JEE (Main) · 2021
JEE (Main) 2021 — 18 March, 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
- Electrostatics3
- Kinematics3
- Work, Energy and Power2
27 easy · 3 medium · 0 hard
Chemistry
30 Q
- Coordination Compounds3
- Hydrocarbons2
- Amines2
22 easy · 8 medium · 0 hard
Mathematics
30 Q
- Matrices and Determinants3
- Permutations and Combinations3
- Integral Calculus3
16 easy · 13 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 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance For the drop to stay at rest, $qE = mg$.
Mass: $m = \frac{4}{3}\pi r^3\rho = \frac{4}{3}\pi(2\times10^{-3})^3(3000) = 1.005\times10^{-4}$ kg, so $mg = 9.86\times10^{-4}$ N.
Charge: $q = \frac{mg}{E} = \frac{9.86\times10^{-4}}{3.55\times10^{5}} = 2.78\times10^{-9}$ C.
Number of electrons: $n = \frac{q}{e} = \frac{2.78\times10^{-9}}{1.6\times10^{-19}} \approx 1.73\times10^{10}$.
Answer: (B). | ||||
| 2 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power With constant power $P$, the work done equals the kinetic energy gained: $Pt = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2P}{m}}\,t^{1/2}$.
Then $x = \int_0^t v\,dt = \sqrt{\frac{2P}{m}}\cdot\frac{2}{3}t^{3/2}$, so $x \propto t^{3/2}$.
Answer: (A). | ||||
| 3 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The current $i$ flows through A and then divides equally among the three identical solenoids B, C and D in parallel, so C carries $\frac{i}{3}$.
Inside a long solenoid $B = \mu_0 n i$, proportional to the current. So $B_C = \frac{1}{3}B_A = \frac{3}{3} = 1$ T.
Answer: (A). | ||||
| 4 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the graph: C is at $P_C = 100\ \text{N/m}^2$, $V_C = 4\ \text{m}^3$; D is at $P_D = 200\ \text{N/m}^2$, $V_D = 3\ \text{m}^3$.
Work done by the gas in an adiabatic process: $W = \frac{P_CV_C - P_DV_D}{\gamma - 1} = \frac{400 - 600}{0.4} = -500$ J (the gas is compressed, so the work done by it is negative).
Answer: (B). | ||||
| 5 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane With constant acceleration $a$: $a(t)$ is a horizontal line, $v(t) = u + at$ is a straight line (rising for $a > 0$), and $x(t) = x_0 + ut + \frac{1}{2}at^2$ is a parabola curving upward.
(A) has $x(t)$ linear, which needs zero acceleration. (B) has $v(t)$ falling while $x(t)$ curves upward with a positive constant $a$: inconsistent. (D) has a linear $x(t)$ and a curved $v(t)$: inconsistent with constant $a$. (C) shows the parabola, the rising straight line and the constant.
Answer: (C). | ||||
| 6 | 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 Take $i = i_0\cos\omega t$, maximum at $t = 0$. It equals $i_{rms} = \frac{i_0}{\sqrt{2}}$ when $\omega t = \frac{\pi}{4}$, i.e. after $t = \frac{T}{8}$.
$T = \frac{1}{50}$ s $= 20$ ms, so $t = \frac{20}{8} = 2.5$ ms.
Answer: (A). | ||||
| 7 | 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 Fringe width: $\beta = \frac{\lambda D}{d} = \frac{5890\times10^{-10}\times0.5}{0.5\times10^{-3}} = 5.89\times10^{-4}$ m.
The first and third bright fringes are two fringe widths apart: $2\beta = 1.178\times10^{-3}$ m $= 1178\times10^{-6}$ m.
Answer: (D). | ||||
| 8 | 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 $\lambda = \frac{h}{mv}$, so $\frac{\lambda_p}{\lambda_e} = \frac{m_ev_e}{m_pv_p}$.
With $v_p = 4v_e$: $2 = \frac{m_e}{4m_p}$, so $m_p = \frac{m_e}{8}$.
Answer: (D). | ||||
| 9 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $g = \frac{4\pi^2 l}{T^2}$, so $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$.
$\frac{\Delta l}{l} = \frac{0.1}{10} = 1\%$. The total time is 100 s with a 1 s resolution, so $\frac{\Delta T}{T} = \frac{1}{100} = 1\%$.
$\frac{\Delta g}{g} = 1\% + 2\% = 3\%$.
Answer: (B). | ||||
| 10 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei In the Bohr model the ground-state energy is $E = -\frac{me^4}{8\varepsilon_0^2h^2}$, proportional to the mass of the orbiting particle.
So the ionization energy becomes $207\times13.6 = 2815.2$ eV, and the ionization potential is 2815.2 V (written in the options as eV).
Answer: (D). | ||||
| 11 | 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 For two independent decay modes the rates add: $\frac{dN}{dt} = -(\lambda_1 + \lambda_2)N$, so $\lambda = \lambda_1 + \lambda_2$.
With $\lambda = \frac{\ln 2}{T_{1/2}}$: $\frac{1}{T_{1/2}} = \frac{1}{T_{1/2}^{(1)}} + \frac{1}{T_{1/2}^{(2)}}$, giving $T_{1/2} = \frac{T_{1/2}^{(1)}T_{1/2}^{(2)}}{T_{1/2}^{(1)} + T_{1/2}^{(2)}}$.
Answer: (A). | ||||
| 12 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter A current loop in a magnetic field settles where its potential energy $U = -\vec{m}\cdot\vec{B}$ is least, i.e. where the flux through it is greatest.
For a flexible wire of fixed length, the greatest area is a circle, and the greatest flux comes with the plane normal to the field (magnetic moment along $\vec{B}$). The forces on the elements pull the loop outward into that shape.
Answer: (B). | ||||
| 13 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\rho = \frac{RA}{l} = \frac{V\pi d^2}{4Il}$, so $\frac{\Delta\rho}{\rho} = \frac{\Delta V}{V} + 2\frac{\Delta d}{d} + \frac{\Delta I}{I} + \frac{\Delta l}{l}$.
The errors are the last-digit uncertainties: $\frac{0.1}{5.0} = 2\%$, $2\times\frac{0.01}{5.00} = 0.4\%$, $\frac{0.01}{2.00} = 0.5\%$, $\frac{0.1}{10.0} = 1\%$.
Total: $2 + 0.4 + 0.5 + 1 = 3.9\%$.
Answer: (A). | ||||
| 14 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation By Kepler's third law $T^2 \propto R^3$, so $T' = T\left(\frac{9R}{R}\right)^{3/2} = 27T$.
Answer: (C). | ||||
| 15 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Resonance frequency $\omega_0 = \frac{1}{\sqrt{LC}}$ does not depend on $R$, so (A) is wrong.
Bandwidth $\Delta\omega = \frac{R}{L}$ increases with $R$: (B) is right.
Quality factor $Q = \frac{\omega_0 L}{R}$ decreases, so (C) and (D) are wrong.
Answer: (B). | ||||
| 16 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Not seeing a distant object clearly is myopia (short-sightedness).
Seeing a uniform mesh of lines as non-uniform and distorted, with lines in different directions not equally sharp, is astigmatism.
Both together: myopia with astigmatism.
Answer: (D). | ||||
| 17 | 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 By the law of equipartition of energy, each degree of freedom carries an average energy of $\frac{1}{2}k_BT$.
Answer: (B). | ||||
| 18 | Communication SystemsLayers of the atmosphere | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on layers of the atmosphere, from Communication Systems. NCERT: No matching NCERT chapter The school-text table of atmospheric layers used for radio propagation gives:
Troposphere up to about 10 km, so (a) 10 km → (iv).
D layer at 65–75 km, counted as part of the stratosphere, so (b) 70 km → (iii).
F$_1$ layer at 170–190 km, counted as part of the mesosphere, so (c) 180 km → (ii).
F$_2$ layer at 250–400 km, in the thermosphere, so (d) 270 km → (i).
Answer: (A). | ||||
| 19 | 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 Magnitude: $E = cB = 3\times10^{8}\times2.0\times10^{-8} = 6.0$ V/m.
Direction: the wave travels along $\vec{E}\times\vec{B}$, i.e. $+\hat{i}$. With $\vec{B}$ along $\hat{k}$, $\vec{E}$ must be along $\hat{j}$, since $\hat{j}\times\hat{k} = \hat{i}$.
$\vec{E} = 6.0\,\hat{j}$ V/m.
Answer: (B). | ||||
| 20 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion No external torque acts about the axis, so angular momentum is conserved:
$Mr^2\omega = (Mr^2 + 2mr^2)\omega'$, giving $\omega' = \omega\frac{M}{M+2m}$.
Answer: (B). | ||||
| 21 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Power gain of a CE amplifier $= \beta^2\frac{R_{out}}{R_{in}}$.
$10^6 = \beta^2\times\frac{10^4}{100} = 100\beta^2$, so $\beta^2 = 10^4$ and $\beta = 100$.
Answer: 100. | ||||
| 22 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Each piece has mass 10 kg. Piece 1 moves along $+y$ at 10 m/s; piece 2 moves at $v_2$ at 30° below the $x$-axis.
$y$-momentum (initially zero): $10\times10 = 10\,v_2\sin30^\circ$, so $v_2 = 20$ m/s.
Check with $x$-momentum: $10\times10\sqrt{3} = 10\times20\cos30^\circ = 100\sqrt{3}$, which agrees.
Answer: 20. | ||||
| 23 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity The 50 $\Omega$ and 20 $\Omega$ resistors join the same two nodes, so they are in parallel: $\frac{50\times20}{70} = \frac{100}{7}\ \Omega$.
Total resistance: $10 + \frac{100}{7} = \frac{170}{7}\ \Omega$; current $I = \frac{170}{170/7} = 7$ A.
Voltage across 10 $\Omega$: $7\times10 = 70$ V.
Answer: 70. | ||||
| 24 | 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 Taking the track as smooth, mechanical energy is conserved from A (height 10 m, starting from rest) to B (height 5 m):
$\frac{1}{2}v^2 = g(10 - 5) = 50$, so $v = 10$ m/s.
Answer: 10. | ||||
| 25 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves From the mean position $x = A\sin\omega t$, with $\omega = \frac{2\pi}{T} = \pi$ rad/s.
$\frac{A}{2} = A\sin\pi t$ gives $\pi t = \frac{\pi}{6}$, so $t = \frac{1}{6}$ s.
Answer: 6. | ||||
| 26 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance At $t = 0$ the capacitor's voltage is $V = \frac{q}{C} = \frac{30\ \mu\text{C}}{3\ \mu\text{F}} = 10$ V.
The initial discharge current is $I = \frac{V}{R} = \frac{10}{5\times10^{6}} = 2\times10^{-6}$ A $= 2\ \mu$A.
Answer: 2. | ||||
| 27 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane To land directly opposite, the swimmer's resultant velocity along the flow must be zero:
$u + 10\cos120^\circ = 0$, so $u = 10\times\frac{1}{2} = 5$ m/s.
Answer: 5. | ||||
| 28 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $\Delta l = \frac{FL}{\pi r^2 Y}$, so with the same $F$ and $Y$, $L \propto \Delta l\,r^2$.
$\frac{a}{b} = \frac{2\times r^2}{4\times(4r)^2} = \frac{2}{64} = \frac{1}{32}$.
Answer: 32. | ||||
| 29 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With a slab of thickness $t$ and dielectric constant $K$: $C = \frac{\varepsilon_0A}{d - t + \frac{t}{K}}$.
$C = \frac{8.85\times10^{-12}\times100}{10 - 5 + 0.5} = \frac{8.85\times10^{-10}}{5.5} = 1.609\times10^{-10}$ F $= 160.9$ pF.
To the nearest integer: 161.
Answer: 161. | ||||
| 30 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $v^2 = u^2 - 2as$: $0 = 100 - 2a(0.5)$, so $a = 100\ \text{m/s}^2$.
$F = ma = 0.1\times100 = 10$ N.
Answer: 10. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Solid StateClose packed structures | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on close packed structures, from Solid State. NCERT: No matching NCERT chapter For $N$ atoms of A in the hcp lattice there are $2N$ tetrahedral voids.
M occupies $\frac{2}{3}\times2N = \frac{4N}{3}$ of them, so M : A $= \frac{4}{3} : 1 = 4 : 3$.
Formula: $\text{M}_4\text{A}_3$.
Answer: (B). | ||||
| 32 | 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 Angular nodes $= l = 0$, so it is an s orbital.
Radial nodes $= n - l - 1 = 2$ gives $n = 3$.
The orbital is 3s.
Answer: (C). | ||||
| 33 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter (a) Deacon's process ($4\text{HCl} + \text{O}_2 \rightarrow 2\text{Cl}_2 + 2\text{H}_2\text{O}$) uses $\text{CuCl}_2$: (ii).
(b) Contact process ($2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$) uses $\text{V}_2\text{O}_5$: (iv).
(c) Cracking of hydrocarbons uses the zeolite ZSM-5: (i).
(d) Hydrogenation of vegetable oils uses finely divided nickel: (iii).
Answer: (D). | ||||
| 34 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties $\text{Na}^+$, $\text{Mg}^{2+}$ and $\text{Al}^{3+}$ are isoelectronic (10 electrons). As the nuclear charge rises, the radius falls: $\text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+}$.
Only 0.72 Å and 0.54 Å are both smaller than 1.02 Å and in decreasing order (these are the accepted values).
Answer: (B). | ||||
| 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 In the electrolytic (Hall-Héroult) extraction, purified alumina is dissolved in molten cryolite, $\text{Na}_3\text{AlF}_6$, which lowers the melting point of the mixture to about 1240 K and improves its conductivity.
Bauxite is the ore itself; kaolinite is a clay; calamine is a zinc ore.
Answer: (D). | ||||
| 36 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter On boiling, $\text{Mg(HCO}_3)_2 \rightarrow \text{Mg(OH)}_2\downarrow + 2\text{CO}_2\uparrow$: magnesium is removed as the hydroxide, not the carbonate. So A is false.
R is false too: by measured values $\mathrm{Mg(OH)_2}$ has the smaller solubility product (about $1.8 \times 10^{-11}$, against about $10^{-5}$ to $10^{-8}$ for $\mathrm{MgCO_3}$), which is why the hydroxide precipitates.
With A and R both false, none of the four options applies.
No option is correct as printed. | ||||
| 37 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter (a) $\text{Ca(OCl)}_2$ is the active part of bleaching powder: bleach, (iii).
(b) $\text{CaSO}_4\cdot\frac{1}{2}\text{H}_2\text{O}$ is plaster of Paris: (iv).
(c) CaO (quicklime) is a main raw material of cement: (ii).
(d) $\text{CaCO}_3$ neutralises stomach acid: antacid, (i).
Answer: (D). | ||||
| 38 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Phosphonic (phosphorous) acid condenses with $\text{PCl}_3$ to give pyrophosphorous acid: $\text{PCl}_3 + 5\text{H}_3\text{PO}_3 \rightarrow 3\text{H}_4\text{P}_2\text{O}_5 + 3\text{HCl}$.
$\text{H}_4\text{P}_2\text{O}_5$ is $\text{HO}-\text{P(=O)(H)}-\text{O}-\text{P(=O)(H)}-\text{OH}$: two hydrogens are on oxygen (ionisable) and two are bonded directly to phosphorus (not ionisable).
Answer: (B). | ||||
| 39 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Chlorophyll is a magnesium complex: (iv). Vitamin $\text{B}_{12}$ (cyanocobalamin) contains cobalt: (iii). The anticancer drug cisplatin is a platinum complex: (ii). Grubbs catalyst (olefin metathesis) is a ruthenium complex: (i).
Answer: (B). | ||||
| 40 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Trans square-planar $[\text{NiBr}_2(\text{PPh}_3)_2]$: the two Br are opposite each other, and so are the two $\text{PPh}_3$. Options (A) and (D) show the cis isomer, so they are out.
In the octahedral drawings the axial positions are top and bottom, and in the square of four the upper-left position is trans to the lower-right, and the upper-right trans to the lower-left.
Mer isomer: the three identical ligands lie in one plane, so two of them are trans to each other.
(B): $\text{NO}_2$ at top, upper-right and lower-left; upper-right and lower-left are trans, so the three $\text{NO}_2$ are meridional (and so are the three $\text{NH}_3$: bottom, upper-left, lower-right).
(C): $\text{NO}_2$ at top, upper-right and lower-right; no two are trans, so this is the facial isomer.
Answer: (B). | ||||
| 41 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter (A) True: methane is a greenhouse gas, and as a hydrocarbon it takes part with nitrogen oxides in the reactions that form photochemical smog (and tropospheric ozone).
(B) True: anaerobic decay in flooded paddy fields releases methane.
(C) True: molecule for molecule, methane traps far more heat than $\text{CO}_2$.
(D) False: reducing (classical) smog is smoke and fog with sulphur dioxide; methane is not a part of it.
Answer: (B). | ||||
| 42 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $\text{C}_3\text{H}_6\text{O}$ includes propanal ($\text{CH}_3\text{CH}_2\text{CHO}$), propanone ($\text{CH}_3\text{COCH}_3$), prop-2-en-1-ol ($\text{CH}_2=\text{CHCH}_2\text{OH}$) and methoxyethene ($\text{CH}_3\text{OCH}=\text{CH}_2$): these differ in functional group, so the formula shows functional group isomerism.
There is no pair of stable isomers with the same functional group at different positions on the same chain, and only one ether, so neither positional isomerism nor metamerism arises.
Answer: (B). | ||||
| 43 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (a) Alcoholic KOH brings about dehydrohalogenation, a $\beta$-elimination: (iii).
(b) Pd on $\text{BaSO}_4$ (partly poisoned) is Lindlar's catalyst: (iv).
(c) BHC, $\text{C}_6\text{H}_6\text{Cl}_6$, is made by adding three molecules of $\text{Cl}_2$ to benzene in UV light: (ii).
(d) Polyacetylene, a conducting polymer, is used for electrodes in batteries: (i).
Answer: (D). | ||||
| 44 | Alcohols, Phenols and EthersPrimary, secondary and tertiary alcohols | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on primary, secondary and tertiary alcohols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers An instant reaction with Lucas reagent means A is a tertiary alcohol.
$\text{C}_8\text{H}_8\text{O}$ that gives a tertiary alcohol with a Grignard reagent is acetophenone, $\text{C}_6\text{H}_5\text{COCH}_3$.
$\text{C}_6\text{H}_5\text{COCH}_3 + \text{C}_2\text{H}_5\text{MgBr}$, then hydrolysis, gives A $= \text{C}_6\text{H}_5\text{C(OH)(CH}_3)\text{C}_2\text{H}_5$ (2-phenylbutan-2-ol, $\text{C}_{10}\text{H}_{14}\text{O}$).
Lucas reagent replaces OH by Cl: B $= \text{C}_6\text{H}_5\text{C(Cl)(CH}_3)\text{C}_2\text{H}_5$, 2-chloro-2-phenylbutane, $\text{C}_{10}\text{H}_{13}\text{Cl}$.
Answer: (D). | ||||
| 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 Acid hydrolysis of a nitrile goes in two stages: $\text{R-C}\equiv\text{N} \xrightarrow{\text{H}_2\text{O, H}^+} \text{R-CONH}_2 \xrightarrow{\text{H}_2\text{O, H}^+, \Delta} \text{R-COOH}$.
With mild conditions the first stage stops at the amide, which on further heating with acid gives cyclohexanecarboxylic acid, as shown. So A is cyclohexanecarboxamide.
Answer: (B). | ||||
| 46 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Alkaline $\text{KMnO}_4$ oxidises a side chain with a benzylic hydrogen all the way to $-\text{COO}^-$; acidification then gives $-\text{COOH}$. The methoxy group is not attacked.
4-Methylanisole therefore gives 4-methoxybenzoic acid.
Answer: (B). | ||||
| 47 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Aniline with $\text{NaNO}_2$ and HCl at 273–278 K is diazotised: X is benzenediazonium chloride, $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$.
The diazonium ion couples with N,N-dimethylaniline at the para position (the ortho positions are hindered and less activated), giving the azo dye Y, 4-(dimethylamino)azobenzene, $\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{N(CH}_3)_2$ (para).
Answer: (C). | ||||
| 48 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Cimetidine (an $\text{H}_2$-receptor blocker) is an antacid: (a)-(ii). Alitame is an artificial sweetener: (b)-(iv). Novestrol (ethynylestradiol) is an antifertility drug: (c)-(i). Valium is a tranquilizer: (d)-(iii).
Answer: (B). | ||||
| 49 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Glucose, fructose and galactose are themselves reducing monosaccharides and do not hydrolyse.
Sucrose is non-reducing, because the glycosidic link joins the reducing carbons of both units (C1 of glucose and C2 of fructose); it hydrolyses to glucose and fructose, both reducing.
Answer: (D). | ||||
| 50 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Nitrite diazotises sulphanilic acid in acetic acid; the diazonium salt couples with 1-naphthylamine to give a red azo dye. This is the test for $\text{NO}_2^-$.
Answer: (A). | ||||
| 51 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\text{C}_2\text{H}_6 + \frac{7}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}$.
3 g of ethane (M = 30) is 0.1 mol, which gives 0.3 mol of water.
Molecules: $0.3\times6.023\times10^{23} = 18.07\times10^{22}$, so $x \approx 18$.
Answer: 18. | ||||
| 52 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure For second-row diatomics the bond orders run: 14 electrons (as in $\text{N}_2$, CO) bond order 3; 15 electrons bond order 2.5; 16 electrons ($\text{O}_2$) bond order 2.
A bond order of 2.5 needs 15 electrons, i.e. one electron in $\pi^*2p$ beyond the 14-electron configuration. The neutral molecule of two second-row atoms with 15 electrons is NO.
Answer: 15. | ||||
| 53 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta_rH = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$.
Broken in $\text{C}_2\text{H}_6$: one C$-$C and six C$-$H. Formed: one C$=$C and four C$-$H in $\text{C}_2\text{H}_4$, and one H$-$H.
Net: $\Delta_rH = 347 + 2(414) - 611 - 436 = 1175 - 1047 = 128$ kJ $\text{mol}^{-1}$.
Answer: 128. | ||||
| 54 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions The freezing point must be $-3.885$°C, so $\Delta T_f = 3.885$ K.
$\Delta T_f = iK_fm$: $3.885 = i\times1.85\times2$, so $i = 1.05$.
For $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$, $i = 1 + \alpha$, so $\alpha = 0.05 = 50\times10^{-3}$.
Answer: 50. | ||||
| 55 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Henderson equation: $\text{pH} = \text{p}K_a + \log\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$.
$5.74 = 4.74 + \log\frac{[\text{salt}]}{1.0}$, so $[\text{salt}] = 10^{1} = 10$ M.
Answer: 10. | ||||
| 56 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Find $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}}$ through free energies: $\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}$, $\Delta G_1^\circ = -3F(-0.036)$; $\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}$, $\Delta G_2^\circ = -2F(-0.440)$.
$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$: $\Delta G^\circ = \Delta G_1^\circ - \Delta G_2^\circ$, so $E^\circ = 3(-0.036) - 2(-0.440) = -0.108 + 0.880 = 0.772$ V.
$E^\circ_{cell} = 0.772 - 0.539 = 0.233$ V, with $n = 2$.
$\Delta_rG^\circ = -nFE^\circ = -2\times96500\times0.233 = -44969$ J $\approx -45$ kJ.
Answer: 45. | ||||
| 57 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let $r = k[\text{NO}]^x[\text{Cl}_2]^y$.
Runs 1 and 2 ($[\text{Cl}_2]$ doubled): $\frac{0.35}{0.18} \approx 2 = 2^y$, so $y = 1$.
Runs 2 and 3 ($[\text{NO}]$ doubled): $\frac{1.40}{0.35} = 4 = 2^x$, so $x = 2$.
Overall order $= 2 + 1 = 3$.
Answer: 3. | ||||
| 58 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Oxalate is $\text{C}_2\text{O}_4^{2-}$; three of them carry $-6$, and with $\text{K}_3$ the complex ion is $[\text{Cr(C}_2\text{O}_4)_3]^{3-}$, so chromium is +3, $3d^3$.
In an octahedral field the three electrons occupy the three $t_{2g}$ orbitals singly, whatever the field strength: 3 unpaired electrons.
Answer: 3. | ||||
| 59 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\text{HOCH}_2\text{CH}_2\text{CHO} \rightarrow \text{CH}_2=\text{CHCHO} + \text{H}_2\text{O}$, one mole to one mole.
Acrolein wanted: $\frac{7.8}{56} = 0.1393$ mol. At 64% yield the hydroxypropanal needed is $\frac{0.1393}{0.64} = 0.2176$ mol.
Mass: $0.2176\times74 = 16.1$ g, about 16.
Answer: 16. | ||||
| 60 | AminesClassification and identification of amines | Class 12: Amines | NumericalEasy | |
A easy numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines $\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 + 3\text{CH}_3\text{Br} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{N}^+(\text{CH}_3)_3\text{Br}^- + 2\text{HBr}$.
Molar mass of $\text{C}_{10}\text{H}_{16}\text{NBr}$: $120 + 16 + 14 + 80 = 230$, so 23 g is 0.1 mol of the salt.
Each mole takes 3 mol of $\text{CH}_3\text{Br}$: $0.3 = 3\times10^{-1}$ mol.
Answer: 3. | ||||
| 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 correctEasy | |
A easy 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 $f$ needs $x\ge0$ and $g$ needs $x\le1$, so $f\pm g$ and $g-f$ are defined on $[0,1]$.
$f/g$ also needs $g(x)\ne0$, i.e. $x\ne1$; $g/f$ needs $f(x)\ne0$, i.e. $x\ne0$.
Common domain: $0<x<1$.
Answer: (B). | ||||
| 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 For a circle the $|z|^2$ term must be present, so $a\ne0$. Dividing by $a$: $|z|^2 + \frac{\overline{\alpha}}{a}z + \frac{\alpha}{a}\overline{z} + \frac{d}{a} = 0$, i.e. $\left|z + \frac{\alpha}{a}\right|^2 = \frac{|\alpha|^2}{a^2} - \frac{d}{a}$.
This is a circle with centre $-\frac{\alpha}{a}$ and radius$^2 = \frac{|\alpha|^2 - ad}{a^2}$, which must be positive: $|\alpha|^2 - ad > 0$.
(A) allows $a = 0$ (a line) and a zero radius; (C) allows a negative radius$^2$; (D) gives $|z|^2 = -\frac{d}{a} < 0$, no locus.
Answer: (B). | ||||
| 63 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Trace is linear. Let $a = \text{Tr}(A)$, $b = \text{Tr}(B)$.
$a + 2b = 1 - 3 + 1 = -1$ and $2a - b = 2 - 1 + 2 = 3$.
Solving: $a = 1$, $b = -1$, so $a - b = 2$.
Answer: (B). | ||||
| 64 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A non-trivial solution needs $\begin{vmatrix}\alpha&\beta&\gamma\\\beta&\gamma&\alpha\\\gamma&\alpha&\beta\end{vmatrix} = 0$.
This determinant equals $-(\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha) = -\frac{1}{2}(\alpha+\beta+\gamma)\left[(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2\right]$.
$\alpha+\beta+\gamma = -a \ne 0$, so for real roots $\alpha = \beta = \gamma$.
Then $a = -3\alpha$ and $b = 3\alpha^2$, so $\frac{a^2}{b} = \frac{9\alpha^2}{3\alpha^2} = 3$.
Answer: (C). | ||||
| 65 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations There are $\frac{4!}{2!} = 12$ distinct numbers.
Fix a place. Digit 1 there: the rest (2, 2, 3) arrange in 3 ways. Digit 3 there: 3 ways. Digit 2 there: the rest (1, 2, 3) arrange in 6 ways.
Sum of digits in each place: $1\times3 + 3\times3 + 2\times6 = 24$.
Total: $24\times1111 = 26664$.
Answer: (B). | ||||
| 66 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Put $x = 1$: $\sum a_k = 4^{20} = 2^{40}$. Put $x = -1$: $\sum(-1)^ka_k = 2^{20}$.
So $a_1 + a_3 + \dots + a_{39} = \frac{2^{40} - 2^{20}}{2} = 2^{19}(2^{20} - 1)$.
$a_{39}$: $x^{39}$ needs nineteen factors $2x^2$ and one factor $x$: $a_{39} = 20\times2^{19}$.
$a_1 + a_3 + \dots + a_{37} = 2^{19}(2^{20} - 1) - 20\times2^{19} = 2^{19}(2^{20} - 21)$.
Answer: (C). | ||||
| 67 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $y$ be the value. The pattern repeats after two levels: $y = 3 + \cfrac{1}{4 + \frac{1}{y}} = 3 + \frac{y}{4y + 1}$.
$y(4y + 1) = 3(4y + 1) + y$, so $4y^2 - 12y - 3 = 0$.
$y = \frac{12 + \sqrt{144 + 48}}{8} = \frac{12 + 8\sqrt{3}}{8} = 1.5 + \sqrt{3}$ (the positive root, since $y > 3$).
Answer: (A). | ||||
| 68 | 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 General term with $2k+1$, $k = 1$ to $100$: $\frac{1}{(2k+1)^2 - 1} = \frac{1}{4k(k+1)} = \frac{1}{4}\left(\frac{1}{k} - \frac{1}{k+1}\right)$.
The sum telescopes: $\frac{1}{4}\left(1 - \frac{1}{101}\right) = \frac{1}{4}\cdot\frac{100}{101} = \frac{25}{101}$.
Answer: (A). | ||||
| 69 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The terms are $(100 - k)(100 + k) = 10^4 - k^2$ for $k = 0, 1, \dots, 99$.
Sum $= 100\times10^4 - \sum_{k=1}^{99}k^2 = 10^6 - \frac{99\times100\times199}{6} = 100^3 - 199\times1650$.
So $\alpha = 3$, $\beta = 1650$, and the slope is $\frac{\beta}{\alpha} = \frac{1650}{3} = 550$.
Answer: (D). | ||||
| 70 | 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 $f$ is even, so it is enough to check $x = 1$, where $f(x) = \frac{1}{x}$ for $x \ge 1$.
Continuity: $a + b = 1$. Equal derivatives: $2a = -\frac{1}{x^2}\Big|_{x=1} = -1$, so $a = -\frac{1}{2}$ and $b = \frac{3}{2}$.
Answer: (B). | ||||
| 71 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\text{cosec}^{-1}x$ needs $|x| \ge 1$.
$\sqrt{x - [x]}$ in the denominator needs $x - [x] > 0$, i.e. $x$ not an integer.
Together: non-integers with $|x| \ge 1$; since $\pm1$ are integers this is all non-integers outside $[-1, 1]$.
Answer: (C). | ||||
| 72 | 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 Near 0: $\sin^{-1}x = x + \frac{x^3}{6} + \dots$ and $\tan^{-1}x = x - \frac{x^3}{3} + \dots$
So $\sin^{-1}x - \tan^{-1}x = \frac{x^3}{2} + \dots$, and $L = \frac{1/2}{3} = \frac{1}{6}$.
$6L + 1 = 2$.
Answer: (C). | ||||
| 73 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $4x^2 - 4x + 6 = (2x - 1)^2 + 5$. Let $t = \sqrt{(2x-1)^2 + 5}$; then $dt = \frac{2(2x-1)}{t}\,dx$.
The integral becomes $\frac{1}{2}\int\cos t\,dt = \frac{1}{2}\sin t + c = \frac{1}{2}\sin\sqrt{(2x-1)^2+5} + c$.
Answer: (B). | ||||
| 74 | Differential EquationsOrder and degree | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on order and degree, from Differential Equations. NCERT: Class 12: Differential Equations Differentiate: $2yy' = 4a$, so $a = \frac{yy'}{2}$.
Substitute: $y^2 = 2yy'\left(x + \frac{yy'}{2}\right) = 2xyy' + y^2(y')^2$.
Divide by $y$: $y(y')^2 + 2xy' - y = 0$.
Answer: (A). | ||||
| 75 | CirclesRadius and centre | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections First circle: centre $(5, 5)$, radius $\sqrt{25 + 25 - 41} = 3$. Second: centre $(11, 5)$, radius $\sqrt{121 + 25 - 137} = 3$.
Distance between centres $= 6 = 3 + 3$: the circles touch externally, at one point $(8, 5)$.
Answer: (C). | ||||
| 76 | CirclesRadius and centre | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Centres: M $(0, 0)$, N $(1, 0)$, O $(1, 1)$, P $(0, 1)$.
MN, NO, OP and PM all have length 1 and meet at right angles: a square (which is also a rectangle, rhombus and parallelogram, but square is the exact description).
Answer: (C). | ||||
| 77 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines Substitute: $3x + 4(mx + 1) = 9$, so $x = \frac{5}{3 + 4m}$.
$x$ is an integer when $3 + 4m \in \{\pm1, \pm5\}$, i.e. $m = -\frac{1}{2}, -1, \frac{1}{2}, -2$.
The integer values are $m = -1$ ($x = -5$) and $m = -2$ ($x = -1$): 2 values.
Answer: (C). | ||||
| 78 | Straight LinesSlope, parallel and perpendicular lines | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines The given line has slope $3\sqrt{2}$. A line of slope $m$ makes the angle when $\left|\frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m}\right| = \sqrt{2}$.
Taking $m - 3\sqrt{2} = \sqrt{2}(1 + 3\sqrt{2}m) = \sqrt{2} + 6m$: $m = -\frac{4\sqrt{2}}{5}$.
Through $(1, 3)$: $y - 3 = -\frac{4\sqrt{2}}{5}(x - 1)$, i.e. $4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$.
(The other sign gives $m = \frac{2\sqrt{2}}{7}$, which is not among the options.)
Answer: (A). | ||||
| 79 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$ give $\begin{vmatrix}1&0&\sin^2x\\-1&1&\cos^2x\\0&-1&1+4\sin2x\end{vmatrix}$.
Expanding along the first row: $1\cdot(1 + 4\sin2x + \cos^2x) + \sin^2x\cdot(1 - 0) = 2 + 4\sin2x$.
$2 + 4\sin2x = 0$ gives $\sin2x = -\frac{1}{2}$; with $0 < 2x < 2\pi$, $2x = \frac{7\pi}{6}, \frac{11\pi}{6}$, so $x = \frac{7\pi}{12}, \frac{11\pi}{12}$.
Answer: (C). | ||||
| 80 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra Rotating the axes does not change the length of the vector: $(3p)^2 + 1 = (p + 1)^2 + 10$.
$9p^2 + 1 = p^2 + 2p + 11$, so $8p^2 - 2p - 10 = 0$, i.e. $4p^2 - p - 5 = 0$, giving $p = \frac{5}{4}$ or $p = -1$.
Of these only $-1$ is an option.
Answer: (B). | ||||
| 81 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $0, z_1, z_2$ form an equilateral triangle exactly when $z_1^2 + z_2^2 = z_1z_2$, i.e. $(z_1 + z_2)^2 = 3z_1z_2$.
With $z_1 + z_2 = -a$ and $z_1z_2 = 12$: $a^2 = 36$, so $|a| = 6$.
Answer: 6. | ||||
| 82 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals From the two conditions, $f(x^2) = 4x^3 - g(4 - x) = 4x^3 + g(x)$.
$f(x^2)$ is even, so $\int_{-4}^{4}f(x^2)\,dx = 2\int_0^4 f(x^2)\,dx = 2\int_0^4\left(4x^3 + g(x)\right)dx$.
$\int_0^4 g(x)\,dx = \int_0^4 g(4 - x)\,dx = -\int_0^4 g(x)\,dx$, so it is 0. And $\int_0^4 4x^3\,dx = 256$.
Value $= 2\times256 = 512$.
Answer: 512. | ||||
| 83 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Divide numerator and denominator by $x^{14}$: $\int\frac{5x^{-6} + 7x^{-8}}{(x^{-5} + x^{-7} + 2)^2}\,dx$.
With $t = x^{-5} + x^{-7} + 2$, $dt = -(5x^{-6} + 7x^{-8})\,dx$, so the integral is $-\int\frac{dt}{t^2} = \frac{1}{t} + C = \frac{x^7}{x^2 + 1 + 2x^7} + C$.
$f(0) = 0$ gives $C = 0$; $f(1) = \frac{1}{4}$, so $K = 4$.
Answer: 4. | ||||
| 84 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The curve is $xy = \pm1$, symmetric under rotation by $90^\circ$ about the origin, so take the square centred at the origin: $A\left(t, \frac{1}{t}\right)$, $B\left(-\frac{1}{t}, t\right)$, $C\left(-t, -\frac{1}{t}\right)$, $D\left(\frac{1}{t}, -t\right)$ (each on $xy = \pm1$).
Midpoint of AB: $\left(\frac{t - 1/t}{2}, \frac{t + 1/t}{2}\right)$; on the curve, $\left(\frac{t^2 - 1/t^2}{4}\right)^2 = 1$, so $\left(t^2 - \frac{1}{t^2}\right)^2 = 16$.
Area $= AB^2 = \left(t + \frac{1}{t}\right)^2 + \left(t - \frac{1}{t}\right)^2 = 2\left(t^2 + \frac{1}{t^2}\right)$.
$\left(t^2 + \frac{1}{t^2}\right)^2 = \left(t^2 - \frac{1}{t^2}\right)^2 + 4 = 20$, so Area$^2 = 4\times20 = 80$.
Answer: 80. | ||||
| 85 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The plane is the perpendicular bisector: normal along $(2-4, 3+3, -5-1) = (-2, 6, -6) \parallel (1, -3, 3)$, through the midpoint $(3, 0, -2)$.
$x - 3y + 3z + d = 0$ with $3 - 0 - 6 + d = 0$: $d = 3$.
All integer forms are $k(1, -3, 3, 3)$; the least sum of squares is at $k = \pm1$: $1 + 9 + 9 + 9 = 28$.
Answer: 28. | ||||
| 86 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The planes are $x - 2y + 2z + d = 0$, with $\frac{|1 - 4 + 6 + d|}{3} = 1$, i.e. $|3 + d| = 3$: $d = 0$ or $d = -6$.
With $a = 1$, $b = -2$, $c = 2$ (any common multiple gives the same ratio): $K = \frac{b - d}{c - a} = -2 - d$.
$d = 0$ gives $K = -2$; $d = -6$ gives $K = 4$. The positive value is 4.
Answer: 4. | ||||
| 87 | StatisticsMean, median and mode | Class 11: Statistics | NumericalEasy | |
A easy numerical question on mean, median and mode, from Statistics. NCERT: Class 11: Statistics Total age before: $25\times40 = 1000$. After: $25\times39 = 975$.
$1000 - 60 + x = 975$, so $x = 35$.
Answer: 35. | ||||
| 88 | 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 Write 0 to 999 as three-digit strings 000 to 999 (adding leading zeros writes no extra 3s, and 0 has none).
Each of the three places takes the digit 3 in $100$ of the $1000$ strings: $3\times100 = 300$.
1000 has no 3. Total: 300.
Answer: 300. | ||||
| 89 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Each quarter of the circle has two outer numbers and one inner entry.
Top-left: 1, 2, inner 1. Bottom-right: 4, 7, inner $3^6$. Bottom-left: 12, 8, inner $4^{24}$.
The base of each inner entry is the difference of its outer pair (2 $-$ 1 = 1, 7 $-$ 4 = 3, 12 $-$ 8 = 4) and the exponent is the factorial of that base ($1! = 1$, $3! = 6$, $4! = 24$): inner $= n^{n!}$.
Top-right: 5 $-$ 3 = 2, so the missing value is $2^{2!} = 4$.
Answer: 4. | ||||
| 90 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin x \ne 0$, so $x \ne 0, \pi, 2\pi$.
If $\cot x \ge 0$: $0 = \frac{1}{\sin x}$, impossible.
If $\cot x < 0$: $-2\cot x = \frac{1}{\sin x}$, so $-2\cos x = 1$, $\cos x = -\frac{1}{2}$: $x = \frac{2\pi}{3}$ or $\frac{4\pi}{3}$.
At $\frac{2\pi}{3}$, $\cot x = -\frac{1}{\sqrt{3}} < 0$: valid. At $\frac{4\pi}{3}$, $\cot x = \frac{1}{\sqrt{3}} > 0$: rejected.
One solution.
Answer: 1. | ||||
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