JEE (Main) · 2021
JEE (Main) 2021 — 31 August, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Laws of Motion2
- Properties of Bulk Matter2
- Kinematics2
18 easy · 12 medium · 0 hard
Chemistry
30 Q
- Hydrocarbons3
- p-Block Elements2
- Haloalkanes and Haloarenes2
20 easy · 10 medium · 0 hard
Mathematics
30 Q
- Integral Calculus3
- Matrices and Determinants2
- Sequences and Series2
8 easy · 20 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 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement From $F = ma = m\frac{V}{T}$, mass is $m = \frac{FT}{V}$.
So $[m] = [\mathrm{FTV^{-1}}]$.
Answer: (D). | ||||
| 2 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The electron is captured into $n = 2$, where the energy is $-\frac{13.6}{4} = -3.4$ eV.
Energy released as the photon: $E = 2.6 - (-3.4) = 6.0$ eV $= 6.0\times1.6\times10^{-19} = 9.6\times10^{-19}$ J.
$\nu = \frac{E}{h} = \frac{9.6\times10^{-19}}{6.6\times10^{-34}} = 1.45\times10^{15}$ Hz $= 1.45\times10^{9}$ MHz.
Answer: (C). | ||||
| 3 | Laws of MotionEquilibrium of concurrent forces | Class 11: Laws of Motion | StatementsEasy | |
A easy statements question on equilibrium of concurrent forces, from Laws of Motion. NCERT: Class 11: Laws of Motion Statement I: $\vec{F_1}+\vec{F_2}+\vec{F_3} = 0$, so three such forces acting at a point (concurrent) have zero resultant and are in equilibrium (triangle law). True.
Statement II: forces forming the sides of a closed triangle taken in the same order add to zero, which is the condition for translatory equilibrium. True.
Answer: (A). | ||||
| 4 | Properties of Bulk MatterHeat transfer | 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 heat transfer, 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 thin shell of radius $r$ and thickness $dr$: $H = -K\,4\pi r^2\frac{d\theta}{dr}$, the same $H$ through every shell.
$H\int_{r_1}^{r_2}\frac{dr}{r^2} = 4\pi K(\theta_2-\theta_1)$, so $H\left(\frac{1}{r_1}-\frac{1}{r_2}\right) = 4\pi K(\theta_2-\theta_1)$.
$H = \frac{4\pi K r_1 r_2(\theta_2-\theta_1)}{r_2-r_1}$ (flowing inward, from the hotter outer shell).
Answer: (A). | ||||
| 5 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | StatementsMedium | |
A medium statements question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Since $\vec{P}\perp\vec{Q}$: $|\vec{P}+\vec{Q}| = |\vec{P}-\vec{Q}| = \sqrt{P^2+Q^2} = a$.
Resultant of two forces of magnitude $a$ at angle $\theta$: $R^2 = 2a^2(1+\cos\theta)$.
$R^2 = 3a^2$: $\cos\theta_1 = \frac{1}{2}$, $\theta_1 = 60^\circ$. $R^2 = 2a^2$: $\cos\theta_2 = 0$, $\theta_2 = 90^\circ$.
So $\theta_1 < \theta_2$ (Statement I true) and the values are $60^\circ$ and $90^\circ$ (Statement II true).
Answer: (A). | ||||
| 6 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation Let $x = \frac{r}{R_E}$. At depth $r$: $g_d = g(1-x)$. At height $r$: $g_h = \frac{g}{(1+x)^2}$.
$\frac{g_d}{g_h} = (1-x)(1+x)^2 = (1-x)(1+2x+x^2) = 1+x-x^2-x^3$.
Answer: (B). | ||||
| 7 | 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 Total moles: $n = \frac{PV}{RT} = \frac{400\times10^3\times500\times10^{-6}}{8.3\times300} \approx 0.080$ mol.
With $m_H + m_O = 0.76$ g and $\frac{m_H}{2}+\frac{m_O}{32} = 0.080$: subtracting $\frac{1}{32}$ of the first, $m_H\left(\frac{1}{2}-\frac{1}{32}\right) = 0.080-0.02375$, so $m_H \approx 0.12$ g and $m_O \approx 0.64$ g.
$\frac{m_O}{m_H} = \frac{0.64}{0.12} = \frac{16}{3}$.
Answer: (D). | ||||
| 8 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Masses $m$ and $2m$. Momentum: $3m\times40 = m\times60 + 2m\,v$, so $v = 30$ m/s.
$K_i = \frac{1}{2}(3m)(40)^2 = 2400m$; $K_f = \frac{1}{2}m(60)^2+\frac{1}{2}(2m)(30)^2 = 1800m+900m = 2700m$.
Fractional change $= \frac{2700m-2400m}{2400m} = \frac{1}{8}$.
Answer: (A). | ||||
| 9 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity The bare wire on the far left joins the top and bottom rails, so the $5\ \Omega$ resistor is short-circuited and the left end of the first $2\ \Omega$ is at the bottom rail's potential.
So the first $2\ \Omega$ (top) is in parallel with the middle vertical $2\ \Omega$: $1\ \Omega$. Adding the second top $2\ \Omega$ in series: $3\ \Omega$.
The two $3\ \Omega$ resistors across A and B are in parallel: $1.5\ \Omega$.
$R_{AB} = 3\parallel1.5 = \frac{3\times1.5}{4.5} = 1\ \Omega$.
Answer: (C). | ||||
| 10 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Buoyancy reduces the effective gravity: $g' = g\left(1-\frac{\rho_l}{\rho}\right) = \frac{3g}{4}$. New length $l' = \frac{4l}{3}$.
$T' = 2\pi\sqrt{\frac{l'}{g'}} = 2\pi\sqrt{\frac{4l/3}{3g/4}} = 2\pi\sqrt{\frac{16l}{9g}} = \frac{4}{3}T$.
Answer: (A). | ||||
| 11 | 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 Load on each column: $F = \frac{50\times10^3\times9.8}{4} = 1.225\times10^5$ N.
Area: $A = \pi(1^2-0.5^2) = 0.75\pi = 2.356\ \mathrm{m^2}$.
Strain $= \frac{F}{AY} = \frac{1.225\times10^5}{2.356\times2.0\times10^{11}} = 2.60\times10^{-7}$.
Answer: (A). | ||||
| 12 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | StatementsEasy | |
A easy statements question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves (a) False: K.E. and P.E. are equal only at $x = \pm\frac{A}{\sqrt{2}}$.
(b) False: over an arbitrary interval the averages differ; they are equal only over whole periods (or half periods).
(c) True: total mechanical energy $\frac{1}{2}kA^2$ is constant.
(d) True: over one period both averages are $\frac{1}{4}kA^2$.
Answer: (D). | ||||
| 13 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves At $t = 0$: $kz = \pi$ for $q_1$ and $kz = 3\pi$ for $q_2$, and $\cos\pi = \cos3\pi = -1$.
So both charges see the same $\vec{B}$ and the same $\vec{E}$ (the wave fields are equal at the two points), and they have the same velocity.
The force $\vec{F} = q(\vec{E}+\vec{v}\times\vec{B})$ is then proportional to $q$: $\frac{F_1}{F_2} = \frac{4\pi}{2\pi} = 2$.
Answer: (C). | ||||
| 14 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Each sphere (solid, as intended): $I = \frac{2}{5}MR^2 + Md^2 = \frac{2}{5}(1.5)(0.5)^2 + 1.5(2.5)^2 = 0.15+9.375 = 9.525\ \mathrm{kg\,m^2}$.
System: $2\times9.525 = 19.05\ \mathrm{kg\,m^2}$.
Answer: (A). | ||||
| 15 | Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current A field parallel to the plane of the coil gives no flux through it, so no current is induced: (C) and (D) are out.
The arrows on the coil show the induced current going anticlockwise as seen, which produces a field coming out of the page inside the coil.
By Lenz's law the induced field supports the original flux when that flux is decreasing, so $\vec{B}$ is outward and decreasing.
Answer: (B). | ||||
| 16 | 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): inputs $V_A$ and $V_B$ feed the output through forward diodes $D_1$, $D_2$ with $R$ to ground. If either input is 5 V, its diode conducts and $V_o$ is high; only with both at 0 V is $V_o$ low. This is an OR gate.
(B): an npn transistor in common-emitter mode with output at the collector. Input high: the transistor saturates and $V_o \approx 0$; input low: cut off and $V_o = 5$ V. This is a NOT gate.
Answer: (D). | ||||
| 17 | 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 Distance of the centroid from each side: $d = \frac{a}{2\sqrt{3}} = \frac{0.09}{2\sqrt{3}} = 0.026$ m.
One side: $B_1 = \frac{\mu_0 I}{4\pi d}(\sin60^\circ+\sin60^\circ) = 10^{-7}\times\frac{1.5}{0.026}\times\sqrt{3} = 1.0\times10^{-5}$ T.
Three sides add: $B = 3\times10^{-5}$ T. A clockwise current gives a field into the page, i.e. 'inside' the plane.
Answer: (A). | ||||
| 18 | Electronic DevicesRectifiers | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | StatementsEasy | |
A easy statements question on rectifiers, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Statement I: a capacitor across the load charges at the peaks and discharges slowly through $R_L$, smoothing the output. True.
Statement II: an inductor in series opposes changes in current and passes dc while blocking the ripple (a choke filter). True.
Answer: (A). | ||||
| 19 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter At the same temperature the momentum is $p = \sqrt{3mkT}\propto\sqrt{m}$; take $\Delta p\sim p$.
$\Delta x \geq \frac{h}{4\pi\Delta p}$, so $\Delta x\propto\frac{1}{\sqrt{m}}$ and $\frac{\Delta x_e}{\Delta x_p} = \sqrt{\frac{m_p}{m_e}}$.
Answer: (C). | ||||
| 20 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | StatementsEasy | |
A easy statements question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance (a) Correct: with the outward normal, entering lines give $\vec{E}\cdot d\vec{A} < 0$.
(b) Correct: by symmetry each face carries $\frac{q}{6\varepsilon_0}$.
(c) Correct: no enclosed charge, so zero net flux.
(d) Incorrect: a field parallel to the surface is perpendicular to $d\vec{A}$, so its flux is zero.
Answer: (D). | ||||
| 21 | 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 Prism angle $A = 60^\circ$. At minimum deviation with $i = A = 60^\circ$, $r = \frac{A}{2} = 30^\circ$, so $\mu = \frac{\sin60^\circ}{\sin30^\circ} = \sqrt{3}$.
$PA = 10\cos30^\circ = 5\sqrt{3}$ cm $= 5\sqrt{3}\times10^{-2}$ m; speed in the prism $v = \frac{c}{\sqrt{3}}$.
$t = \frac{PA}{v} = \frac{5\sqrt{3}\times10^{-2}\times\sqrt{3}}{3\times10^8} = 5\times10^{-10}$ s.
Answer: 5. | ||||
| 22 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $P_2 = P_1\left(\frac{V_1}{V_2}\right)^{\gamma} = 200\times4^{1.5} = 1600$ kPa.
$P_1V_1 = 200\times10^3\times1200\times10^{-6} = 240$ J; $P_2V_2 = 1600\times10^3\times300\times10^{-6} = 480$ J.
$W = \frac{P_1V_1-P_2V_2}{\gamma-1} = \frac{240-480}{0.5} = -480$ J, so $|W| = 480$ J.
Answer: 480. | ||||
| 23 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With the battery connected $V$ stays 200 V and $C$ becomes $2C$.
$\Delta U = \frac{1}{2}(2C-C)V^2 = \frac{1}{2}\times200\times10^{-6}\times(200)^2 = 4$ J (an increase).
Answer: 4. | ||||
| 24 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The two fourth bright fringes are $8\beta$ apart: $8\beta = 2.4$ cm, $\beta = 0.3$ cm $= 3\times10^{-3}$ m.
$\lambda = \frac{\beta d}{D} = \frac{3\times10^{-3}\times0.3\times10^{-3}}{1.5} = 6\times10^{-7}$ m.
$f = \frac{c}{\lambda} = \frac{3\times10^8}{6\times10^{-7}} = 5\times10^{14}$ Hz.
Answer: 5. | ||||
| 25 | Communication SystemsAmplitude modulation and bandwidth | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on amplitude modulation and bandwidth, from Communication Systems. NCERT: No matching NCERT chapter An AM station occupies $2f_m = 2\times6 = 12$ kHz (both sidebands).
Number of stations $= \frac{6\times10^6}{12\times10^3} = 500$.
Answer: 500. | ||||
| 26 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= 1\text{ MSD}-1\text{ VSD} = 1-0.9 = 0.1$ mm.
Reading $= 10+8\times0.1 = 10.8$ mm $= 1.08$ cm. Correcting the positive zero error: $1.08-0.04 = 1.04$ cm (diameter).
Radius $= 0.52$ cm $= 52\times10^{-2}$ cm.
Answer: 52. | ||||
| 27 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $v^2 = u^2+2ax$, so the slope of the $v^2$-$x$ line is $2a$.
From the graph: slope $= \frac{80-20}{30-0} = 2$ (also $\frac{60-40}{20-10} = 2$), so $a = 1\ \mathrm{m/s^2}$.
Answer: 1. | ||||
| 28 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At very high frequency each capacitor ($X_C = \frac{1}{\omega C}\to0$) acts as a plain wire and the inductor ($X_L = \omega L\to\infty$) as a break.
The lower-left branch ($1\ \Omega$ with $20$ H) carries no current. The source current goes through the upper $1\ \Omega$ to the middle junction, which the vertical $0.5$ F capacitor now joins to the lower right branch.
From there the two $2\ \Omega$ resistors (each with a shorted capacitor) are in parallel: $1\ \Omega$.
$Z = 1+1 = 2\ \Omega$.
Answer: 2. | ||||
| 29 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Magnetising field $H = nI$ is unchanged. Magnetisation $M = \chi H$ with $\chi = \mu_r-1$, and the core's moment is $MV$.
Fractional change $= \frac{\chi_2-\chi_1}{\chi_1} = \frac{749-499}{499} = \frac{250}{499}$.
Answer: 250. | ||||
| 30 | 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 $R = \frac{P}{I^2} = \frac{192}{16} = 12\ \Omega$.
With 8 A: $H = I^2Rt = 64\times12\times5 = 3840$ J.
Answer: 3840. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics (A) At constant $T$ and $P$, $\Delta S_{total} = -\frac{\Delta G_{system}}{T}$, so $\frac{\Delta G_{system}}{\Delta S_{total}} = -T$. Correct.
(B) Reversible isothermal work on an ideal gas: $w = -nRT\ln\frac{V_f}{V_i}$. Correct.
(C) $\Delta G^\circ = -RT\ln K$ and $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ give $\ln K = -\frac{\Delta H^\circ - T\Delta S^\circ}{RT}$; the minus sign is missing. Incorrect.
(D) $K = e^{-\Delta G^\circ/RT}$. Correct.
Answer: (C). | ||||
| 32 | ElectrochemistryConductance | Class 12: Electrochemistry | Matching listEasy | |
A easy matching list question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell constant $\frac{l}{A}$: $\mathrm{m^{-1}}$ (iii). Molar conductivity: $\mathrm{S\,cm^2\,mol^{-1}}$ (i). Conductivity $\kappa$: $\mathrm{\Omega^{-1}m^{-1}}$ (iv). Degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$: dimensionless (ii).
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).
Answer: (B). | ||||
| 33 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Iodine oxides are the most stable among halogen oxides (greater polarity of the I-O bond); chlorine oxides come next, helped by multiple bonding involving d orbitals; bromine oxides are the least stable (the middle row anomaly).
Order: I > Cl > Br.
Answer: (A). | ||||
| 34 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter (A) True: H-H (435.88 kJ/mol) has the highest bond enthalpy of any single bond between two atoms of an element.
(B) False: at about 2000 K dihydrogen is only about 0.081% dissociated into atoms, not 8.1%.
(C) True: $\mathrm{Zn + 2HCl \rightarrow ZnCl_2 + H_2}$ and $\mathrm{Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2}$.
(D) True: atomic hydrogen is formed by irradiating $\mathrm{H_2}$ at high temperature with UV light (or in an electric arc).
Answer: (B). | ||||
| 35 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter (A) True: lithium salts are commonly hydrated, e.g. $\mathrm{LiCl\cdot2H_2O}$.
(R) True: the very small $\mathrm{Li^+}$ ion has the highest charge density, hence the highest polarising power in the group.
That high charge density makes $\mathrm{Li^+}$ attract water molecules strongly (highest hydration enthalpy), so (R) explains (A).
Answer: (A). | ||||
| 36 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Sulphurous acid $\mathrm{H_2SO_3}$, $\mathrm{OS(OH)_2}$: one S=O.
Peroxodisulphuric acid $\mathrm{H_2S_2O_8}$, $\mathrm{HO-SO_2-O-O-SO_2-OH}$: two S=O on each S, four in all.
Pyrosulphuric acid $\mathrm{H_2S_2O_7}$, $\mathrm{HO-SO_2-O-SO_2-OH}$: four S=O.
So 1, 4 and 4.
Answer: (B). | ||||
| 37 | 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 Eu (Z = 63): $\mathrm{[Xe]}4f^76s^2$. Removing the two 6s electrons gives $\mathrm{Eu^{2+}}$: $\mathrm{[Xe]}4f^7$, a stable half-filled shell.
In spite of this it readily changes to the common +3 state, so it is a strong reducing agent.
Answer: (A). | ||||
| 38 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctMedium | |
A medium single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions A species can disproportionate only if its element is in an intermediate oxidation state that can both rise and fall.
Cannot: $\mathrm{F_2}$ (fluorine has no positive state), $\mathrm{ClO_4^-}$ (Cl +7, highest), $\mathrm{Cr_2O_7^{2-}}$ (Cr +6, highest), $\mathrm{MnO_4^-}$ (Mn +7, highest).
Can: $\mathrm{ClO_2^-}$ (Cl +3), $\mathrm{Cl_2}$ (0), $\mathrm{Mn^{3+}}$ ($\mathrm{2Mn^{3+} \rightarrow Mn^{2+} + MnO_2}$), and manganate $\mathrm{MnO_4^{2-}}$ (Mn +6, $\mathrm{3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O}$).
Options (A), (B) and (C) each contain $\mathrm{F_2}$, $\mathrm{ClO_4^-}$ or $\mathrm{Cr_2O_7^{2-}}$; only (D) can be all-disproportionating, with its first species read as manganate $\mathrm{MnO_4^{2-}}$.
Answer: (D). | ||||
| 39 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Oxidation state of Fe: $x + 0 - 2 = +1$, so $x = +3$, $3d^5$.
CO is a strong-field ligand, so the complex is low spin: $t_{2g}^5e_g^0$, one unpaired electron.
$\mu = \sqrt{n(n+2)} = \sqrt{3} = 1.73$ BM.
Answer: (B). | ||||
| 40 | 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 Ammonium salts in rain drops reach the ground as wet deposition.
$\mathrm{SO_2}$ is also absorbed directly on solid and liquid ground surfaces and so is deposited as dry deposition.
X = ammonium salts, Y = $\mathrm{SO_2}$.
Answer: (A). | ||||
| 41 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Alcoholic NaOH ($\mathrm{NaOH/C_2H_5OH}$) brings about dehydrohalogenation (E2) rather than substitution.
The C-Cl carbon is allylic (next to the ring double bond). Removing H from the neighbouring CH(CH3) carbon and Cl gives a new double bond conjugated with the old one: 1-methylcyclopenta-1,3-diene.
The conjugated, more substituted diene (Saytzeff product) is the major product.
Answer: (B). | ||||
| 42 | HydrocarbonsConformations of ethane | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on conformations of ethane, from Hydrocarbons. NCERT: Class 11: Hydrocarbons I: the two $\mathrm{CH_3}$ groups are 180° apart (anti), the lowest energy.
III: staggered with the $\mathrm{CH_3}$ groups 60° apart (gauche), slightly higher.
IV: eclipsed with each $\mathrm{CH_3}$ eclipsing an H (partially eclipsed), higher still.
II: fully eclipsed, $\mathrm{CH_3}$ eclipsing $\mathrm{CH_3}$, the highest.
Increasing energy: I < III < IV < II.
Answer: (C). | ||||
| 43 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes 1. $\mathrm{Br_2/Fe}$: bromination gives bromobenzene, $\mathrm{C_6H_5Br}$.
2. Mg/dry ether: phenylmagnesium bromide, $\mathrm{C_6H_5MgBr}$.
3. $\mathrm{CH_3OH}$ has an acidic H, which protonates the Grignard reagent: $\mathrm{C_6H_5MgBr + CH_3OH \rightarrow C_6H_6 + Mg(OCH_3)Br}$.
Answer: (A). | ||||
| 44 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids 4-(Hydroxymethyl)benzaldehyde has no $\alpha$-hydrogen, so hot NaOH brings about the Cannizzaro reaction.
One molecule is reduced to 1,4-benzenedimethanol (a diol); the other is oxidised to 4-(hydroxymethyl)benzoate, which $\mathrm{H_3O^+}$ turns into 4-(hydroxymethyl)benzoic acid, a monocarboxylic acid that also has an alcohol group.
The $\mathrm{CH_2OH}$ group is not oxidised, so no dicarboxylic acid forms.
Answer: (B). | ||||
| 45 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Friedel-Crafts acylation of benzene by succinic anhydride ($\mathrm{AlCl_3}$, 2 eq) opens the anhydride: A = $\mathrm{C_6H_5COCH_2CH_2COOH}$ (4-oxo-4-phenylbutanoic acid).
Clemmensen reduction (Zn/Hg, HCl) reduces the ketone C=O to $\mathrm{CH_2}$ and leaves the COOH: B = $\mathrm{C_6H_5CH_2CH_2CH_2COOH}$ (4-phenylbutanoic acid).
Answer: (A). | ||||
| 46 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Nitration (conc. $\mathrm{HNO_3}$ + conc. $\mathrm{H_2SO_4}$): A = nitrobenzene.
$\mathrm{Cl_2}$/anhyd. $\mathrm{AlCl_3}$: $\mathrm{-NO_2}$ is a meta director, so B = 3-chloronitrobenzene.
Fe/HCl reduces $\mathrm{-NO_2}$ to $\mathrm{-NH_2}$: C = 3-chloroaniline.
Answer: (A). | ||||
| 47 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Aniline with acetic anhydride is acetylated on nitrogen: A = acetanilide, $\mathrm{C_6H_5NHCOCH_3}$ (no Friedel-Crafts on the ring; the amino group is the reactive site).
The $\mathrm{-NHCOCH_3}$ group is a moderate ortho/para director; with $\mathrm{Br_2}$ in $\mathrm{CH_3COOH}$ at room temperature it gives mainly the para monobromo product: B = 4-bromoacetanilide.
Answer: (A). | ||||
| 48 | 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 Glyptal (ethylene glycol + phthalic acid), Dacron (ethylene glycol + terephthalic acid) and PHBV (3-hydroxybutanoic + 3-hydroxypentanoic acids) are polyesters.
Novolac is a phenol-formaldehyde resin (the linear precursor of Bakelite): no ester links.
Answer: (B). | ||||
| 49 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Keratin (hair, wool), myosin (muscle) and collagen are fibrous proteins.
Albumin is a globular protein (soluble in water).
Answer: (B). | ||||
| 50 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter $\mathrm{Mn^{2+}}$: precipitated as MnS by $\mathrm{H_2S}$ in ammoniacal medium, Group IV (iii).
$\mathrm{As^{3+}}$: sulphide soluble in yellow ammonium sulphide, Group IIB (iv).
$\mathrm{Cu^{2+}}$: sulphide insoluble in yellow ammonium sulphide, Group IIA (ii).
$\mathrm{Al^{3+}}$: $\mathrm{Al(OH)_3}$ with $\mathrm{NH_4Cl/NH_4OH}$, Group III (i).
Answer: (C). | ||||
| 51 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{Na_2O + H_2O \rightarrow 2NaOH}$. Moles of $\mathrm{Na_2O} = \frac{20.0}{62} = 0.3226$; NaOH $= 0.6452$ mol.
Molarity $= \frac{0.6452}{0.500} = 1.290$ M $= 12.9\times10^{-1}$ M, nearest integer 13.
Answer: 13. | ||||
| 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 In a ccp array there is one octahedral void per anion.
All octahedral sites filled: cations : anions = 1 : 1, formula AB, so $x = 1$.
Answer: 1. | ||||
| 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 Zn: $\mathrm{[Ar]}3d^{10}4s^2$; $\mathrm{Zn^+}$: $\mathrm{[Ar]}3d^{10}4s^1$.
The outermost electron is in 4s: $l = 0$, so $m_l = 0$.
Answer: 0. | ||||
| 54 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{O_2^{2-}}$ has 18 electrons: $\sigma1s^2\sigma^*1s^2\sigma2s^2\sigma^*2s^2\sigma2p_z^2\pi2p_x^2\pi2p_y^2\pi^*2p_x^2\pi^*2p_y^2$.
Both $\pi^*$ orbitals are full, so there are no unpaired electrons (diamagnetic).
Answer: 0. | ||||
| 55 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions In acetone (monomer): $0.17 = 1.7\times\frac{1.22/M}{0.100}$, so $\frac{1.22}{M} = 0.010$ mol and $M = 122$ g/mol.
In benzene the acid dimerises completely: $i = \frac{1}{2}$.
$\Delta T_b = iK_bm = \frac{1}{2}\times2.6\times\frac{0.010}{0.100} = 0.13$°C $= 13\times10^{-2}$°C.
Answer: 13. | ||||
| 56 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Excess $\mathrm{H^+}$: $50 - 30 = 20$ mmol in $80$ mL, so $[\mathrm{H^+}] = 0.25$ M.
$\mathrm{pH} = -\log0.25 = -\log\frac{2.5}{10} = 1 - 0.3979 = 0.6021 = 6021\times10^{-4}$.
Answer: 6021. | ||||
| 57 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Arrhenius: $\log_{10}k = \log_{10}A - \frac{E_a}{2.303RT}$, so $\frac{E_a}{2.303R} = 2.47\times10^3$ K.
$E_a = 2.303\times8.314\times2470 = 4.73\times10^4$ J/mol $= 47.3$ kJ/mol, nearest integer 47.
Answer: 47. | ||||
| 58 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter $V = kP^{1/n}$. From the two data: $\frac{15}{10} = 2^{1/n}$, so $\frac{1}{n} = \frac{\log1.5}{\log2} = \frac{0.4771-0.3010}{0.3010} = 0.5850$.
At 300 mm: $V = 10\times3^{1/n}$, $\log V = 1 + 0.5850\times0.4771 = 1.2791$.
So $x = 1.2791 = 127.9\times10^{-2}$, nearest integer 128.
Answer: 128. | ||||
| 59 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In the electrolytic refining of copper the anode mud contains antimony, selenium, tellurium, silver, gold and platinum: Sb, Se, Te, Ag, Au, Pt.
Pb and Ru are not among the main impurities listed for the anode mud.
Answer: 6. | ||||
| 60 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Balancing atoms: C gives $x = 2$; H gives $2\times\frac{y}{2} = 7$, so $y = 7$; N gives $z = 1$.
(Check O: $2x+\frac{y}{2} = 4+3.5 = 7.5$ on both sides.)
Answer: 7. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Putting $m = n = 1$ repeatedly: $f(n) = nf(1)$ for every $n \in \mathbf{N}$.
$f(6) = 6f(1) = 18$, so $f(1) = 3$ and $f(n) = 3n$.
$f(2)\cdot f(3) = 6\times9 = 54$.
Answer: (D). | ||||
| 62 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\frac{z-i}{z-1}$ purely imaginary means $\arg$ is $\pm\frac{\pi}{2}$: the segments from $z$ to $i$ and to $1$ are perpendicular, so $z$ lies on the circle with diameter joining $(0,1)$ and $(1,0)$.
Centre $\left(\frac{1}{2},\frac{1}{2}\right)$, radius $\frac{1}{\sqrt{2}}$.
Distance of $(3,3)$ from the centre: $\sqrt{2}\times\frac{5}{2} = \frac{5}{\sqrt{2}}$.
Minimum $= \frac{5}{\sqrt{2}}-\frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$, reached at $z = 1+i$, where $\frac{z-i}{z-1} = \frac{1}{i} = -i$ is indeed purely imaginary.
Answer: (B). | ||||
| 63 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $2\log_4(10-2^{-x}) = \log_2(10-2^{-x})$, so the equation is $\log_2\frac{2^{x+1}(10-2^{-x})}{(3+2^x)^2} = 0$.
With $t = 2^x$: $2t\left(10-\frac{1}{t}\right) = (3+t)^2$, i.e. $20t-2 = t^2+6t+9$, so $t^2-14t+11 = 0$.
Both roots $t = 7\pm\sqrt{38}$ are positive and keep $10-\frac{1}{t} > 0$, so both give real $x$.
$x_1+x_2 = \log_2(t_1t_2) = \log_2 11$.
Answer: (A). | ||||
| 64 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}1&\cos\gamma&\cos\beta\\\cos\gamma&1&\cos\alpha\\\cos\beta&\cos\alpha&1\end{vmatrix} = 1-\cos^2\alpha-\cos^2\beta-\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma$.
With $\gamma = 2\pi-(\alpha+\beta)$, $\cos\gamma = \cos(\alpha+\beta)$, and
$\cos^2\alpha+\cos^2\beta+\cos^2(\alpha+\beta)-2\cos\alpha\cos\beta\cos(\alpha+\beta) = \cos^2\alpha+\cos^2\beta-\cos(\alpha+\beta)\cos(\alpha-\beta) = \cos^2\alpha+\cos^2\beta-(\cos^2\alpha-\sin^2\beta) = 1$.
So $\Delta = 0$. The system is homogeneous, hence consistent, and with $\Delta = 0$ it has infinitely many solutions.
Answer: (A). | ||||
| 65 | 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 $\frac{S_{10}}{S_p} = \frac{10^2}{p^2}$ for $p \neq 10$ means $S_n = kn^2$.
$\frac{10}{2}(2a+9d)\cdot p^2 = 100\cdot\frac{p}{2}(2a+(p-1)d)$ gives $p(2a+9d) = 10(2a+(p-1)d)$, i.e. $(p-10)(2a-d) = 0$, so $d = 2a$.
Then $a_n = a+(n-1)2a = (2n-1)a$ and $\frac{a_{11}}{a_{10}} = \frac{21}{19}$.
Answer: (B). | ||||
| 66 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Intersection: $\frac{x^2}{a^2}+\frac{ab-x^2}{b^2} = 1$ gives $x^2 = \frac{a^2b}{a+b}$, $y^2 = \frac{ab^2}{a+b}$, so $\frac{x^2}{y^2} = \frac{a}{b}$.
Slopes: ellipse $m_1 = -\frac{b^2x}{a^2y}$, circle $m_2 = -\frac{x}{y}$.
$m_1-m_2 = \frac{x}{y}\cdot\frac{a^2-b^2}{a^2}$ and $1+m_1m_2 = 1+\frac{b^2}{a^2}\cdot\frac{x^2}{y^2} = 1+\frac{b}{a} = \frac{a+b}{a}$.
$\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right| = \sqrt{\frac{a}{b}}\cdot\frac{(a-b)(a+b)}{a^2}\cdot\frac{a}{a+b} = \frac{a-b}{\sqrt{ab}}$.
Answer: (A). | ||||
| 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 $\tan^3x-\tan x = \tan x\cdot\frac{\sin^2x-\cos^2x}{\cos^2x} = -\frac{\tan x(\cos x-\sin x)(\cos x+\sin x)}{\cos^2x}$ and $\cos\left(x+\frac{\pi}{4}\right) = \frac{\cos x-\sin x}{\sqrt{2}}$.
$\alpha = \lim_{x\to\pi/4}\left(-\frac{\sqrt{2}\tan x(\cos x+\sin x)}{\cos^2x}\right) = -\frac{\sqrt{2}\cdot1\cdot\sqrt{2}}{1/2} = -4$.
$\beta = e^{\lim_{x\to0}\cot x(\cos x-1)} = e^{\lim_{x\to0}\left(-\frac{\cos x\,(1-\cos x)}{\sin x}\right)} = e^0 = 1$.
Roots $-4$ and $1$: product $-\frac{4}{a} = -4$ gives $a = 1$; sum $-\frac{b}{a} = -3$ gives $b = 3$.
Answer: (C). | ||||
| 68 | 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 Mean value theorem on $[0,1]$ and $[1,2]$: $f'(c_1) = 1$ for some $c_1 \in (0,1)$ and $f'(c_2) = 1$ for some $c_2 \in (1,2)$.
Rolle's theorem for $f'$ on $[c_1, c_2]$: $f''(c) = 0$ for some $c \in (c_1,c_2) \subset (0,2)$. So (C) always holds.
The others need not hold: $f(x) = x$ has $f' = 1$ everywhere and $f'' = 0$, so (A) and (B) fail; $f(x) = x+x(x-1)(x-2)$ passes through the three points with $f''(x) = 6x-6$, not zero throughout, so (D) fails.
Answer: (C). | ||||
| 69 | 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 $[0,1)$: $[x] = 0$ and the factor is $1$: $\int_0^1\sin\frac{\pi x}{2}dx = \frac{2}{\pi}$.
On $[1,2)$: $[x] = 1$ and the factor is $x-1$. With $t = x-1$: $\int_0^1 t\sin\left(\frac{\pi t}{2}+\frac{\pi}{2}\right)dt = \int_0^1 t\cos\frac{\pi t}{2}dt = \left[\frac{2t}{\pi}\sin\frac{\pi t}{2}\right]_0^1-\frac{2}{\pi}\int_0^1\sin\frac{\pi t}{2}dt = \frac{2}{\pi}-\frac{4}{\pi^2}$.
Total $= \frac{4}{\pi}-\frac{4}{\pi^2}$; times $\pi^2$: $4\pi-4 = 4(\pi-1)$.
Answer: (D). | ||||
| 70 | 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 Cancel $2^x$: $\frac{dy}{dx} = \frac{y+2^y}{1+2^y\log_e2}$, so $dx = \frac{1+2^y\log_e2}{y+2^y}dy$.
The numerator is the derivative of the denominator: $x = \log_e(y+2^y)+C$. With $y(0) = 0$: $0 = \log_e1+C$, $C = 0$.
At $y = 1$: $x = \log_e3 \approx 1.099 \in (1, 2)$.
Answer: (C). | ||||
| 71 | 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 Put $v = \frac{y^2}{x^2}$, $y^2 = vx^2$: $2y\frac{dy}{dx} = x^2\frac{dv}{dx}+2vx$, so $y\frac{dy}{dx} = \frac{x^2}{2}\frac{dv}{dx}+vx$.
The equation becomes $\frac{x^2}{2}\frac{dv}{dx} = x\frac{\phi(v)}{\phi'(v)}$, i.e. $\frac{\phi'(v)}{\phi(v)}dv = \frac{2}{x}dx$.
$\ln\phi(v) = 2\ln x+c$, so $\phi\left(\frac{y^2}{x^2}\right) = kx^2$. At $x = 1$, $y = -1$: $v = 1$, $k = \phi(1)$.
$\phi\left(\frac{y^2}{4}\right)$ is the case $x = 2$: $\phi\left(\frac{y^2}{4}\right) = 4\phi(1)$.
Answer: (C). | ||||
| 72 | 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 Area $= \frac{1}{2}|5(2-\beta)+3(\beta-6)+\alpha(6-2)| = \frac{1}{2}|4\alpha-2\beta-8| = |2\alpha-\beta-4| = 12$.
So A is the pair of lines $2x-y = 16$ and $2x-y = -8$.
Their distances from the origin are $\frac{16}{\sqrt{5}}$ and $\frac{8}{\sqrt{5}}$; the least is $\frac{8}{\sqrt{5}}$.
Answer: (A). | ||||
| 73 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Midpoint $(h,k)$ of $(-3,-5)$ and $(x_1,y_1)$: $x_1 = 2h+3$, $y_1 = 2k+5$.
$\frac{(2h+3)^2}{4}+\frac{(2k+5)^2}{9} = 1$ gives $9(4h^2+12h+9)+4(4k^2+20k+25) = 36$.
$36h^2+16k^2+108h+80k+145 = 0$.
Answer: (D). | ||||
| 74 | 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 Both planes pass through the origin; the line's direction is $(2,3,2)\times(1,-2,1) = (7, 0, -7)$, i.e. $(1,0,-1)$.
For $P(-1,2,-2)$: $\vec{OP}\times\hat{d}$ with $\vec{d} = (1,0,-1)$: $(-1,2,-2)\times(1,0,-1) = (-2,-3,-2)$, of length $\sqrt{17}$.
Distance $= \frac{\sqrt{17}}{\sqrt{2}} = \frac{\sqrt{34}}{2}$.
Answer: (C). | ||||
| 75 | 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 Let $|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = k$. $\vec{a}\times(\vec{u}\times\vec{a}) = k\vec{u}-(\vec{a}\cdot\vec{u})\vec{a}$.
Using $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0$, the sum is $k(3\vec{r}-\vec{a}-\vec{b}-\vec{c})-[(\vec{a}\cdot\vec{r})\vec{a}+(\vec{b}\cdot\vec{r})\vec{b}+(\vec{c}\cdot\vec{r})\vec{c}]$.
For a mutually perpendicular set of equal length, $(\vec{a}\cdot\vec{r})\vec{a}+(\vec{b}\cdot\vec{r})\vec{b}+(\vec{c}\cdot\vec{r})\vec{c} = k\vec{r}$.
So $k(2\vec{r}-\vec{a}-\vec{b}-\vec{c}) = \vec{0}$ and $\vec{r} = \frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$.
Answer: (B). | ||||
| 76 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability An onto function from S to S is a bijection: $6! = 720$ of them.
$g(3) = 2g(1)$ allows $(g(1), g(3)) = (1,2), (2,4), (3,6)$: 3 ways; the other four elements map bijectively onto the remaining four values in $4! = 24$ ways.
Probability $= \frac{3\times24}{720} = \frac{1}{10}$.
Answer: (C). | ||||
| 77 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $\sum x = 56$, $\sum x^2 = 7(16+64) = 560$.
Remaining five: $\sum x = 56-14 = 42$, $\sum x^2 = 560-36-64 = 460$.
Variance $= \frac{460}{5}-\left(\frac{42}{5}\right)^2 = 92-\frac{1764}{25} = \frac{536}{25}$.
Answer: (B). | ||||
| 78 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sec^2x = 1+\tan^2x$, so with $t = 32^{\tan^2x}$: $t+32t = 81$, $t = \frac{27}{11}$.
On $\left[0,\frac{\pi}{4}\right]$, $\tan^2x$ rises from 0 to 1, so $t$ rises from 1 to 32 one-to-one.
Since $1 < \frac{27}{11} < 32$, exactly one $x$ works.
Answer: (B). | ||||
| 79 | 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 $\cos^{-1}$: $-1 \leq \frac{x-1}{x+1} \leq 1$. The right inequality gives $-\frac{2}{x+1} \leq 0$, $x > -1$; the left gives $\frac{2x}{x+1} \geq 0$. Together $x \geq 0$.
$\sin^{-1}$ ($x \neq 1$): $\frac{3x^2+x-1}{(x-1)^2} \leq 1$ gives $2x^2+3x-2 \leq 0$, $-2 \leq x \leq \frac{1}{2}$; $\geq -1$ gives $4x^2-x \geq 0$, $x \leq 0$ or $x \geq \frac{1}{4}$.
Intersection: $\{0\}\cup\left[\frac{1}{4},\frac{1}{2}\right]$.
Answer: (C). | ||||
| 80 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $\sim(A\Rightarrow B) \equiv A\wedge\sim B$.
$\sim[(p\vee r)\Rightarrow(q\vee r)] \equiv (p\vee r)\wedge\sim q\wedge\sim r \equiv p\wedge\sim q\wedge\sim r$ (with $\sim r$, $p\vee r$ reduces to $p$).
Answer: (C). | ||||
| 81 | 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 $(I-A)^3 = I-3A+3A^2-A^3$, so the condition is $A^2 = A$.
$A^2 = \begin{pmatrix}a^2&b(a+d)\\0&d^2\end{pmatrix}$: need $a^2 = a$, $d^2 = d$ (so $a, d \in \{0,1\}$) and $b(a+d) = b$.
$a = d = 0$: $b = 0$ (1 matrix). $a = d = 1$: $2b = b$, $b = 0$ (1). $a+d = 1$ (two ways): any $b$ (3 each, 6).
Total $1+1+6 = 8$.
Answer: 8. | ||||
| 82 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations 4-digit numbers: 9000.
Multiples of 3: $3333-333 = 3000$. Multiples of 7: $1428-142 = 1286$. Multiples of 21: $476-47 = 429$.
Multiples of 3 or 7: $3000+1286-429 = 3857$.
Neither: $9000-3857 = 5143$.
Answer: 5143. | ||||
| 83 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem General term: $\frac{10!}{i!\,j!\,k!}a^i(2b)^j(4ab)^k$ with $i+j+k = 10$.
Power of a: $i+k = 7$; power of b: $j+k = 8$. Adding, $10+k = 15$: $k = 5$, $i = 2$, $j = 3$.
Coefficient $= \frac{10!}{2!\,3!\,5!}\cdot2^3\cdot4^5 = 2520\cdot2^{13} = 315\cdot2^3\cdot2^{13} = 315\cdot2^{16}$.
Answer: 315. | ||||
| 84 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalHard | |
A hard numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Numerators 7, 9, 13, 19, … have differences 2, 4, 6, …: $T_n = n^2-n+7$.
With $x = \frac{1}{5}$: $\sum x^n = \frac{x}{1-x} = \frac{1}{4}$, $\sum nx^n = \frac{x}{(1-x)^2} = \frac{5}{16}$, $\sum n^2x^n = \frac{x(1+x)}{(1-x)^3} = \frac{15}{32}$.
$S = \frac{15}{32}-\frac{5}{16}+\frac{7}{4} = \frac{15-10+56}{32} = \frac{61}{32}$.
$160S = 5\times61 = 305$.
Answer: 305. | ||||
| 85 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $f(x) = ax^3+bx^2+cx+d$. $f'$ has a minimum at $-1$: $f''(-1) = -6a+2b = 0$, $b = 3a$ (and $a > 0$).
$f'(1) = 3a+2b+c = 0$: $c = -9a$.
$f(1) = -5a+d = -10$ and $f(-1) = 11a+d = 6$: $a = 1$, $d = -5$.
$f(x) = x^3+3x^2-9x-5$ ($f''(1) = 12 > 0$, a minimum indeed). $f(3) = 27+27-27-5 = 22$.
Answer: 22. | ||||
| 86 | Integral CalculusPartial fractions | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on partial fractions, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Divide by $\cos^3x$: $\int\frac{\tan x\sec^2x}{1+\tan^3x}dx = \int\frac{t\,dt}{1+t^3}$ with $t = \tan x$.
$\frac{t}{(t+1)(t^2-t+1)} = -\frac{1}{3}\cdot\frac{1}{t+1}+\frac{1}{3}\cdot\frac{t+1}{t^2-t+1}$.
$\frac{1}{3}\int\frac{t+1}{t^2-t+1}dt = \frac{1}{6}\ln(t^2-t+1)+\frac{1}{2}\int\frac{dt}{(t-\frac{1}{2})^2+\frac{3}{4}} = \frac{1}{6}\ln(t^2-t+1)+\frac{1}{\sqrt{3}}\tan^{-1}\frac{2t-1}{\sqrt{3}}$.
So $\alpha = -\frac{1}{3}$, $\beta = \frac{1}{6}$, $\gamma = \frac{1}{\sqrt{3}}$; $\alpha+\beta+\gamma^2 = -\frac{1}{3}+\frac{1}{6}+\frac{1}{3} = \frac{1}{6}$.
$18\times\frac{1}{6} = 3$.
Answer: 3. | ||||
| 87 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Area of the region: $\int_0^{3/2}(1+4x-x^2)dx = \frac{3}{2}+\frac{9}{2}-\frac{9}{8} = \frac{39}{8}$.
The line $y = mx$ cuts off the triangle under it from $x = 0$ to $\frac{3}{2}$ (it stays below the curve there): area $\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{3m}{2} = \frac{9m}{8}$.
$\frac{9m}{8} = \frac{39}{16}$ gives $m = \frac{13}{6}$; check: at $x = \frac{3}{2}$ the line is at $3.25$, below the curve's $4.75$.
$12m = 26$.
Answer: 26. | ||||
| 88 | CirclesStandard and general form | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Centre $B(1,-2)$, radius $r = \sqrt{1+4-1} = 2$. $AB = \sqrt{4+9} = \sqrt{13}$.
PQ is the chord of contact, perpendicular to AB. Distance of B from PQ $= \frac{r^2}{AB} = \frac{4}{\sqrt{13}}$; distance of A from PQ $= \sqrt{13}-\frac{4}{\sqrt{13}} = \frac{9}{\sqrt{13}}$.
The triangles share the base PQ: ratio of areas $= \frac{9}{4}$. $8\times\frac{9}{4} = 18$.
Answer: 18. | ||||
| 89 | Conic SectionsParabola | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent to $y^2 = 8x$ at $(x_1,y_1)$: $yy_1 = 4(x+x_1)$; at $(2,-4)$: $-4y = 4(x+2)$, i.e. $x+y+2 = 0$.
Tangent to $x^2+y^2 = a$: distance from the origin $= \frac{2}{\sqrt{2}} = \sqrt{2} = \sqrt{a}$, so $a = 2$.
Answer: 2. | ||||
| 90 | 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 The line's direction is perpendicular to the plane's normal: $\alpha-15-4 = 0$, $\alpha = 19$.
Its point $(2,2,-2)$ lies on the plane: $2+6+4+\beta = 0$, $\beta = -12$.
$\alpha+\beta = 7$.
Answer: 7. | ||||
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