JEE (Main) · 2021
JEE (Main) 2021 — 31 August, Shift 1
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
- Electronic Devices2
- Optics2
- Electromagnetic Induction and Alternating Currents2
18 easy · 11 medium · 1 hard
Chemistry
30 Q
- Some Basic Concepts of Chemistry2
- d- and f-Block Elements2
- Organic Chemistry: Some Basic Principles and Techniques2
21 easy · 7 medium · 2 hard
Mathematics
30 Q
- Limits, Continuity and Differentiability3
- Integral Calculus3
- Matrices and Determinants2
12 easy · 16 medium · 2 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 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Going round the loop, the source drives current forward through the ideal diode, down through R and into the + terminal of the 3 V cell, which opposes it.
So the diode conducts only while $V_i > 3$ V; then the current is $\frac{V_i - 3}{R}$ and the voltage across R is $V_R = V_i - 3$ (largest value $10 - 3 = 7$ V).
While $V_i < 3$ V the diode is reverse biased, no current flows and $V_R = 0$.
So the output appears only over the part of each positive half-cycle where the input is above the 3 V line, and its size is the height of the input above that line.
Option (B) is the only one that shows output solely where the curve rises above the +3 line; (A) is ordinary half-wave rectification ignoring the cell, (C) and (D) are clipper outputs.
None of the four graphs draws this: (A) shows the whole positive half-cycle of Vᵢ, (B) a curve lying above +3 V, (C) and (D) clipped sines.
No option is correct as printed. | ||||
| 2 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Both A and B feed an AND gate, giving $P = A\cdot B$, and also an OR gate, giving $Q = A + B$.
P and Q feed a NAND gate: $Y = \overline{P\cdot Q} = \overline{AB(A+B)} = \overline{AB}$, since $AB(A+B) = AB$.
(0,0): $Y = 1$; (0,1): $Y = 1$; (1,0): $Y = 1$; (1,1): $Y = 0$.
Answer: (A). | ||||
| 3 | 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 For the chain $A \to B \to C$: $\frac{dN_B}{dt} = \lambda_A N_A - \lambda_B N_B$.
At $t = 0$, $N_B = 0$, so $N_B$ first grows smoothly; as A runs out the decay of B wins and $N_B$ falls smoothly back to zero:
$N_B = \frac{\lambda_A N_0}{\lambda_B - \lambda_A}\left(e^{-\lambda_A t} - e^{-\lambda_B t}\right)$, which starts at zero, has one smooth maximum and tends to zero.
(A) does not start at zero, (C) has a straight rise and a sharp corner, and (D) never decreases.
Answer: (B). | ||||
| 4 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{P}$, so equal wavelengths mean equal momenta: $P_p = P_e$.
$K = \frac{P^2}{2m}$, and the proton is much heavier than the electron, so $K_p < K_e$.
Answer: (C). | ||||
| 5 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Put the corner of the mirrors at the origin, $M_1$ along the y-axis and $M_2$ along the x-axis; then P is at $(a, 2a)$.
Image in $M_1$: $I_1 = (-a, 2a)$. Image in $M_2$: $I_2 = (a, -2a)$. The third image (reflection in both) is $I_3 = (-a, -2a)$.
The distance between the two images formed by the mirrors, $I_1$ and $I_2$: $\sqrt{(2a)^2 + (4a)^2} = 2\sqrt{5}\,a = 2\times2.3\,a = 4.6a$.
Answer: (B). | ||||
| 6 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For the concave lens the focal length is $-f$ and the object distance is $u = -f$.
$\frac{1}{v} = \frac{1}{f_{lens}} + \frac{1}{u} = -\frac{1}{f} - \frac{1}{f} = -\frac{2}{f}$, so $v = -\frac{f}{2}$: a virtual image $\frac{f}{2}$ from the optical centre, on the object's side.
$m = \frac{v}{u} = \frac{-f/2}{-f} = \frac{1}{2}$.
Answer: (C). | ||||
| 7 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Each side of the large loop is at distance $\frac{b}{2}$ from the centre and subtends $45^\circ$ on either side:
$B_{side} = \frac{\mu_0 I}{4\pi (b/2)}(\sin45^\circ + \sin45^\circ) = \frac{\mu_0}{4\pi}\cdot\frac{2\sqrt{2}\,I}{b}$.
Four sides: $B = \frac{\mu_0}{4\pi}\cdot\frac{8\sqrt{2}\,I}{b}$, nearly uniform over the small loop since $b \gg a$.
Flux through the small loop: $\phi = B a^2$, so $M = \frac{\phi}{I} = \frac{\mu_0}{4\pi}\,8\sqrt{2}\,\frac{a^2}{b}$.
Answer: (A). | ||||
| 8 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$ (the circuit is at resonance).
Answer: (C). | ||||
| 9 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Galvanometer and shunt are in parallel, so $I_g G = (I - I_g) S$.
$0.02I \times G = 0.98I \times 5$, so $G = \frac{0.98\times5}{0.02} = 245\ \Omega$.
Answer: (B). | ||||
| 10 | 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 The turns are spread evenly between $a$ and $b$: the number of turns between $r$ and $r + dr$ is $\frac{N}{b-a}dr$.
Each acts as a circular loop of radius $r$: $dB = \frac{\mu_0 I}{2r}\cdot\frac{N\,dr}{b-a}$.
$B = \frac{\mu_0 NI}{2(b-a)}\int_a^b \frac{dr}{r} = \frac{\mu_0 NI}{2(b-a)}\log_e\frac{b}{a}$.
Answer: (A). | ||||
| 11 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Torque = force × distance: $ML^2T^{-2}$ (iii).
Impulse = force × time: $MLT^{-1}$ (i).
Tension is a force: $MLT^{-2}$ (iv).
Surface tension = force per length: $MT^{-2}$ (ii).
Answer: (D). | ||||
| 12 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The charges have opposite signs, so the fields cancel only outside the pair, on the side of the smaller charge (beyond B).
Let the point be $x$ cm to the right of B: $\frac{k(20)}{(5+x)^2} = \frac{k(5)}{x^2}$, so $\frac{5+x}{x} = 2$ and $x = 5$ cm.
Answer: (B). | ||||
| 13 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (A) $\frac{s}{\rho r g}$: $\frac{MT^{-2}}{ML^{-3}\cdot L\cdot LT^{-2}} = L$ — a height. Correct.
(B) $\frac{p a^4}{\eta L}$: $\frac{ML^{-1}T^{-2}\cdot L^4}{ML^{-1}T^{-1}\cdot L} = L^3T^{-1}$ — a volume per unit time (Poiseuille's rate of flow), not a volume $L^3$. Incorrect.
(C) Torque × angle has the dimensions of work. Correct.
(D) $\epsilon\frac{\partial E}{\partial t}$ is the displacement current density, $AL^{-2}$. Correct.
Answer: (B). | ||||
| 14 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $\frac{dp}{p} = -a\,dv$ with $p = p_0$ at $v = 0$ gives $p = p_0e^{-av}$.
For one mole, $T = \frac{pv}{R} = \frac{p_0}{R}ve^{-av}$.
$\frac{dT}{dv} = \frac{p_0}{R}e^{-av}(1 - av) = 0$ at $v = \frac{1}{a}$ (a maximum).
$T_{max} = \frac{p_0}{R}\cdot\frac{1}{a}\cdot e^{-1} = \frac{p_0}{aeR}$.
Answer: (B). | ||||
| 15 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{4}$, so $T_2 = \frac{3}{4}T_1$.
After lowering the sink by 58 (a change of 58°C is a change of 58 K): $1 - \frac{T_2 - 58}{T_1} = \frac{1}{2}$, so $T_2 - 58 = \frac{1}{2}T_1$.
$\frac{3}{4}T_1 - \frac{1}{2}T_1 = 58$ gives $T_1 = 232$ K and $T_2 = 174$ K.
Answer: (D). | ||||
| 16 | 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 For a rod hanging under its own weight W, the elongation is $\Delta l = \frac{WL}{2AY}$ (the average tension is $\frac{W}{2}$).
$\Delta l = \frac{10\times0.2}{2\times(100\times10^{-4})\times2\times10^{11}} = \frac{2}{4\times10^{9}} = 5\times10^{-10}$ m.
Answer: (B). | ||||
| 17 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation At the midpoint, distance $\frac{r}{2}$ from each centre, the potential energy is $U = -\frac{GM_1m}{r/2} - \frac{GM_2m}{r/2} = -\frac{2G(M_1+M_2)m}{r}$.
To escape to infinity: $\frac{1}{2}mV^2 = \frac{2G(M_1+M_2)m}{r}$, so $V = \sqrt{\frac{4G(M_1+M_2)}{r}}$.
Answer: (C). | ||||
| 18 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion About the centre of the circle, $\vec{L} = m\vec{r}\times\vec{v}$ has magnitude $mvr$, constant at constant speed, and its direction is always perpendicular to the plane of the circle (the sense of motion does not change). The force is central, so there is no torque about the centre.
Answer: (D). | ||||
| 19 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Time of fall $t = \sqrt{\frac{2h}{g}}$; in that time the packet moves forward $x = vt = v\sqrt{\frac{2h}{g}}$, so the man must be this far ahead horizontally.
Distance from the helicopter: $\sqrt{x^2 + h^2} = \sqrt{\frac{2v^2h}{g} + h^2}$.
Answer: (B). | ||||
| 20 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctHard | |
A hard single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Let both angles be $\theta$, with speeds $v_1$ (mass M) and $v_2$ (mass m) after the collision.
Perpendicular momentum: $Mv_1\sin\theta = mv_2\sin\theta$, so $Mv_1 = mv_2$.
Along the line: $MV_0 = (Mv_1 + mv_2)\cos\theta = 2Mv_1\cos\theta$, so $v_1 = \frac{V_0}{2\cos\theta}$ and $v_2 = \frac{MV_0}{2m\cos\theta}$.
Energy: $MV_0^2 = Mv_1^2 + mv_2^2 = \frac{MV_0^2}{4\cos^2\theta}\left(1 + \frac{M}{m}\right)$, so $\frac{M}{m} = 4\cos^2\theta - 1$.
The largest value, as $\theta \to 0$, is $4 - 1 = 3$.
Answer: (B). | ||||
| 21 | 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 Upper branch: $4\parallel4 = 2\ \Omega$, then $2\ \Omega$, then $15\parallel10 = 6\ \Omega$: total $10\ \Omega$.
Lower branch: $8\parallel8 = 4\ \Omega$ and $12\parallel12 = 6\ \Omega$: total $10\ \Omega$.
The two branches in parallel give $5\ \Omega$; with the $1\ \Omega$ resistor the circuit has $6\ \Omega$, so $I = \frac{12}{6} = 2$ A.
Each branch carries 1 A, so the drop across the $15\parallel10$ pair is $1\times6 = 6$ V.
Answer: 6. | ||||
| 22 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Released from the extreme position, the displacement is $y = A\cos\omega t$.
KE = PE when $\frac{1}{2}ky^2 = \frac{1}{2}\cdot\frac{1}{2}kA^2$, i.e. $y = \frac{A}{\sqrt{2}}$.
$\cos\omega t = \frac{1}{\sqrt{2}}$ first at $\omega t = \frac{\pi}{4}$, so $t = \frac{\pi/4}{2\pi/T} = \frac{T}{8}$.
Answer: 8. | ||||
| 23 | Laws of MotionNewton's laws | Class 11: Laws of Motion | NumericalMedium | |
A medium numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion In the car's frame the bob feels gravity $g$ downward and a pseudo force $ma$ down the incline. Effective gravity $\vec{g}_{eff} = \vec{g} - \vec{a}$.
Horizontal component: $a\cos30^\circ = 10\times\frac{\sqrt{3}}{2} = 5\sqrt{3}$.
Vertical component: $g + a\sin30^\circ = 10 + 5 = 15$.
The string lies along $\vec{g}_{eff}$: $\tan\phi = \frac{5\sqrt{3}}{15} = \frac{1}{\sqrt{3}}$, so $\phi = 30^\circ$ with the vertical.
Answer: 30. | ||||
| 24 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Momentum: $m\times40 = \frac{m}{2}\times60 + \frac{m}{2}v$, so $v = 20$ m/s.
Initial KE $= \frac{1}{2}m(40)^2 = 800m$. Final KE $= \frac{1}{2}\cdot\frac{m}{2}(60^2 + 20^2) = 1000m$.
Fractional change $= \frac{1000m - 800m}{800m} = \frac{1}{4}$, i.e. $1 : 4$.
Answer: 1. | ||||
| 25 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical 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 $B = \frac{\Delta p}{\Delta V/V}$, so $\Delta p = 9.8\times10^8\times0.005 = 4.9\times10^6$ Pa.
$\Delta p = \rho g h$: $h = \frac{4.9\times10^6}{10^3\times9.8} = 500$ m.
Answer: 500. | ||||
| 26 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{900}{9\times10^{-4}}} = 1000$ m/s.
Successive resonant frequencies of a wire fixed at both ends differ by the fundamental: $\frac{v}{2L} = 550 - 500 = 50$ Hz.
$L = \frac{1000}{2\times50} = 10$ m.
Answer: 10. | ||||
| 27 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In the steady state no current flows into the capacitor, so the three $2\ \text{k}\Omega$ resistors carry $I = \frac{6}{6000} = 1$ mA.
The capacitor is across the lowest resistor: $V = 1\ \text{mA}\times2\ \text{k}\Omega = 2$ V, its upper plate at the higher potential (the cell's positive terminal feeds the top of the chain).
$Q = CV = 50\ \mu\text{F}\times2\ \text{V} = 100\ \mu$C on the upper plate (positive).
Answer: 100. | ||||
| 28 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Total resistance of the wire $= 4\times3 = 12\ \Omega$. Bent into a circle, two diametrically opposite points split it into two halves of $6\ \Omega$ each, in parallel.
$R = \frac{6\times6}{6+6} = 3\ \Omega$.
Answer: 3. | ||||
| 29 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalMedium | |
A medium numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Average energy density of the wave (electric and magnetic parts together): $u = \frac{1}{2}\epsilon_0E_0^2 = \frac{1}{2}\times8.8\times10^{-12}\times2500 = 1.1\times10^{-8}$ J/m$^3$.
$V = \frac{5.5\times10^{-12}}{1.1\times10^{-8}} = 5\times10^{-4}$ m$^3 = 500$ cm$^3$.
Answer: 500. | ||||
| 30 | Communication SystemsLine of sight communication | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on line of sight communication, from Communication Systems. NCERT: No matching NCERT chapter Range $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$ with $h_T + h_R = 160$ m.
For a fixed sum, $\sqrt{h_T} + \sqrt{h_R}$ is largest when $h_T = h_R = 80$ m.
$d = 2\sqrt{2\times6.4\times10^6\times80} = 2\sqrt{1.024\times10^9} = 2\times32000$ m $= 64$ km.
Answer: 64. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry For an ideal gas at constant temperature, Boyle's law gives $PV = nRT = $ constant, whatever the pressure.
So the plot of PV against P is a horizontal straight line.
Answer: (D). | ||||
| 32 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $Cr_2O_7^{2-}$ is two $CrO_4$ tetrahedra sharing one corner oxygen. The bridging Cr$-$O$-$Cr bond is bent (angle about $126^\circ$), and the two Cr$-$O bridge bonds are equal, so the bond is non-linear and symmetrical.
Answer: (C). | ||||
| 33 | Solutionsvan't Hoff factor | Class 12: Solutions | Single correctEasy | |
A easy single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = iK_fm$, so at equal concentration the solute giving the most particles wins.
Glycine (a zwitterion), glucose and hydrazine (a very weak base) are essentially undissociated: $i \approx 1$.
$KHSO_4 \rightarrow K^+ + HSO_4^-$, and $HSO_4^-$ ionises further in part, so $i$ is between 2 and 3.
Answer: (C). | ||||
| 34 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Adsorption of a gas is exothermic, so at a given pressure less gas is adsorbed at the higher temperature: the $T_2$ (lower temperature) isotherm lies above the $T_1$ isotherm.
The isotherm $\frac{x}{m} = kP^{1/n}$ rises steeply at low pressure and levels off towards saturation at high pressure, so the curves bend over rather than being straight lines.
Answer: (A). | ||||
| 35 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Assertion and reasonEasy | |
A easy assertion and reason question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties (A) is true: across a period atoms hold their electrons more firmly, so the tendency to lose electrons (metallic character) falls and the tendency to gain them (non-metallic character) rises.
(R) is true: across a period ionisation enthalpy increases and electron gain enthalpy decreases (becomes more negative).
These two trends are exactly why the character changes, so (R) explains (A).
Answer: (A). | ||||
| 36 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter (A) is true: purified alumina is dissolved in molten cryolite and electrolysed (Hall–Héroult process).
(R) is true: in cryolite $Na_3AlF_6$, $3(+1) + x + 6(-1) = 0$ gives $x = +3$.
But cryolite is added to lower the melting point and raise the conductivity of the melt; the oxidation state of Al in it does not explain the extraction.
Answer: (B). | ||||
| 37 | 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 Statement I: producing syn-gas (a mixture of CO and $H_2$) from coal, $C + H_2O \xrightarrow{1270\ K} CO + H_2$, is called coal gasification. True.
Statement II: syn-gas is a mixture of CO and $H_2$; it does not contain $CO_2$ in a 1 : 1 : 1 ratio. False.
Answer: (C). | ||||
| 38 | 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 Portland cement is mainly dicalcium silicate $Ca_2SiO_4$ (about 26%), tricalcium silicate $Ca_3SiO_5$ (about 51%) and tricalcium aluminate $Ca_3Al_2O_6$ (about 11%).
The largest component is tricalcium silicate.
Answer: (C). | ||||
| 39 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Yb: $[Xe]4f^{14}6s^2$, so $Yb^{2+}$ is $[Xe]4f^{14}$ — a full f shell, no unpaired electron: diamagnetic, and the +2 state is well known (stable full shell).
$Nd^{2+}$ ($4f^4$) and $Ce^{2+}$ ($4f^2$) have unpaired electrons; $La^{2+}$ ($5d^1$) has one unpaired electron and La does not show a stable +2 state.
Answer: (B). | ||||
| 40 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Biuret, $H_2N{-}CO{-}NH{-}CO{-}NH_2$, binds a metal ion (as in the biuret test with $Cu^{2+}$) through two donor atoms, forming a chelate ring: it is a bidentate ligand.
Answer: (A). | ||||
| 41 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Clean water has a BOD of less than 5 ppm, whereas highly polluted water has a BOD of 17 ppm or more.
Answer: (B). | ||||
| 42 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Assertion and reasonEasy | |
A easy assertion and reason question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Propan-1-ol boils at about 370 K (97°C) and propanone at about 329 K (56°C): a difference of about 41°C.
Liquids whose boiling points differ sufficiently (more than about 20°C) can be separated by simple distillation, so (R) is true and is the reason (A) is true.
Answer: (A). | ||||
| 43 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The compound is $CH_3{-}C{\equiv}C{-}CH{=}C(Br){-}CH_3$, a six-carbon chain with one double and one triple bond.
Numbering from either end gives the multiple bonds the locants {2, 4}; in a tie the double bond gets the lower number, so number from the bromine end: C2=C3 and C4≡C5, Br on C2: 2-bromohex-2-en-4-yne.
At C2 the higher-priority group is Br; at C3 it is the propynyl group (not H). In the drawing Br points to the right and the alkynyl chain to the left, on opposite sides of the double bond: E.
Answer: (B). | ||||
| 44 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Assertion and reasonMedium | |
A medium assertion and reason question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $Br_2$ adds to propene to give a cyclic bromonium ion. Water, present in large excess, opens it at the more substituted carbon (the one better able to carry positive charge) — the Markovnikov orientation.
So OH goes to C2 and Br to C1: $CH_3CH(OH)CH_2Br$, 1-bromopropan-2-ol. (A) is true and (R) correctly explains it.
Answer: (A). | ||||
| 45 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Acetate in acetic acid is a weak nucleophile in a polar protic solvent: the substitution goes by $S_N1$, so the rate follows the stability of the carbocation formed.
1. Cl on the allylic carbon with $CH_3$ on the far end of the double bond: the allylic cation is delocalised onto a tertiary carbon — the most stable.
2. Cl allylic, with $CH_3$ on the neighbouring saturated carbon: a secondary allylic cation, still resonance-stabilised.
3. Cl on a ring carbon not next to the double bond: a secondary cation with no resonance.
4. $CH_2Cl$ on the ring: a primary cation — the least stable.
Order: 1 > 2 > 3 > 4, which is the order drawn in (A).
Answer: (A). | ||||
| 46 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Protonation of OH and loss of water give the secondary carbocation $(CH_3)_3C{-}\overset{+}{C}H{-}CH_3$.
A 1,2-methyl shift turns it into the tertiary cation $(CH_3)_2\overset{+}{C}{-}CH(CH_3){-}CH_3$.
Loss of the proton that gives the most substituted alkene (Saytzeff) yields 2,3-dimethylbut-2-ene, $(CH_3)_2C{=}C(CH_3)_2$.
Answer: (C). | ||||
| 47 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A: $SOCl_2$ converts acetic acid into acetyl chloride, $CH_3COCl$.
B: Friedel–Crafts acylation of benzene gives acetophenone, $C_6H_5COCH_3$.
C: cyanide adds to the carbonyl group (base catalysed), giving the cyanohydrin $C_6H_5C(OH)(CN)CH_3$.
Answer: (B). | ||||
| 48 | 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 The starting compound is acetone cyanohydrin, $(CH_3)_2C(OH)CN$.
A: $LiAlH_4$ reduces the nitrile to a primary amine: $(CH_3)_2C(OH)CH_2NH_2$.
B: acid hydrolysis of the nitrile gives the α-hydroxy acid $(CH_3)_2C(OH)COOH$, and with $H_2SO_4$ this tertiary alcohol dehydrates to the conjugated acid $CH_2{=}C(CH_3)COOH$ (methacrylic acid).
Answer: (D). | ||||
| 49 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Phenol and formaldehyde first give o- and p-hydroxymethylphenol; the o-hydroxymethylphenol units condense into the linear polymer Novolac (which cross-links further to Bakelite).
(B) is not a known pair, (C) gives Buna-S and (D) is a monomer of PHBV.
Answer: (A). | ||||
| 50 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Lactose: β-D-galactose C1 linked to C4 of D-glucose — a β-C1–C4 link.
Amylose and maltose have α-C1–C4 links; sucrose has an α-C1 (glucose)–β-C2 (fructose) link.
Answer: (A). | ||||
| 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 Molar mass of $H_2C_2O_4\cdot2H_2O = 90 + 36 = 126$ g mol$^{-1}$; moles $= \frac{6.3}{126} = 0.05$.
Molarity $= \frac{0.05}{0.250} = 0.20 = 20\times10^{-2}$ mol L$^{-1}$.
Answer: 20. | ||||
| 52 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom Ge: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\,4p^2$.
Every s orbital has $m_l = 0$: 1s, 2s, 3s, 4s — all full (4).
Each full p or d subshell has one $m_l = 0$ orbital: 2p, 3p, 3d — all full (3).
4p holds 2 electrons in two different orbitals (Hund's rule), so no 4p orbital is completely filled.
$x = 4 + 3 = 7$.
Answer: 7. | ||||
| 53 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics In the diagram, $x$ is the activation energy above A + B, $y$ is the drop from the A + B level to the M + N level, and $z$ is the height of M + N above the axis.
The enthalpy change is the difference between the product and reactant levels: $|\Delta H| = y = 45$ kJ mol$^{-1}$ (the reaction is exothermic).
Answer: 45. | ||||
| 54 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Molar solubility $s = \frac{x}{M}$ mol L$^{-1}$. $A_3B_2 \rightleftharpoons 3A^{2+} + 2B^{3-}$.
$K_{sp} = (3s)^3(2s)^2 = 27\times4\,s^5 = 108\left(\frac{x}{M}\right)^5$.
Answer: 108. | ||||
| 55 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Cd goes from 0 to +2, so $n = 2$.
$\Delta G^\circ = -nFE^\circ = -2\times96487\times4.315 = -832682.8$ J mol$^{-1}$.
$\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T} = \frac{-825200 + 832682.8}{298} = \frac{7482.8}{298} = 25.1$ J K$^{-1}$.
Answer: 25. | ||||
| 56 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For a first-order reaction $t = \frac{1}{k}\ln\frac{[A]_0}{[A]}$.
75% completion: $t_{75} = \frac{\ln4}{k} = \frac{2\ln2}{k}$ (two half-lives). 50% completion: $t_{50} = \frac{\ln2}{k}$.
Ratio $= 2$.
Answer: 2. | ||||
| 57 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Halic(V) acids are $HXO_3$, with the halogen in the +5 state: $HClO_3$, $HBrO_3$ and $HIO_3$ are known.
Fluorine, the most electronegative element, shows no positive oxidation state and forms no $HFO_3$.
Answer: 3. | ||||
| 58 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure In $CuSO_4\cdot5H_2O$, i.e. $[Cu(H_2O)_4]SO_4\cdot H_2O$, four water molecules are coordinated to $Cu^{2+}$ and the fifth is held by hydrogen bonds between the coordinated water and the sulphate ion.
Answer: 1. | ||||
| 59 | AminesClassification and identification of amines | Class 12: Amines | NumericalMedium | |
A medium numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Nitrobenzene is reduced to aniline by a metal in acid — Sn/HCl, Fe/HCl and Zn/HCl — and by catalytic hydrogenation — $H_2$/Pd and $H_2$/Raney nickel.
A metal in neutral or basic medium (Sn/$NH_4OH$) does not carry the reduction through to aniline; it stops at intermediate products such as phenylhydroxylamine or azoxy/azo compounds.
Reagents I, III, IV, V and VI: 5.
Answer: 5. | ||||
| 60 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | NumericalHard | |
A hard numerical question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter In group II analysis, PbS, CuS and CdS dissolve in 50% $HNO_3$, while HgS does not (it needs aqua regia).
$As_2S_3$ is oxidised and dissolved by nitric acid (to arsenic acid, used in the molybdate test), whereas antimony is converted by nitric acid into an insoluble oxide, so $Sb_2S_3$ does not go into solution.
Soluble: PbS, CuS, CdS and $As_2S_3$ — 4.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions (A) $|x - x| = 0 \le 1$ and $|x - y| = |y - x|$: reflexive and symmetric. Correct.
(B) $|x| - |x| = 0 \le 1$: reflexive. $(0, 2)$: $0 - 2 = -2 \le 1$ is in R, but $(2, 0)$: $2 \le 1$ fails, so not symmetric. Correct.
(C) If $|x| - |y| > 0$ then $|y| - |x| < 0$, so never symmetric. $(1.9, 1)$ and $(1, 0.2)$ are in R but $(1.9, 0.2)$ gives $1.7 > 1$: not transitive. Correct.
(D) Symmetric, yes; but $(0, 1)$ and $(1, 0)$ are in R while $(0, 0)$ is not ($|0 - 0| = 0$), so it is not transitive. This statement is not correct.
Answer: (D). | ||||
| 62 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $t = e^x > 0$: $g(t) = t^4 + 2t^3 - t - 6 = 0$, and each positive root $t$ gives exactly one real $x$.
The coefficients $+, +, -, -$ change sign once, so (Descartes' rule) $g$ has exactly one positive root. Indeed $g(1) = -4 < 0$ and $g(2) = 24 > 0$; and since $g(0) = -6$, $g$ is negative on $(0, t_0)$ and, with $g'(t) = 4t^3 + 6t^2 - 1$ positive for $t \ge \frac{1}{2}$, increasing afterwards.
So there is exactly one real root.
Answer: (B). | ||||
| 63 | 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 $a_r = \omega^r$ with $\omega = e^{2\pi i/9}$, so $a_{r+3} = \omega^3a_r$: the second row is $\omega^3$ times the first, and the determinant is 0.
Check the options: $a_9 = \omega^9 = 1$; $a_1a_9 - a_3a_7 = \omega^{10} - \omega^{10} = 0$; $a_2a_6 - a_4a_8 = \omega^8 - \omega^{12} \ne 0$; $a_5 \ne 0$.
Answer: (B). | ||||
| 64 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix} 2&1&1\\1&-1&1\\1&1&a \end{vmatrix} = 2(-a-1) - (a-1) + 2 = 1 - 3a$, which is 0 for $a = \frac{1}{3}$.
With $a = \frac{1}{3}$: $\frac{2}{3}(2x+y+z) - \frac{1}{3}(x-y+z) = x + y + \frac{1}{3}z$, so the third left side is a combination of the first two, whose right side is $\frac{2}{3}(5) - \frac{1}{3}(3) = \frac{7}{3}$.
The system has no solution when $a = \frac{1}{3}$ and $b \ne \frac{7}{3}$.
Answer: (A). | ||||
| 65 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the numbers be $a, ar, ar^2$. Then $a, 2ar, ar^2$ are in AP: $4ar = a + ar^2$, so $r^2 - 4r + 1 = 0$ and $r = 2 \pm \sqrt{3}$.
Fourth term $ar^3 = 3r^2$ gives $a = \frac{3}{r} > 0$, so an increasing GP needs $r > 1$: $r = 2 + \sqrt{3}$, $\frac{1}{r} = 2 - \sqrt{3}$.
$d = 2ar - a = 6 - \frac{3}{r} = 6 - 3(2 - \sqrt{3}) = 3\sqrt{3}$.
$r^2 = 7 + 4\sqrt{3}$, so $r^2 - d = 7 + \sqrt{3}$.
Answer: (B). | ||||
| 66 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The $n$th term is $\frac{2n+1}{n^2(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$.
The sum telescopes: $S_{10} = 1 - \frac{1}{11^2} = \frac{120}{121}$.
Answer: (C). | ||||
| 67 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\sin(\pi\cos^4x) = \sin(\pi - \pi\cos^4x) = \sin\left(\pi(1 - \cos^4x)\right)$, and $1 - \cos^4x \to 0$.
So the limit is $\lim \frac{\pi^2(1 - \cos^4x)^2}{x^4}$.
$1 - \cos^4x = (1 - \cos^2x)(1 + \cos^2x) = \sin^2x(1 + \cos^2x)$, so $\frac{1 - \cos^4x}{x^2} \to 1\times2 = 2$.
Limit $= \pi^2\times2^2 = 4\pi^2$.
Answer: (C). | ||||
| 68 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Left limit: $\frac{\ln(1 + x/a) - \ln(1 - x/b)}{x} \to \frac{1}{a} + \frac{1}{b}$.
Right limit: $\cos^2x - \sin^2x - 1 = \cos2x - 1 = -2\sin^2x$ and $\sqrt{x^2+1} - 1 = \frac{x^2}{\sqrt{x^2+1}+1}$, so the ratio is $-2\frac{\sin^2x}{x^2}(\sqrt{x^2+1}+1) \to -4$.
Continuity: $k = -4$ and $\frac{1}{a} + \frac{1}{b} = -4$.
$\frac{1}{a} + \frac{1}{b} + \frac{4}{k} = -4 - 1 = -5$.
Answer: (A). | ||||
| 69 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $9x^2 - 12x + 4 = (3x - 2)^2 \ge 0$, so $e^{|9x^2-12x+4|} = e^{(3x-2)^2}$ is differentiable everywhere and never zero.
$|x^2 - 2x - 3| = |(x-3)(x+1)|$ has corners at the simple roots $x = 3$ and $x = -1$; multiplied by a non-zero smooth factor, $f$ keeps these corners.
So $f$ is not differentiable at exactly two points.
Answer: (B). | ||||
| 70 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Write the integrand as $\frac{1}{(x+2)^2}\left(\frac{x-1}{x+2}\right)^{-3/4}$.
Put $t = \frac{x-1}{x+2}$: $dt = \frac{3}{(x+2)^2}dx$.
$\int = \frac{1}{3}\int t^{-3/4}dt = \frac{1}{3}\cdot4t^{1/4} + C = \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1/4} + C$.
Answer: (A). | ||||
| 71 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} = \frac{2^x(2^y - 1)}{2^y}$, so $\int\frac{2^y}{2^y - 1}dy = \int2^xdx$.
$\frac{\ln(2^y - 1)}{\ln2} = \frac{2^x}{\ln2} + C$, i.e. $\ln(2^y - 1) = 2^x + C'$.
$y(0) = 1$: $\ln1 = 1 + C'$, $C' = -1$.
At $x = 1$: $\ln(2^y - 1) = 2 - 1 = 1$, so $2^y = 1 + e$ and $y(1) = \log_2(1 + e)$.
Answer: (D). | ||||
| 72 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Differentiating: $\sqrt{1 - (f'(x))^2} = f(x)$, so $(f')^2 = 1 - f^2$ and $f' = \pm\sqrt{1 - f^2}$.
With $f(0) = 0$ and $f \ge 0$, $f$ must increase from 0: $f' = \sqrt{1 - f^2}$, giving $\sin^{-1}f = x$, $f(x) = \sin x$ on [0, 1].
$\lim_{x\to0}\frac{1}{x^2}\int_0^x\sin t\,dt = \lim\frac{1 - \cos x}{x^2} = \frac{1}{2}$.
Answer: (B). | ||||
| 73 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines First line: $p = \frac{|k\cot2\alpha|}{\sqrt{\mathrm{cosec}^2\alpha + \sec^2\alpha}} = |k\cot2\alpha|\,|\sin\alpha\cos\alpha| = \left|k\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{\sin2\alpha}{2}\right| = \frac{|k\cos2\alpha|}{2}$.
Second line: $q = \frac{|k\sin2\alpha|}{\sqrt{\sin^2\alpha + \cos^2\alpha}} = |k\sin2\alpha|$.
$4p^2 + q^2 = k^2\cos^22\alpha + k^2\sin^22\alpha = k^2$.
Answer: (D). | ||||
| 74 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Vertex $(R, 0)$, focus $(S, 0)$: the distance from vertex to focus is $a = S - R$.
Latus rectum $= 4a = 4(S - R)$.
Answer: (D). | ||||
| 75 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Dividing by 60: $\frac{x\cos\theta}{5} + \frac{y\sin\theta}{12} = 1$, the tangent at the point $(5\cos\theta, 12\sin\theta)$ of the ellipse $\frac{x^2}{25} + \frac{y^2}{144} = 1$.
That ellipse is $144x^2 + 25y^2 = 3600$.
Answer: (C). | ||||
| 76 | 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 The plane is $(3x - 2y + 4z - 7) + \lambda(x + 5y - 2z + 9) = 0$.
At (1, 4, −3): $(3 - 8 - 12 - 7) + \lambda(1 + 20 + 6 + 9) = -24 + 36\lambda = 0$, $\lambda = \frac{2}{3}$.
Multiply by 3: $3(3x - 2y + 4z - 7) + 2(x + 5y - 2z + 9) = 11x + 4y + 8z - 3 = 0$, i.e. $-11x - 4y - 8z + 3 = 0$.
$\alpha + \beta + \gamma = -11 - 4 - 8 = -23$.
Answer: (D). | ||||
| 77 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin18^\circ = \frac{\sqrt{5} - 1}{4}$, so $\mathrm{cosec}\,18^\circ = \frac{4}{\sqrt{5} - 1} = \sqrt{5} + 1$.
$x = 1 + \sqrt{5}$: $x^2 - 2x - 4 = (6 + 2\sqrt{5}) - (2 + 2\sqrt{5}) - 4 = 0$.
Answer: (A). | ||||
| 78 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Squaring: $4|\vec{a}|^2 + 12\vec{a}\cdot\vec{b} + 9|\vec{b}|^2 = 9|\vec{a}|^2 + 6\vec{a}\cdot\vec{b} + |\vec{b}|^2$, so $8|\vec{b}|^2 + 6\vec{a}\cdot\vec{b} - 5|\vec{a}|^2 = 0$.
$|\vec{a}| = 8$ and $\vec{a}\cdot\vec{b} = 8|\vec{b}|\cos60^\circ = 4|\vec{b}|$: $8|\vec{b}|^2 + 24|\vec{b}| - 320 = 0$, i.e. $|\vec{b}|^2 + 3|\vec{b}| - 40 = 0$.
$|\vec{b}| = 5$ (the root −8 is rejected).
Answer: (B). | ||||
| 79 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let the height be $h$: the lower part is $\frac{3h}{10}$. If each part subtends $\theta$, then $\tan\theta = \frac{3h/10}{18} = \frac{h}{60}$ and $\tan2\theta = \frac{h}{18}$.
$\frac{2(h/60)}{1 - h^2/3600} = \frac{h}{18}$ gives $\frac{36}{60} = 1 - \frac{h^2}{3600}$, so $h^2 = 1440$ and $h = 12\sqrt{10}$ m.
Answer: (A). | ||||
| 80 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter (B) $(p \wedge \sim q) \Rightarrow (p \vee q)$: whenever the left side is true, p is true, so $p \vee q$ is true. A tautology.
(A) $p = T, q = F$: $(T \wedge T) \Rightarrow (T \wedge F)$ is $T \Rightarrow F$, false.
(C) the same values make it $T \Rightarrow F$, false.
(D) $p = F, q = F$: $(F \vee T) \Rightarrow (F \vee F)$ is $T \Rightarrow F$, false.
Answer: (B). | ||||
| 81 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom{12}{r}\left(\frac{x}{4}\right)^{12-r}\left(-\frac{12}{x^2}\right)^r$; the power of $x$ is $12 - 3r = 0$, so $r = 4$.
$T_5 = \binom{12}{4}\frac{12^4}{4^8} = 495\cdot\frac{3^4\cdot4^4}{4^8} = 495\cdot\frac{3^4}{4^4}$.
$\frac{3^6}{4^4}k = 495\cdot\frac{3^4}{4^4}$ gives $k = \frac{495}{9} = 55$.
Answer: 55. | ||||
| 82 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | NumericalHard | |
A hard numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations The segment from $-2$ to $2$ subtends the angle $\frac{\pi}{4}$ at $z$, so $z$ lies on an arc of a circle through $\pm2$. Chord 4 $= 2R\sin\frac{\pi}{4}$ gives $R = 2\sqrt{2}$, with the centre 2 units from the chord.
For $z = iy$ with large $y > 0$, $\arg(z - 2) - \arg(z + 2) > 0$, so the arc is above the real axis and, the angle being acute, it is the major arc with centre $2i$: $|z - 2i| = 2\sqrt{2}$, $\mathrm{Im}\,z > 0$.
The point $9\sqrt{2} + 2i$ is $9\sqrt{2}$ from the centre $2i$, on the horizontal through it; the nearest point of the circle, $2\sqrt{2} + 2i$, lies on the arc.
Minimum distance $= 9\sqrt{2} - 2\sqrt{2} = 7\sqrt{2}$, and its square is 98.
Answer: 98. | ||||
| 83 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | NumericalEasy | |
A easy numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 2x + a$.
Increasing on [1, 2]: $2x + a \ge 0$ for all $x \in [1, 2]$, i.e. $a \ge -2$; so $R = -2$.
Decreasing on [1, 2]: $2x + a \le 0$ for all $x \in [1, 2]$, i.e. $a \le -4$; so $S = -4$.
$|R - S| = 2$.
Answer: 2. | ||||
| 84 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $[2x] = -1$ on $[-\frac{1}{2}, 0)$, $0$ on $[0, \frac{1}{2})$, $1$ on $[\frac{1}{2}, 1)$: $\int_{-1/2}^1[2x]\,dx = -\frac{1}{2} + 0 + \frac{1}{2} = 0$.
$\int_{-1/2}^1|x|\,dx = \frac{1}{8} + \frac{1}{2} = \frac{5}{8}$.
$8\times\frac{5}{8} = 5$.
Answer: 5. | ||||
| 85 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Differentiate: $\phi(x) + x\phi'(x) = 3x^2 - 2\phi'(x)$, i.e. $(x + 2)\phi'(x) + \phi(x) = 3x^2$, which is $\frac{d}{dx}\left[(x + 2)\phi(x)\right] = 3x^2$.
So $(x + 2)\phi(x) = x^3 + C$; $\phi(0) = 4$ gives $C = 8$.
$\phi(2) = \frac{8 + 8}{4} = 4$.
Answer: 4. | ||||
| 86 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Lying between the circles means the centres (1, 1) and (9, 1) are on opposite sides: $(7 - \alpha)(31 - \alpha) < 0$, so $7 < \alpha < 31$.
No chord on the first circle: $\frac{|7 - \alpha|}{5} \ge 1$, so $\alpha \ge 12$ (in this range).
No chord on the second circle: $\frac{|31 - \alpha|}{5} \ge 2$, so $\alpha \le 21$.
Integral values 12, 13, …, 21: sum $= \frac{10(12 + 21)}{2} = 165$.
Answer: 165. | ||||
| 87 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line: $(1 + 2t, 2 + 3t, -1 + 6t)$. In the plane: $2 + 4t - 2 - 3t - 1 + 6t = 6$, so $7t = 7$, $t = 1$: the point (3, 5, 5).
Distance squared from (−1, −1, 2): $4^2 + 6^2 + 3^2 = 61$.
Answer: 61. | ||||
| 88 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability The units work independently. P(instrument fails) $= 1 - 0.9\times0.8 = 0.28$.
P(first fails and second works) $= 0.1\times0.8 = 0.08$.
$p = \frac{0.08}{0.28} = \frac{2}{7}$, so $98p = 28$.
Answer: 28. | ||||
| 89 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations VOWELS has six different letters: vowels O, E and consonants V, W, L, S.
All arrangements: $6! = 720$. With the four consonants together as one block: $3!\times4! = 144$.
Words in which the consonants are not all together: $720 - 144 = 576$.
Answer: 576. | ||||
| 90 | StatisticsMean, median and mode | Class 11: Statistics | NumericalEasy | |
A easy numerical question on mean, median and mode, from Statistics. NCERT: Class 11: Statistics The $k$th number is $(3k + 4)(2k + 6) = 6k^2 + 26k + 24$.
Sum for $k = 1$ to 10: $6\times385 + 26\times55 + 24\times10 = 2310 + 1430 + 240 = 3980$.
Mean $= \frac{3980}{10} = 398$.
Answer: 398. | ||||
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