JEE (Main) · 2021
JEE (Main) 2021 — 4 August, 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
30 Q
- Oscillations and Waves3
- Atoms and Nuclei3
- Magnetic Effects of Current and Magnetism3
19 easy · 11 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques2
- Hydrocarbons2
- Amines2
22 easy · 7 medium · 1 hard
Mathematics
30 Q
- Applications of Derivatives4
- Matrices and Determinants3
- Trigonometry3
15 easy · 13 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 | 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 Frequency: $f = \frac{v}{\lambda} = \frac{300}{0.75\times10^{-2}} = 4\times10^4$ Hz $= 40$ kHz.
The audible range is about 20 Hz to 20 kHz, so a 40 kHz wave is ultrasonic and therefore not audible. Both statements (B) and (C) hold, which is option (D).
Answer: (D). | ||||
| 2 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The loss of weight is the buoyant force: $V\rho_w g = 196$, so $V = \frac{196}{1000\times9.8} = 0.02\ \mathrm{m^3}$.
Weight in air: $W = V\rho g = 0.02\times7500\times9.8 = 1470$ N.
Answer: (D). | ||||
| 3 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $E_n = -\frac{13.6}{n^2}$ eV, so $\Delta E_n = 13.6\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) = 13.6\,\frac{2n+1}{n^2(n+1)^2}$.
$\left|\frac{\Delta E_n}{E_n}\right| = \frac{2n+1}{(n+1)^2} \approx \frac{2}{n}$ for large $n$.
So the ratio varies as $\frac{1}{n}$.
Answer: (B). | ||||
| 4 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\alpha$ decay: $Z - 2$, $A - 4$: ${}^{180}_{72} \to {}^{176}_{70}$.
$\beta^-$ decay: $Z + 1$: ${}^{176}_{71}$.
$\alpha$ decay: ${}^{172}_{69}$.
$\gamma$ decay changes neither: $\mathrm{A_4}$ is ${}^{172}_{69}$, atomic number 69 and mass number 172.
Answer: (A). | ||||
| 5 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei First excited state: $n = 2$. Bohr radius and speed: $r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m_e e^2}$, $v_n = \frac{e^2}{2\varepsilon_0 h n}$.
For $n = 2$: $r = \frac{4\varepsilon_0 h^2}{\pi m_e e^2}$, $v = \frac{e^2}{4\varepsilon_0 h}$.
The orbiting electron is a current $I = \frac{ev}{2\pi r}$, and the field at the centre is $B = \frac{\mu_0 I}{2r} = \frac{\mu_0 e v}{4\pi r^2}$.
$r^2 = \frac{16\varepsilon_0^2 h^4}{\pi^2 m_e^2 e^4}$, so $B = \frac{\mu_0 e}{4\pi}\cdot\frac{e^2}{4\varepsilon_0 h}\cdot\frac{\pi^2 m_e^2 e^4}{16\varepsilon_0^2 h^4} = \frac{\mu_0 e^7 \pi m_e^2}{256\varepsilon_0^3 h^5}$.
Answer: (C). | ||||
| 6 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The transistors $T_1$ and $T_2$ are in series between the output point y and earth, with the load resistor to $+V_{CC}$.
y is pulled low only when both transistors conduct, that is when A = 1 and B = 1. For any other input at least one transistor is off, no current flows through the load resistor and y stays high at $V_{CC}$.
Output is 0 only for A = B = 1: this is a NAND gate.
Answer: (B). | ||||
| 7 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a prism of angle $A = 60^\circ$: $\mu = \frac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}}$.
In air: $\sin\frac{60^\circ+\delta_1}{2} = 1.6\times\frac{1}{2} = 0.8 = \frac{4}{5}$, so $\frac{60^\circ+\delta_1}{2} = 53^\circ$ and $\delta_1 = 46^\circ$.
In the liquid the relative index is $\frac{1.6}{1.2} = \frac{4}{3}$: $\sin\frac{60^\circ+\delta_2}{2} = \frac{4}{3}\times\frac{1}{2} = \frac{2}{3}$, so $\frac{60^\circ+\delta_2}{2} = 41^\circ$ and $\delta_2 = 22^\circ$.
Change: $46^\circ - 22^\circ = 24^\circ$.
Answer: (D). | ||||
| 8 | Communication SystemsRange of line-of-sight transmission | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on range of line-of-sight transmission, from Communication Systems. NCERT: No matching NCERT chapter Range of a tower of height h: $d = \sqrt{2Rh} = \sqrt{2\times6.4\times10^6\times250} = \sqrt{3.2\times10^9} \approx 5.66\times10^4$ m.
That is about 56 km.
Answer: (B). | ||||
| 9 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Uniform circular motion needs a force of constant magnitude always perpendicular to the velocity.
A constant magnetic field alone gives $\vec{F} = q\vec{v}\times\vec{B}$, perpendicular to $\vec{v}$ and of constant magnitude: a charge moving perpendicular to $\vec{B}$ goes round a circle.
A constant electric field gives a constant force in a fixed direction, which cannot keep turning the particle: with $E \ne 0$ the path is a parabola (B = 0) or a drifting cycloid-like path (B $\ne$ 0), not a circle. With no field the particle moves in a straight line.
So only $E = 0$, $B \ne 0$ is possible.
Answer: (D). | ||||
| 10 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Field at the centre of the coil: $B = \frac{\mu_0 I}{2R} = \frac{4\pi\times10^{-7}\times2}{2\times0.04} = \pi\times10^{-5}$ T, perpendicular to the plane of the coil.
The rod moves in the plane of the coil, perpendicular to its own length, so the motional emf is $\varepsilon = Bvl = \pi\times10^{-5}\times5\times0.1 = 5\pi\times10^{-6}$ V.
$5\pi\times10^{-6}$ V $= \frac{\pi}{200}$ mV.
Answer: (B). | ||||
| 11 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Mass of the block: $m = \ell b h\, d\rho_w$. Pushed down by $x$, the extra buoyant force is $\ell b x\rho_w g$, a restoring force $F = -(\ell b\rho_w g)x$.
$\omega = \sqrt{\frac{\ell b\rho_w g}{\ell b h d\rho_w}} = \sqrt{\frac{g}{dh}}$.
Answer: (A). | ||||
| 12 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At height $\frac{R_e}{3}$: $g_h = \frac{g}{(1 + 1/3)^2} = \frac{9}{16}g$.
At depth d: $g_d = g\left(1 - \frac{d}{R_e}\right)$.
Equal forces: $1 - \frac{d}{R_e} = \frac{9}{16}$, so $d = \frac{7}{16}R_e$.
Answer: (C). | ||||
| 13 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Treat the remaining disc as the whole disc (mass $\propto R^2$, centre at the origin) minus the hole (mass $\propto r^2$, centre at distance a).
$x_{cm} = \frac{R^2\cdot0 - r^2 a}{R^2 - r^2} = -\frac{ar^2}{R^2 - r^2}$.
The centre of mass lies at a distance $\frac{ar^2}{R^2 - r^2}$ from the origin, on the side opposite the hole.
Answer: (A). | ||||
| 14 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The sides parallel to the y axis cut the field. With the left side at x and the right side at $x + \ell$, the motional emfs are $B(x)\ell v_0$ and $B(x+\ell)\ell v_0$, opposing each other round the loop (the sides along x give none).
Net emf: $\ell v_0\left[B(x+\ell) - B(x)\right] = \ell v_0 B_0\frac{\ell}{a} = v_0 B_0\frac{\ell^2}{a}$.
(Equivalently $\frac{d\Phi}{dt}$ with $\Phi = \int_x^{x+\ell}B_0\left(2+\frac{x'}{a}\right)\ell\,dx'$ gives the same.)
Answer: (C). | ||||
| 15 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity In the figure the 4 V cell (branch EF) has its longer, positive plate towards F (bottom), and the 6 V cell (branch AD) has its positive plate towards A (top).
Take $V_C = 0$ (C, D, F joined) and $V_B = V$ (A, B, E joined).
Current up through the 6 V cell into B: $\frac{6 - V}{2}$. Current from B down through the 4 V cell (negative terminal at E to positive at F): $\frac{V + 4}{1}$. Current from B to C through $5\Omega$: $\frac{V}{5}$.
Junction B: $\frac{6 - V}{2} = (V + 4) + \frac{V}{5}$; multiplying by 10: $30 - 5V = 10V + 40 + 2V$, so $V = -\frac{10}{17}$ V.
Current through $5\Omega$ from B to C is $\frac{V}{5} = -\frac{2}{17}$ A, that is $\frac{2}{17}$ A from C to B.
Answer: (C). | ||||
| 16 | Properties of Bulk MatterElasticity | 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 elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The wire would contract by a strain $\alpha\Delta T$ but is held fixed, so the thermal tension is $T = YA\alpha\Delta T$.
$T = 2\times10^{11}\times10^{-6}\times11\times10^{-6}\times10 = 22$ N.
Answer: (A). | ||||
| 17 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The particle moves along the line $y = x$ with displacement $A\sin314t + B\cos314t$, a single SHM of angular frequency $\omega = 314$ rad/s.
$f = \frac{\omega}{2\pi} = \frac{314}{2\times3.14} = 50$ Hz.
Answer: (B). | ||||
| 18 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter First instrument: vernier zero to the right of the main-scale zero, a positive zero error. With the $8^{th}$ vernier division coinciding, the error is $8\times\mathrm{LC} = 8\times0.01 = 0.08$ cm.
Second instrument: vernier zero to the left, a negative zero error, with the $4^{th}$ division coinciding. Counted the same way the paper intends, its size is $4\times0.01 = 0.04$ cm.
So the positive and negative zero errors are 0.08 cm and 0.04 cm.
Answer: (C). | ||||
| 19 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Position of the point of action relative to the reference point: $\vec{r} = (-1-3)\hat{i} + (2-1)\hat{j} + (1-4)\hat{k} = -4\hat{i} + \hat{j} - 3\hat{k}$.
$\vec{\tau} = \vec{r}\times\vec{F} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-4&1&-3\\2&1&2\end{vmatrix} = (2+3)\hat{i} - (-8+6)\hat{j} + (-4-2)\hat{k} = 5\hat{i} + 2\hat{j} - 6\hat{k}$.
Answer: (C). | ||||
| 20 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium 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 S is 5 m from the wall. Let the beam make angle $\theta$ with the perpendicular SO to the wall; the spot is at $y = 5\tan\theta$ along the wall.
$v = \frac{dy}{dt} = 5\sec^2\theta\,\frac{d\theta}{dt}$.
At P the beam makes $30^\circ$ with the wall, so $\theta = 60^\circ$ and $\sec^2\theta = 4$.
$v = 5\times4\times0.1 = 2$ m/s.
Answer: (C). | ||||
| 21 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Wire of radius r: $R_1 = \frac{\rho L}{\pi r^2}$.
The coating runs from r to 2r: area $\pi(4r^2 - r^2) = 3\pi r^2$, resistivity $2\rho$: $R_2 = \frac{2\rho L}{3\pi r^2} = \frac{2}{3}R_1$.
They are in parallel: $R = \frac{R_1\cdot\frac{2}{3}R_1}{R_1 + \frac{2}{3}R_1} = \frac{2}{5}R_1$.
The resistance falls from $R_1$ to $0.4R_1$, a decrease of 60%.
Answer: 60. | ||||
| 22 | Laws of MotionNewton's laws | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion In equilibrium the spring holds both blocks: spring force $= 3mg$ upward on 2m.
Just after the string is cut the spring force is unchanged (a spring cannot change its extension instantly), and the string tension vanishes.
On 2m: $3mg - 2mg = 2m\,a$, so $a = \frac{g}{2} = 5\ \mathrm{m/s^2}$ (upward).
Answer: 5. | ||||
| 23 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The drop starts at the surface, $r_1 = 0.2$ m from the centre, and ends at $r_2 = 0.5$ m.
Electric energy released (like charges repel): $kQq\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = 9\times10^9\times2.5\times10^{-6}\times10^{-9}\times(5 - 2) = 6.75\times10^{-5}$ J.
Gravity: $mgh = 1.8\times10^{-5}\times10\times0.3 = 5.4\times10^{-5}$ J.
$\frac{1}{2}mv^2 = 12.15\times10^{-5}$ J, so $v^2 = \frac{2\times12.15\times10^{-5}}{1.8\times10^{-5}} = 13.5 = \frac{27}{2}$.
$\frac{x^3}{2} = \frac{27}{2}$ gives $x = 3$.
Answer: 3. | ||||
| 24 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rolling from height h: $v^2 = \frac{2gh}{1 + k^2/R^2}$.
Solid cylinder, $\frac{k^2}{R^2} = \frac{1}{2}$: $v_c^2 = \frac{4gh}{3}$. Spherical shell, $\frac{k^2}{R^2} = \frac{2}{3}$: $v_s^2 = \frac{6gh}{5}$.
Equal radii, so $\frac{\omega_s}{\omega_c} = \frac{v_s}{v_c} = \sqrt{\frac{6/5}{4/3}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.
Answer: 10. | ||||
| 25 | ThermodynamicsFirst law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat absorbed: $Q = 1\times2256 = 2256$ J.
Work done by the system: $W = P\Delta V = 10^5\times(1671 - 1)\times10^{-6} = 167$ J.
$\Delta U = Q - W = 2256 - 167 = 2089$ J.
Answer: 2089. | ||||
| 26 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | NumericalEasy | |
A easy numerical question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$ and $v_{rms} = \sqrt{\frac{3RT}{M}}$.
$\frac{v_{avg}}{v_{rms}} = \sqrt{\frac{8}{3\pi}}$, so $x = 3$.
Answer: 3. | ||||
| 27 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Magnesium: Z = 12, molar mass 24 g/mol.
Moles in 0.5 g: $\frac{0.5}{24}$; atoms: $\frac{0.5}{24}\times6.02\times10^{23} = 1.254\times10^{22}$.
Protons: $12\times1.254\times10^{22} = 1.505\times10^{23}$.
Charge: $1.505\times10^{23}\times1.6\times10^{-19} \approx 2.408\times10^4$ C $= 24.08$ kC, about 24 kC.
Answer: 24.08 (24 to the nearest integer). | ||||
| 28 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Field between the plates: $E = \frac{V}{d} = \frac{800}{5\times10^{-3}} = 1.6\times10^5$ V/m.
Undeflected: $eE = evB$, so $B = \frac{E}{v} = \frac{1.6\times10^5}{2\times10^6} = 0.08$ T $= 8\times10^{-2}$ T.
Answer: 8. | ||||
| 29 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Magnetisation: $M = \frac{m}{V} = \frac{5}{25\times10^{-6}} = 2\times10^5$ A/m.
$B = \mu_0(H + M) = 4\pi\times10^{-7}\times(0.5\times10^5 + 2\times10^5) = 4\pi\times10^{-7}\times2.5\times10^5 = 0.1\pi$ T.
$0.1\pi$ T $= 314.16$ mT, about 314 mT.
Answer: 314.16 (314 to the nearest integer). | ||||
| 30 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics First surface (R = +50 cm, object at infinity): $\frac{1.5}{v_1} - \frac{1}{\infty} = \frac{1.5 - 1}{50}$, so $v_1 = 150$ cm from the first surface.
This is 150 - 100 = 50 cm beyond the second surface: a virtual object, $u = +50$ cm, for the second surface (R = -50 cm).
$\frac{1}{v} - \frac{1.5}{50} = \frac{1 - 1.5}{-50} = \frac{1}{100}$, so $\frac{1}{v} = \frac{1}{100} + \frac{3}{100} = \frac{1}{25}$ and $v = 25$ cm.
The image is 25 cm beyond the far (outer) edge of the sphere.
Answer: 25. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 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 Frenkel defect: the smaller cation leaves its site for an interstitial one; ZnS shows it (A→III).
Schottky defect: equal numbers of cation and anion vacancies in ionic solids of similar-sized ions such as CsCl (B→I).
Impurity defect: $\mathrm{Cd^{2+}}$ replacing two $\mathrm{Ag^+}$ in AgCl leaves a cation vacancy (C→IV).
Metal excess defect: LiCl heated in Li vapour has electrons in anion vacancies (F-centres), giving the pink colour (D→II).
Answer: (C). | ||||
| 32 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{B_2}$ (10 electrons): $\ldots\pi2p_x^1\pi2p_y^1$, two unpaired electrons, paramagnetic.
$\mathrm{C_2}$ (12 electrons): $\ldots\pi2p_x^2\pi2p_y^2$, all paired, diamagnetic.
$\mathrm{S_2}$ is valence-isoelectronic with $\mathrm{O_2}$: two unpaired electrons in the $\pi^*$ orbitals, paramagnetic.
NO (15 electrons): one unpaired electron in $\pi^*$, paramagnetic.
$\mathrm{O_2^{2-}}$ (18 electrons): $\pi^*$ orbitals full, diamagnetic.
Paramagnetic set: $\mathrm{B_2}$, $\mathrm{S_2}$, NO.
Answer: (C). | ||||
| 33 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let the partial pressure of A fall by p. Then B rises by 2p and C by p, so the total pressure is $P = P_0 - p + 2p + p = P_0 + 2p$ and $\frac{dP}{dt} = 2\frac{dp}{dt}$.
For an ideal gas $[A] = \frac{p_A}{RT}$, so the rate $= -\frac{d[A]}{dt} = \frac{1}{RT}\frac{dp}{dt} = \frac{1}{2RT}\frac{dP}{dt}$.
Answer: (C). | ||||
| 34 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich: $\frac{x}{m} = kP^{1/n}$, so $\log\frac{x}{m} = \log k + \frac{1}{n}\log P$, a straight line of slope $\frac{1}{n}$.
The figure shows a horizontal line: slope zero, so $\frac{1}{n} = 0$ and $\frac{x}{m} = k$, independent of pressure (the high-pressure limit).
Answer: (B). | ||||
| 35 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across period 2 ionization enthalpy rises, with two exceptions: Be ($2s^2$, filled) is above B ($2p^1$), and N ($2p^3$, half-filled) is above O ($2p^4$).
Values (kJ/mol): N 1402, O 1314, C 1086, Be 899, B 801.
Order: N > O > C > Be > B.
Answer: (C). | ||||
| 36 | Isolation of MetalsRoasting and calcination | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on roasting and calcination, from Isolation of Metals. NCERT: No matching NCERT chapter Siderite is $\mathrm{FeCO_3}$, a carbonate: heating in limited air (calcination) gives the oxide, $\mathrm{FeCO_3 \rightarrow FeO + CO_2}$.
Copper glance is $\mathrm{Cu_2S}$, a sulphide: heating in excess air (roasting) gives the oxide, $\mathrm{2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2}$.
Answer: (C). | ||||
| 37 | 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 heavier isotopes have lower zero-point vibrational energy, so their molecules sit deeper in the potential well and need more energy to dissociate.
Enthalpies of bond dissociation (kJ/mol): $\mathrm{H_2}$ 435.9, $\mathrm{D_2}$ 443.4, $\mathrm{T_2}$ 446.9.
Order: $\mathrm{T_2 > D_2 > H_2}$.
Answer: (B). | ||||
| 38 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter (A) False: superoxides such as $\mathrm{KO_2}$ contain the $\mathrm{O_2^-}$ ion with an unpaired electron and are paramagnetic.
(B) False: $\mathrm{KO_2}$ is paramagnetic but coloured (yellow to orange), not white.
(C) True: the large superoxide ion is stabilised by a large cation; superoxide stability rises down the group, and $\mathrm{KO_2}$ is stable while $\mathrm{LiO_2}$ is not formed under ordinary conditions.
(D) Not the intended statement: the small $\mathrm{Li^+}$ fits the large $\mathrm{O_2^-}$ poorly, and $\mathrm{LiO_2}$ is not a stable superoxide; the statement is taken as false.
Answer: (C). | ||||
| 39 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2}$: boric acid is made from diborane.
$\mathrm{2NaH + B_2H_6 \rightarrow 2NaBH_4}$: sodium borohydride is made from diborane.
Borax can be reached from diborane: the boric acid it gives is neutralised to borax, $\mathrm{4H_3BO_3 + 2NaOH \rightarrow Na_2B_4O_7 + 7H_2O}$.
Methyl derivatives of diborane are made by replacing terminal hydrogens, and methylation stops at tetramethyldiborane; $\mathrm{B_2(CH_3)_6}$ is not a known compound (trimethylborane is monomeric, $\mathrm{B(CH_3)_3}$), so it cannot be prepared from diborane.
Answer: (D). | ||||
| 40 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Gd (64): $[\mathrm{Xe}]4f^7 5d^1 6s^2$, the half-filled f shell kept (A→IV).
Ho (67): $[\mathrm{Xe}]4f^{11} 6s^2$ (B→III).
Np (93): $[\mathrm{Rn}]5f^4 6d^1 7s^2$ (C→II).
Cm (96): $[\mathrm{Rn}]5f^7 6d^1 7s^2$ (D→I).
Answer: (B). | ||||
| 41 | Environmental ChemistryOzone depletion | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ozone depletion, from Environmental Chemistry. NCERT: No matching NCERT chapter UV light breaks the C–Cl bond: $\mathrm{CF_2Cl_2 \xrightarrow{h\nu} {}^{\bullet}CF_2Cl + {}^{\bullet}Cl}$.
The chlorine radical attacks ozone, $\mathrm{{}^{\bullet}Cl + O_3 \rightarrow ClO^{\bullet} + O_2}$, and is regenerated by $\mathrm{ClO^{\bullet} + O \rightarrow {}^{\bullet}Cl + O_2}$; the radical released from the CFC is $\mathrm{{}^{\bullet}Cl}$.
Answer: (B). | ||||
| 42 | 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 the Carius method a known mass of the compound is heated with fuming nitric acid and silver nitrate in a sealed (Carius) tube. Carbon and hydrogen are oxidised and the halogen is precipitated as silver halide, which is weighed.
Answer: (B). | ||||
| 43 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons But-2-yne is reduced to trans-but-2-ene (the two methyl groups on opposite sides).
Sodium in liquid ammonia (Birch-type dissolving-metal reduction) adds the hydrogens anti and gives the trans alkene.
$\mathrm{H_2}$ with poisoned Pd (Lindlar-type, quinoline) gives the cis alkene; $\mathrm{Pt/H_2}$ reduces all the way to butane; $\mathrm{LiAlH_4}$ does not reduce an isolated C≡C.
Answer: (B). | ||||
| 44 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Replacing the $\alpha$-hydrogen of a carboxylic acid by Cl or Br is the Hell–Volhard–Zelinsky reaction: $\mathrm{X_2}$ with red phosphorus, then water.
$\mathrm{SOCl_2}$ gives the acid chloride; NBS brominates allylic/benzylic positions; halogen at 400–500°C gives free-radical substitution without $\alpha$-selectivity.
Answer: (C). | ||||
| 45 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Hot alkaline $\mathrm{KMnO_4}$ oxidises any alkyl side chain bearing a benzylic hydrogen down to $-\mathrm{COOH}$, whatever its length.
Ethylbenzene gives potassium benzoate, and acidification gives benzoic acid, $\mathrm{C_6H_5COOH}$.
Answer: (A). | ||||
| 46 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers List II: I benzaldehyde, II salicylic acid (2-hydroxybenzoic acid), III anisole (methoxybenzene), IV salicylaldehyde (2-hydroxybenzaldehyde).
Kolbe's reaction: sodium phenoxide + $\mathrm{CO_2}$, then acid, gives salicylic acid (A→II).
Gattermann–Koch reaction: benzene + CO + HCl (anhydrous $\mathrm{AlCl_3}$/CuCl) gives benzaldehyde (B→I).
Reimer–Tiemann reaction: phenol + $\mathrm{CHCl_3}$/NaOH gives salicylaldehyde (C→IV).
Williamson synthesis: sodium phenoxide + $\mathrm{CH_3I}$ gives anisole (D→III).
Answer: (C). | ||||
| 47 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Statement I: benzenediazonium chloride is a colourless crystalline solid, so the statement is incorrect.
Statement II: benzenediazonium fluoroborate is insoluble in water and stable at room temperature; 'water soluble' makes the statement incorrect.
Both are incorrect.
Answer: (B). | ||||
| 48 | BiomoleculesProteins | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The biuret test detects peptide (–CO–NH–) linkages, which give a violet colour with alkaline copper sulphate.
Aspartame is the methyl ester of the dipeptide aspartyl-phenylalanine and contains a peptide bond. Saccharin (a cyclic sulphonimide), sucralose (a chlorinated sugar) and sucrose (a sugar) have no peptide bond.
Answer: (A). | ||||
| 49 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules In the pyranose ring the ring oxygen links C-1 and C-5.
C-1, the anomeric carbon, carries the ring oxygen and the OH group: two oxygens.
C-5 carries the ring oxygen and the $\mathrm{CH_2OH}$ carbon (C-6): one oxygen. C-2 and C-3 carry one OH each.
Answer: (A). | ||||
| 50 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Flame colours: copper, green with a blue centre (A→III); strontium, crimson red (B→IV); barium, apple green (C→II); calcium, brick red (D→I).
Answer: (D). | ||||
| 51 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions 1 L of solution weighs 1250 g and holds $0.30\times1250 = 375$ g of $\mathrm{H_2SO_4}$.
Moles: $\frac{375}{98} = 3.83$, so the molarity is 3.83 M; nearest integer 4.
Answer: 4. | ||||
| 52 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $v = 80$ km/h $= \frac{80\times1000}{3600} = 22.22$ m/s; momentum $p = 0.057\times22.22 = 1.267$ kg m/s.
$\lambda = \frac{h}{p} = \frac{6.6\times10^{-34}}{1.267} = 5.21\times10^{-34}$ m; nearest integer 5.
Answer: 5. | ||||
| 53 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Kinetic energy of the bullet: $\frac{1}{2}\times0.018\times500^2 = 2250$ J $= \frac{2250}{4.184} = 537.8$ cal.
Heat capacity of block + bullet: $1\times0.5 + 0.018\times0.03 = 0.50054$ kcal/K $= 500.54$ cal/K.
$\Delta T = \frac{537.8}{500.54} = 1.07$ K (the block's momentum after impact is negligible: with it the energy turned to heat is almost unchanged, 2210 J, $\Delta T = 1.06$ K).
Final temperature $\approx 26.07$°C; nearest integer 26.
Answer: 26. | ||||
| 54 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions 1.0 L of water is 1000 g = 1 kg, so the molality is 1 mol/kg.
$\Delta T_b = K_b m = 1\times K_b$, so $x = 1$.
Answer: 1. | ||||
| 55 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium With solubility s: $[\mathrm{Ca^{2+}}] = 0.2 + s \approx 0.2$ M and $[\mathrm{SO_4^{2-}}] = s$.
$K_{sp} = 0.2\,s = 9.0\times10^{-6}$, so $s = 4.5\times10^{-5}$ M $= 45\times10^{-6}$ M.
Answer: 45. | ||||
| 56 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Dichromate is reduced (cathode) and iodide oxidised (anode): $E^0_{cell} = E^0_{\mathrm{Cr_2O_7^{2-}/Cr^{3+}}} - E^0_{\mathrm{I_2/I^-}}$.
$0.79 = 1.33 - E^0_{\mathrm{I_2/I^-}}$, so $E^0_{\mathrm{I_2/I^-}} = 0.54$ V $= 54\times10^{-2}$ V.
Answer: 54. | ||||
| 57 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions A: $\left[(\mathrm{H_2O})_4(\mathrm{CN})\mathrm{Fe}(\mu\text{-OH})_2\mathrm{Fe}(\mathrm{CN})(\mathrm{H_2O})_4\right]^{2+}$: $2x + 2(-1)_{\mathrm{CN}} + 2(-1)_{\mathrm{OH}} = +2$, so $x = +3$.
B: $3x + (-2)_{\mathrm{O}} + 6(-1)_{\mathrm{CH_3COO}} = +1$, so $3x = 9$ and $x = +3$.
C: $x + (-1)_{\mathrm{CN}} = +1$ (CO, $\mathrm{NH_3}$, $\mathrm{H_2O}$ neutral), so $x = +2$.
Two species (A and B) have Fe in +3.
Answer: 2. | ||||
| 58 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Both contain $\mathrm{Ni^{2+}}$, $3d^8$.
$\left[\mathrm{NiCl_4}\right]^{2-}$: $\mathrm{Cl^-}$ is a weak-field ligand, $sp^3$ tetrahedral, two unpaired electrons.
$\left[\mathrm{Ni(CN)_4}\right]^{2-}$: $\mathrm{CN^-}$ is strong-field, the electrons pair up, $dsp^2$ square planar, no unpaired electrons.
Sum: 2 + 0 = 2.
Answer: 2. | ||||
| 59 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques All four are methylcyclohexanes with an exocyclic double bond at ring carbon C-1 (C-2 at the top, numbering anticlockwise).
Compounds 1 and 2 have $=\mathrm{CH_2}$: one end of the double bond carries two identical H atoms, so no geometrical isomerism about it.
Compound 3 has $=\mathrm{CH{-}CH_3}$ but methyls at C-2, C-3, C-5 and C-6: the two ring paths from C-1 (C-2→C-3 and C-6→C-5) are identical, so C-1 carries two identical groups and there is no cis/trans pair about the double bond.
Compound 4 has $=\mathrm{CH{-}CH_3}$ with methyls at C-2, C-5 and C-6: the paths C-2 (Me)→C-3 (H) and C-6 (Me)→C-5 (Me) differ, so C-1 carries two different groups and the $\mathrm{CH_3}$ can lie towards either side: geometrical isomers exist.
One compound (the fourth) shows geometrical isomerism about the double bond.
Answer: 1. | ||||
| 60 | AminesClassification and identification of amines | Class 12: Amines | NumericalMedium | |
A medium numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Only primary amines give the carbylamine (isocyanide) test.
Primary amines $\mathrm{C_4H_9NH_2}$: butan-1-amine, butan-2-amine, 2-methylpropan-1-amine and 2-methylpropan-2-amine.
That is 4 structural isomers.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | 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 $(3+\sqrt{5})(3-\sqrt{5}) = 4$, so $\frac{3-\sqrt{5}}{4} = \frac{1}{3+\sqrt{5}}$.
Put $t = (3+\sqrt{5})^{|x^2-4|} \ge 1$: the equation is $t + \frac{4}{t} = 6$, i.e. $t^2 - 6t + 4 = 0$, $t = 3 \pm \sqrt{5}$.
$t = 3 - \sqrt{5} < 1$ is impossible; $t = 3 + \sqrt{5}$ gives $|x^2 - 4| = 1$.
$x^2 = 5$ or $x^2 = 3$: $x = \pm\sqrt{5}, \pm\sqrt{3}$, four roots.
Answer: (B). | ||||
| 62 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|\omega| = 1$ means $|z - 1| = 3|z - 2|$: $(x-1)^2 + y^2 = 9\left[(x-2)^2 + y^2\right]$, i.e. $8(x^2 + y^2) - 34x + 35 = 0$.
With $|z| = 1$, $x^2 + y^2 = 1$: $8 - 34x + 35 = 0$, $x = \frac{43}{34} > 1$.
No point of the unit circle has $x > 1$, so the set is empty.
Answer: (A). | ||||
| 63 | 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 $A(\mathrm{adj}\,A) = |A|I$, so $k = |A|$.
$|A| = x(yz - 8) - 3(z - 8) + 2(2 - 2y) = xyz - 8x - 3z + 24 + 4 - 4y = xyz - (8x + 4y + 3z) + 28$.
$= 60 - 20 + 28 = 68$.
Answer: (C). | ||||
| 64 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The coefficient determinant is $a\begin{vmatrix}1&1&1\\1&b&b^2\\1&a&a^2\end{vmatrix} = a(b-1)(a-1)(a-b)$; it must vanish.
If $a = b$, the third equation is $a(x + ay + a^2z) = 1$, i.e. $x + ay + a^2z = \frac{1}{a}$, which must agree with the second, $x + ay + a^2z = 7$: $a = \frac{1}{7}$.
Then the first two equations are independent ($b \ne 1$), the third repeats the second, and there are infinitely many solutions.
Checks: $(2,2)$ gives $x + 2y + 4z = 7$ and $= \frac{1}{2}$, inconsistent; $(1,1)$ gives $x + y + z = 3$ and $= 7$, inconsistent; $\left(\frac{2}{7},\frac{2}{7}\right)$ gives $7$ and $\frac{7}{2}$, inconsistent.
Answer: (D). | ||||
| 65 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The sum is $\sum a_{2k}2^{2k} = \frac{f(2) + f(-2)}{2}$ with $f(x) = (1 + x + x^2)^4$.
$f(2) = 7^4 = 2401$, $f(-2) = 3^4 = 81$.
Sum $= \frac{2401 + 81}{2} = 1241$.
Answer: (C). | ||||
| 66 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctHard | |
A hard single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Pair the term $k$ (1 to 22) with the term $45 - k$ (23 to 44). Its sign is $(-1)^{45-k} = (-1)^{k-1}$, the same as that of the term $k$.
Numerator $= \sum_{k=1}^{22}(-1)^{k-1}\left[\cos k^\circ + \cos(45-k)^\circ\right] = 2\cos22.5^\circ\sum_{k=1}^{22}(-1)^{k-1}\cos(k - 22.5)^\circ$.
Denominator $= \sum_{k=1}^{22}(-1)^{k-1}\left[\sin k^\circ + \sin(45-k)^\circ\right] = 2\sin22.5^\circ\sum_{k=1}^{22}(-1)^{k-1}\cos(k - 22.5)^\circ$.
The common sum cancels: the value is $\cot22.5^\circ = \sqrt{2} + 1$.
Answer: (D). | ||||
| 67 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = 2$ the base is $8 - 4 - 16 + 13 = 1$ and the exponent's denominator is $8 - 12 + 4 = 0$: a $1^\infty$ form.
Limit $= e^{L}$, $L = \lim_{x\to2}\frac{x^3 - x^2 - 8x + 12}{x^3 - 3x^2 + 4}$.
$x^3 - x^2 - 8x + 12 = (x-2)^2(x+3)$ and $x^3 - 3x^2 + 4 = (x-2)^2(x+1)$, so $L = \frac{5}{3}$.
The limit is $e^{5/3}$.
Answer: (C). | ||||
| 68 | 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 For $x < 0$, $y = e^{x}$; at $x = -1$, $y = \frac{1}{e}$ and the slope is $\frac{1}{e}$.
Normal slope $-e$: $y - \frac{1}{e} = -e(x + 1)$; multiplying by $e$: $ey - 1 = -e^2x - e^2$.
$e^2x + ey + (e^2 - 1) = 0$.
Answer: (C). | ||||
| 69 | 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 Replace $x$ by $\pi - x$: $\mathrm{cosec}(\pi - x) = \mathrm{cosec}\,x$, so $I_1 = \int_0^\pi(\pi - x)e^{\mathrm{cosec}\,x}dx = \pi I_2 - I_1$.
$I_1 = \frac{\pi}{2}I_2$, so $I_1 : I_2 = \pi : 2$.
Answer: (A). | ||||
| 70 | 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 At $x = 1$ both limits are 1, so $y = 0$.
Leibniz rule: $y' = \frac{3x^2}{\sqrt{1+x^6}} - \frac{2x}{\sqrt{1+x^4}}$; at $x = 1$, $y' = \frac{3}{\sqrt{2}} - \frac{2}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Tangent: $y = \frac{1}{\sqrt{2}}(x - 1)$, i.e. $x - \sqrt{2}y - 1 = 0$.
Answer: (D). | ||||
| 71 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $\frac{dy}{\sqrt{4 - y^2}} = \frac{e^x dx}{\sqrt{1 - e^{2x}}}$, so $\sin^{-1}\frac{y}{2} = \sin^{-1}(e^x) + C$.
$y(0) = 1$: $\frac{\pi}{6} = \frac{\pi}{2} + C$, $C = -\frac{\pi}{3}$.
At $x = -\log_e2$, $e^x = \frac{1}{2}$: $\sin^{-1}\frac{y}{2} = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6}$, so $y = -1$.
Answer: (B). | ||||
| 72 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines Vertices: $x + y = 4$ and $x - y = 5$ meet at $\left(\frac{9}{2}, -\frac{1}{2}\right)$; $x + y = 4$ and $x - 2y + 2 = 0$ at $(2, 2)$; $x - y = 5$ and $x - 2y + 2 = 0$ at $(12, 7)$.
The line $y = 3$ crosses the triangle between the sides $x - 2y + 2 = 0$ (at $x = 4$) and $x - y = 5$ (at $x = 8$); the side $x + y = 4$ only spans $y \le 2$.
So $(k, 3)$ is interior for $4 < k < 8$.
Answer: (C). | ||||
| 73 | CirclesStandard and general form | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Touching the y-axis, the radius is $|h|$; passing through $(1,1)$: $(1 - h)^2 + (1 - k)^2 = h^2$.
$1 - 2h + (1 - k)^2 = 0$, so $h = \frac{1 + (1 - k)^2}{2}$.
As $k$ varies over $\mathbb{R}$, $h$ takes every value $\ge \frac{1}{2}$: $h \in \left[\frac{1}{2}, \infty\right)$.
Answer: (A). | ||||
| 74 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = 2x$ has $a = \frac{1}{2}$ and directrix $x = -\frac{1}{2}$.
From $P(x_1, y_1)$ on the parabola the foot of the perpendicular is $\left(-\frac{1}{2}, y_1\right)$; the mid-point is $\left(\frac{x_1 - 1/2}{2}, y_1\right) = (h, k)$.
$x_1 = 2h + \frac{1}{2}$, $y_1 = k$; $y_1^2 = 2x_1$ gives $k^2 = 4h + 1$.
Locus: $y^2 = 4x + 1$.
Answer: (B). | ||||
| 75 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections At an intersection point, with $X = x^2$, $Y = y^2$: $X + Y = 4b$ and $\frac{X}{16} + \frac{Y}{b^2} = 1$ give $X = \frac{16b}{b+4}$, $Y = \frac{4b^2}{b+4}$.
Normals: circle $(x, y)$, ellipse $\left(\frac{x}{16}, \frac{y}{b^2}\right)$. Their dot product is $\frac{X}{16} + \frac{Y}{b^2} = 1$.
$|(x,y)| = 2\sqrt{b}$ and $\left|\left(\frac{x}{16}, \frac{y}{b^2}\right)\right|^2 = \frac{X}{256} + \frac{Y}{b^4} = \frac{b^3 + 64}{16b^2(b+4)} = \frac{b^2 - 4b + 16}{16b^2}$.
$\cos\theta = \frac{1}{2\sqrt{b}\cdot\frac{\sqrt{b^2 - 4b + 16}}{4b}} = \frac{2\sqrt{b}}{\sqrt{b^2 - 4b + 16}}$.
$\cos\frac{\pi}{3} = \frac{1}{2}$: $16b = b^2 - 4b + 16$, $b^2 - 20b + 16 = 0$, $b = 10 \pm 2\sqrt{21}$.
$10 + 2\sqrt{21}$ is option (A) (and for it $X = \frac{16b}{b+4} < 16$, so the curves do meet).
Answer: (A). | ||||
| 76 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra $2\vec{a} - \vec{b} + 3\vec{c} = (2 - 4 + 3)\hat{i} + (2 + 2 - 6)\hat{j} + (2 - 3 + 3)\hat{k} = \hat{i} - 2\hat{j} + 2\hat{k}$, of magnitude 3.
A parallel vector of magnitude 6: $2(\hat{i} - 2\hat{j} + 2\hat{k}) = 2\hat{i} - 4\hat{j} + 4\hat{k}$.
Answer: (C). | ||||
| 77 | 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 Probability that none completes: $\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5} = \frac{1}{5}$.
At least one: $1 - \frac{1}{5} = 0.8$.
Answer: (B). | ||||
| 78 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each round: A fails with probability $\frac{2}{3}$, then B gets a head with probability $\frac{1}{2}$; if both fail (probability $\frac{2}{3}\cdot\frac{1}{2} = \frac{1}{3}$) the next round begins.
B wins in round $n$ (after $n$ throws of the die) with probability $\left(\frac{1}{3}\right)^{n-1}\cdot\frac{2}{3}\cdot\frac{1}{2} = \left(\frac{1}{3}\right)^n$.
After 3 or more throws: $\sum_{n\ge3}\left(\frac{1}{3}\right)^n = \frac{1/27}{1 - 1/3} = \frac{1}{18}$.
Answer: (A). | ||||
| 79 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Each pole's top is seen from O at elevation $30^\circ$, so a pole at distance $d$ has height $d\tan30^\circ = \frac{d}{\sqrt{3}}$: the nearest pole is the shortest and the farthest the tallest.
The feet are at $1, 1 + a, \ldots, 1 + 9a$ from O. Tallest: $\frac{1 + 9a}{\sqrt{3}} = 12$, so $1 + 9a = 12\sqrt{3}$.
$a = \frac{12\sqrt{3} - 1}{9}$.
Answer: (A). | ||||
| 80 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $\sim(p\vee q) = \sim p\wedge\sim q$.
With $r = \sim q$: $(\sim p\wedge\sim q)\vee(p\wedge\sim q) = \sim q\wedge(\sim p\vee p) = \sim q = r$.
Others fail: $r = p$ gives $(\sim p\wedge\sim q)\vee p = p\vee\sim q \ne p$; $r = q$ gives $(\sim p\wedge\sim q)\vee(p\wedge q)$, which is true when p, q are both false although $q$ is false; $r = \sim p$ gives $\sim p\wedge\sim q \ne \sim p$.
Answer: (D). | ||||
| 81 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series With ratio $\rho < 0$: $p = \frac{10}{\rho}$, $r = 10\rho$, and $\rho + \frac{1}{\rho} \le -2$, so $p + q + r = 10\left(\rho + \frac{1}{\rho} + 1\right) \le -10$.
Hence $p + q + r = -15$: $\rho + \frac{1}{\rho} = -\frac{5}{2}$, $\rho = -2$ or $-\frac{1}{2}$.
$\rho = -2$ gives $p = -5$, $r = -20$ ($p > r$); $\rho = -\frac{1}{2}$ gives $p = -20$, $r = -5$ ($p < r$). So $\rho = -\frac{1}{2}$.
Fourth term $= r\rho = \frac{5}{2}$; product with the second term: $10\times\frac{5}{2} = 25$.
Answer: 25. | ||||
| 82 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $M = \begin{bmatrix}-1&-3&-8\\3&-1&-5\\5&4&-1\end{bmatrix}$ ($a_{12} = 1 - 4$, $a_{13} = 1 - 9$, $a_{23} = 4 - 9$; $a_{21} = 4 - 1$, $a_{31} = 6 - 1$, $a_{32} = 6 - 2$).
$X^TMX$ is the sum of all entries: $(-1 - 3 - 8) + (3 - 1 - 5) + (5 + 4 - 1) = -12 - 3 + 8 = -7$.
$-X^TMX = 7$.
Answer: 7. | ||||
| 83 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Each value appears on 2 faces.
Sum zero in three throws: the values are (0, 0, 0), or 1, 0, −1 in some order (two 1's would need −2, impossible).
(0, 0, 0): $2\times2\times2 = 8$ ways.
1, 0, −1 in any order: $3! = 6$ orders, each with $2^3 = 8$ face choices: 48 ways.
Total $8 + 48 = 56$.
Answer: 56. | ||||
| 84 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalEasy | |
A easy numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f(x) = |x - 3| + |x + 2| + |x - 5|$ is a sum of distances from $x$ to $-2$, $3$, $5$; it is least at the median point $x = 3$.
$f(3) = 0 + 5 + 2 = 7$.
Answer: 7. | ||||
| 85 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalEasy | |
A easy numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Put $t = x + 1$: $\frac{(t + 4)(t + 1)}{t} = t + \frac{4}{t} + 5$.
$\frac{d}{dt}: 1 - \frac{4}{t^2} = 0$ at $t = \pm2$; the second derivative $\frac{8}{t^3}$ is positive at $t = 2$, a local minimum.
Value: $2 + 2 + 5 = 9$ (at $x = 1$).
Answer: 9. | ||||
| 86 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\sqrt{\tan x} + \sqrt{\cot x} = \frac{\sin x + \cos x}{\sqrt{\sin x\cos x}} = \sqrt{2}\,\frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}}$.
Put $u = \sin x - \cos x$, $du = (\cos x + \sin x)dx$: the integral is $\sqrt{2}\sin^{-1}(\sin x - \cos x) + C$.
Equivalently, $\sqrt{2}\tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2\tan x}}\right) + C$ (the substitution $t = \sqrt{\tan x}$ gives $\int\frac{2(t^2 + 1)}{t^4 + 1}dt = \sqrt{2}\tan^{-1}\frac{t^2 - 1}{\sqrt{2}t}$).
So $a = \sqrt{2}$, $b = 2$: $a^4 + b^5 = 4 + 32 = 36$.
Answer: 36. | ||||
| 87 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Equal angles with the axes need $|l| = |m| = |n| = \frac{1}{\sqrt{3}}$: direction cosines $\left(\pm\frac{1}{\sqrt{3}}, \pm\frac{1}{\sqrt{3}}, \pm\frac{1}{\sqrt{3}}\right)$, 8 directions.
Opposite directions give the same line, so there are $\frac{8}{2} = 4$ lines (the four body diagonals of a cube).
Answer: 4. | ||||
| 88 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical 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 Vectors in the plane: $(2, 3, -1)$ and $(1, 2, 0)$; normal $= (2, 3, -1)\times(1, 2, 0) = (2, -1, 1)$.
Plane: $2(x - 1) - (y + 2) + (z - 1) = 0$, i.e. $2x - y + z - 5 = 0$.
Distance of $(a, -1, 1)$: $\frac{|2a + 1 + 1 - 5|}{\sqrt{6}} = \frac{|2a - 3|}{\sqrt{6}} = \frac{7}{\sqrt{6}}$, so $2a - 3 = \pm7$, $a = 5$ or $-2$.
$a > 0$: $a = 5$.
Answer: 5. | ||||
| 89 | 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 $|\sin x|$ has corners where $\sin x = 0$ and $|\cos x|$ where $\cos x = 0$; at such a point the other term is smooth, so f has a corner at every multiple of $\frac{\pi}{2}$.
For example, at $x = \frac{\pi}{2}$: to the left $f = \sin x + \cos x$, $f' \to -1$; to the right $f = \sin x - \cos x$, $f' \to 1$.
In $(0, 2\pi)$: $x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}$, three points.
Answer: 3. | ||||
| 90 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $4x^2 - 24x + 35 = (2x - 5)(2x - 7) < 0$ for $\frac{5}{2} < x < \frac{7}{2}$; the only integer is 3, so $\tan A = 3$ and $\sin^2A = \frac{9}{10}$.
The bisector of the second quadrant is $y = -x$, slope $-1$: $\sin B = -1$, $\sin^2B = 1$.
$\sin(A + B)\sin(B - A) = \sin^2B - \sin^2A = 1 - \frac{9}{10} = \frac{1}{10}$.
$40\times\frac{1}{10} = 4$.
Answer: 4. | ||||
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