JEE (Main) · 2020
JEE (Main) 2020 — 9 January, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
25 Q
- Work, Energy and Power2
- Motion of System of Particles and Rigid Body2
- Properties of Bulk Matter2
14 easy · 11 medium · 0 hard
Chemistry
25 Q
- Coordination Compounds2
- Hydrocarbons2
- Some Basic Concepts of Chemistry2
15 easy · 9 medium · 1 hard
Mathematics
25 Q
- Integral Calculus4
- Matrices and Determinants2
- Applications of Derivatives2
14 easy · 11 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $[h] = [ML^2T^{-1}]$, $[c^5] = [L^5T^{-5}]$, $[G] = [M^{-1}L^3T^{-2}]$.
$\left[\frac{hc^5}{G}\right] = \frac{[ML^2T^{-1}][L^5T^{-5}]}{[M^{-1}L^3T^{-2}]} = [M^2L^4T^{-4}]$.
So $[f] = [ML^2T^{-2}]$, the dimension of energy.
Answer: (4). | ||||
| 2 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $W = \int \vec{F}\cdot d\vec{r} = \int_1^0 (-x)\,dx + \int_0^1 y\,dy$.
$\int_1^0 (-x)\,dx = \left[-\frac{x^2}{2}\right]_1^0 = \frac{1}{2}$ and $\int_0^1 y\,dy = \frac{1}{2}$.
Each part depends only on its own coordinate, so the path does not matter: $W = \frac{1}{2} + \frac{1}{2} = 1$ J.
Answer: (3). | ||||
| 3 | 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 Momentum: $mu\hat{i} + mu\frac{\hat{i}+\hat{j}}{2} = 2m\vec{v}$, so $\vec{v} = \frac{u}{4}(3\hat{i} + \hat{j})$ and $v^2 = \frac{10u^2}{16}$.
Initial kinetic energy: $\frac{1}{2}mu^2 + \frac{1}{2}m\cdot\frac{u^2}{2} = \frac{3}{4}mu^2$.
Final kinetic energy: $\frac{1}{2}(2m)\frac{10u^2}{16} = \frac{5}{8}mu^2$.
Loss $= \frac{3}{4}mu^2 - \frac{5}{8}mu^2 = \frac{1}{8}mu^2$.
Answer: (1). | ||||
| 4 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium 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 Each sphere about its own centre: $\frac{2}{5}m\left(\frac{d}{2}\right)^2 = \frac{md^2}{10}$.
About the centroid O: each centre is $\frac{d}{\sqrt{3}}$ away, so $I_0 = 3\left(\frac{md^2}{10} + \frac{md^2}{3}\right) = \frac{13}{10}md^2$.
About the centre of A: A contributes $\frac{md^2}{10}$; B and C are each $d$ away: $2\left(\frac{md^2}{10} + md^2\right)$. So $I_A = \frac{3}{10}md^2 + 2md^2 = \frac{23}{10}md^2$.
$\frac{I_0}{I_A} = \frac{13}{23}$.
Answer: (1). | ||||
| 5 | GravitationOrbital motion | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Momentum along the orbit: $mv + \frac{m}{2}\cdot\frac{v}{2} = \frac{3m}{2}V$, so $V = \frac{5}{6}v$.
At radius R the combined body moves tangentially with a speed less than the circular-orbit speed $v$ but greater than zero (and far below the escape speed $\sqrt{2}v$).
A tangential speed below the circular speed at that point gives a bound elliptical orbit with this point as its farthest point (apogee).
Answer: (4). | ||||
| 6 | Properties of Bulk MatterFluid mechanics | 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 fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Continuity: $A_1v_1 = A_2v_2$, so $v_2 = 2v_1$.
Bernoulli: $\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}(1000)(3v_1^2) = 1500v_1^2 = 700$, so $v_1^2 = 0.467$ and $v_1 = 0.683$ m/s.
Flow rate $= A_1v_1 = 40\times10^{-4}\times0.683 = 2.73\times10^{-3}$ m³/s $\approx 2732$ cm³/s, closest to 2720 cm³/s.
Answer: (3). | ||||
| 7 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics In the P-V cycle: 1 → 2 is an adiabatic expansion (V rises, P and T fall); 2 → 3 is at constant pressure with V falling (so T falls, $V \propto T$); 3 → 1 is at constant volume with P rising (so T rises).
On a V-T graph: 3 → 1 is a horizontal line (V constant) towards higher T. 2 → 3 is a straight line ($V \propto T$, pointing towards the origin) with V and T both falling. 1 → 2 is the adiabatic $TV^{\gamma-1} = $ constant, a curve along which V rises as T falls.
Option (1) has exactly this: horizontal 3 → 1, curved 1 → 2 rising to the left, straight 2 → 3 back down. Option (4) draws 1 → 2 as a straight line, (2) draws 2 → 3 curved and nearly vertical, and (3) has 2 → 3 at constant V.
Answer: (1). | ||||
| 8 | 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 Rigid diatomic: 5 degrees of freedom, $C_V^A = \frac{5}{2}R$.
With a vibrational mode (kinetic and potential terms): 7 degrees of freedom, $C_V^B = \frac{7}{2}R$.
Molar specific heats in the ratio 5 : 7 (the molecular masses do not enter the molar values).
Answer: (2). | ||||
| 9 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves With equal amplitudes a, the two waves at $\pm\frac{\pi}{4}$ add to $2a\cos\frac{\pi}{4} = \sqrt{2}a$ in phase with the first.
Resultant amplitude $a(1 + \sqrt{2})$, so $I = (1+\sqrt{2})^2 I_0 = (3 + 2\sqrt{2})I_0 \approx 5.83\,I_0$.
Answer: (4). | ||||
| 10 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Treat the remaining body as the full sphere minus the small sphere (cavity). The cavity touches the top of the big sphere, so its centre is $\frac{R}{2}$ above the centre O; A is O itself and B is the bottom of the big sphere.
At A: the full sphere gives zero at its centre; the cavity sphere gives $\frac{\rho r}{3\varepsilon_0}$ at a point $r = \frac{R}{2}$ from its centre. So $E_A = \frac{\rho R}{6\varepsilon_0}$.
At B: the full sphere gives $\frac{\rho R}{3\varepsilon_0}$; the cavity sphere (charge $\rho\cdot\frac{4}{3}\pi\frac{R^3}{8}$) at distance $\frac{3R}{2}$ gives $\frac{\rho (R/2)^3}{3\varepsilon_0 (3R/2)^2} = \frac{\rho R}{54\varepsilon_0}$ in the same direction, to be subtracted: $E_B = \frac{\rho R}{3\varepsilon_0} - \frac{\rho R}{54\varepsilon_0} = \frac{17\rho R}{54\varepsilon_0}$.
$\frac{E_A}{E_B} = \frac{1/6}{17/54} = \frac{9}{17} = \frac{18}{34}$.
Answer: (1). | ||||
| 11 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $\vec{E} = \frac{1}{4\pi\varepsilon_0}\frac{3(\vec{p}\cdot\hat{r})\hat{r} - \vec{p}}{r^3}$. With $\vec{p}\cdot\vec{r} = 0$ (equatorial point), $\vec{E} = -\frac{\vec{p}}{4\pi\varepsilon_0 r^3}$, antiparallel to $\vec{p}$.
Direction: $-\vec{p} \propto \hat{i} + 3\hat{j} - 2\hat{k}$.
Answer: (2). | ||||
| 12 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Inside ($r < a$): $B = \frac{\mu_0 I r}{2\pi a^2}$, so at $\frac{a}{3}$: $B_1 = \frac{\mu_0 I}{6\pi a}$.
Outside: $B = \frac{\mu_0 I}{2\pi r}$, so at 2a: $B_2 = \frac{\mu_0 I}{4\pi a}$.
$\frac{B_1}{B_2} = \frac{4}{6} = \frac{2}{3}$.
Answer: (3). | ||||
| 13 | 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 P is (0, a) and Q is (2a, 0). Only the electric force does work, and it moves the particle 2a along x.
(A) $qE(2a) = \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2 = \frac{3}{2}mv^2$, so $E = \frac{3mv^2}{4qa}$: true.
(B) At P the velocity $v\hat{i}$ is along the field: power $= qEv = \frac{3}{4}\frac{mv^3}{a}$: true.
(C) At Q the velocity $-2v\hat{j}$ is perpendicular to $E\hat{i}$, and the magnetic force never does work: total power zero, true.
(D) About O: at P, $|\vec{r}\times m\vec{v}| = mva$; at Q, $|2a\hat{i}\times(-2mv\hat{j})| = 4mva$. The difference is $3mav$, not $2mav$: false.
Answer: (4). | ||||
| 14 | 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 For each wave $\vec{B} = \frac{1}{c}\hat{n}\times\vec{E}$ with $\hat{n}$ the direction of travel.
Wave 1 travels along $\hat{i}$: $\vec{B}_1 = \frac{E_0}{c}\hat{i}\times\hat{j}\cos(\omega t - kx) = \frac{E_0}{c}\hat{k}\cos(\omega t - kx)$.
Wave 2 travels along $\hat{j}$: $\vec{B}_2 = \frac{E_0}{c}\hat{j}\times\hat{k}\cos(\omega t - ky) = \frac{E_0}{c}\hat{i}\cos(\omega t - ky)$.
At the origin at t = 0: $\vec{E} = E_0(\hat{j} + \hat{k})$, $\vec{B} = \frac{E_0}{c}(\hat{i} + \hat{k})$.
$\vec{v}\times\vec{B} = 0.8c\hat{j}\times\frac{E_0}{c}(\hat{i} + \hat{k}) = 0.8E_0(-\hat{k} + \hat{i})$.
$\vec{F} = q(\vec{E} + \vec{v}\times\vec{B}) = E_0q(0.8\hat{i} + \hat{j} + 0.2\hat{k})$.
Answer: (2). | ||||
| 15 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Seen from above (near-normal viewing), each layer appears $\frac{\text{thickness}}{\mu}$ deep:
$d = \frac{h}{\sqrt{2}} + \frac{h}{2\sqrt{2}} = \frac{3h}{2\sqrt{2}} = \frac{3}{4}h\sqrt{2}$.
Answer: (2). | ||||
| 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 Limit of resolution: $\theta = \frac{1.22\lambda}{D} = \frac{1.22\times5500\times10^{-10}}{5} = 1.34\times10^{-7}$ rad.
Separation on the moon: $x = \theta d = 1.34\times10^{-7}\times4\times10^{8} \approx 54$ m, close to 60 m.
Answer: (2). | ||||
| 17 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{\sqrt{2mE}}$, so halving $\lambda$ needs four times the kinetic energy: $E + \Delta E = 4E$, $\Delta E = 3E$.
Answer: (3). | ||||
| 18 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon energy: $\frac{12400}{6561} \approx 1.89$ eV.
Fastest electrons: $p = eBr = 1.6\times10^{-19}\times3\times10^{-4}\times10^{-2} = 4.8\times10^{-25}$ kg m/s.
$K_{max} = \frac{p^2}{2m} = \frac{(4.8\times10^{-25})^2}{2\times9.1\times10^{-31}} = 1.27\times10^{-19}$ J $\approx 0.79$ eV.
$\phi = 1.89 - 0.79 \approx 1.1$ eV.
Answer: (2). | ||||
| 19 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Pitch $= \frac{3}{6} = 0.5$ mm. Least count $= \frac{0.5}{50} = 0.01$ mm $= 0.001$ cm.
Answer: (3). | ||||
| 20 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity The wire makes B and D one node. Between A and BD: 1 Ω ∥ 4 Ω = 0.8 Ω; between BD and C: 2 Ω ∥ 3 Ω = 1.2 Ω. Total 2 Ω, so the cell drives 10 A.
Voltage A to BD: 8 V, so 8 A through 1 Ω (into B) and 2 A through 4 Ω (into D).
Voltage BD to C: 12 V, so 6 A through 2 Ω (out of B) and 4 A through 3 Ω (out of D).
At B: 8 A in, 6 A out, so 2 A flows from B to D through the wire.
Answer: (2). | ||||
| 21 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Differentiate: $2xv = 2at + 2b$, so $xv = at + b$.
Again: $v^2 + x\,\alpha = a$, so the acceleration $\alpha = \frac{a - v^2}{x} = \frac{ax^2 - (at+b)^2}{x^3}$.
$ax^2 - (at+b)^2 = a(at^2 + 2bt + c) - (a^2t^2 + 2abt + b^2) = ac - b^2$, a constant.
So $\alpha = \frac{ac - b^2}{x^3} \propto x^{-3}$: n = 3.
Answer: 3. | ||||
| 22 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical 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 centre of mass falls by $\frac{L}{2}\sin30° = 0.25$ m.
Energy: $mg(0.25) = \frac{1}{2}\cdot\frac{mL^2}{3}\omega^2 = \frac{m\omega^2}{6}$, so $\omega^2 = 1.5g = 15$ (with $g = 10$ m s⁻²).
Answer: 15. | ||||
| 23 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Largest tension: $T = \sigma A = 4.8\times10^7\times10^{-6} = 48$ N.
In the space station only the tension supplies the centripetal force: $m\omega^2 L = 48$, so $\omega^2 = \frac{48}{10\times0.3} = 16$, $\omega = 4$ rad s⁻¹.
Answer: 4. | ||||
| 24 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\varepsilon = L\frac{\Delta I}{\Delta t}$: $L = \frac{100\times0.025\times10^{-3}}{0.25} = 0.01$ H $= 10$ mH.
Answer: 10. | ||||
| 25 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The left diode points from the 12.7 V input towards A, and A is returned to the common line through a resistor, so it conducts: $V_A = 12.7 - 0.7 = 12$ V.
The right diode points from the 4 V terminal towards A; with A at 12 V its anode (4 V) is below its cathode, so it is reverse biased and carries no current.
Answer: 12. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Bohr's condition $2\pi r_n = n\lambda$ with $r_n = n^2a_0$ (hydrogen): $\lambda = \frac{2\pi n^2a_0}{n} = 2\pi n a_0$.
For n = 4: $\lambda = 8\pi a_0$.
Answer: (2). | ||||
| 27 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure 1.73 B.M. $= \sqrt{n(n+2)}$ with n = 1: one unpaired electron.
MO theory: $\mathrm{O_2}$ has two unpaired electrons in $\pi^*$; $\mathrm{O_2^+}$ (one $\pi^*$ electron) and $\mathrm{O_2^-}$ (three $\pi^*$ electrons) each have one.
Answer: (3). | ||||
| 28 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Atomisation of the liquid: $\mathrm{Br_2(l)} \rightarrow \mathrm{Br_2(g)} \rightarrow 2\mathrm{Br(g)}$, so $x = \Delta_{vap}H + y$.
The enthalpy of vaporisation is positive, so x > y.
Answer: (2). | ||||
| 29 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Same rate constant and same A: $\frac{E_a}{R(700)} = \frac{E_a - 30}{R(500)}$.
$500E_a = 700E_a - 21000$, so $E_a = 105$ kJ/mol uncatalysed and $105 - 30 = 75$ kJ/mol catalysed.
Answer: (3). | ||||
| 30 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium In 400 mL: $[\mathrm{Pb^{2+}}] = \frac{0.134\times300}{400} = 0.1005$ M; $[\mathrm{Cl^-}] = \frac{0.4\times100}{400} = 0.1$ M.
$Q = (0.1005)(0.1)^2 = 1.0\times10^{-3} > 1.6\times10^{-5}$.
Answer: (1). | ||||
| 31 | Solid StateClassification of solids | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on classification of solids, from Solid State. NCERT: No matching NCERT chapter A low melting point and no conduction in either state mark a molecular solid: carbon tetrachloride.
ZnS is ionic (high melting, conducts when molten), SiC is a covalent network (very high melting) and mercury is a metal (conducts).
Answer: (2). | ||||
| 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 (I), (II) and (III) are the reasons: the 2p electron penetrates less than a 2s electron, is better shielded by the core, and so is easier to remove.
(IV) is false: across the period the radius falls, so B is smaller than Be.
Answer: (1). | ||||
| 33 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter A reduces $\mathrm{BO_2}$ ($\mathrm{A + BO_2 \rightarrow AO_2 + B}$) when $\Delta G°$ for forming $\mathrm{AO_2}$ lies below that for forming $\mathrm{BO_2}$.
On the diagram the B line is the lower one up to the crossing at about 1400 °C and rises steeply above it; beyond 1400 °C the A line is lower.
Answer: (2). | ||||
| 34 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{N_2O_3}$ is acidic (anhydride of nitrous acid), $\mathrm{Li_2O}$ is basic and $\mathrm{Al_2O_3}$ amphoteric.
(1) starts with a basic oxide, (3) ends with acidic $\mathrm{P_4O_{10}}$, and (4) has basic $\mathrm{Na_2O}$ first.
Answer: (2). | ||||
| 35 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions N in $\mathrm{HNO_2}$ (+3), O in $\mathrm{H_2O_2}$ (−1) and S in $\mathrm{H_2SO_3}$ (+4) are in intermediate oxidation states and can go up or down.
P in $\mathrm{H_3PO_4}$ is at +5, its highest state, so it cannot be oxidised and does not act as a reducing agent.
Answer: (4). | ||||
| 36 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Eu: [Xe] $4f^76s^2$; $\mathrm{Eu^{2+}}$ loses the two 6s electrons: [Xe] $4f^7$.
Ce: [Xe] $4f^15d^16s^2$; $\mathrm{Ce^{3+}}$: [Xe] $4f^1$.
Answer: (3). | ||||
| 37 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds 3.83 BM $= \sqrt{15}$: three unpaired electrons, $d^3$, so Cr(III) and n = 3.
Geometrical isomerism needs two chlorides in the sphere (cis/trans $\mathrm{[Cr(H_2O)_4Cl_2]^+}$); the third chloride is ionic (reacts with $\mathrm{AgNO_3}$), and two waters are of crystallisation.
X = $\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O}$, tetraaquadichloridochromium(III) chloride dihydrate.
Answer: (3). | ||||
| 38 | 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 Square planar [Mabcd] has 3 geometrical isomers: n = 3, so the complex is $\mathrm{[Fe(CN)_6]^{3-}}$ with $\mathrm{Fe^{3+}}$ ($d^5$).
$\mathrm{CN^-}$ is a strong field ligand: $t_{2g}^5e_g^0$, one unpaired electron, $\mu = \sqrt{3} = 1.73$ BM.
CFSE $= 5\times(-0.4\Delta_0) = -2.0\Delta_0$ (pairing energy ignored).
Answer: (2). | ||||
| 39 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Aniline's $\mathrm{NH_2}$ and phenol's OH are Lewis bases: they bind $\mathrm{AlCl_3}$, which turns them into strongly deactivating groups and uses up the catalyst, so Friedel–Crafts fails or gives poor yield.
$\mathrm{CONH_2}$ deactivates the ring (and also binds the catalyst).
Chlorobenzene does not deactivate the catalyst and undergoes Friedel–Crafts alkylation/acylation at the ortho and para positions, so it gives the highest yield.
Answer: (4). | ||||
| 40 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A carbanion is a stronger base the weaker its conjugate acid.
$\mathrm{^-CN}$ (HCN, pKa ≈ 9) is the weakest; then $\mathrm{HC\equiv C^-}$ (sp carbon, pKa ≈ 25); then the allyl anion (ii), stabilised by resonance; then $\mathrm{^-CH_3}$ (iv); and the tert-butyl anion (i) is strongest, destabilised by three electron-releasing methyl groups.
(v) < (iii) < (ii) < (iv) < (i).
Answer: (2). | ||||
| 41 | Chemistry in Everyday LifeChemicals in food | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on chemicals in food, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Ninhydrin positive (amino acid unit with free $\mathrm{NH_2}$): aspartame and alitame, so A and D are these two.
C has chlorine but no nitrogen: sucralose.
Nitroprusside positive (sulphur): saccharin and alitame, so B and D are these two.
Hence D = alitame, A = aspartame, B = saccharin, C = sucralose.
Answer: (2). | ||||
| 42 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines 3-Bromoaniline is diazotised at 273–278 K (X = 3-bromobenzenediazonium chloride); Sandmeyer with $\mathrm{Cu_2Br_2}$ gives Y = 1,3-dibromobenzene.
Nitration: both Br are ortho/para directors. C-4 (and its equivalent C-6) is ortho to one Br and para to the other and is not crowded, so it is the main site; C-2 lies between the two Br and is hindered, C-5 is meta to both.
Z = 2,4-dibromo-1-nitrobenzene, option (1).
Answer: (1). | ||||
| 43 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Hydration of 3-methylbut-1-yne ($\mathrm{HgSO_4/H_2SO_4}$, Markovnikov) gives X = 3-methylbutan-2-one, $\mathrm{(CH_3)_2CH{-}CO{-}CH_3}$.
$\mathrm{C_2H_5MgBr}$ then water gives the tertiary alcohol $\mathrm{(CH_3)_2CH{-}C(OH)(CH_3)C_2H_5}$.
Acid-catalysed dehydration gives the most substituted (Saytzeff) alkene, the tetrasubstituted $\mathrm{(CH_3)_2C{=}C(CH_3)CH_2CH_3}$ (2,3-dimethylpent-2-ene).
Answer: (1). | ||||
| 44 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The three are isomeric hexa-2,4-dienes: (a) has both double bonds trans (E,E), (b) one trans and one cis (E,Z), (c) both cis (Z,Z).
Isomers burn to the same products, so the less stable isomer releases more heat. Each cis double bond adds steric strain: stability (a) > (b) > (c).
Heat of combustion: (a) < (b) < (c).
Answer: (4). | ||||
| 45 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The ozonolysis product has, on adjacent ring carbons, a CHO and a $\mathrm{C(CH_3)_2{-}CO{-}CH_3}$ group. So B is the indene in which ring–CH=C(CH₃)–C(CH₃)₂–ring closes a five-membered ring (1,1,2-trimethylindene); ozonolysis of its C=C gives exactly that aldehyde–ketone.
Working back: B forms when the tertiary allylic carbocation from the alcohol attacks the ortho position of the benzene ring (intramolecular Friedel–Crafts, conc. $\mathrm{H_2SO_4}$/Δ).
From (4), $\mathrm{Ph{-}CH{=}C(CH_3){-}CO{-}CH_3}$ (a methyl ketone, iodoform positive): $\mathrm{CH_3MgBr}$ gives $\mathrm{Ph{-}CH{=}C(CH_3){-}C(OH)(CH_3)_2}$; the cation $\mathrm{^+C(CH_3)_2}$ cyclises onto the ring to give the indene with H on C-3, $\mathrm{CH_3}$ on C-2 and two $\mathrm{CH_3}$ on C-1. Ozonolysis then gives the aryl CHO and the $\mathrm{C(CH_3)_2COCH_3}$ group.
(1) would give no methyl at C-2 (an aldehyde CHO on the side chain instead of a ketone); (2) and (3) put $\mathrm{CH_3}$ on the carbon next to the ring and give a ketone, not the CHO.
Answer: (4). | ||||
| 46 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry 1 L of solution: $1000\times1.4 = 1400$ g, of which 63% is $\mathrm{HNO_3}$: 882 g $= \frac{882}{63} = 14$ mol.
Molarity = 14 M.
Answer: 14. | ||||
| 47 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = iK_fm$ with i = 2 for NaCl: $0.2 = 2\times2\times m$, so m = 0.05 mol kg⁻¹.
For 0.6 kg water: 0.03 mol NaCl $= 0.03\times58.5 = 1.755$ g ≈ 1.76 g.
Answer: 1.76 (NTA set this question aside for all candidates; see the note). | ||||
| 48 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry 1 mol Ag needs 1 F. At the anode $\mathrm{2H_2O \rightarrow O_2 + 4H^+ + 4e^-}$: 1 F gives 0.25 mol $\mathrm{O_2}$.
$V = \frac{nRT}{P} = 0.25\times0.08314\times273 \approx 5.67$ L (at 1 bar the molar volume at 273 K is 22.7 L).
Answer: 5.67. | ||||
| 49 | HydrogenHydrogen and its compounds | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter $10^{-3}$ mol $\mathrm{MgSO_4}$ per litre is equivalent to $10^{-3}$ mol $\mathrm{CaCO_3}$ = 0.1 g = 100 mg per litre (about 1000 g of water).
Hardness = 100 ppm.
Answer: 100. | ||||
| 50 | Some Basic Concepts of ChemistryEmpirical and molecular formula | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Histamine, 2-(1H-imidazol-4-yl)ethanamine, is $\mathrm{C_5H_9N_3}$: M = 60 + 9 + 42 = 111 g mol⁻¹.
% N $= \frac{42}{111}\times100 = 37.8$%.
Answer: 37.8. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\int_0^1 f(x)\,dx = a + \frac{b}{2} + \frac{c}{3}$.
$f(0) + f(1) + 4f\left(\frac{1}{2}\right) = a + (a + b + c) + 4\left(a + \frac{b}{2} + \frac{c}{4}\right) = 6a + 3b + 2c$, and one sixth of this is $a + \frac{b}{2} + \frac{c}{3}$.
(Checking the others with f = 1: (1) gives $\frac{2}{3}$, (2) gives 2, (3) gives 10, not 1.)
Answer: (4). | ||||
| 52 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $t = e^x > 0$: $t^4 + t^3 - 4t^2 + t + 1 = 0$. Divide by $t^2$: $\left(t^2 + \frac{1}{t^2}\right) + \left(t + \frac{1}{t}\right) - 4 = 0$.
With $u = t + \frac{1}{t} \ge 2$: $u^2 - 2 + u - 4 = 0$, $u^2 + u - 6 = 0$, u = 2 or u = −3. Only u = 2 is allowed, giving t = 1, x = 0.
One real root.
Answer: (1). | ||||
| 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 $|z - i| = |z + 2i|$: z is equidistant from i and −2i, so it lies on $y = -\frac{1}{2}$.
$|z| = \frac{5}{2}$: $x^2 + \frac{1}{4} = \frac{25}{4}$, $x^2 = 6$.
$|z + 3i| = \sqrt{x^2 + \left(\frac{5}{2}\right)^2} = \sqrt{6 + \frac{25}{4}} = \sqrt{\frac{49}{4}} = \frac{7}{2}$.
Answer: (3). | ||||
| 54 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $|A| = 1(9 + 4) - 1(3 - 4) + 2(-1 - 3) = 13 + 1 - 8 = 6$.
$|B| = |\mathrm{adj}\,A| = |A|^2 = 36$; $|\mathrm{adj}\,B| = |B|^2 = 1296$.
$|C| = 3^3|A| = 162$. Ratio $= \frac{1296}{162} = 8$.
Answer: (2). | ||||
| 55 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A line of solutions needs infinitely many solutions, so the coefficient determinant is zero:
$\begin{vmatrix}1&4&-2\\1&7&-5\\1&5&\alpha\end{vmatrix} = (7\alpha + 25) - 4(\alpha + 5) - 2(5 - 7) = 3\alpha + 9 = 0$, so $\alpha = -3$.
Then the third row is $\frac{2}{3}$(first) + $\frac{1}{3}$(second): $(1, 5, -3) = \frac{2}{3}(1, 4, -2) + \frac{1}{3}(1, 7, -5)$. For consistency the right sides must match: $\frac{2}{3} + \frac{\beta}{3} = 5$, $\beta = 13$.
$\alpha + \beta = 10$.
Answer: (4). | ||||
| 56 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Tens digit fixed as 2. The leading digit cannot be 0 or 2: 8 choices. The remaining three places are filled from the 8 unused digits: $8\times7\times6 = 336$.
Total $= 8\times336$, so k = 8.
Answer: (1). | ||||
| 57 | 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 Powers of 2: $\frac{1}{4} + \frac{2}{16} + \frac{3}{48} + \frac{4}{128} + \ldots = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots$
This geometric series sums to $\frac{1/4}{1 - 1/2} = \frac{1}{2}$, so the product is $2^{\frac{1}{2}}$.
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 Left limit: $\frac{\sin(a+2)x}{x} + \frac{\sin x}{x} \to a + 2 + 1 = a + 3$.
Right limit: $\frac{x^{1/3}\left[(1 + 3x)^{1/3} - 1\right]}{x^{4/3}} = \frac{(1 + 3x)^{1/3} - 1}{x} \to \frac{1}{3}\cdot3 = 1$.
Continuity: $a + 3 = 1 = b$, so a = −2, b = 1 and a + 2b = 0.
Answer: (2). | ||||
| 59 | Applications of DerivativesRate of change | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on rate of change, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Ice volume $V = \frac{4}{3}\pi(r^3 - 10^3)$ with outer radius r, so $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$.
At thickness 5 cm, r = 15: $-50 = 4\pi(225)\frac{dr}{dt}$, so the thickness falls at $\frac{50}{900\pi} = \frac{1}{18\pi}$ cm/min.
Answer: (3). | ||||
| 60 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Mean value theorem: $\frac{f(c) - f(a)}{c - a} = f'(\xi_1)$ with $\xi_1 \in (a, c)$, and $\frac{f(b) - f(c)}{b - c} = f'(\xi_2)$ with $\xi_2 \in (c, b)$.
$f'' < 0$ makes $f'$ decreasing, so $f'(\xi_1) > f'(\xi_2) > 0$.
Hence $\frac{f(c) - f(a)}{f(b) - f(c)} > \frac{c - a}{b - c}$.
Answer: (2). | ||||
| 61 | Integral CalculusIntegral as an antiderivative | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\sec x + \tan x = \frac{1 + \sin x}{\cos x} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$, and $\frac{\pi}{4} + \frac{x}{2} \in \left(0, \frac{\pi}{2}\right)$, so $f'(x) = \frac{\pi}{4} + \frac{x}{2}$.
$f(x) = \frac{\pi x}{4} + \frac{x^2}{4}$ (as f(0) = 0), so $f(1) = \frac{\pi + 1}{4}$.
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 Write the integrand as $\frac{1}{(x + 4)^2}\left(\frac{x - 3}{x + 4}\right)^{-\frac{6}{7}}$.
Let $t = \frac{x - 3}{x + 4}$: $dt = \frac{7}{(x + 4)^2}dx$.
Integral $= \frac{1}{7}\int t^{-\frac{6}{7}}dt = t^{\frac{1}{7}} + C = \left(\frac{x - 3}{x + 4}\right)^{\frac{1}{7}} + C$.
Answer: (2). | ||||
| 63 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $g(x) = \frac{\sin^8 x}{\sin^8 x + \cos^8 x}$; $g(2\pi - x) = g(x)$. With $x \to 2\pi - x$: $I = \int_0^{2\pi}(2\pi - x)g\,dx$, so $I = \pi\int_0^{2\pi}g\,dx$.
g has period $\pi$ and is symmetric, so $\int_0^{2\pi}g = 4\int_0^{\pi/2}g$; and $\int_0^{\pi/2}\frac{\sin^8}{\sin^8 + \cos^8} = \frac{\pi}{4}$ (add it to the same integral with $x \to \frac{\pi}{2} - x$).
$I = \pi\cdot4\cdot\frac{\pi}{4} = \pi^2$.
Answer: (4). | ||||
| 64 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines C = $\left(\frac{3 + 1 + 2}{3}, \frac{-1 + 3 + 4}{3}\right)$ = (2, 2).
P: y = 3x + 1 in x + 3y = 1 gives 10x = −2, so P = $\left(-\frac{1}{5}, \frac{2}{5}\right)$.
Slope CP $= \frac{2 - 2/5}{2 + 1/5} = \frac{8}{11}$; line: $y - 2 = \frac{8}{11}(x - 2)$, i.e. 8x − 11y + 6 = 0.
(−9, −6): −72 + 66 + 6 = 0, on the line. (The others give 2, −1, −1.)
Answer: (4). | ||||
| 65 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Centre (h, 4), radius |h|. Through (2, 0): $(2 - h)^2 + 16 = h^2$, so 20 = 4h, h = 5: centre (5, 4), radius 5.
Distances from (5, 4): 3x − 4y − 24: |15 − 16 − 24|/5 = 5; 3x + 4y − 6: |15 + 16 − 6|/5 = 5; 4x + 3y − 8: |20 + 12 − 8|/5 = 24/5; 4x − 3y + 17: |20 − 12 + 17|/5 = 5.
Only 4x + 3y − 8 = 0 is not at distance 5.
Answer: (3). | ||||
| 66 | Conic SectionsStandard forms of conics | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections $e_1^2 = 1 - \frac{4}{18} = \frac{7}{9}$; $e_2^2 = 1 + \frac{4}{9} = \frac{13}{9}$.
$k = 15\cdot\frac{7}{9} + 3\cdot\frac{13}{9} = \frac{105 + 39}{9} = 16$.
Answer: (2). | ||||
| 67 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each draw is A or B with probability $\frac{1}{2}$, independently (with replacement).
The second A comes before the third B exactly when the first four draws contain at least two A's.
$P = 1 - P(\text{0 or 1 A in 4}) = 1 - \frac{1 + 4}{16} = \frac{11}{16}$.
Answer: (3). | ||||
| 68 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics For $y_i = x_i - 5$: mean 1, variance $\frac{40}{10} - 1^2 = 3$.
$x_i - 3 = y_i + 2$: mean $\mu = 3$, variance unchanged $\lambda = 3$.
Answer: (4). | ||||
| 69 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\cos\frac{3\pi}{8} = \sin\frac{\pi}{8}$ and $\sin\frac{3\pi}{8} = \cos\frac{\pi}{8}$. With $c = \cos\frac{\pi}{8}$, $s = \sin\frac{\pi}{8}$:
$c^3s + s^3c = sc(c^2 + s^2) = sc = \frac{1}{2}\sin\frac{\pi}{4} = \frac{1}{2\sqrt{2}}$.
Answer: (3). | ||||
| 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 $\sim(p \vee q) \equiv (\sim p) \wedge (\sim q)$.
With p: '$\sqrt{5}$ is an integer' and q: '5 is irrational', the negation is '$\sqrt{5}$ is not an integer and 5 is not irrational'.
Answer: (3). | ||||
| 71 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem A term takes $x^2$ from j factors and $x$ from i factors with i + 2j = 4.
j = 0, i = 4: $\binom{10}{4} = 210$. j = 1, i = 2: $\frac{10!}{1!\,2!\,7!} = 360$. j = 2, i = 0: $\binom{10}{2} = 45$.
Coefficient $= 210 + 360 + 45 = 615$.
Answer: 615. | ||||
| 72 | 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 $\log_{1/2}|\sin x| + \log_{1/2}|\cos x| = 2$, so $|\sin x\cos x| = \frac{1}{4}$ (both factors non-zero), i.e. $|\sin 2x| = \frac{1}{2}$.
For $x \in [0, 2\pi]$, $2x \in [0, 4\pi]$; $|\sin\theta| = \frac{1}{2}$ has 4 solutions in each interval of length $2\pi$, so 8 in all.
Answer: 8. | ||||
| 73 | 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 $[\vec{p}\ \vec{q}\ \vec{r}] = \begin{vmatrix}a+1&a&a\\a&a+1&a\\a&a&a+1\end{vmatrix} = 3a + 1 = 0$, so $a = -\frac{1}{3}$.
$\vec{p} = \frac{1}{3}(2, -1, -1)$, $\vec{q} = \frac{1}{3}(-1, 2, -1)$, $\vec{r} = \frac{1}{3}(-1, -1, 2)$.
$\vec{p}\cdot\vec{q} = \frac{-2 - 2 + 1}{9} = -\frac{1}{3}$, so $3(\vec{p}\cdot\vec{q})^2 = \frac{1}{3}$.
$\vec{r}\times\vec{q} = \frac{1}{9}(-3, -3, -3)$, so $|\vec{r}\times\vec{q}|^2 = \frac{27}{81} = \frac{1}{3}$.
$\frac{1}{3} - \frac{\lambda}{3} = 0$: $\lambda = 1$.
Answer: 1. | ||||
| 74 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalEasy | |
A easy numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} - \frac{y}{x + 1} = (x + 1) - \frac{3}{x + 1}$; integrating factor $\frac{1}{x + 1}$.
$\frac{d}{dx}\left(\frac{y}{x + 1}\right) = 1 - \frac{3}{(x + 1)^2}$, so $\frac{y}{x + 1} = x + \frac{3}{x + 1} + C$.
y(2) = 0: 0 = 2 + 1 + C, C = −3. At x = 3: $\frac{y}{4} = 3 + \frac{3}{4} - 3 = \frac{3}{4}$, y = 3.
Answer: 3. | ||||
| 75 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Segment: (1, −3, 8). Line direction: (4, −4, 7), of length $\sqrt{16 + 16 + 49} = 9$.
Projection $= \frac{4 + 12 + 56}{9} = \frac{72}{9} = 8$.
Answer: 8. | ||||
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