JEE (Main) · 2020
JEE (Main) 2020 — 8 January, Shift 1
Which chapter each question came from, 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.
Physics
25 Q
- Electrostatics3
- Optics3
- Motion of System of Particles and Rigid Body2
18 easy · 7 medium · 0 hard
Chemistry
25 Q
- Coordination Compounds2
- Alcohols, Phenols and Ethers2
- Equilibrium1
14 easy · 8 medium · 3 hard
Mathematics
25 Q
- Matrices and Determinants2
- Sequences and Series2
- Limits, Continuity and Differentiability2
11 easy · 12 medium · 2 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Potential is energy per charge: $[V_0] = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$.
Write $V_0 \propto h^a c^b G^d A^e$ with $[h] = [ML^2T^{-1}]$, $[c] = [LT^{-1}]$, $[G] = [M^{-1}L^3T^{-2}]$.
M: $a - d = 1$; L: $2a + b + 3d = 2$; T: $-a - b - 2d = -3$; A: $e = -1$.
From M, $a = 1 + d$; then L gives $b = -5d$, and T gives $-1 + 2d = -3$, so $d = -1$, $a = 0$, $b = 5$.
So $[V_0] = [c^5 G^{-1} A^{-1}]$, with no power of h at all. None of the four options is this combination, which is why NTA awarded full marks to every candidate.
Answer: none of the options (marks to all). | ||||
| 2 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion Let the stretch be $x$; the particle then moves in a circle of radius $l + x$.
The spring force supplies the centripetal force: $kx = m\omega^2(l + x)$.
$x(k - m\omega^2) = ml\omega^2$, so $x = \frac{ml\omega^2}{k - m\omega^2}$.
Answer: (2). | ||||
| 3 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Split the flag into two rectangles of the same areal density.
Pole: from (0, 0) to (1, 3), area 3, centre (0.5, 1.5).
Flag: from (1, 2) to (2, 3), area 1, centre (1.5, 2.5).
$x_{cm} = \frac{3(0.5) + 1(1.5)}{4} = 0.75$ m, $y_{cm} = \frac{3(1.5) + 1(2.5)}{4} = 1.75$ m.
Answer: (1). | ||||
| 4 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single 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 torque about it).
Before: only the velocity component perpendicular to the rod counts: $L = mv\sin\frac{\pi}{4}\cdot\frac{l}{2} = \frac{mvl}{2\sqrt{2}}$.
After: $I = \frac{Ml^2}{12} + m\left(\frac{l}{2}\right)^2 = \frac{4ml^2}{12} + \frac{ml^2}{4} = \frac{7ml^2}{12}$.
$\omega = \frac{mvl/(2\sqrt{2})}{7ml^2/12} = \frac{6}{7\sqrt{2}}\frac{v}{l} = \frac{3\sqrt{2}}{7}\frac{v}{l}$.
Answer: (3). | ||||
| 5 | GravitationNewton's law of gravitation | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation The field of a solid sphere peaks at its surface, where $E = \frac{GM}{R^2}$.
Sphere 1: peak 2 at $R = 1$, so $GM_1 = 2\times1^2 = 2$.
Sphere 2: peak 3 at $R = 2$, so $GM_2 = 3\times2^2 = 12$.
$\frac{M_1}{M_2} = \frac{2}{12} = \frac{1}{6}$.
Answer: (1). | ||||
| 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 correctMedium | |
A medium 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 The sphere floats if the liquid is at least as dense as the sphere's average density.
Mass: $M = \int_0^R \rho_0\left(1 - \frac{r^2}{R^2}\right)4\pi r^2\,dr = 4\pi\rho_0\left(\frac{R^3}{3} - \frac{R^3}{5}\right) = \frac{8\pi\rho_0R^3}{15}$.
Average density: $\frac{M}{\frac{4}{3}\pi R^3} = \frac{8}{15}\cdot\frac{3}{4}\rho_0 = \frac{2\rho_0}{5}$.
Answer: (3). | ||||
| 7 | 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 The cylinder's weight is unchanged, so the buoyant force is the same: $\rho_0 A(0.80) = \rho_4 A(0.79)$.
$\frac{\rho_4}{\rho_0} = \frac{0.80}{0.79} = 1.0127 \approx 1.01$.
Answer: (2). | ||||
| 8 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Read each step off the V-T diagram.
x → y: a straight line through the origin, $V \propto T$, so pressure is constant and the volume grows: an isobaric expansion (horizontal, rightward on P-V).
y → z: horizontal on the V-T diagram, so V is constant while T falls: isochoric, pressure falls (vertical, downward on P-V).
z → x: vertical on the V-T diagram, so T is constant while V falls: isothermal compression, pressure rises along a hyperbola back to x.
Diagram (2) shows exactly this: x → y rightward at constant P, y → z straight down, and a curve from z back up to x.
Answer: (2). | ||||
| 9 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Mean free time $\tau = \frac{\lambda}{v_{avg}}$, where $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$.
For a given gas sample (fixed number density n), $\lambda$ does not depend on T, while $v_{avg} = \sqrt{\frac{8RT}{\pi M}} \propto \sqrt{T}$.
So $\tau \propto \frac{1}{\sqrt{T}}$: a straight line through the origin when plotted against $\frac{1}{\sqrt{T}}$.
Answer: (3). | ||||
| 10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Let $k = \frac{1}{4\pi\epsilon_0}$. From the figure, A is at 30° above the +x axis, C at 30° below it, and B at 150°, all at distance d from O.
A ($-4q$) pulls the field towards itself: $E_A = \frac{4kq}{d^2}$ along 30°, x-part $\frac{4kq}{d^2}\cos30^\circ = \frac{2\sqrt{3}kq}{d^2}$.
C ($-2q$) pulls towards itself: $E_C = \frac{2kq}{d^2}$ along −30°, x-part $\frac{\sqrt{3}kq}{d^2}$.
B ($+2q$) pushes away from itself, i.e. along −30° (opposite to 150°): x-part $\frac{2kq}{d^2}\cos30^\circ = \frac{\sqrt{3}kq}{d^2}$.
Total: $E_x = \frac{4\sqrt{3}kq}{d^2} = \frac{4\sqrt{3}q}{4\pi\epsilon_0 d^2} = \frac{\sqrt{3}q}{\pi\epsilon_0 d^2}$.
Answer: (4). | ||||
| 11 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance At the same voltage, $U = \frac{1}{2}CV^2$, so $C_2 = 4C_1$.
$C_1 + C_2 = 5C_1 = 10$ μF gives $C_1 = 2$ μF, $C_2 = 8$ μF.
In series: $C = \frac{2\times8}{2+8} = 1.6$ μF.
Answer: (4). | ||||
| 12 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Speed: $K = 1$ MeV $= 1.6\times10^{-13}$ J, $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{3.2\times10^{-13}}{1.6\times10^{-27}}} = \sqrt{2}\times10^7$ m/s.
The field is perpendicular to the velocity, so $ma = qvB$:
$B = \frac{ma}{qv} = \frac{1.6\times10^{-27}\times10^{12}}{1.6\times10^{-19}\times1.414\times10^7} = 7.1\times10^{-4}$ T $= 0.71$ mT.
Answer: (3). | ||||
| 13 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Gauss's law is $\oint\vec{E}\cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$.
It reduces to $|\vec{E}||A| = \frac{q_{enc}}{\epsilon_0}$ only if $\vec{E}$ is everywhere normal to the surface (so $\vec{E}\cdot d\vec{A} = E\,dA$; this is what an equipotential surface guarantees) and $|\vec{E}|$ has the same value all over it (so it comes out of the integral).
Both conditions are needed; a constant magnitude alone (4) is not enough, nor is being equipotential alone (2).
Answer: (3). | ||||
| 14 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Area of the loop: the 16 cm × 4 cm rectangle less the two triangular notches, each of base 4 cm and height 2 cm.
$A = 16\times4 - 2\times\frac{1}{2}\times4\times2 = 64 - 8 = 56$ cm² $= 56\times10^{-4}$ m².
$\frac{dB}{dt} = \frac{(1000 - 500)\times10^{-4}\ \mathrm{T}}{5\ \mathrm{s}} = 10^{-2}$ T/s.
$\varepsilon = A\frac{dB}{dt} = 56\times10^{-4}\times10^{-2} = 56\times10^{-6}$ V $= 56$ μV.
Answer: (4). | ||||
| 15 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Refractive index: $n = \sqrt{\epsilon_r\mu_r} = \sqrt{3\times\frac{4}{3}} = 2$.
$\sin C = \frac{1}{n} = \frac{1}{2}$, so $C = 30^\circ$.
Answer: (1). | ||||
| 16 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics In normal adjustment: $f_o + f_e = 60$ cm and $\frac{f_o}{f_e} = 5$.
So $6f_e = 60$, $f_e = 10$ cm ($f_o = 50$ cm).
Answer: (1). | ||||
| 17 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{\sqrt{2mT}}$, so $\lambda_B = 2\lambda_A$ means $T_B = \frac{T_A}{4}$.
$T_A - 1.5 = \frac{T_A}{4}$ gives $T_A = 2$ eV and $T_B = 0.5$ eV.
$\phi_B = 4.50 - 0.5 = 4$ eV.
Answer: (2). | ||||
| 18 | Atoms and NucleiRutherford model and alpha scattering | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on rutherford model and alpha scattering, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Most α-particles pass almost undeflected and only a very few are scattered through large angles; Rutherford's formula gives $N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$.
So the count is very large at small θ and falls steeply, flattening to a small value at large angles: graph (4).
Answer: (4). | ||||
| 19 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The two switches feed the 5 V source through diodes to a common resistor: the transistor's base is high if A or B is closed. This stage is an OR gate.
The transistor in common-emitter mode inverts: when its base is high it conducts and Y is low, otherwise Y is high.
$Y = \overline{A + B} = \overline{A}\cdot\overline{B}$ (a NOR gate).
Answer: (4). | ||||
| 20 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity At the null point no current flows through the cell, so its internal resistance plays no part: 5 V is balanced by 1000 cm of wire.
Potential gradient: $\frac{5}{1000}$ V/cm; across the whole 1200 cm wire: $V = 6$ V.
$R = \frac{V}{I} = \frac{6}{0.06} = 100\ \Omega$.
Answer: (3). | ||||
| 21 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $v_x = \frac{dx}{dt} = 8 - 6t = 2$ m/s at $t = 1$ s (first particle, along x).
$v_y = \frac{dy}{dt} = -24t^2 = -24$ m/s at $t = 1$ s (second particle, along y).
Velocity of the second relative to the first: $(-2, -24)$ m/s; its speed is $\sqrt{4 + 576} = \sqrt{580}$.
Answer: 580. | ||||
| 22 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Momentum (equal masses): $3\hat{i} + 5\hat{j} = 4\hat{i} + 4\hat{j} + \vec{v}_B$, so $\vec{v}_B = -\hat{i} + \hat{j}$.
$K_B = \frac{1}{2}(0.1)(1 + 1) = 0.1$ J $= \frac{1}{10}$ J.
(Check: total kinetic energy before $= \frac{1}{2}(0.1)(9 + 25) = 1.7$ J, after $= \frac{1}{2}(0.1)(32 + 2) = 1.7$ J, so the collision is indeed elastic.)
Answer: 1. | ||||
| 23 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Speed of sound $v = \sqrt{\frac{\gamma P}{\rho}}$; at the same pressure, doubling the density gives $v = \frac{300}{\sqrt{2}} = 212.1$ m/s.
Open pipe: $f_n = \frac{nv}{2L}$, so $f_2 - f_1 = \frac{v}{2L} = \frac{212.1}{2} = 106.07$ Hz.
Answer: 106.07 (NTA's key accepts 106 to 107.2). | ||||
| 24 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity In cyclic order P = 15 Ω, Q = 12 Ω, R = 4 Ω, S = 10 Ω, the bridge balances when $PR = QS$ (opposite arms multiply equally).
$15\times4 = 60$ but $12\times10 = 120$; so S must become $\frac{60}{12} = 5\ \Omega$.
10 Ω in parallel with X gives 5 Ω when $X = 10\ \Omega$.
Answer: 10. | ||||
| 25 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens maker's formula with one flat face: $\frac{1}{f} = (\mu - 1)\frac{1}{R} = 0.5\times\frac{1}{30}$.
$f = 60$ cm (the focal length of a thin lens does not depend on which face meets the light).
Answer: 60. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Read points off the curve: [Y] ≈ 2 mM at [X] = 1 mM, ≈ 1 mM at [X] = 2 mM and ≈ 0.67 mM at [X] = 3 mM.
The product [X][Y] stays at $2\ \mathrm{mM^2}$ along the curve (a rectangular hyperbola), so the salt is XY with $K_{sp} = [X][Y] = (10^{-3})(2\times10^{-3}) = 2\times10^{-6}$.
For $\mathrm{XY_2}$ or $\mathrm{X_2Y}$ the curve would follow $[X][Y]^2$ or $[X]^2[Y]$ = constant, which these points do not fit (e.g. for $\mathrm{XY_2}$ through (1, 2), [Y] at [X] = 2 would be 1.41 mM, not 1 mM).
Read instead from the pure salt XY₂ ([X] = s = 1 mM, [Y] = 2s = 2 mM on the curve): Ksp = (10⁻³)(2×10⁻³)² = 4×10⁻⁹ M³, which matches option (2) including its unit.
Answer: (1) or (2). | ||||
| 27 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics With the same pre-exponential factor, $\frac{k_{enz}}{k} = e^{-(E_{enz} - E)/RT} = 10^6$.
$-(E_{enz} - E) = RT\ln10^6 = 6(2.303)RT$, so $\Delta E_a = -6(2.303)RT$ (the enzyme lowers it).
Answer: (2). | ||||
| 28 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Ferric hydroxide sol is positively charged, so it is coagulated by the anions of the electrolytes.
Hardy-Schulze rule: the greater the charge on the coagulating ion, the greater its power and the smaller its flocculation value.
Anions: $[\mathrm{Fe(CN)_6}]^{3-}$ (charge 3) > $\mathrm{CrO_4^{2-}}$ (charge 2) > $\mathrm{Br^-} = \mathrm{NO_3^-} = \mathrm{Cl^-}$ (charge 1; $\mathrm{AlCl_3}$ acts through $\mathrm{Cl^-}$, its $\mathrm{Al^{3+}}$ has the same charge sign as the sol).
Flocculation values are in the reverse order: $\mathrm{K_3[Fe(CN)_6]} < \mathrm{K_2CrO_4} < \mathrm{KBr} = \mathrm{KNO_3} = \mathrm{AlCl_3}$.
Answer: (4). | ||||
| 29 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Ethyl acetate, $\mathrm{CH_3COOC_2H_5}$, is a polar molecule (the ester group), so dipole-dipole forces act, together with London dispersion forces present in every liquid.
It has no H bonded to O, N or F, so its molecules cannot hydrogen-bond with one another.
Answer: (4). | ||||
| 30 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom (I) True: as $n_2$ grows the lines crowd together towards the series limit, i.e. they converge as the wavelength decreases.
(II) True: the Balmer series ends on $n_1 = 2$.
(III) True: the smallest energy gap, $n_2 = 3 \to 2$ (the H-α line), gives the longest wavelength.
(IV) False: the Balmer limit gives the energy to ionise from n = 2, not from the ground state; the ionisation energy of hydrogen comes from the Lyman series limit.
Answer: (1). | ||||
| 31 | SolutionsRaoult's law | Class 12: Solutions | Single correctEasy | |
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions At any temperature X has the highest vapour pressure and Z the lowest (X reaches a given pressure at the lowest temperature).
A higher vapour pressure means weaker intermolecular forces, so the interactions increase X < Y < Z.
(A) is false, (B) is true, (C) is false (Z has stronger interactions than Y).
Answer: (2). | ||||
| 32 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Ionization energy rises across period 3 with one exception: Al (3p¹) is lower than Mg (filled 3s²), because the 3p electron is higher in energy and shielded by 3s.
So Na 496 < Al 577 < Mg 737 < Si 786: in the order Na, Mg, Al, Si this reads 496, 737, 577, 786.
Answer: (3). | ||||
| 33 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter NaOH absorbs moisture and $\mathrm{CO_2}$, so it is standardised against a primary standard: oxalic acid, whose solution of exactly known strength can be made by weighing.
Concentrated $\mathrm{H_2SO_4}$ is not a primary standard (it is hygroscopic and its strength is not exactly known), ruling out (1) and (2), and a volumetric flask is not a titration vessel.
The usual practice is the alkali in the burette and a measured volume of the standard oxalic acid in the conical flask with phenolphthalein; a pipette is not used to deliver NaOH, which attacks glass and absorbs $\mathrm{CO_2}$, so (4) is less accurate.
Answer: (3). | ||||
| 34 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Gypsum is $\mathrm{CaSO_4\cdot2H_2O}$. On heating to 393 K it loses part of its water and forms plaster of Paris:
$2(\mathrm{CaSO_4\cdot2H_2O}) \to 2(\mathrm{CaSO_4})\cdot\mathrm{H_2O} + 3\mathrm{H_2O}$, i.e. $\mathrm{CaSO_4\cdot0.5H_2O}$.
Above 393 K the anhydrous salt (dead burnt plaster) forms.
Answer: (3). | ||||
| 35 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Peroxodisulphate is $\mathrm{O_3S{-}O{-}O{-}SO_3}^{2-}$: each S is bonded to four O atoms (three terminal and one of the peroxide bridge), so there are $2\times4 = 8$ S–O bonds.
Rhombic sulphur consists of puckered $\mathrm{S_8}$ rings: eight S atoms in a ring joined by 8 S–S bonds.
Answer: (4). | ||||
| 36 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The third electron is removed from the $\mathrm{M^{2+}}$ ion: $\mathrm{Mn^{2+}}$ $3d^5$, $\mathrm{Fe^{2+}}$ $3d^6$, $\mathrm{Co^{2+}}$ $3d^7$, $\mathrm{Ni^{2+}}$ $3d^8$.
For $\mathrm{Fe^{2+}}$ removing one electron gives the stable half-filled $3d^5$ of $\mathrm{Fe^{3+}}$, so its third ionisation enthalpy is the lowest (about 2957 kJ/mol, against about 3248 for Mn, 3232 for Co and 3393 for Ni).
Answer: (3). | ||||
| 37 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Facial and meridional isomers arise in octahedral complexes of the type $\mathrm{Ma_3b_3}$: the three identical ligands on one face (fac) or along a meridian (mer).
$\mathrm{[Co(NH_3)_3(NO_2)_3]}$ is $\mathrm{Ma_3b_3}$. The others ($\mathrm{M(AA)_2b_2}$, $\mathrm{Ma_4b_2}$, square planar $\mathrm{Ma_2b_2}$) show only cis-trans isomerism.
Answer: (1). | ||||
| 38 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Greenhouse gases absorb infrared radiation from the earth: $\mathrm{CO_2}$, water vapour, CFCs, ozone, methane and nitrous oxide.
$\mathrm{O_2}$ (a homonuclear diatomic with no changing dipole) does not absorb infrared and is not a greenhouse gas.
Answer: (3). | ||||
| 39 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Only the $-\mathrm{CO_2H}$ group is reduced to $-\mathrm{CH_2OH}$; the ketone, the amide and the nitrile are untouched.
Diborane reduces carboxylic acids to primary alcohols rapidly and selectively, faster than it attacks ketones, amides or nitriles.
$\mathrm{LiAlH_4}$ would also reduce the ketone, amide and nitrile; $\mathrm{NaBH_4}$ does not reduce acids but reduces the ketone; $\mathrm{H_2/Pd}$ does not reduce an acid.
Answer: (2). | ||||
| 40 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Isohexane (2-methylpentane) boils at 60 °C and 3-methylpentane at 63 °C (the two are isomers whose boiling points differ by only 3 °C).
Boiling points only 3 °C apart need fractional distillation; simple distillation works only for a large difference.
The lower-boiling isohexane distils out first.
Answer: (1). | ||||
| 41 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers In phenol the lone pair of the OH oxygen is delocalised into the ring, giving the C–O bond partial double-bond character (and the carbon is sp²): it is shorter than the C–O single bond of methanol.
In p-ethoxyphenol the electron-donating ethoxy group pushes electron density into the ring and reduces the delocalisation from the OH oxygen, so its C–OH bond has less double-bond character than in phenol, but still more than in methanol.
Order: phenol < p-ethoxyphenol < methanol.
Answer: (2). | ||||
| 42 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Glucose forms an oxime with hydroxylamine (open-chain aldehyde), and exists as α and β crystalline forms: (2) and (4) are true.
Its pentaacetate has no free –OH at C-1 to open the ring, so it does not form an oxime: (1) is true.
Despite having an aldehyde group, glucose does not give Schiff's test (nor the bisulphite addition): it exists almost entirely in the cyclic hemiacetal form. So (3) is not true.
Answer: (3). | ||||
| 43 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The nitrile is 2,3-dimethylbutanenitrile, $\mathrm{(CH_3)_2CH{-}CH(CH_3){-}CN}$; the alkene is but-1-ene.
Peroxide and heat abstract the hydrogen that gives the most stable radical: the tertiary C–H next to CN, because that radical is both tertiary and delocalised onto the nitrile. A = $\mathrm{(CH_3)_2CH{-}\dot{C}(CH_3){-}CN}$.
A adds to the CH₂ end of but-1-ene to give the more stable secondary radical $\mathrm{(CH_3)_2CH{-}C(CH_3)(CN){-}CH_2{-}\dot{C}H{-}CH_2CH_3}$, which takes a hydrogen atom to give B = $\mathrm{(CH_3)_2CH{-}C(CH_3)(CN){-}CH_2CH_2CH_2CH_3}$: a quaternary carbon bearing CN, methyl, isopropyl and n-butyl.
That is option (3); (1) adds to the wrong carbon of the alkene, and (2) and (4) start from the less stable radical on the isopropyl carbon.
Answer: (3). | ||||
| 44 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The rate of an $E_1$ reaction is set by how easily the carbocation forms.
(A) 1-chlorobutane gives a primary carbocation, the least stable: slowest.
(C) 2-chlorobutane gives a secondary carbocation.
(D) 3-chlorobut-1-ene gives a secondary allylic carbocation, stabilised by resonance.
(B) 4-chlorobut-1-ene: as the C–Cl bond ionises, the C=C two carbons away takes part (homoallylic participation), giving the non-classical cyclopropylcarbinyl cation, which is more stable even than an allylic cation; the product is the conjugated buta-1,3-diene.
Order: B > D > C > A.
Answer: (1). | ||||
| 45 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctHard | |
A hard single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The substrate is geraniol, $\mathrm{(CH_3)_2C{=}CH{-}CH_2CH_2{-}C(CH_3){=}CH{-}CH_2OH}$.
In dilute acid the allylic alcohol ionises to an allylic cation (it can isomerise to the cis, neryl form). The remote $\mathrm{C{=}C}$ of the isopropylidene end attacks it intramolecularly, closing a six-membered ring and leaving a tertiary carbocation $\mathrm{{-}C^+(CH_3)_2}$ on the ring.
Water captures this cation, giving α-terpineol: a 1-methylcyclohexene ring carrying $\mathrm{{-}C(CH_3)_2OH}$ at C-4.
Answer: (2). | ||||
| 46 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Iron needed: 10 ppm of 100 kg $= \frac{10}{10^6}\times10^5$ g $= 1$ g.
Molar mass of $\mathrm{FeSO_4\cdot7H_2O} = 55.85 + 32.00 + 64.00 + 7(18.02) = 277.99$ g/mol (with H = 1.008).
Mass of salt $= 1\times\frac{277.99}{55.85} = 4.98$ g (4.97 g with H = 1).
Answer: 4.98 (NTA's key accepts 4.95 to 4.99). | ||||
| 47 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics From the figure, the axes cross at (2, 2): A = (2 m³, 8 Pa), B = (8 m³, 8 Pa), C = (12 m³, 2 Pa).
NTA's key takes the area between the path and the drawn volume axis:
A → B: rectangle $6\times(8 - 2) = 36$ J; B → C: triangle $\frac{1}{2}\times4\times(8 - 2) = 12$ J; total 48 J.
(The full $\int P\,dV$ down to zero pressure would add the strip $10\times2 = 20$ J below the axis, giving 68 J; see the note.)
Answer: 48. | ||||
| 48 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry The half reaction is written as an oxidation, so $E^0_{ox} = -1.23$ V.
Nernst: $E = E^0_{ox} - \frac{0.0591}{4}\log\left([\mathrm{H^+}]^4\,p_{\mathrm{O_2}}\right) = -1.23 - \frac{0.0591}{4}\log(10^{-20})$.
$= -1.23 + 0.0591\times5 = -1.23 + 0.296 = -0.934$ V.
Answer: −0.93 (NTA's key accepts −0.93 to −0.94). | ||||
| 49 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All three chlorides of $\mathrm{[Co(NH_3)_6]Cl_3}$ are ionisable (outside the coordination sphere).
Moles of complex $= \frac{0.3}{267.46} = 1.122\times10^{-3}$; moles of $\mathrm{Cl^-} = 3\times1.122\times10^{-3} = 3.365\times10^{-3}$.
One $\mathrm{Ag^+}$ per $\mathrm{Cl^-}$: $V = \frac{3.365\times10^{-3}}{0.125}$ L $= 26.92$ mL.
Answer: 26.92 (NTA's key accepts 26.60 to 27.0). | ||||
| 50 | Chemistry in Everyday LifeAntibiotics and antiseptics | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on antibiotics and antiseptics, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Penicillin's core (the fused β-lactam–thiazolidine ring system) has three stereocentres: C-2 (carrying the carboxyl group), C-5 (the ring-fusion carbon bonded to S) and C-6 (carrying the acylamino side chain).
The gem-dimethyl carbon is not chiral (two identical methyls), and the common penicillin G side chain (benzyl) adds none.
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Let $y = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}} = \frac{8^{4x} - 1}{8^{4x} + 1}$.
Then $8^{4x} = \frac{1+y}{1-y}$, so $4x = \log_8\frac{1+y}{1-y} = (\log_8 e)\log_e\frac{1+y}{1-y}$.
$f^{-1}(x) = \frac{1}{4}(\log_8 e)\log_e\left(\frac{1+x}{1-x}\right)$.
Answer: (4). | ||||
| 52 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let the roots be $z = \alpha \pm i\beta$. Sum $2\alpha = -b$, product $\alpha^2 + \beta^2 = 45$.
$|z + 1|^2 = (\alpha + 1)^2 + \beta^2 = \alpha^2 + \beta^2 + 2\alpha + 1 = 45 - b + 1 = 40$, so $b = 6$.
(Complex roots need $b^2 < 180$, which holds.) Then $b^2 - b = 36 - 6 = 30$.
Answer: (3). | ||||
| 53 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The coefficient determinant is $1(16 - 20) - 2(12 - 20) + 3(12 - 16) = -4 + 16 - 12 = 0$.
The third row of coefficients is $2\times$(second) $- 2\times$(first): $2(3, 4, 5) - 2(1, 2, 3) = (4, 4, 4)$.
So the system is consistent exactly when $\delta = 2\mu - 2$, and inconsistent otherwise.
(1, 0): $0 = 0$; (3, 4): $4 = 4$; (4, 6): $6 = 6$, all consistent. (4, 3): $2\mu - 2 = 6 \ne 3$, inconsistent.
Answer: (3). | ||||
| 54 | Straight LinesCartesian coordinates and distance formula | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines Area of PAB $= \frac{1}{2}\left|x'(-1 - 2) + 1(2 - y') + 0(y' + 1)\right| = \frac{1}{2}\left|2 - (3x' + y')\right|$.
P lies on the line, so $3x' + y' = 4\lambda$: area $= \frac{1}{2}|2 - 4\lambda| = 5$.
$|2 - 4\lambda| = 10$ gives $\lambda = 3$ or $\lambda = -2$; of these, 3 is offered.
Answer: (4). | ||||
| 55 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Greatest values: $a = {}^{19}C_9 = {}^{19}C_{10}$, $b = {}^{20}C_{10}$, $c = {}^{21}C_{10} = {}^{21}C_{11}$.
$b = {}^{19}C_9 + {}^{19}C_{10} = 2a$.
$c = {}^{21}C_{11} = \frac{21}{11}\,{}^{20}C_{10} = \frac{21}{11}b = \frac{42}{11}a$.
So $\frac{a}{11} = \frac{b}{22} = \frac{c}{42}$.
Answer: (4). | ||||
| 56 | Sequences and SeriesRelation between A.M. and G.M. | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on relation between a.m. and g.m., from Sequences and Series. NCERT: Class 11: Sequences and Series In an A.P. the middle term is the mean: $f(x) = \frac{1}{2}\left[2(2^x + 2^{-x}) + (3^x + 3^{-x})\right]$.
By A.M. ≥ G.M., $2^x + 2^{-x} \ge 2$ and $3^x + 3^{-x} \ge 2$, both with equality at $x = 0$.
Minimum: $\frac{1}{2}(4 + 2) = 3$.
Answer: (3). | ||||
| 57 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability This is a $1^\infty$ form: the limit is $e^L$ with $L = \lim_{x\to0}\frac{1}{x^2}\left(\frac{3x^2 + 2}{7x^2 + 2} - 1\right) = \lim_{x\to0}\frac{1}{x^2}\cdot\frac{-4x^2}{7x^2 + 2} = -2$.
Limit $= e^{-2} = \frac{1}{e^2}$.
Answer: (4). | ||||
| 58 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\cos^{-1}(-t) = \pi - \cos^{-1}t$ and, for $|x| \le \frac{\pi}{2}$, $\cos^{-1}(\sin|x|) = \frac{\pi}{2} - |x|$.
So $f(x) = x\left(\frac{\pi}{2} + |x|\right)$: $f(x) = \frac{\pi}{2}x + x^2$ for $x \ge 0$ and $\frac{\pi}{2}x - x^2$ for $x < 0$.
$f'(x) = \frac{\pi}{2} + 2x$ on $(0, \frac{\pi}{2})$ and $\frac{\pi}{2} - 2x$ on $(-\frac{\pi}{2}, 0)$; both one-sided derivatives at 0 equal $\frac{\pi}{2}$, so f is differentiable with $f'(0) = \frac{\pi}{2}$ ((1) and (2) false).
$f' = \frac{\pi}{2} - 2x$ decreases on $(-\frac{\pi}{2}, 0)$ and $f' = \frac{\pi}{2} + 2x$ increases on $(0, \frac{\pi}{2})$.
Answer: (3). | ||||
| 59 | Applications of DerivativesRate of change | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on rate of change, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Rolle's theorem needs $f(3) = f(4)$: $\frac{9 + \alpha}{21} = \frac{16 + \alpha}{28}$, so $28(9 + \alpha) = 21(16 + \alpha)$ and $\alpha = 12$.
$f(x) = \ln(x^2 + 12) - \ln 7x$, $f'(x) = \frac{2x}{x^2 + 12} - \frac{1}{x}$; $f'(c) = 0$ gives $2c^2 = c^2 + 12$, $c = 2\sqrt{3}$.
$f''(x) = \frac{2(x^2 + 12) - 4x^2}{(x^2 + 12)^2} + \frac{1}{x^2} = \frac{24 - 2x^2}{(x^2 + 12)^2} + \frac{1}{x^2}$.
At $c^2 = 12$: $f''(c) = 0 + \frac{1}{12} = \frac{1}{12}$.
Answer: (2). | ||||
| 60 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t = \sin x$: $I = \int\frac{dt}{t^3(1 + t^6)^{2/3}} = \int\frac{dt}{t^7(1 + t^{-6})^{2/3}}$.
Let $u = 1 + t^{-6}$, $du = -6t^{-7}dt$: $I = -\frac{1}{6}\int u^{-2/3}du = -\frac{1}{2}u^{1/3} + c = -\frac{1}{2}\cdot\frac{(1 + t^6)^{1/3}}{t^2} + c$.
So $f(x) = -\frac{1}{2\sin^2x}$ and $\lambda = 3$.
$\lambda f\left(\frac{\pi}{3}\right) = 3\times\left(-\frac{1}{2\cdot\frac{3}{4}}\right) = 3\times\left(-\frac{2}{3}\right) = -2$.
Answer: (4). | ||||
| 61 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The curves meet at O and $P(a, a)$; chord OP is $y = x$, so $Q = (b, b)$, $R = (b, 0)$ and area OQR $= \frac{b^2}{2} = \frac{1}{2}$, giving $b = 1$.
Whole area between the curves: $\int_0^a\left(\sqrt{ax} - \frac{x^2}{a}\right)dx = \frac{2}{3}a^2 - \frac{a^2}{3} = \frac{a^2}{3}$.
The part from 0 to $b = 1$ must be half of it: $\frac{2}{3}\sqrt{a} - \frac{1}{3a} = \frac{a^2}{6}$.
Multiplying by $6a$: $4a^{3/2} - 2 = a^3$, i.e. $a^3 + 2 = 4a^{3/2}$.
Squaring: $a^6 + 4a^3 + 4 = 16a^3$, so $a^6 - 12a^3 + 4 = 0$: a satisfies $x^6 - 12x^3 + 4 = 0$.
Answer: (2). | ||||
| 62 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $\frac{dy}{\sqrt{1 - y^2}} = -\frac{dx}{\sqrt{1 - x^2}}$, so $\sin^{-1}y = -\sin^{-1}x + C$.
$y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}$: $\frac{\pi}{3} = -\frac{\pi}{6} + C$, $C = \frac{\pi}{2}$. So $\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2}$, i.e. $y = \sqrt{1 - x^2}$.
At $x = -\frac{1}{\sqrt{2}}$: $y = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Answer: (2). | ||||
| 63 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Let the point on the parabola be $(x_1, y_1)$ with $x_1^2 = 4y_1$, and the dividing point $(h, k)$.
Ratio 1 : 2 from $(0, -1)$: $h = \frac{x_1}{3}$, $k = \frac{y_1 + 2(-1)}{3}$, so $x_1 = 3h$, $y_1 = 3k + 2$.
$9h^2 = 4(3k + 2) = 12k + 8$: the locus is $9x^2 - 12y = 8$.
Answer: (3). | ||||
| 64 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Let $P = (x_0, y_0)$. The normal is along the gradient $(4x_0, 2y_0) \propto (2x_0, y_0)$: its points are $(x_0 + 2x_0t,\ y_0 + y_0t)$.
x-intercept ($t = -1$): $x = -x_0 = -\frac{1}{3\sqrt{2}}$, so $x_0 = \frac{1}{3\sqrt{2}}$.
Then $y_0^2 = 1 - 2x_0^2 = 1 - \frac{1}{9} = \frac{8}{9}$, $y_0 = \frac{2\sqrt{2}}{3}$ (first quadrant).
y-intercept ($t = -\frac{1}{2}$): $\beta = \frac{y_0}{2} = \frac{\sqrt{2}}{3}$.
Answer: (1). | ||||
| 65 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Points $A(3, 8, 3)$, $B(-3, -7, 6)$; directions $\vec{d_1} = (3, -1, 1)$, $\vec{d_2} = (-3, 2, 4)$.
$\vec{d_1}\times\vec{d_2} = (-4 - 2,\ -(12 + 3),\ 6 - 3) = (-6, -15, 3)$, of length $\sqrt{270} = 3\sqrt{30}$.
$\vec{AB} = (-6, -15, 3)$, and $\vec{AB}\cdot(\vec{d_1}\times\vec{d_2}) = 36 + 225 + 9 = 270$.
Distance $= \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30}$.
Answer: (2). | ||||
| 66 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Volume $= |[\vec{u}\ \vec{v}\ \vec{w}]| = \left|\begin{vmatrix}1&1&\lambda\\1&1&3\\2&1&1\end{vmatrix}\right| = |1(1 - 3) - 1(1 - 6) + \lambda(1 - 2)| = |3 - \lambda| = 1$, so $\lambda = 2$ or $4$.
$\cos\theta = \frac{\vec{u}\cdot\vec{w}}{|\vec{u}||\vec{w}|} = \frac{3 + \lambda}{\sqrt{2 + \lambda^2}\sqrt{6}}$.
$\lambda = 2$: $\frac{5}{6}$ (not offered); $\lambda = 4$: $\frac{7}{\sqrt{18}\sqrt{6}} = \frac{7}{6\sqrt{3}}$.
Answer: (3). | ||||
| 67 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics New observations $px_i - q$: new mean $20p - q = 10$, new s.d. $2|p| = 1$, so $p = \pm\frac{1}{2}$.
$p = \frac{1}{2}$ gives $q = 0$, which is excluded. $p = -\frac{1}{2}$ gives $-10 - q = 10$, $q = -20$.
Answer: (4). | ||||
| 68 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Independence: $P(A\cap B) = \frac{1}{18}$, and A is independent of B' too.
(1) $P(A\cup B) = \frac{1}{3} + \frac{1}{6} - \frac{1}{18} = \frac{4}{9}$; $P(A \mid A\cup B) = \frac{1/3}{4/9} = \frac{3}{4}$: false.
(2) $P(A \mid B) = P(A) = \frac{1}{3}$: false.
(3) $P(A \mid B') = P(A) = \frac{1}{3}$: true.
(4) $P(A' \mid B') = P(A') = \frac{2}{3}$: false.
Answer: (3). | ||||
| 69 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\sin(\tan^{-1}x) = \frac{x}{\sqrt{1 + x^2}}$ and $\sin(\cot^{-1}x) = \frac{1}{\sqrt{1 + x^2}}$ for every real x, so $f(x) = \frac{(x + 1)^2}{1 + x^2} - 1 = \frac{2x}{1 + x^2}$.
Integrating the given relation: $y = \frac{1}{2}\sin^{-1}\left(\frac{2x}{1 + x^2}\right) + C$.
$y(\sqrt{3}) = \frac{1}{2}\sin^{-1}\frac{\sqrt{3}}{2} + C = \frac{\pi}{6} + C = \frac{\pi}{6}$, so $C = 0$.
$y(-\sqrt{3}) = \frac{1}{2}\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{6}$.
Answer: (1). | ||||
| 70 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter (1) and (2) both simplify to P by absorption, which is false when P is false: not tautologies.
(3) If $P \wedge (P \to Q)$ is true, then P is true and $P \to Q$ is true, so Q is true (modus ponens). The implication is never false: a tautology.
(4) With Q true and P false, $P \wedge (P \to Q)$ is false, so $Q \to (\ldots)$ is false: not a tautology.
Answer: (3). | ||||
| 71 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $2x^2 + (a - 10)x + \frac{33}{2} - 2a = 0$ has real roots when $D = (a - 10)^2 - 8\left(\frac{33}{2} - 2a\right) \ge 0$.
$D = a^2 - 20a + 100 - 132 + 16a = a^2 - 4a - 32 = (a - 8)(a + 4) \ge 0$, so $a \le -4$ or $a \ge 8$.
Least positive value: 8.
Answer: 8. | ||||
| 72 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\mathrm{tr}(AA^T) = \sum_{i,j}a_{ij}^2$, the sum of the squares of all nine entries.
With entries in $\{-1, 0, 1\}$ each square is 0 or 1, so exactly three entries are $\pm1$ and six are 0.
Choose their places: $\binom{9}{3} = 84$; choose their signs: $2^3 = 8$. Total $84\times8 = 672$.
Answer: 672. | ||||
| 73 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\sum_{k=1}^{20}\frac{k(k+1)}{2} = \frac{1}{2}\left(\sum k^2 + \sum k\right) = \frac{1}{2}(2870 + 210) = 1540$.
(Equivalently $\frac{n(n+1)(n+2)}{6} = \frac{20\cdot21\cdot22}{6} = 1540$.)
Answer: 1540. | ||||
| 74 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Differentiate: $2yy' - 6x + y' = 0$, so $m = y' = \frac{6x}{2y + 1}$ at $P(x, y)$.
The normal $Y - y = -\frac{1}{m}(X - x)$ meets $X = 0$ at $Y = y + \frac{x}{m} = y + \frac{2y + 1}{6}$ (for $x \ne 0$).
$y + \frac{2y + 1}{6} = \frac{3}{2}$ gives $8y + 1 = 9$, $y = 1$. On the curve: $1 - 3x^2 + 1 + 10 = 0$, $x = \pm2$.
$m = \frac{6(\pm2)}{3} = \pm4$, so $|m| = 4$.
Answer: 4. | ||||
| 75 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations All ways to draw 4 of the 12 marbles: $\binom{12}{4} = 495$.
Exclude the draws with all four red: $\binom{5}{4} = 5$.
$495 - 5 = 490$.
Answer: 490. | ||||
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