JEE (Main) · 2020
JEE (Main) 2020 — 5 September, 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
25 Q
- Physical World and Measurement2
- Kinematics2
- Motion of System of Particles and Rigid Body2
12 easy · 12 medium · 1 hard
Chemistry
25 Q
- Biomolecules2
- Amines2
- Coordination Compounds2
15 easy · 9 medium · 1 hard
Mathematics
25 Q
- Limits, Continuity and Differentiability3
- Matrices and Determinants2
- Sequences and Series2
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 MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement For $z = a^2 b^{2/3} c^{-1/2} d^{-3}$ the maximum percentage error adds the powers times the individual errors:
$\frac{\Delta z}{z}\times100 = 2(2) + \frac{2}{3}(1.5) + \frac{1}{2}(4) + 3(2.5) = 4 + 1 + 2 + 7.5 = 14.5\%$.
Answer: (B). | ||||
| 2 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane From B the line of sight makes 45° with the vertical: $\tan45^\circ = \frac{d}{h_1}$, so $h_1 = d$.
From C, at distance $d + 2.464d = 3.464d$ from A, the line of sight makes 60° with the vertical: $\tan60^\circ = \frac{3.464d}{h_1+h_2}$, so $h_1 + h_2 = \frac{3.464d}{1.732} = 2d$.
Hence $h_2 = 2d - d = d$.
Answer: (D). | ||||
| 3 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium 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 At height h the helicopter's speed is $u = \sqrt{2gh}$ upwards, and the packet leaves with this speed.
Taking up as positive, the packet reaches the ground when $-h = ut - \frac{1}{2}gt^2$, i.e. $\frac{1}{2}gt^2 - \sqrt{2gh}\,t - h = 0$.
$t = \frac{\sqrt{2gh} + \sqrt{2gh + 2gh}}{g} = (\sqrt{2} + 2)\sqrt{\frac{h}{g}} \approx 3.41\sqrt{\frac{h}{g}}$.
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: $I\omega = (I + 3I)\omega'$, so $\omega' = \frac{\omega}{4}$.
Initial energy $\frac{1}{2}I\omega^2$; final energy $\frac{1}{2}(4I)\frac{\omega^2}{16} = \frac{1}{8}I\omega^2$.
Fractional loss $= \frac{\frac{1}{2} - \frac{1}{8}}{\frac{1}{2}} = \frac{3}{4}$.
Answer: (B). | ||||
| 5 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At height $R/2$: $g_1 = \frac{g}{(1 + \frac{1}{2})^2} = \frac{4g}{9}$.
At depth d: $g\left(1 - \frac{d}{R}\right) = \frac{4g}{9}$, so $\frac{d}{R} = \frac{5}{9}$.
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 Kinetic energy $= \frac{1}{2}(0.005)(210)^2 = 110.25$ J; half of it, 55.125 J, heats the bullet.
$55.125$ J $= \frac{55.125}{4.2} = 13.125$ cal (1 cal = $4.2\times10^7$ erg = 4.2 J).
$\Delta T = \frac{13.125}{5\times0.030} = 87.5^\circ$C.
Answer: (C). | ||||
| 7 | 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 Just submerged: weight of the shell = weight of water displaced by the whole sphere.
$\frac{27}{8}\rho_w\cdot\frac{4}{3}\pi(R^3 - r^3)g = \rho_w\cdot\frac{4}{3}\pi R^3 g$
$R^3 - r^3 = \frac{8}{27}R^3 \Rightarrow r^3 = \frac{19}{27}R^3 \Rightarrow r = \frac{19^{1/3}}{3}R \approx 0.89R$.
The closest option is $\frac{8}{9}R \approx 0.889R$.
Answer: (D). | ||||
| 8 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics A lies on the isotherm $T_2$, and B, C and D all lie on the isotherm $T_1$. The internal energy of an ideal gas depends only on temperature, so $E_{AB} = E_{AC} = E_{AD} = \frac{3}{2}nR(T_1 - T_2)$.
A → B is at constant pressure with the volume increasing: $W_{AB} > 0$.
A → C is at constant volume: $W_{AC} = 0$.
A → D, as drawn, goes up and to the left, so the volume decreases and $W_{AD} < 0$.
Only option (A) has the internal-energy changes right (all three equal), together with $W_{AB} > 0$ and $W_{AC} = 0$; option (B) has the sign of $W_{AD}$ right but wrongly makes $E_{AD}$ larger. The question's central point is that the three changes in internal energy are equal, so (A) is the intended answer.
Answer: (A). | ||||
| 9 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Mean kinetic energy $\frac{3}{2}kT = 4\times10^{-14}$ erg, so $kT = \frac{8}{3}\times10^{-14}$ erg.
Pressure $P = h\rho g = 2\times13.6\times980 = 26656$ dyn/cm$^2$.
$N = \frac{PV}{kT} = \frac{26656\times4}{\frac{8}{3}\times10^{-14}} = 3.998\times10^{18} \approx 4.0\times10^{18}$.
Answer: (D). | ||||
| 10 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Write $s = \Delta p\, v^a \rho^b f^c$. Dimensions: $[L] = [ML^{-1}T^{-2}][LT^{-1}]^a[ML^{-3}]^b[T^{-1}]^c$.
M: $0 = 1 + b \Rightarrow b = -1$. L: $1 = -1 + a - 3b = 2 + a \Rightarrow a = -1$. T: $0 = -2 - a - c \Rightarrow c = -1$.
So $s = \frac{\Delta p}{\rho v f} = \frac{10}{1\times300\times1000} = 3.3\times10^{-5}$ m $\approx \frac{1}{30}$ mm, of the order of $\frac{3}{100}$ mm.
Answer: (D). | ||||
| 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 The remaining sphere (charge Q) acts as a point charge at its centre. The piece starts at distance R and ends at $R + y$.
Energy conservation: $\frac{1}{2}mv^2 = mgy + \frac{qQ}{4\pi\epsilon_0}\left(\frac{1}{R} - \frac{1}{R+y}\right) = mgy + \frac{qQ\,y}{4\pi\epsilon_0 R(R+y)}$.
$v^2 = 2y\left[\frac{qQ}{4\pi\epsilon_0 R(R+y)m} + g\right]$.
Answer: (C). | ||||
| 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 Charges: $CV$ and $2C\cdot2V = 4CV$. Joined positive to negative, the net charge is $4CV - CV = 3CV$ on a total capacitance $3C$.
Common potential $= \frac{3CV}{3C} = V$; final energy $= \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2$.
Answer: (B). | ||||
| 13 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Current $I = \frac{1000}{220} = 4.55$ A. Loss in the line $= I^2R = (4.55)^2\times2 = 41.3$ W.
Efficiency $= \frac{1000}{1000 + 41.3} = 0.96 = 96\%$.
Answer: (A). | ||||
| 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 $I_g$: $I_g(G + R_1) = 1$ and $I_g(G + R_1 + R_2) = 2$.
Dividing, $G + R_1 + R_2 = 2(G + R_1)$, so $R_2 = R_1 + G$.
Answer: (D). | ||||
| 15 | 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 The loop's magnetic moment is $m = I(2a)^2 = 4Ia^2$, along the y-axis (normal to the XZ plane).
Since $b \gg a$, the wire's field over the loop is that at the origin: $B = \frac{\mu_0 I}{2\pi b}$, perpendicular to both the wire (z) and the line joining it to the origin (y), i.e. along x.
$\vec{m}\perp\vec{B}$, so $\tau = mB = 4Ia^2\cdot\frac{\mu_0 I}{2\pi b} = \frac{2\mu_0 I^2 a^2}{\pi b}$ (the torque is along z).
Answer: (A). | ||||
| 16 | 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 $B_0 = \frac{E_0}{c} = \frac{30}{3\times10^8} = 10^{-7}$ T. The wave travels along x with $\vec{E}$ along y, so $\vec{B}$ is along z, perpendicular to the electron's velocity (along y).
$F_{max} = evB_0 = 1.6\times10^{-19}\times3\times10^7\times10^{-7} = 4.8\times10^{-19}$ N.
Answer: (B). | ||||
| 17 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a concave lens with a real object, in magnitudes: $\frac{1}{v} = \frac{1}{u} + \frac{1}{f}$, so $v = \frac{uf}{u + f}$.
For small u, $v \approx u$ (the curve starts along the line u = v); for every u, $v < u$ and $v < f$, and as $u\to\infty$, $v\to f$.
So the curve starts tangent to the line u = v at the origin, stays below it, and approaches $v = f$ from below: graph (4).
Answer: (D). | ||||
| 18 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\ln R = \ln R_0 - \lambda t$, so each line's slope is $-\lambda$.
A: from 6 to 0 in 10 yr, $\lambda_A = 0.6$; B: from 6 to 0 in 5 yr, $\lambda_B = 1.2$; C: from 2 to 0 in 5 yr, $\lambda_C = 0.4$ (per year).
$T_{1/2} \propto \frac{1}{\lambda}$: $\frac{1}{0.6} : \frac{1}{1.2} : \frac{1}{0.4} = 2 : 1 : 3$.
Answer: (D). | ||||
| 19 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A photodiode is operated in reverse bias. As the reverse bias is raised from zero, the electric field across the depletion region sweeps more of the photo-generated electron-hole pairs out before they recombine, so the photocurrent first increases; once essentially all of them are collected, it becomes independent of the bias, i.e. it saturates (its value then depends only on the light intensity).
Answer: (A). | ||||
| 20 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Successive resonances differ in air-column length by $\frac{\lambda}{2}$: $\frac{\lambda}{2} = 24.5 - 17.0 = 7.5$ cm, so $\lambda = 15$ cm.
$f = \frac{v}{\lambda} = \frac{330}{0.15} = 2200$ Hz.
Answer: (D). | ||||
| 21 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalHard | |
A hard numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Let the particle (mass m = 200 MeV/$c^2$) have speed v; the atom (M = 1000 MeV/$c^2$ = 5m) recoils with speed u.
Momentum: $mv = Mu \Rightarrow u = \frac{v}{5}$.
Kinetic energy of the atom: $\frac{1}{2}(5m)\frac{v^2}{25} = \frac{1}{5}\cdot\frac{1}{2}mv^2 = \frac{K}{5}$, where K is the particle's initial kinetic energy.
The energy lost goes into excitation to n = 2: $K - \frac{K}{5} = 10.2$ eV, so $K = 12.75$ eV $= \frac{N}{4}$.
$N = 51$.
Answer: 51. | ||||
| 22 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\vec{r} = (4\hat{i} + 3\hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = 3\hat{i} + \hat{j} - 2\hat{k}$.
$\vec{\tau} = \vec{r}\times\vec{F} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&1&-2\\1&2&3\end{vmatrix} = (3+4)\hat{i} - (9+2)\hat{j} + (6-1)\hat{k} = 7\hat{i} - 11\hat{j} + 5\hat{k}$.
$|\vec{\tau}| = \sqrt{49 + 121 + 25} = \sqrt{195}$.
Answer: 195. | ||||
| 23 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $C_1$ is small, so the field of $C_2$ over it is the field at the centre: $B = \frac{\mu_0 N_2 I}{2R_2}$.
Mutual inductance $M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2R_2} = \frac{4\pi\times10^{-7}\times500\times200\times\pi\times10^{-4}}{2\times0.2} = \pi^2\times10^{-5}$ H.
$\frac{dI}{dt} = 10t - 2 = 8$ A/s at t = 1 s.
emf $= M\frac{dI}{dt} = 8\pi^2\times10^{-5}$ V $= 0.08\pi^2$ mV.
With $\pi^2 \approx 10$: emf $= 0.8$ mV $= \frac{4}{x}$, so $x = 5$.
Answer: 5. | ||||
| 24 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Least strain: the final image is at infinity, so the intermediate image lies at the eyepiece's focus, 5 cm in front of it, i.e. $10 - 5 = 5$ cm from the objective.
Objective: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $v = +5$, $f = +1$: $\frac{1}{u} = \frac{1}{5} - 1 = -\frac{4}{5}$, so $|u| = 1.25$ cm $= \frac{50}{40}$ cm.
$n = 50$.
Answer: 50. | ||||
| 25 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | NumericalMedium | |
A medium numerical question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Taking θ as the glancing angle on atomic planes 1 Å apart, the first maximum satisfies Bragg's condition $2d\sin\theta = \lambda$:
$\lambda = 2\times1\times\sin60^\circ = 1.732$ Å $= 1.732\times10^{-10}$ m.
$E = \frac{h^2}{2m\lambda^2} = \frac{(6.64\times10^{-34})^2}{2\times9.1\times10^{-31}\times3\times10^{-20}} = 8.075\times10^{-18}$ J $= \frac{8.075\times10^{-18}}{1.6\times10^{-19}} = 50.47$ eV.
Answer: 50.47. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Essential amino acids cannot be made by the body and must come from the diet: valine, leucine, isoleucine, lysine, threonine, methionine, phenylalanine, tryptophan, histidine and arginine.
Tyrosine is made in the body from phenylalanine, so it is non-essential.
Answer: (D). | ||||
| 27 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines (A) pyridine, (B) pyrrole, (C) pyrrolidine, (D) imidazole.
Pyrrole (B): the N lone pair is part of the aromatic sextet, so it is almost non-basic (conjugate acid p$K_a \approx -3.8$).
Pyridine (A): the lone pair is in an $sp^2$ orbital, outside the ring's π system (p$K_a \approx 5.2$).
Imidazole (D): the $sp^2$ N is protonated and the cation is stabilised by resonance over both N atoms (p$K_a \approx 7$).
Pyrrolidine (C): a saturated secondary amine with an $sp^3$ lone pair, the most basic (p$K_a \approx 11.3$).
Order: (B) < (A) < (D) < (C).
Answer: (B). | ||||
| 28 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons o-Xylene undergoes Friedel–Crafts acylation with succinic anhydride. Both methyls activate the ring; the attack goes para to one methyl (position 4), the less hindered site, giving A = 4-(3,4-dimethylphenyl)-4-oxobutanoic acid.
Zn–Hg/HCl (Clemmensen) reduces the aryl ketone C=O to $\mathrm{CH_2}$: 4-(3,4-dimethylphenyl)butanoic acid.
$\mathrm{H_3PO_4}$ then brings about intramolecular Friedel–Crafts acylation onto the ring position ortho to the side chain that is not crowded by a methyl, closing a six-membered ring: B = 6,7-dimethyl-3,4-dihydronaphthalen-1(2H)-one (6,7-dimethyl-1-tetralone).
Option (2) wrongly loses the C=O in B; options (1) and (3) place the acyl group next to a methyl.
Answer: (D). | ||||
| 29 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Reducing a nitrile to a primary amine, $\mathrm{R{-}C{\equiv}N \xrightarrow{LiAlH_4} RCH_2NH_2}$, needs a strong hydride donor. $\mathrm{LiAlH_4}$ does it; $\mathrm{NaBH_4}$ and $\mathrm{NaBH_3CN}$ are too mild to reduce a nitrile, and $\mathrm{CaH_2}$ is only a base/drying agent.
Answer: (A). | ||||
| 30 | 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 (A) acetone, (B) $\mathrm{PhCOCH_2COPh}$ (a 1,3-diketone), (C) methyl acetate, (D) N,N-dimethylacetamide.
The more the carbonyl's pull on the α-carbanion is reduced by a donor group on the carbonyl, the weaker the α-acidity. $\mathrm{-NMe_2}$ donates more strongly by resonance than $\mathrm{-OMe}$, so the amide (D) is the least acidic, then the ester (C), then the ketone (A). In (B) the $\mathrm{CH_2}$ lies between two C=O groups and its anion is delocalised over both, so it is by far the most acidic.
Order: (D) < (C) < (A) < (B).
Answer: (B). | ||||
| 31 | 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 (1) is an ester, $\mathrm{CH_3COOR}$; aqueous base hydrolyses it (saponification) to acetate and ROH, so it is unstable in base.
(2) a tert-butyl ether, (3) a benzyl ether and (4) a tetrahydropyranyl ether (an acetal) are all stable to base; the last two are cleaved by acid or hydrogenolysis, not by base.
Answer: (A). | ||||
| 32 | 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 A low level of the neurotransmitter noradrenaline lowers signal transmission and the mood, causing depression. Antidepressants (e.g. MAO inhibitors such as iproniazid, phenelzine) slow its breakdown and raise its level.
Answer: (D). | ||||
| 33 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The sixth period (n = 6) begins with 6s, and by the (n + l) rule the orbitals then fill in the order 6s → 4f → 5d → 6p (Cs to Rn, 32 elements).
Answer: (D). | ||||
| 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 $\Delta_f G^\circ$ of oxides (per mole of $\mathrm{O_2}$) against temperature; it tells which reductant can reduce which oxide at a given temperature. It says nothing about kinetics, and the pH–potential diagram is a Pourbaix diagram.
Answer: (C). | ||||
| 35 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter The water-gas shift reaction converts the CO of water gas into more hydrogen: $\mathrm{CO + H_2O \xrightarrow[\text{iron chromate}]{673\,K} CO_2 + H_2}$.
(1) makes water gas, (2) is steam reforming of methane, (4) makes producer gas.
Answer: (C). | ||||
| 36 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter In the solid state $\mathrm{PCl_5}$ is ionic, $\mathrm{[PCl_4]^+[PCl_6]^-}$: the cation is tetrahedral ($sp^3$) and the anion octahedral ($sp^3d^2$).
Answer: (C). | ||||
| 37 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Gd: [Xe] $4f^7 5d^1 6s^2$; removing three electrons gives $\mathrm{Gd^{3+}}$: [Xe] $4f^7$, with 7 unpaired electrons.
$\mu = \sqrt{n(n+2)} = \sqrt{63} = 7.9$ BM.
Answer: (A). | ||||
| 38 | 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 Octahedral, high spin $d^6$: $t_{2g}^4 e_g^2$, CFSE $= 4(-0.4\Delta_o) + 2(0.6\Delta_o) = -0.4\Delta_o$.
Tetrahedral (always high spin) $d^6$: $e^3 t_2^3$, CFSE $= 3(-0.6\Delta_t) + 3(0.4\Delta_t) = -0.6\Delta_t$.
Answer: (A). | ||||
| 39 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Normal rain water has pH about 5.6 (dissolved $\mathrm{CO_2}$); about 0.03% $\mathrm{CO_2}$ is the normal atmospheric level; clean water has a BOD below about 5 ppm.
Eutrophication — overgrowth of algae from excess nutrients (phosphates, nitrates), which robs a water body of oxygen — is a sign of pollution.
Answer: (D). | ||||
| 40 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter For bcc, Z = 2 molecules per cell. $M = \frac{\rho a^3 N_A}{Z} = \frac{6.17\times(3\times10^{-8})^3\times6\times10^{23}}{2} = \frac{6.17\times27\times10^{-24}\times6\times10^{23}}{2} = 49.98 \approx 50$ g mol$^{-1}$.
200 g is $\frac{200}{50} = 4$ mol, i.e. $4N_A$ molecules.
Answer: (B). | ||||
| 41 | 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 $r_n = 0.529\frac{n^2}{Z}$ Å, so $r_4 - r_3 = 0.529\times\frac{7}{Z}$ Å.
$\frac{\Delta R_1}{\Delta R_2} = \frac{Z_{He}}{Z_{Li}} = \frac{2}{3}$.
Answer: (D). | ||||
| 42 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure As two H atoms approach from far apart, attraction lowers the energy until it reaches a minimum at the bond length (74 pm, the bond energy below the separated atoms); closer still, nuclear repulsion makes the energy rise steeply. Graph (3) shows this minimum and the levelling off at large distance.
Answer: (C). | ||||
| 43 | EquilibriumLe Chatelier's principle | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on le chatelier's principle, from Equilibrium. NCERT: Class 11: Equilibrium (a) The forward reaction is endothermic, so lowering the temperature shifts the equilibrium towards $\mathrm{N_2O_4}$ (reactant).
(b) Adding the inert gas $\mathrm{N_2}$ raises the total pressure only by adding moles at fixed volume; the partial pressures of $\mathrm{N_2O_4}$ and $\mathrm{NO_2}$ are unchanged, so the equilibrium does not shift.
Answer: (C). | ||||
| 44 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $[A] = [A]_0e^{-k_At}$, $[B] = [B]_0e^{-k_Bt}$ with $[A]_0 = [B]_0$: $\frac{[A]}{[B]} = e^{(k_B - k_A)t} = 4$.
$k_B - k_A = \ln2\left(\frac{1}{180} - \frac{1}{300}\right) = \frac{\ln2}{450}$.
$t = \frac{\ln4}{\ln2/450} = 2\times450 = 900$ s.
Answer: (C). | ||||
| 45 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter At the CMC the surface is covered by surfactant molecules with their polar heads in the water and their non-polar tails pointing out into the air, and molecules in the bulk begin to gather into micelles: tails inside, away from water, and polar heads outside, in contact with water.
(A) shows only the surface layer, no micelle. (B) and (C) put the heads above the surface with the tails in the water, the wrong way round. (D) shows tails out of the water at the surface and a micelle with heads outward in the bulk.
Answer: (D). | ||||
| 46 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{EDTA^{4-}}$ binds through its two N atoms and the four carboxylate O atoms: it is hexadentate, with 6 donor sites.
Answer: 6. | ||||
| 47 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$: 5 mol $\mathrm{O_2}$.
$\mathrm{C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O}$: 6.5 mol each, 13 mol for 2 mol.
Total $= 5 + 13 = 18$ mol.
Answer: 18. | ||||
| 48 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Henry's law: 1 mol (44 g) of $\mathrm{CO_2}$ per kg at 30 bar, so at 3 bar the solution holds 0.1 mol per kg, i.e. about 0.1 M (density 1 g/mL).
Treating dissolved $\mathrm{CO_2}$ as $\mathrm{H_2CO_3}$ and taking only the first dissociation: $[\mathrm{H^+}] = \sqrt{K_{a1}C} = \sqrt{4\times10^{-7}\times0.1} = 2\times10^{-4}$ M.
pH $= 4 - \log2 = 4 - 0.3 = 3.7 = 37\times10^{-1}$.
Answer: 37. | ||||
| 49 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\Delta G^\circ = -nFE^\circ_{cell}$, so $E^\circ_{cell} = -\frac{17370}{3\times96500} = -0.06$ V $= -6\times10^{-2}$ V.
(The positive $\Delta G^\circ$ means the reaction is not spontaneous, so $E^\circ_{cell}$ is negative.)
Answer: −6. | ||||
| 50 | BiomoleculesProteins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Isoleucine has two stereocentres: the α-carbon and the β-carbon ($\mathrm{CH(CH_3)CH_2CH_3}$). Arginine has one (its α-carbon) and proline one (its α-carbon in the ring). The peptide bonds do not create or remove any.
Total $= 2 + 1 + 1 = 4$.
Answer: 4. | ||||
| 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 $n(C\cup T) = 73 + 65 - x \le 100$, so $x \ge 38$; and $x \le \min(73, 65) = 65$.
So $38 \le x \le 65$; of the options only 36 lies outside.
Answer: (D). | ||||
| 52 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z = x + iy$. The four points are $(x, y)$, $(x, -y)$, $\bar{z} - 2x = (-x, -y)$ and $z - 2x = (-x, y)$: a rectangle with sides $2|x|$ and $2|y|$.
For a square of side 4: $2|x| = 2|y| = 4$, so $|x| = |y| = 2$ and $|z| = \sqrt{4 + 4} = 2\sqrt{2}$.
Answer: (C). | ||||
| 53 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $t = |x| \ge 0$: $9t^2 - 18t + 5 = 0 \Rightarrow (3t - 1)(3t - 5) = 0$, so $t = \frac{1}{3}$ or $\frac{5}{3}$, both admissible.
The roots are $x = \pm\frac{1}{3}, \pm\frac{5}{3}$; their product is $\left(-\frac{1}{9}\right)\left(-\frac{25}{9}\right) = \frac{25}{81}$.
Answer: (D). | ||||
| 54 | 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 $R_1 \to R_1 - R_2$: the first row becomes $(\cos^2\theta - \sin^2\theta,\ \cos^2\theta - \sin^2\theta,\ 0) = (k, k, 0)$ with $k = \cos2\theta$.
Expanding along it: $f = k\begin{vmatrix}-1-\cos^2\theta & 1\\10 & -2\end{vmatrix} - k\begin{vmatrix}-\cos^2\theta & 1\\12 & -2\end{vmatrix} = k(2\cos^2\theta - 8) - k(2\cos^2\theta - 12) = 4\cos2\theta$.
For $\theta\in\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$, $2\theta\in\left[\frac{\pi}{2}, \pi\right]$ and $\cos2\theta\in[-1, 0]$, so $f\in[-4, 0]$.
(m, M) = (−4, 0).
Answer: (C). | ||||
| 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 $D = \begin{vmatrix}2&-4&\lambda\\1&-6&1\\\lambda&-10&4\end{vmatrix} = 2(-24+10) + 4(4-\lambda) + \lambda(-10+6\lambda) = 6\lambda^2 - 14\lambda - 12 = 2(3\lambda+2)(\lambda-3)$.
For $\lambda \ne 3, -\frac{2}{3}$ there is a unique solution.
$\lambda = 3$: the first two equations add to $3x_1 - 10x_2 + 4x_3 = 3$, which is the third equation, so the system is consistent (infinitely many solutions).
$\lambda = -\frac{2}{3}$: the third left side equals $-\frac{7}{4}(\text{first}) + \frac{17}{6}(\text{second})$ (check: $2(-\frac{7}{4}) + \frac{17}{6} = -\frac{2}{3}$, $-4(-\frac{7}{4}) - 6(\frac{17}{6}) = -10$, $-\frac{2}{3}(-\frac{7}{4}) + \frac{17}{6} = 4$), so the right sides would have to satisfy $3 = -\frac{7}{4} + \frac{34}{6} = \frac{47}{12}$, which is false: inconsistent.
So the system is inconsistent for exactly one value, $\lambda = -\frac{2}{3}$, which is negative.
Answer: (C). | ||||
| 56 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let $t = 3^{2\sin2\alpha}$, so $0 < t \le 9$. The terms are $\frac{t}{3}$, 14, $\frac{81}{t}$, and $\frac{t}{3} + \frac{81}{t} = 28$.
$t^2 - 84t + 243 = 0 \Rightarrow (t - 3)(t - 81) = 0$; since $t \le 9$, $t = 3$.
The terms are 1, 14, 27 with common difference 13; the sixth term is $1 + 5\times13 = 66$.
Answer: (C). | ||||
| 57 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The left side is a G.P. of 11 terms, first term $2^{10}$, ratio $\frac{3}{2}$:
$2^{10}\cdot\frac{(3/2)^{11} - 1}{\frac{3}{2} - 1} = 2^{11}\left(\frac{3^{11}}{2^{11}} - 1\right) = 3^{11} - 2^{11}$.
So $S = 3^{11}$.
Answer: (A). | ||||
| 58 | 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 $p(x) = (x-2)(x+1)$, so $\alpha = 2$ and the denominator is $x - 2$.
$\sqrt{1 - \cos p} = \sqrt{2}\,\left|\sin\frac{p}{2}\right|$. As $x\to2^+$, $p(x)\to0^+$, so $\sqrt{1-\cos p} \approx \sqrt{2}\cdot\frac{p}{2} = \frac{(x-2)(x+1)}{\sqrt{2}}$.
Limit $= \lim_{x\to2^+}\frac{(x+1)}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.
Answer: (C). | ||||
| 59 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Continuity at $\pi$: $-1 = k_2\cos\pi = -k_2$, so $k_2 = 1$.
First derivative: $2k_1(x - \pi)\to0$ and $-k_2\sin x\to0$ at $\pi$; equal for any $k_1$.
Second derivative: $2k_1 = -k_2\cos\pi = k_2 = 1$, so $k_1 = \frac{1}{2}$.
Answer: (A). | ||||
| 60 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives On the ellipse $x^2 = 5 - \frac{5}{4}y^2$ with $-2 \le y \le 2$.
$PQ^2 = x^2 + (y+4)^2 = 5 - \frac{5}{4}y^2 + y^2 + 8y + 16 = 21 + 8y - \frac{y^2}{4}$.
Its derivative $8 - \frac{y}{2} > 0$ on $[-2, 2]$, so the maximum is at $y = 2$ (P = (0, 2)): $PQ^2 = 21 + 16 - 1 = 36$.
Answer: (A). | ||||
| 61 | Integral CalculusIntegral as an antiderivative | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Differentiating the right side: $\left(g'(x) + g(x)(e^x - e^{-x})\right)e^{(e^x+e^{-x})}$.
Try $g(x) = e^x + 1$: $g' + g(e^x - e^{-x}) = e^x + (e^{2x} - 1 + e^x - e^{-x}) = e^{2x} + 2e^x - e^{-x} - 1$, which is the integrand's factor.
So $g(x) = e^x + 1$ and $g(0) = 2$.
Answer: (B). | ||||
| 62 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals With $x \to -x$: $I = \int_{-\pi/2}^{\pi/2}\frac{dx}{1 + e^{-\sin x}} = \int_{-\pi/2}^{\pi/2}\frac{e^{\sin x}}{1 + e^{\sin x}}dx$.
Adding: $2I = \int_{-\pi/2}^{\pi/2}1\,dx = \pi$, so $I = \frac{\pi}{2}$.
Answer: (B). | ||||
| 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 $\frac{dy}{2 + y} = -\frac{e^x}{5 + e^x}dx \Rightarrow \ln|2 + y| = -\ln(5 + e^x) + C$, i.e. $(2 + y)(5 + e^x) = K$.
$y(0) = 1$: $K = 3\times6 = 18$.
At $x = \ln13$: $(2 + y)\times18 = 18$, so $y = -1$.
Answer: (B). | ||||
| 64 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A tangent to $y^2 = 4x$ is $y = mx + \frac{1}{m}$. It touches $x^2 = 4y$ when $x^2 - 4mx - \frac{4}{m} = 0$ has equal roots: $16m^2 + \frac{16}{m} = 0 \Rightarrow m^3 = -1$, $m = -1$.
Common tangent: $y = -x - 1$, i.e. $x + y + 1 = 0$.
Touching the circle: $c = \frac{|1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Answer: (A). | ||||
| 65 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $\frac{x^2}{16} + \frac{y^2}{9} = 1$: $a = 4$, $b = 3$, $ae = \sqrt{16 - 9} = \sqrt{7}$, so A and B are the foci.
For any point on an ellipse the sum of focal distances is $2a = 8$.
Answer: (D). | ||||
| 66 | 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 is $M = (-1 + 2t, 3 - 2t, -t)$. $\vec{PM} = (2t - 2, 1 - 2t, 3 - t)$ must be perpendicular to $(2, -2, -1)$:
$2(2t-2) - 2(1-2t) - (3-t) = 9t - 9 = 0$, so $t = 1$ and the foot is $M = (1, 1, -1)$.
Image $= 2M - P = (1, 0, 1)$, so $a + b + c = 2$.
Answer: (C). | ||||
| 67 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $[\vec{a}\,\vec{b}\,\vec{c}] = \begin{vmatrix}1&1&n\\2&4&-n\\1&n&3\end{vmatrix} = (12 + n^2) - (6 + n) + n(2n - 4) = 3n^2 - 5n + 6$.
$3n^2 - 5n + 6 = 158 \Rightarrow 3n^2 - 5n - 152 = 0 \Rightarrow (n - 8)(3n + 19) = 0$, so $n = 8$ ($n \ge 0$).
Then $\vec{b}\cdot\vec{c} = 2 + 4n - 3n = 2 + n = 10$, while $\vec{a}\cdot\vec{c} = 1 + n + 3n = 33$.
Answer: (C). | ||||
| 68 | 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 Let the other two be $x, y$. Sum: $56 = 42 + x + y \Rightarrow x + y = 14$.
$\sum x_i^2 = 7(16 + 64) = 560$; the five known squares add to $4 + 16 + 100 + 144 + 196 = 460$, so $x^2 + y^2 = 100$.
$2xy = 196 - 100 = 96$, so $(x - y)^2 = 100 - 96 = 4$ and $|x - y| = 2$.
Answer: (B). | ||||
| 69 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions The n-th term is $\tan^{-1}\frac{1}{1 + n(n+1)} = \tan^{-1}\frac{(n+1) - n}{1 + n(n+1)} = \tan^{-1}(n+1) - \tan^{-1}n$.
The sum telescopes: $S = \tan^{-1}11 - \tan^{-1}1$.
$\tan S = \frac{11 - 1}{1 + 11} = \frac{10}{12} = \frac{5}{6}$.
Answer: (A). | ||||
| 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 $x \leftrightarrow \sim y$ is true exactly when $x$ and $y$ have opposite truth values. Its negation is true exactly when $x$ and $y$ have the same truth value: both true or both false, i.e. $(x \wedge y) \vee (\sim x \wedge \sim y)$ (which is $x \leftrightarrow y$).
Answer: (D). | ||||
| 71 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations SYLLABUS has S twice, L twice, and Y, A, B, U once: six different letters.
The alike pair: S or L, 2 ways. The two distinct letters: any 2 of the remaining 5 different letters, $\binom{5}{2} = 10$ ways.
Each choice can be arranged in $\frac{4!}{2!} = 12$ ways.
Total $= 2\times10\times12 = 240$.
Answer: 240. | ||||
| 72 | 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 $T_{r+1} = \binom{22}{r}x^{m(22-r) - 2r}$. Since $\binom{22}{r} = 1540$ and $\binom{22}{3} = \frac{22\cdot21\cdot20}{6} = 1540$, $r = 3$ or $r = 19$.
$r = 3$: $19m - 6 = 1$ gives $m = \frac{7}{19}$, not natural.
$r = 19$: $3m - 38 = 1$ gives $m = 13$.
Answer: 13. | ||||
| 73 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\left[\frac{x}{2}\right]$ jumps by 1 at every even integer $x = 2k$; in $(-10, 10)$ these are $-8, -6, -4, -2, 0, 2, 4, 6, 8$ (9 points).
At $x = 2k \ne 0$, $f$ jumps by $2k \ne 0$: discontinuous.
At $x = 0$: for $-2 \le x < 0$, $f = -x \to 0$; for $0 \le x < 2$, $f = 0$; so $f$ is continuous at 0.
Number of discontinuities $= 9 - 1 = 8$.
Answer: 8. | ||||
| 74 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | NumericalEasy | |
A easy numerical question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines Write the lines as $2x - y + \frac{\alpha}{2} = 0$ and $2x - y + \frac{\beta}{3} = 0$, parallel to $2x - y + 3 = 0$.
$\frac{|3 - \alpha/2|}{\sqrt{5}} = \frac{1}{\sqrt{5}} \Rightarrow \frac{\alpha}{2} = 2$ or 4, so $\alpha = 4$ or 8.
$\frac{|3 - \beta/3|}{\sqrt{5}} = \frac{2}{\sqrt{5}} \Rightarrow \frac{\beta}{3} = 1$ or 5, so $\beta = 3$ or 15.
Sum $= 4 + 8 + 3 + 15 = 30$.
Answer: 30. | ||||
| 75 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability For one die, P(3 or 5) $= \frac{1}{3}$. For one throw of four dice, P(at least two show 3 or 5) $= 1 - \left(\frac{2}{3}\right)^4 - 4\cdot\frac{1}{3}\left(\frac{2}{3}\right)^3 = 1 - \frac{16}{81} - \frac{32}{81} = \frac{33}{81} = \frac{11}{27}$.
Over 27 throws the number of successes is binomial, with mean $27\times\frac{11}{27} = 11$.
Answer: 11. | ||||
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