JEE (Advanced) · 2017
JEE Advanced 2017, 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
36 Q
- Motion of System of Particles and Rigid Body6
- Magnetic Effects of Current and Magnetism5
- Electromagnetic Induction and Alternating Currents4
14 easy · 20 medium · 2 hard
Chemistry
36 Q
- p-Block Elements7
- Hydrocarbons5
- Thermodynamics3
14 easy · 20 medium · 2 hard
Mathematics
36 Q
- Conic Sections5
- Applications of Derivatives5
- Matrices and Determinants4
17 easy · 19 medium · 0 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
No matching NCERT chapter
Class 11: Hydrocarbons
Class 11: Thermodynamics
Class 11: Structure of Atom
Class 12: Aldehydes, Ketones and Carboxylic Acids
Class 12: Solutions
Class 12: Electrochemistry
Class 11: Chemical Bonding and Molecular Structure
Class 12: Amines
Class 12: Coordination Compounds
Class 11: Organic Chemistry – Some Basic Principles and Techniques
No matching NCERT chapter
Class 11: Redox Reactions
No matching NCERT chapter
Class 12: Chemical Kinetics
Class 12: Haloalkanes and Haloarenes
Class 11: Systems of Particles and Rotational Motion
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 11: Thermodynamics
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 11: Oscillations; Class 11: Waves
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 12: Atoms; Class 12: Nuclei
Class 11: Kinetic Theory
Class 12: Dual Nature of Radiation and Matter
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Class 11: Gravitation
Class 11: Units and Measurement
Class 11: Laws of Motion
Class 11: Conic Sections
Class 12: Application of Derivatives
Class 12: Matrices; Class 12: Determinants
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 11: Probability; Class 12: Probability
Class 11: Permutations and Combinations
Class 12: Vector Algebra
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 11: Complex Numbers and Quadratic Equations
Class 11: Complex Numbers and Quadratic Equations
Class 11: Conic Sections
Class 11: Sequences and Series
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 12: Differential Equations
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 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Multiple correctMedium | |
A medium multiple correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory At very low pressure the molecules do not collide with one another near the plate; each one simply bounces off it. Take a molecule speed of order $u$ and plate speed $v \ll u$.
On the leading face the molecules meet the plate with relative speed about $u + v$; both the rate of hits and the momentum given per hit grow with $u + v$, so the pressure there is about $\rho(u + v)^2$ (up to a constant). On the trailing face it is about $\rho(u - v)^2$.
Pressure difference $\Delta p \propto (u + v)^2 - (u - v)^2 = 4uv$: proportional to $uv$. (A) is true.
The resistive force is $\Delta p \times$ area $\propto uv$, and $u$ is fixed by the gas temperature, so the force is proportional to $v$. (B) is true.
The equation of motion is $m\frac{dv}{dt} = F - kv$. As $v$ grows the resistance grows, so the acceleration falls; it is not constant. (C) is false.
The speed tends to the terminal value $v = F/k$, where $F$ is balanced by the resistive force. (D) is true.
Answer: (A), (B), (D). | ||||
| 2 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Multiple correctMedium | |
A medium multiple correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The cut is a quarter circle of radius $R$: the mass starts at its top, a horizontal distance $R$ to the left of the block's right edge, and leaves the block at the bottom of the cut (the right edge), moving horizontally.
No horizontal external force acts, so horizontal momentum is conserved (zero) and the centre of mass of the system does not move horizontally.
Velocities: $mv + MV = 0$, and energy $mgR = \frac{1}{2}mv^2 + \frac{1}{2}MV^2$. With $V = -\frac{m}{M}v$: $mgR = \frac{1}{2}mv^2\left(1 + \frac{m}{M}\right)$, so $v = \sqrt{\frac{2gR}{1 + \frac{m}{M}}}$. (B) is correct.
Then $V = -\frac{m}{M}v = -\frac{m}{M}\sqrt{\frac{2gR}{1 + m/M}}$, not $-\frac{m}{M}\sqrt{2gR}$. (D) is wrong.
Displacements: relative to the block the mass moves $R$ to the right. If the block moves $X$, the mass moves $X + R$, and $m(X + R) + MX = 0$, so $X = -\frac{mR}{M+m}$. (C) is correct.
The mass leaves at the block's right edge, which is now at $x = X = -\frac{mR}{M+m}$; there is no factor $\sqrt{2}$. (A) is wrong.
Answer: (B), (C). | ||||
| 3 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Multiple correctMedium | |
A medium multiple correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Let $\mu$ be the mass per unit length and $y$ the height above $A$. The tension at height $y$ holds up the block and the rope below: $T(y) = Mg + \mu g y$. The wave speed there is $v(y) = \sqrt{T/\mu}$, a function of position only.
(A) The time to cross the rope is $\int_0^L \frac{dy}{v(y)}$ whichever way the pulse travels, so $T_{AO} = T_{OA}$. Correct.
(B) At the midpoint both pulses have the same speed, but Pulse 1 moves down and Pulse 2 moves up: their velocities are opposite, not the same. Wrong.
(C) Pulse 1 runs from high tension (at $O$) to low tension (at $A$), so it slows down. The frequency stays fixed, so $\lambda = v/f$ becomes shorter. Wrong.
(D) $v = \sqrt{T/\mu}$ involves only the tension and the mass per unit length: a string is non-dispersive. Correct.
Answer: (A), (D). | ||||
| 4 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctMedium | |
A medium multiple correct question on 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 Body temperature $T = 310$ K, area $A = 1\ \text{m}^2$.
(A) The energy the body itself radiates each second is $\sigma T^4 A = 460\left(\frac{310}{300}\right)^4 \approx 460 \times 1.14 \approx 525$ J. What is close to $60$ J is the net loss, $\sigma A(T^4 - T_0^4) \approx 4\sigma T_0^3 \Delta T \cdot A = 4 \times 460 \times \frac{10}{300} \approx 61$ W. The statement is about the energy radiated, so it is wrong.
(B) At the same body temperature the body radiates the same $\sigma T^4 A$; it is the energy absorbed from the cooler room that falls, by $4\sigma T_0^3\Delta T_0$ per unit area. The net loss rises by that amount, but the body does not radiate more. Wrong as stated.
(C) Radiated power is proportional to the exposed area, so curling up reduces the loss at the same body temperature. Correct.
(D) By Wien's law $\lambda_{max}T = $ constant, a hotter body peaks at a shorter wavelength. Wrong.
Answer: (C). | ||||
| 5 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Multiple correctMedium | |
A medium multiple correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Twisting the wire into a figure of eight reverses the sense of circulation in one loop relative to the other. Going round the wire, one loop is traversed clockwise and the other anticlockwise, so their emfs oppose.
At angle $\omega t$ the flux through each loop is $B \times$ area $\times \cos\omega t$. Net flux linked with the wire: $\Phi = B(2A - A)\cos\omega t = BA\cos\omega t$.
Net emf $\varepsilon = -\frac{d\Phi}{dt} = BA\omega\sin\omega t$.
(A) It depends on the difference of the areas, not the sum. Wrong.
(B) The amplitude $BA\omega$ equals that of the small loop (area $A$) alone. Correct.
(C) The emf goes as $\sin\omega t$, not $\cos\omega t$. Wrong.
(D) $|d\Phi/dt|$ is largest when $\sin\omega t = \pm1$, i.e. $\omega t = 90^\circ$: the plane of the loops is then perpendicular to the paper. Correct.
Answer: (B), (D). | ||||
| 6 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Multiple correctEasy | |
A easy multiple correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Current and voltage are in phase at resonance, $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-6}\times10^{-6}}} = 10^6$ rad s$^{-1}$.
(A) $10^4$ rad s$^{-1}$ is not the resonant frequency. Wrong.
(B) $\omega_0 = 1/\sqrt{LC}$ does not contain $R$. Correct.
(C) As $\omega \to 0$, $X_C = \frac{1}{\omega C} \to \infty$: the capacitor blocks the current. Correct.
(D) For $\omega \gg 10^6$, $X_L = \omega L$ is large and $X_C$ small: the circuit is inductive, not capacitive. Wrong.
Answer: (B), (C). | ||||
| 7 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At minimum deviation $r_1 = r_2 = \frac{A}{2}$ and $i_1 = \frac{A + \delta_m}{2} = \frac{A + A}{2} = A$.
(A) $i_1 = A$ is the minimum-deviation incidence, where the ray inside an isosceles prism runs parallel to the base. Correct.
(C) There $r_1 = \frac{A}{2} = \frac{i_1}{2}$. Correct.
(B) $\mu = \frac{\sin i_1}{\sin r_1} = \frac{\sin A}{\sin(A/2)} = 2\cos\frac{A}{2}$, so $A = 2\cos^{-1}\frac{\mu}{2}$, not $\frac{1}{2}\cos^{-1}\frac{\mu}{2}$. Wrong.
(D) The emergent ray grazes the second face when $r_2 = \theta_c$, $\sin\theta_c = \frac{1}{\mu}$. Then $r_1 = A - \theta_c$ and
$\sin i_1 = \mu\sin(A - \theta_c) = \mu\sin A\cos\theta_c - \mu\cos A\sin\theta_c = \sin A\sqrt{\mu^2 - 1} - \cos A$.
With $\mu^2 = 4\cos^2\frac{A}{2}$: $i_1 = \sin^{-1}\left[\sin A\sqrt{4\cos^2\frac{A}{2} - 1} - \cos A\right]$. Correct.
Answer: (A), (C), (D). | ||||
| 8 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Volume is conserved: $\frac{4}{3}\pi R^3 = K\cdot\frac{4}{3}\pi r^3$, so $r = R K^{-1/3}$.
New surface area $K\cdot4\pi r^2 = 4\pi R^2 K^{1/3}$, so $\Delta U = S\cdot4\pi R^2\left(K^{1/3} - 1\right)$.
$S\cdot4\pi R^2 = 0.1 \times 10^{-4} = 10^{-5}$ J, so $10^{-3} = 10^{-5}\left(K^{1/3} - 1\right)$, giving $K^{1/3} = 101 \approx 100$.
$K \approx 10^6$, so $\alpha = 6$.
Answer: 6. | ||||
| 9 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei In the Bohr model the potential energy is $V_n = -\frac{27.2\text{ eV}}{n^2} \propto -\frac{1}{n^2}$.
$\frac{V_i}{V_f} = \frac{n_f^2}{n_i^2} = 6.25$, so $\frac{n_f}{n_i} = 2.5 = \frac{5}{2}$.
Both are whole numbers, so the smallest choice is $n_i = 2$, $n_f = 5$.
Answer: 5. | ||||
| 10 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics With parallel interfaces, Snell's law gives the same $n\sin\theta$ in every layer: $n\sin\theta = 1.6 \times \sin30^\circ = 0.8$.
The ray comes out parallel to the interface between slabs $m-1$ and $m$: it grazes that interface, which means the angle of refraction into slab $m$ is $90^\circ$:
$n_m\sin90^\circ = 0.8 \Rightarrow 1.6 - 0.1m = 0.8 \Rightarrow m = 8$.
Answer: 8. | ||||
| 11 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The car is an observer approaching the source: it receives $f_1 = f_0\frac{v + u}{v}$.
It then re-emits $f_1$ as a source approaching the stationary listener: $f_2 = f_1\frac{v}{v - u} = f_0\frac{v + u}{v - u}$.
$f_2 = 492 \times \frac{332}{328} = 498$ Hz.
Beat frequency $= f_2 - f_0 = 498 - 492 = 6$ Hz.
Answer: 6. | ||||
| 12 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Decay constant $\lambda = \frac{\ln2}{8\text{ days}} = \frac{0.7}{192}$ per hour.
After $t = 11.5$ h: $\lambda t = \frac{0.7 \times 11.5}{192} \approx 0.042$, so the total activity is $A = 2.4\times10^5\,e^{-0.042} \approx 2.4\times10^5(1 - 0.042) \approx 2.3\times10^5$ Bq.
The serum is spread uniformly, so activity per ml of blood is $\frac{115}{2.5} = 46$ Bq/ml.
Blood volume $V = \frac{2.3\times10^5}{46} = 5000$ ml $= 5$ litres.
Answer: 5. | ||||
| 13 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Matching listMedium | |
A medium matching list 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 Constant velocity needs zero net force: $\vec{E} + \vec{v}\times\vec{B} = 0$ (true for either sign of charge).
(D) (II) (iii) (S): $\vec{v}\times\vec{B} = \frac{E_0}{B_0}\hat{y}\times B_0\hat{z} = E_0\hat{x}$, and $\vec{E} = -E_0\hat{x}$. They cancel: the electron goes straight along $+y$ at constant speed.
(A) and (C): the proton starts at rest in a non-zero $\vec{E}$, so it accelerates.
(B) (IV) (i) (S): $\vec{v}\times\vec{B} = 2E_0\,\hat{x}\times\hat{z} = -2E_0\hat{y}$, while $\vec{E} = E_0\hat{z}$: no cancellation.
Answer: (D). | ||||
| 14 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Matching listMedium | |
A medium matching list 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 A helix about $+z$ needs $\vec{B}$ along $z$ (circular motion in the $xy$ plane) and a drift along $+z$.
(A) (IV) (i) (S): the proton moves along $x$, perpendicular to $\vec{B} = B_0\hat{z}$, so it circles in the $xy$ plane; $\vec{E} = E_0\hat{z}$ pushes it steadily along $+z$. A helix with its axis along $+z$ (pitch growing with time). Correct.
(B) (II) (ii) (R): $\vec{v}$ is parallel to $\vec{B}$ (both along $y$), so there is no magnetic force; $\vec{E}$ is also along $y$. Straight line.
(C) (III) (iii) (P): proton from rest, $\vec{E}$ along $-x$ and $\vec{B}$ along $-x$: straight-line motion along $-x$.
(D) (IV) (ii) (R): $\vec{B}$ along $y$, so the circling is in the $xz$ plane and the electric push is along $-y$: a helix about the $-y$ direction.
Answer: (A). | ||||
| 15 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Matching listMedium | |
A medium matching list 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 (B) (III) (ii) (R): the proton starts at rest; $\vec{E} = -E_0\hat{y}$ accelerates it along $-y$. Its velocity is then along $\pm\hat{y}$, parallel to $\vec{B} = B_0\hat{y}$, so the magnetic force stays zero. Straight line along $-y$. Correct.
(D) (III) (ii) (P): same start, but $\vec{B} = -B_0\hat{x}$ is perpendicular to the velocity, so the path bends (a cycloid).
(A) (II) (iii) (Q): the electron starts along $+y$ and is deflected by $\vec{v}\times\vec{B} \neq 0$.
(C) (IV) (ii) (S): the proton starts along $+x$ in $\vec{B}$ along $z$; it does not move along a straight line.
Answer: (B). | ||||
| 16 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Matching listEasy | |
A easy matching list question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta U = \Delta Q - P\Delta V$ with a single pressure $P$ is the first law for a process at constant pressure: isobaric, (iii).
The work done on the gas is then $W = -P(V_2 - V_1) = -PV_2 + PV_1$: (II).
The diagram with 1 → 2 a horizontal line is (P).
(A) adiabatic does not fit $-P\Delta V$ at one pressure; (B) pairs isobaric with $W = 0$; (C) pairs it with the vertical (isochoric) graph (S).
Answer: (D). | ||||
| 17 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Matching listEasy | |
A easy matching list question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics An isochoric process (ii) has no volume change, so $W_{1\to2} = 0$ (III), and on a $P$–$V$ diagram it is a vertical line (S). So (III) (ii) (S) is consistent.
(A) $W = -nRT\ln\frac{V_2}{V_1}$ is isothermal work, not isochoric.
(C), (D) $W = -PV_2 + PV_1$ is isobaric work, not adiabatic.
Answer: (B). | ||||
| 18 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Matching listMedium | |
A medium matching list question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Newton took sound in a gas to be isothermal, $v = \sqrt{P/\rho}$; Laplace's correction is that the compressions and rarefactions are adiabatic, $v = \sqrt{\gamma P/\rho}$. So the process is adiabatic (iv).
Work done on the gas in an adiabatic change: $W = \Delta U = nC_V(T_2 - T_1) = \frac{1}{\gamma - 1}(P_2V_2 - P_1V_1)$: (I).
Of the curves, the adiabatic one is the steeper: in (Q) 1 → 2 drops steeply to the lower curve of the cycle, while (R) is the gentler isothermal. So (Q).
(A) and (B) pair it with isochoric; (C) pairs adiabatic with $W = 0$.
Answer: (D). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 19 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics (A) Compressing from $(p_2, V_2)$ back to $(p_1, V_1)$, the work done on the gas is $\int p_{ext}\,dV$ over the volume change. The external pressure can never usefully exceed the final pressure $p_1$ in a one-step irreversible compression, and holding it at $p_1$ throughout gives $p_1(V_2 - V_1)$, which is larger than the reversible work (the gas pressure is below $p_1$ for most of the path). Among the ways listed it is the maximum. Correct.
(B) In a free expansion (into vacuum) $w = 0$ and $q = 0$, so $\Delta U = 0$; for an ideal gas $U$ depends only on $T$, so $T$ does not change. It is adiabatic ($q = 0$) and isothermal at once. Correct.
(C) From the same $V_1$ to the same $V_2$, the reversible adiabatic curve falls below the isotherm (the gas cools), so the area under it, the work done by the gas, is smaller. Correct.
(D) (i) Reversible isothermal: $\Delta U = 0$, true. (ii) Reversible adiabatic expansion: $\Delta U = w < 0$, since the gas does work and receives no heat; $\Delta U$ is negative, not positive. Wrong.
Answer: (A), (B), (C). | ||||
| 20 | SolutionsRaoult's law | Class 12: Solutions | Multiple correctMedium | |
A medium multiple correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions The horizontal axis is $x_M$, running from $1$ at the left to $0$ at the right. At the right end $x_M = 0$, so $x_L = 1$: the point Z is the vapour pressure of pure L, $p_L^\circ$. That rules out (A) and (C).
(B) Near $x_L \to 1$, L is the solvent and its partial pressure follows Raoult's law, $p_L = x_Lp_L^\circ$: the curve meets Z along the Raoult line. Correct.
(D) Elsewhere $p_L$ lies above the straight Raoult line joining $0$ to Z: a positive deviation. L molecules escape more easily from the mixture than Raoult's law predicts, which means L–M attractions are weaker than the L–L and M–M attractions in the pure liquids. Correct.
Answer: (B), (D). | ||||
| 21 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) In $\text{HClO}_4$ chlorine has four $\sigma$ bonds and no lone pair; in HClO it has one $\sigma$ bond and three lone pairs. Four electron domains in each: $sp^3$. Correct.
(B) $\text{ClO}_4^-$ spreads its negative charge over four equivalent oxygens by resonance, while $\text{ClO}^-$ has none to share it with. The more stable anion makes $\text{HClO}_4$ the far stronger acid. Correct.
(C) Chlorine with water disproportionates to HCl and HClO: $\text{Cl}_2 + \text{H}_2\text{O} \to \text{HCl} + \text{HClO}$. $\text{HClO}_4$ is not formed. Wrong.
(D) $\text{HClO}_4$ is a stronger acid than $\text{H}_3\text{O}^+$, so its conjugate base $\text{ClO}_4^-$ is a weaker base than $\text{H}_2\text{O}$. Correct.
Answer: (A), (B), (D). | ||||
| 22 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter The colour comes from absorbing visible light to promote an electron from the highest occupied MO ($\pi^*$) to the lowest unoccupied MO ($\sigma^*$).
Down the group the atoms get larger, overlap weakens, and the $\pi^*$–$\sigma^*$ gap (the HOMO–LUMO gap) shrinks. Smaller gaps absorb longer wavelengths, so the transmitted colour moves from yellow ($\text{F}_2$) through greenish yellow, red-brown to violet ($\text{I}_2$). (C) and (D) both describe this.
(A) The physical state does not set the colour: the vapours show the same trend. Wrong.
(B) Ionisation energy refers to removing an electron, not the visible absorption. Wrong.
Answer: (C), (D). | ||||
| 23 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Multiple correctMedium | |
A medium multiple correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Pink $\text{MCl}_2\cdot6\text{H}_2\text{O}$ with $\mu = 3.87$ B.M. ($3$ unpaired electrons) is cobalt(II) chloride: X is $[\text{Co(H}_2\text{O)}_6]\text{Cl}_2$, high-spin $d^7$.
With ammonia, $\text{NH}_4\text{Cl}$ and air, Co(II) is oxidised to Co(III): Y is $[\text{Co(NH}_3)_6]\text{Cl}_3$, a 1:3 electrolyte, low-spin $d^6$ with no unpaired electrons.
With excess HCl, X gives the blue tetrahedral $[\text{CoCl}_4]^{2-}$ (Z), $d^7$ with $3$ unpaired electrons, $3.87$ B.M.
(A) All three chlorides of Y are outside the sphere: $\text{AgNO}_3$ gives three equivalents of AgCl, not two. Wrong.
(B) Low-spin Co(III) with six $\text{NH}_3$ uses inner $d$ orbitals: $d^2sp^3$. Correct.
(C) $[\text{CoCl}_4]^{2-}$ is tetrahedral. Correct.
(D) $[\text{Co(H}_2\text{O)}_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}$ is endothermic; cooling to $0^\circ$C shifts it to the left, so the solution is pink. Correct.
Answer: (B), (C), (D). | ||||
| 24 | 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 The compound is a benzene ring with $\text{CH}_3$ and Cl para to each other.
(A) As a substituted benzene, the two substituents are cited in alphabetical order (chloro before methyl), and when the locants are $1,4$ either way the first-cited group gets locant $1$: 1-chloro-4-methylbenzene. Correct.
(B) IUPAC keeps toluene as a parent name; the chlorine is then at C-4: 4-chlorotoluene. Correct.
(C) 4-methylchlorobenzene treats chlorobenzene as the parent, which is not an IUPAC parent name (chloro is only a prefix). Wrong.
(D) 1-methyl-4-chlorobenzene breaks alphabetical order and gives locant 1 to the later-cited group. Wrong.
Answer: (A), (B). | ||||
| 25 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Multiple correctMedium | |
A medium multiple correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Bromine adds to alkenes through a cyclic bromonium ion that the bromide opens from the back: anti (trans) addition.
(i) The alkene is trans-2-butene. Anti addition gives meso-2,3-dibromobutane; M and N are both this meso compound (one achiral product).
(ii) The alkene is cis-2-butene. Anti addition gives the pair of enantiomers $(2R,3R)$ and $(2S,3S)$-2,3-dibromobutane: O and P are a racemic pair.
(A) O and P are enantiomers, not identical. Wrong.
(B) The meso compound and either chiral isomer are stereoisomers that are not mirror images: diastereomers. So (M, O) and (N, P) are both diastereomeric pairs. Correct.
(C) A meso compound has no enantiomer among O and P. Wrong.
(D) Both reactions go by anti (trans) addition. Correct.
Answer: (B), (D). | ||||
| 26 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter Cell volume: $a^3 = (400 \times 10^{-10}\text{ cm})^3 = 6.4 \times 10^{-23}\text{ cm}^3$.
Mass of one cell: $8 \times 6.4\times10^{-23} = 5.12 \times 10^{-22}$ g, and an fcc cell holds $4$ atoms.
Number of cells in $256$ g: $\frac{256}{5.12\times10^{-22}} = 5 \times 10^{23}$; atoms $= 4 \times 5\times10^{23} = 2 \times 10^{24}$.
Answer: 2. | ||||
| 27 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Conductivity: $\kappa = G\frac{l}{A} = 5\times10^{-7} \times \frac{120}{1} = 6 \times 10^{-5}$ S cm$^{-1}$.
Molar conductivity: $\Lambda_m = \frac{1000\,\kappa}{c} = \frac{1000 \times 6\times10^{-5}}{0.0015} = 40$ S cm$^2$ mol$^{-1}$.
Degree of dissociation from the pH: $[\text{H}^+] = 10^{-4} = c\alpha$, so $\alpha = \frac{10^{-4}}{1.5\times10^{-3}} = \frac{1}{15}$.
$\Lambda_m^\circ = \frac{\Lambda_m}{\alpha} = 40 \times 15 = 600 = 6 \times 10^2$ S cm$^2$ mol$^{-1}$.
Answer: 6. | ||||
| 28 | 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 Count the central atom's valence electrons, adjust for charge, and subtract those used in bonds.
$[\text{TeBr}_6]^{2-}$: Te has $6 + 2 = 8$; six Te–Br bonds use $6$; $2$ left = $1$ lone pair.
$[\text{BrF}_2]^+$: Br has $7 - 1 = 6$; two bonds use $2$; $4$ left = $2$ lone pairs.
$\text{SNF}_3$: S has $6$; three S–F bonds and an S≡N triple bond use all $6$: $0$ lone pairs.
$[\text{XeF}_3]^-$: Xe has $8 + 1 = 9$; three bonds use $3$; $6$ left = $3$ lone pairs.
Sum $= 1 + 2 + 0 + 3 = 6$.
Answer: 6. | ||||
| 29 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Fill the molecular orbitals and look for unpaired electrons.
$\text{H}_2$ ($\sigma1s^2$): diamagnetic.
$\text{He}_2^+$ (3 electrons, $\sigma1s^2\sigma^*1s^1$): one unpaired, paramagnetic.
$\text{Li}_2$ ($\sigma2s^2$ beyond the core): diamagnetic.
$\text{Be}_2$ ($\sigma2s^2\sigma^*2s^2$): all paired, diamagnetic (bond order $0$).
$\text{B}_2$ ($\pi2p_x^1\pi2p_y^1$): two unpaired, paramagnetic.
$\text{C}_2$ ($\pi2p^4$): diamagnetic.
$\text{N}_2$: diamagnetic.
$\text{O}_2^-$ (17 electrons, $\pi^*2p^3$): one unpaired, paramagnetic.
$\text{F}_2$: diamagnetic.
Diamagnetic: $\text{H}_2$, $\text{Li}_2$, $\text{Be}_2$, $\text{C}_2$, $\text{N}_2$, $\text{F}_2$ = $6$.
Answer: 6. | ||||
| 30 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Aromatic: cyclic, planar, fully conjugated, with $4n + 2$ $\pi$ electrons.
1. Cyclooctatetraene: $8$ $\pi$ electrons, tub-shaped and not aromatic.
2. Cyclopropenyl anion: $4$ $\pi$ electrons, antiaromatic. No.
3. Cyclopropenyl cation: $2$ $\pi$ electrons ($n = 0$). Aromatic.
4. 1,3-Cyclohexadiene: has two $sp^3$ carbons, not fully conjugated. No.
5. Cycloheptatrienyl (tropylium) cation: $6$ $\pi$ electrons. Aromatic.
6. Cyclopentadienyl cation: $4$ $\pi$ electrons, antiaromatic. No.
7. Cyclopentadienyl anion: $6$ $\pi$ electrons. Aromatic.
8. 1,2-Dihydronaphthalene: contains an intact benzene ring. Aromatic compound.
9. Phenanthrene: $14$ $\pi$ electrons, fused benzenoid. Aromatic.
Count $= 5$.
Answer: 5. | ||||
| 31 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Matching listMedium | |
A medium matching list question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Facts used: (i) is the 1s function and (iii) the $2p_z$ function (the $\cos\theta$ and the $e^{-Zr/2a_o}$). Radial nodes $= n - l - 1$: 1s $0$, 2s $1$, $2p_z$ $0$, $3d_{z^2}$ $0$. The $xy$-plane ($\theta = 90^\circ$) is a nodal plane of $2p_z$ only; $3d_{z^2}$ has nodal cones instead. Graph (P) crosses zero once: an s function with one radial node, the 2s. At the nucleus the s orbitals have $|\psi(0)|^2 \propto Z^3/a_o^3$, which is the largest value of the density; p and d orbitals vanish there. (S): $\frac{E_{2\to4}}{E_{2\to6}} = \frac{\frac14 - \frac1{16}}{\frac14 - \frac1{36}} = \frac{3/16}{2/9} = \frac{27}{32}$, independent of $Z$, so true for every hydrogen-like species.
(A) 1s has no radial node. Wrong.
(B) $3d_{z^2}$ has no $xy$ nodal plane, and its density is zero at the nucleus. Wrong.
(C) 2s has one radial node, and (P) is its wave function. Correct.
(D) (iii) is the $2p_z$ function, but (P) has a radial node and is non-zero at $r = 0$; $2p_z$ has neither. Wrong.
Answer: (C). | ||||
| 32 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Matching listMedium | |
A medium matching list question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Facts used: (i) is the 1s function and (iii) the $2p_z$ function (the $\cos\theta$ and the $e^{-Zr/2a_o}$). Radial nodes $= n - l - 1$: 1s $0$, 2s $1$, $2p_z$ $0$, $3d_{z^2}$ $0$. The $xy$-plane ($\theta = 90^\circ$) is a nodal plane of $2p_z$ only; $3d_{z^2}$ has nodal cones instead. Graph (P) crosses zero once: an s function with one radial node, the 2s. At the nucleus the s orbitals have $|\psi(0)|^2 \propto Z^3/a_o^3$, which is the largest value of the density; p and d orbitals vanish there. (S): $\frac{E_{2\to4}}{E_{2\to6}} = \frac{\frac14 - \frac1{16}}{\frac14 - \frac1{36}} = \frac{3/16}{2/9} = \frac{27}{32}$, independent of $Z$, so true for every hydrogen-like species.
(A) 1s with its function (i), and (S) is true for hydrogen. Correct.
(B) (i) is the 1s function, not the 2s. Wrong.
(C) (P) has a radial node; the 1s function has none. Wrong.
(D) 1s is spherical and has no nodal plane. Wrong.
Answer: (A). | ||||
| 33 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Matching listMedium | |
A medium matching list question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Facts used: (i) is the 1s function and (iii) the $2p_z$ function (the $\cos\theta$ and the $e^{-Zr/2a_o}$). Radial nodes $= n - l - 1$: 1s $0$, 2s $1$, $2p_z$ $0$, $3d_{z^2}$ $0$. The $xy$-plane ($\theta = 90^\circ$) is a nodal plane of $2p_z$ only; $3d_{z^2}$ has nodal cones instead. Graph (P) crosses zero once: an s function with one radial node, the 2s. At the nucleus the s orbitals have $|\psi(0)|^2 \propto Z^3/a_o^3$, which is the largest value of the density; p and d orbitals vanish there. (S): $\frac{E_{2\to4}}{E_{2\to6}} = \frac{\frac14 - \frac1{16}}{\frac14 - \frac1{36}} = \frac{3/16}{2/9} = \frac{27}{32}$, independent of $Z$, so true for every hydrogen-like species.
(A) 1s, its function (i), density largest at the nucleus: all correct.
(B) 2s has one radial node, and its density at the nucleus is $\frac{Z^3}{8\pi a_o^3} \propto \frac{1}{a_o^3}$: correct.
(C) (iii) is the $2p_z$ function, not the 1s function: this combination is incorrect.
(D) 1s with (i), and (S) holds for $\text{He}^+$ since the ratio does not depend on $Z$: correct.
Answer: (C). | ||||
| 34 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Matching listEasy | |
A easy matching list question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Acetophenone, $\text{C}_6\text{H}_5\text{COCH}_3$, is a methyl ketone. With NaOH/$\text{Br}_2$ (hypobromite) the methyl group is tribrominated and cleaved: the haloform reaction,
$\text{C}_6\text{H}_5\text{COCH}_3 + 3\text{Br}_2 + 4\text{NaOH} \to \text{C}_6\text{H}_5\text{COONa} + \text{CHBr}_3 + 3\text{NaBr} + 3\text{H}_2\text{O}$; acidifying gives benzoic acid. So (II) (i) (S).
(B) Phenol with $\text{Br}_2$/h$\nu$ is not a condensation and gives no benzoic acid. (C) Toluene is not carboxylated by NaOH/$\text{CO}_2$ (that is the Kolbe reaction of phenol). (D) Benzaldehyde does not react with NaOH/$\text{CO}_2$ by substitution.
Answer: (A). | ||||
| 35 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Matching listMedium | |
A medium matching list question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Benzaldehyde heated with acetic anhydride and potassium acetate undergoes the Perkin condensation:
$\text{C}_6\text{H}_5\text{CHO} + (\text{CH}_3\text{CO})_2\text{O} \xrightarrow{\text{CH}_3\text{COOK}} \text{C}_6\text{H}_5\text{CH=CHCOOH} + \text{CH}_3\text{COOH}$.
It gives two different acids, cinnamic acid and acetic acid. So (III) (iii) (P).
(A) Acetophenone does not react with NaOH/$\text{CO}_2$. (B) Phenol with acetic anhydride is acetylated (an ester), not carboxylated. (D) Toluene is not a methyl ketone; no haloform reaction.
Answer: (C). | ||||
| 36 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Matching listEasy | |
A easy matching list question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\text{Br}_2$ with light ($h\nu$) splits into bromine atoms. With toluene they abstract a benzylic hydrogen, and the benzyl radical takes Br from $\text{Br}_2$: a free-radical chain substitution giving benzyl bromide, $\text{C}_6\text{H}_5\text{CH}_3 \xrightarrow{\text{Br}_2/h\nu} \text{C}_6\text{H}_5\text{CH}_2\text{Br}$. So (I) (ii) (R).
(A) Benzaldehyde with $\text{Br}_2$/h$\nu$ is not a condensation. (B) Phenol with NaOH/$\text{Br}_2$ is ionic electrophilic bromination, not carboxylation. (C) Acetophenone with acetic anhydride/acetate is not a radical substitution.
Answer: (D). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 37 | Conic SectionsHyperbola | Class 11: Conic Sections | Multiple correctEasy | |
A easy multiple correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The line is $y = 2x + 1$: slope $m = 2$, intercept $c = 1$. It touches $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ when $c^2 = a^2m^2 - b^2$:
$1 = 4a^2 - 16$, so $a^2 = \frac{17}{4}$ and $(2a)^2 = 17$.
Test each set: is the square of the largest side the sum of the other two squares?
(A) $a, 4, 1$: $16$ against $\frac{17}{4} + 1 = \frac{21}{4}$. Not right-angled.
(B) $a, 4, 2$: $16$ against $\frac{17}{4} + 4 = \frac{33}{4}$. Not right-angled.
(C) $2a, 8, 1$: $64$ against $17 + 1 = 18$. Not right-angled.
(D) $2a, 4, 1$: $17 = 16 + 1$. Right-angled.
So (A), (B), (C) cannot be sides of a right angled triangle.
Answer: (A), (B), (C). | ||||
| 38 | Conic SectionsParabola | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The chord of $y^2 = 16x$ with midpoint $(h, k)$ is $T = S_1$: $ky - 8(x + h) = k^2 - 16h$, of slope $\frac{8}{k}$.
It must have slope $-2$ (the line $2x + y = p$), so $k = -4$.
Then the chord is $-4y - 8x - 8h = 16 - 16h$, i.e. $2x + y = 2h - 4$. So $p = 2h - 4$.
For a real chord that is not a tangent the midpoint lies inside the parabola: $k^2 < 16h$, i.e. $h > 1$.
(A) $k = -4$, $h = 2$ needs $p = 0$, not $-2$. No.
(B) $k = -3$. No.
(C) $k = -4$, $h = 3$, $p = 2 \cdot 3 - 4 = 2$, and $h > 1$. Yes.
(D) $k = -3$. No.
Answer: (C). | ||||
| 39 | 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 On $[n, n + 1)$, $[x] = n$, so $\cos(\pi(x + n)) = (-1)^n\cos\pi x$ and $f(x) = (-1)^n x\cos\pi x$.
At an integer $n$: the value (from the right) is $(-1)^n \cdot n\cos n\pi = (-1)^n \cdot n(-1)^n = n$.
The left-hand limit uses the piece on $[n - 1, n)$: $(-1)^{n-1}\cdot n\cos n\pi = (-1)^{n-1}(-1)^n n = -n$.
The two agree only when $n = -n$, i.e. $n = 0$. So $f$ is continuous at $x = 0$ and discontinuous at every other integer.
At $x = -1$: right value $-1$, left limit $1$. At $x = 1$: $1$ against $-1$. At $x = 2$: $2$ against $-2$. All discontinuous.
Answer (our solve): (A), (C), (D). | ||||
| 40 | 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 Each function is continuous, so a change of sign between $x = 0$ and $x = 1$ guarantees a zero in $(0, 1)$ (intermediate value theorem). Throughout, $0 < f < 1$.
(A) $g(x) = x^9 - f(x)$: $g(0) = -f(0) < 0$ and $g(1) = 1 - f(1) > 0$. Has a zero. Correct.
(B) $g(x) = x - \int_0^{\frac{\pi}{2} - x}f(t)\cos t\,dt$: $g(0) = -\int_0^{\pi/2}f(t)\cos t\,dt < 0$; $g(1) = 1 - \int_0^{\frac{\pi}{2} - 1}f(t)\cos t\,dt > 1 - \int_0^{\frac{\pi}{2}-1}\cos t\,dt = 1 - \sin\left(\frac{\pi}{2} - 1\right) = 1 - \cos1 > 0$. Has a zero. Correct.
(C) $e^x - \int_0^x f(t)\sin t\,dt$: for $x \in (0, 1)$, $e^x > 1$ while the integral is less than $\int_0^1\sin t\,dt = 1 - \cos1 < 1$. Always positive. Wrong.
(D) $f(x) + \int_0^{\pi/2}f(t)\sin t\,dt$ is a sum of positive terms. Never zero. Wrong.
Answer: (A), (B). | ||||
| 41 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants If $A = M^2$ with $M$ real, then $\det A = (\det M)^2 \ge 0$.
(A) $I = I^2$. It is a square.
(B) $\text{diag}(1, 1, -1)$ has determinant $-1 < 0$. Not a square.
(C) $\text{diag}(1, -1, -1)$: the $2\times2$ block $-I_2$ is the square of the rotation $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, so it equals $M^2$ for $M = \begin{pmatrix}1&0&0\\0&0&1\\0&-1&0\end{pmatrix}$. It is a square.
(D) $-I_3$ has determinant $-1 < 0$. Not a square.
Answer: (B), (D). | ||||
| 42 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Multiple correctEasy | |
A easy multiple correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Since $b = a - 1$: $az + b = a(z + 1) - 1$, so $\frac{az + b}{z + 1} = a - \frac{1}{z + 1}$.
$\frac{1}{z + 1} = \frac{(x + 1) - iy}{(x + 1)^2 + y^2}$, so $Im\left(a - \frac{1}{z + 1}\right) = \frac{y}{(x + 1)^2 + y^2}$.
Setting this equal to $y$ (with $y \neq 0$): $(x + 1)^2 + y^2 = 1$, so $x = -1 \pm \sqrt{1 - y^2}$.
(A) and (B) are these two values; (C) and (D) are not.
Answer: (A), (B). | ||||
| 43 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Multiple correctEasy | |
A easy multiple correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(X \cap Y) = P(Y|X)P(X) = \frac{2}{5}\cdot\frac{1}{3} = \frac{2}{15}$.
$P(Y) = \frac{P(X \cap Y)}{P(X|Y)} = \frac{2/15}{1/2} = \frac{4}{15}$. (A) correct.
$P(X'|Y) = 1 - P(X|Y) = \frac{1}{2}$. (B) correct.
$P(X \cap Y) = \frac{2}{15}$, not $\frac{1}{5}$. (C) wrong.
$P(X \cup Y) = \frac{1}{3} + \frac{4}{15} - \frac{2}{15} = \frac{7}{15}$, not $\frac{2}{5}$. (D) wrong.
Answer: (A), (B). | ||||
| 44 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Centre $(-1, -2)$, radius$^2 = 1 + 4 + p = 5 + p$.
On the $x$-axis: $x^2 + 2x - p = 0$; on the $y$-axis: $y^2 + 4y - p = 0$.
Three common points arise in two ways.
1. The circle passes through the origin (counted once though it is on both axes) and cuts each axis again: $p = 0$. Then the points are $(0, 0)$, $(-2, 0)$, $(0, -4)$: three.
2. The circle touches one axis and cuts the other twice.
Touching the $x$-axis: radius $= 2$, $p = -1$. Then $y^2 + 4y + 1 = 0$ has two real roots: $1 + 2 = 3$ points. Works.
Touching the $y$-axis: radius $= 1$, $p = -4$. Then $x^2 + 2x + 4 = 0$ has no real root: only $1$ point. Fails.
So $p = 0$ or $p = -1$: two values.
Answer: 2. | ||||
| 45 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The integrand is a derivative: $\frac{d}{dt}\left[f(t)\,\text{cosec}\,t\right] = f'(t)\,\text{cosec}\,t - f(t)\,\text{cosec}\,t\cot t$.
So $g(x) = f\left(\frac{\pi}{2}\right)\text{cosec}\frac{\pi}{2} - \frac{f(x)}{\sin x} = 3 - \frac{f(x)}{\sin x}$.
As $x \to 0$: $\frac{f(x)}{\sin x} = \frac{f(x) - f(0)}{x}\cdot\frac{x}{\sin x} \to f'(0) \cdot 1 = 1$.
$\lim_{x\to0}g(x) = 3 - 1 = 2$.
Answer: 2. | ||||
| 46 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Infinitely many solutions need the determinant to vanish:
$\det = 1(1 - \alpha^2) - \alpha(\alpha - \alpha^3) + \alpha^2(\alpha^2 - \alpha^2) = 1 - 2\alpha^2 + \alpha^4 = (1 - \alpha^2)^2$, so $\alpha = \pm1$.
$\alpha = 1$: every equation reads $x + y + z = $ something, with right sides $1, -1, 1$: inconsistent, no solution.
$\alpha = -1$: the equations are $x - y + z = 1$, $-x + y - z = -1$, $x - y + z = 1$: all the same plane, infinitely many solutions.
So $\alpha = -1$ and $1 + \alpha + \alpha^2 = 1 - 1 + 1 = 1$.
Answer: 1. | ||||
| 47 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $x = 10!$ (all ten letters, each once).
For $y$: choose the letter that appears twice ($10$ ways); the other $8$ places hold $8$ distinct letters from the remaining $9$ ($\binom{9}{8} = 9$ ways); arrange the $10$ letters with one pair alike: $\frac{10!}{2!}$.
$y = 10 \times 9 \times \frac{10!}{2} = 45 \times 10!$.
$\frac{y}{9x} = \frac{45 \times 10!}{9 \times 10!} = 5$.
Answer: 5. | ||||
| 48 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the sides be $a - d$, $a$, $a + d$ with $d > 0$; the hypotenuse is $a + d$.
$(a + d)^2 = a^2 + (a - d)^2 \Rightarrow 4ad = a^2 \Rightarrow a = 4d$. The sides are $3d, 4d, 5d$.
Area $= \frac{1}{2}\cdot3d\cdot4d = 6d^2 = 24$, so $d = 2$ and the sides are $6, 8, 10$.
Smallest side $= 6$.
Answer: 6. | ||||
| 49 | Conic SectionsStandard forms of conics | Class 11: Conic Sections | Matching listMedium | |
A medium matching list question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections The slope-form tangents and their points of contact: circle (I): $y = mx + a\sqrt{m^2 + 1}$ (ii), touching at (Q). Ellipse (II), $\frac{x^2}{a^2} + y^2 = 1$: $y = mx + \sqrt{a^2m^2 + 1}$ (iv), at (R). Parabola (III): $y = mx + \frac{a}{m}$, i.e. $my = m^2x + a$ (i), at (P). Hyperbola (IV), $\frac{x^2}{a^2} - y^2 = 1$: $y = mx + \sqrt{a^2m^2 - 1}$ (iii), at (S).
With $a = \sqrt{2}$, the point $(-1, 1)$ lies on the circle $x^2 + y^2 = 2$ (I): $1 + 1 = 2$. (It is not on $x^2 + 2y^2 = 2$, on $y^2 = 4\sqrt2x$, or on $x^2 - 2y^2 = 2$.)
The tangent there is $-x + y = 2$, i.e. $y = x + 2$, slope $m = 1$: $y = mx + a\sqrt{m^2 + 1} = x + \sqrt2\cdot\sqrt2 = x + 2$ (ii), and (Q) gives $\left(\frac{-\sqrt2}{\sqrt2}, \frac{\sqrt2}{\sqrt2}\right) = (-1, 1)$.
So (I) (ii) (Q).
Answer: (B). | ||||
| 50 | Conic SectionsStandard forms of conics | Class 11: Conic Sections | Matching listMedium | |
A medium matching list question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections The slope-form tangents and their points of contact: circle (I): $y = mx + a\sqrt{m^2 + 1}$ (ii), touching at (Q). Ellipse (II), $\frac{x^2}{a^2} + y^2 = 1$: $y = mx + \sqrt{a^2m^2 + 1}$ (iv), at (R). Parabola (III): $y = mx + \frac{a}{m}$, i.e. $my = m^2x + a$ (i), at (P). Hyperbola (IV), $\frac{x^2}{a^2} - y^2 = 1$: $y = mx + \sqrt{a^2m^2 - 1}$ (iii), at (S).
$(8, 16)$ lies on $y^2 = 4ax$ with $a = 8$: $256 = 32 \times 8$. The parabola's tangent of slope $m$ is $y = mx + \frac{a}{m}$; with $m = 1$, $y = x + 8$. Written as in (i): $my = m^2x + a$. Point of contact $\left(\frac{a}{m^2}, \frac{2a}{m}\right) = (8, 16)$, which is (P).
Check the circle: $8^2 + 16^2 = 320$ would need $a = \sqrt{320}$, and then $y = x + a\sqrt2 \ne x + 8$. So (III) (i) (P).
Answer: (C). | ||||
| 51 | Conic SectionsStandard forms of conics | Class 11: Conic Sections | Matching listMedium | |
A medium matching list question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections The slope-form tangents and their points of contact: circle (I): $y = mx + a\sqrt{m^2 + 1}$ (ii), touching at (Q). Ellipse (II), $\frac{x^2}{a^2} + y^2 = 1$: $y = mx + \sqrt{a^2m^2 + 1}$ (iv), at (R). Parabola (III): $y = mx + \frac{a}{m}$, i.e. $my = m^2x + a$ (i), at (P). Hyperbola (IV), $\frac{x^2}{a^2} - y^2 = 1$: $y = mx + \sqrt{a^2m^2 - 1}$ (iii), at (S).
Try the ellipse (II), $\frac{x^2}{a^2} + y^2 = 1$: $\frac{3}{a^2} + \frac{1}{4} = 1$ gives $a^2 = 4$. Its tangent at $\left(\sqrt3, \frac12\right)$ is $\frac{\sqrt3x}{4} + \frac{y}{2} = 1$, i.e. $\sqrt3x + 2y = 4$. It matches.
Slope $m = -\frac{\sqrt3}{2}$: $a^2m^2 + 1 = 4 \cdot \frac34 + 1 = 4$, and $y = mx + \sqrt{a^2m^2 + 1} = -\frac{\sqrt3}{2}x + 2$: the same line (iv).
Point of contact (R): $\left(\frac{-4\cdot(-\sqrt3/2)}{2}, \frac{1}{2}\right) = \left(\sqrt3, \frac12\right)$.
(For the hyperbola (IV), $\frac{3}{a^2} - \frac14 = 1$ gives $a^2 = \frac{12}{5}$, whose tangent there is $\frac{5\sqrt3}{12}x - \frac{y}{2} = 1$, not the given line.)
So (II) (iv) (R).
Answer: (D). | ||||
| 52 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Matching listMedium | |
A medium matching list question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 1 + \frac1x - \ln x - 1 = \frac1x - \ln x$ and $f''(x) = -\frac{1}{x^2} - \frac1x < 0$ for all $x > 0$.
Column 1: (I) $f(1) = 1 > 0$, $f(e^2) = e^2 + 2 - 2e^2 = 2 - e^2 < 0$: a zero in $(1, e^2)$, true. (II) $f'(1) = 1 > 0$, $f'(e) = \frac1e - 1 < 0$: true. (III) on $(0, 1)$, $\frac1x > 0$ and $-\ln x > 0$, so $f' > 0$: false. (IV) $f'' < 0$ everywhere: false.
Column 2: $f(x) = x(1 - \ln x) + \ln x \to -\infty$, so (i) false and (ii) true. $f'(x) = \frac1x - \ln x \to -\infty$: (iii) true. $f''(x) \to 0$: (iv) true.
Column 3: (P) $f' > 0$ on $(0, 1)$: true. (Q) for $x > e$, $\ln x > 1 > \frac1x$, so $f' < 0$: true. (R) $f'' < 0$, so $f'$ is decreasing, not increasing: false. (S) true.
(A) has (i), false. (B) (II), (ii), (Q) are all true. (C) has (III) and (R), false. (D) has (IV), false.
Answer: (B). | ||||
| 53 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Matching listMedium | |
A medium matching list question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 1 + \frac1x - \ln x - 1 = \frac1x - \ln x$ and $f''(x) = -\frac{1}{x^2} - \frac1x < 0$ for all $x > 0$.
Column 1: (I) $f(1) = 1 > 0$, $f(e^2) = e^2 + 2 - 2e^2 = 2 - e^2 < 0$: a zero in $(1, e^2)$, true. (II) $f'(1) = 1 > 0$, $f'(e) = \frac1e - 1 < 0$: true. (III) on $(0, 1)$, $\frac1x > 0$ and $-\ln x > 0$, so $f' > 0$: false. (IV) $f'' < 0$ everywhere: false.
Column 2: $f(x) = x(1 - \ln x) + \ln x \to -\infty$, so (i) false and (ii) true. $f'(x) = \frac1x - \ln x \to -\infty$: (iii) true. $f''(x) \to 0$: (iv) true.
Column 3: (P) $f' > 0$ on $(0, 1)$: true. (Q) for $x > e$, $\ln x > 1 > \frac1x$, so $f' < 0$: true. (R) $f'' < 0$, so $f'$ is decreasing, not increasing: false. (S) true.
(A) has (R), false. (B) (II), (iii), (S) are all true. (C) has (III), false. (D) has (IV) and (i), false.
Answer: (B). | ||||
| 54 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Matching listMedium | |
A medium matching list question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 1 + \frac1x - \ln x - 1 = \frac1x - \ln x$ and $f''(x) = -\frac{1}{x^2} - \frac1x < 0$ for all $x > 0$.
Column 1: (I) $f(1) = 1 > 0$, $f(e^2) = e^2 + 2 - 2e^2 = 2 - e^2 < 0$: a zero in $(1, e^2)$, true. (II) $f'(1) = 1 > 0$, $f'(e) = \frac1e - 1 < 0$: true. (III) on $(0, 1)$, $\frac1x > 0$ and $-\ln x > 0$, so $f' > 0$: false. (IV) $f'' < 0$ everywhere: false.
Column 2: $f(x) = x(1 - \ln x) + \ln x \to -\infty$, so (i) false and (ii) true. $f'(x) = \frac1x - \ln x \to -\infty$: (iii) true. $f''(x) \to 0$: (iv) true.
Column 3: (P) $f' > 0$ on $(0, 1)$: true. (Q) for $x > e$, $\ln x > 1 > \frac1x$, so $f' < 0$: true. (R) $f'' < 0$, so $f'$ is decreasing, not increasing: false. (S) true.
(A) (I), (iii), (P): all true. (B) (II), (iv), (Q): all true. (D) (II), (iii), (P): all true. (C) (III), (i), (R) are all false: the incorrect combination.
Answer: (C). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The mass is fixed: $M = \frac{4}{3}\pi R^3\rho$, so $\rho \propto R^{-3}$.
Take logarithms and differentiate: $\ln\rho = \text{const} - 3\ln R$, so $\frac{1}{\rho}\frac{d\rho}{dt} = -\frac{3}{R}\frac{dR}{dt}$.
This is a constant, say $-k$. Then $v = \frac{dR}{dt} = \frac{k}{3}R$, so $v \propto R$.
Answer: (A). | ||||
| 2 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion When the polygon rests on a side, its centre of mass is at the inradius from the ground: $h$ is the distance from the centre to the middle of a side.
While it tips over the leading vertex, the centre is at a fixed distance from that vertex: the circumradius $R_c$. It is highest when it is directly above the vertex, at height $R_c$.
In a regular $n$-gon the side subtends $\frac{2\pi}{n}$ at the centre, so the half-angle is $\frac{\pi}{n}$ and $h = R_c\cos\frac{\pi}{n}$, i.e. $R_c = \frac{h}{\cos(\pi/n)}$.
So $\Delta = R_c - h = h\left(\frac{1}{\cos(\pi/n)} - 1\right)$.
Answer: (B). | ||||
| 3 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation for the fastest electron: $K = \frac{hc}{\lambda} - \phi_0$, and $\lambda_d = \frac{h}{\sqrt{2mK}}$, so $K = \frac{h^2}{2m\lambda_d^2}$.
So $\frac{h^2}{2m\lambda_d^2} = \frac{hc}{\lambda} - \phi_0$. Differentiate both sides:
$-\frac{h^2}{m\lambda_d^3}\,d\lambda_d = -\frac{hc}{\lambda^2}\,d\lambda$
$\frac{d\lambda_d}{d\lambda} = \frac{mc}{h}\cdot\frac{\lambda_d^3}{\lambda^2}$.
So $\frac{\Delta\lambda_d}{\Delta\lambda} \propto \frac{\lambda_d^3}{\lambda^2}$.
Answer: (D). | ||||
| 4 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The six-pointed star is the outline of two equilateral triangles. Opposite outer vertices are $4a$ apart, so each outer vertex is $2a$ from the centre: that is the circumradius of each triangle, so each triangle's side lies at perpendicular distance $a$ (the inradius) from the centre.
Each side of a triangle has length $2\sqrt3\,a$ and is cut into three equal parts of $\frac{2a}{\sqrt3}$; the middle part is inside the star. So the loop is made of 12 straight pieces, each running along a line at distance $a$ from the centre, from a point $\frac{a}{\sqrt3}$ from the foot of the perpendicular (an inner vertex) to a point $\sqrt3\,a$ from it (an outer vertex).
Seen from the centre, those ends are at angles $\tan^{-1}\frac{1}{\sqrt3} = 30^\circ$ and $\tan^{-1}\sqrt3 = 60^\circ$ from the perpendicular. Field of one piece:
$B_1 = \frac{\mu_0 I}{4\pi a}(\sin60^\circ - \sin30^\circ) = \frac{\mu_0 I}{4\pi a}\cdot\frac{\sqrt3 - 1}{2}$.
All twelve pieces circulate the same way round the centre, so their fields add:
$B = 12B_1 = \frac{\mu_0 I}{4\pi a}\,6(\sqrt3 - 1)$.
Answer: (A). | ||||
| 5 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\vec R = \vec Q - \vec P$ runs from the tip $P$ to the tip $Q$. The point $S$ lies on it at distance $b|\vec R|$ from $P$, so $\overrightarrow{PS} = b\vec R$.
Then $\vec S = \vec P + b\vec R = \vec P + b(\vec Q - \vec P) = (1 - b)\vec P + b\vec Q$.
Answer: (A). | ||||
| 6 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation To escape the system the rocket's total energy at launch must be at least zero:
$\frac{1}{2}mv_S^2 = \frac{GM_em}{R_e} + \frac{GM_sm}{r}$, with $M_s = 3\times10^5M_e$ and $r = 2.5\times10^4R_e$.
$\frac{GM_s}{r} = \frac{3\times10^5}{2.5\times10^4}\cdot\frac{GM_e}{R_e} = 12\frac{GM_e}{R_e}$.
So $v_S^2 = 2\cdot13\frac{GM_e}{R_e} = 13v_e^2$ and $v_S = \sqrt{13}\times11.2 \approx 40.4\ \text{km s}^{-1}$, closest to $42\ \text{km s}^{-1}$.
Answer: (B). | ||||
| 7 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The measured time is the fall plus the sound's return: $T = \sqrt{\frac{2L}{g}} + \frac{L}{v}$.
$\frac{dT}{dL} = \frac{1}{\sqrt{2gL}} + \frac{1}{v} = \frac{1}{\sqrt{2\cdot10\cdot20}} + \frac{1}{300} = \frac{1}{20} + \frac{1}{300} = \frac{16}{300}$ s per m.
So $\delta L = \frac{\delta T}{dT/dL} = 0.01\times\frac{300}{16} = 0.1875$ m, and $\frac{\delta L}{L} = \frac{0.1875}{20} \approx 0.94\%$, closest to $1\%$.
Answer: (B). | ||||
| 8 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctMedium | |
A medium multiple correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter In the field the particle moves on a circle of radius $r = \frac{p}{QB}$. With $\vec v$ along $+x$ and $\vec B$ into the page, $Q\vec v\times\vec B$ points along $+y$, so it curves upward; the centre of the circle is at $(0, -R + r)$, on the $y$-axis.
(A) The path reaches at most $x = r$. It turns back into region 1 (a semicircle) if $r < \frac{3R}{2}$, i.e. $B > \frac{2p}{3QR}$. True.
(B) For the circle to pass through $P_2 = \left(\frac{3R}{2}, 0\right)$: $\left(\frac{3R}{2}\right)^2 + (R - r)^2 = r^2$, so $\frac{9R^2}{4} + R^2 = 2Rr$ and $r = \frac{13R}{8}$, i.e. $B = \frac{8p}{13QR}$. The circle then meets $x = \frac{3R}{2}$ at $y = 0$ and $y = \frac{5R}{4}$; moving up from $P_1$ it reaches $y = 0$ first, so it leaves through $P_2$. True.
(C) The longest path that still returns is the semicircle with $r = \frac{3R}{2}$. At the farthest point from the $y$-axis the momentum has turned from $p\hat i$ to $p\hat j$; the change has magnitude $|p\hat j - p\hat i| = \sqrt2\,p$, not $\frac{p}{\sqrt2}$. False.
(D) The re-entry point is a diameter away: $2r = \frac{2mv}{QB}$, proportional to the mass, not inversely. False.
Answer: (A), (B). | ||||
| 9 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Multiple correctEasy | |
A easy multiple correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Two sinusoids of amplitude $V_0$ that differ in phase by $\phi$ have a difference of amplitude $2V_0\sin\frac{\phi}{2}$.
For $X$ and $Y$, $\phi = \frac{2\pi}{3}$: amplitude $2V_0\sin60^\circ = \sqrt3V_0$, rms $\frac{\sqrt3V_0}{\sqrt2} = V_0\sqrt{\frac32}$.
For $Y$ and $Z$ the phase difference is again $\frac{2\pi}{3}$, so the reading is the same, $V_0\sqrt{\frac32}$ (and likewise for $Z$ and $X$).
(A) True. (B) False: it is $V_0\sqrt{3/2}$. (C) False. (D) True: every pair gives the same reading.
Answer: (A), (D). | ||||
| 10 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The charge sits just above the top of the dome, on its axis, at height $R$ above the centre of the flat face.
The closed surface (dome + flat face) contains no charge, so the total flux is $0$: (B) is false.
Flux through the flat disc: seen from the charge, the disc (radius $R$ at distance $R$) subtends a cone of half-angle $45^\circ$, solid angle $\Omega = 2\pi(1 - \cos45^\circ)$. The field lines through it leave the closed surface (outward normal of the flat face points away from the charge), so $\phi_{flat} = \frac{Q}{4\pi\epsilon_0}\Omega = \frac{Q}{2\epsilon_0}\left(1 - \frac{1}{\sqrt2}\right)$.
Then $\phi_{curved} = 0 - \phi_{flat} = -\frac{Q}{2\epsilon_0}\left(1 - \frac{1}{\sqrt2}\right)$: (A) is true.
(C) The normal component $\frac{kQ\cos\theta}{d^2}$ changes across the disc as the distance $d$ and angle $\theta$ change: false.
(D) Every point of the rim is at the same distance $\sqrt2R$ from the charge, so the rim is an equipotential: true.
Answer: (A), (D). | ||||
| 11 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a circle much larger than $d$, a point at angle $\theta$ from $P_1$ (the perpendicular bisector) has path difference $\Delta x = d\sin\theta$.
At $P_2$ (on the line of the sources) $\Delta x = d = 1.8\times10^{-3}$ m $= \frac{1.8\times10^{-3}}{6\times10^{-7}}\lambda = 3000\lambda$, a whole number of wavelengths: a bright spot, and the largest path difference anywhere, so the highest order.
(A) False: $P_2$ is bright. (B) True.
(C) From $P_1$ ($\Delta x = 0$) to $P_2$ ($\Delta x = 3000\lambda$) the order runs from 0 to 3000, so about 3000 fringes: true.
(D) Bright spots at $d\sin\theta = n\lambda$, so $d\cos\theta\,\Delta\theta = \lambda$ and $\Delta\theta = \frac{\lambda}{d\cos\theta}$, which grows as $\theta$ goes towards $90^\circ$ (towards $P_2$): false.
Answer: (B), (C). | ||||
| 12 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Multiple correctEasy | |
A easy multiple correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The ideal inductors are in parallel, so they always have the same voltage: $L_1\frac{dI_1}{dt} = L_2\frac{dI_2}{dt}$. Both currents start at zero, so integrating, $L_1I_1 = L_2I_2$ at every instant: the ratio $\frac{I_1}{I_2} = \frac{L_2}{L_1}$ is fixed. (C) true.
After a long time the inductors carry steady current with no voltage, so the total is $\frac{V}{R}$, shared in the ratio $I_1 : I_2 = L_2 : L_1$:
$I_1 = \frac{V}{R}\frac{L_2}{L_1 + L_2}$, $I_2 = \frac{V}{R}\frac{L_1}{L_1 + L_2}$. (A) and (B) true.
(D) At $t = 0$ an inductor's current cannot jump, so the current through $R$ is $0$, not $\frac{V}{R}$. False.
Answer: (A), (B), (C). | ||||
| 13 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Multiple correctMedium | |
A medium multiple correct 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 floor is frictionless, so the only forces are gravity and the vertical normal reaction: no horizontal force acts and the bar starts from rest.
(A) The centre of mass has no horizontal acceleration and no initial horizontal velocity, so the midpoint falls straight down. True.
(B) Put the midpoint on the line $x = 0$. At angle $\theta$ the midpoint is at height $\frac{L}{2}\cos\theta$ and the lower end is on the floor, so the top end A is at $x = -\frac{L}{2}\sin\theta$, $y = L\cos\theta$: $\frac{x^2}{(L/2)^2} + \frac{y^2}{L^2} = 1$, an ellipse, not a parabola. False.
(C) About the contact point the normal reaction has no moment; gravity acts at the midpoint, horizontally $\frac{L}{2}\sin\theta$ away: $\tau = mg\frac{L}{2}\sin\theta \propto \sin\theta$. True.
(D) The midpoint starts at height $\frac{L}{2}$ and is at $\frac{L}{2}\cos\theta$: it has moved $\frac{L}{2}(1 - \cos\theta)$. True.
Answer: (A), (C), (D). | ||||
| 14 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Multiple correctHard | |
A hard multiple correct 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 step has height $R$, so its corner $Q$ is level with the wheel's centre $C$, a distance $R$ to its right; $P$ (the leftmost point), $C$ and $Q$ lie on one horizontal line, $S$ is the top and $X$ is on the lower left. While climbing, the wheel turns about the corner $Q$.
(C) A force normal to the rim at $P$ acts along the radius $PC$, i.e. along the line $PCQ$, which passes through $Q$. Its moment about $Q$ is zero, so $\tau = 0$. True. (This force also cannot lift the wheel.)
(A) A tangential force at $P$ is vertical, with lever arm $PQ = 2R$ about $Q$. Since $P$ and $Q$ are both points of the wheel, which turns about $Q$, the distance $PQ$ and the force's line relative to it stay the same while it keeps acting tangentially at $P$: its torque $2RF$ does not decrease (and the weight's opposing torque $MgR\cos\phi$ falls as the wheel rises through angle $\phi$, so the net torque grows). False.
(B) A normal force at $X$ acts through the centre; its own moment about $Q$ keeps its value, but the weight's moment about $Q$, $MgR\cos\phi$, changes as the wheel climbs, so the torque about $Q$ is not constant. False on the key's reading.
(D) A tangential (horizontal) force at the top $S$ has lever arm $R$ about $Q$ (the vertical height of $S$ above $Q$), so $\tau = FR \neq 0$, turning the wheel up the step; with $F > Mg$ it climbs.
The official key gives (C), and also accepts (C) and (D) together.
Answer: (C). | ||||
| 15 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | ComprehensionEasy | |
A easy comprehension question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charging to $V_0$ moves charge $q = CV_0$ through the source, which does work $qV_0 = CV_0^2$.
The capacitor stores $E_C = \frac{1}{2}CV_0^2$, and the rest is dissipated in $R$: $E_D = CV_0^2 - \frac{1}{2}CV_0^2 = \frac{1}{2}CV_0^2$.
So $E_C = E_D$ (whatever the value of $R$).
Answer: (A). | ||||
| 16 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | ComprehensionMedium | |
A medium comprehension question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In a step where a source of voltage $V_2$ charges the capacitor from $V_1$ to $V_2$:
work by the source $= C(V_2 - V_1)V_2$; increase in stored energy $= \frac{1}{2}C(V_2^2 - V_1^2)$;
heat $= C(V_2 - V_1)V_2 - \frac{1}{2}C(V_2 - V_1)(V_2 + V_1) = \frac{1}{2}C(V_2 - V_1)^2$.
Each of the three steps raises the voltage by $\frac{V_0}{3}$, so each dissipates $\frac{1}{2}C\frac{V_0^2}{9}$, and in all
$E_D = 3\cdot\frac{1}{2}C\frac{V_0^2}{9} = \frac{1}{3}\left(\frac{1}{2}CV_0^2\right)$.
Answer: (C). | ||||
| 17 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | ComprehensionHard | |
A hard comprehension question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The ring's centre goes round the axis of the cone on a circle of radius $R - r$ (Figure 2: the small circle of radius $r$ touches the ring internally). Taking the centre to go round at $\omega_0$, its speed is $v = \omega_0(R - r)$.
If the ring rolls without slipping on the small circle as a fixed track, the ring's point in contact is at rest, so the ring spins about its centre at $\omega = \frac{v}{R} = \frac{\omega_0(R - r)}{R}$.
Then $KE = \frac{1}{2}Mv^2 + \frac{1}{2}(MR^2)\omega^2 = \frac{1}{2}M\omega_0^2(R - r)^2 + \frac{1}{2}M\omega_0^2(R - r)^2 = M\omega_0^2(R - r)^2$, which is option (C).
But the paragraph also says the finger itself goes round at $\omega_0$, so the contact point moves at $\omega_0r$ rather than being at rest. Matching the ring's rim to the moving fingertip instead gives a spin of $\omega_0$ and $KE = \frac{1}{2}M\omega_0^2\left[(R - r)^2 + R^2\right]$, which is none of the options. The question is ambiguous, and the official key awards full marks to all. | ||||
| 18 | Laws of MotionFriction | Class 11: Laws of Motion | ComprehensionMedium | |
A medium comprehension question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion The ring's centre moves on a circle of radius $R - r$ at angular speed $\omega_0$. The only horizontal force on the ring is the finger's normal push $N$ on the inner rim, which supplies the centripetal force:
$N = M\omega_0^2(R - r)$.
The weight is held up by friction at the finger, at most $\mu N$:
$\mu M\omega_0^2(R - r) \ge Mg$, so $\omega_0 \ge \sqrt{\frac{g}{\mu(R - r)}}$.
Below this the ring slips down.
Answer: (A). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 19 | SolutionsColligative properties | Class 12: Solutions | Single correctEasy | |
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions Moles of ethanol $= \frac{34.5}{46} = 0.75$; molality $= \frac{0.75}{0.5} = 1.5$ mol kg$^{-1}$.
$\Delta T_f = K_f m = 2\times1.5 = 3$ K, so the solution freezes at $273 - 3 = 270$ K.
On the plot: vapour pressure rises with temperature for ice, water and the solution alike; the solution's curve lies below pure water's (lower vapour pressure), and the freezing point is where a liquid's curve meets the ice curve: 273 K for water and 270 K for the solution.
(A) and (D) draw the liquids' vapour pressure falling with temperature, which is wrong; (B) puts the new freezing point at 271 K. Only (C) has rising curves, the solution below water, and the meeting with the ice curve at 270 K.
Answer: (C). | ||||
| 20 | ElectrochemistryNernst equation | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, with $n = 2$.
Nernst: $E = E^o - \frac{2.303RT}{2F}\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 1.1 - \frac{2.303RT}{2F}\log10 = 1.1 - \frac{2.303RT}{2F}$.
$\Delta G = -nFE = -2F\left(1.1 - \frac{2.303RT}{2F}\right) = 2.303RT - 2.2F$.
Answer: (B). | ||||
| 21 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics For graphite $\rightarrow$ diamond at 1 bar, $\Delta G^o = 2900$ J mol$^{-1}$. At constant temperature $\left(\frac{\partial\Delta G}{\partial P}\right)_T = \Delta V = -2\times10^{-6}$ m$^3$ mol$^{-1}$ (solids: $\Delta V$ taken as constant).
At equilibrium $\Delta G = 0$:
$0 = 2900 + (-2\times10^{-6})(P - 10^5)$, so $P - 10^5 = \frac{2900}{2\times10^{-6}} = 1.45\times10^9$ Pa $= 14500$ bar.
$P = 14500 + 1 = 14501$ bar.
Answer: (A). | ||||
| 22 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Concentrated nitric acid makes iron passive (a thin oxide film); no hydrogen.
(B) Copper reduces concentrated HNO$_3$ to NO$_2$: $\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}$; no hydrogen.
(C) Zinc is amphoteric and dissolves in alkali with hydrogen: $\text{Zn} + 2\text{NaOH} + 2\text{H}_2\text{O} \rightarrow \text{Na}_2[\text{Zn(OH)}_4] + \text{H}_2\uparrow$.
(D) Gold dissolves in cyanide only because air's oxygen oxidises it: $4\text{Au} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{Au(CN)}_2]^- + 4\text{OH}^-$; no hydrogen.
Answer: (C). | ||||
| 23 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions With H = +1 and O = −2:
H$_3$PO$_2$: $3 + x - 4 = 0$, $x = +1$.
H$_3$PO$_3$: $3 + x - 6 = 0$, $x = +3$.
H$_4$P$_2$O$_6$: $4 + 2x - 12 = 0$, $x = +4$.
H$_3$PO$_4$: $3 + x - 8 = 0$, $x = +5$.
Order: H$_3$PO$_4$ (+5) > H$_4$P$_2$O$_6$ (+4) > H$_3$PO$_3$ (+3) > H$_3$PO$_2$ (+1).
Answer: (C). | ||||
| 24 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Step i: NaNO$_2$/HCl at 0 °C diazotises the aromatic amine, giving the aryl diazonium chloride on the ring that carried NH$_2$.
Step ii: in aqueous NaOH the OH becomes phenoxide, which strongly activates its ring at the positions ortho and para to it. The diazonium group, held right beside that ring by the biaryl bond, couples with it intramolecularly (azo coupling), closing a new six-membered ring that contains N=N: a benzo[c]cinnoline.
The ring can close at either carbon ortho to the biaryl bond on the phenol ring. One is para to the oxygen and unhindered; the other lies between the biaryl bond and the OH. Coupling para to the phenoxide is preferred, which leaves the OH on the far side of the new ring from the N=N, as in (C).
(A) would need a Sandmeyer reaction (CuCl); (B) is only the diazonium salt and phenoxide, not the product after coupling; (D) is coupling at the crowded position ortho to OH.
Answer: (C). | ||||
| 25 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines All four are bases because of an imine-type nitrogen; what matters is how well the cation formed on protonation is stabilised by resonance.
IV (guanidine): the guanidinium ion spreads the positive charge over three equivalent nitrogens. Strongest (pK$_a$ of the conjugate acid about 13.6).
I (acetamidine): the amidinium ion shares the charge over two equivalent nitrogens (pK$_a$ about 12.4).
II (2-imidazoline, a cyclic amidine): also an amidinium on protonation, but with less substitution help and ring constraint; a little weaker than I (pK$_a$ about 11).
III (imidazole): the protonated ring is aromatic either way, but the nitrogen's lone pair is sp$^2$ in an aromatic ring and the ring is electron-withdrawing; much weaker (pK$_a$ about 7).
Order: IV > I > II > III.
Answer: (C). | ||||
| 26 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter (A) Adsorbed molecules lose freedom, so $\Delta S < 0$; for adsorption to be spontaneous $\Delta G = \Delta H - T\Delta S < 0$ needs $\Delta H < 0$. True.
(B) A gas with a higher critical temperature is more easily liquefied, its intermolecular forces are stronger, and it is adsorbed more. Ethane (563 K) is adsorbed more than nitrogen (126 K). True.
(C) A cloud is liquid droplets dispersed in a gas: an aerosol, not an emulsion (an emulsion is liquid in liquid). False.
(D) Brownian motion is more vigorous for smaller particles and less viscous media: it depends on size as well. False.
Answer: (A), (B). | ||||
| 27 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\ln K = -\frac{\Delta G^o}{RT} = \frac{\Delta S_{sys}^o}{R} - \frac{\Delta H^o}{RT}$, and the surroundings' entropy change is $\Delta S_{surr} = -\frac{\Delta H}{T}$. The system's entropy term does not depend on $T$; the temperature effect comes entirely through $\Delta S_{surr}$.
Exothermic ($\Delta H < 0$): $\Delta S_{surr} = \frac{|\Delta H|}{T}$ is favourable, and it shrinks as $T$ rises, so $K$ decreases. (D) true; (A) gives the wrong reason (the system's entropy change).
Endothermic ($\Delta H > 0$): $\Delta S_{surr} = -\frac{\Delta H}{T}$ is unfavourable, and its size falls as $T$ rises, so $K$ increases. (B) true; (C) gives the wrong reason.
Answer: (B), (D). | ||||
| 28 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | Multiple correctMedium | |
A medium multiple correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The steric factor is $P = \frac{A_{\text{observed}}}{A_{\text{predicted}}}$, comparing the experimental pre-exponential factor with the one predicted by collision theory for the Arrhenius expression $k = PZe^{-E_a/RT}$.
(A) $P$ multiplies the pre-exponential term only; the activation energy is a separate quantity. True.
(B) $P = 4.5 > 1$, so the experimental frequency factor is higher than the predicted one. True.
(C) $P$ says nothing about whether a catalyst is needed; the reaction proceeds at its rate. False.
(D) The reverse of (B). False.
Answer: (A), (B). | ||||
| 29 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Multiple correctHard | |
A hard multiple correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes I is benzyl bromide, II (bromomethyl)cyclohexane, III tert-butyl bromide, IV 1-bromo-1-phenylethane.
(A) Benzyl bromide gives the resonance-stabilised benzyl cation and tert-butyl bromide a tertiary cation: both can react by S$_N$1. True.
(B) Benzyl bromide is primary and unhindered, and its S$_N$2 transition state is helped by the ring; (bromomethyl)cyclohexane is a primary halide: both react by S$_N$2. True.
(C) IV is a chiral secondary benzylic bromide; under S$_N$2 conditions the nucleophile attacks from the back and the configuration is inverted. True.
(D) The order depends on the conditions: for S$_N$1 the tertiary III reacts faster than the primary I; for S$_N$2 the hindered tertiary III is slowest but primary I is faster than secondary IV. "IV > I > III" is not right for either mechanism in general.
The official key accepts (A), (B), (C) and also (A), (B), (C), (D).
Answer: (A), (B), (C). | ||||
| 30 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) In Al$_2$(CH$_3$)$_6$ two methyl groups bridge the aluminium atoms; a methyl has no lone pair, so each Al–C–Al bridge is a three-centre two-electron bond. True.
(B) In diborane B$_2$H$_6$ the two B–H–B bridges are three-centre two-electron "banana" bonds. True.
(C) In Al$_2$Cl$_6$ each bridging chlorine donates a lone pair: the bridges are ordinary two-electron dative bonds (three-centre four-electron), not 3c–2e. False.
(D) Boron is small and its BCl$_3$ is monomeric with a vacant p orbital; aluminium chloride relieves its electron deficiency by dimerising. BCl$_3$ is the stronger Lewis acid (as the key holds). True.
Answer: (A), (B), (D). | ||||
| 31 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Amphoteric oxides react with both acids and bases.
(A) Cr$_2$O$_3$, BeO, SnO and SnO$_2$ are all amphoteric. True.
(B) CrO (chromium in +2) is basic. False.
(C) NO is neutral and B$_2$O$_3$ is acidic. False.
(D) ZnO, Al$_2$O$_3$, PbO and PbO$_2$ are all amphoteric. True.
Answer: (A), (D). | ||||
| 32 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Multiple correctMedium | |
A medium multiple correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Q, C$_8$H$_8$O, gives Cannizzaro but not haloform: an aldehyde with no α-hydrogen and no CH$_3$CO group, i.e. a methylbenzaldehyde (CH$_3$C$_6$H$_4$CHO).
S, C$_8$H$_8$O, gives haloform but not Cannizzaro: a methyl ketone, which with 8 carbons is acetophenone C$_6$H$_5$COCH$_3$.
So P must be an alkene that ozonolyses to a tolualdehyde (ArCH= with Ar a tolyl group), and R one that gives acetophenone (C$_6$H$_5$C(CH$_3$)= with an unsubstituted phenyl).
(A) 4-Methylstyrene → 4-methylbenzaldehyde + HCHO; α-methylstyrene → acetophenone + HCHO. Suitable.
(B) 1-(3-methylphenyl)propene → 3-methylbenzaldehyde + CH$_3$CHO; C$_6$H$_5$C(CH$_3$)=C(CH$_3$)$_2$ → acetophenone + acetone. Suitable.
(C) The first gives 2-methylbenzaldehyde (fine), but the second, 2-CH$_3$C$_6$H$_4$C(CH$_3$)=CHCH$_3$, gives 2-methylacetophenone, C$_9$H$_{10}$O. Not suitable.
(D) The second, 4-CH$_3$C$_6$H$_4$C(CH$_3$)=CH$_2$, gives 4-methylacetophenone, C$_9$H$_{10}$O. Not suitable.
Answer: (A), (B). | ||||
| 33 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | ComprehensionEasy | |
A easy comprehension question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $2\text{KClO}_3 \xrightarrow{\text{MnO}_2,\ \Delta} 2\text{KCl} + 3\text{O}_2$: W is O$_2$.
In excess oxygen white phosphorus burns to the higher oxide: $\text{P}_4 + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}$ (P$_4$O$_6$ forms only with limited oxygen). X is P$_4$O$_{10}$.
Answer: (C). | ||||
| 34 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | ComprehensionEasy | |
A easy comprehension question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter P$_4$O$_{10}$ is a powerful dehydrating agent. It removes water from pure nitric acid, giving its anhydride, and takes up that water as metaphosphoric acid:
$4\text{HNO}_3 + \text{P}_4\text{O}_{10} \rightarrow 2\text{N}_2\text{O}_5 + 4\text{HPO}_3$.
Y is N$_2$O$_5$ and Z is HPO$_3$.
Answer: (B). | ||||
| 35 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | ComprehensionHard | |
A hard comprehension question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons P is ethyl 3-(4-tert-butylphenyl)propanoate, Ar–CH$_2$CH$_2$–CO$_2$Et.
P → Q: excess CH$_3$MgBr adds twice to the ester, giving the tertiary alcohol Ar–CH$_2$CH$_2$–C(CH$_3$)$_2$OH.
Q → R: cold H$_2$SO$_4$ ionises it to the tertiary carbocation Ar–CH$_2$CH$_2$–C$^+$(CH$_3$)$_2$, which attacks its own ring at the position ortho to the side chain (intramolecular Friedel–Crafts alkylation), closing a five-membered ring: 1,1-dimethyl-6-tert-butylindane (gem-dimethyl next to the ring fusion, tert-butyl on the benzene ring).
R → S: CH$_3$COCl/AlCl$_3$ puts an acetyl group on the benzene ring (Friedel–Crafts acylation), at a free position away from the bulky tert-butyl group.
Only (A) keeps the indane skeleton with the gem-dimethyl group, the tert-butyl group and one COCH$_3$ on the ring. (B) and (C) have a C=C (dehydration products), and (D) has a sulfonic acid group and an open chain.
Answer: (A). | ||||
| 36 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | ComprehensionMedium | |
A medium comprehension question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Q is the tertiary alcohol Ar–CH$_2$CH$_2$–C(CH$_3$)$_2$OH. In cold H$_2$SO$_4$ it loses water to give a tertiary carbocation, which at once alkylates its own aromatic ring to close the indane ring: Q → R is an (intramolecular) Friedel–Crafts alkylation. No alkene is left in R and no sulfonation occurs at 0 °C.
R → S with CH$_3$COCl/AlCl$_3$ is Friedel–Crafts acylation.
Answer: (D). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 37 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A plane perpendicular to both given planes contains both their normals, so its normal is their cross product:
$(2, 1, -2)\times(3, -6, -2) = \big(1\cdot(-2) - (-2)(-6),\ (-2)(3) - 2(-2),\ 2(-6) - 1\cdot3\big) = (-14, -2, -15)$, i.e. $(14, 2, 15)$.
Through $(1, 1, 1)$: $14x + 2y + 15z = 14 + 2 + 15 = 31$.
Answer: (C). | ||||
| 38 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Write $\vec p, \vec q, \vec r, \vec s$ for the position vectors. The first equality:
$\vec p\cdot\vec q + \vec r\cdot\vec s = \vec r\cdot\vec p + \vec q\cdot\vec s$
$\Rightarrow \vec p\cdot(\vec q - \vec r) - \vec s\cdot(\vec q - \vec r) = 0 \Rightarrow (\vec p - \vec s)\cdot(\vec q - \vec r) = 0$.
So $SP \perp QR$. In the same way the second equality gives $(\vec q - \vec s)\cdot(\vec r - \vec p) = 0$, i.e. $SQ \perp RP$.
$S$ lies on two altitudes, so it is the orthocentre.
Answer: (D). | ||||
| 39 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $dy = \frac{dx}{8\sqrt x\,\sqrt{9 + \sqrt x}\,\sqrt{4 + \sqrt{9 + \sqrt x}}}$.
Let $u = \sqrt{4 + \sqrt{9 + \sqrt x}}$. By the chain rule
$\frac{du}{dx} = \frac{1}{2u}\cdot\frac{1}{2\sqrt{9 + \sqrt x}}\cdot\frac{1}{2\sqrt x}$, which is exactly the right-hand side.
So $dy = du$ and $y = \sqrt{4 + \sqrt{9 + \sqrt x}} + c$.
$y(0) = \sqrt{4 + 3} + c = \sqrt7 + c = \sqrt7$, so $c = 0$.
$y(256) = \sqrt{4 + \sqrt{9 + 16}} = \sqrt{4 + 5} = 3$.
Answer: (A). | ||||
| 40 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives By the mean value theorem on $\left[\frac12, 1\right]$ there is $c \in \left(\frac12, 1\right)$ with
$f'(c) = \frac{f(1) - f(1/2)}{1 - 1/2} = \frac{1/2}{1/2} = 1$.
$f'' > 0$ means $f'$ is strictly increasing, and $1 > c$, so $f'(1) > f'(c) = 1$.
Answer: (D). | ||||
| 41 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The $(j, j)$ entry of $M^TM$ is the sum of squares of column $j$ of $M$, so the trace of $M^TM$ is the sum of the squares of all nine entries.
Each square is $0$, $1$ or $4$, and they must add to $5$:
one entry $2$, one entry $1$, seven $0$: $9\times8 = 72$ matrices;
five entries $1$, four $0$: $\binom{9}{5} = 126$ matrices.
Total $72 + 126 = 198$.
Answer: (B). | ||||
| 42 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Every five-element subset of $S$ has some number $k$ of odd elements. $S$ has only four even numbers, so $k = 0$ is impossible; every subset is counted in exactly one $N_k$ with $1 \le k \le 5$.
So the sum is the number of all five-element subsets: $\binom{9}{5} = 126$.
Answer: (D). | ||||
| 43 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability All nonnegative solutions of $x + y + z = 10$: $\binom{12}{2} = 66$, taken as equally likely.
For a given $z$, $x + y = 10 - z$ has $11 - z$ solutions. For $z = 0, 2, 4, 6, 8, 10$: $11 + 9 + 7 + 5 + 3 + 1 = 36$.
Probability $= \frac{36}{66} = \frac{6}{11}$.
Answer: (B). | ||||
| 44 | 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 By the Leibniz rule, $g'(x) = \sin^{-1}(\sin2x)\cdot2\cos2x - \sin^{-1}(\sin x)\cdot\cos x$.
At $x = \frac{\pi}{2}$: $\sin2x = \sin\pi = 0$, so the first term is $\sin^{-1}(0)\cdot2\cos\pi = 0$; and $\cos\frac{\pi}{2} = 0$ kills the second term. So $g'\left(\frac{\pi}{2}\right) = 0$.
At $x = -\frac{\pi}{2}$: likewise $\sin(-\pi) = 0$ and $\cos\left(-\frac{\pi}{2}\right) = 0$, so $g'\left(-\frac{\pi}{2}\right) = 0$.
Both derivatives are $0$, so none of the four options is true. The official key awards full marks to all. | ||||
| 45 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Multiple correctMedium | |
A medium multiple correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Solve for $\cos\beta$: $\cos\beta(2 + \cos\alpha) = 1 + 2\cos\alpha$, so $\cos\beta = \frac{1 + 2\cos\alpha}{2 + \cos\alpha}$.
Then $\tan^2\frac{\beta}{2} = \frac{1 - \cos\beta}{1 + \cos\beta} = \frac{(2 + \cos\alpha) - (1 + 2\cos\alpha)}{(2 + \cos\alpha) + (1 + 2\cos\alpha)} = \frac{1 - \cos\alpha}{3(1 + \cos\alpha)} = \frac{1}{3}\tan^2\frac{\alpha}{2}$.
So $\tan\frac{\alpha}{2} = \pm\sqrt3\tan\frac{\beta}{2}$: for a given pair one of (A) and (C) holds, depending on the signs of the two tangents, while (B) and (D) (with $\sqrt3$ on the wrong angle) are false in general.
Since the condition fixes only $\tan^2$, neither (A) nor (C) is true for every admissible pair, and the half-angle tangents are undefined when $\alpha$ or $\beta$ is an odd multiple of $\pi$. The official key awards full marks to all. | ||||
| 46 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Multiple correctEasy | |
A easy multiple correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $h(x) = e^{-2x}f(x)$. Then $h'(x) = e^{-2x}(f'(x) - 2f(x)) > 0$, so $h$ is increasing, and $h(0) = 1$.
For $x > 0$: $h(x) > 1$, i.e. $f(x) > e^{2x}$. (C) true.
Then $f'(x) > 2f(x) > 2e^{2x} > 0$ on $(0, \infty)$: $f$ is increasing, (A) true and (B) false; and $f'(x) > 2e^{2x} > e^{2x}$, so (D) is false.
Answer: (A), (C). | ||||
| 47 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctMedium | |
A medium multiple correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $x < 1$: $|1 - x| = 1 - x$, so the numerator is $1 - x(2 - x) = (1 - x)^2$ and $f(x) = (1 - x)\cos\frac{1}{1 - x}$. Since $|\cos| \le 1$ and $1 - x \to 0$, $\lim_{x\to1^-}f(x) = 0$. (A) true, (B) false.
For $x > 1$: $|1 - x| = x - 1$, the numerator is $1 - x\cdot x = 1 - x^2$, and $f(x) = \frac{(1 - x)(1 + x)}{x - 1}\cos\frac{1}{1 - x} = -(1 + x)\cos\frac{1}{1 - x}$. As $x \to 1^+$, $-(1 + x) \to -2$ while $\cos\frac{1}{1 - x}$ keeps oscillating between $-1$ and $1$, so the limit does not exist. (C) false, (D) true.
Answer: (A), (D). | ||||
| 48 | 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 Expand along the first row:
$f = \cos2x(\cos^2x + \sin^2x) - \cos2x(-\cos^2x + \sin^2x) + \sin2x(-\sin x\cos x - \sin x\cos x)$
$= \cos2x + \cos2x\cdot\cos2x - \sin^22x = \cos2x + \cos4x$.
$f'(x) = -2\sin2x - 4\sin4x = -2\sin2x(1 + 4\cos2x)$.
In $(-\pi, \pi)$: $\sin2x = 0$ at $x = -\frac{\pi}{2}, 0, \frac{\pi}{2}$; and $\cos2x = -\frac14$ for four values of $2x$ in $(-2\pi, 2\pi)$, so four more values of $x$. Seven zeros in all: (B) true, (A) false.
$f(x) = \cos2x + \cos4x \le 2$, with $f(0) = 2$: the maximum is at $x = 0$. (C) true, (D) false.
Answer: (B), (C). | ||||
| 49 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Area of $R$: $\int_0^1(x - x^3)\,dx = \frac12 - \frac14 = \frac14$.
Half of it to the left of $x = \alpha$: $\int_0^\alpha(x - x^3)\,dx = \frac{\alpha^2}{2} - \frac{\alpha^4}{4} = \frac18$, i.e. $2\alpha^4 - 4\alpha^2 + 1 = 0$. (C) true; (D) is a different equation, false.
$\alpha^2 = \frac{4 - \sqrt{8}}{4} = 1 - \frac{1}{\sqrt2} \approx 0.293$ (the other root exceeds 1), so $\alpha \approx 0.541$, which lies in $\left(\frac12, 1\right)$. (B) true, (A) false.
Answer: (B), (C). | ||||
| 50 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $[k, k+1]$, $x + 1 \ge k + 1$, so $\frac{k+1}{x(x+1)} \le \frac{1}{x}$ (strictly except at one point). Hence
$I < \int_1^{99}\frac{dx}{x} = \log_e99$. (B) true, (A) false.
Also $x \le k + 1$, so $\frac{k+1}{x(x+1)} \ge \frac{1}{x+1}$. Hence
$I > \int_1^{99}\frac{dx}{x+1} = \log_e\frac{100}{2} = \log_e50 \approx 3.9 > \frac{49}{50}$. (D) true, (C) false.
Answer: (B), (D). | ||||
| 51 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | ComprehensionEasy | |
A easy comprehension question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\overrightarrow{OX}$ and $\overrightarrow{OY}$ are unit vectors along $\overrightarrow{QR}$ and $\overrightarrow{RP}$. These two sides meet at $R$, but $\overrightarrow{QR}$ points into $R$ and $\overrightarrow{RP}$ out of it, so the angle between the vectors is $\pi - R$.
$|\overrightarrow{OX}\times\overrightarrow{OY}| = \sin(\pi - R) = \sin R = \sin(\pi - P - Q) = \sin(P + Q)$.
Answer: (A). | ||||
| 52 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | ComprehensionEasy | |
A easy comprehension question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $P + Q = \pi - R$, so $\cos(P + Q) = -\cos R$, and likewise for the others: the sum is $-(\cos P + \cos Q + \cos R)$.
In a triangle $\cos P + \cos Q + \cos R = 1 + 4\sin\frac{P}{2}\sin\frac{Q}{2}\sin\frac{R}{2} \le \frac32$, with equality for the equilateral triangle.
So the minimum of the given sum is $-\frac32$.
Answer: (B). | ||||
| 53 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | ComprehensionEasy | |
A easy comprehension question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha^2 = \alpha + 1$, so $\alpha^{12} = \alpha^{11} + \alpha^{10}$; likewise $\beta^{12} = \beta^{11} + \beta^{10}$.
Multiply by $p$ and $q$ and add: $a_{12} = a_{11} + a_{10}$.
Answer: (B). | ||||
| 54 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | ComprehensionMedium | |
A medium comprehension question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations By the recurrence $a_{n+2} = a_{n+1} + a_n$: $a_2 = a_1 + a_0$, $a_3 = 2a_1 + a_0$, $a_4 = 3a_1 + 2a_0$.
With $\alpha, \beta = \frac{1 \pm \sqrt5}{2}$: $a_0 = p + q$, $a_1 = \frac{p + q}{2} + \frac{(p - q)\sqrt5}{2}$.
$a_4 = \frac{3(p + q)}{2} + \frac{3(p - q)\sqrt5}{2} + 2(p + q) = \frac{7(p + q)}{2} + \frac{3(p - q)}{2}\sqrt5 = 28$.
By the FACT ($a = \frac{7(p+q)}{2} - 28$, $b = \frac{3(p - q)}{2}$ rational): $p - q = 0$ and $p + q = 8$, so $p = q = 4$.
$p + 2q = 4 + 8 = 12$.
Answer: (D). | ||||
Sit JEE Advanced 2017 free.
Timed like the exam or chapter by chapter, with a solution for each question. Run by a non-profit trust.