JEE (Advanced) · 2007

JEE Advanced 2007, question by question

Which chapter each question came from, how the paper was spread, and how to solve each one. Keep the official paper open beside this page.

Sit this paper free All years, chapter-wise

Summary

The heaviest chapters in each subject, and how the questions were spread by level. One year is not a forecast.

Physics
32 Q
1. Electrostatics5
2. Atoms and Nuclei4
3. Laws of Motion3
16 easy · 16 medium · 0 hard
Chemistry
41 Q
1. p-Block Elements6
2. Electrochemistry5
3. Chemical Bonding and Molecular Structure4
27 easy · 14 medium · 0 hard
Mathematics
41 Q
1. Applications of Derivatives5
2. Conic Sections5
3. Integral Calculus4
16 easy · 24 medium · 1 hard

Chapters

Every chapter asked, and the NCERT chapters to read for it.

p-Block Elements6
No matching NCERT chapter
Electrochemistry5
Class 12: Electrochemistry
Chemical Bonding and Molecular Structure4
Class 11: Chemical Bonding and Molecular Structure
Hydrocarbons3
Class 11: Hydrocarbons
Organic Chemistry: Some Basic Principles and Techniques3
Class 11: Organic Chemistry – Some Basic Principles and Techniques
Thermodynamics3
Class 11: Thermodynamics
Coordination Compounds2
Class 12: Coordination Compounds
Redox Reactions2
Class 11: Redox Reactions
Aldehydes, Ketones and Carboxylic Acids2
Class 12: Aldehydes, Ketones and Carboxylic Acids
Solid State2
No matching NCERT chapter
Isolation of Metals1
No matching NCERT chapter
Solutions1
Class 12: Solutions
Surface Chemistry1
No matching NCERT chapter
Polymers1
No matching NCERT chapter
Principles of Qualitative Analysis1
No matching NCERT chapter
Structure of Atom1
Class 11: Structure of Atom
Chemical Kinetics1
Class 12: Chemical Kinetics
Biomolecules1
Class 12: Biomolecules
Practical Organic Chemistry1
No matching NCERT chapter

Applications of Derivatives5
Class 12: Application of Derivatives
Conic Sections5
Class 11: Conic Sections
Integral Calculus4
Class 12: Integrals; Class 12: Application of Integrals
Probability3
Class 11: Probability; Class 12: Probability
Vector Algebra3
Class 12: Vector Algebra
Circles3
Class 11: Conic Sections
Sequences and Series3
Class 11: Sequences and Series
Quadratic Equations2
Class 11: Complex Numbers and Quadratic Equations
Differential Equations2
Class 12: Differential Equations
Complex Numbers2
Class 11: Complex Numbers and Quadratic Equations
Trigonometry2
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Limits, Continuity and Differentiability2
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Straight Lines2
Class 11: Straight Lines
Matrices and Determinants1
Class 12: Matrices; Class 12: Determinants
Permutations and Combinations1
Class 11: Permutations and Combinations
Three Dimensional Geometry1
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry

Questions

Tap Explain on any question for a short summary and our full working.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
1	Current ElectricityMetre bridge	Class 12: Current Electricity	Single correctEasy
A easy single correct question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity At balance in a metre bridge, the ratio of the resistances in the two gaps equals the ratio of the wire lengths on the two sides. Let the unknown resistance be $X > 2\ \Omega$ and let the balance point be at $l$ cm from the end on the side of the 2 $\Omega$ resistance: $\frac{2}{X} = \frac{l}{100 - l}$. Since $X > 2$, $l < 50$. After interchanging, the balance point is at $100 - l$ from the same end, so the shift is $(100 - l) - l = 100 - 2l = 20$, giving $l = 40$ cm. Then $\frac{2}{X} = \frac{40}{60}$, so $X = 3\ \Omega$. Answer: (A).
2	Experimental SkillsOptics experiments	No matching NCERT chapter	Single correctMedium
A medium single correct question on optics experiments, from Experimental Skills. NCERT: No matching NCERT chapter Parallax rule: when the eye moves sideways, the nearer of two objects appears to move opposite to the eye relative to the farther one. The eye moves to the left and the image appears to the right of the pin, so the image has shifted opposite to the eye: the image is nearer to the eye than the pin. The eye is beyond the pin, looking towards the mirror, so the image being nearer to the eye means it is farther from the mirror than the pin: $\|v\| > x$. Also, the image is inverted and real, so the object is beyond the focus ($x > f$). For a concave mirror a real image is farther from the mirror than the object only when the object lies between $f$ and $2f$ (at $2f$ they coincide; beyond $2f$ the image is nearer). Hence $f < x < 2f$. Answer: (B).
3	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 Each half of the string has length $a$. When the masses are $2x$ apart, each half makes an angle $\theta$ with the vertical where $\sin\theta = \frac{x}{a}$ and $\cos\theta = \frac{\sqrt{a^2 - x^2}}{a}$. The string is light, so at the mid-point the force $F$ is balanced by the two tensions: $2T\cos\theta = F$, so $T = \frac{F}{2\cos\theta}$. Each mass is pulled horizontally by $T\sin\theta$ (the vertical part of the tension is balanced by the normal reaction, since $F$ is small and the masses stay on the surface): $m\,a_{\text{acc}} = T\sin\theta = \frac{F}{2}\tan\theta = \frac{F}{2}\cdot\frac{x}{\sqrt{a^2 - x^2}}$. So the acceleration is $\frac{F}{2m}\frac{x}{\sqrt{a^2 - x^2}}$. Answer: (B).
4	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance By Gauss's law, the field in the gap between the cylinders (radius $r$ between the two) depends only on the charge enclosed within radius $r$, i.e. the charge on the axis and on the inner cylinder: $E = \frac{\lambda_{\text{enc}}}{2\pi\varepsilon_0 r}$. A potential difference between the cylinders exists exactly when this field is non-zero. (A) Charge on the inner cylinder: enclosed charge is non-zero, so there is a field in the gap and a potential difference. True. (B) Charge only on the outer cylinder: nothing is enclosed in the gap, the field there is zero, no potential difference. False. (C) A line charge on the axis is enclosed, so there is a field in the gap and a potential difference. The statement says none appears: false. (D) Charge on both: the inner cylinder's charge is still enclosed, so a potential difference appears. False. Answer: (A).
5	ElectrostaticsConductors and dielectrics	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charge is conserved. The sphere is isolated and neutral; the outside charge only induces a separation of charge: negative charge gathers on the near side and an equal positive charge on the far side. The total charge on the sphere stays zero. (A), (B), (C) all describe a net negative charge, which would need charge to flow onto the sphere from somewhere; it is not earthed, so none does. Answer: (D).
6	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The capacitor branch (3 $\mu$F and 6 $\mu$F, junction X) and the resistor branch (3 $\Omega$ and 6 $\Omega$, junction Y) are both connected across the 9 V battery. Take the left end at 9 V and the right end at 0 V. Switch open: the capacitors are in series, $C = \frac{3 \times 6}{9} = 2\ \mu$F, so each carries $Q = 2 \times 9 = 18\ \mu$C. On the plates joined at X: $-18 + 18 = 0$. So X holds zero net charge. Switch closed: X is joined to Y. In steady state the current through the resistors is $\frac{9}{3+6} = 1$ A, so Y (and X) is at $9 - 3 = 6$ V. 3 $\mu$F capacitor: voltage $9 - 6 = 3$ V, charge 9 $\mu$C; its plate at X carries $-9\ \mu$C. 6 $\mu$F capacitor: voltage $6 - 0 = 6$ V, charge 36 $\mu$C; its plate at X carries $+36\ \mu$C. Net charge now at X: $36 - 9 = 27\ \mu$C. It was 0 before, and it can only have arrived through the switch, from Y to X. Answer: (C).
7	OpticsReflection and refraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Below the critical angle there is both partial reflection and refraction, so (A) and (B) are out. The reflected ray makes angle $\theta$ with the normal (in water); the refracted ray makes angle $r$ with the normal (in air), and going from denser to rarer, $r > \theta$. The two rays lie on opposite sides of the surface, and the angle between them is $180^\circ - \theta - r$. Since $r > \theta$, $180^\circ - \theta - r < 180^\circ - 2\theta$. Answer: (C).
8	Atoms and NucleiNuclear fission and fusion	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Both splits conserve mass number and charge: $137 + 97 + 2 = 236$, $53 + 39 = 92$; $140 + 94 + 2 = 236$, $56 + 36 = 92$. Fission of a heavy nucleus such as $^{236}U$ into medium-mass nuclei releases energy, because the fragments have a higher binding energy per nucleon. The energy released is the decrease in rest mass energy, so the rest mass energy of $^{236}U$ is greater than that of the products. (A) says exactly this: true. (B) and (C) say the opposite, and (D) says no energy is released: false. Answer: (A).
9	Atoms and NucleiHydrogen spectrum	Class 12: Atoms; Class 12: Nuclei	Single correctMedium
A medium single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The ultraviolet lines are the Lyman series ($n \to 1$); its longest wavelength is the $2 \to 1$ line: $\frac{1}{122\ \text{nm}} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$, so $\frac{1}{R} = \frac{3 \times 122}{4} = 91.5$ nm. The infrared series are Paschen ($n \to 3$), Brackett, Pfund…; the shortest infrared wavelength is the series limit of Paschen ($\infty \to 3$): $\frac{1}{\lambda} = \frac{R}{9}$, so $\lambda = 9 \times 91.5 = 823.5$ nm, which is about 823 nm (the Brackett limit, $16/R \approx 1464$ nm, is longer). Answer: (B).
10	Work, Energy and PowerWork-energy theorem	Class 11: Work, Energy and Power	Assertion and reasonMedium
A medium assertion and reason question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Statement-1. On the level surface the whole kinetic energy $\frac{1}{2}mv^2$ is lost to friction, so the loss of mechanical energy is $\frac{1}{2}mv^2$. On the incline the block goes up a distance $d$ with deceleration $g(\sin 30^\circ + \mu\cos 30^\circ)$, so $d = \frac{v^2}{2g(\sin 30^\circ + \mu\cos 30^\circ)}$. The mechanical energy lost is only the work against friction, $\mu mg\cos 30^\circ\, d = \frac{1}{2}mv^2 \cdot \frac{\mu\cos 30^\circ}{\sin 30^\circ + \mu\cos 30^\circ} < \frac{1}{2}mv^2$; the rest is stored as potential energy. So Statement-1 is true. Statement-2. The coefficient of friction is a property of the two surfaces in contact and does not depend on the angle of inclination. False. (The smaller loss comes from the smaller normal force and the shorter distance, not from a change in $\mu$.) Answer: (C).
11	Work, Energy and PowerCollisions	Class 11: Work, Energy and Power	Assertion and reasonMedium
A medium assertion and reason question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Statement-2: momentum is conserved in every collision (no external impulse), elastic or not. True. Statement-1: for a head-on (one-dimensional) elastic collision, the coefficient of restitution is 1, so the speed of separation equals the speed of approach: true. It is the conservation of kinetic energy together with momentum that gives this result; momentum conservation alone holds in inelastic collisions too, where the relative speeds differ. So if Statement-1 is read for a head-on collision, both are true but Statement-2 does not explain it: (B). For an oblique elastic collision only the components of relative velocity along the line of impact are equal; the relative speeds before and after need not be. Read in that general sense, Statement-1 is false and the answer is (D). The official key accepts either (B) or (D).
12	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Assertion and reasonEasy
A easy assertion and reason question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Statement-1: the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ is derived for paraxial rays, i.e. when the aperture is small compared with the radius of curvature. For larger mirrors spherical aberration appears and the formula fails. True. Statement-2: the laws of reflection hold at every point of any surface, plane or curved, with the normal at that point. False. Answer: (C).
13	Atoms and NucleiBohr model	Class 12: Atoms; Class 12: Nuclei	Assertion and reasonEasy
A easy assertion and reason question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Statement-1: characteristic X-rays come from transitions between inner energy levels of the target atoms; their wavelengths depend only on the target element, not on the accelerating voltage (which changes the cut-off wavelength of the continuous spectrum). True. Statement-2: only a small part of the electrons' kinetic energy becomes X-rays; most becomes heat in the target. True. Statement-2 does not explain why the characteristic lines are fixed; that comes from the fixed atomic energy levels. Answer: (B).
20	Physical World and MeasurementUnits and dimensions	Class 11: Units and Measurement	Matrix matchMedium
A medium matrix match question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement First reduce each Column II unit to base units. (P) volt × coulomb = joule, so (P) is J m = $\mathrm{kg\,m^3\,s^{-2}}$. (Q) is $\mathrm{kg\,m^3\,s^{-2}}$ as written. (R) is $\mathrm{m^2\,s^{-2}}$ (a speed squared). (S) farad × volt² = (C/V) × V² = C V = J, so (S) is J/kg = $\mathrm{m^2\,s^{-2}}$. So (P) = (Q) and (R) = (S). (A) $GM_eM_s = F r^2$ (from $F = \frac{GM_eM_s}{r^2}$): N m² = $\mathrm{kg\,m^3\,s^{-2}}$. A → P, Q. (B) $\frac{3RT}{M} = v_{\text{rms}}^2$: $\mathrm{m^2\,s^{-2}}$. B → R, S. (C) $F = qvB$ gives $\frac{F}{qB} = v$, so $\frac{F^2}{q^2B^2} = v^2$: $\mathrm{m^2\,s^{-2}}$. C → R, S. (D) $\frac{GM_e}{R_e} = gR_e = v_{\text{orbit}}^2$ near the surface: $\mathrm{m^2\,s^{-2}}$. D → R, S. Official: A → P, Q; B → R, S; C → R, S; D → R, S.
21	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Matrix matchMedium
A medium matrix match question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current (A) The capacitor discharges through the wire. The current and the potential difference fall exponentially (not constant), and the charge on the plates falls too; while the current flows, heat $I^2R$ is produced. A → Q. (B) A wire moving through a magnetic field has a motional emf $Bvl$ across it. With nothing connected, no current flows: the magnetic force pushes electrons to one end until the electric field of the separated charges balances it. So constant charges sit at the ends and a constant potential difference $Bvl$ exists between them; no current, no heat. B → R, S. (C) In a constant external field along the wire, electrons shift until the field of the induced end charges cancels the external field inside the conductor. Constant charges appear at the ends; inside, the net field is zero, so the conductor is an equipotential: no potential difference, no current. C → S. (D) A battery of constant emf drives a constant current $\frac{E}{R}$ (ignoring internal resistance changes), producing heat, with a constant potential difference across the wire. A current-carrying wire is neutral overall and does not hold fixed end charges in the sense of (S). D → P, Q, R. Official: A → Q; B → R, S; C → S; D → P, Q, R.
22	Atoms and NucleiBohr model	Class 12: Atoms; Class 12: Nuclei	Matrix matchEasy
A easy matrix match question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (A) Transitions between atomic energy levels give line spectra: the characteristic X-rays (inner-shell transitions) and the hydrogen spectrum. A → P, R. (B) Electrons are emitted from a material in the photoelectric effect, and in $\beta$-decay an electron is emitted (from the nucleus of an atom of the material). B → Q, S. (C) Moseley's law, $\sqrt{\nu} \propto (Z - b)$, describes the frequencies of characteristic X-rays. C → P. (D) Photon energy turning into kinetic energy of electrons is the photoelectric effect ($K_{\max} = h\nu - \phi$). D → Q. Official: A → P, R; B → Q, S; C → P; D → Q.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
23	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 CO has 14 electrons and bond order 3. $\mathrm{NO^+}$: $7 + 8 - 1 = 14$ electrons, bond order 3. $\mathrm{CN^-}$: $6 + 7 + 1 = 14$ electrons, bond order 3. $\mathrm{N_2}$: 14 electrons, bond order 3. $\mathrm{NO^-}$: $7 + 8 + 1 = 16$ electrons. The two electrons beyond 14 go into the antibonding $\pi^*$ orbitals, so bond order $= \frac{10 - 6}{2} = 2$. Only $\mathrm{NO^-}$ differs. Answer: (A).
24	p-Block ElementsGroup 16 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) $\mathrm{Na_2O_2}$ contains the peroxide ion $\mathrm{O_2^{2-}}$ (18 electrons, all paired): diamagnetic. (B) $\mathrm{O_3}$ has all electrons paired: diamagnetic. (C) $\mathrm{N_2O}$ has 22 electrons, all paired: diamagnetic. (D) $\mathrm{KO_2}$ contains the superoxide ion $\mathrm{O_2^-}$ (17 electrons), with one unpaired electron in a $\pi^*$ orbital: paramagnetic. Answer: (D).
25	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 Zinc blende is ZnS. After concentration by froth flotation it is roasted in air to the oxide: $\mathrm{2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2}$ The oxide is then reduced with coke (carbon) at high temperature: $\mathrm{ZnO + C \rightarrow Zn + CO}$ Self-reduction is used for copper and lead sulphides, not zinc; reduction by another metal (as with Cr or Mn by aluminium) is not used for zinc. Answer: (B).
26	HydrocarbonsElectrophilic substitution and directive influence	Class 11: Hydrocarbons	Single correctMedium
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The substrate is benzanilide, $\mathrm{C_6H_5NH{-}CO{-}C_6H_5}$, nitrated with conc. $\mathrm{HNO_3}$/conc. $\mathrm{H_2SO_4}$ (electrophile $\mathrm{NO_2^+}$). The ring on nitrogen carries $\mathrm{-NHCOPh}$: the nitrogen lone pair is donated into this ring (less strongly than in aniline, since it is also shared with the C=O), so this ring is activated and substitution goes ortho/para; the bulky group makes para the major product. The ring on the carbonyl carries $\mathrm{-CONHPh}$, an electron-withdrawing group, so that ring is deactivated (and would be meta-directing). So nitration occurs on the activated N-phenyl ring, at the para position: 4-nitrobenzanilide, structure (B). Answer: (B).
27	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctEasy
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Making an alkyne from a vicinal dibromide needs two dehydrobrominations. Alcoholic KOH removes the first HBr easily: $\mathrm{BrCH_2CH_2Br \rightarrow CH_2{=}CHBr}$. The second elimination, from a vinyl bromide, is difficult (the C–Br bond has partial double-bond character) and needs a stronger base, sodamide: $\mathrm{CH_2{=}CHBr \xrightarrow{NaNH_2} HC{\equiv}CH}$. (A) alone stops mostly at vinyl bromide. (C) aqueous KOH gives substitution (the glycol), not elimination. (D) Zn removes both Br atoms together and gives ethene, not ethyne. Answer: (B).
28	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The structural isomers of hexane: 1. n-hexane, $\mathrm{CH_3(CH_2)_4CH_3}$ 2. 2-methylpentane 3. 3-methylpentane 4. 2,2-dimethylbutane 5. 2,3-dimethylbutane There are 5. Answer: (C).
29	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 In white phosphorus, $\mathrm{P_4}$ is a tetrahedron. Each P atom forms three P–P single bonds and keeps one lone pair: four electron pairs, so each P is treated as $sp^3$ hybridised. An $sp^3$ orbital is one part s and three parts p, so its p-character is $\frac{3}{4} = 75\%$. Answer: (D).
30	Solutionsvan't Hoff factor	Class 12: Solutions	Single correctEasy
A easy single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Molar mass of $\mathrm{C_{11}H_8O_2}$ $= 11 \times 12 + 8 \times 1 + 2 \times 16 = 172\ \mathrm{g\ mol^{-1}}$. Molality $m = \frac{20/172}{0.050} = 2.326\ \mathrm{mol\ kg^{-1}}$. Expected depression without association: $K_f m = 1.72 \times 2.326 = 4.0$ K. $\Delta T_f = i K_f m$, so $i = \frac{2}{4.0} = 0.5$ (the acid dimerises in benzene through hydrogen bonding). Answer: (A).
31	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta_rG^\circ = \Delta_rH^\circ - T\Delta_rS^\circ = -54070 - 298 \times 10 = -57050\ \mathrm{J\ mol^{-1}}$. $\Delta_rG^\circ = -2.303RT\log_{10}K$, so $\log_{10}K = \frac{57050}{5705} = 10$. Answer: (B).
32	p-Block ElementsGroup 13 elements	No matching NCERT chapter	Assertion and reasonEasy
A easy assertion and reason question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement-1: boron forms only covalent compounds; it does not form a $\mathrm{B^{3+}}$ ion in its compounds. True. Statement-2: a $\mathrm{B^{3+}}$ ion would be extremely small and highly charged; its huge polarising power (Fajans' rules) would distort any anion, so the bond becomes covalent. Together with the very high sum of the first three ionisation enthalpies, this is why boron is covalent. True, and it explains Statement-1. Answer: (A).
33	p-Block ElementsGroup 13 elements	No matching NCERT chapter	Assertion and reasonEasy
A easy assertion and reason question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement-1: $\mathrm{H_3BO_3}$ is a weak monobasic acid ($K_a \approx 10^{-9}$). True. Statement-2: it does not donate its own protons. It is a Lewis acid: boron accepts a hydroxide ion from water, releasing a proton from the water: $\mathrm{B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^- + H_3O^+}$ So it is not a proton donor in the Brønsted sense. False. Answer: (C).
34	Chemical Bonding and Molecular StructureHydrogen bonding	Class 11: Chemical Bonding and Molecular Structure	Assertion and reasonEasy
A easy assertion and reason question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure In $o$-hydroxybenzoic acid (salicylic acid) the OH and COOH groups are close enough to form an intramolecular hydrogen bond, so Statement-2 is true. That ties up the groups within each molecule and reduces association between molecules. The para isomer cannot do this and forms intermolecular hydrogen bonds, so it is more associated and has the higher boiling point. Statement-1 says the opposite: false. Answer: (D).
35	Surface ChemistryEmulsions, surfactants and micelles	No matching NCERT chapter	Assertion and reasonMedium
A medium assertion and reason question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter Statement-1: surfactant ions aggregate into micelles only once their concentration exceeds the CMC (and above the Kraft temperature). True. Statement-2: at the CMC the ions cluster into large, slower micelles that also bind counter-ions, so the (molar) conductivity of the solution falls sharply there; the break in the conductivity plot is how the CMC is measured. Taken as true. The fall in conductivity is a consequence of micelle formation, not its cause, so Statement-2 does not explain Statement-1. Answer: (B).
36	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Argon is a noble gas: chemically inert. In arc welding it is used as a shield around the hot metal to keep out oxygen and nitrogen, which would otherwise oxidise or nitride the molten metal. It does not lower melting points (B), it is not flammable (C) and it is not a fuel (D). Answer: (A).
37	Chemical Bonding and Molecular StructureVSEPR theory	Class 11: Chemical Bonding and Molecular Structure	Single correctEasy
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Xe has 8 valence electrons. In $\mathrm{XeO_3}$ it forms three Xe=O bonds (using 6 electrons) and keeps one lone pair. Three bonding domains and one lone pair: $sp^3$ electron geometry, molecular shape trigonal pyramidal (like $\mathrm{NH_3}$). Answer: (C).
38	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Xenon fluorides hold xenon in a high positive oxidation state (+4 in $\mathrm{XeF_4}$, +6 in $\mathrm{XeF_6}$) and readily give it up: they are powerful oxidising and fluorinating agents, e.g. $\mathrm{XeF_4 + 4KI \rightarrow 2I_2 + 4KF + Xe}$ and $\mathrm{XeF_4 + Pt \rightarrow PtF_4 + Xe}$. They are reactive (they hydrolyse), and not basic (they act as fluoride donors or acceptors). So they are oxidizing: (A). The official key accepts (A) or (B).
39	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	ComprehensionEasy
A easy comprehension question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Moles of NaCl $= 4.0 \times 0.500 = 2.0$ mol, so 2.0 mol of $\mathrm{Cl^-}$. At the anode: $\mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}$. Two chloride ions give one $\mathrm{Cl_2}$, so complete electrolysis gives 1.0 mol of $\mathrm{Cl_2}$. Answer: (B).
40	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	ComprehensionMedium
A medium comprehension question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry With a mercury cathode, sodium is discharged and dissolves in the mercury as sodium amalgam: $\mathrm{Na^+ + e^- + Hg \rightarrow Na(Hg)}$. There are 2.0 mol of $\mathrm{Na^+}$. Taking the amalgam as NaHg (one Hg per Na), its molar mass is $23 + 200 = 223\ \mathrm{g\ mol^{-1}}$. Maximum mass of amalgam $= 2.0 \times 223 = 446$ g. Answer: (D).
41	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	ComprehensionEasy
A easy comprehension question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry 2.0 mol of $\mathrm{Cl^-}$ must be oxidised (and 2.0 mol of $\mathrm{Na^+}$ or water reduced at the cathode); each needs one electron, so 2.0 mol of electrons, i.e. 2 faradays. $Q = 2 \times 96500 = 193000$ C. Answer: (D).
42	Coordination CompoundsIsomerism in complexes	Class 12: Coordination Compounds	Matrix matchMedium
A medium matrix match question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) $\mathrm{[Co(NH_3)_4(H_2O)_2]^{2+}}$: Co is +2 ($d^7$, an odd number of electrons, so at least one unpaired: paramagnetic). An octahedral $\mathrm{MA_4B_2}$ complex has cis and trans isomers. A → P, Q, S. (B) $\mathrm{[Pt(NH_3)_2Cl_2]}$: Pt is +2 ($d^8$), square planar, all electrons paired: diamagnetic. Square planar $\mathrm{MA_2B_2}$ has cis and trans isomers. B → P, R, S. (C) $\mathrm{[Co(H_2O)_5Cl]^+}$: Co is $+2$ ($x - 1 = +1$), $d^7$: paramagnetic. $\mathrm{MA_5B}$ has no geometrical isomers. C → Q, S. (D) $\mathrm{[Ni(H_2O)_6]^{2+}}$: Ni is +2 ($d^8$ octahedral, two unpaired electrons in $e_g$): paramagnetic; $\mathrm{MA_6}$ has no isomers. D → Q, S. Official: A → P, Q, S; B → P, R, S; C → Q, S; D → Q, S.
43	PolymersNylon, Teflon, Bakelite and PVC	No matching NCERT chapter	Matrix matchEasy
A easy matrix match question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter (A) Cellulose is a natural polymer of $\beta$-D-glucose units joined by glycosidic linkages. A → P, S. (B) Nylon-6,6 is a synthetic condensation polymer of hexamethylenediamine and adipic acid, joined by amide links. B → Q, R. (C) Proteins are natural polymers of amino acids joined by peptide (amide) bonds. C → P, R. (D) Sucrose is a disaccharide, not a polymer; its glucose and fructose units are joined by a glycosidic linkage. D → S. Official: A → P, S; B → Q, R; C → P, R; D → S.
44	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	Matrix matchMedium
A medium matrix match question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics (A) Hydrogen at 200 atm and 273 K: at high pressure the volume of the molecules matters, and for $\mathrm{H_2}$ the attraction constant $a$ is very small, so the van der Waals equation reduces to $P(V - nb) = nRT$, giving $Z = 1 + \frac{Pb}{RT} > 1$. A → P, S. (B) Hydrogen at very low pressure behaves ideally: $PV = nRT$. B → R. (C) $\mathrm{CO_2}$ at 1 atm and 273 K is below its critical temperature and easily liquefied; attraction dominates and $Z < 1$. C → P, Q. (D) The key reads this as a gas whose own molecular volume is large, so the excluded-volume term dominates: $P(V - nb) = nRT$ and $Z \neq 1$. D → P, S. Official: A → P, S; B → R; C → P, Q; D → P, S.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
45	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 From the first equation: $\alpha + \beta = p$, $\alpha\beta = r$. From the second: $\frac{\alpha}{2} + 2\beta = q$, i.e. $\alpha + 4\beta = 2q$ (and the product $\frac{\alpha}{2}\cdot 2\beta = \alpha\beta = r$ agrees). Subtracting: $3\beta = 2q - p$, so $\beta = \frac{2q - p}{3}$, and $\alpha = p - \beta = \frac{4p - 2q}{3} = \frac{2(2p - q)}{3}$. $r = \alpha\beta = \frac{2}{9}(2p - q)(2q - p)$. Answer: (D).
46	Differential EquationsLinear differential equations	Class 12: Differential Equations	Single correctMedium
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations The numerator vanishes at $t = x$, so the limit is the derivative with respect to $t$ of $t^2 f(x) - x^2 f(t)$ at $t = x$: $2x f(x) - x^2 f'(x) = 1$. Rewrite: $x^2 f'(x) - 2x f(x) = -1$. Dividing by $x^4$: $\frac{d}{dx}\left(\frac{f(x)}{x^2}\right) = -\frac{1}{x^4}$. Integrate: $\frac{f(x)}{x^2} = \frac{1}{3x^3} + C$, so $f(x) = \frac{1}{3x} + Cx^2$. $f(1) = 1$ gives $\frac{1}{3} + C = 1$, $C = \frac{2}{3}$. $f(x) = \frac{1}{3x} + \frac{2x^2}{3}$. Answer: (A).
47	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 Count circular arrangements of the 10 people. Given event B (each American couple together): treat each American couple as one block. The units are 4 blocks + Indian man + Indian wife = 6 units, arranged in a circle in $5!$ ways, with $2^4$ internal orders: $n(B) = 5! \cdot 2^4$. Event A ∩ B (also the Indian couple together): 5 blocks in a circle, $4!$ ways, $2^5$ internal orders: $n(A \cap B) = 4! \cdot 2^5$. $P(A \mid B) = \frac{4! \cdot 32}{5! \cdot 16} = \frac{2}{5}$. Answer: (C).
48	Applications of DerivativesTangents and normals	Class 12: Application of Derivatives	Single correctMedium
A medium single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Tangent at $(c, e^c)$: $y = e^c(1 + t)$ where $t = x - c$. Chord through $(c - 1, e^{c-1})$ and $(c + 1, e^{c+1})$: slope $\frac{e^{c+1} - e^{c-1}}{2} = e^c\sinh 1$, so $y = e^c\left(e^{-1} + (t + 1)\sinh 1\right)$. Equating: $1 + t = e^{-1} + \sinh 1 + t\sinh 1$, so $t(1 - \sinh 1) = e^{-1} + \sinh 1 - 1 = \cosh 1 - 1$. $\cosh 1 - 1 \approx 0.543 > 0$ and $1 - \sinh 1 \approx -0.175 < 0$, so $t \approx -3.1 < 0$. Geometric view: the chord has slope $e^c\sinh 1 > e^c$, steeper than the tangent; at $x = c$ the chord lies above the tangent (the curve is convex), so going left the steeper chord comes down to meet the tangent: the intersection is to the left of $x = c$. Answer: (A).
49	Integral CalculusFundamental theorem of calculus	Class 12: Integrals; Class 12: Application of Integrals	Single correctEasy
A easy single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals At $x = \frac{\pi}{4}$, $\sec^2 x = 2$, so both numerator and denominator are 0. By L'Hôpital's rule and the fundamental theorem of calculus: numerator's derivative $= f(\sec^2 x)\cdot 2\sec^2 x\tan x \to f(2)\cdot 2 \cdot 2 \cdot 1 = 4f(2)$; denominator's derivative $= 2x \to \frac{\pi}{2}$. Limit $= \frac{4f(2)}{\pi/2} = \frac{8}{\pi}f(2)$ (taking $f$ continuous). Answer: (A).
50	Conic SectionsHyperbola	Class 11: Conic Sections	Single correctEasy
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse: $\frac{x^2}{4} + \frac{y^2}{3} = 1$, $e^2 = 1 - \frac{3}{4} = \frac{1}{4}$, foci $(\pm 1, 0)$. Hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $a = \sin\theta$ and the same foci: $a^2 + b^2 = 1$, so $b^2 = 1 - \sin^2\theta = \cos^2\theta$. Equation: $\frac{x^2}{\sin^2\theta} - \frac{y^2}{\cos^2\theta} = 1$, i.e. $x^2\,\mathrm{cosec}^2\theta - y^2\sec^2\theta = 1$. Answer: (A).
51	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Single correctEasy
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Coplanar means the scalar triple product is zero: $\begin{vmatrix} -\lambda^2 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2 \end{vmatrix} = 0$. For a matrix with every diagonal entry $d$ and every off-diagonal entry 1, the determinant is $(d - 1)^2(d + 2)$. With $d = -\lambda^2$: $(\lambda^2 + 1)^2(2 - \lambda^2) = 0$. $\lambda^2 + 1 \neq 0$ for real $\lambda$, so $\lambda^2 = 2$, $\lambda = \pm\sqrt{2}$: two values. Answer: (C).
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 Take east as the real axis and north as the imaginary axis. North-east is at $45^\circ$: the first step is $3e^{i\pi/4}$. North-west is at $135^\circ = 45^\circ + 90^\circ$: the second step is $4e^{i3\pi/4} = 4i\,e^{i\pi/4}$ (multiplying by $i$ turns by $90^\circ$). $P = 3e^{i\pi/4} + 4ie^{i\pi/4} = (3 + 4i)e^{i\pi/4}$. Answer: (D).
53	TrigonometryTrigonometric identities	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctEasy
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions First equation: $2\sin^2\theta = 1 - 2\sin^2\theta$, so $\sin^2\theta = \frac{1}{4}$, $\sin\theta = \pm\frac{1}{2}$. Second equation: $2(1 - \sin^2\theta) = 3\sin\theta$, i.e. $2\sin^2\theta + 3\sin\theta - 2 = 0$, $(2\sin\theta - 1)(\sin\theta + 2) = 0$, so $\sin\theta = \frac{1}{2}$ ($-2$ is impossible). Both hold when $\sin\theta = \frac{1}{2}$: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ in $[0, 2\pi]$. Two solutions. Answer: (C).
54	ProbabilityBayes' theorem	Class 11: Probability; Class 12: Probability	Assertion and reasonMedium
A medium assertion and reason question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Statement-2: the $H_i$ are mutually exclusive and exhaustive, so their probabilities add to 1. True. Statement-1: $P(H_i \mid E) = \frac{P(E \mid H_i)P(H_i)}{P(E)}$. Since $0 < P(E) < 1$, this is strictly greater than $P(E \mid H_i)P(H_i)$ only when $P(E \mid H_i)P(H_i) > 0$. If $E$ and some $H_i$ are disjoint, $P(E \mid H_i) = 0$ and both sides are 0, so the strict inequality fails. (For example, $n = 2$, $E = H_1$: then $P(H_2 \mid E) = 0 = P(E \mid H_2)P(H_2)$.) So Statement-1 is not true for every $i$ in general: false. Answer: (D).
55	CirclesStandard and general form	Class 11: Conic Sections	Assertion and reasonEasy
A easy assertion and reason question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Statement-2: the locus of points from which perpendicular tangents are drawn to $x^2 + y^2 = r^2$ is the director circle $x^2 + y^2 = 2r^2$. Here $r^2 = 169$, so it is $x^2 + y^2 = 338$. True. Statement-1: $17^2 + 7^2 = 289 + 49 = 338$, so $(17, 7)$ lies on the director circle and the tangents from it are perpendicular. True, and it follows directly from Statement-2. Answer: (A).
56	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Assertion and reasonMedium
A medium assertion and reason question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Going round a regular hexagon, each side turns by $60^\circ$ from the last. Taking $\vec{PQ}$ at angle $0^\circ$: $\vec{QR}$ at $60^\circ$, $\vec{RS}$ at $120^\circ$, $\vec{ST}$ at $180^\circ$. So $\vec{ST}$ is antiparallel to $\vec{PQ}$: $\vec{PQ} \times \vec{ST} = \vec{0}$; and $\vec{RS}$ is at $120^\circ$ to $\vec{PQ}$: $\vec{PQ} \times \vec{RS} \neq \vec{0}$. Statement-1: $\vec{PQ} \times (\vec{RS} + \vec{ST}) = \vec{PQ} \times \vec{RS} \neq \vec{0}$. True. Statement-2 has it the wrong way round. False. Answer: (C).
57	Integral CalculusIntegral as an antiderivative	Class 12: Integrals; Class 12: Application of Integrals	Assertion and reasonEasy
A easy assertion and reason question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Statement-2: $\sin(x + \pi) = -\sin x$, so $\sin^2(x + \pi) = \sin^2 x$. True. Statement-1: $F(x) = \int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$. Then $F(x + \pi) = \frac{x + \pi}{2} - \frac{\sin 2x}{4} + C = F(x) + \frac{\pi}{2} \neq F(x)$. A periodic integrand with non-zero mean over a period does not give a periodic integral. False. Answer: (D).
58	Sequences and SeriesArithmetic progressions	Class 11: Sequences and Series	ComprehensionMedium
A medium comprehension question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $V_r = \frac{r}{2}\left[2r + (r - 1)(2r - 1)\right] = \frac{r}{2}(2r^2 - r + 1) = r^3 - \frac{r^2}{2} + \frac{r}{2}$. $\sum_{r=1}^{n} V_r = \frac{n^2(n + 1)^2}{4} - \frac{n(n + 1)(2n + 1)}{12} + \frac{n(n + 1)}{4}$ $= \frac{n(n + 1)}{12}\left[3n(n + 1) - (2n + 1) + 3\right] = \frac{n(n + 1)}{12}(3n^2 + n + 2)$. Check $n = 1$: $V_1 = 1$, and $\frac{1 \cdot 2 \cdot 6}{12} = 1$. Answer: (B).
59	Sequences and SeriesArithmetic progressions	Class 11: Sequences and Series	ComprehensionMedium
A medium comprehension question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series From $V_r = r^3 - \frac{r^2}{2} + \frac{r}{2}$: $V_{r+1} - V_r = (3r^2 + 3r + 1) - \frac{2r + 1}{2} + \frac{1}{2} = 3r^2 + 2r + 1$. $T_r = 3r^2 + 2r + 1 - 2 = 3r^2 + 2r - 1 = (3r - 1)(r + 1)$. For $r \geq 1$ both factors are at least 2 ($3r - 1 \geq 2$, $r + 1 \geq 2$), so $T_r$ is always composite: $T_1 = 4$, $T_2 = 15$, $T_3 = 32$. It is not always odd (4) or always even (15), and never prime. Answer: (D).
60	Sequences and SeriesArithmetic progressions	Class 11: Sequences and Series	ComprehensionEasy
A easy comprehension question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series With $T_r = 3r^2 + 2r - 1$: $Q_r = T_{r+1} - T_r = 3(2r + 1) + 2 = 6r + 5$. So $Q_1 = 11$, $Q_2 = 17$, $Q_3 = 23, \ldots$: an A.P. with common difference 6. Answer: (B).
61	Conic SectionsParabola	Class 11: Conic Sections	ComprehensionMedium
A medium comprehension question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Intersection: $x^2 + 8x - 9 = 0$, so $x = 1$ ($x = -9$ gives no real $y$). $P = (1, 2\sqrt{2})$, $Q = (1, -2\sqrt{2})$. Tangent to the circle at $P$: $x + 2\sqrt{2}y = 9$, meeting the $x$-axis at $R = (9, 0)$ (by symmetry the tangent at $Q$ meets it there too). Tangent to the parabola at $P$: $2\sqrt{2}\,y = 4(x + 1)$, meeting the $x$-axis at $S = (-1, 0)$. Both triangles have base $PQ$ on the line $x = 1$. Their heights are the distances of $S$ and $R$ from that line: $1 - (-1) = 2$ and $9 - 1 = 8$. Ratio of areas $= 2 : 8 = 1 : 4$. Answer: (C).
62	Conic SectionsParabola	Class 11: Conic Sections	ComprehensionMedium
A medium comprehension question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Intersection: $x^2 + 8x - 9 = 0$, so $x = 1$ ($x = -9$ gives no real $y$). $P = (1, 2\sqrt{2})$, $Q = (1, -2\sqrt{2})$. Tangent to the circle at $P$: $x + 2\sqrt{2}y = 9$, meeting the $x$-axis at $R = (9, 0)$ (by symmetry the tangent at $Q$ meets it there too). Tangent to the parabola at $P$: $2\sqrt{2}\,y = 4(x + 1)$, meeting the $x$-axis at $S = (-1, 0)$. Triangle $PRS$: $SR = 10$, $PS = \sqrt{2^2 + 8} = 2\sqrt{3}$, $PR = \sqrt{8^2 + 8} = 6\sqrt{2}$. Area $= \frac{1}{2} \times SR \times (\text{height of } P) = \frac{1}{2} \times 10 \times 2\sqrt{2} = 10\sqrt{2}$. Circumradius $= \frac{abc}{4\Delta} = \frac{10 \cdot 2\sqrt{3} \cdot 6\sqrt{2}}{40\sqrt{2}} = 3\sqrt{3}$. Answer: (B).
63	Conic SectionsParabola	Class 11: Conic Sections	ComprehensionMedium
A medium comprehension question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Intersection: $x^2 + 8x - 9 = 0$, so $x = 1$ ($x = -9$ gives no real $y$). $P = (1, 2\sqrt{2})$, $Q = (1, -2\sqrt{2})$. Tangent to the circle at $P$: $x + 2\sqrt{2}y = 9$, meeting the $x$-axis at $R = (9, 0)$ (by symmetry the tangent at $Q$ meets it there too). Tangent to the parabola at $P$: $2\sqrt{2}\,y = 4(x + 1)$, meeting the $x$-axis at $S = (-1, 0)$. Triangle $PQR$: $PQ = 4\sqrt{2}$, $PR = QR = \sqrt{64 + 8} = 6\sqrt{2}$. Semi-perimeter $s = \frac{4\sqrt{2} + 12\sqrt{2}}{2} = 8\sqrt{2}$. Area $= \frac{1}{2} \times PQ \times 8 = 16\sqrt{2}$ (height from $R$ to $x = 1$ is 8). Inradius $= \frac{\Delta}{s} = \frac{16\sqrt{2}}{8\sqrt{2}} = 2$. Answer: (D).
64	Matrices and DeterminantsConsistency of linear systems	Class 12: Matrices; Class 12: Determinants	Matrix matchMedium
A medium matrix match question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The determinant of the system is $\Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = -(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$, and $a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]$, which is 0 only when $a = b = c$. (A) $a = b = c \neq 0$: all three equations are $x + y + z = 0$, the same plane. A → R. (B) $\Delta = 0$ but $a, b, c$ not all equal, so the rank is 2 and the solutions form a line. Adding the equations gives $(a + b + c)(x + y + z) = 0$, true for all points; and $x = y = z$ satisfies each equation since $a + b + c = 0$. So the common line is $x = y = z$. B → Q. (C) $\Delta \neq 0$: only the trivial solution, the planes meet at the single point (0, 0, 0). C → P. (D) $a = b = c$ and $a + b + c = 0$ force $a = b = c = 0$: every equation is $0 = 0$, satisfied by the whole space. D → S. Official: A → R; B → Q; C → P; D → S.
65	Limits, Continuity and DifferentiabilityDifferentiability	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Matrix matchMedium
A medium matrix match question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability (A) $x\|x\|$ is $x^2$ for $x \geq 0$ and $-x^2$ for $x < 0$: continuous, derivative $2\|x\|$ exists everywhere (0 at the origin), and it is strictly increasing (derivative $\geq 0$, zero only at a point). A → P, Q, R. (B) $\sqrt{\|x\|}$ is continuous; at 0 the difference quotient $\frac{\sqrt{\|h\|}}{h}$ is unbounded, so not differentiable there. It decreases on $(-1, 0)$. B → P, S. (C) $x + [x]$ is $x - 1$ on $(-1, 0)$ and $x$ on $[0, 1)$: it jumps from $-1$ to 0 at $x = 0$, so it is discontinuous and not differentiable there, yet it is strictly increasing on $(-1, 1)$ (each piece increases and the jump is upward). C → R, S. (D) On $(-1, 1)$, $\|x - 1\| + \|x + 1\| = (1 - x) + (x + 1) = 2$, a constant: continuous and differentiable, not strictly increasing. D → P, Q. Official: A → P, Q, R; B → P, S; C → R, S; D → P, Q.
66	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	Matrix matchEasy
A easy matrix match question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals (A) $[\tan^{-1}x]_{-1}^{1} = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$. A → S. (B) $[\sin^{-1}x]_0^1 = \frac{\pi}{2}$. B → S. (C) $\int \frac{dx}{1 - x^2} = \frac{1}{2}\log\left\|\frac{1 + x}{1 - x}\right\|$. From 2 to 3: $\frac{1}{2}\left[\log\frac{4}{2} - \log\frac{3}{1}\right] = \frac{1}{2}\log\frac{2}{3}$. C → P. (D) $[\sec^{-1}x]_1^2 = \frac{\pi}{3} - 0 = \frac{\pi}{3}$. D → R. Official: A → S; B → S; C → P; D → R.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
1	Experimental SkillsHeat and sound experiments	No matching NCERT chapter	Single correctEasy
A easy single correct question on heat and sound experiments, from Experimental Skills. NCERT: No matching NCERT chapter The fork is held just above the open mouth of the vertical tube, and the sound must travel down the air column. The prongs vibrate in the plane that contains them, so to push air up and down the tube they are held in a vertical plane. (A) is right and (B) is wrong. The tube is closed at the water surface and open at the top, so resonance occurs when the air column (plus the end correction) is an odd number of quarter wavelengths: first resonance $l_1 + e = \frac{\lambda}{4}$, second resonance $l_2 + e = \frac{3\lambda}{4}$. Neither length is close to $\lambda$ or to $\frac{\lambda}{2}$, so (C) and (D) are wrong. Answer: (A).
2	Physical World and MeasurementError analysis	Class 11: Units and Measurement	Single correctMedium
A medium single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $Y = \frac{FL}{A\,\Delta l} = \frac{4MgL}{\pi d^2 \Delta l}$ $Y = \frac{4 \times 1.0 \times 9.8 \times 2}{\pi (0.4 \times 10^{-3})^2 (0.8 \times 10^{-3})} = \frac{78.4}{\pi \times 1.28 \times 10^{-10}} \approx 1.95 \times 10^{11}\ \text{N/m}^2 \approx 2.0 \times 10^{11}\ \text{N/m}^2$. M, g and L are exact, so only $d$ and $\Delta l$ carry error: $\frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta(\Delta l)}{\Delta l} = 2 \times \frac{0.01}{0.4} + \frac{0.05}{0.8} = 0.05 + 0.0625 = 0.1125$ $\Delta Y = 0.1125 \times 2.0 \times 10^{11} \approx 0.225 \times 10^{11}\ \text{N/m}^2$. Rounded to one decimal place this lies between 0.2 and 0.3, which is why the official key accepts both. Answer: (A) or (B), $(2.0 \pm 0.2) \times 10^{11}$ or $(2.0 \pm 0.3) \times 10^{11}\ \text{N/m}^2$.
3	Laws of MotionNewton's laws	Class 11: Laws of Motion	Single correctEasy
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion By Newton's second law the force is the rate of change of momentum: $\vec{F} = \frac{d\vec{p}}{dt} = Ak\left[-\hat{i}\sin(kt) - \hat{j}\cos(kt)\right]$ $\vec{F} \cdot \vec{p} = A^2 k\left[-\sin(kt)\cos(kt) + \sin(kt)\cos(kt)\right] = 0$ Both vectors are non-zero, so they are perpendicular. (The magnitude of $\vec{p}$ is constant, $\|\vec{p}\| = A$, so the force only turns it, as in uniform circular motion.) Answer: (D) $90^\circ$.
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 Rolling without slipping, the kinetic energy is $\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2\left(1 + \frac{K^2}{R^2}\right)$, with $I = mK^2$ and $\omega = \frac{v}{R}$. At the highest point it is momentarily at rest (friction does no work in pure rolling), so $\frac{1}{2}mv^2\left(1 + \frac{K^2}{R^2}\right) = mg \cdot \frac{3v^2}{4g}$ $1 + \frac{K^2}{R^2} = \frac{3}{2}$, so $\frac{K^2}{R^2} = \frac{1}{2}$, i.e. $I = \frac{1}{2}mR^2$. That is a disc (ring 1, solid sphere $\frac{2}{5}$, hollow sphere $\frac{2}{3}$). Answer: (D) disc.
5	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctMedium
A medium single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The section ABCD is a rectangle of width $2R$ (a diameter) and height $h$. Pressure force across it: at depth $y$ the pressure is $P_0 + \rho g y$, acting on a strip of area $2R\,dy$: $F_P = \int_0^h (P_0 + \rho g y)\,2R\,dy = 2P_0Rh + R\rho g h^2$. This pushes the two halves apart. Surface tension: the free surface is cut along the top edge AB, of length $2R$. Across that line each half pulls the other with force $T \times 2R$, which pulls the halves together. Net force of one half on the other: $\left\|2P_0Rh + R\rho g h^2 - 2RT\right\|$. Answer: (B).
6	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Treat the cavity as the full sphere (charge density $\rho$) plus a small sphere of density $-\rho$ filling the cavity. Inside a uniformly charged sphere, Gauss's law gives $\vec{E} = \frac{\rho \vec{r}}{3\varepsilon_0}$, where $\vec{r}$ is measured from its centre. At a point in the cavity, with $\vec{r}_1$ from the big sphere's centre $O$ and $\vec{r}_2$ from the cavity's centre $O'$: $\vec{E} = \frac{\rho \vec{r}_1}{3\varepsilon_0} - \frac{\rho \vec{r}_2}{3\varepsilon_0} = \frac{\rho}{3\varepsilon_0}(\vec{r}_1 - \vec{r}_2) = \frac{\rho}{3\varepsilon_0}\overrightarrow{OO'}$ This is the same vector at every point of the cavity and is not zero, since the cavity is off-centre. Answer: (B) non-zero and uniform.
7	ElectrostaticsElectric dipole	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Both points lie in the plane $z = 0$, which is the perpendicular bisector of the dipole. Every point of that plane is equidistant from $+q$ and $-q$, so the potential there is $V = \frac{kq}{r} - \frac{kq}{r} = 0$. Work done by the field $= q_0(V_i - V_f) = q_0(0 - 0) = 0$. The electrostatic field is conservative, so the path does not matter. Answer: (C) zero.
8	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Single correctMedium
A medium single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter In $a < x < 2a$: $\vec{F} = q\vec{v} \times \vec{B} = qv_0B_0(\hat{i} \times \hat{j}) = qv_0B_0\hat{k}$. The charge starts bending towards $+z$ on a circular arc, so (B) and (D), which bend towards $-z$, are out. In $2a < x < 3a$ the field is reversed, so the curvature reverses: the arc bends back the other way, with the same radius. By symmetry the charge turns back through the same angle it turned through in the first region and leaves at $x = 3a$ moving parallel to $+x$ again, but displaced to positive $z$. The path is a smooth S-shaped curve, still rising throughout. The velocity of a charge in a magnetic field changes direction continuously, so the path cannot have a sharp corner at $x = 2a$, which rules out (C). Answer: (A).
9	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Electron momentum $p = \frac{h}{\lambda}$, so its kinetic energy is $K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$. The shortest X-ray wavelength comes from an electron giving all its energy to one photon: $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$, so $\lambda_0 = \frac{2mc\lambda^2}{h}$. Answer: (A).
10	Motion of System of Particles and Rigid BodyCentre of mass	Class 11: Systems of Particles and Rotational Motion	Assertion and reasonEasy
A easy assertion and reason question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Statement-1: zero torque about the centre of mass does not mean zero external force. A force whose line of action passes through the centre of mass (gravity on a falling ball, for instance) gives no torque about it but accelerates it. The velocity of the centre of mass is constant only when the net external force is zero. False. Statement-2: an isolated system has no external force, so its total linear momentum is conserved. True. Answer: (D).
11	Laws of MotionNewton's laws	Class 11: Laws of Motion	Assertion and reasonEasy
A easy assertion and reason question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Statement-1: if the cloth is jerked out quickly, the friction on the dishes acts for a very short time, so the impulse it gives them is small and they barely move: they stay on the table by their inertia. True. Statement-2: Newton's third law. True. The trick is explained by inertia (the first law) and the small impulse, not by the third law. So both are true, and Statement-2 is not the explanation. Answer: (B).
12	Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Assertion and reasonMedium
A medium assertion and reason question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The alternating current in the coil makes an alternating flux along the iron rod, which links the ring and induces an alternating current in it. Near the ring the field lines spread out from the rod, so the field has a radial (horizontal) component. The force on the ring current is $I\,d\vec{l} \times \vec{B}$; with the ring current tangential and the radial component horizontal, this force is vertical. Because of the ring's inductance, its current lags behind the changing flux, and by Lenz's law the ring opposes the increase of flux through it: averaged over a cycle the ring is repelled by the coil, i.e. pushed upward. At a certain height this average force balances the ring's weight and it floats (the familiar jumping-ring experiment). Statement-1 is true and Statement-2 gives exactly the mechanism. Answer: (A).
13	Behaviour of Perfect Gas and Kinetic TheoryKinetic theory	Class 11: Kinetic Theory	Assertion and reasonEasy
A easy assertion and reason question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Statement-1: kinetic theory gives $PV = \frac{1}{3}Nm\overline{v^2}$, so the total translational kinetic energy is $\frac{1}{2}Nm\overline{v^2} = \frac{3}{2}PV$. True. Statement-2: molecules do collide and their velocities change. True. But Statement-1 comes from the momentum the molecules transfer to the walls; it does not depend on the molecules colliding with each other (the ideal-gas derivation even ignores those collisions). Not the explanation. Answer: (B).
20	Oscillations and WavesSHM	Class 11: Oscillations; Class 11: Waves	Matrix matchMedium
A medium matrix match question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves (A) $v^2 = c_1^2(c_2 - x^2)$. Differentiating, $2v\frac{dv}{dt} = -2c_1^2 x v$, so $a = -c_1^2 x$: simple harmonic motion with $\omega = c_1$ and amplitude $\sqrt{c_2}$. It turns back at both ends, again and again, so not Q or S, and its kinetic energy rises and falls, so not R. A → P. (B) $\frac{dx}{dt} = -kx$ gives $x = x_0e^{-kt}$: the object moves towards the origin, slowing down, and never reaches it. It never changes direction (Q) and its speed $k\|x\|$, hence kinetic energy, keeps decreasing (R). Acceleration $a = k^2x$ points away from the origin, so it is not SHM. B → Q, R. (C) In the elevator's frame the object feels its weight, the spring force and a pseudo-force $ma$ downward. Initially at the old equilibrium, it now has a new equilibrium $\frac{ma}{k_s}$ lower, and oscillates about it: simple harmonic motion, changing direction repeatedly. C → P. (D) The escape speed is $\sqrt{2GM_e/R_e}$ and the launch speed $2\sqrt{GM_e/R_e}$ is larger, so the object escapes: it moves outward for ever (Q), slowing all the time as gravity pulls back, so its kinetic energy keeps decreasing (R). D → Q, R. Official key: A → P; B → Q, R; C → P; D → Q, R, with D → R alone also accepted.
21	Magnetic Effects of Current and MagnetismBiot-Savart law	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Matrix matchMedium
A medium matrix match question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Read the current directions from the arrows in the figures. (A) Parallel wires, currents in the same direction. At the midpoint the two fields are equal in size and opposite in direction (one into the page, one out), so they cancel. Parallel currents attract, so not S. A → Q, R. (B) Coaxial loops with currents in the same sense. On the axis each loop's field points along the axis in the same direction, so they add. Currents in the same sense attract. B → P. (C) Coplanar loops side by side: the left one carries current anticlockwise (down at its left edge), the right one clockwise (down at its right edge). P lies outside both loops. Outside an anticlockwise loop, in its plane, the field points into the page; outside a clockwise loop it points out of the page. The fields are opposite and, by symmetry, equal, so they cancel. The facing sides of the loops both carry current upward, so the loops attract. C → Q, R. (D) Concentric loops: the outer carries current clockwise (up at its left edge), the inner anticlockwise (up at its right edge). At the centre their fields are opposite; they do not cancel, because the inner loop of smaller radius gives the larger field $\frac{\mu_0 I}{2r}$. Antiparallel currents repel, so the loops repel. D → Q, S. Official key: A → Q, R; B → P; C → Q, R; D → Q, with D → Q, S also accepted.
22	Properties of Bulk MatterThermal expansion and calorimetry	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Matrix matchEasy
A easy matrix match 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 (A) A bimetallic strip bends because its two metals expand by different amounts when heated: thermal expansion of solids (S). The bending turns heat into mechanical movement, so energy conversion (Q) is also accepted. A → S (or Q, S). (B) A steam engine turns heat into mechanical work: energy conversion. B → Q. (C) An incandescent lamp gives light because its filament is white hot: radiation from a hot body (P); electrical energy becomes heat and light, so energy conversion (Q) is also accepted. C → P (or P, Q). (D) An electric fuse works by melting when the current is too large (R); electrical energy is converted to heat, so Q is also accepted. D → R (or Q, R). Official key: A → Q, S or S alone; B → Q; C → P, Q or P alone; D → Q, R or R alone.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
23	Redox ReactionsBalancing redox equations	Class 11: Redox Reactions	Single correctEasy
A easy single correct question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Mohr's salt, $\text{FeSO}_4 \cdot (\text{NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O}$, supplies $\text{Fe}^{2+}$, which is oxidised to $\text{Fe}^{3+}$ (one electron each). Dichromate is reduced: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ (two Cr, each +6 to +3, six electrons). $\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$ Six moles of Mohr's salt per mole of dichromate. Answer: (D) 6.
24	Coordination CompoundsValence bond theory	Class 12: Coordination Compounds	Single correctMedium
A medium single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds In a metal carbonyl the metal donates d-electron density back into the empty $\pi^$ orbitals of CO (back bonding). Electrons in $\pi^$ weaken the C–O bond, lowering its bond order. The more electron-rich the metal, the stronger the back bonding. $[\text{V(CO)}_6]^-$, $[\text{Cr(CO)}_6]$ and $[\text{Mn(CO)}_6]^+$ are isoelectronic ($d^6$), and the negative charge on the vanadium complex makes its metal the richest in electrons, while the positive charge on the manganese complex holds its d-electrons tightest. $[\text{Fe(CO)}_5]$ is neutral, like the chromium complex. So back bonding is greatest, and the C–O bond order lowest, in $[\text{V(CO)}_6]^-$ (its C–O stretching frequency is the lowest of the set). Answer: (D).
25	Principles of Qualitative AnalysisCharacteristic precipitates and colours	No matching NCERT chapter	Single correctEasy
A easy single correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter $\text{Hg}^{2+} + 2\text{I}^- \rightarrow \text{HgI}_2$ (scarlet red precipitate) $\text{HgI}_2 + 2\text{I}^- \rightarrow [\text{HgI}_4]^{2-}$ (colourless, soluble; Nessler's reagent is its alkaline solution) $\text{Hg}^{2+} + \text{Co}^{2+} + 4\text{SCN}^- \rightarrow \text{Co}[\text{Hg(SCN)}_4]$ (deep blue crystals), the standard confirmatory test for mercury(II). Checking the others: $\text{Pb}^{2+}$ gives a yellow $\text{PbI}_2$; $\text{Cu}^{2+}$ gives white CuI with brown iodine; $\text{Co}^{2+}$ gives no red iodide. Answer: (B) $\text{Hg}^{2+}$.
26	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 Ozonolysis with reductive work-up (Zn, $\text{H}_2\text{O}$) cleaves the ring C=C and gives two CHO groups on the same chain: E = hexane-1,6-dial, $\text{OHC}-(\text{CH}_2)_4-\text{CHO}$. Aqueous KOH brings about an intramolecular aldol condensation. The enolate at C-2 (next to one CHO) attacks the other CHO (C-6), forming a five-membered ring; dehydration of the aldol gives the double bond conjugated with the remaining CHO: F = cyclopent-1-ene-1-carbaldehyde, the CHO on a carbon of the ring double bond. (B) has the double bond out of conjugation; (C) and (D) are acids, which need an oxidant that is not present. Answer: (A).
27	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	Single correctEasy
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Bromine adds to an alkene anti, through a cyclic bromonium ion. Anti addition to $trans$-2-butene gives only meso-2,3-dibromobutane (attack at either carbon of the bromonium ion gives the same meso compound). The meso compound has a plane of symmetry and is optically inactive. (Anti addition to $cis$-2-butene would instead give the racemic pair.) Answer: (A) 1.
28	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 Compare the four structures by the usual rules: like charges on adjacent atoms are very destabilising; opposite charges close together are better than far apart. In (A) the carbocation sits on the carbon bonded directly to the nitro nitrogen, which itself carries a positive charge: two positive charges on adjacent atoms. That is the worst arrangement. (B) puts the negative charge next to the positive nitrogen (a stabilising attraction) and the positive charge further along. (C) and (D) keep the positive carbon at the far end of the chain, away from the positive nitrogen. So (A) is the least stable. Answer: (A).
29	Structure of AtomAtomic orbitals	Class 11: Structure of Atom	Single correctEasy
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom In positron emission a proton becomes a neutron: $p \rightarrow n + e^+ + \nu$. The mass number is unchanged and the atomic number falls by one. $^{23}_{11}\text{Na} \rightarrow ^{23}_{10}\text{Ne} + e^+$ Mass number : atomic number $= 23/10$. Answer: (C).
30	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics 373 K is the boiling point of water at 1 bar (to the precision of the question), so liquid and vapour are in equilibrium: $\Delta G = 0$. Liquid turning into gas increases disorder: $\Delta S = \frac{\Delta H_{vap}}{T} > 0$. Answer: (A).
31	Chemical KineticsOrder of reaction	Class 12: Chemical Kinetics	Single correctEasy
A easy single correct question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let rate $= k[\text{G}]^x[\text{H}]^y$. Doubling both: $2^{x+y} = 8$, so $x + y = 3$. Doubling G alone: $2^x = 2$, so $x = 1$, and then $y = 2$. Overall order $x + y = 3$. Answer: (D).
32	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Assertion and reasonEasy
A easy assertion and reason question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\text{M} + (x + y)\text{NH}_3 \rightarrow [\text{M(NH}_3)_x]^+ + [e(\text{NH}_3)_y]^-$ Statement-1: the dilute solutions are blue. True. Statement-2: the metal cations are indeed solvated as $[\text{M(NH}_3)_n]^+$. True. The blue colour, however, comes from the ammoniated (solvated) electrons, which absorb in the red region of the spectrum, not from the solvated metal cations. So Statement-2 does not explain Statement-1. Answer: (B).
33	BiomoleculesCarbohydrates	Class 12: Biomolecules	Assertion and reasonEasy
A easy assertion and reason question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Statement-1: glucose, an aldose, reduces Fehling's solution and a reddish-brown precipitate appears. True. Statement-2: the precipitate is copper(I) oxide, $\text{Cu}_2\text{O}$, not CuO (which is black); glucose is oxidised to gluconic acid (as its salt): $\text{RCHO} + 2\text{Cu}^{2+} + 5\text{OH}^- \rightarrow \text{RCOO}^- + \text{Cu}_2\text{O} + 3\text{H}_2\text{O}$ Because it names CuO, Statement-2 is false. Answer: (C).
34	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Assertion and reasonEasy
A easy assertion and reason question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Statement-1 is the definition of chirality. True. Statement-2 is false: a molecule can be chiral without any chiral centre, for example a suitably substituted allene such as penta-2,3-diene ($\text{CH}_3\text{CH}{=}\text{C}{=}\text{CHCH}_3$) or a hindered biphenyl, which are chiral through axial chirality. Answer: (C).
35	Solid StateClassification of solids	No matching NCERT chapter	Assertion and reasonMedium
A medium assertion and reason question on classification of solids, from Solid State. NCERT: No matching NCERT chapter Statement-1: germanium is a semiconductor with a band gap of only about 0.7 eV. True. Statement-2: an energy level of an isolated germanium atom is sharp, its spread infinitesimally small (the official key treats this as true). The small band gap arises when the atoms come together in the crystal: their levels split and broaden into bands, and in germanium the valence and conduction bands come close. The sharpness of the isolated atomic levels does not explain why the gap is small. Both true, Statement-2 not the explanation. Answer: (B).
36	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	Single correctEasy
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry A species is oxidised by an oxidant whose reduction potential is higher than its own couple's. Standard reduction potentials (acid solution): $\text{Mn}^{3+}/\text{Mn}^{2+}$ 1.51 V; $\text{Cl}_2/\text{Cl}^-$ 1.36 V; $\text{O}_2/\text{H}_2\text{O}$ 1.23 V; $\text{Fe}^{3+}/\text{Fe}^{2+}$ 0.77 V; $\text{I}_2/\text{I}^-$ 0.54 V. (A) $\text{O}_2$ (1.23) is weaker than $\text{Cl}_2$ (1.36): it cannot oxidise $\text{Cl}^-$. (B) $\text{I}_2$ (0.54) is weaker than $\text{Fe}^{3+}$ (0.77): it cannot oxidise $\text{Fe}^{2+}$ (the reverse happens: $\text{Fe}^{3+}$ oxidises $\text{I}^-$). (C) $\text{Cl}_2$ (1.36) is stronger than $\text{I}_2$ (0.54): $\text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2$. Correct. (D) $\text{Cl}_2$ (1.36) is weaker than $\text{Mn}^{3+}$ (1.51): it cannot oxidise $\text{Mn}^{2+}$ to $\text{Mn}^{3+}$. Answer: (C).
37	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	Single correctMedium
A medium single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Compare each couple with $\text{O}_2/\text{H}_2\text{O}$, $E^\circ = 1.23$ V in acid. $\text{Mn}^{3+}/\text{Mn}^{2+}$: $E^\circ = 1.51$ V, higher than 1.23 V. So $\text{Mn}^{3+}$ is a strong enough oxidant to oxidise water: $4\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow 4\text{Mn}^{2+} + \text{O}_2 + 4\text{H}^+$ ($E^\circ_{cell} = 0.28$ V $> 0$). That is why $\text{Mn}^{3+}$ does not survive in acid solution. $\text{Fe}^{3+}/\text{Fe}^{2+}$: $E^\circ = 0.77$ V, lower than 1.23 V, so $\text{Fe}^{3+}$ cannot oxidise water and is stable; (C) is false. (A) and (B): $\text{O}_2$ (1.23 V) cannot oxidise $\text{Mn}^{2+}$ to $\text{Mn}^{3+}$ (1.51 V), and in any case making $\text{Mn}^{3+}$ would not explain its instability. Answer: (D).
38	Practical Organic ChemistryDetection of elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on detection of elements, from Practical Organic Chemistry. NCERT: No matching NCERT chapter Lassaigne's test for nitrogen. Fusing aniline with sodium gives NaCN in the extract. $\text{Fe}^{2+} + 6\text{CN}^- \rightarrow [\text{Fe(CN)}_6]^{4-}$ Air (and acid) oxidise part of the iron(II) to iron(III), which combines with hexacyanoferrate(II): $4\text{Fe}^{3+} + 3[\text{Fe(CN)}_6]^{4-} \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3$ (Prussian blue, iron(III) hexacyanoferrate(II)) Check the charges: $4 \times (+3) + 3 \times (-4) = 0$. None of the other formulas is neutral with sensible charges. Answer: (A).
42	Redox ReactionsOxidation number	Class 11: Redox Reactions	Matrix matchMedium
A medium matrix match question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions (A) $2\text{O}_2^- \rightarrow \text{O}_2 + \text{O}_2^{2-}$: oxygen goes from $-\frac{1}{2}$ to 0 and to $-1$, the same species both oxidised and reduced. Redox and disproportionation. A → P, S. (B) $2\text{CrO}_4^{2-} + 2\text{H}^+ \rightarrow \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$: chromium stays +6, so no redox. Dichromate is two $\text{CrO}_4$ tetrahedra sharing one bridging oxygen: a dimeric bridged tetrahedral metal ion. B → R. (C) $2\text{MnO}_4^- + 5\text{NO}_2^- + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{NO}_3^- + 3\text{H}_2\text{O}$: Mn +7 to +2, N +3 to +5, a redox reaction; nitrate, $\text{NO}_3^-$, is trigonal planar. C → P, Q. (D) The brown-ring test: $\text{NO}_3^- + 3\text{Fe}^{2+} + 4\text{H}^+ \rightarrow \text{NO} + 3\text{Fe}^{3+} + 2\text{H}_2\text{O}$, then $[\text{Fe(H}_2\text{O)}_6]^{2+} + \text{NO} \rightarrow [\text{Fe(H}_2\text{O)}_5\text{NO}]^{2+}$ (brown). Redox; no trigonal planar product, no dimer, no disproportionation. D → P. Official key: A → P, S; B → R; C → P, Q; D → P.
43	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	Matrix matchMedium
A medium matrix match question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) Benzaldehyde has a C=O group: it gives an orange 2,4-dinitrophenylhydrazone precipitate (P) and adds HCN to form its cyanohydrin (S). A → P, S. (B) Propyne is a terminal alkyne; its acidic H is replaced by silver, giving a white precipitate of $\text{CH}_3\text{C}{\equiv}\text{CAg}$ with (ammoniacal) silver nitrate. B → Q. (C) $\text{CN}^-$ gives white AgCN with $\text{Ag}^+$ (Q), is a good nucleophile (R), and is the nucleophile that attacks C=O in cyanohydrin formation (S). C → Q, R, S. (D) $\text{I}^-$ gives yellow AgI (Q) and is a good nucleophile (R); it plays no part in cyanohydrin formation. D → Q, R. Official key: A → P, S; B → Q; C → Q, R, S; D → Q, R.
44	Solid StateSeven crystal systems	No matching NCERT chapter	Matrix matchMedium
A medium matrix match question on seven crystal systems, from Solid State. NCERT: No matching NCERT chapter Parameters: cubic $a = b = c$, $\alpha = \beta = \gamma = 90^\circ$; rhombohedral $a = b = c$, $\alpha = \beta = \gamma \ne 90^\circ$; tetragonal $a = b \ne c$, all angles $90^\circ$; hexagonal $a = b \ne c$, $\alpha = \beta = 90^\circ$, $\gamma = 120^\circ$; monoclinic $a \ne b \ne c$, $\alpha = \gamma = 90^\circ$, $\beta \ne 90^\circ$. (A) Simple cubic and fcc are two lattices of the one cubic system: $a = b = c$, $\alpha = \beta = \gamma$. A → P, S. (B) Cubic and rhombohedral are two different systems, and both have $a = b = c$ and $\alpha = \beta = \gamma$. B → P, Q. (C) Cubic and tetragonal are two systems; tetragonal has $a = b \ne c$, so not P; both have three right angles, so not R. C → Q. (D) Hexagonal and monoclinic are two systems, and each has exactly two angles of $90^\circ$. D → Q, R. Official key: A → P, S; B → P, Q; C → Q; D → Q, R.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
45	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 The point inside a triangle that divides it into three triangles of equal area is the centroid. (Area of $OQR$ = one third of area $OPQ$ means R is at one third of P's height above OQ, and likewise for the other two sides; only the centroid does this for all three.) $R = \left(\frac{0 + 3 + 6}{3}, \frac{0 + 4 + 0}{3}\right) = \left(3, \frac{4}{3}\right)$. Check: area $OPQ = \frac{1}{2} \times 6 \times 4 = 12$; area $OQR = \frac{1}{2} \times 6 \times \frac{4}{3} = 4$, one third. Answer: (C).
46	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	Single correctEasy
A easy single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $\|z\| = 1$, $\bar{z} = \frac{1}{z}$. Dividing numerator and denominator by $z$: $\frac{z}{1 - z^2} = \frac{1}{\frac{1}{z} - z} = \frac{1}{\bar{z} - z}$. With $z = \cos\theta + i\sin\theta$, $\bar{z} - z = -2i\sin\theta$, which is non-zero since $z \ne \pm 1$. So $\frac{z}{1 - z^2} = \frac{1}{-2i\sin\theta} = \frac{i}{2\sin\theta}$, purely imaginary. All the values lie on the $y$-axis. Answer: (D).
47	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 $E^c \cap F^c \cap G$ is $G$ with the parts in $E$ or in $F$ removed: $P(E^c \cap F^c \cap G) = P(G) - P(E \cap G) - P(F \cap G) + P(E \cap F \cap G)$. By pairwise independence $P(E \cap G) = P(E)P(G)$ and $P(F \cap G) = P(F)P(G)$; and $P(E \cap F \cap G) = 0$. So $P(E^c \cap F^c \cap G) = P(G)\left[1 - P(E) - P(F)\right]$. $P(E^c \cap F^c \mid G) = \frac{P(E^c \cap F^c \cap G)}{P(G)} = 1 - P(E) - P(F) = P(E^c) - P(F)$. Answer: (C).
48	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 $\frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1}$. Differentiate with respect to $y$, using the chain rule through $x$: $\frac{d^2x}{dy^2} = \frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^{-1}\right]\cdot\frac{dx}{dy} = -\left(\frac{dy}{dx}\right)^{-2}\frac{d^2y}{dx^2}\cdot\left(\frac{dy}{dx}\right)^{-1}$ $= -\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3}$. Answer: (D).
49	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 Separate the variables: $\frac{y\,dy}{\sqrt{1 - y^2}} = dx$. Integrate: $-\sqrt{1 - y^2} = x + c$. Square: $1 - y^2 = (x + c)^2$, i.e. $(x + c)^2 + y^2 = 1$. These are circles of radius 1 with centres $(-c, 0)$ on the $x$-axis. Answer: (C).
50	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Single correctEasy
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Cross $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ with $\vec{a}$: $\vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0}$, so $\vec{a} \times \vec{b} = \vec{c} \times \vec{a}$. Cross it with $\vec{b}$: $\vec{b} \times \vec{a} + \vec{b} \times \vec{c} = \vec{0}$, so $\vec{b} \times \vec{c} = \vec{a} \times \vec{b}$. Hence $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$. Three unit vectors adding to zero form an equilateral triangle, so each pair is at $120^\circ$: $\|\vec{a} \times \vec{b}\| = \sin 120^\circ = \frac{\sqrt{3}}{2} \ne 0$. So (B) holds and (A) fails. (C) would need $\vec{a} \times \vec{c} = \vec{c} \times \vec{a}$, i.e. both zero, contradicting the non-zero value. (D) fails because the three cross products are equal, not perpendicular. Answer: (B).
51	CirclesRadius and centre	Class 11: Conic Sections	Single correctMedium
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections ABCD is a right trapezium with parallel sides $AB = 2x$, $CD = x$ and the perpendicular side AD. A circle touching both parallel sides has diameter equal to the distance between them, so $AD = 2r$. Area: $\frac{1}{2}(2x + x)(2r) = 3xr = 18$, so $xr = 6$. A quadrilateral with an inscribed circle has equal sums of opposite sides: $AB + CD = AD + BC$, so $BC = 3x - 2r$. BC is the slant side: its horizontal run is $AB - CD = x$ and its rise is $2r$, so $BC^2 = x^2 + 4r^2$. $(3x - 2r)^2 = x^2 + 4r^2 \Rightarrow 9x^2 - 12xr = x^2 \Rightarrow 8x^2 = 12xr \Rightarrow x = \frac{3r}{2}$. Then $xr = \frac{3r^2}{2} = 6$, so $r^2 = 4$ and $r = 2$. Answer: (B) 2.
52	Integral CalculusIntegration by substitution	Class 12: Integrals; Class 12: Application of Integrals	Single correctMedium
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f(f(x))$: $[f(x)]^n = \frac{x^n}{1 + x^n}$, so $1 + [f(x)]^n = \frac{1 + 2x^n}{1 + x^n}$ and $f(f(x)) = \frac{f(x)}{(1 + [f(x)]^n)^{1/n}} = \frac{x}{(1 + x^n)^{1/n}}\cdot\frac{(1 + x^n)^{1/n}}{(1 + 2x^n)^{1/n}} = \frac{x}{(1 + 2x^n)^{1/n}}$. Each further application adds $x^n$ in the bracket, so with $f$ applied $n$ times $g(x) = \frac{x}{(1 + nx^n)^{1/n}}$. $\int x^{n-2}g(x)\,dx = \int \frac{x^{n-1}}{(1 + nx^n)^{1/n}}\,dx$. Put $u = 1 + nx^n$, $du = n^2x^{n-1}\,dx$: $= \frac{1}{n^2}\int u^{-1/n}\,du = \frac{1}{n^2}\cdot\frac{u^{1 - 1/n}}{1 - \frac{1}{n}} + K = \frac{1}{n(n-1)}(1 + nx^n)^{1 - \frac{1}{n}} + K$. Answer: (A).
53	Permutations and CombinationsSimple applications	Class 11: Permutations and Combinations	Single correctEasy
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The letters in alphabetical order: C, C, H, I, N, O. Any word starting with H, I, N or O comes after COCHIN, so count only words starting with C. The remaining letters are C, H, I, N, O (all different). Second letter C, H, I or N (all before O): each leaves 4 distinct letters, $4! = 24$ words each, $4 \times 24 = 96$ words. Words starting CO: the remaining letters C, H, I, N in alphabetical order give C-O-C-H-I-N, which is COCHIN itself, the first of them. So 96 words come before COCHIN. Answer: (C).
54	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Assertion and reasonMedium
A medium assertion and reason question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The line of intersection is perpendicular to both normals $\vec{n}_1 = (3, -6, -2)$ and $\vec{n}_2 = (2, 1, -2)$: $\vec{n}_1 \times \vec{n}_2 = \hat{i}\left((-6)(-2) - (-2)(1)\right) - \hat{j}\left((3)(-2) - (-2)(2)\right) + \hat{k}\left((3)(1) - (-6)(2)\right) = 14\hat{i} + 2\hat{j} + 15\hat{k}$. So Statement-2 is true. Statement-1: at $t = 0$ the line passes through $(3, 1, 0)$. In the first plane $3(3) - 6(1) - 2(0) = 3 \ne 15$, so this point is not on the planes and the line given is not the line of intersection (it is only parallel to it). False. A point that does lie on both planes: put $z = 0$, then $3x - 6y = 15$ and $2x + y = 5$ give $x = 3$, $y = -1$. So the line of intersection is $x = 3 + 14t$, $y = -1 + 2t$, $z = 15t$, not the one in Statement-1. Answer: (D).
55	Conic SectionsParabola	Class 11: Conic Sections	Assertion and reasonEasy
A easy assertion and reason question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Complete the square: $y = -\frac{1}{2}(x^2 - 2x) + 1 = -\frac{1}{2}(x - 1)^2 + \frac{3}{2}$, i.e. $(x - 1)^2 = -2\left(y - \frac{3}{2}\right)$. This is a parabola with vertex $\left(1, \frac{3}{2}\right)$ and axis $x = 1$. Statement-2 (a parabola is symmetric about its axis) is true, and it is exactly why the curve is symmetric about $x = 1$: Statement-1 is true and follows from it. Answer: (A).
56	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	Assertion and reasonMedium
A medium assertion and reason question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = -\sin x$, which is zero exactly at the multiples of $\pi$. Any closed interval $[t, t + \pi]$ of length $\pi$ contains a multiple of $\pi$, so Statement-1 is true. Statement-2: $\cos$ has period $2\pi$, so $f(t) = f(t + 2\pi)$. True. Statement-2 with Rolle's theorem only gives a $c$ in $[t, t + 2\pi]$, an interval twice as long, so it does not prove Statement-1. Answer: (B).
57	Straight LinesDistance of a point from a line	Class 11: Straight Lines	Assertion and reasonMedium
A medium assertion and reason question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines $L_1$ and $L_2$ meet at the origin O. On $L_3$ ($y = -2$): $P = (-2, -2)$ from $y = x$, and $Q = (1, -2)$ from $y = -2x$. Angle POQ: $\vec{OP} \cdot \vec{OQ} = (-2)(1) + (-2)(-2) = 2 > 0$, so angle POQ is acute; it is the acute angle between $L_1$ and $L_2$, and its bisector meets PQ at R. By the angle-bisector theorem in triangle OPQ, $\frac{PR}{RQ} = \frac{OP}{OQ} = \frac{\sqrt{8}}{\sqrt{5}} = \frac{2\sqrt{2}}{\sqrt{5}}$. Statement-1 is true. Statement-2 is false: an angle bisector does not in general split a triangle into two similar triangles (it does only in special cases such as an isosceles triangle bisected at its apex). Answer: (C).
61	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	ComprehensionEasy
A easy comprehension question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The meeting points are the roots of $f(x) = ke^x - x = 0$. For $k \le 0$: $f'(x) = ke^x - 1 < 0$ for all $x$, so $f$ is strictly decreasing and has at most one root. As $x \to -\infty$, $ke^x \to 0$ and $-x \to +\infty$, so $f \to +\infty$; as $x \to +\infty$, $f \le -x \to -\infty$. Being continuous, $f$ takes both signs, so it has a root. (For $k = 0$ the root is $x = 0$.) Exactly one point. Answer: (B).
62	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	ComprehensionMedium
A medium comprehension question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives For $k > 0$: $f'(x) = ke^x - 1 = 0$ at $x = -\ln k$, and $f''(x) = ke^x > 0$, so this is the minimum. $f_{min} = f(-\ln k) = k \cdot \frac{1}{k} + \ln k = 1 + \ln k$. As $x \to \pm\infty$, $f \to +\infty$. So $f$ has exactly one root when its minimum just touches zero: $1 + \ln k = 0 \Rightarrow k = \frac{1}{e}$. Answer: (A).
63	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	ComprehensionMedium
A medium comprehension question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives From the previous part, for $k > 0$ the function $f(x) = ke^x - x$ decreases to the minimum $1 + \ln k$ at $x = -\ln k$ and then increases, tending to $+\infty$ at both ends. Two distinct roots need the minimum to be negative (one root on each side of it, by the result in the paragraph): $1 + \ln k < 0 \Rightarrow \ln k < -1 \Rightarrow 0 < k < \frac{1}{e}$. Answer: (A).
64	Quadratic EquationsNature of roots	Class 11: Complex Numbers and Quadratic Equations	Matrix matchMedium
A medium matrix match question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $f(x) = \frac{(x - 1)(x - 5)}{(x - 2)(x - 3)}$, and $f(x) - 1 = \frac{(x^2 - 6x + 5) - (x^2 - 5x + 6)}{(x - 2)(x - 3)} = \frac{-(x + 1)}{(x - 2)(x - 3)}$. (A) $-1 < x < 1$: numerator $(x - 1)(x - 5) > 0$ (both factors negative), denominator $> 0$ (both negative), so $f > 0$. Also $-(x + 1) < 0$ and denominator $> 0$, so $f - 1 < 0$. Thus $0 < f < 1$. A → P, R, S. (B) $1 < x < 2$: numerator $< 0$ ($x - 1 > 0$, $x - 5 < 0$), denominator $> 0$: $f < 0$, hence also $f < 1$. B → Q, S. (C) $3 < x < 5$: numerator $< 0$, denominator $> 0$ (both factors positive): $f < 0 < 1$. C → Q, S. (D) $x > 5$: numerator $> 0$, denominator $> 0$: $f > 0$; and $f - 1 = \frac{-(x + 1)}{(x - 2)(x - 3)} < 0$. So $0 < f < 1$. D → P, R, S. Official key: A → P, R, S; B → Q, S; C → Q, S; D → P, R, S.
65	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Matrix matchHard
A hard matrix match question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Since $\sin^{-1}(ax) = \frac{\pi}{2} - \cos^{-1}(ax)$, the condition becomes $\cos^{-1}(y) + \cos^{-1}(bxy) = \cos^{-1}(ax)$. Take the cosine of both sides: $y \cdot bxy - \sqrt{1 - y^2}\sqrt{1 - b^2x^2y^2} = ax$, i.e. $bxy^2 - ax = \sqrt{(1 - y^2)(1 - b^2x^2y^2)}$, and square. (A) $a = 1$, $b = 0$: $\cos^{-1}(0) = \frac{\pi}{2}$, so $\sin^{-1}x + \cos^{-1}y = 0$. Writing $\cos^{-1}y = \theta \in [0, \pi]$, we need $\sin^{-1}x = -\theta$, which forces $\theta \in [0, \frac{\pi}{2}]$, $x = -\sin\theta$, $y = \cos\theta$. Then $x^2 + y^2 = 1$. A → P. (B) $a = b = 1$: $x^2(y^2 - 1)^2 = (1 - y^2)(1 - x^2y^2)$, so $(1 - y^2)\left[x^2(1 - y^2) - (1 - x^2y^2)\right] = 0$, i.e. $(1 - y^2)(x^2 - 1) = 0$: $(x^2 - 1)(y^2 - 1) = 0$. B → Q. (C) $a = 1$, $b = 2$: $x^2(2y^2 - 1)^2 = (1 - y^2)(1 - 4x^2y^2)$. Left: $4x^2y^4 - 4x^2y^2 + x^2$. Right: $1 - y^2 - 4x^2y^2 + 4x^2y^4$. Cancelling gives $x^2 = 1 - y^2$: $x^2 + y^2 = 1$. C → P. (D) $a = b = 2$: $4x^2(y^2 - 1)^2 = (1 - y^2)(1 - 4x^2y^2)$, so $(1 - y^2)\left[4x^2(1 - y^2) - 1 + 4x^2y^2\right] = (1 - y^2)(4x^2 - 1) = 0$: $(4x^2 - 1)(y^2 - 1) = 0$. D → S. None of the cases forces $y = x$, so R is matched by no row. Official key: A → P; B → Q; C → P; D → S.
66	CirclesIntersection of a line and a circle	Class 11: Conic Sections	Matrix matchMedium
A medium matrix match question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections For two circles, the line through the two centres is normal to both, so every pair of circles has a common normal: Q for A, B and C, and never S. (A) Two intersecting circles have two common (direct) tangents. A → P, Q. (B) Two mutually external circles have four common tangents. B → P, Q. (C) If one circle lies strictly inside the other, any tangent to the inner circle cuts the outer one, so there is no common tangent. C → Q, R. (D) The two branches of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: a tangent to one branch never touches the other (the tangent at $(a\sec\theta, b\tan\theta)$ meets the hyperbola only there; the asymptotes are not tangents), so no common tangent. The transverse axis ($y = 0$) is normal to both branches at the vertices, so there is a common normal. D → Q, R. Official key: A → P, Q; B → P, Q; C → Q, R; D → Q, R.

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