JEE (Advanced) · 2009

JEE Advanced 2009, 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.

Chemistry
39 Q
1. Organic Chemistry: Some Basic Principles and Techniques6
2. p-Block Elements5
3. Some Basic Concepts of Chemistry3
15 easy · 19 medium · 5 hard
Mathematics
39 Q
1. Conic Sections5
2. Trigonometry5
3. Matrices and Determinants5
9 easy · 24 medium · 6 hard
Physics
39 Q
1. Electrostatics4
2. Motion of System of Particles and Rigid Body4
3. Work, Energy and Power4
14 easy · 22 medium · 3 hard

Chapters

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

Organic Chemistry: Some Basic Principles and Techniques6
Class 11: Organic Chemistry – Some Basic Principles and Techniques
p-Block Elements5
No matching NCERT chapter
Some Basic Concepts of Chemistry3
Class 11: Some Basic Concepts of Chemistry
Aldehydes, Ketones and Carboxylic Acids3
Class 12: Aldehydes, Ketones and Carboxylic Acids
Coordination Compounds3
Class 12: Coordination Compounds
Principles of Qualitative Analysis3
No matching NCERT chapter
Chemical Kinetics2
Class 12: Chemical Kinetics
Thermodynamics2
Class 11: Thermodynamics
Surface Chemistry1
No matching NCERT chapter
Solutions1
Class 12: Solutions
Polymers1
No matching NCERT chapter
Solid State1
No matching NCERT chapter
Classification of Elements and Periodicity1
Class 11: Classification of Elements and Periodicity in Properties
Hydrocarbons1
Class 11: Hydrocarbons
Chemical Bonding and Molecular Structure1
Class 11: Chemical Bonding and Molecular Structure
Electrochemistry1
Class 12: Electrochemistry
Biomolecules1
Class 12: Biomolecules
Haloalkanes and Haloarenes1
Class 12: Haloalkanes and Haloarenes
Equilibrium1
Class 11: Equilibrium
d- and f-Block Elements1
Class 12: The d- and f-Block Elements

Conic Sections5
Class 11: Conic Sections
Trigonometry5
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Matrices and Determinants5
Class 12: Matrices; Class 12: Determinants
Applications of Derivatives4
Class 12: Application of Derivatives
Integral Calculus3
Class 12: Integrals; Class 12: Application of Integrals
Probability3
Class 11: Probability; Class 12: Probability
Three Dimensional Geometry2
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Circles2
Class 11: Conic Sections
Complex Numbers2
Class 11: Complex Numbers and Quadratic Equations
Limits, Continuity and Differentiability2
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Differential Equations1
Class 12: Differential Equations
Vector Algebra1
Class 12: Vector Algebra
Permutations and Combinations1
Class 11: Permutations and Combinations
Sequences and Series1
Class 11: Sequences and Series
Straight Lines1
Class 11: Straight Lines
Quadratic Equations1
Class 11: Complex Numbers and Quadratic Equations

Questions

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

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
1	Some Basic Concepts of ChemistryMole concept	Class 11: Some Basic Concepts of Chemistry	Single correctEasy
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry The average atomic mass is the abundance-weighted mean of the isotope masses (taking each mass as its mass number): $A = 0.05 \times 54 + 0.90 \times 56 + 0.05 \times 57$ $= 2.70 + 50.40 + 2.85 = 55.95$ Answer: (B) 55.95.
2	Some Basic Concepts of ChemistryMole concept	Class 11: Some Basic Concepts of Chemistry	Single correctEasy
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry The van der Waals equation is $\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$ The attractions between molecules pull a molecule hitting the wall back into the gas, so the measured pressure is less than the ideal pressure. The term $\frac{an^2}{V^2}$ added to $P$ corrects for these attractive forces. The term $nb$ corrects for the volume of the molecules themselves (repulsion), not the attraction. Answer: (B) $\frac{an^2}{V^2}$.
3	Surface ChemistryColloids and their preparation	No matching NCERT chapter	Single correctEasy
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter $\text{Sb}_2\text{S}_3$ (like $\text{As}_2\text{S}_3$) forms a negatively charged sol, because it adsorbs sulphide ions. A negative sol is coagulated by cations, and by the Hardy–Schulze rule the coagulating power rises steeply with the charge of the ion of opposite sign. Cations available: $\text{Na}^+$ (+1), $\text{Ca}^{2+}$ (+2), $\text{Al}^{3+}$ (+3), $\text{NH}_4^+$ (+1). $\text{Al}^{3+}$ has the highest charge, so $\text{Al}_2(\text{SO}_4)_3$ is the most effective. Answer: (C).
4	SolutionsConcentration terms	Class 12: Solutions	Single correctMedium
A medium single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Henry's law: $p_{\text{N}_2} = K_H\, x_{\text{N}_2}$, where $x_{\text{N}_2}$ is the mole fraction of $\text{N}_2$ in the solution. Partial pressure of $\text{N}_2$ over the water: $p = 0.8 \times 5 = 4$ atm. $x_{\text{N}_2} = \frac{4}{1.0 \times 10^5} = 4.0 \times 10^{-5}$ Since this is tiny, $x \approx \frac{n_{\text{N}_2}}{n_{\text{water}}}$, so $n_{\text{N}_2} = 4.0 \times 10^{-5} \times 10 = 4.0 \times 10^{-4}$ mol. Answer: (A) $4.0 \times 10^{-4}$.
5	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Single correctMedium
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter White phosphorus burns in excess oxygen to give $\text{P}_4\text{O}_{10}$; $\text{P}_4\text{O}_6$ forms only when the supply of oxygen is limited: $\text{P}_4 + 3\text{O}_2 \rightarrow \text{P}_4\text{O}_6$ (A) Dry $\text{O}_2$ alone is not limiting, so oxidation goes on to $\text{P}_4\text{O}_{10}$. (B) A mixture of $\text{O}_2$ and $\text{N}_2$ dilutes the oxygen (this is how $\text{P}_4\text{O}_6$ is made, with a restricted supply of oxygen diluted by nitrogen), so the reaction stops at $\text{P}_4\text{O}_6$. Correct. (C) Moist $\text{O}_2$: the oxides would be hydrolysed to phosphorus oxoacids. (D) In aqueous NaOH, $\text{P}_4$ disproportionates to $\text{PH}_3$ and hypophosphite. Answer: (B).
6	Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctEasy
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Carboxylic acids (pKa about 4–4.5) are far stronger acids than phenols (pKa about 9–10), because the carboxylate ion is stabilised by resonance over two equivalent oxygens. Between the acids: the $\text{CH}_3$ group in (IV) is electron-releasing (+I, hyperconjugation) and destabilises the carboxylate, so (IV) is weaker than benzoic acid (III): III > IV. Between the phenols: Cl in (II) is electron-withdrawing (–I outweighs +R at the para position) and stabilises the phenoxide, so 4-chlorophenol is more acidic than phenol: II > I. Order: (III) > (IV) > (II) > (I). Answer: (A).
7	PolymersNatural rubber and cellulose	No matching NCERT chapter	Single correctEasy
A easy single correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Polymers are classified by the strength of the forces between chains: Nylon: fibre, strong hydrogen bonds between amide groups. Cellulose: extensive hydrogen bonding between OH groups (fibre). Poly(vinyl chloride): thermoplastic, dipole–dipole forces of intermediate strength. Natural rubber (cis-polyisoprene): an elastomer, chains held only by weak van der Waals forces, which is why it stretches and recoils. The weakest intermolecular attraction is in natural rubber. Answer: (D).
8	Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The ring carries OH, CN and Br. The nitrile is the principal group (higher than OH), so the parent is benzonitrile and the carbon bearing CN is C-1. OH and Br become prefixes (hydroxy, bromo). In the drawing, Br is next to CN (ortho) and OH is meta to CN on the other side (OH is para to Br). Numbering from C-1 towards Br gives Br at 2 and OH at 5: locants {2, 5}. Numbering the other way gives OH at 3 and Br at 6: {3, 6}. The lower set is {2, 5}. Prefixes alphabetically: bromo before hydroxy. Name: 2-Bromo-5-hydroxybenzonitrile. (A) and (C) do not use the nitrile as the principal group; (D) uses the higher locant set. Answer: (B).
9	Solid StatePoint defects	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on point defects, from Solid State. NCERT: No matching NCERT chapter (A) False. A Frenkel defect (a small ion, usually the cation, leaving its site for an interstitial site) is favoured by a LARGE difference in the sizes of cation and anion, as in AgCl, AgBr, ZnS. A small difference favours Schottky defects (NaCl, KCl). (B) True. In a Frenkel defect the ion is displaced (dislocated) from its lattice site to an interstitial site, so it is also called a dislocation defect. (C) True. An F-centre is an anion vacancy occupied by an unpaired electron; trapping an electron in the lattice in this way creates the F-centre (and gives colour, e.g. yellow NaCl in Na vapour). (D) False. Schottky defects lower the density of the solid and increase its conductivity, so they do affect physical properties. Answer: (B), (C). The official key also accepts (C) alone.
10	Coordination CompoundsIsomerism in complexes	Class 12: Coordination Compounds	Multiple correctMedium
A medium multiple correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) $[\text{Pt(en)Cl}_2]$: square planar $\text{Pt}^{2+}$. The bidentate en must span two cis positions, so the two Cl are necessarily cis. Only one form: no geometrical isomerism. (B) $[\text{Pt(en)}_2]\text{Cl}_2$: square planar $[\text{Pt(en)}_2]^{2+}$ with two identical symmetric chelates. Only one arrangement: no geometrical isomerism. (C) $[\text{Pt(en)}_2\text{Cl}_2]\text{Cl}_2$: octahedral $\text{Pt}^{4+}$, type $[\text{M(AA)}_2\text{B}_2]$. The two Cl can be cis or trans: shows geometrical isomerism (the cis form is also optically active). (D) $[\text{Pt(NH}_3)_2\text{Cl}_2]$: square planar $[\text{MA}_2\text{B}_2]$, which exists as cis-platin and trans-platin: shows geometrical isomerism. Answer: (C), (D).
11	Classification of Elements and PeriodicityPeriodic trends	Class 11: Classification of Elements and Periodicity in Properties	Multiple correctMedium
A medium multiple correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties On burning in air or oxygen the alkali metals give different oxides: Li gives mainly the monoxide $\text{Li}_2\text{O}$, Na gives mainly the peroxide $\text{Na}_2\text{O}_2$, and K, Rb, Cs give superoxides $\text{MO}_2$. $2\text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O}_2$ (excess air) With a limited supply some $\text{Na}_2\text{O}$ also forms, so $\text{Na}_2\text{O}$ may accompany the peroxide. (A) $\text{Na}_2\text{O}_2$: the main product. Correct. (B) $\text{Na}_2\text{O}$: formed to a small extent; the official key accepts it along with (A). (C) $\text{NaO}_2$: sodium superoxide is not formed by burning in air (it needs oxygen under high pressure). (D) NaOH is not a combustion product (it needs water). Answer: (A). The official key also accepts (A), (B).
12	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Multiple correctHard
A hard multiple correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques X is hex-3-ene-2,5-diol, $\text{CH}_3\text{–CH(OH)–CH=CH–CH(OH)–CH}_3$. It has two stereocentres (C-2 and C-5), which carry identical sets of groups, and a double bond that can be cis or trans. For each geometry of the double bond the two stereocentres give (R,R), (S,S) and (R,S); because the two ends are the same, (R,S) = (S,R) is a meso form (it has a mirror plane in the cis isomer and a centre of symmetry in the trans isomer). So trans: (R,R), (S,S), meso = 3; cis: (R,R), (S,S), meso = 3. (A) Total stereoisomers $= 3 + 3 = 6$. True. (B) The 6 stereoisomers fall into 4 diastereomeric sets (cis pair, cis-meso, trans pair, trans-meso), so the number of diastereomers is not 3. False. (C) With a trans double bond the enantiomers are only the (R,R)/(S,S) pair: 2, not 4. False. (D) With a cis double bond the enantiomers are again the (R,R)/(S,S) pair: 2. True. Answer: (A), (D).
13	Principles of Qualitative AnalysisAnalysis of anions	No matching NCERT chapter	ComprehensionMedium
A medium comprehension question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter The methylene blue test is a test for sulphide. In strongly acidic solution a sulphide releases $\text{H}_2\text{S}$; with p-amino-N,N-dimethylaniline and an oxidant ($\text{Fe}^{3+}$) the sulphur is built into the thiazine ring of methylene blue, giving the blue colour. None of nitrate, chloride or sulphate can supply the sulphur. X is $\text{Na}_2\text{S}$. Answer: (D).
14	Principles of Qualitative AnalysisCharacteristic precipitates and colours	No matching NCERT chapter	ComprehensionMedium
A medium comprehension question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Y must (i) oxidise the sulphide/amine mixture to methylene blue and (ii) give an intense blue precipitate with $\text{K}_4[\text{Fe(CN)}_6]$. Both fit $\text{Fe}^{3+}$: it is the oxidant in the methylene blue test, and $4\text{Fe}^{3+} + 3[\text{Fe(CN)}_6]^{4-} \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3$ (Prussian blue) which dissolves in excess ferrocyanide (as soluble Prussian blue, $\text{KFe}[\text{Fe(CN)}_6]$). $\text{Fe}^{3+}$ with $[\text{Fe(CN)}_6]^{3-}$ gives only a brown colour, as the passage says. $\text{MgCl}_2$ and $\text{ZnCl}_2$ give white precipitates and are not oxidants; $\text{FeCl}_2$ gives a white/pale blue precipitate with ferrocyanide and is not an oxidant. Y is $\text{FeCl}_3$. Answer: (C).
15	Principles of Qualitative AnalysisCharacteristic precipitates and colours	No matching NCERT chapter	ComprehensionMedium
A medium comprehension question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Y is $\text{FeCl}_3$. With potassium hexacyanoferrate(III) the $\text{Fe}^{3+}$ combines with $[\text{Fe(CN)}_6]^{3-}$ to give the brown, soluble ferric ferricyanide: $\text{Fe}^{3+} + [\text{Fe(CN)}_6]^{3-} \rightarrow \text{Fe}[\text{Fe(CN)}_6]$ (brown colour) (A) and (D) contain Mg and Zn, which are not in the system; (C) $\text{Fe}_4[\text{Fe(CN)}_6]_3$ is Prussian blue, the blue precipitate formed with hexacyanoferrate(II), not the brown colour. Z is $\text{Fe}[\text{Fe(CN)}_6]$. Answer: (B).
16	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	ComprehensionHard
A hard comprehension question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Work the whole sequence from the answer choices. P must be a methyl ketone (positive iodoform test). Take P = $\text{Ph–CH=C(CH}_3\text{)–CO–CH}_3$ (3-methyl-4-phenylbut-3-en-2-one, option B). 1. MeMgBr then $\text{H}^+/\text{H}_2\text{O}$: the ketone becomes the tertiary alcohol $\text{Ph–CH=C(CH}_3\text{)–C(OH)(CH}_3)_2$. 2. $\text{H}_2\text{SO}_4$, $\Delta$: loss of water gives the tertiary allylic carbocation $\text{(CH}_3)_2\text{C}^+\text{–C(CH}_3\text{)=CH–Ph}$, which attacks the ortho carbon of the phenyl ring (intramolecular Friedel–Crafts alkylation) and closes a five-membered ring. Q = 1,1,2-trimethyl-1H-indene (the "olefin": its C2=C3 double bond). 3. Ozonolysis of the C2=C3 bond: C3 (CH, attached to the ring) becomes CHO; C2 (bearing CH3) becomes a methyl ketone. R = $o\text{-OHC–C}_6\text{H}_4\text{–C(CH}_3)_2\text{–CO–CH}_3$. 4. Base: the only enolisable position is the COCH3 methyl; its enolate attacks the aldehyde to form a six-membered ring, and heating dehydrates the aldol: S = 1,1-dimethylnaphthalen-2(1H)-one, the ketone conjugated with the new CH=CH. Checking the other choices for P: (A) $\text{Ph–CH=CH–CO–CH}_3$ is a methyl ketone, but the same steps give 1,1-dimethylindene as Q and the dialdehyde $o\text{-OHC–C}_6\text{H}_4\text{–C(CH}_3)_2\text{–CHO}$ as R. Neither aldehyde of this R has an α-hydrogen, so it cannot undergo the intramolecular aldol reaction the passage describes. (C) is an ethyl ketone and gives no iodoform test. (D) $\text{Ph–CO–CH=CH–CH}_3$ is not a methyl ketone ($\text{PhCO–}$), so no iodoform test. Answer: (B).
17	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	ComprehensionHard
A hard comprehension question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Work the whole sequence from the answer choices. P must be a methyl ketone (positive iodoform test). Take P = $\text{Ph–CH=C(CH}_3\text{)–CO–CH}_3$ (3-methyl-4-phenylbut-3-en-2-one, option B). 1. MeMgBr then $\text{H}^+/\text{H}_2\text{O}$: the ketone becomes the tertiary alcohol $\text{Ph–CH=C(CH}_3\text{)–C(OH)(CH}_3)_2$. 2. $\text{H}_2\text{SO}_4$, $\Delta$: loss of water gives the tertiary allylic carbocation $\text{(CH}_3)_2\text{C}^+\text{–C(CH}_3\text{)=CH–Ph}$, which attacks the ortho carbon of the phenyl ring (intramolecular Friedel–Crafts alkylation) and closes a five-membered ring. Q = 1,1,2-trimethyl-1H-indene (the "olefin": its C2=C3 double bond). 3. Ozonolysis of the C2=C3 bond: C3 (CH, attached to the ring) becomes CHO; C2 (bearing CH3) becomes a methyl ketone. R = $o\text{-OHC–C}_6\text{H}_4\text{–C(CH}_3)_2\text{–CO–CH}_3$. 4. Base: the only enolisable position is the COCH3 methyl; its enolate attacks the aldehyde to form a six-membered ring, and heating dehydrates the aldol: S = 1,1-dimethylnaphthalen-2(1H)-one, the ketone conjugated with the new CH=CH. So Q is the indene with gem-dimethyl at C1 and a methyl at C2, and R is the keto-aldehyde with CHO on the ring and $\text{C(CH}_3)_2\text{COCH}_3$ beside it: option (A). (B) has no C2 methyl, and its R is a dialdehyde. (C) has Me and Et at C1, which would need an ethyl Grignard. (D) puts methyls at C1 and C3; its R does not come from P with MeMgBr. Answer: (A).
18	Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro	Class 12: Aldehydes, Ketones and Carboxylic Acids	ComprehensionHard
A hard comprehension question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Work the whole sequence from the answer choices. P must be a methyl ketone (positive iodoform test). Take P = $\text{Ph–CH=C(CH}_3\text{)–CO–CH}_3$ (3-methyl-4-phenylbut-3-en-2-one, option B). 1. MeMgBr then $\text{H}^+/\text{H}_2\text{O}$: the ketone becomes the tertiary alcohol $\text{Ph–CH=C(CH}_3\text{)–C(OH)(CH}_3)_2$. 2. $\text{H}_2\text{SO}_4$, $\Delta$: loss of water gives the tertiary allylic carbocation $\text{(CH}_3)_2\text{C}^+\text{–C(CH}_3\text{)=CH–Ph}$, which attacks the ortho carbon of the phenyl ring (intramolecular Friedel–Crafts alkylation) and closes a five-membered ring. Q = 1,1,2-trimethyl-1H-indene (the "olefin": its C2=C3 double bond). 3. Ozonolysis of the C2=C3 bond: C3 (CH, attached to the ring) becomes CHO; C2 (bearing CH3) becomes a methyl ketone. R = $o\text{-OHC–C}_6\text{H}_4\text{–C(CH}_3)_2\text{–CO–CH}_3$. 4. Base: the only enolisable position is the COCH3 methyl; its enolate attacks the aldehyde to form a six-membered ring, and heating dehydrates the aldol: S = 1,1-dimethylnaphthalen-2(1H)-one, the ketone conjugated with the new CH=CH. In R the aldehyde carbon is attached to the ring and the CO–CH3 sits on the CMe2 carbon. The enolate $\text{–CO–CH}_2^-$ adds to CHO, giving a ring of six atoms: ring C, CMe2, C=O, CH2, CH(OH), ring C. Dehydration gives CH=CH conjugated with both the ring and the C=O. S: a bicyclic enone with the C=O next to the gem-dimethyl carbon (1,1-dimethylnaphthalen-2(1H)-one), option (B). (A), (C) and (D) put the C=O on the carbon bonded to the benzene ring, which in R is the aldehyde carbon; in the aldol that carbon becomes CH(OH) and then part of the C=C, never a ketone. (A) and (D) also lack the gem-dimethyl group. Answer: (B).
19	Chemical Bonding and Molecular StructureMolecular orbital theory	Class 11: Chemical Bonding and Molecular Structure	Matrix matchMedium
A medium matrix match question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Molecular orbital configurations (for $\text{B}_2$, $\text{C}_2$, $\text{N}_2$ the s–p mixing pushes $\sigma 2p_z$ above $\pi 2p$; for O and beyond there is no significant mixing). $\text{B}_2$ (10 e): $\ldots \pi 2p_x^1 \pi 2p_y^1$. Two unpaired electrons: paramagnetic. Bond order $= \frac{6-4}{2} = 1$. s–p mixing present (that is why it is paramagnetic). $\text{N}_2$ (14 e): bond order 3, all paired: diamagnetic; s–p mixing present. $\text{O}_2^-$ (17 e): bond order $\frac{10-7}{2} = 1.5$, one unpaired electron: paramagnetic; no s–p mixing. $\text{O}_2$ (16 e): bond order 2, two unpaired electrons: paramagnetic; no s–p mixing. Every one of these species can lose an electron (oxidation, e.g. $\text{O}_2^- \rightarrow \text{O}_2$, $\text{N}_2 \rightarrow \text{N}_2^+$) and gain an electron (reduction, e.g. $\text{O}_2 \rightarrow \text{O}_2^-$, $\text{N}_2$ to nitride), so all show (Q) and (R). (A) $\text{B}_2$: P, Q, R, T (bond order 1, not S). (B) $\text{N}_2$: Q, R, S, T (diamagnetic). (C) $\text{O}_2^-$: P, Q, R. (D) $\text{O}_2$: P, Q, R, S. Official: A → P, Q, R, T; B → Q, R, S, T; C → P, Q, R; D → P, Q, R, S.
20	Organic Chemistry: Some Basic Principles and TechniquesTypes of organic reactions	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Matrix matchMedium
A medium matrix match question on types of organic reactions, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) Butanenitrile: $\text{H}_2$/Pd–C reduces C≡N to $\text{–CH}_2\text{NH}_2$ (P); $\text{SnCl}_2$/HCl reduces it to the imine, hydrolysed to the aldehyde (Stephen reduction) (Q); DIBAL-H reduces a nitrile to the aldehyde (S); alkaline hydrolysis gives the carboxylate (T). A → P, Q, S, T. (B) Ethyl acetate: DIBAL-H reduces an ester (to aldehyde at low temperature) (S); alkaline hydrolysis (saponification) (T). No C=C or C≡N for catalytic hydrogenation under ordinary conditions, no $\text{SnCl}_2$ reduction, no amine. B → S, T. (C) But-2-en-1-ol: the C=C is hydrogenated by $\text{H}_2$/Pd–C (P). Nothing to hydrolyse, nothing reduced by $\text{SnCl}_2$ or DIBAL-H (the alcohol is already reduced). C → P. (D) Butan-1-amine, a primary amine: carbylamine reaction with $\text{CHCl}_3$/alcoholic KOH gives the foul-smelling isocyanide (R). D → R. Official: A → P, Q, S, T; B → S, T; C → P; D → R.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
21	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctEasy
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A general point of the line: $Q = (1 - 3\mu,\ -1 + \mu,\ 2 + 5\mu)$. $\overrightarrow{PQ} = Q - P = (-2 - 3\mu,\ -3 + \mu,\ -4 + 5\mu)$. $\overrightarrow{PQ}$ is parallel to the plane when it is perpendicular to the plane's normal $(1, -4, 3)$: $(-2 - 3\mu) - 4(-3 + \mu) + 3(-4 + 5\mu) = 0$ $-2 - 3\mu + 12 - 4\mu - 12 + 15\mu = 0$ $8\mu - 2 = 0 \Rightarrow \mu = \frac{1}{4}$ Answer: (A).
22	CirclesEquation from a diameter	Class 11: Conic Sections	Single correctMedium
A medium single correct question on equation from a diameter, from Circles. NCERT: Class 11: Conic Sections The circle has centre $C(3, 2)$ and radius $\sqrt{9 + 4 + 11} = \sqrt{24}$. The radius to a point of contact is perpendicular to the tangent: $\angle CAP = \angle CBP = 90^\circ$. So $A$ and $B$ both lie on the circle with $PC$ as diameter, which is therefore the circumcircle of $PAB$. Circle on the diameter joining $(1, 8)$ and $(3, 2)$: $(x - 1)(x - 3) + (y - 8)(y - 2) = 0$ $x^2 - 4x + 3 + y^2 - 10y + 16 = 0$ $x^2 + y^2 - 4x - 10y + 19 = 0$ Answer: (B).
23	Differential EquationsSeparation of variables	Class 12: Differential Equations	Single correctMedium
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Differentiate both sides with respect to $x$ (fundamental theorem of calculus): $\sqrt{1 - (f'(x))^2} = f(x)$, so $(f'(x))^2 = 1 - f^2(x)$ and $f'(x) = \pm\sqrt{1 - f^2(x)}$. Since $f \ge 0$ and $f(0) = 0$, $f$ cannot decrease from 0, so take $f' = \sqrt{1 - f^2}$. ($f \equiv 0$ is not possible: it would need $\sqrt{1 - 0} = 0$.) Separate: $\frac{df}{\sqrt{1 - f^2}} = dx \Rightarrow \sin^{-1} f = x + c$, and $f(0) = 0$ gives $c = 0$. So $f(x) = \sin x$ on $[0, 1]$ (and $1 < \frac{\pi}{2}$, so this stays valid). For $x > 0$, $\sin x < x$: $f\left(\frac{1}{2}\right) = \sin\frac{1}{2} < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) = \sin\frac{1}{3} < \frac{1}{3}$. Answer: (C).
24	Complex NumbersAlgebra of complex numbers	Class 11: Complex Numbers and Quadratic Equations	Single correctMedium
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z\bar{z}^3 + \bar{z}z^3 = z\bar{z}(\bar{z}^2 + z^2) = \|z\|^2 \cdot 2\,\text{Re}(z^2) = (x^2 + y^2) \cdot 2(x^2 - y^2)$. So $(x^2 + y^2)(x^2 - y^2) = 175$. With $x, y$ integers, $x^2 + y^2$ and $x^2 - y^2$ are a factor pair of 175 of the same parity with $x^2 + y^2 > x^2 - y^2 > 0$: $(175, 1)$, $(35, 5)$, $(25, 7)$. $(175, 1)$: $x^2 = 88$, not a square. $(35, 5)$: $x^2 = 20$, not a square. $(25, 7)$: $x^2 = 16$, $y^2 = 9$. So $x = \pm 4$, $y = \pm 3$: the roots are $(\pm 4, \pm 3)$, a rectangle of sides 8 and 6. Area $= 8 \times 6 = 48$. Answer: (A).
25	Conic SectionsEllipse	Class 11: Conic Sections	Single correctMedium
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The ellipse is $\frac{x^2}{9} + \frac{y^2}{1} = 1$: $a = 3$, $b = 1$. Take $A(3, 0)$, $B(0, 1)$ (other choices of extremities give congruent triangles). Line $AB$: $\frac{x}{3} + y = 1$, i.e. $x = 3 - 3y$. Auxiliary circle: $x^2 + y^2 = 9$. Substituting: $(3 - 3y)^2 + y^2 = 9 \Rightarrow 10y^2 - 18y = 0 \Rightarrow y = 0$ (the point $A$) or $y = \frac{9}{5}$. So $M = \left(-\frac{12}{5}, \frac{9}{5}\right)$. Triangle $OAM$ has base $OA = 3$ on the x-axis and height $\frac{9}{5}$: Area $= \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10}$. Answer: (D).
26	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Single correctHard
A hard single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\|\vec{a} \times \vec{b}\| \le 1$ and $\|\vec{c} \times \vec{d}\| \le 1$ for unit vectors, so the dot product can equal 1 only if both cross products are unit vectors in the same direction. Hence $\vec{a} \perp \vec{b}$, $\vec{c} \perp \vec{d}$, and the planes of $(\vec{a}, \vec{b})$ and $(\vec{c}, \vec{d})$ coincide: all four vectors are coplanar. Also, by the identity $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$: $\frac{1}{2}(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c}) = 1$. (A), (B): all the vectors lie in one plane, so $\vec{a}, \vec{b}, \vec{c}$ and $\vec{b}, \vec{c}, \vec{d}$ are coplanar. Both false. (C): if $\vec{b} = \vec{d}$, then $\vec{a} \cdot \vec{d} = \vec{a} \cdot \vec{b} = 0$ and the identity gives $\frac{1}{2} = 1$; if $\vec{b} = -\vec{d}$, it gives $-\frac{1}{2} = 1$. Both impossible, so $\vec{b}$ and $\vec{d}$ are not parallel. True. (D): if $\vec{a} = \pm\vec{d}$, then $\vec{a} \cdot \vec{c} = \pm\vec{d} \cdot \vec{c} = 0$, contradicting $\vec{a} \cdot \vec{c} = \frac{1}{2}$. False. Answer: (C).
27	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	Single correctMedium
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations By De Moivre, $z^{2m-1} = \cos(2m-1)\theta + i\sin(2m-1)\theta$, so the sum is $S = \sin\theta + \sin 3\theta + \ldots + \sin 29\theta$. Multiply by $2\sin\theta$ and use $2\sin\theta\sin(2m-1)\theta = \cos(2m-2)\theta - \cos 2m\theta$; the sum telescopes: $2\sin\theta \cdot S = \cos 0 - \cos 30\theta = 1 - \cos 30\theta = 2\sin^2 15\theta$ $S = \frac{\sin^2 15\theta}{\sin\theta}$ At $\theta = 2^\circ$: $15\theta = 30^\circ$, $\sin^2 30^\circ = \frac{1}{4}$, so $S = \frac{1}{4\sin 2^\circ}$. Answer: (D).
28	Permutations and CombinationsSimple applications	Class 11: Permutations and Combinations	Single correctMedium
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Seven digits each at least 1 sum to at least 7; the extra 3 must be shared out, each digit getting at most 2 extra (digit 3). Ways to make the extra 3: $2 + 1$ or $1 + 1 + 1$. Case $2 + 1$: one 3, one 2 and five 1s. Arrangements $= \frac{7!}{5!} = 42$. Case $1 + 1 + 1$: three 2s and four 1s. Arrangements $= \binom{7}{3} = 35$. Total $= 42 + 35 = 77$. Answer: (C).
29	Integral CalculusArea under simple curves	Class 12: Integrals; Class 12: Application of Integrals	Multiple correctMedium
A medium multiple correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The region lies between $x = 0$ and $x = 1$ (where $e^x = e$), below $y = e$ and above $y = e^x$: Area $= \int_0^1 (e - e^x)\,dx = e - (e - 1) = 1$. (A) $e - 1 \approx 1.718 \ne 1$. False. (B) Put $u = e + 1 - y$: as $y$ runs from 1 to $e$, $u$ runs from $e$ to 1, $dy = -du$: $\int_1^e \ln(e + 1 - y)\,dy = \int_1^e \ln u\,du = [u\ln u - u]_1^e = (e - e) - (0 - 1) = 1$. True. (C) $e - \int_0^1 e^x\,dx = e - (e - 1) = 1$: the rectangle $[0,1]\times[0,e]$ minus the area under the curve. True. (D) Integrating in $y$: the region is $1 \le y \le e$, $0 \le x \le \ln y$, so Area $= \int_1^e \ln y\,dy = [y\ln y - y]_1^e = 1$. True. Answer: (B), (C), (D).
30	Limits, Continuity and DifferentiabilityLimits of functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Multiple correctMedium
A medium multiple correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Expand by the binomial series: $\sqrt{a^2 - x^2} = a\left(1 - \frac{x^2}{a^2}\right)^{1/2} = a - \frac{x^2}{2a} - \frac{x^4}{8a^3} - \ldots$ Numerator $= \frac{x^2}{2a} + \frac{x^4}{8a^3} + \ldots - \frac{x^2}{4} = x^2\left(\frac{1}{2a} - \frac{1}{4}\right) + \frac{x^4}{8a^3} + \ldots$ For $L$ to be finite the $x^2$ term must vanish: $\frac{1}{2a} = \frac{1}{4} \Rightarrow a = 2$. Then $L = \frac{1}{8a^3} = \frac{1}{64}$. (A) $a = 2$: true. (B) $a = 1$: false. (C) $L = \frac{1}{64}$: true. (D) $L = \frac{1}{32}$: false. Answer: (A), (C).
31	TrigonometryTrigonometric identities	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Multiple correctMedium
A medium multiple correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\cos B + \cos C = 2\cos\frac{B+C}{2}\cos\frac{B-C}{2} = 2\sin\frac{A}{2}\cos\frac{B-C}{2}$. So $2\sin\frac{A}{2}\cos\frac{B-C}{2} = 4\sin^2\frac{A}{2} \Rightarrow \cos\frac{B-C}{2} = 2\sin\frac{A}{2}$. By the sine rule: $\frac{b + c}{a} = \frac{\sin B + \sin C}{\sin A} = \frac{2\sin\frac{B+C}{2}\cos\frac{B-C}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}} = \frac{\cos\frac{B-C}{2}}{\sin\frac{A}{2}} = 2$. So $b + c = 2a$: (B) true, (A) false. $AB + AC = 2a = 2\,BC$ is constant and greater than $BC$, so $A$ moves on an ellipse with foci $B$ and $C$ (excluding the two points on line $BC$): (C) true, (D) false. Answer: (B), (C).
32	TrigonometryTrigonometric identities	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Multiple correctMedium
A medium multiple correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $s = \sin^2 x$, so $\cos^2 x = 1 - s$: $\frac{s^2}{2} + \frac{(1 - s)^2}{3} = \frac{1}{5}$ Multiply by 30: $15s^2 + 10(1 - 2s + s^2) = 6 \Rightarrow 25s^2 - 20s + 4 = 0 \Rightarrow (5s - 2)^2 = 0$. So $\sin^2 x = \frac{2}{5}$, $\cos^2 x = \frac{3}{5}$. (A) $\tan^2 x = \frac{2/5}{3/5} = \frac{2}{3}$. True; (C) false. (B), (D): $\sin^8 x = \frac{16}{625}$, $\cos^8 x = \frac{81}{625}$: $\frac{\sin^8 x}{8} + \frac{\cos^8 x}{27} = \frac{2}{625} + \frac{3}{625} = \frac{5}{625} = \frac{1}{125}$. So (B) true, (D) false. (Alternatively, by Cauchy–Schwarz $\frac{u^2}{2} + \frac{v^2}{3} \ge \frac{(u+v)^2}{5} = \frac{1}{5}$ with equality only when $\frac{u}{2} = \frac{v}{3}$.) Answer: (A), (B).
33	Matrices and DeterminantsTypes of matrices	Class 12: Matrices; Class 12: Determinants	ComprehensionMedium
A medium comprehension question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Counting $\mathscr{A}$: let $d$ be the number of 1s on the diagonal and $k$ the number of symmetric off-diagonal pairs that are 1. Then $d + 2k = 5$ with $d \le 3$, $k \le 3$, so $d$ is odd: $d = 1, k = 2$: $\binom{3}{1}\binom{3}{2} = 9$ matrices; $d = 3, k = 1$: $\binom{3}{1} = 3$ matrices. Total 12. (The 1s off the diagonal come in pairs because the matrix is symmetric, which is why the diagonal must hold an odd number of 1s.) Answer: (A) 12.
34	Matrices and DeterminantsDeterminants of order two and three	Class 12: Matrices; Class 12: Determinants	ComprehensionHard
A hard comprehension question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Counting $\mathscr{A}$: let $d$ be the number of 1s on the diagonal and $k$ the number of symmetric off-diagonal pairs that are 1. Then $d + 2k = 5$ with $d \le 3$, $k \le 3$, so $d$ is odd: $d = 1, k = 2$: $\binom{3}{1}\binom{3}{2} = 9$ matrices; $d = 3, k = 1$: $\binom{3}{1} = 3$ matrices. Total 12. Determinants of the 12 matrices. Write the off-diagonal pairs as (12), (13), (23). Type $d = 3$, $k = 1$ (identity plus one pair): e.g. $\begin{pmatrix}1&1&0\\1&1&0\\0&0&1\end{pmatrix}$ has two equal rows, so all 3 are singular. Type $d = 1$, $k = 2$: with the diagonal 1 at position $i$: if both chosen pairs contain $i$ (e.g. $i = 1$, pairs (12), (13)): $\begin{pmatrix}1&1&1\\1&0&0\\1&0&0\end{pmatrix}$, two equal rows, singular; if one chosen pair is the pair not containing $i$ (e.g. $i = 1$ with (12), (23): $\begin{pmatrix}1&1&0\\1&0&1\\0&1&0\end{pmatrix}$, det $= -1$; with (13), (23): $\begin{pmatrix}1&0&1\\0&0&1\\1&1&0\end{pmatrix}$, det $= -1$): non-singular. So each $i$ gives 2 non-singular and 1 singular matrix: 6 non-singular matrices in all, and 6 singular ones (3 + 3). A unique solution exists exactly when $\det A \ne 0$: 6 matrices, which is at least 4 but less than 7. Answer: (B).
35	Matrices and DeterminantsConsistency of linear systems	Class 12: Matrices; Class 12: Determinants	ComprehensionHard
A hard comprehension question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Counting $\mathscr{A}$: let $d$ be the number of 1s on the diagonal and $k$ the number of symmetric off-diagonal pairs that are 1. Then $d + 2k = 5$ with $d \le 3$, $k \le 3$, so $d$ is odd: $d = 1, k = 2$: $\binom{3}{1}\binom{3}{2} = 9$ matrices; $d = 3, k = 1$: $\binom{3}{1} = 3$ matrices. Total 12. Determinants of the 12 matrices. Write the off-diagonal pairs as (12), (13), (23). Type $d = 3$, $k = 1$ (identity plus one pair): e.g. $\begin{pmatrix}1&1&0\\1&1&0\\0&0&1\end{pmatrix}$ has two equal rows, so all 3 are singular. Type $d = 1$, $k = 2$: with the diagonal 1 at position $i$: if both chosen pairs contain $i$ (e.g. $i = 1$, pairs (12), (13)): $\begin{pmatrix}1&1&1\\1&0&0\\1&0&0\end{pmatrix}$, two equal rows, singular; if one chosen pair is the pair not containing $i$ (e.g. $i = 1$ with (12), (23): $\begin{pmatrix}1&1&0\\1&0&1\\0&1&0\end{pmatrix}$, det $= -1$; with (13), (23): $\begin{pmatrix}1&0&1\\0&0&1\\1&1&0\end{pmatrix}$, det $= -1$): non-singular. So each $i$ gives 2 non-singular and 1 singular matrix: 6 non-singular matrices in all, and 6 singular ones (3 + 3). Only the 6 singular matrices can give an inconsistent system. Check each with right side $(1, 0, 0)$: Identity plus (12): $x + y = 1$, $x + y = 0$: inconsistent. Identity plus (13): $x + z = 1$, $y = 0$, $x + z = 0$: inconsistent. Identity plus (23): $x = 1$, $y + z = 0$ twice: consistent. $a_{11} = 1$ with (12), (13): $x + y + z = 1$, $x = 0$, $x = 0$: consistent. $a_{22} = 1$ with (12), (23): $y = 1$, $x + y + z = 0$, $y = 0$: inconsistent. $a_{33} = 1$ with (13), (23): $z = 1$, $z = 0$, $x + y + z = 0$: inconsistent. So 4 matrices give an inconsistent system: more than 2. Answer: (B).
36	ProbabilityProbability distribution of a random variable	Class 11: Probability; Class 12: Probability	ComprehensionEasy
A easy comprehension question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $X = 3$ means the first two tosses are not six and the third is six (tosses independent): $P(X = 3) = \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} = \frac{25}{216}$. Answer: (A).
37	ProbabilityProbability distribution of a random variable	Class 11: Probability; Class 12: Probability	ComprehensionEasy
A easy comprehension question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $X \ge 3$ means the first two tosses are both not six: $P(X \ge 3) = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$. (Check: $1 - P(X = 1) - P(X = 2) = 1 - \frac{1}{6} - \frac{5}{36} = \frac{25}{36}$.) Answer: (B).
38	ProbabilityAddition and multiplication theorems	Class 11: Probability; Class 12: Probability	ComprehensionMedium
A medium comprehension question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(X \ge n) = \left(\frac{5}{6}\right)^{n-1}$ (the first $n - 1$ tosses are not six). $P(X \ge 6 \mid X > 3) = \frac{P(X \ge 6)}{P(X \ge 4)} = \frac{(5/6)^5}{(5/6)^3} = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$. (The memoryless property: given three failures, needing at least two more failures has probability $(5/6)^2$.) Answer: (D).
39	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	Matrix matchHard
A hard matrix match question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Useful values: $\frac{\pi}{8} \approx 0.39$, $\frac{\pi}{2} \approx 1.57$, $\frac{5\pi}{4} \approx 3.93$, $\pi \approx 3.14$. (A) $\frac{dy}{y} = -\frac{dx}{(x - 3)^2} \Rightarrow \ln\|y\| = \frac{1}{x - 3} + c$, so $y = Ke^{1/(x-3)}$, defined for every $x \ne 3$. An interval is contained in the domain if it does not contain 3. (P), (Q), (S) do not contain 3; (R) and (T) do. A → P, Q, S. (B) Put $t = x - 3$: $\int_{-2}^{2} (t + 2)(t + 1)t(t - 1)(t - 2)\,dt = \int_{-2}^{2} t(t^2 - 1)(t^2 - 4)\,dt = 0$ (odd integrand). Intervals containing 0: (P) and (T); (Q) and (S) are open at 0. B → P, T. (C) $f(x) = \cos^2 x + \sin x$, $f'(x) = -2\sin x\cos x + \cos x = \cos x(1 - 2\sin x)$. $f'$ changes from + to − where $\sin x = \frac{1}{2}$: local maxima at $x = \frac{\pi}{6}, \frac{5\pi}{6}$ (plus $2n\pi$); at $x = \pm\frac{\pi}{2}$ there are minima. $\frac{\pi}{6} \approx 0.52$ lies in (P), (Q), (R) ($0.39 < 0.52$), (T); not in (S). C → P, Q, R, T. (D) $\tan^{-1}$ is increasing, so $\tan^{-1}(\sin x + \cos x)$ increases where $\sin x + \cos x$ does: $\cos x - \sin x = \sqrt{2}\cos\left(x + \frac{\pi}{4}\right) > 0$, i.e. $x \in \left(-\frac{3\pi}{4}, \frac{\pi}{4}\right)$ (mod $2\pi$). Of the intervals only (S) $\left(0, \frac{\pi}{8}\right)$ lies inside. D → S. Official: A → P, Q, S; B → P, T; C → P, Q, R, T; D → S.
40	Conic SectionsStandard forms of conics	Class 11: Conic Sections	Matrix matchMedium
A medium matrix match question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections (P) The distance from the origin to $hx + ky = 1$ equals the radius 2: $\frac{1}{\sqrt{h^2 + k^2}} = 2 \Rightarrow h^2 + k^2 = \frac{1}{4}$, a circle. (Q) The difference of distances from $(-2, 0)$ and $(2, 0)$ is the constant 3, less than the distance 4 between them: a hyperbola (both branches). (R) $\frac{x}{\sqrt{3}} = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$ are $\cos\theta$, $\sin\theta$ with $t = \tan\frac{\theta}{2}$, so $\frac{x^2}{3} + y^2 = 1$: an ellipse. (S) $e \ge 1$: $e = 1$ for a parabola and $e > 1$ for a hyperbola. (T) With $z = x + iy$: $\text{Re}(z + 1)^2 = (x + 1)^2 - y^2$, so $(x + 1)^2 - y^2 = x^2 + y^2 + 1 \Rightarrow 2x = 2y^2 \Rightarrow y^2 = x$: a parabola. A (circle) → P; B (parabola) → S, T; C (ellipse) → R; D (hyperbola) → Q, S. Official: A → P; B → S, T; C → R; D → Q, S.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
41	ElectrostaticsConductors and dielectrics	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance By induction (Gauss's law inside each conductor), the inner surface of a shell carries minus the total charge enclosed, and its outer surface carries the total charge up to and including that shell: Outer surface of shell 1: $Q_1$; of shell 2: $Q_1 + Q_2$; of shell 3: $Q_1 + Q_2 + Q_3$. Equal surface densities: $\frac{Q_1}{4\pi R^2} = \frac{Q_1 + Q_2}{4\pi(2R)^2} = \frac{Q_1 + Q_2 + Q_3}{4\pi(3R)^2}$ So $Q_1 + Q_2 = 4Q_1 \Rightarrow Q_2 = 3Q_1$, and $Q_1 + Q_2 + Q_3 = 9Q_1 \Rightarrow Q_3 = 5Q_1$. $Q_1 : Q_2 : Q_3 = 1 : 3 : 5$. Answer: (B).
42	Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium single correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Sliding begins when $\tan\theta > \mu = \sqrt{3}$, i.e. $\theta > 60^\circ$. Toppling begins when the vertical through the centre of mass passes outside the lower edge of the base: $\tan\theta > \frac{\text{base}/2}{\text{height}/2} = \frac{10}{15} = \frac{2}{3}$, i.e. $\theta > \tan^{-1}\frac{2}{3} \approx 33.7^\circ$. As $\theta$ increases from 0, the toppling angle (about $34^\circ$) is reached long before the sliding angle ($60^\circ$). The block stays at rest up to about $34^\circ$ and then topples. (A) At $30^\circ$, $\tan 30^\circ < \sqrt{3}$: no sliding. (C), (D): the block has already toppled before $60^\circ$. Answer: (B).
43	OpticsReflection and refraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctMedium
A medium single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Speed of the ball after falling $20 - 12.8 = 7.2$ m: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 7.2} = 12$ m/s. For a fish looking up along the normal from water ($\mu = \frac{4}{3}$) at an object in air at height $y$ above the surface, the image appears at height $y' = \mu y$ (objects in the rarer medium look farther away). So $\frac{dy'}{dt} = \mu\frac{dy}{dt}$: apparent speed $= \frac{4}{3} \times 12 = 16$ m/s. Answer: (C).
44	Motion of System of Particles and Rigid BodyCentre of mass	Class 11: Systems of Particles and Rotational Motion	Single correctEasy
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Treat each part as a point mass at its own centre: outer circle $6m$ at $y = 0$; two inner circles $m$ each at $y = a$; vertical line $m$ at $y = 0$; horizontal line $m$ at $y = -a$. Total mass $= 6m + m + m + m + m = 10m$. $y_{cm} = \frac{6m(0) + m(a) + m(a) + m(0) + m(-a)}{10m} = \frac{a}{10}$. Answer: (A).
45	Work, Energy and PowerCollisions	Class 11: Work, Energy and Power	Single correctMedium
A medium single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power In an elastic head-on collision between equal masses the velocities are exchanged. So the motion looks exactly as if the two particles passed through each other: one "speed $v$" motion and one "speed $2v$" motion each carry on round the circle. Approach speed is $v + 2v = 3v$, so they meet each time they have together covered one circumference $2\pi R$: every $\frac{2\pi R}{3v}$. The speed-$v$ motion returns to A after $\frac{2\pi R}{v}$, and the speed-$2v$ motion returns after $\frac{\pi R}{v}$ and again at $\frac{2\pi R}{v}$. So both particles are first back at A together at $t = \frac{2\pi R}{v}$. Meetings in that time: $\frac{2\pi R/v}{2\pi R/(3v)} = 3$; the third happens at A itself (at the moment they return). Collisions other than at A: $3 - 1 = 2$. Answer: (C).
46	Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctMedium
A medium single correct question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The wire runs round the outer square (through c and d), crosses over at the lower right corner, and runs round the inner square (through a and b). Because of the crossing, if the outer square is traversed clockwise the inner square is traversed anticlockwise: the two emfs oppose each other. It is one wire, so $I_1 = I_2$: (A) and (B) are false. The outer square encloses more area, so its emf wins: net emf $= -\frac{dB}{dt}(A_{outer} - A_{inner})$, with the sense set by the outer square. The field into the page is increasing, so by Lenz's law the current in the outer square circulates anticlockwise (its own field out of the page). Along the top of the outer square that is from right to left: from d to c, i.e. $I_2$ is in the direction dc. The inner square is then traversed clockwise, which along its top edge is left to right: from a to b, i.e. $I_1$ is in the direction ab. Answer: (D).
47	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 flux is $\frac{q_{enc}}{\varepsilon_0}$. Find the charge inside the cube $-\frac{a}{2} \le x, y, z \le \frac{a}{2}$. Disk: centre on the face $x = -\frac{a}{2}$, lying in the x-y plane; the face cuts it in half, so half its charge is inside: $3$ C. Rod: from $x = \frac{a}{4}$ to $\frac{5a}{4}$; the part inside is $\frac{a}{4}$ to $\frac{a}{2}$, length $\frac{a}{4}$ out of $a$: $\frac{1}{4} \times 8 = 2$ C. Point charge $-7$ C at $\left(\frac{a}{4}, -\frac{a}{4}, 0\right)$: inside. Point charge $3$ C at $\left(-\frac{3a}{4}, \frac{3a}{4}, 0\right)$: outside. $q_{enc} = 3 + 2 - 7 = -2$ C, so the flux is $\frac{-2\text{C}}{\varepsilon_0}$. Answer: (A).
48	Oscillations and WavesSHM	Class 11: Oscillations; Class 11: Waves	Single correctEasy
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves From the graph: $x = 0$ at $t = 0$ and rising, amplitude 1 cm, and the motion repeats every 8 s (zeros at 0, 4, 8, 12 s). So $x = \sin\left(\frac{2\pi}{8}t\right) = \sin\frac{\pi t}{4}$ cm, $\omega = \frac{\pi}{4}$ rad/s. $a = -\omega^2 x = -\frac{\pi^2}{16}\sin\frac{\pi t}{4}$. At $t = \frac{4}{3}$ s: $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$, so $a = -\frac{\pi^2}{16} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{32}\pi^2$ cm/s². Answer: (D).
49	Laws of MotionConservation of linear momentum	Class 11: Laws of Motion	Multiple correctEasy
A easy multiple correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion (A) $\vec{F}_{ext} = \frac{d\vec{P}}{dt} = 0$, so the total linear momentum is constant. True. (B) Internal forces can do work (an explosion, a stretched spring releasing), so the kinetic energy can change. Not sure. (C) Zero net force does not mean zero net torque: a couple has zero resultant force but non-zero torque, which changes the angular momentum. Not sure. (D) Internal conservative forces can change the potential energy (and external forces summing to zero can still do work). Not sure. Answer: (A).
50	Experimental SkillsOptics experiments	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on optics experiments, from Experimental Skills. NCERT: No matching NCERT chapter For a concave mirror with real object and real image (distances as magnitudes): $\frac{1}{u} + \frac{1}{v} = \frac{1}{f} = \frac{1}{24}$, i.e. $v = \frac{24u}{u - 24}$. (42, 56): $\frac{1}{42} + \frac{1}{56} = \frac{4 + 3}{168} = \frac{1}{24}$. Consistent. (48, 48): $\frac{2}{48} = \frac{1}{24}$. Consistent (object at 2f). (60, 40): $\frac{1}{60} + \frac{1}{40} = \frac{2 + 3}{120} = \frac{1}{24}$. Consistent. (66, 33): expected $v = \frac{24 \times 66}{42} \approx 37.7$ cm; recorded 33 cm, off by about 4.7 cm, far more than the 0.2 cm possible error. Incorrect. (78, 39): expected $v = \frac{24 \times 78}{54} \approx 34.7$ cm; recorded 39 cm, off by about 4.3 cm. Incorrect. (All distances fit on the 1.5 m bench, so length is not the issue.) Answer: (C), (D).
51	Current ElectricityOhm's law	Class 12: Current Electricity	Multiple correctMedium
A medium multiple correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity $R_2 = 6$ kΩ and $R_L = 1.5$ kΩ are in parallel: $\frac{6 \times 1.5}{7.5} = 1.2$ kΩ. In series with $R_1 = 2$ kΩ: total 3.2 kΩ. (A) $I = \frac{24}{3.2\ \text{k}\Omega} = 7.5$ mA. True. (B) Voltage across the parallel pair (and $R_L$) $= 7.5\ \text{mA} \times 1.2\ \text{k}\Omega = 9$ V, not 18 V. False. (C) $P_1 = I^2R_1 = (7.5)^2 \times 2 = 112.5$ mW; $P_2 = \frac{9^2}{6} = 13.5$ mW. Ratio $\approx 8.3$, not 3. False. (D) Interchanged: $R_1 = 6$ kΩ in series, $R_2 = 2$ kΩ parallel to $R_L$: $\frac{2 \times 1.5}{3.5} = \frac{6}{7}$ kΩ. Total $6 + \frac{6}{7} = \frac{48}{7}$ kΩ, $I = \frac{24 \times 7}{48} = 3.5$ mA, $V_L = 3.5 \times \frac{6}{7} = 3$ V. Power in $R_L$ goes from $\frac{9^2}{1.5}$ to $\frac{3^2}{1.5}$: it decreases by a factor of 9. True. Answer: (A), (D).
52	Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom	Class 11: Kinetic Theory	Multiple correctEasy
A easy multiple correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Monatomic ideal gas: $C_v = \frac{3}{2}R$, $C_p = \frac{5}{2}R$. Diatomic ideal gas (rigid, 5 degrees of freedom): $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$. (A) $C_p - C_v = R$ for both. False. (B) $C_p + C_v$: $4R$ (monatomic) against $6R$ (diatomic). True. (C) $\frac{C_p}{C_v}$: $\frac{5}{3} \approx 1.67$ against $\frac{7}{5} = 1.4$; smaller for the diatomic gas. False. (D) $C_p C_v$: $\frac{15}{4}R^2$ against $\frac{35}{4}R^2$. True. Answer: (B), (D).
53	Atoms and NucleiNuclear fission and fusion	Class 12: Atoms; Class 12: Nuclei	ComprehensionEasy
A easy comprehension question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei A plasma is a fully ionised gas. Ionisation happens because at the very high temperature of the core the thermal energy of the particles far exceeds the ionisation energy of the atoms, so collisions strip the electrons off. (A) The strong force acts only at nuclear distances and causes fusion, not ionisation. (B) Coulomb repulsion between deuterons does not ionise anything. (C) The Coulomb force between deuteron and electron is what holds the atom together; it opposes ionisation. (D) The high temperature: correct. Answer: (D).
54	Atoms and NucleiNuclear fission and fusion	Class 12: Atoms; Class 12: Nuclei	ComprehensionMedium
A medium comprehension question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Moving head-on with equal and opposite momenta, the deuterons can turn all their kinetic energy into potential energy at closest approach. Minimum temperature: total kinetic energy = Coulomb energy at $r = 4 \times 10^{-15}$ m. $2 \times 1.5kT = \frac{e^2}{4\pi\varepsilon_0 r}$ $3kT = \frac{1.44 \times 10^{-9}}{4 \times 10^{-15}}\ \text{eV} = 3.6 \times 10^5$ eV $T = \frac{3.6 \times 10^5}{3 \times 8.6 \times 10^{-5}} \approx 1.4 \times 10^9$ K This lies between $1.0 \times 10^9$ K and $2.0 \times 10^9$ K. Answer: (A).
55	Atoms and NucleiNuclear fission and fusion	Class 12: Atoms; Class 12: Nuclei	ComprehensionEasy
A easy comprehension question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Compute the Lawson number $nt_0$ for each and compare with $5 \times 10^{14}\ \text{s/cm}^3$: (A) $2.0 \times 10^{12} \times 5.0 \times 10^{-3} = 1.0 \times 10^{10}$ (B) $8.0 \times 10^{14} \times 9.0 \times 10^{-1} = 7.2 \times 10^{14}$ (C) $4.0 \times 10^{23} \times 1.0 \times 10^{-11} = 4.0 \times 10^{12}$ (D) $1.0 \times 10^{24} \times 4.0 \times 10^{-12} = 4.0 \times 10^{12}$ Only (B) exceeds $5 \times 10^{14}$. Answer: (B).
56	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	ComprehensionEasy
A easy comprehension question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter With $n$ loops between the nodes at 0 and $a$: $n\frac{\lambda}{2} = a \Rightarrow \lambda = \frac{2a}{n}$. de Broglie: $p = \frac{h}{\lambda} = \frac{nh}{2a}$. $E = \frac{p^2}{2m} = \frac{n^2h^2}{8ma^2}$. For fixed $n$, $E \propto a^{-2}$. Answer: (A).
57	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	ComprehensionMedium
A medium comprehension question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter With $n$ loops between the nodes at 0 and $a$: $n\frac{\lambda}{2} = a \Rightarrow \lambda = \frac{2a}{n}$. de Broglie: $p = \frac{h}{\lambda} = \frac{nh}{2a}$. $E = \frac{p^2}{2m} = \frac{n^2h^2}{8ma^2}$. Ground state $n = 1$: $E_1 = \frac{h^2}{8ma^2}$. $\frac{h}{a} = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-9}} = 1.0 \times 10^{-25}$, so $\frac{h^2}{a^2} = 1.0 \times 10^{-50}$. $E_1 = \frac{1.0 \times 10^{-50}}{8 \times 1.0 \times 10^{-30}} = 1.25 \times 10^{-21}$ J $= \frac{1.25 \times 10^{-21}}{1.6 \times 10^{-19}}$ eV $\approx 7.8 \times 10^{-3}$ eV $\approx 8$ meV. Answer: (B).
58	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	ComprehensionEasy
A easy comprehension question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter With $n$ loops between the nodes at 0 and $a$: $n\frac{\lambda}{2} = a \Rightarrow \lambda = \frac{2a}{n}$. de Broglie: $p = \frac{h}{\lambda} = \frac{nh}{2a}$. $E = \frac{p^2}{2m} = \frac{n^2h^2}{8ma^2}$. Speed $v = \frac{p}{m} = \frac{nh}{2ma} \propto n$. Answer: (D).
59	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Matrix matchHard
A hard matrix match question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Read the charges from the figures. (P) Hexagon with alternating + and − charges. Opposite corners carry opposite charges, and the field at M from a + at one corner and a − at the opposite corner point the same way; the three such pairs give equal fields at $120^\circ$ to each other, which add to zero: $E = 0$. Three + and three − at the same distance: $V = 0$. Rotating about the axis through M: all six charges move on the same circle, net charge zero, so the equivalent currents cancel: $B = 0$, $\mu = 0$. P: A, C. (Q) Along the line: −, +, − on one side of M and +, −, + on the other, so each charge has an equal and opposite charge at the mirror point. $V = 0$ (pairs cancel). $E \ne 0$ (the + and − of each pair push the same way along the line). Rotation about PQ: each mirror pair moves on the same circle with opposite charges, so currents cancel: $B = 0$, $\mu = 0$. Q: C only. (R) Outer ring: three + charges at $120^\circ$; inner ring: three − charges at $120^\circ$. Each symmetric ring gives zero field at its centre: $E = 0$. $V = 3kq\left(\frac{1}{R_{out}} - \frac{1}{R_{in}}\right) \ne 0$. Rotating, the rings carry opposite currents $\frac{3q\omega}{2\pi}$ on circles of different radius: $B = \frac{\mu_0 I}{2}\left(\frac{1}{R_{out}} - \frac{1}{R_{in}}\right) \ne 0$ and $\mu = I\pi(R_{out}^2 - R_{in}^2) \ne 0$. R: A, B, D. (S) The four corners carry − and the two mid-points of the long sides carry +. By symmetry about M, $E = 0$. $V$: the + charges are at $\frac{a}{2}$, the − charges at $\frac{\sqrt{5}}{2}a$: $V = kq\left(\frac{4}{a} - \frac{8}{\sqrt{5}a}\right) \ne 0$. Rotation about PQ (the long axis through M): the two + charges form a ring of radius $\frac{a}{2}$ centred at M, the corners form two − rings of radius $\frac{a}{2}$ centred a distance $a$ away. Their fields at M do not cancel (the near ring is much stronger), so $B \ne 0$; magnetic moment $\propto \sum q r^2 = (2q - 4q)\left(\frac{a}{2}\right)^2 \ne 0$. S: A, B, D. (T) Left ring: three + charges; right ring: three − charges, placed so that each + charge is the mirror image of a − charge in PQ. $V = 0$ (same set of distances for + and −). $E \ne 0$ (both rings push a test charge the same way, from left to right). Rotating about PQ, each + charge shares its circle with its mirror − charge: currents cancel, $B = 0$, $\mu = 0$. T: C only. Collecting by Column I: A ($E = 0$): P, R, S. B ($V \ne 0$): R, S. C ($B = 0$): P, Q, T. D ($\mu \ne 0$): R, S. Official: A → P, R, S; B → R, S; C → P, Q, T; D → R, S.
60	Work, Energy and PowerConservation of energy	Class 11: Work, Energy and Power	Matrix matchHard
A hard matrix match question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power (P) Y slides at constant velocity, so the net force on Y is zero: the total force from X (normal reaction plus friction) balances the weight, magnitude $Mg$ (A). X is fixed: its potential energy is constant. Friction turns mechanical energy into heat continuously (C). P is at the foot of the incline, not on the vertical through Y, so the weight of Y has a torque about P. P → A, C. (Q) The lift moves up at constant velocity (an inertial frame). Y is pushed down by its own weight and by the magnetic repulsion of Z (which supports Z's weight, $Mg$), so X pushes Y up with $2Mg$, not $Mg$. X rises steadily, so its potential energy keeps increasing (B). The lift does work on the system: its mechanical energy increases, not decreases. P lies on the common vertical axis through Y, so the weight of Y passes through P: zero torque (D). Q → B, D. (R) Lift going down at constant velocity: X (and everything) loses potential energy steadily with no change in kinetic energy, so the mechanical energy of X + Y keeps decreasing (C). The force of X on the pulley Y balances the pulley's weight and the two string tensions ($m_0 g + $ string forces), not $Mg$. X's potential energy is decreasing, not increasing. P is a point on the table, not on the vertical through Y: non-zero torque. R → C. (S) Non-viscous liquid: Y accelerates down, so the buoyant force from X is less than $Mg$. As Y sinks, liquid moves up to fill the space above it: the potential energy of X increases (B). No viscosity: mechanical energy of X + Y is conserved, not decreasing. P is at the bottom corner of the container, off the line of fall: non-zero torque. S → B. (T) Terminal velocity: no acceleration, so buoyancy plus viscous drag from X equal $Mg$ (A). Liquid is displaced upward as Y falls: potential energy of X increases (B). Viscous drag dissipates energy continuously (C). P is off the line of fall: non-zero torque. T → A, B, C. Collecting by Column I: A: P, T. B: Q, S, T. C: P, R, T. D: Q. Official: A → P, T; B → Q, S, T; C → P, R, T; D → Q.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
1	Chemical KineticsArrhenius equation	Class 12: Chemical Kinetics	Single correctEasy
A easy single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The Arrhenius equation in base-10 logarithms is $\log k = \log A - \frac{E_a}{2.303R}\cdot\frac{1}{T}$. Comparing with $\log k = 6.0 - 2000\cdot\frac{1}{T}$: $\log A = 6.0$, so $A = 1.0 \times 10^{6}\ \mathrm{s^{-1}}$ (a first-order rate constant has unit $\mathrm{s^{-1}}$). $\frac{E_a}{2.303R} = 2000\ \mathrm{K}$, so $E_a = 2000 \times 2.303 \times 8.314\ \mathrm{J\ mol^{-1}} \approx 38.3\ \mathrm{kJ\ mol^{-1}}$. Answer: (D).
2	Coordination CompoundsValence bond theory	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds CO is a neutral ligand, so chromium is in the 0 oxidation state: $\mathrm{Cr}(0)$ is $3d^5 4s^1$, six valence electrons. CO is a strong-field ligand: the electrons pair up in the three $t_{2g}$ orbitals ($t_{2g}^6$), and the empty $3d$ orbitals take part in $d^2sp^3$ bonding with the six CO ligands. Unpaired electrons $n = 0$, so $\mu = \sqrt{n(n+2)} = 0$ BM. Answer: (A).
3	Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctMedium
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The positive charge is on C-3. Groups that can shift to it are on the neighbouring carbons: C-2 carries H, OH and the methyl C-1; C-4 carries H and a methyl (and C-5). A 1,2-shift moves the positive charge to the carbon the group left, so compare the new cations: A shift from C-4 (H or $\mathrm{CH_3}$) leaves a cation at C-4 with only alkyl (hyperconjugative) help. A shift from C-2 leaves a cation at C-2, which carries an OH. The oxygen lone pair delocalises onto C-2 ($\mathrm{C^+-OH \leftrightarrow C=O^+H}$): every atom then has an octet, so this oxocarbenium ion is far more stable (it simply loses $\mathrm{H^+}$ to give a ketone). So the group migrates from C-2. Of the two groups on C-2 that can move, H migrates in preference to $\mathrm{CH_3}$ (hydride has the higher migratory aptitude). Hence the H at C-2 migrates. Answer: (D).
4	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 The four structures are of diazomethane. Rank them by the usual rules: more covalent bonds and complete octets first, then the negative charge on the more electronegative atom, and no sextet on a positive nitrogen. (I) $\mathrm{H_2C=\overset{+}{N}=\overset{-}{N}}$: every atom has an octet, and the negative charge sits on the terminal N (more electronegative than C). Most stable. (III) $\mathrm{H_2\overset{-}{C}-\overset{+}{N}\equiv N}$: every atom has an octet too, but the negative charge is on carbon, less electronegative. Second. (II) $\mathrm{H_2\overset{+}{C}-N=\overset{-}{N}}$: carbon has only a sextet (positive carbon). Third. (IV) $\mathrm{H_2\overset{-}{C}-N=\overset{+}{N}}$: the terminal nitrogen has a sextet and a positive charge, which is worse than a positive carbon with a sextet because N is more electronegative. Least stable. Order: (I) > (III) > (II) > (IV). Answer: (B).
5	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	Multiple correctMedium
A medium multiple correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\mathrm{NO_3^-}$ oxidises a metal M when the cell (nitrate reduced at the cathode, M oxidised at the anode) has a positive emf: $E^0_{cell} = 0.96 - E^0(\mathrm{M^{n+}/M}) > 0$, that is $E^0(\mathrm{M^{n+}/M}) < 0.96$ V. V: $-1.19$ V, oxidised. Fe: $-0.04$ V, oxidised. Hg: $+0.86$ V, oxidised ($E^0_{cell} = +0.10$ V). Au: $+1.40$ V, not oxidised ($E^0_{cell} = -0.44$ V). (A) V and Hg: both oxidised. True. (B) Hg and Fe: both oxidised. True. (C) Fe and Au: Au is not oxidised. False. (D) Fe and V: both oxidised. True. Answer: (A), (B), (D).
6	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	Multiple correctEasy
A easy multiple correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics A state function depends only on the state of the system, not on how it was reached. (A) Internal energy: a state function. True. (B) Irreversible expansion work: work depends on the path. False. (C) Reversible expansion work: still work, still path-dependent (different reversible paths between the same states do different work). False. (D) Molar enthalpy: $H = U + PV$ per mole, a state function (an intensive one). True. Answer: (A), (D).
7	p-Block ElementsGroup 13 elements	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Diborane can be split by a Lewis base in two ways. Unsymmetrical (heterolytic) cleavage gives the ionic product $\mathrm{[BH_2L_2]^+[BH_4]^-}$: two ligands crowd onto one boron. This happens with small bases: $\mathrm{NH_3}$ (which gives $\mathrm{[BH_2(NH_3)_2]^+[BH_4]^-}$), $\mathrm{CH_3NH_2}$ and $\mathrm{(CH_3)_2NH}$. Symmetrical cleavage gives two adducts $\mathrm{L\cdot BH_3}$. Bulky bases do this, because two of them cannot fit on one boron: $\mathrm{(CH_3)_3N}$ gives $\mathrm{2\,(CH_3)_3N\cdot BH_3}$. So (A), (B), (C) are true and (D) is false. Answer: (A), (B), (C).
8	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Draw each skeleton. (A) $\mathrm{N_2O}$: linear N–N–O ($\mathrm{N\equiv N^+-O^-}$). Has an N–N bond. True. (B) $\mathrm{N_2O_3}$: $\mathrm{O=N-NO_2}$. Has an N–N bond. True. (C) $\mathrm{N_2O_4}$: $\mathrm{O_2N-NO_2}$ (the dimer of $\mathrm{NO_2}$). Has an N–N bond. True. (D) $\mathrm{N_2O_5}$: $\mathrm{O_2N-O-NO_2}$, the two N atoms joined through an oxygen. No N–N bond. False. Answer: (A), (B), (C).
9	BiomoleculesCarbohydrates	Class 12: Biomolecules	Multiple correctMedium
A medium multiple correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules X: a glucopyranose ring joined through oxygen to C-2 of a fructofuranose ring. The glucose C-1 has its H up and the glycosidic O down in the Haworth drawing, which is the $\alpha$ configuration. This is sucrose: both anomeric carbons (glucose C-1 and fructose C-2) are used in the link, so there is no free hemiacetal and X is non-reducing. Y: a glucopyranose ring whose C-1 has the glycosidic O up (H down), which is $\beta$, joined to C-4 of a second glucose. The second glucose keeps a free anomeric OH at its C-1 (a hemiacetal that can open to an aldehyde), so Y is reducing (it is cellobiose, $\beta$-1,4). (A) X reducing, Y non-reducing: the reverse of the truth. False. (B) X non-reducing, Y reducing. True. (C) Linkages $\alpha$ in X and $\beta$ in Y. True. (D) $\beta$ in X and $\alpha$ in Y. False. Answer: (B), (C).
10	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Matrix matchMedium
A medium matrix match question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Each metal gives its nitrate; the reduction product of nitric acid depends on the acid's strength and on the metal. (A) $\mathrm{3Cu + 8HNO_3(dil) \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O}$. A → P, S. (B) $\mathrm{Cu + 4HNO_3(conc) \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O}$. B → Q, S. (C) Zinc is a stronger reducing agent; with dilute acid it takes nitrogen further down: $\mathrm{4Zn + 10HNO_3(dil) \rightarrow 4Zn(NO_3)_2 + N_2O + 5H_2O}$. C → R, T. (D) $\mathrm{Zn + 4HNO_3(conc) \rightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O}$. D → Q, T. Official: A → P, S; B → Q, S; C → R, T; D → Q, T.
11	Haloalkanes and HaloarenesSN1 and SN2 mechanisms	Class 12: Haloalkanes and Haloarenes	Matrix matchHard
A hard matrix match question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (A) 3-Bromo-2,3-dihydrobenzofuran. The C–Br carbon is benzylic and secondary, so Br is easily replaced by a nucleophile (P). Loss of HBr across C-2/C-3 (elimination, Q) gives benzofuran, in which the furan ring has become aromatic; the official key also counts this conversion of the dihydro ring into the aromatic ring as dehydrogenation (T). A → P, Q, T. (B) Benzyl alcohol, $\mathrm{C_6H_5CH_2OH}$. The OH can be replaced (with HBr, $\mathrm{SOCl_2}$ and so on: nucleophilic substitution, P); it is acetylated by acetic anhydride to benzyl acetate (S); and a primary alcohol is dehydrogenated over hot copper to the aldehyde, $\mathrm{C_6H_5CH_2OH \rightarrow C_6H_5CHO + H_2}$ (T). It has no β-hydrogen for elimination to an alkene. B → P, S, T. (C) Salicylaldehyde (2-hydroxybenzaldehyde). The CHO undergoes nucleophilic addition (HCN, bisulfite and so on: R); the phenolic OH is acetylated by acetic anhydride (S). C → R, S. (D) 1-Bromo-2-nitrobenzene. An aryl halide does not normally undergo substitution, but the nitro group ortho to Br withdraws electrons and stabilises the anionic (Meisenheimer) intermediate, so Br is replaced by $\mathrm{OH^-}$, $\mathrm{NH_3}$ and so on: nucleophilic aromatic substitution (P). D → P. Official: A → P, Q, T; B → P, S, T; C → R, S; D → P.
12	ThermodynamicsEnthalpy	Class 11: Thermodynamics	NumericalMedium
A medium numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat released = $C\,\Delta T = 2.5 \times (298.45 - 298.0) = 2.5 \times 0.45 = 1.125$ kJ. Moles of gas burnt $= \frac{3.5}{28} = 0.125$ mol. Heat released per mole $= \frac{1.125}{0.125} = 9$ kJ $\mathrm{mol^{-1}}$, so the combustion energy is $-9$ kJ $\mathrm{mol^{-1}}$. At constant volume this is strictly $\Delta U$; the question takes it as the enthalpy of combustion, and the numerical value asked for (the magnitude, as a single digit) is 9. Answer: 9.
13	Some Basic Concepts of ChemistryMole concept	Class 11: Some Basic Concepts of Chemistry	NumericalEasy
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $u_{rms} = \sqrt{\frac{3RT}{M}}$ and $u_{mp} = \sqrt{\frac{2RT}{M}}$. Setting them equal: $\frac{3 \times 400}{40} = \frac{2 \times 60}{M_Y}$, so $30 = \frac{120}{M_Y}$ and $M_Y = 4$. Answer: 4.
14	EquilibriumIonic equilibrium	Class 11: Equilibrium	NumericalEasy
A easy numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium The anion of a weak acid hydrolyses: $\mathrm{A^- + H_2O \rightleftharpoons HA + OH^-}$, with $K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{10^{-4}} = 10^{-10}$. $[\mathrm{OH^-}] = \sqrt{K_b\,c} = \sqrt{10^{-10} \times 10^{-2}} = 10^{-6}$ M. pOH = 6, so pH = 14 − 6 = 8. (Equivalently pH $= 7 + \frac{1}{2}(pK_a + \log c) = 7 + \frac{1}{2}(4 - 2) = 8$.) Answer: 8.
15	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	NumericalEasy
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Only $\alpha$ emission changes the mass number, by 4 each time: $238 - 214 = 24$, so 6 $\alpha$ particles. Six $\alpha$ particles lower the atomic number by 12: $92 - 12 = 80$. The product has 82, so the atomic number must rise by 2, which takes 2 $\beta^-$ emissions (each raises Z by 1). Total $= 6 + 2 = 8$. Answer: 8.
16	d- and f-Block ElementsPotassium dichromate and permanganate	Class 12: The d- and f-Block Elements	NumericalEasy
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Fusing $\mathrm{MnO_2}$ with an alkali (KOH) in air or with an oxidant ($\mathrm{KNO_3}$, $\mathrm{KClO_3}$) gives green potassium manganate: $\mathrm{2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O}$. In $\mathrm{K_2MnO_4}$: $2(+1) + x + 4(-2) = 0$, so $x = +6$. Answer: 6.
17	Coordination CompoundsWerner theory	Class 12: Coordination Compounds	NumericalEasy
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds In crystalline blue vitriol the copper ion is surrounded by four water molecules in a square plane, $\mathrm{[Cu(H_2O)_4]^{2+}}$, with two sulfate oxygens weakly above and below. The fifth water is not on copper: it is held by hydrogen bonds between the coordinated waters and the sulfate ion. The formula is $\mathrm{[Cu(H_2O)_4]SO_4\cdot H_2O}$. Answer: 4.
18	p-Block ElementsGroup 13 elements	No matching NCERT chapter	NumericalMedium
A medium numerical question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Solid aluminium chloride has a layer lattice (like $\mathrm{CrCl_3}$) in which each $\mathrm{Al^{3+}}$ sits in an octahedral hole surrounded by six chloride ions. (In the vapour and in non-polar solvents it is the dimer $\mathrm{Al_2Cl_6}$ with four-coordinate Al, and at high temperature the monomer with three; the question asks about the crystal.) Answer: 6.
19	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	NumericalMedium
A medium numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $\mathrm{C_5H_{10}}$ with a ring has no double bond. List the rings: Five-membered: cyclopentane. 1 Four-membered: methylcyclobutane. 1 Three-membered: ethylcyclopropane (1); 1,1-dimethylcyclopropane (1); 1,2-dimethylcyclopropane, which has cis (a meso form, 1) and trans (chiral, a pair of enantiomers, 2) forms. Total $= 1 + 1 + 1 + 1 + 1 + 2 = 7$. Answer: 7.

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
20	Sequences and SeriesArithmetic progressions	Class 11: Sequences and Series	Single correctEasy
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The $r$th term is $T_r = S_r - S_{r-1} = cr^2 - c(r-1)^2 = c(2r - 1)$. Sum of squares: $\sum_{r=1}^{n} c^2(2r-1)^2 = c^2\sum_{r=1}^{n}(4r^2 - 4r + 1)$ $= c^2\left[\frac{4n(n+1)(2n+1)}{6} - 2n(n+1) + n\right] = c^2 n\left[\frac{2(n+1)(2n+1) - 6(n+1) + 3}{3}\right]$ $= c^2 n\cdot\frac{4n^2 + 6n + 2 - 6n - 6 + 3}{3} = \frac{n(4n^2 - 1)c^2}{3}$. Check with $n = 1$: the only term is $c$, its square $c^2$, and the formula gives $\frac{3c^2}{3} = c^2$. Answer: (C).
21	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctEasy
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Equal angles with the axes and positive direction cosines give $l = m = n = \frac{1}{\sqrt{3}}$. A point at distance $r$ from $P$ along the line is $\left(2 + \frac{r}{\sqrt{3}}, -1 + \frac{r}{\sqrt{3}}, 2 + \frac{r}{\sqrt{3}}\right)$. On the plane: $2\left(2 + \frac{r}{\sqrt{3}}\right) + \left(-1 + \frac{r}{\sqrt{3}}\right) + \left(2 + \frac{r}{\sqrt{3}}\right) = 9$, so $5 + \frac{4r}{\sqrt{3}} = 9$ and $r = \sqrt{3}$. So $PQ = \sqrt{3}$ (Q is (3, 0, 3)). Answer: (C).
22	Conic SectionsEllipse	Class 11: Conic Sections	Single correctMedium
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The ellipse is $\frac{x^2}{16} + \frac{y^2}{4} = 1$: $a = 4$, $b = 2$, $e^2 = 1 - \frac{4}{16} = \frac{3}{4}$, $ae = 2\sqrt{3}$. Take $P(4\cos\theta, 2\sin\theta)$. The normal is $\frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^2 - b^2$, that is $\frac{4x}{\cos\theta} - \frac{2y}{\sin\theta} = 12$. At $y = 0$: $x = 3\cos\theta$, so $Q(3\cos\theta, 0)$. Midpoint $M = \left(\frac{7}{2}\cos\theta, \sin\theta\right)$, so its locus is $\frac{4x^2}{49} + y^2 = 1$. The latus rectums are $x = \pm 2\sqrt{3}$: $\frac{4 \times 12}{49} + y^2 = 1$ gives $y^2 = \frac{1}{49}$, $y = \pm\frac{1}{7}$. Points: $\left(\pm 2\sqrt{3}, \pm\frac{1}{7}\right)$. Answer: (C).
23	Straight LinesCentroid, orthocentre and circumcentre	Class 11: Straight Lines	Single correctMedium
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines The first line meets $y = 0$ at $B(-p, 0)$; the second meets it at $C(-q, 0)$. Vertex $A$ (the meeting point of the first two lines): multiply the first by $q$ and the second by $p$ and subtract: $[q(1+p) - p(1+q)]x = -pq(1+p) + pq(1+q)$, so $(q - p)x = pq(q - p)$ and $x = pq$. Then $(1+p)pq - py + p(1+p) = 0$ gives $y = (1+p)(1+q)$. So $A = (pq, (1+p)(1+q))$. The altitude from $A$ is perpendicular to $BC$ (the $x$-axis): $x = pq$. The altitude from $B$ is perpendicular to the second line, whose slope is $\frac{1+q}{q}$: it is $y = -\frac{q}{1+q}(x + p)$. At $x = pq$: $y = -\frac{q}{1+q}\cdot p(q + 1) = -pq$. Orthocentre $H = (pq, -pq)$: it always lies on $x + y = 0$, a straight line. Answer: (D).
24	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	Multiple correctHard
A hard multiple correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replace $x$ by $-x$: $\frac{\sin(-nx)}{(1 + \pi^{-x})\sin(-x)} = \frac{\pi^x \sin nx}{(1 + \pi^x)\sin x}$. So $I_n = \int_{-\pi}^{\pi}\frac{\pi^x\sin nx}{(1+\pi^x)\sin x}dx$, and adding the two forms, $2I_n = \int_{-\pi}^{\pi}\frac{\sin nx}{\sin x}dx$. The integrand $\frac{\sin nx}{\sin x}$ is even, so $I_n = \int_0^{\pi}\frac{\sin nx}{\sin x}dx$. Now $\sin(n+2)x - \sin nx = 2\cos(n+1)x\,\sin x$, so $I_{n+2} - I_n = 2\int_0^{\pi}\cos(n+1)x\,dx = 0$ for $n \geq 0$. With $I_0 = 0$ and $I_1 = \int_0^{\pi}1\,dx = \pi$: every even $I_{2m} = 0$ and every odd $I_{2m+1} = \pi$. (A) $I_n = I_{n+2}$: shown above. True. (B) $\sum_{m=1}^{10} I_{2m+1} = 10\pi$: ten terms each $\pi$. True. (C) $\sum_{m=1}^{10} I_{2m} = 0$: every term is 0. True. (D) $I_n = I_{n+1}$: $I_0 = 0 \neq \pi = I_1$. False. Answer: (A), (B), (C).
25	Conic SectionsHyperbola	Class 11: Conic Sections	Multiple correctMedium
A medium multiple correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The hyperbola is $\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$, rectangular, so its eccentricity is $\sqrt{2}$ and its foci are $(\pm\frac{1}{\sqrt{2}}\cdot\sqrt{2}, 0) = (\pm 1, 0)$. The ellipse has $e = \frac{1}{\sqrt{2}}$: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b^2 = a^2(1 - e^2) = \frac{a^2}{2}$. Orthogonal intersection. At a common point $(x_1, y_1)$ the slopes are $-\frac{b^2x_1}{a^2y_1}$ (ellipse) and $\frac{x_1}{y_1}$ (hyperbola). Their product is $-1$: $\frac{b^2}{a^2}\cdot\frac{x_1^2}{y_1^2} = 1$, so $x_1^2 = 2y_1^2$. On the hyperbola, $2x_1^2 - 2y_1^2 = 1$ gives $2y_1^2 = 1$, so $y_1^2 = \frac{1}{2}$, $x_1^2 = 1$. On the ellipse, $\frac{1}{a^2} + \frac{1/2}{a^2/2} = 1$ gives $a^2 = 2$, $b^2 = 1$. Ellipse: $\frac{x^2}{2} + y^2 = 1$, that is $x^2 + 2y^2 = 2$; foci $(\pm ae, 0) = (\pm 1, 0)$. (The two curves are confocal, as orthogonal central conics must be.) (A) True. (B) True. (C) $x^2 + 2y^2 = 4$: wrong size. False. (D) Foci $(\pm\sqrt{2}, 0)$: False. Answer: (A), (B).
26	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	Multiple correctMedium
A medium multiple correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = \cos\frac{1}{x} + \frac{1}{x}\sin\frac{1}{x}$. $f''(x) = \frac{1}{x^2}\sin\frac{1}{x} - \frac{1}{x^2}\sin\frac{1}{x} - \frac{1}{x^3}\cos\frac{1}{x} = -\frac{1}{x^3}\cos\frac{1}{x}$. For $x \geq 1$, $0 < \frac{1}{x} \leq 1 < \frac{\pi}{2}$, so $\cos\frac{1}{x} > 0$ and $f''(x) < 0$: $f'$ is strictly decreasing. As $x \to \infty$, $\frac{1}{x} \to 0$, so $f'(x) \to \cos 0 + 0 = 1$. Since $f'$ decreases towards the limit 1, $f'(x) > 1$ for every $x \geq 1$. Then $f(x+2) - f(x) = \int_x^{x+2}f'(t)\,dt > 2$. (A) $f(x+2) - f(x) < 2$ for some $x$: never. False. (B) $\lim_{x\to\infty}f'(x) = 1$. True. (C) $f(x+2) - f(x) > 2$ for all $x \geq 1$. True. (D) $f'$ strictly decreasing on $[1, \infty)$. True. Answer: (B), (C), (D).
27	Conic SectionsParabola	Class 11: Conic Sections	Multiple correctMedium
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Take $P(at^2, 2at)$. Tangent: $ty = x + at^2$, meeting the axis at $T(-at^2, 0)$. Normal: $y = -tx + 2at + at^3$, meeting the axis at $N(2a + at^2, 0)$. Centroid: $h = \frac{at^2 - at^2 + 2a + at^2}{3} = \frac{2a + at^2}{3}$, $k = \frac{2at}{3}$. Eliminate $t = \frac{3k}{2a}$: $3h = 2a + a\cdot\frac{9k^2}{4a^2}$, so $k^2 = \frac{4a}{9}(3h - 2a) = \frac{4a}{3}\left(h - \frac{2a}{3}\right)$. Locus: $y^2 = \frac{4a}{3}\left(x - \frac{2a}{3}\right)$, a parabola with vertex $\left(\frac{2a}{3}, 0\right)$ and $4A = \frac{4a}{3}$, so $A = \frac{a}{3}$. (A) Vertex $\left(\frac{2a}{3}, 0\right)$. True. (B) Directrix: $x = \frac{2a}{3} - \frac{a}{3} = \frac{a}{3}$, not $x = 0$. False. (C) Latus rectum $= 4A = \frac{4a}{3}$, not $\frac{2a}{3}$. False. (D) Focus: $\left(\frac{2a}{3} + \frac{a}{3}, 0\right) = (a, 0)$. True. Answer: (A), (D).
28	TrigonometryTrigonometric functions	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Multiple correctMedium
A medium multiple correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions For angles $\alpha$ and $\beta = \alpha + \frac{\pi}{4}$: $\cot\alpha - \cot\beta = \frac{\sin(\beta - \alpha)}{\sin\alpha\sin\beta}$, so $\operatorname{cosec}\alpha\operatorname{cosec}\beta = \frac{\cot\alpha - \cot\beta}{\sin\frac{\pi}{4}} = \sqrt{2}(\cot\alpha - \cot\beta)$. The sum telescopes: $\sqrt{2}\left[\cot\theta - \cot\left(\theta + \frac{6\pi}{4}\right)\right] = \sqrt{2}\left[\cot\theta - \cot\left(\theta + \frac{3\pi}{2}\right)\right] = \sqrt{2}(\cot\theta + \tan\theta)$. $\cot\theta + \tan\theta = \frac{1}{\sin\theta\cos\theta} = \frac{2}{\sin 2\theta}$, so the equation is $\frac{2\sqrt{2}}{\sin 2\theta} = 4\sqrt{2}$, i.e. $\sin 2\theta = \frac{1}{2}$. With $0 < 2\theta < \pi$: $2\theta = \frac{\pi}{6}$ or $\frac{5\pi}{6}$, so $\theta = \frac{\pi}{12}$ or $\frac{5\pi}{12}$. (A) $\frac{\pi}{4}$: $\sin 2\theta = 1$. False. (B) $\frac{\pi}{6}$: $\sin 2\theta = \frac{\sqrt{3}}{2}$. False. (C) True. (D) True. Answer: (C), (D).
29	TrigonometryTrigonometric functions	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Matrix matchMedium
A medium matrix match question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions (A) $\sin^2 2\theta = 2 - 2\sin^2\theta = 2\cos^2\theta$, so $4\sin^2\theta\cos^2\theta = 2\cos^2\theta$: either $\cos\theta = 0$ ($\theta = \frac{\pi}{2}$) or $\sin^2\theta = \frac{1}{2}$ ($\theta = \frac{\pi}{4}$). Among the listed values, $\frac{\pi}{6}$: $2\cdot\frac{1}{4} + \frac{3}{4} = \frac{5}{4} \neq 2$; $\frac{\pi}{3}$: $\frac{3}{2} + \frac{3}{4} \neq 2$; $\pi$: $0 \neq 2$. A → Q, S. (B) A product of step functions can jump only where $\frac{6x}{\pi}$ or $\frac{3x}{\pi}$ is an integer. Check each value. $x = \frac{\pi}{6}$: $\frac{6x}{\pi} = 1$ jumps (0 to 1), $\left[\frac{3x}{\pi}\right] = [0.5] = 0$ steady; $f$ goes from $0$ to $1\cdot\cos 0 = 1$. Discontinuous (P). $x = \frac{\pi}{4}$: $\frac{6x}{\pi} = 1.5$, $\frac{3x}{\pi} = 0.75$, neither an integer. Continuous. $x = \frac{\pi}{3}$: left value $1\cdot\cos 0 = 1$, value and right value $2\cos 1 \approx 1.08$. Discontinuous (R). $x = \frac{\pi}{2}$: $\frac{6x}{\pi} = 3$ jumps, $\left[\frac{3x}{\pi}\right] = [1.5] = 1$; left $2\cos 1$, right $3\cos 1$. Discontinuous (S). $x = \pi$: left $5\cos 2 \approx -2.08$, right $6\cos 3 \approx -5.94$. Discontinuous (T). B → P, R, S, T. (C) Volume $= \left\|\begin{vmatrix}1&1&0\\1&2&0\\1&1&\pi\end{vmatrix}\right\| = \pi(2 - 1) = \pi$. C → T. (D) $\vec{a} + \vec{b} = -\sqrt{3}\,\vec{c}$, so $\|\vec{a} + \vec{b}\|^2 = 3$: $1 + 1 + 2\cos\phi = 3$, $\cos\phi = \frac{1}{2}$, $\phi = \frac{\pi}{3}$. D → R. Official: A → Q, S; B → P, R, S, T; C → T; D → R.
30	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 (A) Let $g(x) = xe^{\sin x} - \cos x$. $g(0) = -1 < 0$ and $g\left(\frac{\pi}{2}\right) = \frac{\pi}{2}e > 0$. On $\left(0, \frac{\pi}{2}\right)$, $xe^{\sin x}$ increases (both factors positive and increasing) and $\cos x$ decreases, so $g$ is strictly increasing and has exactly one zero. A → P. (B) Three planes through the origin meet in a line (not just the origin) when the determinant vanishes: $\begin{vmatrix}k&4&1\\4&k&2\\2&2&1\end{vmatrix} = k(k - 4) - 4(4 - 4) + (8 - 2k) = k^2 - 6k + 8 = 0$, so $k = 2$ or $4$. For each the three planes are distinct, so they share a line. B → Q, S. (C) Let $S(x) = \|x-1\| + \|x-2\| + \|x+1\| + \|x+2\|$. For $-1 \leq x \leq 1$, $S = 6$; for $\|x\| \geq 2$, $S = 4\|x\|$. At integers: $S(0) = S(\pm1) = 6$, $S(\pm2) = 8$, $S(\pm3) = 12$, $S(\pm4) = 16$, $S(\pm5) = 20$, and so on. So $4k \in \{6, 8, 12, 16, 20, \ldots\}$, i.e. $k \in \{\frac{3}{2}, 2, 3, 4, 5, \ldots\}$. From the list: 2, 3, 4, 5. C → Q, R, S, T. (D) $\frac{dy}{y + 1} = dx$, so $y + 1 = Ce^x$; $y(0) = 1$ gives $C = 2$. $y(\ln 2) = 2\cdot 2 - 1 = 3$. D → R. Official: A → P; B → Q, S; C → Q, R, S, T; D → R.
31	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	NumericalEasy
A easy numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $x^2 - 9x + 20 \leq 0$ means $(x - 4)(x - 5) \leq 0$, so $A = [4, 5]$. $f'(x) = 6x^2 - 30x + 36 = 6(x - 2)(x - 3) > 0$ on $[4, 5]$, so $f$ increases there and is largest at $x = 5$: $f(5) = 250 - 375 + 180 - 48 = 7$. Answer: 7.
32	Matrices and DeterminantsConsistency of linear systems	Class 12: Matrices; Class 12: Determinants	NumericalEasy
A easy numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The second equation gives $z = 3x$. The first then gives $y = 3x - z = 0$. The third: $-3x + 0 + 3x = 0$, always true. So the solutions are $(x, 0, 3x)$. $x^2 + 9x^2 = 10x^2 \leq 100$ gives $x^2 \leq 10$, so the integer $x$ is $-3, -2, -1, 0, 1, 2, 3$: seven points. Answer: 7.
33	TrigonometryTrigonometric functions	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	NumericalMedium
A medium numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $BC = a$. The cosine rule at $B$: $AC^2 = AB^2 + a^2 - 2\cdot AB\cdot a\cos 30°$, so $8 = 16 + a^2 - 4\sqrt{3}a$, i.e. $a^2 - 4\sqrt{3}a + 8 = 0$. $a = 2\sqrt{3} \pm 2$ (both positive): these are $BC$ and $BC'$, the two triangles of the ambiguous case. Area $= \frac{1}{2}\cdot AB\cdot a\sin 30° = \frac{1}{2}\cdot 4 \cdot a \cdot\frac{1}{2} = a$. Difference $= (2\sqrt{3} + 2) - (2\sqrt{3} - 2) = 4$. Answer: 4.
34	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	NumericalMedium
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The limit means $\frac{p(x)}{x^2} \to 1$ as $x \to 0$, so $p$ has no constant or $x$ term and its $x^2$ coefficient is 1: $p(x) = ax^4 + bx^3 + x^2$. $p'(x) = 4ax^3 + 3bx^2 + 2x = x(4ax^2 + 3bx + 2)$ must vanish at $x = 1$ and $x = 2$: $4a + 3b + 2 = 0$ and $16a + 6b + 2 = 0$. Subtracting twice the first from the second: $8a - 2 = 0$, $a = \frac{1}{4}$; then $3b = -3$, $b = -1$. $p(x) = \frac{x^4}{4} - x^3 + x^2$, and $p(2) = 4 - 8 + 4 = 0$. Answer: 0.
35	Integral CalculusFundamental theorem of calculus	Class 12: Integrals; Class 12: Application of Integrals	NumericalEasy
A easy numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Since $f$ is continuous, the right side is differentiable, so $f'(x) = f(x)$; and $f(0) = \int_0^0 f = 0$. Then $\frac{d}{dx}\left(e^{-x}f(x)\right) = e^{-x}(f' - f) = 0$, so $e^{-x}f(x) = f(0) = 0$: $f \equiv 0$. $f(\ln 5) = 0$. Answer: 0.
36	CirclesStandard and general form	Class 11: Conic Sections	NumericalHard
A hard numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Put the centres at $(-3, 0)$ and $(3, 0)$, so $P$ is the origin. A line through $P$ tangent to $C_1$: $y = mx$ at distance 1 from $(-3, 0)$: $\frac{3\|m\|}{\sqrt{1 + m^2}} = 1$, so $m^2 = \frac{1}{8}$. By symmetry the same line is at distance 1 from $(3, 0)$: it is a transverse common tangent of $C_1$ and $C_2$. The circle $C$ touches $C_1$ and $C_2$ externally, so its centre is on the perpendicular bisector: $(0, h)$, radius $r$, and it must also touch the line $y = mx$. Tangent to the line: $\frac{\|h\|}{\sqrt{1 + m^2}} = r$, with $\sqrt{1 + m^2} = \frac{3}{2\sqrt{2}}$, so $h^2 = \frac{9r^2}{8}$. Touching $C_1$ externally: $9 + h^2 = (r + 1)^2$, so $9 + \frac{9r^2}{8} = r^2 + 2r + 1$, which gives $r^2 - 16r + 64 = 0$, $(r - 8)^2 = 0$. $r = 8$. Answer: 8.
37	Quadratic EquationsNature of roots	Class 11: Complex Numbers and Quadratic Equations	NumericalMedium
A medium numerical question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $f(x) = x^2 - 8kx + 16(k^2 - k + 1)$. Both roots real, distinct and $\geq 4$ needs: Discriminant $> 0$: $64k^2 - 64(k^2 - k + 1) > 0$, so $k > 1$. Vertex to the right of 4: $\frac{8k}{2} = 4k > 4$, so $k > 1$. $f(4) \geq 0$: $16 - 32k + 16k^2 - 16k + 16 = 16(k^2 - 3k + 2) = 16(k - 1)(k - 2) \geq 0$, so $k \leq 1$ or $k \geq 2$. Together: $k \geq 2$. The smallest value is 2 (then the roots are 4 and 12). Answer: 2.
38	Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	NumericalEasy
A easy numerical question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f(0) = 0 + e^0 = 1$, so $g(1) = 0$. Differentiating $f(g(x)) = x$: $g'(x) = \frac{1}{f'(g(x))}$. $f'(x) = 3x^2 + \frac{1}{2}e^{x/2}$, so $f'(0) = \frac{1}{2}$, and $g'(1) = \frac{1}{f'(0)} = 2$. Answer: 2.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
39	Oscillations and WavesSimple pendulum and spring systems	Class 11: Oscillations; Class 11: Waves	Single correctMedium
A medium single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The springs are in series between the wall and the mass, so (being light) they carry the same force at every instant: $k_1x_1 = k_2x_2$, where $x_1$, $x_2$ are their extensions. The mass moves by $x_1 + x_2$. At the extreme position $x_1 + x_2 = A$, so $x_1 = \frac{k_2A}{k_1 + k_2}$. P is the junction; its displacement equals the extension of the spring $k_1$ (whose other end is fixed to the wall). Its amplitude is $\frac{k_2A}{k_1 + k_2}$. Answer: (D).
40	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 In the frame of the wire the bead feels gravity $mg$ downwards and a pseudo force $ma$ opposite to the acceleration, besides the normal reaction, which is perpendicular to the wire. For equilibrium the resultant of $mg$ and $ma$ must also be perpendicular to the wire, so the tangent makes angle $\theta$ with the horizontal where $\tan\theta = \frac{a}{g}$. The slope of the parabola is $\frac{dy}{dx} = 2kx$, so $2kx = \frac{a}{g}$ and $x = \frac{a}{2gk}$ (on the side opposite to the acceleration). Answer: (B).
41	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	Single correctMedium
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon energies $E = \frac{1240}{\lambda}$: 550 nm gives 2.25 eV, 450 nm gives 2.76 eV, 350 nm gives 3.54 eV. Plate p (2.0 eV) is ejected by all three wavelengths; q (2.5 eV) by 450 nm and 350 nm; r (3.0 eV) by 350 nm only. Saturation current is set by the number of electrons ejected per second. The three wavelengths have equal intensity, so p receives the most effective photons, then q, then r: $I_{sat}$ is largest for p and smallest for r. Stopping potential is set by the most energetic photon (350 nm): $V_0 = 3.54 - \phi$, i.e. 1.54 V for p, 1.04 V for q, 0.54 V for r. The curve of p starts at the most negative voltage, then q, then r. Graph (A) has the saturation currents in the order p > q > r and the cut-offs with p the most negative and r the least. (B) and (D) give all three the same saturation current; (C) reverses the order of the currents. Answer: (A).
42	Oscillations and WavesSHM	Class 11: Oscillations; Class 11: Waves	Single correctMedium
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Turning the rod through a small angle $\theta$ moves each end by $\frac{L}{2}\theta$ along its spring, so each spring exerts a force $k\frac{L}{2}\theta$ at a lever arm $\frac{L}{2}$. Both torques oppose the rotation: $\tau = -2\cdot k\frac{L}{2}\theta\cdot\frac{L}{2} = -\frac{kL^2}{2}\theta$. Moment of inertia about the centre: $I = \frac{ML^2}{12}$. $\omega^2 = \frac{kL^2/2}{ML^2/12} = \frac{6k}{M}$, so $f = \frac{1}{2\pi}\sqrt{\frac{6k}{M}}$. Answer: (C).
43	Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Multiple correctMedium
A medium multiple correct question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Switching on the current changes the flux through each ring; the induced current (Lenz's law) makes the ring repel the solenoid. The rings have the same shape, so the same induced emf; the induced current is $\propto \frac{1}{R} \propto \frac{1}{\rho}$, and the upward force, and its impulse $J$, are proportional to that current. The ring leaves with speed $v = \frac{J}{m} \propto \frac{1}{\rho m}$ and rises to $h = \frac{v^2}{2g} \propto \frac{1}{(\rho m)^2}$. So $h_A > h_B$ requires $\rho_Am_A < \rho_Bm_B$. (A) $\rho_A > \rho_B$, $m_A = m_B$: $\rho_Am_A > \rho_Bm_B$. False. (B) $\rho_A < \rho_B$, $m_A = m_B$: $\rho_Am_A < \rho_Bm_B$. True. (C) $\rho_A > \rho_B$, $m_A > m_B$: the product is larger for A. False. (D) $\rho_A < \rho_B$, $m_A < m_B$: the product is smaller for A. True. Answer: (B), (D).
44	Experimental SkillsHeat and sound experiments	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on heat and sound experiments, from Experimental Skills. NCERT: No matching NCERT chapter (A) At the first resonance the air column is shorter, so less energy is lost in it and the sound reaching the ear from the open end is louder than at the second resonance. True. (B) The fork is held with its prongs vibrating vertically over the mouth of the tube, i.e. the prongs in a vertical plane, so that they push the air column up and down. Held in a horizontal plane they would not drive the column. False. (C) The prongs of a tuning fork vibrate with an amplitude of a fraction of a millimetre, not about 1 cm. False. (D) The antinode at the open end lies a little beyond the mouth (end correction $e \approx 0.3d$), so $l_1 + e = \frac{\lambda}{4}$ and the length of the column at the first resonance is somewhat shorter than $\frac{\lambda}{4}$. True. Answer: (A), (D).
45	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Multiple correctMedium
A medium multiple correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the figure: A = (2, 3), B = (3, 2), C = (2, 1), D = (1, 2) in the plotted units, and the cycle runs clockwise A → B → C → D → A. (A) $PV$ is 6 at A and at B, but in between, on the circle $(V - 2)^2 + (P - 2)^2 = 1$, at $V = P = 2 + \frac{1}{\sqrt{2}}$ it is about 7.3. $PV$ (and so $T$) is not constant, so A → B is not isothermal. False. (B) From B to D the volume falls from 3 to 1, so the gas has work done on it ($W_{by} < 0$), and $PV$ falls from 6 to 2, so $T$ and $U$ fall ($\Delta U < 0$). Then $Q = \Delta U + W_{by} < 0$: heat flows out. True. (C) Work in A → B → C is the area under the semicircle taken out (V from 2 to 3) and back (3 to 2): their difference is the area of the half disc, $\frac{\pi}{2}$ units, not zero. False. (D) A clockwise cycle on a $P$-$V$ diagram encloses a positive net work done by the gas. True. Answer: (B), (D).
46	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	Multiple correctEasy
A easy multiple correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The Coulomb force on $-q$ always points towards $+Q$: a central force, so its torque about $+Q$ is zero and the angular momentum about $+Q$ is conserved. (A) Angular momentum constant. True. (B) Linear momentum constant: a net force acts, so momentum changes (in direction always, and in size along an ellipse). False. (C) Angular velocity constant: $L = mr^2\omega$ is constant while $r$ changes along the ellipse, so $\omega$ changes. False. (D) Linear speed constant: the force has a component along the velocity except at the ends of the major axis, so the speed changes (fastest nearest the charge). False. Answer: (A).
47	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	Multiple correctEasy
A easy multiple correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rolling without slipping on a fixed surface: with the centre moving at $\vec{v}$, $\vec{V}_A = \vec{0}$, $\vec{V}_B = \vec{v}$, $\vec{V}_C = 2\vec{v}$. (A) $\vec{V}_C - \vec{V}_A = 2\vec{v}$, but $2(\vec{V}_B - \vec{V}_C) = -2\vec{v}$. False. (B) $\vec{V}_C - \vec{V}_B = \vec{v} = \vec{V}_B - \vec{V}_A$. True. (C) $\|\vec{V}_C - \vec{V}_A\| = 2v = 2\|\vec{V}_B - \vec{V}_C\|$. True. (D) $\|\vec{V}_C - \vec{V}_A\| = 2v \neq 4v$. False. Answer: (B), (C).
48	ThermodynamicsFirst law	Class 11: Thermodynamics	Matrix matchMedium
A medium matrix match question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics (P) Charging a capacitor stores energy $\frac{1}{2}CV^2$ in it: its energy increases (A). (Q) Adiabatic compression: the work done by the piston (mechanical energy) goes into the gas as internal energy, the random motion of its molecules (B), and the energy of the gas increases (A). (R) The rigid gas container loses heat to the surroundings: its internal energy decreases, with no work. None of A, B, C, D in the sense intended. (S) Fission: nuclear (internal) energy appears as kinetic energy of the fragments and neutrons (C), and the rest mass of the products is less than that of the nucleus (D). (T) The changing flux drives an induced current in the loop, which heats it: the loop's energy increases (A). By rows: A → P, Q, T; B → Q; C → S; D → S. Official: A → P, Q, T; B → Q; C → S; D → S.
49	OpticsInterference and diffraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Matrix matchHard
A hard matrix match question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The sheet on $S_2$ adds an optical path $(\mu - 1)t$ to the light from $S_2$. The effective path difference at a point is $\Delta = (S_1P - S_2P) - (\mu - 1)t$, the phase difference $\delta = \frac{2\pi}{\lambda}\Delta$, and with equal slits $I = 4I_0\cos^2\frac{\delta}{2}$. Geometric path differences: 0 at $P_0$, $\frac{\lambda}{4}$ at $P_1$, $\frac{\lambda}{3}$ at $P_2$. (A) No sheet. $P_0$: $\delta = 0$ (P), $I = 4I_0$. $P_1$: $\delta = \frac{\pi}{2}$, $I = 2I_0$. $P_2$: $\delta = \frac{2\pi}{3}$, $I = 4I_0\cos^2 60° = I_0$. So $I(P_0) > I(P_1)$ (S); $I(P_2) < I(P_1)$. A → P, S. (B) $(\mu - 1)t = \frac{\lambda}{4}$. $P_0$: $\Delta = -\frac{\lambda}{4}$, $I = 2I_0$. $P_1$: $\Delta = 0$, $\delta = 0$ (Q), $I = 4I_0$, the maximum, so neither S nor T. $P_2$: $\Delta = \frac{\lambda}{12}$, $I < 4I_0$. B → Q. (C) $(\mu - 1)t = \frac{\lambda}{2}$. $P_0$: $\Delta = -\frac{\lambda}{2}$, $I = 0$. $P_1$: $\Delta = -\frac{\lambda}{4}$, $I = 2I_0$. $P_2$: $\Delta = -\frac{\lambda}{6}$, $\delta = \frac{\pi}{3}$, $I = 4I_0\cos^2 30° = 3I_0 > 2I_0$ (T). C → T. (D) $(\mu - 1)t = \frac{3\lambda}{4}$. $P_0$: $\Delta = -\frac{3\lambda}{4}$, $I = 4I_0\cos^2 135° = 2I_0$. $P_1$: $\Delta = -\frac{\lambda}{2}$, $I = 0$ (R). $P_2$: $\Delta = -\frac{5\lambda}{12}$, $\delta = \frac{5\pi}{6}$, $I = 4I_0\cos^2 75° \approx 0.27I_0 > 0$. So $I(P_0) > I(P_1)$ (S) and $I(P_2) > I(P_1)$ (T). D → R, S, T. Official: A → P, S; B → Q; C → T; D → R, S, T.
50	Properties of Bulk MatterHeat transfer	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalMedium
A medium numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Heat flows from P to both ends. With conductivity $K$ and cross-section $A$: to the ice end, $H_1 = \frac{KA(400 - 0)}{\lambda x}$; to the water end, $H_2 = \frac{KA(400 - 100)}{(10 - \lambda)x}$. Equal masses per unit time: $\frac{H_1}{80} = \frac{H_2}{540}$, so $\frac{400}{80\lambda} = \frac{300}{540(10 - \lambda)}$, i.e. $\frac{5}{\lambda} = \frac{5}{9(10 - \lambda)}$. $9(10 - \lambda) = \lambda$, so $\lambda = 9$. Answer: 9.
51	Properties of Bulk MatterFluid mechanics	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalMedium
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Water stops flowing when the pressure at the orifice equals the atmosphere: $P_{air} + \rho g h = P_0$ with $h = 0.2$ m, so $P_{air} = 10^5 - 1000 \times 10 \times 0.2 = 98000$ Pa. The trapped air (temperature constant) obeys Boyle's law. Initially it is at $P_0$ and fills a length $0.5 - H$; finally it fills $0.5 - 0.2 = 0.3$ m: $10^5(0.5 - H) = 98000 \times 0.3 = 29400$, so $0.5 - H = 0.294$ m and $H = 0.206$ m = 206 mm. Fall in level $= 206 - 200 = 6$ mm. Answer: 6.
52	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalMedium
A medium numerical 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 A soap bubble has two surfaces, so the excess pressure inside is $\frac{4T}{r}$. A: $P_A = 8 + \frac{4 \times 0.04}{0.02} = 8 + 8 = 16$ N/m$^2$. B: $P_B = 8 + \frac{4 \times 0.04}{0.04} = 8 + 4 = 12$ N/m$^2$. At the same temperature $n \propto PV \propto Pr^3$: $\frac{n_B}{n_A} = \frac{12 \times 4^3}{16 \times 2^3} = \frac{768}{128} = 6$. Answer: 6.
53	Work, Energy and PowerCollisions	Class 11: Work, Energy and Power	NumericalEasy
A easy numerical question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Elastic collision of A ($m$, 9 m/s) with B ($2m$, at rest): $v_B = \frac{2m_A}{m_A + m_B}u = \frac{2m}{3m}\times 9 = 6$ m/s. B ($2m$, 6 m/s) sticks to C ($m$, at rest): momentum $2m \times 6 = 3m\,v$, so $v = 4$ m/s. Answer: 4.
54	Magnetic Effects of Current and MagnetismBiot-Savart law	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	NumericalMedium
A medium numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $3^2 + 4^2 = 5^2$, so the right angle is at P. Sides PQ and PR pass through P, so they give no field there; only QR contributes. Distance of P from QR: $d = \frac{PQ \cdot PR}{QR} = \frac{12x}{5}$. Let F be the foot of this perpendicular: $QF = \frac{PQ^2}{QR} = \frac{9x}{5}$ and $FR = \frac{PR^2}{QR} = \frac{16x}{5}$. For a straight segment, $B = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)$ with $\sin\alpha_1 = \frac{QF}{PQ} = \frac{3}{5}$ and $\sin\alpha_2 = \frac{FR}{PR} = \frac{4}{5}$. $B = \frac{\mu_0 I}{4\pi}\cdot\frac{5}{12x}\cdot\frac{7}{5} = \frac{7\mu_0 I}{48\pi x}$. $k = 7$. Answer: 7.
55	Work, Energy and PowerWork-energy theorem	Class 11: Work, Energy and Power	NumericalEasy
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Acceleration: $a = \frac{(0.72 - 0.36)g}{0.72 + 0.36} = \frac{g}{3} = \frac{10}{3}$ m/s$^2$; the lighter block rises. Tension: $T = m(g + a) = 0.36 \times \frac{40}{3} = 4.8$ N. Distance risen in 1 s: $s = \frac{1}{2}at^2 = \frac{5}{3}$ m. Work by the string on the 0.36 kg block $= Ts = 4.8 \times \frac{5}{3} = 8$ J. Answer: 8.
56	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	NumericalEasy
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charge within radius $r$: $q(r) = \int_0^r \kappa r'^a\,4\pi r'^2\,dr' = \frac{4\pi\kappa}{a + 3}r^{a+3}$. Gauss's law: $E(r) = \frac{q(r)}{4\pi\varepsilon_0 r^2} \propto r^{a+1}$ for $r \leq R$. $\frac{E(R/2)}{E(R)} = \left(\frac{1}{2}\right)^{a+1} = \frac{1}{8}$, so $a + 1 = 3$ and $a = 2$. Answer: 2.
57	Oscillations and WavesStanding waves and beats	Class 11: Oscillations; Class 11: Waves	NumericalEasy
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Linear density $\mu = \frac{1.0 \times 10^{-3}}{0.20} = 5 \times 10^{-3}$ kg/m. Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{0.5}{5 \times 10^{-3}}} = 10$ m/s. The string vibrates at the driving frequency: $\lambda = \frac{v}{f} = \frac{10}{100} = 0.1$ m = 10 cm. Successive nodes are $\frac{\lambda}{2} = 5$ cm apart (the 20 cm string holds four loops, so it is at resonance). Answer: 5.

Sit JEE Advanced 2009 free.

Timed like the exam or chapter by chapter, with a solution for each question. Run by a non-profit trust.

Start free