JEE (Main) · 2020
JEE (Main) 2020 — 8 January, 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
- Electrostatics3
- Kinematics2
- Motion of System of Particles and Rigid Body2
12 easy · 12 medium · 1 hard
Chemistry
25 Q
- p-Block Elements2
- Coordination Compounds2
- Organic Chemistry: Some Basic Principles and Techniques2
14 easy · 9 medium · 2 hard
Mathematics
25 Q
- Matrices and Determinants2
- Sequences and Series2
- Limits, Continuity and Differentiability2
10 easy · 14 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 $g = \frac{4\pi^2 L}{T^2}$, so $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$.
Length 25.0 cm is read to 0.1 cm: $\frac{\Delta L}{L} = \frac{0.1}{25.0} = 0.004$.
The time for 40 oscillations is 50 s with a 1 s resolution: $\frac{\Delta T}{T} = \frac{1}{50} = 0.02$.
$\frac{\Delta g}{g} = 0.004 + 2(0.02) = 0.044 = 4.40\%$.
Answer: (3). | ||||
| 2 | KinematicsUniform circular motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\vec{v} = \frac{d\vec{r}}{dt} = \omega(-\sin\omega t\,\hat{i} + \cos\omega t\,\hat{j})$, and $\vec{v}\cdot\vec{r} = \omega(-\sin\omega t\cos\omega t + \cos\omega t\sin\omega t) = 0$, so $\vec{v} \perp \vec{r}$.
$\vec{a} = \frac{d\vec{v}}{dt} = -\omega^2(\cos\omega t\,\hat{i} + \sin\omega t\,\hat{j}) = -\omega^2\vec{r}$, which points from the particle towards the origin.
Answer: (3). | ||||
| 3 | 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 In the frame of free fall the two approach each other at the constant relative speed $\sqrt{2gh}$, so they meet after $t_0 = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$.
At that moment the dropped particle has velocity $-gt_0$ (downwards) and the thrown one $\sqrt{2gh} - gt_0$ (upwards). Since $gt_0 = \sqrt{\frac{gh}{2}}$ and $\sqrt{2gh} = 2\sqrt{\frac{gh}{2}}$, the velocities are $-\sqrt{\frac{gh}{2}}$ and $+\sqrt{\frac{gh}{2}}$.
Equal masses sticking together: the combined velocity is $\frac{1}{2}\left(-\sqrt{\frac{gh}{2}} + \sqrt{\frac{gh}{2}}\right) = 0$.
They meet at height $h - \frac{1}{2}gt_0^2 = h - \frac{h}{4} = \frac{3h}{4}$, and the combined mass falls from rest: $\frac{3h}{4} = \frac{1}{2}gt^2$, so $t = \sqrt{\frac{3h}{2g}} = \sqrt{\frac{3}{2}}\sqrt{\frac{h}{g}}$.
Answer: (2). | ||||
| 4 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium 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 From the figure the cavity touches the sphere's surface, so its centre O is at distance $OC = R - 1$ from C, and G, on the cavity's surface on the far side, is at $CG = 1 - (R - 1) = 2 - R$ on the other side of C.
Mass is proportional to volume. Take C as origin and the line CO as the axis: the whole sphere ($R^3$) has its centre of mass at C, so
$0 = (R^3 - 1)(-(2 - R)) + 1\cdot(R - 1)$, i.e. $(R^3 - 1)(2 - R) = R - 1$.
Dividing by $R - 1$ (with $R^3 - 1 = (R - 1)(R^2 + R + 1)$): $(R^2 + R + 1)(2 - R) = 1$.
Answer: (3). | ||||
| 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 For a solid sphere rolling without slipping, $K = \frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{2}{5}mr^2\cdot\frac{v^2}{r^2} = \frac{7}{10}mv^2$.
$K = \frac{7}{10}(0.5)(0.05)^2 = 0.35 \times 0.0025 = 8.75\times10^{-4}$ J.
Answer: (3). | ||||
| 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 wall is 10 m wide; each part is 5 m high, with the lighter liquid $\rho_1$ on top.
Upper part MN: the gauge pressure rises from 0 to $5\rho_1 g$, average $2.5\rho_1 g$; force $F_1 = 2.5\rho_1 g \times (5\times10)$.
Lower part NO: the pressure rises from $5\rho_1 g$ to $5\rho_1 g + 5\rho_2 g = 15\rho_1 g$, average $10\rho_1 g$; force $F_2 = 10\rho_1 g\times(5\times10)$.
$\frac{F_1}{F_2} = \frac{2.5}{10} = \frac{1}{4}$ (the atmosphere acts on both sides of the wall and is left out).
Answer: (4). | ||||
| 7 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics For a reversible engine $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{10}$, so $\frac{T_2}{T_1} = \frac{9}{10}$.
Run backwards as a refrigerator, its coefficient of performance is $\beta = \frac{T_2}{T_1 - T_2} = \frac{1 - \eta}{\eta} = 9$.
Heat taken from the cold reservoir: $Q_2 = \beta W = 9\times10 = 90$ J.
Answer: (3). | ||||
| 8 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Helium (monatomic): $C_V = \frac{3}{2}R$, $C_P = \frac{5}{2}R$. Rigid oxygen (diatomic): $C_V = \frac{5}{2}R$, $C_P = \frac{7}{2}R$.
Mixture: $C_V = \frac{n\cdot\frac{3}{2}R + 2n\cdot\frac{5}{2}R}{3n} = \frac{13}{6}R$; $C_P = \frac{n\cdot\frac{5}{2}R + 2n\cdot\frac{7}{2}R}{3n} = \frac{19}{6}R$.
$\frac{C_P}{C_V} = \frac{19}{13}$.
Answer: (3). | ||||
| 9 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v = \sqrt{\frac{T}{\mu}}$, so $T \propto v^2$. Halving the speed quarters the tension:
$T = \frac{2.06\times10^4}{4} = 5.15\times10^3$ N.
Answer: (2). | ||||
| 10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance For a charged conducting sphere, $E = \frac{kQ}{R^2}$ and $V = \frac{kQ}{R}$ at the surface, so $V = ER$.
$\frac{V_1}{V_2} = \frac{E_1}{E_2}\cdot\frac{R_1}{R_2} = \frac{R_1}{R_2}\cdot\frac{R_1}{R_2} = \left(\frac{R_1}{R_2}\right)^2$.
Answer: (2). | ||||
| 11 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance At distance x from the edge where the gap is d, the gap is $d + \alpha x$. Treat the plate as strips of width dx (area $a\,dx$) in parallel:
$C = \int_0^a \frac{\epsilon_0 a\,dx}{d + \alpha x} = \frac{\epsilon_0 a}{\alpha}\ln\left(1 + \frac{\alpha a}{d}\right)$.
With $\frac{\alpha a}{d}$ small, $\ln(1 + y) \approx y - \frac{y^2}{2}$:
$C \approx \frac{\epsilon_0 a}{\alpha}\left(\frac{\alpha a}{d} - \frac{\alpha^2a^2}{2d^2}\right) = \frac{\epsilon_0 a^2}{d}\left(1 - \frac{\alpha a}{2d}\right)$.
Answer: (2). | ||||
| 12 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctHard | |
A hard single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Put B at the origin, AB along the y-axis (current upwards, from $y = -\infty$ to 0) and BC along the x-axis (current to the right, from 0 to $\infty$). Tangent to both lines, the circle has its centre at $(R, R)$, touching AB at N $(0, R)$ and BC at D $(R, 0)$.
Field of a straight segment: $\frac{\mu_0 I}{4\pi d}(\sin\theta_2 - \sin\theta_1)$, angles measured from the foot of the perpendicular.
AB: $d = R$, foot at N; the segment runs from $-90^\circ$ to $-45^\circ$: $\frac{\mu_0 I}{4\pi R}\left(1 - \frac{1}{\sqrt{2}}\right)$, into the page.
BC: $d = R$, foot at D; the segment runs from $-45^\circ$ to $90^\circ$: $\frac{\mu_0 I}{4\pi R}\left(1 + \frac{1}{\sqrt{2}}\right)$, out of the page.
Net from the straight parts: $\frac{\mu_0 I}{4\pi R}\cdot\frac{2}{\sqrt{2}} = \frac{\mu_0 I}{2\sqrt{2}\pi R}$, out of the page.
The loop D → M → N → D is traversed anticlockwise (up the right side, over the top, down the left): $\frac{\mu_0 I}{2R}$, out of the page.
Total: $\frac{\mu_0 I}{2R} + \frac{\mu_0 I}{2\sqrt{2}\pi R} = \frac{\mu_0 I}{2\pi R}\left(\pi + \frac{1}{\sqrt{2}}\right)$.
Answer: (1). | ||||
| 13 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The force $qE$ is constant, so the acceleration $a = \frac{qE}{m}$ is constant. From rest, $v^2 = 2ax$, so $v = \sqrt{\frac{2qEx}{m}} \propto \sqrt{x}$.
A $\sqrt{x}$ curve starts steeply at the origin and keeps rising with decreasing slope: graph (3).
Answer: (3). | ||||
| 14 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The current grows as $i = \frac{\epsilon}{R}\left(1 - e^{-t/t_c}\right)$ with $t_c = \frac{L}{R}$.
$q = \int_0^{t_c} i\,dt = \frac{\epsilon}{R}\left[t + t_c e^{-t/t_c}\right]_0^{t_c} = \frac{\epsilon}{R}\left(t_c + \frac{t_c}{e} - t_c\right) = \frac{\epsilon t_c}{eR} = \frac{\epsilon L}{eR^2}$.
Answer: (1). | ||||
| 15 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Magnitude: $E = cB = 3\times10^8\times5\times10^{-8} = 15$ V/m.
Direction: $\vec{E}\times\vec{B}$ points along the propagation, $+\hat{k}$. With $\vec{B}$ along $\hat{j}$, $\hat{i}\times\hat{j} = \hat{k}$, so $\vec{E}$ is along $+\hat{i}$.
$\vec{E} = 15\hat{i}$ V/m.
Answer: (1). | ||||
| 16 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Take distances as magnitudes for a real object beyond the focus ($x > f$). The mirror equation gives the image distance $v = \frac{fx}{x - f}$, so $|m| = \frac{v}{x} = \frac{f}{x - f}$.
At $x \to f$, $|m| \to \infty$; at $x = 2f$, $|m| = 1$; as $x \to \infty$, $|m| \to 0$. The magnification falls continuously from a very large value through 1 at $2f$ towards zero: graph (2).
Answer: (2). | ||||
| 17 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Phase difference $\phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{8} = \frac{\pi}{4}$.
$I = I_{max}\cos^2\frac{\phi}{2}$, so $\frac{I}{I_{max}} = \cos^2\frac{\pi}{8} = \frac{1 + \cos\frac{\pi}{4}}{2} = \frac{1 + 0.707}{2} = 0.853$.
Answer: (1). | ||||
| 18 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The force on the electron is $-e\vec{E} = eE_0\hat{k}$, so it gains a velocity $\frac{eE_0t}{m}$ along $\hat{k}$, while the x- and y-components stay $v_0$.
Initial speed $\sqrt{2}v_0$; speed at time t: $\sqrt{2v_0^2 + \frac{e^2E_0^2t^2}{m^2}}$.
$\lambda = \frac{h}{mv}$, so $\frac{\lambda}{\lambda_0} = \frac{\sqrt{2}v_0}{\sqrt{2v_0^2 + \frac{e^2E_0^2t^2}{m^2}}} = \frac{1}{\sqrt{1 + \frac{e^2E_0^2t^2}{2m^2v_0^2}}}$.
This is option (2), where E stands for $E_0$.
Answer: (2). | ||||
| 19 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Input 1 passes through a NOT gate: 0. Both inputs of the top NAND gate come from this point, so its output is $\overline{0\cdot0} = 1$.
That output is fed back to the second input of the bottom NAND gate, whose first input is 0: $\overline{0\cdot1} = 1$.
The middle NAND gate gets the 0 inverted again by the second NOT gate (1) and the bottom gate's output (1): $Y = \overline{1\cdot1} = 0$.
Answer: (1). | ||||
| 20 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter A series resistance $R$ is needed with $I_g(G + R) = V$:
$R = \frac{V}{I_g} - G = \frac{10}{10^{-3}} - 100 = 10000 - 100 = 9900\ \Omega = 9.9$ kΩ.
Answer: (2). | ||||
| 21 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Let the fall take time t. Then $\frac{1}{2}gt^2 = 100$ and, up to half a second before landing, $\frac{1}{2}g\left(t - \frac{1}{2}\right)^2 = 100 - 19 = 81$.
Dividing: $\frac{t}{t - \frac{1}{2}} = \frac{10}{9}$, so $9t = 10t - 5$, $t = 5$ s.
$g = \frac{2\times100}{25} = 8$ ms⁻².
Answer: 8 (NTA set this question aside for all candidates; see the note). | ||||
| 22 | GravitationEscape velocity | Class 11: Gravitation | NumericalMedium | |
A medium numerical question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Energy conservation: $\frac{1}{2}v^2 - \frac{GM}{R} = \frac{1}{2}u^2 - \frac{GM}{10R}$, with $\frac{2GM}{R} = v_e^2$.
$v^2 = u^2 + v_e^2\left(1 - \frac{1}{10}\right) = 144 + 0.9\times125.44 = 144 + 112.9 = 256.9$.
$v = 16.03$ km/s, i.e. 16 km/s to the nearest integer.
Answer: 16. | ||||
| 23 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter With equal specific heat, the final temperature of a mix is the volume-weighted mean. Let the containers be at $\theta_1, \theta_2, \theta_3$.
$\theta_1 + 2\theta_2 = 3\times60 = 180$; $\theta_2 + 2\theta_3 = 90$; $2\theta_1 + \theta_3 = 180$.
From the third, $\theta_3 = 180 - 2\theta_1$; the second gives $\theta_2 = 90 - 2\theta_3 = 4\theta_1 - 270$; the first gives $\theta_1 + 8\theta_1 - 540 = 180$, so $\theta_1 = 80$°C, $\theta_2 = 50$°C, $\theta_3 = 20$°C.
Equal volumes of all three: $\theta = \frac{80 + 50 + 20}{3} = 50$°C.
Answer: 50. | ||||
| 24 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Total emf 20 V; the current is $I = \frac{20}{25 + R_p}$, where $R_p$ is the parallel combination.
Terminal voltage of the 20 Ω cell: $10 - 20I = 0$, so $I = 0.5$ A and $25 + R_p = 40$, $R_p = 15\ \Omega$.
$\frac{30R}{30 + R} = 15$ gives $30R = 450 + 15R$, $R = 30\ \Omega$.
Answer: 30. | ||||
| 25 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right)$.
First member ($n = 3$): $\frac{1}{4} - \frac{1}{9} = \frac{5}{36}$. Second member ($n = 4$): $\frac{1}{4} - \frac{1}{16} = \frac{3}{16}$.
$\lambda_2 = \lambda_1\cdot\frac{5/36}{3/16} = 6561\times\frac{80}{108} = 4860$ Å $= 486$ nm.
Answer: 486. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Assertion and reasonMedium | |
A medium assertion and reason question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Assertion: the activity of metals as hydrogenation catalysts rises from Group 5 to Group 11 with a maximum at Groups 7–9 (e.g. Fe, Co, Ni and their heavier congeners) — a standard observation, so it is true.
Reason: a good catalyst adsorbs the reactants with moderate strength — strongly enough to activate them, but not so strongly that the products cannot leave. Metals to the left of Groups 7–9 adsorb too strongly, those to the right too weakly. The maximum activity is therefore not due to the strongest adsorption; the reason is false.
Answer: (3). | ||||
| 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 $\log k = \log A - \frac{E_a}{2.303R}\cdot\frac{1}{T}$: a plot of $\log k$ against $\frac{1}{T}$ is a straight line of slope $-\frac{E_a}{2.303R}$, so the steeper the line, the larger the activation energy.
From the figure, c is the steepest, then a, then d, and b is the flattest.
$E_c > E_a > E_d > E_b$.
Answer: (3). | ||||
| 28 | EquilibriumpH and buffers | Class 11: Equilibrium | Assertion and reasonEasy | |
A easy assertion and reason question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Breaking O–H bonds to form $\mathrm{H^+}$ and $\mathrm{OH^-}$ needs energy: the dissociation of water is endothermic, so the reason is false.
Being endothermic, the dissociation increases with temperature, $K_w$ rises (e.g. $10^{-14}$ at 25 °C, about $5.5\times10^{-14}$ at 50 °C), $[\mathrm{H^+}]$ rises and the pH of pure water falls (to about 6.6 at 50 °C). The assertion is false too.
Answer: (4). | ||||
| 29 | 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 $r_n = \frac{n^2a_0}{Z}$. For $\mathrm{Li^{2+}}$, $Z = 3$; for the second orbit $n = 2$:
$r_2 = \frac{4a_0}{3}$.
Answer: (2). | ||||
| 30 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Down the halogen group the atom gets larger, the C–X bond gets longer and the orbital overlap poorer, so the bond weakens.
Approximate average bond enthalpies: C–F ≈ 485, C–Cl ≈ 330, C–Br ≈ 275, C–I ≈ 240 kJ/mol.
Order: C–F > C–Cl > C–Br > C–I.
Answer: (2). | ||||
| 31 | Solid StatePoint defects | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on point defects, from Solid State. NCERT: No matching NCERT chapter Schottky defects occur in ionic solids whose cation and anion are of similar size (NaCl, KCl, KBr, CsCl); Frenkel defects in solids with a small cation and a large anion (ZnS, AgCl, AgBr, AgI).
AgBr is the standard example of a solid that shows both: the small $\mathrm{Ag^+}$ ion can move into interstitial sites (Frenkel), and its ions are also missing in pairs (Schottky).
ZnS shows only Frenkel defects; CsCl and KBr only Schottky defects.
Answer: (4). | ||||
| 32 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across period 2 the radius falls with rising nuclear charge: F < O < C (about 64, 66 and 77 pm).
Down group 17 it rises: F < Cl < Br; Cl (99 pm) is larger than C, and Br (114 pm) is the largest.
Order: F < O < C < Cl < Br, i.e. (c) < (b) < (a) < (d) < (e).
Answer: (3). | ||||
| 33 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter (a) The limestone flux decomposes to CaO, which combines with the silica gangue to form the slag $\mathrm{CaSiO_3}$: it occurs.
(b) In the upper, cooler zone (500–800 K) CO reduces $\mathrm{Fe_2O_3}$ stepwise, first to $\mathrm{Fe_3O_4}$: it occurs.
(c) Some FeO combines with silica and passes into the slag as $\mathrm{FeSiO_3}$: it occurs (as a side reaction).
(d) FeO is reduced in the furnace by CO and C (FeO + CO → Fe + $\mathrm{CO_2}$; FeO + C → Fe + CO). Its direct thermal decomposition into Fe and $\mathrm{O_2}$ has a large positive ΔG at furnace temperatures and does not take place.
Only (d) does not occur.
(c) FeO + SiO₂ → FeSiO₃ is the slag-forming step of copper extraction (removing iron from the matte), not a reaction of the iron blast furnace, where CaO is the flux and CaSiO₃ the slag. (d) FeO is reduced by CO or C, not decomposed on its own. So (c) and (d) do not occur.
Answer: (3). | ||||
| 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 The isotopes of hydrogen are protium $\mathrm{^1_1H}$ (0 neutrons), deuterium $\mathrm{^2_1H}$ (1 neutron) and tritium $\mathrm{^3_1H}$ (2 neutrons).
$x + y + z = 0 + 1 + 2 = 3$.
Answer: (3). | ||||
| 35 | 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 The gas that turns $\mathrm{CuSO_4}$ solution deep blue-violet is ammonia: $\mathrm{Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+}}$.
So B is a nitride that is hydrolysed by water. Magnesium burns in nitrogen: $\mathrm{3Mg + N_2 \rightarrow Mg_3N_2}$, and $\mathrm{Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3}$.
Sodium does not combine with nitrogen on heating (among the alkali metals only lithium forms a nitride), and the nitrates in (3) and (4) are not formed from metal and nitrogen.
A = Mg, B = $\mathrm{Mg_3N_2}$.
Answer: (2). | ||||
| 36 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2}$: X is sodium hypophosphite.
$\mathrm{NaH_2PO_2 + HCl \rightarrow H_3PO_2 + NaCl}$: Y is hypophosphorous acid, $\mathrm{H_3PO_2}$.
Its structure has one P–OH and two P–H bonds; only the O–H hydrogen is ionisable, so the acid is monobasic.
Answer: (1). | ||||
| 37 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) $\mathrm{Ni(CO)_4}$: Ni(0), $3d^{10}$ after pairing, tetrahedral ($sp^3$): no unpaired electron, μ = 0.
(B) $\mathrm{[Ni(H_2O)_6]^{2+}}$: Ni(II), $3d^8$, octahedral: two unpaired electrons, μ = $\sqrt{8} = 2.83$ BM.
(C) $\mathrm{[Ni(CN)_4]^{2-}}$: Ni(II), $d^8$ with the strong-field cyanide, square planar ($dsp^2$): μ = 0.
(D) $\mathrm{PdCl_2(PPh_3)_2}$: Pd(II), $4d^8$; the large crystal-field splitting of a 4d metal makes it square planar and low-spin: μ = 0.
So (A) ≈ (C) ≈ (D) < (B).
Answer: (4). | ||||
| 38 | 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 (a) Square planar $\mathrm{Ma_3b}$: every arrangement is the same — no geometrical isomers.
(b) Octahedral $\mathrm{Mab_5}$: no geometrical isomers.
(c) Square planar $\mathrm{Ma_2bc}$: the two $\mathrm{NH_3}$ can be cis or trans — two isomers.
(d) Octahedral $\mathrm{Ma_4bc}$: Cl and Br can be cis or trans — two isomers.
Answer: (3). | ||||
| 39 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Maltose is a disaccharide of two α-D-glucose units joined by an α(1→4) glycosidic link between C1 of one unit and C4 of the other.
(Sucrose is α-D-glucose + β-D-fructose; lactose is β-D-galactose + β-D-glucose.)
Answer: (2). | ||||
| 40 | 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 In Kjeldahl's method the compound is digested with concentrated $\mathrm{H_2SO_4}$, which turns its nitrogen into $\mathrm{(NH_4)_2SO_4}$.
Nitrogen in nitro and azo groups and in a ring (pyridine) is not converted to ammonium sulphate under these conditions, so the method fails for nitrobenzene.
Urea, propanenitrile and aniline are all estimated by it.
Answer: (3). | ||||
| 41 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The reactant is 2-methylcyclopropenone. Its carbonyl oxygen is strongly basic, because protonating it gives a hydroxycyclopropenyl cation: a planar, cyclic, fully conjugated ring with 2 π electrons — aromatic by Hückel's rule (4n + 2 with n = 0).
So with $\mathrm{H_3O^+}$ the major product is the protonated, aromatic cation, 1-hydroxy-2-methylcyclopropenyl cation, rather than any addition product.
Answer: (4). | ||||
| 42 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons X takes up 2 $\mathrm{H_2}$, so it has two C=C bonds; option (2) is a ketone, not a hydrocarbon.
Ozonolysis with Zn/$\mathrm{H_2O}$ cuts each C=C into two C=O; Tollens' reagent then oxidises the aldehyde groups to COOH.
Number the ring of option (3) from the carbon carrying the exocyclic $\mathrm{=CH_2}$ (C1), with the ring double bond at C3=C4. Ozonolysis gives HCHO and, from the ring, C1 as a ketone and C3, C4 as aldehyde groups:
$\mathrm{OHC-CH_2-CO-CH_2-CH_2-CHO}$ (A), which Tollens' reagent oxidises to $\mathrm{HOOC-CH_2-CO-CH_2-CH_2-COOH}$, 3-oxohexanedioic acid (B).
Option (1), with the double bonds conjugated (C1=$\mathrm{CH_2}$, C2=C3), gives $\mathrm{HOOC-CO-CH_2CH_2CH_2-COOH}$, the 2-oxo acid; option (4) gives cyclohexane-1,3-dione and no acid.
Answer: (3). | ||||
| 43 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Phenol reacts with formaldehyde (acid or base catalysed): the ring is attacked at its ortho and para positions by the electrophilic carbon of HCHO, giving o- and p-hydroxymethylphenols — an electrophilic aromatic substitution.
These then link to further phenol rings through $\mathrm{-CH_2-}$ bridges with loss of water (dehydration), giving novolac and, on further cross-linking, Bakelite.
Answer: (1). | ||||
| 44 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers All the structures are $\mathrm{C_9H_{18}O_3}$. Alcohols form intermolecular hydrogen bonds; ethers cannot (they have no O–H). So a triol boils far higher than an isomeric triether.
For A to boil higher than B, A must be a triol and B the trimethoxy ether: option (2).
Options (3) and (4) have the ether as A, so A would boil lower; in option (1) both are triols with three O–H groups each, and there is no clear basis for A boiling higher.
Answer: (2). | ||||
| 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 Hydroboration–oxidation adds water anti-Markovnikov and without rearrangement: OH goes to the less substituted alkene carbon.
A: $\mathrm{CH_3-CH(CH(CH_3)_2)-CH(OH)-CH_2CH_3}$ (4,5-dimethylhexan-3-ol).
Dilute $\mathrm{H_2SO_4}$, heat: loss of water gives a secondary carbocation at the former C–OH carbon. A 1,2-hydride shift from the neighbouring CH (which carries $\mathrm{CH_3}$ and isopropyl) gives a more stable tertiary cation: $\mathrm{CH_3-\overset{+}{C}(CH(CH_3)_2)-CH_2CH_2CH_3}$.
Loss of a proton gives the most substituted (Saytzeff) alkene: removing the H of the isopropyl group gives the tetrasubstituted alkene $\mathrm{CH_3-C(=C(CH_3)_2)-CH_2CH_2CH_3}$.
Answer: (4). | ||||
| 46 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles of $\mathrm{O_2}$: $n = \frac{PV}{RT} = \frac{1\times492}{0.082\times300} = 20$ mol.
The equation (balanced as written) gives 1 mol $\mathrm{O_2}$ per mol $\mathrm{NaClO_3}$, so 20 mol of $\mathrm{NaClO_3}$ is needed.
Molar mass of $\mathrm{NaClO_3} = 23 + 35.5 + 48 = 106.5$ g/mol; mass $= 20\times106.5 = 2130$ g.
Answer: 2130 (NTA's key accepts 2120 to 2140). | ||||
| 47 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta U = nC_V\Delta T$, so $C_V = \frac{5000}{4\times(500 - 300)} = \frac{5000}{800} = 6.25$ J mol⁻¹ K⁻¹.
Answer: 6.25. | ||||
| 48 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $\mathrm{Sn + Pb^{2+} \rightarrow Sn^{2+} + Pb}$, $n = 2$; $E^0_{cell} = -0.13 - (-0.14) = 0.01$ V.
At equilibrium $E_{cell} = 0$: $0 = 0.01 - \frac{0.06}{2}\log\frac{[\mathrm{Sn^{2+}}]}{[\mathrm{Pb^{2+}}]}$, so $\log\frac{[\mathrm{Sn^{2+}}]}{[\mathrm{Pb^{2+}}]} = \frac{1}{3}$.
Ratio $= 10^{1/3} = 2.154 \approx 2.15$.
Answer: 2.15 (NTA's key accepts 2.13 to 2.17). | ||||
| 49 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure "Sum" is the total number of such angles (NTA's key is 20).
Square pyramid (Ni): four basal ligands and one apical. 90° angles: 4 apical–basal + 4 between adjacent basal ligands = 8; 180°: 2 (the trans basal pairs); 120°: none. Total 10 (all $\binom{5}{2} = 10$ pairs).
Trigonal bipyramid (Fe): 90°: 2 axial × 3 equatorial = 6; 120°: 3 between equatorial ligands; 180°: 1 (axial–axial). Total 10.
Sum of the counts: 10 + 10 = 20.
Answer: 20. | ||||
| 50 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons A, the simplest alkyne, is ethyne; passed through a red-hot copper (or iron) tube it trimerises to benzene (B).
Friedel–Crafts methylation with one equivalent of $\mathrm{CH_3Cl}$/anhydrous $\mathrm{AlCl_3}$ gives toluene (C), $\mathrm{C_6H_5CH_3}$.
In toluene the six ring carbons, the five ring hydrogens and the methyl carbon all lie in the ring's plane (12 atoms). The methyl group can rotate so that one of its three hydrogens also lies in that plane.
Maximum number of coplanar atoms: 12 + 1 = 13.
Answer: 13. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions For $1 < x < 2$, $[x] = 1$ and $f(x) = \frac{x}{1 + x^2}$. Its derivative $\frac{1 - x^2}{(1 + x^2)^2} < 0$ for $x > 1$, so f falls from $\frac{1}{2}$ (at $x \to 1$) to $\frac{2}{5}$ (at $x \to 2$), both ends excluded: $\left(\frac{2}{5}, \frac{1}{2}\right)$.
For $2 \le x < 3$, $[x] = 2$ and $f(x) = \frac{2x}{1 + x^2}$, also decreasing: from $\frac{4}{5}$ at $x = 2$ (included) to $\frac{6}{10} = \frac{3}{5}$ as $x \to 3$ (excluded): $\left(\frac{3}{5}, \frac{4}{5}\right]$.
Range: $\left(\frac{2}{5}, \frac{1}{2}\right) \cup \left(\frac{3}{5}, \frac{4}{5}\right]$.
Answer: (3). | ||||
| 52 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha = \omega$, a cube root of unity: $\omega^3 = 1$, $1 + \omega + \omega^2 = 0$.
$b = \sum_{k=0}^{100}(\omega^3)^k = 101$.
$\sum_{k=0}^{100}\omega^{2k} = \frac{1 - \omega^{202}}{1 - \omega^2}$; since $202 = 3\cdot67 + 1$, $\omega^{202} = \omega$, so the sum is $\frac{1 - \omega}{(1 - \omega)(1 + \omega)} = \frac{1}{1 + \omega}$.
$a = (1 + \omega)\cdot\frac{1}{1 + \omega} = 1$.
Sum of roots $102$, product $101$: $x^2 - 102x + 101 = 0$.
Answer: (4). | ||||
| 53 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $t = 3^x > 0$: $t^2 - t + 2 = |t - 1| + |t - 2|$.
$t \ge 2$: $t^2 - t + 2 = 2t - 3$, i.e. $t^2 - 3t + 5 = 0$, discriminant $9 - 20 < 0$: no root.
$1 \le t < 2$: $t^2 - t + 2 = (t - 1) + (2 - t) = 1$, i.e. $t^2 - t + 1 = 0$: no real root.
$0 < t < 1$: $t^2 - t + 2 = (1 - t) + (2 - t) = 3 - 2t$, i.e. $t^2 + t - 1 = 0$, $t = \frac{\sqrt{5} - 1}{2} \approx 0.618$, which lies in $(0, 1)$.
So $3^x = \frac{\sqrt{5} - 1}{2}$ gives exactly one real x: S is a singleton.
Answer: (2). | ||||
| 54 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The characteristic equation of A: $\lambda^2 - (\mathrm{tr}A)\lambda + |A| = 0$ with $\mathrm{tr}A = 6$, $|A| = 8 - 18 = -10$.
By the Cayley–Hamilton theorem $A^2 - 6A - 10I = O$. Multiply by $A^{-1}$: $A - 6I - 10A^{-1} = O$, so $10A^{-1} = A - 6I$.
Answer: (4). | ||||
| 55 | 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 $D = \begin{vmatrix}\lambda&2&2\\2\lambda&3&5\\4&\lambda&6\end{vmatrix} = \lambda(18 - 5\lambda) - 2(12\lambda - 20) + 2(2\lambda^2 - 12) = -\lambda^2 - 6\lambda + 16 = -(\lambda + 8)(\lambda - 2)$.
So the solution is unique for every $\lambda$ except $-8$ and 2; (1) is false and, at $\lambda = 8$, (2) is false.
At $\lambda = 2$: $D_x = \begin{vmatrix}5&2&2\\8&3&5\\10&2&6\end{vmatrix} = 5(8) - 2(-2) + 2(-14) = 16 \ne 0$, so the system is inconsistent: no solution.
Answer: (4). | ||||
| 56 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The odd powers of $\sqrt{x^2 - 1}$ cancel:
$2\left[x^6 + \binom{6}{2}x^4(x^2 - 1) + \binom{6}{4}x^2(x^2 - 1)^2 + (x^2 - 1)^3\right]$
$= 2\left[x^6 + 15x^6 - 15x^4 + 15x^6 - 30x^4 + 15x^2 + x^6 - 3x^4 + 3x^2 - 1\right] = 2[32x^6 - 48x^4 + 18x^2 - 1]$.
$\alpha = -96$, $\beta = 36$; $\alpha - \beta = -132$ (and $\alpha + \beta = -60$).
Answer: (2). | ||||
| 57 | 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 $a + 9d = \frac{1}{20}$, $a + 19d = \frac{1}{10}$: $10d = \frac{1}{20}$, $d = \frac{1}{200}$, $a = \frac{1}{20} - \frac{9}{200} = \frac{1}{200}$.
$S_{200} = \frac{200}{2}\left(2a + 199d\right) = 100\cdot\frac{2 + 199}{200} = \frac{201}{2} = 100\frac{1}{2}$.
Answer: (4). | ||||
| 58 | 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 The limit has the form $\frac{0}{0}$. By L'Hôpital's rule and the fundamental theorem of calculus,
$\lim_{x \to 0}\frac{x\sin(10x)}{1} = 0$.
Answer: (4). | ||||
| 59 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctHard | |
A hard single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $g(x) = x\,(f(x) - f(1))$. g is continuous on [0, 1], differentiable on (0, 1), and $g(0) = 0 = g(1)$. By Rolle's theorem there is $c \in (0, 1)$ with
$g'(c) = f(c) - f(1) + cf'(c) = 0$, i.e. $f(c) - f(1) = -cf'(c)$.
Hence $|f(c) - f(1)| = c|f'(c)| < |f'(c)|$, since $0 < c < 1$ (provided $f'(c) \ne 0$): option (1).
(2) and (3) fail, e.g., for $f(x) = x$: (2) would need $1 - c < 1 - c$; (3) would need $c + 1 < 1 + c$. (4) asks that the chord slope from c to 1 equal the slope at c itself, which fails for $f(x) = x^2$ (it gives $1 + c = 2c$, $c = 1$).
Answer: (1). | ||||
| 60 | 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 Differentiate: $2x + 2y + 2xy' - 6yy' = 0$. At (2, 2): $8 + 4y' - 12y' = 0$, $y' = 1$.
(Indeed the curve is the pair of lines $(x - y)(x + 3y) = 0$, and (2, 2) lies on $y = x$.)
The normal has slope $-1$: $y - 2 = -(x - 2)$, i.e. $x + y - 4 = 0$.
Distance from the origin: $\frac{4}{\sqrt{2}} = 2\sqrt{2}$.
Answer: (3). | ||||
| 61 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $g(x) = 2x^3 - 9x^2 + 12x + 4$. Then $g'(x) = 6(x - 1)(x - 2) < 0$ on (1, 2), so g decreases from $g(1) = 9$ to $g(2) = 8$.
On (1, 2), $\frac{1}{3} < \frac{1}{\sqrt{g(x)}} < \frac{1}{\sqrt{8}}$; integrating over an interval of length 1: $\frac{1}{3} < I < \frac{1}{2\sqrt{2}}$.
Squaring: $\frac{1}{9} < I^2 < \frac{1}{8}$.
Answer: (1). | ||||
| 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 parabola and the line meet where $x^2 = 3 - 2x$: $x^2 + 2x - 3 = 0$, $x = -3, 1$.
Area $= \int_{-3}^{1}(3 - 2x - x^2)\,dx = \left[3x - x^2 - \frac{x^3}{3}\right]_{-3}^{1} = \left(3 - 1 - \frac{1}{3}\right) - (-9 - 9 + 9) = \frac{5}{3} + 9 = \frac{32}{3}$.
Answer: (1). | ||||
| 63 | Differential EquationsOrder and degree | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on order and degree, from Differential Equations. NCERT: Class 12: Differential Equations Differentiate: $2x = 4by'$, so $b = \frac{x}{2y'}$.
Substitute: $x^2 = \frac{2x}{y'}\left(y + \frac{x}{2y'}\right) = \frac{2xy}{y'} + \frac{x^2}{(y')^2}$.
Multiply by $\frac{(y')^2}{x}$: $x(y')^2 = 2yy' + x$.
Answer: (3). | ||||
| 64 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections $L_1$: $\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1$, slope $-1$. Perpendicular to it, $m = 1$: the line is $x - y + c = 0$.
Tangent to the circle with centre (3, 0), radius 1: $\frac{|3 + c|}{\sqrt{2}} = 1$, so $(c + 3)^2 = 2$, i.e. $c^2 + 6c + 7 = 0$.
Answer: (3). | ||||
| 65 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $\frac{x^2}{36} - \frac{y^2}{b^2} = 1$ through (10, 16): $\frac{100}{36} - \frac{256}{b^2} = 1$, so $\frac{256}{b^2} = \frac{64}{36}$, $b^2 = 144$.
Normal at $(x_1, y_1)$: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$: $\frac{36x}{10} + \frac{144y}{16} = 180$, i.e. $3.6x + 9y = 180$, or $2x + 5y = 100$.
Answer: (4). | ||||
| 66 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The plane is the perpendicular bisector of the segment joining the point and its image.
Normal: $\left(1 + \frac{7}{3}, 2 + \frac{4}{3}, 3 + \frac{1}{3}\right) = \frac{10}{3}(1, 1, 1)$. Midpoint: $\left(-\frac{2}{3}, \frac{1}{3}, \frac{4}{3}\right)$.
Plane: $x + y + z = -\frac{2}{3} + \frac{1}{3} + \frac{4}{3} = 1$.
Check: (1, −1, 1) gives 1 ✓; the others give −3, 3 and −1.
Answer: (3). | ||||
| 67 | 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 $\vec{b}\times(\vec{c} - \vec{a}) = \vec{0}$, so $\vec{c} = \vec{a} + \lambda\vec{b}$.
$\vec{c}\cdot\vec{a} = |\vec{a}|^2 + \lambda\,\vec{a}\cdot\vec{b} = 6 + 4\lambda = 0$, so $\lambda = -\frac{3}{2}$.
$\vec{c}\cdot\vec{b} = \vec{a}\cdot\vec{b} + \lambda|\vec{b}|^2 = 4 - \frac{3}{2}\cdot3 = -\frac{1}{2}$.
Answer: (2). | ||||
| 68 | 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 $\sum x = 200$; $\sum x^2 = 20(4 + 10^2) = 2080$.
Corrected: $\sum x = 200 - 9 + 11 = 202$; $\sum x^2 = 2080 - 81 + 121 = 2120$.
Mean $= 10.1$; variance $= \frac{2120}{20} - 10.1^2 = 106 - 102.01 = 3.99$.
Answer: (2). | ||||
| 69 | 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 P(exactly one) $= P(A\cup B) - P(A\cap B)$.
$\frac{2}{5} = \frac{1}{2} - P(A\cap B)$, so $P(A\cap B) = \frac{1}{10} = 0.10$.
Answer: (3). | ||||
| 70 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter (4): $\sim(p \vee \sim q) \equiv \sim p \wedge q$, which implies q and hence $p \vee q$. So the implication is true in every case: a tautology.
The others fail:
(1) p = F, q = F: $p \vee \sim q$ is T but $p \wedge q$ is F.
(2) p = F, q = T: $\sim p \wedge q$ is T but $p \wedge q$ is F.
(3) p = F, q = F: $\sim(p \wedge \sim q)$ is T but $p \vee q$ is F.
Answer: (4). | ||||
| 71 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations EXAMINATION: A, I, N twice each; E, X, M, T, O once — 8 distinct letters.
All four different: $^8P_4 = 1680$.
One pair alike, two different: $3\times\binom{7}{2}\times\frac{4!}{2!} = 3\times21\times12 = 756$.
Two pairs alike: $\binom{3}{2}\times\frac{4!}{2!2!} = 3\times6 = 18$.
Total: $1680 + 756 + 18 = 2454$.
Answer: 2454. | ||||
| 72 | 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 $n(n+1)(2n+1) = 2n^3 + 3n^2 + n$.
For n = 1 to 7: $\sum n^3 = 784$, $\sum n^2 = 140$, $\sum n = 28$.
Sum $= \frac{1}{4}(2\cdot784 + 3\cdot140 + 28) = \frac{1568 + 420 + 28}{4} = \frac{2016}{4} = 504$.
Answer: 504. | ||||
| 73 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'$ is quadratic with a root at $-1$ and its vertex at $x = 1$ (where $f'' = 0$), so its other root is 3: $f'(x) = k(x + 1)(x - 3) = k(x^2 - 2x - 3)$.
$f(x) = k\left(\frac{x^3}{3} - x^2 - 3x\right) + C$; $f(-1) = \frac{5k}{3} + C = 10$, $f(1) = -\frac{11k}{3} + C = -6$. Subtracting, $\frac{16k}{3} = 16$, $k = 3$.
$f''(x) = 3(2x - 2)$: $f''(3) = 12 > 0$, a local minimum (and $f''(-1) < 0$, a maximum).
Answer: 3. | ||||
| 74 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $m^2x^2 = x$ gives $P = \left(\frac{1}{m^2}, \frac{1}{m}\right)$.
Tangent to $y^2 = x$ at $(x_1, y_1)$: $yy_1 = \frac{x + x_1}{2}$; at $y = 0$, $x = -x_1$, so $Q = \left(-\frac{1}{m^2}, 0\right)$.
Area $= \frac{1}{2}\cdot OQ\cdot y_P = \frac{1}{2}\cdot\frac{1}{m^2}\cdot\frac{1}{m} = \frac{1}{2m^3} = 4$, so $m^3 = \frac{1}{8}$, $m = 0.5$.
Answer: 0.5. | ||||
| 75 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sqrt{1 + \cos2\alpha} = \sqrt{2}\cos\alpha$ (cos α > 0), so the first gives $\tan\alpha = \frac{1}{7}$.
$\sqrt{\frac{1 - \cos2\beta}{2}} = \sin\beta = \frac{1}{\sqrt{10}}$, so $\tan\beta = \frac{1}{3}$ and $\tan2\beta = \frac{2/3}{1 - 1/9} = \frac{3}{4}$.
$\tan(\alpha + 2\beta) = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{3}{28}} = \frac{25/28}{25/28} = 1$.
Answer: 1. | ||||
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