JEE (Main) · 2021
JEE (Main) 2021 — 26 February, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Oscillations and Waves5
- Optics3
- Atoms and Nuclei2
15 easy · 14 medium · 1 hard
Chemistry
30 Q
- Chemical Bonding and Molecular Structure2
- Isolation of Metals2
- Hydrogen2
21 easy · 8 medium · 1 hard
Mathematics
30 Q
- Integral Calculus3
- Applications of Derivatives3
- Three Dimensional Geometry2
9 easy · 18 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 | 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 Activity falls exponentially: $\frac{A}{5} = A e^{-\lambda(t_2 - t_1)}$, so $\lambda(t_2 - t_1) = \ln 5$ and $\lambda = \frac{\ln 5}{t_2 - t_1}$.
Average (mean) life $\tau = \frac{1}{\lambda} = \frac{t_2 - t_1}{\ln 5}$.
Answer: (C). | ||||
| 2 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Assertion and reasonMedium | |
A medium assertion and reason question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics A simple microscope (a convex lens held close to the eye) forms a virtual image. The object and its image lie on the same ray through the optical centre, which is practically at the eye, so the angle the image subtends at the eye equals the angle the object itself subtends: A is true.
The gain comes from the fact that, with the lens, the object can be placed much closer to the eye than the near point (25 cm) and still be seen clearly; at that small distance it subtends a much larger angle than it would at 25 cm. So R is true, but it explains why the simple microscope magnifies, not why the angular sizes of object and image are equal (that follows from both lying on the same ray through the lens at the eye). R is therefore not the explanation of A.
Answer: (B). | ||||
| 3 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves With 5 beats/s, fork A is either $340 + 5 = 345$ Hz or $340 - 5 = 335$ Hz.
Filing a fork removes mass from its prongs and raises its frequency. If A were 345 Hz, raising it would increase the beats; if A is 335 Hz, raising it to 338 Hz brings it closer to 340 Hz and the beats fall to 2/s.
So the frequency of A is 335 Hz.
Answer: (A). | ||||
| 4 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The first gate is a NAND with both inputs joined to A, so it gives $\bar{A}$. The OR gate gives $\bar{A} + B$, and the final NOT gives $Y = \overline{\bar{A} + B} = A\bar{B}$.
From the waveforms: A is 1 for $0$–$2$ s, 0 for $2$–$3$ s, 1 for $3$–$5$ s; B is 1 for $1$–$2$ s and $3$–$4$ s, and 0 otherwise.
$Y = A\bar{B}$ is 1 only when A = 1 and B = 0: for $0$–$1$ s and $4$–$5$ s, and 0 in between.
That is the second waveform.
Answer: (B). | ||||
| 5 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | StatementsEasy | |
A easy statements question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves A second's pendulum is one whose time period is 2 s, so Statement I is false.
Going from one extreme position to the other is half an oscillation, which takes $\frac{T}{2} = 1$ s, so Statement II is true.
Answer: (D). | ||||
| 6 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $[V] = \frac{[\text{work}]}{[\text{charge}]} = \frac{ML^2T^{-2}}{IT} = [ML^2T^{-3}I^{-1}]$.
$[C] = \frac{[Q]}{[V]} = \frac{IT}{ML^2T^{-3}I^{-1}} = [M^{-1}L^{-2}T^4I^2]$.
$[\lambda] = \frac{[C]}{[V]} = \frac{M^{-1}L^{-2}T^4I^2}{ML^2T^{-3}I^{-1}} = [M^{-2}L^{-4}I^3T^7]$.
Answer: (C). | ||||
| 7 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current A plane flying horizontally with horizontal wings cuts only the vertical component of the earth's field: $B_V = B\sin 60^\circ = 2.5 \times 10^{-4} \times 0.866 = 2.165 \times 10^{-4}$ T.
$v = 180$ km/h $= 50$ m/s, $l = 10$ m.
$\varepsilon = B_V l v = 2.165 \times 10^{-4} \times 10 \times 50 = 0.10825$ V $= 108.25$ mV.
Answer: (A). | ||||
| 8 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The cord does not slip, so the falling weight's speed is $v = \omega r$. Energy is conserved:
$mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}m\omega^2 r^2$, so $\omega^2 = \frac{2mgh}{I + mr^2}$.
Answer: (C). | ||||
| 9 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane At the point of projection the slope is $\frac{dy}{dx}\Big|_{x=0} = \alpha$, so $\tan\theta = \alpha$ and $\theta = \tan^{-1}\alpha$.
Maximum height where $\frac{dy}{dx} = \alpha - 2\beta x = 0$, i.e. $x = \frac{\alpha}{2\beta}$: $H = \alpha\cdot\frac{\alpha}{2\beta} - \beta\cdot\frac{\alpha^2}{4\beta^2} = \frac{\alpha^2}{4\beta}$.
Answer: (C). | ||||
| 10 | 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 The force F pulls B to the left, away from A, so the spring is stretched. The only horizontal force on A is the spring force, so the spring force is $Ma$ (pulling A towards B, and pulling B towards A).
On B, along the direction of F: $F - Ma = Ma_B$, so $a_B = \frac{F - Ma}{M}$.
Answer: (D). | ||||
| 11 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | StatementsEasy | |
A easy statements question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Statement I: the net charge of a dipole is zero, so by Gauss's law the flux through the sphere is zero. The field of a dipole, however, is not zero at points in the sphere. True.
Statement II: inside a conductor the field is zero, so the field is zero everywhere on the surface of radius $r < R$, and the flux through it, $\oint \vec{E}\cdot d\vec{A}$, is also zero (all the charge sits on the outer surface, so the enclosed charge is zero). The claim that the flux is not zero is false.
Answer: (C). | ||||
| 12 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The top speed reached is $v = a_1 t_1$, and it is brought to zero in time $t_2$: $v = a_2 t_2$.
So $a_1 t_1 = a_2 t_2$ and $\frac{t_1}{t_2} = \frac{a_2}{a_1}$.
Answer: (B). | ||||
| 13 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory For an ideal gas with f degrees of freedom, $U = \frac{f}{2}nRT = \frac{f}{2}PV$ (up to a constant reference value, which is the 4 here).
Comparing with $U = 3PV + 4$: $\frac{f}{2} = 3$, so $f = 6$. Six degrees of freedom (3 translational + 3 rotational) belong to a non-linear polyatomic gas.
Answer: (C). | ||||
| 14 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Photon energy: $E = 13.6\left(1 - \frac{1}{25}\right) = 13.056$ eV $= 13.056 \times 1.6 \times 10^{-19} = 2.089 \times 10^{-18}$ J.
Momentum is conserved: the atom's momentum equals the photon's, $mv = \frac{E}{c}$.
$v = \frac{E}{mc} = \frac{2.089 \times 10^{-18}}{1.67 \times 10^{-27} \times 3 \times 10^8} \approx 4.17$ m/s.
Answer: (B). | ||||
| 15 | Properties of Bulk MatterElasticity | 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 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 With original length L, the extension is proportional to tension: $l_1 = L + kT_1$, $l_2 = L + kT_2$ (k the same for both, $k = \frac{L}{AY}$).
Multiply the first by $T_2$, the second by $T_1$ and subtract: $T_2 l_1 - T_1 l_2 = L(T_2 - T_1)$.
$L = \frac{T_2 l_1 - T_1 l_2}{T_2 - T_1}$.
Answer: (D). | ||||
| 16 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves In SHM $v = \omega\sqrt{A^2 - x^2}$, so $\frac{x^2}{A^2} + \frac{v^2}{\omega^2 A^2} = 1$.
This is an ellipse with semi-axes A and $\omega A$ (a circle only in the special case $\omega = 1$ in those units).
Answer: (C). | ||||
| 17 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics On reflection the component of the ray along the normal reverses while the component along the surface is unchanged.
Component of $\vec{a}$ along the normal: $(\vec{a}\cdot\vec{c})\vec{c}$. So $\vec{b} = \vec{a} - (\vec{a}\cdot\vec{c})\vec{c} - (\vec{a}\cdot\vec{c})\vec{c} = \vec{a} - 2(\vec{a}\cdot\vec{c})\vec{c}$.
Answer: (B). | ||||
| 18 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $V = 30\sin 100t$: $V_0 = 30$ V, $\omega = 100$ rad/s.
$X_L = \omega L = 100 \times 0.1 = 10\ \Omega$; $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-4}} = 100\ \Omega$.
$Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{120^2 + 90^2} = 150\ \Omega$, so $I_0 = \frac{30}{150} = 0.2$ A.
Resonant frequency: $f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1 \times 10^{-4}}} = \frac{1}{2\pi \times 3.16 \times 10^{-3}} \approx 50$ Hz.
Answer: (B). | ||||
| 19 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctHard | |
A hard single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Electric force $qE = 5 \times 10^{-3} \times 200 = 1$ N, horizontal, along the field (to the right in the figure). The incline rises to the right, so the body slides down to the left, and this force has a component $qE\cos 30^\circ = 0.866$ N up the incline and $qE\sin 30^\circ = 0.5$ N pressing into the incline.
Normal reaction: $N = mg\cos 30^\circ + qE\sin 30^\circ = 9.8 \times 0.866 + 0.5 = 8.99$ N; friction $= 0.2 \times 8.99 = 1.80$ N, up the incline.
Along the incline (down): $ma = mg\sin 30^\circ - qE\cos 30^\circ - \mu N = 4.9 - 0.866 - 1.80 = 2.24$ N, so $a = 2.24$ m/s$^2$.
Length of incline $= \frac{1}{\sin 30^\circ} = 2$ m. From rest, $2 = \frac{1}{2}(2.24)t^2$, so $t = \sqrt{1.79} \approx 1.3$ s.
Answer: (B). | ||||
| 20 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity On stretching the volume is constant, so $A \propto \frac{1}{l}$ and $R = \frac{\rho l}{A} \propto l^2$.
$\frac{R'}{R} = (1.25)^2 = 1.5625$, an increase of 56.25%, about 56%.
Answer: (B). | ||||
| 21 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The Zener holds the 5 k$\Omega$ load at 30 V, so the 4 k$\Omega$ series resistor has $90 - 30 = 60$ V across it: $I = \frac{60}{4000} = 15$ mA.
Load current $= \frac{30}{5000} = 6$ mA. Zener current $= 15 - 6 = 9$ mA.
Answer: 9. | ||||
| 22 | 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 Read '$\frac{5}{8}$ oscillation' as $\frac{5}{8}$ of the path of a full oscillation: one oscillation covers 4A, so the bob travels $\frac{5}{8} \times 4A = 2.5A$.
From the mean position, half an oscillation (time $\frac{T}{2}$) covers 2A and brings it back to the mean position. The remaining $\frac{A}{2}$ from the mean position takes time t with $\frac{A}{2} = A\sin\frac{2\pi t}{T}$, so $\frac{2\pi t}{T} = \frac{\pi}{6}$ and $t = \frac{T}{12}$.
Total time $= \frac{T}{2} + \frac{T}{12} = \frac{7}{12}T$, so $\alpha = 7$ ($\beta = 12$).
Answer: 7. | ||||
| 23 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Take one mole. From $V = KT^{2/3}$: $\frac{dV}{V} = \frac{2}{3}\frac{dT}{T}$.
$W = \int P\,dV = \int \frac{RT}{V}dV = \int RT \cdot \frac{2}{3}\frac{dT}{T} = \frac{2}{3}R\,\Delta T = \frac{2}{3}R \times 90 = 60R$.
So $x = 60$.
Answer: 60. | ||||
| 24 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $K_{max} = E - \phi$: in the first case $2\phi - \phi = \phi$, in the second $10\phi - \phi = 9\phi$.
$v \propto \sqrt{K}$, so $\frac{v_1}{v_2} = \sqrt{\frac{\phi}{9\phi}} = \frac{1}{3}$; $x : y = 1 : 3$ and $x = 1$.
Answer: 1. | ||||
| 25 | Communication SystemsAmplitude modulation | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter An AM signal occupies a bandwidth of twice the highest modulating frequency: $2 \times 5 = 10$ kHz per station.
Number of stations $= \frac{90}{10} = 9$.
Answer: 9. | ||||
| 26 | GravitationAcceleration due to gravity | Class 11: Gravitation | NumericalMedium | |
A medium numerical question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation C is 3200 km $= \frac{R}{2}$ above the surface: $g_C = \frac{gR^2}{(1.5R)^2} = \frac{4}{9}g$.
A is inside the earth at distance $r$ from O: $g_A = g\frac{r}{R}$. Setting $g_A = g_C$: $r = \frac{4}{9}R$.
OA $= \frac{4}{9}R$, AB $= R - \frac{4}{9}R = \frac{5}{9}R$, so OA : AB $= 4 : 5$ and $x = 4$.
Answer: 4. | ||||
| 27 | ThermodynamicsFirst law | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta U = Q - W = Q - \frac{Q}{5} = \frac{4Q}{5}$.
For 1 mole, $\Delta U = C_V\Delta T$ and $Q = C\Delta T$, so $C = C_V \cdot \frac{Q}{\Delta U} = \frac{5R}{2} \times \frac{5}{4} = \frac{25R}{8}$.
So $x = 25$.
Answer: 25. | ||||
| 28 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The image S' is 60 cm behind the mirror, opposite its centre. The man sees it only along lines from S' that pass through the mirror, whose edges are 25 cm either side of the centre line.
The man's line is $60 + 120 = 180$ cm from S', three times the 60 cm to the mirror, so by similar triangles the visible stretch is $3 \times 50 = 150$ cm.
Answer: 150. | ||||
| 29 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v = \omega\sqrt{a^2 - y^2}$ and $v_{max} = a\omega$. Setting $v = \frac{a\omega}{2}$: $a^2 - y^2 = \frac{a^2}{4}$, so $y = \frac{\sqrt{3}a}{2}$.
So $x = 3$.
Answer: 3. | ||||
| 30 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Volume is conserved: $R^3 = 27r^3$, so $R = 3r$. Charge is conserved: $Q = 27q$.
The energy of a charged sphere is $U = \frac{Q^2}{8\pi\varepsilon_0 R}$, so $\frac{U_{big}}{U_{small}} = \frac{(27)^2}{3} = 243$.
Answer: 243. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Bond order $= \frac{1}{2}(N_b - N_a)$ from molecular orbital theory:
$Ne_2$ (20 electrons): $\frac{1}{2}(10 - 10) = 0$; $N_2$ (14): $\frac{1}{2}(10 - 4) = 3$; $F_2$ (18): $\frac{1}{2}(10 - 8) = 1$; $O_2$ (16): $\frac{1}{2}(10 - 6) = 2$.
So (a)→(iii), (b)→(iv), (c)→(i), (d)→(ii).
Answer: (D). | ||||
| 32 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter In hot water $FeCl_3$ hydrolyses to a sol of hydrated ferric oxide, $Fe_2O_3 \cdot xH_2O$. The particles adsorb the common ion present in the medium, $Fe^{3+}$, and so carry a positive charge.
(Adding $FeCl_3$ to NaOH instead gives a negative sol, by adsorbing $OH^-$.)
Answer: (A). | ||||
| 33 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The magnitude of the (negative) electron gain enthalpy falls down group 16 from S, because the atoms grow larger: S > Se > Te.
Oxygen is the exception: its very small 2p shell already has strong electron–electron repulsion, so an added electron is less welcome, and O has the smallest value of the four (S $-200$, Se $-195$, Te $-190$, O $-141$ kJ mol$^{-1}$).
Order: S > Se > Te > O.
Answer: (D). | ||||
| 34 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Siderite is $FeCO_3$ (Fe); calamine is $ZnCO_3$ (Zn); malachite is $CuCO_3 \cdot Cu(OH)_2$ (Cu); cryolite is $Na_3AlF_6$ (Al).
So (a)→(iii), (b)→(v), (c)→(i), (d)→(iv).
Answer: (B). | ||||
| 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 Of the isotopes of hydrogen only tritium is radioactive. It decays by emitting a low-energy $\beta^-$ particle: $^3_1H \rightarrow {}^3_2He + \beta^- + \bar{\nu}$ (half-life about 12.3 years).
Answer: (D). | ||||
| 36 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Sodium carbonate is made by the Solvay (ammonia–soda) process; titanium is refined by the van Arkel method (via volatile $TiI_4$); chlorine is made by Deacon's process ($4HCl + O_2 \xrightarrow{CuCl_2} 2Cl_2 + 2H_2O$); sodium hydroxide is made by the Castner–Kellner cell (electrolysis of brine with a mercury cathode).
So (a)→(iv), (b)→(iii), (c)→(i), (d)→(ii).
Answer: (A). | ||||
| 37 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter $B_2O_3$ (non-metal oxide; with water gives boric acid) and $SiO_2$ (reacts with bases, e.g. NaOH to give sodium silicate) are both acidic.
CaO and BaO are basic; $N_2O$ is neutral.
Answer: (A). | ||||
| 38 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Assertion and reasonMedium | |
A medium assertion and reason question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter A: $TlI_3$, like $CsI_3$, is $Tl^+[I_3]^-$: the metal is in the +1 state ($Tl^{3+}$ is a strong oxidant and cannot coexist with $I^-$). True.
R: Tl is $[Xe]4f^{14}5d^{10}6s^26p^1$, so it does have fourteen 4f electrons. True.
But A is explained by the stability of +1 thallium (the inert pair effect) and by $Tl^{3+}$ oxidising iodide to $I_3^-$; merely having 4f electrons is not that explanation. So both are true and R does not explain A.
Answer: (B). | ||||
| 39 | 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 Calgon is sodium hexametaphosphate, $Na_6P_6O_{18}$ (Graham's salt), a water-soluble polymeric phosphate. It softens water by binding $Ca^{2+}$ and $Mg^{2+}$ into soluble complexes, not by precipitating them.
The second most abundant element by weight in the earth's crust is silicon, which Calgon does not contain (it contains Na, P and O; oxygen is the first). So the first statement is the one that is not true.
Answer: (A). | ||||
| 40 | Principles Related to Practical ChemistryDetection of elements and functional groups | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Ceric ammonium nitrate turns from yellow to red with an alcohol (an alcohol–cerium(IV) complex).
$CHCl_3$ with alcoholic KOH is the carbylamine test: a primary amine gives the foul-smelling isocyanide.
Answer: (A). | ||||
| 41 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure C1 and C3 each have one double bond (three σ bonds): $sp^2$. C2, the central allene carbon, has two double bonds (two σ bonds): $sp$. C4 is a methyl carbon with four σ bonds: $sp^3$.
Answer: (C). | ||||
| 42 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The starting ketone is $CH_3CH_2CH(C_2H_5)COCH_2CH_3$ (4-ethylhexan-3-one).
Step 1, Zn/HCl (Clemmensen reduction), turns C=O into $CH_2$: 3-ethylhexane, $CH_3CH_2CH(C_2H_5)CH_2CH_2CH_3$.
Step 2, $Cr_2O_3$ at 773 K and 10–20 atm, is aromatisation: a six-carbon chain closes to a ring and loses hydrogen. The hexane chain closes into a benzene ring and the ethyl group stays as a substituent (every six-carbon chain in 3-ethylhexane leaves a two-carbon side chain), giving ethylbenzene, $C_6H_5CH_2CH_3$.
Answer: (A). | ||||
| 43 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers $SOCl_2$ replaces the OH of an alcohol by Cl. It does not replace a phenolic OH, whose C–O bond has partial double-bond character.
The reactant has a phenolic OH on the aromatic ring, a secondary (benzylic) alcohol on the saturated ring and a primary $CH_2OH$. The two alcoholic OH groups become Cl and $CH_2Cl$; the phenolic OH stays.
That is the third structure (OH on the ring, Cl on the saturated ring, $CH_2Cl$).
Answer: (C). | ||||
| 44 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids With NaOH the dialdehyde undergoes an intramolecular aldol condensation. The carbanion formed at the $\alpha$-carbon of one $CH_2CH_2CHO$ chain attacks the CHO of the other chain.
Counting the ring: $\alpha$-C, $\beta$-C, two aromatic carbons, $\beta'$-C, $\alpha'$-C and the attacked carbonyl carbon — a seven-membered ring fused to benzene. On heating, the aldol dehydrates to the conjugated $\alpha,\beta$-unsaturated aldehyde, with the CHO outside the ring on the C=C.
That is the benzocycloheptene carbaldehyde, the second structure.
Answer: (B). | ||||
| 45 | 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 (i) HCHO, NaOH: crossed Cannizzaro. HCHO is oxidised to formate and 4-methoxybenzaldehyde is reduced to 4-methoxybenzyl alcohol, $CH_3OC_6H_4CH_2OH$.
(ii) NaH then $CH_3CH_2Br$ in DMF: Williamson synthesis gives the ethyl ether $CH_3OC_6H_4CH_2OCH_2CH_3$.
(iii) HI, heat: both ethers are cleaved. The aryl methyl ether gives the phenol ($ArOH + CH_3I$), since I$^-$ cannot attack the aryl carbon. The benzyl ethyl ether is cleaved at the benzylic carbon (stable benzylic cation/easy substitution), giving the benzyl iodide and ethanol.
A is 4-(iodomethyl)phenol, $HOC_6H_4CH_2I$.
Answer: (B). | ||||
| 46 | AminesDiazonium salts | Class 12: Amines | Matching listEasy | |
A easy matching list question on diazonium salts, from Amines. NCERT: Class 12: Amines (a) Benzenediazonium chloride with $Cu_2Cl_2$ gives chlorobenzene: Sandmeyer reaction (ii).
(b) The same change with copper powder and HCl: Gatterman reaction (iv).
(c) Two alkyl halides coupled by sodium in dry ether: Wurtz reaction (i).
(d) Two aryl halides coupled by sodium in dry ether: Fittig reaction (iii).
Answer: (A). | ||||
| 47 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Seliwanoff's test (resorcinol in HCl) gives a quick cherry-red colour with ketoses, which dehydrate faster than aldoses.
The xanthoproteic test (conc. $HNO_3$, yellow colour turning orange with alkali) detects proteins with aromatic amino acids.
Answer: (C). | ||||
| 48 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Sucrose is α-D-glucose + β-D-fructose; lactose is β-D-galactose + β-D-glucose; maltose is two α-D-glucose units.
So (a)→(ii), (b)→(i), (c)→(iii).
Answer: (A). | ||||
| 49 | Principles Related to Practical ChemistryDetection of elements and functional groups | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter 2,4-Dinitrophenylhydrazine condenses with the C=O group of aldehydes and ketones to give a yellow-orange 2,4-dinitrophenylhydrazone precipitate. Of the options, it identifies an aldehyde.
Answer: (B). | ||||
| 50 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines Phenylmethanamine (benzylamine, $C_6H_5CH_2NH_2$) is an alkylamine: its lone pair is not delocalised into the ring, so it is by far the most basic ($pK_b \approx 4.7$).
In the three anilines the lone pair is delocalised into the ring; each N-methyl group pushes electron density onto nitrogen and raises the basicity: N,N-dimethylaniline ($pK_b \approx 8.9$) > N-methylaniline ($\approx 9.2$) > aniline ($\approx 9.4$).
Order: A > B > C > D.
Answer: (A). | ||||
| 51 | 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 Sodium needed: $70.0 \times 50 = 3500$ mg $= 3.5$ g $= \frac{3.5}{23} = 0.1522$ mol.
Moles of $NaNO_3$ are the same; molar mass $= 23 + 14 + 48 = 85$ g mol$^{-1}$. Mass $= 0.1522 \times 85 = 12.93$ g, which is 13 to the nearest integer.
Answer: 13. | ||||
| 52 | Solid StateClose packed structures | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on close packed structures, from Solid State. NCERT: No matching NCERT chapter In a close-packed lattice of N spheres there are N octahedral voids (and 2N tetrahedral voids).
So there is 1 octahedral void per lattice site.
Answer: 1. | ||||
| 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 $\Delta v = 0.05 \times 90 = 4.5$ m s$^{-1}$, $m = 0.01$ kg.
$\Delta x = \frac{h}{4\pi m\Delta v} = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 0.01 \times 4.5} = \frac{6.63 \times 10^{-34}}{0.5652} = 1.17 \times 10^{-33}$ m.
To the nearest integer: 1.
Answer: 1. | ||||
| 54 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Build $SF_6$(g) from its atoms: $S(g) + 6F(g) \rightarrow SF_6(g)$, $\Delta H = -6E_{S-F}$.
By Hess's law, $\Delta H = \Delta_f H(SF_6) - \Delta_f H(S, g) - 6\Delta_f H(F, g) = -1100 - 275 - 480 = -1855$ kJ mol$^{-1}$.
$E_{S-F} = \frac{1855}{6} = 309.17$ kJ mol$^{-1}$, which is 309 to the nearest integer.
Answer: 309. | ||||
| 55 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Molality, if unassociated: $\frac{12.2/122}{0.100} = 1.0$ mol kg$^{-1}$, which would lower the freezing point by $1.86$ K.
Observed $\Delta T_f = 0.93$ K, so $i = \frac{0.93}{1.86} = 0.5$. With 100% association into $(C_6H_5COOH)_n$, $i = \frac{1}{n}$, so $n = 2$ (the dimer).
Answer: 2. | ||||
| 56 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Ammonium phosphate is a salt of a weak acid and a weak base, so its pH does not depend on concentration:
$pH = 7 + \frac{1}{2}(pK_a - pK_b) = 7 + \frac{1}{2}(5.23 - 4.75) = 7 + 0.24 = 7.24$.
Answer: 7.24. | ||||
| 57 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$, $n = 2$; $E^\theta_{cell} = 0.80 - (-0.76) = 1.56$ V.
$E = E^\theta - \frac{0.059}{2}\log\frac{[Zn^{2+}]}{[Ag^+]^2} = 1.56 - 0.0295\log\frac{0.1}{(0.01)^2} = 1.56 - 0.0295 \times 3 = 1.4715$ V.
$1.4715$ V $= 147.15 \times 10^{-2}$ V, so $x = 147$ to the nearest integer.
Answer: 147. | ||||
| 58 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The fraction of molecules with energy at least $E_a$ is $e^{-E_a/RT}$.
$x = \frac{E_a}{RT} = \frac{80900}{8.31 \times 700} = \frac{80900}{5817} = 13.91$, which is 14 to the nearest integer.
Answer: 14. | ||||
| 59 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In neutral or faintly alkaline solution permanganate oxidises thiosulphate to sulphate: $8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$.
A is $SO_4^{2-}$, in which sulphur is +6.
Answer: 6. | ||||
| 60 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The complex is octahedral of type $[M(AA)_2BC]$.
Trans form (Br and $NH_3$ opposite, the two oxalates in one plane): it has a mirror plane, so it is optically inactive — 1 isomer.
Cis form (Br and $NH_3$ adjacent): no mirror plane, so it exists as a pair of enantiomers — 2 isomers.
Total stereoisomers $= 1 + 2 = 3$.
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry For the plane $4x - 5y + 2z - 8 = 0$, the image of $(1, 3, 5)$ is given by $\frac{\alpha - 1}{4} = \frac{\beta - 3}{-5} = \frac{\gamma - 5}{2} = \frac{-2(4 - 15 + 10 - 8)}{16 + 25 + 4} = \frac{18}{45} = \frac{2}{5}$.
$\alpha = 1 + \frac{8}{5} = \frac{13}{5}$, $\beta = 3 - 2 = 1$, $\gamma = 5 + \frac{4}{5} = \frac{29}{5}$.
$5(\alpha + \beta + \gamma) = 13 + 5 + 29 = 47$.
Answer: (D). | ||||
| 62 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium 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 The range of f is $\{2, 4, 6, 8, 10\}$: each odd k goes to the next even number and each even k to itself.
$g(f(k)) = f(k)$ for all k means $g(y) = y$ for every y in the range of f, i.e. g must fix 2, 4, 6, 8, 10.
The values of g at 1, 3, 5, 7, 9 are unrestricted: 10 choices each, giving $10^5$ functions.
Answer: (D). | ||||
| 63 | 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 The curves meet at $x = \frac{\pi}{4}$.
$A_1$ lies between $\cos x$ (above) and $\sin x$ from 0 to $\frac{\pi}{4}$: $A_1 = \int_0^{\pi/4}(\cos x - \sin x)\,dx = \sqrt{2} - 1$.
$A_2$ is the region under both curves down to the x-axis, up to $x = \frac{\pi}{2}$: $A_2 = \int_0^{\pi/4}\sin x\,dx + \int_{\pi/4}^{\pi/2}\cos x\,dx = 2\left(1 - \frac{1}{\sqrt{2}}\right) = 2 - \sqrt{2} = \sqrt{2}(\sqrt{2} - 1)$.
So $A_1 : A_2 = 1 : \sqrt{2}$, and $A_1 + A_2 = 1$ (together they fill the area under $\cos x$ from 0 to $\frac{\pi}{2}$).
Answer: (B). | ||||
| 64 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a + b}{1 - ab} = \frac{\pi}{4}$ (valid as $ab < 1$), so $a + b = 1 - ab$, i.e. $(1 + a)(1 + b) = 2$.
The series is $\left(a - \frac{a^2}{2} + \frac{a^3}{3} - \ldots\right) + \left(b - \frac{b^2}{2} + \ldots\right) = \log_e(1 + a) + \log_e(1 + b) = \log_e 2$.
Answer: (C). | ||||
| 65 | 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 $\frac{dy}{dx} = y^2 + \frac{y}{x}$. Then $\frac{x\,dy - y\,dx}{y^2} = x\,dx$, i.e. $-d\left(\frac{x}{y}\right) = x\,dx$, so $\frac{x}{y} = -\frac{x^2}{2} + C$.
At $x = -2$ the line gives $y = 3$: $-\frac{2}{3} = -2 + C$, $C = \frac{4}{3}$.
At $x = 3$: $\frac{3}{y} = -\frac{9}{2} + \frac{4}{3} = -\frac{19}{6}$, so $y = -\frac{18}{19}$.
Answer: (B). | ||||
| 66 | Sequences and SeriesArithmetic and geometric means | Class 11: Sequences and Series | Single correctHard | |
A hard single correct question on arithmetic and geometric means, from Sequences and Series. NCERT: Class 11: Sequences and Series Put $m = 2n + 1$ (so m = 3, 5, 7, …): $n^2 + 6n + 10 = \frac{m^2 + 10m + 29}{4} = \frac{m(m - 1) + 11m + 29}{4}$.
Sum $= \frac{1}{4}\sum_{m\ \text{odd} \ge 3}\left[\frac{1}{(m - 2)!} + \frac{11}{(m - 1)!} + \frac{29}{m!}\right]$.
With $\sinh 1 = 1 + \frac{1}{3!} + \ldots = \frac{e - e^{-1}}{2}$ and $\cosh 1 = 1 + \frac{1}{2!} + \ldots = \frac{e + e^{-1}}{2}$:
$\sum\frac{1}{(m - 2)!} = \sinh 1$, $\sum\frac{1}{(m - 1)!} = \cosh 1 - 1$, $\sum\frac{1}{m!} = \sinh 1 - 1$.
Sum $= \frac{1}{4}\left[30\sinh 1 + 11\cosh 1 - 40\right] = \frac{1}{4}\left[\frac{41}{2}e - \frac{19}{2}e^{-1} - 40\right] = \frac{41}{8}e - \frac{19}{8}e^{-1} - 10$.
Answer: (D). | ||||
| 67 | 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 At $x = 0$: $f(0) = 1$. Differentiating: $f'(x) = e^x f(x) + e^x = e^x(f(x) + 1)$.
$\frac{df}{f + 1} = e^x\,dx \Rightarrow \ln(f + 1) = e^x + C$; at $x = 0$: $\ln 2 = 1 + C$.
$f(x) + 1 = 2e^{e^x - 1}$, so $f(x) = 2e^{(e^x - 1)} - 1$.
Answer: (A). | ||||
| 68 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Write $xf(a) - af(x) = (x - a)f(a) - a(f(x) - f(a))$.
Limit $= f(a) - a f'(a) = 4 - 2a$.
Answer: (B). | ||||
| 69 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium 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 $g(x) = \frac{(x - 2)(x + 1)}{(x - 2)(2x + 3)} = \frac{x + 1}{2x + 3}$ for $x \ne -\frac{3}{2}$ (with $g(2) = \frac{3}{7}$ by the given definition).
fog needs $-1 \le \frac{x + 1}{2x + 3} \le 1$.
$\frac{x + 1}{2x + 3} \le 1 \iff \frac{x + 2}{2x + 3} \ge 0 \iff x \le -2$ or $x > -\frac{3}{2}$.
$\frac{x + 1}{2x + 3} \ge -1 \iff \frac{3x + 4}{2x + 3} \ge 0 \iff x < -\frac{3}{2}$ or $x \ge -\frac{4}{3}$.
Both together: $x \in (-\infty, -2] \cup \left[-\frac{4}{3}, \infty\right)$.
Answer: (B). | ||||
| 70 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections With M the midpoint of AB, $(1, -\frac{1}{2})$, $PA^2 + PB^2 = 2PM^2 + \frac{AB^2}{2}$, which is greatest when P is the point of the circle farthest from M.
That point lies on the line through M and the centre (1, 1), i.e. on $x = 1$: P = (1, 2).
A and B also lie on $x = 1$, so P, A, B lie on a straight line.
Answer: (D). | ||||
| 71 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra Collinear: $\frac{x}{1} = \frac{-1}{y} = \frac{1}{z}$, so $y = -\frac{1}{x}$, $z = \frac{1}{x}$, and the vector is $x\hat{i} - \frac{1}{x}\hat{j} + \frac{1}{x}\hat{k}$.
With $x = 1$ this is $\hat{i} - \hat{j} + \hat{k}$, whose unit vector is $\frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$.
The other options are impossible: (A) and (D) need $x = 0$ or $z = 0$, and (C) needs $y = z$, i.e. $-\frac{1}{x} = \frac{1}{x}$.
Answer: (B). | ||||
| 72 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $F_1$: take A, B, C all true. Then $A \wedge \sim B$ is F, $\sim C \wedge (A \vee B)$ is F, $\sim A$ is F, so $F_1$ is F. Not a tautology.
$F_2$: $B \rightarrow \sim A \equiv \sim B \vee \sim A$, so $F_2 \equiv A \vee B \vee \sim B \vee \sim A$, which is always true. A tautology.
Answer: (C). | ||||
| 73 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Total arrangements: $\frac{7!}{2!\,3!\,2!} = 210$. Divisible by 2 means the last digit is 4; the rest 3, 3, 4, 4, 5, 5 arrange in $\frac{6!}{2!\,2!\,2!} = 90$ ways.
Probability $= \frac{90}{210} = \frac{3}{7}$ (three of the seven digits are 4).
Answer: (A). | ||||
| 74 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $D = \begin{vmatrix}1&2&-3\\2&6&-11\\1&-2&7\end{vmatrix} = 1(42 - 22) - 2(14 + 11) - 3(-4 - 6) = 20 - 50 + 30 = 0$, so there is never a unique solution.
The rows are dependent: $5R_1 - 2R_2 = (1, -2, 7) = R_3$. So the third equation agrees with the first two exactly when $c = 5a - 2b$, i.e. $5a = 2b + c$; then the system has infinitely many solutions (and none otherwise).
Answer: (C). | ||||
| 75 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives For a triangle inscribed in a circle of radius r, area $= 2r^2\sin A\sin B\sin C$, which (with $A + B + C = \pi$) is greatest when $A = B = C = 60^\circ$: the equilateral triangle.
Its side is $2r\sin 60^\circ = \sqrt{3}r$ and its height $\frac{3r}{2}$ (not $\frac{2r}{3}$).
Answer: (D). | ||||
| 76 | 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 Direction of L: $(1, 2, 1) \times (0, 1, 2) = (3, -2, 1)$. A point on L (z = 0): y = 4, x = −2. So L: $(-2 + 3t, 4 - 2t, t)$.
Foot of the perpendicular from Q(3, 2, 1): $(P - Q)\cdot(3, -2, 1) = 0$: $3(3t - 5) - 2(2 - 2t) + (t - 1) = 14t - 20 = 0$, $t = \frac{10}{7}$.
$P = \left(\frac{16}{7}, \frac{8}{7}, \frac{10}{7}\right)$, so $21(\alpha + \beta + \gamma) = 21 \times \frac{34}{7} = 102$.
Answer: (B). | ||||
| 77 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = 1$: the right-hand limit is $\sin\pi = 0$, so $|a + 1 + b| = 0$, giving $a + b = -1$.
Check at $x = -1$: left-hand limit $2\sin\frac{\pi}{2} = 2$, and $|a - 1 + b| = |-2| = 2$. Continuous.
Answer: (B). | ||||
| 78 | CirclesStandard and general form | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections A point of the circle is $(\cos\theta, \sin\theta)$; the midpoint is $\left(\frac{3 + \cos\theta}{2}, \frac{2 + \sin\theta}{2}\right)$.
So $(2h - 3)^2 + (2k - 2)^2 = 1$, i.e. $\left(h - \frac{3}{2}\right)^2 + (k - 1)^2 = \frac{1}{4}$: a circle of radius $\frac{1}{2}$.
Answer: (C). | ||||
| 79 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $\frac{1}{y} + \frac{1}{z} = \frac{y + z}{yz} = \frac{5}{yz} = \frac{5}{6}$, so $yz = 6$ with $y + z = 5$, $y > z$: $y = 3$, $z = 2$.
Odd divisors are the divisors of $3^3 5^2$: $(3 + 1)(2 + 1) = 12$.
Answer: (C). | ||||
| 80 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals In $f\left(\frac{1}{x}\right) = \int_1^{1/x}\frac{\ln t}{1 + t}dt$ put $t = \frac{1}{u}$: it becomes $\int_1^x \frac{\ln u}{u(1 + u)}du$.
$f(x) + f\left(\frac{1}{x}\right) = \int_1^x \ln t\left[\frac{1}{1 + t} + \frac{1}{t(1 + t)}\right]dt = \int_1^x \frac{\ln t}{t}dt = \frac{(\ln x)^2}{2}$.
At $x = e$: $\frac{1}{2}$.
Answer: (C). | ||||
| 81 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals In $I_{m,n}$ put $x = \frac{t}{1 + t}$ ($1 - x = \frac{1}{1 + t}$, $dx = \frac{dt}{(1 + t)^2}$, t from 0 to $\infty$): $I_{m,n} = \int_0^\infty \frac{t^{m-1}}{(1 + t)^{m+n}}dt$.
Split at 1: $\int_1^\infty \frac{t^{m-1}}{(1 + t)^{m+n}}dt$ with $t = \frac{1}{u}$ becomes $\int_0^1 \frac{u^{n-1}}{(1 + u)^{m+n}}du$.
So $I_{m,n} = \int_0^1 \frac{x^{m-1} + x^{n-1}}{(1 + x)^{m+n}}dx$, and $\alpha = 1$.
Answer: 1. | ||||
| 82 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | NumericalHard | |
A hard numerical question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x + iy$: $z(1 + i) + \bar{z}(1 - i) = 2(x - y) \ge -10$, i.e. $y \le x + 5$. And $|z + 5| \le 4$ is the disc of centre $(-5, 0)$, radius 4.
The line $y = x + 5$ passes through the centre, so the region is the half-disc on the side $y \le x + 5$; its arc is $(-5 + 4\cos\theta, 4\sin\theta)$ with $\sin\theta \le \cos\theta$, i.e. $\theta \in \left[-\frac{3\pi}{4}, \frac{\pi}{4}\right]$.
$|z + 1|^2$ is the squared distance from $(-1, 0)$. On the arc it is $(4\cos\theta - 4)^2 + 16\sin^2\theta = 32(1 - \cos\theta)$, largest at $\theta = -\frac{3\pi}{4}$: $32\left(1 + \frac{1}{\sqrt{2}}\right) = 32 + 16\sqrt{2}$. Points on the straight edge are no farther (the distance along a chord peaks at its ends).
$\alpha = 32$, $\beta = 16$, so $\alpha + \beta = 48$.
Answer: 48. | ||||
| 83 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives A curve all of whose normals pass through one point is a circle with that point as centre. So (a, b) is equidistant from (3, −3) and $(4, -2\sqrt{2})$:
$(3 - a)^2 + (3 + b)^2 = (4 - a)^2 + (2\sqrt{2} + b)^2 \Rightarrow 18 - 6a + 6b = 24 - 8a + 4\sqrt{2}b \Rightarrow a + (3 - 2\sqrt{2})b = 3$.
With $a - 2\sqrt{2}b = 3$: subtracting gives $3b = 0$, so $b = 0$, $a = 3$.
$a^2 + b^2 + ab = 9$.
Answer: 9. | ||||
| 84 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $p'(x) = 10x^4 + 20x^3 + 30x^2 + 20x + 10 = 10(x^2 + x + 1)^2 > 0$, so p is strictly increasing and has exactly one real root.
$p(-1) = -2 + 5 - 10 + 10 - 10 + 10 = 3 > 0$ and $p(-2) = -64 + 80 - 80 + 40 - 20 + 10 = -34 < 0$, so the root lies in $(-2, -1)$.
$a = -2$ and $|a| = 2$.
Answer: 2. | ||||
| 85 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Let $\bar{X}$ be the mean. From the first sum, $\bar{X} - \alpha = 2$.
$\frac{1}{18}\sum(X_i - \beta)^2 = \sigma^2 + (\bar{X} - \beta)^2$, so $5 = 1 + (\bar{X} - \beta)^2$ and $\bar{X} - \beta = \pm 2$.
Since $\beta \ne \alpha$, $\bar{X} - \beta = -2$. Then $\alpha = \bar{X} - 2$, $\beta = \bar{X} + 2$ and $|\alpha - \beta| = 4$.
Answer: 4. | ||||
| 86 | 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 rows and columns 1 and 3 form the block $M = \begin{bmatrix}1&0\\3&-1\end{bmatrix}$, with $M^2 = I$; the middle entry is 2 on its own.
So $M^{20} = I$, $M^{19} = M$, and the block equation reads $I + (\alpha + \beta)M = I$, giving $\alpha + \beta = 0$.
Middle entry: $2^{20} + \alpha\,2^{19} + 2\beta = 4$. With $\beta = -\alpha$: $\alpha(2^{19} - 2) = 4 - 2^{20} = -2(2^{19} - 2)$, so $\alpha = -2$, $\beta = 2$.
$\beta - \alpha = 4$.
Answer: 4. | ||||
| 87 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha$ and $\beta$ are roots of $t^2 - t - 1 = 0$, so $\alpha^{n+1} = \alpha^n + \alpha^{n-1}$ (likewise for $\beta$), and $p_{n+1} = p_n + p_{n-1}$.
$29 = p_n + 11$, so $p_n = 18$ and $p_n^2 = 324$.
Answer: 324. | ||||
| 88 | 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 $\gcd(N, 18) = 3$ means N is divisible by 3 but by neither 2 nor 9: an odd multiple of 3 that is not a multiple of 9.
4-digit multiples of 3: from 1002 to 9999, $3333 - 333 = 3000$, alternately even and odd starting with even, so 1500 are odd.
4-digit multiples of 9: from 1008 to 9999, $1111 - 111 = 1000$, of which 500 are odd.
Count $= 1500 - 500 = 1000$.
Answer: 1000. | ||||
| 89 | 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 The roots of $4x^2 - 9x + 5 = 0$ are 1 and $\frac{5}{4}$. Since $AM \ge GM$, $AM = \frac{5}{4}$ and $GM = 1$.
So the two terms satisfy $T_p + T_q = \frac{5}{2}$ and $T_pT_q = 1$: they are the roots of $t^2 - \frac{5}{2}t + 1 = 0$, i.e. 2 and $\frac{1}{2}$.
The sequence (ratio $-\frac{1}{2}$) runs −16, 8, −4, 2, −1, $\frac{1}{2}$, …: 2 is the 4th term and $\frac{1}{2}$ the 6th.
$p + q = 10$.
Answer: 10. | ||||
| 90 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the circle $x^2 + y^2 = \frac{31}{4}$.
$y = mx + c$ touches the ellipse if $c^2 = 9m^2 + 4$, and the circle if $c^2 = \frac{31}{4}(1 + m^2)$.
$9m^2 + 4 = \frac{31}{4} + \frac{31}{4}m^2 \Rightarrow 36m^2 + 16 = 31 + 31m^2 \Rightarrow m^2 = 3$.
Answer: 3. | ||||
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