JEE (Advanced) · 2023

JEE Advanced 2023, 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
34 Q
1. Probability5
2. Trigonometry4
3. Integral Calculus3
9 easy · 21 medium · 4 hard
Physics
34 Q
1. Oscillations and Waves4
2. Properties of Bulk Matter4
3. Optics3
15 easy · 14 medium · 5 hard
Chemistry
34 Q
1. Aldehydes, Ketones and Carboxylic Acids4
2. Thermodynamics3
3. Amines3
8 easy · 24 medium · 2 hard

Chapters

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

Probability5
Class 11: Probability; Class 12: Probability
Trigonometry4
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Integral Calculus3
Class 12: Integrals; Class 12: Application of Integrals
Complex Numbers3
Class 11: Complex Numbers and Quadratic Equations
Matrices and Determinants3
Class 12: Matrices; Class 12: Determinants
Conic Sections2
Class 11: Conic Sections
Limits, Continuity and Differentiability2
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Three Dimensional Geometry2
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Vector Algebra2
Class 12: Vector Algebra
Differential Equations2
Class 12: Differential Equations
Sets, Relations and Functions1
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Sequences and Series1
Class 11: Sequences and Series
Binomial Theorem1
Class 11: Binomial Theorem
Statistics1
Class 11: Statistics
Applications of Derivatives1
Class 12: Application of Derivatives
Circles1
Class 11: Conic Sections

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	Sets, Relations and FunctionsOne-one, into and onto functions	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	Multiple correctEasy
A easy multiple correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions (A) $S$ is an infinite set and $T$ has 4 elements, so there are infinitely many ways to assign values: for instance, for each $a \in (0,1)$ the function that is $1$ at $x = a$ and $0$ elsewhere is a different function. True. (B) A strictly increasing function is one-one. An infinite set cannot be mapped one-one into the 4-element set $T$, so there is no strictly increasing function at all. False. (C) A continuous function from an interval into the discrete set $T$ takes one value on that interval (by the intermediate value property it cannot jump between integers). $S$ is made of 3 separate intervals, so a continuous function is fixed by choosing one of 4 values on each: $4^3 = 64 \le 120$. True. (D) Every continuous function is constant on each of the three open intervals, so its derivative exists and is $0$ at every point of $S$. True. Answer: (A), (C), (D).
Q.2	Conic SectionsEllipse	Class 11: Conic Sections	Multiple correctMedium
A medium multiple correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections A tangent to $y^2 = 12x$ (here $a = 3$) with slope $m$ is $y = mx + \frac{3}{m}$, touching at $\left(\frac{3}{m^2}, \frac{6}{m}\right)$. It touches $\frac{x^2}{6} + \frac{y^2}{3} = 1$ when $c^2 = 6m^2 + 3$: $\frac{9}{m^2} = 6m^2 + 3 \Rightarrow 2m^4 + m^2 - 3 = 0 \Rightarrow (m^2 - 1)(2m^2 + 3) = 0$, so $m = \pm 1$. $T_1: y = x + 3$ and $T_2: y = -x - 3$. Both meet the $x$-axis at $(-3, 0)$: (C) true, (D) false. Points of contact. Parabola: $m = 1$ gives $A_1 = (3, 6)$; $m = -1$ gives $A_4 = (3, -6)$. Ellipse: the line $y = mx + c$ touches $\frac{x^2}{A} + \frac{y^2}{B} = 1$ at $\left(-\frac{Am}{c}, \frac{B}{c}\right)$. For $m = 1, c = 3$: $A_2 = (-2, 1)$; for $m = -1, c = -3$: $A_3 = (-2, -1)$. $A_1A_4$ (length 12, on $x = 3$) and $A_2A_3$ (length 2, on $x = -2$) are parallel, $5$ apart, so the quadrilateral is a trapezium of area $\frac{1}{2}(12 + 2)\times 5 = 35$. (A) true, (B) false. Answer: (A), (C).
Q.3	Integral CalculusArea under simple curves	Class 12: Integrals; Class 12: Application of Integrals	Multiple correctHard
A hard multiple correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals First the facts about $f$. $f'(x) = x^2 - 2x + \frac{5}{9} = 0$ at $x = \frac{1}{3}$ (and $\frac{5}{3}$, outside). $f(0) = \frac{17}{36} \approx 0.472$, $f\left(\frac{1}{3}\right) = \frac{7}{81} + \frac{17}{36} \approx 0.559$ (maximum), $f(1) = \frac{13}{36} \approx 0.361$ (minimum). So $\frac{1}{4} < f(x) < \frac{2}{3}$ on $[0,1]$. $\int_0^1 f\,dx = \frac{1}{12} - \frac{1}{3} + \frac{5}{18} + \frac{17}{36} = \frac{1}{2}$, so red area $=$ green area $= \frac{1}{2}$. For a line at height $h$: green above $= \int_0^1 (1 - \max(f, h))\,dx$, green below $= \int_0^1 (h - f)^+\,dx$, red below $= \int_0^1 \min(f, h)\,dx$, red above $= \int_0^1 (f - h)^+\,dx$. (A) Needs green above $= \frac{1}{4}$ (half of $\frac{1}{2}$). Green above decreases as $h$ rises, and even at $h = \frac{2}{3}$ (above the whole curve) it is $1 - \frac{2}{3} = \frac{1}{3} > \frac{1}{4}$. Never. False. (B) Needs red below $= \frac{1}{4}$. At $h = \frac{1}{4}$ the line lies below the whole curve, so red below $= \int_0^1 \frac{1}{4}\,dx = \frac{1}{4}$ and red above $= \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$. $h = \frac{1}{4}$ is in the closed interval. True. (C) Green above $-$ red below $= \int_0^1 (1 - \max(f,h) - \min(f,h))\,dx = \int_0^1 (1 - f - h)\,dx = \frac{1}{2} - h$. Zero at $h = \frac{1}{2} \in \left[\frac{1}{4}, \frac{2}{3}\right]$. True. (D) Red above $-$ green below $= \int_0^1 ((f-h)^+ - (h-f)^+)\,dx = \int_0^1 (f - h)\,dx = \frac{1}{2} - h$. Zero at $h = \frac{1}{2}$. True. Answer: (B), (C), (D).
Q.4	Limits, Continuity and DifferentiabilityLimits of functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctMedium
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability If $x \in \left[\frac{1}{n+1}, \frac{1}{n}\right)$ then $nx < 1 \le (n+1)x$, so $\frac{1}{x} - 1 \le n < \frac{1}{x}$, and $f(x) = \sqrt n$. Upper bound: $f(x)g(x) < \sqrt{n}\cdot 2\sqrt{x} = 2\sqrt{nx} < 2$. Lower bound: on $[x^2, x]$, $\sqrt{1 - t} \ge \sqrt{1 - x}$, so $\int_{x^2}^{x} \sqrt{\frac{1-t}{t}}\,dt \ge \sqrt{1-x}\int_{x^2}^{x} \frac{dt}{\sqrt t} = 2\sqrt{1-x}\,(\sqrt x - x)$. Then $f(x)g(x) > \sqrt{\frac{1}{x} - 1}\cdot 2\sqrt{1-x}(\sqrt x - x) = 2(1 - x)(1 - \sqrt x)$, which tends to $2$ as $x \to 0$. By the squeeze theorem $\lim_{x\to 0} f(x)g(x) = 2$. Answer: (C).
Q.5	Three Dimensional GeometrySkew lines and shortest distance	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Single correctMedium
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry By symmetry take the main diagonal through $O(0,0,0)$ with direction $\vec b = (1,1,1)$. A face diagonal that passes through $O$ or $(1,1,1)$ meets it (distance $0$). The others are skew to it; take the one from $(1,0,0)$ to $(0,1,0)$, direction $\vec a = (-1, 1, 0)$. $\vec a \times \vec b = (1\cdot 1 - 0\cdot 1,\ 0\cdot 1 - (-1)\cdot 1,\ -1\cdot 1 - 1\cdot 1) = (1, 1, -2)$, of length $\sqrt6$. Shortest distance $= \frac{\|(1,0,0)\cdot(1,1,-2)\|}{\sqrt6} = \frac{1}{\sqrt 6}$. Every skew face diagonal is carried to this one by a symmetry of the cube that keeps the main diagonal, so all give $\frac{1}{\sqrt6}$ (for example the diagonal from $(0,1,1)$ to $(1,0,1)$: $(1,-1,0)\times(1,1,1) = (-1,-1,2)$, and $(0,1,1)\cdot(-1,-1,2) = 1$, again $\frac{1}{\sqrt6}$). Maximum $= \frac{1}{\sqrt6}$. Answer: (A).
Q.6	ProbabilityProbability of an event	Class 11: Probability; Class 12: Probability	Single correctMedium
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Points of $X$: $y^2 < 5x$ forces $x \ge 1$, and $\frac{x^2}{8} < 1$ forces $x \le 2$. $x = 1$: $y^2 < 5$ and $y^2 < 17.5$, so $y \in \{-2, \dots, 2\}$: 5 points. $x = 2$: $y^2 < 10$ and $y^2 < 10$, so $y \in \{-3, \dots, 3\}$: 7 points. So $\|X\| = 12$ and there are $\binom{12}{3} = 220$ choices. Three points on one vertical line are collinear (area 0). Otherwise two points lie on one line and one on the other; the lines are $1$ apart, so the area is $\frac{1}{2}\cdot\|\Delta y\|\cdot 1$, a positive integer exactly when $\Delta y$ is even. Two on $x = 1$ with even difference: from $\{-2, 0, 2\}$: 3 pairs; from $\{-1, 1\}$: 1 pair; total 4, times 7 choices on $x = 2$: 28. Two on $x = 2$ with even difference: from $\{-3,-1,1,3\}$: 6 pairs; from $\{-2,0,2\}$: 3 pairs; total 9, times 5 choices on $x = 1$: 45. Favourable $= 28 + 45 = 73$. Probability $= \frac{73}{220}$. Answer: (B).
Q.7	Conic SectionsParabola	Class 11: Conic Sections	Single correctMedium
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Take $P = (at^2, 2at)$. The normal there is $y = -tx + 2at + at^3$, of slope $m = -t$, so $t = -m$. It meets $y = 0$ at $Q = (2a + at^2, 0)$. The focus is $F = (a, 0)$, so $FQ = a + at^2 = a(1 + m^2)$. Area of $PFQ = \frac{1}{2}\cdot FQ \cdot \|2at\| = a^2 m(1 + m^2) = 120$. With positive integers: $m = 3$ gives $30a^2 = 120$, $a = 2$. (Check the others: $m = 1$: $2a^2 = 120$, no; $m = 2$: $10a^2 = 120$, no; $m = 4$: $68a^2 = 120$, no.) $(a, m) = (2, 3)$. Answer: (A).
Q.8	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 $\sqrt{1 + \cos 2x} = \sqrt{2\cos^2 x} = \sqrt2\,\|\cos x\|$, so the equation is $\|\cos x\| = \tan^{-1}(\tan x)$. On each interval write $u = \tan^{-1}(\tan x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$; then $x = u + k\pi$ and $\|\cos x\| = \|\cos u\| = \cos u$. So on every one of the three intervals the equation becomes $\cos u = u$ with $u \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. $\cos u = u$ has exactly one root there: for $u \le 0$, $\cos u > 0 \ge u$; on $\left(0, \frac{\pi}{2}\right)$, $\cos u - u$ falls strictly from $1$ to $-\frac{\pi}{2}$, so it crosses zero once ($u \approx 0.739$). That gives one solution in each interval: $x \approx 0.739 - \pi$, $x \approx 0.739$, $x \approx 0.739 + \pi$. Number of solutions $= 3$.
Q.9	Integral CalculusArea under simple curves	Class 12: Integrals; Class 12: Application of Integrals	NumericalEasy
A easy numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Each piece is a straight segment, and $f \ge 0$, so the area is a sum of four triangles. $\left[0, \frac{1}{2n}\right]$: from $f(0) = n$ down to $0$: area $\frac{1}{2}\cdot\frac{1}{2n}\cdot n = \frac{1}{4}$. $\left[\frac{1}{2n}, \frac{3}{4n}\right]$: from $0$ up to $2n\left(\frac{3}{2} - 1\right) = n$: area $\frac{1}{2}\cdot\frac{1}{4n}\cdot n = \frac{1}{8}$. $\left[\frac{3}{4n}, \frac{1}{n}\right]$: from $n$ down to $0$: area $\frac{1}{8}$. $\left[\frac{1}{n}, 1\right]$: from $0$ up to $\frac{n}{n-1}(n - 1) = n$: area $\frac{1}{2}\left(1 - \frac{1}{n}\right) n = \frac{n-1}{2}$. Total $= \frac{1}{2} + \frac{n-1}{2} = \frac{n}{2} = 4$, so $n = 8$. The peaks of the four pieces are all $n$, so the maximum of $f$ is $n = 8$.
Q.10	Sequences and SeriesGeometric progressions	Class 11: Sequences and Series	NumericalMedium
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Write the $r$-th term: $7\overbrace{5\cdots5}^{r}7 = 7\cdot 10^{r+1} + 5(10^r + \cdots + 10) + 7 = 7\cdot10^{r+1} + \frac{5(10^{r+1} - 10)}{9} + 7 = \frac{68\cdot 10^{r+1} + 13}{9}$. (Check: $r = 0$ gives $\frac{693}{9} = 77$; $r = 1$ gives $\frac{6813}{9} = 757$.) $S = \sum_{r=0}^{98} \frac{68\cdot10^{r+1} + 13}{9} = \frac{1}{9}\left(68\cdot\frac{10^{100} - 10}{9} + 13\cdot 99\right) = \frac{68\cdot 10^{100} + 10903}{81}$. The 99-fives number is $N = \frac{68\cdot10^{100} + 13}{9}$, so $68\cdot 10^{100} = 9N - 13$. $S = \frac{9N - 13 + 10903}{81} = \frac{9N + 10890}{81} = \frac{N + 1210}{9}$. So $m = 1210$, $n = 9$ (both below 3000), and $m + n = 1219$.
Q.11	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 Multiply by the conjugate $7 + 3i\cos\theta$: $\frac{(1967 + 1686i\sin\theta)(7 + 3i\cos\theta)}{49 + 9\cos^2\theta} = \frac{(13769 - 5058\sin\theta\cos\theta) + i(5901\cos\theta + 11802\sin\theta)}{49 + 9\cos^2\theta}$. For a real value the imaginary part must vanish: $5901\cos\theta + 11802\sin\theta = 0$, i.e. $\tan\theta = -\frac{1}{2}$. Then $\cos^2\theta = \frac{1}{1 + \tan^2\theta} = \frac{4}{5}$ and $\sin\theta\cos\theta = \frac{\tan\theta}{1 + \tan^2\theta} = -\frac{2}{5}$. Value $= \frac{13769 + 5058\cdot\frac{2}{5}}{49 + 9\cdot\frac{4}{5}} = \frac{13769 + 2023.2}{56.2} = \frac{15792.2}{56.2} = 281$. Every real element of $A$ equals 281, so the one positive integer is $n = 281$.
Q.12	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	NumericalHard
A hard numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra The plane's normal $(\sqrt3, 2, 3)$ has length $\sqrt{3 + 4 + 9} = 4$. Distance $\frac{\|\sqrt3\alpha + 2\beta + 3\gamma - 16\|}{4} = \frac{7}{2}$ gives $\sqrt3\alpha + 2\beta + 3\gamma = 2$ or $30$. For a unit vector the left side is at most $4$, so it is $2$. So $S$ is the circle cut from the unit sphere by the plane $\sqrt3 x + 2y + 3z = 2$, whose distance from the origin is $\frac{2}{4} = \frac{1}{2}$. The circle's radius is $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt3}{2}$. $\vec u, \vec v, \vec w$ are equidistant points on this circle: an equilateral triangle inscribed in it, of side $\frac{\sqrt3}{2}\cdot\sqrt3 = \frac{3}{2}$ and area $\frac{\sqrt3}{4}\cdot\frac{9}{4} = \frac{9\sqrt3}{16}$. The tetrahedron with vertices $O, \vec u, \vec v, \vec w$ has this base and height $\frac{1}{2}$: volume $\frac{1}{3}\cdot\frac{9\sqrt3}{16}\cdot\frac{1}{2} = \frac{3\sqrt3}{32}$. The parallelepiped is 6 times the tetrahedron: $V = \frac{9\sqrt3}{16}$. $\frac{80}{\sqrt3}V = \frac{80\cdot 9}{16} = 45$.
Q.13	Binomial TheoremGeneral and middle term	Class 11: Binomial Theorem	NumericalEasy
A easy numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem First expansion: $T_{r+1} = \binom{4}{r}(ax^2)^{4-r}\left(\frac{70}{27bx}\right)^r$ has power $8 - 3r = 5$, so $r = 1$: coefficient $4a^3\cdot\frac{70}{27b} = \frac{280a^3}{27b}$. Second: $T_{r+1} = \binom{7}{r}(ax)^{7-r}\left(-\frac{1}{bx^2}\right)^r$ has power $7 - 3r = -5$, so $r = 4$: coefficient $\binom{7}{4}\frac{a^3}{b^4} = \frac{35a^3}{b^4}$. Equate ($a \ne 0$): $\frac{280}{27b} = \frac{35}{b^4} \Rightarrow b^3 = \frac{35\cdot 27}{280} = \frac{27}{8} \Rightarrow b = \frac{3}{2}$. $2b = 3$.
Q.14	Matrices and DeterminantsConsistency of linear systems	Class 12: Matrices; Class 12: Determinants	Matching listMedium
A medium matching list question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}1&2&1\\1&0&\alpha\\2&-3&\beta\end{vmatrix} = 1(0 + 3\alpha) - 2(\beta - 2\alpha) + 1(-3) = 7\alpha - 2\beta - 3$. So $\Delta = 0$ exactly when $\beta = \frac{1}{2}(7\alpha - 3)$. When $\Delta = 0$: look for $p E_1 + q E_2 = E_3$. From $x$: $p + q = 2$; from $y$: $2p = -3$, so $p = -\frac{3}{2}$, $q = \frac{7}{2}$; then $z$: $p + q\alpha = \frac{7\alpha - 3}{2} = \beta$, which holds. The right sides give $7p + 11q = -\frac{21}{2} + \frac{77}{2} = 28$. (P) $\gamma = 28$: the third equation repeats the first two, and two independent equations in three unknowns give infinitely many solutions → (3). (Q) $\gamma \ne 28$: the equations contradict → no solution (2). (R) $\Delta \ne 0$: unique solution (1). (It is not (4) or (5): (4) satisfies the third equation only when $\gamma = 28$, and (5) fails the first equation.) (S) $\Delta \ne 0$ with $\alpha = 1$, $\gamma = 28$: the unique solution is $(11, -2, 0)$: $11 - 4 + 0 = 7$, $11 + 0 = 11$, $22 + 6 + 0 = 28$. → (4). Answer: (A).
Q.15	StatisticsMean deviation	Class 11: Statistics	Matching listEasy
A easy matching list question on mean deviation, from Statistics. NCERT: Class 11: Statistics $N = 5 + 2 + 3 + 2 + 4 + 4 = 20$, $\sum f_ix_i = 15 + 16 + 33 + 20 + 20 + 16 = 120$. (P) Mean $= \frac{120}{20} = 6$ → (3). (Q) In order: 3 (×5), 4 (×4), 5 (×4), 8 (×2), 10 (×2), 11 (×3); cumulative 5, 9, 13, 15, 17, 20. The 10th and 11th values are both 5, so the median is 5 → (2). (R) $\sum f_i\|x_i - 6\| = 5\cdot3 + 2\cdot2 + 3\cdot5 + 2\cdot4 + 4\cdot1 + 4\cdot2 = 15 + 4 + 15 + 8 + 4 + 8 = 54$; $\frac{54}{20} = 2.7$ → (4). (S) $\sum f_i\|x_i - 5\| = 5\cdot2 + 2\cdot3 + 3\cdot6 + 2\cdot5 + 0 + 4\cdot1 = 10 + 6 + 18 + 10 + 4 = 48$; $\frac{48}{20} = 2.4$ → (5). Answer: (A).
Q.16	Three Dimensional GeometrySkew lines and shortest distance	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Matching listMedium
A medium matching list question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry If $\ell_2$ is not parallel to $H$ it meets $H$ and $d(H) = 0$. So the largest $d(H)$ comes from the plane through $\ell_1$ parallel to $\ell_2$, whose normal is perpendicular to both directions: $(1,1,1)\times(1,0,1) = (1\cdot1 - 1\cdot0,\ 1\cdot1 - 1\cdot1,\ 1\cdot0 - 1\cdot1) = (1, 0, -1)$. $H_0$ passes through the origin (on $\ell_1$): $x - z = 0$. (P) $d(H_0)$ = distance of $(0, 1, -1)$ (on $\ell_2$) from $H_0$ $= \frac{\|0 + 1\|}{\sqrt2} = \frac{1}{\sqrt2}$ → (5). (Q) $(0,1,2)$: $\frac{\|0 - 2\|}{\sqrt2} = \sqrt2$ → (4). (R) The origin lies on $H_0$: $0$ → (3). (S) $x = 1$, $x = z$ gives $z = 1$, and $y = z = 1$: the point $(1,1,1)$, at distance $\sqrt3$ from the origin → (1). Answer: (B).
Q.17	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	Matching listMedium
A medium matching list question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z = x + iy$ with $y \ne 0$ and $r = \|z\|$. Imaginary part: $2\cdot 2xy - 4y = 0 \Rightarrow 4y(x - 1) = 0 \Rightarrow x = 1$. Real part: $r^3 + 2(x^2 - y^2) + 4x - 8 = 0 \Rightarrow r^3 + 2(1 - y^2) - 4 = 0$. With $y^2 = r^2 - 1$: $r^3 + 2(2 - r^2) - 4 = r^3 - 2r^2 = 0$, so $r = 2$ and $y^2 = 3$. (P) $\|z\|^2 = 4$ → (2). (Q) $\|z - \bar z\|^2 = \|2iy\|^2 = 4y^2 = 12$ → (1). (R) $\|z\|^2 + \|z + \bar z\|^2 = 4 + (2x)^2 = 4 + 4 = 8$ → (3). (S) $\|z + 1\|^2 = (x + 1)^2 + y^2 = 4 + 3 = 7$ → (5). Answer: (B).

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	KinematicsProjectile motion	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Multiple correctMedium
A medium multiple correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane (A) The slide is frictionless, so $\frac{1}{2}mu_0^2 = mgh$ and $u_0 = \sqrt{2gh}$, along $\hat x$. True. Fall from height $3h$: time $t = \sqrt{\frac{6h}{g}}$, vertical speed on landing $v_z = \sqrt{2g\cdot3h} = \sqrt{6gh}$ downward. (C) $\tan\theta = \frac{\sqrt{6gh}}{\sqrt{2gh}} = \sqrt3$, so $\theta = 60^\circ$. True. (B) At the bounce the ground reverses and reduces only the normal (vertical) component: $v_z' = \frac{1}{\sqrt3}\sqrt{6gh} = \sqrt{2gh}$ upward; the horizontal part $\sqrt{2gh}$ is unchanged (smooth ground). So $\vec v = \sqrt{2gh}(\hat x + \hat z)$, not $(\hat x - \hat z)$. False. (D) $h_1 = \frac{v_z'^2}{2g} = h$, and $d = u_0 t = \sqrt{2gh}\sqrt{\frac{6h}{g}} = 2\sqrt3\,h$. So $\frac{d}{h_1} = 2\sqrt3$. True. Answer: (A), (C), (D).
Q.2	OpticsPolarisation	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Multiple correctMedium
A medium multiple correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics (A) No reflected light at all means the light falls at Brewster's angle and its electric field lies in the plane of incidence (only that polarisation is fully transmitted). True. Brewster: $\tan i = \mu_b = \sqrt3$, so $i = 60^\circ$ and the refraction angle is $r_1 = 90^\circ - 60^\circ = 30^\circ$. Deviation $\delta = i + e - A$ gives $60^\circ = 60^\circ + e - A$, so $e = A$. At the second face $r_2 = A - 30^\circ$ and $\sin e = \sqrt3\sin(A - 30^\circ)$. With $e = A$: $\sin A = \frac{3}{2}\sin A - \frac{\sqrt3}{2}\cos A$, so $\tan A = \sqrt3$ and $A = 60^\circ$. (B) The prism angle is $60^\circ$, not $45^\circ$. False. (D) The emergence angle is $e = A = 60^\circ$. True. (C) Red at minimum deviation: $\mu_r = \frac{\sin\frac{A + \delta_{\min}}{2}}{\sin\frac{A}{2}} = \frac{\sin 45^\circ}{\sin 30^\circ} = \sqrt2$. True. Answer: (A), (C), (D).
Q.3	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 Both cells have their positive plate on the left wire. Let $V$ be the potential of the left wire over the right one. Key open. If the 1 A in the 5 V branch flowed out of the cell's positive end, $V = 5 - 1 = 4$ V; the 3 Ω would carry $\frac{4}{3}$ A and the 15 V branch only $\frac{1}{3}$ A, giving $R = 33\,\Omega$, which matches nothing. So the 15 V cell drives 1 A backwards through the 5 V cell: $V = 5 + 1\times1 = 6$ V. Current through the 3 Ω: $\frac{6}{3} = 2$ A. Current from the 15 V branch $= 2 + 1 = 3$ A, so $15 - 3R = 6$ and $R = 3\,\Omega$. (A) True. (B) The 3 Ω resistor carries 2 A, so $\|I_1\| = 2$ A. True (see note on its sign). Key closed. Seen from the capacitor branch, the rest of the circuit is a source of $V_{th} = 6$ V with $R_{th} = 3 \parallel 1 \parallel 3 = \frac{1}{\frac13 + 1 + \frac13} = 0.6\,\Omega$. Charging resistance $3 + 0.6 = 3.6\,\Omega$, time constant $\tau = 3.6 \times 2\,\mu\text{F} = 7.2\,\mu$s. (C) $i(t) = \frac{6}{3.6}e^{-(t - t_0)/\tau}$; at $t - t_0 = 7.2\,\mu$s, $i = \frac{5}{3}\times0.36 = 0.6$ A. True. (D) Finally $q = CV_{th} = 2\,\mu\text{F}\times 6\text{ V} = 12\,\mu$C. True. Answer: (A), (B), (C), (D).
Q.4	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 About the pivot: $I = \frac{1}{3}ML^2 = \frac{1}{3}(1)(0.04) = \frac{0.04}{3}$ kg m², and the impact point is at $\frac{L}{2} = 0.1$ m. Angular momentum about the pivot (the pivot force has no torque): $m u\frac{L}{2} = -m v\frac{L}{2} + I\omega$, i.e. $0.05 = -0.01v + \frac{0.04}{3}\omega$. Elastic collision: the speed of separation at the contact point equals the speed of approach: $\omega\frac{L}{2} + v = u$, i.e. $v = 5 - 0.1\omega$. Substitute: $0.05 + 0.01(5 - 0.1\omega) = \frac{0.04}{3}\omega \Rightarrow 0.1 = \left(\frac{0.04}{3} + 0.001\right)\omega = 0.014\overline{3}\,\omega$, so $\omega = 6.98$ rad s$^{-1}$. $v = 5 - 0.698 = 4.30$ m s$^{-1}$. Answer: (A).
Q.5	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In 10 s the liquid volume is $2500$ cm³; over the base $50 \times 5 = 250$ cm² it stands $10$ cm high. The plates (50 cm × 50 cm, $d = 5$ cm apart) now form two capacitors in parallel: filled part: area $0.5 \times 0.1 = 0.05$ m², with $K = 3$; empty part: area $0.5 \times 0.4 = 0.2$ m². $C = \frac{\epsilon_0}{d}(KA_1 + A_2) = \frac{9\times10^{-12}}{0.05}(3\times0.05 + 0.2) = 1.8\times10^{-10}\times 0.35 = 6.3\times10^{-11}$ F $= 63$ pF. Answer: (B).
Q.6	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Single correctEasy
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $TV^{\gamma - 1}$ is constant, so $T_AV_0^{\gamma-1} = T_f(5V_0)^{\gamma - 1}$ and $T_A = 5^{\gamma - 1}T_f$. Isothermal: the temperature does not change, so $T_B = T_f$. $\frac{T_A}{T_B} = 5^{\gamma - 1}$. Answer: (A).
Q.7	GravitationAcceleration due to gravity	Class 11: Gravitation	Single correctEasy
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g = \frac{GM}{r^2}$ with $r = R + h$, so $\frac{g_P}{g_Q} = \left(\frac{r_Q}{r_P}\right)^2 = \frac{36}{25}$ and $\frac{r_Q}{r_P} = \frac{6}{5}$. $r_P = R + \frac{R}{3} = \frac{4R}{3}$, so $r_Q = \frac{6}{5}\cdot\frac{4R}{3} = \frac{8R}{5}$ and $h_Q = \frac{8R}{5} - R = \frac{3R}{5}$. Answer: (A).
Q.8	Dual Nature of Matter and RadiationPhotoelectric effect	Class 12: Dual Nature of Radiation and Matter	NumericalEasy
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Work function: $\phi = \frac{hc}{\lambda_0} = \frac{1240}{310} = 4$ eV. Photon energy: $E = K_{max} + \phi = 1.95 + 4 = 5.95$ eV. For $n = 4 \to 3$: $E = 13.6Z^2\left(\frac{1}{9} - \frac{1}{16}\right) = 13.6Z^2\cdot\frac{7}{144}$. $Z^2 = \frac{5.95 \times 144}{13.6 \times 7} = \frac{856.8}{95.2} = 9$, so $Z = 3$.
Q.9	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalHard
A hard numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Follow the light that goes first to $M_2$ (focal length $12$ cm). S is $10$ cm in front of it, inside the focus: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ with $u = -10$, $f = -12$ gives $\frac{1}{v} = -\frac{1}{12} + \frac{1}{10} = \frac{1}{60}$, $v = +60$ cm: a virtual image $60$ cm behind $M_2$, i.e. $80$ cm from L. The reflected light now crosses L as if from a point $80$ cm away: $\frac{1}{v} = \frac{1}{10} - \frac{1}{80} = \frac{7}{80}$, $v = \frac{80}{7}$ cm on the $M_1$ side. For the next image to fall on S, the light coming back through L must be parallel, because S lies at the focus of L ($10$ cm). That needs $M_1$ to send out a parallel beam, so the image at $\frac{80}{7}$ cm must sit at the focus of $M_1$ ($f = 10$ cm). L to $M_1$ $= \frac{80}{7} + 10 = \frac{150}{7}$ cm, so $n = 150$.
Q.10	Physical World and MeasurementError analysis	Class 11: Units and Measurement	NumericalEasy
A easy numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement With distances taken as positive for a real image, $\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{10} + \frac{1}{20}$, so $f = \frac{20}{3}$ cm. Differentiating, $\frac{\Delta f}{f^2} = \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2} = \frac{0.1}{100} + \frac{0.2}{400} = 0.0015$. $\frac{\Delta f}{f} = f \times 0.0015 = \frac{20}{3}\times 0.0015 = 0.01$, i.e. $1\%$. $n = 1$.
Q.11	ThermodynamicsFirst law	Class 11: Thermodynamics	NumericalEasy
A easy numerical question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics At constant pressure the work is $W = P\Delta V = n_{total}R\Delta T = 3R\Delta T = 66$ J, so $R\Delta T = 22$ J. $\Delta U = \left(2\cdot\frac{3}{2} + 1\cdot\frac{5}{2}\right)R\Delta T = 5.5 \times 22 = 121$ J.
Q.12	KinematicsRelative velocity	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	NumericalEasy
A easy numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Let the person be at $x$ and the shadow tip at $s$ from the foot of the post. Similar triangles: $\frac{4}{s} = \frac{1.6}{s - x}$, so $4(s - x) = 1.6s$ and $s = \frac{4}{2.4}x = \frac{5}{3}x$. Speed of the tip $= \frac{5}{3}\times 60 = 100$ cm s$^{-1}$; relative to the person, $100 - 60 = 40$ cm s$^{-1}$.
Q.13	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 The centre of mass divides the 10 cm rod so that $30 \times r_1 = 20 \times r_2$ with $r_1 + r_2 = 10$: $r_1 = 4$ cm (30 gm) and $r_2 = 6$ cm (20 gm). $I = 0.03(0.04)^2 + 0.02(0.06)^2 = 4.8\times10^{-5} + 7.2\times10^{-5} = 1.2\times10^{-4}$ kg m². $\omega = \sqrt{\frac{\kappa}{I}} = \sqrt{\frac{1.2\times10^{-8}}{1.2\times10^{-4}}} = 10^{-2} = 10\times10^{-3}$ rad s$^{-1}$. $n = 10$.
Q.14	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	Matching listEasy
A easy matching list question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei In each, the mass number falls by 4, so exactly one $\alpha$ is emitted (lowering $Z$ by 2); $\beta$ particles then fix $Z$ ($\beta^-$ raises it by 1, $\beta^+$ lowers it by 1). (P) $92 \to 90$ after $\alpha$; need $91$: one $\beta^-$ → (4). (Q) $82 \to 80$; need $82$: two $\beta^-$ → (3). (R) $81 \to 79$; need $82$: three $\beta^-$ → (2). (S) $91 \to 89$; need $88$: one $\beta^+$ → (1). Answer: (A).
Q.15	Properties of Bulk MatterHeat transfer	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Matching listMedium
A medium matching list question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Peak wavelengths from $\lambda_m = \frac{2.9\times10^{-3}}{T}$: 2000 K → 1450 nm (infrared); 3000 K → 967 nm (infrared); 5000 K → 580 nm (visible); 10000 K → 290 nm (ultraviolet). (1) Photon energy at 290 nm: $\frac{1240}{290} = 4.28$ eV $> 4$ eV, so only 10000 K emits photoelectrons from a 4 eV metal: S → 1. (2) Only 580 nm is visible: R → 2. (3) The central maximum's width grows with $\lambda$, so the longest peak wavelength (2000 K) gives the widest: P → 3. (4) $\frac{\sigma T^4}{\sigma(6000)^4} = \frac{1}{16}$ needs $T = 3000$ K: Q → 4. (5) Imaging bones needs X-rays; none of these peaks is an X-ray, so (5) is unused. P → 3, Q → 4, R → 2, S → 1. Answer: (C).
Q.16	Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Matching listHard
A hard matching list question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $C = \frac{1}{\omega_0^2L} = \frac{1}{10^{10}\times0.05} = 2\times10^{-9}$ F. At $\omega = 8\times10^4$: $X_L = \omega L = 4000\,\Omega$, $X_C = \frac{1}{\omega C} = \frac{1}{1.6\times10^{-4}} = 6250\,\Omega$, so $\|X\| = 2250\,\Omega$. Current $0.05I_0$ means $Z = 20R$: $R^2 + 2250^2 = 400R^2$, so $R = \frac{2250}{\sqrt{399}} \approx 112.6\,\Omega$ ($\approx 112.5\,\Omega$ taking $\sqrt{399}\approx 20$). (P) $I_0 = \frac{45}{R} \approx 0.400$ A $= 400$ mA → (3). (Q) $Q = \frac{\omega_0L}{R} = \frac{5000}{112.6} \approx 44.4$ → (1). (R) Bandwidth $= \frac{R}{L} = \frac{112.6}{0.05} \approx 2250$ rad s$^{-1}$ → (4). (S) At resonance the current is in phase with the voltage; the peak instantaneous power is $V_0I_0 = 45\times0.4 = 18$ W → (2). Answer: (B).
Q.17	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Matching listMedium
A medium matching list question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current With speed $v$ the emf is $B\ell v$, the current $\frac{B\ell v}{R}$ and the retarding force $\frac{B^2\ell^2v}{R}$. Here $B\ell = 4\times0.25 = 1$ T m, so the force is $\frac{v}{10}$ N. $m\frac{dv}{dt} = mg - \frac{B^2\ell^2}{R}v$ gives $v = v_t(1 - e^{-t/\tau})$ with $v_t = \frac{mgR}{B^2\ell^2} = \frac{0.02\times10\times10}{1} = 2$ m s$^{-1}$ and $\tau = \frac{mR}{B^2\ell^2} = \frac{0.02\times10}{1} = 0.2$ s. At $t = 0.2$ s: $v = 2(1 - 0.4) = 1.2$ m s$^{-1}$. (P) emf $= B\ell v = 1.2$ V → (3). (Q) Force $= \frac{B^2\ell^2v}{R} = \frac{1.2}{10} = 0.12$ N → (4). (R) Power $= \frac{\text{emf}^2}{R} = \frac{1.44}{10} = 0.144 \approx 0.14$ W → (2). (S) $v_t = 2.00$ m s$^{-1}$ → (5). Answer: (D).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	Isolation of MetalsRoasting and calcination	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on roasting and calcination, from Isolation of Metals. NCERT: No matching NCERT chapter (A) Malachite is $\mathrm{CuCO_3\cdot Cu(OH)_2}$. Heating it drives off $\mathrm{CO_2}$ and $\mathrm{H_2O}$ and leaves $\mathrm{CuO}$ (and the process is a calcination); cuprite is $\mathrm{Cu_2O}$. False. (B) Calamine $\mathrm{ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2}$; $\mathrm{ZnO}$ is zincite. True. (C) In the reverberatory furnace the iron of copper pyrites ends up as $\mathrm{FeO}$, which silica removes as slag: $\mathrm{FeO + SiO_2 \to FeSiO_3}$. True. (D) The cyanide process: $\mathrm{4Ag + 8CN^- + 2H_2O + O_2 \to 4[Ag(CN)_2]^- + 4OH^-}$, then $\mathrm{2[Ag(CN)_2]^- + Zn \to [Zn(CN)_4]^{2-} + 2Ag}$. True. Answer: (B), (C), (D).
Q.2	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	Multiple correctHard
A hard multiple correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids P: the nitrile with PhMgBr and acid gives the ketone $\mathrm{CH_3CH_2CH(CH_3)CH_2COPh}$; a second PhMgBr gives the tertiary alcohol $\mathrm{CH_3CH_2CH(CH_3)CH_2C(OH)Ph_2}$. The $\mathrm{CH(CH_3)}$ carbon carries H, $\mathrm{CH_3}$, $\mathrm{C_2H_5}$ and $\mathrm{CH_2C(OH)Ph_2}$: asymmetric. P is chiral. Q: Friedel–Crafts acylation gives acetophenone; PhMgBr gives $\mathrm{Ph_2C(OH)CH_3}$, with two identical phenyls: no asymmetric carbon. R: the cadmium reagent turns the acid chloride into the ketone $\mathrm{CH_3CH_2COCH_2Ph}$; PhMgBr gives $\mathrm{CH_3CH_2C(OH)(Ph)CH_2Ph}$, whose carbinol carbon bears OH, ethyl, phenyl and benzyl: asymmetric. S: $\mathrm{PhCH_2CHO \to PhCH_2CH(OH)Ph \to PhCH_2COPh}$ (oxidation) $\to$ cyanohydrin $\mathrm{PhCH_2C(OH)(CN)Ph}$; hot $\mathrm{H_2SO_4}$ hydrolyses the nitrile and dehydrates the tertiary alcohol to $\mathrm{PhCH{=}C(Ph)COOH}$, which has no asymmetric carbon. (A) Q has none. False. (B) Q has none. False. (C) True. (D) True. Answer: (C), (D).
Q.3	Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids	Class 12: Aldehydes, Ketones and Carboxylic Acids	Multiple correctMedium
A medium multiple correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids P: hydroboration–oxidation adds water anti-Markovnikov: $\mathrm{PhCH_2CH_2OH}$. Q: $\mathrm{CrO_3/H_2SO_4}$ oxidises the primary alcohol to $\mathrm{PhCH_2COOH}$; $\mathrm{Cl_2}$/red P (Hell–Volhard–Zelinsky) then water gives the $\alpha$-chloro acid $\mathrm{PhCHClCOOH}$ (the water hydrolyses the acid chloride intermediate, so Q is not an acid chloride). R: $\mathrm{SOCl_2}$ gives $\mathrm{PhCH_2CH_2Cl}$, NaCN gives $\mathrm{PhCH_2CH_2CN}$, and hot acid hydrolysis gives $\mathrm{PhCH_2CH_2COOH}$ (3-phenylpropanoic acid), not the amide. S: concentrated $\mathrm{H_2SO_4}$ brings about intramolecular Friedel–Crafts acylation, closing a five-membered ring: 1-indanone. (A) R and S wrong. (B) all three right. (C) Q wrong. (D) Q, R, S wrong. Answer: (B).
Q.4	Principles of Qualitative AnalysisCharacteristic precipitates and colours	No matching NCERT chapter	Single correctMedium
A medium single correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter A white hydroxide that $\mathrm{PbO_2}$ in acid oxidises to a coloured solution points to manganese(II): $\mathrm{MnCl_2 + 2NaOH \to Mn(OH)_2}$ (white) $\mathrm{+ 2NaCl}$. $\mathrm{PbO_2}$/$\mathrm{H_2SO_4}$ on heating oxidises $\mathrm{Mn^{2+}}$ to purple permanganate: $\mathrm{2Mn^{2+} + 5PbO_2 + 4H^+ \to 2MnO_4^- + 5Pb^{2+} + 2H_2O}$. So X $= \mathrm{MnO_4^-}$ (manganate $\mathrm{MnO_4^{2-}}$ is not stable in acid; chromate would come from a green, not white, hydroxide). The filtrate holds $\mathrm{Cl^-}$; warming with $\mathrm{MnO(OH)_2}$ and concentrated $\mathrm{H_2SO_4}$ liberates $\mathrm{Cl_2}$, which oxidises $\mathrm{I^-}$ to $\mathrm{I_2}$ and turns starch blue. Y $= \mathrm{Cl_2}$. Answer: (C).
Q.5	ElectrochemistryConductance	Class 12: Electrochemistry	Single correctMedium
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry With degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$, Ostwald's law gives $K_a = \frac{c\alpha^2}{1 - \alpha} = \frac{c\Lambda_m^2}{\Lambda_m^{\circ}(\Lambda_m^{\circ} - \Lambda_m)}$. Rearranging: $K_a\Lambda_m^{\circ 2} - K_a\Lambda_m^{\circ}\Lambda_m = c\Lambda_m^2$; divide by $K_a\Lambda_m^{\circ 2}\Lambda_m$: $\frac{1}{\Lambda_m} = \frac{1}{\Lambda_m^{\circ}} + \frac{c\Lambda_m}{K_a\Lambda_m^{\circ 2}}$. Intercept $P = \frac{1}{\Lambda_m^{\circ}}$, slope $S = \frac{1}{K_a\Lambda_m^{\circ 2}}$, so $\frac{P}{S} = K_a\Lambda_m^{\circ}$. Answer: (A).
Q.6	EquilibriumSolubility product	Class 11: Equilibrium	Single correctMedium
A medium single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Dissolved MX gives $\mathrm{M^+}$ and $\mathrm{X^-}$, part of which is protonated: $s = [\mathrm{X^-}] + [\mathrm{HX}] = [\mathrm{X^-}]\left(1 + \frac{[\mathrm{H^+}]}{K_a}\right)$. With $K_{sp} = s[\mathrm{X^-}]$: $s^2 = K_{sp}\left(1 + \frac{[\mathrm{H^+}]}{K_a}\right)$. Ratio of the two cases: $\left(\frac{10^{-3}}{10^{-4}}\right)^2 = 100 = \frac{1 + 10^{-2}/K_a}{1 + 10^{-7}/K_a}$. If $K_a \gg 10^{-7}$ the denominator is 1 and $\frac{10^{-2}}{K_a} = 99$, $K_a \approx 1.0\times10^{-4}$ (consistent: $10^{-7}/K_a = 10^{-3}$, negligible). $pK_a = 4$. Answer: (B).
Q.7	Chemistry in Everyday LifeDrugs and their classification	No matching NCERT chapter	Single correctEasy
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter HI cleaves a phenyl alkyl ether at the alkyl–O bond: Q is phenol (plus an alkyl iodide). Kolbe's reaction (sodium phenoxide + $\mathrm{CO_2}$, then acid) gives salicylic acid, R. Acetylation with acetic anhydride gives acetylsalicylic acid (aspirin), S. (A) describes antidepressants such as MAO inhibitors. False. (B) Aspirin is a non-narcotic analgesic that works by inhibiting prostaglandin synthesis. True. (C) It is non-narcotic. False. (D) Aspirin is 2-acetoxybenzoic acid ($\mathrm{CH_3COO{-}}$ on the ring), not ortho-acetylbenzoic acid ($\mathrm{CH_3CO{-}}$ on the ring). False. Answer: (B).
Q.8	p-Block ElementsGroup 14 elements	No matching NCERT chapter	NumericalEasy
A easy numerical question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{(CH_3)_2SiCl_2}$: $M = 28 + 2(15) + 71 = 129$ g mol$^{-1}$; $\frac{516}{129} = 4$ mol. Hydrolysis gives $\mathrm{(CH_3)_2Si(OH)_2}$, which condenses; four units close the ring $\mathrm{[(CH_3)_2SiO]_4}$. Unit $\mathrm{(CH_3)_2SiO}$: $28 + 30 + 16 = 74$; tetramer $296$ g mol$^{-1}$. 4 mol of monomer give 1 mol of X at full yield. Mass at 75%: $0.75\times296 = 222$ g.
Q.9	States of MatterGas laws	No matching NCERT chapter	NumericalEasy
A easy numerical question on gas laws, from States of Matter. NCERT: No matching NCERT chapter $Z = \frac{PV_m}{RT}$, so $x = \frac{ZRT}{V_m} = \frac{0.5\times0.08\times800}{0.4} = 80$ atm. Ideal at the same $T$ and $P$: $y = \frac{RT}{P} = \frac{0.08\times800}{80} = 0.8$ dm³ mol$^{-1}$ (equivalently $y = \frac{V_m}{Z}$). $\frac{x}{y} = \frac{80}{0.8} = 100$.
Q.10	Chemical KineticsArrhenius equation	Class 12: Chemical Kinetics	NumericalMedium
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Arrhenius: $\log k = \log A - \frac{E_a}{2.303R}\cdot\frac{1}{T}$. Forward: at $\frac{1}{T} = 0.002$ (500 K), $\log k_f = 9$: $9 = 15 - \frac{E_f}{2.303R}(0.002)$, so $\frac{E_f}{2.303R} = 3000$ K. $K = \frac{k_f}{k_b}$, so at 500 K $\log k_b = 9 - 6 = 3$: $3 = 11 - \frac{E_b}{2.303R}(0.002)$, so $\frac{E_b}{2.303R} = 4000$ K. At 250 K ($\frac{1}{T} = 0.004$): $\log k_b = 11 - 4000\times0.004 = -5$. $\|\log k_b\| = 5$.
Q.11	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	NumericalMedium
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Monoatomic: $\gamma = \frac{5}{3}$, $\gamma - 1 = \frac{2}{3}$. From the figure $V_1 = 10$ m³, $T_1 = 600$ K, $T_2 = 60$ K. A → B adiabatic: $q = 0$ and $T_1V_1^{2/3} = T_2V_2^{2/3}$: $V_2^{2/3} = 10\times10^{2/3}$, so $V_2 = 10^{3/2}\times10 = 10^{5/2}$ m³. B → C isothermal: $q = RT_2\ln\frac{V_3}{V_2}$. The total heat is $RT_2\ln10$, so $\frac{V_3}{V_2} = 10$ and $V_3 = 10^{7/2}$ m³. $2\log V_3 = 2\times\frac{7}{2} = 7$.
Q.12	EquilibriumChemical equilibrium	Class 11: Equilibrium	NumericalMedium
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Start with 6 mol A in 1 L. At $T_1$: $[P] = 4$, $[A] = 2$, $K_1 = 2$. At $T_2$: $[P] = 2$, $[A] = 4$, $K_2 = \frac{1}{2}$. $\Delta G^{\ominus} = -RT\ln K$: $\Delta G_1^{\ominus} = -RT_1\ln2 = -2RT_2\ln2$; $\Delta G_2^{\ominus} = -RT_2\ln\frac{1}{2} = RT_2\ln2$. $\Delta G_2^{\ominus} - \Delta G_1^{\ominus} = 3RT_2\ln2 = RT_2\ln8$. $x = 8$.
Q.13	AminesClassification and identification of amines	Class 12: Amines	NumericalMedium
A medium numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Excess $\mathrm{LiAlH_4}$ reduces all four $\mathrm{C{\equiv}N}$ groups to $\mathrm{CH_2NH_2}$: $\mathrm{(H_2NCH_2)_2CH{-}CH(CH_2NH_2)_2}$, a tetra-amine. Each primary amine condenses with acetophenone (in excess) to an imine $\mathrm{{-}N{=}C(CH_3)Ph}$. P has four such groups and no ring containing nitrogen (non-heterocyclic, as stated). sp² carbons per imine unit: the $\mathrm{C{=}N}$ carbon (1) and the six phenyl carbons (6) = 7. The $\mathrm{CH_3}$, the $\mathrm{CH_2}$ and the two $\mathrm{CH}$ carbons are sp³. Total $= 4\times7 = 28$.
Q.14	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Matching listEasy
A easy matching list question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (P) $\mathrm{P_2O_3 + 3H_2O \to 2H_3PO_3}$ → (2). (Q) White phosphorus disproportionates in hot alkali: $\mathrm{P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2}$ → (3). (R) $\mathrm{PCl_5 + CH_3COOH \to CH_3COCl + POCl_3 + HCl}$ → (4). (S) Hypophosphorous acid is a strong reducing agent: $\mathrm{H_3PO_2 + 2H_2O + 4AgNO_3 \to 4Ag + H_3PO_4 + 4HNO_3}$ → (5). Answer: (D).
Q.15	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 (1) $\mathrm{Fe^{2+}}$, $d^6$, weak-field water, octahedral high spin: $t_{2g}^4e_g^2$. (2) $\mathrm{Mn^{2+}}$, $d^5$, octahedral high spin: $t_{2g}^3e_g^2$. (3) $\mathrm{Co^{3+}}$, $d^6$, $\mathrm{NH_3}$ pairs the electrons (low spin): $t_{2g}^6e_g^0$. (4) $\mathrm{Fe^{3+}}$, $d^5$, tetrahedral: $e^2t_2^3$. (5) $\mathrm{Co^{2+}}$, $d^7$, tetrahedral: $e^4t_2^3$ (not in List-I). P → 3, Q → 2, R → 4, S → 1. Answer: (D).
Q.16	Haloalkanes and HaloarenesSN1 and SN2 mechanisms	Class 12: Haloalkanes and Haloarenes	Matching listMedium
A medium matching list question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (P) In 1-bromo-2-ethylpentane the Br is on $\mathrm{C1}$ ($\mathrm{CH_2Br}$), and the stereocentre is $\mathrm{C2}$. $\mathrm{S_N2}$ at $\mathrm{C1}$ does not touch the bonds of $\mathrm{C2}$, so its configuration is retained → (2). (Q) In 2-bromopentane the Br is on the stereocentre; backside attack gives inversion of configuration → (1). (R) The tertiary stereocentre ionises to a planar carbocation, attacked from both faces: the two enantiomers of 3-methylhexan-3-ol (racemisation) → (3). (S) The molecule has two stereocentres. $\mathrm{S_N1}$ at the C–Br centre scrambles that centre only, while the other keeps its configuration: the two alcohols differ at one centre of two, so they are diastereomers → (5). Answer: (B).
Q.17	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	Matching listMedium
A medium matching list question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Products of List-II: (1) Clemmensen reduction gives ethylbenzene. (2) $\mathrm{KMnO_4}$ oxidises toluene to benzoic acid; $\mathrm{SOCl_2}$ gives benzoyl chloride. (3) Friedel–Crafts methylation gives toluene. (4) Diazotisation gives benzenediazonium chloride. (5) Zinc dust reduces phenol to benzene. (P) Etard reaction ($\mathrm{CrO_2Cl_2}$) turns toluene into benzaldehyde: reactant toluene → (3). (Q) Gattermann reaction (Cu/HX) turns a benzenediazonium salt into a haloarene → (4). (R) Gattermann–Koch reaction (CO, HCl, anhyd. $\mathrm{AlCl_3}$) formylates benzene → (5). (S) Rosenmund reduction ($\mathrm{H_2}$, Pd/$\mathrm{BaSO_4}$) turns benzoyl chloride into benzaldehyde → (2). Answer: (D).

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	Differential EquationsLinear differential equations	Class 12: Differential Equations	Single correctMedium
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Differentiate both sides with respect to $x$ (fundamental theorem of calculus): $3f(x) = f(x) + x f'(x) - x^2$, so $x f'(x) - 2f(x) = x^2$. Divide by $x^3$: $\frac{x f'(x) - 2f(x)}{x^3} = \frac{1}{x}$, and the left side is $\frac{d}{dx}\left(\frac{f(x)}{x^2}\right)$. So $\frac{f(x)}{x^2} = \ln x + C$. At $x = 1$: $f(1) = \frac{1}{3} = C$. (The original equation at $x = 1$ reads $0 = f(1) - \frac{1}{3}$, which agrees.) $f(x) = x^2\left(\ln x + \frac{1}{3}\right)$, so $f(e) = e^2\left(1 + \frac{1}{3}\right) = \frac{4e^2}{3}$. Answer: (C).
Q.2	ProbabilityAddition and multiplication theorems	Class 11: Probability; Class 12: Probability	Single correctMedium
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $p = P(H) = \frac{1}{3}$, $q = P(T) = \frac{2}{3}$, tosses independent. Until it stops, the outcomes alternate. Let $a$ = probability of finally stopping with $HH$ when the last toss was $H$, and $b$ the same when the last toss was $T$. From $H$: the next toss is $H$ (stop with head) or $T$ (move to state $T$): $a = p + q\,b$. From $T$: the next toss is $H$ (move to state $H$) or $T$ (stop with tail): $b = p\,a$. So $a = p + pq\,a$, giving $a = \frac{p}{1 - pq}$ and $b = \frac{p^2}{1 - pq}$. The first toss is $H$ or $T$: $P = p\,a + q\,b = \frac{p^2 + qp^2}{1 - pq} = \frac{p^2(1 + q)}{1 - pq}$. With $p = \frac{1}{3}$, $q = \frac{2}{3}$: $P = \frac{\frac{1}{9}\cdot\frac{5}{3}}{1 - \frac{2}{9}} = \frac{5}{27}\cdot\frac{9}{7} = \frac{5}{21}$. Answer: (B).
Q.3	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctMedium
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $u = \frac{6y}{9 - y^2}$. For $0 < \|y\| < 3$, $9 - y^2 > 0$, so $u$ has the sign of $y$. Case $0 < y < 3$: $u > 0$ and $\cot^{-1}\left(\frac{1}{u}\right) = \tan^{-1}u$. The equation is $2\tan^{-1}u = \frac{2\pi}{3}$, so $u = \tan\frac{\pi}{3} = \sqrt{3}$: $6y = \sqrt{3}(9 - y^2)$, i.e. $y^2 + 2\sqrt{3}\,y - 9 = 0$, $y = -\sqrt{3} \pm 2\sqrt{3}$. Only $y = \sqrt{3}$ is in $(0, 3)$. Case $-3 < y < 0$: $u < 0$ and $\cot^{-1}\left(\frac{1}{u}\right) = \pi + \tan^{-1}u$. The equation is $2\tan^{-1}u + \pi = \frac{2\pi}{3}$, so $\tan^{-1}u = -\frac{\pi}{6}$, $u = -\frac{1}{\sqrt{3}}$: $6\sqrt{3}\,y = -(9 - y^2)$, i.e. $y^2 - 6\sqrt{3}\,y - 9 = 0$, $y = 3\sqrt{3} \pm 6$. Only $y = 3\sqrt{3} - 6 \approx -0.80$ is in $(-3, 0)$. Sum $= \sqrt{3} + 3\sqrt{3} - 6 = 4\sqrt{3} - 6$. Answer: (C).
Q.4	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Single correctEasy
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra (A) $\vec b - \vec a = (2, 4, 8)$, $\vec c - \vec a = \left(\frac{12}{5}, \frac{6}{5}, 12\right)$, $\vec d - \vec a = (1, -1, 6)$. Their scalar triple product is $\frac{1}{5}\begin{vmatrix}2&4&8\\12&6&60\\1&-1&6\end{vmatrix} = \frac{1}{5}\left[2(36 + 60) - 4(72 - 60) + 8(-12 - 6)\right] = \frac{1}{5}(192 - 48 - 144) = 0$. The points are coplanar, so (A) is false. (B) $\frac{\vec b + 2\vec d}{3} = \frac{(3 + 4,\ 6 + 2,\ 3 + 2)}{3} = \left(\frac{7}{3}, \frac{8}{3}, \frac{5}{3}\right)$. The point dividing $PR$ internally in $5 : 4$ is $\frac{5\vec c + 4\vec a}{9} = \frac{(17 + 4,\ 16 + 8,\ 35 - 20)}{9} = \left(\frac{7}{3}, \frac{8}{3}, \frac{5}{3}\right)$. Equal, so (B) is true. (C) The external point would be $\frac{5\vec c - 4\vec a}{1} = (13, 8, 55)$, a different point; false. (D) $\vec b \times \vec d = (6\cdot1 - 3\cdot1,\ 3\cdot2 - 3\cdot1,\ 3\cdot1 - 6\cdot2) = (3, 3, -9)$, whose square magnitude is $9 + 9 + 81 = 99 \neq 95$; false. Answer: (B).
Q.5	Matrices and DeterminantsAdjoint and inverse	Class 12: Matrices; Class 12: Determinants	Multiple correctMedium
A medium multiple correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Row $i = 1$: $1$ divides everything, so row 1 is $(1, 1, 1)$. Row $i = 2$: $j + 1$ even for $j = 1, 3$: $(1, 0, 1)$. Row $i = 3$: $j + 1$ divisible by $3$ for $j = 2$: $(0, 1, 0)$. $M = \begin{pmatrix}1&1&1\\1&0&1\\0&1&0\end{pmatrix}$, $\det M = 1(0 - 1) - 1(0 - 0) + 1(1 - 0) = 0$. (A) $\det M = 0$, so $M$ is not invertible; false. (C) Since $\det M = 0$, $MX = \mathbf 0$ has a nonzero solution (e.g. $X = (1, 0, -1)^T$); true. (B) We need $(M + I)X = \mathbf 0$ with $X \neq \mathbf 0$, i.e. $\det(M + I) = 0$: $\begin{vmatrix}2&1&1\\1&1&1\\0&1&1\end{vmatrix} = 2(0) - 1(1) + 1(1) = 0$. True (e.g. $X = (0, 1, -1)^T$). (D) $\det(M - 2I) = \begin{vmatrix}-1&1&1\\1&-2&1\\0&1&-2\end{vmatrix} = -1(4 - 1) - 1(-2 - 0) + 1(1 - 0) = -3 + 2 + 1 = 0$, so $M - 2I$ is not invertible; false. Answer: (B), (C).
Q.6	Limits, Continuity and DifferentiabilityDifferentiability	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Multiple correctMedium
A medium multiple correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $g(x) = \left(x - \frac{1}{4}\right)^2\left(x - \frac{1}{2}\right)$, so $g'(x) = \left(x - \frac{1}{4}\right)\left(3x - \frac{5}{4}\right)$. On $(0, 1)$, $[4x]$ is $0, 1, 2, 3$ on $\left(0, \frac14\right), \left[\frac14, \frac12\right), \left[\frac12, \frac34\right), \left[\frac34, 1\right)$; only $\frac14, \frac12, \frac34$ need checking. At $x = \frac14$: $g\left(\frac14\right) = 0$ and $g'\left(\frac14\right) = 0$, so $f$ ($= 0$ on the left, $= g$ on the right) is continuous and differentiable there. At $x = \frac12$: $g\left(\frac12\right) = 0$, so $f$ is continuous; left derivative $g'\left(\frac12\right) = \frac14\cdot\frac14 = \frac{1}{16}$, right derivative $2g'\left(\frac12\right) = \frac18$. Continuous but not differentiable. At $x = \frac34$: $g\left(\frac34\right) = \frac14\cdot\frac14 = \frac{1}{16} \neq 0$, so $f$ jumps from $\frac{2}{16}$ to $\frac{3}{16}$: discontinuous. (A) True: the only discontinuity is at $\frac34$. (B) True: $x = \frac12$. (C) False: $f$ fails to be differentiable only at $\frac12$ and $\frac34$, two points. (D) $f < 0$ only on $\left(\frac14, \frac12\right)$, where $f = g$. There $g' = 0$ at $x = \frac{5}{12}$: $g\left(\frac{5}{12}\right) = \left(\frac16\right)^2\left(-\frac{1}{12}\right) = -\frac{1}{432}$. The minimum is $-\frac{1}{432}$, not $-\frac{1}{512}$; false. Answer: (A), (B).
Q.7	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	Multiple correctMedium
A medium multiple correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $h(x) = f(x) - x$. Then $h'' = f'' > 0$ on $(-1, 1)$, so $h'$ is strictly increasing there. (B) If $h$ had three zeros in $(-1, 1)$, Rolle's theorem would give two zeros of $h'$, and then a zero of $h''$ between them, impossible. So $X_f \le 2$; true. (A) $f(x) = x^2 + 1$: $x^2 - x + 1 > 0$ always, so $X_f = 0$; true. (C) $f(x) = 2x^2$: $2x^2 = x$ at $x = 0$ and $x = \frac12$, both in $(-1, 1)$, so $X_f = 2$; true. (D) $f(x) = x^2$: $x^2 = x$ at $x = 0$ and $x = 1$, and $1 \notin (-1, 1)$, so $X_f = 1$. Such a function exists; false. Answer: (A), (B), (C).
Q.8	Integral CalculusFundamental theorem of calculus	Class 12: Integrals; Class 12: Application of Integrals	NumericalEasy
A easy numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The upper limit is $u(x) = x\tan^{-1}x$. Since $\tan^{-1}x$ has the sign of $x$, $u(x) \ge 0$ for all $x$, with $u = 0$ only at $x = 0$. For $t \ge 0$ the integrand $\frac{e^{t - \cos t}}{1 + t^{2023}}$ is positive, so $f(x) = \int_0^{u(x)} (\text{positive})\,dt \ge 0$, with equality exactly when $u(x) = 0$, i.e. at $x = 0$. Minimum value $= f(0) = 0$. Answer: $0$.
Q.9	Differential EquationsLinear differential equations	Class 12: Differential Equations	NumericalEasy
A easy numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Divide by $(x^2 - 5)^2$: $\frac{(x^2 - 5)y' - 2xy}{(x^2 - 5)^2} = -2x$, and the left side is $\frac{d}{dx}\left(\frac{y}{x^2 - 5}\right)$. So $\frac{y}{x^2 - 5} = -x^2 + C$. At $x = 2$: $\frac{7}{-1} = -4 + C$, so $C = -3$. $y = (x^2 - 5)(-x^2 - 3) = -(x^2 - 5)(x^2 + 3)$. With $u = x^2 \ge 0$: $y = -(u^2 - 2u - 15) = 16 - (u - 1)^2$. The maximum is $16$, at $u = 1$ ($x = \pm 1$). Answer: $16$.
Q.10	ProbabilityAddition and multiplication theorems	Class 11: Probability; Class 12: Probability	NumericalMedium
A medium numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability A five-digit number from these digits is a multiple of $5$ only if its last digit is $0$ (there is no $5$). Then the first four digits come from $\{1, 2, 2, 2, 4, 4\}$, and the leading digit is automatically non-zero. Count 4-digit strings from $\{1, 2^3, 4^2\}$ by how many of each are used: without 1: $(2,2,2,4)$: $4$; $(2,2,4,4)$: $6$ — total $10$. with 1: $(1,2,2,2)$: $4$; $(1,2,2,4)$: $12$; $(1,2,4,4)$: $12$ — total $28$. Multiples of $5$: $38$. Multiple of $20$: last digit $0$ and tens digit even, so the number ends in $20$ or $40$; the first three digits come from what is left. Ends in $20$: three digits from $\{1, 2^2, 4^2\}$: $(2,2,4)$ $3$, $(2,4,4)$ $3$, $(1,2,2)$ $3$, $(1,2,4)$ $6$, $(1,4,4)$ $3$ — $18$. Ends in $40$: three digits from $\{1, 2^3, 4\}$: $(2,2,2)$ $1$, $(2,2,4)$ $3$, $(1,2,2)$ $3$, $(1,2,4)$ $6$ — $13$. Multiples of $20$: $31$. All elements equally likely, so $p = \frac{31}{38}$ and $38p = 31$. Answer: $31$.
Q.11	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 Put the circle's centre at the origin of the complex plane with $A_k = 2\omega^k$, $\omega = e^{2\pi i/8}$. The $A_k$ are the roots of $w^8 = 2^8 = 256$, so $\prod_{k}(z - A_k) = z^8 - 256$. For $P = z$ on the circle, $\|z\| = 2$: $PA_1\cdots PA_8 = \|z^8 - 256\|$, where $\|z^8\| = 256$. By the triangle inequality $\|z^8 - 256\| \le \|z^8\| + 256 = 512$, with equality when $z^8 = -256$ (e.g. $P$ midway between two adjacent vertices on the arc). Answer: $512$.
Q.12	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 Expand along the third row: $\det = -5\,(ad - bc)$. The matrix is invertible exactly when $ad \neq bc$. Total choices: $8^4 = 4096$. Count $ad = bc$: Both zero: $ad = 0$ for $8^2 - 7^2 = 15$ pairs $(a, d)$, and likewise $15$ pairs $(b, c)$: $225$. Both non-zero: then all four are primes and, by unique factorisation, $\{b, c\} = \{a, d\}$. If $a = d$: $b = c = a$, $7$ ways. If $a \neq d$: $7\cdot6$ ordered pairs, each with $2$ orders of $(b, c)$: $84$. Total $91$. (A product cannot be zero on one side and non-zero on the other.) Singular: $225 + 91 = 316$. Invertible: $4096 - 316 = 3780$. Answer: $3780$.
Q.13	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 The midpoint $M$ of a common tangent segment $PQ$ has $MP = MQ$, and these are the tangent lengths from $M$ to $C_1$ and $C_2$. So $M$ has equal power with respect to both circles: it lies on their radical axis. Both midpoints do, so the line joining them is the radical axis. $C_1: x^2 + y^2 - 1 = 0$, $C_2: x^2 + y^2 - 8x - 2y + 17 - r^2 = 0$. Subtracting: $8x + 2y = 18 - r^2$. It meets the $x$-axis at $B = \left(\frac{18 - r^2}{8}, 0\right)$. $AB^2 = \left(4 - \frac{18 - r^2}{8}\right)^2 + 1 = 5$, so $4 - \frac{18 - r^2}{8} = \pm 2$, i.e. $\frac{18 - r^2}{8} = 2$ or $6$, giving $r^2 = 2$ or $r^2 = -30$. So $r^2 = 2$ ($r = \sqrt2 \in (1, 3)$; the circles are apart since $OA = \sqrt{17} > 1 + r$, so the common tangents exist). Answer: $2$.
Q.14	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 Let $A$ be the largest angle (obtuse) and $C$ the smallest, so $A = C + \frac{\pi}{2}$ and $B = \pi - A - C = \frac{\pi}{2} - 2C$. The largest side $a$ and smallest $c$ are the ends of the AP: $a + c = 2b$. With circumradius $1$, sides are $2\sin A, 2\sin B, 2\sin C$: $\sin A + \sin C = 2\sin B$, i.e. $\cos C + \sin C = 2\cos 2C = 2(\cos C - \sin C)(\cos C + \sin C)$. So $\cos C - \sin C = \frac12$. Squaring: $1 - \sin 2C = \frac14$, $\sin 2C = \frac34$; then $\cos C + \sin C = \sqrt{1 + \frac34} = \frac{\sqrt7}{2}$ and $\cos 2C = \frac12\cdot\frac{\sqrt7}{2} = \frac{\sqrt7}{4}$. ($2C \approx 48.6^\circ$, so $C < B$ as needed.) Area $a = \frac{(2\sin A)(2\sin B)(2\sin C)}{4R} = 2\cos C\cos 2C\sin C = \sin 2C\cos 2C = \frac34\cdot\frac{\sqrt7}{4} = \frac{3\sqrt7}{16}$. $64a = 12\sqrt7$, so $(64a)^2 = 144\cdot7 = 1008$. Answer: $1008.00$.
Q.15	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 Take $A$ obtuse and largest, $C$ smallest: $A = C + \frac{\pi}{2}$, $B = \frac{\pi}{2} - 2C$, and the AP gives $a + c = 2b$. With $R = 1$: $\cos C + \sin C = 2\cos 2C$, so $\cos C - \sin C = \frac12$, $\sin 2C = \frac34$, $\cos C + \sin C = \frac{\sqrt7}{2}$, $\cos 2C = \frac{\sqrt7}{4}$. Area $\Delta = \frac{abc}{4R} = \sin 2C\cos 2C = \frac{3\sqrt7}{16}$. Semi-perimeter $s = \frac{a + b + c}{2} = \frac{3b}{2} = 3\sin B = 3\cos 2C = \frac{3\sqrt7}{4}$. Inradius $r = \frac{\Delta}{s} = \frac{3\sqrt7/16}{3\sqrt7/4} = \frac14 = 0.25$. Answer: $0.25$.
Q.16	ProbabilityProbability distribution of a random variable	Class 11: Probability; Class 12: Probability	NumericalEasy
A easy numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability A $6 \times 6$ square has $7 \times 7 = 49$ lattice points. Corners: $4$ points with $2$ friends. Other boundary points: $4 \times 5 = 20$ with $3$ friends. Interior: $5 \times 5 = 25$ with $4$ friends. No point has $0$ or $1$ friend. $E(X) = \frac{4\cdot2 + 20\cdot3 + 25\cdot4}{49} = \frac{168}{49} = \frac{24}{7}$. $7E(X) = 24$. Answer: $24.00$.
Q.17	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 Friend pairs are the unit segments of the grid: each of the $7$ rows has $6$, and each of the $7$ columns has $6$, so $84$ pairs. (Check: the friend counts add to $168 = 2 \times 84$.) Pairs of distinct points: $\binom{49}{2} = 1176$. $p = \frac{84}{1176} = \frac{1}{14}$, so $7p = \frac12 = 0.5$. Answer: $0.50$.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	ElectrostaticsElectric dipole	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctEasy
A easy single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance P is about $141$ mm away, far compared with the $4$ mm size of the dipole, so the dipole formula applies: $V = \frac{\vec p\cdot\hat r}{4\pi\varepsilon_0 r^2}$, with $\hat r = \frac{1}{\sqrt2}(1, 1)$. First dipole: $\vec p = q\,(0, 4)$ mm, so $\vec p\cdot\hat r = \frac{4q}{\sqrt2}$. New dipole: $\vec p' = q\left[(-1, 2) - (1, -2)\right] = q\,(-2, 4)$ mm, so $\vec p'\cdot\hat r = \frac{(-2 + 4)q}{\sqrt2} = \frac{2q}{\sqrt2}$. The centre is still the origin, so $r$ is unchanged and $V' = \frac{2}{4}V_0 = \frac{V_0}{2}$. Answer: (B).
Q.2	Physical World and MeasurementDimensional analysis	Class 11: Units and Measurement	Single correctEasy
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $[Y] = [M L^{-1} T^{-2}]$, $[c] = [L T^{-1}]$, $[h] = [M L^2 T^{-1}]$, $[G] = [M^{-1} L^3 T^{-2}]$. M: $\beta - \gamma = 1$. L: $\alpha + 2\beta + 3\gamma = -1$. T: $-\alpha - \beta - 2\gamma = -2$. From M, $\beta = 1 + \gamma$; from T, $\alpha = 2 - \beta - 2\gamma = 1 - 3\gamma$. Into L: $(1 - 3\gamma) + 2(1 + \gamma) + 3\gamma = -1$, so $3 + 2\gamma = -1$, $\gamma = -2$. Then $\beta = -1$, $\alpha = 7$. Answer: (A).
Q.3	Laws of MotionNewton's laws	Class 11: Laws of Motion	Single correctEasy
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $\dot x = \alpha y$ and $\dot y = 2\alpha x$. $a_x = \frac{d}{dt}(\alpha y) = \alpha\dot y = 2\alpha^2 x$, and $a_y = \frac{d}{dt}(2\alpha x) = 2\alpha\dot x = 2\alpha^2 y$. $\vec F = m\vec a = 2m\alpha^2(x\hat x + y\hat y)$. Answer: (A).
Q.4	Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom	Class 11: Kinetic Theory	Single correctEasy
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Equipartition: $U_n = \frac{n}{2}RT$, so at fixed $T$ a larger $n$ means a larger $U_n$. $\gamma = 1 + \frac{2}{n}$ and $\mathrm{v}_n = \sqrt{\frac{\gamma RT}{M}}$; taking the same molar mass $M$, a larger $n$ means a smaller $\gamma$ and so a smaller $\mathrm{v}_n$. (A) $\mathrm{v}_3 < \mathrm{v}_6$ is false. (B) $\mathrm{v}_5 > \mathrm{v}_3$ is false. (C) $\mathrm{v}_5 > \mathrm{v}_7$ and $U_5 < U_7$: both true. (D) $\mathrm{v}_6 < \mathrm{v}_7$ is false. Answer: (C).
Q.5	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 Wave picture: inside the slab the optical path at height $y$ is $n(y)\,d$, with $n(y) = n_1 + \frac{n_2 - n_1}{h}y$. Across the height $h$ the emerging wavefront is therefore retarded at the top by $(n_2 - n_1)d$ more than at the bottom. The emerging plane wavefront is tilted so that the lower part has advanced further: the wave turns towards the higher index, i.e. upward, through an angle $\theta$ with $h\sin\theta = (n_2 - n_1)d$. Ray picture (check): the ray equation gives $n\frac{d\varphi}{dx} = \frac{dn}{dy}$, so after the thin slab $n\sin\varphi = \frac{(n_2 - n_1)d}{h}$, and Snell's law at the exit face gives $\sin\theta = n\sin\varphi = \frac{(n_2 - n_1)d}{h}$. For a thin slab the angle is small and $\theta \approx \tan^{-1}\left[\frac{(n_2 - n_1)d}{h}\right]$: (B) is true and (C) false. The angle depends on $n_2 - n_1$ (with $d$ and $h$) but not on $n_1$ or $n_2$ separately: (D) is true. (A) involves $n_2^2 - n_1^2 = (n_2 - n_1)(n_2 + n_1)$, which depends on the individual values; false. Answer: (B), (D).
Q.6	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	Multiple correctMedium
A medium multiple correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Moment of inertia of the annulus about its axis: $I = \frac{M(a^2 + b^2)}{2}$. $h_m$ is the height at which the disk starts rolling with no friction impulse needed: $v = \frac{\mathcal I_0}{M}$, $\omega = \frac{\mathcal I_0 h}{I}$, and rolling requires $v = \omega b$: $\frac{\mathcal I_0}{M} = \frac{\mathcal I_0 h_m b}{I}$, so $h_m = \frac{I}{Mb} = \frac{a^2 + b^2}{2b}$. (A) $a \to 0$ (solid disk): $h_m = \frac{b}{2}$; true. (B) $a \to b$ (ring): $h_m = \frac{2b^2}{2b} = b$; true. (C) At $h = h_m$, $\omega = \frac{v}{b} = \frac{\mathcal I_0}{Mb}$, independent of $a$; true. (D) With $\mu = 0$ and the impulse through the centre there is no torque at all, so $\omega = 0$ while $v = \frac{\mathcal I_0}{M}$; with no friction nothing ever starts it turning, so it slides without rolling; true. Answer: (A), (B), (C), (D).
Q.7	Electromagnetic WavesProperties of electromagnetic waves	Class 12: Electromagnetic Waves	Multiple correctMedium
A medium multiple correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Wave speed: $v = \frac{f}{1/\lambda} = \frac{5\times10^{14}}{10^7/3} = 1.5\times10^8$ m s$^{-1}$, so $n = \frac{c}{v} = 2$: (D) true. The wave travels along $+z$, so $\vec B = \frac{1}{v}\hat z\times\vec E$. With $\hat z\times\hat x = \hat y$ and $\hat z\times\hat y = -\hat x$: $B_x = -\frac{E_y}{v} = -\frac{30}{1.5\times10^8}\sin[\ldots] = -2\times10^{-7}\sin[\ldots]$: (A) true. $B_y = \frac{E_x}{v} = \frac{60}{1.5\times10^8}\sin[\ldots] = 4\times10^{-7}\sin[\ldots]$, not $2\times10^{-7}$: (B) false. (C) $\vec E$ points along $2\hat x + \hat y$, at $\tan^{-1}\frac12 \approx 26.6^\circ$ to the $x$-axis, not $30^\circ$: false. Answer: (A), (D).
Q.8	Motion of System of Particles and Rigid BodyAngular momentum	Class 11: Systems of Particles and Rotational Motion	NumericalMedium
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Linear: $v = \frac{J}{m} = \frac{\sqrt{\pi/2}\times10^{-2}}{5\times10^{-3}} = 2\sqrt{\frac{\pi}{2}}$ m s$^{-1}$, so $v^2 = 2\pi$. Angular, about a diameter ($I = \frac{mR^2}{4}$): $J r = I\omega$, so $\omega = \frac{4Jr}{mR^2} = \frac{4vr}{R^2} = \frac{4v\cdot\frac{2}{3}\times10^{-2}}{\frac{16}{9}\times10^{-4}} = 150\,v$ rad s$^{-1}$. Time to return: $t = \frac{2v}{g} = \frac{v}{5}$. Rotations: $n = \frac{\omega t}{2\pi} = \frac{150v\cdot v/5}{2\pi} = \frac{30v^2}{2\pi} = \frac{30\cdot2\pi}{2\pi} = 30$. Answer: $30$.
Q.9	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 field of the wire, $B = \frac{\mu_0 I}{2\pi y}$, depends only on the distance $y$ from it, so motion along $x$ does not change the flux. Only $v_y = \frac{v}{2}$ matters. The sides parallel to the wire (length $w = 2$ cm) are at $y_1 = d = 4$ cm and $y_2 = 8$ cm. EMF $= \frac{\mu_0 I}{2\pi}\,w\,v_y\left(\frac{1}{y_1} - \frac{1}{y_2}\right)$. $\frac{\mu_0 I}{2\pi} = 2\times10^{-7}\times10 = 2\times10^{-6}$; $\frac{1}{0.04} - \frac{1}{0.08} = 12.5$ m$^{-1}$. EMF $= 2\times10^{-6}\times0.02\times12.5\times\frac{v}{2} = 2.5\times10^{-7}\,v$ V. Current $= \frac{\text{EMF}}{R} = 2.5\times10^{-6}\,v = 10\times10^{-6}$, so $v = 4$ m s$^{-1}$. Answer: $4$.
Q.10	Oscillations and WavesStanding waves and beats	Class 11: Oscillations; Class 11: Waves	NumericalEasy
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For a string fixed at both ends the harmonics are $f_0, 2f_0, 3f_0, \ldots$; successive ones differ by $f_0$, so $f_0 = 1000 - 750 = 250$ Hz (they are the 3rd and 4th). $f_0 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ with $\mu = 2\times10^{-5}$ kg m$^{-1}$: $\sqrt{\frac{T}{\mu}} = 2\times1\times250 = 500$ m s$^{-1}$. $T = 500^2\times2\times10^{-5} = 5$ N. Answer: $5$.
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	NumericalMedium
A medium numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The piston is weightless, so the trapped air starts at atmospheric pressure $P_0 = 10^5$ Pa. Isothermal compression to $\frac{100}{101}V_0$: $P = \frac{101}{100}P_0 = P_0 + 1000$ Pa. The capillary's top is open to the atmosphere ($P_0$), while the liquid surface in the container is under the compressed air ($P$). Just under the meniscus the pressure is $P - \rho g h$, and across the meniscus (contact angle $0$) $P_0 - (P - \rho g h) = \frac{2T}{r}$. So $\rho g h = \frac{2T}{r} + (P - P_0) = \frac{2\times0.075}{10^{-4}} + 1000 = 1500 + 1000 = 2500$ Pa. $h = \frac{2500}{10^3\times10} = 0.25$ m $= 25$ cm. Answer: $25$.
Q.12	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	NumericalEasy
A easy numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Activity halves every half-life: $A = A_0\,2^{-t/T_{1/2}}$. Both start at $A_0$. After its 3rd half-life $A_1 = \frac{A_0}{2^3}$; after its 7th $A_2 = \frac{A_0}{2^7}$. $\frac{A_1}{A_2} = 2^{7-3} = 16$. Answer: $16$.
Q.13	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	NumericalMedium
A medium numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Cycle I (read from the diagram): $a$ at $4P_0$ from $V_0$ to $2V_0$: $W = 4P_0V_0$. $b$ isothermal from $(2V_0, 4P_0)$ to $(4V_0, 2P_0)$ (both have $PV = 8P_0V_0$): $W = 8P_0V_0\ln 2$. $c$ at $2P_0$ from $4V_0$ to $V_0$: $W = -6P_0V_0$. $d$ isochoric: $0$. $W_I = P_0V_0(8\ln2 - 2) = 2P_0V_0(4\ln2 - 1)$. Cycle II: $a'$ isothermal from $(V_0, 4P_0)$ to $(2V_0, 2P_0)$: $W = 4P_0V_0\ln2$. $b'$ isochoric: $0$. $c'$ at $P_0$ from $2V_0$ to $V_0$: $W = -P_0V_0$. $d'$: $0$. $W_{II} = P_0V_0(4\ln2 - 1)$. $\frac{W_I}{W_{II}} = 2$. Answer: $2$.
Q.14	Oscillations and WavesDoppler effect	Class 11: Oscillations; Class 11: Waves	NumericalMedium
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Take $O$ as origin, $P = (0, -r)$, $Q = (r, 0)$, $R = (0, r)$. Moving anticlockwise, $S_2$ at $Q$ has velocity $4\sqrt2$ m s$^{-1}$ along $+y$. The line from $Q$ to the detector at $P$ points along $\frac{(-1, -1)}{\sqrt2}$. The component of the source velocity along it is $4\sqrt2\times\left(-\frac{1}{\sqrt2}\right) = -4$ m s$^{-1}$: the source moves away from $P$ at $4$ m s$^{-1}$. $f = f_0\frac{v}{v + v_s} = 656\times\frac{324}{328} = 648$ Hz. Answer: $648.00$.
Q.15	Oscillations and WavesDoppler effect	Class 11: Oscillations; Class 11: Waves	NumericalMedium
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $S_2$ at $R$ moves horizontally (anticlockwise at the top means along $-x$), perpendicular to the line $RP$, so it has no velocity component towards the detector: $f_2 = 656$ Hz. $S_1$ moves along $OP$ towards the detector at $4$ m s$^{-1}$: $f_1 = 656\times\frac{324}{324 - 4} = 656\times\frac{324}{320} = 664.2$ Hz. Beat frequency $= f_1 - f_2 = 8.2$ Hz. Answer: $8.20$.
Q.16	Properties of Bulk MatterFluid mechanics	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalHard
A hard numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Hot air density (same pressure, ideal gas): $\rho = \rho_a\frac{T_a}{T} = 1.2\times\frac{300}{360} = 1.0$ kg m$^{-3}$. Take the furnace (wide, so the air in it is nearly at rest) at pressure $P_a$, and the outside air at the chimney's base also at $P_a$. At the chimney top, $h = 9$ m higher, the outside pressure is $P_a - \rho_a g h$, and the hot air leaves at that pressure. Bernoulli along the hot air from the furnace to the exit: $P_a = (P_a - \rho_a g h) + \rho g h + \frac12\rho v^2$, so $\frac12\rho v^2 = (\rho_a - \rho)g h$. $v^2 = \frac{2(1.2 - 1.0)\times10\times9}{1.0} = 36$, $v = 6$ m s$^{-1}$. Mass flow $= \rho\,\frac{\pi d^2}{4}\,v = 1.0\times\frac{3.14\times0.01}{4}\times6 = 0.0471$ kg s$^{-1}$ $= 47.1$ g s$^{-1}$. Answer: $47.10$.
Q.17	Properties of Bulk MatterFluid mechanics	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalMedium
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter With the cap on, the air is at rest. Hot air density $\rho = 1.2\times\frac{300}{360} = 1.0$ kg m$^{-3}$. Below the cap (hot air, starting from $P_a$ at the chimney's base): $P_{\text{bottom}} = P_a - \rho g h$. Above the cap (outside air, starting from $P_a$ at the same base level): $P_{\text{top}} = P_a - \rho_a g h$. $\Delta P = (\rho_a - \rho)g h = 0.2\times10\times9 = 18$ N m$^{-2}$ (the inside pushes up on the cap). Answer: $18.00$.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	Chemical Bonding and Molecular StructureMolecular orbital theory	Class 11: Chemical Bonding and Molecular Structure	Single correctMedium
A medium single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{F_2}$ has $14$ valence electrons. For O$_2$ and F$_2$ there is little $2s$–$2p$ mixing, so the $\sigma$ orbital from $2p_z$ ($2\sigma$) lies below the two $\pi$ orbitals ($1\pi$). Order: $1\sigma < 1\sigma^* < 2\sigma < 1\pi < 1\pi^* < 2\sigma^$. Filling: $1\sigma^2\,1\sigma^{2}\,2\sigma^2\,1\pi^4\,1\pi^{4}$, with $2\sigma^$ empty (bond order $1$, no unpaired electrons). (A) has $1\pi$ below $2\sigma$ (the order for B$_2$, C$_2$, N$_2$). (B) puts $2\sigma^$ below $1\pi^$ and fills it. (D) leaves two electrons in $2\sigma^$ and two unpaired in $1\pi^$. (C) shows the correct order and filling. Answer: (C).
Q.2	Surface ChemistryColloids and their preparation	No matching NCERT chapter	Single correctEasy
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter (I) True: lyophobic sols (metals, sulphides) need special methods such as Bredig's arc or chemical reactions; simple mixing does not give them. (II) True: an emulsion is a liquid dispersed in a liquid. (III) False: micelles form only above the critical micelle concentration and above the Kraft temperature, and in a suitable solvent. (IV) False: the Tyndall effect needs a difference in refractive index between the particles and the medium; with equal indices there is no scattering. Answer: (A).
Q.3	Haloalkanes and HaloarenesSN1 and SN2 mechanisms	Class 12: Haloalkanes and Haloarenes	Single correctMedium
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The starting halide is isobutyl chloride, $\mathrm{(CH_3)_2CHCH_2Cl}$; its Grignard reagent is $\mathrm{(CH_3)_2CHCH_2MgCl}$. P: the Grignard reagent with water gives the alkane isobutane, $\mathrm{(CH_3)_2CHCH_3}$, not an alcohol. (A) false. Q: with $\mathrm{CO_2}$ then acid, 3-methylbutanoic acid; with NaOH, sodium 3-methylbutanoate. Kolbe electrolysis couples two isobutyl radicals: $\mathrm{(CH_3)_2CHCH_2CH_2CH(CH_3)_2}$, 2,5-dimethylhexane, eight carbons. (B) true. R: with $\mathrm{CH_3CHO}$, 4-methylpentan-2-ol; $\mathrm{CrO_3}$ oxidises it to the ketone 4-methylpentan-2-one (six carbons). A ketone with $\alpha$-hydrogens does not undergo the Cannizzaro reaction. (C) false. S: NaCN gives 3-methylbutanenitrile; $\mathrm{H_2/Ni}$ gives 3-methylbutan-1-amine; $\mathrm{CHCl_3/KOH}$ (carbylamine) gives the isocyanide $\mathrm{RNC}$; $\mathrm{LiAlH_4}$ reduces it to the secondary amine $\mathrm{RNHCH_3}$, N-methyl-3-methylbutan-1-amine (six carbons, but secondary). (D) false. Answer: (B).
Q.4	BiomoleculesCarbohydrates	Class 12: Biomolecules	Single correctMedium
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Not oxidised by bromine water: X is a non-reducing sugar, so both anomeric carbons are tied up in the glycosidic link and there is no free hemiacetal OH. Hydrolysis gives a laevorotatory solution: this is the inversion of cane sugar. Sucrose ($+66.5^\circ$) gives glucose ($+52.5^\circ$) and fructose ($-92.4^\circ$), a net laevorotatory mixture. (A) is $\alpha$-D-glucopyranose linked C1–C2 to $\beta$-D-fructofuranose: sucrose. (B) and (C) each keep a free anomeric OH on the right-hand pyranose, so they are reducing and would be oxidised by bromine water. (D) links two glucose units through both anomeric carbons (non-reducing, like trehalose), but its hydrolysis gives only glucose, which is dextrorotatory. Answer: (A).
Q.5	Coordination CompoundsIsomerism in complexes	Class 12: Coordination Compounds	Multiple correctMedium
A medium multiple correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Square-planar $\mathrm{[Pt(NH_3)_2Br_2]}$ ($\mathrm{MA_2B_2}$) shows geometrical (cis–trans) isomerism. (A) $\mathrm{[Pt(en)(SCN)_2]}$: the short en chelate must span two cis positions, so the two SCN are always cis: no geometrical isomers. (Its SCN can show linkage isomerism, a different type.) False. (B) $\mathrm{[Zn(NH_3)_2Cl_2]}$: tetrahedral, all positions adjacent: no geometrical isomers. False. (C) $\mathrm{[Pt(NH_3)_2Cl_4]}$: octahedral $\mathrm{MA_2B_4}$: the two NH$_3$ can be cis or trans. True. (D) $\mathrm{[Cr(en)_2(H_2O)(SO_4)]^+}$: octahedral $\mathrm{M(AA)_2BC}$: H$_2$O and SO$_4$ can be cis or trans. True. Answer: (C), (D).
Q.6	Solid StatePacking in fcc, bcc and hcp	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter Radii in terms of $r_x$: $r_y = \frac{8}{\sqrt3}r_x$, $r_z = \frac{\sqrt3}{2}\cdot\frac{8}{\sqrt3}r_x = 4r_x$. Edges: fcc $L_x = 2\sqrt2\,r_x \approx 2.83\,r_x$; bcc $L_y = \frac{4r_y}{\sqrt3} = \frac{32}{3}r_x \approx 10.67\,r_x$; simple cubic $L_z = 2r_z = 8r_x$. (A) Packing efficiencies are $74\%$ (fcc), $68\%$ (bcc), $52.4\%$ (simple cubic): true. (B) $L_y \approx 10.67\,r_x > L_z = 8r_x$: true. (C) $L_x \approx 2.83\,r_x < L_y$: false. (D) Molar masses: $M_y = \frac{2}{3}M_z = 2M_x$. Density $= \frac{ZM}{N_A L^3}$: $\rho_x \propto \frac{4M_x}{(2\sqrt2)^3 r_x^3} = \frac{4M_x}{22.6\,r_x^3}$ and $\rho_y \propto \frac{2\cdot2M_x}{(32/3)^3 r_x^3} = \frac{4M_x}{1214\,r_x^3}$. So $\rho_x > \rho_y$: true. Answer: (A), (B), (D).
Q.7	Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro	Class 12: Aldehydes, Ketones and Carboxylic Acids	Multiple correctHard
A hard multiple correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids P: alkaline $\mathrm{KMnO_4}$ oxidises the ethyl group (benzylic H) to $-$COOH; the tert-butyl group has no benzylic H and survives. P = 4-tert-butylbenzoic acid (one COOH). Q: NaOH hydrolyses both the acid chloride and the methyl ester; acidification gives benzene-1,3-dicarboxylic acid (isophthalic acid). R: hot aqueous acid hydrolyses the ester and the nitrile; the $\beta$-keto acid loses $\mathrm{CO_2}$, leaving 3-acetylbenzoic acid; chromic acid then oxidises the aryl methyl ketone to $-$COOH. R = isophthalic acid. S: the Grignard reagent of the aryl bromide (the aldehyde protected as the dioxolane) takes up $\mathrm{CO_2}$; acid work-up gives the acid and frees the aldehyde: 4-formylbenzoic acid; Tollens' reagent oxidises CHO to COOH. S = terephthalic acid. (A) Dacron's acid monomer is terephthalic acid and glyptal's is phthalic acid (anhydride); P is neither. False. (B) P has only one COOH. False. (C) Q and R are both isophthalic acid. True. (D) R (isophthalic acid) has no $\alpha$-hydrogen and no carbonyl to enolise, so no aldol; S (terephthalic acid) has no CHO, so no Cannizzaro. True. Answer: (C), (D).
Q.8	Redox ReactionsBalancing redox equations	Class 11: Redox Reactions	NumericalEasy
A easy numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions In acid, permanganate oxidises $\mathrm{H_2S}$ to sulphur: $\mathrm{5H_2S + 2MnO_4^- + 6H^+ \to 5S + 2Mn^{2+} + 8H_2O}$ (as the full equation: $\mathrm{5H_2S + 2KMnO_4 + 3H_2SO_4 \to 5S + 2MnSO_4 + K_2SO_4 + 8H_2O}$). Water: $x = 8$. Electrons: each S goes from $-2$ to $0$, $2$ electrons $\times 5 = 10$ (equal to $2$ Mn $\times 5$). $y = 10$. $x + y = 18$. Answer: $18$.
Q.9	Chemical Bonding and Molecular StructureHybridisation	Class 11: Chemical Bonding and Molecular Structure	NumericalMedium
A medium numerical question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Count bond pairs plus lone pairs (steric number) on the central atom: $\mathrm{[I_3]^+}$: central I, 2 bonds + 2 lone pairs = 4, $sp^3$ (bent). (yes) $\mathrm{[SiO_4]^{4-}}$: 4, $sp^3$. (yes) $\mathrm{SO_2Cl_2}$: 4, $sp^3$. (yes) $\mathrm{SOCl_2}$: 3 bonds + 1 lone pair, $sp^3$. (yes) $\mathrm{Ni(CO)_4}$: Ni(0), tetrahedral, $sp^3$. (yes) $\mathrm{XeF_2}$ (2 + 3), $\mathrm{SF_4}$ (4 + 1), $\mathrm{ClF_3}$ (3 + 2), $\mathrm{XeO_2F_2}$ (4 + 1): $sp^3d$. $\mathrm{XeF_4}$ (4 + 2): $sp^3d^2$. $\mathrm{[PtCl_4]^{2-}}$: square planar, $dsp^2$. Total $sp^3$: $5$. Answer: $5$.
Q.10	Redox ReactionsOxidation number	Class 11: Redox Reactions	NumericalMedium
A medium numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions $\mathrm{Br_3O_8}$: $\mathrm{O_3Br{-}BrO_2{-}BrO_3}$; the terminal Br are $+6$ and the central Br $+4$: none at $0$. $\mathrm{F_2O}$: O is $+2$, F is $-1$: none. $\mathrm{H_2S_4O_6}$ (tetrathionic acid): $\mathrm{HO_3S{-}S{-}S{-}SO_3H}$; the two middle S are $0$: $2$. $\mathrm{H_2S_5O_6}$ (pentathionic acid): three middle S at $0$: $3$. $\mathrm{C_3O_2}$: $\mathrm{O{=}C{=}C{=}C{=}O}$; the outer C are $+2$ and the central C is $0$: $1$. Sum $= 0 + 0 + 2 + 3 + 1 = 6$. Answer: $6$.
Q.11	Structure of AtomBohr model	Class 11: Structure of Atom	NumericalEasy
A easy numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Bohr radii: $r_n = \frac{a\,n^2}{Z} = 26.45\,n^2$ pm for $Z = 2$. So $105.8$ pm is $n = 2$ and $26.45$ pm is $n = 1$. $\Delta E = R_H Z^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 2.2\times10^{-18}\times4\times\frac34 = 6.6\times10^{-18}$ J. $\lambda = \frac{hc}{\Delta E} = \frac{6.6\times10^{-34}\times3\times10^8}{6.6\times10^{-18}} = 3\times10^{-8}$ m $= 30$ nm. Answer: $30$.
Q.12	SolutionsColligative properties	Class 12: Solutions	NumericalMedium
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions First solution: $0.2$ molal means $0.2$ mol ($12$ g) urea per $1000$ g solvent, i.e. $0.2$ mol in $1012$ g of solution. $50$ mL weighs $50\times1.012 = 50.6$ g, so it holds $0.2\times\frac{50.6}{1012} = 0.01$ mol urea. Second solution: $\frac{0.06}{60} = 0.001$ mol. Total $0.011$ mol in $300$ mL (volumes add since $\Delta_{\text{mix}}V = 0$): $C = \frac{0.011}{0.3}$ mol L$^{-1}$. $\pi = CRT = \frac{0.011}{0.3}\times62\times300 = 0.011\times62\times1000 = 682$ Torr. Answer: $682$.
Q.13	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	NumericalMedium
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 4-Methyloct-1-ene is $\mathrm{C_9H_{18}}$, $M = 126$; $\frac{2.52}{126} = 0.02$ mol. With peroxide, HBr adds anti-Markovnikov: the major ($9$ parts) product is the primary bromide 1-bromo-4-methyloctane, the minor ($1$ part) the secondary 2-bromo isomer. Combined yield $50\%$: $0.01$ mol of bromides, of which the primary is $\frac{9}{10}\times0.01 = 0.009$ mol. With diethylamine it gives the ammonium salt; $\mathrm{K_2CO_3}$ frees the tertiary amine S, $\mathrm{CH_3(CH_2)_3CH(CH_3)CH_2CH_2CH_2N(C_2H_5)_2}$ = $\mathrm{C_{13}H_{29}N}$, $M = 156 + 29 + 14 = 199$. Mass $= 0.009\times199 = 1.791$ g $= 1791$ mg. Answer: $1791$.
Q.14	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	NumericalMedium
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics At $600$ K the plot gives $S_T - S_0 = 6$ for $\beta$ and $5$ for $\alpha$. With $S_\beta - S_\alpha = 0$ at $0$ K, $\Delta S(600) = S_\beta - S_\alpha = 6 - 5 = 1$ J mol$^{-1}$ K$^{-1}$. $\frac{d(\Delta S)}{dT} = \frac{\Delta C_p}{T}$ with $\Delta C_p = 1$ J mol$^{-1}$ K$^{-1}$, so $\Delta S(300) = \Delta S(600) + \Delta C_p\ln\frac{300}{600} = 1 - 0.69 = 0.31$ J mol$^{-1}$ K$^{-1}$. Answer: $0.31$.
Q.15	ThermodynamicsEnthalpy	Class 11: Thermodynamics	NumericalMedium
A medium numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics From the plot, $\Delta S(600) = 6 - 5 = 1$ J mol$^{-1}$ K$^{-1}$ (the two phases have equal entropy at $0$ K). At the transition temperature $600$ K the phases are in equilibrium: $\Delta G = 0$, so $\Delta H(600) = T\Delta S = 600\times1 = 600$ J mol$^{-1}$. Kirchhoff: $\Delta H(300) = \Delta H(600) + \Delta C_p(300 - 600) = 600 - 300 = 300$ J mol$^{-1}$. Answer: $300.00$.
Q.16	AminesDiazonium salts	Class 12: Amines	NumericalMedium
A medium numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines Sn/HCl reduces the three $\mathrm{NO_2}$ groups to $\mathrm{NH_2}$: 1,3,5-tris(4-aminophenyl)benzene. Excess $\mathrm{NaNO_2/HCl}$ at $0\,^\circ$C gives the tris-diazonium chloride P. Water hydrolyses each $\mathrm{ArN_2^+Cl^-}$ to ArOH: Q = 1,3,5-tris(4-hydroxyphenyl)benzene, $\mathrm{C_{24}H_{18}O_3}$. Aqueous bromine brominates each phenol ring at its two free ortho positions (para is taken by the link to the central ring): $6$ Br in place of $6$ H. Mass change $6\times(80 - 1) = 474$ g mol$^{-1}$, matching the data. R = $\mathrm{C_{24}H_{12}Br_6O_3}$: heteroatoms $6$ Br $+ 3$ O $= 9$. Answer: $9.00$.
Q.17	AminesDiazonium salts	Class 12: Amines	NumericalMedium
A medium numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines P is the tris-diazonium chloride of 1,3,5-tris(4-aminophenyl)benzene: $\mathrm{C_6H_3(C_6H_4N_2^+Cl^-)_3}$. With excess phenol in base each diazonium group couples at the para position of a phenol: $\mathrm{ArN_2^+Cl^- \to Ar{-}N{=}N{-}C_6H_4OH}$. Each coupling replaces Cl ($35.5$) by $\mathrm{C_6H_4OH}$ ($93$): $+57.5$; three of them give $172.5$ g mol$^{-1}$, matching the data. S = $\mathrm{C_6H_3(C_6H_4{-}N{=}N{-}C_6H_4OH)_3}$: carbon $6 + 3\times12 = 42$; heteroatoms $6$ N $+ 3$ O $= 9$. Total $42 + 9 = 51$. Answer: $51.00$.

Sit JEE Advanced 2023 free.

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

Start free