JEE (Main) · 2021
JEE (Main) 2021 — 24 February, 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
- Oscillations and Waves4
- Electronic Devices3
- Motion of System of Particles and Rigid Body2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- Amines3
- Some Basic Concepts of Chemistry3
- Chemical Bonding and Molecular Structure2
21 easy · 9 medium · 0 hard
Mathematics
30 Q
- Three Dimensional Geometry3
- Applications of Derivatives3
- Integral Calculus3
10 easy · 18 medium · 2 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $g = \frac{4\pi^2 L}{T^2}$, so $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$.
$\frac{\Delta L}{L} = \frac{1\text{ mm}}{1000\text{ mm}} = 0.1\%$ and $2\frac{\Delta T}{T} = 2\times\frac{0.01}{1.95} = 1.03\%$.
$\frac{\Delta g}{g} = 0.1\% + 1.03\% = 1.13\%$.
Answer: (C). | ||||
| 2 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Write $a = v\frac{dv}{dx}$: $mv\,dv = -\alpha x^2\,dx$.
Integrate from $v_0$ at $x = 0$ to $v = 0$ at $x = d$: $-\frac{1}{2}mv_0^2 = -\frac{\alpha d^3}{3}$, so $d = \left(\frac{3mv_0^2}{2\alpha}\right)^{1/3}$.
The options drop the mass (equivalently, take $m = 1$, or read $\alpha$ as force constant per unit mass): $d = \left(\frac{3v_0^2}{2\alpha}\right)^{1/3}$.
Answer: (C). | ||||
| 3 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion From the figure, the disc's centre is at $x = a$ from O and the hole (radius $\frac{a}{2}$) touches the disc's rim on the right, so its centre is at $x = \frac{3a}{2}$.
Take mass proportional to area: disc $\pi a^2$, hole $\frac{\pi a^2}{4}$.
$x_{cm} = \frac{\pi a^2\cdot a - \frac{\pi a^2}{4}\cdot\frac{3a}{2}}{\pi a^2 - \frac{\pi a^2}{4}} = \frac{a - \frac{3a}{8}}{\frac{3}{4}} = \frac{5a}{8}\cdot\frac{4}{3} = \frac{5a}{6}$.
Answer: (B). | ||||
| 4 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At the pole there is no rotational effect: $mg = 49$ N, so $m = 5$ kg.
At the equator the balance reads $m(g - \omega^2 R)$ with $\omega = \frac{2\pi}{86400} = 7.27\times10^{-5}$ rad s⁻¹.
$\omega^2 R = (7.27\times10^{-5})^2 \times 6.4\times10^6 = 0.0338$ m s⁻².
Reading $= 49 - 5\times0.0338 = 49 - 0.17 = 48.83$ N.
Answer: (C). | ||||
| 5 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Let $T$ be the temperature of the isotherm AB, so $P_1V_1 = RT$.
A→B (isothermal, volume $V_1 \to 2V_1$): $W_{AB} = RT\ln\frac{2V_1}{V_1} = RT\ln 2$.
B→C (constant volume): $W_{BC} = 0$.
C→A (adiabatic): $W_{CA} = \frac{P_CV_C - P_AV_A}{\gamma - 1} = \frac{\frac{P_1}{4}\cdot 2V_1 - P_1V_1}{\gamma - 1} = -\frac{P_1V_1}{2(\gamma - 1)} = -\frac{RT}{2(\gamma - 1)}$.
Net work $= RT\left(\ln 2 - \frac{1}{2(\gamma - 1)}\right)$.
Answer: (A). | ||||
| 6 | 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 In kinetic theory the molecules collide elastically with the walls; each collision reverses the normal component of momentum, and the rate of this momentum change per unit area is the pressure. They neither lose energy, stick, nor are attracted by the walls.
Answer: (D). | ||||
| 7 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Gravity along the incline, $Mg\sin\alpha$, is constant: it only shifts the equilibrium position.
Displaced by $x$ along the incline, one spring is stretched and the other compressed by $x$, so the restoring force is $2kx$: the springs act in parallel, $k_{eff} = 2k$.
$f = \frac{1}{2\pi}\sqrt{\frac{2k}{M}}$.
Answer: (C). | ||||
| 8 | 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 In SHM $v = \omega\sqrt{A^2 - x^2}$, so $\frac{x^2}{A^2} + \frac{v^2}{\omega^2A^2} = 1$: an ellipse in the $x$–$v$ plane (a circle only in the special case $\omega = 1$ in the chosen units).
Answer: (B). | ||||
| 9 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves A travelling wave is a function of $(x \pm vt)$ only.
(A) separates into a function of $x$ times a function of $t$: no travel. (B) is not of the form $f(x - vt)$. (D) is a standing wave.
(C) $A\sin(15x - 2t) = A\sin 15\left(x - \frac{2}{15}t\right)$ travels along $+x$ with speed $\frac{2}{15}$.
Answer: (C). | ||||
| 10 | 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 Each electron attracts the proton with $F = \frac{q^2}{4\pi\varepsilon_0(d^2 + x^2)}$. The components along the line cancel; the components back toward the midpoint add:
$F_{net} = 2\cdot\frac{q^2}{4\pi\varepsilon_0(d^2 + x^2)}\cdot\frac{x}{\sqrt{d^2 + x^2}} \approx \frac{q^2}{2\pi\varepsilon_0 d^3}\,x$ for $x \ll d$.
So $m\omega^2 = \frac{q^2}{2\pi\varepsilon_0 d^3}$ and $\omega = \left(\frac{q^2}{2\pi\varepsilon_0 md^3}\right)^{1/2}$.
Answer: (A). | ||||
| 11 | 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 Just after the switch closes, an inductor carries no current (it acts as an open circuit), so both branches containing L carry nothing.
The current flows only through the series R and the plain R branch: $R_{total} = 2 + 2 = 4\ \Omega$.
$i = \frac{9}{4} = 2.25$ A.
Answer: (A). | ||||
| 12 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter In an applied field, domains already aligned with the field grow at the expense of the others, which shrink, and the domains' magnetisation rotates toward the field. So domains may grow or shrink and change orientation.
Answer: (C). | ||||
| 13 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Microwaves are produced by special vacuum tubes such as the magnetron → (ii).
Infrared comes from the vibration of atoms and molecules (hot bodies) → (iv).
Gamma rays come from radioactive decay of nuclei → (i).
X-rays come from transitions of inner-shell electrons (and deceleration of fast electrons) → (iii).
(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).
Answer: (C). | ||||
| 14 | 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 $\beta = \frac{\lambda D}{d}$. Violet light has a shorter wavelength than red, so $\beta$ decreases: consecutive fringes come closer. The central fringe stays bright (zero path difference) and the minima stay dark for equal slits.
Answer: (B). | ||||
| 15 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The shortest wavelength is when an electron's whole energy $eV$ becomes one photon: $\lambda_{min} = \frac{hc}{eV}$.
With $hc = 1240$ eV nm: $\lambda_{min} = \frac{1240}{1.24\times10^6}$ nm $= 10^{-3}$ nm.
Answer: (C). | ||||
| 16 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{mv}$; equal $\lambda$ means equal momentum, so $\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} = \frac{4}{1}$.
Answer: (A). | ||||
| 17 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $h\nu = 13.6Z^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ eV.
2→1: $\frac{3}{4} = 0.75$; 3→2: $\frac{5}{36} \approx 0.139$; 4→3: $\frac{7}{144} \approx 0.049$; 5→4: $\frac{9}{400} \approx 0.023$.
The largest energy gap, hence the highest frequency, is 2→1.
Answer: (A). | ||||
| 18 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Zener breakdown needs a very strong field across the junction, which tears electrons from covalent bonds. Heavy doping on both sides gives a very narrow depletion layer, so even a few volts produce a field of order $10^6$ V m⁻¹ or more.
Answer: (C). | ||||
| 19 | 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 Statement I: two separate diodes joined back to back do not make a transistor. A transistor needs a very thin, lightly doped base shared by both junctions so that carriers injected from the emitter cross the base to the collector; two discrete diodes have no such common thin base. False.
Statement II: the current gain $\beta = \frac{I_C}{I_B}$. True.
Answer: (D). | ||||
| 20 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The circuit is an OR gate with inputs $A$ and $\bar{B}$, followed by a NOT gate: $C = \overline{A + \bar{B}}$.
By De Morgan, $C = \bar{A}\cdot B$.
That is an AND gate fed with $\bar{A}$ and $B$: option (C). (Option (B) has a bubble at the output, giving $\overline{\bar{A}B}$; option (A) gives $\overline{A\bar{B}}$; option (D) is NOR, $\overline{A + B}$.)
Answer: (C). | ||||
| 21 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $K = \frac{p^2}{2m}$; with equal $p$, $\frac{K_A}{K_B} = \frac{m_B}{m_A} = \frac{2}{1}$.
Answer: 2. | ||||
| 22 | 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 $\Delta L = \frac{FL}{AY}$ with $A = \frac{\pi d^2}{4}$. Doubling $L$ doubles $\Delta L$; doubling $d$ makes $A$ four times, quartering $\Delta L$.
New elongation $= 0.04\times\frac{2}{4} = 0.02$ m $= 2$ cm.
Answer: 2. | ||||
| 23 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Each side: mass 1 kg, length $l = 0.4$ m. Its midpoint is at the apothem distance $\frac{\sqrt{3}}{2}l = 0.2\sqrt{3}$ m from the centre.
For one side, by the parallel-axis theorem: $\frac{ml^2}{12} + m\left(\frac{\sqrt{3}}{2}l\right)^2 = \frac{0.16}{12} + 0.12 = 0.1\overline{3}$ kg m².
Six sides: $I = 6\times0.1\overline{3} = 0.8$ kg m² $= 8\times10^{-1}$ kg m².
Answer: 8. | ||||
| 24 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | NumericalEasy | |
A easy numerical question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$ depends only on temperature; the pressure does not enter.
$v_2 = 200\sqrt{\frac{400}{300}} = 200\cdot\frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}}$ m s⁻¹.
Answer: 400. | ||||
| 25 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Speed of each car: $7.2$ km/h $= 2$ m s⁻¹.
Each driver hears his own horn at 676 Hz and the other's horn (source and observer approaching) at
$f' = 676\cdot\frac{340 + 2}{340 - 2} = 676\cdot\frac{342}{338} = 2\times342 = 684$ Hz.
Beat frequency $= 684 - 676 = 8$ Hz.
Answer: 8. | ||||
| 26 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The charge is at height $\frac{a}{2}$ above the centre of a square of side $a$: it is at the centre of a cube of which the square is one face. By symmetry and Gauss's law each face carries one sixth of the total flux.
$\phi = \frac{q}{6\varepsilon_0} = \frac{12\times10^{-6}}{6\times8.85\times10^{-12}} = 2.26\times10^5$ N m² C⁻¹ $= 226\times10^3$ N m²/C.
Answer: 226. | ||||
| 27 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $J = \sigma E = 5\times10^7\times10^{-2} = 5\times10^5$ A m⁻².
$I = J\cdot\pi r^2 = 5\times10^5\times\pi(0.5\times10^{-3})^2 = 0.125\pi$ A $= 125\pi$ mA $= 5^3\pi$ mA.
Answer: 5. | ||||
| 28 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At resonance the impedance is $R$ and the power factor 1: $P = \frac{V^2}{R}$.
$R = \frac{120^2}{16} = \frac{14400}{16} = 900\ \Omega$.
Answer: 900. | ||||
| 29 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalEasy | |
A easy numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves In vacuum $\lambda_0 = \frac{c}{f} = \frac{3\times10^8}{3\times10^9} = 0.1$ m $= 10$ cm.
Refractive index (non-magnetic medium) $n = \sqrt{\varepsilon_r} = 1.5$; the frequency is unchanged, so $\lambda = \frac{10}{1.5} = 6.667$ cm $= 666.7\times10^{-2}$ cm.
Rounded to the nearest integer: 667.
Answer: 666.67 (667 to the nearest integer). | ||||
| 30 | Communication SystemsAttenuation in a transmission medium | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on attenuation in a transmission medium, from Communication Systems. NCERT: No matching NCERT chapter Total gain over 20 km $= -5\times20 = -100$ dB.
$-100 = 10\log_{10}\frac{P_o}{P_i}$, so $\frac{P_o}{P_i} = 10^{-10}$.
$P_o = 100\text{ W}\times10^{-10} = 10^{-8}$ W, so $x = 8$.
Answer: 8. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctMedium | |
A medium single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In the Bohr model $r \propto \frac{n^2}{Z}$ and $v \propto \frac{Z}{n}$.
(A) $KE = \frac{1}{2}mv^2 \propto \frac{Z^2}{n^2}$. True.
(B) $vn \propto Z$, not $Z^2$. False.
(C) Frequency $= \frac{v}{2\pi r} \propto \frac{Z/n}{n^2/Z} = \frac{Z^2}{n^3}$, not $\frac{Z^3}{n^3}$. False.
(D) $F = \frac{kZe^2}{r^2} \propto \frac{Z}{(n^2/Z)^2} = \frac{Z^3}{n^4}$. True.
So (A) and (D) only.
Answer: (B). | ||||
| 32 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure The central I in $\mathrm{I_3^-}$ has 2 bond pairs and 3 lone pairs: 5 electron pairs, trigonal bipyramidal arrangement. The three lone pairs take the equatorial positions, so the two bonded I atoms are axial: the ion is linear with an I−I−I angle of 180°.
Answer: (A). | ||||
| 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 Blood is a negatively charged colloid. It is coagulated by cations, and by the Hardy–Schulze rule the higher the charge of the coagulating ion, the stronger its effect. $\mathrm{Fe^{3+}}$ from $\mathrm{FeCl_3}$ carries the highest charge among the cations offered ($\mathrm{Na^+}$, $\mathrm{Mg^{2+}}$, $\mathrm{Fe^{2+}}$), so $\mathrm{FeCl_3}$ clots blood most effectively.
Answer: (C). | ||||
| 34 | Chemical Bonding and Molecular StructureIonic bonding and lattice enthalpy | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on ionic bonding and lattice enthalpy, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure LiF vs LiCl: the larger $\mathrm{Cl^-}$ is more polarised by the small $\mathrm{Li^+}$ (Fajans' rules), so LiCl is more covalent and melts lower (LiF about 845 °C, LiCl about 605 °C): LiF > LiCl.
MgO vs NaCl: MgO has doubly charged ions of small size and a far larger lattice enthalpy (MgO about 2850 °C, NaCl about 801 °C): MgO > NaCl.
Answer: (B). | ||||
| 35 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Kaolinite, $\mathrm{Al_2(OH)_4Si_2O_5}$, is an aluminium ore → (a)-(iii).
Siderite, $\mathrm{FeCO_3}$, is an iron ore → (b)-(i).
Malachite, $\mathrm{CuCO_3\cdot Cu(OH)_2}$, is a copper ore → (c)-(iv).
Calamine, $\mathrm{ZnCO_3}$, is a zinc ore → (d)-(ii).
Answer: (D). | ||||
| 36 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter A: Hydrogen makes up most of the mass of the Universe (stars, interstellar gas), yet the troposphere is mostly $\mathrm{N_2}$ and $\mathrm{O_2}$ with only traces of $\mathrm{H_2}$. True.
R: Hydrogen is the lightest element. True.
Being the lightest, $\mathrm{H_2}$ molecules move fastest at a given temperature and escape the Earth's gravitational pull, which is why so little remains in the atmosphere. R explains A.
Answer: (A). | ||||
| 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 Characteristic flame lines: Li crimson red 670.8 nm → (a)-(ii); Na yellow 589.2 nm → (b)-(iv); Rb red-violet 780.0 nm → (c)-(iii); Cs blue 455.5 nm → (d)-(i).
Answer: (B). | ||||
| 38 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements (A) V is +4 in $\mathrm{VOSO_4}$ and is readily oxidised to +5: a reducing agent. Correct.
(B) Ru is +8 in $\mathrm{RuO_4}$: a strong oxidising agent. Correct.
(C) $\mathrm{Cr_2O_3}$ dissolves in both acids and alkalis: amphoteric. Correct.
(D) Ruby is $\mathrm{Al_2O_3}$ with a little $\mathrm{Cr^{3+}}$; its red colour is due to $\mathrm{Cr^{3+}}$, not $\mathrm{Co^{3+}}$. Incorrect.
Answer: (D). | ||||
| 39 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Densities (g cm⁻³): Zn 7.14, Cr 7.19, Fe 7.87, Co 8.90, Cu 8.96. Zinc, with its full d-subshell and weak metallic bonding, is the least dense.
Zn < Cr < Fe < Co < Cu.
Answer: (B). | ||||
| 40 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[FeCl_4]^{2-}}$: $\mathrm{Fe^{2+}}$, $d^6$, tetrahedral with weak-field $\mathrm{Cl^-}$, high spin: 4 unpaired, $\mu = \sqrt{4\times6} = 4.90$ BM.
$\mathrm{[Co(C_2O_4)_3]^{3-}}$: $\mathrm{Co^{3+}}$, $d^6$, octahedral; oxalate pairs the electrons of $\mathrm{Co^{3+}}$ ($t_{2g}^6$): 0 unpaired, 0 BM.
$\mathrm{MnO_4^{2-}}$: Mn +6, $d^1$: 1 unpaired, $\mu = \sqrt{3} = 1.73$ BM.
Answer: (A). | ||||
| 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 Statement I: BOD measures the oxygen that microorganisms need to break down the organic matter in water. A high BOD depletes dissolved oxygen and kills aquatic life, so its value matters for their survival. True.
Statement II: Clean water has a BOD below 5 ppm; there is no 'optimum' BOD of 6.5 ppm (polluted water may exceed 17 ppm). The figure of about 6.5 ppm belongs to the dissolved oxygen needed by aquatic life, not to BOD. False.
Answer: (C). | ||||
| 42 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Only a primary aromatic amine gives a stable arenediazonium salt, which couples with alkaline $\beta$-naphthol to give an orange-red azo dye.
(A) aniline: primary aromatic amine. Yes.
(B) benzylamine: the $\mathrm{NH_2}$ is on an alkyl carbon; its diazonium salt is unstable and gives the alcohol and $\mathrm{N_2}$, no dye.
(C) N-methylaniline: secondary amine; gives an N-nitrosamine.
(D) N,N-dimethylaniline: tertiary amine; it does not form a diazonium salt.
Answer: (A). | ||||
| 43 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Hückel's rule: a cyclic, planar, fully conjugated ring with $(4n+2)$ π electrons is aromatic.
(A) Furan: 4 π electrons from the two C=C plus one lone pair of O, 6 in all, fully conjugated. Aromatic.
(B) Cyclobutadiene dication: 2 π electrons ($n = 0$) over a fully conjugated four-membered ring. Aromatic.
(C) Cycloheptatriene: the ring contains an $sp^3$ $\mathrm{CH_2}$, which breaks the cyclic conjugation. Non-aromatic.
(D) Anthracene: 14 π electrons ($n = 3$), planar and fully conjugated. Aromatic.
Answer: (C). | ||||
| 44 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Acid-catalysed hydration ($\mathrm{Hg^{2+}}$) follows Markovnikov's rule: the OH goes to the more substituted carbon of the triple bond and the enol tautomerises to a carbonyl. Only ethyne gives an aldehyde; every other alkyne gives a ketone.
(A) $\mathrm{CH_3CHO}$ from ethyne. Possible.
(B) $\mathrm{CH_3COCH_2CH_3}$ from but-1-yne or but-2-yne. Possible.
(C) Cyclohexyl methyl ketone from ethynylcyclohexane. Possible.
(D) $\mathrm{CH_3CH_2CHO}$, propanal: the only C₃ alkyne, propyne, gives acetone, not propanal. Not possible.
Answer: (D). | ||||
| 45 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (i) chlorobenzene, (ii) 4-nitrochlorobenzene, (iii) 2,4-dinitrochlorobenzene, (iv) 2,4,6-trinitrochlorobenzene.
Nucleophilic substitution on an aryl halide goes through a negatively charged (Meisenheimer) intermediate. Each nitro group at an ortho or para position delocalises that negative charge and speeds the reaction, so the more o/p nitro groups, the more reactive.
(i) < (ii) < (iii) < (iv).
Answer: (C). | ||||
| 46 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The reaction turns the ring C=O of the enone (carvone) into $\mathrm{CH_2}$, leaving both C=C double bonds untouched.
Wolff–Kishner reduction (hydrazine, then a strong base such as sodium ethoxide on heating) does exactly this and does not attack C=C.
$\mathrm{NaBH_4}$ gives the alcohol, not $\mathrm{CH_2}$; $\mathrm{Ni/H_2}$ would also hydrogenate the C=C bonds; red P + $\mathrm{Cl_2}$ is for α-halogenation of acids.
Answer: (B). | ||||
| 47 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines The nitro group is a meta director, so brominate first: $\mathrm{Br_2/Fe}$ gives m-bromonitrobenzene.
$\mathrm{Sn/HCl}$ reduces $\mathrm{NO_2}$ to $\mathrm{NH_2}$: m-bromoaniline.
$\mathrm{NaNO_2/HCl}$ at 0–5 °C gives the diazonium salt.
$\mathrm{CuBr/HBr}$ (Sandmeyer) replaces $\mathrm{N_2^+}$ by Br: m-dibromobenzene.
Answer: (D). | ||||
| 48 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Buna-S is a copolymer of 1,3-butadiene (Bu) and styrene (S), polymerised with sodium (Na) as catalyst in the original process.
Answer: (C). | ||||
| 49 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Valium (diazepam) is a tranquilizer → (iv). Morphine is a narcotic analgesic → (iii). Norethindrone is a synthetic progestogen used as an antifertility drug → (i). Lack of vitamin $\mathrm{B_{12}}$ causes pernicious anaemia → (ii).
(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii).
Answer: (C). | ||||
| 50 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (a) Acid chloride to aldehyde: Rosenmund reduction, $\mathrm{H_2/Pd{-}BaSO_4}$ → (ii).
(b) α-Chlorination of a carboxylic acid: Hell–Volhard–Zelinsky, $\mathrm{Cl_2}$/red P then $\mathrm{H_2O}$ → (iv).
(c) Amide to amine with one carbon fewer: Hofmann bromamide degradation, $\mathrm{Br_2/NaOH}$ → (i).
(d) C=O to $\mathrm{CH_2}$ in acid: Clemmensen reduction, Zn(Hg)/conc. HCl → (iii).
(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).
Answer: (D). | ||||
| 51 | 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 $\mathrm{C_xH_y} + \left(x + \frac{y}{4}\right)\mathrm{O_2} \rightarrow x\,\mathrm{CO_2} + \frac{y}{2}\mathrm{H_2O}$.
By volume (Avogadro): $x = 4$ and $x + \frac{y}{4} = 6$, so $y = 8$ ($\mathrm{C_4H_8}$).
Answer: 8. | ||||
| 52 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $n = \frac{4.75}{26} = 0.1827$ mol; $T = 323$ K; $P = \frac{740}{760} = 0.9737$ atm.
$V = \frac{nRT}{P} = \frac{0.1827\times0.0826\times323}{0.9737} = 5.01$ L.
Answer: 5. | ||||
| 53 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta_rG^\circ = \Delta_fG^\circ(\mathrm{C_6H_6}) - 3\Delta_fG^\circ(\mathrm{C_2H_2}) = -1.24\times10^5 + 6.12\times10^5 = 4.88\times10^5$ J mol⁻¹.
$\log K = -\frac{\Delta_rG^\circ}{2.303RT} = -\frac{4.88\times10^5}{2.303\times8.314\times298} = -\frac{4.88\times10^5}{5706} = -85.5$.
$|\log K| = 85.5 = 855\times10^{-1}$.
Answer: 855. | ||||
| 54 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Molality of butane (M = 58): $m = \frac{10/58}{0.200} = 0.862$ mol kg⁻¹.
$\Delta T_f = K_f m = 5.12\times0.862 = 4.41$ °C.
Freezing point $= 5.5 - 4.41 = 1.09$ °C, which rounds to 1.
Answer: 1.09 (1 to the nearest integer). | ||||
| 55 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium With solubility $s$: $[\mathrm{Pb^{2+}}] = 0.1 + s \approx 0.1$ and $[\mathrm{I^-}] = 2s$.
$K_{sp} = 0.1\times(2s)^2 = 0.4s^2 = 8.0\times10^{-9}$, so $s^2 = 2\times10^{-8}$ and $s = 1.41\times10^{-4} = 141\times10^{-6}$ mol L⁻¹.
Answer: 141. | ||||
| 56 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry $\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$.
$E = E^\circ - \frac{0.059}{5}\log\frac{[\mathrm{Mn^{2+}}]}{[\mathrm{MnO_4^-}][\mathrm{H^+}]^8}$.
With the Mn concentrations unchanged, $\Delta E = \frac{0.059}{5}\times8\times\Delta\log[\mathrm{H^+}] = \frac{0.059\times8}{5}\times4 = 0.3776$ V $= 3776\times10^{-4}$ V.
Answer: 3776. | ||||
| 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 $k = \frac{0.693}{3.33} = 0.2081$ h⁻¹.
$\ln\frac{1}{f} = kt = 0.2081\times9 = 1.873$, so $\log_{10}\frac{1}{f} = \frac{1.873}{2.303} = 0.813 = 81.3\times10^{-2}$.
Answer: 81. | ||||
| 58 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter α- (rhombic) and β- (monoclinic) sulphur are both made of puckered $\mathrm{S_8}$ rings with all electrons paired: diamagnetic.
$\mathrm{S_2}$ (in sulphur vapour at high temperature) has, like $\mathrm{O_2}$, two unpaired electrons in antibonding π* orbitals: paramagnetic.
Number = 1.
Answer: 1. | ||||
| 59 | AminesClassification and identification of amines | Class 12: Amines | NumericalEasy | |
A easy numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Gabriel phthalimide synthesis makes primary amines by $S_N2$ alkylation of potassium phthalimide with an alkyl halide, followed by hydrolysis.
(A) $\mathrm{(CH_3)_2CH{-}CH_2{-}NH_2}$ from isobutyl halide: yes.
(B) $\mathrm{CH_3CH_2NH_2}$ from ethyl halide: yes.
(C) $\mathrm{C_6H_5CH_2NH_2}$ from benzyl halide: yes.
(D) Aniline: no, since aryl halides do not undergo $S_N2$ with the phthalimide anion.
Total = 3.
Answer: 3. | ||||
| 60 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Aniline $\mathrm{C_6H_5NH_2}$, M = 93: $\frac{1.86}{93} = 0.02$ mol, giving 0.02 mol acetanilide $\mathrm{C_6H_5NHCOCH_3}$, M = 135: 2.70 g.
After a 10% loss: $2.70\times0.9 = 2.43$ g $= 243\times10^{-2}$ g.
Answer: 243. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The midpoint $M = \left(\frac{a+20}{2}, \frac{6+b}{2}, -\frac{a}{2}\right)$ of P and its image lies on the line, so for some $t$:
$\frac{a+20}{2} = 3 + 7t$, $\frac{6+b}{2} = 2 + 5t$, $-\frac{a}{2} = 1 - 9t$,
i.e. $a + 14 = 14t$, $b + 2 = 10t$, $a + 2 = 18t$.
From the first and third: $12 = -4t$, so $t = -3$, $a = -56$, $b = -32$.
Check: $\overrightarrow{PQ} = (76, -38, 38)$ and $(76)(7) + (-38)(5) + (38)(-9) = 532 - 190 - 342 = 0$, so PQ is perpendicular to the line.
$|a + b| = 88$.
Answer: (C). | ||||
| 62 | 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 Centroid: $\frac{2a + 2}{3} = \frac{10}{3}$ gives $a = 4$; $\frac{c + 2b}{3} = \frac{7}{3}$ gives $c + 2b = 7$.
AP: $2b = a + c = 4 + c$, so $c = 2b - 4$ and $4b - 4 = 7$: $b = \frac{11}{4}$, $c = \frac{3}{2}$.
The equation is $4x^2 + \frac{11}{4}x + 1 = 0$: $\alpha + \beta = -\frac{11}{16}$, $\alpha\beta = \frac{1}{4}$.
$\alpha^2 + \beta^2 - \alpha\beta = (\alpha + \beta)^2 - 3\alpha\beta = \frac{121}{256} - \frac{3}{4} = \frac{121 - 192}{256} = -\frac{71}{256}$.
Answer: (B). | ||||
| 63 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The parabola lies entirely above the line ($x^2 + 4 - (4x - 1) = (x - 2)^2 + 1 > 0$). The closest point is where the tangent is parallel to the line: $\frac{dy}{dx} = 2x = 4$, so $x = 2$, $y = 8$.
(Equivalently, the vertical gap $(x - 2)^2 + 1$, and hence the distance, is least at $x = 2$.)
Answer: (B). | ||||
| 64 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The point lies on the line: $\frac{3\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = 2\sqrt{3}$.
(B) $x^2 + 9y^2 = 9$: $\frac{27}{4} + \frac{9}{4} = 9$, so the point is on it. Tangent $T = 0$: $\frac{3\sqrt{3}}{2}x + \frac{9}{2}y = 9$, i.e. $3\sqrt{3}x + 9y = 18$, i.e. $x + \sqrt{3}y = 2\sqrt{3}$. This is the given line.
(A) The tangent to the circle at the point is $\frac{3\sqrt{3}}{2}x + \frac{1}{2}y = 7$, a different line. (C) $\frac{1}{6\sqrt{3}}\cdot\frac{3\sqrt{3}}{2} = \frac{1}{4} = y^2$: the point is on it, but its tangent $\frac{y}{2} = \frac{1}{12\sqrt{3}}\left(x + \frac{3\sqrt{3}}{2}\right)$ has slope $\frac{1}{6\sqrt{3}} \ne -\frac{1}{\sqrt{3}}$. (D) $2\cdot\frac{27}{4} - 18\cdot\frac{1}{4} = 9$: on it, but the tangent $3\sqrt{3}x - 9y = 9$ has positive slope.
Answer: (B). | ||||
| 65 | 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 $x^2 - 2x - 2 = (x - 1)^2 - 3$. Put $t = x - 1$: $I = \int_0^2 [t^2 - 3]\,dt = \int_0^2 [t^2]\,dt - 6$.
$\int_0^2 [t^2]\,dt = 0\cdot1 + 1\cdot(\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3}) = 5 - \sqrt{2} - \sqrt{3}$.
$I = 5 - \sqrt{2} - \sqrt{3} - 6 = -\sqrt{2} - \sqrt{3} - 1$.
Answer: (C). | ||||
| 66 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $\sim p \wedge (p \vee q) = (\sim p \wedge p) \vee (\sim p \wedge q) = \sim p \wedge q$.
Its negation, by De Morgan: $\sim(\sim p \wedge q) = p \vee \sim q$.
Answer: (D). | ||||
| 67 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let $M = A^2B^2 - B^2A^2$. Since $A^T = A$ and $B^T = -B$, $(A^2)^T = A^2$ and $(B^2)^T = (B^T)^2 = B^2$.
$M^T = B^2A^2 - A^2B^2 = -M$: M is skew-symmetric of odd order 3, so $\det M = 0$.
A homogeneous system with a singular coefficient matrix has non-trivial solutions, hence infinitely many.
Answer: (D). | ||||
| 68 | 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 Integrating $f'(x) = f'(2 - x)$: $f(x) = -f(2 - x) + C$. At $x = 0$: $C = f(0) + f(2) = 1 + e^2$. So $f(x) + f(2 - x) = 1 + e^2$.
$I = \int_0^2 f(x)\,dx = \int_0^2 f(2 - x)\,dx$, so $2I = \int_0^2 (1 + e^2)\,dx = 2(1 + e^2)$ and $I = 1 + e^2$.
Answer: (B). | ||||
| 69 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $ff'' - (f')^2 = 0$. Since $f' \ne 0$ and $f$ is continuous with $f(0) = 1$, write $\frac{f''}{f'} = \frac{f'}{f}$: $\ln|f'| = \ln|f| + c$, so $f' = kf$.
At $x = 0$: $2 = k\cdot1$, so $f' = 2f$ and $f(x) = e^{2x}$.
$f(1) = e^2 \approx 7.39$, in (6, 9).
Answer: (C). | ||||
| 70 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The curve passes through the origin: $c = 0$. Slope at the origin: $y'(0) = b = 1$. Through (1, 2): $a + b + c = 2$, so $a = 1$.
Answer: (A). | ||||
| 71 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Hockey-stick identity: ${}^2C_2 + {}^3C_2 + \ldots + {}^nC_2 = {}^{n+1}C_3$.
Sum $= {}^{n+1}C_2 + 2\,{}^{n+1}C_3 = \frac{(n+1)n}{2} + \frac{(n+1)n(n-1)}{3} = \frac{n(n+1)}{6}\left[3 + 2(n - 1)\right] = \frac{n(n+1)(2n+1)}{6}$.
(It is $1^2 + 2^2 + \ldots + n^2$, since ${}^{k+1}C_2 + {}^kC_2 = k^2$.)
Answer: (B). | ||||
| 72 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}1&-2&0\\1&-1&k\\0&k&4\end{vmatrix} = 1(-4 - k^2) + 2(4) = 4 - k^2$.
$k \ne \pm2$: $\Delta \ne 0$, unique solution. (A) is true, so (B) and (C) are false.
$k = 2$: subtracting the first equation from the second gives $y + 2z = -3$, while the third gives $2y + 4z = 6$, i.e. $y + 2z = 3$. Inconsistent: no solution. (D) is true.
$k = -2$: the second minus the first gives $y - 2z = -3$ and the third gives $-2y + 4z = 6$, i.e. $y - 2z = -3$: consistent, infinitely many solutions. So (E), which claims infinitely many for $k \ne -2$, is false.
Correct: (A) and (D) only.
Answer: (A). | ||||
| 73 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives f is continuous: at $x = -5$ both pieces give 275, at $x = 4$ both give −400.
$x < -5$: $f' = -55 < 0$, decreasing.
$-5 < x < 4$: $f' = 6x^2 - 6x - 120 = 6(x - 5)(x + 4)$, positive for $-5 < x < -4$ and negative for $-4 < x < 4$.
$x > 4$: $f' = 6x^2 - 6x - 36 = 6(x - 3)(x + 2) > 0$.
So f increases on $(-5, -4) \cup (4, \infty)$.
Answer: (B). | ||||
| 74 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta = \sin^{-1}\frac{\sqrt{63}}{8}$, an acute angle with $\cos\theta = \frac{1}{8}$.
$\cos\frac{\theta}{2} = \sqrt{\frac{1 + \frac{1}{8}}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$ and $\sin\frac{\theta}{2} = \frac{\sqrt{7}}{4}$.
$\tan\frac{\theta}{4} = \frac{\sin(\theta/2)}{1 + \cos(\theta/2)} = \frac{\sqrt{7}/4}{7/4} = \frac{1}{\sqrt{7}}$.
Answer: (A). | ||||
| 75 | 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 Speed $432$ km/h $= 120$ m/s, so the plane flies $120\times20 = 2400$ m.
With height $h$, its horizontal distances from A are $h\cot60^\circ$ and then $h\cot30^\circ$ (flying away from A):
$h(\cot30^\circ - \cot60^\circ) = h\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}} = 2400$, so $h = 1200\sqrt{3}$ m.
Answer: (D). | ||||
| 76 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Planes through the intersection: $(x + y + z - 1) + \lambda(x - 2y + 2) = 0$.
Through (1, 0, 2): $(1 + 0 + 2 - 1) + \lambda(1 - 0 + 2) = 2 + 3\lambda = 0$, so $\lambda = -\frac{2}{3}$.
$3(x + y + z - 1) - 2(x - 2y + 2) = 0$ gives $x + 7y + 3z = 7$, i.e. $\vec{r}\cdot(\hat{i} + 7\hat{j} + 3\hat{k}) = 7$.
Answer: (B). | ||||
| 77 | 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 (a) For the implication to fail we need $\sim p$ false (p true) with $\sim q \wedge (p \rightarrow q)$ true, i.e. q false and $p \rightarrow q$ true; but p true and q false makes $p \rightarrow q$ false. It never fails: a tautology (modus tollens).
(b) For it to fail we need q false with $(p \vee q) \wedge \sim p$ true, i.e. p false and $p \vee q$ true; but p and q both false makes $p \vee q$ false. It never fails: a tautology (disjunctive syllogism).
Answer: (C). | ||||
| 78 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $x\frac{dy}{dx} + y = \frac{d}{dx}(xy) = bx^4$, so $xy = \frac{bx^5}{5} + C$ and $y = \frac{bx^4}{5} + \frac{C}{x}$.
Through (1, 2): $\frac{b}{5} + C = 2$.
$\int_1^2 f(x)\,dx = \frac{b}{25}(32 - 1) + C\ln2 = \frac{31b}{25} + \left(2 - \frac{b}{5}\right)\ln2$.
Setting this equal to $\frac{62}{5}$: $\left(\frac{31b}{25} - \frac{62}{5}\right) = \left(\frac{b}{5} - 2\right)\ln 2$, i.e. $\frac{31}{25}(b - 10) = \frac{1}{5}(b - 10)\ln2$. Since $\frac{31}{25} \ne \frac{\ln 2}{5}$, $b = 10$ (and then $C = 0$, $f(x) = 2x^4$).
Answer: (A). | ||||
| 79 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each subset is chosen independently from the $2^5$ subsets: $4^5 = 2^{10}$ equally likely ordered pairs (A, B).
Each element is in both, only A, only B, or neither. Exactly two in both: choose them in ${}^5C_2 = 10$ ways; each of the other 3 has 3 choices: $3^3 = 27$. Favourable $= 270$.
$P = \frac{270}{2^{10}} = \frac{135}{2^9}$.
Answer: (C). | ||||
| 80 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The curves meet where $5x^2 = 2x^2 + 9$: $x = \pm\sqrt{3}$.
Area $= \int_{-\sqrt{3}}^{\sqrt{3}} (9 - 3x^2)\,dx = 2\left[9x - x^3\right]_0^{\sqrt{3}} = 2(9\sqrt{3} - 3\sqrt{3}) = 12\sqrt{3}$.
Answer: (D). | ||||
| 81 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Nine 1s and one $k$. $\sigma^2 = \frac{9 + k^2}{10} - \left(\frac{9 + k}{10}\right)^2 = \frac{90 + 10k^2 - 81 - 18k - k^2}{100} = \frac{9(k - 1)^2}{100}$.
$\frac{9(k - 1)^2}{100} < 10$ gives $(k - 1)^2 < 111.1$, so $k - 1 \le 10$ and $k \le 11$.
Answer: 11. | ||||
| 82 | CirclesStandard and general form | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections $(x - 5)^2 + y^2 = 9\left[(x + 5)^2 + y^2\right]$
$x^2 - 10x + 25 + y^2 = 9x^2 + 90x + 225 + 9y^2$
$8x^2 + 8y^2 + 100x + 200 = 0$, i.e. $x^2 + y^2 + \frac{25}{2}x + 25 = 0$.
Centre $\left(-\frac{25}{4}, 0\right)$, $r^2 = \frac{625}{16} - 25 = \frac{225}{16}$, $r = \frac{15}{4}$.
$4r^2 = \frac{225}{4} = 56.25$.
Answer: 56.25 (56 to the nearest integer). | ||||
| 83 | CirclesRadius and centre | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Centre (2, 3). The normal at (5, 7) passes through the centre: slope $\frac{4}{3}$, $y - 3 = \frac{4}{3}(x - 2)$, meeting the x-axis at $x = -\frac{1}{4}$.
The tangent is perpendicular: slope $-\frac{3}{4}$, $y - 7 = -\frac{3}{4}(x - 5)$, meeting the x-axis at $x = 5 + \frac{28}{3} = \frac{43}{3}$.
The triangle has base on the x-axis from $-\frac{1}{4}$ to $\frac{43}{3}$, length $\frac{175}{12}$, and height 7 (the y-coordinate of (5, 7)).
$A = \frac{1}{2}\cdot\frac{175}{12}\cdot7 = \frac{1225}{24}$, so $24A = 1225$.
Answer: 1225. | ||||
| 84 | 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 $-1 + i\sqrt{3} = 2e^{i2\pi/3}$, so its 21st power is $2^{21}e^{i14\pi} = 2^{21}$. $(1 - i)^2 = -2i$, so $(1 - i)^{24} = (-2i)^{12} = 2^{12}$. First term: $2^9 = 512$.
$1 + i\sqrt{3} = 2e^{i\pi/3}$, so its 21st power is $2^{21}e^{i7\pi} = -2^{21}$. $(1 + i)^{24} = (2i)^{12} = 2^{12}$. Second term: $-512$.
$k = 0$, so $n = 0$ and the sum runs over $j = 0$ to 5:
$\sum_{j=0}^{5}(j + 5)(j + 4) = 5\cdot4 + 6\cdot5 + 7\cdot6 + 8\cdot7 + 9\cdot8 + 10\cdot9 = 20 + 30 + 42 + 56 + 72 + 90 = 310$.
Answer: 310. | ||||
| 85 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalHard | |
A hard numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The groups are named, so choose C's members, then split the rest between A and B with neither empty ($2^m - 2$ ways for $m$ students).
$|C| = 1$: ${}^{10}C_1(2^9 - 2) = 10\times510 = 5100$.
$|C| = 2$: ${}^{10}C_2(2^8 - 2) = 45\times254 = 11430$.
$|C| = 3$: ${}^{10}C_3(2^7 - 2) = 120\times126 = 15120$.
Total $= 5100 + 11430 + 15120 = 31650$.
Answer: 31650. | ||||
| 86 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem By Vandermonde's identity the first sum is $\binom{25}{k}$ and the second is $\binom{25}{k+1}$; together $\binom{25}{k} + \binom{25}{k+1} = \binom{26}{k+1}$.
With the given convention the sum is defined for every integer $k$, so 'exists' must be read as 'is maximum': $\binom{26}{k+1}$ is greatest at the middle, $k + 1 = 13$.
So $k = 12$.
Answer: 12. | ||||
| 87 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Replace $x$ by $\frac{1}{x}$: $af\left(\frac{1}{x}\right) + \alpha f(x) = \frac{b}{x} + \beta x$.
Add the two equations: $(a + \alpha)\left[f(x) + f\left(\frac{1}{x}\right)\right] = (b + \beta)\left(x + \frac{1}{x}\right)$.
So the expression $= \frac{b + \beta}{a + \alpha} = \frac{2}{1} = 2$.
Answer: 2. | ||||
| 88 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry First line: $\frac{x - \lambda}{1} = \frac{y - \frac{1}{2}}{\frac{1}{2}} = \frac{z}{-\frac{1}{2}}$: through $(\lambda, \frac{1}{2}, 0)$, direction $(2, 1, -1)$.
Second line: $\frac{x}{1} = \frac{y + 2\lambda}{1} = \frac{z - \lambda}{1}$: through $(0, -2\lambda, \lambda)$, direction $(1, 1, 1)$.
$(2, 1, -1)\times(1, 1, 1) = (2, -3, 1)$, of length $\sqrt{14}$.
Joining vector $(-\lambda, -2\lambda - \frac{1}{2}, \lambda)$; its dot product with $(2, -3, 1)$ is $-2\lambda + 6\lambda + \frac{3}{2} + \lambda = 5\lambda + \frac{3}{2}$.
$\frac{|5\lambda + \frac{3}{2}|}{\sqrt{14}} = \frac{\sqrt{7}}{2\sqrt{2}}$ gives $|5\lambda + \frac{3}{2}| = \frac{7}{2}$: $\lambda = \frac{2}{5}$ or $\lambda = -1$.
$\lambda$ is an integer, so $\lambda = -1$ and $|\lambda| = 1$.
Answer: 1. | ||||
| 89 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Terms $a, ar, ar^2, ar^3$. Product of the first three: $(ar)^3 = 1$, so $ar = 1$.
Sum $= a(1 + r + r^2 + r^3)$; sum of reciprocals $= \frac{1 + r + r^2 + r^3}{ar^3}$. Their ratio is $a^2r^3 = \frac{65/12}{65/18} = \frac{3}{2}$.
$a^2r^3 = (ar)^2r = r$, so $r = \frac{3}{2}$ and $a = \frac{2}{3}$. Check: $\frac{2}{3} + 1 + \frac{3}{2} + \frac{9}{4} = \frac{65}{12}$.
$\alpha = ar^2 = \frac{2}{3}\cdot\frac{9}{4} = \frac{3}{2}$, so $2\alpha = 3$.
Answer: 3. | ||||
| 90 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $x \ge 5$: $(x + 1)^2 \ge 36 > \frac{27}{4}$, so no root.
$x < 5$: $(x + 1)^2 + 5 - x = \frac{27}{4}$ gives $x^2 + x - \frac{3}{4} = 0$, so $x = \frac{-1 \pm 2}{2}$: $x = \frac{1}{2}$ or $x = -\frac{3}{2}$, both less than 5.
Number of real roots $= 2$.
Answer: 2. | ||||
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