in the order

a. P > S > R > Q b. S > R > Q > P c. Q > R > S > P d. P > Q > R > S
49. The number of angular and radial nodes in 3p orbital respectively are
a. 3, 1 b. 1, 1 c. 2, 1 d. 2,3
50. The resistance of 0.01 m KCl solution at 298 K is 1500 Ω. If the conductivity of 0.01 m KCl
solution at 298 K is 0.146 × 10−3 S cm−1 . The cell constant of the conductivity cell in cm−1 is
a. 0.219 b. 0.291 c. 0.301 d. 0.194
51. H2 (g) + 2 AgCl(s) ⇌ 2 Ag(s) + 2HCl(aq)
0
Ecell at 25°C for the cell is 0.22 V. The equilibrium constant at 25°C is
a. 2.8 × 107 b. 5.2 × 108 c. 2.8 × 105 d. 5.2 × 104
52. For a reaction A + 2B → Products, when concentration of B alone is increased half-life remains
the same. If concentration of A alone is doubled, rate remains the same. The unit of rate
constant for the reaction is
a. s −1 b. L mol−1 s −1 c. mol L−1 s −1 d. atm−1
53. The third ionisation enthalpy is highest in
a. Alkali metals b. alkaline earth metals
c. Chalcogens d. pnictogens
54. If the rate constant for a first-order reaction is k, the time (t) required for the completion of 99%
of the reaction is given by
4.606 2.303 0.693 6.909
a. t = k
b. t = k
c. t = k
d. t = k
55. The rate of a gaseous reaction is given by the expression k[A][B]2 . If the volume of vessel is
reduced to one half of the initial volume, the reaction rate as compared to original rate is
1 1
a. 16
b. 8 c. 8 d. 16
56. The correct IUPAC name of

a. 4 – Ethyl – 1 – fluoro – 2 – nitrobenzene b. 1 – Ethyl – 4 – Fluoro – 3 – nitrobenzene

c. 3 – Ethyl – 6 – Fluronitrobenzene d. 5 – Ethyl – 2 - Fluronitrobenzene
57. Higher order (> 3) reactions are rare due to
a. Shifting of equilibrium towards reactants due to elastic collisions
b. Loss of active species on collision
c. Low probability of simultaneous collision of all reacting species.
 d. Increases in entropy as more molecules are involved
58. Arrange benzene, n – hexane and ethyne in decreasing order of their acidic behaviour.
a. Benzene > n − Hexane > Ethyne b. n − Hexane > Benzene > Ethyne
c. Ethyne > n − Hexane > Benzene d. Ethyne > Benzene > n − Hexane
59. A colloidal solution is subjected to an electric field than colloidal particles more towards anode.
The amount of electrolytes of BaCl2 , AlCl3 and NaCl required to coagulate the given colloid is in
the order
a. NaCl > BaCl2 > AlCl3 b. BaCl2 > AlCl3 > NaCl
c. AlCl3 = NaCl = BaCl2 d. AlCl3 > BaCl2 > NaCl
60. Which of the following is an incorrect statement?
a. Hydrogen bonding is stronger than dispersion forces
b. Sigma bonds are stronger than π − bonds
c. Ionic bonding is non – directional
d. σ − Electrons are referred to as mobile electrons.

MATHEMATICS

1. The maximum slope of the curve 𝑦 = −𝑥 3 + 3𝑥 2 + 2𝑥 − 27 is

a. 1 b. 23 c. 5 d. -23
𝑥 3 sin(tan−1 (𝑥 4 ))
2. ∫ 1+𝑥 8
𝑑𝑥 is equal to
− cos(tan−1 (𝑥 4 )) cos(tan−1 (𝑥 4 ))
a. 4
+𝐶 b. 4
+𝐶
− cos(tan−1 (𝑥 3 )) sin(tan−1 (𝑥 4 ))
c. 4
+𝐶 d. 4
+𝐶
𝑥 2 𝑑𝑥
3. The value of ∫ is equal to
√𝑥 6 +𝑎6
a. log|𝑥 3 + √𝑥 6 + 𝑎6 |+C b. log|𝑥 3 − √𝑥 6 + 𝑎6 |+C
1 1
c. log|𝑥 3 + √𝑥 6 + 𝑎6 |+C d. log|𝑥 3 − √𝑥 6 + 𝑎6 |+C
3 3
𝑥𝑒 𝑥 𝑑𝑥
4. The value of ∫ (1+𝑥)2 is equal to
𝑒𝑥
a. 𝑒 𝑥 (1 + 𝑥)+C b. 𝑒 𝑥 (1 + 𝑥 2 ) + 𝐶 c. 𝑒 𝑥 (1 + 𝑥)2 + 𝐶 d. 1+𝑥 + 𝑐
1+sin 𝑥
5. The value of ∫ 𝑒 𝑥 [1+cos 𝑥] 𝑑𝑥 is equal to
𝑥
a. 𝑒 𝑥 tan + 𝐶 b. 𝑒 𝑥 tan 𝑥 + 𝐶 c. 𝑒 𝑥 (1 + cos 𝑥) + 𝐶 d. 𝑒 𝑥 (1 + sin 𝑥) + 𝐶
2
𝜋/4
6. If 𝐼𝑛 = ∫0 tan𝑛 𝑥 𝑑𝑥 when 𝑛 is positive integer then 𝐼10 + 𝐼8 is equal to
1 1 1
a. 9 b. 7 c. 8 d. 9
4042 √𝑥𝑑𝑥
7. The value of ∫0 is equal to
√ √4042−𝑥
𝑥+
a. 4042 b. 2021 c. 8084 d. 1010
8. The area of the region bounded by 𝑦 = √16 − and 𝑥 −axis is 𝑥2
a. 8 square units b. 20𝜋 square units c. 16𝜋 square units d. 256𝜋 square units
𝑥2 𝑦2
9. If the area of the Ellipse is + = 1 is 20 𝜋 square units, then 𝜆 is
25 𝜆2
a. ±4 b. ±3 c. ±3 d. ±1
10. Solution of differential equation 𝑥𝑑𝑦 − 𝑦𝑑𝑥 = 0 represents
a. A rectangular Hyperbola b. Parabola whose vertex is at origin
c. Straight line passion through origin d. A circle whose center is origin
𝑑𝑦 𝑦+1
11. The number of solutions of 𝑑𝑥 = 𝑥−1 when 𝑦(1) = 2 is
a. Three b. one c. infinite d. two
12. A vector 𝑎⃑ makes equal acute angles on the coordinate axis. Then the projection of vector 𝑏⃗⃑ =
5𝑖̂ + 7𝑗̂ − 𝑘̂ on 𝑎⃑ is
11 11 4 3
a. 15
b. c. 5 d. 5
√3 √3
13. The diagonals of a parallelogram are the vectors 3𝑖̂ + 6𝑗̂ − 2𝑘̂ and −𝑖̂ − 2𝑗̂ − 8𝑘̂ then the length
of the shorter side of parallelogram is at
a. 2√3 b. √14 c. 3√5 d. 4√3
14. If 𝑎⃑ ∙ 𝑏⃗⃑ = 0 and 𝑎⃑ + 𝑏⃗⃑ makes an angle 60° with 𝑎⃑ then
a. |𝑎⃑| = 2|𝑏⃗⃑| b.2|𝑎⃑| = |𝑏⃗⃑| c. |𝑎⃑| = √3|𝑏⃗⃑| d. √3|𝑎⃑| = |𝑏⃗⃑|
15. If the area of the parallelogram with 𝑎⃑ and 𝑏⃗⃑ as two adjacent sides is 15 sq. units then the area
of the parallelogram having 3𝑎⃑ + 2𝑏⃗⃑ and 𝑎⃑ + 3𝑏⃗⃑ as two adjacent sides in sq. units is
a. 45 b. 75 c. 105 d. 120
16. The equation of the line joining the points (-3, 4, 11) and (1, -2, 7) is
𝑥+3 𝑦−4 𝑧−11 𝑥+3 𝑦−4 𝑧−11
a. 2
= 3
= 4
b.
−2
= 3 = 2
𝑥+3 𝑦+4 𝑧+11 𝑥+3 𝑦+4 𝑧+11
c. −2
= 3
= 4
d. 2 = −3 = 2
√3 1 √3 √3 1 √3
17. The angle between the lines whose direction cosines are ( 4 , 4 , 2 ) and ( 4 , 4 , − 2 ) is
𝜋 𝜋 𝜋
a. 𝜋 b. 2 c. 3 d. 4
18. If a plane meets the coordinate axes at A, B and C in such a way that the centroid of triangle ABC
is at the point (1, 2, 3) then the equation of the plane is
𝑥 𝑦 𝑧 𝑥 𝑦 𝑧 𝑥 𝑦 𝑧 1 𝑥 𝑦 𝑧
a. 1
+2+3 =1 b. 3 + 6 + 9 = 1 c. 1 + 2 + 3 = 3 d. 1 − 2 + 3 = −1
19. The area of the quadrilateral ABCD, when A(0, 4, 1), B(2, 3, -1), C(4, 5, 0) and D(2, 6, 2) is equal
to
a. 9 sq. units b. 18 sq. units c. 27 sq. units d. 81 sq. units
20. The shaded region is the solution set of the inequalities

a. 5𝑥 + 4𝑦 ≥ 20, 𝑥 ≤ 6, 𝑦 ≥ 3, 𝑥 ≥ 0, 𝑦 ≥ 0
b. 5𝑥 + 4𝑦 ≤ 20, 𝑥 ≤ 6, 𝑦 ≤ 3, 𝑥 ≥ 0, 𝑦 ≥ 0
 c. 5𝑥 + 4𝑦 ≥ 20, 𝑥 ≤ 6, 𝑦 ≤ 3, 𝑥 ≥ 0, 𝑦 ≥ 0
d. 5𝑥 + 4𝑦 ≥ 20, 𝑥 ≥ 6, 𝑦 ≤ 3, 𝑥 ≥ 0, 𝑦 ≥ 0
3 1
21. Given that A and B are two events such that 𝑃(𝐵) = 5 , 𝑃(𝐴/𝐵) = 2 and 𝑃(𝐴 ∪ 𝐵) =
4
5
𝑡ℎ𝑒𝑛 𝑃(𝐴) =
3 1 1 3
a. 10
b. 2 c. 5 d. 5
22. If A, B and C are three independent events such that 𝑃(𝐴) = 𝑃(𝐵) = 𝑃(𝐶) = 𝑃 then 𝑃 ( at least
two of A, B, C occur)=
a. 𝑃3 − 3𝑃 b. 3𝑃 − 2𝑃2 c. 3𝑃2 − 2𝑃3 d. 3𝑃2
23. Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 the
probability of getting a sum as 3 is
1 5 1 2
a. 18
b. 18 c. 5 d. 5
24. A car manufacturing factor has two plants X and Y. plant X manufactures 70% of cars and plant Y
manufactures 30% of cars. 80% of cars at plant X and 90% of cars at plant Y are rated as
standard quality. A car is chosen at random and is found to be of standard quality. The
probability that it has come from plant X is
56 56 56 56
a. 73
b. 84 c. 83 d. 79
25. In a certain town 65% families own cellphones, 15000 families own scooter and 15% families
own both. Taking into consideration that the families own at least one of the two, the total
number of families in the town is
a. 20000 b. 30000 c. 40000 d. 50000
26. A and B are non-singleton sets and 𝑛(𝐴 × 𝐵) = 35. If 𝐵 ⊂ 𝐴 then 𝑛(𝐴)𝐶𝑛(𝐵) =
a. 28 b. 35 c. 42 d. 21
𝑥
27. Domain of 𝑓(𝑥) = 1−|𝑥|
is
a. 𝑅 − [−1, 1] b. (−∞, 1) c. (−∞, 1) ∪ (0, 1) d. 𝑅 − {−1, 1}
28. The value of cos 1200° + tan 1485° is
1 3 3 1
a. 2
b. 2 c. − 2 d. − 2
29. The value of tan 1° tan 2° tan 3° … … … … tan 89° is
1
a. 0 b. 1 c. d. -1
2
1+𝑖 𝑥
30. If (1−𝑖) = 1 then
a. 𝑥 = 4𝑛 + 1; 𝑛 ∈ 𝑁 b. 𝑥 = 2𝑛 + 1; 𝑛 ∈ 𝑁
c. 𝑥 = 2𝑛; 𝑛 ∈ 𝑁 d. 𝑥 = 4𝑛; 𝑛 ∈ 𝑁
31. The cost of revenue functions of a product are given by 𝐶(𝑥) = 20𝑥 + 4000 and 𝑅(𝑥) = 60𝑥 +
2000 respectively where 𝑥 the number of items is produced and sold. The value of 𝑥 to earn
Profit is
a. > 50 b. > 60 c. > 80 d. > 40
32. A student has to answer 10 questions, choosing at least 4 from each of the parts A and B. if
there are 6 questions in part A and 7 in part B, then the number of ways can the student choose
10 questions is
 a. 256 b. 352 c. 266 d. 426
33. If the middle term of the A.P is 300 than the sum of its first 51 terms is
a. 15300 b. 14800 c. 16500 d. 14300
34. The equation of straight line which passes through the point (𝑎 cos 𝜃 , 𝑎 sin3 𝜃) and
3

perpendicular to 𝑥 sec 𝜃 + 𝑦 cosec 𝜃 = 𝑎 is

𝑥 𝑦
a. + = 𝑎 cos 𝜃 b. 𝑥 cos 𝜃 − 𝑦 sin 𝜃 = 𝑎 cos 2𝜃
𝑎 𝑎
c. 𝑥 cos 𝜃 + 𝑦 sin 𝜃 = 𝑎 cos 2𝜃 d. 𝑥 cos 𝜃 − 𝑦 sin 𝜃 = −𝑎 cos 2𝜃
35. The main points of the sides of a triangle are (1, 5, -1) (0, 4, -2) and (2, 3, 4) then centroid of the
triangle
1 1
a. (1, 4, 3) b. (1, 4, 3) c. (-1, 4, 3) d. (3 , 2, 4)
36. Consider the following statements:
𝑎𝑥 2 +𝑏𝑥+𝑐
Statement 1 : lim 𝑐𝑥 2 +𝑏𝑥+𝑎 is 1(where a + b + c ≠ 0)
𝑥→1
1 1
+ 1
𝑥 2
Statement 2 : lim is 4
𝑥→−2 𝑥+2
a. Only statement 2 is true b. Only statement 1 is true
c. Both statements 1 and 2 are true d. Both statements 1 and 2 are false
𝑎 𝑏
37. If 𝑎 and 𝑏 are fixed non-zero constant, then the derivative of − + cos 𝑥 is 𝑚𝑎 + 𝑛𝑏 − 𝑝
𝑥4 𝑥2
where
−2 −4 2
a. 𝑚 = 4𝑥 3 ; 𝑛 = 𝑥3
; 𝑝 = sin 𝑥 b. 𝑚 = 𝑥5
;𝑛 = 𝑥 3 ; 𝑝 = sin 𝑥
−4 −2 3 2
c. 𝑚 = 𝑥5
;𝑛 = 𝑥3
; 𝑝 = − sin 𝑥 d. 𝑚 = 4𝑥 ; 𝑛 = 𝑥 3 ; 𝑝 = −sin 𝑥
38. The standard deviation of the numbers 31, 32, 33……….. 46, 47 is
17 472 −1
a. √ b. √ c. 2√6 d. 4√3
12 12
39. If 𝑃(𝐴) = 0.59, 𝑃(𝐵) = 0.30 and 𝑃(𝐴 ∩ 𝐵) = 0.21 then 𝑃(𝐴′ ∩ 𝐵′) =
a. 0.11 b. 0.38 c. 0.32 d. 0.35
2𝑥; 𝑥 > 3
40. 𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = {𝑥 2 ; 1 < 𝑥 ≤ 3 then
3𝑥; 𝑥 ≤ 1
a. 14 b. 9 c. 5 d. 11
2𝑥
41. Let 𝐴 = {𝑥: 𝑥 ∈ 𝑅, 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟} Define 𝑓: 𝐴 → 𝑅 as 𝑓(𝑥) = 𝑥−1, then 𝑓 is
a. Injective but not surjective
b. surjective but not Injective
c. bijective
d. neither injective nor surjective
42. the function 𝑓(𝑥) = √3 sin 2𝑥 − cos 2𝑥 + 4 is one-one in the interval
−𝜋 𝜋 𝜋 −𝜋 −𝜋 𝜋 −𝜋 −𝜋
a. [ , ] b. [ , ] c. [ , ] d. [ , ]
6 3 6 3 2 2 6 3
1
43. Domain of the function 𝑓(𝑥) = where [𝑥] is greatest integer ≤ 𝑥 is
√[𝑥]2 −[𝑥]−6
a. (−∞, −2) ∪ [4, ∞] b. (−∞, −2) ∪ [3, ∞]
b. [−∞, −2] ∪ [4, ∞] d. [−∞, −2] ∪ (3, ∞)
 𝜋
44. cos [cot −1 (−√3) + ] =
6
1
a. 0 b. 1 c. d. -1
√2
1 5𝜋 √3
45. tan−1 [ sin 2 ] sin−1 [cos (sin−1 2 )] =
√3
𝜋 𝜋
a. 0 b. 6 c. 3 d. 𝜋
2 1
1 −2 1
46. If 𝐴 = [ ] 𝐵 = [3 2] then (𝐴𝐵)′ is equal to
2 1 3
1 1
−3 −2 −3 10 −3 7 −3 7
a. [ ] b. [ ] c. [ ] d. [ ]
10 7 −2 7 10 2 10 −2
47. Let 𝑀 be 2 × 2 symmetric matrix with integer entries, then 𝑀 is invertible if
a. The first column of 𝑀 is the transpose of second row of 𝑀
b. The second row of 𝑀 is the transpose of first column of 𝑀
c. 𝑀 is a diagonal matrix with non-zero entries in the principal diagonal
d. The product of entries in the principal diagonal of 𝑀 is the product of entries in the other
diagonal.
48. If 𝐴 and 𝐵 are matrices of order 3 and |𝐴| = 5, |𝐵| = 3 then |3𝐴𝐵| is
a. 425 b. 405 c. 565 d. 585
49. If 𝐴 and 𝐵 are invertible matrices then which of the following is not correct?
a. 𝑎𝑑𝑗 𝐴 = |𝐴|𝐴−1 b. det(𝐴−1 ) = [𝑑𝑒𝑡(𝐴)]−1
b. (𝐴𝐵)−1 = 𝐵−1 𝐴−1 d. (𝐴 + 𝐵)1 = 𝐵−1 + 𝐴−1
cos 𝑥 1 0
50. If 𝑓(𝑥) = | 0 2 cos 𝑥 3 | then lim 𝑓(𝑥) =
𝑥→𝑘
0 1 2 cos 𝑥
a. -1 b. 1 c. 0 d. 3
1 2 3
51. If 𝑥 3 − 2𝑥 2 − 9𝑥 + 18 = 0 and 𝐴 = |4 𝑥 6| then the maximum value of 𝐴 is
7 8 9
a. 96 b. 36 c. 24 d. 120
3
52. At 𝑥 = 1, the function 𝑓(𝑥) = {𝑥 − 1 1 < 𝑥 < ∞ is
𝑥−1 ∞<𝑥 ≥1
a. Continuous and differentiable
b. Continuous and non-differentiable
c. Discontinuous and differentiable
d. Discontinuous and non-differentiable
𝑑𝑦
53. If 𝑦 = (cos 𝑥 2 )2 , then 𝑑𝑥 is equal to
a. −4𝑥 sin 2𝑥 2 b. −𝑥 sin 𝑥 2 c. −2𝑥 sin 2𝑥 2 d. −𝑥 cos 2𝑥 2
54. For constant 𝑎, (𝑥 𝑥 + 𝑥 𝑎 + 𝑎 𝑥 + 𝑎𝑎 ) is
a. 𝑥 𝑥 (1 + log 𝑥) + 𝑎𝑥 𝑎−1
b. 𝑥 𝑥 (1 + log 𝑥) + 𝑎𝑥 𝑎−1 + 𝑎 𝑥 log 𝑎
c. 𝑥 𝑥 (1 + log 𝑥) + 𝑎𝑎 (1 + log 𝑥)
d. 𝑥 𝑥 (1 + log 𝑥) + 𝑎𝑎 (1 + log 𝑥) + 𝑎𝑥 𝑎−1
55. Consider the following statements:
 𝑑𝑦 log10 𝑒 1
Statement 1: if 𝑦 = log10 𝑥 + log 𝑒 𝑥 then = +
𝑑𝑥 𝑥 𝑥
𝑑 log 𝑥 𝑑 log 𝑥
Statement 2: 𝑑𝑥 (log10 𝑥) = log 10 and 𝑑𝑥
(log 𝑒 𝑥) =
log 𝑒
a. Statement 1 is true; statement 2 is false
b. Statement 1 is false; statement 2 is true
c. Both statement 1 and 2 are true
d. Both statement 1 and 2 are false
𝜃
56. If the parametric equation of a curve is given by 𝑥 = cos 𝜃 + log tan 2 and 𝑦 = sin 𝜃, then the
𝑑𝑦
points for which 𝑑𝑥 = 0 are given by
𝑛𝜋 𝜋
a. 𝜃 = 2
,𝑛 ∈𝑧 b. 𝜃 = (2𝑛 + 1) 2 , 𝑛 ∈ 𝑧
c. 𝜃 = (2𝑛 + 1)𝜋, 𝑛 ∈ 𝑧 d. 𝜃 = 𝑛𝜋, 𝑛 ∈ 𝑧
𝑑𝑦
57. If 𝑦 = (𝑥 − 1)2 (𝑥 − 2)3 (𝑥 − 3)5 then 𝑑𝑥 at 𝑥 = 4 is equal to
a. 108 b. 54 c. 36 d. 516
𝑡2 𝑡
58. A particle starts from rest and its angular displacement (in radians) is given by 𝜃 = + . If the
20 5
angular velocity at the end of 𝑡 = 4 is 𝑘, then the value of 5𝑘 is
a. 0.6 b. 5 c. 5𝑘 d. 3
2 3
59. If the parabola 𝑦 = 𝑎𝑥 + 6𝑥 + 𝛽 passes through the point (0, 2) and has its tangent at 𝑥 =
2
parallel to 𝑥 axis, then
a. 𝛼 = 2, 𝛽 = −2 b. 𝛼 = −2, 𝛽 = 2 c. 𝛼 = 2, 𝛽 = 2 d. 𝛼 = −2, 𝛽 = −2
2
60. The function 𝑓(𝑥) = 𝑥 − 2𝑥 is strictly decreasing in the interval
a. (−∞, 1) b. (1, ∞) c. 𝑅 d. (−∞, ∞)

Biology

1. Identify the correct statement's from the following :

I. Cuscuta is a chlorophyllous endoparasite.
II. The human liverfluke needs only one host to complete its life cycle.
III. The life cycle of endoparasite is more complex due to their extreme specialisation.
IV. During the course of evolution the host bird's eggs have evolved to resemble the eggs of
the parasitic bird.
a. I, II, III b. II, IV c. Only III d. I, III and IV
2. Relate Column I with Column II with regard to predatory behaviour.
Column I Column II
(a) Calotropis (i) Invertebrates
(b) Pisaster (ii) Distasteful
(c) Monarch butterfly (iii) Cryptically coloured
(d) Frogs (iv) Cardioglycoside

a. (a)-(iv); (b)-(i); (c)- (iii); (d)-(ii) b. (a)-(iv); (b)-(i); (c)-(ii); (d)-(iii)

b. (a)- (ii); (b)-(iv); (c)-(i): (d)-(iii) d. (a)-(iii); (b)-(i): (c)-(ii); (d)-(iv)
3. Small mammals and birds are rarely found in polar regions. The reason is that
a. they have a larger surface area relative to their volume.
b. they tend to gain heat very fast.
c. they expend less energy to generate body heat.
d. None of the above.
4. Identify the incorrect statement.
a. CAM plants close their stomata during daytime.
b. Seals have a thick layer of fat to reduce body heat.
c. Lizards bask in the sun during winter.
d. Tribes living in high altitude have the same RBC count as people living in the plains.
5. Population size keeps changing depending on different factor/s such as
a. food availability. b. adverse weather.
c. predation pressure. d. All of the above.
6. Identify the incorrect statement.
I. Speciation is generally a function of time.
II. Tropical environment is less seasonal, relatively more constant and predictable.
III. Solar energy contributes to high productivity.
IV. Temperate regions have remained relatively undisturbed for millions of years.
a. I, III, IV b. II, III c. Only IV d. III, IV
7. The correct equation depicting species-area relationship is
a. logS=logC+Zlog A. b. logC=logS+ Zlog A.
c. log A= log C+ZlogS. d. logZ=log C+ Slog A.
8. Match Column I and Column I.
Column I Column II
(a) Narrowly utilitarian argument (i) Conserving biodiversity for major
ecosystem services
(b) Broadly utilitarian argument (ii) Every species has an intrinsic value
and moral duty to pass our
biological legacy in good order to
future generation.
(c) Ethical argument (iii) Receiving benefits like food,
medicine and industrial products.

a. (a)-(i); (b)-(ii); (c)-(iii) b. (a)-(ii); (b)-(iii); (c) - (i)

c. (a)- (iii); (b)-(i). (c)-(ii) d. (a)-(iii), (b) - (ii), (c)-(i)
9. Identify the correct statement/s about er
Advanced ex situ conservation includes
I. cryopreservation of gametes.
II. plant tissue culture method.
III. seed bank.
IV. in vitro fertilisation.
a. Only II b. I and II c. I, II, III, IV d. None of the above
10. The concept of "Contagium vivum fluidum" was given by
a. D. J. Ivanowsky. b. M.W. Beijerinek.
 c. W. M. Stanley. d. R. H. Whittaker.
11. Identify the odd one out.
a. Ustilago b. Alternaria
c. Colletotrichum d. Trichoderma
12. The plant body having holdfast, stipe and frond is a characteris- tic of
a. Laminaria. b. Volvox. c. Gelidium. d. Porphyra.
13. Identify the correct statement/s regarding class aves.
I. Forelimbs are modified into wings and hindlimbs are modified for walking and swimming.
II. Heart is completely four-chambered.
III. They are homeotherms.
IV. They are oviparous and development is direct.
a. Both I and III b. Both I and IV c. I, II and III d. All are correct
14. Epigynous flower is one in which
a. ovary is superior and other floral parts are inferior.
b. ovary is inferior and other floral parts are superior.
c. all the floral parts are at the same level.
d. None of the above.
15. The following type of cell junction is not found in animal tissues.
a. Adhering junction b. Tight junction
c. Gap junction d. Loose junction
16. A bacterial flagellum is composed of
a. filament, hook and basal body.
b. vesicles, tubules and lamellae.
c. pili. Fimbriae and filament.
d. hook, tubules and Fimbriac.
17. Match the compounds of Column I with their functions in Column II
Column I Column II
(a) Trypsin (i) Fights infectious agents
(b) GLUT - 4 (ii) Is an intercellular ground substance
(c) Collagen (iii) Works as an enzyme
(d) Antibody (iv) Enables glucose transport into cells

a. (a) – (iv); (b) – (iii); (c) – (ii); (d) – (i) b. (a) – (iii); (b) – (iv); (c) – (i); (d) – (ii)
b. (a) – (iv); (b) – (iii); (c) – (i); (d) – (ii) d. (a) – (iii); (b) – (iv); (c) – (ii); (d) – (i)
18. The correct sequence of events in prophase I is
a. Synapsis → Crossing over → Chiasmata → Terminalisation
b. Crossing over→ Synapsis Chiasmata→ Terminalisation
c. Chiasmata → Synapsis → Crossing over → Terminalisation.
d. Chiasmata → Crossing over → Synapsis → Terminalisation.
19. The enzyme that is not found is C, plants is
a. ATP synthase b. RUBP carboxylase
c. NADP reductase d. PEP carboxylase
20. Match the location of the cell given in Column I with its function in Column II.
Column I Column II
 (a) Mitochondrial matrix (i) Krebs cycle
(b) Cytoplasm (ii) ETC
(c) 𝐹0 and 𝐹1 (iii) Glycolysis
(d) Inner mitochondrial membrane (iv) ATP synthesis

a. (a)-(i); (b)-(iii); (c) - (iv); (a)-(ii) b. (a)-(ii); (b)-(iv); (c)-(ii); (d) - (iii)
c. (a)-(iii); (b)(ii); (c)-(i); (d)-(iv) d. (a)-(iv); (b)-(i); (c)-(iii); (d)-(ii)
21. Identify the incorrect statement/s.
I. Kinetin is a derivative of Adenine which is a pyrimidine.
II. The technique of decapitation is widely used in tea plantations.
III. Ethylene is a gaseous plant hormone.
IV. Use of G𝐴3 hastens the malting process in brewing industry.
V. ABA is a growth promoter.
a. I. II, III, IV b. Only III c. II, III, IV d. I and IV
22. Calculate the cardiac output of an individual having 70 heart beats/min with a stroke volume of
55 ml.
a. 3750 ml b. 125 ml c. 3850 ml d. None of the above
23. In a standard ECG, one of the following functions of its compo nents is not correctly interpreted.
a. P is the contraction of only left atria
b. QRS complex represents ventrical contraction.
c. T is the end of systole
d. P is the contraction of both atria.
24. Match the hormones of Column I with its functions in Column II.
Column I Column II
(a) Catecholamines (i) Diurnal rhythm
(b) MSH (ii) Immune response
(c) Thymosins (iii) Pigmentation
(d) Melatonin (iv) Stress hormone

a. (a)-(iv); (b)-(iii); (c) - (ii); (d) - (i) b. (a)-(iii); (b)-(ii); (c) – (iv); (d) – (i)
c. (a)-(ii); (b)-(iv); (c) – (iii); (d) – (i) d. (a)-(i); (b)(ii); (c) – (iii); (d) – (iv)
25. How many microsporangia are located at the corners of a typi- cal bilobed anther of
angiosperm?
a. 2 b. 4 3. 8 d. 1
26. In Bryophytes and Pteridophytes the number of male gametes produced is several thousand
times the number of female gametes produced.
Reason: Large number of male gametes fail to reach the female gametes during transport.
a. Assertion is correct but reason in incorrect.
b. Both Assertion and reason are correct.
c. Assertion is incorrect, but reason is correct.
d. Both Assertion and reason are incorrect.
27. In the given diagram identify the parts labelled as a, b, c and d.
 a. a→ Coleoptile, b→ Scutellum, c→ Pericarp, d→ Coleorhiza
b. a → Coleoptile, b→ Scutellum, c→ Coleorhiza, d→ Pericarp
c. a→ Pericarp, b→ Coleorhiza, e→ Scutellum, d → Coleoptile
d. a→ Coleorhiza. b→ Coleoptile, c→ Scutellum, d→ Pericarp
28. Consider the following statements and choose the correct answer from the given options.
Statement 1: Innermost layer of microsporangium is tapetum.
Statement 2: Cells of tapetum possess dense cytoplasm more than one nucleus and nourishes
developing pollen grains.
a. Both Statements 1 and 2 are incorrect.
b. Both Statements 1 and 2 are correct.
c. Statement I is correct and 2 is incorrect.
d. Statement 2 is correct and I is incorrect.
29. Identify the correct statement.
a. Only one megaspore present towards chalazal end remains functional.
b. 3 megaspores present towards chalazal end degenerate gradually.
c. Each megaspore mother cell directly develops into a megaspore.
d. Each female gametophyte is 7-celled and 7-nucleated structure.
30. Which of the following aquatic plant does not show pollination by water?
a. Vallisneria b. Hydrilla c. Water hyacinth d. Zostera
31. Which cell of the female gametophyte is involved in the forma- tion of primary endosperm
nucleus (PEN) after fertilization?
a. Antipodals b. Synergids c. Egg cell d.Central Cell
32. In the given diagram of human sperm, identify the functions of the labelled parts. a, b and c.
 a. a→ Heipe u petietration of sperm into ovum.
b→ Helps in movement of sperm
c→ Provides energy for the movement of sperms into the female reproductive tract.
b. a→ Helps in penetration of sperm into ovum.
b→ Provides energy for the movement of sperm.
c→ Helps in movement of sperm.
c. a→ Helps in movement of sperm.
b→ Helps in penetration of sperm into ovum.
c→ Provides energy for the movement of sperms.
d. a→ Provides energy for the movement of sperm.
b→ Helps in movement of sperm.
c→Helps in penetration of sperm into ovum.
33. Select the correct path of flow of milk during breast feeding.
a. Mammary tubules Mammary duct → Mammary ampulla→ Lactiferous duct→ Alveoli
b. Mammary tubules → Mammary duct → Lactiferous duct→Mammary ampulla → Alveoli
c. Alveoli →Mammary tubules →Mammary ampulla →Mammary duct → Lactiferous duct
d. Alveoli Mammary tubules → Mammary duct → Mammary ampulla→ Lactiferous duct
34. Under the influence of oxytocin which layer of the uterus exhibits strong contractions during
parturition?
a. Endometrium b. Myometrium c. Perimetrium d. Both (1) and (3)
35. Select the incorrect statement about contraceptives.
a. They are regular requirements for the maintenance of reproductive health
b. They have a significant role in checking uncontrolled growth of population
c. They are practised against a natural reproductive events like conception or pregnancy The
possible ill-effects like nausea, abdominal pain,
d. irregular menstrual bleeding or even breast cancer should not be totally ignored
36. The method of directly injecting a sperm into ovum is called
a. GIFT b. ZIFT c. ICSI. d. IVF-ET
37. Match Column I with Column II and find the correct answer.
Column I Column II
(a) Aneuploidy (i) Increase in whole set of chromo- somes
(b) Monoploidy (ii) Loss or gain of a chromosome
 (c) Polyploidy (iii) Two sets of chromosomes
(d) Diploidy (iv) A single set of chromosomes

a. (a)-(i); (b)-(ii); (c)-(iii); (d) - (iv) b. (a)-(iii); (b)- (i); (c)- (ii); (d)-(iv)
c. (a) –(ii); (b)-(iv); (c)-(i);(d)-(iii) d. (a)-(iv); (b) (iii); (c)-(i), (d)-(ii)
38. The genotype of a husband and wife are 𝐼 𝐴 𝐼 𝐵 and 𝐼 𝐴 𝐼 0 Among the blood types of their children,
A - B how many different genotypes and phenotypes are passible?
a. 3 genotypes; 3 phenotypes
b. 4 genotypes, 3 phenotypes
c. 4 phenotypes; 3 genotypes
d. 4 phenotypes; 4 genotypes
39. What is the possible blood group of children whose parents are heterozygous for A and B blood
groups?
a. A. B only b. A. B. A B and O
c. AB only d. A. B and AB only
40. Match the Column I with Column II.
Column I Column II
(a) Autosomal trisomy (i) Turner's Syndrome
(b) Allosomal trisomy (ii) Mendelian disorder
(c) Allosomal Monosomy (iii) Klinefelter's Syndrome
(d) Cystic fibrosis (iv) Down's. Syndrome

a. (a)-(i); (b)-(ii); (c)-(iii); (d) -(iv) b. (a)-(i); (b)-(ii); (c)(iv); (d)-(iii)

b. (a)-(iv); (b)(iii):(c)(ii); (d)-(i) d. (a)-(iv); (b)-(iii) (c)-(i); (d) - (ii)
41. Which among the following characters selected by Mendel in a pea plant is a recessive
character?
a. Inflated (full) pod b. Green pod colour
c. White flower d. Axillary flower
42. Match the scientists of Column 1 with their contributions in Column II.
Column I Column II
(a) Griffith (i) Lac operon
(b) Jacob and Monad (ii) DNA is the genetic material
(c) Meseison and stahl (iii) Transforming principle
(d) Harshey and Chase (iv) DNA replicates semi - conservatively

a. (a) – (i); (b) - (ii); (c) – (iii) ; (d) – (iv) b. (a) – (i); (b) (iv); (c) – (ii) ; (d) – (iii)
b. (a) – (iii); (b) (i); (c) – (iv) ; (d) – (ii) d. (a) – (iii); (b) (ii); (c) – (i) ; (d) – (iv)
43. In which region of the t-RNA molecule is the amino-acid binding site located?
a. 5' end b. anticodon loop c. 3' end d. None of the above
15 14
44. E. Coli fully labelled 𝑁 with is allowed to grow in 𝑁 medium The two strands of DNA
molecule of the first-generation bacteria have
a. same density and resemble with their parent DNA.
b. same density but do not resemble with their parent DNA
c. different density but do not reseple with their parent DNA
 d. different density but resemble with their parent DNA
45. Experiments involving use of radioactive thymidine to detect distribution of newly synthesized
DNA in the chromosome was performed us which plant?
a. Vicia faba b. Pisum sativum
c. Coccus nucifera Antirrhinum
46. If the sequend of nucleotides in a templates stand of DNA is 3’-ATGCTTCCGGAAT-5’. Write the
sequence in the coresponding region of the transcribed m-RNA.
a. 5’-TAC GAA GGC CTT-3′ b. 5’-UAC GAA GGC UUA-3'
b. 3’-UAC GAA GGC UUA-5' d. 3’-TACGAA GGC CTT-5
47. Pneumonia is caused by
a.Streptococcus pneumonia. b. Haemophilus influenzae.
c. Both (1) and (2) d. None
48. The development of quick immune response in a person infected with deadly microbes by
administering preformed anti- bodies is
a. active immunity. b. cell-mediated immunity.
c. innate immunity. d. passive immunization.
49. Which is the most feared property of malignant tumor?
a. Neoplasty b. Metastasis
b. Rapid invasive growth d. Loss of contact inhibition
50. Identify the techniques useful in detecting the cancers of internal organs.
a. CT b. MRI c. Radiography d. All of the above
51. Which among the following plants is a source of drug which is native to America?
a. Papaver Somniferum b. Erythroxylum coca
b. Cannabis sativa d. Atropa belladonna
52. The technology of biogas production was developed in India due to the efforts of
a. KVIC b. IARI c. CDRI d. Both (1) and (2)
53. Which among the following products of microbes is not obtained from fungi?
a. Penicillin b. Statins c. Swiss cheese d. Cyclosporin - A
54. Match the following.
Column I Column II
(a) Cyclosporin-A (i) Clot busters
(b) Streptokinase (ii) Antibiotic
(c) Statins (iii) Immuno suppressive agent
(d) Penicillin (iv) Blood cholesterol lowering agent

a. (a)- (iii); (b)-(i); (c)(iv); (d)-(ii) b. (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)

b. (a)-(i); (b)-(ii); (c)(iii); (d)-(iv) d. (a)-(i); (b)-(ii); (c)-(iv); (d)-(iii)
55. Taq polymerase that finds its application in PCR is obtained from
a. Thermus aquaticus. b. Agrobacterium tumifaciens.
b. Bacillus thuringiensis. d. Salmonella typhimurium.
56. Rop-gene which codes for the proteins involved in the replica- tion of the plasmid pBR322 in E.
coli is located at restriction site of
a. HindIII. d. EcoRI. c. Pvu II. d. BamHI
57. Rapid antigen test and RT-PCR are the two-diagnosis test for Covid-19 virus. PCR, a molecular
diagnostic tool, stands for
a. Polymerase chain reaction. b. Polymerase chain reagent.
b. Physiological chain reaction. d. Physiological chain reagent.
58. Which of the following diagnostic tools allows the detection of very low concentration of
bacterium or viruses by amplifying their nucleic acid?
a. ELISA b. PCR c. Autoradiography d. r-DNA technology
59. Silencing of a gene could be achieved through the use of
a. short interfering RNA (RNAi). b. antisense RNA.
b. by both A and B d. None of the above.
60. 𝛼-1 antitrypsin is
a. an antacid b. an enzyme
c. used to treat emphysema d. used to treat arthritis.

EXPLENATIONS
PHYSICS
1. (D) : work energy and power
Self-inductance and mutual inductance have same unit both measure magnetic flux
per unit current. The magnetic flux ha unit 𝑊𝑏.
𝜙 𝑊𝑏
𝐿=𝑀= =
𝐴 𝐴
2. (b) :
This is a 𝑅𝐿𝐶 circuit, the total impedance is calculated as,
𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2
= √(45Ω)2 − (4Ω − 4Ω)2 = 45Ω
The current in the circuit is expressed as,
𝐸 90𝑉
𝐼= = = 2𝐴
𝑍 45Ω
The current is 2 A.
The potential is calculated as,
𝑉 = (𝑋𝐿 − 𝑋𝐶 )𝐼
= (4Ω − 4Ω) − 2𝐴 = 0𝑉
 𝑡ℎ𝑒 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑖𝑠 0 𝑉.
3. (c) :
At resonance the 𝐿𝐶 circuit is expressed as,
𝑑2 𝐼 𝑙
𝐿 2+ 𝐼=0
𝑑𝑡 𝐶
The spring block system is expressed as,
𝑑2 𝑥
𝑚 2 + 𝑘𝑥 = 0
𝑑𝑥
Comparing both equations, we get
1
𝑚 ∝ 𝐿, 𝑘 ∝
𝐶
4. (b) ;
The change as a function of time is expressed as,
𝑞 = 𝑄 sin 𝜔𝑡
Here, Q is the maximum charge on the capacitor and 𝜔 is the angular frequency of
oscillation.
At time 𝑡 = 0, the charge is 𝑞 = 0.
We know that current is expressed as,
𝑑𝑞 𝑑(𝑄 sin 𝜔𝑡)
𝐼= = = 𝜔𝑄 cos 𝜔𝑡
𝑑𝑡 𝑑𝑡
So, substituting 𝑡 = 0,
𝐼0 = 𝜔𝑄 cos(0) = 𝜔𝑄
𝐼0
⇒ 𝑄 = 𝐼0 √𝐿𝐶
𝜔
= (2𝐴)√(3 × 10−3 𝐻)(2.70 × 10−6 𝐹)
= 1.8 × 10−4 𝐶 = 18 × 10−5 𝐶
5. (c) :
The magnetic field amplitude is expressed as,
𝐸0 120𝑁/𝐶
𝐵0 = = = 400𝑛𝑇
𝑐 3 × 108 𝑚/𝑠
The angular frequency is expressed as,
𝜔 = 2𝜋𝑓 = 2𝜋(50𝑀𝐻𝑧) = 3.14 × 108 𝑟𝑎𝑑𝑠 −1 = 𝜋 × 108 𝑟𝑎𝑑𝑠 −1
The propagation constant is expressed as,
𝜔
𝑘 = = 1.05𝑟𝑎𝑑𝑚−1
𝑐
The wavelength is given as,
𝑐
𝜆 = = 6𝑚
𝑓
6. (b) :
Motion of a particle in circular orbit is accelerated motion, thus source of
electromagnetic wave can be a charge moving in the circular orbit.
7. (b) :
 When a light wave passes from one medium to other, Refractive index charges due to
which its velocity and wave-length charges.
𝑐 𝑐
𝑛= ⇒𝑣=
𝑣 𝑛
𝑐
𝜆 = 𝑣𝑓 =
𝑛
So, charge in light wave velocity and wavelength, wave bents.
8. (d) :
The refractive index from air to glass is,
3
𝑎𝜇𝑔 =
2
The refractive index from air to water is,
4
𝑎𝜇𝑔 =
3
The refractive index from water to glass is,
𝑎𝜇𝑔 3/2 9
𝑤𝜇𝑔 = = =
𝑎𝜇𝑤 4/3 8
The expression for the focal length for water glass medium is,
1 1 1
= (𝑤𝜇𝑔 − 1) ( − )
𝑓𝑤 𝑅1 𝑅2
The expression for the focal length for air, in air-glass medium is
1 1 1
= (𝑎𝜇𝑔 − 1) ( − )
𝑓𝑎 𝑅1 𝑅2
the required ratio is,
𝑓𝑤 (𝑎𝜇𝑔 − 1) 3/2 − 1 4
= = =
𝑓𝑎 (𝑤 − 1) 9/8 − 1 1
𝜇𝑔
9. (d) :
There are three arrangements of the thin biconvex lens. For the first case, two biconvex
lenses the power of the combination is,
1 1 𝑓1 + 𝑓2
𝑃1 = + =
𝑓1 𝑓2 𝑓1 𝑓2
For the second case, first biconvex lens is in contact with the third biconcave lens, the
power of the combination is,
1 1 𝑓3 − 𝑓1
𝑃2 = + =
𝑓1 𝑓3 𝑓1 𝑓3
for the third case, second biconvex lens is in contact with the third biconcave lens, the
power of combination is,
1 1 𝑓3 − 𝑓2
𝑃3 = + =
𝑓2 𝑓3 𝑓3 𝑓2
10. (b) :
The lens formula is,
1 1 1
= −
𝑓 𝑣 𝑢
 1 1 1
⇒ = −
−20𝑐𝑚 𝑣 4𝑐𝑚
⇒ 𝑣 = 5𝑐𝑚
The magnification is given as,
𝑣 5
𝑚= = = 1.25
𝑢 4
The new size of the image is,
ℎ𝑖′
= 1.25
ℎ𝑖′
ℎ𝑖′ = (1.25)(2𝑐𝑚) = 2.5𝑐𝑚
11. (b) :
The principle of diffraction says,
𝑎 sin 𝜃 = 𝑛𝜆
Where 𝑎 is slit width, 𝑛 is order of secondary maxima, and 𝜆 is wavelength.
Given that, 𝜃 = 30°, 𝑛 = 1 and 𝜆 = 650𝑛𝑚. Then, 𝑎 sin 30° = 1(6500 × 10−10 𝑚)
(6500×10−10 𝑚)
⇒𝑎= = 1.3 × 10−10 𝑚 = 1.3𝜇𝑚 = 1.3 micron
sin 30°
12. (a) :
The single slit diffraction pattern is observed when a coherent light is passed through
one slit and the slit size is comparable to wavelength of light. Intensity of the central
fringe is maximum and then it decreases. The fringe pattern is of unequal width and
unequal intensity. The light energy is conserved in ideal case as there is no loss due to
heat or other medium issues.
13. (b):
The expression for the resultant intensity is,
∆𝜙
𝐼𝑅 = 4𝐼0 cos2 ( )
2
Path difference is given by,
𝑥
∆𝑥 = 𝑑
𝐷
The phase difference is given by,
2𝜋∆𝑥
∆𝜙 =
𝜆
𝜙 𝜋∆𝑥 𝑥
= = 𝜋𝑑
2 𝜆 𝜆𝐷
The resultant intensity will be,
𝑥
𝐼𝑅 = 4𝐼0 cos2 (𝜋𝑑 )
𝜆𝐷
14. (d) :
The Einstein photoelectric equation is expressed as,
𝐸 = work function +𝐾𝑚𝑎𝑥
⇒ ℎ𝑓 = 𝜙 + 𝐾𝑚𝑎𝑥
ℎ𝑐 1 2
⇒ = 𝜙 + 𝑚𝑣𝑚𝑎𝑥
𝜆 2
Substitute the given value, we have
 (6.6 × 10−34 𝐽 ∙ 𝑠)(3 × 108 𝑚/𝑠) 1
−10
= (1 × 1.6 × 10−19 𝐽) + (9.1 × 10−31 𝑘𝑔)𝑣 2
(3000 × 10 𝑚) 2
6
⇒ 𝑣 = 1 × 10 𝑚/𝑠
15. (d):
Given that, the kinetic energy of proton is 1/8th of its rest mass energy. That is,
𝑝12 1
𝐾= = 𝑚𝑐 2
2𝑚 8
2 𝑚𝑐
⇒ 𝑝1 = √ 𝑚2 𝑐 2 =
8 2
For second case, we have the kinetic energy of photon is equal to its rest mass energy.
That is,
𝐾 = 𝑝2 𝑐 − 𝑚𝑐 2 = 𝑚𝑐 2
⇒ 𝑝2 𝑐 = 2 𝑚𝑐 2 ⇒ 𝑝2 = 2𝑚𝑐
𝑝 −𝑝
Therefore, the ratio will be 1𝑝 2,
2
𝑝1 − 𝑝2 (𝑚𝑐/2) − 2𝑚𝑐 3
= =
𝑝1 𝑚𝑐/2 4
16. (d) :
The Einstein photoelectric equation is expressed as,
𝐸 =work function +𝐾𝑚𝑎𝑥
⇒ ℎ𝑓 = 𝜙 + 𝐾𝑚𝑎𝑥
⇒ 𝐾𝑚𝑎𝑥 = −𝜙 + ℎ𝑓
The final equation resembles,
𝑦 = 𝑚𝑥 + 𝑐
Here, 𝑦 = 𝐾𝑚𝑎𝑥 , 𝑚 = ℎ, 𝑥 = 𝑓 and 𝐶 = −𝜙.
So, the correct graph between kinetic energy and frequency is

17. (b) :
The energy of an electron in the orbit of the atom is expressed as,
𝑘𝑍 2
𝐸𝑛 = 2
𝑛
Here, 𝑍 iz atomic number, 𝑛 is the orbit in which electron is present, and 𝑘 is constant.
So, the energy of second orbit of hydrogen atom is,
 𝑘(1)2 𝑘
𝐸1 = 𝐸 = =
22 4
Energy of electron in third orbit of 𝐻𝑒 + is,
𝑘(2)2 4𝑘 𝑘 16 16 16𝐸
𝐸3 = 2
= = ( ) = 𝐸( )=
3 9 4 9 9 9
18. (*):
According to Bohr’s second postulate,
𝑛ℎ
𝑚𝑣𝑟 = 2𝜋 (1)
Where 𝑚 be the mass of an electron, 𝑣 be the velocity and 𝑟 be the radius of orbit in
which the electron revolves around the nucleus.
The velocity of an electron is expressed as,
𝑛ℎ
𝑣= (2)
2𝜋𝑚𝑟
Here, the centripetal force is provided by the electrostatic force of attraction between
electron and nucleus.
𝑚𝑣 2 1 𝑒2
=
𝑟 4𝜋𝜀0 𝑟 2
1 𝑒2
⇒ 𝑟 = 4𝜋𝜀 2 (3)
0 𝑚𝑣
From Eqs. (2) and (3), we get
1 𝑒2 4𝜋 2 𝑚2 𝑟 2 𝑒 2
𝑟= =
4𝜋𝜀0 𝑛ℎ 2 4𝜋𝜀0 𝑚𝑛2 ℎ2
𝑚 (2𝜋𝑚𝑟)
𝜀0 𝑛2 ℎ2
⇒𝑟=
𝜋𝑚𝑒 2
19. (b) :
The angular momentum is expressed as,
𝑛ℎ
𝐿=
2𝜋
For 𝑛 = 3, we have
3ℎ
𝐿=
2𝜋
The wavelength is expressed as,
ℎ ℎ ℎ𝑟
𝜆= = =
𝑝 𝑚𝑣 𝑚𝑣𝑟
The velocity of an electron is expressed as,
𝑛ℎ
𝑣=
2𝜋𝑚𝑟
For 𝑛 = 3, we have
3ℎ
𝑣=
2𝜋𝑚𝑟
ℎ𝑟 ℎ𝑟 2𝜋𝑟
So,𝜆 = 3ℎ = 3ℎ/2𝜋 = 3
𝑚𝑟( )
2𝜋𝑚𝑟
The radius is expressed as,
 𝑎0 𝑛2
𝑟=
𝑍
Substitute the value in wavelength equation,
2𝜋 𝑎0 𝑛2
𝜆= ( )
3 𝑍
For 𝑛 = 3 and Z=3, we have
2𝜋 𝑎0 (3)2
𝜆= ( ) = 2𝜋𝑎0
3 3
Comparing from the given equation,

𝑃𝜋𝑎0 = 2𝜋𝑎0
𝑃=2
20. (d):
The potential energy of the two nucleon is in the order of MeV. Thus, the graph in
options (A) and (D) are two possible options. Nuclear force is of attractive nature when
the separation is greater than 1 fm and repulsive if less than 1 fm. So, the option (A) is
not possible because point of stable equilibrium will be negative not positive.
21. (b) :
For elastic collision masses of both must be equal so that they can exchange the
velocities. To slow down the speed of neutron substance should be made up of 1 proton
for perfectly elastic, that is, we need light nuclei not heavy. In heavy nuclei only
direction will change not the speed.
22. (b) :
The given figure can redraw as following figure

Truth table of the given circuit is

𝑋1 𝑋2 𝑌1 𝑌2
0 0 1 0
0 1 1 0
1 0 1 0
1 1 0 1

The truth table shows AND logic gate operation.

23. (d) :
The width of the depletion layer decreases when diode is forward biased and if it is
reversed biased the barrier potential is increased. Photodiode is operated in reverse
bias to operate in photoconductive mode. An LED is a heavily doped P-N junction diode
which emits spontaneous radiation of forward of forward biasing.
24. (c):
 The energy of photon is expressed as,
ℎ𝑐
𝐸 = ℎ𝑓 =
𝜆
Given that, 𝜆 = 600𝑛𝑚. So,
(6.6 × 10−34 𝐽 ∙ 𝑠)(3 × 108 𝑚/𝑠 2 )
𝐸= = 3.3 × 10−19 𝐽
(6000 × 10−10 𝑚)
1 𝑒𝑉
= (3.3 × 10−19 𝐽) = 2.0625𝑒𝑉
1.6 × 10−19 𝐽
The only photodiode that will detect the light is 𝐷2 with 2 eV.
25. (b) :
In the 𝑣 − 𝑡 graph represents the displacement and from the given figure, we have the
area under uniformly accelerated motion is less than the area under the uniform motion
graph.
26. (b):
The area under acceleration time graph gives the maximum velocity of particle,
Therefore, from the given figure, we have
1 1
𝑣 = (base)(height)= (10s)(8m𝑠 −1 ) = 40m𝑠 −1
2 2
27. (c):
The horizontal range in any projectile motion is given by,
𝑢2 sin 2𝜃
𝑅=
𝑔
For maximum range, 𝜃 = 45°,
𝑢2
𝑅=
𝑔
So,
𝑢2
16000𝑚 =
10𝑚𝑠 −1
⇒ 𝑢 = 400𝑚𝑠 −1
28. (d):
The trajectory of the projectile is parabolic in nature because only force that acts on the
motion is gravitational force towards the center of the earth. The air resistance will
affect the projectile and the ideal parabolic path will be altered.
29. (a) :
The angle will be acute for whole second half of the projectile motion but will be
minimum and acute both at same time is at the point where ball just strikes the ground.
So only the point where ball strikes the surface is point of minimum and acute angle
between the velocity and acceleration.
30. (d):
The position of the particle at any given instant of time in vector form is,
1
𝑠=𝑢 ⃗ 𝑡 + 𝑎𝑡 2
2
⃗ = 10𝑗̂𝑚𝑠 −1 , 𝑎 = (8𝑖̂ + 2𝑗̂)𝑚𝑠 −1 and 𝑥 = 16m.
Here, 𝑠 = 𝑥𝑖̂ + 𝑦𝑗̂, 𝑢
 1
6𝑖̂ + 𝑦𝑗̂ = 10𝑡𝑗̂ + (8𝑖̂ + 2𝑗̂)𝑡 2
2
⇒ 6𝑖̂ + 𝑦𝑗̂ = 4𝑡 2 𝑖̂ + (10𝑡 + 𝑡 2 )𝑗̂
Compare the 𝑥 and 𝑦 components of both sides, we get
𝑥 = 4𝑡 2 ⇒ 16𝑚 = 4𝑡 2 ⇒ 𝑡 = 2𝑠
And
𝑦𝑗̂ = 10𝑡 + 𝑡 2
Substitute the value of time, we get
𝑦 = 10(2𝑠) + (2𝑠)2 = 24𝑚
31. (a):
When the coin is just on the verge of slipping, it just over-comes the force of static
friction (𝐹𝑠 ) which acts as centripetal force for being that coin in the circular motion.
So, we have,
𝐹𝑠 = 𝑚𝜔2 𝑟
If we double the 𝜔, then 𝑟, woll also self-adjust to a new 𝑟, since the static friction will
remain same.
We have,
𝑚𝜔2 𝑟 = 𝑚(2𝜔)2 𝑟′
𝑟
⇒ 𝑟 ′ = 4 cm
Here, 𝑟 = 4 cm. Then,
𝑟 4
𝑟 ′ = 4 cm= 4 cm = 1 cm
32. (b) :
The coefficient of restitution (e) is the ratio of the final to initial relative speed between
two objects after they collide. The velocity of mass 1 after collision is expressed as,
(𝑚1 − 𝑒𝑚2 )𝑢1 + (1 + 𝑒)𝑚2 𝑢2
𝑣1 =
𝑚1 + 𝑚2
2 2
(1𝑘𝑔 − 3 (2𝑘𝑔)) (12𝑚𝑠 −1 ) + (1 + 3) (2𝑘𝑔)(−24𝑚𝑠 −1 )
=
1𝑘𝑔 + 2𝑘𝑔
= −28𝑚𝑠 −1
The velocity of mass 2 after collision is expressed as,
(𝑚2 − 𝑒𝑚1 )𝑢1 + (1 + 𝑒)𝑚1 𝑢1
𝑣2 =
𝑚1 + 𝑚2
2 2
(2𝑘𝑔 − 3 (1𝑘𝑔)) (−24𝑚𝑠 −1 ) + (1 + 3) (2𝑘𝑔)(12𝑚𝑠 −1 )
=
1𝑘𝑔 + 2𝑘𝑔
= −4 𝑚𝑠 −1
33. (c) :
Momentum of the system changes only due to external force or impulse. When a ball
hits the floor and gets rebound, then no external force or impulse acts on the ball as
well as the floor, thus total momentum of the “ball + earth” system remains conserved.
 Also, some of the mechanical energy of the ball is always lost in an inelastic collision due
to its deformation; hence it cannot remain conserved in this collision.
34. (c) :
The pure rolling is motion of the round object without any slipping or skidding at the
point of contact between the two bodies. The rolling is superposition of two motion:
translation with respect to the surface, and rotation around its own axis. From the graph
the linear motion shows that the object is wheel or specifically a disc shaped.
35. (b) :
The expression for the gravitational force is,
𝐺𝑚1 𝑚2
𝐹=
𝑟2
The force due to mass 8 kg placed at vertex A,
𝐺(8𝑘𝑔)(2𝑘𝑔)
𝐹1 = = 16𝐺
1𝑚
Similarly, force due to mass 8 kg placed at vertex B,
𝐹2 = 16𝐺
Resultant force is expressed as,
𝐹12 = 2𝐹1 sin 30° = 16𝐺
The force due to 4 kg particle at centroid G,
𝐺(4𝑘𝑔)(2𝑘𝑔) 8𝐺
𝐹3 = 𝑟2
= 𝑟2
Since, the net force is zero. So,
𝐹12 + 𝐹3 = 0
8𝐺
⇒ 16𝐺 + 2 = 0
𝑟
1
⇒𝑟=
√2
36. (b):
The expression for capillary action is expressed as,
2𝑇 cos 𝜃
ℎ=
𝑟𝑑𝑔
1
⇒ℎ∝
𝑟
Given that, the height of water level in capillary tube P is 2/3 of the height in capillary
tube Q. So,
𝑟𝑃 ℎ𝑄 ℎ𝑄 3
= = = =3∶2
𝑟𝑄 ℎ𝑃 2 ℎ 2
3 𝑄
37. (c) :
We know that ideal gas expression is,
1 𝑛𝑅
𝑃𝑉 = 𝑛𝑅𝑇 ⇒ =
𝑇 𝑃𝑉
Since pressure is constant, we have
∆𝑉 𝑛𝑅
𝑃∆𝑉 = 𝑛𝑅∆𝑇 ⇒ ∆𝑇 = 𝑃
= constant
 The volumetric thermal expansion is expressed as,
∆𝑉 𝑛𝑅 1
𝛼𝑉 = = =
𝑉∆𝑇 𝑃𝑉 𝑇
So, the correct graphical representation is,

38. (b) :
The efficiency of Carnot engine working between two temperature is expressed as,
𝑇𝐿 𝑇𝐻 − 𝑇𝐿
𝜂 =1− =
𝑇𝐻 𝑇𝐻
Carnot engine will not work if (𝑇𝐻 < 𝑇𝐿 ).
At, (𝑇𝐻 < 𝑇𝐿 ), the efficiency 𝜂 will be zero.
From this point the increasing 𝑇𝐻 will only increase 𝜂.
The only graph representing such movement is,

39. (c) :
The value of degree of freedom (𝑓) for monoatomic and diatomic gases are 3 and 5,
respectively.
The charge in internal energy for the mixture is,
𝑓
𝑑𝑈 = 𝑛𝑅𝑇
2
So, the total internal energy,
𝑓 𝑓
𝑑𝑈𝑡𝑜𝑡𝑎𝑙 = ( 𝑛𝑅𝑇) + ( 𝑛𝑅𝑇)
2 𝑚𝑜𝑛𝑜 2 𝑑𝑖𝑎
3 5 8
= 𝑛𝑅𝑇 + 𝑛𝑅𝑇 = 𝑛𝑅𝑇
2 2 2
For 𝑛 = 2 moles. We get
8
𝑑𝑈𝑡𝑜𝑡𝑎𝑙 = (2𝑚𝑜𝑙)𝑅𝑇 = 8𝑅𝑇
2
40. (a) :
 The first assumption is correct because it is basic law that we take under consideration
for the ideal environment of the simple pendulum.
The second one is also correct because it is considered that angle of motion should be
negligible, 10° is one negligible angle because sine of 10° is 0.18. Any angle smaller than
will tend do zero effect in charge.
Thus, both (i) and (ii) are correct.
41. (d) :
In longitudinal waves, the vibrations are parallel to the direction of wave travel. In
transverse waves, the vibrations are at right angle to the direction of wave travel. To
possess both type of motion the material must bulk as well as shear moduli. Solid posses
both, therefore both motion is observed in solids, while gases do have longitudinal
waves.
42. (a) :
The equation of motion due to force applied is,
𝐹 = 𝑚𝑎
And electric force due to field is,
𝐹 = 𝑞𝐸 = 𝑒𝐸
The acceleration due to circular motion is,
𝑎 = 𝜔2 𝑥
So, we have
𝑒𝐸 = 𝑚𝜔2 𝑥
Therefore, electric field at the distance 𝑥 from the axis of rotation is,
𝑚𝜔2 𝑥
𝐸=
𝑒
43. (b) :
The electric field due to infinite, straight uniformly charged wire is expressed as,
𝜆
𝐸=
2𝜋𝜀0 𝑟
Where 𝜆 is linear charge density.
1
Therefore, electric is inversely proportional to the distance 𝑟, that is, 𝐸 ∝ 𝑟.
44. (c) :
The electric potential energy is expressed as,
𝑃𝐸 = 𝐸𝑄𝑥
The kinetic energy is,
1
𝐾 = 𝑚𝑣 2
2
Total energy is constant,
1
𝐸𝑄𝑥 = 𝑚𝑣 2 = 𝐾
2
𝐾
⇒𝑄=
𝐸𝑥
Given that, 𝐾 = 0.12𝐽, 𝐸⃗ = (300𝑁𝐶 −1 )𝑖 and 𝑥 = 0.5 m. Then,
 0.12𝐽
𝑄= = 800𝜇𝐶
(300𝑁𝐶 −1 )(0.5𝑚)
45. (b) :
The expression for the separation and dielectric constant between a plate,
1
𝑥 = 𝑡 (1 − )
𝑘
𝑥 1
⇒1− =
𝑡 𝑘
𝑡
⇒𝑘=
𝑡−𝑥
Given that, 𝑥 = 3.5 mm and 𝑡 = 4 mm. So,
4 × 10−3 𝑚
𝑘= = 8.0
4 × 10−3 𝑚 − 3.5 × 10−3 𝑚
46. (a) :
We know that,
Volume of 8 small drop = =Volume of big drop
4 4
8 × 𝜋𝑟 3 = 𝜋𝑅 3
3 3
⇒ 𝑅 = 2𝑟
Let 𝐶 be the capacitance of smaller drop, 𝐶′ be the capacitance of bigger drop and store
in it is 𝑞. So,
𝐶 = 4𝜋𝜀0 𝑟, 𝐶 ′ = 4𝜋𝜀0 𝑅
The ratio of capacitance for smaller drop and bigger drop is,
𝐶′ 𝑅 2𝑟
= = =2
𝐶 𝑟 𝑟
47. (c) :
Polar molecule is molecule having one end slightly positive and other end slightly
negative due to difference is electronegativity of constituent molecules. Centers of
positive and negative charges are separated with or without the presence of electric
field. Due dipole in nature polar molecule has dipole moment permanently, HCl is one
polar molecule with 𝐻 + and 𝐶𝑙 − atoms.
48. (d) :
The number of capacitors in a row,
desired voltage 1500𝑉
𝑛= = =3
voltage across each capacitor 500𝑉
Given that, desired voltage = 1.5 kV and voltage across capacitor = 500 V.
1500𝑉
𝑛= =3
500𝑉
The number of rows is expressed as,
desired capacitance 6𝜇𝐹
𝑚= = =9
total capacitance in each row 2𝜇𝐹/3
Given that, desired capacitance = 6𝜇𝐹 and total capacitance in one row =
1 2
1 1 1 = 3 𝜇𝐹.
+ +
2𝜇𝐹 2𝜇𝐹 2𝜇𝐹
 6𝜇𝐹
𝑚= =9
2𝜇𝐹/3
Therefore, total number of capacitance are,
𝑚 × 𝑛 = 9 × 3 = 27
49. (b) :
The capacitance is expressed as
𝑄
𝐶=
𝑉
120𝜇𝐶
=
10𝑉
= 12𝜇𝐹
Given that, 𝑄 = 120𝜇𝐶 and 𝑉 = 10𝑉. So,
120𝜇𝐶
𝐶= = 12𝜇𝐹
10𝑉
The energy stored in a capacitor is expressed as,
1
𝑈 = 𝑄𝑉
2
1
= (120𝜇𝐶)(10𝑉) = 600𝜇𝐽
2
50. (d) :
The resistance is expressed as,
𝜌𝑙
𝑅= 𝐴
(1)
′ ′
Given that, 𝑙 = 2𝑙 then 𝐴 = 𝐴/2. Since, the length of wire gets doubled, the cross-
sectional area will become half because volume of wires remains constant.
The new resistance is,
𝜌(2𝑙)
𝐸′ = (2)
𝐴/2
from (1)and (2), we get
𝜌(2𝑙)
𝑅′ 𝐴/2
=
𝑅 𝜌𝑙
𝐴
⇒ 𝑅 ′ = 4𝑅
Given that 𝑅 = 3Ω. So,
𝑅 ′ = 4(3Ω) = 12Ω
51. (a) :
The meter bridge works on the principle of Wheatstone bridge. That is,
𝑅1 𝑅2 𝑅1 𝑅𝐴𝐷 𝑥
= ⇒ = =
𝑅𝐴𝐷 𝑅𝐷𝐵 𝑅2 𝑅𝐷𝐵 100 − 𝑥
The balancing length is independent of area of cross section of the wire AB. Thus, there
will be no change no charge in 𝑥 due to doubling of wire cross-section AB.
52. (d) :
The drift velocity relation for electron is,
𝐼
𝑣𝑑 =
𝑛𝑒𝐴
 Here 𝐼 = 1𝐴, 𝑛 = 8 × 1028 𝑚−3 , 𝑒 = 1.6 × 10−19 𝐶 and 𝐴 = 5 × 10−7 𝑚2 . So,
1𝐴 1
𝑣𝑑 = = 𝑚/𝑠
(8 × 10 )(1.6 × 10 𝐶)(5 × 10 𝑚 ) 6.4 × 103
28 −19 −7 2

The time taken by electron take to drift one end to the other end of 1m wire is,
𝐿 1𝑚
𝑇= = = 6.4 × 103 𝑠
𝑣𝑑 ( 1
𝑚/𝑠)
6.4 × 103
53. (b) :
There is electrical conductor connected across the voltage 𝑉. The motion will be
assumed to be conserved. It means loss in potential energy is equal to the gain in kinetic
energy. So,
∆𝐾 = −∆𝑈
Therefore, for the given case there is increase in kinetic energy and decrease in
potential energy,
The potential energy is expressed as,
∆𝑈 = −𝐼𝑉∆𝑡
And ∆𝐾 = −∆𝑈 = 𝐼𝑉∆𝑡
54. (c) :
Due to strong magnetic field on a stationary electron, there will be attraction force to
both external magnetic poles, cancelling each other so the net force on the electron will
always be zero, causing the electron to remain stationary. Magnetic force act on moving
change particles.
55. (a) :
The force exerted by one wire, having current 𝐼1 , on the other parallel wire having
current 𝐼2 , per unit length is expressed as,
𝐹 𝜇0 𝐼1 𝐼2
=
𝐿 2𝜋𝑟
Where, 𝑟 is separation of wire.
Here, 𝜇0 = 4𝜋 × 10−7 𝑁𝐴−2 , 𝐼1 = 𝐼2 = 10𝐴 and 𝑟 = 10 cm. So,
𝐹 (4𝜋 × 10−7 𝑁𝐴−2 )(10𝐴)(10𝐴)
= = 2 × 10−4 𝑁𝑚−1
𝐿 2𝜋(10 × 10−2 𝑚)
The force will be attractive.
56. (d) :
A toroid is a donut-shaped coil closely wound with one continuous wire. So, the
direction of magnetic field due to inner surface is opposite to the direction of magnetic
field due to outer surface.
We have to find magnetic field in three regions I, II and III as shown in the following
figure
 The magnetic field due to inner and outer surface at region II is given by Ampere’s
circuital law, that is,
𝐵𝐼𝐼 = 𝜇0 𝑛𝐼
Where, 𝑛 number of turns per unit length and it is given by
𝑁 𝑁
𝑛= =
2𝜋𝑟 2𝜋 ( 1 + 𝑅2 )
𝑅
2
𝜇0 𝑁𝐼
Then, 𝐵𝐼𝐼 = 𝜋(𝑅 +𝑅 )
1 2
Magnetic field outside the toroid is zero, that is 𝐵𝐼 and 𝐵𝐼𝐼𝐼 = 0
So, the best graph is represents the magnetic field due to toroid is,

57. (b) :
When centripetal force is equal to magnetic force, than particle does not strike the
solenoid. That is,
𝑚𝑣 2 𝑞𝐵𝑅
= 𝑞𝑣𝐵 ⇒ 𝑣 =
𝑅 𝑚
Here 𝑅 = 𝑟/2. So,
𝐵𝑞𝑟
𝑣=
2𝑚
The magnetic field of solenoid is,
𝐵 = 𝜇0 𝑛𝐼
So, the velocity is,
 (𝜇0 𝑛𝐼) 𝜇0 𝑛𝐼𝑞𝑟
𝑣= =
2𝑚 2𝑚
58. (b) :
The horizontal components are known as latitude and vertical components are called
longitude. Equator is latitude and largest horizontal line. At the magnetic poles there is
no latitudes as it is junction of longitudes. So, at magnetic poles Earth’s magnetic field
do not have horizontal component.
59. (c) :
The option (a) is positive electric monopole while option (b) is negative. And monopoles
do not exist for magnetic field. Option (d) shows magnetic field of the magnet having
North and South poles or dipoles, but dipoles do not exist in electric field. The only
option here which is analogous to both electric magnetic and electric field is option (c),
as time varying magnetic field causes time varying electric field.
60. (*) :
The emf induced due to self-inductance is expressed as,
𝑑𝐼
𝐸 = −𝐿
𝑑𝑡
From the given graph, we have
𝐼1 = 4𝐴 at 20 s
And, 𝐼2 = 3𝐴 at 40 s
(3𝐴−4𝐴)
So, 𝐸 = −(6𝑚𝐻) (40𝑠−20𝑠) = 3.0 × 10−4 𝑉
None of the option matches.

CHEMISTRY

1. (c): In colloids, the potential difference between a fixed layer and diffused layer of the opposite
charge is called as Zeta potential.
2. (b): The decomposition reactions of the given compounds are given below:
∆
i. Pb(NO3 )2 → PbO + 2NO2 + O2
∆
ii. NH4 NO3 → N2 O + 2 H2 O
∆
iii. NH4 NO2 → N2 + 2H2 O
∆
iv. 2 NaNO3 → 2NaNO2 + O2
Hence, only NH4 NO3 will decompose to give N2 O.

3. (b, c): Down the group the electronegativity decreases and the atomic size increases, therefore,
the bond length decreases so as the thermal stability decreases in same order.
4. (b): The boiling point is depended on strength of hydrogen bond and the molecular mass.
Although F is more electronegative than O but, H2 O form four hydrogen bonds while HF form
two hydrogen bond.

5. (c): The reaction involved is

XeF6 + H2 O → XeOF4 + 2HF
The hybridization of XeOF4
[V + M + A − C] [8 + 4 + 0 − 0] 12
Hyb. = = = = 6(sp3 d2 )
2 2 2
Hence, the geometry of XeOF4 is octahedral and shape is square pyramidal.

6. (c): The reaction involved is

[o] cooling [H]
Np O → NO2 → N2 O4 → N2 O3
P Q R S
7. (c, d): (i) Paramagnetic behaviour is depend upon the number of unpaired electrons:
μspin = √n(n + 2) BM
Cr 2+ = [Ar]4s 0 3d4 : Number of unpaired electron is 4.
Mn2+ = [Ar]4s 0 3d5 : Number of unpaired electron is 5.
Fe2+ = [Ar]4s 0 3d6 : Number of unpaired electron is 4.
Therefore, the correct order of paramagnetic behaviour is:
Cr 2+ = Fe2+ < Mn2+ . Hence option (c) properties stated is incorrect.
(ii) As moving left to right in 3d series, a decrease is metallic radius is observed and increase in
the atomic mass, therefore, in same order density is decreases. Hence, the correct order of
density is: Sc < Cr < Fe
8. (d): The transition elements have unpaired electrons in +2 oxidation state, therefore, they show
paramagnetic behaviour.
9. (b): From the ideal gas, we have
 pV = nRT
The combined ideal gas law,
p1 V1 p2 V2
=
T1 T2
p2
Given, T1 = 2T2 and p1 =
2
Substituting the values in Eq. (1), we get
p2 V1 p2 V2
=
2 × 2T2 T2
⇒ V1 = 4V2
10. (d): The colour properties of transition elements are due to the unpaired electrons.
Sc 3+ = [Ar]4s 0 3d0 : Number of unpaired electron is 0.
Mn2+ = [Ar]4s 0 3d5 : Number of unpaired electron is 5.
Ni2+ = [Ar]4s 0 3d8 : Number of unpaired electron is 2.
Ti4+ = [Ar]4s 0 3d0 : Number of unpaired electron is 0.
Ti3+ = [Ar]4s 0 3dn : Number of unpaired electron is 1.
Hence, Mn2+ and Ti3+ have unpaired electrons and will show colours in aqueous solution.
11. (b): The CFSE for octahedral complexes is:
CFSE = (0.6 × eng − 0.4 × t n2g )∆o
12. (b): H – I bond is weaker than H – Br bond so that iodine free radicals combine to form iodine
molecules, hence, the peroxide effect is only observed with H – Br
13. (a): The correct IUPAC name for [Co(NH3 )5 (CO3 )]Cl is:
Pentaamminecarbonatecobalt (III) chloride.
14. (a): The compounds having only one type of ligands are known as homoleptic complexes.
15. (a): The absorption of light by a coordination complex is depend upon the strength of the ligand
complex central metal have. Higher is the crystal filed energy lower is the wavelength of
absorption.
Therefore, CN- is the strongest ligand and will absorb lowest wavelength of light.
Hence, the correct order is:
[CoCl(NH3 )5 ]2+ > [Co(NH3 )6 ]3+ > [Co(CN)6 ]3−
16. (b): The benzyl radical is more stable than alkyl radical, therefore, the substitution will be on
benzyl position. The reaction involved is

17. (b): Let the reaction is

A2 + B2 → 2AB
The given bond enthalpy ratio is:
x
A2 : B2 : AB ⇒ x: : x
2
 Now,
1 1
∆f H = A2 + B2 − AB
2 2
x x
⇒ −100 kJmol−1 = + − x [Given, ∆f H = −100 kJmol−1 ]
2 4
2x + x − 4x
⇒ −100 kJmol−1 =
4
⇒ −400 kJmol−1 = −x
Therefore, the bond enthalpy of B2 is:
400
∆bond H = = 200 kJmol−1
2
18. (a): In SN 2 mechanism, the substrate should have less hindrance. Hence, the correct order of
reactivity towards SN 2 mechanism is:
CH3 H H H
| | | |
C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br < C6 H5 − C − Br
| | | |
C6 H5 C6 H5 CH3 H
19. (a): The reaction involved is
HBr
CH2 = CH − CH2 − OH CH3 − CH − CH2 Br
→
Excess
|
Br
20. (a): The reaction involved is
21. (a, b, c): For the given reaction
A(g) + B(g) ⇌ C(g) + D(g); ∆H = −Q kJ
Since, heat is releasing so, the above is an exothermic reaction, hence for such reactions Keq is
directly proportional to temperature. However, change in the concentration of any reactant or
product will affect the position of equilibrium, but equilibrium constant will remain same.
Moreover, there will not be any change in value of Keq by change in pressure. Therefore, option
(1), (2) and (3) are correct.
22. (c): Since, compound Y give iodoform test, therefore Y must be a methyl carbonyl compound or
ethanol. As compound X is oxidized by PCC in DCM to given a methyl carbonyl compound,
therefore, X is an alcohol, that is, ethanol. The reaction involved is
PCC I2 + NaOH
CH3 CH2 OH ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ CH3 CHO CHI3
DCM ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
(X) (Y) Yellow ppt.
23. (b): Since, compound ‘A’ gives characteristics colour with neutral FeCl3 solution, therefore,
compound ‘A’ is a phenol. The reaction involved is
24. (d): In the given reaction sequence, first step is chlorination of acetic acid. Second step is Friedel
Craft’s acylation, that gives acetophenone. The next step is nucleophilic addition of HCN
followed by hydrolysis to give α − hydroxy − carboxylic acid. The reaction involved is

25. (d): According to conjugate acid – base theory, a weak acid produces a strong conjugate base
and vice-versa. That is, smaller the value of Ka, weaker will be the acid. Therefore, HCN
(k a = 4 × 10−10 ) is weakest acid among the given acid. Therefore, it will give strongest
conjugate base that is CN-.
HCN ⇌ H + + CN −
26. (c): Only acetylene CH ≡ CH) gives acetaldehyde through oxymercuration reaction. The
reaction involved is
HgSO4 /H2 SO4
HC ≡ CH CH3 CHO
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
H2 O
(A) (Acetaldehyde)
Acetylene
Ethanol on oxidation with PCC gives CH3 CHO. The reaction involved is
 PCC
CH3 CH2 OH ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ CH3 CHO
(oxidation)
(B) (Acetaldehyde)
Ethanol
Reduction of alkyl nitriles with SnCl2 /HCl is an example of Stephen reduction reaction. The
reaction involved is
(i)SnCl2 /HCl
CH3 CN ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗+ CH3 CHO
(ii)H3 O
(c) (Acetaldehyde)
Methyl nitrile
27. (a): A carboxylic acid can be reduced by LiAlH4/ether. Therefore, the reaction involved is
LiAlH4
CH3 − COOH CH3 CH2 OH
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
ether
28. (b): The reaction involved is
OH
|
(i)CH MgBr
3
CH3 − CHO ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
+
CH3 − C − CH3
(ii)H3 O
|
H
(A)
(Propan – 2 – ol)
Conc.n H2 SO4 ↓ ∆(−H2 O)
(i)B2 H6
CH3 − CH2 − CH2 − OH CH3 − CH = CH2
←
(ii) H2 O2 / OH−
(C) (B)
(Propan – 1 – ol)
The compound (A) and (C) are position isomers as position of −OH group has changed.
29. (d): An oxidation process involves the addition of oxygen atoms, removal of electrons, addition
of electronegative element as well as removal of H – atoms from an atom. Whereas removal of
electronegative atom does not count as oxidation.
30. (c): The given transformation can be carried out by using iodoform regents followed by
acidification. Since, I2 / NaOH is a weak oxidizing agent, therefore the double bond is
unaffected. The reaction involved is
(i) I2 / NaOH
CH3 CH = CHCH2 COCH3 → CH3 CH = CHCH2 CO O− Na+ + CHCl3
(ii) H2 O+
CH3 CH = CHCH2 CO O− Na+ → CH3 CH = CHCH2 COOH
31. (a): The reaction involved is

alc NH3 2 CH3 Cl

C6 H5 CH2 − Cl → C6 H5 CH2 − NH2 → C6 H5 CH2 − N − CH3
|
CH3
 N, N – Dimethyl phenylmethanamine
32. (d): Since, alkyl halides are more reactive than aryl halides, as aryl halides do not undergo
nucleophilic substitution reaction easily. Therefore, in Gabriel phthalimide synthesis only
aliphatic amines are prepared

33. (c): Permanent hardness of water is removed by using washing soda, Calgon’s method and ion
exchange method. However temporary hardness of water is removed by Clark ′ s method.
Ca(OH)2 is Clark’s reagent. It removes hardness of water by converting bicarbonates into
carbonates.
Ca(OH)2 + Ca (HCO3 )2 → 2CaCO3 ↓ +H2 O
34. (b): Compound (A) which gives only one compound on addition with HCl. The possible structure
of A is
HCl
CH3 − CH = CH − CH3 → CH3 − CH − CH2 − CH3
(A) |
Cl
↓ NH3 (1mol)
(i)NaNO2 / HCl
CH3 − CH − CH2 − CH3 ← CH3 − CH − CH2 − CH3
(ii) H2 O

| |

OH NH2

(D)

Compound (D) have chiral centre, therefore, it is an optically active compound.

35. (a): RNA and DNA contains D – ribose and Deoxyribose sugars, which are D − Chiral sugars.
These sugars result in the optical activity of DNA and RNA
36. (c): Moving down the group in alkaline earth metals:
(i) Electropositive of alkaline earth metals increases as size increases.
(ii) Ionization energy decreases as size increases.
(iii) Solubility of their hydroxide in water.
Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2
(iv) Solubility of their sulphate in water.
37. (c): In nuclei acids, the pairing bases are as follows:
G ≡ C and A = T
The complementary sequence of the given chain is given below:
 G≡C
A=T
T=A
G≡C
C≡G
Therefore, CTACG is the complimentary chain.
38. (a): The metal chloride salt is treated with sulphuric acid, which produces hydrogen chloride gas.
MCl + H2 SO4 → MHSO4 + HCl
39. (a): The Buckminster fullerene has 20 six membered ring whereas 12 five membered rings.
40. (b): Oxide ions occupy CCP close packing, therefore no. of oxygen atoms (N)=4
IN CCP, no. of octahedral voids (N) =4
1
Therefore, No. Of Be2+ ion = × 4 = 1
4
No. of tetrahedral voids 2N = 8
1
No. of Al3+ ions 4 × 8 = 2
Hence, the formula of compound is BeAl2 O4
41. (c): (i) Frankel defect is known as dislocation defect. In which a smaller ion (cation) is dislocated
from its normal site to interstitial site. It creates a vacancy defect at its original site and an
interstitial defect at its new location.
(ii) Schottky defect is a vacancy defect in ionic solids.
(iii) Trapping of an electron in the lattice leads to the formation of F − centre is correct.
42. (c): We know
Z×M
ρ= 3 … (1)
a × NA
For BCC, Z = 2
Substituting the values in equation (1), we get
2×M
6.15 gcm−3 =
(3 × 10−8 cm)3 × 6.022 × 1023 /mol
6.15 g cm−3 × 27 × 10−24 cm3 × 6.022 × 1023 /mol
⇒M=
2
⇒ M = 50 gmol−1
Hence molar mass of metal is 50 gmol−1
43. (a): Given,
Mole fraction of N2 = 0.8, pT = 5 atm
From Dalton′ s law, we have
PN2 = xN2 × pN2 = 0.8 × 5 atm
pN2 = 4 atm
From Henry’s law, we have
pN2 = k N2 xN2
Susbtituting the values in eq. (1), we get
nN
4 atm = 1.0 × 105 atm × 2
10
Thus, nN2 = 4 × 10−4
44. (d): Given, Mass of C = 2.4 g
No. of H − atoms = 1.2 × 1023 atoms
Moles of O atoms = 0.2 moles
2.4 g
The No. moles of C = = 0.2 mol
12 gmol−1
Using the number of H – atoms, moles of H atoms are calculated as follows:
We know
1 mole = 6.022 × 1023 atoms
1.2 × 1023 atoms
⇒ Moles of H − atoms = = 0.2 mol
6.022 × 1023 atoms/mol
Therefore, empirical formula = C0.2 H0.2 O0.2
Or C1 H1 O1 ≈ CHO
45. (d): (1) k H value is different for a gas in any solvent, as it depends on the solvent.
(2) According to Dalton’s law: Solubility ∝ pressure,
p = kH ∙ x
Therefore, solubility inversely proportional to k H . So more the value of k H less will be solubility.
(3): k H value increases on increasing the temperature correct.
(4): The gases which liquifies easily has less value of k H , correct
46. (a): According to Henry’s law, we have
p = kH ∙ x
p
or x =
kH
Thus, solubility is inversely proportional to k H . More than value of k H less will be the solubility.
Gases k H Values
I. Argon 40.3
II. Carbon dioxide 1.67
III. Formaldehyde 1.83 × 10−5
IV. Methane 0.413
Hence, the correct increasing order of solubility of gas in liquid is I < II < IV < III
47. (a): Given, poA = 450 mm Hg, poB = 700 mm Hg
pT = 600 mm Hg xA =? xB =?
pT = pA + pB = pA ∙ poA + xB ∙ poB
Or pT = xA ∙ poA + (1 − xA )poB (since, xA + xB = 1)
Thus, 600 = xA × 4500 + 700 − 700 xA
600 − 700 = (450 − 700)xA
100
⇒ 100 = 250xA ⇒ xA = = 0.4
250
Therefore, xB = 1 − xA = 1 − 0.4 = 0.6
Hence, the comoposition of mixture is ∶ xA = 0.4 and xB = 0.6
48. (c): From Nernst equation, we have
0.0591 [P]
Ecell = E 0 − log
n [R]
For the given half-cell reaction,
 Zn → Zn2+ + 2e−
We have
Ecell ∝ [P]
Since in above reaction product is Zn2+ , so more the concentration of Zn2+ ions, more will be
the Ecell . Hence, the correct order of Ecell is:
Q>R>S>P
49. (b): Number of angular nodes = l
Number of radial nodes = n − l − 1
For 3p, n = 3
l = 1,
Therefore, angular nodes (l) = 1
Radial nodes = 3 − 1 − 1 = 1
50. (a): Given, R = 1500 Ω, concentration = 0.01 m
l
Conductivity (k) = 0.146 × 10−3 S cm−1 , Cell constant A =?
We know
A
R = ρ( )
l
1 1 l
⇒ = ∙( )
R ρ A
1 l
or = k ∙ ( ) … (1)
R A
Substituting the values in equation (1), we get
l
⇒ = k ∙ R = 0.146 × 10−3 S cm−1 × 1500 S −1
A
= 219 × 10−1 cm−1 = 0.219 cm−1
51. (a): We know
o 0.0591
Ecell = log k eq
n
From the given chemical reactions,
H2 + 2AgCl ⇌ 2 Ag(s) + 2 HCl
2AgCl + 2e− ⇌ 2Ag(s) + 2Cl−
The change in number of moles n = 2
Substituting the values in eq. (1), we get
0.0591
0.22 = log k eq
2
0.22 × 2
⇒ log k eq = = 7.457
0.0591
⇒ k eq = Antilog(7.457) = 2.8 × 107
52. (a): For the given reaction
A + 2B → P
The rate expression is
Rate = k[A]α [B]β [Where α and β are order w. r. t A and B]
When [B] is increased, half-life remains same, this implies that [B] does not affect t1/2.
 0.693
t1 = [First − order reaction]
2 k
Therefore, α = 0,
Hence, the rate expression becomes
Rate = k[A]0 [B]1
Or Rate = k[B]1
⇒ The overall order of reaction is 1.0
So, the unit of rate constant (k) = s −1
53. (b): Since, the general electronic configuration of alkaline earth metals is ns2. Therefore, after
the two ionization enthalpies, the electronic configuration of alkaline earth metals becomes ns0,
that is, a noble gas configuration, which required higher third ionization energy to extract from
noble gas configuration.
54. (b): We know, For first order reaction:
2.303 [A]0
t= log
k [A]1
For 99% completion: [A]1 = 100 − 99 = 1%
2.303 100
t 99% = log
t 1
2.030
⇒ t 99% = ×2 [Since, log 100 = 2]
k
4.606
⇒ t 99% =
k
55. (c): The rate expression is
Rate0 = k[A][B]2 for gaseous reaction
1 V1
⇒ p ∝ so, when V2 =
V 2
1 2
⇒p= = , pressure becomes double
V1 V1
2
Hence, the rate expression becomes
⇒ Rate1 = k[2A]1 [2B]2
⇒ Rate1 = 8k[A]1 [B]2
⇒ Rate1 = 8Rate0 ⇒⇒
56. (a): The correct IUPAC name of the given compound is

4 – ethyl – 1 – fluoro – 2 – nitrobenzene.

57. (c): The higher order reactions (> 3) are very less because there is very low probability of
effective collisions between the reacting species.
58. (d): Since, the carbon having more s – character will have more acidic hydrogen, therefore,
ethyne have most acidic hydrogen since, it has sp – hybridized carbon atom.
H − C ≡ C − H → HC ≡ C ⊝
↓
(sp carbon) more stable due to more e- negativity than sp2 and sp3 hybridized carbon
atoms.
Benzene is less acidic than ethyne, as its carbon of benzene is sp2 – hybridized.

n – Hexane is least acidic, as its carbon is sp3 – hybridized

All C − sp3 least e- negative, thus least acidic.

Hence, the correct acidic order is: Ethye > Benzene > n - Hexane
59. (a): According to Hardy – Schulze rule, more the charge of flocculating ion, more will be the
flocculating power. However, flocculating power is inversely proportional to flocculating value.
1
Flocculating value ∝
Flocculating Power
NaCl ⇒ Na+ , BaCl2 ⇒ Ba2+ , AlCl3 ⇒ Al3+
Hence, the correct order of the amount of electrolyte required will be:
NaCl > BaCl2 > AlCl3
⇒ That means AlCl3 will be required less compared to BaCl2 and NaCl.
60. (d): 1. H – bonding is stronger than dispersion forces.
2. A sigma bond is formed by head-to-head overlap while π − bond is formed by lateral overlap.
So head to head overlap is more stronger and a sigma bond is stronger than π − bond.
3. Ionic bonds are non – directional as cation and anion (charge species) are involved.

4. π − Electrons are known as mobile electrons. For example, in benzene, the π − electrons are
delocalized over the ring in benzene.
 MATHEMATICS

1. Chapter: Straight Lines

(3) Given that, 𝑦 = −𝑥 3 + 3𝑥 2 + 2𝑥 − 27
Differentiate with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥
= −3𝑥 2 + 6𝑥 + 2
𝑑𝑦
Let slope 𝑚 = 𝑑𝑥 = −3𝑥 2 + 6𝑥 + 2
𝑑𝑚
Then, 𝑑𝑥
= −6𝑥 + 6
𝑑𝑚
For maximum or minimum put 𝑑𝑥
=0
⇒ 6𝑥 + 6 = 0
⇒𝑥=1
𝑑2 𝑚
Now, 𝑑𝑥 2
= −6 < 0
Thus, 𝑚 is maximum for 𝑥 = 1 and the maximum value of 𝑚 is given by
𝑚 = −3 + 6 + 5 = 5
2. Chapter: Integrals
𝑥 3 sin(tan−1 (𝑥 4 ))
(1) Given that, ∫ 1+𝑥 8
𝑑𝑥
4 3
Let 𝑥 = 𝑡 then 4𝑥 𝑑𝑥 = 𝑑𝑡 so,
𝑥 3 sin(tan−1 +𝑥 4 ) 1 sin(tan−1 𝑡)
∫ 𝑑𝑥 = ∫ 𝑑𝑡
1 + 𝑥8 4 1 + 𝑡2
1
Let tan−1 𝑡 = 𝑢 then 1+𝑡 2 𝑑𝑡 = 𝑑𝑢 so,
𝑥 3 sin(tan−1 +𝑥 4 ) 1 sin(tan−1 𝑡) 1
∫ 8
𝑑𝑥 = ∫ 2
𝑑𝑡 = ∫ sin 𝑢 𝑑𝑢
1+𝑥 4 1+𝑡 4
1 −1 −1
= (− cos 𝑢) + 𝐶 = cos(tan−1 𝑡) + 𝐶 = cos(tan−1 𝑥 4 ) + 𝐶
4 4 4
3. (c):
𝑥2
Given that, ∫ √𝑥 6 +𝑎6 𝑑𝑥
Let 𝑥 3 = 𝑡 then 3𝑥 2 𝑑𝑥 = 𝑑𝑡
𝑥2 𝑥2 1 𝑑𝑡
∫ 𝑑𝑥 = ∫ 𝑑𝑥 = ∫ ∙
6
√𝑥 + 𝑎 6 3 2 3
√(𝑥 ) + (𝑎 )2 √𝑡 + (𝑎 ) 3
2 3 2
1 𝑑𝑡 1
= ∫ [log |𝑡 + √𝑡 2 + (𝑎3 )2 | + 𝐶1 ]
3 √𝑡 2 +(𝑎 3 )2 3
1
= 3 log|𝑥 3 + √𝑥 6 + 𝑎6 | + 𝐶
4. (d):
𝑥𝑒 𝑥 𝑑𝑥
Given that, ∫ (1+𝑥)2
𝑑𝑢 𝑑𝑣 1 1
Let 𝑢 = 𝑥𝑒 𝑥 ⇒ 𝑑𝑥 = 𝑥𝑒 𝑥 + 𝑒 𝑥 and 𝑑𝑥 = (1+𝑥)2 ⇒ 𝑣 = − 𝑥+1
Now, using Integration parts, that is,
𝑑𝑣 𝑑𝑢
∫ 𝑢 𝑑𝑢 = 𝑢𝑣 − ∫ 𝑑𝑥 𝑣
 𝑥𝑒 𝑥 𝑑𝑥 −𝑥𝑒 𝑥 −𝑥𝑒 𝑥 −𝑒 𝑥
So, ∫ (1+𝑥)2 = 𝑥+1
−∫ 𝑥+1
𝑑𝑥 ………..(i)
−𝑥𝑒 𝑥 −𝑒 𝑥
Now solve, ∫ 𝑥+1
𝑑𝑥 = − ∫ 𝑒 𝑥 𝑑𝑥 = −𝑒 𝑥 + 𝐶
From Eq.(1), we get
𝑥𝑒 𝑥 𝑑𝑥
−𝑥𝑒 𝑥
∫ (1+𝑥)2 = + 𝑒𝑥 + 𝐶
𝑥+1
−𝑥𝑒 𝑥 +𝑥𝑒 𝑥 +𝑒 𝑥 𝑒𝑥
= 𝑥+1
+ 𝐶 = 𝑥+1 + 𝐶
1+sin 𝑥
5. (a) Given that, ∫ 𝑒 𝑥 (1+cos 𝑥)
𝑥 𝑥 𝑥
We know that, sin 𝑥 = 2 sin (2) cos (2) , sin2 𝜃 + cos2 𝜃 = 1 and1 + cos 𝑥 = 2 cos2 2.
So,
𝑥 𝑥 𝑥 𝑥
1 + sin 𝑥 sin2 (2) + cos2 (2) + 2 sin (2) ∙ cos 2
𝑥 𝑥
𝑒 ( )=𝑒 ( 𝑥 )
1 + cos 𝑥 2cos2 ( )
2
𝑥 𝑥 2 𝑥 𝑥 2
(sin +cos ) 𝑒 𝑥 sin(2)+cos(2)
𝑥 2 2
=𝑒 ( 𝑥 )= 2
( 𝑥 )
2cos2 ( ) cos( )
2 2
𝑥 𝑥 2
𝑒𝑥 sin cos 𝑒𝑥 𝑥 2
2 2
= 2
( 𝑥 + 𝑥 ) = 2
(tan (2) + 1)
cos cos
2 2
𝑒𝑥 𝑥 𝑥
= 2
(tan2 (2) + (1)2 + 2 (tan 2) (1))
𝑒𝑥 𝑥 𝑥
= 2
(tan2 2 + 1 + 2 tan 2)
𝑒𝑥 𝑥 𝑥
= (sec 2 + 2 tan ) [𝑠𝑖𝑛𝑐𝑒, 1 + tan2 𝜃 = sec 2 𝜃]
2 2 2
1 𝑥 2 𝑥
= 𝑒 𝑥 ( ∙ sec 2 + tan )
2 2 2 2
𝑥 1 𝑥
= 𝑒 𝑥 (tan ( ) + ∙ sec 2 ( ))
2 2 2
𝑥 1+sin 𝑥 𝑥 𝑥 1 2 𝑥
Now, ∫ 𝑒 ( ) = ∫ 𝑒 (tan ( ) + sec ( )) 𝑑𝑥
1+cos 𝑥 2 2 2
It is of the form, ∫ 𝑒 𝑥 [𝑓(𝑥) + 𝑓′(𝑥)]𝑑𝑥 = 𝑒 𝑥 𝑓(𝑥) + 𝐶
1+sin 𝑥 𝑥
So, ∫ 𝑒 𝑥 (1+cos 𝑥) = 𝑒 𝑥 tan (2) + 𝐶
6. (d) : Given that,
𝜋/4
𝐼𝑛 = ∫ tan𝑛 𝑥 𝑑𝑥
0
Replacing 𝑛 → 𝑛 + 1, we get
𝜋/4
𝐼𝑛+1 = ∫0 tan𝑛+1 𝑥 𝑑𝑥 (1)
Now, replacing 𝑛 → 𝑛 − 1, we get
𝜋/4
𝐼𝑛−1 = ∫0 tan𝑛−1 𝑥 𝑑𝑥 (2)
Add Eqs. (1) and (2), we get
𝜋/4 𝜋/4
𝐼𝑛+1 + 𝐼𝑛−1 = ∫ tan𝑛+1 𝑥 𝑑𝑥 + ∫ tan𝑛−1 𝑥 𝑑𝑥
0 0
𝜋/4 𝜋/4
∫0 (tan𝑛+1 𝑥 + tan 𝑛−1
𝑥) 𝑑𝑥 = ∫0 tan𝑛−1 𝑥(tan2 𝑥 + 1) 𝑑𝑥
 𝜋/4
= ∫0 (tan𝑛−1 𝑥 (sec 2 𝑥) 𝑑𝑥
Let tan 𝑥 = 𝑡 then sec 2 𝑥 = 𝑑𝑡 so, limits charge from 0 to 1.
𝜋/4 1
∫ (tan𝑛−1 𝑥 (sec 2 𝑥) 𝑑𝑥 = ∫ 𝑡 𝑛−1 𝑑𝑡
0 0
𝑢𝑛 1 1
= | =
𝑛 0 𝑛
𝜋/4 1
Therefore, 𝐼𝑛+1 + 𝐼𝑛−1 = ∫0 ((tan𝑛+1 𝑥 + tan𝑛−1 𝑥)𝑑𝑥 = 𝑛)
1
For 𝑛 = 9, we have 𝐼9+1 + 𝐼9−1 = 𝐼10 + 𝐼8 = 9
7. (b) :
4042 √𝑥
Given that 𝐼 = ∫0 𝑑𝑥 (1)
√𝑥+√4042−𝑥
𝑏
Using property ∫𝑎 𝑓(𝑎 + 𝑏 − 𝑥) 𝑑𝑥, we get
4042
√4042 − 𝑥𝑑𝑥
𝐼=∫
0 √4042 − 𝑥 + √4042 − (4042 − 𝑥)
4042 √4042−𝑥
𝐼 = ∫0 (2)
√4042−𝑥+√𝑥
Add Eqs. (1) and (2), we get
4042 4042
√𝑥 √4042 − 𝑥
𝐼+𝐼 =∫ +∫
0 √𝑥 + √4042 − 𝑥 0 √4042 − 𝑥 + √𝑥
4042 4042
√𝑥 + √4042 − 𝑥
2𝐼 = ∫ =∫ 1
0 √𝑥 + √4042 − 𝑥 0
4042
⇒ 2𝐼 = [𝑥]4042
0 ⇒𝐼= = 2021
2
8. (c) :
Given equation of curve is 𝑦 = √16 − 𝑥 2 and the equation of line is 𝑥 − axis which
means 𝑦 = 0. The equation of curve shows the ellipse having
√16 − 𝑥 2 = 0
⇒ 16 − 𝑥 2 = 0
⇒ 𝑎2 = ±4
So, intersection points are (4, 0) and (-4, 0)
The condition is shown in the following figure:

Therefore, area of curve is given by

4
∫ (16 − 𝑥 2 )1/2 𝑑𝑥
−4
 4
𝑥 42 𝑥
= [ √42 − 𝑥 2 + sin−1 ]
2 2 2 −4
4 8 4 −4 −4
= [ √42 − 42 + sin−1 ] − [ √42 − (−4)2 + 8 sin−1 ( )]
2 2 2 4 2
8 𝜋 𝜋
= [2.0 + ∙ − 0 + 8 ∙ ] = 8𝜋 square unit
2 2 2
9. (a) :
𝑥2 𝑦2
Given that, 25 + 𝜆2
𝑦2 𝑥2
⇒ = −1 −
𝜆2 25
𝜆
⇒ 𝑦 = ± √25 − 𝑥 2
5
Given that area of ellipse is 20 𝜋 square units. So,
𝜆 5
4 [± ∫ √25 − 𝑥 2 𝑑𝑥 ] = 20𝜋
5 0
5
4𝜆 𝑥 2 52 𝑥
⇒± [ √5 − 𝑥 2 + sin−1 ] = 20𝜋
5 2 2 50
4𝜆 52
⇒± [0 + sin−1 1 − 0 − 0] = 20𝜋
5 2
4𝜆 25 𝜋
⇒± × × = 20𝜋
5 2 2
⇒ ±5 𝜆 = 20 ⇒ 𝜆 = ±4
10. (c) :
Given that,
𝑥𝑑𝑦 − 𝑦𝑑𝑥 = 0
⇒ 𝑥𝑑𝑦 = 𝑦𝑑𝑥
𝑑𝑦 𝑑𝑥
⇒ =
𝑦 𝑥
On integrating both side
𝑑𝑦 𝑑𝑥
∫
=∫
𝑦 𝑥
log 𝑦 = log 𝑥 + log 𝐶
⇒ log 𝑦 = log 𝑥 𝐶
⇒ 𝑦 = 𝐶𝑥
The equation shows straight line passing through origin.
11. (b) :
𝑑𝑦 𝑦+1
Given that, =
𝑑𝑥 𝑥−1
𝑑𝑦 𝑑𝑥
⇒ =
𝑦+1 𝑥−1
On integrating both sides, we get
𝑑𝑦 𝑑𝑥
∫ =∫
𝑦+1 𝑥−1
 ⇒ log(𝑦 + 1) = log(𝑥 − 1) + log 𝐶
⇒ (𝑦 + 1) = 𝐶(𝑥 − 1)
𝑦+1
⇒𝐶=
𝑥−1
When 𝑥 = 1 and 𝑦 = 2 then 𝐶 = ∞
So, required solution is 𝑥 − 1 = ∞
Hence, infinite solutions exist.
12. (b):
Given that, 𝑏⃗⃑ = 5𝑖̂ + 7𝑗̂ − 𝑘̂
Let 𝑎⃑ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂
It makes equal angle with coordinates axes, that is,
cos 𝜃𝑥 = cos 𝜃𝑦 = cos 𝜃𝑧
𝑥 𝑦 𝑧
⇒ = =
√𝑥 2 + 𝑦 2 + 𝑧 2 √𝑥 2 + 𝑦 2 + 𝑧 2 √𝑥 2 + 𝑦 2 + 𝑧 2
⇒𝑥=𝑦=𝑧
So, ̂
𝑎⃑ = 𝑥(𝑖̂ + 𝑗̂ + 𝑘 )
Then, |𝑎⃑| = √3
⇒ √𝑥 2 + 𝑥 2 + 𝑥 2 = √3
⇒ 3𝑥 2 = 3 ⇒ 𝑥 = ±1
Therefore, 𝑎⃑ = 𝑖̂ + 𝑗̂ + 𝑘̂ or 𝑎⃑ = (𝑖̂ + 𝑗̂ + 𝑘̂ )
Projection of 𝑏⃗⃑ on 𝑎⃑ is given by
𝑎⃑ ∙ 𝑏⃗⃑ ±(𝑖̂ + 𝑗̂ + 𝑘̂ ) ∙ (5𝑖̂ + 7𝑗̂ − 𝑘̂ ) ±(5 + 7 − 1) 11
𝑎⃑ = = = =±
𝑎⃑ √3 √3 √3
13. (*) :
Given that, 3𝑖̂ + 6𝑗̂ − 2𝑘̂ and −𝑖̂ − 2𝑗̂ − 8𝑘̂.
Let ABCD are the vertices of parallelogram, vertex A be origin and O is mid-point of
diagonals, so
⃗⃗⃗⃗⃗ = 3𝑖̂ + 6𝑗̂ − 2𝑘̂ or 𝐶 = (3, 6, −2)
𝐴𝐶
3 6 −2 3
And 𝑂 = (2 , 2 ,
2
) = (2 , 3, −1)
Let 𝐵 = (𝑎, 𝑏, 𝑐) so,
1 3 1
̅̅̅̅ = (𝐵𝐷
𝑂𝐵 ̅̅̅̅ ) = ( − 𝑎) 𝑖̂ + (3 − 𝑏)𝑗̂ + [−1 − 𝑐]𝑘̂ = (−𝑖̂ − 2𝑗̂ − 8𝑘̂ )
2 2 2
3 −1
⇒ −𝑎 = , 3 − 𝑏 = −1, −1 − 𝑐 = −4
2 2
⇒ 𝑎 = 2, 𝑏 = 4, 𝑐 = 3
Then, 𝐵 = (2, 4, 3)
Therefore, lengths of the sides of parallelogram are
̅̅̅̅| = √(2)2 + (4)2 + (3)2 = √4 + 16 + 9 = √29
|𝐴𝐵
̅̅̅̅ | = √(3 − 2)2 + (6 − 4)2 + (−2 − 3)2 = √30
|𝐵𝐶
Hence, shorter side of parallelogram is √29
14. (d) :
 Given that, 𝑎̅ ∙ 𝑏̅ = 0 and 𝑎̅ + 𝑏̅ make and angle 60° with 𝑎̅. Since, 𝑎̅ ∙ 𝑏̅ = 0 the vectors
𝑎̅ and 𝑏̅ are perpendicular to each other.

|𝑏| ̅ |𝑏| ̅
So, tan 60° = |𝑎̅| ⇒ √3 = |𝑎̅| ⇒ √3|𝑎̅| = |𝑏̅|
15. (c) :
Given that, |𝑎̅ × 𝑏̅| = 15
If the sides are (3𝑎̅ + 2𝑏̅) 𝑎𝑛𝑑 (𝑎̅ + 3𝑏̅) . then area is given by
Area = |(3𝑎̅ + 2𝑏̅) × (𝑎̅ + 3𝑏̅)|
= |2(𝑏̅ × 𝑎̅) + 9(𝑎̅ × 𝑏̅)| = |−2(𝑎̅ × 𝑏̅) + 9(𝑎̅ × 𝑏̅)|
= 7|𝑎̅ × 𝑏̅| = 7(15) = 105 Sq. units.
16. (b) :
We know that the equation of the line passing through the points (𝑥1 , 𝑦1 , 𝑧1 ) and
(𝑥2 , 𝑦2 , 𝑧2 ) is given by
𝑥 − 𝑥1 𝑦 − 𝑦1 𝑧 − 𝑧1
= =
𝑥2 − 𝑥1 𝑦2 − 𝑦1 𝑧2 − 𝑧1
Given that points are (-3, 4, 11) and (1, -2, 7). So, the equation of the line is given by
𝑥+3 𝑦−4 𝑧 − 11
= =
1 + 3 −2 − 4 7 − 11
𝑥 + 3 𝑦 − 4 𝑧 − 11
⇒ = =
4 −6 −4
Clearly the directives of the line are proportional to -2, 3, 2
𝑥+3 𝑦−4 𝑧−11
Therefore, the required equation of line is −2
= 3
= 2
17. (c):
√3 2 √3
Given that, (𝑙1 , 𝑚1 , 𝑛1 )= ( 4 , 4 , 2
) and
√3 1 −√3
(𝑙2 , 𝑚2 , 𝑛2 )= ( 4 ,4, 2 )
So, angle between the lines are given by
|𝑙1 𝑙2 + 𝑚1 𝑚2 + 𝑛1 𝑛2 |
cos 𝜃 =
√𝑙12 + 𝑚12 + 𝑛12 × √𝑙22 + 𝑚22 + 𝑛22
3 1 3
|16 + 16 − 4|
−1
= | =|
√1√1 2
1 𝜋
⇒ cos 𝜃 = ⇒ 𝜃 =
2 3
18. (b) :
 Let the plan meets the coordinate axes at the points A(a, 0, 0), B(0, b, 0) and C(0, 0, c).
the centroid of the triangle
𝑎 𝑏 𝑐
ABC is 𝑂 = (3 , 3 , 3).
Given that, centroid 𝑂 = (1, 2, 3). So,
𝑎 𝑏 𝑐
( , , ) = (1, 2, 3)
3 3 3
⇒ 𝑎 = 3, 𝑏 = 6, 𝑐 = 9
Now, the equation of plane in intercept form is given by
𝑥 𝑦 𝑧
+ + =1
𝑎 𝑏 𝑐
𝑥 𝑦 𝑧
⇒ + + =1
3 6 9
19. (a) :
Given that, ABCD is a quadrilateral, where A = (0, 4, 1), B = (2, 3, -1), C = (4, 5, 0) and D =
(2, 6, 2).
Area of quadrilateral is given by
Area of quadrilateral = (Area of ∆𝐴𝐵𝐷) +(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐵𝐶𝐷)
1 1
= |𝐴𝐵 ̅̅̅̅ × ̅̅̅̅ ̅̅̅̅ × ̅̅̅̅
𝐴𝐷 | + |𝐶𝐵 𝐶𝐷|
2 2
1 1
= |3𝑖̂ − 6𝑗̂ + 6𝑘̂ | + |−3𝑖̂ + 6𝑗̂ − 6𝑘̂ |
2 2
1 1 1 1 1
= 2 √81 + 2 √81 = 2 (9) +2 (9) =2 (18) = 9 sq. unit
20. (c) :
From the given shaded region in the figure, we can see four lines bounded the shaded
region.
The three lines are given as
𝑥 = 6, 𝑦 = 0 and 𝑦 = 3
Other line is given as
−5
𝑦= 𝑥 + 5 ⇒ 5𝑥 + 4𝑦 = 20
4
To make a given shaded region we have following inequalities from there four lines are
5𝑥 + 4𝑦 ≥ 20, 0 ≤ 𝑥 ≤ 6, 0 ≤ 𝑦 ≤ 3
Or 5𝑥 + 4𝑦 ≥ 20, 𝑥 ≤ 6, 𝑦 ≤ 3, 𝑥 ≥ 0, 𝑦 ≥ 0
21. (2) :
3 1 4
Given that, 𝑃(𝐵) = 5 , 𝑃(𝐴/𝐵) = 2 and 𝑃(𝐴 ∪ 𝐵) = 5.
We know that,
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
𝑃(𝐴∩𝐵)
Since, 𝑃(𝐴/𝐵) = 𝑃(𝐵)
⇒ 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴/𝐵) ∙ 𝑃(𝐵)
Then, 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴/𝐵) ∙ 𝑃(𝐵)
4 3 1 3
⇒ = 𝑃(𝐴) + − ( ) ( )
5 5 2 5
1
⇒ 𝑃(𝐴) =
2
22. (c) :
Given that, 𝑃(𝐴) = 𝑃(𝐵) = 𝑃(𝐶) = 𝑃
𝑃(at least 2 of 𝐴, 𝐵, 𝐶 occur)
= 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶 ′ ) + 𝑃(𝐴 ∩ 𝐵′ ∩ 𝐶) + 𝑃(𝐴′ ∩ 𝐵 ∩ 𝐶) + 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶)
𝑃(𝐴)𝑃(𝐵)𝑃(𝐶′) + 𝑃(𝐴)𝑃(𝐵′)𝑃(𝐶) + 𝑃(𝐴′)𝑃(𝐵)𝑃(𝐶) + 𝑃(𝐴)𝑃(𝐵)𝑃(𝐶)
= 𝑃2 (1 − 𝑃) + 𝑃2 (1 − 𝑃) + 𝑃2 (1 − 𝑃) + 𝑃3 = 3𝑃2 − 2𝑃3
23. (c) :
Given that, 𝐸 = {(1, 1)(1, 2)(1,3)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)} 𝑎𝑛𝑑 𝐹 =
{(1, 2)(2, 1)}
So, probability of getting a sum as 3 is
𝐹 2 1
𝑃( ) = =
𝐸 10 5
24. (3) :
Let 𝐸1 be the event of plant 𝑋 manufacturing car, 𝐸2 be the event of plant 𝑌
manufacturing car and 𝐴 be event car is of standard quality. So, the probability when a
car chosen at random it has come from plant 𝑋 is given by
𝑃(𝐸1 ) ∙ 𝑃(𝐴/𝐸1 )
𝑃(𝐸1 /𝐴) =
𝑃(𝐸1 ) ∙ 𝑃(𝐴/𝐸1 ) + 𝑃(𝐸2 ) ∙ 𝑃(𝐴/𝐸2 )
Given that, 𝑃(𝐸1 ) = 0.7, 𝑃(𝐸2 ) = 0.3, 𝑃(𝐴/𝐸1 ) = 0.8 and 𝑃(𝐴/𝐸2 ) = 0.9. so,
𝐸1 0.7 × 0.8 56
𝑃( ) = =
𝐴 0.7 × 0.8 + 0.3 × 0.9 83
25. (b) :
Let the total families in the town be 𝑥, families own cell phones be 𝐴 and families own
scooter be 𝐵.
Given that, 65% of 𝑥 have cell phones, 15000 have scooter and 15% of 𝑥 have both.
So, total number of families is given by
𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵)
65 15
⇒𝑥= 𝑥 + 15,000 − 𝑥
100 100
⇒ 𝑥 = 0.5𝑥 + 15,000
1
⇒ 𝑥 = 15,000 ⇒ 𝑥 = 30,000
2
26. (d) :
Given that, 𝑛(𝐴 × 𝐵) = 35.
If 𝐵 ⊂ 𝐴, then 𝑛(𝐵) = 5, 𝑛(𝐴) = 7.
7×6
Therefore, 𝑛(𝐴)𝐶𝑛(𝐵) = 7𝐶5 = 2
= 21
27. (d) :
𝑥
Given that, 𝑓(𝑥) = 1−|𝑥|
So, |𝑥| ≠ 1
Therefore, domain = 𝑅 − {−1, +1}
28. (a):
Given that, cos 1200° + tan 1485°
 = cos[1080° + 120°] + tan[1440° + 45°]
= cos[(180 × 6)° + 120°] + tan[(180 × 8)° + 45°]
= cos 120° + tan 45° = cos(180° − 60°) + 1
−1 1
= − cos 60° + 1 = +1=
2 2
29. (2) :
Given that, tan 1° ∙ tan 2° … … . tan 89°
tan 1° ∙ tan 2° … tan 45° … . tan(90 − 2)° tan(90 − 1)°
tan 1° ∙ tan 2° … tan 45° … . . cot 2° cot 1° = tan 45° = 1
30. (d) :
1+𝑖 𝑥
Given that, (1−𝑖) = 1
1+𝑖
We know that, (1−𝑖) = 𝑖
⇒ 𝑖 𝑥 = 1 ⇒ 𝑖 4𝑛 = 1
Therefore, 𝑥 = 4𝑛, 𝑛 ∈ ℕ
31. (a):
Given that, 𝐶(𝑥) = 20𝑥 + 4000 and 𝑅(𝑥) = 60𝑥 + 2000.
For earn profit, we want
𝑅(𝑥) − 𝐶(𝑥) > 0
⇒ 60𝑥 + 2000 − 20𝑥 − 4000 > 0
⇒ 40𝑥 > 2000
2000
⇒𝑥> ⇒ 𝑥 > 50
40
32. (c) :
Given that, part 𝐴 have 6 questions and part 𝐵 have 7 questions.
We have the following ways to choose 10 questions from parts 𝐴 and :
A B Number of selections
4 6 6×5
6𝐶4 × 6𝐶6 = ×7
2
= 105
5 5 7×6
6𝐶5 × 7𝐶5 = 6 ×
2
= 126
6 4 7×6×5
6𝐶6 × 7𝐶4 = 1 ×
2
= 35

Therefore, total = 105+126+35=266

33. (a) :
𝑡1 +𝑡𝑛
Given that, middle term 2
= 300 ⇒ 𝑡1 + 𝑡𝑛 = 600
So, sum of first 51 terms is given by
51 51
𝑆51 = (𝑡1 + 𝑡𝑛 ) = × 600 = 51 × 300 = 15300
2 2
34. (b) :
 Given that, point (𝑎 cos2 𝜃 , sin2 𝜃) and 𝑥 sec 𝜃 + 𝑦 cosec 𝜃 = 𝑎. The equation of the
line passing through point (𝑎 cos3 𝜃 , sin3 𝜃) and perpendicular to line 𝑥 sec 𝜃 +
𝑦 cosec 𝜃 = 𝑎 is given by
𝑥 𝑦 𝑎 cos3 𝜃 𝑎 sin3 𝜃
− = =
sin 𝜃 cos 𝜃 sin 𝜃 cos 𝜃
𝑥 cos 𝜃 − 𝑦 sin 𝜃 𝑎 cos4 𝜃 − 𝑎 sin4 𝜃
⇒ =
sin 𝜃 cos 𝜃 sin 𝜃 cos 𝜃
𝑥 cos 𝜃 − 𝑦 sin 𝜃 𝑎(cos 4 𝜃 − sin4 𝜃)
⇒ =
sin 𝜃 cos 𝜃 sin 𝜃 cos 𝜃
𝑎(cos2 𝜃−sin2 𝜃)(cos2 𝜃−sin2 𝜃)
= sin 𝜃 cos 𝜃
𝑥 cos 𝜃 − 𝑦 sin 𝜃 𝑎(cos 2𝜃)(1)
⇒ =
sin 𝜃 cos 𝜃 sin 𝜃 cos 𝜃
𝑥 cos 𝜃 − 𝑦 sin 𝜃 𝑎 cos 2𝜃
⇒ =
sin 𝜃 cos 𝜃 sin 𝜃 cos 𝜃
⇒ 𝑥 cos 𝜃 − 𝑦 sin 𝜃 = 𝑎 cos 2𝜃
35. (b) :
Given that, triangle with vertices (1, 5, -1), (0, 4, -2) and (2, 3, 4). So, centroid of the
triangle
1+0+2 5+4+3 −1−2+4 1
Centroid = ( 3
, 3 , 3 ) = (1, 4, 3)
36. (2) :
𝑎𝑥 2 +𝑏𝑥+𝑐
Given that, statement 1: lim 𝑐𝑥 2 +𝑏𝑥+𝑎 and statement 2:
𝑥→1
2+𝑥
lim
𝑥→−2 2𝑥(𝑥 + 2)
Taking statement 1, we get
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑎 + 𝑏 + 𝑐
lim 2 = =1
𝑥→1 𝑐𝑥 + 𝑏𝑥 + 𝑎 𝑐+𝑏+𝑎
Taking statement 2, we get
2+𝑥 0
lim =
𝑥→−2 2𝑥(𝑥 + 2) 0
So, we use L’Hospitals Rule, that is,
𝑑
(2 + 𝑥) 1 0+1 −1
lim 𝑑𝑥 = lim = lim =
𝑥→−2 𝑑 𝑥→−2 2(𝑥 + 2) + 2𝑥 𝑥→−2 4𝑥 + 4 4
[2𝑥(𝑥 + 2)]
𝑑𝑥
Therefore, only statement 1 is true.
37. (2) :
𝑑 𝑎 𝑏
Given that, 𝑑𝑥 (𝑥 4 − 𝑥 2 + cos 𝑥) = 𝑚𝑎 + 𝑛𝑏 − 𝑝
⇒ (−4𝑎)𝑥 −5 + 2𝑏(𝑥 −3 ) − sin 𝑥 = 𝑚𝑎 + 𝑛𝑏 − 𝑝
Comparing both sides, we get
−4 2
𝑚 = 5 , 𝑛 = 3 ; 𝑝 = sin 𝑥
𝑥 𝑥
38. (c) :
 Standard deviation of number 31, 32, 33,….. 46, 47 is same as the standard deviation of
the number 1, 2, 3,……….17 standard deviation of first 𝑛 natural is given by

𝑛2 − 1
𝜎=√
12
Here 𝑛 = 17. Therefore, standard deviation is

172 − 1 288
𝜎=√ =√ = 2√6
12 12
39. (c) :
Given that, 𝑃(𝐴) = 0.59, 𝑃(𝐵) = 0.03 and 𝑃(𝐴 ∩ 𝐵) = 0.21.
So,
𝑃(𝐴′ ∩ 𝐵′) = 𝑃(𝐴 ∪ 𝐵)′
= 1 − 𝑃(𝐴 ∪ 𝐵)
= 1 − [𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)]
= 1 − 0.68 = 0.32
40. (d):
Given that,
2𝑥; 𝑥 > 3
𝑓(𝑥) = {𝑥 2 ; 1 < 𝑥 ≤ 3
3𝑥; 𝑥 ≤ 1
Then, 𝑓(−2) + 𝑓(3) + 𝑓(4) = 3(−2) + 32 + 2(4)
= −6 + 9 + 8 = 11
41. (a) :
2𝑥
Let 𝐴 = {𝑥: 𝑥 ∈ 𝑅; 𝑥 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟} and define 𝑓: 𝐴 → 𝑅 as 𝑓(𝑥) = 𝑥−1
Check for one-one function,
∀𝑥1 , 𝑥2 ∈ 𝐴
𝑓(𝑥1 ) = 𝑓(𝑥2 )
2𝑥1 2𝑥2
=
𝑥1 − 1 𝑥2 − 1
𝑥1𝑥2 − 𝑥1 = 𝑥1 𝑥2 − 𝑥2
𝑥1 = 𝑥2
So, 𝑓 is one-one function.
Now check for onto function,
2𝑥
𝑦=
𝑥−1
𝑥𝑦 − 𝑦 = 2𝑥
𝑥𝑦 − 2𝑥 = 𝑦
𝑥(𝑦 − 2) = 𝑦
𝑦
𝑥= ∉𝐴
𝑦−2
So, 𝑓 is not onto function.
42. (a) :
 Given that, function 𝑓(𝑥) = √3 sin 2𝑥 − cos 2𝑥 + 4
√3 1
⇒ 𝑓(𝑥) = 2 { sin 2𝑥 − cos 2𝑥} + 4
2 2
𝜋 𝜋
= 2 {sin 2𝑥 cos − cos 2𝑥 sin } + 4
2 2
𝜋
= 2 {sin (2𝑥 − )} + 4
6
−𝜋 𝜋
sin 𝑥 is one-one in [ , ]. So,
2 2

−𝜋 𝜋 𝜋
≤ 2𝑥 − ≤
2 6 2
𝜋 𝜋 𝜋 𝜋
⇒ 6 − 2 ≤ 2𝑥 ≤ 2 + 6

−𝜋 2𝜋
⇒ ≤ 2𝑥 ≤
3 6

−𝜋 𝜋 −𝜋 𝜋
⇒ 6
≤𝑥≤ 3
⇒ 𝑥 ∈ [ 6 , 3]

43. (a) :
1
Given that, function 𝑓(𝑥) =
√[𝑥]2 −[𝑥]−6
So, [𝑥]2 − [𝑥] − 6 > 0

⇒ [𝑥]2 − 3[𝑥] + 2[𝑥] − 6 > 0

⇒ [𝑥]([𝑥] − 3) + 2([𝑥] − 3) > 0

⇒ ([𝑥] − 3)([𝑥] − 2) > 0
⇒ [𝑥] < −2 or ⇒ [𝑥] > 3
Therefore, the domain of the given function is (−∞, −2) ∪ [4, ∞)

44. (d):
𝜋
Given that, cos [cot −1 (−√3) + ]
6
−1
𝜋 𝜋
= cos [cot cot (𝜋 − ) + ]
6 6
𝜋 𝜋
= cos (𝜋 − + ) = cos 𝜋 = −1
6 6
45. (*) :
1 5𝜋 √3
Given that, tan−1 [ sin ] sin−1 [cos (sin−1 )]
√3 3 2
−1 1 𝜋 −1 −1 𝜋
, tan [ 3 sin (2𝜋 + 2 )] sin [cos (sin sin ( 3 ))]
√
1 𝜋 𝜋
tan−1 [ sin ( )] sin−1 [cos ( )]
√3 2 3
1 1
tan−1 [ (1)] sin−1 [ ]
√3 2
 𝜋 𝜋
tan−1 [tan ( )] sin−1 [sin ( )]
6 6
𝜋 𝜋 𝜋2
= ∙ =
6 6 36
46. (b) :
2 1
1 −2 1
Given that, 𝐴 = [ ] and 𝐵 = [3 2]. Then,
2 1 3
1 1
2 1
1 −2 1 2−6+1 1−4+1
𝐴𝐵 = [ ] [ 3 2] = [ ]
2 1 3 4+3+3 2+2+3
1 1
−3 −2
𝐴𝐵 = [ ]
10 7
−3 10
So, (𝐴𝐵)′ = (𝐴𝐵)𝑇 = [ ]
−2 7
47. (c) :
For matrix to be invertible, determinant must not be equal to zero, that is matrix should
be non-singular.
𝑎 ℎ
Let matrix be 𝑀 = ( ) then its determinant is |𝑀| = 𝑎𝑏 − ℎ2 ≠ 0 ⇒ 𝑎𝑏 ≠ ℎ2
ℎ 𝑏
Therefore, 𝑀 is a diagonal matrix with non-zero entries in the main diagonal of 𝑀 is not
the square of an integer.
48. (b) :
Given that, matrix of order 3 and |𝐴| = 5, |𝐵| = 3.
Then, |3𝐴𝐵| = 33 |𝐴||𝐵| (since, |𝑘𝐴| = 𝑘 𝑛 |𝐴|)
= 33 (5)(3) = 27(15) = 405
49. (4) :
Given that, A and B are invertible matrices.
So,(𝐴 + 𝐵)′ = 𝐴′ + 𝐵′ ⇒ (𝐴 + 𝐵)𝑟 = 𝐴𝑟 + 𝐵𝑟
But (𝐴 + 𝐵)−1 = 𝐵−1 + 𝐴−1 is not correct.
50. (a):
cos 𝑥 1 0
Given that, 𝑓(𝑥) = | 0 2 cos 𝑥 3 |
0 1 2 cos 𝑥
Expand along first column, we get
= cos 𝑥 (4 cos2 𝑥 − 3) − 0(0 − 2 cos 𝑥) + 0(3 − 0)
= 4 cos3 𝑥 − 3 cos 𝑥 = cos 3𝑥
So, 𝑓(𝑥) = cos 3𝑥
Therefore, lim 𝑓(𝑥) = cos 3𝜋 = −1
𝑥→𝜋
1 2 3
51. Given that, 𝑥 3 − 2𝑥 2 − 9𝑥 + 18 = 0 and 𝐴 = |4 𝑥 6|.
7 8 9
Now find the value of 𝑥 from given equation
𝑥 3 − 2𝑥 2 − 9𝑥 + 18 = 0
⇒ 𝑥 2 (𝑥 − 2) − 9(𝑥 − 2) = 0
⇒ (𝑥 2 − 9)(𝑥 − 2) = 0
 ⇒ (𝑥 − 3)(𝑥 + 3)(𝑥 − 2) = 0
⇒ 𝑥 = 2, 3, −3
For 𝑥 = 2, we get
1 2 3
𝐴 = | 4 2 6|
7 8 9
= 1[(2 × 9) − (6 × 8)] − 2[(4 × 9) − (6 × 7)] + 3[(4 × 8) − (2 × 7)]
= [18 − 48] − 2[36 − 42] + 3[32 − 14]
= −30 + 12 + 54 = 36
For 𝑥 = 3, we get
1 2 3
𝐴 = | 4 3 6|
7 8 9
1[(3 × 9) − (6 × 8)] − 2[(4 × 9) − (6 × 7)] + 3[(4 × 8) − (3 × 7)]
= [27 − 48] − 2[36 − 42] + 3[32 − 21]
= −21 + 12 + 33 = 24
For 𝑥 = −3, we get
1 2 3
𝐴 == |4 −3 6|
7 8 9
1[(−3 × 9) − (6 × 8)] − 2[(4 × 9) − (6 × 7)] + 3[(4 × 8) − (−3 × 7)]
= [−27 − 48] − 2[36 − 42] + 3[32 + 21]
= −75 + 12 + 159 = 96
So, the maximum value of 𝐴 is 96.
52. (b) :
𝑥3 − 1 1<𝑥>∞
Given that, 𝑓(𝑥) = {
𝑥 − 1, ∞<𝑥≥1
3
At 𝑥 = 1, 𝑓(𝑥)𝑥=1 at LHL = (1) − 1 = 0 and 𝑅𝐻𝐿 = 1 − 1 = 0. So, 𝑅𝐻𝐿 = 𝐿𝐻𝐿.
Therefore, 𝑓(𝑛) is continuous
Now, 𝑅𝐻𝐷 = 𝑅{𝑓′(𝑥)} = 3𝑥 2 = 𝑅{𝑓′(1)} = 3(1)2 = 3
𝐿𝐻𝐷 = 𝐿{𝑓′(1)} = 1 = 𝐿{𝑓′(1)} = 1
𝐿𝐻𝐷 ≠ 𝑅𝐻𝐷
Therefore, 𝑓(𝑛) is not differentiable
53. (c) :
Given that, 𝑦 = (cos 𝑥 2 )2
𝑑𝑦
So, 𝑑𝑥
= 2(cos 𝑥 2 )(− sin 𝑥 2 )2𝑥
= −4𝑥 cos 𝑥 2 sin 𝑥 2 = −2𝑥(2 cos 𝑥 2 sin 𝑥 2 ) = −2𝑥 sin(2𝑥 2 )
54. (b) :
𝑑
Given that, (𝑥 𝑥 + 𝑥 𝑎 + 𝑥 𝑥 + 𝑥 𝑎 )
𝑑𝑥
𝑑 𝑑 𝑑 𝑑
= 𝑑𝑥 (𝑥 𝑥 ) + 𝑑𝑥 (𝑥 𝑎 ) + 𝑑𝑥 (𝑥 𝑥 ) + 𝑑𝑥 (𝑥 𝑎 )
𝑥 (1 𝑎−1 𝑥
=𝑥 + log 𝑥) + 𝑎𝑥 + 𝑎 log 𝑎 + 0
𝑥 (1 𝑎−1
=𝑥 + log 𝑥) + 𝑎𝑥 + 𝑎 𝑥 log 𝑎
55. (a) :
 Given that, statement 1 : 𝑦 = log10 𝑥 + log 𝑒 𝑥 and statement 2 :
𝑑 log 𝑥 𝑑 log 𝑥
(log10 𝑥) = log 10 and 𝑑𝑥 (log 𝑒 𝑥) = log 𝑒.
𝑑𝑥
Now, 𝑦 = log10 𝑥 + log 𝑒 𝑥
log 𝑒 𝑥
= + log 𝑒 𝑥 = log10 𝑒 log 𝑒 𝑥 + log 𝑒 𝑥
log 𝑒 10
𝑑𝑦 log10 𝑒 1
So, 𝑑𝑥
= 𝑥
+𝑥
𝑑 log 𝑥 𝑑 log𝑒 𝑥
But 𝑑𝑥 (log10 𝑥) ≠ log 10
and 𝑑𝑥 (log 𝑒 𝑥) ≠ log 𝑒
.
Therefore, statement 1 is true but statement 2 is false.
56. (d) :
𝜃
Given that, 𝑥 = cos 𝜃 + log tan ( ) and 𝑦 = sin 𝜃.
2
𝑑𝑥 1 𝜃 1
So, 𝑑𝜃
= − sin 𝜃 + tan 𝜃/2 sec 2 2 ∙ 2
𝜃
cos 1 1 1
2
= − sin 𝜃 + 𝜃 𝜃 ∙ 2 = − sin 𝜃 + 𝜃 𝜃
sin cos2 2 sin cos
2 2 2 2

1 1 cos2 𝜃
= − sin 𝜃 + = − sin 𝜃 + =
2𝜃 sin 𝜃 sin 𝜃
sin
2
𝑑𝑦
And 𝑑𝑥
= cos 𝜃
𝑑𝑦 cos 𝜃
Then, 𝑑𝑥
= cos2 𝜃
= tan 𝜃 ⇒ tan 𝜃 = 0
sin 𝜃
𝜃 = 𝑛𝜋, 𝑛𝑒𝑍
57. (d) :
Given that, 𝑦 = (1 − 𝑥)2 (𝑥 − 2)3 (𝑥 − 3)5
⇒ log 𝑦 = log[(1 − 𝑥)2 (𝑥 − 2)3 (𝑥 − 3)5 ]
⇒ log 𝑦 = log(1 − 𝑥)2 + log(𝑥 − 2)3 + log(𝑥 − 3)5
⇒ log 𝑦 = 2 log(𝑥 − 1) + 3 log(𝑥 − 2) + 5 log(𝑥 − 3)
1 𝑑𝑦 2 3 5
So, 𝑦 𝑑𝑥 = 𝑥−1 + 𝑥−2 + 𝑥−3
𝑑𝑦 2 3 5
⇒ = 𝑦[ + + ]
𝑑𝑥 𝑥−1 𝑥−2 𝑥−3
𝑑𝑦 2 3 5
⇒ = (1 − 𝑥)2 (𝑥 − 2)3 (𝑥 − 3)5 [ + + ]
𝑑𝑥 𝑥−1 𝑥−2 𝑥−3
At 𝑥 = 4, we get
𝑑𝑦 2 3 5
(𝑑𝑥 ) = 32 ∙ 23 ∙ 15 [ 3 + 2 + 1 ]
𝑥=4
4 + 9 + 30
= 9 × 8[ ] = 12 × 43 = 516
6
58. (d) :
𝑡2 𝑡
Given that, 𝜃 = 20 + 5
𝑑𝜃 𝑡 1
Then, 𝜔= 𝑑𝑡
= 10 + 5
At 𝑡 = 4, we get 𝜔𝑡=4 = 𝑘
 2 1
⇒ + =𝑘
5 5
3
⇒ 𝑘 = ⇒ 5𝑘 = 3
5
59. (c) :
Given that, 𝑦 = 𝛼𝑥 2 − 6𝑥 + 𝛽 (1)
𝑑𝑦
So, 𝑑𝑥
= 2𝛼𝑥 − 6
3
At 𝑥 = 2, the tangent is parallel to 𝑥 − axis. Then,
𝑑𝑦
( ) = 2𝛼𝑥 − 6 = 0
𝑑𝑥 𝑥=3
2
⇒ 3𝛼 − 6 = 0 ⇒ 𝛼 = 2
If point (0, 2) lies on parabola Eq. (1), we get
2=0−0−𝛽 ⇒𝛽 =2
60. (a) :
Given that, 𝑓(𝑥) = 𝑥 2 − 2𝑥
So, 𝑓 ′ (𝑥) = 2𝑥 − 2
Since, 𝑓 ′ (𝑥) < 0
⇒ 2𝑥 − 2 < 0
⇒𝑥<1
Therefore, the given function is strictly decreasing in the interval 𝑥 ∈ (−∞, 1).

BIOLOGY

1. Chapter: Organism and Populations

(3) : I. Cuscuta plant found on hedge plants which does not have leaves and chlorophyll for
manufacturing its own food, they are endoparasites live in the bodies of their hosts.
II. The human liverfluke require two host to complete its life cycle, it uses snail and fish as
intermediate hosts to facili- tate parasitization of its primary host.
III. Endoparasites live in the bodies of their hosts (e.g., in liver, kidneys, red blood cells, etc.). The
life cycles of endoparasites are more complex, as they are more specialized for their mode of
life. Endoparasites have elaborate adaptive mechanisms like holdfast organs for attachment,
obligatory anaerobic respiration, loss of sensory organs, complex life cycles involving one or
more intermediate hosts, many larval stages in the life cycle, high fecundity, etc.
IV. In the brood parasitism, the parasitic bird's eggs have evolved to resemble the eggs of the
host bird. The eggs of parasites mimic those of their hosts, to deceive hosts to accept, have
harder shells, to avoid rejection by punc- ture, and require slightly shorter incubation times,
causing a size advantage for parasitic nestlings. Such as laying eggs by cuckoo (koel) in crow's
nest.
2. Chapter: Organism and Populations
(2) The weed Calotropis have chemical defenses against herbivory. They produce highly toxic
cardiac glycosides, so are avoided by cattle and hence grows abundant in the fields.
 Pisaster is a genus of pacific sea stars, they feed on many invertebrates most commonly on
mussels.
The caterpillars of Monarch butterflies feed on plants of the milkweed family, they do not break
down the chemicals that protect these plants from most herbivores. Instead, they store them in
fat within their bodies.
Some species frogs are cryptically colored, blending col- oration with the background, escaping
predation. It is referred to as camouflage or cryptic coloration (blending coloration). Such as
cryptic coloration of the canyon tree frog (Hyla arenicolor) allows it to disappear on a granite
background.
3. Chapter: Organism and Populations
(1) Small animals have a large surface area as a result of which they tend to lose heat very fast.
This is the reason why they are not observed in polar regions.
4. Chapter: Organism and Populations
(4) Plants in hot and arid climates have crassulacean acid metabolism (CAM) pathway of
photosynthesis in which the stomata remain closed during the daytime.
In seals found in polar seas, the fat layer or blubber present below the skin minimizes loss of
heat and acts as an insulator. Because of the behavioral adaptations, lizard basks in the sun. The
temperature regulation would require lizard Anolis cristatellus to travel long distances to find an
exposed sunny area. Hence its body temperature conforms to that of the environment.
At the higher altitudes, the oxygen is thin, which influences respiration. At higher altitude
polycythemia, that is, increased number of RBCs, is commonly reported in human beings. That's
why tribes living in high altitude have higher red blood cells count or total haemoglobin than
people living in the plains.
5. Chapter: Organism and Populations
(4) Population size helps in determining the position of a population in a given habitat and is
expressed in terms of population density (N), which is defined as the number of individuals per
unit area, or per unit volume. Most populations grow rapidly if optimal conditions for growth
and reproduction of its individuals exist. This is because the size of populations is dependent on
several factors such as availability of resources, predation and unfavorable environmental
conditions.
6. Chapter: Biodiversity and Conservation
(3) The tropical region harbours more number of species as this area remained stable for
millions of years. Hence, species in tropics got a long time for speciation and evolutionary
diversification. Unlike the tropical regions, the temperate regions were subjected to frequent
glaciations in the past.
7. Chapter: Biodiversity and Conservation
(1) An important concept in ecology, the SAR links area with species richness. The bigger the
area, the more are the number of species found. The first mathematical description of the
species-area relationship was proposed by Arrhenius in 1920 and modified by Gleason in 1922.
S=C𝐴2
log S=log C+ Zlog A
 where S = species richness, A = area, Z = the slope of the line (regression coefficient) and C = the
intercept on the y-axis. The Z value is usually higher for islands (around 0.3) than for the
mainland (less than 0.2). Among continents, it varies from 0.6 to 1.2.
8. Chapter: Biodiversity and Conservation
(3) Narrowly utilitarian means that a species or group of species provides a product that is of
direct value to people. Biodiversity has provided us with many goods that have economic value
and can be directly valued. Such as food Fats, oil, medicines, biopesticides, resins, fibers, dyes,
etc.
Broadly utilitarian refers to the public services that nature and its diversity provide that are
essential or valuable to human life, and would be expensive or impossible to replace by direct
human action. Public service also refers to the idea that species have roles in their ecosystems,
and that some of these are necessary for the persistence of their ecosystems. Such as
maintenance of gaseous con- centration, regulation of global cycle, pollination, nutrition cycle,
etc.
Ethical means the philosophical and spiritual belief that species have the right to exist,
independent of their value to people. Spiritual refers to the way that nature and its diversity has
moved people ever since and nature and its diversity have been written about, an uplifting
experience, often perceived as a religious one. Thus, it becomes our moral obligation to
maintain the wellbeing of these spe- cies, so that the future generations also can appreciate
their value.
9. Chapter: Biodiversity and Conservation
(3) The ex involves the conservation of genetic resources of species away from their area of
origin or development. This strategy is aimed to conserve more genetic material in less area. The
various ways in which ex situ conservation can be achieved are tissue culture, cryopreservation,
gene bank and off-site collection.
10. Chapter: Biological Classification
(3) Martinus Beijerinck (1898), the Dutch microbiologist, determined the infectious nature of the
filtrates. He was the first one to characterize viruses. He called this fluid Contagious vivum
fluidum or infectious living fluid. The term virus which had been earlier refer to infectious
agents. According to the "Contagium vivum fluidum" the virus was used by M. W. Beijerinck to
refer to specific pathogenic (disease-causing) molecules incorporated into cells.
11. Chapter: Biological Classification
(1) Colletotrichum (red rot of sugarcane), Alternaria (early blight of potato and tomato),
Trichophyton and Trichoderma (causes skin diseases) are the example of Fungi Imperfecti or
Deuteromycota. These are called imperfect because there is no sexual stage has been observed
in their life cycles. As a result. they cannot be assigned to a taxonomic group. While Ustilago is
the example of Basidiomycetes (The Club Fungi), cause plant diseases that destroy crops, such
as wheat, corn and coffee.
12. Chapter: Plant Kingdom
(1) Laminaria is brown algae belong to class Phaeophyceae These algae usually have a structure
differentiated into a root- like portion or a holdfast, stem-like stalk called stipe that car- ries the
 leaf-like blades or fronds. The larger forms contain air bladders (or sir sacs) for buoyancy. They
are multicellular and may be branched, filamentous or highly branched.
13. Chapter: Animal Kingdom
(4) Class Aves are homeothermic or warm-blooded like mam- mals and can regulate their body
temperature by the energy released during metabolism. The fore limbs are modified into wings
that help in flying while hind limbs or legs of the birds are covered by the scales and are adapted
for perching, walking or swimming, etc., usually bear four, sometimes three and rarely two toes.
The heart of the birds is four-chambered with two atria and two ventricles separated. They are
oviparous, fertilization takes place internally. The female has a well-developed left oviduct (one
on right side being vestigial) while the male does not have any copulatory organ, paired testes
are present.
14. Chapter: Morphology of Flowering Plants
(2) In Epigynous flower, the margins of the thalamus grow upwards and surround the ovary
covering it to the top and fused with it. The floral whorls are thus located above the ovary.
Hence, the ovary is called inferior, while other floral organs are called superior. The condition is
called epigyny.
For example, cucumber, ray florets of sunflower, guava, etc.
15. Chapter: Structural Organisation in Animals
(4) The cells are held together to form tissues and are tightly joined into functional units known
as cell junctions, they are contact points between the plasma membranes of tissue cells. The
three main types of cell junctions present in animal tissue are tight junctions consist of web-like
strands of transmembrane proteins that fuse together the outer surfaces of adjacent plasma
membranes to seal off passageways between adjacent cells (stomach, intestines, etc), adhering
junctions (adherens june- tions, desmosomes and hemidesmosomes) and gap junction in which
the plasma membranes are not fused together as in tight proteins that fuse together une
membranes to seal off passageways between adjacent cells (stomach, intestines, etc.), adhering
junctions (adherens junc- tions, desmosomes and hemidesmosomes) and gap junction in which
the plasma membranes are not fused together as in tight junctions but are separated by a very
narrow intercellular gap (space).
16. Chapter: Cell: The Unit of Life
(1) Appendages used by bacteria for movement is the flagella (singular: flagellum). Flagella are
like a hollow tube made up of protein subunits called flagellin. A bacterial flagellum con- sists of
three parts: the basal body complex, the filament and the hook that connects them. The basal
body is made up of a protein other than flagellin. The structure of the basal body differs subtly
in Gram negative and positive bacteria.
17. Chapter: Biomolecules
(4) Trypsin is a protein, act as enzyme that catalyzes protein degradation, and also called
proteinase. GLUT-4 is used to transport glucose in the cells. Collagen in bone and other con-
nective tissues forms the intercellular ground substance such as keratin in skin, hair and
fingernails. Antibody aid responses that protect body against foreign substances and invading
pathogens.
18. Chapter: Cell Cycle and Cell Division
 (1) Prophase of the first meiotic division (i.e., prophase I) is typically lengthened in
extraordinary fashion when compared to mitotic prophase. The first meiotic prophase is
complex and is divided into leptotene, zygotene, pachytene, diplotene and diakinesis. Synapsis is
the second stage of prophase I occurs in zygotene, it is marked by the visible association of
homologous chromosomes with one another. Synapsis can be of three types: (1) procentric; (2)
proterminal and (3) intermediate. In pachytene, crossing over (exchange of genetic material)
takes place between non-sister chromatids of the homologous chromosomes. Diplotene formed
Chiasmata by covalent junctions between a chromatid from one homologue and a non-sister
chromatid from the other homologue. In Diakinesis, the final stage the chiasmata terminalize,
the meiotic spindle is assembled, the chromosomes become recompacted and they prepared for
separation.
19. Chapter: Photosynthesis in Higher Plants PEPcase is an enzyme phosphoenol pyruvate
carboxylase that catalyzes the carboxylation of PEP in C4 plants. It uses bicar- bonate ion (HC𝑂3− )
as substrate rather than C𝑂2 . C4 pathway (or HSK pathway) is a closed cycle pathway in which
PEPcase is used to convert PEP and bicarbonate into OAA as the first stable product of C𝑂2
fixation.
20. Chapter: Respiration in Plants
(1) The process of Tricarboxylic acid cycle (TCA) takes place in mitochondrial matrix, also known
as Krebs's cycle. It consists of a series of eight reactions that oxidizes the ace- tyl group of acetyl-
CoA to two molecules of C𝑂2 produces reduced compounds NADH and FAD𝐻2 .
Glycolysis, the first step of anaerobic and aerobic respiration takes place in cytoplasm. The
products of glycolysis are 2 molecules of pyruvic acid, 2ATP from substrate level
phosphorylation and 3 ATP from oxidative phosphorylation of NAD𝐻2 molecules.
ATP synthase is also known as proton-pumping ATP syn- thase and 𝐹1 -𝐹0 −ATPase. This enzyme
becomes active in the presence of proton gradient the higher concentration and enter 𝐹1 .
phosphorylation and 3 ATP from oxidative phosphoryla- tion of NADH, molecules.
ATP synthase is also known as proton-pumping ATP syn- thase and 𝐹1 - 𝐹0 -ATPase. This enzyme
becomes active in the presence of proton gradient, the higher concentration of protons is on the
𝐹0 side than on the 𝐹1 side. Due to this, the protons move from higher to lower concentration
and enter 𝐹1 Passage of protons is coupled to the catalytic site in 𝐹1 which is site for synthesis of
ATP from ADP and Pi.
The terminal oxidation takes place in the two processes- electron transport system (ETS)
followed by oxidative phosphorylation. The electrons carried by NADH + 𝐻 + and FAD𝐻2 then
pass into the mitochondrial electron trans- port system (ETS) consisting of linked electron
carriers that are present in the inner mitochondrial membrane. The electrons from electron
transport chain (ETC) are finally passed to 𝑂2 and 𝐻2 O is produced.
21. Chapter: Plant Growth and Development
(4) Derivatives of adenine are N6-furfurylamino purine, kinetin, etc. Kinetin is the examples of
synthetic cytokinins. The natural cytokinins are generally synthesized in root apices, developing
shoot buds, axillary buds, young leaves and fruits, etc. Abscisic acid (ABA) had an important role
in abscission and dormancy. In other cases, its responses have been found to antagonize those
of gibberellins, auxins and cytokinins (e.g.. seed germination, seed or bud dormancy, ripening of
fruits. senescence, transpiration) and keeps them in check, thus con- trolling growth.
22. Chapter: Body Fluid and Circulations
(3) Cardiac output (CO) is the volume of blood ejected from the left ventricle (or the right
ventricle) into the aorta (or pulmo- nary trunk) each minute. Cardiac output equals the stroke
vol- ume (SV), the volume of blood ejected by the ventricle during each contraction, multiplied
by the heart rate (HR), the number of heart beats per minute: CO = SV X HR. The stroke volume,
SV=55ml and heart rate, HR = 70 beats/min.
Therefore, cardiac output = 55 x 70 = 3850 ml.
23. Chapter: Body Fluid and Circulations
(1) The ECG is a composite record of action potentials produced by all the heart muscle fibers
during each heartbeat. In a typical record, three clearly recognizable waves appear with each
heart. P-wave is first wave, it is a small upward deflection on the ECG. It represents atrial
depolarization or excitation. which spreads from the SAN through contractile fibers in both atria
leads to their contraction. QRS complex is second wave begins as a downward deflection,
continues as a large, upright. triangular wave, and ends as a downward wave. It represents rapid
ventricular depolarization, as the action potential spreads through ventricular contractile fibers.
It marks the beginning of systole And T-wave is the third wave, it is a dome-shaped upward
deflection, and it indicates ventricular repolarization It occurs just as the ventricles are starting
to relax. The systole ends at the end of T-wave
24. Chapter: Chemical Coordination and Integration
(1) The adrenal medulla produces three catecholamine hormones-norepinephrine, epinephrine,
and a small amount of dopamine. In stressful situations and during exercise, impulses from the
hypothalamus stimulate sympathetic pre- ganglionic neurons, which in turn stimulate the
chromaffin cells to secrete epinephrine and norepinephrine. These two hormones greatly
augment the fight-or-flight response.
Melanocyte-stimulating hormone (MSH) increases skin pigmentation in amphibians by
stimulating the dispersion of melanin granules in melanocytes (cells containing melanin).
Continued administration of MSH for several days in
humans does produce a darkening of the skin. Thymosin is a peptide hormone produced by the
thymus helps in antibody production for providing humoral immunity.
The pineal gland secretes melatonin, an amine hormone derived from tryptophan. Melatonin
appears to contribute to the setting of the body's biological clock, which is controlled by the
suprachiasmatic nucleus of the hypothalamus. It also helps maintain body temperature and
sleep-wake cycle.
25. Chapter: Sexual Reproduction in Flowering Plants
(2) The anther of angiosperm is bilobed and the two lobes are connected by sterile connective
tissue called connective. In a bilobed anther, each lobe has two theca is called dithecous. A
longitudinal groove separates the theca. When cut transversely, four microsporangia (two in
each lobe) one at each corner, are clearly observed. Eventually, microsporangia develop into
pollen sacs and carry pollen grains.
26. Chapter: Reproduction in Organisms
 (2) In bryophytes and pteridophytes, water serves as a medium for the transfer of gametes.
Since the probability of reaching the female gamete is very less, the male gametes are produced
in large numbers.
27. Chapter: Sexual Reproduction in Flowering Plants
(1) A series of divisions in embryonic development of mono- cot seed leads to formation of two
regions, one of which forms the upper part of the cotyledon while the other forms the lower
part or plumule. The cotyledon (scutellum) grows faster than the plumule and pushes the
plumule to one side. thus, plumule comes to lie in a depression. At this stage the cells in the
middle undergo divisions to form the hypocotyl and radicle. In few cereals, both radicle and
plumule get covered by sheaths developed from scutellum called coleorhiza and coleoptile
respectively. The outer layer of seed is peri- carp. Hence, (1) Coleoptile, (2) Scutellum, (3)
Pericarp and (4) Coleorhiza.
28. Chapter: Sexual Reproduction in Flowering Plants
(2) Tapetum is the innermost layer of cells. These cells are generally polyploid, multinucleate
and possess dense cytoplasm. In tapetal cells, the nucleus divides but cytokinesis does not take
place, so same cell contains two or more nuclei. Its main function is to provide nourishment to
the developing pollen grains. It also deposits a lipid-rich coat containing Ubisch granules around
each microspore for exine formation.
29. Chapter: Sexual Reproduction in Flowering Plants
(2) In the structure of female gametophyte or megagametophyte the eight nuclei so formed get
arranged in two groups of four each. One group of four nuclei is present at the micropylar end
while the other at the chalazal end. One nucleus from each end then migrates towards the
center. These nuclei are known as polar nuclei. Cellularization then occurs in the nuclei at both
ends, that is, three nuclei at micropylar end and three nuclei at chalazal end. The three cells at
the micropy- lar end form the egg apparatus. The two polar nuclei situated below the egg
apparatus are present in a single large central cell and three cells situated at the chalazal end are
called antipodals.
30. Chapter: Reproduction in Organisms
(3) Eichhornia crassipes (water hyacinth) is the example of vegetative propagation. The
reproduction takes place through offsets. The offset can grow into a new plant if it breaks from
the parent plant. Their internodes are shorter and thicker than those of runners. Leaves are
produced at its apex and adventitious roots at the base.
31. Chapter: Sexual Reproduction in Flowering Plants
(4) The process of double fertilization in angiosperm involve two type of fusion in the embryo
sac. Fusion of sperm cell (male gamete) with egg cell (female gamete) to form a diploid zygots
known as syngamy and fusion of the other sperm cell with two polar nuclei to form primary
endosperm cell containing triploid nucleus also known as primary endosperm nucleus (PEN) in
the central cell of embryo sac known as triple fusion.
32. Chapter: Human Reproduction
(2) The major parts of a sperm are the head, neck, middle piece and tail. (1) The acrosome, a
cap-like vesicle filled with enzymes hyaluronidase and proteases that help a sperm to penetrate
the ovum to bring about fertilization. (2) The middle piece contains numerous mitochondria
 arranged in a spiral, which provide the energy (ATP) for locomotion of sperm to the site of
fertilization and for sperm metabolism. (3) The principal piece is the longest portion of the tail,
and the end piece is the terminal, tapering portion of the tail that helps in movement of sperm.
33. Chapter: Human Reproduction
(4) The functions of the mammary glands are the synthesis, secretion, and ejection of milk.
When milk is being produced, it passes from the alveoli into a series of secondary tubules and
then into the mammary ducts. Near the nipple, the mammary ducts again join together to form
a wider mammary ampulla which expand to form sinuses called lactiferous sinuses, where some
milk may be stored before draining into a lactiferous duct. Each lactiferous duct typically carries
milk from one of the lobes to the exterior.
34. Chapter: Human Reproduction
(2) The myometrium is the middle layer of the uterus. The myometrium consists of three layers
of smooth muscle fibers that are thickest in the fundus and thinnest in the cervix. During labor
and childbirth, coordinated contractions of the myometrium in response to oxytocin from the
posterior pituitary help expel the fetus from the uterus.
35. Chapter: Reproductive Health
(1) Birth control refers to restricting the number of children by various methods designed to
control fertility and prevent conception. Contraceptive methods help in preventing unwanted
pregnancies. The properties of an ideal contraceptive are user- friendly (easy to use), affordable
and readily available, effective, not interfere with the sexual desire or drive or act of the user.
They are not regular requirements for the maintenance of reproductive health they are only
used to control unwanted pregnancy not to maintain any health.
36. Chapter: Reproductive Health
(3) Intra-cytoplasmic sperm injection (ICSI) technique is used to treat male infertility that is low
sperm count, low motility of sperm or intracellular sperm infection and in older couples, etc. In
this technique, the sperms are obtained from ejaculation and a single sperm is injected into a
mature egg with the help of a micro needle.
37. Chapter: Principles of Inheritance and Variation
(3) Aneuploidy refers to a numerical change in part of the genome, usually a change in a single
chromosome. Individuals having an extra chromosome or missing a chromo- some are said to be
aneuploid.
Monoploidy is define as presence of one half of the normal number of chromosomes.
Polyploidy occurs with the presence of extra chromosome sets. A polyploidy organism has more
than two sets of chromosomes (2n diploid) or genomes-for example, trip- loid (3n), tetraploid
(4n), pentaploid (5n), hexaploid (6n), heptaploid (7n), octoploid (8).
The presence of two sets of chromosomes or two copies of genes is known as diploidy.
38. Chapter: Principles of Inheritance and Variation
(2) The genotype of a husband and wife are 𝐼 𝐴 𝐼𝐵 and 𝐼 𝐴 𝐼𝑜 . Hence, four different types of
genotypes are possible of their children that are 𝐼 𝐴 𝐼𝐵 , 𝐼 𝐴 𝐼 𝑂 , 𝐼 𝐵 𝐼 𝑂 , 𝐼 𝐴 𝐼 𝐴 IP. And their possible
phenotypes are blood group A with genotype 𝐼 𝐴 𝐼 𝐴 , 𝐼 𝐴 𝐼 𝑂 , blood group B with gen- otype 𝐼 𝐵 𝐼 𝑂
and blood group AB with genotype 𝐼 𝐴 𝐼𝐵 .
39. Chapter: Principles of Inheritance and Variation
 (2) The A, B and O blood groups are controlled by three different alleles of gene I. These are
𝐼 𝐴 , 𝐼 𝐵 and i. The slight difference in these alleles lies in the composition of sugar obtained from
them. The sugar produced by alleles 𝐼 𝐴 and 𝐼 𝐵 are slightly different, while no sugar is produced
by allele i. Each person can possess two of the three alleles. It has been found that 𝐼 𝐴 and 𝐼 𝐵 are
completely dominant over i because i does not produce any sugar. Thus, in genotype 𝐼 𝐴 i, only
𝐼 𝐴 expresses that is blood group A and in genotype 𝐼 𝐵 i, only 𝐼 𝐵 expresses that is blood group B.
Similarly, in genotype 𝐼 𝐴 𝐼 𝐵 both 𝐼 𝐴 𝐼 𝐵 is expressed that is blood group AB and in genotype ii no
gene is expressed which is known as blood group O. Hence, the blood group of children whose
parents are heterozygous for A and B blood can be A, B, A B or O.
40. Chapter: Principles of Inheritance and Variation
(4) Presence of an extra copy of a chromosome is consider as trisomy and If the trait is
expressed equally in both males and females it is known as autosomal. For example, Down's
syndrome (47, +21) in which chromosome number 21 is in triplicate in human beings.
In some cases, zygote with an extra Y (XYY) develops into a physically normal male who is likely
to be taller than average. The individual has 47 chromosomes (2n+1). This condition is called
Klinefelter syndrome, which is characterized by mental retardation, sterility and the presence of
feminine physical characteristics.
A single homologous chromosome is lost in monosomy The monosomics have 2n-1 number of
chromosomes. For example, Turner syndrome in which 45, X individuals originate from eggs or
sperm that lack a sex chromosome or from loss of a sex chromosome in mitosis sometime after
fertilization.
Mendelian disorders are inherited to the offspring following principle of inheritance. Cystic
fibrosis is an autosomal recessive mutation, this includes excessively salty sweat, lungs, pancreas
and liver become clogged with a thick mucus, leading to chronic infections and eventual
malfunctions of these vital organs.
41. Chapter: Principles of Inheritance and Variation
(3) Gregor Johann Mendel explained principles of heredity and variation through his
hybridization experiments on the garden pea plant, Pisum sativum. Mendel chose 14 pea plants
to focus on seven clearly definable characters including plant height and flower color, each of
which occurred in two allelomorphic traits. The flower colour has purple and white colour that is
dominant and recessive respectively.
42. Chapter: Molecular Basis of Inheritance
(3) The transforming principle was discovered by the Frederick Griffith, while studying
pneumococcal infections in mice in 1928. Transformation is the change in an organ- ism's
characteristics because of the transfer of genetic information.
Francois Jacob and Jacques Monod proposed the operon model based on their studies on the
lac operon in 1961 and were first to elucidate transcription in regulated gene expression. Their
operon model explained the regulation of genes required for lactose utilization in E. coli.
The template mechanism of DNA replication given by Watson and Crick was proven correct by
Matthew Meselson and Franklin Stahl in 1958. They used heavy ( 15𝑁) and light ( 14𝑁) isotopes
(not radioactive) of nitrogen to distinguish between parental and newly synthesized DNA
 strands. The results of Meselson and Stahl's experiment ruled out the possibility of conservative
and dispersive schemes of replication and supported semiconservative DNA replication.
Hershey and Martha Chase conducted experiments to establish that DNA is the genetic material.
They used bacteriophages (viruses that infect bacteria) and proved that the genetic information
of the bacterial virus bacterio- phage 𝑇2 was present in its DNA.
43. Chapter: Molecular Basis of Inheritance
(3) The amino acids are attached to the tRNAs by high-energy bonds between the carboxyl
groups of the amino acids and the 3'-OH termini of the tRNAs. This process is called charging of
tRNA or aminoacylation of tRNA.
44. Chapter: Molecular Basis of Inheritance
(3) In the experiment of Meselson and Stahl's, heavy (15N) and light (14N) isotopes (not
radioactive) of nitrogen are used to distinguish between parental and newly synthesized DNA
strands. After one generation (after 20 min in which E. coli divides), all DNA molecules would be
15
𝑁- 14𝑁 hybrids. Their density would be halfway between that expected for totally heavy and
totally light DNA. After the second generation (40 min), two types of bands would appear in the
gradients-one containing 15𝑁- 14𝑁 hybrids and another containing only 14N.
45. Chapter: Molecular Basis of Inheritance
(1) In 1958, semi-conservative replication was shown for the first time in eukaryotic
chromosomes in root tip cells of Vicia fabu by J. Herbert Taylor, Philip Woods and Walter
Hughes. Vicia faba chromosomes were labeled by growing root tips in medium containing
radioactive 𝐻 3 -thymidine later in non- radioactive alkaloid colchicine. It was observed that both
chromatids of each pair were similarly labeled at the first c-meta-phase but only one of the
chromatids of each pair was radioactive in second c-metaphase.
46. 46. Chapter: Molecular Basis of Inheritance
(2) The two DNA strands are of opposite polarity. In the process of transcription, the template
strand that codes for mRNA have polarity 3-5'and coding strand have polarity 5'-3' that does not
code for anything. Hence, the sequence in the corresponding region of the transcribed m-RNA
will be 5'-UAC GAA GGC UUA-3'.
47. Chapter: Human Health and Diseases
(3) Pneumonia is an infection of the lungs which is usually caused by bacteria such as
Streptococcus pneumoniae or Haemophiles influenzae. The small droplets containing these
bacteria settle in the alveoli and passages of the lung and the bacteria rapidly multiply. The part
of the lung then becomes filled with pus and fluid.
48. Chapter: Human Health and Diseases
(4) Passive immunity produced when ready-made antibodies are introduced into the body. It is
called passive because the host's own immune system does not make antibodies. Such as
antibodies made by a mother's immune system transferred to her offspring or antibodies made
by other hosts introduced into a new host. Hence, the development of quick immune response
in a person infected with deadly microbes by administering pre- formed antibodies is passive
immunity.
49. Chapter: Human Health and Diseases
 (2) Malignant tumor developed through a multi-step process (tumorigenesis) characterized by a
progression of genetic and sometimes by both non-genetic or epigenetic alterations that make
the cells increasingly less responsive to the body's normal regulatory machinery and better able
to invade normal tissues. This tumor starts growing rapidly and spreads to the other organs of
the body through blood and lymph, these cancerous cells continue to divide and develop
secondary tumors called metastasis.
50. Chapter: Human Health and Diseases
(4) Timely detection and diagnosis can prevent cancer dissemination from the origin. Cancer
detection is based on many techniques such as non-invasive in vivo magnetic resonance imaging
(MRI) that uses magnetic field and non-ionizing radiations to study the changes in living tissue.
Radiology can detect large group of cancer cells, but it is not successful to detect a single cancer
cell or even a few. The computed tomography (CT) scan biopsy can detect size, shape location of
the cancerous cells and help to remove the small piece of tissue.

51. Chapter: Human Health and Disease

(2) The leaves of the shrub Erythroxylum coca used to extract an additive drug coca alkaioid or
cocaine. Coke is the cocaine used as the illegal drug. It brings users a strong sense of euphoria
and energy before leading to agitation, depression and paraInoia. It is the second most-used
illegal drug that is native to America.
52. Chapter: Microbes in Human Welfare
(4) The Indian Agricultural Research Institute (IARI) has been a pioneer in developing the
technology for biogas production. After the installation of the first plant in 1941, several models
were developed. The Khadi and Village Industries Commission (KVIC) developed another model
in 1960s for the production of biogas.
53. Chapter: Microbes in Human Welfare
(3) The Swiss cheese is produced by fermentation of lactic acid by the bacterium
Propionibacterium shermanii that produces propionic acid, acetic acid and carbon dioxide. The
acids provide flavor to the cheese and the carbon dioxide, which becomes trapped in the curd,
produces the characteristic holes in the cheese.
54. Chapter: Microbes in Human Welfare
(1) Cyclosporin A is produced by the fungus Trichoderma polysporum and Cylindrocarpon
lucidum. It is used as an immunosuppressive agent that prevents rejection in patients who have
had transplants of kidney, liver, pancreas, etc.
Streptokinase is obtained from Streptomyces sp. works as a tissue plasminogen activator. It
helps in dissolving blood clots inside the blood vessels and has fibrinolytic effect.
Statins are produced by the yeast Monascus purpureus. They are used as agents that lower the
blood-cholesterol. The enzyme 3-hydroxy-3-methyl-glutaryl-CoA reductase (HMG CoA
reductase) is involved in synthesis of cholesterol in the liver. Statin inhibits this enzyme from
catalyzing the reaction for cholesterol synthesis. Penicillin amidase obtained from Bacillus, used
in pharmaceuticals as antibiotics.
55. Chapter: Biotechnology: Principles and Processes
 (1) Taq polymerase is a restriction endonuclease obtained from the bacterium Thermus
aquaticus. It is generally used to carry out the amplification. This organism lives in hot springs,
hence is thermostable and resistant to denaturation by heat treatment.
56. Chapter: Biotechnology: Principles and Processes
(3) Rop gene which codes for the proteins involved in the replication of the plasmid pBR322 in E.
coli is located at restriction site of Pvull. The repressor of primer (Rop) gene codes for Rop
protein that controls number of copies and prevents the replication of pBR322.
57. Chapter: Biotechnology: Principles and Processes
(1) The polymerase chain reaction (PCR) is a technique in which a short region of a DNA
molecule is copied many times in vitro in presence of DNA polymerase enzyme. Hence, this is
the technique that used to detect the genetic material, ribonucleic acid or RNA of virus that is
responsible for COVID- 19 that is SARS-CoV-2. In Rapid antigen test and RT-PCR test for Covid-19
virus PCR, a molecular diagnostic tool, stands for Polymerase Chain Reaction.
58. Chapter: Biotechnology and Its Applications
(2) Polymerase Chain Reaction (PCR) is technique that permits diagnostic tests to be performed
on small amounts of DNA. PCR can generate large amounts of DNA from minuscule samples,
such as that in a single cell. Since the DNA segment of interest is amplified, so very less amount
is sufficient. The use of PCR technique allows the detection of presence of dis- ease-causing
bacteria or viruses much before the disease manifests itself, thereby allowing early detection
and treatment.
59. Chapter: Biotechnology and Its Applications
(3) To prevent the infestation by nematodes in tobacco plants, a technique based on the process
of RNA interference (RNAi) or post transcriptional gene silencing (PTGS) has been adopted. RNA
interference (RNAi) is a process in which a specific mRNA is degraded due to the presence of a
double-stranded small interference RNA (dsRNA) whose sequence is contained within the mRNA
sequence. After entering the cell, dsRNA produced from an infection produce short interfering
RNA (siRNA). siRNA unwinds into two strands-passenger strand (sense) and guide strand
(antisense). Both sense as well as anti- sense RNA are form dsRNA being complementary to each
other. This siRNA binds with the RNA induced silencing complex (RISC). This prevents the
translation of specific mRNA known as gene silencing.
60. Chapter: Biotechnology and Its Applications
(3) A transgenic sheep, Tracy was produced in 1985 that contained a human gene which
expressed high levels of the human protein a-1-antitrypsin deficiency of which can lead to a rare
form of emphysema in humans.