JEE (Main) · 2021
JEE (Main) 2021 — 26 August, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Oscillations and Waves3
- Physical World and Measurement2
- Electronic Devices2
18 easy · 11 medium · 1 hard
Chemistry
30 Q
- Coordination Compounds3
- Hydrocarbons3
- Chemical Bonding and Molecular Structure2
19 easy · 10 medium · 1 hard
Mathematics
30 Q
- Trigonometry4
- Matrices and Determinants3
- Integral Calculus3
7 easy · 19 medium · 4 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Magnetic induction: from $F = qvB$, $[B] = \frac{\mathrm{MLT^{-2}}}{\mathrm{AT\cdot LT^{-1}}} = \mathrm{MT^{-2}A^{-1}}$, i.e. (iii).
Magnetic flux: $[\phi] = [B][\text{area}] = \mathrm{ML^2T^{-2}A^{-1}}$, i.e. (i).
Permeability: from $B = \frac{\mu_0 I}{2\pi r}$, $[\mu] = \frac{[B]\mathrm{L}}{\mathrm{A}} = \mathrm{MLT^{-2}A^{-2}}$, i.e. (iv).
Magnetization: magnetic moment per unit volume, $\frac{\mathrm{AL^2}}{\mathrm{L^3}} = \mathrm{M^0L^{-1}A}$, i.e. (ii).
So (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).
Answer: (C). | ||||
| 2 | Communication SystemsLine of sight communication | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on line of sight communication, from Communication Systems. NCERT: No matching NCERT chapter Maximum LoS distance: $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
$\sqrt{2\times6.4\times10^6\times50} = \sqrt{6.4\times10^8} \approx 25.3$ km.
$\sqrt{2\times6.4\times10^6\times80} = \sqrt{1.024\times10^9} = 32.0$ km.
$d \approx 25.3 + 32.0 = 57.3$ km.
Answer: (D). | ||||
| 3 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $T = 2\pi\sqrt{L/g}$, so $\frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta L}{L} = 0.05\% = 5\times10^{-4}$.
The clock runs slow; in a day it loses $86400\times5\times10^{-4} = 43.2$ s.
Answer: (C). | ||||
| 4 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Let $P = \overline{A+B}$ (the middle gate). The upper gate gives $Q = \overline{A+P}$, the lower gate $R = \overline{B+P}$, and $Y = \overline{Q+R}$.
A = 0, B = 0: P = 1, Q = 0, R = 0, Y = 1.
A = 0, B = 1: P = 0, Q = 1, R = 0, Y = 0.
A = 1, B = 0: P = 0, Q = 0, R = 1, Y = 0.
A = 1, B = 1: P = 0, Q = 0, R = 0, Y = 1.
Y = 1, 0, 0, 1: the circuit is an XNOR gate, the first table.
Answer: (A). | ||||
| 5 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane At release the bomb has the plane's horizontal velocity, and (air resistance neglected) it keeps it, so it stays directly below the plane. Relative to the plane its only motion is the vertical fall under gravity: a vertical straight line.
(To a ground observer the path is a parabola.)
Answer: (B). | ||||
| 6 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctHard | |
A hard single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $N_A = 2N_0e^{-\lambda t}$ where $N_0 = N_B(0)$, and $\frac{dN_B}{dt} = \lambda N_A - \lambda N_B$.
Try $N_B = e^{-\lambda t}f(t)$: $f' = 2\lambda N_0$, so $f = N_0(1+2\lambda t)$ and $\frac{N_B(t)}{N_B(0)} = (1+2\lambda t)e^{-\lambda t}$.
It starts at 1. Its derivative is $\lambda e^{-\lambda t}(2 - 1 - 2\lambda t) = \lambda e^{-\lambda t}(1-2\lambda t)$: it rises until $t = \frac{1}{2\lambda}$, where it reaches $2e^{-1/2} \approx 1.21$, then falls towards zero.
That is the second graph (maximum at $t = \frac{1}{2\lambda}$).
Answer: (B). | ||||
| 7 | 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 In the figure $\vec{B}$ is drawn from the head of $\vec{A}$, and the angle between the two drawn arrows at that point is $120^\circ$; so the angle between the directions of $\vec{A}$ and $\vec{B}$ is $180^\circ - 120^\circ = 60^\circ$.
Resolve $\vec{A}-\vec{B}$ along and across $\vec{A}$: along $\vec{A}$, $A - B\cos60^\circ = A - \frac{B}{2}$; across it, $B\sin60^\circ = \frac{\sqrt{3}B}{2}$ (on the side away from $\vec{B}$).
$\tan\beta = \frac{\sqrt{3}B/2}{A - B/2} = \frac{\sqrt{3}B}{2A-B}$.
Answer: (B). | ||||
| 8 | 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 $\lambda = \frac{h}{\sqrt{2mE}}$, so $E \propto \frac{1}{\lambda^2}$.
With $\lambda' = \frac{3}{4}\lambda$: $E' = \frac{16}{9}E$.
Extra energy $= \frac{16}{9}E - E = \frac{7}{9}E$.
Answer: (C). | ||||
| 9 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion A conical pendulum: with $\theta$ the string's angle to the vertical, $\sin\theta = \frac{r}{L} = \frac{1}{\sqrt{2}}$, so $\theta = 45^\circ$.
$T\cos\theta = mg$ and $T\sin\theta = \frac{mv^2}{r}$ give $v^2 = rg\tan\theta = rg$.
$v = \sqrt{rg}$.
Answer: (A). | ||||
| 10 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium 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 AB is the cylinder's own axis, so $I_{AB} = \frac{1}{2}Mr^2 = \frac{1}{2}M(0.2)^2 = 0.02M$.
CD is parallel to AB at distance $\frac{L}{2} = 0.4$ m: $I_{CD} = 0.02M + M(0.4)^2 = 0.18M$.
$0.18M = 2.7$ gives $M = 15$ kg.
Volume $= \pi r^2L = \pi(0.04)(0.8) = 0.1005\ \mathrm{m^3}$; density $= \frac{15}{0.1005} \approx 149\ \mathrm{kg/m^3} = 1.49\times10^2\ \mathrm{kg/m^3}$.
Answer: (B). | ||||
| 11 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $B_0 = \frac{E_0}{c} = \frac{800}{3\times10^8} = 2.67\times10^{-6}$ T.
Largest force when the velocity is perpendicular to $\vec{B}$: $F = evB_0 = 1.6\times10^{-19}\times3\times10^7\times2.67\times10^{-6} = 1.28\times10^{-17}$ N $= 12.8\times10^{-18}$ N.
Answer: (B). | ||||
| 12 | 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 correctMedium | |
A medium 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 Equal masses, so heat lost = heat gained reduces to specific heats.
$x$ with $y$: $c_x(16-10) = c_y(20-16)$, so $c_y = 1.5c_x$.
$y$ with $z$: $c_y(26-20) = c_z(30-26)$, so $c_z = 1.5c_y = 2.25c_x$.
$x$ with $z$: $c_x(T-10) = 2.25c_x(30-T)$, so $3.25T = 77.5$ and $T = 23.85^\circ$C.
This is 23.84°C as printed (the option truncates $23.846$).
Answer: (B). | ||||
| 13 | 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 Tension in an Atwood machine: $T = \frac{2m_1m_2g}{m_1+m_2} = \frac{2\times3\times5\times10}{8} = 37.5$ N.
The wire must not break: $\frac{T}{\pi r^2} \le \frac{2400}{\pi}$, so $r^2 \ge \frac{37.5}{2400} = 0.015625\ \mathrm{m^2}$.
$r_{min} = 0.125$ m $= 12.5$ cm.
Answer: (A). | ||||
| 14 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Treat it as an ideal (Carnot) refrigerator between $T_2 = 263$ K and $T_1 = 298$ K.
Coefficient of performance $= \frac{T_2}{T_1-T_2} = \frac{263}{35}$.
Heat removed from the cold space per second: $Q_2 = \frac{263}{35}\times35 = 263$ J/s.
(The heat given to the surroundings would be $263 + 35 = 298$ J/s.)
Answer: (B). | ||||
| 15 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Capacitor branch: $X_C = \frac{1}{100\times100\times10^{-6}} = 100\ \Omega$, so $Z_1 = 100 - 100j\ \Omega$ and $I_1 = \frac{200}{100-100j} = 1 + j$ A ($\sqrt{2}$ A, leading by $45^\circ$).
Inductor branch: $X_L = 100\times0.5 = 50\ \Omega$, so $Z_2 = 50 + 50j\ \Omega$ and $I_2 = \frac{200}{50+50j} = 2 - 2j$ A ($2\sqrt{2}$ A, lagging by $45^\circ$).
The branch currents are $90^\circ$ apart, so $I = (1+j)+(2-2j) = 3 - j$ and $|I| = \sqrt{10} \approx 3.16$ A.
None of the options is 3.16 A. Option (B), 4.24 A $= 3\sqrt{2}$, is the plain sum $\sqrt{2} + 2\sqrt{2}$ of the branch currents, which ignores their phase difference.
Answer: none of the options; the current is $\sqrt{10} \approx 3.16$ A. | ||||
| 16 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Dalton's law with the ideal gas equation: $p = \frac{(n_1+n_2)RT}{V} = \frac{3\times8.3\times400}{4.0\times10^{-3}} = 2.49\times10^6$ Pa $= 24.9\times10^5$ Pa.
Answer: (A). | ||||
| 17 | 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 Bulb resistance: $R_b = \frac{V^2}{P} = \frac{100^2}{500} = 20\ \Omega$; at 500 W it carries $I = \frac{500}{100} = 5$ A.
Total resistance needed: $\frac{200}{5} = 40\ \Omega$, so $R = 40 - 20 = 20\ \Omega$.
Answer: (A). | ||||
| 18 | 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 The capacitor splits into two halves of area $\frac{A}{2}$ in parallel.
Empty half: $C_1 = \frac{\varepsilon_0(A/2)}{d} = \frac{\varepsilon_0A}{2d}$.
Filled half: the two slabs of thickness $\frac{d}{2}$ are in series, $C_2 = \frac{\varepsilon_0(A/2)}{\frac{d}{2K_1}+\frac{d}{2K_2}} = \frac{\varepsilon_0A}{d}\cdot\frac{K_1K_2}{K_1+K_2}$.
$C = C_1 + C_2 = \frac{\varepsilon_0A}{d}\left(\frac{1}{2}+\frac{K_1K_2}{K_1+K_2}\right)$.
Answer: (D). | ||||
| 19 | 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 Read each as a parallel pair joined in series with the other two.
(A) $\frac{6\times8}{14} + 2 + 4 = \frac{66}{7}\ \Omega$.
(B) $\frac{2\times6}{8} + 4 + 8 = 13.5\ \Omega$.
(C) $\frac{2\times4}{6} + 6 + 8 = \frac{4}{3} + 14 = \frac{46}{3}\ \Omega$.
(D) $\frac{4\times6}{10} + 2 + 8 = 12.4\ \Omega$.
Answer: (C). | ||||
| 20 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Every point of a ring is at distance $a$ from its own centre and $\sqrt{s^2+a^2}$ from the other ring's centre. With $k = \frac{1}{4\pi\varepsilon_0}$:
$V_1 = \frac{kQ}{a} - \frac{kQ}{\sqrt{s^2+a^2}}$, $V_2 = -\frac{kQ}{a} + \frac{kQ}{\sqrt{s^2+a^2}}$.
$V_1 - V_2 = 2kQ\left[\frac{1}{a}-\frac{1}{\sqrt{s^2+a^2}}\right] = \frac{Q}{2\pi\varepsilon_0}\left[\frac{1}{a}-\frac{1}{\sqrt{s^2+a^2}}\right]$.
Answer: (D). | ||||
| 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 force F acts on the lower 2 kg block; only friction drives the 1 kg block on top.
Largest acceleration friction can give the top block: $a = \mu g = 0.5\times10 = 5\ \mathrm{m\ s^{-2}}$.
Moving together: $F = (1+2)\times5 = 15$ N.
Answer: 15. | ||||
| 22 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The zener holds the load at 10 V (the open-circuit load voltage $24\times\frac{5}{6} = 20$ V exceeds 10 V, so the zener conducts).
Series current: $\frac{24-10}{1\ \mathrm{k\Omega}} = 14$ mA; load current: $\frac{10}{5\ \mathrm{k\Omega}} = 2$ mA; zener current $I_z = 12$ mA.
Power: $P = V_zI_z = 10\times12 = 120$ mW.
Answer: 120. | ||||
| 23 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalMedium | |
A medium numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The energy is expressed through $g$ and $T$ via the length: $E \propto mgl$ with $l = \frac{gT^2}{4\pi^2}$, so $E \propto mg^2T^2$.
Relative error: $\frac{\Delta E}{E} = 2\frac{\Delta g}{g} + 2\frac{\Delta T}{T} = 2(4) + 2(3) = 14\%$.
Answer: 14. | ||||
| 24 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\sin2\pi t + \cos2\pi t = \sqrt{2}\sin\left(2\pi t + \frac{\pi}{4}\right)$, so $x_2 = 5\sqrt{2}\cdot\sqrt{2}\sin\left(2\pi t+\frac{\pi}{4}\right) = 10\sin\left(2\pi t+\frac{\pi}{4}\right)$.
Amplitudes 10 and 5: the ratio is 2.
Answer: 2. | ||||
| 25 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Final speed $v = \frac{c}{6} = \frac{3\times10^8}{6} = 5\times10^7$ m/s; kinetic energy $K = \frac{1}{2}m_pv^2 = \frac{1}{2}\times1.67\times10^{-27}\times2.5\times10^{15} = 2.0875\times10^{-12}$ J (about 13.05 MeV).
The proton crosses the gap twice a revolution, gaining $2eV = 2\times1.6\times10^{-19}\times12000 = 3.84\times10^{-15}$ J (24 keV) per revolution.
$n = \frac{2.0875\times10^{-12}}{3.84\times10^{-15}} \approx 543.62$ revolutions.
Answer: 543.62 (544 to the nearest integer). | ||||
| 26 | 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 Unpolarised light through $P_1$ drops to $\frac{I}{2}$; since the screen still gets $\frac{I}{2}$, $P_2$ is parallel to $P_1$.
Rotating $P_2$ by $\theta$ (Malus's law): $\frac{I}{2}\cos^2\theta = \frac{3I}{8}$, so $\cos^2\theta = \frac{3}{4}$ and $\theta = 30^\circ$.
Answer: 30. | ||||
| 27 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The final image falls on the object only if the rays from the lens strike the mirror normally and retrace their path, i.e. the lens alone would form its image at the mirror's centre of curvature.
Radius of curvature $R = 2f = 30$ cm behind the mirror, so the lens image (without the mirror) is $8 + 30 = 38$ cm from the lens.
Distance from the object: $12 + 38 = 50$ cm.
Answer: 50. | ||||
| 28 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\varepsilon_{max} = NBA\omega = 20\times3.0\times10^{-2}\times\pi(0.08)^2\times50$.
$\pi(0.08)^2 = 2.011\times10^{-2}\ \mathrm{m^2}$, so $\varepsilon_{max} = 0.6\times2.011\times10^{-2}\times50 = 0.603$ V $= 60.3\times10^{-2}$ V.
Nearest integer: 60.
Answer: 60. | ||||
| 29 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Area: $A = \frac{\sqrt{3}}{4}(0.1)^2 = \frac{\sqrt{3}}{4}\times10^{-2}\ \mathrm{m^2}$.
With the plane parallel to $\vec{B}$ the moment is perpendicular to it: $\tau = IAB = 0.2\times\frac{\sqrt{3}}{4}\times10^{-2}\times0.02 = \sqrt{3}\times10^{-5}$ N m.
So $x = 3$.
Answer: 3. | ||||
| 30 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Phase difference: $\phi = kx_0 = 6.28\times3.5 = 2\pi\times3.5 = 7\pi$, an odd multiple of $\pi$, so the waves are in opposite phase.
$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos7\pi} = |A_1 - A_2| = 12 - 5 = 7$ mm.
Answer: 7. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure London (dispersion) forces arise between instantaneous and induced dipoles; their interaction energy varies as $\frac{1}{r^6}$, i.e. $r^{-6}$.
So $x = -6$.
Answer: (D). | ||||
| 32 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{O_2^-}$ has 17 electrons: $\sigma1s^2\sigma^*1s^2\sigma2s^2\sigma^*2s^2\sigma2p_z^2\pi2p_x^2\pi2p_y^2\pi^*2p_x^2\pi^*2p_y^1$.
Bond order $= \frac{10-7}{2} = 1.5$; one unpaired electron in $\pi^*$, so it is paramagnetic.
Answer: (B). | ||||
| 33 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter A precipitate adsorbs the ion of its own lattice that is in excess.
(A) KI into excess $\mathrm{AgNO_3}$: AgI adsorbs $\mathrm{Ag^+}$, a positive sol.
(B) $\mathrm{AgNO_3}$ into excess KI: AgI adsorbs $\mathrm{I^-}$, giving $\mathrm{AgI/I^-}$, a negative sol.
(C) Hydrated ferric oxide from $\mathrm{FeCl_3}$ in hot water adsorbs $\mathrm{Fe^{3+}}$: positive.
(D) Hydrated alumina is a positive sol.
Answer: (B). | ||||
| 34 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter The chalcogens are group 16: O, S, Se, Te, Po.
Tb, Pu and Pm are f-block elements and Ti is a transition metal, so only (C) has all three from group 16.
Answer: (C). | ||||
| 35 | Isolation of MetalsFroth flotation | No matching NCERT chapter | StatementsEasy | |
A easy statements question on froth flotation, from Isolation of Metals. NCERT: No matching NCERT chapter Statement I: sphalerite is ZnS and copper glance is $\mathrm{Cu_2S}$; true.
Statement II: in froth flotation two sulphide ores can be separated by adjusting the oil-to-water ratio or with a depressant (NaCN depresses ZnS in a ZnS–PbS mixture); true.
Answer: (A). | ||||
| 36 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter (A) is true: $\mathrm{D_2O}$ is used to label hydrogen and to study reaction mechanisms (isotope effects).
(R) is false: the O–D bond is stronger (lower zero-point energy), so O–H cleaves faster than O–D, not slower.
Answer: (C). | ||||
| 37 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Assertion and reasonMedium | |
A medium assertion and reason question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter (A) is true: $\mathrm{BaCO_3}$ is practically insoluble in water and is the most thermally stable of the alkaline earth carbonates.
(R) is true: down the group the cation grows larger, polarises the carbonate ion less, and the carbonate decomposes only at higher temperature.
The large $\mathrm{Ba^{2+}}$ is why $\mathrm{BaCO_3}$ is so stable, so (R) explains (A).
Answer: (A). | ||||
| 38 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Complete hydrolysis: $\mathrm{PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl}$.
In $\mathrm{H_3PO_4}$, $\mathrm{OP(OH)_3}$, all three hydrogens are on oxygen and are ionisable; none is bonded to P.
Non-ionisable hydrogens: 0.
Answer: (D). | ||||
| 39 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctHard | |
A hard single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The splitting $\Delta_o$ grows with the ligand's field strength ($\mathrm{F^- < H_2O < NH_3 < en}$) and with the metal's oxidation state (Co(III) splits more than Co(II)).
B: Co(II) with the weak-field water has the smallest splitting.
A: Co(III) with $\mathrm{F^-}$, high-spin $d^6$.
C and D: Co(III) with the strong-field ligands $\mathrm{NH_3}$ and en, low-spin $d^6$ ($t_{2g}^6$, CFSE $= -2.4\Delta_o$), D the larger since en is stronger than $\mathrm{NH_3}$.
Computed in full: [CoF₆]³⁻ is high-spin d⁶, CFSE = −0.4Δo with Δo ≈ 13 000 cm⁻¹, about 5200 cm⁻¹; [Co(H₂O)₆]²⁺ is high-spin d⁷, CFSE = −0.8Δo with Δo ≈ 9300 cm⁻¹, about 7400 cm⁻¹. So A has the smaller CFSE.
Order of increasing CFSE: A < B < C < D.
Answer: (A). | ||||
| 40 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) $[\mathrm{M(AA)_2B_2}]$: cis and trans.
(B) $[\mathrm{Co(CN)_5(NC)}]^{3-}$ is of the type $\mathrm{MA_5B}$ (one isocyanide among five cyanides): only one arrangement, no geometrical isomers (it shows linkage isomerism instead).
(C) $\mathrm{MA_4B_2}$: cis and trans.
(D) $\mathrm{MA_3B_3}$: fac and mer.
Answer: (B). | ||||
| 41 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter (A) is true: photochemical smog cracks rubber and damages plant life.
(R) is true: ozone, nitric oxide, acrolein, formaldehyde and PAN make it an oxidising smog.
The cracking of rubber is caused by these oxidants (chiefly ozone attacking the C=C bonds), so (R) explains (A).
Answer: (A). | ||||
| 42 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Aromatic: cyclic, planar, fully conjugated, with $(4n+2)\pi$ electrons.
(A) Cyclopentadienyl anion: $6\pi$ electrons, aromatic.
(B) Cyclooctatetraene: $8\pi$ electrons; it is tub-shaped, not planar, so it is non-aromatic.
(C) Tropylium cation: $6\pi$ electrons, aromatic.
(D) Phenol: a benzene ring, $6\pi$ electrons, aromatic.
Answer: (B). | ||||
| 43 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques C1 and C2 are stereocentres.
cis-1,2-dimethylcyclopropane has a mirror plane: one meso form.
trans-1,2-dimethylcyclopropane is chiral: a pair of enantiomers.
Total: 1 + 2 = 3 stereoisomers.
Answer: (C). | ||||
| 44 | 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 With $\mathrm{Br_2}$ and the Lewis acid $\mathrm{AlBr_3}$ the electrophile $\mathrm{Br^+}$ attacks the ring (no light or radical conditions for the side chain).
The acyl group $-\mathrm{COC_4H_9}$ withdraws electrons by resonance and deactivates the ortho and para positions most, so substitution is at the meta position.
Product: 1-(3-bromophenyl)pentan-1-one.
Answer: (A). | ||||
| 45 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Matching listMedium | |
A medium matching list question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (a) Acid hydrolysis of ethyl acetate gives acetic acid and ethanol: (ii).
(b) DIBAL-H reduces an ester to the aldehyde: (iii).
(c) Stephen reduction ($\mathrm{SnCl_2}$/HCl, then water) turns a nitrile into the aldehyde: (iv).
(d) One equivalent of $\mathrm{CH_3MgBr}$ adds to the nitrile to give an imine salt, hydrolysed to acetone: (i).
So (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i). (DIBAL-H also reduces nitriles, but $\mathrm{SnCl_2}$/HCl does not reduce esters, so (b) must take (iii).)
Answer: (A). | ||||
| 46 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Cyanohydrin formation needs the nucleophile $\mathrm{CN^-}$; a trace of base (NaOH) generates it from HCN, while acid suppresses it. So X is NaOH.
$\mathrm{LiAlH_4}$ reduces $-\mathrm{C{\equiv}N}$ to $-\mathrm{CH_2NH_2}$ and leaves the OH: Y is $\mathrm{CH_3CH_2CH(CH_3)CH(OH)CH_2NH_2}$, 1-amino-3-methylpentan-2-ol.
That is X - NaOH with the $-\mathrm{CH(OH)CH_2NH_2}$ structure.
Answer: (C). | ||||
| 47 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Acetic anhydride acylates the nucleophilic primary amine of the $-\mathrm{CH_2NH_2}$ group.
The amide nitrogen of $-\mathrm{CONH_2}$ is not nucleophilic (its lone pair is delocalised onto the carbonyl), so it is not acetylated.
P: the benzylamine nitrogen becomes $-\mathrm{CH_2NHCOCH_3}$ and $-\mathrm{CONH_2}$ is unchanged.
Answer: (A). | ||||
| 48 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Chlordiazepoxide (a benzodiazepine, like meprobamate and equanil in use) relieves anxiety and tension: it is a tranquilizer.
Answer: (A). | ||||
| 49 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Assertion and reasonEasy | |
A easy assertion and reason question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules (A) is true: sucrose is a disaccharide and is non-reducing because both anomeric carbons are tied up in the glycosidic bond.
(R) is false: the linkage is between C1 of $\alpha$-D-glucose and C2 of $\beta$-D-fructose, not $\beta$-glucose and $\alpha$-fructose.
Answer: (C). | ||||
| 50 | Practical Organic ChemistryDetection of functional groups | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on detection of functional groups, from Practical Organic Chemistry. NCERT: No matching NCERT chapter In the phthalein test, phthalic anhydride condenses at the position para to the phenolic OH (phenol gives phenolphthalein, pink in alkali; resorcinol gives fluorescein; catechol also gives a coloured product).
In p-cresol the para position is blocked by $\mathrm{CH_3}$, so no phthalein dye forms and no colour appears.
Answer: (D). | ||||
| 51 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles of Na: $\frac{3.45}{23} = 0.15$; each $\mathrm{Na_3PO_4}$ has 3 Na, so $n = 0.05$ mol.
Molarity $= \frac{0.05}{0.100\ \mathrm{L}} = 0.5\ \mathrm{M} = 50\times10^{-2}\ \mathrm{mol\ L^{-1}}$.
Answer: 50. | ||||
| 52 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $\nu = \frac{c}{\lambda} = \frac{3\times10^8}{500\times10^{-9}} = 6.0\times10^{14}$ Hz.
$\frac{1}{2}mv^2 = h(\nu-\nu_0) = 6.63\times10^{-34}\times1.7\times10^{14} = 1.127\times10^{-19}$ J.
$v = \sqrt{\frac{2\times1.127\times10^{-19}}{9.0\times10^{-31}}} = \sqrt{2.505\times10^{11}} = 5.0\times10^5$ m/s.
Answer: 5. | ||||
| 53 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $\mathrm{H_2O(l) \rightarrow H_2O(g)}$: $\Delta n_g = 1$, so $\Delta H = \Delta U + \Delta n_gRT$.
$\Delta U = 41 - \frac{8.3\times373}{1000} = 41 - 3.096 = 37.90\ \mathrm{kJ\ mol^{-1}}$.
Answer: 37.90. | ||||
| 54 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Ethylene glycol $\mathrm{C_2H_6O_2}$: $M = 62$; $n = \frac{83}{62} = 1.339$ mol.
Molality $= \frac{1.339}{0.625} = 2.142$ mol/kg; $\Delta T_f = 1.86\times2.142 = 3.98$ K.
$T_f = 273 - 3.98 = 269.02$ K, nearest integer 269.
Answer: 269. | ||||
| 55 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Initially $Q = 1 < 100$, so the reaction moves forward by $x$: $\frac{(1+x)^2}{(1-x)^2} = 100$.
$\frac{1+x}{1-x} = 10$, so $x = \frac{9}{11} = 0.818$ M.
$[D] = 1 + 0.818 = 1.818\ \mathrm{M} = 181.8\times10^{-2}$ M, nearest integer 182.
Answer: 182. | ||||
| 56 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry The given values are oxidation potentials, so $E^0_{cell} = E^0_{\mathrm{Zn/Zn^{2+}}} - E^0_{\mathrm{Cu/Cu^{2+}}} = 0.76 + 0.34 = 1.10$ V.
$E_{cell} = 1.10 - \frac{0.059}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} = 1.10 - 0.0295\log2 = 1.10 - 0.0089 = 1.091$ V.
$= 109.1\times10^{-2}$ V, nearest integer 109.
Answer: 109. | ||||
| 57 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The two terms are the forward and backward rates: $k_f = 4.8\times10^{-5}$, $k_b = 2.4\times10^{-3}$.
At equilibrium the net rate is zero: $K_c = \frac{k_f}{k_b} = \frac{4.8\times10^{-5}}{2.4\times10^{-3}} = 0.02$ (water, the solvent, is left out).
Nearest integer: 0.
Answer: 0. | ||||
| 58 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The dissociation constant is the reciprocal of the stability constant: $\frac{1}{2.1\times10^{13}} = 4.76\times10^{-14}$.
$y = 4.76$, nearest integer 5.
Answer: 5. | ||||
| 59 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $M(\mathrm{BaSO_4}) = 137 + 32 + 64 = 233$; 233 g of $\mathrm{BaSO_4}$ hold 32 g of S.
% S $= \frac{32}{233}\times\frac{1.44}{0.471}\times100 = 41.99\%$, nearest integer 42.
Answer: 42. | ||||
| 60 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Moles of vapour: $\frac{448}{22400} = 0.02$ mol, so the molar mass is $\frac{1.53}{0.02} = 76.5$ g/mol.
An alkene (it undergoes ozonolysis) with one Cl: $\mathrm{C_nH_{2n-1}Cl}$ has $M = 14n + 34.5$; $14n + 34.5 = 76.5$ gives $n = 3$.
A is $\mathrm{C_3H_5Cl}$, e.g. $\mathrm{CH_2{=}CH{-}CH_2Cl}$, which on ozonolysis gives HCHO and $\mathrm{ClCH_2CHO}$, both aldehydes.
Number of carbon atoms: 3.
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\mathrm{cosec}^{-1}t$ needs $|t| \ge 1$, and $x \ne 0$.
$\left|\frac{1+x}{x}\right| \ge 1 \iff (1+x)^2 \ge x^2 \iff 1 + 2x \ge 0 \iff x \ge -\frac{1}{2}$.
Domain: $\left[-\frac{1}{2}, \infty\right) - \{0\}$.
Answer: (A). | ||||
| 62 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\sqrt{3}+i = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$, so $(\sqrt{3}+i)^{100} = 2^{100}\left(\cos\frac{50\pi}{3} + i\sin\frac{50\pi}{3}\right)$.
$\frac{50\pi}{3} = 16\pi + \frac{2\pi}{3}$: the value is $2^{100}\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 2^{99}(-1 + \sqrt{3}i)$.
So $p = -1$, $q = \sqrt{3}$: sum $\sqrt{3}-1$, product $-\sqrt{3}$.
Equation: $x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0$.
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 $\Delta = \begin{vmatrix}1&1&1\\1&2&3\\1&3&\lambda\end{vmatrix} = (2\lambda - 9) - (\lambda - 3) + (3 - 2) = \lambda - 5$.
Unique solution iff $\lambda \ne 5$: $p = \frac{5}{6}$.
If $\lambda = 5$: subtracting equations, $y + 2z = \mu - 5$ and $y + 2z = 1 - \mu$; these agree only when $\mu = 3$ (infinitely many solutions), otherwise there is no solution.
No solution: $\lambda = 5$, $\mu \ne 3$, i.e. 5 outcomes of 36: $q = \frac{5}{36}$.
Answer: (D). | ||||
| 64 | 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 $A^2 = \begin{pmatrix}1&0&0\\1&1&1\\1&0&0\end{pmatrix}$, $A^3 = \begin{pmatrix}1&0&0\\2&1&1\\1&0&0\end{pmatrix}$; by induction $A^n = \begin{pmatrix}1&0&0\\ n-1&1&1\\1&0&0\end{pmatrix}$ (the row multiplying gives $(n-1) + 0 + 1 = n$ in the next power).
So $A^n = A + (n-1)E$ with $E$ the matrix having 1 in row 2, column 1 and zeros elsewhere.
$A^{2025} - A^{2020} = 5E$, and $A^6 - A = 5E$ as well (while $A^5 - A = 4E$).
Answer: (C). | ||||
| 65 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $n(n+1)x^2 + 2(2n+1)x + 4 = (nx+2)((n+1)x+2)$.
So $\frac{x}{(nx+2)((n+1)x+2)} = \frac{1}{nx+2} - \frac{1}{(n+1)x+2}$, and the sum telescopes to $\frac{1}{x+2} - \frac{1}{10x+2}$.
At $x = 2$: $\frac{1}{4} - \frac{1}{22} = \frac{11-2}{44} = \frac{9}{44}$.
Answer: (C). | ||||
| 66 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $\ln f = x^2(\ln2 - \ln x)$; $\frac{f'}{f} = 2x(\ln2 - \ln x) - x = x(2\ln2 - 2\ln x - 1)$.
$f' = 0$ at $\ln x = \ln2 - \frac{1}{2}$, i.e. $x = \frac{2}{\sqrt{e}}$; $f'$ changes from + to −, a local maximum.
There $\frac{2}{x} = \sqrt{e}$ and $x^2 = \frac{4}{e}$, so $f = (\sqrt{e})^{4/e} = e^{2/e}$.
Answer: (B). | ||||
| 67 | 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 Replace $x$ by $-x$: $I = \int_{-\pi/2}^{\pi/2}\frac{(1+\sin^2x)\pi^{\sin x}}{1+\pi^{\sin x}}dx$.
Adding: $2I = \int_{-\pi/2}^{\pi/2}(1+\sin^2x)dx = \pi + \frac{\pi}{2} = \frac{3\pi}{2}$.
$I = \frac{3\pi}{4}$.
Answer: (B). | ||||
| 68 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} = \frac{1}{x} - \frac{e^y}{2x^2}$. Put $t = e^{-y}$: $\frac{dt}{dx} = -t\frac{dy}{dx} = -\frac{t}{x} + \frac{1}{2x^2}$.
Linear: $\frac{dt}{dx} + \frac{t}{x} = \frac{1}{2x^2}$, integrating factor $x$: $\frac{d}{dx}(tx) = \frac{1}{2x}$, so $xe^{-y} = \frac{1}{2}\ln x + C$.
$y(e) = 1$: $e\cdot e^{-1} = \frac{1}{2} + C$, so $C = \frac{1}{2}$.
At $x = 1$: $e^{-y} = \frac{1}{2}$, so $y(1) = \log_e2$.
Answer: (C). | ||||
| 69 | 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 Let $\theta = \angle PCB$, $\tan\theta = 2$: the distance of P from the pencil is $\sqrt{5}\sin\theta = \sqrt{5}\cdot\frac{2}{\sqrt{5}} = 2$ inches.
For distance 1 the angle $\phi$ between CP and the pencil needs $\sqrt{5}\sin\phi = 1$, so $\tan\phi = \frac{1}{2}$.
Rotate the pencil towards P by $\theta - \phi$: $\tan(\theta-\phi) = \frac{2 - \frac{1}{2}}{1 + 2\cdot\frac{1}{2}} = \frac{3}{4}$.
(The other choice, the angle $\pi - \phi$, needs a rotation of $90^\circ$ since $\theta + \phi = 90^\circ$, which is not acute.)
Answer: (B). | ||||
| 70 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Take the floor as D(0,0,0), C(10,0,0), B(10,10,0), A(0,10,0) and height $h$: H(0,0,h), G(10,0,h).
The diagonals AG and BH meet at their common midpoint P(5,5,h/2).
$\vec{PG} = (5,-5,\frac{h}{2})$, $\vec{PH} = (-5,-5,\frac{h}{2})$: $\cos\angle GPH = \frac{-25+25+\frac{h^2}{4}}{50+\frac{h^2}{4}} = \frac{1}{5}$.
$\frac{5h^2}{4} = 50 + \frac{h^2}{4}$, so $h^2 = 50$ and $h = 5\sqrt{2}$ m.
Answer: (A). | ||||
| 71 | CirclesStandard and general form | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Circles touching $x - 2y = 0$ at (2, 1): $(x-2)^2 + (y-1)^2 + \lambda(x - 2y) = 0$, i.e. $x^2 + y^2 + (\lambda-4)x - (2\lambda+2)y + 5 = 0$.
Common chord with $C_1$ (subtract): $(\lambda-4)x - (2\lambda+4)y + 10 = 0$. It is a diameter of $C_1$, so it passes through its centre (0, −1): $2\lambda + 4 + 10 = 0$, $\lambda = -7$.
C: $x^2 + y^2 - 11x + 12y + 5 = 0$, centre $\left(\frac{11}{2}, -6\right)$, $r^2 = \frac{121}{4} + 36 - 5 = \frac{245}{4}$, $r = \frac{7\sqrt{5}}{2}$.
Diameter $= 7\sqrt{5}$.
Answer: (D). | ||||
| 72 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $e^2 = 1 + \frac{b^2}{a^2} = \frac{5}{4}$ gives $b^2 = \frac{a^2}{4}$; P on the curve: $\frac{24}{a^2} - \frac{12}{a^2} = 1$, so $a^2 = 12$, $b^2 = 3$.
Tangent at P: $\frac{-2\sqrt{6}x}{12} - \frac{\sqrt{3}y}{3} = 1$; at $x = 0$, $y = -\sqrt{3}$: Q(0, −√3).
Normal at P: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$, i.e. $\frac{12x}{-2\sqrt{6}} + \frac{3y}{\sqrt{3}} = 15$; at $x = 0$, $y = 5\sqrt{3}$: R(0, 5√3).
$QR = 5\sqrt{3} + \sqrt{3} = 6\sqrt{3}$.
Answer: (D). | ||||
| 73 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Chord with mid-point $(h,k)$: $T = S_1$, i.e. $hx - ky = h^2 - k^2$, or $y = \frac{h}{k}x - \frac{h^2-k^2}{k}$.
It touches $y^2 = 8x$ ($a = 2$) when $c = \frac{a}{m}$: $-\frac{h^2-k^2}{k} = \frac{2k}{h}$.
$-h(h^2 - k^2) = 2k^2$, so $k^2(h-2) = h^3$.
Locus: $y^2(x-2) = x^3$.
Answer: (C). | ||||
| 74 | 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 On $[k, k+1)$ put $x = k + t$: the integrand is $(2k + t)e^{-t}$.
$\int_0^1(2k+t)e^{-t}dt = 2k\left(1 - \frac{1}{e}\right) + \left(1 - \frac{2}{e}\right)$.
Summing $k = 0$ to 4: $20\left(1 - \frac{1}{e}\right) + 5\left(1 - \frac{2}{e}\right) = 25 - 30e^{-1}$.
$\alpha = -30$, $\beta = 25$ (indeed $5\alpha + 6\beta = 0$); $(\alpha+\beta)^2 = 25$.
Answer: (A). | ||||
| 75 | 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 (S1): $(p\to q) \equiv \sim p \vee q$ and $(\sim q \to p) \equiv q \vee p$; their disjunction contains $\sim p \vee p$, so it is always true: a tautology. True.
(S2): $\sim p \vee q \equiv \sim(p \wedge \sim q)$, so the statement is $r \wedge \sim r$, always false: a fallacy. True.
Answer: (D). | ||||
| 76 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $h(x) = \min\{\{x\}, 1 - \{x\}\}$, the distance from $x$ to the nearest integer: a zig-zag between 0 and $\frac{1}{2}$.
At an integer $n$: from the left $\{x\}\to1$ so $h \to \min(1,0) = 0$; from the right $h \to 0$; and $h(n) = 0$. So h is continuous on $[-2, 2]$.
It has corners at the integers $-1, 0, 1$ and at the half-integers $-\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{3}{2}$: 7 points of non-differentiability in $(-2, 2)$, more than four.
Answer: (C). | ||||
| 77 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\frac{1}{2r^2} = \frac{2}{4r^2} = \frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}$, so $\tan^{-1}\frac{1}{2r^2} = \tan^{-1}(2r+1) - \tan^{-1}(2r-1)$.
The sum telescopes: $p = \tan^{-1}101 - \tan^{-1}1$.
$\tan p = \frac{101 - 1}{1 + 101} = \frac{100}{102} = \frac{50}{51}$.
Answer: (A). | ||||
| 78 | 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 $P(X \ge n) = \left(\frac{5}{6}\right)^{n-1}$ (the first $n-1$ tosses are not six).
$P(X \ge 5 \mid X > 2) = \frac{P(X \ge 5)}{P(X \ge 3)} = \frac{(5/6)^4}{(5/6)^2} = \frac{25}{36}$.
Answer: (B). | ||||
| 79 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Planes through the line: $(x + y + 4z - 16) + \lambda(-x + y + z - 6) = 0$.
Through (1, 2, 3): $-1 + \lambda(-2) = 0$, $\lambda = -\frac{1}{2}$, giving $3x + y + 7z - 26 = 0$.
(−8, 8, 6): $-24 + 8 + 42 - 26 = 0$, on P. (6, −6, 2): $18 - 6 + 14 - 26 = 0$, on P. (4, 2, 2): $12 + 2 + 14 - 26 = 2 \ne 0$, not on P. (3, 3, 2): $9 + 3 + 14 - 26 = 0$, on P.
Answer: (C). | ||||
| 80 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin\frac{7\pi}{8} = \sin\frac{\pi}{8}$, $\sin\frac{5\pi}{8} = \sin\frac{3\pi}{8} = \cos\frac{\pi}{8}$, $\sin\frac{2\pi}{8} = \sin\frac{6\pi}{8} = \frac{1}{\sqrt{2}}$.
Product $= 2\cdot\frac{1}{2}\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8} = \left(\frac{1}{2}\sin\frac{\pi}{4}\right)^2 = \left(\frac{1}{2\sqrt{2}}\right)^2 = \frac{1}{8}$.
Answer: (A). | ||||
| 81 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha$ is common: $3\alpha^2 - 3\alpha + 6\lambda = 0$ and $3\alpha^2 - 10\alpha + 27\lambda = 0$; subtracting, $7\alpha = 21\lambda$, $\alpha = 3\lambda$.
Then $9\lambda^2 - 3\lambda + 2\lambda = 0$, and $\lambda \ne 0$ gives $\lambda = \frac{1}{9}$, $\alpha = \frac{1}{3}$.
$\beta = 1 - \alpha = \frac{2}{3}$, $\gamma = \frac{10}{3} - \alpha = 3$.
$\frac{\beta\gamma}{\lambda} = \frac{2}{1/9} = 18$.
Answer: 18. | ||||
| 82 | 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 For 3×3 matrices: $\det(kM) = k^3\det M$ and $\det(\mathrm{Adj}\,M) = (\det M)^2$. Let $d = \det A$.
$\det(2A) = 8d$; $\det\mathrm{Adj}(2A) = 2^6d^2$; $\det\mathrm{Adj}(\mathrm{Adj}(2A)) = 2^{12}d^4$.
$\det(2\,\mathrm{Adj}(\mathrm{Adj}(2A))) = 2^{15}d^4$; its adjoint has determinant $2^{30}d^8$; doubling gives $2^{33}d^8$.
$2^{33}d^8 = 2^{41}$, so $d^8 = 2^8$ and $\det(A^2) = d^2 = 4$.
Answer: 4. | ||||
| 83 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $\begin{bmatrix}12\\12-k+i\end{bmatrix} = \begin{bmatrix}12\\k-i\end{bmatrix}$, so the first sum is $\sum_i\binom{9}{i}\binom{12}{k-i} = \binom{21}{k}$ (Vandermonde; the zero terms are exactly those outside the range).
Likewise the second sum is $\sum_i\binom{8}{i}\binom{13}{k-i} = \binom{21}{k}$. So $A_k = 2\binom{21}{k}$.
$A_4 - A_3 = 2(5985 - 1330) = 9310 = 190\times49$.
Answer: 49. | ||||
| 84 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Multiples of 11 from 110 to 495 are $11m$, $m = 10, \dots, 45$. Remove those with a digit 1:
Hundreds digit 1: 110–198 (all removed). Then 209, 220, 231 (×), 242, 253, 264, 275, 286, 297, 308, 319 (×), 330, 341 (×), 352, 363, 374, 385, 396, 407, 418 (×), 429, 440, 451 (×), 462, 473, 484, 495.
Kept: 209, 220, 242, 253, 264, 275, 286, 297 (sum 2046); 308, 330, 352, 363, 374, 385, 396 (sum 2508); 407, 429, 440, 462, 473, 484, 495 (sum 3190).
Total $= 2046 + 2508 + 3190 = 7744$.
Answer: 7744. | ||||
| 85 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f'(x) = 6(x-2)(x+1)$: local maximum at $a = -1$, local minimum at $b = 2$.
$f(x) = x(2x^2 - 3x - 12)$ is positive on $(-1, 0)$ and negative on $(0, 2]$ (the other roots are $\frac{3\pm\sqrt{105}}{4}$, outside $[-1, 2]$).
With $F(x) = \frac{x^4}{2} - x^3 - 6x^2$: $F(-1) = -\frac{9}{2}$, $F(0) = 0$, $F(2) = -24$.
$A = [F(0) - F(-1)] + |F(2) - F(0)| = \frac{9}{2} + 24 = \frac{57}{2}$; $4A = 114$.
Answer: 114. | ||||
| 86 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean: $3 + 7 + x + y = 20$, so $x + y = 10$.
Variance: $\frac{9 + 49 + x^2 + y^2}{4} - 25 = 10$, so $x^2 + y^2 = 82$ and $xy = \frac{100 - 82}{2} = 9$: $x = 9$, $y = 1$.
New numbers: 21, 9, 10, 8; mean $= \frac{48}{4} = 12$.
Answer: 12. | ||||
| 87 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Sum: $\vec{s} = (2-\lambda)\hat{i} + 6\hat{j} - 2\hat{k}$.
Projection: $\frac{(2-\lambda) + 12 - 2}{\sqrt{(2-\lambda)^2 + 40}} = 1$, so $(12-\lambda)^2 = (2-\lambda)^2 + 40$.
$144 - 24\lambda = 44 - 4\lambda$, giving $\lambda = 5$ (and $12 - \lambda > 0$ as needed).
Answer: 5. | ||||
| 88 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $2i = 2e^{i\pi/2}$ and $1 - i = \sqrt{2}e^{-i\pi/4}$, so the expression is $2^{\frac{n+2}{2}}e^{i(3n-2)\pi/4}$.
It is a positive real when $\frac{(3n-2)\pi}{4}$ is a multiple of $2\pi$: $3n - 2 \equiv 0 \pmod 8$, i.e. $n \equiv 6 \pmod 8$.
Least: $n = 6$, giving $2^4 = 16$, a positive integer.
Answer: 6. | ||||
| 89 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Normal: $(6,7,8)\times(3,5,7) = (49-40,\ 24-42,\ 30-21) = (9,-18,9) \parallel (1,-2,1)$.
Through (−1, 1, 3): $x - 2y + z = 0$ (it also contains (1, 2, 3), so the lines are coplanar).
$PQ = \frac{|7 + 4 + 13|}{\sqrt{6}} = \frac{24}{\sqrt{6}}$, $PQ^2 = \frac{576}{6} = 96$.
Answer: 96. | ||||
| 90 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $c_2 = a_1 - 3 + 2b_1 = 12$ and $c_3 = a_1 - 6 + 4b_1 = 13$; subtracting, $2b_1 - 3 = 1$, so $b_1 = 2$ and $a_1 = 11$.
$\sum a_k = \frac{10}{2}(22 - 27) = -25$; $\sum b_k = 2(2^{10} - 1) = 2046$.
$\sum c_k = 2046 - 25 = 2021$.
Answer: 2021. | ||||
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