JEE (Main) · 2020
JEE (Main) 2020 — 4 September, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
25 Q
- Kinematics2
- Properties of Bulk Matter2
- Electrostatics2
8 easy · 16 medium · 1 hard
Chemistry
25 Q
- Classification of Elements and Periodicity2
- Coordination Compounds2
- Alcohols, Phenols and Ethers2
13 easy · 10 medium · 2 hard
Mathematics
25 Q
- Integral Calculus3
- Quadratic Equations2
- Matrices and Determinants2
7 easy · 16 medium · 2 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Heat conduction: $\frac{dQ}{dt} = kA\frac{\Delta T}{\Delta x}$, so $k = \frac{(dQ/dt)\,\Delta x}{A\,\Delta T}$.
$[dQ/dt] = \mathrm{ML^2T^{-3}}$, $[\Delta x] = \mathrm{L}$, $[A] = \mathrm{L^2}$, $[\Delta T] = \mathrm{K}$.
$[k] = \frac{\mathrm{ML^2T^{-3}}\cdot\mathrm{L}}{\mathrm{L^2}\cdot\mathrm{K}} = \mathrm{MLT^{-3}K^{-1}}$.
Answer: (B). | ||||
| 2 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Take upward velocity as positive and height y above the floor.
Falling from rest at $y = h$: $v^2 = 2g(h - y)$ with $v < 0$. At $y = h$, $v = 0$; as y falls to 0, $|v|$ grows to $\sqrt{2gh}$. The curve $y = h - \frac{v^2}{2g}$ is a parabola with its axis along the h-axis, so it meets the h-axis perpendicularly at $y = h$ (not tangentially).
After the bounce: $v^2 = 2g(\frac{h}{2} - y)$ with $v > 0$, starting from $v = \sqrt{gh}$ at $y = 0$ and reaching 0 at $y = \frac{h}{2}$; again an arc meeting the axis perpendicularly.
So the graph is a lower arc from $(h, 0)$ to the negative v-axis (motion towards smaller h) and an upper arc from the positive v-axis to $(\frac{h}{2}, 0)$ (motion towards larger h). (A) has the arcs tangent to the axis, (B) has straight lines (v linear in h is wrong), and (D) starts the rise at the origin with zero speed. Only (C) fits.
Answer: (C). | ||||
| 3 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Along x: $u_x = 0$, $a_x = 10$, so $x = \frac{1}{2}(10)t^2 = 5t^2 = 20$, giving $t = 2$ s.
Along y: $u_y = 5$, $a_y = 4$, so $y_0 = 5(2) + \frac{1}{2}(4)(2)^2 = 10 + 8 = 18$ m.
Answer: (C). | ||||
| 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 $mv$ is conserved through every perfectly inelastic collision; the moving lump keeps growing.
After hitting m: mass 2m, speed $\frac{v}{2}$. After 2m: mass 4m, speed $\frac{v}{4}$. After 4m: mass 8m, speed $\frac{v}{8}$. After 8m (the last block starts moving): mass 16m, speed $\frac{v}{16}$.
Final kinetic energy $= \frac{1}{2}(16m)\left(\frac{v}{16}\right)^2 = \frac{1}{16}\cdot\frac{1}{2}mv^2$.
Energy lost $= 1 - \frac{1}{16} = \frac{15}{16} = 93.75\%$, close to 94.
Answer: (D). | ||||
| 5 | GravitationGravitational potential energy | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation $E = -\frac{dV}{dx}$, so $V(x) = -\int_{\infty}^{x} E\,dx = -\int_{\infty}^{x}\frac{Ax\,dx}{(x^2+a^2)^{3/2}}$.
With $\int \frac{x\,dx}{(x^2+a^2)^{3/2}} = -\frac{1}{(x^2+a^2)^{1/2}}$: $V(x) = -\left[-\frac{A}{(x^2+a^2)^{1/2}}\right]_{\infty}^{x} = \frac{A}{(x^2+a^2)^{1/2}}$.
Its magnitude is $\frac{A}{(x^2+a^2)^{1/2}}$.
Answer: (B). | ||||
| 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 Forces on the bubble: buoyancy $\rho Vg$ upward and weight $mg$ downward. Newton's second law: $\rho Vg - mg = ma$.
$m = \frac{\rho Vg}{g + a}$, with $V = \frac{4}{3}\pi(1)^3 = 4.19\ \mathrm{cm^3}$, $\rho = 1$.
$m = \frac{4.19 \times 980}{980 + 9.8} = 4.19 \times \frac{1}{1.01} = 4.15$ gm.
Answer: (C). | ||||
| 7 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Heat given by the water cooling from 25°C to 0°C: $0.2 \times 4200 \times 25 = 21000$ J.
This melts $m = \frac{21000}{3.4\times10^5}$ kg $= 0.0618$ kg $\approx 61.7$ g (less than the 100 g available, so the final state is 0°C).
Answer: (B). | ||||
| 8 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | MatchingEasy | |
A easy matching question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $\gamma = 1 + \frac{2}{f}$ with f degrees of freedom.
Monatomic: $f = 3$, $\gamma = \frac{5}{3}$ (IV). Diatomic rigid: $f = 5$, $\gamma = \frac{7}{5}$ (I). Diatomic non-rigid (with the vibrational mode, kinetic and potential): $f = 7$, $\gamma = \frac{9}{7}$ (II). Triatomic rigid (non-linear): $f = 6$, $\gamma = \frac{8}{6} = \frac{4}{3}$ (III).
A→IV, B→I, C→II, D→III.
Answer: (A). | ||||
| 9 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Two crests are a whole number of wavelengths apart: $5 = n\lambda$. A crest and a trough are an odd number of half-wavelengths apart: $1.5 = (2m+1)\frac{\lambda}{2}$, i.e. $3 = (2m+1)\lambda$.
So $\lambda$ divides both 5 and 3 an integer number of times, with $\frac{3}{\lambda}$ odd. $\lambda = 1$: $n = 5$, $\frac{3}{\lambda} = 3$ (odd) - allowed. $\lambda = \frac{1}{3}$: $n = 15$, $\frac{3}{\lambda} = 9$ - allowed. $\lambda = \frac{1}{5}$: $n = 25$, $15$ - allowed. In general $\lambda = \frac{1}{k}$ with k odd; an even k gives $\frac{3}{\lambda} = 3k$ even, which is not allowed, and wavelengths such as 2, 3 or 5 fail $5 = n\lambda$ and $3 = (2m+1)\lambda$ together.
Possible wavelengths: $\frac{1}{1}, \frac{1}{3}, \frac{1}{5}, \ldots$
Answer: (C). | ||||
| 10 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In the figures the sheet $\sigma_+$ is the horizontal line and $\sigma_-$ the vertical one.
Each sheet gives a uniform field $\frac{|\sigma|}{2\varepsilon_0}$ perpendicular to itself: the positive (horizontal) sheet gives a vertical field pointing away from it, and the negative (vertical) sheet a horizontal field pointing towards it.
In each quadrant the total field is therefore uniform, so the field lines are straight segments running from the positive sheet to the negative sheet. Their slope is $\frac{E_+}{E_-} = \frac{|\sigma_+|}{|\sigma_-|} > 1$: steeper than 45°.
A line leaving the horizontal sheet at distance a from the crossing meets the vertical sheet at a height greater than a, so the lines form diamonds that are taller than they are wide, with arrows from $\sigma_+$ towards $\sigma_-$. Electrostatic field lines cannot form closed loops, which rules out (C) and (D); (B) has the slope less than 45°, which would need $|\sigma_-| > |\sigma_+|$.
Answer: (A). | ||||
| 11 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Let $k = \frac{1}{4\pi\epsilon_0}$. The electrostatic force is conservative, so only the end points matter.
At the origin: distances $\frac{d}{2}$ from both charges. $V_i = k\left(\frac{4q}{d/2} - \frac{q}{d/2}\right) = \frac{6kq}{d}$.
At $x = d$: distance $\frac{3d}{2}$ from 4q and $\frac{d}{2}$ from $-q$. $V_f = k\left(\frac{4q}{3d/2} - \frac{q}{d/2}\right) = k\left(\frac{8q}{3d} - \frac{2q}{d}\right) = \frac{2kq}{3d}$.
Change in energy of q: $q(V_f - V_i) = kq^2\left(\frac{2}{3} - 6\right)\frac{1}{d} = -\frac{16}{3}\cdot\frac{q^2}{4\pi\epsilon_0 d} = -\frac{4q^2}{3\pi\epsilon_0 d}$.
The energy decreases by $\frac{4q^2}{3\pi\epsilon_0 d}$.
Answer: (C). | ||||
| 12 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity The resistor has the terminal voltage across it: $I = \frac{P}{V} = \frac{0.5}{2.5} = 0.2$ A.
Voltage across the internal resistance: $3.0 - 2.5 = 0.5$ V, so the power there is $0.5 \times 0.2 = 0.10$ W.
Answer: (D). | ||||
| 13 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium 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 An arc subtending angle $\theta$ at its centre gives $B = \frac{\mu_0 I}{4\pi r}\theta$ there (the straight radial leads give no field at O).
From the figure, A (radius 2 cm) has a 90° gap, so it subtends $360° - 90° = 270° = \frac{3\pi}{2}$; B (radius 4 cm) has a 60° gap, so it subtends $300° = \frac{5\pi}{3}$.
$\frac{B_A}{B_B} = \frac{I_A\theta_A/r_A}{I_B\theta_B/r_B} = \frac{2\times\frac{3\pi}{2}/2}{3\times\frac{5\pi}{3}/4} = \frac{3\pi/2}{5\pi/4} = \frac{6}{5}$.
Answer: (A). | ||||
| 14 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy 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 $\tau = MB\sin\theta$: $0.018 = M(0.06)(0.5)$, so $M = 0.6\ \mathrm{A\,m^2}$.
From stable ($\theta = 0$, $U = -MB$) to unstable ($\theta = 180°$, $U = +MB$): $W = 2MB = 2(0.6)(0.06) = 0.072$ J $= 7.2\times10^{-2}$ J.
Answer: (B). | ||||
| 15 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At entry (a) the flux through the coil is increasing, so a current flows and G deflects one way.
When the small magnet is completely inside (b) the flux through the coil hardly changes, so the emf and the deflection are zero.
At exit (c) the flux decreases, so the current reverses (Lenz's law) and G deflects the other way.
Deflection one side, zero, deflection the other side: option (B).
Answer: (B). | ||||
| 16 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves In order of decreasing wavelength the spectrum runs: radio waves (metres and more), microwaves (mm to cm), infrared, visible (400-700 nm), ultraviolet, X-rays (about 0.01-10 nm), gamma rays.
So $\lambda_{\text{radio}} > \lambda_{\text{micro}} > \lambda_{\text{visible}} > \lambda_{x\text{-rays}}$. The other options put visible light above radio waves or X-rays above radio waves.
Answer: (A). | ||||
| 17 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics By Malus's law the transmitted intensity is $I_0\cos^2\omega t$; over a full revolution the average of $\cos^2$ is $\frac{1}{2}$.
Average power through the polariser: $\frac{1}{2}I_0 A = \frac{1}{2}(3.3)(3\times10^{-4}) = 4.95\times10^{-4}$ W.
Time for one revolution: $T = \frac{2\pi}{\omega} = \frac{6.28}{31.4} = 0.2$ s.
Energy per revolution $= 4.95\times10^{-4}\times0.2 = 9.9\times10^{-5}$ J $\approx 1.0\times10^{-4}$ J.
Answer: (B). | ||||
| 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 One-dimensional elastic collision with B at rest: $v_A = \frac{m_A - m_B}{m_A + m_B}v_0 = \frac{\frac{m}{2} - \frac{m}{3}}{\frac{m}{2} + \frac{m}{3}}v_0 = \frac{1/6}{5/6}v_0 = \frac{v_0}{5}$.
$\lambda = \frac{h}{m_A v}$, so A's wavelength becomes $5\lambda_0$.
$\Delta\lambda = 5\lambda_0 - \lambda_0 = 4\lambda_0$.
Answer: (D). | ||||
| 19 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The circuit has two zener diodes in series, back to back, across the output (after the resistor $R_1$).
For any polarity one diode is forward biased (small drop) and the other is reverse biased. While $|V_{in}|$ is below the breakdown voltage no current flows through them and $V_o = V_{in}$.
When $|V_{in}|$ exceeds about 6 V, the reverse-biased zener breaks down and holds the output at about 6 V (plus the small forward drop of the other diode), the rest of the voltage dropping across $R_1$.
This happens in both half-cycles because of the symmetric arrangement, so the 10 V peaks are clipped flat on both sides.
Answer: (B). | ||||
| 20 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The minimum energy for ejection is the work function $\phi = hf_0$, where $f_0$ is the threshold frequency, the frequency at which the stopping potential is zero.
The data point B is printed as $(5.5, 0)$: the stopping potential is zero at $f = 5.5\times10^{14}$ Hz, so $f_0 = 5.5\times10^{14}$ Hz.
$\phi = \frac{6.62\times10^{-34}\times5.5\times10^{14}}{1.6\times10^{-19}}$ eV $= \frac{3.641\times10^{-19}}{1.6\times10^{-19}}$ eV $= 2.27$ eV.
Answer: (C). | ||||
| 21 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalHard | |
A hard numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Let the side be a and the mass M. For a uniform equilateral triangular lamina about the perpendicular axis through its centroid, $I_0 = \frac{Ma^2}{12}$.
Triangle ADE is similar with side $\frac{a}{2}$, so its mass is $\frac{M}{4}$ and about its own centroid $G'$: $I' = \frac{M}{4}\cdot\frac{(a/2)^2}{12} = \frac{Ma^2}{192} = \frac{I_0}{16}$.
Height of ABC: $h = \frac{\sqrt3 a}{2}$. G is $\frac{2h}{3}$ below A; $G'$ is $\frac{2}{3}\cdot\frac{h}{2} = \frac{h}{3}$ below A. So $GG' = \frac{h}{3} = \frac{a}{2\sqrt3}$ and $GG'^2 = \frac{a^2}{12}$.
Parallel axes: $I_{ADE,G} = \frac{I_0}{16} + \frac{M}{4}\cdot\frac{a^2}{12} = \frac{I_0}{16} + \frac{I_0}{4} = \frac{5I_0}{16}$.
Remaining: $I_0 - \frac{5I_0}{16} = \frac{11I_0}{16}$, so $N = 11$.
Answer: 11. | ||||
| 22 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $I_1 = \frac{MR^2}{2}$, $I_2 = \frac{M(R/2)^2}{2} = \frac{MR^2}{8}$.
Angular momentum is conserved: $I_1\omega_1 = (I_1 + I_2)\omega_2$, so $\omega_2 = \frac{1/2}{5/8}\omega_1 = \frac{4}{5}\omega_1$.
$\frac{K_f}{K_i} = \frac{(I_1+I_2)\omega_2^2}{I_1\omega_1^2} = \frac{I_1}{I_1+I_2} = \frac{4}{5}$.
Energy lost $= \frac{1}{5} = 20\%$.
Answer: 20. | ||||
| 23 | ThermodynamicsFirst law | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics No heat or work is exchanged with the surroundings (closed, rigid vessel), so the total internal energy $nC_VT$ is conserved; $C_V$ is the same for both portions.
$0.1(200) + 0.05(400) = 0.15\,T$, so $T = \frac{20 + 20}{0.15} = 266.67$ K.
Answer: 266.67. | ||||
| 24 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics With the final image at the near point D = 25 cm, $m = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)$, taking the objective's image distance as the tube length L.
$100 = \frac{20}{1}\left(1 + \frac{25}{f_e}\right)$, so $1 + \frac{25}{f_e} = 5$ and $f_e = \frac{25}{4} = 6.25$ cm.
Answer: 6.25. | ||||
| 25 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.
Lyman ($n_1 = 1$): longest $\lambda = \frac{4}{3R}$ ($n_2 = 2$), shortest $\lambda = \frac{1}{R}$ ($n_2 = \infty$). Difference $= \frac{1}{3R} = 304$ Å, so $\frac{1}{R} = 912$ Å.
Paschen ($n_1 = 3$): longest $\frac{1}{\lambda} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$, $\lambda = \frac{144}{7R}$; shortest $\lambda = \frac{9}{R}$.
Difference $= \left(\frac{144}{7} - 9\right)\frac{1}{R} = \frac{81}{7}\times912 = 10553.14$ Å.
Answer: 10553.14. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | MatchingEasy | |
A easy matching question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Foam is gas dispersed in a liquid: froth (e). Gel is a liquid dispersed in a solid: jellies (c). Aerosol is a solid or liquid dispersed in a gas: smoke (a). Emulsion is a liquid dispersed in a liquid: milk (f).
(Cell fluid is a sol and rubber is a sol of liquid in solid only in some classifications; neither is needed here.)
(i)→(e), (ii)→(c), (iii)→(a), (iv)→(f).
Answer: (A). | ||||
| 27 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry The Daniell cell has $E^\circ_{cell} = 0.34 - (-0.76) = 1.10$ V, opposed by the external voltage $E_{ext}$.
$E_{ext} < 1.1$ V: the cell works normally; Zn dissolves at the anode and Cu deposits at the cathode, electrons flowing from Zn to Cu. (A) is correct.
$E_{ext} = 1.1$ V: the two balance and no current flows. (B) is correct.
$E_{ext} > 1.1$ V: the cell is driven backwards (electrolytic); electrons flow from Cu to Zn (C is correct), Zn is deposited at the Zn electrode and Cu dissolves at the Cu electrode. (D) states the opposite, so it is incorrect.
Answer: (D). | ||||
| 28 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Starting from A, the forward rate (a) is largest at first and falls as A is used up; the reverse rate (b) starts at zero and rises as B builds up.
At equilibrium the two rates become equal and stay constant (dynamic equilibrium), so the curves meet and continue as one horizontal line.
Only (A) shows this; in (B) the curves cross, in (C) they level off at different values, and in (D) they never meet.
Answer: (A). | ||||
| 29 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics (a) True: for an ideal gas U depends only on T, and $H = U + PV = U + RT$ also depends only on T.
(b) False: for an ideal gas $z = \frac{PV}{RT} = 1$ exactly.
(c) True: Mayer's relation $C_{P,m} - C_{V,m} = R$.
(d) True: since U depends only on T, $dU = C_V\,dT$ for any process, not only at constant volume.
So (a), (c) and (d).
Answer: (D). | ||||
| 30 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure On a potential energy curve the depth of the minimum gives the bond enthalpy, the position of the minimum gives the bond length, and the steepness and narrowness of the well give the stiffness of the bond.
A–B has the deepest well (about $-550$ kJ mol$^{-1}$), deeper than A–A (about $-450$), so (A) is false.
A–B's well is also deep and narrow, rising very steeply on both sides, so it has the stiffest bond: (B) is true.
A–D has its minimum at the largest distance (about 145 pm): the longest bond, so (C) is false.
Electronegativity cannot be read from the curves, so (D) is not supported.
Answer: (B). | ||||
| 31 | 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 Balmer lines end on $n = 2$: $\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{n^2}\right)$. They run from 656 nm ($n = 3$) down to the series limit at 365 nm, the familiar visible lines $H_\alpha$, $H_\beta$, $H_\gamma$, $H_\delta$. (Lyman lies in the ultraviolet, Paschen and beyond in the infrared.)
Answer: (B). | ||||
| 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 All four are isoelectronic (10 electrons). The radius falls as the nuclear charge rises: O (Z = 8) > F (9) > Na (11) > Mg (12).
So $\mathrm{O^{2-} > F^- > Na^+ > Mg^{2+}}$ (about 140, 133, 102 and 72 pm).
Answer: (A). | ||||
| 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 (a) True: in the blast furnace $\mathrm{CaCO_3 \rightarrow CaO + CO_2}$, and the CaO removes silica as slag $\mathrm{CaSiO_3}$.
(b) True: silver is leached with cyanide as the anionic complex $\mathrm{[Ag(CN)_2]^-}$, then displaced by zinc.
(c) True: Mond's process, $\mathrm{Ni + 4CO \rightarrow Ni(CO)_4}$ then decomposition on heating.
(d) True: van Arkel method, volatile iodides $\mathrm{ZrI_4}$, $\mathrm{TiI_4}$ decomposed on a hot filament.
All four are correct.
Answer: (B). | ||||
| 34 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter In excess air lithium forms the monoxide $\mathrm{Li_2O}$, sodium the peroxide $\mathrm{Na_2O_2}$, and potassium the superoxide $\mathrm{KO_2}$. Larger cations stabilise the larger peroxide and superoxide anions.
Answer: (B). | ||||
| 35 | 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 $\mathrm{2Pb(NO_3)_2 \xrightarrow{\Delta} 2PbO + 4NO_2 + O_2}$: (A) is the brown gas $\mathrm{NO_2}$.
On cooling it dimerises to colourless $\mathrm{N_2O_4}$ (B).
$\mathrm{N_2O_4 + 2NO \rightarrow 2N_2O_3}$: (C) is dinitrogen trioxide, a blue solid.
In $\mathrm{N_2O_3}$: $2x = 6$, so nitrogen is +3.
Answer: (B). | ||||
| 36 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The actinoids are Z = 90-103, so Z = 101 (mendelevium) is an actinoid.
After lawrencium (103) the 6d series begins: Z = 104 (rutherfordium, $\mathrm{[Rn]5f^{14}6d^27s^2}$) lies below Hf in Group 4.
Answer: (D). | ||||
| 37 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Pt(II) is square planar. The bidentate en must span two cis positions, so the two nitrite ligands are cis to each other: no geometrical isomers, and the planar complex has no optical isomer.
Nitrite is ambidentate (N-bonded nitro, $\mathrm{-NO_2}$, or O-bonded nitrito, $\mathrm{-ONO}$), giving linkage isomers: $\mathrm{[Pt(en)(NO_2)_2]}$, $\mathrm{[Pt(en)(NO_2)(ONO)]}$ and $\mathrm{[Pt(en)(ONO)_2]}$.
Total: 3.
Answer: (C). | ||||
| 38 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Count unpaired electrons (water, chloride, hydroxide and ammonia with these 2+ ions give high-spin complexes):
$\mathrm{[Cr(H_2O)_6]^{2+}}$: $\mathrm{Cr^{2+}}$ $d^4$, 4 unpaired.
$\mathrm{[CoCl_4]^{2-}}$: $\mathrm{Co^{2+}}$ $d^7$ tetrahedral, 3 unpaired.
$\mathrm{[Fe(H_2O)_6]^{2+}}$: $\mathrm{Fe^{2+}}$ $d^6$ high spin, 4 unpaired.
$\mathrm{[Mn(H_2O)_6]^{2+}}$: $\mathrm{Mn^{2+}}$ $d^5$, 5 unpaired (against 4 for Cr(II)).
$\mathrm{[Co(OH)_4]^{2-}}$: $d^7$, 3 unpaired; $\mathrm{[Fe(NH_3)_6]^{2+}}$: $d^6$ high spin, 4 unpaired.
Only (B) pairs equal values: 4 and 4, $\mu = \sqrt{24} = 4.90$ BM each.
Answer: (B). | ||||
| 39 | Alcohols, Phenols and EthersPrimary, secondary and tertiary alcohols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on primary, secondary and tertiary alcohols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Immediate turbidity with $\mathrm{ZnCl_2}$/conc. HCl (Lucas test) shows a tertiary alcohol, which forms the chloride at once through a stable carbocation.
All four options are esters $\mathrm{C_6H_{12}O_2}$. Their alcohol parts: (A) tert-butyl alcohol (tertiary), (B) butan-2-ol (secondary, turbidity in about 5 minutes), (C) 2-methylpropan-1-ol (primary, no turbidity at room temperature), (D) propan-1-ol (primary).
So (A) is tert-butyl acetate, hydrolysed to acetic acid (B) and tert-butyl alcohol (C).
Answer: (A). | ||||
| 40 | 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 A COOH on a ring is named as a cyclopentanecarboxylic acid, with the ring carbon bearing it as C1.
Numbering towards the methyl: C2 has CH3 and C4 has Br, locants {2, 4}. Numbering the other way: Br at C3 and CH3 at C5, locants {3, 5}. The lower set {2, 4} wins.
Name: 4-bromo-2-methylcyclopentane-1-carboxylic acid. (A) and (D) wrongly use '-oic acid' for a ring carbon, and (C) has the higher locant set.
Answer: (B). | ||||
| 41 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $\mathrm{AgNO_3}$ removes chloride as AgCl; the rate follows the stability of the carbocation formed (SN1-like).
(A) 3-chlorocyclopropene gives the cyclopropenyl cation, aromatic ($2\pi$ electrons, Huckel), so it ionises very easily.
(B) is the same ring with OMe on a double-bond carbon; the +M effect of OMe further stabilises the aromatic cation, so (B) > (A).
(C) gives a secondary isopropyl cation; (D) gives a secondary cation next to $\mathrm{CH_2NO_2}$, whose strong -I effect destabilises it, so (C) > (D).
Order: (B) > (A) > (C) > (D).
Answer: (B). | ||||
| 42 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Maltose is two glucose units joined by an $\alpha$-1,4-glycosidic link. C1 of the first glucose is bonded to the ring O and to the O of the link: an acetal. C1 of the second glucose keeps a free OH next to its ring O: a hemiacetal (which is why maltose is a reducing sugar). Glucose is an aldose, so there are no ketals.
Answer: (B). | ||||
| 43 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Sodalime decarboxylates an acid $\mathrm{RCOOH \rightarrow RH}$. Toluene results from a methylbenzoic acid (and also from phenylacetic acid), so P is $\mathrm{C_8H_8O_2}$.
Ring bromination must give only one product.
Phenylacetic acid: $\mathrm{CH_2COOH}$ is ortho/para directing, giving ortho and para isomers.
2- and 3-methylbenzoic acids: the four free ring positions are all different, and the directing effects lead to more than one product.
4-Methylbenzoic acid: the two positions ortho to $\mathrm{CH_3}$ are also meta to COOH, and they are equivalent by symmetry; both groups direct there, so only one monobromo product $\mathrm{C_8H_7O_2Br}$ forms.
Answer: (D). | ||||
| 44 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Neopentyl alcohol $\mathrm{(CH_3)_3CCH_2OH}$ is protonated and loses water; the primary cation rearranges at once by a 1,2-methyl shift to the tertiary cation $\mathrm{(CH_3)_2\overset{+}{C}CH_2CH_3}$.
Loss of a proton from this cation: from the $\mathrm{CH_2}$ gives the more substituted 2-methylbut-2-ene $\mathrm{(CH_3)_2C=CHCH_3}$ (major, A, Saytzeff); from a $\mathrm{CH_3}$ gives 2-methylbut-1-ene $\mathrm{CH_2=C(CH_3)CH_2CH_3}$ (minor, B).
3-Methylbut-1-ene cannot come from this cation.
Answer: (B). | ||||
| 45 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines $\mathrm{CHCl_3}$ with alcoholic KOH is the carbylamine (isocyanide) test, given only by primary amino groups ($\mathrm{-NH_2}$).
Adenine has a primary $\mathrm{-NH_2}$ on its purine ring; lysine has two primary amino groups. Proline is a secondary amine (ring NH), and thymine has no primary amino group (only ring NH and C=O).
So adenine and lysine react.
Answer: (C). | ||||
| 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 $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$.
$\mathrm{N_2}$: $\frac{2800}{28} = 100$ mol; $\mathrm{H_2}$: $\frac{1000}{2} = 500$ mol. 100 mol $\mathrm{N_2}$ needs only 300 mol $\mathrm{H_2}$, so $\mathrm{N_2}$ is limiting.
$\mathrm{NH_3}$ formed: 200 mol $\times$ 17 g/mol = 3400 g.
Answer: 3400. | ||||
| 47 | SolutionsRaoult's law | Class 12: Solutions | NumericalMedium | |
A medium numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions Treat the mixture as ideal (Raoult's law): $P = x_{hex}P^\circ_{hex} + x_{hep}P^\circ_{hep}$.
First solution: $550 = \frac{1}{4}P^\circ_{hex} + \frac{3}{4}P^\circ_{hep}$, i.e. $2200 = P^\circ_{hex} + 3P^\circ_{hep}$.
After adding 1 mol heptane: $560 = \frac{1}{5}P^\circ_{hex} + \frac{4}{5}P^\circ_{hep}$, i.e. $2800 = P^\circ_{hex} + 4P^\circ_{hep}$.
Subtracting: $P^\circ_{hep} = 600$ mm Hg (and $P^\circ_{hex} = 400$ mm Hg).
Answer: 600. | ||||
| 48 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics First order: $kt = 2.303\log\frac{[A]_0}{[A]}$.
75% complete: $k(90) = 2.303\log4 = 2.303(0.60)$.
60% complete: $kt = 2.303\log\frac{100}{40} = 2.303\log2.5 = 2.303(0.40)$.
Dividing: $\frac{t}{90} = \frac{0.40}{0.60}$, so $t = 60$ minutes.
Answer: 60. | ||||
| 49 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions $\mathrm{2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O}$.
$\mathrm{KMnO_4}$: $\frac{0.316}{158} = 2\times10^{-3}$ mol, so $\mathrm{H_2O_2} = \frac{5}{2}\times2\times10^{-3} = 5\times10^{-3}$ mol $= 5\times10^{-3}\times34 = 0.17$ g.
Purity $= \frac{0.17}{0.2}\times100 = 85\%$.
Answer: 85. | ||||
| 50 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalHard | |
A hard numerical question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Starting nitrile: 2-(3-methylcyclohexyl)propanenitrile, $\mathrm{R{-}CH(CH_3){-}C{\equiv}N}$ with R = 3-methylcyclohexyl.
$\mathrm{C_2H_5MgBr}$ then $\mathrm{H_3O^+}$ converts the nitrile into a ketone: [A] = $\mathrm{R{-}CH(CH_3){-}CO{-}C_2H_5}$.
$\mathrm{CH_3MgBr}$ then water gives the tertiary alcohol [B] = $\mathrm{R{-}CH(CH_3){-}C(OH)(CH_3)(C_2H_5)}$.
Chiral centres in [B]:
1. The carbinol carbon: OH, $\mathrm{CH_3}$, $\mathrm{C_2H_5}$ and the $\mathrm{CH(CH_3)R}$ group, all different.
2. The side-chain CH: H, $\mathrm{CH_3}$, the ring and the carbinol group.
3. Ring C1 (bearing the side chain): H, the side chain, and two ring paths that differ because of the 3-methyl group.
4. Ring C3 (bearing $\mathrm{CH_3}$): H, $\mathrm{CH_3}$ and two different ring paths.
Total: 4.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $n(A\cup B) = 63 + 76 - x \le 100$, so $x \ge 39$. Also $x \le n(A) = 63$.
So $39 \le x \le 63$. Of the options only 55 lies in this range.
Answer: (B). | ||||
| 52 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $[x+2] = [x] + 2$. With $n = [x]$: $n^2 + 2n + 4 - 7 = 0$, i.e. $n^2 + 2n - 3 = 0$, so $n = 1$ or $n = -3$.
$[x] = 1$ gives every $x \in [1, 2)$ and $[x] = -3$ every $x \in [-3, -2)$: infinitely many solutions (of which only 1 and $-3$ are integers).
Answer: (D). | ||||
| 53 | 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 $u = \frac{2x + i(2y+1)}{x + i(y-k)} = \frac{[2x + i(2y+1)][x - i(y-k)]}{x^2 + (y-k)^2}$.
Re: $\frac{2x^2 + (2y+1)(y-k)}{x^2+(y-k)^2}$; Im: $\frac{x(2y+1) - 2x(y-k)}{x^2+(y-k)^2} = \frac{x(1+2k)}{x^2+(y-k)^2}$.
The curve Re(u) + Im(u) = 1 is $2x^2 + (2y+1)(y-k) + x(1+2k) = x^2 + (y-k)^2$, i.e. $x^2 + (y-k)(y+1+k) + (1+2k)x = 0$, or $x^2 + y^2 + (1+2k)x + y - k - k^2 = 0$: a circle.
On the y-axis ($x = 0$): $(2y+1)(y-k) = (y-k)^2$, so $(y-k)(y+1+k) = 0$, giving $y = k$ and $y = -(1+k)$.
$PQ = k + (1 + k) = 2k + 1 = 5$, so $k = 2$.
Answer: (B). | ||||
| 54 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Multiplying, $\begin{bmatrix}\cos\alpha & i\sin\alpha\\ i\sin\alpha & \cos\alpha\end{bmatrix}\begin{bmatrix}\cos\beta & i\sin\beta\\ i\sin\beta & \cos\beta\end{bmatrix} = \begin{bmatrix}\cos(\alpha+\beta) & i\sin(\alpha+\beta)\\ i\sin(\alpha+\beta) & \cos(\alpha+\beta)\end{bmatrix}$, so $A^5$ has angle $5\theta = \frac{5\pi}{24}$.
$a = d = \cos\frac{5\pi}{24}$, $b = c = i\sin\frac{5\pi}{24}$.
(A) $a^2 - d^2 = 0$: true. (B) $a^2 - b^2 = \cos^2 + \sin^2 = 1 \ne \frac{1}{2}$: not true. (C) $a^2 + b^2 = \cos^2 - \sin^2 = \cos\frac{5\pi}{12} \approx 0.26$, in $[0, 1]$: true. (D) $a^2 - c^2 = 1$: true.
Answer: (B). | ||||
| 55 | 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 sum is ${}^{30}C_6 + {}^{31}C_6 + \cdots + {}^{50}C_6$.
By the hockey-stick identity $\sum_{m=6}^{N}{}^{m}C_6 = {}^{N+1}C_7$:
$\sum_{m=30}^{50}{}^{m}C_6 = {}^{51}C_7 - {}^{30}C_7$.
Answer: (C). | ||||
| 56 | 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 Let the GP be $\alpha, \alpha r, \alpha r^2, \alpha r^3$.
$\alpha + \beta = \alpha(1+r) = 3$ and $\gamma + \delta = \alpha r^2(1+r) = 6$, so $r^2 = 2$.
$p = \alpha\beta = \alpha^2 r$ and $q = \gamma\delta = \alpha^2 r^5 = p r^4 = 4p$.
$(2q + p) : (2q - p) = 9p : 7p = 9 : 7$.
Answer: (B). | ||||
| 57 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The general bracket is $1 - (2k)^2(2k-1) = 1 - 8k^3 + 4k^2$ for $k = 1, \ldots, 10$.
Sum $= 1 + 10 - 8\sum k^3 + 4\sum k^2 = 11 - 8(3025) + 4(385) = 11 - 24200 + 1540 = 11 - 22660$.
$22660 = 220 \times 103$, so the sum is $11 - 220(103)$: $(\alpha, \beta) = (11, 103)$.
Answer: (C). | ||||
| 58 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives By the mean value theorem on $f'$: $f'(5) - f'(2) = 3f''(c) \ge 12$, so $f'(5) \ge 17$.
Also $f'(x) \ge 5 + 4(x - 2)$ on $[2, 5]$, so $f(5) - f(2) \ge \int_2^5 [5 + 4(x-2)]dx = 15 + 18 = 33$, and $f(5) \ge 41$.
Hence $f(5) + f'(5) \ge 58 \ge 28$, so (B) holds; (A) and (D) are false, and (C) need not hold since $f'(5) \ge 17$ and $f''(5) \ge 4$ give $f'(5) + f''(5) \ge 21$.
Answer: (B). | ||||
| 59 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Differentiate with respect to y:
$-\sqrt2 b\sin x\,\frac{dx}{dy}(a - \sqrt2 b\cos y) + (a + \sqrt2 b\cos x)\sqrt2 b\sin y = 0$.
At $x = y = \frac{\pi}{4}$: $\sqrt2 b\sin\frac{\pi}{4} = \sqrt2 b\cos\frac{\pi}{4} = b$, so $-b(a - b)\frac{dx}{dy} + (a + b)b = 0$.
$\frac{dx}{dy} = \frac{a+b}{a-b}$.
Answer: (C). | ||||
| 60 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $[0, 4]$, $0 \le f(x) = |x-2| \le 2$, so $g(x) = |f(x) - 2| = 2 - |x - 2|$.
$g(x) - f(x) = 2 - 2|x - 2|$.
$\int_0^3 |x-2|\,dx = \int_0^2(2-x)dx + \int_2^3(x-2)dx = 2 + \frac{1}{2} = \frac{5}{2}$.
$\int_0^3(g - f)dx = 2(3) - 2\cdot\frac{5}{2} = 1$.
Answer: (C). | ||||
| 61 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Note $\frac{d}{dx}(x\sin x + \cos x) = x\cos x$. Write the integrand as $x\sec x\cdot\frac{x\cos x}{(x\sin x + \cos x)^2}$.
By parts with $u = x\sec x$ and $dv = \frac{x\cos x}{(x\sin x+\cos x)^2}dx$, $v = -\frac{1}{x\sin x + \cos x}$:
$I = -\frac{x\sec x}{x\sin x + \cos x} + \int\frac{\sec x + x\sec x\tan x}{x\sin x + \cos x}dx$.
$\sec x + x\sec x\tan x = \frac{\cos x + x\sin x}{\cos^2 x}$, so the last integral is $\int\sec^2 x\,dx = \tan x$.
$I = \tan x - \frac{x\sec x}{x\sin x + \cos x} + C$.
Answer: (D). | ||||
| 62 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f(3) - f(1) = \int_1^3\frac{\sqrt x}{(1+x)^2}dx$. Put $x = \tan^2\theta$, $dx = 2\tan\theta\sec^2\theta\,d\theta$; $\theta$ runs from $\frac{\pi}{4}$ to $\frac{\pi}{3}$.
Integrand: $\frac{\tan\theta\cdot2\tan\theta\sec^2\theta}{\sec^4\theta} = 2\sin^2\theta = 1 - \cos2\theta$.
$\int_{\pi/4}^{\pi/3}(1 - \cos2\theta)d\theta = \left[\theta - \frac{\sin2\theta}{2}\right]_{\pi/4}^{\pi/3} = \left(\frac{\pi}{3} - \frac{\sqrt3}{4}\right) - \left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt3}{4}$.
Answer: (B). | ||||
| 63 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Divide by $x^2$: $\frac{xy' - y}{x^2} = \frac{d}{dx}\left(\frac{y}{x}\right) = x\cos x + \sin x = \frac{d}{dx}(x\sin x)$.
So $\frac{y}{x} = x\sin x + C$. $y(\pi) = \pi$ gives $1 = 0 + C$, $C = 1$: $y = x^2\sin x + x$.
$y' = 2x\sin x + x^2\cos x + 1$, $y'' = 2\sin x + 4x\cos x - x^2\sin x$.
At $\frac{\pi}{2}$: $y'' = 2 - \frac{\pi^2}{4}$ and $y = \frac{\pi^2}{4} + \frac{\pi}{2}$. Sum $= 2 + \frac{\pi}{2}$.
Answer: (C). | ||||
| 64 | Straight LinesSlope, parallel and perpendicular lines | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines $\vec{AB} = (2, -1)$, $AB = \sqrt5$. AC is perpendicular to AB, so along $\pm(1, 2)$; to stay in the first quadrant take $(1, 2)$, unit vector $\frac{(1,2)}{\sqrt5}$.
Area $= \frac{1}{2}AB\cdot AC = \frac{\sqrt5}{2}AC = 5\sqrt5$, so $AC = 10$.
$C = (1, 2) + \frac{10}{\sqrt5}(1, 2) = (1 + 2\sqrt5,\ 2 + 4\sqrt5)$. (The other direction would give $1 - 2\sqrt5 < 0$.)
Abscissa: $1 + 2\sqrt5$.
Answer: (A). | ||||
| 65 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $\phi$ is greatest at $t = \frac{1}{2}$: $e = \frac{5}{12} + \frac{1}{2} - \frac{1}{4} = \frac{2}{3}$.
$b^2 = a^2(1 - e^2) = \frac{5a^2}{9}$. Latus rectum $\frac{2b^2}{a} = 10$ gives $b^2 = 5a$.
$\frac{5a^2}{9} = 5a$, so $a = 9$, $b^2 = 45$, and $a^2 + b^2 = 81 + 45 = 126$.
Answer: (B). | ||||
| 66 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Normal at $(x_1, y_1)$: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$. At P(3, 3) through (9, 0): $3a^2 = a^2 + b^2$, so $b^2 = 2a^2$.
P on the curve: $\frac{9}{a^2} - \frac{9}{2a^2} = 1$, so $a^2 = \frac{9}{2}$.
$e^2 = 1 + \frac{b^2}{a^2} = 3$. $(a^2, e^2) = \left(\frac{9}{2}, 3\right)$.
Answer: (D). | ||||
| 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 $f(x) = \begin{vmatrix}x & -2 & 3\\ -2 & x & -1\\ 7 & -2 & x\end{vmatrix} = x(x^2 - 2) + 2(-2x + 7) + 3(4 - 7x) = x^3 - 27x + 26$.
$f'(x) = 3x^2 - 27 = 0$ at $x = \pm3$; $f''(-3) = -18 < 0$, so $x_0 = -3$.
Then $\vec a = (-3, -2, 3)$, $\vec b = (-2, -3, -1)$, $\vec c = (7, -2, -3)$.
$\vec a\cdot\vec b = 6 + 6 - 3 = 9$, $\vec b\cdot\vec c = -14 + 6 + 3 = -5$, $\vec c\cdot\vec a = -21 + 4 - 9 = -26$. Sum $= -22$.
Answer: (B). | ||||
| 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 of all eight $= 80$; the six given sum to 63, so $p + q = 17$.
$\sum x^2 = 8(13.5 + 10^2) = 908$; the six give $25 + 49 + 100 + 144 + 196 + 225 = 739$, so $p^2 + q^2 = 169$.
$2pq = 17^2 - 169 = 120$, $(p - q)^2 = 169 - 120 = 49$, so $|p - q| = 7$ (the two are 12 and 5).
Answer: (C). | ||||
| 69 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let AC = d and let P be at height h, a horizontal distance x from A.
Along BC (from B, height 15 at A, down to C): similar triangles give $\frac{h}{15} = \frac{d - x}{d}$. Along AD (from A up to D, height 10 at C): $\frac{h}{10} = \frac{x}{d}$.
Adding: $\frac{h}{15} + \frac{h}{10} = 1$, so $h = \frac{15\times10}{25} = 6$ m, whatever the distance between the poles.
Answer: (A). | ||||
| 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 $S_1$: take p = T, q = T. Then $q\vee p$ = T, while $p\leftrightarrow\sim q$ is T $\leftrightarrow$ F = F, so the implication is F. Not a tautology: $S_1$ is incorrect.
$S_2$: take p = T, q = F. Then $\sim q$ = T and $\sim p\leftrightarrow q$ is F $\leftrightarrow$ F = T, so the statement is T. Not a fallacy: $S_2$ is incorrect.
Answer: (D). | ||||
| 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 For infinitely many solutions the determinant must vanish: $\begin{vmatrix}1 & -2 & 3\\ 2 & 1 & 1\\ 1 & -7 & a\end{vmatrix} = (a + 7) + 2(2a - 1) + 3(-15) = 5a - 40 = 0$, so $a = 8$.
The third equation must then be a combination of the first two: $\lambda(1, -2, 3) + \mu(2, 1, 1) = (1, -7, 8)$ gives $\lambda + 2\mu = 1$, $-2\lambda + \mu = -7$, so $\mu = -1$, $\lambda = 3$ (and $3\cdot3 - 1 = 8$ checks).
Constants: $9\lambda + b\mu = 24$, i.e. $27 - b = 24$, $b = 3$.
$a - b = 8 - 3 = 5$.
Answer: 5. | ||||
| 72 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Let $P(x) = 2x^2 + 3x + 4$. Then $P\left(\frac{2}{x}\right) = \frac{8}{x^2} + \frac{6}{x} + 4 = \frac{2}{x^2}(4 + 3x + 2x^2) = \frac{2}{x^2}P(x)$.
So $P(x)^{10} = \frac{x^{20}}{2^{10}}P\left(\frac{2}{x}\right)^{10}$, i.e. $\sum a_r x^r = \sum_r a_r 2^{r-10}x^{20-r}$.
Comparing the coefficient of $x^{20-r}$: $a_{20-r} = 2^{r-10}a_r$. With $r = 7$: $a_{13} = 2^{-3}a_7$.
$\frac{a_7}{a_{13}} = 8$.
Answer: 8. | ||||
| 73 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f'(x) = \lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to0}\frac{f(h) + xh^2 + x^2h}{h} = 1 + 0 + x^2$.
$f'(3) = 1 + 9 = 10$.
Answer: 10. | ||||
| 74 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Planes through the line of intersection: $(x + 4y - z + 7) + \lambda(3x + y + 5z - 8) = 0$.
Matching $ax + by + 6z - 15 = 0$: the ratio of the z-coefficient to the constant, $\frac{-1 + 5\lambda}{7 - 8\lambda} = \frac{6}{-15}$, gives $15 - 75\lambda = 42 - 48\lambda$, so $\lambda = -1$.
Plane: $-2x + 3y - 6z + 15 = 0$, i.e. $2x - 3y + 6z = 15$ ($a = 2$, $b = -3$).
Distance of (3, 2, $-1$): $\frac{|6 - 6 - 6 - 15|}{\sqrt{4 + 9 + 36}} = \frac{21}{7} = 3$.
Answer: 3. | ||||
| 75 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Taking the shots as independent, $P(\text{at least one hit in } n) = 1 - (0.9)^n > \frac{1}{4}$, i.e. $(0.9)^n < 0.75$.
$(0.9)^2 = 0.81$, $(0.9)^3 = 0.729$. Least $n = 3$.
Answer: 3. | ||||
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