JEE (Main) · 2020
JEE (Main) 2020 — 2 September, 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
25 Q
- Electrostatics3
- Oscillations and Waves2
- Motion of System of Particles and Rigid Body2
12 easy · 13 medium · 0 hard
Chemistry
25 Q
- Alcohols, Phenols and Ethers3
- Chemical Bonding and Molecular Structure2
- Structure of Atom2
12 easy · 12 medium · 1 hard
Mathematics
25 Q
- Sequences and Series3
- Matrices and Determinants2
- Limits, Continuity and Differentiability2
8 easy · 16 medium · 1 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 MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Energy is $[ML^2T^{-2}]$ and momentum is $[MLT^{-1}]$, so $E = P\times[LT^{-1}]$.
A length is $A^{1/2}$, so $LT^{-1} = A^{1/2}T^{-1}$ and $[E] = [PA^{1/2}T^{-1}]$.
Answer: (C). | ||||
| 2 | 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 The graph starts at the positive extreme at $t=0$, crosses zero at $T/4$ and $3T/4$, and is at the negative extreme at $T/2$: $x = A\cos\omega t$.
(A) At $3T/4$, $x=0$, so the force $-kx$ is zero: true.
(B) At $T$, $x=+A$, an extreme, so the acceleration $\omega^2A$ is maximum: true.
(C) At $T/4$, $x=0$, the mean position, so the speed is maximum: true.
(D) At $T/2$, $x=-A$: the energy is all potential and K.E. is zero, so they are not equal: false.
So (A), (B) and (C).
Answer: (C). | ||||
| 3 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Released from rest, the mass has a constant horizontal acceleration $qE/m$ and a constant vertical acceleration $g$.
Both displacements grow as $\frac{1}{2}at^2$, so $\frac{y}{x} = \frac{g}{qE/m}$, a constant: the path is a straight line through the starting point.
Answer: (C). | ||||
| 4 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Angular momentum is conserved: $L = 0.1\times10 + 0.2\times5 = 2$ kg m$^2$ s$^{-1}$, and $I = 0.3$ kg m$^2$.
$\omega = \frac{2}{0.3} = \frac{20}{3}$ rad/s, and $K = \frac{L^2}{2I} = \frac{4}{0.6} = \frac{20}{3}$ J.
Answer: (D). | ||||
| 5 | 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 height $h$: $g\frac{R^2}{(R+h)^2}$. At depth $h$: $g\left(1-\frac{h}{R}\right)$. With $x = h/R$:
$\frac{1}{(1+x)^2} = 1-x \Rightarrow (1-x)(1+x)^2 = 1 \Rightarrow x - x^2 - x^3 = 0 \Rightarrow x^2 + x - 1 = 0$ (as $x\neq0$).
$x = \frac{\sqrt5-1}{2}$, so $h = \frac{\sqrt5R - R}{2}$.
Answer: (C). | ||||
| 6 | 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 | Single correctEasy | |
A easy single correct 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 Linear strain $\alpha\Delta T = 0.02\%$; volume strain $\gamma\Delta T = 3\alpha\Delta T = 0.06\%$.
Mass is fixed, so $\rho \propto 1/V$ and the density falls by about 0.06%.
Answer: (D). | ||||
| 7 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Each tangent makes the contact angle $\theta$ with the vertical wall; the two tangents meet at $2\theta = 60^\circ$, so $\theta = 30^\circ$.
$h = \frac{2T\cos\theta}{r\rho g} = \frac{2\times0.05\times0.866}{0.15\times10^{-3}\times667\times10} = \frac{0.0866}{1.0005} \approx 0.087$ m.
Answer: (C). | ||||
| 8 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat taken in: $1915 + 125 = 2040$ J. Work in the cycle = net heat $= 1915 - 40 + 125 - Q = 2000 - Q$.
$\eta = \frac{2000 - Q}{2040} = 0.5 \Rightarrow Q = 2000 - 1020 = 980$ J.
Answer: (D). | ||||
| 9 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory In a closed (fixed-volume) container the number density $n$ is fixed, so $\lambda = \frac{1}{\sqrt2\pi d^2 n}$ does not change: (C) true, (A) false.
The mean speed grows as $\sqrt T$, so the mean collision time $\lambda/\bar v$ decreases: (B) true, (D) false.
Answer: (C). | ||||
| 10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Equal $\sigma$ means charge in proportion to area: $q_1 = \frac{Qr^2}{R^2+r^2}$ on the inner shell, $q_2 = \frac{QR^2}{R^2+r^2}$ on the outer.
$V_0 = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{r} + \frac{q_2}{R}\right) = \frac{1}{4\pi\epsilon_0}\frac{(r+R)Q}{R^2+r^2}$.
Answer: (D). | ||||
| 11 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charge is conserved: $10\times50 = (10 + C)\times20 \Rightarrow 10 + C = 25 \Rightarrow C = 15\ \mu$F.
Answer: (A). | ||||
| 12 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Period $T = \frac{2\pi m}{qB} = \frac{2\pi\times1.67\times10^{-27}}{1.6\times10^{-19}\times0.3} = 2.19\times10^{-7}$ s.
Pitch $= v\cos60^\circ\,T = 2\times10^5\times2.19\times10^{-7} = 0.0437$ m.
$l = 10\times0.0437 \approx 0.44$ m.
Answer: (B). | ||||
| 13 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Adding the segment AD twice in opposite directions splits the loop into two rectangles of area $ab$ each, carrying the same current I.
ABCDA lies in the x-y plane and ADEFA in the x-z plane. With the current senses shown, their moments are $abI\,\hat k$ and $abI\,\hat j$.
Total $\vec m = abI(\hat j + \hat k)$: magnitude $\sqrt2\,abI$, along $\frac{\hat j + \hat k}{\sqrt2}$.
Answer: (C). | ||||
| 14 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current A 45$^\circ$ lead means $X_L = R$, so the coil's impedance is $Z = \sqrt2 X_L$.
Taking the 100 $\Omega$ as the coil's impedance (see note): $X_L = \frac{100}{\sqrt2} = 70.7\ \Omega$.
$L = \frac{X_L}{2\pi f} = \frac{70.7}{2\pi\times1000} = 1.13\times10^{-2}$ H $\approx 1.1\times10^{-2}$ H.
Answer: (B). | ||||
| 15 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves The wave travels along $\vec E\times\vec B$: $\hat k\times(2\hat i - 2\hat j) = 2\hat j + 2\hat i$.
Unit vector: $\frac{1}{\sqrt2}(\hat i + \hat j)$.
Answer: (B). | ||||
| 16 | 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 $\beta\propto\lambda$, so the number in a fixed segment goes as $1/\lambda$: $N = 16\times\frac{700}{400} = 28$.
Answer: (C). | ||||
| 17 | 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 $\frac{\lambda_p}{\lambda_e} = \frac{m_ev_e}{m_pv_p} = \frac{m_e}{5m_p} = 1.878\times10^{-4}$.
$m_p = \frac{9.1\times10^{-31}}{5\times1.878\times10^{-4}} = \frac{9.1\times10^{-31}}{9.39\times10^{-4}} \approx 9.7\times10^{-28}$ kg.
Answer: (B). | ||||
| 18 | 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 $\nu\propto\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{2n+1}{n^2(n+1)^2} \approx \frac{2n}{n^4} = \frac{2}{n^3}$ for $n\gg1$.
Answer: (D). | ||||
| 19 | 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 NAND gate: $X = \overline{AB}$. OR gate: $Y = A + B$. AND gate: $W = XY$. NOR gate: $Z = \overline{X + W}$.
Since $W = XY$ is 1 only when $X$ is 1, $X + W = X$, so $Z = \overline{X} = AB$.
For (1,0), (0,0), (1,1), (0,1): Z = 0, 0, 1, 0.
Answer: (C). | ||||
| 20 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity $E_2$ is connected at end P, so it is balanced by the wire from P to the jockey: $100 - 49 = 51$ cm.
With S open no current flows through $E_2$ at balance, so $1.02 = k\times51$ and $k = 0.02$ V/cm.
Answer: (B). | ||||
| 21 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power After the collision $m$ has half its kinetic energy: $v_1 = \frac{u}{\sqrt2}$.
Elastic, so the other half goes to $10m$: $\frac12(10m)v_2^2 = \frac14mu^2 \Rightarrow v_2 = \frac{u}{\sqrt{20}}$.
Momentum perpendicular to x: $mv_1\sin\theta_1 = 10mv_2\sin\theta_2 \Rightarrow \frac{\sin\theta_1}{\sqrt2} = \frac{10}{\sqrt{20}}\sin\theta_2 \Rightarrow \sin\theta_1 = \sqrt{10}\sin\theta_2$.
So n = 10.
Answer: 10. | ||||
| 22 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion With areal density $\sigma$: the disc has mass $\sigma\pi a^2$ at O; the hole has mass $\sigma\frac{a^2}{4}$ with its centre at $\frac{a}{2}$.
$x_{cm} = \frac{0 - \frac{a^2}{4}\cdot\frac a2}{\pi a^2 - \frac{a^2}{4}} = -\frac{a}{8\pi - 2}$.
$X = 8\pi - 2 = 23.13$, nearest integer 23.
Answer: 23. | ||||
| 23 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\rho = 9\times10^{-3}$ kg cm$^{-3}$ $= 9000$ kg m$^{-3}$. Stress $= Y\times$strain $= 9\times10^{10}\times4.9\times10^{-4} = 4.41\times10^7$ N m$^{-2}$.
$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\text{stress}}{\rho}} = \sqrt{\frac{4.41\times10^7}{9000}} = \sqrt{4900} = 70$ m/s.
Fundamental: $f = \frac{v}{2L} = \frac{70}{2} = 35$ Hz.
Answer: 35. | ||||
| 24 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity When charged, no current flows through C, so R5 carries none and the capacitor's right plate is at the battery's right-hand node N.
From the left node L: $R_1 + R_2 = 4\ \Omega$ in parallel with $R_3 = 2\ \Omega$ gives $\frac43\ \Omega$ to the middle node M, then $R_4 = 2\ \Omega$ to N. Total $\frac{10}{3}\ \Omega$, so $I = 3$ A.
$V_L - V_M = 3\times\frac43 = 4$ V, so the $R_1R_2$ branch carries 1 A and the drop across $R_1$ is 2 V.
Taking $V_L = 10$ V and $V_N = 0$: the left plate is at $10 - 2 = 8$ V and the right plate at 0. The capacitor holds 8 V.
Answer: 8. | ||||
| 25 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At entry $\sin r = \frac{\sin60^\circ}{\sqrt3} = \frac12$, $r = 30^\circ$. The chord makes equal angles with the two radii, so the ray meets the far surface at $30^\circ$.
The reflected ray makes $30^\circ$ with the normal inside; the refracted ray leaves at $60^\circ$ outside.
Angle between them $= 180^\circ - 30^\circ - 60^\circ = 90^\circ$.
Answer: 90. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics I to II: [B] doubles and the rate goes up 4 times, so the order in B is 2. I to III: [A] doubles and the rate doubles, so the order in A is 1.
Rate $= k[A][B]^2$; from I, $k = \frac{6\times10^{-3}}{0.1\times0.01} = 6$.
IV: $7.2\times10^{-2} = 6\times X\times0.04 \Rightarrow X = 0.3$.
V: $2.88\times10^{-1} = 6\times0.3\times Y^2 \Rightarrow Y^2 = 0.16 \Rightarrow Y = 0.4$.
Answer: (D). | ||||
| 27 | SolutionsColligative properties | Class 12: Solutions | Single correctEasy | |
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions The salt solution outside is hypertonic to the cell sap, so water passes out of the mango's cells through their semipermeable membranes into the solution, and the mango shrinks. This is osmosis (exosmosis).
Answer: (C). | ||||
| 28 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter (a) Valid: as the surface fills, fewer free sites with unbalanced forces remain, so less heat is released for each further amount adsorbed; $\Delta H$ becomes less negative.
(b) Valid: ammonia is more easily liquefied (higher critical temperature, stronger van der Waals forces) than nitrogen, so it is adsorbed more.
(c) Not valid: adsorption satisfies the residual forces, so they decrease.
(d) Not valid: adsorption is exothermic, so raising the temperature lowers the equilibrium amount adsorbed.
So (a) and (b).
Answer: (A). | ||||
| 29 | 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 S in SF$_4$ has 6 valence electrons: 4 bond pairs and 1 lone pair, 5 electron pairs in all ($sp^3d$).
Counting the lone pair, the arrangement is trigonal bipyramidal (the shape of the atoms alone is see-saw).
Answer: (D). | ||||
| 30 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom For $n = 4$, $l = 0, 1, 2, 3$. $m = -2$ needs $l \ge 2$, so only $l = 2$ (4d) and $l = 3$ (4f) qualify: 2 subshells.
Answer: (A). | ||||
| 31 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | MatchingMedium | |
A medium matching question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Ion-ion (Coulomb) energy varies as $\frac{1}{r}$.
Dipole-dipole energy between fixed (stationary) polar molecules varies as $\frac{1}{r^3}$.
London dispersion energy varies as $\frac{1}{r^6}$.
So (I)-(a), (II)-(c), (III)-(d).
Answer: (B). | ||||
| 32 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across period 3 the oxides change from basic (Na, Mg) through amphoteric (Al) to acidic (Si, P, S, Cl). So the basic-oxide element comes first and the acidic one last: X < Y < Z (for example Mg, Al, S).
Answer: (A). | ||||
| 33 | 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 Pig iron from the blast furnace is remelted with scrap iron and coke to give cast iron. Cast iron in turn is refined to wrought iron (in a reverberatory furnace lined with haematite) and to steel. Pig iron is its raw material, not its product, so the answer is wrought iron and steel.
Answer: (A). | ||||
| 34 | Chemistry in Everyday LifeCleansing agents | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on cleansing agents, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Toilet-cleaning liquids are strongly acidic (they contain hydrochloric acid). The acid is neutralised with a mild base that is safe on skin: aqueous sodium hydrogencarbonate. NaOH and NH$_3$ are themselves corrosive, and vinegar is an acid.
Answer: (C). | ||||
| 35 | 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 Lithium and magnesium are diagonally related and alike: neither forms a solid hydrogencarbonate, and both burn in nitrogen to form nitrides ($\mathrm{Li_3N}$, $\mathrm{Mg_3N_2}$). The other alkali metals form solid hydrogencarbonates and do not form nitrides directly.
Answer: (C). | ||||
| 36 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter [XeF$_5$]$^-$: the electrons about Xe are 8 (Xe) + 5 (one from each F) + 1 (the charge) = 14, i.e. 7 pairs: 5 bond pairs and 2 lone pairs ($sp^3d^3$); the lone pairs sit above and below the plane, so the shape is pentagonal planar.
XeO$_3$F$_2$: 5 $\sigma$ bonds and no lone pair on Xe ($sp^3d$), so it is trigonal bipyramidal (F atoms axial, O atoms equatorial).
Answer: (B). | ||||
| 37 | 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 The spectrochemical series puts $\mathrm{F^-} < \mathrm{NCS^-}$ (N-bonded) $< \mathrm{NH_3}$, so $\Delta_o$ is largest for the ammine and smallest for the fluoride.
$\lambda_{max}\propto\frac{1}{\Delta_o}$: the shortest wavelength (A) is (iii), the middle one (B) is (i), the longest (C) is (ii).
So A-(iii), B-(i), C-(ii).
Answer: (D). | ||||
| 38 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[Ni(NH_3)_2Cl_2]$: Ni(II) with these ligands is tetrahedral ($sp^3$); a tetrahedral $MA_2B_2$ has no geometrical or optical isomers.
$[Ni(NH_3)_4(H_2O)_2]^{2+}$: octahedral $MA_4B_2$, cis and trans.
$[Ni(en)_3]^{2+}$: optical isomers.
$[Pt(NH_3)_2Cl_2]$: square planar, cis and trans.
Answer: (A). | ||||
| 39 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $H_b$ is the carboxylic acid proton (pKa about 4-5): most acidic.
$H_c$ is a phenolic OH para to the nitro group, whose $-R$ effect stabilises the phenoxide: next (pKa about 7).
$H_d$ is a phenolic OH meta to the nitro group, helped only by its $-I$ effect: less acidic than $H_c$ but still a phenol (pKa about 8-9).
$H_a$ is a terminal alkyne C-H (pKa about 25): least acidic.
So b > c > d > a.
Answer: (A). | ||||
| 40 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The substrate is 2-methyl-5-nitrophenol. The OH group is the strongest activator and directs ortho and para: the free positions it activates are C4 (para) and C6 (ortho).
Both are ortho to the existing NO$_2$; C6 is also crowded between OH and NO$_2$, so substitution goes mainly para to OH, at C4.
The major product is 2-methyl-4,5-dinitrophenol.
Answer: (A). | ||||
| 41 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Hydrolysis of sucrose gives glucose and fructose. Seliwanoff's reagent (resorcinol in HCl) quickly gives a red (cherry-red) colour with a ketose; fructose is a ketose, so a red colour forms.
Answer: (A). | ||||
| 42 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Reaction (1) is SN1: rate $= k[\text{t-BuBr}]$ does not contain [OH$^-$], so the base's concentration has no effect on it.
Reaction (2) is E2: rate $= k[\text{t-BuBr}][OH^-]$, so it depends on the base's concentration and on the base itself (an alkoxide changes the rate).
Only the third statement is true.
Answer: (C). | ||||
| 43 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctHard | |
A hard single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers B gives a yellow precipitate with AgNO$_3$ (AgI), so B is an alkyl iodide, not a phenol: the ether must have an alkyl group on O that HI can take as iodide.
C tautomerises, so C is an enol, and D gives the iodoform test, so D has $\mathrm{CH_3CO-}$.
$\mathrm{C_6H_5CH_2-O-CH=CH_2 + HI \rightarrow C_6H_5CH_2I\ (B) + CH_2=CH-OH\ (C)}$, and $\mathrm{CH_2=CHOH \rightarrow CH_3CHO}$ (D), which gives iodoform. Its formula is C$_9$H$_{10}$O.
In the other three the O is bonded to the ring, so HI gives a phenol, which gives no precipitate with AgNO$_3$.
Answer: (B). | ||||
| 44 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Bromide is a far better leaving group than fluoride, so Br leaves from C3 and a $\beta$-H goes from C2 or C4.
The C2-H is made more acidic by the electron-withdrawing F on the same carbon, and removing it gives the more substituted (trisubstituted) alkene.
Major product: $\mathrm{CH_3CH_2CH=C(F)CH_3}$ (2-fluoropent-2-ene).
Answer: (B). | ||||
| 45 | Alcohols, Phenols and EthersPrimary, secondary and tertiary alcohols | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on primary, secondary and tertiary alcohols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers C$_3$H$_6$O with a Grignard reagent: propanal and propanone.
Propanal + CH$_3$MgBr gives butan-2-ol, $\mathrm{CH_3CH_2CH(OH)CH_3}$: secondary (Lucas turbidity in about five minutes) with a $\mathrm{CH_3CH(OH)-}$ group (iodoform positive). This is C.
Propanone + CH$_3$MgBr gives 2-methylpropan-2-ol, $\mathrm{(CH_3)_3COH}$: tertiary (turbidity at once) and iodoform negative. This is D.
Both are alcohols, so both give the ceric ammonium nitrate test.
Answer: (C). | ||||
| 46 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3\times10^8}{300\times10^{-9}} = 6.63\times10^{-19}$ J.
$KE = 6.63\times10^{-19} - 4.41\times10^{-19} = 2.22\times10^{-19}$ J $= 222\times10^{-21}$ J.
Answer: 222. | ||||
| 47 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $\mathrm{C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)}$: $\Delta n_g = 2 - 3 = -1$.
$\Delta U = \Delta H - \Delta n_gRT = -327000 - (-1)(2)(300) = -326400$ cal.
Heat evolved at constant volume: 326400 cal.
Answer: 326400. | ||||
| 48 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Reduction $\mathrm{Cu^+ \to Cu}$ (0.52 V) with oxidation $\mathrm{Cu^+ \to Cu^{2+}}$: $E^0_{cell} = 0.52 - 0.16 = 0.36$ V, with $n = 1$.
$\ln K = \frac{nFE^0}{RT} = \frac{0.36}{0.025} = 14.4 = 144\times10^{-1}$.
Answer: 144. | ||||
| 49 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements K$_2$Cr$_2$O$_7$: $2 + 2x - 14 = 0$, $x = +6$. KMnO$_4$: $1 + y - 8 = 0$, $y = +7$. K$_2$FeO$_4$: $2 + z - 8 = 0$, $z = +6$.
$x + y + z = 19$.
Answer: 19. | ||||
| 50 | Some Basic Concepts of ChemistryEmpirical and molecular formula | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Take 12 g of C: then H is 3 g and O is 16 g. Moles C : H : O $= 1 : 3 : 1$, empirical formula $\mathrm{CH_3O}$.
$\mathrm{CH_3O}$ itself cannot be a stable molecule; the saturated acyclic compound is $\mathrm{C_2H_6O_2}$ (ethane-1,2-diol).
$\mathrm{C_2H_6O_2 + \frac52O_2 \rightarrow 2CO_2 + 3H_2O}$, so 2 mol need 5 mol of O$_2$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series By induction $f(k) = kf(1) = 2k$ for every natural k.
$g(n) = \sum_{k=1}^{n-1}2k = (n-1)n = 20 \Rightarrow n = 5$.
Answer: (B). | ||||
| 52 | 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 Write $f(x) = a(x-3)(x-\beta)$, $a\neq0$.
$f(-1) = a(-4)(-1-\beta) = 4a(1+\beta)$ and $f(2) = a(-1)(2-\beta) = a(\beta-2)$.
$4 + 4\beta + \beta - 2 = 0 \Rightarrow \beta = -\frac25$, which lies in $(-1, 0)$.
Answer: (B). | ||||
| 53 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $2\sqrt{-54} = 6\sqrt6\,i$. Let $(x+iy)^2 = 3 + 6\sqrt6\,i$: $x^2 - y^2 = 3$, $xy = 3\sqrt6$, so $x^2 = 9$, $y^2 = 6$ and $\sqrt{3+6\sqrt6 i} = \pm(3 + \sqrt6 i)$. Likewise $\sqrt{3-6\sqrt6 i} = \pm(3 - \sqrt6 i)$.
Taking each sign independently, the difference $s_1(3+\sqrt6 i) - s_2(3-\sqrt6 i)$ ($s_1, s_2 = \pm1$) has imaginary part $(s_1 + s_2)\sqrt6$: $2\sqrt6$, $0$ or $-2\sqrt6$, depending on the signs chosen.
Of the options only $-2\sqrt6$ is possible (from $-(3+\sqrt6 i) - (3-\sqrt6 i) = -6 - 2\sqrt6 i$).
Answer: (C). | ||||
| 54 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | Single correctHard | |
A hard single correct question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A$ is symmetric, so $A^TA = A^2 = I$. Comparing entries: $a^2 + b^2 + c^2 = 1$ and $ab + bc + ca = 0$, so $(a+b+c)^2 = 1$.
Then $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (a+b+c) = \pm1$.
$2 - 3abc = \pm1 \Rightarrow abc = \frac13$ or $abc = 1$. Of the options, $\frac13$.
Answer: (B). | ||||
| 55 | 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 $|P| = 1(-3+36) - 2(2+4) + 1(-18-3) = 33 - 12 - 21 = 0$, and the first two rows are independent, so the rank is 2.
The solutions of $PX = 0$ form a line through the origin, $X = tX_0$. On the unit sphere this gives $t = \pm\frac{1}{|X_0|}$: exactly two points.
Answer: (C). | ||||
| 56 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Blue lines (neighbours on the circle): $n$. Red lines: ${}^nC_2 - n = \frac{n(n-3)}{2}$.
$\frac{n(n-3)}{2} = 99n \Rightarrow n - 3 = 198 \Rightarrow n = 201$.
Answer: (A). | ||||
| 57 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_{11} = \frac{11}{2}(2a_1 + 10d) = 0 \Rightarrow d = -\frac{a_1}{5}$.
$a_1, a_3, \ldots, a_{23}$ is an A.P. of 12 terms with difference $2d$: sum $= 6(2a_1 + 11\cdot2d) = 6\left(2a_1 - \frac{22a_1}{5}\right) = -\frac{72}{5}a_1$.
Answer: (D). | ||||
| 58 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S = (x + x^2 + \ldots + x^9) + a\sum_{i=0}^{8}(k + 2i) = \frac{x^{10} - x}{x-1} + a(9k + 72)$.
The given $S = \frac{x^{10}-x}{x-1} + 45a$, so $9k + 72 = 45 \Rightarrow k = -3$.
Answer: (A). | ||||
| 59 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability A $1^\infty$ form: the limit is $e^L$ with $L = \lim_{x\to0}\frac{\tan(\frac\pi4 + x) - 1}{x} = \lim\frac{1}{x}\cdot\frac{2\tan x}{1-\tan x} = 2$.
So the limit is $e^2$.
Answer: (D). | ||||
| 60 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives At $x = 0$: $y = 1 + \cos^2 0 = 2$. Also $\cos^2(\sin^{-1}x) = 1 - x^2$.
$y' = (1+x)^{2y}\left[2y'\ln(1+x) + \frac{2y}{1+x}\right] - 2x$; at $(0, 2)$: $y' = 1\cdot[0 + 4] - 0 = 4$.
Normal slope $-\frac14$: $y - 2 = -\frac{x}{4}$, i.e. $x + 4y = 8$.
Answer: (A). | ||||
| 61 | 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 $f'(x) = \frac{g(x)}{x^2}$ with $g(x) = \frac{x}{1+x} - \ln(1+x)$.
$g(0) = 0$ and $g'(x) = \frac{1}{(1+x)^2} - \frac{1}{1+x} = -\frac{x}{(1+x)^2}$: $g$ rises on $(-1, 0)$ up to 0 and falls after 0, so $g(x) < 0$ for all $x\neq0$.
Hence $f' < 0$ on both sides, and $f$ (continuous at 0, as $f(x)\to1$) decreases on $(-1, \infty)$.
Answer: (D). | ||||
| 62 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The region lies between $y = x^2$ and $y = 2x$ for $0\le y\le4$; for a given $y$, $x$ runs from $\frac y2$ to $\sqrt y$.
Total area $= \int_0^4\left(\sqrt y - \frac y2\right)dy = \frac{16}{3} - 4 = \frac43$.
$\int_0^\alpha\left(\sqrt y - \frac y2\right)dy = \frac23\alpha^{3/2} - \frac{\alpha^2}{4} = \frac23 \Rightarrow 8\alpha^{3/2} - 3\alpha^2 = 8$, i.e. $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$.
Answer: (C). | ||||
| 63 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $y = vx$: $v + xv' = v + \frac{v^2}{2} \Rightarrow \frac{2dv}{v^2} = \frac{dx}{x} \Rightarrow -\frac2v = \ln x + C$.
At (1, 2), $v = 2$: $C = -1$. So $\frac{2x}{y} = 1 - \ln x$, $y = \frac{2x}{1 - \ln x}$.
$f\left(\frac12\right) = \frac{1}{1 - \ln\frac12} = \frac{1}{1 + \ln2}$.
Answer: (A). | ||||
| 64 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines For (1, 2): $1 + 2 - 1 = 2 > 0$. So we need $\sin\theta + \cos\theta - 1 > 0$, i.e. $\sqrt2\sin\left(\theta + \frac\pi4\right) > 1$.
With $\theta + \frac\pi4 \in \left(\frac\pi4, \frac{5\pi}{4}\right)$, this needs $\theta + \frac\pi4 \in \left(\frac\pi4, \frac{3\pi}{4}\right)$, so $\theta\in\left(0, \frac\pi2\right)$.
Answer: (D). | ||||
| 65 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections By symmetry the other two vertices are $(h, \pm k)$ with each side from the origin at $30^\circ$ to the axis: $k = h\tan30^\circ = \frac{h}{\sqrt3}$.
On the parabola: $\frac{h^2}{3} = 8h \Rightarrow h = 24$, $k = 8\sqrt3$. Side $= 2k = 16\sqrt3$.
Area $= \frac{\sqrt3}{4}(16\sqrt3)^2 = \frac{\sqrt3}{4}\times768 = 192\sqrt3$.
Answer: (C). | ||||
| 66 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Hyperbola $\frac{x^2}{10} - \frac{y^2}{10\cos^2\theta} = 1$: $e_h^2 = 1 + \cos^2\theta$.
Ellipse $\frac{x^2}{5\cos^2\theta} + \frac{y^2}{5} = 1$, major axis along y ($a^2 = 5$, $b^2 = 5\cos^2\theta$): $e_e^2 = 1 - \cos^2\theta$.
$1 + \cos^2\theta = 5(1 - \cos^2\theta) \Rightarrow \cos^2\theta = \frac23$, so $b^2 = \frac{10}{3}$.
Latus rectum $= \frac{2b^2}{a} = \frac{20/3}{\sqrt5} = \frac{4\sqrt5}{3}$.
Answer: (C). | ||||
| 67 | 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 Normal $= (1, -2, 2)\times(2, 3, -1) = (2 - 6,\ 4 + 1,\ 3 + 4) = (-4, 5, 7)$.
Plane: $-4(x-3) + 5(y-1) + 7(z-1) = 0$. At $(\alpha, -3, 5)$: $-4\alpha + 12 - 20 + 28 = 0 \Rightarrow \alpha = 5$.
Answer: (C). | ||||
| 68 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(E_2^C\cap E_3^C\cap E_1) = P(E_1) - P(E_1\cap E_2) - P(E_1\cap E_3) + P(E_1\cap E_2\cap E_3) = P(E_1)[1 - P(E_2) - P(E_3)]$, using pairwise independence and the zero triple intersection.
Dividing by $P(E_1)$: $1 - P(E_2) - P(E_3) = P(E_3^C) - P(E_2)$.
Answer: (C). | ||||
| 69 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - \frac12\sin^22\theta$, which takes every value in $\left[\frac12, 1\right]$.
So $\lambda = -(\cos^4\theta + \sin^4\theta) \in \left[-1, -\frac12\right]$.
Answer: (B). | ||||
| 70 | 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) is $p\leftrightarrow q$, false when $p$ and $q$ differ.
(B) If $\sim p$ and $p\vee q$ are both true, $p$ is false so $q$ must be true: the implication never fails. Tautology.
(C) With $q$ false, $\sim q$ is true, so the premise is true and the conclusion false: not a tautology.
(D) With $p$ false and $q$ true: $q\to p$ is false and $p\to q$ is true, so both parts are false: not a tautology.
Answer: (B). | ||||
| 71 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The coefficients are binomial coefficients; take ${}^nC_{r-1} : {}^nC_r : {}^nC_{r+1} = 2 : 5 : 12$.
$\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r} = \frac52 \Rightarrow 2n + 2 = 7r$.
$\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1} = \frac{12}{5} \Rightarrow 5n = 17r + 12$.
From the first, $n = \frac{7r-2}{2}$; then $\frac{35r - 10}{2} = 17r + 12 \Rightarrow r = 34$, $n = 118$.
Answer: 118. | ||||
| 72 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $\cos\alpha = \frac35$, $\sin\alpha = \frac45$ ($0 < \alpha < \frac\pi2$). Then $\frac35\cos kx - \frac45\sin kx = \cos(kx + \alpha)$.
Near $x = 0$, $kx + \alpha\in(0, \pi)$, so $\cos^{-1}\cos(kx + \alpha) = kx + \alpha$ and $y = \sum_{k=1}^6 k(kx + \alpha)$.
$\frac{dy}{dx} = \sum_{k=1}^6k^2 = 91$.
Answer: 91. | ||||
| 73 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{OP} = \frac{\lambda\vec{OB} + \vec{OA}}{\lambda + 1}$.
$\vec{OB}\cdot\vec{OP} = \frac{14\lambda + 6}{\lambda + 1}$ (as $|\vec{OB}|^2 = 14$, $\vec{OA}\cdot\vec{OB} = 6$).
$\vec{OA}\times\vec{OP} = \frac{\lambda}{\lambda+1}\vec{OA}\times\vec{OB} = \frac{\lambda}{\lambda+1}(2\hat i - \hat j - \hat k)$, with square magnitude $\frac{6\lambda^2}{(\lambda+1)^2}$.
$\frac{14\lambda + 6}{\lambda+1} - \frac{18\lambda^2}{(\lambda+1)^2} = 6 \Rightarrow (14\lambda+6)(\lambda+1) - 18\lambda^2 = 6(\lambda+1)^2 \Rightarrow 10\lambda^2 - 8\lambda = 0$.
$\lambda > 0$, so $\lambda = 0.8$.
Answer: 0.8. | ||||
| 74 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $[1, \frac43)$, $[3x] = 3$ and $|2x - 3| = 3 - 2x$: $\int = \left[3x - x^2\right]_1^{4/3} = \frac29$.
On $[\frac43, \frac53)$, $[3x] = 4$ and $|2x - 4| = 4 - 2x$: $\int = \left[4x - x^2\right]_{4/3}^{5/3} = \frac13$.
On $[\frac53, 2)$, $[3x] = 5$ and $|2x - 5| = 5 - 2x$: $\int = \left[5x - x^2\right]_{5/3}^{2} = \frac49$.
Total $= \frac29 + \frac39 + \frac49 = 1$.
Answer: 1. | ||||
| 75 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics The variance of an A.P. of n terms with difference d is $\frac{(n^2-1)d^2}{12}$ (the same as for $d, 2d, \ldots, nd$).
$\frac{120d^2}{12} = 10d^2 = 90 \Rightarrow d = 3$ (increasing, so $d > 0$).
Answer: 3. | ||||
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