NEET Previous Year Question Paper 2024 Code S5 with Solutions – Free PDF Download

 

QUESTIONS

1. In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:
(1) BA and DC
(2) AB and DC
(3) BA and CD
(4) AB and CD
Answer (2)
Sol. North of magnet is moving away from solenoid 1 so end B of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end C of solenoid 2 is South.

2. 290/82 X → α → Y → α → Z → β → P → α → Q
In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are
(1) 286,81
(2) 280,81
(3) 286,80
(4) 288,82
Answer (1)
Sol. 290/82 X → α → Y → α → Z → β → P → α → Q
A → 286
Z = 81

3. In a vernier callipers, (N + 1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:
(1) 10(N + 1)
(2) 1/(10N)
(3) 1/[100(N + 1)]
(4) 100N
Answer (3)

6. The output (Y) of the given logic gate is similar to the output of an/a
(1) AND gate
(2) NAND gate
(3) NOR gate
(4) OR gate
Answer (1)
Sol. Y1 = A · A = A
Y2 = B + B = B
Y = Y1 + Y2 = A + B = A · B
= A,B is similar to output of AND Gate

7. The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm². The length of the 400 g rod is nearly:
(1) 72.0 cm
(2) 8.5 cm
(3) 17.5 cm
(4) 20.7 cm
Answer (2)
Sol. Moment of inertia of rod = I = mℓ²/12
⇒ 2400 = 400 ℓ²/12
⇒ 72 = ℓ²
⇒ ℓ = √72 = 8.48 cm = 8.5 cm

8. If the monochromatic source in Young's double slit experiment is replaced by white light, then
(1) All bright fringes will be of equal width
(2) Interference pattern will disappear
(3) There will be a central dark fringe surrounded by a few coloured fringes
(4) There will be a central bright white fringe surrounded by a few coloured fringes
Answer (4)
Sol. At central point on screen, path difference is zero for all wavelength. So, central bright fringe is white and other fringes depend on wavelength as β = λD/d Therefore, other fringes will be coloured.

9. The quantities which have the same dimensions as those of solid angle are:
(1) angular speed and stress
(2) strain and angle
(3) stress and angle
(4) strain and arc
Answer (2)
Sol. Solid angle dΩ = dA/r² has dimensions [M0L0T0]
Strain = Δl/l has dimensions [M0L0T0]
Angle measured in radians is also dimensionless [M0L0T0]
θ = l/r

10. In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8 × 10⁻⁶ kg m². If the magnitude of magnetic moment of the needle is x × 10⁻⁵ Am², then the value of 'x' is:
(1) 1280π²
(2) 5π²
(3) 128π²
(4) 50π²
Answer (1)
Sol. Time period of Oscillation, T = 2π√(I/MB)
⇒ 1/4 = 2π√(9.8 × 10⁻⁶ / (M × 0.049))
⇒ 1/16 = 4π² × 9.8 × 10⁻⁶ / (M × 49 × 10⁻³)

11. Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity ν1 while body B is at rest before collision. The velocity of the system after collision is ν2. The ratio ν1 : ν2 is
(1) 1:4
(2) 1:2
(3) 2:1
(4) 4:1
Answer (3)
Sol. Before collision ⇒ A → ν1 rest
It undergoes completely inelastic collision
Using conservation of linear momentum
Initial momentum = Final momentum
⇒ mν1 = mν2 + mν2
⇒ mν1 = 2mν2
⇒ ν1/ν2 = 2/1

12. A horizontal force 10 N is applied to a block A as shown in figure. The mass of blocks A and B are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is:
(1) 10 N
(2) Zero
(3) 4 N
(4) 6 N
Answer (4)
Sol. F = (M1 + M2)a
a = 10/(2 + 3) = 2 ms⁻²
F = M2(2) = 3 × 2 N = 6 N

13. A logic circuit provides the output Y as per the following truth table:
A B Y
0 0 1
0 1 0
1 0 1
1 1 0
The expression for the output Y is:
(1) B
(2) AB + A
(3) AB + A
(4) B
Answer (4)
Sol. According to given truth table, output is independent on value of A
Output Y = B

14. A bob is whirled in a horizontal plane by means of a string with an initial speed of α rpm. The tension in the string is T. If speed becomes 2α while keeping the same radius, the tension in the string becomes:
(1) √2T
(2) T
(3) 4T
(4) T/4
Answer (3)

15. The mass of a planet is 1/10 that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:
(1) 3.92 m s⁻²
(2) 19.6 m s⁻²
(3) 9.8 m s⁻²
(4) 4.9 m s⁻²
Answer (1)
Sol. g' = GM'/R'² = GM / [10(R/2)²]
= 4GM/(10R²) = 0.4 × 9.8
= 3.92 m s⁻²

16. Given below are two statements: Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges. Statement II: Atoms of each element are stable and emit their characteristic spectrum. In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Statement I is incorrect but Statement II is correct
(2) Both Statement I and Statement II are correct
(3) Both Statement I and Statement II are incorrect
(4) Statement I is correct but Statement II is incorrect
Answer (4)
Sol. Statement I is true as atoms are electrically neutral because they contain equal number of positive and negative charges. Statement II is wrong as atom of most of the elements are stable and emit characteristic spectrum. But this statement is not true for every atom.

17. The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are 8 × 10⁸ N m⁻² and 2 × 10¹¹ N m⁻², is:
(1) 8 mm
(2) 4 mm
(3) 0.4 mm
(4) 40 mm
Answer (2)
Sol. In the case for maximum elongation, Stress = Elastic limit
δmax = σelastic × L / Young's modulus = 8 × 10⁸ × 1 / (2 × 10¹¹) = 4 × 10⁻³ = 4 mm
i.e. maximum elongation is 4 mm

18. The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is:
(1) 10 V
(2) 4 V
(3) 6 V
(4) 8 V
Answer (4)
Sol. Current in circuit i = 10/(4 + 1) = 2A
Terminal voltage = E - iR = 10 - 2 × 1 = 8V

19. A thermodynamic system is taken through the cycle abcd. The work done by the gas along the path bc is:
(1) -60J
(2) Zero
(3) 30J
(4) -90J
Answer (2)
Sol. Path bc is an isochoric process.
Work done by gas along path bc is zero.

20. Consider the following statements A and B and identify the correct answer:
A. For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph.
B. In a reverse biased pn junction diode, the current measured in (μA) is due to majority charge carriers.
(1) Both A and B are incorrect
(2) A is correct but B is incorrect
(3) A is incorrect but B is correct
(4) Both A and B are correct
Answer (2)
Sol. A: Solar cell characteristics
B: In reverse biased pn junction diode, the current measured in (μA) is due to minority charge carrier.

21. Match List-I with List-II.
List-I (Material)
List-II (Susceptibility χ)
A. Diamagnetic
B. Ferromagnetic
C. Paramagnetic
D. Non-magnetic
I. 0 < χ < ε (a small positive number)
Choose the correct answer from the options given below
(1) A-IV, B-III, C-II, D-I
(2) A-II, B-III, C-IV, D-I
(3) A-II, B-I, C-III, D-IV
(4) A-III, B-II, C-I, D-IV
Answer (2)
Sol. (Material) (Susceptibility χ)
Diamagnetic
Ferromagnetic
Paramagnetic
Non-magnetic

22. Match List I with List II.
List I (Spectral Lines of Hydrogen for transitions from)
List II (Wavelengths (nm))
A. n2 = 3 to n1 = 2 I. 410.2
B. n2 = 4 to n1 = 2 II. 434.1
C. n2 = 5 to n1 = 2 III. 656.3
D. n2 = 6 to n1 = 2 IV. 486.1
Choose the correct answer from the options given below:
(1) A-I, B-II, C-III, D-IV
(3) A-III, B-IV, C-II, D-I
Answer (3)

23. At any instant of time t, the displacement of any particle is given by 2t - 1 (SI unit) under the influence of force of 5 N. The value of instantaneous power is (in SI unit):
(1) 6
(2) 10
(3) 5
(4) 7
Answer (2)
Sol. x = 2t - 1
ν = dx/dt = 2 m s⁻¹
P = F,ν = 2 × 5 = 10 W

24. In the following circuit, the equivalent capacitance between terminal A and terminal B is:
(1) 4 μF
(2) 2 μF
(3) 1 μF
(4) 0.5 μF
Answer (2)
Sol. Given circuit is balanced Wheatstone bridge
CAB = 1 + 1 = 2 μF

25. A wire of length r and resistance 100Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
(1) 60Ω
(2) 26Ω
(3) 52Ω
(4) 55Ω
Answer (3)
Sol. Divided into 10 parts R = ρl/A
R' = ρl/(10A) = R/10
RS = 5 × R/10 [series]
RS = 50
RP = R/50 [parallel]
Req = RS + RP = 52Ω

26. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The potential (V) at any axial point, at 2m distance (r) from the centre of the dipole of dipole moment vector P of magnitude, 4 × 10⁻⁶ C m, is ±9 × 10³ V.
(Take 1/(4πε0) = 9 × 10⁹ SI units)
Reason R: V = ± 2P/(4πε0 r²), where r is the distance of any axial point, situated at 2m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
(1) A is false but R is true.
(2) Both A and R are true and R is the correct explanation of A.
(3) Both A and R are true and R is NOT the correct explanation of A.
(4) A is true but R is false.
Answer (4)
Sol. The potential V at any point, at distance r from centre of dipole = KP cosθ/r²
At axial point where θ = 0°, V = KP/r² = 9 × 10⁹ × 4 × 10⁻⁶ / 2² = 9 × 10³ V
At axial point where θ = 180°, V = -KP/r² = -9 × 10³ V

27. If c is the velocity of light in free space, the correct statements about photon among the following are:
A. The energy of a photon is E = hν
B. The velocity of a photon is c.
C. The momentum of a photon, ρ = hν/c
D. In a photon-electron collision, both total energy and total momentum are conserved.
E. Photon possesses positive charge.
Choose the correct answer from the options given below:
(1) A,B,D and E only
(2) A and B only
(3) A,B,C and D only
(4) A,C and D only
Answer (3)
Sol. (A) If c is the velocity of light so, E = hν (Energy of photon)
(B) Velocity of photon is equal to velocity of light i.e. c.
λ = h/ρ
p = h/λ
p = hν/c
(D) In photon-electron collision both total energy and total momentum are conserved.

28. A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π × 10⁻⁷ SI units):
(1) 44T
(2) 44mT
(3) 4.4T
(4) 4.4mT
Answer (4)
Sol. The magnitude of magnetic field due to circular coil of N turns is given by
Bc = μ0 i N / (2R)
= (4π × 10⁻⁷ × 7 × 100) / (2 × 0.1)
= 4.4 × 10⁻³ T
= 4.4 mT

29. A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m-1, then the excess force required to take it away from the surface is
(1) 99N
(2) 19.8mN
(3) 198N
(4) 1.98mN
Answer (2)

30. An unpolarised light beam strikes a glass surface at Brewster's angle. Then
(1) The reflected light will be completely polarised but the refracted light will be partially polarised.
(2) The reflected light will be partially polarised.
(3) The refracted light will be completely polarised.
(4) Both the reflected and refracted light will be completely polarised.
Answer (1)
Sol. According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized.

31. The graph which shows the variation of (1/λ²) and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle):
(1)
(2)
(3)
(4)
Answer (1)

32. A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively)?
(1) Point P has zero speed
(2) Point P moves slower than point Q
(3) Point P moves faster than point Q
(4) Both the points P and Q move with equal speed
Answer (3)
Sol. In the case of pure rolling,
The topmost point will have velocity 2v while point Q i.e. lowest point will have zero velocity. Hence point P moves faster than point Q.

33. If x = 5 sin(πt + π/3) m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
(1) 5m 1s
(2) 5cm 2s
(3) 5m 2s
(4) 5cm 1s
Answer (3)

34. A particle moving with uniform speed in a circular path maintains:
(1) Varying velocity and varying acceleration
(2) Constant velocity
(3) Constant acceleration
(4) Constant velocity but varying acceleration
Answer (1)
Sol. A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.

35. A light ray enters through a right angled prism at point P with the angle of incidence 30° as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is:
(1) √3/2
(2) √5/4
(3) √5/2
(4) √3/4
Answer (3)
Sol. A = 90°
In prism, r1 + c = A
r1 = 90° - c
sin c = 1/μ ⇒ cos c = √(μ² - 1)/μ
Apply Snell's law, on incidence surface
1 · sin 30° = μ sin(r1) ⇒ 1 × 1/2 = μ × sin(90° - c)
1/2 = μ × √(μ² - 1)/μ

36. The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is:
(1) GmM/(3R)
(2) 5GmM/(6R)
(3) 2GmM/(3R)
(4) GmM/(2R)
Answer (2)
Sol. Apply energy conservation,
Ui + Ki = Ui + Ki
⇒ -GmM/R + Ki = -GmM/(3R) + 1/2 mν²
⇒ -GmM/R + Ki = GmM/(3R) + 1/2 × m × GM/(3R)
⇒ Ki = 1/6 GmM/R + GmM/R
Ki = 5/6 GmM/R

37. The property which is not of an electromagnetic wave travelling in free space is that:
(1) They originate from charges moving with uniform speed
(2) They are transverse in nature
(3) The energy density in electric field is equal to energy density in magnetic field
(4) They travel with a speed equal to 1/√(μ0ε0)
Answer (1)
Sol. The EM waves originate from an accelerating charge. The charge moving with uniform velocity produces steady state magnetic field.

38. A force defined by F = αt² + βt acts on a particle at a given time t. The factor which is dimensionless, if α and β are constants, is:
(1) αβ/t
(2) βt/α
(3) αt/β
(4) αβt
Answer (3)

39. If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x/2 times its original time period. Then the value of x is:
(1) 4
(2) √3
(3) √2
(4) 2√3
Answer (3)
Sol. T' = 2π√(ℓ'/g) where ℓ' = ℓ/2
T = 2π√(ℓ/g)
T' = x/2 T
2π√(ℓ/2g) = x/2 2π√(ℓ/g)
1/√2 = x/2 ⇒ x = √2

40. A metallic bar of Young's modulus, 0.5 × 10¹¹ N m⁻² and coefficient of linear thermal expansion 10⁻⁵ °C⁻¹, length 1 m and area of cross-section 10⁻³ m² is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is:
(1) 2 × 10³ N
(2) 5 × 10³ N
(3) 50 × 10³ N
(4) 100 × 10³ N
Answer (3)
Sol. Thermal strain = Longitudinal strain = αΔT
⇒ Longitudinal strain, δ = 10⁻⁵ × 10² = 10⁻³
⇒ Compressive stress = δ × Young's Modulus = 10⁻³ × 0.5 × 10¹¹ = 0.5 × 10⁸
⇒ Compressive force = 0.5 × 10⁸ × 10⁻³ = 0.5 × 10⁵ = 5 × 10⁴ × 10/10 = 50 × 10³ N

41. A 10 μF capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π = 3.14):
(1) 0.35A
(2) 0.58A
(3) 0.93A
(4) 1.20A
Answer (3)
Sol. Capacitive Reactance XC = 1/ωC = 1/(2πfC) = 1/(2 × 3.14 × 50 × 10 × 10⁻⁶) = 1000/3.14
Vrms = 210 V
irms = Vrms/XC = 210/XC
Peak current = √2 irms = √2 × 210/1000 × 3.14 = 0.932 = 0.93 A

42. A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:
(1) 32
(2) 34
(3) 28
(4) 17
Answer (3)
Sol. f0 = 140 cm and fe = 5 cm
For distant object,
m = f0/fe = 140/5 = 28

43. The following graph represents the T - V curves of an ideal gas (where T is the temperature and V the volume) at three pressures P1, P2 and P3 compared with those of Charles's law represented as dotted lines.
Then the correct relation is:
(1) P1 > P2 > P3
(2) P3 > P2 > P1
(3) P1 > P3 > P2
(4) P2 > P1 > P3
Answer (1)

44. An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60° with each other. The magnetic moment of this new magnet is:
(1) M/√3
(2) M
(3) M/2
(4) 2M
Answer (3)
Sol. M = ml
Δl = 2(l/2) sin30° = l/2
M' = ml/2 = M/2

45. The velocity (v) - time (t) plot of the motion of a body is shown below:
The acceleration (a)-time (t) graph that best suits this motion is:
(1)
(2)
(3)
(4)

46. A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:
(1) Displacement current of magnitude greater than I flows but can be in any direction
(2) There is no current
(3) Displacement current of magnitude equal to I flows in the same direction as I
(4) Displacement current of magnitude equal to I flows in a direction opposite to that of I
Answer (3)
Sol. According to modified Ampere's law ∮ B.dl = μ0(Ic + Id)
For Loop L1 Ic ≠ 0 and Id = 0
For Loop L2 Ic = 0 and Id ≠ 0
Due to KCL Ic = Id

47. A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:
A. hold the sheet there if it is magnetic.
B. hold the sheet there if it is non-magnetic.
C. move the sheet away from the pole with uniform velocity if it is conducting.
D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar.
Choose the correct statement(s) from the options given below:
(1) C only
(2) B and D only
(3) A and C only
(4) A, C and D only
Answer (3)
Sol. A. A magnetic pole will repel or attract magnetic sheet so force is need.
B. If sheet is non-magnetic, no force needed.
C. If it is conducting, then there will be eddy current in sheet, which opposes the motion. So forces is needed move sheet with uniform speed.
D. The non-conducting and non-polar sheet do not interact with magnetic field of magnet.

48. Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
(1) 2:3
(2) 1:1
(3) 2:9
(4) 1:2
Answer (3)
Sol. Power Consumed = P = V²/R
PA/PB = RB/RA
RA = 2RB
For Series Combination PS = V²/(3RB)
For Parallel Combination PP = 3V²/(2RB)
PS/PP = 2/9

49. Choose the correct circuit which can achieve the bridge balance.
(1)
(2)
(3)
(4)
Answer (2)
Sol. In option (2),
10/15 = 10/(5 + RD)
The diode can conduct and have resistance RD = 10 Ω because diode have dynamic resistance. In that case bridge will be balanced.

50. If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then A. the charge stored in it, increases. B. the energy stored in it, decreases. C. its capacitance increases. D. the ratio of charge to its potential remains the same. E. the product of charge and voltage increases. Choose the most appropriate answer from the options given below:
(1) A, B and C only
(2) A, B and E only
(3) A, C and E only
(4) B, D and E only
Answer (3)
Sol. Given V = V = Constant
(i) C' = ε0A/d', C = ε0A/d
d' < d
C' > C
Hence, final capacitance greater than initial capacitance.
(ii) U' = 1/2 C'V²
U = 1/2 CV²
U' > U
Hence final energy is greater than initial energy
(iii) Q'/V' = C' and Q/V = C
Q'/V' = Q/V
(iv) Product of charge and voltage X' = Q'V = C'V²
X = QV = CV²
X' > X

CHEMISTRY
SECTION-A

51. For the reaction 2A ⇌ B + C, Kc = 4 × 10⁻³. At a given time, the composition of reaction mixture is: [A] = [B] = [C] = 2 × 10⁻³ M. Then, which of the following is correct?
(1) Reaction has gone to completion in forward direction.
(2) Reaction is at equilibrium.
(3) Reaction has a tendency to go in forward direction.
(4) Reaction has a tendency to go in backward direction.
Answer (4)
Sol. 2A ⇌ B + C, Kc = 4 × 10⁻³
At a given time t, Qc is to be calculated and been compared with Kc.
Qc = [B][C]/[A]² = (2 × 10⁻³)(2 × 10⁻³)/(2 × 10⁻³)²
Qc = 1
As Qc > Kc, so reaction has a tendency to move backward.

52. A compound with a molecular formula of C6H14 has two tertiary carbons. Its IUPAC name is:
(1) 2,2-dimethylbutane
(2) n-hexane
(3) 2-methylpentane
(4) 2,3-dimethylbutane
Answer (4)
Sol. CH3 – CH2 – CH2 – CH2 – CH2 – CH3 has no tertiary carbon (n-Hexane)
H3C – CH2 – CH2 – CH – CH3 has only one tertiary carbon (2-Methylpentane)
H3C – CH – CH – CH3 has two tertiary carbon. (2, 3-Dimethylbutane)
H3C – C – CH2 – CH3 has no tertiary carbon (2, 2-Dimethylbutane)

53. Given below are two statements:
Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane
Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct
(2) Both Statement I and Statement II are correct
(3) Both Statement I and Statement II are incorrect
(4) Statement I is correct but Statement II is incorrect
Answer (2)
Sol. Both statement I and statement II are correct.
Boiling point of n-pentane = 309 K
isopentane = 301 K
neopentane = 282.5
As branching increases molecules attain the shape of a sphere results in smaller area of contact thus weak intermolecular forces between spherical molecules, which are overcome at relatively lower temperature. Leading to decrease in boiling point.

54. The energy of an electron in the ground state (n = 1) for He+ ion is –x J, then that for an electron in n = 2 state for Be3+ ion in J is
(1) -4/9 x
(2) –x
(3) -x/9
(4) –4x
Answer (2)
Sol. En = -RH(Z²/n²) J
For He+ (n = 1),
En = -x = -RH(2²/1²) = -4RH
∴ RH = x/4
For Be3+ (n = 2),
En = -RH(Z²/n²) J
= -x/4 × (4×4)/(2×2) = -x J

55. Among Group 16 elements, which one does NOT show –2 oxidation state?
(1) Po
(2) O
(3) Se
(4) Te
Answer (1)
Sol. Oxygen shows –2, –1, +1 and +2 oxidation states
Selenium shows –2, +2, +4 and +6 oxidation states
Tellurium shows –2, +2, +4 and +6 oxidation states
Polonium shows +2 and +4 oxidation states

56. Match List I with List II.
List I (Complex) List II (Type of isomerism)
A. [Co(NH3)5(NO2)]Cl2 I. Solvate isomerism
B. [Co(NH3)5(SO4)]Br II. Linkage isomerism
C. [Co(NH3)6][Cr(CN)6] III. Ionization isomerism
D. [Co(H2O)6]Cl3 IV. Coordination isomerism
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-III, D-I
(2) A-II, B-III, C-IV, D-I
(3) A-I, B-II, C-IV, D-III
(4) A-I, B-IV, C-II, D-III
Answer (2)
Sol. A. [Co(NH3)5(NO2)]Cl2 II. Linkage isomerism due to 'N' and 'O' linkage by NO2
B. [Co(NH3)5(SO4)]Br III. Ionization isomerism
C. [Co(NH3)6][Cr(CN)6] IV. Coordination isomerism
D. [Co(H2O)6]Cl3 I. Solvate isomerism

57. The reagents with which glucose does not react to give the corresponding tests/products are
A. Tollen's reagent
B. Schiff's reagent
C. HCN
D. NH2OH
E. NaHSO3
Choose the correct options from the given below:
(1) E and D
(2) B and C
(3) A and D
(4) B and E
Answer (4)
Sol. Despite having the aldehyde group glucose does not give Schiff's test and it does not form the hydrogen sulphite addition product with NaHSO3.

58. Given below are two statements:
Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction.
Statement II : Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is incorrect but Statement II is true
(2) Both statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is correct but Statement II is false
Answer (2)
Sol. Aniline does not undergo Friedel-Crafts alkylation reaction due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. Aniline (aromatic primary amine) cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with anion formed by phthalimide.

59. Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follow the order H2O > H2Te > H2Se > H2S
Statement II: On the basis of molecular mass, H2O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (2)
Sol. Statement I is correct, because boiling point of hydrides of group 16 follows the order H2O > H2Te > H2Se > H2S
Statement II due to intermolecular H-bonding H2O shows higher boiling point than respective hydrides of group 16. (Both Statement are true)
Order from H2Te to H2S is due to decreasing molar mass.

60. Fehling's solution 'A' is
(1) aqueous sodium citrate
(2) aqueous copper sulphate
(3) alkaline copper sulphate
(4) alkaline solution of sodium potassium tartrate (Rochelle's salt)
Answer (2)
Sol. Fehling solution 'A' = Aqueous copper sulphate
Fehling solution 'B' = Alkaline sodium potassium tartrate (Rochelle salt)

61. Which one of the following alcohols reacts instantaneously with Lucas reagent?
(1) CH3 – C – OH
(2) CH3 – CH2 – CH2 – CH2OH
(3) CH3 – CH2 – CH – OH
(4) CH3 – CH – CH2OH
Answer (1)
Sol. Tertiary alcohols react instantaneously with Lucas reagent and gives immediate turbidity. In case of tertiary alcohols, they form halides easily with Lucas reagent (conc. HCl and ZnCl2)

62. The most stable carbocation among the following is:
(1)
(2)
(3)
(4)
Answer (1)
Sol. The stability of carbocation can be described by the hyperconjugation. Greater the extent of hyperconjugation, more is the stability of carbocation.
(1) → 7 α-H
(2) → 3 α-H
(3) → 5 α-H
(4) → 1 α-H
Stability order of carbocations = (1) > (3) > (2) > (4)

63. Match List I with List II.
List I (Compound)
List II (Shape/geometry)
A. NH3 B. BrF5 C. XeF4 D. SF6
I. Trigonal Pyramidal II. Square Planar III. Octahedral IV. Square Pyramidal
Choose the correct answer from the options given below:
(1) A-II, B-III, C-IV, D-I
(2) A-I, B-IV, C-II, D-III
(3) A-II, B-IV, C-III, D-I
(4) A-III, B-IV, C-I, D-II
Answer (2)
Sol. NH3 ⇒ sp³ hybridised with 1 lone pair. Structure will be Trigonal Pyramidal. BrF5 ⇒ sp³d² hybridised with 1 lone pair. Structure will be Square Pyramidal. XeF4 ⇒ sp³d² with two lone pairs. Structure will be Square Planar. SF6 ⇒ sp³d² with no lone pair. Structure will be Octahedral. A-I, B-IV, C-II, D-III

64. Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N
Choose the correct answer from the options given below:
(1) Li < Be < N < B < C
(2) Li < Be < B < C < N
(3) Li < B < Be < C < N
(4) Li < Be < C < B < N
Answer (3)
Sol. Increasing order of first ionization enthalpy is Li < B < Be < C < N
Element First ionization enthalpy (ΔH/kJ mol-1)
Li 520
Be 899
B 801
C 1086
N 1402

65. Given below are two statements:
Statement I: Both [Co(NH3)6]³⁺ and [CoF6]³⁻ complexes are octahedral but differ in their magnetic behaviour.
Statement II: [Co(NH3)6]³⁺ is diamagnetic whereas [CoF6]³⁻ is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (2)
Sol. In [Co(NH3)6]³⁺ Co³⁺ ion is having 3d⁶ configuration. Electronic configuration of Co³⁺: 3d 4s 4p In presence of NH3 ligand, pairing of electrons takes place and it becomes diamagnetic complex ion. In presence of NH3 ligand: 3d 4s 4p: [Co(NH3)6]³⁺ is octahedral with d²sp³ hybridisation and it is diamagnetic in nature. In case of [CoF6]³⁻ Co is in +3 oxidation state and it is having 3d⁶ configuration. In presence of weak field F- ligand, pairing does not take place. In presence of F- ligands: 3d 4s 4p 4d: In [CoF6]³⁻ Co³⁺ is sp³d² hybridised with four unpaired electrons, so it is paramagnetic in nature.

66. 1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to
(1) 200 mg
(2) 0 mg
(3) 250 mg
(4) Zero mg
Answer (3)
Sol. M = W×1000/(M2×V(in mL))
W = M×M2×V(in mL)/1000 = 0.75×36.5×25/1000 = 0.684 g (Mass of HCl)
HCl + NaOH → NaCl + H2O
36.5 g HCl reacts with NaOH = 40 g
0.684 g HCl reacts with NaOH = 40/36.5×0.684 = 0.750 g
Amount of NaOH left = 1 g - 0.750 g = 0.250 g = 250 mg

67. The Henry's law constant (KH) values of three gases (A, B, C) in water are 145, 2×10⁻⁵ and 35 kbar, respectively. The solubility of these gases in water follow the order:
(1) A > B > C
(2) B > A > C
(3) B > C > A
(4) A > C > B
Answer (3)
Sol. Value of Henry's law constant α 1/Solubility of gas
Higher the value of KH at a given pressure, lower is the solubility of the gas in the liquid.
KH value of gases (given): A > C > B
Order of solubility of gases in water: B > C > A

68. Identify the correct reagents that would bring about the following transformation.
(1) (i) H2O/H⁺ (ii) PCC
(2) (i) H2O/H⁺ (ii) CrO3
(3) (i) BH3 (ii) H2O2/OH (iii) PCC
(4) (i) BH3 (ii) H2O2/OH (iii) alk.KMnO4 (iv) H3O⊕
Answer (3)

69. 'Spin only' magnetic moment is same for which of the following ions?
A. Ti³⁺ B. Cr²⁺ C. Mn²⁺ D. Fe²⁺ E. Sc³⁺
Choose the most appropriate answer from the options given below.
(1) A and D only
(2) B and D only
(3) A and E only
(4) B and C only
Answer (2)
Sol. Ions No. of unpaired electrons Configuration
Ti³⁺ (2) 3d¹
Cr²⁺ 4 3d⁴
Mn²⁺ 5 3d⁵
Fe²⁺ 4 3d⁶
Sc³⁺ 0 3d⁰
Spin only magnetic moment is given by √n(n+2) BM
∴ Cr²⁺ and Fe²⁺ will have same spin only magnetic moment.

70. In which of the following processes entropy increases?
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from 130K to 0K
C. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
D. Cl2(g) → 2Cl(g)
(1) C and D
(2) A and C
(3) A, B and D
(4) A, C and D
Answer (4)
Sol. When a liquid evaporates to vapour entropy increases.
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
Number of gaseous product molecules increases so entropy increases.
Cl2(g) → 2Cl(g)
1 mole Cl2(g) form 2 mol Cl(g). So entropy increases.

71. Intramolecular hydrogen bonding is present in
(1) HF
(2) o-nitrophenol
(3) p-nitrophenol
(4) m-nitrophenol
Answer (2)
Sol. In o-nitrophenol intramolecular H-bonding is present.

72. In which of the following equilibria, Kp and Kc are NOT equal?
(1) 2BrCl(g) ⇌ Br2(g) + Cl2(g)
(2) PCl5(g) ⇌ PCl3(g) + Cl2(g)
(3) H2(g) + I2(g) ⇌ 2HI(g)
(4) CO2(g) + H2O(g) ⇌ CO2(g) + H2(g)
Answer (2)
Sol. Kp = Kc(RT)Δng
for Kp ≠ Kc
Δng ≠ 0
Δng = np - nr
(1) Δng = 2 - 2 = 0
(2) Δng = 2 - 1 = 1
(3) Δng = 2 - 2 = 0
(4) Δng = 2 - 2 = 0

73. The E° value for the Mn³⁺/Mn²⁺ couple is more positive than that of Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ due to change of
(1) d³ to d⁵ configuration
(2) d⁵ to d⁴ configuration
(3) d⁵ to d² configuration
(4) d⁴ to d⁵ configuration
Answer (4)
Sol. E°(Mn³⁺/Mn²⁺) > E°(Cr³⁺/Cr²⁺) or E°(Fe³⁺/Fe²⁺)
Electronic configuration of Mn³⁺ = [Ar]3d⁴
Electronic configuration of Mn²⁺ = [Ar]3d⁵
Electronic configuration of Cr³⁺ = [Ar]3d³
Electronic configuration of Cr²⁺ = [Ar]3d⁴
As Mn³⁺ from d⁴ configuration goes to more stable d⁵ configuration (Half filled), due to more exchange energy in d⁵ configuration.

74. Activation energy of any chemical reaction can be calculated if one knows the value of
(1) rate constant at two different temperatures
(2) rate constant at standard temperature
(3) probability of collision
(4) orientation of reactant molecules during collision
Answer (1)
Sol. To calculate value of Ea Equation used is log(k2/k1) = Ea/2.303R (1/T1 - 1/T2)
Hence Ea can be calculated if value of rate constant k is known at two different temperatures T1 and T2

75. Match List I with List II.
List I List II (Molecule) (Number and types of bond/s between two carbon atoms)
A. ethane I. one σ-bond and two π-bonds
B. ethene II. two π-bonds
C. carbon molecule, C2 III. one σ-bond
D. ethyne IV. one σ-bond and one π-bond
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-I, D-II
(2) A-I, B-IV, C-II, D-III
(3) A-IV, B-III, C-II, D-I
(4) A-III, B-IV, C-II, D-I
Answer (4)
Sol. (A) Ethane: one (C-C) σ bond
(B) Ethene: one (C-C) σ and one (C-C) π bond
(C) C2: two (C-C) π bonds
(D) Ethyne: two (C-C) π bonds and one (C-C) σ bond

76. On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as
(1) Chromatography
(2) Crystallization
(3) Sublimation
(4) Distillation
Answer (3)
Sol. (1) Chromatography: It is based on separation by using stationary and mobile phase.
(2) Crystallization: It is based on difference in the solubilities of the compound and impurities in a suitable solvent.
(3) Sublimation: It is the purification technique based on principle that on heating, some solid substances change from solid to vapour state without passing through liquid state.
(4) Distillation: It is used to separate volatile liquids from non-volatile impurities and the liquids having sufficient difference in their boiling point.

77. Match List I with List II.
List I List II (Conversion) (Number of Faraday required)
A. 1 mol of H2O to O2 I. 3F
B. 1 mol of MnO4⁻ to Mn²⁺ II. 2F
C. 1.5 mol of Ca from molten CaCl2 III. 1F
D. 1 mol of FeO to Fe2O3 IV. 5F
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-II, D-I
(2) A-II, B-IV, C-I, D-III
(3) A-II, B-IV, C-III, D-I
(4) A-III, B-III, C-II, D-IV
Answer (2)
Sol. 4OH⁻ → 2H2O + O2 + 4e⁻ for 2 mole of H2O = 4F charge is required for 1 mole of H2O = 4F/2 = 2F required
MnO4⁻ → Mn²⁺ for 1 mole MnO4 5F charge is required
Ca²⁺ + 2e⁻ → Ca For 1 mole Ca²⁺ ion required = 2F
1.5 mole Ca²⁺ ion required = 2/1×1.5 = 3F
FeO → Fe2O3 for 1 mole FeO 1F charge is required.

78. Match List I with List II
List I (Quantum Number) (Information provided)
A. m I. Shape of orbital
B. ms II. Size of orbital
C. l III. Orientation of orbital
D. n IV. Orientation of spin of electron
Choose the correct answer from the options given below:
(1) A-II, B-I, C-IV, D-III
(2) A-I, B-II, C-III, D-IV
(3) A-III, B-IV, C-I, D-II
(4) A-II, B-IV, C-III, D-I
Answer (3)
Sol. Magnetic quantum number m informs about orientation of orbital.
Spin quantum number ms informs about orientation of spin of electron.
Azimuthal quantum number (l) informs about shape of orbital
Principal quantum number (n) informs about size of orbital

79. Which reaction is NOT a redox reaction?
(1) BaCl2 + Na2SO4 → BaSO4 + 2NaCl
(2) Zn + CuSO4 → ZnSO4 + Cu
(3) 2KClO3 + I2 → 2KIO3 + Cl2
(4) H2 + Cl2 → 2HCl
Answer (1)
Sol. (1) BaCl2 + Na2SO4 → BaSO4 + 2NaCl
This is not a redox reaction as there is no change in oxidation state.
(2) Zn + CuSO4 → ZnSO4 + Cu, Redox reaction
(3) 2KClO3 + I2 → 2KIO3 + Cl2, Redox reaction
(4) H2 + Cl2 → 2HCl, Redox reaction

80. Match List I with List II.
List I (Reaction)
List II (Reagents/Condition)
A. Cyclohexene → 2 Cyclohexanone
B. Benzene → Benzophenone
C. Cyclohexanol → Cyclohexanone
D. Ethylbenzene → Benzoic acid
I. C6H5COCl/Anhyd. AlCl3
II. CrO3
III. KMnO4/KOH, Δ
IV. (i) O3 (ii) Zn-H2O
Choose the correct answer from the options given below:
(1) A-I, B-IV, C-II, D-III
(2) A-IV, B-I, C-III, D-II
(3) A-III, B-I, C-II, D-IV
(4) A-IV, B-I, C-II, D-III
Answer (4)
Sol. (A) Reductive ozonolysis
(B) Friedel-Crafts acylation reaction.
(C) Secondary alcohols are oxidised to ketones by CrO3
(D) Ethylbenzene → Benzoic acid by KMnO4/KOH, Δ

81. The highest number of helium atoms is in
(1) 2.271098 L of helium at STP
(2) 4 mol of helium
(3) 4 u of helium
(4) 4 g of helium
Answer (2)
Sol. (1) 2.271098 L of He at STP = 2.271/22.71098 = 0.1 mole = 0.1 NA He atom
(2) 4 mol of He = 4 NA He atoms
(3) 4 u of He = 4 u = 1 He atom
(4) 4 g of Helium = 4/4 mole = 1 mole = NA He atom

82. Which plot of ln k vs 1/T is consistent with Arrhenius equation?
(1)
(2)
(3)
(4)
Answer (1)
Sol. The Arrhenius equation is given as k = Ae^(-Ea/RT)
ln k = ln A - Ea/RT
ln k vs 1/T gives a straight line graph with slope = -Ea/R and intercept = ln A
Intercept = ln A, Slope = -Ea/R

83. Match List I with List II.
List-I (Process)
A. Isothermal process
B. Isochoric process
C. Isobaric process
D. Adiabatic process
List-II (Conditions)
I. No heat exchange
II. Carried out at constant temperature
III. Carried out at constant volume
IV. Carried out at constant pressure
Choose the correct answer from the options given below:
(1) A-II, B-III, C-IV, D-I
(2) A-I, B-II, C-III, D-IV
(3) A-III, B-IV, C-I, D-II
(4) A-IV, B-I, C-II, D-III
Answer (1)
Sol. (A) Isothermal process ⇒ Temperature is constant throughout the process
(B) Isochoric process ⇒ Volume is constant throughout the process
(C) Isobaric process ⇒ Pressure is constant throughout the process
(D) Adiabatic process ⇒ No exchange of heat (q) between system and surrounding

84. Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si
Choose the correct answer from the options given below:
(1) F < O < N < C < Si
(2) Si < C < N < O < F
(3) Si < C < O < N < F
(4) O < F < N < C < Si
Answer (2)
Sol. Electronegativity increases across the period on moving left to right. It decreases on moving down the group. The correct option is Si < C < N < O < F

85. The compound that will undergo SN1 reaction with the fastest rate is
(1)
(2)
(3)
(4)
Answer (1)
Sol. Reactivity towards SN1 depends upon stability of carbocation. Order of stability is (1) > (3) > (2) > (4). Hence (1) is most reactive.

SECTION-B

86. The products A and B obtained in the following reactions, respectively, are
3ROH + PCl3 → 3RCl + A
ROH + PCl5 → RCl + HCl + B
(1) H3PO3 and POCl3
(2) POCl3 and H3PO3
(3) POCl3 and H3PO4
(4) H3PO4 and POCl3
Answer (1)
Sol. These reactions are preparation of haloalkanes from alcohols.
3ROH + PCl3 → 3RCl + H3PO3 (A)
ROH + PCl5 → RCl + HCl + POCl3 (B)
A and B are H3PO3 and POCl3 respectively.

87. The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation. Given R = 8.314 J K⁻¹ mol⁻¹, log4 = 0.6021
(1) 3804 kJ/mol
(2) 38.04 kJ/mol
(3) 380.4 kJ/mol
(4) 3.80 kJ/mol
Answer (2)
Sol. log(k2/k1) = Ea/2.303R (1/T1 - 1/T2)
log(4/1) = Ea/2.303R (1/300 - 1/330)
Ea = (log4)×2.303×8.314×300×330 / 30
= 3.804 × 10⁴ J/mol
= 38.04 kJ/mol

88. Identify the major product C formed in the following reaction sequence:
CH3-CH2-CH2-I → NaCN → A → Partial hydrolysis → B → NaOH/Br2 → C
(1) α-bromobutanoic acid
(2) propylamine
(3) butylamine
(4) butanamide
Answer (2)
Sol. CH3CH2CH2-I → NaCN → CH3CH2CH2-CN (A) → Partial hydrolysis → CH3CH2CH2-CONH2 (B) → NaOH/Br2 → CH3CH2CH2-NH2 (C) (Propyl amine)
Step-I is SN2 reaction with CN⁻ nucleophile.
Step-II will give amide.
Step-III is Hoffmann bromamide degradation reaction.

89. Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given: Molar mass of Cu: 63 g mol⁻¹, 1 F = 96487 C)
(1) 0.0315 g
(2) 3.15 g
(3) 0.315 g
(4) 31.5 g
Answer (3)
Sol. Cu²⁺(aq) + 2e⁻ → Cu(s)
Mass of Cu deposited (w) = M×i×t / nF
= 63×9.6487×100 / (2×96487)
= 0.315 g

90. During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of Fe²⁺ ion?
(1) dilute sulphuric acid
(2) dilute hydrochloric acid
(3) concentrated sulphuric acid
(4) dilute nitric acid
Answer (1)
Sol. During the preparation of Mohr's salt, dilute sulphuric acid is added to prevent the hydrolysis of Fe²⁺ ion.

91. The plot of osmotic pressure (Π) vs concentration (mol L⁻¹) for a solution gives a straight line with slope 25.73 L bar mol⁻¹. The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar mol⁻¹ K⁻¹)
(1) 12.05°C
(2) 37°C
(3) 310°C
(4) 25.73°C
Answer (2)

92. Given below are certain cations. Using inorganic qualitative analysis, arrange them in increasing group number from 0 to VI.
A. Al³⁺ B. Cu²⁺ C. Ba²⁺ D. Co²⁺ E. Mg²⁺
Choose the correct answer from the options given below.
(1) E,A,B,C,D
(2) B,A,D,C,E
(3) B,C,A,D,E
(4) E,C,D,B,A
Answer (2)
Sol. Group Cations
Group-II Cu²⁺
Group-III Al³⁺
Group-IV Co²⁺
Group-V Ba²⁺
Group-VI Mg²⁺
The correct order of group number of ions is Cu²⁺ < Al³⁺ < Co²⁺ < Ba²⁺ < Mg²⁺ (B) (A) (D) (C) (E)
The correct order is B,A,D,C,E

93. Consider the following reaction in a sealed vessel at equilibrium with concentrations of N2 = 3.0×10⁻³ M, O2 = 4.2×10⁻³ M and NO = 2.8×10⁻³ M.
2NO(g) ⇌ N2(g) + O2(g)
If 0.1 mol L⁻¹ of NO(g) is taken in a closed vessel, what will be degree of dissociation (α) of NO(g) at equilibrium?
(1) 0.717
(2) 0.00889
(3) 0.0889
(4) 0.8889
Answer (1)
Sol. 2NO(g) ⇌ N2(g) + O2(g)
Kc = [N2][O2]/[NO]²
= (3×10⁻³×4.2×10⁻³)/(2.8×10⁻³×2.8×10⁻³)
= 1.607
t=0: 0.1, 0, 0
At eq: 0.1-0.1α, 0.05α, 0.05α
Kc = (0.05α×0.05α)/(0.1-0.1α)²
Kc = 0.05α×0.05α / 0.01(1-α)²
1.607 = (0.05)²α² / 0.01(1-α)²
α²/(1-α)² = 1.607×(0.1)²/(0.05)²
α/(1-α) = 1.27×0.1/0.05 = 2.54
α = 2.54 - 2.54α
3.54α = 2.54
α = 2.54/3.54 = 0.717

94. Given below are two statements:
Statement I: [Co(NH3)6]³⁺ is a homoleptic complex whereas [Co(NH3)4Cl2]⁺ is a heteroleptic complex.
Statement II: Complex [Co(NH3)6]³⁺ has only one kind of ligands but [Co(NH3)4Cl2]⁺ has more than one kind of ligands.
In the light of the above statements, choose the correct answer from the options given below.
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (2)
Sol. [Co(NH3)6]³⁺ is a homoleptic complex as only one type of ligands (NH3) is coordinated with Co³⁺ ion. While [Co(NH3)4Cl2]⁺ is a heteroleptic complex in which Co³⁺ ion is ligated with more than one type of ligands, i.e., NH3 and Cl⁻.

95. Major products A and B formed in the following reaction sequence, are
Answer (2)
Sol. Reaction sequence as per mechanism.

96. Identify the correct answer.
(1) Three canonical forms can be drawn for CO3²⁻ ion
(2) Three resonance structures can be drawn for ozone
(3) BF3 has non-zero dipole moment
(4) Dipole moment of NF3 is greater than that of NH3
Answer (1)
Sol. (1) CO3²⁻ has three canonical forms.
(2) In ozone, there are two resonating structures.
(3) BF3 has dipole moment = 0
(4) Dipole moment of NF3 is less than that of NH3.

97. A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is: (Given atomic masses of A = 64; B = 40; C = 32 u)
(1) ABC4
(2) ABC2
(3) ABC3
(4) ABC2
Answer (3)
Sol. Element Mass percentage % No. of moles No. of moles/Smallest number Simplest whole number
A 32% 32/64 = 1/2 ×2 = 1
B 20% 20/40 = 1/2 ×2 = 1
C 48% 48/32 = 3/2 ×2 = 3
So, empirical formula of X = A1 : B1 : C3
The correct empirical formula of compound X is ABC3.

98. The pair of lanthanoid ions which are diamagnetic is
(1) Pm³⁺ and Sm³⁺
(2) Ce⁴⁺ and Yb²⁺
(3) Ce³⁺ and Eu²⁺
(4) Gd³⁺ and Eu³⁺
Answer (2)
Sol. Magnetic moment μ = √n(n+2)
n → number of unpaired electron
Ce⁴⁺ ⇒ (Xe)4f⁰, μ = 0
Yb²⁺ ⇒ (Xe)4f¹⁴, μ = 0
Ce³⁺ ⇒ (Xe)4f¹, μ = √3
Eu²⁺ ⇒ (Xe)4f⁷, μ = √63
Gd³⁺ ⇒ (Xe)4f⁷, μ = √63
Eu³⁺ ⇒ (Xe)4f⁶, μ = √48
Diamagnetic: Ce⁴⁺ and Yb²⁺.

101. Which one of the following is not a criterion for classification of fungi?
(1) Fruiting body
(2) Morphology of mycelium
(3) Mode of nutrition
(4) Mode of spore formation
Answer (3)
Sol. The morphology of the mycelium, mode of spore formation and fruiting bodies form the basis for the division of the kingdom fungi into various classes.

102. Given below are two statements:
Statement I: Parenchyma is living but collenchyma is dead tissue.
Statement II: Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (1)
Sol. Collenchyma is also living tissue.
Gymnosperm lack xylem vessels but presence of xylem vessels is the characteristic of angiosperm.

103. A pink flowered Snapdragon plant was crossed with a red flowered Snapdragon plant. What type of phenotype/s is/are expected in the progeny?
(1) Red, Pink as well as white flowered plants
(2) Only red flowered plants
(3) Red flowered as well as pink flowered plants
(4) Only pink flowered plants
Answer (3)
Sol. Pink colour flower in snapdragon have genotype Rr
Red flowered snapdragon have genotype RR when they both are crossed
RR × Rr
Gametes: R, R and R, r
Progeny: RR, Rr, RR, Rr
Phenotype: Red : Pink : White = 2 : 2 : 0
So the progeny that we get are red and pink flowered plants only.

104. Identify the set of correct statements:
A. The flowers of Vallisneria are colourful and produce nectar.
B. The flowers of water lily are not pollinated by water.
C. In most of water-pollinated species, the pollen grains are protected from wetting.
D. Pollen grains of some hydrophytes are long and ribbon like.
E. In some hydrophytes, the pollen grains are carried passively inside water.
Choose the correct answer from the options given below.
(1) B, C, D and E only
(2) C, D and E only
(3) A, B, C and D only
(4) A, C, D and E only
Answer (1)
Sol. Flowers of Vallisneria are not colourful and do not produce nectar. Waterlily is pollinated by insect or wind. In water-pollinated species, pollen grains are protected from wetting by a mucilaginous covering. In some hydrophytes such as Vallisneria pollen grains are carried passively by water current.

105. These are regarded as major causes of biodiversity loss:
A. Over exploitation
B. Co-extinction
C. Mutation
D. Habitat loss and fragmentation
E. Migration
Choose the correct option:
(1) A, B and D only
(2) A, C and D only
(3) A, B, C and D only
(4) A, B and E only
Answer (1)
Sol. Major causes of biodiversity losses are:
(1) Co-extinctions
(2) Habitat loss and fragmentation
(3) Over-exploitation
(4) Alien species invasions
Hence correct option is A, B and D only.

106. Match List I with List II
List I List II
A. Clostridium butylicum I. Ethanol
B. Saccharomyces cerevisiae II. Streptokinase
C. Trichoderma polysporum III. Butyric acid
D. Streptococcus sp. IV. Cyclosporin-A
Choose the correct answer from the options given below:
(1) A-IV, B-I, C-III, D-II
(2) A-III, B-I, C-IV, D-II
(3) A-III, B-IV, C-II, D-I
(4) A-II, B-I, C-IV, D-III
Answer (2)
Sol. A. Clostridium butylicum — Butyric acid
B. Saccharomyces cerevisiae — Ethanol
C. Trichoderma polysporum — Cyclosporin-A
D. Streptococcus sp. — Streptokinase

107. Which of the following are required for the dark reaction of photosynthesis?
A. Light
B. Chlorophyll
C. CO2
D. ATP
E. NADPH
Choose the correct answer from the options given below:
(1) D and E only
(2) A, B and C only
(3) B, C and D only
(4) C, D and E only
Answer (4)
Sol. For dark reaction of photosynthesis there are the requirement of CO2, ATP, NADPH.

108. Match List I with List II
List I List II
A. Two or more alternative forms of a gene I. Back cross
B. Cross of F1 progeny with homozygous recessive parent II. Ploidy
C. Cross of F1 progeny with any of the parents III. Allele
D. Number of chromosome sets in plant IV. Test cross
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-II, D-I
(2) A-II, B-I, C-III, D-IV
(3) A-III, B-IV, C-I, D-II
(4) A-III, B-IV, C-I, D-II
Answer (3)
Sol. A. Two or more alternative forms of gene are called alleles.
B. Cross of F1 progeny with homozygous recessive parent is a test cross.
C. Cross of F1 progeny with any of the parents is a back cross.
D. Number of chromosome sets in plant is called ploidy.

109. Given below are two statements:
Statement I: Chromosomes become gradually visible under light microscope during leptotene stage.
Statement II: The beginning of diplotene stage is recognized by dissolution of synaptonemal complex.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (2)
Sol. During leptotene stage the chromosomes become gradually visible under the light microscope. The beginning of diplotene is recognised by the dissolution of the synaptonemal complex and the tendency of the recombined homologous chromosomes of the bivalents to separate from each other except at the site of crossover. Thus both statement I and II are correct.

110. What is the fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism?
A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism.
B. It may get integrated into the genome of the recipient.
C. It may multiply and be inherited along with the host DNA.
D. The alien piece of DNA is not an integral part of chromosome.
E. It shows ability to replicate.
Choose the correct answer from the options given below:
(1) A and E only
(2) A and B only
(3) D and E only
(4) B and C only
Answer (4)
Sol. Correct answer is option (4) because:
The fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism are:
(B) It may get integrated into the genome of the recipient
(C) It may multiply and be inherited along with the host DNA
This piece of DNA would not be able to multiply itself in the progeny cells of the organism but when gets integrated into the genome of the recipient, it may multiply and be inherited along with the host DNA.

111. Identify the type of flowers based on the position of calyx, corolla and androecium with respect to the ovary from the given figures (a) and (b)
(1) (a) Perigynous; (b) Perigynous
(2) (a) Epigynous; (b) Hypogynous
(3) (a) Hypogynous; (b) Epigynous
(4) (a) Perigynous; (b) Epigynous
Answer (1)
Sol. If gynoecium is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level, it is called perigynous. Both diagram shows perigynous condition.

112. Which of the following is an example of actinomorphic flower?
(1) Sesbania
(2) Datura
(3) Cassia
(4) Pisum
Answer (2)
Sol. Datura shows actinomorphic flower. In Cassia, Pisum and Sesbania, zygomorphic flowers are seen.

113. Which one of the following can be explained on the basis of Mendel's Law of Dominance?
A. Out of one pair of factors one is dominant and the other is recessive.
B. Alleles do not show any expression and both the characters appear as such in F2 generation.
C. Factors occur in pairs in normal diploid plants.
D. The discrete unit controlling a particular character is called factor.
E. The expression of only one of the parental characters is found in a monohybrid cross.
Choose the correct answer from the options given below:
(1) A, B, C, D and E
(2) A, B and C only
(3) A, C, D and E only
(4) B, C and D only
Answer (3)
Sol. According to Law of Dominance:
(2) Characters are controlled by discrete units called factors
(3) Factors occur in pairs
(4) In a dissimilar pair of factors one member of the pair dominates (dominant) the other recessive
The law of dominance is used to explain the expression of only one of the parental characters in a monohybrid cross.
Law of segregation is based on the fact that the alleles do not show any expression and both the characters are recovered as such in F2 generation.

114. The cofactor of the enzyme carboxypeptidase is:
(1) Haem
(2) Zinc
(3) Niacin
(4) Flavin
Answer (2)
Sol. The correct answer is option (2) as the cofactor of the enzyme carboxypeptidase is zinc. Niacin is associated with coenzyme NAD and NADP. Option (1) is incorrect as haem is the prosthetic group in peroxidase and catalase.

115. The equation of Verhulst-Pearl logistic growth is dN/dt = rN[(K-N)/K]
From this equation, K indicates:
(1) Population density
(2) Intrinsic rate of natural increase
(3) Biotic potential
(4) Carrying capacity
Answer (4)
Sol. In the equation dN/dt = rN[(K-N)/K], K represents carrying capacity.

116. Match List I with List II
List-I List-II
A. Nucleolus I. Site of formation of glycolipid
B. Centriole II. Organization like the cartwheel
C. Leucoplasts III. Site for active ribosomal RNA synthesis
D. Golgi apparatus IV. For storing nutrients
Choose the correct answer from the options given below:
(1) A-I, B-II, C-III, D-IV
(2) A-III, B-II, C-IV, D-I
(3) A-II, B-III, C-I, D-IV
(4) A-III, B-IV, C-II, D-I
Answer (2)
Sol. Nucleolus is a site for active ribosomal RNA synthesis
Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like the cartwheel.
Leucoplasts are the colourless plastids of varied shapes and sizes with stored nutrients.
Golgi apparatus is the important site for formation of glycoproteins and glycolipids.

117. Match List I with List II
List-I List-II
A. Rhizopus I. Mushroom
B. Ustilago II. Smut fungus
C. Puccinia III. Bread mould
D. Agaricus IV. Rust fungus
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-II, D-I
(2) A-III, B-II, C-IV, D-I
(3) A-I, B-III, C-II, D-IV
(4) A-III, B-II, C-I, D-IV
Answer (2)
Sol. Rhizopus is a bread mould fungus. Ustilago is a smut fungi. Puccinia is known as rust fungi. Agaricus is commonly called mushroom.
A-III, B-II, C-IV, D-I

118. The lactose present in the growth medium of bacteria is transported to the cell by the action of
(1) Polymerase
(2) Beta-galactosidase
(3) Acetylase
(4) Permease
Answer (4)
Sol. The y gene lac operon codes for permease enzyme, which increase the permeability of cell to β-galactosides. So, the lactose present in the growth medium of bacteria is transported into the cell by the action of permease.

119. List of endangered species was released by
(1) IUCN
(2) GEAC
(3) WWF
(4) FOAM
Answer (1)
Sol. List of endangered species was released by IUCN.

120. A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and down stream end;
(1) Promotor, Structural gene, Terminator
(2) Repressor, Operator gene, Structural gene
(3) Structural gene, Transposons, Operator gene
(4) Inducer, Repressor, Structural gene
Answer (1)
Sol. A transcription unit of DNA is defined primarily by the three regions in the DNA:
(i) A promoter
(ii) The structural gene
(iii) A terminator
The promoter is said to be located towards 5'-end (upstream) of the structural gene (the reference is made with respect to the polarity of coding strand). The terminator is located towards 3'-end (downstream) of the coding strand.

121. Formation of interfascicular cambium from fully developed parenchyma cells is an example for
(1) Maturation
(2) Differentiation
(3) Redifferentiation
(4) Dedifferentiation
Answer (4)
Sol. The phenomenon of formation of interfascicular cambium from fully differentiated parenchyma cells is called dedifferentiation.

122. Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of:
(1) Enzyme activation
(2) Cofactor inhibition
(3) Feedback inhibition
(4) Competitive inhibition
Answer (4)
Sol. Correct answer is option (4) because malonate shows close structural similarity with the substrate and it competes with the substrate for the substrate binding site of the enzyme succinic dehydrogenase. Option (1), (2) and (3) are incorrect as enzyme activation, co-factor inhibition are not showing structural similarity with substrate.

123. Lecithin, a small molecular weight organic compound found in living tissues, is an example of:
(1) Carbohydrates
(2) Amino acids
(3) Phospholipids
(4) Glycerides
Answer (3)
Sol. The correct answer is option (3). Some lipids have phosphorous and a phosphorylated organic compound in them. These are phospholipids. They are found in cell membrane. Lecithin is one example. Option (4) is incorrect as glycerides are another group of lipids in which both glycerol and fatty acids are present. Option (1) and (2) are incorrect as carbohydrates and amino acids are separate groups of biomolecules.

124. Bulliform cells are responsible for
(1) Providing large spaces for storage of sugars.
(2) Inward curling of leaves in monocots.
(3) Protecting the plant from salt stress.
(4) Increased photosynthesis in monocots.
Answer (2)
Sol. In grasses, certain adaxial epidermal cells along the veins modify themselves into large, empty, colourless cells. These are called bulliform cells. When the bulliform cells in the leaves have absorbed water and are turgid, the leaf surface is exposed. When they are flaccid due to water stress, they make the leaves curl inwards to minimise water loss.

125. Spindle fibers attach to kinetochores of chromosomes during
(1) Telophase
(2) Prophase
(3) Metaphase
(4) Anaphase
Answer (3)
Sol. Spindle fibers attach to kinetochores of chromosome in metaphase stage.

126. How many molecules of ATP and NADPH are required for every molecule of CO2 fixed in the Calvin cycle?
(1) 3 molecules of ATP and 2 molecules of NADPH
(2) 2 molecules of ATP and 3 molecules of NADPH
(3) 2 molecules of ATP and 2 molecules of NADPH
(4) 3 molecules of ATP and 3 molecules of NADPH
Answer (1)
Sol. For fixation of 1 molecule of CO2 in Calvin cycle 3 ATP molecules and 2 NADPH molecules are required.

127. Tropical regions show greatest level of species richness because
A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
B. Tropical environments are more seasonal.
C. More solar energy is available in tropics.
D. Constant environments promote niche specialization.
E. Tropical environments are constant and predictable.
Choose the correct answer from the options given below.
(1) A, B and D only
(2) A, C, D and E only
(3) A and B only
(4) A, B and E only
Answer (2)
Sol. Only statement B is incorrect because tropical environments unlike temperate ones, are less seasonal, relatively more constant and predictable. Thus statements A, C, D and E are correct.

128. In the given figure, which component has thin outer walls and highly thickened inner walls?
(1) B
(2) C
(3) D
(4) A
Answer (2)
Sol. Guard cells of stomata have thin outer wall and highly thickened inner walls.

129. Identify the part of the seed from the given figure which is destined to form root when the seed germinates.
(1) D
(2) A
(3) B
(4) C
Answer (4)
Sol. Radicle is destined to form root. In the given diagram 'C' represent radicle.

130. Auxin is used by gardeners to prepare weed-free lawns. But no damage is caused to grass as auxin
(1) can help in cell division in grasses, to produce growth.
(2) promotes apical dominance.
(3) promotes abscission of mature leaves only.
(4) does not affect mature monocotyledonous plants.
Answer (4)
Sol. Auxin does not affect mature monocot plants. In monocots, especially grasses show limited translocation and cause rapid degradation of external auxin.

131. The capacity to generate a whole plant from any cell of the plant is called:
(1) Somatic hybridization
(2) Totipotency
(3) Micropropagation
(4) Differentiation
Answer (2)
Sol. Totipotency is defined as the capacity to generate a whole plant from any cell of the plant.

132. The type of conservation in which the threatened species are taken out from their natural habitat and placed in special setting where they can be protected and given special care is called
(1) Sustainable development
(2) in-situ conservation
(3) Biodiversity conservation
(4) Semi-conservative method
Answer (3)
Sol. The type of conservation in which threatened species are taken out from their natural habitat and placed in special setting where they can be protected and given special care is called ex-situ conservation which is a type of biodiversity conservation.

133. In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). In order to find out the genotype of the black seed plant, with which of the following genotype will you cross it?
(1) BB/Bb
(2) BB
(3) bb
(4) Bb
Answer (3)
Sol. To determine the genotype of a black seed colour at F2 the black seed from F2 is crossed with the white seed colour. This is called a test cross.
To determine the genotype of (BB/Bb) black seed we need to cross them with white seed i.e. bb.

134. Hind II always cuts DNA molecules at a particular point called recognition sequence and it consists of:
(1) 10 bp
(2) 8 bp
(3) 6 bp
(4) 4 bp
Answer (3)
Sol. The correct answer is option (3).
The first restriction endonuclease - Hind II, whose functioning depends on a specific DNA nucleotide sequence was isolated. It was found that Hind II always cut DNA molecules at a particular point by recognising sequence of six base pairs.
Option (1), (2) and (4) are incorrect because they have either more than 6 or less than 6 bp.

135. Given below are two statements:
Statement I: Bt toxins are insect group specific and coded by a gene cry IAc.
Statement II: Bt toxin exists as inactive protein in B. thuringiensis. However, after ingestion by the insect the inactive protein gets converted into active form due to acidic pH of the insect gut.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (4)
Sol. The correct answer is option (4) as specific Bt toxin genes were isolated from Bacillus thuringiensis and incorporated into the several crop plants such as cotton. The choice of genes depends upon the crop and the targeted pest as most Bt toxins are insect-group specific. The toxin is coded by a gene named cry. There are a number of them, for example, the proteins encoded by the genes cry IAc and cry IIAb control the cotton bollworms, that of cry IAb controls corn borer.

136. Read the following statements and choose the set of correct statements:
In the members of Phaeophyceae,
A. Asexual reproduction occurs usually by biflagellate zoospores.
B. Sexual reproduction is by oogamous method only.
C. Stored food is in the form of carbohydrates which is either mannitol or laminarin.
D. The major pigments found are chlorophyll a, c and carotenoids and xanthophyll.
E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin.
Choose the correct answer from the options given below:
(1) A, B, C and E only
(2) A, B, C and D only
(3) B, C, D and E only
(4) A, C, D and E only
Answer (4)
Sol. In members of Phaeophyceae sexual reproduction is by oogamous, isogamous or anisogamous methods. Therefore correct set of statements are A, C, D and E.

137. In an ecosystem if the Net Primary Productivity (NPP) of first trophic level is 100x (kcal m⁻²) yr⁻¹, what would be the GPP (Gross Primary Productivity) of the third trophic level of the same ecosystem?
(1) 100x/3x (kcal m⁻²) yr⁻¹
(2) x/10 (kcal m⁻²) yr⁻¹
(3) x (kcal m⁻²) yr⁻¹
(4) 10x (kcal m⁻²) yr⁻¹
Answer (4)
Sol. NPP at first trophic level would be the GPP for second trophic level. NPP at second trophic level would be GPP for third trophic level. Therefore, 100x (kcal m⁻²) yr would be GPP at second trophic level and 100x × 10% (kcal m⁻²) yr i.e., 10x (kcal m⁻²) yr energy would be GPP at third trophic level.

138. Identify the step in tricarboxylic acid cycle, which does not involve oxidation of substrate.
(1) Isocitrate → α-ketoglutaric acid
(2) Malic acid → Oxaloacetic acid
(3) Succinic acid → Malic acid
(4) Succinyl-CoA → Succinic acid
Answer (4)
Sol. Oxidation involves the loss of electrons (often as part of hydrogen) from a molecule, leaving to an increase in its oxidation state. This process is typically associated with the transfer of electrons to an electron acceptor which is reduced in the process. The conversion of succinyl CoA to succinic acid does not involve oxidation of substrate.

139. Match List-I with List-II
List-I List-II
A. GLUT-4 I. Hormone
B. Insulin II. Enzyme
C. Trypsin III. Intercellular ground substance
D. Collagen IV. Enables glucose transport into cells
Choose the correct answer from the options given below.
(1) A-III, B-IV, C-I, D-II
(2) A-IV, B-I, C-II, D-III
(3) A-I, B-II, C-III, D-IV
(4) A-II, B-III, C-IV, D-I
Answer (2)
Sol. Correct answer is option (2)
A. GLUT-4 IV. Enables glucose transport into cells
B. Insulin I. Hormone
C. Trypsin II. Enzyme
D. Collagen III. Intercellular ground substance

140. Spraying sugarcane crop with which of the following plant growth regulators, increases the length of stem, thus, increasing the yield?
(1) Abscisic acid
(2) Auxin
(3) Gibberellin
(4) Cytokinin
Answer (3)
Sol. Sugarcane store carbohydrate as sugar in their stems. Spraying sugarcane crop with gibberellins increases the length of the stem, thus increasing the yield.

141. Match List I with List II
List I List II (Types of Stamens) (Example)
A. Monoadelphous I. Citrus
B. Diadelphous II. Pea
C. Polyadelphous III. Lily
D. Epiphyllous IV. China-rose
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-IV, B-II, C-I, D-III
(3) A-IV, B-I, C-II, D-III
(4) A-II, B-III, C-IV, D-I
Answer (2)
Sol. In China rose monoadelphous androecium is present. Diadelphous androecium is found in pea plant. Polyadelphous androecium is found in citrus. Epiphyllous androecium is found in lily.

142. Match List I with List II
List I List II
A. Frederick Griffith I. Genetic code
B. Francois Jacob & Jacque Monod II. Semi-conservative mode of DNA replication
C. Har Gobind Khorana III. Transformation
D. Meselson & Stahl IV. Lac operon
Choose the correct answer from the options given below:
(1) A-IV, B-I, C-II, D-III
(2) A-III, B-II, C-I, D-IV
(3) A-III, B-IV, C-I, D-II
(4) A-II, B-III, C-IV, D-I
Answer (3)
Sol. Frederick Griffith series of experiment witness miraculous transformation in the bacteria. The elucidation of Lac operon was a result of a close association between geneticist, Francois Jacob and a biochemist, Jacques Monod. Meselson and Stahl gave semi-conservative mode of DNA replication. Har Gobind Khorana developed chemical method to define combination of bases in genetic code.

143. Which of the following statement is correct regarding the process of replication in E.coli?
(1) The DNA dependent DNA polymerase catalyses polymerization in 5'→3' direction
(2) The DNA dependent DNA polymerase catalyses polymerization in one direction that is 3'→5'
(3) The DNA dependent RNA polymerase catalyses polymerization in one direction, that is 5'→3'
(4) The DNA dependent DNA polymerase catalyses polymerization in 5'→3' as well as 3'→5' direction
Answer (1)
Sol. In Prokaryotes, like E.coli during replication, the DNA dependent DNA polymerase catalyse polymerization only in one direction, that is 5'→3'

144. Given below are two statements:
Statement I: In C3 plants, some O2 binds to RuBisCO, hence CO2 fixation is decreased.
Statement II: In C4 plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (4)
Sol. In C3 plant, some O2 bind to RuBisCO, and hence CO2 fixation is decreased. Statement II is incorrect, photorespiration does not occur in C4 plants as they lack RuBisCO in mesophyll. Hence statement I is the only correct option.

145. Identify the correct description about the given figure:
(1) Compact inflorescence showing complete autogamy
(2) Wind pollinated plant inflorescence showing flowers with well exposed stamens.
(3) Water pollinated flowers showing stamens with mucilaginous covering.
(4) Cleistogamous flowers showing autogamy.
Answer (2)
Sol. The given diagram shows a wind pollinated plant showing compact inflorescence and well exposed stamens. Stamens are exposed so complete autogamy does not occur.

146. Match List I with List II
List I List II
A. Rose I. Twisted aestivation
B. Pea II. Perigynous flower
C. Cotton III. Drupe
D. Mango IV. Marginal placentation
Choose the correct answer from the options given below:
(1) A-II, B-III, C-IV, D-I
(2) A-II, B-IV, C-I, D-III
(3) A-I, B-II, C-III, D-IV
(4) A-IV, B-III, C-II, D-I
Answer (2)
Sol. Rose have half-inferior ovary, thus it is known as Perigynous flower.
In Pea, the placenta form a ridge along the ventral suture of the ovary and ovules are borne on this ridge forming two rows. In Cotton, twisted aestivation is present. In Mango, fruit is drupe.

147. The DNA present in chloroplast is:
(1) Circular, single stranded
(2) Linear, double stranded
(3) Circular, double stranded
(4) Linear, single stranded
Answer (3)
Sol. The DNA present in chloroplast is circular double stranded.

148. Which of the following are fused in somatic hybridization involving two varieties of plants?
(1) Pollens
(2) Callus
(3) Somatic embryos
(4) Protoplasts
Answer (4)
Sol. Protoplast of two varieties of plants are fused in somatic hybridization.

149. Match List I with List II
List I List II
A. Robert May I. Species-Area relationship
B. Alexander von Humboldt II. Long term ecosystem experiment using out door plots
C. Paul Ehrlich III. Global species diversity at about 7 million
D. David Tilman IV. Rivet popper hypothesis
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-II, D-I
(2) A-II, B-III, C-I, D-IV
(3) A-III, B-I, C-IV, D-II
(4) A-I, B-III, C-II, D-IV
Answer (3)
Sol. Robert May places the global species diversity at about 7 million.
Alexander von Humboldt gave species-area relationship.
Paul Ehrlich used an analogy "Rivet popper hypothesis" to explain the role of species in the ecosystem.
David Tilman performed long term ecosystem experiments using out door plots.

150. Match List I with List II
List I List II
A. Citric acid cycle I. Cytoplasm
B. Glycolysis II. Mitochondrial matrix
C. Electron transport system III. Intermembrane space of mitochondria
D. Proton gradient IV. Inner mitochondrial membrane
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-II, D-I
(2) A-I, B-II, C-III, D-IV
(3) A-II, B-I, C-IV, D-III
(4) A-III, B-IV, C-I, D-II
Answer (3)
Sol. Citric acid cycle occurs in mitochondrial matrix.
Glycolysis occurs in cytosol in most of the organism.
Electron transport system is present in the inner mitochondrial membrane.
Proton gradient is formed across the intermembrane space of mitochondria.

151. Which of the following is not a component of Fallopian tube?
(1) Ampulla
(2) Uterine fundus
(3) Isthmus
(4) Infundibulum
Answer (2)
Sol. The correct answer is option (2) as uterine fundus is the upper, dome-shaped part of the uterus, above the opening of fallopian tubes.
Option (3) is incorrect as isthmus is the last and narrow part of the oviduct that links to the uterus. Option (4) is incorrect as infundibulum is the part of oviduct which is closer to the ovary. Option (1) is incorrect as ampulla is the wider part of the oviduct.

152. Match List I with List II:
List I List II
A. Expiratory capacity I. Expiratory reserve volume + Tidal volume + Inspiratory reserve volume
B. Functional residual capacity II. Tidal volume + Expiratory reserve volume
C. Vital capacity III. Tidal volume + Inspiratory reserve volume
D. Inspiratory capacity IV. Expiratory reserve volume + Residual volume
Choose the correct answer from the options given below:
(1) A-I, B-III, C-II, D-IV
(2) A-II, B-IV, C-I, D-III
(3) A-III, B-II, C-IV, D-I
(4) A-II, B-I, C-IV, D-III
Answer (2)
Sol. Expiratory capacity = Tidal volume + Expiratory reserve volume
Functional residual capacity = Expiratory reserve volume + Residual volume
Vital capacity = Expiratory reserve volume + Tidal volume + Inspiratory reserve volume
Inspiratory capacity = Tidal volume + Inspiratory reserve volume

153. Which of the following are Autoimmune disorders?
A. Myasthenia gravis
B. Rheumatoid arthritis
C. Gout
D. Muscular dystrophy
E. Systemic Lupus Erythematosus (SLE)
Choose the most appropriate answer from the options given below:
(1) C, D & E only
(2) A, B & D only
(3) A, B & E only
(4) B, C & E only
Answer (3)

154. Match List I with List II:
List I (Sub Phases of Prophase I) List II (Specific Characters)
A. Diakinesis I. Synaptonemal complex formation
B. Pachytene II. Completion of terminalisation of chiasma
C. Zygotene III. Chromosomes look like thin threads
D. Leptotene IV. Appearance of recombination nodules
Choose the correct answer from the options given below
(1) A-IV, B-III, C-II, D-I
(2) A-IV, B-II, C-III, D-I
(3) A-I, B-II, C-IV, D-III
(4) A-II, B-IV, C-I, D-III
Answer (4)
Sol. (A) Diakinesis - Completion of terminalisation of chiasma
(B) Pachytene - Appearance of recombination nodules
(C) Zygotene - Synaptonemal complex formation
(D) Leptotene - Chromosomes look like thin threads
A-II, B-IV, C-I, D-III

155. In both sexes of cockroach, a pair of jointed filamentous structures called anal cerci are present on
(1) 11th segment
(2) 5th segment
(3) 10th segment
(4) 8th and 9th segment
Answer (3)
Sol. Correct answer is option (3), because in both sexes of cockroach, 10th segment bears a pair of jointed filamentous structures called anal cerci. Options (2), (4) and (1) are incorrect because 5th, 8th and 9th segments do not bear such structures. In adult cockroaches only 10th segments are present in abdomen. 11th abdominal segment is absent.

156. Which one of the following factors will not affect the Hardy-Weinberg equilibrium?
(1) Constant gene pool
(2) Genetic recombination
(3) Genetic drift
(4) Gene migration
Answer (1)
Sol. The correct answer is option (1) as a constant gene pool will not disturb the Hardy-Weinberg equilibrium. Option (3), (4) & (5) will affect the equilibrium leading to evolution.

157. Match List I with List II
List I List II
A. Non-medicated IUD I. Multiload 375
B. Copper releasing IUD II. Progestogens
C. Hormone releasing IUD III. Lippes loop
D. Implants IV. LNG-20
Choose the correct answer from the option given below:
(1) A-III, B-I, C-IV, D-II
(2) A-III, B-I, C-II, D-IV
(3) A-I, B-III, C-IV, D-II
(4) A-IV, B-I, C-II, D-III
Answer (1)
Sol. Correct answer is option (1) because
Lippes loop is a non-medicated IUD. Multiload 375 is a copper releasing IUD. LNG-20 is a hormone releasing IUD. Progestogens are used as implants.

158. Match List I with List II:
List I List II
A. Fibrous joints I. Adjacent vertebrae, limited movement
B. Cartilaginous joints II. Humerus and Pectoral girdle, rotational movement
C. Hinge joints III. Skull, don't allow any movement
D. Ball and socket joints IV. Knee, help in locomotion
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-IV, B-II, C-II, D-III
(3) A-I, B-III, C-II, D-IV
(4) A-II, B-III, C-I, D-IV
Answer (1)
Sol. The correct answer is option no. (1) as
Fibrous joints do not allow any movement. This type of joint is shown by the flat skull bones which fuse end-to-end with the help of dense fibrous connective tissues in the form of sutures.
Cartilaginous joint is present between the adjacent vertebrae in the vertebral column and this permits limited movements.
Hinge joint is a type of synovial joint present in knee, help in locomotion
Ball and socket joint is also a type of synovial joint present between humerus and pectoral girdle and allows rotational movement.

159. Match List I with List II:
List I List II
A. α-I antitrypsin I. Cotton bollworm
B. Cry IAb II. ADA deficiency
C. Cry IAc III. Emphysema
D. Enzyme replacement therapy IV Corn borer
Choose the correct answer form the options given below:
(1) A-II, B-IV, C-I, D-III
(2) A-II, B-I, C-IV, D-III
(3) A-II, B-I, C-III, D-IV
(4) A-II, B-IV, C-I, D-III
Answer (4)
Sol. The correct answer is option (4) as
α-I antitrypsin → Is used for treatment of Emphysema
Cry IAb gene → Controls corn borer
Cry IAc gene → Controls cotton bollworms
Enzyme replacement therapy → Can be used as treatment option in ADA deficiency.

160. Which of the following factors are favourable for the formation of oxyhaemoglobin in alveoli?
(1) Low pCO2 and High temperature
(2) High pO2 and High pCO2
(3) High pO2 and Lesser H⁺ concentration
(4) Low pCO2 and High H⁺ concentration
Answer (3)
Sol. The correct answer is option (3) as
Conditions favourable for formation of oxyhaemoglobin in alveoli are high pO2 less H⁺ concentration low pCO2 and low temperature. Option (1), (2) and (4) are not correct as they do not favour the formation of oxyhaemoglobin.

161. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R:
Assertion A : FSH acts upon ovarian follicles in female and Leydig cells in male.
Reason R : Growing ovarian follicles secrete estrogen in female while interstitial cells secrete androgen in male human being.
In the light of the above statements, choose the correct answer from the options given below:
(1) A is false but R is true
(2) Both A and R are true and R is the correct explanation of A
(3) Both A and R are true but R is NOT the correct explanation of A
(4) A is true but R is false
Answer (1)

162. Match List I with List II:
List I List II
A. Pons I. Provides additional space for Neurons, regulates posture and balance.
B. Hypothalamus II. Controls respiration and gastric secretions.
C. Medulla III. Connects different regions of the brain.
D. Cerebellum IV. Neuro secretory cells
Choose the correct answer from the options given below:
(1) A-II, B-I, C-II, D-IV
(2) A-II, B-II, C-I, D-IV
(3) A-III, B-IV, C-II, D-I
(4) A-II, B-II, C-II, D-IV
Answer (3)
Sol. The correct answer is option (3) as
A. Pons - Part of hindbrain, it connects different regions of the brain.
B. Hypothalamus - Also have neuro secretory cells which secrete hormones.
C. Medulla oblongata - Part of hindbrain which controls respiration and gastric secretions.
D. Cerebellum - Part of hindbrain with convoluted surface which provides additional space for neurons, also regulates posture and balance.

163. Match List I with List II:
List I List II
A. Axoneme I. Centriole
B. Cartwheel pattern II. Cilia and flagella
C. Crista III. Chromosome
D. Satellite IV. Mitochondria
Choose the correct answer from the options given below:
(1) A-II, B-I, C-IV, D-III
(2) A-IV, B-II, C-III, D-I
(3) A-IV, B-II, C-III, D-I
(4) A-II, B-IV, C-III, D-II
Answer (1)
Sol. Axoneme is seen in cilia and flagella
Centriole shows cartwheel appearance
Crista is found in mitochondria
Satellite is present in chromosomes

164. Match List I with List II:
List I List II
A. Pterophyllum I. Hag fish
B. Myxine II. Saw fish
C. Pristis III. Angel fish
D. Exocoetus IV. Flying fish
Choose the correct answer from the options given below:
(1) A-III, B-II, C-II, D-IV
(2) A-II, B-II, C-II, D-IV
(3) A-III, B-II, C-II, D-IV
(4) A-IV, B-II, C-II, D-II
Answer (3)
Sol. The correct option is option no. (3) as
Pterophyllum is the scientific name for Angel fish.
Myxine is the scientific name for Hag fish.
Pristis is the scientific name for Saw fish.
Exocoetus is the scientific name for Flying fish.

165. Which one is the correct product of DNA dependent RNA polymerase to the given template? 3'TACATGGCAATATCCATTCAS'
(1) 5'ATGTACCGTTTATAGGTAAAGT3'
(2) 5'AUGUACCGUUUUAUGGUAAAGU3'
(3) 5'AUGUAAAGUUUAUAGGUAAAGU3'
(4) 5'AUGUACCGUUUUAAGGGAAAGU3'
Answer (2)
Sol. Template DNA is: 3'TACATGGCAATATCCATTCAS'
5'AUGUACCGUUUUAUGGUAAAGU3'm-RNA

166. Given below are two statements:
Statement I: The presence or absence of hymen is not a reliable indicator of virginity.
Statement II: The hymen is torn during the first coitus only.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (4)
Sol. The correct answer is option no. (4) because the presence or absence of hymen is not a reliable indicator of virginity because hymen can also be broken by a sudden jolt, insertion of a vaginal tampon, active participation in some sports and in some women the hymen persists even after coitus.

167. Match List I with List II:
List-I List-II
A. Lipase I. Peptide bond
B. Nuclease II. Ester bond
C. Protease III. Glycosidic bond
D. Amylase IV. Phosphodiester bond
Choose the correct answer from the options given below:
(1) A-IV, B-I, C-III, D-II
(2) A-IV, B-II, C-III, D-I
(3) A-III, B-II, C-I, D-IV
(4) A-II, B-IV, C-I, D-III
Answer (4)
Sol. The correct answer is option (4) as
A. Lipase - Digests ester bond found in lipids.
B. Nuclease - Helps in digestion of phosphodiester bonds found in nucleic acids.
C. Protease - Helps in digestion of peptide bond found in proteins.
D. Amylase - Digests/breaks the glycosidic bonds found in carbohydrates i.e., digest starch into smaller molecules, ultimately yielding maltose, which in turn is cleaved into two glucose molecules by maltase.

168. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:
Assertion A : Breast-feeding during initial period of infant growth is recommended by doctors for bringing a healthy baby.
Reason R : Colostrum contains several antibodies absolutely essential to develop resistance for the new born baby.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) A is not correct but R is correct
(2) Both A and R are correct and R is the correct explanation of A
(3) Both A and R are correct but R is NOT the correct explanation of A
(4) A is correct but R is not correct
Answer (2)
Sol. Correct answer is option (2)
Breast-feeding during initial period of infant growth is recommended by doctors for bringing a healthy baby as colostrum contains several antibodies absolutely essential to develop resistance for the new born baby.

169. The flippers of the Penguins and Dolphins are the example of the
(1) Divergent evolution
(2) Adaptive radiation
(3) Natural selection
(4) Convergent evolution
Answer (4)
Sol. The correct answer is option (4), because the flippers of the Penguins and Dolphins perform similar function but they are not anatomically similar structures. This is example of analogous structures.
Option (2) is incorrect as adaptive radiation is the process of evolution of different species in a given geographical area starting from a point and literally radiating to the other areas of geography. Option (3) is incorrect as natural selection is a key mechanism of evolution. Option (1) is incorrect as divergent evolution results in the formation of homologous structures.

170. Match List I with List II:
List I List II
A. Cocaine I. Effective sedative in surgery
B. Heroin II. Cannabis sativa
C. Morphine III. Erythroxylum
D. Marijuana IV. Papaver somniferum
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-I, D-II
(2) A-IV, B-III, C-I, D-II
(3) A-I, B-III, C-II, D-IV
(4) A-II, B-I, C-III, D-IV
Answer (1)
Sol. The correct option is (1) as
A. Cocaine - Obtained from plant Erythroxylum coca, stimulating action on CNS.
B. Heroin - Formed by the acetylation of morphine which is obtained from plant Papaver somniferum.
C. Morphine - Obtained from Papaver somniferum, is an effective sedative in surgery.
D. Marijuana - Obtained from Cannabis sativa, produces hallucinogenic effect and affects cardiovascular system of the body.

171. Match List I with List II:
List I List II
A. Common cold I. Plasmodium
B. Haemozoin II. Typhoid
C. Widal test III. Rhinoviruses
D. Allergy IV. Dust mites
Choose the correct answer from the options given below:
(1) A-IV, B-II, C-III, D-I
(2) A-II, B-IV, C-III, D-I
(3) A-I, B-III, C-II, D-IV
(4) A-III, B-I, C-II, D-IV
Answer (4)
Sol. Correct answer is option (4) because
Common cold is caused by Rhinoviruses
Haemozoin is released in blood due to ruptured RBCs after Plasmodium infection.
Widal test is used to confirm the typhoid fever.
Allergy is caused due to dust mites.

172. Match List I with List II:
List I List II
A. Down's syndrome I. 11th chromosome
B. α-Thalassemia II. X' chromosome
C. β-Thalassemia III. 21st chromosome
D. Klinefelter's syndrome IV. 16th chromosome
Choose the correct answer from the options given below:
(1) A-IV, B-I, C-II, D-III
(2) A-I, B-II, C-III, D-IV
(3) A-II, B-III, C-IV, D-I
(4) A-III, B-IV, C-I, D-II
Answer (4)
Sol. Down's syndrome is due to presence of an additional copy of chromosome number 21. Klinefelter's syndrome is caused due to presence of an additional copy of X-chromosome. α-Thalassemia is controlled by two closely linked genes on chromosome 16 of each parent. β-Thalassemia is controlled by a single gene HBB on chromosome 11 of each parent.

173. Match List I with List II:
List I List II
A. Typhoid I. Fungus
B. Leishmaniasis II. Nematode
C. Ringworm III. Protozoa
D. Filariasis IV. Bacteria
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-III, D-I
(2) A-I, B-III, C-II, D-IV
(3) A-IV, B-III, C-I, D-II
(4) A-III, B-I, C-IV, D-II
Answer (3)
Sol. The correct answer is option (3) as
Typhoid - Caused by Salmonella typhimurium (Bacteria)
Leishmaniasis - Caused by protozoan i.e., Leishmania donovani
Ringworm - Caused by fungi belonging to the genera Microsporum, Trichophyton and Epidermophyton
Filariasis - Caused by Wuchereria bancrofti and Wuchereria malayi (Nematode)

174. Following are the stages of pathway for conduction of an action potential through the heart
A. AV bundle
B. Purkinje fibres
C. AV node
D. Bundle branches
E. SA node
Choose the correct sequence of pathway from the options given below
(1) E-A-D-B-C
(2) E-C-A-D-B
(3) A-E-C-B-D
(4) B-D-E-C-A
Answer (2)
Sol. Correct answer is option (2) because the correct pathway of conduction of action potential is SA → AV node → AV bundle → Bundle branches → Purkinje fibres

175. Following are the stages of cell division :
A. Gap 2 phase
B. Cytokinesis
C. Synthesis phase
D. Karyokinesis
E. Gap 1 phase
Choose the correct sequence of stages from the options given below :
(1) E-C-A-D-B
(2) C-E-D-A-B
(3) E-B-D-A-C
(4) B-D-E-A-C
Answer (1)
Sol. The correct sequence of stages of cell division is Gap 1 phase → Synthesis phase → Gap 2 phase (E) (C) (A) → Karyokinesis → Cytokinesis (D) (B)
The correct sequence will be → E → C → A → D → B

176. Match List I with List II :
List I List II
A. Pleurobrachia I. Mollusca
B. Radula II. Ctenophora
C. Stomochord III. Osteichthyes
D. Air bladder IV. Hemichordata
Choose the correct answer from the options given below :
(1) A-IV, B-III, C-II, D-I
(2) A-IV, B-II, C-III, D-I
(3) A-II, B-I, C-IV, D-III
(4) A-II, B-IV, C-I, D-III
Answer (3)
Sol. The correct answer is option (3) as
A. Pleurobrachia is a member of phylum Ctenophora.
B. Radula is a rasping feeding organ present in phylum Mollusca.
C. Stomochord Rudimentary structure similar to notochord found in the collar region of members of phylum Hemichordata.
D. Air bladder is found in Osteichthyes which provides them buoyancy.

177. Which of the following is not a steroid hormone?
(1) Glucagon
(2) Cortisol
(3) Testosterone
(4) Progesterone
Answer (1)
Sol. The correct answer is option (1) as glucagon is a proteinaceous hormone secreted from pancreas. Options (2), (3) and (4) are not correct as they are steroid in nature.

178. The "Ti plasmid" of Agrobacterium tumefaciens stands for
(1) Temperature independent plasmid
(2) Tumour inhibiting plasmid
(3) Tumor independent plasmid
(4) Tumor inducing plasmid
Answer (4)
Sol. The correct answer is option (4) as Ti plasmid of Agrobacterium tumefaciens is tumor inducing plasmid, containing T-DNA which causes tumor in several dicot plants. Options (2), (3) and (1) are not correct.

179. Given below are some stages of human evolution. Arrange them in correct sequence. (Past to Recent)
A. Homo habilis
B. Homo sapiens
C. Homo neanderthalensis
D. Homo erectus
Choose the correct sequence of human evolution from the options given below:
(1) A-D-C-B
(2) D-A-C-B
(3) B-A-D-C
(4) C-B-D-A
Answer (1)
Sol. Correct answer is option (1) because the correct sequence of stages of human evolution from past to recent is Homo habilis → Homo erectus → Homo neanderthalensis → Homo sapiens

180. Consider the following statements:
A. Annelids are true coleomates
B. Poriferans are pseudocoelomates
C. Aschelminthes are acoelomates
D. Platyhelminthes are pseudocoelomates
Choose the correct answer from the options given below:
(1) D only
(2) B only
(3) A only
(4) C only
Answer (3)
Sol. The correct answer is option no. (3), because annelids are true coelomate animals. Options (1), (2) and (4) are incorrect because poriferans are acoelomates, aschelminthes are pseudocoelomates and platyhelminthes are acoelomates.

181. The following diagram showing restriction sites in E. coli cloning vector pBR322. Find the role of 'X' and 'Y' genes:
(1) Gene 'X' is responsible for recognitions sites and 'Y' is responsible for antibiotic resistance.
(2) The gene 'X' is responsible for resistance to antibiotics and 'Y' for protein involved in the replication of Plasmid.
(3) The gene 'X' is responsible for controlling the copy number of the linked DNA and 'Y' for protein involved in the replication of Plasmid.
(4) The gene 'X' is for protein involved in replication of Plasmid and 'Y' for resistance to antibiotics.
Answer (3)
Sol. Correct answer is option (3), because
'X' in the given diagram is ori while 'Y' is rop. 'X' which is ori is responsible for controlling the copy number of the linked DNA and 'Y' which is rop codes for protein involved in the replication of plasmid. Options (1), (2) and (4) are incorrect as 'X' and 'Y' are not related to these functions.

182. Which of the following is not a natural/traditional contraceptive method?
(1) Vaults
(2) Coitus interruptus
(3) Periodic abstinence
(4) Lactational amenorrhea
Answer (1)
Sol. The correct answer is option (1) as
Vault is a barrier method of contraception which is made of rubber that is inserted into the female reproductive tract to cover the cervix during the coitus.
Option (2) is incorrect as coitus interruptus is a natural method of contraception in which male partner withdraws his penis from the vagina just before ejaculation so as to avoid insemination. Option (3) is incorrect as periodic abstinence is also a natural method of contraception in which couples avoid coitus during the fertile period.
Option (4) is incorrect as lactational amenorrhea is also a natural method of contraception which is based on the fact that ovulation and therefore the cycle do not occur during the period of intense lactational following parturition.

183. Which of the following statements is incorrect?
(1) Bio-reactors have an agitator system, an oxygen delivery system and foam control system
(2) A bio-reactor provides optimal growth conditions for achieving the desired product
(3) Most commonly used bio-reactors are of stirring type
(4) Bio-reactors are used to produce small scale bacterial cultures
Answer (4)
Sol. Correct answer is option (4)
The statement (4) is incorrect because bioreactors are used for processing of large volumes (100 - 1000 litres) of culture.
Small volume cultures cannot yield appreciable quantities of products. To produce in large quantities the development of bioreactors is required.

184. Three types of muscles are given as a, b and c. Identify the correct matching pair along with their location in human body:
Name of muscle/location
(1) (a) Involuntary - Nose tip (b) Skeletal - Bone (c) Cardiac - Heart
(2) (a) Smooth - Toes (b) Skeletal - Legs (c) Cardiac - Heart
(3) (a) Skeletal - Triceps (b) Smooth - Stomach (c) Smooth - Heart
(4) (a) Skeletal - Biceps (b) Involuntary - Intestine (c) Cardiac - Heart
Answer (3)
Sol. The correct answer is option (3) as
Figure (a) represents skeletal muscle fibres which are closely attached to skeletal bones. In a typical muscle such as triceps and biceps, striated muscle fibres are bundled together in a parallel fashion.
Figure (b) represents smooth muscle fibres which are present in the wall of internal organs such as the blood vessels, stomach and intestine.
Figure (c) represents cardiac muscle fibres which are exclusively present in the heart.

185. Given below are two statements :
Statement I : In the nephron, the descending limb of loop of Henle is impermeable to water and permeable to electrolytes.
Statement II : The proximal convoluted tubule is lined by simple columnar brush border epithelium and increases the surface area for reabsorption.
In the light of the above statements, choose the correct answer from the option given below :
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (3)
Sol. Correct answer is option (3) because
Statement I is false as the descending limb of loop of Henle is permeable to water and almost impermeable to electrolytes.
Statement II is false as proximal convoluted tubule is lined by simple cuboidal brush border epithelium which increases the surface area for reabsorption.

SECTION-B

186. Given below are two statements :
Statement I : Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.
Statement II : Both bone marrow and thymus provide micro environments for the development and maturation of T-lymphocytes.
In the light of above statements, choose the most appropriate answer from the options given below :
(1) Statement I is incorrect but Statement II is correct.
(2) Both Statement I and Statement II are correct.
(3) Both Statement I and Statement II are incorrect.
(4) Statement I is correct but Statement II is incorrect.
Answer (2)
Sol. The correct answer is option no. (2) as both statements I and II are correct.
In humans, the bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.
Both bone-marrow and thymus provide micro-environments for the development and maturation of T-lymphocytes.
Options (1), (3) and (4) are incorrect.

187. Given below are two statements: Statement I: Mitochondria and chloroplasts both double membranes bound organelles. Statement II: Inner membrane of mitochondria is relatively less permeable, as compared chloroplast. In the light of the above statements, choose the mis appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct.
(2) Both Statement I and Statement II are correct.
(3) Both Statement I and Statement II are incorrect.
(4) Statement I is correct but Statement II is incorrect.
Answer (4)
Sol. Both mitochondria and chloroplasts are double membrane bound cell organelles. Transport of ions occurs across the inner membrane of mitochondria. The inner membrane of chloroplast is impermeable to ions and metabolites. Therefore, it is said that inner membrane of mitochondria is relatively more permeable to that of chloroplast.

188. Match List I with List II:
List I List II
A. Mesozoic Era I. Lower invertebrates
B. Proterozoic Era II. Fish & Amphibia
C. Cenozoic Era III. Birds & Reptiles
D. Paleozoic Era IV. Mammals
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-II, B-I, C-III, D-IV
(3) A-III, B-I, C-II, D-IV
(4) A-II, B-II, C-IV, D-III
Answer (1)
Sol. The correct answer is option no. (1)
(A) Mesozoic Era - (III) Birds & Reptiles
(B) Proterozoic Era - (I) Lower invertebrates
(C) Cenozoic Era - (IV) Mammals
(D) Paleozoic Era - (II) Fish & Amphibia

189. Given below are two statements:
Statement I: Gause's competitive exclusion principle states that two closely related species competing for different resources cannot exist indefinitely.
Statement II: According to Gause's principle, during competition, the inferior will be eliminated. This may be true if resources are limiting.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true.
(2) Both Statement I and Statement II are true.
(3) Both Statement I and Statement II are false.
(4) Statement I is true but Statement II is false.
Answer (1)
Sol. Gause's competitive exclusion principle states that two closely related species competing for the same resources cannot exist indefinitely and the competitively inferior one will be eliminated eventually. This may be true if resources are limiting.

190. Regarding catalytic cycle of an enzyme action, select the correct sequential steps :
A. Substrate enzyme complex formation.
B. Free enzyme ready to bind with another substrate.
C. Release of products.
D. Chemical bonds of the substrate broken.
E. Substrate binding to active site.
Choose the correct answer from the options given below :
(1) E, D, C, B, A
(2) E, A, D, C, B
(3) A, E, B, D, C
(4) B, A, C, D, E
Answer (2)
Sol. The correct answer is option (2) which is E, A, D, C, B.
The catalytic cycle of an enzyme action can be described in the following steps.
(1) The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and run through the catalytic cycle once again.
(2) First, the substrate binds to the active site of the enzyme, fitting into the active site.
(3) The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.
(4) The active site of the enzyme, now in close proximity of the substrate breaks the chemical bonds of the substrate and the new enzyme-product complex is formed.
Options (1), (3) and (4) are incorrect as the steps mentioned are in the wrong sequence.

191. Match List I with List II :
List I List II
A. P wave I. Heart muscles are electrically silent.
B. QRS complex II. Depolarisation of ventricles.
C. T wave III. Depolarisation of atria.
D. T-P gap IV. Repolarisation of ventricles.
Choose the correct answer from the options given below :
(1) A-IV, B-II, C-I, D-III
(2) A-I, B-III, C-IV, D-II
(3) A-III, B-II, C-IV, D-I
(4) A-II, B-III, C-I, D-IV
Answer (3)
Sol. The correct answer is option no. (3) as
A. P wave - III. Depolarisation of atria.
B. QRS complex - II. Depolarisation of ventricles.
C. T wave - IV. Repolarisation of ventricles.
D. T-P gap - I. Heart muscles are electrically silent.

192. Match List I with List II:
List I List II
A. RNA polymerase III I. snRNPs
B. Termination of transcription II. Promotor
C. Splicing of Exons III. Rho factor
D. TATA box IV. SnRNAs, tRNA
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-I, D-II
(2) A-II, B-IV, C-I, D-III
(3) A-III, B-II, C-IV, D-I
(4) A-III, B-IV, C-I, D-II
Answer (1)
Sol. In eukaryotes, RNA polymerase III codes for snRNAs, tRNA and 5s rRNA. Splicing of exons is performed by snRNPs. TATA box is present in promoter region of transcription unit. Rho factor is responsible for termination of transcription.

193. Identify the correct option (A), (B), (C), (D) with respect to spermatogenesis.
(1) ICSH, Leydig cells, Sertoli cells, spermatogenesis.
(2) FSH, Leydig cells, Sertoli cells, spermiogenesis.
(3) ICSH, Interstitial cells, Leydig cells, spermiogenesis.
(4) FSH, Sertoli cells, Leydig cells, spermatogenesis.
Answer (2)
Sol. The correct answer is option no. (2) as
(A) is FSH which is a pituitary hormone.
(B) is Leydig cells which are found in the interstitial space outside of the seminiferous tubules.
(C) is Sertoli cells are found inside the seminiferous tubules.
(D) is Spermiogenesis which is a process that helps in transformation of spermatids into spermatozoa.

194. The following are the statements about non-chordates:
A. Pharynx is perforated by gill slits.
B. Notochord is absent.
C. Central nervous system is dorsal.
D. Heart is dorsal if present.
E. Post anal tail is absent.
Choose the most appropriate answer from the options given below:
(1) B, C & D only
(2) A & C only
(3) A, B & D only
(4) B, D & E only
Answer (4)
Sol. The correct answer is option no. (4) as the features of non-chordates among the given statements are:
B. Notochord is absent.
D. Heart is dorsal if present.
E. Post anal tail is absent.
Statements A and C are features of chordates.
Hence, option (4) is correct and options (1), (2) and (3) are incorrect.

195. Choose the correct statement given below regarding juxta medullary nephron.
(1) Juxta medullary nephrons outnumber the cortical nephrons.
(2) Juxta medullary nephrons are located in the columns of Bertini.
(3) Renal corpuscle of juxta medullary nephron lies in the outer portion of the renal medulla.
(4) Loop of Henle of juxta medullary nephron runs deep into medulla.
Answer (4)
Sol. The correct answer is option no. (4) because the length of loop of Henle of juxta medullary nephron is longer than the length of loop of Henle of cortical nephron and runs deep into medulla.
Option (1) is incorrect as juxta medullary nephrons are lesser in number than cortical nephrons.
Option (2) is incorrect as juxta medullary nephron are not present in columns of Bertini.
Option (3) is incorrect because renal corpuscle of juxta medullary nephron lies in inner cortical region.

196. Given below are two statements:
Statement I: The cerebral hemispheres are connected by nerve tract known as corpus callosum.
Statement II: The brain stem consists of the medulla oblongata, pons and cerebrum.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct.
(2) Both Statement I and Statement II are correct.
(3) Both Statement I and Statement II are incorrect.
(4) Statement I is correct but Statement II is incorrect.
Answer (4)
Sol. The correct answer is option (4) as statement I is correct but statement II is incorrect. In human brain, a deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres. The cerebral hemispheres are connected by a tract of nerve fibres called corpus callosum. Three major regions make up the brain stem i.e. mid brain, pons and medulla oblongata. Cerebrum is a part of forebrain which does not form brain stem. Options (1), (2) and (3) are incorrect.

197. Match List I with List II related to digestive system of cockroach.
List I List II
A. The structures used for storing of food I. Gizzard
B. Ring of 6-8 blind tubules at junction of foregut and midgut. II. Gastric Caeca
C. Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut. III. Malpighian tubules
D. The structures used for grinding the food. IV. Crop
Choose the correct answer from the options given below:
(1) A-III, B-II, C-IV, D-I
(2) A-IV, B-II, C-III, D-I
(3) A-I, B-II, C-III, D-IV
(4) A-IV, B-III, C-III, D-I
Answer (2)
Sol. The correct answer is option no. (2) as
The structure used for grinding the food particles - Gizzard
The structure used for storing of food - Crop
Ring of 6-8 blind tubules at junction of foregut and midgut which assists in secretion of digestive juices - Gastric Caeca
Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut which assists in elimination of nitrogenous wastes - Malpighian tubules

198. Match List I with List II:
List I List II
A. Unicellular glandular epithelium I. Salivary glands
B. Compound epithelium II. Pancreas
C. Multicellular glandular epithelium III. Goblet cells of alimentary canal
D. Endocrine glandular epithelium IV. Moist surface of buccal cavity
Choose the correct answer from the options given below:
(1) A-II, B-II, C-IV, D-II
(2) A-II, B-II, C-III, D-IV
(3) A-IV, B-III, C-II, D-II
(4) A-III, B-IV, C-I, D-II
Answer (4)
Sol. The correct answer is option no. (4) as
A. Unicellular glandular epithelium - (III) Goblet cells of alimentary canal
B. Compound epithelium - (IV) Lines moist surface of buccal cavity
C. Multicellular glandular epithelium - (I) Salivary glands
D. Endocrine glandular epithelium - (II) Pancreas

199. As per ABO blood grouping system, the blood group of father is B⁺, mother is A⁺ and child is O⁺. Their respective genotype can be
A. IB/IA/ii
B. IBIB/IAIA/ii
C. IAIB/IAIB/i
D. IA/iIB/iIA
E. iIB/iIA/IAIB
Choose the most appropriate answer from the options given below :
(1) D & E only
(2) A only
(3) B only
(4) C & B only
Answer (2)
Sol. Genotype of father with blood group B⁺ = IB/iIB
Genotype of mother with blood group A⁺ = IA/iIA
Genotype of child with blood group O⁺ = ii
Hence only 'A' is correct.

200. Match List I with List II :
List I List II
A. Exophthalmic goiter I. Excess secretion of cortisol, moon face & hyperglycemia.
B. Acromegaly II. Hypo-secretion of thyroid hormone and stunted growth.
C. Cushing's syndrome III. Hyper secretion of thyroid hormone & protruding eye balls.
D. Cretinism IV. Excessive secretion of growth hormone.
Choose the correct answer from the options given below :
(1) A-III, B-IV, C-I, D-II
(2) A-I, B-III, C-II, D-IV
(3) A-IV, B-II, C-I, D-III
(4) A-III, B-IV, C-II, D-I
Answer (1)

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