NEET (UG) 2026 — Code 50 — Answer Key & Solutions

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Q1.

A particle of mass M moves along a horizontal x-axis from x = 0 to x = L. The coefficient of kinetic friction varies as a function of x as μ_k(x) = μ₀ − αx, with μ_k(L) = 0. If the total work done by the frictional force during the motion is nμ₀MgL (where g is the acceleration due to gravity), the value of n is :

  • A. 3
  • B. 1
  • C. 1/3
  • D. 1/2 ✓

Solution: $n = \tfrac{1}{2}$ On a horizontal surface $N = Mg$, so friction is $f(x) = \mu_k(x)\,Mg$. $$\mu_k(L) = 0 \Rightarrow \alpha = \dfrac{\mu_0}{L}$$ Magnitude of work done by friction $= \displaystyle\int_0^L (\mu_0 - \alpha x)\,Mg\,dx = Mg\left(\mu_0 L - \dfrac{\alpha L^2}{2}\right) = \tfrac{1}{2}\mu_0 MgL$ $$\therefore n = \tfrac{1}{2}$$

Q2.

The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the molecules of gas A is twice the diameter of the molecules of B. If number densities of the gases A and B are n_A and n_B respectively, then the correct option is :

  • A. n_A = n_B
  • B. n_A = 2 n_B
  • C. n_A = (1/4) n_B
  • D. n_A = (1/2) n_B ✓

Solution: $n_A = \tfrac{1}{2}n_B$ Mean free path $\lambda = \dfrac{1}{\sqrt{2}\,\pi d^2 n}$ With $d_A = 2d_B$: $\dfrac{\lambda_A}{\lambda_B} = \dfrac{d_B^2\,n_B}{d_A^2\,n_A} = \dfrac{n_B}{4n_A} = \dfrac{1}{2}$ $$\therefore n_A = \tfrac{1}{2}n_B$$

Q3.

Two identical inductors are connected in two different configurations P and Q, where a time varying current I(t) is flowing, as shown in the figure. The induced emf between points a and b for configuration P is E_P and that for configuration Q is E_Q. The ratio E_P/E_Q is : [Neglect the effect of mutual inductance.]

  • A. 1/4
  • B. 1/2
  • C. 1
  • D. 2 ✓

Solution: $E_P/E_Q = 2$ Emf induced in an inductor $|E| = L\left|\dfrac{dI}{dt}\right|$ In P (series) the current $I(t)$ flows through both coils, but a and b are across only the first coil, so $E_P = L\dfrac{dI}{dt}$. In Q the two identical coils are in parallel between a and b, so $L_{eq} = \dfrac{L}{2}$ and $E_Q = \dfrac{L}{2}\dfrac{dI}{dt}$. $$\therefore \dfrac{E_P}{E_Q} = \dfrac{L}{L/2} = 2$$

Q4.

For sound waves, if the number of nodes for the 5th harmonic of an open-ended pipe is a and that for the 5th harmonic of the same pipe with one of its ends closed is m, then a/m is :

  • A. 5/9
  • B. 9/5
  • C. 5/3 ✓
  • D. 4/5

Solution: $a/m = \tfrac{5}{3}$ An open pipe has $n$ displacement nodes in its $n$th harmonic, so its 5th harmonic has 5 nodes: $a = 5$. A closed pipe has only odd harmonics; in its 5th harmonic the pipe length is $\dfrac{5\lambda}{4}$, giving 3 nodes: $m = 3$. $$\therefore \dfrac{a}{m} = \dfrac{5}{3}$$

Q5.

Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and μ₀ is the permeability of free space, the inductance of the solenoid is :

  • A. μ₀πn²r²l ✓
  • B. μ₀n²r²l
  • C. (μ₀/2π)n²r²l
  • D. 2μ₀πn²r²l

Solution: $L = \mu_0\pi n^2 r^2 l$ Self-inductance of a long solenoid $L = \mu_0 n^2 A l = \mu_0 n^2(\pi r^2)\,l = \mu_0\pi n^2 r^2 l$

Q6.

Consider a particle moving along a straight line, whose position as a function of time is given by s(t) = αt² − βt + γ, where α = 1 m s⁻², β = 6 m s⁻¹ and γ = 5 m. The average speed of the particle, in m s⁻¹, from t = 0 to t = 6 s is :

  • A. 12
  • B. 6
  • C. 3 ✓
  • D. 0

Solution: $3\ \text{m s}^{-1}$ Velocity $v = \dfrac{ds}{dt} = 2t - 6$, which is zero at $t = 3$ s, so the particle turns back there. $s(0) = 5$ m, $s(3) = -4$ m, $s(6) = 5$ m Distance travelled $= 9\ \text{m} + 9\ \text{m} = 18\ \text{m}$ $$\therefore \text{average speed} = \dfrac{18\ \text{m}}{6\ \text{s}} = 3\ \text{m s}^{-1}$$

Q7.

Consider the following nuclear reaction : ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He. Take masses of ²³⁸U, ²³⁴Th and ⁴He as 238.050 u, 234.043 u and 4.003 u, respectively. The Q value for the reaction, in keV, is : (Given : 1 u = 931.5 MeV c⁻²)

  • A. 3726 ✓
  • B. 3730
  • C. 3736
  • D. 3740

Solution: 3726 keV Mass defect $\Delta m = 238.050 - (234.043 + 4.003) = 0.004$ u $\therefore$ Q value $= 0.004 \times 931.5\ \text{MeV} = 3.726\ \text{MeV} = 3726\ \text{keV}$

Q8.

A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time t, then variation of the photocurrent I and magnitude of the stopping potential |V| with time is best represented by :

  • A. (1) ✓
  • B. (2)
  • C. (3)
  • D. (4)

Solution: Graph (1): $I$ falls linearly, $|V|$ stays constant Photocurrent is proportional to the intensity of light, so $I$ falls linearly with time. Stopping potential depends on the frequency of light, not its intensity, so $|V|$ stays constant.

Q9.

Three identical capacitors P, Q and S, each of the capacitance C, are connected to a battery of voltage V, as shown in the figure. If the energy stored in the capacitor P and total energy stored in the system are U_P and U_T, respectively, then the ratio U_P/U_T is :

  • A. 2/3
  • B. 1/3
  • C. 1/2
  • D. 1/6 ✓

Solution: $U_P/U_T = \tfrac{1}{6}$ P and Q in series: $C_{PQ} = \dfrac{C}{2}$; S in parallel with them: $C_{eq} = \dfrac{C}{2} + C = \dfrac{3C}{2}$ Total energy $U_T = \tfrac{1}{2}C_{eq}V^2 = \tfrac{3}{4}CV^2$ P and Q in series share the battery voltage, so $V_P = \dfrac{V}{2}$ and $U_P = \tfrac{1}{2}C\left(\dfrac{V}{2}\right)^2 = \tfrac{1}{8}CV^2$ $$\therefore \dfrac{U_P}{U_T} = \dfrac{1/8}{3/4} = \dfrac{1}{6}$$

Q10.

Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between P and Q is 15 N m⁻². The area of cross-section at P and Q are 40 cm² and 20 cm², respectively. The rate of flow of water through the pipe, in cm³ s⁻¹, is : [Take density of water = 1000 kg m⁻³]

  • A. 100
  • B. 200
  • C. 300
  • D. 400 ✓

Solution: $400\ \text{cm}^3\,\text{s}^{-1}$ Equation of continuity: $A_P v_P = A_Q v_Q \Rightarrow v_Q = 2v_P$ Bernoulli's principle for a horizontal pipe: $P_P - P_Q = \tfrac{1}{2}\rho(v_Q^2 - v_P^2) = \tfrac{3}{2}\rho v_P^2$ $$15 = \tfrac{3}{2} \times 1000 \times v_P^2 \Rightarrow v_P = 0.1\ \text{m s}^{-1} = 10\ \text{cm s}^{-1}$$ $$\therefore \text{rate of flow} = A_P v_P = 40 \times 10 = 400\ \text{cm}^3\,\text{s}^{-1}$$

Q11.

A current I₀ flows through a metallic circular loop of radius r as shown in the figure. Resistance of the segment ABC is half that of ADC. Magnitude of the magnetic field at the center O of the loop is :

  • A. μ₀I₀/12r ✓
  • B. μ₀I₀/4r
  • C. μ₀I₀/2r
  • D. μ₀I₀/2πr

Solution: $B = \dfrac{\mu_0 I_0}{12r}$ Current divides between ABC and ADC inversely as their resistances: $I_{ABC} = \dfrac{2I_0}{3}$, $I_{ADC} = \dfrac{I_0}{3}$ Each semicircle gives a field $\dfrac{\mu_0 I}{4r}$ at O, and the two fields are opposite. $$\therefore B = \dfrac{\mu_0}{4r}\left(\dfrac{2I_0}{3} - \dfrac{I_0}{3}\right) = \dfrac{\mu_0 I_0}{12r}$$

Q12.

In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity (v) with the ratio of density of spherical ball (σ) to density of the liquid (ρ), is best represented by :

  • A. (1)
  • B. (2)
  • C. (3) ✓
  • D. (4)

Solution: Graph (3): a straight line of positive slope, $v = 0$ at $\sigma/\rho = 1$ At terminal velocity, weight equals buoyant force plus viscous drag (Stokes' law): $\tfrac{4}{3}\pi R^3\sigma g = \tfrac{4}{3}\pi R^3\rho g + 6\pi\eta Rv$ $$v = \dfrac{2R^2 g}{9\eta}(\sigma - \rho) = K\left(\dfrac{\sigma}{\rho} - 1\right), \quad K = \dfrac{2R^2 g\rho}{9\eta} > 0$$ $\therefore$ $v$ against $\sigma/\rho$ is a straight line of positive slope, zero at $\sigma/\rho = 1$ and equal to $-K$ at $\sigma/\rho = 0$.

Q13.

Two planets P₁ and P₂ with equal mass have radii R₁ and R₂, respectively, where R₂ = R₁/2. The escape speeds of P₁ and P₂ are v₁ and v₂, respectively. Then v₂/v₁ is :

  • A. 1/√2
  • B. 1
  • C. √2 ✓
  • D. 2

Solution: $v_2/v_1 = \sqrt{2}$ Escape speed $v = \sqrt{\dfrac{2GM}{R}} \propto \dfrac{1}{\sqrt{R}}$ for equal masses $$\therefore \dfrac{v_2}{v_1} = \sqrt{\dfrac{R_1}{R_2}} = \sqrt{2}$$

Q14.

In a solar system, the time-period of revolution of a planet during a circular orbit of radius R is proportional to :

  • A. R^(1/2)
  • B. R^(3/2) ✓
  • C. R²
  • D. R³

Solution: $T \propto R^{3/2}$ Kepler's law of periods: $T^2 \propto R^3$, so $T \propto R^{3/2}$.

Q15.

Two infinitely long parallel conducting wires A and B carry currents I and 2I, respectively, in the same direction. The wire A has uniform mass per unit length λ and lies on an insulated floor. The wire B is kept fixed at a height h above the floor. The minimum magnitude of h so that the wire A does not rise from the floor is : [g is the acceleration due to gravity and μ₀ is the permeability of free space.]

  • A. μ₀I²/2πλg
  • B. μ₀I²/πλg ✓
  • C. 2μ₀I²/πλg
  • D. 4μ₀I²/πλg

Solution: $h_{\min} = \dfrac{\mu_0 I^2}{\pi\lambda g}$ Currents in the same direction attract, so B pulls A up with force per unit length $\dfrac{\mu_0 I(2I)}{2\pi h} = \dfrac{\mu_0 I^2}{\pi h}$ Weight of A per unit length $= \lambda g$; A does not rise if $\dfrac{\mu_0 I^2}{\pi h} \le \lambda g$ $$\therefore h_{\min} = \dfrac{\mu_0 I^2}{\pi\lambda g}$$

Q16.

An ideal Zener diode with breakdown voltage of −3 V is reverse biased with a negative input voltage V_i = −5 V. The magnitude of voltage difference between points B and A is :

  • A. 3 V
  • B. 2 V ✓
  • C. 1 V
  • D. 0 V

Solution: 2 V $|V_i| = 5$ V exceeds the breakdown voltage of 3 V, so the Zener is in breakdown and the voltage across it stays at 3 V. Kirchhoff's loop rule: $V_i = V_Z + V_R$; with B and A across the resistor, $5 = 3 + V_{BA}$ $$\therefore |V_{BA}| = 2\ \text{V}$$

Q17.

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (γ = 5/3) decreases from 60 K to 50 K. The work done by the gas in the process is : (Take the universal gas constant as R = 8.3 J mol⁻¹ K⁻¹)

  • A. 41.5 J
  • B. 83 J
  • C. 124.5 J ✓
  • D. 166 J

Solution: 124.5 J In an adiabatic process $Q = 0$, so work done by the gas $W = -\Delta U = nC_V(T_1 - T_2)$, with $C_V = \tfrac{3}{2}R$ for a monatomic gas $$\therefore W = 1 \times \tfrac{3}{2} \times 8.3 \times (60 - 50)\ \text{J} = 124.5\ \text{J}$$

Q18.

A ray of light with wavelength λ is incident on three different photo-electric cells namely 1, 2 and 3. The threshold wavelengths of these cells are λ₁, λ₂ and λ₃ respectively and the magnitude of stopping potentials are V₁, V₂ and V₃ respectively. The relation between λ and threshold wavelengths are λ₁ < λ, λ₂ > λ and λ₃ ≫ λ. The correct option is :

  • A. V₁ = 0, V₂ < V₃ ✓
  • B. V₁ = 0, V₂ > V₃
  • C. V₁ > V₂, V₃ = 0
  • D. V₁ < V₂, V₃ = 0

Solution: $V_1 = 0$ and $V_2 < V_3$ Einstein's photoelectric equation: $eV_0 = hc\left(\dfrac{1}{\lambda} - \dfrac{1}{\lambda_0}\right)$, with emission only if $\lambda \le \lambda_0$ Cell 1: $\lambda > \lambda_1$, so no emission and $V_1 = 0$. Cells 2 and 3: $\lambda < \lambda_2$ and $\lambda \ll \lambda_3$, so both emit. $\lambda_3 > \lambda_2 \Rightarrow \dfrac{1}{\lambda_3} < \dfrac{1}{\lambda_2}$, $\therefore V_3 > V_2$

Q19.

A photon and an electron, each of 20 eV energy, move in free space. The ratio of the momentum of the electron p_e to that of photon p_ph, i.e. p_e/p_ph, is : (Take speed of light = 3 × 10⁸ m s⁻¹, charge of electron = 1.6 × 10⁻¹⁹ C and mass of electron = 9 × 10⁻³¹ kg)

  • A. 2/450
  • B. 1/250
  • C. 225 ✓
  • D. 275

Solution: $p_e/p_{ph} = 225$ Photon momentum $p_{ph} = \dfrac{E}{c}$; electron momentum (non-relativistic) $p_e = \sqrt{2mE}$ $$\dfrac{p_e}{p_{ph}} = c\sqrt{\dfrac{2m}{E}}$$ $$\dfrac{2m}{E} = \dfrac{2 \times 9 \times 10^{-31}}{20 \times 1.6 \times 10^{-19}} = \dfrac{9}{16} \times 10^{-12}$$ $$\therefore \dfrac{p_e}{p_{ph}} = 3 \times 10^8 \times \tfrac{3}{4} \times 10^{-6} = 225$$

Q20.

Which of the following measurements require 'index correction' ?

  • A. Measurement of resistance of a wire using meter bridge
  • B. Measurement of gravitational acceleration using simple pendulum
  • C. Measurement of focal length of lenses using optical bench ✓
  • D. Measurement of speed of sound using resonance tube

Solution: Focal length of lenses on an optical bench On an optical bench the actual distance between the lens and a needle can differ from the distance read between the index marks of their uprights; index correction removes this error from focal-length readings. (A) A metre bridge needs end correction, not index correction. (B) A simple pendulum needs no index correction; its effective length is measured from the point of suspension to the centre of the bob. (D) A resonance tube needs end correction at its open end, not index correction.

Q21.

A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius R, having uniform positive charge density ρ, as shown in the figure. The initial and final positions of the charge are marked by A and B at distances 2R and 3R respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is ρR²/nε₀. The value of n is : (ε₀ is the permittivity of vacuum)

  • A. 2
  • B. 6
  • C. 9
  • D. 18 ✓

Solution: $n = 18$ A and B lie outside the sphere, so it acts as a point charge $Q = \tfrac{4}{3}\pi R^3\rho$ at its centre. Work on a unit charge $|W| = \dfrac{Q}{4\pi\varepsilon_0}\left(\dfrac{1}{2R} - \dfrac{1}{3R}\right) = \dfrac{\rho R^3}{3\varepsilon_0} \times \dfrac{1}{6R} = \dfrac{\rho R^2}{18\varepsilon_0}$ $$\therefore n = 18$$

Q22.

Consider three media P, Q and R with refractive indices 1, 1.25 and 1.5, respectively. The medium Q having a thickness of 5 cm is placed between extended media P and R as shown in the figure. An object is placed at the centre of medium Q. If viewed from medium P near the normal direction, the apparent depth of O is h₁. For similar observation from medium R, the apparent depth is h₂. The value of |h₁ − h₂|, in cm, is :

  • A. 0
  • B. 1 ✓
  • C. 2
  • D. 3

Solution: 1 cm The object is at the centre of the 5 cm medium Q, so its real depth from each interface is 2.5 cm. Apparent depth $= \text{real depth} \times \dfrac{n_{\text{observer}}}{n_{\text{object}}}$ From P: $h_1 = 2.5 \times \dfrac{1}{1.25} = 2.0$ cm; from R: $h_2 = 2.5 \times \dfrac{1.5}{1.25} = 3.0$ cm $$\therefore |h_1 - h_2| = 1\ \text{cm}$$

Q23.

A frictionless circular wire of unit radius is fixed on the horizontal plane. Two point particles of unit mass start moving simultaneously from point A (θ = π/2) with identical uniform angular speeds in opposite directions, and meet again at point B (θ = −π/2). During this time, which of the following figures schematically represent the magnitude of the total linear momentum P of the system, as a function of θ ?

  • A. (1)
  • B. (2)
  • C. (3) ✓
  • D. (4)

Solution: Figure (3): $|P| = 2\omega|\cos\theta|$, zero at A and B, largest at $\theta = 0$ Taking $\theta$ as the position of particle 1, particle 2 is at $\pi - \theta$. With unit radius, $\vec{v}_1 = (\omega\sin\theta,\ -\omega\cos\theta)$ and $\vec{v}_2 = (-\omega\sin\theta,\ -\omega\cos\theta)$ With unit masses, $\vec{P} = \vec{v}_1 + \vec{v}_2 = (0,\ -2\omega\cos\theta)$, so $|\vec{P}| = 2\omega|\cos\theta|$ $\therefore |\vec{P}|$ is zero at $\theta = \pm\dfrac{\pi}{2}$ (A and B) and largest, $2\omega$, at $\theta = 0$: the inverted-U curve of figure (3).

Q24.

The temperature of a metallic sphere of radius R is raised by a small amount ΔT. If the linear coefficient of thermal expansion of the metal is α, the approximate increase in the volume of the sphere is :

  • A. 2πR³αΔT
  • B. 3πR³αΔT
  • C. 4πR³αΔT ✓
  • D. 6πR³αΔT

Solution: $\Delta V = 4\pi R^3\alpha\Delta T$ Coefficient of volume expansion $\alpha_V = 3\alpha$ $$\Delta V = \alpha_V V\Delta T = 3\alpha \times \tfrac{4}{3}\pi R^3 \times \Delta T = 4\pi R^3\alpha\Delta T$$

Q25.

A cylindrical cork of uniform density floats in a liquid of density ρ. If the cork is depressed slightly and released, it oscillates harmonically with time period T. If the same cork floats in another liquid of density ρ₀, then another oscillation has time period 2T. The value of ρ₀/ρ is :

  • A. 4
  • B. 2
  • C. 1/2
  • D. 1/4 ✓

Solution: $\rho_0/\rho = \tfrac{1}{4}$ Pushed down by $x$, the cork feels an extra buoyant force $\rho_{\text{liquid}}Agx$, so $k = \rho_{\text{liquid}}Ag$ and $m = \rho_{\text{cork}}AL$ ($A$ = cross-section, $L$ = length) $$T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{\rho_{\text{cork}}L}{\rho_{\text{liquid}}\,g}} \propto \dfrac{1}{\sqrt{\rho_{\text{liquid}}}}$$ $$\dfrac{2T}{T} = \sqrt{\dfrac{\rho}{\rho_0}} \Rightarrow \dfrac{\rho}{\rho_0} = 4$$ $$\therefore \dfrac{\rho_0}{\rho} = \tfrac{1}{4}$$

Q26.

One main scale division of a Vernier callipers is equal to 1 mm and the number of divisions on the Vernier scale is 10. When both the jaws touch each other, the zero of the Vernier scale shifts to the left of the zero of the main scale in such a way that 4th Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of the wire to be 1.00 cm, the actual length of the wire is :

  • A. 0.60 cm
  • B. 0.96 cm
  • C. 1.00 cm
  • D. 1.04 cm ✓

Solution: Least count = 1 mm/10 = 0.1 mm. The vernier zero to the left of the main zero is a negative zero error of magnitude 4 × 0.1 = 0.4 mm = 0.04 cm, which is added: actual = 1.00 + 0.04 = 1.04 cm. ⚠ Zero-error convention can vary — verify against the official key.

Q27.

A solid sphere A of radius R and mass M is attached at a point to a smaller solid sphere B of radius r < R and mass m < M, so that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of A is I_A, and that calculated about a vertical axis passing through the centre of B is I_B. The difference I_A − I_B is :

  • A. (M − m)(R + r)²
  • B. (m − M)(R + r)² ✓
  • C. (m − M)(R − r)²
  • D. 0

Solution: $I_A - I_B = (m - M)(R + r)^2$ The spheres touch externally, so the distance between their centres is $R + r$. Theorem of parallel axes: $I_A = \tfrac{2}{5}MR^2 + \tfrac{2}{5}mr^2 + m(R + r)^2$ Similarly $I_B = \tfrac{2}{5}MR^2 + \tfrac{2}{5}mr^2 + M(R + r)^2$ $$\therefore I_A - I_B = (m - M)(R + r)^2$$

Q28.

Consider a spring-mass system executing simple harmonic motion, where the spring constant is k and the mass is m. At any instant the displacement is x and the speed of the particle is v. On the v−x plane, if the graph of v as a function of x is a circle, then the correct option is :

  • A. k = 1/m
  • B. k = m ✓
  • C. k = m²
  • D. k = √m

Solution: $k = m$ In SHM $v = \omega\sqrt{A^2 - x^2}$, i.e. $\dfrac{v^2}{\omega^2} + x^2 = A^2$, an ellipse on the $v$–$x$ plane. It is a circle only when $\omega = \sqrt{\dfrac{k}{m}} = 1$ $$\therefore k = m$$

Q29.

The lens combination as shown in the figure consists of two lenses, L₁ and L₂, of focal lengths +10 cm and −10 cm, respectively. The position of the image formed is :

  • A. 20 cm to the left of the concave lens
  • B. 60 cm to the left of the concave lens ✓
  • C. 30 cm to the right of the concave lens
  • D. 60 cm to the right of the concave lens

Solution: 60 cm to the left of the concave lens Lens formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ $L_1$ ($f_1 = +10$ cm, $u_1 = -30$ cm): $\dfrac{1}{v_1} = \dfrac{1}{10} - \dfrac{1}{30} = \dfrac{2}{30}$, so $v_1 = +15$ cm, a real image 15 cm to the right of $L_1$ This image is a virtual object for $L_2$, which is 3 cm away: $u_2 = 15 - 3 = +12$ cm $L_2$ ($f_2 = -10$ cm): $\dfrac{1}{v_2} = -\dfrac{1}{10} + \dfrac{1}{12} = -\dfrac{1}{60}$, so $v_2 = -60$ cm $\therefore$ the final image is virtual, 60 cm to the left of the concave lens.

Q30.

An ac voltage V = 220 sin(2π × 10⁴ t) Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is : (Given : L = 10 mH, C = 25 μF, R = 100 Ω)

  • A. 2.2 A ✓
  • B. 5.5 A
  • C. 11.0 A
  • D. 22.0 A

Solution: At series resonance Z = R = 100 Ω, so I₀ = 220/100 = 2.2 A. ⚠ The printed L, C, ω do not actually satisfy resonance (give ≈0.35 A) — verify against the official key.

Q31.

A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is L_A and L_B computed about points A and B, respectively, with OB = 2 × OA. The value of L_A/L_B is :

  • A. 1/4
  • B. 1/2
  • C. 1 ✓
  • D. 2

Solution: $L_A/L_B = 1$ The disc spins about its fixed centre O, so its total linear momentum is zero: each mass element moving with $\vec v$ has an identical one on the opposite side moving with $-\vec v$. Angular momentum about any point P: $\vec L_P = \vec L_O + \vec r_{O/P} \times \vec P_{\text{total}}$. Since $\vec P_{\text{total}} = 0$, the second term vanishes and $\vec L_P = \vec L_O$ for every point. $\therefore L_A = L_B = L_O$, whatever OA and OB are, so $L_A/L_B = 1$.

Q32.

A conducting loop of finite resistance lies on the x−y plane. There is a constant magnetic field in the z direction. The area of the loop varies with time as A = A₀(1 + sin t). The figure that correctly indicates the qualitative behaviour of the power P dissipated in the loop as a function of time is :

  • A. (1)
  • B. (2) ✓
  • C. (3)
  • D. (4)

Solution: Graph of $P \propto \cos^2 t$: never negative, period $\pi$ Faraday's law: $\varepsilon = -B\dfrac{dA}{dt} = -BA_0\cos t$. Power dissipated: $P = \dfrac{\varepsilon^2}{R} \propto \cos^2 t$. $\therefore$ P is never negative: identical humps that touch zero, with period $\pi$.

Q33.

A point charge Q is placed inside a cavity within a solid isolated conducting sphere. Consider points A, B and C as shown in the figure, where the magnitudes of the electric fields are E_A, E_B and E_C, respectively. The points B and C are at the same distance from the centre of the solid sphere. The correct option is :

  • A. E_A > 0, E_B = E_C
  • B. E_A = 0, E_B = E_C ✓
  • C. E_A > 0, E_B > E_C
  • D. E_A = 0, E_B > E_C

Solution: $E_A = 0,\ E_B = E_C$ Inside the material of a conductor $E = 0$, so $E_A = 0$. Outside, the induced charge on the outer surface makes the field spherically symmetric. $\therefore$ Points at equal distance from the centre have equal fields: $E_B = E_C$.

Q34.

Consider a fixed uniformly charged insulating sphere with radius R and total charge +Q. A point charge −q (q < Q) with mass m is released from rest at a distance of 3R from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is : (ε₀ is the permittivity of vacuum, neglect gravitational forces)

  • A. √(3Qq/4πε₀mR)
  • B. √(2Qq/3πε₀mR)
  • C. √(Qq/3πε₀mR) ✓
  • D. √(Qq/4πε₀mR)

Solution: $v = \sqrt{\dfrac{Qq}{3\pi\varepsilon_0 mR}}$ Only the electrostatic force acts, so $K_i + U_i = K_f + U_f$, with $K_i = 0$ (released from rest). Outside the sphere its charge acts as a point charge Q at the centre: $U = -\dfrac{Qq}{4\pi\varepsilon_0 r}$. $$\tfrac{1}{2}mv^2 = U_{3R} - U_R = \dfrac{Qq}{4\pi\varepsilon_0 R} - \dfrac{Qq}{12\pi\varepsilon_0 R} = \dfrac{Qq}{6\pi\varepsilon_0 R}$$ $$\therefore v = \sqrt{\dfrac{Qq}{3\pi\varepsilon_0 mR}}$$

Q35.

In Geiger-Marsden's experiment, the number of scattered α-particles N(θ) is plotted as a function of the scattering angle θ. Which of the following options represents the correct plot ?

  • A. (1)
  • B. (2)
  • C. (3) ✓
  • D. (4)

Solution: Graph of $N(\theta) \propto 1/\sin^4(\theta/2)$: very high near 0°, falling steeply, then flattening Rutherford's scattering formula: $N(\theta) \propto \dfrac{1}{\sin^4(\theta/2)}$. At small θ, $\sin(\theta/2)$ is tiny, so $N(\theta)$ is very large (about $10^7$): most α-particles pass nearly straight through. As θ approaches 180°, $\sin(\theta/2) \to 1$ and $N(\theta)$ falls to its smallest value (about $10^1$): only about 1 in 8000 is deflected by more than 90°. $\therefore$ The plot starts very high near 0°, drops sharply, then flattens out.

Q36.

Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, +α, while the other one has a negative temperature coefficient of resistance, −α, as shown in the figure. The current through these circuits are denoted by I_A and I_B. At initial temperature, the resistance of the two resistors is R₀. As the temperature is increased, the correct option that describes the variation of current in these circuits is :

  • A. I_A remains constant while I_B increases ✓
  • B. I_A decreases while I_B increases
  • C. I_A increases while I_B decreases
  • D. both I_A and I_B remain constant

Solution: $I_A$ remains constant while $I_B$ increases In the series circuit $R = R_0(1+\alpha\Delta T) + R_0(1-\alpha\Delta T) = 2R_0$, which is constant, so the current stays constant. In the parallel circuit $R = \dfrac{R_0}{2}(1-\alpha^2\Delta T^2)$, which decreases as the temperature rises, so the current increases.

Q37.

Consider that α, k_B and b represent the Stefan-Boltzmann constant, Boltzmann constant and Wien's displacement law constant, respectively. The dimension of α·k_B³·b is :

  • A. [M⁴L⁷T⁻⁹K⁻⁶] ✓
  • B. [L⁻¹K⁻¹T⁻¹]
  • C. [L⁻²T⁻²K⁻¹]
  • D. [L⁻²T⁻¹K⁻¹]

Solution: $[\text{M}^4\text{L}^7\text{T}^{-9}\text{K}^{-6}]$ Stefan's law: energy radiated per second per unit area $= \alpha T^4$, so $[\alpha] = \dfrac{[\text{ML}^2\text{T}^{-3}]}{[\text{L}^2\text{K}^4]} = [\text{MT}^{-3}\text{K}^{-4}]$. $k_BT$ is an energy, so $[k_B] = [\text{ML}^2\text{T}^{-2}\text{K}^{-1}]$ and $[k_B^3] = [\text{M}^3\text{L}^6\text{T}^{-6}\text{K}^{-3}]$. Wien's law $\lambda_m T = b$, so $[b] = [\text{LK}]$. $$\therefore [\alpha k_B^3 b] = [\text{M}^{1+3}\text{L}^{6+1}\text{T}^{-3-6}\text{K}^{-4-3+1}] = [\text{M}^4\text{L}^7\text{T}^{-9}\text{K}^{-6}]$$

Q38.

An electromagnetic wave travelling in a lossless dielectric medium having dielectric constant ε_r = 9 has the electric field E_x = E₀ sin(kz − 2π×10⁶ t) V m⁻¹, where E₀ is the amplitude and k is the wave vector. Among the following options, the incorrect choice is :

  • A. The speed of the electromagnetic wave inside the medium is 10⁸ ms⁻¹
  • B. The wavelength of the electromagnetic wave inside the medium is 300 m ✓
  • C. The magnetic field is given by the relation B_y = (B₀/v) sin(kz − 2π×10⁶ t), where v is the speed of the electromagnetic wave inside the medium
  • D. The direction of propagation of the electromagnetic wave is along +z

Solution: Wavelength of 300 m is wrong: $\lambda = 100$ m in the medium With $f = \dfrac{\omega}{2\pi} = 10^6\ \text{Hz}$ and $v = 10^8\ \text{m s}^{-1}$, the wavelength in the medium is $\lambda = \dfrac{v}{f} = 100\ \text{m}$, not 300 m (300 m is $c/f$, the wavelength in vacuum). (A) True: $v = \dfrac{c}{\sqrt{\varepsilon_r}} = \dfrac{3\times10^8}{\sqrt{9}} = 10^8\ \text{m s}^{-1}$, taking $\mu_r = 1$. (C) True: $\vec B$ is along y, in phase with $\vec E$, with amplitude $E_0/v$ in the medium. (D) True: a phase of the form $(kz - \omega t)$ means the wave travels along +z.

Q39.

One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is :

  • A. 400 J
  • B. 500 J
  • C. 600 J ✓
  • D. 800 J

Solution: 600 J In a complete cycle the internal energy returns to its starting value: $\Delta U = 0$. By the first law, $Q_{\text{net}} = W_{\text{net}}$, the area enclosed by the cycle on the P–V diagram. The cycle is a rectangle with $\Delta P = 300 - 100 = 200\ \text{N m}^{-2}$ and $\Delta V = 5 - 2 = 3\ \text{m}^3$. $$\therefore Q_{\text{net}} = \Delta P \times \Delta V = 200 \times 3 = 600\ \text{J}$$

Q40.

Consider that an electron is revolving in an excited state of a Hydrogen atom with velocity √25.6 × 10⁵ m s⁻¹. The radius of the orbit is x × 10⁻⁹ m. The value of x is : (e = 1.6×10⁻¹⁹ C, mₑ = 9×10⁻³¹ kg, 1/4πε₀ = 9×10⁹ N m² C⁻²)

  • A. 1 ✓
  • B. 2
  • C. 3
  • D. 4

Solution: $x = 1$ In Bohr's model the electrostatic attraction provides the centripetal force: $\dfrac{mv^2}{r} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}$, so $r = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{mv^2}$. $$r = \dfrac{(9\times10^9)(1.6\times10^{-19})^2}{(9\times10^{-31})(25.6\times10^{10})}\ \text{m} = \dfrac{2.56\times10^{-29}}{25.6\times10^{-21}}\ \text{m} = 1\times10^{-9}\ \text{m}$$ $\therefore x = 1$

Q41.

A car travels on a circular racetrack of radius 50 m, which is banked at an angle θ. If the car travels at a speed of 10 m s⁻¹, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be 10 m s⁻², the value of θ is :

  • A. tan⁻¹(1/√5) ✓
  • B. tan⁻¹(2/√5)
  • C. tan⁻¹(2√5)
  • D. tan⁻¹(√3/2)

Solution: Minimum wear ⇒ no friction ⇒ tan θ = v²/(rg) = 100/(50×10) = 1/5, θ = tan⁻¹(1/5). ⚠ None of the printed options is exactly tan⁻¹(1/5) — verify the option values against the official key.

Q42.

Three identical p-n junction diodes D₁, D₂ and D₃ are connected across a battery as shown in the figure. If the widths of the depletion regions of D₁, D₂ and D₃ are W₁, W₂ and W₃, respectively, then the correct option is :

  • A. W₁ > W₂ > W₃
  • B. W₃ = W₁ > W₂
  • C. W₃ > W₂ > W₁ ✓
  • D. W₂ > W₁ = W₃

Solution: $W_3 > W_2 > W_1$ The 5 V battery's left terminal is positive and its right terminal negative. $D_1$: p-side (triangle) to +, n-side to −, so it is forward biased and $W_1$ decreases (smallest). $D_2$: its branch has an open switch or break, so no voltage is applied across it; it is unbiased and $W_2$ is unchanged (middle). $D_3$: p-side to −, n-side to +, so it is reverse biased and $W_3$ increases (largest). $\therefore W_3 > W_2 > W_1$

Q43.

The following table presents parts of the electromagnetic spectrum (P–S) and their corresponding major applications (I–IV). The correct option is :

  • A. P-I, Q-II, R-III, S-IV
  • B. P-IV, Q-I, R-II, S-III
  • C. P-II, Q-I, R-IV, S-III ✓
  • D. P-II, Q-IV, R-III, S-I

Solution: P-II, Q-I, R-IV, S-III P-II: microwave ovens warm food, as the microwave frequency matches the resonant frequency of water molecules. Q-I: UV lamps kill germs, so they are used in water purifiers. R-IV: gamma rays are used in medicine to destroy cancer cells. S-III: radio waves are used for AM and FM radio communication.

Q44.

Bob B of mass m at rest is hanging vertically from the ceiling via a massless string of length 10 m, as shown in the figure. Point mass A of mass m travelling horizontally with speed 10 m s⁻¹ hits bob B elastically. The bob B rises to a height h after the collision. Taking the acceleration due to gravity g = 10 m s⁻² and neglecting the size of the bob, the value of h is :

  • A. 8
  • B. 7
  • C. 5 ✓
  • D. 2.5

Solution: $h = 5$ m In an elastic collision of equal masses the velocities are exchanged: A stops and B moves off with $10\ \text{m s}^{-1}$. $$\therefore h = \dfrac{v^2}{2g} = \dfrac{(10)^2}{2\times10}\ \text{m} = 5\ \text{m}$$

Q45.

An ideal gas is made of polyatomic molecules. Each of the molecules has three translational and three rotational modes. If the number of vibrational modes is f and the ratio of heat capacities C_P/C_V is 8/7, then the value of f is :

  • A. 4 ✓
  • B. 2
  • C. 3
  • D. 1

Solution: $f = 4$ Each vibrational mode has kinetic and potential energy, so it counts as two degrees of freedom: $d = 3 + 3 + 2f$. $$\gamma = \dfrac{C_P}{C_V} = \dfrac{d+2}{d} = \dfrac{8}{7} \Rightarrow d = 14$$ $$\therefore 6 + 2f = 14 \Rightarrow f = 4$$

Q46.

For the following reaction sequence, choose the correct option :

  • A. If P is the sodium salt of a carboxylic acid, Q is a primary alcohol
  • B. P and Q are aromatic compounds
  • C. If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light ✓
  • D. Both P and Q are carbonyl compounds

Solution: P gives benzoic acid on acidification; Q (chloroform) gives poisonous phosgene in air and light Friedel–Crafts acylation of benzene (CH₃COCl/AlCl₃) gives acetophenone, C₆H₅COCH₃, whose methyl ketone group undergoes the haloform reaction with NaOCl: P = sodium benzoate (C₆H₅COONa), Q = chloroform (CHCl₃). P on acidification gives benzoic acid; Q in air and light is slowly oxidised to the poisonous gas phosgene: 2CHCl₃ + O₂ → 2COCl₂ + 2HCl. (A) Q is chloroform, not a primary alcohol. (B) Chloroform is aliphatic, not aromatic. (D) Chloroform has no carbonyl group, so P and Q are not both carbonyl compounds.

Q47.

Given below are two statements : Statement-I : [Fe(ox)₃]³⁻ is chiral. Statement-II : trans-[Cr(C₂O₄)₂(H₂O)₂]⁻ is chiral. (Given : ox = HOOC-COOH). In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect ✓
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Statement-I is correct but Statement-II is incorrect Statement-I is correct: [Fe(ox)₃]³⁻, an octahedral complex with three didentate ligands, has non-superimposable mirror images (d and l forms), so it is chiral. Statement-II is incorrect: trans-[Cr(C₂O₄)₂(H₂O)₂]⁻ has a plane of symmetry, so it is achiral; only its cis isomer is chiral.

Q48.

The following carbocation is stabilized by the interaction of the empty p orbital with :

  • A. filled σ and filled π orbitals ✓
  • B. empty σ* and empty π* orbitals
  • C. empty σ* and filled π orbitals
  • D. filled σ and empty π* orbitals

Solution: Filled σ and filled π orbitals A carbocation is stabilised by donation of electrons from filled orbitals into its empty p orbital. These are the filled σ (C–H) orbitals (hyperconjugation) and the filled π system (conjugation).

Q49.

In potash alum, the ratio of K⁺ and SO₄²⁻ ions is :

  • A. 1:2 ✓
  • B. 2:1
  • C. 2:3
  • D. 3:2

Solution: 1 : 2 Potash alum is KAl(SO₄)₂·12H₂O, with one K⁺ and two SO₄²⁻ per formula unit, so K⁺ : SO₄²⁻ = 1 : 2.

Q50.

The correct statement about peptides and proteins is :

  • A. Tertiary structure of proteins has two or more polypeptide subunits
  • B. Only the proteins having a quaternary structure are biologically active
  • C. In β-pleated sheet structures, peptide chains are held together by intermolecular hydrogen bonds ✓
  • D. In α-helices, the polypeptide chain is twisted into a left-handed screw through intramolecular hydrogen bonds

Solution: In β-pleated sheets, peptide chains are held together by intermolecular hydrogen bonds In a β-pleated sheet the peptide chains are stretched out and laid side by side, held together by intermolecular hydrogen bonds. (A) Two or more polypeptide subunits make the quaternary structure; the tertiary structure is the overall folding of a polypeptide chain. (B) Many proteins made of a single chain, such as myoglobin, are biologically active without a quaternary structure. (D) The α-helix is a right-handed screw, not a left-handed one.

Q51.

The numbers 17.0145 and 21.0235 were rounded to three figures after the decimal point. The resulting numbers, respectively, are :

  • A. 17.014 and 21.023
  • B. 17.015 and 21.023
  • C. 17.014 and 21.024 ✓
  • D. 17.015 and 21.024

Solution: 17.014 and 21.024 When the digit to be removed is 5, the preceding digit is left unchanged if it is even and increased by one if it is odd. 17.0145: the digit before 5 is 4 (even), so it stays: 17.014. 21.0235: the digit before 5 is 3 (odd), so it becomes 4: 21.024.

Q52.

The correct order of solubility of the given salts in water at 298 K (K_sp: AgBr = 5.0×10⁻¹³, Zn(OH)₂ = 1.0×10⁻¹⁵, Hg₂Cl₂ = 1.3×10⁻¹⁸) is :

  • A. Hg₂Cl₂ > Zn(OH)₂ > AgBr
  • B. Hg₂Cl₂ > AgBr > Zn(OH)₂
  • C. Zn(OH)₂ > Hg₂Cl₂ > AgBr
  • D. Zn(OH)₂ > AgBr > Hg₂Cl₂ ✓

Solution: Zn(OH)₂ > AgBr > Hg₂Cl₂ AgBr ⇌ Ag⁺ + Br⁻: $K_{sp} = s^2$, so $s = \sqrt{5.0\times10^{-13}} = 7.1\times10^{-7}\ \text{mol L}^{-1}$. Zn(OH)₂ ⇌ Zn²⁺ + 2OH⁻: $K_{sp} = 4s^3$, so $s = \left(\dfrac{1.0\times10^{-15}}{4}\right)^{1/3} = 6.3\times10^{-6}\ \text{mol L}^{-1}$. Hg₂Cl₂ ⇌ Hg₂²⁺ + 2Cl⁻: $K_{sp} = 4s^3$, so $s = \left(\dfrac{1.3\times10^{-18}}{4}\right)^{1/3} = 6.9\times10^{-7}\ \text{mol L}^{-1}$. ∴ Zn(OH)₂ > AgBr > Hg₂Cl₂

Q53.

Among the following options, the correct trend in the electron gain enthalpy is :

  • A. F > Cl > Br > I
  • B. Br > Cl > F > I
  • C. Cl > F > Br > I ✓
  • D. I > Br > Cl > F

Solution: Cl > F > Br > I Down group 17 the added electron goes into a shell farther from the nucleus, so electron gain enthalpy becomes less negative: Cl > Br > I. F is the exception: the added electron enters the small n = 2 shell and is strongly repelled by the electrons already there, so F is less negative than Cl. ∴ Order of negative electron gain enthalpy: Cl > F > Br > I (−349, −328, −325 and −295 kJ mol⁻¹).

Q54.

Given below are two statements : Assertion A : For an ideal solution formed by mixing liquids P and Q, ΔmixH = 0 and ΔmixV = 0. Reason R : No interactions occur between P and Q. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both A and R are correct and R is the correct explanation of A
  • B. Both A and R are correct but R is NOT the correct explanation of A
  • C. A is correct but R is not correct ✓
  • D. A is not correct but R is correct

Solution: Assertion A is correct but Reason R is not correct A is correct: an ideal solution obeys Raoult's law over the entire range of concentration, with $\Delta_{\text{mix}}H = 0$ and $\Delta_{\text{mix}}V = 0$. R is not correct: P–Q interactions are present; in an ideal solution they are nearly equal to the P–P and Q–Q interactions, not absent.

Q55.

The amino acid that gives a red-to-blood-red colour on treating with sodium nitroprusside is :

  • A. leucine
  • B. threonine
  • C. methionine ✓
  • D. serine

Solution: methionine The sodium nitroprusside test detects sulphur-containing amino acids, and methionine, CH₃SCH₂CH₂CH(NH₂)COOH, is the only one of the four with sulphur.

Q56.

The standard electrode potential (E°) for the half-cell reaction Fe³⁺ + e⁻ → Fe²⁺ at 298 K is : (Given : E°(Fe³⁺/Fe) = −0.04 V and E°(Fe²⁺/Fe) = −0.44 V at 298 K)

  • A. +0.40 V
  • B. +0.76 V ✓
  • C. −0.48 V
  • D. +0.92 V

Solution: +0.76 V Electrode potentials cannot be added, but Gibbs energies can: $\Delta G^\circ = -nFE^\circ$. Fe³⁺ + 3e⁻ → Fe: $\Delta G_1^\circ = -3F(-0.04) = +0.12F$ Fe²⁺ + 2e⁻ → Fe: $\Delta G_2^\circ = -2F(-0.44) = +0.88F$ Fe³⁺ + e⁻ → Fe²⁺ is the first minus the second: $\Delta G_3^\circ = 0.12F - 0.88F = -0.76F$ $$\therefore E_3^\circ = -\dfrac{\Delta G_3^\circ}{1 \times F} = +0.76\ \text{V}$$

Q57.

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO₄ solution. If the volume of KMnO₄ solution required to reach the end point is 10 mL, the strength of the KMnO₄ solution is :

  • A. 0.10 M ✓
  • B. 0.20 M
  • C. 0.25 M
  • D. 0.15 M

Solution: 0.10 M In acid medium 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O, so 2 mol MnO₄⁻ react with 5 mol oxalate. Oxalic acid taken $= 0.25\ \text{M} \times 10\ \text{mL} = 2.5\ \text{mmol}$, so MnO₄⁻ needed $= \tfrac{2}{5} \times 2.5 = 1.0\ \text{mmol}$. ∴ Molarity of KMnO₄ $= \dfrac{1.0\ \text{mmol}}{10\ \text{mL}} = 0.10\ \text{M}$

Q58.

According to crystal field theory, the correct order of ligands with respect to their decreasing order of field strength is :

  • A. CO > NH₃ > H₂O > Cl⁻ ✓
  • B. Cl⁻ > NH₃ > H₂O > CO
  • C. CO > H₂O > NH₃ > Cl⁻
  • D. Cl⁻ > H₂O > NH₃ > CO

Solution: CO > NH₃ > H₂O > Cl⁻ In the spectrochemical series field strength increases in the order Cl⁻ < H₂O < NH₃ < CO, so the decreasing order is CO > NH₃ > H₂O > Cl⁻.

Q59.

Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K. The values of ΔS_system and ΔS_surroundings are : (R is the universal gas constant)

  • A. ΔS_system = 0, ΔS_surroundings = 0
  • B. ΔS_system = 4.606R, ΔS_surroundings = −4.606R
  • C. ΔS_system = 0, ΔS_surroundings = 4.606R
  • D. ΔS_system = 4.606R, ΔS_surroundings = 0 ✓

Solution: $\Delta S_{\text{sys}} = 4.606R,\ \Delta S_{\text{surr}} = 0$ In free expansion $p_{\text{ext}} = 0$, so $w = 0$; for an ideal gas at constant temperature $\Delta U = 0$, so $q = 0$. No heat is exchanged with the surroundings: $\Delta S_{\text{surr}} = \dfrac{q_{\text{surr}}}{T} = 0$. Entropy is a state function, so $\Delta S_{\text{sys}}$ is found along a reversible isothermal path: $\Delta S_{\text{sys}} = nR\ln\dfrac{V_2}{V_1}$. $$\therefore \Delta S_{\text{sys}} = 2 \times 2.303\,R\log\dfrac{100}{10} = 4.606R$$

Q60.

2A → B is a zero-order reaction, where k = 1.0 mol L⁻¹ min⁻¹. If the initial concentration of A is 2 M, then the time taken to complete 75% of the reaction will be :

  • A. 1.5 min
  • B. 0.75 min ✓
  • C. 1.0 min
  • D. 2.0 min

Solution: 0.75 min For 2A → B, rate $= -\tfrac{1}{2}\dfrac{d[\text{A}]}{dt} = k$ (zero order), so $[\text{A}]_t = [\text{A}]_0 - 2kt$. 75% complete leaves $[\text{A}]_t = 0.25 \times 2 = 0.5\ \text{M}$, so $[\text{A}]_0 - [\text{A}]_t = 1.5\ \text{M}$. $$\therefore t = \dfrac{[\text{A}]_0 - [\text{A}]_t}{2k} = \dfrac{1.5}{2 \times 1.0}\ \text{min} = 0.75\ \text{min}$$

Q61.

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Generally, 3d transition metals have high melting points. Reason R : Involvement of 3d-electrons in addition to 4s-electrons in the interatomic metallic bonding. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both A and R are correct and R is the correct explanation of A ✓
  • B. Both A and R are correct and R is NOT the correct explanation of A
  • C. A is correct but R is not correct
  • D. A is not correct but R is correct

Solution: Both A and R are correct and R is the correct explanation of A The high melting points of transition metals are due to the $(n-1)d$ electrons taking part in interatomic metallic bonding along with the $ns$ electrons. ∴ A and R are both correct, and R explains A.

Q62.

For a salt XY, which is a strong electrolyte, the plot of Λ_m versus √c has a slope of −90.0 S cm² mol⁻³ᐟ² L¹ᐟ² at 298 K. At 0.01 M concentration of XY, the value of Λ_m is 145.0 S cm² mol⁻¹. The limiting molar conductivity Λ°_m of Y⁻ will be : (Given : Λ°_m(X⁺) = 74.0 S cm² mol⁻¹)

  • A. 80.0 ✓
  • B. 100.0
  • C. 90.0
  • D. 76.0

Solution: $80.0\ \text{S cm}^2\ \text{mol}^{-1}$ For a strong electrolyte $\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$; the slope of the plot is $-A$, so $A = 90.0$. $\Lambda_m^\circ$(XY) $= 145.0 + 90.0 \times \sqrt{0.01} = 145.0 + 9.0 = 154.0\ \text{S cm}^2\ \text{mol}^{-1}$ ∴ by Kohlrausch's law, $\Lambda_m^\circ$(Y⁻) $= 154.0 - 74.0 = 80.0\ \text{S cm}^2\ \text{mol}^{-1}$

Q63.

The amount of carbon dioxide evolved upon complete combustion of 116 g of n-butane is : (Given : atomic mass in amu, H = 1, C = 12 and O = 16)

  • A. 352 g ✓
  • B. 322 g
  • C. 176 g
  • D. 362 g

Solution: 352 g 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O, so 1 mol of butane gives 4 mol of CO₂. Moles of butane $= \dfrac{116\ \text{g}}{58\ \text{g mol}^{-1}} = 2\ \text{mol}$ ∴ mass of CO₂ $= 2 \times 4 \times 44\ \text{g} = 352\ \text{g}$

Q64.

For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is 6.64 kJ mol⁻¹ and R = 8.3 J K⁻¹ mol⁻¹, the temperature at which the rate constant becomes e² min⁻¹ is :

  • A. 125 K
  • B. 150 K
  • C. 200 K ✓
  • D. 250 K

Solution: 200 K Arrhenius equation: $\ln k = \ln A - \dfrac{E_a}{RT}$ $\dfrac{E_a}{R} = \dfrac{6640\ \text{J mol}^{-1}}{8.3\ \text{J K}^{-1}\ \text{mol}^{-1}} = 800\ \text{K}$ From the plot's intercept $\ln A = 6$, and $k = e^2\ \text{min}^{-1}$ gives $\ln k = 2$. ∴ $2 = 6 - \dfrac{800}{T}$, so $T = 200\ \text{K}$

Q65.

Given below are two statements : Statement-I : Heating NaCl with concentrated H₂SO₄ and MnO₂ results in oxidation of Mn. Statement-II : Heating NaI with concentrated H₂SO₄ and MnO₂ results in reduction of Mn. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect
  • D. Statement-I is incorrect but Statement-II is correct ✓

Solution: Statement-I is incorrect but Statement-II is correct MnO₂ acts as the oxidising agent: Mn goes from +4 to +2, and the halide ion is oxidised to the halogen. Statement-I is incorrect: 4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂, so Mn is reduced, not oxidised. Statement-II is correct: with NaI, I⁻ is oxidised to I₂ and Mn is reduced from +4 to +2.

Q66.

Among the species given below, the spin-only magnetic moment is highest for : (Given : Atomic number of Ti = 22, Mn = 25, Fe = 26 and Co = 27)

  • A. [Mn(CN)₆]³⁻ ✓
  • B. [Fe(CN)₆]³⁻
  • C. [Co(NH₃)₆]³⁺
  • D. [Ti(H₂O)₆]³⁺

Solution: [Mn(CN)₆]³⁻ Spin-only moment $\mu = \sqrt{n(n+2)}\ \text{BM}$, so it is highest for the largest number of unpaired electrons $n$. [Mn(CN)₆]³⁻: Mn³⁺ is 3d⁴ and strong-field CN⁻ gives $t_{2g}^4$, so $n = 2$ and $\mu = \sqrt{8} = 2.83\ \text{BM}$. (B) [Fe(CN)₆]³⁻: Fe³⁺ is 3d⁵, low spin $t_{2g}^5$, $n = 1$. (C) [Co(NH₃)₆]³⁺: Co³⁺ is 3d⁶, low spin $t_{2g}^6$, $n = 0$. (D) [Ti(H₂O)₆]³⁺: Ti³⁺ is 3d¹, $n = 1$.

Q67.

The lanthanide ion having four unpaired electrons is : (Given : Atomic numbers of Ce = 58, Nd = 60, Tb = 65 and Ho = 67)

  • A. Nd³⁺
  • B. Ce³⁺
  • C. Tb³⁺
  • D. Ho³⁺ ✓

Solution: Ho³⁺ Ho³⁺ is [Xe]4f¹⁰: ten electrons in seven 4f orbitals leave four unpaired. (A) Nd³⁺ is 4f³: three unpaired. (B) Ce³⁺ is 4f¹: one unpaired. (C) Tb³⁺ is 4f⁸: six unpaired.

Q68.

The formula of tetraammineaquachloridocobalt(III) chloride is :

  • A. [Co(NH₃)₄Cl₂]Cl·H₂O
  • B. [Co(NH₃)₄]Cl₃·H₂O
  • C. [Co(NH₃)₄(H₂O)Cl]Cl
  • D. [Co(NH₃)₄(H₂O)Cl]Cl₂ ✓

Solution: [Co(NH₃)₄(H₂O)Cl]Cl₂ Tetraammine, aqua and chlorido mean four NH₃, one H₂O and one Cl⁻ bound to Co³⁺ inside the coordination sphere. Charge on [Co(NH₃)₄(H₂O)Cl] $= +3 + 4(0) + 1(0) + 1(-1) = +2$, so two Cl⁻ ions are needed outside the sphere. (A) has two chlorido ligands and the water outside the sphere. (B) has neither the aqua nor the chlorido ligand inside the sphere. (C) has the right complex ion but only one Cl⁻ for its +2 charge.

Q69.

Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure. w₁, w₂, w₃ and w₄ represent the work done (in calories) in the processes 1, 2, 3 and 4 respectively. ΔU₁ and ΔU₂ are the changes in the internal energy for processes 2 and 4 respectively. [Use R = 2 cal K⁻¹ mol⁻¹]. The correct option is :

  • A. w₁ + w₃ = −2T₁ ln(V₂/V₁) − 2T₂ ln(V₄/V₃) ✓
  • B. w₂ + w₄ = ΔU₂ − ΔU₁
  • C. w₁ + w₂ = 2T₁ ln(V₂/V₁)
  • D. w₁ + w₂ + w₃ + w₄ = 0

Solution: $w_1 + w_3 = -2T_1 \ln(V_2/V_1) - 2T_2 \ln(V_4/V_3)$ Processes 1 and 3 are reversible isothermal: $w = -nRT\ln\dfrac{V_f}{V_i}$, with $n = 1$ and $R = 2\ \text{cal K}^{-1}\ \text{mol}^{-1}$. Processes 2 and 4 are adiabatic: $q = 0$, so $w_2 = \Delta U_1$ and $w_4 = \Delta U_2$. ∴ $w_1 + w_3 = -2T_1\ln\dfrac{V_2}{V_1} - 2T_2\ln\dfrac{V_4}{V_3}$

Q70.

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The first ionization enthalpy of O is lower than that of N and F. Reason R : The loss of an electron from O leads to a stable half-filled p orbital. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both A and R are correct and R is the correct explanation of A
  • B. Both A and R are correct but R is NOT the correct explanation of A ✓
  • C. A is correct but R is not correct
  • D. A is not correct but R is correct

Solution: Both A and R are correct but R is not the correct explanation of A A is correct: O (2p⁴) has two electrons paired in one 2p orbital, and their repulsion makes its first ionization enthalpy lower than that of half-filled N (2p³). F has a higher nuclear charge than O, so its first ionization enthalpy is higher than that of O. R is correct: O⁺ is 2p³, a stable half-filled configuration. But R does not explain why O is lower than F, so R is not the correct explanation of A.

Q71.

Consider the following statements about the solutions formed by mixing two liquids : A. An ideal solution thus formed obeys Raoult's law throughout the composition range. B. Mixture of chloroform and acetone shows negative deviation from Raoult's law. C. Mixture of aniline and phenol shows positive deviation from Raoult's law. The correct option is :

  • A. A and B only ✓
  • B. B and C only
  • C. A only
  • D. A and C only

Solution: Statements A and B only Statement A is true: an ideal solution obeys Raoult's law over the entire range of concentration. Statement B is true: chloroform forms a hydrogen bond with acetone, so the mixture shows negative deviation. Statement C is false: phenol and aniline hydrogen bond with each other more strongly than with their own kind, so the deviation is negative, not positive. (B) and (D) include the false statement C, and (C) leaves out the true statement B.

Q72.

One of the products formed in the following reaction is :

  • A. Dicyclohexylamine
  • B. 2-Aminobiphenyl derivative
  • C. Bicyclohexyl
  • D. Cyclohexane ✓

Solution: Cyclohexane Grignard reagents react with any source of proton, even amines, to give the hydrocarbon. The N–H proton of cyclohexylamine converts cyclohexylmagnesium bromide to cyclohexane: C₆H₁₁MgBr + C₆H₁₁NH₂ → C₆H₁₂ + C₆H₁₁NHMgBr.

Q73.

The correct statement is :

  • A. Boron has a maximum covalency of four ✓
  • B. Beryllium has three valence orbitals
  • C. Magnesium has a maximum covalency of four
  • D. Aluminium has five valence orbitals

Solution: Boron has a maximum covalency of four Boron has only four valence orbitals (2s and three 2p) and no d orbitals, so its maximum covalency is four, as in BF₄⁻. (B) Beryllium has four valence orbitals, 2s and three 2p, not three. (C) Magnesium can use d orbitals and reach a coordination number of six, so four is not its maximum. (D) Aluminium has 3s, three 3p and vacant 3d orbitals, more than five, so its covalency can go beyond four, as in [AlF₆]³⁻.

Q74.

A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to N ⇌ D. At 60 °C, the concentrations of both N and D are equal at equilibrium, and the standard enthalpy change of denaturation is 666 kJ mol⁻¹. The standard entropy change (ΔS°) in kJ K⁻¹ mol⁻¹ of the protein upon denaturation at 60 °C is closest to :

  • A. 2.0 ✓
  • B. 2000.0
  • C. 333.0
  • D. 11.1

Solution: $2.0\ \text{kJ K}^{-1}\ \text{mol}^{-1}$ At equilibrium [N] = [D], so $K = 1$ and $\Delta G^\circ = -RT\ln K = 0$. $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0 \Rightarrow \Delta S^\circ = \dfrac{\Delta H^\circ}{T}$ ∴ $\Delta S^\circ = \dfrac{666\ \text{kJ mol}^{-1}}{(60 + 273)\ \text{K}} = 2.0\ \text{kJ K}^{-1}\ \text{mol}^{-1}$

Q75.

Match the species in List-I with their geometry in List-II and choose the correct answer :

  • A. A-II, B-III, C-I, D-II
  • B. A-III, B-IV, C-I, D-II ✓
  • C. A-III, B-I, C-II, D-IV
  • D. A-III, B-II, C-I, D-IV

Solution: A-III, B-IV, C-I, D-II A-III: PCl₅ has five bond pairs on P (sp³d), so it is trigonal bipyramidal. B-IV: BrF₅ has five bond pairs and one lone pair on Br (sp³d²), so it is square pyramidal. C-I: BF₄⁻ has four bond pairs on B (sp³), so it is tetrahedral. D-II: in [Ni(CN)₄]²⁻, Ni²⁺ (3d⁸) with strong-field CN⁻ is dsp² hybridised, so it is square planar.

Q76.

Given below are two statements : Statement-I : trans-But-2-ene upon treatment with Br₂ in CCl₄ gives the following product. Statement-II : cis-But-2-ene upon treatment with alkaline KMnO₄ gives the following product. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect ✓
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Statement-I is correct but Statement-II is incorrect Br₂ in CCl₄ adds anti, through a cyclic bromonium ion; cold alkaline KMnO₄ (Baeyer's reagent) adds the two –OH groups syn. Statement-I is correct: anti addition to trans-but-2-ene gives meso-2,3-dibromobutane, and the drawn Newman projection is meso (a plane of symmetry with the methyls eclipsed). Statement-II is incorrect: syn addition to cis-but-2-ene should give the meso diol, but the structure drawn is the threo (racemic) form.

Q77.

Consider the following reaction sequence and choose the correct option :

  • A. K and L are geometrical isomers ✓
  • B. K and L are enantiomers
  • C. M and N are geometrical isomers
  • D. M and N are stereoisomers

Solution: K and L are geometrical isomers Geometrical isomers have the same structural formula but differ in the spatial arrangement of groups about a double bond or a ring.

Q78.

The complex which has both facial and meridional isomers is : (Given : py = pyridine and en = H₂N-CH₂-CH₂-NH₂)

  • A. [Cr(py)₃Cl₃] ✓
  • B. [Cr(H₂O)₆]³⁺
  • C. [Co(NH₃)₄(H₂O)₂]³⁺
  • D. [Ni(en)₂(H₂O)₂]²⁺

Solution: [Cr(py)₃Cl₃] Facial and meridional isomers occur in octahedral complexes of the type [Ma₃b₃]; [Cr(py)₃Cl₃] is of this type. (B) [Cr(H₂O)₆]³⁺ has six identical ligands, so it has no geometrical isomers. (C) [Co(NH₃)₄(H₂O)₂]³⁺, of the type [Ma₄b₂], shows only cis and trans isomers. (D) [Ni(en)₂(H₂O)₂]²⁺ shows cis and trans isomers (the cis form is optically active), but no fac and mer forms.

Q79.

Identify the reactions which give aniline as the major product. Choose the correct answer from the options given below :

  • A. A and B only
  • B. B and D only ✓
  • C. A and C only
  • D. C and D only

Solution: Reactions B and D only Reaction B: Hofmann bromamide degradation of benzamide gives aniline. Reaction D: acid hydrolysis of acetanilide gives aniline. Reaction A gives benzylamine, and NaBH₄ does not reduce nitrobenzene to aniline.

Q80.

Match the vitamins in List-I with their sources in List-II and choose the correct answer :

  • A. A-IV, B-III, C-I, D-II
  • B. A-IV, B-I, C-II, D-III ✓
  • C. A-III, B-I, C-II, D-IV
  • D. A-IV, B-II, C-III, D-I

Solution: A-IV, B-I, C-II, D-III A-IV: vitamin A is found in carrots, fish liver oil, butter and milk. B-I: vitamin B₁₂ is found in meat, fish, egg and curd. C-II: vitamin E is found in vegetable oils such as wheat germ oil and sunflower oil. D-III: vitamin K is found in green leafy vegetables.

Q81.

The correct decreasing order of the oxidation state of the underlined atom in each molecule is :

  • A. B₂O₁₀ > SO₃ > H₂O
  • B. N₂O₅ > Al₂O₃ > H₂S ✓
  • C. P₂O₃ > N₂O₃ > SO₃
  • D. P₂O₅ > Cl₂O₇ > AlH₃

Solution: N₂O₅ > Al₂O₃ > H₂S Oxidation states: N in N₂O₅ is +5, Al in Al₂O₃ is +3 and S in H₂S is −2, so +5 > +3 > −2 is a decreasing order. (A) S in SO₃ is +6, higher than the first atom, so the order is not decreasing. (C) P in P₂O₃ and N in N₂O₃ are both +3, and S in SO₃ is +6, so the order rises. (D) Cl in Cl₂O₇ is +7, higher than P in P₂O₅ (+5).

Q82.

The compound that CANNOT be obtained from the aldol condensation reaction shown below is :

  • A. (1) ✓
  • B. (2)
  • C. (3)
  • D. (4)

Solution: Compound (1), with the C=C inside the ring In an aldol condensation the new C=C joins the two units, between the enolate carbon and the former carbonyl carbon; A has its C=C inside the ring, so it cannot form. The enolate forms only at the –CH₂– α-carbon of 2,2-dimethylcyclopentanone: the other α-carbon carries two methyl groups and no H, and benzaldehyde has no α-H. (B) and (D) Crossed condensation of this enolate with benzaldehyde gives the E and Z products. (C) Self-condensation joins this α-carbon by a C=C to the former carbonyl carbon of a second ketone molecule.

Q83.

Among the following, the compound having conjugated double bonds is :

  • A. hepta-1,3-diene ✓
  • B. hepta-1,4-diene
  • C. hepta-1,5-diene
  • D. hepta-1,6-diene

Solution: Hepta-1,3-diene Conjugated double bonds alternate with a single bond: hepta-1,3-diene has C1=C2–C3=C4. (B) Hepta-1,4-diene: one CH₂ group (C3) separates the double bonds, so they are isolated. (C) Hepta-1,5-diene: two CH₂ groups separate the double bonds. (D) Hepta-1,6-diene: three CH₂ groups separate the double bonds.

Q84.

Given below are two statements : Statement-I : Oxidation of p-nitrotoluene with acidic KMnO₄ gives an acid that is stronger than benzoic acid. Statement-II : Reduction of p-nitrotoluene with Sn/HCl followed by neutralization gives an amine that is more basic than aniline. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct ✓
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Both Statement-I and Statement-II are correct Statement-I is correct: acidic KMnO₄ oxidises –CH₃ to –COOH, giving p-nitrobenzoic acid, and the electron-withdrawing –NO₂ group makes it a stronger acid than benzoic acid. Statement-II is correct: Sn/HCl reduces –NO₂ to –NH₂, giving p-toluidine, and the electron-donating –CH₃ group makes it more basic than aniline.

Q85.

The green paramagnetic species formed by heating KMnO₄ at 513 K is :

  • A. K₂MnO₄ ✓
  • B. Mn₃O₄
  • C. MnO
  • D. KO₂

Solution: K₂MnO₄ (potassium manganate) At 513 K, 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂. In K₂MnO₄, Mn is +6 (3d¹), so the manganate ion is green and paramagnetic with one unpaired electron.

Q86.

Consider the following reaction, and choose the correct option :

  • A. On treating compound P with saturated NaHCO₃ solution, brisk effervescence is observed
  • B. Compound P can be prepared by treating benzene with anhydrous AlCl₃ and CH₃COCl
  • C. On treatment with bromine water, compound P gives a white precipitate
  • D. P is obtained by the hydrogenation of benzoyl chloride with Pd on BaSO₄ ✓

Solution: P is obtained by the hydrogenation of benzoyl chloride with Pd on BaSO₄ The Etard reaction (CrO₂Cl₂ in CS₂, then H₃O⁺) converts toluene to benzaldehyde, P. Benzaldehyde is also made by Rosenmund reduction: benzoyl chloride is hydrogenated over Pd on BaSO₄. (A) Benzaldehyde is not a carboxylic acid, so it gives no CO₂ with NaHCO₃. (B) Benzene with CH₃COCl and anhydrous AlCl₃ gives acetophenone, not benzaldehyde. (C) Bromine water gives a white precipitate with phenol or aniline, not with benzaldehyde.

Q87.

A 1:3 electrolyte in an aqueous solution is :

  • A. [CoCl₂(NH₃)₄]Cl
  • B. [CoCl(NH₃)₅]Cl₂
  • C. [Co(NH₃)₆]Cl₃ ✓
  • D. [Co(NH₃)₃(NO₂)₃]

Solution: [Co(NH₃)₆]Cl₃ [Co(NH₃)₆]Cl₃ → [Co(NH₃)₆]³⁺ + 3Cl⁻: one cation and three anions, a 1:3 electrolyte. (A) [CoCl₂(NH₃)₄]Cl gives one Cl⁻ per complex ion: a 1:1 electrolyte. (B) [CoCl(NH₃)₅]Cl₂ gives two Cl⁻: a 1:2 electrolyte. (D) [Co(NH₃)₃(NO₂)₃] is a neutral complex and gives no ions: a non-electrolyte.

Q88.

Consider the following schematic plots of orbital wavefunction (ψ_r) against distance (r) from the nucleus. The figure representing two radial nodes in the orbital is :

  • A. A
  • B. B
  • C. C ✓
  • D. D

Solution: Plot C, which crosses zero twice A radial node is a distance $r$ (other than $r = 0$ or $r \to \infty$) where $\psi_r$ crosses zero and changes sign. C crosses the axis twice (downward, then upward), so it has two radial nodes (3s). (A) never crosses zero: no radial node (1s). (B) crosses once: one radial node (2s). (D) starts at zero at the nucleus and crosses once: one radial node (3p).

Q89.

Arrange the following compounds in the increasing order of polarity : A. CH₃CH₂OCH₂CH₃, B. CH₃CH₂OH, C. CH₃COCH₃, D. CH₃COOH. Choose the correct answer from the options given below :

  • A. A < B < C < D
  • B. C < A < B < D
  • C. C < A < D < B
  • D. A < C < B < D ✓

Solution: Diethyl ether < acetone < ethanol < acetic acid (A < C < B < D) Polarity rises with the polar group present and, most of all, with hydrogen bonding through O–H. A, diethyl ether: weakly polar C–O bonds and no O–H, so it is the least polar. C, acetone: a strongly polar C=O group but no O–H, so it is more polar than the ether but less than an alcohol. B, ethanol: its –OH group forms strong intermolecular hydrogen bonds. D, acetic acid: has both C=O and O–H and forms hydrogen-bonded dimers, so it is the most polar.

Q90.

The highest occupied molecular orbital for Ne₂ is :

  • A. π2p
  • B. σ2p
  • C. π*2p
  • D. σ*2p ✓

Solution: σ*2p For O₂, F₂ and Ne₂ the energy order is $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$. Ne₂ has $2 \times 10 = 20$ electrons, which fill all of them: $2 + 2 + 2 + 2 + 2 + 4 + 4 + 2 = 20$. ∴ the highest occupied molecular orbital is $\sigma^* 2p_z$.

Q91.

The number of vertebrae in a human is :

  • A. 7
  • B. 12
  • C. 26 ✓
  • D. 206

Solution: 26 The vertebral column is formed by 26 serially arranged vertebrae: cervical (7), thoracic (12), lumbar (5), sacral (1, fused) and coccygeal (1, fused).

Q92.

Symbiotic association between fungi and algae are called :

  • A. lichens ✓
  • B. sponges
  • C. mycorrhiza
  • D. chrysophytes

Solution: lichens Fungi form symbiotic associations with algae as lichens and with the roots of higher plants as mycorrhiza.

Q93.

Cell theory was formulated by :

  • A. Schleiden and Schwann ✓
  • B. Robert Brown
  • C. Singer and Nicolson
  • D. Antonie Van Leeuwenhoek

Solution: Schleiden and Schwann Schleiden, a botanist, and Schwann, a zoologist, together formulated the cell theory.

Q94.

Which of the following are characteristics of prokaryotic cells ? (a) Ribosomes are made of 50S and 30S subunits (b) They can have plasmids (c) They contain mesosomes (d) They have peroxisomes. Choose the correct answer from the options given below :

  • A. (b) and (c) only
  • B. (a) and (c) only
  • C. (a), (c) and (d) only
  • D. (a), (b) and (c) only ✓

Solution: (a), (b) and (c) only Prokaryotic cells have 70S ribosomes, can carry plasmids and have mesosomes, but lack membrane-bound organelles, so (a), (b) and (c) are correct and (d) is not. (A) Leaves out (a); the 70S prokaryotic ribosome is made of 50S and 30S subunits. (B) Leaves out (b); many bacteria have plasmids, small circular DNA outside the genomic DNA. (C) Includes (d), but peroxisomes are membrane-bound organelles found only in eukaryotic cells.

Q95.

Which of the following is not a part of the human central neural system ?

  • A. Arachnoid
  • B. Dura mater
  • C. Pia mater
  • D. Pericardium ✓

Solution: pericardium The pericardium is the double-walled membranous bag enclosing the heart; it is not part of the central neural system. (A) Arachnoid is the very thin middle layer of the cranial meninges covering the brain. (B) Dura mater is the outer layer of the cranial meninges. (C) Pia mater is the inner layer, in contact with the brain tissue.

Q96.

Mitochondrial inner membrane encloses :

  • A. matrix ✓
  • B. cytosol
  • C. mucus
  • D. aqueous humor

Solution: matrix The inner membrane of a mitochondrion encloses the inner compartment, called the matrix.

Q97.

Match List-I with List-II and choose the correct answer from the options given below :

  • A. A-III, B-IV, C-I, D-II
  • B. A-II, B-IV, C-I, D-III ✓
  • C. A-II, B-IV, C-III, D-I
  • D. A-IV, B-III, C-I, D-II

Solution: A-II, B-IV, C-I, D-III A-II: Cristae are the infoldings of the inner mitochondrial membrane. B-IV: Cisternae are the flat, disc-shaped sacs of the Golgi apparatus. C-I: Thylakoids are the flattened membranous sacs in the stroma of the chloroplast. D-III: Phospholipids form the lipid bilayer of the cell membrane.

Q98.

The plastid that stores xanthophyll is known as :

  • A. chloroplast
  • B. chromoplast ✓
  • C. aleuroplast
  • D. amyloplast

Solution: chromoplast Chromoplasts contain fat-soluble carotenoid pigments like carotene and xanthophylls, which give a yellow, orange or red colour.

Q99.

Which of the following statements related to pituitary gland are correct ? (a) It is divided anatomically into adenohypophysis and neurohypophysis (b) It secretes follicle stimulating hormone (c) It secretes melanocyte stimulating hormone (d) It does not secrete prolactin. Choose the correct answer from the options given below :

  • A. (a) and (b) only
  • B. (a), (b) and (c) only ✓
  • C. (a) and (c) only
  • D. (b) and (c) only

Solution: (a), (b) and (c) only The pituitary has an adenohypophysis and a neurohypophysis and secretes FSH, MSH and prolactin, so (a), (b) and (c) are correct and only (d) is wrong. (A) Leaves out (c); MSH is secreted by the pars intermedia. (C) Leaves out (b); FSH is secreted by the pars distalis. (D) Leaves out (a); the gland is divided anatomically into an adenohypophysis and a neurohypophysis.

Q100.

Photorespiration reaction catalyzed by RuBisCO is shown below : RuBP + O₂ → 3-Phosphoglycerate + X. Identify 'X' from the given options :

  • A. Phosphoenolpyruvate
  • B. 2-Phosphoglycolate ✓
  • C. Oxaloacetate
  • D. Malate

Solution: 2-phosphoglycolate In photorespiration RuBisCO binds O₂, and RuBP forms one molecule of 3-phosphoglycerate and one of 2-phosphoglycolate.

Q101.

Mad cow disease is caused by :

  • A. prions ✓
  • B. viroids
  • C. Aspergillus sp.
  • D. Mycoplasma sp.

Solution: prions Bovine spongiform encephalopathy (BSE), commonly called mad cow disease, is caused by prions, agents made of abnormally folded protein.

Q102.

Which pigment has absorption peak at 700 nm in the photosynthetic reaction centre PS I (P700) ?

  • A. Chlorophyll b
  • B. Chlorophyll a ✓
  • C. Xanthophylls
  • D. Carotenoids

Solution: chlorophyll a In PS I the reaction centre chlorophyll a has an absorption peak at 700 nm, hence it is called P700.

Q103.

In water, frogs respire using :

  • A. skin ✓
  • B. buccal cavity
  • C. lungs
  • D. trachea

Solution: skin In water the skin acts as the aquatic respiratory organ (cutaneous respiration); dissolved oxygen is exchanged through it by diffusion.

Q104.

Which of the following represents the correct sequence of arrangement of bones in the lower limb of humans ?

  • A. Femur-tibia-patella-tarsal
  • B. Patella-femur-tibia-tarsal
  • C. Femur-patella-tibia-tarsal ✓
  • D. Femur-tarsal-patella-tibia

Solution: femur → patella → tibia → tarsal The femur (thigh bone) comes first, the patella (knee cap) covers the knee ventrally, the tibia forms the leg and the tarsals form the ankle.

Q105.

Phyllotaxy is the pattern of arrangement of :

  • A. leaves ✓
  • B. flowers
  • C. fruits
  • D. sepals

Solution: leaves Phyllotaxy is the pattern of arrangement of leaves on the stem or branch, usually alternate, opposite or whorled.

Q106.

Match List-I with List-II and choose the correct answer from the options given below :

  • A. A-I, B-II, C-IV, D-III
  • B. A-II, B-I, C-IV, D-III ✓
  • C. A-II, B-I, C-III, D-IV
  • D. A-I, B-I, C-III, D-IV

Solution: A-II, B-I, C-IV, D-III A-II: Starch is the main store of energy in plant tissues. B-I: Antibodies fight infectious agents. C-IV: Concanavalin A is a lectin. D-III: GLUT-4 enables glucose transport into cells.

Q107.

Given below are two statements : Statement-I : When any plane passing through the central axis of the body divides the organism into two identical halves, it is called radial symmetry. Statement-II : In phylum Echinodermata, both adults and larvae are radially symmetrical. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect ✓
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Statement-I is correct but Statement-II is incorrect Statement-I is the definition of radial symmetry, but in echinoderms only the adults are radially symmetrical; the larvae are bilaterally symmetrical. (A) Statement-II is incorrect, as echinoderm larvae are bilaterally symmetrical. (B) Statement-I is correct; coelenterates, ctenophores and adult echinoderms show this body plan. (D) It is the reverse: Statement-I is correct and Statement-II is incorrect.

Q108.

Endomembrane system includes :

  • A. endoplasmic reticulum, Golgi complex, lysosomes and vacuole ✓
  • B. endoplasmic reticulum, chloroplast, peroxisomes and vacuole
  • C. mitochondria, chloroplast, peroxisomes and vacuole
  • D. Golgi complex, chloroplast, peroxisomes and vacuole

Solution: endoplasmic reticulum, Golgi complex, lysosomes and vacuole The endomembrane system includes the endoplasmic reticulum, Golgi complex, lysosomes and vacuoles. Mitochondria, chloroplasts and peroxisomes are not part of it, because their functions are not coordinated with these components.

Q109.

How many molecules of pyruvic acid are produced at the end of glycolysis from 206 molecules of glucose ?

  • A. 206
  • B. 309
  • C. 103
  • D. 412 ✓

Solution: 412 molecules Glycolysis breaks one molecule of glucose into two molecules of pyruvic acid. ∴ Pyruvic acid formed $= 2 \times 206 = 412$ molecules.

Q110.

Which of the following plant growth regulators is used as herbicide ?

  • A. 2,4-D ✓
  • B. Kinetin
  • C. Abscisic acid
  • D. Gibberellin

Solution: 2,4-D 2,4-D, a synthetic auxin, is widely used to kill dicotyledonous weeds; it does not affect mature monocotyledonous plants.

Q111.

Given below are two statements : Statement-I : In gymnosperms, the male and female gametophytes remain within the sporangia. Statement-II : In gymnosperms, seeds are not covered. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct ✓
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Both Statement-I and Statement-II are correct Gymnosperm gametophytes remain within the sporangia and their seeds are not covered, so both statements are correct. (B) Neither statement is incorrect. (C) Statement-II is also correct: the ovules are not enclosed by an ovary wall, so the seeds are naked. (D) Statement-I is also correct: unlike bryophytes and pteridophytes, the gametophytes are not free-living.

Q112.

Match List-I with List-II and choose the correct answer from the options given below :

  • A. A-I, B-III, C-II, D-IV
  • B. A-III, B-II, C-I, D-IV
  • C. A-II, B-I, C-IV, D-III
  • D. A-II, B-IV, C-I, D-III ✓

Solution: A-II, B-IV, C-I, D-III A-II: Spherical bacteria are cocci. B-IV: Rod-shaped bacteria are bacilli. C-I: Comma-shaped bacteria are vibrio. D-III: Spiral bacteria are spirilla.

Q113.

Which of the following are characteristic features of Solanaceae family ? (a) Flowers are bisexual and actinomorphic (b) Calyx have five sepals and are united (c) Androecium have five stamens and are epipetalous (d) Ovary is inferior. Choose the correct answer from the options given below :

  • A. (a), (b) and (c) only ✓
  • B. (b) only
  • C. (a) and (b) only
  • D. (b) and (c) only

Solution: (a), (b) and (c) only Solanaceae flowers are bisexual and actinomorphic, with five united sepals and five epipetalous stamens, but the ovary is superior, so (d) is wrong. (B) Leaves out (a) and (c), which are both correct. (C) Leaves out (c); the five stamens are epipetalous. (D) Leaves out (a); the flowers are bisexual and actinomorphic.

Q114.

Select the correct sequence of events that led to a gradual understanding of photosynthesis in green plants.

  • A. Absorption spectra of chlorophyll a and b → production of glucose → release of oxygen → role of air
  • B. Role of air → release of oxygen → production of glucose → absorption spectra of chlorophyll a and b ✓
  • C. Release of oxygen → production of glucose → absorption spectra of chlorophyll a and b → role of air
  • D. Production of glucose → release of oxygen → absorption spectra of chlorophyll a and b → role of air

Solution: role of air → release of oxygen → production of glucose → absorption spectra of chlorophyll a and b Priestley (1770) showed the essential role of air in the growth of green plants. Ingenhousz then showed that green parts give off bubbles of oxygen in bright sunlight. Sachs (1854) gave evidence that plants produce glucose, usually stored as starch. Engelmann later described the first action spectrum, which roughly resembles the absorption spectra of chlorophyll a and b.

Q115.

The number of action potentials generated by sino-atrial node (SAN) in a healthy human per minute is :

  • A. 28-30
  • B. 70-75 ✓
  • C. 100-110
  • D. 120-140

Solution: 70–75 per minute The SAN generates the maximum number of action potentials, 70–75 per minute, and initiates the rhythmic contraction of the heart, so it is called the pacemaker.

Q116.

How many turns of Calvin cycle are required for the formation of three molecules of glucose ?

  • A. 1
  • B. 3
  • C. 6
  • D. 18 ✓

Solution: 18 turns Each turn of the Calvin cycle fixes one CO₂, so one glucose (six carbons) needs 6 turns. ∴ Turns for three glucose $= 3 \times 6 = 18$.

Q117.

Which of the following statements is incorrect ?

  • A. Blood coagulates in response to an injury
  • B. Blood clot consists of fibrins
  • C. Fibrin is produced from fibrinogen
  • D. Fibrinogen is produced from fibrin ✓

Solution: Fibrinogen is produced from fibrin (the incorrect statement) Thrombin converts fibrinogen into fibrin, so the reverse, fibrinogen produced from fibrin, is incorrect. (A) An injury or trauma does start coagulation, by making platelets release clotting factors. (B) A clot is indeed mainly a network of fibrin threads trapping dead and damaged formed elements. (C) Fibrin is indeed formed from fibrinogen, so this statement is correct.

Q118.

Match List-I with List-II and choose the correct answer from the options given below :

  • A. A-II, B-III, C-IV, D-V, E-I
  • B. A-III, B-I, C-II, D-IV, E-V
  • C. A-III, B-II, C-V, D-IV, E-I
  • D. A-III, B-V, C-II, D-IV, E-I ✓

Solution: A-III, B-V, C-II, D-IV, E-I A-III: Mango (Mangifera indica) belongs to the family Anacardiaceae. B-V: Its genus is Mangifera. C-II: Its class is Dicotyledonae. D-IV: Its phylum (division) is Angiospermae. E-I: Its order is Sapindales.

Q119.

Arrange the following taxonomic categories in ascending order : (a) Genus (b) Class (c) Order (d) Phylum (e) Family (f) Kingdom (g) Species. Choose the correct answer from the options given below :

  • A. (g), (a), (e), (c), (b), (d), (f) ✓
  • B. (a), (c), (e), (b), (d), (g), (f)
  • C. (g), (c), (a), (b), (e), (d), (f)
  • D. (g), (a), (c), (b), (e), (d), (f)

Solution: (g), (a), (e), (c), (b), (d), (f) Ascending order runs from the lowest category to the highest: species → genus → family → order → class → phylum → kingdom. ∴ In the given labels this is (g), (a), (e), (c), (b), (d), (f).

Q120.

Match List-I with List-II and choose the correct answer from the options given below :

  • A. A-IV, B-IV, C-I, D-III
  • B. A-IV, B-II, C-III, D-I
  • C. A-IV, B-III, C-II, D-I
  • D. A-IV, B-II, C-I, D-III ✓

Solution: A-IV, B-II, C-I, D-III A-IV: In pea the ovules form two rows on a ridge along the ventral suture (marginal). B-II: In tomato the ovules are attached to an axial placenta in a multilocular ovary (axile). C-I: In Argemone the ovules develop on the inner wall of the ovary (parietal). D-III: In primrose the ovules are borne on a central axis with no septa (free central).

Q121.

Sphenopsida class belongs to :

  • A. bryophytes
  • B. angiosperms
  • C. gymnosperms
  • D. pteridophytes ✓

Solution: pteridophytes Pteridophytes are classified into Psilopsida, Lycopsida, Sphenopsida (Equisetum) and Pteropsida.

Q122.

Which of the following statements regarding photorespiration are correct ? (a) Do not occur in C3 plants (b) CO₂ is consumed and O₂ is generated (c) Phosphoglycolate is formed (d) No synthesis of ATP and NADPH. Choose the correct answer from the options given below :

  • A. (a) and (d) only
  • B. (c) and (d) only ✓
  • C. (b) and (d) only
  • D. (a) and (c) only

Solution: (c) and (d) only In photorespiration RuBP binds O₂ to form one PGA and one phosphoglycolate (c); the pathway makes no ATP or NADPH (d). (A) and (D) include (a), but photorespiration occurs in C3 plants; it does not occur in C4 plants. (C) includes (b), but in photorespiration O₂ is consumed and CO₂ is released.

Q123.

Smooth endoplasmic reticulum :

  • A. has ribosomes attached to its surface
  • B. is the major site for the synthesis of lipids ✓
  • C. is actively involved in protein synthesis
  • D. is a site for the synthesis of carbohydrates

Solution: the major site for the synthesis of lipids The smooth ER is the major site for synthesis of lipids; in animal cells lipid-like steroidal hormones are also made there. Ribosomes on the surface and active protein synthesis are features of the rough ER.

Q124.

Which one of the following statements is incorrect ?

  • A. α-cells of pancreas secrete glucagon
  • B. δ-cells of pancreas secrete insulin ✓
  • C. Glucagon stimulates glycogenolysis
  • D. β-cells of pancreas secrete insulin

Solution: δ-cells of pancreas secrete insulin δ-cells of the pancreas secrete somatostatin, not insulin. (A) α-cells do secrete glucagon. (C) Glucagon does stimulate glycogenolysis in liver cells, raising blood sugar. (D) β-cells do secrete insulin.

Q125.

Genus represents :

  • A. an individual plant or animal
  • B. a population of plants and animals
  • C. a group of closely related species ✓
  • D. a group of closely related families

Solution: a group of closely related species A genus is a group of related species that share more characters with one another than with species of other genera.

Q126.

Which of the following is not a prokaryote ?

  • A. Bacteria
  • B. Blue green algae
  • C. Mycoplasma
  • D. Fungi ✓

Solution: Fungi Fungi are eukaryotes; bacteria, blue-green algae (cyanobacteria) and mycoplasma are prokaryotes of kingdom Monera.

Q127.

Which of the following plant growth regulators promotes internode elongation prior to flowering in cabbage ?

  • A. Abscisic acid
  • B. Gibberellin ✓
  • C. Indole butyric acid
  • D. Ethephon

Solution: Gibberellin Gibberellins promote bolting, the internode elongation just before flowering, in beet, cabbage and other plants with rosette habit.

Q128.

The correct sequence of adult cell cycle phases is :

  • A. G1-G2-S-M
  • B. G1-M-G2-S
  • C. G1-S-G2-M ✓
  • D. S-M-G2-G1

Solution: G1-S-G2-M Interphase passes through G1 (gap 1), S (DNA synthesis) and G2 (gap 2), and is followed by the M phase (mitosis).

Q129.

Match List-I with List-II and choose the correct answer from the options given below :

  • A. A-II, B-III, C-I ✓
  • B. A-II, B-I, C-III
  • C. A-III, B-II, C-I
  • D. A-III, B-I, C-II

Solution: A-II, B-III, C-I A-II: fusion of protoplasms of two gametes is plasmogamy. B-III: fusion of two nuclei is karyogamy. C-I: meiosis in the zygote gives haploid spores.

Q130.

Given below are two statements : Statement-I : The class name Reptilia refers to creeping or crawling mode of locomotion. Statement-II : All organisms belonging to Reptilia have three chambered heart. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect ✓
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Statement-I is correct but Statement-II is incorrect Statement-I is correct: the name comes from the Latin repere or reptum, to creep or crawl. Statement-II is incorrect: the heart of reptiles is usually three-chambered but four-chambered in crocodiles.

Q131.

Given below are two statements : Statement-I : Chromosomes are fully condensed at the end of prophase I. Statement-II : Meiosis I resembles mitosis. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect ✓
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Statement-I is correct but Statement-II is incorrect Statement-I is correct: in diakinesis, the last stage of prophase I, the chromosomes are fully condensed. Statement-II is incorrect: meiosis II, not meiosis I, resembles a normal mitosis; meiosis I separates homologous chromosomes.

Q132.

Which of the following is not a characteristic of chordates ?

  • A. Presence of notochord
  • B. Central nervous system is dorsal
  • C. Presence of gills ✓
  • D. Presence of post anal part

Solution: Presence of gills Chordates have a pharynx perforated by gill slits; gills themselves are found only in some chordates, such as fishes. (A), (B) and (D) are fundamental chordate features: a notochord, a dorsal hollow nerve cord and a post-anal part (tail).

Q133.

Length of the stem at time 0 is 20 cm. The arithmetic growth rate is 30 cm per day. What is the length of the stem at the end of the 7th day ?

  • A. 50 cm
  • B. 170 cm
  • C. 230 cm ✓
  • D. 460 cm

Solution: 230 cm In arithmetic growth the length increases by a constant amount per unit time: $L_t = L_0 + rt$. Length after 7 days, $L_7 = 20\ \text{cm} + (30\ \text{cm day}^{-1}) \times 7\ \text{days}$ $\therefore L_7 = 20 + 210 = 230\ \text{cm}$

Q134.

Arrange the following elements in descending order of their contribution to the percentage weight of the human body : (a) Carbon (b) Oxygen (c) Hydrogen (d) Nitrogen. Choose the correct answer from the options given below :

  • A. (b), (a), (c), (d) ✓
  • B. (c), (a), (b), (d)
  • C. (b), (a), (d), (c)
  • D. (a), (b), (c), (d)

Solution: Oxygen > carbon > hydrogen > nitrogen, i.e. (b), (a), (c), (d) By weight the human body is about 65% oxygen, 18.5% carbon, 9.5% hydrogen and 3.3% nitrogen. Most of the body's weight is water, and one-ninth of the weight of water is hydrogen, so hydrogen outweighs nitrogen.

Q135.

In frogs, the number of pairs of cranial nerves arising from the brain are :

  • A. 6
  • B. 9
  • C. 10 ✓
  • D. 12

Solution: 10 pairs Ten pairs of cranial nerves arise from the brain of a frog.

Q136.

Which of the following is used as a clot buster ?

  • A. Streptokinase ✓
  • B. Penicillin
  • C. Cyclosporin A
  • D. Statins

Solution: Streptokinase Streptokinase, produced by the bacterium Streptococcus and modified by genetic engineering, is used as a clot buster to remove clots from blood vessels after myocardial infarction.

Q137.

The inactive form of Bt toxin is converted to the active form in the insect gut :

  • A. due to alkaline pH ✓
  • B. due to acidic pH
  • C. by proteases
  • D. by nucleases

Solution: due to alkaline pH Bt toxin exists as an inactive protoxin; in the insect gut the alkaline pH solubilises the crystals and converts it into the active toxin.

Q138.

Given below are two statements : Statement-I : Down's syndrome is caused by the absence of one of the X-chromosomes. Statement-II : Turner's syndrome is caused by the presence of an additional copy of the chromosomes. In light of the above statements, choose the correct answer from the options given below :

  • A. Both Statement-I and Statement-II are correct
  • B. Both Statement-I and Statement-II are incorrect ✓
  • C. Statement-I is correct but Statement-II is incorrect
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Both Statement-I and Statement-II are incorrect Statement-I is incorrect: Down's syndrome is caused by an additional copy of chromosome 21 (trisomy 21). Statement-II is incorrect: Turner's syndrome is caused by the absence of one of the X chromosomes (45, X0).

Q139.

Which of the following disease is not sexually transmitted ?

  • A. Syphilis
  • B. Tuberculosis ✓
  • C. Gonorrhoea
  • D. Genital warts

Solution: Tuberculosis Syphilis, gonorrhoea and genital warts are sexually transmitted infections; tuberculosis is a bacterial disease spread by droplets in the air.

Q140.

Sperm motility is due to :

  • A. flagellar movement ✓
  • B. ciliary movement
  • C. amoeboid movement
  • D. muscular movement

Solution: flagellar movement The sperm tail is a flagellum, and flagellar movement helps in the swimming of spermatozoa.

Q141.

Natural selection can lead to : (a) stabilisation (b) genetic drift (c) directional change (d) disruption. Choose the correct answer from the options given below :

  • A. (a) only
  • B. (a), (c) and (d) only ✓
  • C. (a), (b), (c) and (d) only
  • D. (a) and (c) only

Solution: (a), (c) and (d) only Natural selection can lead to stabilisation, directional change or disruption of the distribution of a character. (A) and (D) leave out disruption, and (A) also leaves out directional change. (C) includes genetic drift, which is a change in allele frequency by chance, not by natural selection.

Q142.

The method of directly injecting sperm into ovum in assisted reproductive technology is called :

  • A. Gamete intra fallopian transfer (GIFT)
  • B. Zygote intra fallopian transfer (ZIFT)
  • C. Intra cytoplasmic sperm injection (ICSI) ✓
  • D. Embryo transfer (ET)

Solution: Intra cytoplasmic sperm injection (ICSI) In ICSI a sperm is directly injected into the ovum to form an embryo in the laboratory.

Q143.

Which of the following structure is not a part of the male reproductive system ?

  • A. Rete testis
  • B. Epididymis
  • C. Vasa efferentia
  • D. Infundibulum ✓

Solution: Infundibulum The infundibulum is the funnel-shaped part of the fallopian tube closer to the ovary; rete testis, vasa efferentia and epididymis are male sex accessory ducts.

Q144.

Arrange the following in descending order of number of species in the Amazonian rain forest : (a) Plants (b) Birds (c) Fishes (d) Invertebrates (e) Mammals. Choose the correct answer from the options given below :

  • A. (c) > (b) > (d) > (e) > (a)
  • B. (d) > (a) > (c) > (b) > (e) ✓
  • C. (e) > (b) > (a) > (c) > (d)
  • D. (b) > (a) > (c) > (d) > (e)

Solution: Invertebrates > plants > fishes > birds > mammals, i.e. (d) > (a) > (c) > (b) > (e) The Amazonian rain forest has more than 1,25,000 species of invertebrates, 40,000 of plants, 3,000 of fishes, 1,300 of birds and 427 of mammals.

Q145.

Given below are two statements : Statement-I : Ovulation is caused by LH surge leading to rupture of Graafian follicles. Statement-II : Graafian follicle remaining after ovulation transforms into corpus luteum and secretes large amount of estrogen. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect ✓
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Statement-I is correct but Statement-II is incorrect Statement-I is correct: the LH surge at mid-cycle induces rupture of the Graafian follicle and release of the ovum. Statement-II is incorrect: the corpus luteum formed from the remaining follicle secretes large amounts of progesterone, not estrogen.

Q146.

Which of the following are primary consumers in a food chain ?

  • A. Parasites
  • B. Predators
  • C. Herbivores ✓
  • D. Carnivores

Solution: Herbivores Herbivores feed directly on producers (plants), so they are the primary consumers.

Q147.

A population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele A is 0.1, the frequency of AA is :

  • A. 0.01 ✓
  • B. 0.02
  • C. 0.10
  • D. 0.99

Solution: 0.01 At Hardy-Weinberg equilibrium $p^2 + 2pq + q^2 = 1$, where $p$ is the frequency of allele A and $p^2$ the frequency of AA. $\therefore$ Frequency of AA, $p^2 = (0.1)^2 = 0.01$

Q148.

Match List-I with List-II and choose the correct answer from the options given below :

  • A. A-III, B-IV, C-II, D-I
  • B. A-III, B-IV, C-I, D-II ✓
  • C. A-II, B-IV, C-I, D-III
  • D. A-IV, B-III, C-I, D-II

Solution: A-III, B-IV, C-I, D-II A-III: excess growth hormone in adults causes acromegaly. B-IV: LH induces ovulation of fully mature Graafian follicles. C-I: vasopressin (ADH) stimulates reabsorption of water and electrolytes by the distal tubules of the kidney. D-II: oxytocin stimulates vigorous contraction of the uterus during childbirth.

Q149.

The opening between the right atrium and the right ventricle is guarded by :

  • A. bicuspid valve
  • B. tricuspid valve ✓
  • C. semilunar valve
  • D. sino-atrial node

Solution: tricuspid valve The right atrioventricular opening is guarded by the tricuspid valve, which has three cusps; the bicuspid (mitral) valve guards the left one.

Q150.

Sponges exchange O₂ with CO₂ by :

  • A. simple diffusion over their entire body surfaces ✓
  • B. moist cuticle
  • C. tracheal tubes
  • D. gills

Solution: simple diffusion over their entire body surfaces Lower invertebrates such as sponges, coelenterates and flatworms exchange O₂ with CO₂ by simple diffusion over their entire body surface.

Q151.

How many theca are present in each lobe of a typical bilobed angiosperm anther ?

  • A. 2 ✓
  • B. 6
  • C. 8
  • D. 12

Solution: 2 theca in each lobe A typical angiosperm anther is bilobed, each lobe having two theca, i.e., the anther is dithecous.

Q152.

Muscle contraction is initiated by a signal sent from the central nervous system by the release of :

  • A. acetyl choline ✓
  • B. acetyl coenzyme A
  • C. cyclic guanine monophosphate
  • D. cyclic adenine monophosphate

Solution: acetyl choline A signal from the CNS reaches the neuromuscular junction through a motor neuron and releases the neurotransmitter acetyl choline, which generates an action potential in the sarcolemma.

Q153.

Which of the following statements about lac-operon is correct ?

  • A. Gene i is constitutively expressed ✓
  • B. Lactose activates repressor to bind to the operator
  • C. Genes i, z, y and a share a single common promoter
  • D. Galactose can act as an inducer of lac operon

Solution: Gene i is constitutively expressed The repressor protein is synthesised all the time (constitutively) from the regulatory gene i. (B) Lactose, the inducer, inactivates the repressor, so the repressor cannot bind to the operator. (C) Only the structural genes z, y and a share one promoter; the regulatory gene i has its own. (D) The inducer is lactose (or allolactose), not galactose.

Q154.

Which of the following in the female gametophyte of an angiosperm helps in guiding the pollen tube for fertilizing the egg ?

  • A. Antipodals
  • B. Synergids ✓
  • C. Central cells
  • D. Polar nuclei

Solution: Synergids The synergids have cellular thickenings at the micropylar tip, the filiform apparatus, which guide the pollen tube into the synergid.

Q155.

Which of the following plant produces non-albuminous seeds ?

  • A. Wheat
  • B. Maize
  • C. Barley
  • D. Pea ✓

Solution: Pea Non-albuminous seeds have no residual endosperm, as it is completely consumed during embryo development (e.g., pea); wheat, maize and barley seeds retain part of it and are albuminous.

Q156.

If the diploid chromosome number is 36, what would be the chromosome number in its endosperm ?

  • A. 18
  • B. 36
  • C. 54 ✓
  • D. 72

Solution: 54 The endosperm is triploid ($3n$), formed by triple fusion of a male gamete with the two polar nuclei. $2n = 36 \Rightarrow n = 18$ ∴ Chromosome number in the endosperm $= 3n = 3 \times 18 = 54$

Q157.

Which of the following statements about the reabsorption process in Henle's loop are correct ? (a) The descending limb of Henle's loop is permeable to water but almost impermeable to electrolytes (b) Urine gets concentrated in Henle's loop (c) Reabsorption of Na⁺ and water takes place in Henle's loop (d) Active or passive transport of electrolytes occurs in the ascending limb of Henle's loop. Choose the correct answer from the options given below :

  • A. (a) and (b) only
  • B. (b), (c) and (d) only
  • C. (a), (b) and (d) only
  • D. (a), (b), (c) and (d) only ✓

Solution: All four: (a), (b), (c) and (d) All four are correct: water leaves the descending limb, which is almost impermeable to electrolytes, so the filtrate is concentrated; electrolytes such as Na⁺ leave the ascending limb actively or passively. (A) Leaves out (c) and (d), which are also correct. (B) Leaves out (a), which is also correct. (C) Leaves out (c), which is also correct.

Q158.

Which of the following is the correct order of arrangement of the vertebrae in the vertebral column from the head to toe ?

  • A. Cervical vertebra, thoracic vertebra, sacrum, lumbar vertebra
  • B. Sacrum, lumbar vertebra, thoracic vertebra, cervical vertebra
  • C. Cervical vertebra, lumbar vertebra, thoracic vertebra, sacrum
  • D. Cervical vertebra, thoracic vertebra, lumbar vertebra, sacrum ✓

Solution: Cervical, thoracic, lumbar, sacrum Starting from the skull, the vertebral column has cervical (7), thoracic (12), lumbar (5), sacral (1, fused) and coccygeal (1, fused) regions.

Q159.

Which of the following is not evidence for evolution ?

  • A. Convergent evolution of traits like wings of birds and butterflies
  • B. Paleontological evidence from fossil records
  • C. Embryological support for evolution as proposed by Ernst Haeckel ✓
  • D. Divergent evolution of anatomical structures such as forelimbs

Solution: Embryological support proposed by Ernst Haeckel Haeckel's embryological support was disproved by Karl Ernst von Baer, who noted that embryos never pass through the adult stages of other animals. (A) Wings of birds and butterflies are analogous structures, a result of convergent evolution, and are evidence for evolution. (B) Fossils in rock layers of different ages show that life forms have varied over time. (D) Homologous forelimbs of whales, bats, cheetah and human show divergent evolution and indicate common ancestry.

Q160.

Given below are two statements : Statement-I : Modern Homo sapiens arose in Australia and moved across continents. Statement-II : Homo sapiens arose around 75000 to 10000 years ago. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect
  • D. Statement-I is incorrect but Statement-II is correct ✓

Solution: Statement-I is incorrect but Statement-II is correct Homo sapiens arose in Africa, not Australia, and moved across continents, so Statement-I is incorrect. Modern Homo sapiens arose during the ice age between 75,000 and 10,000 years ago, so Statement-II is correct.

Q161.

Consider a population of 10 million cells. Given the per-capita birth rate is 0.002 (per unit time) and the per-capita death rate is 0.002 (per unit time), the expected number of cells after 10 generations is :

  • A. 1 million
  • B. 5 million
  • C. 10 million ✓
  • D. 100 million

Solution: 10 million Exponential growth: $\dfrac{dN}{dt} = (b - d)N = rN$ Intrinsic rate of natural increase, $r = b - d = 0.002 - 0.002 = 0$ ∴ $\dfrac{dN}{dt} = 0$, so the population stays at 10 million whatever the number of generations.

Q162.

During PCR, primers bind to the DNA strands in the ______ step.

  • A. denaturation
  • B. extension
  • C. annealing ✓
  • D. ligation

Solution: annealing In the annealing step the primers bind to complementary regions of the single-stranded DNA, after denaturation and before extension by DNA polymerase.

Q163.

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The logistic growth model of populations is considered more realistic than the exponential growth model. Reason R : Resources are finite. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both A and R are correct and R is the correct explanation of A ✓
  • B. Both A and R are correct but R is not the correct explanation of A
  • C. A is correct but R is incorrect
  • D. A is not correct but R is correct

Solution: Both A and R are correct and R is the correct explanation of A No population in nature has unlimited resources; resources are finite and become limiting sooner or later. ∴ Growth levels off at the carrying capacity (K), so the logistic model is more realistic and R explains A.

Q164.

Adaptive radiation in placental mammals and Australian Marsupials leading to similarity between distinct species is an example of :

  • A. divergent evolution
  • B. convergent evolution ✓
  • C. founder effect
  • D. genetic drift

Solution: convergent evolution More than one adaptive radiation in an isolated geographical area, producing similar forms such as the placental wolf and the Tasmanian wolf (a marsupial), is convergent evolution.

Q165.

Which of the following are secondary lymphoid organs ? (a) Tonsils (b) Bone marrow (c) Spleen (d) Thymus. Choose the correct answer from the options given below :

  • A. (a) and (b) only
  • B. (a) and (c) only ✓
  • C. (b) and (d) only
  • D. (a) and (d) only

Solution: Tonsils and spleen, (a) and (c) only Mature lymphocytes migrate to the secondary lymphoid organs: spleen, lymph nodes, tonsils, Peyer's patches and appendix. (A) Bone marrow is a primary lymphoid organ. (C) Bone marrow and thymus are both primary lymphoid organs, where immature lymphocytes differentiate. (D) Thymus is a primary lymphoid organ.

Q166.

Which of the following hormone is not secreted by human placenta ?

  • A. hCG
  • B. Estrogen
  • C. Progesterone
  • D. LH ✓

Solution: LH The placenta acts as an endocrine tissue and secretes hCG, hPL, estrogens and progestogens; LH is secreted by the anterior pituitary.

Q167.

Which of the following enzymes synthesizes precursor mRNA ?

  • A. RNA polymerase I
  • B. RNA polymerase II ✓
  • C. RNA polymerase III
  • D. DNA polymerase

Solution: RNA polymerase II RNA polymerase II transcribes the precursor of mRNA, the heterogeneous nuclear RNA (hnRNA).

Q168.

Given below are two statements : Statement-I : Plasmids are autonomously replicating DNA. Statement-II : Plasmids are extrachromosomal DNA. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both Statement-I and Statement-II are correct ✓
  • B. Both Statement-I and Statement-II are incorrect
  • C. Statement-I is correct but Statement-II is incorrect
  • D. Statement-I is incorrect but Statement-II is correct

Solution: Both Statement-I and Statement-II are correct Plasmids are autonomously replicating, circular, extra-chromosomal DNA, so both statements are correct.

Q169.

For a person with blood group O, which of the following is not a possible combination of the parents' blood group genotypes ?

  • A. (1) ✓
  • B. (2)
  • C. (3)
  • D. (4)

Solution: The pair in which a parent has no $i$ allele Blood group O is the genotype $ii$, so the child must get one $i$ allele from each parent. ∴ No pair with an $I^AI^A$, $I^BI^B$ or $I^AI^B$ parent can have an O child, since that parent has no $i$ allele to pass on.

Q170.

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Forelimbs of human and bats are homologous. Reason R : Forelimbs of humans and bats have similar anatomical structure. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Solution: Both A and R are true and R is the correct explanation of A Human and bat forelimbs have the same pattern of bones (humerus, radius, ulna, carpals, metacarpals, phalanges) though they perform different functions. Structures with similar anatomy but different functions are homologous, so R explains A.

Q171.

Colostrum, secreted by the mother during the initial days of lactation, is abundant in :

  • A. IgG
  • B. IgM
  • C. IgA ✓
  • D. IgD

Solution: IgA The yellowish colostrum secreted in the initial days of lactation has abundant IgA antibodies, which protect the infant (passive immunity).

Q172.

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Abingdon tortoise in Galapagos islands became extinct within a decade after goats were introduced. Reason R : Goats were more efficient at browsing than the Abingdon tortoise. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both A and R are correct and R is the correct explanation of A ✓
  • B. Both A and R are correct but R is not the correct explanation of A
  • C. A is correct but R is incorrect
  • D. A is not correct but R is correct

Solution: Both A and R are correct and R is the correct explanation of A Introduced goats competed with the Abingdon tortoise for the same food plants. The goats' greater browsing efficiency led to the tortoise's extinction within a decade, so R explains A.

Q173.

The covering of ovum at ovulation is :

  • A. endometrium
  • B. zona radiata
  • C. zona pellucida ✓
  • D. chorion

Solution: zona pellucida The secondary oocyte forms a new membrane around it, the zona pellucida, which covers the ovum released at ovulation.

Q174.

Match List-I with List-II and choose the correct answer from the options given below :

  • A. A-III, B-IV, C-II, D-I
  • B. A-I, B-II, C-III, D-IV
  • C. A-II, B-I, C-IV, D-III
  • D. A-III, B-I, C-II, D-IV ✓

Solution: A-III, B-I, C-II, D-IV A-III: in competition both species are harmed (−, −). B-I: in predation the predator benefits and the prey is harmed (+, −). C-II: in mutualism both species benefit (+, +). D-IV: in commensalism one species benefits and the other is neither harmed nor benefited (+, 0).

Q175.

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : In an experiment, Mendel observed that the F1 progeny plants are all tall, with none being dwarf. Reason R : Stem height is a contrasting trait, with tall being dominant and dwarf being recessive. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both A and R are correct and R is the correct explanation of A ✓
  • B. Both A and R are correct but R is not the correct explanation of A
  • C. A is correct but R is not correct
  • D. A is not correct but R is correct

Solution: Both A and R are correct and R is the correct explanation of A A cross of tall (TT) and dwarf (tt) plants gives only Tt plants in the F1. Tall (T) is dominant over dwarf (t), so every F1 plant is tall and none is dwarf; R explains A.

Q176.

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : In recombinant DNA technology, lysozyme is used for disrupting bacterial cells while cellulase is for plant cells. Reason R : Isolation of genetic material needs disruption of cells. In light of the above statements, choose the most appropriate answer from the options given below :

  • A. Both A and R are correct and R is the correct explanation of A ✓
  • B. Both A and R are correct but R is not the correct explanation of A
  • C. A is correct but R is not correct
  • D. A is not correct but R is correct

Solution: Both A and R are correct and R is the correct explanation of A DNA is enclosed within membranes, so cells must be broken open to isolate it. This is done with enzymes: lysozyme for bacteria, cellulase for plant cells and chitinase for fungi; R explains A.

Q177.

Which of the following is used as an effective sedative and painkiller for treating post-surgery patients ?

  • A. Interferon
  • B. Antibiotics
  • C. Morphine ✓
  • D. Anti-retroviral drugs

Solution: Morphine Morphine, an opioid extracted from the latex of the poppy plant Papaver somniferum, is a very effective sedative and painkiller, useful in patients who have undergone surgery.

Q178.

Which of the following statements are correct ? (a) Energy flow from producers to consumers is unidirectional (b) Energy pyramid can never be inverted (c) Transfer of energy follows the 1% law. Choose the correct answer from the options given below :

  • A. (a), (b) and (c)
  • B. (a) and (b) only ✓
  • C. (a) and (c) only
  • D. (b) and (c) only

Solution: Statements (a) and (b) only Energy flows one way, from producers to consumers, and the pyramid of energy is always upright because some energy is lost as heat at each step. (A) Statement (c) is wrong: only about 10 per cent of the energy passes to the next trophic level (10 per cent law). (C) Includes the wrong statement (c) and leaves out (b). (D) Includes the wrong statement (c) and leaves out (a).

Q179.

Which of the following statements is correct about Plasmodium ?

  • A. Reproduces sexually in liver cells
  • B. Reproduces sexually in RBCs
  • C. Gametocytes develop in mosquito gut
  • D. Fertilization takes place in mosquito gut ✓

Solution: Fertilisation takes place in the mosquito gut The female Anopheles takes up gametocytes with its blood meal, and fertilisation and development take place in the mosquito's stomach (gut). (A) In liver cells the parasite reproduces asexually. (B) In RBCs the parasite reproduces asexually. (C) Gametocytes, the sexual stages, develop in human RBCs, not in the mosquito gut.

Q180.

Match List-I with List-II and choose the correct answer from the options given below :

  • A. A-II, B-I, C-IV, D-III ✓
  • B. A-I, B-II, C-IV, D-III
  • C. A-III, B-IV, C-I, D-II
  • D. A-IV, B-I, C-III, D-II

Solution: A-II, B-I, C-IV, D-III A-II: transformation is the introduction of a piece of DNA into a host bacterium. B-I: a cloning site is a recognition site for a restriction enzyme, where alien DNA is linked. C-IV: antibiotic resistance genes act as selectable markers to pick out the transformants. D-III: ori is the sequence from where replication starts.

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