NEET (UG) · 2025

NEET 2025, question by question

Which chapter each question came from, how the paper was spread, and how to solve each one. Keep the official paper open beside this page.

Sit this paper free All years, chapter-wise

Summary

The heaviest chapters in each subject, and how the questions were spread by level. One year is not a forecast.

Physics
45 Q
1. Optics4
2. Properties of Bulk Matter3
3. Magnetic Effects of Current and Magnetism3
24 easy · 19 medium · 2 hard
Chemistry
45 Q
1. Hydrocarbons4
2. Organic Chemistry: Some Basic Principles and Techniques4
3. Coordination Compounds3
31 easy · 13 medium · 1 hard
Botany
50 Q
1. Molecular Basis of Inheritance8
2. Sexual Reproduction in Flowering Plants6
3. Plant Kingdom5
37 easy · 12 medium · 1 hard
Zoology
40 Q
1. Biotechnology: Principles and Processes6
2. Human Reproduction5
3. Human Health and Disease5
27 easy · 13 medium · 0 hard

Chapters

Every chapter asked, and the NCERT chapters to read for it.

Molecular Basis of Inheritance8
Class 12: Molecular Basis of Inheritance
Sexual Reproduction in Flowering Plants6
Class 12: Sexual Reproduction in Flowering Plants
Plant Kingdom5
Class 11: Plant Kingdom
Organisms and Populations4
Class 12: Organisms and Populations
Cell: The Unit of Life4
Class 11: Cell: The Unit of Life
Microbes in Human Welfare4
Class 12: Microbes in Human Welfare
Ecosystem3
Class 12: Ecosystem
Principles of Inheritance and Variation3
Class 12: Principles of Inheritance and Variation
Biodiversity and Conservation2
Class 12: Biodiversity and Conservation
Morphology of Flowering Plants2
Class 11: Morphology of Flowering Plants
Plant Growth and Development2
Class 11: Plant Growth and Development
Photosynthesis in Higher Plants2
Class 11: Photosynthesis in Higher Plants
Respiration in Plants1
Class 11: Respiration in Plants
The Living World1
Class 11: The Living World
Anatomy of Flowering Plants1
Class 11: Anatomy of Flowering Plants
Biological Classification1
Class 11: Biological Classification
Cell Cycle and Cell Division1
Class 11: Cell Cycle and Cell Division

Questions

Tap Explain on any question for a short summary and our full working.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
1	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctHard
A hard single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Measure $y$ from the level of the flat liquid surface far from the wall, and take a thin strip of the surface between $x$ and $x + dx$, one unit long in the $z$ direction. Surface tension pulls on each edge of the strip along the surface. Its vertical parts are $S\sin\theta(x+dx)$ upward on the right edge and $S\sin\theta(x)$ downward on the left edge, so the net vertical pull is $S\,[\theta(x+dx) - \theta(x)] = S\frac{d\theta}{dx}dx = S\frac{d^2y}{dx^2}dx$ (using $\theta = \frac{dy}{dx}$). Just under a surface raised by $y$ above the flat level, the liquid pressure is lower than the atmosphere by $\rho g y$ (hydrostatics). So the pressure difference pushes the strip down with a force $\rho g y\,dx$ per unit length. For the strip (massless film) to be in equilibrium: $S\frac{d^2y}{dx^2}dx = \rho g y\,dx$, that is $\frac{d^2y}{dx^2} = \frac{\rho g}{S}y$. (Check: the solution $y = y_0 e^{(x-L)/\ell}$ with $\ell = \sqrt{S/\rho g}$ rises towards the wall and dies away from it, the shape in the figure.) Answer: (2).
2	OpticsOptical instruments	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a compound microscope in normal adjustment (final image at infinity), with the tube length $L$ taken as the distance between the inner foci, $m = \frac{L}{f_o}\cdot\frac{D}{f_e} = \frac{40}{2}\times\frac{25}{4} = 20\times6.25 = 125$. (If the final image were formed at the near point the eyepiece would give $1 + \frac{D}{f_e} = 7.25$ and $m = 145$, which is not offered, so normal adjustment is meant.) Answer: (2) 125.
3	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctEasy
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The magnetic force $q\vec{v}\times\vec{B}$ is perpendicular to both $\vec{v}$ and $\vec{B}$. To cancel it, the electric force $q\vec{E}$ must be equal and opposite, so $\vec{E}$ must be perpendicular to $\vec{B}$ (and to $\vec{v}$), with $qE = qvB \Rightarrow E = vB$. $v = \frac{c}{100} = 3\times10^6$ m/s, so $E = 3\times10^6\times9\times10^{-4} = 2700 = 27\times10^2$ V m$^{-1}$. Answer: (2).
4	Laws of MotionFriction	Class 11: Laws of Motion	Single correctMedium
A medium single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Starting from rest over the same length, $L = \frac{1}{2}at^2$, so $t \propto \frac{1}{\sqrt{a}}$. $t_{rough} = 2t_{smooth} \Rightarrow a_{smooth} = 4a_{rough}$. Smooth: $a = g\sin45°$. Rough: $a = g(\sin45° - \mu_k\cos45°)$. $g\sin45° = 4g(\sin45° - \mu_k\cos45°)$; with $\sin45° = \cos45°$: $1 = 4(1 - \mu_k)$, so $\mu_k = \frac{3}{4} = 0.75$. Answer: (4) 0.75.
5	Work, Energy and PowerWork-energy theorem	Class 11: Work, Energy and Power	Single correctEasy
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power By the work–energy theorem the brakes' work removes all the kinetic energy: $F d = K$, so $F = \frac{K}{d}$. $F_A = \frac{100}{1000} = 0.1$ N, $F_B = \frac{225}{1500} = 0.15$ N. $\frac{F_A}{F_B} = \frac{0.1}{0.15} = \frac{2}{3}$. Answer: (2).
6	Current ElectricityWheatstone bridge	Class 12: Current Electricity	Single correctMedium
A medium single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity Read the circuit: A and F are joined by a wire, and so are C and D. Between them the resistors are A–B 5 Ω, B–C 2.5 Ω (so B–D 2.5 Ω), A–D 6 Ω, F–E 3 Ω (so A–E 3 Ω) and E–D 1.5 Ω. The battery branch runs from B through 1.5 Ω and 5.5 Ω, the 5 V cell and $\frac{1}{3}$ Ω back to E. So the network seen by the battery is between B and E: a bridge with arms B–A 5 Ω, A–E 3 Ω, B–D 2.5 Ω, D–E 1.5 Ω and the 6 Ω between A and D. $\frac{5}{3} = \frac{2.5}{1.5}$, so the bridge is balanced and no current flows in the 6 Ω. $R_{BE} = \frac{(5+3)(2.5+1.5)}{(5+3)+(2.5+1.5)} = \frac{8\times4}{12} = \frac{8}{3}$ Ω. Total resistance $= \frac{8}{3} + 1.5 + 5.5 + \frac{1}{3} = 3 + 7 = 10$ Ω. $I = \frac{5}{10} = 0.5$ A. Answer: (2) 0.5 A.
7	Work, Energy and PowerMotion in a vertical circle	Class 11: Work, Energy and Power	Single correctMedium
A medium single correct question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At P the bob is above the horizontal through O by $l\sin\theta$, so its height above the lowest point is $l(1+\sin\theta)$. Radial equation at P (gravity has a component $mg\sin\theta$ towards O): $T + mg\sin\theta = \frac{mv^2}{l}$. The string goes slack when $T = 0$: $v^2 = gl\sin\theta$. Energy conservation from the bottom: $v_0^2 = v^2 + 2gl(1+\sin\theta) = gl\sin\theta + 2gl + 2gl\sin\theta = gl(2 + 3\sin\theta)$. $\frac{v}{v_0} = \left(\frac{gl\sin\theta}{gl(2+3\sin\theta)}\right)^{1/2} = \left(\frac{\sin\theta}{2+3\sin\theta}\right)^{1/2}$. Answer: (4).
8	Electronic DevicesLogic gates	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctEasy
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The upper gate is a NOR of A and B; the lower gate is a NAND of B and A. Their outputs go into an AND gate: $Y = \overline{A+B}\cdot\overline{AB}$. Truth table (A, B → NOR, NAND → Y): 0,0 → 1,1 → 1; 0,1 → 0,1 → 0; 1,0 → 0,1 → 0; 1,1 → 0,0 → 0. Y is 1 only when both inputs are 0, which is the NOR gate. (Whenever $\overline{A+B} = 1$, both inputs are 0 and $\overline{AB}$ is 1 too, so $Y = \overline{A+B}$.) Answer: (4) NOR.
9	Electromagnetic WavesProperties of electromagnetic waves	Class 12: Electromagnetic Waves	Single correctMedium
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves The phase $5x + 1.5\times10^9 t$ means the wave travels along $-x$, with speed $\frac{\omega}{k} = \frac{1.5\times10^9}{5} = 3\times10^8$ m/s. $B_0 = \frac{E_0}{c} = \frac{60}{3\times10^8} = 2\times10^{-7}$ T, in phase with $E$. $\vec{B}$ is perpendicular to both $\vec{E}$ (along z) and the direction of travel (x), so it is along y. Check the sense: $\vec{E}\times\vec{B}$ must point along $-x$; $\hat{k}\times\hat{j} = -\hat{i}$, so $\vec{B}$ is along $+y$ when $E_z$ is positive. $B_y = 2\times10^{-7}\cos(5x + 1.5\times10^9 t)$ T. Answer: (1).
10	Laws of MotionNewton's laws	Class 11: Laws of Motion	Single correctEasy
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Speed just before impact: $v_1 = \sqrt{2gh_1} = \sqrt{2\times9.8\times40} = \sqrt{784} = 28$ m/s, downward. Speed just after: $v_2 = \sqrt{2\times9.8\times10} = \sqrt{196} = 14$ m/s, upward. Impulse = change in momentum $= m(v_2 - (-v_1)) = 0.5\times(14 + 28) = 21$ N s (upward). Answer: (1) 21 NS.
11	Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Going from A to B along the current: the 1 H inductor drops $L\frac{di}{dt} = 1\times1 = 1$ V (the current is increasing, so the induced emf opposes it). The cell's longer (positive) plate faces A, so going from A to B we go from + to −: a further drop of 5 V. The 2 Ω resistor drops $iR = 2\times2 = 4$ V. $V_A - V_B = 1 + 5 + 4 = 10$ V. Answer: (4) 10 volt.
12	Magnetic Effects of Current and MagnetismMagnetic dipole	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctEasy
A easy single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Magnetic moment of a current loop: $\mu = I A = I\pi r^2$. The current is the same, so $\mu \propto r^2$. $\frac{\mu_1}{\mu_2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$. Answer: (1) 1:4.
13	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For thin lenses in contact the powers add: $P = p + p + p + p = 4p$. In a combination each lens forms an image that is the object of the next, so the total magnification is the product of the separate magnifications: $m\times m\times m\times m = m^4$. Answer: (3) $4p$ and $m^4$.
14	Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation	Class 11: Kinetic Theory	Single correctMedium
A medium single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Gauge pressure is measured above the atmosphere, so the absolute pressure left in the cylinder is $11 + 1 = 12$ atm $= 12\times1.01\times10^5 = 1.212\times10^6$ Pa. Moles left: $n = \frac{PV}{RT} = \frac{1.212\times10^6\times30\times10^{-3}}{\frac{100}{12}\times300} = \frac{36360}{2500} = 14.544$ mol. Moles withdrawn: $18.20 - 14.544 = 3.656$ mol. Mass withdrawn: $3.656\times32 = 117$ g $\approx 0.116$ kg. Answer: (3) 0.116 kg.
15	KinematicsMotion in a straight line	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctMedium
A medium single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\frac{dt}{dx} = 2x + 1$, so $v = \frac{dx}{dt} = \frac{1}{2x+1}$. $a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = \left(-\frac{2}{(2x+1)^2}\right)\cdot\frac{1}{2x+1} = -\frac{2}{(2x+1)^3}$. Check: $a = -2v^3$, and $v^3 = \frac{1}{(2x+1)^3}$, the same. Answer: (2).
16	Electromagnetic Induction and Alternating CurrentsAC circuits	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctEasy
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Net reactance $X = X_L - X_C = 45 - 25 = 20$ Ω. Impedance $Z = \sqrt{R^2 + X^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2} \approx 28.28$ Ω. Current $I = \frac{V}{Z} = \frac{220}{28.28} \approx 7.8$ A. Phase angle: $\tan\phi = \frac{X}{R} = \frac{20}{20} = 1$, so $\phi = 45°$. Answer: (2) 7.8 A and 45°.
17	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	Single correctEasy
A easy single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion With no external torque the angular momentum $I\omega$ is conserved. For a uniform sphere $I = \frac{2}{5}MR^2$, so with the mass unchanged, $I \propto R^2$. Doubling $R$ makes $I$ four times larger, so $\omega$ becomes a quarter and the period $T = \frac{2\pi}{\omega}$ four times longer: $T' = 4\times27 = 108$ days. Answer: (4) 108 days.
18	Magnetic Effects of Current and MagnetismMagnetic dipole	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctMedium
A medium single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The electron moves in a circle of radius $r$ with $evB = \frac{mv^2}{r}$, so $v = \frac{eBr}{m}$. Its current is $I = \frac{e}{2\pi r/v} = \frac{ev}{2\pi r}$, and its magnetic moment $\mu = I\pi r^2 = \frac{evr}{2} = \frac{e^2Br^2}{2m}$. Lowest state, $n = 1$: flux $B\pi r^2 = \frac{h}{e}$, so $r^2 = \frac{h}{\pi eB}$. $\mu = \frac{e^2B}{2m}\cdot\frac{h}{\pi eB} = \frac{he}{2\pi m}$. Answer: (2).
19	Properties of Bulk MatterHeat transfer	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctEasy
A easy single correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Thermal resistance of a rod $= \frac{l}{kA}$. With the same $l$ and $A$, the side rods have resistance $R$ each and the middle rod $2R$ (half the conductivity). Total $4R$. In steady state the same heat current $H$ flows through all three: $H = \frac{3T - T}{4R} = \frac{T}{2R}$. $T_1 = 3T - HR = 3T - \frac{T}{2} = \frac{5T}{2}$. $T_2 = T_1 - H(2R) = \frac{5T}{2} - T = \frac{3T}{2}$ (check: $T_2 - HR = T$, the right end). $\frac{T_1}{T_2} = \frac{5/2}{3/2} = \frac{5}{3}$. Answer: (3).
20	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With slabs of thickness $t_1$, $t_2$ the capacitance is $C = \frac{\varepsilon_0 A}{d - t_1 - t_2 + \frac{t_1}{K_1} + \frac{t_2}{K_2}}$; the air gap left is $d - \frac{3d}{8} - \frac{d}{2} = \frac{d}{8}$. $C = 2C_0$ means the denominator equals $\frac{d}{2}$: $\frac{d}{8} + \frac{3d}{8K_1} + \frac{d}{2K_2} = \frac{d}{2} \Rightarrow \frac{3}{8K_1} + \frac{1}{2K_2} = \frac{3}{8}$. $K_2 = \frac{K_1}{1.25} = 0.8K_1$, so $\frac{1}{2K_2} = \frac{0.625}{K_1}$: $\frac{0.375 + 0.625}{K_1} = \frac{3}{8} \Rightarrow K_1 = \frac{8}{3} \approx 2.66$. (Check: $K_2 = 2.13$; $\frac{d}{8} + \frac{3d}{8\times2.667} + \frac{d}{2\times2.133} = 0.125d + 0.1406d + 0.2344d = 0.5d$.) Answer: (1) 2.66.
21	KinematicsRelative velocity	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctMedium
A medium single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Buses in each direction are spaced $d = vT$ apart, where $v$ is the bus speed. Buses going her way overtake her at relative speed $v - 60$: $\frac{d}{v-60} = \frac{1}{2}$ h. Buses coming the other way meet her at relative speed $v + 60$: $\frac{d}{v+60} = \frac{1}{6}$ h. Dividing: $\frac{v+60}{v-60} = 3 \Rightarrow v + 60 = 3v - 180 \Rightarrow v = 120$ km/h. Then $d = \frac{180}{6} = 30$ km and $T = \frac{d}{v} = \frac{30}{120}$ h $= 15$ min. (Check: $\frac{30}{120-60} = 0.5$ h = 30 min.) Answer: (4) 15 min, 120 km/h.
22	Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium single correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The rod makes 60° with the wall, so 30° with the floor. Forces: weight $mg = 200$ N at the middle, the wall's normal force $N_w$ (horizontal, the wall is smooth), the floor's normal force $N$ and friction $f$. Horizontal balance: $f = N_w$. Torques about the foot of the rod: the weight acts at horizontal distance $\frac{L}{2}\cos30°$, the wall's force at height $L\sin30°$: $N_w L\sin30° = mg\frac{L}{2}\cos30° \Rightarrow N_w = \frac{mg}{2}\cot30° = 100\sqrt{3}$ N. So $f = 100\sqrt{3}$ N (about 173 N). Answer: (2).
23	Oscillations and WavesSimple pendulum and spring systems	Class 11: Oscillations; Class 11: Waves	Single correctMedium
A medium single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Frequency: $\omega = \sqrt{\frac{k}{m}}$. As sand leaks the mass falls, so $\omega$ rises, levelling off when the sand is gone. Amplitude: the sand leaves with the box's own velocity, so it carries away its kinetic energy without pushing the box. The oscillation energy $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$ therefore falls ($\frac{dE}{dt} = \frac{1}{2}\frac{dm}{dt}v^2 < 0$). Since $E = \frac{1}{2}kA^2$ with $k$ fixed, $A$ falls, levelling off when the sand is gone. (For a slow change, the oscillation keeps $A^2\omega$ constant, so $A \propto m^{1/4}$.) So $\omega(t)$ increases and $A(t)$ decreases: graph (2). Answer: (2).
24	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctHard
A hard single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The excess pressure inside the balloon is $\Delta P = \frac{cS}{r}$ ($c$ = 2 or 4, a constant). By Bernoulli's equation the gas leaves with $\frac{1}{2}\rho v^2 = \Delta P$: $v = \sqrt{\frac{2cS}{\rho r}} \propto r^{-1/2}$, so $a = -\frac{1}{2}$. The volume falls at the rate gas flows out: $\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = -Av$, i.e. $4\pi r^2\frac{dr}{dt} = -A\sqrt{\frac{2cS}{\rho}}\,r^{-1/2}$. $r^{5/2}dr = -\frac{A}{4\pi}\sqrt{\frac{2cS}{\rho}}\,dt$. Integrating from $R$ to 0 over time $T$: $\frac{2}{7}R^{7/2} = \frac{A}{4\pi}\sqrt{\frac{2cS}{\rho}}\,T \Rightarrow T \propto S^{-1/2}A^{-1}\rho^{1/2}R^{7/2}$. So $\alpha = -\frac{1}{2}$, $\beta = -1$, $\gamma = \frac{1}{2}$, $\delta = \frac{7}{2}$. (Dimensional check: $\sqrt{\rho/S}$ has units s m$^{-3/2}$; times $R^{7/2}/A$, m$^{3/2}$, gives seconds.) Answer: (3).
25	Experimental SkillsVernier calipers and screw gauge	No matching NCERT chapter	Single correctEasy
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter 1 VSD $= \frac{9}{10}$ MSD $= 0.09$ cm, so the least count is $1$ MSD $- 1$ VSD $= 0.1 - 0.09 = 0.01$ cm. With the jaws closed the vernier zero is at $+0.1$ cm: a positive zero error of 0.1 cm. Reading $= 5 + 8\times0.01 = 5.08$ cm. Corrected diameter $= 5.08 - 0.1 = 4.98$ cm. Answer: (3) 4.98 cm.
26	Electromagnetic WavesDisplacement current	Class 12: Electromagnetic Waves	Single correctEasy
A easy single correct question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Between the plates $E = \frac{\sigma}{\varepsilon_0}$ grows at a constant rate, so there is a steady, uniform displacement current density. Ampère–Maxwell law on a circle of radius $r$ about the axis (plate radius $R$): inside, $r < R$: $B\cdot2\pi r = \mu_0 i_d\frac{r^2}{R^2}$, so $B = \frac{\mu_0 i_d r}{2\pi R^2}$, growing with $r$; outside, $r > R$: $B = \frac{\mu_0 i_d}{2\pi r}$, falling with $r$. So the field is non-zero off the axis, both inside and outside, and is largest at $r = R$, the cylindrical surface joining the edges of the plates. Answer: (3).
27	OpticsPolarisation	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Brewster's law: $\tan i_B = n = 1.73 = \sqrt{3}$, so $i_B = 60°$, and the angle of reflection equals it: 60°. At Brewster's angle the reflected light is completely (plane) polarized; the transmitted light is only partially polarized. The reflected and refracted rays are perpendicular, so the refraction angle is 30°. Answer: (1).
28	ElectrostaticsCoulomb's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Identical spheres in contact share their total charge equally. Third sphere with A: $\frac{q + 0}{2}$, so A keeps $\frac{q}{2}$ and the third sphere carries $\frac{q}{2}$. Third sphere with B: $\frac{q + q/2}{2} = \frac{3q}{4}$ each, so B keeps $\frac{3q}{4}$. $F' = k\frac{(q/2)(3q/4)}{r^2} = \frac{3}{8}\cdot k\frac{q^2}{r^2} = \frac{3F}{8}$. Answer: (4).
29	Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation	Class 11: Kinetic Theory	Single correctMedium
A medium single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The container as a whole is isolated (no heat in, no work done), so the total internal energy is conserved. For the same ideal gas $U = nC_vT = \frac{C_v}{R}pV$, so the total $pV$ is conserved: $p(V_1 + V_2) = p_1V_1 + p_2V_2 = 1\times2 + 2\times3 = 8$ atm L. $p = \frac{8}{5} = 1.6$ atm. (Second way: $T_1 = \frac{2}{5R}$, $T_2 = \frac{6}{4R}$; the common temperature is $T = \frac{5T_1 + 4T_2}{9} = \frac{8}{9R}$, and $p = \frac{9RT}{5} = 1.6$ atm.) Answer: (2) 1.6 atm.
30	Atoms and NucleiBohr model	Class 12: Atoms; Class 12: Nuclei	Single correctMedium
A medium single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Circular orbit: $\frac{mv^2}{r} = F$, so $v = \sqrt{\frac{Fr}{m}} \propto r^{1/2}$. Bohr's condition: $mvr = \frac{nh}{2\pi}$, so $m\sqrt{\frac{Fr}{m}}\,r \propto n \Rightarrow r^{3/2} \propto n \Rightarrow r \propto n^{2/3}$. Then $v \propto r^{1/2} \propto n^{1/3}$. Answer: (3).
31	GravitationKepler's laws	Class 11: Gravitation	Single correctEasy
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Kepler's third law: $T^2 \propto r^3$, so $\frac{T_{Mars}}{T_{Mercury}} = 4^{3/2} = 8$. $T_{Mercury} = \frac{687}{8} \approx 86$ days; the nearest option is 88 earth days (the orbit ratio is only "about" 4). Answer: (1).
32	GravitationAcceleration due to gravity	Class 11: Gravitation	Single correctEasy
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation Gravitational force $\propto \frac{1}{(R+h)^2}$. At $h = \frac{R}{3}$, $R + h = \frac{4R}{3}$: $W' = 48\times\left(\frac{R}{4R/3}\right)^2 = 48\times\frac{9}{16} = 27$ N. Answer: (2) 27 N.
33	Current ElectricityOhm's law	Class 12: Current Electricity	Single correctEasy
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Each piece has resistance $\frac{R}{8}$. Four in parallel: $\frac{R/8}{4} = \frac{R}{32}$. Two such sets in series: $2\times\frac{R}{32} = \frac{R}{16}$. Answer: (3).
34	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Bohr's condition says the orbit holds a whole number of wavelengths: $2\pi r_n = n\lambda$. $r_2 = n^2a_0 = 4\times0.052 = 0.208$ nm. $\lambda = \frac{2\pi r_2}{2} = \pi\times0.208 \approx 0.65$ nm, closest to 0.67 nm. (Second way: $v_2 = \frac{2.19\times10^6}{2} \approx 1.09\times10^6$ m/s; $\lambda = \frac{h}{mv} = \frac{6.63\times10^{-34}}{9.1\times10^{-31}\times1.09\times10^6} \approx 0.67$ nm.) Answer: (2) 0.67 nm.
35	ElectrostaticsElectric dipole	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $U = -pE\cos\theta$, so $\Delta U = pE(\cos0° - \cos60°) = 5\times10^{-6}\times4\times10^5\times\left(1 - \frac{1}{2}\right) = 2\times0.5 = 1.0$ J (an increase). Answer: (2) 1.0 J.
36	Current ElectricityKirchhoff's rules	Class 12: Current Electricity	Single correctMedium
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity The 1 Ω and 3 Ω both join A's side to the C–D link, and the 2 Ω and 4 Ω join the link to B's side. With C and D joined by a wire: $(1\parallel3) = \frac{3}{4}$ Ω, $(2\parallel4) = \frac{8}{6} = \frac{4}{3}$ Ω, total $\frac{3}{4} + \frac{4}{3} = \frac{25}{12}$ Ω. Total current $= \frac{50}{25/12} = 24$ A. Voltage across the first pair $= 24\times\frac{3}{4} = 18$ V: 18 A through 1 Ω (into C), 6 A through 3 Ω (into D). Voltage across the second pair $= 24\times\frac{4}{3} = 32$ V: 16 A through 2 Ω (out of C), 8 A through 4 Ω (out of D). At C: 18 A in, 16 A out, so 2 A flows from C to D (check at D: 6 + 2 = 8 A out through 4 Ω). Answer: (2) 2.0 A.
37	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon: $E = \frac{hc}{\lambda}$, so $\lambda_{photon} = \frac{hc}{E}$. Electron (non-relativistic): $p = \sqrt{2mE}$, so $\lambda_{electron} = \frac{h}{\sqrt{2mE}}$. $\frac{\lambda_{photon}}{\lambda_{electron}} = \frac{hc}{E}\cdot\frac{\sqrt{2mE}}{h} = c\sqrt{\frac{2m}{E}}$. Answer: (3).
38	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	StatementsMedium
A medium statements question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter A: above the threshold frequency, each photon can free at most one electron, so the photocurrent is proportional to the intensity: a straight line through the origin. Correct. B: a current that does not change with intensity is wrong. C: a current rising from zero frequency is wrong: below the threshold frequency there is no emission at all. D: above threshold the (saturation) photocurrent depends on how many photons arrive, not on their frequency; for a given intensity it does not keep rising with frequency (if anything it falls, as fewer photons carry the same energy). So a current growing steadily with frequency is wrong; it is the stopping potential, not the current, that rises linearly with frequency above threshold. Only A is correct. Answer: (1) A only.
39	Motion of System of Particles and Rigid BodyMoment of inertia	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Let the whole sphere (radius $2R$) have mass $M$. The small sphere has $\frac{1}{8}$ of the volume, so mass $\frac{M}{8}$; its centre is at distance $R$ from the Y-axis, which passes through the large sphere's centre. Whole sphere about Y: $\frac{2}{5}M(2R)^2 = \frac{8}{5}MR^2 = \frac{64}{40}MR^2$. Small sphere about Y (parallel axes): $\frac{2}{5}\frac{M}{8}R^2 + \frac{M}{8}R^2 = \frac{7}{40}MR^2$. Remaining part: $\frac{64}{40}MR^2 - \frac{7}{40}MR^2 = \frac{57}{40}MR^2$. Ratio $= \frac{7}{57}$. Answer: (3).
40	Electronic DevicesRectifiers	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctMedium
A medium single correct question on rectifiers, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits At $t = 15$ ms the phase is $100\pi\times0.015 = 1.5\pi$, and $\sin1.5\pi = -1$: the input is in its negative half-cycle (at its negative peak). In a centre-tap full-wave rectifier the two diodes take turns: $D_1$, on the upper end of the secondary, conducts in the positive half-cycle, and $D_2$, on the lower end, in the negative half-cycle. So at 15 ms $D_1$ is reverse biased and $D_2$ is forward biased. Answer: (2).
41	ThermodynamicsFirst law	Class 11: Thermodynamics	Single correctMedium
A medium single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics First law: $Q = \Delta U + W$. Equal $Q$ and equal $\Delta U$ mean equal work $W = p\Delta V$, and the pressures are equal, so the volume changes are equal: $\pi r_A^2\times16 = \pi r_B^2\times9 \Rightarrow \frac{r_A^2}{r_B^2} = \frac{9}{16} \Rightarrow \frac{r_A}{r_B} = \frac{3}{4}$. Answer: (2).
42	Physical World and MeasurementError analysis	Class 11: Units and Measurement	Single correctEasy
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement For a product of powers the maximum relative errors add, each multiplied by its power: $\frac{\Delta P}{P} = 3\frac{\Delta a}{a} + 2\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{1}{2}\frac{\Delta d}{d} = 3(1) + 2(3) + 2 + \frac{1}{2}(4) = 3 + 6 + 2 + 2 = 13\%$. Answer: (3) 13%.
43	OpticsPolarisation	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics After the middle sheet (Malus's law): $I_0\cos^2 22.5°$. Its axis is then at $90° - 22.5° = 67.5°$ to the last polaroid, so after it: $I = I_0\cos^2 22.5°\,\sin^2 22.5° = \frac{I_0}{4}\sin^2 45° = \frac{I_0}{4}\cdot\frac{1}{2} = \frac{I_0}{8}$. Answer: (3).
44	Oscillations and WavesSimple pendulum and spring systems	Class 11: Oscillations; Class 11: Waves	Single correctEasy
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Maximum speed in SHM: $v_{max} = A\omega = A\sqrt{\frac{k}{m}}$. The masses are equal and so are the maximum speeds: $A_P\sqrt{k_1} = A_Q\sqrt{k_2} \Rightarrow \frac{A_Q}{A_P} = \sqrt{\frac{k_1}{k_2}}$. Answer: (4).
45	Oscillations and WavesStanding waves and beats	Class 11: Oscillations; Class 11: Waves	Single correctEasy
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Open pipe of length $L$: $f = \frac{v}{2L}$. Dipped to half its length, it becomes a pipe closed at one end (the water surface) with an air column $\frac{L}{2}$: $f' = \frac{v}{4(L/2)} = \frac{v}{2L} = f$. Answer: (2) $f$.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
46	Structure of AtomBohr model	Class 11: Structure of Atom	Single correctEasy
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Rydberg formula for hydrogen: $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. n = 2 → 3: $\frac{1}{\lambda_1} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$. n = 4 → 6: $\frac{1}{\lambda_2} = R\left(\frac{1}{16} - \frac{1}{36}\right) = R\cdot\frac{9 - 4}{144} = \frac{5R}{144}$. $\frac{\lambda_1}{\lambda_2} = \frac{5R/144}{5R/36} = \frac{36}{144} = \frac{1}{4}$. Answer: (4).
47	Classification of Elements and PeriodicityPeriodic trends	Class 11: Classification of Elements and Periodicity in Properties	StatementsMedium
A medium statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties A: False. Gallium has a very low melting point (about 303 K, it melts in the hand), and so does caesium (about 302 K); Ga is not a high-melting metal. B: False. On the Pauling scale both N and Cl have electronegativity 3.0. C: True. Ar (18), $K^+$ (19 − 1), $Cl^-$ (17 + 1), $Ca^{2+}$ (20 − 2) and $S^{2-}$ (16 + 2) all have 18 electrons. D: False. Mg ($3s^2$, filled subshell) has a higher first ionization enthalpy than Al ($3p^1$); the order is Si > Mg > Al > Na. E: True. Atomic radius grows down group 1, so Cs is larger than both Li and Rb. True: C and E. Answer: (2).
48	Principles Related to Practical ChemistryQualitative salt analysis	No matching NCERT chapter	MatchingEasy
A easy matching question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter In the usual scheme of cation analysis: A. $Co^{2+}$ is precipitated as CoS by $H_2S$ in ammoniacal medium: Group IV → III. B. $Mg^{2+}$ is left to the last group (Group VI, after the carbonates of Group V): → IV. C. $Pb^{2+}$ is precipitated as $PbCl_2$ by dilute HCl: Group I → I. D. $Al^{3+}$ is precipitated as $Al(OH)_3$ by $NH_4OH$ in the presence of $NH_4Cl$: Group III → II. A-III, B-IV, C-I, D-II. Answer: (2).
49	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (i) HBr with benzoyl peroxide adds by the free-radical (anti-Markovnikov, peroxide effect) route: Br goes to the less substituted ring carbon, H to the carbon bearing $CH_3$. Product: 1-bromo-2-methylcyclopentane. (ii) KCN replaces Br by CN ($S_N2$): 2-methylcyclopentane-1-carbonitrile. (iii) Na(Hg)/$C_2H_5OH$ reduces the nitrile to a primary amine (Mendius reduction), $-C\equiv N \rightarrow -CH_2NH_2$. P = (2-methylcyclopentyl)methanamine: a cyclopentane ring with $CH_3$ and $CH_2NH_2$ on adjacent carbons, structure (1). (Structure (2) would come from Markovnikov addition; (3) and (4) are isocyanides, which KCN does not give as the major product.) Answer: (1).
50	Structure of AtomBohr model	Class 11: Structure of Atom	Single correctEasy
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom For a one-electron ion in the first orbit: $E_1 = -R_H Z^2$ and $r_1 = \frac{a_0}{Z}$. $Li^{2+}$ (Z = 3): $E_1 = -2.18\times10^{-18}\times9 = -19.62\times10^{-18}$ J; $r_1 = \frac{52.9}{3} = 17.6$ pm. $He^+$ (Z = 2): $E_1 = -2.18\times10^{-18}\times4 = -8.72\times10^{-18}$ J; $r_1 = \frac{52.9}{2} = 26.45 \approx 26.4$ pm. Answer: (1).
51	Coordination CompoundsValence bond theory	Class 12: Coordination Compounds	StatementsMedium
A medium statements question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A. $[NiCl_4]^{2-}$: Ni(II), $3d^8$; $Cl^-$ is a weak ligand, $sp^3$ tetrahedral, 2 unpaired electrons: paramagnetic. B. $Ni(CO)_4$: Ni(0), $3d^{10}$ after pairing, $sp^3$ tetrahedral: diamagnetic. C. $[Ni(CN)_4]^{2-}$: Ni(II), $3d^8$; strong-field $CN^-$ pairs the electrons, $dsp^2$ square planar: diamagnetic. D. $[Ni(H_2O)_6]^{2+}$: Ni(II), $3d^8$ octahedral; $t_{2g}^6e_g^2$ always has 2 unpaired electrons: paramagnetic. E. $Ni(PPh_3)_4$: Ni(0), $3d^{10}$, tetrahedral: diamagnetic. Paramagnetic: A and D. Answer: (3).
52	p-Block ElementsGroup 15 elements	No matching NCERT chapter	StatementsEasy
A easy statements question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: every group 15 element forms a hydride $EH_3$: $NH_3$, $PH_3$, $AsH_3$ (arsine), $SbH_3$, $BiH_3$. Correct. Statement II: antimony shows the +5 state in its oxide $Sb_2O_5$ (antimony pentoxide); it is bismuth whose +5 oxide is hardly stable. Incorrect. Answer: (3).
53	Classification of Elements and PeriodicityModern periodic table	Class 11: Classification of Elements and Periodicity in Properties	StatementsEasy
A easy statements question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Main group (s- and p-block) elements have their last electron in an s or p subshell. A. $[Ne]3s^1$ is Na, s-block: main group. B. $[Ar]3d^34s^2$ is V, d-block. C. $[Kr]4d^{10}5s^25p^5$ is I, p-block: main group. D. $[Ar]3d^{10}4s^1$ is Cu, d-block (group 11), not a main group element. E. $[Rn]5f^06d^27s^2$ is Th, an actinoid (f-block). Main group: A and C. Answer: (2).
54	Some Basic Concepts of ChemistryLaws of chemical combination	Class 11: Some Basic Concepts of Chemistry	Single correctEasy
A easy single correct question on laws of chemical combination, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Dalton's theory (indivisible atoms combining in whole-number ratios, neither created nor destroyed) accounts for the laws of conservation of mass, constant proportions and multiple proportions. It could not explain Gay-Lussac's law of gaseous volumes (gases combine in simple volume ratios): that needed Avogadro's idea of molecules, since by Dalton's picture atoms would have to split. Answer: (4).
55	Redox ReactionsOxidation number	Class 11: Redox Reactions	Single correctEasy
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions $KO_2$ is potassium superoxide ($K^+O_2^-$): K is an alkali metal, +1 (each O is $-\frac{1}{2}$). $H_2O_2$ is a peroxide: $2(+1) + 2x = 0 \Rightarrow x = -1$. $H_2SO_4$: $2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$. +1, −1, +6. Answer: (1).
56	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	Single correctEasy
A easy single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics First order: $t = \frac{2.303}{k}\log\frac{[A]_0}{[A]}$ with $k = \frac{0.693}{t_{1/2}} = 0.693$ min$^{-1}$. 99.9% complete leaves 0.1%: $\frac{[A]_0}{[A]} = 1000$, $\log1000 = 3$. $t = \frac{2.303\times3}{0.693} \approx 9.97$ min $\approx$ 10 minutes. (Check: after 10 half-lives $\frac{1}{2^{10}} = \frac{1}{1024} \approx 0.1\%$ remains.) Answer: (4).
57	Coordination CompoundsCrystal field theory	Class 12: Coordination Compounds	Single correctMedium
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The light absorbed has energy equal to the crystal-field splitting: the larger the splitting, the shorter the wavelength absorbed. B. $[Co(CN)_6]^{3-}$: Co(III) with the strongest ligand $CN^-$, the largest splitting; absorbs in the ultraviolet / violet end (shortest $\lambda$). A. $[Co(NH_3)_6]^{3+}$: Co(III) with $NH_3$, large splitting; absorbs blue (about 475 nm), looks yellow-orange. D. $[Ti(H_2O)_6]^{3+}$: Ti(III) with water; absorbs blue-green (about 500 nm), looks purple. C. $[Cu(H_2O)_4]^{2+}$: Cu(II), +2 charge and weak water ligands, smallest splitting; absorbs orange-red (about 600–800 nm), looks blue. Increasing wavelength: B < A < D < C. Answer: (2).
58	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Cis-trans isomerism needs two different groups on each of two carbons whose rotation is locked (a double bond or a ring). Pent-1-ene: $CH_2=CH-C_3H_7$, C1 carries two H atoms: no. 2-Methylhex-2-ene: $(CH_3)_2C=CH-C_3H_7$, C2 carries two methyls: no. 1,1-Dimethylcyclopropane: both methyls on one carbon: no. 1,2-Dimethylcyclohexane: the two methyls on adjacent ring carbons can be on the same side (cis) or opposite sides (trans) of the ring: yes. Answer: (4).
59	EquilibriumIonic equilibrium	Class 11: Equilibrium	StatementsMedium
A medium statements question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium A: adding the three steps gives the overall ionization $H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-}$, whose constant is the product $K = K_{a_1}K_{a_2}K_{a_3}$; so $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$. True. B: removing $H^+$ from a neutral molecule is easier than from an anion that already carries negative charge, so $H_3PO_4$ is the strongest of the three acids. True. C: for the same reason $K_{a_1} > K_{a_2} > K_{a_3}$ (about $7\times10^{-3}$, $6\times10^{-8}$, $5\times10^{-13}$). True. D: no such relation exists; with the values above it is plainly false. A, B and C. Answer: (4).
60	HydrocarbonsAromatic hydrocarbons	Class 11: Hydrocarbons	Single correctEasy
A easy single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (1) Sodium benzoate heated with soda lime is decarboxylated: $C_6H_5COONa + NaOH \xrightarrow{CaO,\ \Delta} C_6H_6 + Na_2CO_3$. Gives benzene. (2) n-Hexane over $Mo_2O_3$ (or $Cr_2O_3$, $V_2O_5$) at 773 K and 10–20 atm is cyclised and dehydrogenated (aromatised) to benzene. (3) Ethyne passed through a red-hot iron tube at 873 K trimerises to benzene. (4) Benzenediazonium chloride warmed with water is hydrolysed: $C_6H_5N_2^+Cl^- + H_2O \rightarrow C_6H_5OH + N_2 + HCl$. This gives phenol, not benzene (benzene would need $H_3PO_2$ or ethanol). Answer: (4).
61	ElectrochemistryConductance	Class 12: Electrochemistry	Single correctEasy
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Kohlrausch's law: $\Lambda^\circ_m = \Lambda^\circ_+ + \Lambda^\circ_- = 349.6 + 50.4 = 400$ S cm$^2$ mol$^{-1}$. Degree of dissociation: $\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{90}{400} = 0.225$. Answer: (3).
62	Chemical Bonding and Molecular StructureMolecular orbital theory	Class 11: Chemical Bonding and Molecular Structure	StatementsEasy
A easy statements question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Statement I: bond order $= \frac{1}{2}(N_b - N_a)$. Zero means bonding and antibonding electrons cancel, so there is no net bond and the molecule does not exist as a stable species (e.g. $He_2$). False. Statement II: a higher bond order means a stronger bond and a shorter bond length (bond length is inversely related to bond order). False. Answer: (2).
63	Coordination CompoundsWerner theory	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Conductance depends on the number of ions a compound gives in solution. $[Co(NH_3)_3Cl_3]$: all three chlorides are in the coordination sphere; it is neutral and gives no ions, so its conductance is minimum. (The others, as printed, give ions: $[Co(NH_3)_6]Cl_3$ gives 4 ions, $[Co(NH_3)_5Cl]Cl$ ionises at its outer chloride, and $[Co(NH_3)_4Cl_2]$ as printed is the cation $[Co(NH_3)_4Cl_2]^+$ of a salt.) Answer: (1).
64	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	MatchingEasy
A easy matching question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Count the electron pairs on Xe (8 valence electrons): A. $XeO_3$: 3 bonding domains + 1 lone pair = 4: $sp^3$, pyramidal → II. B. $XeF_2$: 2 bond pairs + 3 lone pairs = 5: $sp^3d$, linear → I. C. $XeOF_4$: 5 bonding domains + 1 lone pair = 6: $sp^3d^2$, square pyramidal → IV. D. $XeF_6$: 6 bond pairs + 1 lone pair = 7: $sp^3d^3$, distorted octahedral → III. A-II, B-I, C-IV, D-III. Answer: (1).
65	ThermodynamicsEnthalpy	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta H = H_P - H_R = -74.8$ kJ mol$^{-1}$: the reaction is exothermic, so the products lie 74.8 kJ mol$^{-1}$ below the reactants, and the 74.8 must be marked between the reactant level and the product level. (1): R high, P lower, 74.8 marked between the R and P levels. Correct. (2) and (4): P above R, an endothermic profile. Wrong. (3): P below R, but the 74.8 is marked from the top of the energy hump to the reactant level (an activation energy), not between R and P. Wrong. Answer: (1).
66	SolutionsConcentration terms	Class 12: Solutions	MatchingEasy
A easy matching question on concentration terms, from Solutions. NCERT: Class 12: Solutions A. Humidity: water (liquid) in air (gas) → II. B. Alloys (e.g. brass): one solid metal in another → I. C. Amalgams: liquid mercury in a solid metal (e.g. sodium) → IV. D. Smoke: solid particles in a gas → III (classed this way in the matching; strictly smoke is a colloid, an aerosol). A-II, B-I, C-IV, D-III. Answer: (2).
67	AminesBasicity of amines	Class 12: Amines	Single correctMedium
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines Aliphatic amines are far stronger bases than aryl amines, whose lone pair is delocalised into the ring. Among the aliphatic ones in water, the secondary amine $(C_2H_5)_2NH$ (N-ethylethanamine, $pK_b$ 3.0) is stronger than the primary $C_2H_5NH_2$ (ethanamine, $pK_b$ 3.3). Among the aryl amines, the electron-releasing methyl makes $C_6H_5NHCH_3$ (N-methylaniline, $pK_b$ 9.3) slightly more basic than $C_6H_5NH_2$ (benzenamine, $pK_b$ 9.38). N-ethylethanamine > ethanamine > N-methylaniline > benzenamine. Answer: (3).
68	Some Basic Concepts of ChemistryMole concept	Class 11: Some Basic Concepts of Chemistry	StatementsEasy
A easy statements question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles of atoms = moles of substance × atoms per formula unit. A. $\frac{212}{106} = 2$ mol $\times$ 6 atoms = 12 mol of atoms. B. $\frac{248}{62} = 4$ mol $\times$ 3 = 12 mol. C. $\frac{240}{40} = 6$ mol $\times$ 3 = 18 mol. D. $\frac{12}{2} = 6$ mol $\times$ 2 = 12 mol. E. $\frac{220}{44} = 5$ mol $\times$ 3 = 15 mol. A, B and D each contain $12N_A$ atoms. Answer: (2).
69	BiomoleculesNucleic acids and vitamins	Class 12: Biomolecules	MatchingEasy
A easy matching question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules A. Vitamin $B_{12}$ deficiency: pernicious anaemia → IV. B. Vitamin D deficiency: rickets in children → III. C. Vitamin $B_2$ (riboflavin) deficiency: cheilosis (cracked lips and corners of the mouth) → I. D. Vitamin $B_6$ (pyridoxine) deficiency: convulsions → II. A-IV, B-III, C-I, D-II. Answer: (2).
70	Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctEasy
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Alkyl groups release electrons (+I effect), which destabilises the carboxylate anion and lowers acidity. The more alkyl groups on the α-carbon, the weaker the acid. HCOOH (no alkyl group, $pK_a$ 3.75) > $CH_3COOH$ (4.76) > $(CH_3)_2CHCOOH$ > $(CH_3)_3CCOOH$ (about 5.0). Answer: (3).
71	d- and f-Block ElementsTransition element properties	Class 12: The d- and f-Block Elements	StatementsMedium
A medium statements question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Statement I: in a ferromagnetic substance the unpaired electrons of many atoms align together in domains, giving a very strong attraction to a magnetic field; it is described as an extreme form of paramagnetism. True. Statement II: Cr (Z = 24) is $[Ar]3d^54s^1$; $Cr^{2+}$ is $[Ar]3d^4$, 4 unpaired electrons. Nd (Z = 60) is $[Xe]4f^46s^2$; $Nd^{3+}$ is $[Xe]4f^3$, 3 unpaired electrons. Not the same. False. Answer: (3).
72	Organic Chemistry: Some Basic Principles and TechniquesPurification methods	Class 11: Organic Chemistry – Some Basic Principles and Techniques	MatchingEasy
A easy matching question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A. Chloroform (b.p. 334 K) and aniline (b.p. 457 K) boil far apart: simple distillation → IV. B. Crude oil is a mixture of many liquids with close boiling ranges: fractional distillation → III. C. Glycerol decomposes at its normal boiling point, so it is recovered from spent-lye by distillation under reduced pressure → I. D. Aniline is steam volatile and immiscible with water: steam distillation → II. A-IV, B-III, C-I, D-II. Answer: (1).
73	EquilibriumChemical equilibrium	Class 11: Equilibrium	Single correctMedium
A medium single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium At equilibrium $K_c = \frac{k_f}{k_b} = \frac{1}{2500} = 4\times10^{-4}$. $K_P = K_c(RT)^{\Delta n}$ with $\Delta n = 2 - 1 = 1$: $K_P = 4\times10^{-4}\times0.0831\times1000 = 4\times10^{-4}\times83.1 = 0.0332 \approx 0.033$. Answer: (3).
74	AminesDiazonium salts	Class 12: Amines	StatementsEasy
A easy statements question on diazonium salts, from Amines. NCERT: Class 12: Amines Statement I: aniline is diazotised with nitrous acid ($NaNO_2$ + HCl) at 273–278 K; the diazonium salt is unstable, is used in solution, and decomposes (explosively) when dry. Correct. Statement II: direct iodination of benzene is reversible and difficult, so iodobenzene is made by warming the diazonium salt with KI: $C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + KCl + N_2$. Correct. Answer: (1).
75	HydrocarbonsAlkanes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The compound is 2-methylbutane, $(CH_3)_2CH-CH_2-CH_3$. It has four kinds of hydrogen: 1. On either of the two equivalent $CH_3$ groups at C2 → 1-chloro-2-methylbutane, $ClCH_2-CH(CH_3)-CH_2CH_3$. C2 now carries H, $CH_3$, $CH_2Cl$ and $C_2H_5$: chiral, a pair of enantiomers (2). 2. On C2 (the CH) → 2-chloro-2-methylbutane: achiral (1). 3. On C3 (the $CH_2$) → 2-chloro-3-methylbutane, $(CH_3)_2CH-CHCl-CH_3$: C3 is chiral, a pair of enantiomers (2). 4. On C4 (the end $CH_3$) → 1-chloro-3-methylbutane: achiral (1). Total $2 + 1 + 2 + 1 = 6$ products. Answer: (4) 6.
76	Chemical Bonding and Molecular StructureHybridisation	Class 11: Chemical Bonding and Molecular Structure	Single correctMedium
A medium single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A C–H bond is stronger the more s character the carbon uses for it (the electrons are held closer to the carbon and the bond is shorter). II: the H is on an sp carbon of the alkyne (50% s): strongest, about 130 kcal mol$^{-1}$. I: the H is on an $sp^2$ ring carbon of benzene (33% s): about 113 kcal mol$^{-1}$. III: the marked H is on the $sp^3$ $CH_2$ of cyclopropene (25% s), and the radical left behind is allylic to the ring double bond: weakest. II > I > III. Answer: (1).
77	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctEasy
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Bromine water is decolourised by compounds that react with bromine quickly at room temperature. (1) Cyclohexane is a saturated alkane: no reaction with bromine water (it needs light for substitution). Does not decolourise. (2) Phenol: the activated ring gives a white precipitate of 2,4,6-tribromophenol. Decolourises. (3) Styrene ($C_6H_5CH=CH_2$): bromine adds across the C=C. Decolourises. (4) Aniline: gives a white precipitate of 2,4,6-tribromoaniline. Decolourises. Answer: (1).
78	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Excess $CH_3MgBr$ attacks both electrophilic groups: the ketone $C_6H_5CO-$ adds $CH_3^-$ to give a magnesium alkoxide, which $H_3O^+$ turns into the tertiary alcohol $C_6H_5C(CH_3)(OH)-$; the nitrile $-C\equiv N$ adds $CH_3^-$ once to give an imine salt $-C(CH_3)=NMgBr$, which cannot react further; on acid hydrolysis it becomes a ketone, $-COCH_3$. Product: $C_6H_5C(CH_3)(OH)CH_2CH_2COCH_3$ (5-hydroxy-5-phenylhexan-2-one), structure (2). (The new ketone appears only during the aqueous work-up, when no Grignard reagent is left, so it is not converted to the diol (3).) Answer: (2).
79	Solutionsvan't Hoff factor	Class 12: Solutions	Single correctEasy
A easy single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Elevation of boiling point $\Delta T_b = iK_bm$: the largest $i\times$ concentration boils highest. Urea: $1\times0.01 = 0.01$. $KNO_3$: $2\times0.01 = 0.02$. $Na_2SO_4$: $3\times0.01 = 0.03$. Glucose: $1\times0.015 = 0.015$. $Na_2SO_4$ gives the most particles. Answer: (3).
80	d- and f-Block ElementsTransition element properties	Class 12: The d- and f-Block Elements	MatchingEasy
A easy matching question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements A. Haber process ($N_2 + 3H_2 \rightarrow 2NH_3$) uses an iron catalyst → I. B. Wacker oxidation of ethene to ethanal uses $PdCl_2$ → II. C. Wilkinson's catalyst for hydrogenating alkenes is $[(PPh_3)_3RhCl]$ → III. D. The Ziegler (Ziegler–Natta) catalyst for polymerising alkenes is $TiCl_4$ with $Al(CH_3)_3$ (a trialkylaluminium) → IV. A-I, B-II, C-III, D-IV. Answer: (3).
81	SolutionsRaoult's law	Class 12: Solutions	Single correctEasy
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Mole fractions: $x_X = \frac{5}{15} = \frac{1}{3}$, $x_Y = \frac{2}{3}$. An ideal solution (Raoult's law) would have $p = \frac{1}{3}\times63 + \frac{2}{3}\times78 = 21 + 52 = 73$ torr. The observed 70 torr is lower, so the solution shows negative deviation (X–Y attractions stronger than in the pure liquids; such solutions contract on mixing, so (4) is also wrong). Answer: (2).
82	BiomoleculesCarbohydrates	Class 12: Biomolecules	Single correctEasy
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules A keto sugar (ketose) among the options can only be fructose; glucose is an aldose and maltose and sucrose are disaccharides. D-Fructose is found in honey and fruits, is laevorotatory (hence "laevulose"), and its cyclic (furanose) form exists as α and β anomers. Answer: (2) D-Fructose.
83	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctEasy
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids An ester has to be reduced only as far as the aldehyde. DIBAL-H, $AlH(iBu)_2$, delivers one hydride to the ester at low temperature; the intermediate is hydrolysed by water to the aldehyde: $C_6H_5COOCH_3 \rightarrow C_6H_5CHO$. $LiAlH_4$ reduces the ester all the way to the alcohol $C_6H_5CH_2OH$; $NaBH_4$ hardly reduces esters; $H_2$/Pd-$BaSO_4$ (Rosenmund) reduces acid chlorides, not esters. Answer: (2).
84	Haloalkanes and HaloarenesSN1 and SN2 mechanisms	Class 12: Haloalkanes and Haloarenes	Assertion and reasonEasy
A easy assertion and reason question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Both compounds are primary alkyl halides with the same carbon chain, differing only in the halogen. In $S_N2$ the C–X bond breaks in the rate-determining step. Iodide is large and highly polarisable, the C–I bond is the longest and weakest, and $I^-$ is the most stable (weakest base) halide ion, so it is the best leaving group: reactivity R–I > R–Br > R–Cl. So A is true, R is true, and R is the reason for A. Answer: (1).
85	ThermodynamicsHess law	Class 11: Thermodynamics	Single correctMedium
A medium single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Crystallisation: $Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$, $\Delta H = -4.5$ kcal/mol. $\Delta H = \Delta_fH(BaSO_4) - \Delta_fH(Ba^{2+}) - \Delta_fH(SO_4^{2-})$: $-4.5 = -349 - \Delta_fH(Ba^{2+}) - (-216)$ $\Delta_fH(Ba^{2+}) = -349 + 216 + 4.5 = -128.5$ kcal/mol. (Check: $-128.5 + (-216) - 4.5 = -349$.) Answer: (1).
86	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctHard
A hard single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Cyclic ethers have the oxygen in the ring. $C_4H_8O$ has one ring (one degree of unsaturation). Five-membered ring: oxolane (tetrahydrofuran): 1. Four-membered ring (oxetane) with one $CH_3$: 2-methyloxetane (C2 chiral, 2 enantiomers) and 3-methyloxetane (achiral): 3. Three-membered ring (oxirane) with two more carbons: ethyloxirane (chiral, 2 enantiomers): 2; 2,2-dimethyloxirane (achiral): 1; 2,3-dimethyloxirane: cis (meso, 1) and trans (a pair of enantiomers, 2): 3. Total $1 + 3 + 2 + 1 + 3 = 10$. Answer: (3) 10.
87	Chemical Bonding and Molecular StructureDipole moment and Fajans' rule	Class 11: Chemical Bonding and Molecular Structure	StatementsMedium
A medium statements question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A: dipole moments $H_2O$ 1.85 D > $NH_3$ 1.47 D > $CHCl_3$ 1.04 D. Correct. B: lone pairs on Xe: $XeF_4$ 2, $XeO_3$ 1, $XeF_2$ 3. The order is $XeF_2 > XeF_4 > XeO_3$, not the one given. Wrong. C: bond lengths O–H ≈ 96 pm, C–H ≈ 109 pm, N–O ≈ 136 pm (single); O–H is the shortest, so the order given is wrong. D: bond enthalpies $N_2$ 946 > $O_2$ 498 > $H_2$ 436 kJ mol$^{-1}$. Correct. A and D. Answer: (1).
88	EquilibriumLe Chatelier's principle	Class 11: Equilibrium	StatementsEasy
A easy statements question on le chatelier's principle, from Equilibrium. NCERT: Class 11: Equilibrium The forward reaction is endothermic ($\Delta H > 0$), so by Le Chatelier's principle a higher temperature shifts the equilibrium towards NO: A true, B false. Adding either reactant, $N_2$ or $O_2$, shifts the equilibrium forward: C and D true. (Pressure has no effect: $\Delta n_g = 0$.) A, C and D. Answer: (4).
89	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	Single correctEasy
A easy single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The unit s$^{-1}$ shows a first order reaction. $t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{0.03}\log\frac{7.2}{0.9} = \frac{2.303}{0.03}\times\log8 = \frac{2.303\times3\times0.301}{0.03} = \frac{2.0796}{0.03} \approx 69.3$ s. (Second way: $\frac{7.2}{0.9} = 8 = 2^3$, three half-lives; $t_{1/2} = \frac{0.693}{0.03} = 23.1$ s, and $3\times23.1 = 69.3$ s.) Answer: (1) 69.3 s.
90	Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In Lassaigne's test the organic compound is fused with sodium metal, which converts N, S and halogens into soluble ionic sodium salts: NaCN (from C and N), $Na_2S$ (from S) and NaX (from a halogen X), tested afterwards in the sodium fusion extract. Heating with copper(II) oxide, $2CuO + C \rightarrow 2Cu + CO_2$, is the test for carbon (the $CO_2$ turns lime water milky), not part of Lassaigne's test. Answer: (4).

Botany questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
91	Respiration in PlantsETS and oxidative phosphorylation	Class 11: Respiration in Plants	Single correctEasy
A easy single correct question on ets and oxidative phosphorylation, from Respiration in Plants. NCERT: Class 11: Respiration in Plants The electron transport chain in the inner mitochondrial membrane has five complexes: - Complex I: NADH dehydrogenase, which takes electrons from NADH. - Complex II: succinate dehydrogenase, which oxidises succinate to fumarate in the Krebs cycle and passes the electrons on through FADH$_2$ to ubiquinone. - Complex III: cytochrome bc$_1$ complex. - Complex IV: cytochrome c oxidase, which hands electrons to oxygen. - Complex V: ATP synthase. So complex II is succinate dehydrogenase. Answer: (2).
95	The Living WorldWhat is living	Class 11: The Living World	Single correctEasy
A easy single correct question on what is living, from The Living World. NCERT: Class 11: The Living World Explaining living processes by breaking them into their molecules and the physical and chemical laws those molecules obey is called the physico-chemical approach. Studying and understanding living organisms this way, by reducing them to physics and chemistry, is called reductionist biology. The concepts and techniques of physics and chemistry are applied to biology. A purely chemical, physiological or behavioural study is not the definition. Answer: (1).
96	Molecular Basis of InheritanceDNA structure	Class 12: Molecular Basis of Inheritance	StatementsEasy
A easy statements question on dna structure, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance Statement I: There is evidence that essential life processes such as metabolism, translation and splicing evolved around RNA. RNA was the first genetic material, and it also works as a catalyst (ribozymes, for instance the peptidyl transferase of the ribosome). Its 2'-OH group makes it reactive and hence unstable. Correct. Statement II: DNA evolved from RNA through chemical modifications (2'-deoxyribose, thymine in place of uracil) that make it more stable. Being double-stranded with complementary strands, DNA resists change by having evolved repair processes. Correct. Answer: (1).
97	Organisms and PopulationsInteractions	Class 12: Organisms and Populations	Single correctEasy
A easy single correct question on interactions, from Organisms and Populations. NCERT: Class 12: Organisms and Populations An epiphyte such as an orchid grows on the branch of a mango tree only for support and better access to light; it makes its own food and takes nothing from the tree. The orchid benefits and the mango is neither harmed nor benefited (+, 0). That is commensalism. Answer: (1).
98	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	StatementsEasy
A easy statements question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life Eukaryotic (cytoplasmic) ribosomes are 80S and prokaryotic ribosomes are 70S (A true). Every ribosome is made of a large and a small subunit (B true). The 80S ribosome has 60S and 40S subunits; the 70S has 50S and 30S (C true). Sedimentation coefficients are not additive, which is why 60S + 40S gives 80S. D and E give wrong small subunits (20S, or 30S for the 80S ribosome). Answer: (1) A, B, C.
99	Biodiversity and ConservationConservation strategies	Class 12: Biodiversity and Conservation	Single correctEasy
A easy single correct question on conservation strategies, from Biodiversity and Conservation. NCERT: Class 12: Biodiversity and Conservation Ex-situ (off-site) conservation takes threatened plants and animals out of their natural habitat and protects them in special settings: zoological parks, botanical gardens, wildlife safari parks, seed banks and cryopreservation. National parks, wildlife sanctuaries and other protected areas conserve species within their own habitat, which is in-situ conservation. Answer: (3).
100	EcosystemProductivity and decomposition	Class 12: Ecosystem	StatementsEasy
A easy statements question on productivity and decomposition, from Ecosystem. NCERT: Class 12: Ecosystem Statement I: Except for the deep-sea hydrothermal ecosystems, the sun is the only source of energy for all ecosystems on Earth; producers trap it in photosynthesis. Correct. Statement II: The rate of production of organic matter during photosynthesis is the gross primary productivity (GPP). Part of it is used by the plants in respiration; what remains, GPP − R, is the net primary productivity (NPP). So the statement describes GPP, not NPP. Incorrect. Answer: (3).
102	Sexual Reproduction in Flowering PlantsPollination and outbreeding devices	Class 12: Sexual Reproduction in Flowering Plants	Assertion and reasonMedium
A medium assertion and reason question on pollination and outbreeding devices, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants Assertion: Flowers pollinated by wind or water are not very colourful and do not produce nectar. True: they have no animal visitor to attract or reward. Reason: Wind- and water-pollinated flowers produce an enormous amount of pollen. True: much of the pollen is lost on the way, and the large output makes up for that chance. Explanation: The lack of colour and nectar follows from not needing animal pollinators, not from the amount of pollen produced. Large pollen output is a separate adaptation to the wasteful transport by wind or water. So R does not explain A. Answer: (2).
103	Microbes in Human WelfareIndustrial and sewage applications	Class 12: Microbes in Human Welfare	Single correctEasy
A easy single correct question on industrial and sewage applications, from Microbes in Human Welfare. NCERT: Class 12: Microbes in Human Welfare Brewer's yeast, Saccharomyces cerevisiae, ferments malted cereals and fruit juices to make ethanol. Depending on the raw material and processing (with or without distillation), different beverages result: wine and beer are produced without distillation, whereas whisky, brandy and rum are produced by distillation of the fermented broth. Answer: (3) Beer.
104	Morphology of Flowering PlantsFlower and fruit	Class 11: Morphology of Flowering Plants	StatementsEasy
A easy statements question on flower and fruit, from Morphology of Flowering Plants. NCERT: Class 11: Morphology of Flowering Plants In a floral formula, $\oplus$ marks an actinomorphic (radially symmetrical) flower and % marks a zygomorphic one. The gynoecium is G; a line drawn below it, $\underline{\text{G}}$, means a superior ovary, and a line above it, $\overline{\text{G}}$, an inferior ovary. Statement I gives both symbols the wrong meaning: incorrect. Statement II gives both correctly: correct. Answer: (4).
105	Microbes in Human WelfareIndustrial and sewage applications	Class 12: Microbes in Human Welfare	Single correctEasy
A easy single correct question on industrial and sewage applications, from Microbes in Human Welfare. NCERT: Class 12: Microbes in Human Welfare Streptokinase, produced by the bacterium Streptococcus and modified by genetic engineering, is used as a clot buster: it dissolves clots in the blood vessels of patients who have had a myocardial infarction (heart attack). Answer: (4).
106	Molecular Basis of InheritanceHuman genome project	Class 12: Molecular Basis of Inheritance	Single correctEasy
A easy single correct question on human genome project, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance Among the salient features found by the Human Genome Project: chromosome 1 has the most genes (2968), and the Y chromosome has the fewest (231). Answer: (3) Chromosome 1.
108	Plant Growth and DevelopmentPlant hormones	Class 11: Plant Growth and Development	Single correctEasy
A easy single correct question on plant hormones, from Plant Growth and Development. NCERT: Class 11: Plant Growth and Development Cytokinins help produce new leaves and chloroplasts, and they promote nutrient mobilisation: nutrients are drawn towards the tissue treated with cytokinin, which keeps the leaf green and delays its senescence. Ethylene and abscisic acid hasten senescence. Answer: (4) Cytokinin.
111	Plant KingdomBryophytes and pteridophytes	Class 11: Plant Kingdom	Single correctEasy
A easy single correct question on bryophytes and pteridophytes, from Plant Kingdom. NCERT: Class 11: Plant Kingdom The main plant body of a pteridophyte is the sporophyte. In its sporangia, spore mother cells undergo meiosis to form spores (B). A spore germinates into a small, free-living, photosynthetic thalloid gametophyte, the prothallus (A). The prothallus bears the male and female sex organs, antheridia and archegonia (D). Water is needed to carry the antherozoids released from the antheridia to the mouth of the archegonium (E), where one fuses with the egg: fertilisation (C), giving a zygote that grows into a new sporophyte. Sequence: B, A, D, E, C. Answer: (1).
113	Microbes in Human WelfareBiocontrol agents and biofertilisers	Class 12: Microbes in Human Welfare	Single correctEasy
A easy single correct question on biocontrol agents and biofertilisers, from Microbes in Human Welfare. NCERT: Class 12: Microbes in Human Welfare Biological nitrogen fixation is done only by prokaryotes, which have the enzyme nitrogenase. A. Azotobacter: a free-living soil bacterium that fixes atmospheric nitrogen (used as a biofertiliser). B, C, E. Oscillatoria, Anabaena and Nostoc: cyanobacteria (blue-green algae) that fix atmospheric nitrogen; Anabaena and Nostoc do it in heterocysts. D. Volvox: a eukaryotic, colonial green alga. Eukaryotes have no nitrogenase, so it cannot fix nitrogen. Answer: (2) D only.
114	Molecular Basis of InheritanceTranscription and translation	Class 12: Molecular Basis of Inheritance	StatementsMedium
A medium statements question on transcription and translation, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance Statement I: During translation, the anticodon of each charged tRNA base-pairs with a codon on the mRNA, and the rRNA of the small subunit binds the mRNA (the ribosome-binding site lines up against rRNA) and holds it in place. So both tRNA and rRNA interact with mRNA. Incorrect. Statement II: RNA interference silences a specific mRNA with a complementary double-stranded RNA that binds to it and prevents its translation. It takes place in all eukaryotic organisms as a method of cellular defence (for example against viruses with RNA genomes and mobile genetic elements). Correct. Answer: (4).
118	Anatomy of Flowering PlantsAnatomy of root, stem and leaf	Class 11: Anatomy of Flowering Plants	Single correctEasy
A easy single correct question on anatomy of root, stem and leaf, from Anatomy of Flowering Plants. NCERT: Class 11: Anatomy of Flowering Plants A monocot stem has a sclerenchymatous hypodermis, a large number of scattered vascular bundles each surrounded by a sclerenchymatous bundle sheath, and a large, conspicuous parenchymatous ground tissue. The vascular bundles are conjoint and closed (no cambium), peripheral bundles are smaller than the central ones, phloem parenchyma is absent, and water-containing cavities are present within the bundles. So the statement that is not correct is that the hypodermis is parenchymatous: in a monocot stem it is sclerenchymatous. (A dicot stem has a collenchymatous hypodermis.) Answer: (1).
119	Plant KingdomBryophytes and pteridophytes	Class 11: Plant Kingdom	Single correctMedium
A medium single correct question on bryophytes and pteridophytes, from Plant Kingdom. NCERT: Class 11: Plant Kingdom In bryophytes the main plant body is the haploid gametophyte, attached to the substratum by rhizoids (B). Its antheridia release biflagellate antherozoids into water (E), which swim to an archegonium where one fuses with the egg (A). The zygote develops into a multicellular sporophyte, which stays attached to the gametophyte and depends on it (D). Some cells of the sporophyte undergo reduction division (meiosis) to produce haploid spores (C), which germinate to form new gametophytes. Sequence: B, E, A, D, C. Answer: (3).
121	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	MatchingEasy
A easy matching question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life A. The centromere holds the two chromatids together; its kinetochores attach the chromosome to spindle fibres, so it works in cell division (II). B. Cilia are hair-like outgrowths that move the cell or the fluid around it: cell movement (III). C. Cristae are the infoldings of the inner membrane of the mitochondrion (I). D. The cell membrane is built on a phospholipid bilayer with proteins in it (fluid mosaic model) (IV). Answer: (4) A-II, B-III, C-I, D-IV.
122	Photosynthesis in Higher PlantsLight reaction	Class 11: Photosynthesis in Higher Plants	MatchingEasy
A easy matching question on light reaction, from Photosynthesis in Higher Plants. NCERT: Class 11: Photosynthesis in Higher Plants Paper chromatography of a leaf extract separates its pigments by colour: - Chlorophyll a: bright or blue-green (III) - Chlorophyll b: yellow-green (I) - Xanthophylls: yellow (II) - Carotenoids (carotenes): yellow to yellow-orange (IV) Answer: (2) A-III, B-I, C-II, D-IV.
124	Sexual Reproduction in Flowering PlantsSeed and fruit development	Class 12: Sexual Reproduction in Flowering Plants	Single correctEasy
A easy single correct question on seed and fruit development, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants In a monocot seed such as maize or wheat, the seed coat is membranous and generally fused with the fruit wall. The endosperm is bulky and stores food. Its outer covering separates it from the embryo as a proteinous layer called the aleurone layer. The embryo has one large scutellum and a short axis with a plumule and radicle, enclosed in sheaths called the coleoptile and the coleorhiza. Answer: (4) Aleurone layer.
127	Sexual Reproduction in Flowering PlantsMegasporogenesis	Class 12: Sexual Reproduction in Flowering Plants	Assertion and reasonEasy
A easy assertion and reason question on megasporogenesis, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants Assertion: The functional megaspore divides three times by free nuclear mitosis to give 8 nuclei. Three cells at the micropylar end form the egg apparatus, three at the chalazal end are the antipodals, and the two remaining polar nuclei sit in the large central cell. So the mature embryo sac is 8-nucleate and 7-celled. True. Reason: The egg apparatus consists of the egg cell and two synergids. The two polar nuclei lie in the central cell, not in the egg apparatus. False. Answer: (3).
128	Cell: The Unit of LifeProkaryotic and eukaryotic cells	Class 11: Cell: The Unit of Life	Single correctEasy
A easy single correct question on prokaryotic and eukaryotic cells, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life The plasma membrane of a prokaryotic cell forms special infoldings called mesosomes, as vesicles, tubules and lamellae. They help in cell wall formation, DNA replication and its distribution to daughter cells, respiration, secretion, and in increasing the surface area of the plasma membrane and its enzymatic content. Answer: (1) Mesosome.
129	Molecular Basis of InheritanceTranscription and translation	Class 12: Molecular Basis of Inheritance	StatementsMedium
A medium statements question on transcription and translation, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance In eukaryotes the primary transcript (hnRNA) is processed in the nucleus: - Splicing: the introns are removed and the exons joined in a defined order (B). - Capping: an unusual nucleotide, methyl guanosine triphosphate, is added to the 5' end of hnRNA (C). - Tailing: 200–300 adenylate residues are added at the 3' end, independent of the template (D). Only the fully processed hnRNA, now mRNA, is transported out of the nucleus for translation, so A (transport before splicing) is wrong. E describes the pairing of a double-stranded RNA with an mRNA in RNA interference, not processing of the transcript. Answer: (2) B, C, D only.
130	Principles of Inheritance and VariationDeviations from Mendelism	Class 12: Principles of Inheritance and Variation	Single correctMedium
A medium single correct question on deviations from mendelism, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation A polygenic trait, such as human skin colour or height, is controlled by three or more genes, each with a small additive effect, and is also influenced by the environment. Instead of falling into the distinct classes Mendel described (tall or dwarf, 3 : 1), the offspring show a continuous range of phenotypes, a bell-shaped distribution. Because the trait does not follow Mendel's simple pattern of distinct alternative forms, its inheritance is described as non-Mendelian. (Each gene by itself still segregates as Mendel described, but the trait's pattern is not Mendelian.) It is not the pattern of a single autosomal dominant or X-linked recessive gene. Answer: (2).
132	Biological ClassificationFive kingdom classification	Class 11: Biological Classification	Single correctMedium
A medium single correct question on five kingdom classification, from Biological Classification. NCERT: Class 11: Biological Classification Identify each kingdom: C. Prokaryotes with a cell wall of polysaccharide and amino acids (peptidoglycan): Monera. E. Unicellular eukaryotes, cellular body organisation: Protista. A. Multicellular (mostly) heterotrophs with chitin cell walls, loose tissue at most: Fungi. D. Eukaryotic autotrophs with tissue/organ level: Plantae. B. Heterotrophs with tissue/organ/organ system level: Animalia. In Whittaker's table, body organisation goes from cellular (Monera, Protista) to multicellular/loose tissue (Fungi), to tissue/organ (Plantae), to tissue/organ/organ system (Animalia). Increasing order: C, E, A, D, B. Answer: (2).
133	Organisms and PopulationsPopulation attributes	Class 12: Organisms and Populations	Single correctEasy
A easy single correct question on population attributes, from Organisms and Populations. NCERT: Class 12: Organisms and Populations Professor Ramdeo Misra is revered as the Father of Ecology in India. He laid the foundations for research and teaching in ecology in the country, and his work on tropical communities, succession, productivity and nutrient cycling led to the first post-graduate course in ecology in India. He was also instrumental in setting up the National Committee for Environmental Planning and Coordination, which later led to the Ministry of Environment and Forests. Answer: (2).
134	Molecular Basis of InheritanceDNA structure	Class 12: Molecular Basis of Inheritance	MatchingEasy
A easy matching question on dna structure, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance A. Hershey and Chase (1952), with bacteriophages labelled with radioactive phosphorus and sulphur, gave unequivocal proof that DNA is the genetic material (IV). B. Euchromatin is the loosely packed, lightly stained chromatin, and is transcriptionally active (III). C. Griffith (1928) did his transformation experiments with Streptococcus pneumoniae (I). D. Heterochromatin is densely packed and stains dark, and is inactive (II). Answer: (3) A-IV, B-III, C-I, D-II.
140	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	Assertion and reasonMedium
A medium assertion and reason question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life Assertion: The Golgi apparatus principally performs the function of packaging materials, to be delivered either to intracellular targets or secreted outside the cell. True. Reason: The Golgi cisternae have two faces: the convex cis (forming) face and the concave trans (maturing) face. Materials to be packaged, in the form of vesicles from the ER, fuse with the cis face and move toward the maturing face; on the way they are modified (glycoproteins and glycolipids are formed) and they are released from the trans face. True. The reason describes how the Golgi receives, modifies and releases the ER's products, which is exactly how it carries out the packaging and delivery stated in the assertion. So R explains A. Answer: (1).
141	Sexual Reproduction in Flowering PlantsSeed and fruit development	Class 12: Sexual Reproduction in Flowering Plants	MatchingEasy
A easy matching question on seed and fruit development, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants A. The embryo of grasses has one cotyledon, called the scutellum (II). B. Non-albuminous seeds have no residual endosperm, because it is used up during embryo development: pea and groundnut (III). C. In some grass embryos a small outgrowth opposite the scutellum, the epiblast, is a rudimentary second cotyledon (IV). D. In some seeds such as black pepper and beet, remnants of the nucellus persist: this persistent nucellus is the perisperm (I). Answer: (1) A-II, B-III, C-IV, D-I.
145	Principles of Inheritance and VariationMendelian genetics	Class 12: Principles of Inheritance and Variation	Single correctEasy
A easy single correct question on mendelian genetics, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation RRYY × rryy gives F1 all RrYy, round yellow. The F1 forms four kinds of gametes, RY, Ry, rY and ry, in equal proportion because the genes assort independently. Selfing gives 16 combinations: - R_Y_ round yellow: 9 - R_yy round green: 3 - rrY_ wrinkled yellow: 3 - rryy wrinkled green: 1 Phenotypic ratio 9 : 3 : 3 : 1, the dihybrid ratio. Answer: (3).
146	Molecular Basis of InheritanceDNA structure	Class 12: Molecular Basis of Inheritance	Single correctEasy
A easy single correct question on dna structure, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance In eukaryotes, negatively charged DNA is wrapped round a set of positively charged basic proteins called histones. A protein takes its charge from its amino acid side chains: histones are rich in the basic amino acids lysine and arginine, whose side chains carry positive charges. Eight histone molecules form the histone octamer around which DNA wraps to make a nucleosome. Answer: (1).
150	EcosystemProductivity and decomposition	Class 12: Ecosystem	Single correctEasy
A easy single correct question on productivity and decomposition, from Ecosystem. NCERT: Class 12: Ecosystem Productivity is the rate of biomass production: the amount produced per unit area over a time period. It is expressed in terms of weight (g m$^{-2}$ yr$^{-1}$) or energy (kcal m$^{-2}$ yr$^{-1}$). Only option (4) contains both an area and a time. Answer: (4).
152	Principles of Inheritance and VariationGenetic disorders	Class 12: Principles of Inheritance and Variation	Single correctHard
A hard single correct question on genetic disorders, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation Find the pattern. Only females are shown as carriers, an affected mother (F0) has carrier daughters with an unaffected father, and a carrier mother has an affected son: the disease is X-linked recessive. Write the normal allele X$^A$ and the mutant X$^a$. The F2 parents of the F3 children: the female is a carrier, X$^A$X$^a$; the male (her cousin, a consanguineous marriage shown by the double line) is affected, X$^a$Y. Cross X$^A$X$^a$ × X$^a$Y: - X$^A$X$^a$: carrier daughter, no disease, 1/4 - X$^a$X$^a$: affected daughter, 1/4 - X$^A$Y: normal son, 1/4 - X$^a$Y: affected son, 1/4 A child with no disease who carries the mutation in one allele is the carrier daughter, X$^A$X$^a$: probability 1/4 of all births. (Sons have a single X, so a son is either normal or affected, never a carrier.) Answer: (1) 1/4.
153	Sexual Reproduction in Flowering PlantsMicrosporogenesis	Class 12: Sexual Reproduction in Flowering Plants	Assertion and reasonMedium
A medium assertion and reason question on microsporogenesis, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants Assertion: The tapetum is the innermost wall layer of the microsporangium. Its cells possess dense cytoplasm and generally have more than one nucleus. True. Reason: The tapetum nourishes the developing pollen, from the sporogenous tissue and microspore mother cells through to the microspores. More nuclei (and the polyploid state of tapetal cells) mean more gene copies and a higher rate of synthesis of the materials the tapetum supplies, so the multinucleate condition does make it a more efficient nourishing layer. True, and it is why the tapetal cells are built this way. Answer: (1).
154	Sexual Reproduction in Flowering PlantsMegasporogenesis	Class 12: Sexual Reproduction in Flowering Plants	Single correctMedium
A medium single correct question on megasporogenesis, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants The megaspore mother cell undergoes one meiotic division to form four megaspores, of which one is functional (monosporic development). The nucleus of the functional megaspore then divides mitotically three successive times (free nuclear divisions): 1 → 2 → 4 → 8 nuclei. Cell walls then form, giving the 8-nucleate, 7-celled embryo sac, the mature female gametophyte. So: 1 meiosis and 3 (rounds of) mitosis. Answer: (3).
155	Morphology of Flowering PlantsFlower and fruit	Class 11: Morphology of Flowering Plants	Single correctEasy
A easy single correct question on flower and fruit, from Morphology of Flowering Plants. NCERT: Class 11: Morphology of Flowering Plants A zygomorphic flower can be divided into two similar halves only along one particular vertical plane. Pea, bean, gulmohur and Cassia have zygomorphic flowers; in the pea the papilionaceous corolla (standard, two wings, keel) is bilaterally symmetrical. Petunia, Datura and chilli belong to Solanaceae, whose flowers are actinomorphic: they can be cut into two equal halves along any radial plane through the centre. Answer: (3) Pea.
157	Organisms and PopulationsInteractions	Class 12: Organisms and Populations	StatementsMedium
A medium statements question on interactions, from Organisms and Populations. NCERT: Class 12: Organisms and Populations Statement II: Each fig species can be pollinated only by its partner wasp species. The female wasp pollinates the fig inflorescence while searching for suitable egg-laying sites, and the fig offers the wasp sites for laying eggs, using some of its developing seeds to nourish the larvae. Both partners gain, so this is mutualism. Correct. Statement I: A fig is a plant product, the fruit of an inflorescence (syconium). A wasp that enters to pollinate and lay eggs dies inside, and its body is broken down by the fig's own enzymes; the offspring leave the fruit. The ripe fruit does not contain wasps as an animal part of the food, and the fig remains a plant food. Calling it a 'non-vegetarian fruit because it has enclosed fig wasps' is not a biological statement and is incorrect. Statement II is also not correct as worded: the wasp develops inside the fig's inflorescence (the syconium), pollinating its flowers there; the 'fig fruit' is not what the wasp pollinates. Answer: (2).
158	Cell Cycle and Cell DivisionMitosis	Class 11: Cell Cycle and Cell Division	Single correctEasy
A easy single correct question on mitosis, from Cell Cycle and Cell Division. NCERT: Class 11: Cell Cycle and Cell Division Spindle fibres, made of microtubules, attach to the kinetochores of the chromosomes and align them at the metaphase plate. In anaphase the centromeres split and the spindle fibres pull the sister chromatids (now daughter chromosomes) to opposite poles. So their main function is to separate the chromosomes equally into the two daughter nuclei. Answer: (1).
159	Plant KingdomGymnosperms	Class 11: Plant Kingdom	Single correctEasy
A easy single correct question on gymnosperms, from Plant Kingdom. NCERT: Class 11: Plant Kingdom Gymnosperms (gymnos: naked, sperma: seeds) are plants in which the ovules are not enclosed by any ovary wall and remain exposed, before and after fertilisation. The seeds that develop after fertilisation are therefore not covered: they are naked. In angiosperms the seeds are enclosed in fruits. Answer: (2).
162	Plant KingdomBryophytes and pteridophytes	Class 11: Plant Kingdom	MatchingEasy
A easy matching question on bryophytes and pteridophytes, from Plant Kingdom. NCERT: Class 11: Plant Kingdom A. Salvinia is a heterosporous aquatic fern, a pteridophyte (IV). B. Polytrichum is a moss, a bryophyte (III). C. Salvia is a flowering plant (of the mint family), an angiosperm (I). D. Ginkgo is a gymnosperm (II). Do not confuse Salvinia (fern) with Salvia (angiosperm). Answer: (2) A-IV, B-III, C-I, D-II.
163	Molecular Basis of InheritanceTranscription and translation	Class 12: Molecular Basis of Inheritance	Single correctEasy
A easy single correct question on transcription and translation, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance George Gamow, a physicist, argued that since there are only 4 bases and they must code for 20 amino acids, the code should be made of a combination of bases. A code of two bases gives only $4^2 = 16$ combinations, too few; three bases give $4^3 = 64$, more than enough. He therefore proposed a triplet code. Har Gobind Khorana's chemical synthesis of RNA and Marshall Nirenberg's cell-free system later helped decipher it. Answer: (1).
164	Biodiversity and ConservationConservation strategies	Class 12: Biodiversity and Conservation	MatchingEasy
A easy matching question on conservation strategies, from Biodiversity and Conservation. NCERT: Class 12: Biodiversity and Conservation A. The 'Evil Quartet' is the name for the four major causes of biodiversity loss: habitat loss and fragmentation, over-exploitation, alien species invasions and co-extinctions (III). B. Ex situ conservation includes cryopreservation of gametes in a viable and fertile condition for long periods, along with zoos, botanical gardens and seed banks (I). C. Lantana camara is an invasive alien weed that threatens native species, an example of alien species invasion (II). D. The dodo of Mauritius is an example of recent extinction (IV). Answer: (2) A-III, B-I, C-II, D-IV.
168	Plant KingdomBryophytes and pteridophytes	Class 11: Plant Kingdom	Single correctEasy
A easy single correct question on bryophytes and pteridophytes, from Plant Kingdom. NCERT: Class 11: Plant Kingdom Liverworts such as Marchantia reproduce asexually by fragmentation of the thallus or by gemmae. Gemmae are green, multicellular, asexual buds that develop in small receptacles called gemma cups on the thallus. They become detached from the parent and germinate to form new individuals. Answer: (2).
170	EcosystemEnergy flow	Class 12: Ecosystem	StatementsEasy
A easy statements question on energy flow, from Ecosystem. NCERT: Class 12: Ecosystem Statement I: Energy fixed by producers from sunlight passes to herbivores and then to carnivores; it does not return. So there is a unidirectional flow of energy from the sun to producers and then to consumers. Correct. Statement II: Ecosystems are not exempt from the second law of thermodynamics. At every transfer some energy is lost as heat, which is why only about 10 per cent passes to the next trophic level, and an ecosystem needs a constant supply of energy to synthesise the molecules it needs and to counteract the universal tendency toward increasing disorder. Incorrect. Answer: (3).
171	Photosynthesis in Higher PlantsCalvin cycle	Class 11: Photosynthesis in Higher Plants	Single correctEasy
A easy single correct question on calvin cycle, from Photosynthesis in Higher Plants. NCERT: Class 11: Photosynthesis in Higher Plants RuBisCO (RuBP carboxylase-oxygenase), the most abundant enzyme in the world, catalyses the first step of the Calvin cycle: carboxylation of ribulose-1,5-bisphosphate (RuBP) by CO$_2$, giving two molecules of 3-PGA. It can also bind O$_2$ (photorespiration), but its active site has a much greater affinity for CO$_2$ when the two are present in equal amounts; the relative concentrations decide which binds. Photolysis of water is done by the water-splitting complex of photosystem II. Answer: (4).
173	Plant Growth and DevelopmentPlant hormones	Class 11: Plant Growth and Development	StatementsMedium
A medium statements question on plant hormones, from Plant Growth and Development. NCERT: Class 11: Plant Growth and Development A. True. Auxins induce parthenocarpy, for example in tomatoes. B. True. Plant growth regulators are broadly divided into promoters (auxins, gibberellins, cytokinins) and inhibitors (abscisic acid); ethylene can do both, and an auxin such as 2,4-D kills dicot weeds. C. True. A differentiated cell that has lost the ability to divide first regains it (dedifferentiation, forming a meristem such as the interfascicular cambium or cork cambium); the cells it produces then lose the ability again and mature into specific functions (redifferentiation). Redifferentiation follows dedifferentiation. D. False. Abscisic acid is a growth inhibitor, the stress hormone that promotes dormancy and closure of stomata. E. False. In apical dominance the growing apical bud inhibits the growth of the lateral (axillary) buds; removing the shoot tip (decapitation) lets them grow. Answer: (1) A, B, C only.
174	Molecular Basis of InheritanceTranscription and translation	Class 12: Molecular Basis of Inheritance	Single correctEasy
A easy single correct question on transcription and translation, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance In bacteria, RNA polymerase associates transiently with the initiation factor sigma ($\sigma$) to start transcription at the promoter, and with the termination factor rho ($\rho$) to stop it at the terminator. The association with these factors changes the specificity of the RNA polymerase to initiate or terminate. Answer: (3) $\rho$ (rho).
177	Microbes in Human WelfareMicrobes in household products	Class 12: Microbes in Human Welfare	StatementsMedium
A medium statements question on microbes in household products, from Microbes in Human Welfare. NCERT: Class 12: Microbes in Human Welfare Household products: - Lactobacillus (lactic acid bacteria) turns milk into curd (B). - Saccharomyces cerevisiae (baker's yeast) ferments dough to make bread (D). - Propionibacterium sharmanii produces CO$_2$ that forms the large holes of Swiss cheese (E). Industrial products: - Aspergillus niger, a fungus, is used to produce citric acid (A). - Trichoderma polysporum, a fungus, produces cyclosporin A, an immunosuppressive agent used in organ transplant patients (C). Not involved in household products: A and C. Answer: (2).
180	Organisms and PopulationsPopulation growth models	Class 12: Organisms and Populations	Single correctEasy
A easy single correct question on population growth models, from Organisms and Populations. NCERT: Class 12: Organisms and Populations With unlimited resources a population grows exponentially, $\dfrac{dN}{dt} = rN$. Resources are finite, so growth slows as the population nears the carrying capacity $K$; the Verhulst-Pearl logistic growth is $\dfrac{dN}{dt} = rN\left(\dfrac{K-N}{K}\right)$ where $N$ is the population density at time $t$ and $r$ the intrinsic rate of natural increase. When $N$ is small the bracket is close to 1 and growth is nearly exponential; as $N \to K$ the bracket tends to 0 and growth stops, giving the sigmoid curve. Answer: (2).

Zoology questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
92	Biotechnology: Principles and ProcessesPCR and bioreactors	Class 12: Biotechnology: Principles and Processes	Single correctEasy
A easy single correct question on pcr and bioreactors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes Each PCR cycle (denaturation, annealing of primers, extension by Taq polymerase) copies every DNA molecule present, so the number of molecules doubles with each cycle. Starting from one molecule, after 1 cycle there are 2, after 2 cycles 4, after 3 cycles 8, and after $n$ cycles $2^n$ copies. The growth is exponential, not linear ($2n + 1$) or a square ($N^2$, $2N^2$). Answer: (2) $2^n$.
93	Reproductive HealthART	Class 12: Reproductive Health	StatementsMedium
A medium statements question on art, from Reproductive Health. NCERT: Class 12: Reproductive Health Go through each statement. A. False. IVF is not a procedure with a high risk of death for the mother. B. True. ART needs highly precise handling by specialists and expensive instruments and reagents, so it is costly and available only in a few centres. C. False. The gametes need not come from the husband and wife; donor sperm or ova can be used. D. True. A couple who could adopt one of the many orphaned children may instead choose costly IVF for a child of their own, so adoption falls. E. False. IVF is available in India, though mostly in private centres in cities. F. True. Not every embryo transferred implants and develops; an early embryo may fail to survive, so a cycle may have to be repeated. The drawbacks are B, D and F. Answer: (1).
94	Structural Organisation in AnimalsFrog: morphology and anatomy	Class 11: Structural Organisation in Animals	Single correctEasy
A easy single correct question on frog: morphology and anatomy, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals In the frog, deoxygenated blood from the body is collected by the venae cavae (two precavals and a postcaval), which open into the sinus venosus and from there into the right atrium. Oxygenated blood from the lungs returns through the pulmonary veins to the left atrium, and blood leaves the heart through the conus arteriosus and the arterial arches (the aortae and pulmocutaneous arteries). Answer: (4) Vena cava.
101	Body Fluids and CirculationCirculatory disorders	Class 11: Body Fluids and Circulation	MatchingEasy
A easy matching question on circulatory disorders, from Body Fluids and Circulation. NCERT: Class 11: Body Fluids and Circulation A. Emphysema is a chronic disorder in which the alveolar walls are damaged, so the respiratory surface decreases (II); cigarette smoking is a major cause. B. Angina pectoris is acute chest pain that appears when not enough oxygen is reaching the heart muscle (III). C. Glomerulonephritis is inflammation of the glomeruli of the kidney (IV). D. Tetany is rapid spasms (wild contractions) of muscle due to low Ca$^{++}$ in body fluid (I). So A-II, B-III, C-IV, D-I. Answer: (4).
107	Structural Organisation in AnimalsFrog: morphology and anatomy	Class 11: Structural Organisation in Animals	Single correctEasy
A easy single correct question on frog: morphology and anatomy, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals Male frogs are told apart from females by vocal sacs and by a copulatory (nuptial) pad on the first digit of each fore limb. The pad helps the male grip the female during amplexus. Females have neither. Answer: (4).
109	Animal KingdomBasis of classification	Class 11: Animal Kingdom	Single correctEasy
A easy single correct question on basis of classification, from Animal Kingdom. NCERT: Class 11: Animal Kingdom A body cavity lined by mesoderm on all sides (against the body wall and around the gut) is a true coelom; animals with it are coelomates, and a schizocoelom is one kind. When the mesoderm is present only as scattered pouches between the ectoderm and the endoderm, so that the cavity is not lined by mesoderm against the gut, the cavity is a pseudocoelom (as in roundworms). Here the mesoderm lies only toward the body wall and not around the alimentary canal, so the animal is a pseudocoelomate. Answer: (2).
110	Human ReproductionGametogenesis	Class 12: Human Reproduction	MatchingEasy
A easy matching question on gametogenesis, from Human Reproduction. NCERT: Class 12: Human Reproduction A human sperm has a head, neck, middle piece and tail. A. The head holds the elongated haploid nucleus, the genetic material (IV). B. The middle piece has many mitochondria, which produce the energy for movement (III). C. The acrosome, a cap over the head, is filled with enzymes that help in fertilisation of the ovum (I). D. The tail gives the sperm its motility (II). Answer: (1) A-IV, B-III, C-I, D-II.
112	Body Fluids and CirculationCardiac cycle	Class 11: Body Fluids and Circulation	StatementsMedium
A medium statements question on cardiac cycle, from Body Fluids and Circulation. NCERT: Class 11: Body Fluids and Circulation The normal activities of the heart are regulated intrinsically, that is auto-regulated by specialised muscles (nodal tissue: SA node, AV node, bundle of His), so the heart is myogenic (A). A special neural centre in the medulla oblongata can moderate the cardiac function through the autonomic nervous system: sympathetic nerves raise the heart rate and the strength of contraction, parasympathetic nerves lower them (B). Adrenal medullary hormones, adrenaline and noradrenaline, also increase the cardiac output (C). The adrenal cortical hormones (glucocorticoids, mineralocorticoids) are not among the regulators of cardiac activity described. Answer: (1) A, B and C only.
115	Biotechnology: Principles and ProcessesRestriction enzymes and vectors	Class 12: Biotechnology: Principles and Processes	Single correctMedium
A medium single correct question on restriction enzymes and vectors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes The EcoRI site in this plasmid lies inside the gene for beta-galactosidase. Inserting the alien DNA there interrupts the gene, so recombinant plasmids no longer make the enzyme: insertional inactivation. The bacteria are grown on a medium containing a chromogenic substrate. A non-recombinant plasmid still makes beta-galactosidase, which acts on the substrate and gives blue colonies. A recombinant plasmid cannot, so its colonies stay white (colourless). White colonies are picked as recombinants. The ampicillin and tetracycline resistance genes are untouched by an insert at EcoRI, so antibiotic plates cannot tell recombinants from non-recombinants (they only show which cells took up a plasmid at all). Answer: (3).
116	Biotechnology and its ApplicationsHuman insulin and vaccine production	Class 12: Biotechnology and its Applications	Single correctEasy
A easy single correct question on human insulin and vaccine production, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications In 1983 the American company Eli Lilly prepared two DNA sequences corresponding to the A and B chains of human insulin and introduced them into plasmids of the bacterium E. coli to produce the insulin chains. The chains were produced separately, extracted and joined by disulphide bonds to form human insulin. Answer: (1) Bacterium.
117	BiomoleculesEnzymes	Class 11: Biomolecules	Single correctEasy
A easy single correct question on enzymes, from Biomolecules. NCERT: Class 11: Biomolecules Enzymes are put in six classes: - Oxidoreductases/dehydrogenases: oxidation–reduction between two substrates. - Transferases: transfer of a group G (other than hydrogen) between a pair of substrates S and S$^\#$: S – G + S$^\#$ $\rightarrow$ S + S$^\#$ – G. - Hydrolases: hydrolysis of ester, ether, peptide and similar bonds. - Lyases: removal of groups from substrates by mechanisms other than hydrolysis, leaving double bonds. - Isomerases: interconversion of optical, geometric or positional isomers. - Ligases: linking together of two compounds. The reaction given moves the group G from S to S$^\#$, which is the defining reaction of a transferase. Answer: (3).
120	Human Health and DiseaseCancer and drug abuse	Class 12: Human Health and Disease	StatementsEasy
A easy statements question on cancer and drug abuse, from Human Health and Disease. NCERT: Class 12: Human Health and Disease A. True. Radiography (X-rays), computed tomography (CT) and magnetic resonance imaging (MRI) are used to detect cancers of the internal organs. B. False. Chemotherapeutic drugs are used to kill cancerous cells; their side effects on normal cells are a drawback, not their purpose. C. True. Patients are given biological response modifiers such as $\alpha$-interferon, which activate their immune system and help in destroying the tumour. D. False. The biological response modifiers are substances like $\alpha$-interferon; chemotherapeutic drugs are a different treatment. E. False. Leukaemia is detected by an increased count of (abnormal) white blood cells in blood and bone marrow tests. Correct: A and C. Answer: (4).
123	Human ReproductionEmbryonic development	Class 12: Human Reproduction	StatementsMedium
A medium statements question on embryonic development, from Human Reproduction. NCERT: Class 12: Human Reproduction The course of human foetal development: - After one month of pregnancy, the embryo's heart is formed (C true). - By the end of the second month, the foetus develops limbs and digits (D true). - By the end of 12 weeks (the first trimester), most of the major organ systems are formed, for example the limbs and external genital organs are well developed (A true, so B, which says 8 weeks, is false). - The first movements of the foetus and the appearance of hair on the head are usually seen during the fifth month (E true). - By the end of 24 weeks (second trimester) the body is covered with fine hair, the eyelids separate and eyelashes form. Correct: A, C, D and E. Answer: (4).
125	Excretory Products and their EliminationUrine formation	Class 11: Excretory Products and their Elimination	Single correctMedium
A medium single correct question on urine formation, from Excretory Products and their Elimination. NCERT: Class 11: Excretory Products and their Elimination In each drawing an arrow from the lumen outward is reabsorption, and an arrow into the lumen is secretion. Proximal convoluted tubule (PCT): nearly all essential nutrients and 70–80 per cent of the electrolytes and water are reabsorbed here, so NaCl, HCO$_3^-$ and H$_2$O leave the lumen. It keeps the pH and ionic balance of body fluids by selective secretion of H$^+$ and ammonia (NH$_3$) into the filtrate and absorption of HCO$_3^-$ from it. Distal convoluted tubule (DCT): conditional reabsorption of Na$^+$ (NaCl) and water takes place, and HCO$_3^-$ is reabsorbed; H$^+$ and K$^+$ (and NH$_3$) are selectively secreted to keep the pH and the sodium–potassium balance. Drawing (2) shows exactly this. Drawings (1) and (4) put water flowing into the tubule (secreted), and (4) also reabsorbs NH$_3$ in the distal tubule; drawing (3) secretes NaCl into the distal tubule. Answer: (2).
126	Biotechnology: Principles and ProcessesPCR and bioreactors	Class 12: Biotechnology: Principles and Processes	Single correctMedium
A medium single correct question on pcr and bioreactors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes In a simple stirred-tank bioreactor, a motor at the top (B) turns a central shaft. Flat-bladed impellers on the shaft, down in the culture broth (A), mix the contents and spread the oxygen. Sterile air is bubbled in at the bottom through a sparger (D). Stirring and aeration make foam at the surface of the broth, so the uppermost bladed disc, just under the lid and above the liquid (C), is the foam braker: it breaks the foam as it rises. Answer: (4) C.
131	BiomoleculesEnzymes	Class 11: Biomolecules	Single correctEasy
A easy single correct question on enzymes, from Biomolecules. NCERT: Class 11: Biomolecules A prosthetic group is an organic compound tightly bound to the apoenzyme. In peroxidase and catalase, which catalyse the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and is part of the active site of the enzyme. Carbonic anhydrase uses zinc as a cofactor, and RuBisCo is activated by Mg$^{2+}$. Answer: (4) Catalase.
135	Human Health and DiseaseCancer and drug abuse	Class 12: Human Health and Disease	StatementsMedium
A medium statements question on cancer and drug abuse, from Human Health and Disease. NCERT: Class 12: Human Health and Disease Malignant tumours are a mass of proliferating cells called neoplastic or tumour cells (A). They grow very rapidly (B), invading and damaging the surrounding normal tissues (C), and through metastasis spread to distant sites. Remaining confined to the original location (D) is the mark of a benign tumour, which causes little damage, so it is not a neoplastic characteristic in this sense. Answer: (2) A, B, C only.
136	Biotechnology: Principles and ProcessesRestriction enzymes and vectors	Class 12: Biotechnology: Principles and Processes	StatementsEasy
A easy statements question on restriction enzymes and vectors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes Statement I: After electrophoresis the separated bands are cut out of the agarose gel and the DNA is extracted from the gel piece (elution). These purified DNA fragments are used in constructing recombinant DNA by joining them with cloning vectors. Correct. Statement II: DNA is negatively charged, so in an electric field the fragments move toward the anode (positive electrode) through the agarose matrix, which acts as a sieve. The smaller the fragment, the farther it moves; so the smallest fragments lie nearest the anode end and the largest remain close to the wells. Correct. Answer: (1).
137	BiomoleculesCarbohydrates, lipids and nucleic acids	Class 11: Biomolecules	MatchingEasy
A easy matching question on carbohydrates, lipids and nucleic acids, from Biomolecules. NCERT: Class 11: Biomolecules Adenine is a nitrogen base, a purine (C-I). Adenine joined to a sugar (ribose) is a nucleoside, adenosine (A-III). A nucleoside with a phosphate esterified to its sugar is a nucleotide, adenylic acid (adenosine monophosphate) (B-II). Alanine is an amino acid, with a methyl group as its R group (D-IV). Answer: (3) A-III, B-II, C-I, D-IV.
138	Human ReproductionGametogenesis	Class 12: Human Reproduction	StatementsMedium
A medium statements question on gametogenesis, from Human Reproduction. NCERT: Class 12: Human Reproduction A. True. Oogenesis begins during embryonic development: primary oocytes enter prophase I of meiosis before birth and stay arrested there. In the male, spermatocytes start meiosis only at puberty. B. True. In the male, a secondary spermatocyte completes meiosis II soon after meiosis I. In the female, the primary oocyte completes meiosis I only just before ovulation (years after birth), and the secondary oocyte completes meiosis II only if a sperm enters it, so the gap is far longer. C. False. The first polar body is formed when the primary oocyte completes meiosis I, giving a large secondary oocyte and the small first polar body. It goes with the formation of the secondary oocyte, not the primary. D. False. The LH surge in mid-cycle induces ovulation. Menstruation follows when the corpus luteum degenerates and progesterone falls, which breaks down the endometrium. Answer: (1) A and B are true.
139	Animal KingdomChordate classes	Class 11: Animal Kingdom	Single correctEasy
A easy single correct question on chordate classes, from Animal Kingdom. NCERT: Class 11: Animal Kingdom All living members of the class Cyclostomata (lampreys, Petromyzon, and hagfishes, Myxine) are ectoparasites on some fishes. They have a sucking, circular mouth without jaws, an elongated body with 6–15 pairs of gill slits, and a cartilaginous cranium and vertebral column. They are marine but migrate to fresh water for spawning, and die a few days after. Answer: (4).
142	Animal KingdomChordate classes	Class 11: Animal Kingdom	Assertion and reasonEasy
A easy assertion and reason question on chordate classes, from Animal Kingdom. NCERT: Class 11: Animal Kingdom Assertion: Phylum Chordata includes Urochordata and Cephalochordata (protochordates), which have a notochord but never a vertebral column, as well as Vertebrata. So every vertebrate is a chordate, but not every chordate is a vertebrate. True. Reason: Vertebrates have a notochord during the embryonic period; in the adult it is replaced by a cartilaginous or bony vertebral column. True. This is the reason given for the assertion: vertebrates qualify as chordates because they possess a notochord (in the embryo), while the protochordates, which keep only a notochord and never form a vertebral column, are chordates but not vertebrates. R explains A. Answer: (1).
143	Human Health and DiseaseImmunity	Class 12: Human Health and Disease	Single correctMedium
A medium single correct question on immunity, from Human Health and Disease. NCERT: Class 12: Human Health and Disease Each antibody molecule has four peptide chains, two small light chains and two longer heavy chains, so it is written H$_2$L$_2$; the chains are held together by disulphide bonds. Each chain has a variable region at its amino (N-terminal) end and a constant region toward its carboxyl (C-terminal) end. The antigen-binding sites are formed by the variable regions at the tips of the two arms of the Y, at the N-terminal ends. So it is wrong to say the antigen-binding site is at the C-terminal region. Answer: (3).
144	Biotechnology and its ApplicationsTransgenic organisms	Class 12: Biotechnology and its Applications	Single correctEasy
A easy single correct question on transgenic organisms, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications RNA interference silences a specific mRNA by a complementary double-stranded RNA (dsRNA) molecule that binds to the mRNA and prevents its translation. The dsRNA may come from an infection by viruses with RNA genomes or from mobile genetic elements that replicate through an RNA intermediate. In making nematode-resistant tobacco, nematode-specific genes were introduced so that they produced both sense and anti-sense RNA; the two strands formed dsRNA that silenced the parasite's mRNA. Answer: (1).
147	Human ReproductionMenstrual cycle	Class 12: Human Reproduction	Single correctEasy
A easy single correct question on menstrual cycle, from Human Reproduction. NCERT: Class 12: Human Reproduction The reproductive cycle in female primates is the menstrual cycle. The first menstruation begins at puberty and is called menarche. In human females, cycles stop around 50 years of age, which is termed menopause. Answer: (2).
148	Chemical Coordination and IntegrationEndocrine glands	Class 11: Chemical Coordination and Integration	MatchingEasy
A easy matching question on endocrine glands, from Chemical Coordination and Integration. NCERT: Class 11: Chemical Coordination and Integration A. The atrial wall of the heart secretes atrial natriuretic factor (ANF), which lowers blood pressure (III). B. The juxtaglomerular cells of the kidney secrete erythropoietin, which stimulates formation of RBCs (I). C. Endocrine cells of the gastro-intestinal tract secrete gastrin, secretin, cholecystokinin and GIP (IV). D. The adrenal cortex (zona glomerulosa) secretes aldosterone, the main mineralocorticoid (II). Answer: (4) A-III, B-I, C-IV, D-II.
149	BiomoleculesEnzymes	Class 11: Biomolecules	Single correctEasy
A easy single correct question on enzymes, from Biomolecules. NCERT: Class 11: Biomolecules Many enzymes need a non-protein constituent, a cofactor, bound to them to be catalytically active. The protein portion of such an enzyme is called the apoenzyme. Cofactors are of three kinds: prosthetic groups (organic, tightly bound, e.g. haem in catalase), coenzymes (organic, transiently bound, e.g. NAD, NADP) and metal ions (e.g. zinc in carboxypeptidase). Removing the cofactor leaves an inactive apoenzyme. Answer: (3).
151	EvolutionEvidences of evolution	Class 12: Evolution	Single correctEasy
A easy single correct question on evidences of evolution, from Evolution. NCERT: Class 12: Evolution The sweet potato is a modified root and the potato a modified stem (tuber). They are structurally and developmentally different, yet both store food and look alike. Organs with different origins that perform the same function are analogous, and they arise by convergent evolution, where different structures evolve toward a similar function. Answer: (1).
156	Human Health and DiseaseImmunity	Class 12: Human Health and Disease	StatementsEasy
A easy statements question on immunity, from Human Health and Disease. NCERT: Class 12: Human Health and Disease The primary lymphoid organs, bone marrow and thymus, are where immature lymphocytes differentiate into antigen-sensitive lymphocytes. After maturation the lymphocytes migrate to the secondary lymphoid organs, the spleen, lymph nodes, tonsils, Peyer's patches of the small intestine and the appendix, which provide the sites where lymphocytes interact with antigens and proliferate into effector cells. Thymus (A) and bone marrow (B) are primary organs. Answer: (4) C, D, E only.
160	Chemical Coordination and IntegrationEndocrine glands	Class 11: Chemical Coordination and Integration	StatementsMedium
A medium statements question on endocrine glands, from Chemical Coordination and Integration. NCERT: Class 11: Chemical Coordination and Integration The adrenal medulla secretes adrenaline and noradrenaline (catecholamines), the emergency or 'fight or flight' hormones. They increase alertness, cause pupillary dilation (so A, constriction, is false), piloerection (raising of hairs, C true) and sweating; they increase the heart beat, the strength of heart contraction (D true) and the rate of respiration; and they stimulate the breakdown of glycogen, raising blood glucose, so they are hyperglycaemic (B true), and also the breakdown of lipids and proteins. Answer: (2) B, C and D only.
161	Biotechnology and its ApplicationsHuman insulin and vaccine production	Class 12: Biotechnology and its Applications	Single correctEasy
A easy single correct question on human insulin and vaccine production, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications Insulin is a protein (a small polypeptide hormone of two chains). Taken by mouth, it would be broken down like any dietary protein by the proteolytic enzymes of the gut (pepsin, trypsin, chymotrypsin and peptidases) into amino acids before it could reach the blood, so it is given by injection. Answer: (2).
165	Chemical Coordination and IntegrationEndocrine glands	Class 11: Chemical Coordination and Integration	Single correctEasy
A easy single correct question on endocrine glands, from Chemical Coordination and Integration. NCERT: Class 11: Chemical Coordination and Integration The posterior pituitary (pars nervosa, neurohypophysis) stores and releases two hormones, oxytocin and vasopressin (anti-diuretic hormone, ADH), which are actually synthesised by neurosecretory cells of the hypothalamus and carried down their axons to the neurohypophysis. LH, FSH and ACTH are made and secreted by the anterior pituitary (adenohypophysis) itself. Answer: (2).
166	Animal KingdomNon-chordate phyla	Class 11: Animal Kingdom	StatementsEasy
A easy statements question on non-chordate phyla, from Animal Kingdom. NCERT: Class 11: Animal Kingdom The most distinctive feature of echinoderms is the water vascular system, which helps in locomotion, capture and transport of food, and respiration. So A (respiration and locomotion) and C (capture and transport of food) are right. Echinoderms have no excretory system, and digestion is done by the alimentary canal, not the water vascular system, so B, D and E are wrong. Answer: (2) A and C only.
167	Human Health and DiseaseImmunity	Class 12: Human Health and Disease	Single correctEasy
A easy single correct question on immunity, from Human Health and Disease. NCERT: Class 12: Human Health and Disease Innate immunity is a non-specific type of defence that is present at the time of birth. It works through barriers: physical (skin, mucus coating), physiological (stomach acid, saliva, tears), cellular (neutrophils, monocytes, natural killer lymphocytes, macrophages) and cytokine (interferons). Acquired immunity, with its humoral (antibody) and cell-mediated (T cell) responses, is pathogen-specific and characterised by memory. Answer: (2).
169	Structural Organisation in AnimalsFrog: morphology and anatomy	Class 11: Structural Organisation in Animals	Single correctEasy
A easy single correct question on frog: morphology and anatomy, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals The frog has two special venous connections: the hepatic portal system, between the liver and the intestine, and the renal portal system, between the kidney and the lower parts of the body. In the renal portal system, veins from the hind limbs and lower body break up into capillaries in the kidneys before the blood returns to the heart. Answer: (4).
172	Biotechnology: Principles and ProcessesRestriction enzymes and vectors	Class 12: Biotechnology: Principles and Processes	StatementsMedium
A medium statements question on restriction enzymes and vectors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes Gene cloning by recombinant DNA technology needs: restriction enzymes (molecular scissors) to cut the DNA of interest and the vector at specific sites (A); DNA ligase to join the DNA fragment into the cut vector (B); and DNA polymerase to copy the DNA, as in PCR amplification of the gene of interest (E). 'DNA mutase' is not an enzyme used in cloning, and a DNA recombinase (which mediates site-specific recombination in cells) is not one of the essential tools either. Not essential: C and D. Answer: (1).
175	Structural Organisation in AnimalsFrog: morphology and anatomy	Class 11: Structural Organisation in Animals	StatementsMedium
A medium statements question on frog: morphology and anatomy, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals Frogs respire on land and in water by two different methods. In water, the skin acts as the aquatic respiratory organ (cutaneous respiration): dissolved oxygen in the water is exchanged through the skin by diffusion. On land, the buccal cavity, skin and lungs act as the respiratory organs (respiration by lungs is pulmonary respiration). So the land part of the statement is right, but the water part wrongly adds the buccal cavity. Answer: (3).
176	Human ReproductionEmbryonic development	Class 12: Human Reproduction	Single correctEasy
A easy single correct question on embryonic development, from Human Reproduction. NCERT: Class 12: Human Reproduction Monozygotic (identical) twins develop from a single zygote that splits, so they have the same genes and are always of the same sex. A boy and a girl must have come from two different zygotes, each formed by a separate ovum fertilised by a separate sperm. Such twins are dizygotic, or fraternal, twins. Being fraternal twins says nothing about IVF, and like ordinary siblings they share on average half their genes, not 75%. Answer: (2).
178	Chemical Coordination and IntegrationEndocrine glands	Class 11: Chemical Coordination and Integration	MatchingMedium
A medium matching question on endocrine glands, from Chemical Coordination and Integration. NCERT: Class 11: Chemical Coordination and Integration C. Melanocyte stimulating hormone (MSH) is secreted by the pars intermedia of the pituitary (I). D. Catecholamines (adrenaline and noradrenaline) are secreted by the adrenal medulla (III). That leaves options (1) and (3), which differ only in A and B. A. After ovulation the ruptured Graafian follicle becomes the corpus luteum, which secretes mainly progesterone (IV). B. During pregnancy, relaxin is secreted by the ovary in the later phase of pregnancy (II). Answer: (1) A-IV, B-II, C-I, D-III.
179	Biotechnology: Principles and ProcessesRestriction enzymes and vectors	Class 12: Biotechnology: Principles and Processes	StatementsEasy
A easy statements question on restriction enzymes and vectors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes In this method the recombinant DNA is inserted within the coding sequence of the enzyme beta-galactosidase, which inactivates the gene (insertional inactivation). Colonies whose plasmid has no insert still make the enzyme and produce a blue colour with the chromogenic substrate. Colonies whose plasmid carries an insert cannot make the enzyme and stay without colour (white): these are identified as recombinant colonies. Statement I says the blue colonies are recombinant: incorrect. Statement II says the colonies without blue colour carry the insert: correct. Answer: (4).

Sit NEET 2025 free.

Timed like the exam or chapter by chapter, with a solution for each question. Run by a non-profit trust.

Start free