JEE (Advanced) · 2012

JEE Advanced 2012, 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
40 Q
1. Motion of System of Particles and Rigid Body7
2. Electrostatics6
3. Optics4
16 easy · 23 medium · 1 hard
Chemistry
40 Q
1. Thermodynamics4
2. Aldehydes, Ketones and Carboxylic Acids4
3. Coordination Compounds3
17 easy · 20 medium · 3 hard
Mathematics
40 Q
1. Limits, Continuity and Differentiability5
2. Applications of Derivatives4
3. Permutations and Combinations3
11 easy · 26 medium · 3 hard

Chapters

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

Motion of System of Particles and Rigid Body7
Class 11: Systems of Particles and Rotational Motion
Electrostatics6
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Optics4
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Magnetic Effects of Current and Magnetism4
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Kinematics3
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Electromagnetic Induction and Alternating Currents3
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Properties of Bulk Matter2
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Atoms and Nuclei2
Class 12: Atoms; Class 12: Nuclei
Physical World and Measurement1
Class 11: Units and Measurement
Behaviour of Perfect Gas and Kinetic Theory1
Class 11: Kinetic Theory
Current Electricity1
Class 12: Current Electricity
Laws of Motion1
Class 11: Laws of Motion
Oscillations and Waves1
Class 11: Oscillations; Class 11: Waves
Dual Nature of Matter and Radiation1
Class 12: Dual Nature of Radiation and Matter
Thermodynamics1
Class 11: Thermodynamics
Experimental Skills1
No matching NCERT chapter
Gravitation1
Class 11: Gravitation

Limits, Continuity and Differentiability5
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Applications of Derivatives4
Class 12: Application of Derivatives
Permutations and Combinations3
Class 11: Permutations and Combinations
Conic Sections3
Class 11: Conic Sections
Matrices and Determinants3
Class 12: Matrices; Class 12: Determinants
Integral Calculus3
Class 12: Integrals; Class 12: Application of Integrals
Three Dimensional Geometry3
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Circles3
Class 11: Conic Sections
Trigonometry3
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Probability3
Class 11: Probability; Class 12: Probability
Sequences and Series2
Class 11: Sequences and Series
Vector Algebra2
Class 12: Vector Algebra
Sets, Relations and Functions1
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Complex Numbers1
Class 11: Complex Numbers and Quadratic Equations
Differential Equations1
Class 12: Differential Equations

Questions

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

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
1	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take the insect's mass as $m$ and the rod's moment of inertia about $O$ as $I_0$. At time $t < T$ the insect is at distance $r = vt$ from $O$, so the angular momentum of the system about the vertical axis is $L = (I_0 + m v^2 t^2)\,\omega$. With $\omega$ constant, the torque needed is $\tau = \frac{dL}{dt} = 2 m v^2 \omega\, t$, which grows linearly from zero. For $t > T$ the insect has stopped at the end, $L$ is constant, so $\tau = 0$. (The insect's outward motion also needs no torque by itself; only the changing moment of inertia does.) So $\|\vec{\tau}\|$ rises as a straight line through the origin up to $t = T$ and then drops to zero: plot (B). Answer: (B).
2	Properties of Bulk MatterHeat transfer	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctEasy
A easy single correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter In steady state the middle plate absorbs as much radiation as it emits. It receives $\sigma(2T)^4$ per unit area from the first plate and $\sigma(3T)^4$ from the third, and, being black on both faces with high conductivity (one temperature $T_2$), it emits $\sigma T_2^4$ from each of its two faces: $2\sigma T_2^4 = \sigma(16T^4 + 81T^4) = 97\sigma T^4$ $T_2 = \left(\frac{97}{2}\right)^{1/4} T$. Answer: (C).
3	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 By Gauss's law, a Gaussian sphere of radius $r < R$ encloses no charge, so $E = 0$ inside the shell. Outside, $E = \frac{Q}{4\pi\varepsilon_0 r^2}$: it jumps to its largest value $\frac{Q}{4\pi\varepsilon_0 R^2}$ just outside $r = R$ and then falls off as $1/r^2$. The potential outside is $V = \frac{Q}{4\pi\varepsilon_0 r}$; since $E = 0$ inside, $V$ is constant there and equal to its surface value $\frac{Q}{4\pi\varepsilon_0 R}$. So $V$ is flat up to $R$ and then decreases as $1/r$, continuously. In the graphs the full line is $\|\vec{E}\|$ and the dashed line is $V$. Only graph (D) has $\|\vec{E}\|$ zero up to $R$, a jump at $R$ and a decay, together with $V$ constant inside and decaying outside. Answer: (D).
4	Physical World and MeasurementError analysis	Class 11: Units and Measurement	Single correctEasy
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Least count of each instrument $= \frac{\text{pitch}}{\text{divisions}} = \frac{0.5}{100} = 0.005$ mm, so $\Delta d = \Delta l = 0.005$ mm. From $Y = \frac{4MLg}{\pi l d^2}$: $\frac{\Delta Y}{Y} = \dots + \frac{\Delta l}{l} + 2\frac{\Delta d}{d}$. Contribution of $l$: $\frac{0.005}{0.25} = 0.02$. Contribution of $d$: $2\times\frac{0.005}{0.5} = 0.02$. The two contributions are the same. Answer: (A).
5	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 Block: it starts at rest at the extreme position $x = 4.9 + 0.2 = 5.1$ m, so $x(t) = 4.9 + 0.2\cos\left(\frac{\pi t}{3}\right)$. At $t = 1$ s: $x = 4.9 + 0.2\cos 60^\circ = 4.9 + 0.1 = 5.0$ m. Pebble: it is launched from $P$ ($x = 10$ m) towards $O$ and must land on the block (at ground level) at $t = 1$ s, at $x = 5$ m, so its range is $5$ m and its time of flight is $1$ s. Time of flight: $\frac{2v\sin 45^\circ}{g} = 1 \Rightarrow v\sin45^\circ = 5 \Rightarrow v = 5\sqrt{2} = \sqrt{50}$ m/s. Check with the range: $\frac{v^2\sin 90^\circ}{g} = \frac{50}{10} = 5$ m. ✓ Answer: (A).
6	OpticsInterference and diffraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Fringe width $\beta = \frac{\lambda D}{d}$ is proportional to the wavelength. $\lambda_R > \lambda_G > \lambda_B$ (red about 700 nm, green about 530 nm, blue about 470 nm), so $\beta_R > \beta_G > \beta_B$. Answer: (D).
7	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion About $O$: the position vector $\vec{r}$ of the mass lies in the $x$-$y$ plane and the velocity $\vec{v}$ is tangent to the circle, also in that plane. $\vec{L}_O = m\,\vec{r}\times\vec{v}$ is along $+z$ with magnitude $m r^2\omega$, fixed in size and direction: constant. About $P$: the position vector from $P$ is $\vec{r}' = \vec{r} - h\hat{k}$ (with $P$ at height $h$ above $O$). Then $\vec{L}_P = m(\vec{r} - h\hat{k})\times\vec{v} = \vec{L}_O - m h\,\hat{k}\times\vec{v}$. The second term lies in the horizontal plane and turns with the mass, so $\vec{L}_P$ has a constant magnitude but a rotating direction: it varies with time. (Equivalently, the net force — tension plus gravity — is the horizontal centripetal force through $O$, so its torque about $O$ is zero but its torque about $P$ is not.) Answer: (C).
8	Behaviour of Perfect Gas and Kinetic TheoryKinetic theory	Class 11: Kinetic Theory	Single correctEasy
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$ depends only on the temperature and the molar mass, not on the number of moles. Both gases are at the same temperature, so $\frac{v_{rms}(\text{He})}{v_{rms}(\text{Ar})} = \sqrt{\frac{M_{Ar}}{M_{He}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$. Answer: (D).
9	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 The proton feels its weight $mg$ downward and the electric force $eE$ horizontally (the plates are vertical). Released from rest, it moves along the net force, which makes $45^\circ$ with the vertical when $eE = mg$. $E = \frac{X}{d}$, so $X = \frac{mgd}{e} = \frac{(1.67\times10^{-27})(10)(10^{-2})}{1.6\times10^{-19}} \approx 1.0\times10^{-9}$ V. Answer: (C).
10	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a thin plano-convex lens, $\frac{1}{f} = (n-1)\frac{1}{R}$. The two lenses are in contact, so their powers add: $\frac{1}{f} = \frac{1.5 - 1}{14} + \frac{1.2 - 1}{14} = \frac{0.7}{14} = \frac{1}{20}$, $f = 20$ cm. Object at $u = -40$ cm: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40}$, so $v = 40.0$ cm. Answer: (B).
11	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Multiple correctMedium
A medium multiple correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance All three charges lie on the $y$-axis inside the cube. (A) True. The arrangement is symmetric under $x \to -x$ (every charge has $x = 0$), so the flux through the face $x = +a/2$ equals that through $x = -a/2$. (B) False. The arrangement is also symmetric under $y \to -y$ (the two $-q$ charges are at $y = \pm a/4$), so the fluxes through $y = +a/2$ and $y = -a/2$ are equal, not one more than the other. (C) True. Net charge enclosed $= -q + 3q - q = q$, so by Gauss's law the total flux out of the cube is $\frac{q}{\varepsilon_0}$. (D) True. The charges lie on the $y$-axis, so the arrangement is unchanged by swapping $x$ and $z$ (a rotation by $90^\circ$ about the $y$-axis). The face $z = +a/2$ maps onto $x = +a/2$, so the fluxes are equal. Answer: (A), (C), (D).
12	Current ElectricityKirchhoff's rules	Class 12: Current Electricity	Multiple correctMedium
A medium multiple correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Call the left junction $A$ and the right junction $B$; the $12$ V battery maintains $V_A - V_B = 12$ V (the current $I_1$ flows up the left wire into $A$). Branches: $A$–$P$ $2\,\Omega$, $P$–$S$ $2\,\Omega$, $S$–$B$ $2\,\Omega$ (top); $A$–$Q$ $4\,\Omega$, $Q$–$T$ $4\,\Omega$, $T$–$B$ $4\,\Omega$ (bottom); bridges $P$–$Q$ $1\,\Omega$ and $S$–$T$ $1\,\Omega$. Try zero current in both bridges. Top path: $6\,\Omega$ carries $\frac{12}{6} = 2$ A; bottom path: $12\,\Omega$ carries $\frac{12}{12} = 1$ A. Taking $V_B = 0$, $V_A = 12$ V: $V_P = 12 - 2\times2 = 8$ V and $V_Q = 12 - 1\times4 = 8$ V; $V_S = 8 - 4 = 4$ V and $V_T = 8 - 4 = 4$ V. $V_P = V_Q$ and $V_S = V_T$, so the bridges indeed carry no current (each half is a balanced Wheatstone bridge, $\frac{2}{4} = \frac{2}{4}$), and this is the solution. (A) True: the current through $PQ$ is zero. (B) True: $I_1 = 2 + 1 = 3$ A. (C) True: $V_S = 4$ V is less than $V_Q = 8$ V. (D) True: $I_2$, the current in $PS$, is $2$ A. Answer: (A), (B), (C), (D).
13	Laws of MotionFriction	Class 11: Laws of Motion	Multiple correctMedium
A medium multiple correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Weight $mg = 0.1\times10 = 1$ N. The incline rises from $P$ (lower end) to $Q$ (upper end) and the horizontal $1$ N force pushes the block towards the vertical side $OQ$, i.e. into the incline. Along the incline: the horizontal force has component $1\cos\theta$ up the incline (towards $Q$); gravity has component $1\sin\theta$ down the incline (towards $P$). (A) True. At $\theta = 45^\circ$ the two components are equal, so the block is in equilibrium with no friction needed: it stays at rest. (B) False and (C) True. For $\theta > 45^\circ$, $\sin\theta > \cos\theta$, so the block tends to slide down towards $P$; friction (if large enough) acts up the incline, towards $Q$, and it can stay at rest. (D) False. For $\theta < 45^\circ$ the block tends to move up towards $Q$, so friction acts towards $P$, not $Q$. Answer: (A), (C).
14	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Multiple correctMedium
A medium multiple 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 Both fields are along $y$. The magnetic force $q\vec{v}\times\vec{B}$ acts only on the velocity component perpendicular to $y$ (the $x$-component $v\cos\theta$), making the charge circle in a plane parallel to $x$-$z$. The electric force $qE_0\hat{j}$ accelerates it uniformly along $y$, and the magnetic force does not affect $v_y$. (A) False. At $\theta = 0^\circ$ the charge does circle about the $y$-direction, but it is also accelerated along $y$, so the path is not confined to the $x$-$z$ plane. (B) False. The motion is helical along $y$, but $v_y$ grows, so the pitch $v_y\cdot\frac{2\pi m}{qB_0}$ increases each turn; it is not constant. (C) True. At $\theta = 10^\circ$ there is a perpendicular component (circling) and a $y$-component that keeps increasing: a helix along $y$ with increasing pitch. (D) True. At $\theta = 90^\circ$ the velocity is along $\vec{B}$, so there is no magnetic force; the electric force along $y$ accelerates it in a straight line along $y$. Answer: (C), (D).
15	Oscillations and WavesWave motion	Class 11: Oscillations; Class 11: Waves	Multiple correctMedium
A medium multiple correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At an open end the pressure is held at atmospheric value (a pressure node), so a pressure pulse reflects with its phase reversed: a high-pressure (compression) pulse returns as a low-pressure (rarefaction) pulse. At a closed end the air cannot move (a displacement node, a pressure antinode), so a pressure pulse reflects without change of phase: a high-pressure pulse returns as a high-pressure pulse. (A) False, (B) True, (C) False, (D) True. Answer: (B), (D).
16	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	NumericalMedium
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Treat the body as a full cylinder of density $\rho$ plus a sphere of density $-\rho$ filling the cavity; $P$ lies on the line through the cavity's centre perpendicular to the axis, at distance $2R$. Full cylinder (Gauss's law, outside): $E_1 = \frac{\rho R^2}{2\varepsilon_0 r} = \frac{\rho R^2}{2\varepsilon_0 (2R)} = \frac{\rho R}{4\varepsilon_0}$, directed away from the axis. Sphere of radius $R/2$ with charge $-\rho\cdot\frac{4}{3}\pi\left(\frac{R}{2}\right)^3$: at distance $2R$, $E_2 = \frac{\rho (R/2)^3}{3\varepsilon_0 (2R)^2} = \frac{\rho R}{96\varepsilon_0}$, directed towards the cavity centre (opposite to $E_1$). $E = \frac{\rho R}{4\varepsilon_0} - \frac{\rho R}{96\varepsilon_0} = \frac{23\rho R}{96\varepsilon_0} = \frac{23\rho R}{16\times6\,\varepsilon_0}$. So $k = 6$. Answer: 6.
17	Magnetic Effects of Current and MagnetismAmpere's law	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	NumericalMedium
A medium numerical question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter From the figure: the cylinder has radius $a$ and centre $O$; the cavity has radius $a/2$ and touches the outer surface, so its centre $C$ is at $a/2$ from $O$; $P$ is on the surface on the far side, on the line $CO$, at distance $a$ from $O$ and $\frac{3a}{2}$ from $C$. Superpose a full cylinder carrying $J$ and a cavity-sized cylinder carrying $-J$. Full cylinder at its surface (Ampère's law): $B_1 = \frac{\mu_0 J \pi a^2}{2\pi a} = \frac{\mu_0 J a}{2}$. Cavity cylinder (radius $a/2$) at distance $\frac{3a}{2}$: $B_2 = \frac{\mu_0 J\pi (a/2)^2}{2\pi(3a/2)} = \frac{\mu_0 J a}{12}$. $P$, $O$ and $C$ are on one line, so both fields at $P$ are along the same line (perpendicular to $PO$), and the cavity's reversed current makes $B_2$ opposite to $B_1$: $B = \frac{\mu_0 J a}{2} - \frac{\mu_0 J a}{12} = \frac{5}{12}\mu_0 a J$. So $N = 5$. Answer: 5.
18	Motion of System of Particles and Rigid BodyMoment of inertia	Class 11: Systems of Particles and Rotational Motion	NumericalMedium
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion From the figure: the big disc has radius $2R$ and centre $O$; the removed disc has radius $R$ and its centre is at distance $R$ from $O$ (it passes through $O$ and touches the rim); $P$ is on the rim of the big disc, at distance $2R$ from $O$ along the direction perpendicular to the line joining the centres. Let the removed disc have mass $m$; the big disc (four times the area) has mass $4m$. About $O$: big disc $\frac{1}{2}(4m)(2R)^2 = 8mR^2$; removed disc $\frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$. $I_O = 8mR^2 - \frac{3}{2}mR^2 = \frac{13}{2}mR^2$. About $P$: big disc $8mR^2 + 4m(2R)^2 = 24mR^2$; removed disc: its centre is at distance $\sqrt{(2R)^2 + R^2} = \sqrt{5}R$ from $P$, so $\frac{1}{2}mR^2 + 5mR^2 = \frac{11}{2}mR^2$. $I_P = 24mR^2 - \frac{11}{2}mR^2 = \frac{37}{2}mR^2$. $\frac{I_P}{I_O} = \frac{37}{13} \approx 2.85$, which is $3$ to the nearest integer. Answer: 3.
19	Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	NumericalMedium
A medium numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Pass current $I$ in the circular loop. On its axis at $z = \sqrt{3}R$: $B = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 I}{16R}$, along $z$. Since $a \ll R$, this field is uniform over the small square. The plane of the square makes $45^\circ$ with the $z$-axis, so its normal also makes $45^\circ$ with $z$. Flux through the two turns: $\Phi = 2\,B a^2\cos45^\circ = \frac{2}{\sqrt{2}}\cdot\frac{\mu_0 I a^2}{16R} = \frac{\mu_0 I a^2}{8\sqrt{2}R}$. $M = \frac{\Phi}{I} = \frac{\mu_0 a^2}{2^{3}\cdot2^{1/2}R} = \frac{\mu_0 a^2}{2^{7/2}R}$. So $\frac{p}{2} = \frac{7}{2}$, $p = 7$. Answer: 7.
20	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	NumericalMedium
A medium numerical question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter At closest approach (head-on) all the initial kinetic energy has become potential energy: $K = \frac{1}{4\pi\varepsilon_0}\frac{Qe}{r} = eV$ with $V = \frac{9\times10^9\times120\times1.6\times10^{-19}}{10\times10^{-15}} = 1.728\times10^{7}$ V. $\lambda = \frac{h}{\sqrt{2m_pK}} = \frac{h}{\sqrt{2m_p eV}} = \frac{h}{e}\sqrt{\frac{e}{2m_pV}}$. $\frac{e}{2m_pV} = \frac{1.6\times10^{-19}}{2\times\frac{5}{3}\times10^{-27}\times1.728\times10^{7}} = \frac{1.6\times10^{-19}}{5.76\times10^{-20}} = \frac{25}{9}$, so its square root is $\frac{5}{3}$. $\lambda = 4.2\times10^{-15}\times\frac{5}{3} = 7.0\times10^{-15}$ m $= 7$ fm. Answer: 7.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
21	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 Allene is $\mathrm{H_2C{=}C{=}CH_2}$. Each terminal carbon forms one double bond and two C–H bonds: three σ-bonds, so $sp^2$. The central carbon forms two double bonds: two σ-bonds and two π-bonds, so $sp$. The carbons are $sp$ and $sp^2$. Answer: (B).
22	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	Single correctMedium
A medium single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics For one mole with $b = 0$, the van der Waals equation $\left(P + \frac{a}{V^2}\right)V = RT$ gives $PV = RT - a\cdot\frac{1}{V}$. So $PV$ against $\frac{1}{V}$ is a straight line with intercept $RT$ ($= 0.082\times300 = 24.6$ L atm, as on the graph) and slope $-a$. From the two marked points $(2.0,\ 21.6)$ and $(3.0,\ 20.1)$: slope $= \frac{20.1 - 21.6}{3.0 - 2.0} = -1.5$. $a = 1.5$ atm L$^2$ mol$^{-2}$. Answer: (C).
23	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Complete ozonolysis cleaves all three C=C bonds, each C=C becoming two C=O groups: $\mathrm{CH_3CH{=}\|{=}CH{-}CH(CH_3){-}CH{=}\|{=}CH{-}CH(CH_3){-}CH{=}\|{=}CHCH_3}$ The fragments are: $\mathrm{CH_3CHO}$ (twice), and $\mathrm{OHC{-}CH(CH_3){-}CHO}$ (twice, one from each former stereocentre). In 2-methylpropanedial the central carbon carries $\mathrm{H}$, $\mathrm{CH_3}$ and two identical $\mathrm{CHO}$ groups, so it is not a stereocentre: the molecule is achiral. Ethanal is achiral too. Number of optically active products $= 0$. Answer: (A).
24	Solid StatePacking in fcc, bcc and hcp	No matching NCERT chapter	Single correctEasy
A easy single correct question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter X (circles) sits at the 8 corners and the 6 face centres: $8\times\frac{1}{8} + 6\times\frac{1}{2} = 4$ X per cell. M (squares) sits at the body centre (1, wholly inside) and at the centres of the four edges shown (each edge centre is shared by 4 cells): $1 + 4\times\frac{1}{4} = 2$ M per cell. $\mathrm{M : X} = 2 : 4$, so the empirical formula is $\mathrm{MX_2}$. Answer: (B).
25	Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The product is pentaerythritol, $\mathrm{C(CH_2OH)_4}$. Ethanal has three α-hydrogens. Each is used in a crossed aldol addition to methanal (which has none and is the acceptor): 1. $\mathrm{CH_3CHO + HCHO \rightarrow HOCH_2CH_2CHO}$ 2. $\mathrm{HOCH_2CH_2CHO + HCHO \rightarrow (HOCH_2)_2CHCHO}$ 3. $\mathrm{(HOCH_2)_2CHCHO + HCHO \rightarrow (HOCH_2)_3CCHO}$ The last aldehyde has no α-hydrogen left; the fourth HCHO reduces it in a crossed Cannizzaro reaction to $\mathrm{C(CH_2OH)_4}$ (HCHO is oxidised to formate). So three aldol reactions occur. Answer: (C).
26	Coordination CompoundsCrystal field theory	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Aqueous $\mathrm{CuSO_4}$ contains $[\mathrm{Cu(H_2O)_6}]^{2+}$, a $d^9$ ion whose d–d transition absorbs in the red/orange-red region (around 600–800 nm). The transmitted (seen) light is the complementary colour, blue (blue-green), which is why the solution looks blue. So the light absorbed is orange-red. Answer: (A).
27	Chemistry in Everyday LifeDrugs and their classification	No matching NCERT chapter	Single correctEasy
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Picric acid is 2,4,6-trinitrophenol: its acidic group is a phenolic –OH. Barbituric acid is a cyclic ureide (pyrimidine-2,4,6-trione): its acidity is from C–H/N–H, with no –COOH. Ascorbic acid (vitamin C) is a lactone whose acidity comes from an enediol –OH. Aspirin is acetylsalicylic acid, 2-acetoxybenzoic acid: it has a –COOH group. Answer: (D).
28	Structure of AtomBohr model	Class 11: Structure of Atom	Single correctMedium
A medium single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Bohr's quantisation: $mvr = \frac{nh}{2\pi}$, and for hydrogen $r_n = n^2a_0$. For $n = 2$: $r = 4a_0$, so $v = \frac{2h}{2\pi m(4a_0)} = \frac{h}{4\pi m a_0}$. $KE = \frac{1}{2}mv^2 = \frac{1}{2}m\cdot\frac{h^2}{16\pi^2m^2a_0^2} = \frac{h^2}{32\pi^2 m a_0^2}$. Answer: (C).
29	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 Oxidation states of nitrogen: $\mathrm{HNO_3}$: $+1 + x - 6 = 0 \Rightarrow x = +5$. $\mathrm{NO}$: $+2$. $\mathrm{N_2}$: $0$. $\mathrm{NH_4Cl}$: in $\mathrm{NH_4^+}$, $x + 4 = +1 \Rightarrow x = -3$. Decreasing order: $\mathrm{HNO_3}\,(+5) > \mathrm{NO}\,(+2) > \mathrm{N_2}\,(0) > \mathrm{NH_4Cl}\,(-3)$. Answer: (B).
30	Coordination CompoundsNomenclature	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Ligands are named in alphabetical order, ignoring the multiplying prefixes: "ammine" ($\mathrm{NH_3}$, spelt with double m) comes before "aqua" ($\mathrm{H_2O}$) because "am" precedes "aq". Two $\mathrm{NH_3}$: diammine; four $\mathrm{H_2O}$: tetraaqua. Cobalt is in the +3 state ($x + 0 = +3$ for the cation balancing three $\mathrm{Cl^-}$). Name: diamminetetraaquacobalt(III) chloride. "Diamine" (single m) in (C) is wrong spelling for the ammonia ligand, and (A)/(B) put aqua first. Answer: (D).
31	Practical Organic ChemistryChemical methods of separation	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on chemical methods of separation, from Practical Organic Chemistry. NCERT: No matching NCERT chapter The scheme needs a mixture that is separated both by aqueous NaOH and by aqueous $\mathrm{NaHCO_3}$: one component must dissolve in each base and the other in neither. NaOH dissolves both carboxylic acids and phenols (as their sodium salts); $\mathrm{NaHCO_3}$ dissolves only the stronger carboxylic acids, not phenols. Benzyl alcohol dissolves in neither. (A) Phenol + benzoic acid: NaOH dissolves both, so NaOH does not separate them. Not correct. (B) Benzoic acid + benzyl alcohol: either base takes the acid into water and leaves the alcohol. Correct. (C) Benzyl alcohol + phenol: $\mathrm{NaHCO_3}$ dissolves neither. Not correct. (D) Benzyl alcohol + phenylacetic acid: either base dissolves the acid only. Correct. Answer: (B), (D).
32	Surface ChemistryColloids and their preparation	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Lyophobic sols are stabilised by charge, not by solvation. (A) Correct. The particles preferentially adsorb ions (commonly their own ions) from the solution and become charged; like charges repel and prevent coagulation. (B) Incorrect. Preferential adsorption of solvent (a solvation layer) is what stabilises lyophilic colloids. (C) Incorrect. Attraction between oppositely charged particles would cause coagulation, not stability. (D) Correct. The fixed layer of adsorbed ions and the diffuse layer of counter-ions form an electrical double layer; the potential difference across it (the zeta potential) keeps the particles apart. Answer: (A), (D).
33	Organic Chemistry: Some Basic Principles and TechniquesElectronic effects	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Multiple correctHard
A hard multiple correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) Cyclohexa-1,3-diene: an ordinary conjugated diene, non-aromatic but stable. Not unstable. (B) Cyclobutadiene: planar, cyclic, conjugated with $4\pi$ electrons — antiaromatic. It cannot be isolated at room temperature (it dimerises at once). Unstable. (C) Cyclopentadienone: the polarised C=O leaves a partial positive charge on carbon, giving the ring cyclopentadienyl-cation character with $4\pi$ electrons — antiaromatic. It dimerises rapidly at room temperature. Unstable. (D) Tropone (cycloheptatrienone): the polarisation C$^{\delta+}$–O$^{\delta-}$ gives the ring tropylium-cation character with $6\pi$ electrons — aromatic, so it is stable. Answer: (B), (C).
34	Principles of Qualitative AnalysisCharacteristic precipitates and colours	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter HCl, HBr and HI give precipitates of AgCl (white), AgBr (pale yellow) and AgI (yellow) with silver nitrate. Thiosulfate forms the very stable complex $[\mathrm{Ag(S_2O_3)_2}]^{3-}$ and dissolves all three (this is why "hypo" is the photographic fixer): $\mathrm{AgX + 2S_2O_3^{2-} \rightarrow [Ag(S_2O_3)_2]^{3-} + X^-}$ HF gives no precipitate, because AgF is freely soluble in water. Answer: (A), (C), (D).
35	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Multiple correctMedium
A medium multiple correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics From the figure, $X \to Y$ is an expansion at constant pressure and $Y \to Z$ is a pressure drop at constant volume; the dashed path goes straight from $X$ to $Z$. (A) Correct. Entropy is a state function, so $\Delta S_{x\to z} = \Delta S_{x\to y} + \Delta S_{y\to z}$. (B) Incorrect. Work is a path function: along the direct path the pressure falls during the expansion, so the work differs from that along $X\to Y\to Z$. (C) Correct. $Y \to Z$ is at constant volume, so $w_{y\to z} = 0$ and $w_{x\to y\to z} = w_{x\to y}$. (D) Incorrect. $\Delta S_{x\to y\to z} = \Delta S_{x\to y} + \Delta S_{y\to z}$, and $\Delta S_{y\to z} = nC_v\ln\frac{T_z}{T_y} \neq 0$ (the temperature falls as the pressure falls at constant volume). Answer: (A), (C).
36	BiomoleculesProteins	Class 12: Biomolecules	NumericalMedium
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules At pH 7 the terminal $\mathrm{NH_3^+}$ and $\mathrm{COO^-}$ cancel, so the net charge comes from the side chains: $\mathrm{(CH_2)_4NH_2}$ (lysine) is protonated to $\mathrm{(CH_2)_4NH_3^+}$: $+1$. $\mathrm{CH_2COOH}$ (aspartic acid) is ionised to $\mathrm{CH_2COO^-}$: $-1$. H, $\mathrm{CH_3}$, $\mathrm{CH_2OH}$, $\mathrm{CH_2CONH_2}$ (amide, not basic): $0$. Net charges: I $0$; II $0$; III $-1$; IV $+1$; V $0$; VI $+2$; VII $-1$; VIII $+1$; IX $+1$. Positively charged: IV, VI, VIII, IX — four peptides. Answer: 4.
37	Classification of Elements and PeriodicityModern periodic table	Class 11: Classification of Elements and Periodicity in Properties	NumericalEasy
A easy numerical question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Balance mass numbers: $63 + 1 = 6(1) + 4 + 2(1) + A \Rightarrow A = 52$. Balance atomic numbers: $29 + 1 = 6(0) + 2 + 2(1) + Z \Rightarrow Z = 26$. $Z = 26$ is iron, $[\mathrm{Ar}]3d^64s^2$: 8 electrons beyond the argon core, so it is in group 8. Answer: 8.
38	BiomoleculesCarbohydrates	Class 12: Biomolecules	NumericalHard
A hard numerical question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules The open chain $\mathrm{C1HO{-}C2H_2{-}C3HOH{-}C4HOH{-}C5HOH{-}C6H_2OH}$ (a 2-deoxyhexose) has stereocentres at C3, C4 and C5. In the pyranose form the ring closes between the C5–OH and C1, so C1 becomes a new stereocentre (the anomeric carbon, α or β). The ring now has stereocentres C1, C3, C4 and C5. The D-configuration fixes C5. The other three, C1, C3 and C4, are free: $2^3 = 8$ stereoisomers. Answer: 8.
39	Some Basic Concepts of ChemistryMole concept	Class 11: Some Basic Concepts of Chemistry	NumericalEasy
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Molarity of the stock: 1000 mL weighs $1250$ g and contains $0.292\times1250 = 365$ g HCl $= \frac{365}{36.5} = 10$ mol. So the stock is $10$ M. Dilution: $M_1V_1 = M_2V_2 \Rightarrow 10\times V_1 = 0.4\times200 \Rightarrow V_1 = 8$ mL. Answer: 8.
40	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	NumericalEasy
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For a first-order reaction, the time to fall to a fraction $f$ of the initial concentration is $t = \frac{2.303}{k}\log\frac{1}{f}$. $t_{1/8} = \frac{2.303}{k}\log 8 = \frac{2.303}{k}(3\times0.3) = \frac{2.303}{k}(0.9)$. $t_{1/10} = \frac{2.303}{k}\log 10 = \frac{2.303}{k}(1)$. $\frac{t_{1/8}}{t_{1/10}}\times10 = 0.9\times10 = 9$. Answer: 9.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
41	Permutations and CombinationsSimple applications	Class 11: Permutations and Combinations	Single correctMedium
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations We count onto maps from 5 distinct balls to 3 distinct persons, by inclusion–exclusion. All distributions: $3^5 = 243$. Subtract those missing a given person: $\binom{3}{1}2^5 = 96$. Add back those missing two given persons: $\binom{3}{2}1^5 = 3$. $243 - 96 + 3 = 150$. (Check by cases: split $3,1,1$ gives $\binom{5}{3}\cdot2\cdot3 = 60$ ways after assigning to persons; split $2,2,1$ gives $\frac{\binom{5}{2}\binom{3}{2}}{2}\cdot3! = 90$; total $150$.) Answer: (B).
42	Limits, Continuity and DifferentiabilityDifferentiability	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctMedium
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = 0$: $\frac{f(h) - f(0)}{h} = h\left\|\cos\frac{\pi}{h}\right\|$, and $\left\|h\cos\frac{\pi}{h}\right\| \le \|h\| \to 0$. So $f'(0) = 0$: differentiable at $0$. At $x = 2$: $\cos\frac{\pi}{2} = 0$, so $f(2) = 0$. Let $g(x) = x^2\cos\frac{\pi}{x}$; then $f = \|g\|$ near $2$ and $g'(x) = 2x\cos\frac{\pi}{x} + \pi\sin\frac{\pi}{x}$, $g'(2) = 0 + \pi = \pi \neq 0$. So $g$ changes sign at $2$ with non-zero slope, and $\|g\|$ has left derivative $-\pi$ and right derivative $+\pi$: not differentiable at $2$. Answer: (B).
43	Sets, Relations and FunctionsOne-one, into and onto functions	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	Single correctEasy
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f'(x) = 6x^2 - 30x + 36 = 6(x - 2)(x - 3)$. On $[0, 2]$, $f' \ge 0$: $f$ increases from $f(0) = 1$ to $f(2) = 16 - 60 + 72 + 1 = 29$. On $[2, 3]$, $f' \le 0$: $f$ decreases to $f(3) = 54 - 135 + 108 + 1 = 28$. Range $= [1, 29]$, equal to the codomain: onto. Every value in $[28, 29)$ is taken once on $[0,2]$ and once on $(2,3]$: not one-one. Answer: (B).
44	Limits, Continuity and DifferentiabilityLimits of functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctEasy
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\frac{x^2 + x + 1}{x + 1} = x + \frac{1}{x + 1}$. So the expression is $(1 - a)x - b + \frac{1}{x + 1}$. For a finite limit the $x$ term must vanish: $a = 1$. Then the limit is $-b = 4$, so $b = -4$. Answer: (B).
45	Complex NumbersAlgebra of complex numbers	Class 11: Complex Numbers and Quadratic Equations	Single correctMedium
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Write $z = x + iy$ with $y \neq 0$. $z^2 + z + 1 = (x^2 - y^2 + x + 1) + i(2xy + y)$. Real means $y(2x + 1) = 0$; as $y \neq 0$, $x = -\frac{1}{2}$. Then $a = \frac{1}{4} - y^2 - \frac{1}{2} + 1 = \frac{3}{4} - y^2 < \frac{3}{4}$ (strictly, since $y \neq 0$). So $a$ can be $-1$, $\frac{1}{3}$ or $\frac{1}{2}$, but not $\frac{3}{4}$. (Alternatively: $z^2 + z + 1 - a = 0$ must have non-real roots, so its discriminant $1 - 4(1 - a) < 0$, i.e. $a < \frac{3}{4}$.) Answer: (D).
46	Conic SectionsEllipse	Class 11: Conic Sections	Single correctMedium
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The rectangle around $E_1$ has sides $x = \pm3$, $y = \pm2$, so its vertices are $(\pm3, \pm2)$. Take $E_2$ with the same axes: $\frac{x^2}{A} + \frac{y^2}{B} = 1$. Through $(0, 4)$: $B = 16$. Through $(3, 2)$: $\frac{9}{A} + \frac{4}{16} = 1 \Rightarrow A = 12$. Since $16 > 12$, the major axis is along $y$: $e^2 = 1 - \frac{12}{16} = \frac{1}{4}$, $e = \frac{1}{2}$. Answer: (C).
47	Matrices and DeterminantsDeterminants of order two and three	Class 12: Matrices; Class 12: Determinants	Single correctEasy
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $b_{ij} = 2^i\cdot a_{ij}\cdot 2^j$, so $Q = DPD$ with $D = \mathrm{diag}(2, 4, 8)$: row $i$ is multiplied by $2^i$ and column $j$ by $2^j$. $\det Q = (\det D)^2\det P = (2\cdot4\cdot8)^2\cdot 2 = (2^6)^2\cdot2 = 2^{13}$. Answer: (D).
48	Integral CalculusIntegration by substitution	Class 12: Integrals; Class 12: Application of Integrals	Single correctHard
A hard single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $t = \sec x + \tan x$. Then $dt = (\sec x\tan x + \sec^2x)\,dx = \sec x\cdot t\,dx$, and since $\sec x - \tan x = \frac{1}{t}$, $\sec x = \frac{1}{2}\left(t + \frac{1}{t}\right)$. $\sec^2x\,dx = \sec x\cdot\frac{dt}{t} = \frac{1}{2}\left(1 + \frac{1}{t^2}\right)dt$. $I = \frac{1}{2}\int\left(t^{-9/2} + t^{-13/2}\right)dt = \frac{1}{2}\left(\frac{t^{-7/2}}{-7/2} + \frac{t^{-11/2}}{-11/2}\right) + K = -\frac{t^{-7/2}}{7} - \frac{t^{-11/2}}{11} + K$ $= -\frac{1}{t^{11/2}}\left\{\frac{1}{11} + \frac{1}{7}t^2\right\} + K$, with $t = \sec x + \tan x$. Answer: (C).
49	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctMedium
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Direction of $QR$: $Q - R = (1, 4, 1)$. Points of the line: $(1 + s,\ -1 + 4s,\ 4 + s)$. $P$: $5(1 + s) - 4(-1 + 4s) - (4 + s) = 1 \Rightarrow 5 - 12s = 1 \Rightarrow s = \frac{1}{3}$, so $P = \left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$. $S$: $(T - R)\cdot(1,4,1) = (1, 2, 0)\cdot(1, 4, 1) = 9$ and $\|(1,4,1)\|^2 = 18$, so $s = \frac{1}{2}$ and $S = \left(\frac{3}{2}, 1, \frac{9}{2}\right)$. $\vec{PS} = \left(\frac{1}{6}, \frac{2}{3}, \frac{1}{6}\right)$, $\|PS\| = \sqrt{\frac{1 + 16 + 1}{36}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$. Answer: (A).
50	CirclesStandard and general form	Class 11: Conic Sections	Single correctMedium
A medium single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Let $(h, k)$ be the mid-point. The chord with this mid-point is $hx + ky = h^2 + k^2$. The chord of contact from $(x_1, y_1)$ is $x_1x + y_1y = 9$. These are the same line, so $x_1 = \frac{9h}{h^2 + k^2}$, $y_1 = \frac{9k}{h^2 + k^2}$. $(x_1, y_1)$ lies on $4x - 5y = 20$: $\frac{36h - 45k}{h^2 + k^2} = 20$, i.e. $20(h^2 + k^2) - 36h + 45k = 0$. Locus: $20(x^2 + y^2) - 36x + 45y = 0$. Answer: (A).
51	TrigonometryTrigonometric identities	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Multiple correctHard
A hard multiple correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Range of $\theta$: $\tan(2\pi - \theta) = -\tan\theta > 0$, so $\tan\theta < 0$; with $\sin\theta < 0$ this puts $\theta$ in the fourth quadrant. And $-1 < \sin\theta < -\frac{\sqrt{3}}{2}$ there gives $\frac{3\pi}{2} < \theta < \frac{5\pi}{3}$, so $0 < \cos\theta < \frac{1}{2}$. Simplify: $\tan\frac{\theta}{2} + \cot\frac{\theta}{2} = \frac{1}{\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \frac{2}{\sin\theta}$, so the right side is $2\sin\theta\cos\varphi - 1$. $2\cos\theta - 2\cos\theta\sin\varphi = 2\sin\theta\cos\varphi - 1 \Rightarrow 2(\sin\theta\cos\varphi + \cos\theta\sin\varphi) = 2\cos\theta + 1$ $\sin(\theta + \varphi) = \cos\theta + \frac{1}{2} \in \left(\frac{1}{2}, 1\right)$. With $\theta \in \left(\frac{3\pi}{2}, \frac{5\pi}{3}\right)$ and $\varphi \in [0, 2\pi]$, $\theta + \varphi \in \left(\frac{3\pi}{2}, \frac{11\pi}{3}\right)$, and the sine lies in $\left(\frac{1}{2}, 1\right)$ there only for $\theta + \varphi \in \left(\frac{13\pi}{6}, \frac{17\pi}{6}\right)$. So $\varphi > \frac{13\pi}{6} - \theta > \frac{13\pi}{6} - \frac{5\pi}{3} = \frac{\pi}{2}$ and $\varphi < \frac{17\pi}{6} - \theta < \frac{17\pi}{6} - \frac{3\pi}{2} = \frac{4\pi}{3}$: $\varphi \in \left(\frac{\pi}{2}, \frac{4\pi}{3}\right)$. (A) $0 < \varphi < \frac{\pi}{2}$: impossible — $\varphi$ cannot satisfy it. Correct. (B) $\frac{\pi}{2} < \varphi < \frac{4\pi}{3}$: possible (e.g. $\theta = \frac{19\pi}{12}$ gives $\varphi \approx 124^\circ$). Not correct. (C) $\frac{4\pi}{3} < \varphi < \frac{3\pi}{2}$: impossible. Correct. (D) $\frac{3\pi}{2} < \varphi < 2\pi$: impossible. Correct. Answer: (A), (C), (D).
52	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	Multiple correctHard
A hard multiple correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $S = \int_0^1 e^{-x^2}dx$ (numerically about $0.747$). (A) True. On $[0, 1]$, $x^2 \le 1$, so $e^{-x^2} \ge e^{-1}$ and $S \ge \frac{1}{e}$. (B) True. On $[0, 1]$, $x^2 \le x$, so $e^{-x^2} \ge e^{-x}$ and $S \ge \int_0^1 e^{-x}dx = 1 - \frac{1}{e}$. (C) False. $\frac{1}{4}\left(1 + \frac{1}{\sqrt{e}}\right) \approx \frac{1}{4}(1.607) \approx 0.40$, which is less than $S$: indeed $S \ge 1 - \frac{1}{e} \approx 0.632$. (D) True. Split at $\frac{1}{\sqrt{2}}$ and bound $e^{-x^2}$ above by its value at the left end of each piece (it is decreasing): on $\left[0, \frac{1}{\sqrt{2}}\right]$, $e^{-x^2} \le 1$; on $\left[\frac{1}{\sqrt{2}}, 1\right]$, $e^{-x^2} \le e^{-1/2}$. So $S \le \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{e}}\left(1 - \frac{1}{\sqrt{2}}\right)$. Answer: (A), (B), (D).
53	ProbabilityBayes' theorem	Class 11: Probability; Class 12: Probability	Multiple correctMedium
A medium multiple correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(X_1) = \frac{1}{2}$, $P(X_2) = P(X_3) = \frac{1}{4}$. Exactly two working: $X_1X_2X_3^c$: $\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{3}{4} = \frac{3}{32}$; $X_1X_2^cX_3$: $\frac{3}{32}$; $X_1^cX_2X_3$: $\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4} = \frac{1}{32}$. Total $\frac{7}{32}$. All three: $\frac{1}{32}$. So $P(X) = \frac{8}{32} = \frac{1}{4}$. (A) False. $P(X_1^c \cap X) = P(X_1^cX_2X_3) = \frac{1}{32}$, so $P(X_1^c\,\|\,X) = \frac{1/32}{1/4} = \frac{1}{8}$, not $\frac{3}{16}$. (B) True. $P(\text{exactly two}\,\|\,X) = \frac{7/32}{1/4} = \frac{7}{8}$. (C) False. Given $E_2$ works, the ship needs at least one of $E_1, E_3$: $P(X\,\|\,X_2) = 1 - \frac{1}{2}\cdot\frac{3}{4} = \frac{5}{8}$, not $\frac{5}{16}$. (D) True. Given $E_1$ works: $P(X\,\|\,X_1) = 1 - \frac{3}{4}\cdot\frac{3}{4} = \frac{7}{16}$. Answer: (B), (D).
54	Conic SectionsHyperbola	Class 11: Conic Sections	Multiple correctMedium
A medium multiple correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Tangents of slope $m = 2$: $y = 2x + c$ with $c^2 = a^2m^2 - b^2 = 9\cdot4 - 4 = 32$, $c = \pm4\sqrt{2}$. The tangent $y = mx + c$ touches $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $\left(-\frac{a^2m}{c}, -\frac{b^2}{c}\right)$. $c = 4\sqrt{2}$: $\left(-\frac{18}{4\sqrt{2}}, -\frac{4}{4\sqrt{2}}\right) = \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$ — option (B). $c = -4\sqrt{2}$: $\left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ — option (A). Check: $\frac{81/8}{9} - \frac{1/2}{4} = \frac{9}{8} - \frac{1}{8} = 1$. ✓ (C) and (D): $\frac{27}{9} - \frac{8}{4} = 1$, so they lie on the hyperbola, but the tangent slope there is $\frac{b^2x}{a^2y} = \frac{4\cdot3\sqrt{3}}{9\cdot(-2\sqrt{2})} \neq 2$. Not points of contact. Answer: (A), (B).
55	Differential EquationsLinear differential equations	Class 12: Differential Equations	Multiple correctMedium
A medium multiple correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor: $e^{-\int\tan x\,dx} = e^{\ln\cos x} = \cos x$. $(y\cos x)' = 2x\sec x\cos x = 2x \Rightarrow y\cos x = x^2 + C$; $y(0) = 0$ gives $C = 0$. $y = x^2\sec x$, $y' = 2x\sec x + x^2\sec x\tan x$. (A) True. $y\left(\frac{\pi}{4}\right) = \frac{\pi^2}{16}\sqrt{2} = \frac{\pi^2}{8\sqrt{2}}$. (B) False. $y'\left(\frac{\pi}{4}\right) = \frac{\pi}{2}\sqrt{2} + \frac{\pi^2}{16}\sqrt{2}$, not $\frac{\pi^2}{18}$. (C) False. $y\left(\frac{\pi}{3}\right) = \frac{\pi^2}{9}\cdot2 = \frac{2\pi^2}{9}$. (D) True. $y'\left(\frac{\pi}{3}\right) = \frac{2\pi}{3}\cdot2 + \frac{\pi^2}{9}\cdot2\cdot\sqrt{3} = \frac{4\pi}{3} + \frac{2\sqrt{3}\pi^2}{9} = \frac{4\pi}{3} + \frac{2\pi^2}{3\sqrt{3}}$. Answer: (A), (D).
56	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	NumericalMedium
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f$ is even, so study $x \ge 0$. $0 \le x \le 1$: $f = x + 1 - x^2$, $f' = 1 - 2x$: rises from $f(0) = 1$ to $f\left(\frac{1}{2}\right) = \frac{5}{4}$, then falls to $f(1) = 1$. $x \ge 1$: $f = x + x^2 - 1$, increasing. So on $x \ge 0$: local minimum at $0$ (by evenness $f$ also rises to the left of $0$), local maximum at $\frac{1}{2}$, local minimum at $1$. By symmetry: local maximum at $-\frac{1}{2}$ and local minimum at $-1$. Points: $-1, -\frac{1}{2}, 0, \frac{1}{2}, 1$ — five. Answer: 5.
57	Sequences and SeriesGeometric progressions	Class 11: Sequences and Series	NumericalMedium
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let $u = \sqrt{4 - \frac{1}{3\sqrt{2}}\sqrt{4 - \frac{1}{3\sqrt{2}}\cdots}}$, so $u = \sqrt{4 - \frac{u}{3\sqrt{2}}}$, $u > 0$. $u^2 + \frac{u}{3\sqrt{2}} - 4 = 0 \Rightarrow 3\sqrt{2}u^2 + u - 12\sqrt{2} = 0$. $u = \frac{-1 + \sqrt{1 + 4\cdot3\sqrt{2}\cdot12\sqrt{2}}}{6\sqrt{2}} = \frac{-1 + \sqrt{289}}{6\sqrt{2}} = \frac{16}{6\sqrt{2}} = \frac{8}{3\sqrt{2}}$. The argument of the logarithm is $\frac{u}{3\sqrt{2}} = \frac{8}{18} = \frac{4}{9} = \left(\frac{3}{2}\right)^{-2}$. Value $= 6 + (-2) = 4$. Answer: 4.
58	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	NumericalMedium
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The least degree is 3, with $p'(x) = k(x - 1)(x - 3) = k(x^2 - 4x + 3)$. $p(x) = k\left(\frac{x^3}{3} - 2x^2 + 3x\right) + C$. $p(3) = k(9 - 18 + 9) + C = C = 2$. $p(1) = k\left(\frac{1}{3} - 2 + 3\right) + 2 = \frac{4k}{3} + 2 = 6 \Rightarrow k = 3$. $k > 0$ makes $p'$ positive before $1$, negative between $1$ and $3$, positive after: maximum at $1$, minimum at $3$, as required. $p'(0) = 3k = 9$. Answer: 9.
59	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	NumericalMedium
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\sum\|\vec{a} - \vec{b}\|^2 = 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 9$, so $\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\frac{3}{2}$. Then $\|\vec{a} + \vec{b} + \vec{c}\|^2 = 3 + 2\left(-\frac{3}{2}\right) = 0$, so $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $\vec{b} + \vec{c} = -\vec{a}$. $2\vec{a} + 5\vec{b} + 5\vec{c} = 2\vec{a} - 5\vec{a} = -3\vec{a}$, of magnitude $3$. Answer: 3.
60	Conic SectionsParabola	Class 11: Conic Sections	NumericalMedium
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Focus of $y^2 = 8x$ ($4a = 8$, $a = 2$): $S = (2, 0)$. Intersections: put $x = \frac{y^2}{8}$ into the circle: $\frac{y^4}{64} + y^2 - \frac{y^2}{4} - 4y = 0 \Rightarrow y\left(\frac{y^3}{64} + \frac{3y}{4} - 4\right) = 0 \Rightarrow y = 0$ or $y^3 + 48y - 256 = 0$. $y = 4$ is a root ($64 + 192 - 256 = 0$), and $y^3 + 48y - 256$ is increasing, so it is the only one. $P = (0, 0)$, $Q = (2, 4)$. Triangle with vertices $(0,0)$, $(2,4)$, $(2,0)$: base $SQ = 4$ (vertical, on $x = 2$), height $2$. Area $= \frac{1}{2}\cdot4\cdot2 = 4$. Answer: 4.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
1	Magnetic Effects of Current and MagnetismMagnetic dipole	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctEasy
A easy single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The loop is a square of side $a$ with a semicircle of diameter $a$ on each of its four sides. Area enclosed: $A = a^2 + 4\cdot\frac{1}{2}\pi\left(\frac{a}{2}\right)^2 = a^2 + \frac{\pi a^2}{2} = \left(\frac{\pi}{2}+1\right)a^2$. The arrow on the loop runs downwards on the left lobe, i.e. the current circulates anticlockwise as seen with $\hat{k}$ out of the paper, so by the right-hand rule the moment points along $+\hat{k}$. $\vec{\mu} = I\vec{A} = \left(\frac{\pi}{2}+1\right)a^2I\hat{k}$. Answer: (B).
2	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 Apply Ampere's law to a circle of radius $r$ coaxial with the cylinder, with $J$ the current density. $r < \frac{R}{2}$: no current is enclosed, so $B = 0$. $\frac{R}{2} \le r \le R$: $B\cdot 2\pi r = \mu_0 J\pi\left(r^2 - \frac{R^2}{4}\right)$, so $B = \frac{\mu_0 J}{2}\left(r - \frac{R^2}{4r}\right)$. This is zero at $r = \frac{R}{2}$ (no jump) and increases, not linearly, up to $r = R$. $r > R$: $B = \frac{\mu_0 I}{2\pi r}$, falling as $\frac{1}{r}$ and continuous with the value at $R$. Only graph (D) starts from zero at $R/2$, rises to a peak at $R$ and then decays. Answer: (D).
3	Properties of Bulk MatterFluid mechanics	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctMedium
A medium single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Let $V$ be the total (outer) volume of the closed shell and $v$ the volume of its material, so the inside holds $V - v$. Let $w$ be the volume of water inside. Measure masses in units of the density of water. Floating half-submerged: weight = buoyancy, so $\rho_c v + w = \frac{V}{2}$, i.e. $w = \frac{V}{2} - \rho_c v$. The shell is more than half-filled when $w > \frac{V - v}{2}$: $\frac{V}{2} - \rho_c v > \frac{V}{2} - \frac{v}{2} \iff \rho_c < 0.5$. So it is more than half-filled if $\rho_c < 0.5$, exactly half-filled if $\rho_c = 0.5$, and less than half-filled if $\rho_c > 0.5$. (A) True. (B) False: $\rho_c > 1$ gives less than half. (C) False: $\rho_c > 0.5$ gives less than half. (D) False: $\rho_c < 0.5$ gives more than half. Answer: (A).
4	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The upper plate of the $4\ \mu F$ capacitor holds $+80\ \mu C$, so by induction its lower plate holds $-80\ \mu C$. The lower plate of the $4\ \mu F$ capacitor and the upper plates of the $2\ \mu F$ and $3\ \mu F$ capacitors form an isolated conductor with zero net charge, so the two upper plates together carry $+80\ \mu C$. The $2\ \mu F$ and $3\ \mu F$ capacitors are in parallel (both lower plates earthed), so they share this charge in proportion to capacitance: $q_3 = \frac{3}{2+3}\times 80 = 48\ \mu C$. Answer: (C).
5	ThermodynamicsFirst law	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics The balloon needs no energy to expand, so the gas inside stays at the outside pressure: the heating is isobaric. Helium is monatomic: $C_p = \frac{5}{2}R$. $Q = nC_p\Delta T = 2 \times \frac{5}{2} \times 8.31 \times 5 = 207.75 \approx 208$ J. Answer: (D).
6	KinematicsRelative velocity	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctHard
A hard single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Work in the frame of the disc. Both pebbles have the same velocity relative to the disc, horizontally along $\overrightarrow{OR}$ (the $+y$ direction) plus a vertical part. In this rotating frame each pebble feels the Coriolis force $-2m\vec{\omega}\times\vec{v}$, which acts sideways to its horizontal motion, and both feel it in the same direction because their relative velocities are equal. If the disc turns anticlockwise ($\vec{\omega} = \omega\hat{k}$, $\hat{k}$ out of the paper): $-2m\omega\hat{k}\times v\hat{j} = +2m\omega v\,\hat{i}$, so both pebbles drift towards $+x$, the unshaded side. If the disc turns clockwise, both drift towards $-x$, the shaded side. (The same follows in the ground frame: e.g. for $Q$, starting a distance $q$ from $O$ along $OR$, the signed distance of its landing point from the turned diameter is $q(\sin\theta - \theta\cos\theta) + R_{range}\sin\theta$ with $\theta = \omega t < \frac{\pi}{4}$, which is positive for anticlockwise rotation.) Either way both pebbles land on the same side, so (A) and (B) are impossible. The figure's arrow for $\omega$ reads as anticlockwise, which gives (C); the official key accepts (C) or (D), since the sense of rotation can be read either way. Answer: (C) or (D).
7	Experimental SkillsHeat and sound experiments	No matching NCERT chapter	Single correctMedium
A medium single correct question on heat and sound experiments, from Experimental Skills. NCERT: No matching NCERT chapter $\lambda = \frac{v}{f} = \frac{336}{512} = 0.65625$ m, so $\frac{\lambda}{4} = 16.4$ cm. At the first resonance the effective length of the air column is $\frac{\lambda}{4} = l + e$, where $e$ is the end correction at the open top: $e = 0.6r = 0.6 \times 2 = 1.2$ cm (i.e. $0.3d$). $l = 16.4 - 1.2 = 15.2$ cm. The scale's zero is at the top of the tube, so the water level reads 15.2 cm. Answer: (B).
8	KinematicsRelative velocity	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctMedium
A medium single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Put the centres on the $x$-axis, the left disc's at $(-d, 0)$ and the right disc's at $(d, 0)$, with $P$ and $Q$ on the line of centres at $t = 0$. The discs turn in opposite senses, so at every instant $Q$'s position is the mirror image of $P$'s in the perpendicular bisector of the line of centres. With $\theta = \omega t$, $P$'s velocity is $R\omega(-\sin\theta, -\cos\theta)$ and $Q$'s is $R\omega(\sin\theta, -\cos\theta)$ (mirror images: equal $y$-parts, opposite $x$-parts). $\vec{v}_P - \vec{v}_Q = (-2R\omega\sin\theta, 0)$, so $v_r = 2R\omega\|\sin\omega t\|$. $v_r = 0$ at $t = 0, \frac{T}{2}, T$ and is largest ($2R\omega$) at $\frac{T}{4}$ and $\frac{3T}{4}$: two arches starting from zero. Answer: (A).
9	OpticsReflection and refraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	ComprehensionEasy
A easy comprehension question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Snell's law $\sin\theta_1 = n\sin\theta_2$ with $n < 0$ gives $\sin\theta_2 < 0$: the refraction angle is measured on the other side of the normal from the usual one. So the refracted ray lies on the same side of the normal as the incident ray, which is never the case for ordinary materials. Only diagram (C) shows the refracted ray in the meta-material on the incident ray's side of the normal. Answer: (C).
10	OpticsReflection and refraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	ComprehensionEasy
A easy comprehension question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics From $n = \frac{c}{v}$ with $n$ negative, the speed (a positive magnitude) is $v = \frac{c}{\|n\|}$. (B) true; (A) and (C) false. The frequency is unchanged on refraction, so $\lambda_m = \frac{v}{f} = \frac{c}{\|n\|f} = \frac{\lambda_{air}}{\|n\|}$, not $\lambda_{air}\|n\|$. (D) false. Answer: (B).
11	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	ComprehensionEasy
A easy comprehension question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The energy released in the decay, $Q \approx 0.8 \times 10^6$ eV, is shared between the electron and the anti-neutrino (the heavy proton takes only a tiny recoil energy). The electron's kinetic energy can be anything from nearly zero up to $0.8 \times 10^6$ eV. When the electron gets almost nothing, the anti-neutrino carries almost all of it. So the anti-neutrino's maximum energy is nearly $0.8 \times 10^6$ eV. Answer: (C).
12	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	ComprehensionMedium
A medium comprehension question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The electron can still be left with (nearly) zero kinetic energy, the rest going to the anti-neutrino and the proton, so the lower limit stays $0$. With a mass of $3\ eV/c^2$, at least $3$ eV of the released energy must go into the anti-neutrino's rest energy. The electron's maximum kinetic energy is therefore about $0.8 \times 10^6 - 3$ eV, strictly less than $0.8 \times 10^6$ eV. Range: $0 \le K < 0.8 \times 10^6$ eV. Answer: (D).
13	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	ComprehensionMedium
A medium comprehension question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion In both cases the disc–stick system is one rigid body turning about the fixed $z$-axis with angular velocity $\omega\hat{k}$. A rigid body has a single angular velocity: every line fixed in the body (for instance $PQ$) turns at $\omega$ about the vertical, whatever the disc's tilt. So the motion is the centre of mass going round the $z$-axis plus a rotation of the body about a vertical axis through the centre of mass, with angular velocity $\omega\hat{k}$. The tilt of the disc only changes how the disc is placed, not the direction of the body's angular velocity. The instantaneous axis through the centre of mass is vertical in both cases (a) and (b). Answer: (A).
14	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	ComprehensionMedium
A medium comprehension question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The angular velocity of a rigid body is the same about every point of it. The whole system turns at $\omega\hat{k}$ about the $z$-axis, so the rotation about the vertical axis through the centre of mass is also at angular speed $\omega$. This holds whether the disc is vertical (case (a)) or tilted at $45^\circ$ (case (b)); the tilt does not create any additional spin. Answer: (D).
15	Electromagnetic Induction and Alternating CurrentsAC circuits	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Multiple correctMedium
A medium multiple correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Two branches are in parallel across the 20 V source. Branch 1: $100\ \mu F$ in series with $100\,\Omega$. $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-4}} = 100\,\Omega$, $Z_1 = 100\sqrt{2}\,\Omega$, $I_1 = \frac{20}{100\sqrt{2}} = \frac{\sqrt{2}}{10}$ A, leading the voltage by $45^\circ$. Branch 2: $0.5$ H in series with $50\,\Omega$. $X_L = \omega L = 50\,\Omega$, $Z_2 = 50\sqrt{2}\,\Omega$, $I_2 = \frac{20}{50\sqrt{2}} = \frac{2\sqrt{2}}{10}$ A, lagging by $45^\circ$. The two currents are $90^\circ$ apart: $I = \sqrt{I_1^2 + I_2^2} = \sqrt{0.02 + 0.08} = \sqrt{0.1} \approx 0.316$ A. (A) $0.316$ A is about $0.3$ A; it is true only as an approximation. (B) $0.3\sqrt{2} \approx 0.42$ A: false. (C) $V_{100} = I_1 \times 100 = 10\sqrt{2}$ V: true. (D) $V_{50} = I_2 \times 50 = 10\sqrt{2}$ V, not 10 V: false. Answer: (C); the official key also accepts (A), (C).
16	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Multiple correctEasy
A easy multiple correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The field lines of the straight wire are circles round it, so the field passes into the plane of the loop on one side of the diameter and out of it on the other. By symmetry the flux through the two halves is equal and opposite: the net flux through the loop is zero whatever the current. The flux stays zero as the current changes, so $\varepsilon = -\frac{d\phi}{dt} = 0$ in every case. (A) True. (B) False. (C) True. (D) False. Answer: (A), (C).
17	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Multiple correctMedium
A medium multiple correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Take $O$ as origin with $D$ along $+x$. Charges: $A(-L, 0)$: $+2q$; $D(L, 0)$: $-2q$; $F$ and $E$ at the top: $+q$ and $-q$; $B$ and $C$ at the bottom: $+q$ and $-q$. Each is at distance $L$ from $O$. Field at $O$: a charge $+Q$ at $\vec{r}$ gives $\frac{Q}{4\pi\varepsilon_0L^2}$ along $-\hat{r}$. $A$ and $D$: each gives $2K$ along $+x$, total $4K$. $F(+q)$ and $C(-q)$ lie on one diagonal: each gives $K$ along $\overrightarrow{FC}$, total $2K$ along $\overrightarrow{FC}$, at $60^\circ$ below $+x$. $B(+q)$ and $E(-q)$: total $2K$ along $\overrightarrow{BE}$, at $60^\circ$ above $+x$. These two add to $2 \times 2K\cos 60^\circ = 2K$ along $+x$. Total $\vec{E} = 6K$ along $OD$. (A) true. Potential at $O$: $\frac{1}{4\pi\varepsilon_0 L}(2q - 2q + q - q + q - q) = 0$. (B) true. $PR$ is the perpendicular bisector of $AD$, $FE$ and $BC$; each $+$ charge has an equal $-$ charge as its mirror image in it, so every point of $PR$ is equidistant from each pair and $V = 0$ all along $PR$. (C) true. $ST$ runs along $AD$, and the field at $O$ (a point of it) is $6K$ along this line, so the potential changes along $ST$. (D) false. Answer: (A), (B), (C).
18	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	Multiple correctEasy
A easy multiple correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For rolling without slipping, $a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$ and $v^2 = \frac{2gh}{1 + \frac{I}{mR^2}}$. $P$'s mass is near the rim, so $I_P > I_Q$. (A) False: $Q$ has the larger acceleration and reaches first. (B) False: $P$ has the smaller acceleration. (C) False: both lose the same $mgh$, but $P$ puts a bigger share into rotation, so $Q$ has more translational kinetic energy. (D) True: $v_Q > v_P$ and $\omega = \frac{v}{R}$ with equal radii. Answer: (D).
19	GravitationEscape velocity	Class 11: Gravitation	Multiple correctMedium
A medium multiple correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{r}} = \sqrt{\frac{2G}{r}\cdot\frac{4}{3}\pi r^3\rho} \propto r$ for a fixed density. Areas $A$ and $4A$ give radii $r$ and $2r$, so $M_Q = 8M_P$ and $M_R = 9M_P$, giving $r_R = 9^{1/3}r \approx 2.08r$. $V_P : V_Q : V_R = 1 : 2 : 9^{1/3} \approx 1 : 2 : 2.08$. (A) False. (B) True: $V_R > V_Q > V_P$. (C) False: $V_R/V_P = 9^{1/3} \approx 2.08$. (D) True: $V_P/V_Q = \frac{1}{2}$. Answer: (B), (D).
20	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	Multiple correctMedium
A medium multiple correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The ring rolls clockwise without slipping, so its centre $O$ moves to the right with $v_O = 3R\omega$: $\vec{v}_O = 3R\omega\hat{i}$. (A) true. The disc shares the centre $O$ and turns anticlockwise at $\frac{\omega}{2}$. $P$ is at $\vec{r} = R(\cos 30^\circ\,\hat{i} + \sin 30^\circ\,\hat{k})$ from $O$. For anticlockwise rotation (in the $x$-$z$ view with $x$ to the right and $z$ up), the velocity of $P$ relative to $O$ is $\frac{\omega}{2}R(-\sin 30^\circ\,\hat{i} + \cos 30^\circ\,\hat{k}) = -\frac{1}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k}$. $\vec{v}_P = 3R\omega\hat{i} - \frac{1}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k} = \frac{11}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k}$. (B) true; (C) and (D) false. Answer: (A), (B).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
21	Coordination CompoundsValence bond theory	Class 12: Coordination Compounds	Single correctMedium
A medium single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $Ni^{2+}$ is $3d^8$. In a four-coordinate complex it can be: tetrahedral ($sp^3$): the eight d electrons keep two unpaired electrons, so the complex is paramagnetic; square planar ($dsp^2$): the electrons pair up in four d orbitals, leaving the $d_{x^2-y^2}$ empty, so the complex is diamagnetic. With mixed ligands (two $Cl^-$, two phosphines) the two geometries lie close in energy and interconvert with temperature, which is why the magnetism changes. Paramagnetic state: tetrahedral; diamagnetic state: square planar. Answer: (C).
22	Isolation of MetalsOres and concentration	No matching NCERT chapter	Single correctEasy
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Argentite ($Ag_2S$) is leached with sodium cyanide solution while air is blown through: $Ag_2S + 4NaCN \rightleftharpoons 2Na[Ag(CN)_2] + Na_2S$. The oxygen of the air oxidises $Na_2S$ (to sulphate/thiosulphate), pulling the equilibrium to the right, so $O_2$ is the oxidising agent. Silver is then displaced from the complex with zinc dust, the reducing agent: $2[Ag(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Ag$. Answer: (B).
23	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$ Phosphorus starts at oxidation state 0 and goes both down, to $-3$ in $PH_3$, and up, to $+1$ in sodium hypophosphite $NaH_2PO_2$: the same element is both oxidised and reduced, so this is a disproportionation reaction. The correct description is "disproportionation; $-3$ and $+1$", which is none of the four options.
24	Chemical Bonding and Molecular StructureVSEPR theory	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Xe has 8 valence electrons. Two Xe=O double bonds use 4 of them and two Xe–F bonds use 2, leaving one lone pair. Electron domains: 4 bonding directions + 1 lone pair = 5, so the arrangement is trigonal bipyramidal ($sp^3d$) with the lone pair in an equatorial position. Four atoms around Xe with one equatorial lone pair give a see-saw shape (the two F axial, the two O and the lone pair equatorial). Answer: (D).
25	SolutionsColligative properties	Class 12: Solutions	Single correctMedium
A medium single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Delta T_b = K_b m$, so $m = \frac{2}{0.76} = 2.63$ mol per kg of water. For a dilute solution the mole fraction of solute is $x_2 \approx \frac{n_2}{n_1} = m \times \frac{18}{1000} = 2.63 \times 0.018 = 0.0474$. The solution boils at its new boiling point under 1 atm, and at the boiling point of water the pure solvent's vapour pressure is 760 mm Hg. Raoult's law: $\frac{p^\circ - p}{p^\circ} = x_2 \Rightarrow p = 760(1 - 0.0474) = 724$ mm Hg. Answer: (A).
26	Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Acids with a carbonyl group $\beta$ to the carboxyl carbon ($\beta$-keto acids, malonic-type acids) lose $CO_2$ on gentle warming through a six-membered cyclic transition state, giving an enol that tautomerises to the ketone. (B) 2-oxocyclohexanecarboxylic acid is a $\beta$-keto acid: the ring C=O is $\beta$ to the COOH carbon. It decarboxylates readily to cyclohexanone. (D) (2-oxocyclohexyl)acetic acid is a $\gamma$-keto acid: the six-membered cyclic path is not available. (A) and (C) are diacids whose carboxyl groups are too far apart (a glutaric-type and a succinic-type acid): neither has a C=O on the carbon $\beta$ to a carboxyl carbon, so they need strong heating. Answer: (B).
27	ThermodynamicsHess law	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Make $C_2H_2$ from gaseous atoms: $2C(g) + 2H(g) \rightarrow C_2H_2(g)$ releases $E_{C\equiv C} + 2E_{C-H}$. By Hess's law: $\Delta H_f = [\Delta H(2C(s) \rightarrow 2C(g)) + \Delta H(H_2 \rightarrow 2H)] - [E_{C\equiv C} + 2E_{C-H}]$ $225 = 1410 + 330 - (E_{C\equiv C} + 700)$ $E_{C\equiv C} = 1740 - 700 - 225 = 815$ kJ mol$^{-1}$. Answer: (D).
28	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 Cyanide adds to the carbonyl of butanone: G is the cyanohydrin $CH_3CH_2C(OH)(CN)CH_3$. Hot 95% $H_2SO_4$ does two things: it hydrolyses $-CN$ (through the amide) to $-COOH$, and it dehydrates the tertiary alcohol. Loss of water towards the $CH_2$ gives the more substituted, conjugated alkene. H: $CH_3-CH=C(CH_3)-COOH$ (2-methylbut-2-enoic acid). (B) keeps the nitrile and (D) stops at the amide; the strong hot acid takes both through to the acid. (C) has not been dehydrated. Answer: (A).
29	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	ComprehensionEasy
A easy comprehension question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Bleaching powder is $Ca(OCl)Cl$ (with $CaCl_2$ and $Ca(OH)_2$); its oxo-salt is the hypochlorite $OCl^-$, from hypochlorous acid $HOCl$. The anhydride of $HOCl$ is $Cl_2O$: $Cl_2O + H_2O \rightarrow 2HOCl$ (chlorine $+1$ in both). $Cl_2O_7$ is the anhydride of $HClO_4$; $ClO_2$ and $Cl_2O_6$ are mixed anhydrides of chlorous/chloric and chloric/perchloric acids. Answer: (A).
30	Principles Related to Practical ChemistryTitrimetric analysis	No matching NCERT chapter	ComprehensionMedium
A medium comprehension question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter $OCl^- + 2I^- + 2H^+ \rightarrow Cl^- + I_2 + H_2O$ $I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$ So $1\ OCl^- \equiv 1\ I_2 \equiv 2\ S_2O_3^{2-}$. (KI, 15 mmol, is in excess.) Thiosulphate used $= 48 \times 0.25 = 12$ mmol, so $I_2 = 6$ mmol and $OCl^- = 6$ mmol. Molarity of bleach $= \frac{6}{25} = 0.24$ M. Answer: (C).
31	ElectrochemistryNernst equation	Class 12: Electrochemistry	ComprehensionMedium
A medium comprehension question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry For the concentration cell (anode on the left, $n = 2$): $E = \frac{0.059}{2}\log\frac{[M^{2+}]_{cathode}}{[M^{2+}]_{anode}} = \frac{0.059}{2}\log\frac{0.001}{s}$ $0.059 = \frac{0.059}{2}\log\frac{10^{-3}}{s} \Rightarrow \log\frac{10^{-3}}{s} = 2 \Rightarrow s = 10^{-5}$ mol dm$^{-3}$. In the saturated solution $[M^{2+}] = s$ and $[X^-] = 2s$: $K_{sp} = s(2s)^2 = 4s^3 = 4 \times 10^{-15}$. Answer: (B).
32	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	ComprehensionEasy
A easy comprehension question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\Delta G = -nFE = -2 \times 96500 \times 0.059 = -11387$ J mol$^{-1} \approx -11.4$ kJ mol$^{-1}$. The negative sign agrees with a cell giving a positive emf spontaneously. Answer: (D).
33	Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro	Class 12: Aldehydes, Ketones and Carboxylic Acids	ComprehensionMedium
A medium comprehension question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $\mathbf{J}$, $C_9H_8O_2$, is a carboxylic acid ($NaHCO_3$ effervescence) with a C=C (decolourises Baeyer's reagent): cinnamic acid, $C_6H_5CH=CHCOOH$. Cinnamic acid is made by the Perkin reaction: an aromatic aldehyde heated with acetic anhydride and sodium acetate. $C_6H_5CHO + (CH_3CO)_2O \xrightarrow{CH_3COONa} C_6H_5CH=CHCOOH + CH_3COOH$ So $\mathbf{I}$ is benzaldehyde. Benzyl alcohol, acetophenone and styrene do not undergo the Perkin condensation. Answer: (A).
34	HydrocarbonsElectrophilic substitution and directive influence	Class 11: Hydrocarbons	ComprehensionMedium
A medium comprehension question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathbf{J}$ is cinnamic acid, $C_6H_5CH=CHCOOH$. (i) $H_2$, Pd/C reduces the C=C: 3-phenylpropanoic acid, $C_6H_5CH_2CH_2COOH$. (ii) $SOCl_2$ gives the acid chloride $C_6H_5CH_2CH_2COCl$. (iii) Anhydrous $AlCl_3$: intramolecular Friedel–Crafts acylation at the ortho position closes a five-membered ring with the C=O attached to the benzene ring. $\mathbf{K}$ is 1-indanone (indan-1-one). Answer: (C).
35	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	Multiple correctMedium
A medium multiple correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics From $(P_1, V_1, T_1)$ the gas expands reversibly to $V_2$: isothermally to $(P_2, V_2, T_2)$ and adiabatically to $(P_3, V_2, T_3)$. (A) True: the isothermal path keeps $T_2 = T_1$. (B) False: in a reversible adiabatic expansion the gas does work at the cost of its internal energy, so it cools: $T_3 < T_1$. (C) The isothermal curve lies above the adiabatic one, so the area under it (the work done by the gas) is larger: $\|w_{iso}\| > \|w_{ad}\|$. Read as work done by the gas, (C) is true. Read with the sign convention $w$ = work done on the gas, both are negative and $w_{iso} < w_{ad}$, so (C) would be false. The official key accepts either reading. (D) True: $\Delta U_{iso} = 0$, while $\Delta U_{ad} = nC_V(T_3 - T_1) < 0$. Answer: (A), (C), (D); the official key also accepts (A), (D).
36	Surface ChemistryPhysisorption and chemisorption	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter I: at constant pressure the amount adsorbed falls steadily as $T$ rises: physisorption (exothermic, weak van der Waals forces, no activation needed). II: at constant pressure the amount adsorbed rises with $T$ (over the range shown): chemisorption, which needs activation energy and so increases with temperature at first. III: amount against pressure, more adsorbed at 200 K than at 250 K at every pressure: the low-temperature, pressure-driven behaviour of physisorption. IV: a potential-energy curve with an activation barrier $E_{ads}$ and a deep well, $\Delta H_{ads} = 150$ kJ mol$^{-1}$, of the size of a chemical bond: chemisorption. (A) True. (B) False: III is physisorption. (C) True. (D) False. Answer: (A), (C).
37	Principles of Qualitative AnalysisCharacteristic precipitates and colours	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter First step: ferricyanide oxidises iodide: $2[Fe(CN)_6]^{3-} + 2I^- \rightarrow 2[Fe(CN)_6]^{4-} + I_2$; with excess $I^-$ the iodine stays as brownish-yellow $I_3^-$. Fe goes from $+3$ to $+2$ and I from $-1$ to 0: a redox reaction. (A) true. With $ZnSO_4$, the ferrocyanide now present gives the white precipitate $K_2Zn_3[Fe(CN)_6]_2$ (zinc ferrocyanide), not the ferricyanide $Zn_3[Fe(CN)_6]_2$. (B) false. The filtrate still holds the iodine (removed by thiosulphate: $I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$, colourless), so it turns starch blue. (C) true. Zinc compounds dissolve in excess NaOH as zincate, $[Zn(OH)_4]^{2-}$; the white precipitate dissolves. (D) true. Answer: (A), (C), (D).
38	p-Block ElementsGroup 14 elements	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) False: diamond's three-dimensional network of strong $sp^3$ C–C bonds makes it the hardest natural substance; graphite's layers slide over each other, so it is soft. (B) True: in graphite each $sp^2$ carbon has one electron in a delocalised $\pi$ system, so it conducts electricity; diamond has no free electrons and is an insulator. (C) False: diamond is an exceptionally good conductor of heat (through lattice vibrations of its rigid network), better than graphite. (D) True: in graphite one $\pi$ bond is shared over three C–C bonds, giving a bond order of about $1\frac{1}{3}$ (bond length 141.5 pm), against 1 in diamond (154 pm). Answer: (B), (D).
39	Alcohols, Phenols and EthersPreparation and properties	Class 12: Alcohols, Phenols and Ethers	Multiple correctMedium
A medium multiple correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers T is a six-membered lactone (cyclic ester) with the methyl on the ring carbon $\beta$ to the C=O: 4-methyltetrahydropyran-2-one, the lactone of 5-hydroxy-3-methylpentanoic acid. (A) True: hot aqueous NaOH hydrolyses the ester to the soluble sodium salt $HOCH_2CH_2CH(CH_3)CH_2COO^-Na^+$. $LiAlH_4$ reduces the lactone to the diol U: $HOCH_2CH_2CH(CH_3)CH_2CH_2OH$ (3-methylpentane-1,5-diol). (B) False: the CH carrying the methyl now has two identical $-CH_2CH_2OH$ arms, so U has no stereocentre and is optically inactive (the stereocentre of T is lost). (C) True: excess acetic anhydride acetylates both OH groups. U is $C_6H_{14}O_2$; each acetylation adds $C_2H_2O$, so W, $CH_3COOCH_2CH_2CH(CH_3)CH_2CH_2OCOCH_3$, is $C_{10}H_{18}O_4$. (D) True: $CrO_3/H^+$ oxidises both primary alcohols to acids: V is 3-methylpentanedioic acid, $HOOCCH_2CH(CH_3)CH_2COOH$, which liberates $CO_2$ from $NaHCO_3$. Answer: (A), (C), (D).
40	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Multiple correctHard
A hard multiple correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques All five are $CH_3-CH(OH)-CH(OH)Cl$ with two stereocentres: C1 $= CH(OH)Cl$ and C2 $= CH(OH)CH_3$. Assign each configuration (CIP priorities: at C1, $Cl > OH > C2 > H$; at C2, $OH > C1 > CH_3 > H$). In a sawhorse the lower-left carbon is the one nearer the viewer; in a Newman projection the bonds meeting at the centre belong to the front carbon. M: front C2 has OH left, H right, $CH_3$ down; back C1 has OH left, H right, Cl down. Working out the sense of rotation of the three highest priorities with H accounted for gives C1 $R$, C2 $S$: M is (1$R$,2$S$). N: the front carbon is now C1 (H left, OH right, Cl down) and the back is C2 (OH left, H right, $CH_3$ down): C1 $R$, C2 $R$. N is (1$R$,2$R$). O: front C2 as in M ($S$); back C1 with Cl up, OH right, H left: $R$. O is (1$R$,2$S$). P (Newman): front C2 with $CH_3$ up, H lower right, OH lower left: $R$; back C1 with OH upper right, Cl down, H upper left: $S$. P is (1$S$,2$R$). Q (Newman): front C2 as in P: $R$; back C1 with OH upper left, H upper right, Cl down: $R$. Q is (1$R$,2$R$). (A) True: M (1$R$,2$S$) and N (1$R$,2$R$) differ at one centre only: diastereomers, i.e. non-mirror-image stereoisomers. (B) True: O has the same configuration as M; it is the same molecule in another conformation. (C) True: P (1$S$,2$R$) has both centres inverted: the enantiomer of M. (D) False: Q (1$R$,2$R$) is a diastereomer of M (it is identical to N). Answer: (A), (B), (C).

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
41	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctMedium
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A plane through the line of intersection: $(x + 2y + 3z - 2) + \lambda(x - y + z - 3) = 0$, i.e. $(1+\lambda)x + (2-\lambda)y + (3+\lambda)z - (2 + 3\lambda) = 0$. At $(3, 1, -1)$ the left side is $(3 + 2 - 3 - 2) + \lambda(3 - 1 - 1 - 3) = -2\lambda$. $\|\vec{n}\|^2 = (1+\lambda)^2 + (2-\lambda)^2 + (3+\lambda)^2 = 3\lambda^2 + 4\lambda + 14$. Distance: $\frac{2\|\lambda\|}{\sqrt{3\lambda^2 + 4\lambda + 14}} = \frac{2}{\sqrt{3}} \Rightarrow 3\lambda^2 = 3\lambda^2 + 4\lambda + 14 \Rightarrow \lambda = -\frac{7}{2}$. Plane: $-\frac{5}{2}x + \frac{11}{2}y - \frac{1}{2}z + \frac{17}{2} = 0$, i.e. $5x - 11y + z = 17$. Answer: (A).
42	TrigonometryTrigonometric identities	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctMedium
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\frac{2\sin P - 2\sin P\cos P}{2\sin P + 2\sin P\cos P} = \frac{1 - \cos P}{1 + \cos P} = \tan^2\frac{P}{2}$. Semi-perimeter $s = \frac{2 + 3.5 + 2.5}{2} = 4$, so $s - a = 2$, $s - b = \frac{1}{2}$, $s - c = \frac{3}{2}$. $\tan\frac{P}{2} = \frac{(s-b)(s-c)}{\Delta} = \frac{\frac{1}{2}\cdot\frac{3}{2}}{\Delta} = \frac{3}{4\Delta}$. The expression is $\left(\frac{3}{4\Delta}\right)^2$. (Here $\Delta = \sqrt{4 \cdot 2 \cdot \frac{1}{2} \cdot \frac{3}{2}} = \sqrt{6}$, but the answer is asked in terms of $\Delta$.) Answer: (C).
43	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Single correctEasy
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Let $\vec{c} = 2\hat{i} + 3\hat{j} + 4\hat{k}$. Then $\vec{a} \times \vec{c} = \vec{c} \times \vec{b} = -\vec{b} \times \vec{c}$, so $(\vec{a} + \vec{b}) \times \vec{c} = \vec{0}$ and $\vec{a} + \vec{b} = \lambda\vec{c}$. $\|\vec{c}\| = \sqrt{29}$, so $\|\lambda\|\sqrt{29} = \sqrt{29}$ and $\lambda = \pm 1$. $(\vec{a} + \vec{b}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k}) = \pm(-14 + 6 + 12) = \pm 4$. A possible value is 4. Answer: (C).
44	Matrices and DeterminantsAlgebra of matrices	Class 12: Matrices; Class 12: Determinants	Single correctEasy
A easy single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Transpose both sides: $P = 2P^T + I = 2(2P + I) + I = 4P + 3I$, so $3P = -3I$ and $P = -I$. Then $PX = -X$ for every column $X$, in particular for any non-zero one. (A) would need $X = 0$; (B) and (C) would need $-X = X$ or $-X = 2X$, again $X = 0$. Answer: (D).
45	Limits, Continuity and DifferentiabilityLimits of functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctMedium
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For small $a$, $(1+a)^k - 1 = ka + O(a^2)$, so the coefficients are $\frac{a}{3}$, $\frac{a}{2}$, $\frac{a}{6}$ to first order. Dividing the equation by $a$ and letting $a \to 0^+$, the roots tend to those of $\frac{x^2}{3} + \frac{x}{2} + \frac{1}{6} = 0$, i.e. $2x^2 + 3x + 1 = 0$. $(2x + 1)(x + 1) = 0$: $x = -\frac{1}{2}$ and $x = -1$. Answer: (B).
46	ProbabilityProbability of an event	Class 11: Probability; Class 12: Probability	Single correctEasy
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Whatever $D_4$ shows, each of $D_1$, $D_2$, $D_3$ independently misses that number with probability $\frac{5}{6}$. P(none of them shows it) $= \left(\frac{5}{6}\right)^3 = \frac{125}{216}$. P(at least one shows it) $= 1 - \frac{125}{216} = \frac{91}{216}$. Answer: (A).
47	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	Single correctMedium
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $g(x) = \ln\frac{\pi + x}{\pi - x}$ is odd ($g(-x) = -g(x)$) and $\cos x$ is even, so $\int_{-\pi/2}^{\pi/2} g(x)\cos x\,dx = 0$. $x^2\cos x$ is even: $I = 2\int_0^{\pi/2} x^2\cos x\,dx$. By parts twice: $\int x^2\cos x\,dx = x^2\sin x + 2x\cos x - 2\sin x$. From 0 to $\frac{\pi}{2}$: $\frac{\pi^2}{4} + 0 - 2 = \frac{\pi^2}{4} - 2$. $I = 2\left(\frac{\pi^2}{4} - 2\right) = \frac{\pi^2}{2} - 4$. Answer: (B).
48	Sequences and SeriesArithmetic progressions	Class 11: Sequences and Series	Single correctEasy
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The reciprocals $\frac{1}{a_n}$ form an AP: $\frac{1}{a_1} = \frac{1}{5}$, $\frac{1}{a_{20}} = \frac{1}{25} = \frac{1}{5} + 19d$, so $d = -\frac{4}{475}$. $a_n < 0 \iff \frac{1}{a_n} < 0 \iff \frac{1}{5} - \frac{4(n-1)}{475} < 0 \iff 4(n - 1) > 95 \iff n - 1 > 23.75$. The least such $n$ is 25. Answer: (D).
49	Permutations and CombinationsFundamental principle of counting	Class 11: Permutations and Combinations	ComprehensionMedium
A medium comprehension question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Every such number starts with 1. Build them digit by digit: a valid $n$-digit number ending in 1 is any valid $(n-1)$-digit number followed by 1, so $b_n = a_{n-1}$; one ending in 0 must have 1 just before it, so $c_n = b_{n-1}$. Hence $a_n = b_n + c_n = a_{n-1} + a_{n-2}$. $a_1 = 1$ (just 1), $a_2 = 2$ (10, 11), $a_3 = 3$, $a_4 = 5$, $a_5 = 8$, $a_6 = 13$. $b_6 = a_5 = 8$. Answer: (B).
50	Permutations and CombinationsFundamental principle of counting	Class 11: Permutations and Combinations	ComprehensionMedium
A medium comprehension question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations From $b_n = a_{n-1}$, $c_n = b_{n-1} = a_{n-2}$ and $a_n = b_n + c_n$: (A) $a_{17} = b_{17} + c_{17} = a_{16} + a_{15}$. True. (B) $c_{17} = a_{15}$ and $c_{16} + c_{15} = a_{14} + a_{13} = a_{15}$, so $c_{17} = c_{16} + c_{15}$. False. (C) $b_{17} = a_{16} = b_{16} + c_{16}$. False. (D) $c_{17} + b_{16} = a_{15} + a_{15} = 2a_{15}$, while $a_{17} = a_{16} + a_{15} > 2a_{15}$. False. Answer: (A).
51	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	ComprehensionMedium
A medium comprehension question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $g'(x) = \left(\frac{2(x-1)}{x+1} - \ln x\right)f(x)$. For $x > 1$, $f(x) = (1-x)^2\sin^2 x + x^2 > 0$. Let $h(x) = \frac{2(x-1)}{x+1} - \ln x$. Then $h(1) = 0$ and $h'(x) = \frac{4}{(x+1)^2} - \frac{1}{x} = \frac{4x - (x+1)^2}{x(x+1)^2} = -\frac{(x-1)^2}{x(x+1)^2} < 0$ for $x > 1$. So $h(x) < 0$ on $(1, \infty)$, hence $g'(x) < 0$ there: $g$ is decreasing on $(1, \infty)$. Answer: (B).
52	Limits, Continuity and DifferentiabilityContinuity	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	ComprehensionMedium
A medium comprehension question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability P: $(1-x)^2\sin^2 x + x^2 + 2x = 2 + 2x^2 \iff (1-x)^2\sin^2 x = x^2 - 2x + 2 = (1-x)^2 + 1 \iff -(1-x)^2\cos^2 x = 1$. The left side is never positive, so there is no solution: P is false. Q: let $k(x) = 2f(x) + 1 - 2x - 2x^2 = 2(1-x)^2\sin^2 x + 1 - 2x$. $k(0) = 1 > 0$ and $k(1) = 0 + 1 - 2 = -1 < 0$. $k$ is continuous, so by the intermediate value theorem it vanishes somewhere in $(0, 1)$: Q is true. Answer: (C).
53	CirclesIntersection of a line and a circle	Class 11: Conic Sections	ComprehensionEasy
A easy comprehension question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections The tangent at $P(\sqrt{3}, 1)$ to $x^2 + y^2 = 4$ is $\sqrt{3}x + y = 4$, of slope $-\sqrt{3}$. $L$ is perpendicular to it, so it has slope $\frac{1}{\sqrt{3}}$: $x - \sqrt{3}y = c$. Tangent to the circle with centre $(3, 0)$, radius 1: $\frac{\|3 - c\|}{2} = 1$, so $c = 1$ or $c = 5$. $L$: $x - \sqrt{3}y = 1$ or $x - \sqrt{3}y = 5$; the first is option (A). (B) and (D) have the wrong slope; (C) has $c = -1$. Answer: (A).
54	CirclesIntersection of a line and a circle	Class 11: Conic Sections	ComprehensionMedium
A medium comprehension question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Circles: centre $(0, 0)$, radius 2 and centre $(3, 0)$, radius 1. The distance between centres, 3, equals $2 + 1$: they touch externally at $(2, 0)$, with common tangent $x = 2$ there. The other two common tangents pass through the external centre of similitude, which divides the centres externally in the ratio $2 : 1$: $(6, 0)$. A line $y = m(x - 6)$ is tangent to $x^2 + y^2 = 4$ when $\frac{6\|m\|}{\sqrt{1 + m^2}} = 2$, i.e. $m^2 = \frac{1}{8}$, $m = \pm\frac{1}{2\sqrt{2}}$. These are $x \pm 2\sqrt{2}y = 6$; option (D) is one. Check: distance from $(3, 0)$ is $\frac{3}{3} = 1$. (A) $x = 4$ is at distance 4 from the origin. (B) $y = 2$ is at distance 2 from $(3, 0)$. (C) is at distance $\frac{1}{2}$ from $(3, 0)$. None is tangent to both. Answer: (D).
55	Limits, Continuity and DifferentiabilityContinuity	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Multiple correctMedium
A medium multiple correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = 2n$: from the left ($x \in (2n-1, 2n)$) $f \to b_n + \cos 2n\pi = b_n + 1$; the value is $a_n + \sin 2n\pi = a_n$. Continuity: $a_n = b_n + 1$, i.e. $a_n - b_n = 1$ for every $n$. At $x = 2n + 1$: the value is $a_n + \sin(2n+1)\pi = a_n$; from the right ($x \in (2n+1, 2n+2)$, the interval of $b_{n+1}$) $f \to b_{n+1} + \cos(2n+1)\pi = b_{n+1} - 1$. Continuity: $a_n - b_{n+1} = -1$ for every $n$, i.e. $a_{n-1} - b_n = -1$. (A) False: $a_{n-1} - b_{n-1} = 1$. (B) True. (C) False: it equals $-1$. (D) True. Answer: (B), (D).
56	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	Multiple correctEasy
A easy multiple correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = e^{x^2}(x - 2)(x - 3)$: positive on $(0, 2)$, negative on $(2, 3)$, positive on $(3, \infty)$. (A) True: $f'$ changes from $+$ to $-$ at 2, a local maximum. (B) True: $f' < 0$ on $(2, 3)$. (C) True: $f'$ is differentiable and $f'(2) = f'(3) = 0$, so by Rolle's theorem $f''(c) = 0$ for some $c \in (2, 3)$. (D) True: $f'$ changes from $-$ to $+$ at 3, a local minimum. Answer: (A), (B), (C), (D).
57	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Multiple correctMedium
A medium multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Line 1 passes through $A(1, -1, 0)$ with direction $(2, k, 2)$; line 2 through $B(-1, -1, 0)$ with direction $(5, 2, k)$. $\overrightarrow{AB} = (-2, 0, 0)$. Coplanar: $\begin{vmatrix} -2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{vmatrix} = -2(k^2 - 4) = 0$, so $k = \pm 2$. $k = 2$: normal $(2, 2, 2) \times (5, 2, 2) = (0, 6, -6)$, plane $y - z = c$ through $(1, -1, 0)$: $y - z = -1$. (C) $k = -2$: normal $(2, -2, 2) \times (5, 2, -2) = (0, 14, 14)$, plane $y + z = -1$. (B) (A) and (D) contain neither pair of lines. Answer: (B), (C).
58	ProbabilityAddition and multiplication theorems	Class 11: Probability; Class 12: Probability	Multiple correctEasy
A easy multiple correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(Y) = \frac{P(X \cap Y)}{P(X\|Y)} = \frac{1/6}{1/2} = \frac{1}{3}$ and $P(X) = \frac{P(X \cap Y)}{P(Y\|X)} = \frac{1/6}{1/3} = \frac{1}{2}$. (A) $P(X \cup Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{2}{3}$. True. (B) $P(X)P(Y) = \frac{1}{6} = P(X \cap Y)$: independent. True; so (C) is false. (D) $P(X^c \cap Y) = P(Y) - P(X \cap Y) = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$. False. Answer: (A), (B).
59	Matrices and DeterminantsAdjoint and inverse	Class 12: Matrices; Class 12: Determinants	Multiple correctEasy
A easy multiple correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3 \times 3$ matrix, $\|\text{adj}\,P\| = \|P\|^{3-1} = \|P\|^2$. $\|\text{adj}\,P\| = 1(1\cdot 3 - 7\cdot 1) - 4(2\cdot 3 - 7\cdot 1) + 4(2\cdot 1 - 1\cdot 1) = -4 + 4 + 4 = 4$. $\|P\|^2 = 4$, so $\|P\| = \pm 2$. Answer: (A), (D).
60	TrigonometryTrigonometric identities	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Multiple correctMedium
A medium multiple correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $2 - \sec^2\theta = 1 - \tan^2\theta$, and $\frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta$ gives $1 - \tan^2\theta = \cos 2\theta\sec^2\theta = \frac{2\cos 2\theta}{1 + \cos 2\theta}$. So $\frac{2}{2 - \sec^2\theta} = \frac{1 + \cos 2\theta}{\cos 2\theta} = 1 + \frac{1}{\cos 2\theta}$. $\cos 4\theta = \frac{1}{3} \Rightarrow 2\cos^2 2\theta - 1 = \frac{1}{3} \Rightarrow \cos 2\theta = \pm\sqrt{\frac{2}{3}}$, and both signs occur for $\theta$ in the given range (one in $(0, \frac{\pi}{4})$, one in $(\frac{\pi}{4}, \frac{\pi}{2})$). $f\left(\frac{1}{3}\right) = 1 \pm \sqrt{\frac{3}{2}}$: the values in (A) and (B).

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