JEE (Main) · 2021
JEE (Main) 2021 — 17 March, 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
- Current Electricity2
- Electromagnetic Induction and Alternating Currents2
14 easy · 15 medium · 1 hard
Chemistry
30 Q
- d- and f-Block Elements2
- Coordination Compounds2
- Hydrocarbons2
25 easy · 5 medium · 0 hard
Mathematics
30 Q
- Integral Calculus4
- Trigonometry2
- Vector Algebra2
6 easy · 20 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 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter An AM wave contains the carrier and the two side bands $f_c \pm f_m$, so its bandwidth is $2f_m$.
$f_m = \frac{\omega_m}{2\pi} = \frac{1.57\times10^{8}}{2\times3.14} = 2.5\times10^{7}$ Hz $= 25$ MHz.
Bandwidth $= 2\times25 = 50$ MHz.
Answer: (A). | ||||
| 2 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity The cells are drawn in the same sense, so their emfs add: $I = \frac{2E + E}{R + r_1 + r_2} = \frac{3E}{R + r_1 + r_2}$.
Terminal p.d. of the first cell: $2E - Ir_1 = 0 \Rightarrow 2E = \frac{3Er_1}{R + r_1 + r_2}$.
$2(R + r_1 + r_2) = 3r_1 \Rightarrow R = \frac{r_1}{2} - r_2$.
Answer: (A). | ||||
| 3 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The to-and-fro distance 6 cm is twice the amplitude, so $A = 3$ cm $= 0.03$ m.
$\omega = 2\pi f = 2\pi\times245 \approx 1.54\times10^{3}\ \mathrm{rad\,s^{-1}}$, i.e. $1.5\times10^{3}$.
$k = \frac{\omega}{v} = \frac{1539}{300} \approx 5.1\ \mathrm{m^{-1}}$.
So $Y = 0.03\sin[5.1x - (1.5\times10^{3})t]$.
Answer: (B). | ||||
| 4 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Orbit radii from the centre: $r_1 = R + 11R = 12R$ (period 24 h, the planet's own period) and $r_2 = R + 2R = 3R$.
By Kepler's third law $T \propto r^{3/2}$: $T_2 = 24\left(\frac{3R}{12R}\right)^{3/2} = 24\times\frac{1}{8} = 3$ h.
Answer: (C). | ||||
| 5 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Taking the collision of C with A as elastic (equal masses), C stops and A moves off with speed v while the spring is still unstretched.
Maximum compression occurs when A and B move with a common velocity: $mv = 2mV \Rightarrow V = \frac{v}{2}$.
Energy: $\frac{1}{2}mv^2 = \frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 + \frac{1}{2}Kx^2 \Rightarrow \frac{1}{2}Kx^2 = \frac{1}{4}mv^2$.
$x = v\sqrt{\frac{m}{2K}}$.
Answer: (B). | ||||
| 6 | 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 $v_{max} = A\omega$ with $\omega = \sqrt{\frac{K}{m}}$.
$A_1\sqrt{\frac{K_1}{m}} = A_2\sqrt{\frac{K_2}{m}} \Rightarrow \frac{A_1}{A_2} = \sqrt{\frac{K_2}{K_1}}$.
Answer: (D). | ||||
| 7 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current (a) Pure resistor: current and voltage are in phase, phase difference zero → (ii).
(b) Pure inductor: current lags the voltage by $\frac{\pi}{2}$ → (iii).
(c) Pure capacitor: current leads the voltage by $\frac{\pi}{2}$ → (i).
(d) Series LCR: $\tan\phi = \frac{X_C - X_L}{R}$ (angle by which the current leads) → (iv).
Answer: (C). | ||||
| 8 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $X_L = 2\pi fL$: halving f halves $X_L$.
$I = \frac{V}{X_L}$: with the same voltage the current doubles.
Answer: (B). | ||||
| 9 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For a solid sphere rolling without slipping on an incline, $a = \frac{g\sin\theta}{1 + \frac{k^2}{R^2}} = \frac{g\sin30^\circ}{1 + \frac{2}{5}} = \frac{5}{14}g$, directed down the incline both going up and coming down.
With $g = 9.8\ \mathrm{ms^{-2}}$: $a = 3.5\ \mathrm{ms^{-2}}$.
Time to stop and come back to A: $t = \frac{2u}{a} = \frac{2\times1}{3.5} \approx 0.57$ s.
Answer: (B). | ||||
| 10 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Rising to $\frac{81}{100}$ of the fall height means the speed after each bounce is $e = 0.9$ times the speed before it.
Distance: $5 + 2(5)(0.81) + 2(5)(0.81)^2 + \dots = 5\cdot\frac{1 + 0.81}{1 - 0.81} = \frac{9.05}{0.19} \approx 47.63$ m.
Time: first fall $t_0 = \sqrt{\frac{2\times5}{10}} = 1$ s; each later up-and-down trip takes $2t_0e^n$. Total $= t_0\frac{1 + e}{1 - e} = \frac{1.9}{0.1} = 19$ s.
Average speed $= \frac{47.63}{19} \approx 2.50\ \mathrm{ms^{-1}}$.
Answer: (B). | ||||
| 11 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $x = \int_0^1 (v_0 + gt + Ft^2)\,dt = v_0 + \frac{g}{2} + \frac{F}{3}$.
Answer: (C). | ||||
| 12 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter P is at the end of each straight wire, at perpendicular distance r. Each semi-infinite wire gives $\frac{\mu_0 I}{4\pi r}$.
The semicircle of radius r gives $\frac{\mu_0 I}{4r}$ at its centre.
Following the current (right along the top wire, round the semicircle, left along the bottom wire) the current circulates the same way about P everywhere, so all three fields point the same way (into the page).
$B = 2\cdot\frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4\pi r}(2 + \pi)$.
Answer: (C). | ||||
| 13 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Lyman lines are ultraviolet; Paschen and Brackett lines are infrared. The Balmer series (transitions to $n = 2$, 656 nm to 365 nm) lies in the visible region.
Answer: (B). | ||||
| 14 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\frac{1}{2}mv_1^2 = hf_1 - \phi$ and $\frac{1}{2}mv_2^2 = hf_2 - \phi$ (same work function).
Subtracting: $v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$.
Answer: (B). | ||||
| 15 | 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 upper path is an AND gate followed by NOT: $\overline{AB}$. The lower path is an OR gate: $A + B$. These feed an AND gate:
$Y = \overline{AB}\cdot(A + B) = (\bar{A} + \bar{B})(A + B) = \bar{A}B + A\bar{B} = A \oplus B$.
Truth table: 00→0, 01→1, 10→1, 11→0, which is XOR.
Answer: (C). | ||||
| 16 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity Take $V_A = 10$ V, $V_C = 0$, and unknown potentials $V_B$, $V_D$. AB = 100 Ω, BC = 10 Ω, AD = 60 Ω, DC = 5 Ω, BD = 15 Ω.
Node B: $\frac{10 - V_B}{100} = \frac{V_B}{10} + \frac{V_B - V_D}{15}$.
Node D: $\frac{10 - V_D}{60} = \frac{V_D}{5} + \frac{V_D - V_B}{15}$.
Multiplying out: $53V_B - 20V_D = 30$ (node B, ×300) and $-4V_B + 17V_D = 10$ (node D, ×60).
Solving: $V_B = \frac{710}{821}$ V $\approx 0.8648$ V and $V_D = \frac{650}{821}$ V $\approx 0.7917$ V.
$I_g = \frac{V_B - V_D}{15} = \frac{60/821}{15} = \frac{4}{821}$ A $\approx 4.87\times10^{-3}$ A $= 4.87$ mA (from B to D).
Answer: (D). | ||||
| 17 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For damping force $-bv$, the amplitude falls as $A = A_0e^{-bt/2m}$.
$6 = 12e^{-b(120)/(2\times1)} \Rightarrow 60b = \ln2 \Rightarrow b = \frac{0.693}{60} = 1.16\times10^{-2}\ \mathrm{kg\,s^{-1}}$.
Option (A) carries the mantissa 1.16 with the exponent printed as $10^{2}$; it is evidently this value with the minus sign of the exponent lost.
Answer: (A). | ||||
| 18 | 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 two modes add 4.
$f = 3 + 3 + 4 = 10$, $C_V = \frac{f}{2}R = 5R$, $C_P = 6R$.
$\beta = \frac{6R}{5R} = 1.2$.
Answer: (B). | ||||
| 19 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics (a) P–V: a vertical line is constant volume, not adiabatic. Wrong.
(b) P–T: the line labelled isothermal is horizontal (constant P); an isothermal on a P–T graph is vertical. Wrong.
(c) V–T: isothermal is the vertical line (T constant); adiabatic $VT^{1/(\gamma-1)} =$ constant falls as T rises. Correct.
(d) P–T: isothermal is the vertical line; adiabatic $P \propto T^{\gamma/(\gamma-1)}$ rises steeply. Correct.
Answer: (D). | ||||
| 20 | Properties of Bulk MatterElasticity | 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 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 Hydraulic stress $= \rho gh = 1000\times9.8\times2000 = 1.96\times10^{7}$ Pa.
Bulk modulus $= \frac{1.96\times10^{7}}{0.0136} \approx 1.44\times10^{9}\ \mathrm{Nm^{-2}}$.
Answer: (D). | ||||
| 21 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Initial charge $Q = 2\times10 = 20\ \mu$C, shared between the capacitors in parallel.
Common potential $V = \frac{20}{2 + 8} = 2$ V; charge on $C_2 = 8\times2 = 16\ \mu$C.
Answer: 16. | ||||
| 22 | Laws of MotionFriction | Class 11: Laws of Motion | NumericalMedium | |
A medium numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion One rope runs from the boy's hands over the pulleys to the wood, so its tension T is the boy's pull F everywhere.
System boy + wood: the rope pulls it up at the boy's hands (taken vertical) with T, and horizontally at the wood with T.
Normal reaction $N = (4 + 5)g - T = 90 - T$.
No sliding: $T \le \mu N = 0.5(90 - T) \Rightarrow 1.5T \le 45 \Rightarrow T \le 30$ N.
Answer: 30. | ||||
| 23 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\lambda_{medium} = \frac{2}{3}\lambda_{air} \Rightarrow \mu = \frac{3}{2}$.
$v = +10$ m and $|u| = \frac{3}{2}\times10 = 15$ m, so $u = -15$ m.
$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$: $\frac{1.5}{10} + \frac{1}{15} = \frac{0.5}{R}$.
$\frac{9 + 4}{60} = \frac{13}{60} = \frac{0.5}{R} \Rightarrow R = \frac{30}{13}$ m, so $x = 30$.
Answer: 30. | ||||
| 24 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalEasy | |
A easy numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Intensity $I = \frac{P}{4\pi r^2} = \frac{1}{2}c\varepsilon_0E_0^2$, so at a fixed distance $E \propto \sqrt{P}$.
$E' = E\sqrt{\frac{60}{100}} = \sqrt{\frac{3}{5}}E$, so $x = 3$.
Answer: 3. | ||||
| 25 | Electromagnetic WavesDisplacement current | Class 12: Electromagnetic Waves | NumericalHard | |
A hard numerical question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves With $E = E_0\sin\omega t$ between the plates: $J_c = \frac{E}{\rho} = \frac{E_0\sin\omega t}{\rho}$ and $J_d = \varepsilon\frac{dE}{dt} = \varepsilon\omega E_0\cos\omega t$.
$\frac{J_c}{J_d} = \frac{\tan\omega t}{\rho\varepsilon\omega}$.
At $t = \frac{1}{800}$ s: $\omega t = 2\pi\times900\times\frac{1}{800} = \frac{9\pi}{4}$, and $\tan\frac{9\pi}{4} = 1$.
$\rho\varepsilon\omega = 0.25\times80\times\frac{1}{4\pi\times9\times10^{9}}\times2\pi\times900 = \frac{0.25\times80\times900}{18\times10^{9}} = 10^{-6}$.
$\frac{J_c}{J_d} = 10^{6}$, so $x = 6$.
Answer: 6. | ||||
| 26 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance A surface parallel to the Y–Z plane has its area vector along $\hat{i}$: $\vec{A} = 0.4\hat{i}$.
$\phi = \vec{E}\cdot\vec{A} = \frac{2}{5}\times4.0\times10^{3}\times0.4 = 640\ \mathrm{Nm^2C^{-1}}$.
Answer: 640. | ||||
| 27 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take a as the radius of the disc. For the quarter disc in the first quadrant, using polar strips: $x_{cm} = \frac{\int_0^{\pi/2}\int_0^a (r\cos\theta)\,r\,dr\,d\theta}{\frac{\pi a^2}{4}} = \frac{\frac{a^3}{3}\cdot1}{\frac{\pi a^2}{4}} = \frac{4a}{3\pi}$, and by symmetry $y_{cm} = \frac{4a}{3\pi}$.
Comparing with $\frac{x}{3}\frac{a}{\pi}$: $x = 4$.
Answer: 4. | ||||
| 28 | Laws of MotionFriction | Class 11: Laws of Motion | NumericalMedium | |
A medium numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Pull at angle $\theta$ above the horizontal: $F\cos\theta = \mu(mg - F\sin\theta) \Rightarrow F = \frac{\mu mg}{\cos\theta + \mu\sin\theta}$.
The denominator is largest, $\sqrt{1 + \mu^2}$, at $\tan\theta = \mu$.
$F_{min} = \frac{\mu mg}{\sqrt{1 + \mu^2}} = \frac{\frac{1}{\sqrt{3}}\times10}{\frac{2}{\sqrt{3}}} = 5$ N.
Answer: 5. | ||||
| 29 | 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 Force $F = -\frac{dU}{dr} = -4U_0r^3$ (attractive), so $\frac{mv^2}{r} = 4U_0r^3 \Rightarrow mv^2 = 4U_0r^4$.
Bohr: $mvr = \frac{nh}{2\pi} \Rightarrow v = \frac{nh}{2\pi mr}$.
$m\frac{n^2h^2}{4\pi^2m^2r^2} = 4U_0r^4 \Rightarrow r^6 \propto n^2 \Rightarrow r \propto n^{1/3}$.
So $\alpha = 3$.
Answer: 3. | ||||
| 30 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | NumericalMedium | |
A medium numerical question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Volume of one drop: $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi\cdot\frac{3}{40\pi}\times10^{-9} = 10^{-10}\ \mathrm{cm^3}$; 100 drops: $10^{-8}\ \mathrm{cm^3}$ of solution.
Oleic acid in it: $0.01\times10^{-8} = 10^{-10}\ \mathrm{cm^3}$.
Thickness $= \frac{10^{-10}}{4} = 2.5\times10^{-11}$ cm $= 2.5\times10^{-13}$ m $= 25\times10^{-14}$ m.
Answer: 25. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $N_3^-$ is isoelectronic with $CO_2$ (16 valence electrons): $[N=N=N]^-$ with an sp central N and no lone pair on it, so it is linear.
$NO_2$ (a lone electron on N), $O_3$ (a lone pair on the central O) and $Cl_2O$ (two lone pairs on O) are all bent.
Answer: (A). | ||||
| 32 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics (A), (B): liquid → solid, the molecules become ordered; entropy of the system falls at either temperature.
(C): 4 mol of gas give 2 mol of gas; entropy falls.
(D): adsorption restricts the freedom of the gas molecules; entropy falls.
(E): dissolving NaCl frees the ions from the lattice; entropy rises.
So (A), (B), (C) and (D).
Answer: (A). | ||||
| 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 negative sol is coagulated by cations, and by the Hardy–Schulze rule the higher the charge of the cation the greater its flocculating power. Of the cations given, $Ba^{2+}$ has the higher charge.
Answer: (A). | ||||
| 34 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Be–Al, B–Si and Li–Mg are pairs with a diagonal relationship (second-period element and the third-period element one group to the right).
Li–Na are members of the same group (group 1), a different relationship.
Answer: (C). | ||||
| 35 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Haematite $Fe_2O_3$ (ii); bauxite $Al_2O_3\cdot xH_2O$ (i); magnetite $Fe_3O_4$ (iv); malachite $CuCO_3\cdot Cu(OH)_2$ (iii).
Answer: (B). | ||||
| 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 Cation exchange resins carry acidic $-SO_3H$ groups whose $H^+$ is exchanged for metal ions: $2RSO_3H + Ca^{2+} \rightarrow (RSO_3)_2Ca + 2H^+$.
Anion exchange resins carry basic amino groups ($-NH_2$, which in water become $-NH_3^+OH^-$) whose $OH^-$ is exchanged for anions.
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 In the Solvay process ammonia is recovered by heating the ammonium chloride solution with slaked lime:
$2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + CaCl_2 + 2H_2O$.
The by-product is $CaCl_2$.
Answer: (C). | ||||
| 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 $N_2O$ and NO are neutral oxides; $N_2O_3$ and $NO_2$ (and $N_2O_5$) are acidic.
Answer: (A). | ||||
| 39 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Z = 24 is chromium, $[Ar]3d^54s^1$. Using the 4s and the five 3d electrons it shows oxidation states +2, +3, +4, +5 and +6 (+3 and +6 being the most common), i.e. +2 to +6.
Answer: (B). | ||||
| 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) Both ions are complexes, so ligands can be exchanged between Co and Cr: co-ordination isomerism (iii).
(b) $NO_2^-$ can bind through N or O: linkage isomerism (i).
(c) Water can be inside or outside the coordination sphere ($[Cr(H_2O)_5Cl]Cl_2\cdot H_2O$ etc.): solvate (hydrate) isomerism (ii).
(d) The cis bis(oxalato) complex has no plane of symmetry: optical isomerism (iv).
Answer: (C). | ||||
| 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 Eutrophication is the nutrient enrichment of water bodies: run-off of fertilisers (A) and phosphate detergents (B) supplies the nutrients, and the resulting dense plant (algal) growth (C) uses up the dissolved oxygen.
A lack of nutrients (D) is the opposite of what causes eutrophication, so (D) is the incorrect reason.
Answer: (D). | ||||
| 42 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Kjeldahl's method does not work for nitro and azo/diazo compounds or for nitrogen in a ring (pyridine), because that nitrogen is not converted to ammonium sulphate on digestion with conc. $H_2SO_4$.
Benzylamine has an amino nitrogen, which is converted to $(NH_4)_2SO_4$, so it can be estimated.
Answer: (B). | ||||
| 43 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes An ambident nucleophile can attack through two different atoms.
$CN^-$ attacks through C (giving nitriles, with KCN) or N (giving isocyanides, with AgCN).
$NO_2^-$ attacks through N (nitroalkanes, with $AgNO_2$) or O (alkyl nitrites, with $KNO_2$).
$RCOO^-$ attacks through either of two equivalent O atoms (same product), and $I^-$ has only one atom. So (A) and (C).
Answer: (C). | ||||
| 44 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Alkanes with a tertiary H are oxidised by $KMnO_4$ at that carbon to the tertiary alcohol: $(CH_3)_2CHCH_2CH_3 \rightarrow (CH_3)_2C(OH)CH_2CH_3$, 2-methylbutan-2-ol. Statement I is correct.
n-Alkanes have no tertiary H and are not easily oxidised by $KMnO_4$. Statement II is incorrect.
Answer: (C). | ||||
| 45 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons A is a primary carbocation; B is a secondary carbocation, stabilised by more hyperconjugation and +I effect, so B is more stable.
The more stable carbocation has the lower-energy transition state and forms faster (Markovnikov addition).
Answer: (D). | ||||
| 46 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Invertase hydrolyses sucrose to glucose and fructose; zymase (from yeast) ferments glucose to ethanol and carbon dioxide.
Answer: (A). | ||||
| 47 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines The product keeps $-OCH_3$ on the ring, so A ($C_7H_7N_2OCl$) is methoxybenzenediazonium chloride, $CH_3O-C_6H_4-N_2^+Cl^-$ (the para isomer in the options).
Ethanol reduces a diazonium salt, replacing $-N_2^+$ by H and being oxidised to ethanal:
$ArN_2^+Cl^- + CH_3CH_2OH \rightarrow ArH + N_2 + CH_3CHO + HCl$.
So X = $CH_3CHO$ and Y = HCl.
Answer: (D). | ||||
| 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 A sulphonyl chloride (Hinsberg's reagent, benzenesulphonyl chloride or p-toluenesulphonyl chloride) gives with a primary amine a sulphonamide soluble in alkali, with a secondary amine one insoluble in alkali, and does not react with a tertiary amine; this separates the three.
Chloroform/KOH (carbylamine test) only detects primary amines.
Answer: (C). | ||||
| 49 | Chemistry in Everyday LifeChemicals in food | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on chemicals in food, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Sucralose is an artificial sweetener (ii); sodium benzoate is a food preservative (iv); bithionol is an antiseptic added to soaps (iii).
That leaves the glyceryl stearate with (i), a non-ionic surfactant; only option (B) has (a)→(ii), (c)→(iv), (d)→(iii).
Answer: (B). | ||||
| 50 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Fructose, $C_6H_{12}O_6$, has six carbons and a keto group at C-2: a ketohexose. (Its cyclic form is a furanose, not a pyranose.)
Answer: (B). | ||||
| 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 STP: 1 bar and 273 K (the bar in R points to this definition). $n = \frac{PV}{RT} = \frac{1\times0.020}{0.083\times273} = \frac{0.020}{22.66} = 8.83\times10^{-4}$ mol of $Cl_2$.
Atoms $= 2\times8.83\times10^{-4}\times6.023\times10^{23} = 1.06\times10^{21}$.
To the nearest integer: 1.
Answer: 1. | ||||
| 52 | Solid StatePoint defects | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on point defects, from Solid State. NCERT: No matching NCERT chapter Each $Sr^{2+}$ replaces two $K^+$ and leaves one cation vacancy.
Moles of KBr in 1 g: $\frac{1}{39.1 + 79.9} = \frac{1}{119} = 8.40\times10^{-3}$ mol.
$10^{-5}$ mole percent = mole fraction $10^{-7}$: moles of $SrBr_2 = 10^{-7}\times8.40\times10^{-3} = 8.40\times10^{-10}$.
Vacancies $= 8.40\times10^{-10}\times6.023\times10^{23} = 5.06\times10^{14}$.
To the nearest integer: 5.
Answer: 5. | ||||
| 53 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Fe: $[Ar]3d^64s^2$; the $3d^6$ set has 4 unpaired electrons.
$\mu = \sqrt{n(n+2)} = \sqrt{24} = 2\sqrt{2}\sqrt{3} = 2\times1.41\times1.73 = 4.88$ BM $= 48.8\times10^{-1}$ BM.
To the nearest integer: 49.
Answer: 49. | ||||
| 54 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $K_4Fe(CN)_6 \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$ (5 ions): $i = 1 + (5-1)(0.4) = 2.6$, so the effective molality is $2.6\times1 = 2.6$.
Same boiling point, so the solution of A has molality 2.6. In 100 g: 18.1 g A in 81.9 g water.
$\frac{18.1/M}{0.0819} = 2.6 \Rightarrow M = \frac{18.1}{2.6\times0.0819} = \frac{18.1}{0.2129} = 85.0$.
To the nearest integer: 85.
Answer: 85. | ||||
| 55 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $K_P = K_C(RT)^{\Delta n}$ with $\Delta n = 2 - 1 = 1$.
$T = \frac{K_P}{K_C R} = \frac{600.1}{20.4\times0.0831} = \frac{600.1}{1.6952} = 354.0$ K.
To the nearest integer: 354.
Answer: 354. | ||||
| 56 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell constant $G^* = \kappa R = 0.14\times4.19 = 0.5866\ \mathrm{m^{-1}}$.
$\kappa_{HCl} = \frac{G^*}{R} = \frac{0.5866}{1.03} = 0.5695\ \mathrm{S\ m^{-1}} = 56.95\times10^{-2}\ \mathrm{S\ m^{-1}}$.
To the nearest integer: 57.
Answer: 57. | ||||
| 57 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Elementary, so rate $= k[A]^2[B_2]$.
Reducing the volume to one third triples every concentration: the rate becomes $3^2\times3 = 27$ times.
Answer: 27. | ||||
| 58 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $FeCl_3 + 3H_2C_2O_4 + 3KOH \rightarrow K_3[Fe(C_2O_4)_3] + 3HCl + 3H_2O$ (the acid is taken up by excess KOH).
Each oxalate is bidentate, so Fe has coordination number $3\times2 = 6$; Werner's secondary valency is the coordination number.
Answer: 6. | ||||
| 59 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalEasy | |
A easy numerical question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Mesityl oxide is 4-methylpent-3-en-2-one, $(CH_3)_2C=CH-CO-CH_3$.
C–C sigma bonds: two $CH_3-C$ bonds on C-4, the C=C double bond (one sigma), $CH-CO$, and $CO-CH_3$: $2 + 1 + 1 + 1 = 5$.
Answer: 5. | ||||
| 60 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Benzoyl chloride $C_7H_5OCl$: M = 140.5; $\frac{0.140}{140.5} = 9.96\times10^{-4}$ mol.
Diphenylamine $C_{12}H_{11}N$: M = 169; $\frac{0.388}{169} = 2.30\times10^{-3}$ mol, in excess, so benzoyl chloride is limiting.
Product $C_6H_5CON(C_6H_5)_2$, $C_{19}H_{15}NO$: M = 273. Theoretical yield $= 9.96\times10^{-4}\times273 = 0.272$ g.
% yield $= \frac{0.210}{0.272}\times100 = 77.2\%$.
To the nearest integer: 77.
Answer: 77. | ||||
| 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 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 For $x \in [-1, 1]$, $x^2 \in [0, 1]$.
If $0 \le x^2 < \frac{2}{3}$: $[x^2 + \frac{1}{3}] = 0$ and $[x^2 - \frac{2}{3}] = -1$, so the left side is $\sin^{-1}0 + \cos^{-1}(-1) = \pi$. But $x^2 < 1 < \pi$: no solution.
If $\frac{2}{3} \le x^2 \le 1$: $[x^2 + \frac{1}{3}] = 1$ and $[x^2 - \frac{2}{3}] = 0$, so the left side is $\frac{\pi}{2} + \frac{\pi}{2} = \pi$, again more than $x^2$: no solution.
Number of solutions: 0.
Answer: (A). | ||||
| 62 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter (A) $(p \wedge q) \wedge (p \vee q)$ is false when p, q are both false. Not a tautology.
(B) $(p \wedge q) \vee (p \rightarrow q)$ is false when p is true and q false (both parts false).
(C) $(p \wedge q) \rightarrow (p \rightarrow q)$: whenever $p \wedge q$ is true, q is true, so $p \rightarrow q$ is true. The implication is never false: a tautology.
(D) $(p \wedge q) \wedge (p \rightarrow q)$ is false whenever p is false.
Answer: (C). | ||||
| 63 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{OP}\cdot\vec{OQ} = -x + 2y - 3x = 0 \Rightarrow y = 2x$.
$\vec{PQ} = (-1 - x)\hat{i} + (2 - 2x)\hat{j} + (3x + 1)\hat{k}$; $|\vec{PQ}|^2 = (1 + x)^2 + (2 - 2x)^2 + (3x + 1)^2 = 14x^2 + 6 = 20 \Rightarrow x^2 = 1$, so $x = 1$, $y = 2$.
Coplanarity: $\begin{vmatrix}1&2&-1\\-1&2&3\\3&z&-7\end{vmatrix} = 1(-14 - 3z) - 2(7 - 9) - 1(-z - 6) = -4 - 2z = 0 \Rightarrow z = -2$.
$x^2 + y^2 + z^2 = 1 + 4 + 4 = 9$.
Answer: (D). | ||||
| 64 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Apply $R_1 \rightarrow R_1 - 2R_2 + R_3$ (the determinant is unchanged): the new first row is $(3 - 8 + 5,\ 4\sqrt{2} - 10\sqrt{2} + k,\ x - 2y + z) = (0,\ k - 6\sqrt{2},\ 0)$, since $x + z = 2y$.
Expanding along it: $\Delta = -(k - 6\sqrt{2})(4z - 5y)$.
With $y = x + d$, $z = x + 2d$: $4z - 5y = 3d - x \ne 0$ because $x \ne 3d$.
So $k = 6\sqrt{2}$ and $k^2 = 72$.
Answer: (D). | ||||
| 65 | 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 $\tan(\pi\cos^2\theta) = \tan(\pi - \pi\sin^2\theta) = -\tan(\pi\sin^2\theta)$.
As $\theta \to 0$: $\frac{-\tan(\pi\sin^2\theta)}{\sin(2\pi\sin^2\theta)} \to \frac{-\pi\sin^2\theta}{2\pi\sin^2\theta} = -\frac{1}{2}$.
Answer: (B). | ||||
| 66 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The integrand has period 1, so the integral is $10\int_0^1 [\sin 2\pi x]e^{-x}dx$.
On $(0, \frac{1}{2})$, $0 < \sin 2\pi x \le 1$ gives $[\sin 2\pi x] = 0$ (except at the single point $x = \frac{1}{4}$, which does not matter); on $(\frac{1}{2}, 1)$, $[\sin 2\pi x] = -1$.
$\int_0^1 [\sin 2\pi x]e^{-x}dx = -\int_{1/2}^1 e^{-x}dx = e^{-1} - e^{-1/2}$.
So the integral is $10e^{-1} - 10e^{-1/2} + 0$: $\alpha = 10$, $\beta = -10$, $\gamma = 0$, and $\alpha + \beta + \gamma = 0$.
Answer: (A). | ||||
| 67 | 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 $x^2 + x^{5/4} = x^{5/4}(x^{3/4} + 1)$ and $x + x^{1/4} = x^{1/4}(x^{3/4} + 1)$, so dividing by $2x^{5/4}(x^{3/4} + 1)$:
$\frac{dy}{dx} - \frac{y}{2x} = \frac{x}{x^{3/4} + 1}$.
Integrating factor $x^{-1/2}$: $\frac{d}{dx}\left(\frac{y}{\sqrt{x}}\right) = \frac{x^{1/2}}{x^{3/4} + 1}$.
Put $x = t^4$: $\int\frac{t^2\cdot4t^3}{t^3 + 1}dt = \frac{4}{3}\int\frac{u}{u + 1}du$ (with $u = t^3 = x^{3/4}$) $= \frac{4}{3}\left(x^{3/4} - \ln(1 + x^{3/4})\right)$.
So $\frac{y}{\sqrt{x}} = \frac{4}{3}\left(x^{3/4} - \ln(1 + x^{3/4})\right) + C$. At $x = 1$: $1 - \frac{4}{3}\ln2 = \frac{4}{3} - \frac{4}{3}\ln2 + C \Rightarrow C = -\frac{1}{3}$.
At $x = 16$ ($\sqrt{x} = 4$, $x^{3/4} = 8$): $y = 4\left(\frac{32}{3} - \frac{4}{3}\ln9 - \frac{1}{3}\right) = 4\left(\frac{31}{3} - \frac{8}{3}\ln3\right)$.
Answer: (A). | ||||
| 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 Write $Q = 3\sin x + \cos x + 3$. Dividing by $\cos x\cdot Q$: $\frac{dy}{dx} - y\tan x = \frac{1}{Q\cos x}$.
Integrating factor $\cos x$: $\frac{d}{dx}(y\cos x) = \frac{1}{3\sin x + \cos x + 3}$.
With $t = \tan\frac{x}{2}$: $3\sin x + \cos x + 3 = \frac{2t^2 + 6t + 4}{1 + t^2}$ and $dx = \frac{2dt}{1 + t^2}$, so $\int\frac{dx}{Q} = \int\frac{dt}{t^2 + 3t + 2} = \ln\frac{t + 1}{t + 2} + C$.
$y(0) = 0$ ($t = 0$): $0 = \ln\frac{1}{2} + C \Rightarrow C = \ln2$, so $y\cos x = \ln\frac{2(t + 1)}{t + 2}$.
At $x = \frac{\pi}{3}$: $t = \frac{1}{\sqrt{3}}$, $\cos x = \frac{1}{2}$: $y = 2\ln\frac{2(1 + \sqrt{3})}{1 + 2\sqrt{3}} = 2\ln\frac{2(1 + \sqrt{3})(2\sqrt{3} - 1)}{11} = 2\ln\frac{2\sqrt{3} + 10}{11}$.
Answer: (D). | ||||
| 69 | 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 For $x > 0$: $f'(x) = 2 - \sin\frac{1}{x} + \frac{1}{x}\cos\frac{1}{x}$.
At $x = \frac{1}{2n\pi}$: $f' = 2 + 2n\pi > 0$; at $x = \frac{1}{(2n + 1)\pi}$: $f' = 2 - (2n + 1)\pi < 0$. So $f'$ changes sign infinitely often near 0, and f is not monotonic on $(0, \infty)$.
For $x < 0$: $f(x) = -x\left(2 - \sin\frac{1}{x}\right)$, and $f(-x) = x\left(2 + \sin\frac{1}{x}\right)$ behaves the same way, so $f'$ again changes sign infinitely often: not monotonic on $(-\infty, 0)$.
Answer: (D). | ||||
| 70 | 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 $kr - 1 < [kr] \le kr$, so summing over $k = 1$ to n: $\frac{rn(n + 1)}{2} - n < \sum[kr] \le \frac{rn(n + 1)}{2}$.
Dividing by $n^2$ both bounds tend to $\frac{r}{2}$; by the sandwich theorem the limit is $\frac{r}{2}$.
Answer: (B). | ||||
| 71 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability We need the four consecutive digits 1, 0, 0, 1, the places being independent.
Starting at an odd place (odd, even, odd, even): $P(1) \cdot P(0) \cdot P(0) \cdot P(1) = \frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{2} = \frac{1}{18}$.
Starting at an even place (even, odd, even, odd): $\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{2}{3} = \frac{1}{18}$.
Either way the probability is $\frac{1}{18}$.
For one fixed starting place the probability is $\frac{1}{18}$ (D). Counting both starting parities as exclusive cases, as in a five-digit string where 1001 can sit at places 1–4 or 2–5 but not both, gives $\frac{1}{18} + \frac{1}{18} = \frac{1}{9}$ (A).
Answer: (A) or (D). | ||||
| 72 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations 14 points in all; subtract the collinear triples on each side:
$^{14}C_3 - {}^3C_3 - {}^5C_3 - {}^6C_3 = 364 - 1 - 10 - 20 = 333$.
Answer: (C). | ||||
| 73 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Tangent at (3, 4): $3x + 4y = 25$, so $P = \left(\frac{25}{3}, 0\right)$, $Q = \left(0, \frac{25}{4}\right)$.
OPQ is right-angled at O with legs $\frac{25}{3}$, $\frac{25}{4}$ and hypotenuse $\frac{125}{12}$. Inradius $= \frac{a + b - c}{2} = \frac{1}{2}\cdot\frac{100 + 75 - 125}{12} = \frac{25}{12}$, so the incentre is $\left(\frac{25}{12}, \frac{25}{12}\right)$.
$r^2 = 2\left(\frac{25}{12}\right)^2 = \frac{1250}{144} = \frac{625}{72}$.
Answer: (B). | ||||
| 74 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x + iy$: $S_1$ is the closed disc of centre (1, 0), radius $\sqrt{2}$; $(1 - i)(x + iy) = (x + y) + i(y - x)$, so $S_2$ is the half-plane $x + y \ge 1$; $S_3$ is $y \le 1$.
The centre (1, 0) lies in all three ($1 + 0 \ge 1$, $0 \le 1$), and so do nearby points such as (1.5, 0) and (2, 0): the intersection is a region with infinitely many points.
Answer: (D). | ||||
| 75 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $F'(x) = f(x)$, so the integral is $\int_0^1 2e^{-x}\sin x\cdot e^x dx = 2\int_0^1\sin x\,dx = 2(1 - \cos1)$.
$1 - \cos1 = \frac{1}{2!} - \frac{1}{4!} + \frac{1}{6!} - \frac{1}{8!} + \ldots$, an alternating series with decreasing terms, so
$2\left(\frac{1}{2} - \frac{1}{24} + \frac{1}{720} - \frac{1}{40320}\right) < 2(1 - \cos1) < 2\left(\frac{1}{2} - \frac{1}{24} + \frac{1}{720}\right) = \frac{331}{360}$.
The lower bound is $\frac{331}{360} - \frac{1}{20160} > \frac{330}{360}$. (Numerically $2(1 - \cos1) \approx 0.9194 = \frac{330.98}{360}$.)
Answer: (A). | ||||
| 76 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $\sum_{r=0}^{6}{}^6C_r\,{}^6C_{6-r}$ is the coefficient of $x^6$ in $(1 + x)^6(1 + x)^6 = (1 + x)^{12}$, i.e. $^{12}C_6 = 924$.
Answer: (A). | ||||
| 77 | 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 A point of the first line: $(-1 + 2t, 3 + t, -2 - t)$. The foot of the perpendicular from (2, 3, 1) satisfies $(2t - 3, t, -3 - t)\cdot(2, 1, -1) = 6t - 3 = 0$, so $t = \frac{1}{2}$, foot $(0, \frac{7}{2}, -\frac{5}{2})$, and the image is $(-2, 4, -6)$.
The second line passes through (2, 1, −1) with direction (3, −2, 1). The vector from (2, 1, −1) to (−2, 4, −6) is (−4, 3, −5).
Normal $= (3, -2, 1)\times(-4, 3, -5) = (10 - 3,\ -4 + 15,\ 9 - 8) = (7, 11, 1)$.
Plane: $7(x - 2) + 11(y - 1) + (z + 1) = 0 \Rightarrow 7x + 11y + z = 24$.
$\alpha + \beta + \gamma = 19$.
Answer: (C). | ||||
| 78 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Centre C(1, 2), radius $\sqrt{1 + 4 - 4} = 1$.
Let the angle between the tangents be $2\theta$, $\tan2\theta = \frac{12}{5}$: $\frac{2t}{1 - t^2} = \frac{12}{5} \Rightarrow 6t^2 + 5t - 6 = 0 \Rightarrow \tan\theta = \frac{2}{3}$; also $\sin2\theta = \frac{12}{13}$.
$PA = \frac{r}{\tan\theta} = \frac{3}{2}$. Area PAB $= \frac{1}{2}PA^2\sin2\theta = \frac{1}{2}\cdot\frac{9}{4}\cdot\frac{12}{13} = \frac{27}{26}$.
$\angle ACB = \pi - 2\theta$, so area CAB $= \frac{1}{2}r^2\sin2\theta = \frac{6}{13}$.
Ratio $= \frac{27/26}{6/13} = \frac{9}{4}$, i.e. 9 : 4.
Answer: (A). | ||||
| 79 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Solutions are the intersections of $y = \tan x$ with the line $y = \frac{1}{2}\left(\frac{\pi}{2} - x\right)$, which falls from $\frac{\pi}{4}$ at $x = 0$ to $-\frac{3\pi}{4}$ at $x = 2\pi$.
On $[0, \frac{\pi}{2})$: $\tan x$ rises from 0 (below $\frac{\pi}{4}$) to $+\infty$: one crossing.
On $(\frac{\pi}{2}, \frac{3\pi}{2})$: $\tan x$ rises from $-\infty$ to $+\infty$ while the line stays finite: one crossing.
On $(\frac{3\pi}{2}, 2\pi]$: $\tan x$ rises from $-\infty$ to 0, while the line is between $-\frac{\pi}{2}$ and $-\frac{3\pi}{4}$: one crossing.
Total 3.
Answer: (B). | ||||
| 80 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Differentiating $y^2 = 4x - 20$: $2yy' = 4$, so at (6, 2) the slope is 1 and L is $y = x - 4$.
$y = mx + c$ touches $\frac{x^2}{a^2} + \frac{y^2}{b} = 1$ when $c^2 = a^2m^2 + b$: $16 = 2(1) + b \Rightarrow b = 14$.
Answer: (B). | ||||
| 81 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On [1, e], $\log|x| = \ln x$. By parts: $I_n = \left[\frac{x^{20}}{20}(\ln x)^n\right]_1^e - \frac{n}{20}I_{n-1}$, so $20I_n = e^{20} - nI_{n-1}$.
$20I_{10} = e^{20} - 10I_9$ and $20I_9 = e^{20} - 9I_8$, so $e^{20} = 20I_9 + 9I_8$.
$20I_{10} = 20I_9 + 9I_8 - 10I_9 = 10I_9 + 9I_8$: $\alpha = 10$, $\beta = 9$, $\alpha - \beta = 1$.
Answer: 1. | ||||
| 82 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | NumericalHard | |
A hard numerical question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines O is the circumcentre, so $OA = OB = OC = R$ and (taking the angles as the directions of OA, OB, OC) $A = R(\cos\alpha, \sin\alpha)$, etc.
With the circumcentre at the origin the orthocentre is $A + B + C$ (the centroid is $\frac{A + B + C}{3}$ and lies one third of the way from O to H). On the y-axis: $\cos\alpha + \cos\beta + \cos\gamma = 0$.
Then $\cos^3\alpha + \cos^3\beta + \cos^3\gamma = 3\cos\alpha\cos\beta\cos\gamma$ (since $a + b + c = 0$).
$\sum\cos3\alpha = 4\sum\cos^3\alpha - 3\sum\cos\alpha = 12\cos\alpha\cos\beta\cos\gamma$.
The ratio is 12 and its square is 144.
Answer: 144. | ||||
| 83 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series AP: $2\log_{10}(4^x - 2) = 1 + \log_{10}\left(4^x + \frac{18}{5}\right) \Rightarrow (4^x - 2)^2 = 10\cdot4^x + 36$.
With $t = 4^x$: $t^2 - 14t - 32 = 0 \Rightarrow t = 16$ (t = −2 impossible), so $x = 2$ (and $4^x - 2 = 14 > 0$).
Determinant at $x = 2$: $\begin{vmatrix}3&1&4\\1&0&2\\2&1&0\end{vmatrix} = 3(0 - 2) - 1(0 - 4) + 4(1 - 0) = -6 + 4 + 4 = 2$.
Answer: 2. | ||||
| 84 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean of all $= \frac{2n(6) + n(3)}{3n} = 5$, so $\sum x^2 = 3n(4 + 25) = 87n$.
Let $S_1$, $S_2$ be the sums of squares of the two groups ($S_1 + S_2 = 87n$; the group sums are 12n and 3n).
New sum of squares $= (S_1 + 2\cdot12n + 2n) + (S_2 - 2\cdot3n + n) = 87n + 26n - 5n = 108n$.
New mean $= \frac{2n(7) + n(2)}{3n} = \frac{16}{3}$.
$k = \frac{108n}{3n} - \frac{256}{9} = 36 - \frac{256}{9} = \frac{68}{9}$, so $9k = 68$.
Answer: 68. | ||||
| 85 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f''(x) = 2a$ is constant, so $2a = \frac{1}{2}$, $a = \frac{1}{4}$.
$f'(-1) = -2a + b = 1 \Rightarrow b = \frac{3}{2}$; $f(-1) = a - b + c = 2 \Rightarrow c = \frac{13}{4}$.
$f'(x) = \frac{x}{2} + \frac{3}{2} > 0$ on [−1, 1], so the maximum is $f(1) = \frac{1}{4} + \frac{3}{2} + \frac{13}{4} = 5$.
Least $\alpha = 5$.
Answer: 5. | ||||
| 86 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = {}^nC_r\,a^r x^{n - 3r}$, so the coefficients of the 3rd, 4th, 5th terms are ${}^nC_2a^2$, ${}^nC_3a^3$, ${}^nC_4a^4$.
$\frac{{}^nC_2}{{}^nC_3\,a} = \frac{3}{a(n - 2)} = \frac{12}{8} \Rightarrow a(n - 2) = 2$; $\frac{{}^nC_3}{{}^nC_4\,a} = \frac{4}{a(n - 3)} = \frac{8}{3} \Rightarrow a(n - 3) = \frac{3}{2}$.
Dividing: $\frac{n - 2}{n - 3} = \frac{4}{3} \Rightarrow n = 6$, $a = \frac{1}{2}$.
Term independent of x: $n - 3r = 0 \Rightarrow r = 2$: ${}^6C_2\left(\frac{1}{2}\right)^2 = \frac{15}{4} = 3.75$.
Answer: 3.75. | ||||
| 87 | 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 $x^2 \le x + 6$ for $-2 \le x \le 3$, so on [−3, −2] $f = x + 6$ and on [−2, 0] $f = x^2$; on [0, 1], $\sqrt{x} \ge x^2$, so $f = \sqrt{x}$. f is non-negative throughout.
$A = \int_{-3}^{-2}(x + 6)dx + \int_{-2}^{0}x^2dx + \int_0^1\sqrt{x}\,dx = \frac{7}{2} + \frac{8}{3} + \frac{2}{3} = \frac{41}{6}$.
$6A = 41$.
Answer: 41. | ||||
| 88 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $\frac{(x + y + z)^2}{3} + \frac{(lx - nz)^2}{l^2 + n^2} + \frac{(x - 2y + z)^2}{6} = 9$ must be $x^2 + y^2 + z^2 = 9$ identically.
$\frac{(x + y + z)^2}{3} + \frac{(x - 2y + z)^2}{6} = \frac{3x^2 + 6y^2 + 3z^2 + 6zx}{6} = \frac{x^2}{2} + y^2 + \frac{z^2}{2} + zx$.
So $\frac{(lx - nz)^2}{l^2 + n^2}$ must equal $\frac{x^2 + z^2}{2} - zx = \frac{(x - z)^2}{2}$, which needs $l = n$.
$l - n = 0$.
Answer: 0. | ||||
| 89 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{x} = \lambda\vec{a} + \mu\vec{b} = (2\lambda + \mu, -\lambda + 2\mu, \lambda - \mu)$.
Perpendicular to (3, 2, −1): $3(2\lambda + \mu) + 2(-\lambda + 2\mu) - (\lambda - \mu) = 3\lambda + 8\mu = 0$.
Projection on $\vec{a}$: $\frac{\vec{x}\cdot\vec{a}}{|\vec{a}|} = \frac{17\sqrt{6}}{2}$ with $|\vec{a}| = \sqrt{6}$, so $\vec{x}\cdot\vec{a} = 51$. Also $\vec{x}\cdot\vec{a} = 6\lambda + \mu(\vec{a}\cdot\vec{b}) = 6\lambda - \mu$.
$\lambda = -\frac{8\mu}{3} \Rightarrow -16\mu - \mu = 51 \Rightarrow \mu = -3$, $\lambda = 8$.
$\vec{x} = (13, -14, 11)$, $|\vec{x}|^2 = 169 + 196 + 121 = 486$.
Answer: 486. | ||||
| 90 | 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 $(A - I)B = 0$ with $B \ne 0$, so $A - I$ is singular: $\begin{vmatrix}a - 1&b\\c&d - 1\end{vmatrix} = 0$.
$(a - 1)(d - 1) - bc = 0 \Rightarrow ad - bc = a + d - 1 = 2020$.
Answer: 2020. | ||||
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