NEET (UG) · 2026

NEET 2026, 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. Electromagnetic Induction and Alternating Currents4
2. Electrostatics4
3. Laws of Motion3
24 easy · 21 medium · 0 hard
Chemistry
45 Q
1. Coordination Compounds6
2. Aldehydes, Ketones and Carboxylic Acids3
3. Organic Chemistry: Some Basic Principles and Techniques3
20 easy · 22 medium · 3 hard
Zoology
42 Q
1. Evolution6
2. Human Reproduction5
3. Locomotion and Movement4
38 easy · 3 medium · 1 hard
Botany
48 Q
1. Cell: The Unit of Life7
2. Biological Classification5
3. Photosynthesis in Higher Plants5
44 easy · 4 medium · 0 hard

Chapters

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

Cell: The Unit of Life7
Class 11: Cell: The Unit of Life
Biological Classification5
Class 11: Biological Classification
Photosynthesis in Higher Plants5
Class 11: Photosynthesis in Higher Plants
Sexual Reproduction in Flowering Plants4
Class 12: Sexual Reproduction in Flowering Plants
Morphology of Flowering Plants3
Class 11: Morphology of Flowering Plants
Plant Growth and Development3
Class 11: Plant Growth and Development
The Living World3
Class 11: The Living World
Principles of Inheritance and Variation3
Class 12: Principles of Inheritance and Variation
Organisms and Populations3
Class 12: Organisms and Populations
Plant Kingdom2
Class 11: Plant Kingdom
Cell Cycle and Cell Division2
Class 11: Cell Cycle and Cell Division
Biodiversity and Conservation2
Class 12: Biodiversity and Conservation
Ecosystem2
Class 12: Ecosystem
Molecular Basis of Inheritance2
Class 12: Molecular Basis of Inheritance
Respiration in Plants1
Class 11: Respiration in Plants
Microbes in Human Welfare1
Class 12: Microbes in Human Welfare

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	Laws of MotionFriction	Class 11: Laws of Motion	Single correctEasy
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion On a horizontal surface the normal force is $Mg$, so the friction force at position $x$ is $f(x) = \mu_k(x)Mg = (\mu_0 - \alpha x)Mg$. The condition $\mu_k(L) = 0$ gives $\mu_0 - \alpha L = 0$, so $\alpha = \frac{\mu_0}{L}$. Magnitude of the work done by friction: $W = \int_0^L (\mu_0 - \alpha x)Mg\,dx = Mg\left(\mu_0 L - \frac{\alpha L^2}{2}\right) = Mg\left(\mu_0 L - \frac{\mu_0 L}{2}\right) = \frac{1}{2}\mu_0 MgL$. Check another way: $\mu_k$ falls linearly from $\mu_0$ to 0, so its average over the path is $\frac{\mu_0}{2}$, and $W = \frac{\mu_0}{2}Mg\cdot L$ again. (The work is negative, since friction opposes the motion; the paper asks for $n$ in its magnitude.) So $n = \frac{1}{2}$. Answer: (4) $\frac{1}{2}$.
2	Behaviour of Perfect Gas and Kinetic TheoryMean free path	Class 11: Kinetic Theory	Single correctEasy
A easy single correct question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Mean free path: $\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}$, so $\lambda \propto \frac{1}{d^2 n}$. $\frac{\lambda_A}{\lambda_B} = \frac{d_B^2\, n_B}{d_A^2\, n_A}$. With $\lambda_A = \frac{1}{2}\lambda_B$ and $d_A = 2d_B$: $\frac{1}{2} = \frac{d_B^2\, n_B}{4d_B^2\, n_A} = \frac{n_B}{4n_A}$, so $4n_A = 2n_B$ and $n_A = \frac{1}{2}n_B$. Check: with $n_A = \frac{1}{2}n_B$, $d_A^2 n_A = 4d_B^2\cdot\frac{1}{2}n_B = 2d_B^2 n_B$, twice that of B, so $\lambda_A$ is half of $\lambda_B$, as given. Answer: (4) $n_A = \frac{1}{2}n_B$.
3	Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctEasy
A easy single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Let each inductor have inductance $L$. Configuration P: the two inductors are in series and $a$, $b$ are the two ends of the upper inductor only. The full current $I(t)$ flows through it, so $E_P = L\frac{dI}{dt}$. Configuration Q: the two identical inductors are in parallel between $a$ and $b$; by symmetry each carries $\frac{I}{2}$. The emf across the pair is $E_Q = L\frac{d}{dt}\left(\frac{I}{2}\right) = \frac{L}{2}\frac{dI}{dt}$. (Equivalently, the parallel pair has $L_{eq} = \frac{L}{2}$ and carries $I$.) $\frac{E_P}{E_Q} = \frac{L}{L/2} = 2$. Answer: (4) 2.
4	Oscillations and WavesStanding waves and beats	Class 11: Oscillations; Class 11: Waves	Single correctMedium
A medium single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Open pipe (antinodes at both ends): the $k$-th harmonic has $k$ half-wavelengths in the pipe, $L = k\frac{\lambda}{2}$, and $k$ displacement nodes between the two end antinodes. For the 5th harmonic, $n = 5$. Closed pipe (node at the closed end, antinode at the open end): only odd harmonics exist, $L = (2p-1)\frac{\lambda}{4}$, and the $(2p-1)$-th harmonic has $p$ nodes (counting the one at the closed end). The 9th harmonic has $2p - 1 = 9$, $p = 5$, so $m = 5$. Check by counting: closed-pipe 1st harmonic has 1 node, 3rd has 2, 5th has 3, 7th has 4, 9th has 5. $\frac{n}{m} = \frac{5}{5} = 1$. Answer: (3) 1.
5	Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctEasy
A easy single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Field inside a long solenoid carrying current $I$: $B = \mu_0 nI$. Flux through each turn: $\mu_0 nI\cdot\pi r^2$. Number of turns: $nl$. Total flux linkage: $N\Phi = nl\cdot\mu_0 nI\pi r^2 = \mu_0\pi n^2 r^2 l\,I$. $L = \frac{N\Phi}{I} = \mu_0\pi n^2 r^2 l$ (that is, $\mu_0 n^2 A l$ with $A = \pi r^2$). Answer: (1) $\mu_0\pi n^2 r^2 l$.
6	KinematicsMotion in a straight line	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy 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 $s(t) = t^2 - 6t + 5$, so $v = \frac{ds}{dt} = 2t - 6$, which is zero at $t = 3$ s: the particle turns back there. $s(0) = 5$ m, $s(3) = 9 - 18 + 5 = -4$ m, $s(6) = 36 - 36 + 5 = 5$ m. Distance travelled: $\|{-4} - 5\| + \|5 - (-4)\| = 9 + 9 = 18$ m. Average speed $= \frac{18}{6} = 3$ ms$^{-1}$. (The displacement is zero, so the average velocity is 0, which is option (4): a trap, since the question asks for speed.) Check: $\int_0^6 \|2t - 6\|\,dt$ is two triangles of base 3 and height 6, area $2\times\frac{1}{2}\times3\times6 = 18$ m. Answer: (3) 3.
7	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Mass defect: $\Delta m = m_U - (m_{Th} + m_{He}) = 238.050 - (234.043 + 4.003) = 238.050 - 238.046 = 0.004$ u. $Q = \Delta m\cdot c^2 = 0.004\times931.5$ MeV $= 3.726$ MeV $= 3726$ keV. Check: $931.5\times4 = 3726$, and $\times10^{-3}$ MeV per u-thousandth gives 3726 keV. Answer: (1) 3726.
8	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The frequency of the light does not change; only its power (intensity) falls. Photocurrent is proportional to the number of photons arriving per second, that is to the intensity. With the power falling linearly in time, $I$ falls linearly in time. The stopping potential depends only on the frequency and the work function: $e\|V\| = h\nu - \phi$. Neither changes, so $\|V\|$ stays constant. The pair "I decreasing linearly, $\|V\|$ constant" is graph (1). Answer: (1).
9	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 From the figure, three branches join the same two vertical wires: P and Q in series, S alone, and the battery. So the series pair P–Q and the capacitor S are each directly across V. Branch P–Q: equivalent $\frac{C}{2}$, charge $q = \frac{C}{2}V$; since P and Q are identical each has $\frac{V}{2}$ across it. $U_P = \frac{1}{2}C\left(\frac{V}{2}\right)^2 = \frac{CV^2}{8}$. Total: $U_T = \frac{1}{2}C_{eq}V^2$ with $C_{eq} = \frac{C}{2} + C = \frac{3C}{2}$, so $U_T = \frac{3CV^2}{4}$. Check by adding: $U_P + U_Q + U_S = \frac{CV^2}{8} + \frac{CV^2}{8} + \frac{CV^2}{2} = \frac{6CV^2}{8} = \frac{3CV^2}{4}$. $\frac{U_P}{U_T} = \frac{1/8}{3/4} = \frac{1}{6}$. Answer: (4) 1/6.
10	Properties of Bulk MatterFluid mechanics	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctMedium
A medium single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Continuity: $A_P v_P = A_Q v_Q$, so $v_Q = \frac{40}{20}v_P = 2v_P$. Bernoulli for a horizontal pipe: $P_P - P_Q = \frac{1}{2}\rho(v_Q^2 - v_P^2) = \frac{1}{2}\rho(4v_P^2 - v_P^2) = \frac{3}{2}\rho v_P^2$. $15 = \frac{3}{2}\times1000\times v_P^2 \Rightarrow v_P^2 = 0.01$, $v_P = 0.1$ m s$^{-1}$ = 10 cm s$^{-1}$. Rate of flow $= A_P v_P = 40\text{ cm}^2\times10\text{ cm s}^{-1} = 400$ cm$^3$s$^{-1}$. Check at Q: $v_Q = 20$ cm s$^{-1}$, $20\times20 = 400$ cm$^3$s$^{-1}$; and $\frac{1}{2}\times1000\times(0.04 - 0.01) = 15$ Pa. Answer: (4) 400.
11	Magnetic Effects of Current and MagnetismBiot-Savart law	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctMedium
A medium single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Current enters at A (top) and leaves at C (bottom), diametrically opposite, so ABC and ADC are the two semicircles, in parallel. The current divides inversely as resistance: $R_{ABC} = \frac{1}{2}R_{ADC}$, so $I_{ABC} = \frac{2}{3}I_0$ and $I_{ADC} = \frac{1}{3}I_0$. A semicircle carrying $I$ gives $\frac{\mu_0 I}{4r}$ at the centre. The two halves carry current from A to C on opposite sides, one clockwise and one anticlockwise about O, so their fields are opposite. The straight leads lie along the diameter through O, so they give no field at O. $B = \frac{\mu_0}{4r}\left(\frac{2}{3}I_0 - \frac{1}{3}I_0\right) = \frac{\mu_0 I_0}{12r}$. Answer: (1) $\frac{\mu_0 I_0}{12r}$.
12	Properties of Bulk MatterViscosity	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctMedium
A medium single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal velocity, weight = buoyancy + viscous drag: $\frac{4}{3}\pi r^3\sigma g = \frac{4}{3}\pi r^3\rho g + 6\pi\eta r v \Rightarrow v = \frac{2r^2 g(\sigma - \rho)}{9\eta} = \frac{2r^2 g\rho}{9\eta}\left(\frac{\sigma}{\rho} - 1\right)$. For fixed $r$ and liquid, $v$ is a straight line in $\frac{\sigma}{\rho}$, zero at $\frac{\sigma}{\rho} = 1$ and rising linearly for $\frac{\sigma}{\rho} > 1$ (the balls that sink). Graph (1) is the only one whose line starts from $v = 0$ at $\frac{\sigma}{\rho} = 1$. Graphs (2) and (3) give a non-zero $v$ at $\frac{\sigma}{\rho} = 1$, and (4) shows no dependence on density. Answer: (1).
13	GravitationEscape velocity	Class 11: Gravitation	Single correctEasy
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Escape speed: $v = \sqrt{\frac{2GM}{R}}$. With equal masses, $v \propto \frac{1}{\sqrt{R}}$. $\frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} = \sqrt{2}$. Answer: (3) $\sqrt{2}$.
14	GravitationKepler's laws	Class 11: Gravitation	Single correctEasy
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation For a circular orbit gravity supplies the centripetal force: $\frac{GMm}{R^2} = \frac{mv^2}{R}$, so $v = \sqrt{\frac{GM}{R}}$. $T = \frac{2\pi R}{v} = 2\pi\sqrt{\frac{R^3}{GM}}$, so $T^2 \propto R^3$ (Kepler's third law) and $T \propto R^{3/2}$. Answer: (2) R$^{3/2}$.
15	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 Parallel currents in the same direction attract, so B (above) pulls A upward. Force per unit length on A: $\frac{F}{l} = \frac{\mu_0 I_A I_B}{2\pi h} = \frac{\mu_0 I(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}$. A stays on the floor while this does not exceed its weight per unit length $\lambda g$: $\frac{\mu_0 I^2}{\pi h} \le \lambda g \Rightarrow h \ge \frac{\mu_0 I^2}{\pi\lambda g}$. The minimum height is $\frac{\mu_0 I^2}{\pi\lambda g}$. Answer: (2).
16	Electronic DevicesSpecial-purpose diodes	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctEasy
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The Zener diode (between C and B) and the resistor R (between B and A) are in series across the 5 V supply, with the diode reverse biased. 5 V exceeds the 3 V breakdown voltage, so the ideal Zener conducts in breakdown and holds exactly 3 V across itself. The rest of the supply appears across the resistor, between B and A: $5 - 3 = 2$ V. Answer: (2) 2 V.
17	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $Q = 0$, so the work done by the gas equals the fall in internal energy. $W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{1\times8.3\times10}{2/3} = 8.3\times15 = 124.5$ J. Check: for a monatomic gas $\Delta U = \frac{3}{2}nR\Delta T = 1.5\times8.3\times10 = 124.5$ J. Answer: (3) 124.5 J.
18	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	Single correctMedium
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Emission happens only if $\lambda$ is shorter than the threshold wavelength $\lambda_0$, and then $eV = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$. Cell 1: $\lambda > \lambda_1$, so no photoelectrons and $V_1 = 0$. Cells 2 and 3: $\lambda < \lambda_2$ and $\lambda < \lambda_3$, so both emit. A larger threshold wavelength means a smaller work function and so a larger stopping potential. Since $\lambda_3 \gg \lambda$ while $\lambda_2$ is merely greater, $\lambda_3 > \lambda_2$ and $V_3 > V_2$. So $V_1 = 0$ and $V_2 < V_3$. Answer: (1).
19	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	Single correctMedium
A medium single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $E = 20$ eV $= 20\times1.6\times10^{-19} = 3.2\times10^{-18}$ J. Photon: $p_{Ph} = \frac{E}{c} = \frac{3.2\times10^{-18}}{3\times10^8} = 1.067\times10^{-26}$ kg m s$^{-1}$. Electron (non-relativistic, since 20 eV is tiny beside 511 keV): $p_e = \sqrt{2mE} = \sqrt{2\times9\times10^{-31}\times3.2\times10^{-18}} = \sqrt{5.76\times10^{-48}} = 2.4\times10^{-24}$ kg m s$^{-1}$. $\frac{p_e}{p_{Ph}} = \frac{2.4\times10^{-24}}{1.067\times10^{-26}} = 225$. Check in closed form: $\frac{p_e}{p_{Ph}} = \frac{\sqrt{2mE}}{E/c} = c\sqrt{\frac{2m}{E}} = 3\times10^8\sqrt{\frac{18\times10^{-31}}{3.2\times10^{-18}}} = 3\times10^8\times\sqrt{5.625\times10^{-13}} = 3\times10^8\times7.5\times10^{-7} = 225$. Answer: (3) 225.
20	Experimental SkillsOptics experiments	No matching NCERT chapter	Single correctEasy
A easy single correct question on optics experiments, from Experimental Skills. NCERT: No matching NCERT chapter Index correction (bench error) belongs to the optical bench. The distance between, say, the object needle and the lens read from the bench scale (the difference of the uprights' positions) need not equal the actual distance between them, because the index marks on the uprights are not exactly in line with the needle tip or the lens centre. An index rod of known length is placed between them to find this error, and every distance read off the scale is corrected by it. A metre bridge needs end corrections, a resonance tube an end correction for the open end, and a simple pendulum none of these. Answer: (3).
21	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The electrostatic field is conservative, so the work done in moving the charge slowly depends only on the end points A and B, not on the path through the tube inside the sphere. Charge of the sphere: $Q = \frac{4}{3}\pi R^3\rho$. Outside it, $V(r) = \frac{Q}{4\pi\epsilon_0 r} = \frac{\rho R^3}{3\epsilon_0 r}$. $V_A = V(2R) = \frac{\rho R^2}{6\epsilon_0}$, $V_B = V(3R) = \frac{\rho R^2}{9\epsilon_0}$. Work by the external agent on a unit charge: $W = q(V_B - V_A) = \frac{\rho R^2}{\epsilon_0}\left(\frac{1}{9} - \frac{1}{6}\right) = -\frac{\rho R^2}{18\epsilon_0}$; the electric force does $+\frac{\rho R^2}{18\epsilon_0}$. The magnitude is $\frac{\rho R^2}{18\epsilon_0}$. Check: $\frac{1}{6} - \frac{1}{9} = \frac{3 - 2}{18} = \frac{1}{18}$. So $n = 18$. Answer: (4) 18.
22	OpticsReflection and refraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics O is at the centre of the 5 cm layer, 2.5 cm from each face. Near-normal viewing across a plane surface: apparent distance $= \text{real distance}\times\frac{n_{\text{observer}}}{n_{\text{object}}}$. From P ($n = 1$): $h_1 = 2.5\times\frac{1}{1.25} = 2$ cm. From R ($n = 1.5$): $h_2 = 2.5\times\frac{1.5}{1.25} = 3$ cm. $\|h_1 - h_2\| = 1$ cm. Answer: (2) 1.
23	Laws of MotionConservation of linear momentum	Class 11: Laws of Motion	Single correctMedium
A medium single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Take the circle's centre as origin, $\theta$ measured from the horizontal diameter, A at the top ($\theta = \frac{\pi}{2}$) and B at the bottom. By symmetry the two particles are always at the same $\theta$, one on each side: positions $(\cos\theta, \sin\theta)$ and $(-\cos\theta, \sin\theta)$, with $\theta$ falling from $\frac{\pi}{2}$ to $-\frac{\pi}{2}$ at rate $\omega$. Velocities (unit radius, unit mass): $\omega(\sin\theta, -\cos\theta)$ and $\omega(-\sin\theta, -\cos\theta)$. The horizontal parts cancel; the vertical parts add: $\vec{P} = (0, -2\omega\cos\theta)$, so $\|P\| = 2\omega\cos\theta$. $\|P\|$ is 0 at A ($\theta = \frac{\pi}{2}$) where the velocities are opposite, rises to its maximum $2\omega$ at $\theta = 0$ where both move straight down together, and falls back to 0 at B. That single smooth hump is graph (3). Graph (1) is constant, (2) has its zero in the middle, and (4) falls steadily. Answer: (3).
24	Properties of Bulk MatterThermal expansion and calorimetry	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 thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Volume expansion coefficient $\gamma = 3\alpha$, so $\Delta V = 3\alpha V\Delta T = 3\alpha\cdot\frac{4}{3}\pi R^3\Delta T = 4\pi R^3\alpha\Delta T$. Check: $\Delta V = \frac{dV}{dR}\Delta R = 4\pi R^2\cdot R\alpha\Delta T$, the same. Answer: (3) $4\pi R^3\alpha\Delta T$.
25	Oscillations and WavesSHM	Class 11: Oscillations; Class 11: Waves	Single correctEasy
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Pushed down a further $x$, the cork of cross-section $A$ feels an extra upthrust $\rho A g x$, a restoring force with $k = \rho Ag$. $T = 2\pi\sqrt{\frac{m}{\rho Ag}}$, so $T \propto \frac{1}{\sqrt{\rho}}$ for the same cork. $\frac{T_2}{T_1} = \sqrt{\frac{\rho_1}{\rho_2}} = 2 \Rightarrow \frac{\rho_2}{\rho_1} = \frac{1}{4}$. Answer: (4) 1/4.
26	Experimental SkillsVernier calipers and screw gauge	No matching NCERT chapter	Single correctMedium
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{1\text{ mm}}{10} = 0.1$ mm $= 0.01$ cm. With the jaws closed the vernier zero lies to the left of the main-scale zero: a negative zero error. One vernier division is $0.9$ mm. If the vernier zero sits $x$ to the left and the 4th vernier division meets a main-scale mark, $-x + 4\times0.9 = 3$ mm, so $x = 0.6$ mm: the zero error is $-0.06$ cm. Actual length = observed reading $-$ zero error $= 1.00 - (-0.06) = 1.06$ cm, which is not an option. (4) 1.04 cm follows only from taking the error as $-4\times0.01$ cm, which is not how a negative zero error is read. No option is correct as printed.
27	Motion of System of Particles and Rigid BodyMoment of inertia	Class 11: Systems of Particles and Rotational Motion	Single correctEasy
A easy 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 The spheres touch at one point, so their centres are $d = R + r$ apart. About the axis through A's centre: $I_A = \frac{2}{5}MR^2 + \left(\frac{2}{5}mr^2 + md^2\right)$ (parallel axes for B). About the axis through B's centre: $I_B = \frac{2}{5}mr^2 + \left(\frac{2}{5}MR^2 + Md^2\right)$. $I_A - I_B = (m - M)d^2 = (m - M)(R + r)^2$, negative since $m < M$. Answer: (2) $(m - M)(R + r)^2$.
28	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 Energy is conserved: $\frac{1}{2}kx^2 + \frac{1}{2}mv^2 = E$, that is $\frac{x^2}{2E/k} + \frac{v^2}{2E/m} = 1$, an ellipse with semi-axes $\sqrt{2E/k}$ along $x$ and $\sqrt{2E/m}$ along $v$. It is a circle only when the two semi-axes are equal: $k = m$ (in the stated units). Equivalently, $x = A\sin\omega t$, $v = A\omega\cos\omega t$ trace a circle only if $\omega = \sqrt{k/m} = 1$. Answer: (2) $k = m$.
29	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctMedium
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics From the figure the object O is 30 cm to the left of the convex lens $L_1$, and the concave lens $L_2$ is 3 cm to the right of $L_1$. $L_1$: $u = -30$ cm, $f = +10$ cm. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}$, so $v = +15$ cm, to the right of $L_1$. This image lies $15 - 3 = 12$ cm to the right of $L_2$: a virtual object for $L_2$, $u = +12$ cm, $f = -10$ cm. $\frac{1}{v} = -\frac{1}{10} + \frac{1}{12} = \frac{-6 + 5}{60} = -\frac{1}{60}$, so $v = -60$ cm. The final image is 60 cm to the left of the concave lens. Check: magnification $m = \frac{15}{-30}\times\frac{-60}{12} = (-0.5)(-5) = 2.5$; the image at 57 cm left of $L_1$ is a magnified virtual image, consistent with the converging light being spread by the diverging lens. Answer: (2).
30	Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctEasy
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\omega = 2\times10^3$ rad s$^{-1}$. $X_L = \omega L = 2000\times0.01 = 20\ \Omega$; $X_C = \frac{1}{\omega C} = \frac{1}{2000\times25\times10^{-6}} = \frac{1}{0.05} = 20\ \Omega$. $X_L = X_C$: the circuit is at resonance, so $Z = R = 100\ \Omega$. Check: $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.01\times25\times10^{-6}}} = \frac{1}{\sqrt{2.5\times10^{-7}}} = \frac{1}{5\times10^{-4}} = 2000$ rad s$^{-1}$. $I_0 = \frac{V_0}{Z} = \frac{220}{100} = 2.2$ A. Answer: (1) 2.2 A.
31	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For any point $X$: $\vec{L}_X = \vec{L}_{cm} + M\,\vec{r}_{cm/X}\times\vec{v}_{cm}$. The disc's centre of mass is the fixed centre O, so $\vec{v}_{cm} = 0$ and the second term vanishes for every point. The angular momentum is the same, $I\omega$ along the axis, about A, about B, or about any point. (Directly: each element $dm$ at $\vec{r}$ from O has $\vec{v}$; about $X$ its angular momentum is $(\vec{r} - \vec{r}_X)\times\vec{v}\,dm$, and $\sum\vec{r}_X\times\vec{v}\,dm = \vec{r}_X\times\vec{p} = 0$ since the total momentum is zero.) $\frac{L_A}{L_B} = 1$. Answer: (3) 1.
32	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Flux: $\Phi = BA_0(1 + \sin t)$. Induced emf: $\varepsilon = -\frac{d\Phi}{dt} = -BA_0\cos t$. Power dissipated in resistance $R$: $P = \frac{\varepsilon^2}{R} = \frac{B^2A_0^2}{R}\cos^2 t$. At $t = 0$ the power is at its maximum; it falls to zero at $t = \frac{\pi}{2}$ and rises again to the maximum at $t = \pi$, repeating, never negative. Graph (2) starts at its maximum, falls to zero and rises again. Graph (1) starts at zero (a $\sin^2 t$ shape), (3) is constant, and (4) grows steadily. Answer: (2).
33	ElectrostaticsConductors and dielectrics	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance A lies inside the cavity, close to Q. There the field is that of Q together with the induced charge $-Q$ on the cavity wall; it is not zero (it is zero only in the conducting material itself). So $E_A \neq 0$. The induced $+Q$ sits on the outer surface of the isolated sphere. The conductor shields the outside from the details of the cavity, so this charge spreads uniformly over the outer spherical surface, and outside the sphere the field is $\frac{Q}{4\pi\epsilon_0 r^2}$, radial from the sphere's centre. B and C are at the same distance from the centre, so $E_B = E_C$, even though Q is off-centre. Answer: (2).
34	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Energy conservation, with $k = \frac{1}{4\pi\epsilon_0}$; the charge is attracted inward and gains kinetic energy equal to the fall in potential energy: $\frac{1}{2}mv^2 = kQq\left(\frac{1}{R} - \frac{1}{3R}\right) = \frac{2kQq}{3R}$. $v^2 = \frac{4kQq}{3mR} = \frac{4Qq}{3\cdot4\pi\epsilon_0 mR} = \frac{Qq}{3\pi\epsilon_0 mR}$. $v = \sqrt{\frac{Qq}{3\pi\epsilon_0 mR}}$. Answer: (3).
35	Atoms and NucleiRutherford model and alpha scattering	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on rutherford model and alpha scattering, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Rutherford's scattering formula gives $N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$: very many particles at small angles and a steep fall as $\theta$ grows, with only a very few scattered back near 180°. Plot (3) has $N(\theta)$ very large at small angles, dropping steeply and levelling off at a small value for large angles. Plot (1) rises with angle, (2) is flat at first, and (4) peaks near 90°; none matches. Answer: (3).
36	Current ElectricityResistivity	Class 12: Current Electricity	Single correctMedium
A medium single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity After a rise $\Delta T$ the resistances are $R_0(1 - \alpha\Delta T)$ and $R_0(1 + \alpha\Delta T)$. Circuit (A), in series: $R_A = R_0(1 - \alpha\Delta T) + R_0(1 + \alpha\Delta T) = 2R_0$, unchanged, so $I_A$ stays constant. Circuit (B), in parallel: $R_B = \frac{R_0^2(1 - \alpha\Delta T)(1 + \alpha\Delta T)}{2R_0} = \frac{R_0}{2}(1 - \alpha^2\Delta T^2)$, which decreases as $\Delta T$ grows, so $I_B$ increases. Answer: (1).
37	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	Single correctMedium
A medium single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\sigma_s$: from $E = \sigma T^4$ (power per area), $[\sigma] = \frac{ML^2T^{-3}}{L^2K^4} = MT^{-3}K^{-4}$. $k_B$: energy per kelvin, $[k_B] = ML^2T^{-2}K^{-1}$. $b$: from $\lambda_m T = b$, $[b] = LK$. $[\sigma k_B^{-1} b] = MT^{-3}K^{-4}\cdot M^{-1}L^{-2}T^2K\cdot LK = L^{-1}T^{-1}K^{-2}$. Check the powers: M: $1 - 1 = 0$; L: $-2 + 1 = -1$; T: $-3 + 2 = -1$; K: $-4 + 1 + 1 = -2$. Answer: (1) $[L^{-1}T^{-1}K^{-2}]$.
38	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 (1) $v = \frac{c}{\sqrt{\epsilon_r}} = \frac{3\times10^8}{3} = 10^8$ m s$^{-1}$ (non-magnetic medium): correct. (2) $\omega = 2\pi\times10^6$ rad s$^{-1}$, so $f = 10^6$ Hz and $\lambda = \frac{v}{f} = \frac{10^8}{10^6} = 100$ m, not 300 m (300 m is the wavelength in vacuum): incorrect. (3) The magnetic field is in phase with $E$, of amplitude $\frac{E_0}{v}$, and along $y$ so that $\vec{E}\times\vec{B}$ points along $+z$ ($\hat{x}\times\hat{y} = \hat{z}$): this is the intended correct statement. (4) The phase $kz - \omega t$ is constant for $z$ increasing with $t$, so the wave travels along $+z$: correct. The incorrect choice is (2). Answer: (2).
39	ThermodynamicsFirst law	Class 11: Thermodynamics	Single correctMedium
A medium single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics From the figure: a (2 m$^3$, 300 N m$^{-2}$) → b (5, 300) → c (5, 100) → d (2, 100) → a, traversed clockwise on the P–V diagram. Over a complete cycle $\Delta U = 0$, so the net heat supplied equals the net work done by the gas, which is the area enclosed (positive for a clockwise cycle): $W = (5 - 2)\times(300 - 100) = 3\times200 = 600$ J. Net heat supplied $= 600$ J. Check by legs: $W_{ab} = 300\times3 = 900$ J, $W_{cd} = 100\times(-3) = -300$ J, the isochoric legs do none: total 600 J. Answer: (3) 600 J.
40	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 The Coulomb force supplies the centripetal force: $\frac{mv^2}{r} = \frac{ke^2}{r^2}$, so $r = \frac{ke^2}{mv^2}$. $v^2 = 25.6\times10^{10}$ m$^2$s$^{-2}$. $r = \frac{9\times10^9\times(1.6\times10^{-19})^2}{9\times10^{-31}\times25.6\times10^{10}} = \frac{9\times10^9\times2.56\times10^{-38}}{9\times10^{-31}\times2.56\times10^{11}} = 10^{9 - 38 + 31 - 11} = 10^{-9}$ m. Check with Bohr's model: $v = \frac{2.19\times10^6}{n}$ m s$^{-1}$ and $\sqrt{25.6}\times10^5 = 5.06\times10^5$, so $n \approx 4.3$, near 4; $r = 0.053n^2$ nm $\approx 1$ nm for $n \approx 4.3$, consistent with $x = 1$. Answer: (4) 1.
41	Laws of MotionCircular motion dynamics	Class 11: Laws of Motion	Single correctEasy
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion Wear is least when no friction is needed: the normal force alone supplies the centripetal force. Then $\tan\theta = \frac{v^2}{rg} = \frac{100}{50\times10} = \frac{1}{5}$. $\theta = \tan^{-1}\left(\frac{1}{5}\right)$. Answer: (1).
42	Electronic Devicesp-n junction diode	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctMedium
A medium single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits From the figure, the 5 V battery (positive plate on the left) connects across the closed loop formed by the branches of $D_1$ and $D_3$; the branch of $D_2$ is broken (open) at the gap drawn after $D_2$, so no voltage is applied to $D_2$. $D_1$ points from the positive side to the negative side: forward biased, so its depletion region is narrower than at equilibrium. $D_3$ points the other way: reverse biased, so its depletion region is wider than at equilibrium. $D_2$ is unbiased: its depletion region keeps its equilibrium width, between the two. $W_3 > W_2 > W_1$. Answer: (3).
43	Electromagnetic WavesEM spectrum	Class 12: Electromagnetic Waves	MatchingEasy
A easy matching question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Microwaves heat food in a microwave oven: P-II. UV rays kill germs and are used to purify water: Q-I. Gamma rays destroy cancer cells in radiotherapy: R-IV. Radio waves carry AM and FM broadcasts: S-III. P-II, Q-I, R-IV, S-III. Answer: (3).
44	Work, Energy and PowerCollisions	Class 11: Work, Energy and Power	Single correctEasy
A easy single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power In a head-on elastic collision between equal masses, the velocities are exchanged: A stops and B moves off at 10 m s$^{-1}$. B then swings up on the string. Energy: $\frac{1}{2}mv^2 = mgh \Rightarrow h = \frac{v^2}{2g} = \frac{100}{20} = 5$ m. Since $h = 5$ m is less than the string's length (10 m), the bob stays below the level of the support and the string remains taut, so all the kinetic energy becomes potential energy at the top of the swing. Answer: (3) 5.
45	Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom	Class 11: Kinetic Theory	Single correctMedium
A medium single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Each translational and rotational mode contributes $\frac{1}{2}R$ to $C_V$; each vibrational mode contributes $R$ (kinetic and potential energy). $C_V = \frac{3}{2}R + \frac{3}{2}R + fR = (3 + f)R$, $C_P = C_V + R = (4 + f)R$. $\gamma = \frac{4 + f}{3 + f} = \frac{8}{7} \Rightarrow 28 + 7f = 24 + 8f \Rightarrow f = 4$. Check: $\frac{4 + 4}{3 + 4} = \frac{8}{7}$. (Counting each vibrational mode as only $\frac{1}{2}R$ would give $f = 8$, not an option.) Answer: (1) 4.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
46	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Step i: Friedel–Crafts acylation. Benzene + CH$_3$COCl / anhydrous AlCl$_3$ gives acetophenone, C$_6$H$_5$COCH$_3$. Step ii: NaOCl with a methyl ketone is the haloform reaction: C$_6$H$_5$COCH$_3$ + 3NaOCl → C$_6$H$_5$COCCl$_3$ → C$_6$H$_5$COONa + CHCl$_3$. So P = sodium benzoate and Q = chloroform. (1) Q is chloroform, not a primary alcohol: wrong. (2) Chloroform is not aromatic: wrong. (3) Sodium benzoate gives benzoic acid on acidification, and chloroform is slowly oxidised by air in light to phosgene, COCl$_2$, a poisonous gas (which is why it is stored in dark bottles filled to the top): correct. (4) Neither is a carbonyl compound in the sense of an aldehyde or ketone: wrong. Answer: (3).
47	Coordination CompoundsIsomerism in complexes	Class 12: Coordination Compounds	StatementsMedium
A medium statements question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Statement-I: $[\text{Fe(ox)}_3]^{3-}$ is an octahedral tris-chelate, M(AA)$_3$. Its three rings form a propeller with no mirror plane and no centre of symmetry, so it exists as a pair of enantiomers ($\Delta$ and $\Lambda$): chiral. Correct. Statement-II: in $trans$-$[\text{Cr(H}_2\text{O)}_2\text{(ox)}_2]^-$ the two oxalates lie in one plane and the two water molecules are above and below it. That plane is a mirror plane, so the trans isomer is achiral (only the cis isomer is chiral). Incorrect. Answer: (3).
48	Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctMedium
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The cation is C$_6$H$_5$–C$^+$H–CH$_3$ (1-phenylethyl cation): the positive carbon carries a benzene ring and a methyl group. An empty orbital is stabilised only by donation of electrons into it, so it must interact with filled orbitals; an empty orbital mixing with another empty orbital gives no stabilisation. Benzene ring: the empty $p$ orbital overlaps with the ring's filled $\pi$ orbitals (resonance), spreading the positive charge over the ortho and para positions. Methyl group: its C–H bonds, filled $\sigma$ orbitals, overlap with the empty $p$ orbital (hyperconjugation). So the stabilisation comes from filled $\sigma$ and filled $\pi$ orbitals. Answer: (1).
49	p-Block ElementsGroup 13 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Potash alum is K$_2$SO$_4\cdot$Al$_2$(SO$_4$)$_3\cdot$24H$_2$O, or KAl(SO$_4$)$_2\cdot$12H$_2$O. Per formula KAl(SO$_4$)$_2\cdot$12H$_2$O there is 1 K$^+$ and 2 SO$_4^{2-}$ (charge check: +1 + 3 = +4 = 2 × 2). K$^+$ : SO$_4^{2-}$ = 1 : 2 (the double formula gives 2 : 4, the same). Answer: (1) 1:2.
50	BiomoleculesProteins	Class 12: Biomolecules	Single correctEasy
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules (1) The arrangement of two or more polypeptide subunits is the quaternary structure, not the tertiary: wrong. (2) Many proteins made of a single chain (with no quaternary structure) are fully active, for example many enzymes: wrong. (3) In the $\beta$-pleated sheet the peptide chains lie side by side, nearly fully extended, and are held together by hydrogen bonds between different chains (intermolecular): correct. (4) The $\alpha$-helix is a right-handed screw, held by hydrogen bonds within the same chain between the –NH of one residue and the C=O of the fourth residue along: wrong because it says left-handed. Answer: (3).
51	Some Basic Concepts of ChemistryMole concept	Class 11: Some Basic Concepts of Chemistry	Single correctMedium
A medium single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry The rule for rounding off when the digit to be dropped is exactly 5: the preceding digit is left unchanged if it is even and increased by one if it is odd. 17.0145: the digit kept last is 4 (even), so it stays: 17.014. 21.0235: the digit kept last is 3 (odd), so it becomes 4: 21.024. Answer: (3) 17.014 and 21.024.
52	EquilibriumSolubility product	Class 11: Equilibrium	Single correctMedium
A medium single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium The salts have different formulas, so compare the molar solubilities $s$, not $K_{sp}$ directly. AgBr ⇌ Ag$^+$ + Br$^-$: $K_{sp} = s^2$, $s = \sqrt{5.0\times10^{-13}} = 7.1\times10^{-7}$ M. Zn(OH)$_2$ ⇌ Zn$^{2+}$ + 2OH$^-$: $K_{sp} = 4s^3$, $s^3 = 2.5\times10^{-16}$, $s = 6.3\times10^{-6}$ M. Hg$_2$Cl$_2$ ⇌ Hg$_2^{2+}$ + 2Cl$^-$: $K_{sp} = 4s^3$, $s^3 = 3.25\times10^{-19}$, $s = 6.9\times10^{-7}$ M. Check the close pair: $(7.1\times10^{-7})^3 = 3.6\times10^{-19} > 3.25\times10^{-19}$, so AgBr is indeed slightly more soluble than Hg$_2$Cl$_2$. Order: Zn(OH)$_2$ > AgBr > Hg$_2$Cl$_2$. Answer: (4).
53	Classification of Elements and PeriodicityElectron gain enthalpy	Class 11: Classification of Elements and Periodicity in Properties	Single correctEasy
A easy single correct question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Electron gain enthalpies of the halogens are negative; their magnitudes are about F −333, Cl −349, Br −325, I −296 kJ mol$^{-1}$. Chlorine has the most negative value. Fluorine's is less negative than chlorine's because its small 2p subshell is crowded and the added electron is strongly repelled. Below chlorine the value becomes less negative as size grows. Order of magnitude: Cl > F > Br > I. Answer: (3).
54	SolutionsRaoult's law	Class 12: Solutions	Assertion and reasonEasy
A easy assertion and reason question on raoult's law, from Solutions. NCERT: Class 12: Solutions Assertion: an ideal solution obeys Raoult's law at all compositions, and for it $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$. Correct. Reason: in an ideal solution P and Q certainly do interact; the point is that the P–Q interactions are nearly the same as the P–P and Q–Q interactions, so mixing changes neither enthalpy nor volume. "No interactions" is wrong. Answer: (3).
55	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 Sodium nitroprusside is the test for sulphur in the sodium fusion extract: sulphur becomes Na$_2$S, which with nitroprusside gives a coloured complex, Na$_4$[Fe(CN)$_5$NOS]. Of the four amino acids only methionine, CH$_3$SCH$_2$CH$_2$CH(NH$_2$)COOH, contains sulphur. Leucine, threonine and serine contain only C, H, N and O. Answer: (3) methionine.
56	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	Single correctMedium
A medium single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Electrode potentials do not add, but $\Delta G° = -nFE°$ does. Fe$^{3+}$ + 3e$^-$ → Fe: $\Delta G_1° = -3F(-0.04) = +0.12F$. Fe$^{2+}$ + 2e$^-$ → Fe: $\Delta G_2° = -2F(-0.44) = +0.88F$. Fe$^{3+}$ + e$^-$ → Fe$^{2+}$ is the first minus the second: $\Delta G° = 0.12F - 0.88F = -0.76F$. $E° = -\frac{\Delta G°}{1\cdot F} = +0.76$ V. Check: $3(-0.04) - 2(-0.44) = -0.12 + 0.88 = 0.76$. Answer: (2) +0.76 V.
57	Principles Related to Practical ChemistryTitrimetric analysis	No matching NCERT chapter	Single correctEasy
A easy single correct question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Balanced reaction: 2MnO$_4^-$ + 5C$_2$O$_4^{2-}$ + 16H$^+$ → 2Mn$^{2+}$ + 10CO$_2$ + 8H$_2$O. Oxalic acid taken: $10\times0.25 = 2.5$ mmol. Permanganate needed: $\frac{2}{5}\times2.5 = 1.0$ mmol. Molarity of KMnO$_4$ $= \frac{1.0\text{ mmol}}{10\text{ mL}} = 0.10$ M. Check with equivalents: oxalic acid (n-factor 2) gives $2.5\times2 = 5$ meq; KMnO$_4$ (n-factor 5) needs $M\times5\times10 = 5$, so $M = 0.10$ M. Answer: (1) 0.10 M.
58	Coordination CompoundsCrystal field theory	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Spectrochemical series (increasing field strength): I$^-$ < Br$^-$ < SCN$^-$ < Cl$^-$ < S$^{2-}$ < F$^-$ < OH$^-$ < C$_2$O$_4^{2-}$ < H$_2$O < NCS$^-$ < edta$^{4-}$ < NH$_3$ < en < CN$^-$ < CO. Decreasing: CO > NH$_3$ > H$_2$O > Cl$^-$. Answer: (1).
59	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Single correctMedium
A medium single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics In a free expansion of an ideal gas, $w = 0$, $q = 0$ and $\Delta U = 0$, so the temperature stays 300 K. Entropy is a state function, so the system's change is found along a reversible isothermal path between the same states: $\Delta S_{system} = nR\ln\frac{V_2}{V_1} = 2R\ln10 = 2\times2.303R = 4.606R$. No heat passes to or from the surroundings, so $\Delta S_{surroundings} = 0$. (The total, $+4.606R$, is positive, as it must be for an irreversible process; option (2) would describe the reversible isothermal expansion.) Answer: (4).
60	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	Single correctMedium
A medium single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For the reaction 2A → B, the rate of reaction is $r = -\frac{1}{2}\frac{d[A]}{dt}$, and for zero order $r = k$. So A is used up at $-\frac{d[A]}{dt} = 2k = 2.0$ mol L$^{-1}$ min$^{-1}$, a constant. 75% of 2 M is 1.5 M of A to be consumed: $t = \frac{1.5}{2.0} = 0.75$ min. Check by the integrated law: $[A] = [A]_0 - 2kt$; with $[A] = 0.5$ M, $0.5 = 2 - 2t$, $t = 0.75$ min. Answer: (2) 0.75 min.
61	d- and f-Block ElementsTransition element properties	Class 12: The d- and f-Block Elements	Assertion and reasonEasy
A easy assertion and reason question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Assertion: the 3d metals (except zinc, and to a degree manganese) have high melting points. Correct. Reason: their strong metallic bonding comes from the large number of electrons, both (n−1)d and ns, that take part in the delocalised bonding; the melting points rise to a maximum around d$^5$, where the number of unpaired d electrons is greatest. Correct, and it is exactly why the melting points are high. Answer: (1).
62	ElectrochemistryConductance	Class 12: Electrochemistry	Single correctMedium
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Kohlrausch's relation for a strong electrolyte: $\Lambda_m = \Lambda_m^0 - A\sqrt{c}$, the slope being $-A = -90.0$. At $c = 0.01$ M, $\sqrt{c} = 0.1$: $145.0 = \Lambda_m^0 - 90.0\times0.1 \Rightarrow \Lambda_m^0 = 154.0$ S cm$^2$ mol$^{-1}$. Kohlrausch's law of independent migration: $\Lambda_m^0 = \lambda^0_{X^+} + \lambda^0_{Y^-}$, so $\lambda^0_{Y^-} = 154.0 - 74.0 = 80.0$ S cm$^2$ mol$^{-1}$. Answer: (1) 80.0.
63	Some Basic Concepts of ChemistryStoichiometry	Class 11: Some Basic Concepts of Chemistry	Single correctEasy
A easy single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry 2C$_4$H$_{10}$ + 13O$_2$ → 8CO$_2$ + 10H$_2$O. Molar mass of butane $= 4\times12 + 10 = 58$ g mol$^{-1}$, so 116 g is 2 mol. 2 mol of butane give 8 mol of CO$_2$: $8\times44 = 352$ g. Check by carbon: 116 g of butane holds $116\times\frac{48}{58} = 96$ g of C = 8 mol C, which becomes 8 mol CO$_2$ = 352 g. Answer: (1) 352 g.
64	Chemical KineticsArrhenius equation	Class 12: Chemical Kinetics	Single correctMedium
A medium single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Arrhenius: $\ln k = \ln A - \frac{E_a}{RT}$. The plot of $\ln k$ against $\frac{1}{T}$ is a straight line whose intercept on the $\ln k$ axis is $\ln A$; from the figure the line meets the axis at 6, so $\ln A = 6$. For $k = e^2$: $2 = 6 - \frac{E_a}{RT} \Rightarrow \frac{E_a}{RT} = 4$. $T = \frac{E_a}{4R} = \frac{6640}{4\times8.3} = \frac{6640}{33.2} = 200$ K. Check: $\frac{6640}{8.3\times200} = \frac{6640}{1660} = 4$. Answer: (3) 200 K.
65	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	StatementsMedium
A medium statements question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Both are the preparation of a halogen: the halide is oxidised and MnO$_2$ (Mn +4) is reduced to Mn$^{2+}$. 4NaCl + MnO$_2$ + 4H$_2$SO$_4$ → MnCl$_2$ + 4NaHSO$_4$ + 2H$_2$O + Cl$_2$. 2NaI + MnO$_2$ + 3H$_2$SO$_4$ → 2NaHSO$_4$ + MnSO$_4$ + 2H$_2$O + I$_2$. In both, Mn goes from +4 to +2: it is reduced. Statement-I says Mn is oxidised: incorrect. Statement-II says Mn is reduced: correct. Answer: (4).
66	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 $[\text{Mn(CN)}_6]^{3-}$: Mn$^{3+}$, d$^4$; CN$^-$ is strong-field, low spin $t_{2g}^4$: 2 unpaired, $\mu = \sqrt{8} = 2.83$ BM. $[\text{Fe(CN)}_6]^{3-}$: Fe$^{3+}$, d$^5$, low spin $t_{2g}^5$: 1 unpaired, 1.73 BM. $[\text{Co(NH}_3)_6]^{3+}$: Co$^{3+}$, d$^6$; NH$_3$ pairs the electrons with Co$^{3+}$, $t_{2g}^6$: 0 unpaired, 0 BM. $[\text{Ti(H}_2\text{O)}_6]^{3+}$: Ti$^{3+}$, d$^1$: 1 unpaired, 1.73 BM. The highest is $[\text{Mn(CN)}_6]^{3-}$. Answer: (1).
67	d- and f-Block ElementsLanthanoids and actinoids	Class 12: The d- and f-Block Elements	Single correctMedium
A medium single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Ln$^{3+}$ ions are [Xe]4f$^n$ with $n = Z - 57$. Ce$^{3+}$: 4f$^1$, 1 unpaired. Nd$^{3+}$: 4f$^3$, 3 unpaired. Tb$^{3+}$: 4f$^8$, 7 orbitals with one pair, 6 unpaired. Ho$^{3+}$: 4f$^{10}$: seven electrons singly, three more paired, $7 - 3 = 4$ unpaired. Answer: (4) Ho$^{3+}$.
68	Coordination CompoundsNomenclature	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Inside the coordination sphere: four NH$_3$ (tetraammine), one H$_2$O (aqua) and one Cl$^-$ (chlorido) around Co(III). Charge of the complex ion: $+3 + 4(0) + 0 + (-1) = +2$, so two chloride ions balance it outside the brackets. Formula: $[\text{Co(NH}_3)_4(\text{H}_2\text{O})\text{Cl}]\text{Cl}_2$. Answer: (4).
69	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	Single correctHard
A hard single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics The cycle is a Carnot cycle: 1 isothermal expansion at $T_1$ ($V_1 \to V_2$), 2 adiabatic ($T_1 \to T_2$), 3 isothermal compression at $T_2$ ($V_3 \to V_4$), 4 adiabatic back ($T_2 \to T_1$). In chemistry $w$ is the work done on the gas, so for a reversible isothermal step $w = -nRT\ln\frac{V_f}{V_i}$. (1) $w_1 = -2T_1\ln\frac{V_2}{V_1}$ and $w_3 = -2T_2\ln\frac{V_4}{V_3}$, so $w_1 + w_3 = -2T_1\ln\frac{V_2}{V_1} - 2T_2\ln\frac{V_4}{V_3}$: correct. (2) For the adiabatic steps $q = 0$, so $w_2 = \Delta U_2 = C_V(T_2 - T_1)$ and $w_4 = \Delta U_4 = C_V(T_1 - T_2)$. Then $w_2 + w_4 = \Delta U_2 + \Delta U_4 = 0$, whereas $\Delta U_2 - \Delta U_4 = 2C_V(T_2 - T_1) \neq 0$: wrong. (3) $w_1 + w_2 = -2T_1\ln\frac{V_2}{V_1} + C_V(T_2 - T_1)$, not $+2T_1\ln\frac{V_2}{V_1}$: wrong. (4) Over the cycle $\Delta U = 0$, so the total work equals minus the net heat, $-(q_1 + q_3) \neq 0$; the enclosed area is not zero: wrong. Answer: (1).
70	Classification of Elements and PeriodicityIonization enthalpy	Class 11: Classification of Elements and Periodicity in Properties	Assertion and reasonMedium
A medium assertion and reason question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Assertion: first ionization enthalpies are N 1402, O 1314, F 1681 kJ mol$^{-1}$, so O is lower than both. Correct. Reason: O is 2p$^4$, with one doubly occupied 2p orbital. Removing an electron gives O$^+$, 2p$^3$, a stable half-filled set; the electron removed is the paired one, pushed out by the repulsion within that orbital. Correct. It explains the assertion: it is why O breaks the general rise across the period (O lower than N). F, beyond O, has the larger nuclear charge and returns to the usual trend, so O lies below both. Answer: (1).
71	SolutionsRaoult's law	Class 12: Solutions	StatementsEasy
A easy statements question on raoult's law, from Solutions. NCERT: Class 12: Solutions A: an ideal solution is defined as one that obeys Raoult's law over the whole range of composition. True. B: chloroform's C–H hydrogen bonds to the oxygen of acetone; the new attractions are stronger than those in the pure liquids, the vapour pressure is lower than Raoult's law predicts: negative deviation. True. C: phenol's –OH hydrogen bonds to aniline's nitrogen, again stronger attractions in the mixture: negative deviation, not positive. False. A and B only. Answer: (1).
72	HydrocarbonsAlkanes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Cyclohexylmagnesium bromide is a Grignard reagent: its carbon is strongly basic. Cyclohexylamine has acidic N–H hydrogens (any compound with H on O, N or S, or a terminal alkyne, destroys a Grignard reagent). C$_6$H$_{11}$MgBr + C$_6$H$_{11}$NH$_2$ → C$_6$H$_{12}$ + C$_6$H$_{11}$NH–MgBr The Grignard carbon simply takes the proton, giving cyclohexane; no C–N or C–C bond forms. Structure (4), cyclohexane, is the product. Answer: (4).
73	p-Block ElementsGroup 13 elements	No matching NCERT chapter	Single correctMedium
A medium single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter (1) Boron has only the 2s and three 2p orbitals in its valence shell, no d orbitals, so it can form at most four bonds (as in BF$_4^-$): its maximum covalency is four. Correct. (2) Beryllium (2s, 2p) has four valence orbitals, one 2s and three 2p: wrong. (3) Magnesium is in the third period; with its 3d orbitals available its maximum covalency is six, not four (as for aluminium, which forms [AlF$_6$]$^{3-}$): wrong. (4) Aluminium's valence shell has 3s, three 3p and five 3d orbitals, nine in all, not five: wrong. Answer: (1).
74	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Single correctMedium
A medium single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Equal concentrations at equilibrium: $K = \frac{[D]}{[N]} = 1$, so $\Delta G° = -RT\ln K = 0$. $\Delta G° = \Delta H° - T\Delta S° = 0 \Rightarrow \Delta S° = \frac{\Delta H°}{T} = \frac{666\text{ kJ mol}^{-1}}{333\text{ K}} = 2.0$ kJ K$^{-1}$mol$^{-1}$. (60 °C = 333 K; using 60 K by mistake gives 11.1, option (4).) Answer: (1) 2.0.
75	Chemical Bonding and Molecular StructureVSEPR theory	Class 11: Chemical Bonding and Molecular Structure	MatchingEasy
A easy matching question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A. PCl$_5$: 5 bond pairs, no lone pair (sp$^3$d): trigonal bipyramidal, III. B. BrF$_5$: 5 bond pairs and 1 lone pair (sp$^3$d$^2$): square pyramidal, IV. C. BF$_4^-$: 4 bond pairs (sp$^3$): tetrahedral, I. D. [Ni(CN)$_4$]$^{2-}$: Ni$^{2+}$ d$^8$ with strong-field CN$^-$, dsp$^2$: square planar, II. A-III, B-IV, C-I, D-II. Answer: (2).
76	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	StatementsHard
A hard statements question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Bromine adds anti (through a bromonium ion); cold alkaline KMnO$_4$ adds the two OH groups syn (through a cyclic manganate ester). trans-But-2-ene + anti addition → meso-2,3-dibromobutane. cis-But-2-ene + syn addition → meso-butane-2,3-diol. Reading the Newman projections (front carbon's bonds meet at the centre: up, lower left, lower right; back carbon's bonds: upper left, upper right, down): Statement I: front C has Br (up), Me (lower right), H (lower left); back C has Br (down), Me (upper right), H (upper left). The Br atoms are anti but the two Me groups are gauche. In the meso compound, with Br anti, the Me groups and H atoms must also be anti (the molecule then has a centre of symmetry). Here they are not, so this is one enantiomer of the chiral (±) dibromide, the product of cis-but-2-ene. Statement I is incorrect. Statement II: front C has OH (up), Me (lower right), H (lower left); back C has OH (upper right), Me (down), H (upper left). Going clockwise both carbons read OH → Me → H, so turning the back carbon by 60° eclipses OH with OH, Me with Me and H with H: the molecule has a mirror plane. It is meso-butane-2,3-diol, the syn product of cis-but-2-ene. Statement II is correct. Answer: (4).
77	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Ph–C≡C–Me (1-phenylpropyne): H$_2$, Pd/C (Lindlar's catalyst) adds H$_2$ syn: K = cis-(Z)-1-phenylprop-1-ene. Na / liquid NH$_3$ (Birch-type reduction) adds H anti: L = trans-(E)-1-phenylprop-1-ene. So K and L are cis/trans, that is geometrical, isomers. (1) correct; they are not mirror images, so (2) is wrong. K + HBr (ionic, Markovnikov): H adds to C2, giving the benzylic cation, so M = Ph–CHBr–CH$_2$CH$_3$ (1-bromo-1-phenylpropane). L + HBr / benzoyl peroxide (radical, anti-Markovnikov): Br$^\cdot$ adds to C1 to give the stable benzylic radical, so N = Ph–CH$_2$–CHBr–CH$_3$ (2-bromo-1-phenylpropane). M and N differ in the position of Br: they are position (structural) isomers, neither geometrical isomers nor stereoisomers. (3) and (4) are wrong. Answer: (1).
78	Coordination CompoundsIsomerism in complexes	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Facial and meridional isomers belong to octahedral complexes of the type [MA$_3$B$_3$]: the three A ligands either occupy one face of the octahedron (fac) or lie in a plane through the metal (mer). $[\text{Cr(py)}_3\text{Cl}_3]$ is MA$_3$B$_3$: it has fac and mer isomers. $[\text{Cr(H}_2\text{O)}_6]^{3+}$ (MA$_6$) has no geometrical isomers; $[\text{Co(NH}_3)_4(\text{H}_2\text{O})_2]^{3+}$ (MA$_4$B$_2$) and $[\text{Ni(en)}_2(\text{H}_2\text{O})_2]^{2+}$ (M(AA)$_2$B$_2$) have cis and trans isomers. Answer: (1).
79	AminesClassification and identification of amines	Class 12: Amines	Single correctMedium
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines A. Benzonitrile, C$_6$H$_5$CN, with LiAlH$_4$ is reduced to benzylamine, C$_6$H$_5$CH$_2$NH$_2$, not aniline: no. B. Benzamide with KOH and Br$_2$ is the Hofmann bromamide degradation: C$_6$H$_5$CONH$_2$ → C$_6$H$_5$NH$_2$, the amine with one carbon fewer: aniline, yes. C. NaBH$_4$ does not reduce an aromatic nitro group; nitrobenzene is reduced to aniline by Sn/HCl, Fe/HCl or H$_2$/Pd: no. D. Acetanilide, C$_6$H$_5$NHCOCH$_3$, heated with HCl and water is hydrolysed to aniline (as its hydrochloride, freed by base) and acetic acid: yes. B and D only. Answer: (2).
80	BiomoleculesNucleic acids and vitamins	Class 12: Biomolecules	MatchingEasy
A easy matching question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Vitamin A: carrots (β-carotene), fish liver oil, butter — IV. Vitamin B$_{12}$: found in animal foods — meat, fish, eggs, curd — I. Vitamin E: vegetable oils such as sunflower and wheat-germ oil — II. Vitamin K: green leafy vegetables — III. A-IV, B-I, C-II, D-III. Answer: (2).
81	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 (1) P in P$_4$O$_{10}$ +5, S in SO$_3$ +6, O in H$_2$O −2: +5 < +6, not decreasing. (2) N in N$_2$O$_5$ +5, Al in Al$_2$O$_3$ +3, S in H$_2$S −2: +5 > +3 > −2, decreasing. Correct. (3) Pb in PbO$_2$ +4, N in N$_2$O$_3$ +3, S in SO$_3$ +6: not decreasing. (4) P in P$_4$O$_6$ +3, Cl in Cl$_2$O$_7$ +7, Al in AlH$_3$ +3: not decreasing. Answer: (2).
82	Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctHard
A hard single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids 2,2-Dimethylcyclopentanone has α-hydrogens only at C5 (C2 carries two methyls). With NaOH it forms the enolate at C5, which can attack either benzaldehyde or a second molecule of the ketone; heating then dehydrates the β-hydroxy ketone by E1cB, which always removes the α-hydrogen and gives the double bond conjugated with C=O. Cross condensation with PhCHO: 5-benzylidene-2,2-dimethylcyclopentanone, as its E isomer (2) or Z isomer (4). Both can be obtained (E being the major). Self condensation: the enolate adds to the C=O of a second ketone; dehydration towards the remaining α-H of the first ring gives the exocyclic, conjugated enone (3). Structure (1) has the new double bond inside the second ring, away from the C=O (a β,γ-unsaturated ketone joined through an sp$^3$ CH). Dehydration in an aldol condensation does not place the double bond there, so (1) is the compound that cannot be obtained. Answer: (1).
83	Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Double bonds are conjugated when they alternate with a single bond: C=C–C=C. Hepta-1,3-diene: CH$_2$=CH–CH=CH–CH$_2$CH$_2$CH$_3$, double bonds at C1–C2 and C3–C4, separated by one single bond: conjugated. In the 1,4-, 1,5- and 1,6-dienes the double bonds are separated by one or more sp$^3$ carbons: isolated. Answer: (1).
84	AminesBasicity of amines	Class 12: Amines	StatementsMedium
A medium statements question on basicity of amines, from Amines. NCERT: Class 12: Amines Statement-I: KMnO$_4$ oxidises the CH$_3$ to COOH, giving $p$-nitrobenzoic acid. The –NO$_2$ group withdraws electrons (–I, –R) and stabilises the carboxylate ion, so $p$-nitrobenzoic acid (p$K_a$ 3.4) is stronger than benzoic acid (4.2). Correct. Statement-II: Sn/HCl reduces –NO$_2$ to –NH$_2$, giving $p$-toluidine after neutralisation. The CH$_3$ group donates electrons (+I and hyperconjugation), raising the electron density on N, so $p$-toluidine (p$K_b$ 8.9) is more basic than aniline (9.4). Correct. Answer: (1).
85	d- and f-Block ElementsPotassium dichromate and permanganate	Class 12: The d- and f-Block Elements	Single correctEasy
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements On heating to 513 K, potassium permanganate decomposes: 2KMnO$_4$ → K$_2$MnO$_4$ + MnO$_2$ + O$_2$. K$_2$MnO$_4$ (potassium manganate) is green; manganese is +6, d$^1$, with one unpaired electron, so it is paramagnetic. (KO$_2$ is paramagnetic but yellow and is not formed here; MnO and Mn$_3$O$_4$ are not green products of this reaction.) Answer: (1) K$_2$MnO$_4$.
86	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Toluene with chromyl chloride in CS$_2$, then hydrolysis, is the Étard reaction: the methyl group is oxidised only as far as the aldehyde. P = benzaldehyde, C$_6$H$_5$CHO. (1) Effervescence with NaHCO$_3$ is given by carboxylic acids; benzaldehyde gives none: wrong. (2) Benzene with CH$_3$COCl / AlCl$_3$ gives acetophenone, not benzaldehyde: wrong. (3) A white precipitate with bromine water is the test for phenol (or aniline): wrong. (4) Benzoyl chloride with H$_2$ over Pd on BaSO$_4$ (Rosenmund reduction) gives benzaldehyde: correct. Answer: (4).
87	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 A 1:3 electrolyte gives one complex ion and three counter-ions. $[\text{Co(NH}_3)_6]\text{Cl}_3$ → $[\text{Co(NH}_3)_6]^{3+}$ + 3Cl$^-$: 1:3. $[\text{CoCl}_2(\text{NH}_3)_4]\text{Cl}$ is 1:1, $[\text{CoCl(NH}_3)_5]\text{Cl}_2$ is 1:2, and $[\text{Co(NH}_3)_3(\text{NO}_2)_3]$ is a non-electrolyte. Answer: (3).
88	Structure of AtomAtomic orbitals	Class 11: Structure of Atom	Single correctEasy
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom A radial node is a distance $r > 0$ at which $\psi_r$ passes through zero and changes sign ($r = 0$ itself does not count). A: never crosses zero, no radial node (like 1s). B: crosses zero once, one radial node (like 2s). C: starts positive, crosses to negative, then crosses back to positive before dying away: two crossings, two radial nodes (like 3s). D: starts at zero at $r = 0$ (a p-type function), rises, then crosses zero once: one radial node (like 3p). Answer: (3) C.
89	Chemical Bonding and Molecular StructureDipole moment and Fajans' rule	Class 11: Chemical Bonding and Molecular Structure	Single correctMedium
A medium single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Polarity here is the usual organic order of functional groups (as in solubility in water or elution in chromatography), set by how polar the group is and how strongly it hydrogen bonds: Ether (A): a weakly polar C–O–C, no O–H; least polar. Ketone (C): the strongly polar C=O, but no O–H. Alcohol (B): O–H, a hydrogen-bond donor and acceptor. Carboxylic acid (D): C=O and O–H together, hydrogen bonding strongly (dimers); most polar. A < C < B < D. Answer: (4).
90	Chemical Bonding and Molecular StructureMolecular orbital theory	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Ne$_2$ would have 20 electrons, 16 of them in the valence shell. For O$_2$ onwards the order is $\sigma_{2s} < \sigma^_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^_{2p_x} = \pi^_{2p_y} < \sigma^_{2p_z}$. Sixteen valence electrons fill every one of these: $\sigma_{2s}^2\sigma_{2s}^{2}\sigma_{2p_z}^2\pi_{2p}^4\pi_{2p}^{4}\sigma_{2p_z}^{2}$. The last orbital filled, the HOMO, is $\sigma^_{2p}$. (Bond order $\frac{10 - 10}{2} = 0$, which is why Ne$_2$ does not exist as a stable molecule.) Answer: (4) $\sigma^*_{2p}$.

Zoology questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
91	Locomotion and MovementSkeletal system	Class 11: Locomotion and Movement	Single correctEasy
A easy single correct question on skeletal system, from Locomotion and Movement. NCERT: Class 11: Locomotion and Movement The human vertebral column is formed by 26 serially arranged vertebrae: 7 cervical, 12 thoracic, 5 lumbar, 1 sacrum (five fused) and 1 coccyx (four fused). 7 is the number of cervical vertebrae, 12 the number of thoracic vertebrae (and of rib pairs), and 206 the number of bones in the whole adult skeleton. Answer: 26.
95	Neural Control and CoordinationCentral nervous system	Class 11: Neural Control and Coordination	Single correctEasy
A easy single correct question on central nervous system, from Neural Control and Coordination. NCERT: Class 11: Neural Control and Coordination The brain (and spinal cord) are covered by the cranial meninges: the outer dura mater, the thin middle arachnoid and the inner pia mater, which is in contact with the brain tissue. The pericardium is the double-walled membranous bag around the heart, not a part of the central neural system. Answer: Pericardium.
99	Chemical Coordination and IntegrationEndocrine glands	Class 11: Chemical Coordination and Integration	StatementsEasy
A easy statements question on endocrine glands, from Chemical Coordination and Integration. NCERT: Class 11: Chemical Coordination and Integration (a) True: the pituitary is divided anatomically into an adenohypophysis and a neurohypophysis. (b) True: the pars distalis of the adenohypophysis secretes FSH (with GH, PRL, TSH, ACTH and LH). (c) True: the pars intermedia secretes melanocyte stimulating hormone (MSH). (d) False: prolactin is secreted by the pars distalis. So (a), (b) and (c) only.
103	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 water, the skin of the frog acts as the aquatic respiratory organ (cutaneous respiration): dissolved oxygen diffuses across the moist, highly vascular skin. On land, the buccal cavity, skin and lungs serve; lungs are used only on land. Frogs have no trachea of the insect kind. Answer: skin.
104	Locomotion and MovementSkeletal system	Class 11: Locomotion and Movement	Single correctEasy
A easy single correct question on skeletal system, from Locomotion and Movement. NCERT: Class 11: Locomotion and Movement Going down the leg: the femur (thigh bone) runs from the hip to the knee; the patella (knee cap) covers the knee joint ventrally; below the knee are the tibia and fibula; then the ankle bones, the tarsals, followed by metatarsals and phalanges. So femur → patella → tibia → tarsal.
106	BiomoleculesProteins and amino acids	Class 11: Biomolecules	MatchingEasy
A easy matching question on proteins and amino acids, from Biomolecules. NCERT: Class 11: Biomolecules A. Starch is the storage polysaccharide of plants, an energy store (II). B. Antibodies are proteins that fight infectious agents (I). C. Concanavalin A is a lectin, a secondary metabolite of plants (IV). D. GLUT-4 is a membrane protein that enables glucose transport into cells (III). So A-II, B-I, C-IV, D-III.
107	Animal KingdomBasis of classification	Class 11: Animal Kingdom	StatementsEasy
A easy statements question on basis of classification, from Animal Kingdom. NCERT: Class 11: Animal Kingdom Statement I is the definition of radial symmetry: any plane through the central axis divides the body into two identical halves (coelenterates, ctenophores, adult echinoderms). It is correct. Statement II is incorrect: adult echinoderms are radially symmetrical, but their larvae are bilaterally symmetrical. So Statement I is correct but Statement II is incorrect.
115	Body Fluids and CirculationCardiac cycle	Class 11: Body Fluids and Circulation	Single correctEasy
A easy single correct question on cardiac cycle, from Body Fluids and Circulation. NCERT: Class 11: Body Fluids and Circulation The SA node, the pacemaker of the heart, generates the maximum number of action potentials, about 70–75 per minute, and so initiates and maintains the rhythmic contraction of the heart: the normal heart rate is 72 beats per minute. Answer: 70 - 75.
117	Body Fluids and CirculationBlood composition	Class 11: Body Fluids and Circulation	Single correctEasy
A easy single correct question on blood composition, from Body Fluids and Circulation. NCERT: Class 11: Body Fluids and Circulation Blood coagulates in response to an injury or trauma, to prevent excessive loss of blood (1 is correct). The clot is a network of threads called fibrins, in which formed elements are trapped (2 is correct). Fibrins are formed from the inactive plasma protein fibrinogen by the enzyme thrombin (3 is correct). Statement (4) reverses this: fibrinogen is not produced from fibrin, so it is the incorrect statement.
124	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 Each Islet of Langerhans has α-cells, which secrete glucagon, and β-cells, which secrete insulin. Glucagon acts on liver cells and stimulates glycogenolysis, raising blood sugar. So the incorrect statement is that α-cells secrete insulin.
130	Animal KingdomChordate classes	Class 11: Animal Kingdom	StatementsEasy
A easy statements question on chordate classes, from Animal Kingdom. NCERT: Class 11: Animal Kingdom Statement I is correct: the class name Reptilia refers to their creeping or crawling mode of locomotion (Latin repere, to creep or crawl). Statement II is incorrect: the reptilian heart is usually three-chambered, but it is four-chambered in crocodiles, so not all reptiles have a three-chambered heart. So Statement I is correct but Statement II is incorrect.
132	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 The fundamental features of chordates are a notochord, a dorsal hollow single nerve cord, paired pharyngeal gill slits and a post-anal part (tail). Having pharyngeal gill slits is a chordate character, so "absence of gills" is not a characteristic of chordates (it is the non-chordates that lack gill slits). Answer: Absence of gills.
134	BiomoleculesCarbohydrates, lipids and nucleic acids	Class 11: Biomolecules	Single correctMedium
A medium single correct question on carbohydrates, lipids and nucleic acids, from Biomolecules. NCERT: Class 11: Biomolecules By weight, the human body is about 65% oxygen, about 18.5% carbon, about 9.5–10% hydrogen and about 3.3% nitrogen. Oxygen is heaviest because most of the body is water, and each water molecule is 16/18 oxygen by mass; hydrogen atoms are the most numerous but light. So oxygen > carbon > hydrogen > nitrogen: (a), (b), (c), (d). Options (2), (3) and (4) put hydrogen or carbon above oxygen, which cannot be right.
135	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 a peripheral nervous system of cranial and spinal nerves: ten pairs of cranial nerves arise from its brain. (Humans and other amniotes have twelve pairs.) Answer: 10.
137	Biotechnology and its ApplicationsBt crops	Class 12: Biotechnology and its Applications	Single correctEasy
A easy single correct question on bt crops, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications Bacillus thuringiensis makes the Bt toxin as an inactive protoxin (crystal). When an insect eats it, the alkaline pH of the insect gut solubilises the crystals and converts the protoxin into the active toxin, which binds to the midgut epithelial cells, creates pores, makes the cells swell and lyse, and kills the insect. Answer: due to alkaline pH.
139	Reproductive HealthSTDs	Class 12: Reproductive Health	Single correctEasy
A easy single correct question on stds, from Reproductive Health. NCERT: Class 12: Reproductive Health Syphilis, gonorrhoea and genital warts (human papilloma virus) are sexually transmitted infections. Tuberculosis, caused by Mycobacterium tuberculosis, spreads through droplets in the air from an infected person and is not an STI. Answer: Tuberculosis.
140	Human ReproductionGametogenesis	Class 12: Human Reproduction	Single correctEasy
A easy single correct question on gametogenesis, from Human Reproduction. NCERT: Class 12: Human Reproduction A sperm has a head, neck, middle piece and tail. The tail is a flagellum, and its beating (powered by the mitochondria of the middle piece) moves the sperm: this is flagellar movement. Ciliary movement moves the ovum along the fallopian tube; amoeboid movement is seen in macrophages and leucocytes; muscular movement is of limbs and organs. Answer: flagellar movement.
141	EvolutionHardy-Weinberg principle	Class 12: Evolution	StatementsEasy
A easy statements question on hardy-weinberg principle, from Evolution. NCERT: Class 12: Evolution Natural selection can lead to stabilisation (more individuals acquire the mean character value), directional change (more individuals acquire a value other than the mean) or disruption (more individuals acquire values at both ends of the distribution). Genetic drift is a separate cause of change in allele frequency, by chance, especially in small populations; it is not a result of natural selection. So (a), (c) and (d) only.
142	Reproductive HealthART	Class 12: Reproductive Health	Single correctEasy
A easy single correct question on art, from Reproductive Health. NCERT: Class 12: Reproductive Health Intra-cytoplasmic sperm injection (ICSI) is a specialised procedure in which a sperm is directly injected into the ovum to form an embryo in the laboratory. GIFT transfers an ovum from a donor into the fallopian tube of a female who cannot produce ova; ZIFT transfers a zygote or early embryo (up to 8 blastomeres) into the fallopian tube; ET is the general transfer of an embryo. Answer: ICSI.
143	Human ReproductionMale and female reproductive systems	Class 12: Human Reproduction	Single correctEasy
A easy single correct question on male and female reproductive systems, from Human Reproduction. NCERT: Class 12: Human Reproduction The male accessory ducts are the rete testis, vasa efferentia, epididymis and vas deferens. The infundibulum is the funnel-shaped part of the fallopian tube (oviduct) closest to the ovary, with finger-like fimbriae; it belongs to the female reproductive system. Answer: Infundibulum.
145	Human ReproductionMenstrual cycle	Class 12: Human Reproduction	StatementsEasy
A easy statements question on menstrual cycle, from Human Reproduction. NCERT: Class 12: Human Reproduction Statement I is correct: the rapid secretion of LH reaching a peak in the middle of the cycle (the LH surge, about day 14) induces rupture of the Graafian follicle and so the release of the ovum (ovulation). Statement II is incorrect: the remaining parts of the Graafian follicle do transform into the corpus luteum, but the corpus luteum secretes large amounts of progesterone, which maintains the endometrium; it is not chiefly an estrogen source. So Statement I is correct but Statement II is incorrect.
147	EvolutionHardy-Weinberg principle	Class 12: Evolution	Single correctEasy
A easy single correct question on hardy-weinberg principle, from Evolution. NCERT: Class 12: Evolution At Hardy–Weinberg equilibrium, with allele frequencies p (A) and q (a), the genotype frequencies are $p^2 + 2pq + q^2 = 1$. The frequency of AA is $p^2 = (0.1)^2 = 0.01$. (0.10 is p itself; the frequency of aa would be $0.9^2 = 0.81$.) Answer: 0.01.
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. Excess growth hormone in adults (especially middle age) causes acromegaly (III). B. LH induces ovulation of fully mature follicles (IV). C. Vasopressin (ADH) acts on the kidney and stimulates reabsorption of water and electrolytes by the distal tubules (I). D. Oxytocin stimulates vigorous contraction of the uterus at child birth (II). So A-III, B-IV, C-I, D-II.
149	Body Fluids and CirculationCardiac cycle	Class 11: Body Fluids and Circulation	Single correctEasy
A easy single correct question on cardiac cycle, from Body Fluids and Circulation. NCERT: Class 11: Body Fluids and Circulation The opening between the right atrium and the right ventricle is guarded by a valve of three muscular flaps or cusps, the tricuspid valve. The bicuspid (mitral) valve guards the left atrio-ventricular opening; semilunar valves guard the openings of the pulmonary artery and aorta; the SA node is the pacemaker tissue, not a valve. Answer: tricuspid valve.
150	Breathing and Exchange of GasesRespiratory organs	Class 11: Breathing and Exchange of Gases	Single correctEasy
A easy single correct question on respiratory organs, from Breathing and Exchange of Gases. NCERT: Class 11: Breathing and Exchange of Gases Lower invertebrates like sponges, coelenterates and flatworms exchange $\mathrm{O_2}$ with $\mathrm{CO_2}$ by simple diffusion over their entire body surface. Earthworms use their moist cuticle, insects a network of tracheal tubes, and most aquatic arthropods and molluscs use gills. Answer: simple diffusion over their entire body surfaces.
152	Locomotion and MovementMuscle contraction	Class 11: Locomotion and Movement	Single correctEasy
A easy single correct question on muscle contraction, from Locomotion and Movement. NCERT: Class 11: Locomotion and Movement Muscle contraction is initiated by a signal sent by the CNS through a motor neuron. When the signal reaches the neuromuscular junction, a neurotransmitter, acetylcholine, is released, which generates an action potential in the sarcolemma; this spreads through the fibre and releases calcium ions into the sarcoplasm. Acetyl coenzyme A is a metabolic intermediate, and cGMP and cAMP are second messengers inside cells. Answer: acetyl choline.
157	Excretory Products and their EliminationUrine formation	Class 11: Excretory Products and their Elimination	StatementsHard
A hard statements question on urine formation, from Excretory Products and their Elimination. NCERT: Class 11: Excretory Products and their Elimination (a) Correct: the descending limb of the loop of Henle is permeable to water but almost impermeable to electrolytes, so water leaves it. (b) Correct as intended: losing water, the filtrate becomes concentrated as it moves down the descending limb; the loop, with the vasa recta, builds the medullary gradient that lets urine be concentrated. (d) Correct: the ascending limb is impermeable to water but allows transport of electrolytes actively or passively, so the filtrate is diluted as it moves up. (c) is the statement left out: reabsorption in Henle's loop is minimal, and water and Na⁺ are handled in separate limbs (water in the descending, electrolytes in the ascending), so it is not the textbook description of the loop. So (a), (b) and (d) only.
158	Locomotion and MovementSkeletal system	Class 11: Locomotion and Movement	Single correctEasy
A easy single correct question on skeletal system, from Locomotion and Movement. NCERT: Class 11: Locomotion and Movement From the skull downwards, the vertebral column has 7 cervical vertebrae (neck), 12 thoracic (chest, carrying the ribs), 5 lumbar (lower back), then the sacrum (fused) and coccyx. So cervical → thoracic → lumbar → sacrum. Option (2) is the right order read upwards, from toe to head.
159	EvolutionEvidences of evolution	Class 12: Evolution	Single correctMedium
A medium single correct question on evidences of evolution, from Evolution. NCERT: Class 12: Evolution Fossils (paleontological evidence) show that life forms changed over time. Homologous organs such as the forelimbs of whales, bats, cheetahs and humans show divergent evolution from a common ancestor; analogous organs such as the wings of birds and butterflies show convergent evolution. Embryological support for evolution was proposed by Ernst Haeckel, from features common to all vertebrate embryos but absent in adults (gill slits, for example); but this proposal was disapproved on careful study by Karl Ernst von Baer, who noted that embryos never pass through the adult stages of other animals. So Haeckel’s embryological argument is the one that is not accepted as evidence.
160	EvolutionHuman evolution	Class 12: Evolution	StatementsEasy
A easy statements question on human evolution, from Evolution. NCERT: Class 12: Evolution Statement I is incorrect: Homo sapiens (modern humans) arose in Africa and moved across continents, developing into distinct races. Statement II is correct as the textbook states it: during the ice age between 75,000 and 10,000 years ago, modern Homo sapiens arose. So Statement I is incorrect but Statement II is correct.
162	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 has three steps: denaturation (heat separates the two DNA strands), annealing (the two primers bind to their complementary sequences on the single strands as the mixture cools) and extension (thermostable DNA polymerase extends the primers). Ligation is not a PCR step. Answer: annealing.
164	EvolutionEvidences of evolution	Class 12: Evolution	Single correctEasy
A easy single correct question on evidences of evolution, from Evolution. NCERT: Class 12: Evolution When more than one adaptive radiation occurs in isolated geographical areas, the placental mammals and the Australian marsupials each radiated into similar forms (placental wolf and Tasmanian wolf, mole and marsupial mole, flying squirrel and flying phalanger). Unrelated (distant) lineages coming to resemble each other is convergent evolution. Divergent evolution is the radiation within one lineage; founder effect and genetic drift are chance changes in allele frequency. Answer: convergent evolution.
165	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, where immature lymphocytes differentiate into antigen-sensitive lymphocytes, are the bone marrow and thymus. After maturation the lymphocytes move to the secondary lymphoid organs: spleen, lymph nodes, tonsils, Peyer’s patches of the small intestine, appendix and MALT. So tonsils and spleen, (b) and (c) only.
166	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 The placenta acts as an endocrine tissue and produces human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens and progestogens; relaxin is secreted by the ovary in later pregnancy. LH is secreted by the anterior pituitary, not the placenta. Answer: LH.
168	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 Plasmids are autonomously replicating circular extra-chromosomal DNA found in bacteria: they replicate independently of the chromosomal DNA (Statement I correct) and lie outside the chromosome (Statement II correct). This is why they are used as cloning vectors. So both statements are correct.
170	EvolutionEvidences of evolution	Class 12: Evolution	Assertion and reasonEasy
A easy assertion and reason question on evidences of evolution, from Evolution. NCERT: Class 12: Evolution Homologous organs have the same basic anatomical structure (and origin) but may perform different functions. The forelimbs of whales, bats, cheetahs and humans share the same bones — humerus, radius, ulna, carpals, metacarpals and phalanges — though they are used for swimming, flying, running and grasping. So A is correct, R is correct, and the similar anatomical structure is exactly why they are called homologous: R explains A.
171	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 The yellowish fluid colostrum, secreted by the mother during the initial days of lactation, has abundant IgA antibodies, which protect the infant (natural passive immunity). IgG crosses the placenta to the foetus during pregnancy; IgM is the first antibody of a primary response; IgD is a B-cell receptor. Answer: IgA.
173	Human ReproductionGametogenesis	Class 12: Human Reproduction	Single correctEasy
A easy single correct question on gametogenesis, from Human Reproduction. NCERT: Class 12: Human Reproduction The secondary oocyte released at ovulation is surrounded by a new membrane, the zona pellucida (with the cells of the corona radiata outside it). The endometrium is the inner lining of the uterus, and the chorion is an extra-embryonic membrane of the embryo, formed after implantation. Answer: zona pellucida.
176	Biotechnology: Principles and ProcessesrDNA technology	Class 12: Biotechnology: Principles and Processes	Assertion and reasonEasy
A easy assertion and reason question on rdna technology, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes To cut DNA with restriction enzymes it must first be isolated in pure form, and since DNA is enclosed within membranes, the cell must be broken open (R is correct). This is done by treating bacterial cells with lysozyme, plant cells with cellulase and fungal cells with chitinase (A is correct). The enzymes are used precisely because isolating the genetic material needs disruption of cells, so R explains A.
177	Human Health and DiseaseCancer and drug abuse	Class 12: Human Health and Disease	Single correctEasy
A easy single correct question on cancer and drug abuse, from Human Health and Disease. NCERT: Class 12: Human Health and Disease Morphine, an opioid obtained from the latex of the poppy plant Papaver somniferum, is a very effective sedative and painkiller and is very useful in patients who have undergone surgery. Interferons are proteins that help cells resist viral infection; antibiotics kill or inhibit bacteria; anti-retroviral drugs are used against HIV. Answer: Morphine.
179	Human Health and DiseasePathogens and diseases	Class 12: Human Health and Disease	Single correctMedium
A medium single correct question on pathogens and diseases, from Human Health and Disease. NCERT: Class 12: Human Health and Disease In humans Plasmodium reproduces only asexually: first in the liver cells, then in the red blood cells. Some parasites in the RBCs differentiate into sexual stages, the gametocytes, so gametocytes develop in human blood, not in the mosquito (3 is wrong). When a female Anopheles bites and takes up the gametocytes, fertilisation and further development take place in the mosquito’s intestine (gut), forming sporozoites that reach its salivary glands. So the correct statement is that fertilisation takes place in the mosquito gut.
180	Biotechnology: Principles and ProcessesRestriction enzymes and vectors	Class 12: Biotechnology: Principles and Processes	MatchingEasy
A easy matching question on restriction enzymes and vectors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes A. Transformation is the procedure by which a piece of DNA is introduced (transferred) into a host bacterium (II). B. A cloning site is a recognition site for a restriction enzyme, where the vector is cut to insert the alien DNA (I). C. Selection of transformants uses selectable markers, commonly genes for resistance to antibiotics such as ampicillin or tetracycline (IV). D. Ori is the origin of replication, the sequence from which replication starts (III). So A-II, B-I, C-IV, D-III.

Botany questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
92	Biological ClassificationViruses and lichens	Class 11: Biological Classification	Single correctEasy
A easy single correct question on viruses and lichens, from Biological Classification. NCERT: Class 11: Biological Classification Lichens are symbiotic associations of algae (the phycobiont, which makes food) and fungi (the mycobiont, which gives shelter and absorbs water and minerals). Mycorrhiza is an association of a fungus with the roots of a higher plant; sponges are animals (phylum Porifera); chrysophytes are diatoms and desmids, protists. Answer: lichens.
93	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 Matthias Schleiden (1838, plants) and Theodor Schwann (1839, animals) together formulated the cell theory: all living organisms are made of cells and their products. (Rudolf Virchow later added that new cells arise from pre-existing cells.) Robert Brown discovered the nucleus, Singer and Nicolson proposed the fluid mosaic model of the membrane, and Leeuwenhoek first saw a live cell. Answer: Schleiden and Schwann.
94	Cell: The Unit of LifeProkaryotic and eukaryotic cells	Class 11: Cell: The Unit of Life	StatementsEasy
A easy statements question on prokaryotic and eukaryotic cells, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life (a) True: prokaryotic ribosomes are 70S, made of a 50S and a 30S subunit. (b) True: many bacteria carry plasmids, small circular DNA outside the genomic DNA (for example, giving antibiotic resistance). (c) True: mesosomes, infoldings of the plasma membrane, are characteristic of prokaryotes. (d) False: peroxisomes are membrane-bound organelles; prokaryotes have no membrane-bound organelles. So (a), (b) and (c) only.
96	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	Single correctEasy
A easy single correct question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life A mitochondrion has two membranes. The outer membrane is its boundary; the inner membrane, folded into cristae, divides the lumen into two compartments: the outer compartment (between the membranes) and the inner compartment, filled with a dense, homogeneous substance called the matrix. Cytosol lies outside the organelle; mucus and aqueous humor are not cell contents at all. Answer: matrix.
97	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. Cristae are the infoldings of the inner mitochondrial membrane (II). B. The Golgi apparatus is a stack of flat, disc-shaped sacs called cisternae (IV). C. Thylakoids are the flattened membranous sacs in the stroma of the chloroplast, stacked into grana (I). D. Phospholipids, arranged as a bilayer, are the main lipids of the cell membrane (III). So A-II, B-IV, C-I, D-III.
98	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	Single correctEasy
A easy single correct question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life Chromoplasts contain fat-soluble carotenoid pigments such as carotene and xanthophylls, which give parts of the plant yellow, orange or red colours. Chloroplasts mainly hold chlorophyll (their carotenoids are accessory pigments, not stored); aleuroplasts store proteins and amyloplasts store starch, both leucoplasts without pigment. Answer: chromoplast.
100	Photosynthesis in Higher PlantsPhotorespiration	Class 11: Photosynthesis in Higher Plants	Single correctEasy
A easy single correct question on photorespiration, from Photosynthesis in Higher Plants. NCERT: Class 11: Photosynthesis in Higher Plants When RuBisCO acts as an oxygenase, RuBP (5 C) combines with $\mathrm{O_2}$ to give one molecule of 3-phosphoglycerate (3 C) and one of phosphoglycolate (2 C). Phosphoenolpyruvate is the $\mathrm{CO_2}$ acceptor of C4 plants, and oxaloacetate and malate are the C4 acids, none of them made by RuBisCO. Answer: 2-Phosphoglycolate.
101	Biological ClassificationViruses and lichens	Class 11: Biological Classification	Single correctEasy
A easy single correct question on viruses and lichens, from Biological Classification. NCERT: Class 11: Biological Classification Prions are infectious agents made of abnormally folded protein. They cause bovine spongiform encephalopathy (BSE), commonly called mad cow disease, and its human analogue, Creutzfeldt–Jakob disease. Viroids are free RNA that infect plants (potato spindle tuber); Aspergillus is a fungus and Mycoplasma a wall-less bacterium. Answer: prions.
102	Photosynthesis in Higher PlantsLight reaction	Class 11: Photosynthesis in Higher Plants	Single correctEasy
A easy single correct question on light reaction, from Photosynthesis in Higher Plants. NCERT: Class 11: Photosynthesis in Higher Plants In each photosystem, all pigments except one molecule of chlorophyll a form the light harvesting complex. That single chlorophyll a molecule is the reaction centre: in PS I it absorbs maximally at 700 nm (P700), and in PS II at 680 nm (P680). Chlorophyll b, xanthophylls and carotenoids are accessory pigments in the antenna. Answer: Chlorophyll a.
105	Morphology of Flowering PlantsRoot, stem and leaf	Class 11: Morphology of Flowering Plants	Single correctEasy
A easy single correct question on root, stem and leaf, from Morphology of Flowering Plants. NCERT: Class 11: Morphology of Flowering Plants Phyllotaxy is the pattern of arrangement of leaves on the stem or branch: alternate (china rose), opposite (guava, Calotropis) or whorled (Alstonia). The arrangement of flowers on the floral axis is inflorescence, and the arrangement of sepals or petals in the bud is aestivation. Answer: leaves.
108	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	Single correctEasy
A easy single correct question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life The endomembrane system is the group of membranous organelles whose functions are coordinated: the endoplasmic reticulum, the Golgi complex, lysosomes and vacuoles. Mitochondria, chloroplasts and peroxisomes work independently of them and are not part of it, which rules out the other three options. Answer: endoplasmic reticulum, Golgi complex, lysosomes and vacuole.
109	Respiration in PlantsGlycolysis	Class 11: Respiration in Plants	Single correctEasy
A easy single correct question on glycolysis, from Respiration in Plants. NCERT: Class 11: Respiration in Plants In glycolysis one molecule of glucose (6 C) is partially oxidised to two molecules of pyruvic acid (3 C each). So 206 molecules of glucose give 206 × 2 = 412 molecules of pyruvic acid.
110	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 2,4-D (2,4-dichlorophenoxyacetic acid), a synthetic auxin, is widely used to kill dicotyledonous weeds; it does not affect mature monocots, so it keeps lawns weed-free. Kinetin is a cytokinin, abscisic acid a growth inhibitor used in dormancy and stress, and gibberellin a growth promoter; none is used as a weedkiller. Answer: 2,4-D.
111	Plant KingdomGymnosperms	Class 11: Plant Kingdom	StatementsEasy
A easy statements question on gymnosperms, from Plant Kingdom. NCERT: Class 11: Plant Kingdom Statement I is correct: in gymnosperms the male and female gametophytes do not have an independent, free-living existence; they remain within the sporangia retained on the sporophyte. Statement II is correct: gymnosperms (gymnos, naked; sperma, seed) have ovules not enclosed by an ovary wall, so the seeds are naked, not covered by a fruit. So both statements are correct.
112	Biological ClassificationMonera and Protista	Class 11: Biological Classification	MatchingEasy
A easy matching question on monera and protista, from Biological Classification. NCERT: Class 11: Biological Classification Bacteria are grouped by shape: spherical bacteria are cocci (A-II), rod-shaped are bacilli (B-IV), comma-shaped are vibrio (C-I), and spiral ones are spirilla (D-III). So A-II, B-IV, C-I, D-III.
113	Morphology of Flowering PlantsPlant families: Malvaceae, Cruciferae, Leguminosae, Compositae, Gramineae	Class 11: Morphology of Flowering Plants	StatementsEasy
A easy statements question on plant families: malvaceae, cruciferae, leguminosae, compositae, gramineae, from Morphology of Flowering Plants. NCERT: Class 11: Morphology of Flowering Plants Solanaceae (the potato family): flowers bisexual and actinomorphic (a true); calyx of five united sepals, persistent (b true); corolla of five united petals; androecium of five stamens, epipetalous (c true); gynoecium bicarpellary, syncarpous, with a superior ovary, obliquely placed (d false). Floral formula: ⊕ ⚥ K(5) C(5) A5 G(2), the ovary superior. So (a), (b) and (c) only.
114	Photosynthesis in Higher PlantsLight reaction	Class 11: Photosynthesis in Higher Plants	Single correctMedium
A medium single correct question on light reaction, from Photosynthesis in Higher Plants. NCERT: Class 11: Photosynthesis in Higher Plants The early experiments in order: 1. Joseph Priestley (1770) showed, with a candle and a mouse under a bell jar with a mint plant, that air plays an essential role in the growth of green plants (role of air). 2. Jan Ingenhousz (1730–1799) showed that sunlight is needed for the plant to purify the air, seen as bubbles of oxygen released from the green parts (release of oxygen). 3. Julius von Sachs (1854) showed that green plants produce glucose, stored as starch, in their green parts (production of glucose). 4. T. W. Engelmann (1843–1909), splitting light with a prism and using aerobic bacteria on Cladophora, described the action spectrum, which roughly resembles the absorption spectra of chlorophyll a and b. So role of air → release of oxygen → production of glucose → absorption spectra of chlorophyll a and b.
116	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 Each turn of the Calvin cycle fixes one molecule of $\mathrm{CO_2}$. One glucose molecule has six carbons, so it needs six turns. Three molecules of glucose need 3 × 6 = 18 turns (fixing 18 $\mathrm{CO_2}$, using 54 ATP and 36 NADPH).
118	The Living WorldTaxonomic categories	Class 11: The Living World	MatchingEasy
A easy matching question on taxonomic categories, from The Living World. NCERT: Class 11: The Living World Mango is classified as: genus Mangifera, family Anacardiaceae, order Sapindales, class Dicotyledonae, and division (phylum) Angiospermae. So A (family)-III, B (genus)-V, C (class)-II, D (phylum)-IV, E (order)-I.
119	The Living WorldTaxonomic categories	Class 11: The Living World	Single correctEasy
A easy single correct question on taxonomic categories, from The Living World. NCERT: Class 11: The Living World The taxonomic hierarchy, from the lowest category to the highest, is species → genus → family → order → class → phylum (division) → kingdom. In the given letters this is (g), (a), (e), (c), (b), (d), (f).
120	Morphology of Flowering PlantsFlower and fruit	Class 11: Morphology of Flowering Plants	MatchingEasy
A easy matching question on flower and fruit, from Morphology of Flowering Plants. NCERT: Class 11: Morphology of Flowering Plants A. Marginal: the placenta forms a ridge along the ventral suture of the ovary, as in pea (IV). B. Axile: the placenta is axial in a multilocular ovary, as in china rose, tomato and lemon (II). C. Parietal: ovules on the inner wall of the ovary, as in mustard and Argemone (I). D. Free central: ovules on a central axis without septa, as in Dianthus and Primrose (III). So A-IV, B-II, C-I, D-III.
121	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 Pteridophytes are divided into four classes: Psilopsida (Psilotum), Lycopsida (Selaginella, Lycopodium), Sphenopsida (Equisetum) and Pteropsida (Dryopteris, Pteris, Adiantum). So Sphenopsida is a class of pteridophytes.
122	Photosynthesis in Higher PlantsPhotorespiration	Class 11: Photosynthesis in Higher Plants	StatementsMedium
A medium statements question on photorespiration, from Photosynthesis in Higher Plants. NCERT: Class 11: Photosynthesis in Higher Plants (a) False: photorespiration occurs in C3 plants; it is C4 plants in which it does not occur, because they concentrate $\mathrm{CO_2}$ at RuBisCO. (b) False: it is the reverse. In photorespiration $\mathrm{O_2}$ is consumed (RuBP is oxygenated) and $\mathrm{CO_2}$ is released. (c) True: oxygenation of RuBP gives one phosphoglycerate and one phosphoglycolate. (d) True: in the photorespiratory pathway there is neither synthesis of sugars, nor of ATP and NADPH; rather it uses ATP and NADPH. So (c) and (d) only.
123	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	Single correctEasy
A easy single correct question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life The smooth endoplasmic reticulum has no ribosomes on its surface and is the major site for the synthesis of lipids; in animal cells, lipid-like steroidal hormones are made there. Ribosomes attached to the surface and active protein synthesis describe the rough ER. Carbohydrates are not made on the SER (the Golgi apparatus is where glycoproteins and glycolipids are formed). Answer: is the major site for the synthesis of lipids.
125	The Living WorldTaxonomic categories	Class 11: The Living World	Single correctEasy
A easy single correct question on taxonomic categories, from The Living World. NCERT: Class 11: The Living World A genus comprises a group of related species which have more characters in common than species of other genera; for example, potato, tomato and brinjal are three species of the genus Solanum. A group of closely related genera makes a family, and a group of related families an order, not a genus. Answer: a group of closely related species.
126	Biological ClassificationFive kingdom classification	Class 11: Biological Classification	Single correctEasy
A easy single correct question on five kingdom classification, from Biological Classification. NCERT: Class 11: Biological Classification Bacteria, blue-green algae (cyanobacteria) and mycoplasma are all prokaryotes, placed in kingdom Monera. Fungi have a well-defined nucleus and membrane-bound organelles: they are eukaryotes, in a kingdom of their own. Answer: Fungi.
127	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 Gibberellins promote bolting, the elongation of the internodes just prior to flowering, in beet, cabbage and many plants with a rosette habit. Abscisic acid inhibits growth; IBA, an auxin, is used for rooting cuttings; ethephon releases ethylene, which ripens fruit and promotes senescence. Answer: Gibberellin.
128	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 The cell cycle has interphase followed by the M phase. Interphase is G1 (gap 1, growth), then S (DNA synthesis), then G2 (gap 2, preparation for division); then comes M (mitosis). So G1 → S → G2 → M.
129	Biological ClassificationFungi	Class 11: Biological Classification	MatchingEasy
A easy matching question on fungi, from Biological Classification. NCERT: Class 11: Biological Classification The sexual cycle of fungi has three steps: fusion of the protoplasms of two motile or non-motile gametes is plasmogamy (A-II); fusion of the two nuclei is karyogamy (B-III); and meiosis in the zygote produces haploid spores (C-I). So A-II, B-III, C-I.
131	Cell Cycle and Cell DivisionMeiosis	Class 11: Cell Cycle and Cell Division	StatementsEasy
A easy statements question on meiosis, from Cell Cycle and Cell Division. NCERT: Class 11: Cell Cycle and Cell Division Statement I is correct: diakinesis, the last stage of prophase I, ends with the chromosomes fully condensed, the meiotic spindle assembled and the nuclear envelope broken down. Statement II is false: meiosis I is the reductional division, with pairing of homologues, crossing over and separation of homologous chromosomes. It is meiosis II that resembles a normal mitosis, separating sister chromatids. So Statement I is correct, but Statement II is false.
133	Plant Growth and DevelopmentGrowth phases	Class 11: Plant Growth and Development	Single correctEasy
A easy single correct question on growth phases, from Plant Growth and Development. NCERT: Class 11: Plant Growth and Development In arithmetic growth, a constant amount is added in each unit of time: $L_t = L_0 + rt$. $L_7 = 20 + 30 \times 7 = 20 + 210 = 230$ cm. (50 cm would be the length after one day only.) Answer: 230 cm.
136	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" to remove clots from the blood vessels of patients who have had a myocardial infarction (heart attack). Penicillin is an antibiotic; cyclosporin A, from the fungus Trichoderma polysporum, is an immunosuppressant used in organ transplants; statins, from the yeast Monascus purpureus, lower blood cholesterol. Answer: Streptokinase.
138	Principles of Inheritance and VariationGenetic disorders	Class 12: Principles of Inheritance and Variation	StatementsEasy
A easy statements question on genetic disorders, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation Statement I is incorrect: Down's syndrome is caused by an additional copy of chromosome 21 (trisomy 21), not by the absence of an X chromosome. Statement II is incorrect: Turner's syndrome is caused by the absence of one X chromosome (45, X0), not by an additional copy. The two causes have been swapped, so both statements are incorrect.
144	Biodiversity and ConservationPatterns of biodiversity	Class 12: Biodiversity and Conservation	Single correctMedium
A medium single correct question on patterns of biodiversity, from Biodiversity and Conservation. NCERT: Class 12: Biodiversity and Conservation The largely tropical Amazonian rain forest in South America is estimated to harbour about 1.25 million species of invertebrates (many still to be discovered), about 40,000 species of plants, 3,000 of fishes, 1,300 of birds and 427 of mammals (with 427 amphibians and 378 reptiles). In descending order: invertebrates > plants > fishes > birds > mammals, that is (d) > (a) > (c) > (b) > (e).
146	EcosystemEnergy flow	Class 12: Ecosystem	Single correctEasy
A easy single correct question on energy flow, from Ecosystem. NCERT: Class 12: Ecosystem Animals that feed directly on the producers (plants) are the primary consumers, the herbivores. Animals that eat the herbivores are carnivores, the secondary consumers; predators are carnivores in this sense, and parasites may feed at any level. Answer: Herbivores.
151	Sexual Reproduction in Flowering PlantsMicrosporogenesis	Class 12: Sexual Reproduction in Flowering Plants	Single correctEasy
A easy single correct question on microsporogenesis, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants A typical angiosperm anther is bilobed, and each lobe has two thecae (it is dithecous); a longitudinal groove often separates the thecae. So each lobe has 2 thecae, and the whole anther is tetrasporangiate, with four microsporangia, one in each corner. Answer: 2.
153	Molecular Basis of InheritanceLac operon	Class 12: Molecular Basis of Inheritance	Single correctMedium
A medium single correct question on lac operon, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance (1) Correct: the repressor of the lac operon is synthesised all the time, constitutively, from the i gene. (2) Incorrect: lactose (the inducer) binds the repressor and inactivates it, so the repressor cannot bind the operator. (3) Incorrect: the structural genes z, y and a share one promoter and operator, but the regulatory gene i has its own promoter. (4) Incorrect: the inducer is lactose (more precisely allolactose, made from it); galactose, one of its products, does not induce the operon. Answer: Gene i is constitutively expressed.
154	Sexual Reproduction in Flowering PlantsMegasporogenesis	Class 12: Sexual Reproduction in Flowering Plants	Single correctEasy
A easy single correct question on megasporogenesis, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants At the micropylar end of the embryo sac the egg apparatus has two synergids and an egg cell. The synergids have special thickenings at the micropylar tip, the filiform apparatus, which guide the pollen tube into the synergid. The antipodals lie at the chalazal end; the central cell with its two polar nuclei gives rise to the endosperm after triple fusion. Answer: Synergids.
155	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 Non-albuminous seeds have no residual endosperm, because it is completely consumed during embryo development; examples are pea and groundnut. Albuminous seeds retain part of the endosperm, as in wheat, maize, barley, castor and sunflower. Answer: Pea.
156	Sexual Reproduction in Flowering PlantsDouble fertilisation	Class 12: Sexual Reproduction in Flowering Plants	Single correctEasy
A easy single correct question on double fertilisation, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants With 2n = 36, n = 18. The endosperm nucleus forms by triple fusion of one male gamete (n) with the two polar nuclei (n + n), so it is triploid: 3n = 3 × 18 = 54.
161	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 The intrinsic rate of natural increase is r = b − d = 0.002 − 0.002 = 0. With $dN/dt = rN = 0$, the population neither grows nor declines: $N_t = N_0 e^{rt} = N_0$. So after 10 generations it is still 10 million cells.
163	Organisms and PopulationsPopulation growth models	Class 12: Organisms and Populations	Assertion and reasonEasy
A easy assertion and reason question on population growth models, from Organisms and Populations. NCERT: Class 12: Organisms and Populations Exponential growth assumes unlimited resources, but no population in nature has unlimited resources. Since resources are finite and become limiting sooner or later, a population grows until it reaches the carrying capacity K, giving the sigmoid curve of the Verhulst–Pearl logistic model. So the logistic model is considered more realistic (A is correct), and the finiteness of resources (R, correct) is exactly why. Both are correct and R explains A.
167	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 eukaryotes there are three RNA polymerases in the nucleus: RNA polymerase I transcribes rRNAs (28S, 18S, 5.8S), RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs, and RNA polymerase II transcribes the precursor of mRNA, the heterogeneous nuclear RNA (hnRNA). DNA polymerase makes DNA, not RNA. Answer: RNA polymerase II.
169	Principles of Inheritance and VariationDeviations from Mendelism	Class 12: Principles of Inheritance and Variation	Single correctEasy
A easy single correct question on deviations from mendelism, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation A person of blood group O has genotype $ii$, so each parent must pass on an $i$ allele. (1) $I^{A}i \times I^{B}i$: both can give $i$, so $ii$ is possible (1/4). (2) $I^{A}i \times I^{A}i$: $ii$ possible (1/4). (3) $I^{B}i \times I^{B}i$: $ii$ possible (1/4). (4) The father $I^{A}I^{B}$ carries no $i$ allele, so he can give only $I^{A}$ or $I^{B}$; the child would be A, AB or B, never O. Answer: Father $I^{A}I^{B}$ and Mother $I^{A}i$.
172	Biodiversity and ConservationConservation strategies	Class 12: Biodiversity and Conservation	Assertion and reasonEasy
A easy assertion and reason question on conservation strategies, from Biodiversity and Conservation. NCERT: Class 12: Biodiversity and Conservation This is the classic example of competition (and of loss of biodiversity through an introduced species): the Abingdon tortoise in the Galapagos Islands became extinct within a decade after goats were introduced on the island, apparently due to the greater browsing efficiency of the goats. A is correct, R is correct, and the goats out-competing the tortoise for food is the reason for its extinction. So R explains A.
174	Organisms and PopulationsInteractions	Class 12: Organisms and Populations	MatchingEasy
A easy matching question on interactions, from Organisms and Populations. NCERT: Class 12: Organisms and Populations A. (−/−) both species lose: competition (III). B. (+/−) one benefits and the other is harmed: predation (I) (parasitism also belongs here). C. (+/+) both benefit: mutualism (II). D. (+/0) one benefits and the other is neither harmed nor benefited: commensalism (IV). So A-III, B-I, C-II, D-IV.
175	Principles of Inheritance and VariationMendelian genetics	Class 12: Principles of Inheritance and Variation	Assertion and reasonEasy
A easy assertion and reason question on mendelian genetics, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation Crossing true-breeding tall (TT) and dwarf (tt) pea plants, Mendel found all the F1 progeny tall (Tt), none dwarf: A is correct. Stem height was one of his seven pairs of contrasting traits, tall being dominant over dwarf: R is correct. Because tall is dominant, the Tt hybrids all show the tall phenotype and dwarfness is masked, so R explains A.
178	EcosystemEnergy flow	Class 12: Ecosystem	StatementsEasy
A easy statements question on energy flow, from Ecosystem. NCERT: Class 12: Ecosystem (a) Correct: energy flows one way, from the sun to producers and then to consumers; it is not recycled. (b) Correct: the pyramid of energy is always upright and can never be inverted, because some energy is always lost as heat at each step. (c) Incorrect: only about 10 per cent of the energy is transferred from one trophic level to the next — the 10 per cent law. (About 1 per cent, or 2–10 per cent, is the share of incident sunlight that plants capture as gross primary productivity, which is a different figure.) So (a) and (b) only.

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