JEE (Advanced) · 2014
JEE Advanced 2014, 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
- Properties of Bulk Matter5
- Optics4
- Electrostatics4
15 easy · 23 medium · 2 hard
Chemistry
40 Q
- Chemical Bonding and Molecular Structure5
- p-Block Elements4
- Aldehydes, Ketones and Carboxylic Acids3
15 easy · 21 medium · 4 hard
Mathematics
40 Q
- Integral Calculus7
- Limits, Continuity and Differentiability6
- Permutations and Combinations3
4 easy · 33 medium · 3 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 11: Permutations and Combinations
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 11: Conic Sections
Class 11: Probability; Class 12: Probability
Class 12: Matrices; Class 12: Determinants
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Class 12: Vector Algebra
Class 12: Application of Derivatives
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 11: Conic Sections
Class 11: Sequences and Series
Class 11: Straight Lines
Class 12: Differential Equations
Class 11: Binomial Theorem
Class 11: Complex Numbers and Quadratic Equations
Class 11: Complex Numbers and Quadratic Equations
Class 11: Chemical Bonding and Molecular Structure
No matching NCERT chapter
Class 12: Aldehydes, Ketones and Carboxylic Acids
Class 11: Hydrocarbons
Class 11: Some Basic Concepts of Chemistry
Class 11: Organic Chemistry – Some Basic Principles and Techniques
Class 12: Alcohols, Phenols and Ethers
Class 12: Amines
Class 11: Thermodynamics
Class 11: Redox Reactions
No matching NCERT chapter
Class 12: Solutions
Class 12: Coordination Compounds
Class 12: Electrochemistry
No matching NCERT chapter
Class 12: Biomolecules
Class 11: Structure of Atom
- Principles Related to Practical Chemistry1
No matching NCERT chapter
Class 12: Chemical Kinetics
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 12: Current Electricity
Class 11: Systems of Particles and Rotational Motion
Class 11: Thermodynamics
No matching NCERT chapter
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Class 11: Work, Energy and Power
Class 11: Laws of Motion
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 11: Oscillations; Class 11: Waves
Class 11: Units and Measurement
Class 12: Dual Nature of Radiation and Matter
Class 12: Atoms; Class 12: Nuclei
Class 11: Gravitation
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 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Multiple correctHard | |
A hard multiple correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current While A is joined to B, the AC source drives the current $I_0\cos\omega t$ straight into the capacitor, so its charge is
$q(t) = \int_0^t I_0\cos\omega t\,dt = \frac{I_0}{\omega}\sin\omega t$.
(A) The largest charge is $\frac{I_0}{\omega} = \frac{1}{500} = 2 \times 10^{-3}$ C, not $1 \times 10^{-3}$ C. False.
(B) At $\omega t = \frac{7\pi}{6}$, $I = I_0\cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$ A: the current has reversed from the initial (clockwise) direction, so it is anticlockwise. False.
At that instant $q = \frac{I_0}{\omega}\sin\frac{7\pi}{6} = -1 \times 10^{-3}$ C, i.e. the capacitor holds $1 \times 10^{-3}$ C with the opposite polarity to the one the initial current was building up. Its voltage is $\frac{q}{C} = \frac{10^{-3}}{20 \times 10^{-6}} = 50$ V.
(C) After switching to D, the capacitor's 50 V is in series with the battery's 50 V and the two add round the right-hand loop, so the initial current is $\frac{50 + 50}{10} = 10$ A. True.
(D) Finally the capacitor is charged to the battery's emf: $q_f = CV = 20 \times 10^{-6} \times 50 = 1 \times 10^{-3}$ C with the battery's polarity. Its charge goes from $-1 \times 10^{-3}$ C to $+1 \times 10^{-3}$ C, so the battery delivers $Q = 2 \times 10^{-3}$ C. True.
Answer: (C) and (D). | ||||
| 2 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctEasy | |
A easy multiple correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Fringe width $\beta = \frac{\lambda D}{d}$ is proportional to $\lambda$.
(A) $\lambda_2 > \lambda_1$, so $\beta_2 > \beta_1$. True.
(B) The number of fringes in a distance $y$ is $m = \frac{y}{\beta}$; the smaller fringe width gives more fringes, so $m_1 > m_2$. True.
(C) 3rd maximum of $\lambda_2$: $y = 3\frac{\lambda_2 D}{d} = 1800\frac{D}{d}$ (nm). 5th minimum of $\lambda_1$: $y = \frac{(2 \times 5 - 1)\lambda_1}{2}\frac{D}{d} = 4.5 \times 400\frac{D}{d} = 1800\frac{D}{d}$. They coincide. True.
(D) Angular fringe separation is $\frac{\lambda}{d}$, smaller for $\lambda_1$. False.
Answer: (A), (B) and (C). | ||||
| 3 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Multiple correctMedium | |
A medium multiple correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For a stationary wave $y = A\sin kx\cos\omega t$ the end $x = 0$ is a node (the fixed end). Two conditions must hold:
1. The wave speed $\frac{\omega}{k}$ must be 100 m s$^{-1}$.
2. The end $x = 3$ m is the vibrating (driven) end, so it cannot be a node; it is an antinode: $\sin 3k = \pm 1$, i.e. $3k = \frac{(2n+1)\pi}{2}$.
(A) $k = \frac{\pi}{6}$, $\omega = \frac{50\pi}{3}$: $\frac{\omega}{k} = 100$; $\sin\frac{3\pi}{6} = 1$. Possible.
(B) $k = \frac{\pi}{3}$, $\omega = \frac{100\pi}{3}$: speed 100, but $\sin\frac{3\pi}{3} = \sin\pi = 0$, a node at the vibrating end. Not possible.
(C) $k = \frac{5\pi}{6}$, $\omega = \frac{250\pi}{3}$: speed 100; $\sin\frac{15\pi}{6} = \sin\frac{5\pi}{2} = 1$. Possible.
(D) $k = \frac{5\pi}{2}$, $\omega = 250\pi$: speed 100; $\sin\frac{15\pi}{2} = -1$. Possible.
Answer: (A), (C) and (D). | ||||
| 4 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctEasy | |
A easy multiple correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The two portions share the same plates, so they are two capacitors in parallel with the same potential difference $V$ and the same gap $d$.
(A), (B) In each portion $E = \frac{V}{d}$, so $E_1 = E_2$ and $\frac{E_1}{E_2} = 1$. (A) true, (B) false.
With plate area $A$: $C_1 = \frac{K\varepsilon_0 (A/3)}{d}$ and $C_2 = \frac{\varepsilon_0 (2A/3)}{d}$.
(C) $\frac{Q_1}{Q_2} = \frac{C_1}{C_2} = \frac{K}{2}$, not $\frac{3}{K}$. False.
(D) $C = C_1 + C_2 = \frac{\varepsilon_0 A}{3d}(K + 2)$, so $\frac{C}{C_1} = \frac{K+2}{K}$. True.
Answer: (A) and (D). | ||||
| 5 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctEasy | |
A easy multiple correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $E_1 = \frac{Q}{4\pi\varepsilon_0 r^2}$, $E_2 = \frac{\lambda}{2\pi\varepsilon_0 r}$, $E_3 = \frac{\sigma}{2\varepsilon_0}$.
(A) $E_1(r_0) = E_3(r_0)$: $\frac{Q}{4\pi r_0^2} = \frac{\sigma}{2}$, so $Q = 2\pi\sigma r_0^2$. False.
(B) $E_2(r_0) = E_3(r_0)$: $\frac{\lambda}{2\pi r_0} = \frac{\sigma}{2}$, so $r_0 = \frac{\lambda}{\pi\sigma}$. False.
(C) Halving $r$ multiplies $E_1$ by 4 and $E_2$ by 2. With $E_1(r_0) = E_2(r_0) = E$: $E_1(r_0/2) = 4E = 2 \times 2E = 2E_2(r_0/2)$. True.
(D) $E_2(r_0/2) = 2E$ while $E_3$ stays $E$, so $E_2(r_0/2) = 2E_3(r_0/2)$. False.
Answer: (C). | ||||
| 6 | Experimental SkillsHeat and sound experiments | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on heat and sound experiments, from Experimental Skills. NCERT: No matching NCERT chapter The first resonance of a tube closed at one end is at length $\frac{\lambda}{4}$ (end correction neglected), so $v = 4fL$.
$v = 4 \times 244 \times 0.350 = 341.6$ m s$^{-1}$; with $L$ from 0.345 to 0.355 m, $v$ lies between 336.7 and 346.5 m s$^{-1}$.
The speed of sound in a gas is $v = \sqrt{\frac{\gamma RT}{M}}$ with $M$ in kg mol$^{-1}$. With $M$ in grams, $M_{kg} = M \times 10^{-3}$, so $v = \sqrt{100\gamma RT}\sqrt{\frac{10}{M}}$.
Monatomic gases (Ne, Ar): $\gamma = \frac{5}{3}$, $100\gamma \approx 167$, so $v = 640\sqrt{\frac{10}{M}}$.
Diatomic gases (N$_2$, O$_2$): $\gamma = \frac{7}{5}$, $100\gamma = 140$, so $v = 590\sqrt{\frac{10}{M}}$.
Neon: $640 \times \frac{7}{10} = 448$ m s$^{-1}$. Nitrogen: $590 \times \frac{3}{5} = 354$ m s$^{-1}$. Oxygen: $590 \times \frac{9}{16} \approx 331.9$ m s$^{-1}$. Argon: $640 \times \frac{17}{32} = 340$ m s$^{-1}$.
Only argon lies within 336.7–346.5 m s$^{-1}$.
Answer: (D). | ||||
| 7 | Current ElectricityResistivity | Class 12: Current Electricity | Multiple correctEasy | |
A easy multiple correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity The same heat $H$ is needed each time, and on the same supply voltage $V$, $H = \frac{V^2 t}{R}$, so $t \propto R$.
Old wire: $R = \frac{\rho L}{\pi d^2/4}$. Each new wire has four times the cross-section, so its resistance is $\frac{R}{4}$.
Series: $\frac{R}{4} + \frac{R}{4} = \frac{R}{2}$, so $t = \frac{4}{2} = 2$ min. (B) true, (C) false.
Parallel: $\frac{R/4}{2} = \frac{R}{8}$, so $t = \frac{4}{8} = 0.5$ min. (D) true, (A) false.
Answer: (B) and (D). | ||||
| 8 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | Multiple correctMedium | |
A medium multiple correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Let the ladder have length $l$. On the point of slipping, the bottom tends to move away from the wall, so friction at the floor is $\mu_2 N_2$ towards the wall and friction at the wall is $\mu_1 N_1$ upward.
Horizontal: $N_1 = \mu_2 N_2$. Vertical: $N_2 + \mu_1 N_1 = mg$.
Torques about the foot: $mg\frac{l}{2}\cos\theta = N_1 l\sin\theta + \mu_1 N_1 l\cos\theta$.
(C) From the force equations: $N_2 + \mu_1\mu_2 N_2 = mg$, so $N_2 = \frac{mg}{1 + \mu_1\mu_2}$. True.
(D) With $\mu_1 = 0$ the torque equation gives $\frac{mg}{2}\cos\theta = N_1\sin\theta$, i.e. $N_1\tan\theta = \frac{mg}{2}$. True.
(A) With $\mu_1 = 0$, $N_2 = mg$, so $N_2\tan\theta = \frac{mg}{2}$ would need $\tan\theta = \frac{1}{2}$; it is not true in general. False.
(B) With $\mu_2 = 0$, $N_1 = \mu_2 N_2 = 0$; then nothing balances the weight's torque about the foot and the ladder cannot be in equilibrium. False.
Answer: (C) and (D). | ||||
| 9 | 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 A thin film of uniform thickness has both its surfaces concentric and practically coincident; refracting into and out of it has no net effect on the focusing, so the system acts as a single convex air–glass surface of radius $R$ (the film's index drops out: $\frac{n_1 - 1}{R} + \frac{n_2 - n_1}{R} = \frac{n_2 - 1}{R}$).
Air to glass, parallel rays ($u = \infty$): $\frac{n_2}{v} = \frac{n_2 - 1}{R}$, so $f_1 = \frac{n_2 R}{n_2 - 1} = \frac{1.5R}{0.5} = 3R$. (A) true, (B) false.
Glass to air, parallel rays inside the glass: $\frac{1}{v} = \frac{n_2 - 1}{R}$ (the surface is convex towards the air, so it converges), giving $f_2 = \frac{R}{n_2 - 1} = \frac{R}{0.5} = 2R$. (C) true, (D) false.
Answer: (A) and (C). | ||||
| 10 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Multiple correctMedium | |
A medium multiple correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity $R_2$ joins two junctions, P (the left wire, between $V_1$ and $R_3$) and M (the right junction, between $R_1$ and $V_2$). Between these junctions there are two other branches: $V_1$ in series with $R_1$ (the upper loop) and $V_2$ in series with $R_3$ (the lower loop). Going round the outer loop the two emfs act in the same sense.
If no current flows in $R_2$, P and M are at the same potential, and a single current $I$ flows round the outer loop: $I = \frac{V_1 + V_2}{R_1 + R_3}$.
Equal potentials at P and M through the upper branch need $V_1 = IR_1$ (and then, through the lower branch, $V_2 = IR_3$). So the condition is
$\frac{V_1}{R_1} = \frac{V_2}{R_3}$, and $R_2$ can be anything.
(A) $V_1 = V_2$, $R_1 = R_3$: satisfied.
(B) $V_1 = V_2$, $R_1 = R_3$ ($= 2R_2$): satisfied.
(C) $V_1 = 2V_2$, $R_3 = 2R_1$: $\frac{V_1}{R_1} = \frac{2V_2}{R_1} = \frac{4V_2}{R_3} \neq \frac{V_2}{R_3}$. Not satisfied.
(D) $V_2 = 2V_1$, $R_3 = 2R_1$: $\frac{V_2}{R_3} = \frac{2V_1}{2R_1} = \frac{V_1}{R_1}$. Satisfied.
Answer: (A), (B) and (D). | ||||
| 11 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The velocity of B relative to A, $\vec{v}_B - \vec{v}_A$, is perpendicular to $\vec{v}_A$, so the component of $\vec{v}_B$ along A's direction equals $v_A$.
The angle between the two velocities is $60^\circ - 30^\circ = 30^\circ$: $v_B\cos 30^\circ = 100\sqrt{3}$, so $v_B = 200$ m s$^{-1}$.
The relative speed is the perpendicular component: $v_B\sin 30^\circ = 100$ m s$^{-1}$.
Seen from A, B comes straight at it along this perpendicular line and has 500 m to cover, so $t_0 = \frac{500}{100} = 5$ s.
Answer: 5. | ||||
| 12 | Experimental SkillsProperties of matter experiments | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on properties of matter experiments, from Experimental Skills. NCERT: No matching NCERT chapter $Y = \frac{FL}{A\,\Delta l}$; the load, length and area are given as exact, so the only measured quantity with an error is the extension $\Delta l$.
First reading: $3.20 \times 10^{-2} + 20 \times 1.0 \times 10^{-5} = 3.220 \times 10^{-2}$ m.
Second reading: $3.20 \times 10^{-2} + 45 \times 1.0 \times 10^{-5} = 3.245 \times 10^{-2}$ m.
Extension: $\Delta l = 25 \times 10^{-5} = 2.5 \times 10^{-4}$ m.
The extension is read with a least count of $1.0 \times 10^{-5}$ m, so its maximum error is $1.0 \times 10^{-5}$ m:
$\frac{\Delta Y}{Y} \times 100 = \frac{1.0 \times 10^{-5}}{2.5 \times 10^{-4}} \times 100 = 4\%$.
Answer: 4. | ||||
| 13 | 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 The three forces run round the triangle in one sense, so they add to zero: the centre stays at rest and only rotation results.
Each side of an equilateral triangle inscribed in a circle of radius $R$ is at distance $\frac{R}{2}$ (the inradius) from the centre, so each force has a lever arm of 0.25 m, and all three turn the disc the same way.
Torque: $\tau = 3 \times 0.5 \times 0.25 = 0.375$ N m.
Moment of inertia about the centre: $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 1.5 \times 0.25 = 0.1875$ kg m$^2$.
$\alpha = \frac{\tau}{I} = 2$ rad s$^{-2}$, so after 1 s, $\omega = 2$ rad s$^{-1}$.
Answer: 2. | ||||
| 14 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Both wires' fields at the charge are perpendicular to the plane. The charge's velocity lies in the plane, so $qu B = \frac{mu^2}{R}$ and $R = \frac{mu}{qB}$: $\frac{R_1}{R_2} = \frac{B_2}{B_1}$.
The charge is at $X_1$ from one wire and $X_0 - X_1 = 2X_1$ from the other.
Same directions: between the wires the two fields oppose: $B_1 = \frac{\mu_0 I}{2\pi}\left(\frac{1}{X_1} - \frac{1}{2X_1}\right) = \frac{\mu_0 I}{2\pi}\cdot\frac{1}{2X_1}$.
Opposite directions: the fields add: $B_2 = \frac{\mu_0 I}{2\pi}\left(\frac{1}{X_1} + \frac{1}{2X_1}\right) = \frac{\mu_0 I}{2\pi}\cdot\frac{3}{2X_1}$.
$\frac{R_1}{R_2} = \frac{B_2}{B_1} = 3$.
Answer: 3. | ||||
| 15 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | NumericalEasy | |
A easy numerical question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Let $d \propto \rho^a S^b f^c$. Dimensions: $[\rho] = $ M L$^{-3}$, $[S] = \frac{\text{power}}{\text{area}} = \frac{\text{M L}^2\text{T}^{-3}}{\text{L}^2} = $ M T$^{-3}$, $[f] = $ T$^{-1}$.
L: $-3a = 1$, so $a = -\frac{1}{3}$.
M: $a + b = 0$, so $b = \frac{1}{3}$.
T: $-3b - c = 0$, so $c = -1$.
So $d \propto S^{1/3}$ and $n = 3$.
Answer: 3. | ||||
| 16 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Voltmeter (series resistance): $0.006(G + 4990) = 30$, so $G + 4990 = 5000$ and $G = 10$ $\Omega$.
Ammeter (shunt $S$): the shunt carries $1.5 - 0.006 = 1.494$ A at the galvanometer's full-scale voltage $0.006 \times 10 = 0.06$ V:
$S = \frac{0.06}{1.494} = \frac{60}{1494} = \frac{10}{249}$ $\Omega$.
$\frac{2n}{249} = \frac{10}{249}$, so $n = 5$.
Answer: 5. | ||||
| 17 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The pulling force is constant in magnitude and direction, so its work is the force times the component of the displacement along it. The displacement from P to Q is the chord PQ itself, of length $\sqrt{3^2 + 4^2} = 5$ m, along the force:
$W_F = 18 \times 5 = 90$ J.
Gravity: the block rises 4 m, $W_g = -mgh = -1 \times 10 \times 4 = -40$ J.
The rail's normal force does no work and there is no friction. Starting from rest at P:
$K_Q = 90 - 40 = 50$ J $= 5 \times 10$ J, so $n = 5$.
Answer: 5. | ||||
| 18 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Work in the rocket's frame. Once thrown, both balls are free, so in this frame each has the same acceleration, 2 m s$^{-2}$ towards $-x$ (the pseudo-acceleration).
Since both have the same acceleration, their relative motion is uniform: they approach each other at $0.3 + 0.2 = 0.5$ m s$^{-1}$ over 4 m, so they meet at $t = \frac{4}{0.5} = 8$ s.
Positions in the rocket frame: $x_1 = 0.3t - t^2$ and $x_2 = 4 - 0.2t - t^2$; $x_1 = x_2$ gives $0.5t = 4$, $t = 8$ s.
The chamber's walls, however, are not out of the way: ball 1's position $0.3t - t^2$ returns to 0, the left wall, at $t = 0.3$ s. If it then stays at the left end, ball 2 reaches it when $4 - 0.2t - t^2 = 0$: $t = \frac{-0.2 + \sqrt{0.04 + 16}}{2} \approx 1.9$ s, about 2 s.
The official key accepts both readings: 8 (balls treated as free throughout) or 2 (ball 1 held at the left end).
Answer: 8 (or 2). | ||||
| 19 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion No external torque acts about the axis, and everything starts at rest, so the total angular momentum about the axis stays zero.
The two balls move in opposite directions on opposite sides of the centre, so their angular momenta are in the same sense: $L_{balls} = 2 \times 0.05 \times 9 \times 0.25 = 0.225$ kg m$^2$ s$^{-1}$.
The platform (a uniform disc) must carry an equal and opposite angular momentum: $I = \frac{1}{2} \times 0.45 \times 0.5^2 = 0.05625$ kg m$^2$.
$\omega = \frac{0.225}{0.05625} = 4$ rad s$^{-1}$.
Answer: 4. | ||||
| 20 | ThermodynamicsFirst law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Path $ia$ is at constant volume, so $W_{iaf} = 0 + 200 = 200$ J.
First law on $iaf$: $U_f - U_i = Q_{iaf} - W_{iaf} = 500 - 200 = 300$ J, so $U_f = 400$ J.
$Q_{ib} = (U_b - U_i) + W_{ib} = 100 + 50 = 150$ J.
$Q_{bf} = (U_f - U_b) + W_{bf} = 200 + 100 = 300$ J.
$\frac{Q_{bf}}{Q_{ib}} = \frac{300}{150} = 2$.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 21 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Multiple correctEasy | |
A easy multiple correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Each option pairs a common name with an IUPAC name; the pair is correct when both name the same alcohol and the IUPAC name is right.
(A) tert-Butanol is (CH$_3$)$_3$C–OH. Its longest chain holding the OH carbon is three carbons with a methyl at C-2 and OH at C-2: 2-methylpropan-2-ol. Correct.
(B) "1,1-dimethylethan-1-ol" describes the same molecule but is not a valid IUPAC name: the longest chain is propane, not ethane. Incorrect.
(C) n-Butanol is CH$_3$CH$_2$CH$_2$CH$_2$OH = butan-1-ol. Correct.
(D) Isobutyl alcohol is (CH$_3$)$_2$CHCH$_2$OH = 2-methylpropan-1-ol. Correct.
Answer: (A), (C) and (D). | ||||
| 22 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Multiple correctMedium | |
A medium multiple correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Z is 3-tert-butylphenol. The strongly activating OH group directs halogens to its ortho and para positions: C-2, C-4 and C-6. This is the electronic effect of the phenolic group (C).
C-2 lies between the OH and the bulky tert-butyl group, and C-4 is next to the tert-butyl group too; C-6 is free. How many of these positions a halogen can occupy depends on the halogen's size: the small Cl atom gets into all three (tri-substitution), the larger Br into two, and the large I into only one. So the steric effect of the halogen (A) and the steric hindrance of the tert-butyl group (B) explain the pattern.
(D) The tert-butyl group's own electronic (weak +I) effect is small and is not what decides the number of substitutions. Not an explanation.
Answer: (A), (B) and (C). | ||||
| 23 | AminesBasicity of amines | Class 12: Amines | Multiple correctMedium | |
A medium multiple correct question on basicity of amines, from Amines. NCERT: Class 12: Amines The substrate has two NH$_2$ groups: a primary benzylic amine (–CH$_2$NH$_2$) and a primary amide (–CONH$_2$).
The amine nitrogen's lone pair is free, so it is a good nucleophile. The amide nitrogen's lone pair is delocalised onto the carbonyl oxygen, so it is a very poor nucleophile.
Acetic anhydride therefore acetylates only the amine: –CH$_2$NH$_2$ + (CH$_3$CO)$_2$O $\rightarrow$ –CH$_2$NHCOCH$_3$ + CH$_3$COOH, leaving the amide untouched.
(B) and (D) acetylate the amide, which does not happen in preference to the amine; (C) acetylates both and wrongly gives water as the by-product (acetylation by an anhydride releases acetic acid).
Answer: (A). | ||||
| 24 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Multiple correctEasy | |
A easy multiple correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics This is a free expansion of an ideal gas in an insulated vessel.
(A) The vessel is thermally insulated, so $q = 0$. True.
Work against zero external pressure is $w = -P_{ext}\Delta V = 0$, so $\Delta U = q + w = 0$.
(B) For an ideal gas $U$ depends only on $T$, so $T_2 = T_1$. True.
(C) At the same temperature, $PV = nRT$ gives $P_2V_2 = P_1V_1$. True.
(D) $PV^{\gamma}$ = constant holds only for a reversible adiabatic process; this expansion is irreversible (and its temperature does not change), so it does not apply. False.
Answer: (A), (B) and (C). | ||||
| 25 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | Multiple correctEasy | |
A easy multiple correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) In ice each water molecule is hydrogen-bonded tetrahedrally to four others, giving an open cage structure less dense than liquid water, so ice floats. True.
(B) In water, a primary ammonium ion RNH$_3^+$ has three N–H hydrogens to hydrogen-bond with water, while R$_3$NH$^+$ has only one; the better solvation stabilises RNH$_3^+$ and makes the primary amine the stronger base in water. True.
(C) Formic acid is stronger than acetic acid because acetic acid's methyl group (+I effect) destabilises the carboxylate; hydrogen bonding is not the reason. False.
(D) In a non-polar solvent such as benzene, acetic acid exists as a cyclic dimer held by two O–H$\cdots$O hydrogen bonds. True.
Answer: (A), (B) and (D). | ||||
| 26 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Multiple correctEasy | |
A easy multiple correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry (A) The salt bridge's electrolyte (KCl, KNO$_3$) is inert: its ions only move to keep the two half-cells neutral and take no part in the cell reaction. True.
(B) It does not stop ions moving; it is the path by which its ions migrate between the half-cells. False.
(C) Without the salt bridge the circuit is open and charge builds up in the half-cells, so the reaction stops at once; the bridge (or some equivalent ionic contact) is needed for the cell reaction to proceed. True.
(D) It is meant to keep the two solutions from mixing while allowing ionic contact. False.
Answer: (A) and (C); the official key also accepts (A) alone. | ||||
| 27 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Cu$_2$S is reduced to copper by a copper compound that holds oxygen (auto-reduction), the sulfur leaving as SO$_2$.
(A) CuFeS$_2$ contains no oxygen; heating it with Cu$_2$S gives no copper. False.
(B) $\text{Cu}_2\text{S} + 2\text{CuO} \rightarrow 4\text{Cu} + \text{SO}_2$. True.
(C) $\text{Cu}_2\text{S} + 2\text{Cu}_2\text{O} \rightarrow 6\text{Cu} + \text{SO}_2$ (the self-reduction step of copper extraction). True.
(D) $\text{Cu}_2\text{S} + \text{CuSO}_4 \rightarrow 3\text{Cu} + 2\text{SO}_2$. True.
Answer: (B), (C) and (D). | ||||
| 28 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) B(OH)$_3$ is not a protic acid; it is a weak monobasic Lewis acid that accepts OH$^-$ from water: $\text{B(OH)}_3 + 2\text{H}_2\text{O} \rightleftharpoons [\text{B(OH)}_4]^- + \text{H}_3\text{O}^+$. Its acidity is not due to self ionization. False.
(B) Ethylene glycol (a cis-diol) forms a stable chelate with [B(OH)$_4$]$^-$, pulling the equilibrium to the right, so the solution becomes more acidic (it can then be titrated with NaOH). True.
(C) Hydrogen bonding joins the planar B(OH)$_3$ units into flat layers: a two-dimensional sheet structure, not a three-dimensional one. False.
(D) It ionises only slightly in water, so it is a weak electrolyte. True.
Answer: (B) and (D). | ||||
| 29 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | Multiple correctEasy | |
A easy multiple correct question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Chlorine goes from +5 (ClO$_3^-$) to $-1$ (Cl$^-$): gains 6 electrons. Iodine goes from $-1$ to 0: each I$^-$ loses 1 electron, so 6 I$^-$ are needed.
$6\text{I}^- + \text{ClO}_3^- + 6\text{H}^+ \rightarrow \text{Cl}^- + 3\text{I}_2 + 3\text{H}_2\text{O}$.
The 6 H$^+$ come from 6 H$_2$SO$_4$, each leaving HSO$_4^-$:
$6\text{I}^- + \text{ClO}_3^- + 6\text{H}_2\text{SO}_4 \rightarrow \text{Cl}^- + 6\text{HSO}_4^- + 3\text{I}_2 + 3\text{H}_2\text{O}$.
(A) The coefficient of HSO$_4^-$ is 6. True.
(B) Iodide is oxidised to iodine. True.
(C) Sulphur stays +6 in H$_2$SO$_4$ and HSO$_4^-$; it is not reduced. False.
(D) Water is formed. True.
Answer: (A), (B) and (D); the official key also accepts (A) and (D). | ||||
| 30 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Multiple correctMedium | |
A medium multiple correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) Sodium dissolves in liquid ammonia to give ammoniated electrons: $\text{Na} + (x+y)\text{NH}_3 \rightarrow [\text{Na(NH}_3)_x]^+ + [e(\text{NH}_3)_y]^-$. The unpaired solvated electron makes the blue solution paramagnetic. True.
(B) Potassium burns in excess oxygen to the superoxide KO$_2$. The O$_2^-$ ion has 17 electrons, one unpaired in a $\pi^*$ orbital: paramagnetic. True.
(C) $3\text{Cu} + 8\text{HNO}_3\text{(dil)} \rightarrow 3\text{Cu(NO}_3)_2 + 2\text{NO} + 4\text{H}_2\text{O}$. NO has an odd number of electrons (15), so it is paramagnetic (and so is Cu$^{2+}$, d$^9$). True.
(D) Air oxidation of 2-ethylanthraquinol gives H$_2$O$_2$ and 2-ethylanthraquinone, both with all electrons paired: diamagnetic. False.
Answer: (A), (B) and (C). | ||||
| 31 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalHard | |
A hard numerical question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A saturated ketone C$_n$H$_{2n}$O has $M = 14n + 16 = 100$, so $n = 6$: C$_6$H$_{12}$O.
The ketones of C$_6$H$_{12}$O:
1. hexan-2-one
2. hexan-3-one
3. 3-methylpentan-2-one (chiral: C-3 carries H, CH$_3$, C$_2$H$_5$ and COCH$_3$), as two enantiomers
4. 4-methylpentan-2-one
5. 3,3-dimethylbutan-2-one
6. 2-methylpentan-3-one
That is 7 ketones counting the two enantiomers of 3.
NaBH$_4$ attacks the flat C=O from either face equally. An achiral ketone whose carbonyl carbon becomes a stereocentre therefore gives a 50:50 pair of enantiomers, a racemate. A chiral ketone (an enantiomer of 3) already has a stereocentre, so the new one gives two diastereomers, not a racemate.
Achiral ketones and the alcohols they give: hexan-2-ol, hexan-3-ol, 4-methylpentan-2-ol, 3,3-dimethylbutan-2-ol and 2-methylpentan-3-ol (C-3 bears H, OH, C$_2$H$_5$ and CH(CH$_3$)$_2$): each is chiral, so each of these 5 ketones gives a racemic product.
Answer: 5. | ||||
| 32 | 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 XeF$_4$: 4 bond pairs + 2 lone pairs (octahedral arrangement), square planar.
SF$_4$: 4 bond pairs + 1 lone pair, see-saw.
SiF$_4$: 4 bond pairs, tetrahedral.
BF$_4^-$: sp$^3$, tetrahedral.
BrF$_4^-$: 4 bond pairs + 2 lone pairs, square planar.
[Cu(NH$_3$)$_4$]$^{2+}$: d$^9$ Cu(II), square planar (dsp$^2$).
[FeCl$_4$]$^{2-}$: Fe(II) with weak-field Cl$^-$, tetrahedral.
[CoCl$_4$]$^{2-}$: Co(II), tetrahedral.
[PtCl$_4$]$^{2-}$: d$^8$ Pt(II), square planar.
Square planar: XeF$_4$, BrF$_4^-$, [Cu(NH$_3$)$_4$]$^{2+}$ and [PtCl$_4$]$^{2-}$.
Answer: 4. | ||||
| 33 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Colours of the sulfides (as precipitated in group analysis):
PbS black, CuS black, HgS black, Ag$_2$S black, NiS black, CoS black.
MnS buff (pink), SnS$_2$ yellow.
Bi$_2$S$_3$ is dark brown (brownish black), and is counted as black by some descriptions.
Black: PbS, CuS, HgS, Ag$_2$S, NiS, CoS = 6 (7 if Bi$_2$S$_3$ is counted as black).
Answer: 6 (the official key accepts 6 or 7). | ||||
| 34 | HydrocarbonsConformations of ethane | Class 11: Hydrocarbons | NumericalHard | |
A hard numerical question on conformations of ethane, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The Fischer projection shows CH$_3$C(Br)(Cl)–C(Br)(Cl)CH$_3$. Upper carbon: Cl at the top, Br on the left, CH$_3$ on the right. Lower carbon: Br on the left, Cl on the right, CH$_3$ at the bottom.
Configuration of the upper carbon (priorities Br > Cl > C(Br)(Cl)CH$_3$ > CH$_3$): Br (left) → Cl (top) → C (bottom) runs clockwise, but the lowest-priority CH$_3$ is on a horizontal bond, so the carbon is S.
Lower carbon (Br > Cl > C(Br)(Cl)CH$_3$ > CH$_3$): Br (left) → Cl (right) → C (top) runs anticlockwise with CH$_3$ on a vertical bond, so it is S.
The compound is (2S,3S): chiral, not meso. So no conformation has a centre of symmetry, and none can have zero dipole moment.
The stable conformers are the three staggered ones about the C2–C3 bond (eclipsed forms are energy maxima). All three have non-zero dipole moment.
Answer: 3. | ||||
| 35 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Acidified K$_2$Cr$_2$O$_7$: $\text{Cr}_2\text{O}_7^{2-} + 6\text{I}^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}$. Yes.
Alkaline KMnO$_4$: oxidises iodide further, to iodate: $2\text{MnO}_4^- + \text{I}^- + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + \text{IO}_3^- + 2\text{OH}^-$. Not iodine.
CuSO$_4$: $2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI} + \text{I}_2$. Yes.
H$_2$O$_2$: $\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{I}_2 + 2\text{H}_2\text{O}$. Yes.
Cl$_2$: $\text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2$. Yes.
O$_3$: $\text{O}_3 + 2\text{I}^- + \text{H}_2\text{O} \rightarrow \text{I}_2 + \text{O}_2 + 2\text{OH}^-$. Yes.
FeCl$_3$: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$. Yes.
HNO$_3$: oxidises I$^-$ to I$_2$. Yes.
Na$_2$S$_2$O$_3$ is a reducing agent (it reduces I$_2$ to I$^-$). No.
Count: 7.
Answer: 7. | ||||
| 36 | BiomoleculesProteins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Break every amide bond (C(=O)–N) and look at the residues.
The two end rings are 2,5-diketopiperazines whose ring carbons are CH$_2$ groups: each ring is two glycine units, so they give glycine (H$_2$NCH$_2$COOH).
The chain residues carry side chains –CH$_2$–(cyclopropyl) (two of them) and –C(CH$_3$)$_3$ (two of them). Their amino acids, 2-amino-3-cyclopropylpropanoic acid and 2-amino-3,3-dimethylbutanoic acid (tert-leucine), are not among the naturally occurring (protein) amino acids.
So the only distinct naturally occurring amino acid obtained is glycine.
Answer: 1. | ||||
| 37 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom For $n = 4$, $l$ can be 0, 1, 2, 3. $|m_l| = 1$ needs $l \geq 1$, and each of $l = 1, 2, 3$ has two orbitals with $m_l = +1$ and $-1$: $3 \times 2 = 6$ orbitals.
Each orbital holds one electron with $m_s = -\frac{1}{2}$, so there are 6 such electrons.
Answer: 6. | ||||
| 38 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $R = N_A k = 6.023 \times 10^{23} \times 1.380 \times 10^{-23} = 8.312$ J K$^{-1}$ mol$^{-1}$ (8.31174 before rounding).
Both factors have 4 significant figures (the trailing zero in 1.380 counts), and a product keeps as many significant figures as the least precise factor: 4.
Answer: 4. | ||||
| 39 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 L of the solution: it holds 3.2 mol of H$_2$X.
With no change in volume on dissolving, the solvent's volume is 1 L = 1000 mL, so its mass is $1000 \times 0.4 = 400$ g = 0.4 kg.
Molality $= \frac{3.2}{0.4} = 8$ mol kg$^{-1}$.
Answer: 8. | ||||
| 40 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\text{MX}_2 \rightleftharpoons \text{M}^{2+} + 2\text{X}^-$: 1 mol gives $(1 - \alpha) + \alpha + 2\alpha = 1 + 2\alpha$ mol of particles.
$i = 1 + 2(0.5) = 2$.
The depression is proportional to the number of particles, so the ratio is $i = 2$.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 41 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Since $M$ and $N$ commute, $M$ commutes with $N^2$, so
$(M + N^2)(M - N^2) = M^2 - MN^2 + N^2M - N^4 = M^2 - N^4 = O$.
$M - N^2 \neq O$. If $M + N^2$ were invertible, multiplying by its inverse would give $M - N^2 = O$, a contradiction. So $\det(M + N^2) = 0$.
(A) $M^2 + MN^2 = M(M + N^2)$, so $\det(M^2 + MN^2) = \det M \cdot \det(M + N^2) = 0$. True.
(B) Take $U = M - N^2 \neq O$: $(M^2 + MN^2)U = M(M + N^2)(M - N^2) = O$. True.
(C) The determinant is 0, not $\geq 1$. False.
(D) (B) gives a non-zero $U$ with $(M^2 + MN^2)U = O$. False.
Answer: (A) and (B). | ||||
| 42 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctMedium | |
A medium multiple correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let the common maximum $m$ be reached by $f$ at $x_1$ and by $g$ at $x_2$. Put $h = f - g$, continuous on [0, 1].
$h(x_1) = m - g(x_1) \geq 0$ and $h(x_2) = f(x_2) - m \leq 0$. By the intermediate value theorem there is $c$ between them with $h(c) = 0$, i.e. $f(c) = g(c)$.
(A) At that $c$ both sides are equal. True.
(D) At that $c$, $(f(c))^2 = (g(c))^2$. True.
(B), (C) Not true for every pair. Take $f = g = 1$: (B) would need $1 + 1 = 1 + 3$, (C) would need $1 + 3 = 1 + 1$, for some $c$; neither ever holds. False.
Answer: (A) and (D). | ||||
| 43 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $\phi(t) = \frac{1}{t}e^{-(t + 1/t)}$. By the Leibniz rule,
$f'(x) = \phi(x) - \phi\left(\frac{1}{x}\right)\cdot\left(-\frac{1}{x^2}\right) = \frac{e^{-(x + 1/x)}}{x} + x e^{-(x + 1/x)}\cdot\frac{1}{x^2} = \frac{2e^{-(x + 1/x)}}{x} > 0$ for all $x > 0$.
(A) $f$ is increasing on the whole of $(0, \infty)$, so on $[1, \infty)$. True.
(B) It is increasing on (0, 1) as well. False.
(C) $f\left(\frac{1}{x}\right) = \int_x^{1/x}\phi(t)\,dt = -\int_{1/x}^{x}\phi(t)\,dt = -f(x)$. True.
(D) $f(2^{-x}) = f\left(\frac{1}{2^x}\right) = -f(2^x)$, so $x \mapsto f(2^x)$ is odd. True.
Answer: (A), (C) and (D). | ||||
| 44 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Multiple correctMedium | |
A medium multiple correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 5x^4 - 5 = 0$ at $x = \pm 1$. $f$ increases on $(-\infty, -1)$, decreases on $(-1, 1)$ and increases on $(1, \infty)$.
Local maximum $f(-1) = -1 + 5 + a = a + 4$; local minimum $f(1) = 1 - 5 + a = a - 4$.
$f$ has three real roots exactly when the maximum is positive and the minimum negative: $a + 4 > 0$ and $a - 4 < 0$, i.e. $-4 < a < 4$. (D) true.
If $a > 4$, the local minimum $a - 4 > 0$, so $f$ crosses zero only once, to the left of $-1$. (B) true, (A) false.
If $a < -4$, the local maximum $a + 4 < 0$, so again only one real root (to the right of 1). (C) false.
Answer: (B) and (D). | ||||
| 45 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctMedium | |
A medium multiple correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Continuity: as $x \to a$, $\int_a^x f \to 0 = g(a)$, and as $x \to b$, $\int_a^x f \to \int_a^b f$, matching the constant beyond $b$. So $g$ is continuous everywhere.
At $a$: left derivative 0 (constant), right derivative $f(a) \geq 1$. They differ, so $g$ is not differentiable at $a$. (A) true.
At $b$: left derivative $f(b) \geq 1$, right derivative 0. Not differentiable at $b$. (C) true.
(B) fails at $a$ and $b$; (D) fails because $g$ is differentiable at neither point.
Answer: (A) and (C). | ||||
| 46 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Multiple correctMedium | |
A medium multiple correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Let $u(x) = \log(\sec x + \tan x)$; on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\sec x + \tan x > 0$.
(A) $\sec(-x) + \tan(-x) = \sec x - \tan x = \frac{1}{\sec x + \tan x}$, so $u(-x) = -u(x)$ and $f(-x) = -f(x)$. Odd. True.
(B) $u'(x) = \frac{\sec x\tan x + \sec^2 x}{\sec x + \tan x} = \sec x > 0$, so $u$ is strictly increasing, and so is $f = u^3$. One-one. True.
(C) As $x \to \frac{\pi}{2}^-$, $\sec x + \tan x \to \infty$ and $u \to \infty$; as $x \to -\frac{\pi}{2}^+$, $\sec x + \tan x = \frac{1 + \sin x}{\cos x} \to 0^+$ and $u \to -\infty$. Being continuous, $u$ takes every real value, and so does $f$. Onto $\mathbb{R}$. True.
(D) An odd function that is not identically zero is not even. False.
Answer: (A), (B) and (C). | ||||
| 47 | 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 First line: points $(t, t, 1)$, direction $(1, 1, 0)$. For the foot $Q$: $(\lambda - t, \lambda - t, \lambda - 1)\cdot(1, 1, 0) = 0$ gives $t = \lambda$, so $Q = (\lambda, \lambda, 1)$.
Second line: points $(s, -s, -1)$, direction $(1, -1, 0)$. For the foot $R$: $(\lambda - s) - (\lambda + s) = 0$ gives $s = 0$, so $R = (0, 0, -1)$.
$\vec{PQ} = (0, 0, 1 - \lambda)$ and $\vec{PR} = (-\lambda, -\lambda, -1 - \lambda)$.
$\angle QPR = 90^\circ$: $\vec{PQ}\cdot\vec{PR} = (1 - \lambda)(-1 - \lambda) = \lambda^2 - 1 = 0$, so $\lambda = \pm 1$.
For $\lambda = 1$, $P = (1, 1, 1)$ lies on the first line, so $Q = P$ and there is no angle QPR. Hence $\lambda = -1$.
Answer: (C). | ||||
| 48 | 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 Each pair has dot product $\sqrt{2}\cdot\sqrt{2}\cos\frac{\pi}{3} = 1$, and each vector has $\vec{v}\cdot\vec{v} = 2$.
$\vec{a} \perp \vec{y} \times \vec{z}$, so $\vec{a}$ lies in the plane of $\vec{y}$ and $\vec{z}$: $\vec{a} = \alpha\vec{y} + \beta\vec{z}$. $\vec{a}\cdot\vec{x} = \alpha + \beta = 0$, so $\vec{a} = \alpha(\vec{y} - \vec{z})$, and $\vec{a}\cdot\vec{y} = \alpha(2 - 1) = \alpha$.
(B) $\vec{a} = (\vec{a}\cdot\vec{y})(\vec{y} - \vec{z})$. True. (D) is its negative, which would make $\vec{a} = \vec{0}$. False.
Likewise $\vec{b} = \gamma\vec{z} + \delta\vec{x}$ with $\vec{b}\cdot\vec{y} = \gamma + \delta = 0$: $\vec{b} = \gamma(\vec{z} - \vec{x})$ and $\vec{b}\cdot\vec{z} = \gamma(2 - 1) = \gamma$.
(A) $\vec{b} = (\vec{b}\cdot\vec{z})(\vec{z} - \vec{x})$. True.
(C) $\vec{a}\cdot\vec{b} = (\vec{a}\cdot\vec{y})(\vec{b}\cdot\vec{z})(\vec{y} - \vec{z})\cdot(\vec{z} - \vec{x}) = (\vec{a}\cdot\vec{y})(\vec{b}\cdot\vec{z})(1 - 1 - 2 + 1) = -(\vec{a}\cdot\vec{y})(\vec{b}\cdot\vec{z})$. True.
Answer: (A), (B) and (C). | ||||
| 49 | CirclesStandard and general form | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Let $S: x^2 + y^2 + 2gx + 2fy + c = 0$. Two circles cut orthogonally when $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
With $x^2 + y^2 - 1 = 0$ ($g = f = 0$, $c = -1$): $0 = c - 1$, so $c = 1$.
With $x^2 + y^2 - 2x - 15 = 0$ ($g = -1$, $f = 0$, $c = -15$): $-2g = c - 15 = -14$, so $g = 7$.
Through $(0, 1)$: $1 + 2f + c = 0$, so $f = -1$.
Centre $(-g, -f) = (-7, 1)$; radius $\sqrt{g^2 + f^2 - c} = \sqrt{49 + 1 - 1} = 7$.
Answer: (B) and (C). | ||||
| 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 Write $M = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ with integers $a, b, c$; $\det M = ac - b^2$.
(A) The first column $\begin{pmatrix} a \\ b \end{pmatrix}$ equals the transpose of the second row $(b \;\; c)$: $a = b$ and $b = c$. Then $\det M = a^2 - a^2 = 0$: not invertible. False.
(B) The second row $(b \;\; c)$ is the transpose of the first column: $b = a$, $c = b$. The same singular matrix. False.
(C) $b = 0$ and $a, c \neq 0$: $\det M = ac \neq 0$. True.
(D) If $ac$ is not the square of an integer, it cannot equal $b^2$, so $\det M \neq 0$. True.
Answer: (C) and (D). | ||||
| 51 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the common ratio be $r = \frac{b}{a}$, a positive integer: $b = ar$, $c = ar^2$.
$\frac{a + ar + ar^2}{3} = ar + 2$ gives $a(r^2 - 2r + 1) = 6$, i.e. $a(r - 1)^2 = 6$.
The only perfect square dividing 6 is 1, so $(r - 1)^2 = 1$; $r = 2$ (since $r = 0$ would make $b = 0$), and $a = 6$.
Check: 6, 12, 24 has mean 14 = 12 + 2.
$\frac{a^2 + a - 14}{a + 1} = \frac{36 + 6 - 14}{7} = \frac{28}{7} = 4$.
Answer: 4. | ||||
| 52 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations There are $\binom{n}{2}$ segments in all. For $n \geq 3$ there are $n$ adjacent pairs (blue), so $\binom{n}{2} - n$ are red. ($n = 2$ gives one blue and no red segment, so it does not qualify.)
$\binom{n}{2} - n = n$: $\frac{n(n-1)}{2} = 2n$, so $n - 1 = 4$ and $n = 5$.
(Check: a pentagon has 5 sides and 5 diagonals.)
Answer: 5. | ||||
| 53 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalHard | |
A hard numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The smallest possible choice is 1, 2, 3, 4, 5, with sum 15; 5 more must be added.
Write $n_i = i + m_i$. Since $n_{i+1} > n_i$ means $n_{i+1} \geq n_i + 1$, we need $0 \leq m_1 \leq m_2 \leq m_3 \leq m_4 \leq m_5$ with $m_1 + \dots + m_5 = 5$.
Each such list is a partition of 5 into at most 5 parts (the non-zero $m_i$ read from the top), and each partition gives exactly one list. The partitions of 5 are
5; 4+1; 3+2; 3+1+1; 2+2+1; 2+1+1+1; 1+1+1+1+1 — seven.
They give (1,2,3,4,10), (1,2,3,5,9), (1,2,3,6,8), (1,2,4,5,8), (1,2,4,6,7), (1,3,4,5,7), (2,3,4,5,6).
Answer: 7. | ||||
| 54 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $x \leq 0$: $f = 1 - x$, and $x^2 \geq -x$ exactly when $x \leq -1$. So $h = x^2 + 1$ on $(-\infty, -1]$ and $h = 1 - x$ on $[-1, 0]$.
For $x > 0$: $f = 1 + x$, and $x^2 \leq x$ exactly when $0 < x \leq 1$. So $h = x^2 + 1$ on $(0, 1]$ and $h = x + 1$ on $[1, \infty)$.
$h$ is continuous (the pieces meet at $x = -1$: 2, $x = 0$: 1, $x = 1$: 2). Check the slopes:
$x = -1$: left $2x = -2$, right $-1$. Not differentiable.
$x = 0$: left $-1$, right $2x = 0$. Not differentiable.
$x = 1$: left $2x = 2$, right 1. Not differentiable.
Everywhere else each piece is smooth. So 3 points.
Answer: 3. | ||||
| 55 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $u = (1 - x^2)^5$, so $u' = -10x(1 - x^2)^4$. Both $u$ and $u'$ vanish at $x = 1$, and $u' = 0$ at $x = 0$.
By parts: $\int_0^1 4x^3 u''\,dx = \left[4x^3u'\right]_0^1 - \int_0^1 12x^2u'\,dx = -\int_0^1 12x^2u'\,dx$.
Again by parts: $-\int_0^1 12x^2u'\,dx = -\left[12x^2u\right]_0^1 + \int_0^1 24xu\,dx = \int_0^1 24x(1 - x^2)^5\,dx$.
With $w = 1 - x^2$, $dw = -2x\,dx$: $\int_0^1 x(1 - x^2)^5\,dx = \frac{1}{2}\int_0^1 w^5\,dw = \frac{1}{12}$.
Value $= 24 \times \frac{1}{12} = 2$.
Answer: 2. | ||||
| 56 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Differentiate implicitly:
$2(y - x^5)(y' - 5x^4) = (1 + x^2)^2 + x\cdot 2(1 + x^2)\cdot 2x$.
At $(1, 3)$: $y - x^5 = 2$, so the left side is $4(y' - 5)$; the right side is $4 + 2 \cdot 2 \cdot 2 = 12$.
$y' - 5 = 3$, so $y' = 8$.
Answer: 8. | ||||
| 57 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Exponent: $\frac{1 - x}{1 - \sqrt{x}} = 1 + \sqrt{x} \to 2$.
Base: put $h = x - 1 \to 0$. The numerator is $-ah + \sin h$ and the denominator $h + \sin h$, so the base is $\frac{-a + \frac{\sin h}{h}}{1 + \frac{\sin h}{h}} \to \frac{1 - a}{2}$.
The limit is $\left(\frac{1 - a}{2}\right)^2 = \frac{1}{4}$, so $1 - a = \pm 1$: $a = 0$ or $a = 2$.
For $a = 2$ the base tends to $-\frac{1}{2}$: near $x = 1$ it is negative, and a negative number raised to the non-integer power $1 + \sqrt{x}$ is not a real number, so the limit does not exist. For $a = 0$ the base tends to $\frac{1}{2}$ and the limit is $\frac{1}{4}$.
The largest admissible value is $a = 0$.
Answer: 0. | ||||
| 58 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $f$ is a zig-zag: $f(x) = x$ on $[0, \pi]$, $2\pi - x$ on $[\pi, 2\pi]$, $x - 2\pi$ on $[2\pi, 3\pi]$, $4\pi - x$ on $[3\pi, 4\pi]$.
The line $y = 1 - \frac{x}{10}$ falls from 1 at $x = 0$ and is negative for $x > 10$ (where it cannot meet $f \geq 0$). Solve piece by piece:
$[0, \pi]$: $x = 1 - \frac{x}{10}$, $x = \frac{10}{11} \approx 0.91$. In range.
$[\pi, 2\pi]$: $2\pi - x = 1 - \frac{x}{10}$, $x = \frac{10(2\pi - 1)}{9} \approx 5.87$. In range (3.14 to 6.28).
$[2\pi, 3\pi]$: $x - 2\pi = 1 - \frac{x}{10}$, $x = \frac{10(2\pi + 1)}{11} \approx 6.62$. In range (6.28 to 9.42).
$[3\pi, 4\pi]$: $4\pi - x = 1 - \frac{x}{10}$, $x = \frac{10(4\pi - 1)}{9} \approx 12.85 > 4\pi \approx 12.57$. Not in range.
So there are 3 solutions.
Answer: 3. | ||||
| 59 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines For $P = (x, y)$ in the first quadrant, $d_1 = \frac{|x - y|}{\sqrt{2}}$ and $d_2 = \frac{x + y}{\sqrt{2}}$.
$d_1 + d_2 = \frac{|x - y| + x + y}{\sqrt{2}} = \frac{2\max(x, y)}{\sqrt{2}} = \sqrt{2}\max(x, y)$.
The condition is $\sqrt{2} \leq \max(x, y) \leq 2\sqrt{2}$: the square $[0, 2\sqrt{2}]^2$ with the square $[0, \sqrt{2})^2$ removed.
Area $= (2\sqrt{2})^2 - (\sqrt{2})^2 = 8 - 2 = 6$.
Answer: 6. | ||||
| 60 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Each pair has dot product $\frac{1}{2}$. Let $V = [\vec{a}\;\vec{b}\;\vec{c}] \neq 0$ (non-coplanar).
Dot both sides with $\vec{a}$: $0 + \vec{a}\cdot(\vec{b} \times \vec{c}) = V = p + \frac{q}{2} + \frac{r}{2}$.
With $\vec{b}$: $0 + 0 = 0 = \frac{p}{2} + q + \frac{r}{2}$.
With $\vec{c}$: $\vec{c}\cdot(\vec{a} \times \vec{b}) + 0 = V = \frac{p}{2} + \frac{q}{2} + r$.
Subtracting the third from the first: $\frac{p - r}{2} = 0$, so $r = p$. The second then gives $q = -p$, and the first gives $p - \frac{p}{2} + \frac{p}{2} = p = V$.
So $p = r = V$, $q = -V$, and $\frac{p^2 + 2q^2 + r^2}{q^2} = \frac{V^2 + 2V^2 + V^2}{V^2} = 4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Falling: $v = gt$, so $K = \frac{1}{2}mg^2t^2$ grows as a parabola in $t$ (curving upward), not linearly.
Collision: the force is like a spring force ($F \propto$ compression), so the ball decelerates smoothly to rest and is pushed back; $K$ falls quickly to zero and rises again to the same value (the surface is smooth and the ball returns to its original height, so no energy is lost). This gives a narrow dip to zero.
Rising: $K$ decreases again as a parabola, mirror image of the fall, reaching zero at the top.
So the sketch must show a parabolic rise, a sharp dip to zero with rounded sides, and a parabolic fall: sketch (B). (A) has the wrong curvature, (C) has no dip to zero, (D) is linear in $t$.
Answer: (B). | ||||
| 2 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion Let $\theta$ be the angle of the bead's radius from the vertical. Starting from rest at the top, energy conservation gives $v^2 = 2gR(1-\cos\theta)$.
Radial equation (towards the centre): $mg\cos\theta - N = \frac{mv^2}{R}$, where $N$ is the force of the wire on the bead taken radially outward.
$N = mg\cos\theta - 2mg(1-\cos\theta) = mg(3\cos\theta - 2)$.
For $\cos\theta > \frac{2}{3}$ (early), $N > 0$: the wire pushes the bead outward, so the bead pushes the wire radially inward.
For $\cos\theta < \frac{2}{3}$ (later, down to B at $\theta = 90^\circ$), $N < 0$: the wire pulls the bead inward, so the bead pushes the wire radially outward. (The bead cannot leave the wire, since the wire passes through it.)
Answer: (D). | ||||
| 3 | Current ElectricityMetre bridge | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity Balance: $\frac{R}{90} = \frac{l}{100-l} = \frac{40}{60}$, so $R = 60\,\Omega$.
Error: $R = 90\frac{l}{100-l}$, so $\frac{\Delta R}{R} = \frac{\Delta l}{l} + \frac{\Delta l}{100-l}$ (the same $\Delta l$ enters both lengths).
With $\Delta l = 0.1$ cm: $\frac{\Delta R}{R} = \frac{0.1}{40} + \frac{0.1}{60} = 0.0025 + 0.00167 = 0.00417$.
$\Delta R = 60 \times 0.00417 = 0.25\,\Omega$.
$R = 60 \pm 0.25\,\Omega$.
Answer: (C). | ||||
| 4 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Use Gauss's law with $k = \frac{1}{4\pi\varepsilon_0}$.
Sphere 1 (radius $R/2$): P is outside, $E_1 = \frac{kQ}{R^2}$.
Sphere 2 (radius $R$): P is on the surface, $E_2 = \frac{k(2Q)}{R^2} = \frac{2kQ}{R^2}$.
Sphere 3 (radius $2R$): P is inside, $E = \frac{kq\,r}{a^3}$ with $q = 4Q$, $a = 2R$, $r = R$: $E_3 = \frac{k(4Q)R}{8R^3} = \frac{kQ}{2R^2}$.
So $E_2 > E_1 > E_3$.
Answer: (C). | ||||
| 5 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Light leaves the top face only inside the cone of the critical angle $C$, so the spot radius is $r = h\tan C$.
$r = \frac{11.54}{2} = 5.77$ mm, $h = 10$ mm, so $\tan C = 0.577 = \frac{1}{\sqrt3}$ and $C = 30^\circ$.
At the critical angle, $\sin C = \frac{\mu_\ell}{\mu_b}$: $\mu_\ell = 2.72 \times \frac{1}{2} = 1.36$.
Answer: (C). | ||||
| 6 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on 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 A sphere of radius $r$ intercepts the parallel beam over its cross-section $\pi r^2$, but radiates (and absorbs from the surroundings) over its whole surface $4\pi r^2$.
Steady state: $I\pi r^2 = \sigma\,4\pi r^2(T^4 - T_0^4)$, so $T^4 - T_0^4 = \frac{I}{4\sigma} = \frac{912}{4 \times 5.7\times10^{-8}} = 4.0\times10^{9}\ \text{K}^4$.
$T_0^4 = 300^4 = 8.1\times10^9$, so $T^4 = 12.1\times10^9$, $T^2 = 1.1\times10^5$, $T \approx 332$ K.
Answer: (A). | ||||
| 7 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon energies: $\frac{1240}{248} = 5.0$ eV and $\frac{1240}{310} = 4.0$ eV.
$\frac{1}{2}mu_1^2 = 5 - \phi$, $\frac{1}{2}mu_2^2 = 4 - \phi$, and $u_1 = 2u_2$ gives $5 - \phi = 4(4 - \phi)$.
$3\phi = 11$, $\phi \approx 3.7$ eV.
Answer: (A). | ||||
| 8 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Moseley's law for the $K_\alpha$ line: $\frac{1}{\lambda} \propto (Z-1)^2$ (one electron left in the K shell screens the nucleus).
$\frac{\lambda_{Cu}}{\lambda_{Mo}} = \frac{(42-1)^2}{(29-1)^2} = \frac{41^2}{28^2} = \frac{1681}{784} \approx 2.14$.
Answer: (B). | ||||
| 9 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation For the same density, $g \propto$ radius, so on the planet's surface $g_s = \frac{10}{10} = 1$ m s$^{-2}$, and $R = 6\times10^5$ m.
Inside the planet $g(r) = g_s\frac{r}{R}$. The wire runs from $r = \frac{4R}{5}$ to $r = R$; the person must hold its whole weight:
$F = \int_{4R/5}^{R} \lambda g_s \frac{r}{R}\,dr = \frac{\lambda g_s}{2R}\left(R^2 - \frac{16R^2}{25}\right) = \frac{9}{50}\lambda g_s R$.
$F = 0.18 \times 10^{-3} \times 1 \times 6\times10^5 = 108$ N.
Answer: (B). | ||||
| 10 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The tube narrows upward (as drawn), so its wall is inclined at $\frac{\alpha}{2}$ to the vertical, leaning inward.
The surface tension acts along the liquid surface at the contact angle $\theta$ measured from the wall; the wall itself is tilted by $\frac{\alpha}{2}$, so this force makes an angle $\theta + \frac{\alpha}{2}$ with the vertical.
Vertical balance of the raised column (taking it as a column of radius $b$, the thin-tube approximation): $2\pi bS\cos\left(\theta + \frac{\alpha}{2}\right) = \pi b^2 h\rho g$.
$h = \frac{2S}{b\rho g}\cos\left(\theta + \frac{\alpha}{2}\right)$.
Answer: (D). | ||||
| 11 | ThermodynamicsFirst law | Class 11: Thermodynamics | ComprehensionMedium | |
A medium comprehension question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics The lower (monatomic) gas is held at constant volume by the fixed partition. The upper (diatomic) gas is under the free piston, so it is at constant pressure.
No heat leaves the container, so heat lost by the lower gas = heat gained by the upper gas:
$2\cdot\frac{3}{2}R(700 - T) = 2\cdot\frac{7}{2}R(T - 400)$
$3(700 - T) = 7(T - 400)$, $2100 + 2800 = 10T$, $T = 490$ K.
Answer: (D). | ||||
| 12 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | ComprehensionMedium | |
A medium comprehension question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Now both gases stay at the constant pressure set by the piston, so each exchanges heat at $C_P$.
$2\cdot\frac{5}{2}R(700 - T) = 2\cdot\frac{7}{2}R(T - 400)$
$5(700 - T) = 7(T - 400)$, $12T = 6300$, $T = 525$ K.
At constant pressure $W = nR\Delta T$ for each gas:
$W = 2R(525 - 700) + 2R(525 - 400) = -350R + 250R = -100R$.
(Check: total heat from outside is zero, and $\Delta U = 2\cdot\frac{3}{2}R(-175) + 2\cdot\frac{5}{2}R(125) = -525R + 625R = 100R = -W$.)
Answer: (D). | ||||
| 13 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | ComprehensionEasy | |
A easy comprehension 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 Continuity (air treated as incompressible): $A_1v_1 = A_2v_2$.
$v_2 = v_1\left(\frac{r_1}{r_2}\right)^2 = 5\ \text{mm s}^{-1} \times \left(\frac{20}{1}\right)^2 = 2000\ \text{mm s}^{-1} = 2\ \text{m s}^{-1}$.
Answer: (C). | ||||
| 14 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | ComprehensionMedium | |
A medium comprehension 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 For a given piston speed the air speed $v$ in the nozzle is fixed. By Bernoulli, the pressure in the nozzle is below atmospheric by about $\frac{1}{2}\rho_a v^2$.
This pressure difference drives the liquid up the tube and out; neglecting the small height, $\frac{1}{2}\rho_\ell u^2 \approx \frac{1}{2}\rho_a v^2$, so the liquid speed is $u = v\sqrt{\frac{\rho_a}{\rho_\ell}}$.
The tube has a fixed cross-section, so the volume rate $= Au \propto \sqrt{\frac{\rho_a}{\rho_\ell}}$.
Answer: (A). | ||||
| 15 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | ComprehensionHard | |
A hard comprehension question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Take the paper as the $xy$-plane, the loop's axis as $z$ (out of the paper, "above"), wire 1 at $x = -d$ and wire 2 at $x = +d$, both along $y$ (P, R at the bottom; Q, S at the top).
Loop: counterclockwise seen from above, so on the axis its field points up: $B_L = \frac{\mu_0 I a^2}{2(a^2+h^2)^{3/2}}$.
A wire at distance $r = \sqrt{d^2+h^2}$ gives $B = \frac{\mu_0 I}{2\pi r}$; its $z$-component is $B\frac{d}{r}$.
If both currents are in the same direction (PQ and RS), their $z$-components are opposite and cancel, leaving a net horizontal field that nothing can cancel. With PQ (along $+y$) and SR (along $-y$), the $z$-components add, and both point down (for PQ at $x=-d$: $\hat{y}\times(d\hat{x} + h\hat{z})$ has $z$-component $-d$; likewise for SR), while the $x$-components cancel.
Zero field: $\frac{\mu_0 I a^2}{2r^3} = 2\cdot\frac{\mu_0 I}{2\pi r}\cdot\frac{d}{r}$, and with $d \approx a$: $\frac{a^2}{2r} = \frac{a}{\pi}$, so $r = \frac{\pi a}{2}$.
$h^2 = r^2 - a^2 = a^2\left(\frac{\pi^2}{4} - 1\right) \approx 1.47a^2$, $h \approx 1.2a$.
Answer: (C). | ||||
| 16 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | ComprehensionMedium | |
A medium comprehension question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter With opposite currents, the two wires' fields at the loop's centre are both perpendicular to the paper and in the same sense:
$B = 2\cdot\frac{\mu_0 I}{2\pi d} = \frac{\mu_0 I}{\pi d}$.
The loop's moment $m = I\pi a^2$ was along this field (normal to the paper); after turning by $30^\circ$ the angle between $\vec m$ and $\vec B$ is $30^\circ$.
$\tau = mB\sin30^\circ = I\pi a^2\cdot\frac{\mu_0 I}{\pi d}\cdot\frac{1}{2} = \frac{\mu_0 I^2 a^2}{2d}$.
Answer: (B). | ||||
| 17 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Matching listMedium | |
A medium matching list question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Let $F_1 = \frac{kqQ}{a^2+b^2}$ (charges at $\pm a$) and $F_2 = \frac{kqQ}{4a^2+b^2}$ (charges at $\pm 2a$). A charge at $(x_0, 0)$ pushes $q$ (if like) along $(-x_0, b)$.
P (all positive): the $x$-components cancel in pairs, all $y$-components are up: $+y$ (3).
Q ($Q_1, Q_2$ positive push $q$ to the right; $Q_3, Q_4$ negative pull it to the right): $y$-components cancel in mirror pairs, net $+x$ (1).
R ($Q_1, Q_4$ positive, $Q_2, Q_3$ negative): symmetric, so $x$ cancels. $y$: repulsion from the far pair $2F_2\frac{b}{\sqrt{4a^2+b^2}}$ up, attraction to the near pair $2F_1\frac{b}{\sqrt{a^2+b^2}}$ down; the near pair is stronger and more nearly vertical, so the net is $-y$ (4).
S ($Q_1, Q_3$ positive, $Q_2, Q_4$ negative): $y$-components cancel ($Q_2$ and $Q_3$ cancel, $Q_1$ and $Q_4$ cancel). $x$: the near pair both push towards $-x$ with $2kqQ\frac{a}{(a^2+b^2)^{3/2}}$; the far pair both push towards $+x$ with $2kqQ\frac{2a}{(4a^2+b^2)^{3/2}}$. The near pair wins when $b$ is not large, giving $-x$ (2).
Only code (A), P→3, Q→1, R→4, S→2, fits.
Answer: (A). | ||||
| 18 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Matching listEasy | |
A easy matching list question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens maker's formula with $\mu = 1.5$: a biconvex lens has $\frac{1}{f} = 0.5\cdot\frac{2}{r} = \frac{1}{r}$; a plano-convex lens $\frac{1}{f} = \frac{0.5}{r}$, $f = 2r$; a plano-concave lens $f = -2r$. Thin lenses in contact add powers.
P: two biconvex lenses: $\frac{1}{F} = \frac{2}{r}$, $F = \frac{r}{2}$ (2).
Q: two plano-convex lenses: $\frac{1}{F} = \frac{1}{2r} + \frac{1}{2r} = \frac{1}{r}$, $F = r$ (4).
R: two plano-concave lenses: $\frac{1}{F} = -\frac{1}{r}$, $F = -r$ (3).
S: biconvex + plano-concave: $\frac{1}{F} = \frac{1}{r} - \frac{1}{2r} = \frac{1}{2r}$, $F = 2r$ (1).
Answer: (B). | ||||
| 19 | Laws of MotionFriction | Class 11: Laws of Motion | Matching listMedium | |
A medium matching list question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion $m_1$ is above $m_2$ and frictionless, so it leans on $m_2$; only $m_2$'s friction can hold both.
At rest: friction on $m_2$ $= (m_1 + m_2)g\sin\theta$, possible only if this is at most $\mu m_2 g\cos\theta$:
$\tan\theta \le \frac{\mu m_2}{m_1 + m_2} = \frac{0.3 \times 2}{3} = 0.2$, i.e. $\theta \le 11.5^\circ$.
P ($5^\circ$) and Q ($10^\circ$): at rest, friction $(m_1 + m_2)g\sin\theta$ (2).
R ($15^\circ$) and S ($20^\circ$): the blocks slide together, and the kinetic friction on $m_2$ is $\mu m_2 g\cos\theta$ (3).
Answer: (D). | ||||
| 20 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Matching listMedium | |
A medium matching list 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 In the lift's frame gravity is replaced by an effective $g_e$ ($g + a$ going up with acceleration, $g - a$ going down, $g$ at constant velocity, $0$ in free fall).
Efflux speed (Torricelli): $v = \sqrt{2g_e h}$, with $h$ the depth of the hole below the water surface. Time to fall height $H$ to the floor: $t = \sqrt{\frac{2H}{g_e}}$.
Range $d = vt = 2\sqrt{hH}$, independent of $g_e$. So for P, Q and R, $d = 1.2$ m (1).
In free fall $g_e = 0$: there is no pressure excess at the hole, so no water leaks out (4).
Answer: (C). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 21 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The ether X is a trityl ether, $\text{Ph}_3\text{C–OR}$. In acid the oxygen is protonated and the C–O bond breaks to give the triphenylmethyl carbocation ($\text{S}_\text{N}1$); the cation is then captured by water to give $\text{Ph}_3\text{COH}$. The rate is set by how stable that carbocation is.
(A) Replacing a phenyl by methyl removes one ring's resonance stabilisation: slower.
(D) X itself: the reference rate.
(B) A para-methoxyphenyl group stabilises the cation strongly by +R donation from the OMe lone pair: faster than X.
(C) Two para-methoxyphenyl groups stabilise the cation still more: fastest.
Answer: (C). | ||||
| 22 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Class I holds the doubly branched isomers (2,3-dimethylbutane and 2,2-dimethylbutane), class II the singly branched ones (2-methylpentane and 3-methylpentane), and class III the straight chain (n-hexane).
More branching makes the molecule more compact and nearly spherical, so the surface area of contact and the van der Waals (London) attractions fall, and the boiling point falls.
So n-hexane boils highest and the doubly branched isomers lowest: III > II > I.
Answer: (B). | ||||
| 23 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The substrate is 5-chloropentan-2-one, $\text{Cl–CH}_2\text{CH}_2\text{CH}_2\text{–CO–CH}_3$.
Step 1: at 0 °C, $\text{CH}_3\text{MgBr}$ adds to the ketone carbonyl (the alkyl chloride reacts much more slowly), giving the tertiary alkoxide $\text{ClCH}_2\text{CH}_2\text{CH}_2\text{C(CH}_3)_2\text{O}^-$.
The alkoxide sits four carbons from the C–Cl, so it displaces chloride intramolecularly ($\text{S}_\text{N}2$), closing a five-membered ring: 2,2-dimethyltetrahydrofuran (2,2-dimethyloxolane).
Step 2: aqueous acid does not open this ether.
(A) would need the chlorine replaced by hydrogen; (B) needs elimination of HCl; (C) is an enol ether with a lost methyl; none forms here.
Answer: (D). | ||||
| 24 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctMedium | |
A medium single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions With periodate: $\text{KIO}_4 + \text{H}_2\text{O}_2 \rightarrow \text{KIO}_3 + \text{H}_2\text{O} + \text{O}_2$. Iodine goes from +7 to +5 (reduced) while the oxygen of $\text{H}_2\text{O}_2$ goes from −1 to 0 in $\text{O}_2$: $\text{H}_2\text{O}_2$ is the reducing agent.
With hydroxylamine: $\text{NH}_2\text{OH} + 2\text{H}_2\text{O}_2 \rightarrow \text{HNO}_2 + 3\text{H}_2\text{O}$ (or further oxidation). Nitrogen goes from −1 upward (oxidised) and the peroxide oxygen goes from −1 to −2: $\text{H}_2\text{O}_2$ is the oxidising agent.
Answer: (A). | ||||
| 25 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Thionyl chloride chlorinates white phosphorus to phosphorus trichloride:
$\text{P}_4 + 8\text{SOCl}_2 \rightarrow 4\text{PCl}_3 + 4\text{SO}_2 + 2\text{S}_2\text{Cl}_2$.
$\text{SO}_2\text{Cl}_2$, $\text{SCl}_2$ and $\text{POCl}_3$ are not the products of this reaction (the sulfur ends up as $\text{SO}_2$ and $\text{S}_2\text{Cl}_2$).
Answer: (A). | ||||
| 26 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Complete hydrolysis: $\text{XeF}_6 + 3\text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6\text{HF}$, so P is $\text{XeO}_3$.
In alkali: $\text{XeO}_3 + \text{OH}^- \rightarrow \text{HXeO}_4^-$, so Q is the hydrogen xenate ion.
Slow disproportionation in alkali: $2\text{HXeO}_4^- + 2\text{OH}^- \rightarrow \text{XeO}_6^{4-} + \text{Xe} + \text{O}_2 + 2\text{H}_2\text{O}$.
The gases released are Xe and $\text{O}_2$: two.
Answer: (C). | ||||
| 27 | Principles Related to Practical ChemistryDetection of elements and functional groups | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter In the dye test, a diazonium salt (from aniline, $\text{NaNO}_2$ and HCl at 0–5 °C) couples with β-naphthol to give an orange-red azo dye.
The coupling partner must be activated: in alkali, β-naphthol becomes the naphthoxide ion, a far stronger electron donor, which couples readily at C-1. In acidic or neutral solution, or in dichloromethane, the coupling hardly occurs.
So an alkaline solution of β-naphthol is needed.
Answer: (D). | ||||
| 28 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate $= k[\text{M}]^n$. Doubling $[\text{M}]$ multiplies the rate by $2^n = 8$, so $n = 3$.
Answer: (B). | ||||
| 29 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Liquid to vapour increases disorder: $\Delta S_{\text{system}} > 0$.
Vaporisation is endothermic, so the surroundings lose heat: $\Delta S_{\text{surroundings}} = -\frac{\Delta H_{\text{vap}}}{T} < 0$.
At 100 °C and 1 atm the two phases are in equilibrium, so the two changes are equal and opposite and $\Delta S_{\text{total}} = 0$.
Answer: (B). | ||||
| 30 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Without s–p mixing, the order of the valence MOs is $\sigma2s < \sigma^*2s < \sigma2p_z < \pi2p_x = \pi2p_y < \pi^*2p < \sigma^*2p_z$.
$\text{Be}_2$ (4 valence electrons): $\sigma2s^2\sigma^*2s^2$, all paired: diamagnetic.
$\text{B}_2$ (6): $\ldots\sigma2p_z^2$, all paired: diamagnetic.
$\text{C}_2$ (8): $\ldots\sigma2p_z^2\,\pi2p_x^1\pi2p_y^1$, two unpaired electrons: paramagnetic.
$\text{N}_2$ (10): $\ldots\sigma2p_z^2\,\pi2p^4$, all paired: diamagnetic.
(With mixing, as actually happens, $\text{B}_2$ would be the paramagnetic one; the question rules that out.)
Answer: (C). | ||||
| 31 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | ComprehensionMedium | |
A medium comprehension question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons M is but-3-yn-1-ol, $\text{HOCH}_2\text{CH}_2\text{C≡CH}$.
1. Excess $\text{NaNH}_2$ removes both acidic protons, giving the dianion $^-\text{OCH}_2\text{CH}_2\text{C≡C}^-$.
2. One equivalent of ethyl iodide alkylates the more nucleophilic carbon end (the acetylide): $^-\text{OCH}_2\text{CH}_2\text{C≡C–CH}_2\text{CH}_3$.
3. One equivalent of methyl iodide then methylates the alkoxide (Williamson): $\text{CH}_3\text{OCH}_2\text{CH}_2\text{C≡CCH}_2\text{CH}_3$.
4. Lindlar's catalyst adds $\text{H}_2$ syn, stopping at the cis alkene.
X = cis-$\text{CH}_3\text{OCH}_2\text{CH}_2\text{CH=CHCH}_2\text{CH}_3$, (Z)-1-methoxyhex-3-ene: structure (A). (B) is the trans isomer; (C) and (D) have the ethyl on oxygen and a methyl on the alkene, the reverse alkylation.
Answer: (A). | ||||
| 32 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | ComprehensionMedium | |
A medium comprehension question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids N is but-1-yne, $\text{CH}_3\text{CH}_2\text{C≡CH}$.
1. Two equivalents of $\text{NaNH}_2$: one deprotonates the alkyne, the other the OH of the bromohydrin added next.
2. The acetylide displaces bromide from $\text{BrCH}_2\text{CH(O}^-)\text{CH}_3$ (directly, or via the epoxide formed from the alkoxide, opened at the less hindered $\text{CH}_2$; the product is the same): $\text{CH}_3\text{CH}_2\text{C≡CCH}_2\text{CH(O}^-)\text{CH}_3$.
3. Mild acid gives the alcohol $\text{CH}_3\text{CH}_2\text{C≡CCH}_2\text{CH(OH)CH}_3$.
4. $\text{H}_2$, Pd/C reduces the triple bond fully: heptan-2-ol.
5. $\text{CrO}_3$ oxidises the secondary alcohol: Y = heptan-2-one, $\text{CH}_3\text{CO(CH}_2)_4\text{CH}_3$, $\text{C}_7\text{H}_{14}\text{O}$.
Y is a methyl ketone, so it gives the iodoform test; as a ketone it does not reduce Tollens' reagent. X (from Q.31) is $\text{C}_7\text{H}_{14}\text{O}$ too, but an ether: same formula, different functional group, so Y is a functional isomer of X (it has no C=C, so it cannot be a geometrical isomer).
Answer: (C). | ||||
| 33 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | ComprehensionMedium | |
A medium comprehension question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A metal ion that gives a tetrahedral complex with one ligand and a square planar one with another must be $d^8$: $\text{Ni}^{2+}$.
With excess HCl: $[\text{NiCl}_4]^{2-}$, tetrahedral (weak-field $\text{Cl}^-$, $sp^3$).
With excess KCN: $[\text{Ni(CN)}_4]^{2-}$, square planar (strong-field $\text{CN}^-$ pairs the electrons, $dsp^2$).
So M1 = $\text{Ni}^{2+}$, Q = HCl, R = KCN.
$\text{Zn}^{2+}$ and $\text{Cd}^{2+}$ ($d^{10}$) give tetrahedral complexes with both reagents (M2 behaves this way), and $\text{Co}^{2+}$ with cyanide does not give a square planar $[\text{Co(CN)}_4]^{2-}$ in this sense.
Answer: (B). | ||||
| 34 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | ComprehensionMedium | |
A medium comprehension question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter M2 forms tetrahedral complexes with both $\text{Cl}^-$ and $\text{CN}^-$ ($[\text{ZnCl}_4]^{2-}$, $[\text{Zn(CN)}_4]^{2-}$): M2 is $\text{Zn}^{2+}$.
With a little KOH: $\text{Zn}^{2+} + 2\text{OH}^- \rightarrow \text{Zn(OH)}_2\downarrow$ (white).
With excess KOH the amphoteric hydroxide dissolves: $\text{Zn(OH)}_2 + 2\text{OH}^- \rightarrow [\text{Zn(OH)}_4]^{2-}$ (tetrahedral).
The ferrocyanide and phosphate precipitates of zinc do not dissolve in excess reagent, and chromate does not give a white precipitate that redissolves.
Answer: (D). | ||||
| 35 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | ComprehensionEasy | |
A easy comprehension question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Graham's law: rate of diffusion $\propto \frac{1}{\sqrt{M}}$, and in the same time the distances covered are in the ratio of the rates.
$\frac{d}{24 - d} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{\frac{40}{10}} = 2$.
$d = 48 - 2d$, $d = 16$ cm.
Answer: (C). | ||||
| 36 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | ComprehensionMedium | |
A medium comprehension question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry X and Y have equal molecular diameters and move through the same inert gas at the same pressure, so their mean free paths are essentially the same: (A) and (B) do not explain a difference.
The lighter X molecules move faster ($\bar{u} \propto \frac{1}{\sqrt{M}}$), so they collide with the inert-gas molecules more often per unit time. Each collision randomises the direction of motion, so X's net progress along the tube is held back more than Graham's simple ratio allows. X therefore covers less distance than predicted, and d comes out smaller.
That is the increased collision frequency of X with the inert gas compared with Y.
Answer: (D). | ||||
| 37 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Matching listHard | |
A hard matching list question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Two things decide the pathway.
(i) Is $\text{CO}_2$ lost at once (concerted, P and Q) or only after $\text{RCO}_2^\bullet$ forms (stepwise, R and S)? Concerted loss happens when $\text{R}^\bullet$ is a stabilised radical: benzyl ($\text{C}_6\text{H}_5\text{CH}_2$, in 1 and 3). With R = phenyl (2 and 4), $\text{C}_6\text{H}_5^\bullet$ is unstable, so the benzoyloxy radical $\text{C}_6\text{H}_5\text{CO}_2^\bullet$ forms first and loses $\text{CO}_2$ afterwards.
(ii) Does the alkoxy radical $\text{R'O}^\bullet$ fragment (β-scission) to a carbonyl compound and a radical $\text{X}^\bullet$ (Q and R)? A tertiary alkoxy radical does: $\text{C}_6\text{H}_5\text{CH}_2\text{C(CH}_3)_2\text{O}^\bullet \rightarrow \text{C}_6\text{H}_5\text{CH}_2^\bullet + \text{acetone}$ (3), and $\text{C}_6\text{H}_5\text{C(CH}_3)_2\text{O}^\bullet \rightarrow \text{CH}_3^\bullet + \text{acetophenone}$ (4). Methoxy radical ($\text{CH}_3\text{O}^\bullet$ in 1 and 2) does not.
So: P (concerted, no scission) → 1; Q (concerted, scission) → 3; R (stepwise, scission) → 4; S (stepwise, no scission) → 2.
Answer: (A). | ||||
| 38 | AminesDiazonium salts | Class 12: Amines | Matching listHard | |
A hard matching list question on diazonium salts, from Amines. NCERT: Class 12: Amines Scheme I: $\text{KMnO}_4$ oxidises a side-chain $\text{CH}_3$ to COOH; $\text{SOCl}_2$ then $\text{NH}_3$ make the amide. From 4-nitrotoluene: 4-nitrobenzoic acid → 4-nitrobenzamide, $\text{C}_7\text{H}_6\text{N}_2\text{O}_3$. S → 1.
Scheme II: Sn/HCl reduces $\text{NO}_2$ to $\text{NH}_2$; acetylation protects it; sulfonation blocks the para position; nitration goes ortho to NHCOCH₃; dilute acid and heat remove the $\text{SO}_3\text{H}$ and hydrolyse the amide; base frees the amine: 2-nitroaniline, $\text{C}_6\text{H}_6\text{N}_2\text{O}_2$. From nitrobenzene. R → 2.
Scheme III: acetylene over red-hot iron trimerises to benzene; vigorous nitration gives 1,3-dinitrobenzene; $\text{H}_2\text{S/NH}_3$ (Zinin) reduces one nitro group to 3-nitroaniline; diazotisation and hydrolysis give 3-nitrophenol, $\text{C}_6\text{H}_5\text{NO}_3$. P → 3.
Scheme IV: resorcinol is sulfonated at the two activated positions 4 and 6, nitration then enters position 2 (between the OH groups), and dilute acid with heat removes the sulfonic acid groups: 2-nitroresorcinol, $\text{C}_6\text{H}_5\text{NO}_4$. Q → 4.
P→3, Q→4, R→2, S→1.
Answer: (C). | ||||
| 39 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Matching listMedium | |
A medium matching list question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds P: $\text{Cr}^{3+}$, $d^3$, three unpaired electrons: paramagnetic. Octahedral $\text{MA}_4\text{B}_2$: cis and trans forms. → 3.
Q: $\text{Ti}^{3+}$, $d^1$: paramagnetic. $\text{MA}_5\text{B}$ has no geometrical isomers, but the Cl inside can swap with a nitrate outside, $[\text{Ti(H}_2\text{O)}_5\text{NO}_3]\text{Cl(NO}_3)$: ionisation isomerism. → 1.
R: $\text{Pt}^{2+}$, $d^8$, square planar: diamagnetic. With a symmetric chelate (en) occupying two cis positions, the other two positions are equivalent in shape, so no cis-trans isomer; Cl inside and $\text{NO}_3^-$ outside can swap: ionisation isomerism. → 4.
S: $\text{Co}^{3+}$, $d^6$ with $\text{NH}_3$: low spin, diamagnetic. $\text{MA}_4\text{B}_2$ octahedral: cis-trans. → 2.
Code: P→3, Q→1, R→4, S→2.
Answer: (B). | ||||
| 40 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Matching listMedium | |
A medium matching list question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Shaded and unshaded lobes are opposite phases. Overlap of like phases is bonding; of unlike phases, antibonding. Overlap along the line joining the nuclei is σ; side-on is π.
P: two d orbitals end to end; the facing lobes are both unshaded (same phase), on the internuclear axis: d–d σ bonding (2).
Q: a p orbital beside a d orbital, side by side; the upper p lobe (shaded) faces a shaded d lobe and the lower (unshaded) faces an unshaded one: p–d π bonding (3).
R: the same arrangement but with the p orbital reversed, so the facing lobes have opposite phase: p–d π antibonding (1).
S: a $d_{z^2}$ orbital end on to a d orbital along the axis; its shaded lobe meets the unshaded lobe of the other: d–d σ antibonding (4).
P→2, Q→3, R→1, S→4.
Answer: (C). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 41 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor: $e^{\int \frac{x}{x^2-1}dx} = e^{\frac{1}{2}\ln|x^2-1|} = \sqrt{1-x^2}$ on $(-1, 1)$.
$\frac{d}{dx}\left(y\sqrt{1-x^2}\right) = x^4 + 2x$, so $y\sqrt{1-x^2} = \frac{x^5}{5} + x^2 + C$, and $f(0) = 0$ gives $C = 0$.
$f(x) = \frac{x^5/5}{\sqrt{1-x^2}} + \frac{x^2}{\sqrt{1-x^2}}$. The first part is odd and integrates to zero over the symmetric interval.
$\int_{-\sqrt3/2}^{\sqrt3/2}\frac{x^2}{\sqrt{1-x^2}}dx = 2\int_0^{\pi/3}\sin^2\theta\,d\theta$ (with $x = \sin\theta$) $= \left[\theta - \frac{\sin2\theta}{2}\right]_0^{\pi/3} = \frac{\pi}{3} - \frac{\sqrt3}{4}$.
Answer: (B). | ||||
| 42 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $\text{cosec}\,x + \cot x = e^u$. Then $\text{cosec}\,x - \cot x = e^{-u}$ (their product is $\text{cosec}^2x - \cot^2x = 1$), so $2\,\text{cosec}\,x = e^u + e^{-u}$.
Differentiating: $(-\text{cosec}\,x\cot x - \text{cosec}^2x)\,dx = e^u\,du$, i.e. $-\text{cosec}\,x\,(\cot x + \text{cosec}\,x)\,dx = e^u du$, so $dx = -\frac{du}{\text{cosec}\,x} = -\frac{2\,du}{e^u + e^{-u}}$.
Limits: $x = \frac{\pi}{2}$: $e^u = 1$, $u = 0$; $x = \frac{\pi}{4}$: $e^u = \sqrt2 + 1$, $u = \log(1+\sqrt2)$.
$I = \int_{\log(1+\sqrt2)}^{0}(e^u + e^{-u})^{17}\cdot\left(-\frac{2}{e^u + e^{-u}}\right)du = \int_0^{\log(1+\sqrt2)} 2(e^u + e^{-u})^{16}\,du$.
Answer: (A). | ||||
| 43 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem A term $x^{2a}x^{3b}x^{4c}$ has coefficient $\binom{4}{a}\binom{7}{b}\binom{12}{c}$; we need $2a + 3b + 4c = 11$ with $a \le 4$.
$3b$ must be odd, so $b = 1$ or $3$.
$b = 1$: $a + 2c = 4$: $(a, c) = (4, 0), (2, 1), (0, 2)$: $7\left[1 + 6\cdot12 + 66\right] = 7 \times 139 = 973$.
$b = 3$: $a + 2c = 1$: $(a, c) = (1, 0)$: $\binom{7}{3}\binom{4}{1} = 35 \times 4 = 140$.
Total $= 973 + 140 = 1113$.
Answer: (C). | ||||
| 44 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals By the fundamental theorem and the chain rule, $F'(x) = f(\sqrt{x^2})\cdot2x = 2xf(x)$ for $x \ge 0$.
So $f'(x) = 2xf(x)$, giving $f(x) = Ce^{x^2}$, and $f(0) = 1$ gives $f(x) = e^{x^2}$.
$F(2) = \int_0^4 f(\sqrt t)\,dt = \int_0^4 e^{t}\,dt = e^4 - 1$.
Answer: (B). | ||||
| 45 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent to $y^2 = 8x$ ($a = 2$): $y = mx + \frac{2}{m}$. It touches the circle of radius $\sqrt2$ when $\frac{|2/m|}{\sqrt{1+m^2}} = \sqrt2$: $\frac{4}{m^2} = 2(1+m^2)$, $m^4 + m^2 - 2 = 0$, $m^2 = 1$.
Tangents: $y = x + 2$ and $y = -x - 2$.
Points on the circle (feet of perpendiculars from O): for $x - y + 2 = 0$, $(-1, 1)$; for $x + y + 2 = 0$, $(-1, -1)$.
Points on the parabola $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$: $(2, 4)$ and $(2, -4)$.
The quadrilateral is a trapezium with parallel vertical sides of lengths 2 (at $x = -1$) and 8 (at $x = 2$), 3 apart:
Area $= \frac{1}{2}(2 + 8)\times3 = 15$.
Answer: (D). | ||||
| 46 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Using $\sin3x = 3\sin x - 4\sin^3x$:
LHS $= \sin x + 4\sin x\cos x - 3\sin x + 4\sin^3 x = 2\sin x\left(-1 + 2\cos x + 2\sin^2x\right) = 2\sin x\left(1 + 2\cos x - 2\cos^2x\right)$.
With $c = \cos x$: $1 + 2c - 2c^2 = \frac{3}{2} - 2\left(c - \frac{1}{2}\right)^2 \le \frac{3}{2}$, and $0 < \sin x \le 1$ on $(0, \pi)$.
So LHS $\le 2\cdot1\cdot\frac{3}{2} = 3$, with equality only if $\sin x = 1$ and $\cos x = \frac{1}{2}$ together, which is impossible. (At $\cos x = \frac{1}{2}$ the LHS is $2\cdot\frac{\sqrt3}{2}\cdot\frac{3}{2} \approx 2.6$.)
Hence LHS $< 3$ throughout: no solution.
Answer: (D). | ||||
| 47 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let the two sides be $a, b$: $a + b = x$, $ab = y$. Then $x^2 - c^2 = y$ gives $a^2 + b^2 + 2ab - c^2 = ab$, i.e. $a^2 + b^2 - c^2 = -ab$, so $\cos C = -\frac{1}{2}$ and $C = 120^\circ$.
Area $\Delta = \frac{1}{2}ab\sin C = \frac{\sqrt3}{4}y$; semi-perimeter $s = \frac{x + c}{2}$.
$r = \frac{\Delta}{s}$ and $R = \frac{abc}{4\Delta}$, so $\frac{r}{R} = \frac{4\Delta^2}{s\cdot abc} = \frac{4\cdot\frac{3y^2}{16}}{\frac{x+c}{2}\cdot yc} = \frac{3y}{2c(x+c)}$.
Answer: (B). | ||||
| 48 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctHard | |
A hard single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Card 1 goes into envelope 2. Cards 2, 3, 4, 5, 6 go into envelopes 1, 3, 4, 5, 6, with cards 3–6 kept out of their own envelopes; card 2 has no restriction (its own envelope is already used).
Count by where card 2 goes.
Card 2 into envelope 1: cards 3–6 must be deranged among envelopes 3–6: $D_4 = 9$.
Card 2 not into envelope 1: treat envelope 1 as card 2's 'forbidden' envelope; then all five cards 2–6 are deranged among envelopes 1, 3, 4, 5, 6: $D_5 = 44$.
Total $= 9 + 44 = 53$.
Answer: (C). | ||||
| 49 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Only the pattern of boys and girls matters: $\binom{5}{2} = 10$ equally likely choices of the girls' positions $i < j$.
The first girl (at $i$) has $i - 1$ boys and no girls ahead: need $i - 1 \ge 1$, $i \ge 2$.
The second girl (at $j$) has one girl and $j - 2$ boys ahead: need $j - 2 \ge 2$, $j \ge 4$.
Pairs: $(2,4), (2,5), (3,4), (3,5), (4,5)$: 5 of the 10.
Probability $= \frac{5}{10} = \frac{1}{2}$.
Answer: (A). | ||||
| 50 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Purely imaginary roots $\pm i\beta$ ($\beta > 0$) mean $p(x) = a(x^2 + \beta^2)$ with $a \ne 0$ real.
$p(p(x)) = 0$ means $p(x) = \pm i\beta$, i.e. $x^2 = -\beta^2 \pm \frac{i\beta}{a}$.
For real $x$, $x^2$ is real and $\ge 0$; for purely imaginary $x$, $x^2$ is real and $\le 0$. Here $x^2$ has a nonzero imaginary part $\pm\frac{\beta}{a}$, so no root is real or purely imaginary.
Answer: (D). | ||||
| 51 | Conic SectionsParabola | Class 11: Conic Sections | ComprehensionMedium | |
A medium comprehension question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections For a focal chord the parameters multiply to $-1$, so $Q = \left(\frac{a}{t^2}, -\frac{2a}{t}\right)$, parameter $-\frac{1}{t}$.
The chord joining parameters $t_1, t_2$ has slope $\frac{2}{t_1 + t_2}$, so slope of $QR = \frac{2}{r - \frac{1}{t}}$.
Slope of $PK = \frac{2at - 0}{at^2 - 2a} = \frac{2t}{t^2 - 2}$.
Parallel: $r - \frac{1}{t} = \frac{t^2 - 2}{t}$, so $r = \frac{t^2 - 1}{t}$.
Answer: (D). | ||||
| 52 | Conic SectionsParabola | Class 11: Conic Sections | ComprehensionMedium | |
A medium comprehension question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent at $P$: $ty = x + at^2$.
Normal at $S$: $y = -sx + 2as + as^3$, and with $s = \frac{1}{t}$: $y = -\frac{x}{t} + \frac{2a}{t} + \frac{a}{t^3}$.
Substitute $x = ty - at^2$: $y = -y + at + \frac{2a}{t} + \frac{a}{t^3}$.
$2y = \frac{a(t^4 + 2t^2 + 1)}{t^3} = \frac{a(t^2+1)^2}{t^3}$, so $y = \frac{a(t^2+1)^2}{2t^3}$.
Answer: (B). | ||||
| 53 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | ComprehensionMedium | |
A medium comprehension question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $g\left(\frac{1}{2}\right) = \int_0^1 \frac{dt}{\sqrt{t(1-t)}}$ (as an improper integral).
Put $t = \sin^2\theta$, $dt = 2\sin\theta\cos\theta\,d\theta$, $\sqrt{t(1-t)} = \sin\theta\cos\theta$:
$g\left(\frac{1}{2}\right) = \int_0^{\pi/2} 2\,d\theta = \pi$.
Answer: (A). | ||||
| 54 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | ComprehensionMedium | |
A medium comprehension question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Substitute $t \rightarrow 1 - t$ in the integral (the limits $h$ and $1-h$ swap):
$g(a) = \int t^{a-1}(1-t)^{-a}\,dt = \int t^{-(1-a)}(1-t)^{(1-a)-1}\,dt = g(1-a)$.
Differentiating $g(a) = g(1-a)$: $g'(a) = -g'(1-a)$. At $a = \frac{1}{2}$: $g'\left(\frac{1}{2}\right) = -g'\left(\frac{1}{2}\right)$, so $g'\left(\frac{1}{2}\right) = 0$.
(Indeed $g(a) = \frac{\pi}{\sin\pi a}$, which is symmetric about $a = \frac{1}{2}$.)
Answer: (D). | ||||
| 55 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | ComprehensionEasy | |
A easy comprehension question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Total outcomes $3 \times 5 \times 7 = 105$. Odd cards: box 1 has 2 (odd) and 1 (even); box 2 has 3 and 2; box 3 has 4 and 3.
The sum is odd when an odd number of the three are odd:
all three odd: $2\cdot3\cdot4 = 24$;
only $x_1$ odd: $2\cdot2\cdot3 = 12$;
only $x_2$ odd: $1\cdot3\cdot3 = 9$;
only $x_3$ odd: $1\cdot2\cdot4 = 8$.
Favourable $= 53$; probability $= \frac{53}{105}$.
Answer: (B). | ||||
| 56 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | ComprehensionEasy | |
A easy comprehension question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability AP means $x_1 + x_3 = 2x_2$: $x_1$ and $x_3$ have the same parity and $x_2 = \frac{x_1 + x_3}{2} \le 5$, i.e. $x_1 + x_3 \le 10$.
$x_1 = 1$: $x_3 \in \{1, 3, 5, 7\}$: 4.
$x_1 = 2$: $x_3 \in \{2, 4, 6\}$: 3.
$x_1 = 3$: $x_3 \in \{1, 3, 5, 7\}$ (sums up to 10): 4.
Favourable $= 11$; probability $= \frac{11}{105}$.
Answer: (C). | ||||
| 57 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Matching listMedium | |
A medium matching list question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations The $z_k$ are the tenth roots of unity other than 1.
P (read as "$z_k \cdot z_j = 1$"): $z_j = z_{10-k}$ is also in the set and $z_kz_{10-k} = 1$. True (1).
Q: $z = \frac{z_k}{z_1}$ always solves $z_1z = z_k$, since $z_1 \ne 0$. The statement is False (2).
R: $1 + x + \cdots + x^9 = \prod_{k=1}^{9}(x - z_k)$; at $x = 1$, $\prod|1 - z_k| = 10$, so the expression is 1 (3).
S: the ten roots sum to 0, so $\sum_{k=1}^{9}\cos\frac{2k\pi}{10} = -1$ and $1 - (-1) = 2$ (4).
P→1, Q→2, R→3, S→4.
Answer: (C). | ||||
| 58 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Matching listMedium | |
A medium matching list question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals P: $f(x) = bx + cx^2$ ($f(0) = 0$) with $\frac{b}{2} + \frac{c}{3} = 1$, i.e. $3b + 2c = 6$ in non-negative integers: $(b, c) = (2, 0), (0, 3)$. Two polynomials (2).
Q: $f = \sqrt2\sin\left(x^2 + \frac{\pi}{4}\right)$ is maximum when $x^2 = \frac{\pi}{4} + 2n\pi$. With $0 \le x^2 \le 13$: $x^2 = \frac{\pi}{4} \approx 0.79$ or $\frac{9\pi}{4} \approx 7.07$ (the next, $\frac{17\pi}{4} \approx 13.35$, is too big). Each gives $x = \pm\sqrt{x^2}$: 4 points (3).
R: with $g(x) = \frac{3x^2}{1+e^x}$, $g(x) + g(-x) = 3x^2\left(\frac{1}{1+e^x} + \frac{e^x}{1+e^x}\right) = 3x^2$, so $\int_{-2}^{2}g = \int_0^2 3x^2\,dx = 8$ (1).
S: $\cos2x$ is even and $\log\frac{1+x}{1-x}$ is odd, so the numerator's integrand is odd on a symmetric interval: numerator 0, while the denominator is positive. Ratio 0 (4).
P→2, Q→3, R→1, S→4.
Answer: (D). | ||||
| 59 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Matching listHard | |
A hard matching list question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability P: let $x = \cos\theta$, so $y = \cos3\theta$. $\frac{dy}{dx} = \frac{3\sin3\theta}{\sin\theta}$, and differentiating again one finds $(1 - x^2)y'' - xy' + 9y = 0$ (the Chebyshev equation; check with $y = 4x^3 - 3x$: $(1-x^2)24x - x(12x^2 - 3) + 9(4x^3 - 3x) = 0$). So $(x^2 - 1)y'' + xy' = 9y$ and the expression is 9 (4).
Q: consecutive position vectors have equal length $\rho$ and angle $\frac{2\pi}{n}$. The $n-1$ cross products all point the same way, magnitude $(n-1)\rho^2\sin\frac{2\pi}{n}$; the dot products sum to $(n-1)\rho^2\cos\frac{2\pi}{n}$. Equal magnitudes: $\left|\tan\frac{2\pi}{n}\right| = 1$, so $\frac{2\pi}{n} = \frac{\pi}{4}$ or $\frac{3\pi}{4}$; the smallest integer $n > 2$ is $n = 8$ (3).
R: the normal at $(x_0, y_0)$ has slope $\frac{a^2y_0}{b^2x_0} = \frac{2y_0}{x_0}$; perpendicular to $x + y = 8$ means slope 1, so $x_0 = 2y_0$. On the ellipse: $\frac{4y_0^2}{6} + \frac{y_0^2}{3} = 1$, $y_0 = \pm1$. At $(2, 1)$ the normal is $y = x - 1$, passing through $(h, 1)$ with $h = 2$ (2). (At $(-2, -1)$ it is $y = x + 1$, giving $h = 0$, not in the list.)
S: for $x > 0$ both arguments are in $(0, 1)$, so $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\frac{A+B}{1-AB}$:
$\frac{A+B}{1-AB} = \frac{(4x+1) + (2x+1)}{(2x+1)(4x+1) - 1} = \frac{6x+2}{8x^2+6x} = \frac{3x+1}{x(4x+3)}$.
Setting equal to $\frac{2}{x^2}$: $x(3x+1) = 2(4x+3)$, $3x^2 - 7x - 6 = 0$, $x = 3$ or $-\frac{2}{3}$. One positive solution (1).
P→4, Q→3, R→2, S→1.
Answer: (A). | ||||
| 60 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Matching listMedium | |
A medium matching list question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f_4(x) = x^2$ for $x < 0$ and $e^{2x} - 1$ for $x \ge 0$.
P: for $x < 0$, $f_4$ takes every value in $(0, \infty)$; for $x \ge 0$, every value in $[0, \infty)$. So it is onto $[0, \infty)$, and not one-one (each positive value is taken twice). It is continuous at 0 but not differentiable there (left derivative 0, right derivative 2). → 1.
Q: $f_3$ is continuous at 0 and its one-sided derivatives there are $\cos0 = 1$ and 1, so it is differentiable everywhere; $\sin x$ repeats values for $x < 0$, so it is not one-one. → 3.
R: $f_2\circ f_1(x) = x^2$ for $x < 0$ and $e^{2x}$ for $x \ge 0$: it jumps from 0 to 1 at $x = 0$ (not continuous), and values above 1 are taken twice (not one-one). → 2.
S: $f_2(x) = x^2$ on $[0, \infty)$ is continuous and strictly increasing, so one-one. → 4.
P→1, Q→3, R→2, S→4.
Answer: (D). | ||||
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