JEE (Main) · 2021
JEE (Main) 2021 — 16 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
- Optics2
- Electromagnetic Induction and Alternating Currents2
- Oscillations and Waves2
18 easy · 12 medium · 0 hard
Chemistry
30 Q
- Amines3
- Hydrocarbons3
- p-Block Elements2
17 easy · 13 medium · 0 hard
Mathematics
30 Q
- Three Dimensional Geometry3
- Limits, Continuity and Differentiability3
- Integral Calculus3
8 easy · 18 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 SystemsLine-of-sight communication | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on line-of-sight communication, from Communication Systems. NCERT: No matching NCERT chapter For line-of-sight communication the maximum separation is $d_m = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
The towers are identical, so $h_T = h_R = h$ and $45 = 2\sqrt{2Rh}$, i.e. $\sqrt{2Rh} = 22.5$ km.
$h = \frac{(22.5)^2}{2 \times 6400}$ km $= \frac{506.25}{12800}$ km $\approx 0.03955$ km $= 39.55$ m.
Answer: (B) 39.55 m. | ||||
| 2 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter For a charge $q$ accelerated through $V$: $\lambda = \frac{h}{\sqrt{2mqV}}$. The electron and proton have the same $|q|$ and the same $V$, so
$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{\frac{1.00727}{0.00055}} = \sqrt{1831.4} \approx 42.8 \approx 43$.
Answer: (A) $43 : 1$. | ||||
| 3 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens-maker's formula in a medium: $\frac{1}{f} = \left(\frac{\mu_{lens}}{\mu_{medium}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
With $\mu_{lens} = \mu_{medium} = 1.4$, the first factor is zero, so $\frac{1}{f} = 0$ and $f$ is infinite: the lens does not deviate light at all.
Answer: (C) Infinite. | ||||
| 4 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Colour is fixed by frequency: red light has a lower frequency than blue. In the same medium both travel at the same speed, so $\lambda = \frac{v}{\nu}$ differs too (red has the longer wavelength).
Answer: (B) Different frequencies and different wavelengths. | ||||
| 5 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Let the left edge of the loop be at $x$; the right edge is at $x + d$. The edges parallel to the $x$-axis move along their own length and give no motional emf.
The two vertical edges (length $d$) each give $Bvd$, in opposition:
$\varepsilon = \left[B(x+d) - B(x)\right] v_o d = \frac{B_o}{a}\left[(x+d) - x\right] v_o d = \frac{B_o v_o d^2}{a}$.
(Equivalently, $\Phi = \int_x^{x+d} \frac{B_o x'}{a} d\,dx'$ and $\frac{d\Phi}{dt} = \frac{B_o d}{a}\,d\,v_o$.)
Answer: (B) $\frac{B_o v_o d^2}{a}$. | ||||
| 6 | 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 $A = A_0 e^{-bt/2m}$.
$\frac{b}{2m} = \frac{20 \text{ g/s}}{2 \times 500 \text{ g}} = 0.02 \text{ s}^{-1}$.
$\frac{A_0}{2} = A_0 e^{-0.02t} \Rightarrow t = \frac{\ln 2}{0.02} = \frac{0.693}{0.02} = 34.65$ s.
Answer: (A) 34.65 s. | ||||
| 7 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $\lambda = \frac{kT}{\sqrt{2}\,\pi d^2 P}$, with $T = 300$ K, $d = 0.3 \times 10^{-9}$ m, $P = 1.01 \times 10^5$ Pa.
Numerator: $1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21}$.
Denominator: $1.414 \times 3.1416 \times 9 \times 10^{-20} \times 1.01 \times 10^5 \approx 4.04 \times 10^{-14}$.
$\lambda \approx 1.02 \times 10^{-7}$ m $= 102$ nm.
Answer: (D) 102 nm. | ||||
| 8 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on 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 Reynolds number $R_e = \frac{\rho v D}{\eta}$ with $v = \frac{Q}{\pi r^2}$ and $D = 2r$, so $R_e = \frac{2\rho Q}{\pi r \eta}$.
$Q_1 = 0.18$ L/min $= 3 \times 10^{-6}$ m$^3$/s: $R_e = \frac{2 \times 10^3 \times 3 \times 10^{-6}}{\pi \times 0.005 \times 10^{-3}} \approx 382$.
$Q_2 = 0.48$ L/min $= 8 \times 10^{-6}$ m$^3$/s: $R_e \approx 1019$.
Flow is steady (streamline) for $R_e < 1000$ and unsteady between 1000 and 2000 (turbulent above 2000). The flow goes from steady ($R_e \approx 382$) to unsteady ($R_e \approx 1019$).
Answer: (A) Steady flow to unsteady flow. | ||||
| 9 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The magnetic force $\vec{F} = Q(\vec{v} \times \vec{B})$ is always perpendicular to $\vec{v}$, hence to the displacement $\vec{dl}$. So $dW = \vec{F}\cdot\vec{dl} = 0$: a magnetic field does no work.
Answer: (B) Zero. | ||||
| 10 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei After 33% decay, 67% remains; after 67% decay, 33% remains. The amount falls from $0.67N_0$ to $0.33N_0$, very nearly a halving.
$t = \frac{T_{1/2}}{\ln 2}\ln\frac{0.67}{0.33} = \frac{20}{0.693} \times 0.708 \approx 20.4$ min $\approx 20$ minutes (one half-life).
Answer: (A) 20 minutes. | ||||
| 11 | Electromagnetic Induction and Alternating CurrentsTransformers | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on transformers, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The load draws 60 W at $i_L = 0.11$ A, so $V_2 = \frac{P}{i_L} = \frac{60}{0.11} \approx 545$ V.
The primary is at 220 V, so $V_2 > V_1$: the transformer steps the voltage up.
Answer: (A) Step-up transformer. | ||||
| 12 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\lambda = \frac{0.693}{2.7 \times 86400 \text{ s}} \approx 2.97 \times 10^{-6}$ s$^{-1}$.
$N = \frac{1.50 \times 10^{-3}}{198} \times 6 \times 10^{23} \approx 4.55 \times 10^{18}$.
$A = \lambda N \approx 1.35 \times 10^{13}$ Bq $= \frac{1.35 \times 10^{13}}{3.7 \times 10^{10}}$ Ci $\approx 365$ Ci.
Of the options only 357 Ci is of this size (the others are 30% or more away); the small difference comes from rounding of the constants.
The printed data give about 365 Ci, which no option equals (the nearest, 357 Ci, is 2% away).
No option is correct as printed. | ||||
| 13 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter In a cold bath both metals contract, but A (higher expansion coefficient) contracts more. A, on the left, becomes the shorter strip, so it lies on the inside (concave side) of the bend: the strip bends towards A, i.e. to the left.
Answer: (B) Bend towards the left. | ||||
| 14 | 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 After the collision, energy is conserved during the swing: $V = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.098} = \sqrt{1.9208} \approx 1.386$ m/s.
Momentum is conserved in the collision: $mv = (M+m)V$, so $v = \frac{6.00}{0.010} \times 1.386 \approx 831.6$ m/s.
This is option (C), 831.4 m/s (the difference is rounding).
Answer: (C) 831.4 m/s. | ||||
| 15 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | StatementsMedium | |
A medium statements question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion Statement I: $v = 7$ km/h $= 1.944$ m/s. Required $\frac{v^2}{r} = \frac{3.78}{2} = 1.89$ m/s$^2$; available $\mu g = 0.2 \times 9.8 = 1.96$ m/s$^2$. Since $1.89 < 1.96$ the cyclist does not slip: true.
Statement II: on a road banked at $45°$ with the same $\mu = 0.2$, $v_{max}^2 = rg\frac{\tan\theta + \mu}{1 - \mu\tan\theta} = 2 \times 9.8 \times \frac{1.2}{0.8} = 29.4$, so $v_{max} \approx 5.42$ m/s $\approx 19.5$ km/h; and $v_{min}^2 = rg\frac{1 - \mu}{1 + \mu} \approx 13.1$, $v_{min} \approx 3.6$ m/s $\approx 13$ km/h. 18.5 km/h ($\approx 5.14$ m/s) lies between them: true.
Answer: (A) Both statements are true. | ||||
| 16 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The acceleration is $\vec{a} = \frac{d\vec{v}}{dt} = t\hat{i} + 3\hat{j}$; at $t = 2$ s, $\vec{a} = 2\hat{i} + 3\hat{j}$, which lies in the $xy$-plane at an angle $\tan^{-1}\left(\frac{2}{3}\right)$ from the $y$-axis.
(The velocity at $t = 2$ s is $2\hat{i} + 6\hat{j} + 9\hat{k}$, whose direction from the $x$-axis is $\cos^{-1}\frac{2}{11}$ and from the $y$-axis $\cos^{-1}\frac{6}{11}$; neither appears among the options.)
Answer: (D) $\tan^{-1}\left(\frac{2}{3}\right)$ from the $y$-axis. | ||||
| 17 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $Y = \frac{MgL}{\pi r^2 \Delta l}$, so $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta L}{L} + 2\frac{\Delta r}{r} + \frac{\Delta(\Delta l)}{\Delta l}$.
$= \frac{1}{1000} + \frac{1}{1000} + 2 \times \frac{0.001}{0.2} + \frac{0.001}{0.5} = 0.001 + 0.001 + 0.010 + 0.002 = 0.014$, i.e. 1.4 %.
Answer: (A) 1.4 %. | ||||
| 18 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Joule heating: $H = I^2 R t$, so at fixed $R$ and $t$, $H \propto I^2$.
Doubling the current multiplies the heat by 4: $H = 4 \times 500 = 2000$ J.
Answer: (D) 2000 J. | ||||
| 19 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The field of an infinite line charge at distance $r$ is $E = \frac{\lambda}{2\pi\varepsilon_0 r}$. The point $(3, 0, 0)$ is at $r = 3$ m.
The 'surface charge density' asked for is the flux density $D = \varepsilon_0 E = \frac{\lambda}{2\pi r} = \frac{8}{2\pi \times 3}$ nC m$^{-2}$ $\approx 0.424$ nC m$^{-2}$.
(For reference, $E = \frac{2 \times 9 \times 10^9 \times 8 \times 10^{-9}}{3} = 48$ N/C, close to option (A)'s number, but (A) has the wrong unit.)
Answer: (C) 0.424 nC m$^{-2}$. | ||||
| 20 | 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 A NAND gate with both inputs joined is a NOT gate. The first two gates give $\bar{A}$ and $\bar{B}$.
The middle NAND gives $\overline{\bar{A}\cdot\bar{B}} = A + B$ (De Morgan).
The last gate (inputs joined) inverts it: $Y = \overline{A + B}$, a NOR gate.
Answer: (D) NOR Gate. | ||||
| 21 | GravitationGravitational potential energy | Class 11: Gravitation | NumericalMedium | |
A medium numerical question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation Build the earth (taken as a uniform sphere of density $\rho$) shell by shell. When the radius is $r$, the mass is $m = \frac{4}{3}\pi r^3 \rho$, and bringing a shell $dm = 4\pi r^2 \rho\,dr$ from infinity releases $\frac{Gm\,dm}{r}$.
Self-energy: $U = -\int_0^R \frac{G}{r}\cdot\frac{4}{3}\pi r^3\rho \cdot 4\pi r^2 \rho\,dr = -\frac{16\pi^2 G\rho^2}{3}\cdot\frac{R^5}{5} = -\frac{3}{5}\frac{GM^2}{R}$.
The energy to be supplied is $\frac{3}{5}\frac{GM^2}{R}$, so $x = 3$.
Answer: 3. | ||||
| 22 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane To land directly opposite, the swimmer's velocity relative to water must cancel the drift: he heads upstream at angle $\alpha$ from the perpendicular with $12\sin\alpha = 6$, so $\alpha = 30°$.
Measured from the direction of flow, his heading is $90° + 30° = 120°$.
Answer: 120. | ||||
| 23 | Laws of MotionNewton's laws | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $\vec{a} = \frac{\vec{F}}{m} = \hat{i} + 1.5\hat{j} + 2.5\hat{k}$ m/s$^2$. From rest at the origin, $\vec{r} = \frac{1}{2}\vec{a}t^2 = 8\vec{a}$ at $t = 4$ s.
$\vec{r} = 8\hat{i} + 12\hat{j} + 20\hat{k}$, consistent with $x = 8$, $z = 20$; so $b = 12$.
Answer: 12. | ||||
| 24 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The force acts at $(2, 0, 0)$. Relative to $(2, 3, 4)$: $\vec{r} = (2-2)\hat{i} + (0-3)\hat{j} + (0-4)\hat{k} = -3\hat{j} - 4\hat{k}$.
$\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&-3&-4\\4&3&4\end{vmatrix} = \hat{i}(-12+12) - \hat{j}(0+16) + \hat{k}(0+12) = -16\hat{j} + 12\hat{k}$.
$|\vec{\tau}| = \sqrt{256 + 144} = 20$.
Answer: 20. | ||||
| 25 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical 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 rolling without slipping, $a = \frac{g\sin\theta}{1 + \frac{I}{ma^2}}$. For a solid disc $I = \frac{1}{2}ma^2$, so $a = \frac{g\sin\theta}{3/2} = \frac{2}{3}g\sin\theta$.
Thus $b = 3$.
Answer: 3. | ||||
| 26 | ThermodynamicsSecond law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Carnot efficiency: $\eta = 1 - \frac{T_2}{T_1}$ with $T_1 = 127 + 273 = 400$ K.
$0.6 = 1 - \frac{T_2}{400} \Rightarrow T_2 = 160$ K $= 160 - 273 = -113$ °C.
Answer: $-113$. | ||||
| 27 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance A slab of thickness $t$ filling the full area: $C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.
$C = \frac{2\varepsilon_0}{1 - 0.5 + \frac{0.5}{3.2}} = \frac{2\varepsilon_0}{0.5 + 0.15625} = \frac{2\varepsilon_0}{0.65625} \approx 3.05\,\varepsilon_0$.
To the nearest integer, 3.
Answer: 3. | ||||
| 28 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity $H = I^2 R t \Rightarrow R = \frac{H}{I^2 t} = \frac{10 \times 10^{-3}}{(2 \times 10^{-3})^2 \times 1} = \frac{10^{-2}}{4 \times 10^{-6}} = 2500\ \Omega$.
Answer: 2500. | ||||
| 29 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For thin prisms, deviation $\delta = (\mu - 1)A$ and angular dispersion $\omega\delta$.
Achromatism (no net dispersion): $\omega_1\delta_1 = \omega_2\delta_2 \Rightarrow 0.02\,\delta_1 = 0.03\,\delta_2 \Rightarrow \delta_2 = \frac{2}{3}\delta_1$.
Net deviation: $\delta_1 - \delta_2 = 2° \Rightarrow \frac{1}{3}\delta_1 = 2° \Rightarrow \delta_1 = 6°$.
Crown prism: $A_1 = \frac{\delta_1}{\mu_1 - 1} = \frac{6°}{0.5} = 12°$.
Answer: 12. | ||||
| 30 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Speed of sound $v = \sqrt{\frac{B}{\rho}}$; equal compressibility means equal $B$, so $\frac{v_2}{v_1} = \sqrt{\frac{\rho_1}{\rho_2}}$.
Closed pipe, first overtone: $f = \frac{3v_1}{4L}$. Open pipe of length $L'$, first overtone: $f = \frac{2v_2}{2L'} = \frac{v_2}{L'}$.
Equal: $L' = \frac{4L}{3}\cdot\frac{v_2}{v_1} = \frac{4}{3}L\sqrt{\frac{\rho_1}{\rho_2}}$, so $x = 4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Buckminsterfullerene $C_{60}$ is a soccer-ball cage of 20 six-membered rings and 12 five-membered rings.
(A) 12 six-membered and 24 five-membered rings: incorrect.
(B) Each six-membered ring is fused to three six- and three five-membered rings: correct.
(C) Each five-membered ring is fused only to six-membered rings: correct.
(D) Each carbon is $sp^2$-like and forms three $\sigma$ bonds: correct.
Answer: (A). | ||||
| 32 | 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) Colloids do show colligative properties (small in value, because the particles are few): correct.
(B) Colloidal particles show Brownian motion: correct.
(C) By the Hardy–Schulze rule, the higher the charge of the coagulating ion, the greater its flocculating power: Al$^{3+}$ > Na$^+$, correct.
(D) Colloidal particles (1–1000 nm) pass through ordinary filter paper; only ultrafilters stop them: incorrect.
Answer: (D). | ||||
| 33 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties $Z = 33$ is arsenic (As), a metalloid; $Z = 53$ is iodine (I), a non-metal; $Z = 83$ is bismuth (Bi), a metal.
Answer: (C). | ||||
| 34 | 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 In the Ellingham diagram the $Al_2O_3$ line lies far below the C/CO line at any practical temperature, so carbon cannot reduce alumina; aluminium is obtained by electrolysis of alumina in molten cryolite (Hall–Héroult).
$Fe_2O_3$ (blast furnace), $ZnO$ (heated with coke at about 1673 K) and $Cu_2O$ are all reduced by coke.
Answer: (D) $Al_2O_3 \rightarrow Al$. | ||||
| 35 | HydrogenHydrogen and its compounds | No matching NCERT chapter | StatementsEasy | |
A easy statements question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter (A) $H_2O_2$ is used in treating domestic and industrial effluents: true.
(B) Oxygen is in the $-1$ state, so it acts both as an oxidant and as a reductant: true.
(C) $H_2O_2$ has an open-book (non-planar) structure; the two O–H bonds lie in different planes (dihedral angle about 111° in the gas phase): false.
(D) $H_2O_2$ is miscible with water in all proportions: true.
Answer: (A) (A), (B) and (D) only. | ||||
| 36 | 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 X: a low first IE (495) and a huge jump to the second (4563) means one valence electron, after which a noble-gas core is broken: an alkali metal, Na.
Y: first and second IEs both moderate (731, 1450), the jump coming only after two electrons: an alkaline earth metal, Mg.
(F would have a very high first IE, about 1680 kJ/mol.)
Answer: (A) X = Na; Y = Mg. | ||||
| 37 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $H_3PO_3$ has two P–OH groups (one H is on P): dibasic. 50 mL × 1 M = 50 mmol needs 100 mmol NaOH = 100 mL of 1 M NaOH.
$H_3PO_2$ has one P–OH group (two H on P): monobasic. 100 mL × 2 M = 200 mmol needs 200 mmol NaOH = 200 mL.
Answer: (C) 100 mL and 200 mL. | ||||
| 38 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements (a) Ce is +4 in $(NH_4)_2[Ce(NO_3)_6]$: $Ce^{4+}$ is $4f^0$, 0 unpaired electrons, $\mu = 0$.
(b) $Gd^{3+}$ is $4f^7$: 7 unpaired, $\mu = \sqrt{7 \times 9} = \sqrt{63} \approx 7.94$ BM.
(c) $Eu^{3+}$ is $4f^6$: 6 unpaired, $\mu = \sqrt{6 \times 8} = \sqrt{48} \approx 6.93$ BM.
Order: (a) < (c) < (b).
Answer: (B). | ||||
| 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 All four iron(II) halides $FeF_2$, $FeCl_2$, $FeBr_2$, $FeI_2$ are known.
Among iron(III) halides, $FeF_3$, $FeCl_3$ and $FeBr_3$ are known, but $FeI_3$ is not a stable compound: $Fe^{3+}$ oxidises $I^-$ ($2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$).
So $x$ = F, Cl, Br, I and $y$ = F, Cl, Br.
Answer: (B). | ||||
| 40 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | StatementsEasy | |
A easy statements question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Greenhouse gases absorb the infrared radiation emitted by the earth: carbon dioxide, methane, water vapour, nitrous oxide, CFCs and ozone.
Oxygen ($O_2$), a homonuclear diatomic, does not absorb in the infrared and is not a greenhouse gas.
So (A), (C) and (D).
Answer: (C). | ||||
| 41 | Practical Organic ChemistryDetection of elements | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on detection of elements, from Practical Organic Chemistry. NCERT: No matching NCERT chapter (a) Lassaigne's (sodium fusion) test detects nitrogen, sulphur, phosphorus and halogens: (iii).
(b) Heating with copper(II) oxide oxidises carbon to $CO_2$ (lime water turns milky): carbon, (i).
(c) Silver nitrate (after boiling with $HNO_3$) gives silver halide precipitates: halogen, (iv).
(d) A black precipitate with acetic acid and lead acetate is $PbS$: sulphur, (ii).
Answer: (B) (a)→(iii), (b)→(i), (c)→(iv), (d)→(ii). | ||||
| 42 | AminesBasicity of amines | Class 12: Amines | StatementsEasy | |
A easy statements question on basicity of amines, from Amines. NCERT: Class 12: Amines Statement I: in NaH hydrogen is present as $H^-$, which readily gives up electrons ($H^- \rightarrow \frac{1}{2}H_2 + e^-$); saline hydrides are strong reducing agents, not oxidising agents. False.
Statement II: the nitrogen lone pair of pyridine is in an $sp^2$ orbital, not part of the aromatic sextet, and is available to accept a proton. True.
Answer: (D). | ||||
| 43 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons A silver mirror with Tollens' reagent means A is a reducing compound: an aldehyde or formic acid.
(A) But-2-yne gives butane-2,3-dione / acetic acid: no silver mirror.
(B) 2,3-Dimethylbut-2-ene, $(CH_3)_2C=C(CH_3)_2$, gives two molecules of acetone: a ketone, no mirror.
(C) But-1-yne, $HC\equiv C-CH_2CH_3$, has a terminal $\equiv CH$; on ozonolysis it gives propanoic acid and formic acid (HCOOH), and formic acid reduces Tollens' reagent to a silver mirror.
(D) Isopropylidenecyclopropane gives acetone and cyclopropanone: ketones only, no mirror.
Answer: (C). | ||||
| 44 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The double bond survives and Cl appears on both allylic carbons (C-3 and C-6) of cyclohexene: substitution, not addition.
Allylic substitution goes by a free-radical chain (the allylic radical is resonance-stabilised), which needs $Cl_2$ with UV light (or high temperature) and a low $Cl_2$ concentration.
$Cl_2$ in the dark with $AlCl_3$ would add across the double bond; HCl (with or without a catalyst) would add HCl.
Answer: (C) A = $Cl_2$; UV light. | ||||
| 45 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The ring $-CH_3$ is oxidised to $-COOH$ while the methoxy group stays. Side-chain oxidation of an alkylbenzene is done with alkaline $KMnO_4$ followed by acidification ($H^+$) to free the acid from its potassium salt.
$LiAlH_4$ and $NaBH_4$ are reducing agents, and Zn–Hg/HCl (Clemmensen) reduces C=O to $CH_2$.
Answer: (B) Alkaline $KMnO_4$, $H^+$. | ||||
| 46 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A Grignard reagent adds once to the C≡N of a nitrile, giving the magnesium salt of an imine, $Ar-C(=NMgBr)-C_6H_5$. Acid hydrolysis converts the imine to a ketone: $Ar-CO-C_6H_5$.
The ether group, $-CH(CH_3)OCH_3$, has no acidic H and is untouched.
So X is the ortho-substituted benzophenone with the $CH(CH_3)OCH_3$ group intact, option (A). (B) and (C) are amines from reduction, not from a Grignard reaction; (D) wrongly replaces the methoxy group by phenyl.
Answer: (A). | ||||
| 47 | AminesBasicity of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines An acyl group withdraws the nitrogen lone pair by resonance into C=O. Diacetamide, $(CH_3CO)_2NH$, has two acyl groups delocalising the one lone pair, so it is the least basic (it is in fact weakly acidic).
$(CH_3CO)NHC_2H_5$ (one acyl group) is next; the alkylamines $(C_2H_5)_2NH$ and $(C_2H_5)_3N$ are strongly basic.
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 $R-X + NH_3 \rightarrow R-NH_3^+X^-$: ammonolysis gives the amine as its ammonium salt (with HX). A strong base liberates the free amine: $R-NH_3^+X^- + NaOH \rightarrow R-NH_2 + H_2O + NaX$.
Of the options, this is removal of the acidic species (HX, the ammonium salt's proton).
Answer: (D) to remove acidic impurities. | ||||
| 49 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Phenol–formaldehyde resins (novolac and its cross-linked form, bakelite) are the thermosetting adhesive and binder used to bond wood veneers into laminated sheets (plywood and decorative laminates), as well as in combs, switches and handles.
cis-Polyisoprene is natural rubber, not a laminating resin.
Answer: (D) Phenol and formaldehyde resin. | ||||
| 50 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The $\alpha$-helix and $\beta$-pleated sheet (secondary structure) are held by hydrogen bonds between the $C=O$ and $N-H$ groups of the peptide backbone. Peptide bonds make the primary structure; glycosidic bonds belong to carbohydrates.
Answer: (C) Hydrogen bonding. | ||||
| 51 | 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 $Pb^{2+}$: $35 \times 0.15 = 5.25$ mmol.
Chromic sulphate $Cr_2(SO_4)_3$: $20 \times 0.12 = 2.4$ mmol, giving $3 \times 2.4 = 7.2$ mmol $SO_4^{2-}$.
$Pb^{2+} + SO_4^{2-} \rightarrow PbSO_4$: $Pb^{2+}$ is limiting, so $5.25$ mmol $= 5.25 \times 10^{-3} = 525 \times 10^{-5}$ mol $PbSO_4$.
Answer: 525. | ||||
| 52 | Solid StateClose packed structures | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on close packed structures, from Solid State. NCERT: No matching NCERT chapter Atoms: $\frac{0.581}{70} \times 6.023 \times 10^{23} = 0.0083 \times 6.023 \times 10^{23} \approx 5.0 \times 10^{21}$.
In a close-packed structure there is 1 octahedral and 2 tetrahedral voids per atom, i.e. 3 voids per atom.
Voids $= 3 \times 5.0 \times 10^{21} = 15 \times 10^{21}$.
Answer: 15. | ||||
| 53 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom For $n = 5$, $l = 0, 1, 2, 3, 4$. $m_l = +2$ needs $l \geq 2$: $l = 2, 3, 4$ (5d, 5f, 5g), one orbital each.
Answer: 3. | ||||
| 54 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $Fe + 2HCl \rightarrow FeCl_2 + H_2$: moles of $H_2$ = moles of Fe $= \frac{50}{55.85} = 0.8953$.
Work done by the gas against constant pressure: $w = P\Delta V = n_{H_2}RT$ (the volume of the solids and liquid is negligible).
$w = 0.8953 \times 8.314 \times 298 \approx 2218$ J.
Answer: 2218. | ||||
| 55 | SolutionsRaoult's law | Class 12: Solutions | NumericalEasy | |
A easy numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions $x_A = \frac{1}{3}$, $x_B = \frac{2}{3}$.
Raoult's law: $P = x_A P_A^\circ + x_B P_B^\circ = \frac{1}{3}(21) + \frac{2}{3}(18) = 7 + 12 = 19$ kPa.
Answer: 19. | ||||
| 56 | EquilibriumIonic equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $Ka_2$ is negligible; the $H^+$ comes from the first step. $Ka_1$ is large, so solve the quadratic:
$\frac{x^2}{0.588 - x} = 0.017 \Rightarrow x^2 + 0.017x - 0.009996 = 0 \Rightarrow x = \frac{-0.017 + \sqrt{0.000289 + 0.039984}}{2} \approx 0.0918$ M.
pH $= -\log 0.0918 \approx 1.04$, which rounds to 1.
(The approximation $x = \sqrt{0.017 \times 0.588} = 0.1$ gives pH 1.0 as well.)
Answer: 1. | ||||
| 57 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Conductivity $\kappa = G \times G^* = 0.55 \text{ mS} \times 1.3 \text{ cm}^{-1} = 0.715$ mS cm$^{-1}$ $= 71.5$ mS m$^{-1}$.
Concentration $c = 5.0$ mmol dm$^{-3}$ $= 5.0$ mol m$^{-3}$.
$\Lambda_m = \frac{\kappa}{c} = \frac{71.5 \text{ mS m}^{-1}}{5.0 \text{ mol m}^{-3}} = 14.3$ mS m$^2$ mol$^{-1}$, i.e. 14.
Answer: 14. | ||||
| 58 | 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 $[A] = [A]_0 e^{-k_A t}$, $[B] = [B]_0 e^{-k_B t}$ with $[A]_0 = [B]_0$, so $\frac{[A]}{[B]} = e^{(k_B - k_A)t} = 16 = 2^4$.
$(k_B - k_A)t = 4\ln 2$, with $k = \frac{\ln 2}{t_{1/2}}$: $\left(\frac{1}{18} - \frac{1}{54}\right)t = 4 \Rightarrow \frac{2}{54}t = 4 \Rightarrow t = 108$ min.
Answer: 108. | ||||
| 59 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $Ti^{3+}$ is $d^1$: the single d–d transition $t_{2g} \rightarrow e_g$ has energy $\Delta_o$.
$\Delta_o = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{498 \times 10^{-9}} \approx 3.99 \times 10^{-19}$ J, i.e. 4.
Answer: 4. | ||||
| 60 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Pressure of dry $N_2$: $758 - 14 = 744$ mm Hg.
Volume at STP: $V_0 = 30 \times \frac{744}{760} \times \frac{273}{287} \approx 27.94$ mL.
Mass of $N_2$: $\frac{28}{22400} \times 27.94 \approx 0.03492$ g.
% N $= \frac{0.03492}{0.1840} \times 100 \approx 18.98$, i.e. 19.
Answer: 19. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry (3, 5, 7) lies on $L_1$: $\frac{3-a}{l} = \frac{5-2}{3} = \frac{7-b}{4} = 1$, so $b = 3$ and $a = 3 - l$.
The perpendicular $(4-3, 3-5, 8-7) = (1, -2, 1)$ is normal to $(l, 3, 4)$: $l - 6 + 4 = 0 \Rightarrow l = 2$, $a = 1$.
$L_1$: through $(1, 2, 3)$ along $(2, 3, 4)$; $L_2$: through $(2, 4, 5)$ along $(3, 4, 5)$.
$(2,3,4) \times (3,4,5) = (15-16,\ 12-10,\ 8-9) = (-1, 2, -1)$, of length $\sqrt{6}$.
Joining vector $(1, 2, 2)$; $(1,2,2)\cdot(-1,2,-1) = -1 + 4 - 2 = 1$.
Shortest distance $= \frac{1}{\sqrt{6}}$.
Answer: (A). | ||||
| 62 | 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 For $x^2 + y^2 + 2gx + 2fy + c = 0$ the intercepts are $2\sqrt{g^2 - c}$ and $2\sqrt{f^2 - c}$. Here $g = \frac{a}{2}$, $f = a$.
$\frac{a^2}{4} - c = 2$ and $a^2 - c = 5$, so $\frac{3a^2}{4} = 3$, $a^2 = 4$, $a = -2$ and $c = -1$.
Circle: $x^2 + y^2 - 2x - 4y - 1 = 0$, centre $(1, 2)$, radius $\sqrt{1 + 4 + 1} = \sqrt{6}$.
A line perpendicular to $x + 2y = 0$: $2x - y + k = 0$. Tangency: $\frac{|2 - 2 + k|}{\sqrt{5}} = \sqrt{6} \Rightarrow |k| = \sqrt{30}$.
Distance from the origin: $\frac{|k|}{\sqrt{5}} = \frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}$ (the same for both tangents).
Answer: (D) $\sqrt{6}$. | ||||
| 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 $\vec{r} \times \vec{a} = \vec{b} \times \vec{r} = -\vec{r} \times \vec{b} \Rightarrow \vec{r} \times (\vec{a} + \vec{b}) = \vec{0}$, so $\vec{r} = t(\vec{a} + \vec{b}) = t(3\hat{i} - \hat{j} + 2\hat{k})$.
$\vec{r}\cdot(\alpha, 2, 1) = t(3\alpha - 2 + 2) = 3\alpha t = 3 \Rightarrow \alpha t = 1$.
$\vec{r}\cdot(2, 5, -\alpha) = t(6 - 5 - 2\alpha) = t - 2\alpha t = t - 2 = -1 \Rightarrow t = 1$, $\alpha = 1$.
$|\vec{r}|^2 = 9 + 1 + 4 = 14$, so $\alpha + |\vec{r}|^2 = 15$.
Answer: (D) 15. | ||||
| 64 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f(x) = 3\ln|x-1| - 3\ln|x+1| - \frac{2}{x-1}$.
$f'(x) = \frac{3}{x-1} - \frac{3}{x+1} + \frac{2}{(x-1)^2} = \frac{6}{(x-1)(x+1)} + \frac{2}{(x-1)^2} = \frac{6(x-1) + 2(x+1)}{(x-1)^2(x+1)} = \frac{4(2x-1)}{(x-1)^2(x+1)}$.
$(x-1)^2 > 0$, so $f'(x) \geq 0$ when $\frac{2x-1}{x+1} \geq 0$: $x < -1$ or $x \geq \frac{1}{2}$ (with $x \neq 1$). $f'$ vanishes only at the single point $\frac{1}{2}$, so $f$ is increasing on $(-\infty, -1)$ and on $\left[\frac{1}{2}, \infty\right) - \{1\}$.
Answer: (B). | ||||
| 65 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections On $y^2 = 3x^2$ the circle gives $4x^2 = 4b$, so $x^2 = b$ and $y^2 = 3b$.
In the ellipse: $\frac{b}{16} + \frac{3b}{b^2} = 1 \Rightarrow \frac{b}{16} + \frac{3}{b} = 1 \Rightarrow b^2 - 16b + 48 = 0 \Rightarrow b = 4$ or $12$.
With $b > 4$, $b = 12$.
Answer: (D) 12. | ||||
| 66 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Reflection in $y = x$ swaps the coordinates, so C is $x^2 = 4y$. P(2, 1) lies on it.
Tangent to $x^2 = 4y$ at $(x_1, y_1)$: $xx_1 = 2(y + y_1)$, i.e. $2x = 2(y + 1)$, or $x - y = 1$.
Answer: (C). | ||||
| 67 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability A 6-digit number uses 6 of the 7 digits, i.e. leaves out one digit.
Total: leave out 0: $6! = 720$; leave out one of 1–6: $6! - 5! = 600$ each (0 cannot lead), $6 \times 600 = 3600$. Total $4320$.
The digit sum is $21 - (\text{omitted digit})$; it is divisible by 3 exactly when the omitted digit is 0, 3 or 6: $720 + 600 + 600 = 1920$.
$P(A) = \frac{1920}{4320} = \frac{4}{9}$.
Answer: (A). | ||||
| 68 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor $e^{\int \tan x\,dx} = \sec x$.
$y\sec x = \int \sin x \sec x\,dx = \int \tan x\,dx = \ln(\sec x) + C$. $y(0) = 0$ gives $C = 0$.
$y = \cos x \ln(\sec x)$; at $\frac{\pi}{4}$: $y = \frac{1}{\sqrt{2}}\ln\sqrt{2} = \frac{1}{2\sqrt{2}}\ln 2$.
Answer: (A). | ||||
| 69 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctHard | |
A hard single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Right of 0: $\{x\} = x \to 0^+$. $\cos^{-1}(1 - x^2) \approx \sqrt{2x^2} = \sqrt{2}\,x$ (since $\cos\theta \approx 1 - \frac{\theta^2}{2}$), $\sin^{-1}(1 - x) \to \frac{\pi}{2}$, and $x - x^3 \approx x$. RHL $= \sqrt{2}\cdot\frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.
Left of 0: $\{x\} = 1 + x = h \to 1^-$. $\cos^{-1}(1 - h^2) \to \cos^{-1}0 = \frac{\pi}{2}$; $\sin^{-1}(1 - h) \approx 1 - h$; $h - h^3 = h(1-h)(1+h) \approx 2(1-h)$. LHL $= \frac{\pi}{2}\cdot\frac{1}{2} = \frac{\pi}{4}$.
LHL $\neq$ RHL, so no value of $\alpha$ makes $f$ continuous at 0.
Answer: (B) no such $\alpha$ exists. | ||||
| 70 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct 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 The plane is $x + y + z = 42$. Put $p = x - 11$, $q = y - 19$, $r = z - 12$; then $p + q + r = 42 - 42 = 0$.
The expression is $3 + \frac{p^3 + q^3 + r^3}{p^2q^2r^2} - \frac{42}{14pqr} = 3 + \frac{p^3 + q^3 + r^3}{p^2q^2r^2} - \frac{3}{pqr}$.
Since $p + q + r = 0$, $p^3 + q^3 + r^3 = 3pqr$, and the middle term is $\frac{3}{pqr}$, which cancels the last.
Value $= 3$.
Answer: (D) 3. | ||||
| 71 | 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 $(a, b) \simeq (4, 3) \iff 3a = 4b \iff (a, b) = (4k, 3k)$ for a positive integer $k$.
Need $2 \leq 3k$ and $4k \leq 30$: $k = 1, 2, \ldots, 7$, giving (4,3), (8,6), (12,9), (16,12), (20,15), (24,18), (28,21).
Answer: (C) 7. | ||||
| 72 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\int_0^1 (x^2 + bx + c)\,dx = \frac{1}{3} + \frac{b}{2} + c = 1 \Rightarrow \frac{b}{2} + c = \frac{2}{3}$.
Remainder theorem: $P(2) = 4 + 2b + c = 5 \Rightarrow 2b + c = 1$.
Subtracting: $\frac{3b}{2} = \frac{1}{3} \Rightarrow b = \frac{2}{9}$, $c = 1 - \frac{4}{9} = \frac{5}{9}$.
$9(b + c) = 9 \times \frac{7}{9} = 7$.
Answer: (A) 7. | ||||
| 73 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Triangles: pick three different sides and one point on each: $\alpha = 5\cdot7\cdot6 + 5\cdot7\cdot9 + 5\cdot6\cdot9 + 7\cdot6\cdot9 = 210 + 315 + 270 + 378 = 1173$.
Quadrilaterals: one point on each of the four sides: $\beta = 5\cdot7\cdot6\cdot9 = 1890$.
$\beta - \alpha = 717$.
Answer: (C). | ||||
| 74 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $[k, k+1)$, $[x] = k$: $\int_k^{k+1} k e^k e^{1-x}\,dx = k e^k\left(e^{1-k} - e^{-k}\right) = k(e - 1)$.
$I = (e - 1)\sum_{k=0}^{9} k = 45(e - 1)$.
Answer: (C). | ||||
| 75 | 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 Triangles PQC and ABC share the angle at C, so $\frac{A_2}{A_1} = \frac{CP}{CA}\cdot\frac{CQ}{CB} = \frac{1}{3}$.
AC: $y = -\frac{x - 2}{3}$. With $y = mx$: $x_P = \frac{2}{3m + 1}$, so $\frac{CP}{CA} = \frac{2 - x_P}{3} = \frac{2m}{3m + 1}$.
BC: $y = 4(x - 2)$. With $y = mx$: $x_Q = \frac{8}{4 - m}$, so $\frac{CQ}{CB} = \frac{x_Q - 2}{1} = \frac{2m}{4 - m}$.
$\frac{4m^2}{(3m+1)(4-m)} = \frac{1}{3} \Rightarrow 12m^2 = -3m^2 + 11m + 4 \Rightarrow 15m^2 - 11m - 4 = 0 \Rightarrow m = 1$ or $-\frac{4}{15}$.
With $m > 0$, $m = 1$ (then P = $(\frac{1}{2}, \frac{1}{2})$ and Q = $(\frac{8}{3}, \frac{8}{3})$ lie on the sides).
Answer: (A) 1. | ||||
| 76 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|5\sqrt{7} + 9i| = \sqrt{175 + 81} = 16$ and $\log_{\sqrt{2}}16 = 8$.
With $r = |z|$, the left side is $2^{\frac{(r+3)(r-1)}{r+1}}$, so the inequality is $\frac{(r+3)(r-1)}{r+1} \geq 3$.
$r^2 + 2r - 3 \geq 3r + 3 \Rightarrow r^2 - r - 6 \geq 0 \Rightarrow (r - 3)(r + 2) \geq 0 \Rightarrow r \geq 3$.
Least $|z| = 3$.
Answer: (C) 3. | ||||
| 77 | 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 $R_1 \to R_1 - R_3$, $R_2 \to R_2 - R_3$:
$f(x) = \begin{vmatrix} 0 & 1 & \cos 2x - \sin 2x \\ 1 & 0 & \cos 2x - \sin 2x \\ \sin^2 x & \cos^2 x & \sin 2x \end{vmatrix}$.
Expanding along $R_1$: $f = -1\left(\sin 2x - (\cos 2x - \sin 2x)\sin^2 x\right) + (\cos 2x - \sin 2x)\cos^2 x = -\sin 2x + (\cos 2x - \sin 2x)(\sin^2 x + \cos^2 x)$.
$f(x) = \cos 2x - 2\sin 2x$, whose maximum is $\sqrt{1^2 + 2^2} = \sqrt{5}$.
Answer: (A) $\sqrt{5}$. | ||||
| 78 | 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 The right side needs $|x| \leq 1$. $x = 0$ is a solution.
For $x \neq 0$, take sines (both sides lie in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ with the same sign): $\frac{3x}{5}\sqrt{1 - \frac{16x^2}{25}} + \frac{4x}{5}\sqrt{1 - \frac{9x^2}{25}} = x \Rightarrow 3\sqrt{25 - 16x^2} + 4\sqrt{25 - 9x^2} = 25$.
The left side decreases as $|x|$ grows, from 35 at $x = 0$ to $3 \cdot 3 + 4 \cdot 4 = 25$ at $|x| = 1$, so $x = \pm 1$ only.
Check $x = 1$: $\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{4}{5} = \frac{\pi}{2} = \sin^{-1}1$; and $x = -1$ by oddness.
Solutions $x = -1, 0, 1$: three.
Answer: (D) 3. | ||||
| 79 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $g(x+1) = \ln x + g(x)$, so $g''(x+1) = g''(x) - \frac{1}{x^2}$.
Applying it for $x = 1, 2, 3, 4$: $g''(5) - g''(1) = -\left(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16}\right) = -\frac{144 + 36 + 16 + 9}{144} = -\frac{205}{144}$.
$|g''(5) - g''(1)| = \frac{205}{144}$.
Answer: (A). | ||||
| 80 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations $C_1$: put $y = vx$: $2v(v + xv') = v^2 - 1 \Rightarrow \frac{2v\,dv}{1 + v^2} = -\frac{dx}{x} \Rightarrow (1 + v^2)x = k \Rightarrow x^2 + y^2 = kx$. Through (1, 1): $x^2 + y^2 = 2x$, the circle of centre (1, 0) and radius 1.
$C_2$: $v + xv' = \frac{2v}{1 - v^2} \Rightarrow \frac{(1 - v^2)\,dv}{v(1 + v^2)} = \frac{dx}{x}$; since $\frac{1 - v^2}{v(1 + v^2)} = \frac{1}{v} - \frac{2v}{1 + v^2}$, $\frac{v}{1 + v^2} = cx \Rightarrow y = c(x^2 + y^2)$. Through (1, 1): $x^2 + y^2 = 2y$, the circle of centre (0, 1) and radius 1.
Two unit circles with centres $\sqrt{2}$ apart meet at (0, 0) and (1, 1). Common area $= 2\left[r^2\cos^{-1}\frac{d}{2r}\right] - \frac{d}{2}\sqrt{4r^2 - d^2} = 2\cdot\frac{\pi}{4} - \frac{\sqrt{2}}{2}\cdot\sqrt{2} = \frac{\pi}{2} - 1$.
(Equivalently, each circle's segment cut off by the chord $y = x$ has area $\frac{\pi}{4} - \frac{1}{2}$.)
Answer: (C). | ||||
| 81 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $t = \tan^{-1}\left(x + \frac{1}{x}\right)$. Then $\frac{dt}{dx} = \frac{1 - \frac{1}{x^2}}{1 + \left(x + \frac{1}{x}\right)^2} = \frac{x^2 - 1}{x^4 + 3x^2 + 1}$.
Split the integrand: $\frac{x^2 - 1}{(x^4 + 3x^2 + 1)\,t} + \frac{1}{x^4 + 3x^2 + 1}$.
First part: $\int \frac{dt}{t} = \ln t$, so $\alpha = 1$.
Second part: $\frac{1}{x^4 + 3x^2 + 1} = \frac{1}{2}\cdot\frac{(x^2 + 1) - (x^2 - 1)}{x^4 + 3x^2 + 1}$.
$\int \frac{x^2 + 1}{x^4 + 3x^2 + 1}dx = \int \frac{1 + \frac{1}{x^2}}{\left(x - \frac{1}{x}\right)^2 + 5}dx = \frac{1}{\sqrt{5}}\tan^{-1}\frac{x^2 - 1}{\sqrt{5}\,x}$, and $\int \frac{x^2 - 1}{x^4 + 3x^2 + 1}dx = \tan^{-1}\frac{x^2 + 1}{x}$.
So the second part is $\frac{1}{2\sqrt{5}}\tan^{-1}\frac{x^2 - 1}{\sqrt{5}\,x} - \frac{1}{2}\tan^{-1}\frac{x^2 + 1}{x}$: $\beta = \frac{1}{2\sqrt{5}}$, $\gamma = \frac{1}{\sqrt{5}}$, $\delta = -\frac{1}{2}$.
$\alpha + \beta\gamma + \delta = 1 + \frac{1}{10} - \frac{1}{2} = \frac{3}{5}$, and $10 \times \frac{3}{5} = 6$.
(Replacing $\beta, \gamma$ by $-\beta, -\gamma$ gives the same function and the same $\beta\gamma$.)
Answer: 6. | ||||
| 82 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Area $= \frac{1}{2}\cdot 12 \cdot 5 \sin A = 30 \Rightarrow \sin A = 1$, so $A = 90°$ and $BC = \sqrt{144 + 25} = 13$.
For a right triangle the hypotenuse is a diameter: $2R = 13$.
$s = \frac{5 + 12 + 13}{2} = 15$, $r = \frac{\Delta}{s} = \frac{30}{15} = 2$.
$2R + r = 15$.
Answer: 15. | ||||
| 83 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The line's direction is $(3, -m, 1)$. From (1, −2, 3) move along it: $(1 + 3t,\ -2 - mt,\ 3 + t)$.
On the plane: $(1 + 3t) + 2(-2 - mt) - 3(3 + t) + 10 = 0 \Rightarrow -2 - 2mt = 0 \Rightarrow t = -\frac{1}{m}$.
Distance $= |t|\sqrt{9 + m^2 + 1} = \frac{\sqrt{10 + m^2}}{|m|} = \sqrt{\frac{7}{2}} \Rightarrow 20 + 2m^2 = 7m^2 \Rightarrow m^2 = 4$.
$|m| = 2$.
Answer: 2. | ||||
| 84 | 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 $\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&-1\\1&2&1\end{vmatrix} = 3\hat{i} - 2\hat{j} + \hat{k}$, and $\vec{c} = \lambda(3\hat{i} - 2\hat{j} + \hat{k})$.
$\vec{c}\cdot(1, 1, 3) = \lambda(3 - 2 + 3) = 4\lambda = 8 \Rightarrow \lambda = 2$.
$\vec{c}\cdot(\vec{a} \times \vec{b}) = 2|\vec{a} \times \vec{b}|^2 = 2 \times 14 = 28$.
Answer: 28. | ||||
| 85 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $x \geq 0$, $f(x) = |x - 1| \geq 0$, so $g(f(x)) = (|x-1| - 1)^2 + b$, continuous there, with $g(f(0)) = b$.
For $x < 0$, $f(x) = x + a$.
Case $a = 0$: $f(x) = x < 0$, $g(f(x)) = x + 1 \to 1$ as $x \to 0^-$. Continuity at 0 needs $b = 1$: $a + b = 1$.
Case $a > 0$: near $0^-$, $f(x) = x + a > 0$ and $g(f(x)) = (x + a - 1)^2 + b \to (a-1)^2 + b$; continuity at 0 needs $a = 1$. Then $f(x) = x + 1$ changes sign at $x = -1$: for $x < -1$, $g(f(x)) = x + 2 \to 1$; for $-1 \leq x < 0$, $g(f(x)) = x^2 + b \to 1 + b$. Continuity at $-1$ needs $b = 0$: $a + b = 1$.
Either way $a + b = 1$.
Answer: 1. | ||||
| 86 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Set I: $\sum x = 20$, $\sum x^2 = 10(2 + 4) = 60$. Set II: $\sum y = 3n$, $\sum y^2 = n(1 + 9) = 10n$.
Combined: $\sigma^2 = \frac{60 + 10n}{10 + n} - \left(\frac{20 + 3n}{10 + n}\right)^2 = \frac{(60 + 10n)(10 + n) - (20 + 3n)^2}{(10 + n)^2} = \frac{n^2 + 40n + 200}{(n + 10)^2}$.
$\frac{n^2 + 40n + 200}{(n + 10)^2} = \frac{17}{9} \Rightarrow 9n^2 + 360n + 1800 = 17n^2 + 340n + 1700 \Rightarrow 8n^2 - 20n - 100 = 0 \Rightarrow 2n^2 - 5n - 25 = 0$.
$n = \frac{5 + 15}{4} = 5$ (the other root is negative).
Answer: 5. | ||||
| 87 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $\sum_{k=0}^{n}(-1)^k\,{}^nC_k\,t^k = (1 - t)^n$, so $A = \left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n + \left(\frac{1}{8}\right)^n + \left(\frac{1}{16}\right)^n + \left(\frac{1}{32}\right)^n$.
This is a G.P. of 5 terms with first term and ratio $\frac{1}{2^n}$: $A = \frac{1}{2^n}\cdot\frac{1 - 2^{-5n}}{1 - 2^{-n}} = \frac{1 - 2^{-5n}}{2^n - 1}$.
$(2^n - 1)A = 1 - 2^{-5n}$; matching $63A = 1 - 2^{-30}$ gives $2^n - 1 = 63$ and $5n = 30$, both $n = 6$.
Answer: 6. | ||||
| 88 | 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 $a_1 = \frac{b_1 - b_2}{\sqrt{3}}$, $a_2 = \frac{b_1 + kb_2}{\sqrt{3}}$.
$a_1^2 + a_2^2 = \frac{2b_1^2 + (1 + k^2)b_2^2 + 2(k - 1)b_1b_2}{3} = \frac{2b_1^2 + 2b_2^2}{3}$
$\Rightarrow (k^2 - 1)b_2^2 + 2(k - 1)b_1b_2 = 0 \Rightarrow (k - 1)\left[(k + 1)b_2^2 + 2b_1b_2\right] = 0$.
The given condition rules out the bracket vanishing, which leaves $k = 1$.
Answer: 1. | ||||
| 89 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series G.P.: $a^2 = \frac{b}{16}$, so $b = 16a^2$. A.P.: $\frac{2}{b} = \frac{1}{a} + 6$.
$\frac{1}{8a^2} = \frac{1}{a} + 6 \Rightarrow 48a^2 + 8a - 1 = 0 \Rightarrow a = \frac{-8 + 16}{96} = \frac{1}{12}$ (positive root), $b = \frac{16}{144} = \frac{1}{9}$.
$72(a + b) = 6 + 8 = 14$.
Answer: 14. | ||||
| 90 | 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 $\log_{a^{1/m}}x = m\log_a x$. The multipliers 2, 3, 6, 11, 18, 27, … have differences 1, 3, 5, 7, 9, …, so the $k$-th is $2 + (k-1)^2$.
$S_n(x) = \log_a x\left[2n + \frac{(n-1)n(2n-1)}{6}\right]$.
$n = 24$: $48 + \frac{23 \cdot 24 \cdot 47}{6} = 48 + 4324 = 4372$, so $4372\log_a x = 1093 \Rightarrow \log_a x = \frac{1}{4}$.
$n = 12$: $24 + \frac{11 \cdot 12 \cdot 23}{6} = 24 + 506 = 530$, so $530\log_a(2x) = 265 \Rightarrow \log_a 2 + \log_a x = \frac{1}{2} \Rightarrow \log_a 2 = \frac{1}{4}$.
$a = 2^4 = 16$.
Answer: 16. | ||||
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