JEE (Main) · 2021
JEE (Main) 2021 — 27 August, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Kinematics2
- Physical World and Measurement2
- Properties of Bulk Matter2
22 easy · 7 medium · 1 hard
Chemistry
30 Q
- Aldehydes, Ketones and Carboxylic Acids3
- p-Block Elements2
- d- and f-Block Elements2
16 easy · 11 medium · 3 hard
Mathematics
30 Q
- Matrices and Determinants3
- Trigonometry3
- Integral Calculus3
8 easy · 19 medium · 3 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Let the interval between drops be $t$. When the first drop lands it has fallen for $2t$ and the second for $t$.
$9.8 = \frac{1}{2}g(2t)^2 = 2gt^2$, so the second drop has fallen $\frac{1}{2}gt^2 = \frac{9.8}{4} = 2.45$ m from the nozzle.
Its height above the floor is $9.8 - 2.45 = 7.35$ m.
Answer: (B). | ||||
| 2 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (a) $\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, so $R_H$ is in $\mathrm{m^{-1}}$: (iii).
(b) $h = E/\nu$: $\mathrm{J\,s} = \mathrm{kg\,m^2s^{-1}}$: (ii).
(c) Energy density is energy per volume: $\mathrm{J/m^3} = \mathrm{kg\,m^{-1}s^{-2}}$: (iv).
(d) $\eta = \frac{F}{A\,(dv/dx)}$: $\mathrm{N\,s\,m^{-2}} = \mathrm{kg\,m^{-1}s^{-1}}$: (i).
Answer: (D). | ||||
| 3 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Angular momentum is conserved: $\omega = \frac{I_1\omega_1+I_2\omega_2}{I_1+I_2}$.
Loss $= \frac{1}{2}I_1\omega_1^2+\frac{1}{2}I_2\omega_2^2-\frac{1}{2}\frac{(I_1\omega_1+I_2\omega_2)^2}{I_1+I_2} = \frac{I_1I_2}{2(I_1+I_2)}(\omega_1-\omega_2)^2$.
Answer: (A). | ||||
| 4 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $h_{max} = \frac{u^2\sin^2\theta}{2g} = \frac{625\times\frac{1}{2}}{20} = 15.625$ m.
Time to the top: $T = \frac{u\sin\theta}{g} = \frac{25\times0.707}{10} = 1.77$ s.
Answer: (C). | ||||
| 5 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion The 8 kg box hangs from a movable pulley held by two segments of the string; the string runs over the fixed pulley to the 2 kg box. If the 8 kg box goes down by $x$, the 2 kg box rises by $2x$, so $a_2 = 2a$.
8 kg box: $80 - 2T = 8a$. 2 kg box: $T - 20 = 2(2a)$, so $T = 20 + 4a$.
$80 - 40 - 8a = 8a$ gives $a = 2.5\ \mathrm{m/s^2}$.
$0.2 = \frac{1}{2}(2.5)t^2$, $t^2 = 0.16$, $t = 0.4$ s.
Answer: (D). | ||||
| 6 | GravitationGravitational potential energy | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation Point mass: $-\frac{G(50)}{25} = -2G$.
Inside the shell the potential equals its surface value: $-\frac{G(100)}{50} = -2G$.
$V = -4G$.
Answer: (C). | ||||
| 7 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter All the potential energy heats the water: $mgh = mc\,\Delta T$.
$\Delta T = \frac{gh}{c} = \frac{9.8\times63}{4200} = 0.147^\circ$C.
Answer: (B). | ||||
| 8 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$; at the same temperature $\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{32}{2}} = 4$.
$v_{H_2} = 4\times160 = 640$ m/s.
Answer: (C). | ||||
| 9 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $C_2$ and $C_3$ are in series between A and B: $C = \frac{6\times12}{18} = 4\ \mu$F, so each carries $4V$ ($\mu$C).
$C_1$ is directly across A and B: $Q_1 = 2V$.
$Q_1 : Q_2 : Q_3 = 2 : 4 : 4 = 1 : 2 : 2$.
Answer: (D). | ||||
| 10 | 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 The arc spans $60^\circ$ on each side of the $x$-axis; the $y$-components cancel.
Linear charge density: $\lambda = \frac{Q}{R(2\pi/3)} = \frac{3Q}{2\pi R}$.
For an arc subtending $2\theta_0$: $E = \frac{2k\lambda\sin\theta_0}{R} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2}{R}\cdot\frac{3Q}{2\pi R}\cdot\frac{\sqrt{3}}{2} = \frac{3\sqrt{3}\,Q}{8\pi^2\varepsilon_0R^2}$.
The charge is negative, so the field at O points towards the arc, along $+\hat{i}$.
Answer: (A). | ||||
| 11 | 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 (i) $x < a$: enclosed current $i_0\frac{x^2}{a^2}$, so $B_1 = \frac{\mu_0 i_0 x}{2\pi a^2}$.
(ii) $a < x < b$: enclosed current $i_0$, so $B_2 = \frac{\mu_0 i_0}{2\pi x}$.
$\frac{B_1}{B_2} = \frac{x^2}{a^2}$.
Answer: (A). | ||||
| 12 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Bands in order: violet (7), green (5), red (multiplier $10^2$), gold ($\pm5\%$).
$R = 75\times10^2 = 7500\ \Omega$; tolerance $5\%$ of 7500 is $375\ \Omega$.
Answer: (D). | ||||
| 13 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Full-scale current: $\frac{50\ \text{div}}{2\ \text{div/mA}} = 25$ mA.
$G = \frac{50\ \text{mV}}{25\ \text{mA}} = 2\ \Omega$.
Answer: (B). | ||||
| 14 | 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 Plano-convex lens: $\frac{1}{f_1} = \frac{\mu_1-1}{R}$. Plano-concave lens: $\frac{1}{f_2} = -\frac{\mu_2-1}{R}$.
In contact: $\frac{1}{F} = \frac{\mu_1-1}{R}-\frac{\mu_2-1}{R} = \frac{\mu_1-\mu_2}{R}$.
$\frac{R}{F} = \mu_1-\mu_2$.
Answer: (A). | ||||
| 15 | 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 $F = MLT^{-2}$ gives $M = FL^{-1}T^{2}$.
Density $= \frac{M}{L^3} = FL^{-4}T^{2}$.
Answer: (C). | ||||
| 16 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At $t = 1$ s the centre is at $x = 0$, so the boundary $x = 0$ cuts the ring along a diameter.
The rate of change of flux is $B\,v\,\ell$, where $\ell$ is the length of the chord on the boundary: $\ell = 2R = 2$ m.
$\varepsilon = 1\times1\times2 = 2$ V.
Answer: (C). | ||||
| 17 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $I_E = I_B + I_C$, so $\beta = \frac{I_C}{I_E - I_C} = \frac{I_C/I_E}{1 - I_C/I_E} = \frac{\alpha}{1-\alpha}$.
Answer: (C). | ||||
| 18 | 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 $I_{max} = (\sqrt{I_1}+\sqrt{I_2})^2 = (2\sqrt{I_0})^2 = 4I_0$ (and $I_{min} = 0$, as stated).
Answer: (C). | ||||
| 19 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon energies with $hc = 1240$ eV nm: $\frac{1240}{670.5} = 1.849$ eV and $\frac{1240}{474.6} = 2.613$ eV.
$eV_2 = eV_1 + (2.613 - 1.849)$ eV $= 0.48 + 0.764 = 1.24$ eV, so $V_2 \approx 1.25$ V.
Answer: (C). | ||||
| 20 | Communication SystemsAntenna height and range | No matching NCERT chapter | Single correctHard | |
A hard single correct question on antenna height and range, from Communication Systems. NCERT: No matching NCERT chapter Range of a transmitting antenna of height $h$: $d = \sqrt{2Rh}$. For $d = 44$ km and $R = 6.4\times10^6$ m:
$h = \frac{d^2}{2R} = \frac{(44\times10^3)^2}{2\times6.4\times10^6} = 151.25$ m, the height of the tower (antenna) needed for this range.
An antenna radiates effectively when its length is $\frac{\lambda}{4}$: $\lambda = 4\times151.25 = 605$ m.
Answer: (D). | ||||
| 21 | 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 Switch open: two branches, each $R + 2R = 3R$, in parallel: $R_{open} = \frac{3R}{2}$.
Switch closed: the middle nodes join; $(R\parallel2R) + (2R\parallel R) = \frac{2R}{3}+\frac{2R}{3} = \frac{4R}{3}$.
$\frac{R_{open}}{R_{closed}} = \frac{3/2}{4/3} = \frac{9}{8}$, so $x = 9$.
Answer: 9. | ||||
| 22 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Old: $\cos\phi_1 = \frac{R}{\sqrt{R^2+9R^2}} = \frac{1}{\sqrt{10}}$.
New: $\cos\phi_2 = \frac{R}{\sqrt{R^2+(3R-2R)^2}} = \frac{1}{\sqrt{2}}$.
$\frac{\cos\phi_2}{\cos\phi_1} = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5} : 1$, so $x = 1$.
Answer: 1. | ||||
| 23 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power To complete the circle the bob needs at the bottom $u = \sqrt{5gl} = \sqrt{5\times10\times0.5} = 5$ m/s.
Momentum (bullet recoils backwards): $0.01v = 1\times5 - 0.01\times100 = 4$.
$v = 400$ m/s.
Answer: 400. | ||||
| 24 | 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 Breaking tensions: $W_1$: $1.25\times10^9\times8\times10^{-7} = 1000$ N; $W_2$: $500$ N.
$W_2$ holds the 10 kg block and the pan load $m$: $(10+m)\times10 \le 500$, so $m \le 40$ kg.
$W_1$ holds everything: $(30+m)\times10 \le 1000$, so $m \le 70$ kg.
The lower wire limits it: $m = 40$ kg.
Answer: 40. | ||||
| 25 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Fundamental of a closed pipe: $f = \frac{v}{4L}$.
$L = \frac{340}{4\times250} = 0.34$ m $= 34$ cm.
Answer: 34. | ||||
| 26 | ThermodynamicsSecond law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Efficiency $= 1 - \frac{240}{300} = 0.2$, and it cannot exceed the Carnot value $1 - \frac{T_2}{T_1}$.
$0.2 \le 1 - \frac{400}{T_1}$ gives $T_1 \ge 500$ K.
Answer: 500. | ||||
| 27 | 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 $y_2 = 5\times2\left(\frac{1}{2}\sin3\pi t + \frac{\sqrt{3}}{2}\cos3\pi t\right) = 10\sin\left(3\pi t + \frac{\pi}{3}\right)$, amplitude 10.
Amplitude of $y_1$ is 10, so the ratio is $1 : 1$ and $x = 1$.
Answer: 1. | ||||
| 28 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalEasy | |
A easy numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $B = \frac{E}{c} = \frac{6}{3\times10^8} = 2\times10^{-8}$ T.
Answer: 2. | ||||
| 29 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Number of lines from level $n$: $\frac{n(n-1)}{2} = \frac{6\times5}{2} = 15$.
Answer: 15. | ||||
| 30 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Largest safe zener current: $I_{max} = \frac{P}{V_Z} = \frac{2}{10} = 0.2$ A.
At the highest input (14 V) the drop across $R_S$ is $4$ V, and with no load all of it flows through the zener: $R_S \ge \frac{4}{0.2} = 20\ \Omega$.
Answer: 20. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter A lyophilic sol owes its stability to two factors, charge and solvation; the extensive solvation layer around each particle is what lyophobic sols lack.
(A) and (D): charge alone (of either sign) is common to both kinds of sol and does not explain the difference. (B): the particles are not uncharged.
Answer: (C). | ||||
| 32 | 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 All five ions are isoelectronic (18 electrons). The radius falls as the nuclear charge rises: P (15) > S (16) > Cl (17) > K (19) > Ca (20).
$\mathrm{P^{3-} > S^{2-} > Cl^- > K^+ > Ca^{2+}}$.
Answer: (B). | ||||
| 33 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Calamine $\mathrm{ZnCO_3}$ (iii); malachite $\mathrm{CuCO_3\cdot Cu(OH)_2}$ (iv); siderite $\mathrm{FeCO_3}$ (ii); sphalerite (zinc blende) $\mathrm{ZnS}$ (i).
Answer: (B). | ||||
| 34 | 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 $\mathrm{PbO_2}$ and $\mathrm{SnO_2}$ are dioxides, not peroxides: they contain no O–O link and give no $\mathrm{H_2O_2}$.
Sodium peroxide reacts with (cold) water directly: $\mathrm{Na_2O_2 + 2H_2O \rightarrow 2NaOH + H_2O_2}$.
Hydrated barium peroxide needs an acid ($\mathrm{BaO_2\cdot8H_2O + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 + 8H_2O}$); with water alone it hardly reacts.
Answer: (C). | ||||
| 35 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter (A) True: LiF, with the very high lattice enthalpy of the small $\mathrm{Li^+}$ and $\mathrm{F^-}$ ions, is the least soluble alkali metal halide.
(B) False: LiF has the most negative enthalpy of formation of the fluorides; the values become less negative down the group.
(C) False: CsI is poorly soluble because of the small hydration enthalpy of its two large ions, not a high lattice enthalpy.
(D) False: for chlorides, bromides and iodides the enthalpy of formation becomes more negative down the group (the reverse of the fluorides).
Answer: (A). | ||||
| 36 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Red phosphorus heated in a sealed tube at 803 K gives $\alpha$-black phosphorus; $\beta$-black phosphorus is made by heating white phosphorus at 473 K under high pressure.
Answer: (B). | ||||
| 37 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mathrm{2KMnO_4 \xrightarrow{513\ K} K_2MnO_4 + MnO_2 + O_2}$.
Potassium manganate contains $\mathrm{MnO_4^{2-}}$ with Mn(+6), $3d^1$: one unpaired electron, so it is paramagnetic, and it is green.
Answer: (C). | ||||
| 38 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Dioxygen difluoride oxidises plutonium to $\mathrm{PuF_6}$: $\mathrm{Pu + 3O_2F_2 \rightarrow PuF_6 + 3O_2}$, which removes plutonium from spent nuclear fuel.
Answer: (D). | ||||
| 39 | Environmental ChemistryOzone depletion | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ozone depletion, from Environmental Chemistry. NCERT: No matching NCERT chapter UV radiation splits $\mathrm{O_2}$ into oxygen atoms, which combine with $\mathrm{O_2}$: $\mathrm{O_2 \xrightarrow{UV} 2O}$, $\mathrm{O + O_2 \rightarrow O_3}$.
Answer: (C). | ||||
| 40 | Practical Organic ChemistryDetection of functional groups | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on detection of functional groups, from Practical Organic Chemistry. NCERT: No matching NCERT chapter Biuret test: alkaline $\mathrm{CuSO_4}$. Barfoed's test: copper(II) acetate in acetic acid. Benedict's test: alkaline copper(II) citrate.
Seliwanoff's test uses resorcinol in concentrated HCl (for ketoses), with no copper.
Answer: (C). | ||||
| 41 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $\mathrm{K_2CO_3}$ turns phenol into phenoxide, which attacks the $\mathrm{CH_2}$ that carries Br in 1-bromo-3-methylbut-2-ene (prenyl bromide).
Direct $\mathrm{S_N2}$ substitution there, with no allylic shift, gives $\mathrm{PhO{-}CH_2{-}CH{=}C(CH_3)_2}$.
(C) would come from attack at the other end of the allyl system, and (A) and (D) need the double bond to move or add.
Answer: (B). | ||||
| 42 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (B) $\mathrm{CH_3CH_2CCl_3}$: alkaline hydrolysis of the $\mathrm{CCl_3}$ group then acid gives $\mathrm{CH_3CH_2COOH}$.
(C) Butan-2-one is a methyl ketone: the iodoform reaction gives $\mathrm{CHI_3}$ and $\mathrm{CH_3CH_2COO^-}$, then $\mathrm{CH_3CH_2COOH}$.
(D) $\mathrm{CH_3CH_2CH_2MgBr + CO_2}$ then $\mathrm{H_3O^+}$ gives $\mathrm{CH_3CH_2CH_2COOH}$, butanoic acid: one carbon more than propionic acid.
(A) Propane has no group that $\mathrm{KMnO_4}$ attacks, so strictly it gives no propionic acid either (see the note); the designed answer is (D).
Answer: (D). | ||||
| 43 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate is a primary bromide, but its $\beta$-hydrogen is benzylic (and tertiary), so it is acidic, and the alkene formed is conjugated with the ring.
Methoxide is a strong base, so E2 elimination wins over substitution: $\mathrm{CH_3CH_2C(C_6H_5){=}CH_2}$.
(A) is the $\mathrm{S_N2}$ product, the minor one; (C) and (D) do not arise from methoxide in methanol.
Answer: (B). | ||||
| 44 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (i) Two moles of $\mathrm{CH_3MgBr}$: one adds to the ketone C=O, the other adds to the C$\equiv$N, giving an imine salt.
(ii) $\mathrm{H_3O^+}$ hydrolyses the imine to a ketone, so CN becomes $\mathrm{COCH_3}$, and gives the tertiary alcohol at C2 (now carrying a $\mathrm{CH_3}$).
(iii) Hot $\mathrm{H_2SO_4}$ dehydrates the tertiary alcohol. The double bond forms towards C1, so that it is conjugated with the benzene ring and tetrasubstituted: $\mathrm{C1{=}C2}$, with $\mathrm{CH_3}$ on C1 and C2 and C3 still $\mathrm{CH(CH_3)}$.
Options (A) and (C) keep CN, which the second Grignard equivalent has consumed; (B) puts the double bond out of conjugation.
Answer: (D). | ||||
| 45 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Ethyl pent-4-ynoate is $\mathrm{HC{\equiv}C{-}CH_2CH_2{-}COOC_2H_5}$.
The ester takes two moles of $\mathrm{CH_3MgBr}$ (ketone, then tertiary alkoxide), so after work-up a $3^\circ$ alcohol, $\mathrm{HC{\equiv}CCH_2CH_2C(OH)(CH_3)_2}$, is formed: Statement I is true.
The terminal alkyne hydrogen is acidic and destroys a third mole of Grignard reagent (giving methane), so one mole of ester uses three moles: Statement II is false.
Answer: (C). | ||||
| 46 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (a) o-Nitrophenol: the OH bonds to the neighbouring $\mathrm{NO_2}$ inside the molecule (intramolecular), so little intermolecular H-bonding.
(b) p-Hydroxyacetanilide (paracetamol): the OH and the N–H/C=O groups point outwards, giving strong intermolecular H-bonds.
(c) The phenolic OH is flanked by two tert-butyl groups; this steric crowding prevents significant intermolecular H-bonding.
Answer: (C). | ||||
| 47 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines $\mathrm{H_2/Pd}$ reduces $\mathrm{NO_2}$ to $\mathrm{NH_2}$: A is p-aminophenol.
With one equivalent of acetic anhydride the more nucleophilic $\mathrm{NH_2}$ is acylated, not the OH: B is p-hydroxyacetanilide (paracetamol).
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 Histamine stimulates the secretion of pepsin and hydrochloric acid in the stomach; antacids such as cimetidine and ranitidine work by blocking its receptors.
(B) is an antidepressant-type hydrazine, (C) a serotonin-like amine and (D) saccharin.
Answer: (A). | ||||
| 49 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Sucrose is $\alpha$-D-glucopyranose linked C1–C2 to $\beta$-D-fructofuranose; hydrolysis gives dextrorotatory D-(+)-glucose and laevorotatory D-(−)-fructose.
Answer: (B). | ||||
| 50 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctHard | |
A hard single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Dilute alkali gives a green gelatinous precipitate. It is usually written $\mathrm{Cr(OH)_3}$, but no definite hydroxide exists: the precipitate is the hydrated oxide $\mathrm{Cr_2O_3\cdot nH_2O}$.
(A) $\mathrm{[Cr(OH)_4]^-}$ (or $\mathrm{[Cr(OH)_6]^{3-}}$) forms only in excess alkali, and it is in solution, not a precipitate, so (D) is wrong too.
Answer: (C). | ||||
| 51 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$. Molar mass of propane $= 44.064$ g/mol.
Propane: $\frac{100}{44.064} = 2.269$ mol; oxygen: $\frac{1000}{32} = 31.25$ mol, in excess.
Products: $\mathrm{CO_2}$ $3\times2.269 = 6.808$ mol; $\mathrm{H_2O}$ $4\times2.269 = 9.078$ mol; $\mathrm{O_2}$ left $31.25 - 11.347 = 19.903$ mol.
Total $= 35.789$ mol; $x_{\mathrm{CO_2}} = \frac{6.808}{35.789} = 0.190 = 19\times10^{-2}$.
Answer: 19. | ||||
| 52 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalHard | |
A hard numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $n_I = \frac{2.8}{28} = 0.1$ mol at 300 K; $n_{II} = \frac{0.2}{28} = 0.00714$ mol at 60 K.
The flasks exchange no heat with the surroundings and do no work, so the internal energy is conserved; with the same gas ($C_v$ the same) the final temperature is
$T = \frac{0.1\times300 + 0.00714\times60}{0.10714} = 284$ K.
$P = \frac{nRT}{V} = \frac{0.10714\times8.31\times284}{3\times10^{-3}} = 8.43\times10^4$ Pa $= 0.843$ bar $= 84\times10^{-2}$ bar.
(Equivalently $P = \frac{R(n_IT_I + n_{II}T_{II})}{V_I+V_{II}}$.)
Answer: 84. | ||||
| 53 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Energy of one photon: $\frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3\times10^8}{10^{-6}} = 1.989\times10^{-19}$ J.
Energy emitted in 0.1 s: $10^{-3}\times0.1 = 10^{-4}$ J.
$N = \frac{10^{-4}}{1.989\times10^{-19}} = 5.03\times10^{14} = 50.3\times10^{13}$, nearest integer 50.
Answer: 50. | ||||
| 54 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{SO_3}$, $\mathrm{NO_3^-}$ and $\mathrm{CO_3^{2-}}$: three bond pairs, no lone pair on the central atom, trigonal planar.
$\mathrm{PCl_3}$: three bond pairs and one lone pair, pyramidal.
Non-pyramidal: 3.
Answer: 3. | ||||
| 55 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta H^\circ = -110.5 - (-266.3) = 155.8$ kJ $\mathrm{mol^{-1}}$.
$\Delta S^\circ = (27.28 + 197.6) - (57.49 + 5.74) = 161.65$ J $\mathrm{mol^{-1}\,K^{-1}}$.
$\Delta G^\circ < 0$ when $T > \frac{155800}{161.65} = 963.8$ K, so the minimum (integer) temperature is 964 K.
Answer: 964. | ||||
| 56 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Molality $= \frac{40/180}{0.200} = 1.111$ mol/kg.
$\Delta T_f = 1.86\times1.111 = 2.07$ K.
$T_f = 273.15 - 2.07 = 271.08$ K, nearest integer 271.
Answer: 271. | ||||
| 57 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)}$. Moles of $\mathrm{NH_4HS} = \frac{5.1}{51} = 0.1$; 20% decomposes, giving 0.02 mol of each gas.
Partial pressure of each: $p = \frac{0.02\times0.082\times300}{2} = 0.246$ atm.
$K_p = p^2 = 0.0605 = 6.05\times10^{-2}$, so $x = 6$.
Answer: 6. | ||||
| 58 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry $\kappa = \frac{G^*}{R} = \frac{1.14}{1500} = 7.6\times10^{-4}$ S $\mathrm{cm^{-1}}$.
$\Lambda_m = \frac{1000\,\kappa}{C} = \frac{1000\times7.6\times10^{-4}}{0.001} = 760$ S $\mathrm{cm^2\,mol^{-1}}$.
Answer: 760. | ||||
| 59 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\log\frac{k_{600}}{k_{700}} = \frac{E_a}{2.303R}\left(\frac{1}{700}-\frac{1}{600}\right) = \frac{209000}{2.303\times8.31}\times(-2.381\times10^{-4}) = -2.60$.
$\log k_{600} = -2.19 - 2.60 = -4.79$, so $k_{600} = 1.62\times10^{-5} = 16.2\times10^{-6}\ \mathrm{s^{-1}}$.
Nearest integer: 16.
Answer: 16. | ||||
| 60 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A tris(bidentate) octahedral complex $\mathrm{[M(AA)_3]}$ is chiral: it exists as a pair of non-superimposable mirror images, the $\Delta$ and $\Lambda$ forms.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $A\cap B$: integer points inside both circles of radius 2 centred at $(2,0)$ and $(0,0)$: $(0,0), (1,0), (1,1), (1,-1), (2,0)$, five points.
$A\cap C$: points inside the circles centred at $(2,0)$ and $(2,2)$: $(1,1), (2,0), (2,1), (2,2), (3,1)$, five points.
The number of relations from a 5-element set to a 5-element set is $2^{5\times5} = 2^{25}$, so $p = 25$.
Answer: (C). | ||||
| 62 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctHard | |
A hard single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $a = 3x^2+4x+3$, $b = 3x^2+4x+2$. The equation is $a^2 - (k+1)ab + kb^2 = (a-b)(a-kb) = 0$.
$a - b = 1 \neq 0$, so $a = kb$: $(3-3k)x^2 + (4-4k)x + (3-2k) = 0$. For $k = 1$ this reads $1 = 0$, impossible.
For $k \neq 1$: $3x^2 + 4x + \frac{3-2k}{1-k} = 0$, real roots when $16 - \frac{12(3-2k)}{1-k} \ge 0$, i.e. $\frac{3(3-2k)}{1-k} \le 4$.
If $k < 1$: $9 - 6k \le 4 - 4k$ gives $k \ge \frac{5}{2}$, a contradiction. If $k > 1$: $9 - 6k \ge 4 - 4k$ gives $k \le \frac{5}{2}$.
So $k \in \left(1, \frac{5}{2}\right]$.
Answer: (A). | ||||
| 63 | 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 $\Delta = \begin{vmatrix}1&1&1\\3&2&5\\9&4&28+[\lambda]\end{vmatrix} = (56 + 2[\lambda] - 20) - (84 + 3[\lambda] - 45) + (12 - 18) = -9 - [\lambda]$.
If $[\lambda] \neq -9$, $\Delta \neq 0$ and there is a unique solution.
If $[\lambda] = -9$ the third equation is $9x + 4y + 19z = -9$. Since $-6(x+y+z) + 5(3x+2y+5z) = 9x + 4y + 19z$ and $-6(4) + 5(3) = -9$, it is the same combination of the first two: the system is consistent, with infinitely many solutions.
So the system has a solution for every real $\lambda$.
Answer: (C). | ||||
| 64 | 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 Let $n = [x]$; then $[x+k] = n + k$ and $A = \begin{pmatrix}n+1&n+2&n+3\\ n&n+3&n+3\\ n&n+2&n+4\end{pmatrix}$.
$R_1 \to R_1 - R_3$, $R_2 \to R_2 - R_3$: rows $(1, 0, -1)$ and $(0, 1, -1)$. Expanding:
$\det A = 1\cdot\left(1\cdot(n+4) - (-1)(n+2)\right) - 1\cdot\left(0\cdot(n+2) - 1\cdot n\right) = (2n+6) + n = 3n + 6$.
$3n + 6 = 192$ gives $n = 62$, so $62 \le x < 63$.
Answer: (C). | ||||
| 65 | 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 $y = \sum_{n\ge1}\frac{n}{n+1}x^{n+1} = \sum_{m\ge2}\left(1 - \frac{1}{m}\right)x^m = \frac{x^2}{1-x} - \left(-\ln(1-x) - x\right)$.
So $y = \frac{x^2}{1-x} + x + \ln(1-x)$. At $x = \frac{1}{2}$: $y = \frac{1}{2} + \frac{1}{2} - \ln 2 = 1 - \ln 2$.
$e^{1+y} = e^{2-\ln2} = \frac{e^2}{2}$.
Answer: (D). | ||||
| 66 | 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 For a finite limit the $x$ terms must cancel: $\sqrt{x^2-x+1} \sim x$, so $a = 1$.
$\sqrt{x^2-x+1} - x = \frac{-x+1}{\sqrt{x^2-x+1}+x} \to \frac{-1}{1+1} = -\frac{1}{2}$.
$(a, b) = \left(1, -\frac{1}{2}\right)$.
Answer: (B). | ||||
| 67 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $x \in \left(\frac{\pi}{2}, \pi\right)$, $\frac{x}{2} \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, so $\sin\frac{x}{2} > \cos\frac{x}{2} > 0$.
$\sqrt{1+\sin x} = \sin\frac{x}{2} + \cos\frac{x}{2}$ and $\sqrt{1-\sin x} = \left|\cos\frac{x}{2} - \sin\frac{x}{2}\right| = \sin\frac{x}{2} - \cos\frac{x}{2}$.
The fraction is $\frac{2\sin\frac{x}{2}}{2\cos\frac{x}{2}} = \tan\frac{x}{2} = \cot\left(\frac{\pi}{2} - \frac{x}{2}\right)$, with $\frac{\pi}{2} - \frac{x}{2} \in \left(0, \frac{\pi}{4}\right)$.
So $y = \frac{\pi}{2} - \frac{x}{2}$ and $\frac{dy}{dx} = -\frac{1}{2}$ everywhere in the interval, including $x = \frac{5\pi}{6}$.
Answer: (B). | ||||
| 68 | 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 $\sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$ ranges over $[1, \sqrt{2}]$ on $\left[0, \frac{\pi}{2}\right]$, and $\tan^{-1}$ is increasing.
$M = \tan^{-1}\sqrt{2}$, $m = \tan^{-1}1$.
$\tan(M - m) = \frac{\sqrt{2} - 1}{1 + \sqrt{2}} = (\sqrt{2}-1)^2 = 3 - 2\sqrt{2}$.
Answer: (C). | ||||
| 69 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $V = x(a-2x)(b-2x) = 4x^3 - 2(a+b)x^2 + abx$.
$V' = 12x^2 - 4(a+b)x + ab = 0$ gives $x = \frac{(a+b) \pm \sqrt{(a+b)^2 - 3ab}}{6} = \frac{a+b \pm \sqrt{a^2+b^2-ab}}{6}$.
The larger root exceeds $\frac{\min(a,b)}{2}$ and is not allowed; at the smaller root $V'' = 24x - 4(a+b) < 0$, a maximum.
$x = \frac{a+b-\sqrt{a^2+b^2-ab}}{6}$.
Answer: (A). | ||||
| 70 | Integral CalculusPartial fractions | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on partial fractions, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $x = t^2$: $I = \int_0^1\frac{2t^2\,dt}{(1+t^2)(1+3t^2)(3+t^2)}$.
With $u = t^2$: $\frac{2u}{(1+u)(1+3u)(3+u)} = \frac{1/2}{1+u} - \frac{3/8}{1+3u} - \frac{3/8}{3+u}$.
$I = \frac{1}{2}\cdot\frac{\pi}{4} - \frac{3}{8}\cdot\frac{1}{\sqrt{3}}\tan^{-1}\sqrt{3} - \frac{3}{8}\cdot\frac{1}{\sqrt{3}}\tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{8} - \frac{3}{8\sqrt{3}}\left(\frac{\pi}{3} + \frac{\pi}{6}\right)$.
$I = \frac{\pi}{8} - \frac{\sqrt{3}\pi}{16} = \frac{\pi}{8}\left(1 - \frac{\sqrt{3}}{2}\right)$.
Answer: (A). | ||||
| 71 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals At $y = 3$: $x = 2$, the point $(2, 3)$. Differentiating, $2(y-2)\frac{dy}{dx} = 1$, slope $\frac{1}{2}$; tangent $x = 2y - 4$, meeting the $x$-axis at $(-4, 0)$.
The region lies between the tangent (left) and the parabola (right) for $0 \le y \le 3$:
Area $= \int_0^3\left[(y-2)^2 + 1 - (2y - 4)\right]dy = \int_0^3 (y-3)^2\,dy = 9$.
Answer: (B). | ||||
| 72 | Differential EquationsOrder and degree | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on order and degree, from Differential Equations. NCERT: Class 12: Differential Equations Latus rectum $= \frac{|3(2) + 4(-3) - 5|}{5} = \frac{11}{5}$.
Parabolas with axis parallel to the $y$-axis: $(x - h)^2 = \frac{11}{5}(y - k)$.
Differentiate: $2(x-h) = \frac{11}{5}\frac{dy}{dx}$; again: $2 = \frac{11}{5}\frac{d^2y}{dx^2}$, i.e. $11\frac{d^2y}{dx^2} = 10$.
Answer: (C). | ||||
| 73 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Treat $x$ as a function of $y$: $\frac{dx}{dy} + \frac{2}{y}x = 10y^2$. Integrating factor $y^2$.
$xy^2 = \int10y^4\,dy = 2y^5 + c$. Through $(0, 1)$: $c = -2$, so $xy^2 = 2y^5 - 2$.
At $(2, \beta)$: $2\beta^2 = 2\beta^5 - 2$, i.e. $\beta^5 - \beta^2 - 1 = 0$.
Answer: (C). | ||||
| 74 | Matrices and DeterminantsArea of triangles using determinants | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on area of triangles using determinants, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Area $= \frac{1}{2}\left|a\left((2b+1) - b\right) + b(b - 0) + 0\right| = \frac{1}{2}\left|a(b+1) + b^2\right| = 1$.
$a(b+1) + b^2 = \pm2$, so $a = \frac{2 - b^2}{b+1}$ or $a = \frac{-2 - b^2}{b+1}$.
Sum $= \frac{-2b^2}{b+1}$.
Answer: (D). | ||||
| 75 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Perpendicular tangents to a parabola meet on its directrix.
$y^2 = 16(x-3)$ has vertex $(3, 0)$ and $4a = 16$, $a = 4$; the directrix is $x = 3 - 4 = -1$.
Locus: $x + 1 = 0$.
Answer: (A). | ||||
| 76 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $n = 2l + 2m$. Then $n(l + m) + lm = 2(l+m)^2 + lm = 2l^2 + 5lm + 2m^2 = (2l + m)(l + 2m) = 0$.
$m = -2l$: $n = -2l$, direction ratios $(1, -2, -2)$. $l = -2m$: $n = -2m$, direction ratios $(-2, 1, -2)$.
Dot product: $-2 - 2 + 4 = 0$, so the lines are perpendicular: $\frac{\pi}{2}$.
Answer: (C). | ||||
| 77 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Plane through the line: $(x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0$.
Parallel to the $x$-axis: coefficient of $x$ is zero, $1 + 2\lambda = 0$, $\lambda = -\frac{1}{2}$.
$2(x+y+z-1) - (2x+3y-z+4) = -y + 3z - 6 = 0$, i.e. $y - 3z + 6 = 0$: $\vec{r}\cdot(\hat{j}-3\hat{k}) + 6 = 0$.
Answer: (D). | ||||
| 78 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability For one person $P(k\ \text{heads}) = \frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \frac{1}{8}$ for $k = 0, 1, 2, 3$.
$P(\text{same}) = \frac{1 + 9 + 9 + 1}{64} = \frac{20}{64} = \frac{5}{16}$.
Answer: (C). | ||||
| 79 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $(p\wedge q) \Rightarrow (p\wedge q\wedge r) \equiv \sim(p\wedge q) \vee (p\wedge q\wedge r) \equiv \sim(p\wedge q) \vee r \equiv \sim p \vee \sim q \vee r$.
(B): $\sim(p\wedge q) \vee (r\wedge q) \equiv \sim p \vee \left(\sim q \vee (r\wedge q)\right) \equiv \sim p \vee \sim q \vee r$. The same.
(C) is a tautology; (A) and (D) are false when, for example, $q, r$ true and $p$ false (A), or $p, r$ true and $q$ false (D), while the given expression is then true.
Answer: (B). | ||||
| 80 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Take $B = (0, 0)$, $A = (0, a)$, $D = (x, 0)$, $C = (x, a+b)$.
From C the angle of depression to A has $\tan\alpha = \frac{b}{x}$ and to B has $\tan\beta = \frac{a+b}{x}$; $\angle ACB = \beta - \alpha$.
$\tan(\beta - \alpha) = \frac{\frac{a+b}{x} - \frac{b}{x}}{1 + \frac{b(a+b)}{x^2}} = \frac{ax}{x^2 + b(a+b)} = \frac{1}{2}$.
$x^2 - 2ax + b(a+b) = 0$.
Answer: (C). | ||||
| 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 With $z = x + iy$: $(x-3)^2 + y^2 = x^2$, i.e. $y^2 = 6x - 9$, a parabola.
For $z_1, z_2$ on it: $y_1^2 - y_2^2 = 6(x_1 - x_2)$. Since $\arg(z_1 - z_2) = \frac{\pi}{4}$, $y_1 - y_2 = x_1 - x_2 \neq 0$.
Dividing: $y_1 + y_2 = 6$.
Answer: 6. | ||||
| 82 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Residues mod 3: $\{3, 6, 9\} \equiv 0$, $\{1, 4\} \equiv 1$, $\{2, 5\} \equiv 2$.
A subset's sum is a multiple of 3 when, from $\{1,4\}$ and $\{2,5\}$, it takes $i$ and $j$ elements with $i + 2j \equiv 0 \pmod 3$: $(0,0)$: 1 way, $(1,1)$: $2\times2 = 4$ ways, $(2,2)$: 1 way, 6 in all; any of the $2^3 = 8$ subsets of $\{3,6,9\}$ can be added.
Subsets (including the empty set) with sum a multiple of 3: $6\times8 = 48$.
Non-empty subsets whose sum is not a multiple of 3: $2^7 - 48 = 80$ (the empty set has sum 0 and is among the 48).
Answer: 80. | ||||
| 83 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $7^3 = 343 = 18\times19 + 1$, so $7^{21} \equiv 1$ and $7^{22} \equiv 7 \pmod{18}$: $3\times7^{22} \equiv 21$.
$10^2 = 100 \equiv 10 \pmod{18}$, so $10^n \equiv 10$ for every $n \ge 1$: $2\times10^{22} \equiv 20$.
Total $\equiv 21 + 20 - 44 = -3 \equiv 15 \pmod{18}$.
Answer: 15. | ||||
| 84 | Integral CalculusIntegral as an antiderivative | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Write $2e^x + 3e^{-x} = \lambda(4e^x + 7e^{-x}) + \mu(4e^x - 7e^{-x})$: $4\lambda + 4\mu = 2$, $7\lambda - 7\mu = 3$.
$\lambda + \mu = \frac{1}{2}$, $\lambda - \mu = \frac{3}{7}$, so $\lambda = \frac{13}{28}$, $\mu = \frac{1}{28}$.
The integral is $\frac{13}{28}x + \frac{1}{28}\ln(4e^x + 7e^{-x}) + C = \frac{1}{14}\left(\frac{13}{2}x + \frac{1}{2}\ln(4e^x + 7e^{-x})\right) + C$.
$u + v = \frac{13}{2} + \frac{1}{2} = 7$.
Answer: 7. | ||||
| 85 | CirclesRadius and centre | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections The centres lie on the normal to the tangent at $(1, 2)$, at distance 5 on either side. The unit normal is $\left(\frac{4}{5}, \frac{3}{5}\right)$.
Centres: $(1 + 4, 2 + 3) = (5, 5)$ and $(1 - 4, 2 - 3) = (-3, -1)$.
$|(5 + 5)(-3 - 1)| = |10\times(-4)| = 40$.
Answer: 40. | ||||
| 86 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections As printed, $(\sec\theta, 2\tan\theta)$ is not on $2x^2 - y^2 = 2$ (it gives $2 - 2\tan^2\theta = 2$). The hyperbola $\frac{x^2}{1} - \frac{y^2}{2} = 1$ has parametric points $(\sec\theta, \sqrt{2}\tan\theta)$; that is taken as intended.
Normal to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(a\sec\theta, b\tan\theta)$: $ax\cos\theta + by\cot\theta = a^2 + b^2$. With $a = 1$, $b = \sqrt{2}$: $x\cos\theta + \sqrt{2}y\cot\theta = 3$.
At B, $\phi = \frac{\pi}{2} - \theta$: $x\sin\theta + \sqrt{2}y\tan\theta = 3$.
Multiply the first by $\sin\theta$ and the second by $\cos\theta$ and subtract: $\sqrt{2}y(\cos\theta - \sin\theta) = 3(\sin\theta - \cos\theta)$, so $\beta = -\frac{3}{\sqrt{2}}$ (for $\theta \neq \frac{\pi}{4}$).
$(2\beta)^2 = 4\times\frac{9}{2} = 18$.
If instead the printed points are kept and the hyperbola is read as $4x^2 - y^2 = 4$ ($a = 1$, $b = 2$), the normals $x\cos\theta + 2y\cot\theta = 5$ and $x\sin\theta + 2y\tan\theta = 5$ meet at $\beta = -\frac{5}{2}$, and $(2\beta)^2 = 25$.
Answer: 18 or 25. | ||||
| 87 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical 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 $\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = -2\cdot\frac{2 - 3 + 4 + 3}{4 + 1 + 1} = -2$, so $S = (-3, 5, 2)$.
R on the plane: $6 - 5 + \gamma + 3 = 0$, $\gamma = -4$, $R = (3, 5, -4)$.
$SR^2 = 6^2 + 0^2 + 6^2 = 72$.
Answer: 72. | ||||
| 88 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $K + 2K + 2K + 3K + K = 9K = 1$, $K = \frac{1}{9}$.
$p = \frac{P(1 < X < 4 \text{ and } X < 3)}{P(X < 3)} = \frac{P(X = 2)}{P(X = 1) + P(X = 2)} = \frac{2K}{3K} = \frac{2}{3}$.
$5p = \frac{10}{3} = \lambda\cdot\frac{1}{9}$, so $\lambda = 30$.
Answer: 30. | ||||
| 89 | 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 $\sin^4\theta + \cos^4\theta = 1 - \frac{1}{2}\sin^2 2\theta$, so the equation is $1 - \frac{1}{2}\sin^2 2\theta - \frac{1}{2}\sin 2\theta = 0$, i.e. $\sin^2 2\theta + \sin 2\theta - 2 = 0$.
$(\sin 2\theta + 2)(\sin 2\theta - 1) = 0$, so $\sin 2\theta = 1$, $\theta = \frac{\pi}{4} + n\pi$.
In $[0, 4\pi]$: $\frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}$, sum $S = 7\pi$.
$\frac{8S}{\pi} = 56$.
Answer: 56. | ||||
| 90 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $20\times12 + 30\mu = 50\times15$ gives $\mu = 17$.
Sum of squares: boys $20(2 + 144) = 2920$; girls $30(2 + 289) = 8730$; total $11650$.
$\sigma^2 = \frac{11650}{50} - 15^2 = 233 - 225 = 8$.
$\mu + \sigma^2 = 17 + 8 = 25$.
Answer: 25. | ||||
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