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Re-NEET 2026 question paper with answers and solutions (code 60)

Updated 9 November 2026By the Testline NEET team

This is the full Re-NEET 2026 question paper (NEET-UG re-exam held on 21 June 2026), booklet code 60, in English. It has all 180 questions with the answer from NTA's final answer key and a worked solution for each.

  • 180 questions, 720 marks. Physics 45, Chemistry 45, Botany 45 and Zoology 45. All questions are compulsory.
  • Marking: +4 for a correct answer, −1 for a wrong one, 0 if left blank.
  • Dropped question: NTA dropped Q43 (Physics) in the final key, so every candidate got 4 marks for it.
  • Solutions are written by the Testline NEET team. The answers follow NTA's final key.

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Physics questions (Q1–45)

Q1. Alternating Current

An ac voltage V=220sin⁡(2×103t) VoltV = 220 \sin(2 \times 10^3 t)\text{ Volt} is applied to a series LCR circuit. Then the current amplitude in this circuit is :
(Given : L=10 mHL = 10\text{ mH}, C=25μFC = 25\mu\text{F}, R=100 ΩR = 100\ \Omega)

  1. 11.0 A11.0\text{ A}
  2. 22.0 A22.0\text{ A}
  3. 2.2 A2.2\text{ A}
  4. 5.5 A5.5\text{ A}
Show answer and solution

Answer: option (3)

ω=2×103 rad s−1\omega=2\times10^3\ \text{rad s}^{-1}, so XL=ωL=20 ΩX_L=\omega L=20\ \Omega and XC=1/(ωC)=20 ΩX_C=1/(\omega C)=20\ \Omega. The circuit is at resonance, Z=R=100 ΩZ=R=100\ \Omega, and I0=V0/Z=220/100=2.2 AI_0=V_0/Z=220/100=2.2\ \text{A}.

Q2. Kinetic Theory

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

  1. nA=14nBn_A = \frac{1}{4} n_B
  2. nA=12nBn_A = \frac{1}{2} n_B
  3. nA=nBn_A = n_B
  4. nA=2nBn_A = 2 n_B
Show answer and solution

Answer: option (2)

Mean free path λ∝1nd2\lambda\propto\frac{1}{n d^2}. With λA=λB/2\lambda_A=\lambda_B/2 and dA=2dBd_A=2d_B: 14nA=12nB\frac{1}{4n_A}=\frac{1}{2n_B}, so nA=12nBn_A=\frac{1}{2}n_B.

Q3. Oscillations

A cylindrical cork of uniform density floats in a liquid of density ρ1\rho_1. 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 ρ2\rho_2, then the similar oscillation has time period 2T. The value of ρ2/ρ1\rho_2 / \rho_1 is :

  1. 1/21/2
  2. 1/41/4
  3. 44
  4. 22
Show answer and solution

Answer: option (2)

For a floating cylinder the restoring force is ρAgx\rho A g x, so T=2πm/(ρAg)∝1/ρT=2\pi\sqrt{m/(\rho A g)}\propto 1/\sqrt{\rho}. Doubling TT needs ρ2=ρ1/4\rho_2=\rho_1/4.

Q4. Oscillations

Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is m kgm\text{ kg} and the spring constant is k Nm−1k\text{ Nm}^{-1}. At a given instant, the extension of the spring is xx meter and the speed of the particle is v ms−1v\text{ ms}^{-1}. On the x−vx - v plane, if the graph of vv as a function of xx is a circle, then the correct option is :

  1. k=m2k = m^2
  2. k=mk = \sqrt{m}
  3. k=1mk = \frac{1}{m}
  4. k=mk = m
Show answer and solution

Answer: option (4)

Energy conservation gives 12kx2+12mv2=E\frac{1}{2}kx^2+\frac{1}{2}mv^2=E, an ellipse in the xx-vv plane. It becomes a circle only when the coefficients are equal, i.e. k=mk=m (numerically).

Q5. Thermodynamics

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (γ=5/3\gamma=5/3) decreases from 60 K60\text{ K} to 50 K50\text{ 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\text{R} = 8.3\text{ J mol}^{-1}\text{ K}^{-1})

  1. 124.5 J124.5\text{ J}
  2. 166 J166\text{ J}
  3. 41.5 J41.5\text{ J}
  4. 83 J83\text{ J}
Show answer and solution

Answer: option (1)

Adiabatic work by the gas W=nR(T1−T2)γ−1=1×8.3×102/3=124.5 JW=\frac{nR(T_1-T_2)}{\gamma-1}=\frac{1\times8.3\times10}{2/3}=124.5\ \text{J}.

Q6. Electromagnetic Waves

The following table presents the part of the electromagnetic spectrum and their corresponding major applications.

Part of the electromagnetic spectrumApplications
P.MicrowaveI.For purifying the water
Q.UV raysII.For warming the food
R.Gamma raysIII.For AM and FM communication systems
S.Radio waveIV.For treating the Cancer cells

The correct option is :

  1. P-II, Q-I, R-IV, S-III
  2. P-II, Q-IV, R-III, S-I
  3. P-I, Q-II, R-III, S-IV
  4. P-I, Q-IV, R-II, S-III
Show answer and solution

Answer: option (1)

Microwaves heat food (P-II), UV rays sterilise/purify water (Q-I), gamma rays are used to destroy cancer cells (R-IV) and radio waves carry AM/FM broadcasts (S-III).

Q7. Electrostatic Potential and Capacitance

A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius RR, having uniform positive charge density ρ\rho, as shown in the figure. The initial and final positions of the charge are marked by AA and BB at distances 2R2R and 3R3R respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is ρR2nε0\frac{\rho R^2}{n \varepsilon_0}. The value of nn is :
(ε0\varepsilon_0 is the permittivity of vacuum)

Figure for question 7: 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 rho, as shown in the figure. The initial and final positions of the charge are marked by A and B at distances 2R and 3R res
  1. 9
  2. 18
  3. 2
  4. 6
Show answer and solution

Answer: option (2)

The field is conservative, so the work equals V(B)−V(A)V(B)-V(A) for a unit charge, whatever the path. Outside the sphere V=ρR33ε0rV=\frac{\rho R^3}{3\varepsilon_0 r}, giving ∣W∣=ρR23ε0(12−13)=ρR218ε0|W|=\frac{\rho R^2}{3\varepsilon_0}\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\rho R^2}{18\varepsilon_0}, so n=18n=18.

Q8. Dual Nature of Radiation and Matter

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 tt, then variation of the photocurrent II and magnitude of the stopping potential ∣V∣|V| with time is best represented by :

Figure for question 8: 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 :

The four options are in the figure above.

Show answer and solution

Answer: option (3)

Photocurrent is proportional to intensity, so it falls linearly as the power falls. The stopping potential depends only on frequency, so ∣V∣|V| stays constant: graph (3).

Q9. Work, Energy and Power

Bob BB of mass mm at rest is hanging vertically from the ceiling via a massless string of length 10 m10\text{ m}, as shown in the figure. Point mass AA of mass mm travelling horizontally with speed 10 ms−110\text{ ms}^{-1} hits bob BB elastically. The bob BB rises hh meter after the collision. Taking the acceleration due to gravity g=10 ms−2g = 10\text{ ms}^{-2} and neglecting the size of the bob, the value of hh is :

Figure for question 9: Bob B of mass m at rest is hanging vertically from the ceiling via a massless string of length 10text{ m}, as shown in the figure. Point mass A of mass m travelling horizontally with speed 10text{ ms}^{-1} hits bob B elastically. The bob B rises h meter after the collision. Taking the acceleration d
  1. 5
  2. 2.5
  3. 8
  4. 7
Show answer and solution

Answer: option (1)

In an elastic head-on collision between equal masses the velocities are exchanged, so B moves off at 10 ms−110\ \text{ms}^{-1}. It rises h=v2/(2g)=100/20=5 mh=v^2/(2g)=100/20=5\ \text{m} (less than the string length, so the string stays taut).

Q10. Kinetic Theory

An ideal gas is made of polyatomic molecules. Each of the molecules has three translational, three rotational and ff number of vibrational modes. If the ratio of heat capacities Cp/CvC_p/C_v of the gas is 8/78/7, then the value of ff is :

  1. 2
  2. 1
  3. 4
  4. 3
Show answer and solution

Answer: option (3)

Each vibrational mode contributes RR to CvC_v, so Cv=(32+32+f)R=(3+f)RC_v=\left(\frac{3}{2}+\frac{3}{2}+f\right)R=(3+f)R and γ=1+13+f=87\gamma=1+\frac{1}{3+f}=\frac{8}{7}, giving 3+f=73+f=7, f=4f=4.

Q11. Mechanical Properties of Fluids

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 (vv) with the ratio of density of spherical ball (σ\sigma) to density of the liquid (ρ\rho), is best represented by :

Figure for question 11: 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 (sigma) to density of the liquid (rho), is best represented by

The four options are in the figure above.

Show answer and solution

Answer: option (3)

Terminal velocity v=2r2(σ−ρ)g9η∝(σρ−1)v=\frac{2r^2(\sigma-\rho)g}{9\eta}\propto\left(\frac{\sigma}{\rho}-1\right): a straight line that is zero at σ/ρ=1\sigma/\rho=1 and rises linearly after it, as in graph (3).

Q12. Gravitation

In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius R\text{R} is proportional to :

  1. R2\text{R}^2
  2. R3\text{R}^3
  3. R1/2\text{R}^{1/2}
  4. R3/2\text{R}^{3/2}
Show answer and solution

Answer: option (4)

By Kepler's third law T2∝R3T^2\propto R^3, so T∝R3/2T\propto R^{3/2}.

Q13. System of Particles and Rotational Motion

A solid sphere AA of radius RR and mass MM is attached at a point to a smaller solid sphere BB of radius r<Rr < R and mass m<Mm < M. Assume 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 AA is IAI_A and that calculated about a vertical axis passing through the centre of BB is IBI_B. The difference IA−IBI_A - I_B is :

Figure for question 13: 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. Assume 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
  1. (m−M)(R−r)2(m - M)(R - r)^2
  2. 00
  3. (M−m)(R+r)2(M - m)(R + r)^2
  4. (m−M)(R+r)2(m - M)(R + r)^2
Show answer and solution

Answer: option (4)

The centres are R+rR+r apart. IA=25MR2+25mr2+m(R+r)2I_A=\frac{2}{5}MR^2+\frac{2}{5}mr^2+m(R+r)^2 and IB=25MR2+M(R+r)2+25mr2I_B=\frac{2}{5}MR^2+M(R+r)^2+\frac{2}{5}mr^2, so IA−IB=(m−M)(R+r)2I_A-I_B=(m-M)(R+r)^2.

Q14. Nuclei

Consider the following nuclear reaction :
238U→ 234Th+ 4He^{238}\text{U} \rightarrow \,{}^{234}\text{Th} + \,{}^4\text{He}
Take masses of 238U^{238}\text{U}, 234Th^{234}\text{Th} and 4He^4\text{He} as 238.050 u238.050\text{ u}, 234.043 u234.043\text{ u} and 4.003 u4.003\text{ u}, respectively. The Q\text{Q} value for the reaction, in keV\text{keV}, is :

[Given : 1 u=931.5 MeV c−21\text{ u} = 931.5\text{ MeV c}^{-2}]

  1. 37363736
  2. 37403740
  3. 37263726
  4. 37303730
Show answer and solution

Answer: option (3)

Mass defect Δm=238.050−(234.043+4.003)=0.004 u\Delta m=238.050-(234.043+4.003)=0.004\ \text{u}, so Q=0.004×931.5 MeV=3.726 MeV=3726 keVQ=0.004\times931.5\ \text{MeV}=3.726\ \text{MeV}=3726\ \text{keV}.

Q15. Electrostatic Potential and Capacitance

Consider a fixed uniformly charged insulating sphere with radius R\mathrm{R} and total charge +Q+\mathrm{Q}. A point charge −q-\mathrm{q} (q≪Q\mathrm{q} \ll \mathrm{Q}) with mass m\mathrm{m} is released from rest at a distance of 3R3\mathrm{R} from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is :

(ϵ0\epsilon_0 is the permittivity of vacuum, neglect gravitational forces).

  1. Qq3πϵ0mR\sqrt{\frac{\mathrm{Qq}}{3\pi\epsilon_0\mathrm{mR}}}
  2. Qq4πϵ0mR\sqrt{\frac{\mathrm{Qq}}{4\pi\epsilon_0\mathrm{mR}}}
  3. 3Qq4πϵ0mR\sqrt{\frac{3\mathrm{Qq}}{4\pi\epsilon_0\mathrm{mR}}}
  4. 2Qq3πϵ0mR\sqrt{\frac{2\mathrm{Qq}}{3\pi\epsilon_0\mathrm{mR}}}
Show answer and solution

Answer: option (1)

Gain in KE = loss in PE: 12mv2=Qq4πϵ0(1R−13R)=Qq6πϵ0R\frac{1}{2}mv^2=\frac{Qq}{4\pi\epsilon_0}\left(\frac{1}{R}-\frac{1}{3R}\right)=\frac{Qq}{6\pi\epsilon_0R}, so v=Qq3πϵ0mRv=\sqrt{\frac{Qq}{3\pi\epsilon_0 mR}}.

Q16. Current Electricity

Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, +α+\alpha, while the other one has a negative temperature of coefficient, −α-\alpha, as shown in the figure. The current through these circuits are denoted by IAI_A and IBI_B. At initial temperature, the resistance of the two resistors is R0R_0. As the temperature is increased, the correct option that describes the variation of current in these circuits is :

Figure for question 16: Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, +alpha, while the other one has a negative temperature of coefficient, -alpha, as shown in the figure. The current through these circuits are denoted by I_A and I_B. At in
  1. IAI_A increases while IBI_B decreases
  2. both IAI_A and IBI_B remain constant
  3. IAI_A remains constant while IBI_B increases
  4. IAI_A decreases while IBI_B increases
Show answer and solution

Answer: option (3)

In series (A) the total resistance R0(1−αΔT)+R0(1+αΔT)=2R0R_0(1-\alpha\Delta T)+R_0(1+\alpha\Delta T)=2R_0 is unchanged, so IAI_A is constant. In parallel (B) Req=R0(1−α2ΔT2)2R_{eq}=\frac{R_0(1-\alpha^2\Delta T^2)}{2} decreases, so IBI_B increases.

Q17. Electromagnetic Induction

A conducting loop of finite resistance lies on the x−yx-y plane. There is a constant magnetic field in the zz direction. The area of the loop varies with time tt, as A=A0(1+sin⁡ t)\mathrm{A}=\mathrm{A}_0(1+\sin\ t) in appropriate units. The figure that correctly indicates the qualitative behaviour of the power PP dissipated in the loop as a function of time is :

Figure for question 17: 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 t, as mathrm{A}=mathrm{A}_0(1+sin t) in appropriate units. The figure that correctly indicates the qualitative behaviour of the power P dissipate

The four options are in the figure above.

Show answer and solution

Answer: option (4)

EMF =B dA/dt=BA0cos⁡t=B\,dA/dt=BA_0\cos t, so P=ε2/R∝cos⁡2tP=\varepsilon^2/R\propto\cos^2 t: maximum at t=0t=0, falling to zero and rising again periodically, as in figure (4).

Q18. Dual Nature of Radiation and Matter

A photon and an electron, each of 20 eV20\text{ eV} energy, move in free space. The ratio of linear momentum of electron pe\mathrm{p}_e to that of photon pPh,pepPh\mathrm{p}_{\mathrm{Ph}}, \frac{\mathrm{p}_e}{\mathrm{p}_{\mathrm{Ph}}} is :
(Take speed of light =3×108 ms−1=3 \times 10^8\text{ ms}^{-1}, charge of electron =−1.6×10−19 C=-1.6 \times 10^{-19}\text{ C} and mass of electron =9×10−31 kg=9 \times 10^{-31}\text{ kg})

  1. 225225
  2. 275275
  3. 2450\frac{2}{450}
  4. 1250\frac{1}{250}
Show answer and solution

Answer: option (1)

Electron: pe=2mE=2×9×10−31×3.2×10−18=2.4×10−24p_e=\sqrt{2mE}=\sqrt{2\times9\times10^{-31}\times3.2\times10^{-18}}=2.4\times10^{-24} SI. Photon: p=E/c=3.2×10−18/(3×108)≈1.07×10−26p=E/c=3.2\times10^{-18}/(3\times10^8)\approx1.07\times10^{-26} SI. Ratio =225=225.

Q19. Units and Measurement

Consider that σs\sigma_{\mathrm{s}}, kBk_{\mathrm{B}}, bb represent Stefan-Boltzmann constant, Boltzmann constant and Wien's displacement law constant, respectively. The dimension of σskB−1b\sigma_{\mathrm{s}} k_{\mathrm{B}}^{-1} b is :

  1. [L−1T−1K−3][\mathrm{L}^{-1}\mathrm{T}^{-1}\mathrm{K}^{-3}]
  2. [L−1T−1K−4][\mathrm{L}^{-1}\mathrm{T}^{-1}\mathrm{K}^{-4}]
  3. [L−1T−1K−2][\mathrm{L}^{-1}\mathrm{T}^{-1}\mathrm{K}^{-2}]
  4. [L−1K−2][\mathrm{L}^{-1}\mathrm{K}^{-2}]
Show answer and solution

Answer: option (3)

[σs]=[MT−3K−4][\sigma_s]=[\text{MT}^{-3}\text{K}^{-4}], [kB]=[ML2T−2K−1][k_B]=[\text{ML}^2\text{T}^{-2}\text{K}^{-1}], [b]=[LK][b]=[\text{LK}]. So [σskB−1b]=[L−1T−1K−2][\sigma_s k_B^{-1} b]=[\text{L}^{-1}\text{T}^{-1}\text{K}^{-2}].

Q20. Moving Charges and Magnetism

Two infinitely long parallel conducting wires A\mathrm{A} and B\mathrm{B} carry currents I\mathrm{I} and 2I2\mathrm{I}, respectively, in the same direction. The wire A\mathrm{A} has uniform mass per unit length λ\lambda and lies on an insulated floor. The wire B\mathrm{B} is kept fixed at a height h\mathrm{h} above the floor. The minimum magnitude of h\mathrm{h} so that the wire A\mathrm{A} does not rise from the floor is :
[g\mathrm{g} is the acceleration due to gravity and μ0\mu_{0} is the permeability of free space.]

  1. 2μ0I2πλg\frac{2\mu_{0}\mathrm{I}^{2}}{\pi \lambda\mathrm{g}}
  2. 4μ0I2πλg\frac{4\mu_{0}\mathrm{I}^{2}}{\pi \lambda\mathrm{g}}
  3. μ0I22πλg\frac{\mu_{0}\mathrm{I}^{2}}{2\pi \lambda\mathrm{g}}
  4. μ0I2πλg\frac{\mu_{0}\mathrm{I}^{2}}{\pi \lambda\mathrm{g}}
Show answer and solution

Answer: option (4)

Parallel currents attract, so B pulls A upward with force per length μ0I(2I)2πh=μ0I2πh\frac{\mu_0 I(2I)}{2\pi h}=\frac{\mu_0I^2}{\pi h}. A stays down if this is at most λg\lambda g, i.e. h≥μ0I2πλgh\ge\frac{\mu_0I^2}{\pi\lambda g}.

Q21. Ray Optics and Optical Instruments

Which of the following measurements require 'index correction'?

  1. Measurement of focal length of lenses using optical bench
  2. Measurement of speed of sound using resonance tube
  3. Measurement of resistance of a wire using meter bridge
  4. Measurement of gravitational acceleration using simple pendulum
Show answer and solution

Answer: option (1)

On an optical bench the scale distance between the uprights differs from the actual distance between the lens and the needle tips; this difference (index error) must be corrected. The other experiments need no such correction.

Q22. Electromagnetic Induction

Two identical inductors are connected in two different configurations P\mathrm{P} and Q\mathrm{Q}, where a time varying current I(t)I(t) is flowing, as shown in the figure. The induced emf between points aa and bb for configuration P\mathrm{P} is EP\mathrm{E}_{\mathrm{P}} and that for configuration Q\mathrm{Q} is EQ\mathrm{E}_{\mathrm{Q}}. The ratio EP/EQ\mathrm{E}_{\mathrm{P}} / \mathrm{E}_{\mathrm{Q}} is :
[Neglect the effect of mutual inductance.]

Figure for question 22: Two identical inductors are connected in two different configurations mathrm{P} and mathrm{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 mathrm{P} is mathrm{E}_{mathrm{P}} and that for configuration mathrm{Q} is mat
  1. 11
  2. 22
  3. 1/41/4
  4. 1/21/2
Show answer and solution

Answer: option (2)

In P, a and b span only one inductor, so EP=L dI/dtE_P=L\,dI/dt. In Q, the inductors are in parallel and each carries I/2I/2, so EQ=L2 dI/dtE_Q=\frac{L}{2}\,dI/dt. Hence EP/EQ=2E_P/E_Q=2.

Q23. Semiconductor Electronics: Materials, Devices and Simple Circuits

Three identical p-n\mathrm{p}\text{-}\mathrm{n} junction diodes D1D_1, D2D_2 and D3D_3 are connected across a battery as shown in the figure. If the width of the depletion regions of D1D_1, D2D_2 and D3D_3 are W1\mathrm{W}_1, W2\mathrm{W}_2 and W3\mathrm{W}_3, respectively, then the correct option is :

Figure for question 23: Three identical mathrm{p}text{-}mathrm{n} junction diodes D_1, D_2 and D_3 are connected across a battery as shown in the figure. If the width of the depletion regions of D_1, D_2 and D_3 are mathrm{W}_1, mathrm{W}_2 and mathrm{W}_3, respectively, then the correct option is :
  1. W3>W2>W1\mathrm{W}_3 > \mathrm{W}_2 > \mathrm{W}_1
  2. W2>W1=W3\mathrm{W}_2 > \mathrm{W}_1 = \mathrm{W}_3
  3. W1>W2>W3\mathrm{W}_1 > \mathrm{W}_2 > \mathrm{W}_3
  4. W3=W1>W2\mathrm{W}_3 = \mathrm{W}_1 > \mathrm{W}_2
Show answer and solution

Answer: option (1)

D1 is forward biased (narrowest depletion region), D2 is in an open branch so it is unbiased (normal width), and D3 is reverse biased (widest). So W3>W2>W1W_3>W_2>W_1.

Q24. Electrostatic Potential and Capacitance

Three identical capacitors PP, QQ and SS, each of the capacitance CC, are connected to a battery of voltage VV as shown in the figure. If the energy stored in the capacitor PP and total energy stored in the system are UP\text{U}_P and UT\text{U}_T, respectively, then the ratio UPUT\frac{\text{U}_P}{\text{U}_T} is :

Figure for question 24: 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 text{U}_P and text{U}_T, respectively, then the ratio frac{text{U}_P}{text{U}_T} is :
  1. 1/21/2
  2. 1/61/6
  3. 2/32/3
  4. 1/31/3
Show answer and solution

Answer: option (2)

P and Q are in series (each gets V/2V/2) and this pair is in parallel with S (gets VV). UP=CV28U_P=\frac{CV^2}{8}, UT=CV28+CV28+CV22=3CV24U_T=\frac{CV^2}{8}+\frac{CV^2}{8}+\frac{CV^2}{2}=\frac{3CV^2}{4}, so UP/UT=1/6U_P/U_T=1/6.

Q25. System of Particles and Rotational Motion

A thin horizontal disc is rotating about a vertical axis passing through its fixed centre OO. Its angular momentum is LAL_A and LBL_B computed about points AA and BB, respectively, with OB=2×OAOB = 2 \times OA. The value of LALB\frac{L_A}{L_B} is :

Figure for question 25: 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 times OA. The value of frac{L_A}{L_B} is :
  1. 11
  2. 22
  3. 14\frac{1}{4}
  4. 12\frac{1}{2}
Show answer and solution

Answer: option (1)

The disc's centre of mass is at rest, so its total linear momentum is zero and its angular momentum is the same about every point (L=IωL=I\omega). Hence LA/LB=1L_A/L_B=1.

Q26. Semiconductor Electronics: Materials, Devices and Simple Circuits

An ideal Zener diode with breakdown voltage of −3 V-3\text{ V} is reverse biased with a negative input voltage Vi=−5 VV_i = -5\text{ V}. The magnitude of voltage difference between point BB and AA is :

Figure for question 26: An ideal Zener diode with breakdown voltage of -3text{ V} is reverse biased with a negative input voltage V_i = -5text{ V}. The magnitude of voltage difference between point B and A is :
  1. 1 V1\text{ V}
  2. 0 V0\text{ V}
  3. 3 V3\text{ V}
  4. 2 V2\text{ V}
Show answer and solution

Answer: option (4)

In breakdown the ideal Zener holds 3 V3\ \text{V} across itself, so the rest, 5−3=2 V5-3=2\ \text{V}, drops across the resistor between B and A.

Q27. Mechanical Properties of Fluids

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 PP and QQ is 15 Nm−215\text{ Nm}^{-2}. The area of cross-section at PP and QQ are 40 cm240\text{ cm}^2 and 20 cm220\text{ cm}^2, respectively. The rate of flow of water through the pipe, in cm3s−1\text{cm}^3\text{s}^{-1}, is :

[Take density of water =1000 kg m−3= 1000\text{ kg m}^{-3}]

Figure for question 27: 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 15text{ Nm}^{-2}. The area of cross-section at P and Q are 40text{ cm}^2 and 20text{ cm}^2, respectively. The rate of flow of water throu
  1. 300300
  2. 400400
  3. 100100
  4. 200200
Show answer and solution

Answer: option (2)

Continuity: vQ=2vPv_Q=2v_P. Bernoulli: ΔP=12ρ(vQ2−vP2)=1500 vP2=15\Delta P=\frac{1}{2}\rho(v_Q^2-v_P^2)=1500\,v_P^2=15, so vP=0.1 ms−1v_P=0.1\ \text{ms}^{-1} and flow rate =40 cm2×10 cm s−1=400 cm3s−1=40\ \text{cm}^2\times10\ \text{cm s}^{-1}=400\ \text{cm}^3\text{s}^{-1}.

Q28. Work, Energy and Power

A particle of mass M\mathrm{M} moves along a horizontal xx axis from x=0x=0 to x=Lx=\mathrm{L}. The coefficient of kinetic friction varies as a function of xx as μk(x)=μ0−αx\mu_k(x)=\mu_0-\alpha x, where μ0,α\mu_0, \alpha are constants of appropriate dimensions, so that μk(L)=0\mu_k(\mathrm{L})=0. The total work done by the frictional force during the motion is nμ0MgLn \mu_0 \mathrm{MgL}, where g\mathrm{g} is the acceleration due to gravity. The value of nn is :

  1. 13\frac{1}{3}
  2. 12\frac{1}{2}
  3. 33
  4. 11
Show answer and solution

Answer: option (2)

μk(L)=0\mu_k(L)=0 gives α=μ0/L\alpha=\mu_0/L. ∣W∣=∫0L(μ0−αx)Mg dx=Mg(μ0L−μ0L2)=12μ0MgL|W|=\int_0^L(\mu_0-\alpha x)Mg\,dx=Mg\left(\mu_0L-\frac{\mu_0L}{2}\right)=\frac{1}{2}\mu_0MgL, so n=12n=\frac{1}{2}.

Q29. Dual Nature of Radiation and Matter

A ray of light with wavelength λ\lambda is incident on three different photoelectric cells namely 1, 2 and 3. The threshold wavelength of these photoelectric cells are λ1,λ2\lambda_1, \lambda_2 and λ3\lambda_3, respectively and the magnitude of stopping potentials of these cells are V1,V2\mathrm{V}_1, \mathrm{V}_2 and V3\mathrm{V}_3, respectively. The relation between λ\lambda and threshold wavelengths are λ1<λ\lambda_1 < \lambda, λ2>λ\lambda_2 > \lambda and λ3≫λ\lambda_3 \gg \lambda. The correct option is :

  1. V1>V2,V3=0\mathrm{V}_1 > \mathrm{V}_2, \mathrm{V}_3=0
  2. V1<V2,V3=0\mathrm{V}_1 < \mathrm{V}_2, \mathrm{V}_3=0
  3. V1=0,V2<V3\mathrm{V}_1=0, \mathrm{V}_2 < \mathrm{V}_3
  4. V1=0,V2>V3\mathrm{V}_1=0, \mathrm{V}_2 > \mathrm{V}_3
Show answer and solution

Answer: option (3)

Cell 1 has λ1<λ\lambda_1<\lambda, so no emission and V1=0V_1=0. For cells 2 and 3, eV=hc(1λ−1λ0)eV=hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) grows with threshold wavelength; since λ3≫λ2\lambda_3\gg\lambda_2, V2<V3V_2<V_3.

Q30. Atoms

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity 25.6×105 ms−1\sqrt{25.6} \times 10^5 \mathrm{~ms}^{-1}. The radius of the orbit is x×10−9 mx \times 10^{-9} \mathrm{~m}. The value of xx is :
[Take the mass of electron to be 9×10−31 kg9 \times 10^{-31} \mathrm{~kg}, charge of electron =−1.6×10−19C=-1.6 \times 10^{-19} \mathrm{C} and 14πε0=9×109 N m2C−2\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}]

  1. 22
  2. 11
  3. 44
  4. 33
Show answer and solution

Answer: option (2)

Coulomb force supplies centripetal force: r=ke2mv2=9×109×(1.6×10−19)29×10−31×25.6×1010=1×10−9 mr=\frac{ke^2}{mv^2}=\frac{9\times10^9\times(1.6\times10^{-19})^2}{9\times10^{-31}\times25.6\times10^{10}}=1\times10^{-9}\ \text{m}, so x=1x=1.

Q31. Laws of Motion

A car travels on a circular racetrack of radius 50 m50\text{ m}, which is banked at an angle θ\theta. If the car travels at a speed 10 ms−110\text{ ms}^{-1}, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be 10 ms−210\text{ ms}^{-2}, the value of θ\theta is :

  1. tan⁡−1(3/2)\tan^{-1}(\sqrt{3}/2)
  2. tan⁡−1(23)\tan^{-1}(2\sqrt{3})
  3. tan⁡−1(15)\tan^{-1}\left(\frac{1}{5}\right)
  4. tan⁡−1(25)\tan^{-1}\left(\frac{2}{5}\right)
Show answer and solution

Answer: option (3)

Tyre wear is minimum when no friction is needed, i.e. at the design speed: tan⁡θ=v2rg=100500=15\tan\theta=\frac{v^2}{rg}=\frac{100}{500}=\frac{1}{5}.

Q32. Thermodynamics

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

Figure for question 32: One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is :
  1. 600 J600\text{ J}
  2. 800 J800\text{ J}
  3. 400 J400\text{ J}
  4. 500 J500\text{ J}
Show answer and solution

Answer: option (1)

For a cycle ΔU=0\Delta U=0, so net heat = net work = area of the rectangle =(5−2)×(300−100)=600 J=(5-2)\times(300-100)=600\ \text{J} (clockwise, so positive).

Q33. System of Particles and Rotational Motion

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=π2)A\left(\theta=\frac{\pi}{2}\right) with identical uniform angular speeds in opposite directions, and meet again at point B(θ=−π2)B\left(\theta=-\frac{\pi}{2}\right). During this time, which of the following figures schematically represent the magnitude of the total linear momentum PP of the system, as a function of θ\theta?

Figure for question 33: A frictionless circular wire of unit radius is fixed on the horizontal plane. Two-point particles of unit mass start moving simultaneously from point Aleft(theta=frac{pi}{2}right) with identical uniform angular speeds in opposite directions, and meet again at point Bleft(theta=-frac{pi}{2}right). Du

The four options are in the figure above.

Show answer and solution

Answer: option (1)

By symmetry the horizontal velocity components cancel and the vertical ones add: P=2v∣cos⁡θ∣P=2v|\cos\theta|. It is zero at θ=±π/2\theta=\pm\pi/2 and maximum at θ=0\theta=0, a single hump as in figure (1).

Q34. Ray Optics and Optical Instruments

The lens combination as shown in the figure, consists of two lenses, L1L_1 and L2L_2, of the focal lengths +10 cm+10\text{ cm} and −10 cm-10\text{ cm}, respectively. The position of the image formed is :

Figure for question 34: The lens combination as shown in the figure, consists of two lenses, L_1 and L_2, of the focal lengths +10text{ cm} and -10text{ cm}, respectively. The position of the image formed is :
  1. 30 cm30\text{ cm} to the right of the concave lens
  2. 60 cm60\text{ cm} to the right of the concave lens
  3. 20 cm20\text{ cm} to the left of the concave lens
  4. 60 cm60\text{ cm} to the left of the concave lens
Show answer and solution

Answer: option (4)

L1L_1: u=−30u=-30, f=10f=10 gives v=+15 cmv=+15\ \text{cm}, i.e. 12 cm beyond L2L_2 (virtual object). L2L_2: 1v=−110+112=−160\frac{1}{v}=-\frac{1}{10}+\frac{1}{12}=-\frac{1}{60}, so v=−60 cmv=-60\ \text{cm}: 60 cm to the left of the concave lens.

Q35. Gravitation

Two planets P1P_1 and P2P_2 with equal mass have radii R1R_1 and R2R_2, respectively, where R2=R12R_2 = \frac{R_1}{2}. The escape speeds of P1P_1 and P2P_2 are v1v_1 and v2v_2, respectively. Then v2v1\frac{v_2}{v_1} is :

  1. 2\sqrt{2}
  2. 22
  3. 12\frac{1}{\sqrt{2}}
  4. 11
Show answer and solution

Answer: option (1)

ve=2GM/R∝1/Rv_e=\sqrt{2GM/R}\propto1/\sqrt{R} for equal masses, so v2/v1=R1/R2=2v_2/v_1=\sqrt{R_1/R_2}=\sqrt{2}.

Q36. Electromagnetic Waves

An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant, ϵr=9\epsilon_r = 9, has the electric field, Ex=E0sin⁡(kz−2π×106t) Vm−1E_x = E_0\sin(kz - 2\pi \times 10^6 t)\text{ Vm}^{-1} where E0E_0 is the amplitude and kk is the wave vector. Among the following options, the incorrect choice is :

  1. The magnetic field is given by the relation By=B0vsin⁡(kz−2π×106t)B_y = \frac{B_0}{v}\sin(kz - 2\pi \times 10^6 t) where vv is the speed of the electromagnetic wave inside the medium
  2. The direction of propagation of the electromagnetic wave is along +z+z
  3. The speed of the electromagnetic wave inside the medium is 108 ms−110^8\text{ ms}^{-1}
  4. The wavelength of the electromagnetic wave inside the medium is 300 m300\text{ m}
Show answer and solution

Answer: option (4)

v=c/ϵr=108 ms−1v=c/\sqrt{\epsilon_r}=10^8\ \text{ms}^{-1} and f=106 Hzf=10^6\ \text{Hz}, so λ=v/f=100 m\lambda=v/f=100\ \text{m}, not 300 m. The wave travels along +z+z (phase kz−ωtkz-\omega t), and with EE along xx the field BB is along yy.

Q37. Motion in a Straight Line

Consider a particle moving along a straight line, whose position as a function of time is given by s(t)=αt2−βt+γs(t) = \alpha t^2 - \beta t + \gamma, where α=1 ms−2\alpha = 1\text{ ms}^{-2}, β=6 ms−1\beta = 6\text{ ms}^{-1} and γ=5 m\gamma = 5\text{ m}. The average speed of the particle, in ms−1\text{ms}^{-1}, from t=0t = 0 to t=6 st = 6\text{ s} is :

  1. 3
  2. 0
  3. 12
  4. 6
Show answer and solution

Answer: option (1)

v=2t−6v=2t-6 changes sign at t=3 st=3\ \text{s}. s(0)=5s(0)=5, s(3)=−4s(3)=-4, s(6)=5s(6)=5, so distance =9+9=18 m=9+9=18\ \text{m} in 6 s: average speed 3 ms−13\ \text{ms}^{-1}.

Q38. Atoms

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

Figure for question 38: In Geiger-Marsden experiment, the number of scattered alpha-particles N(theta) is plotted as a function of scattering angle theta. Which of the following options represents the correct plot ?

The four options are in the figure above.

Show answer and solution

Answer: option (1)

Rutherford scattering gives N(θ)∝1/sin⁡4(θ/2)N(\theta)\propto1/\sin^4(\theta/2): very large at small angles and falling steeply as θ\theta increases, as in plot (1).

Q39. Waves

For sound waves, if the number of nodes for the 5th5^{\text{th}} harmonic of an open-ended pipe is nn and that for the 9th9^{\text{th}} harmonic of the same pipe with one of its ends closed is mm, the ratio nm\frac{n}{m} is :

  1. 11
  2. 35\frac{3}{5}
  3. 59\frac{5}{9}
  4. 95\frac{9}{5}
Show answer and solution

Answer: option (1)

An open pipe in its 5th harmonic has 5 nodes. A closed pipe's 9th harmonic (9λ/4=l9\lambda/4=l) also has 5 nodes (counting the closed end). So n/m=1n/m=1.

Q40. Electromagnetic Induction

Consider a long solenoid of length ll and radius rr. If nn is the number of turns per unit length and μ0\mu_0 is the permeability of free space, the inductance of the solenoid is :

  1. (μ0/2π)n2r2l(\mu_0 / 2\pi)n^2 r^2 l
  2. 2μ0πn2r2l2\mu_0 \pi n^2 r^2 l
  3. μ0πn2r2l\mu_0 \pi n^2 r^2 l
  4. μ0n2r2l\mu_0 n^2 r^2 l
Show answer and solution

Answer: option (3)

B=μ0nIB=\mu_0 n I, flux linkage =(nl)(μ0nI)(πr2)=(nl)(\mu_0 nI)(\pi r^2), so L=μ0πn2r2lL=\mu_0\pi n^2r^2l.

Q41. Moving Charges and Magnetism

A current I0I_0 flows through a metallic circular loop of radius rr as shown in the figure. Resistance of the segment ABC\text{ABC} is half that of ADC\text{ADC}. Magnitude of magnetic field at the center O\text{O} of the loop is :

Figure for question 41: A current I_0 flows through a metallic circular loop of radius r as shown in the figure. Resistance of the segment text{ABC} is half that of text{ADC}. Magnitude of magnetic field at the center text{O} of the loop is :
  1. μ0I02r\frac{\mu_0 I_0}{2r}
  2. μ0I02πr\frac{\mu_0 I_0}{2\pi r}
  3. μ0I012r\frac{\mu_0 I_0}{12r}
  4. μ0I04r\frac{\mu_0 I_0}{4r}
Show answer and solution

Answer: option (3)

Currents divide inversely as resistance: 23I0\frac{2}{3}I_0 through ABC and 13I0\frac{1}{3}I_0 through ADC. Each semicircle gives μ0I4r\frac{\mu_0 I}{4r} in opposite directions, so B=μ04r⋅I03=μ0I012rB=\frac{\mu_0}{4r}\cdot\frac{I_0}{3}=\frac{\mu_0I_0}{12r}.

Q42. Thermal Properties of Matter

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

  1. 4πR3αΔT4\pi R^{3}\alpha\Delta T
  2. 6πR3αΔT6\pi R^{3}\alpha\Delta T
  3. 2πR3αΔT2\pi R^{3}\alpha\Delta T
  4. 3πR3αΔT3\pi R^{3}\alpha\Delta T
Show answer and solution

Answer: option (1)

γ=3α\gamma=3\alpha, so ΔV=3α⋅43πR3ΔT=4πR3αΔT\Delta V=3\alpha\cdot\frac{4}{3}\pi R^3\Delta T=4\pi R^3\alpha\Delta T.

Q43. Units and Measurement

One main scale division of a Vernier calliper is equal to 1 mm1\text{ mm} and the number of divisions on the Vernier scale is 10. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that 4th4^{\text{th}} Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of a wire to be 1 cm1\text{ cm}, the actual length of the wire is :

  1. 1.00 cm1.00\text{ cm}
  2. 1.04 cm1.04\text{ cm}
  3. 0.60 cm0.60\text{ cm}
  4. 0.96 cm0.96\text{ cm}
Show answer and solution

Dropped by NTA. Every candidate got 4 marks for this question.

NTA dropped this question in the final answer key; every candidate gets +4.

Q44. Ray Optics and Optical Instruments

Consider three media P, Q and R with refractive indices 11, 1.251.25, and 1.51.5, respectively. The medium Q having a thickness of 5 cm5\text{ cm} is placed between extended media P and R as shown in the figure. An object O is placed at the center of medium Q. If viewed from medium P near the normal direction, the apparent depth of O is h1\text{h}_1. For similar observation from medium R, the apparent depth is h2\text{h}_2. The value of ∣h1−h2∣|\text{h}_1 - \text{h}_2|, in cm\text{cm}, is :

Figure for question 44: Consider three media P, Q and R with refractive indices 1, 1.25, and 1.5, respectively. The medium Q having a thickness of 5text{ cm} is placed between extended media P and R as shown in the figure. An object O is placed at the center of medium Q. If viewed from medium P near the normal direction, t
  1. 2
  2. 3
  3. 0
  4. 1
Show answer and solution

Answer: option (4)

O is 2.5 cm from each face. From P: h1=2.5×11.25=2 cmh_1=2.5\times\frac{1}{1.25}=2\ \text{cm}. From R: h2=2.5×1.51.25=3 cmh_2=2.5\times\frac{1.5}{1.25}=3\ \text{cm}. So ∣h1−h2∣=1 cm|h_1-h_2|=1\ \text{cm}.

Q45. Electric Charges and Fields

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 EA\text{E}_\text{A}, EB\text{E}_\text{B} and EC\text{E}_\text{C}, respectively. The points B and C are at the same distance from the center of the solid sphere. The correct option is :

Figure for question 45: 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 text{E}_text{A}, text{E}_text{B} and text{E}_text{C}, respectively. The points B and C are at the same distance fro
  1. EA=0,EB>EC\text{E}_\text{A} = 0, \text{E}_\text{B} > \text{E}_\text{C}
  2. EA≠0,EB<EC\text{E}_\text{A} \neq 0, \text{E}_\text{B} < \text{E}_\text{C}
  3. EA=0,EB=EC\text{E}_\text{A} = 0, \text{E}_\text{B} = \text{E}_\text{C}
  4. EA≠0,EB=EC\text{E}_\text{A} \neq 0, \text{E}_\text{B} = \text{E}_\text{C}
Show answer and solution

Answer: option (4)

Inside the cavity the field of Q is present, so EA≠0E_A\neq0. A charge +Q+Q appears uniformly on the outer surface regardless of where Q sits in the cavity, so outside E=Q4πϵ0r2E=\frac{Q}{4\pi\epsilon_0 r^2} and EB=ECE_B=E_C.

Chemistry questions (Q46–90)

Q46. The d- and f-Block Elements

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

  1. Tb3+\text{Tb}^{3+}
  2. Ho3+\text{Ho}^{3+}
  3. Nd3+\text{Nd}^{3+}
  4. Ce3+\text{Ce}^{3+}
Show answer and solution

Answer: option (2)

Ho3+\text{Ho}^{3+} is [Xe]4f10[\text{Xe}]4f^{10}: seven f orbitals hold 10 electrons, leaving 4 unpaired. Tb3+\text{Tb}^{3+} (4f84f^8) has 6, Nd3+\text{Nd}^{3+} (4f34f^3) has 3 and Ce3+\text{Ce}^{3+} (4f14f^1) has 1.

Q47. Electrochemistry

The standard electrode potential (E∘\text{E}^\circ) for the half-cell reaction Fe3++e−→Fe2+\text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+} at 298 K298\text{ K} is
(Given : E∘(Fe3+/Fe)=−0.04 V\text{E}^\circ(\text{Fe}^{3+}/\text{Fe}) = -0.04\text{ V} and E∘(Fe2+/Fe)=−0.44 V\text{E}^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\text{ V} at 298 K298\text{ K})

  1. −0.48 V-0.48\text{ V}
  2. +0.92 V+0.92\text{ V}
  3. +0.40 V+0.40\text{ V}
  4. +0.76 V+0.76\text{ V}
Show answer and solution

Answer: option (4)

Potentials are not additive, but ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ is. Fe3+→Fe\text{Fe}^{3+}\to\text{Fe}: ΔG∘=−3F(−0.04)=+0.12F\Delta G^\circ = -3F(-0.04) = +0.12F; Fe→Fe2+\text{Fe}\to\text{Fe}^{2+}: ΔG∘=+2F(−0.44)=−0.88F\Delta G^\circ = +2F(-0.44) = -0.88F. Sum =−0.76F=−1⋅F⋅E∘= -0.76F = -1\cdot F\cdot E^\circ, so E∘=+0.76 VE^\circ = +0.76\text{ V}.

Q48. The p-Block Elements

Given below are two statements :
Statement-I : Heating NaCl\text{NaCl} with concentrated H2SO4\text{H}_2\text{SO}_4 and MnO2\text{MnO}_2 results in oxidation of Mn\text{Mn}.
Statement-II : Heating NaI\text{NaI} with concentrated H2SO4\text{H}_2\text{SO}_4 and MnO2\text{MnO}_2 results in reduction of Mn\text{Mn}.
In light of the above statements, choose the most appropriate answer from the options given below.

  1. Statement-I is correct but Statement-II is incorrect
  2. Statement-I is incorrect but Statement-II is correct
  3. Both Statement-I and Statement-II are correct
  4. Both Statement-I and Statement-II are incorrect
Show answer and solution

Answer: option (2)

In both cases MnO2\text{MnO}_2 acts as the oxidising agent: halide is oxidised to Cl2\text{Cl}_2 or I2\text{I}_2 while Mn goes from +4+4 in MnO2\text{MnO}_2 to +2+2 in MnSO4\text{MnSO}_4/MnCl2\text{MnCl}_2. So Mn is reduced in both; Statement-I is wrong and Statement-II is right.

Q49. Coordination Compounds

The complex which has facial and meridional isomers is
(Given : py=pyridine\text{py} = \text{pyridine} and en=H2N−CH2−CH2−NH2\text{en} = \text{H}_2\text{N}-\text{CH}_2-\text{CH}_2-\text{NH}_2)

  1. [Co(NH3)4(H2O)2]3+[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{3+}
  2. [Ni(en)2(H2O)2]2+[\text{Ni}(\text{en})_2(\text{H}_2\text{O})_2]^{2+}
  3. [Cr(py)3(Cl)3][\text{Cr}(\text{py})_3(\text{Cl})_3]
  4. [Cr(H2O)6]3+[\text{Cr}(\text{H}_2\text{O})_6]^{3+}
Show answer and solution

Answer: option (3)

fac/mer isomerism arises in octahedral [MA3B3][\text{MA}_3\text{B}_3] complexes. Only [Cr(py)3Cl3][\text{Cr}(\text{py})_3\text{Cl}_3] has this type; MA4B2\text{MA}_4\text{B}_2 and M(AA)2B2\text{M(AA)}_2\text{B}_2 show cis/trans isomerism instead, and MA6\text{MA}_6 has none.

Q50. Organic Chemistry – Some Basic Principles and Techniques

Among the following, the compound having conjugated double bonds is

  1. hepta-1,5-diene
  2. hepta-1,6-diene
  3. hepta-1,3-diene
  4. hepta-1,4-diene
Show answer and solution

Answer: option (3)

Conjugated double bonds alternate with a single bond, i.e. C=C–C=C. In hepta-1,3-diene the double bonds start at C1 and C3, separated by one single bond; the others are isolated dienes.

Q51. Chemical Bonding and Molecular Structure

Match the species in List-I with their geometry in List-II.

List-IList-II
A. PCl5\text{PCl}_5I. Tetrahedral
B. BrF5\text{BrF}_5II. Square Planar
C. BF4−\text{BF}_4^-III. Trigonal bipyramidal
D. [Ni(CN)4]2−[\text{Ni}(\text{CN})_4]^{2-}IV. Square pyramidal

Choose the correct answer from the options given below:

  1. A-III, B-I, C-II, D-IV
  2. A-III, B-II, C-I, D-IV
  3. A-IV, B-III, C-I, D-II
  4. A-III, B-IV, C-I, D-II
Show answer and solution

Answer: option (4)

PCl5\text{PCl}_5 (sp3dsp^3d, 5 bond pairs) is trigonal bipyramidal; BrF5\text{BrF}_5 (5 bond pairs + 1 lone pair) is square pyramidal; BF4−\text{BF}_4^- (sp3sp^3) is tetrahedral; [Ni(CN)4]2−[\text{Ni}(\text{CN})_4]^{2-} (dsp2dsp^2, strong-field CN−\text{CN}^-) is square planar.

Q52. Organic Chemistry – Some Basic Principles and Techniques

The amino acid that gives a red-blood colour on treating its sodium fusion extract with sodium nitroprusside is

  1. methionine
  2. serine
  3. leucine
  4. threonine
Show answer and solution

Answer: option (1)

A colour test with the sodium fusion extract that indicates sulphur (with N, giving the blood-red thiocyanate test) requires an S-containing amino acid. Of the four, only methionine contains sulphur (and nitrogen); serine, leucine and threonine have no S.

Q53. Thermodynamics

A protein undergoes reversible thermal denaturation from its initial state N\textbf{N} to denatured state D\textbf{D} according to N⇌D\textbf{N} \rightleftharpoons \textbf{D}. At 60 ∘C60\,{}^{\circ}\text{C}, the concentrations of both N\textbf{N} and D\textbf{D} are equal at equilibrium, and the standard enthalpy change of denaturation is 666 kJ mol−1666\text{ kJ mol}^{-1}. The standard entropy change (ΔS∘\Delta S^{\circ} in kJ K−1 mol−1\text{kJ K}^{-1}\text{ mol}^{-1}) of the protein upon denaturation at 60 ∘C60\,{}^{\circ}\text{C} is closest to

  1. 333.0
  2. 11.1
  3. 2.0
  4. 2000.0
Show answer and solution

Answer: option (3)

Equal concentrations give K=1K = 1, so ΔG∘=−RTln⁡K=0\Delta G^\circ = -RT\ln K = 0 and ΔH∘=TΔS∘\Delta H^\circ = T\Delta S^\circ. With T=333 KT = 333\text{ K}: ΔS∘=666/333=2.0 kJ K−1 mol−1\Delta S^\circ = 666/333 = 2.0\text{ kJ K}^{-1}\text{ mol}^{-1}.

Q54. Chemical Kinetics

2A→kB2\text{A} \xrightarrow{k} \text{B} is a zero-order reaction, where k=1.0 mol L−1 min−1k = 1.0\text{ mol L}^{-1}\text{ min}^{-1}. If the initial concentration of A is 2 M2\text{ M}, then the time taken to complete 75%75\% of the reaction will be

  1. 1.0 min
  2. 2.0 min
  3. 1.5 min
  4. 0.75 min
Show answer and solution

Answer: option (4)

For 2A→B2\text{A}\to\text{B}, rate =−12d[A]dt=k= -\frac{1}{2}\frac{d[\text{A}]}{dt} = k, so A is consumed at 2k=2 M min−12k = 2\text{ M min}^{-1}. 75% of 2 M is 1.5 M, which takes 1.5/2=0.75 min1.5/2 = 0.75\text{ min}.

Q55. Coordination Compounds

Given below are two statements:
Statement-I : [Fe(ox)3]3−[\text{Fe}(\text{ox})_3]^{3-} is chiral.
Statement-II : trans-[Cr(H2O)2(ox)2]−[\text{Cr}(\text{H}_2\text{O})_2(\text{ox})_2]^- is chiral.
(Given : oxH2=HOOC−COOH\text{oxH}_2 = \text{HOOC}-\text{COOH})
In light of the above statements, choose the most appropriate answer from the options given below:

  1. Statement-I is correct but Statement-II is incorrect
  2. Statement-I is incorrect but Statement-II is correct
  3. Both Statement-I and Statement-II are correct
  4. Both Statement-I and Statement-II are incorrect
Show answer and solution

Answer: option (1)

[M(AA)3][\text{M(AA)}_3] complexes such as tris(oxalato)ferrate(III) have no mirror plane and exist as Δ/Λ\Delta/\Lambda enantiomers, so I is correct. In the trans bis(oxalato) complex the two oxalates lie in one plane, which is a mirror plane, so it is achiral and II is wrong.

Q56. The d- and f-Block Elements

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Generally, 3d3d transition metals have high melting points.
Reason R: Involvement of 3d3d-electrons in addition to 4s4s-electrons in the interatomic metallic bonding.
In light of the above statements, choose the most appropriate answer from the options given below:

  1. A is correct but R is not correct
  2. A is not correct but R is correct
  3. Both A and R are correct and R is the correct explanation of A
  4. Both A and R are correct and R is NOT the correct explanation of A
Show answer and solution

Answer: option (3)

Transition metals have high melting points because, besides the 4s4s electrons, unpaired (n−1)d(n-1)d electrons also take part in metallic bonding, making it stronger. So both are true and R explains A.

Q57. Coordination Compounds

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

  1. [Co(NH3)6]3+[\text{Co}(\text{NH}_3)_6]^{3+}
  2. [Ti(H2O)6]3+[\text{Ti}(\text{H}_2\text{O})_6]^{3+}
  3. [Mn(CN)6]3−[\text{Mn}(\text{CN})_6]^{3-}
  4. [Fe(CN)6]3−[\text{Fe}(\text{CN})_6]^{3-}
Show answer and solution

Answer: option (3)

Unpaired electrons: Co3+\text{Co}^{3+} d6d^6 low spin =0= 0; Ti3+\text{Ti}^{3+} d1=1d^1 = 1; Mn3+\text{Mn}^{3+} d4d^4 low spin (t2g4t_{2g}^4) =2= 2; Fe3+\text{Fe}^{3+} d5d^5 low spin =1= 1. The Mn complex with n=2n = 2 has the highest μ=8≈2.83 BM\mu = \sqrt{8} \approx 2.83\text{ BM}.

Q58. Coordination Compounds

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

  1. Cl−>H2O>NH3>CO\text{Cl}^- > \text{H}_2\text{O} > \text{NH}_3 > \text{CO}
  2. Cl−>NH3>H2O>CO\text{Cl}^- > \text{NH}_3 > \text{H}_2\text{O} > \text{CO}
  3. CO>NH3>H2O>Cl−\text{CO} > \text{NH}_3 > \text{H}_2\text{O} > \text{Cl}^-
  4. CO>H2O>NH3>Cl−\text{CO} > \text{H}_2\text{O} > \text{NH}_3 > \text{Cl}^-
Show answer and solution

Answer: option (3)

In the spectrochemical series Cl−<H2O<NH3<CO\text{Cl}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{CO}, so in decreasing order: CO>NH3>H2O>Cl−\text{CO} > \text{NH}_3 > \text{H}_2\text{O} > \text{Cl}^-.

Q59. The p-Block Elements

In potash alum, the ratio of K+\text{K}^+ and SO42−\text{SO}_4^{2-} ions is

  1. 2 : 3
  2. 3 : 2
  3. 1 : 2
  4. 2 : 1
Show answer and solution

Answer: option (3)

Potash alum is KAl(SO4)2⋅12H2O\text{KAl(SO}_4)_2\cdot 12\text{H}_2\text{O} (i.e. K2SO4⋅Al2(SO4)3⋅24H2O\text{K}_2\text{SO}_4\cdot\text{Al}_2(\text{SO}_4)_3\cdot 24\text{H}_2\text{O}), so K+:SO42−=1:2\text{K}^+ : \text{SO}_4^{2-} = 1 : 2.

Q60. Classification of Elements and Periodicity in Properties

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 stable half-filled pp orbital.
In light of the above statements, choose the most appropriate answer from the options given below:

  1. A is correct but R is not correct
  2. A is not correct but R is correct
  3. Both A and R are correct and R is the correct explanation of A
  4. Both A and R are correct and R is NOT the correct explanation of A
Show answer and solution

Answer: option (3)

O (2p42p^4) has one doubly occupied 2p2p orbital; removing that paired electron relieves repulsion and gives the stable half-filled 2p32p^3 configuration, so O's first IE is lower than N's, and F is higher still. Per the official key, R is taken as the correct explanation of A.

Q61. Amines

Given below are two statements:
Statement-I : Oxidation of pp-nitrotoluene with acidic KMnO4\text{KMnO}_4 gives an acid that is stronger than benzoic acid.
Statement-II : Reduction of pp-nitrotoluene with Sn/HCl\text{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.

  1. Statement-I is correct but Statement-II is incorrect
  2. Statement-I is incorrect but Statement-II is correct
  3. Both Statement-I and Statement-II are correct
  4. Both Statement-I and Statement-II are incorrect
Show answer and solution

Answer: option (3)

Oxidation gives pp-nitrobenzoic acid; the electron-withdrawing −NO2-\text{NO}_2 stabilises the carboxylate, so it is stronger than benzoic acid. Reduction gives pp-toluidine, whose electron-donating −CH3-\text{CH}_3 makes it more basic than aniline. Both statements are correct.

Q62. Organic Chemistry – Some Basic Principles and Techniques

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

Figure for question 62: The following carbocation is stabilized by the interaction of the empty p orbital with
  1. empty σ∗\sigma^* and filled π\pi orbitals
  2. empty σ∗\sigma^* and empty π∗\pi^* orbitals
  3. filled σ\sigma and filled π\pi orbitals
  4. empty σ\sigma and empty π∗\pi^* orbitals
Show answer and solution

Answer: option (3)

The 1-phenylethyl cation is stabilised by hyperconjugation (overlap of filled C–H σ\sigma orbitals of CH3\text{CH}_3 with the empty p orbital) and by resonance with the filled π\pi orbitals of the benzene ring. An empty orbital can only be stabilised by filled ones.

Q63. Structure of Atom

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

Figure for question 63: Consider the following schematic plots of orbital wavefunction (psi_r) against distance (r) from the nucleus. The figure representing two radial nodes in the orbital is
  1. C
  2. D
  3. A
  4. B
Show answer and solution

Answer: option (1)

A radial node is a point (other than r=0r = 0 and ∞\infty) where ψr\psi_r crosses zero. Plot A has no crossing, B one, C two and D three, so C has two radial nodes (e.g. a 3s3s orbital).

Q64. Equilibrium

The correct order of solubility of the given salts in water at 298 K298\text{ K} is

SaltKspK_{sp} at 298 K298\text{ K}
AgBr\text{AgBr}5.0×10−135.0 \times 10^{-13}
Zn(OH)2\text{Zn(OH)}_21.0×10−151.0 \times 10^{-15}
Hg2Cl2\text{Hg}_2\text{Cl}_21.3×10−181.3 \times 10^{-18}
  1. Hg2Cl2>AgBr>Zn(OH)2\text{Hg}_2\text{Cl}_2 > \text{AgBr} > \text{Zn(OH)}_2
  2. Zn(OH)2>AgBr>Hg2Cl2\text{Zn(OH)}_2 > \text{AgBr} > \text{Hg}_2\text{Cl}_2
  3. Hg2Cl2>Zn(OH)2>AgBr\text{Hg}_2\text{Cl}_2 > \text{Zn(OH)}_2 > \text{AgBr}
  4. AgBr>Zn(OH)2>Hg2Cl2\text{AgBr} > \text{Zn(OH)}_2 > \text{Hg}_2\text{Cl}_2
Show answer and solution

Answer: option (2)

AgBr: s=Ksp≈7.1×10−7s = \sqrt{K_{sp}} \approx 7.1\times10^{-7} M. Zn(OH)2\text{Zn(OH)}_2 and Hg2Cl2\text{Hg}_2\text{Cl}_2 (as Hg22++2Cl−\text{Hg}_2^{2+} + 2\text{Cl}^-) both follow Ksp=4s3K_{sp} = 4s^3: s≈6.3×10−6s \approx 6.3\times10^{-6} M and ≈6.9×10−7\approx 6.9\times10^{-7} M. Hence Zn(OH)2>AgBr>Hg2Cl2\text{Zn(OH)}_2 > \text{AgBr} > \text{Hg}_2\text{Cl}_2.

Q65. Coordination Compounds

A 1:31 : 3 electrolyte in an aqueous solution is

  1. [Co(NH3)6]Cl3[\text{Co}(\text{NH}_3)_6]\text{Cl}_3
  2. [Co(NH3)3(NO2)3][\text{Co}(\text{NH}_3)_3(\text{NO}_2)_3]
  3. [CoCl2(NH3)4]Cl[\text{CoCl}_2(\text{NH}_3)_4]\text{Cl}
  4. [CoCl(NH3)5]Cl2[\text{CoCl}(\text{NH}_3)_5]\text{Cl}_2
Show answer and solution

Answer: option (1)

[Co(NH3)6]Cl3[\text{Co}(\text{NH}_3)_6]\text{Cl}_3 ionises into one complex cation [Co(NH3)6]3+[\text{Co}(\text{NH}_3)_6]^{3+} and three Cl−\text{Cl}^-, i.e. a 1 : 3 electrolyte. The others give 1 : 0, 1 : 1 and 1 : 2.

Q66. Biomolecules

The correct statement about peptides and proteins is

  1. In β\beta-pleated sheet structures, peptide chains are held together by intermolecular hydrogen bonds
  2. In α\alpha-helices, the polypeptide chain is twisted into a left-handed screw (helix) through intramolecular hydrogen bonds
  3. Tertiary structure of proteins has two or more polypeptide subunits
  4. Only the proteins having a quaternary structure are biologically active
Show answer and solution

Answer: option (1)

In a β\beta-pleated sheet, neighbouring chains lie side by side and are linked by hydrogen bonds between chains (intermolecular). The α\alpha-helix is right-handed, subunits define quaternary (not tertiary) structure, and many single-chain proteins are active.

Q67. Haloalkanes and Haloarenes

One of the products formed in the following reaction is

Figure for question 67: One of the products formed in the following reaction is

The four options are in the figure above.

Show answer and solution

Answer: option (2)

Cyclohexylamine has acidic N–H protons, so the Grignard reagent simply acts as a strong base: C6H11MgBr+C6H11NH2→C6H12+C6H11NHMgBr\text{C}_6\text{H}_{11}\text{MgBr} + \text{C}_6\text{H}_{11}\text{NH}_2 \rightarrow \text{C}_6\text{H}_{12} + \text{C}_6\text{H}_{11}\text{NHMgBr}. The product shown is cyclohexane (option 2).

Q68. Redox Reactions

In an acidic medium, 10 mL10\text{ mL} of 0.25 M0.25\text{ M} oxalic acid is titrated with KMnO4\text{KMnO}_4 solution. If the volume of KMnO4\text{KMnO}_4 solution required to reach end point is 10 mL10\text{ mL}, the strength of the KMnO4\text{KMnO}_4 solution is

  1. 0.25 M0.25\text{ M}
  2. 0.15 M0.15\text{ M}
  3. 0.10 M0.10\text{ M}
  4. 0.20 M0.20\text{ M}
Show answer and solution

Answer: option (3)

2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}. Oxalic acid =2.5= 2.5 mmol, so KMnO4=25×2.5=1.0\text{KMnO}_4 = \frac{2}{5}\times2.5 = 1.0 mmol in 10 mL, i.e. 0.10 M0.10\text{ M}.

Q69. Hydrocarbons

Consider the following reaction sequences and choose the correct option.

Figure for question 69: Consider the following reaction sequences and choose the correct option.
  1. M and N are geometrical isomers
  2. M and N are stereoisomers
  3. K and L are geometrical isomers
  4. K and L are enantiomers
Show answer and solution

Answer: option (3)

Lindlar reduction of Ph–C≡C–CH3\text{Ph–C}\equiv\text{C–CH}_3 gives cis-1-phenylpropene (K) and Na/liq. NH3\text{NH}_3 gives the trans alkene (L), so K and L are geometrical isomers. HBr adds to give PhCHBrCH2CH3\text{PhCHBrCH}_2\text{CH}_3 (M, via benzylic cation) while HBr/peroxide gives PhCH2CHBrCH3\text{PhCH}_2\text{CHBrCH}_3 (N, via benzylic radical); M and N are position isomers.

Q70. Alcohols, Phenols and Ethers

Arrange the following compounds in the increasing order of polarity
A. CH3CH2OCH2CH3\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3
B. CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}
C. CH3COCH3\text{CH}_3\text{COCH}_3
D. CH3COOH\text{CH}_3\text{COOH}
Choose the correct answer from the options given below.

  1. C < A < B < D
  2. A < C < B < D
  3. A < B < C < D
  4. C < A < D < B
Show answer and solution

Answer: option (2)

Diethyl ether is the least polar (weakly polar C–O–C, no H-bonding), acetone is more polar due to its C=O group, ethanol is a polar protic H-bonding solvent, and acetic acid with both C=O and O–H is the most polar: A < C < B < D.

Q71. Thermodynamics

Consider the reversible processes for 1.0 mol1.0\text{ mol} of an ideal gas as shown in the figure.
Processes 2 and 4 are adiabatic.
w1w_1, w2w_2, w3w_3 and w4w_4 represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; ΔU2\Delta U_2 and ΔU4\Delta U_4 are changes in the internal energy for the processes 2 and 4, respectively.
[use R=2 cal K−1 mol−1R = 2\text{ cal K}^{-1}\text{ mol}^{-1}]
The correct option is

Figure for question 71: Consider the reversible processes for 1.0text{ mol} of an ideal gas as shown in the figure. Processes 2 and 4 are adiabatic. w_1, w_2, w_3 and w_4 represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; Delta U_2 and Delta U_4 are changes in the internal energy for the proces
  1. w1+w2=2T1ln⁡V2V1w_1 + w_2 = 2T_1 \ln\frac{V_2}{V_1}
  2. w1+w2+w3+w4=0w_1 + w_2 + w_3 + w_4 = 0
  3. w1+w3=−2T1ln⁡V2V1−2T2ln⁡V4V3w_1 + w_3 = -2T_1 \ln\frac{V_2}{V_1} - 2T_2 \ln\frac{V_4}{V_3}
  4. w2+w4=ΔU2−ΔU4w_2 + w_4 = \Delta U_2 - \Delta U_4
Show answer and solution

Answer: option (3)

Processes 1 and 3 are reversible isotherms, so w1=−RT1ln⁡V2V1=−2T1ln⁡V2V1w_1 = -RT_1\ln\frac{V_2}{V_1} = -2T_1\ln\frac{V_2}{V_1} and w3=−2T2ln⁡V4V3w_3 = -2T_2\ln\frac{V_4}{V_3}; adding gives option 3. For the adiabats w2=ΔU2w_2 = \Delta U_2 and w4=ΔU4=−ΔU2w_4 = \Delta U_4 = -\Delta U_2, so w2+w4=0w_2 + w_4 = 0, and the cycle's net work is not zero.

Q72. The d- and f-Block Elements

The green paramagnetic species formed by heating KMnO4\text{KMnO}_4 at 513 K513\text{ K} is

  1. MnO\text{MnO}
  2. KO2\text{KO}_2
  3. K2MnO4\text{K}_2\text{MnO}_4
  4. Mn3O4\text{Mn}_3\text{O}_4
Show answer and solution

Answer: option (3)

2KMnO4→513 KK2MnO4+MnO2+O22\text{KMnO}_4 \xrightarrow{513\text{ K}} \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2. Potassium manganate contains Mn6+\text{Mn}^{6+} (d1d^1, one unpaired electron), so it is green and paramagnetic.

Q73. Some Basic Concepts of Chemistry

The numbers 17.014517.0145 and 21.023521.0235 were rounded to three figures after the decimal point. The resulting numbers, respectively, are

  1. 17.01417.014 and 21.02421.024
  2. 17.01517.015 and 21.02421.024
  3. 17.01417.014 and 21.02321.023
  4. 17.01517.015 and 21.02321.023
Show answer and solution

Answer: option (1)

When the digit dropped is exactly 5, the preceding digit is left unchanged if even and raised by one if odd. In 17.0145 the preceding 4 is even, giving 17.014; in 21.0235 the preceding 3 is odd, giving 21.024.

Q74. Hydrocarbons

Given below are two statements :
Statement I : trans-But-2-ene upon treatment with Br2\text{Br}_2 in CCl4\text{CCl}_4 gives the following product.
Statement II : cis-But-2-ene upon treatment with alkaline KMnO4\text{KMnO}_4 gives the following product.
In the light of the above statements, choose the most appropriate answer from the options given below.

Figure for question 74: Given below are two statements : Statement I : trans-But-2-ene upon treatment with text{Br}_2 in text{CCl}_4 gives the following product. Statement II : cis-But-2-ene upon treatment with alkaline text{KMnO}_4 gives the following product. In the light of the above statements, choose the most appropri
  1. Statement I is correct but Statement II is incorrect
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect
Show answer and solution

Answer: option (2)

Anti addition of Br2\text{Br}_2 to trans-but-2-ene gives meso-2,3-dibromobutane, but the Newman projection in Statement I (Br anti, the two Me gauche) is the chiral (±) form, so I is wrong. Syn dihydroxylation of cis-but-2-ene by cold alkaline KMnO4\text{KMnO}_4 gives meso-butane-2,3-diol, which is the structure drawn in Statement II, so II is correct.

Q75. Aldehydes, Ketones and Carboxylic Acids

Consider the following reaction, and choose the correct option.
Toluene→ii. H3O+i. CrO2Cl2, CS2P\text{Toluene} \xrightarrow[\text{ii. H}_3\text{O}^+]{\text{i. CrO}_2\text{Cl}_2,\ \text{CS}_2} \textbf{P}

  1. On treatment with bromine water, compound P gives a white precipitate
  2. Compound P is obtained by the hydrogenation of benzoyl chloride with Pd on BaSO4\text{BaSO}_4
  3. On treating compound P with saturated NaHCO3\text{NaHCO}_3 solution, brisk effervescence is observed
  4. Compound P can be prepared by treating benzene with anhydrous AlCl3\text{AlCl}_3 and CH3COCl\text{CH}_3\text{COCl}
Show answer and solution

Answer: option (2)

The Étard reaction converts toluene into benzaldehyde (P). Benzaldehyde is also made by Rosenmund reduction, i.e. hydrogenation of benzoyl chloride over Pd/BaSO4\text{BaSO}_4. It is not a phenol or acid, and Friedel–Crafts acetylation would give acetophenone.

Q76. Biomolecules

Match the vitamins in List I with their sources in List II

List IList II
A. vitamin AI. meat
B. vitamin B12_{12}II. sunflower oil
C. vitamin EIII. green leafy vegetables
D. vitamin KIV. carrots

Choose the correct answer from the options given below.

  1. A-IV, B-II, C-I, D-III
  2. A-III, B-I, C-IV, D-II
  3. A-II, B-III, C-IV, D-I
  4. A-IV, B-I, C-II, D-III
Show answer and solution

Answer: option (4)

Vitamin A comes from carrots (β\beta-carotene), B12\text{B}_{12} from animal foods such as meat, vitamin E from vegetable oils like sunflower oil, and vitamin K from green leafy vegetables: A-IV, B-I, C-II, D-III.

Q77. Some Basic Concepts of Chemistry

The amount of carbon dioxide evolved upon complete combustion of 116 g116\text{ g} of nn-butane is
(Given: atomic mass in amu H=1\text{H} = 1, C=12\text{C} = 12 and O=16\text{O} = 16)

  1. 176 g176\text{ g}
  2. 362 g362\text{ g}
  3. 352 g352\text{ g}
  4. 322 g322\text{ g}
Show answer and solution

Answer: option (3)

C4H10\text{C}_4\text{H}_{10} has M=58M = 58, so 116 g is 2 mol. Each mole gives 4 mol CO2\text{CO}_2, i.e. 8 mol ×44=352 g\times 44 = 352\text{ g}.

Q78. Aldehydes, Ketones and Carboxylic Acids

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

Figure for question 78: The compound that CANNOT be obtained from the aldol condensation reaction shown below, is

The four options are in the figure above.

Show answer and solution

Answer: option (3)

2,2-Dimethylcyclopentanone has α-H only at C5. Crossed condensation with PhCHO gives 5-benzylidene products (E and Z, options 2 and 4); self-condensation followed by dehydration toward the α-H gives the conjugated exocyclic enone (option 1). Option 3 has the C=C inside the second ring, not conjugated with C=O, which base-catalysed aldol dehydration does not form.

Q79. The p-Block Elements

The correct statement is

  1. Magnesium has a maximum covalency of four.
  2. Aluminium has five valence orbitals.
  3. Boron has a maximum covalency of four.
  4. Beryllium has three valence orbitals.
Show answer and solution

Answer: option (3)

Boron has only 2s2s and 2p2p valence orbitals (four in total, no d orbitals), so its maximum covalency is four, as in [BF4]−[\text{BF}_4]^-. Be also has four valence orbitals, Al has nine (3s,3p,3d3s, 3p, 3d) and Mg can exceed four.

Q80. Coordination Compounds

The formula of tetraammineaquachloridocobalt(III) chloride is

  1. [Co(NH3)4(H2O)Cl]Cl[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})\text{Cl}]\text{Cl}
  2. [Co(NH3)4(H2O)Cl]Cl2[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})\text{Cl}]\text{Cl}_2
  3. [Co(NH3)4Cl2]×H2O[\text{Co}(\text{NH}_3)_4\text{Cl}_2] \times \text{H}_2\text{O}
  4. [Co(NH3)4]Cl3×H2O[\text{Co}(\text{NH}_3)_4]\text{Cl}_3 \times \text{H}_2\text{O}
Show answer and solution

Answer: option (2)

The coordination sphere has 4 NH3\text{NH}_3, 1 H2O\text{H}_2\text{O} and 1 Cl−\text{Cl}^- around Co3+\text{Co}^{3+}, giving a +2+2 charge, balanced by two chloride counter ions: [Co(NH3)4(H2O)Cl]Cl2[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})\text{Cl}]\text{Cl}_2.

Q81. Solutions

Assertion A: For an ideal solution formed by mixing liquids P and Q, ΔmixH=0\Delta_{\text{mix}}H = 0 and ΔmixV=0\Delta_{\text{mix}}V = 0
Reason R: No interactions occur between P and Q
In the light of the above statements, choose the most appropriate answer from the options given below.

  1. A is correct but R is not correct
  2. A is not correct but R is correct
  3. Both A and R are correct and R is the correct explanation of A
  4. Both A and R are correct but R is NOT the correct explanation of A
Show answer and solution

Answer: option (1)

An ideal solution does have ΔmixH=0\Delta_{\text{mix}}H = 0 and ΔmixV=0\Delta_{\text{mix}}V = 0, but this is because P–Q interactions are equal in strength to P–P and Q–Q interactions, not because there are no interactions. So A is true and R is false.

Q82. Amines

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

Figure for question 82: Identify the reactions which give aniline as the major product. Choose the correct answer from the options given below.
  1. A and C only
  2. C and D only
  3. A and B only
  4. B and D only
Show answer and solution

Answer: option (4)

B: Hofmann bromamide degradation of benzamide (Br2\text{Br}_2/KOH) gives aniline. D: acid hydrolysis of acetanilide gives aniline. A: LiAlH4\text{LiAlH}_4 reduces benzonitrile to benzylamine, and C: NaBH4\text{NaBH}_4 does not reduce the nitro group.

Q83. Redox Reactions

The correct decreasing order of oxidation state of the <u>underlined</u> atom in each molecule is

  1. Pb‾O2>N‾2O3>S‾O3\underline{\text{Pb}}\text{O}_2 > \underline{\text{N}}_2\text{O}_3 > \underline{\text{S}}\text{O}_3
  2. P‾4O6>Cl‾2O7>Al‾H3\underline{\text{P}}_4\text{O}_6 > \underline{\text{Cl}}_2\text{O}_7 > \underline{\text{Al}}\text{H}_3
  3. P‾4O10>S‾O3>H2O‾\underline{\text{P}}_4\text{O}_{10} > \underline{\text{S}}\text{O}_3 > \text{H}_2\underline{\text{O}}
  4. N‾2O5>Al‾2O3>H2S‾\underline{\text{N}}_2\text{O}_5 > \underline{\text{Al}}_2\text{O}_3 > \text{H}_2\underline{\text{S}}
Show answer and solution

Answer: option (4)

Oxidation states: (1) +4,+3,+6+4, +3, +6; (2) +3,+7,+3+3, +7, +3; (3) +5,+6,−2+5, +6, -2; (4) N +5+5 > Al +3+3 > S −2-2. Only option 4 is in decreasing order.

Q84. Aldehydes, Ketones and Carboxylic Acids

For the following reaction sequence, choose the correct option
Benzene→ii. NaOCli. CH3COCl, AlCl3P+Q\text{Benzene} \xrightarrow[\text{ii. NaOCl}]{\text{i. CH}_3\text{COCl, AlCl}_3} \text{P} + \text{Q}

  1. If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light
  2. Both P and Q are carbonyl compounds
  3. If P is the sodium salt of a carboxylic acid, Q is a primary alcohol
  4. P and Q are aromatic compounds
Show answer and solution

Answer: option (1)

Friedel–Crafts acylation gives acetophenone; the haloform reaction with NaOCl then gives sodium benzoate (P, gives benzoic acid on acidification) and chloroform (Q). CHCl3\text{CHCl}_3 is oxidised by air in light to poisonous phosgene, COCl2\text{COCl}_2.

Q85. Thermodynamics

Two moles of an ideal gas undergo free expansion from 10 L10\text{ L} to 100 L100\text{ L} at 300 K300\text{ K}. The values of ΔSsystem\Delta S_{\text{system}} and ΔSsurroundings\Delta S_{\text{surroundings}} are
(R is universal gas constant)

  1. ΔSsystem=0\Delta S_{\text{system}} = 0; ΔSsurroundings=4.606 R\Delta S_{\text{surroundings}} = 4.606\text{ R}
  2. ΔSsystem=4.606 R\Delta S_{\text{system}} = 4.606\text{ R}; ΔSsurroundings=0\Delta S_{\text{surroundings}} = 0
  3. ΔSsystem=0\Delta S_{\text{system}} = 0; ΔSsurroundings=0\Delta S_{\text{surroundings}} = 0
  4. ΔSsystem=4.606 R\Delta S_{\text{system}} = 4.606\text{ R}; ΔSsurroundings=−4.606 R\Delta S_{\text{surroundings}} = -4.606\text{ R}
Show answer and solution

Answer: option (2)

Entropy is a state function, so ΔSsys=nRln⁡V2V1=2R×2.303log⁡10=4.606R\Delta S_{\text{sys}} = nR\ln\frac{V_2}{V_1} = 2R \times 2.303\log 10 = 4.606R. In free expansion no heat is exchanged with the surroundings, so ΔSsurr=0\Delta S_{\text{surr}} = 0.

Q86. Chemical Kinetics

For an elementary chemical reaction, the Arrhenius plot is given below.
If the energy of activation is 6.64 kJ mol−16.64\text{ kJ mol}^{-1} and R=8.3 J K−1 mol−1R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}, the temperature at which the rate constant becomes e2 min−1e^2\text{ min}^{-1}, is

Figure for question 86: For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is 6.64text{ kJ mol}^{-1} and R = 8.3text{ J K}^{-1}text{ mol}^{-1}, the temperature at which the rate constant becomes e^2text{ min}^{-1}, is
  1. 200 K200\text{ K}
  2. 250 K250\text{ K}
  3. 125 K125\text{ K}
  4. 150 K150\text{ K}
Show answer and solution

Answer: option (1)

From the plot, the intercept ln⁡A=6\ln A = 6. With ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT} and ln⁡k=2\ln k = 2: EaRT=4\frac{E_a}{RT} = 4, so T=66408.3×4=200 KT = \frac{6640}{8.3 \times 4} = 200\text{ K}.

Q87. Solutions

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.

  1. A only
  2. A and C only
  3. A and B only
  4. B and C only
Show answer and solution

Answer: option (3)

A is the definition of an ideal solution. Chloroform–acetone and phenol–aniline both form hydrogen bonds between the two components, stronger than the original interactions, so both show negative deviation; C is false.

Q88. Chemical Bonding and Molecular Structure

The highest occupied molecular orbital for Ne2\text{Ne}_2 is

  1. π2p∗\pi^*_{2p}
  2. σ2p∗\sigma^*_{2p}
  3. π2p\pi_{2p}
  4. σ2p\sigma_{2p}
Show answer and solution

Answer: option (2)

Ne2\text{Ne}_2 has 20 electrons, which fill all the n = 2 MOs, including π2p∗\pi^*_{2p} and finally σ2p∗\sigma^*_{2p}. The last filled (highest occupied) orbital is therefore σ2p∗\sigma^*_{2p} (bond order 0).

Q89. Electrochemistry

For a salt XY\textbf{XY}, which is a strong electrolyte, the plot of Λm\Lambda_{m} versus c\sqrt{c} has a slope of −90.0 S cm2 mol−3/2 L1/2-90.0\text{ S cm}^2\text{ mol}^{-3/2}\text{ L}^{1/2} at 298 K298\text{ K}. At 0.01 M0.01\text{ M} concentration of XY\textbf{XY}, the value of Λm\Lambda_{m} is 145.0 S cm2 mol−1145.0\text{ S cm}^2\text{ mol}^{-1}. The limiting molar conductivity of Y−\text{Y}^- ion (λY−∘\lambda^\circ_{\text{Y}^-}, in S cm2 mol−1\text{S cm}^2\text{ mol}^{-1}) at 298 K298\text{ K} will be
(Given : λX+∘=74.0 S cm2 mol−1\lambda^\circ_{\text{X}^+} = 74.0\text{ S cm}^2\text{ mol}^{-1})

  1. 90.0
  2. 76.0
  3. 80.0
  4. 100.0
Show answer and solution

Answer: option (3)

Kohlrausch: Λm=Λm∘−Ac\Lambda_m = \Lambda_m^\circ - A\sqrt{c}, so 145.0=Λm∘−90.0×0.1145.0 = \Lambda_m^\circ - 90.0 \times 0.1, giving Λm∘=154.0\Lambda_m^\circ = 154.0. Then λY−∘=154.0−74.0=80.0 S cm2 mol−1\lambda^\circ_{\text{Y}^-} = 154.0 - 74.0 = 80.0\text{ S cm}^2\text{ mol}^{-1}.

Q90. Classification of Elements and Periodicity in Properties

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

  1. Cl>F>Br>I\text{Cl} > \text{F} > \text{Br} > \text{I}
  2. I>Br>Cl>F\text{I} > \text{Br} > \text{Cl} > \text{F}
  3. F>Cl>Br>I\text{F} > \text{Cl} > \text{Br} > \text{I}
  4. Br>Cl>F>I\text{Br} > \text{Cl} > \text{F} > \text{I}
Show answer and solution

Answer: option (1)

Electron gain enthalpy (magnitude, most negative first) is Cl > F > Br > I. F is less negative than Cl because the incoming electron feels strong repulsion in the small, compact 2p2p subshell.

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Botany questions (Q91–135)

Q91. Cell Cycle and Cell Division

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

  1. Statement I is correct, but Statement II is false
  2. Statement I is incorrect, but Statement II is true
  3. Both Statement I and Statement II are true
  4. Both Statement I and Statement II are false
Show answer and solution

Answer: option (1)

By diakinesis at the end of prophase I the chromosomes are fully condensed, so Statement I is true. It is meiosis II, not meiosis I, that resembles mitosis, because meiosis I separates homologous chromosomes and halves the chromosome number.

Q92. Animal Kingdom

Which of the following is not a characteristic of chordates ?

  1. Absence of gills
  2. Presence of post anal part (tail)
  3. Presence of notochord
  4. Central nervous system is dorsal
Show answer and solution

Answer: option (1)

Chordates characteristically have a notochord, a dorsal hollow nerve cord, paired pharyngeal gill slits and a post-anal tail. Absence of gills is therefore not a chordate feature.

Q93. Morphology of Flowering Plants

Phyllotaxy is the pattern of arrangement of ________

  1. fruits
  2. sepals
  3. leaves
  4. flowers
Show answer and solution

Answer: option (3)

Phyllotaxy describes how leaves are arranged on the stem or branch: alternate, opposite or whorled.

Q94. Chemical Coordination and Integration

Which one of the following statements is incorrect ?

  1. Glucagon stimulates glycogenolysis
  2. β-cells of pancreas secrete insulin
  3. α-cells of pancreas secrete glucagon
  4. α-cells of pancreas secrete insulin
Show answer and solution

Answer: option (4)

In the islets of Langerhans, α\alpha-cells secrete glucagon and β\beta-cells secrete insulin. So the statement that α\alpha-cells secrete insulin is wrong.

Q95. Cell: The Unit of Life

The plastid that stores xanthophyll is known as ________ .

  1. aleuroplast
  2. amyloplast
  3. chloroplast
  4. chromoplast
Show answer and solution

Answer: option (4)

Chromoplasts contain fat-soluble carotenoid pigments such as carotene and xanthophyll, giving yellow, orange or red colours.

Q96. Biological Classification

Match List-I with List-II.

List-IList-II
A. Fusion of protoplasms between gametesI. Meiosis
B. Fusion of two nucleiII. Plasmogamy
C. Generation of haploid sporesIII. Karyogamy

Choose the correct answer from the options given below :

  1. A-III, B-II, C-I
  2. A-I, B-III, C-II
  3. A-II, B-III, C-I
  4. A-II, B-I, C-III
Show answer and solution

Answer: option (3)

Fusion of the protoplasms of two gametes is plasmogamy, fusion of their nuclei is karyogamy, and haploid spores are produced by meiosis. So A-II, B-III, C-I.

Q97. Plant Kingdom

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 the light of the above statements, choose the most appropriate answer from the options given below :

  1. Statement I is correct but Statement II is incorrect
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect
Show answer and solution

Answer: option (3)

In gymnosperms the gametophytes are not free-living; they stay within the sporangia on the sporophyte. The ovules are not enclosed in an ovary, so the seeds are naked. Both statements are correct.

Q98. Structural Organisation in Animals

In water, frogs respire using ________.

  1. lungs
  2. trachea
  3. skin
  4. buccal cavity
Show answer and solution

Answer: option (3)

In water a frog breathes through its moist skin (cutaneous respiration), where dissolved oxygen diffuses in. Lungs and the buccal cavity are used on land.

Q99. Biological Classification

Mad cow disease is caused by ________.

  1. Aspergillus sp.
  2. Mycoplasma sp.
  3. prions
  4. viroids
Show answer and solution

Answer: option (3)

Mad cow disease (bovine spongiform encephalopathy) is caused by prions, which are abnormally folded infectious proteins.

Q100. Photosynthesis in Higher Plants

Which of the following statements regarding photorespiration are correct ?
(a) Do not occur in C3 plants
(b) CO2\text{CO}_2 is consumed and O2\text{O}_2 is generated
(c) Phosphoglycolate is formed
(d) No synthesis of ATP and NADPH
Choose the correct answer from the options given below :

  1. (b) and (d) only
  2. (a) and (b) only
  3. (a) and (d) only
  4. (c) and (d) only
Show answer and solution

Answer: option (4)

Photorespiration happens in C3 plants, uses O2\text{O}_2 and releases CO2\text{CO}_2, so (a) and (b) are wrong. RuBisCO acting as an oxygenase forms phosphoglycolate, and the pathway makes no ATP or NADPH, so (c) and (d) are correct.

Q101. Photosynthesis in Higher Plants

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

  1. Release of oxygen →\rightarrow production of glucose →\rightarrow absorption spectra of chlorophyll a and b →\rightarrow role of air
  2. Production of glucose →\rightarrow role of air →\rightarrow release of oxygen →\rightarrow absorption spectra of chlorophyll a and b
  3. Absorption spectra of chlorophyll a and b →\rightarrow production of glucose →\rightarrow release of oxygen →\rightarrow role of air
  4. Role of air →\rightarrow release of oxygen →\rightarrow production of glucose →\rightarrow absorption spectra of chlorophyll a and b
Show answer and solution

Answer: option (4)

Priestley first showed the role of air (1770s), Ingenhousz then showed oxygen release by green parts in light, Sachs showed glucose (starch) production, and later work by Engelmann and others linked the absorption spectra of chlorophyll a and b to photosynthesis.

Q102. Photosynthesis in Higher Plants

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

  1. 1
  2. 18
  3. 6
  4. 3
Show answer and solution

Answer: option (2)

Each turn of the Calvin cycle fixes one CO2\text{CO}_2, so one glucose (6 carbons) needs 6 turns. Three glucose molecules need 3×6=183 \times 6 = 18 turns.

Q103. Cell: The Unit of Life

Mitochondrial inner membrane encloses ____________.

  1. mucus
  2. aqueous humor
  3. matrix
  4. cytosol
Show answer and solution

Answer: option (3)

The inner mitochondrial membrane surrounds the inner compartment called the matrix, which holds enzymes of the Krebs cycle, DNA and ribosomes.

Q104. Body Fluids and Circulation

Which of the following statements is incorrect ?

  1. Fibrin is produced from fibrinogen
  2. Fibrinogen is produced from fibrin
  3. Blood coagulates in response to an injury
  4. Blood clot consists of fibrins
Show answer and solution

Answer: option (2)

During clotting, thrombin converts the soluble plasma protein fibrinogen into insoluble fibrin threads. Fibrinogen is not made from fibrin, so statement (2) is incorrect.

Q105. Plant Kingdom

Sphenopsida class belongs to ____________.

  1. gymnosperms
  2. pteridophytes
  3. bryophytes
  4. angiosperms
Show answer and solution

Answer: option (2)

Pteridophytes are divided into Psilopsida, Lycopsida, Sphenopsida (e.g. Equisetum) and Pteropsida. So Sphenopsida is a pteridophyte class.

Q106. Biological Classification

Match List-I with List-II.

List-IList-II
A. SphericalI. Vibrio
B. RodII. Cocci
C. CommaIII. Spirilla
D. SpirillumIV. Bacilli

Choose the correct answer from the options given below :

  1. A-II, B-I, C-IV, D-III
  2. A-II, B-IV, C-I, D-III
  3. A-I, B-III, C-II, D-IV
  4. A-III, B-II, C-I, D-IV
Show answer and solution

Answer: option (2)

Spherical bacteria are cocci, rod-shaped are bacilli, comma-shaped are vibrio and spiral ones are spirilla. So A-II, B-IV, C-I, D-III.

Q107. Cell: The Unit of Life

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 mesosome
(d) They have peroxisomes

Choose the correct answer from the options given below :

  1. (a), (c) and (d) only
  2. (a), (b) and (c) only
  3. (b) and (c) only
  4. (a) and (c) only
Show answer and solution

Answer: option (2)

Prokaryotes have 70S ribosomes (50S + 30S subunits), may carry plasmids, and have mesosomes formed by infolding of the cell membrane. They lack membrane-bound organelles such as peroxisomes, so (a), (b) and (c) are correct.

Q108. Cell: The Unit of Life

Smooth endoplasmic reticulum ___________ .

  1. is actively involved in protein synthesis
  2. is a site for the synthesis of carbohydrates
  3. has ribosomes attached to its surface
  4. is the major site for the synthesis of lipids
Show answer and solution

Answer: option (4)

Smooth ER has no ribosomes and is the main site of lipid synthesis, including steroid hormones in animal cells. Protein synthesis takes place on rough ER.

Q109. Animal Kingdom

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 the light of the above statements, choose the most appropriate answer from the options given below :

  1. Statement I is correct but Statement II is incorrect
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect
Show answer and solution

Answer: option (1)

Reptilia comes from the creeping or crawling movement of these animals, so Statement I is correct. Most reptiles have a three-chambered heart but crocodiles have four chambers, so Statement II is false.

Q110. Structural Organisation in Animals

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

  1. 10
  2. 12
  3. 6
  4. 9
Show answer and solution

Answer: option (1)

The frog brain gives off 10 pairs of cranial nerves.

Q111. Cell: The Unit of Life

Cell theory was formulated by _______ .

  1. Singer and Nicolson
  2. Antonie Von Leeuwenhoek
  3. Schleiden and Schwann
  4. Robert Brown
Show answer and solution

Answer: option (3)

Schleiden (plants, 1838) and Schwann (animals, 1839) together formulated the cell theory. Virchow later added that new cells arise from pre-existing cells.

Q112. Chemical Coordination and Integration

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 :

  1. (c) and (d) only
  2. (b) and (c) only
  3. (a) and (b) only
  4. (a), (b) and (c) only
Show answer and solution

Answer: option (4)

The pituitary is divided anatomically into the adenohypophysis and neurohypophysis. The anterior part secretes FSH, MSH and prolactin, so (d) is false and (a), (b) and (c) are correct.

Q113. Biological Classification

Which of the following is not a prokaryote ?

  1. Mycoplasma
  2. Fungi
  3. Bacteria
  4. Blue green algae
Show answer and solution

Answer: option (2)

Mycoplasma, bacteria and blue-green algae (cyanobacteria) are prokaryotes in Kingdom Monera. Fungi have a true nucleus and are eukaryotes.

Q114. Plant Growth and Development

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 7th7^{\text{th}} day ?

  1. 230 cm
  2. 460 cm
  3. 50 cm
  4. 170 cm
Show answer and solution

Answer: option (1)

Arithmetic growth follows Lt=L0+rt=20+30×7=230 cmL_t = L_0 + rt = 20 + 30 \times 7 = 230\ \text{cm}.

Q115. Neural Control and Coordination

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

  1. Pia mater
  2. Pericardium
  3. Arachnoid
  4. Dura mater
Show answer and solution

Answer: option (2)

The brain is covered by three meninges: dura mater, arachnoid and pia mater. The pericardium is the membrane around the heart.

Q116. Plant Growth and Development

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

  1. Indole butyric acid
  2. Ethephon
  3. Abscisic acid
  4. Gibberellin
Show answer and solution

Answer: option (4)

Gibberellins cause bolting, which is internode elongation just before flowering in rosette plants such as cabbage and beet.

Q117. Morphology of Flowering Plants

Match List-I with List-II.

List-IList-II
A. Marginal placentationI. _Argemone_
B. Axile placentationII. Tomato
C. Parietal placentationIII. _Primrose_
D. Free central placentationIV. Pea

Choose the correct answer from the options given below :

  1. A-IV, B-III, C-I, D-II
  2. A-IV, B-II, C-I, D-III
  3. A-II, B-IV, C-I, D-III
  4. A-IV, B-II, C-III, D-I
Show answer and solution

Answer: option (2)

Pea has marginal placentation, tomato axile, Argemone parietal and Primrose free central. So A-IV, B-II, C-I, D-III.

Q118. Plant Growth and Development

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

  1. Abscisic acid
  2. Gibberellin
  3. 2,4-D
  4. Kinetin
Show answer and solution

Answer: option (3)

2,4-D is a synthetic auxin used as a weedicide to kill broad-leaved dicot weeds. It does not affect mature monocots.

Q119. Animal Kingdom

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 the light of the above statements, choose the most appropriate answer from the options given below :

  1. Statement I is correct but Statement II is incorrect
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect
Show answer and solution

Answer: option (1)

Statement I is the correct definition of radial symmetry. Adult echinoderms are radially symmetrical but their larvae are bilaterally symmetrical, so Statement II is false.

Q120. Biomolecules

Match List-I with List-II.

List-IList-II
A. StarchI. Fights infection
B. AntibodyII. Energy storage
C. Concanavalin AIII. Glucose transport
D. Glut-4IV. Lectin

Choose the correct answer from the options given below :

  1. A-II, B-I, C-III, D-IV
  2. A-I, B-II, C-III, D-IV
  3. A-I, B-II, C-IV, D-III
  4. A-II, B-I, C-IV, D-III
Show answer and solution

Answer: option (4)

Starch is a storage polysaccharide, antibodies fight infection, concanavalin A is a lectin, and GLUT-4 carries glucose into cells. So A-II, B-I, C-IV, D-III.

Q121. Locomotion and Movement

The number of vertebrae in a human is ________.

  1. 26
  2. 206
  3. 7
  4. 12
Show answer and solution

Answer: option (1)

The adult human vertebral column has 26 vertebrae: 7 cervical, 12 thoracic, 5 lumbar, 1 sacrum and 1 coccyx.

Q122. Photosynthesis in Higher Plants

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

  1. Xanthophylls
  2. Carotenoids
  3. Chlorophyll b
  4. Chlorophyll a
Show answer and solution

Answer: option (4)

The reaction centre of PS I is a chlorophyll a molecule with an absorption peak at 700 nm, hence the name P700.

Q123. The Living World

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 :

  1. (g), (c), (d), (b), (e), (a), (f)
  2. (f), (c), (b), (g), (d), (e), (a)
  3. (g), (a), (e), (c), (b), (d), (f)
  4. (a), (c), (d), (g), (f), (b), (e)
Show answer and solution

Answer: option (3)

From the lowest rank upward the order is species, genus, family, order, class, phylum, kingdom, which is (g), (a), (e), (c), (b), (d), (f).

Q124. The Living World

Match List-I with List-II.

List-IList-II
A. FamilyI. Sapindales
B. GenusII. Dicotyledonae
C. ClassIII. Anacardiaceae
D. PhylumIV. Angiospermae
E. OrderV. _Mangifera_

Choose the correct answer from the options given below :

  1. A-II, B-III, C-V, D-I, E-IV
  2. A-III, B-V, C-II, D-IV, E-I
  3. A-I, B-V, C-II, D-IV, E-III
  4. A-II, B-I, C-III, D-IV, E-V
Show answer and solution

Answer: option (2)

For mango, the family is Anacardiaceae, the genus is Mangifera, the class is Dicotyledonae, the phylum/division is Angiospermae and the order is Sapindales. So A-III, B-V, C-II, D-IV, E-I.

Q125. Biomolecules

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

  1. (b), (c), (d), (a)
  2. (b), (a), (c), (d)
  3. (a), (b), (c), (d)
  4. (c), (a), (b), (d)
Show answer and solution

Answer: option (3)

By percentage weight the human body is about 65% oxygen, 18.5% carbon, 9.5% hydrogen and 3.3% nitrogen, so the order is O > C > H > N.

Q126. Cell: The Unit of Life

Match List-I with List-II.

List-IList-II
A. CristaeI. Flat membrane sacs in stroma of chloroplast
B. CisternaeII. Infoldings in mitochondria
C. ThylakoidsIII. Cell membrane
D. PhospholipidIV. Disc shaped sacs in the Golgi apparatus

Choose the correct answer from the options given below :

  1. A-II, B-IV, C-III, D-I
  2. A-IV, B-III, C-I, D-II
  3. A-III, B-IV, C-I, D-II
  4. A-II, B-IV, C-I, D-III
Show answer and solution

Answer: option (4)

Cristae are infoldings of the inner mitochondrial membrane, cisternae are the disc-shaped sacs of the Golgi apparatus, thylakoids are the flat membrane sacs in the chloroplast stroma, and phospholipids make up the cell membrane. So A-II, B-IV, C-I, D-III.

Q127. Body Fluids and Circulation

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

  1. 100 – 110
  2. 120 – 140
  3. 28 – 30
  4. 70 – 75
Show answer and solution

Answer: option (4)

The SA node is the pacemaker. In a healthy adult it fires about 70–75 action potentials per minute, which sets the normal heart rate.

Q128. Biological Classification

Symbiotic association between fungi and algae are called __________.

  1. mycorrhiza
  2. chrysophytes
  3. lichens
  4. sponges
Show answer and solution

Answer: option (3)

Lichens are mutualistic associations between algae and fungi. Mycorrhiza is a fungus living with plant roots.

Q129. Respiration in Plants

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

  1. 103
  2. 412
  3. 206
  4. 309
Show answer and solution

Answer: option (2)

Glycolysis splits each glucose into 2 pyruvic acid molecules, so 206 glucose give 206×2=412206 \times 2 = 412 pyruvic acid.

Q130. Locomotion and Movement

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

  1. Femur-patella-tibia-tarsal
  2. Femur-tarsal-patella-tibia
  3. Femur-tibia-patella-tarsal
  4. Patella-femur-tibia-tarsal
Show answer and solution

Answer: option (1)

In the lower limb the femur is the thigh bone. The patella covers the knee joint, the tibia (with the fibula) forms the leg, and the tarsals form the ankle.

Q131. The Living World

Genus represents ___________ .

  1. a group of closely related species
  2. a group of closely related families
  3. an individual plant or animal
  4. a population of plants and animals
Show answer and solution

Answer: option (1)

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

Q132. Cell Cycle and Cell Division

The correct sequence of adult cell cycle phases is ___________ .

  1. G1-S-G2-M
  2. S-M-G2-G1
  3. G1-G2-S-M
  4. G1-M-G2-S
Show answer and solution

Answer: option (1)

Interphase runs G1 → S → G2 and is followed by the M phase, so the cycle is G1-S-G2-M.

Q133. Cell: The Unit of Life

Endomembrane system includes ___________

  1. mitochondria, chloroplast, peroxisomes and vacuole
  2. Golgi complex, chloroplast, peroxisomes and vacuole
  3. endoplasmic reticulum, Golgi complex, lysosomes and vacuole
  4. endoplasmic reticulum, chloroplast, peroxisomes and vacuole
Show answer and solution

Answer: option (3)

The endomembrane system is made up of the ER, the Golgi complex, lysosomes and vacuoles, whose functions are coordinated. Mitochondria, chloroplasts and peroxisomes are not included.

Q134. Photosynthesis in Higher Plants

Photorespiration reaction catalyzed by RuBisCo is shown below :
RuBP+O2→3-Phosphoglycerate+X\text{RuBP} + \text{O}_2 \rightarrow \text{3-Phosphoglycerate} + \text{X}
Identify "X" from the given options :

  1. Oxaloacetate
  2. Malate
  3. Phosphoenolpyruvate
  4. 2-Phosphoglycolate
Show answer and solution

Answer: option (4)

When RuBisCO acts as an oxygenase, RuBP + O2\text{O}_2 gives one 3-phosphoglycerate and one 2-phosphoglycolate.

Q135. Morphology of Flowering Plants

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 :

  1. (a) and (b) only
  2. (b), (c) and (d) only
  3. (a), (b) and (c) only
  4. (d) only
Show answer and solution

Answer: option (3)

Solanaceae flowers are bisexual and actinomorphic, with 5 united sepals and 5 epipetalous stamens. The bicarpellary syncarpous ovary is superior, so (d) is false.

Zoology questions (Q136–180)

Q136. Human Reproduction

Sperm motility is due to ________.

  1. amoeboid movement
  2. muscular movement
  3. flagellar movement
  4. ciliary movement
Show answer and solution

Answer: option (3)

A sperm swims by the whip-like beating of its tail, a flagellum powered by mitochondria in the middle piece. So its motility is flagellar movement.

Q137. Biodiversity and Conservation

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 Abingdon tortoise.

In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A is correct but R is not correct
  2. A is not correct but R is correct
  3. Both A and R are correct and R is the correct explanation of A
  4. Both A and R are correct but R is not the correct explanation of A
Show answer and solution

Answer: option (3)

Goats were introduced to the Galapagos and, being far better browsers, outcompeted the Abingdon tortoise for food, driving it extinct within about ten years. Both statements are true and the browsing advantage of goats explains the extinction.

Q138. Organisms and Populations

Match List-I with List-II.

List-IList-II
A. Both species are harmedI. Predation
B. One species is harmed and the other is benefitedII. Mutualism
C. Both species are benefitedIII. Competition
D. One is benefited while the other has no effectIV. Commensalism

Choose the correct answer from the options given below :

  1. A-II, B-I, C-IV, D-III
  2. A-III, B-I, C-II, D-IV
  3. A-III, B-IV, C-II, D-I
  4. A-I, B-II, C-III, D-IV
Show answer and solution

Answer: option (2)

Competition harms both species (−/−), predation benefits one and harms the other (+/−), mutualism benefits both (+/+), and commensalism benefits one with no effect on the other (+/0). So A-III, B-I, C-II, D-IV.

Q139. Human Reproduction

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

  1. Vasa efferentia
  2. Infundibulum
  3. Rete testis
  4. Epididymis
Show answer and solution

Answer: option (2)

The infundibulum is the funnel-shaped end of the fallopian tube in the female tract. Vasa efferentia, rete testis and epididymis are all ducts of the male system.

Q140. Sexual Reproduction in Flowering Plants

Which of the following plant produces non-albuminous seeds?

  1. Barley
  2. Pea
  3. Wheat
  4. Maize
Show answer and solution

Answer: option (2)

In pea the endosperm is used up by the developing embryo, so the mature seed has no endosperm (non-albuminous). Barley, wheat and maize are cereals with persistent endosperm (albuminous).

Q141. Molecular Basis of Inheritance

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

  1. Genes ii, zz, yy and aa share single common promoter
  2. Galactose can act as an inducer of lac operon
  3. Gene ii is constitutively expressed
  4. Lactose activates repressor to bind to the operator
Show answer and solution

Answer: option (3)

The regulator gene ii has its own promoter and is expressed constitutively, making repressor all the time. Only zz, yy, aa share the lac promoter; the inducer is lactose (allolactose), which inactivates the repressor rather than activating it.

Q142. Human Health and Disease

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

  1. Morphine
  2. Anti-retroviral drugs
  3. Interferon
  4. Antibiotics
Show answer and solution

Answer: option (1)

Morphine, an opioid from the poppy latex, is a strong sedative and painkiller given to patients after surgery. The other options do not relieve pain.

Q143. Locomotion and Movement

Which of the following is the correct order of arrangement of vertebrate column from the head to toe?

  1. Cervical vertebra, lumbar vertebra, thoracic vertebra, sacrum
  2. Cervical vertebra, thoracic vertebra, lumbar vertebra, sacrum
  3. Cervical vertebra, thoracic vertebra, sacrum, lumbar vertebra
  4. Sacrum, lumbar vertebra, thoracic vertebra, cervical vertebra
Show answer and solution

Answer: option (2)

The vertebral column runs cervical (7) → thoracic (12) → lumbar (5) → sacral (fused sacrum) → coccygeal from head downward.

Q144. Locomotion and Movement

Muscle contraction is initiated by a signal sent by the central nervous system by the release of _______ .

  1. cyclic guanine monophosphate
  2. cyclic adenine monophosphate
  3. acetyl choline
  4. acetyl coenzyme A
Show answer and solution

Answer: option (3)

A motor neuron signal reaches the neuromuscular junction, where acetylcholine is released and generates an action potential in the sarcolemma, starting contraction.

Q145. Evolution

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 the light of the above statements, choose the most appropriate answer from the options given below.

  1. A is true but R is false
  2. A is false but R is true
  3. Both A and R are correct and R is the correct explanation of A
  4. Both A and R are true, but R is not the correct explanation of A
Show answer and solution

Answer: option (3)

Human and bat forelimbs share the same basic bone plan (humerus, radius-ulna, carpals, metacarpals, phalanges) despite different functions, which is exactly what makes them homologous. So both are true and R explains A.

Q146. Principles of Inheritance and Variation

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 the light of the above statements, choose the correct answer from the options given below :

  1. Statement I is correct but Statement II is incorrect
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect
Show answer and solution

Answer: option (4)

Down's syndrome is trisomy 21 (an extra chromosome 21), and Turner's syndrome is 45,XO (loss of one X). Each statement describes the other's cause, so both are incorrect.

Q147. Excretory Products and their Elimination

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+\text{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 :

  1. (a), (b) and (c) only
  2. (a), (b) and (d) only
  3. (a) and (b) only
  4. (b), (c) and (d) only
Show answer and solution

Answer: option (2)

The descending limb lets water out but not salts, the ascending limb moves electrolytes out actively or passively but not water, and this countercurrent arrangement concentrates the filtrate in the loop. NCERT notes that reabsorption in Henle's loop is minimal, so (c) is excluded and (a), (b), (d) are correct.

Q148. Sexual Reproduction in Flowering Plants

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

  1. 54
  2. 72
  3. 18
  4. 36
Show answer and solution

Answer: option (1)

Here 2n=362n = 36, so n=18n = 18. Endosperm is triploid (3n3n), giving 3×18=543 \times 18 = 54 chromosomes.

Q149. Biodiversity and Conservation

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 :

  1. (e) > (b) > (a) > (c) > (d)
  2. (b) > (a) > (d) > (c) > (e)
  3. (c) > (b) > (d) > (e) > (a)
  4. (d) > (a) > (c) > (b) > (e)
Show answer and solution

Answer: option (4)

NCERT figures for the Amazon: about 1,25,000 invertebrates, 40,000 plants, 3,000 fishes, 1,300 birds and 427 mammals. So invertebrates > plants > fishes > birds > mammals.

Q150. Evolution

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

  1. 0.10
  2. 0.99
  3. 0.01
  4. 0.02
Show answer and solution

Answer: option (3)

At Hardy-Weinberg equilibrium the frequency of AA is p2=(0.1)2=0.01p^2 = (0.1)^2 = 0.01.

Q151. Body Fluids and Circulation

The opening between the right atrium and the right ventricle is guarded by ________.

  1. semilunar valve
  2. sino-atrial node
  3. bicuspid valve
  4. tricuspid valve
Show answer and solution

Answer: option (4)

The right atrioventricular opening is guarded by the three-cusped tricuspid valve; the bicuspid (mitral) valve guards the left side and semilunar valves guard the arterial exits.

Q152. Molecular Basis of Inheritance

Which of the following enzymes synthesizes precursor mRNA?

  1. RNA polymerase III
  2. DNA polymerase
  3. RNA polymerase I
  4. RNA polymerase II
Show answer and solution

Answer: option (4)

In eukaryotes RNA polymerase II transcribes hnRNA, the precursor of mRNA. Pol I makes rRNAs and Pol III makes tRNA, 5S rRNA and snRNAs.

Q153. Biotechnology: Principles and Processes

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 the light of the above statements, choose the most appropriate answer from the options given below :

  1. A is correct but R is not correct
  2. A is not correct but R is correct
  3. Both A and R are correct and R is the correct explanation of A
  4. Both A and R are correct but R is not the correct explanation of A
Show answer and solution

Answer: option (3)

To get DNA out, the cell must first be broken open, so bacterial walls are digested with lysozyme and plant walls with cellulase. Both are true and the need for cell disruption is why these enzymes are used.

Q154. Human Reproduction

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

  1. Progesterone
  2. LH
  3. hCG
  4. Estrogen
Show answer and solution

Answer: option (2)

The placenta acts as an endocrine tissue making hCG, hPL, estrogens and progesterone. LH is made by the anterior pituitary, not the placenta.

Q155. Human Health and Disease

Colostrum, secreted by mother during initial days of lactation, is abundant in ________.

  1. IgA
  2. IgD
  3. IgG
  4. IgM
Show answer and solution

Answer: option (1)

Colostrum is rich in IgA antibodies, which give the newborn passive immunity.

Q156. Organisms and Populations

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

  1. 10 million
  2. 100 million
  3. 1 million
  4. 5 million
Show answer and solution

Answer: option (1)

Net growth rate r=b−d=0.002−0.002=0r = b - d = 0.002 - 0.002 = 0, so the population neither grows nor declines and stays at 10 million.

Q157. Biotechnology: Principles and Processes

Given below are two statements :

Statement I : Plasmids are autonomously replicating DNA.

Statement II : Plasmids are extrachromosomal DNA.

In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Statement I is correct but Statement II is incorrect
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect
Show answer and solution

Answer: option (3)

Plasmids are small circular DNA molecules that lie outside the bacterial chromosome and replicate independently of it. Both statements are true.

Q158. Biotechnology: Principles and Processes

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

  1. annealing
  2. ligation
  3. denaturation
  4. extension
Show answer and solution

Answer: option (1)

PCR cycles are denaturation (strands separate), annealing (primers bind to complementary sites) and extension (Taq polymerase builds new strands). Primer binding is annealing.

Q159. Biotechnology: Principles and Processes

Match List-I with List-II.

List-IList-II
A. TransformationI. Restriction enzyme
B. Cloning siteII. Transfer DNA to host bacteria
C. SelectionIII. Replication
D. OriIV. Antibiotic

Choose the correct answer from the options given below :

  1. A-III, B-IV, C-II, D-I
  2. A-IV, B-I, C-III, D-II
  3. A-II, B-I, C-IV, D-III
  4. A-I, B-II, C-IV, D-III
Show answer and solution

Answer: option (3)

Transformation is getting DNA into host bacteria, a cloning site is a recognition site for a restriction enzyme, selection uses antibiotic-resistance markers, and ori is where replication starts. So A-II, B-I, C-IV, D-III.

Q160. Evolution

Adaptive radiation in placental mammals and Australian Marsupials leading to similarity between distant species is an example of ________.

  1. founder effect
  2. genetic drift
  3. divergent evolution
  4. convergent evolution
Show answer and solution

Answer: option (4)

Placental mammals and Australian marsupials evolved separately yet produced look-alike forms (e.g. wolf and Tasmanian wolf) because they adapted to similar habitats. Similarity in unrelated lines from similar pressures is convergent evolution.

Q161. Human Reproduction

The covering of ovum at ovulation is ________.

  1. zona pellucida
  2. chorion
  3. endometrium
  4. zona radiata
Show answer and solution

Answer: option (1)

The secondary oocyte released at ovulation is surrounded by the zona pellucida, a glycoprotein layer, with corona radiata cells outside it.

Q162. Ecosystem

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 :

  1. (a) and (c) only
  2. (b) and (c) only
  3. (a), (b) and (c)
  4. (a) and (b) only
Show answer and solution

Answer: option (4)

Energy moves one way from producers to consumers and is lost as heat at each step, so an energy pyramid is always upright. Transfer follows the 10% law, not 1%, so only (a) and (b) are correct.

Q163. Sexual Reproduction in Flowering Plants

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

  1. 8
  2. 12
  3. 2
  4. 6
Show answer and solution

Answer: option (3)

A typical anther is bilobed and each lobe has two thecae (dithecous), giving four microsporangia in all. So each lobe holds 2 thecae.

Q164. Human Health and Disease

Which of the following statements is correct about Plasmodium?

  1. Gametocytes develop in mosquito gut
  2. Fertilization takes place in mosquito gut
  3. Reproduces sexually in liver cells
  4. Reproduces sexually in RBCs
Show answer and solution

Answer: option (2)

Male and female gametes of Plasmodium fuse in the stomach (gut) of the female Anopheles mosquito. Gametocytes form in human RBCs, and the liver and RBC stages are asexual.

Q165. Principles of Inheritance and Variation

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

  1. Father : IBiI^Bi and Mother : IBiI^Bi
  2. Father : IAIBI^AI^B and Mother : IAiI^Ai
  3. Father : IAiI^Ai and Mother : IBiI^Bi
  4. Father : IAiI^Ai and Mother : IAiI^Ai
Show answer and solution

Answer: option (2)

A blood group O child is iiii and needs an ii allele from each parent. A father who is IAIBI^AI^B has no ii allele, so option 2 cannot give an O child, while every other pair has ii in both parents.

Q166. Microbes in Human Welfare

Which of the following is used as a clot buster?

  1. Cyclosporin A
  2. Statins
  3. Streptokinase
  4. Penicillin
Show answer and solution

Answer: option (3)

Streptokinase from Streptococcus dissolves blood clots and is used as a clot buster. Cyclosporin A is an immunosuppressant, statins lower cholesterol and penicillin is an antibiotic.

Q167. Reproductive Health

Which of the following disease is not sexually transmitted?

  1. Gonorrhoea
  2. Genital warts
  3. Syphilis
  4. Tuberculosis
Show answer and solution

Answer: option (4)

Gonorrhoea, genital warts and syphilis are STIs; tuberculosis spreads through air droplets, not sexual contact.

Q168. Principles of Inheritance and Variation

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_1 progeny plants are all tall and none are dwarf.

Reason R : Stem height is a contrasting trait, with tall being dominant and dwarf being recessive.

In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A is correct but R is not correct
  2. A is not correct but R is correct
  3. Both A and R are correct and R is the correct explanation of A
  4. Both A and R are correct but R is not the correct explanation of A
Show answer and solution

Answer: option (3)

Crossing true-breeding tall and dwarf plants gave an all-tall F1 because tall is dominant over dwarf for this contrasting trait. Both statements are true and R explains A.

Q169. Reproductive Health

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

  1. Intra cytoplasmic sperm injection (ICSI)
  2. Embryo transfer (ET)
  3. Gamete intra fallopian transfer (GIFT)
  4. Zygote intra fallopian transfer (ZIFT)
Show answer and solution

Answer: option (1)

In ICSI a single sperm is injected straight into the cytoplasm of the ovum in the lab. GIFT, ZIFT and ET transfer gametes or embryos instead.

Q170. Biotechnology and its Applications

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

  1. by proteases
  2. by nucleases
  3. due to alkaline pH
  4. due to acidic pH
Show answer and solution

Answer: option (3)

Bt protoxin crystals dissolve in the alkaline pH of the insect midgut, which converts them to the active toxin that binds midgut cells and makes pores. NCERT credits the activation to the alkaline pH.

Q171. Organisms and Populations

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 the light of the above statements, choose the most appropriate answer from the options given below :

  1. A is correct but R is not correct
  2. A is not correct but R is correct
  3. Both A and R are correct and R is the correct explanation of A
  4. Both A and R are correct but R is not the correct explanation of A
Show answer and solution

Answer: option (3)

Because resources are finite, no population can grow exponentially forever. The logistic model includes a carrying capacity, which makes it more realistic, so R explains A.

Q172. Human Health and Disease

Which of the following are secondary lymphoid organs?

(a) Bone marrow

(b) Tonsils

(c) Spleen

(d) Thymus

Choose the correct answer from the options given below:

  1. (b) and (d) only
  2. (a) and (d) only
  3. (a) and (b) only
  4. (b) and (c) only
Show answer and solution

Answer: option (4)

Bone marrow and thymus are primary lymphoid organs where lymphocytes form and mature. Tonsils and spleen are secondary lymphoid organs where lymphocytes meet antigens.

Q173. Sexual Reproduction in Flowering Plants

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

  1. Central cells
  2. Polar nucleus
  3. Antipodals
  4. Synergids
Show answer and solution

Answer: option (4)

The synergids carry a filiform apparatus at the micropylar end that guides the pollen tube into the embryo sac.

Q174. Human Reproduction

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 transform into corpus luteum and secretes large amount of estrogen.

In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Statement I is correct but Statement II is incorrect
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect
Show answer and solution

Answer: option (1)

The LH surge in mid-cycle ruptures the Graafian follicle and causes ovulation. The leftover follicle becomes the corpus luteum, which mainly secretes large amounts of progesterone, not estrogen, so Statement II is wrong.

Q175. Evolution

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 the light of the above statements, choose the most appropriate answer from the options given below :

  1. Statement I is correct but Statement II is incorrect
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect
Show answer and solution

Answer: option (2)

Modern Homo sapiens arose in Africa, not Australia, and spread across continents. NCERT places this during the ice age about 75,000–10,000 years ago, so only Statement II is correct.

Q176. Ecosystem

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

  1. Herbivores
  2. Carnivores
  3. Parasites
  4. Predators
Show answer and solution

Answer: option (1)

Herbivores eat producers directly and so occupy the second trophic level as primary consumers.

Q177. Evolution

Which of the following is not evidence for evolution?

  1. Embryological support for evolution as proposed by Ernst Heckel
  2. Divergent evolution of anatomical structures such as forelimbs
  3. Convergent evolution of traits like wings of birds and butterflies
  4. Paleontological evidence from fossil records
Show answer and solution

Answer: option (1)

Haeckel's idea that embryos pass through adult stages of ancestors was rejected by Karl Ernst von Baer, so it is not accepted evidence. Homology, analogy and fossils are all valid evidence.

Q178. Animal Kingdom

Sponges exchange O2\text{O}_2 with CO2\text{CO}_2 by ________.

  1. tracheal tubes
  2. gills
  3. simple diffusion over their entire body surfaces
  4. moist cuticle
Show answer and solution

Answer: option (3)

Sponges have no respiratory organs; water flowing through the canal system lets O2\text{O}_2 and CO2\text{CO}_2 diffuse across the body cells.

Q179. Chemical Coordination and Integration

Match List-I with List-II.

List-IList-II
(A) Excess growth hormone(I) Reabsorption of water and electrolytes in kidney
(B) Luteinizing hormone(II) Contraction of uterus during child birth
(C) Vasopressin(III) Acromegaly
(D) Oxytocin(IV) Ovulation

Choose the correct answer from the options given below:

  1. A-II, B-IV, C-I, D-III
  2. A-IV, B-III, C-I, D-II
  3. A-III, B-IV, C-II, D-I
  4. A-III, B-IV, C-I, D-II
Show answer and solution

Answer: option (4)

Excess GH in adults causes acromegaly, LH triggers ovulation, vasopressin (ADH) increases water and electrolyte reabsorption in the kidney, and oxytocin contracts the uterus during birth. So A-III, B-IV, C-I, D-II.

Q180. Evolution

Natural selection can lead to ________.

(a) stabilisation

(b) genetic drift

(c) directional change

(d) disruption

Choose the correct answer from the options given below :

  1. (a), (b), (c) and (d)
  2. (a) and (c) only
  3. (a) only
  4. (a), (c) and (d) only
Show answer and solution

Answer: option (4)

Natural selection can be stabilising, directional or disruptive. Genetic drift is a separate, chance-driven change in allele frequency, not an outcome of selection.

Questions by chapter

How the 180 questions were spread across chapters in this paper.

Physics chapterQuestions
Dual Nature of Radiation and Matter3
Electromagnetic Induction3
Electrostatic Potential and Capacitance3
Ray Optics and Optical Instruments3
System of Particles and Rotational Motion3
Atoms2
Electromagnetic Waves2
Gravitation2
Kinetic Theory2
Mechanical Properties of Fluids2
Moving Charges and Magnetism2
Oscillations2
Semiconductor Electronics: Materials, Devices and Simple Circuits2
Thermodynamics2
Units and Measurement2
Work, Energy and Power2
Alternating Current1
Current Electricity1
Electric Charges and Fields1
Laws of Motion1
Motion in a Straight Line1
Nuclei1
Thermal Properties of Matter1
Waves1
Chemistry chapterQuestions
Coordination Compounds6
Aldehydes, Ketones and Carboxylic Acids3
Organic Chemistry – Some Basic Principles and Techniques3
The d- and f-Block Elements3
The p-Block Elements3
Thermodynamics3
Amines2
Biomolecules2
Chemical Bonding and Molecular Structure2
Chemical Kinetics2
Classification of Elements and Periodicity in Properties2
Electrochemistry2
Hydrocarbons2
Redox Reactions2
Solutions2
Some Basic Concepts of Chemistry2
Alcohols, Phenols and Ethers1
Equilibrium1
Haloalkanes and Haloarenes1
Structure of Atom1
Botany chapterQuestions
Cell: The Unit of Life7
Biological Classification5
Photosynthesis in Higher Plants5
Animal Kingdom3
Morphology of Flowering Plants3
Plant Growth and Development3
The Living World3
Biomolecules2
Body Fluids and Circulation2
Cell Cycle and Cell Division2
Chemical Coordination and Integration2
Locomotion and Movement2
Plant Kingdom2
Structural Organisation in Animals2
Neural Control and Coordination1
Respiration in Plants1
Zoology chapterQuestions
Evolution6
Human Reproduction5
Biotechnology: Principles and Processes4
Human Health and Disease4
Sexual Reproduction in Flowering Plants4
Organisms and Populations3
Principles of Inheritance and Variation3
Biodiversity and Conservation2
Ecosystem2
Locomotion and Movement2
Molecular Basis of Inheritance2
Reproductive Health2
Animal Kingdom1
Biotechnology and its Applications1
Body Fluids and Circulation1
Chemical Coordination and Integration1
Excretory Products and their Elimination1
Microbes in Human Welfare1
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Questions students ask

How many questions were in the Re-NEET 2026 paper?

180 questions for 720 marks: 45 each from Physics, Chemistry, Botany and Zoology. All were compulsory, with +4 for a correct answer and −1 for a wrong one.

Which question was dropped in Re-NEET 2026?

In booklet code 60, NTA dropped Q43 (Physics, units and measurement) in the final answer key. Every candidate got 4 marks for it.

Was Re-NEET 2026 a computer-based test?

No. The 21 June 2026 re-exam was held on pen and paper (OMR). You can still practise this paper on an NTA-style CBT screen on this site.

Who wrote these solutions?

The solutions are written by the Testline NEET team. The answers follow NTA's final answer key.