JEE (Main) · 2021
JEE (Main) 2021 — 3 August, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Magnetic Effects of Current and Magnetism4
- Optics4
- Kinematics3
24 easy · 5 medium · 1 hard
Chemistry
30 Q
- d- and f-Block Elements3
- Amines3
- Some Basic Concepts of Chemistry2
12 easy · 15 medium · 3 hard
Mathematics
30 Q
- Matrices and Determinants4
- Limits, Continuity and Differentiability3
- Three Dimensional Geometry3
10 easy · 20 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter A positive zero error means the gauge reads more than the true value: observed reading = true value + zero error $= 2.675 + 0.02 = 2.695$ mm.
Main scale reading: the largest multiple of 0.5 mm below 2.695 mm is 2.5 mm.
Circular scale part: $2.695 - 2.5 = 0.195$ mm, which is $\frac{0.195}{0.005} = 39$ divisions.
Answer: (C) 39. | ||||
| 2 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | StatementsEasy | |
A easy statements question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $v = \frac{dx}{dt} = 1 - \cos t$ and $a = \frac{dv}{dt} = \sin t$.
A: $1 - \cos t \ge 0$ always, so $v$ is never negative. False.
B: $\sin t < t$ for every $t > 0$, so $x = t - \sin t > 0$. True.
C: $\cos t$ lies in $[-1, 1]$, so $v = 1 - \cos t$ lies in $[0, 2]$. True.
D: $a = \sin t$ is negative for $\pi < t < 2\pi$. False.
E: $a = \sin t$ lies between $-1$ and 1, never between 2 and 3. False.
Answer: (B) B, C only. | ||||
| 3 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Distance in the $n$th second: $s_n = u + \frac{a}{2}(2n - 1)$.
For $n = 20$: $88 = 10 + \frac{a}{2}(39) = 10 + 19.5a$, so $a = \frac{78}{19.5} = 4\ m/s^2$.
Answer: (B) 4 $m/s^2$. | ||||
| 4 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power With constant power $P$, the kinetic energy grows as $\frac{1}{2}mv^2 = Pt$, so $v = \sqrt{\frac{2Pt}{m}} \propto t^{1/2}$.
Distance: $x = \int_0^t v\,dt \propto \int_0^t t^{1/2}dt \propto t^{3/2}$.
Answer: (C) $t^{3/2}$. | ||||
| 5 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics For a Carnot engine $\frac{Q_2}{Q_1} = \frac{T_2}{T_1}$.
$T_2 = 400 \times \frac{300}{500} = 240\ K$.
Answer: (D) 240 $K$. | ||||
| 6 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion Let the tension be $T$. Vertically: $T\cos\alpha = mg$. Horizontally the radius of the circle is $r = l\sin\alpha$, and $T\sin\alpha = m\omega^2 l\sin\alpha$, so $T = m\omega^2 l$.
Then $m\omega^2 l\cos\alpha = mg$, giving $\omega = \sqrt{\frac{g}{l\cos\alpha}}$.
Answer: (A). | ||||
| 7 | ThermodynamicsZeroth law and temperature | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on zeroth law and temperature, from Thermodynamics. NCERT: Class 11: Thermodynamics A constant-volume gas thermometer's pressure varies linearly with Celsius temperature: $P_t = P_0 + \frac{P_{100} - P_0}{100}t$.
$P_{30} = 71 + \frac{29}{100}\times 30 = 71 + 8.7 = 79.7$ cm of Hg.
Answer: (C) 79.7 $cm$ of $Hg$. | ||||
| 8 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion The pulley hangs in the loop of a string whose one end is tied to the ceiling and the other to the ceiling through the spring.
In equilibrium the pulley does not rotate, so the tension $T$ is the same on both sides, and the two vertical segments hold up the pulley: $2T = mg$.
The spring carries the tension $T$: $kx = \frac{mg}{2}$, so $x = \frac{mg}{2k}$ (the moment of inertia plays no part in equilibrium).
Answer: (D). | ||||
| 9 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Take $f_2 = 4\cos\omega t = 4\sin(\omega t + 90°)$. The two phasors 3 (along 0°) and 4 (along 90°) add to amplitude 5 at angle $\phi$ with $\tan\phi = \frac{4}{3} = 1.33$, so $\phi = 53.1°$: the resultant $5\sin(\omega t + 53.1°)$ leads $f_1$ by 53.1°.
Checking the others: (A) gives $6\sin\omega t$, phase 0. (B) gives $\tan\phi = \frac{4\sin 53.1°}{3 + 4\cos 53.1°} = \frac{3.2}{5.4}$, about 30.6°. (C) is $3\sin(\omega t + 36.9°)$, and two equal amplitudes give a phase of 18.45°.
Answer: (D) $f_2(t) = 4\cos(\omega t)$. | ||||
| 10 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctHard | |
A hard single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The masses are on one horizontal level, so the repulsion $F$ is horizontal. Let the threads make angles $\theta_1$ (with $m_1$) and $\theta_2$ (with $m_2$) with the vertical; $\theta_1 + \theta_2 = 45°$.
For each mass, $\tan\theta = \frac{F}{\text{weight}}$: $\tan\theta_1 = \frac{F}{2mg}$, $\tan\theta_2 = \frac{F}{mg}$.
Write $f = \frac{F}{mg}$. Then $\tan 45° = 1 = \frac{\frac{f}{2} + f}{1 - \frac{f}{2}\cdot f}$, so $\frac{3f}{2} = 1 - \frac{f^2}{2}$, i.e. $f^2 + 3f - 2 = 0$.
The positive root is $f = \frac{-3 + \sqrt{9 + 8}}{2} = \frac{\sqrt{17} - 3}{2}$, so $F = \left(\frac{\sqrt{17} - 3}{2}\right)mg$.
Answer: (A). | ||||
| 11 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Joined by a wire, the spheres reach a common potential $V$. At a sphere's surface $E = \frac{kq}{R^2} = \frac{V}{R}$.
$\frac{E_1}{E_2} = \frac{R_2}{R_1} = \frac{12}{15} = \frac{4}{5}$.
Answer: (A) $\frac{4}{5}$. | ||||
| 12 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Total current: $I = \int_0^R \sigma r\cdot 2\pi r\,dr = \frac{2\pi\sigma R^3}{3}$.
Current inside radius $r$: $I_{enc} = \frac{2\pi\sigma r^3}{3} = I\frac{r^3}{R^3}$.
Ampere's law: $B\cdot 2\pi r = \mu_0 I\frac{r^3}{R^3}$, so $B = \frac{\mu_0 I r^2}{2\pi R^3}$.
Answer: (B). | ||||
| 13 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The straight lead-in wire and lead-out wire both lie along radii through O (the figure's dashed lines AO and OB), so neither gives a field at O.
The two arcs subtend 120° and 240° and are in parallel, so their currents are inversely proportional to their lengths: $I_1\theta_1 = I_2\theta_2$ (same resistance per unit length).
An arc of angle $\theta$ gives $B = \frac{\mu_0 I\theta}{4\pi R}$ at the centre; the two arcs carry current in opposite senses around O, so their fields are opposite and equal: $\frac{\mu_0}{4\pi R}(I_1\theta_1 - I_2\theta_2) = 0$.
Answer: (D) Zero. | ||||
| 14 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Full-scale current $I_g = 20$ mA through $G + R$ at 200 V: $R = \frac{V}{I_g} - G = \frac{200}{0.02} - 50 = 10000 - 50 = 9950\ \Omega$.
Answer: (D) 9950 $\Omega$. | ||||
| 15 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics With parallel surfaces, Snell's law at each surface gives $n_1\sin\theta_1 = n_2\sin\theta_2 = n_3\sin\theta_3 = n_4\sin\theta_4$.
AB is in medium $n_1$ and CD in medium $n_4$; CD parallel to AB means $\theta_4 = \theta_1$, so $n_4 = n_1$.
Answer: (D) $n_4 = n_1$. | ||||
| 16 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Reflected and refracted rays perpendicular means incidence at Brewster's angle: $\mu = \tan i_B = \tan 60° = \sqrt{3}$.
Answer: (B) $\sqrt{3}$. | ||||
| 17 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics With the final image at $D = 25$ cm, the eyepiece magnifies $m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6$.
$M = m_o m_e$, so $m_o = \frac{30}{6} = 5$.
Answer: (B) 5. | ||||
| 18 | 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 Light power $= 0.8\times 500 = 400$ W $= N\frac{hc}{\lambda}$ with $N = 8\times 10^{20}$ per second.
$\lambda = \frac{Nhc}{P} = \frac{8\times 10^{20}\times 6.63\times 10^{-34}\times 3\times 10^8}{400} = 3.98\times 10^{-7}$ m $= 398$ nm.
Answer: (D) 398 nm. | ||||
| 19 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | StatementsEasy | |
A easy statements question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Statement I: for the same inputs, OR and AND agree on 00 (both 0) and 11 (both 1) and differ on 01 and 10. So the outputs may be the same or different: true.
Statement II: with B = 0, $Y = A\cdot 0 = 0$ whatever A is, so A cannot affect the output: false.
Answer: (C). | ||||
| 20 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Each input passes through a NOT gate and the two results feed a NOR gate: $Y = \overline{\bar{A} + \bar{B}} = A\cdot B$ (De Morgan), an AND gate.
Before $t = 0$ both inputs are high, so $Y = 1$. Over the six intervals after $t = 0$ the inputs are A: 0, 1, 0, 1, 0, 1 and B: 0, 0, 1, 1, 0, 0.
So $Y = AB$: 0, 0, 0, 1, 0, 0, high only in the fourth interval.
Option (B) is high before $t = 0$, low for three intervals, high for the fourth and low after it.
Answer: (B). | ||||
| 21 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power To complete the loop the car needs $v_{top}^2 \ge gR$ at the top, and starting from rest at height $H$ above the bottom, $v_{top}^2 = 2g(H - 2R)$.
So $H_{min} = 2.5R$. The platform's height above the top of the loop is $h = H - 2R = 0.5R = 0.5\times 760 = 380$ cm.
Answer: 380. | ||||
| 22 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Condensing 80 g of steam at 100°C: $80\times 540 = 43200$ cal.
Cooling 80 g of water from 100°C to 30°C: $80\times 1\times 70 = 5600$ cal (specific heat of water 1 cal/g°C).
Total per minute: $43200 + 5600 = 48800$ cal.
Answer: 48800. | ||||
| 23 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Stretching keeps the volume fixed: $A \propto \frac{1}{L}$, so $R = \frac{\rho L}{A} \propto L^2$.
$\frac{R'}{R} = (1.2)^2 = 1.44$: an increase of 44%.
Answer: 44. | ||||
| 24 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Energy: $\frac{1}{2}LI^2 = 32$ with $I = 8$ A gives $L = \frac{64}{64} = 1$ H.
Heat: $I^2R = 64$ gives $R = 1\ \Omega$.
Time constant $\tau = \frac{L}{R} = 1$ s.
Answer: 1. | ||||
| 25 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon energy $E = \frac{hc}{\lambda} = \frac{1242}{621} = 2$ eV.
$K_{max} = E - \phi = 2 - 1 = 1$ eV.
Answer: 1. | ||||
| 26 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane They meet when $\vec{r_1} + 3\vec{v_1} = \vec{r_2} + 3\vec{v_2}$.
x: $2 + 3 = 5$ and $-10 + 15 = 5$, consistent.
y: $3 - 6 = -3$ and $3 - 3b = -3$, so $b = 2$.
Answer: 2. | ||||
| 27 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Equal original lengths and equal new lengths mean equal expansions: $L\alpha_X(200 - 40) = L\alpha_Y(T - 40)$.
$T - 40 = \frac{\alpha_X}{\alpha_Y}\times 160 = \frac{8}{10}\times 160 = 128$, so $T = 168$ (in °C, as T is defined in the question).
The question asks for T in kelvin: 168 + 273 = 441 K.
Answer: 441 (168 if T is read as the Celsius number defined in the stem). | ||||
| 28 | GravitationKepler's laws | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Kepler's third law: $T \propto r^{3/2}$, so $T' = 10\times 16^{3/2} = 10\times 64 = 640$ hours.
In minutes: $640\times 60 = 38400$.
Answer: 38400. | ||||
| 29 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical 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 Volume $V = \frac{m}{\rho} = \frac{16\times 10^{-3}}{8\times 10^3} = 2\times 10^{-6}\ m^3$.
Intensity of magnetisation $I = \frac{M}{V} = \frac{4}{2\times 10^{-6}} = 2\times 10^6$ A/m.
Answer: 2. | ||||
| 30 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Critical angle: $\sin C = \frac{1}{\sqrt{2}}$, so $r_2 = C = 45°$.
$r_1 + r_2 = A = 75°$, so $r_1 = 30°$.
At the first face: $\sin i = \sqrt{2}\sin 30° = \frac{1}{\sqrt{2}}$, so $i = 45°$.
Answer: 45. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Precision is the closeness of repeated readings to each other; accuracy is their closeness to the true value.
A's readings 5.3 and 5.4 g differ by only 0.1 g (precise), but both are 0.3 to 0.4 g away from 5.0 g (not accurate).
B's readings 4.6 and 4.7 g also differ by 0.1 g (precise) and are 0.3 to 0.4 g away from 5.0 g (not accurate).
Answer: (C). | ||||
| 32 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | StatementsMedium | |
A medium statements question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure In $CH_2F_2$ carbon is $sp^3$, but the four bonds are not equal. The C–F bond pairs lie closer to the more electronegative fluorine, so they are farther from carbon and repel each other less: $\angle FCF$ closes to about 108.3°, below 109.5° (Statement I correct).
The C–H bond pairs are nearer carbon and repel more, so $\angle HCH$ opens to about 111.9°. The two angles are not equal (Statement II incorrect).
Answer: (C). | ||||
| 33 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The second electron is removed from the M$^+$ ion: V$^+$ $3d^4$, Cr$^+$ $3d^5$, Mn$^+$ $3d^5 4s^1$, Fe$^+$ $3d^6 4s^1$.
Cr$^+$ loses an electron from the stable half-filled $3d^5$ set, so Cr is highest (about 1592 kJ/mol).
Mn$^+$ and Fe$^+$ both lose a 4s electron; Fe has the larger nuclear charge, so Fe (about 1561) > Mn (about 1509).
V$^+$ loses a 3d electron from $3d^4$ with the smallest nuclear charge: lowest (about 1414).
Order: Cr > Fe > Mn > V.
Answer: (C). | ||||
| 34 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter An Ellingham diagram plots $\Delta G°$ of oxide formation against temperature. It shows the free energy change, hence the feasibility of a reduction and which metal can reduce which oxide.
It is purely thermodynamic and says nothing about kinetics: it cannot predict the rate of a reaction.
Answer: (B). | ||||
| 35 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Acidified chromate (or dichromate) with $H_2O_2$ gives the blue peroxide $CrO_5$, chromium pentoxide (chromium peroxide, $CrO(O_2)_2$):
$Cr_2O_7^{2-} + 2H^+ + 4H_2O_2 \rightarrow 2CrO_5 + 5H_2O$.
This blue colour is the test for chromate and for $H_2O_2$.
Answer: (B). | ||||
| 36 | 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 A typical Portland cement contains CaO 50–60%, $SiO_2$ 20–25%, $Al_2O_3$ 5–10%, with MgO 2–3% and smaller amounts of $Fe_2O_3$ and $SO_3$.
Order: $CaO > SiO_2 > Al_2O_3 > MgO$.
Answer: (A). | ||||
| 37 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down group 16 the E–H bond becomes longer and weaker, so the hydride gives up hydrogen more easily and is a stronger reducing agent.
Order: $H_2Te > H_2Se > H_2S$.
Answer: (B). | ||||
| 38 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The early actinoids show many oxidation states; the range narrows after the middle of the series.
Th: +4 (and +3); Am: +3, +4, +5, +6; Cm: +3, +4; Fm: +2, +3.
Only americium shows +5.
Answer: (C). | ||||
| 39 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $MnO_4^-$ is a regular tetrahedron (Mn $d^3s$ hybridised), not distorted: (A) false.
Its structure has three Mn=O double bonds and one Mn–O$^-$ bond (the charge delocalised over all four), so it has three Mn–O $\pi$ bonds ($d\pi$–$p\pi$): (B) false, (D) true.
Manganate ($MnO_4^{2-}$, Mn +6) is converted to permanganate (Mn +7) by oxidation, not reduction: (C) false.
Answer: (D). | ||||
| 40 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter 'Blue baby' syndrome (methaemoglobinaemia) comes from excess nitrate in drinking water: A–III.
Eutrophication is driven by phosphate (and nitrate) nutrients: B–IV.
Minamata disease is mercury poisoning: C–I.
Itai-itai disease is cadmium poisoning: D–II.
Answer: (B). | ||||
| 41 | Principles Related to Practical ChemistryDetection of elements and functional groups | No matching NCERT chapter | StatementsMedium | |
A medium statements question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter The Prussian blue colour in Lassaigne's test shows nitrogen present as NaCN in the sodium fusion extract.
Histamine ($C_5H_9N_3$, an imidazole with an aminoethyl chain) contains nitrogen but no sulphur, so it gives the Prussian blue colour: Statement I correct.
Penicillin contains both nitrogen and sulphur. On fusion with the usual amount of sodium they form NaSCN, which with $Fe^{3+}$ gives a blood-red colour, not blue: Statement II incorrect.
Answer: (C). | ||||
| 42 | 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 The substrate is 2-(dimethylaminomethyl)cyclohexanone. Excess MeI methylates the amine to the quaternary salt $-CH_2N^+Me_3\,I^-$.
With NaOH and mild heating the salt undergoes elimination (Hofmann elimination): the hydrogen removed is the acidic one on the ring carbon alpha to C=O, and $NMe_3$ leaves.
The product is the conjugated enone 2-methylenecyclohexanone, an exocyclic C=C on the carbon next to C=O.
Answer: (A). | ||||
| 43 | 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 The alkene is 2-methylbut-1-ene, $CH_2=C(CH_3)CH_2CH_3$.
Oxymercuration–demercuration ($Hg(OAc)_2$, $H_2O$; then $NaBH_4$) adds water with Markovnikov orientation and no rearrangement: OH goes to the more substituted carbon.
Product: 2-methylbutan-2-ol, $(CH_3)_2C(OH)CH_2CH_3$.
Answer: (B). | ||||
| 44 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Nucleophilic substitution of an aryl chloride (addition–elimination) is helped by electron-withdrawing groups ortho or para to Cl, which stabilise the negative intermediate by resonance.
I (2,4-dinitro): two activating $NO_2$ groups, ortho and para: easiest.
IV (4-nitro): one para $NO_2$: next.
II (3-nitro): meta $NO_2$ acts only inductively: harder.
III (4-methyl): the electron-donating $CH_3$ destabilises the intermediate: hardest.
Increasing difficulty: I, IV, II, III.
Answer: (D). | ||||
| 45 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctHard | |
A hard single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Conc. $H_2SO_4$ sulphonates phenol; the sulphonation that the question uses places $SO_3H$ para to OH (the product at higher temperature), giving 4-hydroxybenzenesulphonic acid.
Very dilute $HNO_3$ is a mild nitrating agent for the strongly activated ring; it nitrates once, ortho to OH (para is blocked), and does not displace the sulphonic group.
Product A: 4-hydroxy-3-nitrobenzenesulphonic acid (OH at C-1, $NO_2$ at C-2, $SO_3H$ at C-4).
Answer: (B). | ||||
| 46 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Diazotisation of sulphanilic acid gives the 4-sulphonatobenzenediazonium ion. It couples with N,N-dimethylaniline at the para position of the activated ring (an electrophilic azo coupling).
The product, $^-O_3S$–$C_6H_4$–N=N–$C_6H_4$–$N(CH_3)_2$ with both rings para-substituted, is methyl orange.
(A) is not an azo compound, (B) couples on nitrogen to a quaternary ammonium, and (C) couples ortho; each is also misnamed.
Answer: (D). | ||||
| 47 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Bakelite is a cross-linked (network) thermosetting polymer: A → III.
High-density polythene has linear chains: B → II.
Low-density polythene has branched chains: C → I.
Rayon is regenerated, chemically modified cellulose: semi-synthetic, D → IV.
Answer: (A). | ||||
| 48 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules DNA's sugar is 2-deoxy-D-ribose in its furanose form: no OH at C-2 (two H atoms there), OH at C-3 drawn down in the Haworth projection, and $CH_2OH$ at C-4 drawn up (D series).
β means the anomeric OH at C-1 is on the same side as the $CH_2OH$ group, i.e. up.
(A) has the C-1 OH down (α). (B) has an OH at C-2 (that is ribose). (D) has OH at C-2 and none at C-3 (a 3-deoxy sugar). (C) has C-1 OH up, two H at C-2 and OH down at C-3.
Answer: (C). | ||||
| 49 | AminesClassification and identification of amines | Class 12: Amines | Matching listEasy | |
A easy matching list question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hell-Volhard-Zelinsky: $RCH_2COOH$ with $X_2$/red P gives the α-halo acid: A → II.
Hinsberg's reagent (benzenesulphonyl chloride) distinguishes 1°, 2° and 3° amines: B → IV.
The isocyanide (carbylamine) test, with $CHCl_3$ and KOH, is given only by primary amines: C → I.
Hoffmann bromamide degradation converts an amide into a primary amine with one carbon fewer: D → III.
Answer: (A). | ||||
| 50 | HydrocarbonsConformations of ethane | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on conformations of ethane, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The four Newman projections are of ethane-1,2-diol, $HOCH_2CH_2OH$.
(A) and (D) are eclipsed conformations, raised in energy by torsional strain.
(B) is staggered with the two OH groups anti (180°); (C) is staggered with them gauche (60°).
For ethane-1,2-diol the gauche form is the most stable, because only there the two OH groups are close enough to form an intramolecular hydrogen bond, which outweighs the small gauche repulsion.
Answer: (C). | ||||
| 51 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry For an ideal gas $d = \frac{PM}{RT}$, so $Pd = \frac{P^2M}{RT}$ and $\left(\frac{\partial(Pd)}{\partial P}\right)_T = \frac{2PM}{RT}$.
With $M = 2$ g/mol: $10.0 = \frac{2\times 8.21\times 2}{0.0821\times T}$, so $T = \frac{32.84}{0.821} = 40$ K.
Answer: 40. | ||||
| 52 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalHard | |
A hard numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom de Broglie: $p = \frac{h}{\lambda}$, so $\Delta p = \frac{h}{\lambda^2}\Delta\lambda$.
Minimum uncertainty: $\Delta x\,\Delta p = \frac{h}{4\pi}$, so $\Delta x\cdot\frac{h\Delta\lambda}{\lambda^2} = \frac{h}{4\pi}$, i.e. $\Delta x = \frac{\lambda^2}{4\pi}\cdot\frac{1}{\Delta\lambda}$.
The graph of $\Delta x$ against $\frac{1}{\Delta\lambda}$ is a line through the origin of slope $\tan\theta = 0.318$, so $\frac{\lambda^2}{4\pi} = 0.318$ Å$^2$.
$\lambda^2 = 0.318\times 4\times 3.142 = 3.997$, so $\lambda = 2.0$ Å.
Answer: 2. | ||||
| 53 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics The cycle is a rectangle: A (P, V) → B (P, 2V) → C (2P, 2V) → D (2P, V) → A.
Work done by the gas: A→B, $+P(2V - V) = +PV$; B→C and D→A, zero (constant volume); C→D, $2P(V - 2V) = -2PV$. Net work by the gas $= -PV$ (the cycle runs anticlockwise on the P–V diagram).
In the chemistry sign convention, work done on the gas is positive: $w = -(-PV) = +PV$, so $x = 1$.
Answer: 1. | ||||
| 54 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $-4$°F in Celsius: $\frac{5}{9}(-4 - 32) = -20$°C, so the depression needed is $\Delta T_f = 20$ K.
$m = \frac{\Delta T_f}{K_f} = \frac{20}{1.86} = 10.753$ mol per kg of water.
Mass of ethanol ($M = 46$ g/mol): $10.753\times 46 = 494.6$ g.
Answer: 495. | ||||
| 55 | EquilibriumIonic equilibrium | Class 11: Equilibrium | NumericalHard | |
A hard numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium For a weak acid, $[H^+] = \sqrt{K_aC}$, so $pH = \frac{1}{2}(pK_a - \log C)$.
At $C_1 = 1$ M: $pH_1 = \frac{1}{2}pK_a$.
Doubled: $pH_2 = pK_a = \frac{1}{2}(pK_a - \log C_2)$, so $-\log C_2 = pK_a$ and $C_2 = K_a = 1.8\times 10^{-5}$ M.
$x = \frac{C_1}{C_2} = \frac{1}{1.8\times 10^{-5}} = 55555.6$.
Answer: 55556. | ||||
| 56 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Subtract the second half-reaction from the first: $Pd^{2+} + 4Cl^- \rightarrow PdCl_4^{2-}$ with $E°_{cell} = 0.83 - 0.64 = 0.19$ V and $n = 2$.
$\log K = \frac{nE°}{0.059} = \frac{2\times 0.19}{0.059} = 6.44$ (6.43 with 0.0591).
$6.44 = x\times 10^{-1}$, so $x = 64.4$; nearest integer 64.
Answer: 64. | ||||
| 57 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Experiments 1 and 2 keep [A] fixed and double [B]; the rate rises from 0.20 to 0.80, four times.
$2^n = 4$, so the order with respect to B is $n = 2$. (Experiments 1 and 3 show first order in A.)
Answer: 2. | ||||
| 58 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Acetic acid adsorbed: $(0.85 - 0.75)\times 0.200 = 0.020$ mol $= 0.020\times 60 = 1.2$ g.
Per gram of charcoal: $\frac{1.2}{2} = 0.6$ g $= 6\times 10^{-1}$.
Answer: 6. | ||||
| 59 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds An octahedral cobalt(III) complex needs six ligands. Five water molecules supply five; the sixth must be a chloride, so the compound is $[Co(H_2O)_5Cl]Cl_2$.
In water it gives $[Co(H_2O)_5Cl]^{2+} + 2Cl^-$: three ions.
Answer: 3. | ||||
| 60 | HydrogenHydrogen and its compounds | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter '25 volume' means 1 L of the solution gives 25 L of $O_2$ at STP.
$2H_2O_2 \rightarrow 2H_2O + O_2$: 68 g of $H_2O_2$ give 22.4 L of $O_2$, so 25 L come from $\frac{68\times 25}{22.4} = 75.89$ g.
So the solution holds 75.89 g $H_2O_2$ per litre, i.e. 7.589% (w/v) $= 75.89\times 10^{-1}$; nearest integer 76.
Answer: 76. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $S_1$: $|x|^2 + |x| - 2 = 0$ gives $(|x| + 2)(|x| - 1) = 0$, so $|x| = 1$ and $S_1 = \{-1, 1\}$.
$S_2$: $(x + 2)(x - 1) = 0$, so $S_2 = \{-2, 1\}$.
$S_1 \cup S_2 = \{-2, -1, 1\}$ (3 elements) and $S_1 \cap S_2 = \{1\}$ (1 element).
Answer: (D). | ||||
| 62 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A$ is a cyclic permutation matrix: $A^2 = \begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}$ and $A^3 = I$.
$I + A + A^2$ has a 1 in every position, so $I + A + A^2 = B$.
Then $A^n + A^{n+1} + A^{n+2} = A^n(I + A + A^2) = A^nB$. Each row of $A$ has a single 1, so $AB = B$, and hence $A^nB = B$ for every $n$.
The set is all of $\mathbb{N}$.
Answer: (D). | ||||
| 63 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For order 3: $|kM| = k^3|M|$ and $|\text{adj}\,M| = |M|^2$.
$|4A| = 4^3\cdot 3 = 192 = 2^6\cdot 3$; $|\text{adj}(4A)| = 192^2$.
$|3\,\text{adj}(4A)| = 27\cdot 192^2$; $|\text{adj}(3\,\text{adj}(4A))| = (27\cdot 192^2)^2 = 3^6\cdot 192^4 = 3^6\cdot 2^{24}\cdot 3^4 = 2^{24}\cdot 3^{10}$.
So $(m, n) = (24, 10)$.
Answer: (D). | ||||
| 64 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $D = \begin{vmatrix}1&3&2\\1&\lambda&3\\1&-2&7\end{vmatrix} = (7\lambda + 6) - 3(4) + 2(-2 - \lambda) = 5\lambda - 10$.
If $\lambda \ne 2$, $D \ne 0$ and the system has a unique solution for every $\mu$: (C) is true and (D) is false.
If $\lambda = 2$: subtracting the first equation, $-y + z = 1$ and $-5y + 5z = \mu - 7$. These agree only if $\mu - 7 = 5$, i.e. $\mu = 12$ (then infinitely many solutions); otherwise none. So (A) and (B) are true.
Answer: (D). | ||||
| 65 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = 4$ the base is $16 + 2 - 17 = 1$ and the exponent's denominator is $8 - 2^3 = 0$: the form $1^\infty$.
So $L = e^{\ell}$ with $\ell = \lim_{x\to 4}\frac{2^x + 2^{5-x} - 18}{8 - 2^{3x/4}}$, a $\frac{0}{0}$ form.
L'Hôpital: numerator derivative $\ln 2\,(2^x - 2^{5-x}) \to 14\ln 2$; denominator derivative $-\frac{3}{4}\ln 2\cdot 2^{3x/4} \to -6\ln 2$.
$\ell = -\frac{14}{6} = -\frac{7}{3}$, so $L = e^{-7/3}$.
Answer: (B). | ||||
| 66 | 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 On $(-2, -1)$, $[x] = -2$; on $[2, 3)$, $[x] = 2$; each piece is continuous inside its interval, so only $x = -1$ and $x = 2$ need checking.
At $x = -1$: left limit $1 + 2 + 2a = 3 + 2a$; value $-4 + 5 - b = 1 - b$. So $2a + b = -2$.
At $x = 2$: left limit $8 + 5 - b = 13 - b$; value $12 + 12 - 2a = 24 - 2a$. So $2a - b = 11$.
Adding: $4a = 9$, $a = \frac{9}{4}$; then $b = 2a - 11 = -\frac{13}{2}$.
$8ab = 8\cdot\frac{9}{4}\cdot\left(-\frac{13}{2}\right) = -117$.
Answer: (B). | ||||
| 67 | 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 Differentiating, $\frac{1}{2\sqrt{x}} + \frac{y'}{2\sqrt{y}} = 0$, so the slope at $(x_1, y_1)$ is $-\sqrt{\frac{y_1}{x_1}}$.
The tangent $y - y_1 = -\sqrt{\frac{y_1}{x_1}}(x - x_1)$ simplifies to $\frac{x}{\sqrt{x_1}} + \frac{y}{\sqrt{y_1}} = \sqrt{x_1} + \sqrt{y_1} = 3$.
Intercepts: $3\sqrt{x_1}$ and $3\sqrt{y_1}$; their sum is $3(\sqrt{x_1} + \sqrt{y_1}) = 9$.
Answer: (C). | ||||
| 68 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $\vec{AB} = (2, 3, -1)$, $\vec{AC} = (1, -1, 2)$; normal $\vec{AB}\times\vec{AC} = (5, -5, -5)$, i.e. $(1, -1, -1)$.
The angle $\theta$ between OP, direction $(a, b, c)$, and the plane: $\sin\theta = \frac{|a - b - c|}{\sqrt{3}\sqrt{a^2 + b^2 + c^2}}$.
$(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc = a^2 + b^2 + c^2 - 2[a(b + c) - bc] = a^2 + b^2 + c^2$.
So $\sin\theta = \frac{1}{\sqrt{3}}$, $\theta = \sin^{-1}\frac{1}{\sqrt{3}}$.
Answer: (B). | ||||
| 69 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Write $4x + 13 = 2(2x + 5) + 3$.
$\int\frac{2(2x + 5)}{\sqrt{x^2 + 5x + 6}}dx = 4\sqrt{x^2 + 5x + 6}$.
$\int\frac{3}{\sqrt{(x + \frac{5}{2})^2 - \frac{1}{4}}}dx = 3\log_e\left|x + \frac{5}{2} + \sqrt{x^2 + 5x + 6}\right|$.
So $A = 4$, $B = 3$.
Answer: (D). | ||||
| 70 | 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 $[x^2 \pm 1] = [x^2] \pm 1$, so the integrand is $2[x^2]$.
On $[1, 2]$: $[x^2] = 1$ on $[1, \sqrt{2})$, 2 on $[\sqrt{2}, \sqrt{3})$, 3 on $[\sqrt{3}, 2)$.
$\int_1^2[x^2]dx = (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3}) = 5 - \sqrt{2} - \sqrt{3}$.
The integral is $2(5 - \sqrt{2} - \sqrt{3}) = 10 - 2\sqrt{2} - 2\sqrt{3}$.
Answer: (D). | ||||
| 71 | 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 Let $u = \sin x + \tan^{-1}y$. Then $\frac{du}{dx} = \cos x + \frac{1}{1 + y^2}\frac{dy}{dx}$, the right-hand side.
The equation is $2\sqrt{u}\frac{dy}{dx} = \frac{du}{dx}$, i.e. $\frac{du}{2\sqrt{u}} = dy$, so $\sqrt{u} = y + c$.
$x = 0$ when $y = 0$: $u = 0$, so $c = 0$ and $\sqrt{\sin x + \tan^{-1}y} = y$.
At $y = 1$, $x = b$: $\sin b + \frac{\pi}{4} = 1$, i.e. $4\sin b + \pi = 4$.
Answer: (A). | ||||
| 72 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines The lines are parallel: $x - 2y - a = 0$ and $x - 2y + \frac{7}{3} = 0$. A circle touching both has them as parallel tangents, so the distance between them is its diameter, $\frac{2\sqrt{5}}{3}$.
$\frac{|-a - \frac{7}{3}|}{\sqrt{5}} = \frac{2\sqrt{5}}{3}$ gives $|a + \frac{7}{3}| = \frac{10}{3}$, so $a = 1$ or $a = -\frac{17}{3}$.
Sum: $1 - \frac{17}{3} = -\frac{14}{3}$.
Answer: (D). | ||||
| 73 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Write the line as $y = -\frac{\alpha}{\beta}x + \frac{1}{\beta}$. It touches the hyperbola when $c^2 = a^2m^2 - b^2$: $\frac{1}{\beta^2} = \frac{a^2\alpha^2}{\beta^2} - b^2$.
So $a^2\alpha^2 - b^2\beta^2 = 1$, and the locus is $a^2x^2 - b^2y^2 = 1$.
Answer: (A). | ||||
| 74 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections At $x = 2$: $y^2 = 4\left(1 - \frac{4}{16}\right) = 3$, so the points are $(2, \pm\sqrt{3})$. At $x = 4$, $y_1 = 0$ and the tangent there is $x = 4$.
Tangents at $(2, \pm\sqrt{3})$: $\frac{2x}{16} \pm \frac{\sqrt{3}y}{4} = 1$.
With $x = 4$: $\frac{1}{2} \pm \frac{\sqrt{3}y}{4} = 1$, so $y = \pm\frac{2}{\sqrt{3}}$: A $= (4, \frac{2}{\sqrt{3}})$, B $= (4, -\frac{2}{\sqrt{3}})$.
AB $= \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$.
Answer: (C). | ||||
| 75 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line: $(2t, 3t - 1, 1 - t)$.
In the plane: $4t + (3t - 1) + 6(1 - t) + 1 = t + 6 = 0$, so $t = -6$ and the point is $(-12, -19, 7)$.
Distance from O: $\sqrt{144 + 361 + 49} = \sqrt{554}$.
Answer: (D). | ||||
| 76 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct 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 A parallel plane is $x + 2y + 3z = d$; through (1, 2, 3), $d = 1 + 4 + 9 = 14$.
Check: (2, 6, 1) gives 17; (2, 6, 0) gives $2 + 12 + 0 = 14$; (3, 6, −1) gives 12; (3, 5, 1) gives 16.
Answer: (B). | ||||
| 77 | 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 Variance does not change on shifting by 6. With $d_i = a_i - 6$: $\sigma^2 = \frac{\sum d_i^2}{16} - \left(\frac{\sum d_i}{16}\right)^2 = \frac{40}{16} - \frac{1}{4} = \frac{9}{4}$.
$\sigma = \frac{3}{2}$.
Answer: (B). | ||||
| 78 | 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 An onto function from a 6-element set to itself is a bijection: $6! = 720$ in all.
We need distinct values $f(1), f(2), f(3)$ with $f(3) = f(1) + f(2)$. If $f(1)$ or $f(2)$ were 0, $f(3)$ would equal the other, so both are at least 1, distinct, with sum at most 5.
Ordered pairs: (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (2, 3), (3, 2): 8. The other three elements map onto the remaining three values in $3! = 6$ ways.
Favourable: $8\times 6 = 48$; probability $\frac{48}{720} = \frac{1}{15}$.
Answer: (D). | ||||
| 79 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let the height be $h$. The top and the junction (height $\frac{h}{3}$) are seen at angles $\alpha$, $\beta$ with $\tan\alpha = \frac{h}{40}$, $\tan\beta = \frac{h}{120}$.
$\tan(\alpha - \beta) = \frac{\frac{h}{40} - \frac{h}{120}}{1 + \frac{h^2}{4800}} = \frac{\frac{h}{60}}{1 + \frac{h^2}{4800}} = \frac{1}{2}$.
So $\frac{h}{30} = 1 + \frac{h^2}{4800}$, i.e. $h^2 - 160h + 4800 = 0$, giving $h = 40$ or $h = 120$.
Of the options, 40 m.
Answer: (C). | ||||
| 80 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter If $\square = \wedge$, the statement is false whenever $p$ is false (then $\sim q \wedge p$ is false): not a tautology. So $\square = \vee$.
With $\Diamond = \wedge$: $(\sim q \wedge p) \vee (q \wedge \sim p)$ is true only when $p$ and $q$ differ: not a tautology.
With $\Diamond = \vee$: $(\sim q \wedge p) \vee q \vee \sim p = (p \vee q) \vee \sim p$, always true.
Answer: (D). | ||||
| 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 Dividing by $x$: $x + \frac{1}{x} = 1$. The roots are $x = e^{\pm i\pi/3}$, so $x^n + x^{-n} = 2\cos\frac{n\pi}{3}$.
For $n = 1, \ldots, 5$: $1, -1, -2, -1, 1$. Squares: $1 + 1 + 4 + 1 + 1 = 8$.
Answer: 8. | ||||
| 82 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $t = x^2 + 4x + 6$; then $2x^2 + 8x + 14 = 2t + 2$ and $t(2t + 2) = 12$, i.e. $t^2 + t - 6 = 0$, $t = 2$ or $t = -3$.
$t = 2$: $x^2 + 4x + 4 = 0$, $x = -2$ (real).
$t = -3$: $x^2 + 4x + 9 = 0$, $x = -2 \pm i\sqrt{5}$, each of modulus $\sqrt{4 + 5} = 3$.
Sum of moduli of the non-real roots: $3 + 3 = 6$.
Answer: 6. | ||||
| 83 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations STATISTICS has 10 letters: S × 3, T × 3, I × 2, A, C.
All arrangements: $\frac{10!}{3!\,3!\,2!} = 50400$.
With the three S together (one block): $\frac{8!}{3!\,2!} = 3360$.
Not all together: $50400 - 3360 = 47040 = 3136\times 15 = 56^2\times 15$.
Answer: 15. | ||||
| 84 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $2022 = 6\times 337$, so $2020 \equiv -2 \pmod{337}$ and $2020^{2022} \equiv 2^{2022}$.
337 is prime, so $2^{336} \equiv 1$ (Fermat). $2022 = 6\times 336 + 6$, so $2^{2022} \equiv 2^6 = 64$.
Answer: 64. | ||||
| 85 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series For odd $n = 2m + 1$ the odd cubes carry + and the even cubes −: $S_n = \sum_{k=1}^{n}k^3 - 2\sum_{j=1}^{m}(2j)^3 = \left(\frac{n(n+1)}{2}\right)^2 - 4m^2(m+1)^2$.
With $\frac{n(n+1)}{2} = (2m+1)(m+1)$: $S_n = (m+1)^2[(2m+1)^2 - 4m^2] = (m+1)^2(4m+1)$.
$S_{199}$ ($m = 99$) $= 100^2\cdot 397 = 3970000$; $S_{99}$ ($m = 49$) $= 50^2\cdot 197 = 492500$.
$S_{199} - 4S_{99} = 3970000 - 1970000 = 2\times 10^6$, so $\lambda = 2$.
Answer: 2. | ||||
| 86 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $h(n) = g(n + 2) - g(n + 1)$.
Common difference: $\frac{205 - 53}{101 - 25} = 2$, so $h(n) = 53 + 2(n - 25) = 2n + 3$.
Telescoping: $\sum_{n=1}^{46}h(n) = g(48) - g(2)$.
$\sum_{n=1}^{46}(2n + 3) = 46\cdot 47 + 138 = 2300 = 100k$, so $k = 23$.
Answer: 23. | ||||
| 87 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The curves meet at $x = 0$ and $x = a$. Area $= \left|\int_0^a (ax - x^2)dx\right| = \frac{|a|^3}{6}$.
$\frac{|a|^3}{6} = 36$ gives $|a|^3 = 216$, $|a| = 6$.
Answer: 6. | ||||
| 88 | Matrices and DeterminantsArea of triangles using determinants | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on area of triangles using determinants, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Area $= \frac{1}{2}|0(3 - b) + 2(b - 1) + 4(1 - 3)| = \frac{1}{2}|2b - 10| = |b - 5|$.
$|b - 5| = 6$ gives $b = 11$ or $b = -1$; the positive value is 11.
Answer: 11. | ||||
| 89 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability A $\frac{0}{0}$ form. L'Hôpital: $\lim_{x\to 9}\frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}} = \frac{f'(9)\sqrt{9}}{\sqrt{f(9)}} = \frac{4\cdot 3}{3} = 4$.
Answer: 4. | ||||
| 90 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\cot x$ is undefined at $0$, $\pi$, $2\pi$. On $(0, \pi)$ it decreases continuously from $+\infty$ to $-\infty$ while $x^2$ increases from 0 to $\pi^2$, so $\cot x - x^2$ is strictly decreasing from $+\infty$ to $-\infty$: exactly one root.
On $(\pi, 2\pi)$ the same holds ($\cot x$ falls from $+\infty$ to $-\infty$, $x^2$ rises): exactly one root.
Total: 2.
Answer: 2. | ||||
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