JEE (Advanced) · 2010
JEE Advanced 2010, 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.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level. One year is not a forecast.
Chemistry
47 Q
- Aldehydes, Ketones and Carboxylic Acids5
- Coordination Compounds4
- Thermodynamics4
24 easy · 22 medium · 1 hard
Mathematics
47 Q
- Trigonometry6
- Matrices and Determinants6
- Conic Sections5
9 easy · 29 medium · 9 hard
Physics
47 Q
- Properties of Bulk Matter7
- Optics6
- Current Electricity5
20 easy · 24 medium · 3 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 12: Current Electricity
Class 11: Oscillations; Class 11: Waves
Class 11: Gravitation
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 12: Atoms; Class 12: Nuclei
No matching NCERT chapter
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 11: Laws of Motion
Class 11: Units and Measurement
Class 11: Thermodynamics
Class 11: Kinetic Theory
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 11: Work, Energy and Power
Class 12: Dual Nature of Radiation and Matter
Class 11: Systems of Particles and Rotational Motion
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 12: Matrices; Class 12: Determinants
Class 11: Conic Sections
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 12: Application of Derivatives
Class 12: Integrals; Class 12: Application of Integrals
Class 12: Vector Algebra
Class 11: Probability; Class 12: Probability
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 11: Complex Numbers and Quadratic Equations
Class 11: Conic Sections
Class 11: Sequences and Series
Class 11: Complex Numbers and Quadratic Equations
Class 12: Differential Equations
Class 11: Binomial Theorem
Class 11: Permutations and Combinations
Class 11: Straight Lines
Class 12: Aldehydes, Ketones and Carboxylic Acids
Class 12: Coordination Compounds
Class 11: Thermodynamics
Class 11: Structure of Atom
Class 11: Hydrocarbons
No matching NCERT chapter
Class 11: Chemical Bonding and Molecular Structure
No matching NCERT chapter
Class 12: Biomolecules
Class 12: Alcohols, Phenols and Ethers
Class 12: Chemical Kinetics
Class 11: Equilibrium
Class 12: Electrochemistry
Class 11: Organic Chemistry – Some Basic Principles and Techniques
No matching NCERT chapter
- Principles Related to Practical Chemistry1
No matching NCERT chapter
Class 12: Amines
No matching NCERT chapter
Class 11: Redox Reactions
Class 11: Some Basic Concepts of Chemistry
03
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds EDTA is ethylenediamine, $\mathrm{H_2N{-}CH_2{-}CH_2{-}NH_2}$, in which all four N–H hydrogens are replaced by acetic acid groups $\mathrm{{-}CH_2COOH}$:
$\mathrm{(HOOC{-}CH_2)_2N{-}CH_2{-}CH_2{-}N(CH_2{-}COOH)_2}$.
(A) has a $\mathrm{{-}CH{=}CH{-}}$ bridge, not the saturated ethylene $\mathrm{{-}CH_2{-}CH_2{-}}$ of ethylenediamine.
(B) has the COOH groups bonded straight to N, with no $\mathrm{CH_2}$ between (that would be a carbamic acid, not an acetic acid).
(D) is a different molecule: the acid groups sit on the carbon chain and each N still carries an H.
(C) is the structure above: two N atoms joined by $\mathrm{{-}CH_2{-}CH_2{-}}$, each carrying two $\mathrm{{-}CH_2COOH}$ groups. This hexadentate ligand (two N, four O donors) is what makes EDTA a chelating agent.
Answer: (C). | ||||
| 2 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Ionization isomers have the same formula but exchange a ligand inside the coordination sphere with the counter-ion outside it, so they give different ions in solution.
In $\mathrm{[Cr(H_2O)_4Cl(NO_2)]Cl}$ the ligands are 4 $\mathrm{H_2O}$, one $\mathrm{Cl^-}$ and one $\mathrm{NO_2^-}$; the counter-ion is $\mathrm{Cl^-}$. Swapping the $\mathrm{NO_2^-}$ ligand with the outer $\mathrm{Cl^-}$ gives $\mathrm{[Cr(H_2O)_4Cl_2](NO_2)}$.
(A) and (C) only change how the nitrite binds (through O instead of N): those are linkage isomers, and (A) also loses a Cl ligand. (D) moves a water molecule out of the sphere (a hydrate isomer) and is not even the same set of ions.
Answer: (B). | ||||
| 3 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 3-Octyne is $\mathrm{CH_3CH_2{-}C{\equiv}C{-}CH_2CH_2CH_2CH_3}$: an ethyl group on one side of the triple bond and an n-butyl group on the other.
Sodium amide removes the acidic terminal H of a terminal alkyne, and the acetylide ion then displaces bromide from a primary bromoalkane ($S_N2$):
$\mathrm{CH_3CH_2C{\equiv}CH \xrightarrow{NaNH_2} CH_3CH_2C{\equiv}C^-Na^+ \xrightarrow{BrCH_2CH_2CH_2CH_3} CH_3CH_2C{\equiv}C{-}CH_2CH_2CH_2CH_3}$.
So the alkyne is 1-butyne and the bromoalkane is 1-bromobutane.
(A) 1-bromopentane with 1-butyne gives 3-nonyne; (B) 1-bromopropane with 1-pentyne gives 4-octyne; (C) 1-bromopentane with propyne gives 2-octyne.
Answer: (D). | ||||
| 4 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Ring (a) is six-membered (five carbons and one ring oxygen): a pyranose. Ring (b) is five-membered (four carbons and one ring oxygen): a furanose. That already rules out (B) and (D).
The anomeric carbon of ring (a) is the one bonded to the ring O and to the $\mathrm{{-}OCH_2CH_2O{-}}$ link. In the Haworth drawing its glycosidic oxygen points down, on the side opposite to the $\mathrm{CH_2OH}$ (C-6) group, which points up: that is the $\alpha$-configuration. So ring (a) is a pyranose with an $\alpha$-glycosidic link.
Statement (A) is therefore correct.
(C): in ring (b) the link oxygen at the anomeric carbon is drawn on the same side (down) as the $\mathrm{CH_2OH}$ on the ring carbon at the far end; that cis arrangement is the $\beta$ link of a furanose, so (C) is not correct. (D) fails because ring (b) is not a pyranose.
Answer: (A). | ||||
| 5 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Anisole (methoxybenzene) is an alkyl aryl ether. HBr protonates the ether oxygen, and $\mathrm{Br^-}$ then attacks the methyl carbon by $S_N2$; it cannot attack the aryl carbon, whose C–O bond has partial double-bond character and on which backside attack is impossible.
$\mathrm{C_6H_5{-}O{-}CH_3 + HBr \rightarrow C_6H_5OH + CH_3Br}$
So the products are phenol and methyl bromide. Bromobenzene (B, C) would need the strong aryl C–O bond to break, and HBr does not brominate the ring and release $\mathrm{H_2}$ (A).
Answer: (D). | ||||
| 6 | 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 Arrhenius equation: $k = A\,e^{-E_a/RT}$.
As $T \rightarrow 0$, $e^{-E_a/RT} \rightarrow 0$, so $k$ starts from (essentially) zero; as $T$ rises the exponent becomes less negative and $k$ increases ever faster in the ordinary temperature range. So $k$ rises from zero with an upward-curving (exponential-like) shape.
(B) is a straight line with a nonzero intercept at $T = 0$; (C) and (D) show $k$ falling with temperature. Only (A) has $k$ rising steeply from zero.
Answer: (A). | ||||
| 7 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics By definition, $\Delta_f H^\circ = 0$ only for an element in its standard (most stable) state at 1 bar and 298 K.
(A) Bromine's standard state at 298 K is the liquid, $\mathrm{Br_2(l)}$; $\mathrm{Br_2(g)}$ has $\Delta_f H^\circ$ equal to its enthalpy of vaporisation, about $+31$ kJ mol$^{-1}$.
(B) Chlorine's standard state is the gas $\mathrm{Cl_2(g)}$, so $\Delta_f H^\circ = 0$.
(C) and (D) are compounds, with nonzero enthalpies of formation.
Answer: (B). | ||||
| 8 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics The C–C single bond enthalpy is about 348 kJ mol$^{-1}$. With 1 kcal = 4.184 kJ,
$\frac{348}{4.184} \approx 83$ kcal mol$^{-1}$,
which is of the order of 100 kcal mol$^{-1}$, not 10 or 1000.
Answer: (C). | ||||
| 9 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Temporary hardness is due to the hydrogencarbonates $\mathrm{Ca(HCO_3)_2}$ and $\mathrm{Mg(HCO_3)_2}$.
(B) Clark's method adds a calculated amount of slaked lime:
$\mathrm{Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3\downarrow + 2H_2O}$
$\mathrm{Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3\downarrow + Mg(OH)_2\downarrow + 2H_2O}$
This is the reagent named for temporary hardness.
(A) Calcium phosphate is insoluble and adds no softening action. (D) Sodium hypochlorite is a disinfectant and bleach, not a softener.
(C) Washing soda precipitates $\mathrm{CaCO_3}$ from any soluble calcium salt, so it is the standard reagent for permanent hardness; it would also remove temporary hardness, but the official key gives only (B).
Answer (official key): (B). | ||||
| 10 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Multiple correctMedium | |
A medium multiple correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers In aqueous NaOH phenol is present as phenoxide ion, which is strongly activated at the ortho and para positions. Bromine water then brominates it stepwise to 2,4,6-tribromophenol (as its phenoxide):
phenoxide $\rightarrow$ 4-bromophenoxide (or 2-bromophenoxide) $\rightarrow$ 2,4-dibromophenoxide $\rightarrow$ 2,4,6-tribromophenoxide.
(A) 2,4-dibromophenoxide: on the route, an intermediate.
(C) 4-bromophenoxide: the first substitution at the para position, an intermediate.
(B) 4,4-dibromocyclohexa-2,5-dienone would need two bromines on the same para carbon; it is not on the route to the product.
(D) 3-bromophenoxide has Br at a meta position, which the $\mathrm{{-}O^-}$ group does not activate.
Answer: (A) and (C). | ||||
| 11 | HydrocarbonsConformations of ethane | Class 11: Hydrocarbons | Multiple correctMedium | |
A medium multiple correct question on conformations of ethane, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 2,2-Dimethylbutane is $\mathrm{CH_3{-}C(CH_3)_2{-}CH_2{-}CH_3}$ (C1 to C4, with two methyls on C2). In the projection one carbon carries $\mathrm{H_3C}$, $\mathrm{CH_3}$ and Y (the three bonds at 120° from each other), and the other carries X, H and H.
Look along each C–C bond:
C1–C2: C1 carries H, H, H; C2 carries $\mathrm{CH_3}$, $\mathrm{CH_3}$ and $\mathrm{C_2H_5}$ (C3–C4). So X = H and Y = $\mathrm{C_2H_5}$: option (B).
C2–C3: C2 carries $\mathrm{CH_3}$, $\mathrm{CH_3}$, $\mathrm{CH_3}$ (C1); C3 carries H, H and $\mathrm{CH_3}$ (C4). So X = $\mathrm{CH_3}$ and Y = $\mathrm{CH_3}$: option (D).
C3–C4 and the bonds to the C2 methyls give no carbon with two $\mathrm{CH_3}$ and one more group against a carbon with two H, as drawn.
(A) would need a carbon carrying $\mathrm{CH_3}$, $\mathrm{CH_3}$ and H, which 2,2-dimethylbutane does not have. (C) would need a carbon with H, H and $\mathrm{C_2H_5}$ bonded to a carbon with $\mathrm{CH_3}$, $\mathrm{CH_3}$, H — not this molecule either.
Answer: (B) and (D). | ||||
| 12 | 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 An intensive property does not depend on the amount of substance; an extensive one is proportional to it.
(A) Molar conductivity is conductivity per mole of electrolyte: a per-mole quantity, intensive.
(B) The emf of a cell depends only on the nature of the electrodes, concentrations and temperature, not on the size of the cell: intensive.
(C) Resistance $R = \rho l/A$ depends on the dimensions of the sample: extensive.
(D) Heat capacity doubles when the amount doubles (specific or molar heat capacity would be intensive): extensive.
Answer: (A) and (B). | ||||
| 13 | EquilibriumpH and buffers | Class 11: Equilibrium | Multiple correctMedium | |
A medium multiple correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium A buffer needs a weak acid together with its conjugate base (or a weak base with its conjugate acid).
(A) $\mathrm{HNO_3}$ and $\mathrm{CH_3COOH}$: two acids, no conjugate base present. Not a buffer.
(B) KOH and $\mathrm{CH_3COONa}$: a strong base and a weak base's salt; no weak acid. Not a buffer.
(C) $\mathrm{HNO_3}$ and $\mathrm{CH_3COONa}$: the strong acid converts part of the acetate into acetic acid, $\mathrm{CH_3COO^- + H^+ \rightarrow CH_3COOH}$. With acetate in excess (for example two volumes of acetate to one of nitric acid) the mixture holds both $\mathrm{CH_3COOH}$ and $\mathrm{CH_3COO^-}$: a buffer.
(D) $\mathrm{CH_3COOH}$ and $\mathrm{CH_3COONa}$: the classic acidic buffer.
Answer: (C) and (D). | ||||
| 14 | Isolation of MetalsRoasting and calcination | No matching NCERT chapter | ComprehensionEasy | |
A easy comprehension question on roasting and calcination, from Isolation of Metals. NCERT: No matching NCERT chapter In partial roasting the ore is heated in a limited supply of air. Chalcopyrite breaks up into the sulphides, and the iron sulphide (more readily oxidised than copper sulphide) is converted to its oxide:
$\mathrm{2CuFeS_2 + O_2 \rightarrow Cu_2S + 2FeS + SO_2}$
$\mathrm{2FeS + 3O_2 \rightarrow 2FeO + 2SO_2}$
Copper stays as $\mathrm{Cu_2S}$ because Cu has the greater affinity for sulphur and Fe the greater affinity for oxygen. The products are $\mathrm{Cu_2S}$ and FeO.
Answer: (A). | ||||
| 15 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | ComprehensionEasy | |
A easy comprehension question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter After roasting, the iron is present as FeO (a basic oxide). In the blast or reverberatory furnace silica is added as flux, and FeO combines with it to form a fusible slag that floats on the copper matte and is run off:
$\mathrm{FeO + SiO_2 \rightarrow FeSiO_3}$ (slag).
So iron leaves as iron(II) silicate.
Answer: (D). | ||||
| 16 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | ComprehensionMedium | |
A medium comprehension question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In the Bessemer converter part of the $\mathrm{Cu_2S}$ is oxidised to $\mathrm{Cu_2O}$, which then reacts with the remaining $\mathrm{Cu_2S}$ without any external reducing agent:
$\mathrm{2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2}$
Copper goes from +1 to 0 (reduced). Sulphur goes from −2 in $\mathrm{Cu_2S}$ to +4 in $\mathrm{SO_2}$ (oxidised). The species that gives up the electrons, the reducing agent, is the sulphide ion $\mathrm{S^{2-}}$.
$\mathrm{O^{2-}}$ is unchanged, $\mathrm{SO_2}$ is a product, and elemental S is not present.
Answer: (C). | ||||
| 17 | ElectrochemistryNernst equation | Class 12: Electrochemistry | ComprehensionMedium | |
A medium comprehension question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Left electrode (anode): $\mathrm{M \rightarrow M^+(0.05\ M) + e^-}$; right electrode (cathode): $\mathrm{M^+(1\ M) + e^- \rightarrow M}$. Net: $\mathrm{M^+(1\ M) \rightarrow M^+(0.05\ M)}$.
$E^\circ_{\mathrm{cell}} = 0$ for a concentration cell, so $\Delta G^\circ = -nFE^\circ = 0$; (C) and (D), which give $\Delta G^\circ$ a sign, are wrong.
Nernst: $E_{\mathrm{cell}} = 0 - \frac{2.303RT}{F}\log\frac{0.05}{1} > 0$, since the log is negative. The ions flow from the concentrated to the dilute side spontaneously, so $\Delta G = -nFE_{\mathrm{cell}} < 0$.
Answer: (B). | ||||
| 18 | ElectrochemistryNernst equation | Class 12: Electrochemistry | ComprehensionMedium | |
A medium comprehension question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry $|E_{\mathrm{cell}}| = \frac{2.303RT}{F}\log\frac{1}{c}$ with $c$ the dilute concentration.
With $c = 0.05$: $\log 20 = 1.301$, and $|E| = 70$ mV.
With $c = 0.0025$: $\log 400 = 2\log 20 = 2.602$, exactly twice as large.
So $|E| = 2\times70 = 140$ mV.
(Check: $\frac{2.303RT}{F} = \frac{70}{1.301} \approx 54$ mV, close to the textbook 59 mV at 298 K; the data are a simplified model.)
Answer: (C). | ||||
| 19 | BiomoleculesProteins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The form drawn has three ionisable groups: $\mathrm{{-}NH_3^+}$ on the side chain (the $\varepsilon$-amino group), a free $\mathrm{{-}NH_2}$ on the $\alpha$-carbon, and $\mathrm{{-}COO^-}$.
A basic group is one that can accept a proton.
$\mathrm{{-}NH_2}$ (lone pair on N) can accept $\mathrm{H^+}$: basic.
$\mathrm{{-}COO^-}$ can accept $\mathrm{H^+}$ to become $\mathrm{{-}COOH}$: basic.
$\mathrm{{-}NH_3^+}$ is already protonated; it can only lose a proton, so it is acidic.
Number of basic groups = 2.
Answer: 2. | ||||
| 20 | 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 $\mathrm{C_4H_6}$ has two degrees of unsaturation. A cyclic isomer uses one for a ring and the other for a second ring or a double bond.
Four-membered ring with a double bond: cyclobutene (1).
Three-membered ring with a double bond: 1-methylcyclopropene, 3-methylcyclopropene, and methylenecyclopropane (exocyclic double bond) (3).
Two rings: bicyclo[1.1.0]butane (1).
Total = 1 + 3 + 1 = 5 (no stereoisomers arise: 3-methylcyclopropene's C3 carries H, CH3 and two ring carbons that are equivalent by symmetry).
Answer: 5. | ||||
| 21 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalHard | |
A hard numerical question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Ozonolysis with reductive work-up (Zn, $\mathrm{H_2O}$) cleaves the central C=C of $\Delta^{9,10}$-octalin. The two carbons of that double bond become C=O, and the two rings open into one ten-membered ring:
Y = cyclodecane-1,6-dione.
Y is symmetric. Its $\alpha$-carbons are C2, C10 (next to C1) and C5, C7 (next to C6); by the symmetry of the molecule all four are equivalent, and every enolate can attack only the other carbonyl.
The enolate at C2 attacking C6 closes the ring C2–C3–C4–C5–C6: five-membered. The enolate at C10 attacking C6 closes C10–C9–C8–C7–C6: also five-membered, and it is the mirror image of the same bond-forming pattern, giving the same constitution.
Each gives the same bicyclo[5.3.0]decane skeleton (a five-membered ring fused to a seven-membered ring), which on heating dehydrates to the conjugated enone, bicyclo[5.3.0]dec-1(7)-en-2-one.
So only one intramolecular aldol condensation product is formed.
Answer: 1. | ||||
| 22 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalEasy | |
A easy numerical question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A compound dissolves in aqueous NaOH when it is acidic enough to form a water-soluble sodium salt: carboxylic acids and phenols do; amines, ethers, alcohols, nitro compounds and hydrocarbons do not.
N,N-dimethylaniline: a base, insoluble.
Cyclohexanecarboxylic acid: soluble (carboxylate).
2-Ethoxybenzyl alcohol: an ether and a benzylic alcohol, far too weakly acidic; insoluble.
Phenol: soluble (phenoxide).
Nitrobenzene: no acidic H; insoluble.
4-(N,N-dimethylamino)phenol: has a phenolic OH; soluble.
1,2-Diethylbenzene: hydrocarbon; insoluble.
1-Naphthoic acid: soluble.
Total = 4.
Answer: 4. | ||||
| 23 | EquilibriumIonic equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Red litmus turns blue in a basic solution, i.e. a salt of a strong base and a weak acid, whose anion hydrolyses to give $\mathrm{OH^-}$.
KCN: $\mathrm{CN^- + H_2O \rightleftharpoons HCN + OH^-}$, basic.
$\mathrm{K_2CO_3}$: $\mathrm{CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-}$, basic.
LiCN: as KCN, basic.
$\mathrm{K_2SO_4}$, NaCl: strong acid with strong base, neutral.
$\mathrm{Zn(NO_3)_2}$, $\mathrm{FeCl_3}$, $\mathrm{NH_4NO_3}$: hydrolysis of the cation makes them acidic.
$\mathrm{(NH_4)_2C_2O_4}$: both ions hydrolyse; $K_a(\mathrm{NH_4^+}) \approx 5.6\times10^{-10}$ and $K_b(\mathrm{C_2O_4^{2-}}) \approx 1.6\times10^{-10}$ are comparable, so the solution is nearly neutral (very slightly acidic) and does not turn red litmus blue.
Total = 3.
Answer: 3. | ||||
| 24 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Br has 7 valence electrons; five form Br–F bonds, leaving one lone pair. Six electron pairs take an octahedral arrangement, and with one position taken by the lone pair the molecule is square pyramidal: one apical F and four basal F.
In the ideal square pyramid the 90° F–Br–F angles are:
apical F with each of the four basal F: 4;
each basal F with its two neighbouring basal F: 4 (the two trans pairs make 180°).
That gives 8.
VSEPR also says the lone pair repels the bonding pairs more strongly, pushing the basal F atoms up towards the apical F: the actual angles are about 85°, so strictly no angle is exactly 90°, which gives 0. The official key accepts either 0 or 8.
Answer: 8 (the key accepts 0 or 8). | ||||
| 25 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter This is beryl, a cyclic silicate. The total charge must be zero:
$\mathrm{Be^{2+}}$: $2n$; $\mathrm{Al^{3+}}$: $2\times3 = 6$; $\mathrm{Si^{4+}}$: $6\times4 = 24$; $\mathrm{O^{2-}}$: $18\times(-2) = -36$.
$2n + 6 + 24 - 36 = 0 \Rightarrow 2n = 6 \Rightarrow n = 3$.
Beryl is $\mathrm{Be_3Al_2Si_6O_{18}}$, with the cyclic $\mathrm{Si_6O_{18}^{12-}}$ anion.
Answer: 3. | ||||
| 26 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Average $= \frac{25.2 + 25.25 + 25.0}{3} = \frac{75.45}{3} = 25.15$ mL.
In addition (and averaging), the result cannot be more precise than the least precise reading. 25.2 and 25.0 are known only to the first decimal place, so the average is reported to one decimal place: 25.2 mL (or 25.1 by round-half-to-even; either way one decimal).
25.2 has 3 significant figures.
Answer: 3. | ||||
| 27 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Find the average rate over each interval:
0 to 0.05 min: $\frac{1.0-0.75}{0.05} = 5$ M min$^{-1}$.
0.05 to 0.12 min: $\frac{0.75-0.40}{0.07} = 5$ M min$^{-1}$.
0.12 to 0.18 min: $\frac{0.40-0.10}{0.06} = 5$ M min$^{-1}$.
The rate stays constant while the concentration falls tenfold, so the rate does not depend on [R]: $[R] = [R]_0 - kt$ with $k = 5$ M min$^{-1}$. The reaction is zero order.
Answer: 0. | ||||
| 28 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Fission is started by a neutron: ${}^{235}_{92}\mathrm{U} + {}^{1}_{0}n \rightarrow {}^{142}_{54}\mathrm{Xe} + {}^{90}_{38}\mathrm{Sr} + x\,{}^{1}_{0}n$.
Atomic numbers: $92 = 54 + 38$, balanced.
Mass numbers: $235 + 1 = 142 + 90 + x \Rightarrow 236 = 232 + x \Rightarrow x = 4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 29 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions A, B, C in A.P. with $A + B + C = \pi$ gives $3B = \pi$, so $B = \frac{\pi}{3}$.
By the sine rule, $\frac{a}{c} = \frac{\sin A}{\sin C}$ and $\frac{c}{a} = \frac{\sin C}{\sin A}$. Then
$\frac{a}{c}\sin 2C = \frac{\sin A}{\sin C}\cdot 2\sin C\cos C = 2\sin A\cos C$,
$\frac{c}{a}\sin 2A = \frac{\sin C}{\sin A}\cdot 2\sin A\cos A = 2\sin C\cos A$.
Sum $= 2(\sin A\cos C + \cos A\sin C) = 2\sin(A + C) = 2\sin B = 2\cdot\frac{\sqrt{3}}{2} = \sqrt{3}$.
Answer: (D). | ||||
| 30 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium 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 The two lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ both pass through the origin. The plane containing them has normal
$\vec{n}_1 = (3,4,2)\times(4,2,3) = (4\cdot3 - 2\cdot2,\ 2\cdot4 - 3\cdot3,\ 3\cdot2 - 4\cdot4) = (8, -1, -10)$.
The required plane contains the line through the origin with direction $(2,3,4)$, and, being perpendicular to the first plane, contains the direction $\vec{n}_1$. Its normal is
$(2,3,4)\times(8,-1,-10) = (3(-10) - 4(-1),\ 4\cdot8 - 2(-10),\ 2(-1) - 3\cdot8) = (-26, 52, -26) \parallel (1,-2,1)$.
Through the origin: $x - 2y + z = 0$.
Check: $(2,3,4)$ gives $2 - 6 + 4 = 0$, and $(1,-2,1)\cdot(8,-1,-10) = 8 + 2 - 10 = 0$.
Answer: (C). | ||||
| 31 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability $\omega^r$ depends only on $r \bmod 3$, taking the values $\omega$, $\omega^2$, $1$. Among the faces 1 to 6, each residue mod 3 occurs exactly twice.
Three cube roots of unity (each $1$, $\omega$ or $\omega^2$) add to zero only when they are $1$, $\omega$, $\omega^2$ in some order (any repetition gives $3$, $3\omega$, $3\omega^2$ or $2u + v \ne 0$).
So $r_1, r_2, r_3$ must have three different residues mod 3: $3! = 6$ ways to assign the residues to the three throws, and $2$ faces for each residue: $6\times2^3 = 48$ outcomes.
Probability $= \frac{48}{6^3} = \frac{48}{216} = \frac{2}{9}$.
Answer: (C). | ||||
| 32 | Vector AlgebraAddition of vectors | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on addition of vectors, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{PQ} = (4,0) - (-2,-1) = (6, 1)$, $\vec{SR} = (3,3) - (-3,2) = (6,1)$. Opposite sides are equal and parallel: PQRS is a parallelogram.
$\vec{QR} = (3,3) - (4,0) = (-1, 3)$.
Rectangle? $\vec{PQ}\cdot\vec{QR} = -6 + 3 = -3 \ne 0$: the angle at Q is not a right angle.
Rhombus? $|PQ| = \sqrt{37}$, $|QR| = \sqrt{10}$: adjacent sides differ.
So it is a parallelogram that is neither a rhombus nor a rectangle (and hence not a square).
Answer: (A). | ||||
| 33 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A system of linear equations $AX = B$ has either no solution, exactly one, or infinitely many.
If $X_1 \ne X_2$ are both solutions, then $A(X_1 - X_2) = 0$, and every $X_1 + t(X_2 - X_1)$, $t \in \mathbb{R}$, is also a solution: $A(X_1 + t(X_2 - X_1)) = B + t\cdot0 = B$. So two distinct solutions force infinitely many.
Hence no matrix gives exactly two solutions: the number is 0.
Answer: (A). | ||||
| 34 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Both numerator $\int_0^x \frac{t\ln(1+t)}{t^4+4}dt$ and denominator $x^3$ tend to 0, so use L'Hôpital's rule with the fundamental theorem of calculus:
$\lim_{x\to0}\frac{\frac{x\ln(1+x)}{x^4+4}}{3x^2} = \lim_{x\to0}\frac{\ln(1+x)}{x}\cdot\frac{1}{3(x^4+4)} = 1\cdot\frac{1}{12} = \frac{1}{12}$.
Answer: (B). | ||||
| 35 | Quadratic EquationsForming equations from given roots | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on forming equations from given roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = -p^3 + 3p\,\alpha\beta = q$, so $\alpha\beta = \frac{p^3 + q}{3p}$.
Sum of the roots: $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2}{\alpha\beta} - 2 = \frac{3p^3}{p^3 + q} - 2 = \frac{p^3 - 2q}{p^3 + q}$.
Product of the roots: $1$.
The equation $x^2 - \frac{p^3 - 2q}{p^3 + q}x + 1 = 0$, multiplied by $p^3 + q$ (nonzero since $p^3 \ne -q$):
$(p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0$.
Answer: (B). | ||||
| 36 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Show each function increases on $[0,1]$; then each maximum is the value at $x = 1$.
$f'(x) = 2x(e^{x^2} - e^{-x^2}) \ge 0$ on $[0,1]$.
$g'(x) = e^{x^2}(1 + 2x^2) - 2xe^{-x^2}$. Since $e^{x^2} \ge 1 \ge e^{-x^2}$ and $1 + 2x^2 \ge 2x$ (as $2x^2 - 2x + 1 = 2(x - \frac12)^2 + \frac12 > 0$), $g'(x) > 0$.
$h'(x) = 2xe^{x^2}(1 + x^2) - 2xe^{-x^2} = 2x\left[e^{x^2}(1 + x^2) - e^{-x^2}\right] \ge 0$.
So $a = f(1) = e + \frac{1}{e}$, $b = g(1) = e + \frac{1}{e}$, $c = h(1) = e + \frac{1}{e}$.
$a = b = c$.
Answer: (D). | ||||
| 37 | Conic SectionsParabola | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Take $A = (t_1^2, 2t_1)$ and $B = (t_2^2, 2t_2)$. The circle on AB as diameter has centre $\left(\frac{t_1^2 + t_2^2}{2},\ t_1 + t_2\right)$.
It touches the axis of the parabola (the x-axis), so the distance of the centre from it equals the radius: $|t_1 + t_2| = r$.
Slope of AB $= \frac{2t_1 - 2t_2}{t_1^2 - t_2^2} = \frac{2}{t_1 + t_2} = \pm\frac{2}{r}$.
Both signs occur (reflect the chord in the x-axis).
(A) and (B), $\mp\frac{1}{r}$, would need $|t_1 + t_2| = 2r$, which contradicts tangency.
Answer: (C) and (D). | ||||
| 38 | 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 Cosine rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{\sqrt{3}}{2}$.
Note $ab = (x^2 + x + 1)(x^2 - 1) = x^4 + x^3 - x - 1$, and
$a^2 + b^2 - c^2 = (x^4 + 2x^3 + 3x^2 + 2x + 1) + (x^4 - 2x^2 + 1) - (4x^2 + 4x + 1) = 2x^4 + 2x^3 - 3x^2 - 2x + 1 = 2ab - 3(x^2 - 1)$.
So $\cos C = 1 - \frac{3(x^2 - 1)}{2(x^2 + x + 1)(x^2 - 1)} = 1 - \frac{3}{2(x^2 + x + 1)}$.
Setting this to $\frac{\sqrt{3}}{2}$: $\frac{3}{2(x^2 + x + 1)} = \frac{2 - \sqrt{3}}{2}$, so $x^2 + x + 1 = \frac{3}{2 - \sqrt{3}} = 3(2 + \sqrt{3})$.
$x^2 + x - (5 + 3\sqrt{3}) = 0$. Discriminant $= 1 + 20 + 12\sqrt{3} = (3 + 2\sqrt{3})^2$.
$x = \frac{-1 \pm (3 + 2\sqrt{3})}{2}$, i.e. $x = 1 + \sqrt{3}$ or $x = -(2 + \sqrt{3})$.
Sides must be positive: $c = 2x + 1 > 0$ needs $x > -\frac12$, which rejects $-(2 + \sqrt{3})$ (option A). For $x = 1 + \sqrt{3}$: $c = 3 + 2\sqrt{3}$, $b = 3 + 2\sqrt{3}$, $a = 6 + 3\sqrt{3}$, all positive, and $a < b + c$, so the triangle exists.
(C) $2 + \sqrt{3}$ and (D) $4\sqrt{3}$ do not satisfy the quadratic.
Answer: (B). | ||||
| 39 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Multiple correctMedium | |
A medium multiple correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z - z_1 = t(z_2 - z_1)$ and $z - z_2 = (1 - t)(z_1 - z_2)$ with $0 < t < 1$: z lies strictly between $z_1$ and $z_2$ on the segment joining them.
(A) $|z - z_1| + |z - z_2| = t|z_2 - z_1| + (1 - t)|z_2 - z_1| = |z_1 - z_2|$. True.
(B) $z - z_1$ is a positive multiple of $z_2 - z_1$ and $z - z_2$ a positive multiple of $z_1 - z_2$: the two point in opposite directions, so their arguments differ by $\pi$. False.
(C) The determinant is $(z - z_1)(\bar{z}_2 - \bar{z}_1) - (\bar{z} - \bar{z}_1)(z_2 - z_1)$. With $z - z_1 = t(z_2 - z_1)$ and t real, this is $t|z_2 - z_1|^2 - t|z_2 - z_1|^2 = 0$. True.
(D) $z - z_1 = t(z_2 - z_1)$ with $t > 0$, so both have the same argument. True.
Answer: (A), (C) and (D). | ||||
| 40 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctHard | |
A hard multiple correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f'(x) = \frac{1}{x} + \sqrt{1 + \sin x}$, which exists and is continuous on $(0,\infty)$.
Is $f'$ differentiable? $\sqrt{1 + \sin x} = \sqrt{\left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2} = \left|\sin\frac{x}{2} + \cos\frac{x}{2}\right| = \sqrt{2}\left|\sin\left(\frac{x}{2} + \frac{\pi}{4}\right)\right|$.
This has corners where $\sin x = -1$, i.e. $x = \frac{3\pi}{2} + 2n\pi$ (in $(0,\infty)$), where the left and right derivatives are $\mp\frac{1}{\sqrt{2}}$. So $f''$ does not exist there.
(A) False. (B) True.
(C) $\sqrt{1 + \sin t} \ge 0$ and its average over each period $2\pi$ is a positive constant, so $\int_0^x\sqrt{1 + \sin t}\,dt \rightarrow \infty$ (roughly linearly), and $f(x) \rightarrow \infty$. Meanwhile $|f'(x)| \le \frac{1}{x} + \sqrt{2} \le 1 + \sqrt{2}$ for $x \ge 1$. So for large enough $\alpha > 1$, $|f'(x)| < |f(x)|$ for all $x > \alpha$. True.
(D) $|f(x)| \rightarrow \infty$ as $x \rightarrow \infty$ (and also as $x \rightarrow 0^+$, through $\ln x$), so no bound $\beta$ exists. False.
Answer: (B) and (C). | ||||
| 41 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $x^4(1 - x)^4 = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4$. Divide by $1 + x^2$:
$\frac{x^4(1 - x)^4}{1 + x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1 + x^2}$.
(Check: $(x^6 - 4x^5 + 5x^4 - 4x^2 + 4)(1 + x^2) - 4 = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4$.)
Integrate from 0 to 1:
$\frac{1}{7} - \frac{4}{6} + 1 - \frac{4}{3} + 4 - 4\cdot\frac{\pi}{4} = \frac{1}{7} - 2 + 5 - \pi = \frac{22}{7} - \pi$.
(B) $\frac{2}{105}$ and (D) are other numbers, and (C) is impossible since the integrand is positive on $(0,1)$.
Answer: (A). | ||||
| 42 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | ComprehensionHard | |
A hard comprehension question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\det A = a^2 - bc$.
Skew-symmetric needs diagonal entries $0$ and $c = -b$; with entries in $\{0, \ldots, p-1\}$ that means $a = b = c = 0$, the zero matrix, which is also symmetric. So we only count symmetric matrices: $b = c$.
$\det A = a^2 - b^2 = (a - b)(a + b)$, divisible by the prime p iff $a \equiv b$ or $a \equiv -b \pmod p$.
$a = b$: p matrices ($a = b = 0, 1, \ldots, p-1$).
$a + b = p$ with $a, b \in \{1, \ldots, p-1\}$: $p - 1$ matrices, none with $a = b$ since p is odd. ($a + b = 0$ is $a = b = 0$, already counted.)
Total $= p + (p - 1) = 2p - 1$.
Answer: (D). | ||||
| 43 | 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 Trace $= 2a$. Since p is an odd prime, p divides $2a$ only if $a = 0$. So we need $a \in \{1, \ldots, p-1\}$: $p - 1$ choices.
$\det A = a^2 - bc \equiv 0 \pmod p$ means $bc \equiv a^2 \not\equiv 0$. Then $b \ne 0$, and for each of the $p - 1$ nonzero b there is exactly one c in $\{1, \ldots, p-1\}$ with $bc \equiv a^2$ (b is invertible mod p).
Count $= (p - 1)(p - 1) = (p - 1)^2$.
Answer: (C). | ||||
| 44 | 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 $T_p$ has $p^3$ matrices. Count those with $p \mid \det A = a^2 - bc$.
$a = 0$: need $p \mid bc$, i.e. $b = 0$ or $c = 0$: $p + p - 1 = 2p - 1$ matrices.
$a \ne 0$: from the previous question, $(p - 1)^2$ matrices.
Total with $p \mid \det A$: $2p - 1 + (p - 1)^2 = p^2$.
So the number with $\det A$ not divisible by p is $p^3 - p^2$.
Answer: (D). | ||||
| 45 | Conic SectionsHyperbola | Class 11: Conic Sections | ComprehensionMedium | |
A medium comprehension question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Circle: centre $(4, 0)$, radius 4. A tangent to the hyperbola with slope m is $y = mx \pm\sqrt{9m^2 - 4}$, i.e. $y = mx + c$ with $c^2 = 9m^2 - 4$.
Tangent to the circle: $\frac{|4m + c|}{\sqrt{1 + m^2}} = 4$, so $(4m + c)^2 = 16(1 + m^2)$, i.e. $c^2 + 8mc = 16$.
With $c^2 = 9m^2 - 4$: $8mc = 20 - 9m^2$. Squaring, $64m^2(9m^2 - 4) = (20 - 9m^2)^2$, i.e. $576m^4 - 256m^2 = 400 - 360m^2 + 81m^4$, so $495m^4 + 104m^2 - 400 = 0$, giving $m^2 = \frac{4}{5}$ (the other root is negative).
$m = \frac{2}{\sqrt5}$: $c = \frac{20 - 36/5}{8\cdot2/\sqrt5} = \frac{64/5}{16/\sqrt5} = \frac{4}{\sqrt5}$.
Tangent: $y = \frac{2}{\sqrt5}x + \frac{4}{\sqrt5}$, i.e. $2x - \sqrt5 y + 4 = 0$.
Check option (B): distance from $(4,0)$ is $\frac{|8 + 4|}{\sqrt{4 + 5}} = 4$; and $c^2 = \frac{16}{5} = 9\cdot\frac45 - 4$.
(A) has $c = -4\sqrt5$, with $c^2 = 80 \ne \frac{16}{5}$; (C) and (D) are at distances $4$ and $\frac{20}{5} = 4$ from the centre but are not tangent to the hyperbola ($c^2 \ne 9m^2 - 4$).
Answer: (B). | ||||
| 46 | CirclesEquation from a diameter | Class 11: Conic Sections | ComprehensionEasy | |
A easy comprehension question on equation from a diameter, from Circles. NCERT: Class 11: Conic Sections From the hyperbola, $y^2 = \frac{4x^2}{9} - 4$. Substitute in the circle:
$x^2 + \frac{4x^2}{9} - 4 - 8x = 0 \Rightarrow 13x^2 - 72x - 36 = 0 \Rightarrow (x - 6)(13x + 6) = 0$.
$x = -\frac{6}{13}$ gives $y^2 < 0$, so $x = 6$ and $y^2 = 16 - 4 = 12$: $A, B = (6, \pm2\sqrt3)$.
Circle on AB as diameter: centre $(6, 0)$, radius $2\sqrt3$:
$(x - 6)^2 + y^2 = 12 \Rightarrow x^2 + y^2 - 12x + 24 = 0$.
Answer: (A). | ||||
| 47 | 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 $\tan\theta = \cot5\theta = \tan\left(\frac{\pi}{2} - 5\theta\right) \Rightarrow \theta = \frac{\pi}{2} - 5\theta + k\pi \Rightarrow \theta = \frac{\pi}{12} + \frac{k\pi}{6}$.
In $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$: $\theta = \pm\frac{\pi}{12}, \pm\frac{3\pi}{12}, \pm\frac{5\pi}{12}$. None is a multiple of $\frac{\pi}{5}$, and at each $\cot5\theta$ and $\tan\theta$ are defined.
$\sin2\theta = \cos4\theta = 1 - 2\sin^22\theta \Rightarrow 2\sin^22\theta + \sin2\theta - 1 = 0 \Rightarrow (2\sin2\theta - 1)(\sin2\theta + 1) = 0$.
With $2\theta \in (-\pi, \pi)$: $\sin2\theta = \frac12$ gives $2\theta = \frac{\pi}{6}, \frac{5\pi}{6}$, so $\theta = \frac{\pi}{12}, \frac{5\pi}{12}$; $\sin2\theta = -1$ gives $\theta = -\frac{\pi}{4} = -\frac{3\pi}{12}$.
All three are in the first list. Number of values = 3.
Answer: 3. | ||||
| 48 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Write the denominator in terms of $2\theta$:
$\sin^2\theta + 3\sin\theta\cos\theta + 5\cos^2\theta = \frac{1 - \cos2\theta}{2} + \frac{3}{2}\sin2\theta + \frac{5(1 + \cos2\theta)}{2} = 3 + 2\cos2\theta + \frac32\sin2\theta$.
$2\cos2\theta + \frac32\sin2\theta$ ranges over $\left[-\frac52, \frac52\right]$ since $\sqrt{4 + \frac94} = \frac52$.
Least denominator $= 3 - \frac52 = \frac12 > 0$, so the maximum of the expression is $\frac{1}{1/2} = 2$.
Answer: 2. | ||||
| 49 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\vec a| = 1$, $|\vec b| = 1$, and $\vec a\cdot\vec b = \frac{2 - 2 + 0}{\sqrt{70}} = 0$.
Using $(\vec u\times\vec v)\times\vec w = (\vec u\cdot\vec w)\vec v - (\vec v\cdot\vec w)\vec u$:
$(\vec a\times\vec b)\times\vec a = (\vec a\cdot\vec a)\vec b - (\vec b\cdot\vec a)\vec a = \vec b$,
$(\vec a\times\vec b)\times\vec b = (\vec a\cdot\vec b)\vec b - (\vec b\cdot\vec b)\vec a = -\vec a$.
So $(\vec a\times\vec b)\times(\vec a - 2\vec b) = \vec b + 2\vec a$.
$(2\vec a + \vec b)\cdot(2\vec a + \vec b) = 4|\vec a|^2 + |\vec b|^2 + 4\vec a\cdot\vec b = 4 + 1 + 0 = 5$.
Answer: 5. | ||||
| 50 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The line meets the x-axis at $\left(\frac12, 0\right)$, so the nearest directrix is $x = \frac{a}{e} = \frac12$, i.e. $a = \frac{e}{2}$.
Tangency of $y = -2x + 1$ to the hyperbola: $c^2 = a^2m^2 - b^2$, so $1 = 4a^2 - b^2$, with $b^2 = a^2(e^2 - 1)$:
$1 = a^2(4 - e^2 + 1) = \frac{e^2}{4}(5 - e^2)$.
$e^4 - 5e^2 + 4 = 0 \Rightarrow (e^2 - 1)(e^2 - 4) = 0$. A hyperbola has $e > 1$, so $e^2 = 4$, $e = 2$.
Answer: 2. | ||||
| 51 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Normal to the plane of the two lines: $(2,3,4)\times(3,4,5) = (15 - 16,\ 12 - 10,\ 8 - 9) = (-1, 2, -1)$, i.e. $(1, -2, 1)$.
Through $(1, 2, 3)$: $(x - 1) - 2(y - 2) + (z - 3) = 0 \Rightarrow x - 2y + z = 0$. (The second line's point $(2,3,4)$ also satisfies it: $2 - 6 + 4 = 0$, so the lines do lie in one plane.)
The planes are at a finite distance, hence parallel: $A = 1$. Distance between $x - 2y + z = 0$ and $x - 2y + z = d$ is $\frac{|d|}{\sqrt{1 + 4 + 1}} = \frac{|d|}{\sqrt6} = \sqrt6$.
$|d| = 6$.
Answer: 6. | ||||
| 52 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $[0,1)$, $[x] = 0$ is even: $f(x) = 1 - x$. On $[1,2)$, $[x] = 1$ is odd: $f(x) = x - 1$. In general $f$ has period 2 (shifting x by 2 keeps the parity of $[x]$ and the fractional part), and so does $\cos\pi x$.
So $\int_{-10}^{10} f(x)\cos\pi x\,dx = 10\int_0^2 f(x)\cos\pi x\,dx$.
$\int_0^1 (1 - x)\cos\pi x\,dx = \left[\frac{(1 - x)\sin\pi x}{\pi}\right]_0^1 + \frac1\pi\int_0^1\sin\pi x\,dx = 0 + \frac{1}{\pi}\cdot\frac{2}{\pi} = \frac{2}{\pi^2}$.
$\int_1^2 (x - 1)\cos\pi x\,dx$: put $u = x - 1$, $\cos\pi x = -\cos\pi u$: $= -\int_0^1 u\cos\pi u\,du = -\left(\left[\frac{u\sin\pi u}{\pi}\right]_0^1 - \frac1\pi\int_0^1\sin\pi u\,du\right) = -\left(0 - \frac{2}{\pi^2}\right) = \frac{2}{\pi^2}$.
$\int_0^2 = \frac{4}{\pi^2}$, so $\int_{-10}^{10} = \frac{40}{\pi^2}$, and $\frac{\pi^2}{10}\cdot\frac{40}{\pi^2} = 4$.
Answer: 4. | ||||
| 53 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Add rows 2 and 3 to row 1. Each column sum is $z + 1 + \omega + \omega^2 = z$, so
$\Delta = z\begin{vmatrix} 1 & 1 & 1 \\ \omega & z + \omega^2 & 1 \\ \omega^2 & 1 & z + \omega \end{vmatrix}$.
Subtract column 1 from columns 2 and 3:
$\Delta = z\begin{vmatrix} 1 & 0 & 0 \\ \omega & z + \omega^2 - \omega & 1 - \omega \\ \omega^2 & 1 - \omega^2 & z + \omega - \omega^2 \end{vmatrix} = z\left[(z + \omega^2 - \omega)(z + \omega - \omega^2) - (1 - \omega)(1 - \omega^2)\right]$.
$(z + \omega^2 - \omega)(z + \omega - \omega^2) = z^2 - (\omega - \omega^2)^2 = z^2 - (\omega^2 - 2 + \omega) = z^2 + 3$.
$(1 - \omega)(1 - \omega^2) = 1 - \omega - \omega^2 + 1 = 3$.
So $\Delta = z\cdot z^2 = z^3 = 0$, whose only root is $z = 0$.
Number of distinct z = 1.
Answer: 1. | ||||
| 54 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalHard | |
A hard numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series For $k \ge 2$: $S_k = \frac{(k - 1)/k!}{1 - 1/k} = \frac{k - 1}{k!}\cdot\frac{k}{k - 1} = \frac{1}{(k - 1)!}$. For $k = 1$ the first term is 0, so every term is 0 and $S_1 = 0$.
Write $k^2 - 3k + 1 = (k - 1)(k - 2) - 1$. Then for $k \ge 3$ (where $k^2 - 3k + 1 > 0$):
$(k^2 - 3k + 1)S_k = \frac{(k - 1)(k - 2)}{(k - 1)!} - \frac{1}{(k - 1)!} = \frac{1}{(k - 3)!} - \frac{1}{(k - 1)!}$.
For $k = 2$: $|(4 - 6 + 1)\cdot1| = 1$. For $k = 1$: 0.
$\sum_{k=3}^{100}\left[\frac{1}{(k - 3)!} - \frac{1}{(k - 1)!}\right] = \sum_{j=0}^{97}\frac{1}{j!} - \sum_{j=2}^{99}\frac{1}{j!} = \frac{1}{0!} + \frac{1}{1!} - \frac{1}{98!} - \frac{1}{99!} = 2 - \frac{99 + 1}{99!} = 2 - \frac{100}{99!} = 2 - \frac{100^2}{100!}$.
Whole sum $= 1 + 2 - \frac{100^2}{100!}$, and adding $\frac{100^2}{100!}$ gives 3.
Answer: 3. | ||||
| 55 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalHard | |
A hard numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Multiply the second equation by $yz$: $xyz\sin3\theta = 2z\cos3\theta + 2y\sin3\theta$.
Now all three right sides equal $xyz\sin3\theta$:
(1) $(y + z)\cos3\theta$, (2) $2z\cos3\theta + 2y\sin3\theta$, (3) $(y + 2z)\cos3\theta + y\sin3\theta$.
(1) = (3): $-z\cos3\theta = y\sin3\theta$.
(1) = (2): $(y - z)\cos3\theta = 2y\sin3\theta = -2z\cos3\theta$, so $(y + z)\cos3\theta = 0$.
If $\cos3\theta = 0$, then $\sin3\theta = \pm1$ and $y\sin3\theta = 0$ forces $y = 0$: not allowed. So $y = -z$.
Then $-z\cos3\theta = -z\sin3\theta$, and $z \ne 0$ gives $\tan3\theta = 1$.
With $y = -z$ and $\tan3\theta = 1$, the first equation gives $xyz\sin3\theta = 0$, so $x = 0$; then all three equations hold (for instance (2): $\frac{2}{z}(\sin3\theta - \cos3\theta) = 0$). So every such $\theta$ works.
$3\theta \in (0, 3\pi)$ with $\tan3\theta = 1$: $3\theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}$, i.e. $\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4}$.
Number of values = 3.
Answer: 3. | ||||
| 56 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations The tangent at $(x, y)$ is $Y - y = y'(X - x)$; at $X = 0$ its y-intercept is $y - xy'$. So
$y - xy' = x^3 \Rightarrow y' - \frac{y}{x} = -x^2$.
Integrating factor $\frac1x$: $\frac{d}{dx}\left(\frac{y}{x}\right) = -x \Rightarrow \frac{y}{x} = -\frac{x^2}{2} + C \Rightarrow y = -\frac{x^3}{2} + Cx$.
$f(1) = 1$: $-\frac12 + C = 1$, $C = \frac32$. So $f(x) = \frac{3x - x^3}{2}$, which is differentiable on all of $\mathbb{R}$ and satisfies the condition everywhere (check: $y - xy' = \frac{3x - x^3}{2} - x\cdot\frac{3 - 3x^2}{2} = x^3$).
$f(-3) = \frac{-9 + 27}{2} = 9$.
Answer: 9. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 57 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity The current flows between the two shaded faces, a distance L apart. Each shaded face has area (side L) $\times$ (thickness t) $= Lt$.
$R = \frac{\rho\,(\text{length})}{\text{area}} = \frac{\rho L}{Lt} = \frac{\rho}{t}$.
R is independent of L and inversely proportional to t, so (A), (B) and (D) are false.
Answer: (C). | ||||
| 58 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The ideal-gas model neglects the volume of the molecules and the forces between them.
At low pressure the molecules are far apart, so their own volume is a negligible fraction of the container volume and they rarely come within range of each other's attraction.
At high temperature their kinetic energy is large compared with the energy of intermolecular attraction.
Both conditions make a real gas approach ideal behaviour.
Answer: (D). | ||||
| 59 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity All three bulbs are rated for the same supply voltage V, and at their working temperature $R = \frac{V^2}{P}$: the higher the power, the lower the resistance. So $R_{100} < R_{60} < R_{40}$ when hot.
The filaments are of the same metal and run at similar temperatures, so the hot resistance is roughly the same multiple of the room-temperature resistance for each: the order is the same at room temperature.
$R_{100} < R_{60} < R_{40} \Rightarrow \frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$.
(A) and (B) would need exact relations that the ratings do not give; (C) has the order reversed.
Answer: (D). | ||||
| 60 | Experimental SkillsElectricity experiments | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on electricity experiments, from Experimental Skills. NCERT: No matching NCERT chapter We need a voltmeter across $R_T$ and an ammeter in series with it.
A galvanometer becomes a voltmeter with a high resistance in series: $G_1$ in series with $R_1$, the pair connected in parallel with $R_T$.
A galvanometer becomes an ammeter with a small resistance (shunt) in parallel: $G_2$ with $R_2$ across it, the pair in series with $R_T$ and the source.
Circuit (C) has exactly this: $G_1 + R_1$ across $R_T$, and $G_2$ shunted by $R_2$ in the main line.
(A) uses $R_2$ (small) with $G_1$ across $R_T$ and $R_1$ (large) in series with $G_2$; (B) shunts $G_1$ with $R_1$; (D) swaps the roles of $R_1$ and $R_2$.
Answer: (C). | ||||
| 61 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Impedance of the series RC circuit: $Z = \sqrt{R^2 + \frac{1}{\omega^2C^2}}$.
As $\omega$ increases, the capacitive reactance $\frac{1}{\omega C}$ falls, so Z decreases ((C) and (D) are false).
The current amplitude $\frac{V_0}{Z}$ rises, and the power in the bulb, $I_{\mathrm{rms}}^2R$, rises: the bulb glows brighter.
Answer: (B). | ||||
| 62 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The wire forms a circle of circumference L, radius $r = \frac{L}{2\pi}$.
Force on an element: $d\vec F = I\,d\vec l\times\vec B$. For clockwise current with $\vec B$ into the page, this force points radially outward (at the top the current moves right, and $\hat x\times(-\hat z) = +\hat y$), magnitude $IB\,dl$.
Take a small arc subtending $d\theta$, length $r\,d\theta$. The tension T at its two ends pulls inward with net force $2T\sin\frac{d\theta}{2} \approx T\,d\theta$. Balance:
$T\,d\theta = IB\,r\,d\theta \Rightarrow T = IBr = \frac{IBL}{2\pi}$.
Answer: (C). | ||||
| 63 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Along the incline (up positive), the block is at rest: $P + f - mg\sin\theta = 0$, so $f = mg\sin\theta - P$.
At $P = P_1$: $f = \mu mg\cos\theta$ (positive: friction up the plane, at its limiting value — the block is about to slide down).
As P increases, f decreases linearly with slope $-1$, passing through zero at $P = mg\sin\theta$.
At $P = P_2$: $f = -\mu mg\cos\theta$ (friction down the plane, limiting — about to slide up).
So f falls as a straight line from a positive value at $P_1$ to the equal negative value at $P_2$. That is graph (A).
(B) and (D) show constant friction; (C) shows f increasing with P.
Answer: (A). | ||||
| 64 | GravitationGravitational potential energy | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation From the figure, P is on the axis at distance 4R from the centre.
Surface density: $\sigma = \frac{M}{\pi(16R^2 - 9R^2)} = \frac{M}{7\pi R^2}$.
A ring of radius s and width ds has mass $2\pi\sigma s\,ds$; its potential at P is $-\frac{G\,2\pi\sigma s\,ds}{\sqrt{s^2 + x^2}}$. Integrating from $3R$ to $4R$ with $x = 4R$:
$V = -2\pi G\sigma\left[\sqrt{s^2 + x^2}\right]_{3R}^{4R} = -2\pi G\sigma\left(\sqrt{32}R - 5R\right) = -\frac{2GM}{7R^2}\cdot R\,(4\sqrt2 - 5) = -\frac{2GM}{7R}(4\sqrt2 - 5)$.
The work needed to take a unit mass to infinity (where V = 0) is $0 - V$:
$W = \frac{2GM}{7R}(4\sqrt2 - 5)$, a positive quantity ($4\sqrt2 \approx 5.66$).
Answer: (A). | ||||
| 65 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Field lines start on positive and end on negative charges, and their number is proportional to the charge.
More lines leave $Q_1$ than reach $Q_2$; some of $Q_1$'s lines go off to infinity. So $|Q_1| > |Q_2|$, and lines running from $Q_1$ to $Q_2$ show the charges have opposite signs. (A) true, (B) false.
For two unlike charges the field is zero only on the line joining them, outside the segment, on the side of the smaller charge (there the stronger but farther charge's field can cancel the weaker, nearer one's). Here the smaller charge is $Q_2$, so the null point is at a finite distance to the right of $Q_2$. (D) true, (C) false.
Answer: (A) and (D). | ||||
| 66 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Multiple correctMedium | |
A medium multiple correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $T = \frac{40}{20} = 2$ s. The error in the total time is the least count, 1 s, so the error in one period is $\Delta T = \frac{1}{20} = 0.05$ s. (A) true, (B) false.
$g = \frac{4\pi^2 L}{T^2}$ with L exact, so $\frac{\Delta g}{g} = 2\frac{\Delta T}{T} = 2\times\frac{0.05}{2} = 0.05 = 5\%$. (C) true, (D) false.
Answer: (A) and (C). | ||||
| 67 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Multiple correctMedium | |
A medium multiple correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power For a one-dimensional elastic collision with the 5 kg mass at rest, the 1 kg mass's velocity after is $v_1 = \frac{1 - 5}{1 + 5}u = -\frac{2}{3}u$. With $|v_1| = 2$ m/s, $u = 3$ m/s; and $v_2 = \frac{2\cdot1}{6}u = 1$ m/s.
(A) Total momentum $= 1\times3 = 3$ kg m s$^{-1}$ (check after: $1(-2) + 5(1) = 3$). True.
(B) Momentum of the 5 kg mass $= 5\times1 = 5$ kg m s$^{-1}$, not 4. False.
(C) $v_{cm} = \frac{3}{6} = 0.5$ m/s, KE of the centre of mass $= \frac12\times6\times0.25 = 0.75$ J. True.
(D) Total KE $= \frac12\times1\times3^2 = 4.5$ J (after: $2 + 2.5 = 4.5$ J), not 4 J. False.
Answer: (A) and (C). | ||||
| 68 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctHard | |
A hard multiple correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Angles of the prism: A = 90°, B = 60°, C = 135°, D = 75°. Place AB vertical (A at the bottom) and AD horizontal.
At AB: $\sin r = \frac{\sin60^\circ}{\sqrt3} = \frac12$, $r = 30^\circ$. The normal to AB is horizontal, so inside the prism the ray travels at 30° below the horizontal. Face BC also makes 30° below the horizontal (it meets the vertical AB at 60°), so the ray runs parallel to BC and, starting near B, reaches face CD.
At CD: CD makes 75° with AD, so its normal makes 15° with the horizontal. The ray (30° below horizontal) meets this normal at $30^\circ + 15^\circ = 45^\circ$.
Critical angle: $\sin c = \frac{1}{\sqrt3} \approx 0.577$, $c \approx 35.3^\circ$. Since $45^\circ > c$, the ray is totally internally reflected at CD. (A) true.
After reflection the ray travels at 120° below the horizontal on the far side, i.e. downward and to the left, making 30° with the vertical. It meets AD (horizontal, normal vertical) at incidence 30°, where $\sin e = \sqrt3\sin30^\circ = \frac{\sqrt3}{2}$, $e = 60^\circ$: it comes out through AD. (B) true.
The incident ray travels at 60° below the horizontal, heading right. The emergent ray makes 60° with the vertical, heading down and left, i.e. 30° below the horizontal heading left. The angle between the two directions is $180^\circ - 60^\circ - 30^\circ = 90^\circ$. (C) true, (D) false.
Answer: (A), (B) and (C). | ||||
| 69 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the V–T graph: A is $(T_0, V_0)$, B is $(T_0, 4V_0)$, BC is a straight line through the origin, and CA is at constant volume $V_0$.
(A) A and B are at the same temperature $T_0$, and the internal energy of an ideal gas depends only on T: equal. True.
(B) AB is isothermal: $W = nRT_0\ln\frac{4V_0}{V_0} = P_0V_0\ln4$ (since $nRT_0 = P_0V_0$). True.
(C) On BC, $V \propto T$, so the pressure is constant and equals that at B: $P_B = \frac{P_0V_0}{4V_0} = \frac{P_0}{4}$. So $P_C = \frac{P_0}{4}$. True.
(D) At C, $V = V_0$ and $P = \frac{P_0}{4}$: $T_C = \frac{P_0V_0}{4R} = \frac{T_0}{4}$. True.
Answer: (A), (B), (C) and (D). | ||||
| 70 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | ComprehensionMedium | |
A medium comprehension question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The potential is 0 at the origin and rises to $V_0$, staying there for $|x| \ge X_0$. The particle is trapped (turning points on both sides) only while its energy is below the height of the walls, $E < V_0$; with $E > V_0$ it moves off to infinity with constant speed.
It must also have $E > 0$ to move at all: $E = 0$ means the particle rests at the bottom, and $E < 0$ is impossible since $V \ge 0$.
So periodic motion needs $V_0 > E > 0$.
(B) $E > 0$ is necessary but not enough on its own; the key also accepts it, remarking that (C) implies (B).
Answer: (C) (the official key accepts B, C, or both). | ||||
| 71 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | ComprehensionMedium | |
A medium comprehension question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Near the origin $V = \alpha x^4$, so $[\alpha] = \frac{[\text{energy}]}{[L^4]} = ML^{-2}T^{-2}$. The period can depend only on m, $\alpha$ and A (the motion is not simple harmonic, so A matters).
Let $T \propto m^a\alpha^bA^c$: $M^aM^bL^{-2b}T^{-2b}L^c = T^1$.
Time: $-2b = 1 \Rightarrow b = -\frac12$. Mass: $a + b = 0 \Rightarrow a = \frac12$. Length: $-2b + c = 0 \Rightarrow c = -1$.
$T \propto \frac{1}{A}\sqrt{\frac{m}{\alpha}}$: the period grows as the amplitude shrinks, because the force $4\alpha x^3$ is very weak near the origin.
Answer: (B). | ||||
| 72 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | ComprehensionEasy | |
A easy comprehension question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For $|x| > X_0$ the potential is constant, $V = V_0$. The force is $F = -\frac{dV}{dx} = 0$, so the acceleration is zero: the particle moves there with constant velocity.
Answer: (D). | ||||
| 73 | Current ElectricityResistivity | Class 12: Current Electricity | ComprehensionMedium | |
A medium comprehension question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity From the figure, $T_C(B)$ decreases as B increases. With $B_2 > B_1$, $T_C(B_2) < T_C(B_1)$: in the larger field the resistance drops to zero at a lower temperature.
Above its critical temperature the material is a normal conductor whose resistance is the same curve in either field.
So the dashed ($B_2$) curve must drop to zero at a lower T than the solid ($B_1$) curve, and both must follow the same R–T line above their transitions. Graph (A) shows this.
(C) has the transitions the other way round; (B) and (D) show different normal-state resistances in the two fields and a common transition temperature.
Answer: (A). | ||||
| 74 | Current ElectricityResistivity | Class 12: Current Electricity | ComprehensionMedium | |
A medium comprehension question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $T_C(B)$ falls steadily (monotonically) as B increases, from 100 K at B = 0 to 75 K at 7.5 T.
For B = 5 T, which lies between 0 and 7.5 T, $T_C$ must lie between 75 K and 100 K: (B) can definitely be said.
(A) gives an exact value, 80 K, which cannot be fixed without the exact shape of the curve.
For B = 10 T, beyond 7.5 T, $T_C$ must be below 75 K, so (C) is false; (D) gives an exact value, 70 K, which again cannot be known for certain.
Answer: (B). | ||||
| 75 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a thin lens, $m = \frac{v}{u} = \frac{f}{f + u}$ (sign convention: $u$ negative for a real object).
$u = -25$ cm: $m_{25} = \frac{20}{20 - 25} = -4$.
$u = -50$ cm: $m_{50} = \frac{20}{20 - 50} = -\frac{2}{3}$.
$\frac{m_{25}}{m_{50}} = \frac{-4}{-2/3} = 6$.
Answer: 6. | ||||
| 76 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter After acceleration through V, kinetic energy $= qV$ and momentum $= \sqrt{2mqV}$, so $\lambda = \frac{h}{\sqrt{2mqV}}$.
$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_pq_p}} = \sqrt{4\times2} = \sqrt8 \approx 2.83$.
To the nearest integer: 3.
Answer: 3. | ||||
| 77 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Let each battery have emf E.
Series: emf 2E, internal resistance 2 Ω: $I_1 = \frac{2E}{R + 2}$, $J_1 = \frac{4E^2R}{(R + 2)^2}$.
Parallel: emf E, internal resistance 0.5 Ω: $I_2 = \frac{E}{R + 0.5}$, $J_2 = \frac{E^2R}{(R + 0.5)^2}$.
$J_1 = 2.25J_2$: $\frac{4}{(R + 2)^2} = \frac{2.25}{(R + 0.5)^2} \Rightarrow \frac{2}{R + 2} = \frac{1.5}{R + 0.5}$.
$2R + 1 = 1.5R + 3 \Rightarrow 0.5R = 2 \Rightarrow R = 4$ Ω.
Answer: 4. | ||||
| 78 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy 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 Wien's law: $\lambda_mT = $ constant, so $\frac{T_1}{T_2} = \frac{1500}{500} = 3$.
Stefan's law: power $= \sigma(4\pi r^2)T^4$, so
$\frac{P_A}{P_B} = \left(\frac{r_A}{r_B}\right)^2\left(\frac{T_1}{T_2}\right)^4 = \left(\frac{6}{18}\right)^2\times3^4 = \frac{81}{9} = 9$.
Answer: 9. | ||||
| 79 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The two waves have the same wave number and frequency and differ in phase by $\phi = \frac{\pi}{2}$.
Resultant amplitude: $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi} = \sqrt{16 + 9 + 0} = 5$.
Answer: 5. | ||||
| 80 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The wire acts as a spring of stiffness $k = \frac{YA}{L}$ (force per unit extension).
$\omega^2 = \frac{k}{m} = \frac{YA}{mL} \Rightarrow Y = \frac{\omega^2mL}{A} = \frac{140^2\times0.1\times1}{4.9\times10^{-7}} = \frac{1960}{4.9\times10^{-7}} = 4\times10^9$ N m$^{-2}$.
n = 4.
Answer: 4. | ||||
| 81 | GravitationOrbital motion | Class 11: Gravitation | NumericalMedium | |
A medium numerical question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Both stars go round the centre of mass with the same angular velocity $\omega$, at distances $r_A$, $r_B$ with $m_Ar_A = m_Br_B$: $\frac{r_A}{r_B} = \frac{11}{2.2} = 5$.
$\frac{L_A}{L_B} = \frac{m_Ar_A^2\omega}{m_Br_B^2\omega} = \frac{m_A}{m_B}\left(\frac{r_A}{r_B}\right)^2 = \frac{1}{5}\times25 = 5$.
$\frac{L_A + L_B}{L_B} = 5 + 1 = 6$.
Answer: 6. | ||||
| 82 | GravitationEscape velocity | Class 11: Gravitation | NumericalMedium | |
A medium numerical question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $g = \frac{GM}{R^2} = \frac43\pi G\rho R$, so $R \propto \frac{g}{\rho}$: $\frac{R_p}{R_e} = \frac{\sqrt6/11}{2/3} = \frac{3\sqrt6}{22}$.
$v_e = \sqrt{2gR}$, so $\frac{v_p}{v_e} = \sqrt{\frac{g_p}{g}\cdot\frac{R_p}{R_e}} = \sqrt{\frac{\sqrt6}{11}\cdot\frac{3\sqrt6}{22}} = \sqrt{\frac{18}{242}} = \sqrt{\frac{9}{121}} = \frac{3}{11}$.
$v_p = \frac{3}{11}\times11 = 3$ km s$^{-1}$.
Answer: 3. | ||||
| 83 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter (The "heat capacity" given is the specific heat of ice.)
Heat to warm m grams of ice from −5 °C to 0 °C: $m\times10^{-3}\times2100\times5 = 10.5m$ J.
Heat to melt 1 g: $10^{-3}\times3.36\times10^5 = 336$ J.
$10.5m + 336 = 420 \Rightarrow 10.5m = 84 \Rightarrow m = 8$.
Answer: 8. | ||||
| 84 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves A car approaching at speed v receives $f_0\frac{c + v}{c}$ and, as a moving source, reflects it back to the stationary source as
$f = f_0\frac{c + v}{c - v} \approx f_0\left(1 + \frac{2v}{c}\right)$ for $v \ll c$.
For two cars: $f_1 - f_2 = \frac{2f_0}{c}(v_1 - v_2) = 0.012f_0$.
$v_1 - v_2 = 0.006c = 0.006\times330 = 1.98$ m s$^{-1} = 1.98\times3.6 = 7.128$ km h$^{-1}$.
To the nearest integer: 7.
Answer: 7. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | 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 A spin-only moment $\mu = \sqrt{n(n+2)}$ B.M. of 2.82 B.M. means $n = 2$ unpaired electrons ($\sqrt{8} = 2.83$).
(A) $\mathrm{Ni(CO)_4}$: Ni(0), $3d^{10}$ after pairing, tetrahedral, diamagnetic ($\mu = 0$).
(B) $\mathrm{[NiCl_4]^{2-}}$: $\mathrm{Ni^{2+}}$, $3d^8$; $\mathrm{Cl^-}$ is a weak-field ligand, so the complex is tetrahedral ($sp^3$) and the two unpaired $3d$ electrons stay unpaired: $\mu = \sqrt{8} \approx 2.83$ B.M.
(C) $\mathrm{Ni(PPh_3)_4}$: Ni(0), $d^{10}$, diamagnetic.
(D) $\mathrm{[Ni(CN)_4]^{2-}}$: $\mathrm{Ni^{2+}}$, $d^8$ with the strong-field $\mathrm{CN^-}$: square planar ($dsp^2$), all electrons paired, diamagnetic.
Answer: (B). | ||||
| 2 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Count the electron pairs on the central atom.
(A) $\mathrm{SO_3}$: three bonding regions, no lone pair on S: trigonal planar.
(B) $\mathrm{BrF_3}$: three bond pairs and two lone pairs ($sp^3d$): T-shaped.
(C) $\mathrm{SiO_3^{2-}}$ (the metasilicate unit) is built of $\mathrm{SiO_4}$ tetrahedra sharing corners; as a discrete ion it would be planar like $\mathrm{CO_3^{2-}}$. Not pyramidal.
(D) $\mathrm{OSF_2}$ (thionyl fluoride): S forms three $\sigma$ bonds (to O and two F) and carries one lone pair, so it is $sp^3$ with one lone pair: trigonal pyramidal, like $\mathrm{SOCl_2}$.
Answer: (D). | ||||
| 3 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Step (1): $\mathrm{Br_2/NaOH}$ on a primary amide is the Hofmann bromamide degradation. The amide loses its carbonyl carbon as carbonate and gives the amine with one carbon fewer:
$\mathrm{H_3C{-}C_6H_4{-}CONH_2 \xrightarrow{Br_2/NaOH} H_3C{-}C_6H_4{-}NH_2}$ (p-toluidine).
Step (2): benzoyl chloride acylates the amine nitrogen (benzoylation):
$\mathrm{H_3C{-}C_6H_4{-}NH_2 + C_6H_5COCl \rightarrow H_3C{-}C_6H_4{-}NH{-}CO{-}C_6H_5 + HCl}$.
T is N-(4-methylphenyl)benzamide: the NH is on the tolyl ring and the C=O belongs to the benzoyl group, which is structure (C). (B) has the two rings swapped; (A) and (D) keep the carbonyl carbon that the Hofmann step removes.
Answer: (C). | ||||
| 4 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Where two groups compete, the stronger activating (ortho/para directing) group decides.
P (4-hydroxybenzoic acid): OH is a strong activator, COOH a deactivating meta director. The position ortho to OH is also meta to COOH, so both agree: $\mathrm{NO_2}$ goes ortho to OH (3-nitro-4-hydroxybenzoic acid).
Q (4-methylanisole): $\mathrm{OCH_3}$ is a much stronger activator than $\mathrm{CH_3}$; its para position is taken, so $\mathrm{NO_2}$ enters ortho to $\mathrm{OCH_3}$.
S (phenyl benzoate, $\mathrm{C_6H_5{-}CO{-}O{-}C_6H_5}$): the ring attached to the carbonyl carbon is deactivated by the C=O; the ring attached to the ester oxygen is activated by the O lone pair (ortho/para director). Nitration occurs on the O-phenyl ring, mainly para (the bulky ester group hinders ortho).
These are the three structures of option (C). (A) and (B) nitrate the deactivated benzoyl ring and put Q's nitro group next to $\mathrm{CH_3}$; (D) puts S's nitro group meta.
Answer: (C). | ||||
| 5 | Solid StateClose packed structures | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on close packed structures, from Solid State. NCERT: No matching NCERT chapter The cell is a square of side $L$ with a circle at each corner and one at the centre (a two-dimensional "body-centred" square).
Circles per cell: $4\times\frac{1}{4} + 1 = 2$.
The circles touch along the diagonal: diagonal $= \sqrt{2}L = 4r$, so $r = \frac{\sqrt{2}L}{4}$.
Area occupied: $2\pi r^2 = 2\pi\cdot\frac{2L^2}{16} = \frac{\pi L^2}{4}$.
Packing efficiency $= \frac{\pi L^2/4}{L^2} = \frac{\pi}{4} = 0.7854 = 78.54\%$.
Answer: (D). | ||||
| 6 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{B_2}$ has 10 electrons. With s–p mixing (as for $\mathrm{B_2}$, $\mathrm{C_2}$, $\mathrm{N_2}$) the order is $\sigma1s,\ \sigma^*1s,\ \sigma2s,\ \sigma^*2s,\ (\pi2p_x = \pi2p_y),\ \sigma2p_z,\dots$
Configuration: $\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2$ and then 2 electrons in the degenerate $\pi2p$ pair.
Following Hund's rule they would occupy $\pi2p_x^1\pi2p_y^1$ (paramagnetic). If Hund's rule is violated they pair in one orbital, $\pi2p_x^2$, and no electron is unpaired: diamagnetic.
Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(6 - 4) = 1$ either way.
Answer: (A). | ||||
| 7 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter An acid's basicity is the number of ionisable H atoms, i.e. those on O (O–H), not those bonded directly to the central atom.
$\mathrm{H_3PO_4}$: three P–OH, tribasic.
$\mathrm{H_2SO_4}$: two S–OH, diprotic.
$\mathrm{H_3PO_3}$: $\mathrm{HPO(OH)_2}$, one H on P, two P–OH: diprotic.
$\mathrm{H_2CO_3}$: two OH: diprotic.
$\mathrm{H_2S_2O_7}$ (pyrosulphuric acid, $\mathrm{HO{-}SO_2{-}O{-}SO_2{-}OH}$): two OH: diprotic.
$\mathrm{H_3BO_3}$: a monobasic Lewis acid ($\mathrm{B(OH)_3 + H_2O \rightarrow [B(OH)_4]^- + H^+}$).
$\mathrm{H_3PO_2}$: $\mathrm{H_2PO(OH)}$, one OH: monobasic.
$\mathrm{H_2CrO_4}$: $\mathrm{CrO_2(OH)_2}$: diprotic.
$\mathrm{H_2SO_3}$: $\mathrm{SO(OH)_2}$: diprotic.
Diprotic: $\mathrm{H_2SO_4, H_3PO_3, H_2CO_3, H_2S_2O_7, H_2CrO_4, H_2SO_3}$.
Answer: 6. | ||||
| 8 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Rh(I) is $d^8$, and its four-coordinate complexes are square planar. The complex is of the type $\mathrm{[Mabcd]}$ with four different monodentate ligands.
Fix Cl at one corner; the ligand trans to it can be CO, $\mathrm{PPh_3}$ or $\mathrm{NH_3}$, and the remaining two then occupy the two cis positions, which gives the same arrangement whichever way round (the plane can be flipped). So there are 3 distinct arrangements:
Cl trans to CO, Cl trans to $\mathrm{PPh_3}$, Cl trans to $\mathrm{NH_3}$.
Answer: 3. | ||||
| 9 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions Go through the list.
O: −2 in oxides, −1 in peroxides, +2 in $\mathrm{OF_2}$, +1 in $\mathrm{O_2F_2}$: several.
Cl: −1, +1, +3, +5, +7: several.
F: the most electronegative element, only −1.
N: −3 to +5: several.
P: −3, +3, +5: several.
Sn: +2 and +4.
Tl: +1 and +3.
Na: only +1.
Ti: +2, +3, +4.
Only F and Na show a single non-zero oxidation state.
Answer: 2. | ||||
| 10 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Volume taken up by one atom: $\frac{M}{\rho N_A} = \frac{108}{10.5 \times 6.022\times10^{23}} = 1.71\times10^{-23}\ \mathrm{cm^3}$.
Treat each atom as occupying a small cube of this volume; its edge is $a = (1.71\times10^{-23})^{1/3} = 2.58\times10^{-8}\ \mathrm{cm}$, so each atom covers an area $a^2 = 6.63\times10^{-16}\ \mathrm{cm^2}$ of a surface.
Area given: $10^{-12}\ \mathrm{m^2} = 10^{-12}\times10^4\ \mathrm{cm^2} = 10^{-8}\ \mathrm{cm^2}$.
Number of atoms $= \frac{10^{-8}}{6.63\times10^{-16}} = 1.5\times10^{7}$.
So $y \approx 1.5$ and $x = 7$.
Answer: 7. | ||||
| 11 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Read the points off the graph. The solid curve passes through (0.5, 4), (1, 2), (2, 1), (4, 0.5): $PV = 2$ L atm throughout, so it is an isotherm.
Solid path (isothermal, reversible): $w_s = nRT\ln\frac{V_b}{V_a} = 2\ln\frac{5.5}{0.5} = 2\ln 11 = 2\times2.398 = 4.80$ L atm (work done by the gas).
Dashed path: only the horizontal (constant-pressure) steps do work.
a to (2, 4): $4\times(2 - 0.5) = 6$ L atm.
(2, 1) to (3, 1): $1\times1 = 1$ L atm.
(3, 0.67) to (5.5, 0.67): $\frac{2}{3}\times2.5 = 1.67$ L atm (the last step is at the pressure of the isotherm at V = 3, $\frac{2}{3}$ atm).
$w_d = 6 + 1 + 1.67 = 8.67$ L atm.
$\frac{w_d}{w_s} = \frac{8.67}{4.80} = 1.81$, closest integer 2.
Answer: 2. | ||||
| 12 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | ComprehensionMedium | |
A medium comprehension question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Work back from the lactone. Opening it gives the hydroxy acid $\mathrm{HOCH_2{-}C(CH_3)_2{-}CH(OH){-}COOH}$.
The COOH came from a nitrile (S is a cyanohydrin hydrolysed on acidification and heating), so S is $\mathrm{HOCH_2{-}C(CH_3)_2{-}CH(OH){-}CN}$ and R, the aldehyde that added HCN, is $\mathrm{HOCH_2{-}C(CH_3)_2{-}CHO}$.
R is a crossed aldol: the $\alpha$-carbon of isobutyraldehyde $\mathrm{(CH_3)_2CH{-}CHO}$ (one $\alpha$-H) adds to formaldehyde $\mathrm{HCHO}$ (no $\alpha$-H, the acceptor), giving the $\mathrm{CH_2OH}$ group on the quaternary carbon. Mild base ($\mathrm{K_2CO_3}$) stops it at the aldol.
P = $\mathrm{(CH_3)_2CHCHO}$, Q = HCHO, option (B). Acetaldehyde (A, C) would give a different skeleton, and $\mathrm{(CH_3)_2CHCH_2CHO}$ (C, D) has one carbon too many.
Answer: (B). | ||||
| 13 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | ComprehensionMedium | |
A medium comprehension question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids From the paragraph, R is the crossed aldol of isobutyraldehyde and formaldehyde.
The enolate of $\mathrm{(CH_3)_2CH{-}CHO}$ forms at the $\alpha$-carbon (the CH bearing the two methyls) and attacks HCHO:
$\mathrm{(CH_3)_2CH{-}CHO + HCHO \xrightarrow{K_2CO_3} HOCH_2{-}C(CH_3)_2{-}CHO}$ (3-hydroxy-2,2-dimethylpropanal).
This is the only aldehyde that, after adding HCN and hydrolysis, can close to the given lactone (the $\mathrm{CH_2OH}$ and the new COOH are 1,4-related through the quaternary carbon). It is structure (A).
Answer: (A). | ||||
| 14 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | ComprehensionMedium | |
A medium comprehension question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids HCN adds to the aldehyde carbonyl of R, $\mathrm{HOCH_2{-}C(CH_3)_2{-}CHO}$, to give the cyanohydrin:
$\mathrm{HOCH_2{-}C(CH_3)_2{-}CHO + HCN \rightarrow HOCH_2{-}C(CH_3)_2{-}CH(OH){-}CN}$ = S.
On acidification and heating the CN hydrolyses to COOH, and the acid esterifies with the $\gamma$-$\mathrm{CH_2OH}$ to the five-membered lactone shown (2-hydroxy-3,3-dimethyl-$\gamma$-butyrolactone). This is structure (D).
Answer: (D). | ||||
| 15 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | ComprehensionEasy | |
A easy comprehension question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Spherically symmetric means an s orbital ($l = 0$). The number of radial nodes is $n - l - 1$; one radial node with $l = 0$ gives $n = 2$.
1s has no radial node, 2p is not spherically symmetric, 3s has two radial nodes.
$S_1$ is 2s.
Answer: (B). | ||||
| 16 | Structure of AtomBohr model | Class 11: Structure of Atom | ComprehensionEasy | |
A easy comprehension question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom For a hydrogen-like ion $E_n = E_H\frac{Z^2}{n^2}$, where $E_H = -13.6$ eV is the ground-state energy of hydrogen.
For $\mathrm{Li^{2+}}$, $Z = 3$, and $S_1$ = 2s has $n = 2$:
$E = E_H\cdot\frac{9}{4} = 2.25\,E_H$.
Answer: (C). | ||||
| 17 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | ComprehensionMedium | |
A medium comprehension question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom $S_2$ has energy equal to the hydrogen ground state: $E_H\frac{Z^2}{n^2} = E_H$, so $n^2 = Z^2 = 9$ and $n = 3$.
It has one radial node: $n - l - 1 = 1$, so $l = 3 - 2 = 1$ (a 3p state). This is also consistent with an allowed absorption from 2s ($\Delta l = +1$).
Answer: (B). | ||||
| 18 | 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) Benzenediazonium chloride with phenol in alkali at 0 °C is azo coupling: the diazonium ion attacks the para position of phenoxide, replacing H, an electrophilic aromatic substitution. Coupling reaction (s) and substitution (r). No carbocation, no new stereocentre.
(B) Pinacol with acid is the pinacol–pinacolone rearrangement: protonation and loss of water gives a tertiary carbocation, a methyl shifts, and the oxocarbenium ion loses $\mathrm{H^+}$ to give pinacolone. Carbocation intermediate (t).
(C) $\mathrm{LiAlH_4}$ adds hydride to the C=O of acetophenone (nucleophilic addition, q). The product $\mathrm{C_6H_5CH(OH)CH_3}$ has a new stereocentre; hydride attacks the planar carbonyl equally from both faces, so the product is racemic (p).
(D) The base deprotonates SH; the thiolate displaces chloride intramolecularly (an $\mathrm{S_N2}$ substitution) across the ring, giving the bridged sulphide. Substitution (r).
A→r, s; B→t; C→p, q; D→r. | ||||
| 19 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Matrix matchMedium | |
A medium matrix match question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) $\mathrm{(CH_3)_2SiCl_2 + 2H_2O \rightarrow (CH_3)_2Si(OH)_2 + 2HCl}$; the silanediol condenses to the linear silicone polymer $\mathrm{[{-}Si(CH_3)_2{-}O{-}]_n}$. Hydrogen halide (p) and polymerization (s).
(B) $\mathrm{6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2}$. Xe(+4) disproportionates to Xe(0) and Xe(+6) and oxygen is oxidised to $\mathrm{O_2}$: redox (q), HF formed (p), HF attacks glass ($\mathrm{SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O}$) (r), $\mathrm{O_2}$ formed (t).
(C) $\mathrm{Cl_2 + H_2O \rightarrow HCl + HOCl}$: chlorine disproportionates (0 to −1 and +1), redox (q), with HCl formed (p).
(D) $\mathrm{VCl_5}$ hydrolyses ($\mathrm{VCl_5 + H_2O \rightarrow VOCl_3 + 2HCl}$, and further to vanadium(V) oxo species); V stays +5, so it is only a hydrolysis giving HCl (p).
A→p, s; B→p, q, r, t; C→p, q; D→p. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 20 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Here $A_r = \binom{10}{r}$, $B_r = \binom{20}{r}$, $C_r = \binom{30}{r}$.
$\sum_{r=0}^{10}\binom{10}{r}\binom{20}{r} = \sum_{r=0}^{10}\binom{10}{r}\binom{20}{20-r}$ = coefficient of $x^{20}$ in $(1+x)^{10}(1+x)^{20}$ $= \binom{30}{20} = \binom{30}{10} = C_{10}$.
So $\sum_{r=1}^{10}A_rB_r = C_{10} - 1$ (removing the $r = 0$ term, which is 1).
Similarly $\sum_{r=0}^{10}\binom{10}{r}^2 = \binom{20}{10} = B_{10}$, so $\sum_{r=1}^{10}A_r^2 = B_{10} - 1$.
The sum is $B_{10}(C_{10} - 1) - C_{10}(B_{10} - 1) = -B_{10} + C_{10} = C_{10} - B_{10}$.
Answer: (D). | ||||
| 21 | 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 Count ordered pairs $(X, Y)$ of disjoint subsets first: each of the 4 elements goes into X only, Y only, or neither, so there are $3^4 = 81$ ordered pairs.
An unordered pair $\{X, Y\}$ with $X \ne Y$ is counted twice among these. $X = Y$ together with $X \cap Y = \emptyset$ forces $X = Y = \emptyset$, which is counted once.
Unordered pairs $= \frac{81 - 1}{2} + 1 = 40 + 1 = 41$.
Answer: (D). | ||||
| 22 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f(x) = e^x\left(2 + \int_0^x\sqrt{t^4+1}\,dt\right)$.
At $x = 0$: $f(0) = 2$, so $f^{-1}(2) = 0$.
Differentiate (product rule and the fundamental theorem of calculus):
$f'(x) = e^x\left(2 + \int_0^x\sqrt{t^4+1}\,dt\right) + e^x\sqrt{x^4+1}$, so $f'(0) = 2 + 1 = 3$.
$(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(0)} = \frac{1}{3}$.
Answer: (B). | ||||
| 23 | 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 Distance: $\frac{|1 + 2(-2) - 2(1) - \alpha|}{\sqrt{1+4+4}} = \frac{|-5 - \alpha|}{3} = 5$, so $5 + \alpha = 15$ and $\alpha = 10$ (taking $\alpha > 0$).
The foot is $P + \lambda(1, 2, -2)$ with $\lambda = -\frac{1 - 4 - 2 - 10}{9} = \frac{15}{9} = \frac{5}{3}$.
Foot $= \left(1 + \frac{5}{3},\ -2 + \frac{10}{3},\ 1 - \frac{10}{3}\right) = \left(\frac{8}{3}, \frac{4}{3}, -\frac{7}{3}\right)$.
Check: $\frac{8}{3} + \frac{8}{3} + \frac{14}{3} = 10$.
Answer: (A). | ||||
| 24 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\overrightarrow{AB}| = \sqrt{4 + 100 + 121} = 15$, $|\overrightarrow{AD}| = \sqrt{1+4+4} = 3$, $\overrightarrow{AB}\cdot\overrightarrow{AD} = -2 + 20 + 22 = 40$.
Angle $\theta$ between AB and AD: $\cos\theta = \frac{40}{45} = \frac{8}{9}$.
After rotation in the same plane, AD$'$ is perpendicular to AB, i.e. at $90^\circ$ to AB. Since $\theta < 90^\circ$, AD is turned away from AB through $\alpha = 90^\circ - \theta$ (acute).
$\cos\alpha = \sin\theta = \sqrt{1 - \frac{64}{81}} = \frac{\sqrt{17}}{9}$.
Answer: (B). | ||||
| 25 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability B receives the original colour if both stations are right or both are wrong: $\left(\frac{3}{4}\right)^2 + \left(\frac{1}{4}\right)^2 = \frac{10}{16}$; it receives the other colour with probability $2\cdot\frac{3}{4}\cdot\frac{1}{4} = \frac{6}{16}$.
$P(\text{B green}) = \frac{4}{5}\cdot\frac{10}{16} + \frac{1}{5}\cdot\frac{6}{16} = \frac{40 + 6}{80} = \frac{46}{80}$.
$P(\text{original green} \mid \text{B green}) = \frac{40/80}{46/80} = \frac{40}{46} = \frac{20}{23}$ (Bayes' theorem).
Answer: (C). | ||||
| 26 | CirclesRadius and centre | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections A chord subtending angle $2\phi$ at the centre is at distance $2\cos\phi$ from the centre (radius 2).
The two chords are at distances $2\cos\frac{\pi}{2k}$ and $2\cos\frac{\pi}{k}$.
If they were on the same side of the centre, the gap would be $2\left(\cos\theta - \cos2\theta\right)$ with $\theta = \frac{\pi}{2k}$; with $c = \cos\theta$ this is $2(-2c^2 + c + 1) \le 2\cdot\frac{9}{8} = 2.25 < \sqrt{3} + 1 \approx 2.73$. So they are on opposite sides:
$2\cos\frac{\pi}{2k} + 2\cos\frac{\pi}{k} = \sqrt{3} + 1$, i.e. $\cos\theta + \cos2\theta = \frac{\sqrt{3}}{2} + \frac{1}{2}$.
$\theta = \frac{\pi}{6}$ works ($\frac{\sqrt{3}}{2} + \frac{1}{2}$), and the left side is decreasing on $(0, \frac{\pi}{2})$, so it is the only solution. Then $\frac{\pi}{2k} = \frac{\pi}{6}$, $k = 3$.
$[k] = 3$.
Answer: 3. | ||||
| 27 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Area $= \frac{1}{2}ab\sin C = \frac{1}{2}\cdot6\cdot10\sin C = 30\sin C = 15\sqrt{3}$, so $\sin C = \frac{\sqrt{3}}{2}$; C is obtuse, so $C = 120^\circ$.
Cosine rule: $c^2 = 36 + 100 - 2\cdot6\cdot10\cos120^\circ = 136 + 60 = 196$, $c = 14$.
Semi-perimeter $s = \frac{6 + 10 + 14}{2} = 15$.
$r = \frac{\Delta}{s} = \frac{15\sqrt{3}}{15} = \sqrt{3}$, so $r^2 = 3$.
Answer: 3. | ||||
| 28 | 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 $g(x) = e^{f(x)}$ and $e^u$ is increasing, so g has a local maximum exactly where f does. $g'(x) = f'(x)e^{f(x)}$ has the sign of $f'(x)$.
The factors $(x-2010)^2$ and $(x-2012)^4$ never change sign. The sign of $f'$ changes only at the odd-power factors, $x = 2009$ and $x = 2011$.
For $x < 2009$: $(x-2009) < 0$ and $(x-2011)^3 < 0$, so $f' > 0$.
For $2009 < x < 2011$ (apart from 2010): $f' < 0$.
For $x > 2011$: $f' > 0$.
So f (and g) rises then falls at $x = 2009$: a local maximum; at $x = 2011$ it falls then rises: a local minimum. At 2010 and 2012 there is no change of sign.
Answer: 1. | ||||
| 29 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $a_k - a_{k-1} = a_{k-1} - a_{k-2}$, so the numbers form an A.P.: $a_k = 15 + (k-1)d$.
$\sum_{j=0}^{10}(15 + jd)^2 = 11\cdot225 + 30d\cdot55 + d^2\cdot385 = 2475 + 1650d + 385d^2$.
Setting this equal to $11\times90 = 990$: $385d^2 + 1650d + 1485 = 0$, i.e. $7d^2 + 30d + 27 = 0$, so $d = \frac{-30 \pm 12}{14}$: $d = -3$ or $d = -\frac{9}{7}$.
$27 - 2a_2 > 0$ means $a_2 = 15 + d < 13.5$, i.e. $d < -1.5$. So $d = -3$.
Mean $= \frac{a_1 + a_{11}}{2} = a_6 = 15 + 5(-3) = 0$.
Answer: 0. | ||||
| 30 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3\times3$ matrix, $\det(\operatorname{adj}M) = (\det M)^2$.
B is skew-symmetric ($B^T = -B$) of odd order, so $\det B = 0$ and $\det(\operatorname{adj}B) = 0$.
Expand $\det A$ along the first row:
$(2k-1)\left[(1)(-1) - (-2k)(2k)\right] = (2k-1)(4k^2 - 1)$,
$-2\sqrt{k}\left[(2\sqrt{k})(-1) - (-2k)(-2\sqrt{k})\right] = -2\sqrt{k}\cdot(-2\sqrt{k})(1 + 2k) = 4k(2k+1)$,
$+2\sqrt{k}\left[(2\sqrt{k})(2k) - (1)(-2\sqrt{k})\right] = 2\sqrt{k}\cdot2\sqrt{k}(2k+1) = 4k(2k+1)$.
$\det A = (2k+1)\left[(2k-1)^2 + 8k\right] = (2k+1)(2k+1)^2 = (2k+1)^3$.
So $(2k+1)^6 = 10^6$, $2k + 1 = 10$ ($k > 0$), $k = 4.5$ and $[k] = 4$.
Answer: 4. | ||||
| 31 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | ComprehensionMedium | |
A medium comprehension question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 2 + 6x + 12x^2$ has discriminant $36 - 96 < 0$, so $f'(x) > 0$ for all x: f is strictly increasing and has exactly one real root, and s is that root.
$f\left(-\frac{1}{2}\right) = 1 - 1 + \frac{3}{4} - \frac{1}{2} = \frac{1}{4} > 0$.
$f\left(-\frac{3}{4}\right) = 1 - \frac{3}{2} + \frac{27}{16} - \frac{27}{16} = -\frac{1}{2} < 0$.
So $s \in \left(-\frac{3}{4}, -\frac{1}{2}\right)$.
Answer: (C). | ||||
| 32 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | ComprehensionMedium | |
A medium comprehension question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals From the previous question $t = |s| \in \left(\frac{1}{2}, \frac{3}{4}\right)$. On $[0, t]$, $f(x) > 0$, so the area is
$\int_0^t f(x)\,dx = t + t^2 + t^3 + t^4$, an increasing function of t.
At $t = \frac{1}{2}$: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{15}{16}$.
At $t = \frac{3}{4}$: $\frac{3}{4} + \frac{9}{16} + \frac{27}{64} + \frac{81}{256} = \frac{525}{256} \approx 2.05$.
The area lies in $\left(\frac{15}{16}, \frac{525}{256}\right)$, which is inside $\left(\frac{3}{4}, 3\right)$.
Answer: (A). | ||||
| 33 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | ComprehensionEasy | |
A easy comprehension question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 12x^2 + 6x + 2$, so $f''(x) = 24x + 6$, which is negative for $x < -\frac{1}{4}$ and positive for $x > -\frac{1}{4}$.
Since $t > \frac{1}{2}$, the point $-\frac{1}{4}$ lies inside $(-t, t)$. So $f'$ decreases on $\left(-t, -\frac{1}{4}\right)$ and increases on $\left(-\frac{1}{4}, t\right)$.
Answer: (B). | ||||
| 34 | Conic SectionsEllipse | Class 11: Conic Sections | ComprehensionMedium | |
A medium comprehension question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The chord of contact from $P(3, 4)$ is $\frac{3x}{9} + \frac{4y}{4} = 1$, i.e. $x + 3y = 3$.
Put $x = 3(1 - y)$ in the ellipse: $(1 - y)^2 + \frac{y^2}{4} = 1$, so $-2y + \frac{5y^2}{4} = 0$, giving $y = 0$ or $y = \frac{8}{5}$.
$y = 0$: $x = 3$; $y = \frac{8}{5}$: $x = 3\left(1 - \frac{8}{5}\right) = -\frac{9}{5}$.
A and B are $(3, 0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$. (Check: $x = 3$ is indeed the tangent at $(3, 0)$ and passes through P.)
Answer: (D). | ||||
| 35 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | ComprehensionMedium | |
A medium comprehension question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines $P(3, 4)$, $A(3, 0)$, $B\left(-\frac{9}{5}, \frac{8}{5}\right)$.
PA is the vertical line $x = 3$, so the altitude from B is horizontal: $y = \frac{8}{5}$.
Slope of PB $= \frac{4 - 8/5}{3 + 9/5} = \frac{12/5}{24/5} = \frac{1}{2}$, so the altitude from A has slope $-2$: $y = -2(x - 3)$.
With $y = \frac{8}{5}$: $x - 3 = -\frac{4}{5}$, $x = \frac{11}{5}$.
Orthocentre $\left(\frac{11}{5}, \frac{8}{5}\right)$.
Answer: (C). | ||||
| 36 | Conic SectionsParabola | Class 11: Conic Sections | ComprehensionMedium | |
A medium comprehension question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections AB is $x + 3y - 3 = 0$. The locus is the parabola with focus P and directrix AB:
$(x - 3)^2 + (y - 4)^2 = \frac{(x + 3y - 3)^2}{10}$.
$10(x^2 + y^2 - 6x - 8y + 25) = x^2 + 9y^2 + 9 + 6xy - 6x - 18y$.
$9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0$.
Answer: (A). | ||||
| 37 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Matrix matchHard | |
A hard matrix match question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations (A) z is equidistant from the points $i|z|$ and $-i|z|$, which are mirror images in the real axis. So z lies on their perpendicular bisector, the real axis (z = 0 is also real). The set is the whole real axis: contained in Im z = 0 (q) and hence in $|\mathrm{Im}\,z| \le 1$ (r). It is unbounded, so not in (s) or (t), and it is not an ellipse.
(B) The sum of distances from $\pm4$ is 10: an ellipse with $2a = 10$, focal distance $c = 4$, $e = \frac{4}{5}$ (p). Its semi-axes are 5 and 3, so it is not inside $|\mathrm{Re}\,z| \le 2$, $|\mathrm{Im}\,z| \le 1$ or $|z| \le 3$.
(C) $w = 2e^{i\theta}$: $z = 2e^{i\theta} - \frac{1}{2}e^{-i\theta} = \frac{3}{2}\cos\theta + i\,\frac{5}{2}\sin\theta$. This is the ellipse $\frac{x^2}{(3/2)^2} + \frac{y^2}{(5/2)^2} = 1$, $e = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ (p). $|\mathrm{Re}\,z| \le \frac{3}{2} \le 2$ (s), $|z| \le \frac{5}{2} \le 3$ (t); $|\mathrm{Im}\,z|$ reaches $\frac{5}{2}$, so not (r).
(D) $w = e^{i\theta}$: $z = e^{i\theta} + e^{-i\theta} = 2\cos\theta$, the real segment $[-2, 2]$. It satisfies Im z = 0 (q), $|\mathrm{Im}\,z| \le 1$ (r), $|\mathrm{Re}\,z| \le 2$ (s), $|z| \le 3$ (t).
A→q, r; B→p; C→p, s, t; D→q, r, s, t. | ||||
| 38 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Matrix matchHard | |
A hard matrix match question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry (A) A point of the first line is $P = (2 + \lambda, 1 - 2\lambda, -1 + \lambda)$, of the second $Q = \left(\frac{8}{3} + 2\nu, -3 - \nu, 1 + \nu\right)$. O, P, Q are collinear: $P = kQ$.
Adding twice the third coordinate equation to the second: $(1 - 2\lambda) + 2(-1 + \lambda) = -1 = k(-3 - \nu + 2 + 2\nu) = k(\nu - 1)$.
Subtracting the third from the first: $3 = k\left(\frac{8}{3} + 2\nu - 1 - \nu\right) = k\left(\nu + \frac{5}{3}\right)$.
Dividing: $\frac{\nu + 5/3}{\nu - 1} = -3$, so $\nu = \frac{1}{3}$ and $k = \frac{3}{2}$. Then $Q = \left(\frac{10}{3}, -\frac{10}{3}, \frac{4}{3}\right)$, $P = \frac{3}{2}Q = (5, -5, 2)$ (it is on the first line with $\lambda = 3$).
$\vec{PQ} = \left(-\frac{5}{3}, \frac{5}{3}, -\frac{2}{3}\right)$, $d^2 = \frac{25 + 25 + 4}{9} = 6$ (t).
(B) $\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{3}{4}$. Taking tangents: $\frac{(x+3) - (x-3)}{1 + (x^2 - 9)} = \frac{6}{x^2 - 8} = \frac{3}{4}$, so $x^2 = 16$, $x = \pm4$. For both, $x^2 - 8 > 0$, so the left side lies in $\left(0, \frac{\pi}{2}\right)$ and the equation holds. Values $-4$ and $4$: (p), (r).
(C) $\vec{c} = \frac{\vec{a} - \mu\vec{b}}{4}$, so $\vec{b} + \vec{c} = \frac{\vec{a} + (4 - \mu)\vec{b}}{4}$. With $\vec{a}\cdot\vec{b} = 0$:
$(\vec{b} - \vec{a})\cdot(\vec{a} + (4 - \mu)\vec{b}) = (4 - \mu)|\vec{b}|^2 - |\vec{a}|^2 = 0$, so $|\vec{a}|^2 = (4 - \mu)|\vec{b}|^2$.
$4|\vec{b} + \vec{c}|^2 = |\vec{b} - \vec{a}|^2$: $\frac{|\vec{a}|^2 + (4 - \mu)^2|\vec{b}|^2}{4} = |\vec{a}|^2 + |\vec{b}|^2$, so $(4 - \mu)^2|\vec{b}|^2 - 4|\vec{b}|^2 = 3|\vec{a}|^2 = 3(4 - \mu)|\vec{b}|^2$.
With $u = 4 - \mu$: $u^2 - 3u - 4 = 0$, $u = 4$ or $u = -1$, i.e. $\mu = 0$ or $\mu = 5$. But $|\vec{a}|^2 = u|\vec{b}|^2$ must be positive, which rules out $u = -1$. So $\mu = 0$ (q). (The value 5 comes out of the algebra but gives $|\vec{a}|^2 < 0$; the key also accepts q and s together.)
(D) $\frac{\sin(9x/2)}{\sin(x/2)} = 1 + 2(\cos x + \cos2x + \cos3x + \cos4x)$. Over $[-\pi, \pi]$ each cosine integrates to 0, so $\int_{-\pi}^{\pi}f = 2\pi$ and $\frac{2}{\pi}\cdot2\pi = 4$ (r).
A→t; B→p, r; C→q; D→r. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 39 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter 20 VSD = 16 MSD, so 1 VSD $= \frac{16}{20} = 0.8$ mm.
Least count = 1 MSD − 1 VSD $= 1 - 0.8 = 0.2$ mm.
Answer: (D). | ||||
| 40 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Fundamental of the closed pipe: $f = \frac{v}{4L} = \frac{320}{4\times0.8} = 100$ Hz.
Second harmonic of the string of length $l = 0.5$ m: $f = \frac{2}{2l}\sqrt{\frac{T}{\mu}} = \frac{1}{0.5}\sqrt{\frac{T}{\mu}} = 100$, so the wave speed on the string is $\sqrt{\frac{T}{\mu}} = 50$ m/s.
$\mu = \frac{T}{50^2} = \frac{50}{2500} = 0.02$ kg/m.
Mass $= \mu l = 0.02\times0.5 = 0.01$ kg = 10 g.
Answer: (B). | ||||
| 41 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens first: $u = -30$ cm, $f = 15$ cm: $\frac{1}{v} = \frac{1}{15} - \frac{1}{30} = \frac{1}{30}$, $v = 30$ cm beyond the lens.
The mirror is 10 cm beyond the lens, so the rays converge towards a point 20 cm behind the mirror: a virtual object for the mirror, 20 cm behind it.
A plane mirror images it 20 cm in front of the mirror, i.e. the reflected rays converge towards a point 20 cm from the mirror, which is 10 cm on the far side of the lens (towards the object). The lens intercepts them first.
Lens again (light now travelling back): the rays converge towards a point 10 cm beyond the lens, a virtual object with $u = +10$ cm (sign taken along the new direction of light), $f = 15$ cm:
$\frac{1}{v} = \frac{1}{15} + \frac{1}{10} = \frac{1}{6}$, $v = +6$ cm.
The final image is real, 6 cm from the lens on the object side, i.e. $10 + 6 = 16$ cm from the mirror.
Answer: (B). | ||||
| 42 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion The force is $F = 4 - \frac{4}{3}t$; at $t = 4.5$ s it is $4 - 6 = -2$ N.
Impulse = area under the graph:
0 to 3 s: $\frac{1}{2}\times3\times4 = 6$ N s;
3 to 4.5 s: $-\frac{1}{2}\times1.5\times2 = -1.5$ N s.
Momentum at 4.5 s: $p = 6 - 1.5 = 4.5$ kg m/s (starting from rest).
$K = \frac{p^2}{2m} = \frac{4.5^2}{4} = \frac{20.25}{4} = 5.06$ J.
Answer: (C). | ||||
| 43 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal velocity, Stokes' drag balances the weight (the buoyancy of air is neglected):
$6\pi\eta rv = \frac{4}{3}\pi r^3\rho g$, so $r^2 = \frac{9\eta v}{2\rho g} = \frac{9\times1.8\times10^{-5}\times2\times10^{-3}}{2\times900\times9.8} = 1.837\times10^{-11}\ \mathrm{m^2}$, $r = \frac{3}{7}\times10^{-5}$ m.
Weight: $mg = \frac{4}{3}\pi r^3\rho g = \frac{4}{3}\pi\cdot\frac{27}{343}\times10^{-15}\times900\times9.8 = \frac{4}{3}\pi\cdot\frac{27\times10^{-15}}{343}\times8820$.
Since $\frac{8820}{343} = \frac{180}{7}$: $mg = \frac{4}{3}\pi\times27\times10^{-15}\times\frac{180}{7} = \frac{6480\pi}{7}\times10^{-15}$ N.
Balance in the field: $qE = mg$, so $q = \frac{6480\pi\times10^{-15}/7}{81\pi\times10^5/7} = 80\times10^{-20} = 8.0\times10^{-19}$ C.
Answer: (D). | ||||
| 44 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Each surface element feels an outward electrostatic pressure $\frac{\sigma^2}{2\varepsilon_0}$ (the field just outside is $\frac{\sigma}{\varepsilon_0}$, inside 0, and the element feels the average, $\frac{\sigma}{2\varepsilon_0}$).
The net push on one hemisphere is this pressure times its projected area on the dividing plane:
$F = \frac{\sigma^2}{2\varepsilon_0}\cdot\pi R^2$.
The hemispheres must be pressed together with this force, so $F \propto \frac{1}{\varepsilon_0}\sigma^2R^2$.
Answer: (A). | ||||
| 45 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics For a diatomic ideal gas $\gamma = \frac{7}{5}$, and in an adiabatic process $TV^{\gamma-1}$ is constant.
$T_f = T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1} = T_i\,(32)^{2/5} = T_i\,(2^5)^{2/5} = 4T_i$.
Answer: 4. | ||||
| 46 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The two 2 MΩ resistors in parallel give $R = 1$ MΩ; the two 2 μF capacitors in parallel give $C = 4$ μF.
Time constant $\tau = RC = 10^6\times4\times10^{-6} = 4$ s.
Charging: $V_C = 10\left(1 - e^{-t/4}\right) = 4$, so $e^{-t/4} = 0.6 = \frac{3}{5}$.
$t = 4\ln\frac{5}{3} = 4(1.6 - 1.1) = 2$ s.
Answer: 2. | ||||
| 47 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Convex mirror: $f = +10$ m (images behind the mirror are at positive v). $\frac{1}{u} = \frac{1}{f} - \frac{1}{v}$.
$v = \frac{25}{3}$: $\frac{1}{u} = \frac{1}{10} - \frac{3}{25} = -\frac{1}{50}$, $u = -50$ m.
$v = \frac{50}{7}$: $\frac{1}{u} = \frac{1}{10} - \frac{7}{50} = -\frac{1}{25}$, $u = -25$ m.
The object moves 25 m in 30 s: speed $= \frac{5}{6}$ m/s $= \frac{5}{6}\times3.6 = 3$ km/h.
Answer: 3. | ||||
| 48 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Light escapes through the top face only where its angle of incidence is below the critical angle C: $\sin C = \frac{1}{\mu} = \frac{3}{5}$, so $\tan C = \frac{3}{4}$.
The source is at the bottom face, 8 cm below the top, so the circle has radius $R = 8\tan C = 8\times\frac{3}{4} = 6$ cm.
Answer: 6. | ||||
| 49 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\left|\frac{dN}{dt}\right| = \lambda N_0e^{-\lambda t}$, so $\ln\left|\frac{dN}{dt}\right| = \ln(\lambda N_0) - \lambda t$: a straight line of slope $-\lambda$.
From the graph, the line passes through (4 years, 4) and (6 years, 3): slope $= -\frac{1}{2}$, so $\lambda = 0.5$ per year.
After 4.16 years, $\frac{N_0}{N} = e^{\lambda t} = e^{0.5\times4.16} = e^{2.08} \approx 8$ (since $\ln 8 = 2.079$).
p = 8.
Answer: 8. | ||||
| 50 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | ComprehensionMedium | |
A medium comprehension question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The drop hangs from the rim of the opening, a circle of radius r. Surface tension acts along the drop's surface at the rim, tangent to the sphere of radius R, with total magnitude $T\cdot2\pi r$.
At the rim, the radius of the sphere to the contact circle makes an angle $\theta$ with the vertical where $\sin\theta = \frac{r}{R}$. The tangent there makes the same angle $\theta$ with the horizontal, so the vertical component of the pull is $T\cdot2\pi r\sin\theta$ (the horizontal components cancel round the circle).
$F = 2\pi rT\cdot\frac{r}{R} = \frac{2\pi r^2T}{R}$.
Answer: (C). | ||||
| 51 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | ComprehensionMedium | |
A medium comprehension question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The drop detaches when the weight reaches the surface-tension force:
$\frac{4}{3}\pi R^3\rho g = \frac{2\pi r^2T}{R}$, so $R^4 = \frac{3r^2T}{2\rho g}$.
$R^4 = \frac{3\times25\times10^{-8}\times0.11}{2\times10^3\times10} = \frac{8.25\times10^{-8}}{2\times10^4} = 4.125\times10^{-12}\ \mathrm{m^4}$.
$R = (4.125)^{1/4}\times10^{-3} \approx 1.43\times10^{-3}$ m, i.e. about $1.4\times10^{-3}$ m.
Answer: (A). | ||||
| 52 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | ComprehensionEasy | |
A easy comprehension 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 free drop has one surface of area $4\pi R^2$; its surface energy is $T\cdot4\pi R^2$.
With $R \approx 1.4\times10^{-3}$ m: $E = 0.11\times4\pi\times(1.4\times10^{-3})^2 = 0.11\times4\pi\times1.96\times10^{-6} = 2.7\times10^{-6}$ J.
(With the unrounded $R = 1.43\times10^{-3}$ m it is $2.8\times10^{-6}$ J; the nearest option is still (B).)
Answer: (B). | ||||
| 53 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | ComprehensionEasy | |
A easy comprehension question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Bohr's condition: angular momentum $L = I\omega = \frac{nh}{2\pi}$.
Rotational energy $E = \frac{1}{2}I\omega^2 = \frac{L^2}{2I} = \frac{n^2h^2}{4\pi^2\cdot2I} = n^2\left(\frac{h^2}{8\pi^2I}\right)$.
Answer: (D). | ||||
| 54 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | ComprehensionMedium | |
A medium comprehension question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei With $E_n = \frac{n^2h^2}{8\pi^2I}$ and $n = 0$ not allowed, the ground state is $n = 1$ and the first excited state $n = 2$:
$h\nu = E_2 - E_1 = \frac{3h^2}{8\pi^2I}$, so $I = \frac{3h}{8\pi^2\nu}$.
$I = \frac{3\times2\pi\times10^{-34}}{8\pi^2\times\frac{4}{\pi}\times10^{11}} = \frac{6\pi\times10^{-34}}{32\pi\times10^{11}} = \frac{6}{32}\times10^{-45} = 1.875\times10^{-46}\ \mathrm{kg\,m^2}$.
Answer: (B). | ||||
| 55 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | ComprehensionMedium | |
A medium comprehension question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion About the centre of mass, $I = \mu d^2$ with the reduced mass $\mu = \frac{m_Cm_O}{m_C + m_O} = \frac{12\times16}{28} = \frac{48}{7}$ a.m.u. $= \frac{48}{7}\times\frac{5}{3}\times10^{-27} = \frac{80}{7}\times10^{-27} = 1.143\times10^{-26}$ kg.
Using $I = 1.875\times10^{-46}\ \mathrm{kg\,m^2}$ from the previous question:
$d = \sqrt{\frac{I}{\mu}} = \sqrt{\frac{1.875\times10^{-46}}{1.143\times10^{-26}}} = \sqrt{1.64\times10^{-20}} = 1.28\times10^{-10}$ m $\approx 1.3\times10^{-10}$ m.
Answer: (C). | ||||
| 56 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Matrix matchHard | |
A hard matrix match question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics In every figure the ray comes from the right in $\mu_1$, crosses the lens ($\mu_2$) and leaves into $\mu_3$ on the left. At each surface, a ray entering a denser medium bends towards the normal, and entering a rarer one bends away. No bending at a surface means the indices on its two sides are equal.
First surface ($\mu_1 \to \mu_2$):
(p) convex surface, ray bends towards the axis (converges): towards the normal, so $\mu_2 > \mu_1$.
(r) concave surface, ray bends away from the axis (diverges): for a concave face that is again bending towards the normal, so $\mu_2 > \mu_1$.
(q) convex surface but the ray diverges: $\mu_2 < \mu_1$.
(s), (t) concave surface but the ray converges strongly: $\mu_2 < \mu_1$.
Second surface ($\mu_2 \to \mu_3$):
(p), (r), (t): the ray goes straight on, so $\mu_2 = \mu_3$.
(q), (s): at the flat face the ray bends further away from the normal on leaving, so it enters a rarer medium: $\mu_2 > \mu_3$.
A ($\mu_1 < \mu_2$): p, r. B ($\mu_1 > \mu_2$): q, s, t. C ($\mu_2 = \mu_3$): p, r, t. D ($\mu_2 > \mu_3$): q, s. | ||||
| 57 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Matrix matchHard | |
A hard matrix match question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At 50 Hz: $X_L = 2\pi\times50\times6\times10^{-3} = 1.88\ \Omega$; $X_C = \frac{1}{2\pi\times50\times3\times10^{-6}} = 1061\ \Omega$.
(p) DC, L in series with C: in the steady state no current flows (the capacitor blocks DC), the ideal inductor has no voltage, and the whole V sits on the capacitor: $I = 0$, $V_1 = 0$, $V_2 = V$. Only C.
(q) DC, L with 2 Ω: steady current $I = \frac{V}{2}$; the inductor has $V_1 = 0$ and the resistor $V_2 = 2I = V$. So C ($V_1 = 0$, $V_2 = V$), B ($V_2 > V_1$ with $I \ne 0$) and D ($V_2 \propto I$). A is not counted: $V_1$ is identically zero.
(r) AC, L with 2 Ω: $V_1 = 1.88I$, $V_2 = 2I$. Both proportional to I (A, D) and $V_2 > V_1$ (B).
(s) AC, L with C: $V_1 = 1.88I$, $V_2 = 1061I$: A, B, D.
(t) AC, 1 kΩ with C: $V_1 = 1000I$, $V_2 = 1061I$: A, B, D.
A→r, s, t; B→q, r, s, t; C→p, q; D→q, r, s, t. | ||||
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