JEE (Main) · 2021
JEE (Main) 2021 — 17 March, 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
30 Q
- Motion of System of Particles and Rigid Body3
- Kinematics2
- Laws of Motion2
21 easy · 9 medium · 0 hard
Chemistry
30 Q
- Chemical Bonding and Molecular Structure2
- d- and f-Block Elements2
- Hydrocarbons2
24 easy · 4 medium · 2 hard
Mathematics
30 Q
- Matrices and Determinants3
- Limits, Continuity and Differentiability3
- Circles3
13 easy · 14 medium · 3 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Let the peak speed be $v$, reached after time $t_1$ and lost in time $t_2$: $v = \alpha t_1 = \beta t_2$, and $t_1 + t_2 = t$.
So $v\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) = t$, giving $v = \frac{\alpha\beta t}{\alpha+\beta}$.
The distance is the area of the velocity-time triangle: $s = \frac{1}{2}vt = \frac{\alpha\beta}{2(\alpha+\beta)}t^2$.
Answer: (B). | ||||
| 2 | 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 downward negative lift adds to the weight, so the normal reaction is $N = mg + F_L$.
Friction supplies the centripetal force; at the limit $\mu_s N = \frac{mv^2}{R}$, so $\mu_s(mg + F_L) = \frac{mv^2}{R}$.
$F_L = \frac{mv^2}{\mu_s R} - mg = m\left(\frac{v^2}{\mu_s R} - g\right)$.
Answer: (C). | ||||
| 3 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Efficiency $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{400}{800} = \frac{1}{2}$.
$Q_1 = \frac{W}{\eta} = \frac{1200}{1/2} = 2400$ J.
Answer: (A). | ||||
| 4 | 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 $B = \mu_0\mu_r nI = 4\pi\times10^{-7}\times500\times1000\times5 = 4\pi\times10^{-7}\times2.5\times10^{6} = \pi$ T.
Answer: (D). | ||||
| 5 | Current ElectricityDrift velocity and mobility | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity $I = neAv_d$, so $n = \frac{I}{eAv_d} = \frac{10}{1.6\times10^{-19}\times5\times10^{-6}\times2\times10^{-3}} = \frac{10}{1.6\times10^{-27}} = 6.25\times10^{27} = 625\times10^{25}\ \mathrm{m^{-3}}$.
Answer: (C). | ||||
| 6 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion About A (the centre of the circle) the position vector $\vec{r}$ lies in the horizontal plane and is perpendicular to the velocity, so $\vec{L}_A = M\vec{r}\times\vec{v}$ points along $+z$ with constant magnitude $Mr^2\omega$: it is constant.
About B the position vector runs along the rod, tilted at $\alpha$ to the axis. $\vec{L}_B = M\vec{r}_B\times\vec{v}$ is perpendicular to the rod and to $\vec{v}$; its magnitude $Mlv$ is constant, but its direction turns with the mass, sweeping a cone about the z-axis. So $L_B$ changes direction.
Answer: (B). | ||||
| 7 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current A hot wire ammeter reads the rms value. $I = \sqrt{I_1^2+I_2^2}\,\sin(\omega t+\phi)$, a sinusoid of amplitude $\sqrt{I_1^2+I_2^2}$.
$I_{rms} = \frac{\sqrt{I_1^2+I_2^2}}{\sqrt{2}} = \sqrt{\frac{I_1^2+I_2^2}{2}}$.
Answer: (C). | ||||
| 8 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory A non-linear polyatomic molecule has 3 translational and 3 rotational degrees of freedom; each vibrational mode contributes 2 (kinetic and potential).
$f = 3 + 3 + 2\times24 = 54$.
$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{54} = 1.037 \approx 1.03$.
Answer: (C). | ||||
| 9 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Excess pressures: $P_a = P_0 + \frac{4T}{a}$, $P_b = P_0 + \frac{4T}{b}$. The common film separates the two, so $\frac{4T}{r} = \frac{4T}{a} - \frac{4T}{b}$.
$\frac{1}{r} = \frac{1}{a} - \frac{1}{b} = \frac{b-a}{ab}$, so $r = \frac{ab}{b-a}$.
Answer: (A). | ||||
| 10 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The triangle is equilateral with side 10 cm, O at the origin and Q at (10, 0). So P is at $(5, 5\sqrt{3})$ cm.
About O: $\vec{r} = 5\hat{i} + 5\sqrt{3}\hat{j}$; $\tau_z = xF_y - yF_x = 5(-3) - 5\sqrt{3}(4) = -15 - 20\sqrt{3}$.
About Q: $\vec{r} = -5\hat{i} + 5\sqrt{3}\hat{j}$; $\tau_z = (-5)(-3) - 5\sqrt{3}(4) = 15 - 20\sqrt{3}$.
(Both along $\hat{k}$, in N cm with the lengths in cm.)
Answer: (A). | ||||
| 11 | Properties of Bulk MatterHeat transfer | 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 heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Each wire has length L and area A. In series the thermal resistances add: $\frac{2L}{K A} = \frac{L}{K_1A} + \frac{L}{K_2A}$.
$K = \frac{2K_1K_2}{K_1+K_2}$.
Answer: (B). | ||||
| 12 | 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 $\frac{1}{2}mv'^2 = 0.05\times\frac{1}{2}mv^2$, so $v' = v\sqrt{0.05} = 20\times0.2236 = 4.47\ \mathrm{ms^{-1}}$.
Answer: (B). | ||||
| 13 | 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 Refractive index $\mu = \frac{3\times10^8}{2\times10^8} = 1.5$.
The curved face has sag $h = 0.3$ cm over half-aperture $r = 3$ cm. For a sphere of radius R: $r^2 = h(2R - h)$, so $R = \frac{r^2 + h^2}{2h} = \frac{9 + 0.09}{0.6} \approx 15$ cm.
Lens maker's formula with one plane face: $\frac{1}{f} = (\mu - 1)\frac{1}{R}$, so $f = \frac{15}{0.5} = 30$ cm.
Answer: (A). | ||||
| 14 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei For a one-electron (hydrogen-like) ion, $E_n = -13.6\frac{Z^2}{n^2}$ eV. For carbon $Z = 6$.
Equal to the hydrogen ground state: $\frac{36}{n^2} = 1$, so $n = 6$.
Answer: (C). | ||||
| 15 | 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 Internal energy of $n$ molecules with $F$ degrees of freedom is $\frac{nF}{2}kT$. With no loss of energy:
$\frac{n_1F_1}{2}kT_1 + \frac{n_2F_2}{2}kT_2 = \frac{n_1F_1 + n_2F_2}{2}kT$.
$T = \frac{n_1F_1T_1 + n_2F_2T_2}{n_1F_1 + n_2F_2}$.
Answer: (C). | ||||
| 16 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $K = \frac{1}{2}k(A^2 - x^2)$ and $U = \frac{1}{2}kx^2$. Setting them equal: $A^2 - x^2 = x^2$, so $x = \pm\frac{A}{\sqrt{2}}$.
Answer: (C). | ||||
| 17 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The first two NOR gates have their inputs joined, so each is a NOT gate: they give $\bar{A}$ and $\bar{B}$.
The middle NOR gate gives $\overline{\bar{A}+\bar{B}} = AB$.
The last NOR gate, inputs joined, inverts it: $Y = \overline{AB}$, a NAND gate.
Answer: (C). | ||||
| 18 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Electron: $\lambda_e = \frac{h}{\sqrt{2mE}}$. Photon: $\lambda_p = \frac{hc}{E}$.
$\frac{\lambda_e}{\lambda_p} = \frac{E}{c\sqrt{2mE}} = \frac{1}{c}\left(\frac{E}{2m}\right)^{1/2}$.
Answer: (D). | ||||
| 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 Observed reading $= 8.5 + 6\times0.01 = 8.56$ cm.
A positive zero error of 0.2 mm = 0.02 cm is subtracted: $8.56 - 0.02 = 8.54$ cm.
Answer: (C). | ||||
| 20 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei In the Bohr model $v_n = \frac{Ze^2}{2\varepsilon_0 nh}$ (about $\frac{2.19\times10^6}{n}$ m/s for hydrogen), so $v_n \propto \frac{1}{n}$.
Answer: (D). | ||||
| 21 | Communication SystemsRange of a line-of-sight antenna | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on range of a line-of-sight antenna, from Communication Systems. NCERT: No matching NCERT chapter Range $d = \sqrt{2hR}$, so the service area is $\pi d^2 = 2\pi hR$.
$2\times3.14\times0.03\ \mathrm{km}\times6400\ \mathrm{km} = 3.14\times384 = 1205.76\ \mathrm{km}^2 \approx 1206$.
Answer: 1206. | ||||
| 22 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalMedium | |
A medium numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves For complete absorption at normal incidence the radiation pressure is $\frac{I}{c}$, so $F = \frac{IA}{c}$.
$I = \frac{Fc}{A} = \frac{2.5\times10^{-6}\times3\times10^{8}}{30\ \mathrm{cm}^2} = \frac{750}{30} = 25\ \mathrm{W/cm^2}$. The 40 minutes do not enter.
Answer: 25. | ||||
| 23 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Plate area $= 2\times\frac{3}{2} = 3\ \mathrm{cm}^2$, so each gap forms $C_0 = \frac{3\varepsilon_0}{d}$ (area in $\mathrm{cm}^2$, as the question intends).
B and D are joined, so they are one node. Gaps: A–B ($C_0$), B–C ($C_0$), C–D ($C_0$).
Between C and the B–D node the two capacitors are in parallel: $2C_0$. This is in series with A–B: $C = \frac{C_0\cdot2C_0}{3C_0} = \frac{2}{3}C_0 = \frac{2}{3}\cdot\frac{3\varepsilon_0}{d} = \frac{2\varepsilon_0}{d}$.
Answer: 2. | ||||
| 24 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity $s = R_1 + R_2$, $p = \frac{R_1R_2}{R_1+R_2}$, so $n = \frac{s}{p} = \frac{(R_1+R_2)^2}{R_1R_2}$.
By AM–GM, $(R_1+R_2)^2 \ge 4R_1R_2$, with equality when $R_1 = R_2$. So $n_{min} = 4$.
Answer: 4. | ||||
| 25 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charge stays fixed after the battery is removed. Initial energy $U_1 = \frac{1}{2}CV^2 = \frac{1}{2}\times14\times144 = 1008$ pJ.
With the slab fully in, $C' = 7C$ and $U_2 = \frac{Q^2}{2C'} = \frac{U_1}{7} = 144$ pJ.
The energy lost by the field becomes the slab's kinetic energy as it is pulled in: $1008 - 144 = 864$ pJ, the constant mechanical energy of the oscillation.
Answer: 864. | ||||
| 26 | GravitationEscape velocity | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}}$; at fixed mass $v_e \propto R^{-1/2}$.
For $v_e$ to become 10 times, R must become $\frac{1}{100}$ of itself: $\frac{6400}{100} = 64$ km.
Answer: 64. | ||||
| 27 | KinematicsUniform circular motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $900$ rpm $= 15$ rev/s and $2460$ rpm $= 41$ rev/s.
With uniform acceleration the revolutions equal the average rate times the time: $\frac{15+41}{2}\times26 = 28\times26 = 728$.
Answer: 728. | ||||
| 28 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $T_a = 2\pi\sqrt{\frac{M}{k}}$. Two springs in series have $k_{eq} = \frac{k}{2}$, so $T_b = 2\pi\sqrt{\frac{2M}{k}}$.
$\frac{T_b}{T_a} = \sqrt{2}$, so $x = 2$.
Answer: 2. | ||||
| 29 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rolling down an incline, $a = \frac{g\sin\theta}{1 + \frac{K^2}{R^2}}$; the body with the smallest $\frac{K^2}{R^2}$ arrives first.
Ring 1, disc $\frac{1}{2}$, solid cylinder $\frac{1}{2}$, solid sphere $\frac{2}{5}$. The solid sphere is smallest.
Answer: 4. | ||||
| 30 | Laws of MotionFriction | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Only friction accelerates the upper block, so its largest acceleration is $\mu g = \frac{3}{7}\times9.8 = 4.2\ \mathrm{ms^{-2}}$.
Moving together: $F_{max} = (m + M)\mu g = 5\times4.2 = 21$ N.
Answer: 21. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Three bond pairs and two lone pairs make five electron pairs: trigonal bipyramidal arrangement. The lone pairs take equatorial positions, leaving the three atoms in a T shape (as in $\mathrm{ClF_3}$).
Answer: (B). | ||||
| 32 | 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 Gas dispersed in a solid is a solid foam (pumice, foam rubber). Of the options, foam is that class: an aerosol has a liquid or solid in a gas, a solid sol a solid in a solid, a gel a liquid in a solid.
Answer: (C). | ||||
| 33 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Chlorine has the most negative electron gain enthalpy (about −349 kJ/mol). Fluorine's is less negative (about −328 kJ/mol) because its small, crowded 2p shell repels the added electron. Then Br (−325) and I (−295).
Order of magnitude: Cl > F > Br > I.
Answer: (B). | ||||
| 34 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In an Ellingham diagram, where two lines cross (point A, Al and Mg), the two oxide-formation reactions have equal $\Delta G^\circ$, so the reaction of one metal with the other's oxide has $\Delta G^\circ = 0$.
A sudden increase in slope marks a change of phase (melting or boiling): the entropy change of the oxide-forming reaction becomes more negative, and the slope $-\Delta S^\circ$ jumps.
Answer: (C). | ||||
| 35 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter (A) True: $\mathrm{D_2O}$ slows neutrons in reactors.
(B) True: it is obtained by exhaustive electrolysis of water or as a by-product in some fertilizer industries.
(C) True: it is used in exchange reactions to study reaction mechanisms.
(D) False: the dielectric constant of $\mathrm{D_2O}$ (about 78.1) is slightly lower than that of $\mathrm{H_2O}$ (about 78.4).
Only (D) is incorrect.
Answer: (C). | ||||
| 36 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Smaller alkali-metal ions are more strongly hydrated, so their hydrated ions are larger and move more slowly. Hydration falls from $\mathrm{Na^+}$ to $\mathrm{Cs^+}$, so ionic mobility and conductivity rise: $\mathrm{Cs^+ > Rb^+ > K^+ > Na^+}$.
Answer: (A). | ||||
| 37 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A Lewis base must have a lone pair on the central atom to donate.
$\mathrm{ClF_3}$ (2 lone pairs on Cl), $\mathrm{NF_3}$ (1 on N) and $\mathrm{SF_4}$ (1 on S) have one; in $\mathrm{PCl_5}$ all five valence electrons of P are in bonds, so it has none (it acts as a Lewis acid).
Answer: (B). | ||||
| 38 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Z = 25 is Mn. $\mathrm{Mn^{2+}}$ is $3d^5$; in water (a weak-field ligand) it is high-spin with 5 unpaired electrons.
$\mu = \sqrt{n(n+2)} = \sqrt{35} = 5.92$ BM.
Answer: (C). | ||||
| 39 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Statement I: on heating, $2\mathrm{KMnO_4} \rightarrow \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2}$ (this happens from about 513 K, so at 573 K too). True.
Statement II: both ions are tetrahedral, but $\mathrm{MnO_4^-}$ has Mn(+7), $d^0$, and is diamagnetic; only $\mathrm{MnO_4^{2-}}$ (Mn(+6), $d^1$) is paramagnetic. False.
Answer: (C). | ||||
| 40 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Classical (London) smog forms in cool humid climates from smoke, fog and sulphur dioxide; it is chemically reducing. Photochemical smog, with ozone and acrolein, is oxidising.
Answer: (B). | ||||
| 41 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $R_f = \frac{\text{distance moved by the substance}}{\text{distance moved by the solvent front}}$ is a ratio of two lengths, so it has no unit: Statement I is false.
$R_f$ depends on the solvent (mobile phase) used: Statement II is false.
Answer: (B). | ||||
| 42 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Aromatic: cyclic, planar, fully conjugated, with $(4n+2)$ π electrons.
(A) Cyclopentadiene has an $sp^3$ $\mathrm{CH_2}$: not fully conjugated, non-aromatic.
(B) The cyclopentadienyl anion has 6 π electrons in a fully conjugated ring ($n = 1$): aromatic.
(C) The cyclopentadienyl cation has 4 π electrons: antiaromatic.
(D) The dianion has an $sp^3$ $\mathrm{CH_2}$ between its two charged carbons, so the ring is not fully conjugated: non-aromatic.
Answer: (B). | ||||
| 43 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons HBr in $\mathrm{CCl_4}$ (no peroxide) adds by an ionic, Markovnikov path. Protonation of the vinyl group gives a secondary carbocation $\mathrm{CH_3-\overset{+}{C}H-}$ on the carbon next to the quaternary ring carbon bearing $\mathrm{CH_3}$.
A 1,2-methyl shift from the ring carbon to the cationic carbon gives a more stable tertiary carbocation on the ring carbon, now carrying an isopropyl group.
Bromide adds there: 1-bromo-1-isopropylcyclohexane.
(A) is the anti-Markovnikov (peroxide) product. (B) shows a six-membered ring: shifting a ring bond instead would enlarge the ring to seven members, so the cyclohexane drawn cannot form. (D) needs a shift that does not occur.
Answer: (C). | ||||
| 44 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Chlorobenzene is converted to sodium phenoxide by fusing with aqueous NaOH at 623 K under a pressure of 300 atm (the Dow process); acidifying then gives phenol. No copper catalyst is used.
Answer: (C). | ||||
| 45 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Mesityl oxide, $\mathrm{(CH_3)_2C=CH-CO-CH_3}$, is the dehydrated aldol product of acetone: 4-methylpent-3-en-2-one.
Answer: (A). | ||||
| 46 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hoffmann bromamide degradation: $\mathrm{C_6H_5CONH_2} + \mathrm{Br_2} + 4\mathrm{NaOH} \rightarrow \mathrm{C_6H_5NH_2} + \mathrm{Na_2CO_3} + 2\mathrm{NaBr} + 2\mathrm{H_2O}$; A is aniline.
Carbylamine reaction: $\mathrm{C_6H_5NH_2} + \mathrm{CHCl_3} + 3\mathrm{KOH/NaOH} \rightarrow \mathrm{C_6H_5NC}$; B is phenyl isocyanide.
Answer: (B). | ||||
| 47 | 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 Ethylene glycol with acid converts a ketone into a cyclic ketal (1,3-dioxolane). The ketone carbonyl of ethyl acetoacetate is far more reactive than the ester carbonyl, so only it is protected; the ester stays intact.
Product: ethyl 2-(2-methyl-1,3-dioxolan-2-yl)acetate.
Answer: (D). | ||||
| 48 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Ammonolysis is the cleavage of a C–X bond by ammonia: an alkyl (here benzyl) halide reacts with $\mathrm{NH_3}$ to give the amine. (A) is benzoylation (acylation), (B) is reduction of a nitrile, (D) is salt formation.
Answer: (C). | ||||
| 49 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter A competitive inhibitor binds at the active site, competing with the substrate (A true). An allosteric (non-competitive) inhibitor binds at a different, allosteric site and changes the shape of the active site (B and D true). It does not compete for the active site, so (C) is wrong.
Answer: (C). | ||||
| 50 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Tyrosine is 2-amino-3-(4-hydroxyphenyl)propanoic acid: the α-carbon carries one $\mathrm{CH_2}$ joined to a benzene ring with OH para to it.
Answer: (C). | ||||
| 51 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions 100 molal: 100 mol of solute in 1000 g of water, i.e. $\frac{1000}{18} = 55.56$ mol of water.
$x = \frac{100}{100 + 55.56} = 0.6429 = 64.29\times10^{-2} \approx 64\times10^{-2}$.
Answer: 64. | ||||
| 52 | 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 $n(\mathrm{CH_4}) = \frac{6.4}{16} = 0.4$ mol, $n(\mathrm{CO_2}) = \frac{8.8}{44} = 0.2$ mol; total 0.6 mol.
$P = \frac{nRT}{V} = \frac{0.6\times8.314\times300}{10\times10^{-3}} = 149652$ Pa $= 149.65$ kPa $\approx 150$ kPa.
Answer: 150. | ||||
| 53 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom $|m_l| \le l \le n - 1$, so $m_l = -3$ with $n = 4$ forces $l = 3$ (a 4f orbital).
Radial nodes $= n - l - 1 = 4 - 3 - 1 = 0$.
Answer: 0. | ||||
| 54 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta_rH^\circ = \Delta_fH^\circ(\mathrm{Al_2O_3}) - 3\Delta_fH^\circ(\mathrm{CaO}) = -1675 - 3(-635) = -1675 + 1905 = 230$ kJ (elements have zero enthalpy of formation).
Answer: 230. | ||||
| 55 | SolutionsRaoult's law | Class 12: Solutions | NumericalMedium | |
A medium numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions Henry's law: $p = K_H x$, so $x(\mathrm{O_2}) = \frac{20}{8.0\times10^4} = 2.5\times10^{-4}$.
One $\mathrm{dm^3}$ of solution is 1.0 kg, essentially all water: $\frac{1000}{18} = 55.56$ mol.
Since $x$ is tiny, $n(\mathrm{O_2}) \approx x\cdot n(\mathrm{water}) = 2.5\times10^{-4}\times55.56 = 1.389\times10^{-2}$ mol.
Molar solubility $= 1389\times10^{-5}\ \mathrm{mol\,dm^{-3}}$.
Answer: 1389. | ||||
| 56 | EquilibriumIonic equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium The strong acid fixes $[\mathrm{H^+}] = 0.1$ M (common-ion effect).
$K_a = \frac{[\mathrm{H^+}]\,c\alpha}{c(1-\alpha)} \approx 0.1\,\alpha$, so $\alpha = \frac{2.0\times10^{-6}}{0.1} = 2\times10^{-5}$.
Answer: 2. | ||||
| 57 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions $\mathrm{Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O}$.
Moles of dichromate $= 0.020\times0.03 = 6\times10^{-4}$; moles of $\mathrm{Fe^{2+}} = 6\times6\times10^{-4} = 3.6\times10^{-3}$.
Molarity $= \frac{3.6\times10^{-3}}{0.015} = 0.24$ M $= 24\times10^{-2}$ M.
Answer: 24. | ||||
| 58 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} = \frac{2.303}{570}\log\frac{100}{32}$.
$\log\frac{100}{32} = \log\frac{25}{8} = 2\log5 - 3\log2 = 2(0.699) - 3(0.301) = 0.495$.
$k = \frac{2.303\times0.495}{570} = \frac{1.140}{570} = 2.0\times10^{-3}\ \mathrm{s^{-1}}$.
Answer: 2. | ||||
| 59 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | NumericalHard | |
A hard numerical question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2}$. The product in solution is sodium hypophosphite, $\mathrm{NaH_2PO_2}$ (A).
Hypophosphite (P at +1) is a strong reducing agent and is oxidised to phosphate (+5), giving up 4 electrons:
$\mathrm{H_3PO_2 + 4AgNO_3 + 2H_2O \rightarrow 4Ag + H_3PO_4 + 4HNO_3}$.
So 1 mol of A gives 4 mol of Ag.
Answer: 4. | ||||
| 60 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Benzene: $\frac{3.9}{78} = 0.05$ mol, giving at most 0.05 mol of nitrobenzene $= 0.05\times123 = 6.15$ g.
Yield $= \frac{4.92}{6.15}\times100 = 80\%$.
Answer: 80. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions The statement needs a diagram in which the three sets overlap in pairs but have no region common to all three.
In P the two small circles overlap inside the large circle, so their overlap lies in all three sets. In Q and in R the two outer circles overlap inside the middle circle, again giving a region common to all three.
Every diagram shows a common region of all three sets, so none of them represents 'none play all three games'.
Answer: (D). | ||||
| 62 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\overrightarrow{AB} = iz - z$ and $\overrightarrow{AC} = iz$.
$\overrightarrow{BC} = (z + iz) - iz = z$, of length $|z|$; $\overrightarrow{AC} = iz$ is $z$ turned through $90^\circ$, also of length $|z|$.
So AC and BC are perpendicular and equal: a right isosceles triangle with the right angle at C.
Area $= \frac{1}{2}|z|\cdot|z| = \frac{1}{2}|z|^2$.
Answer: (B). | ||||
| 63 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A^2 = \begin{pmatrix}\sin^2\alpha & 0\\ 0 & \sin^2\alpha\end{pmatrix}$, so $A^2 - \frac{1}{2}I = \left(\sin^2\alpha - \frac{1}{2}\right)I$.
Its determinant is $\left(\sin^2\alpha - \frac{1}{2}\right)^2 = 0$, so $\sin^2\alpha = \frac{1}{2}$; $\alpha = \frac{\pi}{4}$ works.
Answer: (A). | ||||
| 64 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}k&1&1\\1&k&1\\1&1&k\end{vmatrix} = k^3 - 3k + 2 = (k-1)^2(k+2)$. A unique solution exists unless $k = 1$ or $k = -2$.
$k = 1$: all three equations are $x + y + z = 1$: infinitely many solutions.
$k = -2$: adding the equations gives $(k+2)(x+y+z) = 1 + k + k^2$, i.e. $0 = 3$: no solution.
Answer: (D). | ||||
| 65 | 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 Boy-versus-boy matches: $7\times4 = 28$. Girl-versus-girl matches: $6n$.
$28 + 6n = 52$, so $n = 4$.
Answer: (B). | ||||
| 66 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_4 = \binom{7}{3}x^4\left(x^{\log_2 x}\right)^3 = 35\,x^{4 + 3\log_2 x} = 4480$, so $x^{4+3\log_2 x} = 128 = 2^7$.
Put $t = \log_2 x$: $t(4 + 3t) = 7$, i.e. $3t^2 + 4t - 7 = 0$, $(t - 1)(3t + 7) = 0$.
$t = 1$ gives $x = 2$; $t = -\frac{7}{3}$ gives $x = 2^{-7/3}$, not a natural number.
Answer: (B). | ||||
| 67 | 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 Let $y$ be the value. The pattern repeats after two levels: $y = 4 + \frac{1}{5 + \frac{1}{y}} = 4 + \frac{y}{5y + 1}$.
$y(5y + 1) = 4(5y + 1) + y$, so $5y^2 - 20y - 4 = 0$.
$y = \frac{20 \pm \sqrt{400 + 80}}{10} = 2 \pm \frac{4\sqrt{30}}{10} = 2 \pm \frac{2}{5}\sqrt{30}$. The value exceeds 4, so $y = 2 + \frac{2}{5}\sqrt{30}$.
Answer: (A). | ||||
| 68 | 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 $n$th term is $\cot^{-1}(2n^2) = \tan^{-1}\frac{1}{2n^2} = \tan^{-1}\frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)} = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$.
The sum telescopes: $\tan^{-1}201 - \tan^{-1}1 = \tan^{-1}\frac{201 - 1}{1 + 201} = \tan^{-1}\frac{100}{101}$.
So $\cot^{-1}\alpha = \cot^{-1}\frac{101}{100}$ and $\alpha = 1.01$.
Answer: (B). | ||||
| 69 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Take logarithms: $\log y = \log x\cdot\log 5$, so $\log x = \frac{\log y}{\log 5}$.
Then $x = y^{\frac{1}{\log 5}}$, since $\log\left(y^{1/\log 5}\right) = \frac{\log y}{\log 5}$.
Answer: (B). | ||||
| 70 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $0 < x < 1$, $[x] = 0$, so the expression is $\frac{\cos^{-1}x\cdot\sin^{-1}x}{x(1 - x^2)}$.
As $x \to 0^+$: $\cos^{-1}x \to \frac{\pi}{2}$, $\frac{\sin^{-1}x}{x} \to 1$, $1 - x^2 \to 1$. The limit is $\frac{\pi}{2}$.
Answer: (C). | ||||
| 71 | 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 Replace $x$ by $\frac{\pi}{6} + \frac{\pi}{3} - x = \frac{\pi}{2} - x$: $g(\alpha) = \int_{\pi/6}^{\pi/3}\frac{\cos^\alpha x}{\sin^\alpha x + \cos^\alpha x}dx$.
Adding the two forms: $2g(\alpha) = \int_{\pi/6}^{\pi/3}dx = \frac{\pi}{6}$, so $g(\alpha) = \frac{\pi}{12}$ for every $\alpha$: a constant function.
A constant function is even (D is correct), but it is neither strictly increasing (A) nor strictly decreasing (B), and it has no inflection point (C).
Answer: (A), (B) or (C): all three are incorrect statements. | ||||
| 72 | 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 $xy - 1 + x - y = (x - 1)(y + 1)$, so $\frac{dy}{y + 1} = (x - 1)dx$.
$\ln|y + 1| = \frac{x^2}{2} - x + C$; $y(0) = 0$ gives $C = 0$.
$y(1) + 1 = e^{\frac{1}{2} - 1} = e^{-\frac{1}{2}}$, so $y(1) = e^{-\frac{1}{2}} - 1$.
Answer: (A). | ||||
| 73 | 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 The circumcentre lies on every perpendicular bisector. PQ has midpoint $(1, 1)$ and slope $-1$, so its perpendicular bisector is $y = x$.
Meeting $2x - y + 2 = 0$: $2x - x + 2 = 0$, $x = -2$, $y = -2$.
Answer: (D). | ||||
| 74 | 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 The centre lies on the normal at (2, 5), perpendicular to the tangent (slope 2): $y - 5 = -\frac{1}{2}(x - 2)$, i.e. $x + 2y = 12$.
With $x - 2y = 4$: $x = 8$, $y = 2$. Centre (8, 2).
Radius $= \sqrt{(8-2)^2 + (2-5)^2} = \sqrt{45} = 3\sqrt{5}$.
Answer: (D). | ||||
| 75 | CirclesRadius and centre | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections First circle: centre (5, 5), radius $\sqrt{25 + 25 - 41} = 3$. Second: centre (8, 5), radius $\sqrt{64 + 25 - 80} = 3$. Distance between centres = 3.
(A) $0 < 3 < 6$: two intersection points. True.
(B) Each centre is at distance 3, the radius, from the other centre: each circle passes through the other's centre. True.
(C) The centres lie on the circles, not inside them. Incorrect.
(D) Average of radii = 3 = distance. True.
Answer: (C). | ||||
| 76 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A plane containing the y-axis passes through the origin and has no y term: $ax + cz = 0$.
Through (1, 2, 3): $a + 3c = 0$; take $a = 3$, $c = -1$: $3x - z = 0$.
Answer: (D). | ||||
| 77 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter The biconditional is a tautology only if $q * (\sim p) \equiv p \Rightarrow q \equiv \sim p \vee q$. So $*$ is $\vee$ (check: $q \vee \sim p$ matches in every row).
Then $p * (\sim q) = p \vee \sim q \equiv \sim q \vee p \equiv q \Rightarrow p$.
Answer: (C). | ||||
| 78 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{r}\times(\vec{a} - \vec{b}) = \vec{0}$, so $\vec{r} = \lambda(\vec{a} - \vec{b}) = \lambda(-5\hat{i} - 4\hat{j} + 10\hat{k})$.
$\vec{r}\cdot(\hat{i} + 2\hat{j} + \hat{k}) = \lambda(-5 - 8 + 10) = -3\lambda = -3$, so $\lambda = 1$.
$\vec{r}\cdot(2\hat{i} - 3\hat{j} + \hat{k}) = -10 + 12 + 10 = 12$.
Answer: (C). | ||||
| 79 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Count ordered pairs (first, second) with sum $\le 8$ out of 36:
first 1: second 1, 2, 3, 5, 7 (5); first 2: 1, 2, 3, 5 (4); first 3: 1, 2, 3, 5 (4); first 5: 1, 2, 3 (3); first 7: 1 (1); first 11: none.
Total 17, so the probability is $\frac{17}{36}$.
Answer: (C). | ||||
| 80 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctHard | |
A hard single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Taking tangents of both sides, with $\tan\left(\cot^{-1}\frac{1}{x-1}\right) = x - 1$: $\frac{(x+1) + (x-1)}{1 - (x^2 - 1)} = \frac{8}{31}$, so $62x = 16 - 8x^2$, $4x^2 + 31x - 8 = 0$, $x = \frac{1}{4}$ or $x = -8$.
Check each against the original equation, with $\cot^{-1}$ in $(0, \pi)$: for $x < 1$, $\cot^{-1}\frac{1}{x-1} = \pi + \tan^{-1}(x - 1)$.
$x = \frac{1}{4}$: $\tan^{-1}1.25 + \pi + \tan^{-1}(-0.75) \approx 0.896 + 3.142 - 0.644 = 3.394 \ne \tan^{-1}\frac{8}{31} \approx 0.253$. Rejected.
$x = -8$: $\tan^{-1}(-7) + \pi + \tan^{-1}(-9) \approx -1.429 + 3.142 - 1.460 = 0.253$. Valid.
(For $x > 1$ the left side exceeds $\tan^{-1}2$, too large.) Only $x = -8$ remains, so the sum is $-8 = -\frac{32}{4}$.
Answer: (B). | ||||
| 81 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalHard | |
A hard numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives z increases with both x and y in the first quadrant, so the maximum is on the outer boundary of the region.
For $x > 0$, $\frac{75 - 4x}{3} < \frac{100 - 3x}{4}$ (equal at $x = 0$), so only $4x + 3y \le 75$ binds: the boundary is $x = \frac{75 - 3y}{4}$, $0 \le y \le 25$.
On it, $z = y(6x + y) = y\left(\frac{225 - 9y}{2} + y\right) = \frac{y(225 - 7y)}{2}$.
$\frac{dz}{dy} = \frac{225 - 14y}{2} = 0$ at $y = \frac{225}{14}$ (within range; $x = \frac{75}{4} - \frac{675}{56} = \frac{375}{56}$).
$z_{max} = \frac{1}{2}\cdot\frac{225}{14}\cdot\frac{225}{2} = \frac{50625}{56} = 904.02$ (to two decimals).
Answer: 904.02 (nearest integer 904). | ||||
| 82 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\det A = -2$, so $\det(A^4) = 16$.
For a $2\times2$ matrix, $\mathrm{Adj}(2A) = 2\,\mathrm{Adj}A = 2\begin{bmatrix}-1 & -3\\ 0 & 2\end{bmatrix}$.
Both $A$ and $\mathrm{Adj}A$ are upper triangular, so their powers are too, with powered diagonals:
$A^{10}$ has diagonal $(2^{10}, 1)$; $(\mathrm{Adj}(2A))^{10} = 2^{10}(\mathrm{Adj}A)^{10}$ has diagonal $(2^{10}\cdot1, 2^{10}\cdot2^{10})$.
$A^{10} - (\mathrm{Adj}(2A))^{10}$ is upper triangular with diagonal $(0, 1 - 2^{20})$, so its determinant is 0.
Total: $16 + 0 = 16$.
Answer: 16. | ||||
| 83 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $2021 = 17\times118 + 15$, so $2021 \equiv -2 \pmod{17}$ and $2021^{3762} \equiv 2^{3762}$.
$2^4 = 16 \equiv -1$, so $2^8 \equiv 1 \pmod{17}$. $3762 = 8\times470 + 2$, hence $2^{3762} \equiv 2^2 = 4$.
(Binomially: $2^{3762} = 4\cdot(2^8)^{470} = 4(1 + 255)^{470} = 4 + 17k$, as $255 = 17\times15$.)
Answer: 4. | ||||
| 84 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\cos(\sin x) - \cos x = 2\sin\frac{x + \sin x}{2}\,\sin\frac{x - \sin x}{2}$.
As $x \to 0$: $\sin\frac{x + \sin x}{2} \approx x$ and $x - \sin x \approx \frac{x^3}{6}$, so $\sin\frac{x - \sin x}{2} \approx \frac{x^3}{12}$.
The numerator $\approx 2\cdot x\cdot\frac{x^3}{12} = \frac{x^4}{6}$, so $\lim_{x\to0}f(x) = \frac{1}{6}$.
Continuity at 0: $\frac{1}{k} = \frac{1}{6}$, $k = 6$.
Answer: 6. | ||||
| 85 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Since $[x^2]$ is an integer, $\left[[x^2] - \cos x\right] = [x^2] + [-\cos x]$.
On $\left[0, \sqrt{\pi/2}\right]$ ($\sqrt{\pi/2} \approx 1.253 < \frac{\pi}{2}$), $0 < \cos x \le 1$, so $[-\cos x] = -1$.
$[x^2] = 0$ on $[0, 1)$ and $1$ on $[1, \sqrt{\pi/2}]$ (as $x^2 \le \frac{\pi}{2} < 2$).
Integrand: $-1$ on $[0, 1)$, $0$ on $[1, \sqrt{\pi/2}]$. The integral is $-1$, and its absolute value is 1.
Answer: 1. | ||||
| 86 | CirclesRadius and centre | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections First circle: centre (5, 5), radius $\sqrt{50 - 41} = 3$. Second: centre (12, 5), radius $\sqrt{144 + 25 - 160} = 3$.
Distance between centres = 7 > 3 + 3, so the circles are separate. Minimum distance $= 7 - 6 = 1$.
Answer: 1. | ||||
| 87 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Plane: $(2x - 7y + 4z - 3) + \lambda(3x - 5y + 4z + 11) = 0$. At $(-2, 1, 3)$: $(-4 - 7 + 12 - 3) + \lambda(-6 - 5 + 12 + 11) = -2 + 12\lambda = 0$, $\lambda = \frac{1}{6}$.
Multiplying by 6: $12x - 42y + 24z - 18 + 3x - 5y + 4z + 11 = 0$, i.e. $15x - 47y + 28z - 7 = 0$.
So $a = 15$, $b = -47$, $c = 28$, and $2a + b + c - 7 = 30 - 47 + 28 - 7 = 4$.
Answer: 4. | ||||
| 88 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | NumericalHard | |
A hard numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $P(E_1) = x$, $P(E_2) = y$, $P(E_3) = z$. By independence $p = (1-x)(1-y)(1-z)$, $\alpha = p\frac{x}{1-x}$, $\beta = p\frac{y}{1-y}$, $\gamma = p\frac{z}{1-z}$.
Write $u = \frac{x}{1-x}$, $v = \frac{y}{1-y}$, $w = \frac{z}{1-z}$. The first equation becomes $p^2(u - 2v) = p^2uv$, i.e. $u - 2v = uv$, so $u = \frac{2v}{1 - v}$.
With $v = \frac{y}{1-y}$: $\frac{2v}{1-v} = \frac{2y}{1-2y}$, so $\frac{x}{1-x} = \frac{2y}{1-2y}$, giving $x = 2y$.
The second: $p^2(v - 3w) = 2p^2vw$, so $v = \frac{3w}{1 - 2w} = \frac{3z}{1 - 3z}$, i.e. $\frac{y}{1-y} = \frac{3z}{1-3z}$, giving $y = 3z$.
So $x = 6z$ and $\frac{P(E_1)}{P(E_3)} = 6$.
Answer: 6. | ||||
| 89 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $2^x = \tan\theta$, $\theta \in (0, \frac{\pi}{2})$. Then $\frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos2\theta$ with $2\theta \in (0, \pi)$, so $\cos^{-1}(\cos2\theta) = 2\theta$ and $f = \sin2\theta = \frac{2\cdot2^x}{1 + 4^x}$.
$f'(x) = 2\ln2\cdot\frac{2^x(1 + 4^x) - 2^x\cdot2\cdot4^x}{(1 + 4^x)^2} = 2\ln2\cdot\frac{2^x(1 - 4^x)}{(1 + 4^x)^2}$.
At $x = 1$: $f'(1) = 2\ln2\cdot\frac{2(-3)}{25} = -\frac{12}{25}\ln2$, so $\frac{b}{a} = \frac{12}{25}$.
Integers with this ratio are $a = 25k$, $b = 12k$: $|a^2 - b^2| = 481k^2$, least (for $k \ne 0$) at $k = \pm1$: 481.
Answer: 481. | ||||
| 90 | 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 $\vec{a}\cdot\vec{b} = -\alpha\beta - \alpha\beta - 3 = 1$, so $\alpha\beta = -2$.
$\vec{b}\cdot\vec{c} = -\beta + 2\alpha + 1 = -3$, so $\beta = 2\alpha + 4$.
$\alpha(2\alpha + 4) = -2$ gives $(\alpha + 1)^2 = 0$: $\alpha = -1$, $\beta = 2$.
$\vec{a} = (-1, 2, 3)$, $\vec{b} = (-2, 1, -1)$, $\vec{c} = (1, -2, -1)$.
$(\vec{a}\times\vec{b})\cdot\vec{c} = \begin{vmatrix}-1&2&3\\-2&1&-1\\1&-2&-1\end{vmatrix} = -1(-1-2) - 2(2+1) + 3(4-1) = 3 - 6 + 9 = 6$.
$\frac{1}{3}\times6 = 2$.
Answer: 2. | ||||
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