NEET (UG) · 2024

NEET 2024, 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
50 Q
1. Magnetic Effects of Current and Magnetism5
2. Current Electricity4
3. Electrostatics4
39 easy · 10 medium · 1 hard
Chemistry
50 Q
1. Chemical Bonding and Molecular Structure4
2. Thermodynamics3
3. Alcohols, Phenols and Ethers3
39 easy · 10 medium · 1 hard
Botany
52 Q
1. Principles of Inheritance and Variation6
2. Molecular Basis of Inheritance6
3. Biodiversity and Conservation5
40 easy · 12 medium · 0 hard
Zoology
48 Q
1. Biomolecules6
2. Biotechnology: Principles and Processes5
3. Human Health and Disease5
40 easy · 8 medium · 0 hard

Chapters

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

Principles of Inheritance and Variation6
Class 12: Principles of Inheritance and Variation
Molecular Basis of Inheritance6
Class 12: Molecular Basis of Inheritance
Biodiversity and Conservation5
Class 12: Biodiversity and Conservation
Plant Growth and Development5
Class 11: Plant Growth and Development
Cell: The Unit of Life4
Class 11: Cell: The Unit of Life
Morphology of Flowering Plants4
Class 11: Morphology of Flowering Plants
Cell Cycle and Cell Division4
Class 11: Cell Cycle and Cell Division
Anatomy of Flowering Plants3
Class 11: Anatomy of Flowering Plants
Sexual Reproduction in Flowering Plants3
Class 12: Sexual Reproduction in Flowering Plants
Photosynthesis in Higher Plants3
Class 11: Photosynthesis in Higher Plants
Organisms and Populations2
Class 12: Organisms and Populations
Biological Classification2
Class 11: Biological Classification
Respiration in Plants2
Class 11: Respiration in Plants
Microbes in Human Welfare1
Class 12: Microbes in Human Welfare
Plant Kingdom1
Class 11: Plant Kingdom
Ecosystem1
Class 12: Ecosystem

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	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	Single correctEasy
A easy single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For rolling without slipping, $v = \omega R$, and the velocity of any point is the velocity of the centre plus the velocity due to rotation about the centre. Highest point $P$: $v + \omega R = 2v$ (forward). Lowest point $Q$ (in contact with the road): $v - \omega R = 0$. So $P$ moves at $2v$ while $Q$ is momentarily at rest: $P$ moves faster than $Q$. Answer: (4).
2	Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The magnet lies between the solenoids with its N pole facing solenoid-1 and its S pole facing solenoid-2, and it moves to the right. Solenoid-1: the N pole is moving away. By Lenz's law the induced current opposes this, so the face of solenoid-1 nearer the magnet (its right face) becomes an S pole, attracting the receding N pole. Solenoid-2: the S pole is approaching. The induced current opposes the approach, so the face of solenoid-2 nearer the magnet (its left face) becomes an S pole, repelling it. Both solenoids are wound alike. Taking the long slanting lines of each drawn turn as the front of the coil, a current entering at the left-hand terminal ($A$ or $C$) rises on the front of each turn and makes the left end an N pole and the right end an S pole. Solenoid-1 needs an S pole on its right: current enters at $A$, so it flows $A \to B$. Solenoid-2 needs an S pole on its left (so N on its right): the current is reversed, entering at $D$, so it flows $D \to C$. Answer: (3) $AB$ and $DC$.
3	Electronic DevicesSpecial-purpose diodes	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	StatementsEasy
A easy statements question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A: A solar cell supplies power: it drives current opposite to the direction of the applied (positive) voltage, so its I-V curve is drawn in the fourth quadrant (V positive, I negative). A is correct. B: The small reverse-bias current (of the order of $\mu A$) is due to the drift of minority carriers across the junction, not majority carriers. B is incorrect. Answer: (3).
4	Current ElectricityOhm's law	Class 12: Current Electricity	Single correctEasy
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Each part has resistance $\frac{100}{10} = 10\,\Omega$. Five in series: $5 \times 10 = 50\,\Omega$. Five in parallel: $\frac{10}{5} = 2\,\Omega$. In series: $50 + 2 = 52\,\Omega$. Answer: (4).
5	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The network is a bridge: an upper branch of two $2\,\mu F$ capacitors in series, a lower branch of two $2\,\mu F$ capacitors in series, and a $2\,\mu F$ capacitor joining the two middle points. By symmetry the two middle points are at the same potential (each branch splits the voltage equally), so the middle capacitor holds no charge and can be removed. Each branch: $\frac{2 \times 2}{2 + 2} = 1\,\mu F$. Two branches in parallel: $1 + 1 = 2\,\mu F$. Answer: (3).
6	Electronic DevicesLogic gates	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctEasy
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $A$ feeds both inputs of a NAND gate, so its output is $\overline{A}$. $B$ feeds both inputs of a NOR gate, so its output is $\overline{B}$. These go into a NOR gate: $Y = \overline{\overline{A} + \overline{B}}$. By De Morgan's theorem $\overline{\overline{A} + \overline{B}} = A \cdot B$. So the circuit behaves as an AND gate. Answer: (2).
7	Current ElectricityCells, emf and internal resistance	Class 12: Current Electricity	Single correctEasy
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Current: $I = \frac{E}{R + r} = \frac{10}{4 + 1} = 2\,A$. Terminal voltage: $V = E - Ir = 10 - 2 \times 1 = 8\,V$ (check: $IR = 2 \times 4 = 8\,V$). Answer: (1).
8	GravitationAcceleration due to gravity	Class 11: Gravitation	Single correctEasy
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g = \frac{GM}{R^2}$, so $\frac{g_p}{g_e} = \frac{M_p}{M_e}\left(\frac{R_e}{R_p}\right)^2 = \frac{1}{10} \times 2^2 = 0.4$. $g_p = 0.4 \times 9.8 = 3.92\ \text{m s}^{-2}$. Answer: (2).
9	Magnetic Effects of Current and MagnetismMagnetic dipole	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctMedium
A medium single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Period: $T = \frac{5}{20} = 0.25$ s. For a magnetic needle oscillating in a field, $T = 2\pi\sqrt{\frac{I}{mB}}$, so $m = \frac{4\pi^2 I}{T^2 B}$. $m = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{(0.25)^2 \times 0.049} = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{3.0625 \times 10^{-3}} = 4\pi^2 \times 3.2 \times 10^{-3} = 1.28 \times 10^{-2}\,\pi^2$ A m$^2$. $= 1280\pi^2 \times 10^{-5}$ A m$^2$, so $x = 1280\pi^2$. Check: $\frac{9.8 \times 10^{-6}}{0.049} = 2 \times 10^{-4}$ and $\frac{2 \times 10^{-4}}{0.0625} = 3.2 \times 10^{-3}$, as above. Answer: (2).
10	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	StatementsEasy
A easy statements question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter A: $E = h\nu$ — true. B: photons travel at $c$ in free space — true. C: $p = \frac{E}{c} = \frac{h\nu}{c}$ — true. D: in a photon-electron collision total energy and total momentum are conserved — true. E: a photon is electrically neutral — false. So A, B, C and D. Answer: (4).
11	Properties of Bulk MatterElasticity	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 elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Within the elastic limit, maximum strain $= \frac{\text{stress}}{Y} = \frac{8 \times 10^8}{2 \times 10^{11}} = 4 \times 10^{-3}$. Maximum elongation $= 4 \times 10^{-3} \times 1\ \text{m} = 4 \times 10^{-3}$ m $= 4$ mm. Answer: (3).
12	Electronic DevicesLogic gates	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctEasy
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $Y = 1$ exactly when $B = 0$, whatever $A$ is: $Y = \overline{B}$. Check the others: $B$ gives the opposite column; $A.B + \overline{A}$ gives 1 for $A = 1, B = 1$ (table says 0); $A.\overline{B} + \overline{A}$ gives 1 for $A = 0, B = 1$ (table says 0). Answer: (1).
13	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 $I = \frac{ML^2}{12}$, so $L^2 = \frac{12I}{M} = \frac{12 \times 2400}{400} = 72$ cm$^2$. $L = \sqrt{72} \approx 8.49$ cm $\approx 8.5$ cm. Answer: (3).
14	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 $(N + 1)$ VSD $= N$ MSD, so 1 VSD $= \frac{N}{N + 1}$ MSD. Vernier constant $= 1$ MSD $- 1$ VSD $= \frac{1}{N + 1}$ MSD $= \frac{0.1}{N + 1}$ mm. $0.1$ mm $= 0.01$ cm, so the vernier constant $= \frac{0.01}{N + 1}$ cm $= \frac{1}{100(N + 1)}$ cm. Answer: (4).
15	Atoms and NucleiHydrogen spectrum	Class 12: Atoms; Class 12: Nuclei	MatchingEasy
A easy matching question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei These are Balmer lines: $\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{n^2}\right)$. The larger the upper level, the larger the energy gap and the shorter the wavelength. So the order of wavelengths is $3 \to 2$ (longest) $> 4 \to 2 > 5 \to 2 > 6 \to 2$ (shortest). $3 \to 2$: 656.3 nm (III); $4 \to 2$: 486.1 nm (IV); $5 \to 2$: 434.1 nm (II); $6 \to 2$: 410.2 nm (I). Check: $\lambda_{3\to2} = \frac{36}{5R} = \frac{36}{5 \times 1.097 \times 10^{7}} \approx 656$ nm. Answer: (4) A-III, B-IV, C-II, D-I.
16	OpticsInterference and diffraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At the centre the path difference is zero for every wavelength, so all colours interfere constructively there: the central fringe is bright and white. Away from the centre, fringe width $\beta = \frac{\lambda D}{d}$ differs for each colour, so the fringes of different colours separate and a few coloured fringes are seen before they overlap into uniform illumination. Fringe widths are therefore not equal, and the pattern does not vanish. Answer: (1).
17	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$, so $\frac{1}{\lambda^2} = \frac{2mE}{h^2}$. $\frac{1}{\lambda^2}$ is directly proportional to $E$: a straight line through the origin. Answer: (2).
18	Magnetic Effects of Current and MagnetismMagnetic properties of materials	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	MatchingEasy
A easy matching question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Diamagnetic: weakly magnetised opposite to the field, $-1 \leq \chi < 0$ (II). Ferromagnetic: strongly magnetised along the field, $\chi >> 1$ (III). Paramagnetic: weakly magnetised along the field, small positive $\chi$ (IV). Non-magnetic: $\chi = 0$ (I). Answer: (3) A-II, B-III, C-IV, D-I.
19	OpticsPolarisation	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At Brewster's angle ($\tan i_B = \mu$) the reflected and refracted rays are perpendicular. The component of the electric field in the plane of incidence is not reflected at all, so the reflected light is completely plane polarised (vibrations perpendicular to the plane of incidence). The refracted beam carries all of the in-plane component and most of the other one, so it is only partially polarised. Answer: (2).
20	KinematicsUniform circular motion	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy single correct question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane In uniform circular motion the speed is constant but the direction of motion changes continuously, so the velocity varies. The acceleration $\frac{v^2}{r}$ has constant magnitude but always points towards the centre, so its direction changes too: the acceleration also varies. Answer: (2).
21	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	Single correctEasy
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Solid angle $= \frac{\text{area}}{r^2}$ is dimensionless. Strain (change in length / length) and plane angle (arc / radius) are both dimensionless. Arc has dimension [L], angular speed [T$^{-1}$], stress [ML$^{-1}$T$^{-2}$]. Answer: (3).
22	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 The tension provides the centripetal force: $T = m\omega^2 r$. With the same radius and angular speed doubled, $T' = m(2\omega)^2 r = 4T$. Answer: (4).
23	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 Compare with $x = A\sin(\omega t + \phi)$: $A = 5$ m and $\omega = \pi$ rad s$^{-1}$. $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2$ s. Answer: (4).
24	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the P-V diagram, $b$ is at (400 cm$^3$, 100 kPa) and $c$ is at (400 cm$^3$, 300 kPa): the path $bc$ is at constant volume. Work $W = \int P\,dV = 0$ when the volume does not change. (For comparison, the whole cycle encloses $200 \text{ kPa} \times 300 \text{ cm}^3 = 60$ J, but only $bc$ is asked.) Answer: (3).
25	Atoms and NucleiRutherford model and alpha scattering	Class 12: Atoms; Class 12: Nuclei	StatementsEasy
A easy statements question on rutherford model and alpha scattering, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Statement I: an atom has as many electrons as protons, so its net charge is zero — correct. Statement II: atoms of each element are stable, and when excited each element emits its own characteristic line spectrum (the basis of spectroscopic identification) — correct. These are the two experimental facts any atomic model had to explain. Answer: (3).
26	Electromagnetic Induction and Alternating CurrentsTransformers	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctEasy
A easy single correct question on transformers, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current For an ideal transformer $\frac{V_s}{V_p} = \frac{N_s}{N_p} = 2$. So $V_s : V_p = 2 : 1$ (a step-up transformer). Answer: (4).
27	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 Momentum is conserved and the bodies stick together: $m v_1 + 0 = 2m v_2$. $v_2 = \frac{v_1}{2}$, so $v_1 : v_2 = 2 : 1$. Answer: (4).
28	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	Single correctMedium
A medium single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\alpha$ emission: $A$ falls by 4, $Z$ by 2: $Y = (286, 80)$. $e^+$ (positron) emission: $Z$ falls by 1: $Z = (286, 79)$. $\beta^-$ emission: $Z$ rises by 1: $P = (286, 80)$. $e^-$ emission (another $\beta^-$): $Z$ rises by 1: $Q = (286, 81)$. Mass number 286, atomic number 81. Answer: (2).
29	ElectrostaticsElectric dipole	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance On the axis of a short dipole, $V = \pm\frac{1}{4\pi\epsilon_0}\frac{P}{r^2}$ (the sign depending on which side of the dipole the point lies). $V = \pm\frac{9 \times 10^9 \times 4 \times 10^{-6}}{2^2} = \pm 9 \times 10^3$ V. Assertion A is true. Reason R writes $V = \pm\frac{2P}{4\pi\epsilon_0 r^2}$, which would give $\pm 1.8 \times 10^4$ V; the factor 2 belongs to the axial field $E = \frac{2P}{4\pi\epsilon_0 r^3}$, not the potential. R is false. Answer: (1).
30	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 Inside a charged thin spherical shell the electric field is zero, so the potential is the same everywhere inside and equal to the potential on the shell: $V = \frac{kq}{R}$. $C$ is the centre and $P$ is on the shell, so $V_C = V_P = \frac{9 \times 10^9 \times 10^{-6}}{0.03} = 3 \times 10^5$ V. Potential difference $V_C - V_P = 0$. Answer: (2).
31	Properties of Bulk MatterSurface tension	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 surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The water surface pulls on the rim of the disc along its circumference, so the extra force needed is $F = T \times 2\pi r$. $F = 0.07 \times 2\pi \times 0.045 = 0.07 \times 0.2827 = 1.98 \times 10^{-2}$ N $= 19.8$ mN. Answer: (3).
32	OpticsTotal internal reflection and prisms	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctMedium
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The prism is right-angled at $A$, so the refraction angles at the two faces satisfy $r_1 + r_2 = A = 90°$. The ray emerges along face $AC$ (grazing emergence), so at $AC$ the angle inside is the critical angle: $r_2 = \theta_c$, with $\mu\sin\theta_c = 1$. Then $r_1 = 90° - \theta_c$, and Snell's law at $P$: $\sin 30° = \mu\sin r_1 = \mu\cos\theta_c$, so $\mu\cos\theta_c = \frac{1}{2}$. Squaring and adding: $\mu^2(\sin^2\theta_c + \cos^2\theta_c) = 1 + \frac{1}{4}$, so $\mu = \frac{\sqrt{5}}{2}$. Check: $\sin\theta_c = \frac{2}{\sqrt{5}}$, $\cos\theta_c = \frac{1}{\sqrt{5}}$, and $\frac{\sqrt{5}}{2}\cdot\frac{1}{\sqrt{5}} = \frac{1}{2} = \sin 30°$. Answer: (4).
33	Work, Energy and PowerPower	Class 11: Work, Energy and Power	Single correctEasy
A easy single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $v = \frac{dx}{dt} = \frac{d}{dt}(2t - 1) = 2$ m s$^{-1}$. $P = Fv = 5 \times 2 = 10$ W. Answer: (3).
34	Magnetic Effects of Current and MagnetismBiot-Savart law	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctEasy
A easy 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 At the centre of a circular coil of $N$ turns, $B = \frac{\mu_0 N I}{2R}$. $B = \frac{4\pi \times 10^{-7} \times 100 \times 7}{2 \times 0.1} = \frac{8.80 \times 10^{-4}}{0.2} = 4.4 \times 10^{-3}$ T $= 4.4$ mT. Answer: (1).
35	Laws of MotionNewton's laws	Class 11: Laws of Motion	Single correctEasy
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Both blocks move together: $a = \frac{F}{m_A + m_B} = \frac{10}{5} = 2$ m s$^{-2}$. The only horizontal force on $B$ is the push from $A$: $N = m_B a = 3 \times 2 = 6$ N. Check from $A$: $10 - N = 2 \times 2$, so $N = 6$ N. Answer: (1).
36	OpticsOptical instruments	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a telescope in normal adjustment (final image at infinity), $m = \frac{f_o}{f_e} = \frac{140}{5} = 28$. Answer: (4).
37	Current ElectricityOhm's law	Class 12: Current Electricity	Single correctMedium
A medium single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Rated at the supply voltage $V$: $R_A = \frac{V^2}{1000}$, $R_B = \frac{V^2}{2000}$. Series: $P_s = \frac{V^2}{R_A + R_B} = \frac{1}{\frac{1}{1000} + \frac{1}{2000}} = \frac{2000}{3}$ W (equivalently $\frac{P_1P_2}{P_1 + P_2}$). Parallel: $P_p = P_1 + P_2 = 3000$ W. $P_s : P_p = \frac{2000}{3} : 3000 = 2 : 9$. Answer: (4).
38	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 $T = 2\pi\sqrt{\frac{L}{g}}$ does not depend on the mass of the bob. Halving $L$: $T' = \frac{T}{\sqrt{2}} = \frac{\sqrt{2}}{2}T$. So $\frac{x}{2} = \frac{\sqrt{2}}{2}$ and $x = \sqrt{2}$. Answer: (4).
39	Electromagnetic WavesDisplacement current	Class 12: Electromagnetic Waves	Single correctEasy
A easy single correct question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Between the plates there is no conduction current, but the changing electric field gives a displacement current $I_d = \epsilon_0\frac{d\Phi_E}{dt}$. Since $\Phi_E = \frac{q}{\epsilon_0}$, $I_d = \frac{dq}{dt} = I$: the same magnitude as the conduction current, and in the same direction, so that current is continuous around the circuit. Answer: (4).
40	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 The bar is prevented from expanding, so the thermal strain $\alpha\Delta T$ is cancelled by a compressive stress $Y\alpha\Delta T$. $F = YA\alpha\Delta T = 0.5 \times 10^{11} \times 10^{-3} \times 10^{-5} \times 100 = 5 \times 10^4$ N $= 50 \times 10^3$ N. (The length does not enter.) Answer: (4).
41	KinematicsMotion graphs	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The $v$-$t$ graph is a trapezium: $v$ rises at a constant rate, stays constant, then falls at a constant rate to zero. Acceleration is the slope of the $v$-$t$ graph: a constant positive value, then zero, then a constant negative value. That is the graph with a positive rectangle, a gap at zero, then a negative rectangle. Answer: (1).
42	Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation	Class 11: Kinetic Theory	Single correctEasy
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory For a fixed amount of ideal gas at constant pressure, $T = \frac{P}{nR}V$: each isobar on a T-V graph is a line through the origin with slope proportional to $P$. The curve for $P_1$ is the steepest, then $P_2$, then $P_3$. So $P_1 > P_2 > P_3$. Answer: (2).
43	Electromagnetic Induction and Alternating CurrentsAC circuits	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} = \frac{1}{3.14 \times 10^{-3}} \approx 318.5\,\Omega$. 210 V is the rms value: $I_{rms} = \frac{210}{318.5} \approx 0.659$ A. $I_0 = \sqrt{2}\,I_{rms} \approx 1.414 \times 0.659 \approx 0.93$ A. Check: $I_0 = \sqrt{2} \times 210 \times 2\pi f C = 296.98 \times 3.14 \times 10^{-3} \approx 0.933$ A. Answer: (4).
44	Current ElectricityWheatstone bridge	Class 12: Current Electricity	Single correctHard
A hard single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity A bridge can be balanced only if current flows through all four arms, so the diode in the lower right arm must conduct (be forward biased) and must not short-circuit the 5 Ω resistor. The longer plate of the cell is its positive terminal. (1): the diode is forward biased but is connected across the 5 Ω resistor, so it short-circuits that arm (zero resistance): no balance is possible. (2): the cell is reversed, so current enters the bridge at the right; the diode, pointing towards the right-hand junction, is reverse biased and the lower right arm is open. All the current through the 15 Ω arm must then pass through G: no balance. (4): the cell drives current from left to right but the diode points towards the left, so it is reverse biased and the arm is open: no balance. (3): the cell drives current from left to right and the diode, in series with the 5 Ω resistor, points the same way, so it conducts and the lower right arm carries current. Only this circuit can be brought to balance ($\frac{10}{10} = \frac{15}{5 + r_D}$ with the diode's forward resistance $r_D = 10\,\Omega$). Answer: (3).
45	Electromagnetic WavesProperties of electromagnetic waves	Class 12: Electromagnetic Waves	Single correctEasy
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves EM waves travel in free space at $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$, are transverse, and carry equal average energy densities in their electric and magnetic fields. They are produced by accelerated charges; a charge moving with uniform velocity produces steady fields and does not radiate. Answer: (2).
46	GravitationOrbital motion	Class 11: Gravitation	Single correctMedium
A medium single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Orbit radius $r = R + 2R = 3R$. Total energy in orbit: $E_f = -\frac{GMm}{2r} = -\frac{GMm}{6R}$. At rest on the surface (ignoring the earth's rotation): $E_i = -\frac{GMm}{R}$. Energy needed $= E_f - E_i = \frac{GMm}{R} - \frac{GMm}{6R} = \frac{5GMm}{6R}$. Check in parts: raising it costs $GMm\left(\frac{1}{R} - \frac{1}{3R}\right) = \frac{2GMm}{3R}$, and orbital kinetic energy is $\frac{GMm}{6R}$; total $\frac{4 + 1}{6}\frac{GMm}{R}$. Answer: (3).
47	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	StatementsEasy
A easy statements question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With the battery connected, $V$ stays fixed. Bringing the plates closer reduces $d$, so $C = \frac{\epsilon_0 A}{d}$ increases (C true). $Q = CV$ increases (A true). Energy $U = \frac{1}{2}CV^2$ increases (B false). $\frac{Q}{V} = C$ changes (D false). $QV = CV^2$ increases (E true). So A, C and E. Answer: (4).
48	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	Single correctEasy
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Both terms have the dimensions of force: $[\alpha][T^2] = [\beta][T]$. So $\frac{[\alpha][T]}{[\beta]} = 1$: $\frac{\alpha t}{\beta}$ is dimensionless. (Here $[\alpha] = [MLT^{-4}]$ and $[\beta] = [MLT^{-3}]$, so $\frac{\alpha t}{\beta}$ has $[MLT^{-3}]/[MLT^{-3}]$.) Answer: (4).
49	Magnetic Effects of Current and MagnetismMagnetic properties of materials	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	StatementsMedium
A medium statements question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter A: a magnetic sheet is attracted by the pole, so a force is needed to hold it in place — true. B: a non-magnetic sheet feels no magnetic force, so no force is needed to hold it — false. C: moving a conducting sheet through the non-uniform field of the pole induces eddy currents, which oppose the motion (Lenz's law); a force is needed to keep the velocity uniform — true. D: a non-conducting, non-polar sheet has neither eddy currents nor magnetic attraction, so uniform motion needs no force — false. So A and C. Answer: (4).
50	Magnetic Effects of Current and MagnetismMagnetic dipole	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctEasy
A easy single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $M = mL$, with $m$ the pole strength. After bending, the poles stay at the ends of the two arms, each of length $\frac{L}{2}$, with $60°$ between them. The ends and the bend form an equilateral triangle, so the distance between the poles is $\frac{L}{2}$. New moment $M' = m \cdot \frac{L}{2} = \frac{M}{2}$. Answer: (4).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
51	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	MatchingEasy
A easy matching question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Isothermal: constant temperature (II). Isochoric: constant volume (III). Isobaric: constant pressure (IV). Adiabatic: no heat exchange with the surroundings, $q = 0$ (I). Answer: (2) A-II, B-III, C-IV, D-I.
52	HydrocarbonsAlkanes	Class 11: Hydrocarbons	StatementsEasy
A easy statements question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Boiling points: n-pentane 309 K, isopentane (2-methylbutane) 301 K, neopentane (2,2-dimethylpropane) 282.5 K. Statement I is correct. Branching makes the molecule more compact and nearly spherical, reducing the surface area of contact and hence the van der Waals forces, so the boiling point falls. Statement II is correct (and explains I). Answer: (3).
53	Chemical Bonding and Molecular StructureMolecular orbital theory	Class 11: Chemical Bonding and Molecular Structure	MatchingEasy
A easy matching question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Ethane, C–C single bond: one $\sigma$ (III). Ethene, C=C: one $\sigma$ + one $\pi$ (IV). $C_2$: configuration $\sigma 1s^2\sigma^1s^2\sigma 2s^2\sigma^2s^2(\pi 2p_x^2 = \pi 2p_y^2)$; the bond order 2 comes from two $\pi$ bonds and no $\sigma$ bond (II). Ethyne, C≡C: one $\sigma$ + two $\pi$ (I). Answer: (1) A-III, B-IV, C-II, D-I.
54	Alcohols, Phenols and EthersPrimary, secondary and tertiary alcohols	Class 12: Alcohols, Phenols and Ethers	Single correctEasy
A easy single correct question on primary, secondary and tertiary alcohols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Lucas reagent (conc. HCl + anhydrous ZnCl$_2$) converts alcohols to alkyl chlorides through carbocations, so the rate follows carbocation stability: 3° > 2° > 1°. Tertiary alcohols give turbidity at once, secondary in about five minutes, primary not at room temperature. (1) 2-methylpropan-1-ol is 1°, (3) butan-1-ol is 1°, (4) butan-2-ol is 2°, (2) 2-methylpropan-2-ol is 3°. Answer: (2).
55	Structure of AtomQuantum numbers	Class 11: Structure of Atom	MatchingEasy
A easy matching question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom $m_l$ (magnetic): orientation of the orbital in space (III). $m_s$ (spin): orientation of the electron's spin (IV). $l$ (azimuthal): shape of the orbital (I). $n$ (principal): size (and energy) of the orbital (II). Answer: (4) A-III, B-IV, C-I, D-II.
56	d- and f-Block ElementsTransition element properties	Class 12: The d- and f-Block Elements	StatementsEasy
A easy statements question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mu = \sqrt{n(n + 2)}$ BM depends only on the number of unpaired electrons $n$. $Ti^{3+}$: $3d^1$, $n = 1$. $Cr^{2+}$: $3d^4$, $n = 4$. $Mn^{2+}$: $3d^5$, $n = 5$. $Fe^{2+}$: $3d^6$, $n = 4$. $Sc^{3+}$: $3d^0$, $n = 0$. $Cr^{2+}$ and $Fe^{2+}$ both have 4 unpaired electrons: $\mu = \sqrt{24} \approx 4.90$ BM. Answer: (3).
57	Organic Chemistry: Some Basic Principles and TechniquesPurification methods	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Passing directly from solid to vapour (and back to solid on cooling) is sublimation. A sublimable compound (camphor, naphthalene, iodine) is separated from non-sublimable impurities this way. Answer: (4).
58	Structure of AtomBohr model	Class 11: Structure of Atom	Single correctEasy
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom For a hydrogen-like ion $E_n \propto -\frac{Z^2}{n^2}$. $He^+$, $n = 1$: $\frac{Z^2}{n^2} = \frac{4}{1} = 4$, energy $-x$. $Be^{3+}$, $n = 2$: $\frac{Z^2}{n^2} = \frac{16}{4} = 4$, the same ratio. So the energy is $-x$ J. Answer: (3).
59	Chemical KineticsArrhenius equation	Class 12: Chemical Kinetics	Single correctEasy
A easy single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics From the Arrhenius equation, $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. Knowing $k$ at two temperatures gives $E_a$ directly; a single rate constant does not (the pre-exponential factor is unknown). Answer: (2).
60	Chemical KineticsArrhenius equation	Class 12: Chemical Kinetics	Single correctEasy
A easy single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$. A plot of $\ln k$ against $\frac{1}{T}$ is a straight line with negative slope $-\frac{E_a}{R}$ and positive intercept $\ln A$ on the $\ln k$ axis. Only plot (2) falls from a positive intercept. Answer: (2).
61	p-Block ElementsGroup 16 elements	No matching NCERT chapter	StatementsEasy
A easy statements question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Boiling points: $H_2O$ 373 K, $H_2Te$ 271 K, $H_2Se$ 232 K, $H_2S$ 213 K. Statement I is true. Among $H_2S$, $H_2Se$, $H_2Te$ the boiling point rises with molecular mass (stronger van der Waals forces). By that trend $H_2O$ should boil lowest, but extensive intermolecular hydrogen bonding makes it boil highest. Statement II is true. Answer: (3).
62	BiomoleculesCarbohydrates	Class 12: Biomolecules	StatementsEasy
A easy statements question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Glucose reduces Tollens' reagent, forms a cyanohydrin with HCN and an oxime with $NH_2OH$, showing the aldehyde group. But it gives no Schiff's test and forms no hydrogensulphite addition product with $NaHSO_3$: most glucose exists in the cyclic hemiacetal form, with very little free –CHO. This is the evidence for the cyclic structure. So B and E. Answer: (1).
63	Classification of Elements and PeriodicityIonization enthalpy	Class 11: Classification of Elements and Periodicity in Properties	Single correctMedium
A medium single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across period 2 the ionization enthalpy generally increases, with an exception at boron: B loses a $2p$ electron, which is shielded by the $2s$ pair and lies higher in energy, while Be loses an electron from the filled $2s$ subshell. Values (kJ mol$^{-1}$): Li 520, B 801, Be 899, C 1086, N 1402. Order: Li < B < Be < C < N. Answer: (4).
64	Classification of Elements and PeriodicityElectronegativity	Class 11: Classification of Elements and Periodicity in Properties	Single correctEasy
A easy single correct question on electronegativity, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Electronegativity rises across a period and falls down a group. Pauling values: Si 1.8, C 2.5, N 3.0, O 3.5, F 4.0. Order: Si < C < N < O < F. Answer: (3).
65	d- and f-Block ElementsTransition element properties	Class 12: The d- and f-Block Elements	Single correctEasy
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $Mn^{3+}$ is $3d^4$ and $Mn^{2+}$ is $3d^5$. Reduction of $Mn^{3+}$ gives the extra-stable half-filled $d^5$ configuration, so $Mn^{3+}$ is readily reduced and $E°(Mn^{3+}/Mn^{2+})$ is large and positive (+1.57 V). Answer: (1) $d^4$ to $d^5$.
66	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 $NH_3$: 3 bond pairs + 1 lone pair: trigonal pyramidal (I). $BrF_5$: 5 bond pairs + 1 lone pair: square pyramidal (IV). $XeF_4$: 4 bond pairs + 2 lone pairs: square planar (II). $SF_6$: 6 bond pairs: octahedral (III). Answer: (3) A-I, B-IV, C-II, D-III.
67	Coordination CompoundsIsomerism in complexes	Class 12: Coordination Compounds	MatchingEasy
A easy matching question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A: $NO_2^-$ is ambidentate (–NO$_2$ or –ONO): linkage isomerism (II). B: $SO_4^{2-}$ inside and $Br^-$ outside can exchange places, $[Co(NH_3)_5Br]SO_4$: ionization isomerism (III). C: ligands can be exchanged between the two complex ions, e.g. $[Cr(NH_3)_6][Co(CN)_6]$: coordination isomerism (IV). D: water can move between the coordination sphere and the lattice, e.g. $[Co(H_2O)_5Cl]Cl_2\cdot H_2O$: solvate (hydrate) isomerism (I). Answer: (3) A-II, B-III, C-IV, D-I.
68	Chemical Bonding and Molecular StructureHydrogen bonding	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Intramolecular hydrogen bonding needs a donor (–OH) and an acceptor (–NO$_2$) close together in the same molecule, so that a six-membered ring can form. In o-nitrophenol (3) the –OH and –NO$_2$ are on adjacent carbons: the O–H···O=N bond lies within the molecule. m-nitrophenol (1) and p-nitrophenol (4) have the groups too far apart, so they form only intermolecular hydrogen bonds; HF (2) forms intermolecular hydrogen bonds (zig-zag chains). Answer: (3).
69	AminesClassification and identification of amines	Class 12: Amines	StatementsEasy
A easy statements question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Statement I: aniline, a Lewis base, forms a salt with the Lewis acid $AlCl_3$; the $-NH_2$ becomes positively charged and strongly deactivates the ring, so Friedel-Crafts reactions fail. True. Statement II: Gabriel synthesis needs an $S_N2$ attack of phthalimide anion on an alkyl halide; aryl halides do not undergo this substitution, so aromatic primary amines like aniline cannot be made this way. True. Answer: (3).
70	Some Basic Concepts of ChemistryMole concept	Class 11: Some Basic Concepts of Chemistry	Single correctEasy
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry 4 g of He = 1 mol = $N_A$ atoms. 2.271098 L at STP (1 bar, 273.15 K, molar volume 22.71098 L) = 0.1 mol. 4 mol = $4N_A$ atoms. 4 u = the mass of one He atom = 1 atom. The largest is 4 mol. Answer: (3).
71	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	MatchingMedium
A medium matching question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A: cyclohexylidenecyclohexane → 2 cyclohexanone. Cleavage of a C=C into two C=O is ozonolysis: (i) $O_3$, (ii) Zn-$H_2O$ (IV). B: benzene → benzophenone. Friedel-Crafts acylation with benzoyl chloride and anhydrous $AlCl_3$ (I). C: cyclohexanol → cyclohexanone. Oxidation of a secondary alcohol with $CrO_3$ (II). D: ethylbenzene → potassium benzoate. Side-chain oxidation with alkaline $KMnO_4$ and heat; the whole side chain is oxidised to –COOK (III). Answer: (1) A-IV, B-I, C-II, D-III.
72	Alcohols, Phenols and EthersPreparation and properties	Class 12: Alcohols, Phenols and Ethers	Single correctMedium
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers 3-Cyclohexylprop-1-ene ($C_6H_{11}-CH_2-CH=CH_2$) must become 3-cyclohexylpropanal ($C_6H_{11}-CH_2-CH_2-CHO$): the oxygen goes to the terminal carbon (anti-Markovnikov) and stops at the aldehyde. Hydroboration-oxidation ($BH_3$, then $H_2O_2/OH^-$) gives the primary alcohol $C_6H_{11}CH_2CH_2CH_2OH$. PCC oxidises a primary alcohol only as far as the aldehyde. (1) alkaline $KMnO_4$ would go on to the carboxylic acid; (2) and (3) acid-catalysed hydration is Markovnikov, giving a secondary alcohol and then a ketone. Answer: (4).
73	Haloalkanes and HaloarenesSN1 and SN2 mechanisms	Class 12: Haloalkanes and Haloarenes	Single correctEasy
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The $S_N1$ rate depends on the stability of the carbocation formed. (1) bromobenzene: an aryl cation is very unstable; no $S_N1$. (2) 1-bromo-1-phenylethane: gives a secondary benzylic cation, stabilised by resonance with the ring and by the methyl group. (3) (bromomethyl)cyclohexane: a primary cation. (4) bromocyclohexane: a secondary alkyl cation without resonance. The secondary benzylic cation is the most stable, so (2) reacts fastest. Answer: (2).
74	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctEasy
A easy single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Fehling's reagent is made by mixing equal volumes of two solutions just before use: Fehling's A is aqueous copper sulphate, and Fehling's B is alkaline sodium potassium tartrate (Rochelle salt). Answer: (3).
75	Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Carbocation stability: 3° > 2° > 1° (hyperconjugation and inductive effect of alkyl groups). (1) cyclopentylmethyl cation, $C_5H_9-\overset{+}{C}H_2$: primary. (2) 1-methylcyclohexyl cation: tertiary. (3) $CH_3-\overset{+}{C}H-CH(CH_3)-CH_3$: secondary. (4) $CH_3-\overset{+}{C}H-CH_2-CH(CH_3)-CH_3$: secondary. Answer: (2).
76	EquilibriumChemical equilibrium	Class 11: Equilibrium	Single correctEasy
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Reaction quotient $Q_c = \frac{[B][C]}{[A]^2} = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2} = 1$. $Q_c = 1 > K_c = 4 \times 10^{-3}$: there are too many products, so the reaction tends to go in the backward direction. Answer: (1).
77	SolutionsRaoult's law	Class 12: Solutions	Single correctEasy
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Henry's law: $p = K_H x$, so at a given pressure the mole fraction dissolved $x = \frac{p}{K_H}$: the larger $K_H$, the lower the solubility. $K_H$: B ($2 \times 10^{-5}$) < C (35) < A (145). Solubility: B > C > A. Answer: (4).
78	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 A tertiary carbon is bonded to three other carbons. 2,3-dimethylbutane, $(CH_3)_2CH-CH(CH_3)_2$: C-2 and C-3 are each bonded to three carbons — two tertiary carbons. 2,2-dimethylbutane has one quaternary carbon; n-hexane has none; 2-methylpentane has one tertiary carbon. Answer: (1).
79	EquilibriumChemical equilibrium	Class 11: Equilibrium	Single correctEasy
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $K_p = K_c(RT)^{\Delta n_g}$; they are equal only when $\Delta n_g = 0$. (1) $2 - 2 = 0$; (2) $2 - 2 = 0$; (4) $2 - 2 = 0$. (3) $2 - 1 = 1 \neq 0$, so $K_p = K_c RT$. Answer: (3).
80	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 In (2) every element keeps its oxidation number (Ba +2, Cl −1, Na +1, S +6, O −2): it is a double displacement (precipitation) reaction, not redox. (1) H: 0 → +1, Cl: 0 → −1. (3) Zn: 0 → +2, Cu: +2 → 0. (4) I: 0 → +5, Cl: +5 → 0. All redox. Answer: (2).
81	Coordination CompoundsValence bond theory	Class 12: Coordination Compounds	StatementsEasy
A easy statements question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Both have $Co^{3+}$ ($3d^6$) with six ligands: octahedral. $NH_3$ is a strong-field ligand: the $d^6$ electrons pair up ($t_{2g}^6$, $d^2sp^3$), no unpaired electrons — diamagnetic. $F^-$ is a weak-field ligand: high spin ($t_{2g}^4e_g^2$, $sp^3d^2$), four unpaired electrons — paramagnetic. So both statements are true. Answer: (3).
82	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 NaOH: $\frac{1}{40} = 0.025$ mol $= 25$ mmol. HCl: $25 \times 0.75 = 18.75$ mmol. NaOH + HCl → NaCl + H$_2$O, 1 : 1. NaOH left: $25 - 18.75 = 6.25$ mmol $= 6.25 \times 40 = 250$ mg. Check: 18.75 mmol × 40 = 750 mg reacted; 1000 − 750 = 250 mg. Answer: (4).
83	p-Block ElementsGroup 16 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter The stability of the −2 oxidation state falls down group 16 as the electronegativity falls. O, S, Se and Te form compounds in the −2 state (oxides, $H_2Se$, $H_2Te$, tellurides), but polonium, a metal, hardly shows −2. Answer: (2).
84	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	StatementsEasy
A easy statements question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics A: liquid → vapour, much more disorder: entropy increases. B: cooling a crystal towards 0 K decreases its entropy (to zero for a perfect crystal). C: a solid gives gases: entropy increases. D: one mole of gas gives two moles of gas: entropy increases. So A, C and D. Answer: (1).
85	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	MatchingMedium
A medium matching question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry A: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$: 4 F for 2 mol water, so 2 F for 1 mol (II). B: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$: Mn +7 → +2, 5 F (IV). C: $Ca^{2+} + 2e^- \rightarrow Ca$: 1.5 mol × 2 = 3 F (I). D: Fe +2 → +3 in $Fe_2O_3$: 1 F per mol FeO (III). Answer: (3) A-II, B-IV, C-I, D-III.
86	Haloalkanes and HaloarenesElimination reactions	Class 12: Haloalkanes and Haloarenes	Single correctMedium
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $PBr_3$ replaces the –OH of 2-methylcyclohexan-1-ol by –Br: A is 1-bromo-2-methylcyclohexane (the ring and the methyl group are untouched). Alcoholic KOH with heat causes dehydrohalogenation. By Saytzeff's rule the more substituted alkene is the major product: removing the H from the carbon bearing the methyl group gives 1-methylcyclohexene (trisubstituted) rather than 3-methylcyclohexene (disubstituted). So A = 1-bromo-2-methylcyclohexane and B = 1-methylcyclohexene, as in option (3). (1) and (2) keep an –OH in A, which $PBr_3$ would have replaced; (4) shows the less substituted alkene. Answer: (3).
87	EquilibriumChemical equilibrium	Class 11: Equilibrium	Single correctHard
A hard single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $K_c = \frac{[N_2][O_2]}{[NO]^2} = \frac{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}{(2.8 \times 10^{-3})^2} = \frac{12.6}{7.84} = 1.607$. Starting from 0.1 M NO with degree of dissociation $\alpha$: $[NO] = 0.1(1 - \alpha)$, $[N_2] = [O_2] = 0.05\alpha$. $K_c = \frac{(0.05\alpha)^2}{(0.1(1 - \alpha))^2} = \left(\frac{\alpha}{2(1 - \alpha)}\right)^2 = 1.607$. $\frac{\alpha}{2(1 - \alpha)} = \sqrt{1.607} = 1.2677$, so $\alpha = 2.5355(1 - \alpha)$ and $\alpha = \frac{2.5355}{3.5355} = 0.717$. Check: $\frac{0.717}{2 \times 0.283} = 1.267$, squared 1.605 ≈ $K_c$. Answer: (2).
88	Alcohols, Phenols and EthersPreparation and properties	Class 12: Alcohols, Phenols and Ethers	Single correctEasy
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers With $PCl_3$ the three Cl atoms are replaced by three OH groups: $3ROH + PCl_3 \rightarrow 3RCl + H_3PO_3$. So A = $H_3PO_3$ (phosphorous acid). With $PCl_5$: $ROH + PCl_5 \rightarrow RCl + POCl_3 + HCl$. So B = $POCl_3$. Check by atoms: in the first, 3 O and 3 H (from 3 OH) go to $H_3PO_3$. Answer: (2).
89	Principles Related to Practical ChemistryQualitative salt analysis	No matching NCERT chapter	Single correctMedium
A medium single correct question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Group II ($H_2S$ in dilute HCl): $Cu^{2+}$. Group III ($NH_4OH$ with $NH_4Cl$): $Al^{3+}$. Group IV ($H_2S$ in ammoniacal medium): $Co^{2+}$. Group V ($(NH_4)_2CO_3$): $Ba^{2+}$. Group VI: $Mg^{2+}$. Increasing group number: B, A, D, C, E. Answer: (3).
90	Principles Related to Practical ChemistryPreparation of inorganic and organic compounds	No matching NCERT chapter	Single correctEasy
A easy single correct question on preparation of inorganic and organic compounds, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter A few drops of dilute sulphuric acid are added: it suppresses hydrolysis of $Fe^{2+}$ and brings in only sulphate, the salt's own anion. Nitric acid would oxidise $Fe^{2+}$ to $Fe^{3+}$, hydrochloric acid would introduce chloride, and concentrated sulphuric acid is an oxidising, dehydrating agent. Answer: (2).
91	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	Single correctMedium
A medium single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics For a reversible isothermal expansion of an ideal gas, $w = -2.303\,nRT\log\frac{p_1}{p_2}$ (work done on the system; negative because the gas does work on the surroundings). $w = -2.303 \times 1 \times 2.0 \times 298 \times \log\frac{20}{10} = -2.303 \times 596 \times 0.3010 = -413.14$ cal. Check: $nRT\ln 2 = 596 \times 0.6931 = 413.1$ cal in magnitude. Answer: (4) −413.14 calories.
92	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The reactant is 1,2-dicyclohexylethene, $C_6H_{11}-CH=CH-C_6H_{11}$. Hot acidified $KMnO_4$ cleaves the C=C completely. Each =CH– carbon (bearing one H) is oxidised through the aldehyde to a carboxylic acid. So each half becomes cyclohexanecarboxylic acid, $C_6H_{11}-COOH$, and P is option (4). The diol (1) is the product of cold, dilute alkaline $KMnO_4$; the aldehyde (3) would be oxidised further; the diketone (2) is not formed. Answer: (4).
93	SolutionsColligative properties	Class 12: Solutions	Single correctEasy
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Pi = CRT$, so the slope of $\Pi$ against $C$ is $RT$. $T = \frac{25.73}{0.083} = 310$ K $= 310 - 273 = 37$°C. Answer: (3).
94	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: $S_N2$ with cyanide gives butanenitrile, $CH_3CH_2CH_2CN$ (one carbon added). B: partial hydrolysis of the nitrile gives butanamide, $CH_3CH_2CH_2CONH_2$. C: NaOH/$Br_2$ is the Hofmann bromamide degradation, which removes the carbonyl carbon: $CH_3CH_2CH_2NH_2$, propylamine. Answer: (3).
95	Some Basic Concepts of ChemistryEmpirical and molecular formula	Class 11: Some Basic Concepts of Chemistry	Single correctEasy
A easy single correct question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry C makes up $100 - 32 - 20 = 48\%$. Moles in 100 g: A $\frac{32}{64} = 0.5$; B $\frac{20}{40} = 0.5$; C $\frac{48}{32} = 1.5$. Ratio $0.5 : 0.5 : 1.5 = 1 : 1 : 3$. Empirical formula $ABC_3$ (check: $64 + 40 + 96 = 200$, A is $\frac{64}{200} = 32\%$). Answer: (4).
96	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 $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\cdot\frac{T_2 - T_1}{T_1T_2}$ with $T_1 = 300$ K, $T_2 = 330$ K. $0.6021 = \frac{E_a}{2.303 \times 8.314}\cdot\frac{30}{300 \times 330}$. $E_a = \frac{0.6021 \times 2.303 \times 8.314 \times 99000}{30} = 0.6021 \times 19.147 \times 3300 \approx 38044$ J mol$^{-1}$ $\approx 38.04$ kJ mol$^{-1}$. Check with natural logs: $\ln 4 = 1.386$; $E_a = \frac{1.386 \times 8.314 \times 3300}{1} \approx 38030$ J. Answer: (3).
97	Chemical Bonding and Molecular StructureDipole moment and Fajans' rule	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (1) False: in $NH_3$ the lone-pair moment adds to the N–H bond moments, in $NF_3$ it opposes the N–F bond moments; $\mu(NH_3) = 1.47$ D > $\mu(NF_3) = 0.23$ D. (2) True: the double bond in carbonate can be placed on any of the three O atoms, giving three equivalent canonical forms. (3) False: ozone has two resonance structures. (4) False: $BF_3$ is trigonal planar and symmetric, so its dipole moment is zero. Answer: (2).
98	d- and f-Block ElementsLanthanoids and actinoids	Class 12: The d- and f-Block Elements	Single correctEasy
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements A diamagnetic ion has no unpaired electrons: $4f^0$ or $4f^{14}$. $Ce^{4+}$: [Xe] $4f^0$. $Yb^{2+}$: [Xe] $4f^{14}$. Both diamagnetic. The others: $Gd^{3+}$ $4f^7$, $Eu^{3+}$ $4f^6$, $Pm^{3+}$ $4f^4$, $Sm^{3+}$ $4f^5$, $Ce^{3+}$ $4f^1$, $Eu^{2+}$ $4f^7$ — all paramagnetic. Answer: (3).
99	Coordination CompoundsNomenclature	Class 12: Coordination Compounds	StatementsEasy
A easy statements question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A homoleptic complex has the metal bound to only one kind of donor group; a heteroleptic complex has more than one kind. $[Co(NH_3)_6]^{3+}$ has only $NH_3$: homoleptic. $[Co(NH_3)_4Cl_2]^+$ has $NH_3$ and $Cl^-$: heteroleptic. Both statements are true (II is the definition behind I). Answer: (3).
100	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	Single correctEasy
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Charge $Q = It = 9.6487 \times 100 = 964.87$ C $= \frac{964.87}{96487} = 0.01$ F. $Cu^{2+} + 2e^- \rightarrow Cu$: 2 F deposit 1 mol (63 g). Mass $= \frac{0.01}{2} \times 63 = 0.315$ g. Answer: (4).

Botany questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
101	Anatomy of Flowering PlantsTissue systems	Class 11: Anatomy of Flowering Plants	Single correctEasy
A easy single correct question on tissue systems, from Anatomy of Flowering Plants. NCERT: Class 11: Anatomy of Flowering Plants The figure is a stoma seen in surface view. A points to an ordinary epidermal cell, B to the subsidiary cell that rings the stoma, C to a bean-shaped guard cell, and D to the stomatal pore between the two guard cells. Guard cells are the cells whose outer walls (away from the pore) are thin and elastic while the inner walls (lining the pore) are highly thickened. When they swell, the thin outer walls bulge out and pull the thick inner walls apart, opening the pore. So the component is C, the guard cell.
102	Biodiversity and ConservationConservation strategies	Class 12: Biodiversity and Conservation	Single correctEasy
A easy single correct question on conservation strategies, from Biodiversity and Conservation. NCERT: Class 12: Biodiversity and Conservation The International Union for Conservation of Nature and Natural Resources (IUCN) publishes the Red List (Red Data Book), which lists threatened species — critically endangered, endangered and vulnerable — of plants and animals. GEAC is India's Genetic Engineering Appraisal Committee, which judges GM research; WWF is a conservation charity, not the body that issues the Red List. So the answer is IUCN.
103	Biodiversity and ConservationConservation strategies	Class 12: Biodiversity and Conservation	Single correctMedium
A medium single correct question on conservation strategies, from Biodiversity and Conservation. NCERT: Class 12: Biodiversity and Conservation Taking threatened plants and animals out of their natural habitat and keeping them in a special setting — zoological parks, botanical gardens, wildlife safari parks, seed and gene banks — where they are protected and cared for, is ex situ (off-site) conservation. Ex situ is not among the options. In situ conservation is the opposite, protecting species where they live (national parks, sanctuaries, biosphere reserves), so option (3) is wrong. 'Semi-conservative' describes DNA replication, and 'sustainable development' is a policy idea, not a way of keeping species. The only option that still describes the practice is the broad heading it belongs to: biodiversity conservation.
104	Principles of Inheritance and VariationMendelian genetics	Class 12: Principles of Inheritance and Variation	StatementsMedium
A medium statements question on mendelian genetics, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation The law of dominance rests on three ideas: characters are controlled by discrete units called factors (D); factors occur in pairs (C); and when the two factors of a pair are unlike, one — the dominant — expresses itself over the other, the recessive (A). It explains why a monohybrid cross shows only one of the parental characters in F$_1$, and the 3 : 1 ratio in F$_2$ (E). Statement B contradicts dominance: alleles do express themselves, and the recessive character simply reappears in F$_2$ because it was masked, not because 'both characters appear as such'. So A, C, D and E only.
105	Molecular Basis of InheritanceLac operon	Class 12: Molecular Basis of Inheritance	Single correctEasy
A easy single correct question on lac operon, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance The lac operon has three structural genes: z codes for beta-galactosidase, which hydrolyses lactose into glucose and galactose; y codes for permease, which increases the permeability of the cell to lactose and so brings lactose into the cell; a codes for transacetylase. So lactose is taken into the cell by permease.
106	Microbes in Human WelfareIndustrial and sewage applications	Class 12: Microbes in Human Welfare	MatchingEasy
A easy matching question on industrial and sewage applications, from Microbes in Human Welfare. NCERT: Class 12: Microbes in Human Welfare Clostridium butylicum is the bacterium used to produce butyric acid: A-III. Saccharomyces cerevisiae, brewer's yeast, ferments sugar to ethanol: B-I. The fungus Trichoderma polysporum yields cyclosporin A, an immunosuppressant used in organ transplants: C-IV. Streptococcus gives streptokinase, the clot buster: D-II. So A-III, B-I, C-IV, D-II.
107	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 In the logistic equation, $N$ is the population density at time $t$, $r$ is the intrinsic rate of natural increase, and $K$ is the carrying capacity — the largest population the environment's resources can support. As $N$ approaches $K$, the factor $\frac{K-N}{K}$ falls towards zero and growth stops, giving the sigmoid curve. So $K$ is the carrying capacity.
108	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 A transcription unit has three regions: a promoter towards the upstream end (towards 5' of the coding strand), where RNA polymerase binds; the structural gene in the middle; and a terminator at the downstream end (3' of the coding strand), where transcription stops. So promoter, structural gene, terminator.
109	Plant Growth and DevelopmentSeed germination and differentiation	Class 11: Plant Growth and Development	Single correctEasy
A easy single correct question on seed germination and differentiation, from Plant Growth and Development. NCERT: Class 11: Plant Growth and Development Living differentiated cells that have lost the capacity to divide can regain it; this is dedifferentiation. Interfascicular cambium arises in dicot stems when fully differentiated parenchyma cells of the medullary rays between the vascular bundles become meristematic again — dedifferentiation. (Cork cambium is another example.) The cells that cambium later produces, which lose the ability to divide again and mature, are said to redifferentiate.
110	Biological ClassificationFungi	Class 11: Biological Classification	MatchingEasy
A easy matching question on fungi, from Biological Classification. NCERT: Class 11: Biological Classification Rhizopus is the bread mould (Phycomycetes): A-III. Ustilago causes smut disease (Basidiomycetes): B-II. Puccinia causes rust, as in wheat rust: C-IV. Agaricus is the common mushroom: D-I. So A-III, B-II, C-IV, D-I.
112	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 The nucleolus is the site of active ribosomal RNA synthesis: A-III. A centriole is made of nine peripheral triplet microtubules arranged like a cartwheel: B-II. Leucoplasts are colourless plastids that store nutrients — starch, oils, proteins: C-IV. The Golgi apparatus is where glycoproteins and glycolipids are formed: D-I. So A-III, B-II, C-IV, D-I.
113	Morphology of Flowering PlantsFlower and fruit	Class 11: Morphology of Flowering Plants	Single correctEasy
A easy single correct question on flower and fruit, from Morphology of Flowering Plants. NCERT: Class 11: Morphology of Flowering Plants In both drawings the ovary is free, not fused with the thalamus. In (a) the thalamus forms a deep cup with the ovary standing free in its centre; the sepals, petals and stamens arise from the rim of the cup, at about the level of the ovary. The ovary is half-inferior: perigynous. In (b) the thalamus is a shallow saucer with the ovary sitting free on it, and the other parts arise from its rim, again at about the ovary's level, not below a raised ovary and not above a sunken, fused one. This too is perigynous (as in plum and rose). So (a) perigynous, (b) perigynous: (2).
114	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 Metaphase begins after the nuclear envelope has disintegrated: the spindle fibres attach to the kinetochores of the chromosomes, and the chromosomes are brought to the equator, forming the metaphase plate, each with its sister chromatids attached by kinetochores to fibres from opposite poles. In anaphase the attached fibres pull the chromatids apart; in telophase the chromosomes reach the poles. So the attachment is the event of metaphase.
115	Biodiversity and ConservationConservation strategies	Class 12: Biodiversity and Conservation	StatementsEasy
A easy statements question on conservation strategies, from Biodiversity and Conservation. NCERT: Class 12: Biodiversity and Conservation The four major causes of biodiversity loss (the 'evil quartet') are habitat loss and fragmentation, over-exploitation, alien species invasions and co-extinctions. Of the list, A (over-exploitation), B (co-extinction) and D (habitat loss and fragmentation) are among them. Mutation is a source of new variation, not a cause of loss, and migration is a normal movement of organisms. So A, B and D only.
117	Sexual Reproduction in Flowering PlantsPollination and outbreeding devices	Class 12: Sexual Reproduction in Flowering Plants	StatementsMedium
A medium statements question on pollination and outbreeding devices, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants A is false: water-pollinated flowers such as Vallisneria are not colourful and produce no nectar — they need no animal visitors. B is true: waterlily and water hyacinth grow in water but their flowers emerge above it and are pollinated by insects or wind. C is true: the pollen of most water-pollinated species is protected from wetting by a mucilaginous covering. D is true: in seagrasses such as Zostera the pollen grains are long and ribbon-like. E is true: in some hydrophytes (seagrasses) the pollen grains are carried passively inside the water. So B, C, D and E only.
120	Morphology of Flowering PlantsFlower and fruit	Class 11: Morphology of Flowering Plants	Single correctEasy
A easy single correct question on flower and fruit, from Morphology of Flowering Plants. NCERT: Class 11: Morphology of Flowering Plants An actinomorphic flower can be divided into two equal halves by any radial plane through its centre; examples are mustard, Datura and chilli. Pisum (pea) and Sesbania have the papilionaceous flowers of Fabaceae, and Cassia (Caesalpinioideae) also has an irregular corolla: all three are zygomorphic, divisible into equal halves in only one plane. So Datura.
121	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 true: in leptotene, the first stage of prophase I, the chromosomes condense and become gradually visible under the light microscope. Statement II is true: diplotene begins when the synaptonemal complex, formed in zygotene, dissolves and the homologues of each bivalent start to separate, staying joined only at the chiasmata. So both statements are true.
122	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 Flower colour in snapdragon shows incomplete dominance: RR is red, Rr pink, rr white. Pink (Rr) × red (RR): the pink parent gives R or r gametes, the red parent only R. The progeny are RR (red) and Rr (pink) in a 1 : 1 ratio. So red as well as pink flowered plants; no white can appear, since every plant gets an R from the red parent.
123	Anatomy of Flowering PlantsTissue systems	Class 11: Anatomy of Flowering Plants	StatementsEasy
A easy statements question on tissue systems, from Anatomy of Flowering Plants. NCERT: Class 11: Anatomy of Flowering Plants Statement I is false: both parenchyma and collenchyma are living tissues. Collenchyma is living, with walls thickened at the corners by cellulose, hemicellulose and pectin; it is sclerenchyma that is usually dead. Statement II is true: vessels are a characteristic feature of angiosperms, and gymnosperms lack vessels in their xylem (they have tracheids). So Statement I is false but Statement II is true.
125	Biological ClassificationFungi	Class 11: Biological Classification	Single correctEasy
A easy single correct question on fungi, from Biological Classification. NCERT: Class 11: Biological Classification The kingdom Fungi is divided into classes on the basis of the morphology of the mycelium, the mode of spore formation and the fruiting bodies. Mode of nutrition is not used: all fungi are heterotrophic (saprophytic or parasitic), so it cannot tell the classes apart.
126	Plant Growth and DevelopmentSeed germination and differentiation	Class 11: Plant Growth and Development	Single correctEasy
A easy single correct question on seed germination and differentiation, from Plant Growth and Development. NCERT: Class 11: Plant Growth and Development The capacity of a cell or explant to generate a whole plant is totipotency; it is what plant tissue culture relies on. Micropropagation is the technique of raising many plants this way; somatic hybridisation is fusing protoplasts of two varieties; differentiation is a cell becoming specialised.
127	Photosynthesis in Higher PlantsCalvin cycle	Class 11: Photosynthesis in Higher Plants	StatementsEasy
A easy statements question on calvin cycle, from Photosynthesis in Higher Plants. NCERT: Class 11: Photosynthesis in Higher Plants The dark (biosynthetic) reaction is the Calvin cycle in the stroma. It fixes CO$_2$ and reduces it to sugar using the ATP and NADPH made by the light reaction. It does not itself use light or chlorophyll, which are needed only for the light reaction. So CO$_2$, ATP and NADPH: C, D and E only.
128	Biodiversity and ConservationPatterns of biodiversity	Class 12: Biodiversity and Conservation	StatementsMedium
A medium statements question on patterns of biodiversity, from Biodiversity and Conservation. NCERT: Class 12: Biodiversity and Conservation The usual explanations for the high diversity of the tropics are: (A) tropical latitudes have stayed relatively undisturbed for millions of years, unlike temperate regions with repeated glaciations, giving long evolutionary time for diversification; (E) tropical environments are less seasonal, relatively constant and predictable, and (D) such constant environments promote niche specialisation and so more species; and (C) more solar energy is available, contributing to higher productivity. B is false: tropical environments are less seasonal, not more. So A, C, D and E only.
129	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 The figure is a longitudinal section of a monocot (cereal) grain. At the lower end of the embryo a V-shaped sheath encloses the root tip. A points to the coleoptile, the sheath over the shoot tip; B to the plumule (shoot apex and leaf primordia); C into the region inside the lower sheath, the radicle; D to the sheath itself, the coleorhiza. The radicle is the embryonic root and forms the root on germination. So C, which is option (1).
130	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 CO$_2$ fixed by RuBisCO gives two molecules of 3-PGA. Reducing them to triose phosphate takes 2 ATP and 2 NADPH, and regenerating the RuBP acceptor takes 1 more ATP. So every CO$_2$ fixed needs 3 ATP and 2 NADPH (18 ATP and 12 NADPH for one glucose).
131	Principles of Inheritance and VariationMendelian genetics	Class 12: Principles of Inheritance and Variation	Single correctEasy
A easy single correct question on mendelian genetics, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation A plant with the dominant phenotype may be BB or Bb. To find out, it is crossed with the homozygous recessive, bb — a test cross. If all the progeny are black, the plant was BB; if black and white appear 1 : 1, it was Bb. Crossing with BB or Bb would not separate the two cases cleanly, because the dominant allele from that parent would mask the result. So bb.
133	Anatomy of Flowering PlantsAnatomy of root, stem and leaf	Class 11: Anatomy of Flowering Plants	Single correctEasy
A easy single correct question on anatomy of root, stem and leaf, from Anatomy of Flowering Plants. NCERT: Class 11: Anatomy of Flowering Plants In grasses, some adaxial (upper) epidermal cells along the veins are large, empty and colourless: the bulliform cells. When they are turgid the leaf is flat and exposed; when they lose water under water stress they become flaccid and the leaf curls inwards, cutting water loss. So they are responsible for the inward curling (rolling) of leaves in monocots.
134	Principles of Inheritance and VariationMendelian genetics	Class 12: Principles of Inheritance and Variation	MatchingEasy
A easy matching question on mendelian genetics, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation Alternative forms of a gene are alleles: A-III. Crossing the F$_1$ with the homozygous recessive parent is the test cross: B-IV. Crossing the F$_1$ with either parent is a back cross (the test cross is a particular back cross): C-I. The number of chromosome sets in a cell is its ploidy: D-II. So A-III, B-IV, C-I, D-II.
135	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 The synthetic auxin 2,4-D is widely used to kill dicotyledonous weeds. It does not affect mature monocotyledonous plants, and grass is a monocot; so it is used to prepare weed-free lawns, the dicot weeds dying while the grass is unharmed.
136	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 Rose has a perigynous flower, the floral parts on the rim of a cup-shaped thalamus: A-II. Pea has marginal placentation, the ovules along the ridge of a one-chambered ovary: B-IV. Cotton (like China rose and lady's finger) has twisted aestivation: C-I. Mango is a drupe, a fleshy fruit with a stony endocarp: D-III. So A-II, B-IV, C-I, D-III.
137	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 Chloroplasts, like mitochondria, contain their own DNA, which is small, circular and double stranded, together with 70S ribosomes — features resembling a prokaryote's genome. So circular, double stranded.
138	Sexual Reproduction in Flowering PlantsPollination and outbreeding devices	Class 12: Sexual Reproduction in Flowering Plants	Single correctEasy
A easy single correct question on pollination and outbreeding devices, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants The figure shows a compact spike-like inflorescence of a grass with many small flowers whose stamens hang well out on long filaments. Wind-pollinated flowers have well-exposed stamens, so the light, non-sticky pollen is easily dispersed by air currents, and large feathery stigmas to trap it; the flowers are often packed into an inflorescence. So it is a wind-pollinated plant's inflorescence with well-exposed stamens.
139	Biodiversity and ConservationPatterns of biodiversity	Class 12: Biodiversity and Conservation	MatchingEasy
A easy matching question on patterns of biodiversity, from Biodiversity and Conservation. NCERT: Class 12: Biodiversity and Conservation Robert May estimated global species diversity at about 7 million: A-III. Alexander von Humboldt found the species-area relationship from his explorations of South American jungles: B-I. Paul Ehrlich proposed the rivet popper hypothesis, comparing species in an ecosystem to rivets in an aeroplane: C-IV. David Tilman's long-term ecosystem experiments with outdoor plots showed that plots with more species have less year-to-year variation in biomass: D-II. So A-III, B-I, C-IV, D-II.
140	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 Statement I is true: RuBisCO can bind O$_2$ as well as CO$_2$. In C$_3$ plants some O$_2$ binds to it, the RuBP is oxygenated instead of carboxylated, and CO$_2$ fixation decreases — photorespiration. Statement II is false. In C$_4$ plants the mesophyll cells have no RuBisCO (they fix CO$_2$ with PEP carboxylase), so they show no photorespiration at all; the bundle-sheath cells do have RuBisCO but, with CO$_2$ concentrated there, photorespiration does not occur. The statement has the mesophyll cells photorespiring a little, which is wrong. So Statement I is true but Statement II is false.
141	Respiration in PlantsKrebs cycle	Class 11: Respiration in Plants	Single correctMedium
A medium single correct question on krebs cycle, from Respiration in Plants. NCERT: Class 11: Respiration in Plants In the Krebs cycle, the oxidations are: isocitrate to $\alpha$-ketoglutarate (NAD$^+$ reduced, CO$_2$ released), $\alpha$-ketoglutarate to succinyl-CoA (NAD$^+$ reduced), succinate to fumarate (FAD reduced, then fumarate hydrated to malate), and malate to oxaloacetate (NAD$^+$ reduced). The conversion of succinyl-CoA to succinic acid involves no oxidation: the energy of the thioester bond is used to make GTP (ATP) — substrate-level phosphorylation. So succinyl-CoA to succinic acid.
142	Respiration in PlantsETS and oxidative phosphorylation	Class 11: Respiration in Plants	MatchingEasy
A easy matching question on ets and oxidative phosphorylation, from Respiration in Plants. NCERT: Class 11: Respiration in Plants The citric acid (Krebs) cycle runs in the mitochondrial matrix: A-II. Glycolysis takes place in the cytoplasm: B-I. The electron transport system is located in the inner mitochondrial membrane: C-IV. The proton gradient is built up by pumping protons into the intermembrane space: D-III. So A-II, B-I, C-IV, D-III.
143	Plant Growth and DevelopmentSeed germination and differentiation	Class 11: Plant Growth and Development	Single correctEasy
A easy single correct question on seed germination and differentiation, from Plant Growth and Development. NCERT: Class 11: Plant Growth and Development In somatic hybridisation, single cells of two plant varieties have their cell walls digested away by pectinase and cellulase; the naked protoplasts, each surrounded by its plasma membrane, are fused to give a hybrid protoplast, which is then grown into a new plant (a somatic hybrid, such as the pomato). So protoplasts are fused.
145	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 Monoadelphous stamens are united into one bundle, as in China-rose: A-IV. Diadelphous stamens are in two bundles, as in pea (9 + 1): B-II. Polyadelphous stamens are in more than two bundles, as in Citrus: C-I. Epiphyllous stamens are attached to the perianth, as in lily: D-III. So A-IV, B-II, C-I, D-III.
146	Plant KingdomAlgae	Class 11: Plant Kingdom	StatementsMedium
A medium statements question on algae, from Plant Kingdom. NCERT: Class 11: Plant Kingdom In the brown algae (Phaeophyceae): A is true: asexual reproduction is usually by biflagellate, pear-shaped zoospores with two unequal lateral flagella. B is false: sexual reproduction may be isogamous, anisogamous or oogamous, not oogamous only. C is true: food is stored as complex carbohydrates, laminarin or mannitol. D is true: they have chlorophyll a and c, carotenoids and xanthophylls (fucoxanthin, which gives the brown colour). E is true: the vegetative cells have a cellulosic wall, usually covered on the outside by a gelatinous coating of algin. So A, C, D and E only.
147	Molecular Basis of InheritanceDNA structure	Class 12: Molecular Basis of Inheritance	MatchingEasy
A easy matching question on dna structure, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance Frederick Griffith discovered transformation in Streptococcus pneumoniae: A-III. Francois Jacob and Jacques Monod proposed the lac operon model: B-IV. Har Gobind Khorana synthesised RNA molecules of defined sequence and helped crack the genetic code: C-I. Meselson and Stahl proved the semi-conservative mode of DNA replication with $^{15}$N: D-II. So A-III, B-IV, C-I, D-II.
148	Molecular Basis of InheritanceReplication	Class 12: Molecular Basis of Inheritance	Single correctEasy
A easy single correct question on replication, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance The main enzyme of replication is DNA-dependent DNA polymerase. It can polymerise nucleotides in only one direction, 5' to 3', adding each nucleotide to the free 3'-OH end. That is why replication is continuous on one strand (leading) and discontinuous on the other (lagging), whose Okazaki fragments are joined by DNA ligase. Option (4) is about RNA polymerase, an enzyme of transcription, so it does not describe replication. The correct statement is (2).
149	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 Sugarcane stores carbohydrate as sugar in its stems. Spraying the crop with gibberellins increases the length of the stem (by internode elongation), so the yield rises by as much as 20 tonnes per acre. So gibberellin.
150	EcosystemEnergy flow	Class 12: Ecosystem	Single correctMedium
A medium single correct question on energy flow, from Ecosystem. NCERT: Class 12: Ecosystem By Lindeman's 10 per cent law, only about 10% of the energy at one trophic level is transferred to the next. The net primary productivity of the producers, $100x$, is what is available to the second trophic level, which gets $10\%$ of it: $10x$. The third trophic level gets $10\%$ of that: $x$ (kcal m$^{-2}$) yr$^{-1}$. The question calls this the 'GPP' of the third trophic level — strictly, primary productivity belongs only to producers, so it means the gross energy taken in at the third level. So $x$.
151	Principles of Inheritance and VariationGenetic disorders	Class 12: Principles of Inheritance and Variation	MatchingEasy
A easy matching question on genetic disorders, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation Down's syndrome is trisomy of chromosome 21: A-III. $\alpha$-thalassemia is controlled by the two closely linked genes HBA1 and HBA2 on chromosome 16: B-IV. $\beta$-thalassemia is controlled by the single gene HBB on chromosome 11: C-I. Klinefelter's syndrome is an extra X chromosome (47, XXY): D-II. So A-III, B-IV, C-I, D-II.
152	Cell Cycle and Cell DivisionMeiosis	Class 11: Cell Cycle and Cell Division	MatchingEasy
A easy matching question on meiosis, from Cell Cycle and Cell Division. NCERT: Class 11: Cell Cycle and Cell Division Diakinesis is marked by the terminalisation of chiasmata (completed by its end): A-II. In pachytene, recombination nodules appear, the sites of crossing over: B-IV. In zygotene the homologues pair (synapsis) and the synaptonemal complex forms: C-I. In leptotene the chromosomes are long, thin threads that gradually become visible: D-III. So A-II, B-IV, C-I, D-III.
164	Molecular Basis of InheritanceTranscription and translation	Class 12: Molecular Basis of Inheritance	Single correctMedium
A medium single correct question on transcription and translation, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance RNA polymerase reads the template 3'→5' and builds an RNA complementary to it, 5'→3', with U in place of T. Template: 3' T A C A T G G C A A A T A T C C A T T C A 5' RNA: 5' A U G U A C C G U U U A U A G G U A A G U 3' So 5'AUGUACCGUUUAUAGGUAAGU3'.
169	Cell: The Unit of LifeCytoskeleton, cilia and flagella	Class 11: Cell: The Unit of Life	MatchingEasy
A easy matching question on cytoskeleton, cilia and flagella, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life The axoneme is the core of cilia and flagella, with its 9 + 2 microtubules: A-II. A centriole has its nine triplet microtubules arranged in a cartwheel pattern: B-I. Cristae are the infoldings of the inner mitochondrial membrane: C-IV. A satellite is the small fragment beyond the secondary constriction of some chromosomes: D-III. So A-II, B-I, C-IV, D-III.
185	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 A cell cycle runs through interphase — G$_1$ phase, then S (synthesis) phase, when DNA is replicated, then G$_2$ phase — and then the M phase, in which karyokinesis (division of the nucleus) is followed by cytokinesis (division of the cytoplasm). So E (G$_1$) → C (S) → A (G$_2$) → D (karyokinesis) → B (cytokinesis): E-C-A-D-B.
188	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	StatementsMedium
A medium statements question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life Statement I is correct: mitochondria and chloroplasts are both bounded by a double membrane (an outer and an inner membrane). Statement II is incorrect as written. The comparison 'relatively less permeable' belongs to the chloroplast: of the chloroplast's two membranes, the inner one is relatively less permeable than the outer. The statement transfers this to the mitochondrion's inner membrane and compares it with the chloroplast, which is not a valid comparison. So Statement I is correct but Statement II is incorrect.
194	Molecular Basis of InheritanceTranscription and translation	Class 12: Molecular Basis of Inheritance	MatchingEasy
A easy matching question on transcription and translation, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance In eukaryotes, RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs: A-IV. In bacteria, termination of transcription is brought about by the rho factor: B-III. Splicing, which removes introns and joins the exons, is carried out with snRNPs (the spliceosome): C-I. The TATA box is part of the promoter: D-II. So A-IV, B-III, C-I, D-II.
196	Organisms and PopulationsInteractions	Class 12: Organisms and Populations	StatementsMedium
A medium statements question on interactions, from Organisms and Populations. NCERT: Class 12: Organisms and Populations Statement I is false: Gause's competitive exclusion principle says that two closely related species competing for the same resources cannot co-exist indefinitely. Species using different resources do not compete in this way. Statement II is true: according to the principle, the competitively inferior species will be eliminated eventually; this may be true when resources are limiting, but not otherwise. So Statement I is false but Statement II is true.
199	Principles of Inheritance and VariationDeviations from Mendelism	Class 12: Principles of Inheritance and Variation	StatementsMedium
A medium statements question on deviations from mendelism, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation A child of group O has the genotype ii, so it must receive an i allele from each parent. The father, group B, must then be I$^B$i, and the mother, group A, must be I$^A$i. The genotypes, in the order father / mother / child, are I$^B$i / I$^A$i / ii — set A. B fails: I$^B$I$^B$ and I$^A$I$^A$ parents carry no i and cannot have an ii child. C gives the father I$^A$I$^B$ (group AB) and the child I$^B$i (group B). D gives the father group A and the mother group B. E gives the child I$^A$I$^B$, group AB, not O. So A only.

Zoology questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
111	Biotechnology: Principles and ProcessesrDNA technology	Class 12: Biotechnology: Principles and Processes	StatementsMedium
A medium statements question on rdna technology, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes A piece of DNA that carries only the gene of interest has no origin of replication, so on its own it cannot multiply in the progeny cells (A and E are false). It can be kept only if it gets integrated into the genome of the recipient (B); then, being part of a chromosome that can replicate, it multiplies and is inherited along with the host DNA (C). D is false for the same reason — once integrated it is part of the chromosome. This is why genes are cloned into vectors that carry an origin of replication. So B and C only.
116	BiomoleculesCarbohydrates, lipids and nucleic acids	Class 11: Biomolecules	Single correctEasy
A easy single correct question on carbohydrates, lipids and nucleic acids, from Biomolecules. NCERT: Class 11: Biomolecules Lecithin is phosphatidylcholine: glycerol esterified with two fatty acids and with phosphoric acid, to which choline is attached. A lipid that carries a phosphate group is a phospholipid, found in cell membranes. So lecithin is a phospholipid.
118	BiomoleculesEnzymes	Class 11: Biomolecules	Single correctEasy
A easy single correct question on enzymes, from Biomolecules. NCERT: Class 11: Biomolecules Many enzymes need a metal ion as their cofactor; zinc is the cofactor of the proteolytic enzyme carboxypeptidase, forming coordination bonds with side chains at the active site and with the substrate. Haem is the prosthetic group of peroxidase and catalase; niacin is part of the coenzymes NAD and NADP; flavin is part of FAD and FMN.
119	BiomoleculesEnzymes	Class 11: Biomolecules	Single correctEasy
A easy single correct question on enzymes, from Biomolecules. NCERT: Class 11: Biomolecules Malonate closely resembles succinate, the substrate of succinic dehydrogenase, in structure. It competes with succinate for the enzyme's active site and so slows the reaction; adding more succinate overcomes it. An inhibitor that resembles the substrate and competes for the active site is a competitive inhibitor. So it is competitive inhibition.
124	Biotechnology and its ApplicationsBt crops	Class 12: Biotechnology and its Applications	StatementsMedium
A medium statements question on bt crops, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications Statement I: the Bt toxin proteins are specific to insect groups, and they are coded by cry genes; the protein of cryIAc, for example, controls cotton bollworms. Taken as intended, the statement is true. Statement II is false: the toxin is indeed held as an inactive protoxin in Bacillus thuringiensis, but in the insect's gut it is activated by the alkaline pH, which solubilises the crystals. The active toxin binds the midgut epithelium and makes pores that cause the cells to swell and lyse. So Statement I is true but Statement II is false.
132	Biotechnology: Principles and ProcessesRestriction enzymes and vectors	Class 12: Biotechnology: Principles and Processes	Single correctEasy
A easy single correct question on restriction enzymes and vectors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes Hind II, the first restriction endonuclease isolated, always cuts DNA at a particular point by recognising a specific sequence of six base pairs. This is its recognition sequence. So 6 bp.
144	BiomoleculesProteins and amino acids	Class 11: Biomolecules	MatchingEasy
A easy matching question on proteins and amino acids, from Biomolecules. NCERT: Class 11: Biomolecules These are all proteins with different functions. GLUT-4 is the transporter that enables glucose transport into cells: A-IV. Insulin is a hormone: B-I. Trypsin is an enzyme: C-II. Collagen is the intercellular ground substance: D-III. So A-IV, B-I, C-II, D-III.
153	Breathing and Exchange of GasesTransport of gases	Class 11: Breathing and Exchange of Gases	Single correctEasy
A easy single correct question on transport of gases, from Breathing and Exchange of Gases. NCERT: Class 11: Breathing and Exchange of Gases In the alveoli the conditions favour the binding of O$_2$ to haemoglobin: high pO$_2$, low pCO$_2$, lesser H$^+$ concentration (higher pH) and lower temperature. In the tissues the reverse — low pO$_2$, high pCO$_2$, high H$^+$ and higher temperature — favours the release of O$_2$. So high pO$_2$ and lesser H$^+$ concentration.
154	Human Health and DiseasePathogens and diseases	Class 12: Human Health and Disease	MatchingEasy
A easy matching question on pathogens and diseases, from Human Health and Disease. NCERT: Class 12: Human Health and Disease Typhoid is caused by the bacterium Salmonella typhi: A-IV. Leishmaniasis is caused by the protozoan Leishmania: B-III. Ringworm is caused by fungi (Microsporum, Trichophyton, Epidermophyton): C-I. Filariasis is caused by the nematode Wuchereria: D-II. So A-IV, B-III, C-I, D-II.
155	Breathing and Exchange of GasesMechanism and regulation of breathing	Class 11: Breathing and Exchange of Gases	MatchingEasy
A easy matching question on mechanism and regulation of breathing, from Breathing and Exchange of Gases. NCERT: Class 11: Breathing and Exchange of Gases Expiratory capacity (EC) is the air a person can expire after a normal inspiration: TV + ERV: A-II. Functional residual capacity (FRC) is the air left in the lungs after a normal expiration: ERV + RV: B-IV. Vital capacity (VC) is the most air that can be breathed out after a forced inspiration: ERV + TV + IRV: C-I. Inspiratory capacity (IC) is the air that can be inspired after a normal expiration: TV + IRV: D-III. So A-II, B-IV, C-I, D-III.
156	Human Health and DiseaseImmunity	Class 12: Human Health and Disease	StatementsEasy
A easy statements question on immunity, from Human Health and Disease. NCERT: Class 12: Human Health and Disease In an autoimmune disorder the immune system attacks the body's own cells. Myasthenia gravis: antibodies against acetylcholine receptors at the neuromuscular junction cause fatigue and paralysis of skeletal muscle — autoimmune. Rheumatoid arthritis: the immune system attacks the joints — autoimmune. Systemic lupus erythematosus: antibodies against many of the body's own tissues — autoimmune. Gout is inflammation of the joints from deposited uric acid crystals, and muscular dystrophy is a genetic degeneration of skeletal muscle; neither is autoimmune. So A, B & E only.
157	Excretory Products and their EliminationUrine formation	Class 11: Excretory Products and their Elimination	StatementsMedium
A medium statements question on urine formation, from Excretory Products and their Elimination. NCERT: Class 11: Excretory Products and their Elimination Statement I is false: it has the descending limb backwards. The descending limb of the loop of Henle is permeable to water but almost impermeable to electrolytes, so the filtrate becomes concentrated as it goes down. It is the ascending limb that is impermeable to water and lets electrolytes out. Statement II is false: the PCT is lined by simple cuboidal brush border epithelium (not columnar), which does increase the surface area for reabsorption. So both statements are false.
158	Chemical Coordination and IntegrationHormone mechanism	Class 11: Chemical Coordination and Integration	Single correctEasy
A easy single correct question on hormone mechanism, from Chemical Coordination and Integration. NCERT: Class 11: Chemical Coordination and Integration Progesterone, cortisol and testosterone are steroid hormones, derived from cholesterol; being lipid-soluble they enter the cell and act through intracellular receptors. Glucagon, from the alpha cells of the islets of Langerhans, is a peptide hormone. So glucagon is not a steroid.
159	Chemical Coordination and IntegrationEndocrine glands	Class 11: Chemical Coordination and Integration	Assertion and reasonMedium
A medium assertion and reason question on endocrine glands, from Chemical Coordination and Integration. NCERT: Class 11: Chemical Coordination and Integration Assertion A is false. FSH stimulates the growth and development of ovarian follicles in the female, but in the male it acts on the Sertoli cells, stimulating spermatogenesis. The Leydig (interstitial) cells are acted on by LH (also called ICSH), which makes them secrete androgens. Reason R is true: growing ovarian follicles secrete estrogens, and the interstitial (Leydig) cells of the testis secrete androgens such as testosterone. So A is false but R is true.
160	EvolutionHuman evolution	Class 12: Evolution	Single correctEasy
A easy single correct question on human evolution, from Evolution. NCERT: Class 12: Evolution Homo habilis, the first human-like being (about 2 million years ago), came first; then Homo erectus (about 1.5 million years ago); then Neanderthal man, Homo neanderthalensis (about 1,00,000–40,000 years ago); and finally Homo sapiens. So A (habilis) → D (erectus) → C (neanderthalensis) → B (sapiens): A-D-C-B.
161	Structural Organisation in AnimalsAnimal tissues	Class 11: Structural Organisation in Animals	Single correctEasy
A easy single correct question on animal tissues, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals (a) shows long, unbranched, cylindrical fibres with striations and several peripheral nuclei: skeletal muscle, as in the triceps or biceps. (b) shows spindle-shaped, unstriated fibres, each with a single central nucleus: smooth muscle, as in the wall of the stomach and intestine. (c) shows striated, branched fibres joined end to end: cardiac muscle, found only in the heart. Only option (4) names all three correctly: skeletal–triceps, smooth–stomach, cardiac–heart.
162	Human Health and DiseaseCancer and drug abuse	Class 12: Human Health and Disease	MatchingEasy
A easy matching question on cancer and drug abuse, from Human Health and Disease. NCERT: Class 12: Human Health and Disease Cocaine is an alkaloid from the coca plant, Erythroxylum coca: A-III. Heroin (smack) is diacetylmorphine, made from morphine extracted from the latex of the poppy, Papaver somniferum: B-IV. Morphine is a very effective sedative and painkiller, useful in patients who have undergone surgery: C-I. Marijuana comes from the inflorescences of Cannabis sativa: D-II. So A-III, B-IV, C-I, D-II.
163	Biotechnology: Principles and ProcessesRestriction enzymes and vectors	Class 12: Biotechnology: Principles and Processes	Single correctEasy
A easy single correct question on restriction enzymes and vectors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes Agrobacterium tumefaciens, a pathogen of many dicot plants, delivers a piece of DNA (T-DNA) into plant cells and transforms them into tumour cells. The plasmid that does this is the tumour-inducing (Ti) plasmid. It has been modified into a cloning vector that is no longer pathogenic but still delivers genes into plants. So Ti stands for tumour inducing.
165	Animal KingdomChordate classes	Class 11: Animal Kingdom	MatchingEasy
A easy matching question on chordate classes, from Animal Kingdom. NCERT: Class 11: Animal Kingdom Pterophyllum is the angel fish (a bony fish): A-III. Myxine is the hagfish (a cyclostome): B-I. Pristis is the saw fish (a cartilaginous fish): C-II. Exocoetus is the flying fish (a bony fish): D-IV. So A-III, B-I, C-II, D-IV.
166	Locomotion and MovementJoints and disorders	Class 11: Locomotion and Movement	MatchingEasy
A easy matching question on joints and disorders, from Locomotion and Movement. NCERT: Class 11: Locomotion and Movement Fibrous joints, such as the sutures between the flat bones of the skull, allow no movement: A-III. Cartilaginous joints, such as those between adjacent vertebrae, allow limited movement: B-I. The knee is a hinge joint, a synovial joint that helps in locomotion: C-IV. The joint between the humerus and the pectoral girdle is a ball and socket joint, allowing rotational movement: D-II. So A-III, B-I, C-IV, D-II.
167	EvolutionEvidences of evolution	Class 12: Evolution	Single correctEasy
A easy single correct question on evidences of evolution, from Evolution. NCERT: Class 12: Evolution Penguins (birds) and dolphins (mammals) are not closely related, yet both have evolved flippers for swimming. Different lineages coming to have similar structures for the same function — analogous organs — is convergent evolution. Divergent evolution (and adaptive radiation) is the opposite: related forms becoming different, giving homologous organs.
168	Biotechnology and its ApplicationsBt crops	Class 12: Biotechnology and its Applications	MatchingEasy
A easy matching question on bt crops, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications Transgenic animals have been used to produce human $\alpha$-1-antitrypsin, a protein used to treat emphysema: A-III. The protein of cryIAb controls the corn borer: B-IV. The proteins of cryIAc (and cryIIAb) control cotton bollworms: C-I. Enzyme replacement therapy — giving functional ADA by injection — is one treatment of ADA deficiency: D-II. So A-III, B-IV, C-I, D-II.
170	Human ReproductionEmbryonic development	Class 12: Human Reproduction	Assertion and reasonEasy
A easy assertion and reason question on embryonic development, from Human Reproduction. NCERT: Class 12: Human Reproduction Both are true. The milk produced in the first few days of lactation, colostrum, contains several antibodies (IgA) absolutely essential to develop resistance in the newborn baby — passive immunity. That is why doctors recommend breast-feeding during the initial period of infant growth for bringing up a healthy baby. So both A and R are correct and R is the correct explanation of A.
171	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 Each Fallopian tube (oviduct) has three parts: the funnel-shaped infundibulum near the ovary, with fimbriae; the wider ampulla; and the narrow isthmus that joins the uterus. The fundus is the upper dome of the uterus itself, not a part of the Fallopian tube.
172	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 Small volumes of culture can be grown in the laboratory; bioreactors are used to produce products on a large scale — vessels of 100 to 1000 litres in which raw materials are converted into products. So statement (1) is incorrect. The others are correct: a bioreactor provides optimal conditions (temperature, pH, substrate, salts, vitamins, oxygen) for the desired product; it has an agitator system, an oxygen delivery system, a foam control system, and temperature and pH control; and stirred-tank bioreactors are the most commonly used.
173	Neural Control and CoordinationCentral nervous system	Class 11: Neural Control and Coordination	MatchingEasy
A easy matching question on central nervous system, from Neural Control and Coordination. NCERT: Class 11: Neural Control and Coordination The pons consists of fibre tracts that interconnect different regions of the brain: A-III. The hypothalamus contains groups of neurosecretory cells that secrete hypothalamic hormones: B-IV. The medulla contains the centres that control respiration, cardiovascular reflexes and gastric secretions: C-II. The cerebellum, with its very convoluted surface, provides the additional space for many more neurons and regulates posture and balance: D-I. So A-III, B-IV, C-II, D-I.
174	BiomoleculesEnzymes	Class 11: Biomolecules	MatchingEasy
A easy matching question on enzymes, from Biomolecules. NCERT: Class 11: Biomolecules Each enzyme hydrolyses the bond that holds its substrate together. Lipase splits fats, whose fatty acids are joined to glycerol by ester bonds: A-II. Nuclease splits nucleic acids, whose nucleotides are joined by phosphodiester bonds: B-IV. Protease splits proteins at peptide bonds: C-I. Amylase splits starch at glycosidic bonds: D-III. So A-II, B-IV, C-I, D-III.
175	Reproductive HealthContraception	Class 12: Reproductive Health	Single correctEasy
A easy single correct question on contraception, from Reproductive Health. NCERT: Class 12: Reproductive Health Natural (traditional) methods avoid the meeting of ovum and sperm without any device or drug: periodic abstinence (avoiding coitus from day 10 to 17 of the cycle), withdrawal or coitus interruptus, and lactational amenorrhea (no ovulation during intense breast-feeding). Vaults are rubber barriers placed over the cervix by the female — a barrier method, not a natural one.
176	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 action potential starts at the SA node, the pacemaker in the right atrium; it spreads through the atria to the AV node; from there it passes along the AV bundle (bundle of His), which divides into the right and left bundle branches in the interventricular septum; these give rise to the Purkinje fibres throughout the ventricular musculature. So E-C-A-D-B.
177	Animal KingdomNon-chordate phyla	Class 11: Animal Kingdom	MatchingEasy
A easy matching question on non-chordate phyla, from Animal Kingdom. NCERT: Class 11: Animal Kingdom Pleurobrachia, the sea walnut, is a ctenophore: A-II. The radula, a file-like rasping organ in the mouth, is a feature of molluscs: B-I. The stomochord, a rudimentary structure in the collar once taken for a notochord, belongs to hemichordates: C-IV. The air bladder, which regulates buoyancy, is found in bony fishes, Osteichthyes: D-III. So A-II, B-I, C-IV, D-III.
178	Human Health and DiseasePathogens and diseases	Class 12: Human Health and Disease	MatchingEasy
A easy matching question on pathogens and diseases, from Human Health and Disease. NCERT: Class 12: Human Health and Disease The common cold is caused by rhinoviruses: A-III. Haemozoin is the toxin released when red blood cells infected by Plasmodium rupture, causing the chill and high fever of malaria: B-I. The Widal test confirms typhoid fever: C-II. Dust mites (like pollen) are common allergens causing allergy: D-IV. So A-III, B-I, C-II, D-IV.
179	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 A is true: annelids have a true coelom lined by mesoderm. B is false: poriferans are at the cellular level of organisation, with no germ-layer-derived body cavity; they are not pseudocoelomates. C is false: aschelminthes (roundworms) are the pseudocoelomates, their body cavity not fully lined by mesoderm. D is false: platyhelminthes are acoelomates. So A only.
180	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 abdomen of the cockroach has 10 segments. In both sexes the 10th segment bears a pair of jointed filamentous structures, the anal cerci. The males also have a pair of short, unjointed anal styles on the 9th sternum, which females lack. So the 10th segment.
181	Human ReproductionMale and female reproductive systems	Class 12: Human Reproduction	StatementsEasy
A easy statements question on male and female reproductive systems, from Human Reproduction. NCERT: Class 12: Human Reproduction Statement I is true: the hymen is often torn during the first coitus, but it can also be broken by a sudden fall or jolt, insertion of a vaginal tampon, or active participation in some sports such as horse-riding or cycling; in some women it persists even after coitus. So its presence or absence is not a reliable indicator of virginity or sexual experience. Statement II is false for the same reasons: the hymen is not torn only during the first coitus. So Statement I is true but Statement II is false.
182	EvolutionHardy-Weinberg principle	Class 12: Evolution	Single correctEasy
A easy single correct question on hardy-weinberg principle, from Evolution. NCERT: Class 12: Evolution Five factors disturb the Hardy-Weinberg equilibrium: gene migration (gene flow), genetic drift, mutation, genetic recombination and natural selection. A constant gene pool is the equilibrium itself — allele frequencies staying the same from generation to generation — so it does not affect it.
183	Biotechnology: Principles and ProcessesRestriction enzymes and vectors	Class 12: Biotechnology: Principles and Processes	Single correctMedium
A medium single correct question on restriction enzymes and vectors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes pBR322 carries two antibiotic-resistance genes, amp$^R$ (with the Pvu I and Pst I sites) and tet$^R$ (with the BamH I and Sal I sites), together with ori and rop in the segment near Pvu II. The two marked regions X and Y lie in that segment: X is ori, the origin of replication, which also controls the copy number of the linked DNA; Y is rop, which codes for the proteins involved in the replication of the plasmid. So X controls the copy number and Y codes for a replication protein: option (4).
184	Reproductive HealthContraception	Class 12: Reproductive Health	MatchingEasy
A easy matching question on contraception, from Reproductive Health. NCERT: Class 12: Reproductive Health The Lippes loop is a non-medicated IUD: A-III. Multiload 375 (like CuT and Cu7) is a copper-releasing IUD: B-I. LNG-20 (like Progestasert) is a hormone-releasing IUD: C-IV. Implants are placed under the skin and contain progestogens alone or with estrogen: D-II. So A-III, B-I, C-IV, D-II.
186	Excretory Products and their EliminationNephron structure	Class 11: Excretory Products and their Elimination	Single correctEasy
A easy single correct question on nephron structure, from Excretory Products and their Elimination. NCERT: Class 11: Excretory Products and their Elimination In the majority of nephrons the loop of Henle is short and dips only a little into the medulla; these are cortical nephrons. In some nephrons the loop is very long and runs deep into the medulla; these are the juxtamedullary nephrons. Their renal corpuscles lie in the cortex, close to the medulla. So (1) is correct.
187	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 Comparing non-chordates with chordates: non-chordates have no notochord (B true); their central nervous system is ventral, solid and double, not dorsal (C false); their pharynx is not perforated by gill slits (A false); the heart, if present, is dorsal (D true); and there is no post-anal tail (E true). So B, D & E only.
189	BiomoleculesEnzymes	Class 11: Biomolecules	Single correctEasy
A easy single correct question on enzymes, from Biomolecules. NCERT: Class 11: Biomolecules The catalytic cycle of an enzyme: first the substrate binds to the active site of the enzyme (E), fitting into it and forming the enzyme-substrate complex (A); the enzyme alters its shape to fit more tightly round the substrate, and the chemical bonds of the substrate are broken (D); the enzyme releases the products (C); and the free enzyme is ready to bind another molecule of substrate and run the cycle again (B). So E, A, D, C, B.
190	Body Fluids and CirculationECG	Class 11: Body Fluids and Circulation	MatchingEasy
A easy matching question on ecg, from Body Fluids and Circulation. NCERT: Class 11: Body Fluids and Circulation The P wave represents the depolarisation of the atria: A-III. The QRS complex represents the depolarisation of the ventricles: B-II. The T wave represents the repolarisation of the ventricles: C-IV. In the T-P gap, between one beat and the next, the heart muscle is electrically silent (the flat baseline): D-I. So A-III, B-II, C-IV, D-I.
191	Human Health and DiseaseImmunity	Class 12: Human Health and Disease	StatementsMedium
A medium statements question on immunity, from Human Health and Disease. NCERT: Class 12: Human Health and Disease Statement I is correct: the bone marrow is the main (primary) lymphoid organ, where all blood cells, including lymphocytes, are produced. Statement II is correct: the T-lymphocytes arise in the bone marrow and mature in the thymus; both organs provide the micro-environments for the development and maturation of T-lymphocytes. So both statements are correct.
192	Structural Organisation in AnimalsAnimal tissues	Class 11: Structural Organisation in Animals	MatchingEasy
A easy matching question on animal tissues, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals Unicellular glands are isolated glandular cells, such as the goblet cells of the alimentary canal: A-III. Compound (stratified) epithelium, with several layers, protects surfaces such as the moist surface of the buccal cavity and pharynx: B-IV. Multicellular glands are clusters of cells, such as the salivary glands: C-I. Endocrine glandular tissue secretes hormones into the blood — as the pancreas does from its islets: D-II. So A-III, B-IV, C-I, D-II.
193	Neural Control and CoordinationCentral nervous system	Class 11: Neural Control and Coordination	StatementsEasy
A easy statements question on central nervous system, from Neural Control and Coordination. NCERT: Class 11: Neural Control and Coordination Statement I is correct: the two cerebral hemispheres are connected by a tract of nerve fibres, the corpus callosum. Statement II is incorrect: the brain stem is formed by the midbrain, pons and medulla oblongata. The cerebrum is part of the forebrain, not the brain stem. So Statement I is correct but Statement II is incorrect.
195	EvolutionOrigin of life	Class 12: Evolution	MatchingMedium
A medium matching question on origin of life, from Evolution. NCERT: Class 12: Evolution On the geological time scale: The Mesozoic Era is the age of reptiles, in which the birds also arose: A-III. The Proterozoic Era saw the early, lower invertebrates: B-I. The Cenozoic Era is the age of mammals: C-IV. The Paleozoic Era saw the rise of fishes and amphibians: D-II. So A-III, B-I, C-IV, D-II.
197	Chemical Coordination and IntegrationHormonal disorders	Class 11: Chemical Coordination and Integration	MatchingEasy
A easy matching question on hormonal disorders, from Chemical Coordination and Integration. NCERT: Class 11: Chemical Coordination and Integration Exophthalmic goitre (Graves' disease) is hypersecretion of thyroid hormone with protruding eyeballs: A-III. Acromegaly is excess growth hormone in adults: B-IV. Cushing's syndrome is excess cortisol, with a moon face and hyperglycaemia: C-I. Cretinism is hyposecretion of thyroid hormone in childhood, with stunted growth: D-II. So A-III, B-IV, C-I, D-II.
198	Human ReproductionGametogenesis	Class 12: Human Reproduction	Single correctMedium
A medium single correct question on gametogenesis, from Human Reproduction. NCERT: Class 12: Human Reproduction At puberty the hypothalamus releases GnRH, which makes the anterior pituitary secrete two gonadotropins, LH and FSH. LH acts on the Leydig cells and stimulates them to synthesise androgens, which in turn stimulate spermatogenesis: so (B) is Leydig cells. FSH, the other gonadotropin, is (A). It acts on the Sertoli cells, (C), and stimulates the secretion of some factors which help in the process of spermiogenesis, (D) — the transformation of spermatids into sperms. So (A) FSH, (B) Leydig cells, (C) Sertoli cells, (D) spermiogenesis.
200	Structural Organisation in AnimalsFrog: morphology and anatomy	Class 11: Structural Organisation in Animals	MatchingEasy
A easy matching question on frog: morphology and anatomy, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals In the cockroach, the crop, a sac-like part of the foregut, stores food: A-IV. At the junction of the foregut and midgut there is a ring of 6–8 blind tubules, the hepatic or gastric caeca, which secrete digestive juice: B-II. At the junction of the midgut and hindgut there is a ring of 100–150 yellow, thin filaments, the Malpighian tubules, which remove nitrogenous wastes: C-III. The gizzard (proventriculus), with its chitinous teeth, grinds the food: D-I. So A-IV, B-II, C-III, D-I.

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