JEE (Advanced) · 2016
JEE Advanced 2016, 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
- Optics5
- Atoms and Nuclei5
- Properties of Bulk Matter4
8 easy · 23 medium · 5 hard
Chemistry
36 Q
- Aldehydes, Ketones and Carboxylic Acids4
- Coordination Compounds3
- Chemical Kinetics2
12 easy · 21 medium · 3 hard
Mathematics
36 Q
- Matrices and Determinants4
- Limits, Continuity and Differentiability4
- Integral Calculus4
7 easy · 24 medium · 5 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 12: Atoms; Class 12: Nuclei
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 11: Systems of Particles and Rotational Motion
Class 12: Current Electricity
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 12: Dual Nature of Radiation and Matter
Class 11: Oscillations; Class 11: Waves
Class 11: Units and Measurement
Class 11: Laws of Motion
Class 11: Thermodynamics
No matching NCERT chapter
Class 12: Aldehydes, Ketones and Carboxylic Acids
Class 12: Coordination Compounds
Class 12: Chemical Kinetics
No matching NCERT chapter
Class 11: Chemical Bonding and Molecular Structure
No matching NCERT chapter
Class 12: Amines
Class 12: Solutions
Class 12: Biomolecules
Class 11: Equilibrium
Class 11: Structure of Atom
Class 11: Thermodynamics
Class 11: Classification of Elements and Periodicity in Properties
No matching NCERT chapter
Class 12: Alcohols, Phenols and Ethers
Class 11: Some Basic Concepts of Chemistry
Class 12: The d- and f-Block Elements
Class 11: Organic Chemistry – Some Basic Principles and Techniques
Class 12: Electrochemistry
No matching NCERT chapter
No matching NCERT chapter
No matching NCERT chapter
Class 11: Hydrocarbons
Class 12: Matrices; Class 12: Determinants
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 11: Probability; Class 12: Probability
Class 11: Conic Sections
Class 12: Application of Derivatives
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 12: Differential Equations
Class 11: Conic Sections
Class 11: Complex Numbers and Quadratic Equations
Class 11: Complex Numbers and Quadratic Equations
Class 11: Permutations and Combinations
Class 11: Binomial Theorem
Class 11: Sequences and Series
Class 12: Vector Algebra
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 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation: $eV_0 = \frac{hc}{\lambda} - \phi$. Subtracting it for two wavelengths removes the work function: $e(V_1 - V_2) = hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$.
From 0.3 $\mu$m and 0.4 $\mu$m: $\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \left(\frac{1}{0.3} - \frac{1}{0.4}\right) \times 10^6 = \frac{5}{6} \times 10^6$ m$^{-1}$, and $V_1 - V_2 = 1.0$ V.
$h = \frac{e\,\Delta V}{c\,\Delta(1/\lambda)} = \frac{1.6 \times 10^{-19} \times 1.0}{3 \times 10^8 \times \frac{5}{6} \times 10^6} = \frac{1.6 \times 10^{-19}}{2.5 \times 10^{14}} = 6.4 \times 10^{-34}$ J s.
Check with 0.4 and 0.5 $\mu$m: $\Delta(1/\lambda) = 0.5 \times 10^6$ m$^{-1}$, $\Delta V = 0.6$ V, giving $h = \frac{1.6 \times 10^{-19} \times 0.6}{1.5 \times 10^{14}} = 6.4 \times 10^{-34}$ J s again.
Answer: (B). | ||||
| 2 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The stick leans on the top edge of the wall, making $30^\circ$ with the vertical (so $60^\circ$ with the floor). Forces: weight $mg = 16$ N at the middle; the wall's reaction $N$ perpendicular to the stick at the edge; at the floor a normal reaction $R$ (vertical) and friction $f$ (horizontal).
The normal to a stick inclined at $30^\circ$ to the vertical makes $30^\circ$ with the horizontal, so $N$ has a horizontal part $N\cos 30^\circ$ and an upward part $N\sin 30^\circ = N/2$.
Vertical balance: $R + \frac{N}{2} = 16$, and with $R = N$ (given): $\frac{3N}{2} = 16$, $N = \frac{32}{3}$ N.
Horizontal balance: $f = N\cos 30^\circ = \frac{32}{3}\cdot\frac{\sqrt{3}}{2} = \frac{16\sqrt{3}}{3}$ N.
Torques about the bottom end: the edge of the wall is at a distance $\frac{h}{\cos 30^\circ} = \frac{2h}{\sqrt{3}}$ along the stick, and the weight's lever arm is $\frac{l}{2}\sin 30^\circ = \frac{l}{4}$.
$16 \cdot \frac{l}{4} = \frac{32}{3}\cdot\frac{2h}{\sqrt{3}}$, so $4l = \frac{64h}{3\sqrt{3}}$ and $\frac{h}{l} = \frac{12\sqrt{3}}{64} = \frac{3\sqrt{3}}{16}$.
Answer: (D). | ||||
| 3 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Over 3 hours the device puts in $Q = 3000 \times 3 \times 3600 = 3.24 \times 10^7$ J of heat.
The stored water (120 kg) may warm from 10 °C up to 30 °C at most, so it can soak up $120 \times 4200 \times 20 = 1.008 \times 10^7$ J by itself.
The cooler must remove the rest within 3 hours: $P \times 10800 \ge 3.24 \times 10^7 - 1.008 \times 10^7 = 2.232 \times 10^7$ J.
$P_{min} = \frac{2.232 \times 10^7}{10800} \approx 2067$ W.
Answer: (B). | ||||
| 4 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Refraction at PQ: $\sin\alpha = \sqrt{2}\sin r$. With $\alpha = 45^\circ$: $\sin r = \frac{1}{2}$, $r = 30^\circ$.
Critical angle: $\sin C = \frac{1}{\sqrt{2}}$, $C = 45^\circ$.
Take PQ vertical with the ray inside travelling at $30^\circ$ below the horizontal (the normal to PQ). Face PR leaves P at angle $\theta$ to PQ, so its normal makes angle $\theta$ with the horizontal. The angle of incidence on PR is then $r + \theta = 30^\circ + \theta$.
A smaller $\alpha$ gives a smaller $r$ and a smaller angle on PR, so the least $\alpha$ for total internal reflection is the one that makes the angle on PR exactly critical: $30^\circ + \theta = 45^\circ$, $\theta = 15^\circ$.
Answer: (A). | ||||
| 5 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity The material lets the line charge leak out radially to the shell. Let $\lambda(t)$ be the charge per length still on the axis. By Gauss's law the field at radius $r$ is $E = \frac{\lambda(t)}{2\pi\varepsilon r}$, and Ohm's law gives $j = \sigma E = \frac{\sigma\lambda(t)}{2\pi\varepsilon r}$.
The current per unit length leaving the axis is $j \cdot 2\pi r = \frac{\sigma}{\varepsilon}\lambda$, and it drains the line charge: $\frac{d\lambda}{dt} = -\frac{\sigma}{\varepsilon}\lambda$, so $\lambda = \lambda_0 e^{-\sigma t/\varepsilon}$.
Hence $j(t) = \frac{\sigma\lambda_0}{2\pi\varepsilon r}e^{-\sigma t/\varepsilon}$: largest at $t = 0$ and decaying exponentially to zero. That is graph (C). Not linear (A), not flat at first (B), and it does not level off at a non-zero value (D), because the charge leaks away completely.
Answer: (C). | ||||
| 6 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Multiple correctMedium | |
A medium multiple correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Bohr model: $r_n \propto \frac{n^2}{Z}$, $E_n \propto -\frac{Z^2}{n^2}$, $L_n = \frac{nh}{2\pi}$.
Radii: $\frac{r_{n+1} - r_n}{r_n} = \frac{(n+1)^2 - n^2}{n^2} = \frac{2n+1}{n^2} \approx \frac{2}{n}$. $Z$ cancels, so (A) is true, and it varies as $1/n$, so (B) is true.
Energy: $\frac{E_{n+1} - E_n}{E_n} = \frac{1/(n+1)^2 - 1/n^2}{1/n^2}$, of magnitude $\frac{2n+1}{(n+1)^2} \approx \frac{2}{n}$. It varies as $1/n$, not $1/n^3$ (it is the absolute change $\Delta E$ that goes as $1/n^3$). (C) is false.
Angular momentum: $\frac{L_{n+1} - L_n}{L_n} = \frac{1}{n}$. (D) is true.
Answer: (A), (B), (D). | ||||
| 7 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | Multiple correctHard | |
A hard multiple correct question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The car is always equidistant from $M$ and $N$, so its velocity component towards each speaker is the same, $u\cos\phi$, where $u = 60$ km/h $= \frac{50}{3}$ m/s and $\phi$ is the angle between the car's path and the line to a speaker.
For an observer moving towards a source, $f' = f\left(1 + \frac{u\cos\phi}{c}\right)$. So the beat frequency is $\nu = (121 - 118)\left(1 + \frac{u\cos\phi}{c}\right) = 3\left(1 + \frac{u\cos\phi}{c}\right)$, with $\cos\phi$ taken positive while approaching and negative while receding.
(A) At $P$ the car approaches with $\cos\phi = \frac{1800}{\sqrt{1800^2 + 10^2}}$; at $R$ it recedes with the same $|\cos\phi|$; at $Q$, $\cos\phi = 0$. So $\nu_P = 3(1 + k)$, $\nu_R = 3(1 - k)$, $\nu_Q = 3$, and $\nu_P + \nu_R = 2\nu_Q$. True.
(B) $\cos\phi = \frac{x}{\sqrt{x^2 + 10^2}}$, with $x$ the distance from $Q$. Far away it barely changes; near $Q$ it swings from $+1$ to $-1$ over a few tens of metres. The rate of change of $\nu$ is greatest at $Q$. True.
(C) $\nu$ stays nearly constant at its high value far before $Q$, drops steeply through $\nu_Q$ at $Q$, and stays nearly constant at the low value afterwards: the S-shaped curve of plot (C). True.
(D) shows the change slowest at $Q$, the opposite of (B). False.
Answer: (A), (B), (C). | ||||
| 8 | Current ElectricityResistivity | Class 12: Current Electricity | Multiple correctMedium | |
A medium multiple correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Tungsten evaporates unevenly, so some sections become thinner than others.
(A) A thinner section has a larger resistance ($R = \rho l/A$) and, carrying the same current, dissipates more heat ($I^2R$) per unit length; it runs hotter. The temperature is not uniform. False.
(B) As the wire thins, $A$ falls and the resistance of each small section rises with time (and hotter tungsten also has a larger resistivity). False.
(C) Just before it breaks, the thin section is at a very high temperature. By Wien's law the peak of black-body emission moves to higher frequencies, so more light is emitted at the higher band of frequencies. True.
(D) The total resistance grows as the filament thins, and at constant voltage $P = V^2/R$ falls. True.
Answer: (C), (D). | ||||
| 9 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Faint image: the convex surface acts as a convex mirror. A convex mirror always forms a virtual image behind it, so with $u = -30$ cm, $v = +10$ cm: $\frac{1}{f_m} = \frac{1}{v} + \frac{1}{u} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}$, so $f_m = 15$ cm and $R = 2f_m = 30$ cm. (B) is false; the image is virtual and erect, so (C) is false.
Lens: a magnified real image of twice the size has $m = -2$, so $v = 60$ cm. $\frac{1}{f} = \frac{1}{60} + \frac{1}{30} = \frac{1}{20}$, $f = 20$ cm. (D) is true.
Lens maker's formula for a plano-convex lens: $\frac{1}{f} = \frac{n - 1}{R}$, so $\frac{1}{20} = \frac{n - 1}{30}$, $n = 2.5$. (A) is true.
(If the doubled image were taken as virtual, $m = +2$, $v = -60$ cm, then $f = 60$ cm and $n = 1.5$; neither matches an option, so the real image is the one meant.)
Answer: (A), (D). | ||||
| 10 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Multiple correctMedium | |
A medium multiple correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement From Coulomb's law, $\frac{q^2}{\varepsilon\,r}$ is an energy, so $\left[\frac{q^2}{\varepsilon}\right] = $ energy $\times$ length. $k_BT$ is an energy, and $[n] = L^{-3}$.
So $\left[\frac{q^2}{\varepsilon k_B T}\right] = L$.
(A) $\frac{nq^2}{\varepsilon k_BT}$ has $L^{-3}\cdot L = L^{-2}$; its square root is $L^{-1}$. Wrong.
(B) $\frac{\varepsilon k_BT}{nq^2}$ has $L^{-1}\cdot L^{3} = L^2$; its square root is a length. Correct.
(C) $\frac{q^2}{\varepsilon n^{2/3}k_BT}$ has $L \cdot L^{2} = L^3$; its square root is $L^{3/2}$. Wrong.
(D) $\frac{q^2}{\varepsilon n^{1/3}k_BT}$ has $L \cdot L = L^2$; its square root is a length. Correct.
Answer: (B), (D). | ||||
| 11 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Multiple correctHard | |
A hard multiple correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Mutual inductance: find the flux of the wire's field through the triangle. At distance $x$ below the wire the triangle (a right isosceles triangle with its right angle on the wire) is $2x$ wide, so
$\Phi = \int_0^{h}\frac{\mu_0 I}{2\pi x}\,2x\,dx = \frac{\mu_0 I h}{\pi}$, giving $M = \frac{\mu_0 h}{\pi}$ with $h = 0.1$ m.
(A) emf in the wire $= M\frac{dI}{dt} = \frac{\mu_0 \times 0.1}{\pi} \times 10 = \frac{\mu_0}{\pi}$ V. True.
(B) The wire's field depends only on distance from the wire, so turning the loop about the wire leaves the linked flux unchanged: no additional emf. False.
(C) Let the hypotenuse current run left to right (counterclockwise loop with the wire above). A rightward current in the wire puts a field into the page below it, opposite to the loop's own outward normal, so the mutual flux enters with a minus sign; the emf induced in the wire, $-\left(-M\frac{dI}{dt}\right)$, then drives current to the right: the same direction as the hypotenuse current, not opposite. False.
(D) The two legs, which are next to the wire, carry current with a horizontal component to the left, antiparallel to the wire's induced current; antiparallel currents repel, and these near sides outweigh the farther hypotenuse. This is Lenz's law: the force acts to reduce the growing linked flux, pushing them apart. True.
Answer: (A), (D). | ||||
| 12 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Multiple correctEasy | |
A easy multiple correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion At $t = 1$ s: $\vec{r} = \frac{10}{3}\hat{i} + 5\hat{j}$ m.
$\vec{v} = 3\alpha t^2\hat{i} + 2\beta t\hat{j} = 10\hat{i} + 10\hat{j}$ m/s. (A) true.
$\vec{a} = 6\alpha t\hat{i} + 2\beta\hat{j} = 20\hat{i} + 10\hat{j}$ m/s$^2$, so $\vec{F} = m\vec{a} = 2\hat{i} + \hat{j}$ N, not $\hat{i} + 2\hat{j}$. (C) false.
$\vec{L} = m(\vec{r}\times\vec{v}) = 0.1\left(\frac{10}{3}\cdot 10 - 5 \cdot 10\right)\hat{k} = 0.1 \times \left(-\frac{50}{3}\right)\hat{k} = -\frac{5}{3}\hat{k}$ N m s. (B) true.
$\vec{\tau} = \vec{r}\times\vec{F} = \left(\frac{10}{3}\cdot 1 - 5 \cdot 2\right)\hat{k} = -\frac{20}{3}\hat{k}$ N m. (D) true.
Answer: (A), (B), (D). | ||||
| 13 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Treat the slab as many thin layers. Snell's law at every boundary keeps $n\sin\theta$ the same all the way down: $n_1\sin\theta_i = n(z)\sin\theta(z) = n_2\sin\theta_f$. (A) true, so (B) is false.
The lateral displacement $l$ is built up inside the slab: $l = \int_0^d \tan\theta(z)\,dz$, where $\sin\theta(z) = \frac{n_1\sin\theta_i}{n(z)}$. It depends on $n(z)$ (D true) and on $n_1\sin\theta_i$, but not on $n_2$, which only sets the direction of the ray after it has left the slab (C true).
Answer: (A), (C), (D). | ||||
| 14 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter By the Stefan–Boltzmann law with constant emissivity, $P \propto T^4$.
$T_1 = 487 + 273 = 760$ K and $T_2 = 2767 + 273 = 3040$ K $= 4T_1$, so $\frac{P_2}{P_1} = 4^4 = 256 = 2^8$.
$\log_2\frac{P_2}{P_0} = \log_2\frac{P_1}{P_0} + \log_2\frac{P_2}{P_1} = 1 + 8 = 9$.
Answer: 9. | ||||
| 15 | 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 With atomic masses, the $\beta^-$ decay $^{12}\mathrm{B} \to {}^{12}\mathrm{C} + e^- + \bar{\nu}$ releases $Q = (m_B - m_C)c^2$ (the electron masses cancel). The mass of $^{12}\mathrm{C}$ is exactly 12.000 u.
$Q = (12.014 - 12.000) \times 931.5 = 0.014 \times 931.5 = 13.041$ MeV to the ground state.
To the excited state, 4.041 MeV of this stays in the carbon nucleus: $Q^* = 13.041 - 4.041 = 9.000$ MeV.
The $\beta$-particle has its maximum kinetic energy when the antineutrino takes almost nothing (the heavy nucleus' recoil is negligible): $K_{max} \approx 9$ MeV.
Answer: 9. | ||||
| 16 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Photon energy: $E = \frac{hc}{e\lambda} = \frac{1.237 \times 10^{-6}}{970 \times 10^{-10}} \approx 12.75$ eV.
After absorbing it the atom has energy $-13.6 + 12.75 = -0.85$ eV $= -\frac{13.6}{16}$ eV, which is the level $n = 4$.
From $n = 4$ the electron can fall by any route to $n = 1$; the number of distinct lines is $\frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6$.
Answer: 6. | ||||
| 17 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Terminal velocity (Stokes' law): $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$, so $v_t \propto \frac{r^2(\rho - \sigma)}{\eta}$.
P: $r = 0.5$ cm, $\rho - \sigma = 8 - 0.8 = 7.2$, $\eta = 3$: $\frac{0.25 \times 7.2}{3} = 0.6$.
Q: $r = 0.25$ cm, $\rho - \sigma = 8 - 1.6 = 6.4$, $\eta = 2$: $\frac{0.0625 \times 6.4}{2} = 0.2$.
$\frac{v_P}{v_Q} = \frac{0.6}{0.2} = 3$.
Answer: 3. | ||||
| 18 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Each branch is across the battery directly, so the three branches act independently. An inductor's current starts at zero and grows to its steady value $V/r$.
At $t = 0$ only the resistor carries current: $I_{min} = \frac{5}{12}$ A.
In the steady state the inductors are plain resistances: $I_{max} = \frac{5}{3} + \frac{5}{4} + \frac{5}{12} = \frac{20 + 15 + 5}{12} = \frac{40}{12}$ A.
$\frac{I_{max}}{I_{min}} = \frac{40/12}{5/12} = 8$.
Answer: 8. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 19 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom The probability of finding the electron in the thin shell is $P = \psi^2 \cdot 4\pi r^2 dr$.
For 1$s$, $\psi^2 \propto e^{-2r/a_0}$: largest at the nucleus and falling off exponentially. The factor $r^2$ is zero at the nucleus and grows.
So $P \propto r^2 e^{-2r/a_0}$ is zero at $r = 0$, rises to one maximum (at $r = a_0$, the Bohr radius), and then decays smoothly towards zero without ever reaching it at a finite $r$. There is no radial node for 1$s$.
(A) is the shape of $\psi^2$ alone; (B) falls to zero at a finite distance; (C) is non-zero at the nucleus and has an extra dip.
Answer: (D). | ||||
| 20 | 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 Work done by the gas against the constant external pressure: $P_{ext}\Delta V = 3.0 \times (2.0 - 1.0) = 3.0$ L atm $= 303.9$ J.
For an isothermal change of an ideal gas $\Delta U = 0$, so the gas takes in $q = 303.9$ J of heat from the surroundings.
The surroundings (a large reservoir at 300 K) lose this heat reversibly: $\Delta S_{surr} = \frac{-q}{T} = \frac{-303.9}{300} = -1.013$ J K$^{-1}$.
(The value 5.763 is $nR\ln 2$, the entropy change of the gas itself, which is not asked.)
Answer: (C). | ||||
| 21 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Down a group atomic radius normally increases, but gallium is an exception. Between Al and Ga the ten 3$d$ electrons are added; $d$ electrons shield the nucleus poorly, so the effective nuclear charge felt by the outer electrons rises and Ga (about 135 pm) comes out slightly smaller than Al (about 143 pm).
In (about 167 pm) and Tl (about 170 pm) follow in the usual order; Tl is only a little larger than In because of the poorly shielding 4$f$ electrons.
Order: Ga < Al < In < Tl.
Answer: (B). | ||||
| 22 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[\mathrm{Ni(CO)_4}]$: Ni(0), 3$d^{10}$ after pairing with the strong-field CO, tetrahedral; no unpaired electrons. Diamagnetic.
$[\mathrm{NiCl_4}]^{2-}$: Ni$^{2+}$, 3$d^8$, with weak-field Cl$^-$ it is tetrahedral ($sp^3$); two unpaired electrons. Paramagnetic.
$[\mathrm{Co(NH_3)_4Cl_2}]\mathrm{Cl}$: Co$^{3+}$, 3$d^6$, NH$_3$ pairs the electrons (low-spin $t_{2g}^6$). Diamagnetic.
$\mathrm{Na_3[CoF_6]}$: Co$^{3+}$, 3$d^6$ with weak-field F$^-$, high-spin $t_{2g}^4e_g^2$; four unpaired electrons. Paramagnetic.
$\mathrm{Na_2O_2}$: the peroxide ion O$_2^{2-}$ has all electrons paired ($\pi^*$ full). Diamagnetic.
$\mathrm{CsO_2}$: the superoxide ion O$_2^-$ has one unpaired $\pi^*$ electron. Paramagnetic.
Three paramagnetic compounds.
Answer: (B). | ||||
| 23 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Natural rubber is cis-1,4-polyisoprene: $-[\mathrm{CH_2{-}C(CH_3){=}CH{-}CH_2}]_n-$.
Hydrogenating every double bond gives $-[\mathrm{CH_2{-}CH(CH_3){-}CH_2{-}CH_2}]_n-$.
Read the repeat unit as $-\mathrm{CH_2{-}CH(CH_3)}-$ (a propylene unit) followed by $-\mathrm{CH_2{-}CH_2}-$ (an ethylene unit): it is a perfectly alternating ethylene-propylene copolymer.
Vulcanised rubber is rubber cross-linked with sulphur, not hydrogenated; polypropylene and polybutylene have only one kind of unit.
Answer: (A). | ||||
| 24 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | Multiple correctEasy | |
A easy multiple correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Arrhenius equation: $k = Ae^{-E_a/RT}$.
(A) A larger $E_a$ makes $e^{-E_a/RT}$ smaller, so $k$ is smaller: a slow reaction. False.
(B) $e^{-E_a/RT}$ is the fraction of collisions with energy at least $E_a$; it grows with $T$, so $k$ grows. True.
(C) $\frac{d\ln k}{dT} = \frac{E_a}{RT^2}$: the larger $E_a$, the more sharply $k$ changes with temperature. True.
(D) $A$ is the frequency factor, the rate of collisions (with proper orientation) regardless of their energy; the energy condition is carried by the exponential. True.
Answer: (B), (C), (D). | ||||
| 25 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Multiple correctEasy | |
A easy multiple correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics A nucleus with $N/P < 1$ has too many protons for stability (stable nuclei have $N/P \ge 1$, rising above 1 for $Z > 20$). To become stable it must turn a proton into a neutron, raising $N/P$.
(B) Electron capture: $p + e^- \to n + \nu$. Raises $N/P$. Possible.
(D) Positron emission: $p \to n + e^+ + \nu$. Raises $N/P$. Possible.
(A) $\beta^-$ emission turns a neutron into a proton and lowers $N/P$ further. Not possible here.
(C) Neutron emission lowers $N$ and so lowers $N/P$. Not possible here.
Answer: (B), (D). | ||||
| 26 | 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 Borax is $\mathrm{Na_2[B_4O_5(OH)_4]\cdot 8H_2O}$. Its anion is the tetranuclear $[\mathrm{B_4O_5(OH)_4}]^{2-}$ unit. (A) true.
In this unit two boron atoms are trigonal ($\mathrm{BO_3}$, $sp^2$) and two are tetrahedral ($\mathrm{BO_4}$, $sp^3$): equal numbers. (C) true.
The tetrahedral borons make the cage three-dimensional, so the four borons are not in one plane. (B) false.
Each of the four borons carries exactly one terminal $-\mathrm{OH}$ group (four OH for four B). (D) true.
Answer: (A), (C), (D). | ||||
| 27 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Multiple correctEasy | |
A easy multiple correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Lone pairs on the central atom = (valence electrons $-$ electrons used in bonds)/2.
$\mathrm{BrF_5}$: Br has 7; 5 used; one lone pair (square pyramidal).
$\mathrm{ClF_3}$: Cl has 7; 3 used; $4/2 = 2$ lone pairs (T-shaped).
$\mathrm{XeF_4}$: Xe has 8; 4 used; $4/2 = 2$ lone pairs (square planar).
$\mathrm{SF_4}$: S has 6; 4 used; one lone pair (see-saw).
Answer: (B), (C). | ||||
| 28 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter We need a reagent that precipitates the sulphide but leaves the sulphate in solution.
(A) $\mathrm{CuCl_2}$: $\mathrm{Cu^{2+} + S^{2-} \to CuS}$ (black precipitate), while $\mathrm{CuSO_4}$ is soluble. Selective.
(B) $\mathrm{BaCl_2}$ precipitates $\mathrm{BaSO_4}$ (white), while $\mathrm{BaS}$ is soluble: it precipitates the wrong ion. Not selective for sulphide.
(C) Lead acetate gives black $\mathrm{PbS}$, but $\mathrm{PbSO_4}$ is only sparingly soluble and also comes down as a white precipitate, so the separation is not clean.
(D) Sodium nitroprusside gives a violet colour with $\mathrm{S^{2-}}$ ($[\mathrm{Fe(CN)_5NOS}]^{4-}$), a soluble complex: no precipitate.
Answer: (A). | ||||
| 29 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Multiple correctMedium | |
A medium multiple correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Tollens' reagent (ammoniacal silver nitrate) is reduced to a silver mirror by aldehydes and by other easily oxidised groups such as $\alpha$-hydroxy ketones.
(A) Propenal, $\mathrm{CH_2{=}CH{-}CHO}$, is an aldehyde. Positive.
(B) Benzaldehyde is an aldehyde (aromatic aldehydes also reduce Tollens' reagent). Positive.
(C) Benzoin, $\mathrm{PhCH(OH)COPh}$, is an $\alpha$-hydroxy ketone; it is oxidised to benzil, reducing $\mathrm{Ag^+}$ to silver. Positive.
(D) $\mathrm{PhCH{=}CHCOPh}$ is a simple $\alpha,\beta$-unsaturated ketone with no aldehyde or $\alpha$-hydroxy group. Negative.
Answer: (A), (B), (C). | ||||
| 30 | AminesDiazonium salts | Class 12: Amines | Multiple correctMedium | |
A medium multiple correct question on diazonium salts, from Amines. NCERT: Class 12: Amines i) Acetic anhydride/pyridine acetylates aniline to acetanilide, $\mathrm{C_6H_5NHCOCH_3}$. The $-\mathrm{NHCOCH_3}$ group is still ortho/para directing but far less activating than $-\mathrm{NH_2}$, so it is brominated only once.
ii) $\mathrm{KBrO_3/HBr}$ generates $\mathrm{Br_2}$; bromination goes to the less hindered para position: 4-bromoacetanilide.
iii) Acid hydrolysis removes the acetyl group: 4-bromoaniline.
iv) $\mathrm{NaNO_2/HCl}$ at 273–278 K gives 4-bromobenzenediazonium chloride.
v) Cu/HBr (Gattermann reaction) replaces $-\mathrm{N_2^+}$ by Br: 1,4-dibromobenzene.
(Without the protection step, aniline would give 2,4,6-tribromoaniline and finally 1,2,3,5-tetrabromobenzene, (D); the acetylation prevents this.)
Answer: (B). | ||||
| 31 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Multiple correctMedium | |
A medium multiple correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Cumene is oxidised by air to cumene hydroperoxide, which acid splits into phenol and acetone: P is phenol.
Reimer–Tiemann reaction ($\mathrm{CHCl_3/NaOH}$): the major product Q is salicylaldehyde (2-hydroxybenzaldehyde) and the minor product R is 4-hydroxybenzaldehyde.
Q with NaOH and benzyl bromide (Williamson synthesis) gives S, 2-benzyloxybenzaldehyde.
(A) R, the para isomer, has intermolecular hydrogen bonding and is not steam volatile; it is the ortho isomer Q, with intramolecular hydrogen bonding, that distils in steam. False.
(B) Q has a free phenolic $-\mathrm{OH}$ and gives a violet colour with neutral $\mathrm{FeCl_3}$. True.
(C) S still has its $-\mathrm{CHO}$ group and forms a yellow-orange 2,4-dinitrophenylhydrazone precipitate. True.
(D) In S the phenolic OH has become an ether, so there is no $\mathrm{FeCl_3}$ colour. False.
Answer: (B), (C). | ||||
| 32 | SolutionsConcentration terms | Class 12: Solutions | NumericalMedium | |
A medium numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 mol of solution in all: 0.1 mol solute (molar mass $M_A$) and 0.9 mol solvent (molar mass $M_B$).
Mass of solvent $= 0.9M_B$ g; mass of solution $= 0.1M_A + 0.9M_B$ g; volume of solution $= \frac{0.1M_A + 0.9M_B}{2.0}$ mL.
Molality $= \frac{0.1 \times 1000}{0.9M_B}$ and molarity $= \frac{0.1 \times 1000}{V}$. They are equal, so $V$ (in mL) $= 0.9M_B$ (in g):
$\frac{0.1M_A + 0.9M_B}{2} = 0.9M_B$, so $0.1M_A + 0.9M_B = 1.8M_B$ and $0.1M_A = 0.9M_B$.
$\frac{M_A}{M_B} = 9$.
Answer: 9. | ||||
| 33 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $D \propto \lambda\,\bar{u}$.
Mean free path $\lambda = \frac{k_BT}{\sqrt{2}\pi d^2 P} \propto \frac{T}{P}$, and mean speed $\bar{u} = \sqrt{\frac{8RT}{\pi M}} \propto \sqrt{T}$.
So $D \propto \frac{T^{3/2}}{P}$, and $x = \frac{4^{3/2}}{2} = \frac{8}{2} = 4$.
Answer: 4. | ||||
| 34 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In neutral or faintly alkaline solution permanganate is reduced to $\mathrm{MnO_2}$ (Mn: $+7 \to +4$, 3 electrons each), and it oxidises thiosulphate to sulphate (S: average $+2 \to +6$; two S atoms, 8 electrons per $\mathrm{S_2O_3^{2-}}$).
8 mol $\mathrm{MnO_4^-}$ accept $8 \times 3 = 24$ mol electrons, which oxidise $\frac{24}{8} = 3$ mol $\mathrm{S_2O_3^{2-}}$, giving $3 \times 2 = 6$ mol $\mathrm{SO_4^{2-}}$.
Balanced: $8\mathrm{MnO_4^-} + 3\mathrm{S_2O_3^{2-}} + \mathrm{H_2O} \to 8\mathrm{MnO_2} + 6\mathrm{SO_4^{2-}} + 2\mathrm{OH^-}$.
Answer: 6. | ||||
| 35 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalHard | |
A hard numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds L is an unsymmetrical bidentate ligand (binds through N and O), so the complex is of the type $[\mathrm{M(AB)_2X_2}]$, octahedral.
Cl atoms trans: the two chelates lie in the square plane, with the two N atoms either cis or trans to each other: 2 isomers.
Cl atoms cis: the two positions trans to the two Cl atoms are each taken by a donor of a different chelate. They can be N and N, O and O, or N and O: 3 isomers.
Total geometric isomers $= 2 + 3 = 5$.
Answer: 5. | ||||
| 36 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The substrate is one enantiomer of 2-bromopentane, $\mathrm{CH_3{-}CHBr{-}CH_2{-}CH_2{-}CH_3}$ (C1 to C5). Radical bromination at 300 °C can replace an H at any carbon.
C1: 1,2-dibromopentane. C2 keeps its configuration and is still a stereocentre: chiral, 1 product.
C2: 2,2-dibromopentane. C2 now carries two Br: achiral.
C3: 2,3-dibromopentane. C3 becomes a new stereocentre in either configuration, while C2 is unchanged: two diastereomers, both chiral: 2 products.
C4: 2,4-dibromopentane. With C2 fixed, C4 can be R or S: one of them is the (2R,4R)/(2S,4S)-type, chiral; the other is the meso compound, achiral: 1 chiral product.
C5: 1,4-dibromopentane (C2 of the substrate becomes C4): chiral, 1 product.
Chiral products: $1 + 2 + 1 + 1 = 5$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 37 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations First equation: $x = \sec\theta \pm \sqrt{\sec^2\theta - 1}$, and since $\sec^2\theta - 1 = \tan^2\theta$ its roots are $\sec\theta + \tan\theta$ and $\sec\theta - \tan\theta$.
For $-\frac{\pi}{6} < \theta < -\frac{\pi}{12}$, $\tan\theta < 0$ and $\sec\theta > 0$, so the larger root is $\alpha_1 = \sec\theta - \tan\theta$.
Second equation: $x = -\tan\theta \pm \sqrt{\tan^2\theta + 1} = -\tan\theta \pm \sec\theta$. As $\sec\theta > 0$, the smaller root is $\beta_2 = -\tan\theta - \sec\theta$.
$\alpha_1 + \beta_2 = (\sec\theta - \tan\theta) + (-\tan\theta - \sec\theta) = -2\tan\theta$.
Answer: (C). | ||||
| 38 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations No boy: choose 4 girls in $\binom{6}{4} = 15$ ways.
One boy: choose him in $\binom{4}{1} = 4$ ways and 3 girls in $\binom{6}{3} = 20$ ways: $80$ teams.
So $15 + 80 = 95$ teams, and each team chooses its captain in 4 ways: $95 \times 4 = 380$.
Answer: (A). | ||||
| 39 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Multiply by $\sin x\cos x$ (non-zero in $S$): $\sqrt{3}\sin x + \cos x + 2(\sin^2 x - \cos^2 x) = 0$, i.e. $2\sin\left(x + \frac{\pi}{6}\right) - 2\cos 2x = 0$.
So $\sin\left(x + \frac{\pi}{6}\right) = \cos 2x = \sin\left(\frac{\pi}{2} - 2x\right)$.
Case 1: $x + \frac{\pi}{6} = \frac{\pi}{2} - 2x + 2n\pi$, so $x = \frac{\pi}{9} + \frac{2n\pi}{3}$. In $(-\pi, \pi)$: $\frac{\pi}{9}, \frac{7\pi}{9}, -\frac{5\pi}{9}$.
Case 2: $x + \frac{\pi}{6} = \pi - \left(\frac{\pi}{2} - 2x\right) + 2n\pi$, so $x = -\frac{\pi}{3} - 2n\pi$. In $(-\pi, \pi)$: $-\frac{\pi}{3}$.
None of these is $0$ or $\pm\frac{\pi}{2}$.
Sum: $\frac{\pi}{9} + \frac{7\pi}{9} - \frac{5\pi}{9} - \frac{3\pi}{9} = 0$.
Answer: (C). | ||||
| 40 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $P(D \mid T_2) = p$; then $P(D \mid T_1) = 10p$.
Total probability: $0.2 \times 10p + 0.8 \times p = 2.8p = 0.07$, so $p = 0.025$ and $P(D \mid T_1) = 0.25$.
Not defective: from $T_1$, $0.2 \times 0.75 = 0.15$; from $T_2$, $0.8 \times 0.975 = 0.78$. Total $0.93$.
Bayes: $P(T_2 \mid \bar{D}) = \frac{0.78}{0.93} = \frac{78}{93}$.
Answer: (C). | ||||
| 41 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives If $\alpha \le 0$ the left side falls below 1 for large $x$, so $\alpha > 0$. Then we need the minimum of $f(x) = 4\alpha x^2 + \frac{1}{x}$ on $x > 0$ to be at least 1.
$f'(x) = 8\alpha x - \frac{1}{x^2} = 0$ gives $x^3 = \frac{1}{8\alpha}$, $x = \frac{1}{2\alpha^{1/3}}$, a minimum.
$f_{min} = 4\alpha \cdot \frac{1}{4\alpha^{2/3}} + 2\alpha^{1/3} = 3\alpha^{1/3}$.
(Equivalently by AM–GM: $4\alpha x^2 + \frac{1}{2x} + \frac{1}{2x} \ge 3\sqrt[3]{\alpha}$.)
$3\alpha^{1/3} \ge 1$ gives $\alpha \ge \frac{1}{27}$.
Answer: (C). | ||||
| 42 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctMedium | |
A medium multiple correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Coordinates: $O(0,0,0)$, $P(3,0,0)$, $Q(3,3,0)$, $R(0,3,0)$, $T\left(\frac{3}{2}, \frac{3}{2}, 0\right)$, $S\left(\frac{3}{2}, \frac{3}{2}, 3\right)$.
(A) $\vec{OQ} \parallel (1,1,0)$ and $\vec{OS} \parallel (1,1,2)$: $\cos\phi = \frac{2}{\sqrt{2}\sqrt{6}} = \frac{1}{\sqrt{3}}$, which is not $\cos\frac{\pi}{3}$. False.
(B) The plane through $O$ containing $(1,1,0)$ and $(1,1,2)$ has normal $(1,1,0) \times (1,1,2) = (2,-2,0)$, so it is $x - y = 0$. True.
(C) Distance of $P(3,0,0)$ from $x - y = 0$: $\frac{|3 - 0|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$. True.
(D) Line $RS$: through $R(0,3,0)$ with direction $\vec{RS} = \left(\frac{3}{2}, -\frac{3}{2}, 3\right) \parallel (1,-1,2)$. Distance from $O$ $= \frac{|\vec{RO} \times \vec{d}|}{|\vec{d}|}$ with $\vec{OR} = (0,3,0)$: $(0,3,0) \times (1,-1,2) = (6, 0, -3)$, of length $\sqrt{45}$; $|\vec{d}| = \sqrt{6}$. Distance $= \sqrt{\frac{45}{6}} = \sqrt{\frac{15}{2}}$. True.
Answer: (B), (C), (D). | ||||
| 43 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Multiple correctMedium | |
A medium multiple correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $f' + \frac{f}{x} = 2$ is linear with integrating factor $x$: $(xf)' = 2x$, so $xf = x^2 + C$ and $f(x) = x + \frac{C}{x}$.
$f(1) = 1 + C \neq 1$, so $C \neq 0$. Also $f'(x) = 1 - \frac{C}{x^2}$.
(A) $f'\left(\frac{1}{x}\right) = 1 - Cx^2 \to 1$. True.
(B) $xf\left(\frac{1}{x}\right) = x\left(\frac{1}{x} + Cx\right) = 1 + Cx^2 \to 1$, not 2. False.
(C) $x^2f'(x) = x^2 - C \to -C \neq 0$. False.
(D) $f(x) = x + \frac{C}{x}$ is unbounded as $x \to 0^+$ because $C \neq 0$, so $|f(x)| \le 2$ fails on $(0,2)$. False.
Answer: (A). | ||||
| 44 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Multiple correctHard | |
A hard multiple correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\det P = 3(0 + 5\alpha) + 1(0 - 3\alpha) - 2(-10 - 0) = 12\alpha + 20$.
$PQ = kI$ gives $Q = kP^{-1} = \frac{k}{\det P}\,\mathrm{adj}P$, and $\det P \cdot \det Q = k^3$.
The $(2,3)$ entry of $\mathrm{adj}P$ is the cofactor of the $(3,2)$ entry of $P$: $-\begin{vmatrix}3 & -2\\2 & \alpha\end{vmatrix} = -(3\alpha + 4)$.
So $q_{23} = \frac{-k(3\alpha + 4)}{12\alpha + 20} = -\frac{k}{8}$, giving $8(3\alpha + 4) = 12\alpha + 20$, $\alpha = -1$ and $\det P = 8$.
Then $\det Q = \frac{k^3}{8} = \frac{k^2}{2}$ gives $k = 4$.
(A) $\alpha = 0, k = 8$: false.
(B) $4(-1) - 4 + 8 = 0$: true.
(C) $\det(P\,\mathrm{adj}Q) = \det P\,(\det Q)^2 = 8 \times 8^2 = 2^9$ (with $\det Q = \frac{16}{2} = 8$): true.
(D) $\det(Q\,\mathrm{adj}P) = \det Q\,(\det P)^2 = 8 \times 64 = 2^9 \neq 2^{13}$: false.
Answer: (B), (C). | ||||
| 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 Let $s - x = 4\lambda$, $s - y = 3\lambda$, $s - z = 2\lambda$. Adding: $3s - 2s = 9\lambda$, so $s = 9\lambda$ and $x = 5\lambda$, $y = 6\lambda$, $z = 7\lambda$.
Heron: $\Delta = \sqrt{9\lambda \cdot 4\lambda \cdot 3\lambda \cdot 2\lambda} = 6\sqrt{6}\lambda^2$. Inradius $r = \frac{\Delta}{s} = \frac{2\sqrt{6}}{3}\lambda$.
$\pi r^2 = \frac{8\pi}{3}$ gives $\frac{24}{9}\lambda^2 = \frac{8}{3}$, so $\lambda = 1$: sides 5, 6, 7.
(A) $\Delta = 6\sqrt{6}$. True.
(B) $R = \frac{xyz}{4\Delta} = \frac{210}{24\sqrt{6}} = \frac{35}{4\sqrt{6}} = \frac{35\sqrt{6}}{24}$, not $\frac{35\sqrt{6}}{6}$. False.
(C) $r = 4R\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2}$, so the product is $\frac{r}{4R} = \frac{2\sqrt{6}/3}{35/\sqrt{6}} = \frac{12}{105} = \frac{4}{35}$. True.
(D) $\sin^2\frac{X + Y}{2} = \cos^2\frac{Z}{2} = \frac{s(s - z)}{xy} = \frac{9 \cdot 2}{5 \cdot 6} = \frac{3}{5}$. True.
Answer: (A), (C), (D). | ||||
| 46 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Multiple correctHard | |
A hard multiple correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Factor the coefficient: $x^2 + 4x + 4 + y(x + 2) = (x + 2)(x + 2 + y)$. With $X = x + 2$: $X(X + y)\frac{dy}{dX} = y^2$, a homogeneous equation.
Put $y = vX$: $v + X\frac{dv}{dX} = \frac{v^2}{1 + v}$, so $X\frac{dv}{dX} = -\frac{v}{1 + v}$ and $\frac{1 + v}{v}dv = -\frac{dX}{X}$.
Integrating: $\ln v + v = -\ln X + c$, i.e. $\ln y + \frac{y}{x + 2} = c$. Through $(1, 3)$: $c = \ln 3 + 1$.
Curve: $\ln y + \frac{y}{x + 2} = \ln 3 + 1$.
(A), (B) On $y = x + 2$: $\ln(x + 2) + 1 = \ln 3 + 1$, so $x = 1$ only. Exactly one point: (A) true, (B) false.
(C) On $y = (x + 2)^2$: $2\ln(x + 2) + (x + 2) = \ln 3 + 1 \approx 2.10$. For $x > 0$ the left side exceeds $2\ln 2 + 2 \approx 3.39$ and increases, so no solution. False.
(D) On $y = (x + 3)^2$: $2\ln(x + 3) + \frac{(x + 3)^2}{x + 2}$; for $x > 0$ already $2\ln(x + 3) > 2\ln 3 \approx 2.20 > \ln 3 + 1$, and the other term is positive, so no intersection. True.
Answer: (A), (D). | ||||
| 47 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctMedium | |
A medium multiple correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f'(x) = 3x^2 + 3 > 0$, so $f$ is a bijection and $g = f^{-1}$. From $h(g(g(x))) = x$, putting $x = f(f(t))$ gives $h(t) = f(f(t))$.
(A) $f(0) = 2$, so $g(2) = 0$ and $g'(2) = \frac{1}{f'(0)} = \frac{1}{3}$. False.
(B) $h'(1) = f'(f(1))\,f'(1) = f'(6) \cdot f'(1) = 111 \times 6 = 666$. True.
(C) $h(0) = f(f(0)) = f(2) = 8 + 6 + 2 = 16$. True.
(D) $h(g(3)) = f(f(g(3))) = f(3) = 27 + 9 + 2 = 38$, not 36. False.
Answer: (B), (C). | ||||
| 48 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections $x^2 = 2y$ in $x^2 + y^2 = 3$: $y^2 + 2y - 3 = 0$, $y = 1$ (first quadrant), $x = \sqrt{2}$. $P(\sqrt{2}, 1)$.
Tangent at $P$: $\sqrt{2}x + y = 3$.
Centres $(0, k)$ at distance $2\sqrt{3}$ from it: $\frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3}$, $k = 9$ or $-3$. So $Q_2Q_3 = 12$. (A) true.
(B) $R_2, R_3$ are the feet of the perpendiculars from $Q_2, Q_3$ to the tangent, so $R_2R_3$ is the projection of $\vec{Q_2Q_3} = (0, 12)$ on the tangent's direction $\frac{(1, -\sqrt{2})}{\sqrt{3}}$: $12 \cdot \frac{\sqrt{2}}{\sqrt{3}} = 4\sqrt{6}$. True.
(C) The height from $O$ to the tangent is $\frac{3}{\sqrt{3}} = \sqrt{3}$ (the radius of $C_1$). Area $= \frac{1}{2} \cdot 4\sqrt{6} \cdot \sqrt{3} = 6\sqrt{2}$. True.
(D) $Q_2Q_3$ lies on the $y$-axis and $P$ is at distance $\sqrt{2}$ from it: area $= \frac{1}{2} \cdot 12 \cdot \sqrt{2} = 6\sqrt{2}$, not $4\sqrt{2}$. False.
Answer: (A), (B), (C). | ||||
| 49 | CirclesStandard and general form | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Let $P = (\cos\theta, \sin\theta)$. Tangent at $S$: $x = 1$. Tangent at $P$: $x\cos\theta + y\sin\theta = 1$; at $x = 1$, $y = \frac{1 - \cos\theta}{\sin\theta} = \tan\frac{\theta}{2}$. So $Q = \left(1, \tan\frac{\theta}{2}\right)$.
The line through $Q$ parallel to $RS$ (the $x$-axis) is $y = \tan\frac{\theta}{2}$. The normal at $P$ passes through the centre: $y = x\tan\theta$.
With $t = \tan\frac{\theta}{2}$: $E$ has $y = t$ and $x = \frac{t}{\tan\theta} = \frac{t(1 - t^2)}{2t} = \frac{1 - t^2}{2}$.
Locus: $x = \frac{1 - y^2}{2}$, i.e. $y^2 = 1 - 2x$.
$\left(\frac{1}{3}, \pm\frac{1}{\sqrt{3}}\right)$: $y^2 = \frac{1}{3} = 1 - \frac{2}{3}$. On the locus (from $\theta = \pm\frac{\pi}{3}$, allowed points $P$).
$\left(\frac{1}{4}, \pm\frac{1}{2}\right)$: $y^2 = \frac{1}{4}$ but $1 - 2x = \frac{1}{2}$. Not on it.
Answer: (A), (C). | ||||
| 50 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Take $x$ out of column 1 and $x^2$ out of column 2, then split column 3:
$\Delta = x^3\left(\begin{vmatrix}1 & 1 & 1\\2 & 4 & 1\\3 & 9 & 1\end{vmatrix} + x^3\begin{vmatrix}1 & 1 & 1\\2 & 4 & 8\\3 & 9 & 27\end{vmatrix}\right)$.
First determinant: $1(4 - 9) - 1(2 - 3) + 1(18 - 12) = 2$. Second: $1(108 - 72) - 1(54 - 24) + 1(18 - 12) = 12$.
$\Delta = 2x^3 + 12x^6 = 10$, i.e. $6x^6 + x^3 - 5 = 0$, $(6x^3 - 5)(x^3 + 1) = 0$.
$x^3 = \frac{5}{6}$ or $x^3 = -1$: two distinct real $x$.
Answer: 2. | ||||
| 51 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Coefficient of $x^2$: $\sum_{k=2}^{49}\binom{k}{2} + \binom{50}{2}m^2 = \binom{50}{3} + \binom{50}{2}m^2$ (hockey-stick identity).
$\binom{50}{3} = 19600$, $\binom{50}{2} = 1225$, $\binom{51}{3} = 20825$. Dividing $19600 + 1225m^2 = (3n + 1) \cdot 20825$ by 1225: $16 + m^2 = 17(3n + 1)$, so $m^2 = 51n + 1$.
We need the smallest positive $m$ with $m^2 \equiv 1 \pmod{51}$ and $n \ge 1$. $m = 1$ gives $n = 0$ (not positive). Then $m \equiv \pm 1 \pmod{17}$ and $\pmod 3$: $m = 16$ works, $256 = 51 \times 5 + 1$.
$n = 5$.
Answer: 5. | ||||
| 52 | 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 Let $F(x) = \int_0^x \frac{t^2}{1 + t^4}dt - 2x + 1$ on $[0, 1]$.
$F(0) = 1 > 0$. Since $\frac{t^2}{1 + t^4} \le \frac{1}{2}$, $F(1) \le \frac{1}{2} - 1 < 0$. So $F$ has a root in $(0, 1)$.
$F'(x) = \frac{x^2}{1 + x^4} - 2 < 0$, so $F$ is strictly decreasing and the root is unique.
Answer: 1. | ||||
| 53 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Numerator: $x^2\sin\beta x = \beta x^3 + \ldots$
Denominator: $\alpha x - \sin x = (\alpha - 1)x + \frac{x^3}{6} - \ldots$
For a finite non-zero limit the $x$ term must vanish: $\alpha = 1$. Then the limit is $\frac{\beta x^3}{x^3/6} = 6\beta = 1$, so $\beta = \frac{1}{6}$.
$6(\alpha + \beta) = 6 + 1 = 7$.
Answer: 7. | ||||
| 54 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z = \omega$, a complex cube root of unity: $\omega^3 = 1$, $1 + \omega + \omega^2 = 0$.
$P^2 = \begin{bmatrix}\omega^{2r} + \omega^{4s} & \omega^{2s}\left((-\omega)^r + \omega^r\right)\\ \omega^{2s}\left((-\omega)^r + \omega^r\right) & \omega^{4s} + \omega^{2r}\end{bmatrix}$, using $(-\omega)^{2r} = \omega^{2r}$.
Off-diagonal zero: $\omega^r((-1)^r + 1) = 0$, so $r$ is odd: $r = 1$ or $3$.
Diagonal $= -1$: $\omega^{2r} + \omega^{s} = -1$ (as $\omega^{4s} = \omega^s$).
$r = 1$: $\omega^2 + \omega^s = -1$ needs $\omega^s = \omega$, so $s = 1$.
$r = 3$: $1 + \omega^s = -1$ needs $\omega^s = -2$: impossible.
Only $(1, 1)$.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Binding energy (using atomic masses, so electrons cancel): $B = [Z m_H + (A-Z)m_n - M]c^2$.
$B_N = [7(1.007825) + 8(1.008665) - 15.000109]\,\text{u}\,c^2$, $B_O = [8(1.007825) + 7(1.008665) - 15.003065]\,\text{u}\,c^2$.
$B_N - B_O = [(1.008665 - 1.007825) + (15.003065 - 15.000109)]\,\text{u}\,c^2 = (0.000840 + 0.002956) \times 931.5 = 0.003796 \times 931.5 \approx 3.536$ MeV.
Oxygen is less bound only because of its extra Coulomb energy:
$\Delta E = \frac{3}{5}\frac{e^2}{4\pi\varepsilon_0 R}[8\cdot7 - 7\cdot6] = \frac{3}{5}\cdot\frac{1.44 \times 14}{R} = \frac{12.096}{R}$ MeV (R in fm).
Set $\frac{12.096}{R} = 3.536$: $R \approx 3.42$ fm.
Answer: (C). | ||||
| 2 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The activity must fall by a factor of $64 = 2^6$, i.e. through 6 half-lives.
Time $= 6 \times 18 = 108$ days.
Answer: (C). | ||||
| 3 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $P^3V^5 =$ const means $PV^{5/3} =$ const, so for this gas $\gamma = \frac{5}{3}$.
$\Delta U$ depends only on the end states: $\Delta U = \frac{P_fV_f - P_iV_i}{\gamma - 1} = \frac{\frac{10^5}{32}(8\times10^{-3}) - 10^5(10^{-3})}{2/3} = \frac{25 - 100}{2/3} = -112.5$ J.
Work in the two-step path: only the isobaric step does work, $W = P_i(V_f - V_i) = 10^5 \times 7\times10^{-3} = 700$ J.
Heat supplied $Q = \Delta U + W = -112.5 + 700 = 587.5 \approx 588$ J.
Answer: (C). | ||||
| 4 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter In the steady state the same heat current flows through both wires (same cross-section and length taken):
$2K\frac{T - 10}{1} = K\frac{400 - T}{1}$, so $3T = 420$ and the junction is at $T = 140$ °C.
Along PQ the temperature rises linearly from 10 °C to 140 °C, so the average rise above 10 °C is $\frac{0 + 130}{2} = 65$ °C.
Each element $dx$ expands by $\alpha\,\Delta T(x)\,dx$, so $\Delta L = \alpha L\,\overline{\Delta T} = 1.2\times10^{-5} \times 1 \times 65 = 7.8\times10^{-4}$ m $= 0.78$ mm.
Answer: (A). | ||||
| 5 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctHard | |
A hard single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens: $u = -50$, $f = 30$: $\frac{1}{v} = \frac{1}{30} - \frac{1}{50} = \frac{1}{75}$, so $v = 75$ cm. The refracted light converges towards $(75, 0)$, which is 25 cm behind the mirror's pole at $(50, 0)$: a virtual object for the mirror.
The converging beam travels along the $x$-axis and meets the mirror at its pole, where the mirror's normal is at 30° to the beam. Measured along this ray the virtual object is 25 cm behind the mirror. For the convex mirror $f = +50$ cm and the virtual object is at $u = +25$ cm:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{50} - \frac{1}{25} = -\frac{1}{50}$, so $v = -50$ cm: a real image 50 cm in front of the mirror.
It lies along the reflected ray through the pole. A ray along $+x$ meeting a surface whose normal is at 30° is turned through $180° - 60° = 120°$, so the reflected ray goes at 120° to the $x$-axis.
Image: $(50 + 50\cos120°,\ 50\sin120°) = (50 - 25,\ 25\sqrt{3}) = (25, 25\sqrt{3})$.
Answer: (C). | ||||
| 6 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Main scale division (MSD) = 1 mm = 0.1 cm.
$C_1$: 10 VSD = 9 MSD, so 1 VSD = 0.09 cm and the least count is $0.1 - 0.09 = 0.01$ cm. The vernier zero is just past 2.8 cm and the 7th vernier division coincides with a main-scale mark, so the reading is $2.8 + 7 \times 0.01 = 2.87$ cm.
$C_2$: 10 VSD = 11 MSD, so 1 VSD = 0.11 cm. If the zero is at $2.8 + x$ and the $n$th vernier mark meets a main mark, $2.8 + x + 0.11n = 2.8 + 0.1(n+1)$, so $x = 0.1 - 0.01n = 0.01(10 - n)$. In the figure the 7th vernier division coincides, so $x = 0.03$ cm and the reading is $2.83$ cm.
Answer: (B). | ||||
| 7 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Multiple correctHard | |
A hard multiple correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Geometry: the small disc's centre is $l$ from O along the rod and its contact point is $a$ from the centre at right angles to the rod, so the contact point is $\sqrt{l^2 + a^2} = 5a$ from O (and the large disc's is $\sqrt{4l^2 + 4a^2} = 10a$, on the same line). The rod is inclined at $\theta$ to that ground line with $\sin\theta = \frac{a}{5a} = \frac15$, $\cos\theta = \frac{\sqrt{24}}{5}$.
Rolling without slipping: every contact point is at rest, so the instantaneous axis is the ground line through O and the contact points. The total angular velocity is spin $\omega$ along the rod plus precession $\Omega$ about $z$, and it must be horizontal: $\omega\sin\theta - \Omega = 0$ in the vertical direction, so $\Omega = \omega\sin\theta = \frac{\omega}{5}$.
(A) True: the centre of mass goes round the $z$-axis at $\omega/5$.
(B) The centre of mass is at $\frac{m l + 4m(2l)}{5m} = \frac{9l}{5}$ from O along the rod, at horizontal distance $\rho = \frac{9l}{5}\cos\theta$, moving at $v = \Omega\rho$ perpendicular to its position vector. $L = 5m\cdot\frac{9l}{5}\cdot\frac{\omega}{5}\cdot\frac{9l}{5}\cos\theta = \frac{81\times24}{25}\cdot\frac{\sqrt{24}}{5}ma^2\omega \approx 76\,ma^2\omega$, not $81ma^2\omega$. False.
(C) About its own axis the assembly has $I = \frac12ma^2 + \frac12(4m)(2a)^2 = \frac{17}{2}ma^2$, so the spin part of the angular momentum about the centre of mass is $\frac{17}{2}ma^2\omega$. Counted in full, the precession also contributes (the total angular velocity is $\omega\cos\theta$ along the ground line, with components $\omega\cos^2\theta$ along the rod and $\omega\sin\theta\cos\theta$ across it), which makes the magnitude about $9.4\,ma^2\omega$; so (C) holds only when the precession's part is left out.
(D) $L_z$ = orbital part $5m\rho^2\Omega = 5m\cdot\frac{81\cdot24}{25}\cdot\frac{24}{25}a^2\cdot\frac{\omega}{5} \approx 74.6\,ma^2\omega$ plus the spin part's $z$-component (a few $ma^2\omega$); about $77.5\,ma^2\omega$ in all, not $55ma^2\omega$. False.
Answer: (A) (the official key also accepts (A) and (C)). | ||||
| 8 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Multiple correctMedium | |
A medium multiple correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The fastest electrons leave with $K_{max} = \frac{hc}{\lambda_{ph}} - \phi$ and gain $eV$, so at the anode $K = \frac{hc}{\lambda_{ph}} - \phi + eV$ and $\lambda_e = \frac{h}{\sqrt{2mK}}$.
(A) Increasing $\phi$ or $\lambda_{ph}$ lowers $K$ and so increases $\lambda_e$. False.
(B) $d$ does not enter $K$ (the energy gained depends only on $V$). False.
(C) For $eV \gg \phi$ (and the photon energy), $K \approx eV$, $\lambda_e \propto V^{-1/2}$; four times $V$ halves $\lambda_e$. True.
(D) $\lambda_e$ depends on $\lambda_{ph}$ through $\left(\frac{hc}{\lambda_{ph}} - \phi + eV\right)^{-1/2}$, not linearly, so it does not change at the same rate. False.
Answer: (C). | ||||
| 9 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Multiple correctMedium | |
A medium multiple correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (A) $\frac{\Delta r}{r} = \frac{1}{10} = 10\%$. True.
Mean period $\bar T = \frac{0.52 + 0.56 + 0.57 + 0.54 + 0.59}{5} = 0.556 \approx 0.56$ s.
Absolute deviations: 0.036, 0.004, 0.014, 0.016, 0.034; mean deviation $= \frac{0.104}{5} = 0.0208 \approx 0.02$ s (larger than the least count, so it is the error).
(B) $\frac{\Delta T}{T} = \frac{0.02}{0.56} = 3.57\%$. True. (C) False.
$g = \frac{28\pi^2(R - r)}{5T^2}$, so $\frac{\Delta g}{g} = \frac{\Delta R + \Delta r}{R - r} + 2\frac{\Delta T}{T} = \frac{2}{50} + 2(0.0357) = 0.04 + 0.0714 \approx 0.11$.
(D) 11%. True.
Answer: (A), (B), (D). | ||||
| 10 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Multiple correctHard | |
A hard multiple correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Let $I_g$ be the full-scale current of a galvanometer; the range is the voltage (or current) across the whole arrangement when the galvanometer considered reads full scale.
All in series: $V_A = I_g(2R + 2R_C)$.
Arrangement (B): the two galvanometers in parallel ($R_C/2$) carry $2I_g$, in series with $2R$: $V_B = 2I_g\left(2R + \frac{R_C}{2}\right) = I_g(4R + R_C)$.
$V_B - V_A = I_g(2R - R_C) > 0$, so (B) gives the larger voltage range: (A) false, (B) true.
All in parallel: voltage $I_gR_C$; current $= 2I_g + \frac{2I_gR_C}{R}$.
Arrangement (D): galvanometers in series ($2R_C$) carry $I_g$, voltage $2I_gR_C$ across the resistors in parallel ($R/2$): current $= I_g + \frac{4I_gR_C}{R}$.
Difference (all parallel minus (D)) $= I_g\left(1 - \frac{2R_C}{R}\right) > 0$ because $R_C < R/2$. So all in parallel gives the larger current range: (C) true, (D) false.
Answer: (B), (C). | ||||
| 11 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Take the right node at 0 V; the left node (cell's + side) is at 5 V. Two independent branches across the cell:
Upper: 40 μF then 25 kΩ, $\tau_1 = 25\times10^3 \times 40\times10^{-6} = 1$ s. Lower: 50 kΩ then 20 μF, $\tau_2 = 50\times10^3 \times 20\times10^{-6} = 1$ s.
The top node (voltmeter −) is between the 40 μF capacitor and the 25 kΩ resistor: $V_{top} = 5e^{-t}$ (the capacitor's voltage grows from 0).
The bottom node (voltmeter +) is between the 50 kΩ resistor and the 20 μF capacitor: $V_{bottom} = 5(1 - e^{-t})$.
Reading $= V_{bottom} - V_{top} = 5 - 10e^{-t}$.
(A) At $t = 0$: $-5$ V; as $t \to \infty$: $+5$ V. True.
(B) Zero when $e^{-t} = \frac12$, $t = \ln 2$ s. True.
(C) Ammeter current $= \frac{5}{25\,\text{k}}e^{-t} + \frac{5}{50\,\text{k}}e^{-t} \propto e^{-t}$, so after 1 s it is $1/e$ of its initial value. True.
(D) Both capacitors charge fully, so the current falls to zero. True.
Answer: (A), (B), (C), (D). | ||||
| 12 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Multiple correctMedium | |
A medium multiple correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves After sticking, $\omega' = \sqrt{\frac{k}{M + m}}$ in both cases.
Case (i), at $x_0$ the speed is $A\omega$, $\omega = \sqrt{k/M}$. Momentum is conserved: $v' = \frac{MA\omega}{M + m}$. New amplitude $A' = \frac{v'}{\omega'} = \frac{MA}{M + m}\sqrt{\frac{k}{M}}\sqrt{\frac{M + m}{k}} = A\sqrt{\frac{M}{M + m}}$.
Case (ii), at the extreme the block is at rest, so placing $m$ changes nothing about position or velocity: amplitude stays $A$.
(A) True. (B) True: $T' = 2\pi\sqrt{\frac{M + m}{k}}$ in both.
(C) Case (i): energy falls from $\frac12kA^2$ to $\frac12kA'^2$ (an inelastic sticking). Case (ii): still $\frac12kA^2$. False.
(D) Case (i): $v'$ at $x_0$ is $\frac{M}{M + m}$ of the old speed. Case (ii): $A\omega' < A\omega$. Decreases in both. True.
Answer: (A), (B), (D). | ||||
| 13 | 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 The line of the sources is the $y$-axis, perpendicular to the screen. For a point on the screen at angle $\alpha$ from that line, the path difference is $d\cos\alpha$; a fixed path difference is a cone about the $y$-axis, which cuts the screen in a circle centred at O. Only $z > 0$ exists, so the fringes are semicircles centred at O: (D) true, (A) and (C) false.
At O ($\alpha = 0$) the path difference is $d = 0.6003$ mm $= \frac{0.6003\times10^{-3}}{600\times10^{-9}}\lambda = 1000.5\lambda$, an odd number of half-wavelengths: destructive. (B) true.
Answer: (B), (D). | ||||
| 14 | 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 While one edge is in the field (entering, $0 < x < L$; leaving, $3L < x < 4L$) the emf is $B_0Lv$, the current $\frac{B_0Lv}{R}$ and the force on that edge $\frac{B_0^2L^2v}{R}$, always opposing the motion.
$m\frac{dv}{dt} = mv\frac{dv}{dx} = -\frac{B_0^2L^2v}{R}$ gives $\frac{dv}{dx} = -\frac{B_0^2L^2}{mR}$: $v$ falls linearly with $x$ in these two stretches. From $L$ to $3L$ the loop is wholly inside, the flux is constant, there is no current or force and $v$ stays constant. (A) is correct.
Entering, the flux into the page grows, so the induced current opposes it by flowing counter-clockwise: positive, proportional to $v$, so falling linearly. Leaving, the flux falls and the current is clockwise: negative, with magnitude falling linearly. (B) is correct; (C) has the wrong sign while leaving.
The force opposes the motion in both stretches, so $F$ is negative on leaving as well; (D), which shows a positive force from $3L$ to $4L$, is wrong.
Answer: (A), (B). | ||||
| 15 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | ComprehensionMedium | |
A medium comprehension question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion In the rotating frame the only force along the slot is the centrifugal term $m(\vec\omega\times\vec r)\times\vec\omega = m\omega^2x\,\hat{i}$ (the slot is smooth; the Coriolis force and the slot's reaction act along $\hat{j}$).
So $\ddot{x} = \omega^2x$, with $x(0) = \frac{R}{2}$ and $\dot{x}(0) = 0$.
General solution $x = Ae^{\omega t} + Be^{-\omega t}$; the conditions give $A + B = \frac{R}{2}$ and $A - B = 0$, so $A = B = \frac{R}{4}$.
$r = \frac{R}{4}(e^{\omega t} + e^{-\omega t})$.
Answer: (A). | ||||
| 16 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | ComprehensionMedium | |
A medium comprehension question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion From the previous part, $\dot{x} = \frac{R\omega}{4}(e^{\omega t} - e^{-\omega t})$ along $\hat{i}$.
Coriolis term: $2m(\vec{v}_{rot}\times\vec\omega) = 2m\dot{x}\omega(\hat{i}\times\hat{k}) = -2m\omega\dot{x}\,\hat{j}$.
The block does not move along $\hat{j}$ in the rotating frame, so the slot wall must push with $+2m\omega\dot{x}\,\hat{j} = \frac{1}{2}m\omega^2R(e^{\omega t} - e^{-\omega t})\hat{j}$.
Vertically the disc holds the block up against gravity with $mg\hat{k}$.
Net reaction $= \frac{1}{2}m\omega^2R(e^{\omega t} - e^{-\omega t})\hat{j} + mg\hat{k}$.
Answer: (B). | ||||
| 17 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | ComprehensionEasy | |
A easy comprehension question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance On the bottom plate a ball takes the plate's potential and a positive charge, and is repelled upward. On touching the top plate (no bounce, $e = 0$) it comes to that plate's potential $-V_0$: it now carries a negative charge and is repelled by the top plate and attracted by the bottom one, so it travels back down. The cycle repeats.
The motion is under a uniform field (constant force each way), so it is not simple harmonic.
Answer: (C). | ||||
| 18 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | ComprehensionMedium | |
A medium comprehension question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charge on a ball at potential $\pm V_0$: $q \propto V_0$ ($q \approx 4\pi\varepsilon_0 rV_0$ for an isolated sphere; in any case proportional to $V_0$).
Field $E = \frac{2V_0}{h}$, so the force $qE \propto V_0^2$ and the acceleration $a \propto V_0^2$.
Each trip starts from rest (zero restitution): $h = \frac12at^2$, $t = \sqrt{2h/a} \propto \frac{1}{V_0}$.
Each trip carries charge $q$ across the gap, so the average current $\propto \frac{q}{t} \propto V_0 \cdot V_0 = V_0^2$.
Answer: (D). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 19 | ElectrochemistryNernst equation | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $\text{H}_2 + \text{M}^{4+} \rightarrow 2\text{H}^+ + \text{M}^{2+}$, with $n = 2$ and $E^0_{cell} = 0.151 - 0 = 0.151$ V.
With $[\text{H}^+] = 1$ M and $p_{\text{H}_2} = 1$ bar, $Q = \frac{[\text{M}^{2+}]}{[\text{M}^{4+}]} = 10^x$.
$E_{cell} = E^0_{cell} - \frac{0.059}{2}\log Q$: $0.092 = 0.151 - 0.0295x$, so $x = \frac{0.059}{0.0295} = 2$.
Answer: (D). | ||||
| 20 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter KCl is a strong electrolyte whose ions are strongly hydrated and stay away from the surface: the surface tension rises slightly with concentration (II).
Methanol is mildly surface-active: it gathers a little at the surface and lowers the surface tension gradually (I).
Sodium dodecyl sulphate is a surfactant: it lowers the surface tension sharply until the critical micelle concentration, after which extra surfactant forms micelles and the surface tension stays nearly constant (III).
Answer: (D). | ||||
| 21 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter With thiosulphate in excess, $\text{Ag}^+$ forms the soluble complex: $\text{Ag}^+ + 2\text{S}_2\text{O}_3^{2-} \rightarrow [\text{Ag}(\text{S}_2\text{O}_3)_2]^{3-}$ (X, clear solution).
With more $\text{Ag}^+$ the complex is used up and white silver thiosulphate precipitates: $[\text{Ag}(\text{S}_2\text{O}_3)_2]^{3-} + 3\text{Ag}^+ \rightarrow 2\text{Ag}_2\text{S}_2\text{O}_3$ (Y).
On standing it hydrolyses: $\text{Ag}_2\text{S}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow \text{Ag}_2\text{S} + \text{H}_2\text{SO}_4$, the precipitate turning through yellow and brown to black $\text{Ag}_2\text{S}$ (Z).
Answer: (A). | ||||
| 22 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\text{Ni}^{2+}$ ($3d^8$) with ammonia forms $[\text{Ni}(\text{NH}_3)_6]^{2+}$, $sp^3d^2$: octahedral.
$\text{Pt}^{2+}$ ($5d^8$): the large splitting of a 5d ion pairs the electrons, giving $[\text{Pt}(\text{NH}_3)_4]^{2+}$, $dsp^2$: square planar.
$\text{Zn}^{2+}$ ($3d^{10}$): $[\text{Zn}(\text{NH}_3)_4]^{2+}$, $sp^3$: tetrahedral.
Answer: (A). | ||||
| 23 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids An ortho OH hydrogen-bonds to the carboxylate oxygen and stabilises the anion strongly (with the ortho effect). With two ortho OH groups (I, 2,6-dihydroxybenzoic acid) both sides of the carboxylate are bonded, so it is the strongest; salicylic acid (II) with one is next.
A meta OH (III) withdraws electrons by its −I effect without resonance donation to the carboxyl: slightly stronger than benzoic acid.
A para OH (IV) donates electrons by resonance (+R), destabilising the anion: the weakest.
Order: I > II > III > IV.
Answer: (A). | ||||
| 24 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Step i): the ketone $\text{PhCOCH}(\text{CH}_3)_2$ has one α-hydrogen. Its enolate adds to formaldehyde (crossed aldol): $\text{PhCO–C}(\text{CH}_3)_2\text{–CH}_2\text{OH}$.
This β-hydroxy ketone has no α-hydrogen left. With excess formaldehyde and hot alkali, a crossed Cannizzaro-type hydride transfer takes place: formaldehyde (as its hydroxide adduct) gives a hydride to the carbonyl carbon and becomes formate, so the ketone is reduced to the secondary alcohol: $\text{PhCH(OH)–C}(\text{CH}_3)_2\text{–CH}_2\text{OH}$, a 1,3-diol.
Step ii): a 1,3-diol and formaldehyde with a trace of acid form a cyclic acetal, a 1,3-dioxane: ring O–CH₂–O closed through CH(Ph)–C(CH₃)₂–CH₂.
That is structure (A). (B) and (D) keep the ketone, and (C) has a hemiketal OH on the ring carbon, which is not the product of these conditions.
Answer: (A). | ||||
| 25 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Multiple correctMedium | |
A medium multiple correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) $\text{C}_2^{2-}$ has 14 electrons, like $\text{N}_2$: all paired. Diamagnetic. True.
(B) $\text{O}_2^{2+}$ (14 electrons) has bond order 3; $\text{O}_2$ has 2. Higher bond order means a shorter bond. False.
(C) $\text{N}_2^+$ (13 electrons): $\frac{9 - 4}{2} = 2.5$. $\text{N}_2^-$ (15 electrons): $\frac{10 - 5}{2} = 2.5$. Same. True.
(D) $\text{He}_2^+$ has bond order $\frac{2 - 1}{2} = \frac12$: it is bound, so its energy is lower than that of the separated He and He⁺ (and it is not the same as two neutral He atoms, which have one more electron). False.
Answer: (A), (C). | ||||
| 26 | SolutionsRaoult's law | Class 12: Solutions | Multiple correctEasy | |
A easy multiple correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Positive deviation: the A–B attractions are weaker than A–A and B–B, so molecules escape more easily.
(A) $\text{CCl}_4$ breaks up the hydrogen bonding of methanol: positive. True.
(B) $\text{CS}_2$ + acetone: the dipolar attractions of acetone are weakened by the non-polar $\text{CS}_2$: positive (a classic example). True.
(C) Benzene + toluene are alike: nearly ideal. False.
(D) Phenol + aniline form strong O–H···N hydrogen bonds: negative deviation. False.
Answer: (A), (B). | ||||
| 27 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter (A) An atom in the topmost layer has 6 neighbours in its own layer and 3 in the layer below, none above: 9, not 12. False.
(B) ccp (fcc) packing efficiency $= \frac{4\cdot\frac43\pi r^3}{(2\sqrt2r)^3} = \frac{\pi}{3\sqrt2} \approx 74\%$. True.
(C) Per atom there is 1 octahedral void and 2 tetrahedral voids. True.
(D) Atoms touch along a face diagonal: $\sqrt2a = 4r$, so $a = 2\sqrt2r$. True.
Answer: (B), (C), (D). | ||||
| 28 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter (A) The sulphide ore is crushed and concentrated by froth flotation. True.
(B) After roasting, iron is removed in the blast/reverberatory furnace with silica: $\text{FeO} + \text{SiO}_2 \rightarrow \text{FeSiO}_3$ (slag). True.
(C) In the converter, part of $\text{Cu}_2\text{S}$ is oxidised and the rest reduces the oxide: $2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu} + \text{SO}_2$. The escaping $\text{SO}_2$ blisters the copper. True.
(D) Blister copper is refined by poling and electrolytically, not by carbon reduction. False.
Answer: (A), (B), (C). | ||||
| 29 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\text{P}_4\text{O}_{10}$ dehydrates nitric acid: $4\text{HNO}_3 + \text{P}_4\text{O}_{10} \rightarrow 2\text{N}_2\text{O}_5 + 4\text{HPO}_3$. The compound is $\text{N}_2\text{O}_5$.
(A) $\text{P}_4$ with $\text{HNO}_3$ is oxidised to $\text{H}_3\text{PO}_4$ and gives $\text{NO}_2$, not $\text{N}_2\text{O}_5$. False.
(B) $\text{N}_2\text{O}_5$ has all electrons paired. True.
(C) Its structure is $\text{O}_2\text{N–O–NO}_2$: the nitrogens are bridged by oxygen, no N–N bond. False.
(D) $\text{Na} + \text{N}_2\text{O}_5 \rightarrow \text{NaNO}_3 + \text{NO}_2$; $\text{NO}_2$ is a brown gas. True.
Answer: (B), (D). | ||||
| 30 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Multiple correctMedium | |
A medium multiple correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules (A) Invert sugar comes from hydrolysis of sucrose (maltose gives only glucose). False.
(B) Sucrose hydrolyses to equal amounts of D-(+)-glucose and D-(−)-fructose. True.
(C) The D-forms rotate opposite to the given L-forms: D-glucose +52°, D-fructose −92°. Equimolar mixture: $\frac{52 - 92}{2} = -20°$. True.
(D) Bromine water is a mild oxidant: it turns glucose into gluconic acid, not saccharic (glucaric) acid, which needs nitric acid. False.
Answer: (B), (C). | ||||
| 31 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Multiple correctMedium | |
A medium multiple correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Only the aldehyde is to be reduced to a primary alcohol; the carboxylic acid, the ester and the epoxide must survive.
(A) $\text{LiAlH}_4$ also reduces the acid and the ester (and opens epoxides). False.
(B) Borane reduces carboxylic acids quickly (faster than esters), so the COOH would be reduced too. False.
(C) $\text{NaBH}_4$ in ethanol reduces aldehydes but leaves acids, esters and epoxides alone. True.
(D) $\text{H}_2$ over Raney nickel under mild conditions hydrogenates the aldehyde C=O to $\text{CH}_2\text{OH}$ and does not touch the acid or ester. True.
Answer: (C), (D). | ||||
| 32 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Multiple correctMedium | |
A medium multiple correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $tert$-Butylbenzene forms by Friedel–Crafts alkylation through the $tert$-butyl cation $(\text{CH}_3)_3\text{C}^+$.
(A) Sodium ethoxide is a strong base, not a Lewis acid: it eliminates HBr from $tert$-butyl bromide to give isobutylene; benzene is not alkylated. False.
(B) Isobutyl chloride with $\text{AlCl}_3$ gives a primary carbocation-like complex that rearranges by a hydride shift to the $tert$-butyl cation. True.
(C) Isobutylene is protonated by $\text{H}_2\text{SO}_4$ (Markovnikov) to the $tert$-butyl cation. True.
(D) $\text{BF}_3$ makes the OH of isobutyl alcohol a leaving group; the cation formed rearranges to $tert$-butyl. True.
Answer: (B), (C), (D). | ||||
| 33 | EquilibriumChemical equilibrium | Class 11: Equilibrium | ComprehensionEasy | |
A easy comprehension question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium At equilibrium: $\text{X}_2$: $1 - \frac{\beta}{2}$ mol; X: $\beta$ mol; total $1 + \frac{\beta}{2}$ mol, at $P = 2$ bar.
$K_P = \frac{p_X^2}{p_{X_2}} = \frac{\left(\frac{\beta}{1 + \beta/2}\right)^2P^2}{\frac{1 - \beta/2}{1 + \beta/2}P} = \frac{\beta^2P}{(1 + \beta/2)(1 - \beta/2)} = \frac{2\beta^2}{1 - \beta^2/4} = \frac{8\beta^2}{4 - \beta^2}$.
Answer: (B). | ||||
| 34 | EquilibriumChemical equilibrium | Class 11: Equilibrium | ComprehensionMedium | |
A medium comprehension question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $\Delta_rG^\circ > 0$ means $K_P < 1$.
(A) The forward reaction increases the moles of gas, so lowering the pressure shifts it forward (Le Chatelier). Correct statement.
(B) At the start there is no X: $Q = 0$ and $\Delta_rG = \Delta_rG^\circ + RT\ln Q \to -\infty$, so dissociation is spontaneous. Correct statement.
(C) With $\beta = 0.7$: $K_P = \frac{8(0.49)}{4 - 0.49} = \frac{3.92}{3.51} \approx 1.12 > 1$, which would make $\Delta_rG^\circ < 0$. This contradicts the data, so the statement is incorrect.
(D) $K_C = \frac{K_P}{(RT)^{\Delta n}} = \frac{K_P}{0.083 \times 298} = \frac{K_P}{24.7} < 1$. Correct statement.
Answer: (C). | ||||
| 35 | AminesClassification and identification of amines | Class 12: Amines | ComprehensionMedium | |
A medium comprehension question on classification and identification of amines, from Amines. NCERT: Class 12: Amines O is 1,2-dipropylbenzene. Hot acidic $\text{KMnO}_4$ oxidises each side chain to COOH: P is phthalic acid.
Heated with ammonia, P gives phthalamide (Q), $\text{C}_6\text{H}_4(\text{CONH}_2)_2$ (strong heating of it, below, gives the imide).
$\text{Br}_2$/NaOH is the Hofmann bromamide reaction: each $\text{–CONH}_2$ loses its carbonyl carbon and becomes $\text{–NH}_2$. R is benzene-1,2-diamine.
Answer: (A). | ||||
| 36 | BiomoleculesProteins | Class 12: Biomolecules | ComprehensionMedium | |
A medium comprehension question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Strong heating of phthalamide Q expels ammonia to give phthalimide (S).
With KOH, phthalimide forms its potassium salt, which displaces bromide from ethyl 2-bromopropanoate: N-substituted phthalimide carrying $\text{–CH}(\text{CH}_3)\text{COOEt}$ (Gabriel synthesis).
Hydrolysis and acidification free the amine and the acid: $\text{CH}_3\text{CH}(\text{NH}_2)\text{COOH}$, alanine.
Answer: (B). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 37 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Write $P = I + N$ with $N = \begin{bmatrix}0&0&0\\4&0&0\\16&4&0\end{bmatrix}$. Then $N^2 = \begin{bmatrix}0&0&0\\0&0&0\\16&0&0\end{bmatrix}$ and $N^3 = O$.
$I$ and $N$ commute, so by the binomial theorem $P^{50} = I + 50N + \binom{50}{2}N^2 = I + 50N + 1225N^2$.
$Q = P^{50} - I = 50N + 1225N^2$:
$q_{21} = 50 \cdot 4 = 200$, $q_{32} = 50 \cdot 4 = 200$, $q_{31} = 50 \cdot 16 + 1225 \cdot 16 = 20400$.
$\frac{q_{31} + q_{32}}{q_{21}} = \frac{20600}{200} = 103$.
Answer: (B). | ||||
| 38 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The logs are in A.P. with difference $\ln 2$, so the $b_i$ are in G.P. with ratio 2: $b_i = 2^{i-1}b_1$.
$t = b_1(2^{51} - 1)$.
$a_1 = b_1$, $a_{51} = 2^{50}b_1$, so $s = \frac{51}{2}(a_1 + a_{51}) = \frac{51}{2}b_1(1 + 2^{50})$.
$s - t = b_1\left[\frac{51}{2} + \frac{51}{2}2^{50} - 2 \cdot 2^{50} + 1\right] > 0$, so $s > t$.
$a_{101} = a_1 + 100d$ with $50d = a_{51} - a_1$, so $a_{101} = 2a_{51} - a_1 = (2^{51} - 1)b_1$, while $b_{101} = 2^{100}b_1$. Hence $a_{101} < b_{101}$.
Answer: (B). | ||||
| 39 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $A_k = \frac{\pi}{4} + \frac{(k-1)\pi}{6}$, so the second angle is $A_k + \frac{\pi}{6}$.
$\frac{1}{\sin A\sin(A + \frac{\pi}{6})} = \frac{1}{\sin\frac{\pi}{6}}\cdot\frac{\sin((A + \frac{\pi}{6}) - A)}{\sin A\sin(A + \frac{\pi}{6})} = 2\left[\cot A - \cot\left(A + \frac{\pi}{6}\right)\right]$.
The sum telescopes: $2\left[\cot\frac{\pi}{4} - \cot\left(\frac{\pi}{4} + \frac{13\pi}{6}\right)\right]$.
$\frac{\pi}{4} + \frac{13\pi}{6} = 2\pi + \frac{5\pi}{12}$, and $\cot\frac{5\pi}{12} = \cot75° = 2 - \sqrt3$.
Sum $= 2[1 - (2 - \sqrt3)] = 2(\sqrt3 - 1)$.
Answer: (C). | ||||
| 40 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replace $x$ by $-x$: $I = \int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1 + e^{-x}}dx = \int_{-\pi/2}^{\pi/2}\frac{x^2e^x\cos x}{1 + e^x}dx$.
Add the two forms: $2I = \int_{-\pi/2}^{\pi/2}x^2\cos x\,dx = 2\int_0^{\pi/2}x^2\cos x\,dx$.
By parts: $\int x^2\cos x\,dx = x^2\sin x + 2x\cos x - 2\sin x$.
$I = \left[x^2\sin x + 2x\cos x - 2\sin x\right]_0^{\pi/2} = \frac{\pi^2}{4} - 2$.
Answer: (A). | ||||
| 41 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The region lies above $y = \sqrt{|x + 3|}$, below the line $y = \frac{x + 9}{5}$, with $x \le 6$.
Intersections: for $x \ge -3$, $\frac{(x + 9)^2}{25} = x + 3$ gives $x^2 - 7x + 6 = 0$, $x = 1, 6$; for $x < -3$, $(x + 9)^2 = -25(x + 3)$ gives $x^2 + 43x + 156 = 0$, $x = -4$ (or $-39$, where the line is below the axis).
The line is above the curve for $-4 < x < 1$ (at $x = 0$: $1.8 > \sqrt3$); between 1 and 6 the curve is above the line (at $x = 3$: $\sqrt6 > 2.4$).
Area $= \int_{-4}^{1}\frac{x + 9}{5}dx - \int_{-4}^{-3}\sqrt{-x - 3}\,dx - \int_{-3}^{1}\sqrt{x + 3}\,dx = \frac{100 - 25}{10} - \frac23 - \frac23(4)^{3/2} = \frac{15}{2} - \frac23 - \frac{16}{3} = \frac32$.
Answer: (C). | ||||
| 42 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Image of $(3, 1, 7)$ in $x - y + z = 3$: $\frac{x - 3}{1} = \frac{y - 1}{-1} = \frac{z - 7}{1} = -2\cdot\frac{3 - 1 + 7 - 3}{3} = -4$.
$P = (-1, 5, 3)$.
The plane contains the origin and the direction $(1, 2, 1)$ of the line, and $P$. Its normal is $(1, 2, 1) \times (-1, 5, 3) = (6 - 5,\ -1 - 3,\ 5 + 2) = (1, -4, 7)$.
Plane: $x - 4y + 7z = 0$ (and $P$ checks: $-1 - 20 + 21 = 0$).
Answer: (C). | ||||
| 43 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctHard | |
A hard multiple correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Since $\frac{n^n}{n!} = \prod_{r=1}^{n}\frac{n}{r}$, the bracket is $\prod_{r=1}^{n}\frac{\frac{n}{r}\left(x + \frac{n}{r}\right)}{x^2 + \frac{n^2}{r^2}}$. With $t = \frac{r}{n}$ each factor is $\frac{\frac1t(x + \frac1t)}{x^2 + \frac{1}{t^2}} = \frac{1 + xt}{1 + x^2t^2}$.
So $\ln f(x) = \lim \frac{x}{n}\sum_{r=1}^{n}\ln\frac{1 + x\frac{r}{n}}{1 + x^2\frac{r^2}{n^2}} = x\int_0^1\ln\frac{1 + xt}{1 + x^2t^2}dt = \int_0^x\ln\frac{1 + u}{1 + u^2}du$.
Hence $\frac{f'(x)}{f(x)} = \ln\frac{1 + x}{1 + x^2}$, positive for $0 < x < 1$ and negative for $x > 1$: $f$ increases on $(0, 1)$ and decreases after.
(A) $f\left(\frac12\right) < f(1)$. False.
(B) $\frac13 < \frac23 < 1$, so $f\left(\frac13\right) \le f\left(\frac23\right)$. True.
(C) $f'(2) = f(2)\ln\frac35 < 0$. True.
(D) $\ln\frac{4}{10} < \ln\frac{3}{5}$, so $\frac{f'(3)}{f(3)} < \frac{f'(2)}{f(2)}$. False.
Answer: (B), (C). | ||||
| 44 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctMedium | |
A medium multiple correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Cosine is even, so $\cos(|x^3 - x|) = \cos(x^3 - x)$: differentiable everywhere.
$x^3 + x = x(x^2 + 1)$ has the sign of $x$, so $|x^3 + x| = |x|(x^2 + 1)$. For $x \ge 0$: $|x|\sin(|x^3 + x|) = x\sin(x^3 + x)$. For $x < 0$: $(-x)\sin(-(x^3 + x)) = x\sin(x^3 + x)$. So this term is $x\sin(x^3 + x)$ for every $x$: differentiable everywhere.
Therefore $f$ is differentiable at every point for all $a, b$.
(A) True. (B) True. (C) False. (D) False.
Answer: (A), (B). | ||||
| 45 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Multiple correctMedium | |
A medium multiple correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives At $x = 2$ both $f(x)g(x)$ and $f'(x)g'(x)$ tend to 0, so by L'Hôpital's rule
$1 = \lim_{x\to2}\frac{f'g + fg'}{f''g' + f'g''} = \frac{0 + f(2)g'(2)}{f''(2)g'(2) + 0} = \frac{f(2)}{f''(2)}$.
So $f''(2) = f(2) > 0$ (since $f$ takes only positive values).
With $f'(2) = 0$ and $f''(2) > 0$, $f$ has a local minimum at 2: (A) true, (B) false.
(C) $f''(2) = f(2)$, not greater. False.
(D) $f(2) - f''(2) = 0$. True.
Answer: (A), (D). | ||||
| 46 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctHard | |
A hard multiple correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability On $\left[-\frac12, 2\right]$, $x^2 \in [0, 4]$ and $x^2 - 3 \in [-3, 1]$. $[x^2 - 3]$ jumps where $x^2 - 3$ crosses an integer, i.e. $x^2 = 1, 2, 3, 4$: $x = 1, \sqrt2, \sqrt3, 2$. (At $x = 0$, $x^2 = 0$ is a minimum, $x^2 - 3$ stays in $[-3, -2)$ nearby, so $f = -3$ is continuous there; at $x = 2$, $f(2) = 1$ but the left limit is 0.) So $f$ is discontinuous at exactly four points: (A) false, (B) true.
$g(x) = (|x| + |4x - 7|)f(x)$, and $|x| + |4x - 7| > 0$ throughout, so $g$ is discontinuous, hence not differentiable, at $1, \sqrt2, \sqrt3$ in the open interval.
Near $x = 0$: $f = -3$ and $g = -3(|x| + 7 - 4x)$, which has a corner: not differentiable at 0.
Near $x = \frac74$: $\sqrt3 < \frac74 < 2$, where $x^2 \in (3, 4)$ and $f = 0$; so $g = 0$ near $\frac74$ and is differentiable there.
So $g$ fails to be differentiable at exactly $0, 1, \sqrt2, \sqrt3$: four points. (C) true, (D) false.
Answer: (B), (C). | ||||
| 47 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Multiple correctMedium | |
A medium multiple correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z = \frac{a - ibt}{a^2 + b^2t^2}$, so $x = \frac{a}{a^2 + b^2t^2}$, $y = \frac{-bt}{a^2 + b^2t^2}$.
For $a \ne 0$, $b \ne 0$: $x^2 + y^2 = \frac{1}{a^2 + b^2t^2} = \frac{x}{a}$, the circle $x^2 + y^2 - \frac{x}{a} = 0$ with centre $\left(\frac{1}{2a}, 0\right)$ and radius $\frac{1}{2|a|}$.
(A) $a > 0$: radius $\frac{1}{2a}$, centre $\left(\frac{1}{2a}, 0\right)$. True.
(B) $a < 0$: the radius $-\frac{1}{2a}$ is right, but the centre is $\left(\frac{1}{2a}, 0\right)$, not $\left(-\frac{1}{2a}, 0\right)$. False.
(C) $b = 0$: $z = \frac1a$ is real, on the $x$-axis. True.
(D) $a = 0$: $z = \frac{1}{ibt} = -\frac{i}{bt}$, purely imaginary, on the $y$-axis. True.
Answer: (A), (C), (D). | ||||
| 48 | Conic SectionsParabola | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Circle: $(x - 2)^2 + (y - 8)^2 = 4$, centre $S(2, 8)$, radius 2.
The nearest point $P(t^2, 2t)$ of the parabola lies on the normal through $S$. Normal at $t$: $y = -tx + 2t + t^3$. Through $(2, 8)$: $8 = t^3$, $t = 2$, so $P = (4, 4)$.
(A) $SP = \sqrt{4 + 16} = 2\sqrt5$. True.
(B) $SQ = 2$, $QP = 2\sqrt5 - 2$: $SQ : QP = 1 : (\sqrt5 - 1) = (\sqrt5 + 1) : 4$. False.
(C) Normal at $P$: $y = -2x + 12$, meeting the $x$-axis at 6. True.
(D) The radius $SQ$ lies along $SP$, slope $\frac{4 - 8}{4 - 2} = -2$; the tangent at $Q$ is perpendicular to it, slope $\frac12$. True.
Answer: (A), (C), (D). | ||||
| 49 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Multiple correctEasy | |
A easy multiple correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Determinant $= \begin{vmatrix}a&2\\3&-2\end{vmatrix} = -2a - 6$, zero only for $a = -3$.
(B) $a \ne -3$: non-zero determinant, a unique solution for every $\lambda, \mu$. True.
For $a = -3$ the equations are $-3x + 2y = \lambda$ and $3x - 2y = \mu$; adding gives $0 = \lambda + \mu$.
(C) If $\lambda + \mu = 0$ the two equations are the same line: infinitely many solutions. True.
(D) If $\lambda + \mu \ne 0$ they are parallel distinct lines: no solution. True.
(A) Fails when $\lambda + \mu \ne 0$. False.
Answer: (B), (C), (D). | ||||
| 50 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Multiple correctMedium | |
A medium multiple correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\hat{w}$ and $\hat{u} \times \vec{v}$ are both of length 1 and their dot product is 1, so the angle between them is 0: $\hat{u} \times \vec{v} = \hat{w}$.
This needs $\hat{u} \perp \hat{w}$: $u_1 + u_2 + 2u_3 = 0$. If $\vec{v}_0$ is one solution, so is $\vec{v}_0 + \lambda\hat{u}$ for every real $\lambda$ (since $\hat{u} \times \hat{u} = \vec0$).
(A) False. (B) True.
(C) In the $xy$-plane $u_3 = 0$, so $u_1 = -u_2$ and $|u_1| = |u_2|$. True.
(D) In the $xz$-plane $u_2 = 0$, so $u_1 = -2u_3$, i.e. $|u_1| = 2|u_3|$, not $2|u_1| = |u_3|$. False.
Answer: (B), (C). | ||||
| 51 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | ComprehensionEasy | |
A easy comprehension question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability In one game $X - Y$ changes by $+3$ (win, $\frac12$), $0$ (draw, $\frac16$) or $-3$ (loss, $\frac13$). $X > Y$ when $T_1$ wins more games than it loses.
WW: $\frac12\cdot\frac12 = \frac14$. WD or DW: $2\cdot\frac12\cdot\frac16 = \frac16$.
$P(X > Y) = \frac14 + \frac16 = \frac{5}{12}$.
Answer: (B). | ||||
| 52 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | ComprehensionEasy | |
A easy comprehension question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $X = Y$ when wins and losses balance: DD, or one win and one loss.
DD: $\frac16\cdot\frac16 = \frac{1}{36}$. WL or LW: $2\cdot\frac12\cdot\frac13 = \frac13 = \frac{12}{36}$.
$P(X = Y) = \frac{13}{36}$.
Answer: (C). | ||||
| 53 | Conic SectionsEllipse | Class 11: Conic Sections | ComprehensionMedium | |
A medium comprehension question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse: $a^2 = 9$, $b^2 = 8$, $e^2 = 1 - \frac89 = \frac19$, foci $(\pm1, 0)$. So $F_1 = (-1, 0)$, $F_2 = (1, 0)$ and the parabola is $y^2 = 4x$.
Intersection: $\frac{x^2}{9} + \frac{4x}{8} = 1$, $2x^2 + 9x - 18 = 0$, $x = \frac32$ (the other root is negative). $y^2 = 6$: $M = \left(\frac32, \sqrt6\right)$, $N = \left(\frac32, -\sqrt6\right)$.
By symmetry the altitude from $F_1$ is the $x$-axis. The altitude from $M$ is perpendicular to $F_1N$, whose slope is $\frac{-\sqrt6}{5/2} = -\frac{2\sqrt6}{5}$; so its slope is $\frac{5}{2\sqrt6}$.
On it, at $y = 0$: $x - \frac32 = -\sqrt6\cdot\frac{2\sqrt6}{5} = -\frac{12}{5}$, $x = -\frac{9}{10}$.
Orthocentre $\left(-\frac{9}{10}, 0\right)$.
Answer: (A). | ||||
| 54 | Conic SectionsEllipse | Class 11: Conic Sections | ComprehensionMedium | |
A medium comprehension question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections With $M = \left(\frac32, \sqrt6\right)$, $N = \left(\frac32, -\sqrt6\right)$, $F_1 = (-1, 0)$, $F_2 = (1, 0)$ (from the paragraph).
Tangent to the ellipse at $M$: $\frac{x(3/2)}{9} + \frac{y\sqrt6}{8} = 1$; by symmetry the two tangents meet on the $x$-axis, where $\frac{x}{6} = 1$: $R = (6, 0)$.
Normal to $y^2 = 4x$ at $M = (t^2, 2t)$ with $t = \frac{\sqrt6}{2}$: $y = -tx + 2t + t^3$, meeting $y = 0$ at $x = 2 + t^2 = \frac72$: $Q = \left(\frac72, 0\right)$.
Area $MQR = \frac12\cdot QR\cdot\sqrt6 = \frac12\cdot\frac52\cdot\sqrt6 = \frac{5\sqrt6}{4}$.
$MF_1NF_2$ has perpendicular diagonals $F_1F_2 = 2$ and $MN = 2\sqrt6$: area $= \frac12\cdot2\cdot2\sqrt6 = 2\sqrt6$.
Ratio $= \frac{5\sqrt6/4}{2\sqrt6} = \frac58$.
Answer: (C). | ||||
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