JEE (Main) · 2021

JEE (Main) 2021 — 25 July, Shift 2

Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.

Sit this paper free All of 2021

Summary

The heaviest chapters in each subject, and how the questions were spread by level.

Physics
30 Q
1. Kinematics4
2. Current Electricity3
3. Electrostatics2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
1. Aldehydes, Ketones and Carboxylic Acids3
2. Chemical Bonding and Molecular Structure2
3. d- and f-Block Elements2
24 easy · 6 medium · 0 hard
Mathematics
30 Q
1. Binomial Theorem5
2. Vector Algebra3
3. Matrices and Determinants2
14 easy · 16 medium · 0 hard

Questions

Tap Explain on any question for a short summary and our full working.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
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 The argument of a cosine or sine is dimensionless, so $[B] = L^{-1}$ and $[D] = T^{-1}$. Each term has the dimensions of force, so $[A] = MLT^{-2}$. $\left[\frac{AD}{B}\right] = \frac{MLT^{-2}\cdot T^{-1}}{L^{-1}} = ML^2T^{-3}$. Answer: (D).
2	KinematicsMotion in a straight line	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane When dropped, the object shares the balloon's velocity, 10 m/s upwards. Taking up as positive, it reaches the ground when $-75 = 10t - 5t^2$, i.e. $t^2 - 2t - 15 = 0$, so $t = 5$ s. In these 5 s the balloon rises $10\times5 = 50$ m, so its height is $75 + 50 = 125$ m. Answer: (A).
3	KinematicsMotion in a straight line	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane With $\alpha, \beta > 0$ the velocity does not change sign between 1 s and 2 s, so the distance is $\int_1^2 (\alpha t + \beta t^2)\,dt = \frac{\alpha}{2}(4-1) + \frac{\beta}{3}(8-1) = \frac{3}{2}\alpha + \frac{7}{3}\beta$. Answer: (B).
4	KinematicsMotion in a straight line	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctMedium
A medium single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\frac{dt}{dx} = 2mx + n$, so $v = \frac{1}{2mx+n}$. $a = v\frac{dv}{dx} = v\cdot\left(-\frac{2m}{(2mx+n)^2}\right) = -2mv^3$. The acceleration is negative, so the retardation is $2mv^3$. Answer: (A).
5	KinematicsScalars and vectors	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Single correctEasy
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane With $\|\vec{X}\| = \|\vec{Y}\| = a$ and angle $\theta$: $\|\vec{X}-\vec{Y}\|^2 = 2a^2(1-\cos\theta)$ and $\|\vec{X}+\vec{Y}\|^2 = 2a^2(1+\cos\theta)$. $1 - \cos\theta = n^2(1+\cos\theta)$ gives $\cos\theta = \frac{1-n^2}{1+n^2} = \frac{n^2-1}{-n^2-1}$. Answer: (A).
6	Laws of MotionNewton's laws	Class 11: Laws of Motion	Single correctEasy
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $\vec{a} = \vec{F}/m = (8\hat{i} + 2\hat{j})$ m/s$^2$. Starting from rest at the origin, $\vec{r} = \frac{1}{2}\vec{a}t^2 = 50(8\hat{i}+2\hat{j}) = (400\hat{i} + 100\hat{j})$ m. Answer: (A).
7	GravitationAcceleration due to gravity	Class 11: Gravitation	Single correctEasy
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation Equal density means $M \propto R^3$, so $R \propto M^{1/3}$. Then $g = \frac{GM}{R^2} \propto \frac{M}{M^{2/3}} = M^{1/3}$. Doubling the mass multiplies $g$, and so the weight, by $2^{1/3}$: the weight is $2^{1/3}W$. Answer: (D).
8	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctMedium
A medium single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter In vacuum the pressure inside a soap bubble is $P = \frac{4T}{r}$ (two surfaces). At constant temperature the amount of air, $PV$, is conserved: $\frac{4T}{r_1}\cdot\frac{4}{3}\pi r_1^3 + \frac{4T}{r_2}\cdot\frac{4}{3}\pi r_2^3 = \frac{4T}{R}\cdot\frac{4}{3}\pi R^3$, so $R^2 = r_1^2 + r_2^2$. $R = \sqrt{r_1^2 + r_2^2}$. Answer: (D).
9	ThermodynamicsSecond law	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\frac{1}{6} = 1 - \frac{T_2}{T_1}$ and $\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$. Subtracting, $\frac{1}{6} = \frac{62}{T_1}$, so $T_1 = 372$ K $= 99$°C. Answer: (C).
10	Oscillations and WavesSHM	Class 11: Oscillations; Class 11: Waves	Single correctEasy
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At $x = A/2$, $KE = \frac{1}{2}k(A^2 - x^2) = \frac{1}{2}kA^2\left(1 - \frac{1}{4}\right)$, which is $\frac{3}{4}$ of the total energy $\frac{1}{2}kA^2$. Answer: (A).
11	ElectrostaticsConductors and dielectrics	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The field in the dielectric is the free-charge field reduced by $k$: $\frac{q_f - q_b}{A\varepsilon_0} = \frac{q_f}{kA\varepsilon_0}$. So $q_b = q_f\left(1 - \frac{1}{k}\right)$. Answer: (A).
12	ElectrostaticsElectric dipole	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Let $OC = r$. C is on the axis of A, so A gives $E_1 = \frac{2kp_1}{r^3}$ along the axis. C is on the equatorial line of B, so B gives $E_2 = \frac{kp_2}{r^3}$, perpendicular to the axis (antiparallel to $\vec{p}_2$). $\tan 37° = \frac{E_2}{E_1} = \frac{p_2}{2p_1} = \frac{3}{4}$, so $\frac{p_1}{p_2} = \frac{2}{3}$. Answer: (C).
13	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 With the slider in the middle, the wire is two 5 Ω halves. The 2 Ω resistor is joined between A and the slider, so it is in parallel with the half A–slider: $\frac{5\times2}{5+2} = \frac{10}{7}$ Ω. Total resistance $= \frac{10}{7} + 5 = \frac{45}{7}$ Ω; current $= \frac{20\times7}{45} = \frac{28}{9}$ A. Potential drop across the parallel part (and the 2 Ω resistor) $= \frac{28}{9}\times\frac{10}{7} = \frac{40}{9}$ V. Answer: (C).
14	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctEasy
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $r = \frac{mv}{qB}$, so with equal masses $\frac{r_1}{r_2} = \frac{v_1}{v_2}\cdot\frac{q_2}{q_1} = \frac{2}{3}\times\frac{2}{1} = \frac{4}{3}$. Answer: (A).
15	Electromagnetic Induction and Alternating CurrentsAC circuits	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctEasy
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $i = I_0\sin\omega t$. The current is maximum at $\omega t = \frac{\pi}{2}$ and falls to $\frac{I_0}{\sqrt{2}}$ at $\omega t = \frac{3\pi}{4}$. The phase change $\frac{\pi}{4}$ takes $\frac{T}{8} = \frac{1}{8\times50}$ s $= 2.5$ ms. Answer: (A).
16	OpticsTotal internal reflection and prisms	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctMedium
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At the top face, $\sin\theta = \frac{4}{3}\sin\theta'$. The ray meets the adjacent vertical face at angle of incidence $\theta'' = 90° - \theta'$. Total internal reflection needs $\sin\theta'' \ge \frac{3}{4}$, i.e. $\cos\theta' \ge \frac{3}{4}$, so $\sin\theta' \le \frac{\sqrt{7}}{4}$. Then $\sin\theta \le \frac{4}{3}\cdot\frac{\sqrt{7}}{4} = \frac{\sqrt{7}}{3}$: the maximum $\theta$ is $\sin^{-1}\frac{\sqrt{7}}{3}$. Answer: (B).
17	OpticsTotal internal reflection and prisms	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\mu = \frac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}} = \frac{\sin A}{\sin\frac{A}{2}} = 2\cos\frac{A}{2}$ when $\delta_m = A$. So $A = 2\cos^{-1}\left(\frac{\mu}{2}\right)$. Answer: (C).
18	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\frac{hc}{\lambda} - \phi = 4.8e$ and $\frac{hc}{2\lambda} - \phi = 1.6e$. Subtracting: $\frac{hc}{2\lambda} = 3.2e$, so $\frac{hc}{\lambda} = 6.4e$ and $\phi = 6.4e - 4.8e = 1.6e = \frac{1}{4}\cdot\frac{hc}{\lambda}$. Threshold wavelength $\lambda_0 = \frac{hc}{\phi} = 4\lambda$. Answer: (B).
19	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 Equal wavelengths mean equal momenta: $p = mv$. The photon's energy is $pc = mvc$; the electron's kinetic energy is $\frac{1}{2}mv^2$. Ratio $= \frac{\frac{1}{2}mv^2}{mvc} = \frac{v}{2c}$. Answer: (B).
20	Current ElectricityCells, emf and internal resistance	Class 12: Current Electricity	Single correctMedium
A medium single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity With the galvanometer at point (1), only $\varepsilon_1$ lies between A and the galvanometer: $\varepsilon_1 = k(250)$. At point (2), the two cells (drawn in the same sense) are in series and aid: $\varepsilon_1 + \varepsilon_2 = k(400)$. So $\varepsilon_2 = k(150)$ and $\frac{\varepsilon_1}{\varepsilon_2} = \frac{250}{150} = \frac{5}{3}$. Answer: (B).
21	Work, Energy and PowerWork-energy theorem	Class 11: Work, Energy and Power	NumericalEasy
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $W = \int_0^{10}(5y + 20)\,dy = \left[\frac{5y^2}{2} + 20y\right]_0^{10} = 250 + 200 = 450$ J. Answer: 450.
22	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	NumericalEasy
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $I = \frac{1}{2}MR^2 = \frac{1}{2}(10)(0.2)^2 = 0.2$ kg m$^2$. $\omega_0 = 600$ rpm $= 20\pi$ rad/s, so the retardation is $\alpha = \frac{20\pi}{10} = 2\pi$ rad/s$^2$. $\tau = I\alpha = 0.4\pi = 4\pi\times10^{-1}$ Nm. Answer: 4.
23	Behaviour of Perfect Gas and Kinetic TheoryMean free path	Class 11: Kinetic Theory	NumericalEasy
A easy numerical question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Mean free path $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$. With the same $n$, $\frac{\lambda_A}{\lambda_B} = \frac{d_B^2}{d_A^2} = \frac{25}{100} = 0.25 = 25\times10^{-2}$. Answer: 25.
24	Current ElectricityKirchhoff's rules	Class 12: Current Electricity	NumericalMedium
A medium numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Each side is 4 Ω. Across the side AB the loop is 4 Ω in parallel with the other three sides (12 Ω): 3 Ω. With the 1 Ω internal resistance, $I = \frac{9}{4}$ A, and the voltage across AB is $\frac{9}{4}\times3 = 6.75$ V. The path A–D–C–B (12 Ω) carries $\frac{6.75}{12} = 0.5625$ A, a drop of 2.25 V per side. Diagonal AC spans sides AD and DC: $2\times2.25 = 4.5$ V; diagonal DB spans DC and CB: also 4.5 V. $4.5$ V $= 45\times10^{-1}$ V. Answer: 45.
25	Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	NumericalEasy
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Both circuits have the same 220 V source and the same 5 Ω resistance. The powers are equal when the impedance of (b) equals 5 Ω, i.e. at resonance: $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1\times40\times10^{-6}}} = \frac{1}{2\times10^{-3}} = 500$ rad/s. Answer: 500.
26	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	NumericalMedium
A medium numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Photon energy $= \frac{20\times10^{-26}}{500\times10^{-9}} = 4\times10^{-19}$ J $= 2.5$ eV. Maximum KE $= 2.5 - 1.25 = 1.25$ eV $= 2\times10^{-19}$ J. $p = \sqrt{2mK} = \sqrt{2\times9\times10^{-31}\times2\times10^{-19}} = 6\times10^{-25}$ kg m/s. $B = \frac{p}{er} = \frac{6\times10^{-25}}{1.6\times10^{-19}\times0.3} = 1.25\times10^{-5}$ T $= 125\times10^{-7}$ T. Answer: 125.
27	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	NumericalEasy
A easy numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{1}{8} = \left(\frac{1}{2}\right)^3$: three half-lives pass in 30 years, so $T_{1/2} = 10$ years. Answer: 10.
28	Atoms and NucleiNuclear binding energy	Class 12: Atoms; Class 12: Nuclei	NumericalEasy
A easy numerical question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Mass defect $\Delta m = 13(1.00726) + 14(1.00866) - 27.18846 = 13.09438 + 14.12124 - 27.18846 = 0.02716$ u. Energy $= 0.02716x$ J $= 27.16x\times10^{-3}$ J, which rounds to 27. Answer: 27.
29	Electronic DevicesSemiconductors	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	NumericalEasy
A easy numerical question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $n_e n_h = n_i^2$, so $n_e = \frac{(1.5\times10^{16})^2}{4.5\times10^{22}} = \frac{2.25\times10^{32}}{4.5\times10^{22}} = 5\times10^{9}$/m$^3$. Answer: 5.
30	Communication SystemsAmplitude modulation	No matching NCERT chapter	NumericalEasy
A easy numerical question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter Modulation index $\mu = \frac{A_m}{A_c} = \frac{20}{20} = 1$. Answer: 1.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
31	Chemical Bonding and Molecular StructureMolecular orbital theory	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure By molecular orbital theory, removing an electron from $\mathrm{O_2}$ takes it from an antibonding $\pi^$ orbital and adding one puts it into $\pi^$. Bond orders: $\mathrm{O_2^+}$ 2.5, $\mathrm{O_2}$ 2, $\mathrm{O_2^-}$ 1.5, $\mathrm{O_2^{2-}}$ 1. So $\mathrm{O_2^+ > O_2 > O_2^- > O_2^{2-}}$. Answer: (D).
32	Redox ReactionsOxidation number	Class 11: Redox Reactions	Single correctEasy
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions (A) Cr: +6 → +3 for each of two Cr, a change of 6 in all. (B) Mn: +7 → +2, a change of 5. (C) Cr: +6 → +3, a change of 3. (D) C: +3 → +4 for each of two C, a change of 2. Answer: (B).
33	Surface ChemistryColloids and their preparation	No matching NCERT chapter	Matching listEasy
A easy matching list question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Cheese is a gel, liquid dispersed in solid: (iv). Pumice stone is a solid foam, gas in solid: (iii). Hair cream is an emulsion, liquid in liquid: (i). Cloud is an aerosol of liquid droplets in gas: (ii). Answer: (D).
34	Classification of Elements and PeriodicityPeriodic trends	Class 11: Classification of Elements and Periodicity in Properties	Single correctEasy
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties $\mathrm{N^{3-}}$, $\mathrm{O^{2-}}$ and $\mathrm{F^-}$ are isoelectronic (10 electrons). The smaller the nuclear charge, the larger the ion: $\mathrm{N^{3-}}$ (Z = 7) is the largest, about 1.71 Å, bigger than $\mathrm{O^{2-}}$ and $\mathrm{F^-}$ (and far bigger than the N atom). Answer: (B).
35	Isolation of MetalsOres and concentration	No matching NCERT chapter	Matching listEasy
A easy matching list question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Silver ore is concentrated by leaching with dilute NaCN: (a)-(iii). The blast furnace gives pig iron: (b)-(ii). Copper matte is worked in a reverberatory furnace on the way to blister copper: (c)-(i). Froth flotation is used for sulfide ores: (d)-(iv). Answer: (A).
36	HydrogenHydrogen and its compounds	No matching NCERT chapter	Single correctMedium
A medium single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Interstitial (metallic) hydrides are formed by the d-block metals of groups 3, 4 and 5, and in group 6 only by chromium (CrH). The metals of groups 7, 8 and 9 (Mn, Fe, Co) form no hydrides: the hydride gap. Answer: (C).
37	s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium	No matching NCERT chapter	Matching listEasy
A easy matching list question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter $\mathrm{Li_2CO_3}$ alone among the alkali carbonates decomposes easily on heating: (a)-(iv). $\mathrm{NaHCO_3}$ is used in fire extinguishers: (b)-(iii). Potassium is the most abundant cation inside cells: (c)-(ii). CsI, a large cation with a large anion, has the lowest hydration energy relative to its lattice energy and is the least soluble alkali iodide: (d)-(i). Answer: (B).
38	Chemical Bonding and Molecular StructureVSEPR theory	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{CO_3^{2-}}$ has one C=O $\pi$ bond, which can sit on any of the three oxygens: three canonical forms. $\mathrm{SO_3}$ (three S=O) and $\mathrm{SO_2}$ (two S=O) have more than one $\pi$ bond in their usual structures, and $\mathrm{O_2}$ has one $\pi$ bond but no resonance. Answer: (D).
39	d- and f-Block ElementsTransition element properties	Class 12: The d- and f-Block Elements	Single correctEasy
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mathrm{Ti^{3+}}$ is $3d^1$ (n = 1): $\sqrt{3} = 1.73$ BM. $\mathrm{V^{2+}}$ is $3d^3$ (n = 3): $\sqrt{15} = 3.87$ BM. $\mathrm{Sc^{3+}}$ is $3d^0$: 0. Answer: (D).
40	Coordination CompoundsImportance of coordination compounds	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All four are Co(II) complexes with nitrogen donors. Ethylenediamine is a bidentate chelating ligand, and each chelate ring adds stability (the chelate effect). $\mathrm{[Co(en)_3]^{2+}}$ has three chelate rings, the most of the four. Answer: (B).
41	Environmental ChemistryOzone depletion	No matching NCERT chapter	StatementsEasy
A easy statements question on ozone depletion, from Environmental Chemistry. NCERT: No matching NCERT chapter Statement I is false: CFCs are broken down by ultraviolet radiation, not visible light, and they release chlorine free radicals ($\mathrm{Cl^\bullet}$), not chlorine gas. Statement II is false: $\mathrm{O_3 + NO \rightarrow NO_2 + O_2}$; nitrogen gas is not formed. Answer: (B).
42	Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctMedium
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Compare the conjugate bases. Losing the $\mathrm{CH_2}$ proton of cyclopenta-1,3-diene gives the cyclopentadienyl anion: planar, cyclic, with 6 $\pi$ electrons, so aromatic and very stable (pKa about 16). The benzyl anion of toluene is only resonance-stabilised (pKa about 41); the cyclopropenyl anion has 4 $\pi$ electrons and would be antiaromatic; butane's carbanion has no stabilisation at all. Answer: (D).
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 Density rises with the mass of halogen carried: benzene (A, about 0.88 g/mL) < chlorobenzene (B, about 1.1) < dichlorobenzene (C, about 1.29) < bromochlorobenzene (D, about 1.6), since Br is heavier than Cl. Decreasing order: (D) > (C) > (B) > (A). Answer: (A).
44	Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctEasy
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Maleic anhydride is the cyclic anhydride of cis-but-2-enedioic (maleic) acid. On heating, the two COOH groups, held on the same side of the double bond, lose water and close the five-membered ring. The trans acid cannot reach this ring, and alcohol with acid would give an ester, not an anhydride. Answer: (B).
45	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 Excess ethanol with dry HCl converts the aldehyde into its diethyl acetal; the C–Br bond is untouched: A = $\mathrm{BrCH_2CH(OEt)_2}$. Potassium tert-butoxide is a bulky, strong base: it removes the H on the acetal carbon and eliminates HBr (E2) rather than substituting: B = $\mathrm{H_2C{=}C(OEt)_2}$. Answer: (B).
46	AminesDiazonium salts	Class 12: Amines	Single correctMedium
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines With nitrous acid a primary aliphatic amine forms an unstable alkyl diazonium ion, which loses $\mathrm{N_2}$ at once to give a carbocation. Here it is the tertiary 1-methylcycloheptyl cation, already the most stable cation available, so no rearrangement is needed; water captures it. P = 1-methylcycloheptan-1-ol. Answer: (B).
47	AminesDiazonium salts	Class 12: Amines	Single correctMedium
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Sn + HCl reduces nitrobenzene to aniline (A). In acid solution, benzenediazonium chloride couples with aniline at the para position of the ring (C-coupling), giving p-aminobenzene azo compound, 4-aminoazobenzene, a yellow dye. Answer: (A).
48	PolymersBiodegradable polymers	No matching NCERT chapter	Single correctEasy
A easy single correct question on biodegradable polymers, from Polymers. NCERT: No matching NCERT chapter Nylon-2-nylon-6, a biodegradable polyamide, is the copolymer of glycine and 6-aminocaproic acid. Hexamethylenediamine and adipic acid give nylon-6,6, a polyamide that is not biodegradable. Answer: (A).
49	BiomoleculesNucleic acids and vitamins	Class 12: Biomolecules	Single correctEasy
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Cytosine is 4-aminopyrimidin-2(1H)-one: a pyrimidine ring with C=O at C2, N1–H, and the amino group at C4, next to N3. That is structure (A). (B) has the amino group at C5, (C) is a methyl compound, and (D) is thymine. Answer: (A).
50	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctEasy
A easy single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $\mathrm{C_6H_5CN + CH_3MgBr}$ gives an imine salt, which hydrolyses to acetophenone, $\mathrm{C_6H_5COCH_3}$. As a methyl ketone it gives the iodoform test; being a ketone it gives neither Tollens' nor Schiff's test, and it is not an amino acid (ninhydrin). Answer: (C).
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 Leading zeros are not significant; the trailing zero after the decimal point is. The significant digits are 3, 4, 0. Answer: 3.
52	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 At constant volume $\frac{P_1}{T_1} = \frac{P_2}{T_2}$: $T_2 = 300\ \mathrm{K}\times\frac{1.2\times10^6}{3\times10^5} = 1200$ K. $1200 - 273 = 927$°C. Answer: 927.
53	Structure of AtomDual nature and uncertainty principle	Class 11: Structure of Atom	NumericalEasy
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $\Delta v = 5\times10^6\times\frac{0.02}{100} = 1000$ m/s. $\Delta x = \frac{h}{4\pi m\Delta v} = \frac{6.63\times10^{-34}}{4\times3.14\times9.1\times10^{-31}\times1000} = 5.80\times10^{-8}$ m $= 58.0\times10^{-9}$ m. Answer: 58.
54	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	NumericalEasy
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta U = q + w = 150 + (-200) = -50$ J. The magnitude is 50 J. Answer: 50.
55	SolutionsColligative properties	Class 12: Solutions	NumericalEasy
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Delta T_b = K_b\cdot\frac{w/M}{0.100\ \mathrm{kg}}$: $0.60 = 5.0\times\frac{3.00}{0.100M}$, so $M = \frac{5.0\times3.00}{0.060} = 250$ g mol$^{-1}$. Answer: 250.
56	EquilibriumpH and buffers	Class 11: Equilibrium	NumericalEasy
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium $[\mathrm{OH^-}] = 2\times0.005 = 0.01$ M, so $[\mathrm{H_3O^+}] = \frac{10^{-14}}{10^{-2}} = 1\times10^{-12}$ mol L$^{-1}$. Answer: 1.
57	Chemical KineticsRate laws	Class 12: Chemical Kinetics	NumericalEasy
A easy numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Average rate $= \frac{\Delta[\mathrm{B}]}{\Delta t} = \frac{0.2}{0.5\ \mathrm{h}} = 0.4$ mol L$^{-1}$ h$^{-1}$ $= 4\times10^{-1}$. Answer: 4.
58	d- and f-Block ElementsLanthanoids and actinoids	Class 12: The d- and f-Block Elements	NumericalEasy
A easy numerical question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Ho (Z = 67) is $\mathrm{[Xe]4f^{11}6s^2}$. $\mathrm{Ho^{3+}}$ loses the two 6s electrons and one 4f electron: $\mathrm{[Xe]4f^{10}}$. Answer: 10.
59	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 Mass of N $= 0.42\times0.8 = 0.336$ g $= 0.024$ mol, giving 0.024 mol $\mathrm{NH_3}$. $\mathrm{2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4}$, so 0.012 mol $\mathrm{H_2SO_4}$ $= 12$ mL of 1 M acid. Answer: 12.
60	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	NumericalMedium
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Bromine adds anti, through a cyclic bromonium ion. Anti addition to cis-but-2-ene gives 2,3-dibromobutane as the (2R,3R) and (2S,3S) enantiomers in equal amounts, a racemic mixture; the meso isomer comes only from trans-but-2-ene. So P consists of 2 stereoisomers. Answer: 2.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
61	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 With $a = \sin x$ on the diagonal and $b = \cos x$ elsewhere, the determinant is $(a + 2b)(a - b)^2$. $a = b$: $\tan x = 1$, so $x = \frac{\pi}{4}$, which lies in the interval. $a + 2b = 0$: $\tan x = -2$, but on $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ $\tan x \in [-1, 1]$, so there is no root. One distinct root. Answer: (A).
62	Binomial TheoremSimple applications	Class 11: Binomial Theorem	Single correctMedium
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem For $n > 1$, $\left(1+\frac{1}{n}\right)^n = 1 + 1 + \frac{1}{2!}\left(1-\frac{1}{n}\right) + \frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) + \dots$, which is greater than 2 and less than $1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots < 1 + 1 + \frac{1}{2} + \frac{1}{4} + \dots = 3$. With $n = 10^{100}$ the number lies between 2 and 3, so the lowest integer greater than it is 3. Answer: (C).
63	Permutations and CombinationsP(n, r) and C(n, r)	Class 11: Permutations and Combinations	Single correctEasy
A easy single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations ${}^nP_r = {}^nP_{r+1}$: $\frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!}$, so $n - r = 1$. ${}^nC_r = {}^nC_{r-1}$: $r + (r-1) = n$, so $n = 2r - 1$. Then $2r - 1 - r = 1$, giving $r = 2$ (and $n = 3$). Answer: (B).
64	Sequences and SeriesArithmetic progressions	Class 11: Sequences and Series	Single correctMedium
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series For even $n$ the term is $\frac{n}{2}$; for odd $n$ it is $\left[-\frac{n}{2}\right] = -\frac{n+1}{2}$. Even $n = 8, 10, \dots, 100$ give $4 + 5 + \dots + 50 = \frac{47(4+50)}{2} = 1269$. Odd $n = 9, 11, \dots, 99$ give $-(5 + 6 + \dots + 50) = -\frac{46(5+50)}{2} = -1265$. Sum $= 1269 - 1265 = 4$. Answer: (D).
65	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 Let p: the match will be played; q: the weather is good; r: the ground is wet. The statement “p only if (q and not r)” is $p \Rightarrow (q \wedge \sim r)$. Its negation is $p \wedge \sim(q \wedge \sim r) = p \wedge (\sim q \vee r)$: the match will be played and weather is not good or ground is wet. Answer: (D).
66	Binomial TheoremGeneral and middle term	Class 11: Binomial Theorem	Single correctEasy
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = {}^{12}C_r\,2^{(12-r)/3}\,3^{r/4}$. It is rational when $4 \mid r$ and $3 \mid (12 - r)$, i.e. $r$ is a multiple of 12: $r = 0$ or $r = 12$. $r = 0$: $2^4 = 16$. $r = 12$: $3^3 = 27$. Sum $= 43$. Answer: (C).
67	Differential EquationsLinear differential equations	Class 12: Differential Equations	Single correctMedium
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} - \frac{y}{x} = x^2\cos x$, so $\frac{d}{dx}\left(\frac{y}{x}\right) = x\cos x$ and $\frac{y}{x} = x\sin x + \cos x + c$. $y(\pi) = 0$: $0 = 0 - 1 + c$, so $c = 1$ and $y = x(x\sin x + \cos x + 1)$. $y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}\left(\frac{\pi}{2} + 0 + 1\right) = \frac{\pi^2}{4} + \frac{\pi}{2}$. Answer: (D).
68	StatisticsStandard deviation and variance	Class 11: Statistics	Single correctMedium
A medium single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics The second sample has 150 items with mean $\frac{250(15.6) - 100(15)}{150} = \frac{3900 - 1500}{150} = 16$. Sum of squares of the group: $250(13.44 + 15.6^2) = 250(256.8) = 64200$. Of the first sample: $100(9 + 225) = 23400$. Second sample: $\frac{64200 - 23400}{150} = 272$, so its variance is $272 - 16^2 = 16$ and its standard deviation 4. Answer: (D).
69	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	Single correctEasy
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f(x) = \log\left(x + \sqrt{x^2+1}\right)$ is odd: $f(-x) = \log\left(\sqrt{x^2+1} - x\right) = \log\frac{1}{x + \sqrt{x^2+1}} = -f(x)$. The integral of an odd function over $[-1, 1]$ is 0. Answer: (A).
70	ProbabilityProbability distribution of a random variable	Class 11: Probability; Class 12: Probability	Single correctMedium
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Mean $= \sum_{j\ge1} \frac{j}{3^j} = \frac{1/3}{(1 - 1/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}$. P(X positive and even) $= \frac{1}{3^2} + \frac{1}{3^4} + \dots = \frac{1/9}{1 - 1/9} = \frac{1}{8}$. Answer: (A).
71	Binomial TheoremGeneral and middle term	Class 11: Binomial Theorem	Single correctMedium
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The term independent of $x$ is $T_6 = {}^{10}C_5(x\sin\alpha)^5\left(\frac{a\cos\alpha}{x}\right)^5 = {}^{10}C_5\,a^5\,\frac{\sin^5 2\alpha}{32}$. As $\alpha$ varies, $\sin^5 2\alpha$ takes every value in $[-1, 1]$, so the greatest value is ${}^{10}C_5\frac{\|a\|^5}{32}$. Setting it equal to ${}^{10}C_5 = \frac{10!}{(5!)^2}$ gives $\|a\|^5 = 32$, $\|a\| = 2$. $a = 2$ (with $\sin 2\alpha = 1$) and equally $a = -2$ (with $\sin 2\alpha = -1$) give that greatest value. Answer: (B), and (D) is equally correct.
72	Limits, Continuity and DifferentiabilityDifferentiability	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctMedium
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $x > 2$: $f(x) = 5x + \int_0^1(1-t)\,dt + \int_1^x(t-1)\,dt = 5x + \frac{1}{2} + \frac{(x-1)^2}{2}$. As $x \to 2^+$, $f \to 10 + \frac{1}{2} + \frac{1}{2} = 11 = f(2)$: continuous at 2. Right derivative at 2: $5 + \|1 - 2\| = 6$; left derivative: 5. Not differentiable at $x = 2$. (At $x = 1$, $f = 5x + 1$ is differentiable.) Answer: (B).
73	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 $\cot\frac{\theta}{2} = \frac{1 + \cos\theta}{\sin\theta}$ with $\theta = 15°$: $\cos15° = \frac{\sqrt{6}+\sqrt{2}}{4}$, $\sin15° = \frac{\sqrt{6}-\sqrt{2}}{4}$. $\cot7.5° = \frac{4 + \sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{(4 + \sqrt{6} + \sqrt{2})(\sqrt{6} + \sqrt{2})}{4} = \frac{4\sqrt{6} + 4\sqrt{2} + 8 + 4\sqrt{3}}{4} = \sqrt{2} + \sqrt{3} + 2 + \sqrt{6}$. Answer: (B).
74	Quadratic EquationsRoots in real and complex systems	Class 11: Complex Numbers and Quadratic Equations	Single correctEasy
A easy single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\|x\|^2 - \|x\| - 12 = 0$ gives $(\|x\| - 4)(\|x\| + 3) = 0$, so $\|x\| = 4$ ($\|x\| = -3$ is impossible). $x = \pm4$: two real solutions. Answer: (B).
75	Matrices and DeterminantsAlgebra of matrices	Class 12: Matrices; Class 12: Determinants	Single correctEasy
A easy single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $P = I + N$ with $N = \begin{bmatrix}0 & 0\\ \frac{1}{2} & 0\end{bmatrix}$ and $N^2 = 0$. Since $I$ and $N$ commute, $P^{50} = I + 50N = \begin{bmatrix}1 & 0\\ 25 & 1\end{bmatrix}$. Answer: (C).
76	Conic SectionsEllipse	Class 11: Conic Sections	Single correctMedium
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $\frac{x^2}{4} + y^2 = 1$; the tangent at $(2\cos\theta, \sin\theta)$ is $\frac{x\cos\theta}{2} + y\sin\theta = 1$. It meets $x = 2$ at $B\left(2, \frac{1-\cos\theta}{\sin\theta}\right)$ and $x = -2$ at $C\left(-2, \frac{1+\cos\theta}{\sin\theta}\right)$. For the point $(\sqrt{3}, 0)$: $(\sqrt{3}-2)(\sqrt{3}+2) + \frac{(1-\cos\theta)(1+\cos\theta)}{\sin^2\theta} = -1 + 1 = 0$, so it subtends a right angle at BC and lies on the circle for every $\theta$. (It is a focus: $ae = \sqrt{3}$.) Answer: (A).
77	Straight LinesSlope, parallel and perpendicular lines	Class 11: Straight Lines	Single correctMedium
A medium single correct question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines For $ax^2 + 2hxy + by^2 = 0$ the bisectors are $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$. Here $a = 1$, $h = -2$, $b = -5$: $\frac{x^2 - y^2}{6} = \frac{xy}{-2}$, i.e. $x^2 - y^2 = -3xy$, or $x^2 + 3xy - y^2 = 0$. (Check: the lines are $x = 5y$ and $x = -y$, i.e. $y = \frac{x}{5}$ and $y = -x$.) Answer: (B).
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 $\|\vec{a}\times\vec{b}\|^2 + \|\vec{a}\cdot\vec{b}\|^2 = \|\vec{a}\|^2\|\vec{b}\|^2 = 100$, so $\|\vec{a}\cdot\vec{b}\| = \sqrt{100 - 64} = 6$. Answer: (D).
79	Sets, Relations and FunctionsComposition of functions	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	Single correctMedium
A medium single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $(g\circ f)^{-1}$ exists, so $g\circ f: A \rightarrow C$ is one-one and onto. If $f(x_1) = f(x_2)$ then $g(f(x_1)) = g(f(x_2))$, so $x_1 = x_2$: $f$ must be one-one. Every $c \in C$ equals $g(f(a))$ for some $a$, so $g$ must be onto. Neither $f$ onto nor $g$ one-one is forced. Answer: (A).
80	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Single correctEasy
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Coplanar means the scalar triple product is zero: $\begin{vmatrix}a & a & c\\ 1 & 0 & 1\\ c & c & b\end{vmatrix} = a(0 - c) - a(b - c) + c(c - 0) = c^2 - ab = 0$. With $c > 0$, $c = \sqrt{ab}$. Answer: (B).
81	CirclesRadius and centre	Class 11: Conic Sections	NumericalEasy
A easy numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections $\mathrm{Re}(z^2) = x^2 - y^2$, so the circle is $x^2 - y^2 + 2y^2 + 2x = 0$, i.e. $x^2 + y^2 + 2x = 0$, centre $(-1, 0)$. The parabola is $y = (x-3)^2 + 4$, vertex $(3, 4)$. The line through them has slope $\frac{4 - 0}{3 + 1} = 1$: $y = x + 1$, with $y$-intercept 1. Answer: 1.
82	Three Dimensional GeometrySkew lines and shortest distance	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	NumericalEasy
A easy numerical question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The lines pass through $(k, 2, 3)$ and $(-1, -2, -3)$ with directions $(1, 2, 3)$ and $(3, 2, 1)$. They are coplanar when $\begin{vmatrix}k+1 & 4 & 6\\ 1 & 2 & 3\\ 3 & 2 & 1\end{vmatrix} = (k+1)(2-6) - 4(1-9) + 6(2-6) = -4k - 4 + 32 - 24 = 4 - 4k = 0$. $k = 1$. Answer: 1.
83	Binomial TheoremSimple applications	Class 11: Binomial Theorem	NumericalMedium
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $\sum_{r=0}^{n}(2r+1){}^nC_r = 2\sum r\,{}^nC_r + \sum {}^nC_r = 2n\cdot2^{n-1} + 2^n = (n+1)2^n$. $(n+1)2^n = 101\cdot2^{100}$ gives $n = 100$. $2\left[\frac{99}{2}\right] = 2[49.5] = 98$. Answer: 98.
84	Quadratic EquationsRelations between roots and coefficients	Class 11: Complex Numbers and Quadratic Equations	NumericalMedium
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $a^2 + b^2 + c^2 = 1^2 - 2(2) = -3$. $a^2b^2 + b^2c^2 + c^2a^2 = (ab+bc+ca)^2 - 2abc(a+b+c) = 4 - 6 = -2$. $a^4 + b^4 + c^4 = (a^2+b^2+c^2)^2 - 2(a^2b^2+b^2c^2+c^2a^2) = 9 + 4 = 13$. Answer: 13.
85	ProbabilityProbability of an event	Class 11: Probability; Class 12: Probability	NumericalEasy
A easy numerical question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(\text{at least one head}) = 1 - \frac{1}{2^n} \ge 0.9$ requires $2^n \ge 10$. The least such $n$ is 4 ($2^3 = 8 < 10 \le 16$). Answer: 4.
86	Differential EquationsSeparation of variables	Class 12: Differential Equations	NumericalEasy
A easy numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{y} = \frac{2\,dx}{x\log_e x}$ gives $\ln\|y\| = 2\ln\|\log_e x\| + c$, so $y = k(\log_e x)^2$. At $x = 2$: $(\log_e 2)^2 = k(\log_e 2)^2$, so $k = 1$ and $f(x) = (\log_e x)^2$. $f(e) = 1$. Answer: 1.
87	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 The coefficient of $x^r$ is ${}^nC_r\,2^{n-r}3^{-r}$. Equal coefficients of $x^7$ and $x^8$: ${}^nC_7\frac{2^{n-7}}{3^7} = {}^nC_8\frac{2^{n-8}}{3^8}$, so $\frac{{}^nC_8}{{}^nC_7} = \frac{n-7}{8} = 6$, giving $n = 55$. Answer: 55.
88	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	NumericalEasy
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $7\|\vec{a}\|^2 + 16\,\vec{a}\cdot\vec{b} - 15\|\vec{b}\|^2 = 0$ and $7\|\vec{a}\|^2 - 30\,\vec{a}\cdot\vec{b} + 8\|\vec{b}\|^2 = 0$. Subtracting: $46\,\vec{a}\cdot\vec{b} = 23\|\vec{b}\|^2$, so $\vec{a}\cdot\vec{b} = \frac{\|\vec{b}\|^2}{2}$; then $7\|\vec{a}\|^2 = 7\|\vec{b}\|^2$, so $\|\vec{a}\| = \|\vec{b}\|$. $\cos\theta = \frac{\|\vec{b}\|^2/2}{\|\vec{b}\|^2} = \frac{1}{2}$, $\theta = 60°$. Answer: 60.
89	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 the rectangle stand on the base with width $w$. The parts of the base outside it are $\frac{s - w}{2}$ each, and the sides make 60° with the base, so the height is $h = \frac{\sqrt{3}}{2}(s - w)$, with $s = 2\sqrt{2}$. Area $A = \frac{\sqrt{3}}{2}w(s - w)$ is greatest at $w = \frac{s}{2}$: $A_{max} = \frac{\sqrt{3}}{8}s^2 = \frac{\sqrt{3}}{8}\cdot8 = \sqrt{3}$. $A_{max}^2 = 3$. Answer: 3.
90	Limits, Continuity and DifferentiabilityContinuity	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	NumericalMedium
A medium numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $P''$ constant means $P$ is at most quadratic. Continuity at 2 needs the limit to be finite, so $P(2) = 0$, and then $\lim_{x\to2}\frac{P(x)}{\sin(x-2)} = P'(2) = 7$. So $P(x) = (x-2)\left(c(x-2) + 7\right)$. $P(3) = c + 7 = 9$ gives $c = 2$. $P(5) = 3(6 + 7) = 39$. Answer: 39.

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