JEE (Main) · 2020
JEE (Main) 2020 — 6 September, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
25 Q
- Motion of System of Particles and Rigid Body2
- Oscillations and Waves2
- Behaviour of Perfect Gas and Kinetic Theory2
13 easy · 12 medium · 0 hard
Chemistry
25 Q
- Hydrocarbons3
- d- and f-Block Elements3
- Organic Chemistry: Some Basic Principles and Techniques2
15 easy · 9 medium · 1 hard
Mathematics
25 Q
- Matrices and Determinants2
- Sequences and Series2
- Limits, Continuity and Differentiability2
11 easy · 14 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | KinematicsUniform circular motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The tip moves uniformly on a circle of radius $r = 0.1$ m with period $T = 60$ s.
$\omega = \frac{2\pi}{60} \approx 0.105$ rad/s, so the (centripetal) acceleration has magnitude $a = \omega^2 r \approx 0.011\times0.1 \approx 1.1\times10^{-3}\ \mathrm{m\,s^{-2}}$.
This is of the order of $10^{-3}$.
Answer: (B). | ||||
| 2 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Let the radius to the insect make angle $\theta$ with the vertical. Along the surface, the component of gravity is $mg\sin\theta$; the normal reaction is $mg\cos\theta$.
It is on the point of slipping when $mg\sin\theta = \mu mg\cos\theta$, i.e. $\tan\theta = 0.75$, so $\cos\theta = 0.8$.
Height above the bottom: $h = R(1-\cos\theta) = 1\times(1-0.8) = 0.20$ m.
Answer: (B). | ||||
| 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 axis passes through one corner and is parallel to the diagonal that does not pass through that corner.
Distances of the four masses from the axis: the corner on the axis, $0$; the two ends of the parallel diagonal, each $\frac{l}{\sqrt2}$ (half the other diagonal); the opposite corner, $\sqrt2\,l$ (the whole diagonal).
$I = m\left(0 + \frac{l^2}{2} + \frac{l^2}{2} + 2l^2\right) = 3ml^2$.
$L = I\omega = 3ml^2\omega$.
Answer: (B). | ||||
| 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 Take a ring of the surface at slant distance $s$ from the vertex ($0 \le s \le L$, $L$ the slant height). Its radius is $r = \frac{R}{L}s$ and its mass $dm = \frac{M}{\pi RL}\,2\pi r\,ds = \frac{2M}{L^2}s\,ds$.
$I = \int r^2\,dm = \int_0^L \frac{R^2s^2}{L^2}\cdot\frac{2M}{L^2}s\,ds = \frac{2MR^2}{L^4}\cdot\frac{L^4}{4} = \frac{MR^2}{2}$.
Answer: (A). | ||||
| 5 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation At the nearest and farthest points the velocity is perpendicular to the radius, and angular momentum is conserved: $m v_1 r_1 = m v_2 r_2$.
With $v_2 = \frac{v_1}{6}$: $\frac{r_1}{r_2} = \frac{v_2}{v_1} = \frac{1}{6}$.
Answer: (A). | ||||
| 6 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For an extension $x$, the restoring force is $F = \frac{YA}{L}x$, so the wire acts as a spring of constant $k = \frac{YA}{L}$.
$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{YA}{mL}}$.
Answer: (C). | ||||
| 7 | 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 $f = 3 + 2 = 5$. By equipartition, $U = \frac{f}{2}RT = \frac{5}{2}RT$ per mole.
$C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, so $\gamma = \frac{7}{5}$.
Answer: (C). | ||||
| 8 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Equilibrium: $\frac{dU}{dr} = \frac{6A}{r^7} - \frac{12B}{r^{13}} = 0 \Rightarrow r^6 = \frac{2B}{A}$, so $r = \left(\frac{2B}{A}\right)^{1/6}$.
Then $U = -\frac{A}{2B/A} + \frac{B}{(2B/A)^2} = -\frac{A^2}{2B} + \frac{A^2}{4B} = -\frac{A^2}{4B}$.
Answer: (A). | ||||
| 9 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Only the component of the source's velocity along the line to the observer, $v\cos\theta$, matters: $\nu = \frac{\nu_o c}{c - v\cos\theta}$, where $\theta$ is the angle between the velocity and the line from source to observer.
Far away on the approach, $\theta \to 0$ and $\nu$ is highest; as the source comes level with O, $\cos\theta$ falls smoothly to 0 and $\nu \to \nu_o$ (at about $t_0$); as it recedes, $\cos\theta$ becomes negative and $\nu$ falls smoothly below $\nu_o$, levelling off at $\frac{\nu_o c}{c+v}$.
So the frequency decreases smoothly and continuously, crossing $\nu_o$ near $t_0$ — graph (2). The sudden jump of graph (1) is for an observer on the track itself.
Answer: (B). | ||||
| 10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Put O at the origin, A $= (0, x_1)$, B $= (x_2, 0)$.
Field at O: from $Q_1$, $\frac{kQ_1}{x_1^2}$ along $-\hat{j}$; from $Q_2$, $\frac{kQ_2}{x_2^2}$ along $-\hat{i}$. So $\vec{E} = -k\left(\frac{Q_2}{x_2^2}\hat{i} + \frac{Q_1}{x_1^2}\hat{j}\right)$.
The hypotenuse AB has direction $(x_2, -x_1)$. Perpendicular means $\vec{E}\cdot(x_2\hat{i} - x_1\hat{j}) = 0$:
$\frac{Q_2}{x_2^2}x_2 - \frac{Q_1}{x_1^2}x_1 = 0 \Rightarrow \frac{Q_1}{Q_2} = \frac{x_1}{x_2}$.
Answer: (A). | ||||
| 11 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Time constant: $\tau = RC = 10^3\times10\times10^{-9} = 10\ \mu$s.
$0$–$5\ \mu$s (input 5 V): $V_o = 5(1 - e^{-0.5}) \approx 1.97$ V $\approx 2$ V.
$5$–$10\ \mu$s (input 0 V): the capacitor discharges, $V_o = 1.97e^{-0.5} \approx 1.19$ V — it does not fall to zero.
$10$–$15\ \mu$s (input 5 V): $V_o = 5 - (5 - 1.19)e^{-0.5} \approx 2.69$ V, climbing towards 3 V.
So the output builds up step by step and never returns to zero: graphs (2) and (4), which return to 0 at 10 $\mu$s, and (3), which saturates at 2 V, are wrong. Graph (1) shows the build-up from about 2 V towards 3 V.
Answer: (A). | ||||
| 12 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The velocity is perpendicular to $\vec{B}$, so the particle moves on a circle in the X-Y plane of radius $r = \frac{mv}{qB_0}$.
Starting at distance $d$ from the screen and moving straight towards it, the particle can get at most a distance $r$ closer before the circle turns it back (its centre lies on the line $x = d$).
It just grazes the screen when $r = d$, i.e. at $v = \frac{qB_0d}{m}$; for $r < d$ it turns back without reaching the screen.
The critical speed asked for is therefore $\frac{qdB_0}{m}$.
Answer: (A). | ||||
| 13 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The electric force on the electron is $-e\vec{E}$, along $-y$. The magnetic force must be equal and along $+y$.
Magnitude: $evB = eE \Rightarrow B = \frac{E}{v} = \frac{3\times10^4}{6\times10^6} = 5\times10^{-3}$ T.
Direction: $\vec{F} = -e(v\hat{i}\times B\hat{k}) = -evB(-\hat{j}) = +evB\,\hat{j}$, so $\vec{B}$ along $+z$ works.
Answer: (A). | ||||
| 14 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{100}\sqrt{\frac{80\times10^{-3}}{2\times10^{-6}}} = \frac{1}{100}\sqrt{4\times10^4} = \frac{200}{100} = 2$.
Answer: (C). | ||||
| 15 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens: $u = -1$ m, $f = 0.5$ m: $\frac{1}{v} = \frac{1}{0.5} - 1 = 1$, so $v = 1$ m — a real image 1 m behind the lens, i.e. 1 m in front of the mirror.
Mirror: its image is 1 m behind the mirror, i.e. 3 m from the lens. The reflected light goes back through the lens with this as its object, 3 m away.
Lens again: $\frac{1}{v} = \frac{1}{0.5} - \frac{1}{3} = \frac{5}{3}$, $v = 0.6$ m on the far side (the side the object is on), a real image.
Distance from the mirror: $2 + 0.6 = 2.6$ m, real.
Answer: (A). | ||||
| 16 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Path difference $5$ m $= \frac{\lambda}{4}$, i.e. a phase of $90^\circ$. Let each source alone give intensity $I_0$.
At A (on the line, beyond Q): the wave from P travels 5 m farther, losing $90^\circ$, which cancels P's $90^\circ$ lead. Phase difference $0$: $I_A = 4I_0$.
At C (on the line, beyond P): the wave from Q travels 5 m farther and lags another $90^\circ$. Phase difference $180^\circ$: $I_C = 0$.
At B (on the perpendicular bisector): equal paths, phase difference $90^\circ$: $I_B = 2I_0(1+\cos90^\circ) = 2I_0$.
$I_A : I_B : I_C = 4 : 2 : 0 = 2 : 1 : 0$.
Answer: (A). | ||||
| 17 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{\sqrt{2mK}}$: at equal kinetic energy, $\lambda \propto \frac{1}{\sqrt m}$.
$m_e < m_p < m_{He}$, so $\lambda_e > \lambda_p > \lambda_{He^{++}}$.
Answer: (C). | ||||
| 18 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Reaction: ${}^7_3\mathrm{Li} + {}^1_1\mathrm{H} \to 2\,{}^4_2\mathrm{He}$.
Mass defect: $\Delta m = 7.0160 + 1.0079 - 2\times4.0026 = 0.0187$ u.
Energy per reaction, with 1 u $\approx 931.5$ MeV: $Q \approx 0.0187\times931.5 \approx 17.4$ MeV $\approx 2.79\times10^{-12}$ J.
Number of Li nuclei in 20 g: $\frac{20}{7}\times6.02\times10^{23} \approx 1.72\times10^{24}$.
Total energy $\approx 1.72\times10^{24}\times2.79\times10^{-12} \approx 4.80\times10^{12}$ J $= \frac{4.80\times10^{12}}{3.6\times10^6} \approx 1.33\times10^6$ kWh.
Answer: (C). | ||||
| 19 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Reading the input waveforms: A is high from 5 to about 7.5, about 12.5 to 15 and about 17 to 19; B is high from 0 to about 7.5, 10 to 15, and from about 19 on.
Both inputs pass through NOT gates into the final gate. With the final gate read as a NOR (bubbled OR), $Y = \overline{\bar A + \bar B} = A\cdot B$ (De Morgan): Y is high only where A and B are both high — from 5 to about 7.5 and from about 12.5 to 15. That is graph (1).
(Graph (2) is $\overline{A\cdot B}$; graphs (3) and (4) match no combination of the inputs.)
Answer: (A). | ||||
| 20 | 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 Pitch $= 0.5$ mm (displacement per rotation). Least count $= \frac{0.5\ \text{mm}}{50} = 0.01$ mm $= 10\ \mu$m.
With the jaws closed, the circular scale's zero has already gone 4 divisions past the reference line, so the gauge reads $+4\times0.01 = +0.04$ mm when it should read zero: a positive zero error.
Answer: (A). | ||||
| 21 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalEasy | |
A easy numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\rho = \frac{m}{\frac{\pi}{6}d^3}$, so $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta d}{d} = 6.0\% + 3\times1.5\% = 10.5\%$.
$\frac{x}{100} = 10.5 \Rightarrow x = 1050$.
Answer: 1050. | ||||
| 22 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Momentum: $m\vec{v}_1 + m\vec{v}_2 = 2m\vec{V}$. With $|\vec{v}_1| = |\vec{v}_2| = v$ and angle $\theta$ between them, $|\vec{v}_1 + \vec{v}_2| = 2v\cos\frac{\theta}{2}$.
So $V = v\cos\frac{\theta}{2} = \frac{v}{2} \Rightarrow \frac{\theta}{2} = 60^\circ \Rightarrow \theta = 120^\circ$.
Answer: 120. | ||||
| 23 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | NumericalEasy | |
A easy numerical question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Start with $n$ moles. 25% dissociate, each giving two atoms: $0.75n + 2\times0.25n = 1.25n$ moles.
$\frac{P_2}{P_1} = \frac{n_2T_2/V_2}{n_1T_1/V_1} = 1.25\times\frac{2000}{250}\times\frac{1}{2} = 1.25\times8\times0.5 = 5$.
Answer: 5. | ||||
| 24 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current As drawn: P — inductor $L = 50$ mH — cell of 30 V (long plate on the inductor side) — resistor $R = 2\ \Omega$ — Q, with the current I flowing from Q towards P.
Walk from Q to P along the current:
across R (along the current) the potential falls by $IR = 2$ V;
across the cell, from its short (negative) plate to its long (positive) plate, it rises by 30 V;
across the inductor (along the current) it falls by $L\frac{dI}{dt} = 0.05\times(-100) = -5$ V, i.e. it rises by 5 V.
$V_P = V_Q - 2 + 30 + 5 \Rightarrow V_P - V_Q = 33$ V.
Answer: 33. | ||||
| 25 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalEasy | |
A easy numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $I = c\,\epsilon_0E_{rms}^2$, so $E_{rms}^2 = \frac{315/\pi}{3\times10^8\times8.86\times10^{-12}} = \frac{100.27}{2.658\times10^{-3}} \approx 3.772\times10^4$.
$E_{rms} \approx 194.2$ V/m, which is 194 to the nearest integer.
Answer: 194. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines Increasing $\mathrm{pK_b}$ means decreasing basicity.
I: N,N-dimethylaniline with $p$-$\mathrm{OCH_3}$ — the methoxy group's +R effect from the para position raises the electron density on N: the most basic.
II: N,N-dimethylaniline itself.
IV: N-methylaniline with $m$-OH — from the meta position OH acts only by its −I effect, lowering basicity below II.
III: N-methylaniline with $m$-CN — CN is strongly electron-withdrawing (−I): the least basic.
Basicity: I > II > IV > III, so $\mathrm{pK_b}$: I < II < IV < III.
Answer: (A). | ||||
| 27 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate is 3-methylbutan-2-yl mesylate, $\mathrm{(CH_3)_2CH{-}CH(OSO_2CH_3){-}CH_3}$; the mesylate is a good leaving group.
(i) Potassium tert-butoxide is a bulky base: in E2 it removes the more accessible hydrogen of the terminal $\mathrm{CH_3}$, giving the less substituted (Hofmann) alkene, 3-methylbut-1-ene, $\mathrm{(CH_3)_2CH{-}CH{=}CH_2}$.
(ii) Ozonolysis with oxidative work-up ($\mathrm{H_2O_2}$) gives acids: $\mathrm{(CH_3)_2CH{-}COOH}$ and $\mathrm{HCOOH}$ (from the $=\mathrm{CH_2}$ end).
Answer: (A). | ||||
| 28 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons B gives a yellow precipitate with $\mathrm{I_2/NaOH}$ (iodoform test: $\mathrm{CH_3CO-}$ or $\mathrm{CH_3CH(OH)-}$) and a silver mirror (an aldehyde): B is ethanal, $\mathrm{CH_3CHO}$.
C gives no iodoform; its reduction product D gives turbidity with Lucas reagent within 5 minutes, so D is a secondary alcohol and C a ketone without a $\mathrm{CH_3CO-}$ group.
A ($\mathrm{C_7H_{14}}$) minus the $\mathrm{CH_3CH=}$ part leaves five carbons for C: pentan-3-one, $\mathrm{(C_2H_5)_2C{=}O}$, whose reduction gives pentan-3-ol (secondary).
So A is $\mathrm{CH_3CH{=}C(C_2H_5)_2}$, 3-ethylpent-2-ene.
Check the others: (1) gives pentanal (reduced to a primary alcohol); (2) gives acetone (gives iodoform) and butanal; (3) gives pentan-2-one (gives iodoform).
Answer: (D). | ||||
| 29 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The diene is 1-methyl-4-nitrocyclohexa-1,4-diene: each double bond carries the substituent at one end. Each adds one HBr.
$\mathrm{C{=}C}$ bearing $\mathrm{CH_3}$: protonation gives the tertiary carbocation at the methyl-bearing carbon (Markovnikov), so Br goes to the carbon carrying $\mathrm{CH_3}$.
$\mathrm{C{=}C}$ bearing $\mathrm{NO_2}$: a carbocation next to the strongly electron-withdrawing $\mathrm{NO_2}$ is destabilised, so $\mathrm{H^+}$ adds to the carbon bearing $\mathrm{NO_2}$ and the cation forms on its neighbour; Br ends up on the carbon adjacent to the $\mathrm{C{-}NO_2}$ (anti-Markovnikov).
Product: 1-bromo-1-methyl, with Br on the carbon next to the $\mathrm{C{-}NO_2}$ — option (2).
Answer: (B). | ||||
| 30 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Ethene polymerised with a Ziegler–Natta catalyst gives high density polythene (HDPE), which is used for buckets, dustbins, bottles and pipes.
HDPE consists of linear chains that pack closely; it is denser, tougher and chemically inert — which is why it suits such articles. So (R) is true and explains (A).
Answer: (A). | ||||
| 31 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Acid-catalysed ($\mathrm{Hg^{2+}/H^+}$) hydration of the alkyne goes through the more stable vinyl cation. A positive charge on the carbon attached to the $p$-methoxyphenyl ring is stabilised by the +R effect of $\mathrm{OCH_3}$; on the carbon attached to the $p$-nitrophenyl ring it would be destabilised.
So water attacks the carbon next to the anisyl ring, and the enol tautomerises to the ketone $\mathrm{O_2N{-}C_6H_4{-}CH_2{-}CO{-}C_6H_4{-}OCH_3}$. The aryl methyl ether is not cleaved under these conditions.
Answer: (B). | ||||
| 32 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Geometrical isomerism needs two different groups on each doubly bonded carbon.
2-methylpent-1-ene, $\mathrm{CH_2{=}C(CH_3)CH_2CH_2CH_3}$: C1 has two H — no.
2-methylpent-2-ene, $\mathrm{(CH_3)_2C{=}CHCH_2CH_3}$: C2 has two $\mathrm{CH_3}$ — no.
4-methylpent-1-ene, $\mathrm{CH_2{=}CHCH_2CH(CH_3)_2}$: C1 has two H — no.
4-methylpent-2-ene, $\mathrm{CH_3CH{=}CHCH(CH_3)_2}$: each alkene carbon has H and a different alkyl group — cis and trans forms exist.
Answer: (D). | ||||
| 33 | 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 21 Sc, 25 Mn, 42 Mo and 72 Hf are all d-block (transition) elements.
The others contain non-transition elements: 9 F, 17 Cl, 34 Se, 38 Sr; 37 Rb, 50 Sn, 64 Gd (f-block); 32 Ge, 53 I.
Answer: (A). | ||||
| 34 | 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 Bronze is Cu + Sn — correct. German silver is Cu + Zn + Ni — correct. Wrought iron is made from cast iron by oxidising out the impurities in a reverberatory furnace lined with haematite — correct.
Brass is an alloy of copper and zinc, not nickel — incorrect.
Answer: (D). | ||||
| 35 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Down the group the lattice enthalpy of the sulphates changes little, while the hydration enthalpy of the cation falls sharply. The small $\mathrm{Be^{2+}}$ and $\mathrm{Mg^{2+}}$ ions have very high hydration enthalpies, so $\mathrm{BeSO_4}$ and $\mathrm{MgSO_4}$ are readily soluble; $\mathrm{CaSO_4}$, $\mathrm{SrSO_4}$ and $\mathrm{BaSO_4}$ are sparingly soluble.
Answer: (A). | ||||
| 36 | 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 (1) False: liquid $\mathrm{N_2}$ is used in cryosurgery.
(2) True: being inert at ordinary temperatures, $\mathrm{N_2}$ is used to provide an inert atmosphere and as a diluent for reactive chemicals.
(3) False: $\mathrm{N_2}$ combines with $\mathrm{O_2}$ only at very high temperature (electric arc, about 2000 K).
(4) False: $\mathrm{N_2}$ has no unpaired electrons; it is diamagnetic.
Answer: (B). | ||||
| 37 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Ce ($4f^0$ in $\mathrm{Ce^{4+}}$), Tb ($4f^7$ in $\mathrm{Tb^{4+}}$) and Dy show the +4 state.
Eu shows +2 ($4f^7$) and +3, but not +4.
Answer: (B). | ||||
| 38 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mu = \sqrt{n(n+2)}$ BM $= 5.9$ needs $n = 5$ unpaired electrons.
$\mathrm{[MnBr_4]^{2-}}$: $\mathrm{Mn^{2+}}$, $d^5$, high spin in a tetrahedral field: 5 unpaired, $\sqrt{35} \approx 5.92$ BM.
$\mathrm{Ni(CO)_4}$: Ni(0), $d^{10}$ — 0. $\mathrm{[NiCl_4]^{2-}}$: $\mathrm{Ni^{2+}}$, $d^8$ tetrahedral — 2 unpaired (2.83 BM). $\mathrm{[Ni(CN)_4]^{2-}}$: square planar $d^8$ — 0.
Answer: (B). | ||||
| 39 | 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 About 1 ppm of fluoride is safe and beneficial: it converts the hydroxyapatite of tooth enamel into the harder fluorapatite. Above about 2 ppm it causes mottling of teeth, and higher concentrations harm bones.
Answer: (A). | ||||
| 40 | SolutionsConcentration terms | Class 12: Solutions | Single correctMedium | |
A medium single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Mass of solution $= n_1M_1 + n_2M_2$ g; volume $= \frac{n_1M_1 + n_2M_2}{d}$ mL $= \frac{n_1M_1 + n_2M_2}{1000d}$ L.
$C_2 = \frac{n_2}{V} = \frac{1000\,d\,n_2}{n_1M_1 + n_2M_2}$. Divide numerator and denominator by $n_1 + n_2$:
$C_2 = \frac{1000\,d\,x_2}{x_1M_1 + x_2M_2} = \frac{1000\,d\,x_2}{(1-x_2)M_1 + x_2M_2} = \frac{1000\,d\,x_2}{M_1 + x_2(M_2 - M_1)}$.
Answer: (A). | ||||
| 41 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics van 't Hoff: $\ln\frac{K_2}{K_1} = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ with $T_1 = 298$ K, $T_2 = 373$ K.
$\ln 10 = 2.303 = \frac{\Delta H^\circ}{8.314}\times\frac{75}{298\times373} = \frac{\Delta H^\circ}{8.314}\times6.75\times10^{-4}$, so $\Delta H^\circ \approx 28.4$ kJ/mol.
$\Delta G^\circ_1 = -RT_1\ln K_1 = -8.314\times298\times2.303 \approx -5.71$ kJ/mol.
$\Delta G^\circ_2 = -RT_2\ln K_2 = -8.314\times373\times4.605 \approx -14.28$ kJ/mol $\approx -14.29$.
Answer: (B). | ||||
| 42 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Only gases count: $\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}$.
$K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^{-1/2}$, so $K_c = K_p(RT)^{1/2}$.
Answer: (B). | ||||
| 43 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium pOH (at 25 °C): HCl, pH 2, so pOH 12; NaCl, neutral, pOH 7; $\mathrm{CH_3COONa}$, basic by hydrolysis, pOH a little below 7; NaOH, pOH 2.
Decreasing pOH: (A) > (D) > (C) > (B).
Answer: (D). | ||||
| 44 | 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 rate $= k[\text{conc.}]^n$, so $\log[\text{rate}] = \log k + n\log[\text{conc.}]$: the slope of each line is its order.
From the graph the steepest line is [D], then [B], then [A], and [C] is the least steep.
So $d > b > a > c$.
Answer: (C). | ||||
| 45 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter Micelles form only above a particular temperature, the Kraft temperature ($T_k$), and above a particular concentration, the critical micelle concentration.
Answer: (B). | ||||
| 46 | 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 Molar mass of AgBr $= 188$ g/mol. Mass of Br in 1.88 g AgBr $= \frac{80}{188}\times1.88 = 0.80$ g.
% Br $= \frac{0.80}{1.6}\times100 = 50$.
Answer: 50. | ||||
| 47 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Perchloric acid, $\mathrm{HClO_4}$: chlorine is bonded to one OH group and to three oxygen atoms by Cl=O bonds (tetrahedral about Cl).
Answer: 3. | ||||
| 48 | 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 Same amount of gas, same temperature: Boyle's law, $P_1V_1 = P_2V_2$, with $V \propto r^3$.
$P_2 = P_1\left(\frac{r_1}{r_2}\right)^3 = 48\times10^{-3}\times\left(\frac{3}{12}\right)^3 = \frac{48\times10^{-3}}{64} = 0.75\times10^{-3} = 750\times10^{-6}$ bar.
Answer: 750. | ||||
| 49 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\Delta T_b = iK_bm$. For $\mathrm{CaCl_2}$, $i = 3$: $i\,m = 0.15$.
For the complex: $i\times0.10 = 2\times0.15 = 0.30 \Rightarrow i = 3$.
With coordination number 6 and all $x$ $\mathrm{NH_3}$ inside, the complex is $\mathrm{[Cr(NH_3)_xCl_{6-x}]Cl_{x-3}}$, giving $1 + (x-3)$ ions.
$x - 2 = 3 \Rightarrow x = 5$: $\mathrm{[Cr(NH_3)_5Cl]Cl_2}$.
Answer: 5. | ||||
| 50 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Moles of $\mathrm{KClO_3}$ $= \frac{10}{122} = 0.0820$. Each needs 6 F: charge $= 6\times0.0820\times96500 \approx 4.746\times10^4$ C.
Effective current $= 0.60\times2 = 1.2$ A.
$t = \frac{4.746\times10^4}{1.2} \approx 3.955\times10^4$ s $\approx 10.99$ h, i.e. 11 hours.
Answer: 11. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\sqrt{x^2 + y^2} \le 1 + x$. This needs $1 + x \ge 0$, and then squaring is reversible:
$x^2 + y^2 \le 1 + 2x + x^2 \iff y^2 \le 2x + 1 = 2\left(x + \frac{1}{2}\right)$.
(Any point with $y^2 \le 2x + 1$ has $x \ge -\frac{1}{2}$, so $1 + x > 0$ holds automatically.)
Answer: (B). | ||||
| 52 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha + \beta = 64$, $\alpha\beta = 256$; both roots are positive (discriminant $> 0$, sum and product positive).
$\left(\frac{\alpha^3}{\beta^5}\right)^{1/8} = \frac{\alpha^{3/8}}{\beta^{5/8}} = \frac{\alpha}{(\alpha\beta)^{5/8}}$, and similarly the second term is $\frac{\beta}{(\alpha\beta)^{5/8}}$.
Sum $= \frac{\alpha + \beta}{(\alpha\beta)^{5/8}} = \frac{64}{(2^8)^{5/8}} = \frac{64}{32} = 2$.
Answer: (C). | ||||
| 53 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Subtract the first equation from the others: $y + 2z = 3$ and $2y + (\lambda - 1)z = \mu - 2$.
For infinitely many solutions the second must be twice the first: $2y + 4z = 6$. So $\lambda - 1 = 4$ and $\mu - 2 = 6$: $\lambda = 5$, $\mu = 8$.
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 Write $c = \cos^2x$, $s = \sin^2x$, $t = \sin2x$. Apply $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$:
$\begin{vmatrix}0 & 1 & -1\\ 1 & 0 & -1\\ c & s & 1+t\end{vmatrix} = 0 - 1\cdot\big((1+t) + c\big) + (-1)\cdot(s - 0) = -(1 + t + c + s) = -(2 + \sin2x)$.
Since $-1 \le \sin2x \le 1$, the value runs from $-3$ to $-1$: $(m, M) = (-3, -1)$.
Answer: (A). | ||||
| 55 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Treat each family as a block: the 3 blocks can be ordered in $3!$ ways. Inside the blocks: $3!$, $3!$ and $4!$.
Total $= 3!\times3!\times3!\times4! = (3!)^3\cdot(4!)$.
Answer: (A). | ||||
| 56 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $3^{200} = 9^{100} = (1 + 8)^{100} = 1 + 8k$ for an integer $k$ (binomial expansion).
So $\frac{3^{200}}{8} = k + \frac{1}{8}$ and the fractional part is $\frac{1}{8}$.
Answer: (B). | ||||
| 57 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Regroup the left side as a sum of squares:
$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) = (ap - b)^2 + (bp - c)^2 + (cp - d)^2 = 0$.
For real numbers each square is zero: $b = ap$, $c = bp$, $d = cp$, so $\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$.
Hence a, b, c, d are in G.P. (with ratio p).
Answer: (C). | ||||
| 58 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let $f(1) = k$. Then $f(n) = f(1)^n = k^n$ for every natural n, so the sum is a geometric series:
$\sum_{x=1}^{\infty} k^x = \frac{k}{1-k} = 2 \Rightarrow k = \frac{2}{3}$.
$\frac{f(4)}{f(2)} = k^2 = \frac{4}{9}$.
Answer: (D). | ||||
| 59 | 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 Put $h = x - 1$. Since $\int t\cos(t^2)\,dt = \frac{1}{2}\sin(t^2)$:
As printed, the numerator is $\frac{1}{2}\sin(h^4) \approx \frac{h^4}{2}$ and the denominator $h\sin h \approx h^2$, so the limit is $\lim \frac{h^2}{2} = 0$ — which is not among the options.
With the upper limit read as $(x-1)$, the numerator is $\frac{1}{2}\sin(h^2) \approx \frac{h^2}{2}$ and the limit is $\lim_{h\to0}\frac{\frac{1}{2}\sin(h^2)}{h\sin h} = \frac{1}{2}\cdot\frac{h^2}{h^2} = \frac{1}{2}$.
Answer: (C). | ||||
| 60 | Applications of DerivativesRate of change | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on rate of change, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Average speed over $[t_1, t_2]$: $\frac{f(t_2) - f(t_1)}{t_2 - t_1} = \frac{a(t_2^2 - t_1^2) + b(t_2 - t_1)}{t_2 - t_1} = a(t_1 + t_2) + b$.
Instantaneous speed: $f'(t) = 2at + b$ (positive for $t > 0$). Setting $2at + b = a(t_1 + t_2) + b$ gives $t = \frac{t_1 + t_2}{2}$.
Answer: (B). | ||||
| 61 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Integrate $I_2$ by parts with $u = (1 - x^{50})^{101}$, $dv = dx$:
$I_2 = \big[x(1 - x^{50})^{101}\big]_0^1 + \int_0^1 x\cdot101(1 - x^{50})^{100}\cdot50x^{49}\,dx = 5050\int_0^1 x^{50}(1 - x^{50})^{100}\,dx$.
Write $x^{50} = 1 - (1 - x^{50})$: $I_2 = 5050(I_1 - I_2)$, so $5051I_2 = 5050I_1$ and $\alpha = \frac{5050}{5051}$.
Answer: (A). | ||||
| 62 | 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 The region is symmetric in both axes; find the first-quadrant part and multiply by 4.
First quadrant: inside the triangle $x + y \le 1$ (area $\frac{1}{2}$) and $x \le 2y^2$.
Remove the part with $x > 2y^2$: the parabola $x = 2y^2$ meets $x + y = 1$ where $2y^2 + y - 1 = 0$, i.e. $y = \frac{1}{2}$. That part has area $\int_0^{1/2}\big((1 - y) - 2y^2\big)\,dy = \frac{1}{2} - \frac{1}{8} - \frac{1}{12} = \frac{7}{24}$.
First-quadrant area $= \frac{1}{2} - \frac{7}{24} = \frac{5}{24}$; total $= 4\times\frac{5}{24} = \frac{5}{6}$.
Answer: (A). | ||||
| 63 | 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 $1 + x^2 + y^2 + x^2y^2 = (1 + x^2)(1 + y^2)$, so $\frac{y\,dy}{\sqrt{1+y^2}} = -\frac{\sqrt{1+x^2}}{x}\,dx$.
Left: $\int\frac{y\,dy}{\sqrt{1+y^2}} = \sqrt{1+y^2}$.
Right: put $u = \sqrt{1+x^2}$, so $x^2 = u^2 - 1$ and $dx = \frac{u\,du}{x}$: $\int\frac{u}{x}\cdot\frac{u\,du}{x} = \int\frac{u^2}{u^2-1}\,du = u + \frac{1}{2}\ln\frac{u-1}{u+1}$.
So $\sqrt{1+y^2} = -\sqrt{1+x^2} - \frac{1}{2}\ln\frac{u-1}{u+1} + C$, i.e. $\sqrt{1+y^2} + \sqrt{1+x^2} = \frac{1}{2}\log_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right) + C$.
Answer: (C). | ||||
| 64 | 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 The incident ray makes $30^\circ$ with the vertical line $x = 1$, i.e. $60^\circ$ with the horizontal. Going from $(2, 2\sqrt3)$ to $x = 1$ (one unit left), it drops $\tan60^\circ = \sqrt3$: A $= (1, \sqrt3)$. (Rising instead, the reflected ray would go upwards and never meet the x-axis.)
Reflection in a vertical line reverses the horizontal component: the incident direction $(-1, -\sqrt3)$ becomes $(1, -\sqrt3)$.
Line AB: $y - \sqrt3 = -\sqrt3(x - 1)$, i.e. $y = -\sqrt3x + 2\sqrt3$ (it meets the x-axis at B $= (2, 0)$).
At $x = 3$: $y = -\sqrt3$. So AB passes through $(3, -\sqrt3)$; the other points do not satisfy it.
Answer: (D). | ||||
| 65 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The foot of the perpendicular from a focus to any tangent of an ellipse lies on its auxiliary circle $x^2 + y^2 = a^2$; here $a^2 = 4$.
Check: $(-1, \sqrt3)$: $1 + 3 = 4$ ✓; $(1, 2)$: 5; $(-1, \sqrt2)$: 3; $(-2, \sqrt3)$: 7.
Answer: (A). | ||||
| 66 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent of slope m to $y^2 = 4(x+1)$: $y = m(x + 1) + \frac{1}{m}$.
Perpendicular tangent (slope $-\frac{1}{m}$) to $y^2 = 8(x+2)$: $y = -\frac{1}{m}(x + 2) + \frac{2}{-1/m} = -\frac{x+2}{m} - 2m$.
At the intersection: $m(x + 1) + \frac{1}{m} = -\frac{x+2}{m} - 2m$. Multiply by m: $m^2(x + 1) + 1 + x + 2 + 2m^2 = 0 \Rightarrow (m^2 + 1)(x + 3) = 0$.
So $x + 3 = 0$.
Answer: (D). | ||||
| 67 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Line 1: through $P_1(1, -1, 0)$ with direction $\vec{d_1} = (0, -1, 1)$.
Line 2: direction $(1, 1, 1)\times(2, -1, 1) = (2, 1, -3)$; a point with $z = 0$: $x + y = -1$, $2x - y = -3$, so $P_2\left(-\frac{4}{3}, \frac{1}{3}, 0\right)$.
$\vec{d_1}\times\vec{d_2} = (0, -1, 1)\times(2, 1, -3) = (2, 2, 2)$, of length $2\sqrt3$.
$\overrightarrow{P_1P_2} = \left(-\frac{7}{3}, \frac{4}{3}, 0\right)$; dot product with $(2, 2, 2)$: $-\frac{14}{3} + \frac{8}{3} = -2$.
Shortest distance $= \frac{|-2|}{2\sqrt3} = \frac{1}{\sqrt3}$.
Answer: (C). | ||||
| 68 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Variance is unchanged by a shift: with $d_i = x_i - a$,
$\sigma^2 = \frac{\sum d_i^2}{n} - \left(\frac{\sum d_i}{n}\right)^2 = \frac{na}{n} - 1^2 = a - 1$.
$\sigma = \sqrt{a - 1}$.
Answer: (C). | ||||
| 69 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Total selections: $\binom{11}{3} = 165$.
Three numbers in A.P. (arranged in increasing order) with common difference d: the first can be any of the first $11 - 2d$ numbers, $d = 1, \ldots, 5$: $9 + 7 + 5 + 3 + 1 = 25$.
Probability $= \frac{25}{165} = \frac{5}{33}$.
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 By distribution, $p \vee (\sim p \wedge q) = (p \vee \sim p) \wedge (p \vee q) = p \vee q$.
Its negation is $\sim(p \vee q) = \sim p \wedge \sim q$ (De Morgan).
Answer: (A). | ||||
| 71 | Sets, Relations and FunctionsSets and their representation | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical question on sets and their representation, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $2^m - 2^n = 112 = 16\times7 = 2^4(2^3 - 1)$. Since $2^n(2^{m-n} - 1) = 2^4\times7$ with $2^{m-n} - 1$ odd: $n = 4$ and $2^{m-n} = 8$, so $m = 7$.
$m\cdot n = 28$.
Answer: 28. | ||||
| 72 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f'(0) = 0$ from both sides (each piece is $O(x^2)$ near 0).
For $x < 0$: $f'(x) = 5x^4\sin\frac{1}{x} - x^3\cos\frac{1}{x} + 10x$; for $x > 0$: $f'(x) = 5x^4\cos\frac{1}{x} + x^3\sin\frac{1}{x} + 2\lambda x$.
$f''(0^-) = \lim_{h\to0^-}\frac{f'(h) - 0}{h} = 10$ (the other terms are $O(h^2)$); $f''(0^+) = 2\lambda$.
$2\lambda = 10 \Rightarrow \lambda = 5$.
Answer: 5. | ||||
| 73 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalEasy | |
A easy numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let AM $= x$. Then $MD^2 + MC^2 = (x^2 + 64) + \big((10 - x)^2 + 121\big) = 2x^2 - 20x + 285$.
It is least where $4x - 20 = 0$: $x = 5$ m.
Answer: 5. | ||||
| 74 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra With $\theta$ the angle between the unit vectors: $|\vec{a} + \vec{b}| = 2\cos\frac{\theta}{2}$, $|\vec{a} - \vec{b}| = 2\sin\frac{\theta}{2}$ ($0 \le \theta \le \pi$).
Expression $= 2\left(\sqrt3\cos\frac{\theta}{2} + \sin\frac{\theta}{2}\right) = 4\sin\left(\frac{\theta}{2} + \frac{\pi}{3}\right)$, whose greatest value is 4 (at $\theta = \frac{\pi}{3}$).
Answer: 4. | ||||
| 75 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let the height be h. The first point is at horizontal distance h from the foot (elevation $45^\circ$).
After 80 m up the $30^\circ$ slope the observer has risen $40$ m and moved $40\sqrt3$ m horizontally, so is $h - 40\sqrt3$ from the foot and $h - 40$ below the top.
$\tan75^\circ = 2 + \sqrt3 = \frac{h - 40}{h - 40\sqrt3}$.
$h - 40 = (2 + \sqrt3)h - 80\sqrt3 - 120 \Rightarrow (1 + \sqrt3)h = 80 + 80\sqrt3 \Rightarrow h = 80$.
Answer: 80. | ||||
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