JEE (Main) · 2021
JEE (Main) 2021 — 27 July, 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
- Electrostatics4
- Atoms and Nuclei3
- Properties of Bulk Matter3
19 easy · 10 medium · 1 hard
Chemistry
30 Q
- Chemical Bonding and Molecular Structure3
- Coordination Compounds3
- Biomolecules2
23 easy · 7 medium · 0 hard
Mathematics
30 Q
- Integral Calculus5
- Circles3
- Matrices and Determinants3
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 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Nuclei left: $N = N_0e^{-\lambda t}$, so nuclei decayed $N_d = N_0(1-e^{-\lambda t})$.
$f = \frac{N_d}{N_0} = 1-e^{-\lambda t}$.
$\frac{df}{dt} = \lambda e^{-\lambda t}$.
Answer: (B). | ||||
| 2 | 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 Fringe width $\beta = \frac{\lambda D}{d}$. Blue light has a shorter wavelength than orange, so $\beta$ decreases.
The minima stay dark (equal slits) and the centre stays bright (zero path difference).
Answer: (C). | ||||
| 3 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $v = \frac{c}{\sqrt{\mu_r\varepsilon_r}} = \frac{3\times10^8}{\sqrt{81}} = \frac{3\times10^8}{9} = 3.33\times10^7$ m/s.
Answer: (A). | ||||
| 4 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance While charging, $V = V_0(1-e^{-t/RC})$ with $RC = 100\times10^{-6} = 10^{-4}$ s.
$50 = 100(1-e^{-t/RC})$ gives $e^{-t/RC} = \frac{1}{2}$, so $t = RC\ln 2 = 0.69\times10^{-4}$ s.
Answer: (A). | ||||
| 5 | 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 $X_L = 2\pi fL = 2\times\frac{22}{7}\times50\times0.07 = 22\ \Omega$.
$Z = \sqrt{12^2+22^2} = \sqrt{628} \approx 25.06\ \Omega$, so $I = \frac{220}{25.06} \approx 8.8$ A.
$\tan\phi = \frac{X_L}{R} = \frac{22}{12} = \frac{11}{6}$, so $\phi = \tan^{-1}\left(\frac{11}{6}\right)$.
Answer: (A). | ||||
| 6 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single 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 Let the liquid height be $h$ and density $\rho$. Efflux speed $v = \sqrt{2gh}$.
Thrust on the vessel (rate of momentum carried out): $F = \rho a v^2 = 2\rho a g h$.
The vessel is light, so the normal reaction is the weight of the liquid: $N = \rho A h g$.
No sliding needs $\mu N \ge F$: $\mu \ge \frac{2\rho agh}{\rho Agh} = \frac{2a}{A}$.
Answer: (B). | ||||
| 7 | 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 The same current $I$ flows through every cross-section (so the electron current does not change).
$I = neAv_d$: as the area falls from P to Q, $v_d$ increases.
Current density $J = I/A$ increases, so the field $E = \rho J$ increases, not decreases.
Only (A) is true.
Answer: (A). | ||||
| 8 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Total charge $Q = (2C + C)V = 3CV$, which is conserved once the battery is removed.
New capacitance in parallel: $2C + KC = (K+2)C$.
$V' = \frac{3CV}{(K+2)C} = \frac{3V}{K+2}$.
Answer: (D). | ||||
| 9 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The three slabs (K, thickness d; 3K, 2d; 5K, 3d) are in series:
$\frac{1}{C} = \frac{d}{K\varepsilon_0A} + \frac{2d}{3K\varepsilon_0A} + \frac{3d}{5K\varepsilon_0A} = \frac{d}{K\varepsilon_0A}\left(1+\frac{2}{3}+\frac{3}{5}\right) = \frac{34}{15}\frac{d}{K\varepsilon_0A}$.
$C = \frac{15}{34}\frac{K\varepsilon_0A}{d}$.
Answer: (B). | ||||
| 10 | 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 Potential energy $= E - \frac{3E}{4} = \frac{E}{4}$.
$\frac{1}{2}ky^2 = \frac{1}{4}\cdot\frac{1}{2}ka^2$ gives $y^2 = \frac{a^2}{4}$, $y = \frac{a}{2}$.
Answer: (B). | ||||
| 11 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance For each ball: $T\cos\theta = mg$, $T\sin\theta = \frac{q^2}{4\pi\varepsilon_0x^2}$, so $\tan\theta = \frac{q^2}{4\pi\varepsilon_0x^2mg}$.
For a small angle $\tan\theta \approx \sin\theta = \frac{x/2}{l}$.
$\frac{x}{2l} = \frac{q^2}{4\pi\varepsilon_0x^2mg}$ gives $x^3 = \frac{q^2l}{2\pi\varepsilon_0mg}$.
Answer: (B). | ||||
| 12 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic work done by the gas: $W = nC_v(T_{initial} - T_{final})$.
$W_{BC} = C_v(T_1 - T_2)$ and $W_{DA} = C_v(T_2 - T_1)$, so $W_{BC} + W_{DA} = 0$: (A) is false.
$W_{AB} = RT_1\ln\frac{3.5V_0}{V_0} > 0$ and $W_{CD} < 0$ (compression), so (B) is false.
$W_{DC} = RT_2\ln\frac{5.5}{1.5}$ differs from $W_{AB} = RT_1\ln 3.5$ since $T_1 \ne T_2$: (C) is false.
Work along the adiabat taken from A to D is $C_v(T_1 - T_2) = W_{BC}$: (D) is true.
Answer: (D). | ||||
| 13 | 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 Newton's law of cooling with the mean temperature: $\frac{\Delta T}{t} = k(T_{mean} - T_s)$.
First: $\frac{2}{4} = k(60-30)$, so $k = \frac{1}{60}$ per min.
Second: $\frac{2}{t} = \frac{1}{60}(50-30)$, so $t = 6$ min.
Answer: (C). | ||||
| 14 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Kinetic theory: $P = \frac{2}{3}\frac{N}{V}\bar{E}$, so $N = \frac{3PV}{2\bar{E}}$.
$N = \frac{3\times2\times1.013\times10^5\times10^{-3}}{2\times2\times10^{-9}} \approx 1.5\times10^{11}$.
Answer: (B). | ||||
| 15 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane A: write each vector from O. $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$, $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}$, $\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = 2\overrightarrow{AO}$ (D is diametrically opposite A).
Sum $= \overrightarrow{OB} + \overrightarrow{OC} + 3\overrightarrow{AO} + \overrightarrow{AO} = 4\overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{OC}$. A is correct.
R: the polygon law gives $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AD}$, and $\overrightarrow{AD} = 2\overrightarrow{AO}$. R is correct.
A is proved with the triangle law through the centre O; R's chain AB + BC + CD is not the sum in A, so R does not explain A.
Answer: (B). | ||||
| 16 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Larger disc about the perpendicular axis: $I_1 = \frac{MR^2}{2}$.
Same mass per area: smaller disc mass $m = M\frac{r^2}{R^2}$; about a diameter $I_2 = \frac{mr^2}{4} = \frac{Mr^4}{4R^2}$.
$\frac{I_1}{I_2} = \frac{MR^2/2}{Mr^4/(4R^2)} = \frac{2R^4}{r^4}$.
Answer: (C). | ||||
| 17 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion (a) $\frac{ML^2}{12}$: (iii).
(b) $\frac{(2M)L^2}{3} = \frac{2ML^2}{3}$: (iv).
(c) $\frac{M(2L)^2}{12} = \frac{ML^2}{3}$: (ii).
(d) $\frac{(2M)(2L)^2}{3} = \frac{8ML^2}{3}$: (i).
Answer: (B). | ||||
| 18 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Elastic, B at rest: $v_B = \frac{2m_A}{m_A+m_B}u = \frac{2m}{3m}\times9 = 6$ m/s.
B sticks to C: $2m\times6 = 4m\,v$, so $v = 3$ m/s.
Answer: (B). | ||||
| 19 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium 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 $u^2 = 2gh$. At height $\frac{h}{3}$: $\frac{h}{3} = ut - \frac{1}{2}gt^2$, so $t = \frac{u \pm \sqrt{u^2 - \frac{2gh}{3}}}{g} = \frac{u}{g}\left(1 \pm \sqrt{\frac{2}{3}}\right)$.
$\frac{t_1}{t_2} = \frac{1-\sqrt{2/3}}{1+\sqrt{2/3}} = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$.
Answer: (A). | ||||
| 20 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Pitch $= \frac{5\text{ mm}}{5} = 1$ mm.
Least count $= \frac{1\text{ mm}}{50} = 0.02$ mm $= 0.002$ cm, not 0.001 cm, so A is wrong.
R is the correct definition.
Answer: (D). | ||||
| 21 | GravitationGravitational potential energy | Class 11: Gravitation | NumericalHard | |
A hard numerical question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation Neutral point at distance $x$ from the first planet's centre: $\frac{GM}{x^2} = \frac{9GM}{(8R-x)^2}$ gives $8R - x = 3x$, $x = 2R$.
Beyond it the second planet pulls the satellite in, so it need only just reach $x = 2R$.
Potential energy at the surface ($x = R$): $-\frac{GMm}{R} - \frac{9GMm}{7R} = -\frac{16}{7}\frac{GMm}{R}$.
At $x = 2R$: $-\frac{GMm}{2R} - \frac{9GMm}{6R} = -2\frac{GMm}{R}$.
$\frac{1}{2}mv^2 = \left(\frac{16}{7} - 2\right)\frac{GMm}{R} = \frac{2}{7}\frac{GMm}{R}$, so $v^2 = \frac{4}{7}\frac{GM}{R}$.
Answer: 4. | ||||
| 22 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Period $T = \frac{5}{10} = 0.5$ s, and $T = 2\pi\sqrt{\frac{I}{mB}}$.
$B = \frac{4\pi^2I}{mT^2} = \frac{4\times9.85\times5\times10^{-6}}{9.85\times10^{-2}\times0.25} = 8\times10^{-3}$ T $= 8$ mT.
Answer: 8. | ||||
| 23 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Elastic energy $= \frac{1}{2}\frac{YA}{L}x^2$, all given to the stone: $\frac{1}{2}mv^2 = \frac{1}{2}\frac{YA}{L}x^2$.
$v^2 = \frac{0.5\times10^9\times10^{-6}\times(0.04)^2}{0.1\times0.02} = \frac{0.8}{0.002} = 400$, $v = 20$ m/s.
Answer: 20. | ||||
| 24 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Collector current $I_C = \frac{0.6}{1000} = 0.6$ mA $= 600\ \mu$A.
$I_B = \frac{I_C}{\beta} = \frac{600}{24} = 25\ \mu$A.
Answer: 25. | ||||
| 25 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics No bending at BC for every angle needs $n_1 = n_2$.
$1.2 + \frac{10.8\times10^{-14}}{\lambda^2} = 1.45 + \frac{1.8\times10^{-14}}{\lambda^2}$ gives $\frac{9\times10^{-14}}{\lambda^2} = 0.25$.
$\lambda^2 = 3.6\times10^{-13}$ m$^2$, $\lambda = 6\times10^{-7}$ m $= 600$ nm.
Answer: 600. | ||||
| 26 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Both wavelengths are $h/p$.
$\frac{\lambda_{photon}}{\lambda_{particle}} = \frac{p_{particle}}{p_{photon}} = \frac{9.1\times10^{-31}\times10^6}{10^{-27}} = 910$.
Answer: 910. | ||||
| 27 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Activity $A = A_0e^{-t/\tau}$ with mean life $\tau = 30$ ms; discharging charge $q = q_0e^{-t/RC}$.
The ratio is constant only if $RC = \tau$: $R = \frac{30\times10^{-3}}{200\times10^{-6}} = 150\ \Omega$.
Answer: 150. | ||||
| 28 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $I = \frac{e}{T} = \frac{ev}{2\pi r} = \frac{1.6\times10^{-19}\times2.2\times10^6}{2\times\frac{22}{7}\times0.5\times10^{-10}} = 1.12\times10^{-3}$ A.
$1.12$ mA $= 112\times10^{-2}$ mA.
Answer: 112. | ||||
| 29 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current With $T_1$ on $T_2$, the steady current through R and L is $\frac{6}{6} = 1$ A.
On switching to $T_3$, the inductor keeps its current for that instant, now through r: 1 A.
Potential drop across r $= 1\times3 = 3$ V.
Answer: 3. | ||||
| 30 | Communication SystemsAmplitude modulation and bandwidth | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on amplitude modulation and bandwidth, from Communication Systems. NCERT: No matching NCERT chapter Modulation index $\mu = \frac{5}{15} = \frac{1}{3}$. Each side band has amplitude $\frac{\mu A_c}{2} = \frac{A_m}{2} = 2.5$ V.
So $\frac{a}{10} = \frac{b}{10} = 2.5$ V: $a = b = 25$, $\frac{a}{b} = 1$.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Solid StateSeven crystal systems | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on seven crystal systems, from Solid State. NCERT: No matching NCERT chapter All three edges differ ($a \ne b \ne c$), two angles are $90^\circ$ and one ($\beta = 120^\circ$) is not.
That is the monoclinic system ($\alpha = \gamma = 90^\circ$, $\beta \ne 90^\circ$).
Answer: (D). | ||||
| 32 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Statement I: Rutherford's nuclear model says nothing about quantised energies, so it cannot explain the line spectrum. True.
Statement II: Bohr's model gives the electron a fixed orbit (exact position) and an exact momentum at once, which the uncertainty principle forbids. True.
Answer: (A). | ||||
| 33 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate $= k[A]^n$, so $k = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^n} = \mathrm{mol^{1-n}\,L^{n-1}\,s^{-1}}$.
Answer: (D). | ||||
| 34 | 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 NaOH and $\mathrm{Ca(OH)_2}$ are basic: (ii).
$\mathrm{Be(OH)_2}$ and $\mathrm{Al(OH)_3}$ dissolve in both acid and alkali: amphoteric, (iii).
$\mathrm{B(OH)_3}$ (boric acid) is a weak Lewis acid: (i).
Answer: (C). | ||||
| 35 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter An Ellingham diagram plots $\Delta_fG^\circ$ of oxide formation against temperature. It shows which reducing agent can reduce an oxide (A), the free-energy change (C), and kinks where a phase changes (D).
It is purely thermodynamic and says nothing about kinetics, so (B) is incorrect.
Answer: (B). | ||||
| 36 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Anodic oxidation of acidified sulphate (bisulphate) gives peroxodisulphate: $2\mathrm{HSO_4^-} \to \mathrm{H_2S_2O_8} + 2e^-$.
Peroxodisulphuric acid has an O–O bridge: $\mathrm{HO_3S{-}O{-}O{-}SO_3H}$.
Answer: (C). | ||||
| 37 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{Li^+}$ is very small, so its charge density and polarising power are high (Fajans' rules).
It distorts the halide's electron cloud, giving the bond covalent character. A is true and R explains it.
Answer: (A). | ||||
| 38 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions $\mathrm{H_4P_2O_7}$: $4 + 2x - 14 = 0$, $x = +5$.
$\mathrm{H_4P_2O_5}$: $4 + 2x - 10 = 0$, $x = +3$.
$\mathrm{H_4P_2O_6}$: $4 + 2x - 12 = 0$, $x = +4$.
Answer: (C). | ||||
| 39 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Mn is +3: $3d^4$. Chloride is a weak-field ligand, so no pairing: four unpaired electrons, paramagnetic.
The inner 3d orbitals are not free, so the outer 4d orbitals are used: $sp^3d^2$ (high-spin octahedral).
Answer: (C). | ||||
| 40 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[\mathrm{PtCl_2(NH_3)_2}]$, square planar $\mathrm{MA_2B_2}$: cis and trans, 2.
$[\mathrm{Ni(CO)_4}]$, tetrahedral $\mathrm{MA_4}$: one form, 1.
$[\mathrm{Ru(H_2O)_3Cl_3}]$, octahedral $\mathrm{MA_3B_3}$: fac and mer, 2.
$[\mathrm{CoCl_2(NH_3)_4}]^+$, octahedral $\mathrm{MA_4B_2}$: cis and trans, 2.
Answer: (D). | ||||
| 41 | 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 Eutrophication (nutrient enrichment) causes algal blooms; their decay uses up dissolved oxygen, so oxygen falls and conditions turn anaerobic.
(A), (B) and (D) are correct; (C) is not.
Answer: (C). | ||||
| 42 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Barfoed's reagent (copper acetate in dilute acetic acid) is a weak oxidant: reducing monosaccharides give red $\mathrm{Cu_2O}$ within a few minutes, disaccharides only much later.
Seliwanoff's test separates ketoses from aldoses, Tollens' test responds to all reducing sugars, and the iodine test detects starch.
Answer: (B). | ||||
| 43 | HydrocarbonsConformations of ethane | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on conformations of ethane, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Conformers differ only by rotation about the C–C single bond; they are called rotamers.
They are not mirror images or enantiomers.
Answer: (A). | ||||
| 44 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A, benzyl cation: stabilised by resonance with the ring, most stable.
C, ethyl cation: primary, stabilised by hyperconjugation.
B, vinyl cation: positive charge on an $sp^2$ carbon, unstable.
D, ethynyl cation: positive charge on an $sp$ carbon, the most electronegative, least stable.
Order: A > C > B > D.
Answer: (B). | ||||
| 45 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Iodination is reversible because HI reduces $\mathrm{CH_3I}$ back to methane.
An oxidising agent such as concentrated $\mathrm{HIO_3}$ destroys HI as it forms ($5\mathrm{HI} + \mathrm{HIO_3} \to 3\mathrm{I_2} + 3\mathrm{H_2O}$), so the reaction goes to completion.
Answer: (A). | ||||
| 46 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids 2,4-DNP forms coloured hydrazones with aldehydes and ketones only.
(A) acetophenone is a ketone: orange-red 2,4-dinitrophenylhydrazone.
(B) and (C) are esters and (D) is a carboxylic acid; their carbonyl groups are deactivated by resonance with oxygen and do not react.
Answer: (A). | ||||
| 47 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers With $\mathrm{H_3PO_4}$ at $120^\circ$C the alcohol is protonated and loses water; the primary carbocation $\mathrm{C_6H_{11}CH_2^+}$ undergoes a 1,2-hydride shift to the tertiary ring cation.
Loss of a proton gives the more substituted alkene, 1-methylcyclohexene: A.
Hydroboration–oxidation ($\mathrm{(BH_3)_2}$, then $\mathrm{H_2O_2/OH^-}$) adds H and OH anti-Markovnikov (syn): OH goes to the less substituted ring carbon, giving 2-methylcyclohexan-1-ol (trans).
Answer: (C). | ||||
| 48 | AminesBasicity of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines In acetamide the nitrogen lone pair is delocalised onto the strongly electron-withdrawing C=O group, so acetamide is far less basic than aniline (it is almost neutral). Statement I is false.
Statement II correctly explains why aniline is less basic than aliphatic amines. True.
Answer: (D). | ||||
| 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 Furacin is an antiseptic: (iii).
Arsphenamine (salvarsan) was the first antibiotic-type drug, used against syphilis: (i).
Dimetone is a synthetic antihistamine: (iv).
Valium is a tranquilizer: (ii).
Answer: (A). | ||||
| 50 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules A is 5-methyluracil, i.e. thymine (a pyrimidine with a methyl at C-5 and two C=O groups).
In DNA thymine pairs with adenine through two hydrogen bonds.
Answer: (B). | ||||
| 51 | SolutionsConcentration terms | Class 12: Solutions | NumericalMedium | |
A medium numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 L (1000 cm$^3$) of solution: mass $= 1.2\times1000 = 1200$ g.
Taking the volume of the solution as that of the water in it, the water is 1000 cm$^3$ = 1000 g, so NaOH = 200 g = $\frac{200}{40} = 5$ mol.
Molality $= \frac{5}{1\ \text{kg}} = 5$ m.
Answer: 5. | ||||
| 52 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure CO and $\mathrm{NO^+}$ are isoelectronic (14 electrons each).
Bond order $= \frac{10-4}{2} = 3$ for both, so the difference is $0 = \frac{x}{2}$, $x = 0$.
Answer: 0. | ||||
| 53 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\mathrm{H_2O(l) \to H_2O(g)}$: $\Delta n_g = 1$.
$\Delta H - \Delta U = \Delta n_gRT = 1\times8.31\times373 = 3099.6$ J mol$^{-1}$ $\approx 31\times10^2$ J mol$^{-1}$.
Answer: 31. | ||||
| 54 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\pi = \frac{w}{M}\frac{RT}{V}$, so $M = \frac{wRT}{\pi V} = \frac{1.46\times0.083\times300}{2.42\times10^{-3}\times0.1} = 1.502\times10^5$ g mol$^{-1}$.
That is $15.02\times10^4$, nearest integer 15.
Answer: 15. | ||||
| 55 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Let $x$ mol dissociate: $\frac{x^2}{3-x} = 1.844$, so $x^2 + 1.844x - 5.532 = 0$.
$x = \frac{-1.844 + \sqrt{1.844^2 + 4\times5.532}}{2} = \frac{-1.844 + 5.0526}{2} = 1.6043$.
$\mathrm{PCl_5}$ left $= 3 - 1.6043 = 1.3957$ mol $= 1395.7\times10^{-3}$, nearest integer 1396.
Answer: 1396. | ||||
| 56 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry $\Lambda_m = \frac{1000\kappa}{c} = \frac{1000\times2.0\times10^{-5}}{0.001} = 20$ S cm$^2$ mol$^{-1}$.
$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{20}{190} = 0.1053$.
$K_a = \frac{c\alpha^2}{1-\alpha} = \frac{0.001\times0.01108}{0.8947} = 1.238\times10^{-5} = 12.38\times10^{-6}$, nearest integer 12.
Answer: 12. | ||||
| 57 | 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: $\frac{x}{m} = kP^{1/n}$. Doubling P multiplies $\frac{x}{m}$ by $2^{1/n} = 64 = 2^6$.
So $\frac{1}{n} = 6$, $n = \frac{1}{6} = 0.1667 = 16.67\times10^{-2}$, nearest integer 17.
Answer: 17. | ||||
| 58 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[\mathrm{Co(NH_3)_3(NO_2)_3}]$ is octahedral $\mathrm{MA_3B_3}$: fac and mer, $X = 2$.
$[\mathrm{Cr(C_2O_4)_3}]^{3-}$ is $\mathrm{M(AA)_3}$: no geometrical isomers (only a pair of optical isomers), $Y = 0$.
$X + Y = 2$.
Answer: 2. | ||||
| 59 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure N in a tertiary amine is $sp^3$ with one lone pair, so the molecule is pyramidal.
Lone pair–bond pair repulsion squeezes the angle below the tetrahedral $109.5^\circ$, while the bulky alkyl groups open it above ammonia's $107^\circ$: the C–N–C angle is about $108^\circ$.
Answer: 108. | ||||
| 60 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Molar mass of AgCl $= 107.87 + 35.5 = 143.37$.
Mass of Cl $= 0.3849\times\frac{35.5}{143.37} = 0.09531$ g.
% Cl $= \frac{0.09531}{0.5}\times100 = 19.06\%$, nearest integer 19. (The 5.0 g of chlorine is not needed.)
Answer: 19. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | 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 Radius $= OC = \sqrt{13}$. $\overrightarrow{OC} = (2, 3)$, so a unit vector perpendicular to it is $\frac{(-3, 2)}{\sqrt{13}}$.
P, Q $= C \pm \sqrt{13}\cdot\frac{(-3, 2)}{\sqrt{13}} = (2 \mp 3,\ 3 \pm 2)$, i.e. $(-1, 5)$ and $(5, 1)$.
Answer: (C). | ||||
| 62 | 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 $(\vec{a}-\vec{b})\times\vec{b} = \vec{a}\times\vec{b}$ (since $\vec{b}\times\vec{b} = \vec{0}$). $\vec{a}\times\vec{b} = (3-4)\hat{i} - (3+2)\hat{j} + (2+1)\hat{k} = -\hat{i} - 5\hat{j} + 3\hat{k}$.
$\vec{a}\times(\vec{a}\times\vec{b}) = (\vec{a}\cdot\vec{b})\vec{a} - |\vec{a}|^2\vec{b} = 7\vec{a} - 6\vec{b} = (13, -5, -4)$.
$(13, -5, -4)\times(-1, 2, 3) = (-15+8,\ 4-39,\ 26-5) = (-7, -35, 21)$.
$(\vec{a}+\vec{b}) = (0, 3, 5)$; $(0, 3, 5)\times(-7, -35, 21) = (63+175,\ -35-0,\ 0+21) = (238, -35, 21) = 7(34\hat{i} - 5\hat{j} + 3\hat{k})$.
Answer: (D). | ||||
| 63 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem First: $T_{r+1} = \binom{11}{r}x^{22-2r}b^{-r}x^{-r}$; $22 - 3r = 7$ gives $r = 5$: coefficient $\binom{11}{5}b^{-5}$.
Second: $T_{r+1} = \binom{11}{r}x^{11-r}(-1)^rb^{-r}x^{-2r}$; $11 - 3r = -7$ gives $r = 6$: coefficient $\binom{11}{6}b^{-6}$.
$\binom{11}{5} = \binom{11}{6}$, so $b^{-5} = b^{-6}$, $b = 1$.
Answer: (D). | ||||
| 64 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Area $= \int_{1/2}^{2}2^x\,dx - \int_1^2\ln x\,dx$ (the lower curve is 0 for $x < 1$ and $\ln x$ for $x \ge 1$).
$\int_{1/2}^2 2^x dx = \frac{4 - \sqrt{2}}{\ln 2}$; $\int_1^2\ln x\,dx = [x\ln x - x]_1^2 = 2\ln 2 - 1$.
Area $= (4-\sqrt{2})(\ln 2)^{-1} - 2\ln 2 + 1$: $\alpha = 4 - \sqrt{2}$, $\beta = -2$, $\gamma = 1$.
$(\alpha + \beta - 2\gamma)^2 = (-\sqrt{2})^2 = 2$.
Answer: (B). | ||||
| 65 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Characteristic equation: $A^2 - (\text{tr}A)A + |A|I = 0$ with $\text{tr}A = 5$, $|A| = 4 + 2 = 6$: $A^2 - 5A + 6I = 0$.
Multiply by $A^{-1}$: $A - 5I + 6A^{-1} = 0$, so $A^{-1} = \frac{5}{6}I - \frac{1}{6}A$.
$\alpha = \frac{5}{6}$, $\beta = -\frac{1}{6}$, $4(\alpha - \beta) = 4$.
Answer: (B). | ||||
| 66 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Centre C(1, 3), radius 2. $PC = \sqrt{4 + 4} = 2\sqrt{2}$, so $\sin\angle APC = \frac{2}{2\sqrt{2}}$: $\angle APC = 45^\circ$ and $\angle ACB = 90^\circ$.
Chord $AB = 2\cdot2\sin45^\circ = 2\sqrt{2}$, subtending $90^\circ$ at C.
$AD = AB$ means AD also subtends $90^\circ$ at C, so D (not B) is diametrically opposite B.
Then $\angle BAD = 90^\circ$ (angle in a semicircle) with $AB = AD = 2\sqrt{2}$.
Area $= \frac{1}{2}(2\sqrt{2})^2 = 4$.
Answer: (B). | ||||
| 67 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x + iy$: $S_1$ is $(x-3)^2 + (y-2)^2 = 8$; $S_2$ is $x \ge 5$; $S_3$ is $|2y| \ge 8$, i.e. $|y| \ge 4$.
On the circle $y \le 2 + 2\sqrt{2} < 5$ and $y \ge 2 - 2\sqrt{2} > -4$, so $S_3$ needs $y \ge 4$.
Then $(x-3)^2 \ge 4$ and $(y-2)^2 \ge 4$, whose sum is at least 8; equality forces $x = 5$, $y = 4$.
Only $z = 5 + 4i$ lies in all three.
Answer: (B). | ||||
| 68 | 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 The normal is perpendicular to both given normals: $(2, 1, -1)\times(1, -1, -1) = (-1-1,\ -1+2,\ -2-1) = (-2, 1, -3)$.
Plane: $-2(x+1) + y - 3(z+2) = 0$, i.e. $2x - y + 3z + 8 = 0$.
$a + b + c = 2 - 1 + 3 = 4$.
Answer: (C). | ||||
| 69 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha + \beta = -20^{1/4}$, $\alpha\beta = \sqrt{5}$.
$\alpha^2 + \beta^2 = \sqrt{20} - 2\sqrt{5} = 0$.
$\alpha^4 + \beta^4 = 0^2 - 2(\alpha\beta)^2 = -10$.
$\alpha^8 + \beta^8 = (-10)^2 - 2(\alpha\beta)^4 = 100 - 50 = 50$.
Answer: (C). | ||||
| 70 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Left limit: $(1 + |\sin x|)^{3a/|\sin x|} \to e^{3a}$.
Right limit: $\frac{\cot 4x}{\cot 2x} = \frac{\tan 2x}{\tan 4x} \to \frac{2x}{4x} = \frac{1}{2}$, so the limit is $e^{1/2}$.
Continuity: $e^{3a} = e^{1/2} = b$, so $a = \frac{1}{6}$, $b = \sqrt{e}$.
$6a + b^2 = 1 + e$.
Answer: (A). | ||||
| 71 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections A: $x^2 + y^2 - x - y = \frac{1}{2}$, centre $\left(\frac{1}{2}, \frac{1}{2}\right)$, radius 1.
B: $x^2 + y^2 - 4y + \frac{7}{4} = 0$, centre (0, 2), radius $\frac{3}{2}$.
C: $(x-2)^2 + (y-1)^2 \le r^2$, a disc of radius $|r|$ about (2, 1).
The disc contains a circle when $|r| \ge$ (distance between centres) + (that radius).
For A: $\sqrt{\frac{9}{4} + \frac{1}{4}} + 1 = \frac{\sqrt{10}}{2} + 1 \approx 2.58$. For B: $\sqrt{4 + 1} + \frac{3}{2} = \frac{3 + 2\sqrt{5}}{2} \approx 3.74$.
Minimum $|r| = \frac{3 + 2\sqrt{5}}{2}$.
Answer: (D). | ||||
| 72 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean: $60 + a + b = 72$, so $a + b = 12$.
Variance: $\frac{\sum x^2}{8} - 81 = \frac{37}{4}$, so $\sum x^2 = 722$; known squares sum to 642, so $a^2 + b^2 = 80$.
$(a - b)^2 = 2(a^2 + b^2) - (a + b)^2 = 160 - 144 = 16$.
Answer: (A). | ||||
| 73 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} = e^{3x}e^{4y}$, so $e^{-4y}dy = e^{3x}dx$: $-\frac{e^{-4y}}{4} = \frac{e^{3x}}{3} + C$.
$y(0) = 0$: $-\frac{1}{4} = \frac{1}{3} + C$, $C = -\frac{7}{12}$.
At $x = -\frac{2}{3}\ln 2$, $e^{3x} = \frac{1}{4}$: $-\frac{e^{-4y}}{4} = \frac{1}{12} - \frac{7}{12} = -\frac{1}{2}$, so $e^{-4y} = 2$, $y = -\frac{1}{4}\ln 2$.
Answer: (A). | ||||
| 74 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Squaring: $1 + \sin2\theta = \frac{1}{4}$, so $\sin2\theta = -\frac{3}{4}$.
$\cos4\theta = 1 - 2\sin^22\theta = 1 - \frac{9}{8} = -\frac{1}{8}$.
$\sin6\theta = 3\sin2\theta - 4\sin^32\theta = -\frac{9}{4} + \frac{27}{16} = -\frac{9}{16}$.
$16\left(-\frac{12}{16} - \frac{2}{16} - \frac{9}{16}\right) = -23$.
Answer: (B). | ||||
| 75 | 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 $2 \equiv -1 \pmod 3$, so $2^n \equiv (-1)^n$. $2^n - 2 \equiv 0$ needs $(-1)^n \equiv 2 \equiv -1$: n odd.
Two-digit numbers: 90, of which 45 are odd.
Probability $= \frac{45}{90} = \frac{1}{2}$.
Answer: (A). | ||||
| 76 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The reflected ray passes through the image of (2, 1) in the y-axis, (−2, 1), and through (5, 3): slope $\frac{2}{7}$, line $2x - 7y + 11 = 0$.
Nearer focus to its directrix: $\frac{a}{e} - ae = 3a - \frac{a}{3} = \frac{8a}{3} = \frac{8}{\sqrt{53}}$, so $a = \frac{3}{\sqrt{53}}$.
Distance between directrices: $\frac{2a}{e} = 6a = \frac{18}{\sqrt{53}}$.
Other directrix $2x - 7y + c = 0$ with $\frac{|c - 11|}{\sqrt{53}} = \frac{18}{\sqrt{53}}$: $c = 29$ or $c = -7$.
Answer: (C). | ||||
| 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 $(P \vee Q) \wedge \sim P \equiv \sim P \wedge Q$, and $(\sim P \wedge Q) \Rightarrow Q$ is always true: the statement is a tautology.
$\sim(P \Rightarrow Q) \equiv P \wedge \sim Q$, so the biconditional in (C) is also a tautology.
(A) and (B) are false when P is false; (D) is false when both are false.
Answer: (C). | ||||
| 78 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability The form is $\frac{0}{0}$. By L'Hospital's rule: $\lim_{x\to2}\left(2xf(2) - 4f'(x)\right) = 4\cdot4 - 4\cdot1 = 12$.
Answer: (C). | ||||
| 79 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Divide by n: the term is $\frac{\frac{2j-1}{n} + 8}{\frac{2j-1}{n} + 4}$, and $\frac{2j-1}{n} \to 2x$ as $\frac{j}{n} \to x$.
Limit $= \int_0^1\frac{2x + 8}{2x + 4}dx = \int_0^1\left(1 + \frac{4}{2x + 4}\right)dx = 1 + 2[\ln(2x + 4)]_0^1 = 1 + 2\ln\frac{3}{2}$.
Answer: (B). | ||||
| 80 | 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 −x: $x\cos x$ is odd, so $\frac{1}{1 + e^{-x\cos x}} = \frac{e^{x\cos x}}{1 + e^{x\cos x}}$. Adding, $2I = \int_{-\pi/4}^{\pi/4}\frac{dx}{\sin^4x + \cos^4x}$, so $I = \int_0^{\pi/4}\frac{dx}{\sin^4x + \cos^4x}$.
Divide by $\cos^4x$, $t = \tan x$: $I = \int_0^1\frac{1 + t^2}{1 + t^4}dt = \int_0^1\frac{1 + 1/t^2}{(t - 1/t)^2 + 2}dt$.
With $u = t - \frac{1}{t}$ ($-\infty$ to 0): $I = \int_{-\infty}^0\frac{du}{u^2 + 2} = \frac{1}{\sqrt{2}}\cdot\frac{\pi}{2} = \frac{\pi}{2\sqrt{2}}$.
Answer: (A). | ||||
| 81 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The line passes through (2, 3, −2) with direction (−3, 2, 1). The vector to (3, 7, −7) is (1, 4, −5).
Normal: $(-3, 2, 1)\times(1, 4, -5) = (-10-4,\ 1-15,\ -12-2) = (-14, -14, -14) \parallel (1, 1, 1)$.
Plane: $x + y + z = 3$. $d = \frac{3}{\sqrt{3}} = \sqrt{3}$, $d^2 = 3$.
Answer: 3. | ||||
| 82 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$ give rows $(0, -2, -1)$ and $(2, 0, -1)$, with $R_3$ unchanged.
Expanding: $f(x) = 2\left[2(1 + \cos2x) + \sin^2x\right] - 2\cos^2x = 4 + 4\cos2x - 2\cos2x = 4 + 2\cos2x$.
Maximum (at $x = 0$ or $\pi$) $= 6$.
Answer: 6. | ||||
| 83 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $e^xF(x) = \int_3^x(3t^2 + 2t + 4F'(t))dt$, with $F(3) = 0$.
Differentiate: $e^x(F + F') = 3x^2 + 2x + 4F'$, i.e. $(e^x - 4)F' + e^xF = 3x^2 + 2x$, i.e. $\frac{d}{dx}\left[(e^x - 4)F\right] = 3x^2 + 2x$.
So $(e^x - 4)F = x^3 + x^2 - 36$ (using $F(3) = 0$), $F(x) = \frac{x^3 + x^2 - 36}{e^x - 4}$.
$F'(x) = \frac{(3x^2 + 2x)(e^x - 4) - (x^3 + x^2 - 36)e^x}{(e^x - 4)^2}$; at $x = 4$: $\frac{56(e^4 - 4) - 44e^4}{(e^4 - 4)^2} = \frac{12e^4 - 224}{(e^4 - 4)^2}$.
$\alpha = 12$, $\beta = 4$, $\alpha + \beta = 16$.
Answer: 16. | ||||
| 84 | 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 $\vec{a}\times\vec{c} = \vec{a}\times(\vec{a}\times\vec{b}) = (\vec{a}\cdot\vec{b})\vec{a} - |\vec{a}|^2\vec{b} = \vec{a} - 3\vec{b}$.
Directly, $\vec{a}\times\vec{c} = (1,1,1)\times(0,1,-1) = (-2, 1, 1)$, so $\vec{b} = \frac{\vec{a} - (-2,1,1)}{3} = (1, 0, 0)$.
$l = \frac{|\vec{b}\cdot(\vec{a}\times\vec{c})|}{|\vec{a}\times\vec{c}|} = \frac{2}{\sqrt{6}}$, so $3l^2 = 3\cdot\frac{4}{6} = 2$.
Answer: 2. | ||||
| 85 | 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 Need $\log_5(\log_3(\ldots)) > 0$, i.e. $\log_3(18x - x^2 - 77) > 1$, i.e. $18x - x^2 - 77 > 3$: $x^2 - 18x + 80 < 0$, $(x-8)(x-10) < 0$. Domain (8, 10).
With $x \to a + b - x$ the integrand becomes its complement, so $2I = \int_a^b dx = b - a = 2$, $I = 1$.
Answer: 1. | ||||
| 86 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $2\log_3(2^x - 5) = \log_32 + \log_3\left(2^x - \frac{7}{2}\right)$, so with $t = 2^x$: $(t - 5)^2 = 2t - 7$.
$t^2 - 12t + 32 = 0$, $t = 4$ or $8$. The log needs $t > 5$, so $t = 8$, $x = 3$.
Answer: 3. | ||||
| 87 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Coefficient determinant: $1(2 + \alpha) - 1(1 - 2\alpha) - 1(-1 - 4) = 6 + 3\alpha = 0$, so $\alpha = -2$.
Then the third equation must be a combination of the first two: $p(1, 1, -1) + q(1, 2, -2) = (2, -1, 1)$ gives $p = 5$, $q = -3$ (and the z-coefficients agree).
$\beta = 5\cdot2 - 3\cdot1 = 7$. $\alpha + \beta = 5$.
Answer: 5. | ||||
| 88 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalHard | |
A hard numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $f(1) = f(1)^2$ with $f(1) \in S$ gives $f(1) = 1$. The other products inside S are $2\cdot2 = 4$ and $2\cdot3 = 6$: $f(4) = f(2)^2$, $f(6) = f(2)f(3)$.
$f(2)^2 \in S$: $f(2) = 1$ or 2.
$f(2) = 1$: $f(4) = 1$, $f(3)$ any of 7, $f(6) = f(3)$: 7 ways.
$f(2) = 2$: $f(4) = 4$, $2f(3) \in S$ so $f(3) \in \{1, 2, 3\}$: 3 ways.
$f(5)$ and $f(7)$ are free: $7\times7 = 49$.
Total $= (7 + 3)\times49 = 490$.
Answer: 490. | ||||
| 89 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\sin(x+y) + \sin(x-y) = 2\sin x\cos y$, so $\sec y\,y' = 2\sin x\cos y$, i.e. $\sec^2y\,dy = 2\sin x\,dx$.
$\tan y = -2\cos x + C$; $y(0) = 0$ gives $C = 2$: $\tan y = 2 - 2\cos x$.
At $x = \frac{\pi}{2}$: $\tan y = 2$, $\cos^2y = \frac{1}{5}$, and $y' = 2\sin x\cos^2y = \frac{2}{5}$.
$5y'\left(\frac{\pi}{2}\right) = 2$.
Answer: 2. | ||||
| 90 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability With $\{x\} = x - [x]$, $f(x) = \min\{\{x\}, 1 - \{x\}\}$: the distance from x to the nearest integer, a continuous zig-zag. P is empty.
Corners (not differentiable) in (0, 3): at the integers 1, 2 and at 0.5, 1.5, 2.5 — 5 points.
$n(P) + n(Q) = 0 + 5 = 5$.
Answer: 5. | ||||
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