JEE (Main) · 2021
JEE (Main) 2021 — 20 July, 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
- Work, Energy and Power3
- Electrostatics2
- Physical World and Measurement2
21 easy · 9 medium · 0 hard
Chemistry
30 Q
- Coordination Compounds2
- Amines2
- Solid State1
22 easy · 7 medium · 1 hard
Mathematics
30 Q
- Trigonometry2
- Straight Lines2
- Limits, Continuity and Differentiability2
11 easy · 16 medium · 3 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 | 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 left plate is at $+V_2$ and the right plate at $-V_1$, so the potential difference across the plates is $V = V_2-(-V_1) = V_1+V_2$.
The arrangement is two capacitors in series: the dielectric layer of thickness $t$ ($C_1 = \frac{K\varepsilon_0A}{t}$) and the air gap of thickness $d-t$ ($C_2 = \frac{\varepsilon_0A}{d-t}$).
Charge on the plates: $Q = \frac{C_1C_2}{C_1+C_2}(V_1+V_2)$. Potential difference across the air gap: $V_{air} = \frac{Q}{C_2} = \frac{C_1(V_1+V_2)}{C_1+C_2}$.
Field in the air gap (where the bob hangs): $E = \frac{V_{air}}{d-t} = \frac{C_1(V_1+V_2)}{(C_1+C_2)(d-t)}$.
In equilibrium $\tan\theta = \frac{qE}{mg}$, so $\theta = \tan^{-1}\left[\frac{q}{mg}\times\frac{C_1(V_1+V_2)}{(C_1+C_2)(d-t)}\right]$.
Answer: (A). | ||||
| 2 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The AND gate gives $A\cdot B$; B also passes through a NOT gate giving $\bar{B}$. The OR gate gives $Y = A\cdot B + \bar{B}$.
(0,0): $0 + 1 = 1$; (0,1): $0 + 0 = 0$; (1,0): $0 + 1 = 1$; (1,1): $1 + 0 = 1$.
So Y = 1, 0, 1, 1, which is option (B). (Equivalently $Y = A + \bar{B}$.)
Answer: (B). | ||||
| 3 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Impulse = change in momentum: $Mv = \int_0^{2T} F\,dt = F_0\int_0^{2T}\left[1-\frac{(t-T)^2}{T^2}\right]dt$.
$\int_0^{2T}(t-T)^2\,dt = \left[\frac{(t-T)^3}{3}\right]_0^{2T} = \frac{T^3}{3}+\frac{T^3}{3} = \frac{2T^3}{3}$.
So $Mv = F_0\left(2T - \frac{2T}{3}\right) = \frac{4F_0T}{3}$ and $v = \frac{4F_0T}{3M}$.
Answer: (B). | ||||
| 4 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (a) $C = \frac{Q}{V} = \frac{Q^2}{W} = \frac{A^2T^2}{ML^2T^{-2}} = M^{-1}L^{-2}T^4A^2$ → (iii).
(b) $\varepsilon_0 = \frac{q^2}{4\pi Fr^2} = \frac{A^2T^2}{MLT^{-2}L^2} = M^{-1}L^{-3}T^4A^2$ → (ii).
(c) $\mu_0 = \frac{1}{\varepsilon_0c^2} = \frac{1}{M^{-1}L^{-3}T^4A^2\cdot L^2T^{-2}} = MLT^{-2}A^{-2}$ → (iv).
(d) $E = \frac{F}{q} = \frac{MLT^{-2}}{AT} = MLT^{-3}A^{-1}$ → (i).
Answer: (C). | ||||
| 5 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory A non-linear polyatomic molecule has 3 translational and 3 rotational degrees of freedom; each vibrational mode adds 2 (kinetic and potential), so $f = 3 + 3 + 2\times4 = 14$.
$C_V = \frac{f}{2}R = 7R$.
Adiabatic: $Q = 0$, so $W_{by} = -\Delta U = -nC_V\Delta T = -7\times8.314\times10 = -582$ J.
The temperature rises, so 582 J of work is done on the gas.
Answer: (B). | ||||
| 6 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power K.E. $= E_{mech} - U = 8 - U(x)$.
(A) At $x_2$, U is at its minimum (about 0), so K.E. is greatest and the speed is greatest: true.
(B) For $x < x_1$, U = 8 J = $E_{mech}$, so K.E. = 0: the particle is at rest there, not moving at a slowest speed. This statement is incorrect.
(C) For $x > x_4$, U = 6 J is constant, so K.E. = 2 J is constant: true.
(D) At $x_3$, U = 4 J, so K.E. = 4 J: true.
Answer: (B). | ||||
| 7 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei A: true — every element has its own characteristic line spectrum.
B: true — Bohr's postulate: the electron moves in certain stationary (non-radiating) orbits.
C: false — $R = R_0A^{1/3}$ makes the nuclear volume proportional to A, so nuclear density is the same for all nuclei.
D: false — it is the other way round: a free neutron decays (half-life about 10 minutes), a free proton is stable.
E: true — radioactive decay occurs because the nucleus is unstable.
Correct: A, B and E only.
Answer: (A). | ||||
| 8 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal speed, weight = Stokes drag (buoyancy neglected): $\frac{4}{3}\pi R^3\rho_w g = 6\pi\eta R v_t$.
$v_t = \frac{2R^2\rho_w g}{9\eta} = \frac{2\times(0.2\times10^{-3})^2\times1000\times10}{9\times1.8\times10^{-5}} = \frac{8\times10^{-4}}{1.62\times10^{-4}} \approx 4.94$ m/s.
Answer: (B). | ||||
| 9 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Maximum percentage error: $\frac{\Delta y}{y} = 2\frac{\Delta m}{m} + 4\frac{\Delta r}{r} + x\frac{\Delta g}{g} + \frac{3}{2}\frac{\Delta l}{l}$.
$18 = 2(1) + 4(0.5) + xp + \frac{3}{2}(4) = 10 + xp$, so $xp = 8$.
Check the options: $4\times3 = 12$, $5\times2 = 10$, $8\times2 = 16$, $\frac{16}{3}\times\frac{3}{2} = 8$. Only (D) gives 8.
Answer: (D). | ||||
| 10 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Let A reject heat $Q_2$ at T. For a Carnot engine $Q_1 = Q_2\frac{T_1}{T}$, so $W_A = Q_1 - Q_2 = Q_2\frac{T_1-T}{T}$.
B absorbs $\frac{Q_2}{2}$ at T and rejects at $T_3$: $W_B = \frac{Q_2}{2}\left(1-\frac{T_3}{T}\right) = \frac{Q_2}{2}\cdot\frac{T-T_3}{T}$.
$W_A = W_B$: $2(T_1 - T) = T - T_3$, so $T = \frac{2T_1 + T_3}{3} = \frac{2}{3}T_1 + \frac{1}{3}T_3$.
Answer: (A). | ||||
| 11 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Kepler's third law: $T^2 = \frac{4\pi^2}{GM}r^3$, so $M = \frac{4\pi^2}{G}\cdot\frac{r^3}{T^2}$.
$r = 9\times10^6$ m, $r^3 = 7.29\times10^{20}\ \mathrm{m^3}$; $T = 7.5\times3600 = 27000$ s, $T^2 = 7.29\times10^8\ \mathrm{s^2}$.
$M = 6\times10^{11}\times\frac{7.29\times10^{20}}{7.29\times10^{8}} = 6\times10^{11}\times10^{12} = 6.00\times10^{23}$ kg.
Answer: (C). | ||||
| 12 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $x = A\sin\omega t$; at $t = \frac{T}{4}$, $\omega t = \frac{\pi}{2}$ and $x = A$ (extreme position), so all the energy is potential.
$U = \frac{1}{2}m\omega^2A^2$ with $\omega = \frac{2\pi}{0.2} = 10\pi$ rad/s:
$U = \frac{1}{2}\times0.5\times(10\pi)^2\times(0.05)^2 = 0.25\times987\times0.0025 \approx 0.62$ J.
Answer: (C). | ||||
| 13 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power With constant power, the work done in time t is $Pt = \frac{1}{2}mv^2$, so $v = \left(\frac{2P}{m}\right)^{1/2}t^{1/2}$.
$x = \int_0^t v\,dt = \left(\frac{2P}{m}\right)^{1/2}\cdot\frac{2}{3}t^{3/2} = \left(\frac{4}{9}\cdot\frac{2P}{m}\right)^{1/2}t^{3/2} = \left(\frac{8P}{9m}\right)^{1/2}t^{3/2}$.
Answer: (C). | ||||
| 14 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter By Ampere's law, inside a wire of radius R carrying I uniformly: $B = \frac{\mu_0Ir}{2\pi R^2}$ (linear in r); outside: $B = \frac{\mu_0I}{2\pi r}$.
Inside, the slope $\frac{\mu_0I}{2\pi R^2}$ is larger for the thinner wire a.
The peak, at the surface, is $\frac{\mu_0I}{2\pi R}$: higher for a and reached at the smaller radius a.
Outside both wires the field is the same $\frac{\mu_0I}{2\pi r}$, so the curves meet beyond r = b.
This is graph (B).
Answer: (B). | ||||
| 15 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation The two particles are at opposite ends of a diameter, 2R apart.
Gravitational pull supplies the centripetal force: $\frac{G(1)(1)}{(2R)^2} = (1)\omega^2R$.
$\omega^2 = \frac{G}{4R^3}$, so $\omega = \frac{1}{2}\sqrt{\frac{G}{R^3}}$.
Answer: (B). | ||||
| 16 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The second excited state is n = 3, with energy $-\frac{13.6}{9} = -1.51$ eV.
Photon energy = initial energy − final energy $= 3 - (-1.51) = 4.51$ eV.
Work function: $\phi = \frac{hc}{\lambda_0} = \frac{12400\ \mathrm{eV\,Å}}{4000\ \mathrm{Å}} = 3.1$ eV.
$K_{max} = 4.51 - 3.1 = 1.41$ eV.
Answer: (D). | ||||
| 17 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics As i increases from small values, δ first decreases, reaches the minimum deviation $\delta_m$ (when the ray passes symmetrically, $i = e$), and then increases again.
The δ–i curve is therefore U-shaped (not symmetric, but with a single minimum).
Answer: (D). | ||||
| 18 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At resonance $f = \frac{1}{2\pi\sqrt{LC}}$, so $L = \frac{1}{4\pi^2f^2C}$.
$L = \frac{1}{4\times9.87\times3600\times0.1\times10^{-6}} = \frac{1}{1.421\times10^{-2}} \approx 70.36$ H, which is option (B), 70.3 H.
Answer: (B). | ||||
| 19 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $R_T = R_0(1+\alpha T)$: $16 = R_0(1+15\alpha)$, $20 = R_0(1+100\alpha)$.
Dividing: $\frac{20}{16} = \frac{1+100\alpha}{1+15\alpha}$, so $1.25 + 18.75\alpha = 1 + 100\alpha$ and $\alpha = \frac{0.25}{81.25} \approx 0.0031\,°\mathrm{C}^{-1}$.
(Taking 15°C as the reference, $\alpha = \frac{20-16}{16\times85} \approx 0.0029\,°\mathrm{C}^{-1}$; either way about 0.003.)
Answer: (A). | ||||
| 20 | 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 Take O as the origin and $k = \frac{1}{4\pi\varepsilon_0}$. From the figure: B(+q) at $(-l,0)$, G(2q) at $(l,0)$, C(2q) at $(0,l)$, F(q) at $(0,-l)$, A(−q) at $(-l,l)$, D(+q) at $(l,l)$, H(−q) at $(l,-l)$, E(2q) at $(-l,-l)$.
Charges at distance $l$: B gives $\frac{kq}{l^2}$ along +x, G gives $\frac{2kq}{l^2}$ along −x, C gives $\frac{2kq}{l^2}$ along −y, F gives $\frac{kq}{l^2}$ along +y. Sum: $\frac{kq}{l^2}(-1,-1)$.
Charges at distance $\sqrt{2}l$ (each unit charge gives $\frac{kq}{2l^2}$ along the diagonal, components $\frac{kq}{2\sqrt{2}l^2}$): A gives $(-1,+1)$, D gives $(-1,-1)$, H gives $(+1,-1)$, E gives $(+2,+2)$, in units of $\frac{kq}{2\sqrt{2}l^2}$. Sum: $\frac{kq}{2\sqrt{2}l^2}(1,1)$.
Total: $\frac{kq}{l^2}\left(\frac{1}{2\sqrt{2}}-1\right)(1,1)$, magnitude $\frac{kq}{l^2}\sqrt{2}\left(1-\frac{1}{2\sqrt{2}}\right) = \frac{kq}{l^2}\left(\sqrt{2}-\frac{1}{2}\right) = \frac{1}{4\pi\varepsilon_0}\frac{q}{2l^2}(2\sqrt{2}-1)$.
Answer: (C). | ||||
| 21 | Communication SystemsAmplitude modulation and bandwidth | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on amplitude modulation and bandwidth, from Communication Systems. NCERT: No matching NCERT chapter $\mu = \frac{A_{max}-A_{min}}{A_{max}+A_{min}} = \frac{12-3}{12+3} = \frac{9}{15} = 0.6$.
$0.6x = 0.6$, so $x = 1$.
Answer: 1. | ||||
| 22 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity From Fig. 1, V rises linearly from 5 V at t = 3 s to 10 V at t = 4 s (slope 5 V/s), so $V(3.2) = 5 + 5\times0.2 = 6$ V.
In Fig. 2 the source V(t) and the 5 V battery have their positive terminals on the same side, so they oppose each other round the loop.
$I = \frac{V(t) - 5}{R} = \frac{6-5}{1} = 1$ A.
Answer: 1. | ||||
| 23 | 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 $|\varepsilon| = \frac{d\phi_B}{dt} = 20t + 20$ mV; at t = 5 s, $\varepsilon = 120$ mV.
$I = \frac{\varepsilon}{R} = \frac{120\ \mathrm{mV}}{2\ \Omega} = 60$ mA.
Answer: 60. | ||||
| 24 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The $K_\alpha$ photon is emitted when the vacancy moves from K to L, so its energy is $E_K - E_L$.
$E_{K\alpha} = \frac{hc}{\lambda} = \frac{4.14\times10^{-15}\times3\times10^8}{0.071\times10^{-9}} = \frac{1.242\times10^{-6}}{7.1\times10^{-11}} \approx 17.49\times10^3$ eV = 17.49 keV.
$E_L = 27.5 - 17.49 = 10.01$ keV ≈ 10 keV.
Answer: 10. | ||||
| 25 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy 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 Efflux speed $v = \sqrt{2gh}$; the hole is at height $H - h$ above the ground, so the fall time is $\sqrt{\frac{2(H-h)}{g}}$.
Range $R = 2\sqrt{h(H-h)}$, maximum when $h = H - h$, i.e. $h = \frac{H}{2} = 6$ m.
Answer: 6. | ||||
| 26 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The belt does not slip, so the rims have the same linear speed: $\omega_1(3R) = \omega_2R$, i.e. $\omega_2 = 3\omega_1$.
Equal rotational kinetic energy: $\frac{1}{2}I_1\omega_1^2 = \frac{1}{2}I_2\omega_2^2$, so $\frac{I_1}{I_2} = \left(\frac{\omega_2}{\omega_1}\right)^2 = 9$.
Answer: 9. | ||||
| 27 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At angle θ from the vertical, the height is $h = R\cos\theta$ and energy conservation gives $v^2 = 2g(R-h)$.
Radial equation: $mg\cos\theta - N = \frac{mv^2}{R}$. Contact is lost when $N = 0$: $g\frac{h}{R} = \frac{2g(R-h)}{R}$.
$h = 2R - 2h$, so $h = \frac{2R}{3} = 2$ m.
Answer: 2. | ||||
| 28 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Let both speeds be u. The resultant of the river velocity (along the flow) and the swimmer's velocity relative to water must lie along AB.
Resolve perpendicular to AB: the river contributes $u\sin30°$ to one side, the swimmer, heading at θ from AB on the upstream side, contributes $u\sin\theta$ to the other.
For no sideways drift: $u\sin\theta = u\sin30°$, so θ = 30°.
(Along AB both components add, $u\cos30° + u\cos30° > 0$, so the swimmer does move towards B.)
Answer: 30. | ||||
| 29 | 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 At t = 0: $A\sin\phi = 2$ and $v = A\omega\cos\phi = 2\omega$, so $A\cos\phi = 2$.
$A = \sqrt{2^2 + 2^2} = 2\sqrt{2}$ cm, so $x = 2$.
Answer: 2. | ||||
| 30 | OpticsWave optics | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on wave optics, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics In a thickness t: waves in air $= \frac{t}{\lambda/\mu} = \frac{\mu t}{\lambda}$, waves in vacuum $= \frac{t}{\lambda}$.
Difference: $\frac{(\mu-1)t}{\lambda} = 1$, so $t = \frac{\lambda}{\mu-1} = \frac{6\times10^{-7}}{0.0003} = 2\times10^{-3}$ m = 2 mm.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Solid StateClassification of solids | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on classification of solids, from Solid State. NCERT: No matching NCERT chapter (A) True: crystalline solids have a regular arrangement repeated over long distances.
(B) False: crystalline solids are anisotropic (properties differ with direction); amorphous solids are isotropic.
(C) True: amorphous solids are called pseudo solids or super-cooled liquids.
(D) True: they soften gradually over a range of temperatures, having no sharp melting point.
(E) False: with no sharp melting point they have no definite heat of fusion.
Correct: (A), (C), (D) only.
Answer: (A). | ||||
| 32 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In Thomson's model the positive charge is spread thinly through the whole atom with electrons embedded in it; there is no concentrated nucleus.
The heavy, fast α-particles would meet only weak, diffuse electric fields, so they would all pass through, each slightly deflected (by small angles) and losing a little energy (speed) to the atom's charges.
Large-angle scattering (C) and bouncing back (D) need a small dense nucleus, which this model lacks; (A) ignores the small deflections and energy losses that the charges would still cause.
Answer: (B). | ||||
| 33 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Physical adsorption of a gas on charcoal increases with how easily the gas liquefies, which is measured by its critical temperature.
$SO_2$ (critical temperature about 430 K) liquefies far more easily than $H_2$ (about 33 K), so it is adsorbed much more.
Both A and R are true and R explains A.
Answer: (A). | ||||
| 34 | 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 the ionisation enthalpy generally rises, with two exceptions:
Al ($3s^23p^1$) is lower than Mg ($3s^2$), since the 3p electron is easier to remove than a paired 3s electron.
S ($3p^4$) is lower than P ($3p^3$, half-filled and extra stable).
Values (kJ/mol): Al 577 < Mg 738 < S 1000 < P 1012.
Answer: (D). | ||||
| 35 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Copper pyrites contains iron. On roasting the iron sulphide becomes FeO, which silica removes as slag:
$FeO + SiO_2 \rightarrow FeSiO_3$ (iron silicate, slag).
Silica is the flux that converts iron oxide into iron silicate.
Answer: (C). | ||||
| 36 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter The radioactive isotope of hydrogen is tritium, ${}^3_1H$.
Neutrons = 3 − 1 = 2; electrons (neutral atom) = 1.
Answer: (A). | ||||
| 37 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter (a) Lithium, alloyed with lead, is used for white-metal bearings in motor engines → (v).
(b) Liquid sodium is the coolant in fast breeder nuclear reactors → (iii).
(c) Potassium (as KOH, or $KO_2$) absorbs $CO_2$ → (ii).
(d) Caesium, with its very low ionisation enthalpy, is used in photoelectric cells → (i).
Answer: (D). | ||||
| 38 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Chlorous acid $HClO_2$: H–O–Cl=O, one Cl=O.
Chloric acid $HClO_3$: H–O–Cl(=O)$_2$, two Cl=O.
Perchloric acid $HClO_4$: H–O–Cl(=O)$_3$, three Cl=O.
Answer: (C). | ||||
| 39 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter None of these ions forms an insoluble chloride, so nothing precipitates on adding HCl (group I is absent).
$H_2S$ in acid medium gives a low sulphide-ion concentration, enough to precipitate only group II sulphides: here only $Cu^{2+}$ (as black CuS).
$Zn^{2+}$, $Ni^{2+}$ (group IV) need ammoniacal $H_2S$; $Al^{3+}$, $Fe^{3+}$ (group III) need $NH_4OH$; $Ca^{2+}$, $Ba^{2+}$ are group V. ($Fe^{3+}$ is reduced to $Fe^{2+}$ by $H_2S$ with a deposit of sulphur, which is not a cation precipitate.)
Number of cations precipitated = 1.
Answer: (C). | ||||
| 40 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Statement I: $Mn^{3+}$ ($d^4$) and $Fe^{3+}$ ($d^5$) with strong-field $CN^-$ pair up their electrons, leaving two 3d orbitals free: $d^2sp^3$. $[Co(C_2O_4)_3]^{3-}$ ($Co^{3+}$, $d^6$) is a low-spin, diamagnetic complex: also $d^2sp^3$. True.
Statement II: with weak-field $Cl^-$ and $F^-$ the complexes are high spin, $sp^3d^2$: $Mn^{3+}$ $d^4$ has 4 unpaired electrons, $Fe^{3+}$ $d^5$ has 5. Both paramagnetic. True.
Answer: (A). | ||||
| 41 | 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 (a) CO binds haemoglobin to form carboxyhaemoglobin → (iii).
(b) $SO_2$ in the air makes flower buds stiff so that they fall off → (iv).
(c) Polychlorinated biphenyls are carcinogenic → (i).
(d) Oxides of nitrogen are metabolised by plants such as pyrus (they take up $NO_x$) → (ii).
Answer: (A). | ||||
| 42 | Practical Organic ChemistryDetection of elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on detection of elements, from Practical Organic Chemistry. NCERT: No matching NCERT chapter Fusion with sodium converts N, S, halogens and P into ionic NaCN, $Na_2S$, NaX and $Na_3PO_4$, which are tested in the extract.
Carbon and oxygen are not detected this way.
Answer: (B). | ||||
| 43 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Statement I: hyperconjugation operates in the ground state of the molecule all the time; it is a permanent effect. True.
Statement II: in $CH_3-\overset{+}{C}H_2$ the C–H bonds that hyperconjugate belong to the $CH_3$ carbon, which is $sp^3$. The overlap is of a $C_{sp^3}-H_{1s}$ σ bond with the empty 2p orbital of the cationic carbon, not $C_{sp^2}-H$. False.
Answer: (C). | ||||
| 44 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate is $CH_3-CH(OH)-C(=CH_2)-COOCH_3$, a secondary allylic alcohol.
Conc. HBr protonates the OH, water leaves, and an allylic carbocation forms; it is delocalised between the former carbinol carbon and the terminal $CH_2$: $CH_3-\overset{+}{C}H-C(=CH_2)CO_2CH_3 \leftrightarrow CH_3-CH=C(CO_2CH_3)-\overset{+}{C}H_2$.
Bromide attacks the primary end (less hindered, and away from the carbon next to the electron-withdrawing ester), giving the more substituted alkene that is also conjugated with the C=O: $CH_3-CH=C(CH_2Br)-COOCH_3$ (allylic rearrangement).
Options (C) (an O–Br compound) and (D) (ester turned into acid bromide) do not form; (B) is the unrearranged product, which is the minor one.
Answer: (A). | ||||
| 45 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers 1-Phenylpropan-1-ol dehydrates with conc. $H_2SO_4$ on heating through the benzylic carbocation to 1-phenylpropene, conjugated with the ring.
Of the two geometrical isomers, the trans (E) alkene A is more stable (less steric strain between the phenyl and methyl groups) and is the major product; the cis isomer B is minor.
Answer: (A). | ||||
| 46 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids DIBAL-H reduces a nitrile only as far as the imine (as its aluminium complex); hydrolysis then gives the aldehyde: $R-CN \rightarrow R-CH=NH \rightarrow R-CHO$.
So Y is $-CHO$.
Answer: (C). | ||||
| 47 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines The product, 3-chlorophenol, has Cl meta to OH. The meta relation must be set while the nitro group (meta director) is present.
(i) $Cl_2$/$FeCl_3$: nitrobenzene → 3-chloronitrobenzene.
(ii) Fe/HCl: → 3-chloroaniline.
(iii) $NaNO_2$/HCl, 0°C: → 3-chlorobenzenediazonium chloride.
(iv) $H_2O/H^+$ (warm): → 3-chlorophenol.
(B) chlorinates phenol last, giving ortho/para products; (C) chlorinates aniline (o/p); (D) diazotises before the amine exists.
Answer: (A). | ||||
| 48 | 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 This is the Gabriel phthalimide synthesis.
(i) Potassium phthalimide displaces bromide from benzyl bromide ($S_N2$): N-benzylphthalimide.
(ii) Alkaline hydrolysis frees the primary amine (and phthalate): $C_6H_5CH_2NH_2$, benzylamine.
Answer: (B). | ||||
| 49 | Chemistry in Everyday LifeAntibiotics and antiseptics | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on antibiotics and antiseptics, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Statement I: penicillin kills bacteria (it blocks cell-wall synthesis); it is bactericidal, not bacteriostatic. False.
Statement II: the structure drawn — an acylamino group on a β-lactam ring fused to a thiazolidine ring carrying two methyl groups and a COOH — is the general penicillin structure. True.
Answer: (D). | ||||
| 50 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Lactose (milk sugar) is β-D-galactose linked to D-glucose; hydrolysis gives D-galactose and D-glucose.
Maltose and amylose give only glucose; sucrose gives glucose and fructose.
Answer: (B). | ||||
| 51 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry At constant volume and temperature, partial pressure is proportional to moles.
$SO_2$ is limiting: 250 mbar of $SO_2$ uses 125 mbar of $O_2$ and gives 250 mbar of $SO_3$.
Left: $O_2$ = 750 − 125 = 625 mbar, $SO_3$ = 250 mbar, $SO_2$ = 0.
Total = 625 + 250 = 875 mbar.
Answer: 875. | ||||
| 52 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $O_2^{2-}$ has 18 electrons: $\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2\,\sigma2p_z^2\,\pi2p_x^2\pi2p_y^2\,\pi^*2p_x^2\pi^*2p_y^2$.
Bonding orbitals: $\sigma1s$ (2) + $\sigma2s$ (2) + $\sigma2p_z$ (2) + $\pi2p_x,\pi2p_y$ (4) = 10.
Answer: 10. | ||||
| 53 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics $H^+$: $0.4\times0.2\times2 = 0.16$ mol; $OH^-$: $0.6\times0.1 = 0.06$ mol. $OH^-$ is limiting: 0.06 mol of water forms.
Heat released $= 0.06\times57.1 = 3.426$ kJ = 3426 J.
Mass of solution = 1000 mL × 1.0 g/mL = 1000 g.
$\Delta T = \frac{3426}{1000\times4.18} = 0.8196$ K $= 81.96\times10^{-2}$ K ≈ $82\times10^{-2}$ K.
Answer: 82. | ||||
| 54 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Start with 1 mol of HA. 0.5 mol dimerises: $2HA \rightarrow (HA)_2$ gives 0.25 mol of dimer.
The other 0.5 mol dissociates fully: $HA \rightarrow H^+ + A^-$ gives 1.0 mol of ions.
Total particles = 0.25 + 1.0 = 1.25 mol, so $i = 1.25 = 125\times10^{-2}$.
Answer: 125. | ||||
| 55 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Solids do not appear in $K_p$: $K_p = (p_{O_2})^{1/2} = 4$.
$p_{O_2} = 4^2 = 16$ atm.
Answer: 16. | ||||
| 56 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag$, n = 2. $E = E^\circ - \frac{0.059}{2}\log\frac{[Cu^{2+}]}{[Ag^+]^2}$.
Cell 1: $\log\frac{0.1}{(0.01)^2} = \log10^3 = 3$, so $E_1 = E^\circ - 0.0885 = 0.3095$ and $E^\circ = 0.398$ V.
Cell 2: $\log\frac{0.01}{(0.001)^2} = \log10^4 = 4$, so $E_2 = 0.398 - 0.118 = 0.280$ V $= 28\times10^{-2}$ V.
Answer: 28. | ||||
| 57 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The data supplied (ln 2 = 0.69, ln 10 = 2.3 and the two log rules) are exactly those needed if 0.2 mol of A has reacted, leaving 0.8 mol; this is the reading the question is built on.
$k = \frac{1}{t}\ln\frac{[A]_0}{[A]} = \frac{1}{100}\ln\frac{1}{0.8} = \frac{1}{100}(\ln10 - 3\ln2) = \frac{2.3 - 2.07}{100} = 2.3\times10^{-3}\ \mathrm{min^{-1}}$.
$t_{1/2} = \frac{\ln2}{k} = \frac{0.69}{2.3\times10^{-3}} = 300$ min.
(Taken strictly, A → 2B means 0.2 mol of B comes from 0.1 mol of A, leaving 0.9 mol; then $k = \frac{\ln(10/9)}{100} = 1.054\times10^{-3}$ and $t_{1/2} \approx 655$–658 min, a value the given logarithms cannot produce since ln 3 is not supplied.)
As printed: 0.2 mol B means 0.1 mol A used and 0.9 mol A left, so k = (1/100) ln(10/9) ≈ 1.054 × 10⁻³ min⁻¹ and t½ = 0.69/k ≈ 655 min (658 with ln 2 = 0.693). The constants supplied (ln 2, ln 10, no ln 3) only give a clean answer if 0.2 mol of A is taken as used up: k = (1/100) ln(1/0.8) and t½ ≈ 300 min, which is evidently the setter's intent.
Answer: 655 as printed (300 as the setter intended). | ||||
| 58 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds 3 mol of complex give 3 mol of AgCl, so each formula unit has one ionisable $Cl^-$: $[Co(en)_2Cl_2]Cl$.
Inside the sphere: two bidentate en (4 donor atoms) + 2 Cl = 6 donor atoms.
Secondary valency (coordination number) = 6.
Answer: 6. | ||||
| 59 | HydrocarbonsConformations of ethane | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on conformations of ethane, from Hydrocarbons. NCERT: Class 11: Hydrocarbons In a staggered conformation each bond on the front carbon sits midway between two bonds on the back carbon; the dihedral angle between adjacent front and back bonds is 60°.
In $CCl_3-CH_3$ every staggered form is the same: each C–Cl bond is 60° from the two nearest C–H bonds.
Answer: 60. | ||||
| 60 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$: moles of oxalate = $\frac{5}{2}\times$ moles of permanganate.
$MnO_4^-$ used = $10.0\times0.05 = 0.5$ mmol, so oxalic acid = 1.25 mmol in 10.0 mL: molarity 0.125 M.
Oxalic acid dihydrate $H_2C_2O_4\cdot2H_2O$, M = 126 g/mol: strength = $0.125\times126 = 15.75$ g/L = $1575\times10^{-2}$ g/L.
Answer: 1575. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctHard | |
A hard single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $x = y = 0$ gives $2f(0) = 2f(0)^2$; $f(0) = 0$ would make $f \equiv 0$ (put $y = 0$), contradicting $f(\frac{1}{2}) = -1$, so $f(0) = 1$.
With $x = y = \frac{1}{2}$: $f(1) + f(0) = 2f(\frac{1}{2})^2 = 2$, so $f(1) = 1$.
With $y = 1$: $f(x+1) + f(x-1) = 2f(x)$, so $f(n)$ is an arithmetic progression in n with $f(0) = f(1) = 1$; hence $f(k) = 1$ for every integer k. (Equivalently $f(x) = \cos(2\pi x)$ fits all the conditions.)
The sum becomes $\sum_{k=1}^{20}\frac{1}{\sin k\sin(k+1)}$. Since $\frac{\sin((k+1)-k)}{\sin k\sin(k+1)} = \cot k - \cot(k+1)$,
$\sum_{k=1}^{20}\frac{1}{\sin k\sin(k+1)} = \frac{1}{\sin1}(\cot1 - \cot21) = \frac{1}{\sin1}\cdot\frac{\sin(21-1)}{\sin1\sin21} = \mathrm{cosec}^2(1)\,\mathrm{cosec}(21)\sin(20)$.
Answer: (C). | ||||
| 62 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctMedium | |
A medium single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Let $N = 8 + \alpha + \beta$.
Mean 6: $\sum f(x-6) = 4(-4) + 4(0) + 2\alpha + 3\beta = 0$, so $2\alpha + 3\beta = 16$.
Variance 6.8: $\sum f(x-6)^2 = 64 + 4\alpha + 9\beta = 6.8(8+\alpha+\beta)$, so $2.8\alpha - 2.2\beta = 9.6$, i.e. $14\alpha - 11\beta = 48$.
Solving: from the first $\alpha = 8 - 1.5\beta$; then $112 - 21\beta - 11\beta = 48$, $\beta = 2$, $\alpha = 5$, $N = 15$.
Old $\sum fx = 6\times15 = 90$. Changing $x_3$ from 8 to 7 lowers it by $\alpha\times1 = 5$: new mean $= \frac{85}{15} = \frac{17}{3}$.
Answer: (A). | ||||
| 63 | CirclesStandard and general form | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Touching the y-axis at (0, 6), the centre is $(\pm r, 6)$ with radius r.
x-intercept $= 2\sqrt{r^2 - 6^2} = 6\sqrt{5}$, so $r^2 - 36 = 45$ and $r = 9$.
Answer: (C). | ||||
| 64 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines The two sides meet at the origin O, a vertex. The diagonal $11x + 7y = 9$ does not pass through O, so it joins the other two vertices B and D on the two sides.
B: $4x + 5y = 0$ and $11x + 7y = 9$ give $B\left(\frac{5}{3}, -\frac{4}{3}\right)$. D: $7x + 2y = 0$ and $11x + 7y = 9$ give $D\left(-\frac{2}{3}, \frac{7}{3}\right)$.
The diagonals bisect each other: midpoint of BD is $\left(\frac{1}{2}, \frac{1}{2}\right)$. The other diagonal joins O to this point: $y = x$.
Of the options only (2, 2) lies on $y = x$.
Answer: (B). | ||||
| 65 | 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 For $0 \le x \le \frac{\pi}{2}$, sin t increases, so $f(x) = \sin x$; for $\frac{\pi}{2} \le x \le \pi$ the maximum stays 1, so $f(x) = 1$; for $x > \pi$, $f(x) = 2 + \cos x$.
At $x = \frac{\pi}{2}$: both pieces equal 1; left derivative $\cos\frac{\pi}{2} = 0$, right derivative 0. Differentiable.
At $x = \pi$: $f(\pi) = 1$ and $2 + \cos x \to 1$; left derivative 0, right derivative $-\sin\pi = 0$. Differentiable.
Each piece is smooth elsewhere, so f is differentiable everywhere in $(0,\infty)$.
Answer: (D). | ||||
| 66 | 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 Negation swaps the quantifiers and negates the inner statement: $\sim(\forall M\ \exists x: x \ge M) \equiv \exists M\ \forall x: x < M$.
That is: there exists M > 0 such that $x < M$ for all $x \in S$.
Answer: (C). | ||||
| 67 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersections: $x^2 = x + 2$, so $x = -1$ and $x = 2$. The line lies above the parabola between them.
Area $= \int_{-1}^{2}(x + 2 - x^2)\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2} = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{9}{2}$.
Answer: (C). | ||||
| 68 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Rearrange: $x\,dy - y\,dx = x^3dy + x^2y\,dx - 3x^4dx = x^2(x\,dy + y\,dx) - 3x^4dx$.
Divide by $x^2$: $\frac{x\,dy - y\,dx}{x^2} = (x\,dy + y\,dx) - 3x^2dx$, i.e. $d\left(\frac{y}{x}\right) = d(xy) - d(x^3)$.
Integrate: $\frac{y}{x} = xy - x^3 + C$. With $y(3) = 3$: $1 = 9 - 27 + C$, $C = 19$.
At $x = 4$: $\frac{y}{4} = 4y - 64 + 19$, so $\frac{15y}{4} = 45$ and $y = 12$.
Answer: (A). | ||||
| 69 | Straight LinesCartesian coordinates and distance formula | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines Reflection in $y = x$: $(a, b) \rightarrow (b, a)$. Translation: $\rightarrow (b + 2, a)$.
Rotation by $\frac{\pi}{4}$: $(X, Y) \rightarrow \left(\frac{X - Y}{\sqrt{2}}, \frac{X + Y}{\sqrt{2}}\right) = \left(\frac{b + 2 - a}{\sqrt{2}}, \frac{b + 2 + a}{\sqrt{2}}\right)$.
So $b + 2 - a = -1$ and $b + 2 + a = 7$: $a = 4$, $b = 1$.
$2a + b = 9$.
Answer: (C). | ||||
| 70 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The terms simplify: $3^{\log_3\sqrt{25^{x-1}+7}} = \sqrt{25^{x-1}+7}$ and $3^{-\frac{1}{8}\log_3(5^{x-1}+1)} = (5^{x-1}+1)^{-1/8}$.
In increasing powers of the second term, $T_9 = {}^{10}C_8\left(\sqrt{25^{x-1}+7}\right)^2\left((5^{x-1}+1)^{-1/8}\right)^8 = 45\cdot\frac{25^{x-1}+7}{5^{x-1}+1}$.
$T_9 = 180$: $25^{x-1} + 7 = 4(5^{x-1} + 1)$. With $t = 5^{x-1}$: $t^2 - 4t + 3 = 0$, $t = 1$ or 3.
$t = 1$ gives $x = 1$ (and $t = 3$ gives $x = 1 + \log_5 3$, not an option).
Answer: (B). | ||||
| 71 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $S_1$: the disc with centre (2, 0) and radius 1.
$S_2$: with $z = x + iy$, $z(1+i) + \bar{z}(1-i) = 2(x - y) \ge 4$, i.e. $x - y \ge 2$: the half-plane below the line $x - y = 2$, which passes through the centre (2, 0).
So $S_1\cap S_2$ is a half-disc. A boundary point on the circle is $(2 + \cos\theta, \sin\theta)$ with $\cos\theta - \sin\theta \ge 0$, i.e. $\theta\in\left[-\frac{3\pi}{4}, \frac{\pi}{4}\right]$.
$\left|z - \frac{5}{2}\right|^2 = \left(\cos\theta - \frac{1}{2}\right)^2 + \sin^2\theta = \frac{5}{4} - \cos\theta$, largest when $\cos\theta$ is least, at $\theta = -\frac{3\pi}{4}$: $\frac{5}{4} + \frac{1}{\sqrt{2}} = \frac{5 + 2\sqrt{2}}{4}$.
(Points inside or on the diameter are nearer to $\frac{5}{2}$, which lies inside the region.)
Answer: (D). | ||||
| 72 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctHard | |
A hard single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = g(x)$. Between consecutive roots of the five roots of f, Rolle's theorem gives a root of $f' = g$: at least 4 roots of g in (a, b).
Between consecutive roots of g, Rolle's theorem gives a root of $g'$: at least 3 roots of $g'$.
The roots of g and of $g'$ found this way are interlaced and distinct (each root of $g'$ lies strictly between two roots of g), so $g(x)g'(x) = 0$ has at least 4 + 3 = 7 roots in (a, b).
Answer: (C). | ||||
| 73 | 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 Consider $(A^3 + B^3)(A^2 - B^2) = A^5 - A^3B^2 + B^3A^2 - B^5$.
Using $A^5 = B^5$, this is $B^3A^2 - A^3B^2$, and with the given relation $A^3B^2 = A^2B^3$ (the products taken as the question intends, the powers of A and B combining in either order) it is 0.
So $(A^3 + B^3)(A^2 - B^2) = O$. Since $A^2 - B^2$ is invertible, multiplying on the right by its inverse gives $A^3 + B^3 = O$.
Hence $\det(A^3 + B^3) = 0$.
Answer: (A). | ||||
| 74 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Factorise: $x^3 - 3x^2y - xy^2 + 3y^3 = x^2(x - 3y) - y^2(x - 3y) = (x - 3y)(x - y)(x + y)$.
For natural numbers $x + y \ne 0$, so $(x, y)\in R \iff x = y$ or $x = 3y$.
Reflexive: $x = x$. Yes.
Symmetric: $(3, 1)\in R$ but $(1, 3)\notin R$. No.
Transitive: $(9, 3)$ and $(3, 1)$ are in R but $(9, 1)$ is not. No.
Answer: (B). | ||||
| 75 | 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 $8^{2\sin3x}\cdot4^{4\cos3x} = 2^{6\sin3x + 8\cos3x}$, and $6\sin3x + 8\cos3x$ ranges over $[-10, 10]$.
$\alpha = 2^{10}$, $\beta = 2^{-10}$; $\alpha^{1/5} = 4$, $\beta^{1/5} = \frac{1}{4}$.
Sum of roots $\frac{17}{4} = -\frac{b}{8}$, so $b = -34$; product $1 = \frac{c}{8}$, so $c = 8$.
$c - b = 8 + 34 = 42$.
Answer: (A). | ||||
| 76 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For small x, $(1 \mp \sin x)^{1/8} = 1 \mp \frac{\sin x}{8} + O(\sin^2x)$, so the denominator is $-\frac{\sin x}{4} + O(x^3)$ (the square terms cancel).
Limit $= \lim_{x\to0}\frac{x}{-\frac{\sin x}{4}} = -4$.
Answer: (C). | ||||
| 77 | 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 The number of correct guesses X is binomial with n = 8, p = $\frac{1}{2}$: $P(X = r) = \frac{{}^8C_r}{256}$.
$P(X \ge 4) = \frac{70 + 56 + 28 + 8 + 1}{256} = \frac{163}{256} > \frac{1}{2}$.
$P(X \ge 5) = \frac{56 + 28 + 8 + 1}{256} = \frac{93}{256} < \frac{1}{2}$.
Smallest n = 5.
Answer: (C). | ||||
| 78 | 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 General points: $(\alpha + \lambda, 1 + 2\lambda, 1 + 3\lambda)$ and $(4 + \beta\mu, 6 + 3\mu, 7 + 3\mu)$.
Equating y and z: $1 + 2\lambda = 6 + 3\mu$, $1 + 3\lambda = 7 + 3\mu$. Subtracting, $\lambda = 1$; then $\mu = -1$.
The point is $(\alpha + 1, 3, 4)$. On the plane: $\alpha + 1 + 6 - 4 = 8$, so $\alpha = 5$.
x-coordinates: $6 = 4 - \beta$, so $\beta = -2$.
$\alpha - \beta = 7$.
Answer: (C). | ||||
| 79 | 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 $\tan A + \tan B = \frac{\sin(A+B)}{\cos A\cos B}$.
$2x = \tan20° + \tan70° = \frac{\sin90°}{\cos20°\cos70°} = \frac{1}{\cos20°\sin20°} = \frac{2}{\sin40°}$, so $x = \frac{1}{\sin40°}$.
$2y = \tan20° + \tan50° = \frac{\sin70°}{\cos20°\cos50°} = \frac{1}{\cos50°} = \frac{1}{\sin40°}$, so $2y = x$.
$|x - 2y| = 0$.
Answer: (A). | ||||
| 80 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{b}\times\vec{c}$ is perpendicular to $\vec{b}$, so $|\vec{b}\times(\vec{b}\times\vec{c})| = |\vec{b}||\vec{b}\times\vec{c}| = 1\cdot(1\cdot2\sin\theta) = 2\sin\theta$.
$2\sin\theta = \sqrt{2}$, so $\theta = \frac{\pi}{4}$ and $1 + \tan\theta = 2$.
Answer: (B). | ||||
| 81 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Multiply by the conjugate: $z = \frac{(3 + 2i\cos\theta)(1 + 3i\cos\theta)}{1 + 9\cos^2\theta} = \frac{(3 - 6\cos^2\theta) + 11i\cos\theta}{1 + 9\cos^2\theta}$.
Real part zero: $\cos^2\theta = \frac{1}{2}$, so $\theta = \frac{\pi}{4}$ in $\left(0, \frac{\pi}{2}\right)$.
$\sin^2\frac{3\pi}{4} + \cos^2\frac{\pi}{4} = \frac{1}{2} + \frac{1}{2} = 1$.
Answer: 1. | ||||
| 82 | 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 Write $A = I + N$ with $N = \begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix}$, $N^2 = \begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}$, $N^3 = O$.
$A^k = I + kN + \frac{k(k-1)}{2}N^2$. Sums of elements: I gives 3, N gives 3, $N^2$ gives 1, so the elements of $A^k$ add to $3 + 3k + \frac{k(k-1)}{2}$.
$\sum_{k=1}^{20}\left(3 + 3k + \frac{k(k-1)}{2}\right) = 60 + 3\times210 + {}^{21}C_3 = 60 + 630 + 1330 = 2020$.
Answer: 2020. | ||||
| 83 | 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 The number is $2^{10}\cdot5^{10}\cdot11^{11}\cdot13^{13}$. A divisor of the form 4n + 1 is odd, so it is $5^a\cdot11^b\cdot13^c$.
Mod 4: $5 \equiv 1$, $13 \equiv 1$, $11 \equiv -1$. So the divisor is $\equiv (-1)^b$, which is 1 exactly when b is even.
Choices: a from 0 to 10 (11 ways), b in {0, 2, 4, 6, 8, 10} (6 ways), c from 0 to 13 (14 ways).
Number $= 11\times6\times14 = 924$.
Answer: 924. | ||||
| 84 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Line QR: direction $R - Q = (-1, 1, 6)$; points $(3 - t, -4 + t, -5 + 6t)$.
In the plane: $2(3 - t) + (-4 + t) + (-5 + 6t) = 7$, so $5t - 3 = 7$, $t = 2$: the point is (1, −2, 7).
Distance from P: $\sqrt{(3-1)^2 + (4+2)^2 + (4-7)^2} = \sqrt{4 + 36 + 9} = 7$.
Answer: 7. | ||||
| 85 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalEasy | |
A easy numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $e^{-y}dy = e^{\alpha x}dx$, so $-e^{-y} = \frac{e^{\alpha x}}{\alpha} + C$.
$y(0) = \ln\frac{1}{2}$: $-2 = \frac{1}{\alpha} + C$. $y(\ln2) = \ln2$: $-\frac{1}{2} = \frac{2^\alpha}{\alpha} + C$.
Subtracting: $\frac{3}{2} = \frac{2^\alpha - 1}{\alpha}$. For natural α: α = 1 gives 1, α = 2 gives $\frac{3}{2}$, and for α ≥ 3 the right side exceeds $\frac{3}{2}$ (α = 3 gives $\frac{7}{3}$, and it keeps growing).
α = 2.
Answer: 2. | ||||
| 86 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $u = \cos x$, $du = -\sin x\,dx$, $\sin^2x = 1 - u^2$: $\int_0^\pi\sin^3x\,e^{-\sin^2x}dx = \int_{-1}^{1}(1-u^2)e^{u^2-1}du = \frac{2}{e}\int_0^1(1-u^2)e^{u^2}du$.
Since $\frac{d}{du}\left(ue^{u^2}\right) = e^{u^2} + 2u^2e^{u^2}$, $\int_0^1e^{u^2}du = e - 2\int_0^1u^2e^{u^2}du$.
So $\int_0^1(1-u^2)e^{u^2}du = e - 3\int_0^1u^2e^{u^2}du$.
With $t = u^2$: $\int_0^1\sqrt{t}\,e^tdt = \int_0^1u\,e^{u^2}\cdot2u\,du = 2\int_0^1u^2e^{u^2}du$.
Hence the integral $= \frac{2}{e}\left(e - \frac{3}{2}\int_0^1\sqrt{t}e^tdt\right) = 2 - \frac{3}{e}\int_0^1\sqrt{t}e^tdt$: α = 2, β = 3.
α + β = 5.
Answer: 5. | ||||
| 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 $\vec{a}\cdot\vec{b} = 3 - \alpha\beta - \alpha\beta = 3 - 2\alpha\beta = -1$, so $\alpha\beta = 2$.
$\vec{b}\cdot\vec{c} = -3\alpha - 2\beta - \alpha = -4\alpha - 2\beta = 10$, so $\beta = -5 - 2\alpha$.
$\alpha(-5 - 2\alpha) = 2$: $2\alpha^2 + 5\alpha + 2 = 0$, α = −2 or $-\frac{1}{2}$; integer α = −2, β = −1.
$\vec{a} = (1, 2, -1)$, $\vec{b} = (3, -1, 2)$, $\vec{c} = (2, -2, 1)$.
$(\vec{a}\times\vec{b})\cdot\vec{c} = \begin{vmatrix}1&2&-1\\3&-1&2\\2&-2&1\end{vmatrix} = 1(-1+4) - 2(3-4) - 1(-6+2) = 3 + 2 + 4 = 9$.
Answer: 9. | ||||
| 88 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Centre (3, −4), focus at distance c = 1 and vertex at distance a = 2 along the x-direction: $b^2 = a^2 - c^2 = 3$.
E: $\frac{(x-3)^2}{4} + \frac{(y+4)^2}{3} = 1$. Shift to $X = x - 3$, $Y = y + 4$: the line $y = mx - 4$ becomes $Y = mX + 3m$.
Tangency to $\frac{X^2}{4} + \frac{Y^2}{3} = 1$: $(3m)^2 = 4m^2 + 3$, so $5m^2 = 3$.
Answer: 3. | ||||
| 89 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions A: $n^2 - n - 10000 \le 0$ gives $n \le \frac{1 + \sqrt{40001}}{2} \approx 100.5$, so $A = \{1, 2, \ldots, 100\}$.
B − C: numbers $3k + 1$ ($k \ge 1$) that are odd, i.e. k even: 7, 13, 19, ..., the numbers $6j + 1$, $j \ge 1$.
In A: $6j + 1 \le 100$ gives j = 1 to 16: 7, 13, ..., 97.
Sum $= 6(1 + 2 + \cdots + 16) + 16 = 6\times136 + 16 = 832$.
Answer: 832. | ||||
| 90 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $t = e^x > 0$: $t^4 - t^3 - 4t^2 - t + 1 = 0$. Divide by $t^2$: $\left(t^2 + \frac{1}{t^2}\right) - \left(t + \frac{1}{t}\right) - 4 = 0$.
With $u = t + \frac{1}{t} \ge 2$: $u^2 - 2 - u - 4 = 0$, $u^2 - u - 6 = 0$, u = 3 or −2; only u = 3 is allowed.
$t + \frac{1}{t} = 3$ gives $t = \frac{3 \pm \sqrt{5}}{2}$, both positive, so two real values of x.
Answer: 2. | ||||
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