JEE (Main) · 2020
JEE (Main) 2020 — 9 January, 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
25 Q
- Motion of System of Particles and Rigid Body2
- Properties of Bulk Matter2
- Electrostatics2
11 easy · 13 medium · 1 hard
Chemistry
25 Q
- Equilibrium2
- Coordination Compounds2
- Amines2
13 easy · 11 medium · 1 hard
Mathematics
25 Q
- Sequences and Series3
- Limits, Continuity and Differentiability3
- Matrices and Determinants2
9 easy · 15 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Physical World and MeasurementSignificant figures | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on significant figures, from Physical World and Measurement. NCERT: Class 11: Units and Measurement In addition the result keeps as many decimal places as the least precise term.
(i) $24.36 + 0.0724 + 256.2 = 280.6324 \to 280.6$ (one decimal place, from 256.2).
(ii) $24.44 + 16.082 + 240.2 = 280.722 \to 280.7$.
(iii) $25.2 + 19.2812 + 236.183 = 280.6642 \to 280.7$.
(iv) $25 + 236.191 + 19.5 = 280.691 \to 281$ (no decimal place, from 25).
So $A_1+B_1+C_1 < A_2+B_2+C_2 = A_3+B_3+C_3 < A_4+B_4+C_4$, which matches none of the four options; the unrounded sums (280.6324, 280.722, 280.6642, 280.691) give $1 < 3 < 4 < 2$, which matches none either.
Answer: none of the options (NTA awarded full marks to all candidates). | ||||
| 2 | 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 Along y: $u_y = 0$, $a_y = 4$ m/s², so $32 = \frac{1}{2}(4)t^2$, $t^2 = 16$, $t = 4$ s.
Along x: $x = u_x t + \frac{1}{2}a_x t^2 = 3(4) + \frac{1}{2}(6)(16) = 12 + 48 = 60$ m.
Answer: (3). | ||||
| 3 | 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 The string does not slip, so both blocks move with speed $v = \omega R$.
When $m_1$ falls by h, $m_2$ rises by h. Energy conservation:
$m_1gh - m_2gh = \frac{1}{2}(m_1+m_2)\omega^2R^2 + \frac{1}{2}I\omega^2$.
$\omega = \left[\frac{2(m_1-m_2)gh}{(m_1+m_2)R^2 + I}\right]^{1/2}$.
Answer: (2). | ||||
| 4 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At the top the particle moves horizontally with $u\cos 60^\circ = \frac{u}{2}$, at height $H = \frac{u^2\sin^2 60^\circ}{2g} = \frac{3u^2}{8g}$.
Momentum along x in the perfectly inelastic collision: $m\frac{u}{2} + mu = 2mv$, so $v = \frac{3u}{4}$ (horizontal).
The combined mass falls from rest vertically through H: $t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{3u^2}{4g^2}} = \frac{\sqrt{3}}{2}\frac{u}{g}$.
Horizontal distance $= \frac{3u}{4}\cdot\frac{\sqrt{3}u}{2g} = \frac{3\sqrt{3}}{8}\frac{u^2}{g}$.
Answer: (1). | ||||
| 5 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $M = \int_0^L \left(a + \frac{bx^2}{L^2}\right)dx = aL + \frac{bL}{3}$.
$\int_0^L x\left(a + \frac{bx^2}{L^2}\right)dx = \frac{aL^2}{2} + \frac{bL^2}{4}$.
$x_{cm} = \frac{\frac{aL^2}{2} + \frac{bL^2}{4}}{aL + \frac{bL}{3}} = \frac{(2a+b)/4}{(3a+b)/3}L = \frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L$.
Answer: (1). | ||||
| 6 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion With extension x the mass moves on a circle of radius $l + x$; the spring force supplies the centripetal force:
$kx = m\omega^2(l + x)$, so $x = \frac{m\omega^2 l}{k - m\omega^2} \approx \frac{m\omega^2 l}{k}$ for $k \gg m\omega^2$.
Relative change $\frac{x}{l} = \frac{m\omega^2}{k}$.
Answer: (1). | ||||
| 7 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}}$. For B, $\frac{M/2}{R/2} = \frac{M}{R}$, so $v_B = v_A$.
$\frac{v_A}{v_B} = 1 = \frac{n}{4}$, so $n = 4$.
Answer: (4). | ||||
| 8 | 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 Energy per unit volume $u = \frac{\sigma^2}{2Y}$ with stress $\sigma = \frac{F}{\pi d^2/4} \propto \frac{1}{d^2}$, so $u \propto \frac{1}{d^4}$ (same load, same material).
$\frac{u_1}{u_2} = \left(\frac{d_2}{d_1}\right)^4 = \frac{1}{4}$, so $\frac{d_1}{d_2} = 4^{1/4} = \sqrt{2}$.
Answer: (3). | ||||
| 9 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Half immersed, the liquid surface meets the drop along its equator (circumference $2\pi r$), and the surface tension pulls upward with $2\pi rT$.
Weight = buoyancy + surface force: $\frac{4}{3}\pi r^3 dg = \frac{2}{3}\pi r^3\rho g + 2\pi rT$.
$\frac{2}{3}\pi r^3 g(2d - \rho) = 2\pi rT$, so $r^2 = \frac{3T}{(2d-\rho)g}$.
Answer: (2). | ||||
| 10 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | Single correctHard | |
A hard single correct question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Mean free time $\tau = \frac{\lambda}{v_{avg}}$, with $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$ ($d$ = diameter $= 2r$) and $v_{avg} \propto \sqrt{\frac{T}{M}}$.
Same n and T: $\tau \propto \frac{\sqrt{M}}{r^2}$.
$\frac{\tau_{Ar}}{\tau_{Xe}} = \sqrt{\frac{40}{140}}\times\left(\frac{0.1}{0.07}\right)^2 = 0.535 \times 2.04 = 1.09$.
Of the four values, 1.83 is the closest to 1.09.
The ratio is about 1.09, which none of the options gives.
No option is correct as printed. | ||||
| 11 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{540}{6\times10^{-3}}} = \sqrt{90000} = 300$ m/s.
Consecutive resonances of a string fixed at both ends differ by the fundamental: $\frac{v}{2L} = 490 - 420 = 70$ Hz.
$L = \frac{300}{140} = 2.14$ m $\approx 2.1$ m.
Answer: (3). | ||||
| 12 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In both circuits the left plate is positive, so a discharge current would leave the left plate, pass through R and return up the right-hand branch to the negative plate.
In circuit A the diode in the right-hand branch points downwards: it is reverse biased, no current flows, and the charge stays $Q_A = CV$.
In circuit B the diode points upwards: it conducts and the capacitor discharges through R, $Q = CVe^{-t/RC}$. At $t = CR$, $Q_B = \frac{CV}{e}$.
Answer: (4). | ||||
| 13 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Magnetic moment $\mu = I\pi a^2$; restoring torque for a small turn $\theta$: $\tau = -\mu B\theta$.
Moment of inertia of a ring about a diameter: $\frac{ma^2}{2}$.
$\omega^2 = \frac{I\pi a^2 B}{ma^2/2} = \frac{2\pi IB}{m}$, so $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{2\pi IB}} = \sqrt{\frac{2\pi m}{IB}}$.
Answer: (3). | ||||
| 14 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Inside the solenoid $B = \mu_0 nI$ along the axis, and the electron moves perpendicular to it on a circle of radius $r = \frac{mv}{eB}$.
It starts on the axis moving radially, so its circle passes through the axis and reaches at most $2r$ from it. Not hitting the wall: $2r \le R$.
$\frac{2mv}{e\mu_0 nI} \le R$, so $v_{max} = \frac{e\mu_0 nIR}{2m}$.
Answer: (2). | ||||
| 15 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $X_L = \omega L = 314 \times 0.04 = 12.56\ \Omega$; $X_C = \frac{1}{\omega C} = \frac{1}{314 \times 10^{-4}} = 31.85\ \Omega$.
Series LC: net reactance $X_C - X_L = 19.29\ \Omega$ (capacitive), so $I_0 = \frac{10}{19.29} = 0.52$ A.
In a capacitive circuit the current leads the voltage by $\frac{\pi}{2}$: $I = 0.52\sin\left(314t + \frac{\pi}{2}\right) = 0.52\cos 314t$.
Answer: (2). | ||||
| 16 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $\vec{B}$ is perpendicular to both $\vec{E}$ ($\hat{k}$) and the direction of travel, and $\vec{E}\times\vec{B}$ points along the direction of travel.
Try $\hat{b} = \frac{\hat{i}-\hat{j}}{\sqrt{2}}$: $\hat{k}\times\frac{\hat{i}-\hat{j}}{\sqrt{2}} = \frac{\hat{j}+\hat{i}}{\sqrt{2}}$, the direction of travel.
A wave travelling along $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$ has phase $\omega t - \vec{k}\cdot\vec{r}$, i.e. the $\omega t - k(\dots)$ form; option (2) has the wrong sign and the wrong direction, (1) is along the travel and (4) along $\vec{E}$.
Answer: (3). | ||||
| 17 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Light escapes only inside the cone of half-angle C, with $\sin C = \frac{3}{4}$, $\cos C = \frac{\sqrt{7}}{4} = 0.661$.
On a sphere of radius r around the source this cone cuts a cap of height $h = r(1 - \cos C)$ and area $2\pi r^2(1 - \cos C)$.
Fraction $= \frac{2\pi r^2(1-\cos C)}{4\pi r^2} = \frac{1 - 0.661}{2} = 0.17$, i.e. about 17%.
Answer: (4). | ||||
| 18 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Momentum $p = |e|Et$ (from rest), so $\lambda = \frac{h}{|e|Et}$.
$\frac{d\lambda}{dt} = -\frac{h}{|e|Et^2}$.
Answer: (1). | ||||
| 19 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Ionisation energy $= Z^2$ Rydberg $= 9$, so $Z = 3$.
Second excited state is $n = 3$: $\frac{1}{\lambda} = RZ^2\left(1 - \frac{1}{9}\right) = 9R\cdot\frac{8}{9} = 8R$.
$\lambda = \frac{1}{8 \times 1.097\times10^7}$ m $= 1.14\times10^{-8}$ m $= 11.4$ nm.
Answer: (1). | ||||
| 20 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The longer plate of the 9 V cell is on the left, so the current leaves it towards the left, rises through the left wire (as $i$ is drawn) and enters the left corner of the bridge.
The diode D in series with the upper-left 5 Ω points back towards the left corner, so it is reverse biased; the diode D in series with the lower-right 20 Ω points back towards the bottom corner, so it is reverse biased as well. Neither arm conducts.
The only path is left corner → 10 Ω → bottom corner → 5 Ω (vertical) → top corner → 10 Ω → right corner.
Total resistance $= 10 + 5 + 10 + 5 = 30\ \Omega$; $i = \frac{9}{30} = 0.3$ A.
Answer: (2). | ||||
| 21 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $TV^{\gamma-1}$ = constant, so $T_2 = 300 \times 16^{0.4} = 300 \times 3.031 = 909.4$ K.
Isobaric doubling of volume doubles the temperature: $T_3 = 2 \times 909.4 = 1818.9$ K.
Answer: 1819. | ||||
| 22 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance ABCD is the top face $z = 2$ (A(0,0,2), B(3,0,2), C(3,2,2), D(0,2,2)), with normal $\hat{k}$. The field has no z-component, so $\phi_I = 0$.
BCGF is the face $x = 3$, area $2 \times 2 = 4$ m², outward normal $\hat{i}$: $\phi_{II} = 4(3)(4) = 48$ Nm²/C.
$\phi_I - \phi_{II} = 0 - 48 = -48$.
Answer: -48. | ||||
| 23 | Current ElectricityMetre bridge | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity Balance: $\frac{R}{S} = \frac{25}{75} = \frac{1}{3}$.
$R = \frac{\rho L}{\pi d^2/4}$: half the length and half the diameter gives $R' = \frac{1/2}{1/4}R = 2R$, so $\frac{R'}{S} = \frac{2}{3}$.
$\frac{l'}{100 - l'} = \frac{2}{3}$, so $3l' = 200 - 2l'$, $l' = 40$ cm.
Answer: 40. | ||||
| 24 | 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 Fringe width $\beta \propto \lambda$. Same length of screen: $15\beta_1 = 10\beta_2$, so $\lambda = \frac{15}{10}\times500 = 750$ nm.
Answer: 750. | ||||
| 25 | 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 The two identical zener diodes in series hold the output at 8 V, so each drops 4 V.
No load is drawn, so the current through the two 200 Ω resistors in series is $I = \frac{12 - 8}{400} = 10$ mA, and each diode would dissipate $4 \times 10 = 40$ mW.
NTA's key for this question is 12, which this working does not reproduce; see the note.
Answer: 12 (NTA's key). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics The entropy of a substance rises with temperature ($dS = \frac{C\,dT}{T}$), so S depends on T.
The entropy change of a process also depends on the temperature at which it occurs: for a reaction $\Delta S_T = \Delta S_{298} + \int_{298}^{T}\frac{\Delta C_p}{T}dT$, and for a heat transfer $\Delta S = \frac{q_{rev}}{T}$.
Answer: (2). | ||||
| 27 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{Cr(OH)_3 \rightleftharpoons Cr^{3+} + 3OH^-}$: with solubility s, $K_{sp} = s(3s)^3 = 27s^4$.
$[\mathrm{OH^-}] = 3s$, so $[\mathrm{OH^-}]^4 = 81s^4 = 3K_{sp} = 18 \times 10^{-31}$.
$[\mathrm{OH^-}] = (18 \times 10^{-31})^{1/4}$.
Answer: (1). | ||||
| 28 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Conductance comes from dissolved ions. Sea water, well water and saline all carry dissolved salts; distilled water has almost none (only its own slight self-ionisation).
Answer: (4). | ||||
| 29 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Charcoal adsorbs the gases on its surface, so the number of gas molecules in the space, and the pressure, fall with time.
Adsorption slows as the surface fills and reaches equilibrium, so the pressure levels off at a steady value rather than falling to zero.
Answer: (1). | ||||
| 30 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium For $\mathrm{A \rightleftharpoons B}$, $K = \frac{[\mathrm{B}]}{[\mathrm{A}]} = \frac{\text{number of circles}}{\text{number of squares}}$ (same volume).
The figure shows 6 squares and 11 circles, giving $K = \frac{11}{6} \approx 1.8$; the intended count is evidently 12 circles to 6 squares, $K = 2$.
Answer: (1). | ||||
| 31 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Each of the six carbon atoms of benzene is $sp^2$ hybridised and has three $sp^2$ hybrid orbitals.
$6 \times 3 = 18$.
Answer: (3). | ||||
| 32 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The very large jump from the first to the second ionisation enthalpy shows one valence electron: the metal is an alkali metal (496 kJ mol⁻¹ is sodium) and its hydroxide is MOH.
$\mathrm{MOH + HCl \to MCl + H_2O}$: 1 mol HCl.
$\mathrm{2MOH + H_2SO_4 \to M_2SO_4 + 2H_2O}$: 0.5 mol $\mathrm{H_2SO_4}$.
Answer: (4). | ||||
| 33 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{Zn + 2HCl \to ZnCl_2 + H_2}$ and $\mathrm{Zn + 2NaOH \to Na_2ZnO_2 + H_2}$.
Each mole of zinc gives one mole of $\mathrm{H_2}$ in both, so the same mass of zinc gives equal volumes.
Answer: (2). | ||||
| 34 | 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: $\mathrm{Li^+}$ is the smallest alkali-metal ion and has the largest hydration enthalpy.
(b) False: LiCl is largely covalent and dissolves in organic solvents such as pyridine.
(c) True: unlike the other alkali metals, lithium does not form an ethynide with ethyne.
(d) True: lithium (like magnesium, its diagonal partner) reacts only slowly with water.
Answer: (3). | ||||
| 35 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{B_3N_3H_3Cl_3}$ (B-trichloroborazine) is reduced by $\mathrm{LiBH_4}$ to $\mathrm{B_3N_3H_6}$, borazine, the 'inorganic benzene' (B).
Replacing the three B–Cl by B–CH₃ needs a methyl carbanion source, the Grignard reagent MeMgBr (C), giving $\mathrm{H_3N_3B_3(Me)_3}$.
Answer: (2). | ||||
| 36 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds In the octahedral $\mathrm{[Co(NH_3)_4Cl_2]^+}$ the two Cl are adjacent (90°) in the cis isomer and opposite (180°) in the trans isomer. Meridional/facial isomerism needs an $\mathrm{MA_3B_3}$ complex.
Answer: (2). | ||||
| 37 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (I) $\mathrm{Cr^{2+}}$, $d^4$, weak-field water: high spin, 4 unpaired, $\mu = 4.90$ BM.
(II) $\mathrm{Fe^{2+}}$, $d^6$, strong-field cyanide: low spin, 0 unpaired, $\mu = 0$.
(III) $\mathrm{Fe^{3+}}$, $d^5$, with $\Delta_0 > P$: low spin $t_{2g}^5$, 1 unpaired, $\mu = 1.73$ BM.
(IV) $\mathrm{Co^{2+}}$, $d^7$, tetrahedral: $e^4t_2^3$, 3 unpaired, $\mu = 3.87$ BM.
Order: (I) > (IV) > (III) > (II).
Answer: (2). | ||||
| 38 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter BOD is the amount of oxygen used by (aerobic) bacteria to break down the organic matter in a given volume of water, a measure of its organic pollution.
Answer: (3). | ||||
| 39 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines (III) Cyclohexylamine: an $sp^3$ nitrogen with a localised lone pair, a typical aliphatic amine (p$K_b \approx 3.3$), the strongest base.
(II) Pyridine: the lone pair is in an $sp^2$ orbital outside the aromatic ring (p$K_b \approx 8.8$).
(I) Aniline: the lone pair is delocalised into the benzene ring (p$K_b \approx 9.4$).
(IV) Pyrrole: the lone pair is part of the aromatic sextet, so it is hardly basic at all.
Order: (III) > (II) > (I) > (IV).
Answer: (1). | ||||
| 40 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In a vinyl chloride the Cl lone pair is delocalised into the C=C ($\mathrm{Cl^+=CH-CH^-}$), giving the C–Cl bond partial double-bond character.
A $-\mathrm{NO_2}$ group at the other end withdraws this charge ($\mathrm{Cl^+=CH-CH=NO_2^-}$), increasing the delocalisation and the double-bond character, so the C–Cl bond is shortest. $-\mathrm{CH_3}$ and $-\mathrm{OCH_3}$ push electrons into the double bond and lessen it.
Answer: (2). | ||||
| 41 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Molisch's test is positive for all carbohydrates; Barfoed's test is positive only for reducing monosaccharides (in the short time of the test); the biuret test needs peptide bonds, i.e. a protein.
A: carbohydrate but not a monosaccharide, a disaccharide such as lactose.
B: a monosaccharide such as glucose.
C: not a carbohydrate, biuret positive, a protein such as albumin (the single amino acid alanine has no peptide bond and is biuret negative).
Glucose and fructose would both be Barfoed positive, which rules out option (2) for A.
Answer: (1). | ||||
| 42 | PolymersBiodegradable polymers | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biodegradable polymers, from Polymers. NCERT: No matching NCERT chapter PHBV is made from 3-hydroxybutanoic acid and 3-hydroxypentanoic acid; C-3 of each carries H, OH, $\mathrm{CH_2COOH}$ and an alkyl group, so both monomers are chiral.
Chloroprene (neoprene), buta-1,3-diene and acrylonitrile (Buna-N), and hexamethylenediamine and adipic acid (nylon 6,6) are all achiral.
Answer: (4). | ||||
| 43 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctHard | |
A hard single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (1) But-1-ene + HBr gives 2-bromobutane through a planar carbocation: equal amounts of both enantiomers, racemic.
(4) Butan-2-one + HCN: cyanide attacks the planar carbonyl from either face equally, giving a racemic cyanohydrin.
(3) 3-Methylbut-1-ene + HCl: the secondary carbocation rearranges by a 1,2-hydride shift to the tertiary cation, and the main product is 2-chloro-2-methylbutane, $\mathrm{(CH_3)_2CCl-CH_2CH_3}$, which has no chiral centre. The product is achiral, so it is not racemic.
(2) The new chiral centre forms on an achiral, planar cation, but the ring already carries a stereocentre; taking the unspecified starting alkene as a racemate, each diastereomer of the product is itself racemic.
Answer: (3). | ||||
| 44 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Work backwards from cyclopent-1-ene-1-carbaldehyde. It is the intramolecular aldol condensation product of hexane-1,6-dial: the enolate at C-2 attacks C-6, closing a five-membered ring, and dehydration gives the conjugated enal (step v).
Hexane-1,6-dial comes from reductive ozonolysis (steps iii, iv) of cyclohexene, which comes from bromocyclohexane by dehydrobromination with alcoholic KOH (step ii), which comes from cyclohexane by free-radical bromination (step i).
So A is cyclohexane.
Answer: (2). | ||||
| 45 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines [P] gives a coloured azo dye after diazotisation and coupling with β-naphthol, so it is a primary aromatic amine; this rules out the secondary amine (4).
With bromine water it gives $\mathrm{C_7H_6NBr_3}$: a toluidine ($\mathrm{C_7H_9N}$) in which three ring H are replaced by Br. The $-\mathrm{NH_2}$ group directs Br to its two ortho and one para positions, which must all be free.
In m-toluidine positions 2, 4 and 6 (ortho, ortho, para to $\mathrm{NH_2}$) are all free, giving 2,4,6-tribromo-3-methylaniline. In o- and p-toluidine the methyl occupies one of those positions, so only a dibromo product forms.
Answer: (2). | ||||
| 46 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Freezing-point depression: $\Delta T_f = K_f m = 2.0 \times 0.5 = 1.0$ K, so the solution, and the gas, are at $273 - 1 = 272$ K.
Released against the atmosphere (1 atm) at this constant temperature: $V = \frac{nRT}{P} = \frac{0.1 \times 0.08 \times 272}{1} = 2.176$ L $\approx 2.18$ L.
Answer: 2.18. | ||||
| 47 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Mass of 1 L of sea water $= 1000 \times 1.03 = 1030$ g.
ppm $= \frac{10.30\times10^{-3}\text{ g}}{1030\text{ g}} \times 10^6 = 10$.
Answer: 10. | ||||
| 48 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate is inversely proportional to the time taken: $\frac{k_{400}}{k_{300}} = \frac{60}{40} = \frac{3}{2}$.
$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$: $\ln\frac{3}{2} = \frac{E_a}{8.3}\left(\frac{1}{300} - \frac{1}{400}\right) = \frac{E_a}{8.3 \times 1200}$.
Using $\ln\frac{3}{2} = 0.4$ (the data print $\ln\frac{2}{3} = 0.4$, which has the wrong sign): $E_a = 0.4 \times 8.3 \times 1200 = 3984$ J/mol $\approx 3.98$ kJ/mol.
Answer: 3.98 (NTA set this question aside for all candidates; see the note). | ||||
| 49 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Chromate $\mathrm{CrO_4^{2-}}$: tetrahedral Cr bonded to 4 O, so 4 Cr–O links.
Dichromate $\mathrm{Cr_2O_7^{2-}}$: two $\mathrm{CrO_4}$ tetrahedra sharing one corner O; each Cr is bonded to 3 terminal O and the bridging O, so $2 \times 4 = 8$ Cr–O links.
Sum $= 4 + 8 = 12$ (counting each Cr–O linkage, i.e. σ bond, once).
Answer: 12. | ||||
| 50 | Alcohols, Phenols and EthersPrimary, secondary and tertiary alcohols | Class 12: Alcohols, Phenols and Ethers | NumericalMedium | |
A medium numerical question on primary, secondary and tertiary alcohols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Hot copper at 573 K dehydrogenates primary and secondary alcohols but dehydrates a tertiary alcohol to an alkene. So B is the tertiary alcohol that dehydrates to 2-methylbut-2-ene: 2-methylbutan-2-ol, $\mathrm{(CH_3)_2C(OH)CH_2CH_3}$.
B forms from A and $\mathrm{CH_3MgBr}$, so A is butan-2-one, $\mathrm{CH_3COCH_2CH_3}$ ($\mathrm{C_4H_8O}$, M = 72).
% C $= \frac{48}{72} \times 100 = 66.67$.
Answer: 66.67. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $A = (-2, 2)$; $|x - 2| \ge 3$ gives $B = (-\infty, -1] \cup [5, \infty)$.
$A - B = (-1, 2)$, not $[-1, 2)$ since $-1 \in B$: (1) is false.
$B - A = (-\infty, -2] \cup [5, \infty) = \mathbb{R} - (-2, 5)$: (2) is true.
$A \cap B = (-2, -1]$: (3) is false.
$A \cup B = (-\infty, 2) \cup [5, \infty) = \mathbb{R} - [2, 5)$: (4) is false.
Answer: (2). | ||||
| 52 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Repeated root of $ax^2 - 2bx + 5 = 0$: $\alpha = \frac{b}{a}$ and $\alpha^2 = \frac{5}{a}$ (product of roots), so $b\alpha = \frac{b^2}{a} = 5$ (equal roots: $b^2 = 5a$).
$\alpha$ satisfies $x^2 - 2bx - 10 = 0$: $\alpha^2 - 2b\alpha - 10 = 0$, so $\alpha^2 = 10 + 10 = 20$.
Product of roots: $\alpha\beta = -10$, so $\beta^2 = \frac{100}{\alpha^2} = 5$.
$\alpha^2 + \beta^2 = 25$.
Answer: (2). | ||||
| 53 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $|x| + |y| = 4$. Then $|z|^2 = x^2 + y^2$ is least when $|x| = |y| = 2$, giving 8, and greatest at a vertex, giving 16.
So $\sqrt{8} \le |z| \le 4$. $\sqrt{7} < \sqrt{8}$ is impossible; $\sqrt{8}$, $\sqrt{8.5}$ and $\sqrt{10}$ all occur.
Answer: (1). | ||||
| 54 | 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 $R_1 \to R_1 - 2R_2 + R_3$: the new first row is $(a - 2b + c,\ 0,\ 0) = (1, 0, 0)$.
$f(x) = 1 \cdot \begin{vmatrix} x+3 & x+2 \\ x+4 & x+3 \end{vmatrix} = (x+3)^2 - (x+2)(x+4) = 1$.
So $f(x) = 1$ for every x, and $f(50) = 1$.
Answer: (3). | ||||
| 55 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = 7(-24 + 4) - 6(-18 - 2) - 2(-6 - 4) = -140 + 120 + 20 = 0$, so the homogeneous system has infinitely many solutions.
From the third equation $x = 2y + 6z$. In the second: $3(2y + 6z) + 4y + 2z = 10y + 20z = 0$, so $y = -2z$, and then $x = -4z + 6z = 2z$.
Check in the first: $14z - 12z - 2z = 0$.
Answer: (4). | ||||
| 56 | 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 $T_{r+1} = {}^{16}C_r\frac{x^{16-r}}{\cos^{16-r}\theta}\cdot\frac{1}{x^r\sin^r\theta}$; independent of x when $r = 8$:
$T_9 = \frac{{}^{16}C_8}{\sin^8\theta\cos^8\theta} = \frac{{}^{16}C_8\, 2^8}{\sin^8 2\theta}$, least when $\sin 2\theta$ is greatest.
$\frac{\pi}{8} \le \theta \le \frac{\pi}{4}$: $2\theta \in [\frac{\pi}{4}, \frac{\pi}{2}]$, greatest $\sin 2\theta = 1$, so $l_1 = {}^{16}C_8\,2^8$.
$\frac{\pi}{16} \le \theta \le \frac{\pi}{8}$: $2\theta \in [\frac{\pi}{8}, \frac{\pi}{4}]$, greatest $\sin 2\theta = \frac{1}{\sqrt{2}}$, $\sin^8 2\theta = \frac{1}{16}$, so $l_2 = 16\cdot{}^{16}C_8\,2^8$.
$l_2 : l_1 = 16 : 1$.
Answer: (3). | ||||
| 57 | 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 $a_3 + a_5 + \dots + a_{201} = r(a_2 + a_4 + \dots + a_{200})$, so $200 = 100r$, $r = 2$.
$a_1 + a_3 + \dots + a_{199} = \frac{1}{r}(a_2 + a_4 + \dots + a_{200}) = \frac{100}{2} = 50$.
$\sum_{n=1}^{200} a_n = 50 + 100 = 150$.
Answer: (2). | ||||
| 58 | 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 $\frac{4}{x} - 1 < \left[\frac{4}{x}\right] \le \frac{4}{x}$, so $x\left[\frac{4}{x}\right] \to 4$: $A = 4$.
$[x^2]$ jumps where $x^2$ is an integer; f is discontinuous there unless $\sin\pi x = 0$, i.e. unless x is itself an integer.
$\sqrt{A} = 2$, $\sqrt{A+5} = 3$, $\sqrt{A+21} = 5$ are integers, so f is continuous there. $\sqrt{A+1} = \sqrt{5}$: $x^2 = 5$ is an integer but $\sin(\pi\sqrt{5}) \ne 0$, so f jumps from $4\sin\pi\sqrt5$ to $5\sin\pi\sqrt5$.
Answer: (2). | ||||
| 59 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\frac{dx}{d\theta} = 2\cos\theta - 2\cos 2\theta = 4\sin\frac{3\theta}{2}\sin\frac{\theta}{2}$, $\frac{dy}{d\theta} = -2\sin\theta + 2\sin 2\theta = 4\cos\frac{3\theta}{2}\sin\frac{\theta}{2}$.
$\frac{dy}{dx} = \cot\frac{3\theta}{2}$, so $\frac{d^2y}{dx^2} = -\frac{3}{2}\csc^2\frac{3\theta}{2}\Big/\frac{dx}{d\theta}$.
At $\theta = \pi$: $\csc^2\frac{3\pi}{2} = 1$ and $\frac{dx}{d\theta} = -2 - 2 = -4$, so $\frac{d^2y}{dx^2} = \frac{-3/2}{-4} = \frac{3}{8}$.
$\frac{3}{8}$ is none of the options.
Answer: none of the options (NTA awarded full marks to all candidates). | ||||
| 60 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy 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 $f(g(x)) = x$. Differentiating: $f'(g(x))\,g'(x) = 1$.
At $x = a$: $f'(b) \times 5 = 1$, so $f'(b) = \frac{1}{5}$.
Answer: (1). | ||||
| 61 | 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 $F'(x) = x^2 g(x)$ and $g(1) = 0$, so $F'(1) = 0$: x = 1 is a critical point.
$F''(x) = 2x\,g(x) + x^2 g'(x) = 2x\,g(x) + x^2 f(x)$, so $F''(1) = 0 + 1 \times 3 = 3 > 0$.
x = 1 is a point of local minimum.
Answer: (3). | ||||
| 62 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\tan 2\theta + \sec 2\theta = \frac{1 + \sin 2\theta}{\cos 2\theta} = \frac{(\cos\theta + \sin\theta)^2}{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)} = \frac{1 + \tan\theta}{1 - \tan\theta}$.
The integral is $\int\sec^2\theta\,\frac{1 - \tan\theta}{1 + \tan\theta}\,d\theta$. Put $t = \tan\theta$:
$\int\frac{1 - t}{1 + t}dt = \int\left(-1 + \frac{2}{1 + t}\right)dt = -t + 2\ln|1 + t| + C = -\tan\theta + 2\ln|1 + \tan\theta| + C$.
$(\lambda, f(\theta)) = (-1, 1 + \tan\theta)$.
Answer: (1). | ||||
| 63 | 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 On $\frac{1}{2} \le x \le \frac{\sqrt{3}}{2}$, $f(x) = 1 - x$. The curves meet where $1 - x = \left(x - \frac{1}{2}\right)^2$, i.e. $x^2 = \frac{3}{4}$, $x = \frac{\sqrt{3}}{2}$; before that $f > g$.
$\int_{1/2}^{\sqrt3/2}(1 - x)\,dx = \left[x - \frac{x^2}{2}\right]_{1/2}^{\sqrt3/2} = \left(\frac{\sqrt3}{2} - \frac{3}{8}\right) - \frac{3}{8} = \frac{\sqrt3}{2} - \frac{3}{4}$.
$\int_{1/2}^{\sqrt3/2}\left(x - \frac{1}{2}\right)^2dx = \frac{1}{3}\left(\frac{\sqrt3 - 1}{2}\right)^3 = \frac{6\sqrt3 - 10}{24} = \frac{3\sqrt3 - 5}{12}$.
Area $= \frac{\sqrt3}{2} - \frac{3}{4} - \frac{3\sqrt3 - 5}{12} = \frac{3\sqrt3 - 4}{12} = \frac{\sqrt3}{4} - \frac{1}{3}$.
Answer: (4). | ||||
| 64 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $y = vx$: $v + x\frac{dv}{dx} = \frac{v}{1 + v^2}$, so $x\frac{dv}{dx} = -\frac{v^3}{1 + v^2}$.
$\int\frac{1 + v^2}{v^3}dv = -\int\frac{dx}{x}$: $-\frac{1}{2v^2} + \ln v = -\ln x + C$, i.e. $\ln y - \frac{x^2}{2y^2} = C$.
$y(1) = 1$: $C = -\frac{1}{2}$.
$y = e$: $1 - \frac{x^2}{2e^2} = -\frac{1}{2}$, so $x^2 = 3e^2$ and $x = \sqrt{3}\,e$ (taking the positive value).
Answer: (2). | ||||
| 65 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $2b = \frac{4}{\sqrt3}$, so $b^2 = \frac{4}{3}$.
The line is $y = -\frac{x}{6} + \frac{4}{3}$. Tangency to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: $c^2 = a^2m^2 + b^2$, so $\frac{16}{9} = \frac{a^2}{36} + \frac{4}{3}$, $a^2 = 16$.
$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{1}{12} = \frac{11}{12}$, $e = \sqrt{\frac{11}{12}} = \frac{1}{2}\sqrt{\frac{11}{3}}$.
Answer: (2). | ||||
| 66 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = 8x$ has $a = 2$; points $(2t^2, 4t)$. A: $4t = -2$, $t = -\frac{1}{2}$.
For a focal chord $t_1t_2 = -1$, so B has $t = 2$: $B(8, 8)$.
Tangent at t: $ty = x + at^2$, i.e. $2y = x + 8$, or $x - 2y + 8 = 0$.
Answer: (3). | ||||
| 67 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Total probability: $6K^2 + 5K = 1$, $(6K - 1)(K + 1) = 0$, so $K = \frac{1}{6}$ (a probability cannot be negative).
$P(X > 2) = K + 2K + 5K^2 = \frac{1}{2} + \frac{5}{36} = \frac{23}{36}$.
Answer: (3). | ||||
| 68 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctHard | |
A hard single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Total ways: $4^{10}$.
Choose the box for 2 balls (4 ways) and a different box for 3 balls (3 ways); choose its 2 balls ($^{10}C_2 = 45$) and the other box's 3 balls ($^8C_3 = 56$); the remaining 5 balls go into the other two boxes in $2^5$ ways.
Probability $= \frac{4 \times 3 \times 45 \times 56 \times 2^5}{4^{10}} = \frac{967680}{2^{20}} = \frac{945}{2^{10}}$.
Answer: (1). | ||||
| 69 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series For $0 < \theta < \frac{\pi}{4}$, $0 < \tan^2\theta < 1$ and $0 < \cos^2\theta < 1$, so both infinite G.P.s converge.
$x = \frac{1}{1 + \tan^2\theta} = \cos^2\theta$; $y = \frac{1}{1 - \cos^2\theta} = \frac{1}{\sin^2\theta}$.
$y(1 - x) = \frac{\sin^2\theta}{\sin^2\theta} = 1$.
Answer: (2). | ||||
| 70 | 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 An implication is false only when its premise is true and its conclusion false: p = T and $p \wedge \sim q$ = F.
With p = T, $p \wedge \sim q$ = F needs $\sim q$ = F, i.e. q = T.
Answer: (4). | ||||
| 71 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The sum is $\sum_{r=0}^{25}(4r + 1)\,{}^{25}C_r = 4\sum r\,{}^{25}C_r + \sum{}^{25}C_r = 4 \times 25 \times 2^{24} + 2^{25}$.
$= 2^{25}(50 + 1) = 51 \cdot 2^{25}$, so $k = 51$.
Answer: 51. | ||||
| 72 | 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 first A.P. is the numbers $\equiv 3 \pmod 4$ up to 407; the second the numbers $\equiv 2 \pmod 7$ up to 709.
The first common term is 23 ($23 = 4\cdot5 + 3 = 7\cdot3 + 2$), and common terms repeat every $\mathrm{lcm}(4, 7) = 28$: 23, 51, 79, …
$23 + 28(m - 1) \le 407$ gives $m - 1 \le 13.7$, so $m = 14$ terms.
Answer: 14. | ||||
| 73 | CirclesStandard and general form | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections First circle: centre (3, 0), radius $\sqrt{9 - 8} = 1$. Second: centre (0, 4), radius $\sqrt{16 - 16 + k} = \sqrt{k}$.
Distance between centres $= 5$.
External touching: $\sqrt{k} + 1 = 5$, $k = 16$. Internal touching: $\sqrt{k} - 1 = 5$, $k = 36$.
Largest value: 36.
Answer: 36. | ||||
| 74 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The lines through (−1, 3, −1) along (2, 4, 3) and through (−3, −2, 1) along (2, 6, λ) are coplanar when
$\begin{vmatrix} -2 & -5 & 2 \\ 2 & 4 & 3 \\ 2 & 6 & \lambda \end{vmatrix} = -2(4\lambda - 18) + 5(2\lambda - 6) + 2(12 - 8) = 2\lambda + 14 = 0$, so $\lambda = -7$.
Normal: $(2, 4, 3) \times (2, 6, -7) = (-46, 20, 4) \parallel (23, -10, -2)$.
Plane through (−1, 3, −1): $23x - 10y - 2z + 51 = 0$, parallel to the given plane.
Distance $= \frac{|51 - 48|}{\sqrt{23^2 + 10^2 + 2^2}} = \frac{3}{\sqrt{633}}$, so $k = 3$.
Answer: 3. | ||||
| 75 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{b}\cdot\vec{c} = 5|\vec{c}|\cos\frac{\pi}{3} = 10$, so $|\vec{c}| = 4$, and $|\vec{b}\times\vec{c}| = 5 \times 4 \times \sin\frac{\pi}{3} = 10\sqrt3$.
$\vec{a} \perp (\vec{b}\times\vec{c})$, so $|\vec{a}\times(\vec{b}\times\vec{c})| = |\vec{a}|\,|\vec{b}\times\vec{c}| = \sqrt3 \times 10\sqrt3 = 30$.
Answer: 30. | ||||
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