JEE (Main) · 2020
JEE (Main) 2020 — 3 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
- Work, Energy and Power2
- Motion of System of Particles and Rigid Body2
- Properties of Bulk Matter2
18 easy · 6 medium · 1 hard
Chemistry
25 Q
- Some Basic Concepts of Chemistry4
- Classification of Elements and Periodicity2
- Coordination Compounds2
12 easy · 9 medium · 4 hard
Mathematics
25 Q
- Applications of Derivatives3
- Matrices and Determinants2
- Sequences and Series2
11 easy · 13 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 Solar constant $=\frac{\text{energy}}{\text{area}\times\text{time}} = \frac{ML^2T^{-2}}{L^2\,T} = ML^0T^{-3}$.
Answer: (C). | ||||
| 2 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power With constant power $P$ and starting from rest, $\frac12mv^2 = Pt$, so $v=\sqrt{\frac{2Pt}{m}}\propto t^{1/2}$.
Then $s=\int v\,dt \propto t^{3/2}$: the slope $ds/dt$ grows with time, so the graph starts at the origin and bends upward.
Answer: (C). | ||||
| 3 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Momentum is conserved in the collision: $0.1\times20 = 2v \Rightarrow v = 1\ \mathrm{m/s}$.
Kinetic energy just after: $\frac12\times2\times1^2 = 1$ J.
Falling 1 m adds $mgh = 2\times10\times1 = 20$ J, so the kinetic energy at the floor is $1+20 = 21$ J.
Answer: (B). | ||||
| 4 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctHard | |
A hard single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The CM moves in a horizontal circle of radius $\frac l2\sin\theta$, so the pivot supplies $F_H = m\omega^2\frac l2\sin\theta$ (towards the axis) and $F_V = mg$.
Both act at the pivot, a distance $\frac l2$ from the CM. Their torques about the CM: $F_V\cdot\frac l2\sin\theta - F_H\cdot\frac l2\cos\theta$.
Equating to $\frac{ml^2}{12}\omega^2\sin\theta\cos\theta$:
$mg\frac l2\sin\theta - m\omega^2\frac{l^2}{4}\sin\theta\cos\theta = \frac{ml^2}{12}\omega^2\sin\theta\cos\theta$
$\frac g2 = \omega^2 l\cos\theta\left(\frac14+\frac1{12}\right) = \frac{\omega^2l\cos\theta}{3} \Rightarrow \cos\theta = \frac{3g}{2l\omega^2}$.
Answer: (C). | ||||
| 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 Mass within $r$: $M(r) = \int_0^r 4\pi x^2\rho_0\left(1-\frac{x^2}{R^2}\right)dx = 4\pi\rho_0\left(\frac{r^3}{3}-\frac{r^5}{5R^2}\right)$.
Field inside: $E = \frac{GM(r)}{r^2} = 4\pi G\rho_0\left(\frac r3 - \frac{r^3}{5R^2}\right)$.
$\frac{dE}{dr} = 0 \Rightarrow \frac13 = \frac{3r^2}{5R^2} \Rightarrow r^2 = \frac59R^2$, $r=\sqrt{\frac59}R$ (inside the planet; outside, $E$ only falls).
Answer: (B). | ||||
| 6 | Properties of Bulk MatterHeat transfer | 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 heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Newton's law of cooling: $T-20$ falls by the same factor in equal times.
In the first 300 s, $T-20$ goes from 30 to 20: a factor $\frac23$.
In the next 300 s (5 minutes): $T-20 = 20\times\frac23 = 13.3$, so $T\approx33.3$°C.
(The averaged form gives the same: $\frac{10}{300}=k(45-20)$ and $\frac{40-T}{300}=k\left(\frac{40+T}{2}-20\right)$ give $T = 33.3$°C.)
Answer: (C). | ||||
| 7 | 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 Heat taken by water and calorimeter: $(180+20)\times1\times(31-25) = 1200$ cal.
Heat given by the steam: $m(540 + 1\times(100-31)) = 609m$.
$609m = 1200 \Rightarrow m \approx 1.97 \approx 2$ g.
Answer: (A). | ||||
| 8 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $nC_p\times50 = 160 \Rightarrow nC_p = 3.2$; $nC_v\times100 = 240 \Rightarrow nC_v = 2.4$.
$\gamma = \frac{C_p}{C_v} = \frac{3.2}{2.4} = \frac43 = 1+\frac2f \Rightarrow f = 6$.
Answer: (C). | ||||
| 9 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At the equilibrium point the speed is $v = A\omega = A\sqrt{\frac km}$, and the remaining half keeps this speed.
New angular frequency $\omega' = \sqrt{\frac{k}{m/2}} = \sqrt2\,\omega$.
New amplitude $A' = \frac{v}{\omega'} = \frac{A\omega}{\sqrt2\,\omega} = \frac{A}{\sqrt2}$, so $f = \frac1{\sqrt2}$.
Answer: (B). | ||||
| 10 | 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 $qV = \frac12mv^2 \Rightarrow v = \sqrt{\frac{2qV}{m}}$. Both carry charge $e$, so $\frac{v_H}{v_{He}} = \sqrt{\frac{m_{He}}{m_H}} = \sqrt{\frac41} = 2$.
Answer: (D). | ||||
| 11 | 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 The outer shell's charge gives the same potential everywhere inside it, so only $Q_1$ matters:
$V(R)-V(4R) = \frac{Q_1}{4\pi\varepsilon_0}\left(\frac1R-\frac1{4R}\right) = \frac{3Q_1}{16\pi\varepsilon_0R}$.
(Equal densities give $Q_2 = 16Q_1$, so this also equals $\frac{3Q_2}{256\pi\varepsilon_0R}$, which is not among the other options.)
Answer: (B). | ||||
| 12 | 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 Voltmeter in parallel with $400\,\Omega$: $\frac{400\times10000}{10400} = 384.6\,\Omega$.
Reading $= 6\times\frac{384.6}{384.6+800} = 6\times\frac{384.6}{1184.6} \approx 1.95$ V.
Answer: (B). | ||||
| 13 | 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 A perfect diamagnet ($\chi=-1$, a superconductor) expels the field completely: no field lines enter it, so none reach the cavity at its centre. The paramagnetic substance has no field to magnify, and the field in it is zero.
Answer: (C). | ||||
| 14 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Side of the square $= 7.5$ cm, area $= (0.075)^2 = 5.625\times10^{-3}\ \mathrm{m^2}$.
emf $= 0.032\times5.625\times10^{-3} = 1.8\times10^{-4}$ V.
Cross-section of the wire $= \pi(2\times10^{-3})^2 = 1.257\times10^{-5}\ \mathrm{m^2}$; $R = \frac{1.23\times10^{-8}\times0.3}{1.257\times10^{-5}} = 2.94\times10^{-4}\ \Omega$.
$I = \frac{1.8\times10^{-4}}{2.94\times10^{-4}} \approx 0.61$ A.
Answer: (A). | ||||
| 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 $B_0 = \frac{E_0}{c} = E_0\sqrt{\mu_0\epsilon_0}$, in phase with $\vec E$.
$\vec E\times\vec B$ must point along $+\hat i$: $\hat j\times\hat k = \hat i$, so $\vec B$ is along $\hat k$.
At $t=0$: $\vec B = E_0\sqrt{\mu_0\epsilon_0}\cos(kx)\hat k$.
Answer: (A). | ||||
| 16 | OpticsWave optics | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on wave optics, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Optical paths are $n_1L_1$ and $n_2L_2$; phase difference $= \frac{2\pi}{\lambda}(n_1L_1 - n_2L_2)$.
Answer: (B). | ||||
| 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 Photons per second $= \frac{P}{hc/\lambda} = \frac{P\lambda}{hc}$. Same power, so the ratio is $\frac{\lambda_X}{\lambda_V} = \frac{1}{500}$.
Answer: (C). | ||||
| 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 $\rho = \frac{A\times1.67\times10^{-27}}{\frac43\pi(1.3\times10^{-15})^3A} = \frac{1.67\times10^{-27}}{9.2\times10^{-45}} \approx 1.8\times10^{17}\ \mathrm{kg\,m^{-3}}$.
Answer: (C). | ||||
| 19 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $E_g = \frac{hc}{\lambda_{max}} = \frac{6.63\times10^{-34}\times3\times10^8}{400\times10^{-9}} = 4.97\times10^{-19}$ J $= \frac{4.97\times10^{-19}}{1.6\times10^{-19}} \approx 3.1$ eV.
Answer: (A). | ||||
| 20 | Experimental SkillsElectricity experiments | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on electricity experiments, from Experimental Skills. NCERT: No matching NCERT chapter (A) A resistor conducts equally both ways: equal deflection, observed.
(B) A capacitor blocks steady current: in steady state no deflection either way, observed.
(C) A metal wire has almost no resistance, so the meter deflects fully both ways. 'No deflection' is NOT observed.
(D) An LED conducts and glows only when forward biased, and gives no deflection when reversed: observed.
Answer: (C). | ||||
| 21 | Laws of MotionFriction | Class 11: Laws of Motion | NumericalMedium | |
A medium numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Up the plane: retardation $a_1 = g(\sin\theta+\mu\cos\theta)$; down: $a_2 = g(\sin\theta-\mu\cos\theta)$. Over the same distance $d$:
$v_0^2 = 2a_1d$ and $\frac{v_0^2}{4} = 2a_2d$, so $a_1 = 4a_2$.
$\sin\theta+\mu\cos\theta = 4\sin\theta-4\mu\cos\theta \Rightarrow \mu = \frac35\tan30° = \frac{0.6}{\sqrt3} = 0.3464$.
$I = 346.4$, nearest integer 346.
Answer: 346. | ||||
| 22 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Distances from EX: E is on it (0), G is $a$ away, F is $\frac a2$ away.
$I = m a^2 + m\frac{a^2}{4} = \frac54ma^2 = \frac{25}{20}ma^2$, so N = 25.
Answer: 25. | ||||
| 23 | ThermodynamicsSecond law | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat removed from the water: $Q_2 = 100\times80 = 8000$ cal at $T_2 = 273$ K.
Minimum work means a reversible (Carnot) refrigerator: $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$, $T_1 = 300$ K.
$Q_1 = 8000\times\frac{300}{273} = 8791.2$ cal (with 273.15 K and 300.15 K: 8790.8 cal).
Answer: 8791. | ||||
| 24 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter With the plane of the coil parallel to the field the torque is greatest: $\tau = NIAB$.
$B = \frac{1.5}{500\times0.5\times3\times10^{-4}} = \frac{1.5}{0.075} = 20$ T.
Answer: 20. | ||||
| 25 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Mirror formula $\frac1v+\frac1u = \frac1f$; differentiating, $\frac{dv}{dt} = -\frac{v^2}{u^2}\frac{du}{dt}$.
Speed of the image $= \left(\frac{10}{30}\right)^2\times9 = 1\ \mathrm{cm\,s^{-1}}$.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | 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 An ionic micelle needs an ionic surfactant (a long non-polar tail with an ionic head) in water above the critical micelle concentration.
(A) Diethyl ether is not a surfactant: no micelle.
(B) 1,3-Dimethylimidazolium $PF_6^-$ is a small ionic liquid with no long tail: it simply dissolves or separates, no micelle.
(C) Sodium stearate in a non-polar solvent such as toluene can at most form reverse micelles, not ionic micelles in water.
(D) The octyl sulphate anion has a long hydrocarbon tail and an ionic $-SO_4^-$ head; in excess water these aggregate into ionic micelles.
Answer: (D). | ||||
| 27 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium At the start the solution is 0.1 M HCl: pH = 1. The pH rises slowly while HCl is in excess, then steeply near the end.
The total NaOH added (100 mL of 0.1 M = 0.01 mol) exactly equals the HCl (0.01 mol), so the titration stops at the equivalence point, where the solution is NaCl and pH = 7. No excess base is ever added, so the pH never goes above 7.
The curve rises in an S-shape and ends at pH 7: graph (B).
Answer: (B). | ||||
| 28 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) is 2-hydroxybenzoic acid: its $-OH$ and $-COOH$ are next to each other and form an intramolecular hydrogen bond. (B) is 4-hydroxybenzoic acid: its groups are far apart and form intermolecular hydrogen bonds, linking molecules into a network.
(a) The intermolecular network makes (B) pack into a crystal more readily: true.
(b) More intermolecular association gives (B) the higher boiling point: true.
(c) is taken as false in the intended answer: (B)'s strong intermolecular hydrogen bonding holds its molecules together in the solid, which works against dissolving.
So (a) and (b).
Answer: (A). | ||||
| 29 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate $= -\frac12\frac{dn_A}{dt} = -\frac13\frac{dn_B}{dt} = -\frac23\frac{dn_C}{dt}$.
Multiplying by $-2$: $\frac{dn_A}{dt} = \frac23\frac{dn_B}{dt} = \frac43\frac{dn_C}{dt}$.
Answer: (C). | ||||
| 30 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | Single correctHard | |
A hard single correct question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom For $n=1$, $l = 0,1,2$: subshells 1s, 1p, 1d. For $n=2$: 2s, 2p, 2d, 2f.
Filling by the $(n+l)$ rule (lower $n$ first on a tie): 1s (1), 1p (2), 2s (2), 1d (3), 2p (3), 3s (3) ...
Z = 6: $1s^21p^4$, the valence subshell is 1p, not 2p: false.
Z = 9: $1s^21p^62s^1$: one electron in a new s subshell after a filled p, the analogue of lithium: the first alkali metal: true.
Z = 13: $1s^21p^62s^21d^3$: a d subshell is half-filled with 5 electrons, not 3: false.
Z = 8: $1s^21p^6$, a filled s and p set, the analogue of a noble-gas configuration; Z = 9 is alkali-like precisely because it follows this closed core. So (B) and (C) stand or fall together.
Answer: (B) or (C). | ||||
| 31 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Partial pressure is proportional to moles (Dalton's law). Each gas has 1 mol, so each exerts 2 atm.
Total $= 3\times2 = 6$ atm.
Answer: (B). | ||||
| 32 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The jump from 3658 to 25024 kJ $\mathrm{mol^{-1}}$ (about seven times) comes after the third electron: the fourth is taken from an inner shell. So there are 3 valence electrons (this is boron).
Answer: (C). | ||||
| 33 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctMedium | |
A medium single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry 5.6 volume: 1 L gives 5.6 L of $O_2$ at STP $= 0.25$ mol $O_2$.
$2H_2O_2 \rightarrow 2H_2O + O_2$, so $H_2O_2 = 0.5$ mol per litre: 0.5 M.
Mass of $H_2O_2$ per litre $= 0.5\times34 = 17$ g in 1000 g of solution: 1.7%.
Answer: (D). | ||||
| 34 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties (I) Be is above Mg in group 2: smaller radius. True.
(II) IE of Be (899 kJ/mol) is higher than that of Al (577 kJ/mol). True.
(III) Be and Al show a diagonal relationship precisely because their charge/radius ratios are nearly the same; the statement that Be's is greater is taken as false.
(IV) Both form largely covalent compounds ($BeCl_2$, $AlCl_3$) owing to their high polarising power. True.
Answer: (B). | ||||
| 35 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Manganate $MnO_4^{2-}$ has Mn(VI), $d^1$: paramagnetic. Permanganate $MnO_4^-$ has Mn(VII), $d^0$: diamagnetic. So 'both are paramagnetic' is incorrect.
Both are tetrahedral, both have $p\pi$–$d\pi$ bonding, and manganate is green, permanganate purple: these are correct.
Answer: (B). | ||||
| 36 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Molar mass $= 52 + 3\times35 + 6\times18 = 265$.
Loss $= 0.135\times265 = 35.8 \approx 36$ g $= 2H_2O$. Conc. $H_2SO_4$ removes only water of crystallisation, not coordinated water.
So two waters lie outside the coordination sphere: $[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O$ (check: $36/265 = 13.6\%$).
Answer: (B). | ||||
| 37 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Both metals are +2 and $d^6$.
$Ru^{2+}$ is a 4d ion with a strong-field chelate (en): large splitting, low spin, $t_{2g}^6e_g^0$.
$Fe^{2+}$ with weak-field $H_2O$: high spin, $t_{2g}^4e_g^2$.
Answer: (D). | ||||
| 38 | 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 Acid rain corrodes metal pipes (a, correct) and attacks marble and limestone (b, correct). It does cause respiratory ailments (its oxides of sulphur and nitrogen), so (c) is incorrect; it damages trees and plants, so (d) is incorrect.
Answer: (C). | ||||
| 39 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Aldehydes are more reactive than ketones (less steric hindrance, fewer electron-donating alkyl groups). Among aldehydes, benzaldehyde is less reactive than propanal because the ring's resonance reduces the carbonyl carbon's positive charge. Butanone (ethyl) is more hindered and more electron-rich than propanone (methyl).
Increasing order: Butanone < Propanone < Benzaldehyde < Propanal.
Answer: (D). | ||||
| 40 | 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 Chromic anhydride ($CrO_3$) oxidises primary and secondary alcohols: the allylic primary $-OH$ (a) goes to $-CHO$ and the secondary $-OH$ (c) to a ketone.
The tertiary $-OH$ (b) has no H on its carbon and the phenolic $-OH$ (d) is not an alcohol: both survive.
Ceric ammonium nitrate gives a positive test with alcohols and phenols, so P responds because of (b) and (d).
Answer: (B). | ||||
| 41 | AminesClassification and identification of amines | Class 12: Amines | Single correctHard | |
A hard single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines A and C are diazotised, so they are primary aromatic amines (ethylanilines). Hydrolysis gives ethylphenols, and $KMnO_4$ oxidises the ethyl group to $-COOH$: R and S are hydroxybenzoic acids.
R boils lower than S, so R is the ortho acid (salicylic acid, intramolecular hydrogen bond): A is 2-ethylaniline, and C is the meta or para isomer.
B gives an alkali-insoluble sulphonamide with $C_6H_5SO_2Cl$ (Hinsberg test): B is a secondary amine, such as $C_6H_5CH_2NHCH_3$.
Only (D) has A = 2-ethylaniline, B = a secondary amine, C = 4-ethylaniline.
Answer: (D). | ||||
| 42 | 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 Electron-withdrawing nitro groups make the benzylic carbon more electron-poor and stabilise the $S_N2$ transition state, so every dinitro compound reacts faster than benzyl chloride (I).
A nitro group ortho or para withdraws by resonance ($-M$) as well as induction; a meta nitro only by induction.
(II) 2,3-dinitro and (III) 3,4-dinitro each have one o/p nitro and one meta; (IV) 3,5-dinitro has two meta only. (II), with its ortho nitro nearest the reaction centre, is taken as the fastest.
Order: (II) > (III) > (IV) > (I).
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 Ranitidine blocks histamine at the stomach's receptors and cuts acid: antacid (d). Nardil (phenelzine) is a monoamine-oxidase inhibitor: antidepressant (a). Chloramphenicol: broad-spectrum antibiotic (b). Dimetane (brompheniramine): antihistamine (c).
Answer: (B). | ||||
| 44 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids C gives potassium benzoate and benzyl alcohol with conc. KOH (Cannizzaro): C is benzaldehyde, $C_6H_5CHO$.
D gives $(CH_3)_2C=CH-CO-CH_3$ (mesityl oxide) with $Ba(OH)_2$ (aldol condensation): D is acetone.
So B, which ozonolyses to $C_6H_5CHO + (CH_3)_2CO$, is $C_6H_5CH=C(CH_3)_2$.
A with $CH_3MgBr$ then acid/heat must give B: $C_6H_5CH_2COCH_3 \rightarrow C_6H_5CH_2C(OH)(CH_3)_2 \xrightarrow{-H_2O} C_6H_5CH=C(CH_3)_2$ (conjugated with the ring).
Answer: (C). | ||||
| 45 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate is 2-iodo-3,4-dimethylpentane. t-BuOH is a weak, bulky nucleophile and a polar protic solvent; on heating the reaction goes by E1.
Ionisation gives a secondary carbocation at C-2; a 1,2-hydride shift from C-3 gives the more stable tertiary carbocation at C-3.
Loss of the tertiary H on C-4 gives the most substituted (Zaitsev) alkene, tetrasubstituted $CH_3CH_2C(CH_3)=C(CH_3)_2$, 2,3-dimethylpent-2-ene.
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 $6.023\times10^{22}$ molecules $= 0.1$ mol in 10 g, so the molar mass is 100 g/mol.
5 g $= 0.05$ mol in 2 L: $0.025$ M $= 25\times10^{-3}$ M.
Answer: 25. | ||||
| 47 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Isotonic: equal molar concentrations. $\frac{0.73/M_A}{0.25} = \frac{1.65/M_B}{1}$.
$\frac{M_A}{M_B} = \frac{0.73\times4}{1.65} = \frac{2.92}{1.65} = 1.7697 = 176.97\times10^{-2}$, nearest integer 177.
Answer: 177. | ||||
| 48 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Charge $= 2\times8\times60 = 960$ C $= \frac{960}{96000} = 0.01$ F.
6 F give 2 mol $Cr^{3+}$, so 0.01 F gives $\frac{0.01}{3}$ mol $= 3.33\times10^{-3}$ mol $= 0.1733$ g.
Efficiency $= \frac{0.104}{0.1733}\times100 = 60\%$.
Answer: 60. | ||||
| 49 | 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 Equal normalities neutralise in equal volumes: $N_1V_1 = N_2V_2 \Rightarrow 0.1\times V = 0.1\times10$, $V = 10$ mL.
(Phosphinic acid $H_3PO_2$ is monobasic, so here 0.1 N is also 0.1 M; the normality already accounts for the basicity.)
Answer: 10. | ||||
| 50 | BiomoleculesProteins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Two peptide (amide) bonds: 2 C=O.
The C-terminal $-COOH$ of Lys: 1.
The side-chain $-COOH$ of Asp and of Glu: 2. (Lys's side chain is $-(CH_2)_4NH_2$, no C=O.)
Total $= 2+1+2 = 5$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $R_1$: take $a = c = \sqrt{2-\sqrt3}$, $b = 3^{1/4}$. Then $a^2+b^2 = 2$ and $b^2+c^2 = 2$ are rational, so $(a,b),(b,c)\in R_1$, but $a^2+c^2 = 4-2\sqrt3$ is irrational: $(a,c)\notin R_1$. Not transitive.
$R_2$: take $a = 1$, $b = 2^{1/4}$, $c = 1$. $a^2+b^2 = 1+\sqrt2$ and $b^2+c^2 = 1+\sqrt2$ are irrational, so both pairs are in $R_2$, but $a^2+c^2 = 2$ is rational: $(a,c)\notin R_2$. Not transitive.
Neither is transitive.
Answer: (D). | ||||
| 52 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctHard | |
A hard single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $f(x) = (\lambda^2+1)x^2 - 4\lambda x + 2$; $f(0) = 2 > 0$, $f(1) = \lambda^2-4\lambda+3 = (\lambda-1)(\lambda-3)$.
Exactly one root in (0, 1) when $f(0)f(1) < 0$: $(\lambda-1)(\lambda-3) < 0$, $1<\lambda<3$.
Ends: $\lambda = 3$ gives $10x^2-12x+2 = 2(5x-1)(x-1)$, roots $\frac15$ and 1: exactly one root ($\frac15$) in (0, 1), so 3 is included. $\lambda = 1$ gives $2(x-1)^2$, a double root at 1, none in (0, 1), so 1 is excluded.
The set is $(1,3]$.
Answer: (C). | ||||
| 53 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x+iy$: $x = \sqrt{(x-1)^2+y^2} \Rightarrow y^2 = 2x-1$, so both points lie on this parabola.
$\arg(z_1-z_2) = \frac\pi6$: $\frac{y_1-y_2}{x_1-x_2} = \frac1{\sqrt3}$. Since $x = \frac{y^2+1}{2}$, $x_1-x_2 = \frac{(y_1-y_2)(y_1+y_2)}{2}$, so $\frac{2}{y_1+y_2} = \frac1{\sqrt3}$.
$\mathrm{Im}(z_1+z_2) = y_1+y_2 = 2\sqrt3$.
Answer: (C). | ||||
| 54 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $|\mathrm{adj}\,A| = 2(0+4) + 1(1-2) + 1(2-0) = 9$, and $|\mathrm{adj}\,A| = |A|^2$, so $|\lambda| = 3$.
$B = \mathrm{adj}(\mathrm{adj}\,A) = |A|A$, so $|B| = |A|^3|A| = |A|^4 = 81$.
$|(B^{-1})^T| = |B^{-1}| = \frac1{81}$.
Answer: (B). | ||||
| 55 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom9r\left(\frac32x^2\right)^{9-r}\left(-\frac1{3x}\right)^r$; the power of $x$ is $18-3r = 0 \Rightarrow r = 6$.
$k = \binom96\left(\frac32\right)^3\left(\frac13\right)^6 = 84\times\frac{27}{8}\times\frac1{729} = \frac{84}{216} = \frac7{18}$, so $18k = 7$.
Answer: (A). | ||||
| 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 An A.P. with $a = 20$, $d = -\frac25$.
$S_n = \frac n2\left[40 - \frac25(n-1)\right] = 488 \Rightarrow n(202-2n) = 4880 \Rightarrow n^2-101n+2440 = 0 \Rightarrow n = 40$ or $61$.
$T_{40} = 20 - \frac25\times39 = 4.4 > 0$; $T_{61} = 20 - \frac25\times60 = -4 < 0$. So $n = 61$ and the $n^{\text{th}}$ term is $-4$.
Answer: (D). | ||||
| 57 | 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 A $\frac00$ form; by L'Hospital's rule, at $x = a$:
numerator' $= \frac23(3a)^{-2/3} - (3a)^{-2/3} = -\frac13(3a)^{-2/3}$; denominator' $= \frac13(4a)^{-2/3} - \frac43(4a)^{-2/3} = -(4a)^{-2/3}$.
Limit $= \frac13\left(\frac{4a}{3a}\right)^{2/3} = \frac13\left(\frac{16}{9}\right)^{1/3} = \left(\frac{16}{243}\right)^{1/3}$.
And $\left(\frac23\right)\left(\frac29\right)^{1/3} = \left(\frac{8}{27}\cdot\frac29\right)^{1/3} = \left(\frac{16}{243}\right)^{1/3}$.
Answer: (D). | ||||
| 58 | 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 $f'(x) = kx(x-1)(x+1) = k(x^3-x)$, so $f(x) = k\left(\frac{x^4}4 - \frac{x^2}2\right) + f(0)$.
$f(x) = f(0) \Rightarrow \frac{x^2(x^2-2)}{4} = 0 \Rightarrow x = 0, \pm\sqrt2$.
Sum of squares $= 0 + 2 + 2 = 4$.
Answer: (A). | ||||
| 59 | Applications of DerivativesRate of change | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on rate of change, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $S = 6a^2 \Rightarrow \frac{dS}{dt} = 12a\frac{da}{dt} = 3.6 \Rightarrow \frac{da}{dt} = \frac{3.6}{120} = 0.03$ at $a = 10$.
$\frac{dV}{dt} = 3a^2\frac{da}{dt} = 300\times0.03 = 9$.
Answer: (B). | ||||
| 60 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\sin^{-1}\sqrt{\frac{x}{1+x}} = \tan^{-1}\sqrt x$ (an angle with $\sin = \sqrt{\frac x{1+x}}$ has $\tan = \sqrt x$).
Put $x = t^2$: $\int 2t\tan^{-1}t\,dt = t^2\tan^{-1}t - \int\frac{t^2}{1+t^2}dt = t^2\tan^{-1}t - t + \tan^{-1}t$.
$= (x+1)\tan^{-1}\sqrt x - \sqrt x + C$, so $(A, B) = (x+1, -\sqrt x)$.
Answer: (A). | ||||
| 61 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $x = \sin\theta$, $dx = \cos\theta\,d\theta$; the limits become 0 and $\frac\pi6$.
$\int_0^{\pi/6}\frac{\sin^2\theta}{\cos^3\theta}\cos\theta\,d\theta = \int_0^{\pi/6}\tan^2\theta\,d\theta = [\tan\theta-\theta]_0^{\pi/6} = \frac1{\sqrt3} - \frac\pi6 = \frac{2\sqrt3-\pi}{6}$.
$k = 2\sqrt3-\pi$.
Answer: (B). | ||||
| 62 | 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 $(x^3-x^2)dy = (2-x)y\,dx \Rightarrow \frac{dy}{y} = \frac{2-x}{x^2(x-1)}dx$.
Partial fractions: $\frac{2-x}{x^2(x-1)} = -\frac1x - \frac2{x^2} + \frac1{x-1}$.
$\ln y = -\ln x + \frac2x + \ln(x-1) + c \Rightarrow y = K\frac{x-1}{x}e^{2/x}$.
$y(2) = K\cdot\frac12e = e \Rightarrow K = 2$. $y(4) = 2\cdot\frac34e^{1/2} = \frac32\sqrt e$.
Answer: (D). | ||||
| 63 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines Slope of BC $= \frac{-6}{12} = -\frac12$: the altitude from A is $y-7 = 2(x+1)$.
Slope of AC $= \frac{-12}{6} = -2$: the altitude from B is $y-1 = \frac12(x+7)$.
Solving: $2x+9 = \frac{x+9}{2} \Rightarrow x = -3$, $y = 3$.
Answer: (C). | ||||
| 64 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections The latus rectum runs from (1, 2) to (1, −2): length 4, midpoint (1, 0), on the line $x = 1$.
Both centres lie on the perpendicular bisector $y = 0$, at distance $\sqrt{(2\sqrt5)^2 - 2^2} = 4$ from (1, 0): (5, 0) and (−3, 0).
Distance between the centres $= 8$.
Answer: (B). | ||||
| 65 | Conic SectionsStandard forms of conics | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections $e_1^2 = 1-\frac{b^2}{25}$, $e_2^2 = 1+\frac{b^2}{16}$; $e_1^2e_2^2 = 1 \Rightarrow \frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{400} = 0 \Rightarrow b^2 = 9$.
$e_1 = \frac45$, $\alpha = 2\times5\times\frac45 = 8$; $e_2 = \frac54$, $\beta = 2\times4\times\frac54 = 10$.
Answer: (B). | ||||
| 66 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Midpoint (3, 1, 1); direction of the segment $(-2,6,-4) \parallel (1,-3,2)$, the plane's normal.
Plane: $(x-3) - 3(y-1) + 2(z-1) = 0 \Rightarrow x - 3y + 2z = 2$.
$(4,0,-1)$: $4 - 0 - 2 = 2$, on the plane. (The others give 6, 5, −5.)
Answer: (B). | ||||
| 67 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Variance is unchanged by shifting by $p$: $\sigma^2 = \frac{9}{10} - \left(\frac3{10}\right)^2 = 0.9 - 0.09 = 0.81$.
$\sigma = 0.9 = \frac9{10}$.
Answer: (A). | ||||
| 68 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are $9\times10^4$ five-digit numbers.
Two non-zero digits: $\binom92(2^5-2) = 36\times30 = 1080$.
Zero and one non-zero digit $d$: the first digit must be $d$, the other four are 0 or $d$, not all $d$: $9\times(2^4-1) = 135$.
Probability $= \frac{1215}{90000} = \frac{135}{10^4}$.
Answer: (D). | ||||
| 69 | 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 Let the common value be $\lambda$ and $c_1, c_2, c_3$ the three cosines (none zero at $\theta = \frac\pi9$): $a = \frac\lambda{c_1}$, $b = \frac\lambda{c_2}$, $c = \frac\lambda{c_3}$.
Dot product $= ab+bc+ca = \lambda^2\frac{c_1+c_2+c_3}{c_1c_2c_3}$.
$\cos\theta + \cos\left(\theta+\frac{2\pi}3\right) + \cos\left(\theta+\frac{4\pi}3\right) = 0$, so the dot product is 0 and the angle is $\frac\pi2$.
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 An implication is false only when its hypothesis is true and its conclusion false.
$p\wedge q$ true: $p = $ T, $q = $ T. $\sim q\vee r$ false: $\sim q = $ F (fine) and $r = $ F.
Answer: (C). | ||||
| 71 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Twice the first plus the second: $11z = 0$, so $z = 0$ and $x = 2y$. The third, $-14y+14y = 0$, holds.
Solutions $(2t, t, 0)$ with $x^2+y^2+z^2 = 5t^2$: $15\le5t^2\le150 \Rightarrow 3\le t^2\le30 \Rightarrow t^2\in\{4,9,16,25\}$.
$t = \pm2,\pm3,\pm4,\pm5$: 8 solutions.
Answer: 8. | ||||
| 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 Digits $a+b+c = 10$ with $1\le a\le9$, $0\le b,c\le9$. Put $a' = a-1$: $a'+b+c = 9$, non-negative: $\binom{11}{2} = 55$ solutions.
Exclude those with a digit too large: only $a' = 9$ ($a = 10$, $b = c = 0$). $b$ or $c$ cannot exceed 9 here.
$55 - 1 = 54$.
Answer: 54. | ||||
| 73 | Sequences and SeriesArithmetic and geometric means | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic and geometric means, from Sequences and Series. NCERT: Class 11: Sequences and Series G.P.: $3r^4 = 243 \Rightarrow r = 3$; 2nd G.M. $= 3r^2 = 27$.
A.P.: $d = \frac{243-3}{m+1} = \frac{240}{m+1}$; 4th A.M. $= 3+4d = 27 \Rightarrow d = 6 \Rightarrow m+1 = 40$.
$m = 39$.
Answer: 39. | ||||
| 74 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalEasy | |
A easy numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Normal to $y^2 = 4x$ at (1, 2): the tangent slope is $\frac2y = 1$, so the normal is $y-2 = -(x-1)$, meeting the $x$-axis at (3, 0).
Tangent to $y = e^x$ at $c$: $y - e^c = e^c(x-c)$, meeting the $x$-axis at $x = c-1$.
$c - 1 = 3 \Rightarrow c = 4$.
Answer: 4. | ||||
| 75 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Normal $= (\hat i+\hat j)\times(\hat j-\hat k) = -\hat i+\hat j+\hat k$, i.e. $(1,-1,-1)$.
Through (1, 0, 0): $x - y - z = 1$ (check (0, −1, 0): $0+1-0 = 1$, so the second line lies in it too).
Foot from M: $Q = M - \frac{1-0-1-1}{3}(1,-1,-1) = (1,0,1) + \frac13(1,-1,-1) = \left(\frac43,-\frac13,\frac23\right)$.
$3(\alpha+\beta+\gamma) = 3\times\frac53 = 5$.
Answer: 5. | ||||
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