JEE (Main) · 2021

JEE (Main) 2021 — 27 August, Shift 1

Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.

Sit this paper free All of 2021

Summary

The heaviest chapters in each subject, and how the questions were spread by level.

Physics
30 Q
1. Properties of Bulk Matter2
2. Physical World and Measurement2
3. Kinematics2
18 easy · 12 medium · 0 hard
Chemistry
30 Q
1. p-Block Elements2
2. d- and f-Block Elements2
3. Amines2
18 easy · 11 medium · 1 hard
Mathematics
30 Q
1. Integral Calculus3
2. Matrices and Determinants2
3. Sequences and Series2
8 easy · 19 medium · 3 hard

Questions

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

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
1	Properties of Bulk MatterViscosity	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctEasy
A easy single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter An uncharged drop falls at terminal velocity, where the viscous force balances its weight (buoyancy neglected). $F = mg = \frac{4}{3}\pi r^3\rho g = \frac{4}{3}\pi(2.0\times10^{-5})^3(1.2\times10^{3})(9.8)$. $r^3 = 8\times10^{-15}\ \mathrm{m^3}$, so $\frac{4}{3}\pi r^3 = 3.35\times10^{-14}\ \mathrm{m^3}$ and $F = 3.35\times10^{-14}\times1.2\times10^{3}\times9.8 \approx 3.9\times10^{-10}$ N. (The viscosity is not needed: the terminal-velocity condition fixes the force.) Answer: (D).
2	Motion of System of Particles and Rigid BodyMoment of inertia	Class 11: Systems of Particles and Rotational Motion	Single correctEasy
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion About the perpendicular axis through the centre: $I_c = \frac{M(l^2+l^2)}{12} = \frac{Ml^2}{6}$. The corner is at distance $\frac{l}{\sqrt2}$ from the centre, so by the parallel-axis theorem $I = \frac{Ml^2}{6} + M\frac{l^2}{2} = \frac{2}{3}Ml^2$. Answer: (B).
3	Physical World and MeasurementUnits and dimensions	Class 11: Units and Measurement	Single correctMedium
A medium single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement '4s' is 4 seconds of arc: $\theta = \frac{4}{3600}\times\frac{\pi}{180} = 1.94\times10^{-5}$ rad. Radius: $R = \frac{4.4\times9.46\times10^{15}}{1.94\times10^{-5}} = 2.15\times10^{21}$ m. Distance for 4 revolutions: $4\times2\pi R = 5.40\times10^{22}$ m. Speed: $8\times1.5\times10^{11} = 1.2\times10^{12}$ m/s. Time: $t = \frac{5.40\times10^{22}}{1.2\times10^{12}} \approx 4.5\times10^{10}$ s. Answer: (A).
4	KinematicsScalars and vectors	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Resolve each force (with $\sqrt3 = 1.7$, $\frac{1}{\sqrt2} = \frac{\sqrt2}{2} = 0.7$): $\vec{OP}$: 20 N at $30^\circ$ from $y$: $(20\sin30^\circ, 20\cos30^\circ) = (10, 17)$. $\vec{OQ}$: 10 N at $30^\circ$ above $x$: $(10\cos30^\circ, 10\sin30^\circ) = (8.5, 5)$. $\vec{OT}$: 15 N at $60^\circ$ from $y$ towards $-x$: $(-15\sin60^\circ, 15\cos60^\circ) = (-12.75, 7.5)$. $\vec{OS}$: 15 N at $45^\circ$ below $-x$: $(-10.5, -10.5)$. $\vec{OR}$: 20 N at $45^\circ$ below $+x$: $(14, -14)$. $x$: $10+8.5-12.75-10.5+14 = 9.25$; $y$: $17+5+7.5-10.5-14 = 5$. Resultant $= 9.25\hat{i}+5\hat{j}$ N. Answer: (C).
5	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	Single correctEasy
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Quality factor, power factor ($\cos\phi$) and relative permeability ($\mu/\mu_0$) are ratios and have no dimensions. $\mu_0$ has units $\mathrm{T\,m\,A^{-1}}$ (dimensions $[\mathrm{MLT^{-2}A^{-2}}]$). Answer: (D).
6	Electromagnetic WavesProperties of electromagnetic waves	Class 12: Electromagnetic Waves	Single correctEasy
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves E is in V/m and H is in A/m, so $\frac{E}{H}$ is in $\frac{\mathrm{V}}{\mathrm{A}}$ = ohm. (For a wave in vacuum $E/H = 377\ \Omega$, the impedance of free space.) Answer: (B).
7	Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation	Class 11: Kinetic Theory	Single correctMedium
A medium single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $P \propto T^{-3}$ and $PV = nRT$ give $V \propto \frac{T}{P} \propto T^4$. $\gamma = \frac{1}{V}\frac{dV}{dT} = \frac{4}{T}$. Answer: (D).
8	Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation	Class 11: Kinetic Theory	Single correctMedium
A medium single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory At constant volume the load the balloon can carry is proportional to the density of the surrounding air, $\rho \propto \frac{P}{T}$. $W_2 = 185\times\frac{45}{76}\times\frac{300}{266} = 185\times0.5921\times1.1278 \approx 123.54$ kg. Answer: (A).
9	Oscillations and WavesSHM	Class 11: Oscillations; Class 11: Waves	Single correctEasy
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $U = \frac{1}{2}kx^2$ is never negative, and it is zero whenever $x = 0$ (at O, A, B and C) and greatest at the extreme positions. So U against t is a row of positive humps touching zero at O, A, B and C, with twice the frequency of $x$. Answer: (B).
10	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In the steady state no current flows in the capacitor branch, so current flows only through the battery and the $4\ \Omega$ resistor: $I = \frac{5}{4+1} = 1$ A. The terminal voltage is $5 - 1\times1 = 4$ V, and this is across the capacitor branch. In that branch $4\ \mu$F is in series with $2\ \mu\mathrm{F}\parallel2\ \mu\mathrm{F} = 4\ \mu$F, so $C_{eq} = 2\ \mu$F (the $6\ \Omega$ carries no current). $Q = 2\ \mu\mathrm{F}\times4\ \mathrm{V} = 8\ \mu$C, the charge on the $4\ \mu$F capacitor. Answer: (C).
11	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance A ring of radius $r$, width $dr$ carries $dq = \sigma\,2\pi r\,dr$ and gives $dE = \frac{1}{4\pi\varepsilon_0}\frac{dq\,Z}{(r^2+Z^2)^{3/2}}$ along the axis. $E = \frac{\sigma Z}{2\varepsilon_0}\int_0^R\frac{r\,dr}{(r^2+Z^2)^{3/2}} = \frac{\sigma Z}{2\varepsilon_0}\left[\frac{1}{Z} - \frac{1}{\sqrt{Z^2+R^2}}\right] = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{Z}{\sqrt{Z^2+R^2}}\right)$. Answer: (B).
12	Current ElectricityCells, emf and internal resistance	Class 12: Current Electricity	Single correctEasy
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Series: $I_s = \frac{5\times5}{R+5\times1} = \frac{25}{R+5}$. Parallel: $I_p = \frac{5}{R+\frac{1}{5}}$. $\frac{25}{R+5} = \frac{5}{R+0.2}$ gives $25R+5 = 5R+25$, so $R = 1\ \Omega$ (equal to the internal resistance of one cell). Answer: (A).
13	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctEasy
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, so at equal K, $r \propto \frac{\sqrt m}{q}$. Lighter ion: $\frac{\sqrt4}{2} = 1$; heavier ion: $\frac{\sqrt{16}}{3} = \frac43$. The lighter ion moves on the smaller circle, so it is deflected more. Answer: (A).
14	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The flux through the loop changes sharply only while a pole passes through it, so the emf is two short pulses, separated by the time $\frac{l}{v}$ the magnet takes to move its own length. As the N end enters, the flux through the loop grows; as the S end leaves, the same flux falls. By $\varepsilon = -\frac{d\Phi}{dt}$ the two pulses have opposite signs. This rules out (B) and (C). Which pulse is drawn positive depends only on the sign convention chosen for the loop, which the question does not fix: with the loop's normal along the velocity the first pulse is negative (A); with the opposite normal the first is positive (D). Answer: (A) or (D), the two pulses of opposite sign.
15	Electromagnetic WavesProperties of electromagnetic waves	Class 12: Electromagnetic Waves	Single correctEasy
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $v = \frac{\omega}{k} = \frac{10\times10^{10}}{500} = 2\times10^{8}$ m/s $= \frac{2}{3}\times3\times10^{8} = \frac{2}{3}$C. Answer: (A).
16	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctMedium
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The object is beyond C, so the real image lies between F and C: $\|u\| = R + d_1$, $\|v\| = R - d_2$. Mirror formula (all real distances): $\frac{1}{R-d_2} + \frac{1}{R+d_1} = \frac{2}{R}$. $R(2R + d_1 - d_2) = 2(R-d_2)(R+d_1) = 2R^2 + 2R(d_1-d_2) - 2d_1d_2$. So $R(d_1-d_2) = 2d_1d_2$ and $R = \frac{2d_1d_2}{d_1-d_2}$. Answer: (C).
17	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctMedium
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics First lens ($f = +10$ cm, $u = -30$ cm): $\frac1v = \frac1{10} - \frac1{30} = \frac1{15}$, $v = 15$ cm. The second lens is 5 cm further, so this image is a virtual object 10 cm beyond it: $u = +10$ cm, $f = -10$ cm, $\frac1v = -\frac1{10} + \frac1{10} = 0$: the rays leave parallel. The third lens ($f = +30$ cm) focuses parallel rays 30 cm beyond it. Distance from O: $30 + 5 + 10 + 30 = 75$ cm. Answer: (C).
18	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter At fixed frequency, intensity is the number of photons per unit area per unit time. More photons eject more electrons, but each photon still has energy $h\nu$, so the maximum kinetic energy $h\nu - \phi$ is unchanged. Answer: (B).
19	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $N = N_0\left(\frac12\right)^{t/T_{1/2}} = 10^{10}\times\left(\frac12\right)^{1/2} = \frac{10^{10}}{1.414} \approx 7.07\times10^{9} \approx 7\times10^9$. Answer: (C).
20	Electronic DevicesSemiconductors	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctEasy
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits An amplifier needs the output to follow the input linearly, which happens only in the active region (emitter-base forward biased, collector-base reverse biased). Cut-off and saturation are the two states used when the transistor works as a switch. Answer: (D).
21	Electronic Devicesp-n junction diode	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	NumericalMedium
A medium numerical question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Diodes $D_1$ and $D_2$ run from the node at the top of the base resistor to the grounded wire, with their anodes on the node side. The 5 V cell drives current through R towards that node, so the diodes are forward biased and (ideal diodes) hold the node at 0 V. The base of the transistor is therefore at 0 V, the same as its grounded emitter: no base current flows and the transistor is cut off. With no collector current there is no drop across the collector resistor R, so $V_o$ equals the supply, 5 V. Answer: 5.
22	Work, Energy and PowerWork-energy theorem	Class 11: Work, Energy and Power	NumericalEasy
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $F_A d\cos45^\circ = F_B d\cos60^\circ$, so $\frac{F_A}{F_B} = \frac{\cos60^\circ}{\cos45^\circ} = \frac{1/2}{1/\sqrt2} = \frac{1}{\sqrt2}$. So $x = 2$. Answer: 2.
23	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 $v^2 = 5000 + 24x$. Differentiating with respect to time: $2v\frac{dv}{dt} = 24\frac{dx}{dt} = 24v$. So $a = \frac{dv}{dt} = 12\ \mathrm{m/s^2}$ (constant). Answer: 12.
24	GravitationGravitational potential energy	Class 11: Gravitation	NumericalMedium
A medium numerical question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation The two m's are at opposite corners and so are the two (M - m)'s. Each of the four sides joins an m with an M - m; the diagonals ($\sqrt2 d$) join m with m and (M - m) with (M - m). $U = -\frac{G}{d}\left[4m(M-m) + \frac{m^2 + (M-m)^2}{\sqrt2}\right]$. Call the bracket $f(m)$: $\frac{df}{dm} = 4M - 8m + \frac{4m - 2M}{\sqrt2} = (4 - \sqrt2)M - (8 - 2\sqrt2)m$. Setting it to zero: $m = \frac{(4-\sqrt2)M}{2(4-\sqrt2)} = \frac{M}{2}$, and $f'' = -8 + 2\sqrt2 < 0$, so the size of the potential energy is greatest there. So $\frac{M}{m} = 2$, i.e. $x = 2$. Answer: 2.
25	Oscillations and WavesDoppler effect	Class 11: Oscillations; Class 11: Waves	NumericalEasy
A easy numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Source X moves at 36 km/h = 10 m/s and observer Y at 72 km/h = 20 m/s, towards each other. $f' = f\frac{v + v_o}{v - v_s} = f\frac{340+20}{340-10} = f\frac{360}{330}$. $f = 1320\times\frac{330}{360} = 1210$ Hz. Answer: 1210.
26	Current ElectricityCells, emf and internal resistance	Class 12: Current Electricity	NumericalEasy
A easy numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Series: $I = \frac{20}{10n + 10}$. Parallel: $I' = \frac{20}{\frac{10}{n} + 10}$. $\frac{I'}{I} = \frac{10n+10}{\frac{10}{n}+10} = \frac{n+1}{\frac{1+n}{n}} = n$. So $n = 20$. Answer: 20.
27	Magnetic Effects of Current and MagnetismMagnetic dipole	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	NumericalMedium
A medium numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The same wire on the same source carries the same current $I = \frac{V_0}{R}$ in both cases. Triangle: $\frac{24a}{3a} = 8$ turns of area $\frac{\sqrt3}{4}a^2$: $m_1 = 8I\frac{\sqrt3}{4}a^2 = 2\sqrt3\,Ia^2$. Square: $\frac{24a}{4a} = 6$ turns of area $a^2$: $m_2 = 6Ia^2$. $\frac{m_1}{m_2} = \frac{2\sqrt3}{6} = \frac{1}{\sqrt3}$, so $y = 3$. Answer: 3.
28	Electromagnetic Induction and Alternating CurrentsAC circuits	Class 12: Electromagnetic Induction; Class 12: Alternating Current	NumericalMedium
A medium numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $i^2 = 42\sin^2\omega t + 20\sqrt{42}\sin\omega t + 100$. Over a period, $\langle\sin^2\omega t\rangle = \frac12$ and $\langle\sin\omega t\rangle = 0$. $i_{rms}^2 = 21 + 100 = 121$, so $i_{rms} = 11$ A. Answer: 11.
29	Communication SystemsLine of sight communication and antenna height	No matching NCERT chapter	NumericalEasy
A easy numerical question on line of sight communication and antenna height, from Communication Systems. NCERT: No matching NCERT chapter $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$ with $R = 6.4\times10^{6}$ m. $\sqrt{2\times6.4\times10^6\times320} = \sqrt{4.096\times10^{9}} = 64$ km. $\sqrt{2\times6.4\times10^6\times2000} = \sqrt{2.56\times10^{10}} = 160$ km. $d = 64 + 160 = 224$ km. Answer: 224.
30	Properties of Bulk MatterHeat transfer	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalMedium
A medium 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 AB is identical to CD, so its resistance is $10\ \mathrm{K\,W^{-1}}$ and each half (AC, CB) is $5\ \mathrm{K\,W^{-1}}$. At the junction C (temperature T) heat in equals heat out: $\frac{200-T}{5} + \frac{100-T}{5} = \frac{T-125}{10}$. $600 - 4T = T - 125$, so $T = 145^\circ$C. Heat current in CD: $P = \frac{145-125}{10} = 2$ W. Answer: 2.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
31	Some Basic Concepts of ChemistryMole concept	Class 11: Some Basic Concepts of Chemistry	Single correctEasy
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\frac{an^2}{V^2}$ is added to P, so it has the units of pressure: $a = \frac{PV^2}{n^2}$, in $\mathrm{atm\,dm^6\,mol^{-2}}$. (b, by contrast, is in $\mathrm{dm^3\,mol^{-1}}$.) Answer: (B).
32	Solid StateClassification of solids	No matching NCERT chapter	MatchingEasy
A easy matching question on classification of solids, from Solid State. NCERT: No matching NCERT chapter NaCl has all electrons paired: diamagnetic, (a)-(iii). $\mathrm{Fe_3O_4}$ has unequal moments aligned antiparallel, leaving a net moment: ferrimagnetic, (b)-(iv). $\mathrm{O_2}$ has two unpaired electrons: paramagnetic, (c)-(ii). MnO has equal moments aligned antiparallel, cancelling: antiferromagnetic, (d)-(i). Answer: (A).
33	Chemical Bonding and Molecular StructureVSEPR theory	Class 11: Chemical Bonding and Molecular Structure	MatchingEasy
A easy matching question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Xe has 8 valence electrons; each F uses 1 and each double-bonded O uses 2. $\mathrm{XeF_2}$: $8-2 = 6$ left, 3 lone pairs: (a)-(iv). $\mathrm{XeO_2F_2}$: $8-4-2 = 2$, 1 lone pair: (b)-(ii). $\mathrm{XeO_3F_2}$: $8-6-2 = 0$, no lone pair: (c)-(i). $\mathrm{XeF_4}$: $8-4 = 4$, 2 lone pairs: (d)-(iii). Answer: (B).
34	Surface ChemistryColloids and their preparation	No matching NCERT chapter	Single correctEasy
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter The Tyndall effect is the scattering of light by colloidal particles; it is marked when the refractive indices of the dispersed phase and the medium differ greatly. In lyophobic sols they differ much more than in lyophilic sols (whose particles are solvated), so lyophobic colloids show it more strongly. A true solution shows none, and a suspension's particles settle out and do not give the effect. Answer: (C).
35	p-Block ElementsGroup 13 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Back donation ($p\pi$–$p\pi$) from the halogen's filled p orbital into boron's empty 2p orbital is most effective when the orbitals are of the same size: F (2p) with B (2p). It weakens down the group as the halogen orbitals grow (3p, 4p, 5p). So $\mathrm{BF_3}$ has the strongest back bonding (which is why it is the weakest Lewis acid of the four). Answer: (D).
36	Isolation of MetalsSmelting and refining	No matching NCERT chapter	Single correctEasy
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In liquation the impure metal is placed on a sloping hearth and heated just above its melting point: the low-melting metal (tin, lead, bismuth) flows away from the higher-melting impurities. Answer: (B).
37	HydrogenHydrogen and its compounds	No matching NCERT chapter	Single correctEasy
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Deuterium has the same electronic configuration as hydrogen and so the same chemistry, but its greater mass gives a higher bond dissociation enthalpy (D-D 443 kJ/mol against H-H 436 kJ/mol) and slower reactions (the kinetic isotope effect). It is a stable isotope and emits no particles. Answer: (C).
38	s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium	No matching NCERT chapter	Single correctEasy
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Gypsum is $\mathrm{CaSO_4\cdot2H_2O}$ (2); plaster of Paris is $\mathrm{CaSO_4\cdot\frac12H_2O}$ (0.5); heating above 393 K gives anhydrous $\mathrm{CaSO_4}$, dead burnt plaster (0). In the order asked: 2, 0 and 0.5. Answer: (D).
39	p-Block ElementsGroup 16 elements	No matching NCERT chapter	Single correctMedium
A medium single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Polythionic acids are $\mathrm{HO_3S-(S)_{x-2}-SO_3H}$. Each terminal S is bonded to three O (one as OH) and one S: oxidation state +5. The S atoms in the chain are bonded only to sulphur: oxidation state 0. (Check for $x = 3$: average $\frac{10}{3}$, and $\frac{5+5+0}{3} = \frac{10}{3}$.) Answer: (C).
40	d- and f-Block ElementsTransition element properties	Class 12: The d- and f-Block Elements	Single correctEasy
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements For a transition metal the lower oxides are basic and the higher ones acidic. $\mathrm{V_2O_3}$ (V +3) is basic ($\mathrm{V_2O_4}$ amphoteric, $\mathrm{V_2O_5}$ acidic), and CrO (Cr +2) is basic ($\mathrm{Cr_2O_3}$ amphoteric, $\mathrm{CrO_3}$ acidic). Answer: (B).
41	Environmental ChemistryOzone depletion	No matching NCERT chapter	Single correctEasy
A easy single correct question on ozone depletion, from Environmental Chemistry. NCERT: No matching NCERT chapter The gas is a chlorofluorocarbon such as freon-12, $\mathrm{CF_2Cl_2}$: inert, non-toxic and non-flammable, it drifts up to the stratosphere where UV light breaks a C-Cl bond: $\mathrm{CF_2Cl_2} \xrightarrow{h\nu} \dot{\mathrm{C}}\mathrm{l} + \dot{\mathrm{C}}\mathrm{F_2Cl}$. The chlorine radical then attacks ozone ($\mathrm{Cl^\bullet + O_3 \to ClO^\bullet + O_2}$), so ClO is formed later, not initially. Answer: (D).
42	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{NaNH_2}$ removes the terminal acetylenic H: A is $\mathrm{CH_3C\equiv C^-Na^+}$. The acetylide displaces Br from $\mathrm{BrCH_2CH_2CH(OH)CH_3}$ (4-bromobutan-2-ol): B is $\mathrm{CH_3C\equiv C-CH_2CH_2CH(OH)CH_3}$ (hept-5-yn-2-ol). $\mathrm{H_2/Pd\text{-}C}$ saturates the triple bond: C is $\mathrm{CH_3(CH_2)_4CH(OH)CH_3}$ (heptan-2-ol). $\mathrm{CrO_3}$ oxidises the secondary alcohol to the ketone: D is $\mathrm{CH_3CH_2CH_2CH_2CH_2COCH_3}$ (heptan-2-one). Answer: (C).
43	Haloalkanes and HaloarenesNature of the C–X bond	Class 12: Haloalkanes and Haloarenes	Single correctEasy
A easy single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Chlorocyclopentane and Mg in dry ether give the Grignard reagent A, cyclopentylmagnesium chloride. Ethanol has an acidic O-H; the Grignard carbanion takes the proton: $\mathrm{C_5H_9MgCl + C_2H_5OH \to C_5H_{10} + Mg(OC_2H_5)Cl}$. P is cyclopentane. Answer: (B).
44	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The product is but-2-enoic acid, $\mathrm{CH_3CH=CHCOOH}$ (4 carbons). NaOCl followed by acid is the haloform reaction: a methyl ketone $\mathrm{R-CO-CH_3}$ gives $\mathrm{R-COOH}$ (one carbon fewer) and $\mathrm{CHCl_3}$. So P is $\mathrm{CH_3CH=CH-CO-CH_3}$, pent-3-en-2-one, option (A). An aldehyde (B) would be oxidised to an acid with the same carbon count (pent-2-enoic acid), and (C) and (D) have no methyl ketone group. Answer: (A).
45	Alcohols, Phenols and EthersWilliamson synthesis	Class 12: Alcohols, Phenols and Ethers	Assertion and reasonEasy
A easy assertion and reason question on williamson synthesis, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (A) is correct: sodium phenoxide and ethyl bromide give ethyl phenyl ether by $\mathrm{S_N2}$ (Williamson synthesis). (R) is wrong: the C-Br bond of bromobenzene has partial double-bond character and the ring repels the nucleophile, so bromobenzene does not undergo nucleophilic substitution with sodium ethoxide under ordinary conditions. Answer: (C).
46	AminesDiazonium salts	Class 12: Amines	Single correctMedium
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines (i) Ammonia converts the acid chloride to the amide $\mathrm{(CH_3)_2CHCH_2CH_2CONH_2}$. (ii) $\mathrm{Br_2/NaOH}$ (Hofmann bromamide degradation) gives the amine with one carbon fewer: $\mathrm{(CH_3)_2CHCH_2CH_2NH_2}$. (iii), (iv) Nitrous acid turns the primary aliphatic amine into an unstable diazonium salt, which with water gives the alcohol and $\mathrm{N_2}$: $\mathrm{(CH_3)_2CHCH_2CH_2OH}$ (3-methylbutan-1-ol). Answer: (A).
47	AminesClassification and identification of amines	Class 12: Amines	Single correctEasy
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines (A) is true: Gabriel synthesis gives only primary amines. (B) is true: in water, aliphatic $2^\circ$ amines are more basic than $1^\circ$ (e.g. $\mathrm{(C_2H_5)_2NH > C_2H_5NH_2}$). (C) is false: a primary amine has two N-H hydrogens for hydrogen bonding and a secondary amine only one, so association is greater in primary amines. (D) is the textbook statement: primary aliphatic amines give alcohols with nitrous acid, methylamine being the exception (it gives other products, such as methoxymethane). Answer: (C).
48	Chemistry in Everyday LifeDrugs and their classification	No matching NCERT chapter	Single correctMedium
A medium single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter (A) is morphine and (D) is codeine (its methyl ether): both are narcotic analgesics. (B) is diazepam (a benzodiazepine) and (C) is drawn as serotonin (5-hydroxytryptamine); both are classed as tranquillizers. So (A) fails ((B), (C) are not narcotics), (B) fails ((D) is a narcotic), (C) fails ((A), (D) are narcotics), and (D) is correct. Answer: (D).
49	BiomoleculesNucleic acids and vitamins	Class 12: Biomolecules	Single correctEasy
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Uracil in RNA is pyrimidine-2,4(1H,3H)-dione, the diketo (lactam) form with two C=O groups and two N-H groups; N1 is the atom that bonds to ribose. That is structure (A); the others are its enol (lactim) tautomers. Answer: (A).
50	Principles of Qualitative AnalysisCharacteristic precipitates and colours	No matching NCERT chapter	Single correctMedium
A medium single correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter With excess ferrocyanide, $\mathrm{Fe^{3+}}$ and $\mathrm{[Fe(CN)_6]^{4-}}$ combine 1 : 1 with a potassium ion: $\mathrm{Fe^{3+} + K^+ + [Fe(CN)_6]^{4-} \to KFe[Fe(CN)_6]}$, 'soluble' Prussian blue, which stays dispersed as a colloid. With excess $\mathrm{Fe^{3+}}$ the insoluble Prussian blue $\mathrm{Fe_4[Fe(CN)_6]_3}$ precipitates instead. Answer: (C).
51	Structure of AtomBohr model	Class 11: Structure of Atom	NumericalMedium
A medium numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In the $n$th orbit $mvr = \frac{nh}{2\pi}$ and $r_n = n^2a_0$. For $n = 2$: $r = 4a_0$, $mv = \frac{2h}{2\pi\cdot4a_0} = \frac{h}{4\pi a_0}$. $KE = \frac{(mv)^2}{2m} = \frac{h^2}{32\pi^2 m a_0^2}$, so $x = 32\pi^2 = 32\times(3.14)^2 = 315.5072$. $10x = 3155.07 \approx 3155$. Answer: 3155.
52	ThermodynamicsEnthalpy	Class 11: Thermodynamics	NumericalMedium
A medium numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics HCl: $200\times0.2 = 40$ mmol; NaOH: $300\times0.1 = 30$ mmol. 30 mmol (0.030 mol) is neutralised. Heat released: $0.030\times57.1 = 1.713$ kJ $= 1713$ J. Mass of solution: 500 mL, 500 g. $\Delta T = \frac{1713}{500\times4.18} = 0.8196\ ^\circ$C $= 81.96\times10^{-2}$. Nearest integer: 82. Answer: 82.
53	SolutionsColligative properties	Class 12: Solutions	NumericalHard
A hard numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Composition of 1 kg of solution: with $w$ kg of water there are $0.75w$ mol of sucrose (342 g/mol): $1000w + 0.75w\times342 = 1000$, so $w = 0.7959$ kg and sucrose $= 0.5969$ mol. At $-4^\circ$C the remaining solution has $m = \frac{\Delta T_f}{K_f} = \frac{4}{1.86} = 2.151$ mol/kg. Water left liquid: $\frac{0.5969}{2.151} = 0.2776$ kg. Ice separated: $795.9 - 277.6 = 518.3$ g, nearest integer 518. Answer: 518.
54	EquilibriumIonic equilibrium	Class 11: Equilibrium	NumericalMedium
A medium numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Nearly all the silver ends up as the complex: $\mathrm{[Ag(NH_3)_2^+]} \approx 0.80$ M. $K_f = \frac{[\mathrm{Ag(NH_3)_2^+}]}{[\mathrm{Ag^+}][\mathrm{NH_3}]^2}$: $1.0\times10^8 = \frac{0.80}{5.0\times10^{-8}[\mathrm{NH_3}]^2}$, so $[\mathrm{NH_3}]^2 = 0.16$ and free $[\mathrm{NH_3}] = 0.40$ M. Total ammonia: $0.40 + 2\times0.80 = 2.0$ M; in 2 L, 4 mol. Answer: 4.
55	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 In acid, $\mathrm{MnO_4^-}$ takes 5 electrons and $\mathrm{Fe^{2+}}$ gives 1: $\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O}$. $5\times M\times10 = 1\times0.1\times10$, so $M = 0.02$ mol/L. Molar mass of $\mathrm{KMnO_4} = 39 + 55 + 64 = 158$; strength $= 0.02\times158 = 3.16$ g/L $= 316\times10^{-2}$. Answer: 316.
56	Chemical KineticsRate laws	Class 12: Chemical Kinetics	NumericalEasy
A easy numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate $= \frac{1}{2}\frac{d[\mathrm{Cr_2(SO_4)_3}]}{dt} = -\frac{1}{3}\frac{d[\mathrm{C_2H_6O}]}{dt}$. $-\frac{d[\mathrm{C_2H_6O}]}{dt} = \frac{3}{2}\times2.67 = 4.005\ \mathrm{mol\,min^{-1}}$, nearest integer 4. Answer: 4.
57	d- and f-Block ElementsLanthanoids and actinoids	Class 12: The d- and f-Block Elements	NumericalEasy
A easy numerical question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Np (Z = 93): $\mathrm{[Rn]\,5f^4\,6d^1\,7s^2}$. The 5f subshell, the one being filled across the actinoids, holds 4 electrons. Answer: 4.
58	Coordination CompoundsWerner theory	Class 12: Coordination Compounds	NumericalEasy
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Only 1 Cl is ionisable, so 2 Cl are inside the coordination sphere: $\mathrm{[MCl_2L_2]Cl}$. Octahedral means coordination number 6: the two L ligands supply $6 - 2 = 4$ donor atoms, 2 each. L is bidentate. Answer: 2.
59	Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis	Class 11: Organic Chemistry – Some Basic Principles and Techniques	NumericalEasy
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Mass of Br in 0.188 g AgBr $= \frac{80}{188}\times0.188 = 0.080$ g. % Br $= \frac{0.080}{0.2}\times100 = 40$. Answer: 40.
60	Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis	Class 11: Organic Chemistry – Some Basic Principles and Techniques	NumericalMedium
A medium numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques N,N-Dimethylaminopentane is $\mathrm{CH_3(CH_2)_4N(CH_3)_2 = C_7H_{17}N}$, molar mass $84 + 17 + 14 = 115$ g/mol; 57.5 g is 0.5 mol. In the Dumas method: $\mathrm{C_xH_yN_z + (2x + \frac{y}{2})CuO \to xCO_2 + \frac{y}{2}H_2O + \frac{z}{2}N_2 + (2x + \frac{y}{2})Cu}$. For $\mathrm{C_7H_{17}N}$: $2\times7 + \frac{17}{2} = 22.5$ mol CuO per mol. For 0.5 mol: $11.25$ mol $= 1125\times10^{-2}$. Answer: 1125.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
61	Quadratic EquationsNature of roots	Class 11: Complex Numbers and Quadratic Equations	Single correctEasy
A easy single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Complete the square: $(x-2)^2 + 9y^2 = 1$. Since $9y^2 \ge 0$, $(x-2)^2 \le 1$, so $1 \le x \le 3$. Since $(x-2)^2 \ge 0$, $9y^2 \le 1$, so $-\frac13 \le y \le \frac13$. (The curve is an ellipse centred at (2, 0) with semi-axes 1 and $\frac13$.) Answer: (B).
62	Complex NumbersArgand diagram	Class 11: Complex Numbers and Quadratic Equations	Single correctEasy
A easy single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z = x + iy$. $\frac{x + i(y-1)}{x + i(y+2)}$ is real when the imaginary part of $(x + i(y-1))(x - i(y+2))$ vanishes: $x(y-1) - x(y+2) = -3x = 0$, so $x = 0$. Geometrically, $z - i$ and $z + 2i$ are parallel: $z$ lies on the line through $i$ and $-2i$, the imaginary axis (with the point $z = -2i$ itself excluded). So S is a straight line in the complex plane. Answer: (C).
63	Matrices and DeterminantsAlgebra of matrices	Class 12: Matrices; Class 12: Determinants	Single correctMedium
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Trace $= -1$ and $\|A\| = -2K$, so by the Cayley-Hamilton theorem $A^2 + A - 2KI = O$, i.e. $A^2 = -A + 2KI$. $A^4 = (A^2)^2 = A^2 - 4KA + 4K^2I = -(1+4K)A + (2K + 4K^2)I$. $A^4 + 3A = (2 - 4K)A + (2K + 4K^2)I = 2I$. The coefficient of A must vanish: $K = \frac12$; then $2K + 4K^2 = 1 + 1 = 2$, as required. Answer: (C).
64	Sequences and SeriesGeometric progressions	Class 11: Sequences and Series	Single correctEasy
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $y = \log_{10}x\left(1 + \frac13 + \frac19 + \ldots\right) = \log_{10}x\cdot\frac{1}{1 - \frac13} = \frac32\log_{10}x$. The ratio of sums is $\frac{2(1 + 2 + \ldots + y)}{3(1 + 2 + \ldots + y)} = \frac23$. $\frac23 = \frac{4}{\log_{10}x}$ gives $\log_{10}x = 6$, $x = 10^6$, and $y = \frac32\times6 = 9$. Answer: (D).
65	Limits, Continuity and DifferentiabilityLimits of functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctMedium
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $t = x^2 + bx + c = (x-\alpha)(x-\beta) \to 0$ as $x \to \beta$. $e^{2t} - 1 - 2t = \frac{(2t)^2}{2} + \ldots \approx 2t^2$. Limit $= \lim_{x\to\beta}\frac{2(x-\alpha)^2(x-\beta)^2}{(x-\beta)^2} = 2(\beta - \alpha)^2 = 2[(\alpha+\beta)^2 - 4\alpha\beta] = 2(b^2 - 4c)$. Answer: (D).
66	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	Single correctMedium
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Square side $a$, hexagon side $s$: $4a + 6s = 20$, so $a = 5 - \frac32 s$. Area $= a^2 + \frac{3\sqrt3}{2}s^2$. $\frac{dA}{ds} = 2a\left(-\frac32\right) + 3\sqrt3 s = 0$ gives $a = \sqrt3 s$. $5 - \frac32 s = \sqrt3 s$, so $s = \frac{5}{\sqrt3 + \frac32} = \frac{10}{3 + 2\sqrt3}$. $\frac{d^2A}{ds^2} = \frac92 + 3\sqrt3 > 0$, so this is the minimum. Answer: (C).
67	Straight LinesSlope, parallel and perpendicular lines	Class 11: Straight Lines	Single correctMedium
A medium single correct question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines M = (t, 3). Slope of AB $= -\frac{3}{t}$, so the perpendicular bisector is $y - 3 = \frac{t}{3}(x - t)$. At $x = 0$: C = $\left(0, 3 - \frac{t^2}{3}\right)$. P, the mid-point of MC: $x = \frac{t}{2}$, $y = 3 - \frac{t^2}{6}$. With $t = 2x$: $y = 3 - \frac{2x^2}{3}$, i.e. $2x^2 + 3y - 9 = 0$. Answer: (B).
68	Integral CalculusFundamental theorem of calculus	Class 12: Integrals; Class 12: Application of Integrals	Single correctHard
A hard single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\ln U_n = \sum_{r=1}^{n} r\ln\left(1 + \frac{r^2}{n^2}\right)$, so $\frac{-4}{n^2}\ln U_n = -4\cdot\frac1n\sum_{r=1}^{n}\frac{r}{n}\ln\left(1 + \frac{r^2}{n^2}\right) \to -4\int_0^1 x\ln(1+x^2)\,dx$. With $t = 1 + x^2$: $\int_0^1 x\ln(1+x^2)\,dx = \frac12\int_1^2\ln t\,dt = \frac12[t\ln t - t]_1^2 = \frac12(2\ln2 - 1) = \ln2 - \frac12$. The limit is $e^{-4\ln2 + 2} = \frac{e^2}{16}$. Answer: (D).
69	Differential EquationsSeparation of variables	Class 12: Differential Equations	Single correctMedium
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Put $u = y + 2\sin x - 5$: $\frac{du}{dx} = \frac{dy}{dx} + 2\cos x = 2xu$. Separating: $\frac{du}{u} = 2x\,dx$, so $u = u(0)e^{x^2}$, with $u(0) = 7 + 0 - 5 = 2$. At $x = \pi$: $y(\pi) - 5 = 2e^{\pi^2}$ ($\sin\pi = 0$), so $y(\pi) = 2e^{\pi^2} + 5$. Answer: (B).
70	Conic SectionsParabola	Class 11: Conic Sections	Single correctMedium
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = 8x$: focal length 2, directrix $x = -2$. P(2, $-4$) is $(2t^2, 4t)$ with $t = -1$. Tangent: $ty = x + 2t^2$, i.e. $-y = x + 2$; at $x = -2$, $y = 0$: A($-2$, 0). Normal: slope $-t = 1$: $y + 4 = x - 2$, i.e. $y = x - 6$; at $x = -2$, $y = -8$: B($-2$, $-8$). The tangent and normal are perpendicular at P and $PA = PB = 4\sqrt2$, so in square AQBP the diagonals AB and PQ share their mid-point ($-2$, $-4$). Q = $(2(-2) - 2, 2(-4) - (-4)) = (-6, -4)$, so $2a + b = -12 - 4 = -16$. Answer: (B).
71	Differential EquationsLinear differential equations	Class 12: Differential Equations	Single correctEasy
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $f(x) + xf'(x) = \frac{d}{dx}[xf(x)] = x^2$, so $xf(x) = \frac{x^3}{3} + C$. At $(-2, 2)$: $-4 = -\frac83 + C$, $C = -\frac43$. $3xf(x) = x^3 - 4$, i.e. $x^3 - 3xf(x) - 4 = 0$. Answer: (A).
72	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctMedium
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Planes through the line: $(x - y - z - 1) + \lambda(2x + y - 3z + 4) = 0$, i.e. $(1+2\lambda)x + (\lambda-1)y - (1+3\lambda)z + (4\lambda - 1) = 0$. Distance from O: $\frac{(4\lambda-1)^2}{(1+2\lambda)^2 + (\lambda-1)^2 + (1+3\lambda)^2} = \frac{2}{21}$. $21(16\lambda^2 - 8\lambda + 1) = 2(14\lambda^2 + 8\lambda + 3)$, i.e. $308\lambda^2 - 184\lambda + 15 = 0$, so $\lambda = \frac12$ or $\frac{15}{154}$. $\lambda = \frac12$: $2x - \frac12y - \frac52z + 1 = 0$, i.e. $4x - y - 5z + 2 = 0$ (check: $\frac{2}{\sqrt{42}} = \sqrt{\frac{2}{21}}$). The other value gives a plane not among the options. Answer: (D).
73	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctEasy
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The line through $(1, -2, 3)$ with direction cosines $\left(\frac27, \frac37, -\frac67\right)$: a point at distance $r$ is $\left(1 + \frac{2r}{7}, -2 + \frac{3r}{7}, 3 - \frac{6r}{7}\right)$. On the plane: $1 + \frac{2r}{7} + 2 - \frac{3r}{7} + 3 - \frac{6r}{7} = 5$, so $6 - r = 5$, $r = 1$. Answer: (C).
74	TrigonometryTrigonometric identities	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctMedium
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Cross-multiply: $\sin A\sin(C - B) = \sin B\sin(A - C)$. In a triangle $\sin A = \sin(C + B)$ and $\sin B = \sin(A + C)$, and $\sin(X+Y)\sin(X-Y) = \sin^2X - \sin^2Y$: $\sin^2C - \sin^2B = \sin^2A - \sin^2C$, so $2\sin^2C = \sin^2A + \sin^2B$. By the sine rule, $2c^2 = a^2 + b^2$: $b^2, c^2, a^2$ are in A.P. Answer: (B).
75	ProbabilityAddition and multiplication theorems	Class 11: Probability; Class 12: Probability	Single correctMedium
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability A total of 7 comes from (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3): each pair is a face and its opposite. For the biased pair of opposite faces the two orders give $2\left(\frac16 - x\right)\left(\frac16 + x\right) = 2\left(\frac{1}{36} - x^2\right)$; the other two pairs give $4\times\frac{1}{36}$. $P = \frac{6}{36} - 2x^2 = \frac16 - 2x^2 = \frac{13}{96}$, so $2x^2 = \frac{16 - 13}{96} = \frac{1}{32}$, $x^2 = \frac{1}{64}$, $x = \frac18$. Answer: (D).
76	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctMedium
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $s = \sin^{-1}x$, $c = \cos^{-1}x$, with $s + c = \frac{\pi}{2}$. $s^2 - c^2 = (s - c)(s + c) = \frac{\pi}{2}(s - c) = a$, so $s - c = \frac{2a}{\pi}$ and $s = \frac{\pi}{4} + \frac{a}{\pi}$. $2x^2 - 1 = 2\sin^2s - 1 = -\cos2s = -\cos\left(\frac{\pi}{2} + \frac{2a}{\pi}\right) = \sin\left(\frac{2a}{\pi}\right)$. Answer: (B).
77	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	Single correctMedium
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $x^2 - 44x + 484 = (22 - x)^2$, so the integrand is $f(x) = \frac{\ln x^2}{\ln x^2 + \ln(22-x)^2}$. With $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ and $a + b = 22$: $f(22 - x) = \frac{\ln(22-x)^2}{\ln(22-x)^2 + \ln x^2}$. Adding: $2I = \int_6^{16}1\,dx = 10$, so $I = 5$. Answer: (D).
78	Sequences and SeriesGeometric progressions	Class 11: Sequences and Series	Single correctMedium
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The general term is $\frac{2n-1}{n}x^n = \left(2 - \frac1n\right)x^n$ for $n \ge 2$. $\sum_{n\ge2}2x^n = \frac{2x^2}{1-x}$ and $\sum_{n\ge2}\frac{x^n}{n} = -\log_e(1-x) - x$. Sum $= \frac{2x^2}{1-x} + x + \log_e(1-x) = \frac{x^2 + x}{1-x} + \log_e(1-x) = x\left(\frac{1+x}{1-x}\right) + \log_e(1-x)$. Answer: (A).
79	Mathematical ReasoningStatements and logical connectives	No matching NCERT chapter	Single correctEasy
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter The implication can be false only if the antecedent is true and r is false. A true antecedent needs p true; then $p \to q$ forces q true; then $q \to r$ forces r true. So r cannot be false when the antecedent holds: the statement is always true, a tautology (the chain rule of inference). Answer: (A).
80	Binomial TheoremSimple applications	Class 11: Binomial Theorem	Single correctEasy
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $\sum_k {}^{20}C_k\,{}^{20}C_{20-k}$ is the coefficient of $x^{20}$ in $(1+x)^{20}(1+x)^{20} = (1+x)^{40}$, which is ${}^{40}C_{20}$ (and ${}^{20}C_{20-k} = {}^{20}C_k$). Answer: (C).
81	Sets, Relations and FunctionsUnion, intersection and complement	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	NumericalMedium
A medium numerical question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions A: $x < 1$ or $x > 3$. B: $x^2 > 4$, i.e. $x < -2$ or $x > 2$. C: $x \le 2$ or $x \ge 6$. $A \cap B \cap C = (-\infty, -2) \cup [6, \infty)$, so $(A \cap B \cap C)^c = [-2, 6)$. Its integers: $-2, -1, 0, 1, 2, 3, 4, 5$, eight of them, giving $2^8 = 256$ subsets. Answer: 256.
82	Matrices and DeterminantsConsistency of linear systems	Class 12: Matrices; Class 12: Determinants	NumericalMedium
A medium numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The determinant must vanish: $\begin{vmatrix}2&1&-1\\1&-1&-1\\3&3&\beta\end{vmatrix} = 2(-\beta + 3) - (\beta + 3) - (3 + 3) = -3\beta - 3 = 0$, so $\beta = -1$. Then the third equation must be a combination of the first two: $2(2, 1, -1) - (1, -1, -1) = (3, 3, -1)$, so its right side must be $2\times3 - \alpha = 3$, $\alpha = 3$. $\alpha + \beta - \alpha\beta = 3 - 1 + 3 = 5$. Answer: 5.
83	Permutations and CombinationsFundamental principle of counting	Class 11: Permutations and Combinations	NumericalMedium
A medium numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations A six-digit palindrome is $\overline{abccba} = 100001a + 10010b + 1100c$, and $100001 = 11\times9091$, $10010 = 11\times910$, $1100 = 11\times100$: every such number is divisible by 11. Divisibility by 5 needs the last digit $a$ to be 0 or 5; $a \ne 0$ (it is also the first digit), so $a = 5$. b and c are free: $10\times10 = 100$. Answer: 100.
84	Limits, Continuity and DifferentiabilityDerivatives up to order two	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	NumericalHard
A hard numerical question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $t = y^{1/4}$: $t + \frac1t = 2x$, so $t = x \pm \sqrt{x^2 - 1}$ and $y = \left(x \pm \sqrt{x^2-1}\right)^4$. Take $y = (x + \sqrt{x^2-1})^4$: $y' = 4(x + \sqrt{x^2-1})^3\left(1 + \frac{x}{\sqrt{x^2-1}}\right) = \frac{4y}{\sqrt{x^2-1}}$, so $\sqrt{x^2-1}\,y' = 4y$. Squaring: $(x^2-1)y'^2 = 16y^2$. Differentiating: $2xy'^2 + 2(x^2-1)y'y'' = 32yy'$, and dividing by $2y'$: $(x^2-1)y'' + xy' - 16y = 0$. (The other sign gives the same equation.) So $\alpha = 1$, $\beta = -16$, and $\|\alpha - \beta\| = 17$. Answer: 17.
85	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	NumericalMedium
A medium numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f(x) = 3x^4 + 4x^3 - 12x^2 + 4$, $f'(x) = 12x(x+2)(x-1)$: critical points $-2, 0, 1$. $f(-2) = 48 - 32 - 48 + 4 = -28$, $f(0) = 4$, $f(1) = 3 + 4 - 12 + 4 = -1$, and $f \to +\infty$ as $x \to \pm\infty$. The sign changes give one root in each of $(-\infty, -2)$, $(-2, 0)$, $(0, 1)$, $(1, \infty)$: 4 distinct real roots. Answer: 4.
86	CirclesRadius and centre	Class 11: Conic Sections	NumericalHard
A hard numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections $r^2 = \frac{p^2}{4} + \frac{(1-p)^2}{4} - 5 = \frac{2p^2 - 2p + 1}{4} - 5$. $0 < r^2 \le 25$ gives $20 < 2p^2 - 2p + 1 \le 120$, i.e. $9.5 < p^2 - p \le 59.5$. For $p = \sqrt q \ge 0$: $9.5 < q - \sqrt q \le 59.5$ gives $\sqrt q > \frac{1 + \sqrt{39}}{2} \approx 3.62$ and $\sqrt q \le \frac{1 + \sqrt{239}}{2} \approx 8.23$, i.e. $14 \le q \le 67$. For $p = -\sqrt q$: $9.5 < q + \sqrt q \le 59.5$ gives $\sqrt q > \frac{-1 + \sqrt{39}}{2} \approx 2.62$ and $\sqrt q \le \frac{-1 + \sqrt{239}}{2} \approx 7.23$, i.e. $7 \le q \le 52$. Together $7 \le q \le 67$: $67 - 7 + 1 = 61$ integers. Answer: 61.
87	Conic SectionsEllipse	Class 11: Conic Sections	NumericalMedium
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Tangent at $(b\cos\theta, 2a\sin\theta)$: $\frac{x\cos\theta}{b} + \frac{y\sin\theta}{2a} = 1$, with intercepts $\frac{b}{\cos\theta}$ and $\frac{2a}{\sin\theta}$. Area $= \frac12\cdot\frac{b}{\cos\theta}\cdot\frac{2a}{\sin\theta} = \frac{2ab}{\sin2\theta} \ge 2ab$, with equality at $\theta = \frac{\pi}{4}$. So $k = 2$. Answer: 2.
88	StatisticsStandard deviation and variance	Class 11: Statistics	NumericalEasy
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Variance of the first n natural numbers: $\frac{\sum r^2}{n} - \left(\frac{\sum r}{n}\right)^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} = \frac{n^2 - 1}{12}$. $\frac{n^2-1}{12} = 14$ gives $n^2 = 169$, $n = 13$ (odd, as required). Answer: 13.
89	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 $\vec{b}\times\vec{c} = (-9 - 2\beta)\hat{i} + (3 - \beta)\hat{j} + 5\hat{k}$. $(9 + 2\beta)^2 + (3 - \beta)^2 + 25 = 75$ gives $5\beta^2 + 30\beta + 40 = 0$, $\beta = -2$ or $-4$. $\vec{a}\cdot\vec{b} = 1 + 15 + \alpha\beta = 0$, so $\alpha = -\frac{16}{\beta}$: $\alpha = 8$ or $4$. $\|\vec{a}\|^2 = 26 + \alpha^2 = 90$ or $42$. The greatest is 90. Answer: 90.
90	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 With $u = x + \frac12$, $k^2 = \frac34$: $\int\frac{du}{(u^2 + k^2)^2} = \frac{u}{2k^2(u^2 + k^2)} + \frac{1}{2k^3}\tan^{-1}\frac{u}{k}$ (put $u = k\tan\theta$). $\frac{u}{2k^2(u^2+k^2)} = \frac{x + \frac12}{\frac32(x^2+x+1)} = \frac13\cdot\frac{2x+1}{x^2+x+1}$, so $b = \frac13$. $\frac{1}{2k^3} = \frac{1}{2\cdot\frac{3\sqrt3}{8}} = \frac{4}{3\sqrt3}$ and $\frac{u}{k} = \frac{2x+1}{\sqrt3}$, so $a = \frac{4}{3\sqrt3}$. $9(\sqrt3 a + b) = 9\left(\frac43 + \frac13\right) = 15$. Answer: 15.

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