JEE (Main) · 2022
JEE (Main) 2022 — 28 July, 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
- Kinematics3
- Physical World and Measurement2
- Laws of Motion2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- p-Block Elements3
- d- and f-Block Elements2
- Alcohols, Phenols and Ethers2
20 easy · 8 medium · 2 hard
Mathematics
30 Q
- Integral Calculus3
- Conic Sections3
- Matrices and Determinants2
7 easy · 20 medium · 3 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 The argument of a logarithm is dimensionless, so $[\beta x] = [kT]$ = energy, and $[\beta] = \frac{\text{energy}}{\text{length}}$ = force. (A) is true.
$\eta$ and $\sin\theta$ are dimensionless, so $\alpha\beta$ is dimensionless: $[\alpha] = [\text{force}]^{-1}$.
(B) $[\alpha^{-1}x]$ = force × length = energy: true.
(C) $\eta^{-1}\sin\theta$ and $\alpha\beta$ are both dimensionless: true.
(D) $[\alpha] = [\text{force}]^{-1}$ while $[\beta]$ = force: false, so (D) is the incorrect option.
Answer: (D). | ||||
| 2 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\vec{r} = 3t\,\hat{x} + 5t^3\,\hat{y} + 7\,\hat{z}$.
$\vec{v} = 3\hat{x} + 15t^2\hat{y}$ and $\vec{a} = 30t\,\hat{y}$.
At t = 1 s, $\vec{a} = 30\hat{y}$.
Answer: (B). | ||||
| 3 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Mass striking the wall per second: $\rho A v = 1000 \times 10 \times 10^{-4} \times 20 = 20$ kg/s.
The wall stops it, removing momentum $20 \times 20 = 400$ kg m s$^{-2}$ every second.
$F = \rho A v^2 = 400$ N.
Answer: (D). | ||||
| 4 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion With mass per length $\lambda = \frac{m}{L}$, the net driving force is the weight of the extra length: $\lambda g\,[(L - l) - l]$, and the whole mass m accelerates.
$a = \frac{(L - 2l)g}{L} = \frac{g}{2}$ gives $L - 2l = \frac{L}{2}$, so $l = \frac{L}{4}$.
Hence $x = 4$. (Taking $l$ as the longer side gives $l = \frac{3L}{4}$, $x = \frac{4}{3}$, which is not offered.)
Answer: (D). | ||||
| 5 | 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 $\frac{1}{2}(0.2)u^2 = 90$ gives $u = 30$ m/s; $\frac{1}{2}(0.2)v^2 = 40$ gives $v = 20$ m/s after 1 s.
Taking the retardation as uniform: $a = \frac{30 - 20}{1} = 10$ m/s$^2$.
Distance to stop: $s = \frac{u^2}{2a} = \frac{900}{20} = 45$ m.
Answer: (A). | ||||
| 6 | 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 $T \propto \frac{1}{\sqrt{g}}$, so $\frac{g_h}{g} = \left(\frac{4}{6}\right)^2 = \frac{4}{9}$.
$g_h = g\left(\frac{R_E}{R_E + h}\right)^2$, so $\frac{R_E}{R_E + h} = \frac{2}{3}$ and $R_E + h = 1.5R_E$.
$h = 0.5 \times 6400 = 3200$ km.
Answer: (C). | ||||
| 7 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The piston presses on the water with the force $F = 5\times10^5$ N over the whole cross-section, so the pressure at the water surface is $\frac{F}{\pi r^2} = \frac{5\times10^5}{\pi} = 1.59\times10^5$ Pa. Outside the hole the pressure is atmospheric, $P_A = 1.01\times10^5$ Pa.
Bernoulli between the top surface (speed negligible) and the hole, $h = 10$ m lower: $\frac{F}{A} + \rho g h = P_A + \frac{1}{2}\rho v^2$.
$v = \sqrt{\frac{2(F/A - P_A)}{\rho} + 2gh} = \sqrt{\frac{2\times0.582\times10^5}{1000} + 200} = \sqrt{116.3 + 200} = \sqrt{316.3} = 17.8$ m/s.
Answer: (C). | ||||
| 8 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} \propto \sqrt{T}$: doubling it needs $T = 4 \times 300 = 1200$ K, so $\Delta T = 900$ K.
n = 14/28 = 0.5 mol of a diatomic gas, heated in a closed vessel (constant volume): $C_V = \frac{5}{2}R$.
$Q = nC_V\Delta T = 0.5 \times \frac{5}{2} \times 8.32 \times 900 = 9360$ J.
Answer: (C). | ||||
| 9 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$ with $t = \frac{3d}{4}$:
$C = \frac{\varepsilon_0 A}{\frac{d}{4} + \frac{3d}{4K}} = \frac{\varepsilon_0 A}{d}\cdot\frac{4K}{K + 3} = \frac{4KC_0}{3 + K}$.
Answer: (A). | ||||
| 10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Acceleration across the plates: $a = \frac{eE}{m} = 8$ m/s$^2$.
Time between the plates: $t = \frac{1}{2}$ s.
Transverse velocity on leaving: $v_y = at = 4$ m/s, while $v_x = 2$ m/s.
$\tan\theta = \frac{4}{2} = 2$, so $\theta = \tan^{-1}(2)$.
Answer: (B). | ||||
| 11 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Statement I: each part is 20 Ω; four in parallel give $\frac{20}{4} = 5$ Ω. Correct.
Statement II: in parallel the voltage is common, so heat $H = \frac{V^2t}{R} \propto \frac{1}{R}$: $H_{3R} : H_{2R} = 2 : 3$, not 3 : 2. Incorrect.
Answer: (C). | ||||
| 12 | 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 BCD is equilateral, so CD makes 60° with BD, which lies along the field.
$F = IlB\sin 60^\circ = 10 \times 0.05 \times 0.5 \times 0.866 = 0.216$ N.
Answer: (C). | ||||
| 13 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B_1 = \frac{\mu_0 I}{2R}$ and $B_2 = \frac{\mu_0 IR^2}{2(R^2 + x^2)^{3/2}}$ with $x = \sqrt{3}R$.
$(R^2 + 3R^2)^{3/2} = 8R^3$, so $B_2 = \frac{\mu_0 I}{16R} = \frac{B_1}{8}$.
$B_1 : B_2 = 8 : 1$.
Answer: (C). | ||||
| 14 | Electromagnetic Induction and Alternating CurrentsTransformers | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on transformers, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current An ideal transformer passes the same power, 80 kW, on both sides.
Primary: $R_1 = \frac{V_1^2}{P} = \frac{(8000)^2}{80000} = 800$ Ω.
Secondary: $R_2 = \frac{V_2^2}{P} = \frac{(160)^2}{80000} = 0.32$ Ω.
Answer: (C). | ||||
| 15 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves For complete absorption the force is $F = \frac{IA}{c}$, so $I = \frac{Fc}{A}$.
$I = \frac{7.2 \times 10^{-9} \times 3 \times 10^8}{36} = \frac{2.16}{36} = 0.06$ W/cm$^2$ (A in cm$^2$). The 20 minutes do not enter.
Answer: (D). | ||||
| 16 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Taking the power as $P = \frac{1}{f}$ in the medium, the lens maker's formula gives
$1.25 = \left(\frac{1.5}{\mu_m} - 1\right)\left(\frac{1}{0.2} + \frac{1}{0.4}\right) = 7.5\left(\frac{1.5}{\mu_m} - 1\right)$.
$\frac{1.5}{\mu_m} - 1 = \frac{1}{6}$, so $\frac{1.5}{\mu_m} = \frac{7}{6}$ and $\mu_m = \frac{9}{7}$.
Answer: (B). | ||||
| 17 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $K_{max} = E - \phi$: $5\phi - \phi = 4\phi$ and $10\phi - \phi = 9\phi$.
$v_{max} \propto \sqrt{K_{max}}$: ratio $\sqrt{4} : \sqrt{9} = 2 : 3$.
Answer: (C). | ||||
| 18 | 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 After 15 min the fraction left is $1 - \frac{7}{8} = \frac{1}{8} = \left(\frac{1}{2}\right)^3$: three half-lives.
$T_{1/2} = \frac{15}{3} = 5$ min.
Answer: (A). | ||||
| 19 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Voltage gain of the CE amplifier: $A_v = \beta\frac{R_C}{R_B} = 100 \times \frac{10\ \text{k}\Omega}{1\ \text{k}\Omega} = 1000$.
$V_{out} = 1000 \times 1$ mV = 1.0 V.
Answer: (B). | ||||
| 20 | Communication SystemsBandwidth of signals | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on bandwidth of signals, from Communication Systems. NCERT: No matching NCERT chapter Deviation ratio $= \frac{\Delta f}{f_m} = 10$, so the frequency deviation is $\Delta f = 200$ kHz.
FM bandwidth (Carson's rule): $BW = 2(\Delta f + f_m) = 2(200 + 20) = 440$ kHz.
Answer: (D). | ||||
| 21 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Taking upward as positive: displacement in 6 s is $19.6 \times 6 - \frac{1}{2}(9.8)(36) = 117.6 - 176.4 = -58.8$ m, so the tower is 58.8 m high.
Rise above the top: $\frac{u^2}{2g} = \frac{19.6^2}{19.6} = 19.6$ m.
Greatest height above the ground: $58.8 + 19.6 = 78.4$ m $= \frac{k}{5}$, so $k = 392$.
Answer: 392. | ||||
| 22 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $M = \int_0^L \rho_0\left(1 - \frac{x^2}{L^2}\right)dx = \rho_0\left(L - \frac{L}{3}\right) = \frac{2\rho_0 L}{3}$.
$\int_0^L x\rho\,dx = \rho_0\left(\frac{L^2}{2} - \frac{L^2}{4}\right) = \frac{\rho_0 L^2}{4}$.
$x_{cm} = \frac{L^2/4}{2L/3} = \frac{3L}{8}$, so $\alpha = 8$.
Answer: 8. | ||||
| 23 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical 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 At the bottom: $T = mg + \frac{mv^2}{r} = 20 + \frac{2 \times 25}{0.5} = 120$ N.
Stress $= \frac{120}{4 \times 10^{-6}} = 3 \times 10^7$ N/m$^2$.
Strain $= \frac{3 \times 10^7}{10^{11}} = 3 \times 10^{-4} = 30 \times 10^{-5}$.
Answer: 30. | ||||
| 24 | ThermodynamicsFirst law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics With f = 8: $C_V = 4R$ and $C_P = 5R$.
At constant pressure $W = nR\Delta T = 150$ J.
$Q = nC_P\Delta T = 5\,nR\Delta T = 5 \times 150 = 750$ J.
Answer: 750. | ||||
| 25 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $F = -\frac{dU}{dx} = -16\sin 4x \approx -64x$ for small x.
So $k = 64$ N/m and $\omega = \sqrt{\frac{64}{4}} = 4$ rad/s.
$T = \frac{2\pi}{4} = \frac{\pi}{2}$ s, so K = 2.
Answer: 2. | ||||
| 26 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity $R_1 = \frac{220^2}{100} = 484$ Ω and $R_2 = \frac{220^2}{60} = 806.7$ Ω.
$I = \frac{220}{484 + 806.7} = 0.1705$ A.
$P_1 = I^2R_1 = (0.1705)^2 \times 484 = 14.06$ W ≈ 14 W.
Answer: 14. | ||||
| 27 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Just after the switch closes, the inductor carries no current, so the branch with the 2 H inductor and the two 2 Ω resistors carries none.
The battery then drives current through the 2 Ω resistor in its own branch and the 4 Ω resistor: $I = \frac{6}{2 + 4} = 1$ A.
Answer: 1. | ||||
| 28 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics f = −100 cm. After 10 s the object has moved 20 cm, so u = −80 cm.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{100} + \frac{1}{80} = \frac{1}{400}$, so v = +400 cm.
The image is virtual, 400 cm behind the mirror.
Answer: 400. | ||||
| 29 | Experimental SkillsOptics experiments | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on optics experiments, from Experimental Skills. NCERT: No matching NCERT chapter With distances measured from the foci, Newton's formula for a lens is $x_1x_2 = f^2$.
So $f^2 = 225$ and $|f| = 15$ cm.
Answer: 15. | ||||
| 30 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalMedium | |
A medium numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $g = \frac{4\pi^2 l}{T^2}$, so $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$.
100 oscillations take 50 s, measured to 1 s: $\frac{\Delta T}{T} = \frac{1}{50} = 0.02$.
$\frac{\Delta l}{l} = \frac{0.1}{10} = 0.01$.
$\frac{\Delta g}{g} = 0.01 + 0.04 = 0.05 = 5\%$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Overlap of atomic orbitals can be positive (in phase), negative (out of phase) or zero.
An out-of-phase overlap is a negative overlap, not a zero one, so Assertion A is false.
Zero overlap arises when the orbitals approach in a direction or orientation in which the in-phase and out-of-phase parts cancel (for example an s orbital approaching a p orbital along its nodal plane), so Reason R is true.
Answer: (D). | ||||
| 32 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Metallic character falls across a period and rises down a group.
Na > Mg in period 3; Mg > Be down group 2; Be is a metal while Si is a metalloid and P a non-metal, with Si > P across the period.
Order: Na > Mg > Be > Si > P.
Answer: (A). | ||||
| 33 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$. A liquid metal product has higher entropy than a solid one, so $\Delta S$ of the reduction is larger and $\Delta G^\ominus$ more negative.
That is exactly why reduction is easier when the metal is formed as a liquid: A is true and R explains it.
Answer: (A). | ||||
| 34 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Clark's method adds a calculated amount of lime:
$\mathrm{Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3\downarrow + 2H_2O}$
$\mathrm{Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3\downarrow + Mg(OH)_2\downarrow + 2H_2O}$
Magnesium precipitates as the less soluble hydroxide. Products: $\mathrm{CaCO_3}$ and $\mathrm{Mg(OH)_2}$.
Answer: (C). | ||||
| 35 | 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 Statement I: the textbook use of the lithium–magnesium alloy is for armour plates; lithium's alloy with aluminium is the one used for aircraft parts. As worded, Statement I is false.
Statement II: cell-membrane integrity (with neuromuscular function and blood clotting) is a role of calcium ions; magnesium ions act as cofactors of ATP-using enzymes and are part of chlorophyll. Statement II is false.
Answer: (B). | ||||
| 36 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2}$
Answer: (B). | ||||
| 37 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Concentrated nitric acid oxidises iodine to iodic acid and is itself reduced to $\mathrm{NO_2}$:
$\mathrm{I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O}$
Answer: (C). | ||||
| 38 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Count the electrons (Z − charge):
(A) $\mathrm{Sm^{2+}}$: 62 − 2 = 60; $\mathrm{Er^{3+}}$: 68 − 3 = 65. Not isoelectronic.
(B) $\mathrm{Yb^{2+}}$: 68; $\mathrm{Lu^{3+}}$: 68. Isoelectronic (both $4f^{14}$).
(C) $\mathrm{Eu^{2+}}$: 61; $\mathrm{Tb^{4+}}$: 61. Isoelectronic (both $4f^7$).
(D) $\mathrm{Tb^{2+}}$: 63; $\mathrm{Tm^{4+}}$: 65. Also not isoelectronic.
So two pairs, (A) and (D), are not isoelectronic.
Answer: (A) or (D). | ||||
| 39 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Acidified permanganate oxidises HCl to $\mathrm{Cl_2}$ ($\mathrm{2MnO_4^- + 10Cl^- + 16H^+ \rightarrow 2Mn^{2+} + 5Cl_2 + 8H_2O}$).
Part of the permanganate is used up on the acid, so the titre is wrong; that is why HCl is not used. A is true and R explains it.
Answer: (A). | ||||
| 40 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A: Ni(0), $3d^{10}$, tetrahedral: $sp^3$ (I).
B: Ni(II), $3d^8$; strong-field $\mathrm{CN^-}$ pairs the electrons, square planar: $dsp^2$ (IV).
C: Co(III), $3d^6$, strong field, inner-orbital octahedral: $d^2sp^3$ (III).
D: Co(III) with weak-field $\mathrm{F^-}$, outer-orbital octahedral: $sp^3d^2$ (II).
A-I, B-IV, C-III, D-II.
Answer: (B). | ||||
| 41 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{N_2 + O_2 \rightarrow 2NO}$, $\Delta H \approx +180$ kJ mol$^{-1}$: strongly endothermic, and it proceeds only at very high temperature (about 2000 K, as in lightning or an electric arc).
Answer: (D). | ||||
| 42 | 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 The substrate is $\mathrm{(CH_3)_2C{=}CH{-}CH_2CH_2CH_2{-}C(CH_3)_2OH}$.
$\mathrm{H^+}$ protonates the OH and water leaves, giving a tertiary carbocation $\mathrm{{-}CH_2{-}C^+(CH_3)_2}$.
The C=C in the same chain attacks this cation through its CH end, closing a five-membered ring (the cation carbon, three $\mathrm{CH_2}$ and the CH). The positive charge moves to the other alkene carbon, now an exocyclic tertiary centre $\mathrm{{-}C^+(CH_3)_2}$; with heat it loses $\mathrm{H^+}$ to an alkene.
HBr then adds by Markovnikov's rule, putting Br on that tertiary carbon: 1-(2-bromopropan-2-yl)-2,2-dimethylcyclopentane.
Answer: (C). | ||||
| 43 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Electron-donating methyl groups activate the ring: mesitylene (three CH3) > p-xylene (two) > benzene.
Bromine deactivates (−I outweighs +R), and the nitro group deactivates strongly.
Increasing order: nitrobenzene < bromobenzene < benzene < p-xylene < mesitylene, i.e. D < B < E < A < C.
Answer: (B). | ||||
| 44 | 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 I is isopropyl phenyl ether. HI cleaves the alkyl–O bond, never the aryl–O bond (the C–O bond to the ring has partial double-bond character):
$\mathrm{C_6H_5{-}O{-}CH(CH_3)_2 + HI \rightarrow C_6H_5OH + (CH_3)_2CHI}$, so A is phenol.
Heating phenol with zinc dust reduces it to benzene: $\mathrm{C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO}$, so B is benzene.
Answer: (D). | ||||
| 45 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines Direct nitration of aniline gives p-nitroaniline (about 51%), m-nitroaniline (about 47%) and a little o-nitroaniline (about 2%): A is true.
The nitrating mixture is strongly acidic, so most aniline is protonated to the anilinium ion, whose $\mathrm{{-}NH_3^+}$ group is meta-directing; this is why a large meta fraction forms alongside ortho and para. R is true and explains A.
Answer: (A). | ||||
| 46 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter A is neoprene (polychloroprene), an elastomer: III.
B is nylon 6,6, a fibre: II.
C is PVC, a thermoplastic: IV.
D is the phenol–formaldehyde polymer (Bakelite type), thermosetting: I.
A-III, B-II, C-IV, D-I.
Answer: (B). | ||||
| 47 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Statement I: an inhibitor can also bind at an allosteric site, changing the shape of the active site so the substrate cannot bind. Blocking the active site is not the only way: false.
Statement II: some inhibitors form a strong covalent bond with the enzyme that cannot be broken easily, blocking it permanently: true.
Answer: (D). | ||||
| 48 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques TLC separates components by their different adsorption on a stationary adsorbent; the thin layer of silica gel (or alumina) on the glass plate is that adsorbent.
A is true and R explains it.
Answer: (A). | ||||
| 49 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Fructose is $\mathrm{C_6H_{12}O_6}$. HCN adds to the C=O: cyanohydrin $\mathrm{C_7H_{13}NO_6}$. Hydrolysis turns $\mathrm{{-}CN}$ into $\mathrm{{-}COOH}$ (adds $\mathrm{2H_2O}$, loses $\mathrm{NH_3}$): A = $\mathrm{C_7H_{14}O_8}$.
$\mathrm{NaBH_4}$ reduces the keto group to the hexitols (sorbitol/mannitol, $\mathrm{C_6H_{14}O_6}$); HI/P reduces all the OH groups away, giving n-hexane: B = $\mathrm{C_6H_{14}}$.
Answer: (A). | ||||
| 50 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The question shows 2-(cyclohex-1-en-1-yl)acetic acid treated with (1) $\mathrm{I_2/NaHCO_3}$, (2) pyridine, Δ.
Step 1 (iodolactonisation): $\mathrm{I_2}$ forms an iodonium ion on the ring C=C; the carboxylate attacks the ring CH carbon from the opposite face, closing a five-membered γ-lactone. Iodine ends on the ring-fusion carbon that carries the $\mathrm{CH_2}$ of the lactone, anti to the lactone oxygen.
Step 2: pyridine on heating removes HI by E2, which needs an H anti-periplanar to the C–I bond. The H on the other ring-fusion carbon is cis to I (the addition was anti), and the lactone $\mathrm{CH_2}$ hydrogens are held in the five-membered ring, out of line with the C–I bond. An axial H on the neighbouring ring $\mathrm{CH_2}$ is anti to I, so the new C=C forms inside the six-membered ring.
Product: the lactone with the C=C in the cyclohexane ring, from the ring-fusion carbon to the next ring carbon.
Answer: (D). | ||||
| 51 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{H_2SO_4}$: 0.4 mol; NaOH: 0.2 mol (limiting).
$\mathrm{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O}$: 0.2 mol NaOH gives 0.1 mol $\mathrm{Na_2SO_4}$.
Total volume 4 L: $\frac{0.1}{4} = 0.025$ M = 25 millimolar.
Answer: 25. | ||||
| 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 $\rho = \frac{ZM}{N_Aa^3}$ with Z = 4 for fcc.
$a^3 = (4.0 \times 10^{-8})^3 = 6.4 \times 10^{-23}$ cm$^3$.
$M = \frac{\rho N_Aa^3}{Z} = \frac{9.03 \times 6.02 \times 10^{23} \times 6.4 \times 10^{-23}}{4} = 86.98$ g/mol ≈ 87.
Answer: 87. | ||||
| 53 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Kinetic energy of the emitted electron: $K = \frac{h^2}{2m\lambda^2} = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (3.3 \times 10^{-10})^2} = 2.215 \times 10^{-18}$ J.
Minimum energy to escape from the ground state: 13.6 eV $= 2.179 \times 10^{-18}$ J.
Energy absorbed $= 2.179 \times 10^{-18} + 2.215 \times 10^{-18} = 4.394 \times 10^{-18}$ J.
Ratio $= \frac{4.394}{2.179} = 2.02 \approx 2$.
Answer: 2. | ||||
| 54 | SolutionsRaoult's law | Class 12: Solutions | NumericalEasy | |
A easy numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions Raoult's and Dalton's laws: $p_A = x_Ap_A^\circ = y_AP$.
$0.2\,p_A^\circ = 0.5 \times 0.8 = 0.4$, so $p_A^\circ = 2$ atm.
Answer: 2. | ||||
| 55 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{O_2}$ used: 1 − 0.6 = 0.4 mol, so NO used 0.8 mol and $\mathrm{NO_2}$ formed 0.8 mol.
At equilibrium: NO 1.2, $\mathrm{O_2}$ 0.6, $\mathrm{NO_2}$ 0.8; total 2.6 mol at 1 atm.
$K_p = \frac{p_{NO_2}^2}{p_{NO}^2\,p_{O_2}} = \frac{(0.8/2.6)^2}{(1.2/2.6)^2(0.6/2.6)} = \frac{0.64}{1.44} \times \frac{2.6}{0.6} = 1.93$ atm$^{-1}$ ≈ 2.
Answer: 2. | ||||
| 56 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Pressure of dry $\mathrm{N_2}$: 759 − 14.2 = 744.8 mm Hg = 0.980 atm.
$n = \frac{PV}{RT} = \frac{0.980 \times 0.02278}{0.082 \times 280} = 9.72 \times 10^{-4}$ mol, mass $= 9.72 \times 10^{-4} \times 28 = 0.02722$ g.
% N $= \frac{0.02722}{0.125} \times 100 = 21.8 \approx 22$.
Answer: 22. | ||||
| 57 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions $\mathrm{KIO_4 + H_2O_2 \rightarrow KIO_3 + H_2O + O_2}$
Here $\mathrm{H_2O_2}$ acts as a reducing agent; iodine goes from +7 in $\mathrm{KIO_4}$ to +5 in $\mathrm{KIO_3}$.
Answer: 5. | ||||
| 58 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$: the slope is $-\frac{E_a}{R}$.
From the graph the line falls from ln k = 20 at 1/T = 0 to 0 at 1/T = 5: slope $= -\frac{20}{5} = -4$ K.
$E_a = 4 \times 2 = 8$ cal mol$^{-1}$.
Answer: 8. | ||||
| 59 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich: $\frac{x}{m} = kP^{1/n}$, so $\log\frac{x}{m} = \log k + \frac{1}{n}\log P$: a straight line of $\log\frac{x}{m}$ against $\log P$ with a positive intercept $\log k$.
(a) a curve of $\log\frac{x}{m}$ against $\log P$: not in accordance.
(b) a straight line through the origin against P: not in accordance.
(c) a straight line against P: not in accordance.
(d) a straight line against $\log P$ with an intercept: in accordance.
Three curves are not in accordance.
Answer: 3. | ||||
| 60 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics U, V and H depend only on the state of the system: state variables.
Heat q depends on the path: not a state variable.
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Numerator: $|x + 3| - 1 \ge 0$ for $x \le -4$ or $x \ge -2$ (zero at −4 and −2).
Denominator: $|x| - 2 > 0$ for $x < -2$ or $x > 2$, and $< 0$ for $-2 < x < 2$.
Both $\ge 0$ / $> 0$: $x \le -4$ or $x > 2$. Both $\le 0$ / $< 0$: $-4 \le x \le -2$ with $-2 < x < 2$, which is empty.
So $S = [-6, -4] \cup (2, 3]$.
T: $|x|^2 - 7|x| + 9 \le 0$ for $\frac{7 - \sqrt{13}}{2} \le |x| \le \frac{7 + \sqrt{13}}{2}$, i.e. $1.70 \le |x| \le 5.30$: $T = \{\pm 2, \pm 3, \pm 4, \pm 5\}$.
$S \cap T = \{-5, -4, 3\}$: 3 elements.
Answer: (D). | ||||
| 62 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha + \beta = \sqrt{2}$, $\alpha\beta = \sqrt{6}$, so $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = \frac{2 - 2\sqrt{6}}{6} = \frac{1 - \sqrt{6}}{3}$.
$-a = \frac{1 - \sqrt{6}}{3} + 2 = \frac{7 - \sqrt{6}}{3}$.
$b = \frac{1}{(\alpha\beta)^2} + \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\right) + 1 = \frac{1}{6} + \frac{1 - \sqrt{6}}{3} + 1 = \frac{9 - 2\sqrt{6}}{6}$.
$a + b = \frac{-14 + 2\sqrt{6} + 9 - 2\sqrt{6}}{6} = -\frac{5}{6}$.
The equation is $x^2 + \frac{17}{6}x + \frac{7}{6} = 0$, i.e. $6x^2 + 17x + 7 = 0$, with roots $-\frac{1}{2}$ and $-\frac{7}{3}$: real and both negative.
Answer: (B). | ||||
| 63 | 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 $A^T = A$, $B^T = -B$.
(A) $(A^4)^T = A^4$ and $(B^4)^T = (-B)^4 = B^4$: $A^4 - B^4$ is symmetric. True.
(B) $(AB - BA)^T = B^TA^T - A^TB^T = -BA + AB$: symmetric. True.
(C) $(B^5 - A^5)^T = -B^5 - A^5$, which equals $-(B^5 - A^5)$ only if $A^5 = O$. Not true in general.
(D) $(AB + BA)^T = -BA - AB$: skew-symmetric. True.
Answer: (C). | ||||
| 64 | 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 $a + b + c = 3$, $9a + 3b + c = 4$, $4a - 2b + c = \lambda$, $f(0) = c$.
The sum: $c + 3 + \lambda + 4 = 14$ gives $c + \lambda = 7$.
Subtracting the first two: $8a + 2b = 1$.
$\lambda = 4a - 2b + c$ and $c = 3 - a - b$ give $\lambda = 3a - 3b + 3$; with $c + \lambda = 7$: $(3 - a - b) + (3a - 3b + 3) = 7$, so $2a - 4b = 1$.
Solving with $8a + 2b = 1$: $a = \frac{1}{6}$, $b = -\frac{1}{6}$, $c = 3$.
$\lambda = 7 - c = 4$ (check: $4 \cdot \frac{1}{6} + \frac{2}{6} + 3 = 4$).
Answer: (D). | ||||
| 65 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $|x| < 1$, $x^{2n} \to 0$: $f(x) = \cos 2\pi x$.
For $|x| > 1$, divide by $x^{2n}$: $f(x) = \frac{-\sin(x - 1)}{x - 1}$.
At $x = 1$: $f(1) = \frac{1 - 0}{1 + 1 - 1} = 1$; left limit $\cos 2\pi = 1$; right limit $-\lim \frac{\sin(x - 1)}{x - 1} = -1$. Discontinuous.
At $x = -1$: $f(-1) = \frac{1 - \sin(-2)}{1 - 1 - 1} = -(1 + \sin 2)$; right limit $\cos(-2\pi) = 1$. Discontinuous.
Elsewhere f is built from continuous pieces. So f is continuous on $\mathbb{R} - \{-1, 1\}$.
Answer: (B). | ||||
| 66 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = e^{x(1 - x)}\left[1 + x(1 - 2x)\right] = -e^{x(1 - x)}(2x^2 - x - 1) = -e^{x(1 - x)}(2x + 1)(x - 1)$.
$f'(x) > 0$ exactly for $-\frac{1}{2} < x < 1$: f increases on $\left(-\frac{1}{2}, 1\right)$ and decreases outside it.
(B), (C), (D) each contain an interval of the wrong sign.
Answer: (A). | ||||
| 67 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $\sin x - \cos x = \sqrt{2}\sin\left(x - \frac{\pi}{4}\right)$ and $x - \frac{\pi}{4} \in \left[-\frac{\pi}{4}, \frac{3\pi}{4}\right]$, so it ranges over $[-1, \sqrt{2}]$.
$\tan^{-1}$ is increasing: maximum $\tan^{-1}\sqrt{2}$, minimum $\tan^{-1}(-1) = -\frac{\pi}{4}$.
Sum $= \tan^{-1}\sqrt{2} - \frac{\pi}{4}$. An angle with $\tan\theta = \sqrt{2}$ has $\cos\theta = \frac{1}{\sqrt{3}}$, so this is $\cos^{-1}\left(\frac{1}{\sqrt{3}}\right) - \frac{\pi}{4}$.
Answer: (C). | ||||
| 68 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctHard | |
A hard single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $x^2 + y^2 = 8\sin 2t$ and $xy = 8\sin t\cos t\sin 2t = 4\sin^2 2t$, so $(x^2 + y^2)^2 = 16xy$.
At $t = \frac{\pi}{4}$: $x = y = 2$.
Differentiate $(x^2 + y^2)^2 = 16xy$: $(x^2 + y^2)(x + yy') = 4(y + xy')$. At (2, 2): $8(2 + 2y') = 4(2 + 2y')$, so $y' = -1$.
Differentiate again: $(2x + 2yy')(x + yy') + (x^2 + y^2)(1 + y'^2 + yy'') = 4(2y' + xy'')$.
At (2, 2), $y' = -1$: the first term is 0, so $8(2 + 2y'') = 4(-2 + 2y'')$, giving $16 + 16y'' = -8 + 8y''$ and $y'' = -3$.
$\frac{1 + 1}{-3} = -\frac{2}{3}$.
Answer: (D). | ||||
| 69 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Integrate by parts with 1 as the second function:
$I_n = \frac{x}{(x^2 + 5)^n} + 2n\int_0^x \frac{t^2}{(t^2 + 5)^{n+1}}dt = \frac{x}{(x^2 + 5)^n} + 2n(I_n - 5I_{n+1})$.
So $10nI_{n+1} = \frac{x}{(x^2 + 5)^n} + (2n - 1)I_n$.
With n = 5 and $I_5' = \frac{1}{(x^2 + 5)^5}$: $50I_6 = xI_5' + 9I_5$, i.e. $50I_6 - 9I_5 = xI_5'$.
Answer: (A). | ||||
| 70 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $y = \log_e(x + e^2)$ is $x = e^y - e^2$ (rising, through (0, 2) and $(e - e^2, 1)$); $x = \log_e\frac{2}{y}$ is $y = 2e^{-x}$ (falling, through (0, 2) and $(\log_e 2, 1)$). The two meet at (0, 2), and the line $x = \log_e 2$ meets the second curve on y = 1.
The region above y = 1 lies between them for $1 \le y \le 2$:
Area $= \int_1^2\left[\log_e\frac{2}{y} - (e^y - e^2)\right]dy$.
$\int_1^2(\log_e 2 - \log_e y)dy = \log_e 2 - (2\log_e 2 - 1) = 1 - \log_e 2$.
$\int_1^2(e^2 - e^y)dy = e^2 - (e^2 - e) = e$.
Area $= 1 + e - \log_e 2$.
Answer: (B). | ||||
| 71 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor: $e^{\int\frac{dx}{x^2 - 1}} = \left(\frac{x - 1}{x + 1}\right)^{1/2}$.
$\frac{d}{dx}\left[y\left(\frac{x - 1}{x + 1}\right)^{1/2}\right] = \frac{x - 1}{x + 1} = 1 - \frac{2}{x + 1}$, so $y\left(\frac{x - 1}{x + 1}\right)^{1/2} = x - 2\log_e(x + 1) + C$.
At x = 2: $\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = 2 - 2\log_e 3 + C$, so $C = 2\log_e 3 - \frac{5}{3}$.
At x = 8: $y \cdot \frac{\sqrt{7}}{3} = 8 - 2\log_e 9 + 2\log_e 3 - \frac{5}{3} = \frac{19}{3} - 2\log_e 3$.
$\sqrt{7}\,y(8) = 19 - 6\log_e 3$.
Answer: (D). | ||||
| 72 | Differential EquationsOrder and degree | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on order and degree, from Differential Equations. NCERT: Class 12: Differential Equations The centres lie on the perpendicular bisector of the chord, the x-axis: $x^2 + y^2 + 2gx - 4 = 0$ (it passes through (0, ±2)).
So $2g = \frac{4 - x^2 - y^2}{x}$. Differentiating the circle: $2x + 2yy' + 2g = 0$.
$2x + 2yy' + \frac{4 - x^2 - y^2}{x} = 0 \Rightarrow 2x^2 + 2xyy' + 4 - x^2 - y^2 = 0$.
$2xy\frac{dy}{dx} + (x^2 - y^2 + 4) = 0$.
Answer: (A). | ||||
| 73 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Centre C(2, 0), radius r = 1. Tangent length from O: $L = \sqrt{S(0, 0)} = \sqrt{3}$.
OC = 2 and $\angle AOC = 30^\circ$ ($\sin = \frac{1}{2}$), so $\angle AOB = 60^\circ$ and OA = OB = $\sqrt{3}$.
Area $= \frac{1}{2}(\sqrt{3})^2\sin 60^\circ = \frac{3}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}$.
Answer: (B). | ||||
| 74 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Focus of the parabola (ae, 0); its directrix passes through (−ae, 0) and, being perpendicular to the axis through the focus, is $x = -ae$.
Focus-to-directrix distance 2ae, so the parabola's latus rectum is $2 \times 2ae = 4ae$.
$4ae = e \cdot \frac{2b^2}{a}$ gives $b^2 = 2a^2$.
Through $(2\sqrt{2}, -2\sqrt{2})$: $\frac{8}{a^2} - \frac{8}{2a^2} = 1$, so $a^2 = 4$, $b^2 = 8$, $e = \sqrt{1 + 2} = \sqrt{3}$, $ae = 2\sqrt{3}$.
Vertex at the origin: the parabola is $y^2 = 4(2\sqrt{3})x = 8\sqrt{3}x$.
(A) 18 vs 48; (B) $72$ vs $8\sqrt{3} \cdot 3\sqrt{3} = 72$ ✓; (C) 6 vs 24; (D) 72 vs $72\sqrt{2}$.
Answer: (B). | ||||
| 75 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Coplanarity: $\begin{vmatrix}-27 & -20 & -31\\ \lambda & 1 & 2\\ -2 & 3 & \lambda\end{vmatrix} = 0$ (first row: the vector between the given points).
Expanding: $-27(\lambda - 6) + 20(\lambda^2 + 4) - 31(3\lambda + 2) = 20\lambda^2 - 120\lambda + 180 = 20(\lambda - 3)^2 = 0$, so $\lambda = 3$.
Normal: $(3, 1, 2) \times (-2, 3, 3) = (-3, -13, 11)$. Plane through (1, 2, 3): $3x + 13y - 11z + 4 = 0$.
(A) 0 − 26 + 22 + 4 = 0; (B) −15 + 0 + 11 + 4 = 0; (C) 9 − 13 + 0 + 4 = 0; (D) 0 + 52 − 55 + 4 = 1 ≠ 0.
Answer: (D). | ||||
| 76 | 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 Normal: $(-2, 1, -3) \times (-1, 2, -2) = (4, -1, -3)$.
Plane through (2, 2, −2): $4(x - 2) - (y - 2) - 3(z + 2) = 0$, i.e. $4x - y - 3z = 12$.
Intercepts: $\alpha = 3$, $\beta = -12$, $\gamma = -4$; $p = -13$.
$V = \frac{1}{6}|\alpha\beta\gamma| = \frac{144}{6} = 24$.
(V, p) = (24, −13).
Answer: (B). | ||||
| 77 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Let $t = \log_e b > 0$. Acute angle needs $\vec{u}\cdot\vec{v} > 0$: $at^2 - 12 + 6at > 0$, i.e. $a(t^2 + 6t) > 12$, for every $b > 1$.
As $t \to 0^+$ the left side tends to 0 for any fixed a, so the inequality fails for b close to 1 whatever a is.
No real a works: $S = \Phi$.
Answer: (B). | ||||
| 78 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let OP = h. Then $OA = OB = h\cot 15^\circ$, so triangle OAB is isosceles and OC ⊥ AB; $\angle PCO = 45^\circ$ gives OC = h.
$OA^2 = OC^2 + AC^2$: $h^2\cot^2 15^\circ = h^2 + 64$.
$\cot^2 15^\circ = (2 + \sqrt{3})^2 = 7 + 4\sqrt{3}$, so $h^2(6 + 4\sqrt{3}) = 64$.
$h^2 = \frac{32}{3 + 2\sqrt{3}} = \frac{32(2\sqrt{3} - 3)}{3} = \frac{32}{\sqrt{3}}(2 - \sqrt{3})$.
Answer: (B). | ||||
| 79 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(A) = \frac{1/9}{2/5} = \frac{5}{18}$ and $P(B) = \frac{1/9}{1/7} = \frac{7}{9} = \frac{14}{18}$.
(S1) $P(A' \cup B) = 1 - P(A \cap B') = 1 - \left(\frac{5}{18} - \frac{2}{18}\right) = \frac{15}{18} = \frac{5}{6}$: true.
(S2) $P(A \cup B) = \frac{5 + 14 - 2}{18} = \frac{17}{18}$, so $P(A' \cap B') = \frac{1}{18}$: true.
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 "P only if Q" means $P \Rightarrow Q$.
P is p; Q is "he is in his village" ($\sim q$) and "it is Sunday or Saturday" ($r \vee s$).
$p \Rightarrow ((\sim q) \wedge (r \vee s))$.
Answer: (D). | ||||
| 81 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Middle-term coefficients: $\binom{4}{2}\frac{1}{6}\beta^2 = \beta^2$; $2(-3\beta) = -6\beta$; $\binom{6}{3}\left(-\frac{\beta}{2}\right)^3 = -\frac{5}{2}\beta^3$.
A.P.: $2(-6\beta) = \beta^2 - \frac{5}{2}\beta^3$; dividing by $\beta$: $5\beta^2 - 2\beta - 24 = 0$, so $\beta = \frac{12}{5}$ ($\beta > 0$).
$d = -6\beta - \beta^2$, so $\frac{2d}{\beta^2} = -\frac{12}{\beta} - 2 = -5 - 2 = -7$.
$50 - (-7) = 57$.
Answer: 57. | ||||
| 82 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $\binom{b}{3}\binom{g}{2} = 168 = 2^3 \cdot 3 \cdot 7$.
Values of $\binom{b}{3}$: 1, 4, 10, 20, 35, 56, 84, 120, …; of $\binom{g}{2}$: 1, 3, 6, 10, 15, 21, 28, …
The only product equal to 168 is $56 \times 3$: b = 8, g = 3.
b + 3g = 8 + 9 = 17.
Answer: 17. | ||||
| 83 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The major axis is the y-axis ($b^2 = 4 > a^2 = 2$): $e^2 = 1 - \frac{2}{4} = \frac{1}{2}$, foci $(0, \pm\sqrt{2})$, so S = $(0, -\sqrt{2})$.
Tangent at $(\sqrt{2}\cos\theta, 2\sin\theta)$: $\frac{x\cos\theta}{\sqrt{2}} + \frac{y\sin\theta}{2} = 1$. Through R: $\cos\theta + (\sqrt{2} - 1)\sin\theta = 1$.
With $u = \tan\frac{\theta}{2}$: $1 - u^2 + 2(\sqrt{2} - 1)u = 1 + u^2$, so $u = 0$ or $u = \sqrt{2} - 1 = \tan\frac{\pi}{8}$: $\theta = 0$ or $\frac{\pi}{4}$.
P = $(\sqrt{2}, 0)$, Q = $(1, \sqrt{2})$.
$SP^2 = 2 + 2 = 4$, $SQ^2 = 1 + (2\sqrt{2})^2 = 9$. Sum 13.
Answer: 13. | ||||
| 84 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $2 + {}^{49}C_1 + \ldots + {}^{49}C_{49} = 1 + 2^{49}$.
${}^{50}C_0 + {}^{50}C_2 + \ldots + {}^{50}C_{50} = 2^{49}$, so ${}^{50}C_2 + \ldots + {}^{50}C_{50} = 2^{49} - 1$.
The expression is $1 + (2^{49} + 1)(2^{49} - 1) = 2^{98}$: n = 98, m = 1.
n + m = 99.
Answer: 99. | ||||
| 85 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = -\frac{x}{2} = -4ax$ with $a = \frac{1}{8}$; its tangent of slope m is $y = mx - \frac{a}{m}$.
Through (2, 0): $2m = \frac{1}{8m}$, so $m^2 = \frac{1}{16}$, $m = \pm\frac{1}{4}$: the lines $x \mp 4y - 2 = 0$.
Distance from the centre (5, 0): $\frac{|5 - 2|}{\sqrt{17}} = \frac{3}{\sqrt{17}}$, which is the radius $\sqrt{r}$.
$r = \frac{9}{17}$, so 17r = 9.
Answer: 9. | ||||
| 86 | 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 After the first term the terms are $\frac{10 \cdot 2^k}{3^{11 - k}}$ for k = 0 to 10, a G.P. with ratio 6:
$\frac{10}{3^{11}}(1 + 6 + \ldots + 6^{10}) = \frac{10}{3^{11}} \cdot \frac{6^{11} - 1}{5} = 2^{12} - \frac{2}{3^{11}}$.
Adding $\frac{6}{3^{12}} = \frac{2}{3^{11}}$ gives exactly $2^{12}$.
n = 12, m = 1, so m · n = 12.
Answer: 12. | ||||
| 87 | 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 Rearranging: $\tan\theta + \tan 2\theta = \sqrt{5}(1 - \tan\theta\tan 2\theta)$, so $\tan 3\theta = \sqrt{5}$ (if $1 - \tan\theta\tan 2\theta = 0$ the equation would force $\tan\theta + \tan 2\theta = 0$ as well, which is impossible).
$3\theta = \alpha + k\pi$ with $\alpha = \tan^{-1}\sqrt{5} \approx 1.150$, and $3\theta \in [-3\pi, \frac{3\pi}{2})$, i.e. $[-9.42, 4.71)$.
k = −3, −2, −1, 0, 1 give $3\theta \approx -8.27, -5.13, -1.99, 1.15, 4.29$: θ ≈ −2.76, −1.71, −0.66, 0.38, 1.43.
None is an excluded point or makes $\tan\theta$ or $\tan 2\theta$ undefined.
Answer: 5. | ||||
| 88 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Moduli: $|z|^2 = |z| \cdot 2^{1 - |z|}$; $z \neq 0$ (as $b \neq 0$), so $|z| = 2^{1 - |z|}$, whose only solution is $|z| = 1$.
Then $z^2 = \bar{z}$, so $z^3 = z\bar{z} = 1$; with $b \neq 0$, $z = \omega$ or $\omega^2$.
For $z = \omega$: $z + 1 = -\omega^2$, and $\left(\frac{z}{z + 1}\right)^n = (-\omega^{-1})^n = (-1)^n\omega^{2n} = 1$ needs n even and a multiple of 3 (the same for $z = \omega^2$).
Least n = 6.
Answer: 6. | ||||
| 89 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(X = 0) = \frac{\binom{6}{3}}{\binom{10}{3}} = \frac{20}{120}$, $P(1) = \frac{4 \cdot 15}{120} = \frac{60}{120}$, $P(2) = \frac{6 \cdot 6}{120} = \frac{36}{120}$, $P(3) = \frac{4}{120}$.
$E(X) = \frac{60 + 72 + 12}{120} = 1.2$; $E(X^2) = \frac{60 + 144 + 36}{120} = 2$.
$\sigma^2 = 2 - 1.44 = 0.56$, so $100\sigma^2 = 56$.
Answer: 56. | ||||
| 90 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{\sin 6x}{\sin x} = 2(\cos 5x + \cos 3x + \cos x)$ (since $2\sin x\cos kx = \sin(k + 1)x - \sin(k - 1)x$ telescopes).
$\int_0^{\pi/2} 2(\cos 5x + \cos 3x + \cos x)dx = 2\left[\frac{\sin 5x}{5} + \frac{\sin 3x}{3} + \sin x\right]_0^{\pi/2} = 2\left(\frac{1}{5} - \frac{1}{3} + 1\right) = \frac{26}{15}$.
$60 \times \frac{26}{15} = 104$.
Answer: 104. | ||||
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