JEE (Main) · 2021
JEE (Main) 2021 — 22 July, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Motion of System of Particles and Rigid Body3
- Magnetic Effects of Current and Magnetism2
- Optics2
22 easy · 7 medium · 1 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- p-Block Elements2
- Amines2
25 easy · 4 medium · 1 hard
Mathematics
30 Q
- Matrices and Determinants3
- Trigonometry2
- Three Dimensional Geometry2
7 easy · 19 medium · 4 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 | Communication SystemsLine of sight communication and antenna height | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on line of sight communication and antenna height, from Communication Systems. NCERT: No matching NCERT chapter Range of a line-of-sight antenna: $d = \sqrt{2hR_e}$, so $h = \frac{d^2}{2R_e} = \frac{(150\times10^3)^2}{2\times6.5\times10^6} = \frac{2.25\times10^{10}}{1.3\times10^7} \approx 1731$ m.
Area covered: $\pi d^2 = 3.14\times(150)^2 \approx 7.07\times10^4\ \mathrm{km^2}$.
Population covered $= 2000\times7.07\times10^4 \approx 1.413\times10^8 = 1413\times10^5$.
Answer: (C). | ||||
| 2 | Magnetic Effects of Current and MagnetismEarth magnetism | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on earth magnetism, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter In the magnetic meridian, $\tan\delta = \frac{B_V}{B_H}$. In a vertical plane at angle $\theta$ to the meridian only the component $B_H\cos\theta$ of the horizontal field acts, so the apparent dip $\delta'$ satisfies $\tan\delta' = \frac{B_V}{B_H\cos\theta} = \frac{\tan\delta}{\cos\theta} \ge \tan\delta$.
So the true dip is less than the apparent dip (equal only in the magnetic meridian). It is mathematically related, so (D) is wrong, and (A), (B) are wrong.
Answer: (C). | ||||
| 3 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Reflected and refracted rays perpendicular: $r + 90^\circ + r' = 180^\circ$, so $r' = 90^\circ - r$, and $r = i$.
Snell's law from the denser medium (relative index $\mu$): $\mu\sin i = \sin r' = \cos r$, so $\mu = \frac{\cos r}{\sin r} = \cot r$.
Critical angle: $\sin C = \frac{1}{\mu} = \tan r$, so $C = \sin^{-1}(\tan r)$.
Answer: (C). | ||||
| 4 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current In a purely capacitive circuit $q = CE_{g_0}\sin\omega t$ and $I_C = \frac{dq}{dt} = \omega CE_{g_0}\cos\omega t = \omega CE_{g_0}\sin\left(\omega t + \frac{\pi}{2}\right)$.
The current leads the voltage by $90^\circ$: rotating anticlockwise, the $I_C$ phasor is $90^\circ$ ahead of $V_C$. That is the third diagram.
Answer: (C). | ||||
| 5 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory A monoatomic molecule has 3 translational degrees of freedom, each carrying $\frac{1}{2}k_BT$ (equipartition).
Average energy per molecule $= \frac{3}{2}k_BT$.
Answer: (C). | ||||
| 6 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsEasy | |
A easy statements question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Statement I: true. Above the Curie temperature thermal agitation destroys the alignment within the domains and the ferromagnet becomes paramagnetic.
Statement II: false. Heating does not build up the domain structure; as the temperature rises the ordered domains are broken up by thermal agitation, and above the Curie temperature the domains disappear altogether.
Answer: (C). | ||||
| 7 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $\frac{l}{A} = \frac{0.25}{3\times10^{-6}} = 8.33\times10^4\ \mathrm{m^{-1}}$.
$R_{Cu} = 1.7\times10^{-8}\times8.33\times10^4 = 1.417\ \mathrm{m\Omega}$, $R_{Al} = 2.6\times10^{-8}\times8.33\times10^4 = 2.167\ \mathrm{m\Omega}$.
The rods are in parallel between A and B: $R = \frac{1.417\times2.167}{1.417+2.167} = \frac{3.069}{3.583} \approx 0.857\ \mathrm{m\Omega}$, i.e. $0.858\ \mathrm{m\Omega}$.
Answer: (B). | ||||
| 8 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Field of the line charge at distance $r$: $E = \frac{\lambda}{2\pi\varepsilon_0 r}$, radially outward along the $x$-axis.
Net force on the dipole: $F = q\frac{\lambda}{2\pi\varepsilon_0}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$ with $r_1 = 0.010$ m, $r_2 = 0.012$ m.
$\frac{\lambda}{2\pi\varepsilon_0} = 3\times10^{-6}\times1.8\times10^{10} = 5.4\times10^4$; $\frac{1}{0.010} - \frac{1}{0.012} = 100 - 83.33 = 16.67$.
$4 = q\times5.4\times10^4\times16.67 = q\times9\times10^5$, so $q = 4.44\times10^{-6}$ C $= 4.44\ \mu$C.
Answer: (D). | ||||
| 9 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At $t = 0$: $x(0) = B$, and $v = A\omega\cos\omega t - B\omega\sin\omega t$ gives $v(0) = A\omega$.
$C\cos(\omega t - \phi) = C\cos\phi\cos\omega t + C\sin\phi\sin\omega t$, so $C\cos\phi = B = x(0)$ and $C\sin\phi = A = \frac{v(0)}{\omega}$.
Hence $C = \sqrt{x(0)^2 + \frac{v(0)^2}{\omega^2}}$ and $\tan\phi = \frac{v(0)}{x(0)\omega}$.
Answer: (B). | ||||
| 10 | 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 Both gain kinetic energy $eV$ (equal charge magnitude), so $p = \sqrt{2meV}$ and $\lambda = \frac{h}{\sqrt{2meV}}$.
$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$.
Answer: (D). | ||||
| 11 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Vector projection of $\vec{A}$ on $\vec{B}$: $\frac{\vec{A}\cdot\vec{B}}{|\vec{B}|^2}\vec{B} = \frac{1+1}{2}(\hat{i}+\hat{j}) = \hat{i}+\hat{j}$.
(The scalar projection is $\sqrt{2}$, its magnitude.)
Answer: (B). | ||||
| 12 | GravitationEscape velocity | Class 11: Gravitation | Single correctHard | |
A hard single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Projected with escape speed, the total energy is zero: $\frac{1}{2}v^2 = \frac{GM}{r} = \frac{gR_e^2}{r}$, so $v = \frac{dr}{dt} = R_e\sqrt{\frac{2g}{r}}$.
$\int_{R_e}^{R_e+h}\sqrt{r}\,dr = R_e\sqrt{2g}\,t$ gives $\frac{2}{3}\left[(R_e+h)^{3/2} - R_e^{3/2}\right] = R_e\sqrt{2g}\,t$.
$t = \frac{2}{3}\frac{R_e^{3/2}}{R_e\sqrt{2g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2} - 1\right] = \frac{2}{3}\sqrt{\frac{R_e}{2g}}\left[\cdots\right] = \frac{1}{3}\sqrt{\frac{2R_e}{g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2} - 1\right]$.
Answer: (B). | ||||
| 13 | 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 Rolling without slipping from height $h$: $mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$, so $v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$.
$\frac{k^2}{R^2}$: sphere $\frac{2}{5}$, cylinder $\frac{1}{2}$, ring 1. The smaller it is, the larger $v$.
So the sphere is fastest and the ring slowest at the bottom.
Answer: (B). | ||||
| 14 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At constant velocity the porter's force is upward and equal to $mg = 80\times9.8 = 784$ N, while the displacement (0.8 m) is downward.
$W = -784\times0.8 = -627.2$ J.
Answer: (C). | ||||
| 15 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Momentum of the bullet: $0.004\times50 = 0.2\ \mathrm{kg\,ms^{-1}}$. By conservation of momentum the gun receives an equal and opposite impulse, $0.2\ \mathrm{kg\,ms^{-1}}$.
Recoil velocity $= \frac{0.2}{4} = 0.05\ \mathrm{ms^{-1}}$.
Answer: (C). | ||||
| 16 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | MatchingEasy | |
A easy matching question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current (a) $X_L > X_C$: the circuit is inductive and the current lags the emf, (ii).
(b) $X_L = X_C$: the impedance is purely resistive and the current is in phase with the emf, (i).
(c) $X_L < X_C$: capacitive, the current leads the emf, (iv).
(d) At the resonant frequency the impedance is least, so the current is maximum, (iii).
Answer: (B). | ||||
| 17 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $I = \frac{1}{2}\varepsilon_0 cE_0^2$, so $E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} = \sqrt{\frac{0.184}{8.85\times10^{-12}\times3\times10^8}} = \sqrt{69.3} = 8.32$ V/m.
$B_0 = \frac{E_0}{c} = \frac{8.32}{3\times10^8} = 2.77\times10^{-8}$ T.
Answer: (C). | ||||
| 18 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The longest wavelength that creates electron-hole pairs fixes the band gap: $E_g = \frac{hc}{\lambda} = \frac{1240\ \mathrm{eV\,nm}}{621\ \mathrm{nm}} \approx 2$ eV.
Answer: (B). | ||||
| 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 Momentum conservation: the $\alpha$-particle (mass 4) and the daughter (mass 180) have equal and opposite momenta, so the energy divides inversely as the masses.
$K_\alpha = Q\frac{A-4}{A} = 5.5\times\frac{180}{184} = 5.38$ MeV.
Answer: (B). | ||||
| 20 | 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 $T = 2\pi\sqrt{\frac{l}{g}} \propto \sqrt{l}$. With $l' = \frac{l}{16}$, $T' = \frac{T_0}{4}$.
Answer: (B). | ||||
| 21 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Direction of P: $\vec{A}\times\vec{B} = (\hat{i}+\hat{j})\times(\hat{j}+\hat{k}) = \hat{i} - \hat{j} + \hat{k}$.
Direction of Q: $\vec{A}\times\vec{C} = (\hat{i}+\hat{j})\times(-\hat{i}+\hat{j}) = 2\hat{k}$.
$\cos\theta = \frac{(\hat{i}-\hat{j}+\hat{k})\cdot2\hat{k}}{\sqrt{3}\times2} = \frac{1}{\sqrt{3}}$, so $x = 3$.
Answer: 3. | ||||
| 22 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For a semicircular arc, $y_{cm} = \frac{1}{\pi R}\int_0^\pi R\sin\theta\cdot R\,d\theta = \frac{2R}{\pi}$, and $x_{cm} = 0$ by symmetry.
So $|x| = 2$.
Answer: 2. | ||||
| 23 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Newton's law of cooling (average form): $\frac{75-65}{5} = k\left(\frac{75+65}{2} - 25\right) = 45k$, so $k = \frac{2}{45}$ per minute.
Next 5 minutes: $\frac{65-T}{5} = \frac{2}{45}\left(\frac{65+T}{2} - 25\right)$, i.e. $65 - T = \frac{15+T}{9}$, $585 - 9T = 15 + T$, $T = 57^\circ$C.
(The exact exponential law gives the same: the excess over the room falls from 50 to 40 in 5 minutes, ratio 0.8, so the next excess is 32 and $T = 57^\circ$C.)
Answer: 57. | ||||
| 24 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At minimum deviation $r = \frac{A}{2}$, and here $i = 2r = A$.
$\mu = \frac{\sin i}{\sin r} = \frac{\sin A}{\sin(A/2)} = 2\cos\frac{A}{2} = \sqrt{3}$, so $\frac{A}{2} = 30^\circ$ and $A = 60^\circ$.
Answer: 60. | ||||
| 25 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The track is prevented from expanding, so the thermal strain is $\alpha\Delta T = 10^{-5}\times10 = 10^{-4}$.
Energy per unit volume $= \frac{1}{2}Y(\text{strain})^2 = \frac{1}{2}\times10^{11}\times10^{-8} = 500\ \mathrm{J/m^3}$.
Per metre of track: $500\times0.01 = 5$ J/m.
Answer: 5. | ||||
| 26 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Gauss's law in differential form: $\rho = \nabla\cdot\vec{D}$.
$\frac{\partial}{\partial x}(e^{-x}\sin y) = -e^{-x}\sin y$, $\frac{\partial}{\partial y}(-e^{-x}\cos y) = e^{-x}\sin y$, $\frac{\partial}{\partial z}(2z) = 2$.
So $\rho = 2\ \mathrm{C/m^3}$ everywhere, and $Q = \rho\,\Delta V = 2\times2\times10^{-9} = 4\times10^{-9}$ C $= 4$ nC.
Answer: 4. | ||||
| 27 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity $V = \frac{ER}{R+r}$.
$1.25 = \frac{5E}{5+r}$ and $1 = \frac{2E}{2+r}$, so $E = \frac{2+r}{2}$.
Then $1.25(5+r) = \frac{5(2+r)}{2}$: $6.25 + 1.25r = 5 + 2.5r$, $r = 1\ \Omega$, $E = 1.5$ V $= \frac{15}{10}$ V.
Answer: 15. | ||||
| 28 | 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 current is largest when no current is drawn by the load, when the whole current through R passes through the zener.
Voltage across R: $50 - 5 = 45$ V. For the current not to exceed 90 mA: $R_{min} = \frac{45}{0.09} = 500\ \Omega$.
Answer: 500. | ||||
| 29 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rolling without slipping: the point has the translational velocity $v_0$ (horizontal) and the rotational velocity $\omega R = v_0$ (vertical, since the point is level with the centre).
Speed $= \sqrt{v_0^2 + v_0^2} = \sqrt{2}\,v_0$, so $x = 2$.
Answer: 2. | ||||
| 30 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalMedium | |
A medium numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $g = \frac{4\pi^2 l}{T^2}$ with $T = \frac{t}{n}$, so $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta t}{t}$.
Student 1: $\frac{0.1}{64} + 2\times\frac{0.1}{128} = 0.156\% + 0.156\% = 0.31\%$.
Student 2: $\frac{0.1}{64} + 2\times\frac{0.1}{64} = 0.47\%$.
Student 3: $\frac{0.1}{20} + 2\times\frac{0.1}{36} = 0.5\% + 0.56\% = 1.06\%$.
The least error is for student 1.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | MatchingEasy | |
A easy matching question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (a) $\mathrm{SF_4}$: 4 bond pairs + 1 lone pair on S, 5 electron domains, $sp^3d$ (iii).
(b) $\mathrm{IF_5}$: 5 bond pairs + 1 lone pair, 6 domains, $sp^3d^2$ (i).
(c) $\mathrm{NO_2^+}$: linear, O=N=O with no lone pair on N, $sp$ (v).
(d) $\mathrm{NH_4^+}$: tetrahedral, $sp^3$ (iv).
Answer: (B). | ||||
| 32 | Solutionsvan't Hoff factor | Class 12: Solutions | Single correctEasy | |
A easy single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = iK_fm$: at equal concentration the solution with the largest van't Hoff factor freezes lowest.
$\mathrm{Al_2(SO_4)_3} \to 2\mathrm{Al^{3+}} + 3\mathrm{SO_4^{2-}}$, $i = 5$; glucose $i = 1$; KI $i = 2$; $\mathrm{K_2SO_4}$ $i = 3$.
Answer: (A). | ||||
| 33 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Adding $\mathrm{AgNO_3}$ to KI keeps KI in excess while the AgI sol forms. The AgI particles adsorb the common ion in excess, $\mathrm{I^-}$, and the sol is negatively charged: $\mathrm{AgI/I^-}$.
Answer: (A). | ||||
| 34 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Mendeleev published his periodic table in 1869, long before the electron (1897) and the nucleus (1911) were discovered; the structure of the atom was not known then. So (C) is incorrect.
The other statements are true: he wrote a well-known chemistry textbook, he designed an accurate (differential) barometer, and element 101 is mendelevium, named after him.
Answer: (C). | ||||
| 35 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter The soft base $\mathrm{S^{2-}}$ combines with soft acids such as $\mathrm{Pb^{2+}}$ and $\mathrm{Ag^+}$: galena PbS and argentite $\mathrm{Ag_2S}$ are their chief ores.
Al and Mg are hard acids; their ores are oxides, silicates and carbonates (bauxite, magnesite, dolomite).
Answer: (B). | ||||
| 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 Protium and deuterium are stable. Tritium is radioactive and emits low-energy $\beta^-$ particles with a half-life of about 12.3 years.
Answer: (D). | ||||
| 37 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | MatchingEasy | |
A easy matching question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter (a) Ba is in period 6: outer configuration $6s^2$ (ii).
(b) Ca: calcium oxalate is insoluble in water (iii).
(c) Li: its compounds are markedly covalent, and LiCl and similar salts dissolve in organic solvents (i).
(d) Na: NaOH is a very strong monoacidic base (iv).
Answer: (C). | ||||
| 38 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down group 15 the E-H bond becomes longer and weaker, so the hydrides lose hydrogen more easily and their reducing power rises: $\mathrm{NH_3} < \mathrm{PH_3} < \mathrm{AsH_3} < \mathrm{SbH_3} < \mathrm{BiH_3}$.
Answer: (D). | ||||
| 39 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter (a) True: $2\mathrm{NaBH_4} + \mathrm{I_2} \to \mathrm{B_2H_6} + 2\mathrm{NaI} + \mathrm{H_2}$, in which $\mathrm{I_2}$ oxidises the borohydride.
(b) False: each boron is $sp^3$ hybridised.
(c) False: there are two B-H-B bridges, each a 3-centre-2-electron bond.
(d) False: the two bridging H atoms lie above and below the plane of the four terminal H atoms and two B atoms, so the molecule is not planar.
Answer: (D). | ||||
| 40 | 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 Colour and paramagnetism both need unpaired d electrons.
(A) $\mathrm{Cu^{2+}}$ $3d^9$, $\mathrm{Cr^{3+}}$ $3d^3$, $\mathrm{Sc^+}$ $3d^14s^1$: all have unpaired electrons.
(B) $\mathrm{Cu^+}$ and $\mathrm{Zn^{2+}}$ are $3d^{10}$.
(C) $\mathrm{Sc^{3+}}$, $\mathrm{V^{5+}}$, $\mathrm{Ti^{4+}}$ are $3d^0$.
(D) $\mathrm{Mn^{7+}}$ is $3d^0$ and $\mathrm{Hg^{2+}}$ is $5d^{10}$.
Answer: (A). | ||||
| 41 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter The solubility of a gas falls as the temperature rises, so cold water holds the most oxygen; polluted water has its dissolved oxygen used up by the decomposition of organic matter.
Water at $4^\circ$C has the most dissolved $\mathrm{O_2}$.
Answer: (B). | ||||
| 42 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A liquid that decomposes at or below its normal boiling point is distilled under reduced pressure, where it boils at a lower temperature (glycerol from spent lye, for example).
Answer: (D). | ||||
| 43 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) The lone pair on O is conjugated with the C=C of the vinyl ether: resonance.
(B) The $\mathrm{NH_2}$ group is on an $sp^3$ $\mathrm{CH_2}$, not next to the C=C, so its lone pair is not conjugated with the double bond: no resonance.
(C) In the amide the N lone pair is delocalised onto C=O: resonance.
(D) The benzene ring of benzyl alcohol is a resonance hybrid.
Answer: (B). | ||||
| 44 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) 3-Methylhex-1-ene, $\mathrm{CH_2{=}CH{-}CH(CH_3){-}CH_2CH_2CH_3}$: C3 is chiral (H, $\mathrm{CH_3}$, vinyl, propyl), optical isomers.
(B) 4-Methylhex-1-ene, $\mathrm{CH_2{=}CHCH_2{-}CH(CH_3){-}CH_2CH_3}$: C4 is chiral, optical isomers.
(C) 3-Ethylhex-3-ene, $\mathrm{(C_2H_5)_2C{=}CH{-}CH_2CH_3}$: C3 carries two identical ethyl groups, so there is no cis-trans isomerism, and there is no chiral centre.
(D) 3,4-Dimethylhex-3-ene, $\mathrm{CH_3CH_2(CH_3)C{=}C(CH_3)CH_2CH_3}$: each double-bond carbon has two different groups, so it shows geometrical isomerism.
Answer: (C). | ||||
| 45 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers A, $\mathrm{C_6H_6O}$, giving a colour with neutral $\mathrm{FeCl_3}$, is phenol.
Reimer-Tiemann reaction ($\mathrm{CHCl_3}$/KOH, then acid) gives salicylaldehyde, 2-hydroxybenzaldehyde: B.
PCC oxidises a primary alcohol only to the aldehyde, so C is 2-(hydroxymethyl)phenol (salicyl alcohol), which gives salicylaldehyde.
This is the third set.
Answer: (C). | ||||
| 46 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (B) Phenyl acetate is an ester. Excess $\mathrm{CH_3MgBr}$ adds twice: the first addition expels the phenoxide and gives acetone, which takes a second methyl. After hydrolysis: $(\mathrm{CH_3})_3\mathrm{COH}$, a tertiary alcohol (with phenol). This is the textbook route the question tests.
(A) 3-Hydroxyacetophenone: the phenolic OH destroys one equivalent of the Grignard, but with excess reagent the ketone still adds $\mathrm{CH_3}$ to give $\mathrm{HOC_6H_4C(CH_3)_2OH}$, which is also a tertiary alcohol.
(C) 1,4-Naphthoquinone: 1,2-addition of $\mathrm{CH_3}$ to each C=O gives 1,4-dimethyl-1,4-dihydronaphthalene-1,4-diol, a tertiary diol.
(D) The terminal alkyne C-H quenches one equivalent, the ether does not react, and the nitrile gives an imine salt that hydrolyses to a ketone, not an alcohol.
So (A), (B) and (C) all give tertiary alcohols; (B) is the intended answer.
Answer: (B). | ||||
| 47 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Benzenediazonium chloride is reduced to benzene by hypophosphorous acid: $\mathrm{C_6H_5N_2^+Cl^-} + \mathrm{H_3PO_2} + \mathrm{H_2O} \to \mathrm{C_6H_6} + \mathrm{N_2} + \mathrm{H_3PO_3} + \mathrm{HCl}$. So A is $\mathrm{H_3PO_2}$.
Benzene with $\mathrm{CH_3CH_2Cl}$ and anhydrous $\mathrm{AlCl_3}$ (Friedel-Crafts alkylation) gives ethylbenzene. So B is $\mathrm{CH_3CH_2Cl}$.
Answer: (C). | ||||
| 48 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines (A) Aniline is sulphonated by $\mathrm{H_2SO_4}$ (via anilinium hydrogensulphate, on heating) to sulphanilic acid: occurs.
(B) Nitration of aniline does occur, giving p-nitroaniline along with a large share of the m-isomer (from the anilinium ion) and oxidation products.
(C) Aniline does not undergo Friedel-Crafts reaction: the basic $\mathrm{NH_2}$ group forms a salt with the Lewis acid $\mathrm{AlCl_3}$, which deactivates the ring. This reaction does not occur.
(D) Acetylation with acetic anhydride in pyridine gives acetanilide: occurs.
Answer: (C). | ||||
| 49 | PolymersNatural rubber and cellulose | No matching NCERT chapter | MatchingEasy | |
A easy matching question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter (i) is 2-methylbuta-1,3-diene, isoprene. (ii) is 2-chlorobuta-1,3-diene, chloroprene. (iii) is the polymer of chloroprene, neoprene. (iv) is acrylonitrile.
So (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i).
Answer: (D). | ||||
| 50 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Thiamine is vitamin $\mathrm{B_1}$ and pyridoxine is vitamin $\mathrm{B_6}$ (riboflavin is $\mathrm{B_2}$; vitamin E is tocopherol).
Answer: (B). | ||||
| 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 Molar mass of glucose $= 6(12) + 12(1) + 6(16) = 180\ \mathrm{g\,mol^{-1}}$.
Molarity $= \frac{0.72}{180} = 0.004$ M $= 4\times10^{-3}$ M.
Answer: 4. | ||||
| 52 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter CCP (fcc): $Z = 4$. $\rho = \frac{ZM}{N_Aa^3}$, so $M = \frac{\rho N_Aa^3}{Z}$.
$a = 0.4518\ \mathrm{nm} = 4.518\times10^{-8}$ cm, $a^3 = 9.222\times10^{-23}\ \mathrm{cm^3}$.
$M = \frac{7.62\times6.022\times10^{23}\times9.222\times10^{-23}}{4} = \frac{423.2}{4} = 105.8 \approx 106\ \mathrm{g\,mol^{-1}}$.
Answer: 106. | ||||
| 53 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom V: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^3\,4s^2$.
Electrons in p orbitals: $2p^6 + 3p^6 = 12$.
Answer: 12. | ||||
| 54 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics 1 g of carbon is $\frac{1}{12}$ mol.
Heat released $= \frac{248}{12} = 20.67$ kJ $\approx 21$ kJ.
Answer: 21. | ||||
| 55 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $K_p = K_c(RT)^{\Delta n}$ with $\Delta n = 2 - 1 = 1$.
$K_c = \frac{47.9}{0.083\times288} = \frac{47.9}{23.90} = 2.004 \approx 2$.
Answer: 2. | ||||
| 56 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry $n = 2$ and $Q = \frac{[\mathrm{Cu^{2+}}]}{[\mathrm{Ag^+}]^2} = \frac{0.250}{(10^{-3})^2} = 2.5\times10^5$, $\log Q = 5.3979$.
$E_{cell} = E^\ominus_{cell} - \frac{0.0591}{2}\log Q = 2.97 - 0.02955\times5.3979 = 2.97 - 0.160 = 2.81$ V $\approx 3$ V.
Answer: 3. | ||||
| 57 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} = \frac{2.303}{60\ \mathrm{min}}\log\frac{2.40}{1.60} = \frac{2.303}{60}\log1.5$.
$\log1.5 = \log3 - \log2 = 0.477 - (1 - 0.699) = 0.176$.
$k = \frac{2.303\times0.176}{60} = 6.76\times10^{-3}\ \mathrm{min^{-1}} \approx 7\times10^{-3}\ \mathrm{min^{-1}}$.
Answer: 7. | ||||
| 58 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[Co(NH_3)_6]^{2+}}$: $\mathrm{Co^{2+}}$ is $3d^7$. $\mathrm{NH_3}$ does not pair the electrons of $\mathrm{Co^{2+}}$: the complex is high spin, $t_{2g}^5e_g^2$, with 3 unpaired electrons (its magnetic moment is about 4-5 BM).
$\mathrm{[Co(NH_3)_6]^{3+}}$: $\mathrm{Co^{3+}}$ is $3d^6$; with $\mathrm{NH_3}$ it is low spin, $t_{2g}^6$, with 0 unpaired electrons (diamagnetic).
Total $= 3 + 0 = 3$.
Answer: 3. | ||||
| 59 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Acyclic $\mathrm{C_5H_{10}}$:
Straight chain: pent-1-ene, cis-pent-2-ene, trans-pent-2-ene (3).
Branched: 2-methylbut-1-ene, 3-methylbut-1-ene, 2-methylbut-2-ene (3; none shows geometrical isomerism).
Total $= 6$.
Answer: 6. | ||||
| 60 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{C_6H_6}$ (78 g/mol) $\to$ $\mathrm{C_6H_5CH_3}$ (92 g/mol), 1 : 1.
Theoretical yield $= \frac{10}{78}\times92 = 11.79$ g.
Percentage yield $= \frac{9.2}{11.79}\times100 = 78\%$.
Answer: 78. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | 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 Numerator: $0 \le x^2 - x + 1 \le 1$. The left inequality always holds; the right gives $x(x-1) \le 0$, $0 \le x \le 1$.
Denominator: $\sin^{-1}\left(\frac{2x-1}{2}\right) > 0$ needs $0 < \frac{2x-1}{2} \le 1$, i.e. $\frac{1}{2} < x \le \frac{3}{2}$.
Together: $x \in \left(\frac{1}{2}, 1\right]$, so $\alpha + \beta = \frac{3}{2}$.
Answer: (D). | ||||
| 62 | 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 $\sin^7x \le \sin^2x$ and $\cos^7x \le \cos^2x$, since $|\sin x|, |\cos x| \le 1$ (and a negative value makes the seventh power negative). Adding, $\sin^7x + \cos^7x \le 1$, with equality only when each of $\sin x$, $\cos x$ is 0 or 1.
So $(\sin x, \cos x) = (0, 1)$ or $(1, 0)$: $x = 2n\pi$ or $x = 2n\pi + \frac{\pi}{2}$.
In $[0, 4\pi]$: $0, 2\pi, 4\pi, \frac{\pi}{2}, \frac{5\pi}{2}$, five solutions.
Answer: (A). | ||||
| 63 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry First line: $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$ (dividing $3(x-1) = 6(y-2) = 2(z-1)$ by 6), through $A(1, 2, 1)$ with direction $(2, 1, 3)$.
Second line: $\frac{x-2}{1} = \frac{y-\lambda}{2} = \frac{z-3}{4}$, through $B(2, \lambda, 3)$ with direction $(1, 2, 4)$.
$(2,1,3)\times(1,2,4) = (-2, -5, 3)$, of length $\sqrt{38}$.
$\vec{AB} = (1, \lambda-2, 2)$, and $\vec{AB}\cdot(-2,-5,3) = -2 - 5\lambda + 10 + 6 = 14 - 5\lambda$.
$\frac{|14-5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}}$ gives $\lambda = 3$ or $\lambda = \frac{13}{5}$. The integral value is 3.
Answer: (C). | ||||
| 64 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let $u = \begin{pmatrix}1\\1\\1\end{pmatrix}$. Each row sum is 1, so $Au = u$, and then $A^3u = u$.
The sum of all entries of $A^3$ is $u^TA^3u = u^Tu = 3$.
Answer: (C). | ||||
| 65 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctHard | |
A hard single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $(1-\cos2x)^2 = 4\sin^4x \approx 4x^4$.
$\log_e(1+2xe^{-2x}) = 2xe^{-2x} - \frac{(2xe^{-2x})^2}{2} + \cdots = 2x - 4x^2 - 2x^2 + O(x^3) = 2x - 6x^2 + O(x^3)$.
$-2\log_e(1-xe^{-x}) = 2\left(xe^{-x} + \frac{x^2e^{-2x}}{2} + \cdots\right) = 2x - 2x^2 + x^2 + O(x^3) = 2x - x^2 + O(x^3)$.
So the logarithm is $4x + O(x^2)$ and $f(x) \approx \frac{x^3\cdot4x}{4x^4} \to 1$.
For continuity $\alpha = 1$.
Answer: (B). | ||||
| 66 | 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 Use $a \Rightarrow b \equiv \sim a \vee b$; note $\sim q \Rightarrow p \equiv q \vee p$.
(A) $(\sim p \vee q) \vee (q \vee p)$ contains $p \vee \sim p$: tautology.
(B) $(\sim p \vee \sim q) \vee (q \vee p)$: tautology.
(C) $(p \vee q) \vee (q \vee p) = p \vee q$, false when $p$ and $q$ are both false: not a tautology.
(D) $(\sim q \vee p) \vee (q \vee p)$ contains $q \vee \sim q$: tautology.
Answer: (C). | ||||
| 67 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $E_1$: $b^2 = a^2(1-e^2)$.
$E_2$ has its foci at $(0, \pm b)$, so its major axis is along the y-axis: $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ with $B > A$. It passes through $(\pm a, 0)$, so $A = a$; its foci give $Be = b$; and $A^2 = B^2(1-e^2)$.
So $a^2 = \frac{b^2}{e^2}(1-e^2) = \frac{a^2(1-e^2)^2}{e^2}$, i.e. $e = 1 - e^2$.
$e^2 + e - 1 = 0$, $e = \frac{-1+\sqrt{5}}{2}$.
Answer: (A). | ||||
| 68 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $L$: $y = -2x + k$. Tangent to $\frac{x^2}{3} - \frac{y^2}{3} = 1$: $k^2 = 3m^2 - 3 = 12 - 3 = 9$, and $k > 0$ gives $k = 3$.
For $y^2 = \alpha x$ (that is $4a = \alpha$), $y = mx + \frac{\alpha}{4m}$ is a tangent: $\frac{\alpha}{4(-2)} = 3$, so $\alpha = -24$.
Answer: (C). | ||||
| 69 | 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: $(1,-1,2)\times(2,1,-1) = (-1, 5, 3)$.
A point of L with $z = 0$: $x - y = 2$, $2x + y = 2$, so $A\left(\frac{4}{3}, -\frac{2}{3}, 0\right)$.
From $A$ to $Q(1, 2, 0)$: $\left(-\frac{1}{3}, \frac{8}{3}, 0\right)$; its component along L: $t = \frac{\frac{1}{3} + \frac{40}{3}}{35} = \frac{41}{105}$.
$P = A + t(-1, 5, 3)$, so $\alpha + \beta + \gamma = \left(\frac{4}{3} - \frac{2}{3}\right) + t(-1+5+3) = \frac{2}{3} + \frac{287}{105} = \frac{357}{105}$.
$35(\alpha+\beta+\gamma) = \frac{357}{3} = 119$.
Answer: (B). | ||||
| 70 | 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 Divide by $\mathrm{cosec}^2x$: $dy + 2\sin^2x\,dx = (1 + y\cos2x)\,dx$, so $\frac{dy}{dx} = 1 - 2\sin^2x + y\cos2x = (1+y)\cos2x$.
$\int\frac{dy}{1+y} = \int\cos2x\,dx$: $\log_e(1+y) = \frac{\sin2x}{2} + C$.
$y\left(\frac{\pi}{4}\right) = 0$: $0 = \frac{1}{2} + C$, $C = -\frac{1}{2}$.
At $x = 0$: $\log_e(1 + y(0)) = -\frac{1}{2}$, so $(y(0)+1)^2 = e^{-1}$.
Answer: (B). | ||||
| 71 | CirclesStandard and general form | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections $S$: $x^2 + y^2 - 3x + \frac{10}{3}y + \frac{C}{36} = 0$, centre $\left(\frac{3}{2}, -\frac{5}{3}\right)$, $r^2 = \frac{9}{4} + \frac{25}{9} - \frac{C}{36} = \frac{181 - C}{36}$.
Not meeting the axes: $r < \frac{3}{2}$ (distance to the y-axis) and $r < \frac{5}{3}$, so $\frac{181-C}{36} < \frac{9}{4}$, i.e. $C > 100$ (and $C < 181$ for a real circle).
The lines meet at $(2, -1)$. Inside S: $36(4) + 36(1) - 216 - 120 + C < 0$, i.e. $C < 156$.
So $100 < C < 156$.
Answer: (A). | ||||
| 72 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{c} = \vec{a}\times\vec{b}$ is perpendicular to $\vec{a}$ and $\vec{b}$; $\vec{a} = \vec{b}\times\vec{c}$ is perpendicular to $\vec{b}$ and $\vec{c}$. So the three are mutually perpendicular, with $|\vec{c}| = |\vec{a}||\vec{b}|$ and $|\vec{a}| = |\vec{b}||\vec{c}|$, giving $|\vec{b}| = 1$, $|\vec{c}| = 2$.
(A) $[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = |\vec{a}|^2 = 4$, and $[\vec{c}\ \vec{a}\ \vec{b}]$ equals it: sum 8, true.
(B) $(\vec{b}+\vec{c})\times(\vec{b}-\vec{c}) = -2\vec{b}\times\vec{c} = -2\vec{a}$, and $\vec{a}\times(-2\vec{a}) = \vec{0}$: true.
(C) Mutually perpendicular: $|3\vec{a}+\vec{b}-2\vec{c}|^2 = 9(4) + 1 + 4(4) = 53 \ne 51$: not true.
(D) $\frac{\vec{a}\cdot\vec{a}}{|\vec{a}|} = 2$: true.
Answer: (C). | ||||
| 73 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctHard | |
A hard single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Total outcomes: $6^4 = 1296$. Matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$.
All entries different: $6\cdot5\cdot4\cdot3 = 360$.
Singular with distinct entries needs $ad = bc$ with $\{a,d\} \ne \{b,c\}$: the only products of two distinct numbers from 1-6 obtainable in two ways are $6 = 1\cdot6 = 2\cdot3$ and $12 = 2\cdot6 = 3\cdot4$. For each, choose which pair is on the diagonal (2 ways) and order each pair ($2\times2$): 8 matrices, 16 in all.
Favourable: $360 - 16 = 344$. Probability $= \frac{344}{1296} = \frac{43}{162}$.
Answer: (D). | ||||
| 74 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_{10} = 5(2a + 9d) = 530$ gives $2a + 9d = 106$; $S_5 = \frac{5}{2}(2a + 4d) = 140$ gives $2a + 4d = 56$.
So $d = 10$, $a = 8$.
$S_{20} = 10(16 + 190) = 2060$, $S_6 = 3(16 + 50) = 198$.
$S_{20} - S_6 = 1862$.
Answer: (C). | ||||
| 75 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $x > 0$: $f'(x) = -4x^2 + 4x + 3 = -(2x-3)(2x+1) > 0$ for $-\frac{1}{2} < x < \frac{3}{2}$, i.e. on $\left(0, \frac{3}{2}\right)$.
$x < 0$: $f'(x) = 3e^x(1+x) > 0$ for $x > -1$, i.e. on $(-1, 0)$.
$f$ is continuous at 0 (both pieces give 0), so $f$ is increasing on $\left(-1, \frac{3}{2}\right)$.
Answer: (C). | ||||
| 76 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The integrand $\sin^2x\,e^{-\{x/\pi\}}$ has period $\pi$, so $I = 100\int_0^\pi\sin^2x\,e^{-x/\pi}dx$. Put $x = \pi t$: $I = 100\pi\int_0^1\sin^2(\pi t)e^{-t}dt = 50\pi\int_0^1(1 - \cos2\pi t)e^{-t}dt$.
$\int_0^1e^{-t}dt = 1 - e^{-1}$, and $\int_0^1e^{-t}\cos2\pi t\,dt = \left[\frac{e^{-t}(-\cos2\pi t + 2\pi\sin2\pi t)}{1+4\pi^2}\right]_0^1 = \frac{1 - e^{-1}}{1+4\pi^2}$.
$I = 50\pi(1-e^{-1})\left(1 - \frac{1}{1+4\pi^2}\right) = \frac{200\pi^3(1-e^{-1})}{1+4\pi^2}$.
So $\alpha = 200(1 - e^{-1})$.
Answer: (D). | ||||
| 77 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}1&1&1\\3&5&5\\1&2&\lambda\end{vmatrix} = (5\lambda - 10) - (3\lambda - 5) + (6 - 5) = 2\lambda - 4$; no solution needs $\Delta = 0$, $\lambda = 2$.
With $\lambda = 2$: the first two equations give $y + z = 4$ and $x = 2$, and then $x + 2y + 2z = 10$. The third equation is inconsistent exactly when $\mu \ne 10$.
Answer: (B). | ||||
| 78 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z = x + iy$: $x^2 - y^2 + 3x = 0$ and $2xy - 3y = 0$.
$y = 0$: $x^2 + 3x = 0$, $x = 0$ or $-3$.
$x = \frac{3}{2}$: $\frac{9}{4} - y^2 + \frac{9}{2} = 0$, $y = \pm\frac{3\sqrt{3}}{2}$.
So $n = 4$ and $\sum_{k=0}^\infty\frac{1}{4^k} = \frac{1}{1 - \frac{1}{4}} = \frac{4}{3}$.
Answer: (D). | ||||
| 79 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $[e^x + 1] = [e^x] + 1$. With $t = [e^x]$: $t^2 + t - 2 = 0$, $t = 1$ or $t = -2$; $e^x > 0$ rules out $-2$.
$[e^x] = 1$: $1 \le e^x < 2$, so $x \in [0, \log_e2)$.
Answer: (C). | ||||
| 80 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a} = x\vec{b} + y\vec{c}$. $\vec{b}\cdot\vec{d} = 14$, $\vec{c}\cdot\vec{d} = 7$, so $\vec{a}\cdot\vec{d} = 14x + 7y = 0$, $y = -2x$.
$\vec{a} = x(\vec{b} - 2\vec{c}) = x(0, 3, -1)$; $|\vec{a}| = \sqrt{10}$ gives $x = \pm1$.
$[\vec{a}\ \vec{b}\ \vec{c}] = 0$ (coplanar). $[\vec{a}\ \vec{b}\ \vec{d}] + [\vec{a}\ \vec{c}\ \vec{d}] = \vec{a}\cdot\left((\vec{b}+\vec{c})\times\vec{d}\right)$, with $(3, 0, 2)\times(3, 2, 6) = (-4, -12, 6)$.
$\vec{a}\cdot(-4,-12,6) = \pm(0 - 36 - 6) = \mp42$.
A possible value is $-42$ (with $\vec{a} = 3\hat{j} - \hat{k}$).
Answer: (B). | ||||
| 81 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical 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 $f(1) + f(2) + f(3) = 3$ with three distinct values from $A$: only $\{0, 1, 2\}$.
These go to 1, 2, 3 in $3! = 6$ ways; the other 5 elements map bijectively onto the remaining 5 values in $5! = 120$ ways.
Total $= 6\times120 = 720$.
Answer: 720. | ||||
| 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 Let $B = \begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$. $AB$ swaps the first two rows: $\begin{bmatrix}d&e&f\\a&b&c\\g&h&i\end{bmatrix}$; $BA$ swaps the first two columns: $\begin{bmatrix}b&a&c\\e&d&f\\h&g&i\end{bmatrix}$.
Equality: $d = b$, $e = a$, $f = c$, $g = h$. So $B$ is fixed by the five free entries $a, b, c, g, i$.
Number of matrices $= 5^5 = 3125$.
Answer: 3125. | ||||
| 83 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $y^2 = 4(1-x)$ (vertex $(1,0)$, opening left) and $y^2 = 4(x+1)$ (vertex $(-1,0)$, opening right) meet at $(0, 2)$. $y = \frac{1-x^2}{2}$ is a downward parabola through $(\pm1, 0)$ with vertex $(0, \frac{1}{2})$.
The region in the upper half plane lies between the two side parabolas and above $y = \frac{1-x^2}{2}$.
Area between the side parabolas for $0 \le y \le 2$: $\int_0^2\left[\left(1 - \frac{y^2}{4}\right) - \left(\frac{y^2}{4} - 1\right)\right]dy = \int_0^2\left(2 - \frac{y^2}{2}\right)dy = 4 - \frac{4}{3} = \frac{8}{3}$.
Area under $y = \frac{1-x^2}{2}$: $\int_{-1}^1\frac{1-x^2}{2}dx = \frac{2}{3}$.
Required area $= \frac{8}{3} - \frac{2}{3} = 2$.
Answer: 2. | ||||
| 84 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations With five distinct digits no number has more than five digits, so the numbers above 10,000 are the five-digit ones.
First digit non-zero: 4 ways; the remaining four digits in $4! = 24$ ways.
Total $= 4\times24 = 96$.
Answer: 96. | ||||
| 85 | 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 $f$ is continuous everywhere ($f(\pm2) = 0$), with corners at $x = -2, 0, 2$. So $g$ is continuous everywhere: $n = 0$.
$f(x+2)$ has corners at $x = -4, -2, 0$; $f(x-2)$ at $x = 0, 2, 4$.
At $x = 0$: $f(x+2)$ has slopes $-\frac{3}{2}$ (left) and 0 (right); $f(x-2)$ has slopes 0 (left) and $\frac{3}{2}$ (right). $g'(0^-) = -\frac{3}{2} - 0$, $g'(0^+) = 0 - \frac{3}{2}$: equal, so $g$ is differentiable at 0.
At $-4, -2, 2, 4$ only one term has a corner, so $g$ is not differentiable there: $m = 4$.
$n + m = 4$.
Answer: 4. | ||||
| 86 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $2040 = 2^3\cdot3\cdot5\cdot17$, so $n$ must be prime to 2, 3, 5 and 17.
Prime to 30: in each block of 30 the residues are 1, 7, 11, 13, 17, 19, 23, 29 (sum 120). Blocks 1-30, 31-60, 61-90 give $120 + (120 + 240) + (120 + 480) = 1080$; from 91-100 add 91 and 97: total $1080 + 188 = 1268$.
Remove the multiples of 17 among these: $17k \le 100$ with $k$ prime to 30 means $k = 1$, i.e. 17.
Sum $= 1268 - 17 = 1251$.
Answer: 1251. | ||||
| 87 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{k+1} = \binom{10}{k}2^{10-k}x^{r(10-k) - 2k}$.
Constant term: $r(10-k) = 2k$, with coefficient $\binom{10}{k}2^{10-k} = 180$.
$k = 8$: $\binom{10}{8}\cdot2^2 = 45\cdot4 = 180$ (no other $k$ gives 180). Then $2r = 16$, $r = 8$.
Answer: 8. | ||||
| 88 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Dividing by $11^n$, the condition is $\left(\frac{10}{11}\right)^n + \left(\frac{9}{11}\right)^n < 1$, and the left side decreases as $n$ grows.
$n = 4$: $10^4 + 9^4 = 16561 > 11^4 = 14641$, fails. $n = 5$: $10^5 + 9^5 = 159049 < 11^5 = 161051$, holds; so it holds for every $n \ge 5$ and fails for $n = 1, 2, 3, 4$.
Number of elements $= 100 - 4 = 96$.
Answer: 96. | ||||
| 89 | StatisticsMean, median and mode | Class 11: Statistics | NumericalMedium | |
A medium numerical question on mean, median and mode, from Statistics. NCERT: Class 11: Statistics $N = a + b + 26$. Median 14 lies in 12-18: $14 = 12 + \frac{\frac{N}{2} - (a+b)}{12}\times6$, so $\frac{N}{2} - (a + b) = 4$, i.e. $13 - \frac{a+b}{2} = 4$, $a + b = 18$, $N = 44$.
Mean: $\frac{3a + 9b + 15(12) + 21(9) + 27(5)}{44} = \frac{309}{22}$, so $3a + 9b + 504 = 618$, $a + 3b = 38$.
Then $b = 10$, $a = 8$, $(a-b)^2 = 4$.
Answer: 4. | ||||
| 90 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $X = x + 2$, $Y = y + 1$: $\frac{dY}{dX} = e^{Y/X} + \frac{Y}{X}$, homogeneous. With $Y = vX$: $X\frac{dv}{dX} = e^v$, so $e^{-v}dv = \frac{dX}{X}$ and $-e^{-v} = \log_e|X| + C$.
$y(1) = 1$: $X = 3$, $Y = 2$, $v = \frac{2}{3}$, so $C = -e^{-2/3} - \log_e3$ and $e^{-v} = e^{-2/3} - \log_e\frac{X}{3}$ (on the branch $X > 0$ through $X = 3$).
The solution exists while $e^{-v} > 0$ and $X > 0$: $0 < X < 3e^{e^{-2/3}}$, i.e. $-2 < x < 3e^{e^{-2/3}} - 2$.
$|\alpha + \beta| = 3e^{e^{-2/3}} - 4 \approx 3\times1.6709 - 4 = 1.0127 \approx 1.01$.
Answer: 1.01. | ||||
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