JEE (Main) · 2020
JEE (Main) 2020 — 2 September, 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
- Electrostatics2
12 easy · 13 medium · 0 hard
Chemistry
25 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Surface Chemistry2
- Coordination Compounds2
13 easy · 11 medium · 1 hard
Mathematics
25 Q
- Applications of Derivatives3
- Matrices and Determinants2
- Sequences and Series2
9 easy · 16 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 Young's modulus is stress divided by strain, and strain has no dimensions, so $Y$ has the dimensions of stress, force per area: $[Y] = \frac{[F]}{[A]}$.
In terms of F, A and V this is $F^1A^{-1}V^0$ (no power of speed is needed).
Answer: (C). | ||||
| 2 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Convert: 36 km/h = 10 m/s, 72 km/h = 20 m/s, 1.8 km/h = 0.5 m/s.
Take the direction of train A as positive. Train A: $+10$ m/s; the person walks backwards in A, so his ground velocity is $10 - 0.5 = +9.5$ m/s. Train B: $-20$ m/s.
Velocity of the person relative to B: $9.5 - (-20) = 29.5$ m/s.
Answer: (D). | ||||
| 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 Let the 3m mass move off with velocity $(V_x, V_y)$.
Momentum along x: $mu = 3mV_x \Rightarrow V_x = \frac{u}{3}$. Along y: $0 = mv + 3mV_y \Rightarrow V_y = -\frac{v}{3}$.
Kinetic energy is conserved: $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + \frac{1}{2}(3m)\left(\frac{u^2}{9} + \frac{v^2}{9}\right)$, so $u^2 = v^2 + \frac{u^2}{3} + \frac{v^2}{3}$.
Hence $\frac{2u^2}{3} = \frac{4v^2}{3}$, $v^2 = \frac{u^2}{2}$ and $v = \frac{u}{\sqrt{2}}$.
Answer: (C). | ||||
| 4 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion The bead moves in a horizontal circle of radius $a$. With a smooth wire the only forces are gravity $mg$ and the normal reaction $N$, which is perpendicular to the wire.
If the tangent makes angle $\theta$ with the horizontal, the normal makes angle $\theta$ with the vertical: $N\cos\theta = mg$ and $N\sin\theta = m\omega^2 a$, so $\tan\theta = \frac{\omega^2 a}{g}$.
The slope of the wire at P is $\tan\theta = \frac{dy}{dx} = 8Ca$.
So $8Ca = \frac{\omega^2 a}{g}$, $\omega^2 = 8gC$ and $\omega = 2\sqrt{2gC}$.
Answer: (A). | ||||
| 5 | 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 Just as the cylinder starts to rise, it leaves the floor and turns about the edge of the step, so take torques about the edge.
F acts at O perpendicular to the line from the edge to O (length R), so its moment arm is R: torque $FR$.
The centre O is a height $R - a$ above the edge, so its horizontal distance from the edge is $\sqrt{R^2 - (R-a)^2}$; the weight's torque is $Mg\sqrt{R^2 - (R-a)^2}$.
Minimum F: $FR = Mg\sqrt{R^2 - (R-a)^2}$, so $F = Mg\sqrt{1 - \left(\frac{R-a}{R}\right)^2}$.
Answer: (B). | ||||
| 6 | GravitationOrbital motion | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Mass inside radius R: $M(R) = \int_0^R \frac{K}{r}\,4\pi r^2\,dr = 2\pi K R^2$.
By the shell theorem only this mass pulls the star toward the centre. For the circular orbit, $\frac{mv^2}{R} = \frac{GM(R)m}{R^2}$, so $v^2 = \frac{GM(R)}{R} = 2\pi GKR$.
Then $T = \frac{2\pi R}{v}$ and $T^2 = \frac{4\pi^2R^2}{2\pi GKR} = \frac{2\pi R}{GK}$, so $T^2 \propto R$.
Answer: (B). | ||||
| 7 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take torques about B. The rod's weight $mg$ acts at 50 cm from A, i.e. 50 cm from B; the hung weight $2mg$ is 25 cm from B.
$T_A \times 100 = mg \times 50 + 2mg \times 25 = 100\,mg$, so $T_A = mg$.
(Check: $T_B = 3mg - mg = 2mg$.)
Answer: (D). | ||||
| 8 | Properties of Bulk MatterFluid mechanics | 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 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 In the rotating frame the free surface is a surface of constant $p$: along it $\frac{dy}{dx} = \frac{\omega^2 x}{g}$, so $y = \frac{\omega^2x^2}{2g}$ (a paraboloid).
Between the axis and the wall ($x = r = 5$ cm): $h = \frac{\omega^2r^2}{2g} = \frac{25\omega^2}{2g}$, with $r$ in cm (and $g$ in the matching unit, $\mathrm{cm\,s^{-2}}$).
Answer: (B). | ||||
| 9 | 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 oxygen has 5 degrees of freedom: $U_{O_2} = 3\times\frac{5}{2}RT = 7.5RT$.
Monatomic argon has 3: $U_{Ar} = 5\times\frac{3}{2}RT = 7.5RT$.
Total $= 15RT$.
Answer: (C). | ||||
| 10 | 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 Fundamental frequency $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$; the strings are identical (same $L$ and $\mu$), so $f \propto \sqrt{T}$.
$\frac{T_X}{T_Z} = \left(\frac{450}{300}\right)^2 = 1.5^2 = 2.25$.
Answer: (B). | ||||
| 11 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Inside the field the force is $q\vec{E} = -qE\hat{j}$, so the acceleration is $\frac{qE}{m}$ along $-y$ while $v_x = V_0$ stays constant.
At $x = d$ (time $t = \frac{d}{V_0}$): $y_1 = -\frac{qEd^2}{2mV_0^2}$ and $v_y = -\frac{qEd}{mV_0}$, so the slope is $\frac{v_y}{V_0} = -\frac{qEd}{mV_0^2}$.
Beyond $x = d$ there is no field and the particle moves in a straight line: $y = y_1 - \frac{qEd}{mV_0^2}(x - d) = \frac{qEd^2}{2mV_0^2} - \frac{qEd}{mV_0^2}x = \frac{qEd}{mV_0^2}\left(\frac{d}{2} - x\right)$.
Answer: (D). | ||||
| 12 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Room-temperature resistivities (in $\Omega\,\mathrm{m}$): copper $\approx 1.7\times10^{-8}$, aluminium $\approx 2.7\times10^{-8}$, tungsten $\approx 5.6\times10^{-8}$, mercury $\approx 98\times10^{-8}$.
So $\rho_M > \rho_T > \rho_A > \rho_C$. Option (A) puts copper highest, (B) puts aluminium above tungsten, (C) puts aluminium above mercury: all false. (D) $\rho_M > \rho_A > \rho_C$ is true.
Answer: (D). | ||||
| 13 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy 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 Period of the circular part: $T = \frac{2\pi m}{qB}$. Pitch $= v\cos60^\circ \times T = \frac{2\pi m v\cos60^\circ}{qB}$.
$= \frac{2\pi \times 1.67\times10^{-27}\times 2\times10^5}{1.69\times10^{-19}\times0.3} = \frac{2.10\times10^{-21}}{5.07\times10^{-20}} \approx 0.041$ m $\approx 4$ cm.
Answer: (C). | ||||
| 14 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter A permanent magnet must keep its magnetisation (large retentivity) and resist being demagnetised (large coercivity): (A) is right and (B) is not.
A transformer core is taken round its hysteresis loop every cycle, so it needs a thin loop: small retentivity and small coercivity (and high permeability). (C) and (D) both give it large retentivity, so neither fits.
Answer: (A). | ||||
| 15 | 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 The average energy density of a plane wave (electric and magnetic parts together) is $u = \frac{B_0^2}{2\mu_0}$.
$\mu_0 = \frac{1}{\epsilon_0c^2} = \frac{4\pi\times9\times10^9}{9\times10^{16}} = 4\pi\times10^{-7}$ SI units.
$B_0 = \sqrt{2\mu_0u} = \sqrt{2\times4\pi\times10^{-7}\times1.02\times10^{-8}} = \sqrt{2.56\times10^{-14}} \approx 1.6\times10^{-7}$ T $= 160$ nT.
Answer: (A). | ||||
| 16 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The mirror is the silvered part of the sphere, its back (hatched) outside, so it reflects from the inside: a concave mirror with its pole at the 0 mark. The object stands at the 10 cm mark.
The dashed sphere spans the 0 and 16 marks, so its diameter is 16 cm: R = 8 cm and f = −4 cm, with u = −10 cm (the object is beyond C).
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{4} + \frac{1}{10} = -\frac{3}{20}$, so $v = -6.67$ cm and $m = -\frac{v}{u} = -\frac{2}{3}$.
The image is real, inverted and smaller than the object (not magnified).
Answer: (A). | ||||
| 17 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Path difference at a point a distance $y$ from the centre: $\Delta = \frac{yd}{D} = \frac{1.27\times10^{-3}\times1\times10^{-3}}{1} = 1.27\times10^{-6}$ m $= 1.27\ \mu$m.
(This is about $2\lambda = 1.266\ \mu$m, the second bright fringe.)
Answer: (A). | ||||
| 18 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Nuclei in 2 kg: $\frac{2}{235}\times6.023\times10^{26} = 5.13\times10^{24}$.
Energy: $5.13\times10^{24}\times200\times1.6\times10^{-13}$ J $= 1.64\times10^{14}$ J.
Time: $30\times86400 = 2.592\times10^6$ s. Power $= \frac{1.64\times10^{14}}{2.592\times10^6} \approx 6.3\times10^7$ W $\approx 60$ MW.
Answer: (A). | ||||
| 19 | Communication SystemsAmplitude modulation and bandwidth | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on amplitude modulation and bandwidth, from Communication Systems. NCERT: No matching NCERT chapter Expand: $v_m = 5\sin\omega_ct + 3\cos\omega_mt\sin\omega_ct = 5\sin\omega_ct + \frac{3}{2}\sin(\omega_c+\omega_m)t + \frac{3}{2}\sin(\omega_c-\omega_m)t$.
The wave is a carrier of amplitude 5 V with two side bands of amplitude $\frac{mA_c}{2} = \frac{0.6\times5}{2} = \frac{3}{2}$ V each.
So the smallest and largest component amplitudes are $\frac{3}{2}$ V and 5 V.
Answer: (C). | ||||
| 20 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= 1\text{ MSD} - 1\text{ VSD} = 1 - 0.9 = 0.1$ mm $= 0.01$ cm.
Zero error: the vernier zero is to the right of the main-scale zero with the 7th division coinciding, a positive error of $7\times0.01 = +0.07$ cm.
Observed reading: $3.1 + 4\times0.01 = 3.14$ cm. Corrected length $= 3.14 - 0.07 = 3.07$ cm.
Answer: (B). | ||||
| 21 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Let $AC = L$, so $BC = 2L$ and $AB = 3L$. The block starts and ends at rest.
Work by gravity over AB: $mg\sin\theta\times3L$. Work against friction over CA only: $\mu mg\cos\theta\times L$.
Work-energy theorem: $3mgL\sin\theta = \mu mgL\cos\theta$, so $\mu = 3\tan\theta$.
Answer: 3. | ||||
| 22 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics For a rigid diatomic gas $\gamma = \frac{7}{5}$ and $C_V = \frac{5}{2}R$.
Adiabatic: $T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 293\times10^{0.4} = 293\times2.512 \approx 736$ K.
$\Delta U = nC_V\Delta T = 5\times\frac{5}{2}\times8.314\times(736 - 293) \approx 103.9\times443 \approx 4.60\times10^4$ J $\approx 46$ kJ.
Answer: 46. | ||||
| 23 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Connecting the charged capacitor across the uncharged one (the two form a single loop) shares the charge; the energy lost is $\Delta U = \frac{1}{2}\frac{C_1C_2}{C_1+C_2}V^2$.
$\frac{C_1C_2}{C_1+C_2} = \frac{5\times2.5}{7.5} = \frac{5}{3}\ \mu$F, so $\Delta U = \frac{1}{2}\times\frac{5}{3}\times10^{-6}\times220^2 = 0.0403$ J $= \frac{4.03}{100}$ J.
Answer: 4. | ||||
| 24 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Half a rotation ($\pi$ rad) in 0.2 s: $\omega = \frac{\pi}{0.2} = 5\pi$ rad/s.
Flux $\Phi = BA\cos\omega t$, so $\varepsilon_{max} = BA\omega = 3\times10^{-5}\times\pi(0.1)^2\times5\pi = 1.5\times10^{-6}\pi^2 \approx 1.48\times10^{-5}$ V $\approx 14.8\ \mu$V.
Nearest integer: 15.
Answer: 15. | ||||
| 25 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalMedium | |
A medium numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $eV = \frac{hc}{\lambda} - \phi$ and $\frac{eV}{4} = \frac{hc}{3\lambda} - \phi$.
Multiply the second by 4 and equate: $\frac{hc}{\lambda} - \phi = \frac{4hc}{3\lambda} - 4\phi$, so $3\phi = \frac{hc}{3\lambda}$ and $\phi = \frac{hc}{9\lambda}$.
Threshold wavelength $\lambda_0 = \frac{hc}{\phi} = 9\lambda$.
Answer: 9. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | SolutionsRaoult's law | Class 12: Solutions | Single correctEasy | |
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Dissolved glucose (non-volatile) takes up part of the surface, so fewer water molecules escape from the liquid per second: the rate of evaporation, i.e. of molecules leaving the solution, decreases. This is the molecular picture of the lowering of vapour pressure.
The rate at which molecules leave the vapour (condensation) depends on the vapour's own concentration, which does not change at the instant glucose is added; it falls only later as the vapour pressure settles at its new lower value.
Answer: (D). | ||||
| 27 | EquilibriumSolubility product | Class 11: Equilibrium | Assertion and reasonMedium | |
A medium assertion and reason question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium (A) is true: ionic precipitation reactions such as $\mathrm{Cu^{2+} + S^{2-} \to CuS}$ need no bonds to be broken and are practically instantaneous.
(R) is true: the equilibrium constant of precipitation is $K = \frac{1}{K_{sp}}$, very large because $K_{sp}$ of CuS is extremely small.
But an equilibrium constant says how far a reaction goes, not how fast; the speed is a matter of kinetics. So (R) does not explain (A).
Answer: (B). | ||||
| 28 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Bredig's arc method makes metal sols (gold, silver, platinum) by striking an arc between metal electrodes under water: the metal vaporises and condenses into colloidal particles.
Mond process and van Arkel method refine nickel and zirconium/titanium; the Ostwald process makes nitric acid.
Answer: (D). | ||||
| 29 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure With four identical B atoms, a tetrahedral, square planar or rectangular planar arrangement is symmetric: the bond dipoles cancel and the molecule is non-polar.
In a square pyramidal $\mathrm{AB_4}$ (A at the apex above the square of four B atoms, the shape taken when A carries a lone pair opposite the base) all four bond dipoles have a component along the axis in the same sense, so they add: the molecule is polar.
Answer: (D). | ||||
| 30 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | Single correctMedium | |
A medium single correct question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom (A) Black-body radiation curves (a peak that shifts with temperature) could only be explained by Planck's quantisation of energy.
(B) Photoelectron kinetic energy rising linearly with frequency above a threshold that differs for Rb, K and Na is Einstein's photon picture of the photoelectric effect.
(D) A line absorption spectrum shows that atoms absorb only definite energies: quantised energy levels.
(C) The internal energy of argon rising linearly with temperature ($U = \frac{3}{2}nRT$ for a monatomic gas) is classical kinetic theory and needs no quantum idea.
Answer: (C). | ||||
| 31 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry For an ideal gas $PV = \frac{m}{M}RT$, so $d = \frac{PM}{RT}$.
At constant P, $d \propto \frac{1}{T}$: d against T is a falling hyperbola (I is correct) and d against $\frac{1}{T}$ is a straight line through the origin (III is correct). At constant T, $d \propto P$ (IV is correct).
II shows density rising with temperature, which is not correct.
Answer: (B). | ||||
| 32 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across a period the effective nuclear charge rises: electronegativity, ionization enthalpy and (in magnitude) electron gain enthalpy generally increase, while the atomic radius decreases.
Answer: (B). | ||||
| 33 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter An acid-base titration needs a burette on a clamp/stand, a pipette, distilled water, an indicator such as phenolphthalein and a white porcelain tile to see the colour change.
No heating is needed for HCl against NaOH, and volumes are measured with the pipette and burette, not a measuring cylinder.
Answer: (C). | ||||
| 34 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Caesium has the lowest ionization enthalpy (work function) of the stable alkali metals, so visible light ejects electrons from it easily; it is the metal used in photoelectric cells.
Answer: (D). | ||||
| 35 | 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 Gas C, basic and made from B and $\mathrm{H_2}$ over a catalyst, is ammonia; so B is $\mathrm{N_2}$ (Haber process).
$\mathrm{NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O}$; $\mathrm{(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O}$; $\mathrm{2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2}$: all give $\mathrm{N_2}$.
$\mathrm{2Pb(NO_3)_2 \xrightarrow{\Delta} 2PbO + 4NO_2 + O_2}$ gives no $\mathrm{N_2}$ (its $\mathrm{O_2}$ with $\mathrm{H_2}$ gives water, not a basic gas).
Answer: (D). | ||||
| 36 | 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) True: octahedral Mn(II) ($d^5$) with weak-field ligands is high spin, and tetrahedral complexes have so small a splitting ($\Delta_t \approx \frac{4}{9}\Delta_o$) that they are high spin.
(II) True: for the same reason a tetrahedral Ni(II) complex is practically never low spin.
(III) True: strong-field ligands such as $\mathrm{CN^-}$ give low-spin Mn(II), e.g. $\mathrm{[Mn(CN)_6]^{4-}}$ with one unpaired electron.
(IV) False: aqueous $\mathrm{Mn^{2+}}$ is very pale pink, not yellow.
Answer: (B). | ||||
| 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 $\mu = \sqrt{n(n+2)} = 4.90$ BM gives $n = 4$ unpaired electrons: a high-spin $d^6$ ion.
Octahedral options: $-2.4\Delta_o + 2P$ is low-spin $t_{2g}^6e_g^0$, with no unpaired electrons; $-1.6\Delta_o$ is not the CFSE of any $d^6$ arrangement (high-spin octahedral $t_{2g}^4e_g^2$ gives $-0.4\Delta_o$). Neither fits.
Tetrahedral $d^6$ is always high spin, $e^3t_2^3$ with 4 unpaired electrons: CFSE $= 3(-0.6\Delta_t) + 3(+0.4\Delta_t) = -0.6\Delta_t$, with no extra pairing beyond the free ion.
Answer: (B). | ||||
| 38 | Environmental ChemistryOzone depletion | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ozone depletion, from Environmental Chemistry. NCERT: No matching NCERT chapter (A), (B) and (C) are true: stratospheric ozone absorbs UV; CFCs deplete it; ozone is toxic and $\mathrm{NO + O_3 \to NO_2 + O_2}$.
(D) is not true: $\mathrm{Cl^\cdot + O_3 \to ClO^\cdot + O_2}$ gives the chlorine monoxide radical, not chlorine dioxide.
Answer: (D). | ||||
| 39 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The compound is $\mathrm{CH_3-CH(CHO)-CH=CH-CH(CH_3)-COOH}$.
The carboxylic acid is the principal group (suffix -oic acid, C1). The longest chain containing both the COOH carbon and the CHO carbon has six carbons: C1 COOH, C2 CH($\mathrm{CH_3}$), C3=C4, C5 CH($\mathrm{CH_3}$), C6 CHO.
The CHO at the chain end is named as oxo: 2,5-dimethyl-6-oxohex-3-enoic acid.
Answer: (A). | ||||
| 40 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Percentage of bromine: $\frac{0.08}{0.172}\times100 = 46.5\%$.
4-Bromoaniline, $\mathrm{C_6H_6BrN}$ ($M = 172$): $\frac{80}{172} = 46.5\%$. 2,4-Dibromoaniline ($M = 251$): $\frac{160}{251} = 63.7\%$. $\mathrm{CH_3Br}$: $\frac{80}{95} = 84.2\%$. $\mathrm{C_2H_5Br}$: $\frac{80}{109} = 73.4\%$.
Answer: (A). | ||||
| 41 | 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 Protonation of the vinyl group (Markovnikov) gives a secondary carbocation $\mathrm{-CH^+-CH_3}$ on the carbon next to the ring carbon that carries the methyl.
A ring bond migrates to it (ring expansion): the strained five-membered ring becomes a six-membered ring and the charge moves to the former ring carbon, now a tertiary carbocation bearing a methyl, next to the carbon bearing the other methyl.
Loss of a proton from that neighbouring carbon gives the most substituted (tetrasubstituted) alkene, 1,2-dimethylcyclohexene.
Answer: (A). | ||||
| 42 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Addition of HCN (nucleophilic attack of $\mathrm{CN^-}$) is faster the more electron-poor the carbonyl carbon is.
(ii) o-nitrobenzaldehyde: $-NO_2$ withdraws by $-I$ and $-R$ from the ortho position: most reactive. (iv) m-nitrobenzaldehyde: $-I$ only from meta: next.
(i) m-methoxybenzaldehyde: at meta the OMe acts only by its $-I$ effect, a weak pull. (iii) o-methoxybenzaldehyde: the $+R$ effect of OMe at the ortho position feeds electrons to the carbonyl: least reactive.
Increasing order: (iii) < (i) < (iv) < (ii).
Answer: (B). | ||||
| 43 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctHard | |
A hard single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The starting compound is 2-methyl-2H-chromene: a benzene ring fused to a ring $\mathrm{O-CH(CH_3)-CH=CH-}$.
Excess HBr, heat: the alkyl-oxygen bond of the aryl alkyl ether is cleaved (the aryl-O bond is not), giving the phenol and a secondary alkyl bromide; HBr also adds to the C=C conjugated with the ring, Br going to the benzylic carbon. A: $\mathrm{o\text{-}HO-C_6H_4-CHBr-CH_2-CHBr-CH_3}$.
Alcoholic KOH removes both HBr (dehydrohalogenation) and acidification restores the free OH. B: $\mathrm{o\text{-}HO-C_6H_4-CH=CH-CH=CH_2}$.
Ozonolysis with $\mathrm{Zn/H_3O^+}$ cuts each C=C: the aromatic fragment is $\mathrm{o\text{-}HO-C_6H_4-CHO}$, salicylaldehyde (with glyoxal and formaldehyde).
Answer: (C). | ||||
| 44 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes In (C), 1-bromo-2-methylbutane, the bromine is on a $\mathrm{CH_2}$ that is not the stereocentre. Substitution ($S_N2$) occurs at that primary carbon and no bond to the chiral carbon is broken, so its configuration is retained.
(A) is benzylic and reacts largely by $S_N1$ (racemisation); (B), a chiral secondary bromide, undergoes $S_N2$ with inversion; (D) 2-bromopropane is not chiral at all.
Answer: (C). | ||||
| 45 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules (i) The glycoside (acetal) has used the anomeric OH; four OH groups are left: $x = 4$.
(ii) Reduction gives sorbitol (glucitol), with six OH groups: $y = 6$.
(iii) Glucose has five OH groups and forms a pentaacetate: $z = 5$.
Answer: (B). | ||||
| 46 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics 90 g of water is 5 mol. $\Delta H = 5\times41000 = 205000$ J.
$\mathrm{H_2O(l) \to H_2O(g)}$: $\Delta n_g = 5$ mol (liquid volume neglected), so $\Delta U = \Delta H - \Delta n_gRT = 205000 - 5\times8.314\times373 = 205000 - 15505.61 = 189494.39$ J.
Answer: 189494.39. | ||||
| 47 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Cu is oxidised (anode) and $\mathrm{Sn^{2+}}$ reduced (cathode): $E^0_{cell} = E^0_{cathode} - E^0_{anode} = -0.16 - 0.34 = -0.50$ V.
At unit concentrations $\Delta G = \Delta G^0 = -nFE^0_{cell} = -2\times96500\times(-0.50) = +96500$ J.
Answer: 96500. | ||||
| 48 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich isotherm: $\log\frac{x}{m} = \log k + \frac{1}{n}\log p$.
Intercept $\log k = 0.4771$, so $k = 3$; slope $\frac{1}{n} = 2$.
$\frac{x}{m} = kp^2 = 3\times4^2 = 48$.
Answer: 48. | ||||
| 49 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions (A) The complex ion is $\mathrm{[Fe(CN)_5(NOS)]^{4-}}$ (nitroprusside with sulphide); the ligand NOS carries charge $-1$ (NO$^+$ with $\mathrm{S^{2-}}$). $x + 5(-1) + (-1) = -4$, so $x = +2$.
(B) $\mathrm{[FeO_4]^{4-}}$: $y + 4(-2) = -4$, so $y = +4$.
(C) CO is neutral: $z = 0$.
Sum $= 2 + 4 + 0 = 6$.
Answer: 6. | ||||
| 50 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The molecule has the quinine skeleton: a quinuclidine (1-azabicyclo[2.2.2]octane) joined through a CH–O–aryl carbon to a quinoline that carries a $\mathrm{CH(OH)CH_3}$ group.
Chiral carbons: (1) the quinuclidine carbon bearing the methyl substituent; (2) the bridgehead CH, whose three bridges differ because two of them carry substituents; (3) the ring carbon next to N that carries the side chain; (4) the CH bonded to O-aryl, quinolinyl and that ring carbon; (5) the CH(OH)CH$_3$ carbon (H, OH, CH$_3$, aryl).
The aromatic carbons and the $\mathrm{CH_2}$ groups are not stereocentres (the bridgehead N is a stereocentre but not a carbon).
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions The domain of $R^{-1}$ is the range of $R$: the set of $y$ that occur.
$3y^2 \le 8 - x^2 \le 8$ forces $y^2 \le 2$, so $y \in \{-1, 0, 1\}$; each occurs (e.g. with $x = 0$: $3 \le 8$).
Answer: (C). | ||||
| 52 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Each root satisfies $5\alpha^2 + 6\alpha - 2 = 0$; multiply by $\alpha^4$: $5\alpha^6 + 6\alpha^5 - 2\alpha^4 = 0$, and likewise for $\beta$.
Adding: $5S_6 + 6S_5 - 2S_4 = 0$, i.e. $5S_6 + 6S_5 = 2S_4$.
Answer: (B). | ||||
| 53 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $\phi = \frac{\pi}{2} - \frac{2\pi}{9} = \frac{5\pi}{18}$; then $\sin\frac{2\pi}{9} = \cos\phi$ and $\cos\frac{2\pi}{9} = \sin\phi$.
Numerator $1 + \cos\phi + i\sin\phi = 2\cos\frac{\phi}{2}\,e^{i\phi/2}$; the denominator is its conjugate $2\cos\frac{\phi}{2}\,e^{-i\phi/2}$. The ratio is $e^{i\phi}$.
Cubed: $e^{3i\phi} = e^{i\,5\pi/6} = -\frac{\sqrt{3}}{2} + \frac{i}{2} = -\frac{1}{2}(\sqrt{3} - i)$.
Answer: (D). | ||||
| 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 For $A = \begin{pmatrix}a&b\\c&d\end{pmatrix}$ with entries 0 or 1, $|A| = ad - bc \in \{-1, 0, 1\}$.
$|A| = 1$ needs $ad = 1$, $bc = 0$: $a = d = 1$, so $\mathrm{tr}(A) = 2$. (Q) is true.
$A = \begin{pmatrix}1&1\\0&1\end{pmatrix} \ne I_2$ has $|A| = 1$, not $-1$. (P) is false.
Answer: (B). | ||||
| 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 $D = \begin{vmatrix}2&-1&2\\1&-2&\lambda\\1&\lambda&1\end{vmatrix} = 2(-2 - \lambda^2) + (1 - \lambda) + 2(\lambda + 2) = -2\lambda^2 + \lambda + 1$.
$D = 0 \iff 2\lambda^2 - \lambda - 1 = 0 \iff \lambda = 1$ or $\lambda = -\frac{1}{2}$; for every other $\lambda$ the solution is unique.
$\lambda = 1$: first minus third equation gives $x - 2y + z = -2$, contradicting the second ($= -4$): no solution.
$\lambda = -\frac{1}{2}$: twice the third equation is $2x - y + 2z = 8$, contradicting the first ($= 2$): no solution.
So $S = \{1, -\frac{1}{2}\}$ has exactly two elements.
Answer: (C). | ||||
| 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 $T_{r+1} = \binom{10}{r}\alpha^{10-r}\beta^r x^{\frac{10-r}{9} - \frac{r}{6}}$. Independent of $x$: $2(10 - r) = 3r$, $r = 4$, term $= \binom{10}{4}\alpha^6\beta^4 = 210(\alpha^3)^2(\beta^2)^2$.
With $\alpha^3 + \beta^2 = 4$, AM-GM gives $\alpha^3\beta^2 \le \left(\frac{4}{2}\right)^2 = 4$, so $(\alpha^3\beta^2)^2 \le 16$, equality at $\alpha^3 = \beta^2 = 2$.
Maximum $= 210\times16 = 3360 = 10k$, so $k = 336$.
Answer: (C). | ||||
| 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 Write the terms $\frac{a}{r}, a, ar$: product $a^3 = 27$, so $a = 3$ and $S = 3\left(1 + r + \frac{1}{r}\right)$.
$r + \frac{1}{r} \ge 2$ for $r > 0$ and $\le -2$ for $r < 0$, so $S \ge 9$ or $S \le 3(1 - 2) = -3$.
Answer: (B). | ||||
| 58 | 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 The $n$th bracket is $\frac{x^{n+1} - y^{n+1}}{x - y}$.
Sum $= \frac{1}{x - y}\left(\frac{x^2}{1 - x} - \frac{y^2}{1 - y}\right) = \frac{x^2(1 - y) - y^2(1 - x)}{(x - y)(1 - x)(1 - y)}$.
Numerator: $x^2 - y^2 - xy(x - y) = (x - y)(x + y - xy)$. So the sum is $\frac{x + y - xy}{(1 - x)(1 - y)}$.
Answer: (C). | ||||
| 59 | 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 Continuity at $x = 3$: $9c = 9a + 6c$, so $c = 3a$.
Continuity at $x = 1$: $ae + \frac{b}{e} = c = 3a$.
$f'(0) = a - b$ and $f'(2) = 2c\cdot2 = 12a$; so $13a - b = e$, $b = 13a - e$.
Substitute: $ae + \frac{13a - e}{e} = 3a \Rightarrow ae^2 + 13a - e = 3ae \Rightarrow a(e^2 - 3e + 13) = e$.
$a = \frac{e}{e^2 - 3e + 13}$.
Answer: (A). | ||||
| 60 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Slope of the line: $\frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = 1$.
Differentiating, $y' = 1 + \cos y\,y'$, so $y' = \frac{1}{1 - \cos y}$. Setting $y' = 1$ at (a, b) gives $\cos b = 0$, so $\sin b = \pm1$.
The point lies on the curve: $b = a + \sin b = a \pm 1$, so $|b - a| = 1$.
Answer: (A). | ||||
| 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 $p'(x) = k(x - 1)(x - 2) = k(x^2 - 3x + 2)$, so $p(x) = k\left(\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right) + C$.
$p(1) = \frac{5k}{6} + C = 8$ and $p(2) = \frac{2k}{3} + C = 4$. Subtracting, $\frac{k}{6} = 4$, $k = 24$ (positive, so $x = 1$ is indeed the maximum), and $C = 8 - 20 = -12$.
$p(0) = C = -12$.
Answer: (A). | ||||
| 62 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The ellipse has area $\pi\cdot2\cdot3 = 6\pi$.
$\frac{|x|}{2} + \frac{|y|}{3} = 1$ is the rhombus with vertices $(\pm2, 0)$, $(0, \pm3)$, inscribed in the ellipse; its area is $\frac{1}{2}\times4\times6 = 12$.
Required area $= 6\pi - 12 = 6(\pi - 2)$.
Answer: (C). | ||||
| 63 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $\frac{dy}{y + 1} = -\frac{\cos x}{2 + \sin x}dx$, so $\ln(y + 1) = -\ln(2 + \sin x) + C$, i.e. $(y + 1)(2 + \sin x) = K$.
$y(0) = 1$: $K = 2\times2 = 4$. At $x = \pi$: $(y + 1)\times2 = 4$, so $a = 1$.
$\frac{dy}{dx} = -\frac{\cos x\,(y + 1)}{2 + \sin x} = -\frac{(-1)(2)}{2} = 1$ at $x = \pi$, so $b = 1$.
Answer: (B). | ||||
| 64 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $x^2 + 7x + 2 - (3x - 3) = (x + 2)^2 + 1 > 0$, so the curve never meets the line; the nearest point is where the tangent is parallel to it: $2h + 7 = 3$, $h = -2$, $k = 4 - 14 + 2 = -8$.
Normal slope $-\frac{1}{3}$: $y + 8 = -\frac{1}{3}(x + 2)$, i.e. $x + 3y + 26 = 0$.
Answer: (C). | ||||
| 65 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent at $(x_1, y_1)$: $\frac{xx_1}{4} - \frac{yy_1}{2} = 1$, slope $\frac{x_1}{2y_1} = 2$, so $x_1 = 4y_1$.
On the hyperbola: $\frac{16y_1^2}{4} - \frac{y_1^2}{2} = 1 \Rightarrow \frac{7y_1^2}{2} = 1$, $y_1^2 = \frac{2}{7}$, $x_1^2 = \frac{32}{7}$.
$x_1^2 + 5y_1^2 = \frac{32}{7} + \frac{10}{7} = 6$.
Answer: (C). | ||||
| 66 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The line $2x = 3y$, $z = 1$ is $\frac{x}{3} = \frac{y}{2}$, $z = 1$: direction $(3, 2, 0)$.
The plane contains $(2,1,2) - (1,2,1) = (1, -1, 1)$ and $(3, 2, 0)$; normal $= (1, -1, 1)\times(3, 2, 0) = (-2, 3, 5)$.
Plane: $-2(x - 1) + 3(y - 2) + 5(z - 1) = 0$, i.e. $-2x + 3y + 5z = 9$.
Check: $(-2, 0, 1)$ gives $4 + 0 + 5 = 9$. The others give $-9$, $8$, $-8$.
Answer: (B). | ||||
| 67 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctMedium | |
A medium single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics For $1, 2, \ldots, 17$: mean 9, variance $\frac{n^2 - 1}{12} = \frac{288}{12} = 24$.
For Y: variance $a^2\times24 = 216$, $a = 3$ ($a > 0$); mean $9a + b = 17$, $b = -10$.
$a + b = -7$.
Answer: (C). | ||||
| 68 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Primes up to 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 (ten), so P(non-prime | I) $= \frac{20}{30} = \frac{2}{3}$.
Primes from 31 to 50: 31, 37, 41, 43, 47 (five), so P(non-prime | II) $= \frac{15}{20} = \frac{3}{4}$.
Bayes: $P(I \mid \text{non-prime}) = \frac{\frac{1}{2}\cdot\frac{2}{3}}{\frac{1}{2}\cdot\frac{2}{3} + \frac{1}{2}\cdot\frac{3}{4}} = \frac{8}{8 + 9} = \frac{8}{17}$.
Answer: (A). | ||||
| 69 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions The argument is positive, so only $\frac{|x| + 5}{x^2 + 1} \le 1$ is needed: $|x| + 5 \le |x|^2 + 1$, i.e. $t^2 - t - 4 \ge 0$ with $t = |x| \ge 0$.
So $t \ge \frac{1 + \sqrt{17}}{2}$, and the domain is $|x| \ge \frac{1 + \sqrt{17}}{2}$.
Answer: (B). | ||||
| 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 The contrapositive of $p \to q$ is $\sim q \to \sim p$: "If I will not catch the train, then I do not reach the station in time."
(B) is the inverse and (C) the converse; (A) is neither.
Answer: (D). | ||||
| 71 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Letters in order: E, H, M, O, R, T.
Words beginning with E or H: $2\times5! = 240$.
Beginning with M, second letter E or H: $2\times4! = 48$ (total 288).
MO, third letter E, H or R: $3\times3! = 18$ (total 306).
MOT, fourth letter E: $2! = 2$ (total 308). Then MOTH: MOTHER is the first of MOTHER, MOTHRE.
Position $= 308 + 1 = 309$.
Answer: 309. | ||||
| 72 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Write the numerator as $\sum_{k=1}^{n}(x^k - 1)$; each $\frac{x^k - 1}{x - 1} \to k$.
Limit $= 1 + 2 + \ldots + n = \frac{n(n+1)}{2} = 820$, so $n^2 + n - 1640 = 0$ and $n = 40$.
Answer: 40. | ||||
| 73 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $[0, 1]$: $|x - 1| - |x| = 1 - 2x$, so the integrand is $|1 - 2x|$ and $\int_0^1|1 - 2x|\,dx = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
On $[1, 2]$: $|x - 1| - |x| = (x - 1) - x = -1$, integrand 1, integral 1.
Total $= \frac{3}{2} = 1.5$.
Answer: 1.5. | ||||
| 74 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Circle: centre (1, 2), radius $\sqrt{1 + 4 - 4} = 1$.
Two points when the distance from the centre is less than 1: $\frac{|3 + 8 - k|}{5} < 1 \Rightarrow |k - 11| < 5 \Rightarrow 6 < k < 16$.
Integers $k = 7, 8, \ldots, 15$: 9 values.
Answer: 9. | ||||
| 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 $|\vec{a} - \vec{b}|^2 + |\vec{a} - \vec{c}|^2 = 4 - 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}) = 8$, so $\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = -2$.
Each dot product of unit vectors is at least $-1$, so both equal $-1$: $\vec{b} = \vec{c} = -\vec{a}$.
$|\vec{a} + 2\vec{b}|^2 + |\vec{a} + 2\vec{c}|^2 = 2(1 + 4 + 4\vec{a}\cdot\vec{b}) = 2(5 - 4) = 2$.
Answer: 2. | ||||
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