1. For a simple pendulum, having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by:
(1) graph
(2) graph
(3) graph
(4) graph
Answer (4)
Sol. Kinetic energy, K = 1/2 mv²
= 1/2 mA²ω² cos²(ωt + φ)
∴ K ∝ cos²(ωt + φ)
2. A room heater is rated 400 W, 220 V. If the supply voltage drops to 200 V, what will be the power consumed (approximately)?
(1) 400 W
(2) 121 W
(3) 331 W
(4) 200 W
Answer (3)
Sol. Rated power of heater
P0 = V0²/R ...(i)
Consumed power
Pc = V²/R ...(ii) [V → applied voltage]
From (i) and (ii)
Pc/P0 = V²/V0²
Pc = [V/V0]² P0
Pc = [200/220]² × 400
Pc = 331 W
3. The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is:
(1) 600
(2) 300
(3) 900
(4) 150
Answer (4)
Sol. α = (ω2 - ω1)/Δt = ((1200 - 600)/10)(2π/60) = 2π rad/s²
Use equation of motion, ω2² = ω1² + 2αθ
⇒ ω2 = 1200 × 2π/60 = 40π
ω1 = 600 × 2π/60 = 20π
(40π)² = (20π)² + 2 × 2πθ
⇒ 1200π² = 4πθ
θ = 300π radian
Number of revolution θ/2π = 150
4. The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately:
Consider mass of the bob = 20 g
(1) 2.0 m/s
(2) 0.2 m/s
(3) 14.1 m/s
(4) 1.41 m/s
Answer (4)
Sol. At equilibrium position
Total energy = K.E
1/2 mv² = 0.02
1/2 × 20 × v² × 10⁻³ = 2 × 10⁻²
v = √2
v = 1.41 m/s
5. A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 × 10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively:
(Take μ0 = 4π × 10⁻⁷ T m/A)
(1) 2.5 A, 20 A m²
(2) 2 A, 4 A m²
(3) 2.5 A, 2 A m²
(4) 2 A, 10 A m²
Answer (3)
6. A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface?
(Consider the density of water = 1000 kg m⁻³, 1 atm = 1 × 10⁵ Pa and gravitational acceleration g = 10 m/s²)
(1) 9900 m
(2) 990 m
(3) 9000 m
(4) 99 m
Answer (2)
Sol. P = P0 + ρgh
⇒ 100 × 10⁵ = 10⁵ + 10³ × 10 × h
⇒ 10⁷ = 10⁵ + 10⁴h
⇒ 10³ = 10 + h
⇒ h = 1000 - 10 = 990 m
7. Match List I with List II.
List-I
A. E = hν
B. Diffraction and Interference
C. λ = h/p
D. Compton effect
List-II
I. de Broglie wavelength
II. Particle nature of light
III. Wave nature of light
IV. Energy of photon
Choose the correct answer from the options given below.
(1) A-IV, B-III, C-II, D-I
(2) A-IV, B-III, C-I, D-II
(3) A-I, B-IV, C-III, D-II
(4) A-IV, B-I, C-II, D-III
Answer (2)
Sol. A. E = hν is energy of photon
B. Diffraction and interference confirm wave nature of light
C. λ = h/p is de Broglie wavelength of particle.
D. Compton effect confirms particle nature of light.
8. Match List I with List II:
List I
A. Young's Modulus
B. Compressibility
C. Bulk Modulus
D. Poisson's Ratio
List II
I. Δd/ΔL (L/d)
II. FL/A(ΔL)
III. 1/ΔP (ΔV/V)
IV. -P(ΔV/V)
Choose the correct answer from the options given below:
(1) A-III, B-II, C-I, D-IV
(2) A-II, B-III, C-IV, D-I
(3) A-I, B-IV, C-II, D-III
(4) A-IV, B-I, C-II, D-III
Answer (2)
Sol. A. Young's Modulus = Stress/Strain = FL/AΔL
B. Compressibility = 1/Bulk modulus = 1/ΔP (ΔV/V)
C. Bulk Modulus = -P(V/ΔV)
D. Poisson's Ratio = Lateral strain/Longitudinal strain = Δd/ΔL (L/d)
9. Five capacitors of capacitances C1 = C2 = C3 = C4 = 10 μF and C5 = 2.5 μF are connected as shown, along with a battery of 50 V
The equivalent capacitance and the charges on each capacitor respectively are:
(1) 4 μF, 250 μC on C1 to C4 and 125 μC on C5
(2) 5 μF, 250 μC on all capacitors
(3) 5 μF, 125 μC on C1 to C4 and 25 μC on C5
(4) 5 μF, 125 μC on all capacitors
Answer (4)
10. The amount of work done to raise a mass 'm' from the surface of the Earth to a height equal to the radius of the Earth 'R' will be
(1) mgR/2
(2) mgR
(3) mgR/4
(4) 2mgR
Answer (1)
Sol. W.D. = U2 - U1
= -GMm/(R+R) - (-GMm/R)
= -GMm/2R + GMm/R
= GMm/2R = mgR/2
11. When a ruler falls vertically, 5 different persons catch it with different reaction times. (g = 9.8 m s⁻²)
A. Person A has reaction time of 0.20 s.
B. Person B has reaction time of 0.22 s.
C. Person C has reaction time of 0.18 s.
D. Person D has reaction time of 0.19 s.
E. Person E has reaction time of 0.21 s.
What is the correct order of the distance travelled by the ruler for each person?
(1) B > E > A > C > D
(2) C > D > A > E > B
(3) C > D > A > B > E
(4) B > E > A > D > C
Answer (4)
Sol. There will be large distance for large reaction time
Descending order of reaction time ⇒ tB > tE > tA > tD > tC
Descending order of distance covered ⇒ SB > SE > SA > SD > SC
12. The power of a crane, which lifts a mass of 1000 kg to a height of 20 m in 10 s is: (g = 9.8 m/s²)
(1) 19.6 kW
(2) 19.6 W
(3) 39.2 kW
(4) 39.2 W
Answer (1)
Sol. Power = Work/Time = mgh/t = (10³ × 9.8 × 20)/10 = 19.6 kW
13. Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:
(1) 0.5 J
(2) 1.0 × 10⁻⁶ J
(3) 0.5 × 10⁻⁶ J
(4) 1.0 J
Answer (3)
Sol. Energy loss = 1/2 C1C2/(C1 + C2) V²
= 1/2 (200 × 200)/(400) × 10⁻¹²(100)²
= 1/2 × 10⁶ × 10⁻¹² = 0.5 × 10⁻⁶ J
14. An ac circuit contains a resistance of 1 kΩ, a capacitor of 0.1 μF and an inductor of 1 mH connected in series. The resonance frequency of the circuit is approximately:
(1) 15.9 kHz
(2) 20.7 kHz
(3) 10.1 kHz
(4) 13.5 kHz
Answer (1)
Sol. Resonance frequency f = 1/(2π√LC) = 1/(2π√(1 × 10⁻⁷ × 10⁻³)) = 15.9 kHz
15. The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B) with distance (r) from the axis of the conductor in the region is:
Answer (4)
Sol. For a long straight solid wire carrying steady current, which is uniformly distributed across its cross-section, the variation of magnetic field (B) with distance (r) from axis will be
B = μ0Ir/(2πa²) ⇒ B ∝ r, for r < a
B = μ0I/(2πr) ⇒ B ∝ 1/r, for r > a
16. A galvanometer of resistance 100 Ω gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0 - 10 A. The shunt required is:
(1) 0.001 Ω
(2) 0.10 Ω
(3) 1.0 Ω
(4) 0.01 Ω
Answer (4)
Sol. iG = 1 mA = 0.001 A
iS = 10 - iG ≈ 10 A
Both shunt resistance and galvanometer are in parallel connection
∴ iS rS = iG RG
⇒ 10 × rS = 0.001 × 100
⇒ rS = 0.01 Ω
17. A thin wire of length L' and linear mass density m' is bent into a circular ring (in x-y plane) with centre C' as shown in figure. The moment of inertia of the ring about an axis yy' will be:
(1) 3mL³/(8π)
(2) 3mL²/(8π²)
(3) 3mL³/(8π²)
(4) 3mL²/(8π)
Answer (3)
Sol. Mass of thin wire = (Linear mass density) × (Length)
⇒ M = mL
L = 2πr, where r = radius of circular ring = L/(2π)
Using parallel axis theorem,
Iyy' = ICM + Mr² = Mr²/2 + Mr² = 3/2 Mr²
⇒ Iyy' = 3/2 × (mL) × (L/(2π))² = 3mL³/(8π²)
18. In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ, is K units. The intensity of light at a point where the path difference is λ/3 will be
(1) K
(2) 2K
(3) K/2
(4) K/4
Answer (4)
Sol. I = I0 cos²(Δφ/2) [I0 → maximum intensity]
I = I0 cos²(kΔx/2)
K = I0 cos²((2π/λ) × (λ/2))
K = I0
K1 = I0 cos²((2π/λ) × (λ/(3×2))) = I0 cos²(π/3)
K1 = I0/4
K/K1 = 4 ⇒ K1 = K/4
19. Four statements are given (A is mass number):
A. The volume of a nucleus is proportional to A^(1/3).
B. The volume of a nucleus is proportional to A.
C. The difference in mass of an atom and its nucleus is called the mass defect.
D. The difference in mass of a nucleus and its constituents is called the mass defect.
Choose the correct answer from the options given below:
(1) A and D are true, but B and C are false
(2) B and D are true, but A and C are false
(3) B and C are true, but A and D are false
(4) A and C are true, but B and D are false
Answer (2)
Sol. As we know,
Size of nucleus, r = r0(A)^(1/3)
⇒ r³ = r0³ · A
⇒ V ∝ A (∵ V = 4/3 πr³)
So, option A → wrong and B → correct
while the difference between the actual mass of nucleus and its constituents is called the mass defect.
∴ option C → wrong and D → correct
20. In interference and diffraction, the light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe.
A. As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy.
B. Diffraction and interference are characteristics exhibited only by light waves.
Choose the correct answer from the options given below:
(1) A is false, but B is true
(2) A is true, but B is false
(3) A is true and B is also true
(4) Both A and B are false
Answer (2)
Sol. In interference and diffraction there is no loss of energy, the energy gets redistributed. Interference and diffraction both are exhibited in light as well as sound waves.
21. A resistor is connected to a battery of 12 V emf and internal resistance 2 Ω. If the current in the circuit is 0.6 A, the terminal voltage of the battery is:
(1) 12 V
(2) 1.2 V
(3) 10 V
(4) 10.8 V
Answer (4)
Sol. Circuit can be drawn as,
⇒ Terminal voltage of battery V = E - ir
= 12 - 0.6 × 2
= 10.8 V
22. In a metre bridge experiment (see figure), the positions of the cell, E, and galvanometer, G, are interchanged. We shall observe in the galvanometer:
(1) Only the right-sided deflection
(2) Only the left-sided deflection
(3) There will be no deflection irrespective of the position of the jockey
(4) Both right-sided and left-sided deflection and at balance point, no deflection
Answer (4)
Sol. Position of null point will not change when galvanometer (G) and the cell (E) are interchanged. There will be no deflection in galvanometer only at balance point.
In an unbalanced meter bridge, if E and G are interchanged mutually, then the deflection in galvanometer may be towards left-side or right-side.
23. Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as:
(Take π² = 9.8 and g = 9.8 m/s²)
(1) 0.75 m
(2) 1 m
(3) 1.5 m
(4) 2 m
Answer (2)
Sol. Time taken for 30 oscillations = 60 s
Time period of simple pendulum = Time taken for 1 oscillation
⇒ T = 60/30 = 2 s
T = 2π√(l/g) ⇒ l = gT²/(4π²) = (9.8 × 2 × 2)/(4 × 9.8) = 1 m
24. Which of the following statements are correct?
A. Inside a conductor, the electrostatic field is zero.
B. Electric field at the surface of a charged conductor does not depend on its surface charge density.
C. The interior of a charged conductor can have no excess charge in the static situation.
D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point.
E. The electrostatic potential is zero everywhere inside a charged conductor.
Choose the correct answer from the options given below:
(1) A, C and D only
(2) A, C and E only
(3) C, D and E only
(4) A, B and D only
Answer (1)
Sol. A. Electrostatic field is zero inside a conductor.
B. Electric field at the surface of a charged conductor = σ/ε0 n̂, depends on surface charge density (σ)
C. The interior of a charged conductor cannot have any excess charge in the static situation.
D. At the surface of a charged conductor, the electrostatic field ⊥ surface.
E. The electrostatic potential is constant and can be non-zero everywhere inside a charged conductor.
25. Two statements are given below:
A. When the forward bias voltage across a p-n junction diode increases above a certain threshold voltage, the diode current increases significantly.
B. This current is called reverse saturation current.
Choose the correct answer from the options given below:
(1) Both Statements A and B are true
(2) Both Statements A and B are false
(3) Statement A is true, but Statement B is false
(4) Statement A is false, but Statement B is true
Answer (3)
Sol. V-I characteristics of a forward-biased junction diode
When forward bias voltage increases beyond threshold voltage, diode current increases significantly. This is not reverse saturation current.
26. In a concave lens, a ray of light emanating from the object parallel to the principal axis of the lens after refraction:
(1) passes through the second principal focus.
(2) appears to diverge from the first principal focus.
(3) passes through 2F, which is the radius of curvature of the lens.
(4) emerges parallel to the principal axis.
Answer (2*)
Sol. F2 is the second principal focus ⇒ It is the virtual image position for object at infinity.
F1 is the first principal focus ⇒ It is the virtual object position for which image is formed at infinity.
The best appropriate answer is option (2), although it should be second principal focus.
27. An unknown nucleus has a nuclear density of 2.29 × 10¹⁷ kg/m³ and mass of 19.926 × 10⁻²⁷ kg. Its mass number A is approximately:
(Take R0 = 1.2 × 10⁻¹⁵ m, 4π = 12.56)
(1) 16
(2) 20
(3) 12
(4) 19
Answer (3)
Sol. Given
ρ = 2.29 × 10¹⁷ kg/m³, mass m = 19.926 × 10⁻²⁷ kg
R0 = 1.2 × 10⁻¹⁵ m and R = R0A^(1/3)
Now use volume = Mass/Density
⇒ 4/3 πR³ = M/ρ
⇒ 4/3 π[R0(A)^(1/3)]³ = M/ρ
⇒ 4/3 πR0³A = M/ρ
⇒ A = M/ρ × 3/(4πR0³) = (19.926 × 10⁻²⁷ × 3)/(2.29 × 10¹⁷ × 12.56 × (1.2 × 10⁻¹⁵)³) ≈ 12
28. A galvanometer of resistance 100 Ω gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0 - 10 A. The shunt required is:
(1) 0.001 Ω
(2) 0.10 Ω
(3) 1.0 Ω
(4) 0.01 Ω
Answer (4)
Sol. iG = 1 mA = 0.001 A
iS = 10 - iG ≈ 10 A
Both shunt resistance and galvanometer are in parallel connection
∴ iS rS = iG RG
⇒ 10 × rS = 0.001 × 100
⇒ rS = 0.01 Ω
29. A thin wire of length L' and linear mass density m' is bent into a circular ring (in x-y plane) with centre C' as shown in figure. The moment of inertia of the ring about an axis yy' will be:
(1) 3mL³/(8π)
(2) 3mL²/(8π²)
(3) 3mL³/(8π²)
(4) 3mL²/(8π)
Answer (3)
Sol. Mass of thin wire = (Linear mass density) × (Length)
⇒ M = mL
L = 2πr, where r = radius of circular ring = L/(2π)
Using parallel axis theorem,
Iyy' = ICM + Mr² = Mr²/2 + Mr² = 3/2 Mr²
⇒ Iyy' = 3/2 × (mL) × (L/(2π))² = 3mL³/(8π²)
30. For a travelling harmonic wave y(x,t) = 2.0 cos 2π(10t - 0.0080x + 0.35), where x and y are in cm and t in s. The phase difference between oscillatory motion of two points separated by a distance of 0.5 m is:
(1) 0.08π rad
(2) 0.008π rad
(3) 0.8π rad
(4) 8π rad
Answer (3)
Sol. y(x,t) = 2.0 cos 2π(10t - 0.008x + 0.35)
Total phase
φ = 20π - 2π × 8 × 10⁻³x + 2π × 0.35
Δφ = kΔx
Δφ = 2π × 8 × 10⁻³Δx
= 2π × 8 × 10⁻³ × (100/2)
= 8π × 10⁻¹
= 0.8π
31. A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 0.12. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in m s⁻² is:
(g = 10 m/s²)
(1) 1.8
(2) 1.2
(3) 1.5
(4) 2.1
Answer (2)
Sol. a = μg = 0.12 × 10 = 1.2 m/s²
32. In the circuit shown below, the voltage appearing across the diode D will be of the form:
Answer (4)
Sol. Voltage drop will be across the diode, when it will be in reverse bias
In positive half cycle it will be in reverse biased
33. A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixture is 27°C. The ratio of root mean square speed of the molecules of the two gases (V_rms^Ar / V_rms^Cl) is:
(Atomic mass of argon = 40.0 u and molecular mass of chlorine = 70.0 u)
(1) 7/4
(2) 2/√7
(3) √7/2
(4) 7/2
Answer (3)
Sol. v_rms = √(3RT/M)
For same temperature, v_rms ∝ 1/√M
V_rms^Ar / V_rms^Cl = √(M_Cl/M_Ar) = √(70/40) = √(7/2)
34. Match List I with List II:
List-I (Electromagnetic wave)
A. Microwave
B. Visible light
C. Gamma rays
D. Infra-red rays
List-II (Production)
I. Electrons in atoms emit light when they move from a higher energy level to a lower energy level
II. Radioactive decay of nucleus
III. Vibration of atoms and molecules
IV. Klystron valve or magnetron valve
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-II, D-I
(2) A-III, B-IV, C-I, D-II
(3) A-III, B-I, C-II, D-IV
(4) A-IV, B-I, C-II, D-III
Answer (4)
Sol.
(A-IV) Microwave - Klystron valve or magnetron valve
(B-I) Visible light - Electrons in atoms emit light when they move from a higher energy level to a lower energy level
(C-II) Gamma rays - Radioactive decay of nucleus
(D-III) Infra-red rays - Vibration of atoms and molecules
35. The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it, are respectively:
(1) 2 m s⁻² tan⁻¹(3/4) with 8 N force
(2) 2 m s⁻² tan⁻¹(4/3) with 8 N force
(3) 2 m s⁻² tan⁻¹(3/4) with 6 N force
Answer (1)
Sol. F_net = √(6² + 8²) = 10
a = F_net/m = 10/5 = 2 m/s²
tan θ = 6/8
θ = tan⁻¹(3/4) from 8 N
36. The current I in the circuit shown below is:
(All diodes are ideal and identical)
(1) 1/3 A
(2) 5/3 A
(3) 5/9 A
(4) 15/2 A
Answer (2)
37. For a metal of work function 6.6 eV, which of the following wavelengths of incident radiation does not give rise to the photoelectric effect?
(Take Planck's constant as 6.6 × 10⁻³⁴ J s)
(1) 200 nm
(2) 100 nm
(3) 50 nm
(4) 150 nm
Answer (1)
Sol. For incident radiation having wavelength (λ), photoelectric effect doesn't occur when hc/λ < work-function
⇒ λ > hc/W0
⇒ λ > (6.6 × 10⁻³⁴ × 3 × 10⁸)/(6.6 × 1.6 × 10⁻¹⁹)
⇒ λ > (3 × 10⁻⁷)/1.6
⇒ λ > 300/1.6 nm
⇒ λ = 187.5 nm
∴ Option (1) 200 nm is correct.
38. The speed of light in vacuum is taken as unity. If light takes 6 min 40 s to reach the Earth from the Sun, the distance between the Sun and the Earth in new unit is:
(1) 3 × 10⁸
(2) 500
(3) 3 × 10¹⁰
(4) 400
Answer (4)
Sol. Time, t = 6 min 40 s = 360 + 40 = 400 s
Distance in new system d = vt = 1 × 400 = 400
39. A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is 2 cm s⁻¹, in a direction normal to the shorter side of the loop, will be:
(1) 1.8 × 10⁻⁴ volt
(2) 1.3 × 10⁻⁴ volt
(3) 1.2 × 10⁻⁴ volt
(4) 4.8 × 10⁻⁴ volt
Answer (1)
Sol. Induced emf across the shorter side
E_induced = BV [∵ B ⊥ V ⊥ I]
= 0.3 × 2 × 10⁻² × 3 × 10⁻²
= 1.8 × 10⁻⁴ V
= 1.8 × 10⁻⁴ volt
40. The following plots show variation of velocity (v) with time (t) of a ball thrown vertically upward, and falling back. Which of the following plots is/are correct?
(1) B only
(2) A and E only
(3) C only
(4) D only
Answer (3)
Sol. During the whole journey, acceleration due to gravity is vertically downward.
Therefore, slope of velocity vs time curve should be negative throughout the journey.
Statement (C) is correct
41. In a vernier calliper, 20 VSD coincide with 16 MSD (each division of length 1 mm). The least count of the vernier callipers is:
(1) 0.2 cm
(2) 0.1 cm
(3) 0.02 cm
(4) 0.01 cm
Answer (3)
Sol. Least count of vernier callipers
L.C. = 1 MSD - 1 VSD ...(i)
20 VSD = 16 MSD
1 VSD = 16/20 MSD
From (i)
L.C. = 1 MSD - 16/20 MSD
L.C. = 4/20 MSD
42. Each side of a metallic cube of mass 5.580 kg is measured to the 9.0 cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as X × 10³ kg m⁻³ where the value of X is:
(1) 7.654
(2) 7.7
(3) 7.65
(4) 7.6
Answer (2)
Sol. Density = Mass/Volume = 5.580 kg/(9.0 cm)³ = 5.580 × 10⁶ kg/729 m³ = 0.007654321 × 10⁶ kg/m³ = 7.654 × 10³ kg/m³ = 7.7 × 10³ kg/m³
Hence, on comparing X = 7.7
43. A ray of monochromatic light is passing through an equilateral prism (ABC) as shown in the figure. The refracted ray (QR) is parallel to its base (BC) and the angle of incidence (i) is 50°. Then the angle of deviation (δ) is:
(1) 45°
(2) 55°
(3) 35°
(4) 40°
Answer (4)
Sol. i = e
Equation of prism
i + e = A + δ
2i - A = δ
2(50) - 60 = δ
δ = 40°
44. In the first excited state of hydrogen atom, the energy of its electron is -3.4 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately:
(Take 1 eV = 1.6 × 10⁻¹⁹ J, e = 1.6 × 10⁻¹⁹ C and 1/(4πε0) = 9 × 10⁹ N m²/C²)
(1) 2.1 × 10⁻⁸ m
(2) 2.1 × 10⁻¹⁰ m
(3) 2.1 × 10⁻¹¹ m
(4) 2.1 × 10⁻⁹ m
Answer (2)
Sol. KQ²/(2r) = 3.4 eV
(9 × 10⁹ × (1.6 × 10⁻¹⁹)²)/(2 × 3.4 × 1.6 × 10⁻¹⁹) = r
2.1176 × 10⁻¹⁰ = r
r = 2.1 × 10⁻¹⁰ m
45. A uniform metallic wire having resistance 4 Ω is bent to form a square loop (ABCD) (see figure). A resistance of 2 Ω is connected between points B and D and a battery of 2 V is connected across points A and C as shown in the figure. Now the value of current (I) is:
(1) 4 A
(2) 8 A
(3) 4.5 A
(4) 2 A
Answer (4)
Sol. Each side will have resistance of 1 Ω
It is balanced wheat stone bridge
No current in the resistance of 2 Ω
[R_effective]_AC = 1 Ω
I = E/R_eff = 2/1 = 2 A
46. Match List I with List II:
List I (Complex/ion)
A. [Pt(Cl2)(NH3)2]
B. [Co(NH3)6]Cl3
C. [NiCl4]²⁻
D. [Fe(CO)5]
List II (Shape/geometry)
I. Octahedral
II. Trigonal bipyramidal
III. Square planar
IV. Tetrahedral
Choose the correct answer from the options given below:
(1) A-I, B-III, C-IV, D-II
(2) A-III, B-IV, C-I, D-II
(3) A-III, B-I, C-IV, D-II
(4) A-IV, B-I, C-III, D-II
Answer (3)
Sol. In [Pt(Cl2)(NH3)2]; Pt has dsp² hybridisation, so shape of complex is square planar. In [Co(NH3)6]Cl3; Co has dsp³ hybridisation so shape of complex is octahedral. In [NiCl4]²⁻; Ni has sp³ hybridisation so shape of complex ion is tetrahedral. In [Fe(CO)5]; shape of complex is trigonal bipyramidal.
47. Calculate emf of the half cell given below:
Pt (s) | H2(g, 2 atm) | HCl (aq, 0.02 M)
E°_(H2/H+) = 0 V
(Given: 2.303RT/F = 0.059, log 2 = 0.3010)
(1) -0.109 V
(2) 0.109 V
(3) 0.035 V
(4) -0.035 V
Answer (2)
Sol. H2(g) → 2H+(aq) + 2e⁻
∴ E = E° - (2.303RT/nF) log([H+]²/P_H2)
[H+] = 0.02 M (from HCl)
P_H2 = 2 atm
2.303RT/F = 0.059
48. At 298 K, a certain buffer solution contains equal concentrations of X⁻ and HX, Kb for X⁻ is 10⁻¹⁰. What is the pH of this buffer solution?
(1) 10
(2) 4
(3) 2
(4) 6
Answer (2)
Sol. HX ⇌ H⁺ + X⁻
Ka × Kb = Kw
Ka = Kw/Kb = 10⁻¹⁴/10⁻¹⁰ = 10⁻⁴
pH = pKa + log([X⁻]/[HX])
Given that: [X⁻] = [HX]
= 4 + log(1) = 4 + 0 = 4
49. Given below are certain reactions. Identify the reaction for which Kp ≠ Kc.
(1) N2(g) + O2(g) ⇌ 2NO(g)
(2) H2O(g) + CO(g) ⇌ H2(g) + CO2(g)
(3) N2(g) + 3H2(g) ⇌ 2NH3(g)
(4) H2(g) + I2(g) ⇌ 2H(g)
Answer (3)
Sol. Kp = Kc(RT)^Δng
For Kp = Kc (Δng = 0)
Kp ≠ Kc (Δng ≠ 0)
(1) N2(g) + O2(g) ⇌ 2NO(g)
Δng = 0 (Kp = Kc)
(2) H2O(g) + CO(g) ⇌ H2(g) + CO2(g)
Δng = 0 (Kp = Kc)
(3) N2(g) + 3H2(g) ⇌ 2NH3(g)
Δng = -2 (Kp ≠ Kc)
(4) H2(g) + I2(g) ⇌ 2H(g)
Δng = 0 (Kp = Kc)
50. For a certain reaction R → Product, the plot of concentration [R] vs time has a negative slope as shown. The order of reaction is:
(1) 1
(2) 2
(3) 2.5
(4) 0
Answer (4)
Sol. For zero order
[R] = [R0] - kt
So -ve slope shows zero order reaction
51. Consider the following reaction:
2A(g) + B(g) → 2D(g)
ΔU° = -10 kJ mol⁻¹ and ΔS° = -44 kJ K⁻¹ at 298 K.
Identify the correct option with ΔG° for the reaction and spontaneity of the reaction at 298 K. (Given: R = 8.31 J mol⁻¹ K⁻¹)
(1) +0.63568 kJ mol⁻¹ non-spontaneous
(2) -1.635 kJ mol⁻¹ spontaneous
(3) -0.63568 kJ mol⁻¹ spontaneous
(4) +1.635 kJ mol⁻¹ non-spontaneous
Answer (1)
Sol. 2A(g) + B(g) → 2D(g)
ΔU° = -10 kJ/mol
ΔS° = -44 J/K
ΔH° = ΔU° + ΔngRT
ΔH° = -10 - (1 × 298 × 8.31)/1000 = -10 - 2.48 = -12.48 kJ/mol
ΔG° = ΔH° - TΔS°
= -12.48 - (298 × (-44))/1000
= -12.48 + 13.112
= +0.632 kJ/mol
Since ΔG° comes out to be positive, so given process is non-spontaneous.
52. Given below is an expression for the rate constant of a first-order reaction occurring at a certain temperature, T (K).
ln k = 14.34 - (1.25 × 10⁴)/T
The energy of activation in kcal mol⁻¹ for the reaction is:
(Given: k in s⁻¹, R = 1.987 cal mol⁻¹ K⁻¹)
(1) 12.42
(2) 18.63
(3) 14.34
(4) 24.84
Answer (4)
Sol. From Arrhenius equation
k = A e^(-Ea/RT)
ln k = ln A - Ea/RT
Ea/R = 1.25 × 10⁴
Ea = 1.25 × 10⁴ × 1.987
= 24.84 kcal mol⁻¹
53. Select the reagents that reduce nitriles to primary amines.
A. (i) LiAlH4; (ii) H2O
B. Sn + HCl
C. H2/Ni
D. Na(Hg)/C2H5OH
E. Br2/aq. NaOH
Choose the correct answer from the options given below.
(1) A, B and C only
(2) A, D and E only
(3) A, C and D only
(4) B, D and E only
Answer (3)
Sol. R-CN → Na(Hg)/C2H5OH → R-CH2-NH2
R-CN → (i) LiAlH4 → R-CH2-NH2
R-CN → Sn + HCl → R-CHO
R-CN → H2/Ni → R-CH2-NH2
54. The correct statement with regard to the secondary structure of DNA/RNA is
(1) DNA possesses a double strand helix structure and contains thymine as one of the four bases
(2) DNA possesses a single strand helix structure and contains uracil as one of the four bases
(3) RNA possesses a double strand helix structure and contains uracil as one of the four bases
(4) RNA possesses a single strand helix structure and contains thymine as one of the four bases
Answer (1)
Sol. RNA is typically single standard but it contains uracil, not thymine. DNA is secondary structure has a double strand helix consisting of two polynucleotide chains. Its four nitrogenous bases are adenine (A), guanine (G), cytosine (C) and thymine (T).
55. During Lassaigne's test, the elements present in an organic compound are converted from:
(1) Ionic form to ionic form
(2) Covalent form to ionic form
(3) Ionic form to covalent form
(4) Covalent form to covalent form
Answer (2)
Sol. During Lassaigne's test, the element present in an organic compound are converted from covalent form to ionic form.
56. Mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law due to:
(1) Stronger intermolecular forces between chloroform molecules than those between chloroform and acetone molecules.
(2) Formation of hydrogen bonding between acetone and chloroform
(3) Repulsive forces.
(4) Increase in escaping tendency of molecules of each component.
Answer (2)
Sol. Acetone and chloroform show negative deviation from Raoult's law due to stronger H-bonding between acetone and chloroform mixture.
Hence, escaping tendency decreases
Vapour pressure decreases
Boiling point increases.
57. In a test tube containing a salt, a few drops of dilute H2SO4 was added, which gave colourless vapours having the smell of vinegar. The vapours turned the blue litmus paper red. Identify the correct anion from the following:
(1) Carbonate, CO3²⁻
(2) Sulphide, S²⁻
(3) Acetate, CH3COO⁻
(4) Sulphate, SO4²⁻
Answer (3)
Sol. Salt + dil. H2SO4 → CH3COOH (Acidic in Nature)
It turns Blue litmus paper in red
So, correct Answer is: Acetate, CH3COO⁻
58. Identify the correct statement about ClF3 from the following options:
(1) It has T-shaped geometry with three lone pairs on Cl atom.
(2) It has T-shaped geometry with two lone pairs on Cl atom.
(3) It has a trigonal pyramidal geometry with two lone pairs on Cl atom.
(4) It has a planar trigonal geometry with two lone pairs on Cl atom.
Answer (2)
Sol. ClF3 has two lone pairs of electrons on Cl atom with bent T-shape structure (geometry)
59. Match List I with List II:
List-I (Complex)
A. [Pt(NH3)2Cl2]
B. [Co(en)3]³⁺
C. [Co(NH3)5NO2]Cl2
D. [Cr(H2O)6]Cl3
List-II (Type of isomerism)
I. Optical
II. Solvate
III. Geometrical
IV. Linkage
Choose the correct answer from the options given below:
(1) A-I, B-III, C-II, D-IV
(2) A-II, B-IV, C-III, D-I
(3) A-III, B-I, C-IV, D-II
(4) A-III, B-I, C-II, D-IV
Answer (3)
60. The number of chlorine atoms present in the organic products X and Y of the following reactions, respectively, are:
(1) 3 and 3
(2) 6 and 3
(3) 3 and 6
(4) 6 and 6
Answer (2)
Sol.
61. Identify the incorrect statement from the following:
(1) Phosphorus, arsenic and antimony show catenation property.
(2) P(C2H5)3 and As(C6H5)3 form dπ-dπ bond with transition metals.
(3) Nitrogen can form dπ-pπ bond with oxygen.
(4) Nitrogen can form pπ-pπ multiple bonds with itself.
Answer (3)
Sol. Both nitrogen and oxygen do not contain d-orbitals so it cannot form dπ-pπ bond.
Phosphorus and Arsenic can form dπ-dπ bond with transition metals. Since both have vacant d-orbitals by which it can interact with transition metals and can involve in dπ-dπ interaction. Phosphorous, Arsenic and Antimony show catenation property.
62. At a certain temperature, T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:
(1) 700 J
(2) 400 J
(3) 300 J
(4) 500 J
Answer (3)
Sol. From first law of thermodynamics
ΔU = q + w
q = +500 J
w = -200 J
ΔU = 500 - 200
ΔU = 300 J
63. Compound P(C8H8O) gives a red orange precipitate with 2,4-DNP reagent and it does not reduce Fehling's reagent. On drastic oxidation with chromic acid, P gives an aromatic product Q that produces effervescence on treating with aq. NaHCO3. Compounds P and Q, respectively, are:
Answer (1 or 2)
Sol. Degree of unsaturation = (8 + 1) - (8/2) = 9 - 4 = 5
In question it is mentioned that oxidation under drastic condition, therefore it must be ketone
P = Acetophenone gives positive 2,4-DNP test
Chromic acid under drastic condition → Benzoic acid
Benzoic acid + NaHCO3 → Sodium benzoate + H2O + CO2↑
64. The correct IUPAC name of the following compound is:
CH3-CH2-CH(CH3)-CH2-CH(CH3)-CH2-CH3
(1) 3-ethyl-5-methylheptane
(2) 3-methyl-5-ethylheptane
(3) 2,4-diethylhexane
(4) 3,5-diethylhexane
Answer (1)
Sol. CH3-CH2-CH(CH3)-CH2-CH(CH3)-CH2-CH3
IUPAC name: 3-ethyl-5-methylheptane
- Numbering of parent chain should follow lowest locant rule.
- Prefixes should be written in alphabetical order.
65. In the following reaction sequence, X and Z respectively are:
CH3CH2CH2-OH + PCl5 → CH3CH2CH2Cl + X + HCl
(1) X = POCl3; Z = CH3-CH(Br)-CH3
(2) X = POCl3; Z = CH3CH2CH2-Br
(3) X = H3PO3; Z = CH3-CH(Br)-CH3
(4) X = H3PO3; Z = CH3CH2CH2-Br
Answer (2)
Sol. CH3CH2CH2-OH → PCl5 → CH3CH2CH2Cl + POCl3 + HCl (X)
CH3CH2CH2-Br ← HBr/(C6H5CO)2O2 ← CH3CH=CH2 (Y)
So, X = POCl3
Z = CH3CH2CH2Br
66. When 1 dm³ of CO2 gas is passed over hot coke, the volume of gaseous mixture after complete reaction at STP becomes 1.4 dm³. The composition of the gaseous mixture at STP is:
(1) 0.6 dm³ of CO, 0.8 dm³ of CO2
(2) 0.8 dm³ of CO, 0.8 dm³ of CO2
(3) 0.6 dm³ of CO, 0.4 dm³ of CO2
(4) 0.8 dm³ of CO, 0.6 dm³ of CO2
Answer (4)
Sol. CO2 + C(s) → 2CO
1
1 - x 2x
1 - x + 2x = 1 + x = 1.4
x = 0.4 dm³
Volume of CO2 = 1 - 0.4 = 0.6 dm³
Volume of CO = 2 × 0.4 = 0.8 dm³
67. The correct formal charges on oxygen atoms numbered 2, 1 and 3 respectively are:
(1) 0,0,0
(2) +1,0,-1
(3) -1,0,+1
(4) 0,+1,-1
Answer (4)
Sol. Formal charge = Valence electrons - 1/2(shared electrons) - (non-bonding electrons)
(V) (S) (L)
Formal charge on 2nd oxygen atom = 6 - 1/2(4) - 4 = 0
Formal charge on 1st oxygen atom = 6 - 1/2(6) - 2 = +1
Formal charge on 3rd oxygen atom = 6 - 1/2(2) - 6 = -1
68. Match List I with List II:
List-I (Order of reaction)
A. Zero order
B. First order
C. Second order
D. Third order
List-II (Unit of rate constant)
I. mol⁻¹ L s⁻¹
II. mol⁻² L² s⁻¹
III. s⁻¹
IV. mol L⁻¹ s⁻¹
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-II, D-I
(2) A-IV, B-III, C-I, D-II
(3) A-I, B-II, C-III, D-IV
(4) A-IV, B-II, C-I, D-III
Answer (2)
Sol. Unit for rate constant of nth order reaction = (mol/L)^(1-n) s⁻¹
For zero order reaction ⇒ n = 0 unit ⇒ mol L⁻¹ s⁻¹
For first order reaction ⇒ n = 1 unit ⇒ s⁻¹
For second order reaction ⇒ n = 2 unit ⇒ mol L⁻¹ s⁻¹
For third order reaction ⇒ n = 2 unit ⇒ mol L⁻² L² s⁻¹
69. The correct order of increasing metallic character of Na, Be, P, Mg and Si is
(1) Be < Si < P < Mg < Na
(2) P < Si < Na < Mg < Be
(3) P < Si < Be < Mg < Na
(4) P < Mg < Be < Si < Na
Answer (3)
Sol. On moving left to right in a period, metallic character decreases and on moving top to bottom in a group, metallic character increases. So, the correct order of increasing metallic character is P < Si < Be < Mg < Na
Electronegativity ∝ 1/Metallic character
Electronegativity (On Pauling scale) 0.9 Na Mg Be Si P 1.2 1.5 1.8 2.1
70. The number of hydrogen atoms present in 5.4 g of urea is:
(Given: Molar mass of urea : 60 g mol⁻¹, NA : 6.022 × 10²³ particles mol⁻¹)
(1) 2.168 × 10²²
(2) 2.168 × 10²³
(3) 1.084 × 10²²
(4) 1.084 × 10²³
Answer (2)
Sol. Structure of urea: H2N-CO-NH2
Mole of urea = 5.4/60 = 0.09
Number of hydrogen atoms = 0.09 × 4 × 6.022 × 10²³ = 2.1679 × 10²³ = 2.168 × 10²³
71. In a qualitative analysis, Bi³⁺ is detected by appearance of precipitate of BiO(OH)(s). Calculate pH when the following equilibrium exists at 298 K.
BiO(OH)(s) ⇌ BiO⁺(aq) + OH⁻(aq), K = 4 × 10⁻¹⁰
(Given: log2 = 0.3010)
(1) 4.699
(2) 5.286
(3) 8.714
(4) 9.301
Answer (4)
Sol. BiO(OH)(s) ⇌ BiO⁺(aq) + OH⁻(aq)
K = [BiO⁺][OH⁻]/[BiO(OH)(s)]
K = [BiO⁺][OH⁻]/1
K = s × s = s²
s = √K = √(4 × 10⁻¹⁰) = 2 × 10⁻⁵ M
[H⁺] = Kw/[OH⁻] = 1/2 × 10⁻⁹, pH = -log[H⁺] = 9 + log2 = 9.301
72. Match List I with List II:
List I
A. H3C-CH(CH3)-OH → OH
B. CH3COOH → CH3CH2OH
C. CH3CH2CH2OH → CH3-CH(OH)-CH3
D. Benzene → Phenol
List II
I. (i) Oleum; (ii) NaOH, Δ; (iii) H⁺
II. (i) O2; (ii) H2O/H⁺
III. (i) CH3OH, H⁺; (ii) H2, catalyst
IV. (i) conc. H2SO4, Δ; (ii) H⁺/H2O
Choose the correct answer from the options given below:
(1) A-I, B-III, C-IV, D-II
(2) A-II, B-IV, C-III, D-I
(3) A-II, B-III, C-IV, D-I
(4) A-II, B-III, C-I, D-IV
Answer (3)
Sol.
A. Cumene → (i) O2 (ii) H2O/H⁺ → Phenol
B. CH3COOH → (i) CH3OH, H⁺ (ii) H2, catalyst → CH3CH2OH
C. CH3CH2CH2OH → conc. H2SO4/Δ → CH3CH=CH2 → H⁺/H2O → CH3CH(OH)CH3
D. Benzene → (i) Oleum (ii) NaOH, Δ (iii) H⁺ → Phenol
73. A bulb is rated at 150 watt, converting 8% energy into light. If energy of one photon is 4.42 × 10⁻¹⁹ J, how many photons are emitted by the bulb per second?
(1) 27.2 × 10¹⁹
(2) 4.06 × 10¹⁹
(3) 1.35 × 10¹⁹
(4) 2.71 × 10¹⁹
Answer (4)
Sol. Energy = Power × time = 150 watt × 1 s = 150 J
Energy converted to light = 150 × 8/100 = 12 J
E = nhν
Energy of one photon = 4.42 × 10⁻¹⁹
So, n = number of photons = E/hν = 12/(4.42 × 10⁻¹⁹) = 2.715 × 10¹⁹
74. The pair of molecules that are metamers among the following is:
(1) CH3OCH2CH2CH3 and CH3CH2OCH2CH3
(2) CH3CH2CH2OH and CH3-CH(OH)-CH3
(3) CH3CH2CH2CH2CH3 and (CH3)2CHCH2CH3
(4) CH3COCH3 and CH3CH2CHO
Answer (1)
Sol. Metamerism arises due to different alkyl chains on either side of the functional group in the molecule pair of molecules CH3OCH2CH2CH3 and CH3CH2OCH2CH3 are metamers of each other.
75. Identify the correct statements:
(A) The molality of 2.5 g of ethanoic acid (Molar mass : 60 g mol⁻¹) in 75 g of benzene solution is 0.556 m.
(B) The molality of a solution containing 5 g of NaOH (molar mass : 40 g mol⁻¹) in 450 mL of solution is 0.278 M at 298 K.
(C) Aquatic species are more comfortable in cold water.
(D) The solubility of gas increases with decrease in pressure.
(E) For a binary mixture of A and B, the number of moles of A and B are nA and nB respectively. The mole fraction of B will be xB = nA/(nA + nB).
Choose the correct answer from the options given below:
(1) A,B and C only
(2) A,D and E only
(3) A and B only
(4) A and C only
Answer (1)
Sol. (A) Molality = 2.5/60 × 1000/75 = 0.556 mol
(B) Molarity = 5/40 × 1000/450 = 0.278 M
(C) Aquatic species are more comfortable in cold water
Henry's Law KH ∝ Temp ∝ 1/solubility
(D) According to Henry's Law,
P = KH X
P ∝ X
P ∝ Solubility
So as P ↑, solubility ↑
(E) XB = nB/(nA + nB)
76. Which one of the following is an ambidentate ligand?
(1) Oxalate
(2) Ethane-1,2-diamine
(3) Thiocyanate
(4) Ethylenediaminetetraacetate ion
Answer (3)
Sol. An ambidentate ligand is a ligand which has two different donor atoms and either of the two ligates in complex.
(1) Oxalate → didentate
(2) Ethane-1,2-diamine → didentate
(3) Thiocyanate (SCN)⁻ → Ambidentate ligand.
(4) Ethylenediamine tetraacetate ion → hexadentate
77. The functional group that can be identified through phthalein dye test is:
(1) Carboxylic acid
(2) Alcohol
(3) Aldehyde
(4) Phenolic
Answer (4)
Sol. Phthalein Dye test: Phenol on heating with phthalic anhydride in presence of concentrated sulphuric acid forms a colourless condensation compound called phenolphthalein. On further reaction with NaOH it gives pink colour.
So phenolic group is correct answer.
78. Match List I with List II:
List I
A. C2H4
B. C2H2
C. CH4
D. NH3
List II
I. 3 σ bonds, 2 π bonds
II. 3 σ bonds, one lone pair
III. 4 σ bonds
IV. 5 σ bonds, 1 π bond
Choose the correct answer from the options given below:
(1) A-IV, B-I, C-III, D-II
(2) A-III, B-IV, C-II, D-I
(3) A-II, B-III, C-I, D-IV
(4) A-I, B-II, C-IV, D-III
Answer (1)
Sol. C2H4 ⇒ H-C=C-H
Number of σ bonds = 5
Number of π bond = 1
C2H2 ⇒ H-C≡C-H
Number of σ-bonds = 3
Number of π-bonds = 2
CH4 ⇒ H-C-H
Number of σ-bonds = 4
NH3 ⇒ H-N-H
Number of σ-bonds = 3
Number of lone pair = one
A-IV, B-I, C-III, D-II
79. A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is:
(Given: Molar mass of Cu = 63 g mol⁻¹, 1 F = 96487 C mol⁻¹)
(1) 0.2938 g
(2) 0.5876 g
(3) 2.4036 g
(4) 1.7018 g
Answer (1)
Sol. CuSO4 → Cu²⁺ + SO4²⁻
2e⁻ + Cu²⁺ → Cu(s)
W = zit = E × i × t/96500 = (63 × 1.5 × 10 × 60)/(2 × 96500) = 56,700/193000 = 0.2938 g
80. Match List I with List II:
List I (Quantum Numbers)
A. n l
B. 2 1
C. 4 0
D. 5 3
E. 3 2
List II (Orbital)
I. 3d
II. 2p
III. 4s
IV. 5f
Choose the correct answer from the options given below.
(1) A-II, B-III, C-IV, D-I
(2) A-I, B-II, C-III, D-IV
(3) A-II, B-III, C-I, D-IV
(4) A-IV, B-II, C-III, D-I
Answer (1)
Sol. l represents the subshell, for which the values are as following:
l Subshell
0 s
1 p
2 d
3 f
81. The major product Z formed in the following sequence of reactions is
C2H6 → Cl2/UV light → X (monochlorinated product) → NH3 → Y → (i) NaNO2/HCl (ii) H2O → Z
(1) C2H5-N=N-OH
(2) C2H5OH
(3) C2H5NO2
(4) C2H5NH2
Answer (2)
Sol. C2H6 + Cl2 → UV → C2H5Cl + HCl (X)
NH3 → -HCl → C2H5NH2 (Y)
NaNO2/HCl → [C2H5-N2⁺Cl⁻] → H2O → C2H5OH + N2 + HCl (Z)
82. Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:
(1) After losing one more electron, it acquires 4f¹⁴ electronic configuration.
(2) Its nearest inert gas is Radon.
(3) After losing one more electron, it acquires 4f⁰ electronic configuration.
(4) Its atomic number is 61.
Answer (3)
Sol. Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because after losing one more electron it acquires 4f⁰ electronic configuration.
83. Methane reacts with steam at 1273 K in the presence of nickel catalyst to form:
(1) CO and H2
(2) CO and H2O
(3) CO2 and H2
(4) CO2 and H2O
Answer (1)
Sol. CH4(g) + H2O(g) → Ni → CO + 3H2
(Used for industrial preparation of dihydrogen gas)
84. Identify the incorrect statement from the following:
(1) Oxygen exhibits only -2 oxidation state.
(2) The order of catenation property of Group 14 elements is C >> Si > Ge ≈ Sn.
(3) Carbon has the ability to form pπ-pπ multiple bond with itself.
(4) ECl3 (E = B and Al) is a monomer when E = B and a dimer when E = Al.
Answer (1)
Sol. (1) Oxygen exhibits oxidation state of 0 in O2, -2 in oxides, -1 in peroxides. So this statement is incorrect.
(2) Catenation property of group 14 C >> Si > Ge ≈ Sn.
(3) C can form pπ-pπ multiple bond with itself. It is observed when it forms C=C and C≡C
(4) ECl3 (E = B and Al) BCl3 does not form dimer. AlCl3 can form dimer in Al2Cl6.
85. Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at an alkaline pH close to the equivalence point during this titration is:
(1) Pink to colourless
(2) Pinkish red to yellow
(3) Colourless to pink
(4) Yellow to pinkish red
Answer (3)
Sol. Colour of Phenolphthalein before the end point = colourless.
Colour of Phenolphthalein close to equivalence point = Pink.
∴ Colour change = Colourless to pink.
86. Identify the incorrect statement from the following:
(1) The largest and the smallest species among Mg, Mg²⁺, Al and Al³⁺ are Al and Mg²⁺ respectively.
(2) The IUPAC name of the element with atomic number 107 is Unnisleptium.
(3) The similarity in behaviour of Li with Mg is referred to as 'diagonal relationship'
(4) The oxidation state and covalency of Al in [AlCl(H2O)5]²⁺ are 3 and 6 respectively.
Answer (1)
Sol. The largest species is Mg and the smallest one is Al³⁺ among Mg, Mg²⁺, Al and Al³⁺.
Unnisleptium element has atomic number 107
Li and Mg are diagonally related and have similar properties.
The oxidation state and covalency of Al in [AlCl(H2O)5]²⁺ are +3 and 6 respectively.
Statement A is incorrect
87. The following two reactions give the same foul smelling product Z.
C2H5Cl → X → Z
C2H5CNH2 → Br2, NaOH → Y → CHCl3/ethanolic KOH/Δ → Z
X and Z, respectively, are:
(1) X = AgCN; Z = C2H5CN
(2) X = AgCN; Z = C2H5NC
(3) X = KCN; Z = C2H5CN
(4) X = KCN; Z = C2H5NC
Answer (2)
Sol. C2H5Cl → AgCN → C2H5NC(Z) (foul smell compound)
C2H5CNH2 → Br2 + NaOH → C2H5NH2 → CHCl3/ethanolic KOH/Δ → C2H5NC
So, X = AgCN
Z = C2H5NC
88. The calculated 'spin-only' magnetic moment of Ti²⁺ (3d²) is:
(1) 3.87 BM
(2) 4.90 BM
(3) 2.84 BM
(4) 5.92 BM
Answer (3)
Sol. Spin only magnetic moment μ = √(n(n+2)) B.M
n = Number of unpaired e⁻(s)
Electronic configuration of Ti²⁺ ⇒ [Ar]4s⁰3d²
μ = √(2 × 4) = 2.84 BM
89. Match List I with List II:
List I (Transition metal/compound/complex)
A. V2O5
B. Fe
C. PdCl2
D. Ni complex
List II (Catalytic Role)
I. Preparation of ammonia from N2/H2 mixture
II. Polymerisation of alkynes
III. Preparation of H2SO4 and SO2
IV. Oxidation of ethyne to ethanol
Choose the correct answer from the options given below.
(1) A-III, B-IV, C-I, D-II
(2) A-III, B-I, C-IV, D-II
(3) A-III, B-I, C-IV, D-II
(4) A-IV, B-I, C-III, D-II
Answer (2)
Sol. (A) V2O5 → Catalyses the oxidation of SO2 into SO3 in the manufacture of H2SO4 (III)
(B) Fe → Act as catalysts in preparation of ammonia from N2/H2 mixture. (I)
(C) PdCl2 → Oxidation of ethyne to ethanol. (IV)
(D) Ni complex → Polymerisation of alkynes. (II)
A → III, B → I, C → IV, D → II
90. Two products X and Y are formed in the following reaction sequence.
Benzene + CH3Cl → Anhydr. AlCl3 → W → dil. HNO3 + dil. H2SO4 warm → X + Y
The suitable method that can be used for the separation of products X and Y is:
(1) Continuous extraction
(2) Differential extraction
(3) Sublimation
(4) Fractional distillation
Answer (4)
Sol. Benzene + CH3Cl → Anhyd. AlCl3 → Toluene (W) → dil. HNO3 + dil. H2SO4 warm → ortho-isomer (X) + para-isomer (Y)
The ortho and para isomers are separated by fractional distillation under reduced pressure.
o-isomer, (M.P. → -4°C, B.P. → 222°C)
p-isomer, (M.P. → 54°C, B.P. → 238°C)
91. Match List I with List II:
List-I
A. Genetically modified organism
B. Thermostable DNA polymerase
C. Ti plasmid
D. pBR322
List-II
I. Agrobacterium tumefaciens
II. Bt cotton
III. Thermus aquaticus
IV. Escherichia coli
Choose the correct answer from the options given below:
(1) A-II, B-I, C-IV, D-III
(2) A-II, B-III, C-I, D-IV
(3) A-I, B-IV, C-III, D-II
(4) A-I, B-II, C-IV, D-III
Answer (2)
Sol. Genetically modified organism - Bt cotton
Thermostable DNA polymerase - Thermus aquaticus.
Ti plasmid - Agrobacterium tumefaciens
pBR322 - Escherichia coli
92. Exploring molecular, genetic and species-level diversity for products of economic importance is called
(1) Bioprospecting
(2) Biofortification
(3) Biomagnification
(4) Bioremediation
Answer (1)
Sol. Exploring molecular, genetic and species-level diversity for products of economic importance is called as bioprospecting.
93. Which of the following statements are true with reference to the sex-determination in honeybees?
A. An offspring formed from the union of a sperm and an egg, develops as a female (queen or worker).
B. An unfertilized egg develops as a male by parthenogenesis.
C. A male has half the number of chromosomes than that of a female.
D. Males produce sperms by meiosis.
E. Honeybees have a haploidploid sex-determination system.
Choose the correct answer from the options given below:
(1) A,B,C and E only
(2) A,B,C and D only
(3) B,C,D and E only
(4) A,B,D and E only
Answer (1)
94. Match List-I with List-II
List-I (Growth Regulator)
A. 2,4-D
B. GA3
C. Kinetin
D. ABA
List-II (Function/Effect)
I. Brewing industry
II. Stimulation of stomatal closure
III. Herbicide
IV. Nutrient mobilisation
Choose the correct answer from the options given below:
(1) A-I, B-II, C-IV, D-III
(2) A-I, B-IV, C-III, D-II
(3) A-IV, B-III, C-II, D-I
(4) A-III, B-I, C-IV, D-II
Answer (4)
Sol. List-I (Growth Regulator) / List-II (Function/Effect)
A. 2,4-D (Auxin) / III. Herbicide
B. GA3 (Gibberellic Acid) / I. Brewing industry
C. Kinetin (Cytokinin) / IV. Nutrient mobilisation
D. ABA (Abscisic Acid) / II. Stimulation of stomatal closure
95. In racemose inflorescence,
(1) The growth is limited
(2) Flowers are solitary
(3) Flowers are borne in an acropetal succession
(4) The main axis terminates in a flower
Answer (3)
Sol. In racemose type of inflorescence, the main axis continues to grow and the flowers are borne laterally in an acropetal succession. On the contrary, in cymose type of inflorescence, the main axis terminates into a flower and hence, is limited in growth.
96. Since the origin and diversification of life on Earth, there have been five episodes of mass extinction of species. How is the sixth extinction, which is in progress, different from the previous episodes?
(1) The current species extinction rates are far lower than those in previous episodes.
(2) The present species extinction rates are 100 to 1000 times faster than in the pre-human times.
(3) The present net species extinction rate is zero.
(4) The current species extinction rate is nearly 10 times faster than in previous episodes.
Answer (2)
Sol. The current, sixth episode of mass extinction is estimated to be 100 to 1000 times faster than the ones in the pre-human times and our activities are responsible for the faster rates.
97. Alpha-helix is found in which level of protein structure?
(1) Secondary structure
(2) Primary structure
(3) Tertiary structure
(4) Quaternary structure
Answer (1)
Sol. Alpha-helix is shown by secondary structure of protein as it has right handed helices. Primary structure is linear and tertiary structure is a hollow ball-like structure. Quaternary structures are formed by more than one polypeptide chains.
98. The enzyme required for carboxylation in the Calvin cycle is
(1) PEP carboxylase
(2) RuBP carboxylase-oxygenase
(3) Carboxypeptidase
(4) Hexokinase
Answer (2)
Sol. RuBisCO (RuBP carboxylase-oxygenase) is the enzyme required for carboxylation in the calvin cycle.
99. Arrange the following in the correct developmental sequence related to microprogression:
A. Microspore tetrads
B. Sporogenous tissue
C. Pollen grains
D. Pollen mother cells
Choose the correct answer from the options given below:
(1) A, D, C, B
(2) D, A, C, B
(3) B, D, C, A
(4) B, D, A, C
Answer (4)
Sol. The process of formation of microspores from a pollen mother cell (PMC) through meiosis is called microprogression. The correct developmental sequence related to microprogression will be - Sporogenous tissue (B) → Pollen mother cell (D) → Microspore tetrads (A) → Pollen grains (C)
100. Which of the following statements are not true regarding restriction endonucleases?
A. They are called molecular scissors.
B. These are the enzymes responsible for restricting the growth of bacteriophages in E. coli.
C. They cut the DNA only at the centre of the palindromic sites.
D. They remove nucleotides only from the ends of DNA fragments.
E. They recognise specific palindromic base-pair sequences.
Choose the answer from the options given below:
(1) C and D only
(2) A and E only
(3) D and E only
(4) A and B only
Answer (1)
Sol. Statements C and D are incorrect. Restriction endonucleases usually cut the DNA slightly away from the centre of palindromic sites. They cannot remove nucleotides from the ends of the DNA fragment, which is function of restriction exonuclease.
101. In the lac operon, the z gene codes for
(1) the repressor of lac operon
(2) transacetylase
(3) permease
(4) beta-galactosidase
Answer (4)
Sol. In lac operon, i gene codes for - regulator protein
z gene codes for - beta-galactosidase
y gene codes for - Permease
a gene codes for - transacetylase
102. Match List I with List II:
List-I (Phase of cell cycle)
A. G1 phase
B. S phase
C. G2 Phase
D. M phase
List-II (Activity)
I. Actual cell division occurs
II. Cell is metabolically active and continuously grows but does not replicate its DNA
III. Synthesis of DNA occurs and the amount of DNA per cell doubles
IV. Proteins are synthesized while cell growth continues
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-I, D-II
(2) A-IV, B-I, C-II, D-III
(3) A-I, B-II, C-III, D-IV
(4) A-II, B-III, C-IV, D-I
Answer (4)
103. 2(C51H98O6) + 145O2 → 102CO2 + 98H2O + energy
The Respiratory Quotient (RQ) of a biomolecule used for respiration, as per the above equation would be:
(1) Less than 0.5
(2) Between 1.25 and 2
(3) 1.0
(4) Between 0.5 and 0.95
Answer (4)
Sol. RQ = Volume of CO2 evolved / Volume of O2 consumed
As per the given equation,
2(C51H98O6) + 145O2 → 102CO2 + 98H2O + Energy
Hence the RQ can be calculated as,
RQ = 102/145 = 0.7
This value lies between 0.5 and 0.95.
104. Which one of the following is not a characteristic of plant cells in the phase of elongation?
(1) New cell wall deposition
(2) Cell enlargement
(3) Large conspicuous nuclei
(4) Increased vacuolation
Answer (3)
Sol. Phase of elongation is characterised by new cell wall deposition, increased vacuolation and enlargement of cell. Presence of large conspicuous nuclei is a feature of cells in the meristematic phase.
105. Arrange the following steps of somatic hybridisation in a correct sequence.
A. Digestion of cell walls.
B. Isolation of naked protoplasts.
C. Fusion of protoplasts to get hybrid protoplast.
D. Isolation of single cells from two different varieties of plants.
E. Growing of hybrid protoplast to form a new plant.
Choose the correct answer from the options given below:
(1) D, B, A, E, C
(2) E, A, B, C, D
(3) E, B, A, D, C
(4) D, A, B, C, E
Answer (4)
Sol. Scientist have even isolated single cell protoplasts from two different varieties of plants each having a desirable character and after digesting their cell walls have been able to isolate naked protoplasts (surrounded by plasma membrane) that can be fused to get hybrid protoplast, which can be further grown to form a new plant. These hybrids are called somatic hybrids, while the process is called somatic hybridization.
So the correct sequence for the formation of somatic hybrids is D, A, B, C, E.
106. Match List-I with List-II:
List-I
A. Conjunctive tissue
B. Casparian strips
C. Subsidiary cells
D. Starch sheath
List-II
I. Specialised cells in the vicinity of guard cells
II. Endodermal cells rich in starch
III. Tissue between xylem and phloem
IV. Endodermal cells with suberin deposition
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-II, D-I
(2) A-III, B-IV, C-I, D-II
(3) A-IV, B-III, C-I, D-II
(4) A-III, B-IV, C-II, D-I
Answer (2)
Sol. (A) Conjunctive tissue is the tissue between xylem and phloem.
(B) Casparian strips are found in endodermal cells, they are suberin depositions in the cell wall.
(C) Subsidiary cells are specialised cells in the vicinity of guard cells.
(D) Starch sheath is another name for endodermal cells rich in starch.
107. In angiosperms, root hairs arise from which one of the following regions of the root?
(1) The region of elongation
(2) The region of meristematic activity
(3) The region of maturation
(4) The root cap zone
Answer (3)
Sol. From the region of maturation, some epidermal cells form very fine and delicate, thread like structures called root hairs.
108. Which of the following floral formula is the correct floral formula of Solanaceae family?
(1) ⊕ ♀ K(5) C(5) A5 G(2)
(2) ⊕ ♀ K5 C5 A5 G(2)
(3) ⊕ ♀ K(5) C(5) A5 G(2)
(4) ⊕ ♀ K5 C(5) A5 G(2)
Answer (3)
Sol. ⊕ ♀ K(5) C(5) A5 G(2)
Is the floral formula for Solanaceae.
It shows actinomorphic, bisexual, pentamerous flower with epipetalous condition
Generally in Solanaceae, calyx (K) and Corolla (C) shows fusion in sepals and petals respectively. Hence the floral formula must exhibit gamosepalous and gamopetalous condition.
109. Which one of the following is a triploid cell?
(1) Synergid
(2) Primary endosperm cell
(3) Central cell
(4) Zygote
Answer (2)
Sol. Synergid is haploid, Zygote is diploid, Central cell initially contains two polar nuclei which fuse just before fertilization to form a secondary nucleus (2n). Primary endosperm cell (PEC) is triploid.
110. Match List I with List II:
List-I
A. Decomposition
B. Detritus
C. Mineralisation
D. Humification
List-II
I. Accumulation of dark coloured amorphous colloidal substance
II. Release of inorganic nutrients by the activity of microbes in soil
III. Breaking down of complex organic matter into inorganic substances.
IV. Dead remains of plants and animals including fecal matter
Choose the correct answer from the options given below:
(1) A-III, B-II, C-I, D-IV
(2) A-IV, B-III, C-I, D-II
(3) A-I, B-II, C-III, D-IV
(4) A-III, B-IV, C-II, D-I
Answer (4)
Sol. Decomposition is the process of breaking down of complex organic matter into inorganic substance. Detritus includes dead remains of plants and animals including fecal matter and acts as the raw material for decomposition. Degradation of humus by activity of microbes leading to release of inorganic nutrients is called Mineralisation. Accumulation of the dark coloured amorphous substance called humus, is called humification.
111. The main criteria used for Five Kingdom Classification proposed By R.H. Whittaker (1969) included:
A. Cell structure
B. Body organization
C. Presence of flagellum
D. Reproduction
E. Phylogenetic relationships
Choose the correct answer from the options given below:
(1) A, B and E only
(2) A, B, C, D and E
(3) B, C and D only
(4) A, B, D and E only
Answer (4)
Sol. The main criteria for five kingdom classification used by (R.H. Whittaker) includes cell structure, body organization, mode of nutrition, reproduction and phylogenetic relationships.
112. "The Evil Quartet" of biodiversity loss includes which of the following?
(1) Over-exploitation; Alien species invasions; Soil pollution; Co-extinctions
(2) Habitat loss and fragmentation; Air pollution; Water pollution; Co-extinctions
(3) Habitat loss and fragmentation; over-exploitation; Alien species invasions; Co-extinctions
(4) Over-exploitation; Alien species invasions; Air pollution; Co-extinctions
Answer (3)
Sol. "The Evil Quartet" is the Sobriquet used to describe the four major causes of biodiversity loss which includes Habitat loss and fragmentation, over-exploitation; Alien species invasions and Co-extinctions.
113. Arrange the following steps of DNA fingerprinting in a correct sequence.
A. Isolation of DNA and its digestion by restriction endonucleases.
B. Hybridisation using a labelled VNTR probe.
C. Transferring of separated DNA fragments to synthetic membranes.
D. Detection of hybridised DNA fragments by autoradiography.
E. Separation of DNA fragments by electrophoresis.
Choose the correct answer from the options given below:
(1) A, E, B, C, D
(2) A, D, B, E, C
(3) A, B, D, C, E
(4) A, E, C, B, D
Answer (4)
Sol. The following is the correct sequence of steps of DNA fingerprinting.
A. Isolation of DNA and its digestion by restriction endonucleases.
B. Probes made complementary to the VNTR locus are allowed to hybridise with the DNA fragments.
C. The separated DNA fragments are transferred to synthetic membranes made of nylon or nitrocellulose.
D. Finally, the hybridised DNA fragments are detected under X-rays in a technique called autoradiography.
E. DNA fragments are separated based on their size by the technique of gel electrophoresis.
114. Which of the following statements are correct with reference to a transcription unit?
A. A transcription unit in DNA is defined primarily by three regions: promoter, structural gene and terminator.
B. The promoter is said to be located towards the 5'-end of the structural gene.
C. The promoter is a DNA sequence that provides binding site for RNA polymerase.
D. The promoter defines the template and coding strands.
E. The terminator is located towards the 3'-end of the coding strand and it defines the end of the process of transcription.
Choose the correct answer from the options given below:
(1) B, C, D and E only
(2) A, B, C, D and E
(3) A, B, C and D only
(4) A, C, D and E only
Answer (2)
Sol. The promoter and terminator flank the structural gene in a transcription unit. The promoter is said to be located towards 5'-end of the structural gene. It is a DNA sequence that provides binding site for RNA polymerase and it is the presence of promoter in a transcription unit that also defines the template and coding strands. The terminator is located towards 3'-end of the coding strand and it usually defines the end of transcription.
115. Which one of the following types of pollination brings genetically different types of pollen grains to the stigma?
(1) Geitonogamy
(2) Xenogamy
(3) Cleistogamy
(4) Autogamy
Answer (2)
Sol. Transfer of pollen grains from anther to stigma of a different plant is known as Xenogamy. This is the only type of pollination which brings genetically different types of pollen grains to stigma. Cleistogamy flowers are invariably autogamous.
116. Which of the following is an in situ conservation method?
(1) Seed Banks
(2) Sacred Groves
(3) Botanical Gardens
(4) Wildlife Safari Parks
Answer (2)
Sol. In-situ conservation is exemplified by sacred groves. Wildlife Safari parks, Botanical gardens, seed banks are examples of ex-situ conservation.
117. Heterophyllous development in response to environment is an example of which of the following phenomena?
(1) Redifferentiation
(2) Dedifferentiation
(3) Elasticity
(4) Plasticity
Answer (4)
Sol. Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity, e.g., heterophylly in cotton, coriander and larkspur.
118. Match List I with List II:
List I
A. Productivity
B. Net primary productivity
C. Gross primary productivity
D. Secondary productivity
List II
I. Gross primary productivity minus respiration losses
II. Rate of formation of new organic matter by consumers
III. Rate of biomass production
IV. Rate of production of organic matter during photosynthesis
Choose the correct answer from the options given below:
(1) A-III, B-I, C-II, D-IV
(2) A-I, B-II, C-III, D-IV
(3) A-III, B-I, C-IV, D-II
(4) A-I, B-III, C-IV, D-II
Answer (3)
Sol. Productivity is the rate of biomass production.
Net primary productivity is the Gross primary productivity (GPP) minus respiration losses (R).
Gross primary productivity is the rate of production of organic matter during photosynthesis.
Secondary productivity is the rate of formation of new organic matter by consumers.
119. Which of the following statements are correct regarding amino acids?
A. They are substituted methanes.
B. Serine is an aromatic amino acid.
C. Valine is a neutral amino acid.
D. Lysine is an acidic amino acid.
Choose the correct answer from the options given below:
(1) A and B only
(2) C and D only
(3) B and C only
(4) A and C only
Answer (4)
Sol. Statements A and C are correct while statements B and D are incorrect. Serine is an alcoholic amino acid and lysine is a basic amino acid.
120. In which one of the following, the ovules are not enclosed by an ovary wall and remain exposed?
(1) Pinus
(2) Wolffia
(3) Funaria
(4) Selaginella
Answer (1)
Sol. The gymnosperms are plants in which the ovules are not enclosed by any ovary wall and remain exposed, both before and after fertilisation. Pinus is a gymnosperm. Funaria is a moss. Selaginella is a pteridophyte.
121. Which of the following statements are correct with reference to packaging of DNA helix?
A. Histones are organized to form a unit of eight molecules called histone octamer.
B. Histones are negatively charged basic proteins.
C. Histones are rich in the basic amino acid residues - lysine and arginine.
D. The positively charged DNA is wrapped around the histone octamer to form nucleosome.
E. The packaging of chromatin at higher levels requires an additional set of proteins called non-histone chromosomal proteins.
Choose the correct answer from the options given below:
(1) B, D and E only
(2) A, B and D only
(3) C, D and E only
(4) A, C and E only
Answer (4)
Sol. Eight Histones are combined to form a histone octamer. Histones are positively charged basic proteins that are rich in basic amino acids (lysine and arginine). The negatively charged DNA is wrapped around the histone octamer to form nucleosome. The packaging of chromatin at higher levels require NHC (non histone chromosomal) proteins.
122. Match List I with List II:
List I (Placentation)
A. Marginal
B. Axile
C. Parietal
D. Basal
List II (Example)
I. Mustard
II. Pea
III. Marigold
IV. Lemon
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-I, D-III
(2) A-IV, B-II, C-I, D-III
(3) A-III, B-I, C-IV, D-II
(4) A-I, B-III, C-II, D-IV
Answer (1)
Sol. Marginal placentation is found in pea, axile placentation is found in lemon, parietal placentation is found in mustard and basal placentation is found in marigold.
123. Which one of the following is the site for active ribosomal RNA synthesis?
(1) Nucleolus
(2) Kinetochore
(3) Centrosome
(4) Chromatin
Answer (1)
Sol. Nucleolus is the site for active ribosomal RNA synthesis.
124. The main function of bulliform cells in grasses is:
(1) to perform photosynthesis.
(2) to minimize water loss during water stress.
(3) to make the leaf impermeable to fungal spores.
(4) to transport water.
Answer (2)
Sol. Bulliform cells are large empty colourless cells that lose water and become flaccid in water scarce condition. Hence they curl the leaf inwards to minimise water loss by reducing the exposed surface area.
125. Which of the following statements are correct?
A. The Amazon rainforest being cut and cleared for cultivation of soyabeans is an example of habitat loss.
B. Steller's sea cow and passenger pigeon became extinct due to over-exploitation by humans.
C. The Nile perch introduced into Lake Victoria in East Africa helped in population growth of cichlid fish in the lake.
D. Water hyacinth is an invasive species.
E. When a species becomes extinct, the plant and animal species associated with it are not affected.
Choose the correct answer from the options given below:
(1) A, B and D only
(2) B, C and D only
(3) A, B and E only
(4) C, D and E only
Answer (1)
Sol. The Nile Perch introduced into Lake Victoria in East Africa led eventually to extinction of an ecologically unique assemblage of more than 200 species of cichlid fish in the lake.
When a species become extinct, the plant and animal species associated with it in an obligatory way also become, extinct.
Statements, A, B and D are correct.
126. Which one of the following statements is not true about the universal rules of binomial nomenclature?
(1) The first word in the biological name represents the specific epithet, while the second component denotes the genus
(2) The specific epithet in the biological name starts with a small letter
(3) Both the words in a biological name, when handwritten, are separately underlined or printed in italics
(4) Biological names are generally in Latin
Answer (1)
Sol. According to universal rules of nomenclature, the first word denoting the genus starts with a capital letter with the second components denotes the specific epithet and starts with small letter.
127. Which one of the following disorders is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule?
(1) Phenylketonuria
(2) Haemophilia
(3) Sickle-cell anaemia
(4) Thalassemia
Answer (3)
Sol. Sickle cell anaemia is an autosome linked recessive trait that can be transmitted from parents to offsprings when both partners are carrier for the gene. The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at sixth position of beta globin chain of haemoglobin molecule.
128. Find the incorrect statement(s) about photosynthesis from the following:
A. The water splitting complex is associated with PS I.
B. C4 plants use the C3 pathway of CO2 fixation as the main biosynthetic pathway.
C. In C4 plants, photorespiration does not occur.
D. C3 plants exhibit 'Kranz' anatomy.
E. ATP synthesis in chloroplast occurs through chemiosmosis.
Choose the answer from the options given below:
(1) B and C only
(2) B and E only
(3) B only
(4) A and D only
Answer (4)
Sol. The water splitting complex is associated with PS-II. C3 pathway is the main biosynthetic pathway for CO2 fixation in both C3 and C4 plants. C3 plants do not exhibit 'kranz' anatomy. ATP synthesis in chloroplast (Photophosphorylation) occurs by chemiosmosis.
129. Match List I with List II:
List I
A. Trypsin
B. Morphine
C. Concanavalin A
D. Collagen
List II
I. Intercellular ground substance
II. Lectin
III. Enzyme
IV. Alkaloid
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-II, D-I
(2) A-IV, B-III, C-II, D-I
(3) A-I, B-II, C-II, D-IV
(4) A-III, B-II, C-IV, D-I
Answer (1)
Sol. Trypsin is a proteolytic enzyme.
Morphine is a secondary metabolite that belongs to the category of alkaloid. Concanavalin A is a lectin. Collagen acts as an intercellular ground substance.
130. Identify the correct statements about biomolecules.
A. Lipids are generally water soluble.
B. Proteins are polypeptides.
C. Polysaccharides are long chains of sugars.
D. Adenine and guanine are substituted pyrimidines.
E. Almost all enzymes are proteins.
Choose the correct answer from the options given below:
(1) A, B and C only
(2) B, D and E only
(3) B, C and E only
(4) C, D and E only
Answer (3)
Sol. Statements B, C and E are correct. Statements A and D are not true. Lipids are not water soluble. Adenine and guanine are substituted purines.
131. Match List I with List II:
List-1
A. Incomplete dominance
B. Co-dominance
C. Pleiotropy
D. Polygenic inheritance
List-2
I. Human skin colour
II. Inheritance of flower colour in Antirrhinum sp.
III. Phenylketonuria disease in humans
IV. ABO blood groups
Choose the correct answer from the options given below:
(1) A-I, B-IV, C-III, D-II
(2) A-I, B-III, C-II, D-IV
(3) A-II, B-IV, C-III, D-I
(4) A-II, B-I, C-III, D-IV
Answer (3)
Sol. Inheritance of flower colour in snapdragon (Antirrhinum sp.) is an example of incomplete dominance.
ABO blood groups exhibit codominance in case of individuals having AB blood group (IAIB)
Phenylketonuria in humans is an example of pleiotropy since the gene responsible for it leads to multiple phenotypic effects.
Human skin colour is controlled by three pairs of non allelic genes, hence it is an example of polygenic inheritance.
132. Identify the correct sequence of steps in each cycle of Polymerase Chain Reaction:
(1) Denaturation → Extension → Annealing
(2) Denaturation → Annealing → Extension
(3) Annealing → Denaturation → Extension
(4) Extension → Annealing → Denaturation
Answer (2)
Sol. The correct sequence of steps in PCR is Denaturation → Annealing → Extension
133. How many ATP and NADPH molecules are required to make one molecule of glucose through the Calvin pathway?
(1) 12 ATP and 18 NADPH
(2) 18 ATP and 12 NADPH
(3) 6 ATP and 12 NADPH
(4) 24 ATP and 18 NADPH
Answer (2)
Sol. Each turn of Calvin pathway utilizes 3ATP and NADPH + H+ molecules for fixation of 1 CO2 molecule. So for 1 Glucose 6 turns are required, hence 18 ATP and 12 NADPH + H+ are required for glucose synthesis.
134. Match List-I with List-II:
List-I (Process)
A. Glycolysis
B. ETS
C. Accumulation of protons
D. Krebs' cycle
List-II (Location)
I. Inner mitochondrial membrane
II. Mitochondrial matrix
III. Cytoplasm
IV. Intermembrane space
Choose the correct answer from the options given below:
(1) A-IV, B-II, C-I, D-III
(2) A-I, B-IV, C-III, D-II
(3) A-II, B-III, C-IV, D-I
(4) A-III, B-I, C-IV, D-II
Answer (4)
Sol. The site of glycolysis is the cytoplasm in all living organisms. Electron transport system is localized in inner mitochondrial membrane. Accumulation of protons occur in intermembrane space and Krebs's Cycle takes place in mitochondrial matrix.
135. Which of the following statements are correct with respect to DNA separation, isolation and visualization?
A. The cutting of DNA is done by molecular scissors.
B. The DNA fragments separate according to their size in an agarose gel, upon electrophoresis.
C. The separated DNA fragments can be seen without staining when exposed to UV light.
D. The separated DNA fragments, when stained with ethidium bromide, can be seen in visible light.
Choose the correct answer from the options given below:
(1) A and B only
(2) B and D only
(3) A and D only
(4) B and C only
Answer (1)
Sol. The cutting of DNA is possible by the use of restriction enzymes that results in the fragments of DNA. These fragments of DNA can be separated by a technique known as gel electrophoresis. The DNA fragments separate (resolve) according to their size through sieving effect provided by the agarose gel. The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation.
136. What is the probability of having children with 'O' blood group, where both mother and father are heterozygous for 'A' and 'B' blood group, respectively?
(1) 0%
(2) 50%
(3) 25%
(4) 75%
Answer (3)
Sol. Parent: IAi × IBi
gametes: IA, i and IB, i
F1: IAIB (AB), IAi (A), IBi (B), ii (O)
Out of four children, one is with blood group 'O'.
The probability of having children with 'O' blood group will be 25%
137. Match List I with List II:
List I
A. Molluscs
B. Reptiles
C. Adult amphibians
D. Amoeba
List II
I. Pulmonary respiration only
II. Branchial respiration
III. Cellular respiration
IV. Pulmonary and cutaneous respiration
Choose the correct answer from the options given below:
(1) A-III, B-II, C-II, D-IV
(2) A-II, B-I, C-II, D-IV
(3) A-II, B-I, C-IV, D-III
(4) A-II, B-I, C-IV, D-III
Answer (4)
Sol. (A) Molluscs → (II) Perform branchial respiration by using feather-like gills
(B) Reptiles → (I) Perform pulmonary respiration only via lungs
(C) Adult amphibians → (IV) Perform pulmonary and cutaneous respiration via lungs and moist skin, respectively
(D) Amoeba → (III) Performs cellular respiration to generate ATP for survival
Thus, (A)-II, (B)-I, (C)-IV, (D)-III
138. Insertion of a foreign DNA at BamHI site in an E.coli cloning vector pBR322 results in the loss of antibiotic resistance towards:
(1) Ampicillin and tetracycline
(2) Ampicillin
(3) Tetracycline
(4) Gentamycin
Answer (3)
Sol. If one ligate a foreign DNA at the BamHI site of tetracycline resistance gene in the vector pBR322, the recombinant plasmid will lose tetracycline resistance due to insertion of foreign DNA.
139. What is the reason behind production of large holes in 'Swiss Cheese'?
(1) The production of large amount of CO2 and H2 by Trichoderma polysporum
(2) The production of large amount of CO2 and H2 by lactic acid bacteria called Lactobacillus
(3) The production of large amount of CO2 by Propionibacterium sharmanii
(4) The production of large amount of CO2 by Clostridium butylicum
Answer (3)
Sol. The large holes in swiss cheese are due to production of large amount of CO2 by the bacterium Propionibacterium sharmanii. Clostridium butylicum is commercially utilised for butyric acid production. Curd is formed by Lactobacillus.
140. Which of the following is not an example of convergent evolution?
(1) Fore limbs of whales and bats
(2) Flippers of penguins and dolphins
(3) Eyes of octopuses and mammals
(4) Wings of butterflies and birds
Answer (1)
Sol. Fore limbs of whales and bats are examples of divergent evolution that show homology.
141. Non-membrane bound cell organelles found in both prokaryotic and eukaryotic cells are ______.
(1) Lysosomes
(2) Centrosomes
(3) Mitochondria
(4) Ribosomes
Answer (4)
Sol. Ribosome is a non-membrane bound cell organelle, found in both prokaryotic and eukaryotic cells.
142. Ecological pyramids represent the relationship between the organisms at different trophic levels and they are generally inverted for:
(1) Pyramid of number in grassland
(2) Pyramid of energy in pond ecosystem
(3) Pyramid of biomass in grassland
(4) Pyramid of biomass in sea
Answer (4)
Sol. • Pyramid of number in grassland ecosystem is upright
• Pyramid of energy in pond ecosystem is upright
• Pyramid of biomass in grassland is upright
• Pyramid of biomass in sea is inverted
143. Arrange the following events occuring in Renin-Angiotensin mechanism in the correct order:
A. Increase in blood pressure and Glomerular filtration rate
B. Reabsorption of Na+ and water from distal parts of tubule due to Aldosterone
C. Fall in Glomerular filtration rate
D. Vasoconstriction by Angiotensin II and release of Aldosterone.
E. Renin converts Angiotensinogen into Angiotensin I, followed by Angiotensin II.
Choose the correct answer from the options given below:
(1) A, C, E, B, D
(2) C, A, B, D, E
(3) A, D, B, E, C
(4) C, E, D, B, A
Answer (4)
Sol. The JGA plays a complex regulatory role
A fall in glomerular blood flow/GFR can activate the JG cells to release renin which converts angiotensinogen in blood to angiotensin I and further to angiotensin II
Angiotensin II, being a powerful vasoconstrictor, increases the glomerular blood pressure and thereby GFR.
Angiotensin II activates the adrenal cortex to release aldosterone.
Aldosterone causes reabsorption of Na+ and H2O from the distal parts of the tubule. This leads to an increase in blood pressure and GFR.
144. Choose the correct statements regarding population interactions between two species.
A. In both parasitism and commensalism, only one species benefits and the other species is harmed.
B. Both species benefit in mutualism.
C. Both species benefit in commensalism.
D. In parasitism, only one species benefits and the other species is harmed.
E. In amensalism, one species is harmed and the other is unaffected.
Choose the correct answer from the options given below:
(1) A and B only
(2) B and E only
(3) B, D and E only
(4) A and D only
Answer (3)
Sol. In parasitism, one species is benefitted and the other is harmed, whereas, in commensalism one species gets benefitted and the other remains unaffected.
145. In which animal do haploid cells divide mitotically to produce gametes?
(1) Male honeybees
(2) Male grasshoppers
(3) Male earthworms
(4) Male frogs
Answer (1)
Sol. The male honeybee is haploid and females are diploid. The gamete formation in female honey bee is by meiosis, whereas male honeybee form gametes by mitosis. Thus haploid cell undergoes mitosis in male honeybees.
146. In humans, respiration occurs in the following steps. Arrange these steps in the correct order.
A. Diffusion of O2 and CO2 between blood and tissues
B. Diffusion of O2 and CO2 across alveolar membrane
C. Pulmonary ventilation by which atmospheric air is drawn in and CO2 rich alveolar air is released out
D. Cellular respiration
E. Transport of gases by the blood
Choose the correct answer from the options given below
(1) A,B,C,D,E
(2) C,A,B,E,D
(3) C,B,E,A,D
(4) E,A,C,D,B
Answer (3)
Sol. Respiration involves the following steps:
(i) Breathing or pulmonary ventilation by which atmospheric air is drawn in and CO2 rich alveolar air is released out.
(ii) Diffusion of gases across alveolar membrane.
(iii) Transport of gases by the blood
(iv) Diffusion of O2 and CO2 between blood and tissues
(v) Utilisation of O2 by the cells for catabolic reactions and resultant release of CO2
147. The following reaction depicts the activity of a particular class of enzymes:
X-Y + C=C --E--> X + Y-C=C
(Substrate) (Product) (Product)
Identify the enzymes class 'E' from the following options:
(1) Isomerases
(2) Ligases
(3) Transferases
(4) Lyases
Answer (4)
Sol. Lyases are the enzymes that catalyse removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds.
- Transferases are the enzymes that catalyse a transfer of a group between a pair of substrates. Isomerases catalyse inter-conversion of optical, geometric or positional isomers. Ligases catalyse the linking together of 2 compounds.
148. Match List I with List II:
List I (Bioactive molecules)
A. Streptokinase
B. Statins
C. Lipases
D. Cyclosporin A
List II (Importance)
I. Immunosuppressive agent
II. Removal of clots from the blood vessels
III. Blood cholesterol-lowering agent
IV. Detergent formulations
Choose the correct answer from the options given below:
(1) A-II, B-III, C-IV, D-I
(2) A-IV, B-III, C-II, D-I
(3) A-II, B-III, C-IV, D-I
(4) A-II, B-III, C-IV, D-I
Answer (3)
Sol. Streptokinase → Used as 'clot buster' for removing clots from the blood vessels
Statins → Blood cholesterol lowering agent, produced by Monascus purpureus
Lipases → Used in detergent formulation
Cyclosporin A → Used as immunosuppressive agent in organ transplant patients and produced by Trichoderma polysporum
149. Which one of the following is an appropriate example of sexual deceit?
(1) Sea anemone and clown fish
(2) Ophrys and bumblebee
(3) Female wasp and fig
(4) Cuckoo and crow
Answer (2)
Sol. Female wasp and fig - Mutualism
Ophrys and bumblebee - Sexual deceit
Sea anemone and clown fish - Commensalism
Cuckoo and crow - Brood Parasitism
150. Match List I with List II related to muscular/skeletal system:
List I
A. Tetany
B. Arthritis
C. Myasthenia gravis
D. Muscular dystrophy
List II
I. Inflammation of joints
II. Autoimmune disorder affecting neuromuscular junction
III. Wild contraction in muscle due to low Ca++ in body fluid
IV. Progressive degeneration of skeletal muscle
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-II, D-I
(2) A-III, B-I, C-II, D-IV
(3) A-I, B-II, C-III, D-IV
(4) A-III, B-II, C-I, D-IV
Answer (2)
Sol. Arthritis is an inflammation of joints.
- Tetanus is rapid spasms (wild contractions) in muscle due to low Ca++ in body fluids.
- Muscular dystrophy is progressive degeneration of skeletal muscle mostly due to genetic disorder.
- Myasthenia gravis is an autoimmune disorder affecting neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscle.
Hence, A-III, B-I, C-II, D-IV is the correct match.
151. Select the correct statements regarding cell membrane in eukaryotic cell.
A. Membrane of human RBCs has approximately 52% protein.
B. Major phospholipids are arranged in a bilayer.
C. Extensions of the plasma membrane into the cell form mesosomes.
D. Tails towards the inner part of lipids are hydrophobic and thus protected from aqueous medium.
E. Glycocalyx is present on the outer surface of the plasma membrane.
Choose the correct answer from the options given below:
(1) C, D and E only
(2) B, C and E only
(3) A, B and D only
(4) A, C and E only
Answer (3)
Sol. In prokaryotes, extensions of plasma membrane into the cell form mesosomes. Eukaryotes lack such structure. Glycocalyx is present on the outer surface of the plasma membrane in prokaryotes. Eukaryotes do not have glycocalyx.
Hence, only statement A, B and D are correct.
152. Choose the correct statements regarding cell organelles and their inclusions.
A. The endomembrane system includes Golgi complex, endoplasmic reticulum and mitochondria.
B. Rough endoplasmic reticulum bears ribosomes on its surface.
C. Both mitochondria and plastids have circular DNA.
D. A network of microtubules, microfilaments and intermediate filaments present in the cytoplasm is called cytoskeleton.
E. Mitochondrion is a single membrane-bound structure.
Choose the correct answer from the options given below:
(1) A, B and C only
(2) C, D and E only
(3) A and B only
(4) B, C and D only
Answer (4)
Sol. The endomembrane system does not include mitochondria. Mitochondria is a double membrane bound cell organelle.
153. The toxin proteins isolated from Bacillus thuringiensis, coded by which of the following genes would control cotton bollworms and corn borer, respectively?
(1) cry1Ac and cry2Ab
(2) cry1Ac and cry2Ab
(3) cry1Ac and cry1Ab
(4) cry2Ab and cry1Ac
Answer (3)
Sol. Specific Bt toxin genes were isolated from Bacillus thuringiensis and incorporated into the several crop plants such as cotton. The choice of genes depends upon the crop and the targeted pest, as most Bt toxins are insect-group specific. The toxin is coded by a gene named cry. The proteins encoded by the genes cry1Ac and cry1Ab control the cotton bollworms, that of cry1Ab controls corn borer. So, the correct answer is cry1Ac for cotton bollworms and cry1Ab for corn borer.
154. The JGA (Juxta Glomerular Apparatus) is a special sensitive region formed by cellular modifications in related to the same nephron.
(1) Proximal convoluted tubule and efferent renal arteriole
(2) Distal convoluted tubule and efferent renal arteriole
(3) Distal convoluted tubule and afferent renal arteriole
(4) Proximal convoluted tubule and afferent renal arteriole
Answer (3)
Sol. JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact.
155. Choose the correct statements regarding frog's anatomy:
A. Hepatic portal system is the special venous connection between liver and intestine.
B. There are twelve pairs of cranial nerves arising from the brain.
C. The ureters and oviducts open separately into the cloaca in female frogs.
D. Hind-brain consists of cerebellum, medulla oblongata and optic lobes.
E. Sinus venosus joins the right atrium of heart.
Choose the correct answer from the options given below:
(1) B and D only
(2) A, B and C only
(3) B and C only
(4) A, C and E only
Answer (4)
Sol. A. Correct → In frogs, the special venous connection between liver and intestine is called the hepatic portal system.
B. Incorrect → There are ten pairs of cranial nerves arising from the brain of frog.
C. Correct → In female frogs, the uterus and oviduct open separately in the cloaca.
D. Incorrect → In frogs, the midbrain consists of the optic lobes. The hindbrain consists of cerebellum, and medulla oblongata.
E. Correct → In frogs, a triangular structure called sinus venosus joins the right atrium.
Thus, correct statements are (A), (C) and (E)
156. Match List I with List II:
List-I
A. Cortisol
B. Aldosterone
C. Cholecystokinin
D. Progesterone
List-II
I. Stimulates the formation of alveoli in mammary glands
II. Produces anti-inflammatory reactions
III. Stimulates reabsorption of Na+ and water from renal tubule
IV. Stimulates secretion of pancreatic enzymes and bile juice
Choose the correct answer from the options given below:
(1) A-II, B-III, C-IV, D-I
(2) A-II, B-III, C-I, D-IV
(3) A-IV, B-II, C-I, D-III
(4) A-III, B-II, C-IV, D-I
Answer (1)
Sol. The correct answer is option (1) as
- Cortisol produces anti-inflammatory reactions and suppresses the immune response.
- Aldosterone acts mainly at the renal tubules and stimulates the reabsorption of Na+ and water and excretion of K+ and phosphate ions.
- Cholecystokinin (CCK) acts on both pancreas and gall bladder and stimulates the secretion of pancreatic enzymes and bile juice, respectively.
- Progesterone acts on the mammary glands and stimulates the formation of alveoli.
157. The sixth mutant codon of beta globin gene causing polymerization of Haemoglobin and change in RBC shape is ______.
(1) CAG
(2) AUG
(3) GUG
(4) GAG
Answer (3)
Sol. Sickle cell anaemia is an autosomal recessive disorder which is caused by the substitution of Glutamic acid by Valine, at the sixth position of the beta globin chain of the haemoglobin molecule. The substitution of amino acid in the globin protein results due to the single base substitution from GAG to GUG. Hence, GUG is responsible for the change in the shape of RBC.
158. Male frogs can be distinguished from female frogs due to the presence of
A. Bulging eyes
B. Vocal sacs
C. Webbed digits in feet
D. Copulatory pad on first digit of fore limbs
E. Olive green-coloured skin with dark irregular spots
Choose the correct answer from the options given below
(1) A and B only
(2) C and E only
(3) B and D only
(4) B and C only
Answer (3)
Sol. Male frogs can be distinguished from female frogs due to the presence of vocal sacs and copulatory pad on first digit of fore limbs.
Bulging eyes, webbed digit in feet and olive green-coloured skin with dark irregular spots are common in both male and female frogs.
159. The human protein named α-1-antitrypsin, obtained from transgenic animals, is used for the treatment of ______.
(1) Alzheimer's disease
(2) Rheumatoid arthritis
(3) Emphysema
(4) Cystic fibrosis
Answer (3)
Sol. The correct answer is option (3).
The human protein named α-1-antitrypsin, obtained from transgenic animals, is used for the treatment of emphysema.
Transgenic models exist for the study of other diseases, such as Alzheimer's disease, rheumatoid arthritis and cystic fibrosis.
160. Match List I with List II:
List I (Drug)
A. Nicotine
B. Morphine
C. Heroin
D. Cocaine
List II (Effect)
I. Causes sense of euphoria and increased energy
II. Stimulates adrenal gland to release catecholamines into blood circulation
III. Effective sedative and painkiller
IV. A depressant; slows down body function
Choose the correct answer from the options given below:
(1) A-II, B-III, C-I, D-IV
(2) A-III, B-II, C-IV, D-I
(3) A-III, B-II, C-I, D-IV
(4) A-II, B-III, C-IV, D-I
Answer (4)
Sol. The correct answer is option (4) as
Nicotine is present in tobacco and it activates adrenal medulla to release catecholamines into blood circulation, so (A) → II
Morphine acts as an effective sedative and painkiller. It is an opioid, so (B) → III
Heroin acts as a depressant and slows down body function, so (C) → IV
Cocaine acts as a stimulant and causes a sense of euphoria and increased energy, so (D) → I
Thus, (A) → II, (B) → III, (C) → IV, (D) → I
161. The WBC count of a person's blood sample is 8000/cu mm. How many eosinophils and lymphocytes would be in the same blood sample approximately?
(1) 300 - 500/cu mm and 500 - 700/cu mm respectively
(2) 300 - 500/cu mm and 1200 - 1500/cu mm respectively
(3) 100 - 120/cu mm and 160 - 200/cu mm respectively
(4) 160 - 240/cu mm and 1600 - 2000/cu mm respectively
Answer (4)
Sol. Eosinophils constitute 2 - 3% of total WBCs. Hence its value is approximately 2 to 3% of 8000/cu mm = 160 - 240/cu mm
Lymphocytes constitute 20 - 25% of total WBCs. Hence its value is approximately 20 to 25% of 8000/cu mm = 1600 - 2000/cu mm
162. Match List I with List II with respect to chronology of evolution of life forms
List-I
A. About 65 mya
B. About 500 mya
C. About 350 mya
D. About 320 mya
List-II
I. Jawless fish probably evolved
II. The dinosaurs suddenly disappeared from the earth
III. Seaweeds and few plants probably existed
IV. Invertebrates were formed and became active
Choose the correct answer from the options given below:
(1) A(II), B(IV), C(I), D(III)
(2) A(II), B(IV), C(III), D(I)
(3) A(II), B(IV), C(I), D(III)
(4) A(I), B(II), C(III), D(IV)
Answer (3)
Sol. About 65 mya - The dinosaurs suddenly disappeared from the earth.
About 500 mya - Invertebrates were formed and became active.
About 350 mya - Jawless fish probably evolved.
About 320 mya - Seaweeds and few plants probably existed.
163. Match List I and List II
List-I
A. Progestasert
B. Multiload 375
C. Diaphragm
D. Saheli
List-II
I. Barrier made of rubber used by females
II. Oral contraceptive
III. Hormone releasing IUD
IV. Copper releasing IUD
Choose the correct answer from the options given below:
(1) A(IV), B(II), C(I), D(III)
(2) A(IV), B(II), C(I), D(III)
(3) A(III), B(IV), C(I), D(II)
(4) A(III), B(IV), C(I), D(II)
164. The following are the stages of life cycle of Plasmodium. Arrange the stages in the proper order.
A. The parasites reproduce asexually in RBCs, bursting the cells.
B. The parasites reproduce asexually in liver cells, bursting the cells and releasing into blood.
C. Gametocytes develop in RBCs.
D. Sporozoites reach the liver through the blood.
E. Female mosquito injects sporozoites into humans during bite.
Choose the correct answer from the options given below:
(1) A, B, C, D, E
(2) E, C, D, B, A
(3) E, D, B, A, C
(4) C, A, B, D, E
Answer (3)
Sol. Plasmodium enters the human body as sporozoites through the bite of an infected female Anopheles mosquito. The parasites initially multiply asexually within the liver cells and then attack the RBCs resulting in their rupture. Sexual stages (gametocytes) develop in red blood cells. When a female Anopheles mosquito bites an infected person, these parasites enter the mosquito's body and undergo further development.
165. Match List I with List II related to embryonic development at various months of pregnancy:
List-I
A. The foetus movement starts and hair appears on the head
B. The foetus develops limbs and digits
C. The foetus develops external genital organs
D. The foetus body is covered with fine hair; eyelids separate and eyelashes are formed
List-II
I. 24 weeks of pregnancy
II. 20 weeks of pregnancy
III. 8 weeks of pregnancy
IV. 12 weeks of pregnancy
Choose the correct answer from the options given below:
(1) A-II, B-IV, C-III, D-I
(2) A-II, B-III, C-IV, D-I
(3) A-IV, B-II, C-III, D-I
(4) A-II, B-III, C-IV, D-I
Answer (2)
Sol. By the end of second month of pregnancy (8 weeks), the foetus develops limbs and digits By the end of 12 weeks (first trimester) of pregnancy, most of the major organ system are formed. The limbs and external genital organs are also well developed. During the fifth month (20 weeks) of pregnancy, the first movements of the foetus and appearance of hair on the head are usually observed. By the end of about 24 weeks (ends of second trimester) of pregnancy, the body is covered with fine hair, eye lids separate and eyelashes are formed. So, the correct match is A-II, B-III, C-IV, D-I
166. The flightless bird with forelimbs modified as paddle-like structures suited for swimming is known as:
(1) Psittacula
(2) Aptenodytes
(3) Neophron
(4) Struthio
Answer (2)
Sol. Neophron is vulture and Psittacula is a parrot. Both perform flight.
Struthio is ostrich and Aptenodytes is penguin. Both are flightless birds. In penguins, forelimbs are modified into flippers (paddle like structure) and are used for swimming. In ostrichs, forelimbs are small and used for balance while running, not for swimming.
167. Select the incorrect statements from the following:
A. Digestive system in Platyhelminthes is incomplete.
B. Bilateral symmetry is a characteristic feature of adult Echinoderms.
C. Pseudocoelom is possessed by Aschelminthes.
D. Notochord is persistent throughout life in the class Chondrichthyes.
E. Members of class Reptilia maintain a constant body temperature.
Choose the answer from the options given below:
(1) A and C only
(2) B and E only
(3) B and D only
(4) C and D only
Answer (2)
Sol. The correct answer is option (2)
(A) → correct → Platyhelminthes have an incomplete digestive system.
(B) → incorrect → Bilateral symmetry is a characteristic feature of larvae of echinoderms. In adult echinoderms, radial symmetry is seen.
(C) → correct → Aschelminthes are characterised by the presence of pseudocoelom.
(D) → correct → Notochord is persistent throughout life in the Chondrichthyes.
(E) → incorrect → Reptiles are cold-blooded organisms and thus, they cannot maintain a constant body temperature.
Warm-blooded organisms like birds and mammals can maintain a constant body temperature.
Thus, as the incorrect statements are indicated by (B) and (E) only, the correct answer is option (2)
168. A group of researchers procured some fish like animals and upon investigation the following characters were observed:
A. Endoskeleton was made of cartilage.
B. Ectoparasitic; as they were found attached on fish skin with their circular sucking mouth.
C. Paired fins and scales were absent, but 7 pairs of gill slits were present.
Which of the following species of animals did they consider to fit best with these characters?
(1) Scoliodon sp.
(2) Exocoetus sp.
(3) Petromyzon sp.
(4) Branchiostoma sp.
Answer (3)
Sol. Petromyzon sp. have cartilaginous endoskeleton. They have circular sucking mouth. They are ectoparasites on some fishes. Their body is devoid of scales and paired fins. They have 7 pairs (6-15 pairs) of gill slits.
Scoliodon and Exocoetus are not parasites.
169. Choose the correct statements regarding muscle contraction.
A. A motor neuron carries a signal sent by the Central Nervous System (CNS) to the sarcolemma of the muscle fibre.
B. The neural signal generates an action potential which causes the release of Ca++ into sarcoplasm.
C. Increase in Ca++ inactivates the actin for breaking cross bridges.
D. Actin binds to the myosin head to form a cross bridge.
E. Shortening of sarcomere takes place, by pulling actin filaments towards the centre of 'A' band.
Choose the correct answer from the options given below:
(1) A and B only
(2) C and E only
(3) C and D only
(4) A,B,D and E only
Answer (4)
Sol. Statements A, B, D and E are correct while statement C is incorrect.
A neural signal reaching neuromuscular junction releases a neurotransmitter (Acetylcholine) which generates an action potential in the sarcolemma. This spreads through muscle fibre and causes the release of Ca++ into the sarcoplasm. This Ca++ binds to the subunit of troponin on actin filaments and thereby remove the masking of active site for myosin on actin and hence facilitates the formation of cross bridge.
170. Choose the correct statement regarding GIFT to overcome infertility.
(1) Ova collected from a female donor are transferred to the uterus of an infertile female.
(2) Early embryos with up to 8 blastomeres are transferred into the fallopian tube of an infertile female.
(3) Early embryos with up to 8 blastomeres are transferred to the uterus of an infertile female.
(4) It is the transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce ovum but can provide suitable environment for fertilization and development.
Answer (4)
Sol. GIFT is an in-vivo technique used to assist infertility.
GIFT: Gamete Intra Fallopian Transfer technique facilitates the transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one, but can provide suitable environment for fertilization and further development.
ZIFT: Zygote Intra Fallopian Transfer is an in-vitro technique in which zygote or early embryo upto 8 blastomeres is transferred in fallopian tube.
171. Select the incorrect statement with reference to Rh grouping.
A. Erythroblastosis foetalis is a condition observed having foetus with Rh-ve blood and mother with Rh-ve blood.
B. Rh antigen is observed on RBCs in the majority of human beings.
C. Before blood transfusion, Rh group should also be matched.
D. Rh incompatibility is observed when a pregnant mother is Rh-ve and the foetus is Rh-ve.
E. Erythroblastosis foetalis can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the second child.
Choose the answer from the options given below:
(1) A and B only
(2) C and D only
(3) A and E only
(4) B and C only
Answer (3)
Sol. (A) Incorrect → A special case of Rh incompatibility has been observed between the Rh-ve blood of a pregnant mother with Rh-ve blood of the foetus.
(B) Correct → Rh antigen is observed on the surface of RBCs of majority (nearly 80 percent) of humans.
(C) Correct → Before blood transfusion, Rh group should also be matched to avoid severe problems of destruction of RBCs.
(D) Correct → Rh incompatibility (Erythroblastosis foetalis) is observed when a pregnant mother is Rh-ve and the foetus is Rh-ve.
(E) Incorrect → Erythroblastosis foetalis can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.
Thus, the incorrect statements are (A) and (E).
172. Select the set of fishes which belong to the class Osteichthyes:
(1) Saw fish, Fighting fish and Dog fish
(2) Devil fish, Cuttlefish and Hagfish
(3) Starfish, Hagfish and Cuttlefish
(4) Flying fish, Angel fish and Fighting fish
Answer (4)
Sol. The correct answer is option (4).
Flying fish (Exocoetus) is a marine bony fish. Angel fish (Pterophyllum) and fighting fish (Betta) are aquarium bony fishes. Option (2) is incorrect as:- Devil fish (Octopus) and cuttlefish (Sepia) are molluscs. Hag fish (Myxine) is a cyclostome. Option (1) is incorrect as:- Saw fish (Pristis) and dog fish (Scoliodon) are cartilaginous fishes. Option (3) is incorrect as:- Star fish (Asterias) is an echinoderm.
173. In a population of a grasshopper species, the chromosome number of some members is 23 and some other members possess 24 chromosomes. The 23 and 24 chromosome-bearing members in this species are
(1) females and males, respectively
(2) males and females, respectively
(3) all males
(4) all females
Answer (2)
Sol. In grasshopper sex determination is XX - XO type, in which, males have only one X-chromosome besides the autosome (XO), whereas females have a pair of X-chromosome, besides autosomes (XX). Therefore, the individual with 23 chromosomes is a male grasshopper and the one with 24 chromosomes is a female grasshopper.
174. Evolution of human appears parallel to the progressive development of brain and language skills. As such, the evolution of individual species in the sequence of their appearance is:
(1) Ramapithecus → Homo habilis → Homo erectus → Neanderthal → Homo sapiens
(2) Neanderthal → Ramapithecus → Homo habilis → Homo erectus → Homo sapiens
(3) Homo habilis → Homo erectus → Ramapithecus → Neanderthal → Homo sapiens
(4) Homo sapiens → Ramapithecus → Homo habilis → Neanderthal → Homo erectus
Answer (1)
Sol. Evolution of human appears parallel to the progressive development of brain and language skills.
The correct chronological order in which human evolution took place is:
Ramapithecus → Australopithecines → Homo habilis → Homo erectus → Neanderthalensis → Homo sapiens
175. The specific receptors for neurotransmitters in a synapse are present on
(1) Pre-synaptic membrane
(2) Post-synaptic membrane
(3) Myelin sheath
(4) Schwann cell
Answer (2)
Sol. The specific receptors for neurotransmitters in a synapse are present on post-synaptic membrane.
176. Match List-I with List-II.
List-I (Respiratory Volume)
A. ERV (Expiratory Reserve Volume)
B. RV (Residual Volume)
C. IRV (Inspiratory Reserve Volume)
D. TV (Tidal Volume)
List-II (Capacity in mL)
I. 2500 – 3000 mL
II. 500 mL
III. 1000 – 1100 mL
IV. 1100 – 1200 mL
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-I, D-II
(2) A-III, B-I, C-IV, D-II
(3) A-I, B-II, C-III, D-IV
(4) A-I, B-III, C-II, D-IV
Answer (1)
Sol. ERV (Expiratory Reserve Volume) - 1000-1100 mL
RV (Residual Volume) - 1100-1200 mL
IRV (Inspiratory Reserve Volume) - 2500-3000 mL
TV (Tidal Volume) - 500 mL
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