JEE (Advanced) · 2011
JEE Advanced 2011, 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
43 Q
- Coordination Compounds4
- Structure of Atom3
- Principles of Qualitative Analysis3
19 easy · 18 medium · 6 hard
Physics
43 Q
- Oscillations and Waves7
- Electrostatics4
- Properties of Bulk Matter4
14 easy · 25 medium · 4 hard
Mathematics
43 Q
- Conic Sections5
- Integral Calculus4
- Matrices and Determinants4
10 easy · 29 medium · 4 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 11: Oscillations; Class 11: Waves
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 11: Laws of Motion
Class 11: Systems of Particles and Rotational Motion
Class 11: Thermodynamics
Class 12: Atoms; Class 12: Nuclei
Class 12: Current Electricity
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Class 11: Units and Measurement
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 11: Gravitation
No matching NCERT chapter
Class 11: Work, Energy and Power
Class 12: Dual Nature of Radiation and Matter
Class 11: Conic Sections
Class 12: Integrals; Class 12: Application of Integrals
Class 12: Matrices; Class 12: Determinants
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 11: Complex Numbers and Quadratic Equations
Class 12: Vector Algebra
Class 11: Complex Numbers and Quadratic Equations
Class 11: Sequences and Series
Class 11: Probability; Class 12: Probability
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 12: Differential Equations
Class 11: Conic Sections
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Class 12: Application of Derivatives
Class 11: Straight Lines
Class 12: Coordination Compounds
Class 11: Structure of Atom
No matching NCERT chapter
Class 11: Redox Reactions
Class 11: Organic Chemistry – Some Basic Principles and Techniques
No matching NCERT chapter
Class 12: Solutions
Class 12: Electrochemistry
Class 12: Aldehydes, Ketones and Carboxylic Acids
Class 12: Amines
Class 11: Some Basic Concepts of Chemistry
Class 12: Biomolecules
Class 12: The d- and f-Block Elements
No matching NCERT chapter
No matching NCERT chapter
Class 11: Chemical Bonding and Molecular Structure
Class 11: Hydrocarbons
Class 12: Alcohols, Phenols and Ethers
Class 12: Haloalkanes and Haloarenes
Class 12: Chemical Kinetics
No matching NCERT chapter
Class 11: Equilibrium
No matching NCERT chapter
Class 11: Thermodynamics
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 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Very pure nitrogen is made by the thermal decomposition of sodium or barium azide, which gives only the metal and nitrogen:
$\mathrm{Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2}$
(A) $\mathrm{2NH_3 + 3CuO \rightarrow 3Cu + N_2 + 3H_2O}$ gives nitrogen mixed with water vapour and unreacted ammonia.
(B) $\mathrm{NH_4NO_3 \rightarrow N_2O + 2H_2O}$ gives nitrous oxide, not nitrogen.
(C) $\mathrm{(NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + 4H_2O}$ gives nitrogen with water vapour (and some oxides of nitrogen and ammonia), so it is not extra pure.
Answer: (D). | ||||
| 2 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{Ni^{2+}}$ is $3d^8$.
With $\mathrm{Cl^-}$ (weak field, large ligand): $\mathrm{[NiCl_4]^{2-}}$, the 3d electrons stay unpaired and Ni uses $sp^3$ — tetrahedral (paramagnetic).
With $\mathrm{CN^-}$ (strong field): the 3d electrons pair up, one 3d orbital is free and Ni uses $dsp^2$ — $\mathrm{[Ni(CN)_4]^{2-}}$ is square planar (diamagnetic).
With $\mathrm{H_2O}$: $\mathrm{[Ni(H_2O)_6]^{2+}}$, $sp^3d^2$ — octahedral.
Tetrahedral, square planar, octahedral.
Answer: (B). | ||||
| 3 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Balance mass number and atomic number at each step ($\alpha = {}^4_2\mathrm{He}$).
(i) ${}^{27}_{13}\mathrm{Al} + {}^4_2\mathrm{He} \rightarrow {}^{30}_{14}\mathrm{Si} + \mathrm{X}$: mass $31 - 30 = 1$, charge $15 - 14 = 1$, so X is ${}^1_1\mathrm{H}$, a proton.
(ii) ${}^{27}_{13}\mathrm{Al} + {}^4_2\mathrm{He} \rightarrow {}^{30}_{15}\mathrm{P} + \mathrm{Y}$: mass $1$, charge $0$, so Y is ${}^1_0n$, a neutron.
${}^{30}_{15}\mathrm{P} \rightarrow {}^{30}_{14}\mathrm{Si} + \mathrm{Z}$: mass $0$, charge $+1$, so Z is ${}^{\;0}_{+1}e$, a positron.
Proton, neutron, positron.
Answer: (A). | ||||
| 4 | SolutionsConcentration terms | Class 12: Solutions | Single correctEasy | |
A easy single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Moles of urea $= \frac{120}{60} = 2$.
Mass of solution $= 120 + 1000 = 1120$ g, so its volume $= \frac{1120}{1.15} = 973.9$ mL $= 0.9739$ L.
Molarity $= \frac{2}{0.9739} = 2.05$ M.
Answer: (C). | ||||
| 5 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Before the equivalence point: $\mathrm{K^+ + Cl^- + Ag^+ + NO_3^- \rightarrow AgCl\downarrow + K^+ + NO_3^-}$. Each $\mathrm{Cl^-}$ removed as AgCl is replaced by an $\mathrm{NO_3^-}$, whose ionic conductance is nearly the same, so the conductance stays almost constant.
After the equivalence point: the added $\mathrm{Ag^+}$ and $\mathrm{NO_3^-}$ stay in solution, so the number of ions and the conductance rise steadily.
The graph that is flat and then rises is (S). (P) falls first, which would need a fast ion (such as $\mathrm{H^+}$ or $\mathrm{OH^-}$) being replaced; (Q) and (R) do not show a flat first part followed by a rise.
Answer: (D). | ||||
| 6 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The three carboxylic acids are far stronger than $p$-nitrophenol ($\mathrm{p}K_a \approx 7.1$).
Among the acids, $o$-hydroxybenzoic (salicylic) acid is the strongest ($\mathrm{p}K_a \approx 3.0$): in its anion the ortho OH forms an intramolecular hydrogen bond to $\mathrm{COO^-}$, stabilising the conjugate base (ortho effect).
$p$-Hydroxybenzoic acid ($\mathrm{p}K_a \approx 4.6$) and $p$-toluic acid ($\mathrm{p}K_a \approx 4.4$) are weakened by the electron-donating OH (+R) and $\mathrm{CH_3}$ (+I) groups.
Answer: (C). | ||||
| 7 | 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 (i) KOH removes the acidic N–H of phthalimide, giving potassium phthalimide, a good nitrogen nucleophile (the first step of the Gabriel synthesis).
(ii) The phthalimide anion attacks the benzylic $\mathrm{CH_2Cl}$ carbon by $S_N2$; benzylic chloride is very reactive, while the aryl C–Br bond does not undergo nucleophilic substitution under these conditions. Alkylation occurs at nitrogen.
The product is N-(4-bromobenzyl)phthalimide, structure (A). (B) and (D) would need substitution at the aryl carbon, and (C) and (D) are O-alkylated products.
Answer: (A). | ||||
| 8 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Cassiterite (tinstone) is $\mathrm{SnO_2}$.
(A) True: after concentration it is reduced with carbon: $\mathrm{SnO_2 + 2C \rightarrow Sn + 2CO}$.
(B) False: it is an oxide ore, not a sulphide, and self-reduction is used for sulphide ores of Cu, Pb, Hg.
(C) False: copper is not the impurity removed in this extraction.
(D) True: the chief impurity is wolframite, $\mathrm{FeWO_4}$/$\mathrm{MnWO_4}$, which is magnetic and is removed by magnetic separation — the removal of the iron impurity.
Answer: (A), (D). | ||||
| 9 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter (A) True: the adsorbed gas loses freedom, so $\Delta S < 0$; since adsorption is spontaneous ($\Delta G = \Delta H - T\Delta S < 0$), $\Delta H$ must be negative.
(B) True: a gas physisorbed at low temperature can, on heating, acquire the activation energy to form chemical bonds with the surface (for example $\mathrm{H_2}$ on nickel).
(C) False: physisorption decreases with rising temperature (it is exothermic and weak); chemisorption first increases (activation energy) and then decreases.
(D) True: enthalpy of chemisorption is about 80–240 kJ/mol against 20–40 kJ/mol for physisorption, and it needs an activation energy, so it can be slow.
Answer: (A), (B), (D). | ||||
| 10 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Multiple correctMedium | |
A medium multiple correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry (A) True: a postulate of the kinetic theory is that collisions between molecules and with the walls are perfectly elastic.
(B) Not a statement of the theory: the momentum delivered in a collision is $2mv$, which depends on both mass and speed; at a given temperature all gases have the same mean kinetic energy, so lighter molecules move faster. The key does not accept it.
(C) True: by the Maxwell distribution, only a small fraction of molecules has very high speeds; most are near the most probable speed.
(D) True: there are no forces between molecules except during collisions, so between collisions they move in straight lines at constant velocity.
Answer: (A), (C), (D). | ||||
| 11 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Multiple correctMedium | |
A medium multiple correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) Buta-1,3-diene: rotation about the central C2–C3 single bond gives s-cis and s-trans (planar) but also skew conformations in which the two vinyl groups are not coplanar. Not planar in all conformations.
(B) $\mathrm{H{-}C{\equiv}C{-}CH{=}CH_2}$: the $sp$ carbons and their H lie on one straight line, which is the axis of the only single bond that can rotate. Rotation about it moves nothing out of the plane of the vinyl group, so all atoms stay in one plane in every conformation. True.
(C) Ketene $\mathrm{H_2C{=}C{=}O}$: the $sp^2$ $\mathrm{CH_2}$ lies in the plane, and C=C=O is linear on that plane. All atoms coplanar, and there is no rotatable single bond. True.
(D) Allene $\mathrm{H_2C{=}C{=}CH_2}$: the two $\pi$ bonds at the central $sp$ carbon are perpendicular, so the two terminal $\mathrm{CH_2}$ planes are perpendicular. Not planar.
Answer: (B), (C). | ||||
| 12 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | ComprehensionEasy | |
A easy comprehension question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter A light blue solution means hydrated $\mathrm{Cu^{2+}}$. A metal rod displacing a less reactive metal from a colourless solution: $\mathrm{Cu + 2AgNO_3 \rightarrow Cu(NO_3)_2 + 2Ag}$.
The excess $\mathrm{Ag^+}$ then gives white AgCl with NaCl, which dissolves in ammonia, while $\mathrm{Cu^{2+}}$ gives the intense blue $\mathrm{[Cu(NH_3)_4]^{2+}}$.
Fe, Ni and Co would give pale green or pink solutions, not blue.
M is copper.
Answer: (B). | ||||
| 13 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | ComprehensionEasy | |
A easy comprehension question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Copper can displace only a metal below it in the electrochemical series: of the four, only silver ($E^\circ_{\mathrm{Ag^+/Ag}} = +0.80$ V against $+0.34$ V for Cu). Zn, Al and Pb are more reactive than Cu and are not displaced.
The white precipitate O with NaCl is AgCl (from the excess $\mathrm{Ag^+}$), and it dissolves in aqueous ammonia: $\mathrm{AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-}$. ($\mathrm{PbCl_2}$, also white, does not dissolve in ammonia.)
N is $\mathrm{AgNO_3}$.
Answer: (A). | ||||
| 14 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | ComprehensionEasy | |
A easy comprehension question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds From Q.12 and Q.13, M = Cu and N = $\mathrm{AgNO_3}$. The blue solution holds $\mathrm{Cu^{2+}}$ and the remaining $\mathrm{Ag^+}$; AgCl is precipitated by NaCl.
Aqueous ammonia: $\mathrm{AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-}$ and $\mathrm{Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+}}$ (intense blue).
The final solution contains $\mathrm{[Ag(NH_3)_2]^+}$ and $\mathrm{[Cu(NH_3)_4]^{2+}}$.
Answer: (C). | ||||
| 15 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | ComprehensionHard | |
A hard comprehension question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Work backwards. Two acetones from ozonolysis means the alkene before ozonolysis was $\mathrm{(CH_3)_2C{=}C(CH_3)_2}$ (2,3-dimethylbut-2-ene).
That alkene comes from acid dehydration of the alcohol Q; a six-carbon alcohol that gives it after a methyl shift is 3,3-dimethylbutan-2-ol, $\mathrm{(CH_3)_3C{-}CH(OH){-}CH_3}$: the secondary carbocation rearranges by a 1,2-methyl shift to the tertiary cation $\mathrm{(CH_3)_2C^+{-}CH(CH_3)_2}$, which loses $\mathrm{H^+}$ to give the tetrasubstituted alkene.
Q comes from $\mathrm{NaBH_4}$ reduction of the ketone $\mathrm{(CH_3)_3C{-}CO{-}CH_3}$, which is the Markovnikov hydration product ($\mathrm{HgSO_4}$/dil. $\mathrm{H_2SO_4}$) of the terminal alkyne $\mathrm{(CH_3)_3C{-}C{\equiv}CH}$.
P is 3,3-dimethylbut-1-yne, $\mathrm{C_6H_{10}}$.
Check others: (A) would give hexan-2-ol and then hex-2-ene, whose ozonolysis gives aldehydes; (B) and (C) give ketones whose alcohols do not dehydrate to an alkene giving only acetone.
Answer: (D). | ||||
| 16 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | ComprehensionMedium | |
A medium comprehension question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers From Q.15, P is $\mathrm{(CH_3)_3C{-}C{\equiv}CH}$.
(i) dil. $\mathrm{H_2SO_4/HgSO_4}$: Markovnikov hydration of the terminal alkyne gives the methyl ketone $\mathrm{(CH_3)_3C{-}CO{-}CH_3}$.
(ii) $\mathrm{NaBH_4}$, (iii) dil. acid: reduction to the secondary alcohol $\mathrm{(CH_3)_3C{-}CH(OH){-}CH_3}$ (3,3-dimethylbutan-2-ol).
Check: with conc. $\mathrm{H_2SO_4}$ it forms a secondary cation, a methyl shift gives the tertiary cation, and loss of $\mathrm{H^+}$ gives $\mathrm{(CH_3)_2C{=}C(CH_3)_2}$, whose ozonolysis gives two acetones.
Answer: (B). | ||||
| 17 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions In the tetrathionate ion $\mathrm{[O_3S{-}S{-}S{-}SO_3]^{2-}}$ the two central S atoms are bonded only to sulphur, so their oxidation number is 0.
For the whole ion, the four S atoms together carry $-2 - 6 \times (-2) = +10$. The two bridging S contribute 0, so each terminal $\mathrm{SO_3}$ sulphur is $+5$.
Difference $= 5 - 0 = 5$.
Answer: 5. | ||||
| 18 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Bromine disproportionates: Br goes from 0 to $-1$ in NaBr (gain of 1 electron) and from 0 to $+5$ in $\mathrm{NaBrO_3}$ (loss of 5). So 5 NaBr form for every $\mathrm{NaBrO_3}$:
$\mathrm{3Br_2 + 3Na_2CO_3 \rightarrow 5NaBr + NaBrO_3 + 3CO_2}$
Check: Br $6 = 5 + 1$; Na $6 = 5 + 1$; C 3; O $9 = 3 + 6$.
Answer: 5. | ||||
| 19 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom The shell $n = 3$ has $n^2 = 9$ orbitals (one 3s, three 3p, five 3d), holding $2n^2 = 18$ electrons.
Each orbital holds one electron with $m_s = +\frac{1}{2}$ and one with $m_s = -\frac{1}{2}$, so 9 electrons can have $m_s = -\frac{1}{2}$.
Answer: 9. | ||||
| 20 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Photon energy $E = \frac{hc}{\lambda} = \frac{1240\ \mathrm{eV\,nm}}{300\ \mathrm{nm}} \approx 4.13$ eV.
Photoemission needs $\phi < 4.13$ eV: Li (2.4), Na (2.3), K (2.2) and Mg (3.7). Cu (4.8), Ag (4.3), Fe (4.7), Pt (6.3) and W (4.75) all have larger work functions.
Number of metals $= 4$.
Answer: 4. | ||||
| 21 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalHard | |
A hard numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry The piston keeps the total pressure at 1 atm.
If all 1.0 mol of the compound were vapour, its partial pressure would be $\frac{1.0}{1.1} \times 1 = 0.91$ atm, more than its vapour pressure 0.68 atm. So it cannot all evaporate: liquid remains and the vapour is saturated at 0.68 atm.
The partial pressure of He is then $1 - 0.68 = 0.32$ atm, and He fills the whole gas volume:
$V = \frac{nRT}{p_{\mathrm{He}}} = \frac{0.1 \times 0.0821 \times 273}{0.32} = \frac{2.24}{0.32} = 7.0$ L.
Answer: 7. | ||||
| 22 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | NumericalHard | |
A hard numerical question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes C3 carries Br, a cyclopentyl group, an ethyl group (C1–C2) and a propyl group (C4–C6). There are three kinds of $\beta$-hydrogen.
1. From C2: $\mathrm{CH_3{-}CH{=}C(C_5H_9){-}CH_2CH_2CH_3}$ (3-cyclopentylhex-2-ene). C2 has H and $\mathrm{CH_3}$, C3 has cyclopentyl and propyl: cis and trans — 2 alkenes.
2. From C4: $\mathrm{CH_3CH_2{-}C(C_5H_9){=}CH{-}CH_2CH_3}$ (3-cyclopentylhex-3-ene). C3 has ethyl and cyclopentyl, C4 has H and ethyl: cis and trans — 2 alkenes.
3. From the ring C–H: (hexan-3-ylidene)cyclopentane, with an exocyclic double bond. The ring carbon carries two equivalent ring arms, so no geometrical isomers — 1 alkene.
Total $= 2 + 2 + 1 = 5$.
Answer: 5. | ||||
| 23 | BiomoleculesProteins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules A decapeptide has 9 peptide bonds, and hydrolysis adds one water molecule to each.
Mass of the hydrolysed products $= 796 + 9 \times 18 = 958$.
Mass of glycine $= 0.47 \times 958 = 450.3$, so the number of glycine units $= \frac{450.3}{75} \approx 6$.
Answer: 6. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 24 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v_s = 36$ km/h $= 10$ m/s, $v = 320$ m/s.
The building receives (source moving towards it): $f' = f\frac{v}{v - v_s} = 8 \times \frac{320}{310}$ kHz.
The building re-emits $f'$; the driver moves towards it as an observer: $f'' = f'\frac{v + v_s}{v} = 8 \times \frac{330}{310} = 8.52$ kHz $\approx 8.50$ kHz.
Answer: (A). | ||||
| 25 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $n = \frac{5.6}{22.4} = \frac{1}{4}$ mol; helium is monatomic, $\gamma = \frac{5}{3}$, $C_V = \frac{3}{2}R$.
Adiabatic: $T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma - 1} = T_1 (8)^{2/3} = 4T_1$.
Work done on the gas $= \Delta U = nC_V(T_2 - T_1) = \frac{1}{4}\cdot\frac{3}{2}R\cdot 3T_1 = \frac{9}{8}RT_1$ (the work done by the gas is $-\frac{9}{8}RT_1$).
Answer: (A). | ||||
| 26 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The shaded rectangle has corners $(0,0,0)$, $(0,a,0)$, $(a,a,a)$ and $(a,0,a)$. Two edges from the origin are $\vec{u} = a\hat{j}$ and $\vec{w} = a\hat{i} + a\hat{k}$.
Area vector $\vec{A} = \vec{u}\times\vec{w} = a\hat{j}\times(a\hat{i} + a\hat{k}) = a^2(-\hat{k} + \hat{i})$, of magnitude $\sqrt{2}a^2$.
Flux $= \vec{E}\cdot\vec{A} = E_0a^2$. (Equivalently, only the projection of the area on the $yz$-plane, an $a \times a$ square, counts.)
Answer: (C). | ||||
| 27 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.
Hydrogen, first Balmer line ($3 \to 2$): $\frac{1}{6561} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$.
$\mathrm{He^+}$ ($Z = 2$), second Balmer line ($4 \to 2$): $\frac{1}{\lambda} = 4R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{4}$.
$\lambda = 6561 \times \frac{5/36}{3/4} = 6561 \times \frac{5}{27} = 1215$ Å.
Answer: (A). | ||||
| 28 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion A conical pendulum: the string makes angle $\theta$ with the vertical and the circle has radius $r = L\sin\theta$.
Horizontal: $T\sin\theta = m\omega^2 L\sin\theta$, so $T = m\omega^2L$, whatever the angle.
$\omega_{max} = \sqrt{\frac{T_{max}}{mL}} = \sqrt{\frac{324}{0.5 \times 0.5}} = \sqrt{1296} = 36$ rad/s.
Answer: (D). | ||||
| 29 | Current ElectricityMetre bridge | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity X is in the left gap (end A) and 10 Ω in the right gap (end B). Adding the end corrections to the balancing lengths:
$\frac{X}{10} = \frac{52 + 1}{(100 - 52) + 2} = \frac{53}{50}$
$X = 10.6\ \Omega$.
Answer: (B). | ||||
| 30 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In position 1 the $2\,\mu$F capacitor charges to $V$: $U_i = \frac{1}{2}(2)V^2 = V^2$ (in $\mu$J), charge $Q = 2V$.
In position 2 it shares charge with the uncharged $8\,\mu$F capacitor: common potential $V' = \frac{2V}{2 + 8} = \frac{V}{5}$.
$U_f = \frac{1}{2}(10)\left(\frac{V}{5}\right)^2 = \frac{V^2}{5}$.
Dissipated: $V^2 - \frac{V^2}{5} = \frac{4}{5}V^2$, i.e. $80\%$ (in general $\frac{C_2}{C_1 + C_2} = \frac{8}{10}$).
Answer: (D). | ||||
| 31 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance After connection both are at one potential: $\frac{kQ_A}{R_A} = \frac{kQ_B}{R_B}$, so $\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$.
(A) True: the field inside a conductor (and inside a charged shell with no charge within) is zero.
(B) True: $R_A > R_B$, so $Q_A > Q_B$.
(C) True: $\frac{\sigma_A}{\sigma_B} = \frac{Q_A/R_A^2}{Q_B/R_B^2} = \frac{R_A}{R_B}\cdot\frac{R_B^2}{R_A^2} = \frac{R_B}{R_A}$.
(D) True: $E = \frac{\sigma}{\varepsilon_0}$ at the surface, so $\frac{E_A}{E_B} = \frac{R_B}{R_A} < 1$.
Answer: (A), (B), (C), (D). | ||||
| 32 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctMedium | |
A medium multiple 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 Entering perpendicular to the boundary and to $\vec{B}$, each particle moves on a semicircle (radius $r = \frac{mv}{|q|B}$) and leaves the field region moving opposite to its entry velocity.
(A) False: both describe a semicircle and come out.
(B) True: both come out with velocity $-\vec{v}$, so their paths are parallel (they turn in opposite senses because their charges are opposite).
(C) and (D): the time inside is half a period, $t = \frac{\pi m}{|q|B}$. With equal $|q|$ but $m_p \approx 1836\,m_e$, the electron comes out much sooner, so (D) is true and (C) false.
The official key accepts BC, BD or BCD.
Answer: (B), (D) by our solution; the key's first set is (B), (C). | ||||
| 33 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctMedium | |
A medium multiple correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter From the figure (width $w$ into the page): A is $1L$ thick and $4L$ high with $2K$; B, C, D are $4L$ long (from $1L$ to $5L$) with heights $1L$, $2L$, $1L$ and conductivities $3K$, $4K$, $5K$; E is $1L$ thick and $4L$ high with $6K$.
Thermal resistance $R = \frac{\ell}{kA}$:
$R_A = \frac{L}{2K \cdot 4Lw} = \frac{1}{8Kw}$, $R_E = \frac{L}{6K\cdot 4Lw} = \frac{1}{24Kw}$,
$R_B = \frac{4L}{3K\cdot Lw} = \frac{4}{3Kw}$, $R_C = \frac{4L}{4K\cdot 2Lw} = \frac{1}{2Kw}$, $R_D = \frac{4L}{5K\cdot Lw} = \frac{4}{5Kw}$.
B, C, D are in parallel: conductances $\frac{3}{4} + 2 + \frac{5}{4} = 4$ (in $Kw$), so their combination is $\frac{1}{4Kw}$; this is in series with A and E.
(A) True: in series, the same heat current flows through A, the middle group and E.
(B) False: E carries the same heat current as A, not more.
(C) True: with the same current, $\Delta T \propto R$; $R_E = \frac{1}{24Kw}$ is the smallest of $R_A$, $R_{BCD}$ and $R_E$.
(D) True: the middle current divides in proportion to conductance $\frac{3}{4} : 2 : \frac{5}{4}$, i.e. B $= \frac{3}{16}Q$, C $= \frac{1}{2}Q$, D $= \frac{5}{16}Q$, and $\frac{3}{16} + \frac{5}{16} = \frac{1}{2}$.
Answer: (A), (C), (D). | ||||
| 34 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Multiple correctMedium | |
A medium multiple correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Restoring torque about the pivot at angle $\theta$: $\tau = \left(m\frac{L}{2} + ML\right)g\sin\theta$ in both cases, since it depends only on the masses and positions. (A) true, (B) false.
Moment of inertia about the pivot:
Case A (disc fixed to the rod, so it rotates with it): $I_A = \frac{mL^2}{3} + ML^2 + \frac{MR^2}{2}$.
Case B (disc free on a frictionless axle): no torque acts on the disc about its centre, so it does not rotate; it moves like a point mass, $I_B = \frac{mL^2}{3} + ML^2$.
$\omega = \sqrt{\frac{(mL/2 + ML)g}{I}}$ and $I_A > I_B$, so $\omega_A < \omega_B$. (C) false, (D) true.
Answer: (A), (D). | ||||
| 35 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | ComprehensionMedium | |
A medium comprehension question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Take the ground as $x = 0$ and up as positive. The ball starts at $x = 0$ with momentum $+p_0$.
Going up, $x > 0$ and $p$ decreases to 0 at the top ($x = H$). Coming down, $x$ stays positive and $p$ becomes negative, reaching $-p_0$ when the ball is back at $x = 0$.
From $p^2 = p_0^2 - 2m^2gx$, the curve is a parabola lying entirely in $x \ge 0$, traced from $(0, +p_0)$ through $(H, 0)$ to $(0, -p_0)$ — the half-loop on the positive position side with clockwise arrows, figure (D).
Answer: (D). | ||||
| 36 | 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 The point where a phase-space circle cuts the position axis is the amplitude (momentum zero there). From the figure, $A_1 = 2a$ and $A_2 = a$.
For the same oscillator $E = \frac{1}{2}m\omega^2A^2$, so $\frac{E_1}{E_2} = \left(\frac{2a}{a}\right)^2 = 4$.
Answer: (C). | ||||
| 37 | 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 Two features fix the diagram.
Direction: start at the right extreme, $x > 0$, $p = 0$. The spring force points left, so $p$ becomes negative while $x$ decreases: the point moves down and to the left — clockwise in the position–momentum plane.
Damping: the water's viscous drag removes energy each cycle, so the amplitude (and the size of the loop) keeps shrinking — the curve spirals inwards.
Only (B) starts at the outer point on the positive position axis, goes clockwise (down first) and closes the turn at a smaller radius.
Answer: (B). | ||||
| 38 | 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 From Coulomb's law $F = \frac{e^2}{4\pi\varepsilon_0 r^2}$, the combination $\frac{e^2}{\varepsilon_0}$ has the dimensions of force $\times$ length$^2$: $[\mathrm{M L^3 T^{-2}}]$.
$N$ is a number per volume, $[\mathrm{L^{-3}}]$. Then
$\left[\frac{Ne^2}{m\varepsilon_0}\right] = \frac{[\mathrm{L^{-3}}][\mathrm{M L^3 T^{-2}}]}{[\mathrm{M}]} = [\mathrm{T^{-2}}]$,
so $\sqrt{\frac{Ne^2}{m\varepsilon_0}}$ has the dimensions of angular frequency. Options with $e$ to the first power do not work, and (D) is its inverse.
Answer: (C). | ||||
| 39 | 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 $\omega_p = \sqrt{\frac{Ne^2}{m\varepsilon_0}} = \sqrt{\frac{4\times10^{27}\times(1.6\times10^{-19})^2}{10^{-30}\times10^{-11}}} = \sqrt{1.024\times10^{31}} \approx 3.2\times10^{15}$ rad/s.
$\lambda = \frac{2\pi c}{\omega_p} = \frac{2\pi\times3\times10^8}{3.2\times10^{15}} \approx 5.9\times10^{-7}$ m $\approx 600$ nm.
Answer: (B). | ||||
| 40 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalHard | |
A hard numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The stick touches the ring at the end of a horizontal diameter and pushes horizontally with normal force $N = 2$ N; friction from the stick, $f_s = \mu N$, acts vertically at that point. Ground friction $f$ acts horizontally at the bottom.
Translation: $N - f = ma = 2\times0.3 = 0.6$, so $f = 1.4$ N (opposing the push).
Rotation about the centre, $\alpha = \frac{a}{R} = 0.6$ rad/s$^2$, $I = mR^2 = 0.5$ kg m$^2$: net torque $= I\alpha = 0.3$ N m in the rolling sense.
Ground friction gives $fR = 1.4\times0.5 = 0.7$ N m in the rolling sense, so the stick's friction must give $0.7 - 0.3 = 0.4$ N m against it: $f_sR = 0.4$, $f_s = 0.8$ N.
$\mu = \frac{f_s}{N} = \frac{0.8}{2} = 0.4 = \frac{P}{10}$, so $P = 4$.
Answer: 4. | ||||
| 41 | Laws of MotionFriction | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Force (along the incline) to just push it up: $F_1 = mg(\sin\theta + \mu\cos\theta)$.
Force to just prevent sliding down: $F_2 = mg(\sin\theta - \mu\cos\theta)$.
At $45^\circ$, $\sin\theta = \cos\theta$: $1 + \mu = 3(1 - \mu)$, so $\mu = \frac{1}{2}$ and $N = 10\mu = 5$.
Answer: 5. | ||||
| 42 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Equilibrium of the charges: the electrostatic repulsion, $\propto \frac{q^2}{4\pi\varepsilon_0 a^2}$, is balanced by the surface-tension force of the film, $\propto \gamma a$ (tension per unit length times a length).
$\frac{q^2}{a^2} \propto \gamma a \Rightarrow a^3 \propto \frac{q^2}{\gamma}$, so $a = k\left[\frac{q^2}{\gamma}\right]^{1/3}$ (with $\varepsilon_0$ and numerical factors in $k$).
$N = 3$.
Answer: 3. | ||||
| 43 | 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 thermal contraction cancels the stretch due to the load:
$\frac{mgL}{AY} = L\alpha\Delta T$, so $mg = AY\alpha\Delta T$.
$A = \pi(10^{-3})^2 = \pi\times10^{-6}\ \mathrm{m^2}$: $mg = \pi\times10^{-6}\times10^{11}\times10^{-5}\times10 = 10\pi \approx 31.4$ N.
With $g = 10\ \mathrm{m/s^2}$, $m = 3.14$ kg $\approx 3$.
Answer: 3. | ||||
| 44 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Activity $A = \lambda N = \frac{N}{\tau}$, so $N = A\tau = 10^{10}\times10^9 = 10^{19}$ atoms.
Mass $= 10^{19}\times10^{-25} = 10^{-6}$ kg $= 1$ mg.
Answer: 1. | ||||
| 45 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalHard | |
A hard numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The current flows round the curved surface, so the tube acts as a solenoid with current per unit length $\frac{I}{10}$: inside, $B = \mu_0\frac{I}{10}$ along the axis.
Flux through the loop: $\Phi = B\pi r^2 = \mu_0\frac{I}{10}\pi(0.1)^2 = \pi\times10^{-3}\mu_0I$.
EMF: $\varepsilon = -\frac{d\Phi}{dt} = \pi\times10^{-3}\mu_0I_0\times300\sin(300t) = 0.3\pi\mu_0I_0\sin(300t)$.
Current: $i = \frac{\varepsilon}{R} = \frac{0.3\pi}{0.005}\mu_0I_0\sin(300t) = 60\pi\mu_0I_0\sin(300t)$.
Magnetic moment: $i\pi r^2 = 60\pi\times\pi(0.01)\mu_0I_0\sin(300t) = 0.6\pi^2\mu_0I_0\sin(300t)$.
$N = 0.6\pi^2 \approx 5.9 \approx 6$ (self-inductance of the loop neglected).
Answer: 6. | ||||
| 46 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Radius $r = \frac{\sqrt5}{2}$ cm, $r^2 = \frac{5}{4}$ cm$^2 = 1.25\times10^{-4}$ m$^2$.
Each sphere about a diameter: $\frac{2}{5}mr^2 = 0.4\times0.5\times1.25\times10^{-4} = 0.25\times10^{-4}$ kg m$^2$.
Two spheres lie on the diagonal: each contributes $0.25\times10^{-4}$.
The other two are at half the other diagonal from the axis: $d = \frac{4\sqrt2}{2} = 2\sqrt2$ cm, $d^2 = 8\times10^{-4}$ m$^2$. Each contributes $0.25\times10^{-4} + 0.5\times8\times10^{-4} = 4.25\times10^{-4}$.
Total $= 2(0.25) + 2(4.25) = 9\times10^{-4}$ kg m$^2$, so $N = 9$.
Answer: 9. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 47 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Take logarithms. Write $X = \ln x$, $Y = \ln y$, $a = \ln 2$, $b = \ln 3$.
First equation: $a(a + X) = b(b + Y)$.
Second equation: $bX = aY$, so $Y = \frac{bX}{a}$.
Substitute: $a^2 + aX = b^2 + \frac{b^2X}{a}$, i.e. $a^3 + a^2X = ab^2 + b^2X$, so $X(a^2 - b^2) = -a(a^2 - b^2)$.
Since $a \ne b$, $X = -a$: $\ln x_0 = -\ln 2$, $x_0 = \frac{1}{2}$ (and then $y_0 = \frac{1}{3}$).
Answer: (C). | ||||
| 48 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t = x^2$, $dt = 2x\,dx$: $I = \frac{1}{2}\int_{\ln 2}^{\ln 3} \frac{\sin t}{\sin t + \sin(\ln 6 - t)}\,dt = \frac{1}{2}J$.
Since $\ln 2 + \ln 3 = \ln 6$, replacing $t$ by $\ln 6 - t$ keeps the limits and gives $J = \int_{\ln 2}^{\ln 3} \frac{\sin(\ln 6 - t)}{\sin(\ln 6 - t) + \sin t}\,dt$.
Adding: $2J = \int_{\ln 2}^{\ln 3} 1\,dt = \ln\frac{3}{2}$, so $J = \frac{1}{2}\ln\frac{3}{2}$ and $I = \frac{1}{4}\ln\frac{3}{2}$.
Answer: (A). | ||||
| 49 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{v} = \lambda\vec{a} + \mu\vec{b} = (\lambda + \mu)\hat{i} + (\lambda - \mu)\hat{j} + (\lambda + \mu)\hat{k}$: its $\hat{i}$ and $\hat{k}$ components are equal.
Projection on $\vec{c}$: $\frac{\vec{v}\cdot\vec{c}}{|\vec{c}|} = \frac{(\lambda + \mu) - (\lambda - \mu) - (\lambda + \mu)}{\sqrt3} = \frac{\mu - \lambda}{\sqrt3} = \frac{1}{\sqrt3}$, so $\mu - \lambda = 1$.
Only (C) has equal $\hat{i}$ and $\hat{k}$ components: $\lambda + \mu = 3$, $\lambda - \mu = -1$ give $\lambda = 1$, $\mu = 2$, and $\mu - \lambda = 1$. ✓
Answer: (C). | ||||
| 50 | 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 $P$: $\sin\theta = (\sqrt2 + 1)\cos\theta$. Here $\cos\theta \ne 0$ (otherwise $\sin\theta = 0$ too), so $\tan\theta = \sqrt2 + 1$.
$Q$: $\cos\theta = (\sqrt2 - 1)\sin\theta$. Here $\sin\theta \ne 0$, so $\tan\theta = \frac{1}{\sqrt2 - 1} = \sqrt2 + 1$.
Both sets are exactly $\{\theta : \tan\theta = \sqrt2 + 1\}$, so $P = Q$.
Answer: (D). | ||||
| 51 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Total area $R_1 + R_2 = \int_0^1 (1 - x)^2\,dx = \frac{1}{3}$. With $R_1 - R_2 = \frac{1}{4}$: $R_1 = \frac{1}{2}\left(\frac{1}{3} + \frac{1}{4}\right) = \frac{7}{24}$.
$R_1 = \int_0^b (1 - x)^2\,dx = \frac{1 - (1 - b)^3}{3} = \frac{7}{24}$, so $(1 - b)^3 = \frac{1}{8}$, $1 - b = \frac{1}{2}$, $b = \frac{1}{2}$.
Answer: (B). | ||||
| 52 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha^2 = 6\alpha + 2$, so $\alpha^{10} = 6\alpha^9 + 2\alpha^8$; likewise $\beta^{10} = 6\beta^9 + 2\beta^8$.
Subtracting: $a_{10} = 6a_9 + 2a_8$, so $a_{10} - 2a_8 = 6a_9$ and $\frac{a_{10} - 2a_8}{2a_9} = 3$.
Answer: (C). | ||||
| 53 | Straight LinesSlope, parallel and perpendicular lines | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines $\sqrt3x + y = 1$ has slope $-\sqrt3$, inclination $120^\circ$. A line at $60^\circ$ to it has inclination $120^\circ \pm 60^\circ$: $60^\circ$ (slope $\sqrt3$) or $180^\circ$ (horizontal).
The horizontal line through $(3, -2)$ is $y = -2$, which never meets the $x$-axis. So the slope is $\sqrt3$:
$y + 2 = \sqrt3(x - 3)$, i.e. $y - \sqrt3x + 2 + 3\sqrt3 = 0$.
Answer: (B). | ||||
| 54 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Multiple correctMedium | |
A medium multiple correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Let $\vec{a} = (1,1,2)$, $\vec{b} = (1,2,1)$, $\vec{c} = (1,1,1)$. A vector in the plane of $\vec{a}, \vec{b}$ is perpendicular to $\vec{a}\times\vec{b}$; being also perpendicular to $\vec{c}$, it is parallel to $(\vec{a}\times\vec{b})\times\vec{c}$.
$\vec{a}\times\vec{b} = (1\cdot1 - 2\cdot2,\ 2\cdot1 - 1\cdot1,\ 1\cdot2 - 1\cdot1) = (-3, 1, 1)$.
$(-3,1,1)\times(1,1,1) = (1 - 1,\ 1 + 3,\ -3 - 1) = (0, 4, -4) \parallel \hat{j} - \hat{k}$.
(A) $\hat{j} - \hat{k}$: true. (D) $-\hat{j} + \hat{k}$: true (opposite direction).
(B) $-\hat{i} + \hat{j}$ and (C) $\hat{i} - \hat{j}$: perpendicular to $\vec{c}$, but $[(-1,1,0), \vec{a}, \vec{b}] = -1(1 - 4) - 1(1 - 2) + 0 = 4 \ne 0$, so not coplanar. False.
Answer: (A), (D). | ||||
| 55 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The premise cannot hold: for a $3\times3$ skew-symmetric $M$, $\det M = \det M^T = \det(-M) = (-1)^3\det M$, so $\det M = 0$. No $3\times3$ skew-symmetric matrix is non-singular, which is why the question was dropped.
Working formally (as if the inverses existed): $M^T = -M$, $N^T = -N$, $(N^{-1})^T = -N^{-1}$.
$(M^TN)^{-1} = N^{-1}(M^T)^{-1} = -N^{-1}M^{-1}$ and $(MN^{-1})^T = (N^{-1})^TM^T = (-N^{-1})(-M) = N^{-1}M$.
So the expression is $M^2N^2(-N^{-1}M^{-1})(N^{-1}M) = -M^2NM^{-1}N^{-1}M$. With $MN = NM$, $NM^{-1} = M^{-1}N$, so $NM^{-1}N^{-1}M = I$ and the expression is $-M^2$, option (C).
The official key gives marks to all. | ||||
| 56 | Conic SectionsHyperbola | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse $\frac{x^2}{4} + y^2 = 1$: $e = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt3}{2}$, foci $(\pm\sqrt3, 0)$.
Hyperbola: $e = \frac{2}{\sqrt3}$, so $b^2 = a^2(e^2 - 1) = \frac{a^2}{3}$. Through $(\sqrt3, 0)$: $\frac{3}{a^2} = 1$, $a^2 = 3$, $b^2 = 1$.
Hyperbola: $\frac{x^2}{3} - y^2 = 1$, i.e. $x^2 - 3y^2 = 3$.
(A) False ($b^2 = 1$, not 2). (B) True: foci $(\pm ae, 0) = (\pm2, 0)$. (C) False: $e = \frac{2}{\sqrt3}$. (D) True.
Answer: (B), (D). | ||||
| 57 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctMedium | |
A medium multiple correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Putting $x = y = 0$: $f(0) = 0$. For any $x$,
$f'(x) = \lim_{h\to0}\frac{f(x + h) - f(x)}{h} = \lim_{h\to0}\frac{f(h)}{h} = \lim_{h\to0}\frac{f(h) - f(0)}{h} = f'(0)$.
So $f$ is differentiable everywhere with $f'(x) = f'(0)$, a constant; $f(x) = kx$.
(A) False: it is differentiable on all of $\mathbb{R}$, not only on a finite interval.
(B) True: differentiable everywhere, hence continuous everywhere.
(C) True: $f'(x) = f'(0)$ for all $x$.
(D) Read literally, "except at finitely many points" includes zero exceptions, so it is not false; the key accepts BC and BCD.
Answer: (B), (C) (the key also accepts (B), (C), (D)). | ||||
| 58 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | ComprehensionEasy | |
A easy comprehension question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Multiplying out (E) column by column:
$a + 8b + 7c = 0$, $9a + 2b + 3c = 0$, $7a + 7b + 7c = 0$.
The third gives $c = -a - b$; then the first gives $a + 8b - 7a - 7b = 0$, so $b = 6a$ and $c = -7a$. (The second checks: $9a + 12a - 21a = 0$.)
So $(a, b, c) = a(1, 6, -7)$.
On the plane: $2a + 6a - 7a = a = 1$, so $(a, b, c) = (1, 6, -7)$.
$7a + b + c = 7 + 6 - 7 = 6$.
Answer: (D). | ||||
| 59 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | ComprehensionMedium | |
A medium comprehension question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Multiplying out (E) column by column:
$a + 8b + 7c = 0$, $9a + 2b + 3c = 0$, $7a + 7b + 7c = 0$.
The third gives $c = -a - b$; then the first gives $a + 8b - 7a - 7b = 0$, so $b = 6a$ and $c = -7a$. (The second checks: $9a + 12a - 21a = 0$.)
So $(a, b, c) = a(1, 6, -7)$.
With $a = 2$: $b = 12$, $c = -14$.
Using $\omega^3 = 1$: $\frac{3}{\omega^2} = 3\omega$, $\frac{1}{\omega^{12}} = 1$, $\frac{3}{\omega^{-14}} = 3\omega^{14} = 3\omega^2$.
Sum $= 1 + 3(\omega + \omega^2) = 1 - 3 = -2$.
Answer: (A). | ||||
| 60 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | ComprehensionEasy | |
A easy comprehension question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Multiplying out (E) column by column:
$a + 8b + 7c = 0$, $9a + 2b + 3c = 0$, $7a + 7b + 7c = 0$.
The third gives $c = -a - b$; then the first gives $a + 8b - 7a - 7b = 0$, so $b = 6a$ and $c = -7a$. (The second checks: $9a + 12a - 21a = 0$.)
So $(a, b, c) = a(1, 6, -7)$.
With $b = 6$: $a = 1$, $c = -7$, so the equation is $x^2 + 6x - 7 = 0$.
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-6}{-7} = \frac{6}{7}$.
$\sum_{n=0}^{\infty}\left(\frac{6}{7}\right)^n = \frac{1}{1 - 6/7} = 7$.
Answer: (B). | ||||
| 61 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | ComprehensionMedium | |
A medium comprehension question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Head: one ball moves to $U_2$, which then has 2 balls, one of them the original white.
P(white | H) $= \frac{3}{5}\cdot1 + \frac{2}{5}\cdot\frac{1}{2} = \frac{4}{5}$.
Tail: two balls move to $U_2$, which then has 3 balls. The expected number of whites moved is $2\cdot\frac{3}{5} = \frac{6}{5}$, so
P(white | T) $= \frac{1 + 6/5}{3} = \frac{11}{15}$. (By cases: WW $\frac{3}{10}$ gives 1, WR $\frac{6}{10}$ gives $\frac{2}{3}$, RR $\frac{1}{10}$ gives $\frac{1}{3}$: $\frac{3}{10} + \frac{4}{10} + \frac{1}{30} = \frac{22}{30} = \frac{11}{15}$.)
P(white) $= \frac{1}{2}\left(\frac{4}{5} + \frac{11}{15}\right) = \frac{1}{2}\cdot\frac{23}{15} = \frac{23}{30}$.
Answer: (B). | ||||
| 62 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | ComprehensionMedium | |
A medium comprehension question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability From Q.61: P(H and white) $= \frac{1}{2}\cdot\frac{4}{5} = \frac{2}{5} = \frac{12}{30}$ and P(white) $= \frac{23}{30}$.
By Bayes' theorem, P(H | white) $= \frac{12/30}{23/30} = \frac{12}{23}$.
Answer: (D). | ||||
| 63 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = 8x$: $a = 2$, points $(2t^2, 4t)$. Latus rectum ends $L(2, 4)$, $t = 1$, and $L'(2, -4)$, $t = -1$; $P\left(\frac{1}{2}, 2\right)$ has $t = \frac{1}{2}$.
$\Delta_1 = \frac{1}{2}\times8\times\left(2 - \frac{1}{2}\right) = 6$.
Tangents at $t_1$, $t_2$ meet at $(at_1t_2,\ a(t_1 + t_2))$:
$P, L$: $(1, 3)$; $P, L'$: $(-1, -1)$; $L, L'$: $(-2, 0)$.
$\Delta_2 = \frac{1}{2}\left|1(-1 - 0) + (-1)(0 - 3) + (-2)(3 + 1)\right| = \frac{1}{2}|-1 + 3 - 8| = 3$.
$\frac{\Delta_1}{\Delta_2} = 2$ (in general the triangle of three points has twice the area of the triangle of their tangents).
Answer: 2. | ||||
| 64 | 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 With common difference $d$: $S_p = \frac{p}{2}[6 + (p - 1)d]$, so
$\frac{S_{5n}}{S_n} = \frac{5[6 + (5n - 1)d]}{6 + (n - 1)d} = \frac{5[(6 - d) + 5nd]}{(6 - d) + nd}$.
This is independent of $n$ when $d = 0$ (ratio 5) or when $6 - d = 0$ (ratio 25).
$d = 6$: $a_2 = 9$. $d = 0$: $a_2 = 3$.
The intended answer is $d = 6$, $a_2 = 9$; the constant progression also satisfies the condition, and the official key accepts 3, 9, or both.
Answer: 9 (3 also accepted). | ||||
| 65 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta = \frac{\pi}{n}$. Then $\frac{1}{\sin\theta} - \frac{1}{\sin3\theta} = \frac{1}{\sin2\theta}$.
Left side: $\frac{\sin3\theta - \sin\theta}{\sin\theta\sin3\theta} = \frac{2\cos2\theta\sin\theta}{\sin\theta\sin3\theta} = \frac{2\cos2\theta}{\sin3\theta}$.
So $2\sin2\theta\cos2\theta = \sin3\theta$, i.e. $\sin4\theta = \sin3\theta$.
With $0 < \theta < \frac{\pi}{3}$: $4\theta = \pi - 3\theta$, so $\theta = \frac{\pi}{7}$ and $n = 7$.
Answer: 7. | ||||
| 66 | 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 Differentiate: $6f(x) = 3f(x) + 3xf'(x) - 3x^2$, so $xf'(x) - f(x) = x^2$, i.e. $\frac{d}{dx}\left(\frac{f(x)}{x}\right) = 1$.
Then $\frac{f(x)}{x} = x + C$; $f(1) = 2$ gives $C = 1$, so $f(x) = x^2 + x$ and $f(2) = 6$.
But the data are inconsistent: at $x = 1$ the given equation reads $0 = 3f(1) - 1 = 5$. No function satisfies it with $f(1) = 2$ (the equation itself forces $f(1) = \frac{1}{3}$, outside the codomain $[2, \infty)$). That is why the question was dropped.
The value obtained by differentiating is 6. | ||||
| 67 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z$ lies in the disc with centre $(3, 2)$ and radius 2.
$|2z - 6 + 5i| = 2\left|z - \left(3 - \frac{5}{2}i\right)\right|$, twice the distance from $z$ to $\left(3, -\frac{5}{2}\right)$.
That point is $2 + \frac{5}{2} = 4.5$ from the centre, outside the disc, so the least distance is $4.5 - 2 = 2.5$.
Minimum $= 2\times2.5 = 5$.
Answer: 5. | ||||
| 68 | Sequences and SeriesRelation between A.M. and G.M. | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on relation between a.m. and g.m., from Sequences and Series. NCERT: Class 11: Sequences and Series Write the sum as eight positive terms: $a^{-5} + a^{-4} + a^{-3} + a^{-3} + a^{-3} + 1 + a^8 + a^{10}$.
Their product is $a^{-5 - 4 - 9 + 8 + 10} = a^0 = 1$.
By AM ≥ GM, the sum $\ge 8\sqrt[8]{1} = 8$, with equality when all terms are equal, at $a = 1$.
Answer: 8. | ||||
| 69 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $\phi = \tan^{-1}\left(\frac{\sin\theta}{\sqrt{\cos2\theta}}\right)$. Then $\sin\phi = \frac{\sin\theta}{\sqrt{\sin^2\theta + \cos2\theta}} = \frac{\sin\theta}{\sqrt{\cos^2\theta}} = \frac{\sin\theta}{\cos\theta}$ (as $\cos\theta > 0$ on the interval).
So $f(\theta) = \tan\theta$, and $\frac{d f}{d(\tan\theta)} = 1$.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions Haematite is $\mathrm{Fe_2O_3}$: $2x + 3(-2) = 0$, so every iron is $+3$ (III).
Magnetite is $\mathrm{Fe_3O_4}$, which is $\mathrm{FeO\cdot Fe_2O_3}$: one $\mathrm{Fe^{2+}}$ and two $\mathrm{Fe^{3+}}$ (average $+\frac{8}{3}$). So it contains iron in both the II and III states.
Haematite: III; magnetite: II, III.
Answer: (D). | ||||
| 2 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Count the unpaired electrons on each metal ion.
K: $\mathrm{Fe^{3+}}$, $d^5$, with strong-field $\mathrm{CN^-}$: $t_{2g}^5$, one unpaired electron — paramagnetic.
L: $\mathrm{Co^{3+}}$, $d^6$, with $\mathrm{NH_3}$: low spin $t_{2g}^6$ — diamagnetic.
M: $\mathrm{Co^{3+}}$, $d^6$, with oxalate: $\mathrm{Co^{3+}}$ has a large crystal-field splitting and $[\mathrm{Co(ox)_3}]^{3-}$ is low spin $t_{2g}^6$ — diamagnetic.
N: $\mathrm{Ni^{2+}}$, $d^8$, octahedral: $t_{2g}^6e_g^2$, two unpaired electrons — paramagnetic.
O: $\mathrm{Pt^{2+}}$, $d^8$, square planar with $\mathrm{CN^-}$: all eight electrons paired — diamagnetic.
P: $\mathrm{Zn^{2+}}$, $d^{10}$ — diamagnetic.
Diamagnetic: L, M, O, P.
Answer: (C). | ||||
| 3 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter In acid solution the $\mathrm{H^+}$ suppresses the ionisation of $\mathrm{H_2S}$, so $[\mathrm{S^{2-}}]$ is very small. Only sulphides with extremely low solubility products exceed their $K_{sp}$: these are the group II sulphides, such as $\mathrm{CuS}$ (black) and $\mathrm{HgS}$ (black).
$\mathrm{MnS}$ and $\mathrm{NiS}$ (group IV) have much larger $K_{sp}$ and precipitate only in ammoniacal solution, where $[\mathrm{S^{2-}}]$ is high.
Precipitate: CuS and HgS.
Answer: (A). | ||||
| 4 | ElectrochemistryNernst equation | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Two Fe atoms each lose 2 electrons, so $n = 4$.
Reaction quotient: $Q = \frac{[\mathrm{Fe^{2+}}]^2}{P_{\mathrm{O_2}}[\mathrm{H^+}]^4} = \frac{(10^{-3})^2}{0.1 \times (10^{-3})^4} = \frac{10^{-6}}{10^{-13}} = 10^{7}$.
Nernst equation at 298 K:
$E = E^\circ - \frac{0.0591}{n}\log Q = 1.67 - \frac{0.0591}{4} \times 7 = 1.67 - 0.103 = 1.567 \approx 1.57\ \mathrm{V}$.
Answer: (D). | ||||
| 5 | Solutionsvan't Hoff factor | Class 12: Solutions | Single correctMedium | |
A medium single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\mathrm{K_3[Fe(CN)_6]} \rightarrow 3\mathrm{K^+} + [\mathrm{Fe(CN)_6}]^{3-}$, so $i = 4$ (complete dissociation).
Moles of solute $= \frac{0.1}{329} = 3.04 \times 10^{-4}$ mol; molality $m = \frac{3.04 \times 10^{-4}}{0.1\ \mathrm{kg}} = 3.04 \times 10^{-3}$ mol/kg.
$\Delta T_f = iK_fm = 4 \times 1.86 \times 3.04 \times 10^{-3} = 2.26 \times 10^{-2}$ K.
Freezing point $= 0 - 0.0226 \approx -2.3 \times 10^{-2}\ ^\circ$C.
Answer: (A). | ||||
| 6 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines An azo dye with $\beta$-naphthol needs an arenediazonium salt, and only a primary aromatic amine ($\mathrm{Ar{-}NH_2}$) gives a stable one with $\mathrm{NaNO_2}$/HCl at 0–5 °C.
(A) $\mathrm{C_6H_5N(CH_3)_2}$ is a tertiary amine: it undergoes ring nitrosation, not diazotisation.
(B) $\mathrm{C_6H_5NHCH_3}$ is secondary: it gives an N-nitrosamine.
(C) p-Toluidine ($\mathrm{CH_3C_6H_4NH_2}$) is a primary aromatic amine: it gives p-methylbenzenediazonium chloride, which couples with alkaline $\beta$-naphthol to a brilliant orange-red azo dye.
(D) Benzylamine $\mathrm{C_6H_5CH_2NH_2}$ is a primary aliphatic amine: its diazonium ion decomposes at once (to the alcohol and $\mathrm{N_2}$), so no dye.
Answer: (C). | ||||
| 7 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The reactant is 3,4-dihydro-2H-pyran, a cyclic enol ether ($\mathrm{C{=}C{-}O}$ in the ring).
Anhydrous acid protonates the double bond at the carbon away from oxygen, giving an oxocarbenium ion on the carbon next to the ring oxygen (stabilised by the oxygen lone pair).
The alcohol $\mathrm{RCH_2OH}$ adds to that carbon, and loss of $\mathrm{H^+}$ gives 2-(alkoxy)tetrahydropyran (a THP ether).
That carbon now carries two OR groups (the ring O and $\mathrm{OCH_2R}$) and an H: it is an acetal carbon. (Without water no hemiacetal stays; there is no carbonyl for an ester.)
Answer: (B), an acetal. | ||||
| 8 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules The drawing is a six-membered ring of five carbons and one oxygen, so the sugar is a pyranose, not a furanose: (C) is out.
One ring carbon outside the ring carries $\mathrm{CH_2OH}$ (C-5), and the carbon next to the ring oxygen on the other side (C-1, the anomeric carbon) carries H and OH. A carbon bearing H, OH and the ring O is a hemiacetal carbon made from an aldehyde group, so the open-chain form is an aldehyde with six carbons: an aldohexose (glucose here). A ketohexose would have no H on its anomeric carbon. So (B) is true and (A) false.
At C-1 the OH is drawn equatorial (in the mean plane of the chair), on the same side as the $\mathrm{CH_2OH}$: that is the $\beta$ anomer ($\beta$-D-glucopyranose), so (D) $\alpha$-pyranose is not correct.
Answer: (B). | ||||
| 9 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Multiple correctMedium | |
A medium multiple correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Mn in $\mathrm{MnO_4^-}$ is $+7$.
(A) True. In neutral solution: $\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$, Mn $+7 \to +4$: 3 electrons.
(B) False. Five electrons ($\to \mathrm{Mn^{2+}}$) needs acid.
(C) True. In (weakly) alkaline solution permanganate also goes to $\mathrm{MnO_2}$: 3 electrons, e.g. $2\mathrm{KMnO_4} + \mathrm{KI} + \mathrm{H_2O} \rightarrow 2\mathrm{MnO_2} + \mathrm{KIO_3} + 2\mathrm{KOH}$. (Only in strongly alkaline solution with excess reductant is it 1 electron, to manganate.)
(D) True. In acid: $\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$: 5 electrons.
Answer: (A), (C), (D). | ||||
| 10 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Multiple correctHard | |
A hard multiple correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In water $\mathrm{Cu^+}$ disproportionates because the strongly hydrated $\mathrm{Cu^{2+}}$ is more stable. The equilibrium moves to the left when something removes $\mathrm{Cu^+}$ from solution as a very stable form (lowering the free $\mathrm{Cu^+}$ makes more $\mathrm{Cu^+}$ form from $\mathrm{Cu} + \mathrm{Cu^{2+}}$), i.e. when Cu(I) is stabilised.
(A) $\mathrm{NO_3^-}$: forms no insoluble salt or stable complex with $\mathrm{Cu^+}$. No shift.
(B) $\mathrm{Cl^-}$: gives insoluble CuCl ($K_{sp} \approx 10^{-6}$) and $[\mathrm{CuCl_2}]^-$; indeed $\mathrm{Cu} + \mathrm{Cu^{2+}} + 2\mathrm{Cl^-} \rightarrow 2\mathrm{CuCl}$ in chloride solution. Shifts left.
(C) $\mathrm{SCN^-}$: gives insoluble CuSCN. Shifts left.
(D) $\mathrm{CN^-}$: gives CuCN and the very stable $[\mathrm{Cu(CN)_4}]^{3-}$ (cyanide even reduces $\mathrm{Cu^{2+}}$ to Cu(I)). Shifts left.
Answer: (B), (C), (D). | ||||
| 11 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Multiple correctEasy | |
A easy multiple correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics First order: $[A] = [A]_0e^{-kt}$ and $t_{1/2} = \frac{\ln 2}{k}$.
(A) True: the concentration falls exponentially.
(B) True: $k$ rises with temperature (Arrhenius), so $t_{1/2} = \frac{0.693}{k}$ falls.
(C) False: a first-order half-life does not depend on the initial concentration.
(D) True: after 8 half-lives the fraction left is $\left(\frac{1}{2}\right)^8 = \frac{1}{256} = 0.0039$, i.e. 0.39 % left, so the reaction is $100 - 0.39 \approx 99.6\ \%$ complete.
Answer: (A), (B), (D). | ||||
| 12 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Step (ii) condenses the product of step (i) with adipic acid, $\mathrm{HOOC(CH_2)_4COOH}$, on heating to a condensation polymer. The intended partner is a diamine, $\mathrm{H_2N{-}R{-}NH_2}$, giving a polyamide (nylon type). So step (i) must turn $\mathrm{X{-}(CH_2)_4{-}X}$ into a diamine.
(A) $\mathrm{X = COOCH_3}$, $\mathrm{H_2/Ni}$/heat: an ester is not reduced to an amine (and is not normally reduced by $\mathrm{H_2/Ni}$ at all); no diamine. Not correct.
(B) $\mathrm{X = CONH_2}$, $\mathrm{H_2/Ni}$/heat: amides are not reduced by catalytic hydrogenation under these conditions. Not correct.
(C) $\mathrm{X = CONH_2}$, $\mathrm{Br_2/NaOH}$: Hofmann bromamide degradation, $\mathrm{RCONH_2 \rightarrow RNH_2}$, gives $\mathrm{H_2N(CH_2)_4NH_2}$, which with adipic acid gives nylon-4,6. Correct.
(D) $\mathrm{X = CN}$, $\mathrm{H_2/Ni}$/heat: $\mathrm{RCN \rightarrow RCH_2NH_2}$, giving hexamethylenediamine $\mathrm{H_2N(CH_2)_6NH_2}$, which with adipic acid gives nylon-6,6. Correct.
Answer: (C), (D). | ||||
| 13 | 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 $\mathrm{PCl_5}$ gives $\mathrm{POCl_3}$ when it takes an oxygen atom from an oxide or an OH compound:
$\mathrm{SO_2}$: $\mathrm{PCl_5 + SO_2 \rightarrow POCl_3 + SOCl_2}$.
$\mathrm{H_2O}$ (limited): $\mathrm{PCl_5 + H_2O \rightarrow POCl_3 + 2HCl}$.
$\mathrm{H_2SO_4}$: $\mathrm{2PCl_5 + H_2SO_4 \rightarrow SO_2Cl_2 + 2POCl_3 + 2HCl}$.
$\mathrm{P_4O_{10}}$: $\mathrm{6PCl_5 + P_4O_{10} \rightarrow 10POCl_3}$.
$\mathrm{O_2}$ does not convert $\mathrm{PCl_5}$ (it is $\mathrm{PCl_3}$ that is oxidised by $\mathrm{O_2}$ to $\mathrm{POCl_3}$).
The official count is 4: $\mathrm{SO_2}$, $\mathrm{H_2O}$, $\mathrm{H_2SO_4}$ and $\mathrm{P_4O_{10}}$, with $\mathrm{CO_2}$ not counted.
Answer: 4. | ||||
| 14 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds In $\mathrm{[Cr(H_2O)_5Cl]Cl_2}$ the Cl inside the bracket is bonded to Cr and is not precipitated; the two outside are free ions.
Moles of complex $= 0.030 \times 0.01 = 3 \times 10^{-4}$; free $\mathrm{Cl^-} = 2 \times 3 \times 10^{-4} = 6 \times 10^{-4}$ mol.
$\mathrm{Ag^+}$ needed $= 6 \times 10^{-4}$ mol, so volume $= \frac{6 \times 10^{-4}}{0.1}\ \mathrm{L} = 6 \times 10^{-3}\ \mathrm{L} = 6$ mL.
Answer: 6. | ||||
| 15 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium CuCl is far more soluble than AgCl, so it controls $[\mathrm{Cl^-}]$. Only about $10^{-3}$ mol of the 0.1 mol dissolves, so solid CuCl remains and the solution is saturated with it:
$[\mathrm{Cu^+}][\mathrm{Cl^-}] = 10^{-6}$, and with $[\mathrm{Cu^+}] \approx [\mathrm{Cl^-}]$ (the $\sim 10^{-5}$ M from AgCl is negligible), $[\mathrm{Cl^-}] \approx 10^{-3}$ M.
Then $[\mathrm{Ag^+}] = \frac{K_{sp}(\mathrm{AgCl})}{[\mathrm{Cl^-}]} = \frac{1.6 \times 10^{-10}}{10^{-3}} = 1.6 \times 10^{-7}$ M.
So $x = 7$.
Answer: 7. | ||||
| 16 | Solid StateClassification of solids | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on classification of solids, from Solid State. NCERT: No matching NCERT chapter A truncated octahedron is made by cutting off the 6 vertices of a regular octahedron.
Each of the 8 triangular faces loses its three corners and becomes a regular hexagon; each of the 6 cut vertices leaves a square face.
So it has 8 hexagonal faces and 6 square faces (14 faces, 36 edges, 24 vertices; Euler: $24 - 36 + 14 = 2$).
Answer: 8. | ||||
| 17 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The compound is 3-methylpentane, $\mathrm{CH_3CH_2CH(CH_3)CH_2CH_3}$. It has four kinds of H: C-1 (and C-5), C-2 (and C-4), C-3, and the methyl on C-3.
Cl at C-1: $\mathrm{ClCH_2CH_2CH(CH_3)CH_2CH_3}$. C-3 now has H, $\mathrm{CH_3}$, $\mathrm{C_2H_5}$ and $\mathrm{CH_2CH_2Cl}$: one stereocentre, 2 enantiomers.
Cl at C-2: $\mathrm{CH_3CHClCH(CH_3)CH_2CH_3}$ (2-chloro-3-methylpentane). C-2 and C-3 are both stereocentres and the molecule is unsymmetrical: $2^2 = 4$ stereoisomers.
Cl at C-3: $\mathrm{CH_3CH_2CCl(CH_3)CH_2CH_3}$: C-3 carries two ethyl groups, achiral: 1.
Cl on the methyl: $\mathrm{(CH_3CH_2)_2CHCH_2Cl}$: achiral: 1.
Total $= 2 + 4 + 1 + 1 = 8$.
Answer: 8. | ||||
| 18 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Hyperconjugation uses each C–H bond on a carbon directly attached to the positive carbon ($\alpha$-hydrogens). Each such C–H gives one no-bond resonance structure.
The cationic carbon carries:
$\mathrm{CH_3}$: 3 $\alpha$-H;
$\mathrm{CH_2CH_3}$: the $\mathrm{CH_2}$ has 2 $\alpha$-H;
cyclohexyl: its attachment carbon (CH) has 1 $\alpha$-H.
Total $= 3 + 2 + 1 = 6$.
Answer: 6. | ||||
| 19 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Matrix matchMedium | |
A medium matrix match question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics (A) Sublimation of dry ice: a phase transition (P), it absorbs heat, $\Delta H > 0$ (R), and solid → gas raises entropy, $\Delta S > 0$ (S). A → P, R, S.
(B) Decomposition of limestone: endothermic, $\Delta H > 0$ (R); a gas is formed from a solid, $\Delta S > 0$ (S). It is a chemical change, not a phase transition. B → R, S.
(C) Two H atoms combine: a bond forms, so heat is released, $\Delta H < 0$; two particles become one, $\Delta S < 0$ (T). C → T.
(D) White → red phosphorus: a change between allotropes (Q), which is also a transition between solid phases (P). Red P is the more stable (polymeric) form, so $\Delta H < 0$; the polymeric chain is more ordered than discrete $\mathrm{P_4}$ molecules, $\Delta S < 0$ (T). D → P, Q, T.
Official: A → P, R, S; B → R, S; C → T; D → P, Q, T. | ||||
| 20 | Organic Chemistry: Some Basic Principles and TechniquesTypes of organic reactions | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Matrix matchHard | |
A hard 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) $\mathrm{C_6H_5COCH_2CH_2COCH_3}$ with aq NaOH gives 3-phenylcyclopent-2-en-1-one. Hydroxide removes an $\alpha$-H of the methyl ketone to give an enolate (carbanion, T); it attacks the other C=O intramolecularly (nucleophilic addition, S) to close the five-membered ring; the $\beta$-hydroxy ketone loses water to the conjugated enone (dehydration, R). A → R, S, T.
(B) $\mathrm{C_6H_5COCH_2CH_2CH_2Cl}$ with $\mathrm{CH_3MgI}$: the Grignard adds to the C=O (nucleophilic addition, S) to give a tertiary alkoxide, which displaces $\mathrm{Cl^-}$ from the end carbon intramolecularly (nucleophilic substitution, P), closing 2-methyl-2-phenyltetrahydrofuran. B → P, S.
(C) $\mathrm{C_6H_5COCH_2CH_2CH_2{}^{18}OH}$ with $\mathrm{H_2SO_4}$: the $^{18}$OH adds to the protonated C=O (nucleophilic addition, S) to give a cyclic hemiketal containing $^{18}$O in the ring; loss of the original carbonyl oxygen as water (dehydration, R) gives 5-phenyl-2,3-dihydrofuran with $^{18}$O in the ring. C → R, S.
(D) $\mathrm{C_6H_5CH_2CH_2CH_2C(CH_3)_2OH}$ with $\mathrm{H_2SO_4}$: the tertiary alcohol loses water (dehydration, R) to a tertiary carbocation, which attacks the benzene ring ortho to the chain (intramolecular Friedel–Crafts alkylation, electrophilic substitution, Q), giving 1,1-dimethyltetralin. D → Q, R.
Official: A → R, S, T; B → P, S; C → R, S; D → Q, R. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 21 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Going from glass to air, the light meets a less dense medium, so there is a critical angle $\theta_c$ ($\sin\theta_c = \frac{1}{\mu}$, about $42^\circ$ for glass).
For $\theta < \theta_c$: most light is transmitted and a little is reflected; R grows slowly and T falls slowly with $\theta$. Energy is conserved, so $R + T = 100\%$ at every angle: since R is not zero even at $\theta = 0$, T must start below 100 %.
For $\theta > \theta_c$: total internal reflection, so $R = 100\%$ and $T = 0$ all the way to $90^\circ$.
(A) has no critical angle and T reaching zero only at $90^\circ$: wrong.
(B) has the jump at $\theta_c$, but T starts at 100 % while R is already non-zero, so $R + T > 100\%$ near $\theta = 0$: wrong.
(C) has T starting below 100 % (so $R + T = 100\%$), the jump at $\theta_c$, then $R = 100\%$ and $T = 0$: correct.
(D) shows T and R continuing beyond $\theta_c$ with T non-zero: wrong.
Answer: (C). | ||||
| 22 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation In a circular orbit of radius $r$: $\frac{mV^2}{r} = \frac{GMm}{r^2}$, so $V^2 = \frac{GM}{r}$.
To just escape from that point the total energy must be zero:
$\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0 \Rightarrow \frac{1}{2}mv_e^2 = \frac{GMm}{r} = mV^2$.
Answer: (B). | ||||
| 23 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The $N$ turns are spread evenly over the radial width $b - a$, so between $r$ and $r + dr$ there are $dN = \frac{N}{b-a}dr$ turns.
Each turn of radius $r$ gives at the centre $\frac{\mu_0 I}{2r}$ along Z, so
$B = \int_a^b \frac{\mu_0 I}{2r}\cdot\frac{N\,dr}{b-a} = \frac{\mu_0 NI}{2(b-a)}\ln\frac{b}{a}$.
Answer: (A). | ||||
| 24 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $x_1 + x_2 = A\left[\sin\omega t + \sin\left(\omega t + \frac{2\pi}{3}\right)\right] = 2A\cos\frac{\pi}{3}\sin\left(\omega t + \frac{\pi}{3}\right) = A\sin\left(\omega t + \frac{\pi}{3}\right)$.
For rest, $x_3 = -A\sin\left(\omega t + \frac{\pi}{3}\right) = A\sin\left(\omega t + \frac{\pi}{3} + \pi\right) = A\sin\left(\omega t + \frac{4\pi}{3}\right)$.
(As phasors: three equal amplitudes at $0$, $\frac{2\pi}{3}$, $\frac{4\pi}{3}$ add to zero.)
So $B = A$, $\phi = \frac{4\pi}{3}$.
Answer: (B). | ||||
| 25 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Magnetic field lines never begin or end (there are no magnetic monopoles, $\oint\vec{B}\cdot d\vec{A} = 0$), so a valid magnetic pattern must be made of closed lines.
(A) and (B): lines radiate out of, or into, a point. That is the field of a point charge, but it would need a magnetic monopole: not valid for B.
(D): the lines all point away from the central region, upward at the top and downward at the bottom, so the centre acts as a source of lines: not a possible magnetic field.
(C): concentric closed circles. This is the magnetic field round a long straight current; it is also the induced (non-conservative) electric field produced by a changing magnetic flux through the circles, by Faraday's law $\oint\vec{E}\cdot d\vec{l} = -\frac{d\Phi_B}{dt}$. Valid for both.
Answer: (C). | ||||
| 26 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Both leave the post top horizontally and fall 5 m: $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1$ s.
Horizontal speeds after the collision: ball $\frac{20}{1} = 20$ m/s, bullet $\frac{100}{1} = 100$ m/s.
Horizontal momentum is conserved in the collision:
$0.01V = 0.2 \times 20 + 0.01 \times 100 = 4 + 1 = 5 \Rightarrow V = 500$ m/s.
Answer: (D). | ||||
| 27 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{0.5\ \mathrm{mm}}{50} = 0.01$ mm.
Diameter $d = 2.5 + 20 \times 0.01 = 2.70$ mm, with error $\Delta d = 0.01$ mm, so $\frac{\Delta d}{d} = \frac{0.01}{2.70} = 0.37\%$.
$\rho = \frac{m}{\frac{\pi}{6}d^3}$, so $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta d}{d} = 2\% + 3 \times 0.37\% = 3.1\%$.
Answer: (C). | ||||
| 28 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The field adds a constant force $QE$ along the spring axis. With $x$ measured from the old equilibrium:
$m\ddot{x} = -kx + QE = -k\left(x - \frac{QE}{k}\right)$.
This is SHM about the new mean position $x_0 = \frac{QE}{k}$, with the same $\omega = \sqrt{\frac{k}{m}}$: the frequency is unchanged and the mean position is shifted.
Answer: (A). | ||||
| 29 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctMedium | |
A medium multiple correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Let each volume be $V$ and the tension $T > 0$. A is above, held down by the string; B is below, held up by it.
A: buoyancy $= $ weight $+ T$: $d_FVg = d_AVg + T$, so $d_A < d_F$. (A) true, (C) false.
B: weight $=$ buoyancy $+ T$: $d_BVg = d_FVg + T$, so $d_B > d_F$. (B) true.
Adding the two equations: $(d_A + d_B)Vg = 2d_FVg$, so $d_A + d_B = 2d_F$. (D) true.
Answer: (A), (B), (D). | ||||
| 30 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Multiple correctMedium | |
A medium multiple correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current With the dielectric, $C_B = 4C_A$, so $X_C = \frac{1}{\omega C}$ becomes a quarter.
Current: $I = \frac{V}{\sqrt{R^2 + X_C^2}}$; a smaller $X_C$ gives a larger current, so $I_R^A < I_R^B$. (B) true, (A) false.
Capacitor voltage: $V_C = IX_C = \frac{VX_C}{\sqrt{R^2 + X_C^2}} = \frac{V}{\sqrt{\frac{R^2}{X_C^2} + 1}}$, which falls as $X_C$ falls. So $V_C^A > V_C^B$. (C) true, (D) false.
Answer: (B), (C). | ||||
| 31 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance (A) False. Gauss's law rests on the inverse-square law: for $E \propto r^{-2}$ the flux $E \cdot 4\pi r^2$ through any sphere round the charge is the same. With $E \propto r^{-2.5}$ the flux $\propto r^{-0.5}$ would depend on the surface, so the law would fail.
(B) False. Gauss's law is always true, but it gives the field only where symmetry makes $E$ constant over a Gaussian surface; a dipole's field has no such symmetry.
(C) True. Between two charges the fields of unlike charges both point the same way (from + to −) and cannot cancel; the field can be zero between them only if the charges have the same sign.
(D) True. Moving a unit positive charge slowly (no change in kinetic energy) from A to B, the external work equals the rise in potential energy: $W = q(V_B - V_A) = V_B - V_A$.
Answer: (C), (D). | ||||
| 32 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Multiple correctHard | |
A hard multiple correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take right as positive. The ring moves left: $v = -1$ m/s, momentum $2 \times (-1) = -2$ kg m/s. The ball moves right at 20 m/s and leaves vertically, so it loses all its horizontal momentum: $0.1 \times 20 = 2$ kg m/s.
Horizontal impulse on the ring from the ball $= +2$ N s. (The ground's friction impulse during the very short collision is taken as negligible; the vertical impulse on the ring is balanced by the ground's normal impulse.)
Ring after: $2v' = -2 + 2 = 0$, so $v' = 0$: the centre of mass is at rest.
But the ring was spinning ($\omega = \frac{v}{R} = 2$ rad/s, sense of rolling to the left), and the ball's impulse acts at a point only $0.25$ m above the centre and on the ring's surface, so it cannot remove all that spin. Its torque impulse about the centre is small: the point of impact is at $(-0.433, 0.25)$ m from the centre, impulse on the ring $(+2, -1)$ N s, so the angular impulse is $(-0.433)(-1) - (0.25)(2) = -0.067$ N m s and $\Delta\omega = \frac{-0.067}{mR^2} = \frac{-0.067}{0.5} \approx -0.13$ rad/s. The ring still spins at about 1.9 rad/s.
(A) True: the ring rotates about a stationary CM (pure rotation).
(B) False: it still rotates.
(C) With the CM at rest and the ring spinning in the sense it had when rolling to the left, the bottom point now slides to the right over the ground, so kinetic friction on the ring acts to the left. True for the motion just after the collision; the key accepts A alone or A with C.
(D) False once the ring slides on the ground.
Official: (A) or (A), (C). | ||||
| 33 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Work in the train's frame. The ball starts with 10 m/s at $60^\circ$ relative to the train; its acceleration relative to the train is $g$ downward and $a$ backward.
Time to return to the same height: $T = \frac{2 \times 10\sin 60^\circ}{g} = \frac{2 \times 5\sqrt{3}}{10} = \sqrt{3}$ s.
Forward displacement relative to the train: $x = (10\cos 60^\circ)T - \frac{1}{2}aT^2 = 5\sqrt{3} - \frac{3}{2}a$.
This equals 1.15 m: $8.66 - 1.5a = 1.15 \Rightarrow a = \frac{7.51}{1.5} \approx 5$ m/s².
Answer: 5. | ||||
| 34 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The block moves towards the wall, compressing the spring by 0.06 m before it first stops. Work-energy theorem:
$\frac{1}{2}mV^2 = \frac{1}{2}kx^2 + \mu mgx$
$= \frac{1}{2}(2)(0.06)^2 + (0.1)(0.18)(10)(0.06) = 0.0036 + 0.0108 = 0.0144$ J.
$\frac{1}{2}(0.18)V^2 = 0.0144 \Rightarrow V^2 = 0.16 \Rightarrow V = 0.4$ m/s $= \frac{4}{10}$.
Answer: 4. | ||||
| 35 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity The two cells are in parallel between A and B with the same polarity (each has its positive plate towards A): 6 V with 1 Ω and 3 V with 2 Ω.
With no external load, the equivalent emf of parallel cells is
$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{6 + 1.5}{1 + 0.5} = 5$ V.
Check: the loop current is $\frac{6 - 3}{1 + 2} = 1$ A; across AB: $6 - 1 \times 1 = 5$ V $= 3 + 1 \times 2$.
Answer: 5. | ||||
| 36 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalHard | |
A hard numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Take downward (the direction of light) as positive.
Curved air–oil surface (centre of curvature below, in the oil): $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$ with $\mu_1 = 1$, $\mu_2 = \frac{7}{4}$, $u = -24$ cm, $R = +6$ cm:
$\frac{7}{4v} + \frac{1}{24} = \frac{3/4}{6} = \frac{1}{8} \Rightarrow \frac{7}{4v} = \frac{1}{12} \Rightarrow v = 21$ cm below the (thin) oil layer, i.e. below the water surface.
Flat oil–water surface: $\frac{4/3}{v'} = \frac{7/4}{21} \Rightarrow v' = \frac{4}{3} \times 12 = 16$ cm below the water surface.
The tank is 18 cm deep, so the image is $18 - 16 = 2$ cm above the bottom.
Answer: 2. | ||||
| 37 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $Z^2 = R^2 + X_C^2 = 1.25R^2 \Rightarrow X_C = 0.5R$.
$\frac{1}{\omega C} = \frac{R}{2} \Rightarrow RC = \frac{2}{\omega} = \frac{2}{500}$ s $= 4$ ms.
Answer: 4. | ||||
| 38 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalMedium | |
A medium numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon energy: $\frac{hc}{\lambda} = \frac{1240\ \mathrm{eV\,nm}}{200\ \mathrm{nm}} = 6.2$ eV. Maximum kinetic energy $= 6.2 - 4.7 = 1.5$ eV.
Emission stops when the sphere's potential reaches the stopping potential, 1.5 V.
$V = \frac{kQ}{r} \Rightarrow Q = \frac{Vr}{k} = \frac{1.5 \times 0.01}{9 \times 10^9} = 1.67 \times 10^{-12}$ C.
Number of electrons $= \frac{Q}{e} = \frac{1.67 \times 10^{-12}}{1.6 \times 10^{-19}} = 1.04 \times 10^{7}$.
So $Z = 7$.
Answer: 7. | ||||
| 39 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Matrix matchMedium | |
A medium matrix match question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the graph: A = (3V, 3P), B = (1V, 3P), C = (1V, 1P), D = (9V, 1P). $PV$ (so $T$) is 9, 3, 1, 9 units: D → A has $PV$ constant, an isothermal curve.
A → B: isobaric compression; $T$ falls (9 → 3), so $U$ decreases (P); $W$ on gas $> 0$ (T); $Q = \Delta U - W_{on} < 0$, heat lost (R). A → P, R, T.
B → C: isochoric; $P$ and $T$ fall, $U$ decreases (P); no work; $Q = \Delta U < 0$, heat lost (R). B → P, R.
C → D: isobaric expansion; $T$ rises (1 → 9), $U$ increases (Q); heat gained (S). C → Q, S.
D → A: isothermal compression; $\Delta U = 0$; work done on the gas (T); $Q = -W_{on} < 0$, heat lost (R). D → R, T.
Official: A → P, R, T; B → P, R; C → Q, S; D → R, T. | ||||
| 40 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Matrix matchEasy | |
A easy matrix match question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Sound in a pipe is a longitudinal wave; waves on a stretched wire are transverse.
(A) Closed at one end: node at the closed end, antinode at the open end, $L = \frac{\lambda_f}{4}$, so $\lambda_f = 4L$. A → P, T.
(B) Open at both ends: antinodes at both ends, $L = \frac{\lambda_f}{2}$, $\lambda_f = 2L$. B → P, S.
(C) Wire fixed at both ends: nodes at both ends, $L = \frac{\lambda_f}{2}$, $\lambda_f = 2L$. C → Q, S.
(D) Wire also clamped at the middle: a node there, so each half $\frac{L}{2}$ holds one loop: $\frac{L}{2} = \frac{\lambda_f}{2}$, $\lambda_f = L$. D → Q, R.
Official: A → P, T; B → P, S; C → Q, S; D → Q, R. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 41 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The normal at $(x_1, y_1)$ is $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$.
At $P(6, 3)$, putting $y = 0$: $x = \frac{6(a^2 + b^2)}{a^2} = 9$, so $\frac{a^2 + b^2}{a^2} = \frac{3}{2}$.
But $e^2 = 1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2} = \frac{3}{2}$, so $e = \sqrt{\frac{3}{2}}$.
(Check: $\frac{b^2}{a^2} = \frac{1}{2}$ and $\frac{36}{a^2} - \frac{9}{b^2} = 1$ give $a^2 = 18$, $b^2 = 9$, consistent.)
Answer: (B). | ||||
| 42 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $\alpha$ be the common root. Subtracting the equations: $(b - 1)\alpha - 1 - b = 0$, so $\alpha = \frac{b+1}{b-1}$ ($b \neq 1$; for $b = 1$ the equations $x^2 + x - 1 = 0$ and $x^2 + x + 1 = 0$ share no root).
Substitute in $\alpha^2 + \alpha + b = 0$ and multiply by $(b-1)^2$:
$(b+1)^2 + (b+1)(b-1) + b(b-1)^2 = 0$
$b^2 + 2b + 1 + b^2 - 1 + b^3 - 2b^2 + b = b^3 + 3b = 0$.
So $b = 0$ or $b^2 = -3$, i.e. $b = \pm i\sqrt{3}$. Of the options, $-i\sqrt{3}$ is one.
Answer: (B). | ||||
| 43 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Expand along the first row:
$\Delta = 1(1 - c\omega) - a(\omega - c\omega^2) + b(\omega^2 - \omega^2) = 1 - c\omega - a\omega + ac\omega^2 = (1 - a\omega)(1 - c\omega)$.
Non-singular means $a\omega \neq 1$ and $c\omega \neq 1$, i.e. $a \neq \omega^2$ and $c \neq \omega^2$ (since $\omega^2 \cdot \omega = 1$ while $\omega \cdot \omega = \omega^2 \neq 1$).
So $a = c = \omega$, and $b$ may be $\omega$ or $\omega^2$: 2 matrices.
Answer: (A). | ||||
| 44 | CirclesStandard and general form | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Touching the $y$-axis at $(0, 2)$: centre $(h, 2)$, radius $|h|$.
Through $(-1, 0)$: $(-1 - h)^2 + 4 = h^2 \Rightarrow 1 + 2h + 4 = 0 \Rightarrow h = -\frac{5}{2}$.
Circle: $\left(x + \frac{5}{2}\right)^2 + (y - 2)^2 = \frac{25}{4}$.
(A) $\left(-\frac{3}{2}, 0\right)$: $1 + 4 = 5 \neq \frac{25}{4}$. (B) $\left(-\frac{5}{2}, 2\right)$ is the centre. (C) $\left(-\frac{3}{2}, \frac{5}{2}\right)$: $1 + \frac{1}{4} \neq \frac{25}{4}$.
(D) $(-4, 0)$: $\frac{9}{4} + 4 = \frac{25}{4}$. ✓
Answer: (D). | ||||
| 45 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\lim_{x \to 0}(1 + kx)^{1/x} = e^k$, so the limit is $e^{\ln(1 + b^2)} = 1 + b^2$.
Hence $\sin^2\theta = \frac{1 + b^2}{2b}$. By AM–GM, $1 + b^2 \geq 2b$, so the right side is $\geq 1$, while $\sin^2\theta \leq 1$.
Equality forces $b = 1$ and $\sin^2\theta = 1$, so $\theta = \pm\frac{\pi}{2}$ (both in $(-\pi, \pi]$).
Answer: (D). | ||||
| 46 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Since $f \geq 0$, $R_2 = \int_{-1}^{2}f(x)\,dx$.
Use $\int_a^b g(x)\,dx = \int_a^b g(a + b - x)\,dx$ with $a + b = 1$:
$R_1 = \int_{-1}^{2}(1 - x)f(1 - x)\,dx = \int_{-1}^{2}(1 - x)f(x)\,dx = R_2 - R_1$.
So $2R_1 = R_2$.
Answer: (C). | ||||
| 47 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $(g \circ g \circ f)(x) = \sin(\sin x^2)$ and $(f \circ g \circ g \circ f)(x) = \sin^2(\sin x^2)$.
Let $u = \sin(\sin x^2)$: $u^2 = u$, so $u = 0$ or $u = 1$.
$u = 1$ needs $\sin x^2 = \frac{\pi}{2} + 2k\pi$, impossible as $|\sin x^2| \leq 1 < \frac{\pi}{2}$.
$u = 0$ needs $\sin x^2 = k\pi$; with $|\sin x^2| \leq 1$ only $k = 0$: $\sin x^2 = 0$, so $x^2 = n\pi$ with $n = 0, 1, 2, \ldots$
Hence $x = \pm\sqrt{n\pi}$, $n \in \{0, 1, 2, \ldots\}$.
Answer: (A). | ||||
| 48 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections By the section formula, $P = \left(\frac{x}{4}, \frac{y}{4}\right)$. Call it $(h, k)$: $x = 4h$, $y = 4k$.
$y^2 = 4x \Rightarrow 16k^2 = 16h \Rightarrow k^2 = h$.
Locus: $y^2 = x$.
Answer: (C). | ||||
| 49 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctMedium | |
A medium multiple correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability (A) At $x = -\frac{\pi}{2}$: left value $\frac{\pi}{2} - \frac{\pi}{2} = 0$; right limit $-\cos\left(-\frac{\pi}{2}\right) = 0$. Continuous. True.
(B) At $x = 0$: left $-\cos 0 = -1$, right limit $0 - 1 = -1$, so continuous; left derivative $\sin 0 = 0$, right derivative $1$. Not differentiable. True.
(C) At $x = 1$: left $1 - 1 = 0$, right $\ln 1 = 0$; left derivative $1$, right derivative $\frac{1}{x} = 1$. Differentiable. True.
(D) $-\frac{3}{2} > -\frac{\pi}{2} \approx -1.571$, so $x = -\frac{3}{2}$ lies inside $\left(-\frac{\pi}{2}, 0\right)$ where $f = -\cos x$, which is differentiable. True.
Answer: (A), (B), (C), (D). | ||||
| 50 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Multiple correctMedium | |
A medium multiple correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $P(E) = p$, $P(F) = q$; independence gives $P(E \cap F) = pq$.
Exactly one: $p + q - 2pq = \frac{11}{25}$. None: $(1 - p)(1 - q) = 1 - (p + q) + pq = \frac{2}{25}$.
With $s = p + q$, $t = pq$: $s - 2t = \frac{11}{25}$ and $s - t = \frac{23}{25}$, so $t = \frac{12}{25}$, $s = \frac{35}{25} = \frac{7}{5}$.
$p, q$ are the roots of $z^2 - \frac{7}{5}z + \frac{12}{25} = 0$, i.e. $25z^2 - 35z + 12 = 0 \Rightarrow z = \frac{4}{5}, \frac{3}{5}$.
So $\{P(E), P(F)\} = \left\{\frac{4}{5}, \frac{3}{5}\right\}$ in either order: (A) and (D). (B) and (C) give $s = \frac{3}{5}$, wrong.
Answer: (A), (D). | ||||
| 51 | Conic SectionsParabola | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Normal to $y^2 = 4ax$ ($a = 1$) with slope $m$: $y = mx - 2m - m^3$.
Through $(9, 6)$: $6 = 9m - 2m - m^3 \Rightarrow m^3 - 7m + 6 = 0 \Rightarrow (m - 1)(m - 2)(m + 3) = 0$.
$m = 1$: $y = x - 3$, i.e. $y - x + 3 = 0$ — (A).
$m = 2$: $y = 2x - 12$, i.e. $y - 2x + 12 = 0$ — (D).
$m = -3$: $y = -3x + 33$, i.e. $y + 3x - 33 = 0$ — (B).
(C) $y + x - 15 = 0$ has slope $-1$, not a root.
Answer: (A), (B), (D). | ||||
| 52 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Multiple correctHard | |
A hard multiple correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f'(x) = \frac{-(1 - bx) + b(b - x)}{(1 - bx)^2} = \frac{b^2 - 1}{(1 - bx)^2} < 0$, so $f$ is strictly decreasing and one-one on $(0, 1)$.
Its range is $(f(1^-), f(0^+)) = (-1, b)$, not the whole codomain $\mathbb{R}$. So $f : (0, 1) \rightarrow \mathbb{R}$ is not onto and therefore not invertible. (A) true.
(B), (C): $f^{-1}$ does not exist as a function on $(0, 1)$ (and $f$ is not even defined at $0$, so $f'(0)$ is not available), so neither statement can be made. (The formula $\frac{b - x}{1 - bx}$ is its own inverse algebraically, but that does not make $f$ invertible here.) False.
(D) $f^{-1}$ does not exist on $(0, 1)$, so it cannot be differentiable there. False.
Answer: (A). | ||||
| 53 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalHard | |
A hard numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations The question was meant with $\omega$ a non-real cube root of unity ($\omega = e^{2i\pi/3}$, so $1 + \omega + \omega^2 = 0$ and $\bar{\omega} = \omega^2$).
In that case $|x|^2 + |y|^2 + |z|^2 = x\bar{x} + y\bar{y} + z\bar{z}$. Expanding, each $|a|^2$, $|b|^2$, $|c|^2$ appears three times, and every cross term such as $a\bar{b}$ appears with coefficient $1 + \omega^2 + \omega = 0$. So the sum is $3(|a|^2 + |b|^2 + |c|^2)$ and the ratio is 3.
As printed, $\omega = e^{i\pi/3}$ is a sixth root of unity, $1 + \omega + \omega^2 = 1 + i\sqrt{3} \neq 0$, and the ratio is not fixed: $a = 1$, $b = c = 0.001$ gives about 3, while $a = b = c = 1$ gives $x = 3$, $y = z = 1 + i\sqrt{3}$ and ratio $\frac{9 + 4 + 4}{3} = \frac{17}{3}$. That is why the question was given marks to all.
Intended answer: 3. | ||||
| 54 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $f(x) = x^4 - 4x^3 + 12x^2 + x - 1$.
$f''(x) = 12x^2 - 24x + 24 = 12\left[(x - 1)^2 + 1\right] > 0$ for all $x$, so $f'$ is strictly increasing and $f'$ has exactly one zero: $f$ decreases and then increases, so $f$ has at most 2 real roots.
$f(0) = -1 < 0$, and $f(x) \rightarrow +\infty$ as $x \rightarrow \pm\infty$, so $f$ crosses zero once on each side of $0$: at least 2 real roots.
Exactly 2 distinct real roots.
Answer: 2. | ||||
| 55 | 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 A linear equation with integrating factor $e^{\int g'\,dx} = e^{g(x)}$:
$\frac{d}{dx}\left(ye^{g}\right) = gg'e^{g}$.
With $t = g$: $\int te^{t}\,dt = (t - 1)e^{t}$, so $ye^{g} = (g - 1)e^{g} + C$.
At $x = 0$: $0 = (0 - 1)e^0 + C \Rightarrow C = 1$. Hence $y = g - 1 + e^{-g}$.
At $x = 2$: $y(2) = 0 - 1 + e^0 = 0$.
Answer: 0. | ||||
| 56 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let the columns of $M$ be $C_1, C_2, C_3$, so $Me_j = C_j$.
$C_2 = (-1, 2, 3)^T$.
$C_1 - C_2 = (1, 1, -1)^T \Rightarrow C_1 = (0, 3, 2)^T$.
$C_1 + C_2 + C_3 = (0, 0, 12)^T \Rightarrow C_3 = (0, 0, 12)^T - (0, 3, 2)^T - (-1, 2, 3)^T = (1, -5, 7)^T$.
Diagonal: $M_{11} = 0$, $M_{22} = 2$, $M_{33} = 7$. Trace $= 0 + 2 + 7 = 9$.
Answer: 9. | ||||
| 57 | 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 $(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$, so $\vec{r} = \vec{c} + \lambda\vec{b}$.
$\vec{r} \cdot \vec{a} = 0$: $\vec{c} \cdot \vec{a} + \lambda\,\vec{b} \cdot \vec{a} = 0$, with $\vec{c} \cdot \vec{a} = -1 - 3 = -4$ and $\vec{b} \cdot \vec{a} = 1$. So $\lambda = 4$.
$\vec{r} = (1, 2, 3) + 4(-1, 1, 0) = (-3, 6, 3)$.
$\vec{r} \cdot \vec{b} = 3 + 6 + 0 = 9$.
Answer: 9. | ||||
| 58 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections The centre $(0, 0)$ gives $2x - 3y - 1 = -1 < 0$, so the larger part (containing the centre) is where $2x - 3y - 1 < 0$; the smaller part is $2x - 3y - 1 > 0$ inside the circle.
$\left(2, \frac{3}{4}\right)$: $4 - \frac{9}{4} - 1 = \frac{3}{4} > 0$; $4 + \frac{9}{16} < 6$. Inside the smaller part.
$\left(\frac{5}{2}, \frac{3}{4}\right)$: $\frac{25}{4} + \frac{9}{16} > 6$. Outside the circle.
$\left(\frac{1}{4}, -\frac{1}{4}\right)$: $\frac{1}{2} + \frac{3}{4} - 1 = \frac{1}{4} > 0$; inside the circle. Inside the smaller part.
$\left(\frac{1}{8}, \frac{1}{4}\right)$: $\frac{1}{4} - \frac{3}{4} - 1 < 0$. In the larger part.
Count: 2.
Answer: 2. | ||||
| 59 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Matrix matchHard | |
A hard matrix match question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals (A) $\vec{a} + \vec{b} = 2\sqrt{3}\hat{k} = \vec{c}$, so the sides run $\vec{a}$ then $\vec{b}$, closed by $\vec{c}$. The angle between the vectors $\vec{a}$ and $\vec{b}$: $\cos\phi = \frac{-1 + 3}{2 \cdot 2} = \frac{1}{2}$, $\phi = \frac{\pi}{3}$. At the vertex where $\vec{a}$ ends and $\vec{b}$ begins, the internal angle is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$. A → Q.
(B) For all $a$, $b$: $\int_a^b(f(x) - 3x)\,dx = -\int_a^b 2x\,dx$, so $f(x) - 3x = -2x$, $f(x) = x$ and $f\left(\frac{\pi}{6}\right) = \frac{\pi}{6}$. B → P. (The key also accepts P, Q, R, S, T together for this row.)
(C) $\int\sec(\pi x)\,dx = \frac{1}{\pi}\ln|\sec\pi x + \tan\pi x|$. At $x = \frac{5}{6}$: $\sec = -\frac{2}{\sqrt{3}}$, $\tan = -\frac{1}{\sqrt{3}}$, $|\cdot| = \sqrt{3}$. At $x = \frac{7}{6}$: $\sec = -\frac{2}{\sqrt{3}}$, $\tan = \frac{1}{\sqrt{3}}$, $|\cdot| = \frac{1}{\sqrt{3}}$.
So the integral $= \frac{1}{\pi}\left[\ln\sqrt{3} - \ln\frac{1}{\sqrt{3}}\right] = \frac{\ln 3}{\pi}$, and $\frac{\pi^2}{\ln 3} \cdot \frac{\ln 3}{\pi} = \pi$. C → S.
(D) $z = e^{i\theta}$, $0 < \theta < 2\pi$: $1 - z = -2i\sin\frac{\theta}{2}\,e^{i\theta/2}$, with $\sin\frac{\theta}{2} > 0$, so $\arg(1 - z) = \frac{\theta}{2} - \frac{\pi}{2} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, and $\left|\arg\frac{1}{1 - z}\right| = \left|\frac{\pi}{2} - \frac{\theta}{2}\right| < \frac{\pi}{2}$. Its least upper bound is $\frac{\pi}{2}$ (approached as $z \rightarrow 1$). D → T.
Official: A → Q; B → P (or P, Q, R, S, T); C → S; D → T. | ||||
| 60 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Matrix matchHard | |
A hard matrix match question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives (A) $z = e^{i\theta}$: $\frac{2iz}{1 - z^2} = \frac{2i}{\bar{z} - z} = \frac{2i}{-2i\sin\theta} = -\frac{1}{\sin\theta}$, real, with $\sin\theta \in [-1, 1] \setminus \{0\}$. The set is $(-\infty, -1] \cup [1, \infty)$. A → S.
(B) Let $t = 3^{x-1} > 0$; then $8 \cdot 3^{x-2} = \frac{8t}{3}$ and the argument is $\frac{8t}{3(1 - t^2)}$ ($t \neq 1$). Need it in $[-1, 1]$.
If $t < 1$ it is positive: $8t \leq 3 - 3t^2 \Rightarrow 3t^2 + 8t - 3 \leq 0 \Rightarrow (3t - 1)(t + 3) \leq 0 \Rightarrow t \leq \frac{1}{3} \Rightarrow x \leq 0$.
If $t > 1$ it is negative: $8t \leq 3t^2 - 3 \Rightarrow (3t + 1)(t - 3) \geq 0 \Rightarrow t \geq 3 \Rightarrow x \geq 2$.
Domain $(-\infty, 0] \cup [2, \infty)$. B → T.
(C) Expanding along the first row: $1(1 + \tan^2\theta) - \tan\theta(-\tan\theta + \tan\theta) + 1(\tan^2\theta + 1) = 2\sec^2\theta$. For $0 \leq \theta < \frac{\pi}{2}$ this takes every value in $[2, \infty)$. C → R.
(D) $f'(x) = \frac{3}{2}x^{1/2}(3x - 10) + 3x^{3/2} = \frac{x^{1/2}}{2}(9x - 30 + 6x) = \frac{15}{2}x^{1/2}(x - 2)$, which is $\geq 0$ for $x \geq 2$ (and $< 0$ on $(0, 2)$). Increasing on $[2, \infty)$. D → R.
Official: A → S; B → T; C → R; D → R. | ||||
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