NEET 2021 Previous Year Question Paper P2 with Complete Solutions – Free PDF Download

1. The test is of 3 hours duration and Test Booklet contains 200 multiple choice questions (Four options with a single correct answer). There are two sections in each subject, i.e. Section-A & Section-B. You have to attempt all 35 questions from Section-A & only 10 questions from Section-B out of 15. (Candidates are advised to read all 15 questions in each subject of Section-B before they start attempting the question paper. In the event of a candidate attempting more than ten questions, the first ten questions answered by the candidate shall be evaluated.)

Important Instructions :

1. The test is of 3 hours duration and Test Booklet contains 200 multiple choice questions (Four options with a single correct answer). There are two sections in each subject, i.e. Section-A & Section-B. You have to attempt all 35 questions from Section-A & only 10 questions from Section-B out of 15. (Candidates are advised to read all 15 questions in each subject of Section-B before they start attempting the question paper. In the event of a candidate attempting more than ten questions, the first ten questions answered by the candidate shall be evaluated.)2. Each question carries 4 marks. For each correct response, the candidate will get 4 marks. For every wrong response 1 mark shall be deducted from the total score. Unanswered / unattempted questions will be given no marks. The maximum marks are 720.3. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.4. Rough work is to be done in the space provided for this purpose in the Test Booklet only.5. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.6. The CODE for this Booklet is P2.7. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the Test Booklet/Answer Sheet. Use of white fluid for correction is NOT permissible on the Answer Sheet.8. Each candidate must show on demand his/her Admission Card to the Invigilator.9. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her seat.10. Use of Electronic/Manual Calculator is prohibited.11. The candidates are governed by all Rules and Regulations of the examination with regard to their conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of this examination.12. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.13. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet in the Attendance Sheet.

NEET (UG)-2021 (Code-P2)

PHYSICS

SECTION-A

1. If E and G respectively denote energy and gravitational constant, then E/G has the dimensions of

(1) [M²] [L⁻²] [T⁻¹]
(2) [M²] [L⁻¹] [T⁰]
(3) [M] [L⁻¹] [T⁻¹]
(4) [M] [L⁰] [T⁰]

Answer (2)

Sol. Dimensional formula of energy
[E] = [M¹ L² T⁻²] ...(I)
Dimensional formula of gravitational constant
[G] = [M⁻¹ L³ T⁻²] ...(II)
From (I) & (II)
[E]/[G] = [M¹ L² T⁻²]/[M⁻¹ L³ T⁻²]
= [M² L⁻¹ T⁰]
Hence, dimensions of [E/G] = [M² L⁻¹ T⁰]
So, correct option is (2)

2. An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance 'R' are connected in series to an ac source of potential difference 'V' volts as shown in figure.
Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10√2 A. The impedance of the circuit is

(1) 5 Ω
(2) 4√2 Ω
(3) 5/√2 Ω
(4) 4 Ω

Answer (1)

Sol. V_L = 40 volt
V_R = 40 volt
V_C = 10 volt
Now, V_RMS = √(V_R² + (V_L - V_C)²)
= √(40² + (40 - 10)²) = 50 V
I_RMS = I₀/√2 = 10√2/√2 = 10 A
∴ V_RMS = I_RMS × Z
∴ Z = V_RMS/I_RMS = 50/10 = 5 Ω

3. A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is

(1) 4n
(2) n
(3) 2n
(4) 3n

Answer (3)

Sol. Equation of displacement of particle executing SHM is given by x = A sin(ωt + φ) ...(I)
Potential energy of particle executing SHM is given by
U = 1/2 kx²
= 1/2 kA² sin²(ωt + φ) ...(II)
From I and II, it is clear that
Time period of x = A sin(ωt + φ) is
T₁ = 2π/ω ⇒ frequency n₁ = ω/2π
while time period of x² = A² sin²(ωt + φ) is
T₂ = π/ω ⇒ frequency n₂ = ω/π
Hence n₂ = 2n₁

4. A thick current carrying cable of radius 'R' carries current 'I' uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance 'r' from the axis of the cable is represented by

(1) [Graph showing B increasing and then becoming constant]
(2) [Graph showing B increasing exponentially]
(3) [Graph showing B increasing linearly]
(4) [Graph showing B increasing linearly up to R then decreasing hyperbolically]

Answer (4)

Sol. From Ampere's circuital law
B = (μ₀I/2πR²)·r if r < R ⇒ B_inside ∝ r
B = μ₀I/2πr if r ≥ R ⇒ B_outside ∝ 1/r
Hence the correct plot of magnetic field B with distance r from axis of cable is given as
[Graph showing straight line up to R then hyperbola]

5. A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is

(1) 216 MeV
(2) 0.9 MeV
(3) 9.4 MeV
(4) 804 MeV

Answer (1)

Sol. Mass number of reactant = 240
BE per nucleon = 7.6 MeV
Mass number of products = 120
BE per nucleon of product = 8.5 MeV
Total gain in BE = (BE) of products - (BE) of reactants
= [120 + 120] × 8.5 - [240] × 7.6
= (240) × 8.5 - 240 × 7.6
= (2040 - 1824) MeV
Gain in BE = 216 MeV

6. A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is 'd' and the area of each plate is 'A', the energy stored in the capacitor is
(ε₀ = permittivity of free space)

(1) E²Ad/ε₀
(2) 1/2 ε₀E²
(3) ε₀Ed
(4) 1/2 ε₀E²Ad

Answer (4)

Sol. Energy density associated with electric field is given by
u = dU/dV = 1/2 ε₀E²
⇒ dU = 1/2 ε₀E²dV
Total energy stored in the space between the capacitor will be
U = ∫dU = ∫ 1/2 ε₀E²dV
= 1/2 ε₀E²∫dV [E is constant]
= 1/2 ε₀E²V = 1/2 ε₀E²Ad [V = Ad]

7. The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3 × 10⁻³ watt will be (h = 6.6 × 10⁻³⁴ J s)

(1) 10¹⁵
(2) 10¹⁸
(3) 10¹⁷
(4) 10¹⁶

Answer (4)

Sol. The power of a source is given as
(Here n/t is number of photons emitted per second)
⇒ n/t = (3.3 × 10⁻³ × 6 × 10⁻⁷)/(6.6 × 10⁻³⁴ × 3 × 10⁸) = 10¹⁶ photons per second

8. Polar molecules are the molecules

(1) Having a permanent electric dipole moment
(2) Having zero dipole moment
(3) Acquire a dipole moment only in the presence of electric field due to displacement of charges
(4) Acquire a dipole moment only when magnetic field is absent

Answer (1)

Sol. In polar molecules, the centre of positive charges does not coincide with the centre of negative charges.
Hence, these molecules have a permanent electric dipole moment of their own.

9. The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be

(1) 1/3√2
(2) 1/2
(3) 1/2√2
(4) 2/3

Answer (3)

Sol. The activity of a radioactive substance is given as
A = A₀(1/2)^(t/T₁/₂)
Now, A/A₀ = (1/2)^(t/T₁/₂)
⇒ A/A₀ = (1/2)^(150/100)
⇒ A/A₀ = (1/2)^(3/2)
⇒ A/A₀ = 1/2√2

10. A capacitor of capacitance 'C', is connected across an ac source of voltage V, given by
V = V₀ sin ωt
The displacement current between the plates of the capacitor, would then be given by

(1) I_d = V₀ωC sin ωt
(2) I_d = V₀ωC cos ωt
(3) I_d = V₀/ωC cos ωt
(4) I_d = V₀/ωC sin ωt

Answer (2)

Sol. Given V = V₀ sin ωt ...(1)
Now displacement current I_d is given by
I_d = C dV/dt
= C d/dt(V₀ sin ωt) (using equation 1)
= C(V₀ω) cos ωt
I_d = V₀ωC cos ωt

11. A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is

(1) 0.052 cm
(2) 0.52 cm
(3) 0.026 cm
(4) 0.26 cm

Answer (1)

Sol. Here, pitch of the screw gauge, P = 1 mm
Number of circular division, n = 100
Thus least count LC = P/n = 1/100 = 0.01 mm
= 0.001 cm

12. Find the value of the angle of emergence from the prism. Refractive index of the glass is √3.

(1) 90°
(2) 60°
(3) 30°
(4) 45°

Answer (2)

Sol. From the ray diagram shown in the figure.
At point P from Snell's law
sin i/sin r = μ_air/μ_Prism
sin 30°/sin e = 1/√3 (∠r = ∠e emergence)
sin e = √3 · 1/2
e = 60°

13. In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?

(1) 62 cm
(2) 60 cm
(3) 21.6 cm
(4) 64 cm

Answer (2)

Sol. From the application of potentiometer to compare two cells of emfs E₁ and E₂ by balancing lengths ℓ₁ and ℓ₂
E₁/E₂ = ℓ₁/ℓ₂
⇒ ℓ₂ = ℓ₁(E₂/E₁) = (36 cm)(2.5V/1.5V)
= 60 cm

14. A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively

(1) S/4, √(3gS/2)
(2) S/4, 3gS/2
(3) S/4, √(3gS/2)
(4) S/2, √(3gS/2)

Answer (1)

Sol. Let required height of body is y
When body from rest falls through height (S - y)
Then under constant acceleration
v² = 0² + 2g(S - y)
v = √(2g(S - y)) ...(1)
When body is at height y above ground. Potential energy of body of mass m
U = mgy
As per given condition kinetic energy, K = 3U
1/2 m(v)² = 3 × mg(y)
1/2 × m × 2g(S - y) = 3 × mgy (using (1))
S - y = 3y
∴ y = S/4
∴ v = √(2 × g(S - S/4)) = √(3gS/2)

15. The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series?

(1) 4 Ω
(2) 0.25 Ω
(3) 0.5 Ω
(4) 1 Ω

Answer (1)

Sol. All the wires are identical and of same material so they will have same value of resistance. Let it be R. When these are (four) connected in parallel.
Given R_P = 0.25 Ω
∴ 0.25 = R/4
∴ R = 1 Ω
Now these four resistances are arranged in series
R_S = R + R + R + R = 4R
∴ R_S = 4 × 1 = 4 Ω

16. Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine?
(g = 10 m/s²)

(1) 7.0 kW
(2) 10.2 kW
(3) 8.1 kW
(4) 12.3 kW

Answer (3)

Sol. Incident power on turbine = d(mgh)/dt
Now, losses are 10%
∴ power generated = (1 - 10/100) × 9000
= 8100 W
= 8.1 kW

17. An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10⁵ m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

(1) 8 × 10⁻²⁰ N
(2) 4 × 10⁻²⁰ N
(3) 8π × 10⁻²⁰ N
(4) 4π × 10⁻²⁰ N

Answer (1)

Sol. Magnetic field produced due to current carrying wire at point 'A'
B = μ₀/4π · 2I/r
B = (10⁻⁷ × 2 × 5)/(20 × 10⁻²) = 1/2 × 10⁻⁵ (Tesla), upward to the
Now, force acting on electron due to this field
F = q(v × B)
|F| = 1.6 × 10⁻¹⁹ × 10⁵ × 1/2 × 10⁻⁵
= 0.8 × 10⁻¹⁹ N
|F| = 8 × 10⁻²⁰ N

18. Consider the following statements (A) and (B) and identify the correct answer.
(A) A zener diode is connected in reverse bias, when used as a voltage regulator.
(B) The potential barrier of p-n junction lies between 0.1 V to 0.3 V

(1) (A) is incorrect but (B) is correct.
(2) (A) and (B) both are correct.
(3) (A) and (B) both are incorrect
(4) (A) is correct and (B) is incorrect.

Answer (4)

Sol. In reverse biased, after breakdown, voltage across the zener diode becomes constant. Therefore zener diode is connected in reverse biased when used as voltage regulator. Potential barrier of silicon diode is nearly 0.7 V statement A is correct and statement B is incorrect.

19. A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is

(1) 0.628 s
(2) 0.0628 s
(3) 6.28 s
(4) 3.14 s

Answer (1)

Sol. For a spring, kx = F
given x = 5 cm, F = 10 N
⇒ k(5 × 10⁻²) = 10
⇒ k = 1000/5 = 200 N/m
Now, for spring-mass system undergoing SHM
T = 2π√(m/k)
given, m = 2 kg
⇒ T = 2π√(2/200) = 2π/10 = 0.628 s

20. The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.

(1) No current will flow in p-type, current will only flow in n-type
(2) Current in n-type = current in p-type
(3) Current in p-type > current in n-type
(4) Current in n-type > current in p-type.

Answer (4)

Sol. The current through a semiconductor is
I = neAv_d
I = neAμE
I_n/I_p = (n_e eAμ_e E)/(n_h eAμ_h E)
I_n/I_p = μ_e/μ_h
∴ μ_e > μ_h
⇒ I_n > I_p

21. A dipole is placed in an electric field as shown. In which direction will it move?

(1) Towards the right as its potential energy will increase.
(2) Towards the left as its potential energy will increase.
(3) Towards the right as its potential energy will decrease.
(4) Towards the left as its potential energy will decrease.

Answer (3)

Sol. Potential energy of electric dipole in external electric field U = -P·E
Angle between electric field and electric dipole is 180°
U = -PE cos θ
U = -PE cos 180°
U = +PE
On moving towards right electric field strength decrease therefore potential energy decrease.
Net force on electric dipole is towards right and net torque acting on it is zero.
So, it will more towards right.

22. A convex lens 'A' of focal length 20 cm and a concave lens 'B' of focal length 5 cm are kept along the same axis with a distance 'd' between them. If a parallel beam of light falling on 'A' leaves 'B' as a parallel beam, then the distance 'd' in cm will be

(1) 30
(2) 25
(3) 15
(4) 50

Answer (3)

Sol. Parallel beam of light after refraction from convex lens converge at the focus of convex lens. In question it is given light after refraction pass through concave lens becomes parallel. Therefore light refracted from convex lens virtually meet at focus of concave lens.
According to above ray diagram d = f_A - f_B
= 20 - 5 = 15 cm

23. The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is

(1) 4v
(2) v
(3) 2v
(4) 3v

Answer (1)

Sol. Escape velocity from the Earth's surface
v/v₁ = R/4R
v₁ = 4v

24. An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λ_d, then

(1) λ = (2h/mc)λ_d²
(2) λ = (2m/hc)λ_d²
(3) λ_d = (2mc/h)λ²
(4) λ = (2mc/h)λ_d²

Answer (4)

Sol. As per Einstein's photoelectric equation
hc/λ = φ₀ + k
φ₀ : work function
k = maximum kinetic energy of photoelectrons
As per question, φ → 0
∴ hc/λ = k = P²/2m ⇒ P = √(2mhc/λ)
Now De-broglie wavelength,
λ_d = h/P = h/√(2mhc/λ)
⇒ √λ = λ_d√(2mc/h)
⇒ λ = (2mc/h)λ_d²

25. A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since

(1) A large aperture contributes to the quality and visibility of the images.
(2) A large area of the objective ensures better light gathering power.
(3) A large aperture provides a better resolution.
(4) All of the above

Answer (4)

Sol. With larger aperture of objective lens, the light gathering power in telescope is high.
Also, the resolving power or the ability to observe two objects distinctly also depends on the diameter of the objective. Thus objective of large diameter is preferred.
Also, with large diameters fainter objects can be observed. Hence it also contributes to the better quality and visibility of images.
Hence, all options are correct.

26. Two charged spherical conductors of radius R₁ and R₂ are connected by a wire. Then the ratio of surface charge densities of the spheres (σ₁/σ₂) is

(1) R₁²/R₂²
(2) R₁/R₂
(3) R₂/R₁
(4) √(R₁/R₂)

Answer (3)

Sol. When two conductors are connected by a conducting wire, then the two conductors should have same potential.
so, V₁ = V₂
∴ 1/4πε₀ · Q₁/R₁ = 1/4πε₀ · Q₂/R₂
⇒ 1/4πε₀ · Q₁/R₁ × R₁/R₁ = 1/4πε₀ · Q₂/R₂ × R₂/R₂
⇒ Q₁R₁/4πR₁²ε₀ = Q₂R₂/4πR₂²ε₀
⇒ σ₁R₁/ε₀ = σ₂R₂/ε₀
⇒ σ₁/σ₂ = R₂/R₁

27. For a plane electromagnetic wave propagating in x-direction, which one of the following combination gives the correct possible directions for electric field (E) and magnetic field (B) respectively?

(1) -ĵ + k̂, -ĵ + k̂
(2) ĵ + k̂, ĵ + k̂
(3) -ĵ + k̂, -ĵ - k̂
(4) ĵ + k̂, -ĵ - k̂

Answer (3)

Sol. Direction of propagation of electromagnetic waves is along E × B
Given that direction of propagation is along x-axis
(-ĵ + k̂) × (-ĵ + k̂) = 0
(ĵ + k̂) × (ĵ + k̂) = 0
(-ĵ + k̂) × (-ĵ - k̂) = 2î
(ĵ + k̂) × (-ĵ + k̂) = 0
∴ Option (3) is correct.

28. A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is

(1) 5/13 t
(2) 13/5 t
(3) 13/5 t
(4) ...

Answer (3)

Sol. From Average form of Newton's law of cooling
-((T₁ + T₂)/2 - T_s)K = (T₁ - T₂)/Δt
T₁ and T₂ are initial and final temperature and T_s is surrounding temperature.
⇒ -K[(90 + 80)/2 - 20] = (90 - 80)/t

30. If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.

(1) [F][A⁻¹][T]
(2) [F][A][T]
(3) [F][A][T²]
(4) [F][A][T⁻¹]

Answer (3)

Sol. Energy, E ∝ F^a A^b T^c
[E] = [F^a][A^b][T^c]
⇒ [ML²T⁻²] = [ML²T⁻²]^a [LT⁻²]^b [T]^c
[ML²T⁻²] = [M^a L^(a+b) T^(-2a-2b+c)]
Comparing dimensions on both sides.
⇒ a = 1; a + b = 2 and -2 = -2a - 2b + c
⇒ b = 1 ⇒ -2 = -2 - 2 + c
⇒ c = 2
[E] = [FAT²]

31. A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let S_n be the distance travelled by the block in the interval t = n - 1 to t = n. Then, the ratio S_n/S_(n+1) is

(1) 2n/(2n - 1)
(2) (2n - 1)/2n
(3) (2n - 1)/(2n + 1)
(4) (2n + 1)/(2n - 1)

Answer (3)

Sol. Suppose θ is inclination of inclined plane acceleration along inclined plane a = g sin θ
S_n = distance travelled by object during nth second.
Initial speed u = 0
By equation of uniformly accelerated motion
S_n = u + a/2(2n - 1)
S_n = 0 + (g sin θ)/2 (2n - 1) = (g sin θ)/2 (2n - 1) ...(i)

32. The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be

(1) 2Mg
(2) Mg/2
(3) Mg
(4) 3/2 Mg

Answer (2)

Sol. Let F_v be the viscous force and F_B be the Bouyant force acting on the ball.
Then, when body moves with constant velocity
Mg = F_B + F_v [a = 0]
F_v = Mg - F_B
= dVg - d/2 · Vg (M = dVg) V = volume of ball.
= d/2 Vg
F_v = M/2 g

33. Column-I gives certain physical terms associated with flow of current through a metallic conductor.
Column-II gives some mathematical relations involving electrical quantities. Match Column-I and Column-II with appropriate relations.

Column-I
(A) Drift Velocity
(B) Electrical Resistivity
(C) Relaxation Period
(D) Current Density

Column-II
(P) m/(ne²ρ)
(Q) nev_d
(R) eEτ/m
(S) E/J

(1) (A)-(R), (B)-(Q), (C)-(S), (D)-(P)
(2) (A)-(R), (B)-(S), (C)-(P), (D)-(Q)
(3) (A)-(R), (B)-(S), (C)-(Q), (D)-(P)
(4) (A)-(R), (B)-(P), (C)-(S), (D)-(Q)

Answer (2)

Sol. Drift velocity, v_d = eEτ/m
Electrical resistivity, ρ = 1/σ = E/J
Relaxation period, τ = m/(ne²ρ)
Current density, J = I/A = nev_d
(A)-(R), (B)-(S), (C)-(P), (D)-(Q)

34. A radioactive nucleus ᴬ_Z X undergoes spontaneous decay in the sequence
ᴬ_Z X → _(Z-1) B → _(Z-3) C → _(Z-2) D, where Z is the atomic number of element X. The possible decay particles in the sequence are

(1) β⁻, α, β⁺
(2) α, β⁻, β⁺
(3) α, β⁺, β⁻
(4) β⁺, α, β⁻

Answer (4)

Sol. On β⁺ decay atomic number decreases by 1
On β⁻ decay atomic number increases by 1
On α decay atomic number decreases by 2
ᴬ_Z X →(β⁺ decay) _(Z-1) B →(α decay) _(Z-3) C →(β⁻ decay) _(Z-2) D
Hence correct order of decay are β⁺, α, β⁻

35. Match Column-I and Column-II and choose the correct match from the given choices.

Column-I
(A) Root mean square speed of gas molecules
(B) Pressure exerted by ideal gas
(C) Average kinetic energy of a molecule
(D) Total internal energy of 1 mole of a diatomic gas

Column-II
(P) 1/3 nm v̄²
(Q) √(3RT/M)
(R) 5/2 RT
(S) 3/2 k_B T

(1) (A)-(R), (B)-(Q), (C)-(P), (D)-(S)
(2) (A)-(R), (B)-(P), (C)-(S), (D)-(Q)
(3) (A)-(Q), (B)-(R), (C)-(S), (D)-(P)
(4) (A)-(Q), (B)-(P), (C)-(S), (D)-(R)

Answer (4)

Sol. Root mean square speed of gas molecule
= √(3RT/M)
Pressure exerted by ideal gas 1/3 nm v̄²
Average kinetic energy of a molecule 3/2 k_B T
Total internal energy of a gas is (U) = 1/2 nRT
Here, n = 1
f = 5
U = 5/2 RT
Hence, (A)-(Q), (B)-(P), (C)-(S), (D)-(R)

SECTION-B

36. In the product
F = q(v × B)
= qv × (Bî + Bĵ + B₀k̂)
For q = 1 and v = 2î + 4ĵ + 6k̂ and
F = 4î - 20ĵ + 12k̂
What will be the complete expression for B?
(1) 6î + 6ĵ - 8k̂
(2) -8î - 8ĵ - 6k̂
(3) -6î - 6ĵ - 8k̂
(4) 8î + 8ĵ - 6k̂

Answer (3)

Sol. F = q(v × B)
= qv × (Bî + Bĵ + B₀k̂)
Given, q = 1 v = 2î + 4ĵ + 6k̂ and
F = 4î - 20ĵ + 12k̂
⇒ (4î - 20ĵ + 12k̂) = -1 × [(2î + 4ĵ + 6k̂) × (Bî + Bĵ + B₀k̂)]
Thus, calculating values of RHS,
| î ĵ k̂ |
| 2 4 6 |
| B B B₀ |
⇒ î(4B₀ - 6B) - ĵ(2B₀ - 6B) + k̂(2B - 4B)
Comparing L.H.S and R.H.S,
4B₀ - 6B = 4 ⇒ 2B₀ - 3B = 2 ...(1)
-(2B₀ - 6B) = -20 ⇒ B₀ - 3B = 10 ...(2)
2B - 4B = 12 ⇒ B = -6 ...(3)
From (2) and (3)
B = -6 and B₀ = -8
Hence, B = -6î - 6ĵ - 8k̂

37. From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times 'MR²'. Then the value of 'K' is

(1) 1/8
(2) 3/4
(3) 7/8
(4) 1/4

Answer (2)

Sol. Given that,
Mass of Ring = M
Radius of Ring = R
Now 90° arc is removed from circular ring, then mass removed = M/4
Mass of remaining portion = 3M/4
Moment of inertia of remaining part = ∫dmr²
⇒ I = R²∫dm (∵ r = R)
⇒ I = 3MR²/4
So the value of K is 3/4

38. Three resistors having resistances r₁, r₂ and r₃ are connected as shown in the given circuit. The ratio i₃/i₁ of currents in terms of resistances used in the circuit is

(1) r₂/(r₁ + r₃)
(2) r₁/(r₂ + r₃)
(3) r₂/(r₂ + r₃)
(4) r₁/(r₁ + r₂)

Answer (3)

Sol. In parallel combination of resistances r₂ and r₃ potential difference will be equal across both resistance.
So, i₂r₂ = i₃r₃ ⇒ i₂ = i₃r₃/r₂ ...(1)
As per Kirchhoff's first law
⇒ i₁ = i₂ + i₃
⇒ i₁ = (r₃/r₂ + 1)i₃ (from equation 1)
⇒ i₃/i₁ = r₂/(r₂ + r₃)

39. A ball of mass 0.15 kg is dropped from a height 10 m strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s²) nearly

(1) 1.4 kg m/s
(2) 0 kg m/s
(3) 4.2 kg m/s
(4) 2.1 kg m/s

Answer (3)

Sol. Given that:
Mass of ball = 0.15 kg
Height from which ball is dropped = 10 m
Impulse, I = Change in linear momentum = ΔP
I = P_f - P_i
Velocity of ball at ground (v) = √(2gh)
= √(2 × 10 × 10) = 10√2 m/s
I = 0.15 × 10√2(-ĵ) - 0.15 × 10√2(ĵ)
I = 2 × 0.15 × 10√2(-ĵ) = 4.2(-ĵ)
⇒ magnitude of impulse = 4.2 kg m/s

40. A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?

(1) 4 A
(2) 0.2 A
(3) 0.4 A
(4) 2 A

Answer (2)

Sol. In ideal transformer:
Input power = Output power
⇒ V_p I_p = V_s I_s = Given power
⇒ 220 × I_p = 44
⇒ I_p = 0.2 A

41. A uniform conducting wire of length 12a and resistance R is wound up as a current carrying coil in the shape of,
(i) an equilateral triangle of side 'a'
(ii) a square of side 'a'
The magnetic dipole moments of the coil in each case respectively are

(1) 4Ia² and 3Ia²
(2) √3 Ia² and 3Ia²
(3) 3Ia² and Ia²
(4) 3Ia² and 4Ia²

Answer (2)

Sol. Current in the loop will be V/R = I which is same for both loops.
Now magnetic moment of Triangle loop = NIA
M₁ = (12a/3a) · I · (√3/4)a² = √3 Ia²
and magnetic moment of square loop = N'IA'
= (12a/4a) · I · a²
M₂ = 3Ia²

42. A particle of mass 'm' is projected with a velocity v = kV_e (k < 1) from the surface of the earth.
(V_e = escape velocity)
The maximum height above the surface reached by the particle is

(1) Rk²/(1 - k²)
(2) R(k/(1 + k))²
(3) R(k/(1 + k))²
(4) ...

Answer (1)

Sol. given v = kV_e
where, k < 1
Thus, v < V_e
From conservation of mechanical energy,
1/2 mv² - GmM/R = -GmM/(R + h)
⇒ v²/2 = GM/R - GM/(R + h) = h/(R(R + h)) GM
⇒ 1/2 k²V_e² = Gmh/(R(R + h))
We know, V_e = √(2GM/R)
⇒ 1/2 k²(2GM/R) = Gmh/(R(R + h))
k² = h/(R + h)
Rk² + hk² = h
Rk² = h(1 - k²)
∴ h = Rk²/(1 - k²)

43. A series LCR circuit containing 5.0 H inductor, 80 μF capacitor and 40 Ω resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be

(1) 42 rad/s and 58 rad/s
(2) 25 rad/s and 75 rad/s
(3) 50 rad/s and 25 rad/s
(4) 46 rad/s and 54 rad/s

Answer (4)

Sol. The resonance frequency of LCR series circuit is
given as ω₀ = 1/√(LC) = 1/√(5 × 80 × 10⁻⁶) = 50 rad/s
Now half power frequencies are given as
ω = ω₀ ± R/(2L)

44. A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle 'θ' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, θ, is then given by :

(1) θ = sin⁻¹(2gT²/(π²R))^(1/2)
(2) θ = cos⁻¹(π²R/(gT²))^(1/2)
(3) θ = sin⁻¹(π²R/(gT²))^(1/2)
(4) ...

Answer (1)

Sol. To complete a circular path of radius R, time period is T.
so speed of particle (U) = 2πR/T ...(1)
Now the particle is projected with same speed at angle θ to horizontal.
So Maximum Height (H) = U²sin²θ/(2g)
Given that : H = 4R
⇒ U²sin²θ/(2g) = 4R
⇒ sin²θ = 8gR/U² ...(2)
⇒ sin²θ = 8gRT²/(4π²R²) = 2gT²/(π²R) (using equation 1)
⇒ θ = sin⁻¹(2gT²/(π²R))^(1/2)

45. Two conducting circular loops of radii R₁ and R₂ are placed in the same plane with their centres coinciding. If R₁ >> R₂, the mutual inductance M between them will be directly proportional to

(1) R₂²/R₁
(2) R₁²/R₂
(3) R₁/R₂
(4) ...

Answer (1)

Sol. Two concentric coils are of radius R₁ and R₂ as shown
Let current in outer loop be i
Magnetic field at centre = B = μ₀i/(2R₁)
Magnetic flux through inner coil = B × πR₂²
φ = μ₀i/(2R₁) × πR₂²
φ = μ₀i/2 × πR₂²/R₁
as per definition, φ = Mi
⇒ M = (μ₀π/2) R₂²/R₁
∴ M ∝ R₂²/R₁

46. Twenty seven drops of same size are charged at 200 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

(1) 1980 V
(2) 660 V
(3) 1320 V
(4) 1520 V

Answer (1)

Sol. Electric potential due to a charged sphere = kQ/R
k = 9 × 10⁹ N-m²/C²
Q : charge on sphere
R : Radius of sphere
Let charge and radius of smaller drop is q and r respectively
For smaller drop, V = kq/r = 220V
Let R be radius of bigger drop,
As volume remains the same
(4/3 πr³) × 27 = 4/3 πR³
⇒ R = ∛27 r = 3r
Now, using charge conservation,
⇒ Q = 27q
V_bigdrop = kQ/R = k(27q)/(3r) = 9(kq/r)
= 9 × 220 = 1980V

47. A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass 'm' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of 'm' such that the rod is in equilibrium. (g = 10 m/s²)

(1) 1/12 kg
(2) 1/2 kg
(3) 1/3 kg
(4) 1/6 kg

Answer (1)

Sol. Given that
Mass of rod = 500 g
Length of rod = 200 cm
Rod will be in equilibrium, when net torque about point O will be zero.
Torque at point O due to 2 kg mass
τ = r × F = rF sin θ(n̂)
τ₁ = 20 × 20 × 10⁻² × sin 90°(k̂) = 4 N m(k̂)
Torque due to mass of rod :
τ₂ = 5 × 60 × 10⁻² × sin 90°(-k̂) = 3 N m(-k̂)
Torque due to mass m
τ₃ = mg × 120 × 10⁻² × sin 90°(-k̂) = 12m N m(-k̂)
Net torque about point O will be zero
So τ₁ + τ₂ + τ₃ = 0
⇒ 4 - 3 - 12m = 0
⇒ 12m = 1
m = 1/12 kg

48. A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of

(1) 20 cm from the plane mirror, it would be a virtual image
(2) 20 cm from the lens, it would be a real image
(3) 30 cm from the lens, it would be a real image
(4) 30 cm from the plane mirror, it would be a virtual image

Answer (1)

Sol. Using lens formula for first refraction from convex lens
1/v₁ - 1/u = 1/f
v₁ = ? u = -60 cm, f = 30 cm
⇒ 1/v₁ + 1/60 = 1/30 ⇒ v₁ = 60 cm
I₁ here is first image by lens
The plane mirror will produce an image at distance 20 cm to left of it.
For second refraction from convex lens,
u = -20 cm, v = ? f = 30 cm
1/v - 1/u = 1/f ⇒ 1/v + 1/20 = 1/30
⇒ 1/v = 1/30 - 1/20 ⇒ v = -60 cm
Thus the final image is virtual and at a distance, 60 - 40 = 20 cm from plane mirror.

49. A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s?
(Take g = 10 m/s²)

(1) 20√2 m/s, 10 m/s²
(2) 20 m/s, 5 m/s²
(3) 20 m/s, 0
(4) 20√2 m/s, 0

Answer (1)

Sol. Initial velocity of car = 0
Acceleration of car = 5 m/s²
Velocity of car at t = 4 s; v = u + at
⇒ v = 0 + 5 × 4 = 20 ms⁻¹
At t = 4 s, A ball is dropped out of a window so velocity of ball at this instant is 20 ms⁻¹ along horizontal.
After 2 seconds of motion :
Horizontal velocity of ball = 20 ms⁻¹ ∴ a_x = 0
Vertical velocity of ball (v_y) = u_y + a_y t
v_y = 0 + 10 × 2 = 20 ms⁻¹ (∵ a_y = g = 10 m/s²)
So magnitude of velocity of ball
(v) = √(v_x² + v_y²) = 20√2 m/s
Acceleration of ball at t = 6 s is g = 10 m/s²
As ball is under free fall.

50. For the given circuit, the input digital signals are applied at the terminals A, B and C. What would be the output at the terminal y?

(1) [Output waveform]
(2) [Output waveform]
(3) [Output waveform]
(4) [Output waveform]

Answer (3)

Sol. Output of combination of logic gates is given as
y = A·B + B̄·C̄

Input Signals | Output Signal
Time duration | A | B | C | A·B | B̄·C̄ | y = A·B + B̄·C̄
0-t₁ | 0 | 0 | 1 | 0 | 1 | 1
t₁-t₂ | 1 | 0 | 1 | 0 | 0 | 0
t₂-t₃ | 0 | 1 | 0 | 0 | 0 | 0
t₃-t₄ | 1 | 1 | 0 | 1 | 0 | 1
t₄-t₅ | 0 | 1 | 1 | 0 | 0 | 0
t₅-t₆ | 1 | 0 | 1 | 0 | 1 | 1

So the output y is high (1) that is v₀ = 5V

51. The incorrect statement among the following is :
(1) Actinoids are highly reactive metals, especially when finely divided.
(2) Actinoid contraction is greater for element to element than lanthanoid contraction
(3) Most of the trivalent Lanthanoid ions are colorless in the solid state
(4) Lanthanoids are good conductors of heat and electricity

Answer (3)

Sol. Actinoids are highly reactive metals, especially when finely divided
Actinoid contraction is greater from element to element than lanthanoid contraction resulting from poor shielding by 5f electrons
Many trivalent lanthanoids ions are coloured both in the solid state and in aqueous solutions.
Lanthanoids have typical metallic structure and are good conductors of heat and electricity

52. Given below are two statements :
Statement I :
Aspirin and Paracetamol belong to the class of narcotic analgesics.
Statement II :
Morphine and Heroin are non-narcotic analgesics.
In the light of the above statements, choose the correct answer from the options given below.

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

Answer (3)

Sol. Aspirin and paracetamol belong to the class of non-narcotic analgesics
Morphine and Heroin are Narcotic analgesics
Both statement I and statement II are false

53. Statement I : Acid strength increases in the order given as HF ≪ HCl ≪ HBr ≪ HI.
Statement II : As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases.
In the light of the above statements, choose the correct answer from the options given below.

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

Answer (2)

Sol. In the modern periodic table, moving down the group as the size of halogen atom increases, the H-X bond length also increases as a result the bond enthalpy decreases. Hence, The acidic strength also increases.
So, the correct order of acidic strength is
HI > HBr > HCl > HF

54. Which one among the following is the correct option for right relationship between C_p and C_v for one mole of ideal gas?

(1) C_v = RC_p
(2) C_p + C_v = R
(3) C_p - C_v = R
(4) C_p = RC_v

Answer (3)

Sol. At constant volume, q_v = C_v ΔT = ΔU
At constant pressure, q_p = C_p ΔT = ΔH
For a mole of an ideal gas,
ΔH = ΔU + Δ(PV)
= ΔU + Δ(RT)
= ΔU + RΔT
On putting the values of ΔH and ΔU, we have
C_p ΔT = C_v ΔT + RΔT
C_p = C_v + R
C_p - C_v = R

55. The correct option for the number of body centred unit cells in all 14 types of Bravais lattice unit cells is :

(1) 3
(2) 7
(3) 5
(4) 2

Answer (1)

Sol. In 14 types of Bravais lattices, body centred unit cell is present in cubic, tetragonal and orthorhombic crystal systems. Hence, body centred possible variation is present in three crystal systems.

56. Among the following alkaline earth metal halides, one which is covalent and soluble in organic solvents is :

(1) Beryllium chloride
(2) Calcium chloride
(3) Strontium chloride
(4) Magnesium chloride

Answer (1)

Sol. Except for beryllium chloride all other chloride of alkaline earth metals are ionic in nature. Due to small size of Be, Beryllium chloride is essentially covalent and soluble in organic solvents.

57. Tritium, a radioactive isotope of hydrogen, emits which of the following particles?

(1) Neutron
(2) Beta (β⁻)
(3) Alpha (α)
(4) Gamma (γ)

Answer (2)

Sol. Hydrogen has three isotopes : protium, ¹H deuterium, ²H or D and tritium ³H or T. Of these isotopes, only tritium is radioactive and emits low energy β⁻ particles (t₁/₂, 12.33 years).

58. The maximum temperature that can be achieved in blast furnace is :

(1) Upto 5000 K
(2) Upto 1200 K
(3) Upto 2200 K
(4) Upto 1900 K

Answer (3)

Sol. Maximum temperature that can be achieved in blast furnace is upto 2200 K.
(As per NCERT text: 2170 K maximum temperature is given in the figure of blast furnace)

59. BF₃ is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are :

(1) sp² and 8
(2) sp³ and 4
(3) sp³ and 6
(4) sp² and 6

Answer (4)

Sol. Number of electrons around boron atom is 6. Hybridization of B is sp². Shape is trigonal planar.

60. The major product of the following chemical reaction is :
CH₃-CH(CH₃)-CH=CH₂ + HBr → (C₆H₅CO)₂O₂ → ?

(1) CH₃-CH(CH₃)-CBr-CH₂-CH₃
(2) CH₃-CH(CH₃)-CH₂-CH₂-Br
(3) CH₃-CH(CH₃)-CH₂-CH₂-O-COC₆H₅
(4) CH₃-CH(CH₃)-CH(Br)-CH₃

Answer (2)

Sol. Mechanism : Peroxide effect proceeds via free radical chain mechanism.
(i) C₆H₅-C(=O)-O-O-C(=O)-C₆H₅ → Homolysis → 2C₆H₅-C(=O)-O· → 2C₆H₅· + CO₂
(ii) C₆H₅· + H-Br → Homolysis → C₆H₆ + Br·
(iii) CH₃-CH(CH₃)-CH=CH₂ + Br· → CH₃-CH(CH₃)-CH·-CH₂-Br (More Stable secondary free radical)
(iv) CH₃-CH(CH₃)-CH·-CH₂-Br + H-Br → Homolysis → CH₃-CH(CH₃)-CH₂-CH₂-Br (Major product)

61. The molar conductance of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm² mol⁻¹ respectively. The molar conductance of CH₃COOH at infinite dilution is. Choose the right option for your answer.

(1) 540.48 S cm² mol⁻¹
(2) 201.28 S cm² mol⁻¹
(3) 390.71 S cm² mol⁻¹
(4) 698.28 S cm² mol⁻¹

Answer (3)

Sol. According to Kohlrausch law of independent migration of ions.
Λ°_m(CH₃COOH)
= Λ°_m(CH₃COONa) + Λ°_m(HCl) - Λ°_m(NaCl)
= 91.0 S cm² mol⁻¹ + 426.16 S cm² mol⁻¹ - 126.45 S cm² mol⁻¹
= 390.71 S cm² mol⁻¹

62. A particular station of All India Radio, New Delhi broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light c = 3.0 × 10⁸ ms⁻¹]

(1) 21.92 cm
(2) 219.3 m
(3) 219.2 m
(4) 2192 m

Answer (2)

Sol. Energy of electromagnetic radiation (E)
So, c/λ = γ ⇒ λ = c/γ
λ = (3 × 10⁸)/(1368 × 10³) = 219.3 m

63. Which of the following reactions is the metal displacement reaction? Choose the right option.

(1) 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂↑
(2) 2KClO₃ →Δ→ 2KCl + 3O₂
(3) Cr₂O₃ + 2Al →Δ→ Al₂O₃ + 2Cr
(4) Fe + 2HCl → FeCl₂ + H₂↑

Answer (3)

Sol. Both reactions (1) and (2) are examples of decomposition reactions. Reactions (3) and (4), both are examples of displacement reactions, while reaction (3) is an example of metal displacement reaction.

64. The right option for the statement "Tyndall effect is exhibited by", is :

(1) Urea solution
(2) NaCl solution
(3) Glucose solution
(4) Starch solution

Answer (4)

Sol. Tyndall effect is exhibited by colloidal solution only. Among the given options, Urea, NaCl and Glucose solutions are true solutions, so cannot show Tyndall effect. Starch solution is a colloidal solution therefore can show Tyndall effect.

65. The compound which shows metamerism is :

(1) C₄H₁₀O
(2) C₅H₁₂
(3) C₃H₈O
(4) C₃H₆O

Answer (1)

Sol. Compounds with formula C₄H₁₀O can be ethers which may exhibit metamerism. For example
CH₃-CH₂-O-CH₂-CH₃, CH₃-O-CH(CH₃)-CH₃ and CH₃-O-CH₂-CH₂-CH₃ are metamers as structure of alkyl chains are different around the functional group.

66. Match List-I with List-II.
List-I
(a) PCl₅
(b) SF₆
(c) BrF₅
(d) BF₃
List-II
(i) Square pyramidal
(ii) Trigonal planar
(iii) Octahedral
(iv) Trigonal bipyramidal
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)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(4) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

Answer (2)

Sol. (a) PCl₅ : sp³d hybridised and trigonal bipyramidal in shape
(b) SF₆ : sp³d² hybridised and octahedral in shape
(c) BrF₅ : sp³d² hybridised and square pyramidal in shape
(d) BF₃ : sp² hybridised and trigonal planar in shape

67. Which one of the following polymers is prepared by addition polymerization?

(1) Dacron
(2) Teflon
(3) Nylon-66
(4) Novolac

Answer (2)

Sol. Dacron, Nylon-66 and Novolac are prepared by condensation polymerisation.
Teflon is an addition polymer. Monomer of teflon is tetrafluoroethene.
nCF₂=CF₂ →Catalyst/High pressure→ (-CF₂-CF₂-)ₙ Teflon
Tetrafluoroethene

68. The RBC deficiency is deficiency disease of :

(1) Vitamin B₂
(2) Vitamin B₁₂
(3) Vitamin B₆
(4) Vitamin B₁

Answer (2)

Sol. Deficiency of vitamin B₂ (Riboflavin) causes cheilosis, digestive disorders and burning sensation of the skin.
Deficiency of vitamin B₁₂ causes Pernicious anaemia which is RBC deficiency in haemoglobin.
Deficiency of vitamin B₆ (Pyridoxine) causes Convulsions.
Deficiency of vitamin B₁ (Thiamine) causes Beri-Beri (loss of appetite and retarded growth).

69. Identify the compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali.

(1) CH₃-CH₂-N(CH₃)-CH₂-CH₃
(2) CH₃-CH₂-NO₂
(3) CH₃-CH₂-NH-CH₃
(4) CH₃-CH₂-NH₂

Answer (4)

Sol. Benzenesulphonyl chloride (C₆H₅SO₂Cl) is also known as Hinsberg's reagent.
The reaction of Hinsberg's reagent (C₆H₅SO₂Cl) with primary amine (CH₃CH₂NH₂) yields N-ethylbenzene sulphonamide.
C₆H₅-SO₂-Cl + H-NH-C₂H₅ → C₆H₅-SO₂-NH-C₂H₅ + HCl
N-Ethylbenzene sulphonamide (Soluble in alkali)
The reaction of Hinsberg's reagent (C₆H₅SO₂Cl) with secondary amine (C₂H₅NHCH₃) gives, N-Ethyl-N-Methyl benzene sulphonamide
3° amine do not react with Hinsberg reagent

70. Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of :

(1) Having similar chemical properties
(2) Belonging to same group
(3) Diagonal relationship
(4) Lanthanoid contraction

Answer (4)

Sol. The cumulative effect of the contraction of the lanthanoid series, known as lanthanoid contraction, causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. The almost identical radii of Zr (160 pm) and Hf (159 pm) is a consequence of the lanthanoid contraction.

71. Ethylene diaminetetraacetate (EDTA) ion is :

(1) Tridentate ligand with three "N" donor atoms
(2) Hexadentate ligand with four "O" and two "N" donor atoms
(3) Unidentate ligand
(4) Bidentate ligand with two "N" donor atoms

Answer (2)

Sol. Ethylene diaminetetraacetate (EDTA) ion is a hexadentated ligand having four donor oxygen atoms and two donor nitrogen atoms

72. The major product formed in dehydrohalogenation reaction of 2-Bromopentane is Pent-2-ene. This product formation is based on?

(1) Huckel's Rule
(2) Saytzeff's Rule
(3) Hund's Rule
(4) Hofmann Rule

Answer (2)

Sol. Major product formed in dehydrohalogenation reaction of 2-bromopentane is pent-2-ene because according to Saytzeff's rule, in dehydrohalogenation reactions, the preferred product is that alkene which has greater number of alkyl group(s) attached to the doubly bonded carbon atoms.
CH₃-CH₂-CH₂-CH(Br)-CH₃ →OH⁻→ CH₃-CH₂-CH=CH-CH₃ (Pent-2-ene 81%) + CH₃-CH₂-CH₂-CH=CH₂ (Pent-1-ene 19%)

73. An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]

(1) CH₄
(2) CH
(3) CH₂
(4) CH₃

Answer (4)

Sol. Element | Mass percentage | No. of mole | Mole ratio
C | 78% | 78/12 = 6.5 | 6.5/6.5 = 1
H | 22% | 22/1 = 22 | 22/6.5 = 3.38 = 3
Based on above calculation, possible empirical formula is CH₃

74. The pK_b of dimethylamine and pK_a of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethylammonium acetate solution is :

(1) 6.25
(2) 8.50
(3) 5.50
(4) 7.75

Answer (4)

Sol. Dimethylammonium acetate is a salt of weak acid and weak base whose pH can be calculated as
pH = 7 + 1/2(pK_a - pK_b)
= 7 + 1/2(4.77 - 3.27)
= 7.75

75. The structures of beryllium chloride in solid state and vapour phase, are :

(1) Chain in both
(2) Chain and dimer, respectively
(3) Linear in both
(4) Dimer and Linear, respectively

Answer (2)

Sol. Beryllium chloride has a chain structure in the solid state as shown below.
In vapour phase Beryllium chloride tends to form a chloro-bridged dimer.

76. Which one of the following methods can be used to obtain highly pure metal which is liquid at room temperature?

(1) Zone refining
(2) Electrolysis
(3) Chromatography
(4) Distillation

Answer (4)

Sol. Distillation method is generally used for the purification of metals having low boiling point such as Hg, Zn etc.

77. Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are :

(1) 12, 6
(2) 8, 4
(3) 6, 12
(4) 2, 1

Answer (1)

Sol. Number of octahedral and tetrahedral voids formed by N closed packed atoms are N and 2N respectively. Each hexagonal unit cell contains 6 atoms therefore, number of tetrahedral and octahedral voids are 12 and 6 respectively.

78. The correct structure of 2,6-Dimethyl-dec-4-ene is

(1) [Structure]
(2) [Structure]
(3) [Structure]
(4) [Structure]

Answer (2)

Sol. 2,6-Dimethyldec-4-ene

79. Choose the correct option for graphical representation of Boyle's law, which shows a graph of pressure vs. volume of a gas at different temperatures :

(1) [Graph showing P vs V curves]
(2) [Graph showing P vs V curves]
(3) [Graph showing P vs V horizontal line]
(4) [Graph showing P vs V straight lines]

Answer (1)

Sol. According to Boyle's law
P ∝ 1/V ⇒ P = k/V ⇒ PV = k
where k is proportionality constant and equal to nRT.
Graph between P vs. V should be rectangular hyperbola and product of PV increases with increase in temperature.

80. The following solutions were prepared by dissolving 10 g of glucose (C₆H₁₂O₆) in 250 ml of water (P₁), 10 g of urea (CH₄N₂O) in 250 ml of water (P₂) and 10 g of sucrose (C₁₂H₂₂O₁₁) in 250 ml of water (P₃). The right option for the decreasing order of osmotic pressure of these solutions is :

(1) P₃ > P₁ > P₂
(2) P₂ > P₁ > P₃
(3) P₁ > P₂ > P₃
(4) P₂ > P₃ > P₁

Answer (2)

Sol. Osmotic pressure (π) = iCRT where C is molar concentration of the solution
With increase in molar concentration of solution osmotic pressure increases.
Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.
Order of molar mass of solute decreases as Sucrose > Glucose > Urea
So, correct order of osmotic pressure of solution is P₃ < P₁ < P₂

81. The correct sequence of bond enthalpy of 'C-X' bond is :

(1) CH₃-Cl > CH₃-F > CH₃-Br > CH₃-I
(2) CH₃-F < CH₃-Cl < CH₃-Br < CH₃-I
(3) CH₃-F > CH₃-Cl > CH₃-Br > CH₃-I
(4) CH₃-F < CH₃-Cl > CH₃-Br > CH₃-I

Answer (3)

Sol. The size of halogen atom increases from F to I hence bond length from C-F to C-I increases
Bond enthalpy from CH₃-F to CH₃-I decreases

C-X Bond | Bond dissociation enthalpies/kJ mol⁻¹
CH₃-F | 452
CH₃-Cl | 351
CH₃-Br | 293
CH₃-I | 234

82. Dihedral angle of least stable conformer of ethane is :

(1) 0°
(2) 120°
(3) 180°
(4) 60°

Answer (1)

Sol. Ethane has two conformers (i) Eclipsed (ii) Staggered
Eclipsed conformer is least stable while staggered conformer is most stable. In eclipsed conformer the dihedral angle is 0°

83. Noble gases are named because of their inertness towards reactivity. Identify an incorrect statement about them.

(1) Noble gases have large positive values of electron gain enthalpy
(2) Noble gases are sparingly soluble in water
(3) Noble gases have very high melting and boiling points
(4) Noble gases have weak dispersion forces

Answer (3)

84. What is the IUPAC name of the organic compound formed in the following chemical reaction?
Acetone (i) C₂H₅MgBr/dry Ether → Product (ii) H₂O, H⁺

(1) 2-methylbutan-2-ol
(2) 2-methylpropan-2-ol
(3) pentan-2-ol
(4) pentan-3-ol

Answer (1)

Sol. CH₃-C(=O)-CH₃ →(i) C₂H₅MgBr/Dry ether→ CH₃-C(OMgBr)(C₂H₅)-CH₃ →(ii) H₂O/H⁺→ CH₃-C(OH)(C₂H₅)-CH₃
Product (2-methylbutan-2-ol)

85. For a reaction A → B enthalpy of reaction is -4.2 kJ mol⁻¹ and enthalpy of activation is 9.6 kJ mol⁻¹. The correct potential energy profile for the reaction is shown in option.

(1) [Graph]
(2) [Graph]
(3) [Graph]
(4) [Graph]

Answer (3)

Sol. ΔH_reaction = (E_a)_f - (E_a)_b
-4.2 = (E_a)_f - (E_a)_b
-4.2 = 9.6 - (E_a)_b
(E_a)_b = 9.6 + 4.2 = 13.8 kJ mol⁻¹
Since reaction is exothermic, so possible graph is (3) only.
Also (E_a)_f < (E_a)_b so answer is option (3).

SECTION-B

86. Match List-I with List-II.
List-I
(a) 2SO₂(g) + O₂(g) → 2SO₃(g)
(b) HOCl(g) →hv→ OH + Cl
(c) CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
(d) NO₂(g) →hv→ NO(g) + O(g)
List-II
(i) Acid rain
(ii) Smog
(iii) Ozone depletion
(iv) Tropospheric pollution
Choose the correct answer from the options given below.
(1) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
(2) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(3) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(4) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

Answer (4)

87. CH₃CH₂COO⁻Na⁺ →NaOH, +?→ CH₃CH₃ + Na₂CO₃
Consider the above reaction and identify the missing reagent/chemical.

(1) DIBAL-H
(2) B₂H₆
(3) Red Phosphorus
(4) CaO

Answer (4)

Sol. Alkane is produced by heating sodium salt of carboxylic acid with sodium (NaOH and CaO in the ratio of 3:1)
CH₃CH₂COO⁻Na⁺ →NaOH + CaO→ CH₃CH₃ + Na₂CO₃

88. The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3:2 is :
[At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

(1) 350 mm of Hg
(2) 160 mm of Hg
(3) 168 mm of Hg
(4) 336 mm of Hg

Answer (4)

Sol. Given: n_C₆H₆ : n_C₈H₁₈ = 3:2
So, χ_C₆H₆ = 3/5, χ_C₈H₁₈ = 2/5
p_s = p°_C₆H₆ χ_C₆H₆ + p°_C₈H₁₈ χ_C₈H₁₈
= 280 × 3/5 + 420 × 2/5
= 168 + 168 = 336 mm of Hg

89. Match List-I with List-II.
List-I
(a) [Benzene] →CO, HCl/Anhyd. AlCl₃/CuCl→
(b) R-C(=O)-CH₃ + NaOX →
(c) R-CH₂-OH + R'COOH →Conc. H₂SO₄→
(d) R-CH₂COOH →(i) X₂/Red P (ii) H₂O→
List-II
(i) Hell-Volhard-Zelinsky reaction
(ii) Gattermann-Koch reaction
(iii) Haloform reaction
(iv) Esterification
Choose the correct answer from the options given below.
(1) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(2) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
(3) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
(4) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)

Answer (1)

Sol. Gattermann-Koch reaction:
[Benzene] →CO, HCl/Anhyd. AlCl₃/CuCl→ [Benzaldehyde]
Haloform reaction:
R-C(=O)-CH₃ + NaOX → R-C(=O)ONa + CHX₃
Esterification:
R-CH₂-OH + R'-C(=O)-OH →Conc. H₂SO₄→ R'-C(=O)-OCH₂-R
Hell-Volhard-Zelinsky reaction:
R-CH₂COOH →(i) X₂/Red P (ii) H₂O→ R-CH(X)-COOH

90. The intermediate compound 'X' in the following chemical reaction is:
[Benzene with CH₃] + CrO₂Cl₂ →CS₂→ X →H₃O⁺→ [Benzaldehyde]

(1) [Structure with CHCl]
(2) [Structure with CH(OCrOHCl)₂]
(3) [Structure with CH(OCOCH₃)₂]
(4) [Structure with CHCl₂]

Answer (2)

Sol. Etard's reaction
[Toluene] + CrO₂Cl₂ →CS₂→ [Benzaldehyde]

91. The product formed in the following chemical reaction is:
[Cyclohexanone with CH₂COOCH₃ and CH₃] →NaBH₄/C₂H₅OH→ ?

(1) [Cyclohexanol with CH₂COOCH₃ and CH₃]
(2) [Cyclohexanol with CH₂CH(OH)COOCH₃ and CH₃]
(3) [Cyclohexanone with CH₂CH₂OH and CH₃]
(4) [Cyclohexanol with CH₂CH(OH)CH₃ and CH₃]

Answer (1)

Sol. NaBH₄ is a reducing agent. It reduces carbonyl group into alcohols but does not reduce esters.

92. The slope of Arrhenius plot (ln k v/s 1/T) of first order reaction is -5 × 10³ K. The value of E_a of the reaction is. Choose the correct option for your answer.
[Given R = 8.314 J K⁻¹ mol⁻¹]

(1) -83 kJ mol⁻¹
(2) 41.5 kJ mol⁻¹
(3) 83.0 kJ mol⁻¹
(4) 166 kJ mol⁻¹

Answer (2)

Sol. Arrhenius equation
k = Ae^(-E_a/RT)
ln k = ln A + ln e^(-E_a/RT)
ln k = ln A - (E_a/R)(1/T) → (1)
Slope of ln k vs 1/T curve,
m = -E_a/R
-5 × 10³ = -E_a/R
E_a = 5 × 10³ × 8.314 J/mol
= 41.57 × 10³ J/mol
= 41.5 kJ/mol

93. The reagent 'R' in the given sequence of chemical reaction is:
[2,4,6-tribromoaniline] →NaNO₂/HCl/0-5°C→ [2,4,6-tribromobenzenediazonium chloride] →R→ [1,3,5-tribromobenzene]

(1) CuCN/KCN
(2) H₂O
(3) CH₃CH₂OH
(4) HI

Answer (3)

Sol. Reagent R is C₂H₅OH with diazonium salt.

94. For irreversible expansion of an ideal gas under isothermal condition, the correct option is:

(1) ΔU ≠ 0, ΔS_total = 0
(2) ΔU = 0, ΔS_total = 0
(3) ΔU ≠ 0, ΔS_total ≠ 0
(4) ΔU = 0, ΔS_total ≠ 0

Answer (4)

Sol. For a spontaneous process, ΔS_total > 0 and since irreversible process is always spontaneous therefore ΔS_total > 0
Since ΔU = nC_V ΔT and ΔT = 0 for isothermal process therefore ΔU = 0

95. From the following pairs of ions which one is not an iso-electronic pair?

(1) Fe²⁺, Mn²⁺
(2) O²⁻, F⁻
(3) Na⁺, Mg²⁺
(4) Mn²⁺, Fe³⁺

Answer (1)

Sol. Isoelectronic species have same number of electrons.

Species | Number of electrons
Fe²⁺ | 26 - 2 = 24
Mn²⁺ | 25 - 2 = 23
O²⁻ | 8 + 2 = 10
F⁻ | 9 + 1 = 10
Na⁺ | 11 - 1 = 10
Mg²⁺ | 12 - 2 = 10
Fe³⁺ | 26 - 3 = 23

96. The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. What is the dissociation constant of acetic acid? Choose the correct option.
[Λ°_H⁺ = 350 S cm² mol⁻¹]
[Λ°_CH₃COO⁻ = 50 S cm² mol⁻¹]

(1) 2.50 × 10⁻⁵ mol L⁻¹
(2) 1.75 × 10⁻⁴ mol L⁻¹
(3) 2.50 × 10⁻⁴ mol L⁻¹
(4) 1.75 × 10⁻⁵ mol L⁻¹

Answer (4)

Sol. Λ_m = 20 S cm² mol⁻¹
Λ°_m CH₃COOH = Λ°_CH₃COO⁻ + Λ°_m H⁺
= 50 + 350 = 400 S cm² mol⁻¹
α = Λ_m/Λ°_m = 20/400 = 1/20
K_a = Cα²/(1 - α) = Cα² = 7 × 10⁻³ × (1/20)²
= 7 × 10⁻³ × 1/4 × 10⁻²
= 1.75 × 10⁻⁵ mol L⁻¹

97. Which of the following molecules is non-polar in nature?

(1) NO₂
(2) CH₂O
(3) SbCl₅
(4) ...

Answer (4)

Sol. SbCl₅ : Net vector summation of bond moments will be zero so SbCl₅ is a non-polar molecule.
NO₂ : polar molecule.
POCl₃ : polar molecule.
CH₂O : polar molecule.

98. Match List-I with List-II.
List-I
(a) [Fe(CN)₆]³⁻
(b) [Fe(H₂O)₆]³⁺
(c) [Fe(CN)₆]⁴⁻
(d) [Fe(H₂O)₆]²⁺
List-II
(i) 5.92 BM
(ii) 0 BM
(iii) 4.90 BM
(iv) 1.73 BM
Choose the correct answer from the options given below.
(1) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
(2) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
(3) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(4) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

Answer (1)

Sol. Magnetic moment, μ = √(n(n + 2)) BM (where n = number of unpaired electrons)
Complex | No. of unpaired electron(s) | μ (BM)
(a) [Fe(CN)₆]³⁻ | 1 | 1.73
(b) [Fe(H₂O)₆]³⁺ | 5 | 5.92
(c) [Fe(CN)₆]⁴⁻ | 0 | 0
(d) [Fe(H₂O)₆]²⁺ | 4 | 4.90

99. Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O₂ and 2 g H₂ confined in a total volume of one litre at 0°C is :
Given R = 0.082 L atm mol⁻¹ K⁻¹, T = 273 K

(1) 26.02
(2) 2.518
(3) 2.602
(4) 25.18

Answer (4)

Sol. n_O₂ = 4/32 = 1/8
n_H₂ = 2/2 = 1
n_t = 1/8 + 1 = 9/8
P_t V = n_t RT
P_t = 9/8 × 0.082 × 273

Answer (4)

100. In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it?

(1) CO₂ < SiO₂ < SnO₂ < PbO₂ : Increasing oxidizing power
(2) HF < HCl < HBr < HI : Increasing acidic strength
(3) H₂O < H₂S < H₂Se < H₂Te : Increasing pK_a values
(4) NH₃ < PH₃ < AsH₃ < SbH₃ : Increasing acidic character

Answer (3)

Sol. Stronger is the acid, lower is the value of pK_a. On moving down the group, bond dissociation enthalpy of hydrides of group 16 elements decreases hence acidity increases and pK_a value decreases. Correct order of pK_a value will be
H₂O > H₂S > H₂Se > H₂Te

101. Mutations in plant cells can be induced by:

(1) Zeatin
(2) Kinetin
(3) Infrared rays
(4) Gamma rays

Answer (4)

Sol. Several kinds of radiation like gamma rays, X-rays, UV-rays cause mutation. These are physical mutagens. Such induced mutation in plants is done to develop improved varieties. The first natural cytokinin was isolated from unripe maize grain known as zeatin. The cytokinin that was obtained from degraded product of autoclaved herring sperm DNA was kinetin (N⁶-furfuryl aminopurine). Infrared rays cause heating effect.

102. Which of the following is an incorrect statement?

(1) Nuclear pores act as passages for proteins and RNA molecules in both directions between nucleus and cytoplasm
(2) Mature sieve tube elements possess a conspicuous nucleus and usual cytoplasmic organelles
(3) Microbodies are present both in plant and animal cells
(4) The perinuclear space forms a barrier between the materials present inside the nucleus and that of the cytoplasm

Answer (2)

Sol. A mature sieve tube elements possess a peripheral cytoplasm and a large central vacuole but lacks a nucleus. Rest of other statements are correct.

103. The term used for transfer of pollen grains from anthers of one plant to stigma of a different plant which, during pollination, brings genetically different types of pollen grains to stigma, is :

(1) Cleistogamy
(2) Xenogamy
(3) Geitonogamy
(4) Chasmogamy

Answer (2)

Sol. Xenogamy refers to the transfer to pollen grains from anthers of one plant to stigma of a different plant which during pollination, brings genetically different types of pollen grains to stigma. Cleistogamy is a condition in which flower does not open. Geitonogamy refers to the transfer of pollen grain from anther to stigma of another flower of the same plant. Chasmogamy is a condition in which flowers remain open.

104. The factor that leads to Founder effect in a population is :

(1) Genetic drift
(2) Natural selection
(3) Genetic recombination
(4) Mutation

Answer (1)

Sol. Change in gene frequency in a small population by chance is known as genetic drift. Genetic drift has two ramifications, one is bottle neck effect and another is founder's effect. When accidentally a few individuals are dispersed and act as founders of a new isolated population, founder's effect is said to be observed. Crossing over which occurs during gamete formation results in genetic recombination. Mutations are random and directionless.

105. Genera like Selaginella and Salvinia produce two kinds of spores. Such plants are known as:

(1) Heterosporous
(2) Homosporus
(3) Heterosporus
(4) Homosporous

Answer (1)

Sol. Plants like Selaginella and Salvinia produce two kinds of spore i.e., microspores and macrospores. They are known as heterosporous.
Most of the pteridophytes produce single type of spores and are called homosporous
Sorus are brownish or yellowish cluster of spore-producing structures located on the lower surface of fern leaves.

106. The production of gametes by the parents, formation of zygotes, the F₁ and F₂ plants, can be understood from a diagram called :

(1) Net square
(2) Bullet square
(3) Punch square
(4) Punnett square

Answer (4)

Sol. The production of gametes (n) by the parents (2n), the formation of the zygote (2n), the F₁ and F₂ plants can be understood from a diagram called Punnett square.

107. Diadelphous stamens are found in

(1) China rose and citrus
(2) China rose
(3) Citrus
(4) Pea

Answer (4)

Sol. Stamens are said to be diadelphous when these are united in two bundles e.g. Pea. China rose has monoadelphous stamens while, Citrus has polyadelphous stamens. Monoadelphous stamens are grouped in single bundle whereas polyadelphous stamens occur in more than two bundles.

108. Match List-I with List-II.
List-I
(a) Cohesion
(b) Adhesion
(c) Surface tension
(d) Guttation
List-II
(i) More attraction in liquid phase
(ii) Mutual attraction among water molecules
(iii) Water loss in liquid phase
(iv) Attraction towards polar surfaces
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (ii) (i) (iv) (iii)
(2) (ii) (iv) (i) (iii)
(3) (iv) (iii) (ii) (i)
(4) (iii) (i) (iv) (ii)

Answer (2)

Sol. (a) Cohesion is mutual attraction among water molecules.
(b) Adhesion is attraction towards polar surfaces.
(c) Surface tension explains water molecules are more attracted in liquid phase than gaseous phase.
(d) Guttation is loss of water is liquid form from the leaf margins.

109. Gemmae are present in

(1) Some Liverworts
(2) Mosses
(3) Pteridophytes
(4) Some Gymnosperms

Answer (1)

Sol. Gemmae are green, multicellular asexual buds that are produced by some liverworts like Marchantia. Mosses reproduce vegetatively by fragmentation and budding of protonema. Pteridophytes and Gymnosperms normally do not reproduce asexually

110. A typical angiosperm embryo sac at maturity is:

(1) 8-nucleate and 8-celled
(2) 8-nucleate and 7-celled
(3) 7-nucleate and 8-celled
(4) 7-nucleate and 7-celled

Answer (2)

Sol. A typical angiosperm embryo sac has seven cells that are three antipodals, one central cell, one egg cell and two synergids. The central cell has two polar nuclei, hence the embryo sac is eight nucleated.

111. Inspite of interspecific competition in nature, which mechanism the competing species might have evolved for their survival?

(1) Predation
(2) Resource partitioning
(3) Competitive release
(4) Mutualism

Answer (2)

Sol. Inspite of interspecific competition the competing species may co-exist by doing resource partitioning. In mutualism two organisms are equally benefitted. In predation one organism (Predator) eats the another one (Prey). In competition release there occurs dramatical increase in population of a less distributed species when its superior competitor is removed.

112. During the purification process for recombinant DNA technology, addition of chilled ethanol precipitates out :

(1) Polysaccharides
(2) RNA
(3) DNA
(4) Histones

Answer (3)

Sol. Various enzymes like protease, RNase, etc. are added to break down substances like proteins, RNA, etc. Once all these substances are broken down, DNA is left which is precipitated out by adding chilled ethanol.
Histones are basic proteins that help condense DNA in a cell.

113. DNA strands on a gel stained with ethidium bromide when viewed under UV radiation, appear as

(1) Bright blue bands
(2) Yellow bands
(3) Bright orange bands
(4) Dark red bands

Answer (3)

Sol. After the bands are stained, they are viewed in UV light. The bands appear bright orange in colour. Ethidium bromide is the intercalating agent that stacks in between the nitrogenous bases.

114. Which of the following algae produce Carrageen?

(1) Blue-green algae
(2) Green algae
(3) Brown algae
(4) Red algae

Answer (4)

Sol. The cell wall of red algae is composed of agar, carrageen and funori along with cellulose. In brown algae cell wall contains align while in green algae it is composed of cellulose and pectin. In blue green algae cell wall is composed of mucopeptides.

115. Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called

(1) Maturity
(2) Elasticity
(3) Flexibility
(4) Plasticity

Answer (4)

Sol. Plants show plasticity which means the ability of plant to follow different pathways and produce different structures in response to environment.

116. Which of the following plants is monoecious?

(1) Cycas circinalis
(2) Carica papaya
(3) Chara
(4) Marchantia polymorpha

Answer (3)

Sol. When male and female sex organs are present on same plant body, such plants are said to be monoecious. Most of the species of Chara are monoecious. Cycas circinalis, Carica papaya and Marchantia polymorpha are dioecious.

117. The site of perception of light in plants during photoperiodism is

(1) Leaf
(2) Shoot apex
(3) Stem
(4) Axillary bud

Answer (1)

Sol. The site of perception of light in plants during photoperiodism is leaf. The site of perception of low temperature stimulus during vernalisation is shoot apex and embryo. Axillary bud are not sites of perception of photoperiod.

118. The amount of nutrients, such as carbon, nitrogen, phosphorus and calcium present in the soil at any given time, is referred as :

(1) Standing crop
(2) Climax
(3) Climax community
(4) Standing state

Answer (4)

Sol. Amount of all the inorganic substances or nutrients, such as carbon, nitrogen, phosphorus and calcium present in soil at any given time, is referred as standing state. Amount of living material present in different trophic levels at a given time, is referred as standing crop. Climax community is the last community in biotic succession which is relatively stable and is in near equilibrium with the environment of that area.

119. Match List-I with List-II.
List-I
(a) Lenticels
(b) Cork cambium
(c) Secondary cortex
(d) Cork
List-II
(i) Phellogen
(ii) Suberin deposition
(iii) Exchange of gases
(iv) Phelloderm
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (ii) (i) (iii)
(2) (iv) (i) (iii) (ii)
(3) (iii) (i) (iv) (ii)
(4) (ii) (iii) (iv) (i)

Answer (3)

Sol. Lenticels are meant for exchange of gases.
Phellogen is also known as cork cambium.
Phelloderm is also called secondary cortex because it is the cortex that develops during secondary growth.
Cork has deposition of suberin in their cell walls when they get mature.

120. Which of the following stages of meiosis involves division of centromere?

(1) Telophase II
(2) Metaphase I
(3) Metaphase II
(4) Anaphase II

Answer (4)

Sol. Division of centromere occurs in anaphase II.
Telophase II is the last stage of meiosis II. During this phase, the chromatids reach the poles and start uncoiling. Chromosomes form two parallel plates in metaphase I and one plate in metaphase II.

121. The first stable product of CO₂ fixation in Sorghum is

(1) Phosphoglyceric acid
(2) Pyruvic acid
(3) Oxaloacetic acid
(4) Succinic acid

Answer (3)

Sol. Sorghum is a C₄ plant. The first stable product of CO₂ fixation in Sorghum is oxaloacetic acid.
The first stable product in C₃ cycle is 3-phosphoglyceric acid. Pyruvic acid is the end product of glycolysis. Succinic acid is an intermediate product in krebs cycle.

122. Which of the following statements is not correct?

(1) Pyramid of numbers in a grassland ecosystem is upright.
(2) Pyramid of biomass in sea is generally inverted.
(3) Pyramid of biomass in sea is generally upright.
(4) Pyramid of energy is always upright.

Answer (3)

Sol. Pyramid of biomass in sea is inverted. For example, biomass of zooplankton is higher than that of phytoplankton as life span of former is longer and the latter multiply much faster though having shorter life span.
Small standing crop of phytoplankton supports large standing crop of zooplankton

123. Which of the following algae contains mannitol as reserve food material?

(1) Ulothrix
(2) Ectocarpus
(3) Gracilaria
(4) Volvox

Answer (2)

Sol. Ectocarpus is a brown alga belongs to the class Phaeophyceae. Members of this class have mannitol and laminarin as stored food material.
Ulothrix and Volvox belong to Chlorophyceae (green algae). Members of this class have starch as reserve food material. Gracilaria is a member of red algae (Rhodophyceae). This class is characterised by having floridean starch as stored food material.

124. When gene targeting involving gene amplification is attempted in an individual's tissue to treat disease, it is known as :

(1) Safety testing
(2) Biopiracy
(3) Gene therapy
(4) Molecular diagnosis

Answer (3)

Sol. Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in a child/embryo. Biopiracy is the term used to refer to the use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment. Molecular diagnosis refers to the act or process of determining the nature and cause of a disease.

125. Match List-I with List-II.
List-I
(a) Cristae
(b) Thylakoids
(c) Centromere
(d) Cisternae
List-II
(i) Primary constriction in chromosome
(ii) Disc-shaped sacs in Golgi apparatus
(iii) Infoldings in mitochondria
(iv) Flattened membranous sacs in stroma of plastids
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (ii) (iii) (iv) (i)
(2) (iv) (iii) (ii) (i)
(3) (i) (iv) (iii) (ii)
(4) (iii) (iv) (i) (ii)

Answer (4)

Sol. The inner membrane of mitochondria forms infoldings called cristae.
Thylakoids are flattened membranous sacs in stroma of plastids.
Cisternae are disc shaped sacs in Golgi apparatus.
Primary constriction in chromosome that holds two chromatids together is called centromere.
Hence correct option is (4)- a(iii), b(iv), c(i), d(ii)

126. Match List-I with List-II.
List-I
(a) Protoplast fusion
(b) Plant tissue culture
(c) Meristem culture
(d) Micropropagation
List-II
(i) Totipotency
(ii) Pomato
(iii) Somaclones
(iv) Virus free plants
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (iii) (ii) (i)
(2) (iii) (iv) (ii) (i)
(3) (ii) (i) (iv) (iii)
(4) (iii) (iv) (i) (ii)

Answer (3)

Sol. Pomato is obtained as a result of protoplast fusion.
Totipotency is a property of explant to develop into whole plant body during plant tissue culture.
Virus free plants can be obtained through meristem culture.
Somaclones are obtained by the process of micropropagation.

127. Complete the flow chart on central dogma.
(a) DNA (b) mRNA (c) (d)
(1) (a)-Transduction; (b)-Translation; (c)-Replication; (d)-Protein
(2) (a)-Replication; (b)-Transcription; (c)-Transduction; (d)-Protein
(3) (a)-Translation; (b)-Replication; (c)-Transcription; (d)-Transduction
(4) (a)-Replication; (b)-Transcription; (c)-Translation; (d)-Protein

Answer (4)

Sol. Formation of DNA from DNA is replication.
Formation of mRNA from DNA is called Transcription.
Formation of protein from mRNA is called Translation.
So, (a) is Replication
(b) is Transcription
(c) is Translation
(d) is Protein
Transduction is transfer of genetic material from one bacterium to another with the help of virus or a bacteriophage.

128. In the equation GPP - R = NPP
R represents :
(1) Respiration losses
(2) Radiant energy
(3) Retardation factor
(4) Environmental factor

Answer (1)

Sol. In the equation, GPP - R = NPP
R refers to respiratory loss
GPP is gross primary productivity
NPP is net primary productivity

129. When the centromere is situated in the middle of two equal arms of chromosomes, the chromosome is referred as :

(1) Acrocentric
(2) Metacentric
(3) Telocentric
(4) Sub-metacentric

Answer (2)

Sol. When the centromere is situated in the middle of two equal arms of chromosomes, the chromosome is referred as Metacentric.
When the centromere is present slightly away from the middle, it is called sub-metacentric chromosome.
When the centromere is present very close to one end of the chromosome, it is called acrocentric chromosome.
When the centromere is present at terminal position, the chromosome is called telocentric.

130. The plant hormone used to destroy weeds in a field

(1) IBA
(2) IAA
(3) NAA
(4) 2,4-D

Answer (4)

Sol. Some synthetic auxins are used as weedicides. 2,4-D is widely used to remove broad leaved weeds or dicotyledonous weeds in cereal crops or monocotyledonous plants.
IAA and IBA are natural auxins.
NAA is a synthetic auxin.

131. Which of the following are not secondary metabolites in plants?

(1) Rubber, gums
(2) Morphine, codeine
(3) Amino acids, glucose
(4) Vinblastin, curcumin

Answer (3)

Sol. The correct option is (3)
Amino acids and glucose are included under the category of primary metabolites as they have identifiable functions and play known roles in normal physiological processes.
Rubber, gums, morphine, codeine, vinblastin and curcumin are included under the category of secondary metabolites as their role or functions in host organisms is not known yet. However, many of them are useful to human welfare.

132. Amensalism can be represented as:

(1) Species A (+) Species B 0
(2) Species A (-) Species B 0
(3) Species A (+) Species B (+)
(4) Species A (-) Species B (-)

Answer (2)

Sol. Amensalism is an interaction between two organisms of different species in which one species inhibits the growth of other species by secreting certain chemicals. The first species is neither get benefited nor harmed.
(+):(0) interaction is observed in commensalism
(+):(+) interaction is observed in mutualism.
(-):(-) interaction is seen in competition

133. Which of the following is not an application of PCR (Polymerase Chain Reaction)?

(1) Detection of gene mutation
(2) Molecular diagnosis
(3) Gene amplification
(4) Purification of isolated protein

Answer (4)

Sol. PCR is Polymerase Chain Reaction. It is used for making multiple copies of the gene. Hence PCR is used for Gene amplification. PCR-based assays have been developed that detect the presence of gene sequences of the infectious agents. It is also used in detecting mutations. Protein is not the target of PCR. Hence, plays no role in its purification.

134. Which of the following is a correct sequence of steps in a PCR (Polymerase Chain Reaction)?

(1) Annealing, Denaturation, Extension
(2) Denaturation, Annealing, Extension
(3) Denaturation, Extension, Annealing
(4) Extension, Denaturation, Annealing

Answer (2)

Sol. The first step in the polymerase chain reaction is denaturation during which strands of dsDNA separate. This requires temperature around 94°C.
This is followed by annealing in which primers anneal to 3' end of template DNA strand.
Annealing is followed by extension in which Taq polymerase adds nucleotides to 3'OH end of primers.

135. Match List-I with List-II.
List-I
(a) Cells with active cell division capacity
(b) Tissue having all cells similar in structure and function
(c) Tissue having different types of cells
(d) Dead cells with highly thickened walls and narrow lumen
List-II
(i) Vascular tissues
(ii) Meristematic tissue
(iii) Sclereids
(iv) Simple tissue
Select the correct answer from the options given below.
(a) (b) (c) (d)
(1) (ii) (iv) (i)
(2) (ii) (iv) (i) (iii)
(3) (iv) (iii) (ii) (i)
(4) (i) (ii) (iii) (iv)

Answer (2)

Sol. (a) Meristematic tissues are those tissues which have cells with active cell division capacity.
(b) Simple tissues are those tissues which have all the cells similar in structure and function.
(c) Vascular tissues are complex permanent tissues hence they have different types of cells.
(d) Sclereids are sclerenchymatous cells which are dead with highly thickened walls and narrow lumen.

SECTION-B

136. Which of the following statements is correct?

(1) Some of the organisms can fix atmospheric nitrogen in specialized cells called sheath cells
(2) Fusion of two cells is called Karyogamy
(3) Fusion of protoplasm between two motile on non-motile gametes is called plasmogamy
(4) Organisms that depend on living plants are called saprophytes

Answer (3)

Sol. In some blue-green algae specialised cells called heterocyst fixes atmospheric nitrogen into ammonia.
Fusion of two nuclei is called Karyogamy. Organisms that depend on living plants are parasites, saprophytes grow on dead material. Fusion of protoplasts of two cells is called plasmogamy.

137. In the exponential growth equation N_t = N₀e^(rt), e represents

(1) The base of geometric logarithms
(2) The base of number logarithms
(3) The base of exponential logarithms
(4) The base of natural logarithms

Answer (4)

Sol. In the exponential growth equation N_t = N₀e^(rt), e represents the base of natural logarithms
N_t = Population density after time t
N₀ = Population density at time zero
r = Intrinsic rate of natural increase called biotic potential.

138. In some members of which of the following pairs of families, pollen grains retain their viability for months after release?

(1) Rosaceae; Leguminosae
(2) Poaceae; Rosaceae
(3) Poaceae; Leguminosae
(4) Poaceae; Solanaceae

Answer (1)

Sol. In members of some plant families like Solanaceae, Rosaceae and Leguminosae the pollen grains retain their viability for several months. In cereals (Poaceae) pollen grains retain viability for around 30 minutes.

139. Match Column-I with Column-II.
Column-I | Column-II
(a) Nitrococcus | (i) Denitrification
(b) Rhizobium | (ii) Conversion of ammonia to nitrite
(c) Thiobacillus | (iii) Conversion of nitrite to nitrate
(d) Nitrobacter | (iv) Conversion of atmospheric nitrogen to ammonia

Choose the correct answer from options given below.
(a) (b) (c) (d)
(1) (iv) (iii) (ii) (i)
(2) (ii) (iv) (i) (iii)
(3) (i) (ii) (iii) (iv)
(4) (iii) (i) (iv) (ii)

Answer (2)

Sol. Nitrogen fixation is conversion of atmospheric N₂ to NH₃ (ammonia). It is carried out by N₂ fixers such as Rhizobium.
NH₃ is converted to NO₂ (nitrite) by nitrifying bacteria such as Nitrococcus.
Then NO₂ is converted to NO₃ (nitrate) by nitrifying bacteria called Nitrobacter.
Thiobacillus carries out denitrification, a process where NO₂⁻/NO₃⁻ is converted to N₂

140. Match List-I with List-II.
List-I
(a) S phase
(b) G2 phase
(c) Quiescent stage
(d) G1 phase
List-II
(i) Proteins are synthesized
(ii) Inactive phase
(iii) Interval between mitosis and initiation of DNA replication
(iv) DNA replication
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (ii) (iv) (iii) (i)
(2) (iii) (ii) (i) (iv)
(3) (iv) (iii) (ii)
(4) (iv) (i) (ii) (iii)

Answer (4)

Sol. In S phase DNA replication takes place.
In G₂ phase there is synthesis of proteins, RNA etc.
Quiescent stage is inactive stage of cell cycle but cells remain metabolically active in this stage.
G₁ phase is the interval between mitosis and initiation of DNA replication.

141. Match List-I with List-II.
List-I
(a) Protein
(b) Unsaturated fatty acid
(c) Nucleic acid
(d) Polysaccharide
List-II
(i) C = C double bonds
(ii) Phosphodiester bonds
(iii) Glycosidic bonds
(iv) Peptide bonds
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (iii) (i) (ii)
(2) (iv) (i) (ii) (iii)
(3) (i) (iv) (iii) (ii)
(4) (ii) (i) (iv) (iii)

Answer (2)

Sol. In a polypeptide or a protein, amino acids are linked by a peptide bond which is formed when the carboxyl (-COOH) group of one amino acid reacts with amino (-NH₂) group of the next amino acid with the elimination of a water moiety.
Unsaturated fatty acids are with one or more C=C double bonds.
In nucleic acids, a phosphate moiety links the 3'-carbon of one sugar of one nucleotide to the 5'-carbon of the sugar of the succeeding nucleotide. The bond between the phosphate and hydroxyl group is an ester bond. As there is one such ester bond on either side, it is called phosphodiester bond.
In a polysaccharide, the individual monosaccharides are linked by a glycosidic bond.

142. Match Column-I with Column-II
Column-I | Column-II
(a) % C₅ C₁₊₂₊(₂) A₍₃₎ G₁ | (i) Brassicaceae
(b) C₅ C₍₅₎ C₍₅₎ C₍₂₎ | (ii) Liliaceae
(c) C₍₃₊₃₎ A₍₃₊₃₎ G₍₃₎ | (iii) Fabaceae
(d) C₂₊₂ C₄ A₂₋₄ G₍₂₎ | (iv) Solanaceae
Select the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (ii) (i) (iii)
(2) (iii) (iv) (ii) (i)
(3) (i) (ii) (iii) (iv)
(4) (ii) (iii) (iv) (i)

Answer (2)

Sol. The floral formula of
Brassicaceae family - % K₂₊₂ C₄ A₂₊₄ G₍₂₎
Solanaceae family - ⊕ K₍₅₎ C₍₅₎ A₅ G₍₂₎
Fabaceae family - % K₍₅₎ C₁₊₂₊(₂) A₍₉₎₊₁ G₁
Liliaceae family - ⊕ P₍₃₊₃₎ A₃₊₃ G₍₃₎
So a(ii), b(iv), c(ii), d(i) is correct matching.

143. What is the role of RNA polymerase III in the process of transcription in eukaryotes?

(1) Transcribes only snRNAs
(2) Transcribes rRNAs (28S, 18S and 5.8S)
(3) Transcribes tRNA, 5s rRNA and snRNA
(4) Transcribes precursor of mRNA

Answer (3)

Sol. RNA polymerase III transcribes tRNA, ScRNA, 5S rRNA and SnRNA. RNA polymerase I transcribes 5.8S, 18S and 28S rRNA. RNA polymerase II transcribes hnRNA which is precursor of mRNA

144. Which of the following statements is incorrect?

(1) Cyclic photophosphorylation involves both PS I and PS II
(2) Both ATP and NADPH + H⁺ are synthesized during non-cyclic photophosphorylation
(3) Stroma lamellae have PS I only and lack NADP reductase
(4) Grana lamellae have both PS I and PS II

Answer (1)

Sol. Cyclic photophosphorylation involves only PS I. Both PS I and PS II are involved in non-cyclic photophosphorylation where both ATP and NADPH + H⁺ are synthesized. Both PS I and PS II are found on grana lamellae whereas stroma lamellae have PS I only and lack NADP reductase.

145. Now a days it is possible to detect the mutated gene causing cancer by allowing radioactive probe to hybridise its complimentary DNA in a clone of cells, followed by its detection using autoradiography because :

(1) mutated gene does not appear on photographic film as the probe has complementarity with it
(2) mutated gene partially appears on a photographic film
(3) mutated gene completely and clearly appears on a photographic film
(4) mutated gene does not appear on a photographic film as the probe has no complementarity with it

Answer (4)

Sol. Autoradiography allows the detection/localisation of radioactive isotope within a biological sample.
Probe is a radiolabelled ss DNA or ss RNA depending on the technique. To identify the mutated gene probe is allowed to hybridise to its complementary DNA in a clone of cells followed by detection using autoradiography. The mutated gene will not appear on the photographic film, because the probe does not have complementarity with the mutated gene.

146. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence, called as

(1) Polymorphic DNA
(2) Satellite DNA
(3) Repetitive DNA
(4) Single nucleotides

Answer (3)

Sol. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA. The basis of DNA fingerprinting is VNTR a satellite DNA as probe that show very high degree of polymorphism. Polymorphism is the variation at genetic level. Allelic sequence variation has traditionally been described as a DNA polymorphism.

147. Which of the following statements is incorrect?

(1) Oxidation-reduction reactions produce proton gradient in respiration
(2) During aerobic respiration, role of oxygen is limited to the terminal stage
(3) In ETC (Electron Transport Chain), one molecule of NADH + H⁺ gives rise to 2 ATP molecules, and one FADH₂ gives rise to 3 ATP molecules
(4) ATP is synthesized through complex V

Answer (3)

Sol. During respiration, process of ATP synthesis is explained by chemiosmotic model. It says that a proton gradient is required for ATP synthesis that is established by oxidation-reduction reactions. In ETC, one NADH + H⁺ produces 3 ATP while one FADH₂ produces 2 ATP molecules. ATP is synthesised via complex V. In ETS, oxygen acts as terminal electron acceptor.

148. Plasmid pBR322 has PstI restriction enzyme site within gene ampR that confers ampicillin resistance. If this enzyme is used for inserting a gene for β-galactoside production and the recombinant plasmid is inserted in an E.coli strain

(1) It will be able to produce a novel protein with dual ability
(2) It will not be able to confer ampicillin resistance to the host cell
(3) The transformed cells will have the ability to resist ampicillin as well as produce β-galactoside
(4) It will lead to lysis of host cell

Answer (2)

Sol. pBR322 is a commonly used cloning vector. When the gene for β-galactoside is inserted in the ampicillin resistance gene by using Pst I, the recombinant E.coli will lose ampicillin resistance due to insertional inactivation of the antibiotic resistance gene.
The host (recombinant) cell will produce β-galactoside which is not a novel protein nor does it have dual ability.
The transformed cells cannot resist ampicillin as they have lost ampicillin resistance.
A recombinant E. coli is produced and the host cell will not undergo lysis due to insertion of β-galactoside gene.

149. Identify the correct statement.

(1) Split gene arrangement is characteristic of prokaryotes
(2) In capping, methyl guanosine triphosphate is added to the 3' end of hnRNA
(3) RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria
(4) The coding strand in a transcription unit is copied to an mRNA

Answer (3)

Sol. Split gene arrangement is characteristic of eukaryotes.
In capping 5-methyl guanosine triphosphate is added at 5' end of hnRNA.
At 3' end poly-A tail is added.
The non coding or template strand is copied to an mRNA. RNA polymerase associate with ρ factor (Rho factor) and it alters the specificity of the RNA polymerase to terminate the processes.

150. Select the correct pair.

(1) Loose parenchyma cells - Spongy parenchyma rupturing the epidermis and forming a lens shaped opening in bark
(2) Large colorless empty - Subsidiary cells cells in the epidermis of grass leaves
(3) In dicot leaves, vascular - Conjunctive bundles are surrounded tissue by large thick-walled cells
(4) Cells of medullary rays - Interfascicular that form part of cambium ring

Answer (4)

Sol. When the cells of medullary rays differentiated, they give rise to the new cambium called interfascicular cambium.
Loose parenchyma cells rupturing the epidermis and forming a lens-shaped opening in bark are called complementary cells.
Large colourless empty cells in the epidermis of grass leaves are called bulliform cells.
In dicot leave, vascular bundles are surrounded by large thick walled cells called bundle sheath cells.

151. Match List-I with List-II
List-I
(a) Aspergillus niger
(b) Acetobacter aceti
(c) Clostridium butylicum
(d) Lactobacillus
List-II
(i) Acetic Acid
(ii) Lactic Acid
(iii) Citric Acid
(iv) Butyric Acid
Choose the correct answer from the options given below
(a) (b) (c) (d)
(1) (iv) (ii) (i) (iii)
(2) (iii) (i) (iv) (ii)
(3) (i) (ii) (iii) (iv)
(4) (ii) (iii) (i) (iv)

Answer (2)

Sol. Aspergillus niger is involved in production of citric acid.
Acetobacter aceti is involved in production of acetic acid.
Clostridium butylicum is involved in production of butyric acid whereas Lactobacillus is involved in the production of lactic acid.
So a(iii), b(i), c(iv), d(ii) is correct matching.

152. Succus entericus is referred to as:

(1) Chyme
(2) Pancreatic juice
(3) Intestinal juice
(4) Gastric juice

Answer (3)

Sol. Option (3) is correct because succus entericus is referred to as intestinal juice. Chyme is name given to acidic food present in stomach. Exocrine secretion of pancreatic acini is called pancreatic juice. Secretion of gastric glands present in stomach is called gastric juice.

153. Receptors for sperm binding in mammals are present on :

(1) Zona pellucida
(2) Corona radiata
(3) Vitelline membrane
(4) Perivitelline space

Answer (1)

Sol. Option (1) is correct because zona pellucida has receptors for sperm binding (ZP3 receptors) in mammals. Corona radiata is a layer of radially arranged cells of membrana granulosa. Perivitelline space is present in between vitelline membrane and zona pellucida.

154. The fruit fly has 8 chromosomes (2n) in each cell. During interphase of Mitosis if the number of chromosomes at G₁ phase is 8, what would be the number of chromosomes after S phase?

(1) 32
(2) 8
(3) 16
(4) 4

Answer (2)

Sol. In S phase there is duplication of DNA. So amount of DNA increases but not the chromosome number. So, if the number of chromosomes at G₁ phase is 8 in fruit fly then the number of chromosomes will be same in S phase that is 8 only.

155. Select the favourable conditions required for the formation of oxyhaemoglobin at the alveoli.

(1) Low pO₂, low pCO₂, more H⁺, higher temperature
(2) High pO₂, low pCO₂, less H⁺, lower temperature
(3) Low pO₂, high pCO₂, more H⁺, higher temperature
(4) High pO₂, high pCO₂, less H⁺, higher temperature

Answer (2)

Sol. The factors favourable for the formation of oxyhaemoglobin at the alveolar level are; high pO₂, low pCO₂, less H⁺ concentration and lower temperature. The conditions favourable for the dissociation of oxygen from oxyhaemoglobin at the tissue level are; low pO₂, high pCO₂, high H⁺ concentration and high temperature.

156. Match the following:
List-I
(a) Physalia
(b) Limulus
(c) Ancylostoma
(d) Pinctada
List-II
(i) Pearl oyster
(ii) Portuguese Man of War
(iii) Living fossil
(iv) Hookworm
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (i) (iv) (iii) (ii)
(2) (ii) (iii) (i) (iv)
(3) (iv) (i) (iii) (ii)
(4) (ii) (iii) (iv) (i)

Answer (4)

Sol. Option 4 is correct because Physalia is commonly known as Portuguese man of war. Limulus is considered as a living fossil and commonly known as king crab. Ancylostoma is a roundworm and commonly known as hookworm. Pinctada is commonly known as pearl oyster, included in phylum Mollusca.

157. Which stage of meiotic prophase shows terminalisation of chiasmata as its distinctive feature?

(1) Pachytene
(2) Leptotene
(3) Zygotene
(4) Diakinesis

Answer (4)

Sol. In meiosis I, chiasmata (X shaped structure) is formed in diplotene stage while it terminalise in diakinesis stage. Bivalents are formed in zygotene stage and crossing over takes place in pachytene stage. Compaction of chromosomal material occurs in leptotene stage.

158. If Adenine makes 30% of the DNA molecule, what will be the percentage of Thymine, Guanine and Cytosine in it?

(1) T:20; G:25; C:25
(2) T:20; G:30; C:20
(3) T:20; G:20; C:30
(4) T:30; G:20; C:20

Answer (4)

Sol. According to Chargaff's rule, for a double stranded DNA,
[A] = [T],
∵ [A] = 30%, ⇒ [T] = 30%
∴ Since [C] = [G]
∴ 100 - [A + T]
= 100 - [30 + 30]
= 100 - 60 = 40%

159. The partial pressures (in mm Hg) of oxygen (O₂) and carbon dioxide (CO₂) at alveoli (the site of diffusion) are:

(1) pO₂ = 159 and pCO₂ = 0.3
(2) pO₂ = 104 and pCO₂ = 40
(3) pO₂ = 40 and pCO₂ = 45
(4) pO₂ = 95 and pCO₂ = 40

Answer (2)

Sol. Option (2) is correct because pO₂ in alveoli is 104 mm Hg and pCO₂ in alveoli is 40 mm Hg. In atmosphere, pO₂ is 159 mm Hg and pCO₂ is 0.3 mm Hg. In deoxygenated blood, pO₂ is 40 mm Hg and pCO₂ is 45 mm Hg. In oxygenated blood, pO₂ is 95 mm Hg and pCO₂ is 40 mm Hg.

160. Read the following statements
(a) Metagenesis is observed in Helminthes.
(b) Echinoderms are triploblastic and coelomate animals.
(c) Round worms have organ-system level of body organization.
(d) Comb plates present in ctenophores help in digestion.
(e) Water vascular system is characteristic of Echinoderms.
Choose the correct answer from the options given below.

(1) (b), (c) and (e) are correct
(2) (c), (d) and (e) are correct
(3) (a), (b) and (c) are correct
(4) (a), (d) and (e) are correct

Answer (1)

Sol. Metagenesis (alternation of generation) is observed in members of phylum Coelenterata (Cnidaria). Echinoderms are triploblastic and coelomate animals as true coelom is observed in them. Roundworms (Aschelminthes) have organ system level of organization. Comb plates present in ctenophores help in locomotion. Water vascular system is seen in echinoderms, which helps in locomotion, capture and transport of food and respiration.

161. In a cross between a male and female, both heterozygous for sickle cell anaemia gene, what percentage of the progeny will be diseased?

(1) 100%
(2) 50%
(3) 75%
(4) 25%

Answer (4)

Sol. According to given question;
HbA HbS × HbA HbS (Parents)
HbS HbS, HbA HbS, HbA HbS, HbA HbA (Progenies)
Total number of affected progenies = 1
Percentage of diseased/affected progenies
= 1/4 × 100 = 25%

162. Which of the following statements wrongly represents the nature of smooth muscle?

(1) These muscles are present in the wall of blood vessels
(2) These muscle have no striations
(3) They are involuntary muscles
(4) Communication among the cells is performed by intercalated discs

Answer (4)

Sol. Option 4 is incorrect because intercalated discs are found only in cardiac muscle tissue. Smooth muscle fibres are non-striated and involuntary in nature and are present in the wall of blood vessels, uterus, gall bladder, alimentary canal etc.

163. Which one of the following belongs to the family Muscidae?

(1) House fly
(2) Fire fly
(3) Grasshopper
(4) Cockroach

Answer (1)

Sol. Option 1 is correct because housefly belongs to the family Muscidae, class Insecta and phylum Arthropoda. Fire flies are placed in family Lampyridae of class insecta. Grasshopper is also an insect placed in family Acrididae. Cockroach is also an insect placed in family Blattidae.

164. During the process of gene amplification using PCR, if very high temperature is not maintained in the beginning, then which of the following steps of PCR will be affected first?

(1) Ligation
(2) Annealing
(3) Extension
(4) Denaturation

Answer (4)

Sol. Option 4 is correct. High temperature about 94°C is required for the process of denaturation which is the first step of PCR. Ligation of DNA fragments is performed with the help of an enzyme called DNA ligase. Annealing is performed at 50°-60°C which is the second step that can get affected. Addition of nucleotides to the primer, synthesizing a new DNA strand using only the template sequences with the help of enzyme DNA polymerase is called primer extension/polymerisation.

165. Match List-I with List-II
List-I
(a) Metamerism
(b) Canal system
(c) Comb plates
(d) Cnidoblasts
List-II
(i) Coelenterata
(ii) Ctenophora
(iii) Annelida
(iv) Porifera
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (i) (ii) (iii)
(2) (iv) (iii) (i) (ii)
(3) (iii) (iv) (i) (ii)
(4) (iii) (iv) (ii) (i)

Answer (4)

Sol. Metamerism is commonly seen in the members of phylum Annelida where the body is externally and internally divided into segments with a serial repetition of atleast some organs.
Water canal system is present in the members of phylum Porifera.
The body of ctenophores bears 8 external rows of ciliated comb plates which help in locomotion.
Cnidoblasts or cnidocytes are characteristic feature of cnidarians (coelentrata).

166. Which one of the following is an example of Hormone releasing IUD?

(1) Multiload 375
(2) CuT
(3) LNG 20
(4) Cu 7

Answer (3)

Sol. LNG-20 is a hormone releasing IUD which makes the uterus unsuitable for implantation and the cervix hostile to sperms. Multiload 375, CuT and Cu7 are copper releasing IUDs which suppress sperm motility and the fertilizing capacity of sperms.

167. The centriole undergoes duplication during:

(1) G₂ phase
(2) S-phase
(3) Prophase
(4) Metaphase

Answer (2)

Sol. During S phase of cell cycle replication of DNA takes place. In animal cells during S phase, centriole duplicates in the cytoplasm.
In G₂ phase there is duplication of mitochondria, chloroplast and Golgi bodies. Tubulin protein is also synthesized during this phase.
During prophase, condensation of chromatin starts.
During metaphase, chromosomes get aligned at equator to form metaphasic plate.

168. Which of the following characteristics is incorrect with respect to cockroach?

(1) 10th abdominal segment in both sexes, bears a pair of anal cerci
(2) A ring of gastric caeca is present at the junction of midgut and hind gut
(3) Hypopharynx lies within the cavity enclosed by the mouth parts
(4) In females, 7th-9th sterna together form a genital pouch

Answer (2)

Sol. Option (2) is incorrect because a ring of gastric caeca is present at the junction of foregut and midgut. At the junction of midgut and hindgut, malpighian tubules are present. Hypopharynx lies within the cavity enclosed by mouthparts. In female cockroach, the 7th sternum is boat shaped and together with the 8th and 9th sterna forms a genital pouch. 10th abdominal segment in both sexes, bears a pair of anal cerci and 9th sternum only in male cockroach, bears a pair of chitinous anal style.

169. Dobson units are used to measure thickness of:

(1) Troposphere
(2) CFCs
(3) Stratosphere
(4) Ozone

Answer (4)

Sol. The thickness of the ozone in a column of air from the ground to the top of atmosphere is measured in term of Dobson unit (1 DU = 1 ppb)
The lowermost layer of atmosphere is called troposphere.
CFCs are ozone depleting substances. Ozone found in upper part of atmosphere (the stratosphere) is called good ozone.

170. Venereal diseases can spread through :
(a) Using sterile needles
(b) Transfusion of blood from infected person
(c) Infected mother to foetus
(d) Kissing
(e) Inheritance
Choose the correct answer from the option given below
(1) (a) and (c) only
(2) (a), (b) and (c) only
(3) (b), (c) and (d) only
(4) (b) and (c) only

Answer (4)

Sol. Venereal diseases or sexually transmitted diseases or infections are transmitted by sharing of infected needles, surgical instruments with infected person, transfusion of blood or from an infected mother to foetus. Venereal diseases are not transmitted through kissing or inheritance.

171. Which one of the following organisms bears hollow and pneumatic long bones?

(1) Ornithorhynchus
(2) Neophron
(3) Hemidactylus
(4) Macropus

Answer (2)

Sol. Hollow and pneumatic long bones are present in animals that belong to class Aves e.g., Neophron (vulture). Ornithorhynchus (Platypus) and Macropus (Kangaroo) belong to class Mammalia. Hemidactylus (Wall lizard) is a member of class Reptilia.

172. Persons with 'AB' blood group are called as "Universal recipients". This is due to :

(1) Absence of antibodies, anti-A and anti-B, in plasma
(2) Absence of antigens A and B on the surface of RBCs
(3) Absence of antigens A and B in plasma
(4) Presence of antibodies, anti-A and anti-B, on RBCs

Answer (1)

Sol. Option (1) is correct because persons with 'AB' blood group contain antigens 'A' and 'B' but lack antibodies anti-A and anti-B in plasma. So, persons with 'AB' blood group can accept blood from persons with AB as well as the other groups of blood due to lack of antibodies in their blood. Therefore, such persons are called "Universal recipients".

173. The organelles that are included in the endomembrane system are

(1) Golgi complex, Endoplasmic reticulum, Mitochondria and Lysosomes
(2) Endoplasmic reticulum, Mitochondria, Ribosomes and Lysosomes
(3) Endoplasmic reticulum, Golgi complex, Lysosomes and Vacuoles
(4) Golgi complex, Mitochondria, Ribosomes and Lysosomes

Answer (3)

Sol. Endomembrane system consist of endoplasmic reticulum, Golgi complex, vacuoles and lysosomes. Mitochondria is semi-autonomous cell organelle. Ribosome is non-membranous cell organelle.

174. A specific recognition sequence identified by endonucleases to make cuts at specific positions within the DNA is:

(1) Poly(A) tail sequences
(2) Degenerate primer sequence
(3) Okazaki sequences
(4) Palindromic Nucleotide sequences

Answer (4)

Sol. Each restriction endonuclease recognizes a specific palindromic nucleotide sequence in the DNA. Once it finds its specific recognition sequence it bind to DNA and cuts each of the two strands of DNA. During post transcriptional modification in eukaryotes, poly(A) tail (200-300 adenylate residues) are added at 3' end of hnRNA. During DNA replication Okazaki fragments are synthesized discontinuously and joined by DNA ligase. A PCR primer sequence is termed degenerate if some of its position have several possible bases.

175. Identify the incorrect pair

(1) Drugs - Ricin
(2) Alkaloids - Codeine
(3) Toxin - Abrin
(4) Lectins - Concanavalin A

Answer (1)

Sol. Option (1) is incorrect because ricin is a toxin obtained from Ricinus plant. Vinblastin and curcumin are drugs. Morphine and codeine are alkaloids. Abrin is also a toxin obtained by plant Abrus. Concanavalin A is a lectin.

176. With regard to insulin choose correct options.
(a) C-peptide is not present in mature insulin.
(b) The insulin produced by rDNA technology has C-peptide.
(c) The pro-insulin has C-peptide
(d) A-peptide and B-peptide of insulin are interconnected by disulphide bridges.
Choose the correct answer from the options given below
(1) (a) and (d) only
(2) (b) and (c) only
(3) (b) and (d) only
(4) (a), (c) and (d) only

Answer (4)

Sol. Insulin is synthesized as a pro-hormone which contains A-chain, B-chain and an extra stretch called the C-peptide. C-peptide is not present in mature insulin called humulin. Chains A and B are connected by interchain disulphide bridges.

177. Chronic auto immune disorder affecting neuro muscular junction leading to fatigue, weakening and paralysis of skeletal muscle is called as:

(1) Gout
(2) Arthritis
(3) Muscular dystrophy
(4) Myasthenia gravis

Answer (4)

Sol. Option (4) is correct because myasthenia gravis is a chronic auto immune disorder affecting neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscle. Gout is caused due to deposition of uric acid crystals in joints leading to its inflammation. Inflammation of joints is commonly known as arthritis. Muscular dystrophy is a genetic disorder which results in progressive degeneration of skeletal muscle.

178. Which of the following RNAs is not required for the synthesis of protein?

(1) siRNA
(2) mRNA
(3) tRNA
(4) rRNA

Answer (1)

Sol. siRNA are small interfering RNA also called silencing RNA. It is a class of double-stranded RNA, non-coding RNA molecules. mRNA is messenger RNA that carries genetic information provided by DNA. tRNA carries amino acids to the mRNA during translation. rRNA is structural RNA that forms ribosomes which are involved in translation.

179. Which of the following is not an objective of Biofortification in crops?

(1) Improve micronutrient and mineral content
(2) Improve protein content
(3) Improve resistance to diseases
(4) Improve vitamin content

Answer (3)

Sol. Biofortification improves vitamin content, protein content and micronutrient and mineral content. It does not create resistance in plants against diseases.

180. Erythropoietin hormone which stimulates R.B.C. formation is produced by:

(1) Juxtaglomerular cells of the kidney
(2) Alpha cells of pancreas
(3) The cells of rostral adenohypophysis
(4) The cells of bone marrow

Answer (1)

Sol. Option (1) is correct because Juxtaglomerular cells of kidney secrete erythropoietin hormone which stimulates RBC formation. Alpha cells of pancreas produce hormone glucagon. The cells of rostral adenohypophysis synthesizes hormones of anterior lobe of pituitary. The cells of bone marrow are responsible for formation of formed elements.

181. Which enzyme is responsible for the conversion of inactive fibrinogens to fibrins?

(1) Thromboxinase
(2) Thrombin
(3) Renin
(4) Epinephrine

Answer (2)

Sol. During coagulation of blood, an enzyme complex thrombokinase helps in the conversion of prothrombin (present in plasma) into thrombin.
Thrombin further helps in the conversion of inactive fibrinogens into fibrins which form network of threads.
Renin is secreted by JG cells in response to fall in glomerular blood flow, which converts angiotensinogen in blood to angiotensin-I
Epinephrine or adrenaline is secreted by adrenal medulla in response to stress of any kind and during emergency.

182. For effective treatment of the disease, early diagnosis and understanding its pathophysiology is very important. Which of the following molecular diagnostic techniques is very useful for early detection?

(1) Hybridization Technique
(2) Western Blotting Technique
(3) Southern Blotting Technique
(4) ELISA Technique

Answer (3/4*)

Sol. ELISA can be used for early detection of an infection either by detecting the presence of pathogenic antigen or by detecting the antibodies synthesized against the pathogen. Option (3) Southern blotting is used to detect a specific DNA sequence in the given sample and can be detected prior to antibody formation. One can detect presence of pathogenic DNA/RNA. In hybridization technique a ssDNA/ssRNA tagged with a radioactive molecule (probe) is allowed to hybridize its complementary DNA in a clone of cells followed by detection using autoradiography. It is used to find a mutated gene. Western blotting technique is used to detect a specific protein molecule among a mixture of proteins.

183. Sphincter of oddi is present at:

(1) Junction of jejunum and duodenum
(2) Ileo-caecal junction
(3) Junction of hepato-pancreatic duct and duodenum
(4) Gastro-oesophageal junction

Answer (3)

Sol. The bile duct and the pancreatic duct open together into the duodenum as the common hepato-pancreatic duct which is guarded by a sphincter called the sphincter of Oddi. Ileo-caecal valve is present at the junction of ileum and caecum to prevent the backflow of faecal matter into the ileum in humans. Gastro-oesophageal sphincter regulates the opening of oesophagus into stomach.

184. Which is the "Only enzyme" that has "Capability" to catalyse Initiation, Elongation and Termination in the process of transcription in prokaryotes?

(1) DNase
(2) DNA dependent DNA polymerase
(3) DNA dependent RNA polymerase
(4) DNA Ligase

Answer (3)

Sol. In prokaryotes, the DNA dependent RNA polymerase is a holoenzyme that is made of polypeptides (α₂ββ'ω)σ. It is responsible for initiation, elongation and termination during transcription. DNase degrades DNA. DNA dependent DNA polymerase is involved in replication of DNA. DNA ligase joins the discontinuously synthesised fragments of DNA.

185. Match List-I with List-II.
List-I
(a) Vaults
(b) IUDs
(c) Vasectomy
(d) Tubectomy
List-II
(i) Entry of sperm through Cervix is blocked
(ii) Removal of Vas deferens
(iii) Phagocytosis of sperms within the Uterus
(iv) Removal of fallopian tube
Choose the correct answer from the option given below
(a) (b) (c) (d)
(1) (iii) (i) (iv) (ii)
(2) (iv) (ii) (i) (iii)
(3) (i) (iii) (ii) (iv)
(4) (ii) (iv) (iii) (i)

Answer (3)

Sol. Diaphragms, cervical caps and vaults are barrier methods of contraception for female which works by blocking the entry of sperms through the cervix.
IUDs increase phagocytosis of sperms within the uterus.
Vasectomy is a surgical method of contraception in males in which a small part of the vas deferens is removed or tied up through a small incision on the scrotum.
Tubectomy is a surgical method of contraception in females where a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through vagina.

SECTION-B

186. Match List-I with List-II
List-I
(a) Scapula
(b) Cranium
(c) Sternum
(d) Vertebral column
List-II
(i) Cartilaginous joints
(ii) Flat bone
(iii) Fibrous joints
(iv) Triangular flat bone
Choose the correct answer from the options given below
(a) (b) (c) (d)
(1) (iv) (iii) (ii) (i)
(2) (i) (iii) (ii) (iv)
(3) (ii) (iii) (iv) (i)
(4) (iv) (ii) (iii) (i)

Answer (1)

Sol. The correct option is (1).
Scapula is a large triangular flat bone situated in the dorsal part of the thorax between the second and the seventh ribs. Fibrous joint is shown by the flat skull bones which fuse end-to-end with the help of dense fibrous connective tissues in the form of sutures, to form the cranium. Sternum is a flat bone on the ventral midline of thorax. Cartilaginous joints between the adjacent vertebrae in the vertebral column permits limited movements.

187. Following are the statements with reference to 'lipids'.
(a) Lipids having only single bonds are called unsaturated fatty acids
(b) Lecithin is a phospholipid.
(c) Trihydroxy propane is glycerol.
(d) Palmitic acid has 20 carbon atoms including carboxyl carbon.
(e) Arachidonic acid has 16 carbon atoms.
Choose the correct answer from the options given below.
(1) (b) and (e) only
(2) (a) and (b) only
(3) (c) and (d) only
(4) (b) and (c) only

Answer (4)

Sol. The correct option is (4) because lipids having only single bonds are called saturated fatty acids and lipids having one or more C=C double bonds are called unsaturated fatty acids. Palmitic acid has 16 carbon atoms including carboxyl carbon. Arachidonic acid has 20 carbon atoms including the carboxyl carbon. Lecithin is a phospholipid found in cell membrane. Glycerol has 3 carbons, each bearing a hydroxyl (-OH) group.

188. Match List-I with List-II
List-I
(a) Allen's Rule
(b) Physiological adaptation
(c) Behavioural adaptation
(d) Biochemical adaptation
List-II
(i) Kangaroo rat
(ii) Desert lizard
(iii) Marine fish at depth
(iv) Polar seal
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (iii) (ii) (i)
(2) (iv) (ii) (iii) (i)
(3) (iv) (i) (iii) (ii)
(4) (iv) (i) (ii) (iii)

Answer (4)

Sol. Polar seal generally has shorter ears and limbs (extremities) to minimise heat loss. This is with reference to Allen's rule. Kangaroo rat exhibits physiological adaptation. Desert lizard shows behavioural adaptation. They lack the physiological ability to cope-up with extreme temperature but manage the body temperature by behavioural means. Marine fishes at depth are adapted biochemically to survive in great depths in ocean.

189. During muscular contraction which of the following events occur?
(a) 'H' zone disappears
(b) 'A' band widens
(c) 'I' band reduces in width
(d) Myosine hydrolyzes ATP, releasing the ADP and Pi.
(e) Z-lines attached to actins are pulled inwards.
Choose the correct answer from the options given below:
(1) (b), (d), (e), (a) only
(2) (a), (c), (d), (e) only
(3) (a), (b), (c), (d) only
(4) (b), (c), (d), (e) only

Answer (2)

Sol. The correct option is (2) because the length of A-band is retained. During muscle contraction, the following events occur:
(1) The globular head of myosin acts as ATPase and hydrolyses ATP molecule and eventually leads to the formation of cross bridge.
(2) This pulls the actin filament towards the centre of 'A-band'.
(3) The Z-line attached to these actins are also pulled inwards thereby causing a shortening of the sarcomere.
(4) The thin myofilaments move past the thick myofilaments due to which the H-zone narrows. This reduces the length of I-band but retains the length of A-band.
(5) The myosin then releases ADP+Pi, and goes back to its relaxed state.

190. Match List-I with List-II
List-I
(a) Adaptive radiation
(b) Convergent evolution
(c) Divergent evolution
(d) Evolution by anthropogenic action
List-II
(i) Selection of resistant varieties due to excessive use of herbicides and pesticides
(ii) Bones of forelimbs in Man and Whale
(iii) Wings of Butterfly and Bird
(iv) Darwin Finches
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (i) (iv) (iii) (ii)
(2) (iv) (iii) (ii) (i)
(3) (iii) (ii) (i) (iv)
(4) (ii) (i) (iv) (iii)

Answer (2)

Sol. The correct option is (2)
Adaptive radiation is the process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography, for example : Darwin's finches.
Analogous organs which are not anatomically similar structures though they perform similar functions, are a result of convergent evolution, for example: Wings of butterfly and of birds.
Homologous organs which are anatomically similar structures but perform different functions according to their needs, are a result of divergent evolution, for example: Bones of forelimbs in man and whale.
Evolution by anthropogenic action means evolution due to human interference, for example: Antibiotic resistant microbes, herbicides resistant varieties and pesticide resistant varieties.

191. Assertion (A): A person goes to high altitude and experiences 'altitude sickness' with symptoms like breathing difficulty and heart palpitations.
Reason (R): Due to low atmospheric pressure at high altitude, the body does not get sufficient oxygen.
In the light of the above statements, choose the correct answer from the options given below
(1) (A) is false but (R) is true
(2) Both (A) and (R) are true and (R) is the correct explanation of (A)
(3) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(4) (A) is true but (R) is false

Answer (2)

Sol. Altitude sickness can be experienced at high altitude where body does not get enough oxygen due to low atmospheric pressure and causes nausea, fatigue and heart palpitations.
Hence correct option is (2) as [R] is correct explanation of [A].

192. Which of these is not an important component of initiation of parturition in humans?

(1) Release of Prolactin
(2) Increase in estrogen and progesterone ratio
(3) Synthesis of prostaglandins
(4) Release of Oxytocin

Answer (1)

Sol. At the end of gestation, the completely developed foetus is expelled out. This process is called parturition.
Parturition is controlled by a complex neuroendocrine mechanism. Estrogen and progesterone ratio increases as estrogen levels rise significantly. Prostaglandins, which stimulate uterine contractions are also produced that act on myometrium. Oxytocin, the main hormone, also called as birth hormone is released by maternal pituitary, which brings about strong uterine contractions. Prolactin is a lactation hormone that has no role in initiation of parturition.

193. Which of the following secretes the hormone, relaxin, during the later phase of pregnancy?

(1) Uterus
(2) Graafian follicle
(3) Corpus luteum
(4) Foetus

Answer (3)

Sol. The hormone relaxin is produced in the later phase of pregnancy. It is produced by the ovary.
Graafian follicle is not formed when the woman is pregnant. Uterus and foetus do not produce relaxin. Relaxin is produced by the corpus luteum present in the ovary. Ruptured Graafian follicle is called corpus luteum, which has endocrine function.

194. Match List-I with List-II
List-I
(a) Filariasis
(b) Amoebiasis
(c) Pneumonia
(d) Ringworm
List-II
(i) Haemophilus influenzae
(ii) Trichophyton
(iii) Wuchereria bancrofti
(iv) Entamoeba histolytica
Choose the correct answer from the options given below
(a) (b) (c) (d)
(1) (ii) (iii) (i) (iv)
(2) (iv) (i) (iii) (ii)
(3) (iii) (iv) (i) (ii)
(4) (i) (ii) (iv) (iii)

Answer (3)

Sol. The correct option is (3).
Filariasis is the disease caused by Wuchereria bancrofti, filarial worm. Amoebiasis/Amoebic dysentery is caused by a protozoan parasite Entamoeba histolytica in the large intestine of human. Pneumonia is caused by bacteria like Streptococcus pneumoniae and Haemophilus influenzae. Ringworm is caused by fungi belonging to genera Microsporum, Trichophyton and Epidermophyton.

195. Which of the following is not a step in Multiple Ovulation Embryo Transfer Technology (MOET)?

(1) Fertilized eggs are transferred to surrogate mothers at 8-32 cell stage
(2) Cow is administered hormone having LH like activity for super ovulation
(3) Cow yields about 6-8 eggs at a time
(4) Cow is fertilized by artificial insemination

Answer (2)

Sol. Multiple Ovulation Embryo Transfer Technology is used for herd improvement in short time.
Cows are administered hormones, with FSH-like activity for superovulation. 8-32 celled embryos are transferred to surrogate mothers. 6-8 eggs are produced per cycle. Cows can be fertilised by artificial insemination.

196. The Adenosine deaminase deficiency results into

(1) Addison's disease
(2) Dysfunction of Immune system
(3) Parkinson's disease
(4) Digestive disorder

Answer (2)

Sol. Adenosine deaminase (ADA) enzyme is crucial for the immune system to function. Hence, its deficiency results in the dysfunction of immune system.
Hyposecretion of hormones of the adrenal cortex causes Addison's disease. Parkinson's disease is a long-term degenerative disorder of the central nervous system. Disorders which affect GIT & associated glands are called digestive disorders.

197. Which one of the following statements about Histones is wrong?

(1) Histones carry positive charge in the side chain
(2) Histones are organized to form a unit of 8 molecules
(3) The pH of histones is slightly acidic
(4) Histones are rich in amino acids - Lysine and Arginine

Answer (3)

Sol. Histones are rich in basic amino acids residue lysine and arginine with charged side chain. There are five types of histone proteins i.e., H₁, H₂A, H₂B, H₃ and H₄. Four of them occur in pairs to produce a unit of 8 molecules (histone octamer). The pH of histones is basic.

198. Statement I: The codon 'AUG' codes for methionine and phenylalanine.
Statement II: 'AAA' and 'AAG' both codons code for the amino acid lysine.
In the light of the above statements, choose the correct answer from the options given below.
(1) Statement I is incorrect but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is correct but Statement II is false

Answer (1)

Sol. AUG has dual functions, it codes for methionine. It also acts as initiator codon. AUG does not code for phenylalanine. Statement II is true.

199. Identify the types of cell junctions that help to stop the leakage of the substances across a tissue and facilitation of communication with neighbouring cells via rapid transfer of ions and molecules.

(1) Adhering junctions and Gap junctions, respectively
(2) Gap junctions and Adhering junctions, respectively
(3) Tight junctions and Gap junctions, respectively
(4) Adhering junctions and Tight junctions, respectively.

Answer (3)

Sol. Three types of junctions are found in tissues
Tight junctions stop leakage of substances from leaking across a tissue. Adhering junctions cement and keep neighbouring cells together. Gap junctions or communication junctions facilitate communication between cells by connecting the cytoplasm of adjoining cells.

200. Following are the statements about prostomium of earthworm.
(a) It serves as a covering for mouth.
(b) It helps to open cracks in the soil into which it can crawl.
(c) It is one of the sensory structures.
(d) It is the first body segment.
Choose the correct answer from the options given below.
(1) (b) and (c) are correct
(2) (a), (b) and (c) are correct
(3) (a), (b) and (d) are correct
(4) (a), (b), (c) and (d) are correct

Answer (2)

Sol. The anterior end of the earthworm has mouth which has covering called prostomium. Prostomium acts as a wedge to force open cracks in the soil. Prostomium has receptors, so it is sensory in function. The first body segment of earthworm is the peristomium.