JEE (Main) · 2020

JEE (Main) 2020 — 3 September, Shift 1

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 2020

Summary

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

Physics
25 Q
1. Motion of System of Particles and Rigid Body3
2. Properties of Bulk Matter3
3. Magnetic Effects of Current and Magnetism2
15 easy · 10 medium · 0 hard
Chemistry
25 Q
1. Organic Chemistry: Some Basic Principles and Techniques4
2. Chemical Bonding and Molecular Structure2
3. Solutions2
14 easy · 9 medium · 2 hard
Mathematics
25 Q
1. Limits, Continuity and Differentiability3
2. Matrices and Determinants2
3. Sequences and Series2
14 easy · 11 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	Experimental SkillsVernier calipers and screw gauge	No matching NCERT chapter	Single correctEasy
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{\text{pitch}}{\text{number of divisions}} = \frac{0.1}{50} = 0.002$ cm. A reading must be a whole multiple of the least count, so its last (thousandths) digit must be even (it ends in a multiple of 0.002 cm). Of the options only 2.124 cm $= 1062\times0.002$ cm qualifies. Answer: (B).
2	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 $\vec{F} = q\,\vec{v}\times\vec{B}$. $\vec{v}\times\vec{B}$ (without the $10^{-3}$) $= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\5&3&-6\end{vmatrix} = \hat{i}(-18-12) - \hat{j}(-12-20) + \hat{k}(6-15) = -30\hat{i}+32\hat{j}-9\hat{k}$. Force $= 10^{-6}\times10^{-3}\,(-30\hat{i}+32\hat{j}-9\hat{k}) = (-30\hat{i}+32\hat{j}-9\hat{k})\times10^{-9}$ N. Answer: (C).
3	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The block strikes the lower end of the rod, a distance $l$ below O. Angular momentum about O is conserved in the collision (the pivot force has no torque about O). $mvl = I\omega$ with $I = \frac{Ml^2}{3} + ml^2 = \frac{2}{3} + 1 = \frac{5}{3}$ kg m$^2$, so $\omega = \frac{1\times6\times1}{5/3} = 3.6$ rad/s. Kinetic energy after the collision: $\frac{1}{2}I\omega^2 = \frac{1}{2}\times\frac{5}{3}\times12.96 = 10.8$ J. Rising to angle $\theta$: the rod's centre rises $\frac{l}{2}(1-\cos\theta)$ and the block rises $l(1-\cos\theta)$, so the gain in potential energy is $\left(Mg\frac{l}{2} + mgl\right)(1-\cos\theta) = (10 + 10)(1-\cos\theta) = 20(1-\cos\theta)$. $20(1-\cos\theta) = 10.8 \Rightarrow \cos\theta = 0.46 \Rightarrow \theta \approx 62.6^\circ \approx 63^\circ$. Answer: (C).
4	Motion of System of Particles and Rigid BodyMoment of inertia	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium 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 For fixed mass and density the volume is fixed: $\pi R^2L = V$, so $R^2 = \frac{V}{\pi L}$. $I = M\left(\frac{V}{4\pi L} + \frac{L^2}{12}\right)$. Setting $\frac{dI}{dL} = 0$: $-\frac{V}{4\pi L^2} + \frac{L}{6} = 0 \Rightarrow \frac{V}{\pi} = \frac{2L^3}{3}$. So $R^2L = \frac{2L^3}{3}$, i.e. $\frac{L^2}{R^2} = \frac{3}{2}$ and $\frac{L}{R} = \sqrt{\frac{3}{2}}$ (the second derivative is positive, so this is a minimum). Answer: (A).
5	GravitationOrbital motion	Class 11: Gravitation	Single correctMedium
A medium single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Orbital speed at $R_e$: $v_0^2 = \frac{GM}{R_e}$. New speed $v$ with $v^2 = \frac{3}{2}\frac{GM}{R_e}$, still perpendicular to the radius, so $R_e$ is the nearest point (perigee). At the farthest point R the velocity $u$ is again perpendicular to the radius. Angular momentum: $vR_e = uR$. Energy: $\frac{1}{2}v^2 - \frac{GM}{R_e} = \frac{1}{2}u^2 - \frac{GM}{R}$. Substituting $u = \frac{vR_e}{R}$: $\frac{1}{2}v^2\left(1-\frac{R_e^2}{R^2}\right) = GM\left(\frac{1}{R_e}-\frac{1}{R}\right)$, which gives $v^2 = \frac{2GM R}{R_e(R+R_e)}$. $\frac{3}{2}\frac{GM}{R_e} = \frac{2GMR}{R_e(R+R_e)} \Rightarrow 3(R+R_e) = 4R \Rightarrow R = 3R_e$. Answer: (A).
6	Properties of Bulk MatterSurface tension	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 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 Excess pressure in a soap bubble: $\Delta P = \frac{4T}{r}$, so $r \propto \frac{1}{\Delta P}$. $\Delta P_1 = 0.01$ atm and $\Delta P_2 = 0.02$ atm, so $\frac{r_1}{r_2} = 2$ and $\frac{V_1}{V_2} = 2^3 = 8$. Answer: (D).
7	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics The bursting is sudden: the gas expands at once against the outside air, far from equilibrium, so the process is irreversible. It is so fast that no appreciable heat is exchanged with the surroundings, so it is adiabatic. Answer: (C).
8	Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom	Class 11: Kinetic Theory	Single correctEasy
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory A rigid non-linear (triangular) molecule has 3 translational and 3 rotational degrees of freedom, and no vibration (rigid rods): $f = 6$. $U = \frac{f}{2}RT = 3RT$ per mole. Answer: (A).
9	Oscillations and WavesWave motion	Class 11: Oscillations; Class 11: Waves	Single correctMedium
A medium single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Tension at the bottom: $T_1 = 2g$; at the top it also supports the rope: $T_2 = (2+6)g = 8g$. The frequency does not change along the rope, and $v = \sqrt{T/\mu}$, so $\lambda \propto v \propto \sqrt{T}$. $\lambda_2 = 6\times\sqrt{\frac{8g}{2g}} = 6\times2 = 12$ cm. Answer: (D).
10	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 Total charge $= 12 - 3 = 9$ $\mu$C, shared so that the potentials are equal: $\frac{kq_1}{r_1} = \frac{kq_2}{r_2}$, so $q \propto r$. $q_1 : q_2 = \frac{2}{3}R : \frac{1}{3}R = 2 : 1$, giving $q_1 = 6$ $\mu$C and $q_2 = 3$ $\mu$C. Answer: (A).
11	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $C_2$ and $C_3$ are in parallel, and that pair is in series with $C_1$. The total charge 750 $\mu$C flows through $C_1$ and divides between $C_2$ and $C_3$. $C_3$ has the same 20 V as $C_2$: $Q_3 = 8\times20 = 160$ $\mu$C. $Q_2 = 750 - 160 = 590$ $\mu$C. Answer: (C).
12	Current ElectricityCells, emf and internal resistance	Class 12: Current Electricity	Single correctMedium
A medium single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity The internal resistance is that of the electrolyte between the cylinders, with current flowing radially. A shell of radius $r$ and thickness $dr$ has $dr_{int} = \frac{\rho\,dr}{2\pi r l}$. $r_{int} = \int_a^b \frac{\rho\,dr}{2\pi l r} = \frac{\rho}{2\pi l}\ln\frac{b}{a}$. Power in the external resistance is maximum when $R$ equals the internal resistance: $R = \frac{\rho}{2\pi l}\ln\left(\frac{b}{a}\right)$. Answer: (B).
13	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 Each side ($a = 0.1$ m) is at distance $d = \frac{\sqrt{3}}{2}a$ from the centre and subtends $30^\circ$ on each side of the perpendicular. Field of one side: $B_1 = \frac{\mu_0I}{4\pi d}(\sin30^\circ + \sin30^\circ) = \frac{\mu_0I}{4\pi d}$. For 6 sides and 50 turns: $B = 300\times\frac{\mu_0I}{4\pi d} = \frac{75}{d}\cdot\frac{\mu_0I}{\pi}$, with $d = \frac{\sqrt{3}}{2}\times0.1 = \frac{\sqrt{3}}{20}$ m. $B = \frac{75\times20}{\sqrt{3}}\cdot\frac{\mu_0I}{\pi} = 500\sqrt{3}\,\frac{\mu_0I}{\pi}$. Answer: (D).
14	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The loop (area $\pi ab$) lies in the $xy$-plane with $B$ along $z$. Rotating about the $x$-axis, the flux is $\phi = B\pi ab\cos\omega t$. $\varepsilon = -\frac{d\phi}{dt} = B\pi ab\,\omega\sin\omega t$, so the power is $\frac{\varepsilon^2}{R} = \frac{(\pi abB\omega)^2}{R}\sin^2\omega t$. The average of $\sin^2\omega t$ is $\frac{1}{2}$: $\langle P\rangle = \frac{\pi^2a^2b^2B^2\omega^2}{2R}$. Answer: (A).
15	Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\omega = 2\pi\times750 = 4712$ rad/s. $X_L = \omega L = 4712\times0.1803 \approx 849.6$ $\Omega$; $X_C = \frac{1}{\omega C} = \frac{1}{4712\times10^{-5}} \approx 21.2$ $\Omega$. $Z = \sqrt{100^2 + (849.6-21.2)^2} \approx 834.4$ $\Omega$, so $I_{rms} = \frac{20}{834.4} \approx 0.02397$ A. Power in R: $I_{rms}^2R \approx 0.0574$ W. Heat needed: $2\times10 = 20$ J. $t = \frac{20}{0.0574} \approx 348$ s. Answer: (A).
16	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 Amplitude: $E_0 = cB_0 = 3\times10^8\times3\times10^{-8} = 9$ V/m. The phase $(y+ct)$ means the wave travels along $-\hat{j}$. $\vec{E}$ is perpendicular to $\vec{B}$ (along $\hat{i}$) and to the direction of travel, so along $\pm\hat{k}$, and $\vec{E}\times\vec{B}$ must point along $-\hat{j}$. $(-\hat{k})\times\hat{i} = -\hat{j}$, so when $\vec{B}$ is along $+\hat{i}$, $\vec{E}$ is along $-\hat{k}$: $\vec{E} = -9\sin[200\pi(y+ct)]\hat{k}$ V/m. Answer: (B).
17	OpticsInterference and diffraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Angular fringe width $\theta = \frac{\lambda}{d} = \frac{500\times10^{-9}}{0.05\times10^{-3}} = 0.01$ rad. In degrees: $0.01\times\frac{180}{\pi} \approx 0.57^\circ$. Answer: (A).
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 Photon energies ($hc \approx 1240$ eV nm): $E_1 = \frac{1240}{500} = 2.48$ eV, $E_2 = \frac{1240}{200} = 6.2$ eV. $6.2 - \phi = 3(2.48 - \phi) \Rightarrow 2\phi = 7.44 - 6.2 = 1.24 \Rightarrow \phi = 0.62$ eV. The closest option is 0.61 eV. Answer: (B).
19	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{N}{N_0} = e^{-\lambda t} = \frac{9}{16}$. After $t/2$: $e^{-\lambda t/2} = \sqrt{\frac{9}{16}} = \frac{3}{4}$. Answer: (B).
20	Electronic Devicesp-n junction diode	Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits	Single correctEasy
A easy single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Voltage across the resistor: $1.5 - 0.5 = 1.0$ V. For $I \le 10$ mA: $R \ge \frac{1.0}{0.01} = 100$ $\Omega$. 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 Kinetic energy at launch must lift the ball 20 m: $\frac{1}{2}mv^2 = mgh = 0.15\times10\times20 = 30$ J. This energy is the work done by the constant force over 0.2 m: $F\times0.2 = 30$, so $F = 150$ N. Answer: 150.
22	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	NumericalEasy
A easy numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Treat the platform as a uniform disc: $I_p = \frac{1}{2}\times200\times R^2 = 100R^2$. The person on the rim adds $80R^2$; at the centre, nothing. Angular momentum is conserved: $(100R^2 + 80R^2)\times5 = 100R^2\times n$, so $n = 9$ rpm. Answer: 9.
23	Properties of Bulk MatterThermal expansion and calorimetry	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalEasy
A easy numerical question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The unfilled volume stays constant if the beaker's capacity and the mercury expand by the same amount: $V_b\gamma_b\,\Delta T = V_m\gamma_m\,\Delta T$. $V_m = \frac{500\times6\times10^{-6}}{1.5\times10^{-4}} = 20$ cc. Answer: 20.
24	OpticsReflection and refraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalMedium
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The jar's diameter is 30 cm. Empty jar: the line of sight from the hole (height 45 cm) to the point at height 15 cm on the far wall drops 30 cm over 30 cm, so it makes $45^\circ$ with the vertical. With liquid to 30 cm, the same ray in air drops 15 cm to the surface, meeting it 15 cm across. In the liquid it must reach the far bottom edge: 15 cm across and 30 cm down. Angle in air: $\sin i = \frac{1}{\sqrt{2}}$. Angle in liquid: $\sin r = \frac{15}{\sqrt{15^2+30^2}} = \frac{1}{\sqrt{5}}$. $\mu = \frac{\sin i}{\sin r} = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{2.5} \approx 1.58$, so $N = 158$. Answer: 158.
25	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalEasy
A easy numerical 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 $h = \frac{2T\cos\theta}{r\rho g}$ with $\theta = 0$: $T = \frac{hr\rho g}{2}$. $T = \frac{0.15\times1.5\times10^{-4}\times900\times10}{2} = 0.10125$ N/m $= 101.25$ mN/m. Closest integer: 101. Answer: 101.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
26	Chemical KineticsOrder of reaction	Class 12: Chemical Kinetics	Single correctEasy
A easy single correct question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For an elementary (single-step) reaction the order equals the molecularity, which is at least 1; no single collision step can have a rate independent of all concentrations. So a zero order reaction is necessarily a complex, multistep reaction (A false, B true). First order reactions need not be elementary (many are multistep), and second order reactions may be elementary (bimolecular single steps), so (C) and (D) are false. Answer: (B).
27	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 Electron counts and MO bond orders: $\mathrm{NO^+}$ (14 e) 3; NO (15 e) 2.5; $\mathrm{NO^{2+}}$ (13 e) 2.5; $\mathrm{NO^-}$ (16 e) 2 (the extra electron enters an antibonding $\pi^*$ orbital). The lowest bond order, and so the weakest bond, is in $\mathrm{NO^-}$. Answer: (C).
28	SolutionsRaoult's law	Class 12: Solutions	Single correctMedium
A medium single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Henry's law: $p = K_H\,x$. A larger $K_H$ means a smaller mole fraction at a given pressure, so $\alpha$ (largest $K_H$) is the least soluble: (A) is false. The table gives values at 298 K only, so it implies nothing about 308 K: (B) is not implied. A 55.5 molal solution has 55.5 mol of gas per kg (55.5 mol) of water, so $x = \frac{55.5}{55.5+55.5} = 0.5$. For $\gamma$: $p = 2\times10^{-5}\times0.5$ kbar $= 0.01$ bar, not 1 bar: (C) is false. For $\delta$: $p = 0.5\times0.5 = 0.25$ kbar $= 250$ bar: (D) is true. Answer: (D).
29	ElectrochemistryConductance	Class 12: Electrochemistry	Single correctMedium
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry (A) NaCl is very soluble (about 6 M when saturated) while $\mathrm{BaSO_4}$ is sparingly soluble ($\sim10^{-5}$ M), so the saturated NaCl solution conducts far better: true. (D) Ionic mobilities increase with temperature because the viscosity of water falls: true. (B) For $\mathrm{BaSO_4}$ both the solubility and the mobilities rise with temperature, so the conductance rises: true. (C) The solubility of NaCl rises only slightly with temperature, but the mobilities rise, so the conductance of the saturated solution also rises: true. Every statement is true, so no option is false. Answer: none of the options (question treated as dropped).
30	EquilibriumpH and buffers	Class 11: Equilibrium	Single correctEasy
A easy single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium (B): 10 mmol HCl reacts with 10 of the 20 mmol $\mathrm{CH_3COONa}$, leaving 10 mmol $\mathrm{CH_3COOH}$ and 10 mmol $\mathrm{CH_3COONa}$: a weak acid with its salt, an acidic buffer. (A) is a strong acid with its salt (no buffer). (C) leaves excess NaOH. (D) is exact neutralisation, giving only $\mathrm{CH_3COONa}$. Answer: (B).
31	Surface ChemistryColloids and their preparation	No matching NCERT chapter	Single correctEasy
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Light is scattered strongly (Tyndall effect) when the particle size is not much smaller than the wavelength of the light, as with colloidal particles; the condition is usually stated as the diameter being not much smaller than (comparable to) the wavelength. The effect also needs a large difference in refractive index between the phases, in either direction, so (D) as stated is not the condition. Answer: (C).
32	Classification of Elements and PeriodicityModern periodic table	Class 11: Classification of Elements and Periodicity in Properties	Single correctEasy
A easy single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties IUPAC roots: un = 1, nil = 0, enn = 9. Unnilennium is element 109. Answer: (B).
33	Chemical Bonding and Molecular StructureHydrogen bonding	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Water's high boiling point is due to strong intermolecular hydrogen bonding. $\mathrm{H_2S}$ has no significant hydrogen bonding, only weak van der Waals forces, and is a gas at room temperature (boiling point about 213 K). Answer: (A).
34	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Pyrophosphoric acid is $\mathrm{H_4P_2O_7}$: $\mathrm{(HO)_2P(=O)-O-P(=O)(OH)_2}$. It has 4 P$-$OH groups, 2 P=O bonds and 1 P$-$O$-$P bridge. Answer: (D).
35	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{Au + 4H^+ + NO_3^- + 4Cl^- \to AuCl_4^- + NO + 2H_2O}$ (similarly for Pt, giving $\mathrm{PtCl_6^{2-}}$). Nitrate is reduced to nitric oxide, NO. Answer: (B).
36	Coordination CompoundsIsomerism in complexes	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds An octahedral $\mathrm{[M(AA)_2X_2]}$ complex in the cis form has no plane or centre of symmetry and exists as two non-superimposable mirror images, so cis-$\mathrm{[CrCl_2(ox)_2]^{3-}}$ is optically active. The trans form has a mirror plane through the two oxalates. $\mathrm{[MA_2B_4]}$ complexes (cis or trans) have mirror planes and are inactive. Answer: (B).
37	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 $\mathrm{Ti^{3+}}$ is $d^1$; the single absorption is the $t_{2g}\to e_g$ transition, so $\Delta_o = 20300$ $\mathrm{cm^{-1}} = \frac{20300}{83.7} = 242.5$ kJ $\mathrm{mol^{-1}}$. CFSE for $t_{2g}^1$ $= -0.4\Delta_o = -0.4\times242.5 = -97$ kJ $\mathrm{mol^{-1}}$, of magnitude 97. Answer: (D).
38	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 Burning coal releases $\mathrm{SO_2}$ and nitrogen oxides, which form sulphuric and nitric acids in the atmosphere: acid rain. Eutrophication and blue baby syndrome come from fertiliser run-off (phosphates, nitrates), and ozone depletion mainly from CFCs. Answer: (A).
39	Organic Chemistry: Some Basic Principles and TechniquesElectronic effects	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctMedium
A medium single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Approximate $pK_a$ values: propyne (terminal alkyne C$-$H) about 25; acetone ($\alpha$-H) about 19; malononitrile $\mathrm{CH_2(CN)_2}$ about 11; trimethyl methanetricarboxylate $\mathrm{HC(COOCH_3)_3}$ about 7. In (D) the C$-$H is flanked by three ester carbonyls, so its carbanion is delocalised over three C=O groups and is the most stable; that hydrogen is the most acidic. Answer: (D).
40	Organic Chemistry: Some Basic Principles and TechniquesPurification methods	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Glycerol boils at about 563 K and decomposes near its boiling point, so the spent lye from soap making is distilled under reduced pressure, where glycerol boils at a lower temperature without decomposing. Answer: (C).
41	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctMedium
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In each compound the stereocentre carries H, $\mathrm{CH_3}$ and two carbon chains; hydrogenation saturates the C=C, and the product is inactive only if the two chains become identical. (A) propyl and $\mathrm{-CH=C(CH_3)_2}$: after hydrogenation propyl and isobutyl, still different (chiral). (B) isopropyl and $\mathrm{-CH=CH-CH_3}$: after hydrogenation isopropyl and propyl, different (chiral). (C) ethyl and $\mathrm{-CH=CH-CH_3}$: after hydrogenation ethyl and propyl (3-methylhexane), chiral. (D) isopropyl and isopropenyl $\mathrm{-C(CH_3)=CH_2}$: after hydrogenation both groups are isopropyl, so the product, 2,3,4-trimethylpentane, has no stereocentre and is optically inactive. Answer: (D).
42	Alcohols, Phenols and EthersReactions of phenols	Class 12: Alcohols, Phenols and Ethers	Single correctMedium
A medium single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Novestrol (ethynylestradiol) contains a phenolic OH, a tertiary alcohol and a terminal C$\equiv$C, and no carbonyl group. Bromine water reacts with the phenol ring and the alkyne; $\mathrm{ZnCl_2}$/HCl (Lucas reagent) reacts with the tertiary alcohol; $\mathrm{FeCl_3}$ gives the phenol colour test. There is no carbonyl for alcoholic HCN, and no methyl ketone (or $\mathrm{CH_3CH(OH)}$ group) for NaOCl (haloform). Answer: (B).
43	Alcohols, Phenols and EthersPrimary, secondary and tertiary alcohols	Class 12: Alcohols, Phenols and Ethers	Single correctMedium
A medium single correct question on primary, secondary and tertiary alcohols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Since oxidising the alcohol gives the acid, [C] must be a primary alcohol $\mathrm{RCH_2OH}$ and [B] the acid RCOOH with the same R: [A] is $\mathrm{RCOOCH_2R}$, and it must have 10 carbons. (A) Butyl butanoate fits the pattern but is $\mathrm{C_8H_{16}O_2}$, not $\mathrm{C_{10}H_{20}O_2}$: not possible. (B) 2-methylbutanoic acid and 2-methylbutan-1-ol (C5 + C5); the alcohol oxidises to 2-methylbutanoic acid: possible. (C) The alcohol part is butan-2-ol, a secondary alcohol, which oxidises to butanone, not to the acid (3-methylpentanoic acid): not possible. (D) Pivalic acid $\mathrm{(CH_3)_3CCOOH}$ and neopentyl alcohol $\mathrm{(CH_3)_3CCH_2OH}$ (C5 + C5); the alcohol oxidises to pivalic acid: possible. The structure the question tests is (C), the ester of a secondary alcohol; (A) is excluded only by its carbon count. Answer: (C).
44	Haloalkanes and HaloarenesSN1 and SN2 mechanisms	Class 12: Haloalkanes and Haloarenes	Single correctEasy
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (a) The nucleophile is not involved in the slow step, so weak nucleophiles (which do not force an $\mathrm{S_N2}$ path) favour $\mathrm{S_N1}$: correct. (b) Bulky alkyl substituents stabilise the carbocation (inductive effect, hyperconjugation and relief of steric strain on ionisation): correct. (c) The planar carbocation is attacked from both faces, giving racemization: correct. (d) Ionisation needs polar (protic) solvents to stabilise the ions: incorrect. Answer: (C).
45	Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctMedium
A medium single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Products: (a) aniline; (b) benzylamine $\mathrm{C_6H_5CH_2NH_2}$; (c) Stephen reduction gives phenylacetaldehyde $\mathrm{C_6H_5CH_2CHO}$; (d) benzenediazonium chloride. Kjeldahl's method converts amine nitrogen to ammonium sulphate and works for (a) and (b). It fails for (c), which contains no nitrogen at all, and for (d), a diazonium compound, whose nitrogen is lost as $\mathrm{N_2}$ and is not converted to ammonium sulphate (as with azo and nitro compounds). Answer: (B).
46	SolutionsConcentration terms	Class 12: Solutions	NumericalEasy
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 mol in all: 0.1 mol glucose ($0.1\times180 = 18$ g) and 0.9 mol water ($0.9\times18 = 16.2$ g). Mass % of water $= \frac{16.2}{18+16.2}\times100 = \frac{16.2}{34.2}\times100 = 47.37\%$, to the nearest integer 47. Answer: 47.
47	Solid StatePacking in fcc, bcc and hcp	No matching NCERT chapter	NumericalMedium
A medium numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter $Z = \frac{\rho N_A a^3}{M} = \frac{2.7\times10^3\times6.022\times10^{23}\times(405\times10^{-12})^3}{2.7\times10^{-2}} = \frac{2.7\times10^3\times6.022\times10^{23}\times6.643\times10^{-29}}{2.7\times10^{-2}} \approx 4$, so the cell is face-centred cubic. In fcc the atoms touch along the face diagonal: $4r = \sqrt{2}a$, so $r = \frac{a}{2\sqrt{2}} = \frac{405}{2.828} = 143.2$ pm. Answer: 143.
48	ElectrochemistryNernst equation	Class 12: Electrochemistry	NumericalHard
A hard numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $\mathrm{H_2 + 2AgCl \to 2Ag + 2H^+ + 2Cl^-}$, $E^0 = 0.22$ V, $n = 2$. $E = 0.22 - \frac{0.06}{2}\log([\mathrm{H^+}]^2[\mathrm{Cl^-}]^2) = 0.22 - 0.06\log([\mathrm{H^+}][\mathrm{Cl^-}])$. In HCl $[\mathrm{H^+}] = [\mathrm{Cl^-}] = 10^{-pH}$, so $E = 0.22 + 0.12\,pH$. At pH 1, $E = 0.34$ V, which just stops the electrons from Na: $h\nu = 2.3 + 0.34 = 2.64$ eV (same light in both cases). For K the stopping voltage needed is $2.64 - 2.25 = 0.39$ V: $0.22 + 0.12\,pH = 0.39 \Rightarrow pH = 1.417 = 141.7\times10^{-2}$, to the nearest integer 142. Answer: 142.
49	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 $\mathrm{2H_2O_2 \to 2H_2O + O_2}$: 1 L of 8.9 M solution gives $\frac{8.9}{2} = 4.45$ mol $\mathrm{O_2}$. Volume at 273 K and 1 atm: $V = nRT/P = 4.45\times0.0821\times273 = 99.74$ L. So 1 L of solution gives about 99.74 L of $\mathrm{O_2}$: volume strength 99.74, to the nearest integer 100. Answer: 100.
50	HydrocarbonsAlkanes	Class 11: Hydrocarbons	NumericalHard
A hard numerical question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The simplest optically active alkene is 3-methylpent-1-ene, $\mathrm{CH_2=CH-CH(CH_3)-CH_2CH_3}$ ($\mathrm{C_6H_{12}}$). Hydrogenation gives 3-methylpentane, $\mathrm{CH_3CH_2CH(CH_3)CH_2CH_3}$, which has four kinds of hydrogen. 1-halo-3-methylpentane, $\mathrm{XCH_2CH_2CH(CH_3)CH_2CH_3}$: C3 becomes a stereocentre, 2 stereoisomers. 2-halo-3-methylpentane, $\mathrm{CH_3CHXCH(CH_3)CH_2CH_3}$: two stereocentres, 4 stereoisomers. 3-halo-3-methylpentane: achiral, 1. 1-halo-2-ethylbutane, $\mathrm{XCH_2CH(CH_2CH_3)_2}$: achiral, 1. Total $= 2 + 4 + 1 + 1 = 8$. Answer: 8.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
51	Sets, Relations and FunctionsUnion, intersection and complement	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	Single correctEasy
A easy single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Real roots: $(m+1)^2 - 4(m+4) \ge 0 \Rightarrow m^2 - 2m - 15 \ge 0 \Rightarrow (m-5)(m+3) \ge 0$, so $A = (-\infty, -3]\cup[5, \infty)$. $A\cup B = \mathbf{R}$ (true); $A\cap B = \{-3\}$ (true, since 5 is not in B); $B - A = (-3, 5)$ (true). $A - B = (-\infty, -3)\cup[5, \infty)$, which contains 5, so (C) is not true. Answer: (C).
52	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 $\alpha+\beta = -p$, $\alpha\beta = 2$. $\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right) = \alpha\beta + \frac{1}{\alpha\beta} - \frac{\alpha^2+\beta^2}{\alpha\beta} = 2 + \frac{1}{2} - \frac{p^2-4}{2} = \frac{9-p^2}{2}$. $\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right) = \alpha\beta + 2 + \frac{1}{\alpha\beta} = 2 + 2 + \frac{1}{2} = \frac{9}{2}$. Product $= \frac{9}{4}(9-p^2)$. (The second equation only gives $p = 2q$, which turns this into $\frac{9}{4}(9-4q^2)$, not an option.) Answer: (A).
53	Permutations and CombinationsP(n, r) and C(n, r)	Class 11: Permutations and Combinations	Single correctMedium
A medium single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations ${}^nP_{n-1} = n!$, so the $k$-th term of the first sum is $(-1)^{k+1}(k+1)\cdot k! = (-1)^{k+1}(k+1)!$: the first sum is $2! - 3! + 4! - \ldots + 52!$ (the 51st term is $+52!$). The second sum is $1! - 2! + 3! - \ldots + 51!$. Adding, $2!$ cancels $-2!$, $-3!$ cancels $+3!$, and so on up to $51!$; what remains is $1! + 52! = 1 + (52)!$. Answer: (A).
54	Matrices and DeterminantsDeterminants of order two and three	Class 12: Matrices; Class 12: Determinants	Single correctMedium
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $R_2\to R_2-R_1$, $R_3\to R_3-R_1$: rows become $(x-1,\ x-1,\ x-1)$ and $(2x-3,\ 3x-5,\ 7x-13)$. Then $R_3\to R_3-2R_2$: $(-1,\ x-3,\ 5x-11)$. Take $(x-1)$ out of $R_2$ and use $C_2\to C_2-C_1$, $C_3\to C_3-C_1$: $\Delta = (x-1)\begin{vmatrix}x-2&x-1&2x-2\\1&0&0\\-1&x-2&5x-10\end{vmatrix} = -(x-1)\left[(x-1)(5x-10) - (2x-2)(x-2)\right] = -(x-1)^2(x-2)(5-2) = -3(x-1)^2(x-2)$. $-3(x^2-2x+1)(x-2) = -3x^3 + 12x^2 - 15x + 6$, so $B = 12$, $C = -15$ and $B + C = -3$. Answer: (A).
55	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 $T_{r+1} = {}^nC_r\,3^{\frac{n-r}{2}}\,5^{\frac{r}{8}}$ is an integer when $r$ is a multiple of 8 and $n-r$ is even; since $r$ is then even, $n$ must be even. For even $n$ the integral terms are $r = 0, 8, 16, \ldots$, numbering $\left\lfloor\frac{n}{8}\right\rfloor + 1 = 33$, so $\left\lfloor\frac{n}{8}\right\rfloor = 32$, $256 \le n \le 263$. The least such even $n$ is 256. Answer: (C).
56	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 $S_{25} = S_{40} - S_{25}$, so $2S_{25} = S_{40}$. $2\times\frac{25}{2}(6+24d) = \frac{40}{2}(6+39d) \Rightarrow 150 + 600d = 120 + 780d \Rightarrow d = \frac{30}{180} = \frac{1}{6}$. Answer: (A).
57	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 At $x = 0$: $y^2 = y$, so $y(0) = 0$ or 1. Differentiating: $2yy' - 2\tan x = y'$, i.e. $y'(2y-1) = 2\tan x$, so $y'(0) = 0$. Again: $y''(2y-1) + 2y'^2 = 2\sec^2x$; at $x = 0$: $y''(0)(2y(0)-1) = 2$, so $y''(0) = 2$ (if $y(0) = 1$) or $-2$ (if $y(0) = 0$). Either way $\|y''(0)\| = 2$ and $\|y'(0)\| + \|y''(0)\| = 2$. Answer: (D).
58	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 The numerator tends to 1. As $x\to0^+$, $[x] = 0$ and the denominator tends to $\lambda$; as $x\to0^-$, $[x] = -1$ and it tends to $\lambda - 1$. For the limit to exist: $\frac{1}{\|\lambda\|} = \frac{1}{\|\lambda-1\|} \Rightarrow \lambda = \frac{1}{2}$. Then $L = \frac{1}{1/2} = 2$. Answer: (A).
59	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	Single correctEasy
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f(x) = 3x^{5/3} - 7x^{2/3}$, $f'(x) = 5x^{2/3} - \frac{14}{3}x^{-1/3} = \frac{15x - 14}{3x^{1/3}}$. $f'(x) > 0$ when numerator and denominator have the same sign: $x > \frac{14}{15}$ or $x < 0$. Answer: (C).
60	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 The integrand is even and $\pi - \|x\| \ge 0$ on $[-\pi, \pi]$: $2\int_0^\pi(\pi - x)\,dx = 2\left[\pi x - \frac{x^2}{2}\right]_0^\pi = 2\times\frac{\pi^2}{2} = \pi^2$. Answer: (C).
61	Integral CalculusArea under simple curves	Class 12: Integrals; Class 12: Application of Integrals	Single correctMedium
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $x^2 + 1 \le x + 1$ for $0 \le x \le 1$, so the upper boundary is $x^2+1$ on $\left[\frac{1}{2}, 1\right]$ and $x+1$ on $[1, 2]$. Area $= \int_{1/2}^1(x^2+1)\,dx + \int_1^2(x+1)\,dx = \left(\frac{1}{3} - \frac{1}{24} + \frac{1}{2}\right) + \left(4 - \frac{3}{2}\right) = \frac{19}{24} + \frac{5}{2} = \frac{79}{24}$. Answer: (D).
62	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 Separate: $\left(1 + \frac{1}{y^2}\right)dy = \frac{dx}{1+e^{-x}} = \frac{e^x\,dx}{e^x+1}$. Integrate: $y - \frac{1}{y} = \log_e(1+e^x) + C$. At (0, 1): $0 = \log_e2 + C$, so $C = -\log_e2$. $y - \frac{1}{y} = \log_e\left(\frac{1+e^x}{2}\right) \Rightarrow y^2 = 1 + y\log_e\left(\frac{1+e^x}{2}\right)$. Answer: (A).
63	Conic SectionsParabola	Class 11: Conic Sections	Single correctMedium
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Take $P = (3t^2, 6t)$, so $N = (3t^2, 0)$ and $M = (3t^2, 3t)$. The line $y = 3t$ meets the parabola where $9t^2 = 12x$: $Q = \left(\frac{3t^2}{4}, 3t\right)$. Slope of NQ $= \frac{3t}{\frac{3t^2}{4} - 3t^2} = -\frac{4}{3t}$; its $y$-intercept is $-\frac{4}{3t}(0 - 3t^2) = 4t = \frac{4}{3}$, so $t = \frac{1}{3}$. MQ $= 3t^2 - \frac{3t^2}{4} = \frac{9t^2}{4} = \frac{1}{4}$, and PN $= 6t = 2$. Answer: (B).
64	Conic SectionsHyperbola	Class 11: Conic Sections	Single correctEasy
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$: $c^2 = 4 - 3 = 1$, foci $(\pm1, 0)$. Hyperbola: $2a = \sqrt{2}$, $a^2 = \frac{1}{2}$, $b^2 = c^2 - a^2 = \frac{1}{2}$: $x^2 - y^2 = \frac{1}{2}$. Check: (A) $\frac{1}{2}$ yes; (B) $1 - \frac{1}{2} = \frac{1}{2}$ yes; (C) $\frac{3}{2} - \frac{1}{2} = 1$ no; (D) $\frac{3}{2} - 1 = \frac{1}{2}$ yes. Answer: (C).
65	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctEasy
A easy 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 Direction of the line: $(0, 3, -3) \parallel (0, 1, -1)$. A general point: $(1, -2+s, 3-s)$. From (4, 2, 3) to it: $(-3, s-4, -s)$, perpendicular to $(0, 1, -1)$: $(s-4) + s = 0 \Rightarrow s = 2$. Foot $= (1, 0, 1)$. Check: $x+2y-z = 0$; $x-2y+z = 2$; $x-y-2z = -1$; $2x+y-z = 1$. Answer: (D).
66	Three Dimensional GeometrySkew lines and shortest distance	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctEasy
A easy single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Points: $(1+2l, -1, l)$ and $(2+m, -1+m, -m)$. Equal $y$: $-1 = -1 + m \Rightarrow m = 0$. Equal $x$: $1 + 2l = 2 \Rightarrow l = \frac{1}{2}$. Then $z$: $l = \frac{1}{2}$ but $-m = 0$: contradiction. The lines never meet (they are skew). Answer: (D).
67	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 All values lie in the interval $(0, 10]$, whose length is less than 10. For data in an interval of length $R$, the variance is at most $\frac{R^2}{4}$ (attained only with half the weight at each end), so $\sigma \le \frac{R}{2} < 5$. So the standard deviation cannot be 6; 4, 2 and 1 can all be attained by suitable values and frequencies. Answer: (A).
68	ProbabilityAddition and multiplication theorems	Class 11: Probability; Class 12: Probability	Single correctEasy
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Sum 4: (1,3), (2,2), (3,1); sum 8: (2,6), (3,5), (4,4), (5,3), (6,2); sum 12: (6,6). That is 9 equally likely outcomes. Only (4,4) contains a 4. Probability $= \frac{1}{9}$. Answer: (B).
69	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctEasy
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $a = \sin^{-1}\frac{4}{5}$, $b = \sin^{-1}\frac{5}{13}$ (both acute). $\cos(a+b) = \frac{3}{5}\cdot\frac{12}{13} - \frac{4}{5}\cdot\frac{5}{13} = \frac{16}{65}$, and $a+b$ is acute (its sine $\frac{63}{65} > 0$). So $a + b = \cos^{-1}\frac{16}{65} = \frac{\pi}{2} - \sin^{-1}\frac{16}{65}$, and the bracket equals $\frac{\pi}{2}$. $2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$. Answer: (B).
70	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 $\sim(p\wedge\sim q) \equiv \sim p \vee q$. $p \to (\sim p\vee q) \equiv \sim p \vee (\sim p \vee q) \equiv \sim p \vee q$. Answer: (C).
71	Complex NumbersAlgebra of complex numbers	Class 11: Complex Numbers and Quadratic Equations	NumericalEasy
A easy numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\frac{1+i}{1-i} = \frac{(1+i)^2}{2} = i$ and $\frac{1+i}{i-1} = -i$. $i^{m/2} = 1$ needs $\frac{m}{2}$ to be a multiple of 4: least $m = 8$. $(-i)^{n/3} = 1$ needs $\frac{n}{3}$ to be a multiple of 4: least $n = 12$. $\gcd(8, 12) = 4$. Answer: 4.
72	Matrices and DeterminantsAlgebra of matrices	Class 12: Matrices; Class 12: Determinants	NumericalEasy
A easy numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A^2 = \begin{bmatrix}x^2+1&x\\x&1\end{bmatrix}$, $A^4 = (A^2)^2$ has $a_{11} = (x^2+1)^2 + x^2$ and $a_{22} = x^2 + 1$. With $u = x^2$: $u^2 + 3u + 1 = 109 \Rightarrow u^2 + 3u - 108 = 0 \Rightarrow u = 9$ (u cannot be $-12$). $a_{22} = 9 + 1 = 10$. Answer: 10.
73	Sequences and SeriesGeometric progressions	Class 11: Sequences and Series	NumericalEasy
A easy numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{1}{3} + \frac{1}{3^2} + \ldots = \frac{1/3}{1-1/3} = \frac{1}{2}$. $0.16 = \left(\frac{2}{5}\right)^2 = (2.5)^{-2}$, so the value is $(2.5)^{-2\log_{2.5}(1/2)} = \left(\frac{1}{2}\right)^{-2} = 4$. Answer: 4.
74	Limits, Continuity and DifferentiabilityLimits of functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	NumericalMedium
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability The bracket factorises: $\left(1 - \cos\frac{x^2}{2}\right)\left(1 - \cos\frac{x^2}{4}\right)$. Using $1 - \cos\theta \approx \frac{\theta^2}{2}$: $\frac{x^4}{8}\cdot\frac{x^4}{32} = \frac{x^8}{256}$. The limit is $\frac{1}{256} = 2^{-8}$, so $k = 8$. Answer: 8.
75	CirclesRadius and centre	Class 11: Conic Sections	NumericalEasy
A easy numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Centre $(h, 2-h)$ with $0 < h < 2$ (first quadrant). Distance to $x = 3$: $3 - h$; distance to $y = 2$: $\|2 - (2-h)\| = h$. Equal radii: $3 - h = h \Rightarrow h = \frac{3}{2}$, $r = \frac{3}{2}$. Diameter $= 3$. Answer: 3.

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