JEE (Main) · 2021

JEE (Main) 2021 — 26 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. Electronic Devices2
2. Properties of Bulk Matter2
3. Optics2
19 easy · 11 medium · 0 hard
Chemistry
30 Q
1. d- and f-Block Elements3
2. Principles Related to Practical Chemistry3
3. Hydrocarbons3
24 easy · 6 medium · 0 hard
Mathematics
30 Q
1. Limits, Continuity and Differentiability3
2. Integral Calculus3
3. Complex Numbers2
12 easy · 15 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	Experimental SkillsVernier calipers and screw gauge	No matching NCERT chapter	Single correctEasy
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{0.5\ \text{mm}}{50} = 0.01$ mm. With the jaws closed the 5th division is on the reference line, so the zero error is $+5\times0.01 = +0.05$ mm. Observed reading $= 5 + 20\times0.01 = 5.20$ mm. True reading $= 5.20 - 0.05 = 5.15$ mm. Answer: (C).
2	Behaviour of Perfect Gas and Kinetic TheoryKinetic theory	Class 11: Kinetic Theory	Single correctEasy
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$, so at the same temperature the lighter gas is faster. Molar masses: $\mathrm{H_2} = 2$, $\mathrm{O_2} = 32$, $\mathrm{CO_2} = 44$ g/mol, hence $V_H > V_O > V_C$. Answer: (C).
3	Electronic DevicesLogic gates	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctEasy
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Each input goes to a NAND gate with both its inputs joined, which acts as a NOT gate: $X = \bar{A}$, $Y = \bar{B}$. The AND gate gives $Z = \bar{A}\cdot\bar{B} = \overline{A + B}$ (De Morgan), which is the NOR operation. Answer: (D).
4	ThermodynamicsFirst law	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat supplied per second $= \frac{6000}{60} = 100$ W; work done by the system per second $= 90$ W. By the first law, $\frac{dU}{dt} = 100 - 90 = 10$ W. Time $= \frac{2.5\times10^3}{10} = 250$ s $= 2.5\times10^2$ s. Answer: (B).
5	Atoms and NucleiHydrogen spectrum	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\nu \propto \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. For $3\to1$: $1 - \frac{1}{9} = \frac{8}{9}$; for $2\to1$: $1 - \frac{1}{4} = \frac{3}{4}$. $\nu_{21} = 2.92\times10^{15}\times\frac{3/4}{8/9} = 2.92\times10^{15}\times\frac{27}{32} = 2.46\times10^{15}$ Hz. Answer: (D).
6	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctMedium
A medium single correct question on 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 Capillary rise $h = \frac{2T\cos\theta}{r\rho g}$, so the difference is $\Delta h = \frac{2T}{\rho g}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$ with $r_1 = 2.5$ mm, $r_2 = 4.0$ mm. $\frac{2T}{\rho g} = \frac{2\times7.3\times10^{-2}}{10^3\times10} = 1.46\times10^{-5}\ \mathrm{m^2}$. $\frac{1}{r_1} - \frac{1}{r_2} = 400 - 250 = 150\ \mathrm{m^{-1}}$. $\Delta h = 1.46\times10^{-5}\times150 = 2.19\times10^{-3}$ m $= 2.19$ mm. Answer: (D).
7	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 $eV_0 = \frac{hc}{\lambda} - \phi$, so $\Delta V_0 = \frac{hc}{e}\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$. $\frac{hc}{e} = \frac{6.63\times10^{-34}\times3\times10^8}{1.6\times10^{-19}} = 1.243\times10^{-6}$ V m $= 1243$ V nm. $\Delta V_0 = 1243\left(\frac{1}{280} - \frac{1}{400}\right) = 1243\times\frac{120}{112000} \approx 1.33$ V. (At 400 nm the photon energy is about 3.1 eV, above the 2.5 eV work function, so emission still occurs.) Answer: (A).
8	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 rear-view mirror of a car is convex: $f = +10$ cm, object distance $u = -190$ cm. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{10} + \frac{1}{190} = \frac{20}{190}$, so $v = 9.5$ cm. Speed of image $= \left(\frac{v}{u}\right)^2\times$ speed of object $= \left(\frac{9.5}{190}\right)^2\times40 = \frac{40}{400} = 0.1\ \mathrm{ms^{-1}}$. Answer: (A).
9	Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctEasy
A easy single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Energy: $\frac{1}{2}LI^2 = 64 \Rightarrow L = \frac{2\times64}{64} = 2$ H. Power: $I^2R = 640 \Rightarrow R = \frac{640}{64} = 10\ \Omega$. Time constant $\tau = \frac{L}{R} = \frac{2}{10} = 0.2$ s. Answer: (A).
10	Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctEasy
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Average power is maximum at resonance, $X_C = X_L = 250\pi\ \Omega$. $C = \frac{1}{2\pi f X_C} = \frac{1}{2\pi\times50\times250\pi} = \frac{1}{25000\pi^2} = \frac{1}{25000\times10} = 4\times10^{-6}$ F $= 4\ \mu$F. Answer: (D).
11	Magnetic Effects of Current and MagnetismBiot-Savart law	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctMedium
A medium single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter On the axis, $B = \frac{\mu_0 I a^2}{2(a^2 + r^2)^{3/2}} = B_0\left(1 + \frac{r^2}{a^2}\right)^{-3/2}$, where $B_0 = \frac{\mu_0 I}{2a}$ is the field at the centre. For $r \ll a$: $B \approx B_0\left(1 - \frac{3}{2}\frac{r^2}{a^2}\right)$. Fractional change $\frac{B_0 - B}{B_0} = \frac{3}{2}\frac{r^2}{a^2}$. Answer: (D).
12	Current ElectricityCells, emf and internal resistance	Class 12: Current Electricity	Single correctMedium
A medium single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity The left side of the network is one node A and the right side one node B. Between A and B there are four paths: - the top 4 $\Omega$; - the bottom 8 $\Omega$; - the diagonal from the top-left corner: 4 $\Omega$ then 8 $\Omega$ (the wires cross without joining, shown by the hump), 12 $\Omega$; - the diagonal from the bottom-left corner: 2 $\Omega$ then 4 $\Omega$, 6 $\Omega$. $\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{6} = \frac{6 + 3 + 2 + 4}{24} = \frac{15}{24}$, so $R = 1.6\ \Omega$. Current $I = \frac{2.2}{1.6 + 0.6} = 1$ A. Power dissipated in the whole circuit (including the internal resistance) $= EI = 2.2\times1 = 2.2$ W. Answer: (A).
13	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Inside the solid metal sphere ($r < R$) the field is zero. For $R < r < a$: by Gauss's law $E = \frac{q}{4\pi\epsilon_0 r^2}$, falling with $r$. Inside the metal of the shell ($a < r < b$) the field is zero (charge $-q$ is induced on its inner surface). For $r > b$: the shell's outer surface carries $+q$, so $E = \frac{q}{4\pi\epsilon_0 r^2}$ again. Only the third graph shows zero, fall, zero, fall. Answer: (C).
14	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 $C = \frac{K\epsilon_0 A}{d}$ and $R = \frac{\rho d}{A}$, so $RC = \rho K\epsilon_0$. $R = \frac{\rho K\epsilon_0}{C} = \frac{200\times50\times8.85\times10^{-12}}{2\times10^{-12}} = 4.425\times10^4\ \Omega$. Leakage current $I = \frac{V}{R} = \frac{40}{4.425\times10^4} \approx 9.0\times10^{-4}$ A $= 0.9$ mA. Answer: (B).
15	GravitationGravitational potential energy	Class 11: Gravitation	StatementsEasy
A easy statements question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation Inside a uniform shell the field is zero everywhere, so (a) and (c) are true. The potential is constant inside, $V = -\frac{GM}{R}$, which is the same everywhere (d) but not zero, so (b) is false and (e) is false. Answer: (C).
16	Laws of MotionConservation of linear momentum	Class 11: Laws of Motion	Single correctEasy
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Thrust minus weight gives the acceleration: $u\frac{dm}{dt} - mg = ma$. $\frac{dm}{dt} = \frac{m(g + a)}{u} = \frac{1000\times(10 + 20)}{500} = 60\ \mathrm{kg\ s^{-1}}$. Answer: (A).
17	Current ElectricityResistivity	Class 12: Current Electricity	Single correctMedium
A medium single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Area $A = \pi(10^{-3})^2 = 3.14\times10^{-6}\ \mathrm{m^2}$; $\rho_1 = 12\times10^{-8}\ \Omega$ m, $\rho_2 = 51\times10^{-8}\ \Omega$ m. In parallel, $R = \frac{L}{A}\cdot\frac{\rho_1\rho_2}{\rho_1 + \rho_2} = \frac{L}{A}\times\frac{12\times51}{63}\times10^{-8} = \frac{L}{A}\times9.71\times10^{-8}$. $L = \frac{3\times3.14\times10^{-6}}{9.71\times10^{-8}} \approx 97$ m. Answer: (B).
18	KinematicsScalars and vectors	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctMedium
A medium single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Take each magnitude as 1. $\vec{OA}$ at $30^\circ$ above +x: $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$. $\vec{OB}$ at $60^\circ$ below +x: $\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$. $\vec{OC}$ at $45^\circ$ above −x: $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. $\vec{OA} + \vec{OB} - \vec{OC} = \left(\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}},\ \frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\right) = \frac{1}{2}\left(1 + \sqrt{3} + \sqrt{2},\ 1 - \sqrt{3} - \sqrt{2}\right)$. Angle with x-axis: $\tan^{-1}\frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}}$. Answer: (A).
19	Electronic DevicesSemiconductors	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	StatementsEasy
A easy statements question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Statement I: a pentavalent dopant donates an extra electron per atom, so the electron density rises. True. Statement II: each donated electron leaves a fixed positive donor ion, so the n-type crystal as a whole is electrically neutral. False. Answer: (C).
20	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	Single correctEasy
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $[E] = ML^2T^{-2}$, $[L]^2 = (ML^2T^{-1})^2 = M^2L^4T^{-2}$, $[M]^{-5} = M^{-5}$, $[G]^{-2} = (M^{-1}L^3T^{-2})^{-2} = M^2L^{-6}T^4$. Powers: M: $1 + 2 - 5 + 2 = 0$; L: $2 + 4 - 6 = 0$; T: $-2 - 2 + 4 = 0$. P is dimensionless, $[M^0L^0T^0]$. Answer: (D).
21	Work, Energy and PowerConservation of energy	Class 11: Work, Energy and Power	NumericalEasy
A easy numerical question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Mass per metre = 1 kg/m. Initially 1 m (1 kg) hangs, centre 0.5 m below the table: $U_i = -1\times10\times0.5 = -5$ J. Finally all 3 m hang, centre 1.5 m below: $U_f = -3\times10\times1.5 = -45$ J. The table is smooth, so $k = U_i - U_f = 40$ J. Answer: 40.
22	Motion of System of Particles and Rigid BodyMoment of inertia	Class 11: Systems of Particles and Rotational Motion	NumericalMedium
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Handle: a rod of length $6r$ from A; its centre is $3r$ from A, i.e. $2.5r$ from the axis. $I_{rod} = \frac{M(6r)^2}{12} + M(2.5r)^2 = 3Mr^2 + 6.25Mr^2 = 9.25Mr^2$. Ring: diameter $2r$ (radius $r$), centre at $6r + r = 7r$ from A, i.e. $6.5r$ from the axis. About a diameter (the axis is in the plane of the ring) $I = \frac{Mr^2}{2}$, so $I_{ring} = \frac{Mr^2}{2} + M(6.5r)^2 = 0.5Mr^2 + 42.25Mr^2 = 42.75Mr^2$. Total $I = 9.25Mr^2 + 42.75Mr^2 = 52Mr^2$. Answer: 52.
23	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalEasy
A easy numerical 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 Excess pressure inside the outer bubble: $\frac{4T}{6}$; inside the inner bubble, a further $\frac{4T}{3}$. Total excess over atmosphere: $4T\left(\frac{1}{3} + \frac{1}{6}\right) = \frac{4T}{2}$. An equivalent single bubble has $\frac{4T}{R} = \frac{4T}{2}$, so $R = 2$ cm. Answer: 2.
24	Oscillations and WavesStanding waves and beats	Class 11: Oscillations; Class 11: Waves	NumericalEasy
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Nodes are where $\cos(1.57x) = 0$, i.e. $1.57x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$ With $1.57 = \frac{\pi}{2}$, the first node for $x > 0$ is at $x = 1$ cm. Answer: 1.
25	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 $m_1$ at O points perpendicular to OP, so P lies on its equatorial line. Field at P: $B = \frac{\mu_0}{4\pi}\frac{m_1}{r^3} = 10^{-7}\times\frac{1}{1^3} = 10^{-7}$ T, antiparallel to $m_1$. $m_2$ points along OP, perpendicular to this field, so $\tau = m_2B\sin90^\circ = 1\times10^{-7}$ Nm. Answer: 1.
26	Electromagnetic WavesProperties of electromagnetic waves	Class 12: Electromagnetic Waves	NumericalMedium
A medium numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Intensity $I = \frac{1}{2}\epsilon_0 cE_0^2$. For a perfect reflector the pressure is $P = \frac{2I}{c} = \epsilon_0E_0^2$. $P = 8.85\times10^{-12}\times(200)^2 = 3.54\times10^{-7}\ \mathrm{N/m^2} = \frac{354}{10^9}\ \mathrm{N/m^2}$. (The area and the 10 minute exposure do not change the pressure.) Answer: 354.
27	OpticsInterference and diffraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalEasy
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics First bright fringe: $y = \frac{\lambda D}{d}$, so $\lambda = \frac{yd}{D}$. $\lambda_v = \frac{2.0\times10^{-3}\times0.3\times10^{-3}}{1.5} = 400$ nm; $\lambda_r = \frac{3.5\times10^{-3}\times0.3\times10^{-3}}{1.5} = 700$ nm. Difference $= 300$ nm. Answer: 300.
28	Communication SystemsAmplitude modulation and bandwidth	No matching NCERT chapter	NumericalEasy
A easy numerical question on amplitude modulation and bandwidth, from Communication Systems. NCERT: No matching NCERT chapter The modulating signal is $\cos(\omega_m t)$ with $\omega_m = 12560$ rad/s. $f_m = \frac{12560}{2\pi} = \frac{12560}{6.28} = 2000$ Hz $= 2$ kHz. Answer: 2.
29	KinematicsMotion in a straight line	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	NumericalMedium
A medium numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Let the first ball be thrown at $t = 0$ and the second at $t = 3$ s. $y_1 = 35t - 5t^2$, $y_2 = 35(t - 3) - 5(t - 3)^2$. $y_1 - y_2 = 150 - 30t$. Treating the balls as points, they meet when this is zero: $t = 5$ s, at $y = 35\times5 - 5\times25 = 50$ m. (Allowing for the 5 cm radii, the centres are 10 cm apart at contact, $t = 4.9967$ s, centres at 50.05 m and 49.95 m; the point of contact is still at 50 m.) Answer: 50.
30	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 and detector each move away at 20 m/s: $f' = f\,\frac{v - v_d}{v + v_s} = f\times\frac{340 - 20}{340 + 20} = f\times\frac{320}{360}$. $f = 1800\times\frac{360}{320} = 2025$ Hz. Answer: 2025.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
31	Solid StatePoint defects	No matching NCERT chapter	StatementsEasy
A easy statements question on point defects, from Solid State. NCERT: No matching NCERT chapter Statement I: in a Frenkel defect an ion (usually the smaller cation) leaves its site, creating a vacancy, and occupies an interstitial site. True. Statement II: F-centres are anion vacancies occupied by electrons, produced in metal-excess (non-stoichiometric) defects, e.g. NaCl heated in Na vapour. A Frenkel defect does not create F-centres or colour. False. Answer: (C).
32	Structure of AtomBohr model	Class 11: Structure of Atom	StatementsEasy
A easy statements question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In Bohr's model $v = 2.18\times10^6\,\frac{Z}{n}$ m/s. Statement I: $v \propto Z$, so the speed decreases when the nuclear charge decreases. False. Statement II: $v \propto \frac{1}{n}$, so the speed increases as $n$ decreases. True. Answer: (D).
33	ElectrochemistryConductance	Class 12: Electrochemistry	StatementsEasy
A easy statements question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Statement I: limiting molar conductivities are additive over ions. For KCl, Λ°m = λ°(K⁺) + λ°(Cl⁻) ≈ 73.5 + 76.3 ≈ 149.9 S cm² mol⁻¹. For CH₃COOH, Λ°m = λ°(H⁺) + λ°(CH₃COO⁻) ≈ 349.6 + 40.9 ≈ 390.5 S cm² mol⁻¹, because H⁺ is exceptionally mobile. So acetic acid's limiting molar conductivity is the higher one, and Statement I is false. (Being a weak electrolyte lowers its molar conductivity at ordinary concentrations, not its limiting value.) Statement II: on dilution, molar conductivity increases for every electrolyte (more ions per mole for a weak one, less interionic drag for a strong one). So it decreases with increase in concentration, and Statement II is false. Answer: (B).
34	Surface ChemistryPhysisorption and chemisorption	No matching NCERT chapter	Single correctEasy
A easy single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Gas molecules lose freedom of motion when they are held on the surface, so $\Delta S < 0$. For adsorption to be spontaneous, $\Delta G = \Delta H - T\Delta S < 0$, which with $\Delta S < 0$ requires $\Delta H < 0$ (adsorption is exothermic). Answer: (D).
35	Isolation of MetalsSmelting and refining	No matching NCERT chapter	StatementsMedium
A medium statements question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Statement I: the Ellingham diagram plots $\Delta_f G^\circ$ of oxides against temperature; a metal (or C, CO) whose line lies lower can reduce the oxide above it, so it is used to choose the reducing agent. True. Statement II: the slope of each line is $-\Delta S$; $\Delta S$ is roughly constant along a line (it changes only at a phase change) and does not increase from left to right with temperature. False. Answer: (C).
36	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 Clark's method (slaked lime) removes only temporary hardness; Calgon and permutit (zeolite) exchange or sequester $\mathrm{Ca^{2+}}$ and $\mathrm{Mg^{2+}}$ but leave $\mathrm{Na^+}$ ions in the water. Synthetic ion-exchange resins (a cation exchanger in the $\mathrm{H^+}$ form followed by an anion exchanger in the $\mathrm{OH^-}$ form) remove all cations and anions, giving deionised water. Answer: (B).
37	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 $\mathrm{Ca(OH)_2 + CO_2 \rightarrow CaCO_3\downarrow + H_2O}$ (lime water turns milky). With excess $\mathrm{CO_2}$: $\mathrm{CaCO_3 + CO_2 + H_2O \rightarrow Ca(HCO_3)_2}$ (soluble, milkiness disappears). Answer: (A).
38	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Interhalogens are more reactive than the halogens (except fluorine) because the X–X' bond is weaker than the X–X bond of the parent halogen; so ClF is more reactive than $\mathrm{Cl_2}$, and (A) is incorrect. (B) $\mathrm{F_2}$, with its very weak F–F bond, is more reactive than ClF: correct. (C) $\mathrm{ClF + H_2O \rightarrow HOCl + HF}$ (the more electronegative F goes to H): correct. (D) $\mathrm{F_2}$ ($E^\circ = 2.87$ V) is a stronger oxidant than $\mathrm{Cl_2}$ (1.36 V): correct. Answer: (A).
39	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 $\mathrm{Zn^{2+}}$ ($3d^{10}$) and $\mathrm{Cu^+}$ ($3d^{10}$) have no unpaired d electrons and give colourless species; $\mathrm{Cu_2Cl_2}$ and AgCl are moreover insoluble in water. (The nitrogen atmosphere prevents Cu(I) from being oxidised by air.) $\mathrm{Cu^{2+}}$ ($3d^9$) shows d–d transitions, so $\mathrm{CuCl_2}$ gives a blue-green solution. Answer: (B).
40	Environmental ChemistryAir and water pollution	No matching NCERT chapter	Single correctEasy
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Tooth enamel is hydroxyapatite, $[3\mathrm{Ca_3(PO_4)_2}\cdot\mathrm{Ca(OH)_2}]$. Fluoride ions in water convert it into the harder fluorapatite, $[3\mathrm{Ca_3(PO_4)_2}\cdot\mathrm{CaF_2}]$. Answer: (C).
41	Principles Related to Practical ChemistryDetection of elements and functional groups	No matching NCERT chapter	Single correctEasy
A easy single correct question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter In the test for sulphur, sulphide ion reacts with sodium nitroprusside: $\mathrm{S^{2-} + [Fe(CN)_5NO]^{2-} \rightarrow [Fe(CN)_5NOS]^{4-}}$, which is violet. $\mathrm{Fe_4[Fe(CN)_6]_3}$ is Prussian blue, $[\mathrm{Fe(CN)_6}]^{4-}$ is pale yellow, and the iron(III) thiocyanate complexes are blood red. Answer: (B).
42	HydrocarbonsAlkanes	Class 11: Hydrocarbons	Single correctEasy
A easy single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Free-radical bromination is highly selective for the most stable radical; isobutane gives the tertiary radical $\mathrm{(CH_3)_3C^\bullet}$. With excess isobutane, monobromination predominates, giving 2-bromo-2-methylpropane, $\mathrm{(CH_3)_3CBr}$ (more than 99%). Answer: (B).
43	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The diene is isoprene, $\mathrm{CH_2{=}C(CH_3){-}CH{=}CH_2}$, used in excess so only one HBr adds. Protonation at C-1 gives the most stable cation, tertiary and allylic: $\mathrm{(CH_3)_2\overset{+}{C}{-}CH{=}CH_2 \leftrightarrow (CH_3)_2C{=}CH{-}\overset{+}{C}H_2}$. Bromide attack at the tertiary end gives the 1,2-adduct 3-bromo-3-methylbut-1-ene; attack at the primary end gives the 1,4-adduct 1-bromo-3-methylbut-2-ene, which has the more substituted (trisubstituted) double bond and is the more stable, thermodynamic product. Under ordinary conditions the 1,4-adduct is the major product: $\mathrm{(CH_3)_2C{=}CH{-}CH_2Br}$. Answer: (B).
44	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 Aryl halides (the ring Cl or Br in I, II, III and IV) are not hydrolysed by aqueous NaOH because the C–X bond has partial double-bond character. Only IV has a benzylic C–I bond, which is hydrolysed readily: $\mathrm{ArCH_2I + OH^- \rightarrow ArCH_2OH + I^-}$. After acidifying with dil. $\mathrm{HNO_3}$, $\mathrm{Ag^+}$ gives a yellow precipitate of AgI. Answer: (D).
45	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctEasy
A easy single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Acetophenone is a methyl ketone, so $\mathrm{Br_2/KOH}$ (hypobromite) gives the haloform reaction. The methyl group is tribrominated to $\mathrm{C_6H_5COCBr_3}$, which is cleaved by $\mathrm{OH^-}$ to give $\mathrm{C_6H_5COO^-K^+}$ (A, potassium benzoate) and $\mathrm{CHBr_3}$ (B, bromoform). Answer: (A).
46	Alcohols, Phenols and EthersReactions of phenols	Class 12: Alcohols, Phenols and Ethers	Single correctEasy
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers In water phenol is partly ionised and strongly activated, so excess bromine water substitutes every ortho and para position: A = 2,4,6-tribromophenol (white precipitate). In the non-polar solvent $\mathrm{CS_2}$ at low temperature the reaction is controlled and gives monobromophenols, mainly the para isomer: B = 4-bromophenol. Answer: (C).
47	Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $\mathrm{SOCl_2}$ added to methanol generates HCl (and methyl chlorosulphite), the standard way of esterifying an amino acid: the COOH becomes $\mathrm{CO_2CH_3}$. The HCl formed protonates the basic $\alpha$-amino group, so the product is isolated as the ester hydrochloride, $\mathrm{NH_2\cdot HCl}$. The indole N–H is not basic (its lone pair is part of the aromatic sextet), so it is not protonated, and the aromatic ring is not chlorinated. Answer: (B).
48	HydrocarbonsElectrophilic substitution and directive influence	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The COOH and $\mathrm{NO_2}$ groups are meta to each other, so the carboxyl group (a meta director) must be in place before nitration. (B): benzene $\rightarrow$ bromobenzene $\rightarrow$ benzonitrile (cyanide replaces Br) $\rightarrow$ benzoic acid (hydrolysis) $\rightarrow$ nitration meta to COOH gives 3-nitrobenzoic acid. (A) fails: a Grignard reagent cannot be made from an aryl halide that carries a nitro group, which reacts with it. (C) and (D) fail: nitrating bromobenzene puts $\mathrm{NO_2}$ ortho/para to Br, so the final acid would be the 2- or 4-nitro isomer (and (D) also has the nitro–Grignard problem). Answer: (B).
49	PolymersNylon, Teflon, Bakelite and PVC	No matching NCERT chapter	Single correctEasy
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Novolac is the linear phenol–formaldehyde polymer; heating it with more formaldehyde cross-links the chains into the thermosetting resin Bakelite. Answer: (B).
50	Principles Related to Practical ChemistryTitrimetric analysis	No matching NCERT chapter	StatementsEasy
A easy statements question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Statement I: a strong acid–weak base titration has its equivalence point in the acidic range (pH about 4–6), where methyl orange (3.1–4.4) changes colour. True. Statement II: a weak acid–strong base titration has its equivalence point at pH about 8–9, which matches phenolphthalein (8.3–10), so phenolphthalein is suitable. False. Answer: (C).
51	SolutionsConcentration terms	Class 12: Solutions	NumericalEasy
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 kg of water: it holds 3.30 mol KCl $= 3.30\times74.5 = 245.85$ g. Mass of solution $= 1245.85$ g; volume $= \frac{1245.85}{1.20} = 1038.2$ mL. Molarity $= \frac{3.30}{1.0382} = 3.18$ mol $\mathrm{L^{-1}}$, nearest integer 3. Answer: 3.
52	Chemical Bonding and Molecular StructureVSEPR theory	Class 11: Chemical Bonding and Molecular Structure	NumericalEasy
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A T-shaped $\mathrm{AB_3}$ interhalogen (e.g. $\mathrm{ClF_3}$) has 5 electron pairs on A ($sp^3d$, trigonal bipyramidal): 3 bond pairs and 2 lone pairs in the equatorial positions. Answer: 2.
53	Chemical Bonding and Molecular StructureIonic bonding and lattice enthalpy	Class 11: Chemical Bonding and Molecular Structure	NumericalEasy
A easy numerical question on ionic bonding and lattice enthalpy, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\Delta_f H = \Delta_{sub}H + \Delta_i H + \frac{1}{2}\Delta_{bond}H + \Delta_{eg}H + \Delta_{lattice}H$ (formation of the lattice from gaseous ions). $-436.7 = 89.2 + 419.0 + 121.5 - 348.6 + \Delta_{lattice}H$ $\Delta_{lattice}H = -436.7 - 281.1 = -717.8$ kJ $\mathrm{mol^{-1}}$. Magnitude 717.8, nearest integer 718. Answer: 718.
54	Solutionsvan't Hoff factor	Class 12: Solutions	NumericalEasy
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Depression of freezing point $\propto i\times$ concentration. Ethanol is a non-electrolyte, $i = 1$. Taking each salt as a fully dissociated electrolyte: $\mathrm{Ba_3(PO_4)_2}$ $i = 5$, $\mathrm{Na_2SO_4}$ $i = 3$, KCl $i = 2$, $\mathrm{Li_3PO_4}$ $i = 4$. All four have $i > 1$ at the same molarity, so all four freeze below the ethanol solution. Answer: 4.
55	EquilibriumpH and buffers	Class 11: Equilibrium	NumericalEasy
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Millimoles: $\mathrm{NH_4^+} = 5.0\times0.0504 = 0.252$; $\mathrm{NH_3} = 2\times0.0210 = 0.042$. Basic buffer: $[\mathrm{OH^-}] = K_b\frac{n(\mathrm{NH_3})}{n(\mathrm{NH_4^+})} = 1.8\times10^{-5}\times\frac{0.042}{0.252} = 1.8\times10^{-5}\times\frac{1}{6} = 3\times10^{-6}$ M. Answer: 3.
56	Chemical KineticsOrder of reaction	Class 12: Chemical Kinetics	NumericalEasy
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics From (A) to (B), [NO] is tripled at constant $[\mathrm{H_2}]$ and the rate goes from $7\times10^{-9}$ to $2.1\times10^{-8}$, also tripled: $3^x = 3$, so $x = 1$. (From (B) to (C), $[\mathrm{H_2}]$ is quadrupled and the rate is quadrupled, so the reaction is also first order in $\mathrm{H_2}$.) Answer: 1.
57	d- and f-Block ElementsTransition element properties	Class 12: The d- and f-Block Elements	NumericalMedium
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements For a metal, $E^\circ(\mathrm{M^{n+}/M})$ is fixed by the cycle $\mathrm{M(s) \rightarrow M(g) \rightarrow M^{n+}(g) \rightarrow M^{n+}(aq)}$: sublimation enthalpy, ionisation enthalpy and hydration enthalpy. Electron gain enthalpy does not enter this cycle. So three of the listed properties, (A), (B) and (C), affect the reduction potential. Answer: 3.
58	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 Gd (Z = 64): $\mathrm{[Xe]4f^7 5d^1 6s^2}$; the half-filled $4f^7$ is especially stable. Removing the two 6s electrons gives $\mathrm{Gd^{2+}}$: $\mathrm{[Xe]4f^7 5d^1}$. Number of 4f electrons = 7. Answer: 7.
59	Principles Related to Practical ChemistryPreparation of inorganic and organic compounds	No matching NCERT chapter	NumericalEasy
A easy numerical question on preparation of inorganic and organic compounds, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Mohr's salt: $\mathrm{FeSO_4\cdot(NH_4)_2SO_4\cdot6H_2O}$, 6 water molecules. Potash alum: $\mathrm{KAl(SO_4)_2\cdot12H_2O}$, 12 water molecules per formula unit. Ratio $= \frac{6}{12} = 0.5 = 5\times10^{-1}$. Answer: 5.
60	BiomoleculesProteins	Class 12: Biomolecules	NumericalMedium
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules At pH 12.5 every acidic group is deprotonated and the N-terminal amino group is neutral ($\mathrm{NH_2}$). Negative charges: the C-terminal $\mathrm{COO^-}$ (Tyr) 1, the side-chain $\mathrm{COO^-}$ of Glu 1, the side-chain $\mathrm{COO^-}$ of Asp 1, and the phenoxide of Tyr (side-chain pKa about 10.1) 1. Total = 4. Answer: 4.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
61	Sets, Relations and FunctionsUnion, intersection and complement	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	Single correctEasy
A easy single correct 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 Let H and L be the percentages: $n(H) = 89$, $n(L) = 98$, $n(H\cup L) \le 100$. $K = n(H\cap L) = n(H) + n(L) - n(H\cup L) \ge 89 + 98 - 100 = 87$, and $K \le n(H) = 89$. So $87 \le K \le 89$. (A) contains 88, (B) contains 89, (D) contains 87, so K can lie in each of them. No element of {79, 81, 83, 85} lies in [87, 89], so K cannot belong to this set. Answer: (C).
62	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	Single correctMedium
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z = x + iy$. $\frac{z-1}{z+1} = \frac{(x-1+iy)(x+1-iy)}{(x+1)^2+y^2}$, whose real part is proportional to $x^2 + y^2 - 1$ and imaginary part to $2y$. Argument $\frac{\pi}{4}$ means real part = imaginary part > 0: $x^2 + y^2 - 1 = 2y$ with $y > 0$. $x^2 + (y-1)^2 = 2$: centre (0, 1), radius $\sqrt{2}$ (the arc above the real axis). 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 $AA^T = \frac{1}{5}\begin{pmatrix}5 & 0\\ 0 & 5\end{pmatrix} = I$, so A is orthogonal. $Q^2 = A^TBAA^TBA = A^TB^2A$, and in general $Q^{2021} = A^TB^{2021}A$. Hence $AQ^{2021}A^T = B^{2021}$. $B = I + N$ with $N = \begin{pmatrix}0 & 0\\ i & 0\end{pmatrix}$, $N^2 = 0$, so $B^{2021} = I + 2021N = \begin{pmatrix}1 & 0\\ 2021i & 1\end{pmatrix}$. Its inverse is $I - 2021N = \begin{pmatrix}1 & 0\\ -2021i & 1\end{pmatrix}$. Answer: (A).
64	Matrices and DeterminantsConsistency of linear systems	Class 12: Matrices; Class 12: Determinants	Single correctMedium
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a non-trivial solution the determinant must vanish. Add columns 2 and 3 to column 1: each entry of column 1 becomes $2 + 4\sin3\theta$ (since $\cos^2\theta + \sin^2\theta = 1$). Taking it out and subtracting row 1 from rows 2 and 3: $\Delta = (2 + 4\sin3\theta)\begin{vmatrix}1 & \sin^2\theta & 4\sin3\theta\\ 0 & 1 & 0\\ 0 & 0 & 1\end{vmatrix} = 2 + 4\sin3\theta$. $\Delta = 0 \Rightarrow \sin3\theta = -\frac{1}{2}$. With $3\theta \in \left(0, \frac{3\pi}{2}\right)$: $3\theta = \frac{7\pi}{6}$, so $\theta = \frac{7\pi}{18}$. Answer: (B).
65	Sequences and SeriesGeometric progressions	Class 11: Sequences and Series	Single correctHard
A hard single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Use $\frac{1}{x-1} - \frac{1}{x+1} = \frac{2}{x^2-1}$, then $\frac{2}{x^2-1} - \frac{2}{x^2+1} = \frac{4}{x^4-1}$, and in general $\frac{2^k}{x^{2^k}-1} - \frac{2^k}{x^{2^k}+1} = \frac{2^{k+1}}{x^{2^{k+1}}-1}$. Adding for $k = 0, 1, \ldots, 100$, the sum telescopes: $S = \frac{1}{x-1} - \frac{2^{101}}{x^{2^{101}} - 1}$. At $x = 2$: $S = 1 - \frac{2^{101}}{2^{2^{101}} - 1}$. This has exactly the form of option (A), $1 - \frac{2^{101}}{(\ldots) - 1}$ with numerator $2^{101}$; option (A) writes the last power of x as $4^{101}$. Answer: (A).
66	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 $\sum r^2\,{}^nC_r = \sum r(r-1)\,{}^nC_r + \sum r\,{}^nC_r = n(n-1)2^{n-2} + n2^{n-1} = n(n+1)2^{n-2}$. With $n = 20$: $20\times21\times2^{18} = 420\times2^{18}$. Answer: (D).
67	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 $\frac{a}{1-r} = 15$ and $\frac{a^2}{1-r^2} = 150$. Dividing, $\frac{a}{1+r} = 10$. So $\frac{1+r}{1-r} = \frac{3}{2}$, giving $r = \frac{1}{5}$ and $a = 12$. $ar^2 + ar^4 + \ldots = \frac{ar^2}{1-r^2} = \frac{12\times\frac{1}{25}}{\frac{24}{25}} = \frac{1}{2}$. Answer: (B).
68	Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctMedium
A medium single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $\cot^{-1}\sqrt{\frac{1-x}{x}} = \alpha$, $\alpha \in \left(0, \frac{\pi}{2}\right)$: $\csc^2\alpha = 1 + \frac{1-x}{x} = \frac{1}{x}$, so $\sin\alpha = \sqrt{x}$. Let $\tan^{-1}\sqrt{x} = \beta$: $f(x) = \cos2\beta = \frac{1 - \tan^2\beta}{1 + \tan^2\beta} = \frac{1-x}{1+x}$. $f'(x) = \frac{-(1+x) - (1-x)}{(1+x)^2} = \frac{-2}{(1+x)^2}$. $(1-x)^2f'(x) = \frac{-2(1-x)^2}{(1+x)^2} = -2(f(x))^2$, so $(1-x)^2f'(x) + 2(f(x))^2 = 0$. (Check the others: $(1+x)^2f'(x) = -2 \ne \mp2f^2$ in general.) Answer: (C).
69	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 The bracket is a perfect square: $\left(\frac{x+1}{x-1} - \frac{x-1}{x+1}\right)^2 = \left(\frac{4x}{x^2-1}\right)^2$. So the integrand is $\left\|\frac{4x}{x^2-1}\right\| = \frac{4\|x\|}{1-x^2}$ on $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$, an even function. $I = 2\int_0^{1/\sqrt{2}}\frac{4x}{1-x^2}dx = 2\left[-2\ln(1-x^2)\right]_0^{1/\sqrt{2}} = -4\ln\frac{1}{2} = 4\ln2 = \log_e16$. Answer: (C).
70	Integral CalculusFundamental theorem of calculus	Class 12: Integrals; Class 12: Application of Integrals	Single correctMedium
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{1}{n}\sum_{r=0}^{2n-1}\frac{1}{1 + 4\left(\frac{r}{n}\right)^2}$ is a Riemann sum with $x = \frac{r}{n}$ running from 0 to 2. Limit $= \int_0^2\frac{dx}{1+4x^2} = \frac{1}{2}\left[\tan^{-1}2x\right]_0^2 = \frac{1}{2}\tan^{-1}4$. Answer: (B).
71	TrigonometryTrigonometric functions	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctHard
A hard single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\frac{\cos x}{1+\sin x} = \frac{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}}{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$, which is positive on the interval. Case $\tan2x > 0$: $\tan2x = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \Rightarrow \frac{5x}{2} = \frac{\pi}{4} + k\pi \Rightarrow x = \frac{\pi}{10} + \frac{2k\pi}{5}$. In the interval: $x = \frac{\pi}{10}$ and $x = -\frac{3\pi}{10}$ (for both, $\tan2x > 0$). Case $\tan2x < 0$: $\tan(-2x) = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \Rightarrow -\frac{3x}{2} = \frac{\pi}{4} + k\pi \Rightarrow x = -\frac{\pi}{6} - \frac{2k\pi}{3}$. In the interval: $x = -\frac{\pi}{6}$ (and $\tan(-\frac{\pi}{3}) < 0$). Sum $= \frac{\pi}{10} - \frac{3\pi}{10} - \frac{\pi}{6} = -\frac{\pi}{5} - \frac{\pi}{6} = -\frac{11\pi}{30}$. Answer: (B).
72	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 C lies on the bisector: $C = \left(c, \frac{7c-1}{4}\right)$. The midpoint of AC, $\left(\frac{c-3}{2}, \frac{7c+3}{8}\right)$, lies on the median through B: $(c - 3) + \frac{7c+3}{8} - 3 = 0 \Rightarrow 15c = 45 \Rightarrow c = 3$. So $C = (3, 5)$. Slope of CA $= \frac{5-1}{3+3} = \frac{2}{3}$; slope of the bisector $= \frac{7}{4}$. $\tan\frac{\theta}{2} = \left\|\frac{\frac{7}{4} - \frac{2}{3}}{1 + \frac{7}{4}\cdot\frac{2}{3}}\right\| = \frac{13/12}{26/12} = \frac{1}{2}$. $\tan\theta = \frac{2\cdot\frac{1}{2}}{1 - \frac{1}{4}} = \frac{4}{3}$. Answer: (C).
73	Conic SectionsParabola	Class 11: Conic Sections	Single correctMedium
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Circle: $x^2 + y^2 + 30x + \frac{675}{4} = 0$, centre (−15, 0), $r^2 = 225 - \frac{675}{4} = \frac{225}{4}$, $r = 7.5$. Tangent to $y^2 = 4ax$ with $a = 7.5$: $y = mx + \frac{7.5}{m}$. Through (−30, 0): $-30m + \frac{7.5}{m} = 0 \Rightarrow m^2 = \frac{1}{4}$, $m = \pm\frac{1}{2}$. Take $y = \frac{x}{2} + 15$, i.e. $x - 2y + 30 = 0$ (the other is its mirror image). Distance from the centre: $\frac{\|-15 + 30\|}{\sqrt{5}} = 3\sqrt{5}$. Chord length $= 2\sqrt{\frac{225}{4} - 45} = 2\sqrt{\frac{45}{4}} = 3\sqrt{5}$. Answer: (B).
74	Conic SectionsEllipse	Class 11: Conic Sections	Single correctMedium
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $a^2 = 8$, $b^2 = 4$, $e^2 = 1 - \frac{4}{8} = \frac{1}{2}$, foci $(\pm2, 0)$. The tangent is perpendicular to $x + 2y = 0$ (slope $-\frac{1}{2}$), so its slope is 2: $y = 2x \pm\sqrt{8\cdot4 + 4} = 2x \pm 6$. The point of contact of $y = mx + c$ is $\left(-\frac{a^2m}{c}, \frac{b^2}{c}\right)$; for the second quadrant take $c = 6$: $P = \left(-\frac{8}{3}, \frac{2}{3}\right)$. $A = \frac{1}{2}\times SS'\times\|y_P\| = \frac{1}{2}\times4\times\frac{2}{3} = \frac{4}{3}$. $(5 - e^2)A = \frac{9}{2}\times\frac{4}{3} = 6$. Answer: (D).
75	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 Plane through the line: $(x + 2y + 3z + 1) + \lambda(x - y - z - 6) = 0$, normal $(1+\lambda, 2-\lambda, 3-\lambda)$. Perpendicular to $(-2, 1, 1)$: $-2(1+\lambda) + (2-\lambda) + (3-\lambda) = 0 \Rightarrow 3 - 4\lambda = 0 \Rightarrow \lambda = \frac{3}{4}$. Plane: $\frac{7}{4}x + \frac{5}{4}y + \frac{9}{4}z - \frac{7}{2} = 0$, i.e. $7x + 5y + 9z = 14$. Check: (2, −1, 1) gives 18; (0, 1, 1) gives 14 ✓; (1, 0, 1) gives 16; (−1, 1, 2) gives 16. Answer: (B).
76	Differential EquationsLinear differential equations	Class 12: Differential Equations	Single correctMedium
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Rearranged: $\tan x\frac{dy}{dx} + y(1 + \tan^2x) = -\tan^2x$, i.e. $\tan x\frac{dy}{dx} + y\sec^2x = -\tan^2x$. The left side is $\frac{d}{dx}(y\tan x)$, so $y\tan x = -\int(\sec^2x - 1)dx = x - \tan x + C$. $xy = \frac{x(x + C)}{\tan x} - x \to C$ as $x \to 0^+$, so $C = 1$. $y = \frac{x + 1}{\tan x} - 1$; at $x = \frac{\pi}{4}$: $y = \frac{\pi}{4} + 1 - 1 = \frac{\pi}{4}$. Answer: (D).
77	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Single correctEasy
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}\cdot(\vec{b}\times\vec{c}) = [\vec{a}\ \vec{b}\ \vec{c}] = \vec{b}\cdot(\vec{c}\times\vec{a}) = -\vec{b}\cdot(\vec{a}\times\vec{c}) = -\vec{b}\cdot\vec{b} = -\|\vec{b}\|^2 = -2$. (The data are consistent: $\vec{a}\cdot\vec{b} = 0$; solving gives $\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.) Answer: (A).
78	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 An implication is false only when its hypothesis is true and its conclusion false. $\sim r$ true $\Rightarrow$ r is F. Then $q \rightarrow r$ true needs q = F. $p \vee q$ true then needs p = T. Check the conclusion: $p \wedge q = T \wedge F = F$. So p, q, r = T, F, F. Answer: (D).
79	ProbabilityAddition and multiplication theorems	Class 11: Probability; Class 12: Probability	Single correctEasy
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability P(exactly one) $= p(1 - 2p) + 2p(1 - p) = 3p - 4p^2 = \frac{5}{9}$. $36p^2 - 27p + 5 = 0 \Rightarrow p = \frac{27 \pm 3}{72} = \frac{5}{12}$ or $\frac{1}{3}$. Both keep $2p \le 1$; the largest is $\frac{5}{12}$. Answer: (C).
80	StatisticsStandard deviation and variance	Class 11: Statistics	Single correctEasy
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Wrong sum $= 200$; correct sum $= 200 - 25 + 35 = 210$, so $\alpha = 10.5$. Wrong $\sum x^2 = 20(2.5^2 + 10^2) = 20\times106.25 = 2125$; correct $\sum x^2 = 2125 - 625 + 1225 = 2725$. $\beta = \frac{2725}{20} - 10.5^2 = 136.25 - 110.25 = 26$. Answer: (C).
81	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 Units digit 0: hundreds from the other 5 digits, tens from the remaining 4: $5\times4 = 20$. Units digit 4 or 6: hundreds from 4 digits (not 0, not the units digit), tens from the remaining 4: $2\times4\times4 = 32$. Total $= 52$. Answer: 52.
82	Limits, Continuity and DifferentiabilityContinuity	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	NumericalEasy
A easy numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f(0) = a\sin\left(-\frac{\pi}{2}\right) = -a$. $\tan u - \sin u = \tan u(1 - \cos u) \approx u\cdot\frac{u^2}{2}$, so with $u = 2x$: $\lim_{x\to0^+}\frac{\tan2x - \sin2x}{bx^3} = \frac{(2x)^3/2}{bx^3} = \frac{4}{b}$. Continuity: $\frac{4}{b} = -a$, so $ab = -4$ and $10 - ab = 14$. Answer: 14.
83	Quadratic EquationsNature of roots	Class 11: Complex Numbers and Quadratic Equations	NumericalMedium
A medium numerical question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\frac{2}{x-1} - \frac{1}{x-2} = \frac{x-3}{(x-1)(x-2)} = \frac{2}{k}$ gives $2x^2 - (6+k)x + (4 + 3k) = 0$ (with $x \ne 1, 2$). x = 1 would need $2k = 0$ and x = 2 would need $k = 0$, both excluded, so the excluded points never are roots. No real roots: $(6+k)^2 - 8(4+3k) < 0 \Rightarrow k^2 - 12k + 4 < 0 \Rightarrow 6 - 4\sqrt{2} < k < 6 + 4\sqrt{2}$, i.e. $0.34 < k < 11.66$. Integers 1 to 11; sum $= 66$. Answer: 66.
84	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	NumericalEasy
A easy numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Square side $\frac{36-k}{4}$, circle radius $\frac{k}{2\pi}$. Total area $A = \frac{(36-k)^2}{16} + \frac{k^2}{4\pi}$. $\frac{dA}{dk} = -\frac{36-k}{8} + \frac{k}{2\pi} = 0 \Rightarrow k\left(\frac{1}{2\pi} + \frac{1}{8}\right) = \frac{36}{8} \Rightarrow k = \frac{36\pi}{4 + \pi}$ (a minimum, since $A'' > 0$). $\left(\frac{4}{\pi} + 1\right)k = \frac{4+\pi}{\pi}\cdot\frac{36\pi}{4+\pi} = 36$. Answer: 36.
85	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 Boundaries: $y = \frac{3x^2}{4}$ and $y = \frac{6x + 24}{4}$; they meet where $3x^2 = 6x + 24$, i.e. $x = -2, 4$. Area $= \frac{1}{4}\int_{-2}^{4}(6x + 24 - 3x^2)dx = \frac{1}{4}\left[3x^2 + 24x - x^3\right]_{-2}^{4} = \frac{1}{4}(80 + 28) = 27$. Answer: 27.
86	Limits, Continuity and DifferentiabilityDerivatives up to order two	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	NumericalMedium
A medium numerical question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At x = 0: $\ln y = 0$, so y = 1. Differentiate: $1 + y' = (x + y)(4y + 4xy')$. At (0, 1): $1 + y' = 4$, so $y' = 3$. Differentiate again: $y'' = (1 + y')(4y + 4xy') + (x + y)(8y' + 4xy'')$. At (0, 1): $y'' = 4\times4 + 1\times24 = 40$. Answer: 40.
87	Permutations and CombinationsP(n, r) and C(n, r)	Class 11: Permutations and Combinations	NumericalMedium
A medium numerical question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations ${}^kP_k = k!$, and $k\cdot k! = (k+1)! - k!$. The sum telescopes: $\sum_{k=1}^{15}\left[(k+1)! - k!\right] = 16! - 1 = {}^{16}P_{16} - 1$. So q = 16, r = 16, s = 1, and ${}^{q+s}C_{r-s} = {}^{17}C_{15} = {}^{17}C_2 = 136$. Answer: 136.
88	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	NumericalHard
A hard numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The given line meets the plane where $(1 + 2t) - 2(3 + t) - (4 + 2t) = 3$, i.e. $t = -6$: point $(-11, -3, -8)$. Foot of the perpendicular from (1, 3, 4) to the plane: $1 - 6 - 4 - 3 = -12$, so foot $= (1, 3, 4) - \frac{-12}{6}(1, -2, -1) = (3, -1, 2)$. L passes through (3, −1, 2) and (−11, −3, −8): direction $(14, 2, 10) \parallel (7, 1, 5)$. From (3, −1, 2) to (0, 0, 6): $(-3, 1, 4)$. Cross product with (7, 1, 5): $(1, 43, -10)$, magnitude² $= 1950$. $d^2 = \frac{1950}{49 + 1 + 25} = \frac{1950}{75} = 26$. Answer: 26.
89	Complex NumbersAlgebra of complex numbers	Class 11: Complex Numbers and Quadratic Equations	NumericalMedium
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z = e^{-i\pi/3}$, so $z^k + z^{-k} = 2\cos\frac{k\pi}{3}$ and $(z^k + z^{-k})^3 = 8\cos^3\frac{k\pi}{3} = 6\cos\frac{k\pi}{3} + 2\cos k\pi$. $\sum_{k=1}^{21}\cos\frac{k\pi}{3}$: the terms repeat every 6 with sum 0; the last three ($k = 19, 20, 21$) give $\frac{1}{2} - \frac{1}{2} - 1 = -1$. $\sum_{k=1}^{21}\cos k\pi = -11 + 10 = -1$. Sum of cubes $= 6(-1) + 2(-1) = -8$; required value $= 21 - 8 = 13$. Answer: 13.
90	CirclesStandard and general form	Class 11: Conic Sections	NumericalEasy
A easy numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections $x^2 + y^2 + (x-1)^2 + y^2 + x^2 + (y-1)^2 + (x-1)^2 + (y-1)^2 = 18$ $\Rightarrow 4x^2 + 4y^2 - 4x - 4y + 4 = 18 \Rightarrow x^2 + y^2 - x - y - \frac{7}{2} = 0$. Radius² $= \frac{1}{4} + \frac{1}{4} + \frac{7}{2} = 4$, radius 2, diameter 4, $d^2 = 16$. Answer: 16.

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