NEET Previous Year Question Paper 2023 with Solutions
1. A bullet is fired from a gun at the speed of 280 m s⁻¹ in the direction 30° above the horizontal. The maximum height attained by the bullet is (g = 9.8 m s⁻², sin 30° = 0.5)
(1) 3000 m
(2) 2800 m
(3) 2000 m
(4) 1000 m
Answer (4)
Sol. H = u² sin²θ / 2g
H = (280)²(sin² 30) / (2 × 9.8) = (280 × 280 × 0.5 × 0.5) / (2 × 9.8)
H = 1000 m
2. An electric dipole is placed at an angle of 30° with an electric field of intensity 2 × 10⁵ N C⁻¹. It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.
(1) 2 mC
(2) 8 mC
(3) 6 mC
(4) 4 mC
Answer (1)
Sol. E = 2 × 10⁵ N/C
l = 2 cm
τ = 4 Nm
τ = p × E
4 = pE sin θ
4 = p × 2 × 10⁵ × sin 30°
p = 4 × 10⁻⁵ cm
q = p/l = (4 × 10⁻⁵)/0.02 = 2 mC
3. The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = 0.03 N m⁻¹)
(1) 50.1 × 10⁻⁴ J
(2) 30.16 × 10⁻⁴ J
(3) 5.06 × 10⁻⁴ J
(4) 3.01 × 10⁻⁴ J
Answer (4)
Sol. Amount of energy required = [S × ΔA] × 2
⇒ Energy required = [0.03 × 4 × π × 4 × 10⁻⁴] × 2
= 3.015 × 10⁻⁴ J
4. Let a wire be suspended from the ceiling (rigid support) and stretched by a weight W attached at its free end. The longitudinal stress at any point of cross-sectional area A of the wire is
(1) Zero
(2) 2W/A
(3) W/A
(4) W/2A
Answer (3)
Sol. Longitudinal stress = Internal restoring force / Area = F_ext / Area
Stress = W/A
5. Light travels a distance x in time t₁ in air and 10x in time t₂ in another denser medium. What is the critical angle for this medium?
(1) sin⁻¹(10t₁/t₂)
(2) sin⁻¹(10t₂/t₁)
Answer (1)
Sol. μ₂ sin i_c = μ₁
sin i_c = μ₁/μ₂
μ = c/V
So sin i_c = μ₁/μ₂ = V₂/V₁
sin i_c = 10xt₁ / t₂x
i_c = sin⁻¹(10t₁/t₂)
6. For Young's double slit experiment, two statements are given below:
Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains constant.
Statement II : If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true.
(2) Both Statement I and Statement II are true.
(3) Both Statement I and Statement II are false.
(4) Statement I is true but Statement II is false.
Answer (4)
Sol. For YDSE, angular fringe width is given by α = λ/d
It does not depend on the distance of screen from the slit, so statement I is correct.
Angular fringe width α ∝ λ
If λ ↑ → angular separation of fringes increases
So, statement I is true and statement II is false.
7. Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (G = gravitational constant)
(1) 20Gm/R
(2) 8Gm/R
(3) 12Gm/R
(4) 16Gm/R
Answer (4)
Sol. Let electric field at point Q be zero
So, Gm/x² = G(9m)/(R - x)²
(R - x)²/x² = 9
8. The temperature of a gas is -50°C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?
(1) 223 K
(2) 669°C
(3) 3295°C
(4) 3097 K
Answer (3)
Sol. v_rms = √(3RT/m)
T₁ = 273 - 50
v_rms ∝ √T
= 223 K
v_rms is increased by 3 times
T₂ = ?
So, final rms speed = v + 3v = 4v
v₁/v₂ = √(T₁/T₂)
v/4v = √(223/T₂) ⇒ 1/16 = 223/T₂
T₂ = 3568 K
T₂ = 3568 - 273 = 3295°C
9. In a series LCR circuit, the inductance L is 10 mH, capacitance C is 1 μF and resistance R is 100 Ω. The frequency at which resonance occurs is
(1) 1.59 kHz
(2) 15.9 rad/s
(3) 15.9 kHz
(4) 1.59 rad/s
Answer (1)
Sol. For resonance frequency = 1/(2π√(LC))
10. A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is
(1) Along south-west
(2) Along eastward
(3) Along northward
(4) Along north-east
Answer (4)
Sol. ΔP = P_f - P_i
P_f = mu î
P_i = mu(-ĵ)
ΔP = mu î - mu(-ĵ)
ΔP = mu(î + ĵ)
F = ΔP/Δt
Direction of change of momentum and direction of force acting on the player will be same, so correct answer is North east direction
11. The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are
(1) Random errors
(2) Instrumental errors
(3) Personal errors
(4) Least count errors
Answer (1)
Sol. The errors which cannot be associated with any systematic or constant cause are called random errors. These errors can arise due to unpredictable fluctuations in experimental conditions. e.g., random change in pressure, temperature, voltage supply etc.
12. The ratio of radius of gyration of a solid sphere of mass M and radius R about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is
(1) 5:2
(2) 3:5
(3) 5:3
(4) 2:5
Answer (2*)
Sol. Radius of gyration of solid sphere about its own axis = √(2/5) R
Radius of gyration of hollow sphere about its own axis = √(2/3) R
⇒ Required ratio = √(2/5) × √(3/2) = √(3/5)
* None of the option is correct (correct answer is √(3/5))
13. If ∮_s E · dS = 0 over a surface, then
(1) The electric field inside the surface is necessarily uniform
(2) The number of flux lines entering the surface must be equal to the number of flux lines leaving it
(3) The magnitude of electric field on the surface is constant
(4) All the charges must necessarily be inside the surface
Answer (2)
Φ_net = ∮_s E · dS = 0
⇒ Net flux through surface is zero.
⇒ Therefore, the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
14. A Carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is
(1) 200°C
(2) 27°C
(3) 15°C
(4) 100°C
Answer (2)
Sol. Efficiency η = 50/100 = 1/2
Efficiency of Carnot engine
η = 1 - T₂/T₁
η = 1 - T₂/600
1/2 = 1 - T₂/600
T₂/600 = 1/2 ⇒ T₂ = 300 K
T₂ = 300 - 273 = 27°C
15. A 12 V, 60 W lamp is connected to the secondary of a step-down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?
(1) 0.37 A
(2) 0.27 A
(3) 2.7 A
(4) 3.7 A
Answer (2)
16. Given below are two statements:
Statement I: Photovoltaic devices can convert optical radiation into electricity.
Statement II: Zener diode is designed to operate under reverse bias in breakdown region.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Statement I is incorrect but Statement II is correct
(2) Both Statement I and Statement II are correct
(3) Both Statement I and Statement II are incorrect
(4) Statement I is correct but Statement II is incorrect
Answer (2)
Sol. Both Statements are correct.
I: Photovoltaic devices convert optical radiation into electricity.
II: Zener diode is designed to operate under reverse bias in breakdown region. e.g., Zener diode as a voltage regulator.
17. If the galvanometer G does not show any deflection in the circuit shown, the value of R is given by
(1) 400 Ω
(2) 200 Ω
(3) 50 Ω
(4) 100 Ω
Answer (4)
18. The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is
(1) 3:1
(2) 1:2
(3) 2:1
(4) 1:3
Answer (3)
Sol. f₀ = f_open pipe = v/2l
f_c = f_closed pipe = v/4l
f₀/f_c = (v/2l) × (4l/v)
f₀:f_c = 2:1
19. The angular acceleration of a body, moving along the circumference of a circle, is
(1) Along the axis of rotation
(2) Along the radius, away from centre
(3) Along the radius towards the centre
(4) Along the tangent to its position
Answer (1)
Sol. Angular acceleration of a body, moving along the circumference of a circle is along the axis of rotation.
20. The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to
(1) V²
(2) √V
(3) 1/V
(4) 1/√V
Answer (3)
Sol. eV = hc/λ_min
λ_min = hc/eV
λ_min ∝ 1/V
21. The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14 eV, 2.30 eV and 2.75 eV respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?
(1) Na only
(2) Cs only
(3) Both Na and K
(4) K only
Answer (2)
Sol. Energy of incident radiation = 2.80 eV
Work function of Cs → 2.14 eV
Work function of K → 2.30 eV
Work function of Na → 2.75 eV
Since the work function of potassium and sodium are more than energy of incident radiation hence photons may be emitted from caesium.
22. The net magnetic flux through any closed surface is
(1) Negative
(2) Zero
(3) Positive
(4) Infinity
Answer (2)
Sol. ∮ B · dS = zero
Magnetic monopole doesn't exist.
Hence net magnetic flux through any closed surface is zero.
23. A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output?
(1) Load resistance
(2) A centre-tapped transformer
(3) p-n junction diodes
(4) Capacitor
Answer (4)
Sol. Capacitor removes the ac ripple from rectified output.
24. An ac source is connected to a capacitor C. Due to decrease in its operating frequency
(1) Capacitive reactance remains constant
(2) Capacitive reactance decreases
(3) Displacement current increases
(4) Displacement current decreases
Answer (4)
Sol. X_C = 1/ωC
Since ω decreasing X_C will increases hence current will decreases also
conduction current = displacement current
Therefore displacement current will decrease.
25. In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ Hz and amplitude 48 V m⁻¹. Then the amplitude of oscillating magnetic field is (Speed of light in free space = 3 × 10⁸ m s⁻¹)
(1) 1.6 × 10⁻⁶ T
(2) 1.6 × 10⁻⁹ T
(3) 1.6 × 10⁻⁸ T
(4) 1.6 × 10⁻⁷ T
Answer (4)
Sol. From the properties of electromagnetic wave we know that,
C = E₀/B₀
E₀ ⇒ Amplitude of oscillating electric field
B₀ ⇒ Amplitude of oscillating magnetic field
⇒ B₀ = 48/(3 × 10⁵) = 1.6 × 10⁻⁷ T
26. Resistance of a carbon resistor determined from colour codes is (22000 ± 5%) Ω. The colour of third band must be
(1) Yellow
(2) Red
(3) Green
(4) Orange
Answer (4)
Sol. Resistance = (22 × 10³) Ω ± 5%
Third band corresponds to decimal multiplier.
Decimal multiplier = 10³
Colour → Orange
27. A vehicle travels half the distance with speed v and the remaining distance with speed 2v. Its average speed is
(1) 3v/4
(2) v/3
(3) 2v/3
(4) 4v/3
Answer (4)
Sol. v_avg = 2v₁v₂/(v₁ + v₂)
= 2 × v × 2v / (v + 2v)
= 4v/3
28. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be
(1) 16U
(2) 2U
(3) 4U
(4) 8U
Answer (1)
Sol. Potential energy stored in spring U = 1/2 Kx²
U = 1/2 K(2)² where x = 2 cm
U = 1/2 (K)(4)
U = 2K ...(i)
U' = 1/2 K(8)²
U' = 1/2 K × 64 = 32K ...(ii)
On dividing (i) by (ii)
U/U' = 2K/32K = 1/16
U' = 16U
29. In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the Bracket series is
(1) 16λ
(2) 2λ
(3) 4λ
(4) 9λ
Answer (3)
Sol. 1/λ = R[1/n₂² - 1/n₁²]
For Balmer [n₂ = 2, n₁ = ∞]
1/λ = R[1/4 - 1/∞]
λ = 4/R ...(1)
For Bracket, (n₂ = 4, n₁ = ∞)
1/λ' = R[1/16 - 1/∞]
λ' = 16/R ...(2)
Eqⁿ(1)/Eqⁿ(2)
λ' = 4λ
30. The equivalent capacitance of the system shown in the following circuit is
(1) 9 μF
(2) 2 μF
(3) 3 μF
(4) 6 μF
Answer (2)
Sol. For parallel grouping
C₁ = 3 + 3 = 6 μF
For series grouping
C_eq = C₁C₂/(C₁ + C₂) = (3 × 6)/(3 + 6) = 18/9
C_eq = 2 μF
31. A metal wire has mass (0.4 ± 0.002) g radius (0.3 ± 0.001) mm and length (5 ± 0.02) cm. The maximum possible percentage error in the measurement of density will nearly be
(1) 1.4%
(2) 1.2%
(3) 1.3%
(4) 1.6%
Answer (4)
Sol. We know, ρ = Mass/Volume = M/(πr²ℓ)
Using the concept of errors we know,
Δρ/ρ = ΔM/M + 2Δr/r + Δℓ/ℓ
= (0.002/0.4 + (2 × 0.001)/0.3 + 0.02/5)
Δρ/ρ = 0.0156
Δρ/ρ % = 1.56% ≈ 1.6%
32. The venturi-meter works on
(1) The principle of perpendicular axes
(2) Huygen's principle
(3) Bernoulli's principle
(4) The principle of parallel axes
Answer (3)
Sol. Venturi-meter works on the Bernoulli's principle.
33. The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to (1/16)th of its initial value?
(1) 80 minutes
(2) 20 minutes
(3) 40 minutes
(4) 60 minutes
Answer (1)
34. The magnetic energy stored in an inductor of inductance 4 μH carrying a current of 2 A is
(1) 8 μJ
(2) 4 μJ
(3) 4 mJ
(4) 8 mJ
Answer (1)
Sol. Magnetic energy stored in an inductor U = 1/2 Li²
= 1/2 × 4 × 10⁻⁶ × (2)²
= 8 × 10⁻⁶ J
U = 8 μJ
35. The magnitude and direction of the current in the following circuit is
(1) 1.5 A from B to A through E
(2) 0.2 A from B to A through E
(3) 0.5 A from A to B through E
(4) 5/9 A from A to B through E
Answer (3)
Sol. Using Kirchhoff's law
i = (10 - 5)/10 = 5/10
i = 0.5 A
In clock-wise direction (from A to B)
36. In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)?
(1) -50 cm
(2) 40 cm
(3) -40 cm
(4) -100 cm
Answer (4)
Sol. Effective focal length ⇒ f_eff
1/f_eff = 1/f₁ + 1/f₂ + 1/f₃
Also, 1/f = (μ - 1)(1/R₁ - 1/R₂)
1/f₁ = (1.6 - 1)(1/∞ - 1/20) = -0.6/20
1/f₂ = (1.5 - 1)(1/20 - 1/-20) = 0.5/10
1/f₃ = (1.6 - 1)(1/-20 - 1/∞) = -0.6/20
1/f_eff = -0.6/20 + 0.5/10 - 0.6/20
1/f_eff = -0.6/10 + 0.5/10 - -0.1/10 = -1/100
∴ f_eff = -100 cm
37. Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 (g = 10 m s⁻²).
(1) 50 m s⁻²
(2) 1.2 m s⁻²
(3) 150 m s⁻²
(4) 1.5 m s⁻²
Answer (4)
Sol. w.r.t. car
a_b = 0
ma_max = μ_s mg
a_max = μ_s g = 0.15 × 10 = 1.5 m/s²
38. The net impedance of circuit (as shown in figure) will be
(1) 25 Ω
(2) 10√2 Ω
(3) 15 Ω
(4) 5√5 Ω
Answer (4)
Sol. L = 50/π mH
X_L = 2π × 50 × 50/π × 10⁻³ = 5 Ω
C = 10³/π × 10⁻⁶
X_C = 1 × π / (2π × 50 × 10³ × 10⁻⁶) = 10³/100 = 10 Ω
Z = √((X_C - X_L)² + R²)
Z = √((10 - 5)² + 10²) = √125 = 5√5 Ω
39. An electric dipole is placed as shown in the figure.
The electric potential (in 10² V) at point P due to the dipole is (ε₀ = permittivity of free space and 1/4πε₀ = K)
(1) (8/3)qK
(2) (3/8)qK
(3) (5/8)qK
Answer (2)
Sol. Electrostatic potential due to a point charge is given by Kq/r
V_net at point P = Kq/(2 × 10⁻²) - Kq/(8 × 10⁻²)
= Kq × 10²/2 (1 - 1/4)
= (3/8 Kq) × 10² V = 3/8 qK
40. For the following logic circuit, the truth table is
(1)
A B Y
0 0 0
0 1 0
1 0 0
1 1 1
(3)
A B Y
0 0 0
0 1 1
1 0 1
1 1 1
Answer (3)
Sol. Y = A̅ · B̅ = A + B
It is OR gate.
A B Y
0 0 0
0 1 1
1 0 1
1 1 1
41. 10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is
(1) 1000
(2) 10
(3) 100
(4) 1
Answer (3)
Sol. In series combination
R_eq = 10R
i = E/10R
In parallel combination
R_eq = R/10
i' = E/R = 10E/R
i' = 10 × 10i = 100i
42. The resistance of platinum wire at 0°C is 2 Ω and 6.8 Ω at 80°C. The temperature coefficient of resistance of the wire is
(1) 3 × 10⁻¹ °C⁻¹
(2) 3 × 10⁻⁴ °C⁻¹
(3) 3 × 10⁻³ °C⁻¹
(4) 3 × 10⁻² °C⁻¹
Answer (4)
Sol. Using R = R₀(1 + αΔT)
where α is the thermal coefficient of resistance
6.8 = 2(1 + α(80 - 0))
6.8/2 - 1 = α × 80
α = (3.4 - 1)/80 = 2.4/80 = 0.03
∴ α = 3 × 10⁻² °C⁻¹
43. A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s⁻¹. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take g = 10 m s⁻²)
(1) 68 m
(2) 56 m
(3) 60 m
(4) 64 m
Answer (4)
Sol. s = ut - 1/2 gt²
= 16 - 1/2 × 10 × 16
= -64 m
Height of bridge above water surface = 64 m
44. A bullet from a gun is fired on a rectangular wooden block with velocity u. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes u/3. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is
(1) 30 cm
(2) 27 cm
(3) 24 cm
(4) 28 cm
Answer (2)
Sol. between 1 to 2
(u/3)² = u² - 2a × 24
⇒ 2a(24) = 8u²/9 ...(I)
between 2 to 3
0 = (u/3)² - 2as ...(II)
From equation (I) and (II)
2as = 2a(24)/8
s = 3 cm
Length of wooden block is 24 + 3 = 27 cm
45. A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity 3π/Gd represents
(1) √T
(2) T
(3) T²
(4) T³
Answer (3)
Sol. Time period of satellite above earth surface
T = 2π√(R³/GM) = 2π√(R³/(Gd(4/3)πR³))
T = 2π√(3/4πGd)
T = √(3π/Gd) ⇒ T² = 3π/Gd
46. Two thin lenses are of same focal lengths (f), but one is convex and the other one is concave. When they are placed in contact with each other, the equivalent focal length of the combination will be
(1) Infinite
(2) Zero
(3) f/4
(4) f/2
Answer (1)
Sol. Convex lens f₁ > 0, concave lens f₂ < 0
1/f_eq = 1/f₁ + 1/f₂ = 1/f - 1/f = 0
f_eq = ∞
47. A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by
(1) μ₀I/4R [1 - 2/π] pointed into the page
(2) μ₀I/4R [1 - 1/π] pointed away from the page
(3) μ₀I/4R pointed away from the page
Answer (4)
Sol. B_P due to wire 1 = μ₀I/4πR ⊗
B_P due to wire 3 = μ₀I/4πR ⊗
B_P due to wire 2 = μ₀I/4R ⊙
B_net = -μ₀I/2πR + μ₀I/4R = μ₀I/4R [-2/π + 1] = μ₀I/4R [1 - 2/π]
Pointed away from page.
48. The x-t graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at t = 2 s is
(1) π²/16 m s⁻²
(2) π²/8 m s⁻²
(3) -π²/8 m s⁻²
(4) -π²/16 m s⁻²
Answer (1)
Sol. Position of particle as function of time
x = A sin ωt
From figure,
49. A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field B = (2î + 3ĵ - 4k̂) T. The magnitude of the magnetic force acting on the wire is
(1) √3 IL
(2) √5 IL
Answer (4)
Sol. Magnetic force acting on a current carrying wire is
F = iI × B
= iLî × (2î + 3ĵ - 4k̂) = 3Lk̂ + 4Lîĵ
Magnitude of force
|F| = √((3L)² + (4L)²)
= 5L
50. The radius of inner most orbit of hydrogen atom is 5.3 × 10⁻¹¹ m. What is the radius of third allowed orbit of hydrogen atom?
(1) 4.77 A
(2) 0.53 A
(3) 1.06 A
(4) 1.59 A
Answer (1)
Sol. r_n ∝ n²/Z
r₁/r₂ = (1/3)²
r₂ = 9r₁ = 5.3 × 10⁻¹¹ × 9
= 47.7 × 10⁻¹¹
= 4.77 A
51. Match List-1 with List-1.
List-1 List-1
A. Coke I. Carbon atoms are sp³ hybridised
B. Diamond II. Used as a dry lubricant
C. Fullerene III. Used as a reducing agent
D. Graphite IV. Cage like molecules
Choose the correct answer from the options given below :
(1) A-III, B-IV, C-I, D-II
(2) A-III, B-IV, C-I, D-II
(3) A-IV, B-I, C-II, D-II
(4) A-III, B-I, C-IV, D-II
Answer (4)
Sol. Coke is largely used as a reducing agent in metallurgy.
In diamond, each carbon atom undergoes sp³ hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion.
Buckminsterfullerene contains six membered and five membered rings and hence is a cage like molecule.
Graphite is very soft and slippery. Hence, it is used as a dry lubricant in machines running at high temperature.
52. Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.
Reason R : The deep blue solution is due to the formation of amide.
In the light of the above statements, choose the correct answer from the options given below :
(1) A is false but R is true
(2) Both A and R are true and R is the correct explanation of A
(3) Both A and R are true but R is NOT the correct explanation of A
(4) A is true but R is false
Answer (4)
Sol. On dissolving alkali metal (sodium) in liquid ammonia, a deep blue solution is developed due to ammoniated electron which absorbs energy in visible region of light and imparts blue colour. Due to unpaired electron, solution is paramagnetic.
M + (x + y)NH₃ → [M(NH₃)₃]⁺ + [e(NH₃)₃]⁻
So, assertion statement is correct but reason is incorrect.
53. The given compound
CH=CH-CH-CH₂CH₃
X
is an example of
(1) Vinylic halide
(2) Benzylic halide
(3) Aryl halide
(4) Allylic halide
Answer (4)
Sol. α-carbon is sp³ carbon which is right next to C=C<
This α-position is known as allylic position
Hence,
CH=CH-CH-CH₂-CH₃
X
is allylic halide
54. Some tranquilizers are listed below. Which one from the following belongs to barbiturates?
(1) Veronal
(2) Chlordiazepoxide
(3) Meprobamate
(4) Valium
Answer (1)
Sol. Veronal is the derivative of Barbituric acid and considered as barbiturate.
Meprobamate, valium and chlordiazepoxide are other tranquilizers.
55. Identify the product in the following reaction:
(1)
(2)
(3)
(4)
Answer (4)
56. Complete the following reaction
[C] is
(1)
(2)
(3)
(4)
Answer (1)
Sol. [A] + HCN → [B]
[B] → conc. H₂SO₄/Δ → [C]
57. The stability of Cu²⁺ is more than Cu⁺ salts in aqueous solution due to
(1) Second ionisation enthalpy
(2) First ionisation enthalpy
(3) Enthalpy of atomization
(4) Hydration energy
Answer (4)
Sol. The stability of Cu²⁺(aq) is more than Cu⁺(aq) is due to the much more negative Δ_hyd H⁺ of Cu²⁺(aq) than Cu⁺(aq), which more than compensates for second ionisation enthalpy of Cu
Δ_hyd H⁺ of Cu²⁺(aq) = -2121 kJ mol⁻¹
Δ_H⁺ H₁⁺ of Cu = +745 kJ mol⁻¹
Δ_H⁺ H₂⁺ of Cu = +1960 kJ mol⁻¹
58. The right option for the mass of CO₂ produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40)
[CaCO₃] → 1200K → CaO + CO₂
(1) 1.32 g
(2) 1.12 g
(3) 1.76 g
(4) 2.64 g
Answer (3)
Sol. CaCO₃ → 1200K → CaO + CO₂
From 100 g CaCO₃ → 44 g CO₂ produced
As CaCO₃ is 20% pure
So, mass of pure CaCO₃ = 20 × 20/100 = 4 g
So, 100 g CaCO₃ → 44 g CO₂
4 g CaCO₃ → 44/100 × 4 g CO₂
= 1.76 g CO₂
59. The conductivity of centimolar solution of KCl at 25°C is 0.0210 ohm⁻¹ cm⁻¹ and the resistance of the cell containing the solution at 25°C is 60 ohm. The value of cell constant is
(1) 3.34 cm⁻¹
(2) 1.34 cm⁻¹
(3) 3.28 cm⁻¹
(4) 1.26 cm⁻¹
Answer (4)
Sol. Conductivity = conductance × cell constant
k = G G*
G* = k × R = 0.0210 × 60 = 1.26 cm⁻¹
60. Which amongst the following options are correct graphical representation of Boyle's law?
(1)
(2)
(3)
(4)
Answer (3)
Sol. According to Boyle's law,
PV = nRT
P = nRT(1/V)
P versus (1/V) gives straight line graph with slope nRT.
61. Which amongst the following molecules on polymerization produces neoprene?
H₂C = C - CH = CH₂
Cl
H₂C = C - CH = CH₂
Answer (3)
Sol. Neoprene is formed by free radical polymerisation of chloroprene.
nCH₂ = C(Cl) - CH = CH₂ → Polymerisation → [CH₂ - C(Cl) = CH - CH₂]ₙ
2-Chloro-1, 3-butadiene
(Chloroprene)
62. Which of the following reactions will NOT give primary amine as the product?
(1) CH₃CONH₂ → (i) LiAlH₄ (ii) H₂O⁺ → Product
(2) CH₃CONH₂ → Br₂/KOH → Product
(3) CH₃CN → (i) LiAlH₄ (ii) H₂O⁺ → Product
(4) CH₃NC → (i) LiAlH₄ (ii) H₂O⁺ → Product
Answer (4)
Sol. (1) CH₃-C(=O)-NH₂ → (i) LiAlH₄ (ii) H₂O⁺ → CH₃-CH₂-NH₂ (Primary amine)
(2) CH₃-C(=O)-NH₂ → Br₂/KOH → CH₃-NH₂ (Primary amine)
(3) CH₃-C≡N → (i) LiAlH₄ (ii) H₂O⁺ → CH₃-CH₂-NH₂ (Primary amine)
(4) CH₃NC → (i) LiAlH₄ (ii) H₂O⁺ → CH₃-NH-CH₃ (Secondary amine)
63. Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : A reaction can have zero activation energy.
Reasons R : The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy.
In the light of the above statements, choose the correct answer from the options given below :
(1) A is false but R is true
(2) Both A and R are true and R is the correct explanation of A
(3) Both A and R are true and R is NOT the correct explanation of A
(4) A is true but R is false
Answer (3)
Sol. Few reactions can have zero activation energy for example radical reactions.
Activation energy is defined as the minimum amount of extra energy absorbed by reactants to achieve threshold energy.
64. Which one is an example of heterogenous catalysis?
(1) Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron
(2) Oxidation of sulphur dioxide into sulphur trioxide in the presence of oxides of nitrogen
(3) Hydrolysis of sugar catalysed by H⁺ ions
(4) Decomposition of ozone in presence of nitrogen monoxide
Answer (1)
65. Amongst the given options which of the following molecules/ion acts as a Lewis acid?
(1) OH⁻
(2) NH₃
(3) H₂O
(4) BF₃
Answer (4)
Sol. Lewis acids are the one which accepts lone pair of electron due to presence of vacant orbital in outermost shell.
H₂O: → Lewis base
BF₃: → Lewis acid
H⁻: → Lewis base
NH₃: → Lewis base
66. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : In equation ΔG = -nFE_cell value of ΔG depends on n.
Reasons R : Ecell is an intensive property and ΔG is an extensive property.
In the light of the above statements, choose the correct answer from the options given below
(1) A is false but R is true
(2) Both A and R are true and R is the correct explanation of A
(3) Both A and R are true and R is NOT the correct explanation of A
(4) A is true but R is false
Answer (3)
Sol. The value of ΔG depends on n value as per the equation ΔG = -nFE_cell
Where E is the emf of the cell and nF is the amount of charge passed.
So, assertion statement is correct
Ecell is an intensive property while ΔG is an extensive thermodynamic property
So, reason is correct but not explaining the assertion
67. The element expected to form largest ion to achieve the nearest noble gas configuration is
(1) Na
(2) O
(3) F
(4) N
Answer (4)
Sol. For isoelectronic species, as the charge on anion increases, ionic size increases
So, N forms N³⁻ anion with largest ionic size
68. A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is AₓBᵧ, then the value of x + y is in option
(1) 2
(2) 5
(3) 4
(4) 3
Answer (2)
Sol. Number of atoms of element B is N
So, the number of atoms of element A is 1/3 × 2N
The formula of the compound is A₂B_N = A₂B₃
So, x = 2
y = 3
∴ x + y = 5
69. Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Helium is used to dilute oxygen in diving apparatus.
Reason R : Helium has high solubility in O₂
In the light of the above statements, choose the correct answer from the options given below
(1) A is false but R is true
(2) Both A and R are true and R correct explanation of A
(3) Both A and R are true and R is NOT the correct explanation of A
(4) A is true but R is false
Answer (3)
Sol. Helium is used as diluent for oxygen in modern diving apparatus because of its very low solubility in blood. Gases diffuses easily with each other.
70. Which of the following statements are NOT correct?
A. Hydrogen is used to reduce heavy metal oxides to metals.
B. Heavy water is used to study reaction mechanism.
C. Hydrogen is used to make saturated fats from oils.
D. The H-H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any elements.
E. Hydrogen reduces oxides of metals that are more active than iron.
Choose the most appropriate answer from the options given below:
(1) A, B, C only
(2) B, C, D, E only
(3) B, D only
(4) D, E only
Answer (4)
Sol. Statement A, B, C are correct
(D) H-H bond dissociation energy is maximum as compared to single bond between two atom of any element.
(E) Hydrogen reduces oxides of metal that are less active than iron.
71. Which one of the following statements is correct?
(1) Mg plays roles in neuromuscular function and interneuronal transmission
(2) The daily requirement of Mg and Ca in the human body is estimated to be 0.2-0.3 g
(3) All enzymes that utilise ATP in phosphate transfer require Ca as the cofactor
(4) The bone in human body is an inert and unchanging substance
Answer (2)
Sol. All enzymes that utilize ATP in phosphate transfer require Mg as the co-factor. Bone in human body is not an inert and unchanging substance but is continuously being solubilised and redeposited. Ca plays important role in neuromuscular function, interneuronal transmission, cell membrane integrity and blood coagulation. The daily requirement of Mg and Ca in the human body is estimated to be 200 - 300 mg (0.2 - 0.3 g).
72. For a certain reaction, the rate = k[A]²[B], when the initial concentration of A is tripled keeping concentration of B constant, the initial rate would
(1) Increase by a factor of three
(2) Decrease by a factor of nine
(3) Increase by a factor of six
(4) Increase by a factor of nine
Answer (4)
Sol. Rate (r) = k[A]²[B]
When concentration of A is tripled [A'] = [3A]
New rate, r' = k[A']²[B] = k[3A]²[B] = 9k[A]²[B]
r' = 9r
73. Amongst the following the total number of species NOT having eight electrons around central atom in its outermost shell, is
NH₃, AlCl₃, BeCl₂, CCl₄, PCl₅ :
(1) 1
(2) 3
(3) 2
(4) 4
Answer (2)
Sol. NH₃ → 8 e⁻ in nitrogen
AlCl₃ → 6 e⁻ in Al
BeCl₂ → 4 e⁻ in Be
CCl₄ → 8 e⁻ in 'C'
PCl₅ → 10 e⁻ in 'P'
74. In Lassaigne's extract of an organic compound, both nitrogen and sulphur are present, which gives blood red colour with Fe³⁺ due to the formation of
(1) [Fe(SCN)]²⁺
(2) Fe₄[Fe(CN)₆]₃·xH₂O
(3) NaSCN
(4) [Fe(CN)₅NOS]⁴⁻
Answer (1)
Sol. In case, nitrogen and sulphur both are present in organic compound, sodium thiocyanate is formed. It gives blood red colour and no Prussian blue since there are no free cyanide ions.
Na + C + N + S → NaSCN
Fe³⁺ + SCN⁻ → [Fe(SCN)]²⁺ (Blood red)
75. The correct order of energies of molecular orbitals of N₂ molecule, is
(1) σ1s < σ*1s < σ2s < σ*2s < (π2pₓ = π2pᵧ) < (π*2pₓ = π*2pᵧ) < σ2p_z < σ*2p_z
(2) σ1s < σ*1s < σ2s < σ*2s < (π2pₓ = π2pᵧ) < σ2p_z < (π*2pₓ = π*2pᵧ) < σ*2p_z
(3) σ1s < σ*1s < σ2s < σ*2s < σ2p_z < (π2pₓ = π2pᵧ) < (π*2pₓ = π*2pᵧ) < σ*2p_z
(4) σ1s < σ*1s < σ2s < σ*2s < σ2p_z < σ*2p_z < (π2pₓ = π2pᵧ) < (π*2pₓ = π*2pᵧ)
Answer (2)
Sol. For molecules like B₂, C₂, N₂ etc. the increasing order of energies of various molecular orbitals is
σ1s < σ*1s < σ2s < σ*2s < (π2pₓ = π2pᵧ) < σ2p_z < (π*2pₓ = π*2pᵧ) < σ*2p_z
76. Homoleptic complex from the following complexes is
(1) Triamminetriquachromium (III) chloride
(2) Potassium trioxalatoaluminate (III)
(3) Diamminechloridonitrito-N-platinum (II)
(4) Pentaamminecarbonatocobalt (III) chloride
Answer (2)
Sol. Complexes in which a metal is bound to only one kind of donor groups are called as homoleptic complexes
Potassium trioxalatoaluminate (III) K₃[Al(ox)₃]
It is a homoleptic complex
77. Intermolecular forces are forces of attraction and repulsion between interacting particles that will include:
A. dipole - dipole forces
B. dipole - induced dipole forces
C. hydrogen bonding
D. covalent bonding
E. dispersion forces
Choose the most appropriate answer from the options given below:
(1) A, C, D, E are correct
(2) B, C, D, E are correct
(3) A, B, C, D are correct
(4) A, B, C, E are correct
Answer (4)
Sol. Intermolecular forces are the forces of attraction and repulsion between interacting molecules. This term does not include covalent bonds as covalent bond holds atoms of a molecule together. Hence, dipole - dipole forces, dipole - induced dipole forces, hydrogen bonding and dispersion forces are intermolecular forces.
78. Select the correct statements from the following
A. Atoms of all elements are composed of two fundamental particles.
B. The mass of the electron is 9.10939 × 10⁻³¹ kg
C. All the isotopes of a given element show same chemical properties:
D. Protons and electrons are collectively known as nucleons.
E. Dalton's atomic theory, regarded the atom as an ultimate particles of matter
Choose the correct answer from the options given below
(1) B, C and E only
(2) A, B and C only
(3) C, D and E only
(4) A and E only
Answer (1)
Sol. Atoms consist of three fundamental particles: Electrons, protons and neutrons
The mass of the electron is 9.10939 × 10⁻³¹ kg
All the isotopes of a given element show same chemical properties.
Protons and neutrons present in the nucleus are collectively called as nucleons.
Dalton's atomic theory, regarded the atom as the ultimate particle of matter
So, the correct statements are B, C, E only
79. Consider the following reaction and identify the product (P).
CH₃-CH(CH₃)-CH(OH)-CH₃ → HBr → Product (P)
3-Methylbutan-2-ol
(1) CH₃-C(CH₃)(Br)-CH₂Br
(2) CH₃-C(Br)(CH₃)-CH₂-CH₃
(3) CH₃CH=CH-CH₃
(4) CH₃-CH(CH₃)-CH(Br)-CH₃
Answer (2)
Sol. CH₃-CH(CH₃)-CH(OH)-CH₃ →H⊕/HBr→ CH₃-CH(CH₃)-CH(OH₂⁺)-CH₃
→-H₂O→ CH₃-CH(CH₃)-CH⊕-CH₃ (2° carbocation)
Carbocation rearrangement →H⁻ shift→ CH₃-C⊕(CH₃)-CH₂-CH₃
→Br⊖→ CH₃-C(Br)(CH₃)-CH₂-CH₃
80. Identify product (A) in the following reaction:
(1)
(2)
(3)
(4)
Answer (3)
81. Taking stability as the factor, which one of the following represents correct relationship?
(1) TlI > TlI₃
(2) TlCl₃ > TlCl
(3) InI₃ > InI
(4) AlCl > AlCl₃
Answer (1)
Sol. As we move down the group, due to poor shielding effect of intervening d and f orbitals, the increased effective nuclear charge holds ns electrons tightly and therefore restricting their participation in bonding.
So, the relative stability of +1 O.S increases for heavier elements.
E° for In³⁺/In° = -0.16 V
E° for Tl³⁺/Tl° = +1.6 V
Hence, TlI is more stable than TlI₃
82. Given below are two statements:
Statement I: A unit formed by the attachment of a base to 1' position of sugar is known as nucleoside.
Statement II: When nucleoside is linked to phosphorous acid at 5'-position of sugar moiety, we get nucleotide.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (4)
Sol. A unit formed by the attachment of a base to 1' position of sugar is known as nucleoside. In nucleosides, the sugar carbons are numbered as 1', 2', 3', etc. in order to distinguish these from the bases. When nucleoside is linked to phosphoric acid at 5'-position of sugar moiety, we get a nucleotide.
83. Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is :
(1) 18
(2) 16
(3) 32
(4) 30
Answer (3)
Sol. This reaction is called soda lime decarboxylation
CH₃-C(=O)-ONa⁺ + NaOH → CH₄(g) + Na₂CO₃(s)
Methane
Sodium ethanoate
Molar mass of CH₄ = 16 g/mol
Weight of 2 moles of CH₄ = 16 × 2 = 32 g
84. The relation between n_m (n_m = the number of permissible values of magnetic quantum number (m)) for a given value of azimuthal quantum number (l), is
(1) n_m = 2l² + 1
(2) l = (n_m - 1)/2
(3) l = 2n_m + 1
(4) n_m = l + 2
Answer (2)
Sol. Values of n_m (magnetic quantum number) for given azimuthal quantum number can be calculated as following
n_m = 2l + 1
l = (n_m - 1)/2
85. The number of σ bonds, π bonds and lone pair of electrons in pyridine, respectively are:
(1) 12, 2, 1
(2) 11, 2, 0
(3) 12, 3, 0
(4) 11, 3, 1
Answer (4)
Sol. Pyridine
No. of σ bonds = 11
No. of π bonds = 3
No. of lone pair of e⁻ = 1
86. Identify the major product obtained in the following reaction:
(1)
(2)
(3)
(4)
Answer (4)
Sol. Ammoniacal silver nitrate forms silver metal in alkaline medium. Aldehyde is oxidised to corresponding carboxylate ion.
87. Pumice stone is an example of
(1) Foam
(2) Sol
(3) Gel
(4) Solid sol
Answer (4)
Sol. Pumice stone is a solid sol.
Dispersed phase : Gas
Dispersed medium : Solid
88. Consider the following reaction :
Identify products A and B.
(1) A = CH₃ and B = I
(2) A = CH₃ and B = OH
(3) A = CH₂OH and B = I
(4) A = CH₂I and B = OH
Answer (4)
Sol.
(Stable carbocation due to extended conjugation)
89. The reaction that does NOT take place in a blast furnace between 900 K to 1500 K temperature range during extraction of iron is :
(1) CaO + SiO₂ → CaSiO₃
(2) Fe₂O₃ + CO → 2FeO + CO₂
(3) FeO + CO → Fe + CO₂
(4) C + CO₂ → 2CO
Answer (2)
Sol. At 900 - 1500 K (higher temperature range in the blast furnace)
Reaction which take place are :-
C + CO₂ → 2CO
FeO + CO → Fe + CO₂
CaO + SiO₂ → CaSiO₃ (Slag formation)
Fe₂O₃ + CO → 2FeO + CO₂ → This take place at 500-800 K.
90. Which of the following statements are INCORRECT?
A. All the transition metals except scandium form MO oxides which are ionic.
B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc₂O₃ to Mn₂O₇
C. Basic character increases from V₂O₃ to V₂O₄ to V₂O₅
D. V₂O₄ dissolves in acids to give VO₄³⁻ salts.
E. CrO is basic but Cr₂O₃ is amphoteric.
Choose the correct answer from the options given below:
(1) B and C only
(2) A and E only
(3) B and D only
(4) C and D only
Answer (4)
Sol. All transitions metals except Sc form MO oxides which are ionic.
The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc₂O₃ to Mn₂O₇
Acidic character increases from V₂O₃ to V₂O₄ to V₂O₅
V₂O₄ dissolves in acids to give VO²⁺
CrO is basic but Cr₂O₃ is amphoteric.
91. Which complex compound is most stable?
(1) [Co(NH₃)₆][SO₄]₃
(2) [Co(NH₃)₄(H₂O)Br][NO₃]₂
(3) [Co(NH₃)₃(NO₃)₃]
(4) [CoCl₂(en)₂]NO₃
Answer (4)
Sol. Chelating ligands in general form more stable complexes than their monodentate analogs
The most stable complex is [CoCl₂(en)₂]NO₃
92. What fraction of one edge centred octahedral void lies in one unit cell of fcc?
(1) 1/12
(2) 1/2
(3) 1/3
(4) 1/4
Answer (4)
Sol. The total number of octahedral voids in FCC are four
Octahedral voids in FCC = Edge centres + Body centre
The contribution of edge centre = 1/4 : Fraction of one edge centred octahedral void in one unit cell of FCC = 1/4
93. Match List-1 with List-1I:
List-1 (Oxoacids of Sulphur) List-1I (Bonds)
A. Peroxodisulphuric acid I. Two S-OH, Four S=O, One S-O-S
B. Sulphuric acid II. Two S-OH, One S=O
C. Pyrosulphuric acid III. Two S-OH, Four S=O, One S-O-O-S
D. Sulphurous acid IV. Two S-OH, Two S=O
Choose the correct answer from the options given below.
(1) A-III, B-IV, C-II, D-I
(2) A-I, B-III, C-II, D-IV
(3) A-III, B-IV, C-I, D-II
(4) A-I, B-III, C-IV, D-II
Answer (3)
Sol. Peroxodisulphuric acid (H₂S₂O₈)
Sulphuric acid (H₂SO₄)
Pyrosulphuric acid (H₂S₂O₇)
Sulphurous acid (H₂SO₃)
94. Which amongst the following will be most readily dehydrated under acidic conditions?
(1)
(2)
(3)
(4)
Answer (3)
95. The equilibrium concentrations of the species in the reaction A + B ⇌ C + D are 2, 3, 10 and 6 mol L⁻¹, respectively at 300 K. ΔG° for the reaction is (R = 2 cal/mol K)
(1) -13.73 cal
(2) 1372.60 cal
(3) -137.26 cal
(4) -1381.80 cal
Answer (4)
Sol. A + B ⇌ C + D
at equilibrium 2 3 10 6
k_eq = [C][D]/[A][B]
k_eq = (10 × 6)/(2 × 3) = 10
ΔG° = -RT ln K
= -2.303 RT log K
= -2.303 × 2 × 300 × log 10 = -1381.8 cal
96. Given below are two statements:
Statement I: The nutrient deficient water bodies lead to eutrophication
Statement II: Eutrophication leads to decrease in the level of oxygen in the water bodies.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is incorrect but Statement II is true.
(2) Both Statement I and Statement II are true.
(3) Both Statement I and Statement II are false.
(4) Statement I is correct but Statement II is false.
Answer (1)
Sol. Nutrient enriched water bodies support a dense plant population, which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity. This process is called as eutrophication.
97. Consider the following compounds/species:
i. ii. iii. iv. v. vi. vii.
The number of compounds/species which obey Huckel's rule is
(1) 5
(2) 4
(3) 6
(4) 2
Answer (2)
Sol. Criteria for Huckel's rule:
(i) Planarity
(ii) Complete delocalisation of π electrons
(iii) Presence of (4n + 2)π electrons in ring where n is an integer (n = 0, 1, 2, ...)
The compounds which follow Huckel's rule are:
98. Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
(1) ΔH + ΔU = ΔnR
(2) ΔH = ΔU - Δn_gRT
(3) ΔH = ΔU + Δn_gRT
(4) ΔH - ΔU = -ΔnRT
Answer (3)
Sol. Correct relation between change in enthalpy and change in internal energy is
ΔH = ΔU + Δn_gRT
99. Identify the final product [D] obtained in the following sequence of reactions.
(1) HCIO
(2) Na/dry ether
(3)
(4) C₄H₁₀
Answer (2)
Sol. CH₃CHO →(i) LiAlH₄ (ii) H₂O⁺→ CH₃CH₂OH →H₂SO₄→ CH₂=CH₂ →Bromination→ CH₃CH₂Br →Na/dry ether→ Wurtz-Fittig Reaction→ [C]
100. On balancing the given redox reaction,
aCr₂O₇²⁻ + bSO₃²⁻(aq) + cH⁺(aq) → 2aCr³⁺(aq) + bSO₄²⁻(aq) + c/2 H₂O(l)
the coefficients a, b and c are found to be, respectively-
(1) 8, 1, 3
(2) 1, 3, 8
(3) 3, 8, 1
(4) 1, 8, 3
Answer (2)
Sol. Using Ion electron method :
Reduction Half reaction : Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺
Oxidation Half reaction : SO₃²⁻ → SO₄²⁻ + 2e⁻ × 3
Overall reaction : Cr₂O₇²⁻ + 3SO₃²⁻ → 2Cr³⁺ + 3SO₄²⁻
To balance 'O' atoms, adding H₂O on LHS
Cr₂O₇²⁻ + 3SO₃²⁻ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
To balance 'H' atoms, adding H⁺ on RHS
Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
a = 1, b = 3, c = 8
101. Movement and accumulation of ions across a membrane against their concentration gradient can be explained by
(1) Active Transport
(2) Osmosis
(3) Facilitated Diffusion
(4) Passive Transport
Answer (1)
Sol. Movement and accumulation of ions across a membrane against their concentration gradient can be explained by active transport. It uses energy to transport molecules from lower concentration to a higher concentration.
102. Among 'The Evil Quartet', which one is considered the most important cause driving extinction of species?
(1) Co-extinctions
(2) Habitat loss and fragmentation
(3) Over exploitation for economic gain
(4) Alien species invasions
Answer (2)
Sol. Habitat loss and fragmentation is the most important cause driving animals and plants to extinction.
103. Expressed Sequence Tags (ESTs) refers to
(1) Certain important expressed genes.
(2) All genes that are expressed as RNA.
(3) All genes that are expressed as proteins.
(4) All genes whether expressed or unexpressed.
Answer (2)
Sol. All the genes that are expressed as RNA are referred to as Expressed Sequence Tags (ESTs).
104. The phenomenon of pleiotropism refers to
(1) More than two genes affecting a single character
(2) Presence of several alleles of a single gene controlling a single crossover
(3) Presence of two alleles, each of the two genes controlling a single trait
(4) A single gene affecting multiple phenotypic expression
Answer (4)
Sol. When a single gene affects multiple phenotypic expression, the gene is called pleiotropic gene and the phenomenon is called pleiotropism.
105. In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This phenomenon may be called as
(1) Senescence
(2) Differentiation
(3) Dedifferentiation
(4) Development
Answer (3)
Sol. In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This phenomenon may be called as dedifferentiation. Dedifferentiation is a phenomenon by which the living differentiated plant cells, that by now have lost the capacity to divide can regain the capacity of division under certain conditions.
106. Given below are two statements:
Statement I: Endarch and exarch are the terms often used for describing the position of secondary xylem in the plant body.
Statement II: Exarch condition is the most common feature of the root system.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is incorrect but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is correct but Statement II is false
Answer (1)
Sol. Endarch and exarch are the terms often used for describing the position of primary xylem in the plant body. Primary xylem is of two types protoxylem and metaxylem. On the basis of relative position of protoxylem and metaxylem in the organ the arrangement of primary xylem can be endarch or exarch. Exarch type of primary xylem is seen in roots.
Therefore, Statement I is false and Statement II is true.
107. Given below are two statements: One labelled as Assertion A and the other labelled as Reason R:
Assertion A: The first stage of gametophyte in the life cycle of moss is protonema stage.
Reason R: Protonema develops directly from spores produced in capsule.
In the light of the above statements, choose the most appropriate answer from options given below:
(1) A is not correct but R is correct
(2) Both A and R are correct and R is the correct explanation of A
(3) Both A and R are correct but R is NOT the correct explanation of A
(4) A is correct but R is not correct
Answer (2)
Sol. The predominant stage of the life cycle of a moss is the gametophyte which consists of two stages. The first stage is the protonema stage, which develops directly from a spore. Capsule of the sporophyte contains spore which gives rise to protonema. Thus, reason correctly explains the assertion.
108. Which of the following stages of meiosis involves division of centromere?
(1) Telophase
(2) Metaphase I
(3) Metaphase II
(4) Anaphase II
Answer (4)
Sol. Splitting of centromere occurs during anaphase of mitosis or anaphase II of meiosis. During Metaphase I and II, chromosomes align at the equator. During telophase, chromosomes reach the respective poles.
109. Upon exposure to UV radiation, DNA stained with ethidium bromide will show
(1) Bright orange colour
(2) Bright red colour
(3) Bright blue colour
(4) Bright yellow colour
Answer (1)
Sol. Option (1) is the correct answer because in recombinant DNA technology the separated DNA fragments can be visualised only after staining the DNA with a substance known as ethidium bromide followed by exposure to U.V. radiation. You can see bright orange coloured bands of DNA in an ethidium bromide stained gel exposed to U.V. light.
110. The process of appearance of recombination nodules occurs at which sub stage of prophase I in meiosis?
(1) Diakinesis
(2) Zygotene
(3) Pachytene
(4) Diplotene
Answer (3)
Sol. The process of recombination occurs at Pachytene stage of prophase I. This stage is characterised by the appearance of recombination nodules.
111. Cellulose does not form blue colour with Iodine because
(1) It breaks down when iodine reacts with it
(2) It is a disaccharide
(3) It is a helical molecule
(4) It does not contain complex helices and hence cannot hold iodine molecules
Answer (4)
Sol. Option (4) is the correct answer because cellulose does not contain complex helices and hence cannot hold iodine molecules. Option (1), (2) and (3) are not correct as cellulose is a polysaccharide.
112. Family Fabaceae differs from Solanaceae and Liliaceae. With respect to the stamens, pick out the characteristics specific to family Fabaceae but not found in Solanaceae or Liliaceae.
(1) Epiphyllous and Dithecous anthers
(2) Diadelphous and Dithecous anthers
(3) Polyadelphous and epipelatous stamens
(4) Monoadelphous and Monothecous anthers
Answer (2)
Sol. Fabaceae → Diadelphous and dithecous anther.
Solanaceae → Polyandrous, epipetalous and dithecous anther.
Liliaceae → Polyandrous, epiphyllous and dithecous anther.
113. The thickness of ozone in a column of air in the atmosphere is measured in terms of :
(1) Kilobase
(2) Dobson units
(3) Decibels
(4) Decameter
Answer (2)
Sol. The thickness of the ozone in a column of air from the ground to the top of the atmosphere is measured in terms of Dobson units (DU). Noise is measured in decibels.
114. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion A : ATP is used at two steps in glycolysis. Reason R : First ATP is used in converting glucose into glucose-6-phosphate and second ATP is used in conversion of fructose-6-phosphate into fructose-1, 6-diphosphate. In the light of the above statements, choose the correct answer from the options given below : (1) A is false but R is true. (2) Both A and R are true and R is the correct explanation of A. (3) Both A and R are true but R is NOT the correct explanation of A. (4) A is true but R is false. Answer (2)
115. How many ATP and NADPH2 are required for the synthesis of one molecule of Glucose during Calvin cycle?
(1) 18 ATP and 16 NADPH2
(2) 12 ATP and 12 NADPH2
(3) 18 ATP and 12 NADPH2
(4) 12 ATP and 16 NADPH2
Answer (3)
Sol. For every CO₂ molecule entering the Calvin cycle, 3 molecules of ATP and 2 of NADPH2 are required. To make one molecule of glucose, 6 turns of the cycle are required. Thus, ATP and NADPH2 molecules required for synthesis of one molecule of glucose during Calvin cycle will be → 6 × [3 ATP / 2 NADPH2] = 18 ATP / 12 NADPH2
116. During the purification process for recombinant DNA technology, addition of chilled ethanol precipitates out
(1) Polysaccharides
(2) RNA
(3) DNA
(4) Histones
Answer (3)
Sol. Option (3) is the correct answer as, during isolation of the genetic material, purified DNA ultimately precipitates out after the addition of chilled ethanol. Option (4) is not the answer as, proteins can be removed by treatment with proteases. Option (2) is not the answer as RNA can be removed by treatment with ribonuclease.
117. Spraying of which of the following phytohormone on juvenile conifers helps hastening the maturity period, that leads early seed production?
(1) Abscisic Acid
(2) Indole-3-butyric Acid
(3) Gibberellic Acid
(4) Zeatin
Answer (3)
Sol. Spraying juvenile conifers with gibberellins (GAs) hastens the maturity period, thus leading to early seed production.
118. In the equation GPP - R = NPP
GPP is Gross Primary Productivity
NPP is Net Primary Productivity
R here is
(1) Reproductive allocation
(2) Respiratory quotient
(3) Respiratory loss
Answer (3)
Sol. A considerable amount of GPP is utilised by plants in respiration. Gross primary productivity minus respiration losses (R), is the net primary productivity. So R = Respiratory loss
119. Axile placentation is observed in
(1) China rose, Petunia and Lemon
(2) Mustard, Cucumber and Primrose
(3) China rose, Beans and Lupin
(4) Tomato, Dianthus and Pea
Answer (1)
Sol. China rose, Tomato, Petunia and Lemon show axile placentation. Dianthus and Primrose show free central placentation. Pea, Lupin and Beans show marginal placentation. Cucumber and mustard show parietal placentation.
120. Unequivocal proof that DNA is the genetic material was first proposed by
(1) Wilkins and Franklin
(2) Frederick Griffith
(3) Alfred Hershey and Martha Chase
(4) Avery, Macleoid and McCarthy
Answer (3)
Sol. The unequivocal proof that DNA is the genetic material came from the experiment of Alfred Hershey and Martha Chase. Avery, Macleoid and McCarthy gave the biochemical characterisation of Transforming Principle. The transformation experiments by using Pneumococcus was conducted by Frederick Griffith. Wilkins and Franklin produced X-ray diffraction data of DNA.
121. What is the function of tassels in the corn cob?
(1) To protect seeds
(2) To attract insects
(3) To trap pollen grains
(4) To disperse pollen grains
Answer (3)
Sol. Tassels in the corn cob represents stigma and style which wave in the wind to trap pollen grains.
122. Which micronutrient is required for splitting of water molecule during photosynthesis?
(1) Copper
(2) Manganese
(3) Molybdenum
(4) Magnesium
Answer (2)
Sol. Manganese plays a major role in the splitting of water to liberate oxygen during photosynthesis.
Copper is essential for the overall metabolism in plants.
Molybdenum is included in nitrogen metabolism.
Magnesium activates several enzymes involved in photosynthesis and respiration.
123. Given below are two statements: Statement I : The forces generated transpiration can lift a xylem-sized column of water over 130 meters height. Statement II : Transpiration cools leaf surfaces sometimes 10 to 15 degrees evaporative cooling. In the light of the above statements, choose the most appropriate answer from the options given below : (1) Statement I is incorrect but Statement II is correct (2) Both Statement I and Statement II are correct (3) Both Statement I and Statement II are incorrect (4) Statement I is correct but Statement II is incorrect Answer (2)
Sol. Statement I is correct as measurements reveal that the forces generated by transpiration can create pressures sufficient to lift a xylem sized column of water up to 130 meters high. Statement II is also correct as transpiration cools leaf surfaces, sometimes 10 to 15 degrees, by evaporative cooling.
124. Among eukaryotes, replication of DNA takes place in :
(1) G₂ phase
(2) M phase
(3) S phase
(4) G₁ phase
Answer (3)
Sol. Replication of DNA takes place in S-phase of cell cycle in eukaryotes. Most of the cell organelles duplicate in G₁ phase.
125. In gene gun method used to introduce alien DNA into host cells, microparticles of metal are used.
(1) Silver
(2) Copper
(3) Zinc
(4) Tungsten or gold
Answer (4)
Sol. Option (4) is the correct answer because in gene gun method, microparticles of tungsten or gold are used. Gold or tungsten are inert in nature so they do not alter the chemical composition of cells.
126. The reaction centre in PS II has an absorption maxima at
(1) 780 nm
(2) 680 nm
(3) 700 nm
(4) 660 nm
Answer (2)
Sol. In PS-I, the reaction centre chlorophyll a has an absorption peak at 700 nm, while in PS-II, reaction centre has an absorption maxima at 680 nm.
127. Frequency of recombination between gene pairs on same chromosome as a measure of the distance between genes to map their position on chromosome, was used for the first time by
(1) Henking
(2) Thomas Hunt Morgan
(3) Sutton and Boveri
(4) Alfred Sturtevant
Answer (4)
Sol. Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and 'mapped' their position on the chromosome. Sutton and Boveri proposed chromosomal theory of inheritance. Henking discovered X-chromosome. Thomas Hunt Morgan proved chromosomal theory of inheritance and proposed the concept of linkage.
128. The historic Convention on Biological Diversity, 'The Earth Summit' was held in Rio de Janeiro in the year
(1) 2002
(2) 1985
(3) 1992
(4) 1986
Answer (3)
Sol. The historic convention on Biological Diversity, "The Earth Summit" was held in Rio de Janeiro in the year 1992. It called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilisation of its benefits.
129. Identify the correct statements:
A. Detritivores perform fragmentation.
B. The humus is further degraded by some microbes during mineralization.
C. Water soluble inorganic nutrients go down into the soil and get precipitated by a process called leaching.
D. The detritus food chain begins with living organisms.
E. Earthworms break down detritus into smaller particles by a process called catabolism.
Choose the correct answer from the options given below:
(1) D, E, A only
(2) A, B, C only
(3) B, C, D only
(4) C, D, E only
Answer (2)
Sol. The detritus food chain begins with detritus that is dead organic matter. The saprotrophic bacteria and fungi breakdown detritus into simpler inorganic substances by a process called catabolism.
130. Identify the pair of heterosporous pteridophytes among the following:
(1) Equisetum and Salvinia
(2) Lycopodium and Selaginella
(3) Selaginella and Salvinia
(4) Psilotum and Salvinia
Answer (3)
Sol. Selaginella and Salvinia are heterosporous pteridophytes. They produces two different kind of spores. Psilotum, Lycopodium and Equisetum are homosporous pteridophytes.
131. In angiosperm, the haploid, diploid and triploid structures of a fertilized embryo sac sequentially are :
(1) Synergids, antipodals and Polar nuclei
(2) Synergids, Primary endosperm nucleus and zygote
(3) Antipodals, synergids, and primary endosperm nucleus
(4) Synergids, Zygote and Primary endosperm nucleus
Answer (4)
132. What is the role of RNA polymerase III in the process of transcription in Eukaryotes?
(1) Transcription of only snRNAs
(2) Transcription of rRNAs (28S, 18S and 5.8S)
(3) Transcription of tRNA, 5S rRNA and snRNA
(4) Transcription of precursor of mRNA
Answer (3)
Sol. In eukaryotes there are three major types of RNA polymerases.
RNA polymerase I transcribes : 5.8S, 18S, 28S rRNAs
RNA polymerase II transcribes : hnRNAs (precursor of mRNA)
RNA polymerase III transcribes : tRNAs, ScRNA, 5S rRNA and snRNA
133. Which hormone promotes internode/petiole elongation in deep water rice?
(1) 2,4-D
(2) GA₃
(3) Kinetin
(4) Ethylene
Answer (4)
Sol. Ethylene promotes rapid internode/petiole elongation in deep water rice plants.
134. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Late wood has fewer xylary elements with narrow vessels. Reason R : Cambium is less active in winters. In the light of the above statements, choose the correct answer from the options given below : (1) A is false but R is true (2) Both A and R are true and R is the correct explanation of A (3) Both A and R are true but R is NOT the correct explanation of A (4) A is true but R is false Answer (2)
Sol. In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels, and this wood is called autumn wood or late wood.
135. Large, colourful, fragrant flowers with nectar are seen in
(1) Wind pollinated plants
(2) Insect pollinated plants
(3) Bird pollinated plants
(4) Bat pollinated plants
Answer (2)
Sol. Large, colourful, fragrant flowers with nectar attract biotic pollinators (insects), thus, they are seen in insect pollinated plants.
136. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion A : A flower is defined as modified shoot wherein the shoot apical meristem changes to floral meristem. Reason R : Internode of the shoot gets condensed to produce different floral appendages laterally at successive node instead of leaves. In the light of the above statements, choose the correct answer from the options given below : (1) A is false but R is true (2) Both A and R are true and R is the correct explanation of A (3) Both A and R are true but R is NOT the correct explanation of A (4) A is true but R is false Answer (2)
Sol. A flower is a modified shoot wherein the shoot apical meristem changes to floral meristem. Internodes do not elongate and the axis gets condensed. The apex produces different kinds of floral appendages laterally at the successive nodes instead of leaves. Therefore, both A and R are true and R is correct explanation of A.
137. Which of the following combinations is required for chemiosmosis?
(1) Proton pump, electron gradient, NADP synthase
(2) Membrane, proton pump, proton gradient, ATP synthase
(3) Membrane, proton pump, proton gradient, NADP synthase
(4) Proton pump, electron gradient, ATP synthase
Answer (2)
Sol. Chemiosmosis requires a membrane, a proton pump, a proton gradient and ATP synthase.
138. How many different proteins does the ribosome consist of?
(1) 20
(2) 80
(3) 60
(4) 40
Answer (2)
Sol. The ribosome consists of structural RNAs and about 80 different proteins.
139. Given below are two statements : One labelled as Assertion A and the other labelled as Reason R : Assertion A : In gymnosperms the pollen grains are released from the microsporangium and carried by air currents. Reason R : Air currents carry the pollen grains to the mouth of the archegonia where the male gametes are discharged and pollen tube is not formed. In the light of the above statements, choose the correct answer from the options given below : (1) A is false but R is true (2) Both A and R are true and R is the correct explanation of A (3) Both A and R are true but R is NOT the correct explanation of A (4) A is true but R is false Answer (4)
Sol. Assertion is correct but reason is false as in gymnosperms the pollen grains are released from the microsporangium and they are carried in air currents. They come in contact with the opening of the ovules borne on megasporophylls. The pollen tube carrying the male gametes grows towards archegonia in the ovules and discharge their contents near the mouth of the archegonia.
140. Match List I with List II:
List I List II
A. Cohesion I. More attraction in liquid phase
B. Adhesion II. Mutual attraction among water molecules
C. Surface tension III. Water loss in liquid phase
D. Guttation IV. Attraction towards polar surfaces
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-II, B-IV, C-I, D-III
(3) A-IV, B-II, C-III, D-I
(4) A-II, B-I, C-IV, D-III
Answer (2)
Sol. Cohesion represents mutual attraction between water molecules. Adhesion represents attraction of water molecules to polar surfaces. Surface tension represents water molecules are attracted to each other in the liquid phase more than to water in the gas phase. Guttation represent loss of water in liquid phase. Thus, option (2) is correct.
141. Which one of the following statements is NOT correct?
(1) The amount of some toxic substances of industrial waste water increases in the organisms at successive trophic levels
(2) The micro-organisms involved in biodegradation of organic matter in a sewage polluted water body consume a lot of oxygen causing the death of aquatic organisms
(3) Algal blooms caused by excess of organic matter in water improve water quality and promote fisheries
(4) Water hyacinth grows abundantly in eutrophic water bodies and leads to an imbalance in the ecosystem dynamics of the water body
Answer (3)
Sol. Algal bloom imparts a distinct colour to the water bodies. It causes deterioration of the water quantity and fish mortality.
142. Which of the following statements are correct about Klinefelter's Syndrome?
A. This disorder was first described by Langdon Down (1866).
B. Such an individual has overall masculine development. However, the feminine development is also expressed.
C. The affected individual is short statured.
D. Physical, psychomotor and mental development is retarded.
E. Such individuals are sterile.
Choose the correct answer from the options given below:
(1) A and E only
(2) A and B only
(3) C and D only
(4) B and E only
Answer (4)
Sol. Klinefelter's syndrome is caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY. Such an individual has overall masculine development, however, the feminine development is also expressed. Such individuals are sterile. Thus, statement B and E are correct regarding Klinefelter's syndrome.
Statement A, C and D are incorrect w.r.t. Klinefelter's syndrome as they are associated with Down's syndrome.
143. Given below are two statements: Statement I : Gause's 'Competitive Exclusion Principle' states that two closely related species competing for the same resources cannot co-exist indefinitely and competitively inferior one will be eliminated eventually. Statement II : In general, carnivores are more adversely affected by competition than herbivores. In the light of the above statements, choose the correct answer from the options given below: (1) Statement I is incorrect but Statement II is true. (2) Both Statement I and Statement II are true. (3) Both Statement I and Statement II are false. (4) Statement I is correct Statement II is false. Answer (4)
Sol. Gause's 'Competitive Exclusion Principle' states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually. Thus, statement I is correct.
Statement II is incorrect as in general, herbivores and plants appear to be more adversely affected by competition than carnivores.
144. Match List I with List II :
List I List II
A. M Phase I. Proteins are synthesized
B. G₂ Phase II. Inactive phase
C. Quiescent stage III. Interval between mitosis and initiation of DNA replication
D. G₁ Phase IV. Equational division
Choose the correct answer from the options given below :
(1) A-II, B-IV, C-I, D-III
(2) A-II, B-III, C-IV, D-I
(3) A-IV, B-II, C-I, D-III
(4) A-IV, B-I, C-II, D-III
Answer (4)
Sol. M phase or mitosis is the phase where the actual cell division occurs. Mitosis is also called equational division.
During G₂ phase DNA synthesis stops but cell synthesis RNA, proteins, etc. for next phase.
Quiescent stage is inactive phase in which non-dividing cells enters.
G₁ phase is the interval between mitosis and initiation of DNA replication.
Therefore, option (4) is correct.
145. Identify the correct statements:
A. Lenticels are the lens-shaped openings permitting the exchange of gases.
B. Bark formed early in the season is called hard bark.
C. Bark is a technical term that refers to all tissues exterior to vascular cambium.
D. Bark refers to periderm and secondary phloem.
E. Phellogen is single-layered in thickness.
Choose the correct answer from the options given below:
(1) B and C only
(2) B, C and E only
(3) A and D only
(4) A, B and D only
Answer (3)
Sol. Lenticels are lens shaped opening permitting exchange of gases between the outer atmosphere and internal tissue of the stem. Bark that is formed early in the season is called early or soft bark. Towards the end of the season late or hard bark is formed. Bark is non-technical term that refer to all tissues exterior to vascular cambium. Bark refers to a number of tissue types, viz periderm and secondary phloem. Phellogen is couple of layers thick Therefore, only statement A and D are correct.
146. Main steps in the formation of Recombinant DNA are given below. Arrange these steps in a correct sequence.
A. Insertion of recombinant DNA into the host cell
B. Cutting of DNA at specific location by restriction enzyme
C. Isolation of desired DNA fragment
D. Amplification of gene of interest using PCR
Choose the correct answer from the options given below :
(1) B, D, A, C
(2) B, C, D, A
(3) C, A, B, D
(4) C, B, D, A
Answer (2)
Sol. The correct answer is option (2) because recombinant DNA technology involves several steps in specific sequence such as isolation of DNA, fragmentation of DNA by restriction endonucleases, isolation of desired DNA fragment, ligation of the DNA fragment into a vector, transferring the recombinant DNA into the host, culturing the host cells in a medium at large scale and extraction of the desired product.
147. Match List I with List II :
List I List II (Interaction) (Species A and B)
A. Mutualism I. +(A), 0(B)
B. Commensalism II. -(A), 0(B)
C. Amensalism III. +(A), -(B)
D. Parasitism IV. +(A), +(B)
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-IV, B-II, C-I, D-III
(3) A-IV, B-I, C-II, D-III
(4) A-IV, B-II, C-I, D-III
Answer (3)
Sol. (+, +) Mutualism : In this interaction, both the interacting species are benefitted.
(+, 0) Commensalism : Only one species is benefitted and the other species remains unharmed. (-, 0) Amensalism : Neither species is benefitted. One remains unharmed and the other is harmed. (+, -) Parasitism : One species is benefitted and other is negatively effected.
148. Melonate inhibits the growth of pathogenic bacteria by inhibiting the activity of
(1) Dinitrogenase
(2) Succinic dehydrogenase
(3) Amylase
(4) Lipase
Answer (2)
Sol. Option (2) is correct answer of this question because malonate is a competitive inhibitor of enzyme succinate dehydrogenase. Inhibition of succinic dehydrogenase by malonate occurs due to close resemblance of malonate with substrate succinate in structure. Competitive inhibitors are often used in the control of bacterial pathogens.
149. Match List I with List II:
List I List II
A. Iron I. Synthesis of auxin
B. Zinc II. Component of nitrate reductase
C. Boron III. Activator of catalase
D. Molybdenum IV. Cell elongation and differentiation
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-I, D-III
(2) A-III, B-II, C-I, D-IV
(3) A-II, B-III, C-IV, D-I
(4) A-III, B-I, C-IV, D-II
Answer (4)
Sol. Iron activates catalase enzyme.
Zinc is needed in the synthesis of auxin. Boron is required for cell elongation and cell differentiation. Molybdenum is component of nitrogenase and nitrate reductase enzyme. Therefore, option (4) is correct.
150. Match List I with List II:
List I List II
A. Oxidative decarboxylation I. Citrate synthase
B. Glycolysis II. Pyruvate dehydrogenase
C. Oxidative phosphorylation III. Electron transport system
D. Tricarboxylic acid cycle IV. EMP pathway
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-III, D-I
(2) A-II, B-IV, C-I, D-III
(3) A-I, B-IV, C-II, D-III
(4) A-III, B-I, C-II, D-IV
Answer (1)
Sol. Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalyzed by pyruvate dehydrogenase.
The scheme of glycolysis was given by Gustav Embden, Otto Meyrhoff and J. Parnas, and is often referred to as the EMP pathway.
In electron transport system, the energy of oxidation-reduction is utilized for the production of proton gradient required for phosphorylation, thus, this process is also called oxidative phosphorylation.
The TCA (tricarboxylic acid cycle) starts with the condensation of acetyl group with oxaloacetic acid (OAA) and water to yield citric acid. The reaction is catalysed by the enzyme citrate synthase. Thus, option (1) is correct.
151. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Nephrons are of two types: Cortical & Juxta medullary, based on their relative position in cortex and medulla. Reason R: Juxta medullary nephrons have short loop of Henle whereas, cortical nephrons have longer loop of Henle. In the light of the above statements, choose the correct answer from the options given below: (1) A is false but R is true. (2) Both A and R are true and R is the correct explanation of A. (3) Both A and R are true but R is NOT the correct explanation of A. (4) A is true but R is false. Answer (4)
Sol. The correct answer is option (4) because Assertion is true as there are two types of nephrons, i.e., cortical nephrons and juxtamedullary nephrons based on their relative position in the cortex and medulla. Reason is not correct as loop of Henle in juxtamedullary nephrons is very long and runs deep into the medulla. Therefore, Assertion is true but Reason is false.
152. Match List I with List II with respect to human eye.
List I List II
A. Fovea I. Visible coloured portion of eye that regulates diameter of pupil.
B. Iris II. External layer of eye formed of dense connective tissue.
C. Blind spot III. Point of greatest visual acuity or resolution.
D. Sclera IV. Point where optic nerve leaves the eyeball and photoreceptor cells are absent.
Choose the correct answer from the options given below:
(1) A-II, B-III, C-II, D-IV
(2) A-III, B-I, C-IV, D-II
(3) A-IV, B-I, C-II, D-I
(4) A-II, B-IV, C-I, D-II
Answer (2)
Sol. Option (2) is the correct answer because
(i) Fovea is the point of greatest visual acuity or resolution.
(ii) Iris is the visible coloured portion of the eye that regulates diameter of pupil.
(iii) Blind spot is the point where optic nerve leaves the eye-ball and photoreceptor cells are absent.
(iv) Sclera is the external layer of eye formed of dense connective tissue.
153. Which of the following functions is carried out by cytoskeleton in a cell?
(1) Transportation
(2) Nuclear division
(3) Protein synthesis
(4) Motility
Answer (4)
Sol. An elaborate network of filamentous proteinaceous structures consisting of microtubules, microfilaments and intermediate filaments present in cytoplasm is collectively referred to as the cytoskeleton. It is involved in many functions such as mechanical support, motility, maintenance of the shape of the cell.
154. Given below are two statements:
Statement I: Ligaments are dense irregular tissue.
Statement II: Cartilage is dense regular tissue.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Answer (3)
Sol. Option (3) is the correct answer because ligament is an example of dense regular connective tissue so Statement I is incorrect and cartilage is an example of specialised connective tissue and not dense regular tissue. Therefore Statement II is also incorrect.
155. Match List I with List II.
List I List II
A. Heroin I. Effect on cardiovascular system
B. Marijuana II. Slow down body function
C. Cocaine III. Painkiller
D. Morphine IV. Interfere with transport of dopamine
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-I, B-II, C-III, D-IV
(4) A-IV, B-III, C-II, D-I
Answer (2)
Sol. The correct answer is option (2) as
Heroin belongs to the category of opioids and it is a depressant that slows down body functions. Marijuana is known for its effect on the cardiovascular system of the body. Cocaine interferes with the transport of the neurotransmitter dopamine. Morphine is used is a sedative and painkiller.
156. Which of the following is not a cloning vector?
(1) Probe
(2) BAC
(3) YAC
(4) pBR322
Answer (1)
Sol. Option (1) is correct answer because a single stranded DNA or RNA tagged with a radioactive molecule is called a probe and it helps in the detection of mutated gene.
Option (2), (3) and (4) are not correct because YAC, BAC, pBR322 are vectors.
157. Which of the following are NOT considered as the part of endomembrane system?
A. Mitochondria
B. Endoplasmic reticulum
C. Chloroplasts
D. Golgi complex
E. Peroxisomes
Choose the most appropriate answer from the options given below:
(1) A, D and E only
(2) B and D only
(3) A, C and E only
(4) A and D only
Answer (3)
Sol. The endomembrane system include endoplasmic reticulum (ER), golgi complex, lysosomes and vacuoles.
Since the functions of the mitochondria, chloroplast and peroxisomes are not coordinated with the above components, these are not considered as part of endomembrane system.
158. Which of the following statements are correct regarding female reproductive cycle?
A. In non-primate mammals cyclical changes during reproduction are called oestrus cycle.
B. First menstrual cycle begins at puberty and is called menopause.
C. Lack of menstruation may be indicative of pregnancy.
D. Cyclic menstruation extends between menarche and menopause.
Choose the most appropriate answer from the options given below.
(1) A, C and D only
(2) A and D only
(3) A and B only
(4) A, B and C only
Answer (1)
Sol. The correct answer is option (1) as first menstrual cycle that begins at puberty is called menarche. Cyclic menstruation is an indicator of normal reproductive phase and extends between menarche and menopause. In primates, cyclical changes during reproduction are called menstrual cycle.
159. Which one of the following techniques does not serve the purpose of early diagnosis of a disease for its early treatment?
(1) Enzyme Linked Immuno-Sorbent Assay (ELISA) technique
(2) Recombinant DNA Technology
(3) Serum and Urine analysis
(4) Polymerase Chain Reaction (PCR) technique
Answer (3)
Sol. The correct answer is option (3) because using conventional methods of diagnosis like serum and urine analysis, etc, do not help in early diagnosis. Recombinant DNA technology, Polymerase Chain Reaction [PCR] and Enzyme Linked Immuno-Sorbent Assay (ELISA) are some of the techniques that serve the purpose of early diagnosis.
160. Match List I with List II.
List I List II
A. Ringworm I. Haemophilus influenzae
B. Filariasis II. Trichophyton
C. Malaria III. Wuchereria bancrofti
D. Pneumonia IV. Plasmodium vivax
Choose the correct answer from the options given below:
(1) A-III, B-II, C-IV, D-I
(2) A-II, B-III, C-IV, D-I
(3) A-II, B-III, C-I, D-IV
(4) A-III, B-II, C-I, D-IV
Answer (2)
Sol. Option (2) is the correct answer because:
(i) Ringworm is caused by Trichophyton.
(ii) Filariasis is caused by Wuchereria bancrofti.
(iii) Malaria is caused by Plasmodium species.
(iv) Pneumonia is caused by Haemophilus influenzae.
161. Given below are two statements: one is labelled as Assertion A and other is labelled as Reason R. Assertion A: Amniocentesis for sex determination is one of the strategies of Reproductive and Child Health Care Programme. Reason R: Ban on amniocentesis checks increasing menace of female foeticide. In the light of the above statements, choose the correct answer from the options given below. (1) A is false but R is true. (2) Both A and R are true and R is the correct explanation of A. (3) Both A and R are true and R is NOT the correct explanation of A. (4) A is true but R is false. Answer (1)
Sol. The correct answer is option (1) as 'Reproductive and Child Health Care (RCH) programme' deals with creating awareness among people about various reproduction related aspects and providing facilities and support for building up a reproductively healthy society. Amniocentesis is basically used to test for the presence of certain genetic disorders such as Down's syndrome, haemophilia, etc., to determine the survivability of the foetus. Amniocentesis is not a sex determination technique in India and is not a strategy of RCH.
162. Given below are two statements: Statement I: RNA mutates at a faster rate. Statement II: Viruses having RNA genome and shorter life span mutate and evolve faster. In the light of the above statements, choose the correct answer from the options given below: (1) Statement I is false but Statement II is true. (2) Both Statement I and Statement II are true. (3) Both Statement I and Statement II are false. (4) Statement I is true but Statement II is false. Answer (2)
Sol. RNA being unstable, mutate at a faster rate. Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster.
163. Given below are two statements: Statement I: In prokaryotes, the positively charged DNA is held with some negatively charged proteins in a region called nucleoid. Statement II: In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form nucleosome. In the light of the above statements, choose the correct answer from the options given below: (1) Statement I is incorrect but Statement II is true. (2) Both Statement I and Statement II are true. (3) Both Statement I and Statement II are false. (4) Statement I is correct but Statement II is false. Answer (1)
Sol. In prokaryotes, the negatively charged DNA is held with some positively charged proteins in a region termed as nucleoid. In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome.
164. Match List I with List II.
List I List II
A. Taenia I. Nephridia
B. Paramoecium II. Contractile vacuole
C. Periplaneta III. Flame cells
D. Pheretima IV. Urecosse gland
Choose the correct answer from the options given below:
(1) A-III, B-II, C-IV, D-I
(2) A-II, B-I, C-IV, D-III
(3) A-I, B-II, C-III, D-IV
(4) A-IV, B-III, C-II, D-I
Answer (1)
Sol. Option (1) is the correct answer because protonephridia or flame cells are the excretory structures in platyhelminthes. Nephridia are the tubular excretory structures of earthworms (Pheretima) and other annelids. Single celled organisms like Paramoecum have contractile vacuoles for excretion Urecosse glands are present in cockroach.
165. Match List I with List II.
List I List II
A. Gene 'a' I. β-galactosidase
B. Gene 'y' II. Transacetylase
C. Gene 'i' III. Permease
D. Gene 'z' IV. Repressor protein
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-I, D-II
(2) A-III, B-II, C-IV, D-II
(3) A-II, B-III, C-IV, D-I
(4) A-I, B-II, C-IV, D-III
Answer (3)
Sol. In a lac operon,
Gene a codes for enzyme transacetylase.
Gene y codes for enzyme permease.
Gene i codes for repressor protein
Gene z codes for enzyme β-galactosidase.
166. Match List I with List II.
List I List II
A. Vasectomy I. Oral method
B. Coitus interruptus II. Barrier method
C. Cervical caps III. Surgical method
D. Saheli IV. Natural method
Choose the correct answer from the options given below:
(1) A-IV, B-II, C-I, D-III
(2) A-III, B-I, C-IV, D-II
(3) A-III, B-IV, C-II, D-I
(4) A-II, B-III, C-I, D-IV
Answer (3)
Sol. Option (3) is the correct answer because
(i) Vasectomy is a surgical method of contraception
(ii) Coitus interruptus is a natural method of contraception
(iii) Cervical cap is a barrier method of contraception
(iv) Saheli is an oral method of contraception which is a non-steroidal pill
167. Broad palm with single palm crease is visible in a person suffering from
(1) Thalassemia
(2) Down's syndrome
(3) Turner's syndrome
(4) Klinefelter's syndrome
Answer (2)
Sol. Down's syndrome is caused by an additional copy of chromosome number 21. Its symptoms include-
a. Broad palm with characteristic palm crease
b. Short stature with small round head
c. Furrowed tongue and partially open mouth, etc.
168. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Endometrium is necessary for implantation of blastocyst. Reason R: In the absence of fertilization, the corpus luteum degenerates that causes disintegration of endometrium. In the light of the above statements, choose the correct answer from the options given below: (1) A is false but R is true. (2) Both A and R are true and R is the correct explanation of A. (3) Both A and R are true but R is NOT the correct explanation of A. (4) A is true but R is false. Answer (3)
Sol. Option (3) is the correct answer because both Assertion and Reason are true. Implantation is embedding of the blastocyst into endometrium of uterus. Correct explanation of reason is Corpus luteum secretes large amount of progesterone which is essential for maintenance of endometrium of uterus. In absence of fertilisation, the corpus luteum degenerates hence the decrease in the level of progesterone hormone will cause disintegration of endometrium leading to menstruation.
169. Vital capacity of lung is
(1) IRV + ERV + TV
(2) IRV + ERV
(3) IRV + ERV + TV + RV
(4) IRV + ERV + TV - RV
Answer (1)
Sol. Option (1) is the correct answer because vital capacity is the maximum volume of air a person can breathe in after forced expiration. This includes ERV, TV and IRV.
170. Given below are two statements: Statement I: A protein is imagined as a line, the left and represented by first amino acid (C-terminal) and the right end represented by last amino acid (N-terminal). Statement II: Adult human haemoglobin, consists of 4 subunits (two subunits of α type and two subunits of β type.) In the light of the above statements, choose the correct answer from the options given below: (1) Statement I is false but Statement II is true. (2) Both Statement I and Statement II are true. (3) Both Statement I and Statement II are false. (4) Statement I is true but Statement II is false. Answer (1)
Sol. The correct answer is option (1) as a protein is imagined as a line, the left end represented by the first amino acid and the right end is represented by the last amino acid. The first amino acid is also called N-terminal amino acid. The last amino acid is called the C-terminal amino acid.
171. Which one of the following common sexually transmitted diseases is completely curable when detected early and treated properly?
(1) HIV Infection
(2) Genital herpes
(3) Gonorrhoea
(4) Hepatitis-B
Answer (3)
Sol. The correct answer is option (3) because except for hepatitis-B, genital herpes and HIV infection other STIs are completely curable if detected early and treated properly. Gonorrhoea is a bacterial disease which can be treated and cured completely, other diseases mentioned are viral diseases.
172. Given below are two statements: Statement I: Vas deferens receives a duct from seminal vesicle and opens into urethra as the ejaculatory duct. Statement II: The cavity of the cervix is called cervical canal which along with vagina forms birth canal. In the light of the above statements, choose the correct answer from the options given below: (1) Statement I is incorrect but Statement II is true. (2) Both Statement I and Statement II are true. (3) Both Statement I and Statement II are false. (4) Statement I is correct but Statement II is false. Answer (2)
Sol. Option (2) is the correct answer to this question because statement I and statement II both are correct. Vas deferens receives a duct from seminal vesicle and opens into urethra as the ejaculatory duct. The cavity of cervix is called cervical canal which along with vagina forms the birth canal.
173. Match List I with List II.
List I List II
A. P- wave I. Beginning of systole
B. Q- wave II. Repolarisation of ventricles
C. QRS complex III. Depolarisation of atria
D. T- wave IV. Depolarisation of ventricles
Choose the correct answer from the options given below :
(1) A-I, B-II, C-III, D-IV
(2) A-III, B-I, C-IV, D-II
(3) A-IV, B-III, C-II, D-I
(4) A-II, B-IV, C-I, D-III
Answer (2)
Sol. The correct answer is option (2) as in a standard ECG, P-wave represents the electrical excitation (or depolarisation) of the atria which leads to the contraction of both the atria. QRS complex represents the depolarisation of ventricles which initiates the ventricular contraction. T-wave represents the return of the ventricles from excited to normal state.
174. Radial symmetry is NOT found in adults of phylum
(1) Echinodermata
(2) Ctenophora
(3) Hemichordata
(4) Coelenterata
Answer (3)
Sol. Option (3) is the correct answer because hemichordates are bilaterally symmetrical animals. Option (4) is not the answer because coelenterates are radially symmetrical organisms. Option (1) is not the answer because adult echinoderms are radially symmetrical in adult stage Option (2) is not the answer because ctenophores are radially symmetrical organisms.
175. Which one of the following symbols represents mating between relatives in human pedigree analysis?
(1)
(2)
(3)
(4)
Answer (3)
Sol. The symbol representing mating between relatives (consanguineous mating) in human pedigree analysis is (4)
176. Select the correct group/set of Australian Marsupials exhibiting adaptive radiation.
(1) Lemur, Anteater, Wolf
(2) Tasmanian wolf, Bobcat, Marsupial mole
(3) Numbat, Spotted cuscus, Flying phalanger
(4) Mole, Flying squirrel, Tasmanian tiger cat
Answer (3)
Sol. Option (3) is the correct answer because numb, spotted cuscus and flying phalanger are Australian marsupials exhibiting adaptive radiation. Option (4) is incorrect because mole and flying squirrel are placental mammals. Option (1) is incorrect because lemur and wolf are placental mammals. Option (2) is incorrect because bobcat is a placental mammal.
177. Match List I with List II.
List I List II
(Interacting species) (Name of interaction)
A. A Leopard and a Lion in a forest/grassland I. Competition
B. A Cuckoo laying egg in a Crow's nest II. Brood parasitism
C. Fungi and root of a higher plant in Mycorrhizae III. Mutualism
D. A cattle egret and a Cattle in a field IV. Commensalism
Choose the correct answer from the options given below.
(1) A-II, B-III, C-I, D-IV
(2) A-I, B-II, C-III, D-IV
(3) A-I, B-II, C-IV, D-III
(4) A-II, B-IV, C-I, D-III
Answer (2)
Sol. A leopard and a lion in a forest/grassland exemplify competition where both the species are competing for the same resources.
A cuckoo laying egg in a crow's nest is brood parasitism where cuckoo is the parasitic bird that lays its egg in the nest of crow (host bird).
Fungi and root of a higher plant in mycorrhizae exemplify mutualism where both the species are benefitted. The fungi help the plant in the absorption of essential nutrients from the soil while the plant in turn provides the fungi with energy yielding carbohydrates.
A cattle egret and a cattle in a field exemplify commensalism where one species benefits and the other remains unaffected.
The egrets always forage close to where cattle are grazing because the cattle, as they move, stir up and flush out insects from the vegetation that otherwise might be difficult for the egrets to find and catch.
178. Match List I with List II
List I (Cells)
A. Peptic cells
B. Goblet cells
C. Oxynitic cells
D. Hepatic cells
List II (Secretion)
I. Mucus
II. Bile juice
III. Proenzyme pepsinogen
IV. HCl and intrinsic factor for absorption of vitamin B12
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-IV, B-III, C-II, D-I
(3) A-II, B-I, C-III, D-IV
(4) A-I, B-II, C-IV, D-III
Answer (1)
Sol. Option (1) is the correct answer because gastric glands have three major types of cells namely
(i) Mucus neck cells which secrete mucus
(ii) Peptic or chief cells which secrete the proenzyme pepsinogen
(iii) Parietal or oxynitic cells which secrete HCl and intrinsic factor for absorption of vitamin B12.
179. Match List I with List II.
List I List II
A. CCK I. Kidney
B. GIP II. Heart
C. ANF III. Gastric gland
D. ADH IV. Pancreas
Choose the correct answer from the options given below :
(1) A-IV, B-II, C-III, D-I
(2) A-IV, B-III, C-II, D-I
(3) A-III, B-II, C-IV, D-I
(4) A-II, B-IV, C-I, D-III
Answer (2)
Sol. The correct answer is option (2) as Cholecystokinin (CCK) acts on both gall bladder and pancreas and stimulates the secretion of bile juice and pancreatic enzymes respectively. GIP inhibits gastric secretion and motility. Atrial Natriuretic Factor (ANF) is released from the atrial wall of our heart. Anti-diuretic hormone (ADH) acts mainly on the kidney and stimulates resorption of water and electrolytes by the distal tubules.
180. Once the undigested and unabsorbed substances enter the caecum, their backflow is prevented by
(1) Pyloric sphincter
(2) Sphincter of Oddi
(3) Ileo-caecal valve
(4) Gastro-oesophageal sphincter
Answer (3)
Sol. Option (3) is the correct answer because the undigested food (faeces) enters into caecum of the large intestine through ileo-caecal valve, which prevents the backflow of the faecal matter.
Option (4) is not the answer because a muscular sphincter i.e., the gastro-oesophageal sphincter regulates the opening of oesophagus into the stomach.
Option (1) is not the answer because pyloric sphincter regulates the opening in between stomach and duodenum.
Option (2) is not the answer because the opening of common hepato-pancreatic duct is guarded by sphincter of Oddi.
181. Match List I with List II.
List I (Type of Joint) List II (Found between)
A. Cartilaginous Joint I. Between flat skull bones
B. Ball and Socket Joint II. Between adjacent vertebrae in vertebral column
C. Fibrous Joint III. Between carpal and metacarpal of thumb
D. Saddle Joint IV. Between Humerus and Pectoral girdle
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-III, D-I
(2) A-III, B-I, C-II, D-IV
(3) A-II, B-IV, C-I, D-III
(4) A-I, B-IV, C-III, D-II
Answer (3)
Sol. Option (3) is the correct answer because cartilaginous joint is present in between the adjacent vertebrae in the vertebral column.
Option (4) is not the answer because cartilaginous joint is not present between flat skull bones.
Option (1) is not the answer because fibrous joint is not present in between the carpal and metacarpal of thumb.
Option (2) is not the answer because saddle joint is not present in between humerus and pectoral girdle.
182. Which of the following statements is correct?
(1) Algal Bloom decreases fish mortality
(2) Eutrophication refers to increase in domestic sewage and waste water in lakes.
(3) Biomagnification refers to increase in concentration of the toxicant at successive trophic levels.
(4) Presence of large amount of nutrients in water restricts 'Algal Bloom'
Answer (3)
Sol. Increase in the concentration of the toxicant at successive trophic level is called biomagnification.
Large amount of nutrients in water promotes growth of algal bloom. Algal bloom increases fish mortality.
Eutrophication refers to the natural aging of a lake by nutrient enrichment of its water.
183. Given below are two statements:
Statement I: Electrostatic precipitator is most widely used in thermal power plant
Statement II: Electrostatic precipitator in thermal power plant removes ionising radiations
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct.
(2) Both Statement I and Statement II are correct.
(3) Both Statement I and Statement II are incorrect.
(4) Statement I is correct but Statement II is incorrect.
Answer (4)
Sol. Electrostatic precipitator is most widely used in thermal power plant
It can remove over 99 percent particulate matter present in the exhaust from a thermal power plant.
184. In which blood corpuscles, the HIV undergoes replication and produces progeny viruses?
(1) Eosinophils
(2) T_H cells
(3) B-lymphocytes
(4) Basophils
Answer (2)
Sol. The correct answer is option (2) because HIV enters into helper T-lymphocytes (T_H), replicates and produces progeny viruses. The progeny viruses released into blood attack other helper lymphocytes.
185. Given below are two statements :
Statement I : Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat.
Statement II : When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor.
In the light of the above statements, choose the correct answer from the options given below :
(1) Statement I is false but Statement II is true.
(2) Both Statement I and Statement II are true.
(3) Both Statement I and Statement II are false.
(4) Statement I is true but Statement II is false.
Answer (2)
Sol. The correct answer is option (2) as low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat.
Competitive inhibitor due to its close structural similarity with the substrate, competes with the substrate for the substrate-binding site of the enzyme.
186. Select the correct statements with reference to chordates.
A. Presence of a mid-dorsal, solid and double nerve cord.
B. Presence of closed circulatory system.
C. Presence of paired pharyngeal gill slits.
D. Presence of dorsal heart
E. Triploblastic pseudocoelomate animals.
Choose the correct answer from the options given below:
(1) C, D and E only
(2) A, C and D only
(3) B and C only
(4) B, D and E only
Answer (3)
Sol. Option (3) is the correct answer because statements B and C only are correct. Option (2), (3) and (4) are not correct. The chordate characters are presence of closed circulatory system and presence of pharyngeal gill slits. Nerve cord is dorsal, hollow and single. Heart is ventral. They are triploblastic and coelomate.
187. Which of the following statements are correct regarding skeletal muscle?
A. Muscle bundles are held together by collagenous connective tissue layer called fascicle.
B. Sarcoplasmic reticulum of muscle fibre is a store house of calcium ions.
C. Striated appearance of skeletal muscle fibre is due to distribution pattern of actin and myosin proteins.
D. M line is considered as functional unit of contraction called sarcomere.
Choose the most appropriate answer from the options given below:
(1) C and D only
(2) A, B and C only
(3) B and C only
(4) A, C and D only
Answer (3)
Sol. Option (3) is the correct answer because statements B and C are only correct statements while A and D are incorrect statements.
Muscle bundles are held together by collagenous connective tissue layer called fascia. Muscle bundles are called fascicles. The portion of the myofibril between two successive 'Z' lines is considered as functional unit of contraction called sarcomere.
188. Which one of the following is the sequence on corresponding coding strand, if the sequence on mRNA formed is as follows 5'AUCGAUCGAUCGAUCGAUCGAUCGAUCG 3'?
(1) 3'ATCGATCGATCGATCGATCGATCGATCG 5'
(2) 5'UAGCUAUCUAUCUAUCUAUCUAUCUAUCUAUCUAUCUAUCUAUCUA
(3) 3'UAGCUAUCUAUCUAUCUAUCUAUCUAUCUAUCUAUCU
(4) 5'ATCGATCGATCGATCGATCGATCGATCG 3'
Answer (4)
Sol. The sequence of coding strand is same as RNA except thymine at the place of uracil.
Template strand → 3'-TAGCTAGCTAGCTAGCTAGCTAGCTAGC-5'
Coding strand → 5'-ATCGATCGATCGATCGATCGATCGATCG-3'
↓ Transcription
mRNA → 5'AUCGAUCGAUCGAUCGAUCGAUCGAUCG 3'
189. Given below are two statements:
Statement I : During G₀ phase of cell cycle, the cell is metabolically inactive.
Statement II : The centrosome undergoes duplication during S phase of interphase.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct.
(2) Both Statement I and Statement II are correct
(3) Both Statement I and Statement II are incorrect.
(4) Statement I is correct but Statement II is incorrect.
Answer (1)
Sol. Cells in the G₀ stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism. In animal cells, during the S-phase, DNA replication begins in the nucleus, and the centriole duplicates in the cytoplasm.
190. Which of the following are NOT under the control of thyroid hormone?
A. Maintenance of water and electrolyte balance
B. Regulation of basal metabolic rate
C. Normal rhythm of sleep-wake cycle
D. Development of immune system
E. Support the process of RBCs formation
Choose the correct answer from the options given below:
(1) D and E only
(2) A and D only
(3) B and C only
(4) C and D only
Answer (4)
Sol. Option (4) is the correct answer because thyroid hormones play an important role in the regulation of basal metabolic rate, maintenance of water and electrolyte balance and support the process of RBCs formation, whereas this hormone is not involved in regulating normal rhythm of sleep-wake cycle and development of immune system.
191. In cockroach, excretion is brought about by-
A. Phallic gland
B. Urecos gland
C. Nephrocytes
D. Fat body
E. Collateral glands
Choose the correct answer from the options given below:
(1) B and D only
(2) A and E only
(3) A, B and E only
(4) B, C and D only
Answer (4)
Sol. Option (4) is the answer because, In cockroach, excretion is brought about by Malpighian tubules, fat body, nephrocytes and urecos glands. Urecos glands are present in male cockroach of some species. They synthesise uric acid. Nephrocytes are large, colourless, ovoid, binucleate cells attached to the dorsal diaphragm in the body cavity. Fat body accumulates, produces and stores uric acid. Phallic gland is the structure of male reproductive system of cockroach and it secretes the outer layer of spermatophore. Collateral gland is the structure of female reproductive system of cockroach and it secretes the hard egg-case or ootheca around fertilised eggs.
192. Match List I with List II.
List I List II
A. Mast cells I. Ciliated epithelium
B. Inner surface of bronchiolle II. Areolar connective tissue
C. Blood III. Cuboidal epithelium
D. Tubular parts of nephron IV. Specialised connective tissue
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-II, D-I
(2) A-I, B-II, C-IV, D-III
(3) A-II, B-I, C-IV, D-III
(4) A-II, B-I, C-III, D-IV
Answer (4)
Sol. Option (4) is the correct answer because,
Areolar connective tissue contains fibroblasts (cells that produce and secrete fibres), macrophages and mast cells.
Inner surface of bronchioles is lined by ciliated epithelium.
Blood is a specialised connective tissue.
Tubular parts of nephron are lined by cuboidal epithelium.
193. Which of the following is characteristic feature of cockroach regarding sexual dimorphism?
(1) Presence of anal cerci
(2) Dark brown body colour and anal cerci
(3) Presence of anal styles
(4) Presence of sclerites
Answer (3)
Sol. Option (3) is the correct answer because anal styles are present in male cockroaches and absent in female cockroaches. Option (1), (2) and (4) are not the correct answers because sclerites, anal cerci and dark brown body colour are common features of both male and female cockroaches.
194. Select the correct statements.
A. Tetrad formation is seen during Leptotene.
B. During Anaphase, the centromeres split and chromatids separate.
C. Terminalization takes place during Pachytene.
D. Nucleolus, Golgi complex and ER are reformed during Telophase.
E. Crossing over takes place between sister chromatids of homologous chromosome.
Choose the correct answer from the options given below:
(1) B and E only
(2) A and C only
(3) B and D only
(4) A, C and E only
Answer (3)
Sol. Tetrad formation is seen during zygotene stage
During Anaphase, the centromeres split and chromatids separate. Terminalisation of chiasmata takes place during diakinesis.
Nucleolus, golgi complex and ER are reformed during telophase.
Crossing over takes place between non-sister chromatids of homologous chromosomes.
195. Which one of the following is NOT an advantage of inbreeding?
(1) It decreases the productivity of inbred population, after continuous inbreeding.
(2) It decreases homozygosity.
(3) It exposes harmful recessive genes but are eliminated by selection.
(4) Elimination of less desirable genes and accumulation of superior genes takes place due to it.
Answer (1)
Sol. Option (1) is the correct answer because decreasing the productivity of inbred population is not an advantage of inbreeding. Options (3) and (4) are not the answers because they are the advantages of inbreeding. Option (2) is an incorrect statement.
196. The unique mammalian characteristics are:
(1) pinna, monoconodylc skull and mammary glands
(2) hairs, tympanic membrane and mammary glands
(3) hairs, pinna and mammary glands
(4) hairs, pinna and indirect development
197. The parts of human brain that helps in regulation of sexual behaviour, expression of excitement, pleasure, rage, fear etc. are:
(1) Corpus callosum and thalamus
(2) Limbic system and hypothalamus
(3) Corpora quadrigemina and hippocampus
(4) Brain stem and epithalamus
Answer (2)
Sol. Option (2) is the correct answer because limbic system along with hypothalamus regulate the sexual behaviour, expression of excitement, pleasure, rage, fear, etc.
Option (1), (3) and (4) are not correct because corpora quadrigemina is a part of the midbrain and consists of four round swellings. Corpus callosum is a tract of nerve fibres that connects right and left cerebral hemispheres. Thalamus is a major coordinating centre in the forebrain for sensory and motor signalling. Midbrain, pons and medulla oblongata together form the brain stem.
198. Match List I with List II.
List I List II
A. Logistic growth I. Unlimited resource availability condition
B. Exponential growth II. Limited resource availability condition
C. Expanding age pyramid III. The percent individuals of pre-reproductive age is largest followed by reproductive and post reproductive age groups
D. Stable age pyramid IV. The percent individuals of pre-reproductive and reproductive age group are same
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-III, D-I
(2) A-II, B-I, C-III, D-IV
(3) A-II, B-III, C-I, D-IV
(4) A-II, B-IV, C-I, D-III
Answer (2)
Sol. Logistic growth occurs when there is limited resource availability condition.
Exponential growth occurs when there is unlimited resource availability condition.
Expanding age pyramid reflects growing population where the percent individuals of pre-reproductive age is largest followed by reproductive and post-reproductive age groups.
Stable age pyramid shows stable population where the percent individuals of pre-reproductive and reproductive age group are same.
199. Which of the following statements are correct?
A. Basophils are most abundant cells of the total WBCs
B. Basophils secrete histamine, serotonin and heparin
C. Basophils are involved in inflammatory response
D. Basophils have kidney shaped nucleus
E. Basophils are agranulocytes
Choose the correct answer from the options given below:
(1) A and B only
(2) D and E only
(3) C and E only
(4) B and C only
Answer (4)
Sol. Option (4) is the answer because, basophils secrete histamine, serotonin, heparin etc. and are involved in inflammatory response. Option (3) is not the answer because, basophils are granulocytes. Option (1) is not the answer because, neutrophils are the most abundant cells (60-65 %) of the total WBCs whereas basophils are least (0.5-1%) abundant of all WBCs. Option (2) is not the answer because, monocytes have a kidney-shaped nucleus.
200. Which of the following statements are correct?
A. An excessive loss of body fluid from the body switches off osmoreceptors.
B. ADH facilitates water reabsorption to prevent diuresis.
C. ANF causes vasodilation.
D. ADH causes increase in blood pressure.
E. ADH is responsible for decrease in GFR.
Choose the correct answer from the options given below:
(1) C, D and E only
(2) A and B only
(3) B, C and D only
(4) A, B and E only
Answer (3)
Sol. Option (3) is the correct answer because statements B, C and D are true statements. ADH facilitates water reabsorption from DCT of nephron to prevent diuresis, which causes increase in blood pressure. ANF which is secreted by the heart is a vasodilator. Options (1), (2) and (4) are not correct because statements A and E are false. Excessive loss of body fluid from the body switches on the osmoreceptors.
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