JEE (Main) · 2021

JEE (Main) 2021 — 24 February, 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. Gravitation3
2. Properties of Bulk Matter3
3. Optics3
20 easy · 10 medium · 0 hard
Chemistry
30 Q
1. Classification of Elements and Periodicity2
2. Isolation of Metals2
3. Redox Reactions2
19 easy · 10 medium · 1 hard
Mathematics
30 Q
1. Matrices and Determinants3
2. Trigonometry3
3. Applications of Derivatives3
13 easy · 15 medium · 2 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	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	Single correctMedium
A medium single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The exponent must be dimensionless, so $[\alpha][kT] = [x^2]$. Since $kT$ is an energy, $[\alpha] = \frac{L^2}{ML^2T^{-2}} = [M^{-1}T^2]$. Then $[\beta^2] = \frac{[W]}{[\alpha]} = \frac{ML^2T^{-2}}{M^{-1}T^2} = [M^2L^2T^{-4}]$, so $[\beta] = [MLT^{-2}]$. Answer: (C).
2	GravitationOrbital motion	Class 11: Gravitation	Single correctEasy
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Mass m is at distance $r = \frac{2d}{3}$ from the centre of mass. For it: $m\omega^2\cdot\frac{2d}{3} = \frac{G(m)(2m)}{d^2}$, so $\omega^2 = \frac{3Gm}{d^3}$. $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{d^3}{3Gm}}$. Answer: (B).
3	GravitationNewton's law of gravitation	Class 11: Gravitation	Single correctMedium
A medium single correct question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation The particles sit at the corners of a square inscribed in the circle: side $\sqrt{2}$ m, diagonal 2 m. Force towards the centre on one particle: the two neighbours each pull with $\frac{G}{(\sqrt{2})^2} = \frac{G}{2}$ at $45^\circ$ to the radius, and the opposite particle pulls with $\frac{G}{4}$ along it. $F = 2\cdot\frac{G}{2}\cdot\frac{1}{\sqrt{2}} + \frac{G}{4} = \frac{G(2\sqrt{2}+1)}{4}$. With $m = 1$, $r = 1$: $v^2 = F r / m = \frac{G(1+2\sqrt{2})}{4}$, so $v = \frac{\sqrt{(1+2\sqrt{2})G}}{2}$. Answer: (A).
4	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 $I_1 = \frac{1}{2}MR^2$ (ring about a diameter), $I_2 = \frac{1}{2}MR^2$, $I_3 = \frac{1}{2}MR^2$, $I_4 = \frac{2}{5}MR^2$. So $I_1 = I_2 = I_3 = 0.5MR^2 > I_4 = 0.4MR^2$. (Check (A): $I_1 + I_2 = MR^2$ but $I_3 + \frac{5}{2}I_4 = 1.5MR^2$; (B): $MR^2 < 0.9MR^2$ is false.) Answer: (D).
5	GravitationOrbital motion	Class 11: Gravitation	Single correctEasy
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation $\omega = \frac{2\pi}{T}$, so $\frac{\omega_1}{\omega_2} = \frac{T_2}{T_1} = \frac{8}{1}$. Answer: (A).
6	Properties of Bulk MatterThermal expansion and calorimetry	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 thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Each edge becomes $a(1 + \alpha\Delta T)$, so the enclosed volume becomes $a^3(1+\alpha\Delta T)^3 \approx a^3(1 + 3\alpha\Delta T)$. Increase $= 3a^3\alpha\Delta T$ (a hollow box expands as if it were solid). Answer: (D).
7	Properties of Bulk MatterElasticity	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 elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter From $Y = 3K(1-2\sigma)$ and $Y = 2\eta(1+\sigma)$, eliminating $\sigma$ gives $\frac{9}{Y} = \frac{1}{K} + \frac{3}{\eta}$, i.e. $Y = \frac{9K\eta}{\eta + 3K}$. Rearranging: $Y\eta + 3KY = 9K\eta \Rightarrow K(9\eta - 3Y) = Y\eta \Rightarrow K = \frac{Y\eta}{9\eta - 3Y}$. (A) and (B) have wrong denominators; (D) should read $\eta = \frac{3YK}{9K - Y}$. Answer: (C).
8	KinematicsMotion graphs	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane From A to M the velocity falls linearly, so the acceleration is a negative constant; from M to B it rises linearly, so the acceleration is a positive constant. The a–t graph is a negative step followed by a positive step. Answer: (A).
9	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Single correctMedium
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $W_{AB} = nRT\ln\frac{V_2}{V_1} = nRT\ln 2$. $W_{BC} = P_2(V_1 - 2V_1) = -P_2V_1$. At B, $P_2(2V_1) = nRT$, so $P_2V_1 = \frac{nRT}{2}$ and $W_{BC} = -\frac{nRT}{2}$. $W_{CA} = 0$ (isochoric). Total $= nRT\left(\ln 2 - \frac{1}{2}\right)$. Answer: (D).
10	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Isothermal: temperature constant (ii). Isochoric: volume constant (iii). Adiabatic: no heat exchanged, heat content constant (iv). Isobaric: pressure constant (i). Answer: (D).
11	Oscillations and WavesSimple pendulum and spring systems	Class 11: Oscillations; Class 11: Waves	Single correctMedium
A medium single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At equilibrium M moves with its maximum speed $v = A\omega$, $\omega = \sqrt{k/M}$. Placing m on it conserves momentum: $Mv = (M+m)v'$, so $v' = \frac{M}{M+m}A\sqrt{\frac{k}{M}}$. New angular frequency $\omega' = \sqrt{\frac{k}{M+m}}$, and the new amplitude $A' = \frac{v'}{\omega'} = \frac{M}{M+m}A\sqrt{\frac{M+m}{M}} = A\sqrt{\frac{M}{M+m}}$. Answer: (B).
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 In the figure the two cells are joined in a single loop with their positive terminals facing opposite ways (E1's positive plate towards P, E2's towards Y), so their emfs oppose. $I = \frac{6 - 4}{2 + 8} = 0.2$ A, driven by $E_1$, so it flows through $E_2$ against its emf. $V_{XY}$ = p.d. across $E_2$ = $4 + 0.2 \times 8 = 5.6$ V (equivalently across $E_1$: $6 - 0.2\times 2 = 5.6$ V, since Y is joined to P by a plain wire). Answer: (C).
13	Current ElectricityOhm's law	Class 12: Current Electricity	Single correctEasy
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity $q = \int_0^{15}(20t + 8t^2)\,dt = 10t^2 + \frac{8}{3}t^3\Big\|_0^{15} = 2250 + 9000 = 11250$ C. Answer: (C).
14	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Series: $\frac{C}{2}$. Parallel: $2C$. Ratio $= \frac{C/2}{2C} = \frac{1}{4}$. Answer: (D).
15	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Eight equal charges +Q at the vertices give zero field at the centre. The given arrangement is that plus a charge $-2Q$ at the origin. Distance from the origin to the centre $= \frac{\sqrt{3}a}{2}$, so the field of $-2Q$ has magnitude $\frac{1}{4\pi\varepsilon_0}\cdot\frac{2Q}{3a^2/4} = \frac{2Q}{3\pi\varepsilon_0 a^2}$, directed from the centre towards the origin, i.e. along $-\frac{\hat{x}+\hat{y}+\hat{z}}{\sqrt{3}}$. $\vec{E} = \frac{-2Q}{3\sqrt{3}\pi\varepsilon_0 a^2}(\hat{x}+\hat{y}+\hat{z})$. Answer: (C).
16	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 $\Delta I_B = \Delta I_E - \Delta I_C = 4 - 3.5 = 0.5$ mA. $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{3.5}{0.5} = 7$. Answer: (A).
17	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 A runs from the continuum ($n = \infty$) to $n = 1$: the series limit of the Lyman series. B runs from $n = 5$ to $n = 2$: Balmer members are $3\to2$ (first), $4\to2$ (second), $5\to2$ (third), so B is the third member. C runs from $n = 5$ to $n = 3$: Paschen members are $4\to3$ (first), $5\to3$ (second), so C is the second member. Answer: (D).
18	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter For a photon $p = \frac{h}{\lambda}$, so equal momenta mean equal wavelengths: Statement I is true. $p = \frac{h}{\lambda}$ and $E = \frac{hc}{\lambda}$ both increase when $\lambda$ decreases: Statement II is false. Answer: (C).
19	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a spherical mirror (paraxial rays) the focus lies midway between the pole and the centre of curvature, $f = \frac{r}{2}$. For a convex mirror, with the Cartesian sign convention, both $f$ and $r$ are positive (behind the mirror), so $f = +\frac{1}{2}r$. Answer: (D).
20	OpticsInterference and diffraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Amplitudes are in the ratio $3 : 1$. $\frac{I_{max}}{I_{min}} = \frac{(3+1)^2}{(3-1)^2} = \frac{16}{4} = 4$. Answer: (A).
21	Laws of MotionFriction	Class 11: Laws of Motion	NumericalEasy
A easy numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion The horizontal force F is the normal reaction. Friction must hold the weight: $\mu F \ge mg$. $F_{min} = \frac{mg}{\mu} = \frac{0.5 \times 10}{0.2} = 25$ N. Answer: 25.
22	OpticsPolarisation	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalEasy
A easy numerical question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Taking the first reading with the analyzer's axis parallel to the polarizer's (the usual reading), the light reaching the analyzer has intensity 100 lumens. After turning the analyzer by $30^\circ$, by Malus's law $I = 100\cos^2 30^\circ = 100 \times \frac{3}{4} = 75$ lumens. Answer: 75.
23	Work, Energy and PowerCollisions	Class 11: Work, Energy and Power	NumericalEasy
A easy numerical question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The balls move off on opposite sides of the original direction, each at $30^\circ$. Momentum perpendicular to the original direction was zero, so $mv_1\sin 30^\circ = mv_2\sin 30^\circ$, giving $v_1 = v_2$. The ratio is $1 : 1$, so $x = 1$. (Along the original line: $9 = 2v\cos 30^\circ$, $v = 3\sqrt{3}$ m/s each.) Answer: 1.
24	Properties of Bulk MatterFluid mechanics	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalMedium
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Pascal's law: $\frac{mg}{a} = \frac{Mg}{A}$, so $M = m\frac{A}{a} = m\frac{D^2}{d^2}$. Making $D \to 4D$ and $d \to \frac{d}{4}$ multiplies $\frac{D^2}{d^2}$ by $16 \times 16 = 256$. New load $= 100 \times 256 = 25600$ kg. Answer: 25600.
25	Laws of MotionFriction	Class 11: Laws of Motion	NumericalMedium
A medium numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion A block on a slope of angle $\theta$ stays at rest while $\tan\theta \le \mu$. Here $\tan\theta = \frac{dy}{dx} = \frac{x}{2}$, so $\frac{x}{2} \le 0.5 \Rightarrow x \le 1$ m. Maximum height $y = \frac{1^2}{4} = 0.25$ m $= 25$ cm. Answer: 25.
26	Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	NumericalEasy
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $Q = \frac{\omega L}{R} = \frac{2\pi f L}{R} = \frac{2 \times 3.14 \times 10^7 \times 2\times10^{-4}}{6.28} = 2000$. Answer: 2000.
27	Communication SystemsAmplitude modulation	No matching NCERT chapter	NumericalEasy
A easy numerical question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter Modulation index $\mu = \frac{A_m}{A_c} = \frac{20}{80} = 0.25$, i.e. 25 %. Answer: 25.
28	Electronic DevicesSpecial-purpose diodes	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	NumericalMedium
A medium numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The 10 V cell's positive terminal feeds the 1 k$\Omega$ resistor and the top node, so the 5 V Zener (cathode at the top) is reverse biased, in parallel with the 2 k$\Omega$ resistor. Without the Zener the 2 k$\Omega$ resistor would get $10 \times \frac{2}{3} \approx 6.7$ V > 5 V, so the Zener breaks down and holds 5 V across it. $I_{2k} = \frac{5}{2000} = 2.5 \times 10^{-3}$ A $= 25 \times 10^{-4}$ A. (The 1 k$\Omega$ resistor carries 5 mA, so the Zener carries 2.5 mA.) Answer: 25.
29	Electromagnetic WavesProperties of electromagnetic waves	Class 12: Electromagnetic Waves	NumericalEasy
A easy numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $v = \frac{c}{\sqrt{\varepsilon_r\mu_r}} = \frac{3\times10^8}{\sqrt{4}} = 1.5\times10^8 = 15\times10^7$ m/s. Answer: 15.
30	Electromagnetic Induction and Alternating CurrentsTransformers	Class 12: Electromagnetic Induction; Class 12: Alternating Current	NumericalEasy
A easy numerical question on transformers, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Treating the secondary output as 12 V (ignoring rectifier losses and the peak/rms distinction, as the question intends): $\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{220}{12}$. $N_p = 24 \times \frac{220}{12} = 440$. Answer: 440.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
31	Chemical Bonding and Molecular StructureVSEPR theory	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A: $\mathrm{SO_4^{2-}}$ and $\mathrm{CrO_4^{2-}}$ are both tetrahedral — isostructural. B: $\mathrm{SiCl_4}$ and $\mathrm{TiCl_4}$ are both tetrahedral — isostructural. C: $\mathrm{NH_3}$ is trigonal pyramidal (one lone pair), $\mathrm{NO_3^-}$ is trigonal planar — not. D: $\mathrm{BCl_3}$ is trigonal planar, $\mathrm{BrCl_3}$ is T-shaped (two lone pairs on Br) — not. Answer: (A).
32	Surface ChemistryFreundlich adsorption isotherm	No matching NCERT chapter	Single correctEasy
A easy single correct question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich: $\frac{x}{m} = kP^{1/n}$, so $\log\frac{x}{m} = \log k + \frac{1}{n}\log P$. The plot of $\log\frac{x}{m}$ against $\log P$ is a straight line of slope $\frac{1}{n}$, with $\frac{1}{n}$ between 0 and 1. Answer: (D).
33	Classification of Elements and PeriodicityIonization enthalpy	Class 11: Classification of Elements and Periodicity in Properties	Single correctEasy
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across period 3 ionization enthalpy rises, with two exceptions: Al < Mg (Al loses a 3p electron, Mg a paired 3s) and S < P (P has a stable half-filled 3p). Approximate values (kJ/mol): Al 577, Mg 738, Si 786, S 1000, P 1012. Order: Al < Mg < Si < S < P. Answer: (A).
34	Isolation of MetalsFroth flotation	No matching NCERT chapter	Single correctMedium
A medium single correct question on froth flotation, from Isolation of Metals. NCERT: No matching NCERT chapter In the froth flotation of a mixed sulphide ore of ZnS (sphalerite) and PbS, NaCN is added as a depressant: it forms a complex at the surface of ZnS and keeps it from coming into the froth, so the two sulphides are separated. Calamine ($\mathrm{ZnCO_3}$), malachite ($\mathrm{CuCO_3\cdot Cu(OH)_2}$) and siderite ($\mathrm{FeCO_3}$) are not concentrated with a cyanide. Answer: (D).
35	Redox ReactionsOxidation number	Class 11: Redox Reactions	Single correctEasy
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions In both reactions the oxygen of $\mathrm{H_2O_2}$ goes from −1 to 0 in $\mathrm{O_2}$: $\mathrm{H_2O_2}$ is oxidised. (A): Cl goes from +1 (HOCl) to −1 ($\mathrm{Cl^-}$). (B): I goes from 0 ($\mathrm{I_2}$) to −1 ($\mathrm{I^-}$). So $\mathrm{H_2O_2}$ is the reducing agent in both. Answer: (B).
36	Isolation of MetalsOres and concentration	No matching NCERT chapter	Single correctEasy
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Leaching of bauxite: $\mathrm{Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]}$ (X). Passing $\mathrm{CO_2}$ (Y) neutralises the solution and precipitates hydrated alumina: $\mathrm{2Na[Al(OH)_4] + CO_2 \rightarrow Al_2O_3\cdot xH_2O + 2NaHCO_3}$ (Z). Answer: (D).
37	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 $E^\circ(\mathrm{M^{2+}/M})$: Fe −0.44 V, Co −0.28 V, Zn −0.76 V, Cu +0.34 V. Copper is the only 3d metal with a positive value. Answer: (D).
38	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 Gun metal is a bronze: about 88 % Cu, 10 % Sn and 2 % Zn. (Cu, Zn, Ni is German silver; Al, Cu, Mg, Mn is duralumin.) Answer: (A).
39	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 Anaerobic decomposition of vegetation (as in paddy fields and marshes) releases methane, a greenhouse gas; it adds to global warming. Of the options only (B) names global warming. Answer: (B).
40	Alcohols, Phenols and EthersReactions of phenols	Class 12: Alcohols, Phenols and Ethers	Single correctMedium
A medium single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The phthalein test: a phenol condenses with phthalic anhydride through the position para to its OH, giving a phenolphthalein-type dye that is pink in NaOH. It needs a free para position. (A) 4-propylphenol: para position occupied — no phthalein. (B) 2-propylphenol: para position free — gives the pink phthalein. (C) and (D): every position para to an OH carries the methyl or propyl group — no phenolphthalein-type dye (a meta-dihydroxy compound such as (C) gives a fluorescein-type product instead). Answer: (B).
41	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{H^+}$ adds to the terminal $\mathrm{CH_2}$ (Markovnikov), giving the secondary carbocation $\mathrm{(CH_3)_3C{-}\overset{+}{C}H{-}CH_3}$. A 1,2-methyl shift turns it into the more stable tertiary carbocation $\mathrm{(CH_3)_2\overset{+}{C}{-}CH(CH_3){-}CH_3}$. $\mathrm{I^-}$ attacks it: $\mathrm{(CH_3)_2CI{-}CH(CH_3)_2}$, 2-iodo-2,3-dimethylbutane. Answer: (B).
42	HydrocarbonsAlkanes	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Controlled oxidation of an alkane to an aldehyde is done with oxygen over molybdenum oxide ($\mathrm{Mo_2O_3}$) on heating — as $\mathrm{CH_4 + O_2 \xrightarrow{Mo_2O_3,\ \Delta} HCHO + H_2O}$. Copper at high temperature and pressure converts methane to methanol; manganese acetate gives carboxylic acids; $\mathrm{KMnO_4}$ oxidises a tertiary C–H to an alcohol. Answer: (B).
43	Haloalkanes and HaloarenesNature of the C–X bond	Class 12: Haloalkanes and Haloarenes	Single correctHard
A hard single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes With excess Mg in dry ether, each C–Br bond reacts: Mg inserts into the C–Br bond to give an organomagnesium bromide. The first product is therefore the di-Grignard reagent $\mathrm{CH_3CH_2{-}CH(MgBr){-}CH_2{-}CH(MgBr){-}CH_3}$. (A) and (C) are Wurtz-type dimers, not formed with Mg; (B), 1-ethyl-2-methylcyclopropane, is what a 1,3-dihalide finally gives on intramolecular ring closure, a later step. Answer: (D).
44	Haloalkanes and HaloarenesSN1 and SN2 mechanisms	Class 12: Haloalkanes and Haloarenes	Single correctMedium
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes HCl protonates the OH and water leaves, giving a secondary carbocation on the side-chain carbon ($\mathrm{\overset{+}{C}H{-}CH_3}$ attached to the ring). A 1,2-hydride shift from the ring carbon next to it gives a tertiary carbocation on that ring carbon (which now carries an ethyl group). $\mathrm{Cl^-}$ attacks the tertiary carbocation: 1-chloro-1-ethyl-2-methylcyclohexane. Answer: (D).
45	Alcohols, Phenols and EthersPreparation and properties	Class 12: Alcohols, Phenols and Ethers	Single correctMedium
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Cold dilute $\mathrm{KMnO_4}$ (273 K) adds two OH groups across the double bond (syn dihydroxylation): A is 1-methylcyclopentane-1,2-diol. $\mathrm{CrO_3}$ oxidises the secondary alcohol to a ketone and leaves the tertiary alcohol untouched: B is 2-hydroxy-2-methylcyclopentan-1-one. Answer: (B).
46	AminesBasicity of amines	Class 12: Amines	Single correctEasy
A easy single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines In the strongly acidic nitrating mixture, aniline is largely protonated to the anilinium ion $\mathrm{C_6H_5NH_3^+}$. The $\mathrm{-NH_3^+}$ group is electron-withdrawing and meta-directing, so a large share (47 %) of meta-nitroaniline forms besides the para product. Answer: (C).
47	AminesDiazonium salts	Class 12: Amines	Single correctEasy
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines $\mathrm{NaNO_2/HCl}$ diazotises aniline to $\mathrm{C_6H_5N_2^+Cl^-}$; KCN then replaces the diazo group by CN: A is benzonitrile, $\mathrm{C_6H_5CN}$. $\mathrm{SnCl_2/HCl}$ followed by hydrolysis is Stephen reduction: the nitrile goes through the imine to the aldehyde. B is benzaldehyde, $\mathrm{C_6H_5CHO}$. Answer: (D).
48	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 Caprolactam → Nylon 6 (iii); chloroprene → neoprene (iv); isoprene → natural rubber (i); acrylonitrile (with 1,3-butadiene) → Buna-N (ii). Answer: (D).
49	BiomoleculesProteins	Class 12: Biomolecules	Single correctEasy
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The $\alpha$-helix is held by hydrogen bonds between the C=O of one residue and the N–H of the residue four places along the chain. Answer: (D).
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 Borax bead test for copper. In the non-luminous (oxidising) flame, $\mathrm{CuSO_4 + B_2O_3 \rightarrow CuO + SO_3}$ and $\mathrm{CuO + B_2O_3 \rightarrow Cu(BO_2)_2}$: cupric metaborate, which is blue-green, not colourless. In the luminous (reducing) flame, cupric metaborate is reduced to colourless cuprous metaborate (and to red Cu). Statement I calls cupric metaborate colourless: false. Statement II says the non-luminous flame gives cuprous metaborate; it gives cupric metaborate: false. Answer: (B).
51	SolutionsConcentration terms	Class 12: Solutions	NumericalEasy
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Moles $= \frac{4.5}{90} = 0.05$. Molarity $= \frac{0.05}{0.250} = 0.2$ M $= 2\times10^{-1}$ M. Answer: 2.
52	Solid StatePacking in fcc, bcc and hcp	No matching NCERT chapter	NumericalEasy
A easy numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter In bcc the body-centre atom touches the 8 corner atoms; every atom is equivalent, so the coordination number is 8. Answer: 8.
53	Structure of AtomDual nature and uncertainty principle	Class 11: Structure of Atom	NumericalEasy
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $\lambda = \frac{h}{\sqrt{2mqV}}$, so $\frac{\lambda_{Li}}{\lambda_p} = \sqrt{\frac{m_p\,e}{m_{Li}\cdot 3e}} = \sqrt{\frac{1}{8.3 \times 3}} = \sqrt{\frac{1}{24.9}} = 0.2004$. $= 2.004\times10^{-1}$, so $x \approx 2$. Answer: 2.
54	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	NumericalEasy
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta_rG^\circ = -RT\ln K = -R \times 300 \times \ln 100 = -R \times 300 \times 2 \times 2.3 = -1380R$. So $x = 1380$. Answer: 1380.
55	Solutionsvan't Hoff factor	Class 12: Solutions	NumericalMedium
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Molar mass of $\mathrm{ClCH_2COOH}$ = 94.5 g/mol, so 9.45 g is 0.1 mol; in 500 mL (0.5 kg) of water, $m = 0.2$ mol/kg. $\Delta T_f = iK_fm$: $0.5 = i \times 1.86 \times 0.2$, so $i = 1.344$ and $\alpha = i - 1 = 0.344$. Taking the concentration as 0.2 M (dilute solution): $K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.2 \times (0.344)^2}{0.656} = 0.0361 = 36.1\times10^{-3}$. So $x = 36$. Answer: 36.
56	EquilibriumChemical equilibrium	Class 11: Equilibrium	NumericalEasy
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Equal numbers of particles: each partial pressure is 0.5 atm. $K_p = \frac{p_{Cl}^2}{p_{Cl_2}} = \frac{(0.5)^2}{0.5} = 0.5 = 5\times10^{-1}$. Answer: 5.
57	Redox ReactionsBalancing redox equations	Class 11: Redox Reactions	NumericalMedium
A medium numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Sulphur disproportionates: reduction $\mathrm{S^0 \rightarrow S^{2-}}$ (gain 2 e⁻ per S); oxidation $\mathrm{S^0 \rightarrow S^{+2}}$ in $\mathrm{S_2O_3^{2-}}$ (loss 2 e⁻ per S, 4 per thiosulphate). Electron balance: $2b = 4c$, so $b = 2c$. Sulphur: $b + 2c = 8$, so $c = 2$, $b = 4$. Charge: $-a = -2b - 2c = -12$, so $a = 12$, and $d = 6$ balances H and O. $\mathrm{S_8 + 12OH^- \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O}$. Answer: 12.
58	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	NumericalEasy
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics 40 % reacted leaves 60 %: $t = \frac{1}{k}\ln\frac{100}{60} = \frac{0.5108}{3.3\times10^{-4}} = 1548$ s. $1548 / 60 = 25.8$ min, rounded to 26. Answer: 26.
59	Classification of Elements and PeriodicityPeriodic trends	Class 11: Classification of Elements and Periodicity in Properties	NumericalEasy
A easy numerical question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Beryllium's oxide and hydroxide are amphoteric (they dissolve in both acids and alkalis): BeO and $\mathrm{Be(OH)_2}$. BaO and $\mathrm{Sr(OH)_2}$ are basic. Answer: 2.
60	Coordination CompoundsImportance of coordination compounds	Class 12: Coordination Compounds	NumericalMedium
A medium numerical question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Overall formation constant $\beta_4 = K_1K_2K_3K_4 = 10^4 \times 1.58\times10^3 \times 5\times10^2 \times 10^2 = 7.9\times10^{11}$. Dissociation constant $= \frac{1}{\beta_4} = \frac{1}{7.9\times10^{11}} = 1.27\times10^{-12}$. So $x \approx 1$. Answer: 1.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
61	Sets, Relations and FunctionsComposition of functions	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	Single correctEasy
A easy single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f(g(x)) = 2\cdot\frac{x-\frac{1}{2}}{x-1} - 1 = \frac{2x-1}{x-1} - 1 = \frac{x}{x-1}$, for $x \in \mathbb{R}-\{1\}$, into $\mathbb{R}$. One-one: $\frac{x_1}{x_1-1} = \frac{x_2}{x_2-1} \Rightarrow x_1 = x_2$. Not onto: $\frac{x}{x-1} = 1$ has no solution, so 1 is not in the range. Answer: (A).
62	Quadratic EquationsForming equations from given roots	Class 11: Complex Numbers and Quadratic Equations	Single correctMedium
A medium single correct question on forming equations from given roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $pq = s$. Then $p^2 + q^2 = 4 - 2s$ and $p^4 + q^4 = (4-2s)^2 - 2s^2 = 2s^2 - 16s + 16$. $2s^2 - 16s + 16 = 272 \Rightarrow s^2 - 8s - 128 = 0 \Rightarrow s = 16$ or $s = -8$. Since p and q are to be positive, $pq > 0$, so $pq = 16$, and p, q are roots of $x^2 - 2x + 16 = 0$. Answer: (B).
63	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 $\Delta = \begin{vmatrix}3&-2&-k\\2&-4&-2\\1&2&-1\end{vmatrix} = 3(4+4) + 2(-2+2) - k(4+4) = 24 - 8k$, zero when $k = 3$. For $k \neq 3$ the system has a unique solution. With $k = 3$, the first row is the sum of the other two: $(2,-4,-2) + (1,2,-1) = (3,-2,-3)$. The right sides must match the same way: $6 + 5m = 10$, i.e. $m = \frac{4}{5}$, for consistency. So the system is inconsistent when $k = 3$ and $m \neq \frac{4}{5}$. Answer: (C).
64	Binomial TheoremSimple applications	Class 11: Binomial Theorem	Single correctMedium
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem First part: $\sum_{r=1}^{15}(-1)^r r\,{}^{15}C_r = 15\sum_{r=1}^{15}(-1)^r\,{}^{14}C_{r-1} = -15(1-1)^{14} = 0$. Second part: ${}^{14}C_1 + {}^{14}C_3 + \cdots + {}^{14}C_{13} = 2^{13}$; the sum stops at ${}^{14}C_{11}$, so it is $2^{13} - {}^{14}C_{13} = 2^{13} - 14$. Total $= 2^{13} - 14$. Answer: (B).
65	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 For $0 < x < \frac{\pi}{2}$, $0 < \cos^2x < 1$, so $\cos^2x + \cos^4x + \cdots = \frac{\cos^2x}{1-\cos^2x} = \cot^2x$, and the expression is $2^{\cot^2x}$. $t^2 - 9t + 8 = 0$ gives $t = 1$ or $8$. $2^{\cot^2x} = 1$ needs $\cot x = 0$ (not allowed); $2^{\cot^2x} = 8$ gives $\cot^2x = 3$, $\cot x = \sqrt{3}$, $x = \frac{\pi}{6}$. $\frac{2\sin x}{\sin x + \sqrt{3}\cos x} = \frac{2}{1 + \sqrt{3}\cot x} = \frac{2}{1+3} = \frac{1}{2}$. Answer: (A).
66	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 Both numerator and denominator tend to 0. By L'Hospital's rule and the fundamental theorem of calculus: $\lim\frac{\sin\sqrt{x^2}\cdot 2x}{3x^2} = \lim\frac{2\sin\|x\|}{3x}$. As $x \to 0^+$ this is $\frac{2}{3}$. Answer: (A).
67	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	Single correctMedium
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 2x^2 - x - 2\cos x + 2\cos x - (2x-1)\sin x = (2x-1)(x - \sin x)$. For $x > \frac{1}{2}$: $2x - 1 > 0$ and $x - \sin x > 0$, so $f' > 0$: f increases on $\left[\frac{1}{2}, \infty\right)$. On $\left(-\infty, \frac{1}{2}\right]$ the sign changes at 0 (f' > 0 for x < 0, f' < 0 for 0 < x < 1/2), so (C) and (D) are false; (B) is false by the above. Answer: (A).
68	Permutations and CombinationsSimple applications	Class 11: Permutations and Combinations	Single correctEasy
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations With i Indians and 2i foreigners, $i \ge 2$ and $2i \le 8$, so $i = 2, 3, 4$. $^6C_2\,^8C_4 + {}^6C_3\,^8C_6 + {}^6C_4\,^8C_8 = 15\cdot70 + 20\cdot28 + 15\cdot1 = 1050 + 560 + 15 = 1625$. Answer: (B).
69	Limits, Continuity and DifferentiabilityContinuity	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctMedium
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $[x-1]$ jumps only at integers. At an integer n, $\cos\frac{(2n-1)\pi}{2} = 0$, so $f(n) = 0$, and both one-sided limits are (finite integer) $\times\ \cos\frac{(2n-1)\pi}{2} = 0$. So f is continuous at every integer, and elsewhere it is a product of continuous functions. Answer: (D).
70	Integral CalculusIntegration by substitution	Class 12: Integrals; Class 12: Application of Integrals	Single correctEasy
A easy single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $u = \sin x + \cos x$, $du = (\cos x - \sin x)\,dx$, and $u^2 = 1 + \sin 2x$, so $8 - \sin 2x = 9 - u^2$. $\int\frac{du}{\sqrt{9 - u^2}} = \sin^{-1}\frac{u}{3} + c$. So $a = 1$, $b = 3$. Answer: (B).
71	Integral CalculusArea under simple curves	Class 12: Integrals; Class 12: Application of Integrals	Single correctMedium
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersection: $x^2 + 9x - 36 = 0 \Rightarrow x = 3$, $y = \pm3\sqrt{3}$. Area inside both (the region $y^2 \le 9x$ within the circle): $2\left[\int_0^3 3\sqrt{x}\,dx + \int_3^6\sqrt{36 - x^2}\,dx\right] = 2\left[6\sqrt{3} + \left(9\pi - \frac{9\sqrt{3}}{2} - 3\pi\right)\right] = 12\pi + 3\sqrt{3}$. Outside the parabola: $36\pi - (12\pi + 3\sqrt{3}) = 24\pi - 3\sqrt{3}$. Answer: (B).
72	Differential EquationsSeparation of variables	Class 12: Differential Equations	Single correctEasy
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dP}{dt} = 0.5(P - 900)$, so $P - 900 = (850 - 900)e^{t/2} = -50e^{t/2}$. $P = 0$: $50e^{t/2} = 900 \Rightarrow e^{t/2} = 18 \Rightarrow t = 2\log_e 18$. Answer: (D).
73	Straight LinesVarious forms of the equation of a line	Class 11: Straight Lines	Single correctEasy
A easy single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines Line: $\frac{x}{a} + \frac{y}{b} = 1$ with $\frac{1}{2}\left(\frac{1}{a} + \frac{1}{b}\right) = \frac{1}{4}$, i.e. $\frac{2}{a} + \frac{2}{b} = 1$. So every such line passes through (2, 2). It would pass through (1, 1) only if $\frac{1}{a}+\frac{1}{b} = 1$, and through (4, 4) only if $\frac{1}{a}+\frac{1}{b} = \frac{1}{4}$; here $\frac{1}{a}+\frac{1}{b} = \frac{1}{2}$, so neither holds. Only B is on the path. Answer: (B).
74	Conic SectionsParabola	Class 11: Conic Sections	Single correctEasy
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Focus $(a, 0)$, point $(at^2, 2at)$. Mid-point: $h = \frac{a(1+t^2)}{2}$, $k = at$. Eliminating t: $k^2 = a^2t^2 = a(2h - a) = 2a\left(h - \frac{a}{2}\right)$. This is $Y^2 = 4\cdot\frac{a}{2}\left(X - \frac{a}{2}\right)$: vertex $\left(\frac{a}{2}, 0\right)$, focal length $\frac{a}{2}$, so the directrix is $x = \frac{a}{2} - \frac{a}{2} = 0$. Answer: (C).
75	Applications of DerivativesTangents and normals	Class 12: Application of Derivatives	Single correctMedium
A medium single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Tangent: $y = 3t^2x - 2t^3$. With $y = x^3$: $x^3 - 3t^2x + 2t^3 = (x - t)^2(x + 2t) = 0$, so $Q = (-2t, -8t^3)$. Point dividing PQ in 1 : 2: $y = \frac{2\cdot t^3 + 1\cdot(-8t^3)}{3} = -2t^3$. Answer: (D).
76	Three Dimensional GeometryDirection ratios and direction cosines	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctEasy
A easy single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The normal is perpendicular to both given normals: $(3, 1, -2)\times(2, -5, -1) = (-1-10,\ -4+3,\ -15-2) = (-11, -1, -17)$, i.e. $(11, 1, 17)$. Plane: $11(x-1) + (y-2) + 17(z+3) = 0 \Rightarrow 11x + y + 17z + 38 = 0$. Answer: (B).
77	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 A point of the line: $(3+\lambda, 4+2\lambda, 5+2\lambda)$. On the plane: $12 + 5\lambda = 17 \Rightarrow \lambda = 1$, giving $(4, 6, 7)$. Distance from (1, 1, 9): $\sqrt{9 + 25 + 4} = \sqrt{38}$. Answer: (C).
78	ProbabilityProbability distribution of a random variable	Class 11: Probability; Class 12: Probability	Single correctMedium
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each roll gives odd or even with probability $\frac{1}{2}$. For n rolls: $^nC_2\left(\frac{1}{2}\right)^n = {}^nC_3\left(\frac{1}{2}\right)^n \Rightarrow n = 5$. $P(\text{odd number of odd results}) = \frac{^5C_1 + {}^5C_3 + {}^5C_5}{2^5} = \frac{16}{32} = \frac{1}{2}$. Answer: (D).
79	TrigonometryTrigonometric functions	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctEasy
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Heights h and 3h, each 75 m from the observer. $\tan\theta = \frac{h}{75}$ and $\tan(90^\circ - \theta) = \cot\theta = \frac{3h}{75}$. Multiplying: $1 = \frac{3h^2}{75^2} \Rightarrow h^2 = 1875 \Rightarrow h = 25\sqrt{3}$ m. Answer: (D).
80	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 (A) and (B) both reduce to A (absorption), false when A is false. (C): if $A \wedge (A \rightarrow B)$ is true, then A is true and so B is true; the implication can never be false — a tautology. (D): with B true and A false the right side is false, so the implication is false. Answer: (C).
81	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	NumericalHard
A hard numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z = x + iy$. Imaginary part: $y + 2 = 0$, so $y = -2$. Real part: $x + \alpha\sqrt{(x-1)^2 + 4} = 0$. Squaring: $\alpha^2(x^2 - 2x + 5) = x^2$, i.e. $(\alpha^2 - 1)x^2 - 2\alpha^2x + 5\alpha^2 = 0$. For a real x the discriminant must be $\ge 0$: $4\alpha^4 - 20\alpha^2(\alpha^2 - 1) \ge 0 \Rightarrow \alpha^2(5 - 4\alpha^2) \ge 0 \Rightarrow -\frac{\sqrt{5}}{2} \le \alpha \le \frac{\sqrt{5}}{2}$. So $p = -\frac{\sqrt{5}}{2}$, $q = \frac{\sqrt{5}}{2}$ and $4(p^2 + q^2) = 4\left(\frac{5}{4} + \frac{5}{4}\right) = 10$. Answer: 10.
82	ProbabilityAddition and multiplication theorems	Class 11: Probability; Class 12: Probability	NumericalHard
A hard numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $P(B_1) = x$, $P(B_2) = y$, $P(B_3) = z$. By independence, $\frac{\alpha}{p} = \frac{x}{1-x} = a$, $\frac{\beta}{p} = \frac{y}{1-y} = b$, $\frac{\gamma}{p} = \frac{z}{1-z} = c$. Dividing the equations by $p^2$: $a - 2b = ab$ and $b - 3c = 2bc$. So $a = \frac{2b}{1-b}$ and $b = \frac{3c}{1-2c}$. Then $x = \frac{a}{1+a} = \frac{2b}{1+b}$, and $1 + b = \frac{1+c}{1-2c}$ gives $x = \frac{6c}{1+c}$. Also $z = \frac{c}{1+c}$. $\frac{P(B_1)}{P(B_3)} = \frac{x}{z} = 6$. Answer: 6.
83	Matrices and DeterminantsAdjoint and inverse	Class 12: Matrices; Class 12: Determinants	NumericalMedium
A medium numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\|P\| = 3(0 + 5\alpha) + 1(0 - 3\alpha) - 2(-10) = 12\alpha + 20$. $Q = kP^{-1} = \frac{k}{\|P\|}\operatorname{adj}P$, so $\|Q\| = \frac{k^3}{\|P\|} = \frac{k^2}{2} \Rightarrow k = \frac{12\alpha + 20}{2} = 6\alpha + 10$. $q_{23} = \frac{k}{\|P\|}C_{32}$, where $C_{32} = -\begin{vmatrix}3&-2\\2&\alpha\end{vmatrix} = -(3\alpha + 4)$. $-\frac{k(3\alpha+4)}{12\alpha+20} = -\frac{k}{8} \Rightarrow 24\alpha + 32 = 12\alpha + 20 \Rightarrow \alpha = -1$, and then $k = 4$. $\alpha^2 + k^2 = 1 + 16 = 17$. Answer: 17.
84	Matrices and DeterminantsAlgebra of matrices	Class 12: Matrices; Class 12: Determinants	NumericalMedium
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\operatorname{tr}(M^TM)$ is the sum of the squares of all nine entries; each square is 0, 1 or 4. Sum 7: either seven 1s and two 0s, $^9C_7 = 36$ ways; or one 2, three 1s and five 0s, $9 \times {}^8C_3 = 9 \times 56 = 504$ ways. Total $= 540$. Answer: 540.
85	Sequences and SeriesArithmetic progressions	Class 11: Sequences and Series	NumericalMedium
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Three-digit numbers $\equiv 2 \pmod 9$: 101, 110, …, 992 — 100 terms, sum $= 50(101 + 992) = 54650$. Required total $= 274 \times 400 = 109600$, so C must add $109600 - 54650 = 54950$ (l = 2 would add nothing). Three-digit numbers $\equiv l \pmod 9$ for $l \ne 2$: 100 terms from $99 + l$ to $990 + l$, sum $= 50(1089 + 2l) = 54450 + 100l$. $54450 + 100l = 54950 \Rightarrow l = 5$. Answer: 5.
86	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 Let $s = \sin x \in (0, 1)$, $f(s) = \frac{4}{s} + \frac{1}{1-s}$. $f'(s) = -\frac{4}{s^2} + \frac{1}{(1-s)^2} = 0 \Rightarrow 2(1-s) = s \Rightarrow s = \frac{2}{3}$. $f\left(\frac{2}{3}\right) = 6 + 3 = 9$, and $f \to \infty$ at both ends, so the range is $[9, \infty)$ and the least $\alpha$ is 9. Answer: 9.
87	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	NumericalMedium
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\int_{-a}^{a}\|x\|\,dx = a^2$ and $\int_{-a}^{a}\|x-2\|\,dx = \frac{(a+2)^2}{2} + \frac{(a-2)^2}{2} = a^2 + 4$. $2a^2 + 4 = 22 \Rightarrow a = 3$. $\int_3^{-3}(x + [x])\,dx = -\int_{-3}^{3}x\,dx - \int_{-3}^{3}[x]\,dx = 0 - (-3 - 2 - 1 + 0 + 1 + 2) = 3$. Answer: 3.
88	CirclesRadius and centre	Class 11: Conic Sections	NumericalEasy
A easy numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections The given circle has centre (1, 3) and radius $\sqrt{1 + 9 - 6} = 2$. The diameter is a chord of C with mid-point (1, 3), so the line from (2, 1) to (1, 3) is perpendicular to it; that distance is $\sqrt{1 + 4} = \sqrt{5}$. $R^2 = (\sqrt{5})^2 + 2^2 = 9 \Rightarrow R = 3$. Answer: 3.
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 Write $\vec{c} = \lambda\vec{a} + \mu\vec{b}$. Here $\|\vec{a}\|^2 = 3$, $\|\vec{b}\|^2 = 5$, $\vec{a}\cdot\vec{b} = -1$. $\vec{b}\cdot\vec{c} = 0$: $-\lambda + 5\mu = 0$. $\vec{a}\cdot\vec{c} = 7$: $3\lambda - \mu = 7$. So $\mu = \frac{1}{2}$, $\lambda = \frac{5}{2}$. $\vec{c} = \frac{5}{2}(-1, 1, 1) + \frac{1}{2}(2, 0, 1) = \left(-\frac{3}{2}, \frac{5}{2}, 3\right)$. $\vec{a} + \vec{b} + \vec{c} = \left(-\frac{1}{2}, \frac{7}{2}, 5\right)$, $\|\cdot\|^2 = \frac{1}{4} + \frac{49}{4} + 25 = \frac{75}{2}$. $2\|\vec{a} + \vec{b} + \vec{c}\|^2 = 75$. Answer: 75.
90	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	NumericalEasy
A easy numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\tan^{-1}\frac{1}{1 + r(r+1)} = \tan^{-1}(r+1) - \tan^{-1}r$. The sum telescopes to $\tan^{-1}(n+1) - \tan^{-1}1 \to \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$. $\tan\frac{\pi}{4} = 1$. Answer: 1.

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