JEE (Main) · 2021
JEE (Main) 2021 — 25 July, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Atoms and Nuclei2
- Kinematics2
- Dual Nature of Matter and Radiation2
20 easy · 10 medium · 0 hard
Chemistry
30 Q
- Amines3
- Coordination Compounds2
- Aldehydes, Ketones and Carboxylic Acids2
22 easy · 7 medium · 1 hard
Mathematics
30 Q
- Conic Sections3
- Sequences and Series3
- Matrices and Determinants3
7 easy · 22 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | 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 Number of nuclei: $N = \frac{2\times10^{-3}}{198}\times6.02\times10^{23} = 6.08\times10^{18}$.
Decay constant: $\lambda = \frac{0.693}{3\times86400\ \mathrm{s}} = 2.67\times10^{-6}\ \mathrm{s^{-1}}$.
Activity: $A = \lambda N = 2.67\times10^{-6}\times6.08\times10^{18} \approx 1.62\times10^{13} = 16.2\times10^{12}$ disintegrations per second.
This is option (A); (B) is $\lambda$ alone and (C) is $N$ alone, (D) is twice the activity.
Answer: (A). | ||||
| 2 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Thin slabs of thickness $dx$ are in series, so $\frac{1}{C} = \int_0^d \frac{dx}{\varepsilon(x)A}$.
First half: $\int_0^{d/2}\frac{dx}{\varepsilon_0+kx} = \frac{1}{k}\ln\frac{\varepsilon_0+kd/2}{\varepsilon_0} = \frac{1}{k}\ln\frac{2\varepsilon_0+kd}{2\varepsilon_0}$.
The second half is the mirror image and gives the same value.
So $\frac{1}{C} = \frac{2}{kA}\ln\frac{2\varepsilon_0+kd}{2\varepsilon_0}$ and $C = \frac{kA}{2\ln\left(\frac{2\varepsilon_0+kd}{2\varepsilon_0}\right)}$.
Answer: (A). | ||||
| 3 | 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 After 4 s the first droplet has fallen $\frac{1}{2}\times9.8\times4^2 = 78.4$ m.
The next droplet is 34.3 m behind it, so it has fallen $78.4 - 34.3 = 44.1$ m: $4.9t^2 = 44.1$, $t = 3$ s.
The droplets leave the tap 1 s apart: 1 drop per second.
Answer: (A). | ||||
| 4 | Communication SystemsAmplitude modulation and bandwidth | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation and bandwidth, from Communication Systems. NCERT: No matching NCERT chapter Message frequency $f_m = 10^5$ Hz $= 100$ kHz; carrier $f_c = 10^7$ Hz.
Side bands at $f_c \pm f_m$, so the bandwidth is $(f_c+f_m)-(f_c-f_m) = 2f_m = 200$ kHz.
Answer: (A). | ||||
| 5 | 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 A quarter decayed: $N = \frac{3}{4}N_0$, so $t_1 = \frac{1}{\lambda}\ln\frac{4}{3}$.
Half decayed: $t_2 = \frac{1}{\lambda}\ln2$.
$t_2 - t_1 = \frac{1}{\lambda}\ln\left(2\times\frac{3}{4}\right) = \frac{\ln\frac{3}{2}}{\lambda}$.
Answer: (A). | ||||
| 6 | 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 $\lambda = \frac{h}{\sqrt{2mK}}$; at equal kinetic energy $\lambda \propto \frac{1}{\sqrt{m}}$.
$m_e < m_p < m_\alpha$, so $\lambda_e > \lambda_p > \lambda_\alpha$.
Answer: (C). | ||||
| 7 | 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 Momentum is conserved from rest, so the two fragments have equal and opposite momenta $p$.
$\lambda = \frac{h}{p}$ is the same for both: ratio 1 : 1.
Answer: (A). | ||||
| 8 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Fringe width $\beta = \frac{\lambda D}{d}$; $d$ ranges from $d_0 - a_0$ to $d_0 + a_0$.
$\beta_{max} - \beta_{min} = \lambda D\left(\frac{1}{d_0-a_0} - \frac{1}{d_0+a_0}\right) = \frac{2\lambda Da_0}{d_0^2-a_0^2}$.
Answer: (D). | ||||
| 9 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Critical angle: $\sin C = \frac{1}{2.42} = 0.413$, so $C \approx 24.4^\circ$.
The angle of incidence, $30^\circ$, exceeds $C$; Snell's law would need $\sin r = 2.42\times0.5 = 1.21 > 1$.
The ray is totally internally reflected: refraction is not possible.
Answer: (D). | ||||
| 10 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves From the wave: $k = 1.8\ \mathrm{m^{-1}}$, $\omega = 5.4\times10^6\ \mathrm{rad/s}$.
(A) $f = \frac{\omega}{2\pi} = 8.6\times10^5$ Hz, not $54\times10^4$ Hz: false.
(D) $\lambda = \frac{2\pi}{k} = 3.49$ m, not 5.4 m: false.
(C) A perfectly reflecting wall transmits nothing: false.
(B) The reflected wave travels in $-z$ with the same amplitude, $k$ and $\omega$: $3.1\cos[(1.8)z + (5.4\times10^6)t]\hat{i}$. This is the intended answer.
Answer: (B). | ||||
| 11 | Current ElectricityMetre bridge | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity Wire resistance: $1000\ \mathrm{cm}\times0.1\ \Omega/\mathrm{cm} = 100\ \Omega$.
Driver current: $I = \frac{6}{100+20} = 0.05$ A.
Potential across the whole wire: $0.05\times100 = 5$ V, the largest emf that can be balanced.
Answer: (C). | ||||
| 12 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Each input gate is a NAND with its inputs joined, i.e. a NOT gate: outputs $\bar{A}$ and $\bar{B}$.
These feed a NOR gate: $Y = \overline{\bar{A}+\bar{B}} = A\cdot B$ (De Morgan).
The circuit is an AND gate.
Answer: (A). | ||||
| 13 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $C_P - C_V = R$ holds exactly for an ideal gas. A real gas departs from it (here $1.10R$) when it departs from ideal behaviour, which happens at lower temperature.
So in state P the gas is closer to ideal, i.e. hotter: $T_P > T_Q$.
Answer: (A). | ||||
| 14 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Same heating rate $P$ and same mass: $P = mc\frac{dT}{dt}$, so $c \propto \frac{1}{dT/dt}$.
Slopes: A rises 120 °C in 3 s (40 °C/s); B rises 90 °C in 6 s (15 °C/s).
$\frac{c_A}{c_B} = \frac{15}{40} = \frac{3}{8}$.
Answer: (D). | ||||
| 15 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Assertion and reasonEasy | |
A easy assertion and reason question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For a disc, $I_x = I_y = \frac{MR^2}{4}$ and $I_z = \frac{MR^2}{2}$ (perpendicular axes).
Radii of gyration: $K_x = K_y = \frac{R}{2}$ but $K_z = \frac{R}{\sqrt{2}}$ — not all the same, so A is false.
R is the definition of a rigid body: true.
Answer: (D). | ||||
| 16 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Only the component of velocity normal to the wall (along X) reverses.
Ball a strikes head-on: impulse $2mu$.
Ball b strikes at $45^\circ$: normal component $u\cos45^\circ$, impulse $2mu\cos45^\circ = \sqrt{2}\,mu$.
Ratio $= \frac{2mu}{\sqrt{2}mu} = \sqrt{2} : 1$.
Answer: (D). | ||||
| 17 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Matching listEasy | |
A easy matching list question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Read each triangle head-to-tail.
(i) $\vec{A}$ (up) then $\vec{C}$ (right) reach the head of $\vec{B}$: $\vec{B} = \vec{A}+\vec{C}$, i.e. (c).
(ii) $\vec{A}$, $\vec{B}$, $\vec{C}$ go round the triangle and close: $\vec{A}+\vec{B}+\vec{C} = 0$, i.e. (d).
(iii) $\vec{C}$ then $\vec{B}$ end where $\vec{A}$ ends: $\vec{A} = \vec{C}+\vec{B}$, i.e. (b).
(iv) $\vec{A}$ then $\vec{B}$ end where $\vec{C}$ ends: $\vec{C} = \vec{A}+\vec{B}$, i.e. (a).
So (a)→(iv), (b)→(iii), (c)→(i), (d)→(ii).
Answer: (A). | ||||
| 18 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The same tension $F$ acts in both. Extensions add: $\frac{FL}{AY_1} + \frac{FL}{AY_2} = \frac{F(2L)}{AY}$.
$\frac{2}{Y} = \frac{1}{Y_1}+\frac{1}{Y_2}$, so $Y = \frac{2Y_1Y_2}{Y_1+Y_2}$.
Answer: (C). | ||||
| 19 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Angular momentum about the Sun is conserved; at the nearest and farthest points the velocity is perpendicular to the radius.
The minimum speed $v_0$ is at $x_2$: $v_{max}x_1 = v_0x_2$, so $v_{max} = \frac{v_0x_2}{x_1}$.
Answer: (C). | ||||
| 20 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics For an adiabatic process $TV^{\gamma-1} = $ constant, with $\gamma = \frac{5}{3}$ for a monoatomic gas and $V \propto l$.
$T_1l_1^{2/3} = T_2l_2^{2/3}$, so $\frac{T_1}{T_2} = \left(\frac{l_2}{l_1}\right)^{2/3}$.
Answer: (D). | ||||
| 21 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power In a one-dimensional elastic collision with a body at rest: $v_1 = \frac{m_1-m_2}{m_1+m_2}u$.
$\frac{2-m_2}{2+m_2} = \frac{1}{4}$ gives $8 - 4m_2 = 2 + m_2$, $m_2 = 1.2$ kg.
$v_{cm} = \frac{2\times4}{2+1.2} = 2.5$ m/s $= \frac{25}{10}$ m/s.
Answer: 25. | ||||
| 22 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The springs $k$ and $4k$ are in series: $k_{eq} = \frac{k\cdot4k}{5k} = \frac{4k}{5} = 16$ N/m.
Two free bodies oscillate with the reduced mass $\mu = \frac{0.2\times0.8}{0.2+0.8} = 0.16$ kg.
$\omega = \sqrt{\frac{k_{eq}}{\mu}} = \sqrt{\frac{16}{0.16}} = 10$ rad/s.
Answer: 10. | ||||
| 23 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Rise in height: $h = L(1-\cos60^\circ) = 0.5\times0.5 = 0.25$ m.
$v^2 = u^2 - 2gh = 9 - 2\times10\times0.25 = 4$, so $v = 2$ m/s.
Answer: 2. | ||||
| 24 | 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 Steady current: $I_0 = \frac{20}{10^4} = 2$ mA. Time constant: $\tau = \frac{L}{R} = \frac{10^{-2}}{10^4} = 10^{-6}$ s $= 1\ \mu$s.
With the battery removed the current decays through the resistor: $I = I_0e^{-t/\tau} = 2\times e^{-1} = 2\times0.37 = 0.74$ mA $= \frac{74}{100}$ mA.
Answer: 74. | ||||
| 25 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\vec{v} = 20\alpha t\hat{i} + 5\beta\hat{j}$.
$\vec{L} = m\vec{r}\times\vec{v} = m\left[10\alpha t^2\cdot5\beta - 5\beta(t-5)\cdot20\alpha t\right]\hat{k} = 50m\alpha\beta\left(t^2 - 2t^2 + 10t\right)\hat{k} = 50m\alpha\beta(10t - t^2)\hat{k}$.
At $t = 0$, $\vec{L} = 0$; it is zero again when $10t - t^2 = 0$, i.e. $t = 10$ s.
Answer: 10. | ||||
| 26 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Flux: $\phi = B\pi r^2 = 4\times10^{-3}\left(1-\frac{t}{100}\right)$ Wb, so $\varepsilon = \left|\frac{d\phi}{dt}\right| = 4\times10^{-5}$ V, constant.
Power: $P = \frac{\varepsilon^2}{R} = \frac{16\times10^{-10}}{2\times10^{-6}} = 8\times10^{-4}$ W.
The field reaches zero at $t = 100$ s: $E = 8\times10^{-4}\times100 = 0.08$ J $= 80$ mJ.
Answer: 80. | ||||
| 27 | 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 Bulb: $R_b = \frac{V^2}{P} = \frac{100^2}{200} = 50\ \Omega$, rated current $\frac{200}{100} = 2$ A.
On 200 V the total resistance must be $\frac{200}{2} = 100\ \Omega$, so $R = 100 - 50 = 50\ \Omega$.
Answer: 50. | ||||
| 28 | 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 Displaced by $x$ along the line towards one charge (each fixed charge 1 m away):
$F = \frac{kq^2}{(1-x)^2} - \frac{kq^2}{(1+x)^2} \approx kq^2\cdot4x$ (restoring).
$\omega = \sqrt{\frac{4kq^2}{m}} = \sqrt{\frac{4\times9\times10^9\times10}{10^{-6}}} = \sqrt{3.6\times10^{17}} = 6\times10^8$ rad/s $= 6000\times10^5$ rad/s.
Answer: 6000. | ||||
| 29 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{0.1}{100} = 0.001$ cm.
Gauge A reads 5 divisions with the jaws closed: zero error $+0.005$ cm. Gauge B reads 92: the zero lies 8 divisions short, zero error $-0.008$ cm.
A shows $0.322 + 0.005 = 0.327$ cm: main scale 0.3 cm, circular reading 27.
B shows $0.322 - 0.008 = 0.314$ cm: main scale 0.3 cm, circular reading 14.
$|27 - 14| = 13$.
Answer: 13. | ||||
| 30 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B = \mu_0(1+\chi)H$, so the field rises by the fraction $\chi$.
Percentage increase $= \chi\times100 = 2.2\times10^{-3}\ \% = \frac{22}{10^4}$.
Answer: 22. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics (a) Rate constant with time: rate independent of concentration, zero order.
(b) $t_{1/2}$ proportional to initial concentration: $t_{1/2} = \frac{[A]_0}{2k}$, zero order.
(c) Concentration falling exponentially with time: $[A] = [A]_0e^{-kt}$, first order.
(d) Concentration constant with time: no reaction at all; it fits neither order.
(e) Rate proportional to concentration: rate $= k[A]$, first order.
So (a) and (b) are zero order, (c) and (e) first order.
Answer: (A). | ||||
| 32 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter In oil (a non-polar medium) the micelle is reversed: the polar $\mathrm{-COO^-Na^+}$ heads cluster at the centre, away from the oil, and the non-polar hydrocarbon tails $\mathrm{CH_3(CH_2)_{16}-}$ point outwards into the oil.
Such reverse micelles are spherical.
Answer: (A). | ||||
| 33 | 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 $\mathrm{Na^+}$, $\mathrm{Mg^{2+}}$, $\mathrm{Al^{3+}}$ are isoelectronic (10 electrons): the radius falls as the nuclear charge rises, $\mathrm{Al^{3+} < Mg^{2+} < Na^+}$.
$\mathrm{K^+}$ has an extra shell and is the largest.
Order: $\mathrm{Al^{3+} < Mg^{2+} < Na^+ < K^+}$.
Answer: (C). | ||||
| 34 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Bauxite carries $\mathrm{SiO_2}$, iron oxides and $\mathrm{TiO_2}$ as impurities. Hot concentrated NaOH dissolves $\mathrm{Al_2O_3}$ as $\mathrm{Na[Al(OH)_4]}$, and silica also dissolves as sodium silicate.
$\mathrm{Fe_2O_3}$ and $\mathrm{TiO_2}$ stay behind as residue; ZnO is not a constituent of bauxite.
Answer: (A). | ||||
| 35 | 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 Bond dissociation enthalpies at 298 K: $\mathrm{H_2}$ 435.88 kJ/mol, $\mathrm{D_2}$ 443.35 kJ/mol (the heavier isotope has a lower zero-point energy, so its bond is slightly stronger).
$E_D - E_H \approx 7.5$ kJ/mol, so $E_H \simeq E_D - 7.5$.
Answer: (B). | ||||
| 36 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | StatementsEasy | |
A easy statements question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Statement I is false: $\mathrm{Be(OH)_2}$ is amphoteric and dissolves in alkali as $\mathrm{[Be(OH)_4]^{2-}}$.
Statement II is true: the lattice enthalpy falls faster than the hydration enthalpy down the group, so solubility rises from $\mathrm{Be(OH)_2}$ to $\mathrm{Ba(OH)_2}$.
Answer: (D). | ||||
| 37 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Silicon can expand its coordination to six only with small ligands such as F. Six large chloride ions cannot fit around the small Si atom, and the lone pairs of Cl interact weakly with Si, so $\mathrm{[SiCl_6]^{2-}}$ is not known.
$\mathrm{[SiF_6]^{2-}}$, $\mathrm{[GeCl_6]^{2-}}$ and $\mathrm{[Sn(OH)_6]^{2-}}$ all exist.
Answer: (B). | ||||
| 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 Oxidation numbers: Cr in $\mathrm{CrO_3}$ +6, V in $\mathrm{V_2O_5}$ +5, Mn in $\mathrm{MnO_2}$ +4, Fe in $\mathrm{Fe_2O_3}$ +3, Cu in $\mathrm{Cu_2O}$ +1.
Order: (a) > (d) > (c) > (b) > (e).
Answer: (B). | ||||
| 39 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A species responds (is attracted) when it has unpaired electrons.
$\mathrm{[Fe(H_2O)_6]^{3+}}$: $\mathrm{Fe^{3+}}$ $3d^5$ with weak-field water, high spin, 5 unpaired electrons — paramagnetic.
$\mathrm{[Ni(CN)_4]^{2-}}$: $\mathrm{Ni^{2+}}$ $3d^8$, square planar ($dsp^2$), all paired.
$\mathrm{[Co(CN)_6]^{3-}}$: $\mathrm{Co^{3+}}$ $3d^6$, strong field, $t_{2g}^6$, all paired.
$\mathrm{[Ni(CO)_4]}$: Ni(0) $3d^{10}$, all paired.
Answer: (A). | ||||
| 40 | Environmental ChemistryGreen chemistry | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on green chemistry, from Environmental Chemistry. NCERT: No matching NCERT chapter Dry-cleaning uses non-aqueous solvents: chlorinated solvents such as tetrachloroethene $\mathrm{Cl_2C=CCl_2}$ (and carbon tetrachloride as a cleaning fluid), and, as the greener replacement, liquid $\mathrm{CO_2}$ with a suitable detergent.
$\mathrm{H_2O_2}$ is used for bleaching clothes in laundry (with water), not as a dry-cleaning agent.
Answer: (D). | ||||
| 41 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The molecule is 1-nitrobuta-1,3-diene, $\mathrm{CH_2=CH-CH=CH-NO_2}$. The nitro group withdraws electrons by resonance ($-R$), so the $\pi$ electrons of the C3=C4 bond move towards C4 and the nitro group.
(B) $\mathrm{C3^{+}}$, $\mathrm{C4^{-}}$ next to $\mathrm{N^+}$: polarisation in the direction of the $-R$ effect; valid.
(D) The full shift onto oxygen with the charge relayed to the terminal carbon: $\mathrm{^+CH_2-CH=CH-CH=N^+(O^-)_2}$; valid.
(C) Heterolysis of the terminal C=C, leaving the C3=C4 bond and nitro group intact; a valid (minor) form.
(A) puts a positive charge on C4, directly bonded to the already positive nitrogen, with the negative charge pushed away from the electron-withdrawing group — opposite to the $-R$ effect, with like charges on adjacent atoms. This is the incorrect structure.
Answer: (A). | ||||
| 42 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons B, $\mathrm{C_3H_6O}$, formed both by ozonolysis and by acidic permanganate, must be a ketone that survives oxidation: acetone, $\mathrm{(CH_3)_2C=O}$.
So A has the unit $\mathrm{(CH_3)_2C=}$: 2-methylpropene, $\mathrm{(CH_3)_2C=CH_2}$. Its other fragment is HCHO on ozonolysis and $\mathrm{CO_2 + H_2O}$ with $\mathrm{KMnO_4/H^+}$.
But-2-ene would give acetaldehyde/acetic acid; the cycloalkanes are not cleaved.
Answer: (C). | ||||
| 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 The substrate is but-3-en-2-one, $\mathrm{CH_2=CH-CO-CH_3}$.
A Grignard reagent, a hard nucleophile, adds directly to the carbonyl carbon (1,2-addition): $\mathrm{CH_2=CH-C(OMgBr)(CH_3)-C_2H_5}$.
Aqueous acid work-up ($\mathrm{H_2O}$, HCl) only protonates the alkoxide: $\mathrm{CH_2=CH-C(OH)(CH_3)-CH_2CH_3}$, 3-methylpent-1-en-3-ol.
Answer: (D). | ||||
| 44 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Step 1 (crossed aldol): the only enolisable $\alpha$-carbon of 2,2-dimethylcyclopentanone is C5. Its enolate adds to acetaldehyde, giving P = 5-(1-hydroxyethyl)-2,2-dimethylcyclopentanone, which carries a $\mathrm{CH_3CH(OH)-}$ group.
Step 2 (iodoform): $\mathrm{I_2/NaOH}$ oxidises $\mathrm{CH_3CH(OH)R}$ to $\mathrm{CH_3COR}$ and cleaves it: $\mathrm{CHI_3}$ precipitates (filtered off) and the sodium carboxylate stays in the filtrate.
Step 3: HCl liberates the acid, X = 3,3-dimethyl-2-oxocyclopentane-1-carboxylic acid.
Answer: (A). | ||||
| 45 | 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 Phenol is strongly activated; in a polar solvent (bromine water) it forms 2,4,6-tribromophenol at once.
Monobromination to 4-bromophenol needs a non-polar solvent at low temperature: $\mathrm{Br_2}$ in $\mathrm{CS_2}$ or $\mathrm{CHCl_3}$ at 273 K.
A Lewis acid ($\mathrm{FeBr_3}$) is unnecessary and would promote polybromination.
Answer: (B). | ||||
| 46 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hinsberg's reagent (benzenesulphonyl chloride) forms sulphonamides with primary and secondary amines only.
(A) $\mathrm{LiAlH_4}$ reduces CN to $\mathrm{CH_2NH_2}$ (and CHO to $\mathrm{CH_2OH}$): a primary amine, reacts.
(B) $\mathrm{SnCl_2/HCl}$ (Stephen reduction) gives the imine salt, hydrolysed to benzaldehyde: no amine, does not react.
(C) $\mathrm{H_2/Ni}$ gives 3-methylbenzylamine, a primary amine: reacts.
(D) Na/Hg in ethanol (Mendius reduction) gives a benzylamine, a primary amine: reacts.
Answer: (B). | ||||
| 47 | AminesClassification and identification of amines | Class 12: Amines | Assertion and reasonEasy | |
A easy assertion and reason question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Gabriel synthesis needs the phthalimide anion to displace a halide by nucleophilic substitution.
Aryl halides do not undergo this substitution (partial double-bond character of C–X, $sp^2$ carbon), so aromatic primary amines cannot be made this way.
A is true, R is true and explains A.
Answer: (A). | ||||
| 48 | 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 The unit is phenol rings joined at the ortho positions by $\mathrm{-CH_2-}$ bridges: the linear phenol–formaldehyde polymer, Novolac (the precursor of Bakelite).
Neoprene is polychloroprene, Buna-N a butadiene–acrylonitrile copolymer, Acrilan polyacrylonitrile.
Answer: (C). | ||||
| 49 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Fibrous proteins (myosin, fibrin, collagen) are insoluble in water. Globular proteins such as albumin are water soluble.
Answer: (D). | ||||
| 50 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines Only an acid can release $\mathrm{CO_2}$ from bicarbonate.
Acetamide is practically neutral, methylamine and tetramethylammonium hydroxide are bases.
Trimethylammonium chloride, $\mathrm{(CH_3)_3NH^+Cl^-}$, is the salt of a weak base and a strong acid; its cation is the only acidic species: $\mathrm{(CH_3)_3NH^+ + HCO_3^- \rightarrow (CH_3)_3N + H_2O + CO_2}$.
Answer: (D). | ||||
| 51 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter $\mathrm{Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O}$.
Dichromate used: $15\times0.02 = 0.3$ mmol, so $\mathrm{Fe^{2+}} = 6\times0.3 = 1.8$ mmol in 10 mL.
Molarity $= \frac{1.8}{10} = 0.18$ M $= 18\times10^{-2}$ M.
Answer: 18. | ||||
| 52 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $V = 4.00\times10^3\ \mathrm{m^3} = 4.00\times10^6$ L.
$n = \frac{PV}{RT} = \frac{1.0\times4.00\times10^6}{0.083\times300} = 1.606\times10^5$ mol.
Mass $= 16\times1.606\times10^5 = 25.70\times10^5$ g; nearest integer 26.
Answer: 26. | ||||
| 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 Power $= \frac{1000}{10} = 100$ W.
Photon energy $= \frac{hc}{\lambda} = \frac{6.626\times10^{-34}\times3\times10^8}{400\times10^{-9}} = 4.97\times10^{-19}$ J.
Photons per second $= \frac{100}{4.97\times10^{-19}} = 2.01\times10^{20}$; one electron per photon gives $2\times10^{20}$ electrons per second.
Answer: 2. | ||||
| 54 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure The molecule is $\mathrm{CH_3-CH=CH-C\equiv CH}$ ($\mathrm{C_5H_6}$).
C–H bonds: 3 + 1 + 1 + 1 = 6, all $\sigma$.
C–C bonds: 4, each containing one $\sigma$ bond.
Total $\sigma$ bonds $= 6 + 4 = 10$ (plus 3 $\pi$ bonds).
Answer: 10. | ||||
| 55 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics By Hess's law, $\Delta_{sub}H = \Delta_{fus}H + \Delta_{vap}H = 2.8 + 98.2 = 101.0\ \mathrm{kJ\,mol^{-1}}$.
Answer: 101. | ||||
| 56 | SolutionsRaoult's law | Class 12: Solutions | NumericalEasy | |
A easy numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions Henry's law: $x_{CO_2} = \frac{p}{K_H} = \frac{0.835}{1670} = 5.0\times10^{-4}$.
Water in 0.9 L: $\frac{900}{18} = 50$ mol. Since $x_{CO_2}$ is tiny, $n_{CO_2} \approx 5.0\times10^{-4}\times50 = 0.025$ mol $= 25$ mmol.
Answer: 25. | ||||
| 57 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Initially $Q = \frac{1^2}{1\times1} = 1 < K$, so the reaction moves forward.
At equilibrium $[A] = [B] = 1 - y$, $[C] = 1 + 2y$: $\frac{(1+2y)^2}{(1-y)^2} = 100$, so $\frac{1+2y}{1-y} = 10$, $y = 0.75$.
$[C] = 1 + 1.5 = 2.5$ M $= 25\times10^{-1}$ M.
Answer: 25. | ||||
| 58 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $\mathrm{Zn + 2Fe^{3+} \rightarrow Zn^{2+} + 2Fe^{2+}}$, $n = 2$, $E^0_{cell} = 0.77 + 0.76 = 1.53$ V.
$E = E^0 - \frac{0.059}{2}\log\frac{[\mathrm{Zn^{2+}}][\mathrm{Fe^{2+}}]^2}{[\mathrm{Fe^{3+}}]^2}$ with $[\mathrm{Zn^{2+}}] = 1$: $1.500 = 1.53 - 0.059\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}$.
$\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]} = \frac{0.03}{0.059} = 0.508$, so $\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]} = 3.22$.
Fraction as $\mathrm{Fe^{3+}} = \frac{1}{1+3.22} = 0.237 \approx 24\times10^{-2}$.
Answer: 24. | ||||
| 59 | 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 $\mathrm{2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O}$: 1 mol butane gives 5 mol water.
72.0 g water $= 4$ mol, so butane $= 0.8$ mol $= 0.8\times58 = 46.4$ g $= 464\times10^{-1}$ g.
Answer: 464. | ||||
| 60 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All three chlorides precipitate as AgCl, so all three are ionisable (primary valency).
The six octahedral positions (secondary valency) are filled by 3 $\mathrm{NH_3}$ and 3 $\mathrm{H_2O}$: $\mathrm{[Cr(NH_3)_3(H_2O)_3]Cl_3}$.
No chloride satisfies the secondary valency.
Answer: 0. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $e^2 = \frac{1}{3}$ gives $b^2 = a^2(1-e^2) = \frac{2}{3}a^2$. The point gives $\frac{3}{2a^2} + \frac{3}{2a^2} = 1$, so $a^2 = 3$, $b^2 = 2$.
Focus: $\alpha = ae = 1$, $F(1, 0)$.
For a point $(x, y)$ on E, $PF = a - ex = \sqrt{3} - \frac{x}{\sqrt{3}}$. Setting this equal to $\frac{2}{\sqrt{3}}$: $3 - x = 2$, $x = 1$.
Then $y^2 = 2\left(1 - \frac{1}{3}\right) = \frac{4}{3}$, $y = \pm\frac{2}{\sqrt{3}}$, so $PQ = \frac{4}{\sqrt{3}}$ and $PQ^2 = \frac{16}{3}$.
Answer: (C). | ||||
| 62 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Put $u = y - x$: $\frac{du}{dx} = xe^u$, so $e^{-u}du = x\,dx$ and $-e^{-u} = \frac{x^2}{2} + C$. With $u(0) = 0$, $C = -1$: $e^{-u} = 1 - \frac{x^2}{2}$.
So $y = x - \log_e\left(1 - \frac{x^2}{2}\right)$ and $y' = 1 + \frac{2x}{2-x^2} = \frac{2 + 2x - x^2}{2-x^2}$.
$y' = 0$ at $x^2 - 2x - 2 = 0$, $x = 1 \pm \sqrt{3}$; only $x = 1 - \sqrt{3}$ lies in the interval, and $y'$ changes from negative to positive there, a minimum.
$1 - \frac{(1-\sqrt{3})^2}{2} = 1 - (2 - \sqrt{3}) = \sqrt{3} - 1$, so $y_{min} = (1 - \sqrt{3}) - \log_e(\sqrt{3} - 1)$.
Answer: (A). | ||||
| 63 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The curves meet where $2x^2 = 4 - 2x$, i.e. $x^2 + x - 2 = 0$, $x = 1$ (with $x \ge 0$).
Area $= \int_0^1 (4 - 2x - 2x^2)\,dx = 4 - 1 - \frac{2}{3} = \frac{7}{3}$.
Answer: (A). | ||||
| 64 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability An integral of a bounded function is continuous in its upper limit, so $f$ is continuous everywhere on $[0, \infty)$.
Between integers $f'(x) = [x]$. At an integer $n \ge 1$ the left derivative is $n - 1$ and the right derivative is $n$, so $f$ is not differentiable there.
So $f$ is continuous everywhere and differentiable except at the integer points.
Answer: (B). | ||||
| 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 $x < 2$: $x^2 - 5x + 6 = (x-2)(x-3) > 0$, and $5x - x^2 - 6 = -(x^2-5x+6)$, so $f(x) = -\frac{\lambda}{\mu}$.
For $x$ just above 2, $[x] = 2$: $f(x) = e^{\frac{\tan(x-2)}{x-2}} \to e$.
Continuity: $\mu = e$ and $-\frac{\lambda}{\mu} = e$, so $\lambda = -e^2$.
$\lambda + \mu = e - e^2 = e(-e+1)$.
Answer: (D). | ||||
| 66 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctHard | |
A hard single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $g$ cycles each block $\{3n+1, 3n+2, 3n+3\}$: $3n+1 \to 3n+2 \to 3n+3 \to 3n+1$, and has no fixed point.
(A) $g\circ g\circ g$ is the identity, not $g$: false.
(B) $f(g(x)) = f(x)$ with $g(x) \ne x$ makes $f$ take equal values at two different points, so $f$ cannot be one-one: false.
(C) Take $f$ constant on each block: $f(3n+1) = f(3n+2) = f(3n+3) = n+1$. Then $f\circ g = f$ and $f$ is onto $\mathbf{N}$: true.
(D) $g(f(x)) = f(x)$ needs $f(x)$ to be a fixed point of $g$; there are none: false.
Answer: (C). | ||||
| 67 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = \cos x\left(12\sin^3x + 30\sin^2x + 12\sin x\right) = 6\sin x\cos x(2\sin x + 1)(\sin x + 2)$.
On $\left(-\frac{\pi}{6}, 0\right)$: $\sin x \in \left(-\frac{1}{2}, 0\right)$, so $\sin x < 0$ while $\cos x$, $2\sin x + 1$, $\sin x + 2$ are positive: $f' < 0$, decreasing.
On $\left(0, \frac{\pi}{2}\right)$ every factor is positive: increasing.
Answer: (D). | ||||
| 68 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Vertex $(2, 0)$, focus $(4, 0)$: $a = 2$, so P is $y^2 = 8(x - 2)$.
Tangent $y = mx$ from O: $m^2x^2 - 8x + 16 = 0$ must have equal roots, $64 - 64m^2 = 0$, $m = \pm1$, and then $x = 4$, $y = \pm4$.
S(4, 4), R(4, −4): area $= \frac{1}{2}\times8\times4 = 16$.
Answer: (A). | ||||
| 69 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of L is $(t, 0, -t)$.
N: $(t-1, -2, -t+1)\cdot(1, 0, -1) = 2t - 2 = 0$, $t = 1$, N(1, 0, −1), $\vec{PN} = (0, -2, 0)$.
Q: $\vec{PQ} = (s-1, -2, -s+1)$ must be perpendicular to the plane's normal $(1, 1, 2)$: $(s-1) - 2 + 2(1-s) = -s - 1 = 0$, $s = -1$, $\vec{PQ} = (-2, -2, 2)$.
$\cos\alpha = \frac{|\vec{PN}\cdot\vec{PQ}|}{|\vec{PN}||\vec{PQ}|} = \frac{4}{2\times2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Answer: (B). | ||||
| 70 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{1}{a - kb} = \frac{1}{a}\left(1 - \frac{kb}{a}\right)^{-1} \approx \frac{1}{a}\left(1 + \frac{kb}{a} + \frac{k^2b^2}{a^2}\right)$.
Summing over $k = 1$ to $n$: $\frac{n}{a} + \frac{b}{a^2}\cdot\frac{n(n+1)}{2} + \frac{b^2}{a^3}\cdot\frac{n(n+1)(2n+1)}{6}$.
The $n^3$ term comes only from the last sum: $\frac{b^2}{a^3}\cdot\frac{2n^3}{6}$, so $\gamma = \frac{b^2}{3a^3}$.
Answer: (D). | ||||
| 71 | 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 Call the vectors $\vec{u}, \vec{v}, \vec{w}$. Coplanarity is unchanged by replacing $\vec{u}$ with $\vec{u} - \vec{v} = (1+a, a+c, -c)$ and $\vec{w}$ with $\vec{w} - \vec{v} = (1, 0, 1)$.
$\begin{vmatrix} 1+a & a+c & -c \\ 1+b & 2b & -b \\ 1 & 0 & 1 \end{vmatrix} = \left[-b(a+c) + 2bc\right] + \left[2b(1+a) - (a+c)(1+b)\right]$
$= -ab + bc + 2b + 2ab - a - ab - c - bc = 2b - a - c$.
Setting it to zero: $2b = a + c$.
Answer: (A). | ||||
| 72 | 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 Put $t = e^x > 0$: $P(t) = t^6 - t^4 - 2t^3 - 12t^2 + t + 1 = 0$, and each positive root $t$ gives exactly one real $x$.
The signs of the coefficients, $+, -, -, -, +, +$, change twice, so $P$ has at most 2 positive roots (Descartes' rule).
$P(0) = 1 > 0$, $P(1) = 1 - 1 - 2 - 12 + 1 + 1 = -12 < 0$, and $P(t) \to +\infty$: there is a root in $(0, 1)$ and one in $(1, \infty)$.
So there are exactly 2 real roots.
Answer: (B). | ||||
| 73 | 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 The tangents from the eye make $30^\circ$ with the line to the centre, so the distance to the centre is $d = \frac{16}{\sin30^\circ} = 32$ m.
Height of the centre $= 32\sin75^\circ = 32\cdot\frac{\sqrt{6}+\sqrt{2}}{4} = 8(\sqrt{6}+\sqrt{2})$.
The top-most point is 16 m higher: $8(\sqrt{6}+\sqrt{2}) + 16 = 8\left(\sqrt{6}+\sqrt{2}+2\right)$.
Answer: (B). | ||||
| 74 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each ball goes to $B_3$ with probability $\frac{1}{4}$, independently.
$P = \binom{9}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^6 = 84\cdot\frac{3^6}{4^9} = \frac{84}{27}\left(\frac{3}{4}\right)^9$.
$k = \frac{28}{9} \approx 3.11$, which lies in $(2, 4) = \{x : |x - 3| < 1\}$; it is outside $[1, 3]$ and $[4, 6]$.
Answer: (C). | ||||
| 75 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Completing squares: $16(x+1)^2 - 9(y-2)^2 = 144$, i.e. $\frac{(x+1)^2}{9} - \frac{(y-2)^2}{16} = 1$; $a = 3$, $b = 4$, $c = 5$, foci $(4, 2)$ and $(-6, 2)$.
Centroid $(h, k)$ of P$(x, y)$ and the foci: $h = \frac{x - 2}{3}$, $k = \frac{y + 4}{3}$, so $x + 1 = 3(h+1)$, $y - 2 = 3(k-2)$.
Substituting: $(h+1)^2 - \frac{9(k-2)^2}{16} = 1$, i.e. $16(h+1)^2 - 9(k-2)^2 = 16$.
Expanding: $16h^2 - 9k^2 + 32h + 36k - 36 = 0$.
Answer: (D). | ||||
| 76 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_{3n} = 3S_{2n}$: $\frac{3n}{2}\left[2a + (3n-1)d\right] = 3n\left[2a + (2n-1)d\right]$, so $2a + (3n-1)d = 4a + (4n-2)d$, i.e. $2a = -(n-1)d$.
$\frac{S_{4n}}{S_{2n}} = \frac{2n\left[2a + (4n-1)d\right]}{n\left[2a + (2n-1)d\right]} = \frac{2\left[-(n-1) + (4n-1)\right]d}{\left[-(n-1) + (2n-1)\right]d} = \frac{2\cdot3n}{n} = 6$.
Answer: (C). | ||||
| 77 | 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 If $p$ is true, $p \Rightarrow q$ forces $q$ true, and then $q \Rightarrow \sim p$ is false: the expression is false.
If $p$ is false, $p \Rightarrow q$ is true and $q \Rightarrow \sim p$ is true ($\sim p$ is true): the expression is true.
So it has the truth values of $\sim p$.
Answer: (C). | ||||
| 78 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replace $x$ by $\frac{\pi}{24} + \frac{5\pi}{24} - x = \frac{\pi}{4} - x$; then $\tan2x$ becomes $\cot2x$.
$I = \int \frac{dx}{1 + \sqrt[3]{\cot2x}} = \int \frac{\sqrt[3]{\tan2x}}{\sqrt[3]{\tan2x} + 1}dx$.
Adding: $2I = \int_{\pi/24}^{5\pi/24} dx = \frac{4\pi}{24} = \frac{\pi}{6}$, so $I = \frac{\pi}{12}$.
Answer: (C). | ||||
| 79 | 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 $\Delta = \begin{vmatrix} 2 & 3 & 6 \\ 1 & 2 & a \\ 3 & 5 & 9 \end{vmatrix} = 2(18 - 5a) - 3(9 - 3a) + 6(5 - 6) = 3 - a$.
For no solution, $\Delta = 0$: $a = 3$. Then the third left side is the sum of the first two ($3x + 5y + 9z$), so the system is consistent only if $b = 8 + 5 = 13$.
No solution: $a = 3$, $b \ne 13$.
Answer: (B). | ||||
| 80 | 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 $(\sin x + \sin4x) + (\sin2x + \sin3x) = 2\sin\frac{5x}{2}\left(\cos\frac{3x}{2} + \cos\frac{x}{2}\right) = 4\sin\frac{5x}{2}\cos x\cos\frac{x}{2}$.
$\sin\frac{5x}{2} = 0$: $x = \frac{2k\pi}{5}$, $k = 0, 1, \ldots, 5$; sum $= \frac{2\pi}{5}\cdot15 = 6\pi$.
$\cos x = 0$: $x = \frac{\pi}{2}, \frac{3\pi}{2}$; sum $2\pi$.
$\cos\frac{x}{2} = 0$: $x = \pi$.
All values are distinct; total $6\pi + 2\pi + \pi = 9\pi$.
Answer: (B). | ||||
| 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 $\frac{x+1}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$ (sum of cubes) and $\frac{x-1}{x - x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = 1 + x^{-1/2}$.
The expression is $\left(x^{1/3} - x^{-1/2}\right)^{10}$, with general term $\binom{10}{r}(-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$.
$\frac{10-r}{3} = \frac{r}{2}$ gives $r = 4$: the term is $\binom{10}{4} = 210$.
Answer: 210. | ||||
| 82 | 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 $\vec{p} + \vec{q} = (3, 5, 2)$, $\vec{p} - \vec{q} = (1, 1, 0)$; their cross product is $(-2, 2, -2)$.
So $\vec{r} = \lambda(-1, 1, -1)$ with $|\vec{r}| = \sqrt{3}|\lambda| = \sqrt{3}$, $\lambda = \pm1$.
$|\alpha| + |\beta| + |\gamma| = 3$.
Answer: 3. | ||||
| 83 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Middle term of $(1+x)^{20}$: coefficient $\binom{20}{10}$.
Two middle terms of $(1+x)^{19}$: $\binom{19}{9} + \binom{19}{10} = \binom{20}{10}$ (Pascal's rule).
Ratio $= 1$.
Answer: 1. | ||||
| 84 | 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 Base: $1 + \sum_{n\ge1}\frac{4n-2}{3^n}$. With $\sum_{n\ge1}\frac{n}{3^n} = \frac{1/3}{(1-1/3)^2} = \frac{3}{4}$ and $\sum_{n\ge1}\frac{1}{3^n} = \frac{1}{2}$: base $= 1 + 3 - 1 = 3$.
Exponent: $\log_{0.25}\frac{1}{2} = \frac{1}{2}$, since $(0.25)^{1/2} = 0.5$.
$l = 3^{1/2}$, so $l^2 = 3$.
Answer: 3. | ||||
| 85 | 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 Put $u = e^y$: $\frac{du}{dx} - 2u\sin x = -\sin x\cos^2x$, a linear equation with integrating factor $e^{2\cos x}$.
$\frac{d}{dx}\left(ue^{2\cos x}\right) = -\sin x\cos^2x\,e^{2\cos x}$; with $t = \cos x$ the right side integrates to $\int t^2e^{2t}dt = e^{2t}\left(\frac{t^2}{2} - \frac{t}{2} + \frac{1}{4}\right)$.
$u = \frac{\cos^2x}{2} - \frac{\cos x}{2} + \frac{1}{4} + Ce^{-2\cos x}$. At $x = \frac{\pi}{2}$, $u = 1$: $C = \frac{3}{4}$.
At $x = 0$: $u = \frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{3}{4}e^{-2}$, so $\alpha = \frac{1}{4}$, $\beta = \frac{3}{4}$ and $4(\alpha + \beta) = 4$.
Answer: 4. | ||||
| 86 | StatisticsMean, median and mode | Class 11: Statistics | NumericalMedium | |
A medium numerical question on mean, median and mode, from Statistics. NCERT: Class 11: Statistics $\alpha + 110 + 54 + 30 + \beta = 584$, so $\alpha + \beta = 390$.
$\frac{N}{2} = 292$; the median 45 lies in 40 – 50, with cumulative frequency before it $\alpha + 164$ and frequency 30.
$45 = 40 + \frac{292 - (\alpha + 164)}{30}\times10$ gives $128 - \alpha = 15$, $\alpha = 113$, $\beta = 277$.
$|\alpha - \beta| = 164$.
Answer: 164. | ||||
| 87 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The condition for every matrix means $A^n = I$ where $A = \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix}$.
$A^2 = \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix} = iI$, so $A^{2m} = i^mI$, which is $I$ exactly when $4 \mid m$; odd powers are $i^mA \ne I$.
So $S$ is the multiples of 8. Two-digit ones: 16, 24, …, 96, which is 11 numbers.
Answer: 11. | ||||
| 88 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $ad - bc = 15$ with $ad \le 9$ and $bc \ge -9$, so $ad \ge 6$. The products of two entries from $\{0, \pm1, \pm2, \pm3\}$ that are at least 6 are 6 and 9.
$ad = 9$, $bc = -6$: $(a, d) = (3, 3), (-3, -3)$ — 2 ways; $(b, c) = (2, -3), (-2, 3), (3, -2), (-3, 2)$ — 4 ways; 8 matrices.
$ad = 6$, $bc = -9$: 4 ways for $(a, d)$, 2 ways for $(b, c)$; 8 matrices.
Total 16.
Answer: 16. | ||||
| 89 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Let $x, y, z$ be the numbers from classes 10, 11, 12: $x, y \ge 2$, $x + y \le 5$, $z = 10 - x - y \le 8$, so $x + y \in \{4, 5\}$.
$(2, 2, 6)$: $\binom{5}{2}\binom{6}{2}\binom{8}{6} = 10\cdot15\cdot28 = 4200$.
$(2, 3, 5)$: $10\cdot20\cdot56 = 11200$. $(3, 2, 5)$: $10\cdot15\cdot56 = 8400$.
Total $= 23800 = 100k$, $k = 238$.
Answer: 238. | ||||
| 90 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Each root satisfies $x^{n+2} = -5\sqrt{2}x^{n+1} - 10x^n$, so $P_{n+2} + 5\sqrt{2}P_{n+1} = -10P_n$.
Numerator: $P_{17}\left(P_{20} + 5\sqrt{2}P_{19}\right) = -10P_{17}P_{18}$.
Denominator: $P_{18}\left(P_{19} + 5\sqrt{2}P_{18}\right) = -10P_{18}P_{17}$.
The ratio is 1.
Answer: 1. | ||||
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