JEE (Advanced) · 2022

JEE Advanced 2022, question by question

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

Sit this paper free All years, chapter-wise

Summary

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

Mathematics
36 Q
1. Trigonometry4
2. Integral Calculus4
3. Matrices and Determinants4
3 easy · 24 medium · 9 hard
Physics
36 Q
1. Electrostatics6
2. Optics6
3. Motion of System of Particles and Rigid Body4
7 easy · 21 medium · 8 hard
Chemistry
36 Q
1. p-Block Elements6
2. Thermodynamics2
3. Electrochemistry2
10 easy · 22 medium · 4 hard

Chapters

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

Electrostatics6
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Optics6
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Motion of System of Particles and Rigid Body4
Class 11: Systems of Particles and Rotational Motion
Atoms and Nuclei3
Class 12: Atoms; Class 12: Nuclei
Current Electricity3
Class 12: Current Electricity
Electromagnetic Induction and Alternating Currents2
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Kinematics2
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Properties of Bulk Matter2
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Thermodynamics2
Class 11: Thermodynamics
Gravitation1
Class 11: Gravitation
Physical World and Measurement1
Class 11: Units and Measurement
Oscillations and Waves1
Class 11: Oscillations; Class 11: Waves
Dual Nature of Matter and Radiation1
Class 12: Dual Nature of Radiation and Matter
Experimental Skills1
No matching NCERT chapter
Magnetic Effects of Current and Magnetism1
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter

Trigonometry4
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Integral Calculus4
Class 12: Integrals; Class 12: Application of Integrals
Matrices and Determinants4
Class 12: Matrices; Class 12: Determinants
Probability3
Class 11: Probability; Class 12: Probability
Complex Numbers3
Class 11: Complex Numbers and Quadratic Equations
Sequences and Series3
Class 11: Sequences and Series
Conic Sections3
Class 11: Conic Sections
Limits, Continuity and Differentiability2
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Permutations and Combinations2
Class 11: Permutations and Combinations
Circles2
Class 11: Conic Sections
Three Dimensional Geometry2
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Differential Equations2
Class 12: Differential Equations
Quadratic Equations1
Class 11: Complex Numbers and Quadratic Equations
Vector Algebra1
Class 12: Vector Algebra

Questions

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

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	NumericalMedium
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta = \cos^{-1}\sqrt{\frac{2}{2+\pi^2}}$, so $\theta \in (0, \frac{\pi}{2})$, $\cos\theta = \frac{\sqrt2}{\sqrt{2+\pi^2}}$ and $\tan\theta = \frac{\pi}{\sqrt2} \approx 2.22$, which puts $\theta \approx 1.148$, more than $\frac{\pi}{4}$. Second term: $\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{\sqrt2\pi}{1+\frac{\pi^2}{2}} = \frac{2\sqrt2\pi}{2+\pi^2}$. Since $2\theta \in (\frac{\pi}{2}, \pi)$, the principal value is $\sin^{-1}(\sin 2\theta) = \pi - 2\theta$. Third term: $\tan^{-1}\frac{\sqrt2}{\pi} = \tan^{-1}\frac{1}{\tan\theta} = \frac{\pi}{2} - \theta$. Sum: $\frac{3}{2}\theta + \frac{1}{4}(\pi - 2\theta) + \frac{\pi}{2} - \theta = \frac{3\pi}{4} \approx 2.356$. Answer: 2.36 (2.35 if truncated).
Q.2	Limits, Continuity and DifferentiabilityLimits of functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	NumericalMedium
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Put $t = \sqrt{x} - \sqrt{\alpha} \to 0^+$. Then $e^{\sqrt x} - e^{\sqrt\alpha} = e^{\sqrt\alpha}(e^t - 1)$, so $g(x) = \frac{2\ln t}{\sqrt\alpha + \ln(e^t - 1)} = \frac{2\ln t}{\sqrt\alpha + \ln t + \ln\frac{e^t-1}{t}}$. As $t \to 0^+$, $\ln t \to -\infty$ while $\ln\frac{e^t - 1}{t} \to 0$ and $\sqrt\alpha$ is fixed; dividing through by $\ln t$ gives $g(x) \to \frac{2}{1} = 2$. $f$ is continuous, so the limit is $f(2) = \sin\frac{2\pi}{12} = \sin\frac{\pi}{6} = \frac{1}{2}$. Answer: 0.5.
Q.3	ProbabilityProbability of an event	Class 11: Probability; Class 12: Probability	NumericalEasy
A easy numerical question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Write $F, C, B$ for the three symptoms. Pairs: $n(F\cap C) = 190 + 220 - 330 = 80$, $n(C \cap B) = 220 + 220 - 350 = 90$, $n(F \cap B) = 190 + 220 - 340 = 70$; $n(F\cap C\cap B) = 30$. At least one symptom: $190 + 220 + 220 - 80 - 90 - 70 + 30 = 420$, so $900 - 420 = 480$ have none. Exactly one: fever only $190 - 80 - 70 + 30 = 70$, cough only $220 - 80 - 90 + 30 = 80$, breathing only $220 - 90 - 70 + 30 = 90$; total $240$. At most one symptom: $480 + 240 = 720$, probability $\frac{720}{900} = 0.8$. Answer: 0.80.
Q.4	Complex NumbersAlgebra of complex numbers	Class 11: Complex Numbers and Quadratic Equations	NumericalMedium
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\frac{2 + 3z + 4z^2}{2 - 3z + 4z^2} = 1 + \frac{6z}{2 - 3z + 4z^2}$, so the ratio is real exactly when $\frac{z}{2 - 3z + 4z^2}$ is real, i.e. when its reciprocal $\frac{2}{z} - 3 + 4z$ is real. Write $z = re^{i\theta}$ with $\sin\theta \neq 0$. Then $\operatorname{Im}\left(4z + \frac{2}{z}\right) = 4r\sin\theta - \frac{2}{r}\sin\theta = 0$, so $4r = \frac{2}{r}$ and $r^2 = \frac{1}{2}$. Answer: $\|z\|^2 = 0.5$.
Q.5	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	NumericalMedium
A medium numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Rearrange: $(1 - i)\bar z = (1 + i)z^2$, so $\bar z = \frac{1+i}{1-i}z^2 = i z^2$. $z = 0$ is one root. Otherwise take moduli: $\|z\| = \|z\|^2$, so $\|z\| = 1$. Put $z = e^{i\theta}$: $e^{-i\theta} = e^{i(\pi/2 + 2\theta)}$, so $3\theta = -\frac{\pi}{2} + 2k\pi$, giving three distinct values of $\theta$ in $[0, 2\pi)$: $\theta = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$ (that is $z = i$, $-\frac{\sqrt3}{2} - \frac{i}{2}$, $\frac{\sqrt3}{2} - \frac{i}{2}$). Total: $1 + 3 = 4$ distinct roots. Answer: 4.
Q.6	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 Let $D_i = A_{i+1} - A_i = (l_i + d_1)(w_i + d_2) - l_iw_i = d_2l_i + d_1w_i + d_1d_2$. Then $D_{i+1} - D_i = d_2d_1 + d_1d_2 = 2d_1d_2 = 20$: the differences $D_i$ form an AP with common difference 20. $D_{50} = A_{51} - A_{50} = 1000$, so $D_j = 1000 + 20(j - 50)$. $A_{100} - A_{90} = \sum_{j=90}^{99} D_j = 10 \times 1000 + 20\sum_{j=90}^{99}(j - 50) = 10000 + 20(40 + 41 + \dots + 49) = 10000 + 20 \times 445 = 18900$. Answer: 18900.
Q.7	Permutations and CombinationsFundamental principle of counting	Class 11: Permutations and Combinations	NumericalMedium
A medium numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Digits may repeat (nothing in the question forbids it). Count by the first digit, which can only be 2, 3 or 4. First digit 2: all $6^3 = 216$ numbers $2abc$, less those below 2022. Those are $2000, 2002, 2003, 2004, 2006, 2007$ and $2020$: 7 numbers. That leaves $209$. First digit 3: $6^3 = 216$, all inside the interval. First digit 4: second digit 0, 2 or 3 gives $3 \times 36 = 108$; second digit 4 gives $44bc$, and since every allowed third digit is below 8, all $36$ are at most 4482 (the largest is 4477). Second digit 6 or 7 is too large. So $144$. Total: $209 + 216 + 144 = 569$. Answer: 569.
Q.8	CirclesStandard and general form	Class 11: Conic Sections	NumericalMedium
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Place $A$ at the origin, $B = (1, 0)$, $C = (0, 3)$. The circumcircle of a right triangle has the hypotenuse as diameter: centre $(\frac12, \frac32)$, radius $R = \frac{\sqrt{10}}{2}$. A circle touching both legs has centre $(r, r)$. Touching the circumcircle internally means the distance between centres is $R - r$: $(r - \frac12)^2 + (r - \frac32)^2 = (\frac{\sqrt{10}}{2} - r)^2$ $2r^2 - 4r + \frac{5}{2} = \frac{5}{2} - \sqrt{10}\,r + r^2$ $r^2 - (4 - \sqrt{10})r = 0$, so $r = 4 - \sqrt{10} \approx 0.8377$. Answer: 0.84 (0.83 if truncated).
Q.9	Integral CalculusIntegration by substitution	Class 12: Integrals; Class 12: Application of Integrals	Multiple correctMedium
A medium multiple correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $u = (\ln x)^{3/2}$, so $du = \frac{3}{2}\frac{(\ln x)^{1/2}}{x}dx$; $u$ runs from 0 to 1. $\int_0^1 \frac{2}{3}\frac{du}{(a - u)^2} = \frac{2}{3}\left[\frac{1}{a - u}\right]_0^1 = \frac{2}{3}\left(\frac{1}{a - 1} - \frac{1}{a}\right) = \frac{2}{3a(a-1)}$. (The integrand has no singularity because $a \notin [0, 1]$.) Setting it equal to 1: $3a^2 - 3a - 2 = 0$, $a = \frac{3 \pm \sqrt{33}}{6}$, i.e. $a \approx 1.457$ (in $(1, \infty)$) and $a \approx -0.457$ (in $(-\infty, 0)$). Both are allowed. (A) False: two values satisfy it. (B) False: neither value is an integer. (C) True: both are irrational. (D) True: there are two. Answer: (C), (D).
Q.10	Sequences and SeriesArithmetic progressions	Class 11: Sequences and Series	Multiple correctMedium
A medium multiple correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $a_k = 8k - 1$, so $\sum_{k=1}^{m} a_k = 4m(m+1) - m = 4m^2 + 3m$. $T_n = 3 + \sum_{k=1}^{n-1}a_k = 3 + 4(n-1)^2 + 3(n-1) = 4n^2 - 5n + 4$. (A) $T_{20} = 1600 - 100 + 4 = 1504 \neq 1604$. False. (B) $\sum_{k=1}^{20}T_k = 4\cdot\frac{20\cdot21\cdot41}{6} - 5\cdot\frac{20\cdot21}{2} + 4\cdot20 = 11480 - 1050 + 80 = 10510$. True. (C) $T_{30} = 3600 - 150 + 4 = 3454$. True. (D) $\sum_{k=1}^{30}T_k = 4\cdot\frac{30\cdot31\cdot61}{6} - 5\cdot\frac{30\cdot31}{2} + 120 = 37820 - 2325 + 120 = 35615 \neq 35610$. False. Answer: (B), (C).
Q.11	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Multiple correctHard
A hard multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The two faces on $P_1$ and $P_2$ share an edge on their line of intersection $L$. Every other edge is either a line in $P_1$ or in $P_2$ that meets $L$ (not parallel to it), or the edge joining the two remaining vertices, one on $P_1$ and one on $P_2$, both off $L$. So a line can be an edge if it lies in one plane and crosses $L$, or meets $P_1$ and $P_2$ at two different points not on $L$. Normals $n_1 = (10, 15, 12)$, $n_2 = (-2, 5, 4)$; $L$ has direction $n_1 \times n_2 = (0, -64, 80) \parallel (0, -4, 5)$. The point $(0, 4, 0)$ lies on both planes, so on $L$. (A) Direction $(0,0,5)$: $n_1\cdot d = 60$, $n_2 \cdot d = 20$, so it is parallel to neither plane. Through $(1,1,z)$ it meets $P_1$ at $z = \frac{35}{12}$ and $P_2$ at $z = \frac{17}{4}$: two different points, neither on $L$. Can be an edge. (B) Direction $(-5, 2, 3)$: $n_1\cdot d = 16$, $n_2\cdot d = 32$. $(6,0,0)$ lies on $P_1$ but not $P_2$; the line meets $P_2$ at $t = 1$, the point $(1, 2, 3)$, which is not on $P_1$. Can be an edge. (C) Through $(0, 4, 0)$, a point of $L$, and not lying in either plane ($n_1\cdot d = 103$, $n_2 \cdot d = 45$). It meets both planes only at that one point of $L$, so it cannot join a vertex on $P_1$ to one on $P_2$ off $L$, nor lie in a face. Cannot. (D) Direction $(1, -2, 3)$: $n_2\cdot d = 0$ and it passes through $(0,4,0)$ on $P_2$, so it lies in $P_2$. It is not parallel to $L$ and crosses $L$ at $(0, 4, 0)$, so it can be the second edge of the face on $P_2$. Can be an edge. Answer: (A), (B), (D).
Q.12	Three Dimensional GeometryCoordinates of a point in space	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Multiple correctMedium
A medium multiple correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry On the plane $x = -t - p$, $y = t$, $z = 1 + p$, so $x + y + z = 1$ for all $t, p$: the plane is $x + y + z = 1$, normal $(1,1,1)$. For $Q(10, 15, 20)$: $x + y + z - 1 = 44$. The image is $S = Q - \frac{2 \cdot 44}{3}(1,1,1) = \left(-\frac{58}{3}, -\frac{43}{3}, -\frac{28}{3}\right)$. (A) $3(\alpha + \beta) = -58 - 43 = -101$. True. (B) $3(\beta + \gamma) = -43 - 28 = -71$. True. (C) $3(\gamma + \alpha) = -28 - 58 = -86$. True. (D) $3(\alpha + \beta + \gamma) = -129 \neq -121$. False. Answer: (A), (B), (C).
Q.13	Conic SectionsParabola	Class 11: Conic Sections	Multiple correctMedium
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $S = (1, 0)$. A tangent $y = mx + \frac1m$ passes through $(-2, 1)$ when $1 = -2m + \frac1m$, i.e. $2m^2 + m - 1 = 0$, $m = \frac12$ or $-1$. The points of contact $(\frac{1}{m^2}, \frac{2}{m})$ are $(4, 4)$ and $(1, -2)$. Line $S \to (4,4)$ has unit direction $(\frac35, \frac45)$. $\vec{SP} = (-3, 1)$; its component along the line is $-\frac{9}{5} + \frac{4}{5} = -1$, so the foot is at distance 1 from $S$ and $Q = (1,0) - (\frac35, \frac45) = (\frac25, -\frac45)$; the perpendicular has length $\sqrt{\|SP\|^2 - 1} = \sqrt{10 - 1} = 3$. Line $S \to (1,-2)$ is $x = 1$; the foot from $P$ is $(1, 1)$, at distance 1 from $S$, and the perpendicular has length 3. So whichever point is called $P_1$: $SQ_1 = SQ_2 = 1$ and $PQ_1 = PQ_2 = 3$. (A) $SQ_1 = 1 \neq 2$. False. (B) $Q_1Q_2$ = distance between $(\frac25, -\frac45)$ and $(1, 1)$ $= \sqrt{\frac{9}{25} + \frac{81}{25}} = \frac{\sqrt{90}}{5} = \frac{3\sqrt{10}}{5}$. True. (C) $PQ_1 = 3$. True. (D) $SQ_2 = 1$. True. Answer: (B), (C), (D).
Q.14	Matrices and DeterminantsDeterminants of order two and three	Class 12: Matrices; Class 12: Determinants	Multiple correctHard
A hard multiple correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants First determinant, with $s = \sin\theta$: $1(1 + s^2) - s(-s + s) + 1(s^2 + 1) = 2 + 2s^2$; half of it is $1 + \sin^2\theta$. Second determinant: its diagonal is $\sin\pi = 0$, $-\cos\frac{\pi}{2} = 0$, $\tan\pi = 0$, and the off-diagonal entries are negatives of each other: $\sin(\theta - \frac{\pi}{4}) = -\cos(\theta + \frac{\pi}{4})$, $\cot(\theta + \frac{\pi}{4}) = \tan(\frac{\pi}{4} - \theta) = -\tan(\theta - \frac{\pi}{4})$, $\ln\frac{\pi}{4} = -\ln\frac{4}{\pi}$. A $3\times3$ skew-symmetric determinant is 0. So $f(\theta) = 1 + \sin^2\theta$ and $g(\theta) = \|\sin\theta\| + \|\cos\theta\| = \sin\theta + \cos\theta$ on $[0, \frac{\pi}{2}]$, whose maximum is $\sqrt2$ (at $\frac{\pi}{4}$) and minimum is 1 (at the ends). $p(x) = k(x - 1)(x - \sqrt2)$ and $p(2) = k(2 - \sqrt2) = 2 - \sqrt2$, so $k = 1$: $p < 0$ strictly between 1 and $\sqrt2 \approx 1.414$, and $p > 0$ outside $[1, \sqrt 2]$. (A) $\frac{3 + \sqrt2}{4} \approx 1.104$, inside: $p < 0$. True. (B) $\frac{1 + 3\sqrt2}{4} \approx 1.311$, inside: $p < 0$, not $> 0$. False. (C) $\frac{5\sqrt2 - 1}{4} \approx 1.518$, above $\sqrt2$: $p > 0$. True. (D) $\frac{5 - \sqrt2}{4} \approx 0.896$, below 1: $p > 0$, not $< 0$. False. Answer: (A), (C).
Q.15	TrigonometryTrigonometric functions	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Matching listMedium
A medium matching list question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions (I) $\sqrt2\sin(x + \frac{\pi}{4}) = 1$: $x = 2k\pi$ or $x = \frac{\pi}{2} + 2k\pi$. In $[-\frac{2\pi}{3}, \frac{2\pi}{3}]$: $0, \frac{\pi}{2}$. Two elements → (P). (II) $\tan 3x = \frac{1}{\sqrt3}$: $x = \frac{\pi}{18} + \frac{k\pi}{3}$. In $[-\frac{5\pi}{18}, \frac{5\pi}{18}]$: $\frac{\pi}{18}$ and $-\frac{5\pi}{18}$ (the next, $\frac{7\pi}{18}$, is outside). Two → (P). (III) $\cos 2x = \frac{\sqrt3}{2}$: $x = \pm\frac{\pi}{12} + k\pi$. In $[-\frac{6\pi}{5}, \frac{6\pi}{5}]$ (that is $\|x\| \le 1.2\pi$): $\pm\frac{\pi}{12}, \pm\frac{11\pi}{12}, \pm\frac{13\pi}{12}$. Six → (T). (IV) $\sqrt2\sin(x - \frac{\pi}{4}) = 1$: $x = \frac{\pi}{2} + 2k\pi$ or $x = \pi + 2k\pi$. In $[-\frac{7\pi}{4}, \frac{7\pi}{4}]$: $\frac{\pi}{2}, -\frac{3\pi}{2}, \pi, -\pi$. Four → (R). I→P, II→P, III→T, IV→R. Answer: (B).
Q.16	ProbabilityAddition and multiplication theorems	Class 11: Probability; Class 12: Probability	Matching listMedium
A medium matching list question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability In one round: $P(x > y) = \frac{15}{36} = \frac{5}{12}$ (win, W), $P(x = y) = \frac16$ (draw, D), $P(x < y) = \frac{5}{12}$ (loss, L). A win changes $X - Y$ by $+5$, a draw by 0, a loss by $-5$, so $X_i > Y_i$ exactly when $P_1$ has more wins than losses, and $X_i = Y_i$ when they are equal. Two rounds: $P(X_2 > Y_2) = P(WW) + P(WD \text{ or } DW) = \frac{25}{144} + 2\cdot\frac{5}{12}\cdot\frac16 = \frac{25 + 20}{144} = \frac{5}{16}$. $P(X_2 = Y_2) = P(DD) + P(WL \text{ or } LW) = \frac{1}{36} + \frac{50}{144} = \frac{54}{144} = \frac38$. So $P(X_2 \ge Y_2) = \frac{5}{16} + \frac{3}{8} = \frac{11}{16}$. Three rounds: $P(X_3 = Y_3) = P(DDD) + 3!\,P(W)P(L)P(D) = \frac{1}{216} + 6\cdot\frac{25}{144}\cdot\frac16 = \frac{2}{432} + \frac{75}{432} = \frac{77}{432}$. By symmetry between the players, $P(X_3 > Y_3) = \frac12\left(1 - \frac{77}{432}\right) = \frac{355}{864}$. I→Q, II→R, III→T, IV→S. Answer: (A).
Q.17	Matrices and DeterminantsConsistency of linear systems	Class 12: Matrices; Class 12: Determinants	Matching listHard
A hard matching list question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let $\frac1p = a + 9d$, $\frac1q = a + 99d$, $\frac1r = a + 999d$. Dividing the third equation by $pqr$: $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$, i.e. $a(x + y + z) + d(9x + 99y + 999z) = 0$. On solutions of the first two equations, $x + y + z = 1$ and $9x + 99y + 999z = (10x + 100y + 1000z) - (x + y + z) = -1$, so the third equation becomes $a - d = 0$. So: if $a = d$ the third equation adds nothing and the system (two independent equations in three unknowns) has infinitely many solutions; if $a \neq d$ it has no solution. $\frac{q}{r} = \frac{a + 999d}{a + 99d} = 10 \iff a = d$; $\frac{p}{q} = \frac{a + 99d}{a + 9d} = 10 \iff a = d$; $\frac{p}{r} = 100 \iff a + 999d = 100a + 900d \iff a = d$. When $a = d$ the solutions are those of $x + y + z = 1$, $x + 10y + 100z = 0$; both $(0, \frac{10}{9}, -\frac19)$ and $(\frac{10}{9}, -\frac19, 0)$ satisfy them, so (P), (Q), (R), (T) all hold. (I) $\frac{q}{r} = 10$: (P), (Q), (R), (T). (II) $\frac{p}{r} \neq 100$: (S). (III) $\frac{p}{q} \neq 10$: (S). (IV) $\frac{p}{q} = 10$: (P), (Q), (R), (T). Only option (B), I→Q, II→S, III→S, IV→R, is consistent with this. Answer: (B).
Q.18	Conic SectionsEllipse	Class 11: Conic Sections	Matching listMedium
A medium matching list question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The auxiliary circle is $x^2 + y^2 = 4$, so $F = (2\cos\phi, 2\sin\phi)$ and $\alpha = 2\cos\phi$. $E = (\alpha, y_E)$ on the ellipse. Tangent at $E$: $\frac{\alpha x}{4} + \frac{y_E y}{3} = 1$ meets the $x$-axis at $G = (\frac{4}{\alpha}, 0)$, independent of $y_E$. Triangle $FGH$ is right-angled at $H$: base $GH = \frac{4}{\alpha} - \alpha = \frac{4 - \alpha^2}{\alpha}$, height $FH = 2\sin\phi$. Area $= \frac12 \cdot \frac{4 - 4\cos^2\phi}{2\cos\phi}\cdot 2\sin\phi = \frac{2\sin^3\phi}{\cos\phi}$. (I) $\phi = \frac{\pi}{4}$: $\frac{2 \cdot \frac{1}{2\sqrt2}}{\frac{1}{\sqrt2}} = 1$ → (Q). (II) $\phi = \frac{\pi}{3}$: $\frac{2\cdot\frac{3\sqrt3}{8}}{\frac12} = \frac{3\sqrt3}{2}$ → (T). (III) $\phi = \frac{\pi}{6}$: $\frac{2\cdot\frac18}{\frac{\sqrt3}{2}} = \frac{1}{2\sqrt3}$ → (S). (IV) $\phi = \frac{\pi}{12}$: $\frac{2\sin^3\phi}{\cos\phi} = 2\tan\phi\sin^2\phi = 2(2 - \sqrt3)\cdot\frac{2 - \sqrt3}{4} = \frac{(2 - \sqrt3)^2}{2}$, and $(\sqrt3 - 1)^4 = (4 - 2\sqrt3)^2 = 4(2 - \sqrt3)^2$, so this equals $\frac{(\sqrt3 - 1)^4}{8}$ → (P). I→Q, II→T, III→S, IV→P. Answer: (C).

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	GravitationEscape velocity	Class 11: Gravitation	NumericalMedium
A medium numerical question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Let the common initial radius be $R$. $A$ keeps its density and its radius halves, so its new mass is $\frac{M_A}{8}$ and the mass lost is $\frac{7M_A}{8}$. That mass, at density $\rho_A$, occupies $\frac78$ of $A$'s original volume $\frac43\pi R^3$. As a shell on $B$ (radius $R$), the outer radius $R_B'$ satisfies $R_B'^3 = R^3 + \frac78R^3 = \frac{15}{8}R^3$, so $R_B' = \frac{15^{1/3}}{2}R$. $B$'s new mass is $2M_A + \frac78M_A = \frac{23}{8}M_A$. Escape speed $v = \sqrt{\frac{2GM}{r}}$: $\frac{v_B^2}{v_A^2} = \frac{\frac{23}{8}M_A / \frac{15^{1/3}R}{2}}{\frac{M_A}{8} / \frac{R}{2}} = \frac{23}{15^{1/3}}$. So $10n = 23$, $n = 2.3$. Answer: 2.30.
Q.2	Atoms and NucleiNuclear binding energy	Class 12: Atoms; Class 12: Nuclei	NumericalMedium
A medium numerical question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Q-value: $Q = (16.006 + 4.003 - 1.008 - 19.003)\,u\,c^2 = -0.002 \times 930 = -1.86$ MeV, so the reaction needs energy. Not all of the alpha's kinetic energy is available: momentum must be conserved, so the products keep the centre-of-mass motion. The threshold is $K_{th} = -Q\left(1 + \frac{m_\alpha}{m_N}\right) = 1.86 \times \left(1 + \frac{4.003}{16.006}\right) = 1.86 \times 1.2501 \approx 2.325$ MeV. (The exact relativistic expression $K_{th} = \frac{(\sum m_f)^2 - (\sum m_i)^2}{2m_N}c^2 = \frac{20.011^2 - 20.009^2}{2 \times 16.006} \times 930$ MeV gives the same 2.325 MeV.) Answer: 2.33 (2.32 if truncated).
Q.3	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	NumericalEasy
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Read the figure node by node. The whole bottom wire is one node, $G$. The top junction $J$ (the apex where $C_2$, $C_3$, $C_4$ meet) is joined by plain wire along the top to the upper ends of $C_5$ and of the 2 V cell, whose lower ends are on $G$. So the 2 V cell is connected directly across $J$ and $G$, and $C_3$ (between $J$ and $G$) has exactly 2 V across it, whatever the 6 V cell and $C_1$ do (they only fix the charge on $C_1$). $Q_3 = C_3V = 4\,\mu F \times 2\,V = 8\,\mu C$. Answer: 8.
Q.4	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalMedium
A medium numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The rod's foot is on the axis at $u = -\frac{40}{3}$ cm. It leans away from the lens (angle $\frac{2\pi}{3}$ with the axis, as drawn), so its top end is $2\cos\frac{\pi}{3} = 1$ cm further left and $2\sin\frac{\pi}{3} = \sqrt3$ cm high: $u = -\frac{43}{3}$ cm. Foot: $\frac1v = \frac{1}{10} - \frac{3}{40} = \frac{1}{40}$, $v = 40$ cm. Top: $\frac1v = \frac1{10} - \frac{3}{43} = \frac{13}{430}$, $v = \frac{430}{13}$ cm; magnification $\frac{v}{u} = -\frac{30}{13}$, so its image is $\frac{30\sqrt3}{13}$ cm below the axis, matching the given height. The image runs from $(\frac{430}{13}, -\frac{30\sqrt3}{13})$ to $(40, 0)$: horizontal extent $40 - \frac{430}{13} = \frac{90}{13}$. $\tan\alpha = \frac{30\sqrt3/13}{90/13} = \frac{1}{\sqrt3}$, so $\alpha = \frac{\pi}{6}$ and $n = 6$. Answer: 6.
Q.5	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	NumericalMedium
A medium 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 Starting from rest: at $t = \sqrt\pi$, $\omega = \alpha t = \frac{2\sqrt\pi}{3}$ rad/s and the angle turned is $\theta = \frac12\alpha t^2 = \frac{\pi}{3}$. The stone started at the bottom; after the disk turns through $\theta$ it is at height $R(1 - \cos\theta) = 1 - \frac12 = \frac12$ m. Rolling without slipping, its velocity is $\omega \times$ (vector from the contact point to the stone), so its components are $v_x = \omega R(1 - \cos\theta)$ and $v_y = \omega R\sin\theta$. Here $v_y = \frac{2\sqrt\pi}{3}\cdot\frac{\sqrt3}{2} = \sqrt{\frac{\pi}{3}}$ m/s, upward. It then rises a further $\frac{v_y^2}{2g} = \frac{\pi/3}{20} = \frac{\pi}{60}$ m. Maximum height $= \frac12 + \frac{\pi}{60} = \frac12 + \frac{x}{10}$, so $x = \frac{\pi}{6} \approx 0.5236$. Answer: 0.52.
Q.6	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 two 1 N forces are equal and opposite, so their net force is zero; they form a couple. As drawn, the down-slope force acts along a line $r$ below the centre (towards the incline) and the up-slope force along a line $r$ above it. Their moment is $2Fr = 2 \times 1 \times 0.5 = 1$ N m, and its sense turns the sphere the way it would turn rolling UP the slope, so it opposes rolling down. Take $a$ down the slope, friction $f$ up the slope, $I = \frac25mR^2 = 0.4$ kg m², and $\alpha = \frac{a}{R} = a$. Translation: $ma = mg\sin\theta - f$, i.e. $a = 5 - f$. Rotation about the centre: $I\alpha = fR - 1$, i.e. $0.4a = f - 1$. Adding: $1.4a = 4$, $a = \frac{20}{7} \approx 2.857$ m s$^{-2}$. Answer: 2.86.
Q.7	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	NumericalMedium
A medium numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The induced emf is constant: $\varepsilon = A\frac{dB}{dt} = 1 \times 0.04 = 0.04$ V. The circuit is then a constant emf driving an LC loop that starts with $q = 0$, $i = 0$: $L\frac{di}{dt} + \frac{q}{C} = \varepsilon$. Solution: $q = C\varepsilon(1 - \cos\omega t)$, $i = C\varepsilon\omega\sin\omega t$, with $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-4}}} = 100$ rad/s. $i_{max} = C\varepsilon\omega = 10^{-3} \times 0.04 \times 100 = 4 \times 10^{-3}$ A $= 4$ mA. Answer: 4.00.
Q.8	KinematicsProjectile motion	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	NumericalEasy
A easy numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Up to the highest point nothing changes: horizontal distance $\frac{d}{2}$, height $H = \frac{v^2\sin^2\theta}{2g}$, horizontal speed $v\cos\theta$. The fall from $H$ under $g' = \frac{g}{0.81}$ takes $\sqrt{\frac{2H}{g'}} = \sqrt{0.81}\sqrt{\frac{2H}{g}} = 0.9\,\frac{v\sin\theta}{g}$, i.e. 0.9 of the original descent time. So the second half covers $0.9 \times \frac{d}{2}$, and $d' = \frac{d}{2} + 0.45d = 0.95d$. Answer: 0.95.
Q.9	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 The battery stays connected, so the plate voltage stays $V$. Initially $C_a = \frac{K\varepsilon_0A}{d}$. Finally the slab ($d$, constant $K$) is in series with air of total thickness $d$: $C_b = \frac{\varepsilon_0A}{\frac{d}{K} + d} = \frac{K\varepsilon_0A}{d(K + 1)} = \frac{C_a}{K + 1}$. (A) Field in the dielectric: initially $\frac{V}{d}$; finally $\frac{Q_b}{K\varepsilon_0A} = \frac{V}{d(K+1)}$. It falls by a factor $K + 1$, not $2K$. False. (B) The capacitance becomes $\frac{1}{K+1}$ of its first value, which is what the statement means. True. (C) The voltage is held at $V$ by the battery. False. (D) The work done by the agent is $\Delta U - W_{battery} = \frac12(C_b - C_a)V^2 - (C_b - C_a)V^2 = \frac12(C_a - C_b)V^2 = \frac{\varepsilon_0AV^2}{2d}\cdot\frac{K^2}{K + 1}$, which depends on $K$. False. Answer: (B).
Q.10	Current ElectricityKirchhoff's rules	Class 12: Current Electricity	Multiple correctHard
A hard multiple correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Nodes: left $L$, top $T$, right $Rt$, bottom $B$, centre $O$ (take $V_O = 0$). Branches: $L$–$T$ ($R_6$), $T$–$Rt$ ($R_7$), $L$–$B$ ($R_5$), $B$–$Rt$ ($R_8$), $T$–$O$ ($R_2$), $O$–$B$ ($R_4$), $L$–$O$ ($\mathcal{E}_2$, $R_3$), $O$–$Rt$ ($\mathcal{E}_1$, $R_1$). Both cells have the positive (long) plate on the left, so going left to right through either cell the potential drops by its emf. By the up–down symmetry $V_T = V_B = V_M$. Node $T$: $(V_L - V_M) + (V_{Rt} - V_M) + (0 - V_M) = 0$, so $V_L + V_{Rt} = 3V_M$. Node $L$: $2(V_L - V_M) + (V_L - 6 - 0) = 0$, so $3V_L - 2V_M = 6$. Node $Rt$: current from $O$ to $Rt$ through $R_1$ is $(0 - 12 - V_{Rt})$; KCL: $2(V_{Rt} - V_M) - (-12 - V_{Rt}) = 0$, so $3V_{Rt} - 2V_M = -12$. Adding the last two: $3(V_L + V_{Rt}) - 4V_M = -6$, so $5V_M = -6$, $V_M = -1.2$ V; then $V_L = 1.2$ V, $V_{Rt} = -4.8$ V. $I(R_1) = \|0 - 12 - (-4.8)\| = 7.2$ A. (A) True. $I(R_2) = \|V_M - 0\| = 1.2$ A. (B) True. $I(R_3) = \|V_L - 6 - 0\| = 4.8$ A. (C) True. $I(R_5) = \|V_L - V_M\| = 2.4$ A. (D) True. Answer: (A), (B), (C), (D).
Q.11	Properties of Bulk MatterFluid mechanics	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Multiple correctHard
A hard 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 Adiabatic ideal gas with $\gamma = 2$: $P \propto \rho^\gamma = \rho^2$ and $T \propto \rho^{\gamma - 1} = \rho$. $T$ halves (300 K → 150 K), so $\rho_2 = 0.1$ kg m$^{-3}$ and $P_2 = 600 \times \left(\frac12\right)^2 = 150$ Pa. (A) $P_2 = 150$ Pa, not 300. False. (D) $\rho_2 = 0.1$, not 0.05. False. (B) Mass flow $\rho Av = 0.8$: $v_1 = \frac{0.8}{0.2 \times 0.1} = 40$ m/s, $v_2 = \frac{0.8}{0.1 \times 0.4} = 20$ m/s. True. (C) Steady adiabatic flow conserves enthalpy + kinetic + potential energy per unit mass: $\frac{\gamma}{\gamma - 1}\frac{P}{\rho} + \frac{v^2}{2} + gz$. Lower end: $2 \times \frac{600}{0.2} + 800 = 6800$; upper end: $2 \times \frac{150}{0.1} + 200 + 10h = 3200 + 10h$. So $h = 360$ m, not 590. False. Answer: (B).
Q.12	OpticsReflection and refraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Multiple correctHard
A hard multiple correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Put the base on the $x$-axis from $(0, 0)$ to $(L, 0)$, hole at $x = h$; $\theta$ is the angle between the ray and the base. (A) $\theta = 30^\circ$: the ray, the base and the side it heads for form a triangle with angles $30^\circ$ (at the hole) and $60^\circ$ (at the corner), so it meets that side at $90^\circ$ and is reflected straight back to the hole, for any $l$. True. (B) Take $\theta = 60^\circ$, i.e. the ray parallel to one side. Following it (slope-$60^\circ$ ray to the right side, reflection makes it horizontal, then off the left side it goes down parallel to the right side) it returns to the base at $x = L - h$. For $l = \frac{L}{2}$ that is the hole itself, after two reflections: the path through the three midpoints. True. (C) For $\theta = 60^\circ$ and any other $h$, the ray reaches the base at $L - h$, reflects (third reflection) into the starting direction, and the same two side-reflections bring it back to $L - (L - h) = h$, the hole. It comes out, so 'NEVER' is false. False. (D) By the count in (C), for $\theta = 60^\circ$ and $l \ne \frac{L}{2}$ the ray leaves after five reflections (side, side, base, side, side), not six. False. Answer: (A), (B).
Q.13	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Multiple correctMedium
A medium multiple correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The distance from O to a side is $\frac{\sqrt3}{2}a$; the side bisects the perpendicular from each charge, so every charge is at distance $r = \sqrt3a$ from O, the six spaced $60^\circ$ apart. (A) Six equal charges symmetrically placed: their fields cancel. True. (B) $x = -q$ is the symmetric set of six $q$ plus an extra $-2q$ at that position. Field $= \frac{2q}{4\pi\varepsilon_0(3a^2)} = \frac{q}{6\pi\varepsilon_0a^2}$. True. (C) $V = \frac{(5q + 2q)}{4\pi\varepsilon_0\sqrt3a} = \frac{7q}{4\sqrt3\pi\varepsilon_0a}$. True. (D) $V = \frac{5q - 3q}{4\pi\varepsilon_0\sqrt3a} = \frac{2q}{4\sqrt3\pi\varepsilon_0a}$, positive, not $-\frac{3q}{4\sqrt3\pi\varepsilon_0a}$. False. Answer: (A), (B), (C).
Q.14	Atoms and NucleiNuclear binding energy	Class 12: Atoms; Class 12: Nuclei	Multiple correctHard
A hard multiple correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Nuclear forces treat protons and neutrons alike, so the difference between proton and neutron binding comes from the Coulomb repulsion, which only protons feel and which lowers their binding. For $Z$ protons spread uniformly through a sphere of radius $R = R_0A^{1/3}$, the pairwise Coulomb energy is $E_C = \frac{3}{5}\frac{Z(Z-1)e^2}{4\pi\varepsilon_0R} \propto \frac{Z(Z - 1)}{A^{1/3}}$. The question attributes this Coulomb term to the proton binding, so $E_b^p - E_b^n = -E_C$. (A) $\propto Z(Z - 1)$. True. (B) $\propto A^{-1/3}$ (through $R$). True. (C) The Coulomb term reduces proton binding, so $E_b^p - E_b^n$ is negative. False. (D) Positron emission turns a proton into a neutron, $Z \to Z - 1$, so the Coulomb repulsion falls and $E_b^p$ increases. True. Answer: (A), (B), (D).
Q.15	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Matching listHard
A hard matching list question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Field inside the solenoid: $\vec B = B_0\hat n$ with $B_0 = \mu_0mI$. The loop's area vector is $A\hat k$, so the flux is $\Phi = B_0An_z$. Induced current (taking $\hat k$ as the positive sense): $i = -\frac{1}{R}\frac{d\Phi}{dt} = -\frac{B_0A}{R}\dot n_z$, giving the loop a moment $\vec\mu = iA\hat k$. Torque: $\vec\tau = \vec\mu \times \vec B = -\frac{B_0^2A^2}{R}\dot n_z(\hat k \times \hat n) = -\frac{2\alpha}{\omega}\dot n_z(\hat k \times \hat n)$. At $\omega t = \frac{\pi}{6}$: $\sin^2 = \frac14$, $\cos^2 = \frac34$. (I) $n_z = \frac{\cos\omega t}{\sqrt2}$, $\dot n_z = -\frac{\omega\sin\omega t}{\sqrt2}$, $\hat k \times \hat n = -\frac{\sin\omega t}{\sqrt2}\hat i$: $\vec\tau = -\alpha\sin^2\omega t\,\hat i = -\frac{\alpha}{4}\hat i$ → (Q). (II) $n_z = 0$: no flux change, $\vec\tau = 0$ → (P). (III) $\dot n_z = -\frac{\omega\sin\omega t}{\sqrt2}$, $\hat k \times \hat n = \frac{\sin\omega t}{\sqrt2}\hat j$: $\vec\tau = \alpha\sin^2\omega t\,\hat j = \frac{\alpha}{4}\hat j$ → (S). (IV) $n_z = \frac{\sin\omega t}{\sqrt2}$, $\dot n_z = \frac{\omega\cos\omega t}{\sqrt2}$, $\hat k \times \hat n = -\frac{\cos\omega t}{\sqrt2}\hat i$: $\vec\tau = \alpha\cos^2\omega t\,\hat i = \frac{3\alpha}{4}\hat i$ → (R). I→Q, II→P, III→S, IV→R. Answer: (C).
Q.16	KinematicsRelative velocity	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Matching listMedium
A medium matching list question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane (I) Both move on the same circle with speed $1$ m/s, $90^\circ$ apart, so their velocities are always perpendicular: $\|\vec v_A - \vec v_B\| = \sqrt{1 + 1} = \sqrt2$ → (S). (II) Time of flight $= \frac{2v\sin45^\circ}{g} = \frac{2 \times 5\pi/2}{10} = \frac{\pi}{2}$ s $\approx 1.57$ s, so both are in the air at $t = \frac{\pi}{3}$ s. As drawn they are fired towards each other, so the horizontal components ($v\cos45^\circ = \frac{5\pi}{2}$ each) are opposite: relative horizontal speed $5\pi$. Vertically, $v_{Ay} - v_{By} = -gt + g(t - 0.1) = -1$ m/s. Magnitude $\sqrt{25\pi^2 + 1}$ → (T). (III) $v_A = \cos t$, $v_B = \cos(t + \frac{\pi}{2}) = -\sin t$; relative speed $\|\cos t + \sin t\|$ at $t = \frac{\pi}{3}$: $\frac12 + \frac{\sqrt3}{2} = \frac{\sqrt3 + 1}{2}$ → (P). (IV) $A$'s velocity is horizontal, magnitude 1; $B$'s is vertical, 3. Relative speed $\sqrt{1 + 9} = \sqrt{10}$ → (R). I→S, II→T, III→P, IV→R. Answer: (C).
Q.17	ThermodynamicsFirst law	Class 11: Thermodynamics	Matching listMedium
A medium matching list question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics (I) $Q = mL = 10^{-3} \times 2250 = 2.25$ kJ; work by the system $P\Delta V = 10^5(10^{-3} - 10^{-6}) \approx 0.1$ kJ. $\Delta U = 2.25 - 0.1 \approx 2.15$ kJ $\approx 2$ kJ → (P). (II) Isobaric, $V \to 3V$: $T$ goes 500 K → 1500 K. Rigid diatomic, $C_V = \frac52R$: $\Delta U = 0.2 \times \frac52 \times 8 \times 1000 = 4000$ J $= 4$ kJ → (R). (III) Monatomic, $\gamma = \frac53$: $TV^{2/3}$ constant, so $T$ rises by $8^{2/3} = 4$. $\Delta U = \frac32nR(T_f - T_i) = \frac32P_iV_i(4 - 1) = \frac32 \times 2000 \times \frac13 \times 3 = 3000$ J $= 3$ kJ → (T). (IV) Diatomic with vibration: $C_V = \frac72R$, $C_P = \frac92R$. At constant pressure $\Delta U = Q\frac{C_V}{C_P} = 9 \times \frac79 = 7$ kJ → (Q). I→P, II→R, III→T, IV→Q. Answer: (C).
Q.18	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Matching listMedium
A medium matching list question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Use $\frac1v - \frac1u = \frac1f$ with distances measured from each lens, positive to the right. Lens 1 with $f = +10$, $u = -20$: $v = +20$, i.e. 15 cm to the right of lens 2: a virtual object $u_2 = +15$ for lens 2. (I) $f_2 = +15$: $\frac1v = \frac1{15} + \frac1{15}$, $v = 7.5$ cm right of lens 2 → (P). (II) $f_2 = -10$: $\frac1v = -\frac1{10} + \frac1{15} = -\frac1{30}$, $v = -30$: 30 cm to the left of lens 2 → (R). (III) $f_2 = -20$: $\frac1v = -\frac1{20} + \frac1{15} = \frac1{60}$, $v = 60$ cm right → (Q). (IV) Lens 1 $f = -20$, $u = -20$: $\frac1v = -\frac1{20} - \frac1{20}$, $v = -10$ (10 cm left of lens 1), so $u_2 = -15$. Lens 2 $f = +10$: $\frac1v = \frac1{10} - \frac1{15} = \frac1{30}$, $v = 30$ cm right → (T). I→P, II→R, III→Q, IV→T. Answer: (A).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	ThermodynamicsEnthalpy	Class 11: Thermodynamics	NumericalMedium
A medium numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Reaction in the bomb: $\mathrm{2Hg(g) + O_2(g) \to 2HgO(s)}$. Heat released at constant volume: $q = C\Delta T = 20.00 \times 14.8 = 296$ kJ, so $\Delta U = -296$ kJ for 2 mol. $\Delta n_g = 0 - 3 = -3$, so $\Delta H = \Delta U + \Delta n_gRT = -296 + (-3)(8.3 \times 10^{-3})(298) = -296 - 7.42 = -303.42$ kJ. This is $2\Delta_fH(\mathrm{HgO}) - 2\Delta_fH(\mathrm{Hg,g})$ (O$_2$ is zero): $\Delta_fH(\mathrm{HgO}) = \frac{-303.42 + 2 \times 61.32}{2} = \frac{-180.78}{2} = -90.39$ kJ mol$^{-1}$. $\|X\| = 90.39$. Answer: 90.39.
Q.2	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	NumericalEasy
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Potentials do not add, but free energies $\Delta G^0 = -nFE^0$ do. $\mathrm{MnO_4^- \to MnO_2}$ ($+7 \to +4$): $n = 3$, $nE = 3 \times 1.68 = 5.04$. $\mathrm{MnO_2 \to Mn^{2+}}$ ($+4 \to +2$): $n = 2$, $nE = 2 \times 1.21 = 2.42$. $\mathrm{Mn^{2+} \to Mn}$ ($+2 \to 0$): $n = 2$, $nE = 2 \times (-1.03) = -2.06$. Overall $n = 7$: $E^0 = \frac{5.04 + 2.42 - 2.06}{7} = \frac{5.40}{7} = 0.771$ V. Answer: 0.77.
Q.3	EquilibriumpH and buffers	Class 11: Equilibrium	NumericalMedium
A medium numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium The strong base reacts first, with the strongest acid present: $\mathrm{H_2CO_3 + NaOH \to NaHCO_3 + H_2O}$ uses up all 0.01 mol of each. Left: $\mathrm{HCO_3^-}$ $0.01 + 0.01 = 0.02$ mol and $\mathrm{CO_3^{2-}}$ 0.01 mol, a buffer of the second dissociation. $p\mathrm{H} = p\mathrm{K_{a2}} + \log\frac{[\mathrm{CO_3^{2-}}]}{[\mathrm{HCO_3^-}]} = 10.32 + \log\frac{0.01}{0.02} = 10.32 - 0.30 = 10.02$. Answer: 10.02.
Q.4	Some Basic Concepts of ChemistryStoichiometry	Class 11: Some Basic Concepts of Chemistry	NumericalMedium
A medium numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $M(\mathrm{Cu(NO_3)_2}) = 63 + 2 \times 62 = 187$; $\frac{3.74}{187} = 0.02$ mol. $\mathrm{2Cu^{2+} + 4I^- \to 2CuI\downarrow + I_2}$: the white precipitate is CuI, and 0.01 mol $\mathrm{I_2}$ is set free; with excess $\mathrm{I^-}$ it stays in solution as brown $\mathrm{I_3^-}$. $\mathrm{H_2S}$ reduces the iodine and is oxidised to sulphur: $\mathrm{I_2 + H_2S \to 2HI + S\downarrow}$. X is sulphur, 0.01 mol. Mass $= 0.01 \times 32 = 0.32$ g. Answer: 0.32.
Q.5	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 White phosphorus is $\mathrm{P_4}$ (124 g/mol): $\frac{1.24}{124} = 0.01$ mol. $\mathrm{P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2}$: Q is phosphine, 0.01 mol. Phosphine is absorbed by copper sulphate: $\mathrm{3CuSO_4 + 2PH_3 \to Cu_3P_2 + 3H_2SO_4}$, so 0.015 mol $\mathrm{CuSO_4}$ is needed. $M(\mathrm{CuSO_4}) = 63 + 32 + 64 = 159$; mass $= 0.015 \times 159 = 2.385$ g. Answer: 2.38 (truncated) or 2.39 (rounded).
Q.6	Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis	Class 11: Organic Chemistry – Some Basic Principles and Techniques	NumericalEasy
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Red phosphorus and bromine give $\mathrm{PBr_3}$, which converts the alcohol to the bromide; the aryl Br is untouched. R is 4-bromobenzyl bromide, $\mathrm{BrC_6H_4CH_2Br}$ ($\mathrm{C_7H_6Br_2}$). $M = 84 + 6 + 160 = 250$; 1.00 g is 0.004 mol and holds 0.008 mol Br. In the Carius method all of it ends up as AgBr ($M = 188$): $0.008 \times 188 = 1.504$ g. Answer: 1.50.
Q.7	Alcohols, Phenols and EthersReactions of phenols	Class 12: Alcohols, Phenols and Ethers	NumericalEasy
A easy numerical question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Step 1 (Dow process): chlorobenzene with NaOH at 623 K and 300 atm gives sodium phenoxide, i.e. phenol. Step 2: sulphonation with conc. $\mathrm{H_2SO_4}$ followed by conc. $\mathrm{HNO_3}$ replaces the sulphonic groups by nitro groups and gives picric acid, 2,4,6-trinitrophenol, $\mathrm{C_6H_3N_3O_7}$ (the route used because direct nitration of phenol oxidises it). $M = 72 + 3 + 42 + 112 = 229$. Hydrogen: $\frac{3}{229} \times 100 = 1.31\%$. Answer: 1.31.
Q.8	Alcohols, Phenols and EthersPreparation and properties	Class 12: Alcohols, Phenols and Ethers	NumericalMedium
A medium numerical question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers A: three acetylenes trimerise to benzene in a red-hot iron tube. 15 mol gives at most 5 mol; at 80%, 4 mol benzene. B: Friedel-Crafts alkylation with 1-chloropropane; the primary carbocation rearranges to the isopropyl cation, so B is cumene. 50%: 2 mol. C: the cumene process (air oxidation to the hydroperoxide, then acid) gives phenol and acetone (the acetone is the $-\mathrm{CH_3COCH_3}$ shown). 50%: 1 mol phenol. D: acetylation with $\mathrm{CH_3COCl}$/pyridine gives phenyl acetate, $\mathrm{CH_3COOC_6H_5}$ ($\mathrm{C_8H_8O_2}$, $M = 136$). 100%: 1 mol. Mass of D $= 136$ g. Answer: 136.
Q.9	Chemical Bonding and Molecular StructureMolecular orbital theory	Class 11: Chemical Bonding and Molecular Structure	Multiple correctMedium
A medium multiple correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Two $2p_z$ orbitals give $\sigma$/$\sigma^$ when they overlap end-on (molecular axis along $z$) and $\pi$/$\pi^$ when they overlap side-on (axis along $x$ or $y$, the lobes along $z$). (A) The $\sigma(2p_z)$ bonding orbital keeps the two nodal surfaces of the atomic $p_z$ orbitals, one through (near) each nucleus, perpendicular to the axis: two nodal planes in the usual description. True. (B) $\sigma^$ has its nodes perpendicular to the axis (including the $xy$-plane through the centre); no nodal plane contains the molecular axis ($xz$ contains the $z$ axis). False. (C) A bonding $\pi$ orbital has its nodal plane containing the axis (the $xy$-plane for $p_z$ lobes); it has no node perpendicular to the axis through the centre. False. (D) In side-on overlap of $p_z$ orbitals the molecular axis lies in the $xy$-plane, which is the nodal plane of both $p_z$ orbitals; $\pi^$ has that node (plus one perpendicular to the axis). True. Answer: (A), (D).
Q.10	Surface ChemistryPhysisorption and chemisorption	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter (A) Chemisorption involves chemical bonds to surface sites, so it stops at a single (unimolecular) layer. True. (B) Physisorption enthalpies are small, about 20 to 40 kJ mol$^{-1}$; 100 to 140 kJ mol$^{-1}$ is typical of chemisorption (80 to 240). False. (C) Adsorption lowers entropy, so to be spontaneous it must be exothermic; chemisorption is exothermic (only a few unusual cases differ). False. (D) Physisorption is exothermic and needs no activation energy, so lower temperature favours it. True. Answer: (A), (D).
Q.11	Isolation of MetalsOres and concentration	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Bauxite is first purified by leaching (Bayer process), then the alumina is electrolysed (Hall-Heroult process). (A) Alumina is not reduced by carbon; that would need impractically high temperatures, which is why electrolysis is used. False. (B) The sodium aluminate solution is neutralised with $\mathrm{CO_2}$, precipitating hydrated alumina: $\mathrm{2Na[Al(OH)_4] + CO_2 \to Al_2O_3 \cdot xH_2O\downarrow + Na_2CO_3 + H_2O}$. True. (C) Leaching: $\mathrm{Al_2O_3 + 2NaOH + 3H_2O \to 2Na[Al(OH)_4]}$ in hot concentrated NaOH. True. (D) Pure alumina dissolved in molten cryolite $\mathrm{Na_3AlF_6}$ is electrolysed; Al forms at the cathode, and oxygen at the carbon anodes burns them to CO and $\mathrm{CO_2}$. True. Answer: (B), (C), (D).
Q.12	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Galena is PbS. Nitric acid oxidises it, and the gas evolved is nitric oxide: $\mathrm{3PbS + 8HNO_3 \to 3Pb(NO_3)_2 + 3S + 2NO + 4H_2O}$. (A) NO has 11 valence electrons, one unpaired: paramagnetic. True. (B) A diatomic molecule is linear, not bent. False. (C) NO is a neutral oxide. False. (D) NO is colourless (it turns brown only when it meets air and forms $\mathrm{NO_2}$). True. Answer: (A), (D).
Q.13	AminesClassification and identification of amines	Class 12: Amines	Multiple correctMedium
A medium multiple correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Step 1 is the Hell-Volhard-Zelinsky reaction: P $= \mathrm{CH_3CHBrCOOH}$ (2-bromopropanoic acid). Step 2 is the Gabriel synthesis: phthalimide anion displaces Br, and hydrolysis releases phthalic acid and the amine. Q $= \mathrm{CH_3CH(NH_2)COOH}$, alanine. (A) $\mathrm{NaBH_4}$ does not reduce carboxylic acids. False. (B) Excess concentrated ammonia also substitutes the $\alpha$-bromine ($\mathrm{CH_3CHBrCOOH + 2NH_3 \to CH_3CH(NH_2)COO^-NH_4^+}$...), and acidification gives alanine. True. (C) Q is a primary aliphatic amine; with $\mathrm{HNO_2}$ it forms an unstable diazonium ion that loses $\mathrm{N_2}$ (the basis of the Van Slyke amino-nitrogen estimation). True. (D) The $-I$ effect of the $\alpha$-Br stabilises the carboxylate, so P is a stronger acid than propanoic acid. True. Answer: (B), (C), (D).
Q.14	AminesDiazonium salts	Class 12: Amines	Multiple correctMedium
A medium multiple correct question on diazonium salts, from Amines. NCERT: Class 12: Amines A consistent route: P reduces $\mathrm{NO_2}$ to $\mathrm{NH_2}$ (Q = p-toluidine); R is nitrous acid ($\mathrm{NaNO_2/HCl}$) at low temperature, giving S = 4-methylbenzenediazonium chloride; warming with water gives T = p-cresol. U must remove the diazo group (reduction with $\mathrm{H_3PO_2}$ or ethanol gives toluene) and then oxidise the methyl group ($\mathrm{KMnO_4}$/KOH, heat, then acid work-up) to give benzoic acid. (A) $\mathrm{H_2/Pd}$ reduces the nitro group, $\mathrm{NaNO_2/HCl}$ diazotises, $\mathrm{H_3PO_2}$ then $\mathrm{KMnO_4}$-KOH gives benzoic acid. True. (B) Sn/HCl also reduces the nitro group, $\mathrm{HNO_2}$ diazotises, and S is the diazonium chloride drawn. True. (C) S and T are as above, and ethanol, like $\mathrm{H_3PO_2}$, replaces $\mathrm{N_2^+}$ by H before the oxidation. True. (D) If Q were 4-nitrobenzoic acid, reducing it gives 4-aminobenzoic acid, which could not lead to p-cresol (the methyl group is gone). False. Answer: (A), (B), (C).
Q.15	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	Matching listHard
A hard matching list question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics (II) $[\mathrm X] \ll \mathrm{X_s}$: rate $\approx \frac{k}{\mathrm{X_s}}[\mathrm X]$, first order. Half-life independent of the initial concentration → (Q) (also $\ln[\mathrm X]$ linear in time). (III) $[\mathrm X] \gg \mathrm{X_s}$: rate $\approx k$, zero order. $[\mathrm X]$ falls linearly with time → (S). (IV) $[\mathrm X] \gg \mathrm{X_s}$: rate $\approx k[\mathrm X]$, first order. $\ln[\mathrm X]$ falls linearly with time → (T). (I) Over all concentrations, integrate $-\frac{d[\mathrm X]}{dt} = \frac{k[\mathrm X]}{\mathrm{X_s} + [\mathrm X]}$: $\mathrm{X_s}\ln\frac{[\mathrm X]_0}{[\mathrm X]} + ([\mathrm X]_0 - [\mathrm X]) = kt$. At half-life: $t_{1/2} = \frac{\mathrm{X_s}\ln2}{k} + \frac{[\mathrm X]_0}{2k}$, a straight line in $[\mathrm X]_0$ with a positive intercept → (P). (The rate-against-concentration curve (R) also describes it, but only option (A) is consistent with II, III and IV.) I→P, II→Q, III→S, IV→T. Answer: (A).
Q.16	Principles Related to Practical ChemistryPreparation of inorganic and organic compounds	No matching NCERT chapter	Matching listEasy
A easy matching list question on preparation of inorganic and organic compounds, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter The products of each reaction: (P) $\mathrm{Mg(HCO_3)_2 + 2Ca(OH)_2 \to Mg(OH)_2\downarrow + 2CaCO_3\downarrow + 2H_2O}$ (Clark's method): $\mathrm{Mg(OH)_2}$ and $\mathrm{CaCO_3}$. (Q) $\mathrm{BaO_2 + H_2SO_4 \to BaSO_4\downarrow + H_2O_2}$: $\mathrm{H_2O_2}$. (R) $\mathrm{Ca(OH)_2 + MgCl_2 \to Mg(OH)_2\downarrow + CaCl_2}$: $\mathrm{Mg(OH)_2}$. (S) $\mathrm{BaO_2 + 2HCl \to BaCl_2 + H_2O_2}$: $\mathrm{BaCl_2}$ and $\mathrm{H_2O_2}$. (T) $\mathrm{Ca(HCO_3)_2 + Ca(OH)_2 \to 2CaCO_3\downarrow + 2H_2O}$: $\mathrm{CaCO_3}$. Check each option: (A) has IV→R, but (R) gives no $\mathrm{CaCO_3}$. (B) and (C) have I→T, but (T) gives no $\mathrm{H_2O_2}$. (D) I→Q, II→R, III→S, IV→P: every pairing is a real formation. Answer: (D).
Q.17	Coordination CompoundsCrystal field theory	Class 12: Coordination Compounds	Matching listMedium
A medium matching list question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (I) $\mathrm{[Cr(CN)_6]^{4-}}$: Cr(II), $d^4$; cyanide is a strong-field ligand, so low spin $t_{2g}^4e_g^0$, 2 unpaired electrons. Properties: P, R, T. (II) $\mathrm{[RuCl_6]^{2-}}$: Ru(IV), $4d^4$; 4d metals have large splitting and are low spin: $t_{2g}^4$. Properties: P, R, S, T. (III) $\mathrm{[Cr(H_2O)_6]^{2+}}$: Cr(II), $d^4$, weak field, high spin $t_{2g}^3e_g^1$, 4 unpaired: $\mu = \sqrt{4 \times 6} = 4.9$ BM. Properties: Q, T. (IV) $\mathrm{[Fe(H_2O)_6]^{2+}}$: Fe(II), $d^6$, high spin $t_{2g}^4e_g^2$, 4 unpaired, 4.9 BM. Properties: P, Q. Option (A), I→R, T; II→P, S; III→Q, T; IV→P, Q, uses only true pairings. (B) has I→S (Cr is +2); (C) has III→R (it is high spin); (D) has I→Q (only 2 unpaired). Answer: (A).
Q.18	Principles Related to Practical ChemistryDetection of elements and functional groups	No matching NCERT chapter	Matching listMedium
A medium matching list question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter (I) Aniline contains N: the Lassaigne extract gives Prussian blue (P). Bromine water gives a white precipitate of 2,4,6-tribromoaniline (S). No sulphur, so not (Q). → P, S. (II) o-Cresol is a phenol: violet colour with neutral $\mathrm{FeCl_3}$ (T). It is too weak an acid to liberate $\mathrm{CO_2}$ from bicarbonate, and has no N or S. (III) Cysteine, $\mathrm{HSCH_2CH(NH_2)COOH}$, contains S, so the fusion extract (sulphide) responds to sodium nitroprusside (Q), and its carboxylic acid group is taken to release $\mathrm{CO_2}$ from bicarbonate (R). → Q, R. (IV) Caprolactam, a cyclic amide, contains N: Prussian blue (P); no acidic or phenolic group. → P. Only option (D), I→P, S; II→T; III→Q, R; IV→P, has these. Answer: (D).

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	TrigonometryTrigonometric identities	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	NumericalHard
A hard numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Pair the terms. $\frac{\sin\alpha}{\cos\beta} + \frac{\sin\beta}{\cos\alpha} = \frac{\sin\alpha\cos\alpha + \sin\beta\cos\beta}{\cos\alpha\cos\beta}$ and $\frac{\cos\beta}{\sin\alpha} + \frac{\cos\alpha}{\sin\beta} = \frac{\sin\beta\cos\beta + \sin\alpha\cos\alpha}{\sin\alpha\sin\beta}$. Common numerator: $\sin\alpha\cos\alpha + \sin\beta\cos\beta = \frac{1}{2}(\sin 2\alpha + \sin 2\beta) = \sin(\alpha+\beta)\cos(\alpha-\beta)$. Sum of the two denominators' reciprocals: $\frac{1}{\cos\alpha\cos\beta} + \frac{1}{\sin\alpha\sin\beta} = \frac{\cos(\alpha-\beta)}{\sin\alpha\cos\alpha\sin\beta\cos\beta} = \frac{4\cos(\alpha-\beta)}{\sin 2\alpha\,\sin 2\beta}$. So the bracket is $E = \frac{4\sin(\alpha+\beta)\cos^2(\alpha-\beta)}{\sin 2\alpha\sin 2\beta}$. Now $\sin 2\alpha\sin 2\beta = \frac{1}{2}[\cos(2\alpha - 2\beta) - \cos(2\alpha + 2\beta)]$, with $\cos 2(\alpha-\beta) = 2\cdot\frac{4}{9} - 1 = -\frac{1}{9}$ and $\cos 2(\alpha+\beta) = 1 - 2\cdot\frac{1}{9} = \frac{7}{9}$. So $\sin 2\alpha\sin 2\beta = \frac{1}{2}\left(-\frac{1}{9} - \frac{7}{9}\right) = -\frac{4}{9}$ (negative, as it must be with $\beta < 0 < \alpha$). $E = \frac{4\cdot\frac{1}{3}\cdot\frac{4}{9}}{-\frac{4}{9}} = -\frac{4}{3}$, so $E^2 = \frac{16}{9} \approx 1.78$. Greatest integer $\le 1.78$ is $1$. Answer: $1$.
Q.2	Differential EquationsSeparation of variables	Class 12: Differential Equations	NumericalEasy
A easy numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $\frac{dy}{y(y-4)} = \frac{dx}{x}$. Partial fractions: $\frac{1}{y(y-4)} = \frac{1}{4}\left(\frac{1}{y-4} - \frac{1}{y}\right)$. Integrate: $\frac{1}{4}\ln\left\|\frac{y-4}{y}\right\| = \ln x + C$. At $x = 1$, $y = 2$: $\left\|\frac{-2}{2}\right\| = 1$, so $C = 0$ and $\left\|\frac{y-4}{y}\right\| = x^4$. The slope $\frac{y^2 - 4y}{x}$ is never zero, so $y$ never reaches $0$ or $4$; starting at $2$ it stays in $(0, 4)$, where $\frac{y-4}{y} < 0$. Hence $\frac{4 - y}{y} = x^4$, i.e. $y = \frac{4}{1 + x^4}$. $y(\sqrt{2}) = \frac{4}{1 + 4} = \frac{4}{5}$, so $10\,y(\sqrt{2}) = 8$. Answer: $8$.
Q.3	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	NumericalMedium
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The two integrands are inverse functions of each other: if $y = \log_2(x^3 + 1)$ then $x = (2^y - 1)^{1/3}$. In the second integral put $t = (2^x - 1)^{1/3}$, so $x = \log_2(t^3 + 1)$. Limits: $x = 1 \Rightarrow t = 1$; $x = \log_2 9 \Rightarrow t = 2$. $I_2 = \int_1^2 t\, d\big(\log_2(t^3+1)\big) = \big[t\log_2(t^3+1)\big]_1^2 - \int_1^2 \log_2(t^3+1)\,dt = 2\log_2 9 - 1\cdot\log_2 2 - I_1$. So $I_1 + I_2 = 2\log_2 9 - 1 = 4\log_2 3 - 1 \approx 4(1.585) - 1 = 5.34$. Greatest integer: $5$. Answer: $5$.
Q.4	Quadratic EquationsRelations between roots and coefficients	Class 11: Complex Numbers and Quadratic Equations	NumericalMedium
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Take $\log_5$ of both sides and write $t = \log_5 x$: $(16t^3 - 68t)\,t = -16$, i.e. $16t^4 - 68t^2 + 16 = 0$, or $4t^4 - 17t^2 + 4 = 0$. As a quadratic in $t^2$: $t^2 = \frac{17 \pm 15}{8} = 4$ or $\frac{1}{4}$, so $t = \pm 2, \pm\frac{1}{2}$ — four distinct values, each giving one positive $x = 5^t$. Product of the $x$ values $= 5^{t_1 + t_2 + t_3 + t_4} = 5^{0} = 1$. Answer: $1$.
Q.5	Limits, Continuity and DifferentiabilityLimits of functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	NumericalMedium
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability The denominator $x\sin^2 x \sim x^3$, so expand the numerator to order $x^3$. $e^{x^3} = 1 + x^3 + O(x^6)$; $(1 - x^3)^{1/3} = 1 - \frac{x^3}{3} + O(x^6)$; difference $= \frac{4}{3}x^3 + O(x^6)$. $(1 - x^2)^{1/2} - 1 = -\frac{x^2}{2} + O(x^4)$ and $\sin x = x + O(x^3)$, so the product is $-\frac{x^3}{2} + O(x^5)$. Numerator $= \left(\frac{4}{3} - \frac{1}{2}\right)x^3 + \dots = \frac{5}{6}x^3 + \dots$, so $\beta = \frac{5}{6}$. $6\beta = 5$. Answer: $5$.
Q.6	Matrices and DeterminantsDeterminants of order two and three	Class 12: Matrices; Class 12: Determinants	NumericalMedium
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Factor: $A^7 - (\beta-1)A^6 - \beta A^5 = A^5\big(A^2 - (\beta - 1)A - \beta I\big) = A^5 (A - \beta I)(A + I)$. $\det A = \beta(1\cdot(-2) - (-2)\cdot 1) - 0 + 1\cdot(2\cdot 1 - 1\cdot 3) = 0 - 1 = -1 \neq 0$, so $A^5$ is not singular. $\det(A + I) = \det\begin{pmatrix}\beta+1 & 0 & 1\\ 2 & 2 & -2\\ 3 & 1 & -1\end{pmatrix} = (\beta+1)(-2 + 2) + 1\cdot(2 - 6) = -4 \neq 0$. So $\det(A - \beta I) = 0$: $\det\begin{pmatrix}0 & 0 & 1\\ 2 & 1-\beta & -2\\ 3 & 1 & -2-\beta\end{pmatrix} = 1\cdot\big(2\cdot 1 - 3(1 - \beta)\big) = 3\beta - 1 = 0$. $\beta = \frac{1}{3}$, $9\beta = 3$. Answer: $3$.
Q.7	Conic SectionsHyperbola	Class 11: Conic Sections	NumericalHard
A hard numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Here $a = 10$, $b = 8$. $P$ is on the right branch, nearer $S$, so $S_1P - SP = 2a = 20$: $SP = \beta - 20$. The tangent at $P$ bisects $\angle SPS_1$ internally. The line through $S$ parallel to it makes angle $\frac{\alpha}{2}$ with $SP$ (alternate angles) and angle $\frac{\alpha}{2}$ with line $S_1P$ (corresponding angles). So triangle $PSP_1$ has equal base angles $\frac{\alpha}{2}$ at $S$ and $P_1$: it is isosceles with $PP_1 = PS$, and the distance of $P$ from the base $SP_1$ is $\delta = PS\sin\frac{\alpha}{2} = (\beta - 20)\sin\frac{\alpha}{2}$. In triangle $SPS_1$, with $SS_1 = 2c$ and $c^2 = a^2 + b^2$: $4c^2 = \beta^2 + (\beta-20)^2 - 2\beta(\beta-20)\cos\alpha = 400 + 2\beta(\beta-20)(1 - \cos\alpha)$ (using $(\beta - (\beta-20))^2 = 400$). So $2\beta(\beta-20)(1-\cos\alpha) = 4c^2 - 4a^2 = 4b^2 = 256$. With $1 - \cos\alpha = 2\sin^2\frac{\alpha}{2}$: $\beta(\beta-20)\sin^2\frac{\alpha}{2} = b^2 = 64$. Then $\frac{\beta\delta}{9}\sin\frac{\alpha}{2} = \frac{\beta(\beta-20)\sin^2\frac{\alpha}{2}}{9} = \frac{64}{9} \approx 7.11$, the same for every $P$. Greatest integer: $7$. Answer: $7$.
Q.8	Integral CalculusArea under simple curves	Class 12: Integrals; Class 12: Application of Integrals	NumericalMedium
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Both functions are even, so find the area for $0 \le x \le \frac{3}{4}$ and double it. They meet where $x^2 + \frac{5}{12} = 2 - \frac{8x}{3}$, i.e. $12x^2 + 32x - 19 = 0$, $x = \frac{-32 + 44}{24} = \frac{1}{2}$. For $0 \le x \le \frac{1}{2}$ the smaller is $f$ (at $0$: $\frac{5}{12} < 2$); for $\frac{1}{2} \le x \le \frac{3}{4}$ it is $g$. $\int_0^{1/2}\left(x^2 + \frac{5}{12}\right)dx = \frac{1}{24} + \frac{5}{24} = \frac{1}{4}$. $\int_{1/2}^{3/4} g\,dx$ is a triangle with base $\frac{1}{4}$ and height $g(\frac{1}{2}) = \frac{2}{3}$: area $\frac{1}{12}$. Half the area $= \frac{1}{4} + \frac{1}{12} = \frac{1}{3}$, so $\alpha = \frac{2}{3}$ and $9\alpha = 6$. Answer: $6$.
Q.9	TrigonometryTrigonometric functions	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Multiple correctHard
A hard multiple correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Triangle $PQR$: $\angle PQR = 70°$, $\angle PRQ = \angle QRS - \angle PRS = 30°$, so $\angle QPR = 80°$. Sine rule: $\alpha = PQ = \frac{QR\sin 30°}{\sin 80°} = \frac{1}{2\sin 80°}$. Triangle $QRS$: $\angle RQS = 70° - 15° = 55°$, $\angle QRS = 70°$, so $\angle QSR = 55°$ and $RS = QR = 1$. Triangle $PRS$: by the sine rule $\frac{PS}{\sin\angle PRS} = \frac{RS}{\sin\angle RPS}$, so $\beta\sin\theta° = RS\sin 40° = \sin 40°$. $4\alpha\beta\sin\theta° = \frac{4\sin 40°}{2\sin 80°} = \frac{2\sin 40°}{2\sin 40°\cos 40°} = \sec 40° \approx 1.305$. (A) $(0, \sqrt{2}) = (0, 1.414)$: contains $1.305$. TRUE. (B) $(1, 2)$: contains it. TRUE. (C) $(\sqrt{2}, 3)$: $1.305 < 1.414$. FALSE. (D) $(2\sqrt{2}, 3\sqrt{2})$: FALSE. Answer: (A), (B).
Q.10	Sequences and SeriesRelation between A.M. and G.M.	Class 11: Sequences and Series	Multiple correctMedium
A medium multiple correct question on relation between a.m. and g.m., from Sequences and Series. NCERT: Class 11: Sequences and Series $\sin^2\frac{\pi}{6} = \frac{1}{4}$, so $\alpha = \sum_{k\ge1}\left(\frac{1}{4}\right)^k = \frac{1/4}{1 - 1/4} = \frac{1}{3}$ and $g(x) = 2^{x/3} + 2^{(1-x)/3}$. AM–GM: $g(x) \ge 2\sqrt{2^{x/3}\cdot 2^{(1-x)/3}} = 2\cdot 2^{1/6} = 2^{7/6}$, with equality only when $x = 1 - x$, i.e. $x = \frac{1}{2}$. Put $u = 2^{x/3} \in [1, 2^{1/3}]$: $g = u + \frac{2^{1/3}}{u}$ is decreasing then increasing in $u$, so the maximum is at the ends $u = 1$ and $u = 2^{1/3}$, i.e. $x = 0$ and $x = 1$, both giving $1 + 2^{1/3}$. (A) Minimum $2^{7/6}$. TRUE. (B) Maximum $1 + 2^{1/3}$. TRUE. (C) Maximum at two points, $x = 0$ and $x = 1$. TRUE. (D) Minimum only at $x = \frac{1}{2}$. FALSE. Answer: (A), (B), (C).
Q.11	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	Multiple correctHard
A hard multiple correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Write $z = re^{i\phi}$. Then $(\bar z)^2 = r^2e^{-2i\phi}$ and $\frac{1}{z^2} = \frac{1}{r^2}e^{-2i\phi}$, so $(\bar z)^2 + \frac{1}{z^2} = \left(r^2 + \frac{1}{r^2}\right)e^{-2i\phi}$. If its real and imaginary parts are integers $a, b$, then $\left(r^2 + \frac{1}{r^2}\right)^2 = a^2 + b^2$, i.e. $r^4 + \frac{1}{r^4} + 2$ must be a positive integer that is a sum of two squares; conversely any such value is reached by choosing $\phi$. (A) $r^4 = \frac{43 + 3\sqrt{205}}{2}$, $\frac{1}{r^4} = \frac{2(43 - 3\sqrt{205})}{43^2 - 9\cdot205} = \frac{43 - 3\sqrt{205}}{2}$. $r^4 + r^{-4} + 2 = 45 = 6^2 + 3^2$. POSSIBLE. (B) $r^4 = \frac{7 + \sqrt{33}}{4}$, $r^{-4} = \frac{7 - \sqrt{33}}{4}$; sum $+2 = \frac{11}{2}$, not an integer. NOT possible. (C) $r^4 = \frac{9 + \sqrt{65}}{4}$, $r^{-4} = \frac{9 - \sqrt{65}}{4}$; sum $+2 = \frac{13}{2}$. NOT possible. (D) $r^4 = \frac{7 + \sqrt{13}}{6}$, $r^{-4} = \frac{7 - \sqrt{13}}{6}$; sum $+2 = \frac{13}{3}$. NOT possible. Answer: (A).
Q.12	CirclesRadius and centre	Class 11: Conic Sections	Multiple correctMedium
A medium multiple correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections The centres of the small circles are at distance $R + r$ from the centre of $G$ and form a regular $n$-gon whose side is $2r$ (neighbours touch). Half the central angle is $\frac{\pi}{n}$, so $\sin\frac{\pi}{n} = \frac{r}{R + r}$, i.e. $R = r\left(\frac{1}{\sin(\pi/n)} - 1\right)$. (A) $n = 4$: $R = r(\sqrt{2} - 1)$ exactly, so $(\sqrt{2} - 1)r < R$ is FALSE. (B) $n = 5$: $\sin 36° \approx 0.588$, $R \approx 0.70r$, so $r < R$ is FALSE. (C) $n = 8$: $\sin 22.5° \approx 0.383$, $R \approx 1.61r > (\sqrt{2} - 1)r$. TRUE. (D) $n = 12$: $\sin 15° = \frac{\sqrt{3} - 1}{2\sqrt{2}}$, so $R + r = \frac{2\sqrt{2}}{\sqrt{3} - 1}r = \sqrt{2}(\sqrt{3} + 1)r$ and $R = \sqrt{2}(\sqrt{3}+1)r - r < \sqrt{2}(\sqrt{3}+1)r$. TRUE. Answer: (C), (D).
Q.13	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Multiple correctHard
A hard multiple correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra The left side is $\vec c \times \vec b$ (row 1: $c_2b_3 - c_3b_2$, and so on). The right side is $\vec a - \vec c$. So $\vec c \times \vec b = \vec a - \vec c$. $\vec a\cdot\vec b = 0$ gives $3 + b_2 - b_3 = 0$, i.e. $b_3 = b_2 + 3$. Dot with $\vec b$: $0 = \vec a\cdot\vec b - \vec c\cdot\vec b$, so $\vec b\cdot\vec c = 0$. (B) TRUE. Dot with $\vec c$: $0 = \vec a\cdot\vec c - \|\vec c\|^2$, so $\vec a\cdot\vec c = \|\vec c\|^2$. If this were $0$ then $\vec c = \vec 0$ and the equation would give $\vec a = \vec 0$, which is false. So $\vec a\cdot\vec c \ne 0$. (A) FALSE. Cross $\vec b$ with the equation: $\vec b\times(\vec c\times\vec b) = \|\vec b\|^2\vec c - (\vec b\cdot\vec c)\vec b = \|\vec b\|^2\vec c$, and $\vec b\times(\vec a - \vec c) = \vec b\times\vec a + \vec c\times\vec b = \vec b\times\vec a + \vec a - \vec c$. Hence $(1 + \|\vec b\|^2)\vec c = \vec a + \vec b\times\vec a$. Since $\vec a \perp \vec b$, $\|\vec b\times\vec a\| = \|\vec a\|\|\vec b\|$ and $\vec a \perp \vec b\times\vec a$: $(1 + \|\vec b\|^2)^2\|\vec c\|^2 = 11 + 11\|\vec b\|^2$, so $\|\vec c\|^2 = \frac{11}{1 + \|\vec b\|^2} < 11$. (D) TRUE. $\|\vec b\|^2 = 1 + b_2^2 + (b_2 + 3)^2$ with $b_2b_3 > 0$: either $b_2 > 0$ (then $(b_2+3)^2 > 9$, $b_2^2 > 0$) or $b_2 < -3$ (then $b_2^2 > 9$, $(b_2 + 3)^2 > 0$). Either way $\|\vec b\|^2 > 10$. (C) TRUE. Answer: (B), (C), (D).
Q.14	Differential EquationsLinear differential equations	Class 12: Differential Equations	Multiple correctMedium
A medium multiple correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Linear equation with integrating factor $e^{12x}$. Write $\omega = \frac{\pi}{12}$. A particular solution is $\frac{12\cos\omega x + \omega\sin\omega x}{144 + \omega^2}$; adding $Ce^{-12x}$ and using $y(0) = 0$ gives $y(x) = \frac{12\cos\omega x + \omega\sin\omega x - 12e^{-12x}}{144 + \omega^2}$. For large $x$ the exponential dies out and $y$ behaves like $\frac{\sqrt{144 + \omega^2}\cos(\omega x - \varphi)}{144 + \omega^2}$, which oscillates up and down with period $24$. (A), (B) An oscillating function is neither increasing nor decreasing on $\mathbb{R}$. FALSE. (C) Take $\beta = 0$ (or any level strictly inside the band of oscillation): for large $x$ the curve crosses it twice in every period, so infinitely many times. TRUE. (D) The term $e^{-12x}$ is not periodic and does not cancel, so $y$ is not periodic. FALSE. Answer: (C).
Q.15	Permutations and CombinationsSimple applications	Class 11: Permutations and Combinations	Single correctMedium
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations From one box, count the ways to take $k$ balls with at least one red and one blue: $k = 2$ (1R1B): $3\cdot2 = 6$; $k = 3$ (2R1B or 1R2B): $3\cdot2 + 3\cdot1 = 9$; $k = 4$ (3R1B or 2R2B): $1\cdot2 + 3\cdot1 = 5$; $k = 5$: $1$. Each box gives at least $2$, so $8$ balls are taken by the minimum and $2$ more are spread over the boxes. Both extra in one box (sizes $4, 2, 2, 2$): $4\times 5\times 6^3 = 4320$. One extra in each of two boxes (sizes $3, 3, 2, 2$): $\binom{4}{2}\times 9^2\times 6^2 = 6\times81\times36 = 17496$. Total $= 4320 + 17496 = 21816$. Answer: (A).
Q.16	Matrices and DeterminantsAlgebra of matrices	Class 12: Matrices; Class 12: Determinants	Single correctEasy
A easy single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Write $M = I + N$ with $N = \begin{pmatrix}\frac{3}{2} & \frac{3}{2}\\ -\frac{3}{2} & -\frac{3}{2}\end{pmatrix}$. Then $N^2 = \frac{9}{4}\begin{pmatrix}1-1 & 1-1\\ -1+1 & -1+1\end{pmatrix} = 0$. Since $I$ and $N$ commute and $N^2 = 0$, the binomial expansion stops: $M^n = I + nN = \begin{pmatrix}1 + \frac{3n}{2} & \frac{3n}{2}\\ -\frac{3n}{2} & 1 - \frac{3n}{2}\end{pmatrix}$. For $n = 2022$: $\frac{3n}{2} = 3033$, so $M^{2022} = \begin{pmatrix}3034 & 3033\\ -3033 & -3032\end{pmatrix}$. Answer: (A).
Q.17	ProbabilityBayes' theorem	Class 11: Probability; Class 12: Probability	Single correctMedium
A medium single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Box-I has $16$ balls; Box-II $48$; Box-III $16$; Box-IV $32$. P(at least one green): $b$ red, then green from Box-II: $\frac{8}{16}\cdot\frac{15}{48} = \frac{5}{32}$; $b$ blue, then green from Box-III: $\frac{3}{16}\cdot\frac{12}{16} = \frac{9}{64}$; $b$ green (already green): $\frac{5}{16}$. Total $= \frac{10}{64} + \frac{9}{64} + \frac{20}{64} = \frac{39}{64}$. P(white and at least one green): white is only in Box-IV, reached only when $b$ is green: $\frac{5}{16}\cdot\frac{6}{32} = \frac{15}{256}$. Conditional probability $= \frac{15/256}{39/64} = \frac{15}{156} = \frac{5}{52}$. Answer: (C).
Q.18	Integral CalculusFundamental theorem of calculus	Class 12: Integrals; Class 12: Application of Integrals	Single correctHard
A hard single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The $k$-th fraction is $\frac{16k + (9 - 4k)n - 3n^2}{4kn + 3n^2}$ (check $k = 1, 2, 3$). The last has denominator $7n^2 = 4kn + 3n^2$, so $k = n$; its numerator is then $16n + 9n - 4n^2 - 3n^2 = 25n - 7n^2$, as printed. So there are $n$ fractions. Each fraction $= -1 + \frac{16k + 9n}{4kn + 3n^2}$ (because $-(4kn + 3n^2) + 16k + 9n$ is the numerator). The $n$ copies of $-1$ cancel the leading $n$: $f(n) = \sum_{k=1}^{n}\frac{16k + 9n}{n(4k + 3n)} = \frac{1}{n}\sum_{k=1}^{n}\frac{16\frac{k}{n} + 9}{4\frac{k}{n} + 3}$. As a Riemann sum, $\lim f(n) = \int_0^1 \frac{16x + 9}{4x + 3}dx = \int_0^1\left(4 - \frac{3}{4x + 3}\right)dx = 4 - \frac{3}{4}\ln\frac{7}{3}$. Answer: (B).

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	NumericalEasy
A easy numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The force $-k(x\hat i + y\hat j)$ always points towards the $z$-axis and has no $z$ component, so its torque about the $z$-axis is $xF_y - yF_x = x(-ky) - y(-kx) = 0$. So $L_z = m(xv_y - yv_x)$ is conserved, whatever the value of $z$ (the $z$-motion is uniform and does not affect it). At $t = 0$: $xv_y - yv_x = \frac{1}{\sqrt2}\cdot\sqrt2 - \sqrt2\cdot(-\sqrt2) = 1 + 2 = 3$. Answer: $3$ m$^2$ s$^{-1}$.
Q.2	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 Only $\alpha$ decay changes the mass number, by $4$ each time: $230 - 214 = 16$, so $4$ $\alpha$ particles. Four $\alpha$ decays lower $Z$ by $8$, from $90$ to $82$. Each $\beta^-$ raises $Z$ by $1$; to end at $84$ we need $2$ $\beta^-$ decays. Ratio $= 4 : 2 = 2$. Answer: $2$.
Q.3	Current ElectricityWheatstone bridge	Class 12: Current Electricity	NumericalMedium
A medium numerical question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity Balanced bridge: $\frac{R_1}{R_2} = \frac{R_{AM}}{R_{MB}}$, where $M$ is the midpoint of the wire. For a wire whose radius changes linearly, $r = r_A + (r_B - r_A)\frac{x}{L}$, the resistance of a piece is $\int\frac{\rho\,dx}{\pi r^2} = \frac{\rho L}{\pi(r_B - r_A)}\left[\frac{1}{r_{start}} - \frac{1}{r_{end}}\right]$. Radius at $M$: $0.6$ mm. So $R_{AM} \propto \frac{1}{0.2} - \frac{1}{0.6} = 5 - \frac{5}{3} = \frac{10}{3}$ and $R_{MB} \propto \frac{1}{0.6} - \frac{1}{1} = \frac{2}{3}$ (same constant). $\frac{X}{1} = \frac{10/3}{2/3} = 5$. Answer: $5$.
Q.4	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	NumericalMedium
A medium numerical question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Use M, L, T and charge Q as base dimensions. $[e] = Q$; $[m_e] = M$; $[h] = ML^2T^{-1}$; $[k] = [F r^2/q^2] = ML^3T^{-2}Q^{-2}$. $[B]$ from $F = qvB$: $\frac{MLT^{-2}}{Q\cdot LT^{-1}} = MT^{-1}Q^{-1}$. Match powers: M: $\beta + \gamma + \delta = 1$; L: $2\gamma + 3\delta = 0$; T: $-\gamma - 2\delta = -1$; Q: $\alpha - 2\delta = -1$. From L, $\gamma = -\frac{3}{2}\delta$; then T gives $\frac{3}{2}\delta - 2\delta = -1$, so $\delta = 2$, $\gamma = -3$. Then $\beta = 1 + 3 - 2 = 2$ and $\alpha = -1 + 4 = 3$. $\alpha + \beta + \gamma + \delta = 3 + 2 - 3 + 2 = 4$. Answer: $4$.
Q.5	OpticsReflection and refraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalMedium
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics In every unit the ray enters the air layer at $60°$ (each air layer restores the angle, since $\mu\sin$ of the angle is the same in all layers: $1\cdot\sin60° = \frac{\sqrt3}{2}$). Air: sideways shift $h\tan60° = \frac{1}{3}\sqrt3 = \frac{1}{\sqrt3}$ cm. Layer $\mu_1$: $\sin r_1 = \frac{\sqrt3/2}{\sqrt{3/2}} = \frac{1}{\sqrt2}$, $r_1 = 45°$, shift $d\tan45° = d$. Layer $\mu_2$: $\sin r_2 = \frac{\sqrt3/2}{\sqrt3} = \frac{1}{2}$, $r_2 = 30°$, shift $\frac{d}{\sqrt3}$. Per unit: $\frac{1}{\sqrt3} + d\left(1 + \frac{1}{\sqrt3}\right) = \frac{1}{\sqrt3} + \frac{\sqrt3 - 1}{2}\cdot\frac{\sqrt3 + 1}{\sqrt3} = \frac{1}{\sqrt3} + \frac{1}{\sqrt3} = \frac{2}{\sqrt3}$ cm. $n\cdot\frac{2}{\sqrt3} = \frac{8}{\sqrt3}$ gives $n = 4$. Answer: $4$.
Q.6	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	NumericalEasy
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The charge sits at the centre of the flat circle shared by the cone and the hemisphere. Total flux through the closed surface is $\frac{q}{\epsilon_0}$ (Gauss's law). The hemisphere, centred on the charge, subtends exactly half of the full solid angle, so it carries $\frac{q}{2\epsilon_0}$. The cone carries the rest: $\frac{q}{\epsilon_0} - \frac{q}{2\epsilon_0} = \frac{q}{2\epsilon_0} = \frac{3q}{6\epsilon_0}$. Answer: $n = 3$.
Q.7	Oscillations and WavesSimple pendulum and spring systems	Class 11: Oscillations; Class 11: Waves	NumericalEasy
A easy numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Each side of the natural length is half of a simple harmonic motion with its own spring constant. Stretched side: $\omega_1 = \sqrt{k_1/m} = \sqrt{0.09} = 0.3$ s$^{-1}$, time spent $\frac{\pi}{\omega_1} = \frac{10\pi}{3}$ s. Compressed side: $\omega_2 = \sqrt{0.16} = 0.4$ s$^{-1}$, time spent $\frac{\pi}{\omega_2} = 2.5\pi$ s. $T = \left(\frac{10}{3} + \frac{5}{2}\right)\pi = \frac{35}{6}\pi \approx 5.83\pi$ s. Closest integer to $n$: $6$. Answer: $6$.
Q.8	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalMedium
A medium numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Mirror formula with $u = -30$ cm, $f = -10$ cm: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{10} + \frac{1}{30} = -\frac{1}{15}$, so $v = -15$ cm (real). Magnification $m = -\frac{v}{u} = -\frac{1}{2}$. For motion along the axis, the image velocity relative to the mirror is $-m^2$ times the object velocity relative to the mirror. Take the direction from object to mirror as positive. Object: $+15$. Mirror: $V$. Object relative to mirror: $15 - V$; image relative to mirror: $-\frac{1}{4}(15 - V)$. Image in the lab: $V - \frac{1}{4}(15 - V) = 0 \Rightarrow \frac{5V}{4} = \frac{15}{4} \Rightarrow V = 3$ cm s$^{-1}$. Answer: $3$.
Q.9	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 Charge in A: $Q_A = \int_0^1 kr\cdot4\pi r^2dr = \pi k$. Charge in B (from $r = 1$ to $r_B$): $Q_B = \int_1^{r_B}\frac{2k}{r}4\pi r^2dr = 4\pi k(r_B^2 - 1)$. Total $Q = \pi k + 4\pi k(r_B^2 - 1)$, all positive. Outside, the field and potential are those of a point charge $Q$. (A) $r_B = \sqrt{3/2}$: $Q = \pi k + 2\pi k = 3\pi k \neq 0$, so the field outside is not zero. FALSE. (B) $r_B = \frac{3}{2}$: $Q = \pi k + 5\pi k = 6\pi k$; $V = \frac{Q}{4\pi\epsilon_0 r_B} = \frac{6\pi k}{4\pi\epsilon_0\cdot1.5} = \frac{k}{\epsilon_0}$. TRUE. (C) $r_B = 2$: $Q = \pi k + 12\pi k = 13\pi k$, not $15\pi k$. FALSE. (D) $r_B = \frac{5}{2}$: $Q = \pi k + 21\pi k = 22\pi k$; $E = \frac{22\pi k}{4\pi\epsilon_0(25/4)} = \frac{22k}{25\epsilon_0}$, not $\frac{13\pi k}{\epsilon_0}$. FALSE. Answer: (B).
Q.10	Current ElectricityOhm's law	Class 12: Current Electricity	Multiple correctMedium
A medium multiple correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Circuit-1: from A the current passes $R_1$ to a junction; from there $R_2 + R_3 = 5\ \Omega$ in parallel with $\frac{R_1}{2} = 0.5\ \Omega$ to B. Switch S$_1$ joins A's side directly to that junction, i.e. it shorts $R_1$. Open: $R = 1 + \frac{5\times0.5}{5.5} = 1 + \frac{5}{11} = \frac{16}{11}\ \Omega$. Closed: $R = \frac{5}{11}\ \Omega$. Circuit-2: $R_1, R_2, R_3$ in parallel, plus $2R_3 = 6\ \Omega$ through S$_2$. Open: $\frac{1}{R} = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}$, $R = \frac{6}{11}\ \Omega$. Closed: $\frac{1}{R} = \frac{11}{6} + \frac{1}{6} = 2$, $R = \frac{1}{2}\ \Omega$. (A) $6$ V source, $P = V^2/R$: $P_1 = \frac{36\cdot11}{16} = 24.75$ W, $P_2 = \frac{36\cdot11}{6} = 66$ W. $P_1 < P_2$. TRUE. (B) $2$ A source, $P = I^2R$: $P_1 = 4\cdot\frac{16}{11} \approx 5.82$ W, $P_2 = 4\cdot\frac{6}{11} \approx 2.18$ W. $P_1 > P_2$. TRUE. (C) $6$ V in Circuit-1: $Q_1 = \frac{36\cdot11}{5} = 79.2$ W $> P_1$. TRUE. (D) $2$ A: $Q_1 = 4\cdot\frac{5}{11} \approx 1.82$ W, $Q_2 = 4\cdot\frac{1}{2} = 2$ W. $Q_2 < Q_1$ is FALSE. Answer: (A), (B), (C).
Q.11	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Multiple correctHard
A hard multiple correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Read the bubble as a soap bubble in air: it has two surfaces, so the gas pressure is $P = P_a + \frac{4S}{r}$, and the volume is $V \propto r^3$. (A) Insulating surface: adiabatic, $PV^{\gamma} = $ const with $V^{\gamma} \propto r^{5}$: $\left(P_{a1} + \frac{4S}{r_1}\right)r_1^5 = \left(P_{a2} + \frac{4S}{r_2}\right)r_2^5$, so $\left(\frac{r_1}{r_2}\right)^5 = \frac{P_{a2} + 4S/r_2}{P_{a1} + 4S/r_1}$. The option has $2S$ in place of $4S$. FALSE. (B) The atmosphere does work $-P_a\,dV$ on the bubble as it changes size, so the gas's internal energy plus the surface energy changes with $P_a$. FALSE. (C) Conducting surface, temperature fixed: isothermal, $PV = $ const: $\left(P_{a1} + \frac{4S}{r_1}\right)r_1^3 = \left(P_{a2} + \frac{4S}{r_2}\right)r_2^3$, which is the stated relation. TRUE. (D) Adiabatic: $T^{\gamma}P^{1-\gamma} = $ const, i.e. $T^{5/3} \propto P^{2/3}$, $T^{5/2} \propto P$: $\left(\frac{T_2}{T_1}\right)^{5/2} = \frac{P_{a2} + 4S/r_2}{P_{a1} + 4S/r_1}$. TRUE. Answer: (C), (D).
Q.12	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Multiple correctHard
A hard multiple correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Potential energy: $U(z) = \frac{q\sigma}{2\epsilon_0}\left[\sqrt{R^2 + z^2} - z + \beta z\right] = \frac{q\sigma}{2\epsilon_0}u(z)$, with $u(0) = R$ and $u'(z) = \frac{z}{\sqrt{R^2+z^2}} - 1 + \beta$. $\beta = \frac14$: $u' = 0$ at $\frac{z}{\sqrt{R^2+z^2}} = \frac34$, i.e. $z_m = \frac{3R}{\sqrt7} \approx 1.13R$. $u$ falls on $[0, z_m]$ and rises beyond: a well at $z_m$. Starting from rest at $z_0 > z_m$ the particle moves down; it reaches $z = 0$ only if $u(z_0) \ge u(0) = R$. $u(z) = R$: $\sqrt{R^2+z^2} = R + \frac{3z}{4} \Rightarrow \frac{7z^2}{16} = \frac{3Rz}{2} \Rightarrow z = \frac{24R}{7}$. So it reaches the origin for $z_0 \ge \frac{24R}{7}$. (A) $z_0 = \frac{25R}{7} > \frac{24R}{7}$: reaches the origin. TRUE. (B) $z_0 = \frac{3R}{7} < z_m$: $u'(z_0) < 0$, the net force is upward, the particle moves away from the disk and oscillates about $z_m$, never reaching $0$. FALSE. (C) $z_0 = \frac{R}{\sqrt3}$: $u'(z_0) = \frac{1}{2} - \frac{3}{4} < 0$, so it starts upward, turns back at the point beyond $z_m$ where $u = u(z_0)$, and returns to $z_0$ (it oscillates between them). TRUE. (D) $\beta > 1$: $u'(z) = \frac{z}{\sqrt{R^2+z^2}} + (\beta - 1) > 0$ for all $z \ge 0$, so the force is always downward and the particle always reaches the origin. TRUE. Answer: (A), (C), (D).
Q.13	OpticsInterference and diffraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Multiple correctMedium
A medium multiple correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The slit on the interface is lit directly; the light reaching the lower slit has first refracted into medium 2 and travels there at angle $\theta$ to the line joining the slits (the normal). Optical path from the incident wavefront to the lower slit exceeds that to the upper slit by $n_2 d\cos\theta$ (projection of $d$ on the refracted direction). Towards the detector, also at angle $\theta$ to the same line and on the far side, the ray from the upper slit travels the extra $d\cos\theta$ in medium 2, i.e. extra optical path $n_2d\cos\theta$. The two extras cancel: net path difference $0$, phase difference $0$, for every $d$, $n_1$, $n_2$ and angle of incidence. (A) Independent of $d$. TRUE. (B) Zero phase difference: constructive interference. TRUE. (C) It does not depend on $n_1$ (or on $n_2$). FALSE. (D) It vanishes for all values, not only certain ones. FALSE. Answer: (A), (B).
Q.14	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Multiple correctMedium
A medium multiple correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the graph: $A$ is $(0.80\ \text{m}^3, 100$ kPa$)$, $B$ is at $300$ kPa, $C$ is at $V = 0.80$ m$^3$. Adiabatic $A \to B$: $V_B = V_A\left(\frac{P_A}{P_B}\right)^{1/\gamma} = 0.8\left(\frac13\right)^{0.6} = 0.4$ m$^3$. $\|W_{AB}\| = \frac{P_BV_B - P_AV_A}{\gamma - 1} = \frac{120 - 80}{2/3}$ kJ $= 60$ kJ (work done on the gas). Isothermal $B \to C$: $W_{BC} = P_BV_B\ln\frac{V_C}{V_B} = 120\ln2 = 84$ kJ (done by the gas). (A) Total $A \to B \to C$: $-60 + 84 = 24$ kJ, not $144$. FALSE. (B) $84$ kJ. TRUE. (C) $60$ kJ. TRUE. (D) $C \to A$ is at constant volume ($0.80$ m$^3$), so no work. TRUE. Answer: (B), (C), (D).
Q.15	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	Single correctMedium
A medium single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The centre of B moves on a circle of radius $2R$: $v_{cm} = 2R\omega$. Rolling without slipping on the fixed disk A: the contact point of B is at rest, so B's spin $\omega_B$ satisfies $v_{cm} = \omega_BR$, $\omega_B = 2\omega$ (same sense). $L = Mv_{cm}(2R) + I_{cm}\omega_B = M(2R\omega)(2R) + \frac{1}{2}MR^2(2\omega) = 4M\omega R^2 + M\omega R^2 = 5M\omega R^2$. Answer: (B) $n = 5$.
Q.16	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	Single correctEasy
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation in volts: $V_0 = \frac{hc}{e\lambda} - \frac{\phi}{e}$. Intensity does not affect the stopping potential. $6.0 = \frac{1.24\times10^{-6}}{\lambda} - \phi$ and $0.6 = \frac{1.24\times10^{-6}}{4\lambda} - \phi$ (with $\phi$ in eV). Subtract: $5.4 = \frac{3}{4}\cdot\frac{1.24\times10^{-6}}{\lambda}$, so $\frac{1.24\times10^{-6}}{\lambda} = 7.2$ eV and $\lambda = \frac{1.24\times10^{-6}}{7.2} = 1.72\times10^{-7}$ m. $\phi = 7.2 - 6.0 = 1.20$ eV. Answer: (A).
Q.17	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 One rotation moves the screw by $2 \times 0.5 = 1.0$ mm, so the least count is $\frac{1.0}{100} = 0.01$ mm per circular division. Zero error: $+4$ divisions $= +0.04$ mm. Attempt 1: $4\times0.5 + 20\times0.01 - 0.04 = 2.00 + 0.20 - 0.04 = 2.16$ mm. Attempt 2: $2.00 + 0.16 - 0.04 = 2.12$ mm. Mean diameter $2.14$ mm; mean absolute deviation $\frac{0.02 + 0.02}{2} = 0.02$ mm. So $d = 2.14 \pm 0.02$ mm. Radius $r = 1.07 \pm 0.01$ mm. Area $= \pi r^2 = \pi(1.1449)$ mm$^2$ with $\Delta(r^2) = 2r\Delta r = 2(1.07)(0.01) \approx 0.02$. So area $= \pi(1.14 \pm 0.02)$ mm$^2$. Answer: (C).
Q.18	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 Take O as origin, $\hat i$ right, $\hat j$ up, $\hat k$ out of the page. Vertical wire of length $L$ at $x = -L$, current up, ending at $(-L, 0)$: O is at perpendicular distance $L$ from it, level with its top end. $B = \frac{\mu_0I}{4\pi L}(\sin0° + \sin45°) = \frac{\mu_0I}{4\sqrt2\pi L}$; for an upward current, the field to its right is into the page ($-\hat k$). Horizontal piece from $(-L,0)$ to $(-\frac{L}{2},0)$, the straight piece from $(\frac{L}{2},0)$ to $(\frac{L}{4},0)$ and the vertical wire of length $\frac{3L}{4}$ below O all lie on lines through O: no field. Semicircle of radius $\frac{L}{2}$, traversed clockwise: $\frac{\mu_0I}{4(L/2)} = \frac{\mu_0I}{2L}$, into the page. Quarter circle of radius $\frac{L}{4}$, clockwise: $\frac{\mu_0I}{8(L/4)} = \frac{\mu_0I}{2L}$, into the page. Total: $\vec B = -\frac{\mu_0I}{L}\left(\frac12 + \frac12 + \frac{1}{4\sqrt2\pi}\right)\hat k = \frac{-\mu_0I}{L}\left(1 + \frac{1}{4\sqrt2\pi}\right)\hat k$. Answer: (C).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	EquilibriumSolubility product	Class 11: Equilibrium	NumericalHard
A hard numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium First ionisation is complete: $1$ M $\mathrm{H^+}$ and $1$ M $\mathrm{HSO_4^-}$. The added $\mathrm{SO_4^{2-}}$ ($0.018$ M) partly takes up $\mathrm{H^+}$: $\mathrm{SO_4^{2-} + H^+ \rightleftharpoons HSO_4^-}$. Let $y$ M of sulfate be converted: $K_{a2} = \frac{[\mathrm{H^+}][\mathrm{SO_4^{2-}}]}{[\mathrm{HSO_4^-}]} = \frac{(1 - y)(0.018 - y)}{1 + y} = 0.012$. Since $y \ll 1$: $0.018 - y \approx 0.012$, so $[\mathrm{SO_4^{2-}}] \approx 1.2\times10^{-2}$ M (solving exactly gives $\approx 1.22\times10^{-2}$ M). The tiny amount of $\mathrm{PbSO_4}$ that dissolves does not change this. Solubility $s = \frac{K_{sp}}{[\mathrm{SO_4^{2-}}]} = \frac{1.6\times10^{-8}}{1.2\times10^{-2}} \approx 1.3\times10^{-6}$ M. So $\mathrm{X} \approx 1.3$, $\mathrm{Y} = 6$. Answer: $6$.
Q.2	Solutionsvan't Hoff factor	Class 12: Solutions	NumericalMedium
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Relative lowering of vapour pressure $= \frac{60.000 - 59.724}{60.000} = \frac{0.276}{60} = 0.0046 =$ mole fraction of solute particles. Water: $\frac{1800}{18} = 100$ mol. Solute particles: $0.1i$ mol. $\frac{0.1i}{100 + 0.1i} = 0.0046 \Rightarrow 0.1i(1 - 0.0046) = 0.46 \Rightarrow i \approx 4.62$ (the dilute form $\frac{0.1i}{100} = 0.0046$ gives $i = 4.6$). With $n$ ions per formula unit and degree of dissociation $0.9$: $i = 1 + 0.9(n - 1)$. $0.9(n - 1) \approx 3.6 \Rightarrow n - 1 = 4$, $n = 5$. Answer: $5$.
Q.3	ElectrochemistryConductance	Class 12: Electrochemistry	NumericalMedium
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry From the straight line $\Lambda = \Lambda^0 - A\sqrt c$: $339 = \Lambda^0 - 0.01A$ and $336 = \Lambda^0 - 0.04A$, so $A = 100$ and $\Lambda^0(\mathrm{Z_mX_n}) = 340$ S cm$^2$ mol$^{-1}$. Kohlrausch's law: $\Lambda^0 = \nu_+\lambda^0_+ + \nu_-\lambda^0_-$. $\mathrm{Z_mX_n}$: $50m + 80n = 340$. $\mathrm{V_mX_n}$: $100m + 80n = 440$. Subtract: $50m = 100$, $m = 2$, then $n = 3$. $\mathrm{U_mY_p}$: $25m + 100p = 250 \Rightarrow 50 + 100p = 250 \Rightarrow p = 2$. $m + n + p = 2 + 3 + 2 = 7$. Answer: $7$.
Q.4	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	NumericalMedium
A medium numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Dioxygen difluoride is a very strong fluorinating agent; with xenon it gives xenon tetrafluoride: $\mathrm{Xe + 2O_2F_2 \rightarrow XeF_4 + 2O_2}$. So P $= \mathrm{XeF_4}$. Complete hydrolysis of $\mathrm{XeF_4}$: $\mathrm{6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2}$. $24$ mol HF per $6$ mol $\mathrm{XeF_4}$, i.e. $4$ mol HF per mol of P. Answer: $4$.
Q.5	Chemical Bonding and Molecular StructureMolecular orbital theory	Class 11: Chemical Bonding and Molecular Structure	NumericalMedium
A medium numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{2AgNO_3 \xrightarrow{\Delta} 2Ag + 2NO_2 + O_2}$. Both $\mathrm{NO_2}$ (one unpaired electron) and $\mathrm{O_2}$ (two unpaired electrons) are paramagnetic; $\mathrm{O_2}$ has more. $\mathrm{O_2}$ (16 electrons): $\sigma1s^2\,\sigma^1s^2\,\sigma2s^2\,\sigma^2s^2\,\sigma2p_z^2\,\pi2p^4\,\pi^2p^2$. Antibonding electrons: $\sigma^1s^2 + \sigma^2s^2 + \pi^2p^2 = 6$. Answer: $6$.
Q.6	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	NumericalHard
A hard numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Start: cyclohexene carrying a but-2-ynyl group ($\mathrm{-CH_2-C\equiv C-CH_3}$) on the allylic ring carbon (next to the ring double bond). 1. Na, liquid $\mathrm{NH_3}$ reduces the alkyne to the trans alkene: $\mathrm{-CH_2-CH=CH-CH_3}$. 2. Excess $\mathrm{Br_2}$ adds to both C=C: ring $\mathrm{C1, C2}$ dibromide (substituent on C3) and side chain $\mathrm{-CH_2-CHBr-CHBr-CH_3}$. 3. Alcoholic KOH removes two HBr from each vicinal dibromide. Without forming an $sp$ carbon (no alkyne, no allene) each Br must leave with an H from the carbon on its outer side: ring: Br on C1 with H on C6, Br on C2 with H on C3, giving the 1,3-diene $\mathrm{C6=C1-C2=C3}$ with the substituent on C3; chain: $\mathrm{-CH=CH-CH=CH_2}$. The product is a single constitution, a cyclohexa-1,3-diene bearing a buta-1,3-dienyl group on a diene end (a tetraene, fully conjugated). Stereoisomers: the ring double bonds cannot be E/Z; the terminal $\mathrm{=CH_2}$ has no E/Z; there is no stereocentre left. Only the side-chain $\mathrm{ring-CH=CH-}$ double bond can be E or Z. Number of tetraenes $= 2$. Answer: $2$.
Q.7	HydrocarbonsAromatic hydrocarbons	Class 11: Hydrocarbons	NumericalMedium
A medium numerical question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The alkene drawn is oct-4-ene, $\mathrm{CH_3CH_2CH_2CH=CHCH_2CH_2CH_3}$. 1. Reductive ozonolysis: $2$ molecules of butanal, $\mathrm{CH_3CH_2CH_2CHO}$. 2. $\mathrm{KMnO_4}$ oxidises them to butanoic acid, $\mathrm{CH_3CH_2CH_2COOH}$. 3. NaOH, then electrolysis of sodium butanoate (Kolbe): $\mathrm{2CH_3CH_2CH_2COO^- \rightarrow CH_3(CH_2)_4CH_3 + 2CO_2}$, i.e. n-hexane. 4. $\mathrm{Cr_2O_3}$ at $770$ K, $20$ atm: aromatisation of n-hexane to benzene. Benzene has no $\mathrm{-CH_2-}$ group. Answer: $0$.
Q.8	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 P is a chain with six C=C bonds; each tertiary carbinol carbon $\mathrm{C(OH)(CH_3)}$ sits between two of them. Ozonolysis cuts every C=C and leaves seven pieces. Reading from left to right: 1. $\mathrm{CH_3CHO}$ (end) — achiral. 2. $\mathrm{CH_3CO-C(OH)(CH_3)-COCH_3}$ — the central carbon has two identical acetyl groups: achiral. 3. $\mathrm{CH_3CO-C(OH)(CH_3)-CHO}$ — four different groups (OH, $\mathrm{CH_3}$, $\mathrm{COCH_3}$, CHO): chiral. 4. $\mathrm{CH_3CO-C(OH)(CH_3)-COCH_3}$ — achiral. 5. $\mathrm{OHC-C(OH)(CH_3)-COCH_3}$ — chiral. 6. $\mathrm{CH_3CO-C(OH)(CH_3)-COCH_3}$ — achiral. 7. $\mathrm{CH_3CHO}$ (end) — achiral. Chiral molecules formed: pieces 3 and 5, i.e. $2$. Answer: $2$.
Q.9	Some Basic Concepts of ChemistryLaws of chemical combination	Class 11: Some Basic Concepts of Chemistry	Multiple correctMedium
A medium multiple correct question on laws of chemical combination, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Mass of Q per unit mass of P: compound 1: $1$; compound 2: $\frac{55.6}{44.4} \approx 1.25$; compound 3: $1.5$. (A) Compound 3 $= \mathrm{P_3Q_4}$: $\frac{4a_Q}{3a_P} = 1.5 \Rightarrow \frac{a_Q}{a_P} = \frac{9}{8}$. Compound 2 then has $\frac{n}{m} = \frac{1.25}{9/8} = \frac{10}{9}$, i.e. $\mathrm{P_9Q_{10}}$, not $\mathrm{P_3Q_5}$ (which needs $\frac{5}{3}\cdot\frac{9}{8} = 1.875$). FALSE. (B) Compound 3 $= \mathrm{P_3Q_2}$ with $a_P = 20$: mass of P $= 60$, mass of Q $= 1.5\times60 = 90 = 2a_Q$, so $a_Q = 45$. TRUE. (C) Compound 2 $= \mathrm{PQ}$: $\frac{a_Q}{a_P} = 1.25$. Compound 1: $\frac{n}{m}\cdot1.25 = 1 \Rightarrow \frac{n}{m} = \frac{4}{5}$, i.e. $\mathrm{P_5Q_4}$. TRUE. (D) $a_P = 70$, $a_Q = 35$: compound 1 has moles $\frac{50}{70} : \frac{50}{35} = 1 : 2$, i.e. $\mathrm{PQ_2}$, not $\mathrm{P_2Q}$. FALSE. Answer: (B), (C).
Q.10	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Multiple correctMedium
A medium multiple correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics (A) $\Delta G = -nFE_{cell}$ and $\Delta S = -\frac{d\Delta G}{dT} = nF\frac{dE_{cell}}{dT}$. Here $n = 2$, so $\Delta S = 2F\cdot\frac{R}{F} = 2R$, not $R$. FALSE. (B) Concentration cell: the net change is $\mathrm{H^+}(0.1\ \text{M}) \rightarrow \mathrm{H^+}(0.01\ \text{M})$. $E = 0.059\log\frac{0.1}{0.01} > 0$, so it is spontaneous, while $\Delta H = 0$ (same species on both sides). $\Delta G = -T\Delta S < 0$: it is driven by entropy. TRUE. (C) Racemisation turns one enantiomer into an equal mixture of two; $\Delta H \approx 0$ and the entropy of mixing makes $\Delta S = R\ln2 > 0$ per mole. TRUE. (D) $\mathrm{[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O}$: $4$ particles become $7$, so $\Delta S > 0$ (the chelate effect). TRUE. Answer: (B), (C), (D).
Q.11	p-Block ElementsGroup 13 elements	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Boron heated in ammonia: $\mathrm{2B + 2NH_3 \rightarrow 2BN + 3H_2}$. TRUE. (B) Diborane with ammonia gives $\mathrm{B_2H_6\cdot2NH_3}$, which on heating gives borazine and, on stronger heating, boron nitride. TRUE. (C) Boron oxide heated with ammonia: $\mathrm{B_2O_3 + 2NH_3 \rightarrow 2BN + 3H_2O}$. TRUE. (D) $\mathrm{HBF_4}$ with ammonia only forms the salt $\mathrm{NH_4BF_4}$; no BN. FALSE. Answer: (A), (B), (C).
Q.12	Isolation of MetalsSmelting and refining	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter (A) Limestone decomposes to CaO, which combines with the silica impurity: $\mathrm{CaO + SiO_2 \rightarrow CaSiO_3}$ (slag). TRUE. (B) Pig iron from the blast furnace contains about $4\%$ carbon. TRUE. (C) In the furnace, coke reduces carbon dioxide: $\mathrm{C + CO_2 \rightarrow 2CO}$. TRUE. (D) The exhaust gases are mainly $\mathrm{N_2}$, CO and $\mathrm{CO_2}$, not $\mathrm{NO_2}$. FALSE. Answer: (A), (B), (C).
Q.13	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	Multiple correctMedium
A medium multiple correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids P: Friedel–Crafts acylation by succinic anhydride gives 4-oxo-4-phenylbutanoic acid, $\mathrm{C_6H_5COCH_2CH_2COOH}$. Q: Clemmensen reduction of the ketone: 4-phenylbutanoic acid, $\mathrm{C_6H_5CH_2CH_2CH_2COOH}$. R: the acid chloride $\mathrm{C_6H_5(CH_2)_3COCl}$. S: intramolecular Friedel–Crafts acylation closes a six-membered ring: $\alpha$-tetralone (3,4-dihydronaphthalen-1(2H)-one). Clemmensen then gives tetralin. (A) P and Q both carry COOH. TRUE. (B) S is an aryl ketone with no C=C outside the benzene ring; it does not decolorise bromine water. FALSE. (C) P and S both contain a ketone C=O, which forms an oxime with $\mathrm{NH_2OH}$. TRUE. (D) An acid chloride with a dialkylcadmium gives a ketone, not a tertiary alcohol (organocadmiums do not attack the ketone further). FALSE. Answer: (A), (C).
Q.14	PolymersNylon, Teflon, Bakelite and PVC	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter (A) Chloroprene polymerises to neoprene, a synthetic rubber; natural rubber is cis-polyisoprene. FALSE. (B) Teflon is made by heating tetrafluoroethene with a free-radical persulphate catalyst at high pressure. TRUE. (C) PVC softens on heating and hardens on cooling repeatedly: a thermoplastic. TRUE. (D) Ethene at $350-570$ K and $1000-2000$ atm with a peroxide initiator gives low density polythene (branched, by free radicals); high density polythene is made with a Ziegler–Natta catalyst at low pressure. FALSE. Answer: (B), (C).
Q.15	Solid StatePacking in fcc, bcc and hcp	No matching NCERT chapter	Single correctMedium
A medium single correct question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter Per fcc cell: $4$ lattice atoms + $4$ atoms in alternate tetrahedral voids (half of $8$) $= 8$ atoms of X. All are the same atom, so the atoms touch along the corner-to-tetrahedral-void line: that distance is $\frac{\sqrt3a}{4} = 2r$, so $r = \frac{\sqrt3a}{8}$. Packing fraction $= \frac{8\cdot\frac43\pi r^3}{a^3} = \frac{32\pi}{3}\cdot\frac{3\sqrt3}{512} = \frac{\sqrt3\pi}{16} \approx 0.34$. About $34\%$, closest to $35$. Answer: (B).
Q.16	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Chloric acid with hydrochloric acid gives chlorine dioxide (an odd-electron, paramagnetic molecule): $\mathrm{2HClO_3 + 2HCl \rightarrow 2ClO_2 + Cl_2 + 2H_2O}$. $\mathrm{ClO_2}$ is oxidised by ozone to dichlorine hexoxide: $\mathrm{2ClO_2 + 2O_3 \rightarrow Cl_2O_6 + 2O_2}$. Answer: (C).
Q.17	p-Block ElementsGroup 14 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{Pb^{2+} + 2Cl^- \rightarrow PbCl_2}\downarrow$ (white). In more concentrated HCl the extra chloride forms the soluble complex: $\mathrm{PbCl_2 + 2Cl^- \rightarrow [PbCl_4]^{2-}}$. Lead stays $+2$; $\mathrm{PbCl_4}$ and $\mathrm{[PbCl_6]^{2-}}$ are lead(IV) species, which chloride does not form from $\mathrm{Pb^{2+}}$. Answer: (C).
Q.18	BiomoleculesCarbohydrates	Class 12: Biomolecules	Single correctMedium
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules In dilute base D-glucose enolises to an enediol (C1–C2). The enediol can re-ketonise three ways: back to D-glucose, to its C2 epimer D-mannose, or to D-fructose (C=O at C2). C3, C4, C5 are untouched. So the mixture is D-glucose, D-mannose and D-fructose, all with C3–OH on the left and C4–, C5–OH on the right in the Fischer projection. (A) contains D-galactose (C4–OH on the left) and an L-sugar (C5–OH on the left): no. (B) the second sugar has C3–OH on the right (not glucose or mannose), the third is an L-sugar: no. (C) D-glucose (C2 right, C3 left), D-fructose (C=O at C2, C3 left), D-mannose (C2 left, C3 left), all with C4 and C5 on the right: yes. (D) the first has C4–OH on the left (D-galactose): no. Answer: (C).

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