JEE (Advanced) · 2018

JEE Advanced 2018, 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
36 Q
1. Properties of Bulk Matter4
2. Electrostatics4
3. Physical World and Measurement4
19 easy · 16 medium · 1 hard
Chemistry
36 Q
1. Coordination Compounds4
2. Thermodynamics4
3. p-Block Elements3
6 easy · 21 medium · 9 hard
Mathematics
36 Q
1. Differential Equations5
2. Trigonometry4
3. Permutations and Combinations3
8 easy · 22 medium · 6 hard

Chapters

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

Differential Equations5
Class 12: Differential Equations
Trigonometry4
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Permutations and Combinations3
Class 11: Permutations and Combinations
Integral Calculus3
Class 12: Integrals; Class 12: Application of Integrals
Circles3
Class 11: Conic Sections
Complex Numbers2
Class 11: Complex Numbers and Quadratic Equations
Three Dimensional Geometry2
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Sets, Relations and Functions2
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Vector Algebra2
Class 12: Vector Algebra
Probability2
Class 11: Probability; Class 12: Probability
Matrices and Determinants2
Class 12: Matrices; Class 12: Determinants
Conic Sections2
Class 11: Conic Sections
Applications of Derivatives1
Class 12: Application of Derivatives
Sequences and Series1
Class 11: Sequences and Series
Binomial Theorem1
Class 11: Binomial Theorem
Limits, Continuity and Differentiability1
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability

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
Q.1	Laws of MotionCircular motion dynamics	Class 11: Laws of Motion	Multiple correctEasy
A easy multiple correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion The force is $F = -\frac{dV}{dr} = -kr$: an attraction towards $O$ of magnitude $kr$. On the circular orbit this force supplies the centripetal force: $\frac{mv^2}{R} = kR$, so $v^2 = \frac{k}{m}R^2$ and $v = \sqrt{\frac{k}{m}}\,R$. (B) is true and (A) is false. Angular momentum about $O$: $L = mvR = m\sqrt{\frac{k}{m}}\,R \cdot R = \sqrt{mk}\,R^2$. (C) is true and (D) is false. Answer: (B), (C).
Q.2	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 With $m = 1$ kg, $\vec{a} = \vec{F}/m = t\,\hat{i} + \hat{j}$. Integrating from rest: $\vec{v} = \frac{t^2}{2}\hat{i} + t\,\hat{j}$; integrating again from the origin: $\vec{r} = \frac{t^3}{6}\hat{i} + \frac{t^2}{2}\hat{j}$. At $t = 1$ s: $\vec{r} = \frac{1}{6}\hat{i} + \frac{1}{2}\hat{j}$, $\vec{v} = \frac{1}{2}\hat{i} + \hat{j} = \frac{1}{2}(\hat{i} + 2\hat{j})$, $\vec{F} = \hat{i} + \hat{j}$. Torque: $\vec{\tau} = \vec{r} \times \vec{F} = \left(\frac{1}{6}\cdot 1 - \frac{1}{2}\cdot 1\right)\hat{k} = -\frac{1}{3}\hat{k}$ N m. (A) $\|\vec\tau\| = \frac{1}{3}$ N m: true. (B) The torque points along $-\hat{k}$, not $+\hat{k}$: false. (C) $\vec{v} = \frac{1}{2}(\hat{i} + 2\hat{j})$ m s$^{-1}$: true. (D) $\|\vec{r}\| = \sqrt{\frac{1}{36} + \frac{1}{4}} = \frac{\sqrt{10}}{6}$ m, not $\frac{1}{6}$ m: false. Answer: (A), (C).
Q.3	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Multiple correctEasy
A easy multiple correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Balance the upward surface-tension pull on the water column against its weight: $2\pi r\sigma\cos\theta = \pi r^2 h\rho g$, so $h = \frac{2\sigma\cos\theta}{r\rho g}$. (A) For a given tube material ($\theta$ fixed), $h \propto \frac{1}{r}$: $h$ decreases as $r$ increases. True. (B) $h \propto \sigma$, so it is not independent of $\sigma$. False. (C) In a lift accelerating upward at $a$, the effective gravity is $g + a$, so $h = \frac{2\sigma\cos\theta}{r\rho(g + a)}$ is smaller. True. (D) $h \propto \cos\theta$, not $\theta$. False. Answer: (A), (C).
Q.4	Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Multiple correctMedium
A medium multiple correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The middle wire joins the two loops, so each loop is an independent series $LR$ circuit with its own battery. Left loop ($R$, $L$): $i_1 = \frac{V}{R}\left(1 - e^{-Rt/L}\right)$. With the left battery’s positive plate at the top, $i_1$ flows along the top towards the middle and down the middle wire. Right loop ($R$, $2L$): $i_2 = \frac{V}{R}\left(1 - e^{-Rt/2L}\right)$. The right battery is reversed (positive plate at the bottom), so $i_2$ flows along the bottom towards the middle and up the middle wire. So the middle-wire current is $I = i_1 - i_2 = \frac{V}{R}\left(e^{-Rt/2L} - e^{-Rt/L}\right)$. Put $x = e^{-Rt/2L}$: $I = \frac{V}{R}(x - x^2)$, which is largest at $x = \frac{1}{2}$, where $I_{max} = \frac{V}{R}\cdot\frac{1}{4} = \frac{V}{4R}$. $e^{-R\tau/2L} = \frac{1}{2}$ gives $\tau = \frac{2L}{R}\ln 2$. So (B) and (D) are true; (A) and (C) are not. Answer: (B), (D).
Q.5	Magnetic Effects of Current and MagnetismBiot-Savart law	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Multiple correctMedium
A medium multiple correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Fields at the origin. Wire at $x = +R$ (current $I_1$ along $+\hat{j}$): the origin lies on its $-\hat{i}$ side, and $\hat{j}\times(-\hat{i}) = +\hat{k}$, so it gives $+\frac{\mu_0 I_1}{2\pi R}\hat{k}$. Wire at $x = -R$ (current $I_2$): the origin lies on its $+\hat{i}$ side, $\hat{j}\times\hat{i} = -\hat{k}$, so it gives $-\frac{\mu_0 I_2}{2\pi R}\hat{k}$. Together the wires give $\frac{\mu_0 (I_1 - I_2)}{2\pi R}\hat{k}$. Loop: clockwise seen from above means its field on the axis points along $-\hat{k}$. At the origin, a distance $\sqrt{3}R$ below the centre: $B = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 I}{16R}$, along $-\hat{k}$. (A) $I_1 = I_2$: the wires cancel and only the loop’s $\frac{\mu_0 I}{16R}$ remains, so $\vec{B} \neq 0$. True. (B) $I_1 > 0$, $I_2 < 0$: $I_1 - I_2 > 0$, the wires’ field is along $+\hat{k}$ and can cancel the loop’s $-\hat{k}$ field for suitable magnitudes. True. (C) $I_1 < 0$, $I_2 > 0$: the wires’ field is along $-\hat{k}$, the same way as the loop’s, so the total cannot vanish. False. (D) At the loop’s centre $(0, 0, \sqrt{3}R)$ each wire is at distance $2R$. For $I_1 = I_2$ their fields have equal and opposite $z$-components (by the mirror symmetry $x \to -x$) and only $x$-components survive. The loop gives $\frac{\mu_0 I}{2R}$ along $-\hat{k}$ at its centre. So $B_z = -\frac{\mu_0 I}{2R}$. True. Answer: (A), (B), (D).
Q.6	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Multiple correctEasy
A easy multiple correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Read each process off the $T$–$V$ graph. (A) Process I is a curve along which $V$ changes (it decreases a little as $T$ rises), so it is not isochoric (that would be a vertical line). False. (B) Process II is horizontal: $T$ constant while $V$ increases, an isothermal expansion. $\Delta U = 0$ and the gas does positive work, so $Q = W > 0$: heat is absorbed. True. (C) Process IV is horizontal at the lower temperature with $V$ decreasing: isothermal compression, $\Delta U = 0$, $W < 0$, so $Q < 0$: heat is released. True. (D) An isobaric process has $V \propto T$, a straight line through the origin on this graph. I and III are curves that do not do this (in each, $T$ and $V$ change in opposite senses), so neither is isobaric. True. Answer: (B), (C), (D).
Q.7	KinematicsScalars and vectors	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	NumericalEasy
A easy numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Both vectors have magnitude $a$ and the angle between them is $\omega t$. $\|\vec{A} + \vec{B}\| = 2a\left\|\cos\frac{\omega t}{2}\right\|$ and $\|\vec{A} - \vec{B}\| = 2a\left\|\sin\frac{\omega t}{2}\right\|$. The condition gives $\tan\frac{\omega t}{2} = \frac{1}{\sqrt{3}}$, first at $\frac{\omega t}{2} = \frac{\pi}{6}$, i.e. $\omega\tau = \frac{\pi}{3}$. $\tau = \frac{\pi/3}{\pi/6} = 2$ s. Answer: $2.00$.
Q.8	Oscillations and WavesDoppler effect	Class 11: Oscillations; Class 11: Waves	NumericalMedium
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Equidistant means the stationary man is directly above the midpoint: each walker is $5$ m away horizontally and $12$ m below, so $13$ m away, and the line of sight makes $\cos\theta = \frac{5}{13}$ with the walking line. The man behind (speed $2$) is approaching the point below the listener: velocity component towards him $2 \times \frac{5}{13} = \frac{10}{13}$ m s$^{-1}$. The man in front (speed $1$) is moving away: component away $1 \times \frac{5}{13} = \frac{5}{13}$ m s$^{-1}$. $f_1 = 1430\cdot\frac{330}{330 - \frac{10}{13}} = 1430\cdot\frac{4290}{4280}$, $f_2 = 1430\cdot\frac{330}{330 + \frac{5}{13}} = 1430\cdot\frac{4290}{4295}$. Beat frequency $= f_1 - f_2 = 1430 \times 4290 \times \frac{4295 - 4280}{4280 \times 4295} = \frac{92\,020\,500}{18\,382\,600} \approx 5.006$ Hz. Answer: $5.01$ (about $5$ Hz).
Q.9	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	NumericalMedium
A medium numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rolling without slipping down an incline: $a = \frac{g\sin\theta}{1 + I/(mR^2)}$. Ring ($I = mR^2$): $a_r = \frac{g\sin\theta}{2}$. Disc ($I = \frac{1}{2}mR^2$): $a_d = \frac{2g\sin\theta}{3}$. Path length $s = \frac{h}{\sin 60^\circ} = \frac{2h}{\sqrt{3}}$, and $t = \sqrt{\frac{2s}{a}}$. $\frac{s}{g\sin\theta} = \frac{h}{g\sin^2 60^\circ} = \frac{4h}{3g} = \frac{2h}{15}$. Ring: $t_r = \sqrt{4\cdot\frac{2h}{15}} = \sqrt{\frac{8h}{15}}$. Disc: $t_d = \sqrt{3\cdot\frac{2h}{15}} = \sqrt{\frac{2h}{5}} = \sqrt{\frac{6h}{15}}$. $t_r - t_d = \sqrt{\frac{h}{15}}\left(\sqrt{8} - \sqrt{6}\right) = \sqrt{\frac{2h}{15}}\,(2 - \sqrt{3})$. Setting this equal to $\frac{2 - \sqrt{3}}{\sqrt{10}}$: $\frac{2h}{15} = \frac{1}{10}$, so $h = \frac{15}{20} = 0.75$ m. Answer: $0.75$.
Q.10	Work, Energy and PowerCollisions	Class 11: Work, Energy and Power	NumericalMedium
A medium numerical question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Elastic collision of $m_1 = 1$ kg at $u = 2$ m s$^{-1}$ with $m_2 = 2$ kg at rest: $v_1 = \frac{m_1 - m_2}{m_1 + m_2}u = -\frac{2}{3}$ m s$^{-1}$ (the 1 kg block bounces back), $v_2 = \frac{2m_1}{m_1 + m_2}u = \frac{4}{3}$ m s$^{-1}$. The 2 kg block now compresses the spring and performs SHM with $\omega = \sqrt{k/m} = \sqrt{2/2} = 1$ rad s$^{-1}$. It is back at the unstretched position after half a period, $\frac{T}{2} = \frac{\pi}{\omega} = \pi$ s, i.e. back where the collision happened. In that time the 1 kg block has moved away at $\frac{2}{3}$ m s$^{-1}$: distance $\frac{2}{3}\pi \approx 2.094$ m. So the blocks are $\frac{2\pi}{3} \approx 2.09$ m apart. Answer: $2.09$.
Q.11	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	NumericalMedium
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Step 1 ($S_1$ closed, $S_2$ open): only $C_3$ is across the cell, so it charges to $Q_0 = C_3V_0 = 1 \times 8 = 8\ \mu$C. Step 2 ($S_1$ open, $S_2$ closed): $C_3$ is now connected across the series pair $C_1$ ($= \epsilon_r\ \mu$F) and $C_2$ ($= 1\ \mu$F), whose combined capacitance is $C^{\prime} = \frac{\epsilon_r}{\epsilon_r + 1}\ \mu$F. Charge is shared until the voltages are equal. Final charge on $C_3$ is $5\ \mu$C, so the common voltage is $V = \frac{5}{1} = 5$ V. Charge conservation: the series pair holds $8 - 5 = 3\ \mu$C at $5$ V, so $C^{\prime} = \frac{3}{5} = 0.6\ \mu$F. $\frac{\epsilon_r}{\epsilon_r + 1} = 0.6 \Rightarrow \epsilon_r = 0.6\epsilon_r + 0.6 \Rightarrow \epsilon_r = 1.5$. Answer: $1.50$.
Q.12	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	NumericalMedium
A medium numerical 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 Radius $r = \frac{mv}{qB}$ and time for a half circle $\frac{\pi m}{qB}$. Let $r_1 = \frac{mv_0}{qB_1}$; then $r_2 = \frac{r_1}{4}$. At the origin $\vec{v} = v_0\hat{j}$ and the force $q\vec{v}\times\vec{B} \propto \hat{j}\times\hat{k} = \hat{i}$: the particle curves towards $+x$. A semicircle of radius $r_1$ brings it back to the $x$-axis at $x = 2r_1$, moving along $-\hat{j}$, after time $\frac{\pi r_1}{v_0}$. Below the axis the force is along $(-\hat{j})\times\hat{k} = -\hat{i}$: a semicircle of radius $r_2 = \frac{r_1}{4}$ curving back towards $-x$. It crosses the axis from below at $x = 2r_1 - 2r_2 = 1.5r_1$, after a further $\frac{\pi r_2}{v_0} = \frac{\pi r_1}{4v_0}$. So $T = \frac{\pi r_1}{v_0}\left(1 + \frac{1}{4}\right) = \frac{5\pi r_1}{4v_0}$. Distance covered along the $x$-axis: $2r_1$ out and then $2r_2 = 0.5r_1$ back, total $2.5r_1$. Average speed along $x$ $= \frac{2.5r_1}{5\pi r_1/(4v_0)} = \frac{2v_0}{\pi} = \frac{2\pi}{\pi} = 2$ m s$^{-1}$. Answer: $2.00$.
Q.13	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalEasy
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Parallel sunlight converges to the focus at $20$ cm and then spreads out again as a cone. At $22$ cm the beam is $2$ cm beyond the focus, so by similar triangles its radius is $\frac{2}{20} = \frac{1}{10}$ of the lens aperture radius, and its area $\frac{1}{100}$ of the aperture area. With no loss, the same power passes through a $100$ times smaller area: $I = 1.3 \times 100 = 130$ kW m$^{-2}$. Answer: $130$.
Q.14	Properties of Bulk MatterHeat transfer	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalEasy
A easy numerical 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 series the same heat current $H$ flows through both cylinders: $H = \frac{K_1 A_1 (300 - 200)}{L} = \frac{K_2 A_2 (200 - 100)}{L}$. The temperature drops are equal ($100$ K each), so $K_1A_1 = K_2A_2$. $A_2 = \pi(2r)^2 = 4A_1$, hence $K_1 = 4K_2$ and $K_1/K_2 = 4$. Answer: $4.00$.
Q.15	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	ComprehensionEasy
A easy comprehension question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The Lorentz force $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$ requires $E$ and $vB$ to have the same dimensions (equivalently, in an electromagnetic wave $E = cB$). So $[E] = [B][v] = [B][L][T]^{-1}$. Answer: (C).
Q.16	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	ComprehensionEasy
A easy comprehension question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The speed of light satisfies $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$, so $[\mu_0][\epsilon_0] = [c]^{-2} = [L]^{-2}[T]^2$. Hence $[\mu_0] = [\epsilon_0]^{-1}[L]^{-2}[T]^2$. Answer: (D).
Q.17	Physical World and MeasurementError analysis	Class 11: Units and Measurement	ComprehensionMedium
A medium comprehension question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Here the numerator and denominator both depend on the one measured quantity $a$, so they are not independent: expand $r$ itself to first order. $\frac{dr}{da} = \frac{-(1+a) - (1-a)}{(1+a)^2} = -\frac{2}{(1+a)^2}$. So $\Delta r = \left\|\frac{dr}{da}\right\|\Delta a = \frac{2\Delta a}{(1+a)^2}$. (Applying the paragraph’s rule of adding the relative errors of $1 - a$ and $1 + a$ gives the same: $\frac{\Delta r}{r} = \frac{\Delta a}{1-a} + \frac{\Delta a}{1+a} = \frac{2\Delta a}{1-a^2}$, and multiplying by $r = \frac{1-a}{1+a}$ gives $\frac{2\Delta a}{(1+a)^2}$.) Answer: (B).
Q.18	Physical World and MeasurementError analysis	Class 11: Units and Measurement	ComprehensionMedium
A medium comprehension question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Nuclei left after $t = 1$ s: $N = 3000 - (1000 \pm 40) = 2000 \mp 40$. $N = N_0e^{-\lambda t}$ gives $\lambda t = \ln\frac{N_0}{N} = \ln\frac{3000}{2000(1 \mp 0.02)} = \ln\frac{3}{2} - \ln(1 \mp 0.02)$. With $\ln(1 + x) \approx x$: $\lambda = \ln\frac{3}{2} \pm 0.02$ s$^{-1}$ (for $t = 1$ s). So $\Delta\lambda = 0.02$ s$^{-1}$. Answer: (C).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Ammonium nitrate on gentle heating (below about $300\,^\circ$C) gives nitrous oxide, not nitrogen: NH$_4$NO$_3 \to$ N$_2$O + 2H$_2$O. (N$_2$ forms only on strong heating or detonation, above this range.) Not chosen. (B) Ammonium dichromate decomposes on warming (around $180$–$200\,^\circ$C): (NH$_4$)$_2$Cr$_2$O$_7 \to$ N$_2$ + Cr$_2$O$_3$ + 4H$_2$O. Gives N$_2$. (C) Barium azide decomposes on gentle heating to very pure nitrogen: Ba(N$_3$)$_2 \to$ Ba + 3N$_2$. Gives N$_2$. (D) Magnesium nitride is thermally stable; it does not give N$_2$ on heating (with water it gives NH$_3$). Not chosen. Answer: (B), (C).
Q.2	Coordination CompoundsValence bond theory	Class 12: Coordination Compounds	Multiple correctMedium
A medium multiple correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) Fe(0) has $8$ valence electrons ($3d^6 4s^2$) and five CO donate $10$: total $18$. Ni(0) has $10$ ($3d^8 4s^2$) and four CO donate $8$: total $18$. Not $16$. False. (B) CO is a strong-field ligand; the metal electrons are paired (Fe(CO)$_5$ and Ni(CO)$_4$ are diamagnetic). True. (C) A lower oxidation state leaves more electron density on the metal, which increases back-donation into the CO $\pi^$ orbitals; the M–C bond gains double-bond character and strengthens. True. (D) A higher oxidation state reduces back-donation, so less electron density enters the CO $\pi^$ orbital and the C–O bond becomes stronger, not weaker. False. Answer: (B), (C).
Q.3	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Basic character of oxides increases down the group as metallic character increases; N$_2$O$_5$ is strongly acidic, while bismuth’s oxide is far more basic. True. (B) N is a small non-metal close to F in electronegativity, so N–F bonds are covalent; Bi is a large, metallic atom and BiF$_3$ is largely ionic. So NF$_3$ is more covalent. True. (C) NH$_3$ molecules are hydrogen bonded; PH$_3$ molecules are not. NH$_3$ boils at about $-33\,^\circ$C and PH$_3$ at about $-88\,^\circ$C, so PH$_3$ boils lower. True. (D) The N–N single bond is weak because the small N atoms bring their lone pairs close together and these repel; the P–P single bond is stronger. False. Answer: (A), (B), (C).
Q.4	Haloalkanes and HaloarenesSN1 and SN2 mechanisms	Class 12: Haloalkanes and Haloarenes	Multiple correctHard
A hard multiple correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Each of the three steps is an S$_N$2 substitution at the ring carbon that carries the functional group, and each inverts that carbon: 1) PBr$_3$ turns the secondary alcohol into the bromide with inversion. 2) NaI in acetone (Finkelstein) replaces Br by I with inversion. 3) NaN$_3$ in DMF replaces I by N$_3$ with inversion. Three inversions give a net inversion; the methyl-bearing carbon is untouched. So in X the OH must lie on the opposite face to the N$_3$ of the product, on the same carbon. The product, drawn with the ring apex at the top, has Me on a wedge and N$_3$ on a hash (trans). X, drawn the same way, therefore has Me on a wedge and OH on a wedge (cis): that is structure (B). (A) is the trans alcohol: three inversions would give the cis azide. Not X. (C) and (D) are drawn with the ring turned upside down in the plane of the paper while the wedges are kept. Flipping the drawing top-to-bottom while keeping wedges as wedges is a reflection, so (D) is the mirror image of (B) and (C) the mirror image of (A). (D) would give the enantiomer of the given, enantiomerically pure product. Not X. Answer: (B).
Q.5	HydrocarbonsAromatic hydrocarbons	Class 11: Hydrocarbons	Multiple correctMedium
A medium multiple correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A) Three molecules of acetone undergo acid-catalysed aldol condensations and cyclise with loss of three H$_2$O to mesitylene (1,3,5-trimethylbenzene). Yes. (B) Alkynes trimerise in a red-hot iron tube at 873 K; propyne gives 1,3,5-trimethylbenzene. Yes. (C) Br$_2$/NaOH is the haloform reaction: each COCH$_3$ becomes COO$^-$, and acidification gives benzene-1,3,5-tricarboxylic acid. Sodalime then removes all three carboxyl groups (decarboxylation), giving benzene, not mesitylene. No. (D) Clemmensen reduction (Zn/Hg, HCl) reduces each CHO to CH$_3$, giving 1,3,5-trimethylbenzene. Yes. Answer: (A), (B), (D).
Q.6	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 Read the processes: A→B is vertical (same $T_1$): isothermal, $V_1 \to V_2$. B→C joins two states at the same pressure $P_2$ (a straight line through the origin on a $V$–$T$ plot): isobaric, $T_1 \to T_2$. C→A is horizontal (same $V_1$): isochoric, $T_2 \to T_1$. For an ideal gas $U$ and $H$ depend only on $T$: any path from $T_1$ to $T_2$ has $\Delta U = nC_V(T_2 - T_1)$ and $\Delta H = nC_P(T_2 - T_1)$. Work is taken as done on the gas ($\Delta U = q + w$). $q_{AC}$: the isochoric step (reversed C→A) has $w = 0$, so $q_{AC} = \Delta U_{AC} = nC_V(T_2 - T_1) = \Delta U_{BC}$. True. $w_{AB}$: isothermal, $w_{AB} = -nRT_1\ln\frac{V_2}{V_1}$, not $P_2(V_2 - V_1)$. So (A) fails. $w_{BC}$: isobaric at $P_2$ from $V_2$ to $V_1$: $w_{BC} = -P_2(V_1 - V_2) = P_2(V_2 - V_1)$. True. $q_{BC} = \Delta H_{BC} = nC_P(T_2 - T_1) = \Delta H_{AC}$. True. So (B) holds. $\Delta H_{CA} = nC_P(T_1 - T_2)$ and $\Delta U_{CA} = nC_V(T_1 - T_2)$. Since $T_1 < T_2$ and $C_P > C_V$, $\Delta H_{CA}$ is the more negative: $\Delta H_{CA} < \Delta U_{CA}$. So (C) holds (with $q_{AC} = \Delta U_{BC}$ shown above), and (D) fails. Answer: (B), (C).
Q.7	Coordination CompoundsCrystal field theory	Class 12: Coordination Compounds	NumericalMedium
A medium numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Count unpaired electrons in each: H atom: $1s^1$, one unpaired. Paramagnetic. NO$_2$: $23$ valence electrons, an odd-electron molecule. Paramagnetic. O$_2^-$: one unpaired electron in $\pi^$. Paramagnetic. S$_2$ (vapour): like O$_2$, two unpaired electrons in $\pi^$. Paramagnetic. Mn$_3$O$_4$ (Mn$^{2+}$ $d^5$ and Mn$^{3+}$ $d^4$): unpaired electrons. Paramagnetic. (NH$_4$)$_2$[FeCl$_4$]: Fe$^{2+}$ $d^6$, tetrahedral, weak field: $4$ unpaired. Paramagnetic. (NH$_4$)$_2$[NiCl$_4$]: Ni$^{2+}$ $d^8$, tetrahedral: $2$ unpaired. Paramagnetic. K$_2$MnO$_4$: Mn(VI), $d^1$. Paramagnetic. K$_2$CrO$_4$: Cr(VI), $d^0$. Diamagnetic. Only K$_2$CrO$_4$ is diamagnetic. Answer: $1$.
Q.8	Some Basic Concepts of ChemistryStoichiometry	Class 11: Some Basic Concepts of Chemistry	NumericalMedium
A medium numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry (NH$_4$)$_2$SO$_4$ + Ca(OH)$_2$ → CaSO$_4$·2H$_2$O (gypsum) + 2NH$_3$. Molar mass of (NH$_4$)$_2$SO$_4$ $= 2(18) + 96 = 132$; $\frac{1584}{132} = 12$ mol. So $12$ mol gypsum (molar mass $40 + 96 + 36 = 172$): $12 \times 172 = 2064$ g; and $24$ mol NH$_3$. NiCl$_2$·6H$_2$O: $59 + 71 + 108 = 238$; $\frac{952}{238} = 4$ mol. NiCl$_2$·6H$_2$O + 6NH$_3$ → [Ni(NH$_3$)$_6$]Cl$_2$ + 6H$_2$O. $4$ mol need exactly $24$ mol NH$_3$, so all the ammonia is used. [Ni(NH$_3$)$_6$]Cl$_2$: $59 + 6(17) + 71 = 232$; $4 \times 232 = 928$ g. Combined: $2064 + 928 = 2992$ g. Answer: $2992$.
Q.9	Solid StateClose packed structures	No matching NCERT chapter	NumericalEasy
A easy numerical question on close packed structures, from Solid State. NCERT: No matching NCERT chapter The wording (a central anion, face-centred and corner cations) fixes the NaCl cell with M at the corners and face centres and X at the edge centres and the body centre. Start: M $= 8\times\frac{1}{8} + 6\times\frac{1}{2} = 4$; X $= 12\times\frac{1}{4} + 1 = 4$. (i) Keep only the central X: X $= 1$. (ii) Face-centred M become X: X $= 1 + 3 = 4$; M left: corners only, $1$. (iii) Remove the corner M: M $= 0$. (iv) The central X becomes M: X $= 3$, M $= 1$. Ratio anions : cations $= 3/1 = 3$. Answer: $3.00$.
Q.10	ElectrochemistryNernst equation	Class 12: Electrochemistry	NumericalEasy
A easy numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: Mg + Cu$^{2+}$ → Mg$^{2+}$ + Cu, $n = 2$, $Q = \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} = x$. Nernst: $E = E^\circ - \frac{RT}{2F}\ln x$, so $2.70 - 2.67 = \frac{300}{2 \times 11500}\ln x$. $\ln x = \frac{0.03 \times 23000}{300} = 2.30 = \ln 10$, so $x = 10$. Answer: $10.00$.
Q.11	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	NumericalMedium
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Moles from $n = \frac{PV}{RT}$: $n_A \propto \frac{5 \times 1}{400} = \frac{1}{80}$, $n_B \propto \frac{1 \times 3}{300} = \frac{1}{100}$. With a sliding, heat-conducting partition, at equilibrium both sides have the same pressure and the same temperature, so $\frac{V_A}{V_B} = \frac{n_A}{n_B} = \frac{100}{80} = \frac{5}{4}$ (whatever the common final temperature is). The total volume stays $4$ m$^3$: $V_A = 4 \times \frac{5}{9} = \frac{20}{9} \approx 2.22$ m$^3$. Answer: $2.22$.
Q.12	SolutionsRaoult's law	Class 12: Solutions	NumericalEasy
A easy numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions Equimolar: $\frac{1}{2}(p_A^\circ + p_B^\circ) = 45$, so $p_B^\circ = 90 - 20 = 70$ Torr. New solution: $20x_A + 70(1 - x_A) = 22.5 \Rightarrow 50x_A = 47.5 \Rightarrow x_A = 0.95$, $x_B = 0.05$. $\frac{x_A}{x_B} = \frac{0.95}{0.05} = 19$. Answer: $19.00$.
Q.13	EquilibriumSolubility product	Class 11: Equilibrium	NumericalMedium
A medium numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium AB ⇌ A$^+$ + B$^-$, and in acid B$^-$ + H$^+$ ⇌ HB. If $s$ is the solubility, $[\text{A}^+] = s = [\text{B}^-] + [\text{HB}] = [\text{B}^-]\left(1 + \frac{[\text{H}^+]}{K_a}\right) = [\text{B}^-](1 + 10^{5})$. $K_{sp} = [\text{A}^+][\text{B}^-] = \frac{s^2}{1 + 10^5}$, so $s^2 = 2\times10^{-10} \times (1 + 10^5) \approx 2 \times 10^{-5}$. $s = \sqrt{20}\times 10^{-3} \approx 4.47 \times 10^{-3}$ mol L$^{-1}$. Answer: $4.47$.
Q.14	Solutionsvan't Hoff factor	Class 12: Solutions	NumericalHard
A hard numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions From the graph (at 760 mmHg): X boils at 360 K and its NaCl solution at 362 K, so $\Delta T_b = 2$ K; Y boils at 367 K and its NaCl solution at 368 K, so $\Delta T_b = 1$ K. Same molality $m$ and $i = 2$ for NaCl in both: $K_b(X)\cdot 2m = 2$ and $K_b(Y)\cdot 2m = 1$, so $K_b(X) = 2K_b(Y)$. For S dimerising ($2$S ⇌ S$_2$) with degree $\alpha$: $i = 1 - \frac{\alpha}{2}$. Same molality $m^{\prime}$ of S in both. $\Delta T_X = 3\Delta T_Y$: $K_b(X)\,m^{\prime}\left(1 - \frac{\alpha_X}{2}\right) = 3K_b(Y)\,m^{\prime}\left(1 - \frac{0.7}{2}\right)$. $2\left(1 - \frac{\alpha_X}{2}\right) = 3 \times 0.65 = 1.95$, so $1 - \frac{\alpha_X}{2} = 0.975$ and $\alpha_X = 0.05$. Answer: $0.05$.
Q.15	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 Benzene + CO/HCl with anhydrous AlCl$_3$/CuCl (Gattermann–Koch) gives benzaldehyde. Benzaldehyde + Ac$_2$O/NaOAc (Perkin condensation) gives cinnamic acid, PhCH=CH–COOH: this is X. Br$_2$ adds across the C=C; with Na$_2$CO$_3$ the carboxylate of the dibromide loses CO$_2$ and Br$^-$ together, giving $\beta$-bromostyrene, PhCH=CHBr. Moist KOH at 473 K eliminates HBr from the vinyl bromide: PhC≡CH, phenylacetylene. This is Y. (A), (B) and (D) still carry the three-carbon acid chain or its bromides; none survives the decarboxylation and elimination. Answer: (C).
Q.16	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 X is cinnamic acid, PhCH=CH–COOH (see the previous question). H$_2$/Pd-C reduces the C=C (not the ring or the COOH): 3-phenylpropanoic acid, PhCH$_2$CH$_2$COOH. Heating with H$_3$PO$_4$ (a strong acid) generates the acylium ion, which attacks the ortho position of its own ring: intramolecular Friedel–Crafts acylation closes a five-membered ring. Z is 1-indanone: a benzene ring fused to a cyclopentanone, with CH$_2$CH$_2$ and C=O in the new ring. (B) would need an extra C=C, (C) an extra OH, and (D) has lost the carbonyl; none of these steps does that. Answer: (A).
Q.17	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	ComprehensionHard
A hard comprehension question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Identify P. Oxidation to the dibasic acid that makes dacron with ethylene glycol means terephthalic acid: P is a para-disubstituted benzene. Ozonolysis giving an aliphatic ketone means a C=C bearing two alkyl groups on one end. P = 4-(2-methylprop-1-enyl)benzoic acid, HOOC–C$_6$H$_4$–CH=C(CH$_3$)$_2$: C$_{11}$H$_{12}$O$_2$ (checks), and ozonolysis gives acetone. P → Q: H$_2$/Pd-C saturates the side chain (4-isobutylbenzoic acid); SOCl$_2$ gives the acid chloride; MeMgBr with CdCl$_2$ (dimethylcadmium) gives the methyl ketone 4-isobutylacetophenone; NaBH$_4$ gives the secondary alcohol Q, 1-(4-isobutylphenyl)ethanol. Q → R: HCl gives the chloride; Mg/Et$_2$O the Grignard reagent; CO$_2$ then H$_3$O$^+$ adds one carbon as COOH at that carbon. R = 2-(4-isobutylphenyl)propanoic acid (ibuprofen). (B) and (D) are ortho, not para; (C) has a sec-butyl rather than an isobutyl group. Answer: (A).
Q.18	AminesClassification and identification of amines	Class 12: Amines	ComprehensionHard
A hard comprehension question on classification and identification of amines, from Amines. NCERT: Class 12: Amines P is 4-(2-methylprop-1-enyl)benzoic acid (see the previous question). 1) H$_2$/Pd-C: 4-isobutylbenzoic acid. 2) NH$_3$, heat: the amide, 4-isobutylbenzamide. 3) Br$_2$/NaOH (Hofmann bromamide degradation): loses the carbonyl carbon, giving 4-isobutylaniline. 4) CHCl$_3$, KOH, heat (carbylamine reaction): the primary amine becomes the isocyanide Ar–N≡C. 5) H$_2$/Pd-C reduces the isocyanide to the N-methyl secondary amine Ar–NH–CH$_3$. S = N-methyl-4-isobutylaniline. (A) and (C) are benzylamines with a sec-butyl group; (D) is the ortho isomer. Answer: (B).

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	Multiple correctMedium
A medium multiple correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations The question asks for the FALSE statements. (A) $-1 - i$ lies in the third quadrant: $\arg(-1 - i) = -\frac{3\pi}{4}$, not $\frac{\pi}{4}$. The statement is false, so (A) is chosen. (B) $-1 + it$ has negative real part. As $t \to 0^+$ it approaches $-1$ from above and $\arg \to \pi$; as $t \to 0^-$ it approaches from below and $\arg \to -\pi$. $f$ jumps at $t = 0$, so it is not continuous everywhere. False, chosen. (C) $\arg\frac{z_1}{z_2} = \arg z_1 - \arg z_2 + 2k\pi$ for some integer $k$ (the correction that brings the value back into $(-\pi, \pi]$). So the expression equals $2k\pi$. True, not chosen. (D) The argument being $\pi$ means the cross ratio $\frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}$ is a negative real number. A real cross ratio puts $z$ on the circle through $z_1, z_2, z_3$ (a line only when the three are collinear); in general the locus is an arc of that circle, not a straight line. False, chosen. Answer: (A), (B), (D).
Q.2	TrigonometryTrigonometric functions	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Multiple correctMedium
A medium multiple correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Cosine rule at $Q$: $PR^2 = (10\sqrt{3})^2 + 10^2 - 2\cdot 10\sqrt{3}\cdot 10\cos 30^\circ = 300 + 100 - 300 = 100$, so $PR = 10$. So $PR = QR = 10$: the triangle is isosceles with $\angle QPR = \angle PQR = 30^\circ$ and $\angle QRP = 120^\circ$. (A) $\angle QPR = 30^\circ$, not $45^\circ$. False. (B) Area $= \frac{1}{2}\cdot PQ\cdot QR\sin 30^\circ = \frac{1}{2}\cdot 10\sqrt{3}\cdot 10\cdot\frac{1}{2} = 25\sqrt{3}$, and $\angle QRP = 120^\circ$. True. (C) $s = \frac{10 + 10 + 10\sqrt{3}}{2} = 10 + 5\sqrt{3}$; $r = \frac{\Delta}{s} = \frac{25\sqrt{3}}{5(2 + \sqrt{3})} = 5\sqrt{3}(2 - \sqrt{3}) = 10\sqrt{3} - 15$. True. (D) Circumradius $R = \frac{QR}{2\sin\angle QPR} = \frac{10}{2\cdot\frac{1}{2}} = 10$, so the circumcircle has area $100\pi$. True. Answer: (B), (C), (D).
Q.3	Three Dimensional GeometryDirection ratios and direction cosines	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Multiple correctEasy
A easy multiple correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Normals: $\vec{n}_1 = (2, 1, -1)$, $\vec{n}_2 = (1, 2, 1)$. The line of intersection is along $\vec{n}_1\times\vec{n}_2 = (1\cdot1 - (-1)\cdot2,\ (-1)\cdot1 - 2\cdot1,\ 2\cdot2 - 1\cdot1) = (3, -3, 3) \propto (1, -1, 1)$. (A) $(1, 2, -1)$ is not parallel to $(1, -1, 1)$. False. (B) Rewrite the line: $\frac{x - 4/3}{3} = \frac{y - 1/3}{-3} = \frac{z}{3}$, direction $(3, -3, 3) \propto (1, -1, 1)$. It is parallel to the line of intersection, not perpendicular. False. (C) $\cos\theta = \frac{\|\vec{n}_1\cdot\vec{n}_2\|}{\|\vec{n}_1\|\|\vec{n}_2\|} = \frac{2 + 2 - 1}{\sqrt{6}\sqrt{6}} = \frac{1}{2}$, so $\theta = 60^\circ$. True. (D) $P_3$ has normal $(1, -1, 1)$ and passes through $(4, 2, -2)$: $x - y + z = 4 - 2 - 2 = 0$. Distance of $(2, 1, 1)$: $\frac{\|2 - 1 + 1\|}{\sqrt{3}} = \frac{2}{\sqrt{3}}$. True. Answer: (C), (D).
Q.4	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	Multiple correctHard
A hard multiple correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Since $\|f(0)\| \le 2$, $(f^{\prime}(0))^2 \ge 85 - 4 = 81$, so $\|f^{\prime}(0)\| \ge 9$. (A) $f^{\prime}$ is continuous (f is twice differentiable) and $f^{\prime}(0) \neq 0$, so $f^{\prime}$ keeps one sign on some interval around $0$; there $f$ is strictly monotonic and hence one-one. True. (B) Mean value theorem on $[-4, 0]$: $f^{\prime}(x_0) = \frac{f(0) - f(-4)}{4}$ for some $x_0 \in (-4, 0)$, and $\|f(0) - f(-4)\| \le 4$, so $\|f^{\prime}(x_0)\| \le 1$. True. (C) Not forced. For example $f(x) = 2\sin\left(\frac{\sqrt{85}}{2}x\right)$ maps into $[-2, 2]$, has $f(0)^2 + f^{\prime}(0)^2 = 0 + 85 = 85$, and has no limit at infinity. False. (D) Let $g(x) = f(x)^2 + f^{\prime}(x)^2$, so $g(0) = 85$. By (B) there is $x_0 \in (-4, 0)$ with $\|f^{\prime}(x_0)\| \le 1$, so $g(x_0) \le 4 + 1 = 5$; likewise on $[0, 4]$ there is $x_1 \in (0, 4)$ with $g(x_1) \le 5$. $g$ is continuous on $[x_0, x_1]$, so it has a maximum there; since $g(0) = 85$ exceeds both end values, the maximum is at an interior point $\alpha \in (x_0, x_1) \subset (-4, 4)$ and $g^{\prime}(\alpha) = 2f^{\prime}(\alpha)\left(f(\alpha) + f^{\prime\prime}(\alpha)\right) = 0$. At $\alpha$, $g(\alpha) \ge 85$ and $f(\alpha)^2 \le 4$, so $f^{\prime}(\alpha)^2 \ge 81$: $f^{\prime}(\alpha) \neq 0$. Hence $f(\alpha) + f^{\prime\prime}(\alpha) = 0$. True. Answer: (A), (B), (D).
Q.5	Differential EquationsSeparation of variables	Class 12: Differential Equations	Multiple correctMedium
A medium multiple correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $e^{-f}f^{\prime} = e^{-g}g^{\prime}$. Integrating, $-e^{-f(x)} = -e^{-g(x)} - C$, i.e. $e^{-f(x)} - e^{-g(x)} = C$ for all $x$. At $x = 1$: $e^{-1} - e^{-g(1)} = C$. At $x = 2$: $e^{-f(2)} - e^{-1} = C$. Subtracting: $e^{-f(2)} + e^{-g(1)} = 2e^{-1}$. Both terms are positive, so $e^{-f(2)} < \frac{2}{e}$, i.e. $-f(2) < \ln 2 - 1$, so $f(2) > 1 - \ln 2$: (B) true, (A) false. In the same way $e^{-g(1)} < \frac{2}{e}$ gives $g(1) > 1 - \ln 2$: (C) true, (D) false. Answer: (B), (C).
Q.6	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 Write $f(x) = 1 - 2x + e^x\int_0^x e^{-t}f(t)\,dt$; then $f(0) = 1$. Differentiate: $f^{\prime}(x) = -2 + e^x\int_0^x e^{-t}f(t)\,dt + f(x) = -2 + (f(x) - 1 + 2x) + f(x)$, so $f^{\prime} - 2f = 2x - 3$. Try $f = ax + b$: $a - 2ax - 2b = 2x - 3$ gives $a = -1$, $b = 1$. General solution $f = 1 - x + Ce^{2x}$, and $f(0) = 1$ gives $C = 0$. So $f(x) = 1 - x$. (A) $f(1) = 0 \neq 2$. False. (B) $f(2) = -1$. True. On $[0, 1]$, $1 - x \le \sqrt{1 - x^2}$, so the area is $\int_0^1\sqrt{1 - x^2}\,dx - \int_0^1(1 - x)\,dx = \frac{\pi}{4} - \frac{1}{2} = \frac{\pi - 2}{4}$. (C) true, (D) false. Answer: (B), (C).
Q.7	Sets, Relations and FunctionsComposition of functions	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	NumericalEasy
A easy numerical question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Let $a = \log_2 9$. Then $a^{\frac{1}{\log_2 a}} = a^{\log_a 2} = 2$, so the first factor is $\left(a^{\frac{1}{\log_2 a}}\right)^2 = 2^2 = 4$. Second factor: $(\sqrt{7})^{\frac{1}{\log_4 7}} = 7^{\frac{1}{2}\log_7 4} = 7^{\log_7 2} = 2$. Product $= 4 \times 2 = 8$. Answer: $8.00$.
Q.8	Permutations and CombinationsFundamental principle of counting	Class 11: Permutations and Combinations	NumericalEasy
A easy numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations A number is divisible by $4$ exactly when its last two digits form a number divisible by $4$. With digits from $\{1, \dots, 5\}$ the possible endings are $12, 24, 32, 44, 52$: $5$ choices. The first three digits are free: $5^3 = 125$ choices. Total $= 125 \times 5 = 625$. Answer: $625$.
Q.9	Sequences and SeriesArithmetic progressions	Class 11: Sequences and Series	NumericalMedium
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $X$: $1 + 5k$, last term $1 + 5\cdot2017 = 10086$. $Y$: $9 + 7k$, last term $9 + 7\cdot2017 = 14128$. Common terms satisfy $n \equiv 1 \pmod 5$ and $n \equiv 2 \pmod 7$; the smallest is $16$ (which is in both), and they repeat every $35$: $16, 51, 86, \dots$. They must also be $\le 10086$: $16 + 35m \le 10086 \Rightarrow m \le 287.7$, so $m = 0, \dots, 287$: $288$ common terms. $\|X \cup Y\| = 2018 + 2018 - 288 = 3748$. Answer: $3748$.
Q.10	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	NumericalHard
A hard numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions For $\|x\| < \frac{1}{2}$ all the geometric series converge. Left argument: $\sum x^{i+1} = \frac{x^2}{1 - x}$ and $x\sum\left(\frac{x}{2}\right)^i = \frac{x^2}{2 - x}$, so $A = x^2\left(\frac{1}{1 - x} - \frac{1}{2 - x}\right) = \frac{x^2}{(1 - x)(2 - x)}$. Right argument: $\sum\left(-\frac{x}{2}\right)^i = -\frac{x}{2 + x}$ and $\sum(-x)^i = -\frac{x}{1 + x}$, so $B = \frac{x}{1 + x} - \frac{x}{2 + x} = \frac{x}{(1 + x)(2 + x)}$. Since $\frac{\pi}{2} - \cos^{-1}B = \sin^{-1}B$, the equation is $\sin^{-1}A = \sin^{-1}B$, i.e. $A = B$ (both lie well inside $[-1, 1]$ here). $x = 0$ is a solution. Otherwise divide by $x$: $x(1 + x)(2 + x) = (1 - x)(2 - x)$, i.e. $x^3 + 3x^2 + 2x = x^2 - 3x + 2$, so $h(x) = x^3 + 2x^2 + 5x - 2 = 0$. $h^{\prime}(x) = 3x^2 + 4x + 5 > 0$ (discriminant $16 - 60 < 0$), so $h$ has exactly one real root. $h(0) = -2 < 0$ and $h\left(\frac{1}{2}\right) = \frac{1}{8} + \frac{1}{2} + \frac{5}{2} - 2 = \frac{9}{8} > 0$, so that root lies in $\left(0, \frac{1}{2}\right)$. Total: $2$ solutions. Answer: $2.00$.
Q.11	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	NumericalMedium
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $y_n = \left(\prod_{k=1}^{n}\frac{n + k}{n}\right)^{1/n}$, so $\ln y_n = \frac{1}{n}\sum_{k=1}^{n}\ln\left(1 + \frac{k}{n}\right)$. As $n \to \infty$ this Riemann sum tends to $\int_0^1\ln(1 + x)\,dx = \left[(1 + x)\ln(1 + x) - x\right]_0^1 = 2\ln 2 - 1$. $L = e^{2\ln 2 - 1} = \frac{4}{e} \approx 1.47$, so $[L] = 1$. Answer: $1$.
Q.12	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	NumericalEasy
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}$, $\vec{b}$ and $\vec{a}\times\vec{b}$ are mutually perpendicular unit vectors. $\vec{c}\cdot\vec{a} = x = \|\vec{c}\|\cos\alpha = 2\cos\alpha$, and likewise $y = 2\cos\alpha$. $\|\vec{c}\|^2 = x^2 + y^2 + 1 = 8\cos^2\alpha + 1 = 4$. So $8\cos^2\alpha = 3$. Answer: $3.00$.
Q.13	TrigonometryTrigonometric functions	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	NumericalMedium
A medium numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Both roots satisfy the equation; subtract the two: $\sqrt{3}a(\cos\alpha - \cos\beta) + 2b(\sin\alpha - \sin\beta) = 0$. $-2\sqrt{3}a\sin\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2} + 4b\cos\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2} = 0$. $\alpha \neq \beta$ and both lie in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\sin\frac{\alpha - \beta}{2} \neq 0$; divide by it: $-\sqrt{3}a\sin\frac{\pi}{6} + 2b\cos\frac{\pi}{6} = 0$. $-\frac{\sqrt{3}}{2}a + \sqrt{3}b = 0$, so $\frac{b}{a} = \frac{1}{2}$. Answer: $0.50$.
Q.14	Integral CalculusArea under simple curves	Class 12: Integrals; Class 12: Application of Integrals	NumericalEasy
A easy numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Area of $\Delta PQR = \frac{1}{2}\times 2\times 1 = 1$. Side $PQ$ is $y = x$ for $0 \le x \le 1$, and $y = x^n$ ($n > 1$) lies below it there, meeting it at $P$ and $Q$. Area between them: $\int_0^1(x - x^n)\,dx = \frac{1}{2} - \frac{1}{n + 1}$. $\frac{1}{2} - \frac{1}{n + 1} = 0.3 \Rightarrow \frac{1}{n + 1} = 0.2 \Rightarrow n = 4$. Answer: $4.00$.
Q.15	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 If tangents at the ends of a chord meet at $(h, k)$, the chord is the chord of contact $hx + ky = 4$. $E_1E_2$: $y = 1$, i.e. $0\cdot x + 4y = 4$, so $E_3 = (0, 4)$. $F_1F_2$: $x = 1$, i.e. $4x + 0\cdot y = 4$, so $F_3 = (4, 0)$. $G_1G_2$: slope $-1$ through $(1, 1)$: $x + y = 2$, i.e. $2x + 2y = 4$, so $G_3 = (2, 2)$. All three satisfy $x + y = 4$. (They are not on (B): $(0,4)$ gives $16 + 0 = 16$ but $(2,2)$ gives $8$; nor on (C) or (D), which $(0, 4)$ fails.) Answer: (A).
Q.16	CirclesStandard and general form	Class 11: Conic Sections	ComprehensionEasy
A easy comprehension question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Take $P = (2\cos\theta, 2\sin\theta)$ with $0 < \theta < \frac{\pi}{2}$. The tangent is $x\cos\theta + y\sin\theta = 2$. It meets the axes at $M = \left(\frac{2}{\cos\theta}, 0\right)$ and $N = \left(0, \frac{2}{\sin\theta}\right)$; the midpoint is $(h, k) = \left(\frac{1}{\cos\theta}, \frac{1}{\sin\theta}\right)$. $\cos^2\theta + \sin^2\theta = 1$ gives $\frac{1}{h^2} + \frac{1}{k^2} = 1$, i.e. $h^2 + k^2 = h^2k^2$. Answer: (D).
Q.17	ProbabilityProbability of an event	Class 11: Probability; Class 12: Probability	ComprehensionEasy
A easy comprehension question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability All $5! = 120$ seatings are equally likely. $S_1$ sits in $R_1$, and the other four must all miss their own seats: a derangement of $4$, $D_4 = 4!\left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}\right) = 9$. Probability $= \frac{9}{120} = \frac{3}{40}$. Answer: (A).
Q.18	ProbabilityProbability of an event	Class 11: Probability; Class 12: Probability	ComprehensionHard
A hard comprehension question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Count arrangements of $S_1, \dots, S_5$ in a row in which no pair $(S_i, S_{i+1})$ sits together, by inclusion–exclusion over the $4$ "bad" pairs $A_i$ = "$S_i$ and $S_{i+1}$ adjacent". Choosing $k$ of the pairs to be adjacent: the chosen pairs merge into blocks. If the $k$ pairs form $j$ runs of consecutive pairs, there are $5 - k$ units, each run-block can be read in $2$ directions, so $(5 - k)!\,2^{j}$ arrangements. $k = 1$: $4$ choices, each $4!\cdot2 = 48$: $192$. $k = 2$: $3$ adjacent pairs of pairs (one run of 3 students, $3!\cdot2 = 12$ each, $36$) and $3$ separated pairs (two runs, $3!\cdot4 = 24$ each, $72$): $108$. $k = 3$: $\{1,2,3\},\{2,3,4\}$ one run ($2!\cdot2 = 4$ each, $8$); $\{1,2,4\},\{1,3,4\}$ two runs ($2!\cdot4 = 8$ each, $16$): $24$. $k = 4$: one run of all five: $1!\cdot2 = 2$. Good arrangements $= 120 - 192 + 108 - 24 + 2 = 14$. Probability $= \frac{14}{120} = \frac{7}{60}$. Answer: (C).

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	Work, Energy and PowerWork-energy theorem	Class 11: Work, Energy and Power	Multiple correctEasy
A easy multiple correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Integrate from rest: $K = \frac{1}{2}\gamma t^2$. So $\frac{1}{2}mv^2 = \frac{1}{2}\gamma t^2$, giving $v = \sqrt{\frac{\gamma}{m}}\,t$. (A) $a = \frac{dv}{dt} = \sqrt{\frac{\gamma}{m}}$ is constant, so $F = ma = \sqrt{m\gamma}$ is constant. True. (B) $v \propto t$. True. (C) $x = \frac{1}{2}\sqrt{\frac{\gamma}{m}}\,t^2$ grows as $t^2$, not linearly. False. (D) A constant force along a fixed direction does work $\int F\,dx = F(x_2 - x_1)$ that depends only on the end points, so it is conservative. True. Answer: (A), (B), (D).
Q.2	Properties of Bulk MatterViscosity	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Multiple correctEasy
A easy multiple correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Because $h$ is small, the velocity in the liquid falls linearly from $u_0$ at the plate to $0$ at the floor, so the velocity gradient is $\frac{u_0}{h}$ everywhere in the layer. Shear stress (Newton's law of viscosity): $\tau = \eta\frac{u_0}{h}$, the same on the plate and on the floor. Force on the plate: $F = \tau A = \frac{\eta A u_0}{h}$. (A) $F \propto \frac{1}{h}$. True. (B) $F \propto A$, so it depends on the area. False. (C) Stress on the floor $= \frac{\eta u_0}{h}$, which increases with $u_0$. True. (D) Stress on the plate $= \frac{\eta u_0}{h}$, linear in $\eta$. True. Answer: (A), (C), (D).
Q.3	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 Only the charge of the wire inside the shell counts for the flux (Gauss's law); the shell itself is not said to carry charge. The wire is the chord $PQ$ of a great circle; the chord subtends $120^\circ$ at $O$, so its length is $PQ = 2R\sin 60^\circ = \sqrt{3}R$. Enclosed charge $= \sqrt{3}R\lambda$, so the flux is $\Phi = \frac{\sqrt{3}R\lambda}{\epsilon_0}$. (A) True. (C) $\sqrt{2}R\lambda/\epsilon_0$: false. (B) The field of an infinite straight line charge is everywhere perpendicular to the line, i.e. perpendicular to the $z$-axis, so its $z$-component is zero at every point, including every point of the shell. True. (D) The field points radially away from the wire, not from $O$, so it is not normal to the sphere in general. False. Answer: (A), (B).
Q.4	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Multiple correctHard
A hard multiple correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics From the figure: the vertical side of the triangle stands at distance $\frac{f}{2}$ from the mirror with height $\frac{f}{2}$ (the hypotenuse makes $45^\circ$ with the axis and meets the axis at the focus, distance $f$). The base lies on the axis from $\frac{f}{2}$ to $f$. Take a point of the object at distance $x$ in front of the mirror ($\frac{f}{2} \le x \le f$). Mirror formula with real-is-positive distances: the image is virtual, behind the mirror, at distance $v = \frac{xf}{f - x}$, and the magnification is $m = \frac{f}{f - x}$ (erect). Vertical side ($x = \frac{f}{2}$): $v = f$ behind the mirror, $m = 2$, so its image is an erect vertical line of height $f$ at distance $f$ behind the mirror. Hypotenuse: a point at distance $x$ has height $f - x$ (the $45^\circ$ line through the focus). Its image height is $(f - x)\cdot\frac{f}{f - x} = f$, the same for every point, while $v$ runs from $f$ to $\infty$ as $x$ goes from $\frac{f}{2}$ to $f$. So the hypotenuse images to a horizontal line at height $f$ running from the top of the vertical image away from the mirror to infinity. Base on the axis: images onto the axis from $f$ to $\infty$ behind the mirror. The image is a vertical line joined at its top to a horizontal line that goes off to infinity, which is figure (D). The slanted line of (B) and the triangles of (A) and (C) do not match. Answer: (D).
Q.5	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	Multiple correctEasy
A easy multiple correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Only $\alpha$ decay changes the mass number, by 4 each: $N_\alpha = \frac{232 - 212}{4} = 5$. Five $\alpha$ decays lower $Z$ by 10, to $90 - 10 = 80$. Each $\beta^-$ decay raises $Z$ by 1, and the final $Z$ is 82, so $N_\beta = 82 - 80 = 2$. (A) True, (B) false, (C) true, (D) false. Answer: (A), (C).
Q.6	Experimental SkillsHeat and sound experiments	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on heat and sound experiments, from Experimental Skills. NCERT: No matching NCERT chapter Successive resonances of a tube closed at one end differ by $\frac{\lambda}{2}$: $\frac{\lambda}{2} = 83.9 - 50.7 = 33.2$ cm, so $\lambda = 66.4$ cm. (C) True. (A) $v = f\lambda = 500 \times 0.664 = 332$ m s$^{-1}$. True. Resonance condition with end correction $e$: $L + e = \frac{(2n - 1)\lambda}{4}$, with $\frac{\lambda}{4} = 16.6$ cm. $50.7$ cm is not near $16.6$ cm (the fundamental), so (D) is false. The nearest odd multiple is $\frac{3\lambda}{4} = 49.8$ cm, giving $e = 49.8 - 50.7 = -0.9$ cm (and with $83.9$: $\frac{5\lambda}{4} = 83.0$, again $e = -0.9$ cm). An end correction of $+0.9$ cm would need resonances at $48.9$ cm and $82.1$ cm, so (B) is false. Answer: (A), (C).
Q.7	KinematicsMotion in a straight line	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	NumericalEasy
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Impulse gives the initial momentum: $mv_0 = 1.0$, so $v_0 = \frac{1.0}{0.4} = 2.5$ m/s. $x(\tau) = \int_0^{\tau} v_0e^{-t/\tau}dt = v_0\tau\left(1 - e^{-1}\right) = 2.5 \times 4 \times (1 - 0.37) = 10 \times 0.63 = 6.3$ m. Answer: $6.30$.
Q.8	KinematicsProjectile motion	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	NumericalEasy
A easy numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Before the bounce: $H = \frac{u^2\sin^2 45^\circ}{2g} = \frac{u^2}{4g} = 120$, so $u^2 = 480g$. The ball lands with speed $u$. Losing half its kinetic energy, it leaves with $v^2 = \frac{u^2}{2} = 240g$. After the bounce: $H' = \frac{v^2\sin^2 30^\circ}{2g} = \frac{240g \times \frac{1}{4}}{2g} = 30$ m. Answer: $30.00$.
Q.9	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	NumericalEasy
A easy numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $a(t) = \frac{qE_0}{m}\sin\omega t = \frac{1.0 \times 1.0}{10^{-3}}\sin\omega t = 10^3\sin\omega t$ m s$^{-2}$. Starting from rest: $v(t) = \int_0^t 10^3\sin\omega t'\,dt' = \frac{10^3}{\omega}(1 - \cos\omega t) = 1 - \cos\omega t$ (with $\omega = 10^3$). The maximum is when $\cos\omega t = -1$: $v_{max} = 2$ m s$^{-1}$. Answer: $2.00$.
Q.10	Magnetic Effects of Current and MagnetismMoving coil galvanometer	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	NumericalMedium
A medium numerical question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Full-scale current: magnetic torque balances the torsional torque, $NI_gAB = k\theta$. $I_g = \frac{k\theta}{NAB} = \frac{10^{-4}\times0.2}{50\times2\times10^{-4}\times0.02} = \frac{2\times10^{-5}}{2\times10^{-4}} = 0.1$ A. Shunt: the galvanometer and shunt share the same voltage, $I_gG = (I - I_g)S$. $S = \frac{I_gG}{I - I_g} = \frac{0.1\times50}{1.0 - 0.1} = \frac{5}{0.9} = 5.555\ldots\ \Omega$. Answer: $5.56$ (truncated, $5.55$).
Q.11	Properties of Bulk MatterElasticity	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalMedium
A medium numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Extra extension: $\Delta l = \frac{FL}{AY}$ with $F = 1.2\times10 = 12$ N, $L = 1.0$ m. $A = \pi r^2 = 3.2\times(0.25\times10^{-3})^2 = 3.2\times6.25\times10^{-8} = 2\times10^{-7}$ m$^2$. $\Delta l = \frac{12\times1.0}{2\times10^{-7}\times2\times10^{11}} = \frac{12}{4\times10^4} = 3\times10^{-4}$ m $= 0.3$ mm. Vernier least count $= 1$ MSD $- 1$ VSD $= 1 - 0.9 = 0.1$ mm. The vernier moves $0.3$ mm $= 3$ least counts, so its 3rd division now coincides with a main scale division. Answer: $3$.
Q.12	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	NumericalEasy
A easy numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $TV^{\gamma - 1}$ = constant, with $\gamma = \frac{5}{3}$ for a monatomic gas. $T_f = T_i\left(\frac{V_i}{V_f}\right)^{2/3} = 100\times\left(\frac{1}{8}\right)^{2/3} = 100\times\frac{1}{4} = 25$ K. Decrease in internal energy: $nC_V\Delta T = 1\times\frac{3}{2}\times8.0\times(100 - 25) = 12\times75 = 900$ J. Answer: $900$.
Q.13	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	NumericalMedium
A medium numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon energy is just the work function: $6.25$ eV $= 6.25\times1.6\times10^{-19} = 10^{-18}$ J. Photons (and so electrons) per second: $N = \frac{200}{10^{-18}} = 2\times10^{20}$ s$^{-1}$. Each electron starts at rest and reaches the anode with kinetic energy $500$ eV $= 8\times10^{-17}$ J. Momentum: $p = \sqrt{2m_eK} = \sqrt{2\times9\times10^{-31}\times8\times10^{-17}} = \sqrt{1.44\times10^{-46}} = 1.2\times10^{-23}$ kg m/s. Absorbed on impact, so the force is the momentum delivered per second: $F = Np = 2\times10^{20}\times1.2\times10^{-23} = 2.4\times10^{-3}$ N $= 24\times10^{-4}$ N. Answer: $n = 24$.
Q.14	Atoms and NucleiBohr model	Class 12: Atoms; Class 12: Nuclei	NumericalEasy
A easy numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $E_{2\to1} = 13.6Z^2\left(1 - \frac{1}{4}\right) = 13.6Z^2\cdot\frac{3}{4}$ and $E_{3\to2} = 13.6Z^2\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6Z^2\cdot\frac{5}{36}$. Difference: $13.6Z^2\left(\frac{27}{36} - \frac{5}{36}\right) = 13.6Z^2\cdot\frac{22}{36} = 74.8$. $Z^2 = \frac{74.8\times36}{13.6\times22} = \frac{2692.8}{299.2} = 9$, so $Z = 3$. Answer: $3$.
Q.15	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Matching listMedium
A medium matching list question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance (1) Point charge: $E = \frac{Q}{4\pi\epsilon_0d^2} \propto \frac{1}{d^2}$. (2) Short dipole on its axis: $E = \frac{2p}{4\pi\epsilon_0d^3} \propto \frac{1}{d^3}$. (3) Infinite line charge: $E = \frac{\lambda}{2\pi\epsilon_0d} \propto \frac{1}{d}$. (4) Two opposite line charges $2l$ apart (a "line dipole"): $E = \frac{\lambda}{2\pi\epsilon_0}\left(\frac{1}{d - l} - \frac{1}{d + l}\right) = \frac{\lambda}{2\pi\epsilon_0}\cdot\frac{2l}{d^2 - l^2} \approx \frac{\lambda l}{\pi\epsilon_0d^2} \propto \frac{1}{d^2}$. (5) Infinite plane: $E = \frac{\sigma}{2\epsilon_0}$, independent of $d$. So P→5, Q→3, R→1, 4, S→2. Answer: (B).
Q.16	GravitationOrbital motion	Class 11: Gravitation	Matching listEasy
A easy matching list question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Circular orbit: $v = \sqrt{\frac{GM}{R}}$, independent of the satellite's mass. (P) $\frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}} = \sqrt{4} = 2$ → 3. (Q) $L = mvR$: $\frac{L_1}{L_2} = 2\times2\times\frac{1}{4} = 1$ → 2. (R) $K = \frac{1}{2}mv^2$: $\frac{K_1}{K_2} = 2\times2^2 = 8$ → 4. (S) $T \propto R^{3/2}$: $\frac{T_1}{T_2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$ → 1. Answer: (B).
Q.17	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Matching listEasy
A easy matching list question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Read the graph: Process II is horizontal at $3P_0$ from $V_0$ to $3V_0$: isobaric, work $= 3P_0(3V_0 - V_0) = 6P_0V_0$ → 4. Process III is vertical at $3V_0$: isochoric, work zero → 1. Process IV runs from $(3V_0, P_0)$ to $(V_0, 3P_0)$: $PV = 3P_0V_0$ at both ends, and it is the gentler of the two curves, so it is the isothermal, temperature unchanged → 2. Process I is the steeper curve ending at $(V_0, 3P_0)$: the adiabatic ($PV^\gamma$ = constant is steeper than $PV$ = constant), no heat exchanged → 3. So P→3, Q→4, R→1, S→2. Answer: (C).
Q.18	Work, Energy and PowerConservation of energy	Class 11: Work, Energy and Power	Matching listMedium
A medium matching list question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power In every case the force is zero or conservative, so $E$ is conserved (5) throughout. (P) $\vec{v} = \alpha\hat{i} + \beta\hat{j}$ is constant: no force, so $\vec{p}$, $K$, $U$ are constant. $\vec{L} = m\,\vec{r}\times\vec{v} = m t(\alpha\hat{i} + \beta\hat{j})\times(\alpha\hat{i} + \beta\hat{j}) = 0$, constant. P→1, 2, 3, 4, 5. (Q) $\vec{a} = -\omega^2\vec{r}$: a central force, so $\vec{L}$ is conserved; $\vec{p}$ changes. $v^2 = \omega^2(\alpha^2\sin^2\omega t + \beta^2\cos^2\omega t)$ varies because $\alpha \neq \beta$, so $K$ and hence $U$ vary. Q→2, 5. (R) Uniform circular motion: speed $\alpha\omega$ constant, so $K$ constant and $U = E - K$ constant; central force, so $\vec{L}$ constant; $\vec{p}$ changes direction. R→2, 3, 4, 5. (S) $\vec{v} = \alpha\hat{i} + \beta t\hat{j}$: constant force $m\beta\hat{j}$. $\vec{p}$, $K$, $U$ change. $\vec{r}\times\vec{v} = \left(\alpha\beta t^2 - \frac{\alpha\beta t^2}{2}\right)\hat{k} = \frac{\alpha\beta t^2}{2}\hat{k}$ changes, so $\vec{L}$ is not conserved. S→5. Answer: (A).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	Coordination CompoundsIsomerism in complexes	Class 12: Coordination Compounds	Multiple correctMedium
A medium multiple correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The complex is of type $[\mathrm{M(AA)b_3c}]$, octahedral, Co(III), $d^6$. (A) The two ends of en must be cis. Place en in the equatorial plane: the remaining sites are two equatorial (cis to each other) and two axial. Either H$_2$O sits trans to an en nitrogen (equatorial) or trans to an NH$_3$ (axial). These are the only two arrangements, and each has a mirror plane (no optical isomers), so there are exactly two geometrical isomers. True. (B) Replacing en by two CN$^-$ gives $[\mathrm{Co(CN)_2(NH_3)_3(H_2O)}]$, type $[\mathrm{Ma_2b_3c}]$. Isomers: the two CN trans (one isomer); the two CN cis with H$_2$O trans to a CN (one) or trans to an NH$_3$ (one). Three geometrical isomers. True. (C) en, NH$_3$ and H$_2$O give Co(III) ($d^6$) a large splitting: low-spin $t_{2g}^6$, no unpaired electrons, diamagnetic. False. (D) In $[\mathrm{Co(en)(NH_3)_4}]^{3+}$ the H$_2$O is replaced by the stronger-field NH$_3$, so $\Delta_o$ is larger there. The aqua complex has the smaller $\Delta_o$ and absorbs at a longer wavelength. True. Answer: (A), (B), (D).
Q.2	Principles of Qualitative AnalysisGroup analysis of cations	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter (A) The green (bluish-green) flame is the test for copper, not manganese; Mn$^{2+}$ gives no characteristic flame colour. False. (B) In acidic medium the sulphide ion concentration is low, and only the very insoluble CuS (group II) precipitates; MnS (group IV) does not. So only Cu$^{2+}$ gives a precipitate. True. (C) In faintly basic medium the sulphide concentration is high and both CuS and MnS precipitate, so "only Mn$^{2+}$" is wrong. False. (D) $E^\circ(\mathrm{Cu^{2+}/Cu}) = +0.34$ V and $E^\circ(\mathrm{Mn^{2+}/Mn}) = -1.18$ V, so Cu$^{2+}$/Cu has the higher reduction potential. True. Answer: (B), (D).
Q.3	AminesDiazonium salts	Class 12: Amines	Multiple correctHard
A hard multiple correct question on diazonium salts, from Amines. NCERT: Class 12: Amines In mixed acid most aniline is protonated to the meta-directing anilinium ion, so nitration gives para (P, 51%), meta (Q, 47%) and ortho (R, 2%) nitroanilines. R is o-nitroaniline. R → S: 1) Ac$_2$O: o-nitroacetanilide (protects NH$_2$). 2) Br$_2$/AcOH: bromination para to the NHAc group: 4-bromo-2-nitroacetanilide. 3) H$_3$O$^+$: 4-bromo-2-nitroaniline. 4), 5) Diazotisation and EtOH replace the amino group by H: S is 1-bromo-3-nitrobenzene (the Br and NO$_2$ end up meta). S → product: 1) Sn/HCl: 3-bromoaniline. 2) Excess Br$_2$/H$_2$O brominates every free position ortho and para to NH$_2$ (positions 2, 4, 6): 2,3,4,6-tetrabromoaniline. 3), 4) Diazotisation and H$_3$PO$_2$ remove the amino group: 1,2,3,5-tetrabromobenzene (three adjacent Br and one more Br meta to the outer ones). That is structure (D). (A) is 1,2,4,5-, (B) 1,2,3,4-tetrabromobenzene and (C) 1,3,5-tribromobenzene. Answer: (D).
Q.4	BiomoleculesCarbohydrates	Class 12: Biomolecules	Multiple correctMedium
A medium multiple correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules All four options are Haworth rings drawn with the ring O at the back right, C-1 on the right, then C-2, C-3, C-4 going round, and C-5 at the back left. $\beta$-D-glucopyranose in this orientation: CH$_2$OH up at C-5; OH down at C-4, up at C-3, down at C-2; anomeric OH up at C-1 (same side as CH$_2$OH). $\beta$-L-glucopyranose is its mirror image, so every group changes side: CH$_2$OH down at C-5; OH up at C-4, down at C-3, up at C-2; anomeric OH down at C-1 (still on the same side as CH$_2$OH, which is what $\beta$ means). (A) C-5 CH$_2$OH down, C-4 OH up, C-3 OH down, C-2 OH up, but C-1 OH up: this is $\alpha$-L-glucopyranose. Not correct. (B) C-2 has OH down and C-1 OH up: not L-glucose (wrong at C-2) and not $\beta$. Not correct. (C) CH$_2$OH up, so a D sugar; C-2 OH up, C-3 OH up, C-4 OH down: $\beta$-D-mannopyranose. Not correct. (D) C-5 CH$_2$OH down, C-4 OH up, C-3 OH down, C-2 OH up, C-1 OH down: $\beta$-L-glucopyranose. Correct. Answer: (D).
Q.5	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	Multiple correctMedium
A medium multiple correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let $x$ be the pressure of A used up. Then $p_A = P_0 - x$ and $P_t = (P_0 - x) + 2x + x = P_0 + 2x$, so $x = \frac{P_t - P_0}{2}$ and $p_A = P_0 - \frac{P_t - P_0}{2} = \frac{3P_0 - P_t}{2}$. First order: $\ln p_A = \ln P_0 - kt$, so $\ln(3P_0 - P_t) = \ln(2P_0) - kt$: a straight line with negative slope against time. (A) True. (C) $P_0 - P_t$ is negative ($P_t > P_0$), so its logarithm is not even defined. False. (B) $t_{1/3} = \frac{\ln 3}{k}$ is independent of $[\mathrm{A}]_0$; the plot should be horizontal, not falling. False. (D) The rate constant does not depend on $[\mathrm{A}]_0$: a horizontal line. True. Answer: (A), (D).
Q.6	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Multiple correctHard
A hard multiple correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics From the graphs, at equilibrium $[\mathrm{P}] > 5 > [\mathrm{A}]$ at both temperatures (starting from $[\mathrm{A}] = 10$), so $K = \frac{[\mathrm{P}]}{[\mathrm{A}]} > 1$ and $\Delta G^\ominus = -RT\ln K < 0$ at both. Also $[\mathrm{P}]_{eq}$ is larger at $T_1$ than at $T_2$, so $K$ falls as $T$ rises: the reaction is exothermic, $\Delta H^\ominus < 0$. Entropy: $\ln K = -\frac{\Delta H^\ominus}{RT} + \frac{\Delta S^\ominus}{R}$, so $T\ln K = -\frac{\Delta H^\ominus}{R} + \frac{T\Delta S^\ominus}{R}$. Given $\frac{\ln K_1}{\ln K_2} > \frac{T_2}{T_1}$ with $\ln K_2 > 0$: $T_1\ln K_1 > T_2\ln K_2$, i.e. $\frac{T_1\Delta S^\ominus}{R} > \frac{T_2\Delta S^\ominus}{R}$, i.e. $(T_1 - T_2)\Delta S^\ominus > 0$. Since $T_2 > T_1$, $\Delta S^\ominus < 0$. So $\Delta H^\ominus < 0$, $\Delta S^\ominus < 0$, $\Delta G^\ominus < 0$. (A) True. (B) $\Delta H^\ominus > 0$: false. (C) True. (D) $\Delta S^\ominus > 0$: false. Answer: (A), (C).
Q.7	p-Block ElementsGroup 15 elements	No matching NCERT chapter	NumericalMedium
A medium numerical question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter A bridging oxo group is an O joined to two other atoms (X–O–X). $\mathrm{N_2O_3}$: O=N–NO$_2$, the two N atoms bonded directly. No bridge. $\mathrm{N_2O_5}$: O$_2$N–O–NO$_2$. Bridge. $\mathrm{P_4O_6}$: six P–O–P bridges. Bridge. $\mathrm{P_4O_7}$: six P–O–P bridges plus one terminal P=O. Bridge. $\mathrm{H_4P_2O_5}$ (pyrophosphorous acid): H(HO)P(O)–O–P(O)(OH)H. Bridge. $\mathrm{H_5P_3O_{10}}$ (triphosphoric acid): two P–O–P bridges. Bridge. $\mathrm{H_2S_2O_3}$ (thiosulphuric acid): (HO)$_2$S(=O)=S, the second S bonded only to the central S; no S–O–S. No bridge. $\mathrm{H_2S_2O_5}$ (disulphurous acid): HO–S(O)–S(O)$_2$–OH with an S–S bond. No bridge. Total: 5. Answer: $5$.
Q.8	Isolation of MetalsRoasting and calcination	No matching NCERT chapter	NumericalMedium
A medium numerical question on roasting and calcination, from Isolation of Metals. NCERT: No matching NCERT chapter Partial roasting: $\mathrm{2PbS + 3O_2 \rightarrow 2PbO + 2SO_2}$. Self-reduction: $\mathrm{PbS + 2PbO \rightarrow 3Pb + SO_2}$. Adding them: $\mathrm{3PbS + 3O_2 \rightarrow 3Pb + 3SO_2}$, i.e. overall $\mathrm{PbS + O_2 \rightarrow Pb + SO_2}$: one mole of Pb for each mole of O$_2$. Mass of Pb per mass of O$_2$: $\frac{207}{32} = 6.469$. Answer: $6.47$.
Q.9	Principles Related to Practical ChemistryTitrimetric analysis	No matching NCERT chapter	NumericalMedium
A medium numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Every Mn of MnCl$_2$ becomes one MnO$_4^-$. Oxalic acid: $\frac{225}{90} = 2.5$ mmol (H$_2$C$_2$O$_4$, $M = 90$). $\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$ (Mn changes by 5, each oxalate gives 2 electrons). MnO$_4^-$ $= \frac{2}{5}\times2.5 = 1$ mmol, so MnCl$_2$ $= 1$ mmol. Molar mass of MnCl$_2$ $= 55 + 71 = 126$, so the mass is $126$ mg. Answer: $126$.
Q.10	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	NumericalHard
A hard numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Fixed by the drawing: the two ring C–OH centres (wedges), the ring carbon carrying the upper side chain (wedge), the upper C=C and the upper CH(OH) centre (wedge). Not fixed (wavy bonds): (i) the ring carbon carrying the lower side chain, (ii) the geometry of the lower C=C, (iii) the lower CH(OH) centre. Three two-way choices: $2^3 = 8$ stereoisomers. Check for a meso form. The ring is C-1(OH), C-2(upper chain), C-3(lower chain), C-4(OH), C-5(H$_2$). Both OH groups are on the same face. A mirror plane through C-5 and the midpoint of C-2–C-3 exists if the lower chain is on the same face as the upper one, has the same (E) double bond, and its CH(OH) centre is the mirror image of the upper one. Exactly one of the eight combinations is that one, and it is achiral (meso). Every other combination has no symmetry element and is chiral. Optically active stereoisomers: $8 - 1 = 7$. Answer: $7$.
Q.11	AminesClassification and identification of amines	Class 12: Amines	NumericalMedium
A medium numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines A: haloform reaction of the methyl ketone, then acidification: benzoic acid. $10\times0.6 = 6$ mol. B: NH$_3$ and heat: benzamide. $6\times0.5 = 3$ mol. C: Br$_2$/KOH (Hofmann bromamide degradation): aniline. $3\times0.5 = 1.5$ mol. D: Br$_2$ (3 equiv): 2,4,6-tribromoaniline, $\mathrm{C_6H_4Br_3N}$. $1.5$ mol. Molar mass of D: $6(12) + 4(1) + 3(80) + 14 = 72 + 4 + 240 + 14 = 330$. Mass: $1.5\times330 = 495$ g. Answer: $495$.
Q.12	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	NumericalHard
A hard numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Subtract the second reaction from the first: $\mathrm{2Cu + H_2O \rightarrow Cu_2O + H_2}$, $\Delta G^\ominus = -78000 - (-178000) = +100000$ J mol$^{-1}$. With pure solids, $\Delta G = \Delta G^\ominus + RT\ln\frac{p_{\mathrm{H_2}}}{p_{\mathrm{H_2O}}}$. Oxidation is prevented when $\Delta G \ge 0$; the minimum $p_{\mathrm{H_2}}$ makes $\Delta G = 0$: $\ln\frac{p_{\mathrm{H_2}}}{p_{\mathrm{H_2O}}} = -\frac{\Delta G^\ominus}{RT} = -\frac{100000}{8\times1250} = -10$. Water vapour is 1 mole % of 1 bar: $p_{\mathrm{H_2O}} = 0.01$ bar, $\ln 0.01 = -2\times2.3 = -4.6$. $\ln p_{\mathrm{H_2}} = -10 - 4.6 = -14.6$. Answer: $-14.60$.
Q.13	Chemical KineticsArrhenius equation	Class 12: Chemical Kinetics	NumericalMedium
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $K = \frac{k_f}{k_b} = \frac{A_f e^{-E_f/RT}}{A_b e^{-E_b/RT}} = \frac{A_f}{A_b}e^{(E_b - E_f)/RT} = 4e^{2RT/RT} = 4e^2$. $\ln K = \ln 4 + 2 = 1.4 + 2 = 3.4$. $\Delta G^\ominus = -RT\ln K = -2500\times3.4 = -8500$ J mol$^{-1}$. Absolute value: $8500$. Answer: $8500$.
Q.14	ElectrochemistryNernst equation	Class 12: Electrochemistry	NumericalMedium
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction per mole of B formed: $\mathrm{2A(s) + B^{2n+}(aq) \rightarrow 2A^{n+}(aq) + B(s)}$. Reaction quotient: $Q = \frac{[\mathrm{A^{n+}}]^2}{[\mathrm{B^{2n+}}]} = \frac{2^2}{1} = 4$. emf zero means $\Delta G = 0 = \Delta G^\ominus + RT\ln Q$, so $\Delta G^\ominus = -RT\ln 4$. $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$ with $\Delta H^\ominus = 2\Delta G^\ominus$: $T\Delta S^\ominus = \Delta G^\ominus$, so $\Delta S^\ominus = \frac{\Delta G^\ominus}{T} = -R\ln 4$. $\Delta S^\ominus = -8.3\times2\times0.7 = -11.62$ J K$^{-1}$ mol$^{-1}$. Answer: $-11.62$.
Q.15	Coordination CompoundsValence bond theory	Class 12: Coordination Compounds	Matching listMedium
A medium matching list question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (1) $[\mathrm{FeF_6}]^{4-}$: Fe$^{2+}$ $d^6$ with weak-field F$^-$, high spin, outer orbitals: $sp^3d^2$ → R. (2) $[\mathrm{Ti(H_2O)_3Cl_3}]$: Ti$^{3+}$ $d^1$; two empty inner $3d$ orbitals are free: $d^2sp^3$ → S. (3) $[\mathrm{Cr(NH_3)_6}]^{3+}$: Cr$^{3+}$ $d^3$; two inner $3d$ orbitals free: $d^2sp^3$ → S. (4) $[\mathrm{FeCl_4}]^{2-}$: tetrahedral, Fe$^{2+}$ $d^6$ high spin: $sp^3$ → Q. (5) $\mathrm{Ni(CO)_4}$: Ni(0), $3d^{10}$ after pairing, tetrahedral: $sp^3$ → Q. (6) $[\mathrm{Ni(CN)_4}]^{2-}$: Ni$^{2+}$ $d^8$, strong-field CN$^-$ pairs the electrons, square planar: $dsp^2$ → P. So P→6, Q→4, 5, R→1, S→2, 3. Answer: (C).
Q.16	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	Matching listHard
A hard matching list question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids X is 2,2-diphenylpropanoic acid, $\mathrm{Ph_2C(Me)COOH}$. It can come from the methyl ketone $\mathrm{Ph_2C(Me)COMe}$ by the haloform reaction (I$_2$/NaOH or NaOBr), or from the aldehyde $\mathrm{Ph_2C(Me)CHO}$ by oxidation (Tollens' or Fehling's reagent). (P) Pinacol rearrangement. The OH leaves from the carbon giving the more stable cation, $\mathrm{Ph_2C^+}$. From the neighbouring $\mathrm{C(OH)Me_2}$ only methyl can migrate: $\mathrm{Ph_2C(Me){-}C(=O)Me}$. Haloform → X. P→1, 5. (Q) HNO$_2$ diazotises the amine; N$_2$ leaves to give $\mathrm{Ph_2C^+}$. From $\mathrm{CH(OH)Me}$, methyl (alkyl > H) migrates: $\mathrm{Ph_2C(Me){-}CHO}$. Oxidation by Tollens' or Fehling's → X. Q→2, 3. (R) Symmetrical pinacol: cation $\mathrm{PhMeC^+}$; from $\mathrm{C(OH)PhMe}$, phenyl migrates (aryl > alkyl): $\mathrm{Ph_2C(Me){-}C(=O)Me}$. Haloform → X. R→1, 5. (S) Ag$^+$ removes Br$^-$ to give $\mathrm{Ph_2C^+}$; methyl migrates as in Q: $\mathrm{Ph_2C(Me)CHO}$. S→2, 3. HCHO/NaOH (crossed Cannizzaro) would give an alcohol from the aldehyde, not the acid. Answer: (D).
Q.17	Alcohols, Phenols and EthersWilliamson synthesis	Class 12: Alcohols, Phenols and Ethers	Matching listMedium
A medium matching list question on williamson synthesis, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (P) A tertiary halide with a strong, bulky base: E2 elimination, not substitution. Products: isobutylene (4), and the alkoxide takes the proton to become tert-butanol (1). P→1, 4. (Q) A tert-butyl methyl ether is cleaved by HBr by the S$_\mathrm{N}$1 route at the tertiary carbon: $\mathrm{(CH_3)_3CBr + CH_3OH}$. Q→2. (R) A tertiary halide with a strong base (methoxide): elimination dominates, giving isobutylene. R→4. (S) Williamson synthesis the right way round, a tertiary alkoxide with a methyl halide (S$_\mathrm{N}$2 at methyl): $\mathrm{(CH_3)_3COMe}$. S→3. Answer: (B).
Q.18	EquilibriumIonic equilibrium	Class 11: Equilibrium	Matching listMedium
A medium matching list question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium (P) 1 mmol NaOH + 2 mmol CH$_3$COOH → 1 mmol CH$_3$COONa + 1 mmol CH$_3$COOH: a buffer. $[\mathrm{H^+}] = K_a\frac{[\text{acid}]}{[\text{salt}]}$ depends only on the ratio, which dilution does not change. P→1. (Q) 2 mmol each → 2 mmol CH$_3$COONa in 40 mL, diluted from 40 to 80 mL, so its concentration $c$ halves. Salt of a weak acid and a strong base: $[\mathrm{OH^-}] = \sqrt{K_hc}$, so $[\mathrm{H^+}] = \frac{K_w}{\sqrt{K_hc}} \propto \frac{1}{\sqrt{c}}$. Halving $c$ multiplies $[\mathrm{H^+}]$ by $\sqrt{2}$. Q→5. (R) 2 mmol each → NH$_4$Cl, concentration halved by the dilution. Salt of a weak base and a strong acid: $[\mathrm{H^+}] = \sqrt{K_hc} \propto \sqrt{c}$, so it becomes $\frac{1}{\sqrt{2}}$ times. R→4. (S) Solid Ni(OH)$_2$ is still present, so the solution stays saturated: $[\mathrm{OH^-}]$ and hence $[\mathrm{H^+}]$ are unchanged. S→1. Answer: (D).

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Multiple correctMedium
A medium multiple correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Each term telescopes: $\tan^{-1}\frac{(x + j) - (x + j - 1)}{1 + (x + j)(x + j - 1)} = \tan^{-1}(x + j) - \tan^{-1}(x + j - 1)$ (both arguments positive). So $f_n(x) = \tan^{-1}(x + n) - \tan^{-1}x$. The options use $f_j(0)$ and $f_j'(0)$; the formula extends continuously to $x = 0$: $f_j(0) = \tan^{-1}j$. (A) $\tan^2(f_j(0)) = j^2$, and $\sum_{j=1}^{5}j^2 = 1 + 4 + 9 + 16 + 25 = 55$. True. (B) $f_j'(x) = \frac{1}{1 + (x + j)^2} - \frac{1}{1 + x^2}$, so $1 + f_j'(0) = \frac{1}{1 + j^2}$; and $\sec^2(f_j(0)) = 1 + j^2$. Each product is $1$, and the sum of ten of them is $10$. True. (C) $\tan(f_n(x)) = \frac{(x + n) - x}{1 + x(x + n)} = \frac{n}{1 + x^2 + nx} \rightarrow 0$ as $x \rightarrow \infty$, not $\frac{1}{n}$. False. (D) $\sec^2(f_n(x)) = 1 + \tan^2(f_n(x)) \rightarrow 1$. True. Answer: (A), (B), (D).
Q.2	CirclesEquation from a diameter	Class 11: Conic Sections	Multiple correctHard
A hard multiple correct question on equation from a diameter, from Circles. NCERT: Class 11: Conic Sections Locus $E_1$: the common tangent at $M$ meets $T$ at a point $N$. Tangent lengths from $N$ give $NP = NM$ (circle $S_1$) and $NM = NQ$ (circle $S_2$), so $N$ is the midpoint of $PQ$ and $M$ is at distance $\frac{PQ}{2}$ from it: $M$ lies on the circle with diameter $PQ$. Centre $(0, 1)$, radius $\sqrt{4 + 36} = \sqrt{40}$: $x^2 + (y - 1)^2 = 40$. The points $P$ and $Q$ themselves cannot be $M$ (a circle tangent to $T$ at $P$ cannot touch the other circle at $P$), so $E_1$ is this circle without $P$ and $Q$. (A) $(-2, 7)$ is $P$, which is excluded. False. (D) $(0, \frac{3}{2})$: $0 + \frac{1}{4} \ne 40$, not on the circle, so not in $E_1$. True. Locus $E_2$: the midpoint $H$ of a chord through $R$ satisfies $CH \perp HR$ with $C = (0, 1)$, so $H$ lies on the circle with diameter $CR$: $x(x - 1) + (y - 1)^2 = 0$. (C) $(\frac{1}{2}, 1)$: $\frac{1}{2}\cdot(-\frac{1}{2}) + 0 = -\frac{1}{4} \ne 0$. Not in $E_2$. False. (B) $(\frac{4}{5}, \frac{7}{5})$: $\frac{4}{5}\cdot(-\frac{1}{5}) + \frac{4}{25} = 0$, so it is on that circle. But it is the midpoint of the chord through $R$ and $P$: the line $RP$ has direction $(-3, 6) \parallel (1, -2)$, and the foot of the perpendicular from $C$ to it is at $(1 + t, 1 - 2t)$ with $(1 + t) + 4t = 0$, $t = -\frac{1}{5}$: $(\frac{4}{5}, \frac{7}{5})$. That chord has the excluded point $P$ as an end, so it is not in $F_2$ and $(\frac{4}{5}, \frac{7}{5})$ is not in $E_2$. True. Answer: (B), (D).
Q.3	Matrices and DeterminantsConsistency of linear systems	Class 12: Matrices; Class 12: Determinants	Multiple correctHard
A hard multiple correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants In the given system the $y$-column is $-2$ times the $x$-column, so the coefficient matrix has rank 2. Find the dependence of its rows: $a(-1, 2, 5) + b(2, -4, 3) + c(1, -2, 2) = 0$ gives $-a + 2b + c = 0$ and $5a + 3b + 2c = 0$, so $b = 7a$, $c = -13a$. The system is consistent exactly when $b_1 + 7b_2 - 13b_3 = 0$; that plane is $S$. (A) $\begin{vmatrix}1&2&3\\0&4&5\\1&2&6\end{vmatrix} = 1(24 - 10) - 2(0 - 5) + 3(0 - 4) = 12 \ne 0$: solvable for every right side. True. (B) $\begin{vmatrix}1&1&3\\5&2&6\\-2&-1&-3\end{vmatrix} = 0 + 3 - 3 = 0$. Row dependence: $(1,1,3) + (5,2,6) + 3(-2,-1,-3) = 0$, so it needs $b_1 + b_2 + 3b_3 = 0$, which does not follow from $b_1 + 7b_2 - 13b_3 = 0$ (e.g. $(13, 0, 1) \in S$ gives $16 \ne 0$). False. (C) All three rows are multiples of $(1, -2, 5)$: it needs $b_2 = -2b_1$ and $b_3 = -b_1$, not true for all of $S$ (e.g. $(13, 0, 1)$). False. (D) $\begin{vmatrix}1&2&5\\2&0&3\\1&4&-5\end{vmatrix} = 1(0 - 12) - 2(-10 - 3) + 5(8 - 0) = 54 \ne 0$. True. Answer: (A), (D).
Q.4	Conic SectionsEllipse	Class 11: Conic Sections	Multiple correctMedium
A medium multiple correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Tangent to $y^2 = 4x$: $y = mx + \frac{1}{m}$. Tangent to the circle: $\frac{\|1/m\|}{\sqrt{1 + m^2}} = \frac{1}{\sqrt{2}}$, so $m^2(1 + m^2) = 2$, $m^2 = 1$, $m = \pm1$. Lines $y = x + 1$ and $y = -x - 1$ meet at $Q(-1, 0)$, so $a = OQ = 1$ along the $x$-axis, and $2b = \sqrt{2}$, $b = \frac{1}{\sqrt{2}}$. Ellipse: $x^2 + 2y^2 = 1$. $e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2}$, $e = \frac{1}{\sqrt{2}}$; latus rectum $\frac{2b^2}{a} = 1$. (A) true, (B) false. Area between $x = \frac{1}{\sqrt{2}}$ and $x = 1$: $2\int_{1/\sqrt{2}}^{1}\sqrt{\frac{1 - x^2}{2}}\,dx = \sqrt{2}\int_{1/\sqrt{2}}^{1}\sqrt{1 - x^2}\,dx$. $\int\sqrt{1 - x^2}\,dx = \frac{x\sqrt{1 - x^2}}{2} + \frac{\sin^{-1}x}{2}$: at $1$ it is $\frac{\pi}{4}$, at $\frac{1}{\sqrt{2}}$ it is $\frac{1}{4} + \frac{\pi}{8}$. Difference $\frac{\pi}{8} - \frac{1}{4} = \frac{\pi - 2}{8}$. Area $= \frac{\sqrt{2}(\pi - 2)}{8} = \frac{\pi - 2}{4\sqrt{2}}$. (C) true, (D) false. Answer: (A), (C).
Q.5	Complex NumbersAlgebra of complex numbers	Class 11: Complex Numbers and Quadratic Equations	Multiple correctMedium
A medium multiple correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Take the conjugate of the equation: $\bar{t}z + \bar{s}\bar{z} + \bar{r} = 0$. Eliminating $\bar{z}$ between the two: $(\|s\|^2 - \|t\|^2)z = t\bar{r} - \bar{s}r$. If $\|s\| \ne \|t\|$ there is exactly one solution. If $\|s\| = \|t\|$, the equation, written in $x$ and $y$, is one real linear condition (a line) or inconsistent: $L$ is a line (infinitely many points) or empty. (A) Exactly one element forces $\|s\| \ne \|t\|$. True. (B) With $\|s\| = \|t\|$, $L$ may be empty (e.g. $s = t = 1$, $r = i$: $2x + i = 0$ has no real solution). False. (C) $L$ is a point, a line or empty; a line meets the circle $\|z - 1 + i\| = 5$ in at most 2 points. True. (D) More than one element means $\|s\| = \|t\|$ and $L$ is a whole line. True. Answer: (A), (C), (D).
Q.6	Differential EquationsLinear differential equations	Class 12: Differential Equations	Multiple correctHard
A hard multiple correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations The limit is the derivative in $t$ of $f(x)\sin t - f(t)\sin x$ at $t = x$: $f(x)\cos x - f'(x)\sin x = \sin^2 x$. So $\frac{f'(x)\sin x - f(x)\cos x}{\sin^2 x} = -1$, i.e. $\frac{d}{dx}\left(\frac{f(x)}{\sin x}\right) = -1$, and $f(x) = (c - x)\sin x$. $f\left(\frac{\pi}{6}\right) = \frac{1}{2}\left(c - \frac{\pi}{6}\right) = -\frac{\pi}{12}$ gives $c = 0$: $f(x) = -x\sin x$. (A) $f\left(\frac{\pi}{4}\right) = -\frac{\pi}{4\sqrt{2}} \ne \frac{\pi}{4\sqrt{2}}$. False. (B) $-x\sin x < \frac{x^4}{6} - x^2 \iff \sin x > x - \frac{x^3}{6}$ (dividing by $x > 0$), which holds for every $x > 0$ (the function $\sin x - x + \frac{x^3}{6}$ is $0$ at $0$ and has positive derivative $\cos x - 1 + \frac{x^2}{2}$ for $x > 0$). True. (C) $f'(x) = -\sin x - x\cos x$; $f'\left(\frac{\pi}{2}\right) = -1 < 0$ and $f'(x) \rightarrow \pi > 0$ as $x \rightarrow \pi$, so $f'$ vanishes somewhere in $\left(\frac{\pi}{2}, \pi\right)$ (where $\tan x = -x$). True. (D) $f''(x) = -2\cos x + x\sin x$: $f''\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$ and $f\left(\frac{\pi}{2}\right) = -\frac{\pi}{2}$; the sum is $0$. True. Answer: (B), (C), (D).
Q.7	Integral CalculusIntegration by substitution	Class 12: Integrals; Class 12: Application of Integrals	NumericalMedium
A medium numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\left((x + 1)^2(1 - x)^6\right)^{1/4} = (1 + x)^{1/2}(1 - x)^{3/2} = (1 - x)^2\sqrt{\frac{1 + x}{1 - x}}$. Put $u = \frac{1 + x}{1 - x}$: $du = \frac{2}{(1 - x)^2}dx$, and $u$ runs from $1$ to $3$. $\int_0^{1/2}\frac{dx}{(1 - x)^2\sqrt{u}} = \int_1^3\frac{du}{2\sqrt{u}} = \left[\sqrt{u}\right]_1^3 = \sqrt{3} - 1$. The integral is $(1 + \sqrt{3})(\sqrt{3} - 1) = 3 - 1 = 2$. Answer: $2.00$.
Q.8	Matrices and DeterminantsDeterminants of order two and three	Class 12: Matrices; Class 12: Determinants	NumericalMedium
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\det P = \sum \pm p_{1\sigma(1)}p_{2\sigma(2)}p_{3\sigma(3)}$ has six terms, each $0$ or $\pm1$, so $\det P \le 6$. If some entry is $0$, the two terms containing it vanish and at most four terms remain, so $\det P \le 4$. If every entry is $\pm1$, all six terms are $\pm1$. Each entry appears in exactly two of the six products, so the product of the six unsigned products is $\left(\prod p_{ij}\right)^2 = 1$; three terms carry a minus sign, so the product of the six signed terms is $(-1)^3 = -1$. Hence they cannot all be $+1$: at least one is $-1$, and $\det P \le 5 - 1 = 4$. The value $4$ is reached: $\begin{vmatrix}1&1&1\\1&-1&1\\1&1&-1\end{vmatrix} = 1(1 - 1) - 1(-1 - 1) + 1(1 + 1) = 0 + 2 + 2 = 4$. Answer: $4$.
Q.9	Permutations and CombinationsSimple applications	Class 11: Permutations and Combinations	NumericalMedium
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations One-one maps from 5 elements into 7: $\alpha = {}^7P_5 = 7\cdot6\cdot5\cdot4\cdot3 = 2520$. Onto maps from 7 elements onto 5: split $Y$ into 5 non-empty blocks and assign them to $X$. Block sizes are $3,1,1,1,1$ or $2,2,1,1,1$: $\binom{7}{3} = 35$ partitions of the first kind, and $\frac{1}{2}\binom{7}{2}\binom{5}{2} = 105$ of the second, so $140$ partitions in all. $\beta = 140\times5! = 16800$. $\frac{\beta - \alpha}{5!} = \frac{16800 - 2520}{120} = \frac{14280}{120} = 119$. Answer: $119$.
Q.10	Differential EquationsSeparation of variables	Class 12: Differential Equations	NumericalMedium
A medium numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} = 25y^2 - 4$. Separate: $\frac{dy}{25y^2 - 4} = dx$, and $\frac{1}{25y^2 - 4} = \frac{1}{4}\left(\frac{1}{5y - 2} - \frac{1}{5y + 2}\right)$. $\frac{1}{20}\ln\left\|\frac{5y - 2}{5y + 2}\right\| = x + c$; at $x = 0$, $y = 0$ gives $\ln 1 = 0$, so $c = 0$. $\left\|\frac{5y - 2}{5y + 2}\right\| = e^{20x}$, and with $y$ between $-\frac{2}{5}$ and $\frac{2}{5}$ (it starts at $0$ and cannot cross the constant solutions $\pm\frac{2}{5}$): $\frac{2 - 5y}{2 + 5y} = e^{20x}$. As $x \rightarrow -\infty$, $e^{20x} \rightarrow 0$, so $2 - 5y \rightarrow 0$: $y \rightarrow \frac{2}{5}$. Answer: $0.40$.
Q.11	Differential EquationsSeparation of variables	Class 12: Differential Equations	NumericalMedium
A medium numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Put $x = y = 0$: $1 = 2f(0)f'(0) = 2f'(0)$, so $f'(0) = \frac{1}{2}$. Put $y = 0$: $f(x) = f(x)f'(0) + f'(x)f(0) = \frac{1}{2}f(x) + f'(x)$, so $f'(x) = \frac{1}{2}f(x)$. With $f(0) = 1$: $f(x) = e^{x/2}$ (and it does satisfy the equation: $e^{x/2}\cdot\frac{1}{2}e^{y/2} + \frac{1}{2}e^{x/2}e^{y/2} = e^{(x+y)/2}$). $\log_e f(4) = 2$. Answer: $2.00$.
Q.12	Three Dimensional GeometryCoordinates of a point in space	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	NumericalMedium
A medium numerical question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Let $P = (a, b, c)$ and $Q = (0, 0, k)$ on the $z$-axis. $PQ$ is along the plane's normal $(1, 1, 0)$, so $c = k$ and $a = b$. The midpoint $\left(\frac{a}{2}, \frac{a}{2}, c\right)$ is on $x + y = 3$: $a = 3$. So $P = (3, 3, c)$. Distance from the $x$-axis: $\sqrt{b^2 + c^2} = \sqrt{9 + c^2} = 5$, so $c = 4$ (first octant). $R = (3, 3, -4)$, and $PR = 2c = 8$. Answer: $8.00$.
Q.13	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 $P = (1, 0, 0)$, $Q = (0, 1, 0)$, $R = (0, 0, 1)$, $T = (1, 1, 1)$. From $S$: $\vec{p} = \frac{1}{2}(1, -1, -1)$, $\vec{q} = \frac{1}{2}(-1, 1, -1)$, $\vec{r} = \frac{1}{2}(-1, -1, 1)$, $\vec{t} = \frac{1}{2}(1, 1, 1)$. $\vec{p}\times\vec{q} = \frac{1}{4}\left((-1)(-1) - (-1)(1),\ (-1)(-1) - (1)(-1),\ (1)(1) - (-1)(-1)\right) = \frac{1}{4}(2, 2, 0) = \left(\frac{1}{2}, \frac{1}{2}, 0\right)$. $\vec{r}\times\vec{t} = \frac{1}{4}\left((-1)(1) - (1)(1),\ (1)(1) - (-1)(1),\ (-1)(1) - (-1)(1)\right) = \frac{1}{4}(-2, 2, 0) = \left(-\frac{1}{2}, \frac{1}{2}, 0\right)$. Their cross product: $\left(0,\ 0,\ \frac{1}{2}\cdot\frac{1}{2} - \frac{1}{2}\cdot\left(-\frac{1}{2}\right)\right) = \left(0, 0, \frac{1}{2}\right)$. Magnitude $\frac{1}{2}$. Answer: $0.50$.
Q.14	Binomial TheoremSimple applications	Class 11: Binomial Theorem	NumericalMedium
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $r\,{}^{10}C_r = 10\,{}^9C_{r-1}$, so $X = 10\sum_{r=1}^{10}{}^9C_{r-1}\,{}^{10}C_r = 10\sum_{r=1}^{10}{}^9C_{r-1}\,{}^{10}C_{10-r}$. By Vandermonde's identity the sum is the coefficient of $t^9$ in $(1 + t)^9(1 + t)^{10}$: ${}^{19}C_9 = 92378$. $X = 923780$, and $\frac{X}{1430} = 646$. Answer: $646$.
Q.15	Sets, Relations and FunctionsOne-one, into and onto functions	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	Matching listMedium
A medium matching list 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 $E_1$: $\frac{x}{x - 1} > 0$ for $x < 0$ or $x > 1$. Let $u = \frac{x}{x - 1} = 1 + \frac{1}{x - 1}$. For $x < 0$, $u$ takes every value in $(0, 1)$; for $x > 1$, every value in $(1, \infty)$. So $u$ ranges over $(0, 1)\cup(1, \infty)$ and $f = \ln u$ over $(-\infty, 0)\cup(0, \infty)$. P→4. $E_2$: need $-1 \le \ln u \le 1$, i.e. $\frac{1}{e} \le u \le e$. For $x > 1$: $u \le e \iff x \ge e(x - 1) \iff x \ge \frac{e}{e - 1}$. For $x < 0$: $u \ge \frac{1}{e} \iff ex \le x - 1$ (multiplying by $e(x - 1) < 0$) $\iff x \le \frac{1}{1 - e}$. Domain of $g$: $\left(-\infty, \frac{1}{1 - e}\right] \cup \left[\frac{e}{e - 1}, \infty\right)$. S→1. Range of $g$: $\ln u$ covers $[-1, 0)\cup(0, 1]$, so $g$ covers $\left[-\frac{\pi}{2}, 0\right)\cup\left(0, \frac{\pi}{2}\right]$, which contains $(0, 1)$ but not $\left[-\frac{1}{2}, \frac{1}{2}\right]$ (no $0$). Q→2. Domain of $f$ is $(-\infty, 0)\cup(1, \infty)$. It contains set 1 (since $\frac{1}{1 - e} < 0$ and $\frac{e}{e - 1} > 1$), but not 5 or 6 (they include points in $[0, 1]$). R→1. Answer: (A).
Q.16	Permutations and CombinationsSimple applications	Class 11: Permutations and Combinations	Matching listMedium
A medium matching list question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations (P) $\alpha_1 = \binom{6}{3}\binom{5}{2} = 20\times10 = 200$ → 4. (Q) $\alpha_2 = \sum_{k=1}^{5}\binom{6}{k}\binom{5}{k} = 6\cdot5 + 15\cdot10 + 20\cdot10 + 15\cdot5 + 6\cdot1 = 30 + 150 + 200 + 75 + 6 = 461$ → 6. (R) $\alpha_3 = \binom{11}{5} - \binom{6}{5} - \binom{5}{1}\binom{6}{4} = 462 - 6 - 75 = 381$ → 5. (S) 4 members with at least 2 girls: $\binom{5}{2}\binom{6}{2} + \binom{5}{3}\binom{6}{1} + \binom{5}{4} = 150 + 60 + 5 = 215$. Those containing both $M_1$ and $G_1$: 2 girls 2 boys, $\binom{4}{1}\binom{5}{1} = 20$; 3 girls 1 boy (the boy is $M_1$), $\binom{4}{2} = 6$. Total 26. $\alpha_4 = 215 - 26 = 189$ → 2. Answer: (C).
Q.17	Conic SectionsHyperbola	Class 11: Conic Sections	Matching listEasy
A easy matching list question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $L = (0, b)$, $M = (0, -b)$, $N = (a, 0)$. The angle $LNM = 60^\circ$ is bisected by the $x$-axis: $\tan 30^\circ = \frac{b}{a}$, so $a = \sqrt{3}b$. Area $= \frac{1}{2}\cdot2b\cdot a = ab = \sqrt{3}b^2 = 4\sqrt{3}$: $b = 2$, $a = 2\sqrt{3}$. (P) Conjugate axis $2b = 4$ → 4. (Q) $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{3}} = \frac{2}{\sqrt{3}}$ → 3. (R) $2ae = 2\cdot2\sqrt{3}\cdot\frac{2}{\sqrt{3}} = 8$ → 1. (S) $\frac{2b^2}{a} = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}}$ → 2. Answer: (B).
Q.18	Limits, Continuity and DifferentiabilityDifferentiability	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Matching listMedium
A medium matching list question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability (P) $f_1$ is continuous. Near $0$, $1 - e^{-x^2} \approx x^2$, so $f_1(x) \approx \sin\|x\| \approx \|x\|$: $\frac{f_1(h) - f_1(0)}{h} \rightarrow \pm1$ from the two sides. Continuous, not differentiable → 2. (Q) As $x \rightarrow 0^+$, $\frac{\|\sin x\|}{\tan^{-1}x} \rightarrow 1$; as $x \rightarrow 0^-$, $\|\sin x\| > 0$ and $\tan^{-1}x < 0$, so the ratio $\rightarrow -1$. Not continuous → 1. (R) For $x \in (-1, e^{\pi/2} - 2)$, $x + 2 \in (1, e^{\pi/2})$, so $\ln(x + 2) \in (0, \frac{\pi}{2})$ and $\sin$ of it lies in $(0, 1)$. Hence $f_3 \equiv 0$: differentiable with continuous derivative → 4. (S) $f_4'(0) = \lim_{h\rightarrow0}h\sin\frac{1}{h} = 0$, so it is differentiable; for $x \ne 0$, $f_4'(x) = 2x\sin\frac{1}{x} - \cos\frac{1}{x}$, which has no limit at $0$. Derivative not continuous → 3. So P→2, Q→1, R→4, S→3. Answer: (D).

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