NEET Previous Year Question Paper 2024 Code R1 with Solutions – Free PDF Download

NEET Previous Year Question Paper 2024 with Solutions

QUESTIONS

1. 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 moves faster than point Q
(2) Point P has zero speed
(3) Point P moves slower than point Q
(4) Both the points P and Q move with equal speed
Answer (1)
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.

2. 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-III, B-IV, C-II, D-I
(2) A-IV, B-III, C-I, D-II
(3) A-I, B-II, C-III, D-IV
(4) A-II, B-I, C-IV, D-III
Answer (1)

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

4. 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) 6 V
(2) 8 V
(3) 10 V
(4) 4 V
Answer (2)

5. In an ideal transformer, the turns ratio is N_P/N_S = 1/2. The ratio V_S : V_P is equal to (the symbols carry their usual meaning):
(1) 2:1
(2) 1:1
(3) 1:4
(4) 1:2
Answer (1)
Sol. According to transformer ratio, V_S/V_P = N_S/N_P = 2:1

6. 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) sqrt(5)/2
(2) sqrt(3)/4
(3) sqrt(3)/2
(4) sqrt(5)/4
Answer (1)
Sol. A = 90°. In prism, r1 + c = A, r1 = 90° - c. sin c = 1/μ ⇒ cos c = sqrt(μ^2 - 1)/μ. Apply Snell's law: 1·sin 30° = μ sin(r1) ⇒ 1 × 1/2 = μ × sin(90° - c). 1/2 = μ × sqrt(μ^2 - 1)/μ.

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

8. 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) 198 N
(2) 1.98 mN
(3) 99 N
(4) 19.8 mN
Answer (4)
Sol. Excess force = T × 2πR = (7/100) × 2 × 3.14 × (4.5/100) = 197.82 × 10^-4 = 19.8 × 10^-3 N = 19.8 mN.

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

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 m^2. If the magnitude of magnetic moment of the needle is x × 10^-5 Am^2, then the value of 'x' is:
(1) 128π^2
(2) 50π^2
(3) 1280π^2
(4) 5π^2
Answer (3)
Sol. Time period of Oscillation, T = 2π sqrt(I/MB). ⇒ 1/4 = 2π sqrt(9.8 × 10^-6 / (M × 0.049)). ⇒ 1/16 = 4π^2 × 9.8 × 10^-6 / (M × 49 × 10^-3). ⇒ M = 4π^2 × 9.8 × 10^-6 × 16 / (49 × 10^-3) = 12.8π^2 × 10^-3 × 10^-2 × 10^2 = 1280π^2 × 10^-5 Am^2.

11. If the monochromatic source in Young's double slit experiment is replaced by white light, then
(1) There will be a central dark fringe surrounded by a few coloured fringes
(2) There will be a central bright white fringe surrounded by a few coloured fringes
(3) All bright fringes will be of equal width
(4) Interference pattern will disappear
Answer (2)

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

13. 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 × 10^8 N m^-2 and 2 × 10^11 N m^-2, is:
(1) 0.4 mm
(2) 40 mm
(3) 8 mm
(4) 4 mm
Answer (4)
Sol. In the case for maximum elongation, Stress = Elastic limit. δmax = σelastic × L / Young's modulus = 8 × 10^8 × 1 / (2 × 10^11) = 4 × 10^-3 = 4 mm. i.e. maximum elongation is 4 mm.

14. 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) A is incorrect but B is correct
(2) Both A and B are correct
(3) Both A and B are incorrect
(4) A is correct but B is incorrect
Answer (4)

15. A particle moving with uniform speed in a circular path maintains:
(1) Constant acceleration
(2) Constant velocity but varying acceleration
(3) Varying velocity and varying acceleration
(4) Constant velocity
Answer (3)
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.

16. 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, p = 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, C and D only
(2) A, C and D only
(3) A, B, D and E only
(4) A and B only
Answer (1)
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. (C) λ = h/p, p = h/λ, p = hν/c. (D) In photon-electron collision both total energy and total momentum are conserved.

17. 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) 2:1
(2) 4:1
(3) 1:4
(4) 1:2
Answer (1)

18. The graph which shows the variation of (1/λ^2) and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle):
Answer (3)
Sol. de-Broglie wavelength λ = h/P = h/mv = h/sqrt(2mE) where E = 1/2 mv^2. Squaring both sides, λ^2 = h^2/(4m^2E). ⇒ 1/λ^2 = (constant)E. Graph passes through origin with constant slope.

19. An unpolarised light beam strikes a glass surface at Brewster's angle. Then
(1) The refracted light will be completely polarised.
(2) Both the reflected and refracted light will be completely polarised.
(3) The reflected light will be completely polarised but the refracted light will be partially polarised.
(4) The reflected light will be partially polarised.
Answer (3)

20. 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) 5
(2) 7
(3) 6
(4) 10
Answer (4)
Sol. x = 2t - 1, v = dx/dt = 2 m s^-1, P = F·v = 2 × 5 = 10 W.

21. 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) 4.4 T
(2) 4.4 mT
(3) 44 T
(4) 44 mT
Answer (2)
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.

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

23. 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) 4T
(2) T/4
(3) sqrt(2T)
(4) T
Answer (1)
Sol. T = m(α)^2, T' = m(2α)^2 = 4T.

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

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

26. 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) 4 N
(2) 6 N
(3) 10 N
(4) Zero
Answer (2)
Sol. F = (M1 + M2)a, a = 10/(2 + 3) = 2 m s^-2, F = M2(2) = 3 × 2 N = 6 N.

27. 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) 286,80
(2) 288,82
(3) 286,81
(4) 280,81
Answer (3)

28. 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) 1/[100(N + 1)]
(2) 100N
(3) 10(N + 1)
(4) 1/(10N)
Answer (1)
Sol. V.C = MSD - VSD. Given: (N + 1) VSD = N MSD. VSD = (N/(N + 1)) MSD. From (1) and (2): V.C = (MSD) - N/(N + 1)(MSD) = MSD/(N + 1) = 0.01/(N + 1) = 1/[100(N + 1)].

29. 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) 5 m, 2 s
(2) 5 cm, 1 s
(3) 5 m, 1 s
(4) 5 cm, 2 s
Answer (1)
Sol. x = 5 sin(πt + π/3) m. Amplitude = 5 m. ω = π = 2π/T. T = 2π/π = 2 s.

30. 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 CD
(2) BA and DC
(3) AB and DC
(4) AB and CD
Answer (4)
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.

31. 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) A̅B̅ + A̅
(2) B̅
(3) B
(4) A̅B̅ + A̅
Answer (2)
Sol. According to given truth table, output is independent on value of A. Output Y = B̅.

32. A wire of length l and resistance 100 Ω 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) 52 Ω
(2) 55 Ω
(3) 60 Ω
(4) 26 Ω
Answer (1)
Sol. R = ρl/A. R' = ρl/(10A) = R/10. RS = 5 × R/10 = 50. RP = R/50. Req = RS + RP = 52 Ω.

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

34. A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is: (Take 1/(4πε0) = 9 × 10^9 SI units)
(1) 1 × 10^5
(2) 3 × 10^5
(3) Zero
(4) 0.5 × 10^5
Answer (3)
Sol. For uniformly charged spherical shell, V = kq/R (For r ≤ R). ∴ V_C = V_P. V_C - V_P = Zero.

35. 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) 9.8 m s^-2
(2) 4.9 m s^-2
(3) 3.92 m s^-2
(4) 19.6 m s^-2
Answer (3)
Sol. g' = GM'/R'^2 = GM/[10(R/2)^2] = 4GM/(10R^2) = 0.4 × 9.8 = 3.92 m s^-2.

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) 2GmM/(3R)
(2) GmM/(2R)
(3) GmM/(3R)
(4) 5GmM/(6R)
Answer (4)
Sol. Apply energy conservation, Ui + Ki = Ui + Ki. ⇒ -GmM/R + Ki = -GmM/(3R) + 1/2 mν^2. ⇒ -GmM/R + Ki = GmM/(3R) + 1/2 × m × GM/(3R). ⇒ Ki = 1/6 GmM/R + GmM/R = 5/6 GmM/R.

37. 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) 28
(2) 17
(3) 32
(4) 34
Answer (1)
Sol. f0 = 140 cm and fe = 5 cm. For distant object, m = f0/fe = 140/5 = 28.

38. 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:
Answer (4)
Sol. Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially. After that, the slope of v-t curve is constant and positive. After some time, velocity becomes constant and acceleration is zero. After that, the slope of v-t curve is constant and negative.

39. 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) 0.93 A
(2) 1.20 A
(3) 0.35 A
(4) 0.58 A
Answer (1)
Sol. Capacitive Reactance Xc = 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 = 0.93 A.

40. A force defined by F = αt^2 + βt acts on a particle at a given time t. The factor which is dimensionless, if α and β are constants, is:
(1) αt/β
(2) αβt
(3) αβ/α
(4) βt/α
Answer (1)
Sol. From principle of homogeneity, [F] = [αt^2] = [βt]. [α] = [F]/[t^2] and [β] = [F]/[t]. ∴ [α][t] = [β]. ∴ αt/β = dimensionless.

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) 0.93 A
(2) 1.20 A
(3) 0.35 A
(4) 0.58 A
Answer (1)
Sol. Xc = 1/(2πfC) = 1000/3.14. Vrms = 210 V. irms = Vrms/Xc = 210/Xc. Peak current = √2 irms = √2 × 210/1000 × 3.14 = 0.93 A.

42. A metallic bar of Young's modulus, 0.5 × 10^11 N m^-2 and coefficient of linear thermal expansion 10^-5 °C^-1, length 1 m and area of cross-section 10^-3 m^2 is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is:
(1) 50 × 10^3 N
(2) 100 × 10^3 N
(3) 2 × 10^3 N
(4) 5 × 10^3 N
Answer (1)

43. 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 equal to I flows in the same direction as I
(2) Displacement current of magnitude equal to I flows in a direction opposite to that of I
(3) Displacement current of magnitude greater than I flows but can be in any direction
(4) There is no current
Answer (1)

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

45. 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) A and C only
(2) A, C and D only
(3) C only
(4) B and D only
Answer (1)
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.

46. 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, C and E only
(2) B, D and E only
(3) A, B and C only
(4) A, B and E only
Answer (1)
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'V^2. U = 1/2 CV^2. U' > U. Hence final energy is greater than initial energy.

47. 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 equal to I flows in the same direction as I
(2) Displacement current of magnitude equal to I flows in a direction opposite to that of I
(3) Displacement current of magnitude greater than I flows but can be in any direction
(4) There is no current
Answer (1)

48. 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) sqrt(2)
(2) 2sqrt(3)
(3) 4
(4) sqrt(3)
Answer (1)
Sol. T' = 2π sqrt(l'/g) where l' = l/2. T = 2π sqrt(l/g). T' = x/2 T. 2π sqrt(l/2g) = x/2 2π sqrt(l/g). 1/sqrt(2) = x/2 ⇒ x = sqrt(2).

49. 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/2
(2) 2M
(3) M/sqrt(3)
(4) M
Answer (1)
Sol. M = ml. Δl = 2(l/2) sin 30° = l/2. M' = ml/2 = M/2.

50. 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) sqrt(2)
(2) 2sqrt(3)
(3) 4
(4) sqrt(3)
Answer (1)
Sol. T' = 2π sqrt(l'/g) where l' = l/2. T = 2π sqrt(l/g). T' = x/2 T. 2π sqrt(l/2g) = x/2 2π sqrt(l/g). 1/sqrt(2) = x/2 ⇒ x = sqrt(2).

51. 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) A and D
(2) B and E
(3) E and D
(4) B and C
Answer (2)
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.

52. 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) -x/9
(2) -4x
(3) -4/9 x
(4) -x
Answer (4)
Sol. En = -RH(Z^2/n^2) J. For He+ (n = 1), En = -x = -RH(2^2/1^2) = -4RH. ∴ RH = x/4. For Be3+ (n = 2), En = -RH(Z^2/n^2) J = -x/4 × (4×4)/(2×2) = -x J.

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

54. Match List I with List II.
List-I (Process)
List-II (Conditions)
A. Isothermal process I. No heat exchange
B. Isochoric process II. Carried out at constant temperature
C. Isobaric process III. Carried out at constant volume
D. Adiabatic process IV. Carried out at constant pressure
Choose the correct answer from the options given below:
(1) A-IV, B-II, C-III, D-I
(2) A-II, B-III, C-IV, D-I
(3) A-I, B-II, C-III, D-IV
(4) A-IV, B-III, C-II, D-I
Answer (2)
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.

55. 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 a tendency to go in forward direction.
(2) Reaction has a tendency to go in backward direction.
(3) Reaction has gone to completion in forward direction.
(4) Reaction is at equilibrium.
Answer (2)
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 = 1. As Qc > Kc, so reaction has a tendency to move backward.

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-I, B-III, C-IV, D-II
(2) A-I, B-IV, C-III, D-II
(3) A-II, B-IV, C-III, D-I
(4) A-II, B-III, C-IV, D-I
Answer (4)
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. In which of the following processes entropy increases?
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from 130 K to 0 K.
C. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
D. Cl2(g) → 2Cl(g)
Choose the correct answer from the options given below:
(1) A, B and D
(2) A, C and D
(3) C and D
(4) A and C
Answer (2)
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.

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

59. Match List I with List II.
List I (Reaction)
List II (Reagents/Condition)
A. B. C. D.
Choose the correct answer from the options given below:
(1) A-III, B-II, C-II, D-IV
(2) A-IV, B-II, C-II, D-II
(3) A-II, B-IV, C-II, D-II
(4) A-IV, B-II, C-II, D-II
Answer (2)

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

61. Which one of the following alcohols reacts instantaneously with Lucas reagent?
(1) CH3 - CH2 - CH - OH | CH3
(2) CH3 - CH - CH2OH | CH3
(3) CH3 - C - OH | CH3
(4) CH3 - CH2 - CH2 - CH2OH
Answer (3)
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. 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) Both Statement I and Statement II are incorrect
(2) Statement I is correct but Statement II is incorrect
(3) Statement I is incorrect but Statement II is correct
(4) Both Statement I and Statement II are correct
Answer (4)
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.

63. 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) Both Statement I and Statement II are false
(2) Statement I is correct but Statement II is false
(3) Statement I is incorrect but Statement II is true
(4) Both statement I and Statement II are true
Answer (4)
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.

64. The E° value for the Mn3+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to change of
(1) d5 to d2 configuration
(2) d4 to d5 configuration
(3) d3 to d5 configuration
(4) d5 to d4 configuration
Answer (2)
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.

65. 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) Sublimation
(2) Distillation
(3) Chromatography
(4) Crystallization
Answer (1)
Sol. (1) 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. (2) Distillation: It is used to separate volatile liquids from non-volatile impurities and the liquids having sufficient difference in their boiling point. (3) Chromatography: It is based on separation by using stationary and mobile phase. (4) Crystallization: It is based on difference in the solubilities of the compound and impurities in a suitable solvent.

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

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

68. 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.

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

70. Match List I with List II.
List I (Conversion)
List II (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-I, D-II
(2) A-III, B-III, C-I, D-IV
(3) A-III, B-IV, C-II, D-I
(4) A-III, B-IV, C-I, D-II
Answer (4)
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.

71. Among Group 16 elements, which one does NOT show -2 oxidation state?
(1) Se
(2) Te
(3) Po
(4) O
Answer (3)
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.

72. '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 E only
(2) B and C only
(3) A and D only
(4) B and D only
Answer (4)
Sol. Ions No. of unpaired electrons Configuration
Ti3+ 1 3d1
Cr2+ 4 3d4
Mn2+ 5 3d5
Fe2+ 4 3d6
Sc3+ 0 3d0
Spin only magnetic moment is given by sqrt(n(n + 2)) BM. Cr2+ and Fe2+ will have same spin only magnetic moment.

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

74. 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) B > C > A
(2) A > C > B
(3) A > B > C
(4) B > A > C
Answer (1)
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.

75. The compound that will undergo SN1 reaction with the fastest rate is
Answer (3)

76. The most stable carbocation among the following is:
(1) CH3-CH-CH2-CH-CH3 | CH3
(2) cyclopentyl-CH2+
(3) cyclohexyl-CH3+
(4) H3C-CH2-CH-CH3 | CH3
Answer (3)
Sol. The stability of carbocation can be described by the hyperconjugation. Greater the extent of hyperconjugation, more is the stability of carbocation.
(1) → 5 α-H
(2) → 1 α-H
(3) → 7 α-H
(4) → 3 α-H
Stability order of carbocations = (3) > (1) > (4) > (2).

77. 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) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true

78. 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) 250 mg
(2) Zero mg
(3) 200 mg
(4) 750 mg
Answer (1)
Sol. M = W×1000/(M×V(in mL)). W = M×M×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.

79. 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) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true

80. Identify the major product C formed in the following reaction sequence:
CH3-CH2-CH2-I → NaCN → A → OH- → B → NaOH/Br2 → C
(1) butylamine
(2) butanamide
(3) α-bromobutanoic acid
(4) propylamine
Answer (4)

81. 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) 0.315 g
(2) 31.5 g
(3) 0.0315 g
(4) 3.15 g
Answer (1)
Sol. Cu2+(aq) + 2e- → Cu(s). Mass of Cu deposited (w) = M×i×t/nF = 63×9.6487×100/(2×96487) = 0.315 g.

82. Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si
Choose the correct answer from the options given below:
(1) Si < C < O < N < F
(2) O < F < N < C < Si
(3) F < O < N < C < Si
(4) Si < C < N < O < F
Answer (4)
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.

83. Match List I with List II.
List I (Quantum Number)
A. m
B. ms
C. l
D. n
List II (Information provided)
I. Shape of orbital
II. Size of orbital
III. Orientation of orbital
IV. Orientation of spin of electron
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-I, D-II
(2) A-III, B-IV, C-II, D-II
(3) A-III, B-I, C-IV, D-II
(4) A-I, B-III, C-II, D-IV
Answer (1)
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.

84. Match List I with List II.
List I (Compound)
A. NH3
B. BrF5
C. XeF4
D. SF6
List II (Shape/geometry)
I. Trigonal Pyramidal
II. Square Planar
III. Octahedral
IV. Square Pyramidal
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-III, D-II
(2) A-III, B-IV, C-I, D-II
(3) A-III, B-III, C-IV, D-II
(4) A-I, B-IV, C-II, D-III
Answer (4)

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

SECTION-B

86. The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is (Given R = 2.0 cal K^-1 mol-1)
(1) -413.14 calories
(2) 413.14 calories
(3) 100 calories
(4) 0 calorie

87. Identify the correct answer.
(1) BF3 has non-zero dipole moment
(2) Dipole moment of NF3 is greater than that of NH3
(3) Three canonical forms can be drawn for CO3- ion
(4) Three resonance structures can be drawn for ozone
Answer (3)
Sol. (1) BF3 i.e., F-B; Dipole moment = 0. (2) (0.80 × 10^-30 Cm) < (4.90 × 10^-30 Cm). (3) CO3^2- has three canonical forms. (4) In ozone; there are two resonating structures.

88. Major products A and B formed in the following reaction sequence, are
Answer (2)

89. The pair of lanthanoid ions which are diamagnetic is
(1) Ce3+ and Eu2+
(2) Gd3+ and Eu3+
(3) Pm3+ and Sm3+
(4) Ce4+ and Yb2+
Answer (4)
Sol. Magnetic moment μ = sqrt(n(n + 2)). n → number of unpaired electron. Ce4+ ⇒ (Xe)4f0, μ = 0, Diamagnetic. Yb2+ ⇒ (Xe)4f14, μ = 0, Diamagnetic. Ce3+ ⇒ (Xe)4f1, μ = sqrt(3), Paramagnetic. Eu2+ ⇒ (Xe)4f7, μ = sqrt(63), Paramagnetic. Gd3+ ⇒ (Xe)4f7, μ = sqrt(63), Paramagnetic. Eu3+ ⇒ (Xe)4f6, μ = sqrt(48), Paramagnetic. Pm3+ ⇒ (Xe)4f4, μ = sqrt(24), Paramagnetic. Sm3+ ⇒ (Xe)4f5, μ = sqrt(35), Paramagnetic. Hence Ce4+ and Yb2+ are only diamagnetic.

90. 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) ABC3
(2) AB2C2
(3) ABC4
(4) AB2C2
Answer (1)
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 = A/1 : B/1 : C/3. The correct empirical formula of compound X is ABC3.

91. 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) B,C,A,D,E
(2) E,C,D,B,A
(3) E,A,B,C,D
(4) B,A,D,C,E
Answer (4)
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+. The correct order is B,A,D,C,E.

92. 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) 0.0889
(2) 0.8889
(3) 0.717
(4) 0.00889
Answer (3)
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.
2NO(g) ⇌ N2(g) + O2(g)
t = 0: 0.1, 0, 0
0.1-0.1α, 0.05α, 0.05α
Kc = 0.05α × 0.05α / (0.1-0.1α)^2 = 0.05α × 0.05α / 0.01(1-α)^2.
1.607 = (0.05)^2 α^2 / 0.01(1-α)^2.
α^2/(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.

93. 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) 380.4 kJ/mol
(2) 3.80 kJ/mol
(3) 3804 kJ/mol
(4) 38.04 kJ/mol
Answer (4)
Sol. log(k2/k1) = Ea/2.303R (1/T1 - 1/T2). log(4/1) = Ea/2.303R (1/300 - 1/330). Ea = (log(4) × 2.303 × 8.314 × 300 × 330)/30 = 3.804 × 10^4 J/mol = 38.04 kJ/mol.

94. 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) concentrated sulphuric acid
(2) dilute nitric acid
(3) dilute sulphuric acid
(4) dilute hydrochloric acid
Answer (3)
Sol. During the preparation of Mohr's salt, dilute sulphuric acid is added to prevent the hydrolysis of Fe2+ ion.

95. Identify the major product C formed in the following reaction sequence:
CH3-CH2-CH2-I → NaCN → A → OH- → B → NaOH/Br2 → C
(1) butylamine
(2) butanamide
(3) α-bromobutanoic acid
(4) propylamine
Answer (4)

96. 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) 0.315 g
(2) 31.5 g
(3) 0.0315 g
(4) 3.15 g
Answer (1)
Sol. Cu2+(aq) + 2e- → Cu(s). Mass of Cu deposited (w) = M×i×t/nF = 63×9.6487×100/(2×96487) = 0.315 g.

97. For the given reaction:
Answer (2)

98. 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) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true
Answer (4)
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-.

99. 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) 310°C
(2) 25.73°C
(3) 12.05°C
(4) 37°C
Answer (4)
Sol. Π = CRT. Slope = RT. 25.73 = 0.083 × T. T = 25.73/0.083 = 309.47 = 310 K. Temperature in °C = 310 - 273 = 37°C.

100. The products A and B obtained in the following reactions, respectively, are
3ROH + PCl3 → 3RCl + A
ROH + PCl5 → RCl + HCl + B
(1) POCl3 and H3PO4
(2) H3PO4 and POCl3
(3) H3PO3 and POCl3
(4) POCl3 and H3PO3
Answer (3)
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.

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

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

103. 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-II, B-I, C-II, D-IV
(2) A-II, B-IV, C-I, D-II
(3) A-IV, B-II, C-II, D-II
(4) A-I, B-II, C-II, D-IV
Answer (2)
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.

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) A,B,C and D only
(2) A,C,D and E only
(3) B,C,D and E only
(4) C,D and E only
Answer (3)
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. List of endangered species was released by
(1) WWF
(2) FOAM
(3) IUCN
(4) GEAC
Answer (3)
Sol. List of endangered species was released by IUCN.

106. 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) D and E only
(2) B and C only
(3) A and E only
(4) A and B only
Answer (2)
Sol. Correct answer is option (2) 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.

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) B, C and D only
(2) C, D and E only
(3) D and E only
(4) A, B and C only
Answer (2)
Sol. For dark reaction of photosynthesis there are the requirement of CO2, ATP, NADPH.

108. 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) Biodiversity conservation
(2) Semi-conservative method
(3) Sustainable development
(4) in-situ conservation
Answer (1)
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.

109. 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) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true
Answer (2)
Sol. The correct answer is option (2) 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.

110. 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) Structural gene, Transposons, Operator gene
(2) Inducer, Repressor, Structural gene
(3) Promotor, Structural gene, Terminator
(4) Repressor, Operator gene, Structural gene
Answer (3)
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.

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

112. Hind II always cuts DNA molecules at a particular point called recognition sequence and it consists of:
(1) 6 bp
(2) 4 bp
(3) 10 bp
(4) 8 bp
Answer (1)
Sol. The correct answer is option (1). 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 (2), (3) and (4) are incorrect because they have either more than 6 or less than 6 bp.

113. 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) Hypogynous; (b) Epigynous
(2) (a) Perigynous; (b) Epigynous
(3) (a) Perigynous; (b) Perigynous
(4) (a) Epigynous; (b) Hypogynous
Answer (3)
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.

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

115. Which one of the following is not a criterion for classification of fungi?
(1) Mode of nutrition
(2) Mode of spore formation
(3) Fruiting body
(4) Morphology of mycelium
Answer (1)
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.

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

117. 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, C, D and E only
(2) B, C and D only
(3) A, B, C, D and E
(4) A, B and C only
Answer (1)
Sol. According to Law of Dominance: (1) Characters are controlled by discrete units called factors. (2) Factors occur in pairs. (3) 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.

118. 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-I, B-III, C-II, D-IV
(2) A-III, B-II, C-I, D-IV
(3) A-IV, B-III, C-II, D-I
(4) A-III, B-II, C-IV, D-I
Answer (4)
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.

119. Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of:
(1) Feedback inhibition
(2) Competitive inhibition
(3) Enzyme activation
(4) Cofactor inhibition
Answer (2)
Sol. Correct answer is option (2) 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), (3) and (4) are incorrect as enzyme activation, co-factor inhibition are not showing structural similarity with substrate.

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

121. 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 flowered as well as pink flowered plants
(2) Only pink flowered plants
(3) Red, Pink as well as white flowered plants
(4) Only red flowered plants
Answer (1)
Sol. Pink colour flower in snapdragon have genotype Rr. Red flowered snapdragon have genotype RR when they both are crossed. So the progeny that we get are red and pink flowered plants only.

122. 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
(2) Bb
(3) BB/Bb
(4) BB
Answer (1)
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.

123. 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-II, B-IV, C-II, D-I
(2) A-II, B-I, C-IV, D-II
(3) A-IV, B-I, C-II, D-II
(4) A-II, B-I, C-II, D-I
Answer (2)
Sol. A. Clostridium butylicum - Butyric acid. B. Saccharomyces cerevisiae - Ethanol. C. Trichoderma polysporum - Cyclosporin-A. D. Streptococcus sp. - Streptokinase.

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

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

126. 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.

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

130. 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) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true
Answer (4)
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.

131. 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) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true
Answer (3)
Sol. Collenchyma is also living tissue. Gymnosperm lack xylem vessels but presence of xylem vessels is the characteristic of angiosperm.

132. 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,C and D only
(2) A,B and E only
(3) A,B and D only
(4) A,C and D only
Answer (3)
Sol. Major causes of biodiversity losses are (1) Over-exploitation (2) Alien species invasions (3) Co-extinctions (4) Habitat loss and fragmentation. Hence correct option is A, B and D only.

133. The lactose present in the growth medium of bacteria is transported to the cell by the action of
(1) Acetylase
(2) Permease
(3) Polymerase
(4) Beta-galactosidase
Answer (2)
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.

134. Bulliform cells are responsible for
(1) Protecting the plant from salt stress.
(2) Increased photosynthesis in monocots.
(3) Providing large spaces for storage of sugars.
(4) Inward curling of leaves in monocots.
Answer (4)
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.

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

SECTION-B

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) B, C, D and E only
(2) A, C, D and E only
(3) A, B, C and E only
(4) A, B, C and D only
Answer (2)
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. 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-I, C-IV, D-II
(2) A-II, B-III, C-II, D-IV
(3) A-III, B-IV, C-II, D-I
(4) A-II, B-III, C-II, D-IV
Answer (1)

138. 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) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true
Answer (2)
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.

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

140. 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) x (kcal m^-2) yr^-1
(2) 10x (kcal m^-2) yr^-1
(3) 100x/3x (kcal m^-2) yr^-1
(4) x/10 (kcal m^-2) yr^-1
Answer (2)
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/m²/yr) would be GPP at second trophic level and 100x × 10% (kcal/m²/yr) i.e., 10x (kcal/m²/yr) energy would be GPP at third trophic level.

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

142. 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-II, B-I, C-IV, D-III
(2) A-III, B-IV, C-I, D-II
(3) A-IV, B-III, C-II, D-I
(4) A-I, B-II, C-III, D-IV
Answer (1)
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.

143. Match List I with List II
List I
A. Frederick Griffith
B. Francois Jacob & Jacque Monod
C. Har Gobind Khorana
D. Meselson & Stahl
List II
I. Genetic code
II. Semi-conservative mode of DNA replication
III. Transformation
IV. Lac operon
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-I, D-II
(2) A-II, B-III, C-IV, D-I
(3) A-IV, B-I, C-II, D-III
(4) A-III, B-II, C-I, D-IV
Answer (1)

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

145. Identify the correct description about the given figure:
(1) Water pollinated flowers showing stamens with mucilaginous covering.
(2) Cleistogamous flowers showing autogamy.
(3) Compact inflorescence showing complete autogamy
(4) Wind pollinated plant inflorescence showing flowers with well exposed stamens.
Answer (4)

146. 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-I, B-II, C-II, D-IV
(2) A-II, B-III, C-IV, D-I
(3) A-III, B-IV, C-I, D-II
(4) A-IV, B-I, C-II, D-III
Answer (4)
Sol. Correct answer is option (4). A. GLUT-4 IV. Enables glucose transport into cells. B. Insulin I. Hormone. C. Trypsin II. Enzyme. D. Collagen III. Intercellular ground substance.

147. Identify the step in tricarboxylic acid cycle, which does not involve oxidation of substrate.
(1) Succinic acid → Malic acid
(2) Succinyl-CoA → Succinic acid
(3) Isocitrate → α-ketoglutaric acid
(4) Malic acid → Oxaloacetic acid
Answer (2)
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.

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

149. 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-I, B-II, C-III, D-IV
(2) A-IV, B-III, C-II, D-I
(3) A-II, B-III, C-IV, D-I
(4) A-II, B-IV, C-I, D-III
Answer (4)
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.

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

151. 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-IV, C-II, D-I
(2) A-II, B-II, C-II, D-IV
(3) A-II, B-II, C-II, D-IV
(4) A-II, B-II, C-II, D-IV
Answer (1)
Sol. The correct answer is option (1) 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.

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

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

154. 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-III, B-II, C-IV, D-I
(2) A-II, B-I, C-IV, D-II
(3) A-II, B-II, C-II, D-IV
(4) A-II, B-IV, C-II, D-II
Answer (4)
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.

155. 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) Both A and R are true but R is NOT the correct explanation of A
(2) A is true but R is false
(3) A is false but R is true
(4) Both A and R are true and R is the correct explanation of A
Answer (3)
Sol. The correct answer is option (3) as FSH is a gonadotropin affects ovarian follicles in females and causes their growth but in males LH affects Leydig cells leading to secretion of androgens. Growing ovarian follicles secrete estrogen in females while interstitial cells secrete androgen in male human being. Hence, Assertion is false and Reason is true.

156. 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-III, B-II, C-II, D-IV
(2) A-II, B-IV, C-II, D-II
(3) A-IV, B-I, C-II, D-II
(4) A-IV, B-II, C-II, D-I
Answer (2)

162. 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) Skeletal - Triceps (b) Smooth - Stomach (c) Cardiac - Heart
(2) (a) Skeletal - Biceps (b) Involuntary - Intestine (c) Smooth - Heart
(3) (a) Involuntary - Nose tip (b) Skeletal - Bone (c) Cardiac - Heart
(4) (a) Smooth - Toes (b) Skeletal - Legs (c) Cardiac - Heart
Answer (1)
Sol. The correct answer is option (1) 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.

163. 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 chiasmata
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-I, B-II, C-IV, D-III
(2) A-II, B-IV, C-I, D-III
(3) A-IV, B-III, C-II, D-I
(4) A-IV, B-II, C-III, D-I
Answer (2)

164. 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-II, B-II, C-IV, D-I
(2) A-II, B-IV, C-I, D-II
(3) A-IV, B-I, C-II, D-II
(4) A-II, B-II, C-II, D-IV
Answer (2)
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.

165. Which of the following statements is incorrect?
(1) Most commonly used bio-reactors are of stirring type
(2) Bio-reactors are used to produce small scale bacterial cultures
(3) Bio-reactors have an agitator system, an oxygen delivery system and foam control system
(4) A bio-reactor provides optimal growth conditions for achieving the desired product
Answer (2)
Sol. Correct answer is option (2). The statement (2) 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.

166. Match List I with List II:
List I List II
A. Pterophyllum
B. Myxine
C. Pristis
D. Exocoetus
Choose the correct answer from the options given below:
(1) A-II, B-II, C-II, D-IV
(2) A-IV, B-I, C-II, D-II
(3) A-II, B-II, C-I, D-IV
(4) A-II, B-I, C-II, D-IV
Answer (1)

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

168. 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) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true
Answer (2)
Sol. The correct answer is option no. (2) 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.

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

170. 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) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true
Answer (1)
Sol. Correct answer is option (1) 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.

171. 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-B-D-A-C
(2) B-D-E-A-C
(3) E-C-A-D-B
(4) C-E-D-A-B
Answer (3)
Sol. The correct sequence of stages of cell division is Gap 1 phase → Synthesis phase → Gap 2 phase → Karyokinesis → Cytokinesis (E) (C) (A) (D) (B). The correct sequence will be → E → C → A → D → B.

172. In both sexes of cockroach, a pair of jointed filamentous structures called anal cerci are present on
(1) 10th segment
(2) 8th and 9th segment
(3) 11th segment
(4) 5th segment
Answer (1)
Sol. Correct answer is option (1), because in both sexes of cockroach, 10th segment bears a pair of jointed filamentous structures called anal cerci. Options (2), (3) and (4) 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.

173. 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-I, B-III, C-II, D-IV
(2) A-II, B-III, C-I, D-IV
(3) A-III, B-I, C-IV, D-II
(4) A-IV, B-II, C-III, D-I
Answer (3)
Sol. The correct answer is option no. (3) 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.

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

175. 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) A-E-C-B-D
(2) B-D-E-C-A
(3) E-A-D-B-C
(4) E-C-A-D-B
Answer (4)
Sol. Correct answer is option (4) because the correct pathway of conduction of action potential is SA → AV node → AV bundle → Bundle branches → Purkinje fibres.

176. 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-I, B-III, C-IV, D-II
(2) A-IV, B-I, C-II, D-III
(3) A-III, B-I, C-IV, D-II
(4) A-III, B-I, C-II, D-IV
Answer (3)
Sol. Correct answer is option (3) 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.

177. Which of the following is not a natural/traditional contraceptive method?
(1) Periodic abstinence
(2) Lactational amenorrhea
(3) Vaults
(4) Coitus interruptus
Answer (3)
Sol. The correct answer is option (3) 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 (1) is incorrect as periodic abstinence is also a natural method of contraception in which couples avoid coitus during the fertile period. Option (2) 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. Option (4) 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.

178. Which one is the correct product of DNA dependent RNA polymerase to the given template?
3'TACATGGCAATATCCATTCAS
(1) 5'AUGUAAAGUUUAUAGGUAGU3'
(2) 5'AUGUACCGUUUUAUAGGGAAUGU3'
(3) 5'ATGTTACCGTTTATAGGTAAGT3'
(4) 5'AUGUACCGUUUUAUAGGUAAUGU3'
Answer (4)

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

180. 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-I, B-III, C-II, D-IV
(2) A-II, B-I, C-III, D-IV
(3) A-III, B-IV, C-I, D-II
(4) A-IV, B-III, C-I, D-II
Answer (3)
Sol. The correct option is (3) 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.

181. 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-IV, B-III, C-I, D-II
(2) A-III, B-I, C-IV, D-II
(3) A-II, B-IV, C-III, D-I
(4) A-I, B-III, C-II, D-IV
Answer (1)

182. 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-III, B-I, C-II, D-IV
(2) A-III, B-IV, C-I, D-II
(3) A-II, B-IV, C-I, D-III
(4) A-II, B-I, C-IV, D-III
Answer (2)
Sol. The correct answer is option (2) 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.

183. 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) Both A and R are correct but R is NOT the correct explanation of A
(2) A is correct but R is not correct
(3) A is not correct but R is correct
(4) Both A and R are correct and R is the correct explanation of A
Answer (4)
Sol. Correct answer is option (4). 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.

184. The flippers of the Penguins and Dolphins are the example of the
(1) Natural selection
(2) Convergent evolution
(3) Divergent evolution
(4) Adaptive radiation
Answer (2)

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

SECTION-B

186. 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-IV, B-III, C-I, D-II
(2) A-III, B-IV, C-I, D-II
(3) A-II, B-I, C-IV, D-III
(4) A-II, B-I, C-III, D-IV
Answer (2)
Sol. The correct answer is option no. (2) 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.

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

188. Identify the correct option (A), (B), (C), (D) with respect to spermatogenesis.
(1) ICSH, Interstitial cells, Leydig cells, spermiogenesis.
(2) FSH, Sertoli cells, Leydig cells, spermatogenesis.
(3) ICSH, Leydig cells, Sertoli cells, spermatogenesis.
(4) FSH, Leydig cells, Sertoli cells, spermiogenesis.
Answer (4)
Sol. The correct answer is option no. (4) 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.

189. 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-III, B-II, C-IV, D-I
(2) A-II, B-II, C-I, D-IV
(3) A-IV, B-II, C-I, D-II
(4) A-II, B-II, C-IV, D-II
Answer (1)

190. 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) Both Statement I and Statement II are incorrect.
(2) Statement I is correct but Statement II is incorrect.
(3) Statement I is incorrect but Statement II is correct.
(4) Both Statement I and Statement II are correct.
Answer (4)
Sol. The correct answer is option no. (4) 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), (2) and (3) are incorrect.

191. 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) Both Statement I and Statement II are false.
(2) Statement I is true but Statement II is false.
(3) Statement I is false but Statement II is true.
(4) Both Statement I and Statement II are true.
Answer (3)
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.

192. Match List I with List II:

194. 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) B only
(2) C & B only
(3) D & E only
(4) A only
Answer (4)
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.

195. 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-III, B-II, C-IV, D-I
(2) A-III, B-IV, C-I, D-II
(3) A-IV, B-III, C-I, D-II
(4) A-II, B-IV, C-I, D-III
Answer (3)
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.

196. 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-II, D-IV
(2) A-I, B-II, C-IV, D-III
(3) A-III, B-I, C-IV, D-II
(4) A-II, B-I, C-III, D-IV
Answer (3)
Sol. The correct answer is option no. (3)
(A) Mesozoic Era – (III) Birds & Reptiles
(B) Proterozoic Era – (I) Lower invertebrates
(C) Cenozoic Era – (IV) Mammals
(D) Paleozoic Era – (II) Fish & Amphibia

197. 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) Both Statement I and Statement II are incorrect.
(2) Statement I is correct but Statement II is incorrect.
(3) Statement I is incorrect but Statement II is correct.
(4) Both Statement I and Statement II are correct.
Answer (2)
Sol. The correct answer is option (2) 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), (3) and (4) are incorrect.

198. 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) A, E, B, D, C
(2) B, A, C, D, E
(3) E, D, C, B, A
(4) E, A, D, C, B
Answer (4)
Sol. The correct answer is option (4) which is E, A, D, C, B. The catalytic cycle of an enzyme action can be described in the following steps. (1) The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate. (2) 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. (3) 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. (4) First, the substrate binds to the active site of the enzyme, fitting into the active site. Options (1), (2) and (3) are incorrect as the steps mentioned are in the wrong sequence.

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

200. 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) A, B & D only
(2) B, D & E only
(3) B, C & D only
(4) A & C only
Answer (2)
Sol. The correct answer is option no. (2) 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 (2) is correct and options (1), (3) and (4) are incorrect.

जी, प्रश्न 200 यह है:

200. 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) A, B & D only
(2) B, D & E only
(3) B, C & D only
(4) A & C only

Answer (2)

Sol. The correct answer is option no. (2) 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 (2) is correct and options (1), (3) and (4) are incorrect.

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