JEE (Main) · 2021
JEE (Main) 2021 — 26 February, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Laws of Motion4
- Gravitation3
- Properties of Bulk Matter3
20 easy · 10 medium · 0 hard
Chemistry
30 Q
- Chemical Bonding and Molecular Structure2
- d- and f-Block Elements2
- Organic Chemistry: Some Basic Principles and Techniques2
23 easy · 6 medium · 1 hard
Mathematics
30 Q
- Integral Calculus4
- Three Dimensional Geometry3
- Differential Equations3
16 easy · 14 medium · 0 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 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The exponent must be dimensionless, so $[\beta][x^2] = [kT]$, an energy: $[\beta] = \frac{ML^2T^{-2}}{L^2} = [MT^{-2}]$.
Then $[\alpha^2] = \frac{[W]}{[\beta]} = \frac{ML^2T^{-2}}{MT^{-2}} = [L^2]$, so $[\alpha] = [M^0LT^0]$.
Answer: (B). | ||||
| 2 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Assertion and reasonEasy | |
A easy assertion and reason question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power In a head-on elastic collision with Q at rest, conservation of both momentum and kinetic energy gives Q's speed $v_Q = \frac{2M}{M+m}u$.
For $m \ll M$ this tends to $2u$, its largest possible value. So A is correct, R is correct, and A follows directly from R (both conservation laws are what give the formula).
Answer: (A). | ||||
| 3 | GravitationKepler's laws | Class 11: Gravitation | StatementsEasy | |
A easy statements question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Gravity is a central force, so the planet's angular momentum about the sun is conserved and its areal velocity $\frac{dA}{dt} = \frac{L}{2m}$ is constant (Kepler's second law): E is true, and C and D are false because the areal velocity does not change while the speed does.
The speed changes along the orbit (A is false) and is greatest, not least, nearest the sun (B is false).
Answer: (D) E only. | ||||
| 4 | GravitationNewton's law of gravitation | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation The sphere acts on outside points as a point mass M at its centre Y. So the force equals the force of the ring on a point mass M on its axis at distance $x = \sqrt{8}R$:
$F = \frac{GmMx}{(R^2 + x^2)^{3/2}} = \frac{GmM\sqrt{8}R}{(R^2 + 8R^2)^{3/2}} = \frac{GmM\sqrt{8}R}{27R^3} = \frac{\sqrt{8}}{27}\cdot\frac{GmM}{R^2}$.
Answer: (A). | ||||
| 5 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The axis is a side of the square, so it passes through the centres of two spheres: each contributes $\frac{2}{5}ma^2$.
The other two spheres have their centres a distance b from the axis; by the parallel-axis theorem each contributes $\frac{2}{5}ma^2 + mb^2$.
Total: $4\cdot\frac{2}{5}ma^2 + 2mb^2 = \frac{8}{5}ma^2 + 2mb^2$.
Answer: (B). | ||||
| 6 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter With n drops, volume is conserved: $V = \frac{4}{3}\pi R^3 = n\cdot\frac{4}{3}\pi r^3$.
Energy released $= T\,\Delta A = T(4\pi r^2 n - 4\pi R^2) = TV\left(\frac{3}{r} - \frac{3}{R}\right)$, using $4\pi r^2 n = \frac{3V}{r}$ and $4\pi R^2 = \frac{3V}{R}$.
Heat per unit volume $= \frac{3T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$.
Answer: (B). | ||||
| 7 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $K = \frac{P}{|\Delta V|/V}$, so $\frac{|\Delta V|}{V} = \frac{P}{K}$. Mass is fixed, so $\rho V$ is constant and $\frac{\Delta\rho}{\rho} = \frac{|\Delta V|}{V}$ for small changes.
$\Delta\rho = \frac{\rho P}{K}$.
Answer: (A). | ||||
| 8 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation Inside a uniform earth, gravity at distance r from the centre is $\frac{g}{R}r$ towards the centre. Its component along the tunnel, with x measured from the chord's midpoint, is $\frac{g}{R}r\cdot\frac{x}{r} = \frac{g}{R}x$, a restoring force proportional to x whatever the chord's distance from the centre.
So $\omega^2 = \frac{g}{R}$ and $T = 2\pi\sqrt{\frac{R}{g}}$.
Answer: (B). | ||||
| 9 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on 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 In the steady state the same heat current passes through both sheets: $\frac{\theta_2 - \theta}{R_2} = \frac{\theta - \theta_1}{R_1}$.
$R_1\theta_2 - R_1\theta = R_2\theta - R_2\theta_1$, so $\theta = \frac{\theta_1R_2 + \theta_2R_1}{R_1 + R_2}$.
Answer: (A). | ||||
| 10 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion The force supplies the centripetal force: $\frac{mv^2}{R} = \frac{k}{R^3}$, so $v^2 \propto \frac{1}{R^2}$ and $v \propto \frac{1}{R}$.
$T = \frac{2\pi R}{v} \propto R^2$.
Answer: (C). | ||||
| 11 | 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 In series, $\frac{1}{K} = \frac{1}{K_1} + \frac{1}{K_1}$, so $K = \frac{K_1}{2}$: a factor $\frac{1}{2}$.
$T = 2\pi\sqrt{\frac{m}{K}}$, so T changes by $\sqrt{2}$ for the same mass.
Answer: (A). | ||||
| 12 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance For a uniformly charged rod of length L, the field on its perpendicular bisector at distance a is $E = \frac{1}{4\pi\varepsilon_0}\cdot\frac{Q}{a\sqrt{a^2 + L^2/4}}$.
With $a = \frac{\sqrt{3}}{2}L$: $a^2 + \frac{L^2}{4} = L^2$, so $E = \frac{Q}{4\pi\varepsilon_0\cdot\frac{\sqrt{3}}{2}L\cdot L} = \frac{Q}{2\sqrt{3}\pi\varepsilon_0L^2}$.
Answer: (D). | ||||
| 13 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Parallel: $C_1 + C_2$. Series: $\frac{C_1C_2}{C_1 + C_2}$. The condition is $\frac{(C_1 + C_2)^2}{C_1C_2} = \frac{15}{4}$.
With $x = \frac{C_2}{C_1}$: $\frac{(1 + x)^2}{x} = \frac{15}{4}$, i.e. $4x^2 - 7x + 4 = 0$, whose discriminant $49 - 64$ is negative: no real ratio exists.
Indeed $\frac{(1 + x)^2}{x} = x + 2 + \frac{1}{x} \ge 4$ for every positive x, so the parallel value is always at least 4 times the series value, never $\frac{15}{4}$ times. Checking the options: they give $\frac{(1+x)^2}{x} \approx 4.45,\ 4.10,\ 6.02,\ 4.11$, none equal to $\frac{15}{4}$.
No option can be correct; the question is treated as dropped. | ||||
| 14 | 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 $i = \sqrt{i_1^2 + i_2^2}\,\sin(\omega t + \phi)$, a sinusoid of amplitude $i_0 = \sqrt{i_1^2 + i_2^2}$.
$i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{1}{\sqrt{2}}\left(i_1^2 + i_2^2\right)^{\frac{1}{2}}$.
Answer: (C). | ||||
| 15 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity A is joined by a wire to the centre O, and B by a wire to C. So the resistance wanted is that between O and C.
The resistors are OD, OE, DC, EC and DE, all R. From O to C there are two paths O–D–C and O–E–C, each 2R, bridged by DE. By symmetry D and E are at the same potential, so DE carries no current (a balanced bridge).
$R_{AB} = \frac{2R\cdot 2R}{2R + 2R} = R$.
Answer: (C). | ||||
| 16 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The image is erect (same orientation) and diminished (25 cm from 100 cm, $m = +\frac{1}{4}$). A single mirror forms an erect image only when it is virtual, and a concave mirror's virtual image is always enlarged; an erect diminished image comes only from a convex mirror.
A virtual image of a mirror lies behind it, on the opposite side to the object. (Check: $m = -\frac{v}{u} = \frac{1}{4}$ with $f = +40$ cm gives $u = -120$ cm, $v = +30$ cm, behind the mirror.)
Answer: (C). | ||||
| 17 | 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 $\beta = \frac{\lambda D}{d} = \frac{500\times10^{-9}\times 1}{2\times10^{-3}} = 2.5\times10^{-4}$ m $= 0.25$ mm.
Answer: (D). | ||||
| 18 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Assertion and reasonEasy | |
A easy assertion and reason question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The smallest resolvable separation of a microscope is proportional to the wavelength used. Electrons accelerated through kilovolts have de Broglie wavelengths $\lambda = \frac{h}{\sqrt{2meV}}$ of the order of picometres, far below visible wavelengths (400–700 nm).
So the electron microscope resolves far finer detail: A is true, R is true, and R is the reason for A.
Answer: (A). | ||||
| 19 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Third Lyman line: $n = 4 \to 1$, $\frac{1}{\lambda_1} = R\left(1 - \frac{1}{16}\right) = \frac{15R}{16}$.
First Paschen line: $n = 4 \to 3$, $\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$.
$\frac{\lambda_1}{\lambda_2} = \frac{7R/144}{15R/16} = \frac{7\times16}{144\times15} = \frac{7}{135}$.
Answer: (B). | ||||
| 20 | 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 $\lambda = \frac{hc}{E_g} = \frac{6.63\times10^{-34}\times3\times10^8}{1.9\times1.6\times10^{-19}} = \frac{1.989\times10^{-25}}{3.04\times10^{-19}} \approx 6.54\times10^{-7}$ m $= 654$ nm.
654 nm lies in the red part of the visible spectrum (about 620–750 nm).
Answer: (A). | ||||
| 21 | Laws of MotionFriction | Class 11: Laws of Motion | NumericalMedium | |
A medium numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion The force F pushes on the right face, directed down and to the left at $60^\circ$ to the horizontal (as drawn). Its horizontal part is $F\cos60^\circ = \frac{F}{2}$ and its downward part $F\sin60^\circ = \frac{\sqrt{3}F}{2}$.
Normal reaction: $N = mg + \frac{\sqrt{3}F}{2} = 10\sqrt{3} + \frac{\sqrt{3}F}{2}$.
No sliding: $\frac{F}{2} \le \mu N = \frac{1}{3\sqrt{3}}\left(10\sqrt{3} + \frac{\sqrt{3}F}{2}\right) = \frac{10}{3} + \frac{F}{6}$, so $\frac{F}{3} \le \frac{10}{3}$, $F \le 10$ N.
The critical force is 10 N $= 3x$, so $x = \frac{10}{3} \approx 3.33$.
Answer: 3.33. | ||||
| 22 | Laws of MotionNewton's laws | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $a_x = \frac{F_x}{m} = \frac{20}{2} = 10$ m/s$^2$, starting from rest.
$x = \frac{1}{2}a_xt^2 = \frac{1}{2}\times10\times10^2 = 500$ m.
Answer: 500. | ||||
| 23 | Laws of MotionNewton's laws | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion The person's mass is 60 kg. In a lift accelerating downward at a, the balance reads $N = m(g - a) = 60(10 - 1.8) = 60\times8.2 = 492$ N.
Answer: 492. | ||||
| 24 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\mu = 0.135$ g/cm $= 0.0135$ kg/m. From the wave, $k = 1$ m$^{-1}$ and $\omega = 30$ s$^{-1}$, so $v = \frac{\omega}{k} = 30$ m/s.
$T = \mu v^2 = 0.0135\times900 = 12.15$ N $= 1215\times10^{-2}$ N.
Answer: 1215. | ||||
| 25 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | NumericalMedium | |
A medium numerical question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory For ideal gases, $PV = nRT$ in each chamber. When the partition is removed in the insulated container, the total internal energy is unchanged; for the same gas (same $C_v$) this gives $(n_1 + n_2)T = n_1T_1 + n_2T_2$.
Then $P(V_1 + V_2) = R(n_1T_1 + n_2T_2) = P_1V_1 + P_2V_2$ (this is also what an isothermal mixing gives).
$P = \frac{2.0\times4.5 + 3.0\times5.5}{4.5 + 5.5} = \frac{9 + 16.5}{10} = 2.55$ atm $= 25.5\times10^{-1}$ atm.
Answer: 25.5. | ||||
| 26 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalMedium | |
A medium numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Radiated power $= 0.0125\times1000 = 12.5$ W, spread uniformly over a sphere of radius 2 m:
$I = \frac{12.5}{4\pi(2)^2} = \frac{12.5}{16\pi} \approx 0.2487$ W/m$^2$.
$I = \frac{1}{2}c\varepsilon_0E_0^2$, so $E_0^2 = \frac{2I}{c\varepsilon_0} = \frac{0.4974}{3\times10^8\times8.85\times10^{-12}} = \frac{0.4974}{2.655\times10^{-3}} \approx 187.3$ and $E_0 \approx 13.69$ V/m $= 136.9\times10^{-1}$ V/m.
Rounded: 137.
Answer: 137. | ||||
| 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 The potential difference is the work done per unit charge in carrying charge between the terminals: $V = \frac{W}{q}$.
So the work done by the battery in driving 20 C through a p.d. of 15 V is $W = qV = 20\times15 = 300$ J (the time taken does not matter).
Answer: 300. | ||||
| 28 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The battery and the 120 $\Omega$ resistor form the bottom branch; the D1–130 $\Omega$ branch and the D2–100 $\Omega$ branch are in parallel between the left and right wires.
The battery's longer (positive) plate is on the left, so current leaves it to the left, rises through the left wire and flows left to right across the upper branches. D1 points left to right and conducts; D2 points right to left and is reverse biased, so its branch carries nothing.
Circuit resistance: $50 + 130 + 120 = 300\ \Omega$, so $I = \frac{6}{300} = 0.02$ A $= 20$ mA.
Answer: 20. | ||||
| 29 | Communication SystemsAmplitude modulation | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter $\mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}} = \frac{16 - 8}{16 + 8} = \frac{1}{3} \approx 0.3333 = 33.33\times10^{-2}$.
Answer: 33.33. | ||||
| 30 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $Q = \frac{\omega_0L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$.
With $L \to 2L$ and $R \to \frac{R}{2}$ (C unchanged): $Q' = \frac{2}{R}\sqrt{\frac{2L}{C}} = 2\sqrt{2}\,Q = 2\sqrt{2}\times100 \approx 282.84$.
Answer: 282.84. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Angular nodes $= l = 2$, so it is a d orbital. Radial nodes $= n - l - 1 = 2$ gives $n = 5$.
Check the others: 3p has 1 radial and 1 angular; 4d has 1 radial and 2 angular; 4f has 0 radial and 3 angular.
Answer: (D) 5d. | ||||
| 32 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | Assertion and reasonMedium | |
A medium assertion and reason question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A is false: a hydrogen bond is not just a dipole–dipole attraction; it also has a partly covalent (charge-transfer) character and needs a hydrogen on a small, highly electronegative atom, so dipole–dipole forces are not the only interaction at work.
R: fluorine is the most electronegative element (true), and in the strongest hydrogen bonds of HF, such as the $[\mathrm{F{\cdots}H{\cdots}F}]^-$ ion of $\mathrm{KHF_2}$, the hydrogen sits symmetrically between the two fluorines. R is taken as true.
Answer: (D). | ||||
| 33 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Matching listEasy | |
A easy matching list question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The configurations are Be (a), O (b), N (c) and B (d).
Order of first ionization enthalpy: B < Be < O < N. B (801) is lowest since its 2p electron is easier to remove than Be's paired 2s electron (899); N's half-filled 2p shell (1402) is more stable than O's (1314), which loses a paired electron.
(a)→(ii), (b)→(iii), (c)→(iv), (d)→(i).
Answer: (C). | ||||
| 34 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Kernite is $\mathrm{Na_2B_4O_7\cdot4H_2O}$ (boron); cassiterite is $\mathrm{SnO_2}$ (tin); calamine is $\mathrm{ZnCO_3}$ (zinc); cryolite is $\mathrm{Na_3AlF_6}$ (fluorine, among the elements listed).
(a)→(ii), (b)→(i), (c)→(iv), (d)→(iii).
Answer: (A). | ||||
| 35 | HydrogenHydrogen and its compounds | No matching NCERT chapter | StatementsEasy | |
A easy statements question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter A: true; $\mathrm{D_2O}$ is used as a tracer in exchange reactions to study reaction mechanisms.
B: true; it is obtained by exhaustive (prolonged) electrolysis of water, where $\mathrm{H_2O}$ is decomposed preferentially and $\mathrm{D_2O}$ accumulates.
C: true; b.p. of $\mathrm{D_2O}$ is 374.4 K against 373 K.
D: false; $\mathrm{D_2O}$ is more viscous than $\mathrm{H_2O}$.
Answer: (D) A, B and C only. | ||||
| 36 | 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 These are the steps of the Solvay (ammonia–soda) process:
$\mathrm{2NH_3 + H_2O + CO_2 \rightarrow (NH_4)_2CO_3}$, so A is $\mathrm{H_2O}$;
$\mathrm{(NH_4)_2CO_3 + H_2O + CO_2 \rightarrow 2NH_4HCO_3}$, so B is $\mathrm{CO_2}$;
$\mathrm{NH_4HCO_3 + NaCl \rightarrow NH_4Cl + NaHCO_3}$, so C is $\mathrm{NaHCO_3}$, which precipitates because it is sparingly soluble.
Answer: (C). | ||||
| 37 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{PbO_2}$ (lead in the +4 state, made unstable by the inert-pair effect) is a strong oxidising agent, it is amphoteric (dissolves in acids and in alkalis as plumbate), and it is the cathode material of the lead storage battery.
PbO is amphoteric but not a strong oxidant; $\mathrm{Pb_3O_4}$ is a mixed oxide; $\mathrm{PbSO_4}$ is the discharge product, not an oxidant.
Answer: (B). | ||||
| 38 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements An oxide $\mathrm{MO_2}$ needs the +4 state. Ce, Pr, Nd, Tb and Dy show +4 (Pr, Nd and Dy only in their oxides or fluorides).
Yb has the configuration $4f^{14}6s^2$; its common states are +2 ($4f^{14}$) and +3, and it shows no +4 state, so it forms no $\mathrm{YbO_2}$.
Answer: (D) Yb. | ||||
| 39 | 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 Ozone in the stratosphere shields us from UV radiation; ozone in the troposphere is a pollutant.
It forms there from nitrogen oxides and hydrocarbons in sunlight and is a main component of photochemical smog; it is also a greenhouse gas rather than a protection against the greenhouse effect.
Answer: (C). | ||||
| 40 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsEasy | |
A easy statements question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques I: chloroform boils at 334 K and aniline at 457 K. Liquids whose boiling points differ this widely are separated by simple distillation: true.
II: in steam distillation the mixture boils when the sum of the vapour pressures of aniline and water reaches atmospheric pressure, so aniline distils below its own boiling point: true.
Answer: (A). | ||||
| 41 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The Carius method (heating with fuming $\mathrm{HNO_3}$ in a Carius tube) estimates halogens and sulphur, so (A) is true.
Nitrogen is estimated by Dumas' or Kjeldahl's method, not by Carius: (B) is false and (C) is true.
For phosphorus, the phosphoric acid formed is precipitated with magnesia mixture as $\mathrm{MgNH_4PO_4}$, which on ignition gives $\mathrm{Mg_2P_2O_7}$; (D) is accepted as true.
Answer: (B). | ||||
| 42 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Step 1: sodamide, a strong base, removes HBr from the vinyl bromide (dehydrohalogenation): $\mathrm{CH_3CH{=}CHBr \rightarrow CH_3C{\equiv}CH}$ (propyne).
Step 2: an alkyne passed through a red-hot iron tube at 873 K trimerises to a benzene ring. Three propyne molecules give 1,3,5-trimethylbenzene (mesitylene).
Answer: (D). | ||||
| 43 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $\mathrm{Br_2}$ in UV light brominates by a free-radical chain, not on the ring. The hydrogen removed is the one giving the most stable radical: the benzylic $\mathrm{CH_2}$ of the ethyl group, whose radical is stabilised by resonance with the ring.
The major monobrominated product is $\mathrm{3{-}(1{-}bromoethyl)benzonitrile}$, $\mathrm{C_6H_4(CN){-}CH(Br)CH_3}$.
Answer: (C). | ||||
| 44 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids B forms an oxime with hydroxylamine, so it is a carbonyl compound; it fails Tollens' test, so it is a ketone, not an aldehyde. The only $\mathrm{C_4H_8O}$ ketone is butan-2-one.
A gem-dichloride hydrolyses to a carbonyl compound; to give butan-2-one the two Cl must sit on C-2: A is 2,2-dichlorobutane, $\mathrm{CH_3CCl_2CH_2CH_3 \rightarrow CH_3COCH_2CH_3}$.
Answer: (D). | ||||
| 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: phenol with bromine in a non-polar solvent ($\mathrm{CS_2}$) at low temperature gives monobromophenols, mainly the para isomer, p-bromophenol.
B: $\mathrm{CHCl_3}$/NaOH then acid is the Reimer–Tiemann reaction, which puts a –CHO group mainly ortho to –OH: salicylaldehyde (o-hydroxybenzaldehyde).
Answer: (D). | ||||
| 46 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | StatementsEasy | |
A easy statements question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure I: in o-nitrophenol the –OH is hydrogen-bonded to the neighbouring $\mathrm{NO_2}$ within the same molecule, so molecules do not associate; it is volatile in steam: true.
II: for the same reason its intermolecular forces are weak and its melting point is low (about 318 K, against about 387 K for p-nitrophenol, which is held by intermolecular hydrogen bonds): false.
Answer: (C). | ||||
| 47 | 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 sulphonamide insoluble in alkali has no N–H left on the nitrogen, so the amine is secondary (Hinsberg test). This rules out ethylamine (D), which is primary.
It is made by ammonolysis of ethyl chloride, i.e. an N-ethyl group is introduced by attack of an amine on $\mathrm{C_2H_5Cl}$. Of the secondary amines offered, only $\mathrm{CH_3CH_2CH_2NH{-}CH_2CH_3}$ (N-ethylpropan-1-amine) carries an ethyl group on nitrogen: $\mathrm{CH_3CH_2CH_2NH_2 + C_2H_5Cl \rightarrow CH_3CH_2CH_2NHC_2H_5 + HCl}$.
Answer: (B). | ||||
| 48 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Neoprene is the addition polymer of chloroprene (2-chlorobuta-1,3-diene): $\mathrm{n\,CH_2{=}C(Cl){-}CH{=}CH_2 \rightarrow [CH_2{-}C(Cl){=}CH{-}CH_2]_n}$.
The others are Buna-N's unit (A), melamine–formaldehyde (C) and polyacrylonitrile (D).
Answer: (B). | ||||
| 49 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Of the vitamins listed, the one tied to blood clotting is vitamin K: it is needed to make the clotting factors (such as prothrombin) in the liver, and its deficiency increases the blood-clotting time.
Vitamins B and C have no role in clotting at this level (B deficiency causes beri-beri, anaemia and so on; C deficiency scurvy).
Read literally, a vitamin that delays clotting is vitamin E, which has a mild anticoagulant action (it opposes platelet aggregation).
Answer: (D) Vitamin K or (C) Vitamin E. | ||||
| 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 A sulphite with warm dilute $\mathrm{H_2SO_4}$ gives $\mathrm{SO_2}$ (X). $\mathrm{SO_2}$ reduces acidified dichromate to green chromium(III) sulphate (Y):
$\mathrm{K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O}$.
Answer: (D). | ||||
| 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 The power of ten does not affect the count. In 50000.020 every digit counts: the zeros between 5 and 2 are between non-zero digits, and the final zero after the decimal point is significant.
Digits 5, 0, 0, 0, 0, 0, 2, 0: 8 significant figures.
Answer: 8. | ||||
| 52 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $PV_m = RT + Pb$, so $Z = \frac{PV_m}{RT} = 1 + \frac{Pb}{RT}$.
$\left(\frac{\partial Z}{\partial P}\right)_T = \frac{b}{RT}$, so $x = 1$.
Answer: 1. | ||||
| 53 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Spontaneous when $\Delta G = \Delta H - T\Delta S < 0$: $80000 - T(2T) < 0$, i.e. $T^2 > 40000$, $T > 200$ K.
Answer: 200. | ||||
| 54 | SolutionsColligative properties | Class 12: Solutions | NumericalHard | |
A hard numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions NaOH: $0.1\times0.1 = 0.01$ mol. $\mathrm{SO_2}$: $n = \frac{PV}{RT} = \frac{1\times0.224}{0.0821\times298} \approx 0.0092$ mol (about 0.01 mol), more than the 0.005 mol needed.
$\mathrm{SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O}$: NaOH is limiting, giving 0.005 mol of $\mathrm{Na_2SO_3}$, which dissociates into 3 ions ($i = 3$).
Water: $\frac{36}{18} = 2$ mol. For a dilute solution, $\Delta P = P^\circ\cdot\frac{i\,n_{solute}}{n_{water}} = 24\times\frac{3\times0.005}{2} = 0.18$ mm Hg $= 18\times10^{-2}$ mm Hg.
Answer: 18. | ||||
| 55 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium If $\alpha$ mol of $\mathrm{AB_2}$ dissociates, the moles are $1 - \alpha$, $\alpha$, $2\alpha$; total $1 + 2\alpha$.
$n_{total} = \frac{PV}{RT} = \frac{1.9\times25}{0.08206\times300} = \frac{47.5}{24.618} = 1.9295$, so $\alpha = 0.4647$.
Each partial pressure is $n_i\frac{RT}{V}$ with $\frac{RT}{V} = 0.98472$ atm: $K_p = \frac{p_Ap_B^2}{p_{AB_2}} = \frac{4\alpha^3}{1 - \alpha}\left(\frac{RT}{V}\right)^2 = \frac{4(0.4647)^3}{0.5353}\times0.9697 \approx 0.727$.
$K_p \approx 72.7\times10^{-2}$; to the nearest integer, $x = 73$.
Answer: 73. | ||||
| 56 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Each mole of $\mathrm{MnO_4^-}$ takes 5 mol of electrons, i.e. 5 F. For 5 mol: $5\times5 = 25$ F.
Answer: 25. | ||||
| 57 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\Delta E = E_{a,f} - E_{a,b}$, so $E_{a,b} = E_{a,f} - \Delta E = 30 - (-20) = 50$ kJ mol$^{-1}$.
Answer: 50. | ||||
| 58 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter $n(\mathrm{O_2}) = \frac{3.12}{32} = 0.0975$ mol. $V = \frac{nRT}{P} = 0.0975\times0.0821\times300 = 2.401$ L.
Per gram of platinum: $\frac{2.401}{1.2} = 2.001 \approx 2.00$ L.
Answer: 2. | ||||
| 59 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In base dichromate changes to chromate: $\mathrm{Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O}$. This is not a redox change.
In $\mathrm{CrO_4^{2-}}$: $x + 4(-2) = -2$, so $x = +6$.
Answer: 6. | ||||
| 60 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{Mn_2(CO)_{10}}$ is two $\mathrm{Mn(CO)_5}$ units joined by a direct Mn–Mn bond; all ten CO ligands are terminal (five on each Mn).
Number of bridging CO = 0.
Answer: 0. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | 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 $\vec{a}\times(\vec{a}\times\vec{b}) = (\vec{a}\cdot\vec{b})\vec{a} - |\vec{a}|^2\vec{b} = -|\vec{a}|^2\vec{b}$, since $\vec{a}\cdot\vec{b} = 0$.
Then $\vec{a}\times\left(-|\vec{a}|^2\vec{b}\right) = -|\vec{a}|^2(\vec{a}\times\vec{b})$, and once more $\vec{a}\times\left(-|\vec{a}|^2(\vec{a}\times\vec{b})\right) = -|\vec{a}|^2\left(-|\vec{a}|^2\vec{b}\right) = |\vec{a}|^4\vec{b}$.
Answer: (C). | ||||
| 62 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Vectors from $(1, 5, 35)$ to the other three points: $(6, 0, -30)$, $(0, \lambda - 5, -28)$, $(2\lambda - 1, -4, -33)$. Coplanarity needs their scalar triple product to vanish:
$\begin{vmatrix}6&0&-30\\0&\lambda-5&-28\\2\lambda-1&-4&-33\end{vmatrix} = 6\left[-33(\lambda - 5) - 112\right] - 30\left[0 - (\lambda - 5)(2\lambda - 1)\right]$
$= -198\lambda + 318 + 30(2\lambda^2 - 11\lambda + 5) = 60\lambda^2 - 528\lambda + 468 = 0$, i.e. $5\lambda^2 - 44\lambda + 39 = 0$.
Its discriminant $44^2 - 4\cdot5\cdot39 = 1156 > 0$, so both roots are real; their sum is $\frac{44}{5}$.
Answer: (D). | ||||
| 63 | Straight LinesCartesian coordinates and distance formula | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines Vertices: $x = y$ with $x + 2y = 3$ gives $(1, 1)$; $x = y$ with $2x + y = 6$ gives $(2, 2)$; $x + 2y = 3$ with $2x + y = 6$ gives $(3, 0)$.
Sides: $(1,1)$–$(2,2)$: $\sqrt{2}$; $(1,1)$–$(3,0)$: $\sqrt{5}$; $(2,2)$–$(3,0)$: $\sqrt{5}$. Two sides are equal, and $2 + 5 \ne 5$, so there is no right angle.
Answer: (B) isosceles. | ||||
| 64 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Apply $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$ (the value is unchanged):
$R_1 = (-2(a+2),\ -1,\ 0)$, $R_2 = (-2(a+3),\ -1,\ 0)$, $R_3 = ((a+3)(a+4),\ a+4,\ 1)$.
Expanding along the third column: $1\cdot\left[(-2(a+2))(-1) - (-1)(-2(a+3))\right] = 2(a+2) - 2(a+3) = -2$.
Answer: (C). | ||||
| 65 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dN}{dt} = rN$ gives $N = 1000e^{rt}$.
$1200 = 1000e^{2r}$, so $r = \frac{1}{2}\log_e\frac{6}{5}$.
$2000 = 1000e^{rt}$ gives $t = \frac{\log_e 2}{r} = \frac{2\log_e 2}{\log_e(6/5)}$, so $k = 2\log_e 2$ and $\left(\frac{k}{\log_e 2}\right)^2 = 4$.
Answer: (B). | ||||
| 66 | CirclesEquation from a diameter | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on equation from a diameter, from Circles. NCERT: Class 11: Conic Sections The circle passes through O and B and is symmetric about the x-axis, so OB is a diameter: centre $(6.5, 0)$, radius $6.5$, equation $x(x - 13) + y^2 = 0$.
At $x = 1$ (the point A): $y^2 = 1\times12 = 12$, so $AP = AQ = 2\sqrt{3}$ and $PQ = 4\sqrt{3}$.
$AB = 13 - 1 = 12$ is the height of triangle PQB on base PQ: area $= \frac{1}{2}\times4\sqrt{3}\times12 = 24\sqrt{3}$.
Answer: (B). | ||||
| 67 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Numerator: $\sqrt{3}\sin\theta - \cos\theta = 2\sin\left(\theta - \frac{\pi}{6}\right)$; with $\theta = \frac{\pi}{6} + h$ it is $2\sin h$.
Limit $= \lim\limits_{h\to0} 2\cdot\frac{2\sin h}{\sqrt{3}h(\sqrt{3}\cos h - \sin h)} = \frac{4}{\sqrt{3}}\cdot1\cdot\frac{1}{\sqrt{3}} = \frac{4}{3}$.
Answer: (B). | ||||
| 68 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Slope $m = \frac{dy}{dx} = 2x^3 - 15x^2 + 36x - 19$.
$\frac{dm}{dx} = 6x^2 - 30x + 36 = 6(x - 2)(x - 3)$; $\frac{d^2m}{dx^2} = 12x - 30$ is $-6 < 0$ at $x = 2$, so the slope has its (local) maximum at $x = 2$.
$y(2) = 8 - 40 + 72 - 38 = 2$: the point is $(2, 2)$.
Answer: (B). | ||||
| 69 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $I$ be the integral. Replacing x by $-x$: $I = \int_{-\pi/2}^{\pi/2}\frac{3^x\cos^2x}{1 + 3^x}\,dx$.
Adding: $2I = \int_{-\pi/2}^{\pi/2}\cos^2x\,dx = \frac{\pi}{2}$, so $I = \frac{\pi}{4}$.
Answer: (C). | ||||
| 70 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Seven digits each at least 1 add to at least 7; the extra 3 comes from a 2s (extra 1 each) and b 3s (extra 2 each) with $a + 2b = 3$.
$b = 0, a = 3$: three 2s and four 1s, $\frac{7!}{3!\,4!} = 35$ ways.
$b = 1, a = 1$: one 3, one 2, five 1s, $\frac{7!}{5!} = 42$ ways.
Total $35 + 42 = 77$.
Answer: (C). | ||||
| 71 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = {}^{10}C_r\left(tx^{1/5}\right)^{10-r}\left(\frac{(1-x)^{1/10}}{t}\right)^r$ is free of t when $10 - 2r = 0$, $r = 5$:
$T_6 = {}^{10}C_5\,x(1 - x)^{1/2}$.
Maximise $f(x) = x\sqrt{1 - x}$: $f'(x) = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} = 0$ gives $x = \frac{2}{3}$, a maximum, with $f = \frac{2}{3}\cdot\frac{1}{\sqrt{3}} = \frac{2}{3\sqrt{3}}$.
Maximum value $= \frac{10!}{(5!)^2}\cdot\frac{2}{3\sqrt{3}} = \frac{2\cdot10!}{3\sqrt{3}(5!)^2}$.
Answer: (B). | ||||
| 72 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $(1, -1)$ is at distance $\sqrt{2}$ from the origin. Its equivalence class is every point at distance $\sqrt{2}$: $x^2 + y^2 = 2$.
Answer: (B). | ||||
| 73 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let $A = \begin{pmatrix}a&b\\b&c\end{pmatrix}$. Then $A^2$ has diagonal entries $a^2 + b^2$ and $b^2 + c^2$, so $a^2 + 2b^2 + c^2 = 1$.
With integers, $b = 0$ and exactly one of a, c is $\pm1$, the other 0: $(a, c) = (\pm1, 0)$ or $(0, \pm1)$, 4 matrices.
Answer: (B). | ||||
| 74 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $x \ne y$: $\left|\frac{f(x) - f(y)}{x - y}\right| \le |x - y|$. Letting $y \to x$, $f'(x)$ exists and equals 0 for every x.
So f is constant, $f(x) = f(0) = 1 > 0$ for all x.
Answer: (A). | ||||
| 75 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability With n tosses, $P(k\text{ heads}) = \frac{{}^nC_k}{2^n}$. ${}^nC_7 = {}^nC_9$ gives $n = 16$.
$P(2\text{ heads}) = \frac{{}^{16}C_2}{2^{16}} = \frac{120}{2^{16}} = \frac{15}{2^{13}}$.
Answer: (C). | ||||
| 76 | 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 Let each ratio be k. Then $\sin^{-1}x + \cos^{-1}x = (a + b)k = \frac{\pi}{2}$, so $k = \frac{\pi}{2(a + b)}$.
$\tan^{-1}y = ck = \frac{\pi c}{2(a + b)}$, so $\frac{\pi c}{a + b} = 2\tan^{-1}y$.
$\cos(2\tan^{-1}y) = \frac{1 - y^2}{1 + y^2}$.
Answer: (C). | ||||
| 77 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series With first term a and ratio r: $ar^2\cdot ar^4 = (ar^3)^2 = 25$, so $ar^3 = 5$ (terms positive, the series increasing).
$ar + ar^5 = \frac{5}{r^2} + 5r^2 = \frac{25}{2}$ gives $r^2 + \frac{1}{r^2} = \frac{5}{2}$, so $r^2 = 2$ or $\frac{1}{2}$; increasing means $r^2 = 2$.
$T_4 + T_6 + T_8 = ar^3(1 + r^2 + r^4) = 5(1 + 2 + 4) = 35$.
Answer: (D). | ||||
| 78 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series After the first term, the numerators $2, 7, 12, \ldots$ are $5n - 3$: $S = 1 + \sum_{n\ge1}\frac{5n - 3}{3^n}$.
$\sum_{n\ge1}\frac{n}{3^n} = \frac{1/3}{(1 - 1/3)^2} = \frac{3}{4}$ and $\sum_{n\ge1}\frac{1}{3^n} = \frac{1}{2}$.
$S = 1 + 5\cdot\frac{3}{4} - 3\cdot\frac{1}{2} = 1 + \frac{15}{4} - \frac{3}{2} = \frac{13}{4}$.
Answer: (C). | ||||
| 79 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Normals: $P_1$: $(3, 15, 21) = 3(1, 5, 7)$; $P_2$: $(1, -3, -1)$; $P_3$: $(2, 10, 14) = 2(1, 5, 7)$.
$P_1$ and $P_3$ have parallel normals and are distinct ($x + 5y + 7z = 3$ against $x + 5y + 7z = \frac{5}{2}$), so they are parallel; $P_2$ is parallel to neither.
Answer: (B). | ||||
| 80 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $[n-1, n)$, $[x] = n - 1$, so $\int_{n-1}^{n}e^{x-(n-1)}\,dx = \int_0^1 e^t\,dt = e - 1$.
Summing 100 equal terms: $100(e - 1)$.
Answer: (D). | ||||
| 81 | Differential EquationsOrder and degree | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on order and degree, from Differential Equations. NCERT: Class 12: Differential Equations One constant, so the order is 1. Differentiating: $2yy' = a$.
Substituting: $y^2 = 2yy'\left(x + \frac{\sqrt{2yy'}}{2}\right)$, i.e. $y - 2xy' = y'\sqrt{2yy'}$.
Squaring to clear the radical: $(y - 2xy')^2 = 2yy'^3$. The highest power of $y'$ is 3, so the degree is 3.
Degree − order $= 3 - 1 = 2$.
Answer: 2. | ||||
| 82 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $x^3 - 2x^2 + 2x - 1 = (x - 1)(x^2 - x + 1)$. The roots are 1 and $e^{\pm i\pi/3}$ (that is, $-\omega^2$ and $-\omega$).
$\left(e^{\pm i\pi/3}\right)^{162} = e^{\pm 54\pi i} = 1$, and $1^{162} = 1$.
Sum $= 1 + 1 + 1 = 3$.
Answer: 3. | ||||
| 83 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $y = ||x - 1| - 2|$ is zero at $x = -1$ and $x = 3$, rises linearly to $y = 2$ at $x = 1$ between them, and grows again outside them.
The bounded region it encloses with the x-axis is the triangle with vertices $(-1, 0)$, $(3, 0)$, $(1, 2)$: area $= \frac{1}{2}\times4\times2 = 4$.
Answer: 4. | ||||
| 84 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $u = e^{\sin y}$, so $\frac{du}{dx} = e^{\sin y}\cos y\frac{dy}{dx}$. The equation becomes $\frac{du}{dx} + u\cos x = \cos x$, linear with integrating factor $e^{\sin x}$.
$ue^{\sin x} = \int\cos x\,e^{\sin x}\,dx = e^{\sin x} + C$, so $u = 1 + Ce^{-\sin x}$.
$y(0) = 0$ gives $u(0) = 1$, so $C = 0$ and $e^{\sin y} = 1$: $\sin y = 0$, and the solution through the origin is $y(x) = 0$.
The expression is $1 + 0 + 0 + 0 = 1$.
Answer: 1. | ||||
| 85 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations The logarithms need $x > 3$. $\log_4(x - 1) = \frac{1}{2}\log_2(x - 1)$, so $\log_2(x - 1) = 2\log_2(x - 3)$, i.e. $x - 1 = (x - 3)^2$.
$x^2 - 7x + 10 = 0$: $x = 2$ or $x = 5$. Only $x = 5$ satisfies $x > 3$.
Answer: 1. | ||||
| 86 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $3\sin x + 4\cos x$ takes every value in $[-5, 5]$ (amplitude $\sqrt{3^2 + 4^2} = 5$).
So $-5 \le k + 1 \le 5$, i.e. $-6 \le k \le 4$: 11 integers.
Answer: 11. | ||||
| 87 | 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 Normal: $(-1 + 2, -16 + 21, 23 - 29) = (1, 5, -6)$. Plane: $(x - 4) + 5(y + 2) - 6(z - 2) = 0$.
At $(\lambda, 2, 1)$: $\lambda - 4 + 20 + 6 = 0$, so $\lambda = -22$ and $\frac{\lambda}{11} = -2$.
$(-2)^2 - 4(-2) - 4 = 4 + 8 - 4 = 8$.
Answer: 8. | ||||
| 88 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $|\sin 2x|$ has period $\frac{\pi}{2}$, so the integral is $2\int_0^{\pi/2}\sin 2x\,dx = 2\left[-\frac{\cos 2x}{2}\right]_0^{\pi/2} = 2\times1 = 2$.
Answer: 2. | ||||
| 89 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions With $c = \cos x$: $\sqrt{3}c^2 - \sqrt{3}c + c - 1 = 0$, i.e. $\sqrt{3}c(c - 1) + (c - 1) = (c - 1)(\sqrt{3}c + 1) = 0$.
$c = 1$ gives $x = 0$, in the interval; $c = -\frac{1}{\sqrt{3}} < 0$ is impossible for $x \in \left[0, \frac{\pi}{2}\right]$.
Answer: 1. | ||||
| 90 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The sum is $\sum_{k=0}^{29}(30 - k)\binom{30}{k}$. Using $\binom{30}{k} = \binom{30}{30-k}$ and putting $j = 30 - k$: $\sum_{j=1}^{30}j\binom{30}{j} = 30\cdot2^{29}$.
$30\cdot2^{29} = 15\cdot2^{30}$ with 15 odd, so $n = 15$, $m = 30$ and $n + m = 45$.
Answer: 45. | ||||
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