JEE (Main) · 2021
JEE (Main) 2021 — 3 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
- Atoms and Nuclei4
- Oscillations and Waves3
- Gravitation3
24 easy · 6 medium · 0 hard
Chemistry
30 Q
- d- and f-Block Elements4
- Haloalkanes and Haloarenes2
- Organic Chemistry: Some Basic Principles and Techniques2
16 easy · 13 medium · 1 hard
Mathematics
30 Q
- Matrices and Determinants3
- Integral Calculus3
- Quadratic Equations2
15 easy · 15 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 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Starting at the mean position, $x = A\sin\omega t$.
At $x = \frac{A}{\sqrt{2}}$: $\sin\omega t = \frac{1}{\sqrt{2}}$, so $\omega t = \frac{\pi}{4}$.
$t = \frac{\pi}{4}\cdot\frac{T}{2\pi} = \frac{T}{8}$.
Answer: (A). | ||||
| 2 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The block is released from rest at P, a distance A from the equilibrium point O, so P is an extreme position and the amplitude is A. Q is $\frac{A}{2}$ beyond O.
P to O is a quarter oscillation: $\frac{T}{4}$.
O to Q: $\frac{A}{2} = A\sin\omega t$ gives $\omega t = \frac{\pi}{6}$, i.e. $t = \frac{T}{12}$.
Total time $= \frac{T}{4} + \frac{T}{12} = \frac{T}{3} = \frac{1}{3}\cdot 2\pi\sqrt{\frac{m}{K}} = \frac{2\pi}{3}\sqrt{\frac{m}{K}}$.
Answer: (B). | ||||
| 3 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Heat supplied $Q = 100 \times 30 = 3000$ J, all at constant volume (sealed chamber, thermally isolated).
Argon is monatomic: $C_V = \frac{3}{2}R$.
$\Delta T = \frac{Q}{nC_V} = \frac{3000}{1 \times 1.5 \times 8.314} = \frac{3000}{12.471} \approx 240.5$ K.
Answer: (B). | ||||
| 4 | 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 rod lies along a tangent to the sphere, so its axis is at distance $R$ from the sphere's centre.
Parallel axes: $I_{\text{sphere}} = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2 = 1.4\,MR^2$.
The rod's own moment of inertia about its own axis is negligible because its diameter is negligible.
Total $= 1.4\,MR^2$.
Answer: (B). | ||||
| 5 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The disc hangs horizontally and twists about the vertical wire through its centre, so $I = \frac{1}{2}MR^2 = \frac{1}{2}(0.1)(0.04)^2 = 8\times10^{-5}\ \mathrm{kg\,m^2}$.
$T = 2\pi\sqrt{\frac{I}{k}}$ gives $k = \frac{4\pi^2 I}{T^2} = \frac{40 \times 8\times10^{-5}}{0.04} = 8\times10^{-2}\ \mathrm{kg\,m^2\,s^{-2}}$.
Answer: (B). | ||||
| 6 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation At the mid-point each fixed mass is $\frac{l}{2}$ away. For a particle of mass $\mu$ the potential energy there is $U = -2\cdot\frac{Gm\mu}{l/2} = -\frac{4Gm\mu}{l}$.
To escape, $\frac{1}{2}\mu v^2 + U = 0$: $v^2 = \frac{8Gm}{l}$, so $v = \sqrt{\frac{8Gm}{l}} = 2\sqrt{\frac{2Gm}{l}}$.
Answer: (A). | ||||
| 7 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Let friction $f$ act backwards at the contact point.
Translation: $F - f = Ma$. Rotation about the centre: $fR = \frac{2}{5}MR^2\alpha$, and rolling gives $a = R\alpha$, so $f = \frac{2}{5}Ma$.
$F = Ma + \frac{2}{5}Ma = \frac{7}{5}Ma$, so $a = \frac{5F}{7M}$.
Answer: (D). | ||||
| 8 | 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 $x = 3t$ and $y = 4t - 5t^2$. Put $t = \frac{x}{3}$: $y = \frac{4x}{3} - \frac{5x^2}{9}$.
Multiply by 9: $9y = 12x - 5x^2$.
Answer: (A). | ||||
| 9 | 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 $V$ is positive for $0 < t < 2$ and negative for $2 < t < 4$ (it reverses at $t = 2$ s).
Displacement $= \int_0^4 V_0(2-t)\,dt = V_0\left[2t - \frac{t^2}{2}\right]_0^4 = V_0(8 - 8) = 0$.
Distance $= \int_0^2 V_0(2-t)\,dt + \left|\int_2^4 V_0(2-t)\,dt\right| = 2V_0 + 2V_0 = 4V_0$.
Answer: (A). | ||||
| 10 | 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 Young's modulus is the slope of the stress–strain line, i.e. the tangent of its angle with the strain axis.
$Y_A : Y_B = \tan 60^\circ : \tan 30^\circ = \sqrt{3} : \frac{1}{\sqrt{3}} = 3 : 1$.
Answer: (C). | ||||
| 11 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Distance of A from the origin: $\sqrt{2+2} = 2$. Distance of B: 2.
Both points are at the same distance from the point charge, so $V_A = V_B = \frac{kq}{2}$ and $V_A - V_B = 0$.
Answer: (B). | ||||
| 12 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The nucleus has $Z$ protons and $A - Z$ neutrons. The mass defect is the mass of the separate nucleons minus the mass of the nucleus:
$\Delta M = [Zm_p + (A-Z)m_n] - M$.
Answer: (D). | ||||
| 13 | Magnetic Effects of Current and MagnetismEarth magnetism | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on earth magnetism, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter A freely suspended needle settles along the earth's field, so the angle of dip is $45^\circ$ and $B_V = B_H\tan45^\circ = 20\ \mu T$.
Held horizontal, only $B_V$ gives a torque about the suspension point: $\tau = MB_V = (q_m l)B_V = 2.4 \times 0.12 \times 20\times10^{-6} = 5.76\times10^{-6}$ N m.
A vertical force $F$ at one end acts at $\frac{l}{2} = 0.06$ m from the middle: $F \times 0.06 = 5.76\times10^{-6}$, so $F = 9.6\times10^{-5}$ N.
Answer: (D). | ||||
| 14 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Since $b \gg a$, the outer coil's field is uniform over the inner coil: $B = \frac{\mu_0 i}{2b}$ at the centre.
Flux through the inner coil: $\phi = \frac{\mu_0 i}{2b}\cdot\pi a^2$.
Charge circulated $q = \frac{\Delta\phi}{R} = \frac{\mu_0 i\pi a^2}{2Rb}$.
Answer: (D). | ||||
| 15 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a simple microscope with the final image at infinity, $m = \frac{D}{f}$.
$5 = \frac{25}{f}$ gives $f = 5$ cm $= 0.05$ m, so $P = \frac{1}{0.05} = 20$ D.
Answer: (C). | ||||
| 16 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The man (long-sighted) should see an object at the normal near point, 25 cm, through the lens as an image at his own near point, 50 cm, on the same side.
$u = -25$ cm, $v = -50$ cm: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = -\frac{1}{50} + \frac{1}{25} = \frac{1}{50}$.
$f = +50$ cm (a convex lens).
Answer: (B). | ||||
| 17 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Each gate is an OR gate with both inputs inverted, which is a NAND gate ($\bar{X} + \bar{Y} = \overline{XY}$).
$Y_1 = \overline{AB}$, $Y_2 = \overline{A\,Y_1}$, $Y_3 = \overline{Y_1 B}$, $Y = \overline{Y_2 Y_3} = A\,Y_1 + Y_1 B = (A + B)\overline{AB}$.
$(A+B)(\bar{A}+\bar{B}) = A\bar{B} + \bar{A}B$, the exclusive-OR.
Truth table: 00 → 0, 01 → 1, 10 → 1, 11 → 0.
Answer: (B). | ||||
| 18 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $6.25\% = \frac{1}{16} = \left(\frac{1}{2}\right)^4$: four half-lives.
Time $= 4T$ years.
Answer: (B). | ||||
| 19 | Communication SystemsPropagation of electromagnetic waves | No matching NCERT chapter | Assertion and reasonMedium | |
A medium assertion and reason question on propagation of electromagnetic waves, from Communication Systems. NCERT: No matching NCERT chapter A: TV signals (VHF/UHF) are above the critical frequency, so the ionosphere does not reflect them back; they travel as space waves and are received only within the line of sight. A is true.
R: radio-broadcast waves below the critical frequency are reflected by the ionosphere (sky waves) and reach distant receiving antennas. R is true.
R explains why lower-frequency radio reaches beyond the horizon; it does not explain why TV signals cannot, which depends on their frequency being above the critical value. So R is not the correct explanation of A.
Answer: (B). | ||||
| 20 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Along the incline (pull parallel to it): $F - mg\sin30^\circ = ma$, so $F = 4(9.8 \times 0.5 + 1) = 4 \times 5.9 = 23.6$ N.
After 3.0 s from rest, $v = at = 3.0$ m/s.
$P = Fv = 23.6 \times 3.0 = 70.8$ W.
Answer: (A). | ||||
| 21 | 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 A 120 m length of beam takes $t = \frac{120}{3\times10^8} = 4\times10^{-7}$ s to pass a point.
Energy $= Pt = 4\times10^{-3} \times 4\times10^{-7} = 1.6\times10^{-9}$ J $= 16\times10^{-10}$ J.
Answer: 16. | ||||
| 22 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $e = -\frac{d\phi}{dt} = -(15t^2 - 100)$.
At $t = 2$ s: $e = -(60 - 100) = 40$ V.
Answer: 40. | ||||
| 23 | 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 Brackett series: transitions to $n = 4$.
Shortest wavelength ($\infty \to 4$): $\frac{1}{\lambda_{\min}} = \frac{R}{16}$.
Longest wavelength ($5 \to 4$): $\frac{1}{\lambda_{\max}} = R\left(\frac{1}{16} - \frac{1}{25}\right) = \frac{9R}{400}$.
$\frac{\lambda_{\min}}{\lambda_{\max}} = \frac{9R/400}{R/16} = \frac{9}{25}$, so $x = 9$.
Answer: 9. | ||||
| 24 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Take the moment the train starts, with the man 5 m behind it. The man covers $6t$; the train $\frac{1}{2}(2)t^2 = t^2$.
He catches it when $6t = 5 + t^2$, i.e. $t^2 - 6t + 5 = 0$, $t = 1$ s or $5$ s.
He first reaches the train at $t = 1$ s.
Answer: 1. | ||||
| 25 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{1}{16} = \left(\frac{1}{2}\right)^4$: four half-lives in 48 years.
Half-life $= \frac{48}{4} = 12$ years.
Answer: 12. | ||||
| 26 | GravitationOrbital motion | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation The two particles stay diametrically opposite, a distance $2R$ apart.
$\frac{Gm^2}{(2R)^2} = \frac{mv^2}{R}$ gives $v^2 = \frac{Gm}{4R}$, so $v = \sqrt{\frac{1}{4}\cdot\frac{Gm}{R}}$.
$x = 4$.
Answer: 4. | ||||
| 27 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The fringe width $\beta = \frac{\lambda D}{d}$ is proportional to $\lambda$; the field of view is fixed, so $N\lambda$ is constant.
$N_2 = 58 \times \frac{6000}{4000} = 87$.
Answer: 87. | ||||
| 28 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Let the junction be at potential $V$. With S closed the 2 $\Omega$ branch goes to earth (0 V).
Kirchhoff's current law at the junction: $\frac{10 - V}{4} + \frac{5 - V}{2} = \frac{V - 0}{2}$.
Multiply by 4: $10 - V + 10 - 2V = 2V$, so $V = 4$ V.
Current through S $= \frac{4}{2} = 2$ A.
Answer: 2. | ||||
| 29 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $E_x = -\frac{dV}{dx} = \frac{1000}{x^2} + \frac{3000}{x^3} + \frac{1500}{x^4}$.
At $x = 1$ m: $E = 1000 + 3000 + 1500 = 5500$ V/m along $\hat{i}$.
Answer: 5500. | ||||
| 30 | GravitationAcceleration due to gravity | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At latitude $\lambda$ the effective gravity is $g' = g - \omega^2R\cos^2\lambda$.
If the earth stops, the weight rises by $m\omega^2R\cos^2\lambda = 20 \times \omega^2 R \times \frac{1}{2} = 10R\omega^2$.
$x = 10$.
Answer: 10. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | StatementsMedium | |
A medium statements question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mathrm{Cr_2O_7^{2-}}$ is two $\mathrm{CrO_4}$ tetrahedra sharing one oxygen. The six terminal Cr–O bonds are equal (about 163 pm), but the two bridging Cr–O bonds are longer (about 179 pm). Statement I is false.
The Cr–O–Cr bond angle is about $126^\circ$, larger than the H–O–H angle in water ($104.5^\circ$). Statement II is false.
Answer: (B). | ||||
| 32 | Solutionsvan't Hoff factor | Class 12: Solutions | Single correctEasy | |
A easy single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = iK_fm$; the largest $i\times m$ gives the lowest freezing point (500 mL water taken as 0.5 kg).
NaCl: $\frac{5.85}{58.5} = 0.1$ mol, $m = 0.2$, $i = 2$: $im = 0.4$.
Urea: $\frac{6}{60} = 0.1$ mol, $m = 0.2$, $i = 1$: $0.2$.
Glucose: $\frac{18}{180} = 0.1$ mol, $m = 0.2$, $i = 1$: $0.2$.
$\mathrm{MgCl_2}$: $\frac{9.5}{95} = 0.1$ mol, $m = 0.2$, $i = 3$: $0.6$.
$\mathrm{MgCl_2}$ lowers the freezing point most.
Answer: (D). | ||||
| 33 | 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 $\mathrm{Ca(OH)_2}$, an alkaline earth hydroxide of a large dipositive ion, is the strongest base here.
Among the lanthanoid hydroxides basic character falls with the lanthanoid contraction: $\mathrm{Ce(OH)_3 > Lu(OH)_3}$.
$\mathrm{Al(OH)_3}$ is amphoteric, the least basic (the small, highly charged $\mathrm{Al^{3+}}$ polarises the O–H bond most).
Order: $\mathrm{Ca(OH)_2 > Ce(OH)_3 > Lu(OH)_3 > Al(OH)_3}$.
Answer: (C). | ||||
| 34 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In the electrolytic reduction of alumina, cryolite ($\mathrm{Na_3AlF_6}$) and some $\mathrm{CaF_2}$ are added. A is true.
Pure alumina is a poor conductor (and melts above 2000 °C). The fluorides lower the melting point to about 1200 K and raise the conductivity of the melt. R is true, and it is the reason the fluoride is added.
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: first ionisation enthalpies are Li 520, H 1312, F 1681 kJ mol$^{-1}$, so Li < H < F. Correct.
B: of protium, deuterium and tritium only tritium is radioactive. Incorrect.
C: halogens are far more reactive than hydrogen (H–H bond enthalpy is high). Correct.
D: the bare proton has a radius of about $1.5\times10^{-3}$ pm, far below 0.1 pm. Correct.
Only B is incorrect.
Answer: (B). | ||||
| 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 X = $\mathrm{CaCO_3}$: $\mathrm{CaCO_3 \xrightarrow{\Delta} CaO + CO_2}$, so Y = CaO and Z = $\mathrm{CO_2}$.
B = $\mathrm{Ca(OH)_2}$: $\mathrm{Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O}$ forms X.
Y gives X and B: $\mathrm{CaO + CO_2 \rightarrow CaCO_3}$ and $\mathrm{CaO + H_2O \rightarrow Ca(OH)_2}$.
Answer: (B). | ||||
| 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 On hydrolysis $\mathrm{(CH_3)_3SiCl}$ gives $\mathrm{(CH_3)_3SiOH}$, which has only one OH and so can condense at one end only: it caps the chain.
$\mathrm{(CH_3)_2SiCl_2}$ gives the chain (two OH), $\mathrm{CH_3SiCl_3}$ gives cross-links (three OH), and $\mathrm{Si(CH_3)_4}$ has no hydrolysable Cl.
Answer: (B). | ||||
| 38 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Colour needs unpaired d electrons for d–d transitions.
$\mathrm{V^{3+}}$ $3d^2$, $\mathrm{Mn^{3+}}$ $3d^4$, $\mathrm{Cu^{2+}}$ $3d^9$: all coloured.
$\mathrm{Sc^{3+}}$ and $\mathrm{Ti^{4+}}$ are $3d^0$: colourless. Every other set contains one of these.
Answer: (C). | ||||
| 39 | 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 The +3 state is the most stable for lanthanoids.
(A) $\mathrm{Ce^{4+}}$ readily returns to $\mathrm{Ce^{3+}}$ (it is a strong oxidant), so it is not more stable. False.
(B) $\mathrm{Eu^{2+}}$ ($4f^7$) exists but is readily oxidised to $\mathrm{Eu^{3+}}$; $\mathrm{Eu^{3+}}$ is the more stable. False.
(C) $\mathrm{Ce^{4+}}$ is an oxidant ($E^\circ \approx +1.74$ V) and $\mathrm{Eu^{2+}}$ a reducing agent. True.
(D) $\mathrm{La^{3+}}$ is $4f^0$: colourless and diamagnetic. False.
Answer: (C). | ||||
| 40 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[Fe(NCS)(NH_3)_5]^{2+}}$ (sulfate is $-2$): $x - 1 + 0 = +2$, so $x = +3$.
$\mathrm{[Fe(S_2O_3)_3]^{3-}}$: $x + 3(-2) = -3$, so $x = +3$.
$\mathrm{[Fe(CO)_5]}$: CO is neutral, so $x = 0$.
Answer: (B). | ||||
| 41 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter A is true: fluoride above about 2 ppm causes brown mottling of teeth (and higher levels harm bones).
R is true: fluoride (about 1 ppm) converts hydroxyapatite, $\mathrm{[3Ca_3(PO_4)_2\cdot Ca(OH)_2]}$, into the harder fluorapatite, $\mathrm{[3Ca_3(PO_4)_2\cdot CaF_2]}$.
R describes the beneficial effect of fluoride at low levels; it does not explain the mottling caused by excess fluoride.
Answer: (B). | ||||
| 42 | Practical Organic ChemistryDetection of elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on detection of elements, from Practical Organic Chemistry. NCERT: No matching NCERT chapter The sodium fusion extract contains NaCN. With $\mathrm{FeSO_4}$: $\mathrm{6CN^- + Fe^{2+} \rightarrow [Fe(CN)_6]^{4-}}$.
On boiling and acidifying, some $\mathrm{Fe^{2+}}$ is oxidised to $\mathrm{Fe^{3+}}$, which gives Prussian blue: $\mathrm{3[Fe(CN)_6]^{4-} + 4Fe^{3+} \rightarrow Fe_4[Fe(CN)_6]_3\cdot xH_2O}$.
Answer: (C). | ||||
| 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 Hydrolysis of a haloarene by $\mathrm{OH^-}$ goes by nucleophilic aromatic substitution (addition–elimination). Nitro groups ortho and para to Cl stabilise the negatively charged intermediate and speed the reaction.
D (2,4,6-trinitro) > C (2,4-dinitro) > B (4-nitro) > A (chlorobenzene, which needs about 623 K and 300 atm).
Answer: (C). | ||||
| 44 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques 2,4-Dimethylhept-3-ene: $\mathrm{CH_3-CH(CH_3)-CH=C(CH_3)-CH_2CH_2CH_3}$. C3 carries H and the isopropyl group (C1, C2); C4 carries $\mathrm{CH_3}$ and the propyl group (C5–C7).
cis means the main-chain parts, isopropyl on C3 and propyl on C4, are on the same side of the double bond.
(A) has isopropyl across from propyl: trans. (B) has isopropyl and propyl on the same side: cis. (C) puts $\mathrm{CH_3}$ and isopropyl on the same carbon (a different compound), and (D) is a different skeleton.
Answer: (B). | ||||
| 45 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $\mathrm{S_N1}$ rate follows carbocation stability.
B (1-chlorocyclohex-1-ene) is a vinylic chloride: a vinyl cation is very unstable and the C–Cl bond has partial double-bond character. Least reactive.
A (chlorocyclohexane) gives a plain secondary cation.
D (3-chlorocyclohex-1-ene) gives a secondary allylic, resonance-stabilised cation.
C (3-chloro-3-methylcyclohex-1-ene) gives a tertiary allylic cation, the most stable.
B < A < D < C.
Answer: (B). | ||||
| 46 | 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 (A) Friedel–Crafts methylation gives toluene; the Etard reaction ($\mathrm{CrO_2Cl_2}$, then hydrolysis) gives benzaldehyde. Suitable.
(B) Toluene with $\mathrm{Cl_2}$/Fe in the dark substitutes the ring (o- and p-chlorotoluene), not the methyl group; hydrolysis cannot then give benzaldehyde. (Side-chain chlorination to $\mathrm{C_6H_5CHCl_2}$ needs light or heat without a Lewis acid.) Not suitable.
(C) Gatterman–Koch formylation (CO, HCl, anhydrous $\mathrm{AlCl_3}$) gives benzaldehyde directly. Suitable.
(D) Toluene with $\mathrm{CrO_3}$ in acetic anhydride gives benzylidene diacetate, hydrolysed to benzaldehyde. Suitable.
Answer: (B). | ||||
| 47 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines The product is picric acid (2,4,6-trinitrophenol), made by nitrating phenol, so B is phenol.
Phenol is made from A by $\mathrm{NaNO_2/HCl}$ (diazotisation) and then warming the diazonium salt in water, so A is aniline.
Answer: (C). | ||||
| 48 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Borax bead: coloured metaborate beads of transition-metal ions such as $\mathrm{Fe^{3+}}$ (A–III).
Charcoal cavity: arsenic compounds give a white incrustation with a garlic odour (B–I).
Flame test: $\mathrm{Sr^{2+}}$ gives a crimson flame (C–IV).
Lake test: $\mathrm{Al(OH)_3}$ adsorbs blue litmus to form a blue lake, for $\mathrm{Al^{3+}}$ (D–II).
Answer: (A). | ||||
| 49 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Classified by source: protein, starch and cellulose are all natural polymers.
Nylon 6, polythene and Teflon are synthetic; nylon 6,6, Dacron and Buna-N are synthetic; rayon (semi-synthetic), caprolactam (the monomer of nylon 6) and Buna-S belong to the man-made group.
By source the group of natural polymers, (A), is the odd one. (C) is odd in another way: it is the only group that is not all polymers (caprolactam is the monomer of nylon 6).
Answer: (A) or (C). | ||||
| 50 | 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 In a dinucleotide the 3'-OH of one nucleotide's sugar is joined by a phosphodiester bond to the 5'-OH of the next sugar ($3'\rightarrow5'$ link), and the first nucleotide keeps its 5'-phosphate.
(A) shows exactly this: 5'-phosphate on the first ribose, 3'-O–P–O-5' to the second ribose.
(B) joins 5' to 5' through one phosphate; (C) bonds phosphorus directly to a ring carbon and links through the 2'-position; (D) joins the two 5' ends through a pyrophosphate. None is a nucleic-acid linkage.
Answer: (A). | ||||
| 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 Moles of $\mathrm{Cl_2}$: $n = \frac{PV}{RT} = \frac{10^5 \times 2.24\times10^{-3}}{8.314 \times 298} = \frac{224}{2477.6} = 0.0904$ mol.
1 mol $\mathrm{MnO_2}$ (87 g) gives 1 mol $\mathrm{Cl_2}$, so pure $\mathrm{MnO_2} = 0.0904 \times 87 = 7.87$ g.
Purity $= \frac{7.87}{10}\times100 = 78.7\% \approx 79$.
Answer: 79. | ||||
| 52 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Total kinetic energy $= \frac{3}{2}nRT = 1.5 \times 10 \times 8.314 \times 298 = 37\,164$ J $\approx 37.2$ kJ.
Nearest integer: 37.
Answer: 37. | ||||
| 53 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In the Bohr model $r_n = 0.529\,\frac{n^2}{Z}$ Å, so $r \propto n^2$.
$r_1 : r_3 = 1 : 9$, so $x = 9$.
Answer: 9. | ||||
| 54 | 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 A → B: isobaric expansion at $P_0$ from $V_0$ to $2V_0$; work done by the gas $= P_0V_0$.
B → C: reversible isothermal expansion from $2V_0$ to $4V_0$; $nRT = P_0(2V_0)$, so work by the gas $= 2P_0V_0\ln2 = 1.38\,P_0V_0$. At C, $P = \frac{2P_0V_0}{4V_0} = \frac{P_0}{2}$.
C → D: isobaric compression at $\frac{P_0}{2}$ from $4V_0$ to $2V_0$; work by the gas $= -\frac{P_0}{2}(2V_0) = -P_0V_0$.
Total work done by the gas $= P_0V_0 + 1.38P_0V_0 - P_0V_0 = 1.38\,P_0V_0$.
In the chemistry sign convention ($w = -\int P\,dV$, work done on the system) this is $w = -1.38\,P_0V_0$, so $x = -1.38$ and $-x = 1.38 \approx 1$.
Answer: 1. | ||||
| 55 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{NH_3}$: 15 mmol; HCl: 5 mmol. After reaction: 10 mmol $\mathrm{NH_3}$ and 5 mmol $\mathrm{NH_4^+}$, a basic buffer.
$\mathrm{pOH} = \mathrm{p}K_b + \log\frac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} = 4.7 + \log\frac{5}{10} = 4.7 - 0.30 = 4.4$.
$\mathrm{pH} = 14 - 4.4 = 9.6 \approx 10$.
Answer: 10. | ||||
| 56 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry For the concentration cell (anode in the dilute solution): $E = 0.059\log\frac{0.025}{10^{-8}} = 0.059\log(2.5\times10^6)$.
$\log 2.5 = \log10 - 2\log2 = 1 - 0.60 = 0.40$, so $\log(2.5\times10^6) = 6.40$.
$E = 0.059 \times 6.40 = 0.3776$ V $= 3776\times10^{-4}$ V.
Answer: 3776. | ||||
| 57 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics First order: $kt = 2.303\log\frac{[A]_0}{[A]}$.
In 1 min, 98% remains: $k = 2.303\log\frac{100}{98}$.
For 75% inactivation, 25% remains: $kt = 2.303\log 4$.
$t = \frac{\log 4}{\log(100/98)} = \frac{0.6021}{0.008774} = 68.6$ min $\approx 69$.
Answer: 69. | ||||
| 58 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich: $\log\frac{x}{m} = \log k + \frac{1}{n}\log P$.
Slope $\frac{1}{n} = \tan45^\circ = 1$; intercept $\log k = 0.699 = \log 5$, so $k = 5$.
$\frac{x}{m} = 5 \times 0.5 = 2.5$ g per g $= 2500\times10^{-3}$.
Answer: 2500. | ||||
| 59 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The nitronium ion $\mathrm{NO_2^+}$, the acylium ion $\mathrm{CH_3C^+{=}O}$ and the nitrosonium ion $\mathrm{NO^+}$ are electron-deficient cations: electrophiles.
$\mathrm{H_2O}$ has lone pairs on oxygen and acts as a nucleophile.
Number of electrophiles: 3.
Answer: 3. | ||||
| 60 | AminesClassification and identification of amines | Class 12: Amines | NumericalEasy | |
A easy numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines $\mathrm{CHCl_3}$ + KOH reacts only with primary amines (carbylamine reaction).
$\mathrm{C_3H_9N}$ isomers: propan-1-amine and propan-2-amine (primary), N-methylethanamine (secondary), N,N-dimethylmethanamine (tertiary).
Two primary amines.
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 correctEasy | |
A easy 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 Commutativity: $1 * 2 = 1 + 4 + \frac{1}{2} = 5.5$ but $2 * 1 = 2 + 2 + 2 = 6$. Not commutative.
Associativity: $(1 * 1) * 1 = (1 + 2 + 1) * 1 = 4 * 1 = 4 + 2 + 4 = 10$, while $1 * (1 * 1) = 1 * 4 = 1 + 8 + \frac{1}{4} = 9.25$. Not associative.
Answer: (D). | ||||
| 62 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $x^2 - 3x - 4 = (x-4)(x+1)$: roots $4$ and $-1$.
Common root $-1$: $2 - k - 5 = 0$, so $k = -3$.
Common root $4$: $32 + 4k - 5 = 0$, so $k = -\frac{27}{4}$ (not an option).
Answer: (C). | ||||
| 63 | 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 $A = \begin{pmatrix}7&49&343\\49&49&343\\343&343&343\end{pmatrix}$.
$R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$: $\begin{pmatrix}-42&0&0\\-294&-294&0\\343&343&343\end{pmatrix}$, a lower-triangular matrix.
$\det A = (-42)(-294)(343) = (6\cdot7)(6\cdot7^2)(7^3) = 6^2\cdot7^6$.
Answer: (C). | ||||
| 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 Adding a multiple of one row to another leaves the determinant unchanged; multiplying a row by 3 multiplies it by 3.
$|B| = 3|A| = 3(-1) = -3$.
Answer: (B). | ||||
| 65 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A homogeneous system has a non-trivial solution when its determinant is zero.
With $t = k - 3$: $\begin{vmatrix}t&1&1\\1&t&1\\1&1&t\end{vmatrix} = (t+2)(t-1)^2$.
So $(k-1)(k-4)^2 = 0$: $k = 1$ or $k = 4$, two distinct values.
Answer: (B). | ||||
| 66 | 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 $T_{r+1} = \binom{33}{r}(2x^{3/2})^{33-r}(3y^{1/16})^r$: the powers are $x^{\frac{3(33-r)}{2}}$ and $y^{\frac{r}{16}}$.
A positive integral power of $y$ needs $r = 16$ or $32$. Then $33 - r = 17$ or $1$, both odd, so the power of $x$ is $\frac{51}{2}$ or $\frac{3}{2}$, not an integer.
No term qualifies.
Answer: (A). | ||||
| 67 | Sequences and SeriesArithmetic and geometric means | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic and geometric means, from Sequences and Series. NCERT: Class 11: Sequences and Series $l, G_1, G_2, G_3, n$ are in G.P. with ratio $r$, $r^4 = \frac{n}{l}$: $G_k = lr^k$.
$G_1^4 + 2G_2^4 + G_3^4 = l^4(r^4 + 2r^8 + r^{12}) = l^4r^4(1 + r^4)^2 = l^4\cdot\frac{n}{l}\cdot\frac{(l+n)^2}{l^2} = ln(l+n)^2$.
With $l + n = 2m$: $= 4lm^2n$.
Answer: (B). | ||||
| 68 | 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 The form is $1^\infty$: the limit is $e^L$ with $L = \lim_{x\to0}\frac{\ln(1+x)}{e^x - 1} = \lim_{x\to0}\frac{\ln(1+x)}{x}\cdot\frac{x}{e^x-1} = 1\cdot1 = 1$.
The limit is $e$.
Answer: (C). | ||||
| 69 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = \frac{\pi}{4}$: left value $\frac{\pi\alpha}{4} + \sqrt{2}a\cdot\frac{1}{\sqrt{2}} = \frac{\pi\alpha}{4} + a$; right value $\frac{\pi\alpha}{2}\cot\frac{\pi}{4} - c\sin\frac{\pi}{2} = \frac{\pi\alpha}{2} - c$. So $a + c = \frac{\pi\alpha}{4}$.
At $x = \frac{\pi}{2}$: left limit $\pi\alpha\cot\frac{\pi}{2} - c\sin\pi = 0$; right value $-8ac\cos\pi - \pi\alpha b = 8ac - \pi\alpha b$. So $\pi\alpha = \frac{8ac}{b}$.
Then $a + c = \frac{2ac}{b}$, i.e. $\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$: $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.
Answer: (B). | ||||
| 70 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives By the mean value theorem, $f(1) - f(0) = f'(c)$ for some $c \in (0, 1)$, and $|f'(c)| \le 3$.
So $|f(1) - 2| \le 3$, i.e. $-1 \le f(1) \le 5$.
Answer: (C). | ||||
| 71 | 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 $f'(x) = x^2 - 2(2a+3b)x + 24ab$ has roots $-4$ and $4$ (the leading coefficient is positive, so the smaller root is the maximum, as required).
Sum of roots: $2(2a+3b) = 0$, so $a = -\frac{3b}{2}$. Product: $24ab = -16$, so $ab = -\frac{2}{3}$.
$-\frac{3b^2}{2} = -\frac{2}{3}$ gives $b = \pm\frac{2}{3}$, $a = \mp1$.
$4a + 3b = -6b + 3b = -3b = \mp2$. A value is $2$ (with $a = 1$, $b = -\frac{2}{3}$).
Answer: (B). | ||||
| 72 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Rationalise: $\frac{1}{\sqrt{2x+5} - \sqrt{2x+3}} = \frac{\sqrt{2x+5} + \sqrt{2x+3}}{(2x+5) - (2x+3)} = \frac{\sqrt{2x+5} + \sqrt{2x+3}}{2}$.
$\int\sqrt{2x+k}\,dx = \frac{(2x+k)^{3/2}}{3}$.
Integral $= \frac{1}{2}\cdot\frac{1}{3}\left[(2x+5)^{3/2} + (2x+3)^{3/2}\right] + C = \frac{1}{6}\left[(2x+5)^{3/2} + (2x+3)^{3/2}\right] + C$.
Answer: (C). | ||||
| 73 | 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 The normal at $(x, y)$: $(X - x) + \frac{dy}{dx}(Y - y) = 0$. It passes through $(2, 0)$: $(2 - x) - y\frac{dy}{dx} = 0$.
$y\,dy = (2 - x)\,dx$ gives $\frac{y^2}{2} = -\frac{(x-2)^2}{2} + C$, i.e. $(x-2)^2 + y^2 = 2C$.
Through $(-1, 4)$: $9 + 16 = 25$.
$(x-2)^2 + y^2 = 25$.
Answer: (A). | ||||
| 74 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a^2 - b^2 = 9$.
Through $(4, 1)$: $\frac{16}{a^2} + \frac{1}{a^2 - 9} = 1$, so $16(a^2 - 9) + a^2 = a^2(a^2 - 9)$, i.e. $a^4 - 26a^2 + 144 = 0$, $a^2 = 18$ or $8$.
$a^2 > 9$, so $a^2 = 18$, $a = 3\sqrt{2}$, and $e = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Answer: (C). | ||||
| 75 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Intersection: $(x-2)^2 = 2 - x$, so $(x-2)(x-1) = 0$: points $(2, 0)$ and $(1, -1)$.
Tangent to $y^2 = 2 - x$ at $(x_1, y_1)$: $yy_1 = 2 - \frac{x + x_1}{2}$.
At $(2, 0)$: $0 = 2 - \frac{x+2}{2}$, i.e. $x = 2$.
At $(1, -1)$: $-y = 2 - \frac{x+1}{2}$, i.e. $y = \frac{x+1}{2} - 2$.
At $x = 2$: $y = \frac{3}{2} - 2 = -\frac{1}{2}$. The tangents meet at $\left(2, -\frac{1}{2}\right)$.
Answer: (B). | ||||
| 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 Both lines pass through the origin, so their directions are $\vec{OP} = (b-c, c-a, a-b)$ and $\vec{OQ} = \left(\frac{1}{l}, \frac{1}{m}, \frac{1}{n}\right)$.
With first term $A$ and common difference $d$: $\frac{1}{l} = A + (a-1)d$, $\frac{1}{m} = A + (b-1)d$, $\frac{1}{n} = A + (c-1)d$.
$\vec{OP}\cdot\vec{OQ} = (A - d)[(b-c) + (c-a) + (a-b)] + d[a(b-c) + b(c-a) + c(a-b)] = 0 + 0 = 0$.
The lines are perpendicular: the angle is $\frac{\pi}{2}$.
Answer: (B). | ||||
| 77 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Let the plane be $px + qy + rz = s$.
Its trace on $x = 0$ is $qy + rz = s$, the same line as $2y - 3z = 5$: $q = 2t$, $r = -3t$, $s = 5t$.
Its trace on $z = 0$ is $px + qy = s$, the same line as $7x + 4y = 10$: with $q = 2t$, $s = 5t$ this is $\frac{7t}{2}x + 2ty = 5t$, so $p = \frac{7t}{2}$.
Plane: $7x + 4y - 6z = 10$.
Distance of $(1, -2, -1)$: $\frac{|7 - 8 + 6 - 10|}{\sqrt{49 + 16 + 36}} = \frac{5}{\sqrt{101}}$.
Answer: (C). | ||||
| 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 A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}: 6 outcomes.
$A \cap B$ = {(2,5), (5,2)}: 2 outcomes.
$P(B|A) = \frac{2}{6} = \frac{1}{3}$.
Answer: (B). | ||||
| 79 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $P = \prod\cos\theta_i$. Since $\prod\cot\theta_i = 1$, $\prod\sin\theta_i = P$ as well.
$P^2 = \prod\sin\theta_i\cos\theta_i = \prod\frac{\sin2\theta_i}{2} = \frac{1}{2^{10}}\prod\sin2\theta_i \le \frac{1}{2^{10}}$.
So $P \le \frac{1}{2^5} = \frac{1}{32}$, with equality when every $\theta_i = \frac{\pi}{4}$ (which satisfies the condition).
Answer: (D). | ||||
| 80 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $p \leftrightarrow \sim q$ is true exactly when $p$ and $q$ have opposite truth values. Its negation is true exactly when $p$ and $q$ have the same truth value, which is $p \leftrightarrow q$.
Check with $p$ = T, $q$ = T: $\sim(T \leftrightarrow F) = T$, and $p \leftrightarrow q = T$; (C) and (D) differ elsewhere ($p = q$ = F makes (C) false; $p = q$ = T makes (D) false).
Answer: (A). | ||||
| 81 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Case $x \ge 0$: $(x - 2)(2x + 3) = 1$, so $2x^2 - x - 7 = 0$, $x = \frac{1 \pm \sqrt{57}}{4}$; only $\frac{1 + \sqrt{57}}{4}$ is non-negative.
Case $-\frac{3}{2} \le x < 0$: $(-x - 2)(2x + 3) = 1$, so $2x^2 + 7x + 7 = 0$, discriminant $49 - 56 < 0$: none.
Case $x < -\frac{3}{2}$: $(-x - 2)(-(2x + 3)) = 1$, so $2x^2 + 7x + 5 = 0$, $x = -1$ or $-\frac{5}{2}$; only $-\frac{5}{2}$ lies in this range.
Two elements.
Answer: 2. | ||||
| 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 Choose the letter used three times: 6 ways. Choose the three other distinct letters from the remaining 5: $\binom{5}{3} = 10$.
Arrange the six letters (one repeated thrice): $\frac{6!}{3!} = 120$.
Total $= 6 \times 10 \times 120 = 7200$.
Answer: 7200. | ||||
| 83 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The 3-digit multiples of 9 are $108, 117, \ldots, 999$: $\frac{999 - 108}{9} + 1 = 100$ terms.
Sum $= \frac{100}{2}(108 + 999) = 50 \times 1107 = 55350$.
$41k = 55350$, so $k = 1350$.
Answer: 1350. | ||||
| 84 | 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 $|[x] - x| = x - [x] = \{x\}$, the fractional part, which has period 1 and $\int_0^1\{x\}\,dx = \frac{1}{2}$.
The interval $[-2, 4]$ has length 6: the integral is $6 \times \frac{1}{2} = 3$.
Answer: 3. | ||||
| 85 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $y = 5x^2$ and $y = 2x^2 + b$ meet where $3x^2 = b$, $x = \pm\alpha$ with $\alpha = \sqrt{\frac{b}{3}}$. Between them $2x^2 + b \ge 5x^2$.
Area $= 2\int_0^\alpha(b - 3x^2)\,dx = 2(b\alpha - \alpha^3) = 2\alpha\cdot\frac{2b}{3} = \frac{4b}{3}\sqrt{\frac{b}{3}} = \frac{4b^{3/2}}{3\sqrt{3}}$.
$\frac{4b^{3/2}}{3\sqrt{3}} = 12\sqrt{3}$ gives $b^{3/2} = 27$, so $b = 9$.
Answer: 9. | ||||
| 86 | CirclesRadius and centre | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections First circle: centre $C_1(-1, -2)$, radius $r_1 = \sqrt{1 + 4 + 4} = 3$. Second: centre $C_2(2, -2)$, radius $r_2 = \sqrt{8 - k}$.
$C_1C_2 = 3$. Internal contact: $|r_2 - r_1| = 3$, so $r_2 = 6$ ($r_2 = 0$ is impossible).
The first circle lies inside the second; the contact point is on line $C_2C_1$ ($y = -2$), 6 units from $C_2$ through $C_1$: $(2 - 6, -2) = (-4, -2)$ (and 3 units from $C_1$, as it must be).
$5(a^2 + b^2) = 5(16 + 4) = 100$.
Answer: 100. | ||||
| 87 | 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 $\frac{dy}{dx} = \frac{y + 3x^4}{x}$, i.e. $\frac{dy}{dx} - \frac{y}{x} = 3x^3$. Integrating factor $\frac{1}{x}$: $\frac{d}{dx}\left(\frac{y}{x}\right) = 3x^2$, so $\frac{y}{x} = x^3 + C$.
$y(1) = -1$: $-1 = 1 + C$, $C = -2$. So $y = x^4 - 2x$.
$y' = 4x^3 - 2$, $y'' = 12x^2$. At $x = 2$: $y'' = 48$, $y' = 30$, $y = 12$.
$48 - 30 + 12 = 30$.
Answer: 30. | ||||
| 88 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Sum $= 7 \times 7 = 49$; the five known values sum to 40, so $x + y = 9$.
$\frac{\sum x_i^2}{7} - 7^2 = 18$ gives $\sum x_i^2 = 469$; the known squares sum to $4 + 16 + 100 + 121 + 169 = 410$, so $x^2 + y^2 = 59$.
$xy = \frac{(x+y)^2 - (x^2+y^2)}{2} = \frac{81 - 59}{2} = 11$.
Answer: 11. | ||||
| 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 From A: $\sin A = \frac{OP}{AP} = \frac{10\sqrt{3}}{20\sqrt{3}} = \frac{1}{2}$, so the elevation is $30^\circ$.
From C: $\tan C = \frac{10\sqrt{3}}{10} = \sqrt{3}$, so the elevation is $60^\circ$.
In A.P., the elevation from B is $45^\circ$: $BP = \frac{OP}{\sin45^\circ} = 10\sqrt{3}\cdot\sqrt{2} = 10\sqrt{6}$.
$k^2 = 600$.
Answer: 600. | ||||
| 90 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}\times\vec{b} = (\alpha\gamma + 5)\hat{i} + (\beta - 2\gamma)\hat{j} + (-10 - \alpha\beta)\hat{k}$.
So $\alpha\gamma = 21$, $\beta - 2\gamma = -11$, $\alpha\beta = 9$.
$\frac{\beta}{\gamma} = \frac{9}{21} = \frac{3}{7}$, so $\beta = \frac{3\gamma}{7}$; then $\frac{3\gamma}{7} - 2\gamma = -11$ gives $\gamma = 7$, $\beta = 3$, $\alpha = 3$.
$\alpha - \beta + \gamma = 3 - 3 + 7 = 7$.
Answer: 7. | ||||
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