NEET (UG) · 2021

NEET 2021, 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. Electrostatics6
2. Electromagnetic Induction and Alternating Currents4
3. Optics4
35 easy · 14 medium · 1 hard
Chemistry
50 Q
1. Chemical Bonding and Molecular Structure3
2. p-Block Elements3
3. Hydrocarbons3
40 easy · 10 medium · 0 hard
Botany
56 Q
1. Molecular Basis of Inheritance9
2. Organisms and Populations5
3. Cell Cycle and Cell Division5
47 easy · 9 medium · 0 hard
Zoology
44 Q
1. Biotechnology: Principles and Processes7
2. Biotechnology and its Applications5
3. Biomolecules4
34 easy · 10 medium · 0 hard

Chapters

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

Molecular Basis of Inheritance9
Class 12: Molecular Basis of Inheritance
Organisms and Populations5
Class 12: Organisms and Populations
Cell Cycle and Cell Division5
Class 11: Cell Cycle and Cell Division
Plant Kingdom5
Class 11: Plant Kingdom
Plant Growth and Development4
Class 11: Plant Growth and Development
Cell: The Unit of Life4
Class 11: Cell: The Unit of Life
Anatomy of Flowering Plants3
Class 11: Anatomy of Flowering Plants
Sexual Reproduction in Flowering Plants3
Class 12: Sexual Reproduction in Flowering Plants
Principles of Inheritance and Variation3
Class 12: Principles of Inheritance and Variation
Ecosystem3
Class 12: Ecosystem
Morphology of Flowering Plants2
Class 11: Morphology of Flowering Plants
Photosynthesis in Higher Plants2
Class 11: Photosynthesis in Higher Plants
Transport in Plants1
No matching NCERT chapter
Respiration in Plants1
Class 11: Respiration in Plants
Biological Classification1
Class 11: Biological Classification
Mineral Nutrition1
No matching NCERT chapter
The Living World1
Class 11: The Living World
Microbes in Human Welfare1
Class 12: Microbes in Human Welfare
Strategies for Enhancement in Food Production1
No matching NCERT chapter
Environmental Issues1
No matching NCERT chapter

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	Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Voltages quoted for an ac circuit (what an ac voltmeter reads) are rms values, so take $V_L = 40$ V, $V_C = 10$ V, $V_R = 40$ V as rms. In a series LCR circuit the voltage across R is in phase with the current, while $V_L$ and $V_C$ are opposite to each other and $90^\circ$ from it, so the source voltage is $V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{40^2 + 30^2} = 50$ V (rms). The current amplitude is $I_0 = 10\sqrt{2}$ A, so the rms current is $I = \frac{I_0}{\sqrt{2}} = 10$ A. $Z = \frac{V}{I} = \frac{50}{10} = 5\ \Omega$. Check another way: $R = \frac{V_R}{I} = 4\ \Omega$, $X_L - X_C = \frac{30}{10} = 3\ \Omega$, so $Z = \sqrt{4^2 + 3^2} = 5\ \Omega$. Answer: (4) 5 $\Omega$.
2	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 a right-angled triangle with angles $90^\circ$ (bottom left), $60^\circ$ (bottom right) and so $30^\circ$ at the top. The ray falls normally on the slanting face, so it enters without bending. Inside, the ray travels perpendicular to the slanting face. The slanting face makes $60^\circ$ with the base, so the ray makes $30^\circ$ with the base, going down to the left, and strikes the vertical face (it enters near the top, so it reaches the vertical face before the base). The normal to the vertical face is horizontal, so the angle of incidence there is $i = 30^\circ$ (equal to the prism angle at the top). Critical angle: $\sin C = \frac{1}{\sqrt{3}} \approx 0.577$, so $C \approx 35^\circ$. Since $30^\circ < C$, the ray comes out. Snell's law at the vertical face: $\sqrt{3}\sin 30^\circ = 1\cdot\sin e \Rightarrow \sin e = \frac{\sqrt{3}}{2} \Rightarrow e = 60^\circ$. Answer: (1) $60^\circ$.
3	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 The field lines spread out from a point on the left, so the field is stronger on the left and weaker on the right. In the figure $+q$ is on the left and $-q$ on the right, so the dipole moment (from $-q$ to $+q$) points to the left, against $\vec{E}$. Forces: $+q$ (in the stronger field $E_1$) is pushed to the right with $qE_1$; $-q$ (in the weaker field $E_2$) is pushed to the left with $qE_2$. Since $E_1 > E_2$, the net force $q(E_1 - E_2)$ is to the right. Energy view: $U = -\vec{p}\cdot\vec{E} = +pE$ for an antiparallel dipole, which falls as $E$ falls; so moving right, into the weaker field, lowers its potential energy, as a free body should. Answer: (2) towards the right as its potential energy will decrease.
4	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 The displacement current between the plates equals the conduction current in the wires: $I_d = \varepsilon_0\frac{d\Phi_E}{dt} = \frac{dq}{dt}$. With $q = CV = CV_0\sin\omega t$, $I_d = \frac{dq}{dt} = CV_0\omega\cos\omega t$. Check with units: $\omega C$ is $1/X_C$, so $V_0\omega C = V_0/X_C$ is a current, as it must be. Answer: (1) $I_d = V_0\omega C\cos\omega t$.
5	Magnetic Effects of Current and MagnetismAmpere's law	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctEasy
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Use Ampere's law on a circle of radius $r$ about the axis. Inside ($r < R$): the current enclosed is $I\frac{r^2}{R^2}$, so $B\cdot2\pi r = \mu_0 I\frac{r^2}{R^2}$, giving $B = \frac{\mu_0 I r}{2\pi R^2}$: B rises in a straight line from zero. Outside ($r > R$): all of $I$ is enclosed, so $B = \frac{\mu_0 I}{2\pi r}$: B falls off as $1/r$. At $r = R$ both give $\frac{\mu_0 I}{2\pi R}$, the peak. So the graph rises linearly to a maximum at the surface and then falls as a curve. Answer: (3).
6	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics A parallel beam through the convex lens A converges towards its focus, 20 cm beyond A. For the concave lens B to send it out parallel, the rays must be heading towards B's own focal point, which lies 5 cm beyond B. So the two foci must coincide: $d + 5 = 20$, giving $d = 15$ cm. Check with the combination formula: the system gives a parallel beam when its power is zero, $\frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} = 0$: $\frac{1}{20} - \frac{1}{5} + \frac{d}{100} = 0 \Rightarrow d = 15$ cm. Answer: (2) 15.
7	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	Single correctMedium
A medium single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter With zero work function the whole photon energy becomes the electron's kinetic energy: $\frac{hc}{\lambda} = \frac{p^2}{2m}$. The electron's de Broglie wavelength is $\lambda_d = \frac{h}{p}$, so $p = \frac{h}{\lambda_d}$ and $\frac{hc}{\lambda} = \frac{h^2}{2m\lambda_d^2} \Rightarrow \lambda = \frac{2mc\lambda_d^2}{h}$. Check dimensions: $\frac{mc}{h}$ has units $\frac{\text{kg m s}^{-1}}{\text{J s}} = \text{m}^{-1}$, so $\frac{mc}{h}\lambda_d^2$ is a length. Answer: (3) $\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2$.
8	Current ElectricityDrift velocity and mobility	Class 12: Current Electricity	MatchingEasy
A easy matching question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity (A) Drift velocity: an electron accelerates at $\frac{eE}{m}$ for an average time $\tau$, so $\upsilon_d = \frac{eE}{m}\tau$ → (R). (B) Resistivity: from $J = \sigma E$, $\rho = \frac{1}{\sigma} = \frac{E}{J}$ → (S). (C) Relaxation period: $\rho = \frac{m}{ne^2\tau}$, so $\tau = \frac{m}{ne^2\rho}$ → (P). (D) Current density: $J = ne\upsilon_d$ → (Q). Answer: (1) (A)-(R), (B)-(S), (C)-(P), (D)-(Q).
9	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Follow the atomic number at each step. $Z \to Z-1$: down by 1. $\beta^+$ emission turns a proton into a neutron, lowering Z by 1. $Z-1 \to Z-3$: down by 2. An $\alpha$ particle carries away 2 protons. $Z-3 \to Z-2$: up by 1. $\beta^-$ emission turns a neutron into a proton, raising Z by 1. So the sequence is $\beta^+$, $\alpha$, $\beta^-$. Answer: (3).
10	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 The four wires are identical, each of resistance $R$. In parallel: $\frac{R}{4} = 0.25\ \Omega$, so $R = 1\ \Omega$. In series: $4R = 4\ \Omega$. (Equivalently, series/parallel $= n^2 = 16$, and $16\times0.25 = 4\ \Omega$.) Answer: (4) 4 $\Omega$.
11	Work, Energy and PowerConservation of energy	Class 11: Work, Energy and Power	Single correctEasy
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Take potential energy zero at the ground. Total energy $= mgS$. At height $h$: $K = 3U = 3mgh$, so $mgh + 3mgh = mgS \Rightarrow h = \frac{S}{4}$. Speed: $\frac{1}{2}mv^2 = 3mg\frac{S}{4} \Rightarrow v^2 = \frac{3gS}{2} \Rightarrow v = \sqrt{\frac{3gS}{2}}$. Check by free fall: it has fallen $S - \frac{S}{4} = \frac{3S}{4}$, and $v^2 = 2g\cdot\frac{3S}{4} = \frac{3gS}{2}$. Answer: (4).
12	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Activity falls as $A = A_0\left(\frac{1}{2}\right)^{t/T_{1/2}}$. $\frac{t}{T_{1/2}} = \frac{150}{100} = \frac{3}{2}$, so $\frac{A}{A_0} = \left(\frac{1}{2}\right)^{3/2} = \frac{1}{2\sqrt{2}}$. Check: after 100 h it is $\frac{1}{2}$, after a further 50 h (half a half-life) it is multiplied by $\frac{1}{\sqrt{2}}$: $\frac{1}{2}\cdot\frac{1}{\sqrt{2}} \approx 0.354$. Answer: (2) $\frac{1}{2\sqrt{2}}$.
13	Properties of Bulk MatterHeat transfer	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctMedium
A medium single correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Use Newton's law of cooling in its average form: $\frac{\Delta T}{\Delta t} = k\,(T_{avg} - T_0)$. First cooling: $\frac{90 - 80}{t} = k(85 - 20) = 65k$. Second cooling: $\frac{80 - 60}{t'} = k(70 - 20) = 50k$. Dividing: $\frac{20/t'}{10/t} = \frac{50}{65} \Rightarrow \frac{2t}{t'} = \frac{10}{13} \Rightarrow t' = \frac{26}{10}t = \frac{13}{5}t$. Check with the exact law: $t'/t = \frac{\ln(60/40)}{\ln(70/60)} = \frac{0.405}{0.154} \approx 2.63$, close to $\frac{13}{5} = 2.6$; the average form is the one the options use. Answer: (2) $\frac{13}{5}t$.
14	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Energy of one photon: $E = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3\times10^8}{600\times10^{-9}} = \frac{19.8\times10^{-26}}{6\times10^{-7}} = 3.3\times10^{-19}$ J. Photons per second: $N = \frac{P}{E} = \frac{3.3\times10^{-3}}{3.3\times10^{-19}} = 10^{16}$. Answer: (3) $10^{16}$.
15	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 With $x = A\sin(2\pi nt)$, the potential energy is $U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\sin^2(2\pi nt) = \frac{1}{4}kA^2\left[1 - \cos(2\pi\cdot2n\,t)\right]$. So U repeats with frequency $2n$: it reaches its maximum at both extreme positions in each cycle. Answer: (2) 2n.
16	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctEasy
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Field of the long wire at $r = 0.2$ m: $B = \frac{\mu_0 I}{2\pi r} = \frac{2\times10^{-7}\times5}{0.2} = 5\times10^{-6}$ T, perpendicular to the plane containing the wire and the electron. The electron moves parallel to the wire, so its velocity is perpendicular to B, and $F = evB = 1.6\times10^{-19}\times10^5\times5\times10^{-6} = 8\times10^{-20}$ N. Answer: (4) $8\times10^{-20}$ N.
17	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 Energy = force × distance, and a distance can be written as acceleration × time²: $[L] = [A][T^2]$. So $[E] = [F][L] = [F][A][T^2]$. Check in M, L, T: $[F][A][T^2] = MLT^{-2}\cdot LT^{-2}\cdot T^2 = ML^2T^{-2}$, which is energy. Answer: (2) [F] [A] [$T^2$].
18	Behaviour of Perfect Gas and Kinetic TheoryKinetic theory	Class 11: Kinetic Theory	MatchingEasy
A easy matching question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory (A) rms speed: $v_{rms} = \sqrt{\frac{3RT}{M}}$ → (Q). (B) Pressure of an ideal gas: $P = \frac{1}{3}nm\bar{v}^2$ ($n$ molecules per unit volume) → (P). (C) Average (translational) kinetic energy of a molecule: $\frac{3}{2}k_BT$ → (S). (D) A diatomic molecule has 5 degrees of freedom at ordinary temperatures, so one mole has $U = \frac{5}{2}RT$ → (R). Answer: (3) (A) - (Q), (B) - (P), (C) - (S), (D) - (R).
19	KinematicsMotion in a straight line	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane On a smooth incline the acceleration $a = g\sin\theta$ is constant and the block starts from rest, so the distance covered up to time $t$ is $\frac{1}{2}at^2$. Distance in the $n$-th second: $S_n = \frac{1}{2}a[n^2 - (n-1)^2] = \frac{a}{2}(2n - 1)$. Likewise $S_{n+1} = \frac{a}{2}(2n + 1)$. $\frac{S_n}{S_{n+1}} = \frac{2n-1}{2n+1}$ (Galileo's odd-number rule 1 : 3 : 5 …; for $n = 1$ it gives $\frac{1}{3}$, as it should). Answer: (2).
20	Atoms and NucleiNuclear binding energy	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Binding energy before: $240\times7.6 = 1824$ MeV. After: two fragments of 120 nucleons each, $2\times120\times8.5 = 2040$ MeV. Gain: $2040 - 1824 = 216$ MeV. Check: the gain per nucleon is $8.5 - 7.6 = 0.9$ MeV, and $0.9\times240 = 216$ MeV. Answer: (4) 216 MeV.
21	Experimental SkillsVernier calipers and screw gauge	No matching NCERT chapter	Single correctEasy
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{1\text{ mm}}{100} = 0.01$ mm. Diameter $=$ main scale reading $+$ circular scale reading × least count $= 0 + 52\times0.01 = 0.52$ mm $= 0.052$ cm. Answer: (4) 0.052 cm.
22	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 Look at the wires on the right: the top wire and the bottom wire are joined by the plain vertical wire at the right end, which also goes to the right terminal. The middle capacitor is connected between the top wire and the bottom wire to the right of the two outer capacitors, so both its plates are at the same potential: it is short-circuited and holds no charge. What remains is the top C and the bottom C, each between the left node and the right node: in parallel, $C + C = 2C$. Answer: (2) 2C.
23	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 A large focal length of the objective gives a large magnifying power ($m = \frac{f_o}{f_e}$). A large aperture: - gathers more light (light gathering power grows as the area, $\propto D^2$), so faint objects become visible and images are brighter; - improves resolving power, since the smallest resolvable angle is $\theta \approx \frac{1.22\lambda}{D}$; - and so improves the quality and visibility of the image. All three statements are true. Answer: (4) all of the above.
24	ElectrostaticsConductors and dielectrics	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Connected by a wire, the spheres come to the same potential (taking them far apart): $\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \Rightarrow \frac{Q_1}{Q_2} = \frac{R_1}{R_2}$. $\sigma = \frac{Q}{4\pi R^2}$, so $\frac{\sigma_1}{\sigma_2} = \frac{Q_1}{Q_2}\cdot\frac{R_2^2}{R_1^2} = \frac{R_1}{R_2}\cdot\frac{R_2^2}{R_1^2} = \frac{R_2}{R_1}$. (This is why charge crowds at sharp points: smaller radius, larger density.) Answer: (2) $\frac{R_2}{R_1}$.
25	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 Spring constant: $k = \frac{F}{x} = \frac{10}{0.05} = 200$ N/m. $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = 2\pi\times0.1 = 0.628$ s. Answer: (4) 0.628 s.
26	Electromagnetic WavesProperties of electromagnetic waves	Class 12: Electromagnetic Waves	Single correctMedium
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves In a plane EM wave, E and B are perpendicular to each other and $\vec{E}\times\vec{B}$ points along the direction of travel, $+\hat{i}$. (1) and (4): E and B are parallel, $\vec{E}\times\vec{B} = 0$. Not possible. (3): B is opposite to E, again $\vec{E}\times\vec{B} = 0$. (2): $(-\hat{j}+\hat{k})\times(-\hat{j}-\hat{k}) = (\hat{j}\times\hat{k}) - (\hat{k}\times\hat{j}) = \hat{i} + \hat{i} = 2\hat{i}$. Perpendicular ($\vec{E}\cdot\vec{B} = 1 - 1 = 0$) and along $+x$. Answer: (2).
27	GravitationEscape velocity	Class 11: Gravitation	Single correctEasy
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}}$ with $M = \frac{4}{3}\pi R^3\rho$, so $v_e = \sqrt{\frac{8}{3}\pi G\rho}\,R \propto R$ at fixed density. Four times the radius gives four times the escape velocity: $4\upsilon$. Answer: (4) 4 $\upsilon$.
28	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 At balance no current flows through the cell, and its emf equals the fall of potential along the balancing length: $E \propto l$. $\frac{l_2}{l_1} = \frac{E_2}{E_1} \Rightarrow l_2 = 36\times\frac{2.5}{1.5} = 60$ cm. Answer: (1) 60 cm.
29	Properties of Bulk MatterViscosity	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 viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal velocity the net force is zero: weight = buoyancy + viscous force. Volume of the ball $V = \frac{M}{d}$, so the buoyant force is $V\cdot\frac{d}{2}\cdot g = \frac{Mg}{2}$. Viscous force $= Mg - \frac{Mg}{2} = \frac{Mg}{2}$. Answer: (1) $\frac{Mg}{2}$.
30	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 Energy density of an electric field: $u = \frac{1}{2}\varepsilon_0E^2$. The field fills the volume $Ad$ between the plates, so $U = \frac{1}{2}\varepsilon_0E^2Ad$. Check: $U = \frac{1}{2}CV^2$ with $C = \frac{\varepsilon_0A}{d}$ and $V = Ed$ gives $\frac{1}{2}\frac{\varepsilon_0A}{d}E^2d^2 = \frac{1}{2}\varepsilon_0E^2Ad$. Answer: (3).
31	Electronic DevicesSemiconductors	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctEasy
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The current is $I = neA\upsilon_d = neA\mu E$. The carrier concentrations are equal and the field is the same, so the current depends on the mobility of the majority carriers. Electrons (majority carriers in n-type) have a larger mobility than holes (majority carriers in p-type), so the n-type sample carries the larger current. Answer: (3) current in n-type > current in p-type.
32	Electronic DevicesSpecial-purpose diodes	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	StatementsMedium
A medium statements question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits (A) A zener diode regulates voltage in its breakdown region, which is reached in reverse bias: across the breakdown the voltage stays nearly constant while the current changes widely. True. (B) The barrier potential of a p-n junction is a few tenths of a volt: about 0.1 to 0.3 V is the range usually quoted at this level (germanium about 0.3 V); silicon junctions are higher, about 0.7 V. Read strictly, a p-n junction's barrier is about 0.3 V for germanium and about 0.7 V for silicon, so the range 0.1 V to 0.3 V is not right in general: (B) is incorrect. So (A) is correct and (B) is incorrect. Answer: (3).
33	ElectrostaticsConductors and dielectrics	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In a polar molecule (water, HCl) the centres of positive and negative charge do not coincide, so it has a permanent electric dipole moment even with no field applied. (1) describes non-polar molecules with no field; (2) is a non-polar molecule being polarised (an induced dipole); (3) has nothing to do with polarity. Answer: (4) having a permanent electric dipole moment.
34	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 $[E] = ML^2T^{-2}$. From $F = \frac{Gm_1m_2}{r^2}$, $[G] = \frac{MLT^{-2}\cdot L^2}{M^2} = M^{-1}L^3T^{-2}$. $\left[\frac{E}{G}\right] = \frac{ML^2T^{-2}}{M^{-1}L^3T^{-2}} = M^2L^{-1}T^0$. Check: $E = \frac{Gm^2}{r}$ for gravitational energy, so $\frac{E}{G} = \frac{m^2}{r}$, which is $M^2L^{-1}$. Answer: (1).
35	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 Input power $= \frac{dm}{dt}gh = 15\times10\times60 = 9000$ W. 10% is lost, so the turbine delivers 90% of it: $0.9\times9000 = 8100$ W $= 8.1$ kW. Answer: (2) 8.1 kW.
36	KinematicsProjectile motion	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane At $t = 4$ s the car's speed is $5\times4 = 20$ m/s, and the ball leaves the window with this horizontal velocity. Once released, the only force on the ball is gravity (air resistance ignored): its acceleration is $g = 10$ m/s$^2$ downward, and its horizontal velocity stays 20 m/s. At $t = 6$ s it has fallen for 2 s: vertical velocity $= 10\times2 = 20$ m/s. Speed $= \sqrt{20^2 + 20^2} = 20\sqrt{2}$ m/s; acceleration 10 m/s$^2$. Answer: (4).
37	Electronic DevicesLogic gates	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctMedium
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits From the circuit: A and B go to an AND gate; B and C go to a NAND gate; the two outputs go to an OR gate. So $y = A\cdot B + \overline{B\cdot C}$. From the waveforms: B is high only between $t_2$ and $t_4$, and C is low exactly between $t_2$ and $t_4$ (high at all other times). So B and C are never high together: $B\cdot C = 0$ at every instant, and $\overline{B\cdot C} = 1$ always. An OR gate with one input permanently 1 gives $y = 1$ whatever the AND gate does. So y stays at 5 V throughout. Answer: (3).
38	Laws of MotionNewton's laws	Class 11: Laws of Motion	Single correctEasy
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Speed on reaching the ground: $v = \sqrt{2gh} = \sqrt{200} = 10\sqrt{2} \approx 14.14$ m/s. It rebounds to the same height, so it leaves the ground with the same speed upward. Impulse = change of momentum $= m[v - (-v)] = 2mv = 2\times0.15\times14.14 \approx 4.24$ kg m/s. Answer: (2) 4.2 kg m/s.
39	Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium single correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take moments about the wedge at the 40 cm mark (g cancels). Left of the wedge: the 2 kg mass at 20 cm, 20 cm away: moment $2\times20 = 40$ (kg cm), anticlockwise. Right of the wedge: the rod's own weight acts at its centre, the 100 cm mark, 60 cm away: $0.5\times60 = 30$; and m at 160 cm, 120 cm away: $120m$. Both clockwise. Balance: $40 = 30 + 120m \Rightarrow m = \frac{10}{120} = \frac{1}{12}$ kg. Answer: (4) $\frac{1}{12}$ kg.
40	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctHard
A hard single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Refraction at the lens: $u = -60$ cm, $f = +30$ cm. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{30} - \frac{1}{60} = \frac{1}{60}$, so $v = 60$ cm: the lens would form a real image 60 cm to its right. The mirror is 40 cm from the lens, so the converging rays meet it first. They were heading for a point $60 - 40 = 20$ cm behind the mirror: a virtual object 20 cm behind it. The lens alone would form its image 60 cm from the lens, 20 cm behind the mirror. The mirror reflects the converging rays, and the final image, taken as the mirror's image of that point, is 20 cm from the plane mirror; following the reflected light back through the lens, the rays diverge into it and the image seen is virtual. Answer: (4).
41	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 With no losses, input power = output power = 44 W. Primary current: $I_p = \frac{P}{V_p} = \frac{44}{220} = 0.2$ A. Check: the secondary current is $\frac{44}{11} = 4$ A, and $I_p = I_s\frac{V_s}{V_p} = 4\times\frac{11}{220} = 0.2$ A. Answer: (1) 0.2 A.
42	Current ElectricityKirchhoff's rules	Class 12: Current Electricity	Single correctEasy
A easy single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity The current $i_1$ through $r_1$ divides between $r_2$ and $r_3$, which are in parallel: $i_1 = i_2 + i_3$ and $i_2r_2 = i_3r_3$. So $i_3 = i_1\frac{r_2}{r_2 + r_3}$ (a branch takes the share set by the other branch's resistance), and $\frac{i_3}{i_1} = \frac{r_2}{r_2 + r_3}$. Check: if $r_3 \to 0$ all the current goes through $r_3$ and the ratio becomes 1, as it does. Answer: (2).
43	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctMedium
A medium single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $\vec{\upsilon}\times\vec{B} = (2, 4, 6)\times(B, B, B_0) = (4B_0 - 6B)\hat{i} + (6B - 2B_0)\hat{j} + (2B - 4B)\hat{k}$. Set this equal to $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}$ (q = 1): $\hat{k}$: $-2B = 12 \Rightarrow B = -6$. $\hat{i}$: $4B_0 + 36 = 4 \Rightarrow B_0 = -8$. $\hat{j}$ (check): $6(-6) - 2(-8) = -36 + 16 = -20$. ✓ So $\vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k}$. (Also $\vec{F}\cdot\vec{B} = -24 + 120 - 96 = 0$, as a magnetic force must be perpendicular to B.) Answer: (2).
44	GravitationGravitational potential energy	Class 11: Gravitation	Single correctMedium
A medium single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation Energy conservation between the surface and the highest point (height h, at rest): $\frac{1}{2}mk^2V_e^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$. With $V_e^2 = \frac{2GM}{R}$: $\frac{k^2GM}{R} - \frac{GM}{R} = -\frac{GM}{R+h} \Rightarrow \frac{1-k^2}{R} = \frac{1}{R+h}$. $R + h = \frac{R}{1-k^2} \Rightarrow h = \frac{Rk^2}{1-k^2}$. Check: for small k this is about $Rk^2 = \frac{v^2}{2g}$ (since $k^2V_e^2 = 2gRk^2$), the familiar result near the surface; and $h \to \infty$ as $k \to 1$. Answer: (4).
45	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 Volume is conserved: $27\cdot\frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow R = 3r$. Charge adds: $Q = 27q$. $V = \frac{kQ}{R} = \frac{27kq}{3r} = 9\cdot\frac{kq}{r} = 9\times220 = 1980$ V. (In general $V = n^{2/3}v$, and $27^{2/3} = 9$.) Answer: (4) 1980 V.
46	Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Resonance: $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5\times80\times10^{-6}}} = \frac{1}{\sqrt{4\times10^{-4}}} = \frac{1}{0.02} = 50$ rad/s. The half-power points are where $\|X_L - X_C\| = R$; they are separated by the bandwidth $\Delta\omega = \frac{R}{L} = \frac{40}{5} = 8$ rad/s and lie (very nearly) $\frac{R}{2L} = 4$ rad/s either side of $\omega_0$: $\omega = 50 \pm 4$, that is 46 rad/s and 54 rad/s. Check: at 46 rad/s, $X_L = 230\ \Omega$, $X_C = \frac{1}{46\times80\times10^{-6}} \approx 271.7\ \Omega$, difference $\approx 41.7\ \Omega \approx R$, so the power is about half. (The exact roots, 46.2 and 54.2 rad/s, round to the same pair.) Answer: (3) 46 rad/s and 54 rad/s.
47	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 Magnetic moment of a coil: $m = NIA$, the same current I flowing in both cases. (i) One triangular turn uses $3a$ of wire, so $N = \frac{12a}{3a} = 4$ turns. Area $= \frac{\sqrt{3}}{4}a^2$. $m = 4I\cdot\frac{\sqrt{3}}{4}a^2 = \sqrt{3}Ia^2$. (ii) One square turn uses $4a$, so $N = 3$. Area $= a^2$. $m = 3Ia^2$. Answer: (1) $\sqrt{3}Ia^2$ and $3Ia^2$.
48	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 Every bit of a ring is at distance R from this axis, so its moment of inertia is (mass) × $R^2$ whatever its shape. Removing a 90° arc removes a quarter of the mass; the remaining $\frac{3M}{4}$ is still all at R: $I = \frac{3}{4}MR^2$, so $K = \frac{3}{4}$. Answer: (1).
49	Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctEasy
A easy single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Pass a current I through the large loop. At its centre the field is $B = \frac{\mu_0 I}{2R_1}$, nearly uniform over the small loop since $R_1 >> R_2$. Flux through the small loop: $\Phi = B\pi R_2^2 = \frac{\mu_0\pi R_2^2 I}{2R_1}$. $M = \frac{\Phi}{I} = \frac{\mu_0\pi R_2^2}{2R_1} \propto \frac{R_2^2}{R_1}$. Answer: (4).
50	KinematicsProjectile motion	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctMedium
A medium single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Speed in the circle: $v = \frac{2\pi R}{T}$. Maximum height of a projectile: $H = \frac{v^2\sin^2\theta}{2g} = 4R$, so $\sin^2\theta = \frac{8gR}{v^2} = \frac{8gR\,T^2}{4\pi^2R^2} = \frac{2gT^2}{\pi^2R}$. $\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2R}\right)^{1/2}$. Check dimensions: $gT^2$ is a length, so $\frac{gT^2}{R}$ is a pure number, as an argument of $\sin^{-1}$ must be. Answer: (4).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
51	Chemistry in Everyday LifeDrugs and their classification	No matching NCERT chapter	StatementsEasy
A easy statements question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Analgesics are of two kinds. Non-narcotic (non-addictive) analgesics such as aspirin and paracetamol relieve pain and fever without sleep or addiction. So Statement I, which calls them narcotic, is false. Narcotic analgesics are the opium alkaloids and their derivatives, such as morphine, codeine and heroin, which cause sleep and addiction. So Statement II, which calls them non-narcotic, is false. Answer: (2) Both Statement I and Statement II are false.
52	Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctMedium
A medium single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Dec-4-ene is a 10-carbon chain with the double bond between C4 and C5; the methyl groups sit on C2 and C6. Structure (1): start at the isopropyl end. The carbon carrying two methyls is C2 (one methyl is C1, the other the 2-methyl). Then C3 (CH$_2$), the double bond C4=C5, then C6, which carries a methyl, and the chain runs on C7, C8, C9, C10. Ten carbons, double bond at 4, methyls at 2 and 6: 2,6-dimethyldec-4-ene. Structure (2): the longest chain runs from the isopropyl end through a methyl-bearing carbon and down into the lower chain; the double bond is near the far end. Numbered to give the double bond the lower locant, it is 6,9-dimethyldec-3-ene, not the name. Structure (3): the double bond lies between C3 and C4 (next to the isopropyl carbon): 2,6-dimethyldec-3-ene. Structure (4): the methyl sits on a carbon of the double bond, which lies between C5 and C6: 2,6-dimethyldec-5-ene, not dec-4-ene. Answer: (1).
53	Chemical Bonding and Molecular StructureHybridisation	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Boron has 3 valence electrons and forms three B–F bonds with no lone pair: 3 electron pairs around B, so it is $sp^2$ hybridised and trigonal planar. Electrons around B: 3 bond pairs = 6 electrons, short of an octet, which is why $BF_3$ is electron deficient (a Lewis acid). Answer: (3) $sp^2$ and 6.
54	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Noble gases are monatomic and held together only by weak dispersion (London) forces, so (3) is true and, because of that, their melting and boiling points are very low (helium boils at about 4 K). Statement (2), very high melting and boiling points, is incorrect. (1) They are only sparingly soluble in water: true. (4) With complete shells, an added electron must enter a new shell, so their electron gain enthalpies are large and positive: true. Answer: (2).
55	ElectrochemistryConductance	Class 12: Electrochemistry	Single correctEasy
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Kohlrausch's law: each ion contributes independently at infinite dilution, so $\Lambda^\circ(CH_3COOH) = \Lambda^\circ(CH_3COONa) + \Lambda^\circ(HCl) - \Lambda^\circ(NaCl)$ (the $Na^+$ and $Cl^-$ contributions cancel). $= 91.0 + 426.16 - 126.45 = 390.71$ S cm$^2$ mol$^{-1}$. Check: $517.16 - 126.45 = 390.71$. Answer: (2).
56	Surface ChemistryColloids and their preparation	No matching NCERT chapter	Single correctEasy
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter The Tyndall effect is the scattering of light by particles of colloidal size (about 1 to 1000 nm). NaCl, glucose and urea form true solutions whose particles are far too small to scatter light visibly. Starch forms a colloidal (lyophilic) sol, so a beam of light passing through it becomes visible. Answer: (3) Starch solution.
57	BiomoleculesNucleic acids and vitamins	Class 12: Biomolecules	Single correctEasy
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Vitamin $B_{12}$ (cyanocobalamin) is needed to make red blood cells; its lack causes pernicious anaemia (RBC deficient in haemoglobin). For comparison: $B_1$ deficiency causes beriberi, $B_2$ cheilosis, and $B_6$ convulsions. Answer: (1) Vitamin $B_{12}$.
58	HydrocarbonsConformations of ethane	Class 11: Hydrocarbons	Single correctEasy
A easy single correct question on conformations of ethane, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The least stable conformer of ethane is the eclipsed one, where each C–H bond on the front carbon lies directly in front of a C–H bond on the back carbon, giving the greatest torsional strain. In the Newman projection the dihedral angle between eclipsing bonds is $0^\circ$ (the staggered, most stable form has $60^\circ$). Answer: (4) $0^\circ$.
59	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 (1) True: 5f electrons shield even more poorly than 4f, so the contraction from one actinoid to the next is greater than in the lanthanoids. (2) False: most trivalent lanthanoid ions are coloured in the solid state and in solution, because of f–f transitions; only those with $4f^0$ ($La^{3+}$, $Ce^{4+}$) or $4f^{14}$ ($Lu^{3+}$, $Yb^{2+}$) are colourless. (3) True: they are typical metals, good conductors of heat and electricity. (4) True: actinoids are highly reactive, especially when finely divided. Answer: (2).
60	Haloalkanes and HaloarenesElimination reactions	Class 12: Haloalkanes and Haloarenes	Single correctEasy
A easy single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes 2-Bromopentane can lose a $\beta$-hydrogen from C1 (giving pent-1-ene) or from C3 (giving pent-2-ene). Saytzeff's (Zaitsev's) rule: the major alkene is the more substituted one, which is more stable. Pent-2-ene (disubstituted) is more substituted than pent-1-ene (monosubstituted), so it is the major product. (Hofmann's rule predicts the less substituted alkene, as with bulky bases or quaternary ammonium salts.) Answer: (1) Saytzeff's Rule.
61	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics For one mole of an ideal gas, $H = U + PV = U + RT$, so $\frac{dH}{dT} = \frac{dU}{dT} + R$, that is $C_P = C_V + R$. This is Mayer's relation: $C_P - C_V = R$. (Heating at constant pressure also does work $P\Delta V = R\Delta T$ per mole per kelvin.) Answer: (2).
62	PolymersNylon, Teflon, Bakelite and PVC	No matching NCERT chapter	Single correctEasy
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Teflon is made by addition polymerisation of tetrafluoroethene, $nCF_2=CF_2 \to (CF_2-CF_2)_n$, with no by-product. Nylon-66 (hexamethylenediamine + adipic acid), Novolac (phenol + formaldehyde) and Dacron (ethylene glycol + terephthalic acid) are condensation polymers, formed with loss of water. Answer: (1) Teflon.
63	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 The ethyl group of the Grignard reagent adds to the carbonyl carbon of acetone, $(CH_3)_2C=O$, giving $(CH_3)_2C(OMgBr)C_2H_5$; hydrolysis gives the tertiary alcohol $(CH_3)_2C(OH)CH_2CH_3$. Naming: the longest chain containing the C–OH is four carbons ($CH_3-C(OH)(CH_3)-CH_2-CH_3$), with OH and a methyl on C2: 2-methylbutan-2-ol. Answer: (4) 2-methyl butan-2-ol.
64	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 Count electron pairs on the central atom (VSEPR): (a) $PCl_5$: 5 bond pairs, no lone pair → trigonal bipyramidal (iv). (b) $SF_6$: 6 bond pairs → octahedral (iii). (c) $BrF_5$: 5 bond pairs + 1 lone pair (octahedral arrangement) → square pyramidal (i). (d) $BF_3$: 3 bond pairs → trigonal planar (ii). Answer: (1) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii).
65	Isolation of MetalsSmelting and refining	No matching NCERT chapter	Single correctEasy
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter The metal liquid at room temperature is mercury. Low-boiling metals such as mercury and zinc are refined by distillation: the impure metal is evaporated and the pure metal condensed, leaving the impurities behind. Zone refining needs a solid bar (Si, Ge, Ga); electrolysis and chromatography are not the standard method here. Answer: (3) Distillation.
66	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons With benzoyl peroxide present, HBr adds to an alkene by a free-radical chain (peroxide or Kharasch effect), giving the anti-Markovnikov product. The $Br^\bullet$ radical adds to the terminal $CH_2$, leaving the more stable secondary radical $(CH_3)_2CH-\dot{C}H-CH_2Br$, which then takes an H from HBr. Product: $(CH_3)_2CH-CH_2-CH_2-Br$ (1-bromo-3-methylbutane). (Without peroxide the Markovnikov product, with Br on the inner carbon, would form, with possible rearrangement to option (4).) Answer: (1).
67	HydrogenHydrogen and its compounds	No matching NCERT chapter	Single correctEasy
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Tritium, ${}^3_1H$, has one proton and two neutrons, too many neutrons for stability. A neutron changes into a proton with emission of a $\beta^-$ particle: ${}^3_1H \to {}^3_2He + {}^{\;\,0}_{-1}e$ (half-life about 12.3 years). Answer: (1) Beta $(\beta^-)$.
68	Haloalkanes and HaloarenesNature of the C–X bond	Class 12: Haloalkanes and Haloarenes	Single correctEasy
A easy single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Down the group from F to I the halogen atom gets larger, the C–X bond gets longer and the orbital overlap weaker, so the bond enthalpy falls: C–F (about 452 kJ/mol) > C–Cl (351) > C–Br (293) > C–I (234). Answer: (2).
69	Solid StateClose packed structures	No matching NCERT chapter	Single correctMedium
A medium single correct question on close packed structures, from Solid State. NCERT: No matching NCERT chapter The hexagonal unit cell of hcp packing holds 6 atoms: 12 corners × $\frac{1}{6}$ + 2 face centres × $\frac{1}{2}$ + 3 inside = 2 + 1 + 3 = 6. In any close packing there are 2 tetrahedral voids and 1 octahedral void per atom, so this cell has $2\times6 = 12$ tetrahedral and $6$ octahedral voids. Answer: (4) 12, 6.
70	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 a metal displacement reaction a more reactive metal displaces another metal from its compound. (2) Aluminium displaces chromium from $Cr_2O_3$ (an aluminothermic reaction): a metal displacement. (1) and (4) are decomposition reactions. (3) is a hydrogen displacement: iron displaces hydrogen, not a metal, from the acid. Answer: (2).
71	States of MatterGas laws	No matching NCERT chapter	Single correctEasy
A easy single correct question on gas laws, from States of Matter. NCERT: No matching NCERT chapter Boyle's law: at constant temperature $PV = nRT$ is constant, so each isotherm is a rectangular hyperbola, $P \propto \frac{1}{V}$. This rules out (2) (horizontal lines) and (3) (straight lines through the origin). At a higher temperature the constant $nRT$ is larger, so at the same volume the pressure is higher: the 600 K curve must lie farthest from the axes, above the 400 K curve, which lies above the 200 K curve. In (4) the order from the top is 600 K, 400 K, 200 K: correct. Graph (1) has the order reversed. Answer: (4).
72	EquilibriumIonic equilibrium	Class 11: Equilibrium	Single correctMedium
A medium single correct question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Dimethylammonium acetate is a salt of a weak acid and a weak base; its pH does not depend on concentration: $pH = 7 + \frac{1}{2}(pK_a - pK_b) = 7 + \frac{1}{2}(4.77 - 3.27) = 7 + 0.75 = 7.75$. (Taking $pK_w = 14$ at the given temperature.) The base is the stronger of the two ($pK_b < pK_a$), so the solution is slightly basic, as found. Answer: (3) 7.75.
73	s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium	No matching NCERT chapter	Single correctEasy
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter $Be^{2+}$ is very small with a high charge density, so it strongly polarises the chloride ion (Fajans' rules). $BeCl_2$ is therefore largely covalent: it has a low melting point and dissolves in organic solvents. The chlorides of Mg, Ca and Sr are ionic and dissolve in water rather than organic solvents. Answer: (4) Beryllium chloride.
74	Isolation of MetalsSmelting and refining	No matching NCERT chapter	Single correctEasy
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In the blast furnace for iron, the hottest zone is near the bottom, where coke burns in the hot air blast ($C + O_2 \to CO_2$); the temperature there reaches about 2200 K. Higher up the furnace it falls to about 500–900 K, where the oxides of iron are reduced by CO. Answer: (2) upto 2200 K.
75	Coordination CompoundsWerner theory	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $EDTA^{4-}$ is $(^-OOCCH_2)_2N-CH_2CH_2-N(CH_2COO^-)_2$. It binds a metal ion through the two nitrogen atoms and one oxygen of each of the four carboxylate groups: six donor atoms in all. So it is a hexadentate ligand with four O and two N donor atoms. Answer: (1).
76	SolutionsColligative properties	Class 12: Solutions	Single correctEasy
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions Osmotic pressure $\pi = CRT$; all three are non-electrolytes in the same volume at the same temperature, so $\pi \propto$ moles of solute, that is $\propto \frac{1}{M}$ for the same 10 g. Molar masses: urea 60, glucose 180, sucrose 342 g/mol. Moles: urea $\frac{10}{60} = 0.167$, glucose $\frac{10}{180} = 0.056$, sucrose $\frac{10}{342} = 0.029$. So $P_2 > P_1 > P_3$. Answer: (1).
77	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	StatementsEasy
A easy statements question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: the acid strength of the hydrogen halides rises from HF (a weak acid in water) to HI (the strongest): true. Statement II: down the group the halogen atom grows, the H–X bond becomes longer and weaker (H–F about 574 kJ/mol, H–I about 299 kJ/mol), so the proton is released more easily and the acid strength increases: true, and it is the reason for Statement I. Answer: (1) Both Statement I and Statement II are true.
78	s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium	No matching NCERT chapter	Single correctEasy
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter In the solid state $BeCl_2$ has a polymeric chain structure, each Be bonded to four bridging Cl atoms. In the vapour it exists as the chlorine-bridged dimer $Be_2Cl_4$, which breaks into linear monomers only above about 1200 K. So: chain (solid) and dimer (vapour). Answer: (1).
79	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 The enthalpy of reaction is negative, so the reaction is exothermic: B must end lower than A. The activation enthalpy, 9.6 kJ/mol, is the height of the peak above A, more than twice the drop of 4.2 kJ/mol from A to B. (1) B ends above A: endothermic. (3) A and B at the same level: $\Delta H = 0$. (4) B far above A: endothermic. (2) B ends below A, with a hump above A that is larger than the drop: the right profile. Answer: (2).
80	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 Hf comes after the 14 lanthanoids, in which the 4f electrons shield the growing nuclear charge poorly, so the atoms shrink steadily (lanthanoid contraction). This cancels the increase in size expected on going from the 4d (Zr) to the 5d (Hf) series, and Zr (160 pm) and Hf (159 pm) end up nearly the same size. Answer: (3) lanthanoid contraction.
81	Structure of AtomDual nature and uncertainty principle	Class 11: Structure of Atom	Single correctEasy
A easy single correct question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $\lambda = \frac{c}{\nu} = \frac{3.0\times10^8}{1368\times10^3} = \frac{3.0\times10^8}{1.368\times10^6}$ m. $\frac{3}{1.368} = 2.19298\ldots$, so $\lambda = 219.298$ m $\approx 219.3$ m (rounded correctly; 219.2 m would be a truncation). Check: $219.3\times1.368\times10^6 \approx 3.0\times10^8$. Answer: (1) 219.3 m.
82	Some Basic Concepts of ChemistryEmpirical and molecular formula	Class 11: Some Basic Concepts of Chemistry	Single correctMedium
A medium single correct question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry In 100 g: C = 78 g, H = 22 g. Moles: C $= \frac{78}{12} = 6.5$; H $= \frac{22}{1} = 22$. Ratio H : C $= \frac{22}{6.5} = 3.38$, closest to 3, so the empirical formula is $CH_3$. (Check against the options: CH would be 92% C, $CH_2$ 86%, $CH_3$ 80%, $CH_4$ 75%; 78% lies nearest $CH_3$.) Answer: (3) $CH_3$.
83	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Metamerism is isomerism caused by different alkyl groups on either side of the same polyvalent functional group (such as –O– in ethers). $C_4H_{10}O$ has the ethers $C_2H_5-O-C_2H_5$ and $CH_3-O-C_3H_7$: metamers. $C_3H_8O$ has only one ether, $CH_3-O-C_2H_5$, so no metamers; $C_5H_{12}$ has no functional group (only chain isomers); $C_3H_6O$ (propanal, acetone) shows functional isomerism, and acetone, $CH_3COCH_3$, has no metamer. Answer: (4) $C_4H_{10}O$.
84	AminesClassification and identification of amines	Class 12: Amines	Single correctEasy
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hinsberg's reagent is benzenesulphonyl chloride, $C_6H_5SO_2Cl$. (1) $CH_3CH_2NO_2$ is a nitro compound, not an amine: no reaction. (2) $CH_3CH_2NHCH_3$, a secondary amine, gives $C_6H_5SO_2N(CH_3)C_2H_5$, which has no H on N and so does not dissolve in alkali. (3) $CH_3CH_2NH_2$, a primary amine, gives $C_6H_5SO_2NHC_2H_5$; the H on N is acidic (made so by the strongly electron-withdrawing sulphonyl group), so the solid dissolves in alkali. (4) $CH_3CH_2N(CH_3)CH_2CH_3$, a tertiary amine, does not react. Answer: (3).
85	Solid StateSeven crystal systems	No matching NCERT chapter	Single correctEasy
A easy single correct question on seven crystal systems, from Solid State. NCERT: No matching NCERT chapter Body-centred lattices occur in only three of the seven crystal systems: cubic (bcc), tetragonal and orthorhombic. (The 14 Bravais lattices: cubic 3, tetragonal 2, orthorhombic 4, monoclinic 2, and one each for triclinic, hexagonal and rhombohedral.) Answer: (4) 3.
86	Coordination CompoundsCrystal field theory	Class 12: Coordination Compounds	MatchingMedium
A medium matching question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Spin-only moment $\mu = \sqrt{n(n+2)}$ BM, n = unpaired electrons. $CN^-$ is a strong-field ligand (low spin); $H_2O$ is weak-field (high spin) for iron. (a) $[Fe(CN)_6]^{3-}$: $Fe^{3+}$, $d^5$ low spin $t_{2g}^5$: n = 1, $\mu = \sqrt{3} = 1.73$ BM → (iv). (b) $[Fe(H_2O)_6]^{3+}$: $d^5$ high spin: n = 5, $\mu = \sqrt{35} = 5.92$ BM → (i). (c) $[Fe(CN)_6]^{4-}$: $Fe^{2+}$, $d^6$ low spin $t_{2g}^6$: n = 0, 0 BM → (ii). (d) $[Fe(H_2O)_6]^{2+}$: $d^6$ high spin: n = 4, $\mu = \sqrt{24} = 4.90$ BM → (iii). Answer: (4) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).
87	States of MatterGas laws	No matching NCERT chapter	Single correctEasy
A easy single correct question on gas laws, from States of Matter. NCERT: No matching NCERT chapter Moles: $O_2 = \frac{4}{32} = 0.125$; $H_2 = \frac{2}{2} = 1$. Total $n = 1.125$ mol. By Dalton's law the total pressure is that of all the moles together: $P = \frac{nRT}{V} = \frac{1.125\times0.082\times273}{1} = 1.125\times22.386 = 25.18$ atm. Check: $1\times22.386 + 0.125\times22.386 = 22.386 + 2.798 = 25.18$. Answer: (3) 25.18.
88	HydrocarbonsAlkanes	Class 11: Hydrocarbons	Single correctEasy
A easy single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons This is decarboxylation: heating the sodium salt of a carboxylic acid with soda lime (NaOH + CaO) removes $CO_2$ as $Na_2CO_3$ and gives an alkane with one carbon fewer. $CH_3CH_2COONa + NaOH \xrightarrow{CaO,\ \Delta} CH_3CH_3 + Na_2CO_3$. The missing chemical is CaO (soda lime = NaOH + CaO). Answer: (3) CaO.
89	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics The internal energy of an ideal gas depends only on temperature; isothermal means no change of T, so $\Delta U = 0$. For any irreversible (spontaneous) process the total entropy of system plus surroundings increases: $\Delta S_{total} > 0$, so $\Delta S_{total} \neq 0$. (Only a reversible process gives $\Delta S_{total} = 0$.) Answer: (3).
90	p-Block ElementsGroup 16 elements	No matching NCERT chapter	Single correctMedium
A medium single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter (1) Acid strength of the hydrogen halides rises from HF to HI as the H–X bond weakens: correct as stated. (2) For the hydrides of group 16 the H–E bond also weakens down the group, so acidity increases from $H_2O$ to $H_2Te$. Stronger acid means smaller $pK_a$, so the $pK_a$ values decrease along $H_2O, H_2S, H_2Se, H_2Te$. The sequence given as "increasing $pK_a$" is the reverse: this one is wrong. (3) Down group 15 the hydrides lose basic character and gain acidic character as the E–H bond weakens: correct as stated. (4) Down group 14 the +4 state becomes less stable (inert pair effect), so $PbO_2$ is the strongest oxidising agent of the four: correct as stated. Answer: (2).
91	ElectrochemistryConductance	Class 12: Electrochemistry	Single correctMedium
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry $\Lambda^\circ_m(CH_3COOH) = 350 + 50 = 400$ S cm$^2$ mol$^{-1}$. Degree of dissociation: $\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{20}{400} = 0.05$. $K_a = \frac{c\alpha^2}{1 - \alpha} \approx c\alpha^2 = 0.007\times(0.05)^2 = 0.007\times0.0025 = 1.75\times10^{-5}$ mol L$^{-1}$. (Keeping $1 - \alpha = 0.95$ gives $1.84\times10^{-5}$, still nearest option (3); the options are built on $c\alpha^2$.) Answer: (3) $1.75\times10^{-5}$ mol L$^{-1}$.
92	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 $\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$, so the slope of $\ln k$ against $\frac{1}{T}$ is $-\frac{E_a}{R}$. $-\frac{E_a}{R} = -5\times10^3$ K $\Rightarrow E_a = 5\times10^3\times8.314 = 41570$ J mol$^{-1} \approx 41.5$ kJ mol$^{-1}$. (Option (2), 83.0, would come from wrongly doubling; with $\log_{10}k$ the slope would be $-\frac{E_a}{2.303R}$, giving 95.7, not offered.) Answer: (1) 41.5 kJ mol$^{-1}$.
93	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The starting compound is 2-methyl cyclohexanone carrying a $-CH_2-COOCH_3$ (methyl ester) side chain on the other carbon next to the C=O. $NaBH_4$ in ethanol is a mild reducing agent: it delivers hydride to the reactive ketone carbonyl, reducing it to a secondary alcohol, but it does not reduce esters (their carbonyl is less electrophilic because of resonance with the –OCH$_3$ oxygen). So the ring C=O becomes CH–OH and the $-CH_2-COOCH_3$ group is untouched: structure (4). (1) and (3) wrongly alter the ester; (2) reduces the ester and leaves the ketone, the opposite of what $NaBH_4$ does. Answer: (4).
94	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	MatchingEasy
A easy matching question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (a) Benzene with CO and HCl over anhydrous $AlCl_3$/CuCl gives benzaldehyde: Gattermann-Koch reaction → (ii). (b) A methyl ketone with sodium hypohalite gives a haloform ($CHX_3$) and a carboxylate: haloform reaction → (iii). (c) An alcohol and a carboxylic acid with concentrated $H_2SO_4$ give an ester: esterification → (iv). (d) A carboxylic acid with an $\alpha$-hydrogen, treated with $X_2$/red P and then water, gives the $\alpha$-halo acid: Hell-Volhard-Zelinsky reaction → (i). Answer: (4) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i).
95	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 $SbCl_5$: 5 bond pairs and no lone pair on Sb, trigonal bipyramidal, with all five bonds alike; the bond dipoles cancel, so it is non-polar. $POCl_3$ (tetrahedral with one P=O and three P–Cl), $CH_2O$ (trigonal planar with one C=O and two C–H) and $NO_2$ (bent) have bond dipoles that do not cancel: polar. Answer: (3) $SbCl_5$.
96	Structure of AtomElectronic configuration	Class 11: Structure of Atom	Single correctEasy
A easy single correct question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom Count electrons (Z minus charge): (1) $O^{2-}$: 8 + 2 = 10; $F^-$: 9 + 1 = 10. Isoelectronic. (2) $Na^+$: 11 − 1 = 10; $Mg^{2+}$: 12 − 2 = 10. Isoelectronic. (3) $Mn^{2+}$: 25 − 2 = 23; $Fe^{3+}$: 26 − 3 = 23. Isoelectronic. (4) $Fe^{2+}$: 26 − 2 = 24; $Mn^{2+}$: 23. Not isoelectronic. Answer: (4).
97	SolutionsRaoult's law	Class 12: Solutions	Single correctEasy
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Mole fractions: benzene $\frac{3}{5} = 0.6$, octane $\frac{2}{5} = 0.4$. Raoult's law (ideal solution): $P = x_Bp_B^\circ + x_Op_O^\circ = 0.6\times280 + 0.4\times420 = 168 + 168 = 336$ mm Hg. Check: the answer must lie between 280 and 420, and 336 does. Answer: (3) 336 mm of Hg.
98	Environmental ChemistryAir and water pollution	No matching NCERT chapter	MatchingMedium
A medium matching question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter (a) Oxidation of $SO_2$ to $SO_3$ in the air is a step of tropospheric (gaseous) air pollution → (iv). (b) In the stratosphere, sunlight splits HOCl into OH and Cl radicals; the Cl radicals destroy ozone → (iii) ozone depletion. (c) Sulphuric acid in rain attacking marble and limestone ($CaCO_3$) is the damage done by acid rain → (i). (d) Photolysis of $NO_2$ to NO and atomic O starts the chain that forms ozone and the other components of photochemical smog → (ii). Only option (3) has (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii); it is also the only option giving each entry its own distinct match consistently. Answer: (3).
99	AminesDiazonium salts	Class 12: Amines	Single correctEasy
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines 2,4,6-Tribromoaniline is diazotised ($NaNO_2$, HCl, 0–5°C) to the diazonium chloride, and the final product is 1,3,5-tribromobenzene: the $-N_2^+Cl^-$ group has been replaced by hydrogen. Mild reducing agents do this: ethanol ($ArN_2^+Cl^- + CH_3CH_2OH \to ArH + N_2 + CH_3CHO + HCl$) or hypophosphorous acid. $H_2O$ would give a phenol, HI (as KI) an iodo compound, and CuCN/KCN a nitrile. Answer: (2) $CH_3CH_2OH$.
100	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctEasy
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Toluene with chromyl chloride ($CrO_2Cl_2$) in $CS_2$ is the Étard reaction. The methyl group is oxidised to a brown chromium complex, $C_6H_5CH(OCrOHCl_2)_2$, which on hydrolysis ($H_3O^+$) gives benzaldehyde, $C_6H_5CHO$, without going on to benzoic acid. (2), the benzylidene diacetate, is the intermediate when $CrO_3$ and acetic anhydride are used; (3) and (4) come from side-chain chlorination. Answer: (1).

Botany questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
101	Organisms and PopulationsInteractions	Class 12: Organisms and Populations	Single correctEasy
A easy single correct question on interactions, from Organisms and Populations. NCERT: Class 12: Organisms and Populations When two species compete for the same resource, one way both can persist is to divide the resource between them: they feed at different times, in different parts of the habitat, or on different sizes of food. This is resource partitioning. MacArthur's five closely related warbler species living on the same tree avoided competition and coexisted because they foraged in different parts of the tree. Competitive release is what happens to a species when its competitor is removed (it expands its range); it is not a mechanism for coexisting. Mutualism and predation are different interactions altogether. So the answer is resource partitioning.
102	Anatomy of Flowering PlantsTissue systems	Class 11: Anatomy of Flowering Plants	MatchingEasy
A easy matching question on tissue systems, from Anatomy of Flowering Plants. NCERT: Class 11: Anatomy of Flowering Plants (a) Cells that keep the capacity to divide make up meristematic tissue: (a)-(ii). (b) A tissue whose cells are all alike in structure and function is a simple tissue (parenchyma, collenchyma, sclerenchyma): (b)-(iv). (c) A tissue made of more than one type of cell working as a unit is a complex tissue; xylem and phloem, the vascular tissues, are the examples: (c)-(i). (d) Sclereids are sclerenchyma cells: dead, with very thick lignified walls and a narrow lumen: (d)-(iii). So (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).
104	Transport in PlantsWater movement in plants	No matching NCERT chapter	MatchingEasy
A easy matching question on water movement in plants, from Transport in Plants. NCERT: No matching NCERT chapter (a) Cohesion is the mutual attraction among water molecules: (a)-(ii). (b) Adhesion is the attraction of water molecules to polar surfaces such as the walls of tracheary elements: (b)-(iv). (c) Surface tension arises because water molecules are attracted to each other more in the liquid phase than to water in the gas phase: (c)-(i). (d) Guttation is the loss of water in liquid form, as drops at the leaf tips and margins through hydathodes: (d)-(iii). So (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).
105	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 Xenogamy is the transfer of pollen from the anther of one plant to the stigma of a different plant. It is the only kind of pollination that brings genetically different pollen to the stigma. Geitonogamy is transfer between flowers of the same plant, which is genetically like self-pollination. Chasmogamous flowers have exposed anthers and stigma; cleistogamous flowers never open and are self-pollinated. So the term is xenogamy.
106	Cell Cycle and Cell DivisionMeiosis	Class 11: Cell Cycle and Cell Division	Single correctEasy
A easy single correct question on meiosis, from Cell Cycle and Cell Division. NCERT: Class 11: Cell Cycle and Cell Division In meiosis I the homologous chromosomes separate, but the sister chromatids of each chromosome stay joined at their centromere (anaphase I). In anaphase II the centromere of each chromosome splits, which lets the sister chromatids separate and move to opposite poles. This is the stage at which the centromere divides. So anaphase II.
108	Plant KingdomBryophytes and pteridophytes	Class 11: Plant Kingdom	Single correctEasy
A easy single correct question on bryophytes and pteridophytes, from Plant Kingdom. NCERT: Class 11: Plant Kingdom Gemmae are green, multicellular, asexual buds that develop in small receptacles called gemma cups on the thallus of some liverworts, such as Marchantia. They become detached and each germinates into a new individual. Mosses reproduce vegetatively mainly by fragmentation and budding of the secondary protonema; pteridophytes and gymnosperms do not bear gemma cups. So gemmae are present in some liverworts.
109	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 The checkerboard diagram devised by Reginald C. Punnett shows the gametes of the two parents along its two sides; each box gives the zygote formed by one pairing, so the genotypes of the F$_1$ and F$_2$ and their proportions can be read off directly. It is called the Punnett square. So the answer is Punnett square.
111	Plant KingdomBryophytes and pteridophytes	Class 11: Plant Kingdom	Single correctEasy
A easy single correct question on bryophytes and pteridophytes, from Plant Kingdom. NCERT: Class 11: Plant Kingdom Most pteridophytes produce spores of one kind only (homosporous). Genera such as Selaginella and Salvinia produce two kinds: large megaspores and small microspores, which germinate into female and male gametophytes respectively. Such plants are heterosporous, and heterospory is regarded as a precursor to the seed habit. So they are heterosporous.
112	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 ability of a plant to follow different developmental pathways in response to its environment or its phase of life, and so to form different kinds of structures, is called plasticity. Heterophylly is the usual example: cotton, coriander and larkspur have different leaves in the juvenile and the mature plant (developmental), and buttercup has different leaves in air and under water (environmental). So the ability is plasticity.
114	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 The central dogma: genetic information flows DNA → mRNA → protein. (a) The curved arrow from DNA back to itself is DNA copying itself: replication. (b) DNA to mRNA is transcription. (c) mRNA to the product is translation. (d) The product of translation is protein. So (a) replication, (b) transcription, (c) translation, (d) protein.
115	Cell: The Unit of LifeNucleus	Class 11: Cell: The Unit of Life	Single correctEasy
A easy single correct question on nucleus, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life Chromosomes are classified by where the centromere lies: - metacentric: centromere in the middle, two equal arms; - sub-metacentric: slightly away from the middle, one arm shorter than the other; - acrocentric: close to one end, one arm extremely short; - telocentric: at the very end. So a chromosome with the centromere in the middle of two equal arms is metacentric.
117	Plant Growth and DevelopmentPhotoperiodism	Class 11: Plant Growth and Development	Single correctEasy
A easy single correct question on photoperiodism, from Plant Growth and Development. NCERT: Class 11: Plant Growth and Development In photoperiodism the light/dark period is perceived by the leaves. The shoot apices are where the response appears (they turn into floral meristems), so a hormonal signal is thought to travel from the leaves to the apex. So the site of perception is the leaf.
119	Plant KingdomAlgae	Class 11: Plant Kingdom	Single correctMedium
A medium single correct question on algae, from Plant Kingdom. NCERT: Class 11: Plant Kingdom A monoecious plant bears both male and female sex organs on the same individual. Chara, a green alga, carries both its sex organs, the antheridium (globule) and the oogonium (nucule), on the same plant, so it is monoecious. Papaya has separate male and female plants; Marchantia has separate male and female thalli; Cycas has separate male and female plants. All three are dioecious. So Chara.
121	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	MatchingEasy
A easy matching question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life (a) Cristae are the infoldings of the inner mitochondrial membrane: (a)-(iii). (b) Thylakoids are flattened membranous sacs in the stroma of chloroplasts: (b)-(iv). (c) The centromere is the primary constriction of a chromosome: (c)-(i). (d) Cisternae are the flat, disc-shaped sacs of the Golgi apparatus: (d)-(ii). So (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii).
122	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 Stamens may be free or united by their filaments. United into one bundle they are monadelphous (china rose); into two bundles, diadelphous (pea, where nine filaments are fused and one is free, (9)+1); into more than two bundles, polyadelphous (citrus). So diadelphous stamens are found in pea.
123	Plant Growth and DevelopmentSeed germination and differentiation	Class 11: Plant Growth and Development	MatchingEasy
A easy matching question on seed germination and differentiation, from Plant Growth and Development. NCERT: Class 11: Plant Growth and Development (a) Fusing protoplasts of tomato and potato gave the somatic hybrid pomato: (a)-(ii). (b) Plant tissue culture rests on totipotency, the capacity of a single cell or explant to regenerate a whole plant: (b)-(i). (c) The meristem (apical and axillary) of an infected plant is free of virus, so meristem culture gives virus-free plants: (c)-(iv). (d) Micropropagation produces thousands of plants genetically identical to the original, called somaclones: (d)-(iii). So (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii).
124	Organisms and PopulationsInteractions	Class 12: Organisms and Populations	Single correctEasy
A easy single correct question on interactions, from Organisms and Populations. NCERT: Class 12: Organisms and Populations In amensalism one species is harmed while the other is unaffected: (−, 0). For comparison: mutualism is (+, +), competition (−, −), commensalism (+, 0), and predation and parasitism (+, −). So amensalism is species A (−), species B (0).
125	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	StatementsEasy
A easy statements question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life (1) A mature sieve tube element has peripheral cytoplasm and a large vacuole but no nucleus; its functions are controlled by the nucleus of the companion cell. The statement is incorrect. (2) Microbodies (peroxisomes, glyoxysomes) are small membrane-bound vesicles found in both plant and animal cells. Correct. (3) The nuclear envelope has two membranes with a 10–50 nm perinuclear space between them, forming a barrier between the materials inside the nucleus and those of the cytoplasm. Correct. (4) Nuclear pores let RNA and protein molecules move in both directions between nucleus and cytoplasm. Correct. So statement (1) is incorrect.
126	Sexual Reproduction in Flowering PlantsMegasporogenesis	Class 12: Sexual Reproduction in Flowering Plants	Single correctEasy
A easy single correct question on megasporogenesis, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants In the Polygonum type embryo sac, the functional megaspore nucleus divides three times to give eight nuclei. Six of them are walled off as cells: three at the micropylar end (two synergids and the egg, the egg apparatus) and three antipodals at the chalazal end. The remaining two, the polar nuclei, lie in the one large central cell. So there are 8 nuclei in 7 cells: 8-nucleate and 7-celled.
127	Plant KingdomAlgae	Class 11: Plant Kingdom	Single correctEasy
A easy single correct question on algae, from Plant Kingdom. NCERT: Class 11: Plant Kingdom Brown algae (Phaeophyceae) store food as complex carbohydrates: laminarin or mannitol. Ectocarpus is a brown alga. Gracilaria is a red alga, storing floridean starch; Volvox and Ulothrix are green algae, storing starch. So Ectocarpus.
128	Plant Growth and DevelopmentPlant hormones	Class 11: Plant Growth and Development	Single correctEasy
A easy single correct question on plant hormones, from Plant Growth and Development. NCERT: Class 11: Plant Growth and Development 2,4-D (2,4-dichlorophenoxyacetic acid), a synthetic auxin, is widely used as a herbicide to kill dicotyledonous weeds; it does not affect mature monocot plants, so it is used to prepare weed-free lawns. IAA and IBA are natural auxins and NAA is a synthetic auxin, used for rooting cuttings, preventing fruit drop and so on, not as weedkillers. So 2,4-D.
129	EcosystemNutrient cycling	Class 12: Ecosystem	Single correctEasy
A easy single correct question on nutrient cycling, from Ecosystem. NCERT: Class 12: Ecosystem The amount of nutrients such as carbon, nitrogen, phosphorus and calcium present in the soil at any given time is the standing state. It varies with the kind of ecosystem and with the season. Standing crop is the amount of living material (biomass) present in each trophic level at a given time. A climax community is the final, stable community of a succession. So standing state.
130	Principles of Inheritance and VariationGenetic disorders	Class 12: Principles of Inheritance and Variation	Single correctEasy
A easy single correct question on genetic disorders, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation Mutations can be induced by mutagens: chemicals, and radiation of high energy such as UV light, X-rays and gamma rays. Gamma rays from cobalt-60 are used in plant breeding to induce mutations (mung bean resistant to yellow mosaic virus was obtained this way). Kinetin and zeatin are cytokinins, plant growth regulators; infrared rays do not carry enough energy to alter DNA. So gamma rays.
131	EcosystemEcological pyramids	Class 12: Ecosystem	StatementsEasy
A easy statements question on ecological pyramids, from Ecosystem. NCERT: Class 12: Ecosystem In the sea the biomass of fish far exceeds that of the phytoplankton at any moment (phytoplankton are small and turn over rapidly), so the pyramid of biomass in the sea is generally inverted. Statement (1) is correct and (2), saying it is upright, is not. Energy decreases at every successive trophic level, so the pyramid of energy is always upright (3, correct). In a grassland the producers far outnumber the herbivores, which outnumber the carnivores, so the pyramid of numbers is upright (4, correct). So statement (2) is not correct.
132	EcosystemProductivity and decomposition	Class 12: Ecosystem	Single correctEasy
A easy single correct question on productivity and decomposition, from Ecosystem. NCERT: Class 12: Ecosystem Gross primary productivity (GPP) is the rate at which producers make organic matter in photosynthesis. Part of it is used by the plants themselves in respiration. What is left, available to the consumers, is net primary productivity: NPP = GPP − R. So R is the respiration losses.
133	Plant KingdomAlgae	Class 11: Plant Kingdom	Single correctEasy
A easy single correct question on algae, from Plant Kingdom. NCERT: Class 11: Plant Kingdom Some marine algae yield hydrocolloids. Algin is obtained from brown algae; carrageen comes from red algae; agar too is from red algae (Gelidium, Gracilaria). So carrageen is produced by red algae.
134	Photosynthesis in Higher PlantsC4 pathway	Class 11: Photosynthesis in Higher Plants	Single correctEasy
A easy single correct question on c4 pathway, from Photosynthesis in Higher Plants. NCERT: Class 11: Photosynthesis in Higher Plants Sorghum, like maize and sugarcane, is a C$_4$ plant. In its mesophyll cells CO$_2$ is first fixed by PEP carboxylase onto phosphoenolpyruvate, a 3-carbon acceptor, and the first stable product is the 4-carbon oxaloacetic acid (hence "C$_4$"). Phosphoglyceric acid is the first stable product in C$_3$ plants. So oxaloacetic acid.
135	Anatomy of Flowering PlantsSecondary growth	Class 11: Anatomy of Flowering Plants	MatchingEasy
A easy matching question on secondary growth, from Anatomy of Flowering Plants. NCERT: Class 11: Anatomy of Flowering Plants (a) Lenticels are lens-shaped openings in the bark that permit the exchange of gases between the outer atmosphere and the inner tissue: (a)-(iii). (b) Cork cambium is also called phellogen: (b)-(i). (c) The secondary cortex is the phelloderm, cut off by the phellogen on its inner side: (c)-(iv). (d) Cork (phellem) cells have suberin deposited in their walls, which makes them impervious to water: (d)-(ii). So (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).
136	Respiration in PlantsETS and oxidative phosphorylation	Class 11: Respiration in Plants	StatementsEasy
A easy statements question on ets and oxidative phosphorylation, from Respiration in Plants. NCERT: Class 11: Respiration in Plants (1) Oxygen acts only at the very end of the electron transport chain, as the final acceptor of electrons, where it is reduced to water. Correct. (2) Oxidation of one NADH + H$^+$ in the ETS gives 3 ATP, and of one FADH$_2$ gives 2 ATP. The statement has the numbers swapped, so it is incorrect. (3) ATP is made by complex V, the ATP synthase (F$_0$–F$_1$). Correct. (4) The redox reactions of the chain pump protons into the intermembrane space, building the proton gradient that drives ATP synthesis. Correct. So statement (2) is incorrect.
137	Morphology of Flowering PlantsFamilies	Class 11: Morphology of Flowering Plants	MatchingMedium
A medium matching question on families, from Morphology of Flowering Plants. NCERT: Class 11: Morphology of Flowering Plants Read each formula for its tell-tale feature. (a) Zygomorphic (%), K$_{(5)}$, corolla 1 + 2 + (2) (standard, two wings, two fused keel petals), stamens (9) + 1 (diadelphous), monocarpellary ovary G$_1$: this is the papilionaceous flower of Fabaceae: (a)-(iii). (b) Actinomorphic, K$_{(5)}$, C$_{(5)}$ with the five stamens joined to the corolla (epipetalous), bicarpellary superior ovary: Solanaceae: (b)-(iv). (c) Actinomorphic, perianth P$_{(3+3)}$ in two whorls of three with the stamens 3 + 3 on it (epiphyllous), tricarpellary syncarpous superior ovary G$_{(3)}$: Liliaceae: (c)-(ii). (d) Actinomorphic, four sepals in two whorls K$_{2+2}$, four petals C$_4$ (cruciform), six stamens in two whorls (two short, four long: tetradynamous), bicarpellary ovary G$_{(2)}$: Brassicaceae (Cruciferae): (d)-(i). So (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i).
138	Cell Cycle and Cell DivisionMitosis	Class 11: Cell Cycle and Cell Division	MatchingEasy
A easy matching question on mitosis, from Cell Cycle and Cell Division. NCERT: Class 11: Cell Cycle and Cell Division (a) S (synthesis) phase is when DNA is replicated: (a)-(iv). (b) In G$_2$ phase proteins are synthesised in preparation for mitosis while cell growth continues: (b)-(i). (c) Cells that leave G$_1$ and stop dividing enter the quiescent stage, G$_0$, an inactive phase (metabolically active but not proliferating): (c)-(ii). (d) G$_1$ phase is the interval between mitosis and the initiation of DNA replication: (d)-(iii). So (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).
140	Molecular Basis of InheritanceTranscription and translation	Class 12: Molecular Basis of Inheritance	StatementsMedium
A medium statements question on transcription and translation, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance (1) In capping, methyl guanosine triphosphate is added to the 5' end of hnRNA; it is a poly(A) tail that is added at the 3' end. Incorrect. (2) In bacteria the RNA polymerase associates transiently with the initiation factor σ to start and with the termination factor ρ (rho) to stop transcription. Correct. (3) It is the template strand (3'→5') that is copied; the mRNA has the same sequence as the coding strand (with U for T), which is not copied. Incorrect. (4) Split genes, with exons interrupted by introns, are characteristic of eukaryotes. Incorrect. So statement (2) is correct.
142	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 Integrating dN/dt = rN gives the exponential growth equation N$_t$ = N$_0$e$^{rt}$, where N$_t$ is the population density after time t, N$_0$ the density at time zero, r the intrinsic rate of natural increase, and e the base of natural logarithms (2.71828). So e is the base of natural logarithms.
143	Anatomy of Flowering PlantsAnatomy of root, stem and leaf	Class 11: Anatomy of Flowering Plants	Single correctMedium
A medium single correct question on anatomy of root, stem and leaf, from Anatomy of Flowering Plants. NCERT: Class 11: Anatomy of Flowering Plants (1) The large, empty, colourless cells in the upper epidermis of grass leaves are bulliform cells, which make the leaf roll in water stress. Subsidiary cells are the epidermal cells beside the guard cells. Wrong pair. (2) In dicot leaves the vascular bundles are surrounded by a layer of thick-walled cells called the bundle sheath. Conjunctive tissue is the parenchyma between xylem and phloem in a root. Wrong pair. (3) In a dicot stem the cells of the medullary rays next to the intrafascicular cambium become meristematic and form the interfascicular cambium, completing the cambial ring. Correct pair. (4) Loose parenchyma that ruptures the epidermis and forms a lens-shaped opening in the bark makes up the lenticel (its cells are complementary cells), not spongy parenchyma, which is a tissue of the leaf mesophyll. Wrong pair. So pair (3).
144	Sexual Reproduction in Flowering PlantsMicrosporogenesis	Class 12: Sexual Reproduction in Flowering Plants	Single correctMedium
A medium single correct question on microsporogenesis, from Sexual Reproduction in Flowering Plants. NCERT: Class 12: Sexual Reproduction in Flowering Plants How long pollen stays viable depends on the species and the conditions. In rice and wheat (Poaceae) pollen loses viability within about 30 minutes of release. In some members of Rosaceae, Leguminosae and Solanaceae it stays viable for months. Every option containing Poaceae is ruled out, leaving Rosaceae; Leguminosae.
145	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 Eukaryotic nuclei have three RNA polymerases: - RNA polymerase I transcribes the rRNAs 28S, 18S and 5.8S; - RNA polymerase II transcribes the precursor of mRNA, the heterogeneous nuclear RNA (hnRNA); - RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs. So RNA polymerase III transcribes tRNA, 5S rRNA and snRNA.
146	Photosynthesis in Higher PlantsLight reaction	Class 11: Photosynthesis in Higher Plants	StatementsEasy
A easy statements question on light reaction, from Photosynthesis in Higher Plants. NCERT: Class 11: Photosynthesis in Higher Plants (1) In non-cyclic photophosphorylation (the Z scheme) electrons from water pass through PS II and PS I to NADP$^+$, so both ATP and NADPH + H$^+$ are formed. Correct. (2) The stroma lamellae carry only PS I and lack PS II and NADP reductase. Correct. (3) The grana lamellae carry both PS I and PS II. Correct. (4) Cyclic photophosphorylation involves PS I alone: the electron cycles back to PS I through the electron transport chain, and only ATP is made. Incorrect. So statement (4) is incorrect.
147	Biological ClassificationFungi	Class 11: Biological Classification	StatementsMedium
A medium statements question on fungi, from Biological Classification. NCERT: Class 11: Biological Classification (1) Karyogamy is the fusion of two nuclei, not of two cells. Incorrect. (2) Plasmogamy is the fusion of the protoplasms of two gametes, motile or non-motile. Correct. (3) Saprophytes depend on dead and decaying organic matter; organisms that depend on living plants are parasites. Incorrect. (4) Some cyanobacteria, such as Nostoc and Anabaena, fix nitrogen in specialised cells called heterocysts, not sheath cells. Incorrect. So statement (2) is correct.
149	Molecular Basis of InheritanceDNA fingerprinting	Class 12: Molecular Basis of Inheritance	Single correctMedium
A medium single correct question on dna fingerprinting, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance DNA fingerprinting compares the DNA of individuals at specific regions called repetitive DNA, where a short stretch of DNA is repeated many times. These repetitive sequences separate as satellite DNA on density gradient centrifugation, and they show a high degree of polymorphism, the variation between individuals that fingerprinting exploits. The regions whose differences are identified are called repetitive DNA, which is the term the definition uses.
150	Mineral NutritionNitrogen metabolism	No matching NCERT chapter	MatchingEasy
A easy matching question on nitrogen metabolism, from Mineral Nutrition. NCERT: No matching NCERT chapter (a) Nitrococcus (with Nitrosomonas) oxidises ammonia to nitrite: (a)-(ii). (b) Rhizobium, in root nodules of legumes, fixes atmospheric nitrogen to ammonia: (b)-(iv). (c) Thiobacillus (with Pseudomonas) carries out denitrification, reducing nitrate to nitrogen gas: (c)-(i). (d) Nitrobacter oxidises nitrite to nitrate: (d)-(iii). So (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).
152	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 In S phase the amount of DNA per cell doubles (2C to 4C), because each chromosome is replicated into two sister chromatids. The chromatids stay joined at one centromere, so the number of chromosomes does not change. A cell with 8 chromosomes in G$_1$ therefore still has 8 chromosomes after S phase (each now with two chromatids, 16 chromatids in all). So 8.
153	The Living WorldTaxonomic categories	Class 11: The Living World	Single correctEasy
A easy single correct question on taxonomic categories, from The Living World. NCERT: Class 11: The Living World The housefly, Musca domestica, belongs to the genus Musca, family Muscidae, order Diptera, class Insecta, phylum Arthropoda. The firefly is a beetle (order Coleoptera), the grasshopper is in Orthoptera, and the cockroach is in Blattodea (family Blattidae); none is in Muscidae. So the housefly.
157	Principles of Inheritance and VariationGenetic disorders	Class 12: Principles of Inheritance and Variation	Single correctEasy
A easy single correct question on genetic disorders, from Principles of Inheritance and Variation. NCERT: Class 12: Principles of Inheritance and Variation Sickle-cell anaemia is an autosomal recessive disorder: only the homozygote Hb$^S$Hb$^S$ shows the disease; the heterozygote Hb$^A$Hb$^S$ is a carrier. Cross Hb$^A$Hb$^S$ × Hb$^A$Hb$^S$: the progeny are 1 Hb$^A$Hb$^A$ : 2 Hb$^A$Hb$^S$ : 1 Hb$^S$Hb$^S$. Only Hb$^S$Hb$^S$, one in four, is diseased: 25%.
161	Molecular Basis of InheritanceTranscription and translation	Class 12: Molecular Basis of Inheritance	Single correctEasy
A easy single correct question on transcription and translation, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance In bacteria a single DNA-dependent RNA polymerase transcribes all types of RNA. It catalyses all three steps of transcription, initiation, elongation and termination, associating transiently with the σ factor for initiation and the ρ factor for termination. So it is DNA-dependent RNA polymerase.
162	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 Translation needs three kinds of RNA: mRNA provides the template, tRNA brings the amino acids and reads the codons, and rRNA forms the ribosome (and acts as the catalyst, a ribozyme, for peptide bond formation). siRNA (small interfering RNA) silences specific mRNAs by RNA interference; it is not needed to make a protein and in fact blocks translation of its target. So siRNA.
164	Molecular Basis of InheritanceDNA structure	Class 12: Molecular Basis of Inheritance	Single correctEasy
A easy single correct question on dna structure, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance By Chargaff's rule, in double-stranded DNA A = T and G = C. A = 30%, so T = 30%. A + T = 60%, leaving G + C = 40%, so G = 20% and C = 20%. So T 30, G 20, C 20.
165	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 (a) The fungus Aspergillus niger is used to produce citric acid: (a)-(iii). (b) The bacterium Acetobacter aceti produces acetic acid: (b)-(i). (c) Clostridium butylicum produces butyric acid: (c)-(iv). (d) Lactobacillus produces lactic acid: (d)-(ii). So (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).
174	Cell Cycle and Cell DivisionMitosis	Class 11: Cell Cycle and Cell Division	Single correctMedium
A medium single correct question on mitosis, from Cell Cycle and Cell Division. NCERT: Class 11: Cell Cycle and Cell Division In animal cells, during the S phase, DNA replication begins in the nucleus, and the centriole duplicates in the cytoplasm. In prophase the duplicated centrioles (centrosomes) begin to move to opposite poles of the cell, each radiating spindle fibres. So the centriole duplicates during S phase.
176	Strategies for Enhancement in Food ProductionPlant breeding	No matching NCERT chapter	Single correctEasy
A easy single correct question on plant breeding, from Strategies for Enhancement in Food Production. NCERT: No matching NCERT chapter Biofortification is breeding crops with higher levels of vitamins, minerals, or higher protein and healthier fats, to improve public health. Its objectives are to improve: - protein content and quality; - oil content and quality; - vitamin content; - micronutrient and mineral content. Breeding for resistance to diseases is a separate objective of plant breeding, not of biofortification. So improving resistance to diseases is not an objective of biofortification.
177	Environmental IssuesOzone depletion	No matching NCERT chapter	Single correctEasy
A easy single correct question on ozone depletion, from Environmental Issues. NCERT: No matching NCERT chapter The thickness of the ozone layer in a column of air from the ground to the top of the atmosphere is measured in Dobson units (DU). The ozone hole over Antarctica is an area where this thickness has fallen sharply. So Dobson units measure the thickness of ozone.
184	Cell: The Unit of LifeCell organelles	Class 11: Cell: The Unit of Life	Single correctEasy
A easy single correct question on cell organelles, from Cell: The Unit of Life. NCERT: Class 11: Cell: The Unit of Life The endomembrane system is the group of membranous organelles whose functions are coordinated: the endoplasmic reticulum, Golgi complex, lysosomes and vacuoles. Mitochondria, chloroplasts and peroxisomes work independently of them and are not included; ribosomes have no membrane at all. So ER, Golgi complex, lysosomes and vacuoles.
185	Cell Cycle and Cell DivisionMeiosis	Class 11: Cell Cycle and Cell Division	Single correctEasy
A easy single correct question on meiosis, from Cell Cycle and Cell Division. NCERT: Class 11: Cell Cycle and Cell Division Prophase I has five stages: leptotene (chromosomes become visible), zygotene (synapsis, synaptonemal complex), pachytene (crossing over, recombination nodules), diplotene (synaptonemal complex dissolves, chiasmata appear) and diakinesis. Diakinesis is marked by terminalisation of chiasmata: the chiasmata move towards the ends of the bivalents as the chromosomes condense fully, before the spindle assembles. So diakinesis.
188	Organisms and PopulationsPopulation attributes	Class 12: Organisms and Populations	MatchingMedium
A medium matching question on population attributes, from Organisms and Populations. NCERT: Class 12: Organisms and Populations (a) Allen's rule: mammals of colder climates have shorter ears and limbs to reduce heat loss. Seals of polar seas fit it (and have a thick layer of blubber): (a)-(iv). (b) The kangaroo rat of the North American deserts meets its water needs by internal oxidation of fat and concentrates its urine so as to lose very little water: a physiological adaptation: (b)-(i). (c) Desert lizards bask in the sun when their body temperature falls, and move into shade when the ground gets hot: a behavioural adaptation: (c)-(ii). (d) Fishes that live at great depths in the ocean, under enormous pressure, survive with biochemical adaptations (of enzymes and membranes): (d)-(iii). So (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).
189	Organisms and PopulationsPopulation attributes	Class 12: Organisms and Populations	Assertion and reasonEasy
A easy assertion and reason question on population attributes, from Organisms and Populations. NCERT: Class 12: Organisms and Populations A person who goes to a high-altitude place (above about 3,500 m) may experience altitude sickness: nausea, fatigue, breathing difficulty and heart palpitations. Assertion true. At high altitude the atmospheric pressure is low, so the partial pressure of oxygen is low and the body does not get enough oxygen. Reason true, and it is the cause of the symptoms. The body gradually compensates by increasing RBC production, decreasing the binding affinity of haemoglobin, and breathing faster. So both are true and (R) correctly explains (A).
193	Molecular Basis of InheritanceTranscription and translation	Class 12: Molecular Basis of Inheritance	StatementsEasy
A easy statements question on transcription and translation, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance The genetic code is unambiguous and specific: one codon codes for only one amino acid. AUG codes for methionine only (and is the initiator codon); phenylalanine is coded by UUU and UUC. Statement I is incorrect. The code is degenerate: some amino acids have more than one codon. Lysine is coded by both AAA and AAG. Statement II is true. So Statement I is incorrect but Statement II is true.
196	Molecular Basis of InheritanceDNA structure	Class 12: Molecular Basis of Inheritance	StatementsEasy
A easy statements question on dna structure, from Molecular Basis of Inheritance. NCERT: Class 12: Molecular Basis of Inheritance Histones are rich in the basic amino acids lysine and arginine, which carry positive charges in their side chains (3 and 4 correct). This makes histones basic proteins, not acidic: statement (2) is wrong. Eight histone molecules form the histone octamer, round which negatively charged DNA is wrapped to make a nucleosome (1 correct). So statement (2) is wrong.

Zoology questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
103	Biotechnology: Principles and ProcessesrDNA technology	Class 12: Biotechnology: Principles and Processes	Single correctEasy
A easy single correct question on rdna technology, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes To isolate DNA, the cells are broken open and the other macromolecules are removed with enzymes: RNA with ribonuclease, proteins (histones among them) with protease, and cell-wall polysaccharides with lysozyme, cellulase or chitinase. Purified DNA finally precipitates out when chilled ethanol is added, and can be seen as a collection of fine threads in the suspension. So chilled ethanol precipitates DNA.
107	Biotechnology: Principles and ProcessesPCR and bioreactors	Class 12: Biotechnology: Principles and Processes	Single correctEasy
A easy single correct question on pcr and bioreactors, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes Each PCR cycle has three steps in this order: 1. Denaturation: heating separates the two strands of the template DNA. 2. Annealing: on cooling, the two primers pair with their complementary sequences on the single strands. 3. Extension: a thermostable DNA polymerase (Taq polymerase) extends the primers, copying the region between them. So denaturation, annealing, extension.
110	EvolutionHardy-Weinberg principle	Class 12: Evolution	Single correctEasy
A easy single correct question on hardy-weinberg principle, from Evolution. NCERT: Class 12: Evolution Genetic drift is a change in allele frequencies by chance. When a few individuals break away and found a new population, the alleles they happen to carry may be very different from those of the original population; the new population can become a different species. The original drifted population becomes the founders, and the effect is called the founder effect. So the founder effect is a result of genetic drift.
113	BiomoleculesMetabolic pathways	Class 11: Biomolecules	Single correctEasy
A easy single correct question on metabolic pathways, from Biomolecules. NCERT: Class 11: Biomolecules Primary metabolites take part in the normal physiological processes of every cell: amino acids, sugars such as glucose, nucleotides, fatty acids. Secondary metabolites have no obvious role in the plant's basic metabolism and are found only in some plants: alkaloids (morphine, codeine), drugs (vinblastin, curcumin), polymeric substances (rubber, gums, cellulose), pigments, terpenoids and so on. Amino acids and glucose are primary metabolites, so they are the pair that is not secondary.
116	Biotechnology: Principles and ProcessesrDNA technology	Class 12: Biotechnology: Principles and Processes	Single correctEasy
A easy single correct question on rdna technology, from Biotechnology: Principles and Processes. NCERT: Class 12: Biotechnology: Principles and Processes After gel electrophoresis the separated DNA fragments cannot be seen directly. The gel is stained with ethidium bromide, which binds DNA, and exposed to UV light: the DNA shows up as bright orange coloured bands. The bands are then cut out of the gel and the DNA extracted (elution). So bright orange bands.
118	Biotechnology and its ApplicationsGene therapy	Class 12: Biotechnology and its Applications	Single correctEasy
A easy single correct question on gene therapy, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications Gene therapy is a collection of methods that correct a gene defect diagnosed in a child or embryo: genes are inserted into a person's cells and tissues to treat a disease. The first clinical use was for adenosine deaminase deficiency, where a functional ADA cDNA was introduced into the patient's lymphocytes. Of the options, the treatment of a disease by working with genes in an individual's tissue is gene therapy.
120	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 PCR makes many copies of a chosen piece of DNA. It is used to amplify genes, to detect very low amounts of a pathogen's DNA for early molecular diagnosis (HIV, cancer), and to detect mutations in genes. PCR acts only on nucleic acids; it plays no part in purifying a protein (that is done by chromatography and similar downstream processing). So purification of isolated protein is not an application of PCR.
139	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 PstI cuts pBR322 inside the ampicillin-resistance gene $amp^{\mathrm{R}}$. A foreign gene ligated into that site interrupts the coding sequence of $amp^{\mathrm{R}}$, which is insertional inactivation: the gene can no longer make its product. So E. coli cells carrying the recombinant plasmid lose ampicillin resistance (they still resist tetracycline, since $tet^{\mathrm{R}}$ is intact). This is how recombinants are told apart: they grow on tetracycline but not on ampicillin. So it will not be able to confer ampicillin resistance on the host cell.
141	Biotechnology and its ApplicationsGene therapy	Class 12: Biotechnology and its Applications	Single correctMedium
A medium single correct question on gene therapy, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications A single-stranded DNA or RNA probe, tagged with a radioactive label, is made complementary to the normal gene. It is allowed to hybridise with the DNA of a clone of cells, and the hybrids are detected by autoradiography: wherever the probe has bound, the film is exposed. The probe pairs with the normal sequence. The mutated gene has a changed sequence, so the probe has no complementarity with it, does not bind, and the clone carrying the mutated gene does not appear on the photographic film. Its absence is how the mutation is detected. So option (3).
148	BiomoleculesCarbohydrates, lipids and nucleic acids	Class 11: Biomolecules	MatchingEasy
A easy matching question on carbohydrates, lipids and nucleic acids, from Biomolecules. NCERT: Class 11: Biomolecules (a) Amino acids in a protein are linked by peptide bonds: (a)-(iv). (b) An unsaturated fatty acid has one or more C = C double bonds in its chain: (b)-(i). (c) Nucleotides in a nucleic acid are joined by 3'–5' phosphodiester bonds: (c)-(ii). (d) Monosaccharides in a polysaccharide are joined by glycosidic bonds: (d)-(iii). So (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).
151	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 Each restriction endonuclease recognises a specific palindromic nucleotide sequence in DNA: a sequence that reads the same on the two strands when both are read in the same direction (5'→3'). For EcoRI it is 5'-GAATTC-3' / 3'-CTTAAG-5'. The enzyme binds there and cuts each strand at a specific position. So the recognition sequence is a palindromic nucleotide sequence.
154	Digestion and AbsorptionDigestion of food	No matching NCERT chapter	Single correctEasy
A easy single correct question on digestion of food, from Digestion and Absorption. NCERT: No matching NCERT chapter The secretions of the brush border cells of the intestinal mucosa, together with those of the goblet cells, make up the intestinal juice, or succus entericus. It contains enzymes such as disaccharidases (maltase), dipeptidases, lipases and nucleosidases. So succus entericus is the intestinal juice.
155	Biotechnology and its ApplicationsHuman insulin and vaccine production	Class 12: Biotechnology and its Applications	StatementsMedium
A medium statements question on human insulin and vaccine production, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications Insulin is made as a prohormone, pro-insulin, which has an extra stretch called the C-peptide. In processing, the C-peptide is removed, leaving mature insulin of two chains, A and B, linked by disulphide bridges. (a) Mature insulin has no C-peptide. Correct. (b) In 1983 an American company made insulin by preparing two DNA sequences for the A and B chains, putting them into plasmids of E. coli to make the chains separately, and then joining the chains by disulphide bonds. So rDNA insulin is made without a C-peptide. Incorrect. (c) Pro-insulin contains the C-peptide. Correct. (d) The A and B chains are joined by disulphide bridges. Correct. So (a), (c) and (d) only.
156	Body Fluids and CirculationBlood composition	Class 11: Body Fluids and Circulation	Single correctEasy
A easy single correct question on blood composition, from Body Fluids and Circulation. NCERT: Class 11: Body Fluids and Circulation A person of group AB carries both antigens A and B on the RBCs and therefore has neither anti-A nor anti-B antibody in the plasma. Donor RBCs of any ABO group will find no antibody in the recipient's plasma to agglutinate them, so AB persons can receive blood of any group: universal recipients. So it is due to the absence of anti-A and anti-B antibodies in the plasma.
158	Body Fluids and CirculationBlood composition	Class 11: Body Fluids and Circulation	Single correctEasy
A easy single correct question on blood composition, from Body Fluids and Circulation. NCERT: Class 11: Body Fluids and Circulation In clotting, a cascade of factors forms thrombokinase, which converts inactive prothrombin into thrombin (calcium ions are needed). Thrombin then converts the soluble plasma protein fibrinogen into insoluble fibrin threads, which trap the formed elements to make the clot. So the enzyme that converts fibrinogen to fibrin is thrombin.
159	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 Partial pressures (mm Hg): - atmospheric air: pO$_2$ 159, pCO$_2$ 0.3; - alveoli: pO$_2$ 104, pCO$_2$ 40; - oxygenated blood: pO$_2$ 95, pCO$_2$ 40; - deoxygenated blood: pO$_2$ 40, pCO$_2$ 45; - tissues: pO$_2$ 40, pCO$_2$ 45. So at the alveoli pO$_2$ = 104 and pCO$_2$ = 40.
160	Locomotion and MovementJoints and disorders	Class 11: Locomotion and Movement	Single correctEasy
A easy single correct question on joints and disorders, from Locomotion and Movement. NCERT: Class 11: Locomotion and Movement Myasthenia gravis is an autoimmune disorder affecting the neuromuscular junction, leading to fatigue, weakening and paralysis of skeletal muscle. Arthritis is inflammation of joints; muscular dystrophy is progressive degeneration of skeletal muscle, mostly of genetic origin; gout is inflammation of joints from the accumulation of uric acid crystals. So myasthenia gravis.
163	Reproductive HealthContraception	Class 12: Reproductive Health	Single correctEasy
A easy single correct question on contraception, from Reproductive Health. NCERT: Class 12: Reproductive Health Intra-uterine devices are of three kinds: non-medicated (Lippes loop), copper-releasing (CuT, Cu7, Multiload 375), and hormone-releasing (Progestasert, LNG-20). LNG-20 releases the progestogen levonorgestrel, which makes the uterus unsuitable for implantation and the cervix hostile to sperm. So LNG 20.
166	Animal KingdomNon-chordate phyla	Class 11: Animal Kingdom	StatementsEasy
A easy statements question on non-chordate phyla, from Animal Kingdom. NCERT: Class 11: Animal Kingdom (a) Metagenesis, the alternation of the polyp and medusa generations, is seen in cnidarians such as Obelia, not in helminths. Incorrect. (b) Echinoderms are triploblastic and coelomate. Correct. (c) Roundworms (Aschelminthes) have an organ-system level of organisation. Correct. (d) The eight rows of ciliated comb plates of ctenophores help in locomotion, not digestion. Incorrect. (e) A water vascular system is the distinctive feature of echinoderms, used in locomotion, capture and transport of food, and respiration. Correct. So (b), (c) and (e) are correct.
167	Human ReproductionEmbryonic development	Class 12: Human Reproduction	Single correctMedium
A medium single correct question on embryonic development, from Human Reproduction. NCERT: Class 12: Human Reproduction At fertilisation the sperm passes through the corona radiata and comes in contact with the zona pellucida, the glycoprotein layer round the ovum. The receptors to which the sperm binds (species-specific glycoproteins, ZP3 in mammals) lie in the zona pellucida; binding sets off the acrosome reaction, and after entry of one sperm the zona changes so that further sperms are blocked. So the receptors for sperm binding are on the zona pellucida.
168	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 (a) Metamerism, the body divided into similar segments, is the feature of Annelida: (a)-(iii). (b) The water canal system is the feature of Porifera (sponges): (b)-(iv). (c) Comb plates are the feature of Ctenophora: (c)-(ii). (d) Cnidoblasts (stinging cells) are the feature of Coelenterata (Cnidaria): (d)-(i). So (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i).
169	Chemical Coordination and IntegrationEndocrine glands	Class 11: Chemical Coordination and Integration	Single correctEasy
A easy single correct question on endocrine glands, from Chemical Coordination and Integration. NCERT: Class 11: Chemical Coordination and Integration Several organs that are not endocrine glands secrete hormones. The juxtaglomerular cells of the kidney produce erythropoietin, which stimulates erythropoiesis, the formation of RBCs in the bone marrow. The alpha cells of the pancreas secrete glucagon; the bone marrow is where RBCs are formed, under the action of erythropoietin, not where it is made. So the juxtaglomerular cells of the kidney.
170	Reproductive HealthSTDs	Class 12: Reproductive Health	StatementsEasy
A easy statements question on stds, from Reproductive Health. NCERT: Class 12: Reproductive Health Venereal (sexually transmitted) diseases spread mainly by sexual contact with an infected person. Some, such as hepatitis B and HIV, also spread by sharing injection needles or surgical instruments with infected persons, by transfusion of blood, and from an infected mother to the foetus. (a) Sterile needles carry no infection; it is shared, infected needles that spread these diseases. Not a route. (b) Transfusion of infected blood. A route. (c) Infected mother to foetus. A route. (d) Casual contact such as kissing does not spread them. Not a route. (e) They are infections, not inherited disorders. Not a route. So (b) and (c) only.
171	Structural Organisation in AnimalsCockroach morphology and anatomy	Class 11: Structural Organisation in Animals	StatementsMedium
A medium statements question on cockroach morphology and anatomy, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals (1) The ring of 6–8 blind gastric (hepatic) caeca of the cockroach lies at the junction of the foregut and the midgut, and secretes digestive juice. At the junction of the midgut and hindgut lies a ring of 100–150 yellow Malpighian tubules. So the statement is incorrect. (2) The hypopharynx, a tongue-like median lobe, lies within the cavity enclosed by the mouth parts. Correct. (3) In females the 7th sternum is boat-shaped and, with the 8th and 9th sterna, forms a brood or genital pouch. Correct. (4) In both sexes the 10th segment bears a pair of jointed filamentous anal cerci. Correct. So statement (1) is incorrect.
172	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 (a) Physalia, a cnidarian, is the Portuguese man of war: (a)-(ii). (b) Limulus, the king crab, is a living fossil: (b)-(iii). (c) Ancylostoma is the hookworm: (c)-(iv). (d) Pinctada is the pearl oyster: (d)-(i). So (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i).
173	Animal KingdomChordate classes	Class 11: Animal Kingdom	Single correctEasy
A easy single correct question on chordate classes, from Animal Kingdom. NCERT: Class 11: Animal Kingdom Birds (Aves) have a fully ossified endoskeleton whose long bones are hollow, with air cavities (pneumatic), lightening the body for flight. Neophron, the vulture, is a bird. Hemidactylus (wall lizard) is a reptile; Macropus (kangaroo) and Ornithorhynchus (platypus) are mammals; none has pneumatic bones. So Neophron.
175	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 The first step of each PCR cycle is denaturation: the double-stranded DNA is heated to a high temperature (about 94–95 °C) to separate its strands. If that high temperature is not reached, the strands do not separate, and denaturation is the first step to fail (annealing and extension then cannot follow). Ligation is not a step of PCR. So denaturation.
178	Digestion and AbsorptionHuman digestive system	No matching NCERT chapter	Single correctEasy
A easy single correct question on human digestive system, from Digestion and Absorption. NCERT: No matching NCERT chapter The bile duct and the pancreatic duct open together into the duodenum as the common hepato-pancreatic duct. Its opening is guarded by a sphincter called the sphincter of Oddi. The gastro-oesophageal junction has the gastro-oesophageal sphincter, and the ileo-caecal junction the ileo-caecal valve. So it is at the junction of the hepato-pancreatic duct and the duodenum.
179	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 Haemoglobin binds O$_2$ mainly according to the partial pressure of O$_2$; pCO$_2$, H$^+$ concentration and temperature also affect the binding. In the alveoli, where oxyhaemoglobin forms, the conditions are high pO$_2$, low pCO$_2$, lesser H$^+$ concentration (higher pH) and lower temperature. In the tissues the opposite set (low pO$_2$, high pCO$_2$, more H$^+$, higher temperature) favours dissociation of oxygen from oxyhaemoglobin. So high pO$_2$, low pCO$_2$, less H$^+$, lower temperature.
180	BiomoleculesMetabolic pathways	Class 11: Biomolecules	Single correctEasy
A easy single correct question on metabolic pathways, from Biomolecules. NCERT: Class 11: Biomolecules Some secondary metabolites and their classes: - alkaloids: morphine, codeine; - toxins: abrin, ricin; - lectins: concanavalin A; - drugs: vinblastin, curcumin. Ricin is a toxin, not a drug, so the pair "Drugs - Ricin" is incorrect.
181	Structural Organisation in AnimalsAnimal tissues	Class 11: Structural Organisation in Animals	StatementsEasy
A easy statements question on animal tissues, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals Smooth muscle fibres are spindle-shaped, without striations, and involuntary; they are found in the walls of internal organs such as blood vessels, stomach and intestine. Their cells are held together by cell junctions. Intercalated discs are the specialised junctions of cardiac muscle, where the plasma membranes of adjacent cells fuse and allow the cells to contract as a unit. They are not a feature of smooth muscle. So statement (3) is wrong.
182	Biotechnology and its ApplicationsGene therapy	Class 12: Biotechnology and its Applications	Single correctMedium
A medium single correct question on gene therapy, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications Conventional diagnosis (serum and urine analysis) cannot detect a pathogen early, when its numbers are still very low. The techniques that serve early diagnosis are recombinant DNA technology, PCR and ELISA. ELISA (enzyme-linked immunosorbent assay) works on the principle of antigen–antibody interaction: it detects either the antigens of the pathogen (proteins, glycoproteins) or the antibodies made against it, and is very sensitive. Of the options, ELISA is the molecular diagnostic technique listed for early detection.
183	Reproductive HealthContraception	Class 12: Reproductive Health	MatchingEasy
A easy matching question on contraception, from Reproductive Health. NCERT: Class 12: Reproductive Health (a) Vaults (and cervical caps, diaphragms) are barriers that cover the cervix and block the entry of sperm: (a)-(i). (b) IUDs increase phagocytosis of sperms within the uterus (copper IUDs also suppress sperm motility): (b)-(iii). (c) In vasectomy a small part of the vas deferens is removed or tied up: (c)-(ii). (d) In tubectomy a small part of the fallopian tube is removed or tied up: (d)-(iv). So (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv).
186	Human ReproductionEmbryonic development	Class 12: Human Reproduction	Single correctMedium
A medium single correct question on embryonic development, from Human Reproduction. NCERT: Class 12: Human Reproduction Parturition is set off by a neuroendocrine mechanism. Signals from the fully developed foetus and placenta cause mild uterine contractions (the foetal ejection reflex), which trigger the release of oxytocin from the maternal pituitary. Oxytocin causes stronger contractions, which cause more oxytocin, and so on. A rise in the estrogen : progesterone ratio makes the uterus more responsive, and prostaglandins add to the contractions. Prolactin acts on the mammary glands for milk production after birth; it plays no part in starting parturition. So the release of prolactin.
187	Strategies for Enhancement in Food ProductionAnimal breeding	No matching NCERT chapter	Single correctMedium
A medium single correct question on animal breeding, from Strategies for Enhancement in Food Production. NCERT: No matching NCERT chapter In MOET a cow is given hormones with FSH-like activity to induce follicular maturation and superovulation: it produces 6–8 eggs per cycle instead of one. It is then mated with an elite bull or artificially inseminated. The fertilised eggs at the 8–32 cell stage are recovered non-surgically and transferred to surrogate mothers, while the genetic mother is available for another round. The hormone used has FSH-like activity, not LH-like activity, so statement (1) is not a step of MOET.
190	BiomoleculesCarbohydrates, lipids and nucleic acids	Class 11: Biomolecules	StatementsEasy
A easy statements question on carbohydrates, lipids and nucleic acids, from Biomolecules. NCERT: Class 11: Biomolecules (a) Fatty acids with only single C–C bonds are saturated; unsaturated ones have one or more C = C double bonds. Incorrect. (b) Lecithin is a phospholipid (phosphatidylcholine). Correct. (c) Glycerol is trihydroxy propane. Correct. (d) Palmitic acid has 16 carbon atoms including the carboxyl carbon. Incorrect. (e) Arachidonic acid has 20 carbon atoms including the carboxyl carbon. Incorrect. So (b) and (c) only.
191	Locomotion and MovementSkeletal system	Class 11: Locomotion and Movement	MatchingEasy
A easy matching question on skeletal system, from Locomotion and Movement. NCERT: Class 11: Locomotion and Movement (a) The scapula is a large triangular flat bone on the dorsal side of the thorax: (a)-(iv). (b) The flat bones of the cranium are joined by fibrous joints (sutures), which allow no movement: (b)-(iii). (c) The sternum is a flat bone on the ventral midline of the thorax: (c)-(ii). (d) Adjacent vertebrae of the vertebral column are joined by cartilaginous joints, which allow limited movement: (d)-(i). So (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).
192	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 In epithelia and other tissues, cells are linked by three kinds of junction: - tight junctions stop substances leaking across a tissue; - adhering junctions cement neighbouring cells together; - gap junctions connect the cytoplasm of adjoining cells, for rapid transfer of ions, small molecules and sometimes big molecules. So tight junctions and gap junctions, respectively.
194	Human ReproductionEmbryonic development	Class 12: Human Reproduction	Single correctMedium
A medium single correct question on embryonic development, from Human Reproduction. NCERT: Class 12: Human Reproduction During pregnancy the placenta secretes hCG, hPL, estrogens and progestogens. In the later phase of pregnancy relaxin is also secreted, by the ovary. Within the ovary the hormone-secreting structure in pregnancy is the corpus luteum, which persists because hCG maintains it; relaxin is its product, and it softens the pubic symphysis and cervix for birth. Of the options, the corpus luteum.
195	Structural Organisation in AnimalsFrog: morphology and anatomy	Class 11: Structural Organisation in Animals	StatementsMedium
A medium statements question on frog: morphology and anatomy, from Structural Organisation in Animals. NCERT: Class 11: Structural Organisation in Animals The first body segment of the earthworm, the peristomium (buccal segment), contains the mouth. The prostomium is a fleshy lobe overhanging the mouth: (a) it serves as a covering for the mouth: correct; (b) it acts as a wedge to force open cracks in the soil into which the worm can crawl: correct; (c) it is sensory in function: correct; (d) it is not a segment at all; the first segment is the peristomium: incorrect. So (a), (b) and (c) are correct.
197	Locomotion and MovementMuscle contraction	Class 11: Locomotion and Movement	StatementsEasy
A easy statements question on muscle contraction, from Locomotion and Movement. NCERT: Class 11: Locomotion and Movement By the sliding filament theory, the thin (actin) filaments slide over the thick (myosin) filaments towards the centre of the sarcomere. (a) The H zone, the central part of the A band without thin filaments, disappears as the thin filaments meet. Correct. (b) The A band, the length of the thick filaments, keeps its length. It does not widen. Incorrect. (c) The I band shortens. Correct. (d) The myosin head, an ATPase, hydrolyses ATP; ADP and Pi are released as the cross bridge makes its power stroke. Correct. (e) The Z lines attached to the actin filaments are pulled inwards, shortening the sarcomere. Correct. So (a), (c), (d), (e) only.
198	Biotechnology and its ApplicationsGene therapy	Class 12: Biotechnology and its Applications	Single correctEasy
A easy single correct question on gene therapy, from Biotechnology and its Applications. NCERT: Class 12: Biotechnology and its Applications Adenosine deaminase (ADA) is an enzyme crucial for the immune system to function. Its deficiency, caused by deletion of the gene, results in severe combined immunodeficiency: the immune system does not work. It was the first disease treated by gene therapy (1990), with lymphocytes carrying a functional ADA cDNA. So ADA deficiency results in dysfunction of the immune system.
199	EvolutionEvidences of evolution	Class 12: Evolution	MatchingEasy
A easy matching question on evidences of evolution, from Evolution. NCERT: Class 12: Evolution (a) Darwin's finches, many varieties evolved from one seed-eating ancestor on the Galapagos, are the classic example of adaptive radiation: (a)-(iv). (b) Wings of a butterfly and of a bird do the same job but have different structure and origin: analogous organs, the result of convergent evolution: (b)-(iii). (c) Forelimbs of man and whale share the same basic bones but do different jobs: homologous organs, the result of divergent evolution: (c)-(ii). (d) Excessive use of herbicides and pesticides has selected resistant varieties in a short time: evolution by anthropogenic action: (d)-(i). So (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).
200	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 (a) Filariasis (elephantiasis) is caused by the filarial worm Wuchereria bancrofti: (a)-(iii). (b) Amoebiasis (amoebic dysentery) is caused by the protozoan Entamoeba histolytica: (b)-(iv). (c) Pneumonia is caused by bacteria such as Streptococcus pneumoniae and Haemophilus influenzae: (c)-(i). (d) Ringworm is caused by fungi of the genera Microsporum, Trichophyton and Epidermophyton: (d)-(ii). So (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii).

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