JEE (Advanced) · 2015
JEE Advanced 2015, question by question
Which chapter each question came from, how the paper was spread, 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. One year is not a forecast.
Physics
40 Q
- Atoms and Nuclei6
- Optics6
- Electrostatics4
11 easy · 25 medium · 4 hard
Chemistry
40 Q
- Hydrocarbons5
- Organic Chemistry: Some Basic Principles and Techniques3
- Coordination Compounds3
16 easy · 20 medium · 4 hard
Mathematics
40 Q
- Integral Calculus8
- Conic Sections6
- Trigonometry3
7 easy · 27 medium · 6 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Conic Sections
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 11: Probability; Class 12: Probability
Class 12: Application of Derivatives
Class 12: Vector Algebra
Class 11: Permutations and Combinations
Class 12: Matrices; Class 12: Determinants
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 12: Differential Equations
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Class 11: Complex Numbers and Quadratic Equations
Class 11: Sequences and Series
Class 11: Complex Numbers and Quadratic Equations
Class 12: Atoms; Class 12: Nuclei
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 11: Systems of Particles and Rotational Motion
Class 11: Units and Measurement
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 11: Gravitation
Class 11: Oscillations; Class 11: Waves
Class 12: Current Electricity
No matching NCERT chapter
Class 11: Kinetic Theory
Class 12: Dual Nature of Radiation and Matter
Class 11: Work, Energy and Power
Class 11: Thermodynamics
Class 11: Hydrocarbons
Class 11: Organic Chemistry – Some Basic Principles and Techniques
Class 12: Coordination Compounds
No matching NCERT chapter
Class 11: Chemical Bonding and Molecular Structure
Class 12: Electrochemistry
Class 12: Aldehydes, Ketones and Carboxylic Acids
Class 12: Amines
No matching NCERT chapter
Class 11: Redox Reactions
Class 11: Equilibrium
Class 11: Thermodynamics
Class 11: Some Basic Concepts of Chemistry
Class 11: Structure of Atom
Class 12: Solutions
Class 12: Biomolecules
Class 12: The d- and f-Block Elements
No matching NCERT chapter
Class 12: Alcohols, Phenols and Ethers
No matching NCERT chapter
No matching NCERT chapter
03
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The field of an infinite line charge is radial, in planes perpendicular to the line. So the flux through the strip is the fraction of the total flux that leaves the line through the angle the strip subtends at it.
Total flux from a length $L$ of the line: $\frac{\lambda L}{\varepsilon_0}$, spread uniformly over the full angle $2\pi$ around the line.
The strip spans $x$ from $-\frac{a}{2}$ to $\frac{a}{2}$ at a perpendicular distance $\frac{\sqrt{3}}{2}a$ below the line. Each half subtends $\tan\theta = \frac{a/2}{\sqrt{3}a/2} = \frac{1}{\sqrt{3}}$, so $\theta = 30^\circ$, and the whole strip subtends $60^\circ = \frac{2\pi}{6}$.
Flux through $ABCD$ $= \frac{1}{6}\cdot\frac{\lambda L}{\varepsilon_0}$, so $n = 6$.
Answer: 6. | ||||
| 2 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Photon energy: $E = \frac{hc}{\lambda} = \frac{1242}{90} = 13.8$ eV.
Energy needed to ionize from level $n$: $13.8 - 10.4 = 3.4$ eV.
For hydrogen this is $\frac{13.6}{n^2}$ eV, so $n^2 = \frac{13.6}{3.4} = 4$ and $n = 2$.
Answer: 2. | ||||
| 3 | GravitationEscape velocity | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Gravity falls as $\frac{1}{r^2}$, so $g' = \frac{g}{4}$ means the bullet stops at $r = 2R$ (planet radius $R$).
Energy conservation: $\frac{1}{2}v^2 - \frac{GM}{R} = -\frac{GM}{2R}$, so $v^2 = \frac{GM}{R}$.
Escape velocity: $v_{esc}^2 = \frac{2GM}{R} = 2v^2$, so $v_{esc} = v\sqrt{2}$ and $N = 2$.
Answer: 2. | ||||
| 4 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion A uniform disc rolling without slipping has kinetic energy $\frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{1}{2}mR^2\cdot\frac{v^2}{R^2} = \frac{3}{4}mv^2$. Rolling without slipping does no work against friction, so mechanical energy is conserved.
Disc 1: $\frac{3}{4}v_B^2 = \frac{3}{4}v_1^2 + g(30)$.
Disc 2: $\frac{3}{4}v_D^2 = \frac{3}{4}v_2^2 + g(27)$.
With $v_B = v_D$: $\frac{3}{4}(v_2^2 - v_1^2) = g(30 - 27) = 30$, so $v_2^2 - 9 = 40$, $v_2^2 = 49$ and $v_2 = 7$ m/s.
Answer: 7. | ||||
| 5 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Stefan's law: $P = \sigma\cdot4\pi R^2 T^4$, so $\frac{P_A}{P_B} = \left(\frac{R_A}{R_B}\right)^2\left(\frac{T_A}{T_B}\right)^4$.
$10^4 = (400)^2\left(\frac{T_A}{T_B}\right)^4 = 1.6\times10^5\left(\frac{T_A}{T_B}\right)^4$, so $\left(\frac{T_A}{T_B}\right)^4 = \frac{1}{16}$ and $\frac{T_A}{T_B} = \frac{1}{2}$.
Wien's law: $\lambda_{max}T$ = constant, so $\frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A} = 2$.
Answer: 2. | ||||
| 6 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The power available is proportional to the activity, which halves every half-life $T$.
The village needs 12.5% $= \frac{1}{8}$ of the initial power. The available power falls to $\frac{1}{8}$ of its initial value after $\left(\frac{1}{2}\right)^n = \frac{1}{8}$, i.e. $n = 3$ half-lives.
So the plant meets the need for $3T$ years.
Answer: 3. | ||||
| 7 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics From the figure, $S_1$ is a distance $d$ above the water surface (its light travels in air) and $S_2$ a distance $d$ below it (its light travels in water). Both are at the same geometric distance $\sqrt{x^2 + d^2}$ from a point on the surface at distance $x$.
Optical paths: from $S_1$, $1\cdot\sqrt{x^2+d^2}$; from $S_2$, $\frac{4}{3}\sqrt{x^2+d^2}$.
Path difference $= \frac{1}{3}\sqrt{x^2+d^2}$. For a maximum it equals $m\lambda$: $\sqrt{x^2+d^2} = 3m\lambda$, so $x^2 = 9m^2\lambda^2 - d^2$.
Comparing, $p^2 = 9$ and $p = 3$.
Answer: 3. | ||||
| 8 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Mirror (focal length 10 cm, object 15 cm): $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{10} + \frac{1}{15} = -\frac{1}{30}$, so the image is real, 30 cm in front of the mirror, with $m = -\frac{v}{u} = -2$.
This image is $50 - 30 = 20$ cm from the lens and acts as its object.
In air the lens has $f = 10$ cm: $\frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}$, $v = 20$ cm, $m = \frac{v}{u} = \frac{20}{-20} = -1$. So $M_1 = (-2)(-1) = 2$ (erect).
In the medium the mirror is unchanged, but the lens focal length changes: $\frac{1}{f} \propto \left(\frac{1.5}{7/6} - 1\right) = \frac{2}{7}$ against $(1.5 - 1) = \frac{1}{2}$ in air, so $f' = 10\times\frac{1/2}{2/7} = 17.5$ cm.
Lens: $\frac{1}{v} = \frac{1}{17.5} - \frac{1}{20} = \frac{1}{140}$, $v = 140$ cm, $m = \frac{140}{-20} = -7$. So $M_2 = (-2)(-7) = 14$.
$\left|\frac{M_2}{M_1}\right| = \frac{14}{2} = 7$.
Answer: 7. | ||||
| 9 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Vernier callipers: 1 MSD $= \frac{1}{8}$ cm $= 1.25$ mm. 5 VSD = 4 MSD, so 1 VSD $= 0.8$ MSD and the least count is $1 - 0.8 = 0.2$ MSD $= 0.25$ mm.
Screw gauge: one rotation moves it two linear-scale divisions, so pitch = 2 × (linear-scale division), and least count = pitch/100.
(A), (B): pitch $= 2\times0.25 = 0.5$ mm, so least count $= \frac{0.5}{100} = 0.005$ mm. (B) is true, (A) is false.
(C), (D): a linear-scale division $= 2\times0.25 = 0.5$ mm, so pitch $= 1$ mm and least count $= \frac{1}{100} = 0.01$ mm. (C) is true, (D) is false.
Answer: (B), (C). | ||||
| 10 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Multiple correctMedium | |
A medium multiple correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Dimensions: $[h] = ML^2T^{-1}$, $[c] = LT^{-1}$, $[G] = M^{-1}L^3T^{-2}$.
Mass: try $h^a c^b G^d$. $M$: $a - d = 1$; $L$: $2a + b + 3d = 0$; $T$: $-a - b - 2d = 0$. Adding the $L$ and $T$ equations: $a + d = 0$, so $a = \frac{1}{2}$, $d = -\frac{1}{2}$, $b = \frac{1}{2}$. $M = \sqrt{\frac{hc}{G}}$.
Length: $M$: $a - d = 0$; $L$: $2a + b + 3d = 1$; $T$: $-a - b - 2d = 0$. With $d = a$: $5a + b = 1$ and $-3a - b = 0$, so $a = \frac{1}{2}$, $b = -\frac{3}{2}$. $L = \sqrt{\frac{hG}{c^3}}$.
(A) $M \propto \sqrt{c}$: true. (B) $M \propto \frac{1}{\sqrt{G}}$, not $\sqrt{G}$: false. (C) $L \propto \sqrt{h}$: true. (D) $L \propto \sqrt{G}$: true.
Answer: (A), (C), (D). | ||||
| 11 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Multiple correctHard | |
A hard multiple correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For SHM of mass $m$ and amplitude $A$: $p_{max} = m\omega A$, and $E = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}\omega\,(A\,p_{max})$.
Oscillator 1: $A = a$, $p_{max} = b$, so $\omega_1 = \frac{b}{ma}$ and $\frac{E_1}{\omega_1} = \frac{ab}{2}$.
Oscillator 2: $A = R$, $p_{max} = R$, so $\omega_2 = \frac{1}{m}$ and $\frac{E_2}{\omega_2} = \frac{R^2}{2}$.
$\frac{\omega_2}{\omega_1} = \frac{a}{b} = n^2$: (B) is true.
With $b = \frac{a}{n^2}$ and $R = \frac{a}{n}$: $ab = \frac{a^2}{n^2} = R^2$, so $\frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$: (D) is true.
Then $E_1\omega_1 = E_2\omega_2$ would need $\omega_1^2 = \omega_2^2$, i.e. $n = 1$: (A) is false in general.
$\omega_1\omega_2 = \frac{b}{m^2a} = \frac{1}{m^2n^2}$, not $n^2$ (and not even dimensionless): (C) is false.
Answer: (B), (D). | ||||
| 12 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Multiple correctMedium | |
A medium multiple correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion There is no external torque about the axis, so angular momentum is conserved. Initially both masses are at O: $L = MR^2\omega$.
Later, with the masses at $r_1$ and $r_2$: $\left[MR^2 + \frac{M}{8}(r_1^2 + r_2^2)\right]\frac{8}{9}\omega = MR^2\omega$.
So $MR^2 + \frac{M}{8}(r_1^2 + r_2^2) = \frac{9}{8}MR^2$, giving $r_1^2 + r_2^2 = R^2$.
With $r_1 = \frac{3}{5}R$: $r_2^2 = R^2 - \frac{9}{25}R^2 = \frac{16}{25}R^2$, so $r_2 = \frac{4}{5}R$, option (D).
(A) $\frac{2}{3}R$ and (B) $\frac{1}{3}R$ do not satisfy $r_1^2 + r_2^2 = R^2$.
The official key accepts (C) OR (D) OR (C) and (D); the marked option follows the key's first-listed set, (C) — see the note.
Answer (by the working): $\frac{4}{5}R$, (D); official key: (C) or (D) or both. | ||||
| 13 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Midway between two equal line charges the fields cancel, so each point charge is in equilibrium. Let the lines be at $x = \pm d$ and displace the charge to small $x$.
Net field: $E = \frac{\lambda}{2\pi\varepsilon_0}\left(\frac{1}{d + x} - \frac{1}{d - x}\right) \approx -\frac{\lambda}{\pi\varepsilon_0 d^2}x$, i.e. pointing back towards the centre.
For $+q$ the force $qE \propto -x$ is restoring, so it executes SHM.
For $-q$ the force is $+\frac{q\lambda}{\pi\varepsilon_0 d^2}x$, pushing it further the way it was displaced: it keeps moving in the direction of its displacement.
So only (C) is right; (A), (B) and (D) each get one of the two charges wrong.
Answer: (C). | ||||
| 14 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Refraction at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$, light travelling left to right.
Face of $S_1$ (glass to air, convex towards the air, so its centre is in the glass: $R = -10$ cm), $u = -50$ cm: $\frac{1}{v} + \frac{1.5}{50} = \frac{1 - 1.5}{-10} = 0.05$, so $\frac{1}{v} = 0.05 - 0.03 = 0.02$ and $v = +50$ cm. The rays converge to a point 50 cm beyond the face of $S_1$.
Face of $S_2$ (air to glass, convex towards the air, centre in the glass: $R = +10$ cm). For rays parallel inside $S_2$, $v = \infty$: $0 - \frac{1}{u} = \frac{1.5 - 1}{10}$, so $u = -20$ cm. The object must be 20 cm in front of $S_2$'s face.
The image from $S_1$ is that object: $d = 50 + 20 = 70$ cm.
Answer: (B). | ||||
| 15 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctMedium | |
A medium multiple 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 In a uniform field the force on any current-carrying wire equals the force on the straight wire joining its ends: $\vec{F} = I\,\vec{l}_{eff}\times\vec{B}$.
Here the ends lie on the $x$-axis a distance $L + R + R + L = 2(L + R)$ apart, so $\vec{l}_{eff} = 2(L + R)\hat{x}$.
(A) $\vec{B} = B\hat{z}$: $F = 2I(L + R)B \propto (L + R)$. True.
(B) $\vec{B} = B\hat{x}$: $\hat{x}\times\hat{x} = 0$, $F = 0$. True.
(C) $\vec{B} = B\hat{y}$: $\hat{x}\times\hat{y} = \hat{z}$, $F = 2I(L + R)B \propto (L + R)$. True.
(D) contradicts (A): false.
Answer: (A), (B), (C). | ||||
| 16 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Multiple correctMedium | |
A medium multiple correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory (A) Internal energy: H$_2$ (diatomic) $\frac{5}{2}RT$, He $\frac{3}{2}RT$; total $4RT$ for 2 mol, so $2RT$ per mole. True.
(B) Mixture: $C_V = \frac{\frac{5}{2}R + \frac{3}{2}R}{2} = 2R$, $C_P = 3R$, $\gamma = \frac{3}{2}$; mean molar mass $\frac{2 + 4}{2} = 3$ g/mol. Speed of sound $v = \sqrt{\frac{\gamma RT}{M}}$.
$\frac{v_{mix}}{v_{He}} = \sqrt{\frac{\gamma_{mix}/M_{mix}}{\gamma_{He}/M_{He}}} = \sqrt{\frac{3/2 \div 3}{5/3 \div 4}} = \sqrt{\frac{1/2}{5/12}} = \sqrt{\frac{6}{5}}$. True.
(C), (D) $v_{rms} = \sqrt{\frac{3RT}{M}}$: $\frac{v_{He}}{v_{H_2}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$. (D) true, (C) false.
Answer: (A), (B), (D). | ||||
| 17 | Current ElectricityResistivity | Class 12: Current Electricity | Multiple correctMedium | |
A medium multiple correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Between P and Q the Al and Fe parts run side by side over the same length $l = 50$ mm $= 0.05$ m, so they are in parallel.
Fe: area $2\times2 = 4\ \text{mm}^2 = 4\times10^{-6}\ \text{m}^2$; $R_{Fe} = \frac{1.0\times10^{-7}\times0.05}{4\times10^{-6}} = 1.25\times10^{-3}\ \Omega = 1250\ \mu\Omega$.
Al: area $49 - 4 = 45\ \text{mm}^2$; $R_{Al} = \frac{2.7\times10^{-8}\times0.05}{45\times10^{-6}} = 3\times10^{-5}\ \Omega = 30\ \mu\Omega$.
Parallel: $R = \frac{1250\times30}{1250 + 30} = \frac{37500}{1280} = \frac{1875}{64}\ \mu\Omega$.
Answer: (B). | ||||
| 18 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Multiple correctEasy | |
A easy multiple correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation: $eV_0 = \frac{hc}{\lambda} - \phi$, so $V_0 = \frac{hc}{e}\cdot\frac{1}{\lambda} - \frac{\phi}{e}$.
Against $\frac{1}{\lambda}$: a straight line of positive slope $\frac{hc}{e}$, cutting the axis at the threshold $\frac{1}{\lambda_0}$. That is (C); (D), a curve, is wrong.
Against $\lambda$: $V_0$ falls as $\frac{1}{\lambda}$ — a curve, steep at small $\lambda$ and flattening, reaching zero at the threshold $\lambda_0$. That is (A); (B), a straight line, is wrong.
Answer: (A), (C). | ||||
| 19 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Matrix matchMedium | |
A medium matrix match question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (A) Fusion: hydrogen burning in stars (R); the p–p chain also emits neutrinos ($p + p \rightarrow d + e^+ + \nu_e$), so (T) as well. The key accepts R alone or R and T.
(B) Fission in a reactor: ${}^{235}$U absorbs thermal neutrons (P); heavy water is used as the moderator (S).
(C) $\beta$-decay: ${}^{60}$Co is a $\beta^-$ emitter (Q), and every $\beta$-decay emits a (anti)neutrino (T).
(D) $\gamma$-ray emission: the key gives R (stellar fusion releases $\gamma$ rays).
Official: A→R or R,T; B→P,S; C→Q,T; D→R. | ||||
| 20 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Matrix matchHard | |
A hard matrix match question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $F = -\frac{dU}{dx}$.
(A) $U_1' = -\frac{2U_0}{a^2}x\left(1 - \frac{x^2}{a^2}\right)$: zero at $0, \pm a$ (P, Q, R). For $0 < x < a$, $F = \frac{2U_0}{a^2}x(1 - x^2/a^2) > 0$, away from 0 — repulsive, so not S. $U_1(\pm a) = 0$ are minima and $U_1(0) = \frac{U_0}{2}$ a maximum; with $E = \frac{U_0}{4} < \frac{U_0}{2}$ the particle is trapped near $x = -a$ and oscillates (T). A→P,Q,R,T.
(B) $U_2' = \frac{U_0}{a^2}x$: zero only at 0 (Q); $F = -\frac{U_0}{a^2}x$ always points to 0 (S). Its only minimum is at 0, so not T. B→Q,S.
(C) $U_3' = \frac{U_0}{a^2}x\left(1 - \frac{x^2}{a^2}\right)e^{-x^2/a^2}$: zero at $0, \pm a$ (P, Q, R). For $|x| < a$ the force has the sign of $-x$: attractive (S). $x = \pm a$ are maxima ($U = \frac{U_0}{2e}$), so no oscillation about $-a$. C→P,Q,R,S.
(D) $U_4' = \frac{U_0}{2a}\left(1 - \frac{x^2}{a^2}\right)$: zero at $\pm a$ (P, R), not at 0. For $|x| < a$, $F < 0$ everywhere — towards $-x$, not towards 0 — so not S. $x = -a$ is a minimum with $U = -\frac{U_0}{3}$; the barrier at $x = a$ is $U = \frac{U_0}{3} > \frac{U_0}{4}$ and $U \rightarrow \infty$ as $x \rightarrow -\infty$, so a particle of energy $\frac{U_0}{4}$ oscillates about $-a$ (T). D→P,R,T.
Official: A→P,Q,R,T; B→Q,S; C→P,Q,R,S; D→P,R,T. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 21 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques M is camphor: 1,7,7-trimethylbicyclo[2.2.1]heptan-2-one. Its stereocentres are the two bridgehead carbons C1 and C4.
On paper two stereocentres would allow $2^2 = 4$ stereoisomers, but in a bicyclo[2.2.1] system the one-carbon bridge (C7) must span C1 and C4 on the same face. The configurations of the two bridgeheads are therefore locked together: inverting only one would need the bridge on opposite faces, which is impossible.
Only the pair of enantiomers exists (the (1R,4R) and (1S,4S) forms), so the number of stereoisomers is 2.
Answer: 2. | ||||
| 22 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques NaOH removes the phenolic proton, so N is the 2-naphthoxide ion.
Count the structures by where the negative charge sits.
Charge on O: the naphthalene ring keeps its three Kekulé arrangements: 3 structures.
Charge on carbon (with C2=O): the charge can be delocalised to C1, C3, C4a, C6 and C8 (alternate positions reached through the ring's double bonds).
– C1: the remaining eight carbons C3–C8a can pair their double bonds in two ways (C3=C4, C4a=C8a, C5=C6, C7=C8, or C3=C4, C4a=C5, C6=C7, C8=C8a): 2 structures.
– C3: C1=C8a, C4=C4a, C5=C6, C7=C8: 1 structure.
– C4a: C1=C8a, C3=C4, C5=C6, C7=C8: 1 structure.
– C8: C1=C8a, C3=C4, C4a=C5, C6=C7: 1 structure.
– C6: C1=C8a, C3=C4, C4a=C5, C7=C8: 1 structure.
Total $3 + 2 + 1 + 1 + 1 + 1 = 9$.
Answer: 9. | ||||
| 23 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure N$_2$O$_3$ is $\mathrm{O{=}N{-}NO_2}$: an N=O group bonded through N to an NO$_2$ group.
Left N (two bonds: =O and –N): 1 lone pair. Its O (double bond): 2 lone pairs.
Right N (in NO$_2$, bonded to N, to =O and to –O$^-$, formally N$^+$): no lone pair. Its =O: 2 lone pairs; its –O$^-$: 3 lone pairs.
Total $1 + 2 + 0 + 2 + 3 = 8$.
Check: N$_2$O$_3$ has $2\times5 + 3\times6 = 28$ valence electrons, i.e. 14 pairs; the bonds use 6 pairs (N–N 1, N=O 2, N=O 2, N–O 1), leaving $14 - 6 = 8$ lone pairs.
Answer: 8. | ||||
| 24 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Fe$^{3+}$ is $3d^5$.
Thiocyanato-S (SCN$^-$) is a weak-field ligand: high spin, $t_{2g}^3e_g^2$, 5 unpaired electrons, $\mu = \sqrt{5\cdot7} = \sqrt{35} \approx 5.92$ BM.
CN$^-$ is a strong-field ligand: low spin, $t_{2g}^5$, 1 unpaired electron, $\mu = \sqrt{1\cdot3} = \sqrt{3} \approx 1.73$ BM.
Difference $\approx 5.92 - 1.73 = 4.19 \approx 4$.
Answer: 4. | ||||
| 25 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Go through each species.
BeCl$_2$: sp, linear. N$_3^-$: central N sp, linear. N$_2$O: central N sp, linear. NO$_2^+$: sp, linear. All four are linear with no $d$ contribution.
O$_3$: sp$^2$, bent. SCl$_2$: sp$^3$, bent. Not linear.
ICl$_2^-$, I$_3^-$, XeF$_2$: central atom has 3 lone pairs and 2 bonds, sp$^3$d, linear — but the hybridisation uses a $d$ orbital.
Count: 4.
Answer: 4. | ||||
| 26 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom In the one-electron H atom the energy depends only on $n$, so $n = 3$ ($3s$, three $3p$, five $3d$) is 9-fold degenerate.
H$^-$ has two electrons; electron–electron repulsion makes the energy depend on $l$ as well, so orbital energies go $1s < 2s < 2p < \ldots$
Ground state: $1s^2$. First excited state: $1s^1 2s^1$. Second excited state: $1s^1 2p^1$.
The $2p$ electron can be in any of the three $2p$ orbitals, so the degeneracy (ignoring spin) is 3.
Answer: 3. | ||||
| 27 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry For the oxidation $\mathrm{M^+ \rightarrow M^{3+} + 2e^-}$ with $E^0 = -0.25$ V: $\Delta G^0 = -nFE^0 = -2\times96500\times(-0.25) = +48250$ J $= 48.25$ kJ per mole of M$^+$.
Energy available from 1 mol of X: 193 kJ.
Moles of M$^+$ oxidised $= \frac{193}{48.25} = 4$.
Answer: 4. | ||||
| 28 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = iK_fm$: $0.0558 = i\times1.86\times0.01$, so $i = 3$.
The complex gives 3 ions: one complex cation and two Cl$^-$ outside the sphere, i.e. $\mathrm{[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^-}$.
Co(III) needs three chlorides in all; two are ionisable, so one is in the coordination sphere.
Answer: 1. | ||||
| 29 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Multiple correctMedium | |
A medium multiple correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Hydrogenation adds H$_2$ to the C=C and leaves the C–Br stereocentre untouched. The product is optically inactive only if that carbon stops being a stereocentre.
(A) CH$_3$CH=CH–CHBr–CH$_3$ → CH$_3$CH$_2$CH$_2$–CHBr–CH$_3$ (2-bromopentane): the CHBr carbon carries H, Br, CH$_3$ and propyl — still chiral, optically active.
(B) CH$_2$=CH–CHBr–CH$_2$CH$_3$ → CH$_3$CH$_2$–CHBr–CH$_2$CH$_3$ (3-bromopentane): two identical ethyl groups, achiral, optically inactive.
(C) CH$_2$=C(CH$_3$)–CHBr–CH$_3$ → (CH$_3$)$_2$CH–CHBr–CH$_3$ (2-bromo-3-methylbutane): H, Br, CH$_3$ and isopropyl — chiral, active.
(D) the enantiomer of (B); it also gives achiral 3-bromopentane.
Answer: (B), (D). | ||||
| 30 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Multiple correctMedium | |
A medium multiple correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The starting dione is 2-(3-oxopentyl)cyclohexanone: ring C=O at C1, and on C2 the chain CH$_2$–CH$_2$–C(=O)–CH$_2$–CH$_3$.
KOH forms an enolate; the one that can close a six-membered ring is at the CH$_2$ of the ethyl group (α to the side-chain C=O). It attacks the ring carbonyl C1: the new ring is C1–C2–CH$_2$–CH$_2$–C(=O)–CH(CH$_3$)–, six atoms (intramolecular aldol, as in a Robinson annulation).
H$^+$ and heat dehydrate the β-hydroxy ketone; the C=C forms between C1 (the ring-fusion carbon) and the CH(CH$_3$) carbon, conjugated with the C=O.
Product: a bicyclic α,β-unsaturated ketone with the CH$_3$ on the alkene carbon next to the C=O — option (A).
(B) has the double bond not conjugated with the C=O (not the dehydration product). (C) and (D) need four-membered rings, which aldol closure here does not form.
Answer: (A). | ||||
| 31 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Multiple correctMedium | |
A medium multiple correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons HBr protonates a terminal carbon of the diene so as to give the most stable cation. Protonating C1 (the CH$_2$ on the methyl-bearing carbon) gives a tertiary allylic cation:
$\mathrm{(CH_3)_2\overset{+}{C}{-}CH{=}CH_2 \leftrightarrow (CH_3)_2C{=}CH{-}\overset{+}{C}H_2}$.
Br$^-$ can add at the tertiary end (1,2-addition, (B)) or at the primary end (1,4-addition, (D)). The 1,4-product $\mathrm{(CH_3)_2C{=}CH{-}CH_2Br}$ has the more substituted (trisubstituted) double bond and is the thermodynamically favoured, major product.
(A) and (C) come from the less stable cations and are not major.
Answer: (D). | ||||
| 32 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Multiple correctEasy | |
A easy multiple correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules L-glucose is the enantiomer (mirror image) of D-glucose, so every stereocentre is inverted: each OH moves to the other side of the Fischer projection.
D-glucose (C2 to C5): right, left, right, right. L-glucose: left, right, left, left.
(A) left, right, left, left — matches.
(B) right, left, right, left — it differs from D-glucose only at C5, so it is L-idose, not L-glucose.
(C) left, left, right, left and (D) left, left, left, right do not match either.
Answer: (A). | ||||
| 33 | AminesDiazonium salts | Class 12: Amines | Multiple correctHard | |
A hard multiple correct question on diazonium salts, from Amines. NCERT: Class 12: Amines NaNO$_2$/HCl at 0 °C diazotises the primary amine: R–NH$_2$ → R–N$_2^+$. An aliphatic diazonium ion loses N$_2$ at once.
Here the neighbouring carboxylate attacks the carbon from the back as N$_2$ leaves (first inversion), forming a three-membered α-lactone. Water then opens it by attack at the same carbon from the back (second inversion).
Two inversions = overall retention: the OH takes the place of the NH$_2$ with the same spatial arrangement, and the CO$_2$H stays.
Product: 2-hydroxy-4-methylpentanoic acid with OH on the wedge, as in (C).
(B) has the inverted configuration. (A) and (D) lose the CO$_2$H and keep NH$_2$, which this reaction does not do.
Answer: (C). | ||||
| 34 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Multiple correctEasy | |
A easy multiple correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Cr$^{2+}$: [Ar]$3d^4$. Mn$^{3+}$: [Ar]$3d^4$. (C) is true.
(A) Cr$^{2+}$ readily loses an electron to Cr$^{3+}$ ($3d^3$, half-filled $t_{2g}$, very stable): a reducing agent. True.
(B) Mn$^{3+}$ readily gains an electron to Mn$^{2+}$ ($3d^5$, half-filled $d$): an oxidising agent. True.
(D) As a reducing agent Cr$^{2+}$ becomes Cr$^{3+}$, which is $d^3$, not $d^5$. False.
Answer: (A), (B), (C). | ||||
| 35 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In electrolytic refining the impure (blister) copper is the anode, a thin strip of pure copper the cathode, and acidified CuSO$_4$ solution the electrolyte.
At the anode Cu dissolves as Cu$^{2+}$; at the cathode pure Cu deposits. Less reactive impurities (Ag, Au, Pt) do not dissolve and settle below the anode as anode mud.
(A) is false (impure Cu is the anode); (B), (C) and (D) are true.
Answer: (B), (C), (D). | ||||
| 36 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Multiple correctMedium | |
A medium multiple correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions H$_2$O$_2$ can act as an oxidant or a reductant; which one depends on the medium.
In basic solution it reduces Fe$^{3+}$: $\mathrm{2Fe^{3+} + H_2O_2 + 2OH^- \rightarrow 2Fe^{2+} + O_2 + 2H_2O}$. (A) is true.
Na$_2$O$_2$ in water gives NaOH and H$_2$O$_2$: $\mathrm{Na_2O_2 + 2H_2O \rightarrow 2NaOH + H_2O_2}$ — the same alkaline H$_2$O$_2$, so it also reduces Fe$^{3+}$. (B) is true.
In acid, H$_2$O$_2$ is an oxidant: it oxidises Fe$^{2+}$ to Fe$^{3+}$ ($\mathrm{2Fe^{2+} + H_2O_2 + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O}$) rather than reducing Fe$^{3+}$. (C) is false, and so is (D), since Na$_2$O$_2$ with H$_2$SO$_4$ is just acidic H$_2$O$_2$.
Answer: (A), (B). | ||||
| 37 | EquilibriumLe Chatelier's principle | Class 11: Equilibrium | Multiple correctEasy | |
A easy multiple correct question on le chatelier's principle, from Equilibrium. NCERT: Class 11: Equilibrium Two effects of raising the temperature from $T_1$ to $T_2$:
Rate: every reaction is faster at higher temperature, so the yield rises faster and equilibrium is reached sooner — the $T_2$ curve is steeper at the start.
Equilibrium: the forward reaction is exothermic ($\Delta H < 0$), so by Le Chatelier's principle a higher temperature shifts the equilibrium back towards N$_2$ and H$_2$ — the $T_2$ curve levels off at a lower yield.
Only (B) shows a faster rise and a lower final yield.
Answer: (B). | ||||
| 38 | Solid StateClose packed structures | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on close packed structures, from Solid State. NCERT: No matching NCERT chapter A ccp (fcc) unit cell has 4 oxide ions, 4 octahedral holes and 8 tetrahedral holes.
With $m$ of the octahedral holes holding Al$^{3+}$ and $n$ of the tetrahedral holes holding Mg$^{2+}$: $4m$ Al$^{3+}$ and $8n$ Mg$^{2+}$ per 4 O$^{2-}$.
Electrical neutrality: $3(4m) + 2(8n) = 2\times4 = 8$, i.e. $12m + 16n = 8$.
Check the options: (A) $m = \frac{1}{2}, n = \frac{1}{8}$: $6 + 2 = 8$ ✓ — formula Mg$_1$Al$_2$O$_4$, the spinel MgAl$_2$O$_4$.
(B) $12 + 4 = 16$ ✗. (C) $6 + 8 = 14$ ✗. (D) $3 + 2 = 5$ ✗.
Answer: (A). | ||||
| 39 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Matrix matchEasy | |
A easy matrix match question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Siderite FeCO$_3$; malachite CuCO$_3\cdot$Cu(OH)$_2$; bauxite AlO$_x$(OH)$_{3-2x}$ (Al$_2$O$_3\cdot x$H$_2$O); calamine ZnCO$_3$; argentite Ag$_2$S.
(A) Carbonate: siderite, malachite, calamine — P, Q, S.
(B) Sulphide: argentite — T.
(C) Hydroxide: malachite, bauxite — Q, R.
(D) Oxide: bauxite — R.
Official: A→P,Q,S; B→T; C→Q,R; D→R. | ||||
| 40 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Matrix matchMedium | |
A medium matrix match question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics (A) Freezing at the normal freezing point is an equilibrium change: $\Delta G = 0$ (T); liquid → solid lowers the entropy, $\Delta S_{sys} < 0$ (R). Heat is released ($q \neq 0$), so $\Delta U \neq 0$; there is a small volume change, so $w \neq 0$.
(B) Free expansion in isolation: no heat ($q = 0$), no work against a vacuum ($w = 0$), so $\Delta U = 0$ (P, Q, S). Entropy rises and $\Delta G < 0$.
(C) Mixing ideal gases in an isolated container: again $q = 0$, $w = 0$, $\Delta U = 0$ (P, Q, S); $\Delta S > 0$.
(D) The gas returns to its initial state, so every state function is unchanged: $\Delta U = 0$, $\Delta G = 0$ (S, T). Along the same reversible path at 1 atm the heat taken in heating equals that given out in cooling ($nC_P\Delta T$ each way), and likewise the $P\Delta V$ work, so net $q = 0$ and $w = 0$ (P, Q).
Official: A→R,T; B→P,Q,S; C→P,Q,S; D→P,Q,S,T. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 41 | 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 $\cos^4x + \sin^4x = 1 - 2\sin^2x\cos^2x = 1 - \frac{1}{2}\sin^2 2x$.
$\cos^6x + \sin^6x = 1 - 3\sin^2x\cos^2x = 1 - \frac{3}{4}\sin^2 2x$.
The equation becomes $\frac{5}{4}\cos^2 2x + 2 - \frac{5}{4}\sin^2 2x = 2$, i.e. $\cos^2 2x - \sin^2 2x = 0$, i.e. $\cos 4x = 0$.
$4x = \frac{\pi}{2} + k\pi$. For $x \in [0, 2\pi]$, $4x \in [0, 8\pi]$: $4x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots, \frac{15\pi}{2}$ — 8 values.
Answer: 8. | ||||
| 42 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Reflection in $x + y + 4 = 0$ maps $(x, y)$ to $(-y - 4, -x - 4)$ (it is its own inverse).
A point $(X, Y)$ lies on $C$ exactly when its reflection $(-Y - 4, -X - 4)$ lies on $y^2 = 4x$: $(-X - 4)^2 = 4(-Y - 4)$, i.e. $C: (X + 4)^2 = -4(Y + 4)$.
On $Y = -5$: $(X + 4)^2 = 4$, so $X = -2$ or $X = -6$.
$A = (-2, -5)$, $B = (-6, -5)$, $AB = 4$.
Answer: 4. | ||||
| 43 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability With $n$ tosses, $P(\text{at least two heads}) = 1 - P(0) - P(1) = 1 - \frac{1 + n}{2^n}$.
We need $1 - \frac{n + 1}{2^n} \ge 0.96$, i.e. $\frac{n + 1}{2^n} \le 0.04 = \frac{1}{25}$, i.e. $2^n \ge 25(n + 1)$.
$n = 7$: $128 \ge 200$? No. $n = 8$: $256 \ge 225$? Yes.
So the minimum is 8.
Answer: 8. | ||||
| 44 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $n$: treat the 5 girls as one block. Arrange the block and 5 boys: $6!$; arrange the girls inside: $5!$. $n = 6!\cdot5!$.
$m$: arrange the 5 boys: $5!$, leaving 6 gaps (ends included). Choose the girl who stands apart: 5 ways; arrange the other four inside their block: $4!$. Put the block and the single girl in two different gaps (so the single girl is not next to the block): $6\times5 = 30$ ways.
$m = 5!\cdot5\cdot4!\cdot30 = 5!\cdot3600$.
$\frac{m}{n} = \frac{5!\cdot3600}{720\cdot5!} = 5$.
Answer: 5. | ||||
| 45 | Conic SectionsParabola | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections For $y^2 = 4x$ ($a = 1$) the latus rectum ends are $(1, 2)$ and $(1, -2)$, parameters $t = \pm1$.
Normal at $t$: $y = -tx + 2t + t^3$. $t = 1$: $x + y - 3 = 0$. $t = -1$: $x - y - 3 = 0$.
Distance of the centre $(3, -2)$ from $x + y - 3 = 0$: $\frac{|3 - 2 - 3|}{\sqrt{2}} = \sqrt{2}$. From $x - y - 3 = 0$: $\frac{|3 + 2 - 3|}{\sqrt{2}} = \sqrt{2}$. Both equal the radius.
$r^2 = 2$.
Answer: 2. | ||||
| 46 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Split $[-1, 2]$ where $f(x^2)$ or $f(x + 1)$ changes.
$-1 < x < 1$: $0 \le x^2 < 1$, so $f(x^2) = [x^2] = 0$; the integrand is 0.
$1 \le x \le \sqrt{2}$: $1 \le x^2 \le 2$, so $f(x^2) = 1$. And $x + 1 \in [2, 1 + \sqrt{2}]$: $f(x + 1) = 0$ for $x + 1 > 2$ (and $= 2$ only at the single point $x = 1$). The integrand is $\frac{x}{2}$.
$\sqrt{2} < x \le 2$: $x^2 > 2$, so $f(x^2) = 0$.
$I = \int_1^{\sqrt{2}}\frac{x}{2}dx = \frac{1}{4}\left[x^2\right]_1^{\sqrt{2}} = \frac{1}{4}$.
$4I - 1 = 0$.
Answer: 0. | ||||
| 47 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let the inner radius be $r$ and the inner height $h$, so $V = \pi r^2h$.
Material: wall $\pi\left[(r + 2)^2 - r^2\right]h = \pi(4r + 4)h$; bottom $\pi(r + 2)^2\cdot2$.
With $h = \frac{V}{\pi r^2}$: $M(r) = \frac{(4r + 4)V}{r^2} + 2\pi(r + 2)^2 = V\left(\frac{4}{r} + \frac{4}{r^2}\right) + 2\pi(r + 2)^2$.
$M'(r) = -V\left(\frac{4}{r^2} + \frac{8}{r^3}\right) + 4\pi(r + 2) = -\frac{4V(r + 2)}{r^3} + 4\pi(r + 2)$.
$M'(r) = 0$ gives $V = \pi r^3$ (and $M'$ changes from negative to positive there, so it is a minimum).
At $r = 10$: $V = 1000\pi$, so $\frac{V}{250\pi} = 4$.
Answer: 4. | ||||
| 48 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The area is $\int_0^a f(x)\,dx = F'(a) + 2$ for every $a \in \left[0, \frac{1}{2}\right]$. Differentiating in $a$: $f(a) = F''(a)$, so $f(0) = F''(0)$.
By the Leibniz rule, $F'(x) = 2\cos^2\left(x^2 + \frac{\pi}{6}\right)\cdot2x - 2\cos^2x$.
$F''(x) = 4\cos^2\left(x^2 + \frac{\pi}{6}\right) + 4x\cdot\frac{d}{dx}\cos^2\left(x^2 + \frac{\pi}{6}\right) + 2\sin 2x$.
At $x = 0$: $F''(0) = 4\cos^2\frac{\pi}{6} + 0 + 0 = 4\cdot\frac{3}{4} = 3$.
(Consistency: $F'(0) + 2 = -2 + 2 = 0$, the area at $a = 0$.)
Answer: 3. | ||||
| 49 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Use $X^T = -X$, $Y^T = -Y$, $Z^T = Z$, so $(X^k)^T = (-1)^kX^k$ and $(Z^k)^T = Z^k$.
(A) $(Y^3Z^4 - Z^4Y^3)^T = Z^4(-Y^3) - (-Y^3)Z^4 = Y^3Z^4 - Z^4Y^3$: symmetric, so not skew-symmetric in general.
(B) $(X^{44} + Y^{44})^T = X^{44} + Y^{44}$: symmetric.
(C) $(X^4Z^3 - Z^3X^4)^T = Z^3X^4 - X^4Z^3 = -(X^4Z^3 - Z^3X^4)$: skew-symmetric.
(D) $(X^{23} + Y^{23})^T = -X^{23} - Y^{23}$: skew-symmetric.
Answer: (C), (D). | ||||
| 50 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The $(i, j)$ entry is $(i + j\alpha)^2 = i^2\cdot1 + 2i\cdot j\alpha + 1\cdot j^2\alpha^2$, so the matrix is the product
$\begin{pmatrix}1&2&1\\4&4&1\\9&6&1\end{pmatrix}\begin{pmatrix}1&1&1\\\alpha&2\alpha&3\alpha\\\alpha^2&4\alpha^2&9\alpha^2\end{pmatrix}$.
First determinant: $1(4 - 6) - 2(4 - 9) + 1(24 - 36) = -2 + 10 - 12 = -4$.
Second: $\alpha\cdot\alpha^2\begin{vmatrix}1&1&1\\1&2&3\\1&4&9\end{vmatrix} = \alpha^3\cdot(2-1)(3-1)(3-2) = 2\alpha^3$.
So the determinant is $-8\alpha^3$, and $-8\alpha^3 = -648\alpha$ gives $\alpha(\alpha^2 - 81) = 0$: $\alpha = 0, \pm9$.
Of the options, 9 and $-9$ work; $\pm4$ do not.
Answer: (B), (C). | ||||
| 51 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctMedium | |
A medium multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Planes through the line $P_1 \cap P_2$: $(x + z - 1) + \lambda y = 0$ (this family omits only $P_1$ itself; $\lambda = 0$ gives $P_2$, so $\lambda \neq 0$).
Distance of $(0, 1, 0)$: $\frac{|-1 + \lambda|}{\sqrt{2 + \lambda^2}} = 1$, so $(\lambda - 1)^2 = 2 + \lambda^2$, $-2\lambda + 1 = 2$, $\lambda = -\frac{1}{2}$.
$P_3$: $x - \frac{y}{2} + z - 1 = 0$, i.e. $2x - y + 2z - 2 = 0$.
Distance of $(\alpha, \beta, \gamma)$: $\frac{|2\alpha - \beta + 2\gamma - 2|}{3} = 2$, so $2\alpha - \beta + 2\gamma - 2 = \pm6$:
$2\alpha - \beta + 2\gamma - 8 = 0$ (D) or $2\alpha - \beta + 2\gamma + 4 = 0$ (B).
(A) and (C) have the wrong signs on $\beta$ or $\gamma$.
Answer: (B), (D). | ||||
| 52 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctHard | |
A hard multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Every point of $L$ keeps a constant distance from each plane, so $L$ is parallel to both planes: its direction is perpendicular to both normals $(1, 2, -1)$ and $(2, -1, 1)$.
$(1, 2, -1)\times(2, -1, 1) = (2 - 1, -2 - 1, -1 - 4) = (1, -3, -5)$. So $L$: $(t, -3t, -5t)$.
Foot of the perpendicular from $(t, -3t, -5t)$ to $P_1$: the value of $x + 2y - z + 1$ there is $t - 6t + 5t + 1 = 1$, and $|n|^2 = 6$, so the foot is $(t, -3t, -5t) - \frac{1}{6}(1, 2, -1) = \left(t - \frac{1}{6}, -3t - \frac{1}{3}, -5t + \frac{1}{6}\right)$.
(A) $t = \frac{1}{6}$: $\left(0, -\frac{1}{2} - \frac{1}{3}, -\frac{5}{6} + \frac{1}{6}\right) = \left(0, -\frac{5}{6}, -\frac{2}{3}\right)$. On $M$.
(B) $t = 0$: $\left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$. On $M$.
(C) $x = -\frac{5}{6}$ needs $t = -\frac{2}{3}$, giving $y = 2 - \frac{1}{3} = \frac{5}{3} \neq 0$. Not on $M$.
(D) $x = -\frac{1}{3}$ needs $t = -\frac{1}{6}$, giving $y = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \neq 0$. Not on $M$.
Answer: (A), (B). | ||||
| 53 | Conic SectionsParabola | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Write points of $y^2 = 2x$ as $\left(\frac{t^2}{2}, t\right)$: $P$ with $t_1 > 0$, $Q$ with $t_2$.
O on the circle with diameter $PQ$ means $\angle POQ = 90^\circ$: $\frac{t_1^2t_2^2}{4} + t_1t_2 = 0$, so $t_1t_2 = -4$.
Area $= \frac{1}{2}\left|x_1y_2 - x_2y_1\right| = \frac{1}{2}\left|\frac{t_1^2t_2 - t_2^2t_1}{2}\right| = \frac{|t_1t_2|\,|t_1 - t_2|}{4} = |t_1 - t_2|$.
With $t_2 = -\frac{4}{t_1}$: $t_1 + \frac{4}{t_1} = 3\sqrt{2}$, so $t_1^2 - 3\sqrt{2}t_1 + 4 = 0$ and $t_1 = \frac{3\sqrt{2} \pm \sqrt{2}}{2} = 2\sqrt{2}$ or $\sqrt{2}$.
$P = (4, 2\sqrt{2})$ or $(1, \sqrt{2})$: (A) and (D). (B) and (C) lie on the parabola but give the wrong area.
Answer: (A), (D). | ||||
| 54 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Multiple correctMedium | |
A medium multiple correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations The left side is an exact derivative: $\frac{d}{dx}\left[(1 + e^x)y\right] = 1$, so $(1 + e^x)y = x + C$.
$y(0) = 2$: $2\cdot2 = C$, $C = 4$. So $y = \frac{x + 4}{1 + e^x}$.
(A) $y(-4) = 0$. True. (B) $y(-2) = \frac{2}{1 + e^{-2}} \neq 0$. False.
Critical point: $y' = \frac{(1 + e^x) - (x + 4)e^x}{(1 + e^x)^2} = 0$ when $(x + 3)e^x = 1$.
$g(x) = (x + 3)e^x$ is continuous with $g(-1) = \frac{2}{e} \approx 0.74 < 1$ and $g(0) = 3 > 1$, so $g(x) = 1$ somewhere in $(-1, 0)$. (C) true, (D) false.
Answer: (A), (C). | ||||
| 55 | Differential EquationsOrder and degree | Class 12: Differential Equations | Multiple correctHard | |
A hard multiple correct question on order and degree, from Differential Equations. NCERT: Class 12: Differential Equations Circles: $(x - a)^2 + (y - a)^2 = r^2$, two parameters, so a second-order equation.
Differentiate: $(x - a) + (y - a)y' = 0$, so $a = \frac{x + yy'}{1 + y'}$ and $y - a = \frac{y - x}{1 + y'}$.
Differentiate again: $1 + (y')^2 + (y - a)y'' = 0$, i.e. $1 + (y')^2 + \frac{(y - x)y''}{1 + y'} = 0$.
Multiply by $1 + y'$: $(y - x)y'' + (1 + y')(1 + (y')^2) = 0$, i.e. $(y - x)y'' + \left(1 + y' + (y')^2\right)y' + 1 = 0$.
So $P = y - x$ and $Q = 1 + y' + (y')^2$.
(A) false; (B) true; (C) $P + Q = 1 - x + y + y' + (y')^2$: true; (D) $P - Q = y - x - 1 - y' - (y')^2$, not the expression given: false.
Answer: (B), (C). | ||||
| 56 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctHard | |
A hard multiple correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability (A) $\frac{f(t) - f(0)}{t} = \frac{\operatorname{sgn}(t)\,g(t)}{t}$, and $\frac{g(t)}{t} \to g'(0) = 0$, so the quotient tends to 0. $f'(0) = 0$: true.
(B) $h(x) = e^{|x|}$ has one-sided derivatives $+1$ and $-1$ at 0: not differentiable. False.
(C) $h(x) = e^{|x|} \ge 1 > 0$, so $f(h(x)) = g(e^{|x|})$. Its right derivative at 0 is $g'(1)\cdot1$ and its left derivative $g'(1)\cdot(-1)$; these differ because $g'(1) \neq 0$. False.
(D) $h(f(x)) = e^{|f(x)|} = e^{|g(x)|}$. $\frac{e^{|g(t)|} - 1}{t} \approx \frac{|g(t)|}{t} \to 0$ since $\frac{g(t)}{t} \to 0$. Differentiable with derivative 0: true.
Answer: (A), (D). | ||||
| 57 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Multiple correctMedium | |
A medium multiple correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions (A) $\frac{\pi}{2}\sin x$ runs over $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\sin\left(\frac{\pi}{2}\sin x\right)$ runs over $[-1, 1]$, $\frac{\pi}{6}$ times that over $\left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$, and its sine over $\left[-\frac{1}{2}, \frac{1}{2}\right]$. True.
(B) $f(g(x))$: $g(x)$ runs over $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, on which $\sin$ still takes every value in $[-1, 1]$, so the same chain gives $\left[-\frac{1}{2}, \frac{1}{2}\right]$. True.
(C) Near 0: $f(x) \approx \frac{\pi}{6}\cdot\frac{\pi}{2}x$ and $g(x) \approx \frac{\pi}{2}x$, so $\frac{f(x)}{g(x)} \to \frac{\pi}{6}$. True.
(D) $g(f(x)) = \frac{\pi}{2}\sin(f(x))$ with $|f(x)| \le \frac{1}{2}$, so $|g(f(x))| \le \frac{\pi}{2}\sin\frac{1}{2} \approx 1.571\times0.479 \approx 0.75 < 1$. Never 1. False.
Answer: (A), (B), (C). | ||||
| 58 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Multiple correctMedium | |
A medium multiple correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Going round the triangle, $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
$\vec{a} = -(\vec{b} + \vec{c})$: $144 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b}\cdot\vec{c} = 48 + |\vec{c}|^2 + 48$, so $|\vec{c}|^2 = 48$.
(A) $\frac{48}{2} - 12 = 12$. True. (B) $24 + 12 = 36 \neq 30$. False.
(D) $\vec{c} = -(\vec{a} + \vec{b})$: $48 = 144 + 48 + 2\vec{a}\cdot\vec{b}$, so $\vec{a}\cdot\vec{b} = -72$. True.
(C) $\vec{c}\times\vec{a} = -(\vec{a} + \vec{b})\times\vec{a} = -\vec{b}\times\vec{a} = \vec{a}\times\vec{b}$, so the vector is $2\,\vec{a}\times\vec{b}$. $|\vec{a}\times\vec{b}|^2 = 144\cdot48 - 72^2 = 6912 - 5184 = 1728$, $|\vec{a}\times\vec{b}| = 24\sqrt{3}$, and twice that is $48\sqrt{3}$. True.
Answer: (A), (C), (D). | ||||
| 59 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Matrix matchMedium | |
A medium matrix match question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra (A) Projection: $\frac{|\sqrt{3}\alpha + \beta|}{2} = \sqrt{3}$, so $\sqrt{3}\alpha + \beta = \pm2\sqrt{3}$. With $\beta = \frac{\alpha - 2}{\sqrt{3}}$: $3\alpha + \alpha - 2 = \pm6$, so $\alpha = 2$ or $-1$; $|\alpha| = 2, 1$ — P, Q.
(B) Continuity at 1: $-3a - 2 = b + a^2$. Equal derivatives: $-6a = b$. So $-3a - 2 = -6a + a^2$, $a^2 - 3a + 2 = 0$, $a = 1, 2$ — P, Q.
(C) Let $z = 3 - 3\omega + 2\omega^2$. Then $\omega z = 3\omega - 3\omega^2 + 2 = 2 + 3\omega - 3\omega^2$ and $\omega^2z = 3\omega^2 - 3 + 2\omega$. The sum is $z^{4n+3}\left(1 + \omega^{4n+3} + \omega^{2(4n+3)}\right)$, which is 0 exactly when $4n + 3$ is not a multiple of 3, i.e. $n$ not a multiple of 3: $n = 1, 2, 4, 5$ — P, Q, S, T.
(D) AP: $q = 10 - a$, $b = 15 - 2a$. HM: $\frac{2ab}{a + b} = 4$ gives $2a(15 - 2a) = 4(15 - a)$, $2a^2 - 17a + 30 = 0$, $a = 6$ or $\frac{5}{2}$. $a = 6$: $q = 4$, $b = 3$, $|q - a| = 2$. $a = \frac{5}{2}$: $q = \frac{15}{2}$, $b = 10$, $|q - a| = 5$ — Q, T.
Official: A→P,Q; B→P,Q; C→P,Q,S,T; D→Q,T. | ||||
| 60 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Matrix matchHard | |
A hard matrix match question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions (A) In any triangle $\frac{\sin(X - Y)}{\sin(X + Y)} = \frac{a^2 - b^2}{c^2}$, and $\sin Z = \sin(X + Y)$, so $\lambda = \frac{a^2 - b^2}{c^2} = \frac{1}{2}$. $\cos\frac{n\pi}{2} = 0$ for odd $n$: 1, 3, 5 — P, R, S.
(B) $1 + \cos2X = 2\cos^2X = 2 - 2\sin^2X$ and $2\cos2Y = 2 - 4\sin^2Y$, so the condition is $-2\sin^2X + 4\sin^2Y = 2\sin X\sin Y$, i.e. $(\sin X + 2\sin Y)(\sin X - \sin Y) = 0$. Sines are positive, so $\sin X = \sin Y$ and $\frac{a}{b} = 1$ — P.
(C) $OX$ makes $30^\circ$ and $OY$ $60^\circ$ with the $x$-axis; the acute-angle bisector is $y = x$. Distance of $(\beta, 1 - \beta)$: $\frac{|2\beta - 1|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$, so $\beta = 2$ or $-1$; $|\beta| = 2, 1$ — P, Q.
(D) $\alpha = 0$: $y = 3$; above $y = 2\sqrt{x}$ on $[0, 2]$ ($2\sqrt{2} < 3$). $F(0) = \int_0^2(3 - 2\sqrt{x})dx = 6 - \frac{8\sqrt{2}}{3}$, so $F + \frac{8\sqrt{2}}{3} = 6$ (T).
$\alpha = 1$: $y = 3 - x$ on $[0, 1]$ and $1 + x$ on $[1, 2]$, never below $2\sqrt{x}$ (they touch at $x = 1$). $F(1) = \int_0^1(3 - x)dx + \int_1^2(1 + x)dx - \frac{8\sqrt{2}}{3} = \frac{5}{2} + \frac{5}{2} - \frac{8\sqrt{2}}{3}$, so $F + \frac{8\sqrt{2}}{3} = 5$ (S). D→S, T.
Official: A→P,R,S; B→P; C→P,Q; D→S,T. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | GravitationNewton's law of gravitation | Class 11: Gravitation | NumericalMedium | |
A medium numerical question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation With zero tension the two point masses must have the same acceleration under gravity alone. Take the direction towards $M$ as positive.
Near mass (distance $3l$ from $M$): pulled towards $M$ by $M$ and away from $M$ by the other $m$:
$a_1 = \frac{GM}{9l^2} - \frac{Gm}{l^2}$.
Far mass (distance $4l$): pulled towards $M$ by both:
$a_2 = \frac{GM}{16l^2} + \frac{Gm}{l^2}$.
Setting $a_1 = a_2$: $GM\left(\frac{1}{9} - \frac{1}{16}\right) = 2Gm$, so $m = \frac{7M}{144\times2} = \frac{7M}{288}$.
Answer: $k = 7$. | ||||
| 2 | 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 Take logarithms: $\ln E = 2\ln A - \alpha t$.
Differentiate and add the errors: $\frac{\Delta E}{E} = 2\frac{\Delta A}{A} + \alpha\,\Delta t = 2\frac{\Delta A}{A} + \alpha t\,\frac{\Delta t}{t}$.
At $t = 5$ s, $\alpha t = 0.2\times5 = 1$, so
$\frac{\Delta E}{E} = 2(1.25\%) + 1\times(1.50\%) = 2.5\% + 1.5\% = 4\%$.
Answer: 4. | ||||
| 3 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion A thin shell of radius $r$ and mass $dm = 4\pi r^2\rho\,dr$ has moment of inertia $\frac{2}{3}r^2\,dm$ about a diameter, so
$I = \frac{8\pi}{3}\int_0^R \rho(r)\,r^4\,dr$.
$I_A \propto \frac{1}{R}\int_0^R r^5\,dr = \frac{R^5}{6}$, and $I_B \propto \frac{1}{R^5}\int_0^R r^9\,dr = \frac{R^5}{10}$ (same constant $\frac{8\pi k}{3}$).
$\frac{I_B}{I_A} = \frac{1/10}{1/6} = \frac{6}{10}$.
Answer: $n = 6$. | ||||
| 4 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Let each wave have amplitude $a$ (so $I_0 \propto a^2$) and add them as phasors.
The waves with phases 0 and $\pi$ are opposite and cancel.
The waves with phases $\pi/3$ and $2\pi/3$ differ by $\pi/3$; their resultant amplitude is $2a\cos\frac{\pi}{6} = \sqrt3\,a$.
Intensity $\propto (\sqrt3 a)^2 = 3a^2$, i.e. $3I_0$.
Answer: $n = 3$. | ||||
| 5 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $A = \lambda N = A_0e^{-\lambda t}$, so $R = -\frac{dA}{dt} = \lambda A_0 e^{-\lambda t}$.
Both start with the same $A_0$.
$P$: $\lambda = \frac{1}{\tau}$, $R_P = \frac{A_0}{\tau}e^{-2\tau/\tau} = \frac{A_0}{\tau}e^{-2}$.
$Q$: $\lambda = \frac{1}{2\tau}$, $R_Q = \frac{A_0}{2\tau}e^{-2\tau/2\tau} = \frac{A_0}{2\tau}e^{-1}$.
$\frac{R_P}{R_Q} = 2e^{-1} = \frac{2}{e}$.
Answer: $n = 2$. | ||||
| 6 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalHard | |
A hard numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics With prism angle $A = 60^\circ$ and incidence $60^\circ$: $\sin r_1 = \frac{\sin60^\circ}{n}$, $r_2 = A - r_1$, $\sin\theta = n\sin(A - r_1)$.
At $n = \sqrt3$: $\sin r_1 = \frac12$, $r_1 = r_2 = 30^\circ$, $\sin\theta = \sqrt3\cdot\frac12$, $\theta = 60^\circ$.
Differentiate $\sin r_1 = \frac{\sin60^\circ}{n}$: $\cos r_1\frac{dr_1}{dn} = -\frac{\sin60^\circ}{n^2}$, so $\frac{dr_1}{dn} = -\frac{\sqrt3/2}{3\cdot\sqrt3/2} = -\frac13$.
Differentiate $\sin\theta = n\sin(A - r_1)$: $\cos\theta\frac{d\theta}{dn} = \sin(A - r_1) - n\cos(A - r_1)\frac{dr_1}{dn}$
$= \sin30^\circ + \sqrt3\cos30^\circ\cdot\frac13 = \frac12 + \frac12 = 1$.
With $\cos\theta = \frac12$: $\frac{d\theta}{dn} = 2$.
Answer: $m = 2$. | ||||
| 7 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity Name the junctions: $A$ (top, right of $R$), $B$ (top right), $O$ (centre of the diagonals), $C$ (left, between 6 Ω and 12 Ω), $D$ (right, between 2 Ω and 4 Ω), and the bottom wire $E$.
From $A$ to $D$: via $B$ the resistances are 1 Ω then 2 Ω; via $O$ they are 2 Ω then 4 Ω. Since $\frac12 = \frac24$ the bridge is balanced, $B$ and $O$ are at the same potential, and the 8 Ω carries no current. $R_{AD} = (1+2)\parallel(2+4) = 3\parallel6 = 2$ Ω.
Now from $A$ to $E$: via $C$ we have 6 Ω then 12 Ω; via $D$ we have 2 Ω then 4 Ω. Again $\frac{6}{12} = \frac{2}{4}$, so the 10 Ω between $C$ and $D$ carries no current. $R_{AE} = 18\parallel6 = 4.5$ Ω.
Total resistance $= 2 + 4.5 = 6.5$ Ω, so $I = \frac{6.5}{6.5} = 1$ A.
Answer: 1. | ||||
| 8 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Angular momentum $\frac{nh}{2\pi}$ gives $n = 3$. For Li$^{2+}$, $Z = 3$.
Orbit radius $r_n = \frac{n^2a_0}{Z} = \frac{9a_0}{3} = 3a_0$.
Bohr's condition $2\pi r_n = n\lambda$ gives $\lambda = \frac{2\pi\cdot3a_0}{3} = 2\pi a_0$.
Answer: $p = 2$. | ||||
| 9 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctMedium | |
A medium multiple correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $L_1$ floats on $L_2$, so $\sigma_1 < \sigma_2$. With the string taut, $P$ is pulled down by it, so its buoyancy exceeds its weight: $\rho_1 < \sigma_1$. $Q$ is held up by the string, so $\rho_2 > \sigma_2$. For the pair (equal volumes $V$) in equilibrium: $(\rho_1 + \rho_2)Vg = (\sigma_1 + \sigma_2)Vg$, so $\sigma_2 - \rho_1 = \rho_2 - \sigma_1$.
$P$ alone in $L_2$: $\rho_1 < \sigma_1 < \sigma_2$, so it rises, with $|\vec V_P| = \frac{2r^2g(\sigma_2 - \rho_1)}{9\eta_2}$.
$Q$ alone in $L_1$: $\rho_2 > \sigma_2 > \sigma_1$, so it sinks, with $|\vec V_Q| = \frac{2r^2g(\rho_2 - \sigma_1)}{9\eta_1}$.
Using $\sigma_2 - \rho_1 = \rho_2 - \sigma_1$: $\frac{|\vec V_P|}{|\vec V_Q|} = \frac{\eta_1}{\eta_2}$. (A) true, (B) false.
One moves up and the other down, so the velocities are antiparallel: $\vec V_P\cdot\vec V_Q < 0$. (D) true, (C) false.
Answer: (A), (D). | ||||
| 10 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Multiple correctMedium | |
A medium multiple correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\sqrt{\mu_0/\varepsilon_0}$ has the dimensions of resistance (the impedance of free space, about 377 Ω), i.e. of $V/I$. Also $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$, so $\varepsilon_0c = \sqrt{\varepsilon_0/\mu_0}$ has dimensions $I/V$ and $\mu_0c = \sqrt{\mu_0/\varepsilon_0}$ has dimensions $V/I$.
(A) $\frac{\mu_0}{\varepsilon_0} = \left[\frac{V}{I}\right]^2$, so $\mu_0I^2$ and $\varepsilon_0V^2$ match. True.
(B) would need $\frac{\varepsilon_0}{\mu_0} = \frac{V}{I}$, but it is $\left[\frac{I}{V}\right]^2$. False.
(C) $\varepsilon_0cV$ has dimensions $\frac{I}{V}\cdot V = I$. True.
(D) $\mu_0cI$ has dimensions of $V$, while $\varepsilon_0V$ does not. False.
Answer: (A), (C). | ||||
| 11 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Treat the cavity as the full sphere of charge density $\rho$ plus a sphere of density $-\rho$ filling the cavity. Inside a uniform sphere, Gauss's law gives $\vec E = \frac{\rho\vec r}{3\varepsilon_0}$ measured from its centre.
At a point $\vec r$ in the cavity (with $\vec a = \vec{OP}$): $\vec E = \frac{\rho\vec r}{3\varepsilon_0} - \frac{\rho(\vec r - \vec a)}{3\varepsilon_0} = \frac{\rho\vec a}{3\varepsilon_0}$.
The field is uniform, along $\vec a$, with magnitude $\frac{\rho a}{3\varepsilon_0}$: both magnitude and direction depend on $\vec a$ and not on $\vec r$.
(A), (B) false: the direction does not depend on $\vec r$. (C) false: the magnitude is proportional to $a$. (D) true.
Answer: (D). | ||||
| 12 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctEasy | |
A easy multiple correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Stress is along the horizontal axis here. $P$'s curve reaches a larger stress before breaking than $Q$'s, so $P$ has the greater tensile (breaking) strength. (A) true.
$P$ shows a long stretch of large strain beyond its linear part (plastic flow) before fracture, much more than $Q$, so $P$ is more ductile and less brittle. (B) true, (C) false.
Young's modulus is stress/strain, i.e. the reciprocal of the slope on this plot. $P$'s line is steeper (more strain per unit stress), so $P$ has the smaller Young's modulus. (D) false.
Answer: (A), (B). | ||||
| 13 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctMedium | |
A medium multiple correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Inside the sphere, $g(r) = \frac{4}{3}\pi G\rho r$. Hydrostatic balance: $\frac{dP}{dr} = -\rho g(r) = -\frac43\pi G\rho^2 r$.
With $P(R) = 0$: $P(r) = \frac23\pi G\rho^2(R^2 - r^2)$, so $P \propto R^2 - r^2$.
(A) $P(0) = \frac23\pi G\rho^2R^2 \ne 0$. False.
(B) $\frac{1 - 9/16}{1 - 4/9} = \frac{7/16}{5/9} = \frac{63}{80}$. True.
(C) $\frac{1 - 9/25}{1 - 4/25} = \frac{16}{21}$. True.
(D) $\frac{1 - 1/4}{1 - 1/9} = \frac{3/4}{8/9} = \frac{27}{32} \ne \frac{20}{27}$. False.
Answer: (B), (C). | ||||
| 14 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $C_1 = \frac{\varepsilon_0S}{d}$. The space splits into two halves of area $S/2$ side by side (in parallel).
Lower half: dielectric $\varepsilon_1 = 2$ fills the whole gap $d$: $C = \frac{2\varepsilon_0(S/2)}{d} = C_1$.
Upper half: $\varepsilon_2 = 4$ over thickness $d/2$ in series with $\varepsilon_1 = 2$ over $d/2$:
$C_a = \frac{4\varepsilon_0(S/2)}{d/2} = 4C_1$, $C_b = \frac{2\varepsilon_0(S/2)}{d/2} = 2C_1$, in series $\frac{4\cdot2}{4+2}C_1 = \frac43C_1$.
$C_2 = C_1 + \frac43C_1 = \frac73C_1$.
Answer: (D). | ||||
| 15 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Let the piston area be $A$ and spring constant $k$. The spring is relaxed at the start, so the pressure balancing the piston at the start is $P_1$, and later $P = P_1 + \frac{kx}{A}$ with $V = V_1 + Ax$. Pressure rises linearly with volume: $P - P_1 = \frac{k}{A^2}(V - V_1)$.
Case $V_2 = 2V_1$, $T_2 = 3T_1$: $P_2V_2 = 3P_1V_1$, so $P_2 = \frac32P_1$. Spring energy $= \frac12kx^2 = \frac12(P_2 - P_1)A\cdot x = \frac12\cdot\frac{P_1}{2}\cdot V_1 = \frac14P_1V_1$. (A) true.
$\Delta U = \frac32nR\Delta T = \frac32(P_2V_2 - P_1V_1) = \frac32(2P_1V_1) = 3P_1V_1$. (B) true.
Case $V_2 = 3V_1$, $T_2 = 4T_1$: $P_2 = \frac43P_1$. Work by the gas is the area under the straight line: $W = \frac12(P_1 + P_2)(V_2 - V_1) = \frac12\cdot\frac73P_1\cdot2V_1 = \frac73P_1V_1$. (C) true.
$\Delta U = \frac32(4P_1V_1 - P_1V_1) = \frac92P_1V_1$, so $Q = \Delta U + W = \frac92P_1V_1 + \frac73P_1V_1 = \frac{41}{6}P_1V_1 \ne \frac{17}{6}P_1V_1$. (D) false.
Answer: (A), (B), (C). | ||||
| 16 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Multiple correctMedium | |
A medium multiple correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Mass number: $236 = 140 + 94 + A_x + A_y$, so $A_x + A_y = 2$. Charge: $92 = 54 + 38 + Z_x + Z_y$, so $Z_x + Z_y = 0$. Two neutrons satisfy both; $p + e^-$ has mass number 1 and $p + n$ has charge $+1$. So $x = y = n$; (B) and (C) are out.
Energy released: $Q = (140 + 94)(8.5) - 236(7.5) = 1989 - 1770 = 219$ MeV. The neutrons carry $2 + 2 = 4$ MeV, leaving $K_{Xe} + K_{Sr} = 215$ MeV.
Momentum: the two fragments carry (nearly) equal and opposite momenta, so kinetic energy is shared inversely as mass: $K_{Sr} = 215\cdot\frac{140}{234} \approx 129$ MeV, $K_{Xe} = 215\cdot\frac{94}{234} \approx 86$ MeV. The lighter Sr gets more, so (D) is the wrong way round.
Answer: (A). | ||||
| 17 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctMedium | |
A medium multiple 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 carriers drift with speed $v = \frac{I}{neA}$, with cross-section $A = wd$. In equilibrium $eE = evB$, so the Hall field along $z$ is $E = vB$. The faces are a width $w$ apart along $z$, so
$V = Ew = \frac{IB}{ne\,wd}\,w = \frac{IB}{ned}$.
The voltage depends on $I$, $B$, $n$ and the thickness $d$ (the dimension along $\vec B$), but not on $w$.
Same material, so the same $n$; same $I$ and $B$: $V \propto \frac1d$.
$d_1 = 2d_2$ gives $V_2 = 2V_1$: (A) true, (B) false.
With $d_1 = d_2$ the width does not matter: $V_2 = V_1$. (D) true, (C) false.
Answer: (A), (D). | ||||
| 18 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctEasy | |
A easy multiple 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 carriers drift with speed $v = \frac{I}{neA}$, with cross-section $A = wd$. In equilibrium $eE = evB$, so the Hall field along $z$ is $E = vB$. The faces are a width $w$ apart along $z$, so
$V = Ew = \frac{IB}{ne\,wd}\,w = \frac{IB}{ned}$.
The voltage depends on $I$, $B$, $n$ and the thickness $d$ (the dimension along $\vec B$), but not on $w$.
Same $I$ and $d$: $V \propto \frac{B}{n}$.
$B_1 = B_2$, $n_1 = 2n_2$: $\frac{V_2}{V_1} = \frac{n_1}{n_2} = 2$. (A) true, (B) false.
$B_1 = 2B_2$, $n_1 = n_2$: $\frac{V_2}{V_1} = \frac{B_2}{B_1} = \frac12$. (C) true, (D) false.
Answer: (A), (C). | ||||
| 19 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctHard | |
A hard multiple correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At the entry face, $n_0\sin i = n_1\sin\theta$, where $n_0$ is the outside medium. The ray then meets the core wall at $90^\circ - \theta$, which must be at least the critical angle: $\sin(90^\circ - \theta) \ge \frac{n_2}{n_1}$, i.e. $\cos\theta \ge \frac{n_2}{n_1}$. At the limit $n_1\sin\theta = \sqrt{n_1^2 - n_2^2}$, so
$\mathrm{NA} = \sin i_m = \frac{\sqrt{n_1^2 - n_2^2}}{n_0}$.
$S_1$: $n_1^2 - n_2^2 = \frac{45}{16} - \frac{36}{16} = \frac{9}{16}$, so $\mathrm{NA}_1 = \frac{3}{4n_0}$.
$S_2$: $n_1^2 - n_2^2 = \frac{64 - 49}{25} = \frac{15}{25}$, so $\mathrm{NA}_2 = \frac{\sqrt{15}}{5n_0}$.
(A) $S_1$ in water: $\frac34\cdot\frac34 = \frac{9}{16}$. $S_2$ in $\frac{16}{3\sqrt{15}}$: $\frac{\sqrt{15}}{5}\cdot\frac{3\sqrt{15}}{16} = \frac{45}{80} = \frac{9}{16}$. Equal: true.
(B) $S_1$ in $\frac{6}{\sqrt{15}}$: $\frac34\cdot\frac{\sqrt{15}}{6} = \frac{\sqrt{15}}{8}$. $S_2$ in water: $\frac{\sqrt{15}}{5}\cdot\frac34 = \frac{3\sqrt{15}}{20}$. Not equal: false.
(C) $S_1$ in air: $\frac34$. $S_2$ in $\frac{4}{\sqrt{15}}$: $\frac{\sqrt{15}}{5}\cdot\frac{\sqrt{15}}{4} = \frac34$. Equal: true.
(D) $\frac34$ against $\frac{3\sqrt{15}}{20} \approx 0.58$: false.
Answer: (A), (C). | ||||
| 20 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics A ray is guided through the combined structure only if it is guided in both parts.
If light enters the part with $NA_2$ first, only rays with $\sin i \le NA_2$ are trapped at all.
If it enters the part with $NA_1$ first, rays with $NA_2 < \sin i \le NA_1$ are trapped in the first part, but they travel at too steep an angle for the second part, where they are no longer totally reflected and leak out.
Either way, only rays within the smaller acceptance cone pass all the way, so the combination's numerical aperture is the smaller one, $NA_2$.
Answer: (D). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 21 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalHard | |
A hard numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The starting compound is a bicyclic (octalin) secondary alcohol: the CH–OH carbon is next to a carbon carrying two methyl groups, and the other ring already has one C=C.
H$^+$/heat: the OH is protonated and lost, giving a secondary carbocation. A 1,2-methyl shift from the neighbouring gem-dimethyl carbon gives a more stable tertiary carbocation, which loses a proton from the ring-fusion carbon to give the most substituted (tetrasubstituted) alkene. So **P** is a diene: the new C=C plus the original ring C=C.
Cold, dilute, excess KMnO$_4$ (Baeyer's reagent) adds two OH groups across every C=C (syn dihydroxylation). Two double bonds give two diol units.
**Q** therefore has $2\times2 = 4$ hydroxyl groups.
Answer: 4. | ||||
| 22 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalEasy | |
A easy numerical question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids I. Benzene with CO and HCl in the presence of anhydrous AlCl$_3$/CuCl is the Gattermann–Koch reaction: it gives benzaldehyde.
II. Benzal chloride, C$_6$H$_5$CHCl$_2$, hydrolysed with water at 100 °C gives the gem-diol C$_6$H$_5$CH(OH)$_2$, which loses water to benzaldehyde.
III. Benzoyl chloride with H$_2$ over Pd–BaSO$_4$ is the Rosenmund reduction: benzaldehyde.
IV. Methyl benzoate with DIBAL-H in toluene at $-78$ °C, then water, is reduced only as far as the aldehyde: benzaldehyde.
All four give benzaldehyde.
Answer: 4. | ||||
| 23 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Read the ligands off the name: one acetyl (CH$_3$CO$^-$, bound through its carbonyl carbon), one bromido (Br$^-$), two carbonyl (CO, bound through C) and two triethylphosphine (bound through P).
The complex is [Fe(COCH$_3$)Br(CO)$_2$(PEt$_3$)$_2$].
Fe–C bonds: 1 (acetyl) + 2 (carbonyls) = 3. The ethyl carbons of PEt$_3$ are bonded to P, not to Fe.
Answer: 3. | ||||
| 24 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All six are octahedral.
[Co(en)$_2$Cl$_2$]$^+$, type M(AA)$_2$B$_2$: the two Cl can be cis or trans. Yes.
[CrCl$_2$(ox)$_2$]$^{3-}$, also M(AA)$_2$B$_2$: cis and trans. Yes.
[Fe(H$_2$O)$_4$(OH)$_2$]$^+$, type MA$_4$B$_2$: cis and trans. Yes.
[Fe(NH$_3$)$_2$(CN)$_4$]$^-$, type MA$_4$B$_2$: cis and trans. Yes.
[Co(en)$_2$(NH$_3$)Cl]$^{2+}$, type M(AA)$_2$BC: NH$_3$ and Cl can be cis or trans. Yes.
[Co(NH$_3$)$_4$(H$_2$O)Cl]$^{2+}$, type MA$_4$BC: H$_2$O and Cl can be cis or trans. Yes.
All six show cis–trans isomerism.
Answer: 6. | ||||
| 25 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Diborane is solvolysed by methanol to trimethyl borate:
B$_2$H$_6$ + 6CH$_3$OH → 2B(OCH$_3$)$_3$ + 6H$_2$.
Each mole of B$_2$H$_6$ gives 2 moles of B(OCH$_3$)$_3$, so 3 moles give 6 moles.
Answer: 6. | ||||
| 26 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry For a weak acid, $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$. The limiting conductivities are equal ($\lambda^0_{H^+}$ is common and $\lambda^0_{X^-} \approx \lambda^0_{Y^-}$), so $\alpha_{HX} = \frac{\alpha_{HY}}{10}$.
With $\alpha \ll 1$, $K_a = c\alpha^2$:
$\frac{K_a(HX)}{K_a(HY)} = \frac{0.01\,\alpha_{HY}^2/100}{0.10\,\alpha_{HY}^2} = 10^{-3}$.
p$K_a$(HX) $-$ p$K_a$(HY) $= -\log 10^{-3} = 3$.
Answer: 3. | ||||
| 27 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Mass number falls by $238 - 206 = 32$, so the decay emits $32/4 = 8$ alpha particles. (Charge: $92 - 2\times8 = 76$, so 6 beta particles bring it up to 82.)
Each alpha particle becomes a helium atom, a gas: 1 mol U gives 8 mol He. Uranium and lead are solids and add nothing to the pressure.
Gas: initially 1 mol (air); finally $1 + 8 = 9$ mol, at the same volume and temperature.
$\frac{p_f}{p_i} = \frac{9}{1} = 9$.
Answer: 9. | ||||
| 28 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions The complex is [Fe(H$_2$O)$_2$(C$_2$O$_4$)$_2$]$^{2-}$. Both Fe$^{2+}$ and the oxalate are oxidised:
Fe$^{2+}$ → Fe$^{3+}$ loses 1 e$^-$; 2C$_2$O$_4^{2-}$ → 4CO$_2$ loses 4 e$^-$. Total 5 e$^-$ per complex.
MnO$_4^-$ → Mn$^{2+}$ gains 5 e$^-$. So they react 1 : 1:
[Fe(H$_2$O)$_2$(C$_2$O$_4$)$_2$]$^{2-}$ + MnO$_4^-$ + 8H$^+$ → Fe$^{3+}$ + Mn$^{2+}$ + 4CO$_2$ + 6H$_2$O.
(Check: O on the left $2 + 8 + 4 = 14$, right $8 + 6 = 14$; H: $4 + 8 = 12$ both sides; charge $-2 - 1 + 8 = +5 = 3 + 2$.)
8 H$^+$ are used for each MnO$_4^-$, so the rate of change of [H$^+$] is 8 times that of [MnO$_4^-$].
Answer: 8. | ||||
| 29 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Multiple correctHard | |
A hard multiple correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Ozonolysis with Zn/H$_2$O cleaves the C=C of the five-membered ring of the methylindene. The CH= carbon joined to the benzene ring becomes an aryl –CHO, and the other alkene carbon becomes –CHO on the CH$_2$ that stays on the ring. **R** is the dialdehyde 2-(2-formyl-methylphenyl)acetaldehyde: a benzene ring carrying –CHO and –CH$_2$CHO on adjacent carbons, with the methyl where it was.
NH$_3$ condenses with both carbonyl groups (two imine-forming steps with loss of water), closing a six-membered ring containing N: C(=N)H from the aryl CHO, then N, then CH=CH from the CH$_2$CHO. The ring is aromatised to an **isoquinoline** (N is two atoms from the ring junction, not next to it), which rules out the quinolines (B) and (D).
Position of the methyl: in the indene it sits on C-5, next to C-4, which is next to the ring-junction carbon C-3a that carries the alkene CH (C-3). That CH becomes the CH=N carbon C-1 of the isoquinoline, C-3a becomes C-8a, C-4 becomes C-8, and the methyl ends on C-7: 7-methylisoquinoline, drawn in (A). (C) has it on C-6.
Answer: (A). | ||||
| 30 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Multiple correctEasy | |
A easy multiple correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Benzene + propene with acid at high pressure and heat is a Friedel–Crafts alkylation through the secondary (isopropyl) carbocation, so **T** is cumene, C$_6$H$_5$CH(CH$_3$)$_2$, not the n-propyl product.
With a radical initiator and O$_2$, the benzylic tertiary C–H of cumene is the weakest; the cumyl radical takes up O$_2$ and then H, giving cumene hydroperoxide, C$_6$H$_5$C(CH$_3$)$_2$OOH. This is the first stage of the cumene process for phenol and acetone.
(A) puts the hydroperoxide on the wrong (n-propyl) skeleton; (C) and (D) contain C=C bonds that are not formed.
Answer: (B). | ||||
| 31 | AminesDiazonium salts | Class 12: Amines | Multiple correctEasy | |
A easy multiple correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Aniline with NaNO$_2$/HCl at 0 °C is diazotised: **V** is benzenediazonium chloride, C$_6$H$_5$N$_2^+$Cl$^-$.
In NaOH, 2-naphthol is present as the naphthoxide ion, which is strongly activated. The diazonium ion is a weak electrophile and couples at C-1, the position ortho to the O$^-$ that keeps the other ring aromatic in the intermediate (the most reactive site of 2-naphthol).
**W** is 1-phenylazo-2-naphthol, an orange-red dye: structure (A).
Answer: (A). | ||||
| 32 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Structures (H on O in every case): HOCl; HO–Cl=O; HO–Cl(=O)$_2$; HO–Cl(=O)$_3$. Cl is $sp^3$ in all four, with $3, 2, 1, 0$ lone pairs respectively.
(A) Cl=O bonds: 1 in HClO$_2$ + 2 in HClO$_3$ = 3, not two. False.
(B) Lone pairs on Cl: 2 in HClO$_2$ + 1 in HClO$_3$ = 3. True.
(C) HClO$_4$: four σ bonds, no lone pair, $sp^3$ (tetrahedral). True.
(D) Acid strength rises with the number of terminal O atoms (more delocalised anion); HClO$_4$ is the strongest and HClO the weakest. False.
Answer: (B), (C). | ||||
| 33 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter In dilute HCl the sulphide ion concentration is kept very low, so only the least soluble sulphides (group II: Hg$^{2+}$, Pb$^{2+}$, Cu$^{2+}$, Bi$^{3+}$, Cd$^{2+}$, As, Sb, Sn) precipitate.
(A) Ba$^{2+}$ gives no sulphide precipitate at all; ZnS needs ammoniacal medium (group IV). False.
(B) Bi$^{3+}$ precipitates as Bi$_2$S$_3$, but Fe$^{3+}$ is only reduced to Fe$^{2+}$ (with a deposit of sulphur); FeS does not form in acid. False.
(C) CuS and PbS both precipitate. True.
(D) HgS and Bi$_2$S$_3$ both precipitate. True.
Answer: (C), (D). | ||||
| 34 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Hydrolysis turns each Si–Cl into Si–OH, and the silanols condense to Si–O–Si links.
(CH$_3$)$_2$SiCl$_2$ gives (CH$_3$)$_2$Si(OH)$_2$, with two links per Si: a linear chain –[Si(CH$_3$)$_2$–O]$_n$–.
(CH$_3$)$_3$SiCl gives (CH$_3$)$_3$SiOH, with only one link: it caps (terminates) a chain.
CH$_3$SiCl$_3$ and SiCl$_4$ give three or four links and cross-link the polymer; Si(CH$_3$)$_4$ has no Si–Cl and does not react.
Answer: (B). | ||||
| 35 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Electron transfer from the metal means a chemical bond forms between O$_2$ and the surface: this is chemisorption, not physisorption. (A) false.
Adsorption is exothermic (bonds form, and the adsorbed molecule loses freedom, so $\Delta S < 0$ and $\Delta H$ must be negative for it to happen). (B) true.
The extra electrons go into O$_2$'s lowest empty level, the antibonding $\pi^*_{2p}$ orbitals, so their occupancy rises. (C) true.
More antibonding electrons lower the bond order (towards superoxide or peroxide), so the O–O bond lengthens. (D) true.
Answer: (B), (C), (D). | ||||
| 36 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Multiple correctMedium | |
A medium multiple correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Compare with van der Waals' equation $\left(p + \frac{a}{V^2}\right)(V - b) = RT$. Here $a = 0$ but $b \ne 0$.
$b$ is the excluded volume: the atoms have a finite size and cannot come closer than a certain distance, so the potential is infinitely repulsive (a vertical wall) at short range.
$a = 0$ means there is no attraction at all, so beyond contact the potential is zero.
This is the hard-sphere potential: a vertical wall at small $r$ and $V(r) = 0$ beyond, graph (C).
(A) and (D) contain an attractive well ($a \ne 0$); (B) has no repulsive core ($b = 0$, an ideal gas).
Answer: (C). | ||||
| 37 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Multiple correctEasy | |
A easy multiple correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons C$_8$H$_6$ that adds water under Hg$^{2+}$ catalysis is an alkyne: phenylacetylene, C$_6$H$_5$C≡CH.
H$_2$ over Pd–BaSO$_4$ (a poisoned, Lindlar-type catalyst) stops at the alkene: C$_8$H$_8$ is styrene, C$_6$H$_5$CH=CH$_2$.
Hydroboration–oxidation adds H and OH anti-Markovnikov: the OH goes to the terminal CH$_2$. **X** is 2-phenylethanol, C$_6$H$_5$CH$_2$CH$_2$OH.
(B) would be the Markovnikov product; (A) and (D) are carbonyl compounds, not formed here.
Answer: (C). | ||||
| 38 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Multiple correctMedium | |
A medium multiple correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Hg$^{2+}$-catalysed hydration of phenylacetylene follows Markovnikov's rule; the enol tautomerises to the ketone: C$_8$H$_8$O is acetophenone, C$_6$H$_5$COCH$_3$.
EtMgBr adds to the C=O; after water work-up the product is the tertiary alcohol 2-phenylbutan-2-ol, C$_6$H$_5$C(OH)(CH$_3$)CH$_2$CH$_3$.
H$^+$/heat dehydrates it through the tertiary benzylic carbocation. The proton is lost to give the most substituted alkene that is also conjugated with the ring: C$_6$H$_5$C(CH$_3$)=CHCH$_3$, 2-phenylbut-2-ene (trisubstituted, conjugated).
(C) is the less substituted (disubstituted) alkene; (A) and (B) have the wrong carbon skeleton (a straight chain from a benzylic CH$_2$ or CH).
Answer: (D). | ||||
| 39 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Expt. 1: 0.1 mol H$^+$ is neutralised, releasing $0.1\times57.0 = 5.70$ kJ. This raises the beaker and contents by 5.7 °C, so the whole system's heat capacity is $\frac{5700}{5.7} = 1000$ J K$^{-1}$.
Expt. 2: NaOH (0.1 mol) is the limiting reagent, so 0.1 mol acetic acid is neutralised. The same system rises 5.6 °C: heat released $= 1000\times5.6 = 5.60$ kJ, i.e. 56.0 kJ per mole.
The acetic acid must first dissociate, which absorbs heat: $\Delta H_{diss} = 57.0 - 56.0 = 1.0$ kJ mol$^{-1}$.
(Treating the calorimeter as the 200 g of solution alone would give $200\times4.2\times5.7 = 4788$ J, which is not 5.70 kJ; the experiment is used to find the constant, so the total heat capacity is taken from it.)
Answer: (A). | ||||
| 40 | EquilibriumpH and buffers | Class 11: Equilibrium | Multiple correctEasy | |
A easy multiple correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Acetic acid: $0.1\times2.0 = 0.2$ mol. NaOH: $0.1\times1.0 = 0.1$ mol.
After neutralisation: 0.1 mol CH$_3$COOH and 0.1 mol CH$_3$COO$^-$ remain in the same solution, a buffer with equal amounts of acid and salt.
pH $= \mathrm{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]} = \mathrm{p}K_a = -\log(2.0\times10^{-5}) = 5 - 0.30 = 4.7$.
Answer: (B). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 41 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha_k = e^{ik\pi/7}$ are points on the unit circle, each consecutive pair separated by the angle $\frac{\pi}{7}$.
$|\alpha_{k+1} - \alpha_k| = |e^{ik\pi/7}|\,|e^{i\pi/7} - 1| = 2\sin\frac{\pi}{14}$, the same for every $k$.
In the denominator, $\alpha_{4k-1}$ and $\alpha_{4k-2}$ are also consecutive, so each term is again $2\sin\frac{\pi}{14}$.
Ratio $= \frac{12\cdot2\sin(\pi/14)}{3\cdot2\sin(\pi/14)} = 4$.
Answer: 4. | ||||
| 42 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_7 = \frac72(2a + 6d) = 7(a + 3d)$ and $S_{11} = \frac{11}{2}(2a + 10d) = 11(a + 5d)$.
$\frac{7(a + 3d)}{11(a + 5d)} = \frac{6}{11}$ gives $7a + 21d = 6a + 30d$, so $a = 9d$.
Seventh term $a + 6d = 15d$, with $130 < 15d < 140$, i.e. $8.67 < d < 9.33$. All terms are natural numbers, so $d$ is an integer (the difference of two natural numbers) and $d = 9$ (then $a = 81$, $a_7 = 135$).
Answer: 9. | ||||
| 43 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Each factor $(1 + x^r)$ contributes either $1$ or $x^r$, so a term $x^9$ arises once for every way of writing 9 as a sum of distinct positive integers (all at most 100).
List them: $9$; $8+1$; $7+2$; $6+3$; $5+4$; $6+2+1$; $5+3+1$; $4+3+2$. Four distinct parts would need at least $1+2+3+4 = 10 > 9$, so there are no more.
That is 8 ways, so the coefficient is 8.
Answer: 8. | ||||
| 44 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse: $a^2 = 9$, $b^2 = 5$, $c = \sqrt{9 - 5} = 2$. So $f_1 = 2$, $f_2 = -2$.
$P_1$: vertex $(0,0)$, focus $(2,0)$: $y^2 = 8x$. Its tangents are $y = mx + \frac{2}{m}$. Through $(2f_2, 0) = (-4, 0)$: $0 = -4m_1 + \frac{2}{m_1}$, so $m_1^2 = \frac12$.
$P_2$: focus $(-4, 0)$: $y^2 = -16x$. Its tangents are $y = mx - \frac{4}{m}$. Through $(2, 0)$: $0 = 2m_2 - \frac{4}{m_2}$, so $m_2^2 = 2$.
$\frac{1}{m_1^2} + m_2^2 = 2 + 2 = 4$.
Answer: 4. | ||||
| 45 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $e^{\cos(\alpha^n)} - e = e\left(e^{\cos(\alpha^n) - 1} - 1\right)$.
As $\alpha \to 0$, $\cos(\alpha^n) - 1 \approx -\frac{\alpha^{2n}}{2} \to 0$, and $e^u - 1 \approx u$, so the numerator $\approx -\frac{e}{2}\alpha^{2n}$.
The limit $\lim \frac{-\frac{e}{2}\alpha^{2n}}{\alpha^m}$ equals $-\frac{e}{2}$ only when $m = 2n$ (if $m < 2n$ it is 0, if $m > 2n$ it is infinite).
$\frac{m}{n} = 2$.
Answer: 2. | ||||
| 46 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $u = 9x + 3\tan^{-1}x$. Then $\frac{du}{dx} = 9 + \frac{3}{1 + x^2} = \frac{12 + 9x^2}{1 + x^2}$, exactly the second factor.
So $\alpha = \int e^u\,du = \left[e^{9x + 3\tan^{-1}x}\right]_0^1 = e^{9 + 3\pi/4} - 1$.
$1 + \alpha = e^{9 + 3\pi/4} > 0$, so $\log_e|1 + \alpha| - \frac{3\pi}{4} = 9$.
Answer: 9. | ||||
| 47 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f$ is odd, so $F(1) = \int_{-1}^1 f = 0$. Also $f(f(-t)) = f(-f(t)) = -f(f(t))$, so $|f(f(t))|$ is even and $t|f(f(t))|$ is odd: $G(1) = 0$. The limit is of the form $\frac00$.
By L'Hôpital's rule and the fundamental theorem of calculus:
$\lim_{x\to1}\frac{F(x)}{G(x)} = \frac{F'(1)}{G'(1)} = \frac{f(1)}{1\cdot|f(f(1))|} = \frac{1/2}{|f(1/2)|}$.
Setting this equal to $\frac{1}{14}$: $|f(\frac12)| = 7$.
Sign: an odd function has $f(0) = 0$, and that is its only zero. $f$ is continuous and never zero on $(0, \infty)$, so it keeps the sign of $f(1) = \frac12 > 0$ there. Hence $f(\frac12) = 7$ (and $G'(1) \ne 0$, so L'Hôpital applies).
Answer: 7. | ||||
| 48 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra Read “components” as the coefficients when $\vec s$ is written in each basis: $\vec s = 4\vec p + 3\vec q + 5\vec r$, and also
$\vec s = x(-\vec p + \vec q + \vec r) + y(\vec p - \vec q + \vec r) + z(-\vec p - \vec q + \vec r)$.
The three vectors are non-coplanar, so compare coefficients:
$\vec p$: $-x + y - z = 4$; $\vec q$: $x - y - z = 3$; $\vec r$: $x + y + z = 5$.
Adding the first two: $-2z = 7$, $z = -\frac72$. Adding the last two: $2x = 8$, $x = 4$. Then $y = 5 - 4 + \frac72 = \frac92$.
$2x + y + z = 8 + \frac92 - \frac72 = 9$.
Answer: 9 (the official key; marks were later given to all candidates). | ||||
| 49 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Multiple correctMedium | |
A medium multiple correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Distinct real roots: discriminant $1 - 4\alpha^2 > 0$, so $|\alpha| < \frac12$.
$|x_1 - x_2| = \frac{\sqrt{1 - 4\alpha^2}}{|\alpha|} < 1$ gives $1 - 4\alpha^2 < \alpha^2$, i.e. $\alpha^2 > \frac15$, $|\alpha| > \frac{1}{\sqrt5}$.
So $S = \left(-\frac12, -\frac{1}{\sqrt5}\right)\cup\left(\frac{1}{\sqrt5}, \frac12\right)$.
(A) and (D) are exactly these pieces: subsets. (B) and (C) contain values with $|\alpha| < \frac{1}{\sqrt5}$, where the roots are more than 1 apart: not subsets.
Answer: (A), (D). | ||||
| 50 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Multiple correctMedium | |
A medium multiple correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\frac12 < \frac{6}{11} < \frac{1}{\sqrt2}$, so $\frac{\pi}{6} < \sin^{-1}\frac{6}{11} < \frac{\pi}{4}$ and $\frac{\pi}{2} < \alpha < \frac{3\pi}{4}$. Hence $\cos\alpha < 0$. (D) true.
$0 < \frac49 < \frac12$, so $\frac{\pi}{3} < \cos^{-1}\frac49 < \frac{\pi}{2}$ and $\pi < \beta < \frac{3\pi}{2}$ (third quadrant). Hence $\sin\beta < 0$ and $\cos\beta < 0$. (B) true, (A) false.
Adding: $\frac{3\pi}{2} < \alpha + \beta < \frac{9\pi}{4}$, which lies in the fourth quadrant or the first (past $2\pi$); cosine is positive throughout $\left(\frac{3\pi}{2}, \frac{5\pi}{2}\right)$. So $\cos(\alpha + \beta) > 0$. (C) true.
Answer: (B), (C), (D). | ||||
| 51 | Conic SectionsEllipse | Class 11: Conic Sections | Multiple correctHard | |
A hard multiple correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Circle: centre $(0, 1)$, radius $\sqrt2$. Its distance from $x + y = 3$ is $\frac{|0 + 1 - 3|}{\sqrt2} = \sqrt2$, so the line is tangent, at the foot of the perpendicular $P = (1, 2)$.
Points on the line at distance $\frac{2\sqrt2}{3}$ from $P$: move $\pm\frac23$ in $x$ and $\mp\frac23$ in $y$: $\left(\frac53, \frac43\right)$ and $\left(\frac13, \frac83\right)$.
For an ellipse $\frac{x^2}{A} + \frac{y^2}{B} = 1$, the tangent at $(x_0, y_0)$ is $\frac{xx_0}{A} + \frac{yy_0}{B} = 1$. Matching $\frac{x}{3} + \frac{y}{3} = 1$: $A = 3x_0$, $B = 3y_0$.
$E_1$ (major axis along $x$, $A > B$) needs $x_0 > y_0$: $Q = \left(\frac53, \frac43\right)$, $A = 5$, $B = 4$, $e_1^2 = 1 - \frac45 = \frac15$.
$E_2$ (major axis along $y$, $B > A$): $R = \left(\frac13, \frac83\right)$, $A = 1$, $B = 8$, $e_2^2 = 1 - \frac18 = \frac78$.
(A) $\frac15 + \frac78 = \frac{43}{40}$. True.
(B) $e_1e_2 = \sqrt{\frac{7}{40}} = \frac{\sqrt7}{2\sqrt{10}}$. True.
(C) $|e_1^2 - e_2^2| = \frac{27}{40} \ne \frac58$. False.
(D) False, by (B).
Answer: (A), (B). | ||||
| 52 | Conic SectionsHyperbola | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent to $H$ at $P$: $xx_1 - yy_1 = 1$; with $y = 0$, $M = \left(\frac{1}{x_1}, 0\right)$.
The circle touches $H$ at $P$, so its centre lies on the normal to $H$ at $P$. The normal has slope $-\frac{y_1}{x_1}$: $y - y_1 = -\frac{y_1}{x_1}(x - x_1)$; with $y = 0$, $x = 2x_1$. So $N = (2x_1, 0)$.
Centroid: $l = \frac{x_1 + \frac{1}{x_1} + 2x_1}{3} = x_1 + \frac{1}{3x_1}$, $m = \frac{y_1}{3} = \frac{\sqrt{x_1^2 - 1}}{3}$.
(A) $\frac{dl}{dx_1} = 1 - \frac{1}{3x_1^2}$. True; (C) false.
(B) $\frac{dm}{dx_1} = \frac{x_1}{3\sqrt{x_1^2 - 1}}$. True.
(D) $\frac{dm}{dy_1} = \frac13$. True.
Answer: (A), (B), (D). | ||||
| 53 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $g(t) = \sin^6 at + \cos^4 at$. Both $\sin^6$ and $\cos^4$ have period $\pi$, so $g$ has period $\frac{\pi}{a}$; for $a = 2$ or $4$, $\pi$ is a whole number of periods, so $g(t + \pi) = g(t)$.
Split $\int_0^{4\pi} = \sum_{k=0}^{3}\int_{k\pi}^{(k+1)\pi}$ and put $t = u + k\pi$: $\int_{k\pi}^{(k+1)\pi}e^tg(t)\,dt = e^{k\pi}\int_0^\pi e^ug(u)\,du$.
So $L = 1 + e^\pi + e^{2\pi} + e^{3\pi} = \frac{e^{4\pi} - 1}{e^\pi - 1}$, the same for $a = 2$ and $a = 4$ (the denominator is not zero since $g > 0$).
(A) and (C) true; (B) and (D) give a different value.
Answer: (A), (C). | ||||
| 54 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Multiple correctMedium | |
A medium multiple correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $h = f - 3g$. From the table: $h(-1) = 3 - 0 = 3$, $h(0) = 6 - 3 = 3$, $h(2) = 0 + 3 = 3$.
Rolle's theorem on $[-1, 0]$ and on $[0, 2]$: $h'$ vanishes at least once in each of $(-1, 0)$ and $(0, 2)$.
$h''$ never vanishes in either interval, so (being a derivative, it has the intermediate value property) it keeps one sign there; $h'$ is strictly monotonic in each interval and vanishes at most once.
So $f' - 3g' = 0$ has exactly one solution in $(-1, 0)$ and exactly one in $(0, 2)$: two in all.
(B), (C) true; (A) and (D) false.
Answer: (B), (C). | ||||
| 55 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Factor: $f(x) = 7\tan^6x(\tan^2x + 1) - 3\tan^2x(\tan^2x + 1) = (7\tan^6x - 3\tan^2x)\sec^2x$.
With $u = \tan x$: $\int f\,dx = \tan^7x - \tan^3x$. So $\int_0^{\pi/4}f\,dx = (1 - 1) - 0 = 0$. (B) true, (D) false.
By parts: $\int_0^{\pi/4}xf\,dx = \left[x(\tan^7x - \tan^3x)\right]_0^{\pi/4} - \int_0^{\pi/4}(\tan^7x - \tan^3x)\,dx = 0 - \int_0^{\pi/4}\tan^3x(\tan^4x - 1)\,dx$.
$\tan^4x - 1 = (\tan^2x - 1)\sec^2x$, so $\int_0^{\pi/4}\tan^3x(\tan^2x - 1)\sec^2x\,dx = \int_0^1(u^5 - u^3)\,du = \frac16 - \frac14 = -\frac{1}{12}$.
Hence $\int_0^{\pi/4}xf\,dx = \frac{1}{12}$. (A) true, (C) false.
Answer: (A), (B). | ||||
| 56 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Since $2 \le 2 + \sin^4\pi x \le 3$: $64x^3 \le f'(x) \le 96x^3$ for $x \ge 0$.
Integrate from $\frac12$ to $x$ (with $f(\frac12) = 0$), for $x \in [\frac12, 1]$:
$16\left(x^4 - \frac{1}{16}\right) \le f(x) \le 24\left(x^4 - \frac{1}{16}\right)$.
$\int_{1/2}^1\left(x^4 - \frac{1}{16}\right)dx = \frac15\left(1 - \frac{1}{32}\right) - \frac{1}{32} = \frac{31}{160} - \frac{5}{160} = \frac{13}{80}$.
So $2.6 = \frac{16\cdot13}{80} \le \int_{1/2}^1 f\,dx \le \frac{24\cdot13}{80} = 3.9$.
Only (D), $1 \le I \le 12$, contains this range. (A) has $m = 13 > 3.9$; (B) has $M = \frac12 < 2.6$; (C) has $M = 0$ though the integral is positive.
Answer: (D). | ||||
| 57 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Multiple correctEasy | |
A easy multiple correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $p_1 = \frac{n_1}{n_1 + n_2}$ and $p_2 = \frac{n_3}{n_3 + n_4}$ be the chances of red from each box. Each box is chosen with probability $\frac12$, so by Bayes' theorem
$P(\text{II}\mid\text{red}) = \frac{p_2}{p_1 + p_2} = \frac13$, i.e. $p_1 = 2p_2$.
(A) $p_1 = \frac36 = \frac12$, $p_2 = \frac{5}{20} = \frac14$. True.
(B) $p_1 = \frac39 = \frac13$, $p_2 = \frac{10}{60} = \frac16$. True.
(C) $p_1 = \frac{8}{14} = \frac47$, $p_2 = \frac{5}{25} = \frac15$; $\frac47 \ne \frac25$. False.
(D) $p_1 = \frac{6}{18} = \frac13$, $p_2 = \frac15$; $\frac13 \ne \frac25$. False.
Answer: (A), (B). | ||||
| 58 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Multiple correctEasy | |
A easy multiple correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $N = n_1 + n_2$. After one ball leaves box I, the second draw is from $N - 1$ balls.
$P(\text{red}) = \frac{n_1}{N}\cdot\frac{n_1 - 1}{N - 1} + \frac{n_2}{N}\cdot\frac{n_1}{N - 1} = \frac{n_1(n_1 - 1 + n_2)}{N(N - 1)} = \frac{n_1}{N}$.
(Removing a ball at random does not change the chance that a random remaining ball is red.)
So $\frac{n_1}{n_1 + n_2} = \frac13$, i.e. $n_2 = 2n_1$.
(C) $10, 20$ and (D) $3, 6$ satisfy this. (A) gives $\frac{4}{10}$ and (B) gives $\frac25$: false.
Answer: (C), (D). | ||||
| 59 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Multiple correctMedium | |
A medium multiple correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = F(x) + xF'(x)$.
(A) $f'(1) = F(1) + F'(1) = F'(1) < 0$. True.
(B) $F$ is strictly decreasing on $(\frac12, 3)$ and $F(1) = 0$, so $F(2) < 0$ and $f(2) = 2F(2) < 0$. True.
(C) For $x \in (1, 3)$: $F(x) < F(1) = 0$ and $xF'(x) < 0$, so $f'(x) < 0$; it never vanishes. True.
(D) contradicts (C). False.
Answer: (A), (B), (C). | ||||
| 60 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals By parts: $\int_1^3x^2F'\,dx = \left[x^2F\right]_1^3 - 2\int_1^3xF\,dx = 9(-4) - 0 - 2\int_1^3f\,dx$.
Setting this to $-12$: $-36 - 2\int_1^3f\,dx = -12$, so $\int_1^3f\,dx = -12$. (D) true, (B) false.
By parts again: $\int_1^3x^3F''\,dx = \left[x^3F'\right]_1^3 - 3\int_1^3x^2F'\,dx = 27F'(3) - F'(1) + 36 = 40$, so $27F'(3) - F'(1) = 4$.
Now $f'(x) = F(x) + xF'(x)$: $f'(3) = -4 + 3F'(3)$ and $f'(1) = F'(1)$.
$9f'(3) - f'(1) = -36 + 27F'(3) - F'(1) = -36 + 4 = -32$, so $9f'(3) - f'(1) + 32 = 0$. (C) true.
(A) would need $9f'(3) + f'(1) = 32$, i.e. $27F'(3) + F'(1) = 68$, which does not follow from the data (with $27F'(3) - F'(1) = 4$ it would force $F'(1) = 32 > 0$, contradicting $F' < 0$). False.
Answer: (C), (D). | ||||
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