JEE (Main) · 2021
JEE (Main) 2021 — 18 March, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Gravitation3
- Work, Energy and Power2
- Motion of System of Particles and Rigid Body2
22 easy · 7 medium · 1 hard
Chemistry
30 Q
- Chemical Bonding and Molecular Structure2
- Amines2
- Aldehydes, Ketones and Carboxylic Acids2
21 easy · 8 medium · 1 hard
Mathematics
30 Q
- Conic Sections3
- Trigonometry2
- Vector Algebra2
14 easy · 13 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 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$ and $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
$\frac{v_{rms}}{v_{avg}} = \sqrt{\frac{3RT}{M}\cdot\frac{\pi M}{8RT}} = \sqrt{\frac{3\pi}{8}}$; the temperature and molar mass cancel.
Answer: (A). | ||||
| 2 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Let $m_1$ arrive with speed $u$ and both leave with speed $v$: $m_1$ backwards, $m_2$ forwards.
Momentum: $m_1 u = m_2 v - m_1 v$.
Taking the collision as elastic, the speed of separation equals the speed of approach: $v + v = u$, so $u = 2v$.
Then $2m_1 v = (m_2 - m_1)v$, so $m_2 = 3m_1$ and $m_2 : m_1 = 3 : 1$.
Answer: (D). | ||||
| 3 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation The force is $F = -\frac{dU}{dr} = -\frac{C}{r^2}$, attractive, of magnitude $\frac{C}{r^2}$.
For a circular orbit $\frac{mv^2}{r} = \frac{C}{r^2}$, so $r = \frac{C}{mv^2}$.
So $r \propto \frac{1}{v^2}$: r falls steeply as v grows, the decreasing curve of option (C).
Answer: (C). | ||||
| 4 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctHard | |
A hard single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The string leaves the far side of the cylinder parallel to the incline (angle $60^\circ$) and is tied to the wall above; friction acts at the contact point.
Normal reaction: $N = mg\cos60^\circ = \frac{mg}{2}$.
For the cylinder to stay at rest, torques about its centre need $T = f$, and forces along the incline need $T + f = mg\sin60^\circ$, so $f = \frac{\sqrt{3}}{4}mg \approx 0.43\,mg$.
But the largest friction available is $\mu_s N = 0.4 \times \frac{mg}{2} = 0.2\,mg$, which is less than $0.43\,mg$. So the cylinder cannot stay at rest: the contact point slides and friction takes its limiting value.
$f = \mu N = 0.4 \times \frac{mg}{2} = \frac{mg}{5}$.
Answer: (A). | ||||
| 5 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation Bodies at the equator float when the effective gravity there is zero: $\omega^2 R = g$.
$\omega = \sqrt{\frac{10}{6.4\times10^6}} = 1.25\times10^{-3}\ \mathrm{rad/s}$.
$T = \frac{2\pi}{\omega} = \frac{6.28}{1.25\times10^{-3}} = 5024\ \mathrm{s} \approx 83.7$ minutes $\approx 84$ minutes.
Answer: (D). | ||||
| 6 | 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 The semicircle has radius r with $\pi r = L$, so $r = \frac{L}{\pi}$.
Every element of the wire is at distance r from the axis through the centre, so $I = Mr^2 = \frac{ML^2}{\pi^2}$.
Answer: (A). | ||||
| 7 | 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 Resolving power is inversely proportional to the wavelength used, and $\lambda = \frac{h}{mv}$.
At the same speed, $\frac{RP_p}{RP_e} = \frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} \approx 1837$.
Answer: (A). | ||||
| 8 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane From the graph, $v = v_0\left(1 - \frac{x}{x_0}\right)$.
$a = v\frac{dv}{dx} = v_0\left(1-\frac{x}{x_0}\right)\left(-\frac{v_0}{x_0}\right) = -\frac{v_0^2}{x_0} + \frac{v_0^2}{x_0^2}x$.
This is a straight line with negative intercept $-\frac{v_0^2}{x_0}$ at $x = 0$, rising to zero at $x = x_0$: option (C).
Answer: (C). | ||||
| 9 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $PV^\gamma = $ constant. Taking logarithms and differentiating, $\frac{dP}{P} + \gamma\frac{dV}{V} = 0$.
$\frac{dP}{P} = -\gamma\frac{dV}{V}$.
Answer: (A). | ||||
| 10 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation In time dt the radius vector sweeps $dA = \frac{1}{2}|\vec{r}\times\vec{v}\,dt|$.
$\frac{dA}{dt} = \frac{1}{2}|\vec{r}\times\vec{v}| = \frac{|\vec{r}\times M\vec{v}|}{2M} = \frac{L}{2M}$.
Answer: (B). | ||||
| 11 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Entropy is an extensive quantity: the entropy of a system is the sum of the entropies of its parts.
So the total entropy is $S_1 + S_2$ (the options offer no 'greater than' choice; product, difference and ratio are not even dimensionally entropies).
Answer: (C). | ||||
| 12 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves A period of $\frac{\pi}{\omega}$ means an angular frequency of $2\omega$.
(A) $\sin\omega t + \cos\omega t = \sqrt{2}\sin(\omega t + \frac{\pi}{4})$: SHM, but with period $\frac{2\pi}{\omega}$. Not this.
(B) $\sin^2\omega t = \frac{1}{2} - \frac{1}{2}\cos 2\omega t$: it satisfies $\frac{d^2x}{dt^2} = -4\omega^2\left(x - \frac{1}{2}\right)$, an SHM about the mean position $\frac{1}{2}$ with period $\frac{\pi}{\omega}$.
(C) $3\cos\left(\frac{\pi}{4} - 2\omega t\right)$: SHM of amplitude 3 and angular frequency $2\omega$, period $\frac{\pi}{\omega}$.
(D) is a sum of three different frequencies: periodic (period $\frac{2\pi}{\omega}$) but not simple harmonic.
Answer: (C), the pure sinusoid; (B) is also an SHM of period $\frac{\pi}{\omega}$ about a shifted mean position. | ||||
| 13 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{36 + 36} = 6\sqrt{2}\ \Omega$.
Power factor $\cos\phi = \frac{R}{Z} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Answer: (A). | ||||
| 14 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter (A) False: electric monopoles (isolated charges) exist; magnetic monopoles are not known to exist.
(B) True: at the ends and outside a finite solenoid the lines spread out and curve round; they cannot be straight and confined.
(C) True: for an ideal toroid the field is wholly inside the core and zero outside.
(D) False as intended: inside a bar magnet (as inside a solenoid) the lines run nearly parallel from S to N.
(E) A perfect diamagnet (a superconductor) has $\chi = -1$, which is also a true statement.
Of the options, the one built from the intended true statements about field lines is (B) and (C) only.
Answer: (B). | ||||
| 15 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $I_E = I_B + I_C$. Dividing by $I_C$: $\frac{1}{\alpha} = \frac{1}{\beta} + 1 = \frac{1+\beta}{\beta}$.
So $\alpha = \frac{\beta}{1+\beta}$ (equivalently $\beta = \frac{\alpha}{1-\alpha}$).
Answer: (D). | ||||
| 16 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.
For the proton (m, e) and the $\alpha$-particle (4m, 2e): $\frac{r_p}{r_\alpha} = \frac{\sqrt{2mK_p}/e}{\sqrt{8mK_\alpha}/2e} = \sqrt{\frac{K_p}{K_\alpha}}$.
$\sqrt{\frac{K_p}{K_\alpha}} = 2$, so $K_p : K_\alpha = 4 : 1$.
Answer: (B). | ||||
| 17 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The rays meet PQ normally and go straight on to PR. The prism as drawn is an isosceles right-angled prism, so the angle of incidence on PR is $45^\circ$.
A ray emerges only if $45^\circ$ is less than the critical angle, i.e. $\mu < \frac{1}{\sin45^\circ} = \sqrt{2} \approx 1.414$.
Red (1.27) emerges; green (1.42) and blue (1.49) exceed 1.414 and are totally internally reflected.
Answer: (D). | ||||
| 18 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $U = \frac{1}{2}Li^2$, so $U = \frac{1}{4}U_{max}$ when $i = \frac{1}{2}i_0$.
$i = i_0\left(1 - e^{-Rt/L}\right) = \frac{i_0}{2}$ gives $e^{-Rt/L} = \frac{1}{2}$, so $t = \frac{L}{R}\ln2$.
Answer: (A). | ||||
| 19 | 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 $\vec{E}$ and $\vec{B}$ are perpendicular to each other and to the direction of propagation (y), so they lie along x and z.
The pairs are $E_x, B_z$ or $E_z, B_x$.
Answer: (C). | ||||
| 20 | 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 A free proton cannot turn into a neutron, since the neutron is heavier.
Inside a nucleus the energy can come from the change in nuclear binding energy, and $p \to n + e^+ + \nu$ ($\beta^+$ decay) does occur.
So the conversion is possible only inside the nucleus.
Answer: (B). | ||||
| 21 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The horizontal component of velocity is unchanged; the vertical component reverses, from $+u\sin45^\circ$ to $-u\sin45^\circ$.
$|\Delta p| = 2mu\sin45^\circ = 2\times0.005\times5\sqrt{2}\times\frac{1}{\sqrt{2}} = 0.05 = 5\times10^{-2}\ \mathrm{kg\,m\,s^{-1}}$.
Answer: 5. | ||||
| 22 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance All the forces act along the y-axis in the same direction.
$F = kq_1q_2\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \cdots\right) = 9\times10^9\times10^{-6}\times1\times\frac{1}{1-\frac{1}{4}}$.
$F = 9\times10^3\times\frac{4}{3} = 12\times10^3$ N.
Answer: 12. | ||||
| 23 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Let each wire have length l and area A: $R_1 = \frac{6l}{A}$, $R_2 = \frac{3l}{A}$.
In parallel, $R = \frac{R_1R_2}{R_1+R_2} = \frac{18l/A}{9} = \frac{2l}{A}$.
The combination is a conductor of length l and area 2A, so $\rho_{eff}\frac{l}{2A} = \frac{2l}{A}$, giving $\rho_{eff} = 4\ \Omega$ cm.
Answer: 4. | ||||
| 24 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At maximum compression all the kinetic energy is stored in the spring: $\frac{1}{2}\times4\times10^2 = \frac{1}{2}\times100\times d^2$.
$d^2 = 4$, so the compression is $d = 2$ m.
The compressed length is $8 - 2 = 6$ m.
Answer: 6. | ||||
| 25 | 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 In the active region, a change of base current of $10\ \mu A$ (e.g. 20 to 30 $\mu A$) changes the collector current by 2 mA.
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{2\times10^{-3}}{10\times10^{-6}} = 200$ (the d.c. ratio, e.g. $\frac{8\ \mathrm{mA}}{40\ \mu A}$, gives the same).
Answer: 200. | ||||
| 26 | Current ElectricityMetre bridge | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity At balance, $\frac{12}{6} = \frac{R_{AP}}{R_{PB}} = \frac{x}{72 - x}$.
$x = 2(72 - x)$, so $3x = 144$ and $x = 48$ cm.
Answer: 48. | ||||
| 27 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For $v \ll c$: $\Delta\lambda = \lambda\frac{v}{c} = 630\times10^{-9}\times\frac{286\times10^3}{3\times10^8}$.
$\Delta\lambda = 6.006\times10^{-10}$ m, so $x \approx 6$.
Answer: 6. | ||||
| 28 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical 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 The load adds a pressure $\frac{mg}{A} = \frac{24\times10}{0.4} = 600$ Pa at the surface.
The tank is so much wider than the opening that the surface speed is negligible. Bernoulli between the surface and B:
$\frac{1}{2}\rho v^2 = \frac{mg}{A} + \rho gh = 600 + 1000\times10\times0.4 = 4600$.
$v^2 = \frac{2\times4600}{1000} = 9.2$, so $v \approx 3.03\ \mathrm{m/s}$, i.e. 3.
Answer: 3. | ||||
| 29 | 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 $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
(i) $d_1 = \sqrt{2R\times20}$. (ii) $d_2 = \sqrt{2R\times20} + \sqrt{2R\times5}$.
Increase $= \frac{\sqrt{2R\times5}}{\sqrt{2R\times20}}\times100\% = \frac{1}{2}\times100\% = 50\%$.
Answer: 50. | ||||
| 30 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalEasy | |
A easy numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $V = \frac{4}{3}\pi r^3$, so $\frac{\Delta V}{V} = 3\frac{\Delta r}{r} = 3\times\frac{0.85}{7.50} = 0.34$.
Percentage error $= 34\%$.
Answer: 34. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Solid StateClassification of solids | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on classification of solids, from Solid State. NCERT: No matching NCERT chapter Ionic solids conduct when molten; metallic solids conduct in both states; molecular solids are soft and melt low.
A hard, high-melting solid that does not conduct in either state is a covalent (network) solid, like diamond or quartz.
Answer: (B). | ||||
| 32 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Statement I: $Li^+$ has two electrons; Bohr's theory applies only to one-electron species ($H$, $He^+$, $Li^{2+}$), so it does not account for $Li^+$. False.
Statement II: Bohr's theory cannot explain the Zeeman effect (splitting in a magnetic field). True.
Answer: (D). | ||||
| 33 | 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 Metal sulphide sols such as CdS (and $As_2S_3$) are negatively charged; metal oxide and hydroxide sols such as $TiO_2$ are positively charged.
Answer: (B). | ||||
| 34 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties First ionization enthalpies (kJ/mol): Mg 738; Ar 1521, Cl 1251, Ne 2081 (higher); Na 496, Li 520 (lower).
X and Y must exceed Mg and Z must be below it: argon, chlorine and sodium fit. Lithium (520) is below Mg, ruling out (B) and (C); in (D) chlorine is not below Mg.
Answer: (A). | ||||
| 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 Mercury, a low-boiling metal, is refined by distillation (ii).
Copper is refined electrolytically (iii).
Silicon for semiconductors is refined by zone refining (iv).
Nickel is refined by the Mond process, a vapour phase refining via Ni(CO)4 (i).
Answer: (A). | ||||
| 36 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter In basic medium $H_2O_2$ oxidises $Mn^{2+}$: $Mn^{2+} + H_2O_2 \to Mn^{4+} + 2OH^-$. So (A) occurs.
In basic medium it also acts as a reducing agent: $I_2 + H_2O_2 + 2OH^- \to 2I^- + 2H_2O + O_2$. So (B) occurs.
The oxidation $PbS + 4H_2O_2 \to PbSO_4 + 4H_2O$ is the textbook example of its oxidising action in acidic medium, so (C) is not counted.
Answer: (C). | ||||
| 37 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Beryllium, transparent to X-rays, is used for the windows of X-ray tubes (iv).
Magnesium powder burns with a brilliant flame: incendiary bombs and signals (iii).
Calcium is used in the extraction of metals from their oxides (ii).
Radium is used in radiotherapy, the treatment of cancer (i).
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 NO: +2; $NO_2$: +4; $N_2O$: +1; $NO_3^-$: +5.
Order: $NO_3^- (+5) > NO_2 (+4) > NO (+2) > N_2O (+1)$.
Answer: (D). | ||||
| 39 | 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 $SiO_2$, MgO and $Na_2O$ have no unpaired electrons and are diamagnetic.
$Mn_3O_4$ ($Mn^{2+}Mn^{3+}_2O_4$, a spinel) contains manganese ions with unpaired d electrons and is magnetic (ferrimagnetic).
Answer: (B). | ||||
| 40 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Blue vitriol is $[Cu(H_2O)_4]SO_4\cdot H_2O$: four water molecules are coordinated to $Cu^{2+}$ (secondary valency, i.e. coordination number, 4).
The fifth water molecule is not coordinated; it is held by hydrogen bonds between the sulphate ion and coordinated water.
Answer: (A). | ||||
| 41 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Statement I: thermal power plants produce fly ash, a non-biodegradable industrial waste. True.
Statement II: biodegradable detergents (and the phosphates in them) feed the growth of bacteria and algae; as these grow and decay they use up the dissolved oxygen and choke aquatic life, the process of eutrophication. True.
Answer: (A). | ||||
| 42 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Statement I: ethanol is itself a nucleophile (and gives ethoxide with a base); AgCN gives the cyanide nucleophile. True.
Statement II: AgCN is largely covalent, so the cyanide attacks through nitrogen and alkyl halides give isocyanides (isonitriles), not nitriles; KCN gives nitriles. False.
Answer: (C). | ||||
| 43 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Carbon a is the methyl carbon, with four single bonds: sp3.
Carbon b is a carbon of the C=C double bond: sp2.
Carbon c is a benzene ring carbon: sp2.
Answer: (A). | ||||
| 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 HI protonates the ether oxygen; iodide then attacks the methyl carbon (SN2), since the aryl C-O bond has partial double bond character and is not broken.
Products: 1-naphthol and CH3I.
Answer: (D). | ||||
| 45 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hofmann bromamide degradation: $RCONH_2 + Br_2 + 4NaOH \to RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.
The isocyanate formed in the rearrangement is hydrolysed in the strongly alkaline solution, and its carbon leaves as carbonate, $CO_3^{2-}$.
Answer: (C). | ||||
| 46 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Dilute NaOH: the enolate of one cyclopentanone adds to the C=O of another (aldol addition), giving the $\beta$-hydroxy ketone X, 2-(1-hydroxycyclopentyl)cyclopentanone.
Acid and heat: X loses water to give the $\alpha,\beta$-unsaturated ketone, the double bond forming between the $\alpha$-carbon and the carbinol carbon so that it is conjugated with C=O: Y is 2-cyclopentylidenecyclopentanone.
(A) and (B) show a non-conjugated Y (and (B) the wrong X); (C) shows an extra ring double bond that cannot form.
Answer: (D). | ||||
| 47 | 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 Hinsberg test: a primary amine gives an N-substituted sulphonamide with an acidic N-H, which dissolves in alkali; a secondary amine gives a sulphonamide with no N-H, insoluble in alkali; a tertiary amine does not react.
(A) and (C) are secondary, (B) is tertiary; only (D), 1-phenylethylamine, is primary.
Answer: (D). | ||||
| 48 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons In the strongly acidic nitrating mixture much of the aniline is protonated to the anilinium ion, $C_6H_5NH_3^+$, which is meta-directing.
Direct nitration of aniline gives about 51% p-nitroaniline (C), 47% m-nitroaniline (B) and 2% o-nitroaniline (A).
Order: C > B > A.
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 Norethindrone is an antifertility drug (a-iii); Salvarsan, the arsenic drug for syphilis, is an antibiotic (b-iv); meprobamate is a tranquilizer (c-i); alitame is an artificial sweetener (d-ii).
Answer: (C). | ||||
| 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 Vitamin K is needed to make the clotting factors, so its deficiency increases the blood clotting time. Cheilosis comes from vitamin B2 deficiency and fragile RBCs from vitamin E deficiency.
Answer: (A). | ||||
| 51 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter By convention the concordant reading is taken: 5.0 mL.
$Na_2CO_3 + 2HCl \to 2NaCl + H_2O + CO_2$, so moles of HCl = 2 × moles of $Na_2CO_3$.
$2\times M\times10.0 = 0.2\times5.0$, so $M = 0.05$ M = 50 mM.
Answer: 50. | ||||
| 52 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Lone pairs on the central atom: $SF_4$ 1, $BF_4^-$ 0, $ClF_3$ 2, $AsF_3$ 1, $PCl_5$ 0, $BrF_5$ 1, $XeF_4$ 2, $SF_6$ 0.
Two lone pairs: $ClF_3$ and $XeF_4$.
Answer: 2. | ||||
| 53 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalHard | |
A hard numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium For a gas reaction $\Delta G^\circ$ gives $K_p$ (pressures in bar): $\ln K_p = -\frac{\Delta G^\circ}{RT} = -\frac{25200}{8.3\times400} = -7.59$.
$\log K_p = -\frac{7.59}{2.3} = -3.3$, so $K_p = 0.501\times10^{-3} = 5.01\times10^{-4}\ \mathrm{bar^{-1}}$.
$K_p = K_C(RT)^{\Delta n}$ with $\Delta n = 1 - 2 = -1$, so $K_C = K_p\,RT$.
With $R = 0.083\ \mathrm{L\,bar\,mol^{-1}K^{-1}}$: $RT = 33.2$, and $K_C = 5.01\times10^{-4}\times33.2 = 1.66\times10^{-2}$.
To the nearest integer, 2.
Answer: 2. | ||||
| 54 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\Delta T_b = iK_bm$: $0.52 = i\times0.52\times2$, so $i = \frac{1}{2}$.
For dimerisation $2A \rightleftharpoons A_2$: $i = 1 - \frac{\alpha}{2}$, so $\frac{1}{2} = 1 - \frac{\alpha}{2}$ and $\alpha = 1$.
The association is 100%.
Answer: 100. | ||||
| 55 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $K_{sp} = s^2 = (8.0\times10^{-4})^2 = 6.4\times10^{-7}$.
In 0.01 M $H_2SO_4$, $[SO_4^{2-}] \approx 0.01$ M (taking the sulphuric acid as fully dissociated to sulphate, as the question intends): $s' = \frac{6.4\times10^{-7}}{0.01} = 6.4\times10^{-5} = 64\times10^{-6}\ \mathrm{mol\,L^{-1}}$.
Answer: 64. | ||||
| 56 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Kohlrausch's law: $\Lambda^\circ_m(BaSO_4) = \Lambda^\circ_m(BaCl_2) + \Lambda^\circ_m(H_2SO_4) - 2\Lambda^\circ_m(HCl)$.
$= 280 + 860 - 2\times426 = 288\ \mathrm{S\,cm^2\,mol^{-1}}$.
Answer: 288. | ||||
| 57 | 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 Taking the reaction as first order (the half-life is quoted as a constant): $k = \frac{\ln2}{t_{1/2}} = 0.69\ \mathrm{min^{-1}}$.
$t = \frac{1}{k}\ln\frac{100}{0.1} = \frac{\ln1000}{0.69} = \frac{3\times2.3}{0.69} = 10$ min.
Answer: 10. | ||||
| 58 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Partial hydrolysis of $XeF_6$ gives $XeO_2F_2$: $XeF_6 + 2H_2O \to XeO_2F_2 + 4HF$. So A is $XeF_6$.
Lone pairs in $XeF_6$: one on Xe and three on each of the six F atoms, $1 + 18 = 19$.
Answer: 19. | ||||
| 59 | 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 Molar mass of benzoic acid $C_7H_6O_2$ = 122 g/mol, so 6.1 g = 0.05 mol.
Molar mass of m-bromobenzoic acid $C_7H_5BrO_2$ = 84 + 5 + 80 + 32 = 201 g/mol; the theoretical yield is $0.05\times201 = 10.05$ g.
Percentage yield $= \frac{7.8}{10.05}\times100 = 77.6\% \approx 78$.
Answer: 78. | ||||
| 60 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalEasy | |
A easy numerical question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \to RCOO^- + 2Ag + 2H_2O + 4NH_3$.
The aldehyde carbon goes from +1 to +3, giving up 2 electrons, which reduce two $Ag^+$ ions to Ag.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | 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 Let $s = \sin^2\alpha$: $15s^2 + 10(1-s)^2 = 6$, i.e. $25s^2 - 20s + 4 = 0$, i.e. $(5s - 2)^2 = 0$.
So $\sin^2\alpha = \frac{2}{5}$, $\cos^2\alpha = \frac{3}{5}$.
$27\sec^6\alpha = 27\left(\frac{5}{3}\right)^3 = 125$ and $8\mathrm{cosec}^6\alpha = 8\left(\frac{5}{2}\right)^3 = 125$.
Sum $= 250$.
Answer: (D). | ||||
| 62 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $(P\Rightarrow Q)\wedge\sim Q$ is true only when Q is false and P is false (modus tollens).
In that case $\sim P$ is true, so $((P\Rightarrow Q)\wedge\sim Q)\Rightarrow\sim P$ is always true: a tautology.
In the same case P, Q and $P\wedge Q$ are all false, so (A), (B) and (D) are false there.
Answer: (C). | ||||
| 63 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Equal angles of elevation from A, B and C mean the foot of the pole is equidistant from the three corners: it is the circumcentre, at distance R = 2 from each.
$h = R\tan\frac{\pi}{3} = 2\sqrt{3}$.
Answer: (B). | ||||
| 64 | 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 The original mean is 0 and the original standard deviation is $\sqrt{\frac{n a^2 + n a^2}{2n}} = |a|$.
Adding b shifts the mean to b and leaves the standard deviation unchanged: b = 5, |a| = 20.
$a^2 + b^2 = 400 + 25 = 425$.
Answer: (B). | ||||
| 65 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(1) = 5pq^4 = 0.4096$, $P(2) = 10p^2q^3 = 0.2048$.
Dividing: $\frac{2p}{q} = \frac{1}{2}$, so $q = 4p$ and $p = \frac{1}{5}$, $q = \frac{4}{5}$ (check: $5\cdot\frac{1}{5}\cdot\frac{256}{625} = 0.4096$).
$P(3) = 10p^3q^2 = 10\cdot\frac{1}{125}\cdot\frac{16}{25} = \frac{160}{3125} = \frac{32}{625}$.
Answer: (D). | ||||
| 66 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra The projection is $|\vec{AB}|\cos A$.
$\cos A = \frac{AB^2 + AC^2 - BC^2}{2\cdot AB\cdot AC} = \frac{100 + 49 - 64}{140} = \frac{85}{140}$.
Projection $= 10\times\frac{85}{140} = \frac{85}{14}$.
Answer: (B). | ||||
| 67 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines In an equilateral triangle the centroid is also the incentre, so r is the distance from the origin to $x + y = 3$: $r = \frac{3}{\sqrt{2}}$.
Also $R = 2r$, so $R + r = 3r = \frac{9}{\sqrt{2}}$.
Answer: (B). | ||||
| 68 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Tangent: $\frac{x\cos\theta}{3\sqrt{3}} + y\sin\theta = 1$. Intercepts $3\sqrt{3}\sec\theta$ and $\mathrm{cosec}\,\theta$.
$S(\theta) = 3\sqrt{3}\sec\theta + \mathrm{cosec}\,\theta$; $S'(\theta) = 3\sqrt{3}\sec\theta\tan\theta - \mathrm{cosec}\,\theta\cot\theta = 0$.
$3\sqrt{3}\frac{\sin\theta}{\cos^2\theta} = \frac{\cos\theta}{\sin^2\theta}$, so $\tan^3\theta = \frac{1}{3\sqrt{3}}$ and $\tan\theta = \frac{1}{\sqrt{3}}$.
$\theta = \frac{\pi}{6}$; S tends to infinity at both ends of the interval, so this is the minimum.
Answer: (B). | ||||
| 69 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $u = y + 1$: $u' + xu = u^2e^{x^2/2}$, a Bernoulli equation. With $v = \frac{1}{u}$: $v' - xv = -e^{x^2/2}$.
Integrating factor $e^{-x^2/2}$: $\frac{d}{dx}\left(ve^{-x^2/2}\right) = -1$, so $ve^{-x^2/2} = -x + C$.
At $x = 2$, $u = 1$, $v = 1$: $e^{-2} = -2 + C$, so $C = 2 + e^{-2}$ and $v = e^{x^2/2}(2 - x + e^{-2})$.
At $x = 1$: $v = e^{1/2}(1 + e^{-2})$, so $u = \frac{e^{-1/2}}{1 + e^{-2}} = \frac{e^{3/2}}{e^2 + 1}$.
$\frac{dy}{dx} = u\left(ue^{1/2} - 1\right) = \frac{e^{3/2}}{e^2+1}\left(\frac{e^2}{e^2+1} - 1\right) = \frac{-e^{3/2}}{(e^2+1)^2}$.
Answer: (D). | ||||
| 70 | 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 Real points need $(4 - x)(x - 2) \ge 0$, i.e. $2 \le x \le 4$; the closed curve is symmetric about the x-axis with $|y| = \frac{x}{2}\sqrt{(4-x)(x-2)}$.
Area $= 2\int_2^4 \frac{x}{2}\sqrt{(4-x)(x-2)}\,dx = \int_2^4 x\sqrt{1 - (x-3)^2}\,dx$.
With $t = x - 3$: $\int_{-1}^{1}(t + 3)\sqrt{1 - t^2}\,dt = 0 + 3\cdot\frac{\pi}{2} = \frac{3\pi}{2}$ (the odd part vanishes; the even part is a half unit disc).
Answer: (D). | ||||
| 71 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $g(3) = \int_0^1 f + \int_1^3 f$.
$\frac{1}{3} \le \int_0^1 f \le 1$ and $0 \le \int_1^3 f \le \frac{1}{2}\times2 = 1$.
So $\frac{1}{3} \le g(3) \le 2$.
Answer: (A). | ||||
| 72 | 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: $\lim_{x\to0^-}\frac{\sin(a+1)x + \sin2x}{2x} = \frac{(a+1) + 2}{2} = \frac{a+3}{2}$.
Right limit: $\frac{\sqrt{x}\left(\sqrt{1 + bx^2} - 1\right)}{bx^{5/2}} = \frac{\sqrt{1+bx^2} - 1}{bx^2} = \frac{1}{\sqrt{1+bx^2} + 1} \to \frac{1}{2}$.
Continuity: $b = \frac{1}{2}$ and $\frac{a+3}{2} = \frac{1}{2}$, so $a = -2$.
$a + b = -\frac{3}{2}$.
Answer: (D). | ||||
| 73 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_2 - S_1 = \frac{4n}{2}[2a + (4n-1)d] - \frac{2n}{2}[2a + (2n-1)d] = n[2a + (6n-1)d] = 1000$.
$S_{6n} = \frac{6n}{2}[2a + (6n-1)d] = 3\times1000 = 3000$.
Answer: (C). | ||||
| 74 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Let S have centre C and radius r. Touching $S_1$ internally: $CO = 3 - r$; touching $S_2$ externally: $CC_2 = 1 + r$.
So $CO + CC_2 = 4$: an ellipse with foci (0, 0) and (2, 0), $2a = 4$.
Centre (1, 0), $a = 2$, $c = 1$, $b^2 = 3$: $\frac{(x-1)^2}{4} + \frac{y^2}{3} = 1$.
$\left(2, \pm\frac{3}{2}\right)$: $\frac{1}{4} + \frac{9/4}{3} = \frac{1}{4} + \frac{3}{4} = 1$. On the locus. The other points fail: (0, ±√3) gives $\frac{5}{4}$, (1, ±2) gives $\frac{4}{3}$, $\left(\frac{1}{2}, \pm\frac{\sqrt5}{2}\right)$ gives $\frac{23}{48}$.
Answer: (A). | ||||
| 75 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A homogeneous system has a non-trivial solution exactly when its determinant is zero:
$\begin{vmatrix}4&\lambda&2\\2&-1&1\\\mu&2&3\end{vmatrix} = 4(-3-2) - \lambda(6-\mu) + 2(4+\mu) = \lambda\mu - 6\lambda + 2\mu - 12 = (\lambda+2)(\mu-6)$.
It vanishes when $\mu = 6$ (for every $\lambda$) or $\lambda = -2$ (for every $\mu$). Of the options, $\mu = 6, \lambda\in\mathbb{R}$.
Answer: (D). | ||||
| 76 | 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 $y = \frac{x-2}{x-3}$ gives $x = \frac{3y-2}{y-1}$, so $f^{-1}(x) = \frac{3x-2}{x-1}$; $g^{-1}(x) = \frac{x+3}{2}$.
$\frac{3x-2}{x-1} + \frac{x+3}{2} = \frac{13}{2}$ gives $2(3x-2) + (x+3)(x-1) = 13(x-1)$, i.e. $x^2 - 5x + 6 = 0$.
x = 2 or 3, both allowed ($x\ne1$). Sum = 5.
Answer: (B). | ||||
| 77 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|w| = 2$, so $|z| = \frac{1}{2}$.
Oz and Ow are at right angles, so the area is $\frac{1}{2}|z||w| = \frac{1}{2}\times\frac{1}{2}\times2 = \frac{1}{2}$.
Answer: (A). | ||||
| 78 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Reflexive: $IAI^{-1} = A$.
Symmetric: if $PAP^{-1} = B$ then $A = P^{-1}BP = (P^{-1})B(P^{-1})^{-1}$.
Transitive: if $PAP^{-1} = B$ and $QBQ^{-1} = C$ then $(QP)A(QP)^{-1} = C$, with QP non-singular.
So R (similarity of matrices) is an equivalence relation.
Answer: (D). | ||||
| 79 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $\frac{x^2}{4} - \frac{y^2}{2} = 1$: $a^2 = 4$, $b^2 = 2$, $e^2 = \frac{3}{2}$, $ae = \sqrt{6}$. F = $(\sqrt{6}, 0)$.
Tangent at P: $\frac{4x}{4} - \frac{\sqrt{6}y}{2} = 1$, i.e. $x - \frac{\sqrt{6}}{2}y = 1$. Q = (1, 0).
On the latus rectum $x = \sqrt{6}$: $y_1 = \frac{2(\sqrt{6} - 1)}{\sqrt{6}}$.
Area $= \frac{1}{2}\cdot QF\cdot y_1 = \frac{1}{2}(\sqrt{6} - 1)\cdot\frac{2(\sqrt{6}-1)}{\sqrt{6}} = \frac{(\sqrt{6}-1)^2}{\sqrt{6}} = \frac{7 - 2\sqrt{6}}{\sqrt{6}} = \frac{7}{\sqrt{6}} - 2$.
Answer: (B). | ||||
| 80 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}| = |\vec{a}|^2 = |\vec{a}|$, so $|\vec{a}| = |\vec{b}| = 1$, and $\vec{a}, \vec{b}, \vec{a}\times\vec{b}$ are mutually perpendicular unit vectors.
$\vec{v} = \vec{a} + \vec{b} + \vec{a}\times\vec{b}$ has $|\vec{v}| = \sqrt{3}$ and $\vec{v}\cdot\vec{a} = 1$.
$\cos\theta = \frac{1}{\sqrt{3}}$, $\theta = \cos^{-1}\frac{1}{\sqrt{3}}$.
Answer: (A). | ||||
| 81 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations The roots of $x^2 + x + 1$ are $\omega$ and $\omega^2$, with $\omega^3 = 1$.
$P(\omega) = f(1) + \omega g(1) = 0$ and $P(\omega^2) = f(1) + \omega^2 g(1) = 0$.
Subtracting: $(\omega - \omega^2)g(1) = 0$, so $g(1) = 0$, and then $f(1) = 0$.
$P(1) = f(1) + g(1) = 0$.
Answer: 0. | ||||
| 82 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants tr P = −1, det P = −6 + 5 = −1, so by the Cayley-Hamilton theorem $P^2 + P - I = O$, i.e. $P^2 = I - P$.
$P^3 = P - P^2 = 2P - I$; $P^4 = 2P^2 - P = 2I - 3P$; $P^5 = 2P - 3P^2 = 5P - 3I$; $P^6 = 5P^2 - 3P = 5I - 8P$.
n = 6.
Answer: 6. | ||||
| 83 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $\frac{x+1}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$ (sum of cubes), and $\frac{x-1}{x - x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = 1 + x^{-1/2}$.
The bracket is $x^{1/3} - x^{-1/2}$.
$T_{r+1} = \binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r$; the power of x is $\frac{10-r}{3} - \frac{r}{2} = 0$, so r = 4.
Term $= \binom{10}{4} = 210$.
Answer: 210. | ||||
| 84 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalHard | |
A hard numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $(r+1)(r+2)(r+3) = r^3 + 6r^2 + 11r + 6$, so $r^3 + 6r^2 + 2r + 5 = (r+1)(r+2)(r+3) - 9r - 1$.
$r!\left(r^3 + 6r^2 + 2r + 5\right) = (r+3)! - 9\,r\cdot r! - r! = (r+3)! - 9[(r+1)! - r!] - r! = (r+3)! - 9(r+1)! + 8\,r!$.
Summing r = 1 to 10: $\sum_{4}^{13}k! - 9\sum_{2}^{11}k! + 8\sum_{1}^{10}k!$.
Collecting: $13! + 12! + 11!(1 - 9) + \sum_{4}^{10}k!(1 - 9 + 8) - 9(2! + 3!) + 8(1! + 2! + 3!) = 13! + 12! - 8\cdot11! - 72 + 72$.
$= 11!(156 + 12 - 8) = 160\cdot11!$.
Answer: 160. | ||||
| 85 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $P'(x) = k(x^2 - 1)$ with $k > 0$ (minimum at 1, maximum at −1), so $P(x) = k\left(\frac{x^3}{3} - x\right) + c$.
$P(-3) = k(-9 + 3) + c = 0$ gives $c = 6k$.
$\int_{-1}^{1}P\,dx = 2c = 18$ (the odd part integrates to 0), so c = 9 and $k = \frac{3}{2}$.
$P(x) = \frac{x^3}{2} - \frac{3x}{2} + 9$; the sum of coefficients is $P(1) = \frac{1}{2} - \frac{3}{2} + 9 = 8$.
Answer: 8. | ||||
| 86 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $y = vx$: $x\,dy - y\,dx = x^2dv$, so $x^2dv = x\sqrt{1 - v^2}\,dx$, i.e. $\frac{dv}{\sqrt{1-v^2}} = \frac{dx}{x}$.
$\sin^{-1}v = \ln x + C$; $y(1) = 0$ gives C = 0. So $y = x\sin(\ln x)$.
For $1 \le x \le e^\pi$, $\ln x\in[0, \pi]$ and $y \ge 0$. Area $= \int_1^{e^\pi}x\sin(\ln x)\,dx$; with $x = e^t$: $\int_0^\pi e^{2t}\sin t\,dt$.
$\int e^{2t}\sin t\,dt = \frac{e^{2t}(2\sin t - \cos t)}{5}$, so the area is $\frac{e^{2\pi}(0 + 1) - (0 - 1)}{5} = \frac{e^{2\pi} + 1}{5}$.
$\alpha = \beta = \frac{1}{5}$, and $10(\alpha + \beta) = 4$.
Answer: 4. | ||||
| 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 Normal $\vec{n} = (3, 4, 2)\times(4, -3, 7) = (28 + 6,\ 8 - 21,\ -9 - 16) = (34, -13, -25)$.
Plane through (1, −6, −5): $34(x - 1) - 13(y + 6) - 25(z + 5) = 0$.
At $(1, -1, \alpha)$: $0 - 13(5) - 25(\alpha + 5) = 0$, so $25\alpha = -190$ and $\alpha = -7.6$.
$|5\alpha| = 38$.
Answer: 38. | ||||
| 88 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The segment joining the point and its image is parallel to the normal (2, −1, 1): $(-3 - 1, 5 - 3, 2 - a) = (-4, 2, 2 - a) = -2(2, -1, 1)$ gives $2 - a = -2$, so a = 4.
The midpoint (−1, 4, 3) lies on the plane: $2(-1) - 4 + 3 = b$, so b = −3.
|a + b| = 1.
Answer: 1. | ||||
| 89 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem With n = 10: $\sum_{k=0}^{10}{}^{10}C_k = 2^{10}$ and $\sum_{k=0}^{10}k\,{}^{10}C_k = 10\cdot2^9 = 5\cdot2^{10}$.
$\sum(4 + 3k)\,{}^{10}C_k = 4\cdot2^{10} + 15\cdot2^{10} = 19\cdot2^{10}$.
Comparing with $\alpha\cdot3^{10} + \beta\cdot2^{10}$: $\alpha = 0$, $\beta = 19$, so $\alpha + \beta = 19$.
The answer form α·3¹⁰ + β·2¹⁰ shows the setter meant the term 2ᵏ + 3k (the printed 2² is a misprint): then Σ2ᵏ·¹⁰Cₖ = 3¹⁰ and Σ3k·¹⁰Cₖ = 15·2¹⁰, so α = 1, β = 15 and α + β = 16.
Answer: 16 (19 if the printed 2² is taken literally). | ||||
| 90 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Putting x = y = 0: $f(0) = f(0)^2$, and $f(0) \ne 0$, so f(0) = 1.
$\lim_{h\to0}\frac{f(h) - 1}{h} = \lim_{h\to0}\frac{f(h) - f(0)}{h} = f'(0) = 3$.
Answer: 3. | ||||
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