JEE (Main) · 2021
JEE (Main) 2021 — 25 February, 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
- Electromagnetic Induction and Alternating Currents4
- Gravitation3
- Experimental Skills2
15 easy · 15 medium · 0 hard
Chemistry
30 Q
- Aldehydes, Ketones and Carboxylic Acids3
- Chemical Bonding and Molecular Structure2
- p-Block Elements2
17 easy · 11 medium · 2 hard
Mathematics
30 Q
- Limits, Continuity and Differentiability3
- Integral Calculus3
- Matrices and Determinants3
11 easy · 18 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 | GravitationEscape velocity | Class 11: Gravitation | Assertion and reasonEasy | |
A easy assertion and reason question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Escape velocity $v_e = \sqrt{\frac{2GM}{R}}$. Two planets of unequal mass can have the same escape velocity if $\frac{M_1}{R_1} = \frac{M_2}{R_2}$, so A is correct.
R requires $M_1R_1 = M_2R_2$, which is not the condition (it is the ratio $M/R$ that must be equal), so R is not correct.
Answer: (C). | ||||
| 2 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $h = E/\nu$: $[ML^2T^{-2}]/[T^{-1}] = [ML^2T^{-1}]$ → (ii).
Kinetic energy: $[ML^2T^{-2}]$ → (iii).
Potential $V = W/q$: $[ML^2T^{-2}]/[IT] = [ML^2I^{-1}T^{-3}]$ → (iv).
Momentum $p = mv$: $[MLT^{-1}]$ → (i).
Answer: (C). | ||||
| 3 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $T = 2\pi\sqrt{\frac{L}{g}}$ gives $g = \frac{4\pi^2 L}{T^2} = \frac{4\pi^2 \times 2}{2^2} = 2\pi^2\ ms^{-2}$.
Answer: (D). | ||||
| 4 | Experimental SkillsHeat and sound experiments | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on heat and sound experiments, from Experimental Skills. NCERT: No matching NCERT chapter $\lambda = \frac{v}{f} = \frac{336}{504} = \frac{2}{3}$ m $= 66.7$ cm.
At first resonance the effective length is $\frac{\lambda}{4} = 16.67$ cm, which equals the air column length plus the end correction $e = 0.3d = 0.3 \times 6 = 1.8$ cm.
Air column (the water level read from the top) $= 16.67 - 1.8 \approx 14.8$ cm.
Answer: (C). | ||||
| 5 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane For a regular octagon with centre O, $\sum \overrightarrow{OX} = \vec{0}$ over all eight vertices.
Each $\overrightarrow{AX} = \overrightarrow{AO} + \overrightarrow{OX}$. Summing over X = B, …, H (seven vertices):
$\sum \overrightarrow{AX} = 7\overrightarrow{AO} + (\vec{0} - \overrightarrow{OA}) = 7\overrightarrow{AO} + \overrightarrow{AO} = 8\overrightarrow{AO}$.
$8(2\hat{i} + 3\hat{j} - 4\hat{k}) = 16\hat{i} + 24\hat{j} - 32\hat{k}$.
Answer: (D). | ||||
| 6 | GravitationOrbital motion | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation $T = 2\pi\sqrt{\frac{r^3}{GM}}$, independent of the satellite's mass. $GM = 6.67\times10^{-11} \times 6\times10^{24} = 4.0\times10^{14}\ m^3s^{-2}$.
$r_A = 7000$ km $= 7\times10^6$ m: $T_A = 2\pi\sqrt{\frac{3.43\times10^{20}}{4.0\times10^{14}}} \approx 5.82\times10^3$ s.
$r_B = 8000$ km $= 8\times10^6$ m: $T_B = 2\pi\sqrt{\frac{5.12\times10^{20}}{4.0\times10^{14}}} \approx 7.11\times10^3$ s.
$T_B - T_A \approx 1.29\times10^3$ s, nearest to $1.33\times10^3$ s; the other options are far off.
Answer: (A). | ||||
| 7 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits With X positive, current leaves X up the left wire. $D_1$ (pointing right, with the 10 Ω resistor) is forward biased; $D_2$ (pointing left, with the 5 Ω resistor) is reverse biased and carries no current.
A silicon diode drops about 0.7 V when conducting, so $I = \frac{5 - 0.7}{10} = 0.43$ A.
Answer: (B). | ||||
| 8 | Communication SystemsAmplitude modulation | No matching NCERT chapter | StatementsEasy | |
A easy statements question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter In amplitude modulation the side bands are at $f_c \pm f_m = 1000 \pm 2$ kHz, i.e. 1002 kHz and 998 kHz, so Statement II is true.
The bandwidth is $2f_m = 4$ kHz, so Statement I is true.
Answer: (A). | ||||
| 9 | 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 | Assertion and reasonMedium | |
A medium assertion and reason 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 Thermal stress arises only when the expansion of a body is prevented. A rod lying freely can expand without restraint, so no stress develops: A is true.
A rod does lengthen on heating: R is true.
But the reason no stress develops is that the expansion is unrestrained, not merely that the length increases (a clamped rod also tends to lengthen, and is then stressed). R does not explain A.
Answer: (B). | ||||
| 10 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Call P the left node and Q the right node. The battery E (current $i$) sits between the junction T of the two 5 Ω resistors and the junction S of the 1 Ω and 4 Ω resistors. The inductor L joins P and Q.
At t = 0 the inductor carries no current (open). From T to S there are two paths: 5 Ω + 1 Ω = 6 Ω through P and 5 Ω + 4 Ω = 9 Ω through Q, in parallel: $\frac{6 \times 9}{15} = \frac{18}{5}$ Ω. So $i = \frac{5E}{18}$.
At t = ∞ the inductor is a plain wire, so P and Q are one node: $(5 \parallel 5) + (1 \parallel 4) = 2.5 + 0.8 = 3.3$ Ω. So $i = \frac{E}{3.3} = \frac{10E}{33}$.
Answer: (D). | ||||
| 11 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{1\ \text{mm}}{100} = 0.01$ mm.
Zero of the circular scale below the reference line means a positive zero error: $+8 \times 0.01 = +0.08$ mm.
Observed reading $= 1$ mm $+ 72 \times 0.01$ mm $= 1.72$ mm.
Diameter $= 1.72 - 0.08 = 1.64$ mm, so radius $= 0.82$ mm.
Answer: (C). | ||||
| 12 | 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 $\lambda = \frac{h}{\sqrt{2mqV}}$, so $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4 \times 2} = 2\sqrt{2} \approx 2.8$.
Answer: (A). | ||||
| 13 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics With $I_1 = 2x\,I_2$: $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$, $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{4\sqrt{I_1I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{2x}\,I_2}{(2x+1)I_2} = \frac{2\sqrt{2x}}{2x+1}$.
Answer: (B). | ||||
| 14 | 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 Let the train's length be $L$ and acceleration $a$. Then $v^2 = u^2 + 2aL$.
When the middle point passes the post the train has moved $L/2$: $v_m^2 = u^2 + 2a\frac{L}{2} = u^2 + \frac{v^2 - u^2}{2} = \frac{u^2 + v^2}{2}$.
$v_m = \sqrt{\frac{v^2+u^2}{2}}$.
Answer: (C). | ||||
| 15 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter With equal momentum $p$, the speed is $v = p/m \propto \frac{1}{m}$: masses 1 : 2 : 4 give speeds $1 : \frac{1}{2} : \frac{1}{4} = 4 : 2 : 1$.
Force $F = qvB \propto \frac{q}{m}$: $\frac{1}{1} : \frac{1}{2} : \frac{2}{4} = 2 : 1 : 1$ (taking the velocity perpendicular to the field, the same for all three).
Answer: (B). | ||||
| 16 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter On the axis $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$, so $\frac{B_1}{B_2} = \left(\frac{R^2 + 0.04}{R^2 + 0.0025}\right)^{3/2} = 8$.
Hence $\frac{R^2 + 0.04}{R^2 + 0.0025} = 4$, so $R^2 + 0.04 = 4R^2 + 0.01$, $3R^2 = 0.03$, $R = 0.1$ m.
Answer: (A). | ||||
| 17 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The series current follows from the resistor: $I = \frac{V_R}{R} = \frac{15}{60} = 0.25$ A.
Inductor: $X_L = \frac{20}{0.25} = 80\ \Omega$, so $L = \frac{X_L}{\omega} = \frac{80}{100} = 0.8$ H.
Capacitor: $X_C = \frac{10}{0.25} = 40\ \Omega$, so $C = \frac{1}{\omega X_C} = \frac{1}{100 \times 40} = 2.5\times10^{-4}$ F $= 250\ \mu F$.
Answer: (D). | ||||
| 18 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics At constant pressure: $dQ = nC_P\,dT$, $dU = nC_V\,dT$, $dW = nR\,dT$.
$dU : dQ : dW = \frac{5}{2} : \frac{7}{2} : 1 = 5 : 7 : 2$.
Answer: (A). | ||||
| 19 | 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 Three half lives of Y are six half lives of X.
$N_1\left(\frac{1}{2}\right)^6 = N_2\left(\frac{1}{2}\right)^3$, so $\frac{N_1}{N_2} = 2^3 = \frac{8}{1}$.
Answer: (C). | ||||
| 20 | GravitationNewton's law of gravitation | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation $F_1 = \frac{GMm}{9R^2}$.
The cavity of radius $R/2$ touches the surface on the side facing the particle, so its centre B is at $R/2$ from O and at $3R - \frac{R}{2} = \frac{5R}{2}$ from the particle. Its mass would be $\frac{M}{8}$.
$F_2 = \frac{GMm}{9R^2} - \frac{G(M/8)m}{(5R/2)^2} = \frac{GMm}{R^2}\left(\frac{1}{9} - \frac{1}{50}\right) = \frac{GMm}{R^2}\cdot\frac{41}{450}$.
$F_1 : F_2 = \frac{50}{450} : \frac{41}{450} = 50 : 41$.
Answer: (B). | ||||
| 21 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | NumericalMedium | |
A medium numerical question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The ordered kinetic energy of the gas becomes internal energy. For n moles of molar mass M:
$\frac{1}{2}(nM)v^2 = n\cdot\frac{3}{2}R\,\Delta T$, so $\Delta T = \frac{Mv^2}{3R}$.
With $M = 4.0$ g mol⁻¹ $= 4\times10^{-3}$ kg mol⁻¹ and $v = 30$ m/s: $Mv^2 = 4\times10^{-3}\times900 = 3.6$.
$\Delta T = \frac{3.6}{3R}$, so $x = 3.6$.
Answer: 3.6 | ||||
| 22 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Volume is conserved: $R = n^{1/3}r = 8r$. Charge adds: $Q = 512q$.
$V = \frac{kQ}{R} = \frac{512}{8}\cdot\frac{kq}{r} = 64 \times 2 = 128$ V.
Answer: 128 | ||||
| 23 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current With no resistance, $L\frac{di}{dt} = 3t$, so $i = \frac{3t^2}{2L} = \frac{3t^2}{4}$ (with $i = 0$ at $t = 0$).
At $t = 4$ s: $i = 12$ A.
Energy $= \frac{1}{2}Li^2 = \frac{1}{2}\times2\times144 = 144$ J.
Answer: 144 | ||||
| 24 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Let $v$ be the speed at the top. At the bottom $v_b^2 = v^2 + 4gl$.
$T_{max} = \frac{mv_b^2}{l} + mg = \frac{mv^2}{l} + 5mg$, $T_{min} = \frac{mv^2}{l} - mg$.
With $l = 1$ m, $g = 10$: $\frac{v^2 + 50}{v^2 - 10} = 5$, so $v^2 + 50 = 5v^2 - 50$, $v^2 = 25$, $v = 5$ m/s.
Answer: 5 | ||||
| 25 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At equilibrium $\frac{dU}{dr} = 0$: $-\frac{10\alpha}{r^{11}} + \frac{5\beta}{r^6} = 0$, so $r^5 = \frac{2\alpha}{\beta}$ and $r = \left(\frac{2\alpha}{\beta}\right)^{1/5}$.
So $\frac{a}{b} = \frac{1}{5}$, i.e. $a = 1$ (with $b = 5$).
Answer: 1 | ||||
| 26 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance A surface parallel to the y–z plane has its normal along $\hat{i}$: $\phi_1 = \frac{3}{5}E_0 \times 0.2 = 0.12E_0$.
A surface parallel to the x–z plane has its normal along $\hat{j}$: $\phi_2 = \frac{4}{5}E_0 \times 0.3 = 0.24E_0$.
$\phi_1 : \phi_2 = 1 : 2$, so $a = 1$.
Answer: 1 | ||||
| 27 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $f = \frac{c}{\lambda} = \frac{3\times10^8}{960} = 3.125\times10^5$ Hz.
$f = \frac{1}{2\pi\sqrt{LC}}$ gives $L = \frac{1}{4\pi^2f^2C} = \frac{1}{4\pi^2 \times 9.766\times10^{10} \times 2.56\times10^{-6}} = \frac{1}{\pi^2\times10^6}$ H.
$L = 1.013\times10^{-7}$ H $\approx 10\times10^{-8}$ H (exactly 10 with $\pi^2 = 10$).
Answer: 10 | ||||
| 28 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a lens $m = \frac{f}{f + u}$ (u negative). The images are the same size, one real (u = −20) and one virtual (u = −10), so $m_1 = -m_2$:
$\frac{f}{f - 20} = -\frac{f}{f - 10}$, so $f - 20 = -(f - 10)$, $2f = 30$, $f = 15$ cm.
Answer: 15 | ||||
| 29 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $P = kV^3$ is the polytropic process $PV^{x} = $ constant with $x = -3$.
$W = \frac{nR(T_2 - T_1)}{1 - x} = \frac{nR \times 200}{1 + 3} = 50\,nR$.
(Directly: $W = \int kV^3\,dV = \frac{k(V_2^4 - V_1^4)}{4} = \frac{P_2V_2 - P_1V_1}{4} = \frac{nR\Delta T}{4}$.)
Answer: 50 | ||||
| 30 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity The wire AB is ten 1 m lengths joined end to end, running alternately left-to-right and right-to-left from A.
$J_1$ is on the 4th length, which runs right to left; it is 20 cm from the left end, i.e. 80 cm along that length. Balancing length $l_1 = 3 + 0.8 = 3.8$ m.
$J_2$ is on the 8th length, which also runs right to left; it is 60 cm from the right end. Balancing length $l_2 = 7 + 0.6 = 7.6$ m.
$\frac{E_1}{E_2} = \frac{l_1}{l_2} = \frac{3.8}{7.6} = \frac{1}{2}$, so $a = 1$.
Answer: 1 | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom An orbital has $n - l - 1$ radial nodes, so its radial distribution function $4\pi r^2R^2_{n,l}$ has $n - l$ maxima.
3s: $n = 3$, $l = 0$, so 2 radial nodes and 3 maxima, with the largest maximum farthest out (about 7 Å for hydrogen). Only plot (D) has three maxima.
((A) is 1s: one peak near 0.5 Å; (B), two peaks, is 2s; (C), one peak at larger r, is 2p.)
Answer: (D). | ||||
| 32 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Bond orders: $O_2^{2-}$ (18 electrons) = 1; $He_2^-$ (3 electrons: $\sigma1s^2\sigma^*1s^1$) = 0.5; $He_2^+$ (3 electrons) = 0.5; $Be_2$ (8 electrons: $\sigma1s^2\sigma^*1s^2\sigma2s^2\sigma^*2s^2$) = 0.
A bond order of zero means the species does not exist: $Be_2$.
Answer: (C). | ||||
| 33 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctHard | |
A hard single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Dissolved silver $[Ag^+] = s$ is shared between $CN^-$ and HCN: $s = [CN^-] + [HCN] = [CN^-]\left(1 + \frac{[H^+]}{K_a}\right)$.
$K_{sp} = [Ag^+][CN^-] = \frac{s^2}{1 + [H^+]/K_a}$.
$\frac{[H^+]}{K_a} = \frac{10^{-3}}{6.2\times10^{-10}} = 1.61\times10^6$, so $s^2 = 2.2\times10^{-16}\times1.61\times10^6 = 3.55\times10^{-10}$.
$s = 1.9\times10^{-5}$ M.
Answer: (D). | ||||
| 34 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich isotherm: $\frac{x}{m} = kP^{1/n}$ ($n > 1$), valid at intermediate pressure; at low pressure the dependence tends to $P^1$ and at high pressure to $P^0$.
At moderate pressure the exponent is $\frac{1}{n}$.
Answer: (D). | ||||
| 35 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter An Ellingham diagram plots the standard Gibbs energy of formation of oxides (or sulphides, halides) against temperature, $\Delta G^\circ$ vs T.
Answer: (A). | ||||
| 36 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter $H_2O_2$ acts as an oxidant when its oxygen (−1) is reduced to −2 in $H_2O$; it acts as a reductant when it is oxidised to $O_2$ (0).
In (A), (B) and (D) $O_2$ is released: $H_2O_2$ reduces $IO_4^-$, $I_2$ and $Cl_2$.
In (C) iodide is oxidised to $I_2$ and $H_2O_2$ becomes water: this shows its oxidising nature.
Answer: (C). | ||||
| 37 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) False: the B–H–B angle at a bridging hydrogen is about 97° (and the H–B–H angles differ too).
(B) False: the two three-centre, two-electron B–H–B bridges are identical.
(C) True: each boron is roughly $sp^3$; the orbitals used for the two strong terminal B–H bonds (H–B–H about 120°) have more s-character, so those used for the weaker, longer bridging bonds have more p-character. The terminal bonds therefore have less p-character.
(D) False: $BH_3$ is electron-deficient, a Lewis acid.
Answer: (C). | ||||
| 38 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | StatementsMedium | |
A medium statements question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Statement I: cerium(IV) has a strong tendency to go to the stable +3 state, so Ce(IV) compounds such as $CeO_2$ are oxidising agents and are used in organic oxidations, including of aldehydes and ketones. True.
Statement II: europium(II) ($4f^7$) readily goes to the common +3 state, so $Eu^{2+}$ in aqueous $EuSO_4$ is a strong reducing agent. True.
Answer: (A). | ||||
| 39 | 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 $V^{2+}$: $[Ar]3d^3$; $Cr^+$: $[Ar]3d^5$; $Mn^{2+}$: $[Ar]3d^5$; $Ni^{2+}$: $[Ar]3d^8$; $Cu^+$: $[Ar]3d^{10}$; $Fe^{2+}$: $[Ar]3d^6$; $Co^+$: $[Ar]3d^8$.
$Cr^+$ and $Mn^{2+}$ are both $3d^5$.
Answer: (B). | ||||
| 40 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Both contain a $d^5$ ion ($Mn^{2+}$ and $Fe^{3+}$) with the strong-field ligand $CN^-$, so the electrons pair up in the $t_{2g}$ set: $t_{2g}^5$, one unpaired electron.
Two inner d orbitals are free, so the hybridisation is $d^2sp^3$ (inner orbital), and both are paramagnetic.
Answer: (A). | ||||
| 41 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | StatementsMedium | |
A medium statements question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Statement I: ozone, the allotrope of oxygen, is formed in and characteristic of photochemical smog, which is oxidising. Reducing (classical) smog is smoke, fog and sulphur dioxide and has no ozone intermediate. False.
Statement II: photochemical smog forms from nitrogen oxides and hydrocarbons in sunlight; the oxides of nitrogen (with those of sulphur, as the statement groups them) are the tropospheric pollutant gases it refers to. Taken as true.
Answer: (D). | ||||
| 42 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Carbon: $2.64\times\frac{12}{44} = 0.72$ g. Hydrogen: $1.08\times\frac{2}{18} = 0.12$ g.
Oxygen: $1.80 - 0.72 - 0.12 = 0.96$ g, i.e. $\frac{0.96}{1.80}\times100 = 53.33\%$.
Answer: (B). | ||||
| 43 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The reactant is 2-methylhexane, $CH_3CH_2CH_2CH_2CH(CH_3)CH_3$ (seven carbons, a six-carbon chain with a methyl branch).
Heating an alkane of six or more carbons over an oxide catalyst (oxides of Cr, V or Mo) at 773 K and 10–20 atm brings about aromatisation: the six-carbon chain cyclises and dehydrogenates to a benzene ring, and the branch stays as a substituent.
2-methylhexane therefore gives toluene (methylbenzene).
Answer: (B). | ||||
| 44 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate is a cyclohexene with $OCH_3$ on the allylic carbon at one end of the double bond and $NO_2$ on the allylic carbon at the other end.
HCl adds through the more stable carbocation. A cation next to the carbon bearing $NO_2$ is strongly destabilised by its −I effect; the cation next to the carbon bearing $OCH_3$ is less destabilised. So $H^+$ adds to the carbon next to the $NO_2$ carbon and $Cl^-$ ends up on the carbon next to the $OCH_3$ carbon: A is 2-chloro-1-methoxy-4-nitrocyclohexane.
NaI in dry acetone (Finkelstein reaction) replaces the chloride by iodide ($S_N2$; NaCl precipitates); the methoxy group is untouched. B has I where A had Cl.
Answer: (B). | ||||
| 45 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids 1) Acid hydrolysis of the nitrile: $CH_3CH_2CN \rightarrow CH_3CH_2COOH$.
2) $SOCl_2$: $CH_3CH_2COOH \rightarrow CH_3CH_2COCl$.
3) $H_2$ over Pd/$BaSO_4$ (Rosenmund reduction) stops at the aldehyde: $CH_3CH_2CHO$.
Answer: (D). | ||||
| 46 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) Wacker oxidation of ethene gives acetaldehyde.
(B) Chromic acid ($CrO_3$–$H_2SO_4$, Jones reagent) oxidises a primary alcohol through the aldehyde to the carboxylic acid: acetic acid, not acetaldehyde.
(C) DIBAL-H reduces a nitrile to the imine, hydrolysed to the aldehyde: acetaldehyde.
(D) Dehydrogenation of ethanol over Cu at 573 K gives acetaldehyde.
Answer: (B). | ||||
| 47 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A compound liberates $CO_2$ from $NaHCO_3$ only if it is a stronger acid than carbonic acid ($pK_a \approx 6.4$).
A, 2,4,6-triaminophenol: the electron-donating $NH_2$ groups make it weaker than phenol ($pK_a \approx 10$): no reaction.
B, benzoic acid ($pK_a \approx 4.2$): liberates $CO_2$.
C, 2,4,6-trinitrophenol (picric acid, $pK_a \approx 0.4$): liberates $CO_2$.
Answer: (C). | ||||
| 48 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines A: Sn/HCl reduces nitrobenzene to aniline; $HNO_2$ (cold) gives benzenediazonium chloride; coupling with aniline gives p-aminoazobenzene.
B: $NaBH_4$ does not reduce an aromatic nitro group to the amine, so no aniline (and no diazonium salt) is formed; this route does not give p-aminoazobenzene.
C: aniline with $HNO_2$ gives benzenediazonium chloride, which couples with aniline in HCl to give p-aminoazobenzene.
Answer: (C). | ||||
| 49 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter (A) False: Buna-N (butadiene–acrylonitrile) is a synthetic rubber.
(B) False: Buna-S is a synthetic elastomer, not a thermosetting polymer.
(C) False: neoprene is an addition homopolymer of chloroprene, used for hoses, gaskets and conveyor belts, not for buckets.
(D) Buna-S is made by free-radical copolymerisation of 1,3-butadiene and styrene, started by a peroxide initiator, which supplies the oxygen-centred radicals; this is the statement taken as correct.
Answer: (D). | ||||
| 50 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Lactose is β-D-galactose joined through its C-1 to C-4 of β-D-glucose (a β-1,4 glycosidic link).
Answer: (A). | ||||
| 51 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Phenolphthalein end point: all the NaOH and the first step $Na_2CO_3 \rightarrow NaHCO_3$.
Methyl orange (second) end point: $NaHCO_3 \rightarrow CO_2$, which uses as many moles of HCl as there were moles of $Na_2CO_3$.
Moles of $Na_2CO_3 = 1.5\times10^{-3}\ \text{L}\times0.1 = 1.5\times10^{-4}$ mol, mass $= 1.5\times10^{-4}\times106 = 0.0159$ g.
Percentage $= \frac{0.0159}{0.4}\times100 = 3.975 \approx 4$.
Answer: 4 | ||||
| 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 At constant volume (a rigid tyre) $\frac{P_1}{T_1} = \frac{P_2}{T_2}$: $T_2 = 300\times\frac{40}{35} = 342.86$ K.
In °C: $342.86 - 273 = 69.86 \approx 70$.
Answer: 70 | ||||
| 53 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta H = \Delta U + \Delta n_g RT$, counting only gases.
$\Delta n_g = (1 + 1) - \frac{3}{2} = +\frac{1}{2}$ (the product gas written $O_2$ is evidently $CO_2$; the count is the same either way).
$\Delta H = -742.24 + \frac{1}{2}\times8.314\times10^{-3}\times298 = -742.24 + 1.24 = -741.00$ kJ $mol^{-1}$.
Magnitude $= 741$.
Answer: 741 | ||||
| 54 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $A_2B_3 \rightarrow 2A^{3+} + 3B^{2-}$ gives 5 ions: $i = 1 + \alpha(5 - 1) = 1 + 0.6\times4 = 3.4$.
$\Delta T_b = iK_bm = 3.4\times0.52\times1 = 1.768$ K.
Boiling point $= 373.15 + 1.768 = 374.92$ K $\approx 375$ K (373 + 1.77 = 374.77 also rounds to 375).
Answer: 375 | ||||
| 55 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Cr: +6 in $CrO_4^{2-}$ → +3 in $Cr(OH)_4^-$, gaining 3 electrons.
S: +2 in $S_2O_3^{2-}$ → +6 in $SO_4^{2-}$, so each $S_2O_3^{2-}$ loses $2\times4 = 8$ electrons.
Electrons lost by thiosulphate: $40\times0.25\times8 = 80$ mmol. Chromate needed: $\frac{80}{3}$ mmol.
Volume $= \frac{80}{3\times0.154} = 173.2$ mL $\approx 173$.
Answer: 173 | ||||
| 56 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\log k = \log A - \frac{E_a}{2.303R}\cdot\frac{1}{T}$, and the slope is −10000 K.
$\log\frac{k_2}{k_1} = -10000\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$: $\log\frac{10^{-4}}{10^{-5}} = 1 = 10000\left(\frac{1}{500} - \frac{1}{T_2}\right)$.
$\frac{1}{T_2} = 0.002 - 0.0001 = 0.0019$, so $T_2 = 526.3$ K $\approx 526$.
Answer: 526 | ||||
| 57 | Chemical Bonding and Molecular StructureIonic bonding and lattice enthalpy | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on ionic bonding and lattice enthalpy, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure From gaseous atoms: $Na_{(g)} \rightarrow Na^+_{(g)}$ (+495.8), $Br_{(g)} \rightarrow Br^-_{(g)}$ (−325.0), $Na^+_{(g)} + Br^-_{(g)} \rightarrow NaBr_{(s)}$ (−728.4).
Total $= 495.8 - 325.0 - 728.4 = -557.6$ kJ $mol^{-1} = -5576\times10^{-1}$ kJ $mol^{-1}$.
Answer: 5576 | ||||
| 58 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter $BF_3$ (empty p orbital on B), $SiCl_4$ (vacant d orbitals on Si) and $PCl_5$ are all hydrolysed by water.
In $SF_6$ the sulphur is coordinatively saturated and sterically shielded by six fluorines, so water cannot attack: it is inert to hydrolysis.
Number $= 1$.
Answer: 1 | ||||
| 59 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 1) Ethyne trimerises in a red-hot iron tube at 873 K to benzene.
2) CO and HCl with anhydrous $AlCl_3$ (Gattermann–Koch reaction) formylate benzene to benzaldehyde, $C_6H_5CHO$.
All six ring carbons and the carbonyl carbon are $sp^2$: 7.
Answer: 7 | ||||
| 60 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $R_f = \frac{\text{distance moved by the compound}}{\text{distance moved by the solvent}}$, both from the base line.
A is 2 cm above the base line; the solvent front is 5 cm above it.
$R_f = \frac{2}{5} = 0.4 = 4\times10^{-1}$.
Answer: 4 | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions By induction $f(n) = nf(1) = cn$ with $c = f(1) \in \mathbb{N}$.
(A) True: if $f$ is onto, 1 is a value, so $c = 1$ and $f(n) = n$.
(B) True: $cn_1 = cn_2 \Rightarrow n_1 = n_2$.
(C) Not true: $(fog)(n) = c\,g(n)$ is one-one only if $g$ is one-one, and an onto $g$ need not be one-one (e.g. $g(1) = g(2) = 1$, $g(n) = n - 1$ for $n \ge 2$).
(D) True: if $g(n_1) = g(n_2)$ then $fog(n_1) = fog(n_2)$, so a one-one $fog$ forces $g$ one-one.
Answer: (C). | ||||
| 62 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $z = x + iy$.
Line 1: $(2 - i)(x + iy) = (2x + y) + i(2y - x)$ and $(2 + i)(x - iy) = (2x + y) + i(x - 2y)$. Equal when $2y - x = x - 2y$: $x = 2y$.
Line 2: $(2 + i)(x + iy) + (i - 2)(x - iy) = 2i(x + 2y)$, so $2i(x + 2y) - 4i = 0$: $x + 2y = 2$.
Normals meet at the centre: $(1, \frac{1}{2})$.
Tangent: $i(x + iy) + (x - iy) + 1 + i = (x - y + 1)(1 + i) = 0$: $x - y + 1 = 0$.
Radius $= \frac{|1 - \frac{1}{2} + 1|}{\sqrt{2}} = \frac{3}{2\sqrt{2}}$.
Answer: (C). | ||||
| 63 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations An upward quadratic is positive for all real x when its discriminant is negative:
$4(3k - 1)^2 - 4(8k^2 - 7) < 0 \Rightarrow 9k^2 - 6k + 1 - 8k^2 + 7 < 0 \Rightarrow k^2 - 6k + 8 < 0 \Rightarrow 2 < k < 4$.
The only integer is $k = 3$.
Answer: (B). | ||||
| 64 | 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 Each is an infinite G.P. with ratio between 0 and 1:
$x = \frac{1}{1 - \cos^2\theta}$, so $\cos^2\theta = 1 - \frac{1}{x}$; $y = \frac{1}{1 - \sin^2\phi}$, so $\sin^2\phi = 1 - \frac{1}{y}$.
$\frac{1}{z} = 1 - \cos^2\theta\sin^2\phi = 1 - \left(1 - \frac{1}{x}\right)\left(1 - \frac{1}{y}\right) = \frac{1}{x} + \frac{1}{y} - \frac{1}{xy} = \frac{x + y - 1}{xy}$.
So $xy = z(x + y - 1)$, i.e. $xy + z = (x + y)z$.
Answer: (B). | ||||
| 65 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f(1) = f(2)$: $1 - a + b - 4 = 8 - 4a + 2b - 4$, so $3a - b = 7$.
$f'(x) = 3x^2 - 2ax + b$; $f'\left(\frac{4}{3}\right) = \frac{16}{3} - \frac{8a}{3} + b = 0$, so $8a - 3b = 16$.
With $b = 3a - 7$: $8a - 9a + 21 = 16$, $a = 5$, $b = 8$.
Answer: (A). | ||||
| 66 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $H_n = 1 + \frac{1}{2} + \dots + \frac{1}{n}$. Then $\frac{H_n}{n^2} \to 0$, so the limit has the form $1^\infty$ and equals $e^{L}$ with $L = \lim n\cdot\frac{H_n}{n^2} = \lim\frac{H_n}{n}$.
Since $H_n \le 1 + \ln n$, $\frac{H_n}{n} \to 0$. So $L = 0$ and the limit is $e^0 = 1$.
Answer: (D). | ||||
| 67 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{\sin\theta\sin2\theta}{1 - \cos2\theta} = \frac{2\sin^2\theta\cos\theta}{2\sin^2\theta} = \cos\theta$. Put $t = \sin\theta$, $dt = \cos\theta\,d\theta$:
$I = \int (t^6 + t^4 + t^2)\sqrt{2t^4 + 3t^2 + 6}\,dt = \int (t^5 + t^3 + t)\sqrt{2t^6 + 3t^4 + 6t^2}\,dt$ (taking $t \ge 0$).
With $u = 2t^6 + 3t^4 + 6t^2$, $du = 12(t^5 + t^3 + t)\,dt$: $I = \frac{1}{12}\int\sqrt{u}\,du = \frac{1}{18}u^{3/2} + c$.
So $I = \frac{1}{18}\left[2\sin^6\theta + 3\sin^4\theta + 6\sin^2\theta\right]^{3/2} + c$.
Writing $s = \sin^2\theta = 1 - \cos^2\theta$: $2s^3 + 3s^2 + 6s = 11 - 18\cos^2\theta + 9\cos^4\theta - 2\cos^6\theta$.
Answer: (D). | ||||
| 68 | 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 For $-1 \le x < 0$, $x^3 \in [-1, 0)$ so $[x^3] = -1$; for $0 \le x < 1$, $[x^3] = 0$.
$\int_{-1}^{0} x^2e^{-1}\,dx + \int_{0}^{1} x^2\,dx = \frac{1}{3e} + \frac{1}{3} = \frac{e + 1}{3e}$.
Answer: (D). | ||||
| 69 | 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 $\frac{dy}{dx} - \frac{y}{x - 2} = \frac{(x - 2)^2 + 4}{x - 2} = (x - 2) + \frac{4}{x - 2}$.
Integrating factor $\frac{1}{x - 2}$: $\frac{d}{dx}\left(\frac{y}{x - 2}\right) = 1 + \frac{4}{(x - 2)^2}$, so $\frac{y}{x - 2} = x - \frac{4}{x - 2} + C$.
At (0, 0): $0 = 0 + 2 + C$, $C = -2$. Then $y = (x - 2)^2 - 4 = x^2 - 4x$.
At $x = 5$: $y = 5$. So (5, 5) lies on the curve; (4, 5), (5, 4) and (4, 4) do not ($y(4) = 0$).
Answer: (B). | ||||
| 70 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines Image $(h, k)$: $\frac{h - 3}{1} = \frac{k - 5}{-1} = -2\cdot\frac{3 - 5 + 1}{2} = 1$, so $(h, k) = (4, 4)$.
Check: $(4 - 2)^2 + (4 - 4)^2 = 4$, so it lies on $(x - 2)^2 + (y - 4)^2 = 4$. (The others give 36, 0 and 8.)
Answer: (B). | ||||
| 71 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The line $2x + y = 1$ has slope −2, so the tangent has slope $m = \frac{1}{2}$.
For $y^2 = 4ax$ with $a = \frac{3}{2}$, the tangent of slope m is $y = mx + \frac{a}{m} = \frac{x}{2} + 3$.
(0, 3): 3 ✓. (−6, 0): 0 ✓. (4, 5): 5 ✓. (5, 4): $\frac{5}{2} + 3 = 5.5 \ne 4$.
Answer: (C). | ||||
| 72 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives At an intersection point $(x, y)$ the normals are along $\left(\frac{x}{a}, \frac{y}{b}\right)$ and $\left(\frac{x}{c}, \frac{y}{d}\right)$. Orthogonality: $\frac{x^2}{ac} + \frac{y^2}{bd} = 0$.
Subtracting the two equations: $x^2\frac{c - a}{ac} + y^2\frac{d - b}{bd} = 0$.
Substituting $\frac{x^2}{ac} = -\frac{y^2}{bd}$: $\frac{y^2}{bd}\left[-(c - a) + (d - b)\right] = 0$, so $c - a = d - b$, i.e. $a - b = c - d$.
Answer: (B). | ||||
| 73 | 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 $n = l + m$, so $l^2 + m^2 = (l + m)^2$, giving $lm = 0$.
$l = 0$: $m = n$, direction (0, 1, 1). $m = 0$: $l = n$, direction (1, 0, 1).
$\cos\alpha = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2}$, $\alpha = 60°$.
$\sin^4\alpha + \cos^4\alpha = \frac{9}{16} + \frac{1}{16} = \frac{5}{8}$.
Answer: (A). | ||||
| 74 | 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 Foot of the perpendicular from A(0, 1, 2): a point of the line is $P(1 + 2t, -1 + 3t, 1 - 2t)$, and $\overrightarrow{AP} = (1 + 2t, -2 + 3t, -1 - 2t)$.
$\overrightarrow{AP}\cdot(2, 3, -2) = 2 + 4t - 6 + 9t + 2 + 4t = 17t - 2 = 0$, so $t = \frac{2}{17}$ and $\overrightarrow{AP} = \frac{1}{17}(21, -28, -21) \parallel (-3, 4, 3)$.
The line is $\frac{x}{-3} = \frac{y - 1}{4} = \frac{z - 2}{3}$. (It is also the only option whose direction is perpendicular to (2, 3, −2).)
Answer: (B). | ||||
| 75 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability For one missile: P(hit) = P(not intercepted) × P(hit | not intercepted) $= \frac{2}{3}\times\frac{3}{4} = \frac{1}{2}$ (an intercepted missile does not hit).
Three independent missiles: $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Answer: (B). | ||||
| 76 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Equal roots need $b^2 = 4ac$ with $a, b, c \in \{1, \dots, 6\}$, so b is even.
$b = 2$: $ac = 1$: (1, 1). $b = 4$: $ac = 4$: (1, 4), (4, 1), (2, 2). $b = 6$: $ac = 9$: (3, 3).
5 favourable outcomes out of $6^3 = 216$: $\frac{5}{216}$.
Answer: (D). | ||||
| 77 | 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 $\sin2\theta + \tan2\theta = \tan2\theta(\cos2\theta + 1) = 2\cos^2\theta\tan2\theta$.
Since $2\cos^2\theta \ge 0$ (and must be non-zero), the expression is positive exactly when $\tan2\theta > 0$ and $\cos\theta \ne 0$.
$\tan2\theta > 0$ for $2\theta \in (0, \frac{\pi}{2}) \cup (\pi, \frac{3\pi}{2}) \cup (2\pi, \frac{5\pi}{2}) \cup (3\pi, \frac{7\pi}{2})$, i.e. $\theta \in (0, \frac{\pi}{4}) \cup (\frac{\pi}{2}, \frac{3\pi}{4}) \cup (\pi, \frac{5\pi}{4}) \cup (\frac{3\pi}{2}, \frac{7\pi}{4})$; none of these contains $\frac{\pi}{2}$ or $\frac{3\pi}{2}$.
Answer: (D). | ||||
| 78 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $24 = 2^3\cdot3$. Distribute the exponent 3 of 2 among x, y, z: $\binom{5}{2} = 10$ ways. Distribute the single 3: 3 ways.
Total $= 10\times3 = 30$.
Answer: (B). | ||||
| 79 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let the tower's height be h. A is at distance $h\cot30° = h\sqrt{3}$ from the base, B at $h\cot45° = h$.
Speed $v = \frac{h(\sqrt{3} - 1)}{20}$. Time from B to the base $= \frac{h}{v} = \frac{20}{\sqrt{3} - 1} = 10(\sqrt{3} + 1)$ s.
Answer: (B). | ||||
| 80 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $A \rightarrow (B \rightarrow A) \equiv \sim A \vee \sim B \vee A$, which is a tautology.
$A \rightarrow (A \vee B) \equiv \sim A \vee A \vee B$, also a tautology, so the two are equivalent.
The others are not tautologies: with A true and B false, $A \rightarrow B$, $A \wedge B$ and $A \leftrightarrow B$ are all false, so (A), (C) and (D) are false there.
Answer: (B). | ||||
| 81 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let $t = \tan\frac{\theta}{2}$. $I - A = \begin{bmatrix}1 & t\\ -t & 1\end{bmatrix}$, with determinant $1 + t^2$ and inverse $\frac{1}{1 + t^2}\begin{bmatrix}1 & -t\\ t & 1\end{bmatrix}$.
$(I + A)(I - A)^{-1} = \frac{1}{1 + t^2}\begin{bmatrix}1 & -t\\ t & 1\end{bmatrix}\begin{bmatrix}1 & -t\\ t & 1\end{bmatrix} = \frac{1}{1 + t^2}\begin{bmatrix}1 - t^2 & -2t\\ 2t & 1 - t^2\end{bmatrix}$.
So $a = \cos\theta$, $b = \sin\theta$, $a^2 + b^2 = 1$ and $13(a^2 + b^2) = 13$.
Answer: 13 | ||||
| 82 | 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 A is symmetric, so $A^2 = AA^T = I$: the rows are orthonormal. Hence $x^2 + y^2 + z^2 = 1$ and $xy + yz + zx = 0$.
Then $(x + y + z)^2 = 1$, and $x + y + z > 0$ gives $x + y + z = 1$.
$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 1\cdot1 = 1$.
$x^3 + y^3 + z^3 = 1 + 3\times2 = 7$.
Answer: 7 | ||||
| 83 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The z-coefficients are twice the y-coefficients in every equation, so put $w = y + 2z$; the coefficient determinant is 0 for every k.
$3x - w = 2$ and $-2x - 2w = 3$ give $w = 3x - 2$, $-8x + 4 = 3$, $x = \frac{1}{8}$, $w = -\frac{13}{8}$.
The system is consistent (and then has infinitely many solutions, since z is free) when the first equation also holds: $\frac{k}{8} - \frac{13}{8} = 1$, so $k = 21$.
Answer: 21 | ||||
| 84 | 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 Three-digit numbers with distinct digits from {1, 2, 3, 4, 5}.
Divisible by 3: digit sum a multiple of 3: {1, 2, 3}, {1, 3, 5}, {2, 3, 4}, {3, 4, 5}, each in 3! = 6 orders: 24.
Divisible by 5: last digit 5, the other two from four digits: $4\times3 = 12$.
Both: sets {1, 3, 5} and {3, 4, 5} with 5 last: 2 each, 4.
Total $= 24 + 12 - 4 = 32$.
Answer: 32 | ||||
| 85 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Side of $A_n$ $= \sqrt{2}\times$ side of $A_{n+1}$, so the sides form a G.P. with ratio $\frac{1}{\sqrt{2}}$ and the areas a G.P. with ratio $\frac{1}{2}$.
Area of $A_n = \frac{144}{2^{n-1}} < 1 \Rightarrow 2^{n-1} > 144 \Rightarrow n - 1 \ge 8$ (since $2^7 = 128$, $2^8 = 256$).
Smallest $n = 9$.
Answer: 9 | ||||
| 86 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Candidates: $x = -\frac{1}{2}$, $x = -2$, and the zeros of $x^2 + x - 2 = (x + 2)(x - 1)$: $x = -2, 1$.
At $x = -\frac{1}{2}$: only $|2x + 1|$ has a corner: not differentiable.
At $x = 1$: only $|x^2 + x - 2|$ has a corner (slope ±3): not differentiable.
At $x = -2$: $-3|x + 2| + |x + 2||x - 1| = |x + 2|(|x - 1| - 3)$, and near −2, $|x - 1| - 3 = -(x + 2)$, so this part is $-|x + 2|(x + 2)$, which is differentiable at −2 (derivative 0). The rest is smooth there.
So f is not differentiable at 2 points.
Answer: 2 | ||||
| 87 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The limit forces $f(x) = x^6 + ax^5 + bx^4 + x^3$.
$f'(x) = 6x^5 + 5ax^4 + 4bx^3 + 3x^2$.
$f'(1) = 9 + 5a + 4b = 0$ and $f'(-1) = -3 + 5a - 4b = 0$: adding, $10a + 6 = 0$, $a = -\frac{3}{5}$; then $b = -\frac{3}{2}$.
$f(2) = 64 + 32a + 16b + 8 = 72 - 19.2 - 24 = 28.8$, so $5f(2) = 144$.
Answer: 144 | ||||
| 88 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Consecutive intersections: $x = \frac{\pi}{4}$ and $\frac{5\pi}{4}$.
$A = \int_{\pi/4}^{5\pi/4}(\sin x - \cos x)\,dx = \left[-\cos x - \sin x\right]_{\pi/4}^{5\pi/4} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$.
$A^4 = 64$.
Answer: 64 | ||||
| 89 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections From the lines: $k(\sqrt{3}x + y) = 4\sqrt{3}$ and $\sqrt{3}x - y = 4\sqrt{3}k$.
Multiplying: $(\sqrt{3}x + y)(\sqrt{3}x - y) = 48$, i.e. $3x^2 - y^2 = 48$, or $\frac{x^2}{16} - \frac{y^2}{48} = 1$.
$e^2 = 1 + \frac{48}{16} = 4$, so $e = 2$.
Answer: 2 | ||||
| 90 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $(\vec{r} - \vec{c})\times\vec{a} = \vec{0}$, so $\vec{r} = \vec{c} + \lambda\vec{a}$.
$\vec{r}\cdot\vec{b} = \vec{c}\cdot\vec{b} + \lambda\,\vec{a}\cdot\vec{b} = 2 - \lambda = 0$, so $\lambda = 2$.
$\vec{r}\cdot\vec{a} = \vec{c}\cdot\vec{a} + 2|\vec{a}|^2 = (1 - 2 + 1) + 2\times6 = 12$.
Answer: 12 | ||||
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