JEE (Main) · 2020
JEE (Main) 2020 — 6 September, Shift 2
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
- Motion of System of Particles and Rigid Body2
- Properties of Bulk Matter2
- Electrostatics2
11 easy · 14 medium · 0 hard
Chemistry
25 Q
- Organic Chemistry: Some Basic Principles and Techniques2
- Isolation of Metals2
- p-Block Elements2
21 easy · 4 medium · 0 hard
Mathematics
25 Q
- Matrices and Determinants2
- Sequences and Series2
- Limits, Continuity and Differentiability2
13 easy · 11 medium · 1 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 MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The readings are taken to two decimal places of a millimetre (0.01 mm), so neither the mean nor its uncertainty can be stated more finely than 0.01 mm.
Mean $5.5375$ mm rounds to $5.54$ mm; the deviation $0.07395$ mm rounds to $0.07$ mm.
So the diameter is recorded as $(5.54 \pm 0.07)$ mm.
Answer: (C). | ||||
| 2 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Let the rain fall vertically with speed $u$. Relative to a car moving with speed $w$, the rain has a horizontal component $w$ and vertical component $u$, so the angle $\theta$ with the horizontal satisfies $\tan\theta = u/w$.
At speed $v$: $\tan 60^\circ = u/v$, so $u = \sqrt{3}\,v$.
At speed $(1+\beta)v$: $\tan 45^\circ = \dfrac{u}{(1+\beta)v} = 1$, so $1+\beta = \sqrt{3}$.
$\beta = \sqrt{3}-1 \approx 0.73$.
Answer: (D). | ||||
| 3 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion With mass $m$: $a_x = \dfrac{k v_y}{m}$, $a_y = \dfrac{k v_x}{m}$.
$\vec{v}\times\vec{a} = (v_x a_y - v_y a_x)\hat{k} = \dfrac{k}{m}(v_x^2 - v_y^2)\hat{k}$.
$\dfrac{d}{dt}(v_x^2 - v_y^2) = 2v_x a_x - 2v_y a_y = \dfrac{2k}{m}(v_x v_y - v_y v_x) = 0$, so $\vec{v}\times\vec{a}$ is constant — (D) is true.
$\vec{v}\cdot\vec{a} = \dfrac{2k}{m}v_x v_y$, which changes as the velocity changes, so (B) is false; since $\vec{F}\cdot\vec{v} = 2k v_x v_y \neq 0$ in general, the force does work and the kinetic energy changes, so (C) is false; a magnetic force is always perpendicular to $\vec{v}$ and does no work, so (A) is false.
Answer: (D). | ||||
| 4 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Momentum conservation with $m_1 = 2m_2$: $2m_2(\sqrt{3}\hat{i}+\hat{j}) = 2m_2(\hat{i}+\sqrt{3}\hat{j}) + m_2\vec{V}_2$.
$\vec{V}_2 = 2(\sqrt{3}-1)\hat{i} + 2(1-\sqrt{3})\hat{j} = 2(\sqrt{3}-1)(\hat{i}-\hat{j})$, which points at $-45^\circ$ to the $x$-axis.
$\vec{V}_1 = \hat{i}+\sqrt{3}\hat{j}$ points at $+60^\circ$.
Angle between them $= 60^\circ + 45^\circ = 105^\circ$.
Answer: (C). | ||||
| 5 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $M = \int_0^L \lambda_0\left(1+\frac{x}{L}\right)dx = \lambda_0\left(L + \frac{L}{2}\right) = \frac{3}{2}\lambda_0 L$.
$I_A = \int_0^L \lambda_0\left(1+\frac{x}{L}\right)x^2\,dx = \lambda_0\left(\frac{L^3}{3}+\frac{L^3}{4}\right) = \frac{7}{12}\lambda_0 L^3$.
$I_A = \frac{7}{12}\lambda_0L^3 = \frac{7}{12}\cdot\frac{2M}{3L}\cdot L^3 = \frac{7}{18}ML^2$.
Answer: (B). | ||||
| 6 | GravitationGravitational potential energy | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation The body need only just pass the neutral point, where the two pulls balance; beyond it the smaller planet pulls it in.
At distance $x$ from the smaller planet's centre: $\dfrac{GM}{x^2} = \dfrac{16GM}{(10a-x)^2} \Rightarrow 10a - x = 4x \Rightarrow x = 2a$ (8a from the larger planet).
Potential energy at the larger planet's surface (2a from its centre, 8a from the smaller): $U_1 = -\dfrac{16GMm}{2a} - \dfrac{GMm}{8a} = -\dfrac{65}{8}\dfrac{GMm}{a}$.
At the neutral point: $U_2 = -\dfrac{16GMm}{8a} - \dfrac{GMm}{2a} = -\dfrac{20}{8}\dfrac{GMm}{a}$.
$\frac{1}{2}mv^2 = U_2 - U_1 = \dfrac{45}{8}\dfrac{GMm}{a} \Rightarrow v^2 = \dfrac{45}{4}\dfrac{GM}{a} \Rightarrow v = \dfrac{3}{2}\sqrt{\dfrac{5GM}{a}}$.
Answer: (B). | ||||
| 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 Bernoulli's equation for a horizontal pipe: $P + \frac{1}{2}\rho v^2 = \frac{P}{2} + \frac{1}{2}\rho V^2$.
$\frac{1}{2}\rho V^2 = \frac{P}{2} + \frac{1}{2}\rho v^2 \Rightarrow V^2 = \frac{P}{\rho} + v^2$.
$V = \sqrt{\dfrac{P}{\rho}+v^2}$.
Answer: (A). | ||||
| 8 | Properties of Bulk MatterHeat transfer | 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 heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter In steady state the same heat current flows through each rod; with equal area and length, $K \times (\text{temperature drop})$ is the same for all three.
Drops: rod 1: $100-70 = 30$; rod 2: $70-20 = 50$; rod 3: $20-0 = 20$ (in $^\circ$C).
$30K_1 = 50K_2 = 20K_3$.
$K_1 : K_3 = 20 : 30 = 2 : 3$ and $K_2 : K_3 = 20 : 50 = 2 : 5$ (also $K_1 : K_2 = 5 : 3$, so neither ordering in (A) or (B) holds).
Answer: (D). | ||||
| 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 time between collisions $\tau = \dfrac{\lambda}{\bar{v}}$.
Mean free path $\lambda = \dfrac{k_BT}{\sqrt{2}\pi d^2 P}$, so at the given pressure P, $\lambda \propto T$.
Mean speed $\bar{v} \propto \sqrt{T}$.
$\tau \propto \dfrac{T}{\sqrt{T}} = \sqrt{T}$.
Answer: (C). | ||||
| 10 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $y = y_0\sin^2\omega t = \dfrac{y_0}{2} - \dfrac{y_0}{2}\cos 2\omega t$: SHM about $y = \dfrac{y_0}{2}$ with angular frequency $2\omega$.
The mean position is the equilibrium extension: $\dfrac{mg}{k} = \dfrac{y_0}{2} \Rightarrow \dfrac{k}{m} = \dfrac{2g}{y_0}$.
$2\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{2g}{y_0}} \Rightarrow \omega = \dfrac{1}{2}\sqrt{\dfrac{2g}{y_0}} = \sqrt{\dfrac{g}{2y_0}}$.
Answer: (C). | ||||
| 11 | 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 By Gauss's law the field of a uniformly charged shell is zero inside ($r < R$) and equals that of a point charge Q at the centre outside ($r > R$).
So $F = 0$ for $r < R$ (ruling out (A), (B) and (D)) and $F = \dfrac{1}{4\pi\epsilon_0}\dfrac{Qq}{r^2}$ for $r > R$.
Answer: (C). | ||||
| 12 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Dipole 1 sits on the axis of dipole 2, where the field is $E = \dfrac{1}{4\pi\epsilon_0}\dfrac{2p_2}{a^3}$ along $\vec{p}_2$, i.e. along $-\hat{i}$.
Interaction energy $U = -\vec{p}_1\cdot\vec{E} = +\dfrac{1}{4\pi\epsilon_0}\dfrac{2p^2}{a^3}$ (positive: anti-parallel coaxial dipoles repel).
Momentum conservation gives equal speeds $v$; energy: $2\cdot\frac{1}{2}mv^2 = \dfrac{2p^2}{4\pi\epsilon_0 a^3}$.
$v^2 = \dfrac{p^2}{2\pi\epsilon_0 m a^3} \Rightarrow v = \dfrac{p}{a}\sqrt{\dfrac{1}{2\pi\epsilon_0 ma}}$.
Answer: (A). | ||||
| 13 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Call the top junction T and the bottom junction B; take $V_B = 0$, $V_T = x$. Three branches join T and B:
left: B → 2 Ω → 20 V cell (negative to positive) → 5 Ω → T, current into T $= \dfrac{20 - x}{7}$;
middle: 10 Ω, current into T $= \dfrac{0 - x}{10}$;
right: B → 4 Ω → 10 V cell (negative to positive, positive plate at the top) → T, current into T $= \dfrac{10 - x}{4}$.
Kirchhoff's junction rule at T: $\dfrac{20-x}{7} - \dfrac{x}{10} + \dfrac{10-x}{4} = 0$. Multiplying by 140: $400 - 20x - 14x + 350 - 35x = 0 \Rightarrow x = \dfrac{750}{69} \approx 10.87$ V.
Current in the right branch $= \dfrac{10 - 10.87}{4} \approx -0.217$ A: about 0.21 A flows from T down into the cell's positive terminal and out of its negative terminal, i.e. from positive to negative terminal inside the 10 V cell.
Answer: (A). | ||||
| 14 | 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 Radius $r = \dfrac{mv}{qB}$; current $I = \dfrac{qv}{2\pi r}$.
$|\mu| = I\pi r^2 = \dfrac{qvr}{2} = \dfrac{mv^2}{2B}$.
Whatever the sign of q, the circulating charge produces a moment opposite to $\vec{B}$ (the field of the loop opposes the applied field), so $\vec{\mu} = -\dfrac{mv^2}{2B}\hat{B} = -\dfrac{mv^2\vec{B}}{2B^2}$.
Answer: (B). | ||||
| 15 | 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 The loop's magnetic moment has magnitude $M = I(2a)^2 = 4Ia^2$, along the $y$-axis (normal to the $xz$ plane).
Since $b \gg a$, the wire's field over the loop is that at the origin: $B = \dfrac{\mu_0 I}{2\pi b}$, along the $x$-axis (perpendicular to the line from the wire to the origin).
$\vec{M}\perp\vec{B}$, so the torque (about the $z$-axis) is $\tau = MB = 4Ia^2\cdot\dfrac{\mu_0I}{2\pi b} = \dfrac{2\mu_0I^2a^2}{\pi b}$.
Answer: (A). | ||||
| 16 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves The phase $(kx + \omega t)$ means the wave travels along $-\hat{i}$; $\omega/k = 1.5\times10^{11}/0.5\times10^3 = 3\times10^8$ m/s, as it must. E has the same phase.
$E_0 = cB_0 = 3\times10^8\times1.2\times10^{-7} = 36$ V/m.
$\vec{E}$ is perpendicular to both $\hat{k}$ and the direction of travel, so it lies along $\pm\hat{j}$, and $\vec{E}\times\vec{B}$ must point along $-\hat{i}$: $(-\hat{j})\times\hat{k} = -\hat{i}$, so $\vec{E}$ is along $-\hat{j}$.
$\vec{E} = -36\sin(0.5\times10^3x + 1.5\times10^{11}t)\hat{j}$ V/m.
Answer: (B). | ||||
| 17 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens maker's formula: double convex, $P = (\mu-1)\left(\frac{1}{R}+\frac{1}{R}\right) = \dfrac{2(\mu-1)}{R}$.
Plano-convex with curved radius $R'$: $P' = \dfrac{\mu-1}{R'} = 1.5P = \dfrac{3(\mu-1)}{R}$.
$R' = \dfrac{R}{3}$.
Answer: (A). | ||||
| 18 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (A) A free proton (1.0072 u) cannot decay into a heavier neutron (1.0087 u) plus a positron: energy is not conserved.
(B) A pair annihilating into a single photon cannot conserve momentum: in the centre-of-mass frame the total momentum is zero, but one photon always carries momentum.
(C) $m_n + m_p = 2.0159$ u $> m_d = 2.0141$ u; the 0.0018 u (about 2.2 MeV) is carried off by the photon, with the deuteron recoiling to balance momentum. Allowed.
(D) Two neutrons carry no charge, but a deuterium atom holds a proton and an electron; the process could only happen through a weak decay that also emits an antineutrino, so as written it does not conserve charge and lepton number.
Answer: (C). | ||||
| 19 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $v_{rms} = \sqrt{\dfrac{3k_BT}{m}} = \sqrt{\dfrac{3\times1.38\times10^{-23}\times400}{4.64\times10^{-26}}} = \sqrt{3.57\times10^5} \approx 597$ m/s.
$p = mv = 4.64\times10^{-26}\times597 \approx 2.77\times10^{-23}$ kg m/s.
$\lambda = \dfrac{h}{p} = \dfrac{6.63\times10^{-34}}{2.77\times10^{-23}} \approx 2.4\times10^{-11}$ m $= 0.24$ Å.
Answer: (D). | ||||
| 20 | Experimental SkillsElectricity experiments | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on electricity experiments, from Experimental Skills. NCERT: No matching NCERT chapter The ammeter must carry the same current as the resistor, so it goes in series; the voltmeter must read the potential difference across the resistor, so it goes in parallel with it.
Answer: (B). | ||||
| 21 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For a solid hemisphere the centre of mass lies on the symmetry axis at $\dfrac{3R}{8}$ from the centre of the flat face.
$x = \dfrac{3\times8}{8} = 3$ cm.
Answer: 3. | ||||
| 22 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the figure: A $(V_0, P_0)$ → B $(V_0, 3P_0)$ → C $(2V_0, 3P_0)$ → D $(2V_0, P_0)$ → A, clockwise.
Net work = enclosed area $= (3P_0 - P_0)(2V_0 - V_0) = 2P_0V_0$.
Heat is absorbed on AB and BC (temperature rises on both). With $nC_V = \frac{3}{2}nR$, $nC_P = \frac{5}{2}nR$ and $nR\,\Delta T = \Delta(PV)$:
AB (isochoric): $Q = \frac{3}{2}(3P_0V_0 - P_0V_0) = 3P_0V_0$.
BC (isobaric): $Q = \frac{5}{2}(6P_0V_0 - 3P_0V_0) = 7.5P_0V_0$.
Heat in $= 10.5P_0V_0$; efficiency $= \dfrac{2}{10.5}\times100 = 19.05\%$.
Answer: 19.05. | ||||
| 23 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $P = \dfrac{V^2}{Z}\cos\phi$ with $\cos\phi = \dfrac{R}{Z}$, so $P = \dfrac{V^2\cos^2\phi}{R}$: $R = \dfrac{250^2\times0.64}{400} = 100\ \Omega$.
$Z = \dfrac{R}{\cos\phi} = 125\ \Omega$, so $X_L = \sqrt{125^2-100^2} = 75\ \Omega$.
Unity power factor needs $X_C = X_L$: $\dfrac{1}{2\pi\times50\times C} = 75 \Rightarrow C = \dfrac{1}{7500\pi}$ F $= \dfrac{10^6}{7500\pi}\ \mu$F $= \dfrac{400}{3\pi}\ \mu$F.
$n = 400$.
Answer: 400. | ||||
| 24 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $I = I_{max}\cos^2\dfrac{\phi}{2}$, with phase difference $\phi = \dfrac{2\pi}{\lambda}\times$ path difference.
Path difference $\lambda$: $\phi = 2\pi$, $I = I_{max} = K$.
Path difference $\lambda/6$: $\phi = \pi/3$, $I = K\cos^2 30^\circ = \dfrac{3K}{4} = \dfrac{9K}{12}$.
$n = 9$.
Answer: 9. | ||||
| 25 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $\beta_{ac} = \left(\dfrac{\Delta I_C}{\Delta I_B}\right)_{V_{CE}}$, taken between the two characteristics on either side of $I_C = 4.0$ mA at $V_{CE} = 10$ V.
Read from the figure at $V_{CE} = 10$ V: the $I_B = 30\ \mu$A curve is at about 4.3 mA and the $I_B = 20\ \mu$A curve at about 3.1 mA.
$\beta_{ac} = \dfrac{(4.3 - 3.1)\times10^{-3}}{(30-20)\times10^{-6}} = \dfrac{1.2\times10^{-3}}{10\times10^{-6}} = 120$.
Answer: 120. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Lactose (milk sugar) is a disaccharide of $\beta$-D-galactose and D-glucose, joined by a $\beta$-glycosidic link from $\mathrm{C_1}$ of galactose to $\mathrm{C_4}$ of glucose.
(A) It is a disaccharide and carries 8 –OH groups (4 on the galactose unit, 4 on the glucose unit): true as intended (the formula of lactose is $\mathrm{C_{12}H_{22}O_{11}}$; see note).
(B) Hydrolysis gives D(+)-glucose and D(+)-galactose: true.
(C) The glucose unit keeps a free hemiacetal at $\mathrm{C_1}$, so lactose is reducing and gives Fehling's test: true.
(D) The link is $\beta$, not $\alpha$: this statement is not true.
Answer: (D). | ||||
| 27 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Natural rubber is cis-polyisoprene: (a) - (IV).
Neoprene is polychloroprene: (b) - (III).
Buna-N is a copolymer of 1,3-butadiene and acrylonitrile: (c) - (II).
Buna-S is a copolymer of 1,3-butadiene and styrene: (d) - (I).
Answer: (D). | ||||
| 28 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Gabriel synthesis: potassium phthalimide is alkylated by an alkyl halide ($\mathrm{S_N2}$) and the N-alkylphthalimide is hydrolysed to a primary amine.
(A) is an amide, not an amine.
(B) Benzylamine: benzyl chloride is an excellent $\mathrm{S_N2}$ substrate, so $\mathrm{C_6H_5CH_2NH_2}$ forms in good yield.
(C) is a secondary amine; Gabriel synthesis gives only primary amines.
(D) Aniline would need substitution at an aryl halide, which does not undergo $\mathrm{S_N2}$ with phthalimide anion.
Answer: (B). | ||||
| 29 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Benzene + CO, HCl, anhydrous $\mathrm{AlCl_3}$ (Gattermann–Koch formylation) gives benzaldehyde: (I) - (R).
Benzonitrile + $\mathrm{SnCl_2}$/HCl, then hydrolysis (Stephen reduction) gives benzaldehyde: (II) - (P).
Benzoyl chloride + $\mathrm{H_2}$ over Pd–$\mathrm{BaSO_4}$ poisoned with S/quinoline (Rosenmund reduction) gives benzaldehyde: (III) - (Q).
Answer: (A). | ||||
| 30 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (a) But-1-ene with HBr in the presence of benzoyl peroxide adds anti-Markovnikov (peroxide effect): A = 1-bromobutane.
(b) 2-Methylpropene with HBr adds by Markovnikov's rule: B = 2-bromo-2-methylpropane (tert-butyl bromide).
(c) But-2-ene with HBr: C = 2-bromobutane.
All three are $\mathrm{C_4H_9Br}$; boiling point falls with branching (smaller surface area, weaker dispersion forces): tertiary < secondary < primary.
So B (about 73 °C) < C (about 91 °C) < A (about 101 °C).
Answer: (B). | ||||
| 31 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The principal group is –CHO, so the parent is benzaldehyde and the ring carbon bearing CHO is C1.
Numbering towards the nitro group gives substituent locants {2 (NO$_2$), 4 (CH$_2$OH), 5 (NH$_2$)}; numbering the other way gives {3, 4, 6}. The first set is lower.
Prefixes in alphabetical order: amino, hydroxymethyl, nitro: 5-amino-4-hydroxymethyl-2-nitrobenzaldehyde.
(C) has the right locants but not in alphabetical order; (D) wrongly makes nitro the parent.
Answer: (B). | ||||
| 32 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Lucas test (classifying alcohols): conc. HCl + anhydrous $\mathrm{ZnCl_2}$: (i)-(d).
Dumas method (nitrogen as $\mathrm{N_2}$): heating with CuO in an atmosphere of $\mathrm{CO_2}$: (ii)-(c).
Kjeldahl's method: digestion with conc. $\mathrm{H_2SO_4}$ to ammonium sulphate: (iii)-(e).
Hinsberg test (classifying amines): benzenesulphonyl chloride with aq. KOH: (iv)-(a).
Answer: (D). | ||||
| 33 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Distillation refines low-boiling metals such as zinc and mercury: the metal is vaporised and condensed, leaving impurities behind.
Nickel is refined by the Mond process, tin by liquation, and gallium by zone refining.
Answer: (A). | ||||
| 34 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter High-purity (> 99.95%) dihydrogen is made by electrolysing warm aqueous barium hydroxide between nickel electrodes.
Electrolysis of acidified water and of brine, and Zn with dilute HCl, give dihydrogen of ordinary purity (brine electrolysis gives it as a by-product of chlorine and NaOH manufacture).
Answer: (D). | ||||
| 35 | 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 Slaked lime $\mathrm{Ca(OH)_2}$ is used for white wash: (I)-(B).
NaCl is the raw material of the Solvay process for washing soda: (II)-(D).
Plaster of Paris $\mathrm{CaSO_4\cdot\frac{1}{2}H_2O}$ is used for casts of statues: (III)-(A).
$\mathrm{CaCO_3}$ is used as an antacid: (IV)-(C).
Answer: (C). | ||||
| 36 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter At 250 K: $\mathrm{2NO + N_2O_4 \rightarrow 2N_2O_3}$ (dinitrogen trioxide, a blue solid at low temperature).
Answer: (B). | ||||
| 37 | 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 X = $\mathrm{Na_2SO_3}$: $\mathrm{Na_2SO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + SO_2}$, so Y = $\mathrm{SO_2}$.
$\mathrm{SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O}$ regenerates X.
$\mathrm{Na_2SO_3 + SO_2 + H_2O \rightarrow 2NaHSO_3}$, so Z = $\mathrm{NaHSO_3}$.
Answer: (B). | ||||
| 38 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Mischmetal is about 95% lanthanoid metals (mainly cerium and lanthanum) with about 5% iron and traces of S, C, Ca and Al; it is used in lighter flints.
Answer: (A). | ||||
| 39 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds When $\Delta_o < P$ (weak field), the fourth electron enters $e_g$ rather than pairing: $t_{2g}^3e_g^1$ (high spin).
When $\Delta_o > P$ (strong field) it pairs in $t_{2g}$: $t_{2g}^4e_g^0$. So (A) and (C) pair the wrong configuration with the wrong condition, and (D) fills $e_g$ before $t_{2g}$.
Answer: (B). | ||||
| 40 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Let the fraction of $^{35}$Cl be $x$: $35x + 37(1-x) = 35.5 \Rightarrow 2x = 1.5 \Rightarrow x = 0.75$.
Ratio $^{35}$Cl : $^{37}$Cl $= 0.75 : 0.25 = 3 : 1$.
Answer: (C). | ||||
| 41 | Solid StateClose packed structures | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on close packed structures, from Solid State. NCERT: No matching NCERT chapter For $N$ oxide ions in ccp there are $N$ octahedral and $2N$ tetrahedral voids. Take $N = 4$ (one fcc cell).
$\mathrm{M_1}$: $50\%$ of 4 = 2; $\mathrm{M_2}$: $12.5\%$ of 8 = 1. Formula $\mathrm{(M_1)_2M_2O_4}$.
Charge balance: $2x + y = 8$. Only (A) fits: $2(+2) + 4 = 8$. (B) gives 10, (C) gives 5, (D) gives 7.
Answer: (A). | ||||
| 42 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter The oxide is stable (its formation is spontaneous) while $\Delta G < 0$. On the $\Delta G$ vs T (Ellingham) line, $\Delta G$ rises with temperature (gas is consumed, $\Delta S < 0$) and crosses zero at some temperature; above it the oxide decomposes.
So the required temperature is where $\Delta G$ changes from negative to positive. A change of slope (from a phase change of M) does not mark the stability limit.
Answer: (C). | ||||
| 43 | SolutionsRaoult's law | Class 12: Solutions | Single correctEasy | |
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Relative lowering of vapour pressure $= $ mole fraction of solute $= \dfrac{n_{solute}}{n_{solute}+n_{water}}$, with $n_{water} = 10$ mol.
$n_A = 0.1$, $n_B = 0.05$, $n_C = 0.001$ mol: the more moles, the larger the lowering.
So A > B > C.
Answer: (A). | ||||
| 44 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Reversing the reaction inverts K; halving the coefficients takes the square root.
$K_c' = \left(\dfrac{1}{64}\right)^{1/2} = \dfrac{1}{8}$.
Answer: (C). | ||||
| 45 | ElectrochemistryNernst equation | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Concentration cell: anode $\mathrm{Cu \rightarrow Cu^{2+}(C_1) + 2e^-}$, cathode $\mathrm{Cu^{2+}(C_2) + 2e^- \rightarrow Cu}$; $E^\circ_{cell} = 0$.
$E_{cell} = \dfrac{0.059}{2}\log\dfrac{C_2}{C_1}$, and $\Delta G = -nFE_{cell} < 0$ requires $E_{cell} > 0$, i.e. $C_2 > C_1$.
Only $C_2 = \sqrt{2}\,C_1$ satisfies this ((A) gives zero; (B) and (D) give $C_2 < C_1$).
Answer: (C). | ||||
| 46 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | NumericalEasy | |
A easy numerical question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties IUPAC roots: nil = 0, un = 1. Un-nil-un-ium = 1, 0, 1.
Atomic number 101 (mendelevium).
Answer: 101. | ||||
| 47 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{AB_2 \rightleftharpoons A^{2+} + 2B^-}$: $K_{sp} = s(2s)^2 = 4s^3$.
$s^3 = \dfrac{3.20\times10^{-11}}{4} = 8\times10^{-12} \Rightarrow s = 2\times10^{-4}$ mol $\mathrm{L^{-1}}$.
Answer: 2. | ||||
| 48 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln\dfrac{k_{313}}{k_{303}} = \dfrac{E_a}{R}\left(\dfrac{1}{303} - \dfrac{1}{313}\right) = \dfrac{E_a}{R}\cdot\dfrac{10}{303\times313}$.
$E_a = \dfrac{R\ln 3.555\times303\times313}{10} = \dfrac{8.314\times1.268\times94839}{10} = 99981$ J $\mathrm{mol^{-1}}$.
$E_a \approx 99.98$ kJ $\mathrm{mol^{-1}}$ (about 100).
Answer: 99.98. | ||||
| 49 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter $\log\dfrac{x}{m} = \log k + \dfrac{1}{n}\log p$: intercept $\log k = 0.4771 \Rightarrow k = 3$; slope $\dfrac{1}{n} = 2$.
$\dfrac{x}{m} = 3\times(0.04)^2 = 3\times0.0016 = 0.0048$ g $= 48\times10^{-4}$ g.
Answer: 48. | ||||
| 50 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | NumericalMedium | |
A medium numerical question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Phenol + $\mathrm{CHCl_3}$ + aq. NaOH is the Reimer–Tiemann reaction; after work-up the major product P is salicylaldehyde (2-hydroxybenzaldehyde), $\mathrm{C_7H_6O_2}$.
Molar mass $= 84 + 6 + 32 = 122$ g $\mathrm{mol^{-1}}$.
%C $= \dfrac{84}{122}\times100 = 68.85\% \approx 69$.
Answer: 69. | ||||
| 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 $f(f(x)) = \dfrac{a - \frac{a-x}{a+x}}{a + \frac{a-x}{a+x}} = \dfrac{a^2 + ax - a + x}{a^2 + ax + a - x} = \dfrac{(a^2-a) + (a+1)x}{(a^2+a) + (a-1)x}$.
For this to equal $x$ for all admissible $x$: the coefficient of $x^2$ in $(a^2+a)x + (a-1)x^2 = (a^2-a) + (a+1)x$ forces $a = 1$; then both sides are $2x$ and $0$ constant terms match.
With $a = 1$: $f\left(-\frac{1}{2}\right) = \dfrac{1+\frac{1}{2}}{1-\frac{1}{2}} = 3$.
Answer: (A). | ||||
| 52 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations The equation is $4x^2 + 2x - 1 = 0$, so $\alpha + \beta = -\frac{1}{2}$ and $4\alpha^2 = 1 - 2\alpha$, i.e. $2\alpha^2 = \frac{1}{2} - \alpha$.
$-2\alpha(\alpha+1) = -2\alpha^2 - 2\alpha = -\frac{1}{2} + \alpha - 2\alpha = -\frac{1}{2} - \alpha = \beta$.
Answer: (B). | ||||
| 53 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z^2 = x^2 - y^2 + 2ixy$ and $i|z|^2 = i(x^2+y^2)$.
Real parts: $x^2 - y^2 = 0$. Imaginary parts: $2xy = x^2 + y^2 \Rightarrow (x-y)^2 = 0 \Rightarrow y = x$.
So $z$ lies on the line $y = x$ (with $z \neq 0$).
Answer: (C). | ||||
| 54 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A$ is a rotation matrix, so $A^4 = \begin{bmatrix}\cos4\theta & \sin4\theta\\ -\sin4\theta & \cos4\theta\end{bmatrix}$.
$B = \begin{bmatrix}\cos\theta+\cos4\theta & \sin\theta+\sin4\theta\\ -(\sin\theta+\sin4\theta) & \cos\theta+\cos4\theta\end{bmatrix}$.
$\det B = (\cos\theta+\cos4\theta)^2 + (\sin\theta+\sin4\theta)^2 = 2 + 2\cos3\theta$.
With $\theta = \pi/5$: $2 + 2\cos\frac{3\pi}{5} = 2 - 2\cos\frac{2\pi}{5} = 2 - 2(0.309) \approx 1.38$, which lies in (1, 2).
Answer: (D). | ||||
| 55 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = {}^{10}C_r(\sqrt{x})^{10-r}\left(-\dfrac{k}{x^2}\right)^r = {}^{10}C_r(-k)^r x^{\frac{10-r}{2} - 2r}$.
Constant term: $\frac{10-r}{2} = 2r \Rightarrow r = 2$.
${}^{10}C_2 k^2 = 45k^2 = 405 \Rightarrow k^2 = 9 \Rightarrow |k| = 3$.
Answer: (B). | ||||
| 56 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series For the $a$-sequence: $60d = a_{100} - a_{40} = -240 \Rightarrow d = -4$; $a_{70} = a_{40} + 30d = -159 - 120 = -279$.
The $b$-sequence has common difference $-4 + 2 = -2$: $b_{100} = b_1 + 99(-2) = -279 \Rightarrow b_1 = -279 + 198 = -81$.
Answer: (B). | ||||
| 57 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Rolle's theorem on $f$ over [0, 1] ($f(0) = f(1)$): $f'(c) = 0$ for some $c \in (0, 1)$.
Rolle's theorem on $f'$ over $[0, c]$ ($f'(0) = f'(c) = 0$): $f''(x) = 0$ for some $x \in (0, c) \subset (0, 1)$. So (C) always holds and (A) never does.
(B) and (D) can fail: $f(x) = x^2(x-1) = x^3 - x^2$ satisfies the conditions, with $f''(x) = 6x - 2$, so $f''(0) = -2 \neq 0$ and $f''$ is not identically zero.
Answer: (C). | ||||
| 58 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $x^2 \ge x$ for $x \le 0$ or $x \ge 1$; $x \ge x^2$ for $0 \le x \le 1$. So $f(x) = x^2$ for $x \le 0$, $x$ for $0 \le x \le 1$, $x^2$ for $x \ge 1$.
At $x = 0$: left derivative $2x = 0$, right derivative $1$ — not differentiable.
At $x = 1$: left derivative $1$, right derivative $2x = 2$ — not differentiable.
Elsewhere $f$ is a polynomial, so $S = \{0, 1\}$.
Answer: (B). | ||||
| 59 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Slope of the segment: $\dfrac{e - 0}{e - 1} = \dfrac{e}{e-1}$.
$f'(x) = \log_e x + 1$, so $\log_e c + 1 = \dfrac{e}{e-1} \Rightarrow \log_e c = \dfrac{1}{e-1} \Rightarrow c = e^{\frac{1}{e-1}}$.
Answer: (A). | ||||
| 60 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctHard | |
A hard single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f(x) = \sin^2x(\lambda + \sin x)$, so $f'(x) = 2\sin x\cos x(\lambda + \sin x) + \sin^2x\cos x = \sin x\cos x(2\lambda + 3\sin x)$.
On $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\cos x > 0$, so the critical points are $\sin x = 0$ and $\sin x = -\frac{2\lambda}{3}$.
Two distinct sign changes of $f'$ (one maximum, one minimum) need $-\frac{2\lambda}{3} \in (-1, 1)$ and $\neq 0$: $\lambda \in \left(-\frac{3}{2}, \frac{3}{2}\right)$, $\lambda \neq 0$.
If $\lambda = 0$, $f'(x) = 3\sin^2x\cos x \ge 0$: no extremum at all. If $|\lambda| \ge \frac{3}{2}$, only $x = 0$ is critical and there is just one extremum.
Answer: (C). | ||||
| 61 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\dfrac{d}{dx}\left(e^x x^x\right) = e^x x^x + e^x x^x(1 + \log_e x) = e^x x^x(2 + \log_e x)$.
So the integral $= \left[e^x x^x\right]_1^2 = 4e^2 - e = e(4e-1)$.
Answer: (C). | ||||
| 62 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The curves meet where $x^2 - 1 = 1 - x^2$, i.e. $x = \pm1$; between them $1 - x^2$ is above.
Area $= \int_{-1}^{1}\left[(1-x^2) - (x^2-1)\right]dx = \int_{-1}^{1}(2 - 2x^2)\,dx = 4 - \frac{4}{3} = \frac{8}{3}$.
Answer: (C). | ||||
| 63 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $y\sin x = \dfrac{2}{\pi}x - 1$. Differentiating: $y'\sin x + y\cos x = \dfrac{2}{\pi}$.
Dividing by $\sin x$: $y' + y\cot x = \dfrac{2}{\pi}\mathrm{cosec}\,x$, so $p(x) = \cot x$.
Answer: (D). | ||||
| 64 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines L: $\dfrac{x}{3} + y = 1$, i.e. $x + 3y - 3 = 0$.
Image of $(x_1, y_1)$: $\dfrac{x - x_1}{1} = \dfrac{y - y_1}{3} = \dfrac{-2(x_1 + 3y_1 - 3)}{1^2 + 3^2} = \dfrac{-2(-1 - 12 - 3)}{10} = \dfrac{16}{5}$.
$x = -1 + \frac{16}{5} = \frac{11}{5}$, $y = -4 + \frac{48}{5} = \frac{28}{5}$.
Answer: (D). | ||||
| 65 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections The centre lies on the normal to $y = x^2$ at (2, 4). The tangent slope there is 4, so the normal has direction $(-4, 1)$: centre $(2 - 4t, 4 + t)$.
Equal distances to (2, 4) and (0, 1): $16t^2 + t^2 = (2-4t)^2 + (3+t)^2 \Rightarrow 17t^2 = 13 - 10t + 17t^2 \Rightarrow t = \frac{13}{10}$.
Centre $= \left(2 - \frac{26}{5}, 4 + \frac{13}{10}\right) = \left(-\frac{16}{5}, \frac{53}{10}\right)$.
Answer: (C). | ||||
| 66 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$; end of latus rectum $\left(ae, \dfrac{b^2}{a}\right)$.
Normal there: $\dfrac{a^2x}{ae} - \dfrac{b^2y}{b^2/a} = a^2 - b^2$, i.e. $\dfrac{ax}{e} - ay = a^2e^2$.
Through $(0, -b)$: $ab = a^2e^2 \Rightarrow b = ae^2$. (Through $(0, b)$ would need $-ab = a^2e^2$, impossible.)
$b^2 = a^2(1 - e^2) = a^2e^4 \Rightarrow e^4 + e^2 - 1 = 0$.
Answer: (D). | ||||
| 67 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry With A$(a,0,0)$, B$(0,b,0)$, C$(0,0,c)$, the centroid is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) = (1, 1, 2)$, so $a = 3$, $b = 3$, $c = 6$.
Plane: $\dfrac{x}{3} + \dfrac{y}{3} + \dfrac{z}{6} = 1$, i.e. $2x + 2y + z = 6$, with normal $(2, 2, 1)$.
Line: $\dfrac{x-1}{2} = \dfrac{y-1}{2} = \dfrac{z-2}{1}$.
Answer: (D). | ||||
| 68 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability P(A∩B) = P(A) + P(B) − P(A∪B) = 0.6 + 0.4 − 0.8 = 0.2.
$\alpha$ = 0.6 + 0.4 + 0.5 − 0.2 − 0.3 − $\beta$ + 0.2 = 1.2 − $\beta$.
$0.85 \le 1.2 - \beta \le 0.95 \Rightarrow 0.25 \le \beta \le 0.35$.
Answer: (C). | ||||
| 69 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let the height be $h$; with 45° elevation the summit is $h$ km away horizontally.
After climbing 1 km at 30°, the climber is at horizontal distance $\frac{\sqrt{3}}{2}$ and height $\frac{1}{2}$.
Elevation 60°: $\dfrac{h - \frac{1}{2}}{h - \frac{\sqrt{3}}{2}} = \sqrt{3} \Rightarrow h - \frac{1}{2} = \sqrt{3}h - \frac{3}{2} \Rightarrow h(\sqrt{3}-1) = 1$.
$h = \dfrac{1}{\sqrt{3}-1}$ km.
Answer: (B). | ||||
| 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 The contrapositive of 'if p then q' is 'if not q then not p'.
Here p: $n^3 - 1$ is even; q: n is odd. Not q: n is even (for an integer); not p: $n^3 - 1$ is odd.
Contrapositive: if n is even, then $n^3 - 1$ is odd. (B) is the inverse and (C) the converse.
Answer: (A). | ||||
| 71 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A homogeneous system has non-zero solutions exactly when its coefficient determinant is zero.
$\Delta = \begin{vmatrix}\lambda-1 & 3\lambda+1 & 2\lambda\\ \lambda-1 & 4\lambda-2 & \lambda+3\\ 2 & 3\lambda+1 & 3(\lambda-1)\end{vmatrix}$. Subtract row 1 from row 2: $R_2 \to (0,\ \lambda-3,\ 3-\lambda)$.
$\Delta = (\lambda-3)\begin{vmatrix}\lambda-1 & 3\lambda+1 & 2\lambda\\ 0 & 1 & -1\\ 2 & 3\lambda+1 & 3\lambda-3\end{vmatrix}$. Add column 3 to column 2: column 2 becomes $(5\lambda+1,\ 0,\ 6\lambda-2)$.
Expanding along row 2 (only the $-1$ in column 3 remains): $\Delta = (\lambda-3)\cdot(-1)\cdot(-1)^{2+3}\begin{vmatrix}\lambda-1 & 5\lambda+1\\ 2 & 6\lambda-2\end{vmatrix} = (\lambda-3)\left[(\lambda-1)(6\lambda-2) - 2(5\lambda+1)\right]$.
$(\lambda-1)(6\lambda-2) - 10\lambda - 2 = 6\lambda^2 - 18\lambda = 6\lambda(\lambda-3)$, so $\Delta = 6\lambda(\lambda-3)^2$.
$\Delta = 0$ for $\lambda = 0$ or $\lambda = 3$; the sum of the distinct values is $0 + 3 = 3$.
Answer: 3. | ||||
| 72 | 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 LETTER has L, E, T, T, E, R: two E's and two T's. All arrangements: $\dfrac{6!}{2!\,2!} = 180$.
Arrangements with the two E's together (treat EE as one block): $\dfrac{5!}{2!} = 60$.
Vowels never together: $180 - 60 = 120$.
Answer: 120. | ||||
| 73 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $f(i) = f(1)^i = 3^i$ for positive integers $i$.
$\sum_{i=1}^n 3^i = \dfrac{3(3^n - 1)}{2} = 363 \Rightarrow 3^n - 1 = 242 \Rightarrow 3^n = 243 \Rightarrow n = 5$.
Answer: 5. | ||||
| 74 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\vec{x}+\vec{y}|^2 = |\vec{x}|^2 \Rightarrow 2\vec{x}\cdot\vec{y} + |\vec{y}|^2 = 0 \Rightarrow 2\vec{x}\cdot\vec{y} = -|\vec{y}|^2$.
$(2\vec{x} + \lambda\vec{y})\cdot\vec{y} = 0 \Rightarrow 2\vec{x}\cdot\vec{y} + \lambda|\vec{y}|^2 = 0 \Rightarrow (\lambda - 1)|\vec{y}|^2 = 0 \Rightarrow \lambda = 1$.
Answer: 1. | ||||
| 75 | StatisticsMean, median and mode | Class 11: Statistics | NumericalMedium | |
A medium numerical question on mean, median and mode, from Statistics. NCERT: Class 11: Statistics The values are $x_0 = 0$ and $x_k = 2^k$ for $k = 1, \ldots, n$, with frequencies ${}^nC_k$; total frequency $2^n$.
$\sum f_kx_k = \sum_{k=1}^n {}^nC_k2^k = (1+2)^n - 1 = 3^n - 1$.
Mean $= \dfrac{3^n - 1}{2^n} = \dfrac{728}{2^n} \Rightarrow 3^n = 729 \Rightarrow n = 6$.
Answer: 6. | ||||
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