NEET Previous Year Question Paper 2025 Code 45 with Solutions – Free PDF Download

 NEET Previous Year Question Paper 2025 with Solutions

 

1. Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θ₀ (θ₀ << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is:
(take θ(x) = sin θ(x) = tan θ(x) = dy/dx, g is the acceleration due to gravity)
(1) d²y/dx² = ρg/S x
(2) d²y/dx² = ρg/S
(3) d²y/dx² = ρg/S
(4) d²y/dx² = ρg/√S
Answer (2)
Sol. ROC = Radius of curvature at point A
Curvature = 1/ROC = |d²y/dx²| / (1 + (dy/dx)²)^(3/2) = |d²y/dx²| / (1 + 0)^(3/2) = d²y/dx² [∵ dy/dx = tan θ = 0]
ΔP = S × curvature
⇒ ρgy = S d²y/dx²
∴ d²y/dx² = ρgy/S

2. A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is
(1) 100
(2) 125
(3) 150
(4) 250
Answer (2)
Sol. m = L/f₀ × D/f_e
40/2 × 25/4
m = 125

3. An electron (mass 9 × 10⁻³¹ kg and charge 1.6 × 10⁻¹⁹ C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B of magnitude 9 × 10⁻⁴ T perpendicular to its direction of motion. We wish to apply an uniform electric field E together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 × 10⁸ ms⁻¹)
(1) E is perpendicular to B and its magnitude is 27 × 10⁴ V m⁻¹
(2) E is perpendicular to B and its magnitude is 27 × 10² V m⁻¹
(3) E is parallel to B and its magnitude is 27 × 10² V m⁻¹
(4) E is parallel to B and its magnitude is 27 × 10⁴ V m⁻¹
Answer (2)
Sol. For no deflection of electron, F_B = F_E
-e(v × B) = -eE
⇒ E = v × B ⇒ E ⊥ B
E = vB = (c/100) × 9 × 10⁻⁴
= (3 × 10⁸/100) × 9 × 10⁻⁴
= 27 × 10² V m⁻¹

4. There are two inclined surfaces of equal length (L) and same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (μ_k) between the object and the rough surface is close to
(1) 0.25
(2) 0.40
(3) 0.5
(4) 0.75
Answer (4)
Sol. t_rough = 2 t_smooth
a_smooth = g sin θ
t ∝ 1/√a ⇒ t_smooth ∝ 1/√(g sin θ)
a_rough = g sin θ - μ_k g cos θ
t_rough/t_smooth = √(sin θ)/(√(sin θ) - μ_k cos θ) = 2
Squaring both sides

5. The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If F_A and F_B are the forces applied by the breaks on cars A and B respectively, then the ratio of F_A/F_B is
(1) 3/2
(2) 2/3
(3) 1/3
(4) 1/2
Answer (2)
Sol. By work-energy theorem,
F S = ΔK.E
⇒ -F S = k_f - k_i
⇒ F S = k_i - k_f
⇒ F_A/F_B = (k_A/k_B) × (S_B/S_A)
= (100/225) × (1500/1000)
= 150/225 = 2/3

6. The current passing through the battery in the given circuit, is:
(1) 2.0A
(2) 0.5A
(3) 2.5A
(4) 1.5A
Answer (2)
Sol. Balanced wheatstone bridge:
Its equivalent R' = (4 × 8)/12 = 8/3 Ω
Circuit can be redrawn as
Req = 8/3 + 1/3 + 1.5 + 5.5
= 10 Ω
I = V/Req = 5/10 = 0.5 A

7. A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v₀ as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v₀ is:
(1) (sin θ)^(1/2)
(2) (cos θ/(2 + 3 sin θ))^(1/2)
(3) (sin θ/(3 sin θ + 2))^(1/2)
(4) (sin θ/(3 sin θ + 2))^(1/2)
Answer (4)
Sol. At Point P, mg sin θ = mv²/l ...(1)
By conservation of mechanical energy at point P & Q
1/2 mv₀² = 1/2 mv² + mg(l + l sin θ)
v₀²/2 = v²/2 + gl(1 + sin θ)
Put gl = v²/sin θ using (1)
v₀²/2 = v²/2 + v²/sin θ (1 + sin θ)
v₀²/2 = v²/2 + v²/sin θ + v²
v₀²/2 = 3/2 v² + 2v²/(2 sin θ)
v₀² = v² [3 + 2/sin θ]
v/v₀ = (sin θ/(3 sin θ + 2))^(1/2)

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

9. The electric field in a plane electromagnetic wave is given by
E_z = 60 cos(5x + 1.5 × 10⁹ t) V/m.
Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
(1) B_y = 2 × 10⁻⁷ cos(5x + 1.5 × 10⁹ t) T
(2) B_x = 2 × 10⁻⁷ cos(5x + 1.5 × 10⁹ t) T
(3) B_z = 60 cos(5x + 1.5 × 10⁹ t) T
(4) B_y = 60 sin(5x + 1.5 × 10⁹ t) T
Answer (1)
Sol. In electromagnetic wave, E and B are in same phase and B₀ = E₀/c; their planes are perpendicular to each other.
∴ B_y = (60/c) cos(5x + 1.5 × 10⁹ t) T
= (60/(3 × 10⁸)) cos(5x + 1.5 × 10⁹ t) T
B_y = 2 × 10⁻⁷ cos(5x + 1.5 × 10⁹ t) T

10. A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s²)
(1) 21 NS
(2) 7 NS
(3) 0
(4) 84 NS
Answer (1)
Sol. v₁ = √(2gh₁)
= √(2 × 9.8 × 40)
v₁ = √784 = 28 m s⁻¹
and v₂ = √(2gh₂) = √(2 × 9.8 × 10)
= √196 = 14 m s⁻¹
Impulse = Δp = m(v_f - v_i) = m(v₂ - v₁)
= 1/2 (14 - (-28))
= 21 NS

11. AB is a part of an electrical circuit (see figure). The potential difference V_A - V_B, at the instant when current i = 2 A and is increasing at a rate of 1 amp/second is:
(1) 5 volt
(2) 6 volt
(3) 9 volt
(4) 10 volt
Answer (4)
Sol. Given, i = 2A and di/dt = +1 A/s
V_A - L(di/dt) - 5 - i × 2 = V_B
⇒ V_A - 1 × 1 - 5 - 2 × 2 = V_B
⇒ V_A - V_B = 10 volt

12. A 2 amp current is flowing through two different small circular copper coils having radii ratio 1 : 2. The ratio of their respective magnetic moments will be
(1) 1:4
(2) 1:2
(3) 2:1
(4) 4:1
Answer (1)
Sol. Magnetic moment of current carrying circular loop = IA
M = IA
M ∝ A [I - Same]
M₁/M₂ = A₁/A₂ = πr₁²/πr₂² = (1/2)² = 1/4

13. In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact. Then the power of the combination and the total magnification in comparison to the power (p) and magnification (m) for each lens will be, respectively
(1) 4p and 4m
(2) p⁴ and 4m
(3) 4p and m⁴
(4) p⁴ and m⁴
Answer (3)
Sol. For series combination of lens
p_eff = p₁ + p₂ + p₃ + p₄ = 4p
m_eff = m₁ × m₂ × m₃ × m₄ = m⁴

14. An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to:
[Given, R = 100/12 J mol⁻¹ K⁻¹, and molecular mass of O₂ = 32, 1 atm pressure = 1.01 × 10⁵ N/m²]
(1) 0.125 kg
(2) 0.144 kg
(3) 0.116 kg
(4) 0.156 kg
Answer (3)
Sol. Number of moles left
n = pV/RT = (12 × 1.01 × 10⁵ N/m² × 30 × 10⁻³ m³)/((100/12) × 300)
n = (12 × 1.01 × 12)/10 = 14.54 moles
Moles removed = 18.2 - 14.54 = 3.656 moles
Mass removed = 3.656 × 32 = 116.99 g = 0.116 kg

15. In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x² + x. The acceleration of the particle is
(1) 2/(x + 2)³
(2) 2/(2x + 1)³
(3) 2/(x + 1)³
(4) 2/(2x + 1)
Answer (2)
Sol. t = x² + x
dt/dx = 2x + 1
v = dx/dt = 1/(2x + 1)
dv/dx = -2/(2x + 1)²
a = v dv/dx = 1/(2x + 1)[-2/(2x + 1)²]
= -2/(2x + 1)³

16. To an ac power supply of 220 V at 50 Hz, a resistor of 20 Ω, a capacitor of reactance 25 Ω and an inductor of reactance 45 Ω are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively
(1) 7.8 A and 30°
(2) 7.8 A and 45°
(3) 15.6 A and 30°
(4) 15.6 A and 45°
Answer (2)
Sol. X_L = 45 Ω, X_C = 25 Ω, R = 20 Ω
I = 220/√((X_L - X_C)² + R²) = 220/√((45 - 25)² + 20²)
= 220/(2√2) = 11/√2 = 7.779 A
tan φ = (X_L - X_C)/R = (45 - 25)/20 = 1
φ = 45°

17. The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.
(1) 100 days
(2) 105 days
(3) 115 days
(4) 108 days
Answer (4)
Sol. Assuming the Sun to be a solid sphere, I = 2/5 mR²
Using conservation of angular momentum, I'ω' = Iω
⇒ 2/5 m(2R)² × 2π/T' = 2/5 mR² × 2π/T
⇒ T' = 4T = 4 × 27 = 108 days

18. A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(h/e) where n is an integer, h is Planck's constant and e is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (m is the mass of the electron)
(1) he/πm
(2) he/2πm
(3) heB/πm
(4) heB/2πm
Answer (2)
Sol. Magnetic force = mv²/r
evB = mv²/r
v = eBr/m
φ = BA
nh/e = Bπr²
Br² = nh/(eπ)
μ = IA
= e/T πr²
= e × v/(2πr) πr²
μ = evr/2
= 1/2 e × eBr/m r
μ = 1/2 e² Br²/m
μ = 1/2 e² nh/(eπm)
μ = neh/(2πm)
for n = 1
μ = eh/(2πm)

19. Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature 3T and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is T₁ and that at the right junction is T₂. The ratio T₁/T₂ is
(1) 3/2
(2) 4/3
(3) 5/3
(4) 5/4
Answer (3)
Sol. In series, Req = R₁ + R₂ + R₃
= l/(2KA) + l/(KA) + l/(2KA)
= 4l/(2KA)
Req = 2l/(KA)
In series rate of heat flow is same
∴ (3T - T₁)/R₁ = (3T - T)/(Req)
(3T - T₁)(2KA)/l = (2T)KA/(2l)
⇒ 6T - 2T₁ = T
⇒ 2T₁ = 5T
⇒ T₁ = 5T/2 ...(1)
Now, equate heat flow rate in 3rd section & total section
(T₂ - T)/R₃ = (3T - T)/Req
⇒ (T₂ - T)(2KA)/l = 2T(KA)/(2l)
⇒ 2T₂ - 2T = T
⇒ T₂ = 3T/2 ...(2)
By equation (1) and equation (2)
T₁/T₂ = (5T × 2)/(2 × 3T) = 5/3

20. The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K₁ and K₂ with thickness 3d/8 and d/2, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K₁ = 1.25K₂ the value of K₁ is:
(1) 2.66
(2) 2.33
(3) 1.60
(4) 1.33
Answer (1)

21. Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of the bus service and the speed (assumed constant) of the buses.
(1) 9 min, 40 km/h
(2) 15 min, 120 km/h
(3) 10 min, 90 km/h
(4) 10 min, 90 km/h
Answer (2)
Sol. X → Y
Let velocity of bus = v km/hr
Relative velocity of bus w.r.t. scooty = (v - 60)
Distance between 2 consecutive buses = vT
(v - 60)30 = vT
Y → X
(v + 60)10 = vT
Equating (1) and (2)
(v - 60)30 = (v + 60)10
∴ v = 120 km/hr
T = 15 min

22. A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take g = 10 m/s²)
(1) 100 N
(2) 100√3 N
(3) 200 N
(4) 200√3 N
Answer (2)
Sol. For translational equilibrium
N₁ = Mg
N₂ = f
For rotational equilibrium
Torque about A, Mg(l/2) cos θ = N₂ l sin θ
(Mg/2) cot θ = N₂ = f
(Mg/2) cot 30° = f
(Mg/2) √3 = N₂
100√3 = f

23. In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency ω(t) and average amplitude A(t) of the system change with time t. Which one of the following options schematically depicts these changes correctly?
(1) ω increases, A decreases
(2) ω increases, A decreases
(3) ω decreases, A increases
(4) ω decreases, A decreases
Answer (2)
Sol. At any point of time, time period is given by
T = 2π√(m/k)
Here m is decreasing, so time period T will be decreasing
Since ω = 2π/T

24. A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a gas of density ρ and takes a spherical shape of radius R. When the gas is allowed to flow freely out of it, its radius r changes from R to 0 (zero) in time T. If the speed v(r) of gas coming out of the balloon depends on r as r^a and T ∝ S^a A^b ρ^c R^d then
(1) α = 1/2, β = -1, γ = +1, δ = 3/2
(2) α = -1/2, β = -1, γ = 1/2, δ = 7/2
(3) α = -1/2, β = -1, γ = 1/2, δ = 7/2
(4) α = -1/2, β = -1, γ = 1/2, δ = 7/2
Answer (3)
Sol. T ∝ S^a A^b ρ^c R^d
M⁰L⁰T¹ = K(M T⁻²)^a (L²)^b (M L⁻³)^c L^d
M⁰L⁰T¹ = K[M^(a+c) L^(2b - 3c + d) T^(-2a)]
-2α = 1 ⇒ α = -1/2
α + γ = 0 ⇒ γ = 1/2
2β - 3γ + δ = 0
2β - 3(1/2) + δ = 0

25. Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at x = 0.1 cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is M = 5 cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is
(1) 5.18 cm
(2) 5.08 cm
(3) 4.98 cm
(4) 5.00 cm
Answer (3)
Sol. Least count = 1 MSD - 1 VSD
= 1 MSD - 9/10 MSD
= 1/10 MSD
= 1/10 × 0.1 cm = 0.01 cm
Zero error = +0.1 cm

26. A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
(1) Zero at all places
(2) Constant between the plates and zero outside the plates
(3) Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
(4) Zero between the plates and non-zero outside
Answer (3)
Sol. Let the surface charge density be σ = q/A
dσ/dt = constant
∴ d/dt(q/A) = constant ⇒ I/A = constant
It means displacement current is constant.
This system will act like a cylindrical wire.
The graph of magnetic field B vs r is

27. An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then
(1) Reflected light is completely polarized and the angle of reflection is close to 60°
(2) Reflected light is partially polarized and the angle of reflection is close to 30°
(3) Both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to 60° and 30°, respectively
(4) Transmitted light is completely polarized with angle of refraction close to 30°
Answer (1)
Sol. Using Brewster law
1.73 = tan θ_P
√3 = tan θ_P
θ_P = 60°
At this polarising angle, reflected light is perfectly polarized and transmitted light is partially polarised.

28. Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:
(1) 3F/5
(2) 2F/3
(3) F/2
(4) 3F/8
Answer (4)
Sol. A/q B/q r F = Kqq/r²
A q/2 B 3q/4 r F' = K(q/2)(3q/4)/r² = 3F/8

29. A container has two chambers of volumes V₁ = 2 litres and V₂ = 3 litres separated by a partition made of a thermal insulator. The chambers contain n₁ = 5 and n₂ = 4 moles of ideal gas at pressures p₁ = 1 atm and p₂ = 2 atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of
(1) 1.3 atm
(2) 1.6 atm
(3) 1.4 atm
(4) 1.8 atm
Answer (2)
Sol. P₁V₁ + P₂V₂ = P(V₁ + V₂)
1(2) + 2(3) = P(2 + 3)
8/5 = P
⇒ 1.6 atm

30. A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the nth orbit and the particle's speed v in the orbit depend on n as
(1) r ∝ n^(1/3); v ∝ n^(1/3)
(2) r ∝ n^(1/3); v ∝ n^(2/3)
(3) r ∝ n^(2/3); v ∝ n^(1/3)
(4) r ∝ n^(4/3); v ∝ n^(-1/3)
Answer (3)
Sol. Given, force is constant

31. The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?
(1) 88 earth days
(2) 225 earth days
(3) 172 earth days
(4) 124 earth days
Answer (1)
Sol. Applying Kepler's 3rd law: T² ∝ R³
Radius of Martian orbit, R' = 4R
(T'/T)² = (R'/R)³ = (4R/R)³ = 4³ = 64 ⇒ T'/T = 8
Length of 1 year on Mercury = T = T'/8 = 687/8 = 85.88 days

32. A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:
(1) 16 N
(2) 27 N
(3) 32 N
(4) 36 N
Answer (2)
Sol. W = mg and g = GM/R²
g_h = GM/(R + h)²
⇒ W_h/W = mg_h/mg = g_h/g = R²/(R + h)² (h = R/3)
⇒ W_h/W = R²/(R + R/3)² = R²/(4R/3)² = 9/16
⇒ W_h = 9/16 W = 9/16 × 48 [W = 48 N]
= 27 N

33. A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
(1) R/64
(2) R/32
(3) R/16
(4) R/8
Answer (3)

34. De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)
(1) 0.067 nm
(2) 0.67 nm
(3) 1.67 nm
(4) 2.67 nm
Answer (2)
Sol. r = 0.052 n²
For n = 2
r = 0.052 × 4 = 0.208 nm
Mvr = nh/(2π)
λ = h/(Mv) = πr
= 3.14 × 0.208 nm
= 0.65317 nm
≈ 0.67 nm

35. An electric dipole with dipole moment 5 × 10⁻⁵ cm is aligned with the direction of a uniform electric field of magnitude 4 × 10⁵ N/C. The dipole is then rotated through an angle of 60° with respect to the electric field. The change in the potential energy of the dipole is:
(1) 0.8 J
(2) 1.0 J
(3) 1.2 J
(4) 1.5 J
Answer (2)
Sol. Given

36. A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:
(1) 1.5 A
(2) 2.0 A
(3) 2.5 A
(4) 3.0 A
Answer (2)
Sol. R_AB = (1Ω || 3Ω) in series with (2Ω || 4Ω)
= (3 × 1)/(3 + 1) + (2 × 4)/(2 + 4)
= 3/4 + 8/6 = (9 + 16)/12 = 25/12 Ω
Now total current through cell
I = 50/(25/12) = 24 A
I_1Ω = 3/4 × 24 = 18 A, I_3Ω = 1/4 × 24 = 6 A
I_2Ω = 4/6 × 24 = 16 A, I_4Ω = 2/6 × 24 = 8 A
Using junction rule at C, I_CD = 18 - 16 = 2 A (From C to D)

37. A photon and an electron (mass m) have the same energy E. The ratio (λ_photon/λ_electron) of their de Broglie wavelengths is: (c is the speed of light)
(1) √(E/2m)
(2) c√(2mE)
(3) c√(2m/E)
(4) (1/c)√(E/2m)
Answer (3)
Sol. For photon, E = hc/λ_ph ⇒ λ_ph = hc/E

38. Which of the following options represent the variation of photoelectric current with property of light shown on the x-axis?
(1) A only
(2) A and C
(3) A and D
(4) B and D
Answer (1)
Sol. Photoelectric current is directly proportional to intensity of light.

39. A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:
(1) 7/8
(2) 7/40
(3) 7/57
(4) 7/64
Answer (3)
Sol. For larger solid sphere about diameter Y-axis,
I_whole = 2/5 M(2R)² = 8/5 MR²
Density of sphere is uniform
⇒ M/V_whole = M_smaller/V_smaller ⇒ M/(4/3 π(2R)³) = M'/(4/3 πR³)
⇒ M' = M/8
Using parallel axis theorem for smaller sphere,
I' = I_cm + M'R² = 2M'R²/5 + M'R² = 7/40 MR²
∴ Ratio = I_smaller/I_remaining = I'/(I_whole - I') = (7/40 MR²)/((8/5 - 7/40)MR²) = 7/(64 - 7) = 7/57

40. A full wave rectifier circuit with diodes (D₁) and (D₂) is shown in the figure. If input supply voltage V_in = 220 sin(100πt) volt, then at t = 15 msec
(1) D₁ is forward biased, D₂ is reverse biased
(2) D₁ is reverse biased, D₂ is forward biased.
(3) D₁ and D₂ both are forward biased
(4) D₁ and D₂ both are reverse biased
Answer (2)
Sol. V_in = 220 sin(100πt) volt

41. A physical quantity P is related to four observations a, b, c and d as follows:
P = a³b²/(c√d)
The percentage errors of measurement in a, b, c and d are 1%, 3%, 2% and 4% respectively. The percentage error in the quantity P is
(1) 10%
(2) 2%
(3) 13%
(4) 15%
Answer (3)
Sol. Maximum % error in P = ΔP/P × 100 = 3(Δa/a × 100) + 2(Δb/b × 100) + (Δc/c × 100) + 1/2(Δd/d × 100)
= 3 × (1) + 2 × (3) + (2) + 1/2 × (4)
= 13%

42. The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at 22.5° from the polarization axis of one of the polaroids, is I₀ (I₀ is the intensity of polarised light after passing through the first polaroid):
(1) I₀/2
(2) I₀/4
(3) I₀/8
(4) I₀/16
Answer (3)
Sol. I₁ = I₀ cos²(45/2)
I₂ = I₁ cos²(90 - 45/2)
= I₀ cos²(45/2) sin²(45/2)
= I₀/4 (4 cos²(45/2) sin²(45/2))
= I₀/4 sin² 45° = I₀/8

43. Two identical point masses P and Q suspended from two separate massless springs of spring constants k₁ and k₂ respectively, oscillate vertically. If their maximum speeds are the same, the ratio (A_Q/A_P) of the amplitude A_Q of mass Q to the amplitude A_P of mass P is
(1) k₂/k₁
(2) k₁/k₂
(3) √(k₂/k₁)
(4) √(k₁/k₂)
Answer (4)
Sol. Maximum velocity V = Aω
v_P = v_Q
A_P ω_P = A_Q ω_Q
A_Q/A_P = ω_P/ω_Q
= √(k_P/m_P × m_Q/k_Q)
= √(k₁/k₂)

44. A pipe open at both ends has a fundamental frequency f in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:
(1) f/2
(2) f
(3) 3f/2
(4) 2f
Answer (2)
Sol. Fundamental frequency of open pipe (at both ends) f = v/(2L) ...(i)
Now immersed in water open pipe behaves as closed pipe.
f' = v/(4(L/2)) = v/(2L) ...(ii)
f = f'

45. The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 → n = 3 and n = 4 → n = 6 transitions, respectively, is
(1) 1/36
(2) 1/16
(3) 1/9
(4) 1/4
Answer (4)
Sol. ΔE = hc/λ = E_final - E_initial (E_n = -R_H/n²)
ΔE_2→3 = hc/λ_2→3 = E_3 - E_2 = -R_H/3² - (-R_H/2²)
∴ λ_2→3 = hc·36/(R_H·5)
ΔE_4→6 = E_6 - E_4 = -R_H/36 + R_H/16 = R_H × 20/(36 × 16)
hc/λ_4→6 = R_H × 20/(36 × 16)
λ_4→6 = hc × 36 × 16/(R_H·20)
λ_2→3/λ_4→6 = (hc·36/(R_H·5))/(hc × 36 × 16/(R_H·20)) = 20/(5 × 16) = 1/4

46. Which of the following statements are true?
A. Unlike Ga that has a very high melting point, Cs has a very low melting point.
B. On Pauling scale, the electronegativity values of N and Cl are not the same.
C. Ar, K⁺, Cl⁻, Ca²⁺ and S²⁻ are all isoelectronic species.
D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na.
E. The atomic radius of Cs is greater than that of Li and Rb.
Choose the correct answer from the options given below:
(1) A, B, and E only
(2) C and E only
(3) C and D only
(4) A, C, and E only
Answer (2)
Sol. Both Ga and Cs have low melting points.
Element Melting point/K
Ga 303 Cs 302
On Pauling scale, the electronegativity value of N and Cl have same (3.0).
Ar, K⁺, Cl⁻, Ca²⁺ and S²⁻ have 18 electrons. So these are isoelectronic species.
The correct order of first ionization enthalpy is Si > Mg > Al > Na
First ionisation enthalpy of Mg is higher than Al because the penetration of a 3s-electron to the nucleus is more than that of a 2p-electron.
Generally down the group atomic radii increases
Atom Atomic radius/pm
Li 152 Rb 244 Cs 262

47. Match List I with List II
List-I (Ion)
A. Co²⁺
B. Mg²⁺
C. Pb²⁺
D. Al³⁺
List-II (Group Number in Cation Analysis)
I. Group-I
II. Group-III
III. Group-IV
IV. Group-VI
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-II, D-I
(2) A-III, B-IV, C-I, D-II
(3) A-III, B-II, C-IV, D-I
(4) A-III, B-II, C-I, D-IV
Answer (2)
Sol. Ion / Group number in Cation Analysis
A. Co²⁺ / Group-IV
B. Mg²⁺ / Group-VI
C. Pb²⁺ / Group-I
D. Al³⁺ / Group-III

48. Predict the major product 'P' in the following sequence of reactions
(i) HBr, benzoyl peroxide
(ii) KCN
(iii) Na(Hg)/C₂H₅OH
(Major)
(1)
(2)
(3)
(4)
Answer (2)
Sol.

49. Energy and radius of first Bohr orbit of He⁺ and Li²⁺ are
[Given R_H = 2.18 × 10⁻¹⁸ J, a₀ = 52.9 pm]
(1) E_n(Li²⁺) = -19.62 × 10⁻¹⁸ J; r_n(Li²⁺) = 17.6 pm; E_n(He⁺) = -8.72 × 10⁻¹⁸ J; r_n(He⁺) = 26.4 pm
(2) E_n(Li²⁺) = -8.72 × 10⁻¹⁸ J; r_n(Li²⁺) = 26.4 pm; E_n(He⁺) = -19.62 × 10⁻¹⁸ J; r_n(He⁺) = 17.6 pm
(3) E_n(Li²⁺) = -19.62 × 10⁻¹⁸ J; r_n(Li²⁺) = 17.6 pm; E_n(He⁺) = -8.72 × 10⁻¹⁸ J; r_n(He⁺) = 26.4 pm
(4) E_n(Li²⁺) = -8.72 × 10⁻¹⁸ J; r_n(Li²⁺) = 17.6 pm; E_n(He⁺) = -19.62 × 10⁻¹⁸ J; r_n(He⁺) = 17.6 pm
Answer (1)
Sol. E_n = -2.18 × 10⁻¹⁸ × Z²/n² J; r_n = 52.9 × n²/Z pm
For He⁺
E_He⁺ = -2.18 × 10⁻¹⁸ × 4 = -8.72 × 10⁻¹⁸ J
r_He⁺ = 52.9 × 1/2 = 26.45 pm
For Li²⁺
E_Li²⁺ = -2.18 × 10⁻¹⁸ × 9 = 19.62 × 10⁻¹⁸ J
r_Li²⁺ = 52.9 × 1/3 = 17.63 pm

50. Which of the following are paramagnetic?
A. [NiCl₄]²⁻
B. Ni(CO)₄
C. [Ni(CN)₄]²⁻
D. [Ni(H₂O)₆]²⁺
E. Ni(PPh₃)₄
Choose the correct answer from the options given below:
(1) A and C only
(2) B and E only
(3) A and D only
(4) A, D and E only
Answer (3)
Sol.
A. [NiCl₄]²⁻; Ni²⁺; 3d⁸; sp³ hybridisation; 2 unpaired electrons; paramagnetic
B. Ni(CO)₄; Ni; 3d⁸4s²; sp³ hybridisation; Zero unpaired electron; diamagnetic
C. [Ni(CN)₄]²⁻; Ni²⁺; 3d⁸; dsp² hybridisation; Zero unpaired electron; diamagnetic
D. [Ni(H₂O)₆]²⁺; Ni²⁺; 3d⁸; sp³d² hybridisation; Two unpaired electron; paramagnetic
E. Ni(PPh₃)₄; Ni; 3d⁸4s²; sp³ hybridisation; zero unpaired electron; Diamagnetic

51. Given below are two statements:
Statement I: Like nitrogen that can form ammonia, arsenic can form arsine.
Statement II: Antimony cannot form antimony pentoxide.
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 correct
(2) Both Statement I and Statement II are incorrect
(3) Statement I is correct but Statement II is incorrect
(4) Statement I is incorrect but Statement II is correct
Answer (3)
Sol. All the elements of group 15 form hydrides of EH₃ type. Nitrogen forms ammonia (NH₃) while Arsenic forms Arsine (AsH₃)
All the elements of group 15 form two types of oxides: E₂O₃ and E₂O₅
Antimony forms antimony pentoxide Sb₂O₅
Hence, statement I is correct and statement II is incorrect

52. Which among the following electronic configurations belong to main group elements?
A. [Ne]3s¹
B. [Ar]3d⁴4s²
C. [Kr]4d¹⁰5s²5p⁵
D. [Ar]3d¹⁰4s¹
E. [Rn]5f⁶6d²7s²
Choose the correct answer from the option given below:
(1) B and E only
(2) A and C only
(3) D and E only
(4) A, C and D only
Answer (2)

53. Dalton's Atomic theory could not explain which of the following?
(1) Law of conservation of mass
(2) Law of constant proportion
(3) Law of multiple proportion
(4) Law of gaseous volume
Answer (4)
Sol. Dalton's theory could explain the laws of chemical combination. However, it could not explain the laws of gaseous volumes.

54. Consider the following compounds:
KO₂, H₂O₂ and H₂SO₄
The oxidation state of the underlined elements in them are, respectively,
(1) +1, -1, and +6
(2) +2, -2, and +6
(3) +1, -2, and +4
(4) +4, -4, and +6
Answer (1)
Sol. KO₂ → Alkali metal always shows +1 oxidation state. Therefore oxidation state of K is +1.
H₂O₂ → Oxidation state of oxygen in H₂O₂ is -1.
H₂SO₄ → Oxidation state of sulphur in H₂SO₄ is +6.

55. If the half-life (t₁/₂) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:
(1) 2 minutes
(2) 4 minutes
(3) 5 minutes
(4) 10 minutes
Answer (4)
Sol. For 1st order reaction
kt = 2.303 log(A₀/A_t) A₀ = initial concentration
A_t = Final concentration
t_99.9% = 10 t₁/₂
t_99.9% = 10 × 1 minute = 10 minutes

56. The correct order of the wavelength of light absorbed by the following complexes is,
A. [Co(NH₃)₆]³⁺
B. [Co(CN)₆]³⁻
C. [Cu(H₂O)₄]²⁺
D. [Ti(H₂O)₆]³⁺
Choose the correct answer from the options given below:
(1) B < D < A < C
(2) B < A < D < C
(3) C < D < A < B
(4) C < A < D < B
Answer (2)
Sol. λ ∝ 1/strength of ligand
λ ∝ 1/splitting
A. [Co(NH₃)₆]³⁺ : 475 nm
B. [Co(CN)₆]³⁻ : 310 nm
C. [Cu(H₂O)₄]²⁺ : 600 nm
D. [Ti(H₂O)₆]³⁺ : 498 nm
Order of λ = C > D > A > B

57. Which one of the following compounds can exist as cis-trans isomers?
(1) Pent-1-ene
(2) 2-Methylhex-2-ene
(3) 1,1-Dimethylcyclopropane
(4) 1,2-Dimethylcyclohexane
Answer (4)
Sol. Cis-trans isomers shown by:
Condition: Restricted rotation around double bond
Or
Different group around double bond
CH₃-CH₂-CH₂-CH=CH₂ no cis-trans
CH₃-CH₂-CH₂-CH=C(CH₃)-CH₃ no cis-trans

58. Phosphoric acid ionizes in three steps with their ionization constant values K_a1, K_a2 and K_a3 respectively, while K is the overall ionization constant. Which of the following statements are true?
A. log K = log K_a1 + log K_a2 + log K_a3
B. H₃PO₄ is a stronger acid than H₂PO₄⁻ and HPO₄²⁻
C. K_a1 > K_a2 > K_a3
D. K_a1 = (K_a2 + K_a3)/2
Choose the correct answer from the options given below:
(1) A and B only
(2) A and C only
(3) B, C and D only
(4) A, B and C only
Answer (4)
Sol. H₃PO₄ is a stronger acid than H₂PO₄⁻ and HPO₄²⁻
H₃PO₄(aq) ⇌ H⁺(aq) + H₂PO₄⁻(aq) K_a1 = 7.5 × 10⁻³
H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq) K_a2 = 6.2 × 10⁻⁸
HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq) K_a3 = 1.7 × 10⁻¹²
K_a1 > K_a2 > K_a3
log K = log K_a1 + log K_a2 + log K_a3
Ans. (A), (B) and (C) only

59. Which one of the following reactions does NOT give benzene as the product?
(1)
(2)
(3)
(4)
Answer (4)
Sol.
1. HC≡CH red hot Iron Tube at 873 K → Benzene
2. H₂O at 873 K → OH + N₂ + HCl Phenol
3. O Na sodalime → Benzene
4. MoO₃ n-hexane 773K, 10-20 atm. → Benzene

60. If the molar conductivity (Λ_m) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be
[Assume Λ_+° = 349.6 S cm² mol⁻¹ and Λ_-° = 50.4 S cm² mol⁻¹]
(1) 0.115
(2) 0.125
(3) 0.225
(4) 0.215
Answer (3)
Sol. Degree of dissociation (α) is given as
α = Λ_m/Λ_m°
Λ_m° = Λ_+° + Λ_-°
= 349.6 + 50.4
= 400 S cm² mol⁻¹
α = Λ_m/Λ_m° = 90/400 = 0.225

61. Given below are two statements:
Statement I: A hypothetical diatomic molecule with bond order zero is quite stable.
Statement II: As bond order increases, the bond length increases.
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 true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Answer (2)
Sol. A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule.
When bond order increases, the bond length decreases.

62. Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?
(1) [Co(NH₃)₃Cl₃]
(2) [Co(NH₃)₄Cl₂]
(3) [Co(NH₃)₃Cl₃]
(4) [Co(NH₃)₅Cl]Cl
Answer (1, 2)
Sol. Conductance of any complex depends on the following factor.
(1) Number of ions produced by complex.
(2) If number of ions are same then we will check charge on complex unit.
(1) [Co³⁺(NH₃)₃Cl₃] Both complex units have no charge. Therefore both complex units have same conductance.
(2) [Co³⁺(NH₃)₄Cl₂]
(3) [Co(NH₃)₆]Cl₃ → [Co(NH₃)₆]³⁺ + 3Cl⁻
(4) [Co(NH₃)₅Cl]Cl → [Co(NH₃)₅Cl]⁺ + Cl⁻

63. Match List-I with List-II
List-I
A. XeO₃
B. XeF₂
C. XeOF₄
D. XeF₆
List-II
I. sp³d; linear
II. sp³; pyramidal
III. sp³d³; distorted octahedral
IV. sp³d²; square pyramidal
Choose the correct answer from the options given below:
(1) A-II, B-I, C-IV, D-III
(2) A-II, B-I, C-III, D-IV
(3) A-IV, B-II, C-III, D-I
(4) A-IV, B-II, C-I, D-III
Answer (1)
Sol.
Molecule / Hybridisation / Shape
XeO₃ / sp³ / Pyramidal
XeF₂ / sp³d / Linear
XeOF₄ / sp³d² / Square Pyramidal
XeF₆ / sp³d³ / Distorted Octahedral

64. C(s) + 2H₂(g) → CH₄(g); ΔH = -74.8 kJ mol⁻¹
Which of the following diagrams gives an accurate representation of the above reaction? [R → reactants; P → products]
(1)
(2)
(3)
(4)
Answer (3)
Sol. ΔH = -74.8 kJ mol⁻¹, it is an exothermic reaction.
So, accurate representation is

65. Match List-I with List-II
List-I (Example)
A. Humidity
B. Alloys
C. Amalgams
D. Smoke
List-II (Type of Solution)
I. Solid in solid
II. Liquid in gas
III. Solid in gas
IV. Liquid in solid
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-I, D-III
(2) A-II, B-I, C-IV, D-III
(3) A-III, B-I, C-IV, D-II
(4) A-III, B-II, C-I, D-IV
Answer (2)
Sol.
Humidity is a solution of liquid in gas
Alloy is a solution of solid in solid
Amalgam is a solution of liquid in solid
Smoke is a solution of solid in gas

66. The correct order of decreasing basic strength of the given amines is:
(1) N-methylaniline > benzenamine > ethanamine > N-ethylethanamine
(2) N-ethylethanamine > ethanamine > benzenamine > N-methylaniline
(3) N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
(4) benzenamine > ethanamine > N-methylaniline > N-ethylethanamine
Answer (3)
Sol. Lower is the value of pK_b higher is the basicity Also aliphatic amines are stronger bases than aromatic amines.
pK_b: Benzenamine > N-Methylaniline > Ethanamine > N-Ethylethanamine
Basic strength: N-Ethylethanamine > Ethanamine > N-Methylaniline > Benzenamine

67. Among the following, choose the ones with equal number of atoms.
A. 212 g of Na₂CO₃(s) [molar mass = 106 g]
B. 248 g of Na₂O(s) [molar mass = 62 g]
C. 240 g of NaOH(s) [molar mass = 40 g]
D. 12 g of H₂(g) [molar mass = 2 g]
E. 220 g of CO₂(g) [molar mass = 44 g]
Choose the correct answer from the options given below:
(1) A, B, and C only
(2) A, B, and D only
(3) B, C, and D only
(4) B, D, and E only
Answer (2)
Sol. Number of atoms = given mass × atomicity × N_A / molar mass
A. 212/106 × 6 × N_A = 12 N_A
B. 248/62 × 3 × N_A = 12 N_A
C. 240/40 × 3 × N_A = 18 N_A
D. 12/2 × N_A × 2 = 12 N_A
E. 220/44 × N_A × 3 = 15 N_A
A, B and D have same number of atoms

68. Match List-I with List-II.
List-I (Name of Vitamin)
A. Vitamin B12
B. Vitamin D
C. Vitamin B2
D. Vitamin B6
List-II (Deficiency disease)
I. Cheilosis
II. Convulsions
III. Rickets
IV. Pernicious anaemia
Choose the correct answer from the options given below:
(1) A-I, B-III, C-II, D-IV
(2) A-IV, B-III, C-I, D-II
(3) A-II, B-III, C-I, D-IV
(4) A-IV, B-III, C-II, D-I
Answer (2)
Sol.
List-I (Name of Vitamin) / List-II (Deficiency disease)
A. Vitamin B12 / Pernicious anaemia
B. Vitamin D / Rickets
C. Vitamin B2 / Cheilosis
D. Vitamin B6 / Convulsions

69. The correct order of decreasing acidity of the following aliphatic acids is
(1) (CH₃)₃CCOOH > (CH₃)₂CHCOOH > CH₃COOH > HCOOH
(2) CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃COOH > HCOOH
(3) HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃COOH
(4) HCOOH > (CH₃)₃CCOOH > (CH₃)₂CHCOOH > CH₃COOH
Answer (3)
Sol. Electron donating group decreases the acidity of carboxylic acids.
So correct order is
HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃COOH

70. Given below are two statements:
Statement I: Ferromagnetism is considered as an extreme form of paramagnetism.
Statement II: The number of unpaired electrons in a Cr²⁺ ion (Z = 24) is the same as that of a Nd³⁺ ion (Z = 60)
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Answer (3)
Sol. Substances which are attracted very strongly in applied magnetic field are termed as ferromagnetic. In fact, ferromagnetism is an extreme form of paramagnetism. Hence statement I is correct.
Cr²⁺ = 3d⁴4s⁰, unpaired electrons = 4
Nd³⁺ = 4f³6s⁰, unpaired electrons = 3
Hence, Statement II is incorrect

71. Match List I with List II
List-I (Mixture)
A. CHCl₃ + C₆H₅NH₂
B. Crude oil in petroleum industry
C. Glycerol from spent-lye
D. Aniline-water
List-II (Method of separation)
I. Distillation under reduced pressure
II. Steam distillation
III. Fractional distillation
IV. Simple distillation
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-I, D-II
(2) A-IV, B-III, C-II, D-I
(3) A-III, B-IV, C-I, D-II
(4) A-III, B-IV, C-II, D-I
Answer (1)

72. For the reaction A(g) ⇌ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 L atm mol⁻¹ K⁻¹]
K_P for the reaction at 1000 K is
(1) 83.1
(2) 2.077 × 10⁵
(3) 0.033
(4) 0.021
Answer (3)
Sol. K_C = k_f/k_b = 1/2500
K_P = K_C(RT)^Δn_g (Δn_g = 2 - 1 = 1)
= 1/2500 × 0.0831 × 1000
= 0.033

73. Given below are two statements:
Statement-I: Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273-278 K. It decomposes easily in the dry state.
Statement-II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.
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 correct
(2) Both Statement I and Statement II are incorrect
(3) Statement I is correct but Statement II is incorrect
(4) Statement I is incorrect but Statement II is correct
Answer (1)
Sol. Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at 273-278 K. Nitrous acid is produced in the reaction mixture by reaction of NaNO₂ with HCl.
C₆H₅NH₂ + NaNO₂ + 2HCl → 273-278K → C₆H₅N₂Cl + NaCl + 2H₂O
Benzenediazonium chloride decomposes easily in the dry state.
Iodobenzene is prepared by shaking benzene diazonium salt with KI because direct insertion of iodine into benzene ring is difficult.

74. How many products (including stereoisomers) are expected from monochlorination of the following compound?
(1) 2
(2) 6
(3) 5
(4) 6
Answer (4)
Sol. Possible monochlorination products:
Total 6 isomers.

75. Among the given compounds I-III, the correct order of bond dissociation energy of C-H bond marked with * is:
(1) II > I > III
(2) II > III > I
(3) III > II > I
(4) I > II > III
Answer (1)
Sol.
I. carbon of this bond is sp² hybridised
II. carbon of this bond is sp hybridised
III. carbon of this bond is sp³ hybridised
Higher the percentage s character, stronger is C-H bond.
Correct order of bond dissociation energy of C-H bond: II > I > III

76. Which one of the following compounds does not decolorize bromine water?
(1) cyclohexane
(2) phenol
(3) styrene
(4) aniline
Answer (1)
Sol. Test for unsaturation i.e. Bromine water Reddish orange colour of bromine solution in CCl₄ will discharge when bromine adds to an unsaturation site.

77. The major product of the following reaction is
(1)
(2)
(3)
(4)
Answer (2)

78. Which of the following aqueous solution will exhibit highest boiling point?
(1) 0.01M Urea
(2) 0.01M KNO₃
(3) 0.01M Na₂SO₄
(4) 0.015M C₆H₁₂O₆
Answer (3)
Sol. ΔT_b = i K_b × m
ΔT_b ∝ i × m
By considering molarity same as molality
(1) 0.01M Urea: i × m = 1 × 0.01 = 0.01
(2) 0.01M KNO₃: i × m = 2 × 0.01 = 0.02
(3) 0.01M Na₂SO₄: i × m = 3 × 0.01 = 0.03
(4) 0.015M C₆H₁₂O₆: i × m = 1 × 0.015 = 0.015
T_b = T_b° + ΔT_b
Higher the value of (i × m) more will be the boiling point.

79. Match List-I with List-II.
List-I
A. Haber process
B. Wacker oxidation
C. Wilkinson catalyst
D. Ziegler catalyst
List-II
I. Fe catalyst
II. PdCl₂
III. [(PPh₃)₃RhCl]
IV. TiCl₄ with Al(CH₃)₃
Choose the correct answer from the options given below:
(1) A-I, B-II, C-IV, D-III
(2) A-II, B-III, C-I, D-IV
(3) A-I, B-II, C-III, D-IV
(4) A-I, B-IV, C-III, D-II
Answer (3)

80. 5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?
(1) The solution shows positive deviation.
(2) The solution shows negative deviation.
(3) The solution is ideal.
(4) The solution has volume greater than the sum of individual volumes.
Answer (2)
Sol. P_total = X_x P_x° + X_y P_y°
= 5/15 × 63 + 10/15 × 78
= 21 + 52
= 73 torr
Observed total pressure of solution is 70 torr.
It is less than calculated total pressure.
Hence, it shows negative deviation.

81. Sugar 'X'
A. is found in honey
B. is a keto sugar
C. exists in α and β-anomeric forms.
D. Is laevorotatory.
'X' is:
(1) D-Glucose
(2) D-Fructose
(3) Maltose
(4) Sucrose
Answer (2)

82. Identify the suitable reagent for the following conversion.
(1) (i) LiAlH₄, (ii) H⁺/H₂O
(2) (i) DIBAL-H, (ii) H₂O
(3) (i) NaBH₄, (ii) H⁺/H₂O
(4) (i) AlH(Bu)₂, (ii) H₂O
Answer (2)
Sol. Esters are reduced to aldehydes with DIBAL-H

83. Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): I undergoes SN2 reaction faster than Cl.
Reason (R): Iodine is a better leaving group because of its large size.
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 the correct explanation of A
(2) Both A and R are true but R is not the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Answer (1)
Sol. Rate of SN2 reaction of I is faster than Cl.
Because iodine is a good leaving group due to large size of iodine. Which stabilises the I⁻ ion.

84. The standard heat of formation, in kcal/mol of Ba²⁺ is:
[Given: standard heat of formation of SO₄²⁻ ion (aq) = -216 kcal/mol, standard heat of crystallisation of BaSO₄(s) = -4.5 kcal/mol, standard heat of formation of BaSO₄(s) = -349 kcal/mol]
(1) -128.5
(2) -133.0
(3) +133.0
(4) +220.5
Answer (1)
Sol. S + 2O₂ + 2e⁻ → SO₄²⁻ ΔH_f = -216 kcal/mol ...(1)
Ba²⁺(g) + SO₄²⁻(g) → BaSO₄(s) ΔH_crystallisation = -4.5 kcal/mol ...(2)
Ba + S + 2O₂ → BaSO₄(s) ΔH_f(BaSO₄) = -349 kcal/mol ...(3)
Ba(s) → Ba²⁺(g) + 2e⁻ ...(4)
From equation (1), (2) and (3) we get equation (4).
Applying equation (3) - (1) - (2)
So, -349 - (-4.5) - (-216)
So - 349 + 4.5 + 216
= -349 + 220.5
= -128.5 kcal/mol

85. Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C₄H₈O is:
(1) 6
(2) 8
(3) 10
(4) 11
Answer (3)
Sol. For cyclic ethers O should be in ring; * carbon here is chiral
(1) 2(d, l pair)
(2) 1
(3) 1
(4) CH₃ CH₃ Meso compound
(5) 2(d, l pair)
(6) 1
(7) 2(d, l pair)
Total number of isomers = 2 + 1 + 1 + 1 + 2 + 1 + 2 = 10

86. Identify the correct orders against the property mentioned
A. H₂O > NH₃ > CHCl₃ - dipole moment
B. XeF₄ > XeO₃ > XeF₂ - number of lone pairs on central atom
C. O-H > C-H > N-O - bond length
D. N₂ > O₂ > H₂ - bond enthalpy
Choose the correct answer from the options given below:
(1) A, D only
(2) B, D only
(3) A, C only
(4) B, C only
Answer (1)
Sol. μ(D)
A. H₂O 1.85, NH₃ 1.47, CHCl₃ 1.04
B. XeF₄: 2 lone pairs of electron, XeO₃: 1 lone pair of electron, XeF₂: 3 lone pairs of electron
C. Order of Bond length: N-O > C-H > O-H
D. N₂ Bond order is 3, H₂ Bond order is 1, O₂ Bond order is 2

87. Higher yield of NO in N₂(g) + O₂(g) ⇌ 2NO(g) can be obtained at
[ΔH of the reaction = +180.7 kJ mol⁻¹]
A. Higher temperature
B. Lower temperature
C. Higher concentration of N₂
D. Higher concentration of O₂
Choose the correct answer from the options given below:
(1) A, D only
(2) B, C only
(3) B, C, D only
(4) A, C, D only
Answer (4)
Sol. Yield of the product generally depends on
Temperature Concentration of reactant(s) and product(s) Pressure
As this is an endothermic reaction (ΔH = +180.7 kJ mol⁻¹), so, increase in temperature will shift equilibrium in forward direction to increase yield of NO.
Increase in concentration of reactants (N₂ and O₂) also shifts the equilibrium in forward direction and increase the yield of NO.
Hence, (A), (C) and (D) only will increase yield of NO.

88. If the rate constant of a reaction is 0.03 s⁻¹, how much time does it take for 7.2 mol L⁻¹ concentration of the reactant to get reduced to 0.9 mol L⁻¹?
(Given: log 2 = 0.301)
(1) 69.3 s
(2) 23.1 s
(3) 210 s
(4) 21.0 s
Answer (1)
Sol. k = 0.03 s⁻¹
t = 2.303/k log(a/(a - x))
= 2.303/0.03 log(7.2/0.9)
= 2.303/0.03 log 8
= 2.303/0.03 × 3 × log 2
= 2.303/0.03 × 3 × 0.301
= 69.3 s

89. Which one of the following reactions does NOT belong to "Lassaigne's test"?
(1) Na + C + N → NaCN
(2) 2Na + S → Na₂S
(3) Na + X → NaX
(4) 2CuO + C → 2Cu + CO₂
Answer (4)
Sol. Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by "Lassaigne's test".
Na + C + N → NaCN
2Na + S → Na₂S
Na + X → NaX (X = Cl, Br, I)

90. The complex II of mitochondrial electron transport chain is also known as
(1) Cytochrome bc1
(2) Succinate dehydrogenase
(3) Cytochrome c oxidase
(4) NADH dehydrogenase
Answer (2)
Sol. Complex II of mitochondrial electron transport chain is also known as succinate dehydrogenase. Cytochrome c oxidase (complex IV), NADH dehydrogenase (complex I), cytochrome bc1 (complex II).

91. Polymerase chain reaction (PCR) amplifies DNA following the equation.
(1) N²
(2) 2ⁿ
(3) 2n + 1
(4) 2N²
Answer (2)
Sol. PCR i.e., polymerase chain reaction amplifies DNA as per the equation 2ⁿ, where 'n' refers to number of cycles. Thus, say, if 3 PCR cycles will run, then 2³ i.e., 2 × 2 × 2 ⇒ 8 DNA fragments will be formed.

92. What are the potential drawbacks in adoption of the IVF method?
A. High fatality risk to mother
B. Expensive instruments and reagents
C. Husband/wife necessary for being donors
D. Less adoption of orphans
E. Not available in India
F. Possibility that the early embryo does not survive
Choose the correct answer from the options given below:
(1) B, D, F only
(2) A, C, D, F only
(3) A, B, C, D only
(4) A, B, C, E, F only
Answer (1)
Sol. Statements B, D and F are correct while statements A, C and E are incorrect. Husband/wife is not necessary for being donors. IVF is available in India.

93. What is the name of the blood vessel that carries deoxygenated blood from the body to the heart in a frog?
(1) Aorta
(2) Pulmonary artery
(3) Pulmonary vein
(4) Vena cava
Answer (4)
Sol. Frog's heart is a muscular structure with three chambers. It receives deoxygenated blood from body parts through the major veins called vena cava. Vena cava carries deoxygenated blood. Aorta and pulmonary vein carries oxygenated blood. Whereas, pulmonary artery will carry deoxygenated blood towards the lungs.

94. Which one of the following statements refers to Reductionist Biology?
(1) Physico-chemical approach to study and understand living organisms
(2) Physiological approach to study and understand living organisms
(3) Chemical approach to study and understand living organisms
(4) Behavioural approach to study and understand living organisms
Answer (1)
Sol. The physico-chemical approach to study and understand living organisms is called 'Reductionist Biology'.

95. Given below are two statements:
Statement I: In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable.
Statement II: DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism.
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 correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct but statement II is incorrect
(4) Statement I is incorrect but statement II is correct
Answer (1)
Sol. In RNA world, RNA was the first genetic material as there are enough evidences to suggest that essential life processes (such as metabolism, translation, splicing, etc) evolved around RNA. RNA used to act as a genetic material as well as catalyst (there are some important biochemical reaction in living systems that are catalyzed by RNA catalysts not by protein enzymes) so, statement I is correct statement II is also correct as DNA being double stranded and having complementary strands further resists changes by evolving a process of repair.

96. Epiphytes that are growing on a mango branch is an example of which of the following?
(1) Commensalism
(2) Mutualism
(3) Predation
(4) Amensalism
Answer (1)
Sol. Commensalism is the type of interaction in which one-species benefits and another is neither harmed nor benefited. An orchid growing as an epiphyte on a mango branch is an example of commensalism.

97. From the statements given below choose the correct option:
A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S.
B. Each ribosome has two sub-units.
C. The two sub-units of 80S ribosome are 60S and 40S while that of 70S are 50S and 30S.
D. The two sub-units of 80S ribosome are 60S and 20S and that of 70S are 50S and 20S.
E. The two sub-units of 80S are 60S and 30S and that of 70S are 50S and 30S.
(1) A, B, C are true
(2) A, B, D are true
(3) A, B, E are true
(4) B, D, E are true
Answer (1)
Sol. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S type. Each ribosome has two sub-units. The two sub-units of 80S ribosome are 60S and 40S while that of 70S are 50S and 30S.

98. Which one of the following is an example of ex-situ conservation?
(1) National Park
(2) Wildlife Sanctuary
(3) Zoos and botanical gardens
(4) Protected areas
Answer (3)
Sol. Zoological parks (Zoos), botanical gardens and wildlife safari parks are examples of ex-situ conservation. Sacred groves, biosphere reserves, national parks and wildlife sanctuaries are examples of in-situ conservation.

99. Given below are two statements:
Statement I: The primary source of energy in an ecosystem is solar energy.
Statement II: The rate of production of organic matter during photosynthesis in an ecosystem is called net primary productivity (NPP).
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 correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct but statement II is incorrect
(4) Statement I is incorrect but statement II is correct
Answer (3)
Sol. Primary source of energy in the ecosystem is solar energy.
Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. Hence, statement I is correct but statement II is incorrect.

100. Match List-I with List-II.
List-I
A. Emphysema
B. Angina Pectoris
C. Glomerulonephritis
D. Tetany
List-II
I. Rapid spasms in muscle due to low Ca in body fluid
II. Damaged alveolar walls and decreased respiratory surface
III. Acute chest pain when net enough oxygen is reaching to heart muscle
IV. Inflammation of glomeruli of kidney
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-III, B-I, C-II, D-IV
(3) A-II, B-IV, C-III, D-I
(4) A-II, B-III, C-IV, D-I
Answer (4)
Sol. Emphysema - Damaged alveolar walls and decreased respiratory surface
Angina pectoris - Acute chest pain when net enough oxygen is reaching to heart muscle
Glomerulonephritis - Inflammation of glomeruli of kidney
Tetany - Rapid spasms in muscle due to low Ca in body fluid

101. Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Both wind and water pollinated flowers are not very colourful and do not produce nectar.
Reason (R): The flowers produce enormous amount of pollen grains in wind and water pollinated flowers.
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 the correct explanation of A
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Answer (2)
Sol. Both wind and water pollinated flowers are not very colourful and do not produce nectar, this is because they rely on wind and water to carry their pollen. Wind and water pollinated flower do not need to attract insect, so they did not evolve to produce bright coloured flower.

102. Which of the following is an example of non-distilled alcoholic beverage produced by yeast?
(1) Whisky
(2) Brandy
(3) Beer
(4) Rum
Answer (3)
Sol. Wine and beer are produced without distillation whereas whisky, brandy and rum are produced by distillation of fermented broth.

103. Given below are two statements:
Statement I: In a floral formula ⊕ stands for zygomorphic nature of the flower, and G stands for inferior ovary.
Statement II: In a floral formula ⊕ stands for actinomorphic nature of the flower and G stands for superior ovary.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are correct
(2) Both Statement I and Statement II are incorrect
(3) Statement I is correct but Statement II is incorrect
(4) Statement I is incorrect but Statement II is correct
Answer (4)
Sol. The floral formula symbol ⊕ is used for actinomorphic flower, while % is used for zygomorphic flower. The symbol G represents gynoecium and G symbol represent superior ovary, while inferior ovary is represented by G. Thus, statement I is incorrect and Statement II is correct.

104. Streptokinase produced by bacterium Streptococcus is used for
(1) Curd production
(2) Ethanol production
(3) Liver disease treatment
(4) Removing clots from blood vessels
Answer (4)
Sol. Streptokinase produced by the bacterium Streptococcus and modified by genetic engineering is used as a 'clot buster' for removing clots from blood vessels of patients who have undergone myocardial infarction leading to heart attack. Curd production is done by Lactobacillus and ethanol production is done by Saccharomyces.

105. Which chromosome in the human genome has the highest number of genes?
(1) Chromosome X
(2) Chromosome Y
(3) Chromosome 1
(4) Chromosome 10
Answer (3)
Sol. In human genome, Chromosome 1 has the highest number of genes, i.e., 2968.

106. Which of the following statement is correct about location of the male frog copulatory pad?
(1) First and Second digit of fore limb
(2) First digit of hind limb
(3) Second digit of fore limb
(4) First digit of the fore limb
Answer (4)
Sol. In male frogs, copulatory pad is present on the first digit of the forelimbs which are absent in female frogs.

107. Which one of the following phytohormones promotes nutrient mobilization which helps in the delay of leaf senescence in plants?
(1) Ethylene
(2) Abscisic acid
(3) Gibberellin
(4) Cytokinin
Answer (4)
Sol. Cytokinins help to overcome apical dominance. They promote nutrient mobilisation which helps in the delay of leaf senescence.

108. While trying to find out the characteristic of a newly found animal, a researcher did the histology of adult animal and observed a cavity with presence of mesodermal tissue towards the body wall but no mesodermal tissue was observed towards the alimentary canal. What could be the possible coelome of that animal?
(1) Acoelomate
(2) Pseudocoelomate
(3) Schizocoelomate
(4) Spongocoelomate
Answer (2)
Sol. In pseudocoelomates, the body cavity is not entirely lined with mesoderm, instead, mesodermal tissue is present along the body wall but not towards the gut.
Schizocoelomates are animals whose coelom or body cavity develops middle from a split in the mesoderm, the middle germ layer of the embryo. In acoelomates, coelom is absent. Spongocoel is a central cavity found in Sponges.

109. Match List - I with List - II.
List - I
A. Head
B. Middle piece
C. Acrosome
D. Tail
List - II
I. Enzymes
II. Sperm motility
III. Energy
IV. Genetic material
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-I, D-II
(2) A-IV, B-III, C-II, D-I
(3) A-III, B-IV, C-II, D-I
(4) A-III, B-II, C-I, D-IV
Answer (1)
Sol. The sperm head contains elongated nucleus which possesses the genetic material. The middle piece possesses numerous mitochondria, which produce energy for movement. Acrosome is a cap-like structure filled with enzymes that help in fertilization of ovum. The tail of sperm facilitates sperm motility essential for fertilisation.

110. Given below are the stages in the life cycle of pteridophytes. Arrange the following stages in the correct sequence.
A. Prothallus stage
B. Meiosis in spore mother cells
C. Fertilisation
D. Formation of archegonia and antheridia in gametophyte.
E. Transfer of anthozooids to the archegonia in presence of water.
Choose the correct answer from the options given below:
(1) B, A, D, E, C
(2) B, A, E, C, D
(3) D, E, C, A, B
(4) E, D, C, B, A
Answer (1)
Sol. In a pteridophytes life cycle, the correct sequence of stages will be given as follows:
B → Meiosis in spore mother cells
A → Prothallus stage
D → Formation of archegonia and antheridia in gametophyte
E → Transfer of anthozooids to the archegonia in presence of water
C → Fertilisation will occur
So, the correct sequence is B → A → D → E → C

111. Cardiac activities of the heart are regulated by:
A. Nodal tissue
B. A special neural centre in the medulla oblongata
C. Adrenal medullary hormones
D. Adrenal cortical hormones
Choose the correct answer from the options given below:
(1) A, B and C Only
(2) A, B, C and D
(3) A, C and D Only
(4) A, B and D Only
Answer (1)
Sol. Normal cardiac activities of the heart are regulated intrinsically, i.e., auto regulated by specialised muscles (nodal tissue), hence the heart is called myogenic. A special neural centre in the medulla oblongata can moderate the cardiac function through autonomic nervous system. Sympathetic nervous system can increase the rate of heartbeat, ventricular contraction and thereby cardiac output. Parasympathetic neural signals decrease the rate of heartbeat, speed of conduction of action potential and thereby the cardiac output. Adrenal medullary hormones can also increase the cardiac output.

112. Which of following organisms cannot fix nitrogen?
A. Azotobacter
B. Oscillatoria
C. Anabaena
D. Volvox
E. Nostoc
Choose the correct answer from the options given below:
(1) A only
(2) D only
(3) B only
(4) E only
Answer (2)
Sol. Azotobacter, Oscillatoria, Anabaena and Nostoc can fix nitrogen but Volvox cannot fix nitrogen.

113. Given below are two statements:
Statement I: Transfer RNAs and ribosomal RNA do not interact with mRNA.
Statement II: RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.
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 correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct but statement II is incorrect
(4) Statement I is incorrect but statement II is correct
Answer (4)
Sol. Both transfer RNAs and ribosomal RNA interact with mRNA.
RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.

114. In the above represented plasmid, an alien piece of DNA is inserted at EcoRI site. Which of the following strategies will be chosen to select the recombinant colonies?
(1) Using ampicillin & tetracycline containing medium plate.
(2) Blue color colonies will be selected.
(3) White color colonies will be selected.
(4) Blue color colonies grown on ampicillin plates can be selected.
Answer (3)
Sol. The correct answer is that white-colored colonies will be selected.
Since an alien piece of DNA is being inserted at EcoRI site, the gene β-galactosidase present here will undergo insertional inactivation.
This gene is responsible for producing blue-colored colonies, but since it has been insertionaly inactivated, white colored colonies will be produced.
Ampicillin and tetracycline resistance genes present in the given DNA will remain intact. Thus, the given DNA will show ampR and tetR.

115. Which of the following genetically engineered organisms was used by Eli Lilly to prepare human insulin?
(1) Bacterium
(2) Yeast
(3) Virus
(4) Phage
Answer (1)
Sol. The correct answer is bacterium.
In 1983, Eli Lilly, an American company, prepared two DNA sequences corresponding to 'A' and 'B' chains of human insulin and introduced them in plasmids of E.coli (a gram negative bacterium) to produce insulin chains.

116. Name the class of enzyme that usually catalyze the following reaction:
S - G + S' → S + S' - G
Where, G → a group other than hydrogen
S → a substrate
S' → another substrate
(1) Hydrolase
(2) Lyase
(3) Transferase
(4) Ligase
Answer (3)
Sol. Enzymes catalysing a transfer of G group, (other than hydrogen) between a pair of substrates, S and S' are known as transferases.
S - G + S' → S + S' - G
- Ligases catalyse the linking together of 2 compounds such as C-O, C-S, C-N bonds etc
- Lyases catalyse removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds
Hydrolases are enzymes that catalyse hydrolysis of ester, ether, peptide, glycosidic, C-C, C-halide or P-N bonds.

117. Find the statement that is NOT correct with regard to the structure of monocot stem.
(1) Hypodermis is parenchymatous.
(2) Vascular bundles are scattered.
(3) Vascular bundles are conjoint and closed.
(4) Phloem parenchyma is absent.
Answer (1)
Sol. In monocot stem, hypodermis is sclerenchymatous.

118. The correct sequence of events in the life cycle of bryophytes is
A. Fusion of anterozoid with egg.
B. Attachment of gametophyte to substratum.
C. Reduction division to produce haploid spores.
D. Formation of sporophyte.
E. Release of anterozoids into water.
Choose the correct answer from the options given below:
(1) D, E, A, C, B
(2) B, E, A, C, D
(3) B, E, A, D, C
(4) D, E, A, B, C
Answer (3)
Sol. The correct sequence of events in the life cycle of bryophytes is
Attachment of gametophyte to substratum.
Release of anterozoids into water.
Fusion of anterozoid with egg.
Formation of sporophyte.
Reduction division to produce haploid spores.

119. Which are correct:
A. Computed tomography and magnetic resonance imaging detect cancers of internal organs.
B. Chemotherapeutic drugs are used to kill non-cancerous cells.
C. α-interferon activate the cancer patients' immune system and helps in destroying the tumour.
D. Chemotherapeutic drugs are biological response modifiers.
E. In the case of leukaemia blood cell counts are decreased.
Choose the correct answer from the options given below:
(1) B and D only
(2) D and E only
(3) C and D only
(4) A and C only
Answer (4)
Sol. Statements A and C are correct while statements B, D and E are incorrect. Chemotherapeutic drugs are used to kill cancerous cells. In case of leukaemia, blood cell counts are increased. α-interferons are biological response modifiers.

120. Match List-1 with List-II.
List-1
A. Centromere
B. Cilium
C. Cristae
D. Cell membrane
List-II
I. Mitochondrion
II. Cell division
III. Cell movement
IV. Phospholipid Bilayer
Choose the correct answer from the options given below:
(1) A-I, B-II, C-III, D-IV
(2) A-II, B-I, C-IV, D-III
(3) A-IV, B-II, C-III, D-I
(4) A-II, B-III, C-I, D-IV
Answer (4)
Sol. Centromere - Helps in cell division
Cilium - Helps in cell movement
Cristae - Finger like structures of mitochondria
Cell membrane - Is a phospholipid bilayer

121. Match List I with List II:
List-I
A. Chlorophyll a
B. Chlorophyll b
C. Xanthophylls
D. Carotenoids
List-II
I. Yellow-green
II. Yellow
III. Blue-green
IV. Yellow to Yellow-orange
Choose the option with all correct matches.
(1) A-III, B-IV, C-II, D-I
(2) A-III, B-I, C-II, D-IV
(3) A-I, B-II, C-IV, D-III
(4) A-I, B-IV, C-III, D-II
Answer (2)
Sol. A chromatographic separation of the leaf pigments shows that the colour that we see in leaves is not due to single pigment but due to four pigments.
Chlorophyll a - Bright or blue-green in the chromatogram
Chlorophyll b - Yellow-green
Xanthophylls - Yellow
Carotenoids - Yellow to Yellow-orange

122. Find the correct statement:
(A) In human pregnancy, the major organ systems are formed at the end of 12 weeks.
(B) In human pregnancy the major organ systems are formed at the end of 8 weeks.
(C) In human pregnancy heart is formed after one month of gestation.
(D) In human pregnancy, limbs and digits develop by the end of second month.
(E) In human pregnancy the appearance of hair is usually observed in the fifth month.
Choose the correct answer from the options given below:
(1) A and E only
(2) B and C only
(3) B, C, D and E only
(4) A, C, D and E only
Answer (4)
Sol. In a human female's pregnancy.
By the end of 12 weeks (1st trimester), most of major organ systems are formed (not by end of 8 weeks). After one month of pregnancy, the embryo's heart is formed. By the end of second month of pregnancy, the foetus develops limbs and digits. The first movements of foetus and appearance of hair on head are usually observed during the fifth month.

123. In the seeds of cereals, the outer covering of endosperm separates the embryo by a protein-rich layer called:
(1) Coleoptile
(2) Coleorhiza
(3) Integument
(4) Aleurone layer
Answer (4)
Sol. In monocot seeds, the outer covering of endosperm separates the embryo by a proteinous layer called aleurone layer.

124. Which of the following diagrams is correct with regard to the proximal (P) and distal (D) tubule of the Nephron.
(1)
(2)
(3)
(4)
Answer (2)
Sol. During urine formation, the tubular cells secrete substances like H⁺, K⁺ and ammonia into the filtrate. Tubular secretion is also an important step in urine formation as it helps in the maintenance of ionic and acid base balance of body fluids.
PCT → Selective secretion of H⁺, ammonia and K⁺ into the filtrate.
DCT → Capable of reabsorption of HCO₃⁻ and selective secretion of H⁺, K⁺ and NH₃.

125. Identify the part of a bio-reactor which is used as a foam breaker from the given figure.
(1) A
(2) B
(3) D
(4) C
Answer (4)
Sol. B - Motor
C - Foam breaker
A - Flat bladed impeller
D - Sterile air
Part labelled as C is foam breaker.

126. Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): A typical unfertilised, angiosperm embryo sac at maturity is 8 nucleate and 7-celled.
Reason (R): The egg apparatus has 2 polar nuclei.
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 the correct explanation of A
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Answer (3)
Sol. A typical Angiosperm embryo sac, at maturity is 7-celled and 8 nucleate. Polar nuclei are situated below the egg apparatus in the large central cell. Three cells are grouped together at micropylar end and constitute the egg apparatus. Hence, A is true but R is false.

127. A specialised membranous structure in a prokaryotic cell which helps in cell wall formation, DNA replication and respiration is
(1) Mesosome
(2) Chromatophores
(3) Cristae
(4) Endoplasmic Reticulum
Answer (1)
Sol. Mesosome is membranous extension in bacterial cell that helps in cell wall formation, DNA replication and contains enzymes for respiration.

128. Which of the following are the post-transcriptional events in an eukaryotic cell?
A. Transport of pre-mRNA to cytoplasm prior to splicing.
B. Removal of introns and joining of exons.
C. Addition of methyl group at 5' end of hnRNA.
D. Addition of adenine residues at 3' end of hnRNA.
E. Base pairing of two complementary RNAs.
Choose the correct answer from the options given below:
(1) A, B, C only
(2) B, C, D only
(3) B, C, E only
(4) C, D, E only
Answer (2)
Sol. The process of copying genetic information from one strand of the DNA into RNA is known as transcription. It occurs in the cytoplasm with the help of transcriptin enzyme.
Transport of pre-mRNA to cytoplasm prior to splicing is a part of transcription.
The primary transcript is converted into functional mRNA after post transcriptional processing involves 3 steps as follows-
Modification of 5' end by capping, Tailing, Splicing.
Base pairing of two complementary RNA is not on event of post-transcription.
Hence, statements B, C, D are post-transcriptional modification events in eukaryotic cell.

129. What is the pattern of inheritance for polygenic trait?
(1) Mendelian inheritance pattern
(2) Non-mendelian inheritance pattern
(3) Autosomal dominant pattern
(4) X-linked recessive inheritance pattern
Answer (2)
Sol. Polygenic inheritance refers to the inheritance of a trait controlled by two or more genes. When human disorders are determined by mutation in the single gene then they are transmitted to the offspring as per Mendelian principle. Polygenic trait shows non-Mendelian inheritance pattern.

130. Which one of the following enzymes contains 'Haem' as the prosthetic group?
(1) RuBisCo
(2) Carbonic anhydrase
(3) Succinate dehydrogenase
(4) Catalase
Answer (4)
Sol. In peroxidase and catalase, which catalyze the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and it is part of the active site of the enzymes. Zinc is the cofactor in enzyme carbonic anhydrase.
Zinc is the cofactor in enzyme carbonic anhydrase.
RuBisCo is the most abundant protein in whole of the biosphere.
Succinate is the substrate of enzyme succinic dehydrogenase.

131. Each of the following characteristics represent a Kingdom proposed by Whittaker. Arrange the following in increasing order of complexity of body organization.
A. Multicellular heterotrophs with cell wall made of chitin.
B. Heterotrophs with tissue/organ/organ system level of body organization.
C. Prokaryotes with cell wall made of polysaccharides and amino acids.
D. Eukaryotic autotrophs with tissue/organ level of body organization.
E. Eukaryotes with cellular body organization.
Choose the correct answer from the options given below:
(1) A, C, E, B, D
(2) C, E, A, D, B
(3) A, C, E, D, B
(4) C, E, A, B, D
Answer (2)

132. Who is known as the father of Ecology in India?
Answer: Ramdeo Misra is known as the father of Ecology in India.

133. Match List I with List II:
List-I
A. Alfred Hershey and Martha Chase
B. Euchromatin
C. Frederick Griffith
D. Heterochromatin
List-II
I. Streptococcus pneumoniae
II. Densely packed and dark-stained
III. Loosely packed and light-stained
IV. DNA as genetic material confirmation
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-I, D-III
(2) A-IV, B-II, C-I, D-III
(3) A-IV, B-III, C-I, D-III
(4) A-III, B-II, C-IV, D-I
Answer (3)
Sol. The unequivocal proof that DNA is the genetic material came from the experiment of Alfred Hershey and Martha Chase.
Euchromatin are lightly stained region with loosely packed chromatin fibre.
Frederick Griffith performed series of experiments by selecting the different strains of Streptococcus pneumoniae.
Heterochromatin are darkly stained region with tightly packed chromatin fibre.

134. Neoplastic characteristics of cells refer to:
A. A mass of proliferating cell
B. Rapid growth of cells
C. Invasion and damage to the surrounding tissue
D. Those confined to original location
Choose the correct answer from the options given below:
(1) A, B only
(2) A, B, C only
(3) A, B, D only
(4) B, C, D only
Answer (2)
Sol. The correct answer will include: A, B and C only.
A neoplasm is a general term for any abnormal growth of tissue.
Neoplastic characteristics of cells refer to
(1) A mass of proliferating cell.
(2) Rapid growth of cells.
(3) Invasion and damage to the surrounding tissue.
Cancer specifically refers to malignant neoplasms, which are cancerous and invasive.
Benign tumours remain confined to their original location. Thus, D is not included in the answer.
The malignant tumours, on the other hand are a mass of proliferating cells called neoplastic or tumour cells. These cells grow very rapidly, invading and damaging the surrounding normal tissues.

135. Given below are two statements:
Statement I: The DNA fragments extracted from gel electrophoresis can be used in construction of recombinant DNA.
Statement II: Smaller size DNA fragments are observed near anode while larger fragments are found near the wells in an agarose gel.
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 correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct but statement II is incorrect
(4) Statement I is incorrect but statement II is correct
Answer (1)
Sol. The cutting of DNA by restriction endonucleases results in the fragments of DNA. These fragments can be separated by a technique known as gel electrophoresis.
The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. This step is known as elution. The DNA fragments purified in this way are used in constructing rDNA by joining them with cloning vectors.
In gel electrophoresis, the DNA fragments separate (resolve) according to their size through sieving effect provided by the agarose gel. Hence, the smaller the fragment size, the farther it moves from cathode towards anode.

136. Match List I with List II:
List-I
A. Adenosine
B. Adenylic acid
C. Adenine
D. Alanine
List-II
I. Nitrogen base
II. Nucleotide
III. Nucleoside
IV. Amino acid
Choose the option with all correct matches.
(1) A-II, B-IV, C-II, D-I
(2) A-II, B-II, C-IV, D-I
(3) A-III, B-II, C-I, D-IV
(4) A-III, B-II, C-I, D-IV
Answer (3)
Sol. The correct answer is A-III, B-II, C-I, D-IV
Adenosine- It is a nucleoside which is composed of nitrogen base and sugar only.
Adenylic acid- It is a nucleotide which is composed of nitrogen base, sugar and a phosphate group is esterified to the sugar.
Adenine- Nitrogen base (Purine)
Alanine- An amino acid that contains a methyl group as the 'R' group.

137. Consider the following:
A. The reductive division for the human female gametogenesis starts earlier than that of the male gametogenesis.
B. The gap between the first meiotic division and the second meiotic division is much shorter for males compared to females.
C. The first polar body is associated with the formation of the primary oocyte.
D. Luteinizing Hormone (LH) surge leads to disintegration of the endometrium and onset of menstrual bleeding.
Choose the correct answer from the options given below:
(1) A and B are true
(2) A and C are true
(3) B and D are true
(4) B and C are true
Answer (1)
Sol. Statements A and B are true while statements C and D are false.
The first polar body is associated with the formation of the secondary oocyte LH surge leads to ovulation. Decreased levels of progesterone during late luteal phase leads to degeneration of the endometrium and onset of menstrual bleeding.

138. All living members of the class Cyclostomata are:
(1) Free living
(2) Endoparasite
(3) Symbiotic
(4) Ectoparasite
Answer (4)
Sol. All living members of class Cyclostomata are ectoparasites.

139. Given below are two statements: one is labelled as Assertion (A), and the other is labelled as Reason (R).
Assertion (A): The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver it to intracellular targets and outside the cell.
Reason (R): Vesicles containing materials made by the endoplasmic reticulum fuse with the cis face of the Golgi apparatus, and they are modified and released from the trans face of the Golgi apparatus.
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 the correct explanation of A
(2) Both A and R are true but R is not the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Answer (2)
Sol. The primary function of Golgi apparatus is to package the materials made by endoplasmic reticulum and deliver it to intracellular targets and outside the cell, this statement is correct and the reason statement is also correct. Golgi apparatus remains in close association with endoplasmic reticulum. Here, assertion and reason statements both are correct but reason is not correctly explaining assertion.

140. Match List I with List II:
List I
A. Scutellum
B. Non-albuminous seed
C. Epiblast
D. Perisperm
List II
I. Persistent nucleolus
II. Cotyledon of Monocot seed
III. Groundnut
IV. Rudimentary cotyledon
Choose the option with all correct matches.
(1) A-II, B-III, C-IV, D-I
(2) A-IV, B-III, C-II, D-I
(3) A-IV, B-III, C-I, D-II
(4) A-II, B-IV, C-III, D-I
Answer (1)
Sol. Scutellum is cotyledon of monocot seed. Groundnut seed is non-albuminous seed. Epiblast is rudimentary cotyledon in monocot seed. Perisperm is persistent nucleolus.

141. Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): All vertebrates are chordates but all chordates are not vertebrate.
Reason (R): The members of subphylum vertebrate possess notochord during the embryonic period, the notochord is replaced by cartilaginous or bony vertebral column in adults.
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 the correct explanation of (A)
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) (A) is true but (R) is false
(4) (A) is false but (R) is true
Answer (1)
Sol. Both (A) and (R) are true and (R) is the correct explanation of (A).
The members of subphylum vertebrate possess notochord during the embryonic period. The notochord is replaced by a cartilaginous or bony vertebral column in the adult. Thus, all vertebrates are chordates but all chordates are not vertebrates.

142. Identify the statement that is NOT correct.
(1) Each antibody has two light and two heavy chains.
(2) The heavy and light chains are held together by disulfide bonds.
(3) Antigen binding site is located at C-terminal region of antibody molecules.
(4) Constant region of heavy and light chains are located at C-terminus of antibody molecules
Answer (3)
Sol. Each antibody molecule has four peptide chains, two small called light chains and two longer called heavy chains. Hence, an antibody is represented as H₂L₂. In an antibody molecule, antigen binding site is located at N-terminal region.

143. Silencing of specific mRNA is possible via RNAi because of
(1) Complementary dsRNA
(2) Inhibitor ssRNA
(3) Complementary tRNA
(4) Non-complementary ssRNA
Answer (1)
Sol. RNAi (RNA interference) takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to a complementary dsRNA molecule that binds to and prevents translation of the mRNA.

144. Genes R and Y follow independent assortment. If RY produce round yellow seeds and rry produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?
(1) Phenotypic ratio - 1 : 2 : 1
(2) Phenotypic ratio - 3 : 1
(3) Phenotypic ratio - 9 : 3 : 3 : 1
(4) Phenotypic ratio - 9 : 7
Answer (3)
Sol. A classical dihybrid cross performed by Mendel involves.
A cross which was made between a pure round yellow seeded pea plant (RRYY) with wrinkled green seeded plant (rryy). Yellow colour is dominant over green and round seed shape over wrinkled seed shape.
Phenotypic ratio in F2 generation
9 : 3 : 3 : 1
Round yellow : Round green : Wrinkled yellow : Wrinkled green

145. Histones are enriched with -
(1) Lysine & Arginine
(2) Leucine & Lysine
(3) Phenylalanine & Leucine
(4) Phenylalanine & Arginine
Answer (1)
Sol. In eukaryotes, packaging of DNA is much more complex. There is a set of positively charged, basic proteins called histones.
Histones are organised to form a unit of light molecules called histone octamer.
They are rich in the basic amino acid residues lysine and arginine.

146. The first menstruation is called:
(1) Menopause
(2) Menarche
(3) Diapause
(4) Ovulation
Answer (2)
Sol. The first menstruation begins at puberty and is called menarche.
Ovulation is the process that deals with the release of secondary oocyte from the mature Graafian follicle. In human beings, menstrual cycles ceases around 50 years of age; that is termed as menopause. Diapause is a state of dormancy or developmental arrest in an organism.

147. Match List-1 with List-II.
List-1
A. Heart
B. Kidney
C. Gastro-intestinal tract
D. Adrenal Cortex
List-II
I. Erythropoietin
II. Aldosterone
III. Atrial natriuretic factor
IV. Secretin
Choose the correct answer from the options given below:
(1) A-II, B-I, C-II, D-IV
(2) A-IV, B-II, C-II, D-I
(3) A-I, B-II, C-IV, D-II
(4) A-II, B-I, C-IV, D-II
Answer (4)

148. The protein portion of an enzyme is called:
(1) Cofactor
(2) Coenzyme
(3) Apoenzyme
(4) Prosthetic group
Answer (3)
Sol. There are number of cases in which non-protein constituents called co-factors are bound to the enzyme to make the enzyme catalytically active. In these instances, the protein portion of the enzymes is called the apoenzyme. Three kinds of co-factors are identified prosthetic groups, co-enzymes and metal ions. Prosthetic groups are organic compounds and they are tightly bound with apoenzyme. Co-enzymes are also organic compounds but their association with apoenzyme is only transient.

149. Which of the following is the unit of productivity of an Ecosystem?
(1) gm-2
(2) KCal m-2
(3) KCal m-3
(4) (KCal m-3)yr-1
Answer (4)
Sol. The rate of biomass production is called productivity. It is expressed in terms of gm-2yr-1 or (KCal m-2)yr-1 to compare the productivity of different ecosystems.

150. Sweet potato and potato represent a certain type of evolution. Select the correct combination of terms to explain the evolution.
(1) Analogy, convergent
(2) Homology, divergent
(3) Homology, convergent
(4) Analogy, divergent
Answer (1)
Sol. Sweet potato is a root modification while potato is a stem modification but both of them have same function. Analogous structures are not anatomically similar structures though they perform similar functions. Analogous structures are the result of convergent evolution.
Homologous organs are anatomically similar but they do not perform similar function. Homologous organs are the result of divergent evolution.

151. With the help of given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in F3 generation.
(1) 1/4
(2) 1/2
(3) 1/8
(4) Zero
Answer (1)
Sol. As in the F1 generation the carrier female and non-affected (normal, not carrier) had affected male child that means the genetic disorder is sex-linked recessive.
The consanguineous mating between female (X^c X) and male (X^c Y)
Out of 4 child only one is carrier i.e. 1/4.

152. Given below are two statements: One is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A): Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus.
Reason (R): Presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microspore mother cells.
In light of the above statements, choose the most appropriate answer from the options given below:
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Answer (3)
Sol. Cell of the tapetum possess dense cytoplasm and generally have more than one nucleus because the presence of more than one nucleus in the tapetal cells increases the efficiency of nourishing the developing pollen grains.

153. How many meiotic and mitotic divisions need to occur for the development of a mature female gametophyte from the megaspore mother cell in an angiosperm plant?
(1) 2 Meiosis and 3 Mitosis
(2) 1 Meiosis and 2 Mitosis
(3) 1 Meiosis and 3 Mitosis
(4) No Meiosis and 2 Mitosis
Answer (3)
Sol. Development of a mature female gametophyte, i.e., embryo sac from a megaspore mother cell in an angiosperm plant requires 1 meiotic and 3 mitotic divisions.

154. Which of the following is an example of a zygomorphic flower?
(1) Petunia
(2) Datura
(3) Pea
(4) Chili
Answer (3)
Sol. Zygomorphic flowers can be divided into two equal halves by only a single vertical plane and shows bilateral symmetry. Pea possess zygomorphic flowers. Chili, Petunia and Datura possess actinomorphic flowers.

155. After maturation, in primary lymphoid organs, the lymphocytes migrate for interaction with antigens to secondary lymphoid organ(s)/tissue(s) like
A. thymus
B. bone marrow
C. spleen
D. lymph nodes
E. Peyer's patches
Choose the correct answer from the options given below
(1) B, C, D only
(2) A, B, C only
(3) E, A, B only
(4) C, D, E only
Answer (4)
Sol. The primary lymphoid organs are bone marrow and thymus where immature lymphocytes differentiate into antigen-sensitive lymphocytes. After maturation, the lymphocytes migrate into secondary lymphoid organs like spleen, lymph nodes, Peyer's patches of small intestine and appendix. These secondary lymphoid organ provide the sites for interaction of lymphocytes with the antigen.

156. Given below are two statements:
Statement I: Fig fruit is a non-vegetarian fruit as it has enclosed fig wasps in it.
Statement II: Fig wasp and fig tree exhibit mutual relationship as fig wasp completes its life cycle in fig fruit and fig fruit gets pollinated by fig wasp.
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 correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct but statement II is incorrect
(4) Statement I is incorrect but statement II is correct
Answer (2)
Sol. Fig fruit is a vegetarian fruit as it only gets pollinated by wasp. Fig tree and fig wasps shows mutualism in which both species are benefitted. So, statement I is incorrect. Statement II is also not correct as fig inflorescence/flower gets pollinated by fig wasp.

157. What is the main function of the spindle fibers during mitosis?
(1) To separate the chromosomes
(2) To synthesize new DNA
(3) To repair damaged DNA
(4) To regulate cell growth
Answer (1)
Sol. During mitosis, spindle fiber get attach to the keretochores of the chromosome and help in the separation of the chromosome.

158. Which one of the following is the characteristic feature of gymnosperms?
(1) Seeds are enclosed in fruits
(2) Seeds are naked
(3) Seeds are absent
(4) Gymnosperms have flowers for reproduction
Answer (2)
Sol. The gymnosperms (Gymnos: naked, sperma: seed) are plants in which the ovules are not enclosed by an ovary wall and remains exposed, both before and after fertilization. The seeds that develop post-fertilization, are not covered, i.e., naked.

159. Consider the following statements regarding function of adrenal medullary hormones:
(A) It causes pupillary constriction.
(B) It is a hyperglycemic hormone.
(C) It causes piloerection.
(D) It increases strength of heart contraction.
Choose the correct answer from the options given below:
(1) C and D only
(2) B, C and D only
(3) A, C and D only
(4) D only
Answer (2)
Sol. Adrenal medulla secretes two hormones called adrenaline or epinephrine and noradrenaline or norepinephrine (also called emergency hormones). Both the hormones -
Cause pupillary dilation (not constriction)
Stimulate breakdown of glycogen resulting in increased concentration of glucose in blood i.e., cause hyperglycemia.
Cause piloerection (raising of hair).
Increase strength of heart contraction i.e., heartbeat.

160. Why can't insulin be given orally to diabetic patients?
(1) Human body will elicit strong immune response
(2) It will be digested in Gastro-intestinal (GI) tract
(3) Because of structural variation
(4) Its bioavailability will be increased
Answer (2)
Sol. Insulin can't be administered orally to diabetic patients as being the proteinaceous molecule, it will be digested in gastro-intestinal tract.

161. Match List-I with List-II.
List-I
A. Pteridophyte
B. Bryophyte
C. Angiosperm
D. Gymnosperm
List-II
I. Salvia
II. Ginkgo
III. Polytrichum
IV. Salvinia
Choose the option with all correct matches.
(1) A-III, B-IV, C-II, D-I
(2) A-IV, B-III, C-I, D-II
(3) A-III, B-IV, C-I, D-II
(4) A-IV, B-III, C-II, D-I
Answer (2)
Sol. Pteridophyte - Salvinia
Bryophyte - Polytrichum
Angiosperm - Salvia
Gymnosperm - Ginkgo

162. Who proposed that the genetic code for amino acids should be made up of three nucleotides?
(1) George Gamow
(2) Francis Crick
(3) Jacque Monod
(4) Franklin Stahl
Answer (1)
Sol. George Gamow, a physicist proposed that genetic code for amino acids should be made up of three nucleotides.

163. Match List I with List II:
List I
A. The Evil Quartet
B. Ex situ conservation
C. Lantana camara
D. Dodo
List II
I. Cryopreservation
II. Alien species invasion
III. Causes of biodiversity losses
IV. Extinction
Choose the option with all correct matches.
(1) A-III, B-II, C-I, D-IV
(2) A-III, B-I, C-II, D-IV
(3) A-III, B-IV, C-II, D-I
(4) A-III, B-II, C-IV, D-I
Answer (2)
Sol. The Evil Quartet Causes of biodiversity losses
Ex situ conservation Cryopreservation
Lantana camara Alien species invasion
Dodo Extinction

164. Which of the following hormones released from the pituitary is actually synthesized in the hypothalamus?
(1) Luteinizing hormone (LH)
(2) Anti-diuretic hormone (ADH)
(3) Follicle-stimulating hormone (FSH)
(4) Adrenocorticotropic hormone (ACTH)
Answer (2)
Sol. Neurohypophysis i.e., posterior pituitary (Pars nervosa) stores and releases two hormones called oxytocin and vasopressin (Also called ADH i.e., antidiuretic hormone) which are actually synthesized by hypothalamus and are transported axonally to neurohypophysis. The pars distalis (anterior pituitary) produces follicle stimulating hormone (FSH), adrenocorticotropic hormone (ACTH) and luteinizing hormone (LH).

165. Role of the water vascular system in Echinoderms is:
A. Respiration and Locomotion
B. Excretion and Locomotion
C. Capture and transport of food
D. Digestion and Respiration
E. Digestion and Excretion
Choose the correct answer from the options given below:
(1) A and B Only
(2) A and C Only
(3) B and C Only
(4) B, D and E Only
Answer (2)
Sol. Water vascular system in Echinoderms helps in locomotion, capture and transport of food and respiration. Excretory system is absent in echinoderms. Excretion takes place through general body surface.

166. Which of the following type of immunity is present at the time of birth and is a non-specific type of defence in the human body?
(1) Acquired Immunity
(2) Innate Immunity
(3) Cell-mediated Immunity
(4) Humoral Immunity
Answer (2)
Sol. Innate immunity is non-specific type of defence, that is present at the time of birth. This is accomplished by providing different types of barriers to the entry of the foreign agents into our body. Acquired immunity is pathogen specific, characterised by memory cells. Immune response mediated by B-lymphocytes is humoral immunity and other immune response mediated by T-lymphocytes is called cell-mediated immunity.

167. In bryophytes, the gemmae help in which one of the following?
(1) Sexual reproduction
(2) Asexual reproduction
(3) Nutrient absorption
(4) Gaseous exchange
Answer (2)
Sol. Gemmae are green, multicellular, asexual buds which develop in small receptacles called gemma cups and help in asexual reproduction in bryophytes.

168. In frog, the Renal portal system is a special venous connection that acts to link:
(1) Liver and intestine
(2) Liver and kidney
(3) Kidney and intestine
(4) Kidney and lower part of body
Answer (4)
Sol. In frogs, special venous connection between liver and intestine as well as the kidney and lower parts of the body are present in frogs. The former is called hepatic portal system and the latter is called renal portal system.

169. Given below are two statements:
Statement I: In ecosystem, there is unidirectional flow of energy of sun from producers to consumers.
Statement II: Ecosystems are exempted from 2nd law of thermodynamics.
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 correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct but statement II is incorrect
(4) Statement I is incorrect but statement II is correct
Answer (3)
Sol. Sun is the only source of energy for all ecosystems on Earth, except for deep sea-hydro-thermal ecosystem. The energy flow is unidirectional from the sun to producers and then to consumers. Ecosystems are not exempted from the second law of thermodynamics. They need a constant supply of energy to synthesise the molecules they require to counteract the universal tendency towards increasing disorderliness.

170. Which of the following statements about RuBisCO is true?
(1) It is active only in the dark
(2) It has higher affinity for oxygen than carbon dioxide
(3) It is an enzyme involved in the photolysis of water
(4) It catalyzes the carboxylation of RuBP
Answer (4)
Sol. Carboxylation is the most crucial step of the Calvin cycle where CO₂ is utilised for the carboxylation of RuBP. This reaction is catalysed by enzyme RuBP carboxylase. Since this enzyme also has an oxygenase activity, RuBisCO has higher affinity for carbon dioxide than oxygen.

171. Which of the following enzyme(s) are NOT essential for gene cloning?
A. Restriction enzymes
B. DNA ligase
C. DNA mutase
D. DNA recombinase
E. DNA polymerase
Choose the correct answer from the options given below:
(1) C and D only
(2) A and B only
(3) D and E only
(4) B and C only
Answer (1)
Sol. Gene cloning is a process where a specific gene or DNA sequence is isolated and replicated, creating multiple identical copies. In gene cloning, restriction enzymes, DNA ligase and DNA polymerase are primarily used.

172. Read the following statements on plant growth and development.
(A) Parthenocarpy can be induced by auxins.
(B) Plant growth regulators can be involved in promotion as well as inhibition of growth.
(C) Dedifferentiation is a pre-requisite for re-differentiation.
(D) Abscisic acid is a plant growth promoter.
(E) Apical dominance promotes the growth of lateral buds.
Choose the option with all correct statements.
(1) A, B, C only
(2) A, C, E only
(3) A, D, E only
(4) B, D, E only
Answer (1)
Sol. ABA is a plant growth inhibitor and an inhibitor of plant metabolism. Apical dominance promotes growth of apical bud. Statements A, B and C are correct.

173. Which factor is important for termination of transcription?
(1) α (alpha)
(2) σ (sigma)
(3) ρ (rho)
(4) γ (gamma)
Answer (3)
Sol. In prokaryotes the RNA polymerase is only capable of catalysing the process of elongation. It associates transiently with initiation factor (σ) and termination factor (ρ) to initiate and terminate the transcription respectively.

174. Frogs respire in water by skin and buccal cavity and on land by skin, buccal cavity and lungs. Choose the correct answer from the following:
(1) The statement is true for water but false for land
(2) The statement is true for both the environment
(3) The statement is false for water but true for land
(4) The statement is false for both the environment
Answer (3)
Sol. In water, frogs respire through skin and not through buccal cavity i.e., undergo cutaneous respiration only. On land, the buccal cavity, skin and lungs act as respiratory organs i.e., undergo buccopharyngeal, cutaneous and pulmonary respiration.

175. Twins are born to a family that lives next door to you. The twins are a boy and a girl. Which of the following must be true?
(1) They are monozygotic twins.
(2) They are fraternal twins.
(3) They were conceived through in vitro fertilization.
(4) They have 75% identical genetic content.
Answer (2)
Sol. Fraternal twins or dizygotic twins are 2 separate fertilized eggs, they usually develop 2 separate amniotic sacs, placentas and supporting structures. If twins are a boy and a girl, this indicates they are fraternal twins.

176. Which of the following microbes is NOT involved in the preparation of household products?
A. Aspergillus niger
B. Lactobacillus
C. Trichoderma polysporum
D. Saccharomyces cerevisiae
E. Propionibacterium sharmanii
Choose the correct answer from the options given below:
(1) A and B only
(2) A and C only
(3) C and D only
(4) C and E only
Answer (2)
Sol. Lactobacillus is used for production of curd.
Saccharomyces cerevisiae is used for the fermentation of palm sap to obtain toddy drink.
Propionibacterium sharmanii is used for production of swiss cheese.
Aspergillus niger is used for the commercial production of citric acid.
Trichoderma polysporum is used for the production of cyclosporin A and also act as a biocontrol agent.
A, C are used in industrial production of citric acid and cyclosporin-A.

177. Match List-1 with List-II.
List-1
A. Progesterone
B. Relaxin
C. Melanocyte stimulating hormone
D. Catecholamines
List-II
I. Pars intermedia
II. Ovary
III. Adrenal Medulla
IV. Corpus luteum
Choose the correct answer from the options given below:
(1) A-IV, B-II, C-I, D-III
(2) A-IV, B-II, C-III, D-I
(3) A-II, B-IV, C-I, D-III
(4) A-III, B-II, C-IV, D-I
Answer (1)
Sol. The correct answer is [A-IV, B-II, C-I, D-III]
Progesterone - A steroidal hormone which is secreted by the corpus luteum
Relaxin - A proteinaceous hormone which is secreted by the ovaries in the later stage of pregnancy
Melanocyte stimulating hormone - A proteinaceous hormone released by the pars intermedia
Catecholamines - An amino-acid derived hormone released from the adrenal medulla during emergency conditions

178. The blue and white selectable markers have been developed which differentiate recombinant colonies from nonrecombinant colonies on the basis of their ability to produce colour in the presence of a chromogenic substrate. Given below are two statements about this method:
Statement I: The blue coloured colonies have DNA insert in the plasmid and they are identified as recombinant colonies.
Statement II: The colonies without blue colour have DNA insert in the plasmid and are identified as recombinant colonies.
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 correct
(2) Both Statement I and Statement II are incorrect
(3) Statement I is correct but Statement II is incorrect
(4) Statement I is incorrect but Statement II is correct
Answer (4)
Sol. Statement I is incorrect but statement II is correct as a recombinant DNA is inserted within the coding sequence of an enzyme, β-galactosidase. This results into inactivation of the gene for synthesis of this enzyme. Thus, presence of insert results into insertional inactivation of the β-galactosidase gene and the colonies do not produce any colour and identified as recombinant colonies. Whereas non-recombinant transformants will produce blue colour in presence of chromogenic substrate.

179. Which one of the following equations represents the Verhulst-Pearl Logistic Growth of population?
Answer (2)
Sol. Logistic growth is described by Verhulst-Pearl logistic growth equation dN/dt = rN((K - N)/K).

180. Which one of the following enzymes contains 'Haem' as the prosthetic group?
(1) RuBisCo
(2) Carbonic anhydrase
(3) Succinate dehydrogenase
(4) Catalase
Answer (4)
Sol. In peroxidase and catalase, which catalyze the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and it is part of the active site of the enzymes. Zinc is the cofactor in enzyme carbonic anhydrase. RuBisCo is the most abundant protein in whole of the biosphere. Succinate is the substrate of enzyme succinic dehydrogenase.

 

NEET Previous Year Question Paper Free PDF Download lInk

NEET Previous Year Question Paper 2025 Code -46

NEET Previous Year Question Paper 2025 Code-47

NEET Previous Year Question Paper 2025 Code -48

NEET Previous Year Question Paper 2024 Code-Q1