JEE (Main) · 2020
JEE (Main) 2020 — 4 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
- Motion of System of Particles and Rigid Body2
- Properties of Bulk Matter2
- Electrostatics2
14 easy · 10 medium · 1 hard
Chemistry
25 Q
- Chemical Bonding and Molecular Structure2
- Coordination Compounds2
- Haloalkanes and Haloarenes2
13 easy · 9 medium · 3 hard
Mathematics
25 Q
- Matrices and Determinants2
- Differential Equations2
- Trigonometry2
11 easy · 14 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 | 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 $[I] = ML^2$, $[F] = MLT^{-2}$, $[v^2] = L^2T^{-2}$, $[W] = ML^2T^{-2}$, $[L^4] = L^4$.
$[x] = \frac{ML^2\cdot MLT^{-2}\cdot L^2T^{-2}}{ML^2T^{-2}\cdot L^4} = ML^{-1}T^{-2}$.
Energy per unit volume is $\frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$, the same. (Force constant is $MT^{-2}$, viscosity $ML^{-1}T^{-1}$, Planck's constant $ML^2T^{-1}$.)
Answer: (B). | ||||
| 2 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion On the way up both gravity and the drag act downward: $v\frac{dv}{dh} = -(g + kv^2)$.
$H = \int_0^u \frac{v\,dv}{g + kv^2} = \frac{1}{2k}\ln\frac{g+ku^2}{g} = \frac{1}{2k}\ln\left(1+\frac{ku^2}{g}\right)$.
Answer: (C). | ||||
| 3 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Work = area under the force-distance graph.
First 15 m: $200\times15 = 3000$ J. Next 15 m, force falling linearly from 200 N to 100 N: $\frac{200+100}{2}\times15 = 2250$ J.
Total $= 5250$ J.
Answer: (D). | ||||
| 4 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Same density and thickness: $M = \rho t\pi R^2$, so $I = \frac12MR^2 \propto R^4$.
$\frac{I_1}{I_2} = \frac{1}{\alpha^4} = \frac{1}{16} \Rightarrow \alpha = 2$.
Answer: (B). | ||||
| 5 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion About the centre, perpendicular to the sheet: $I_O = \frac{M(a^2+b^2)}{12}$ with $a = 60$ cm, $b = 80$ cm.
The corner is at distance $\frac12\sqrt{a^2+b^2}$ from O, so $I_{O'} = \frac{M(a^2+b^2)}{12} + \frac{M(a^2+b^2)}{4} = \frac{M(a^2+b^2)}{3}$.
$\frac{I_O}{I_{O'}} = \frac{1}{4}$.
Answer: (C). | ||||
| 6 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Orbital speed at radius R: $v_o = \sqrt{\frac{GM}{R}}$. Escape speed: $v_e = \sqrt{\frac{2GM}{R}}$.
$\frac{v_o}{v_e} = \frac{1}{\sqrt2}$.
Answer: (C). | ||||
| 7 | Properties of Bulk MatterElasticity | 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 elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $\frac{\Delta V}{V} = \frac{P}{B} = \frac{4\times10^9}{8\times10^{10}} = 0.05 = 5\%$.
For a cube $\frac{\Delta V}{V} = 3\frac{\Delta l}{l}$, so $\frac{\Delta l}{l} = \frac{5}{3}\% \approx 1.67\%$.
Answer: (C). | ||||
| 8 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter A column of height x has mass $dSx$ with its centre at $x/2$: potential energy $\frac12 dSgx^2$.
Initially $U_i = \frac12 dSg(x_1^2+x_2^2)$. Finally both heights are $\frac{x_1+x_2}{2}$: $U_f = 2\times\frac12 dSg\left(\frac{x_1+x_2}{2}\right)^2 = \frac14 dSg(x_1+x_2)^2$.
$U_i - U_f = \frac14 dSg\left[2x_1^2+2x_2^2-(x_1+x_2)^2\right] = \frac14 gdS(x_2-x_1)^2$ (a decrease, lost to viscous dissipation).
Answer: (D). | ||||
| 9 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: no heat exchange, $\Delta Q = 0$ (B). Isothermal (ideal gas): $\Delta U = 0$ (D). Isochoric: no volume change, $\Delta W = 0$ (A). Isobaric: heat, work and internal energy all change (C).
Answer: (B). | ||||
| 10 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The wall receives $f\frac{c}{c-v}$ and reflects it; the moving driver hears $f' = f\frac{c+v}{c-v}$.
$490(330 - v) = 420(330 + v) \Rightarrow 910v = 70\times330 \Rightarrow v = 25.4$ m s$^{-1}$.
$25.4\times3.6 = 91.4$ km h$^{-1}\approx 91$ km h$^{-1}$.
Answer: (C). | ||||
| 11 | 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 Work done by the field from 0 to x: $W = \int_0^x qE_0(1-ax^2)\,dx = qE_0\left(x - \frac{ax^3}{3}\right)$.
KE is zero again when $W = 0$: $x\left(1-\frac{ax^2}{3}\right) = 0 \Rightarrow x = \sqrt{\frac3a}$.
Answer: (D). | ||||
| 12 | 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 $CV_0$ is shared: common voltage $V = \frac{CV_0}{3C/2} = \frac{2V_0}{3}$.
$U_i = \frac12CV_0^2$; $U_f = \frac12\cdot\frac{3C}{2}\cdot\frac{4V_0^2}{9} = \frac13CV_0^2$.
Loss $= \frac12CV_0^2 - \frac13CV_0^2 = \frac16CV_0^2$.
Answer: (D). | ||||
| 13 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctHard | |
A hard 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 The loss of magnetic potential energy $U = -mB\cos\theta$ becomes rotational kinetic energy.
Taking the intended working, the energy released over the 60$^\circ$ turn is $mB(1 - \cos60^\circ) = 20\times4\times\frac12 = 40$ J.
$\frac12I\omega^2 = 40 \Rightarrow \omega^2 = \frac{80}{0.8} = 100 \Rightarrow \omega = 10$ rad s$^{-1}$.
Answer: (B). | ||||
| 14 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Curie's law: $M \propto \frac{B}{T}$.
$M_2 = 6\times\frac{0.3}{0.4}\times\frac{4}{24} = 6\times0.75\times\frac16 = 0.75$ A/m.
Answer: (D). | ||||
| 15 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $U = \frac12Li^2$, so $U = \frac{U_{max}}{n}$ when $i = \frac{I_0}{\sqrt n}$.
$I_0(1 - e^{-Rt/L}) = \frac{I_0}{\sqrt n} \Rightarrow e^{-Rt/L} = \frac{\sqrt n - 1}{\sqrt n}$.
$t = \frac{L}{R}\ln\left(\frac{\sqrt n}{\sqrt n - 1}\right)$.
Answer: (C). | ||||
| 16 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves The wave travels along $+\hat{z}$, and $\vec{B} = \frac{1}{c}\hat{z}\times\vec{E}$, in phase with $\vec{E}$.
$\hat{z}\times(\hat{x}+\hat{y}) = \hat{y} - \hat{x}$, so $\vec{B} = \frac{E_0}{c}(-\hat{x}+\hat{y})\sin(kz-\omega t)$.
Check: $\vec{E}\times\vec{B} \propto (\hat{x}+\hat{y})\times(-\hat{x}+\hat{y}) = 2\hat{z}$, along the propagation.
Answer: (C). | ||||
| 17 | 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 $eV = \frac{hc}{\lambda} - \phi$: the stopping potential depends only on the wavelength and the work function. The slope $\frac{hc}{e}$ and the intercept (threshold) do not depend on intensity, which changes only the photocurrent.
Answer: (A). | ||||
| 18 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei 50 protons: $50\times1.00783 = 50.3915$ u; 70 neutrons: $70\times1.00867 = 70.6069$ u; total 120.9984 u.
Mass defect $= 120.9984 - 119.902199 = 1.096201$ u; binding energy $= 1.096201\times931 \approx 1020.6$ MeV.
Per nucleon: $\frac{1020.6}{120} \approx 8.5$ MeV.
Answer: (C). | ||||
| 19 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Each input passes through a NOR gate with its inputs joined, which acts as NOT: the final NOR gate receives $\bar A$, $\bar B$, $\bar C$.
Output $= \overline{\bar A + \bar B + \bar C} = A\cdot B\cdot C$ (De Morgan), an AND operation.
Answer: (B). | ||||
| 20 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity The 8 V cell (no internal resistance shown) is connected straight across A and C, so $V_A - V_C = 8$ V whatever the rest of the network does.
The A-C branch is the 4 $\Omega$ and 4 $\Omega$ resistors in series (it crosses the B-D wire without a junction), 8 $\Omega$ in all.
$i_1 = \frac{8}{8} = 1$ A.
Answer: (A). | ||||
| 21 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The speed rises linearly from 0 at $t = 0$ to 8 m s$^{-1}$ at $t = 5$ s (the marked point).
Distance = area under the graph $= \frac12\times5\times8 = 20$ m.
Answer: 20. | ||||
| 22 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | NumericalMedium | |
A medium numerical question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Constant T: $PV = $ const, $|\Delta V| = \frac{V}{P}|\Delta P|$. Constant P: $\frac{V}{T} = $ const, $|\Delta V| = \frac{V}{T}|\Delta T|$.
Equal changes: $|\Delta T| = \frac{T}{P}|\Delta P| = \frac{300}{2}|\Delta P|$, so $C = 150$ K/atm.
Answer: 150. | ||||
| 23 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Take $V_A = 40$ V, $V_C = 0$.
Path ABC: $40 + 60 = 100\ \Omega$, current 0.4 A, $V_B = 40 - 0.4\times40 = 24$ V.
Path ADC: $90 + 110 = 200\ \Omega$, current 0.2 A, $V_D = 40 - 0.2\times90 = 22$ V.
$V_B - V_D = 2$ V.
Answer: 2. | ||||
| 24 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Displacement method: $f = \frac{D^2 - d^2}{4D} = \frac{100^2 - 40^2}{4\times100} = \frac{8400}{400} = 21$ cm.
(Check: object distances 30 cm and 70 cm; $\frac{1}{70}+\frac{1}{30} = \frac{1}{21}$.)
$P = \frac{100}{21} = 4.76$ D $= \frac{476}{100}$ D, so $N = 476$.
Answer: 476. | ||||
| 25 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Minima: $a\sin\theta = n\lambda$, so $\sin\theta = \frac{n\lambda}{a} = \frac{n\times6\times10^{-7}}{6\times10^{-5}} = \frac{n}{100}$.
A minimum needs $\sin\theta < 1$: $n = 1, 2, \ldots, 99$ (n = 100 would be at $\theta = 90^\circ$, along the screen, and is not formed on it).
99 on each side: $2\times99 = 198$.
Answer: 198. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Egg white (albumin) is a lyophilic colloid. It forms a protective layer round the dispersed dye particles and stops them coagulating: it acts as a protective colloid and keeps the ink stable. Eosin is the colouring matter, HCHO a preservative and water the medium.
Answer: (B). | ||||
| 27 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\mathrm{Au^+}$ has the higher reduction potential, so it is reduced first.
Charge passed $= 1\times15\times60 = 900$ C $= \frac{900}{96500} = 9.3\times10^{-3}$ mol of electrons.
$\mathrm{Au^+}$ present $= 0.1\times0.250 = 0.025$ mol, more than $9.3\times10^{-3}$ mol, so the gold is never used up and silver does not begin to deposit.
Answer: (D). | ||||
| 28 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Adding the two reactions gives $\mathrm{A \rightleftharpoons P}$, so the constants multiply: $K = \frac{[B][C]}{[A]}\cdot\frac{[P]}{[B][C]} = \frac{[P]}{[A]} = K_{eq}^{(1)}K_{eq}^{(2)}$.
Answer: (C). | ||||
| 29 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics In free expansion the external pressure is zero, so $w = -p_{ext}\Delta V = 0$.
Answer: (A). | ||||
| 30 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{[Ni(CN)_4]^{2-}}$: square planar, $dsp^2$, one d orbital.
$\mathrm{BrF_5}$ and $\mathrm{XeF_4}$: $sp^3d^2$, two d orbitals. $\mathrm{[CrF_6]^{3-}}$: $d^2sp^3$, two d orbitals.
Answer: (A). | ||||
| 31 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom H, Lyman, shortest ($\infty \to 1$): $\frac{1}{\lambda_1} = R$.
He$^+$ ($Z = 2$), Balmer, longest ($3 \to 2$): $\frac{1}{\lambda} = 4R\left(\frac14 - \frac19\right) = \frac{5R}{9}$.
$\lambda = \frac{9}{5R} = \frac{9\lambda_1}{5}$.
Answer: (C). | ||||
| 32 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Adding an electron to H releases energy ($\Delta_{eg}H \approx -73$ kJ mol$^{-1}$): exothermic.
A noble gas does not want an extra electron (positive $\Delta_{eg}H$), adding an electron to $\mathrm{O^-}$ works against repulsion (positive), and ionisation always needs energy.
Answer: (A). | ||||
| 33 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter $\mathrm{BeSO_4}$ is soluble (high hydration enthalpy of the small $\mathrm{Be^{2+}}$), $\mathrm{Be(OH)_2}$ is insoluble, and BeO is a very stable oxide with the wurtzite (4:4) structure, not rock salt.
MgO, CaO and SrO all have the rock-salt structure; $\mathrm{CaSO_4}$ and $\mathrm{SrSO_4}$ are sparingly soluble.
Answer: (B). | ||||
| 34 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{XeF_4 + SbF_5 \rightarrow [XeF_3]^+[SbF_6]^-}$: Xe goes from $sp^3d^2$ (4 bonds, 2 lone pairs) to $sp^3d$ (3 bonds, 2 lone pairs) - changed.
$\mathrm{H_2SO_4 \rightarrow NaHSO_4}$: S stays $sp^3$. $\mathrm{NH_3 \rightarrow NH_4^+}$: N stays $sp^3$. $\mathrm{H_3PO_2 \rightarrow PH_3 + H_3PO_3}$: P stays $sp^3$.
Answer: (A). | ||||
| 35 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements (a) Correct: down a group of the d-block the higher oxidation states become more stable, so W(VI) is more stable than Cr(VI) (which is a strong oxidant).
(b) Incorrect: permanganate oxidises HCl to chlorine, so titrations in HCl give high, unreliable results.
(c) Correct: lanthanoid oxides are used as phosphors (for example in television screens and fluorescent coatings).
Answer: (B). | ||||
| 36 | 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 Co(III) is $d^6$. With $\Delta_0 < P$ the complex is high spin: $t_{2g}^4e_g^2$.
CFSE $= 4(-0.4\Delta_0) + 2(0.6\Delta_0) = -0.4\Delta_0$.
The one electron pair in $t_{2g}$ is also present in the free ion, so no extra pairing energy is counted.
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 $\mathrm{[Pd(gly)_2]}$: Pd(II) $d^8$, square planar, 0 unpaired electrons.
$\mathrm{[Fe(en)(bpy)(NH_3)_2]^{2+}}$: Fe(II) $d^6$ with strong-field N donors, low spin $t_{2g}^6$, 0 unpaired.
$\mathrm{[Co(OX)_2(OH)_2]^-}$: Co(III) $d^6$ with $\Delta_0 > P$, low spin, 0 unpaired.
$\mathrm{[Ti(NH_3)_6]^{3+}}$: Ti(III) $d^1$, 1 unpaired electron - the only paramagnetic one.
Answer: (D). | ||||
| 38 | Isolation of MetalsRoasting and calcination | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on roasting and calcination, from Isolation of Metals. NCERT: No matching NCERT chapter Calcination decomposes carbonates and gives off $\mathrm{CO_2}$, a greenhouse gas (global warming).
Roasting of sulphide ores gives off $\mathrm{SO_2}$, which forms acid rain.
Answer: (A). | ||||
| 39 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes In vinyl chloride ($\mathrm{CH_2=CH-Cl}$) the C$-$Cl carbon is $sp^2$, and a lone pair of Cl is delocalised into the double bond, giving the C$-$Cl bond partial double-bond character.
In the other three the C$-$Cl carbon is $sp^3$ and the bond is a plain single bond.
Answer: (C). | ||||
| 40 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The precipitate (AgBr) forms fastest from the compound that ionises most easily ($\mathrm{S_N1}$).
In (A), (B) and (C) the Br is on a carbon next to a lone-pair donor, so loss of $\mathrm{Br^-}$ gives a resonance-stabilised iminium or oxocarbenium ion. Nitrogen is a better donor than oxygen, so (C) is slower than the amines.
In (A) the nitrogen lone pair is partly delocalised into the phenyl ring; in (B) the piperidine nitrogen is a plain trialkyl amine with its lone pair fully available, so $\mathrm{CH_2=N^+R_2}$ forms most readily.
In (D) the Br is two carbons from N, so no such direct stabilisation arises.
Answer: (B). | ||||
| 41 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers HI cleaves the ether: iodide goes to the smaller alkyl group ($\mathrm{S_N2}$ at ethyl), giving $\mathrm{CH_3CH_2I}$ and the alcohol [A] $= \mathrm{CH_3CH_2CH(CH_3)CH_2OH}$ (2-methylbutan-1-ol).
With conc. $\mathrm{H_2SO_4}$ and heat the primary carbocation (or the protonated alcohol) undergoes a 1,2-hydride shift to the tertiary carbocation $\mathrm{CH_3CH_2\overset{+}{C}(CH_3)_2}$.
Loss of a proton by Saytzeff's rule gives the more substituted alkene $\mathrm{CH_3-CH=C(CH_3)-CH_3}$ (2-methylbut-2-ene).
Answer: (A). | ||||
| 42 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines p-Toluidine $\xrightarrow{\mathrm{NaNO_2/HCl},\ 0-5^\circ C}$ diazonium salt $\xrightarrow{\mathrm{Cu_2Cl_2/HCl}}$ [A] = p-chlorotoluene (Sandmeyer).
$\mathrm{Cl_2}/h\nu$ substitutes the side chain (free radical): [B] = p-chlorobenzyl chloride, $\mathrm{Cl-C_6H_4-CH_2Cl}$.
Na / dry ether (Wurtz coupling): the reactive benzylic C$-$Cl couples, the aryl C$-$Cl does not, giving $\mathrm{Cl-C_6H_4-CH_2-CH_2-C_6H_4-Cl}$.
Answer: (D). | ||||
| 43 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Terfenadine is an antihistamine: it binds to the histamine receptor and blocks it, so histamine cannot act. It does not change how much histamine is secreted.
Answer: (D). | ||||
| 44 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons [P]: the acetylide displaces Br: $\mathrm{HC{\equiv}C-CH(CH_3)-CH(CH_3)_2}$ (3,4-dimethylpent-1-yne).
Hydration (Markovnikov) gives the methyl ketone $\mathrm{CH_3CO-CH(CH_3)-CH(CH_3)_2}$; $\mathrm{NaBH_4}$ reduces it to [Q] $= \mathrm{CH_3CH(OH)-CH(CH_3)-CH(CH_3)_2}$.
Conc. $\mathrm{H_2SO_4}$: the secondary carbocation at C-2 undergoes a 1,2-hydride shift to the tertiary C-3. Loss of the proton on C-4 gives the tetrasubstituted, most stable alkene $\mathrm{CH_3CH_2-C(CH_3)=C(CH_3)_2}$ (2,3-dimethylpent-2-ene).
Answer: (C). | ||||
| 45 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{CH_2=CH-CHO \xrightarrow{NaBH_4} CH_2=CH-CH_2OH \xrightarrow{SOCl_2}}$ [A] $= \mathrm{CH_2=CH-CH_2Cl}$.
Friedel-Crafts alkylation of benzene: [B] $= \mathrm{C_6H_5-CH_2-CH=CH_2}$ (allylbenzene).
DBr adds ionically: $\mathrm{D^+}$ goes to the terminal carbon, giving the secondary cation $\mathrm{C_6H_5-CH_2-\overset{+}{C}H-CH_2D}$. A 1,2-hydride shift from the benzylic $\mathrm{CH_2}$ gives the more stable, resonance-stabilised benzylic cation $\mathrm{C_6H_5-\overset{+}{C}H-CH_2-CH_2D}$.
$\mathrm{Br^-}$ captures it: [C] $= \mathrm{C_6H_5-CHBr-CH_2-CH_2D}$.
Answer: (C). | ||||
| 46 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Equivalents $= 0.1\times0.100 = 0.01$. $\mathrm{Na_2CO_3}$ has n-factor 2, so moles $= 0.005$.
Molar mass $= \frac{1.43}{0.005} = 286 = 106 + 18x \Rightarrow x = 10$.
Answer: 10. | ||||
| 47 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Osmotic pressure is proportional to the total concentration of particles; at the same temperature the contributions simply add in proportion to volume.
$\pi = \frac{0.10\times1 + 0.20\times2}{3} = \frac{0.50}{3} = 0.1667$ atm $= 166.7\times10^{-3}$ atm.
Nearest integer: 167.
Answer: 167. | ||||
| 48 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The fraction of molecules above the threshold is $e^{-E_a/RT}$, so $\ln\frac{f_2}{f_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$, with $T_1 = 300$ K, $T_2 = 315$ K.
$\frac{1}{300} - \frac{1}{315} = \frac{15}{94500} = \frac{1}{6300}$.
$E_a = 1.6094\times8.314\times6300 = 84297.5$ J mol$^{-1}$ (about 84.3 kJ mol$^{-1}$).
Answer: 84297.48. | ||||
| 49 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Basic medium: $\mathrm{2Fe^{2+} + H_2O_2 \rightarrow 2Fe^{3+} + 2OH^-}$, so $x = 2$, $y = 2$.
Acidic medium: $\mathrm{2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2}$, so $x' = 2$, $y' = 8$, $z' = 5$ (check O: $8 + 10 = 18 = 8 + 10$; H: $6 + 10 = 16$).
Sum $= 2 + 2 + 2 + 8 + 5 = 19$.
Answer: 19. | ||||
| 50 | BiomoleculesProteins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Threonine is $\mathrm{CH_3-CH(OH)-CH(NH_2)-COOH}$.
C-2 (bearing $\mathrm{NH_2}$, H, COOH and the side chain) and C-3 (bearing OH, H, $\mathrm{CH_3}$ and the rest) each carry four different groups.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | 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 Count the memberships two ways. In the $X_i$: $50\times10 = 20|T| \Rightarrow |T| = 25$.
In the $Y_i$: $5n = 6|T| = 150 \Rightarrow n = 30$.
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 $\alpha$ satisfies both: $3\alpha^2 - 3\alpha + 6\lambda = 0$ and $3\alpha^2 - 10\alpha + 27\lambda = 0$. Subtracting, $7\alpha = 21\lambda$, $\alpha = 3\lambda$.
Then $9\lambda^2 - 3\lambda + 2\lambda = 0 \Rightarrow \lambda = \frac19$ (as $\lambda \neq 0$), $\alpha = \frac13$.
$\beta = 1 - \alpha = \frac23$; $\alpha\gamma = 9\lambda = 1 \Rightarrow \gamma = 3$.
$\frac{\beta\gamma}{\lambda} = \frac{2}{1/9} = 18$.
Answer: (C). | ||||
| 53 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha = \omega$, with $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
$(2+\omega)^2 = 4 + 4\omega + \omega^2 = 3 + 3\omega = -3\omega^2$.
$(2+\omega)^4 = 9\omega^4 = 9\omega$, so $a = 0$, $b = 9$ and $a + b = 9$.
Answer: (A). | ||||
| 54 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}1&1&1\\2&4&-1\\3&2&\lambda\end{vmatrix} = (4\lambda+2) - (2\lambda+3) + (4-12) = 2\lambda - 9 = 0 \Rightarrow \lambda = \frac92$.
The third equation must be a combination $a(\text{eq 1}) + b(\text{eq 2})$: $a + 2b = 3$, $a + 4b = 2 \Rightarrow b = -\frac12$, $a = 4$; then the z-coefficient is $a - b = \frac92 = \lambda$, and $\mu = 2a + 6b = 8 - 3 = 5$.
$2\lambda + \mu = 9 + 5 = 14$.
Answer: (D). | ||||
| 55 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Put the columns together: $A[x_1\ x_2\ x_3] = [b_1\ b_2\ b_3]$.
$\det[x_1\ x_2\ x_3] = \begin{vmatrix}1&0&0\\1&2&0\\1&1&1\end{vmatrix} = 2$; $\det[b_1\ b_2\ b_3] = 1\times2\times2 = 4$.
$\det A = \frac42 = 2$.
Answer: (B). | ||||
| 56 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Let $N = n + 5$ and the coefficients be $^NC_{r-1}, ^NC_r, ^NC_{r+1}$.
$\frac{^NC_r}{^NC_{r-1}} = \frac{N-r+1}{r} = 2 \Rightarrow N = 3r - 1$; $\frac{^NC_{r+1}}{^NC_r} = \frac{N-r}{r+1} = \frac75 \Rightarrow 5N = 12r + 7$.
$15r - 5 = 12r + 7 \Rightarrow r = 4$, $N = 11$ (n = 6). Check: $^{11}C_3 : ^{11}C_4 : ^{11}C_5 = 165 : 330 : 462 = 5 : 10 : 14$.
Largest coefficient $= ^{11}C_5 = 462$.
Answer: (C). | ||||
| 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 $1 + (n-1)d = 300 \Rightarrow (n-1)d = 299 = 13\times23$. With $14 \le n - 1 \le 49$ and d an integer, $n - 1 = 23$, $d = 13$, $n = 24$.
$a_{20} = 1 + 19\times13 = 248$; $S_{20} = \frac{20}{2}(1 + 248) = 2490$.
Answer: (A). | ||||
| 58 | 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 The limit is the derivative in t of $t^2f^2(x) - x^2f^2(t)$ at $t = x$: $2xf^2(x) - 2x^2f(x)f'(x) = 0$.
As $f > 0$: $f(x) = xf'(x) \Rightarrow \frac{f'}{f} = \frac1x \Rightarrow f(x) = cx$; $f(1) = e$ gives $f(x) = ex$.
$ex = 1 \Rightarrow x = \frac1e$.
Answer: (B). | ||||
| 59 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = 1$: $f(1) = \frac\pi4 + \frac\pi4 = \frac\pi2$ but the right-hand limit is $\frac12(1-1) = 0$: discontinuous, so not differentiable.
At $x = -1$: $f(-1) = \frac\pi4 - \frac\pi4 = 0$ and the left-hand limit is $\frac12(1-1) = 0$: continuous. Left derivative of $\frac12(-x-1)$ is $-\frac12$; right derivative $\frac{1}{1+x^2} = \frac12$: not differentiable.
So f is continuous on $R - \{1\}$ and differentiable on $R - \{-1, 1\}$.
Answer: (C). | ||||
| 60 | 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 AM $\ge$ GM: $2^{\sin x} + 2^{\cos x} \ge 2\sqrt{2^{\sin x + \cos x}} \ge 2\cdot2^{-\frac{\sqrt2}{2}} = 2^{1-\frac{1}{\sqrt2}}$, since $\sin x + \cos x \ge -\sqrt2$.
Both equalities hold at $x = \frac{5\pi}{4}$ ($\sin x = \cos x = -\frac{1}{\sqrt2}$), so this is the minimum.
Answer: (A). | ||||
| 61 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{d}{dx}\left(\tan^4x\sin^43x\right) = 4\tan^3x\sec^2x\sin^43x + 12\tan^4x\sin^33x\cos3x = 2\tan^3x\sin^23x\,(2\sec^2x\sin^23x + 3\tan x\sin6x)$.
So the integrand is $\frac12\frac{d}{dx}(\tan^4x\sin^43x)$.
At $\frac\pi3$: $\sin\pi = 0$, value 0. At $\frac\pi6$: $\tan^4\frac\pi6 = \frac19$, $\sin\frac\pi2 = 1$, value $\frac19$.
Integral $= \frac12\left(0 - \frac19\right) = -\frac{1}{18}$.
Answer: (D). | ||||
| 62 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives By symmetry take C, D at $(\pm a, a^2 - 1)$, $0 < a < 1$. Area $A(a) = 2a(1 - a^2)$.
$A'(a) = 2 - 6a^2 = 0 \Rightarrow a = \frac{1}{\sqrt3}$ (a maximum, as $A'' = -12a < 0$).
$A = \frac{2}{\sqrt3}\cdot\frac23 = \frac{4}{3\sqrt3}$.
Answer: (C). | ||||
| 63 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Put $u = y + 3x$: $\frac{du}{dx} = \frac{dy}{dx} + 3 = \frac{u}{\ln u}$.
$\frac{\ln u}{u}du = dx \Rightarrow \frac12(\ln u)^2 = x + c$, i.e. $x - \frac12(\log_e(y + 3x))^2 = C$.
Answer: (C). | ||||
| 64 | Straight LinesSlope, parallel and perpendicular lines | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines Midpoint $\left(\frac{1+k}{2}, \frac72\right)$; slope of PQ $= \frac{-1}{k-1}$, so the bisector has slope $k - 1$.
Bisector: $y - \frac72 = (k-1)\left(x - \frac{k+1}{2}\right)$. At $x = 0$: $y = \frac72 - \frac{k^2-1}{2} = \frac{8 - k^2}{2} = -4 \Rightarrow k^2 = 16$, $k = \pm4$.
Answer: (D). | ||||
| 65 | CirclesStandard and general form | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Family: $x^2 + y^2 - 6x + \lambda(x^2 + y^2 - 4y) = 0$, centre $\left(\frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda}\right)$.
On the line: $\frac{6}{1+\lambda} - \frac{6\lambda}{1+\lambda} + 12 = 0 \Rightarrow 18 + 6\lambda = 0 \Rightarrow \lambda = -3$.
Circle: $-2(x^2+y^2) - 6x + 12y = 0$, i.e. $x^2 + y^2 + 3x - 6y = 0$.
$(-3, 6)$: $9 + 36 - 9 - 36 = 0$, on the circle. (The other points give 31, 10 and $-8$.)
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 $\frac ae = 4$, $e = \frac12 \Rightarrow a = 2$, $b^2 = a^2(1 - e^2) = 3$: $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
$\frac14 + \frac{\beta^2}{3} = 1 \Rightarrow \beta = \frac32$.
Normal at $(x_1, y_1)$: $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$: $4x - \frac{3y}{3/2} = 1$, i.e. $4x - 2y = 1$.
Answer: (B). | ||||
| 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 Line through the point along (2, 3, $-6$), whose length is 7: $(1 + 2t, -2 + 3t, 3 - 6t)$.
In the plane: $(1 + 2t) - (-2 + 3t) + (3 - 6t) = 5 \Rightarrow 6 - 7t = 5 \Rightarrow t = \frac17$.
Distance $= |t|\times7 = 1$.
Answer: (B). | ||||
| 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 $P(\text{total }6) = \frac{5}{36}$, $P(\text{total }7) = \frac{6}{36} = \frac16$.
A wins on his first throw, or after a round in which both fail, and so on: $P = \frac{5}{36}\left[1 + r + r^2 + \ldots\right]$ with $r = \frac{31}{36}\cdot\frac56 = \frac{155}{216}$.
$P = \frac{5/36}{1 - 155/216} = \frac{5}{36}\cdot\frac{216}{61} = \frac{30}{61}$.
Answer: (D). | ||||
| 69 | 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 cloud be h m above the lake and d its horizontal distance from P. Its image is h m below the lake.
$\tan30^\circ = \frac{h - 200}{d}$, $\tan60^\circ = \frac{h + 200}{d}$. Dividing: $\frac{h+200}{h-200} = 3 \Rightarrow h = 400$.
$d = \frac{200}{\tan30^\circ} = 200\sqrt3$, and $PC = \frac{d}{\cos30^\circ} = \frac{200\sqrt3}{\sqrt3/2} = 400$.
Answer: (C). | ||||
| 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 The contrapositive of $p \Rightarrow q$ is $\sim q \Rightarrow \sim p$.
Here p: f is differentiable at a; q: f is continuous at a. So: if f is not continuous at a, then it is not differentiable at a.
Answer: (C). | ||||
| 71 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Class marks 15, 25, 35 with frequencies 2, x, 2: symmetric, so the mean is 25.
Variance $= \frac{2\times10^2 + x\times0 + 2\times10^2}{4 + x} = \frac{400}{4 + x} = 50 \Rightarrow x = 4$.
Answer: 4. | ||||
| 72 | 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 Choose the 4 correctly answered questions: $^6C_4 = 15$ ways. Each of the other 2 has 3 wrong choices: $3^2 = 9$.
Total $= 15\times9 = 135$.
Answer: 135. | ||||
| 73 | 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 $\int_0^n\{x\}dx = n\int_0^1x\,dx = \frac n2$; $\int_0^n[x]dx = 0 + 1 + \ldots + (n-1) = \frac{n(n-1)}{2}$.
G.P.: $\left(\frac{n(n-1)}{2}\right)^2 = \frac n2\cdot10n(n-1) \Rightarrow \frac{n-1}{4} = 5 \Rightarrow n = 21$.
Answer: 21. | ||||
| 74 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines $P = (3\cos\theta, 3\sin\theta)$, $Q = -P$. Let $s = 3(\cos\theta + \sin\theta)$, so $s^2 \in [0, 18]$.
$\alpha = \frac{|s - 2|}{\sqrt2}$, $\beta = \frac{|-s - 2|}{\sqrt2}$, $\alpha\beta = \frac{|s^2 - 4|}{2}$.
$s^2 - 4$ ranges over $[-4, 14]$, so the maximum of $\alpha\beta$ is $\frac{14}{2} = 7$ (at $\theta = \frac\pi4$).
Answer: 7. | ||||
| 75 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\hat{i}\times(\vec{a}\times\hat{i}) = \vec{a} - (\vec{a}\cdot\hat{i})\hat{i}$, whose square length is $|\vec{a}|^2 - a_1^2$; similarly for $\hat{j}$, $\hat{k}$.
Sum $= 3|\vec{a}|^2 - (a_1^2 + a_2^2 + a_3^2) = 2|\vec{a}|^2 = 2\times9 = 18$.
Answer: 18. | ||||
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