JEE (Advanced) · 2020
JEE Advanced 2020, 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
- Properties of Bulk Matter7
- Electrostatics6
- Motion of System of Particles and Rigid Body3
6 easy · 21 medium · 9 hard
Chemistry
36 Q
- Organic Chemistry: Some Basic Principles and Techniques5
- p-Block Elements3
- Equilibrium3
11 easy · 19 medium · 6 hard
Mathematics
36 Q
- Limits, Continuity and Differentiability5
- Integral Calculus3
- Probability3
4 easy · 25 medium · 7 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 11: Systems of Particles and Rotational Motion
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 11: Units and Measurement
Class 12: Current Electricity
Class 11: Kinetic Theory
Class 11: Laws of Motion
Class 11: Thermodynamics
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 12: Atoms; Class 12: Nuclei
Class 11: Oscillations; Class 11: Waves
Class 12: Dual Nature of Radiation and Matter
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Class 11: Organic Chemistry – Some Basic Principles and Techniques
No matching NCERT chapter
Class 11: Equilibrium
Class 11: Some Basic Concepts of Chemistry
Class 12: Biomolecules
Class 11: Thermodynamics
Class 11: Hydrocarbons
Class 12: Chemical Kinetics
Class 11: Chemical Bonding and Molecular Structure
Class 12: The d- and f-Block Elements
Class 12: Amines
No matching NCERT chapter
Class 12: Coordination Compounds
No matching NCERT chapter
- Principles Related to Practical Chemistry1
No matching NCERT chapter
Class 12: Electrochemistry
Class 11: Classification of Elements and Periodicity in Properties
No matching NCERT chapter
Class 12: Solutions
Class 11: Structure of Atom
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Probability; Class 12: Probability
Class 12: Application of Derivatives
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Class 11: Conic Sections
Class 12: Matrices; Class 12: Determinants
Class 11: Complex Numbers and Quadratic Equations
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 11: Sequences and Series
Class 12: Vector Algebra
Class 11: Permutations and Combinations
Class 11: Complex Numbers and Quadratic Equations
Class 11: Conic Sections
Class 12: Differential Equations
Class 11: Binomial Theorem
03
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | 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 ball touches the plank only at the rim of the hole. In the plane of the figure the two contact points are the ends of a chord of length $2r$, and the ball's centre lies on the perpendicular bisector of that chord, at height $d = \sqrt{R^2 - r^2}$ above the plank.
The ball can stay at rest only while the vertical line through its centre (the line of action of its weight) falls between the two contact points. When the plank is tilted by $\theta$, that vertical line is turned by $\theta$ from the normal to the plank, so it meets the plank at a distance $d\tan\theta$ from the midpoint of the chord, down the slope.
The ball is about to roll over the lower edge when $d\tan\theta = r$:
$\tan\theta = \frac{r}{\sqrt{R^2 - r^2}}$, i.e. $\sin\theta = \frac{r}{R}$.
(Geometrically: the line from the centre to the lower edge makes angle $\theta$ with the normal and has length $R$, with side $r$ along the plank.)
Answer: (A). | ||||
| Q.2 | Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current As the magnet moves, the flux through the part of the disc under it changes, so eddy currents are induced in the aluminium.
By Lenz's law these currents oppose the cause, which is the relative motion between the magnet and the disc. The magnetic force on the eddy currents therefore drags the disc along with the magnet, trying to reduce the relative motion (the principle of the induction motor and of the eddy-current speedometer).
So the light disc starts rotating in the same direction as the magnet revolves (always lagging behind it, so some eddy current, and some heating, remains).
Answer: (B). | ||||
| Q.3 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The roller (radius $10$ cm) rolls without slipping on the floor, so if its centre moves with speed $v$ its angular speed is $\omega = \frac{v}{10\text{ cm}}$.
The top of the axle (radius $5$ cm) moves at $v + \omega\cdot 5\text{ cm} = v + \frac{v}{2} = \frac{3v}{2}$.
The scale does not slip on the axle, so it moves with the top of the axle: $1.5$ times as far as the roller.
When the roller has moved $50$ cm, the scale has moved $75$ cm over the floor, i.e. $75 - 50 = 25$ cm forward relative to the roller.
At the start the edge of the scale (its $100$ cm end) was on top of the axle. Now the scale has advanced $25$ cm past the axle: the axle is under the $75$ cm mark of the scale, and $25$ cm of the scale sticks out beyond it. That is figure (B) ($x = 0$ the far end of the scale, the axle at $x = 75$ cm).
Answer: (B). | ||||
| Q.4 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter While a current $i$ flows, the coil has magnetic moment $\mu = N i\,\pi R^2$, perpendicular to its plane, and so perpendicular to $B_o$ (which lies in the plane of the coil).
The torque on it is $\tau = \mu B_o = N i \pi R^2 B_o$.
The coil hardly turns during the discharge, so the angle stays $90^\circ$ and the angular impulse is
$L = \int \tau\,dt = N\pi R^2 B_o \int i\,dt = N\pi R^2 B_o Q$.
So the angular momentum gained is $\pi N Q B_o R^2$.
Answer: (B). | ||||
| Q.5 | OpticsWave optics | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on wave optics, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Light travels more slowly in glass, so the more glass a part of the wavefront crosses, the further it lags behind.
The left face is flat. The right face has an outward bulge at the top, an inward hollow in the middle and an outward bulge at the bottom.
Top and bottom bulges: the glass is thickest at the middle of each bulge, so the wavefront lags most there. Each of these parts acts as a converging lens, and its piece of wavefront is curved like "(" (middle behind, ends ahead).
Middle hollow: the glass is thinnest at its middle, so the wavefront leads there. This part acts as a diverging lens, and its piece of wavefront is curved like ")" (middle ahead).
So the emergent wavefront, from top to bottom, is "(" then ")" then "(": the middle bulges forward (to the right) and the two outer parts bend back. That is the shape in (A); (B) is its mirror image, and (C), (D) have only one bend.
Answer: (A). | ||||
| Q.6 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Take the level of the kerosene–water interface in the left arm. Pressures at that level in the two arms are equal:
$\rho_k g (0.1) = \rho_w g\,\Delta$, where $\Delta$ is the height of the water in the right arm above the interface.
$\Delta = \frac{800 \times 0.1}{1000} = 0.08$ m.
The water's volume is unchanged and the tube is uniform, so if the left water level falls by $x$ the right one rises by $x$: $2x = 0.08$, $x = 0.04$ m.
Water level on the left: $0.29 - 0.04 = 0.25$ m; on the right: $0.29 + 0.04 = 0.33$ m.
$h_1 = 0.25 + 0.1 = 0.35$ m, $h_2 = 0.33$ m, so $\frac{h_1}{h_2} = \frac{35}{33}$.
Answer: (B). | ||||
| Q.7 | 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 The force is $-\frac{dV}{dr} = -F$: a constant pull of size $F$ towards the origin.
Circular orbit: $\frac{mv^2}{R} = F$. Bohr quantisation: $mvR = \frac{nh}{2\pi}$.
Square the second: $m^2v^2R^2 = \frac{n^2h^2}{4\pi^2}$, and put $mv^2 = FR$: $mFR^3 = \frac{n^2h^2}{4\pi^2}$, so
$R = \left(\frac{n^2h^2}{4\pi^2mF}\right)^{1/3} \propto n^{2/3}$, and $v = \sqrt{\frac{FR}{m}} \propto R^{1/2} \propto n^{1/3}$.
Energy: $E = \frac{1}{2}mv^2 + FR = \frac{1}{2}FR + FR = \frac{3}{2}FR$, and $FR = F\left(\frac{n^2h^2}{4\pi^2mF}\right)^{1/3} = \left(\frac{n^2h^2F^2}{4\pi^2m}\right)^{1/3}$.
So $E = \frac{3}{2}\left(\frac{n^2h^2F^2}{4\pi^2m}\right)^{1/3}$.
(A) has the powers the wrong way round: FALSE. (B) TRUE. (C) TRUE. (D) has 2 in place of $\frac{3}{2}$: FALSE.
Answer: (B), (C). | ||||
| Q.8 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctMedium | |
A medium multiple correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter (A) Stefan's law: $P = \sigma A T^4 = 5.67\times10^{-8} \times 64\times10^{-6} \times (2500)^4$. With $(2500)^4 = 3.906\times10^{13}$: $P = 3.629\times10^{-12} \times 3.906\times10^{13} \approx 141.8$ W. Not 642–645 W: FALSE.
(B) A point source spreads $P$ over a sphere of radius $d = 100$ m. The pupil collects the fraction $\frac{\pi a^2}{4\pi d^2} = \frac{(3\times10^{-3})^2}{4\times10^4} = 2.25\times10^{-10}$.
Power into one eye $= 141.75 \times 2.25\times10^{-10} \approx 3.19\times10^{-8}$ W, inside $3.15$–$3.25 \times10^{-8}$ W: TRUE.
(C) Wien: $\lambda_m = \frac{2.90\times10^{-3}}{2500} = 1.16\times10^{-6}$ m $= 1160$ nm: TRUE.
(D) Energy of one photon at $1740$ nm: $\frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3\times10^8}{1.74\times10^{-6}} = 1.143\times10^{-19}$ J.
Photons per second into the eye $= \frac{3.19\times10^{-8}}{1.143\times10^{-19}} \approx 2.79\times10^{11}$, inside $2.75$–$2.85\times10^{11}$: TRUE.
Answer: (B), (C), (D). | ||||
| Q.9 | 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 Let time have dimension $[X^t]$ and mass $[X^\mu]$ in this system. Then
speed = position/time: $\beta = \alpha - t$;
acceleration = speed/time: $p = \beta - t = \alpha - 2t$;
momentum = mass × speed: $q = \mu + \beta$;
force = mass × acceleration: $r = \mu + p$.
(A) $\alpha + p = 2\alpha - 2t = 2(\alpha - t) = 2\beta$: TRUE.
(B) $p + q - r = p + (\mu + \beta) - (\mu + p) = \beta$: TRUE.
(C) $p - q + r = p - \mu - \beta + \mu + p = 2p - \beta = \alpha - 3t$, which is not $\alpha$ in general: FALSE.
(D) $p + q + r = 2p + 2\mu + \beta$, not $\beta$ in general: FALSE.
Answer: (A), (B). | ||||
| Q.10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The force on the positive charge is along $-\hat{y}$, so the particle moves like a projectile with a constant downward acceleration
$a = \frac{qE}{m} = 10^{10} \times 400\sqrt{3} = 4\sqrt{3}\times10^{12}\ \text{ms}^{-2}$.
Gravity is negligible beside this. T is on the same horizontal line as the entry point, so the horizontal range must be 5 m:
$\frac{u^2\sin 2\theta}{a} = 5$, with $u^2 = 40\times10^{12}$: $\sin 2\theta = \frac{5 \times 4\sqrt{3}\times10^{12}}{40\times10^{12}} = \frac{\sqrt{3}}{2}$.
So $2\theta = 60^\circ$ or $120^\circ$: $\theta = 30^\circ$ or $60^\circ$. (A) FALSE ($45^\circ$ gives a range of $\frac{10}{\sqrt3} \approx 5.77$ m); (B) TRUE.
Time of flight $t = \frac{2u\sin\theta}{a} = \frac{2\times2\sqrt{10}\times10^6\sin\theta}{4\sqrt{3}\times10^{12}} = \sqrt{\frac{10}{3}}\sin\theta\ \mu\text{s}$.
$\theta = 30^\circ$: $t = \frac{1}{2}\sqrt{\frac{10}{3}} = \sqrt{\frac{5}{6}}\ \mu$s. $\theta = 60^\circ$: $t = \frac{\sqrt3}{2}\sqrt{\frac{10}{3}} = \sqrt{\frac{5}{2}}\ \mu$s.
(C) TRUE. (D) $\sqrt{\frac{5}{3}}\ \mu$s is neither: FALSE.
Answer: (B), (C). | ||||
| Q.11 | Current ElectricityResistivity | Class 12: Current Electricity | Multiple correctHard | |
A hard multiple correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity (A) Split the strip into thin semicircular rings of radius $r$ and width $dr$. Each has length $\pi r$, cross-section $t\,dr$, and the full voltage $V_0$ across it, so they are in parallel:
$dG = \frac{t\,dr}{\rho\pi r}$, $G = \frac{t}{\pi\rho}\int_{R_1}^{R_2}\frac{dr}{r} = \frac{t}{\pi\rho}\ln\frac{R_2}{R_1}$.
$I = GV_0 = \frac{V_0t}{\pi\rho}\ln\left(\frac{R_2}{R_1}\right)$: TRUE.
(B), (C) The electrons drift along curved (semicircular) paths, so they need a centripetal force towards the centre. With no magnetic force, this can only come from a radial electric field. The force on an electron is $-e\vec{E}$, so for it to point inwards $\vec{E}$ must point outwards, from the inner surface to the outer one. The field points from higher to lower potential, so the outer surface is at the lower voltage. (B) FALSE, (C) TRUE.
(D) The required field is $E = \frac{mv_d^2}{er}$, so $\Delta V = \int E\,dr \propto v_d^2$. The drift speed is proportional to the current ($I = neAv_d$ for each ring, with all rings scaled together by $V_0$), so $\Delta V \propto I^2$: TRUE.
Answer: (A), (C), (D). | ||||
| Q.12 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Multiple correctHard | |
A hard multiple correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Treat the dissolved molecules as an ideal gas: partial pressure $p = nk_BT$.
(A) The pressure difference across the tube is $\Delta p = \Delta n\,k_BT$, acting on area $S$, so the force pushing the molecules in the tube is $\Delta n\,k_BTS$: TRUE.
(B) The tube holds about $n_1Sl$ molecules (to first order in $\Delta n$), each moving at steady speed $v$ against a drag $\beta v$. Steady drift: total drag = driving force,
$n_1Sl\,\beta v = \Delta n\,k_BTS$, i.e. $n_1\beta vl = \Delta n\,k_BT$: TRUE.
(C) Molecules crossing per second $= n_1vS = n_1S\cdot\frac{\Delta n\,k_BT}{n_1\beta l} = \left(\frac{\Delta n}{l}\right)\left(\frac{k_BT}{\beta}\right)S$: TRUE.
(D) As molecules move across, $n_1$ falls and $n_2$ rises, so $\Delta n$ and with it the rate decrease with time: FALSE.
Answer: (A), (B), (C). | ||||
| Q.13 | Laws of MotionFriction | Class 11: Laws of Motion | NumericalHard | |
A hard numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Let the fingers be at distances $x_L$ and $x_R$ from the centre of mass (the 50 cm mark). Torques about the centre of mass give $N_Lx_L = N_Rx_R$: the finger nearer the centre carries more load.
The scale moves slowly, so the horizontal frictional forces on it balance.
Start: $x_L = 50$, $x_R = 40$, so $N_L < N_R$ and the left finger slips first (kinetic friction $0.32N_L$ on it, static friction on the right).
Stage 1 ends when the right finger's static friction reaches its limit: $0.32N_L = 0.40N_R$, so $\frac{N_L}{N_R} = \frac{x_R}{x_L} = \frac{40}{x_L} = \frac{0.40}{0.32}$, giving $x_L = 32$ cm.
Stage 2: now the right finger slips (kinetic friction $0.32N_R$) and the left one holds (static). The right finger stops when $0.32N_R = 0.40N_L$: $\frac{N_R}{N_L} = \frac{x_L}{x_R} = \frac{32}{x_R} = \frac{0.40}{0.32}$, so $x_R = 32 \times 0.8 = 25.6$ cm.
$x_R = 25.60$ cm. | ||||
| Q.14 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The pressure of water at the bottom of the disc (above atmospheric) is the hydrostatic $\rho gh$.
The edge of the disc is a semicircle of diameter $h$, i.e. radius $\frac{h}{2}$ (the other curvature, of the rim of the glass, is negligible). A curved surface of radius $R$ can hold an excess pressure of $\frac{T}{R}$, so the edge withstands at most $\frac{T}{h/2} = \frac{2T}{h}$.
Water starts flowing when $\rho gh = \frac{2T}{h}$:
$h = \sqrt{\frac{2T}{\rho g}} = \sqrt{\frac{2\times0.07}{10^3\times10}} = \sqrt{1.4\times10^{-5}}$ m $= 3.74\times10^{-3}$ m.
$h \approx 3.74$ mm. | ||||
| Q.15 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalHard | |
A hard numerical question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance On the axis of a dipole the field at distance $x$ is $E = \frac{2Kp}{x^3}$ ($K = \frac{1}{4\pi\varepsilon_0}$), pointing away from the dipole since $\vec p$ points towards $q$. The force on $q$ is outward: $F(x) = \frac{2Kpq}{x^3}$.
Equilibrium (spring of zero natural length): $kl = \frac{2Kpq}{l^3}$.
For a small displacement $s$ along the axis (the natural direction of the oscillation), the net restoring force is
$-ks + \frac{dF}{dx}\Big|_l s = -ks - \frac{6Kpq}{l^4}s = -ks - \frac{3}{l}\cdot\frac{2Kpq}{l^3}s = -ks - 3ks = -4ks$.
So $\omega = \sqrt{\frac{4k}{m}} = 2\sqrt{\frac{k}{m}}$ and $f = \frac{\omega}{2\pi} = \frac{1}{\pi}\sqrt{\frac{k}{m}}$.
$\delta = \pi \approx 3.14$. | ||||
| Q.16 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Helium is monatomic: $\gamma = \frac{5}{3}$, $C_V = \frac{3}{2}R$. Let the initial temperature be $T_1$.
Isothermal, $V_1 \to 4V_1$: $W_{iso} = RT_1\ln 4 = 2RT_1\ln 2$.
Adiabatic, $4V_1 \to 32V_1$ starting at $T_1$: $TV^{\gamma-1}$ is constant, so $T_2 = T_1\left(\frac{4}{32}\right)^{2/3} = T_1\left(\frac{1}{8}\right)^{2/3} = \frac{T_1}{4}$.
$W_{adia} = C_V(T_1 - T_2) = \frac{3}{2}R\cdot\frac{3T_1}{4} = \frac{9}{8}RT_1$.
$\frac{W_{iso}}{W_{adia}} = \frac{2RT_1\ln2}{\frac{9}{8}RT_1} = \frac{16}{9}\ln 2$.
$f = \frac{16}{9} = 1.78$ (1.777…). | ||||
| Q.17 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The fork moves along the axis of the pipe, at its open end, so the air column receives a Doppler-shifted frequency.
Fork moving towards the pipe: $f' = f\frac{v}{v - v_s} = f\frac{320}{318}$. Fork moving away: $f' = f\frac{320}{322}$.
For a given mode the resonant length is inversely proportional to the frequency, $L \propto \frac{1}{f}$:
towards: $L' = L\frac{318}{320}$; away: $L' = L\frac{322}{320}$. Either way $\frac{|\Delta L|}{L} = \frac{2}{320} = 0.00625$.
Percentage change $= 0.625\% \approx 0.62$. | ||||
| Q.18 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance By Gauss's law, flux is proportional to the charge enclosed. The charge within radius $a$ of the centre is
$q(a) = \int_0^a\sigma_0\left(1 - \frac{r}{R}\right)2\pi r\,dr = 2\pi\sigma_0\left(\frac{a^2}{2} - \frac{a^3}{3R}\right)$.
Whole disc ($a = R$): $q = 2\pi\sigma_0R^2\left(\frac12 - \frac13\right) = \frac{\pi\sigma_0R^2}{3}$.
Inside $\frac{R}{4}$: $q = 2\pi\sigma_0R^2\left(\frac{1}{32} - \frac{1}{192}\right) = 2\pi\sigma_0R^2\cdot\frac{5}{192} = \frac{5\pi\sigma_0R^2}{96}$.
$\frac{\phi_0}{\phi} = \frac{1/3}{5/96} = \frac{96}{15} = 6.4$.
Answer: 6.40. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctMedium | |
A medium single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry The distribution drawn is not the lopsided Maxwell curve: it is a narrow, symmetric bell centred on one speed.
For a curve symmetric about its peak, the mean equals the peak, so the most probable speed equals the average speed.
The rms speed is $\sqrt{\bar{u}^2 + \sigma^2}$, where $\sigma$ is the spread. Here the spread is small beside the central speed (the bell is narrow and sits far from the origin), so $u_{rms} \approx \bar{u}$.
So $u_{mp} : u_{av} : u_{rms} = 1 : 1 : 1$.
(C) $1 : 1.128 : 1.224$ is the Maxwell–Boltzmann ratio, which would apply only to the usual skewed curve, not to this symmetric one.
Answer: (A). | ||||
| Q.2 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Potassium superoxide contains the $\text{O}_2^-$ ion, which disproportionates in water and gives off oxygen:
$4\text{KO}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{KOH} + 3\text{O}_2$.
The peroxides $\text{Na}_2\text{O}_2$ and $\text{Li}_2\text{O}_2$ give hydrogen peroxide on hydrolysis (cold water): $\text{Na}_2\text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\text{O}_2$; they do not liberate $\text{O}_2$ as the hydrolysis product.
$\text{Pb}_3\text{O}_4$ ($2\text{PbO}\cdot\text{PbO}_2$) is not hydrolysed to give oxygen.
Answer: (B). | ||||
| Q.3 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter White chlorides insoluble in cold water: $\text{AgCl}$, $\text{PbCl}_2$ (and $\text{Hg}_2\text{Cl}_2$). Cd and Zn chlorides are soluble, so they give no precipitate.
$\text{PbCl}_2$ dissolves in hot water; $\text{AgCl}$ does not. So the precipitate is partly soluble in hot water.
Residue P = $\text{AgCl}$: dissolves in aq. $\text{NH}_3$ as $[\text{Ag}(\text{NH}_3)_2]^+$ and in excess thiosulfate as $[\text{Ag}(\text{S}_2\text{O}_3)_2]^{3-}$.
Solution Q contains $\text{Pb}^{2+}$, which gives yellow $\text{PbI}_2$ with KI.
So X and Y are Ag and Pb.
Answer: (A). | ||||
| Q.4 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In each projection the three bonds meeting at the centre belong to the front carbon and the three starting at the circle belong to the back carbon. Write out what each carbon carries.
P: front C has $\text{CH}_3$, $\text{CH}_3$, $\text{C}_2\text{H}_5$; back C has OH, $\text{CH}_3$, $\text{CH}_3$. Molecule: $(\text{CH}_3)_2\text{C(OH)}{-}\text{C}(\text{CH}_3)_2\text{C}_2\text{H}_5$, 2,3,3-trimethylpentan-2-ol.
Q: front C has $\text{C}(\text{CH}_3)_2\text{OH}$, H, $\text{C}_2\text{H}_5$; back C has H, H, $\text{CH}_3$ (an ethyl group). Molecule: $(\text{CH}_3)_2\text{C(OH)}{-}\text{CH}(\text{C}_2\text{H}_5)_2$, 3-ethyl-2-methylpentan-2-ol.
R: front C has $\text{CH}_3$, OH, $\text{CH}_3$; back C has $\text{C}_2\text{H}_5$, H, $\text{C}_2\text{H}_5$. Molecule: $(\text{CH}_3)_2\text{C(OH)}{-}\text{CH}(\text{C}_2\text{H}_5)_2$, again 3-ethyl-2-methylpentan-2-ol.
S: front C has $\text{CH}_3$, H, H (an ethyl group); back C has $\text{C}_2\text{H}_5$, $\text{CH}(\text{CH}_3)_2$, OH. Molecule: $(\text{C}_2\text{H}_5)_2\text{C(OH)}{-}\text{CH}(\text{CH}_3)_2$, 3-ethyl-2-methylpentan-3-ol.
All four are $\text{C}_8\text{H}_{18}\text{O}$ isomers, but only Q and R are the same compound (it has no stereocentre, so they are identical, just viewed along different bonds).
Answer: (C). | ||||
| Q.5 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Build the structure from the name. Parent: hex-3-en-5-ynoic acid, $\text{C1}$ = COOH, $\text{C2}$–$\text{C3}$, double bond $\text{C3}{=}\text{C4}$, triple bond $\text{C5}{\equiv}\text{C6}$.
Substituents: OH on C2, an ethynyl on C3, a methyl on C4.
So: $\text{HOOC}{-}\text{CH(OH)}{-}\text{C}(\text{C}{\equiv}\text{CH}){=}\text{C}(\text{CH}_3){-}\text{C}{\equiv}\text{CH}$.
The carbon bearing $\text{CH(OH)CO}_2\text{H}$ must also carry an ethynyl, and the other alkene carbon must carry a methyl and an ethynyl.
(A) has an ethyl, not a methyl: wrong. (B) has the ethyl and the acid fragment on different carbons: wrong. (C) has the methyl on C3 and both ethynyls on C4: that is 4-ethynyl-2-hydroxy-3-methylhex-3-en-5-ynoic acid. (D) matches: ethynyl and $\text{CH(OH)CO}_2\text{H}$ on one alkene carbon, ethynyl and methyl on the other.
Answer: (D). | ||||
| Q.6 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Matching listHard | |
A hard matching list question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Assign R/S to C2 and C3 of each and compare with D-erythrose.
D-erythrose (Fischer, both OH on the right): at C2 the priorities are OH > CHO > C3 > H. With H on a horizontal (front) bond, OH → CHO → C3 runs anticlockwise, so the true label is the reverse: R. C3 in the same way (OH > C2 > $\text{CH}_2\text{OH}$ > H): R. D-erythrose is (2R,3R).
P: at C2, H is hashed (behind); OH (down) → CHO (up-left) → C3 (up-right) runs clockwise: R. At C3, H is wedged (in front); OH (hashed, up-left) → C2 (down-left) → $\text{CH}_2\text{OH}$ (down-right) runs anticlockwise, reversed for a front H: R. P = (2R,3R): identical → 2.
R: C2 as in P: R. At C3 the H and OH are swapped (H hashed, OH wedged): S. R = (2R,3S): diastereomer → 1.
Q: at C2 OH is hashed and H wedged; working through the priorities gives S. At C3 H is hashed and OH wedged: R. Q = (2S,3R): diastereomer (L-threose-type) → 1.
S: C2 as in Q: S. At C3 OH is hashed, H wedged: S. S = (2S,3S): the mirror image, the enantiomer (L-erythrose) → 3.
P→2, Q→1, R→1, S→3.
Answer: (C). | ||||
| Q.7 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics The integrand $\frac{RT}{V-b} - \frac{a}{V^2}$ is the pressure of one mole of a van der Waals gas. Writing it in place of $P_{\text{ext}}$ means the external pressure equals the gas's own pressure at every stage, which is the definition of a reversible process for a van der Waals system.
(A) TRUE: the system obeys the van der Waals equation.
(B) TRUE: a reversible isothermal process fits (T constant inside the integral).
(C) TRUE: a reversible adiabatic process also has $P_{\text{ext}} = P_{\text{gas}}$ throughout; T then changes with V, and the integral still holds with T = T(V).
(D) FALSE: in an irreversible process at constant external pressure, $P_{\text{ext}}$ is a constant and not the gas's equation-of-state pressure, so $w = -P_{\text{ext}}\Delta V$.
Answer: (A), (B), (C). | ||||
| Q.8 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Multiple correctMedium | |
A medium multiple correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The compounds are I triphenylmethane, II benzene, III methane, IV cyclopenta-1,3-diene and V ethyne; the acidic H is the one drawn.
(A) TRUE: the triphenylmethyl anion spreads its negative charge over the three rings, which is why $\text{Ph}_3\text{CH}$ is fairly acidic (p$K_a \approx 31$–$33$).
(B) TRUE: the cyclopentadienyl anion is planar, cyclic and fully conjugated with 6 π electrons: aromatic.
(C) TRUE: a nitro group withdraws electrons (–I, –R) and stabilises the aryl anion, so nitrobenzene's ring C–H is more acidic than benzene's.
(D) FALSE: approximate p$K_a$ values are IV ≈ 16, V ≈ 25, I ≈ 31–33, II ≈ 43, III ≈ 50. Cyclopentadiene (aromatic anion) is more acidic than triphenylmethane, so the order is IV > V > I > II > III (in any case IV > I), not I > IV > V > II > III.
Answer: (A), (B), (C). | ||||
| Q.9 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Multiple correctHard | |
A hard multiple correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons P is 6-tert-butyl-1,1-dimethylindane. The free ring positions are C4 (next to the ring $\text{CH}_2$), C5 (next to tert-butyl) and C7 (between tert-butyl and the $\text{CMe}_2$ ring carbon).
Step 1, Friedel–Crafts acylation with succinic anhydride/$\text{AlCl}_3$: the bulky acylium complex cannot go next to tert-butyl (C5, C7), so it enters C4, ortho to the ring alkyl and away from tert-butyl. Q is the 4-oxobutanoic acid at C4: $\text{Ar}{-}\text{CO}{-}\text{CH}_2\text{CH}_2{-}\text{CO}_2\text{H}$. That is structure (B): TRUE.
Step 2, Zn-Hg/HCl (Clemmensen) reduces the ketone to $\text{CH}_2$: $\text{Ar}{-}(\text{CH}_2)_3{-}\text{CO}_2\text{H}$. Then $\text{H}_3\text{PO}_4$ effects intramolecular acylation onto the only free ortho position, C5. R is the tetralone in which the new C=O is bonded to C5 (beside tert-butyl) and the $\text{CH}_2$ chain to C4. Structure (C) has the C=O on the carbon next to the five-membered ring instead: FALSE.
Step 3, $\text{CH}_3\text{MgBr}$ then $\text{H}_3\text{O}^+$ turns the ketone into a tertiary alcohol, and $\text{H}_2\text{SO}_4$/heat dehydrates it to the alkene conjugated with the ring: the carbon next to C5 bears $\text{CH}_3$ and is double-bonded to the next ring carbon. That is (D): TRUE. (A) puts the methyl alkene on the side next to the five-membered ring, which would follow from (C): FALSE.
Answer: (B), (D). | ||||
| Q.10 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Multiple correctMedium | |
A medium multiple correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) $[\text{FeCl}_4]^-$: $\text{Fe}^{3+}$ with four weak-field $\text{Cl}^-$ ligands, $sp^3$, tetrahedral: TRUE.
(B) $[\text{Co(en)(NH}_3)_2\text{Cl}_2]^+$ is of the type $[\text{M(AA)a}_2\text{b}_2]$. Its geometrical isomers: the two Cl trans (NH$_3$ cis), the two NH$_3$ trans (Cl cis), and both pairs cis (this one is also optically active). That is 3, not 2: FALSE.
(C) $\text{Fe}^{3+}$ ($d^5$) in the weak field of $\text{Cl}^-$ has 5 unpaired electrons, $\mu = \sqrt{35} \approx 5.92$ BM. $\text{Co}^{3+}$ ($d^6$) with en and $\text{NH}_3$ is low spin, 0 unpaired electrons, $\mu = 0$. TRUE.
(D) Low-spin $\text{Co}^{3+}$ uses inner $d$ orbitals: $d^2sp^3$, not $sp^3d^2$: FALSE.
Answer: (A), (C). | ||||
| Q.11 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) The acid strengths are $\text{HOCl} < \text{HClO}_3 < \text{HClO}_4$, so the conjugate base strengths are reversed: $\text{ClO}^-$ is the strongest. TRUE.
(B) $\text{ClO}^-$ is diatomic, so it is linear whatever the lone pairs; $\text{ClO}_3^-$ has one lone pair on Cl, which makes it pyramidal instead of trigonal planar; $\text{ClO}_4^-$ has no lone pair on Cl (tetrahedral). Only the chlorate shape is influenced by a lone pair: TRUE.
(C) $3\text{ClO}^- \rightarrow 2\text{Cl}^- + \text{ClO}_3^-$, but $4\text{ClO}_3^- \rightarrow 3\text{ClO}_4^- + \text{Cl}^-$. The sets of ions are different: FALSE.
(D) $\text{ClO}^- + \text{SO}_3^{2-} \rightarrow \text{Cl}^- + \text{SO}_4^{2-}$: hypochlorite oxidises sulfite. TRUE.
Answer: (A), (B), (D). | ||||
| Q.12 | Solid StateIonic radii and radius ratio | No matching NCERT chapter | Multiple correctHard | |
A hard multiple correct question on ionic radii and radius ratio, from Solid State. NCERT: No matching NCERT chapter Reading the figure: M (filled) sits at the centres of the left and right faces; X (open) sits at the midpoints of the four edges that run left–right.
(A) Count per cell: M $= 2 \times \frac12 = 1$; X $= 4 \times \frac14 = 1$. Formula MX: TRUE.
Shift the origin by half an edge along the left–right direction: M then sits at the corners and X at the body centre of a cube of edge $a$. This is the CsCl arrangement.
(B) In CsCl each ion has 8 neighbours of the other kind at the corners of a cube: both have cubic (8-fold) coordination. FALSE.
(C) The M–X distance is half the body diagonal, $\frac{\sqrt3}{2}a$, so the ratio is $0.866$: TRUE.
(D) 0.414 is the limiting radius ratio for octahedral (6-fold) coordination; for the 8-fold coordination of this structure the radius ratio is at least 0.732. FALSE.
Answer: (A), (C). | ||||
| Q.13 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter The first readings are rough; the concordant titre is 9.0 mL (three identical readings).
Oxalic acid is diprotic: $\text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O}$.
Oxalic acid taken $= 5.00 \times 0.10 = 0.50$ mmol, so NaOH needed $= 1.00$ mmol.
$M_{\text{NaOH}} = \frac{1.00}{9.0} = 0.111$ M.
Answer: 0.11. | ||||
| Q.14 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium With $p_A = 1$ bar held fixed, $K_p = \frac{p_B}{p_A} = p_B$ at equilibrium.
From the plot: at 1000 K, $p_B = 10$ bar, so $K_{1000} = 10$; at 2000 K the new steady value is $p_B = 100$ bar (the 115 bar spike is a transient), so $K_{2000} = 100$.
$\Delta G^\circ = -RT\ln K$:
$\frac{\Delta G^\circ_{1000}}{\Delta G^\circ_{2000}} = \frac{1000\ln 10}{2000\ln 100} = \frac{1000\ln10}{2000 \times 2\ln10} = \frac{1}{4}$.
Answer: 0.25. | ||||
| Q.15 | ElectrochemistryBatteries and fuel cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on batteries and fuel cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $E^0_{\text{cell}} = 1.23 - 0.00 = 1.23$ V; each mole of $\text{H}_2$ transfers $n = 2$ mol electrons.
Maximum work per mole of $\text{H}_2$: $-\Delta G^0 = nFE^0 = 2 \times 96500 \times 1.23 = 237390$ J.
For $1.0\times10^{-3}$ mol at 70% efficiency: $W = 0.70 \times 237.39 = 166.17$ J.
This work compresses 1.00 mol of a monatomic ideal gas adiabatically ($q = 0$), so $\Delta U = W$:
$nC_V\Delta T = W$, $\Delta T = \frac{166.17}{1.00 \times \frac{3}{2} \times 8.314} = \frac{166.17}{12.471} = 13.32$ K.
Answer: 13.32. | ||||
| Q.16 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $2\text{Al} + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3\text{H}_2$.
Al: $\frac{5.4}{27.0} = 0.20$ mol. $\text{H}_2\text{SO}_4$: $0.0500 \times 5.0 = 0.25$ mol.
0.20 mol Al would need 0.30 mol acid, so the acid is limiting. $\text{H}_2$ formed = moles of acid = 0.25 mol.
$V = \frac{nRT}{P} = \frac{0.25 \times 0.082 \times 300}{1.0} = 6.15$ L.
Answer: 6.15. | ||||
| Q.17 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Mass number falls by $238 - 206 = 32$, so each nucleus emits $\frac{32}{4} = 8$ alpha particles (and 6 beta particles to fix the charge: $92 - 16 + 6 = 82$).
Initial U atoms: $\frac{68\times10^{-6}}{238} \times 6.022\times10^{23} = 2.857\times10^{-7} \times 6.022\times10^{23} = 1.721\times10^{17}$.
After three half-lives $\frac{1}{8}$ remains, so $\frac{7}{8}$ has decayed: $\frac78 \times 1.721\times10^{17} = 1.506\times10^{17}$ nuclei.
Alpha particles $= 8 \times 1.506\times10^{17} = 1.20\times10^{18}$.
$Z = 1.20$. | ||||
| Q.18 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques P is methyl 2,4,6-triphenylbenzoate, $\text{C}_{26}\text{H}_{20}\text{O}_2$. Its ester is too hindered for ordinary hydrolysis; in conc. $\text{H}_2\text{SO}_4$ it ionises to the acylium ion $\text{ArC}{\equiv}\text{O}^+$ (the ionic intermediate), losing methanol.
The acylium ion attacks the neighbouring ortho phenyl ring (intramolecular Friedel–Crafts acylation), closing a five-membered ring: Q is 1,3-diphenylfluoren-9-one, a yellow (coloured) ketone.
Q = P − $\text{CH}_3\text{OH}$ = $\text{C}_{25}\text{H}_{16}\text{O}$.
Degree of unsaturation $= \frac{2C + 2 - H}{2} = \frac{50 + 2 - 16}{2} = 18$.
Check by counting: fluorenone has 3 rings + 6 C=C + 1 C=O = 10, and each phenyl adds 4 (1 ring + 3 C=C): $10 + 8 = 18$.
Answer: 18. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Expand and group: the expression $= a^2(c + d) - a(c^2 + d^2) + b^2(c + d) - b(c^2 + d^2) = (c + d)(a^2 + b^2) - (a + b)(c^2 + d^2)$.
From $x^2 + 20x - 2020$: $a + b = -20$, $ab = -2020$, so $a^2 + b^2 = 400 + 4040 = 4440$.
From $x^2 - 20x + 2020$: $c + d = 20$, $cd = 2020$, so $c^2 + d^2 = 400 - 4040 = -3640$.
Value $= 20 \times 4440 - (-20)(-3640) = 88800 - 72800 = 16000$.
Answer: (D). | ||||
| Q.2 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $|x|$ is even and $x - \sin x$ is odd, so $f$ is odd: $f(-x) = -f(x)$, with $f(0) = 0$.
For $x > 0$: $f(x) = x(x - \sin x) = x^2 - x\sin x$, and $f'(x) = 2x - \sin x - x\cos x = (x - \sin x) + x(1 - \cos x) > 0$. So $f$ is strictly increasing on $[0, \infty)$, and by oddness on all of $\mathbb{R}$: one-one.
As $x \to \infty$, $f(x) \geq x(x - 1) \to \infty$; by oddness $f(x) \to -\infty$ as $x \to -\infty$. $f$ is continuous, so it takes every real value: onto.
Answer: (C). | ||||
| Q.3 | 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 For $x < 1$: $|x - 1| = 1 - x$, so $f(x) = e^{x-1} - e^{x-1} = 0$. For $x \geq 1$: $f(x) = e^{x-1} - e^{1-x}$.
$g(x) = \cosh(x - 1) > 0$ everywhere, so $g$ lies above $f$ until they meet.
Meeting point ($x \geq 1$): $e^{x-1} - e^{1-x} = \frac12(e^{x-1} + e^{1-x})$, i.e. $e^{x-1} = 3e^{1-x}$, $e^{2(x-1)} = 3$, $x = 1 + \frac12\ln 3$.
Area $= \int_0^1 g\,dx + \int_1^{1+\frac12\ln3}(g - f)\,dx$.
$\int_0^1 \frac12(e^{x-1} + e^{1-x})dx = \frac12[(1 - e^{-1}) + (e - 1)] = \frac12(e - e^{-1})$.
With $u = x - 1$: $g - f = -\frac12e^u + \frac32e^{-u}$, and $\int_0^{\frac12\ln3}\left(-\frac12e^u + \frac32e^{-u}\right)du = \left[-\frac12e^u - \frac32e^{-u}\right]_0^{\frac12\ln3} = \left(-\frac{\sqrt3}{2} - \frac{3}{2\sqrt3}\right) + 2 = 2 - \sqrt3$.
Area $= (2 - \sqrt3) + \frac12(e - e^{-1})$.
Answer: (A). | ||||
| Q.4 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Take $P = (\lambda, 2\lambda)$ (the other end gives the same result by symmetry).
Parabola: $2yy' = 4\lambda$, so the slope at $P$ is $\frac{2\lambda}{2\lambda} = 1$.
Ellipse: slope $= -\frac{b^2x}{a^2y} = -\frac{b^2\lambda}{a^2 \cdot 2\lambda} = -\frac{b^2}{2a^2}$.
Perpendicular: $1 \cdot \left(-\frac{b^2}{2a^2}\right) = -1$, so $b^2 = 2a^2$.
Now $b > a$, so the major axis is along the y-axis and $e^2 = 1 - \frac{a^2}{b^2} = 1 - \frac12 = \frac12$.
$e = \frac{1}{\sqrt2}$. (The ellipse through $P$ just fixes the size: $\frac{\lambda^2}{a^2} + \frac{4\lambda^2}{2a^2} = 1$, $a^2 = 3\lambda^2$.)
Answer: (A). | ||||
| Q.5 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Real and equal roots: $\alpha^2 = 4\beta$, with $\alpha, \beta \in \{0, 1, 2\}$. The possibilities are $(\alpha, \beta) = (0, 0)$ and $(2, 1)$.
$\alpha$: $P(0) = \frac19$, $P(1) = \frac49$, $P(2) = \frac49$. $\beta$: $P(0) = \frac49$, $P(1) = \frac49$, $P(2) = \frac19$.
The coins are independent: $P = \frac19\cdot\frac49 + \frac49\cdot\frac49 = \frac{4}{81} + \frac{16}{81} = \frac{20}{81}$.
Answer: (B). | ||||
| Q.6 | 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 The curve $y = 2\sin 2x$ is symmetric about $x = \frac{\pi}{4}$. For a given height, the widest rectangle is symmetric, with corners on the curve at $x$ and $\frac{\pi}{2} - x$ ($0 \leq x \leq \frac{\pi}{4}$): width $\frac{\pi}{2} - 2x$, height $2\sin 2x$.
Perimeter $p(x) = 2\left(\frac{\pi}{2} - 2x\right) + 4\sin 2x$.
$p'(x) = -4 + 8\cos 2x = 0 \Rightarrow \cos 2x = \frac12 \Rightarrow x = \frac{\pi}{6}$; $p'' = -16\sin 2x < 0$, a maximum.
Width $= \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$, height $= 2\sin\frac{\pi}{3} = \sqrt3$.
Area $= \frac{\pi\sqrt3}{6} = \frac{\pi}{2\sqrt3}$.
Answer: (C). | ||||
| Q.7 | 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 $f(x) = x^2(x - 1) + (x - 1)\sin x = (x - 1)(x^2 + \sin x)$, so $f(1) = 0$ and
$(fg)'(1) = \lim_{x\to1}\frac{f(x)g(x) - 0}{x - 1} = \lim_{x\to1}(x^2 + \sin x)\,g(x)$, provided this limit exists. Note $1 + \sin 1 \neq 0$.
(A) If $g$ is continuous at 1, the limit is $(1 + \sin1)g(1)$: $fg$ is differentiable. TRUE.
(B) Take $g(x) = 0$ for $x \neq 1$ and $g(1) = 1$. Then $fg \equiv 0$ (since $f(1) = 0$), which is differentiable, but $g$ is not continuous at 1. FALSE.
(C) Differentiable implies continuous, so this follows from (A). TRUE.
(D) Take $g(x) = |x - 1|$: $(fg)'(1) = \lim (x^2 + \sin x)|x - 1| = 0$ exists, but $g$ is not differentiable at 1. FALSE.
Answer: (A), (C). | ||||
| Q.8 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3\times3$ matrix, $\text{adj}(\text{adj}\,M) = (\det M)^{3-2}M = (\det M)\,M$.
So $M^{-1} = (\det M)M$, i.e. $M^2 = \frac{1}{\det M}I$. Taking determinants: $(\det M)^2 = (\det M)^{-3}$, so $(\det M)^5 = 1$ and, $\det M$ being real, $\det M = 1$.
(B) TRUE. Then $M^2 = I$: (C) TRUE.
(D) $\text{adj}\,M = (\det M)M^{-1} = M^{-1}$, so $(\text{adj}\,M)^2 = M^{-2} = (M^2)^{-1} = I$: TRUE.
(A) Not always: $M = \text{diag}(1, -1, -1)$ has $\det M = 1$, $M^2 = I$ and $\text{adj}(\text{adj}\,M) = M = M^{-1}$, but $M \neq I$. FALSE.
Answer: (B), (C), (D). | ||||
| Q.9 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Multiple correctMedium | |
A medium multiple correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $w = z + \frac12$: $z^2 + z + 1 = w^2 + \frac34$, so $S$ is $\left|w^2 + \frac34\right| = 1$.
(C) By the triangle inequality $|w|^2 = |w^2| \geq 1 - \frac34 = \frac14$, so $|w| \geq \frac12$: TRUE.
(A) $w = \frac{i\sqrt7}{2}$ gives $w^2 + \frac34 = -1$, modulus 1, so it is in $S$, but $|w| = \frac{\sqrt7}{2} > \frac12$: FALSE.
(B) $|w|^2 \leq 1 + \frac34 = \frac74$, so $|w| \leq \frac{\sqrt7}{2}$ and $|z| \leq |w| + \frac12 \leq \frac{\sqrt7 + 1}{2} \approx 1.82 < 2$: TRUE.
(D) $w^2$ runs over the whole circle $|u + \frac34| = 1$, so $S$ is a curve with infinitely many points: FALSE.
Answer: (B), (C). | ||||
| Q.10 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Multiple correctHard | |
A hard multiple correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $s = \frac{x + y + z}{2}$ and $\Delta$ be the area. Then $\tan\frac{X}{2} = \frac{\Delta}{s(s - x)}$, $\tan\frac{Z}{2} = \frac{\Delta}{s(s - z)}$.
Sum $= \frac{\Delta}{s}\cdot\frac{(s - z) + (s - x)}{(s - x)(s - z)} = \frac{\Delta\,y}{s(s - x)(s - z)}$.
The condition says this equals $\frac{2y}{2s} = \frac{y}{s}$, so $\Delta = (s - x)(s - z)$.
Then $\tan\frac{Y}{2} = \frac{\Delta}{s(s - y)}$, and since $\Delta^2 = s(s - x)(s - y)(s - z)$, $s(s - y) = \frac{\Delta^2}{(s - x)(s - z)} = \Delta$. So $\tan\frac{Y}{2} = 1$, $Y = \frac{\pi}{2}$.
(A) $2Y = \pi \neq X + Z = \frac{\pi}{2}$: FALSE.
(B) $Y = \frac{\pi}{2} = X + Z$: TRUE.
(C) With $Y = 90^\circ$: $x = y\sin X$, $z = y\cos X$, so $\frac{x}{y + z} = \frac{\sin X}{1 + \cos X} = \tan\frac{X}{2}$: TRUE.
(D) Pythagoras gives $x^2 + z^2 - y^2 = 0$, not $xz$: FALSE.
Answer: (B), (C). | ||||
| Q.11 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctMedium | |
A medium multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Both lines pass through $(1, 0, 1)$, their point of intersection. Directions: $\vec d_1 = (1, -1, 3)$, $\vec d_2 = (-3, -1, 1)$, both of length $\sqrt{11}$.
$\vec d_1\cdot\vec d_2 = -3 + 1 + 3 = 1 > 0$, so the angle between $\vec d_1$ and $\vec d_2$ is acute and its bisector is along $\vec d_1 + \vec d_2 = (-2, -2, 4) \parallel (1, 1, -2)$. (This lies in the plane of the two lines.)
$L$ has direction $(l, m, -2)$, so $l = 1$, $m = 1$: $l + m = 2$. (B) TRUE, (D) FALSE.
$L$ passes through $(1, 0, 1)$: $\frac{1 - \alpha}{1} = \frac{0 - 1}{1} = \frac{1 - \gamma}{-2}$, so $\alpha = 2$, $\gamma = -1$, and $\alpha - \gamma = 3$. (A) TRUE, (C) FALSE.
Answer: (A), (B). | ||||
| Q.12 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Integrate by parts ($\sin 1 \approx 0.8415$, $\cos 1 \approx 0.5403$).
(A) $\int_0^1 x\cos x\,dx = [x\sin x + \cos x]_0^1 = \sin1 + \cos1 - 1 \approx 0.3818 \geq 0.375$: TRUE.
(B) $\int_0^1 x\sin x\,dx = [-x\cos x + \sin x]_0^1 = \sin1 - \cos1 \approx 0.3012 \geq 0.3$: TRUE.
(C) $\int_0^1 x^2\cos x\,dx = [x^2\sin x + 2x\cos x - 2\sin x]_0^1 = 2\cos1 - \sin1 \approx 0.2391 < 0.5$: FALSE.
(D) $\int_0^1 x^2\sin x\,dx = [-x^2\cos x + 2x\sin x + 2\cos x]_0^1 = \cos1 + 2\sin1 - 2 \approx 0.2232 \geq 0.2222$: TRUE.
(Without decimals: $\sin x \geq x - \frac{x^3}{6}$ and $\cos x \geq 1 - \frac{x^2}{2}$ give (A) $\geq \frac12 - \frac18 = \frac38$, (B) $\geq \frac13 - \frac1{30} = \frac{3}{10}$, (D) $\geq \frac14 - \frac1{36} = \frac29$.)
Answer: (A), (B), (D). | ||||
| Q.13 | Sequences and SeriesRelation between A.M. and G.M. | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on relation between a.m. and g.m., from Sequences and Series. NCERT: Class 11: Sequences and Series AM–GM: $3^{y_1} + 3^{y_2} + 3^{y_3} \geq 3\sqrt[3]{3^{y_1+y_2+y_3}} = 3\cdot3^3 = 81$, with equality at $y_1 = y_2 = y_3 = 3$. So $m = \log_3 81 = 4$.
AM–GM: $x_1x_2x_3 \leq \left(\frac{x_1 + x_2 + x_3}{3}\right)^3 = 27$, equality at $x_i = 3$. So $M = \log_3 27 = 3$.
$\log_2(m^3) + \log_3(M^2) = \log_2 64 + \log_3 9 = 6 + 2 = 8$.
Answer: 8. | ||||
| Q.14 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $a_1 + \cdots + a_n = \frac{n}{2}[2c + 2(n - 1)] = n(c + n - 1)$ and $b_1 + \cdots + b_n = c(2^n - 1)$.
The equality: $2n(c + n - 1) = c(2^n - 1)$, i.e. $c(2^n - 1 - 2n) = 2n(n - 1)$.
$n = 1$: $-c = 0$, no positive $c$. $n = 2$: $-c = 4$, no.
$n = 3$: $c \cdot 1 = 12$, so $c = 12$.
$n = 4$: $7c = 24$, not an integer. $n = 5$: $21c = 40$, no. $n = 6$: $51c = 60$, no.
For $n \geq 7$: $2^n - 1 - 2n > 2n(n - 1)$ (e.g. $113 > 84$ at $n = 7$, and the left side roughly doubles each step while the right grows slowly), so $0 < c < 1$: no integer.
Only $c = 12$ works (check: $a$: 12, 14, 16 sum 42, doubled 84; $b$: 12, 24, 48 sum 84).
Answer: 1. | ||||
| Q.15 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalHard | |
A hard numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $t = \pi x - \frac{\pi}{4}$. Then $2\pi x = 2t + \frac{\pi}{2}$, so $\sin 2\pi x = \cos 2t$; and $3\pi x + \frac{\pi}{4} = 3t + \pi$, so $\sin\left(3\pi x + \frac{\pi}{4}\right) = -\sin 3t$.
$f = (3 - \cos 2t)\sin t + \sin 3t$. With $s = \sin t$: $\cos 2t\sin t = s - 2s^3$ and $\sin 3t = 3s - 4s^3$, so
$f = 3s - s + 2s^3 + 3s - 4s^3 = 5s - 2s^3 = s(5 - 2s^2)$.
Since $5 - 2s^2 \geq 3 > 0$, $f \geq 0 \iff \sin t \geq 0$.
For $x \in [0, 2]$, $t \in \left[-\frac{\pi}{4}, \frac{7\pi}{4}\right]$, and $\sin t \geq 0$ exactly for $t \in [0, \pi]$, i.e. $x \in \left[\frac14, \frac54\right]$.
$\beta - \alpha = \frac54 - \frac14 = 1$.
Answer: 1. | ||||
| Q.16 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Going round the triangle, $\vec a + \vec b + \vec c = \vec 0$, so $\vec c = -(\vec a + \vec b)$. Let $k = \vec a\cdot\vec b$.
Numerator: $\vec a\cdot\vec c - \vec a\cdot\vec b = (-9 - k) - k = -9 - 2k$.
Denominator: $\vec c\cdot\vec a - \vec c\cdot\vec b = (-9 - k) - (-k - 16) = 7$.
So $\frac{-9 - 2k}{7} = \frac{3}{3 + 4} = \frac37$, giving $-9 - 2k = 3$, $k = -6$.
$|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2 = 144 - 36 = 108$.
Answer: 108. | ||||
| Q.17 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | NumericalHard | |
A hard numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Lower bound. Take $f \in S$, $f \not\equiv 0$. Since $(x - 1)^2$ and $(x + 1)^2$ divide $f$, both $-1$ and $1$ are roots of $f$ and of $f'$. By Rolle's theorem ($f(-1) = f(1) = 0$), $f'$ has a root in $(-1, 1)$. So $f'$ has at least 3 distinct real roots, $m_{f'} \geq 3$.
Between consecutive roots of $f'$, Rolle's theorem gives a root of $f''$: at least 2 distinct roots, $m_{f''} \geq 2$. So $m_{f'} + m_{f''} \geq 5$.
Attained: $f = (x^2 - 1)^2$ (take $a_0 = 1$, the rest 0). $f' = 4x(x^2 - 1)$ has roots $-1, 0, 1$ (3); $f'' = 12x^2 - 4$ has roots $\pm\frac{1}{\sqrt3}$ (2). Sum 5.
(The zero polynomial, with $a_i$ all 0, has infinitely many roots and is not a candidate for the minimum.)
Answer: 5. | ||||
| Q.18 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $(1 - x)^{1/x} = e^{\frac{\ln(1 - x)}{x}} = e^{\frac{-x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots}{x}} = e^{-1}\,e^{-\frac{x}{2} - \frac{x^2}{3} - \cdots} = e^{-1}\left(1 - \frac{x}{2} + O(x^2)\right)$.
So the numerator is $-\frac{e^{-1}}{2}x + O(x^2)$, of order exactly $x$.
$\frac{\text{numerator}}{x^a}$ tends to a finite nonzero number only for $a = 1$, when the limit is $-\frac{1}{2e}$ (for $a < 1$ it tends to 0, for $a > 1$ it is unbounded).
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalHard | |
A hard numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Seen from above, the container is a square whose corner points at the observer; the line of sight runs along the diagonal through the wire W, 12 cm from the corner. The two walls meeting at the corner each make $45^\circ$ with the line of sight, and the observer sees W once through each wall: hence two images, one on each side of the line of sight.
Take the corner as origin and the line of sight as the x-axis, with W at $(-12, 0)$; one wall runs along $y = -x$ ($x < 0$).
A ray reaching a distant eye leaves that wall parallel to the line of sight, at $45^\circ$ to the wall's normal. Inside the water $\sin r = \frac{3}{4}\sin 45^\circ = 0.530$, so $r = 32.0^\circ$.
Inside, the ray runs from W to the exit point $P = (-u, u)$ at $45^\circ - r = 13.0^\circ$ to the line of sight, so $\frac{u}{12 - u} = \tan 13.0^\circ = 0.230$ and $u = 2.25$ cm.
The emergent ray, traced back, runs parallel to the line of sight 2.25 cm from it, and the image lies on it. The other wall gives the mirror image, so the separation is $2 \times 2.25 \approx 4.5$ cm for a distant eye (about 4.0 to 4.4 cm for an eye 5 to 25 cm from the corner).
As a single-digit integer, the separation is 4.
Answer: 4. | ||||
| Q.2 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Work in the train's frame, where the flow is steady. Air ahead of the train comes towards it with speed $v_t$ through the full area $S_0$ at pressure $p_0$, and all of it passes back through the gap of area $S_0 - S_t = 3S_t$ with speed $v$.
Continuity: $S_0 v_t = (S_0 - S_t)v \Rightarrow 4S_t v_t = 3S_t v \Rightarrow v = \frac{4}{3}v_t$.
Bernoulli along a streamline (horizontal): $p_0 + \frac{1}{2}\rho v_t^2 = p + \frac{1}{2}\rho v^2$.
$p_0 - p = \frac{1}{2}\rho\left(\frac{16}{9} - 1\right)v_t^2 = \frac{7}{18}\rho v_t^2$.
Comparing with $\frac{7}{2N}\rho v_t^2$: $2N = 18$, so $N = 9$. | ||||
| Q.3 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The drop floats when the electric force balances its weight: $qE = mg$.
Field between the discs: $E = \frac{V}{d} = \frac{200}{0.01} = 2\times10^{4}$ V/m.
Mass: $m = \rho\cdot\frac{4}{3}\pi r^3 = 900\times\frac{4}{3}\pi(8\times10^{-7})^3 = 900\times4.19\times5.12\times10^{-19} \approx 1.93\times10^{-15}$ kg.
$q = \frac{mg}{E} = \frac{1.93\times10^{-15}\times10}{2\times10^4} \approx 9.65\times10^{-19}$ C.
Number of electrons: $\frac{q}{e} = \frac{9.65\times10^{-19}}{1.6\times10^{-19}} \approx 6.03$, so the drop carries 6 electrons.
Answer: 6. | ||||
| Q.4 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At equilibrium the buoyant force equals the weight: $\rho(h)Vg = Mg$.
At 100 m: $\rho_0 e^{-100/6000}V = 480$.
At 150 m: $\rho_0 e^{-150/6000}V = 480 - N$.
Dividing: $\frac{480 - N}{480} = e^{-50/6000} = e^{-1/120}$.
$N = 480\left(1 - e^{-1/120}\right) \approx 480\times\frac{1}{120} = 4$ (more precisely $480\times0.0083 = 3.98$).
Answer: 4. | ||||
| Q.5 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalHard | |
A hard numerical question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance From the figure: the string (length $l$) makes angle $\alpha$ with the vertical, and the dipole sits on the vertical through the support, a distance $l$ below it, i.e. exactly where q hung at first. The dipole points along the line to q, which is $r = 2l\sin\frac{\alpha}{2}$ long, and q is now a height $h$ above the dipole.
Forces on q: weight $mg$ (vertical), tension $T$ (along the string) and the electric force $F$ (along the dipole-to-charge line). They are parallel to the three sides of the isosceles triangle support-charge-dipole, whose sides are $l$, $l$ and $2l\sin\frac{\alpha}{2}$. So
$\frac{mg}{l} = \frac{T}{l} = \frac{F}{2l\sin\frac{\alpha}{2}} \Rightarrow F = 2mg\sin\frac{\alpha}{2}$.
For a charge on the axis of a dipole, $F = \frac{2kpq}{r^3}$ and the potential energy is $U = \frac{kpq}{r^2} = \frac{Fr}{2}$.
$U = \frac{1}{2}\cdot2mg\sin\frac{\alpha}{2}\cdot2l\sin\frac{\alpha}{2} = 2mgl\sin^2\frac{\alpha}{2} = mgl(1 - \cos\alpha)$.
The charge has risen by $h = l - l\cos\alpha = l(1 - \cos\alpha)$, so $U = mgh$.
The dipole is brought in slowly, so the work done equals the gain in energy of the system: electrostatic $mgh$ plus gravitational $mgh$.
$W = 2mgh$, so $N = 2$. | ||||
| Q.6 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | NumericalMedium | |
A medium numerical question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Let the partition settle at distance $x$ from the top, with cross-section $A$. The partition conducts heat, so both gases share one temperature $T$, taken as 300 K.
Pressures: top $p_1 = \frac{nRT}{Ax}$, bottom $p_2 = \frac{nRT}{A(8 - x)}$.
Equilibrium of the partition: $(p_2 - p_1)A = mg$:
$nRT\left(\frac{1}{8 - x} - \frac{1}{x}\right) = mg$.
$nRT = 0.1\times8.3\times300 = 249$ J and $mg = 83$ N, so $\frac{1}{8 - x} - \frac{1}{x} = \frac{83}{249} = \frac{1}{3}$.
$3x - 3(8 - x) = x(8 - x) \Rightarrow x^2 - 2x - 24 = 0 \Rightarrow (x - 6)(x + 4) = 0$.
$x = 6$ m. | ||||
| Q.7 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctHard | |
A hard multiple correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics (A) Geometry of a spherical cap of radius $R$, half-width $r$ and depth (sagitta) $h$: $(R - h)^2 + r^2 = R^2 \Rightarrow R = \frac{h^2 + r^2}{2h}$. True.
(B) $\frac{3r^2}{2h}$ does not follow from that geometry. False.
For (C) and (D): the rotating surface is the paraboloid $y = \frac{\omega^2x^2}{2g}$, so $h = \frac{\omega^2r^2}{2g}$ and, with $h \ll r$, $R \approx \frac{r^2}{2h} = \frac{g}{\omega^2}$. The surface is concave upwards, with its centre of curvature above it (on the air side).
Refraction at a spherical surface, light going up from the bottom (in water, $n_1 = \frac{4}{3}$) into air ($n_2 = 1$), distances measured upwards from the vertex, object at $u = -H$ (since $h \ll H$), $R = +\frac{g}{\omega^2}$:
$\frac{1}{v} - \frac{4/3}{-H} = \frac{1 - 4/3}{R} \Rightarrow \frac{1}{v} = -\frac{4}{3H} - \frac{1}{3R}$.
$|v| = \frac{3HR}{4R + H} = \frac{3H}{4}\left(1 + \frac{H}{4R}\right)^{-1} = \frac{3H}{4}\left(1 + \frac{\omega^2H}{4g}\right)^{-1}$.
(D) True; (C) False.
Answer: (A), (D). | ||||
| Q.8 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Multiple correctMedium | |
A medium multiple correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion The semicircle lies in the plane of the ramp; y is its highest point and x, z (the ends of the diameter, which runs horizontally across the ramp) are its lowest points.
At y the centre of the circle lies straight down the slope, so the force towards the centre that the weight supplies is its component along the ramp, $mg\sin30^\circ = \frac{mg}{2}$ (the normal reaction balances the other component).
$\frac{mv_y^2}{R} = \frac{mg}{2} \Rightarrow v_y^2 = \frac{gR}{2}$.
Energy: $v_0^2 = v_y^2 + 2gh \Rightarrow v_0^2 - 2gh = \frac{1}{2}gR$.
(A) True; (B) False.
(C) At x and z the skater still moves on the circle with non-zero speed, so $\frac{mv^2}{R} \neq 0$. False.
(D) x and z are the lowest points of the turn, so the speed, and with it the required centripetal force $\frac{mv^2}{R}$, is largest there. True.
Answer: (A), (D). | ||||
| Q.9 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Multiple correctMedium | |
A medium multiple correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Angular momentum about the pivot is conserved (the pivot force has no moment about it):
$mvx = \left(\frac{mL^2}{3} + mx^2\right)\omega \Rightarrow \omega = \frac{3vx}{L^2 + 3x^2}$. (A) True, (B) False.
Maximise: $\frac{d\omega}{dx} = 3v\frac{(L^2 + 3x^2) - 6x^2}{(L^2 + 3x^2)^2} = 0 \Rightarrow x_M = \frac{L}{\sqrt{3}}$. (C) True.
$\omega_M = \frac{3v\cdot\frac{L}{\sqrt{3}}}{L^2 + L^2} = \frac{\sqrt{3}\,v}{2L}$. (D) True.
Answer: (A), (C), (D). | ||||
| Q.10 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Multiple correctMedium | |
A medium multiple correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Cut-off wavelength: an electron gives at most all its energy $eV$ to one photon, so $\lambda_{min} = \frac{hc}{eV}$. Doubling $V$ halves it; $d$ plays no part.
Characteristic lines come from transitions between inner shells of the target atoms, so their wavelengths depend only on the target material: unchanged.
Intensity: halving the filament current halves the number of electrons striking the target per second, so fewer X-ray photons are produced; the intensities decrease.
(A) True. (B) False (the cut-off changes). (C) True. (D) False (the cut-off becomes half, not double).
Answer: (A), (C). | ||||
| Q.11 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In air each sphere is in equilibrium under weight $mg$, tension $T$ and the electric force $F$: $\tan\frac{\alpha}{2} = \frac{F}{mg}$, $T = \frac{mg}{\cos(\alpha/2)}$.
In the liquid the electric force becomes $\frac{F}{21}$ (dielectric constant 21), and the effective weight becomes $(\rho - 800)Vg$ (buoyancy), where $\rho$ is the spheres' density.
Same angle: $\frac{F}{\rho Vg} = \frac{F/21}{(\rho - 800)Vg} \Rightarrow 21(\rho - 800) = \rho \Rightarrow \rho = 840\ \text{kg m}^{-3}$.
(A) False: the force falls to $\frac{1}{21}$ of its value. (B) True.
(C) True.
(D) False: $T = \frac{\text{effective weight}}{\cos(\alpha/2)}$ and the effective weight drops from $840Vg$ to $40Vg$, so the tension decreases.
Answer: (B), (C). | ||||
| Q.12 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Multiple correctMedium | |
A medium multiple correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane On the path $y = \frac{x^2}{2}$: $v_y = x v_x$ and $a_y = v_x^2 + x a_x$.
(C) At the origin the slope $\frac{dy}{dx} = x = 0$, so the velocity points along the x-axis. True. Hence at $t = 0$, $v_x = 1\ \text{ms}^{-1}$, $v_y = 0$.
(A) At the origin $x = 0$, $v_x = 1$: $a_y = 1^2 + 0 = 1\ \text{ms}^{-2}$ (whatever $a_x$ is). True.
(B) If $a_x = 0$, $v_x$ stays $1\ \text{ms}^{-1}$, so $a_y = v_x^2 + 0 = 1\ \text{ms}^{-2}$ at all times. True.
(D) With $a_x = 0$: $x = t$, so at $t = 1$ s, $x = 1$ m, $v_x = 1$, $v_y = xv_x = 1$; the velocity makes $45^\circ$ with the x-axis. True.
Answer: (A), (B), (C), (D). | ||||
| Q.13 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalHard | |
A hard numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics The work asked for is the work done in compressing the bubble beyond what the surrounding water at $p_0$ does, i.e. $W = \int (p - p_0)\,|dV|$; it is of order $a^2$, as the given form says.
When the radius is $R - s$: adiabatic $pV^\gamma = \text{const}$ gives $p = p_0\left(\frac{R}{R - s}\right)^{3\gamma} \approx p_0\left(1 + \frac{3\gamma s}{R}\right)$, so $p - p_0 \approx \frac{3\gamma p_0}{R}s$.
$|dV| = 4\pi(R - s)^2ds \approx 4\pi R^2 ds$.
$W = \int_0^a \frac{3\gamma p_0}{R}s\cdot4\pi R^2\,ds = 4\pi p_0R\cdot\frac{3\gamma}{2}a^2$.
$X = \frac{3\gamma}{2} = \frac{3}{2}\cdot\frac{41}{30} = \frac{41}{20} = 2.05$. | ||||
| Q.14 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity The 50 V cell is across P (top) and Q (bottom). Two paths join P to Q: P–$R_1$–S–$R_3$–Q and P–$R_2$–T–$R_4$–Q; the voltmeter reads S against T.
Balanced: $\frac{R_1}{R_3} = \frac{60}{300} = \frac{R_2}{R_4} = \frac{100}{500}$.
After heating: $R_3' = 300(1 + 0.0004\times100) = 312\ \Omega$.
Left path: current $\frac{50}{60 + 312} = \frac{50}{372}$ A, drop across $R_1$: $V_P - V_S = \frac{60\times50}{372} = 8.065$ V.
Right path: $V_P - V_T = \frac{100\times50}{600} = 8.333$ V.
$V_S - V_T = 8.333 - 8.065 = 0.269$ V (the voltmeter is ideal, drawing no current).
Answer: about 0.27 V. | ||||
| Q.15 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalMedium | |
A medium numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Series: $C = \frac{C_1C_2}{C_1 + C_2} = \frac{2000\times3000}{5000} = 1200$ pF.
From $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$: $\frac{\Delta C}{C^2} = \frac{\Delta C_1}{C_1^2} + \frac{\Delta C_2}{C_2^2}$.
$\frac{\Delta C}{C} = 1200\left(\frac{10}{2000^2} + \frac{15}{3000^2}\right) = 1200(2.5 + 1.667)\times10^{-6} = 0.005 = 0.5\%$.
Energy $U = \frac{1}{2}CV^2$: $\frac{\Delta U}{U} = \frac{\Delta C}{C} + 2\frac{\Delta V}{V} = 0.5\% + 2\times\frac{0.02}{5.00}\times100\% = 0.5\% + 0.8\% = 1.3\%$.
Answer: 1.30. | ||||
| Q.16 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Extra pressure on the ocean floor: $\Delta P = \rho gh = 10^3\times10\times5000 = 5\times10^7$ Pa.
Volume strain: $\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{5\times10^7}{7\times10^{10}} = 7.14\times10^{-4}$.
For a cube $V = l^3$, so $\frac{\Delta l}{l} = \frac{1}{3}\frac{\Delta V}{V} = 2.38\times10^{-4}$.
$\Delta l = 2.38\times10^{-4}$ m $= 0.238$ mm (a decrease).
Answer: 0.24. | ||||
| Q.17 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalHard | |
A hard numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The work done by the batteries against the induced emfs is stored as the magnetic energy of the coupled inductors at steady state:
$W = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + MI_1I_2$.
Steady currents: $I_1 = \frac{V_1}{R_1} = \frac{5}{5} = 1$ A, $I_2 = \frac{V_2}{R_2} = \frac{20}{10} = 2$ A.
$\frac{1}{2}L_1I_1^2 = \frac{1}{2}\times10\times1^2 = 5$ mJ; $\frac{1}{2}L_2I_2^2 = \frac{1}{2}\times20\times2^2 = 40$ mJ; $MI_1I_2 = 5\times1\times2 = 10$ mJ (all inductances in mH).
$W = 5 + 40 + 10 = 55$ mJ. | ||||
| Q.18 | 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 Power absorbed: $700\times0.05 = 35$ W.
Newton's law of cooling with constant $k = 0.001\ \text{s}^{-1}$: $\frac{dT}{dt} = -k\Delta T$, so the heat lost per second is $mck\Delta T = 1\times4200\times0.001\times\Delta T = 4.2\,\Delta T$ W.
After a long time (steady state), gain = loss: $35 = 4.2\,\Delta T$.
$\Delta T = \frac{35}{4.2} = 8.33\ ^\circ$C. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | NumericalEasy | |
A easy numerical question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Look for the pattern of jumps.
$n + 2$: a small $I_1$ (496) and a huge jump to $I_2$ (4562): one valence electron, an alkali metal.
$n + 3$: $I_1$ and $I_2$ small (738, 1451), then a big jump to $I_3$ (7733): two valence electrons, an alkaline earth metal.
$n + 1$: the highest $I_1$ (2081): a noble gas; $n$ with $I_1 = 1681$ is the halogen before it.
With $n < 10$: $n + 2 = 11$ (Na, 496), $n + 3 = 12$ (Mg, 738), $n + 1 = 10$ (Ne, 2081), $n = 9$ (F, 1681).
Answer: 9. | ||||
| Q.2 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure The figure shows the stream bent towards the comb: a liquid made of polar molecules (non-zero dipole moment) is pulled by the comb's non-uniform field.
Polar: HF, H$_2$O, NH$_3$, H$_2$O$_2$ (open-book, non-planar), CHCl$_3$, C$_6$H$_5$Cl: 6.
Non-polar (dipole moment zero by symmetry): O$_2$, CCl$_4$ (tetrahedral), C$_6$H$_6$ (planar hexagonal).
Answer: 6. | ||||
| Q.3 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In a neutral or weakly basic medium permanganate is reduced to MnO$_2$ (Mn: +7 to +4), gaining 3 electrons:
$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$.
Iodide is oxidised: $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$.
Balancing electrons: $2\text{MnO}_4^- + 6\text{I}^- + 4\text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 3\text{I}_2 + 8\text{OH}^-$.
2 mol KMnO$_4$ give 3 mol I$_2$, so 4 mol give 6 mol.
Answer: 6. | ||||
| Q.4 | 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 Acidified chromate with H$_2$O$_2$ gives the blue chromium peroxide CrO$_5$ (extracted into the alcohol):
$\text{Cr}_2\text{O}_7^{2-} + 2\text{H}^+ + 4\text{H}_2\text{O}_2 \rightarrow 2\text{CrO}_5 + 5\text{H}_2\text{O}$.
In CrO$_5$ chromium (+6) carries one Cr=O double bond and two peroxo groups, each $\text{O}{-}\text{O}$ bonded to Cr by two single bonds ('butterfly' structure).
Oxygens bonded to Cr by single bonds only: the 4 peroxo oxygens.
Answer: 4. | ||||
| Q.5 | BiomoleculesProteins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The peptide is Tyr–Asp–Lys: N-terminal tyrosine (4-hydroxybenzyl side chain), aspartic acid in the middle (CH$_2$CO$_2$H side chain), C-terminal lysine ((CH$_2$)$_4$NH$_2$ side chain).
Ionisable groups (typical p$K_a$): C-terminal COOH (~2), Asp side-chain COOH (~4), N-terminal NH$_3^+$ (~9), Tyr phenol OH (~10), Lys side-chain NH$_3^+$ (~10.5).
pH 2: both amino groups protonated (+2), carboxyls essentially un-ionised: $z_1 = +2$.
pH 6: two NH$_3^+$ (+2), two COO$^-$ ($-2$): $z_2 = 0$.
pH 11: both carboxylates ($-2$), phenoxide ($-1$), amino groups neutral: $z_3 = -3$.
$|z_1| + |z_2| + |z_3| = 2 + 0 + 3 = 5$. | ||||
| Q.6 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Colour with neutral FeCl$_3$: a phenol, so an OH on a benzene ring. C$_8$H$_{10}$O$_2$ minus C$_6$H$_4$(OH) leaves C$_2$H$_5$O for a second substituent (degree of unsaturation 4, all used by the ring).
Optical activity needs a chiral carbon. The only C$_2$H$_5$O group with a stereocentre is $-\text{CH(OH)CH}_3$ (the other choices $-\text{CH}_2\text{CH}_2\text{OH}$, $-\text{OCH}_2\text{CH}_3$, $-\text{CH}_2\text{OCH}_3$ are achiral). With three substituents on the ring (two OH and an ethyl, or OH, CH$_3$ and CH$_2$OH, etc.) no carbon is chiral either.
So the compound is HO–C$_6$H$_4$–CH(OH)CH$_3$: ortho, meta and para (3 structures), each as a pair of enantiomers.
Total isomers: $3\times2 = 6$. | ||||
| Q.7 | 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 paramagnetic substance is drawn into the field (pan goes down, figure III); a diamagnetic substance is pushed out of it (pan goes up, figure II).
(A) H$_2$O: all electrons paired, diamagnetic: upwards. True.
(B) K$_4$[Fe(CN)$_6$]: Fe$^{2+}$ (d$^6$) with the strong-field CN$^-$ is low spin, $t_{2g}^6$, no unpaired electrons: diamagnetic, upwards. True.
(C) O$_2$: two unpaired electrons in the $\pi^*$ orbitals (molecular orbital theory): paramagnetic, downwards. True.
(D) C$_6$H$_6$: all electrons paired, diamagnetic: upwards, not downwards. False.
Answer: (A), (B), (C). | ||||
| Q.8 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Multiple correctMedium | |
A medium multiple correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics P is tert-butyl bromide. With NaOH it is substituted by the S$_\text{N}$1 path: the slow step is ionisation to the tert-butyl cation, so rate $= k[\text{P}]$, first order in P.
(A) For a first-order reaction $t_{1/2} = \frac{0.693}{k}$, independent of $[\text{P}]_0$: a horizontal line. True.
(B) Initial rate $= k[\text{P}]_0$ rises in proportion to $[\text{P}]_0$, not flat. False.
(C) $\frac{[\text{Q}]}{[\text{P}]_0} = 1 - e^{-kt}$: it rises quickly at first and levels off towards 1 (concave down), not the upward-curving graph shown. False.
(D) $\ln\frac{[\text{P}]}{[\text{P}]_0} = -kt$: a straight line through the origin with negative slope; the plot shows a positive slope. False.
Answer: (A). | ||||
| Q.9 | 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) Bauxite is leached with NaOH to sodium aluminate; bubbling CO$_2$ through it neutralises the solution and hydrated alumina precipitates: $2\text{Na[Al(OH)}_4] + \text{CO}_2 \rightarrow \text{Al}_2\text{O}_3\cdot x\text{H}_2\text{O}\downarrow + \text{Na}_2\text{CO}_3 + \ldots$ True.
(B) Cryolite Na$_3$AlF$_6$ is added to lower the melting point of alumina (about 2300 K to about 1250 K) and raise conductivity. True.
(C) In the Hall–Héroult cell the oxygen released at the graphite anodes burns them, giving CO and CO$_2$. True.
(D) The cell is a steel vessel lined with carbon, the lining serving as the cathode. True.
Answer: (A), (B), (C), (D). | ||||
| Q.10 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Sn(II) is readily oxidised to the more stable Sn(IV), so SnCl$_2\cdot$2H$_2$O is a common reducing agent (it reduces HgCl$_2$, Fe$^{3+}$). True.
(B) SnO$_2$ is amphoteric: $\text{SnO}_2 + 2\text{KOH} + 2\text{H}_2\text{O} \rightarrow \text{K}_2[\text{Sn(OH)}_6]$. True.
(C) PbCl$_2$ dissolves in HCl (as chloro complexes such as [PbCl$_4$]$^{2-}$, in equilibrium with Pb$^{2+}$); the solution contains Pb$^{2+}$ and Cl$^-$ ions. True.
(D) Pb$_3$O$_4$ behaves as 2PbO·PbO$_2$: $\text{Pb}_3\text{O}_4 + 4\text{HNO}_3 \rightarrow 2\text{Pb(NO}_3)_2 + \text{PbO}_2 + 2\text{H}_2\text{O}$. The PbO part dissolves as Pb(II) and PbO$_2$ is left as Pb(IV); no oxidation state changes, so it is not a redox reaction. False.
Answer: (A), (B), (C). | ||||
| Q.11 | AminesBasicity of amines | Class 12: Amines | Multiple correctMedium | |
A medium multiple correct question on basicity of amines, from Amines. NCERT: Class 12: Amines I aniline, II N,N-dimethylaniline, III 2,4,6-trinitroaniline, IV N,N-dimethyl-2,4,6-trinitroaniline.
I and II: the nitrogen lone pair is delocalised into the ring; the two methyls (+I) make II slightly more basic (p$K_b$ about 8.9 against 9.4 for aniline).
III: the lone pair is delocalised into the ring and on into three nitro groups (two ortho, one para): extremely weak base.
IV: the bulky NMe$_2$ flanked by two ortho nitro groups is twisted out of the ring plane (steric inhibition of resonance), so its lone pair is not delocalised and IV is far more basic than III.
Order: II > I > IV > III.
(A) II > I > III > IV: False (IV is above III).
(B) The p$K_b$ gap I–II is small (about 0.5); the gap III–IV is very large (several units). False.
(C) Resonance (delocalisation of the lone pair) is greater in III than in IV. True.
(D) The steric effect (inhibition of resonance) makes IV more basic than III. True.
Answer: (C), (D). | ||||
| Q.12 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Multiple correctHard | |
A hard multiple correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons P (C$_9$H$_{12}$, 4 degrees of unsaturation) gives propylcyclohexane with Pt/H$_2$: a six-membered ring carrying a three-carbon chain, with three more unsaturations (a ring C=C and a C≡C). NaNH$_2$ then a ketone (R branch) needs a terminal alkyne, so P is a (prop-2-ynyl)cyclohexene.
Q branch: X then hot acidic KMnO$_4$ cleaves every multiple bond to carboxylic acids (the terminal carbon going to CO$_2$).
P as in (A), 4-(prop-2-ynyl)cyclohex-1-ene: cleaving the ring C=C and the side chain gives HOOC–CH$_2$–CH(CH$_2$COOH)–CH$_2$CH$_2$–COOH (C$_8$H$_{12}$O$_6$), but the branching carbon carries two identical CH$_2$COOH groups: achiral. Not the optically active Q.
P as in (C), 3-(prop-2-ynyl)cyclohex-1-ene: cleavage gives HOOC–CH(CH$_2$COOH)–CH$_2$CH$_2$CH$_2$–COOH (C$_8$H$_{12}$O$_6$), whose CH carries H, COOH, CH$_2$COOH and (CH$_2$)$_3$COOH: chiral, an optically active acid. So P is (C). (C) True, (A) False.
(B) X = Pd-C/quinoline/H$_2$ (Lindlar) reduces the alkyne to CH$_2$CH=CH$_2$; KMnO$_4$/H$_2$SO$_4$/heat then cleaves this and the ring C=C to give the same Q (CH$_2$COOH side chain, CO$_2$ lost). It is a valid X. True.
(D) R: the acetylide adds to acetophenone, and acid with heat dehydrates the tertiary propargylic alcohol to the enyne CH$_2$=C(Ph)–C≡C–CH$_2$–(cyclohexenyl). It must carry P's ring, a cyclohex-2-enyl; the structure drawn has the ring of (A) (cyclohex-3-enyl). False.
Answer: (B), (C). | ||||
| Q.13 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium The steep drop (equivalence point) is at 6 mL of HA. At half-equivalence, 3 mL, half the base has become BH$^+$: $[\text{B}] = [\text{BH}^+]$, a buffer with $\text{pOH} = \text{p}K_b$.
From the graph, at 3 mL the pH is 11, so pOH $= 14 - 11 = 3$.
p$K_b = 3$.
Check against the start: with $K_b = 10^{-3}$, 0.1 M B gives $[\text{OH}^-] \approx \sqrt{10^{-3}\times0.1} = 10^{-2}$, pH about 12, matching the beginning of the curve.
Answer: 3.00. | ||||
| Q.14 | SolutionsRaoult's law | Class 12: Solutions | NumericalEasy | |
A easy numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions Raoult's law: $P = x_Ap_A^0 + (1 - x_A)p_B^0$.
$0.25p_A^0 + 0.75p_B^0 = 0.3 \Rightarrow p_A^0 + 3p_B^0 = 1.2$.
$0.5p_A^0 + 0.5p_B^0 = 0.4 \Rightarrow p_A^0 + p_B^0 = 0.8$.
Subtracting: $2p_B^0 = 0.4 \Rightarrow p_B^0 = 0.2$ bar (and $p_A^0 = 0.6$ bar).
Answer: 0.20. | ||||
| Q.15 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalHard | |
A hard numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom At $d_0$ the two atoms are effectively separate: no electron–electron or nucleus–nucleus repulsion, so $E_0$ is the potential energy of two ground-state H atoms.
For a hydrogen atom in its ground state (Bohr model), total energy $= -13.6$ eV and kinetic energy $= +13.6$ eV, so the potential energy is $2\times(-13.6) = -27.2$ eV (zero for electron and nucleus infinitely apart).
Two atoms: $-54.4$ eV per molecule.
Per mole: $-54.4\times1.602\times10^{-19}\ \text{J}\times6.023\times10^{23} = -5.249\times10^{6}$ J mol$^{-1}$.
$E_0 \approx -5248.97$ kJ mol$^{-1}$. | ||||
| Q.16 | AminesDiazonium salts | Class 12: Amines | NumericalEasy | |
A easy numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines P is aniline (C$_6$H$_5$NH$_2$, $M = 93$ g mol$^{-1}$). NaNO$_2$/HCl at 0–5 °C gives benzenediazonium chloride, which couples with 2-naphthol in NaOH at C-1; acidification gives Q, 1-phenylazo-2-naphthol (an orange-red dye), C$_{16}$H$_{12}$N$_2$O.
$M(\text{Q}) = 16\times12 + 12\times1 + 2\times14 + 16 = 248$ g mol$^{-1}$.
P taken: $9.3$ mL $\times 1.00$ g mL$^{-1}$ $= 9.3$ g $= 0.1$ mol.
Q at 75% yield: $0.075$ mol $\times 248 = 18.6$ g.
Answer: 18.60. | ||||
| Q.17 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Reaction: $\text{SnO}_2(s) + \text{C}(s) \rightarrow \text{Sn}(s) + \text{CO}_2(g)$.
$\Delta H^0 = -394.0 - (-581.0) = +187.0$ kJ mol$^{-1}$.
$\Delta S^0 = (52.0 + 210.0) - (56.0 + 6.0) = 200.0$ J K$^{-1}$mol$^{-1}$.
The reduction becomes feasible when $\Delta G^0 = \Delta H^0 - T\Delta S^0 \le 0$: $T \ge \frac{187000}{200} = 935$ K.
Answer: 935.00. | ||||
| Q.18 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium No precipitate while $[\text{Zn}^{2+}][\text{S}^{2-}] \le K_\text{sp}$:
$[\text{S}^{2-}]_{max} = \frac{1.25\times10^{-22}}{0.05} = 2.5\times10^{-21}$ M.
From H$_2$S: $\frac{[\text{H}^+]^2[\text{S}^{2-}]}{[\text{H}_2\text{S}]} = K_\text{NET}$, so $[\text{H}^+]^2 = \frac{K_\text{NET}[\text{H}_2\text{S}]}{[\text{S}^{2-}]} = \frac{10^{-21}\times0.1}{2.5\times10^{-21}} = 0.04$.
$[\text{H}^+]_{min} = 0.2$ M.
Answer: 0.20. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z = 0$ satisfies the equation but has real part 0. For $z = re^{i\theta}$, $r > 0$:
$r^4(e^{4i\theta} - 1) = 4ir^2e^{2i\theta} \Rightarrow r^2(e^{2i\theta} - e^{-2i\theta}) = 4i \Rightarrow r^2\cdot2i\sin2\theta = 4i \Rightarrow r^2\sin2\theta = 2$.
With $z = x + iy$: $r^2\sin2\theta = 2xy$, so $xy = 1$: the rectangular hyperbola.
$\text{Re}(z_1) > 0$ puts $z_1$ on the branch $x, y > 0$; $\text{Re}(z_2) < 0$ puts $z_2$ on $x, y < 0$.
The closest points of the two branches are their vertices $(1, 1)$ and $(-1, -1)$ (for $z_1 = (a, \frac{1}{a})$, $z_2 = (-c, -\frac{1}{c})$: $|z_1 - z_2|^2 = (a + c)^2 + (\frac{1}{a} + \frac{1}{c})^2 \ge 4ac + \frac{4}{ac} \ge 8$, equality at $a = c = 1$).
Minimum $|z_1 - z_2|^2 = 2^2 + 2^2 = 8$. | ||||
| Q.2 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability With $n$ missiles, the number of hits is binomial with $p = \frac{3}{4}$, $q = \frac{1}{4}$. We need $P(\text{at most 2 hits}) \le 0.05$.
$P(\le 2) = q^n + npq^{n-1} + \binom{n}{2}p^2q^{n-2} = \frac{1 + 3n + 9\binom{n}{2}}{4^n}$.
$n = 5$: $\frac{1 + 15 + 90}{1024} = \frac{106}{1024} \approx 0.104 > 0.05$.
$n = 6$: $\frac{1 + 18 + 135}{4096} = \frac{154}{4096} \approx 0.038 \le 0.05$.
Minimum number: 6. | ||||
| Q.3 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Since $OP = OQ$, the perpendicular bisector of $PQ$ passes through $O$: it is the line through $O$ perpendicular to $2x + 4y = 5$, i.e. $y = 2x$. The circumcentre $C$ lies on it and on $x + 2y = 4$: $x + 4x = 4$, so $C = \left(\frac{4}{5}, \frac{8}{5}\right)$, and $OC^2 = \frac{16 + 64}{25} = \frac{16}{5}$.
Foot of the perpendicular from $O$ to $PQ$ (midpoint $M$ of $PQ$): $M = t(2, 4)$ with $20t = 5$, $M = \left(\frac{1}{2}, 1\right)$, $OM = \frac{\sqrt{5}}{2}$.
$CM = OC - OM = \frac{4}{\sqrt{5}} - \frac{\sqrt{5}}{2} = \frac{3}{2\sqrt{5}}$, $CM^2 = \frac{9}{20}$.
$C$ is the circumcentre, so $CP = CO$: $CP^2 = CM^2 + PM^2 \Rightarrow PM^2 = \frac{16}{5} - \frac{9}{20} = \frac{11}{4}$.
$r^2 = OP^2 = OM^2 + PM^2 = \frac{5}{4} + \frac{11}{4} = 4$, so $r = 2$. | ||||
| Q.4 | 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 For a $2\times2$ matrix with eigenvalues $\lambda_1, \lambda_2$: $\text{tr}A = \lambda_1 + \lambda_2 = 3$, $\det A = \lambda_1\lambda_2$, and $\text{tr}A^3 = \lambda_1^3 + \lambda_2^3$.
(Without eigenvalues: Cayley–Hamilton $A^2 = 3A - dI$ gives $A^3 = 3A^2 - dA = (9 - d)A - 3dI$, so $\text{tr}A^3 = 3(9 - d) - 6d = 27 - 9d$.)
$\lambda_1^3 + \lambda_2^3 = (\lambda_1 + \lambda_2)^3 - 3\lambda_1\lambda_2(\lambda_1 + \lambda_2) = 27 - 9d$.
$27 - 9d = -18 \Rightarrow d = 5$. | ||||
| Q.5 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability On $(-1, 1)$: $f(t) = 2$ for $|t| \le \frac{1}{2}$ and $f(t) = 4|t|$ for $\frac{1}{2} < |t| < 1$.
$g(x) = x + 1$ on $(-1, 0)$ and $g(x) = x$ on $[0, 1)$; its values lie in $[0, 1)$.
So $f(g(x))$: on $(-1, -\frac{1}{2}]$: 2; on $(-\frac{1}{2}, 0)$: $4(x + 1)$; on $[0, \frac{1}{2}]$: 2; on $(\frac{1}{2}, 1)$: $4x$.
At $x = 0$: left limit $4(0 + 1) = 4$, value $f(0) = 2$: discontinuous. Everywhere else it is continuous ($4(x + 1) = 2$ and $4x = 2$ at $x = \mp\frac{1}{2}$). $c = 1$.
Not differentiable: at $x = -\frac{1}{2}$ (slope 0 to 4), at $x = \frac{1}{2}$ (slope 0 to 4), and at $x = 0$ (discontinuous). $d = 3$.
$c + d = 4$. | ||||
| Q.6 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = \frac{\pi}{2}$ both numerator and denominator vanish.
Numerator: $4\sqrt{2}(\sin3x + \sin x) = 4\sqrt{2}\cdot2\sin2x\cos x = 16\sqrt{2}\sin x\cos^2x$.
Denominator: $2\sin2x\sin\frac{3x}{2} = \cos\frac{x}{2} - \cos\frac{7x}{2}$, and $\sqrt{2}(1 + \cos2x) = 2\sqrt{2}\cos^2x$. So it is
$\left(\cos\frac{x}{2} - \cos\frac{3x}{2}\right) + \left(\cos\frac{5x}{2} - \cos\frac{7x}{2}\right) - 2\sqrt{2}\cos^2x = 2\sin x\sin\frac{x}{2} + 2\sin3x\sin\frac{x}{2} - 2\sqrt{2}\cos^2x$
$= 2\sin\frac{x}{2}(\sin3x + \sin x) - 2\sqrt{2}\cos^2x = 4\sin\frac{x}{2}\sin2x\cos x - 2\sqrt{2}\cos^2x = 2\cos^2x\left(4\sin\frac{x}{2}\sin x - \sqrt{2}\right)$.
The factor $\cos^2x$ cancels:
limit $= \lim \frac{16\sqrt{2}\sin x}{2\left(4\sin\frac{x}{2}\sin x - \sqrt{2}\right)} = \frac{16\sqrt{2}}{2\left(4\cdot\frac{1}{\sqrt{2}} - \sqrt{2}\right)} = \frac{16\sqrt{2}}{2\sqrt{2}} = 8$. | ||||
| Q.7 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Multiple correctMedium | |
A medium multiple correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $\frac{f'}{f} = \frac{1}{b^2 + x^2}$, so $\ln f(x) = \frac{1}{b}\tan^{-1}\frac{x}{b} + C$; $f(0) = 1$ gives $C = 0$.
$f(x) = \exp\left(\frac{1}{b}\tan^{-1}\frac{x}{b}\right) > 0$ for every $x$ (also clear because $f$ cannot reach 0 from $f(0) = 1$).
(A) $f' = \frac{f}{b^2 + x^2} > 0$ for every $x$, whatever the sign of $b$: $f$ is increasing. True.
(B) For $b < 0$, $f$ is still increasing, not decreasing. False.
(C) $\frac{1}{b}\tan^{-1}\frac{x}{b}$ is odd in $x$, so $f(x)f(-x) = e^0 = 1$. True.
(D) $f(x) - f(-x) = 0$ would make $f$ even, but $f$ is strictly increasing (e.g. $f(x) = 1/f(-x) \neq f(-x)$ for $x \neq 0$). False.
Answer: (A), (C). | ||||
| Q.8 | Conic SectionsHyperbola | Class 11: Conic Sections | Multiple correctHard | |
A hard multiple correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Let $P = (x_1, y_1)$. Tangent: $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$; through $(1, 0)$: $x_1 = a^2$.
Normal: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$, with intercepts $\frac{(a^2 + b^2)x_1}{a^2}$ and $\frac{(a^2 + b^2)y_1}{b^2}$. Equal intercepts: $\frac{x_1}{a^2} = \frac{y_1}{b^2}$, so $y_1 = b^2$.
On the hyperbola: $\frac{a^4}{a^2} - \frac{b^4}{b^2} = 1 \Rightarrow b^2 = a^2 - 1$.
$e^2 = 1 + \frac{b^2}{a^2} = 2 - \frac{1}{a^2}$, which lies in $(1, 2)$ for $a > 1$: $1 < e < \sqrt{2}$. (A) True, (B) False.
Triangle: the tangent meets the x-axis at $(1, 0)$, the normal at $(a^2 + b^2, 0)$, and both pass through $P = (a^2, b^2)$.
$\Delta = \frac{1}{2}(a^2 + b^2 - 1)\cdot b^2 = \frac{1}{2}(2b^2)b^2 = b^4$ (using $a^2 - 1 = b^2$). (D) True; (C) False since $a \ne b$.
Answer: (A), (D). | ||||
| Q.9 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctHard | |
A hard multiple correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f(0) = 0\cdot g(0) = 0$, and $f'(0) = \lim_{h\to0}\frac{f(h)}{h} = \lim_{h\to0}g(h) = 1$. (D) True.
For any $x$: $\frac{f(x + h) - f(x)}{h} = \frac{f(h)(1 + f(x))}{h} \to 1 + f(x)$. So $f$ is differentiable everywhere with $f' = 1 + f$. (A) True.
Solving $f' = 1 + f$, $f(0) = 0$: $f(x) = e^x - 1$. So $f'(1) = e \ne 1$. (C) False.
(B) $g(x) = \frac{e^x - 1}{x}$ for $x \ne 0$; with $g(0) = 1$, $g(x) = \sum_{k\ge0}\frac{x^k}{(k + 1)!}$, a power series, so $g$ is differentiable everywhere. True.
Answer: (A), (B), (D). | ||||
| Q.10 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctEasy | |
A easy multiple correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The segment joining a point and its image is perpendicular to the plane and bisected by it.
Direction $(3 - 1, 2 - 0, -1 + 1) = (2, 2, 0)$, so $(\alpha, \beta, \gamma) = k(1, 1, 0)$: $\gamma = 0$, $\alpha = \beta$. With $\alpha + \gamma = 1$: $\alpha = \beta = 1$.
Midpoint $(2, 1, -1)$ lies on the plane: $\delta = 2 + 1 + 0 = 3$.
(A) $\alpha + \beta = 2$. True. (B) $\delta - \gamma = 3$. True. (C) $\delta + \beta = 4$. True. (D) $\alpha + \beta + \gamma = 2 \ne 3 = \delta$. False.
Answer: (A), (B), (C). | ||||
| Q.11 | 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 Area: $|\overrightarrow{PQ}\times\overrightarrow{PS}| = |a(-b) - b(a)| = 2ab = 8$, so $ab = 4$.
$|\vec{u}| = \frac{|\vec{w}\cdot\overrightarrow{PQ}|}{|\overrightarrow{PQ}|} = \frac{a + b}{\sqrt{a^2 + b^2}}$, $|\vec{v}| = \frac{|a - b|}{\sqrt{a^2 + b^2}}$.
$|\vec{u}| + |\vec{v}| = \frac{a + b + |a - b|}{\sqrt{a^2 + b^2}} = \frac{2\max(a, b)}{\sqrt{a^2 + b^2}} = \sqrt{2}$.
Squaring: $4\max^2 = 2(a^2 + b^2) \Rightarrow \max(a,b)^2 = \min(a,b)^2 \Rightarrow a = b$; with $ab = 4$, $a = b = 2$.
(A) $a + b = 4$. True. (B) $a - b = 0 \ne 2$. False.
(C) $\overrightarrow{PR} = \overrightarrow{PQ} + \overrightarrow{PS} = 2a\hat{i} = 4\hat{i}$, length 4. True.
(D) $\overrightarrow{PQ} = 2\hat{i} + 2\hat{j}$ is itself along $\vec{w}$; the bisectors of $\overrightarrow{PQ}$ and $\overrightarrow{PS} = 2\hat{i} - 2\hat{j}$ are along $\hat{i}$ and $\hat{j}$. False.
Answer: (A), (C). | ||||
| Q.12 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Multiple correctHard | |
A hard multiple correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Simplify each term of $f$:
$\binom{n+i}{p}\binom{p+n}{p-i} = \frac{(n+i)!}{p!\,(n+i-p)!}\cdot\frac{(n+p)!}{(p-i)!\,(n+i)!} = \frac{(n+p)!}{p!\,n!}\cdot\frac{n!}{(p-i)!\,(n+i-p)!} = \binom{n+p}{p}\binom{n}{p-i}$.
So $f(m, n, p) = \binom{n+p}{p}\sum_i\binom{m}{i}\binom{n}{p-i} = \binom{n+p}{p}\binom{m+n}{p}$ (Vandermonde's identity).
Then $g(m, n) = \sum_{p=0}^{m+n}\binom{m+n}{p} = 2^{m+n}$.
(A) $g(n, m) = 2^{n+m} = g(m, n)$. True.
(B) $g(m, n+1) = 2^{m+n+1} = g(m+1, n)$. True.
(C) $g(2m, 2n) = 4^{m+n}$, while $2g(m, n) = 2^{m+n+1}$: equal only if $m + n = 1$, impossible for positive integers. False.
(D) $(g(m, n))^2 = 4^{m+n} = g(2m, 2n)$. True.
Answer: (A), (B), (D). | ||||
| Q.13 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Choose 4 of the 15 days with no two consecutive. Place the 11 non-visit days in a row; they leave 12 gaps (between them and at the two ends), and each visit day goes into a different gap.
Number of ways $= \binom{15 - 4 + 1}{4} = \binom{12}{4} = 495$.
Answer: 495.00. | ||||
| Q.14 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations With 6 persons in 4 rooms, each holding 1 or 2, the occupancies must be 2, 2, 1, 1.
Choose the two rooms that get two persons: $\binom{4}{2} = 6$.
Distribute the 6 distinct persons into rooms of sizes 2, 2, 1, 1 (rooms distinct): $\frac{6!}{2!\,2!\,1!\,1!} = 180$.
Total $= 6\times180 = 1080$.
Answer: 1080.00. | ||||
| Q.15 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Sums from 2 to 12: primes 2, 3, 5, 7, 11; perfect squares 4, 9 (none is both).
$P(4) = \frac{3}{36}$, $P(9) = \frac{4}{36}$.
The process stops at the first roll whose sum is a prime or a square. Given that it stopped at a square, the stopping square is 4 or 9 in the ratio of their probabilities (the rolls before it do not matter):
$p = \frac{P(9)}{P(4) + P(9)} = \frac{4}{3 + 4} = \frac{4}{7}$.
$14p = 8$.
Answer: 8.00. | ||||
| Q.16 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f(1 - x) = \frac{4^{1-x}}{4^{1-x} + 2} = \frac{4}{4 + 2\cdot4^x} = \frac{2}{4^x + 2}$, so $f(x) + f(1 - x) = \frac{4^x + 2}{4^x + 2} = 1$.
Pair $\frac{k}{40}$ with $\frac{40-k}{40}$ for $k = 1, \ldots, 19$: 19 pairs, each summing to 1. The middle term $f\left(\frac{20}{40}\right) = f\left(\frac{1}{2}\right)$ is left over and is cancelled by $-f\left(\frac{1}{2}\right)$.
Value $= 19$.
Answer: 19.00. | ||||
| Q.17 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals By parts:
$\int_0^\pi f'(x)\cos x\,dx = \left[f(x)\cos x\right]_0^\pi + \int_0^\pi f(x)\sin x\,dx = -f(\pi) - f(0) + \int_0^\pi f(x)\sin x\,dx$.
$\int_0^\pi F(x)\cos x\,dx = \left[F(x)\sin x\right]_0^\pi - \int_0^\pi f(x)\sin x\,dx = -\int_0^\pi f(x)\sin x\,dx$ (using $F' = f$).
Adding: $-f(\pi) - f(0) = 2 \Rightarrow 6 - f(0) = 2 \Rightarrow f(0) = 4$.
Answer: 4.00. | ||||
| Q.18 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $t = \sin2\theta$. Then $(\sin\theta + \cos\theta)^2 = 1 + t$ and $(\sin\theta - \cos\theta)^4 = (1 - t)^2$, so $f = t^2 - t + 2$.
$f'(\theta) = (2t - 1)\cdot2\cos2\theta$. Critical points in $(0, \pi)$:
$t = \frac{1}{2}$: $2\theta = \frac{\pi}{6}, \frac{5\pi}{6}$, i.e. $\theta = \frac{\pi}{12}, \frac{5\pi}{12}$. Here $f = \frac{7}{4}$, the least value of $t^2 - t + 2$: local minima.
$\cos2\theta = 0$: $\theta = \frac{\pi}{4}$ ($t = 1$, $f = 2$) and $\theta = \frac{3\pi}{4}$ ($t = -1$, $f = 4$). Near $\frac{\pi}{4}$, $t$ falls below 1 on both sides and $f$ increases with $t$ there, so $f$ is a local maximum; near $\frac{3\pi}{4}$, $t$ rises above $-1$ and $f$ decreases with $t$ there, so again a local maximum.
Local minima at $\frac{\pi}{12}$ and $\frac{5\pi}{12}$: $\lambda_1 + \lambda_2 = \frac{1}{12} + \frac{5}{12} = \frac{1}{2}$.
Answer: 0.50. | ||||
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