JEE (Main) · 2021
JEE (Main) 2021 — 25 February, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Oscillations and Waves4
- Electrostatics4
- Dual Nature of Matter and Radiation3
18 easy · 12 medium · 0 hard
Chemistry
30 Q
- Aldehydes, Ketones and Carboxylic Acids3
- p-Block Elements2
- d- and f-Block Elements2
18 easy · 9 medium · 3 hard
Mathematics
30 Q
- Integral Calculus4
- Conic Sections4
- Matrices and Determinants3
13 easy · 15 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 $\frac{e^2}{4\pi\varepsilon_0 r}$ is an energy, so $\frac{e^2}{4\pi\varepsilon_0}$ has dimensions of energy × length, $[ML^3T^{-2}]$.
$\hbar$ has dimensions of energy × time, $[ML^2T^{-1}]$, so $\hbar c$ has $[ML^3T^{-2}]$.
The ratio (the fine-structure constant) is dimensionless: $[M^0L^0T^0]$.
Answer: (C). | ||||
| 2 | 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 Take $g = 10\ \mathrm{m\,s^{-2}}$. After falling 5 m the first stone has speed $v = \sqrt{2 \times 10 \times 5} = 10\ \mathrm{m\,s^{-1}}$.
Let both reach the ground a time t later, the building's height being H.
First stone: $H - 5 = 10t + 5t^2$. Second stone (from rest): $H - 25 = 5t^2$.
Subtracting: $20 = 10t$, so $t = 2$ s and $H = 25 + 5(4) = 45$ m.
Answer: (A). | ||||
| 3 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium 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 A solid sphere rolling without slipping has kinetic energy $\frac{1}{2}mv_0^2 + \frac{1}{2}\cdot\frac{2}{5}ma^2\cdot\frac{v_0^2}{a^2} = \frac{7}{10}mv_0^2$.
Rolling without slipping, friction does no work, so all of it becomes potential energy at the highest point: $\frac{7}{10}mv_0^2 = mgh$, $h = \frac{7v_0^2}{10g}$.
Distance along the incline: $s = \frac{h}{\sin\theta} = \frac{7v_0^2}{10g\sin\theta}$.
No option prints this value; option (C) prints its reciprocal coefficient, $\frac{10v_0^2}{7g\sin\theta}$, the only option built from the rolling factor $\frac{7}{10}$. It is the evidently intended answer.
Answer: (C). | ||||
| 4 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Angular speed $\omega = \frac{\pi/6}{0.1} = \frac{5\pi}{3}\ \mathrm{rad\,s^{-1}}$.
At M, P is at the extreme position, where the restoring force per unit mass is $\omega^2 A = \frac{25\pi^2}{9}\times 0.36 = \pi^2 \approx 9.87$ N (per kg).
Answer: (B). | ||||
| 5 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Along the curve $PV^{1/2} = \text{constant}$, so $P \propto V^{-1/2}$.
For one mole of ideal gas $T = \frac{PV}{R} \propto V^{-1/2}\cdot V = V^{1/2}$.
$\frac{T_2}{T_1} = \sqrt{\frac{V_2}{V_1}} = \sqrt{2}$.
Answer: (B). | ||||
| 6 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | StatementsMedium | |
A medium statements question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Statement I: in thermal equilibrium every quadratic energy term, rotational as well as translational, is distributed according to the Maxwell–Boltzmann law. True.
Statement II: by equipartition a diatomic molecule has 3 translational and 2 rotational degrees of freedom, so its mean translational energy is $\frac{3}{2}kT$ and its mean rotational energy $kT$. They are not equal. False.
Answer: (C). | ||||
| 7 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves A displacement x stretches one spring and compresses the other by x, so both push the block back: restoring force $(2k + 2k)x = 4kx$.
$T = 2\pi\sqrt{\frac{m}{4k}} = \pi\sqrt{\frac{m}{k}}$.
Answer: (C). | ||||
| 8 | 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 At t = 0: $\sin\phi_0 = \frac{1}{2}$, so $\phi_0 = \frac{\pi}{6}$ or $\frac{5\pi}{6}$.
Velocity $\frac{dY}{dt} = A\omega\cos\phi_0$ must be negative, so $\cos\phi_0 < 0$, giving $\phi_0 = \frac{5\pi}{6}$.
Answer: (B). | ||||
| 9 | 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 Place seven more cubes so that the charge sits at the centre of a cube of twice the edge. Its 24 outer faces, each a face of a small cube not touching the charge, share the flux $\frac{q}{\varepsilon_0}$ equally, so each such face of the small cube carries $\frac{q}{24\varepsilon_0}$. The three faces through the charge carry none.
The shaded area is a triangle on the top face and a triangle on the side face; each is half of a far face, and the paper treats each half as carrying half that face's flux: $2 \times \frac{1}{2}\cdot\frac{q}{24\varepsilon_0} = \frac{q}{24\varepsilon_0}$.
(The top triangle is cut by the diagonal through the point above the charge, so it carries exactly $\frac{q}{48\varepsilon_0}$ by symmetry; see the note on the side triangle.)
Answer: (D). | ||||
| 10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The field between the plates is perpendicular to them, so the velocity component parallel to the plates is unchanged: $v_1\cos\alpha = v_2\cos\beta$.
$\frac{K_1}{K_2} = \frac{v_1^2}{v_2^2} = \frac{\cos^2\beta}{\cos^2\alpha}$.
Answer: (D). | ||||
| 11 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct 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 Below the Curie temperature, exchange interaction aligns all the atomic dipoles within a domain in the same direction, so each domain is spontaneously magnetised to saturation; the domains themselves may point different ways, which is why an unmagnetised sample has no net moment.
Answer: (B). | ||||
| 12 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Without C: $\tan 45^\circ = \frac{X_L}{R}$, so $X_L = R$. Without L: $\tan 45^\circ = \frac{X_C}{R}$, so $X_C = R$.
With both, $X_L = X_C$: the circuit is at resonance and $Z = R = 110\ \Omega$.
$I_{rms} = \frac{220}{110} = 2$ A (taking 220 V as the rms supply voltage).
Answer: (C). | ||||
| 13 | 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 $\frac{hc}{\lambda} = \phi + eV_0$, so $\frac{hc}{\lambda_2} = \frac{hc}{\lambda_1} + e(V_2 - V_1)$.
With $hc = 1240$ eV nm: $\frac{1240}{491} = 2.526$ eV; adding $1.43 - 0.710 = 0.720$ eV gives 3.246 eV.
$\lambda_2 = \frac{1240}{3.246} \approx 382$ nm.
Answer: (C). | ||||
| 14 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The angular spread of the diffraction pattern is of order $\frac{\lambda}{d}$, so a wider pinhole gives a smaller pattern.
A wider pinhole also lets more light through, and that light is spread over a smaller pattern, so the intensity increases.
Answer: (C). | ||||
| 15 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{mv}$; at the same speed $\frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} = 1836$.
Answer: (B). | ||||
| 16 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $E = 13.6\left(1 - \frac{1}{4}\right) = 10.2$ eV.
$\lambda = \frac{1240}{10.2} \approx 121.6$ nm, closest to 121.8 nm (the Lyman-alpha line).
Answer: (A). | ||||
| 17 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter In amplitude modulation the radiated wave is the carrier, whose amplitude varies with the message; it travels at the carrier frequency $f_c$ (with sidebands at $f_c \pm f_m$ close to it).
Its wavelength in air is $\lambda = \frac{c}{f_c}$.
Answer: (C). | ||||
| 18 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits More acceptors put more holes in a p-type semiconductor, and its Fermi level moves down towards the valence band.
More donors put more electrons in an n-type semiconductor, and its Fermi level moves up towards the conduction band.
Answer: (B). | ||||
| 19 | Electronic DevicesRectifiers | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | MatchingEasy | |
A easy matching question on rectifiers, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A rectifier converts a.c. to d.c. (ii); a stabilizer keeps the output voltage constant (iv); a transformer steps a.c. voltage up or down (i); a filter removes the ripple (iii).
Answer: (B). | ||||
| 20 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Each input goes straight to one AND gate and, through a NOT gate, crosses to the other. The upper AND gate gets A and $\bar{B}$; the lower gets $\bar{A}$ and B.
The NOR gate gives $Y = \overline{A\bar{B} + \bar{A}B} = \overline{A \oplus B}$, the XNOR function.
Y = 1 for (0, 0), 0 for (0, 1), 0 for (1, 0), 1 for (1, 1): the second table.
Answer: (B). | ||||
| 21 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $p = \sqrt{2mK}$; at equal K, $\frac{p_1}{p_2} = \sqrt{\frac{4}{16}} = \frac{1}{2}$.
So n : 2 = 1 : 2 and n = 1.
Answer: 1. | ||||
| 22 | GravitationEscape velocity | Class 11: Gravitation | NumericalMedium | |
A medium numerical question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Energy conservation from r = R to r = 11R: $\frac{1}{2}v_i^2 = GM\left(\frac{1}{R} - \frac{1}{11R}\right) = \frac{10}{11}\cdot\frac{GM}{R}$.
With $v_e^2 = \frac{2GM}{R}$: $v_i^2 = \frac{10}{11}v_e^2$, so $v_i = \sqrt{\frac{10}{11}}\,v_e$ and x = 10.
Answer: 10. | ||||
| 23 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v = \sqrt{\frac{T}{\mu}} \propto \sqrt{T}$, so for a small change $\frac{\Delta v}{v} = \frac{1}{2}\frac{\Delta T}{T} = 2\%$.
Exactly: $\sqrt{1.04} - 1 = 0.0198$, an increase of 1.98%, which is 2% to the nearest integer.
Answer: 1.98 (2 to the nearest integer). | ||||
| 24 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\vec{Q}\times\vec{P} = -\vec{P}\times\vec{Q}$, so the condition gives $2\,\vec{P}\times\vec{Q} = 0$: $PQ\sin\theta = 0$.
For non-zero vectors $\sin\theta = 0$; with $0^\circ < \theta < 360^\circ$ the only value is $\theta = 180^\circ$.
Answer: 180. | ||||
| 25 | ThermodynamicsSecond law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{4}$, so $T_2 = \frac{3}{4}T_1$.
Then $1 - \frac{T_2 - 52}{T_1} = \frac{1}{2}$, so $T_2 - 52 = \frac{T_1}{2}$.
$\frac{3}{4}T_1 - 52 = \frac{1}{2}T_1$, giving $T_1 = 208$ K.
Answer: 208. | ||||
| 26 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Each thread makes angle $\theta$ with the vertical, $\sin\theta = \frac{0.10}{0.5} = 0.2$, $\tan\theta = \frac{0.2}{\sqrt{0.96}} = 0.2041$.
Equilibrium: $\tan\theta = \frac{F}{mg}$, with $mg = 10\times10^{-6}\times 10 = 10^{-4}$ N, so $F = 2.041\times10^{-5}$ N.
$F = \frac{kq^2}{r^2}$: $q^2 = \frac{2.041\times10^{-5}\times 0.04}{9\times10^9} = 9.07\times10^{-17}$, $q = 9.52\times10^{-9}$ C.
$\frac{a}{21}\times10^{-8} = 0.952\times10^{-8}$, so $a = 20.0$.
Answer: 20. | ||||
| 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 After contact the total charge $2.0$ nC is shared equally: 1.0 nC each.
$F = \frac{9\times10^9 \times (10^{-9})^2}{(0.5)^2} = 36\times10^{-9}$ N.
Answer: 36. | ||||
| 28 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalMedium | |
A medium numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Power radiated: $0.10 \times 8 = 0.8$ W. Intensity at 10 m: $I = \frac{0.8}{4\pi(10)^2} = \frac{0.002}{\pi}\ \mathrm{W\,m^{-2}}$.
$I = \frac{E_0^2}{2\mu_0 c}$, so $E_0 = \sqrt{2\mu_0 c I} = \sqrt{0.004}\sqrt{\frac{\mu_0 c}{\pi}} = 0.0632\sqrt{\frac{\mu_0 c}{\pi}}$.
$\frac{x}{10} = 0.0632$, so $x = \sqrt{0.4} \approx 0.63$.
Answer: 0.63 (1 to the nearest integer). | ||||
| 29 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity From P to R there are two parallel paths: PR directly (2 $\Omega$, current $i_2$) and P→Q→R (4 $\Omega$, current $i_1$).
Current divides inversely as resistance: $i_1 = 6\times\frac{2}{2+4} = 2$ A (and $i_2 = 4$ A).
Answer: 2. | ||||
| 30 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | NumericalMedium | |
A medium numerical question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $mc^2 = \frac{hc}{\lambda}$, so $m = \frac{h}{c\lambda} = \frac{h}{3\times10^8 \times 10^{-9}} = \frac{h}{0.3} = \frac{10}{3}h$ kg (h in SI units).
Answer: 10. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{XeF_4}$ (square planar), $\mathrm{SiF_4}$ and $\mathrm{BF_4^-}$ (tetrahedral) have all bonds equivalent.
$\mathrm{SF_4}$ has 4 bond pairs and 1 lone pair (trigonal bipyramidal arrangement, see-saw shape): the two axial S–F bonds are longer than the two equatorial ones.
Answer: (C). | ||||
| 32 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-}$: with solubility s, $K_{sp} = s(2s)^2 = 4s^3$.
$s^3 = \frac{5.5\times10^{-6}}{4} = 1.375\times10^{-6}$, so $s = 1.11\times10^{-2}$ mol L$^{-1}$.
Answer: (A). | ||||
| 33 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Hydrophilic (lyophilic) sols are stabilised by solvation, not by electrolytes; they are reversible; they are hard to coagulate.
Because the dispersed particles are heavily solvated, their viscosity is much higher than that of water, so (C) is false.
Answer: (C). | ||||
| 34 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Bond dissociation enthalpies (kJ mol$^{-1}$): $\mathrm{Cl_2}$ 242.6, $\mathrm{Br_2}$ 192.8, $\mathrm{F_2}$ 158.8, $\mathrm{I_2}$ 151.1.
$\mathrm{F_2}$ is anomalously low because of strong repulsion between the lone pairs on the small F atoms.
Order: $\mathrm{Cl_2 > Br_2 > F_2 > I_2}$.
Answer: (C). | ||||
| 35 | 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 Zone refining is used to obtain semiconductor-grade elements of very high purity, such as Si, Ge, B, Ga and In.
Answer: (C). | ||||
| 36 | 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 Steam gives CO with the fuels: $\mathrm{CH_4 + H_2O \xrightarrow{Ni,\ 1270\ K} CO + 3H_2}$; $\mathrm{C + H_2O \xrightarrow{1270\ K} CO + H_2}$ (water gas); $\mathrm{C_3H_8 + 3H_2O \xrightarrow{Ni} 3CO + 7H_2}$.
$\mathrm{CO_2}$ merely dissolves to give carbonic acid, $\mathrm{CO_2 + H_2O \rightleftharpoons H_2CO_3}$: no CO.
Answer: (C). | ||||
| 37 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: rhombic ($\alpha$) sulphur changes to monoclinic ($\beta$) sulphur above 369 K and back below it, reversibly, on slow heating or cooling. True.
Statement II: at room temperature the stable form is rhombic ($\alpha$) sulphur, not monoclinic. False.
Answer: (C). | ||||
| 38 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements German silver (nickel silver) is an alloy of copper, zinc and nickel (about Cu 50–60 %, Zn 20–25 %, Ni 15–20 %); it contains no silver.
Answer: (B). | ||||
| 39 | 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 (i) $\mathrm{Fe^{3+}}$, $d^5$, weak-field $\mathrm{F^-}$: high spin, 5 unpaired, $\mu = 5.92$ BM.
(ii) $\mathrm{Co^{3+}}$, $d^6$, strong-field $\mathrm{NH_3}$: low spin $t_{2g}^6$, 0 unpaired, $\mu = 0$.
(iii) $\mathrm{Ni^{2+}}$, $d^8$, tetrahedral with $\mathrm{Cl^-}$: 2 unpaired, $\mu = 2.83$ BM.
(iv) $\mathrm{Cu^{2+}}$, $d^9$: 1 unpaired, $\mu = 1.73$ BM.
Order: (i) > (iii) > (iv) > (ii).
Answer: (A). | ||||
| 40 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | StatementsEasy | |
A easy statements question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Normal rain water dissolves atmospheric $\mathrm{CO_2}$ and has pH about 5.6. True.
When oxides of sulphur and nitrogen lower the pH below 5.6, the rain is called acid rain. True.
Answer: (A). | ||||
| 41 | Practical Organic ChemistryDetection of elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on detection of elements, from Practical Organic Chemistry. NCERT: No matching NCERT chapter The sodium extract is boiled with dilute nitric acid to decompose any $\mathrm{NaCN}$ and $\mathrm{Na_2S}$ present, which would otherwise give precipitates with $\mathrm{AgNO_3}$; HCl cannot be used because it adds chloride itself.
Answer: (C). | ||||
| 42 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The alkene is 3-nitrobut-1-ene, $\mathrm{CH_2{=}CH{-}CH(NO_2){-}CH_3}$. Sulphuric acid protonates the double bond at the terminal carbon (Markovnikov), giving the secondary cation $\mathrm{CH_3{-}\overset{+}{C}H{-}CH(NO_2){-}CH_3}$.
A hydride shift would put the positive charge on the carbon carrying the electron-withdrawing $\mathrm{NO_2}$ group, which is less stable, so no rearrangement occurs.
Benzene is alkylated by this cation (Friedel–Crafts): $\mathrm{C_6H_5{-}CH(CH_3){-}CH(NO_2){-}CH_3}$, drawn in option (B).
Answer: (B). | ||||
| 43 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $\mathrm{H_2/CO}$ with a rhodium catalyst is hydroformylation (the oxo process): H and CHO add across the double bond, the CHO going mainly to the terminal carbon.
$\mathrm{CH_3CH_2CH{=}CH_2 + CO + H_2 \rightarrow CH_3CH_2CH_2CH_2CHO}$ (pentanal), one carbon more than the alkene.
Answer: (A). | ||||
| 44 | 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 Target: ethyl at C1, $\mathrm{NO_2}$ ortho to it (C2), Br para to it (C4).
(B): Friedel–Crafts acylation gives acetophenone; Clemmensen reduction (Zn-Hg/HCl) gives ethylbenzene; bromination of the o,p-directing ethylbenzene gives mainly 4-bromoethylbenzene; nitration then goes ortho to the more strongly activating ethyl group, giving 4-bromo-2-nitroethylbenzene.
(A) brominates acetophenone, whose acyl group directs meta; (C) and (D) also place the groups wrongly, and Zn/HCl (in A, C) would reduce the nitro group; in (D) nitrobenzene does not undergo Friedel–Crafts acylation.
Answer: (B). | ||||
| 45 | 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 Carboxylic acids are much stronger than phenol, so I is the weakest.
Among the benzoic acids, the electron-withdrawing para-$\mathrm{NO_2}$ (II) increases acidity and the electron-donating para-$\mathrm{CH_3}$ (IV) decreases it.
Order: II > III > IV > I.
Answer: (C). | ||||
| 46 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctHard | |
A hard single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Heated with oxalic acid, a 1,2-diol forms its formate esters (oxalic acid giving formic acid and $\mathrm{CO_2}$); on stronger heating the diformate of the two adjacent carbons loses $\mathrm{CO_2}$ and $\mathrm{HCOOH}$ (as $\mathrm{CO_2 + H_2}$), leaving a C=C between those carbons.
This is the reaction that turns glycerol into allyl alcohol at about 210–260 °C. Applied to ethylene glycol, both carbons carry the OH groups that are removed, so the product is ethylene, $\mathrm{CH_2{=}CH_2}$.
Answer: (C). | ||||
| 47 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines In the strongly acidic nitrating mixture much of the aniline is protonated to the anilinium ion, $\mathrm{C_6H_5NH_3^+}$, which is meta-directing; so the meta product (B) forms in large amount (about 47 %).
The free aniline that remains is nitrated mainly at the para position, and p-nitroaniline (A) is the major product (about 51 %), with about 2 % of the ortho isomer (C).
Answer: (D). | ||||
| 48 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines $\mathrm{C_6H_5NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_6H_5NC + 3KCl + 3H_2O}$: phenyl isocyanide, recognised by its foul smell.
Answer: (C). | ||||
| 49 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Maltose is two D-glucopyranose units joined by an $\alpha(1\to4)$ glycosidic bond. In the Haworth drawing of D-glucose the OH groups at C2 and C4 point down and the OH at C3 points up; the $\alpha$-anomer has the anomeric OH (C1 of the reducing unit) down, trans to $\mathrm{CH_2OH}$.
Options (A) and (D) draw the C3 OH down, which is not glucose. Option (B) has the anomeric OH of the right-hand unit up ($\beta$).
Option (C) has the glucose configuration in both rings, the $\alpha(1\to4)$ link, and the right-hand C1 OH down: the $\alpha$-anomer.
Answer: (C). | ||||
| 50 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | StatementsMedium | |
A medium statements question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Statement I: in ammoniacal solution $\mathrm{Ni^{2+}}$ gives a rosy-red precipitate of nickel dimethylglyoximate with dimethylglyoxime. True.
Statement II: dimethylglyoxime, $\mathrm{(CH_3)_2C_2(NOH)_2}$, is a neutral molecule that binds through its two N atoms, a bidentate ligand. True as a description of the reagent.
(In the nickel complex each molecule has lost one proton and coordinates as the uninegative dimethylglyoximate ion.)
Answer: (A). | ||||
| 51 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter The concordant burette reading is 4.4 mL (three equal readings); NaOH is in the burette.
Oxalic acid is dibasic: $M_{NaOH}\times 4.4 = 2\times 1.25\times 10.0$, so $M_{NaOH} = \frac{25}{4.4} = 5.68$ M.
Rounded off: 6. (The mean of all five readings, 4.44 mL, gives 5.63 M, also 6.)
Answer: 6. | ||||
| 52 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter An fcc cell holds Z = 4 atoms. $\rho = \frac{ZM}{N_A a^3} = \frac{4\times 63.54\times10^{-3}}{6.022\times10^{23}\times(3.596\times10^{-10})^3}$.
$a^3 = 4.650\times10^{-29}\ \mathrm{m^3}$, so $\rho = \frac{0.25416}{2.8003\times10^{-5}} = 9076.27\ \mathrm{kg\,m^{-3}}$ (about 9076).
Answer: 9076.27 (9076 to the nearest integer). | ||||
| 53 | 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 Energy per photon: $E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3.00\times10^8}{663\times10^{-9}} = 3.00\times10^{-19}$ J.
Per mole: $3.00\times10^{-19}\times6.02\times10^{23} = 1.806\times10^5$ J = 180.6 kJ mol$^{-1}$, which rounds to 181.
Answer: 181. | ||||
| 54 | 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 Isothermal ideal gas: $\Delta U = 0$, so $q = -w = P_{ext}(V_2 - V_1)$.
$V_2 - V_1 = nRT\left(\frac{1}{P_2} - \frac{1}{P_1}\right) = 5\times8.314\times293\times\left(\frac{1}{1.3\times10^6} - \frac{1}{2.1\times10^6}\right) = 12180\times2.930\times10^{-7} = 3.569\times10^{-3}\ \mathrm{m^3}$.
$q = 4.3\times10^6\times3.569\times10^{-3} = 1.535\times10^4$ J = 15.35 kJ, which rounds to 15.
Answer: 15. | ||||
| 55 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\mathrm{AB \rightarrow A^+ + B^-}$: $i = 1 + \alpha = 1.75$.
$\Delta T_b = iK_bm$: $m = \frac{2.5}{1.75\times0.52} = 2.75$ molal, which rounds to 3.
Answer: 3. | ||||
| 56 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalHard | |
A hard numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Let $c = [\mathrm{HNO_3}] = [\mathrm{H^+}] = [\mathrm{NO_3^-}]$, and take the gas pressures as 1 bar.
$\mathrm{NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O}$: $E_1 = 0.96 - \frac{0.059}{3}\log\frac{1}{c^5} = 0.96 + \frac{5\times0.059}{3}\log c$.
$\mathrm{NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O}$: $E_2 = 0.79 - 0.059\log\frac{1}{c^3} = 0.79 + 0.177\log c$.
With the same copper half-cell, equal tendency means $E_1 = E_2$: $0.96 + 0.0983x = 0.79 + 0.177x$, so $0.17 = 0.0787x$ and $x = 2.16$.
$2x = 4.32$, which rounds to 4.
Answer: 4. | ||||
| 57 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ with $T_1 = 300$ K, $T_2 = 325$ K.
$\frac{1}{300} - \frac{1}{325} = \frac{25}{97500} = 2.564\times10^{-4}\ \mathrm{K^{-1}}$.
$E_a = \frac{8.314\times\ln 5}{2.564\times10^{-4}} = \frac{8.314\times1.609}{2.564\times10^{-4}} = 52.2$ kJ mol$^{-1}$, which rounds to 52.
Answer: 52. | ||||
| 58 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Caesium, with the lowest ionisation enthalpy of the alkali metals, emits electrons readily under visible light and is the metal used for the electrodes of photoelectric cells.
Lithium and sodium need light of higher energy, and rubidium is not the metal used in practice.
Answer: 1. | ||||
| 59 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Z = 29 is copper; $\mathrm{Cu^{2+}}$ is $[\mathrm{Ar}]3d^9$, with 1 unpaired electron.
$\mu = \sqrt{n(n+2)} = \sqrt{3} = 1.73$ BM.
Answer: 1.73 (2 to the nearest integer). | ||||
| 60 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalEasy | |
A easy numerical question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Sulphanilic acid is 4-aminobenzenesulphonic acid ($\mathrm{-SO_3H}$); picric acid is 2,4,6-trinitrophenol (phenolic OH); ascorbic acid is a cyclic lactone whose acidity comes from an enol OH.
Only aspirin, 2-acetoxybenzoic acid, has a $\mathrm{-COOH}$ group.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $AA^T = \begin{bmatrix}1+\alpha^2 & \alpha-\alpha\beta\\ \alpha-\alpha\beta & \alpha^2+\beta^2\end{bmatrix} = I_2$.
$1 + \alpha^2 = 1$ gives $\alpha = 0$; then $\alpha^2 + \beta^2 = 1$ gives $\beta^2 = 1$.
$\alpha^4 + \beta^4 = 0 + 1 = 1$.
Answer: (B). | ||||
| 62 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\det(2A) = 2^3\det A = 32$.
$R_2 \to 2R_2 + 5R_3$ multiplies the determinant by 2 (adding a multiple of another row changes nothing).
$\det B = 2\times32 = 64$.
Answer: (C). | ||||
| 63 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}2&3&2\\3&2&2\\1&-1&4\end{vmatrix} = 2(8+2) - 3(12-2) + 2(-3-2) = 20 - 30 - 10 = -20 \ne 0$, so the solution is unique.
Subtracting the first two equations: $y = x$; then the third gives $z = 2$ and the first $5x + 4 = 9$, so $x = y = 1$.
Check (D): $1 + 1 + 8 = 10 \ne 12$, so (D) is false.
Answer: (B). | ||||
| 64 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $I_n + I_{n+2} = \int_{\pi/4}^{\pi/2}\cot^n x\,\mathrm{cosec}^2 x\,dx = \left[-\frac{\cot^{n+1}x}{n+1}\right]_{\pi/4}^{\pi/2} = \frac{1}{n+1}$.
So $I_2+I_4 = \frac{1}{3}$, $I_3+I_5 = \frac{1}{4}$, $I_4+I_6 = \frac{1}{5}$; their reciprocals 3, 4, 5 are in A.P.
Answer: (A). | ||||
| 65 | 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 $f(2-x) = \frac{5^{2-x}}{5^{2-x}+5} = \frac{25}{25 + 5\cdot5^x} = \frac{5}{5^x+5}$, so $f(x) + f(2-x) = 1$.
Pair $f\left(\frac{k}{20}\right)$ with $f\left(\frac{40-k}{20}\right)$ for k = 1, …, 19: 19 pairs, each 1. The middle term is $f(1) = \frac{5}{10} = \frac{1}{2}$.
Sum $= 19 + \frac{1}{2} = \frac{39}{2}$.
Answer: (C). | ||||
| 66 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha^2 = 6\alpha + 2$, so $\alpha^{10} = 6\alpha^9 + 2\alpha^8$; likewise for $\beta$. Subtracting: $a_{10} = 6a_9 + 2a_8$.
$\frac{a_{10} - 2a_8}{3a_9} = \frac{6a_9}{3a_9} = 2$.
Answer: (C). | ||||
| 67 | Sequences and SeriesRelation between A.M. and G.M. | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on relation between a.m. and g.m., from Sequences and Series. NCERT: Class 11: Sequences and Series Let $t = a^{a^x} > 0$. Then $f = t + \frac{a}{t} \ge 2\sqrt{t\cdot\frac{a}{t}} = 2\sqrt{a}$ (A.M. ≥ G.M.).
Equality needs $t = \sqrt{a}$, i.e. $a^x = \frac{1}{2}$, which some real x gives when $a \ne 1$; for a = 1, f = 2 = $2\sqrt{1}$.
Answer: (C). | ||||
| 68 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $e^{k\log_e u} = u^k$: the integrand is $\frac{8x^3 + 20x^2}{x^4 + 5x^3 - 7x^2} = \frac{4(2x+5)}{x^2 + 5x - 7}$.
With $u = x^2 + 5x - 7$, $du = (2x+5)dx$: the integral is $4\log_e|x^2 + 5x - 7| + c$.
Answer: (B). | ||||
| 69 | 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 With real coefficients the other root is $1 + 2i$.
Sum of roots $= 2 = -\alpha$, so $\alpha = -2$; product $= 1 + 4 = 5 = \beta$.
$\alpha - \beta = -7$.
Answer: (D). | ||||
| 70 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Solve: $y = 1 - x$ gives $x^2 + 2(1-x)^2 = 2$, $3x^2 - 4x = 0$, so P = (0, 1) and Q = $\left(\frac{4}{3}, -\frac{1}{3}\right)$.
$\vec{OP}\cdot\vec{OQ} = -\frac{1}{3} < 0$, so the angle is obtuse; $\cos\theta = \frac{-1/3}{1\cdot\frac{\sqrt{17}}{3}} = -\frac{1}{\sqrt{17}}$.
OP lies along the y-axis and OQ makes an angle $\tan^{-1}\frac{1}{4}$ below the x-axis, so $\theta = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$.
Answer: (A). | ||||
| 71 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A point of the parabola is $\left(t, \frac{t^2}{2}\right)$; its distance from $x - y - 1 = 0$ is $\frac{\left|t - \frac{t^2}{2} - 1\right|}{\sqrt{2}} = \frac{(t-1)^2 + 1}{2\sqrt{2}}$.
It is least at t = 1: $\frac{1}{2\sqrt{2}}$ (the line does not meet the curve).
Answer: (B). | ||||
| 72 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse: $e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$, foci $(\pm3, 0)$.
Hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ through $(\pm3, 0)$: a = 3. Its eccentricity is $\frac{5}{3}$, so $b^2 = a^2(e^2 - 1) = 9\cdot\frac{16}{9} = 16$.
$\frac{x^2}{9} - \frac{y^2}{16} = 1$.
Answer: (A). | ||||
| 73 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $\vec{AB} = (1, 1, -2)$, $\vec{AC} = (1, 2, -1)$; normal $\vec{n} = \vec{AB}\times\vec{AC} = (3, -1, 1)$.
$\vec{OP} = (2, -1, 1)$: $|\vec{OP}|^2 = 6$ and its component along $\vec{n}$ is $\frac{6 + 1 + 1}{\sqrt{11}} = \frac{8}{\sqrt{11}}$.
Projection on the plane: $\sqrt{6 - \frac{64}{11}} = \sqrt{\frac{2}{11}}$.
Answer: (B). | ||||
| 74 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The sum is $\sum_{r=0}^{n-1}\frac{n}{(n+r)^2} = \frac{1}{n}\sum_{r=0}^{n-1}\frac{1}{\left(1+\frac{r}{n}\right)^2}$ (the first term is r = 0).
As $n \to \infty$ it tends to $\int_0^1\frac{dx}{(1+x)^2} = \left[-\frac{1}{1+x}\right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2}$.
Answer: (B). | ||||
| 75 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Expected sufferers: smoker non-vegetarian $160\times0.35 = 56$; smoker vegetarian $100\times0.20 = 20$; non-smoker vegetarian $140\times0.10 = 14$; total 90.
By Bayes' theorem, $P = \frac{56}{90} = \frac{28}{45}$.
Answer: (C). | ||||
| 76 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability |A|: 7 in the thousands place: $9^3 = 729$; 7 in one of the other three places: $3\times8\times9^2 = 1944$ (leading digit not 0 or 7). Total 2673.
Remainder 2 on division by 5 means the last digit is 2 or 7.
Last digit 7: the other digits avoid 7: $8\times9\times9 = 648$.
Last digit 2: exactly one 7 among the first three digits: $9\times9 + 2\times8\times9 = 81 + 144 = 225$.
Favourable 873; probability $\frac{873}{2673} = \frac{97}{297}$.
Answer: (C). | ||||
| 77 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $2\cos\frac{x+y}{2}\cos\frac{x-y}{2} - \left(2\cos^2\frac{x+y}{2} - 1\right) = \frac{3}{2}$. With $c = \cos\frac{x+y}{2}$: $4c^2 - 4c\cos\frac{x-y}{2} + 1 = 0$.
For real c the discriminant must be non-negative: $16\cos^2\frac{x-y}{2} - 16 \ge 0$, so $\cos\frac{x-y}{2} = \pm1$; in the range given, x = y, and then $c = \frac{1}{2}$, $\frac{x+y}{2} = \frac{\pi}{3}$.
$x = y = \frac{\pi}{3}$: $\sin x + \cos y = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1+\sqrt{3}}{2}$.
Answer: (D). | ||||
| 78 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $x = {}^5P_3 = 60$. $A\times B$ has 15 elements, so $y = {}^{15}P_3 = 15\cdot14\cdot13 = 2730$.
$\frac{y}{x} = \frac{2730}{60} = \frac{91}{2}$, so $2y = 91x$.
Answer: (A). | ||||
| 79 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta = \cot^{-1}5$, $\tan\theta = \frac{1}{5}$: $\tan2\theta = \frac{2/5}{1 - 1/25} = \frac{5}{12}$, so $\sin2\theta = \frac{5}{13}$, $\cos2\theta = \frac{12}{13}$.
Let $\phi = \cos^{-1}\frac{4}{5}$: $\sin\phi = \frac{3}{5}$.
$\sin(2\theta + \phi) = \frac{5}{13}\cdot\frac{4}{5} + \frac{12}{13}\cdot\frac{3}{5} = \frac{56}{65}$, so the cosec is $\frac{65}{56}$.
Answer: (C). | ||||
| 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 The contrapositive of $p \Rightarrow q$ is $\sim q \Rightarrow \sim p$.
p: you will work; q: you will earn money. Contrapositive: if you will not earn money, you will not work.
Answer: (C). | ||||
| 81 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability On [−2, 2]: $|x| = 2 - x^2$ at $|x| = 1$. For $|x| < 1$, $f = |x|$; for $1 < |x| \le 2$, $f = 2 - x^2$.
Corners at x = 0 (|x|) and x = ±1 (slopes ±1 meet ∓2): not differentiable.
At x = ±2, $f = 2 - 4 = -2$, but for $2 < |x| < 3$, $f = [|x|] = 2$: f is discontinuous, so not differentiable.
Elsewhere in (−3, 3) f is smooth. Points: −2, −1, 0, 1, 2: five.
Answer: 5. | ||||
| 82 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Rewrite: $x\,dy - y\,dx = -2xy^2\,dx$. Dividing by $-y^2$: $\frac{y\,dx - x\,dy}{y^2} = 2x\,dx$, i.e. $d\left(\frac{x}{y}\right) = 2x\,dx$.
So $\frac{x}{y} = x^2 + C$.
The lines meet at (2, 1): $2 = 4 + C$, C = −2.
At x = 1: $\frac{1}{y} = 1 - 2 = -1$, y(1) = −1, $|y(1)| = 1$.
Answer: 1. | ||||
| 83 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $7 \equiv -3 \pmod{10}$, so $3^n + 7^n \equiv 3^n + (-3)^n$.
For odd n this is 0 (mod 10). For even n it is $2\cdot3^n$, never divisible by 5.
So n must be odd: the two-digit odd numbers 11, 13, …, 99, which are 45.
Answer: 45. | ||||
| 84 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Denominator $\approx 4ax^2$. Numerator $= ax - 4x - 8x^2 - \ldots$
For a finite limit the x term must vanish: a = 4. Then the limit is $\frac{-8x^2}{16x^2} = -\frac{1}{2} = b$.
$a - 2b = 4 + 1 = 5$.
Answer: 5. | ||||
| 85 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Slopes: for $x = y^4$, $\frac{dy}{dx} = \frac{1}{4y^3}$; for $xy = k$, $\frac{dy}{dx} = -\frac{y}{x}$.
Perpendicular: $\frac{1}{4y^3}\cdot\left(-\frac{y}{x}\right) = -1$, so $4xy^2 = 1$. With $x = y^4$: $4y^6 = 1$, $y^6 = \frac{1}{4}$.
$k = xy = y^5$, so $(4k)^6 = 4^6y^{30} = 4^6\left(\frac{1}{4}\right)^5 = 4$.
Answer: 4. | ||||
| 86 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $3x^2 - 3x - 6 = 3(x-2)(x+1)$: positive on [−2, −1], negative on (−1, 2).
With $F(x) = x^3 - \frac{3}{2}x^2 - 6x$: $F(-2) = -2$, $F(-1) = \frac{7}{2}$, $F(2) = -10$.
Integral $= [F(-1) - F(-2)] - [F(2) - F(-1)] = \frac{11}{2} + \frac{27}{2} = 19$.
Answer: 19. | ||||
| 87 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem x is odd, so 2020 + x is odd. The square of any odd number is 1 more than a multiple of 8: $(2m+1)^2 = 4m(m+1) + 1$.
$(2020 + x)^{2022} = \left[(2020 + x)^2\right]^{1011} \equiv 1^{1011} = 1 \pmod 8$.
Answer: 1. | ||||
| 88 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalHard | |
A hard numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Directions: $l_1$ (1, 2, 2), $l_2$ (2, 2, 1). l is along their cross product $(-2, 3, -2)$: $\vec{r} = \lambda(-2, 3, -2)$.
Meeting $l_1$: $3 + t = -2\lambda$, $-1 + 2t = 3\lambda$, $4 + 2t = -2\lambda$ give t = −1, λ = −1: the point (2, −3, 2).
A point of $l_2$: $(3+2s, 3+2s, 2+s)$; distance² from (2, −3, 2): $(1+2s)^2 + (6+2s)^2 + s^2 = 17$, i.e. $9s^2 + 28s + 20 = 0$, s = −2 or $-\frac{10}{9}$.
s = −2 gives (−1, −1, 0), not in the first octant; $s = -\frac{10}{9}$ gives $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$.
$18(a+b+c) = 18\cdot\frac{22}{9} = 44$.
Answer: 44. | ||||
| 89 | Conic SectionsParabola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent to the parabola: $y = mx + \frac{1}{m}$. Tangent to the circle: $\frac{|3m + 1/m|}{\sqrt{1+m^2}} = 3$, so $(3m^2+1)^2 = 9m^2(1+m^2)$, giving $m^2 = \frac{1}{3}$.
For contact in the first quadrant, $m = \frac{1}{\sqrt{3}}$: the line $x - \sqrt{3}y + 3 = 0$.
Parabola contact: $\left(\frac{1}{m^2}, \frac{2}{m}\right) = (3, 2\sqrt{3})$.
Circle contact: foot of the perpendicular from (3, 0): $(3, 0) - \frac{6}{4}(1, -\sqrt{3}) = \left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$.
$2(a + c) = 2\left(3 + \frac{3}{2}\right) = 9$.
Answer: 9. | ||||
| 90 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}\times\vec{b} = (4\alpha, 8, -4\alpha)$, so $|\vec{a}\times\vec{b}|^2 = 32\alpha^2 + 64 = 192$, giving $\alpha^2 = 4$.
$\vec{a}\cdot\vec{b} = 3 - \alpha^2 + 3 = 2$.
Answer: 2. | ||||
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