JEE (Advanced) · 2025

JEE Advanced 2025, 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
32 Q
1. Limits, Continuity and Differentiability4
2. Vector Algebra3
3. Conic Sections3
4 easy · 21 medium · 7 hard
Physics
32 Q
1. Oscillations and Waves4
2. Electrostatics4
3. Electromagnetic Induction and Alternating Currents3
6 easy · 21 medium · 5 hard
Chemistry
32 Q
1. p-Block Elements3
2. Aldehydes, Ketones and Carboxylic Acids3
3. Coordination Compounds2
11 easy · 17 medium · 4 hard

Chapters

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

Limits, Continuity and Differentiability4
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Vector Algebra3
Class 12: Vector Algebra
Conic Sections3
Class 11: Conic Sections
Probability2
Class 11: Probability; Class 12: Probability
Matrices and Determinants2
Class 12: Matrices; Class 12: Determinants
Sets, Relations and Functions2
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Complex Numbers2
Class 11: Complex Numbers and Quadratic Equations
Differential Equations2
Class 12: Differential Equations
Applications of Derivatives2
Class 12: Application of Derivatives
Integral Calculus2
Class 12: Integrals; Class 12: Application of Integrals
Trigonometry2
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Quadratic Equations1
Class 11: Complex Numbers and Quadratic Equations
Three Dimensional Geometry1
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Permutations and Combinations1
Class 11: Permutations and Combinations
Sequences and Series1
Class 11: Sequences and Series
Statistics1
Class 11: Statistics
Binomial Theorem1
Class 11: Binomial Theorem

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	Quadratic EquationsNature of roots	Class 11: Complex Numbers and Quadratic Equations	Single correctMedium
A medium single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Look at the difference: $f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3$ (the $x^4$ terms cancel). The condition $f(x) \neq g(x)$ for every real $x$ means this polynomial has no real root. A real polynomial of odd degree always has a real root, so it cannot be a cubic: $a_3 = b_3$. (It then has to be a quadratic with no real root, $a_2 \neq b_2$ and $49 < 4(a_1-b_1)(a_2-b_2)$, but that is not needed here.) Now the $x^3$ coefficient of $h$. In $f(x+1)$: from $x^4$ we get $(x+1)^4 \to 4x^3$, and from $a_3x^3$ we get $a_3x^3$; lower powers give no $x^3$. So the coefficient is $4 + a_3$. In $g(x+2)$: from $(x+2)^4 \to \binom{4}{1}2\,x^3 = 8x^3$, and $b_3x^3$. So the coefficient is $8 + b_3$. Coefficient of $x^3$ in $h$ $= (4 + a_3) - (8 + b_3) = -4 + (a_3 - b_3) = -4$. Answer: (C) $-4$.
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 $p_1, p_2, p_3$ be the probabilities that $S_1, S_2, S_3$ solve the problem, the students working independently. $V$ is "$S_1$ solves, given $S_2$ and $S_3$ fail". By independence this conditional probability is just $p_1$, so $p_1 = \frac{1}{10}$. $W$: $p_2(1 - p_3) = \frac{1}{12}$. $U$: $1 - (1-p_1)(1-p_2)(1-p_3) = \frac{1}{2}$, so $(1-p_1)(1-p_2)(1-p_3) = \frac{1}{2}$, and with $1 - p_1 = \frac{9}{10}$: $(1-p_2)(1-p_3) = \frac{5}{9}$. Expand: $(1-p_3) - p_2(1-p_3) = \frac{5}{9}$, so $1 - p_3 = \frac{5}{9} + \frac{1}{12} = \frac{20 + 3}{36} = \frac{23}{36}$. Hence $P(T) = p_3 = \frac{13}{36}$. Answer: (A).
Q.3	Limits, Continuity and DifferentiabilityDifferentiability	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctMedium
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Derivative at $0$: $f'(0) = \lim_{x\to 0}\frac{f(x) - 2}{x} = \lim_{x\to 0}\left(-2x - x\sin\frac{1}{x}\right) = 0$, since $\|x\sin\frac{1}{x}\| \le \|x\|$. So $f$ is differentiable at $0$ and (A) is false. For $x \neq 0$: $f'(x) = -4x - 2x\sin\frac{1}{x} + \cos\frac{1}{x}$. As $x \to 0$ the first two terms go to $0$, but $\cos\frac{1}{x}$ keeps swinging between $-1$ and $1$. So in every interval $(0, \delta)$, and in every interval $(-\delta, 0)$, there are points where $f' > 0$ and points where $f' < 0$ (take $\frac{1}{x} = 2k\pi$ and $\frac{1}{x} = (2k+1)\pi$ for large $\|k\|$). Hence $f$ is neither increasing nor decreasing on any such interval. (B) is false, and (C) is true: on no $(-\delta, 0)$ is $f$ increasing. (D): $f(x) - f(0) = -x^2\left(2 + \sin\frac{1}{x}\right) < 0$ for $x \neq 0$, because $2 + \sin\frac{1}{x} \ge 1$. So $f(0) = 2$ is the largest nearby value: $x = 0$ is a local MAXIMUM, not a minimum. (D) is false. Answer: (C).
Q.4	Matrices and DeterminantsTypes of matrices	Class 12: Matrices; Class 12: Determinants	Single correctHard
A hard single correct question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $Q^{-1} = Q^T$ means $QQ^T = I$: every row of $Q$ is a unit vector. With integer entries, a row whose squares add to $1$ has exactly one entry $\pm 1$ and the rest $0$. The rows are also mutually perpendicular, so the $\pm1$'s sit in different columns: $Q$ is a signed permutation matrix. Now $PQ = QP$. Write $Q = (q_{ij})$. $(PQ)_{ij} = p_i q_{ij}$ and $(QP)_{ij} = q_{ij} p_j$ with $p = (2, 2, 3)$. So $(p_i - p_j)q_{ij} = 0$: $q_{ij} = 0$ whenever $p_i \neq p_j$, i.e. $q_{13} = q_{23} = q_{31} = q_{32} = 0$. So $Q = \begin{pmatrix} A & 0 \\ 0 & c \end{pmatrix}$ with $A$ a $2\times 2$ signed permutation matrix and $c = \pm 1$. Number of $2\times2$ signed permutation matrices: $2$ permutations $\times\ 2^2$ sign choices $= 8$. Choices for $c$: $2$. Total $= 8 \times 2 = 16$. Answer: (C).
Q.5	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Multiple correctMedium
A medium 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 Direction of $L_1$ = cross product of the plane normals: $(2,3,1)\times(1,2,1) = (3\cdot1 - 1\cdot2,\ 1\cdot1 - 2\cdot1,\ 2\cdot2 - 3\cdot1) = (1, -1, 1)$. $L_2$: $(2 + t, -1 - t, 3 + t)$. Put into $M$: $2(2+t) + (-1-t) - 2(3+t) = 6 \Rightarrow -3 - t = 6 \Rightarrow t = -9$. So $Q = (-7, 8, -6)$. Foot $R$: $2(2) + (-1) - 2(3) - 6 = -9$, and $\|n\|^2 = 9$, so $R = P - \frac{-9}{9}(2, 1, -2) = (4, 0, 1)$. Check: $8 + 0 - 2 = 6$. Also $PR = 3$. (A) $PQ = \|t\|\,\|(1,-1,1)\| = 9\sqrt{3}$. TRUE. (B) $QR$: $R - Q = (11, -8, 7)$, $QR^2 = 121 + 64 + 49 = 234$, so $QR = \sqrt{234} \approx 15.3 \neq 15$. FALSE. (C) $PR$ is perpendicular to the plane and $QR$ lies in it, so the triangle is right-angled at $R$: area $= \frac{1}{2}\cdot PR \cdot QR = \frac{1}{2}\cdot 3\cdot\sqrt{234} = \frac{3}{2}\sqrt{234}$. TRUE. (Check: $PR^2 + QR^2 = 9 + 234 = 243 = PQ^2$.) (D) $\cos\angle QPR = \frac{PR}{PQ} = \frac{3}{9\sqrt{3}} = \frac{1}{3\sqrt{3}}$, not $\frac{1}{2\sqrt{3}}$. FALSE. Answer: (A), (C).
Q.6	Sets, Relations and FunctionsOne-one, into and onto functions	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	Multiple correctMedium
A medium multiple correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f$ ($\mathbb{N} = \{1, 2, 3, \dots\}$): odd $n = 1, 3, 5, \dots$ give $1, 2, 3, \dots$; even $n = 2, 4, 6, \dots$ give $1, 0, -1, -2, \dots$. So $f(1) = f(2) = 1$: $f$ is not one-one. Every integer is reached ($k \ge 1$ from odd $n$, $k \le 1$ from even $n$): $f$ is onto. $g$: $n \ge 0$ gives $3, 5, 7, \dots$; $n < 0$ gives $2, 4, 6, \dots$. Different inputs give different outputs, so $g$ is one-one; but $1$ is never reached, so $g$ is not onto. The range of $g$ is $\{2, 3, 4, \dots\}$. (A) $g\circ f$: $g(f(1)) = g(f(2))$, so not one-one; its values lie in the range of $g$, which misses $1$, so not onto. TRUE. (B) $f\circ g$: $g$ is one-one onto $\{2, 3, 4, \dots\}$. On this set $f$ sends odd $m \ge 3$ to $2, 3, 4, \dots$ and even $m \ge 2$ to $1, 0, -1, \dots$, all different, and together they cover $\mathbb{Z}$. So $f\circ g$ is one-one AND onto. "NOT one-one" is false: (B) FALSE. (C) $g$ is one-one but not onto: FALSE. (D) $f$ is not one-one but is onto: TRUE. Answer: (A), (D).
Q.7	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	Multiple correctEasy
A easy multiple correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Square both sides: $(x-1)^2 + (y-2)^2 = 4\left[x^2 + (y-3)^2\right]$. $x^2 - 2x + 1 + y^2 - 4y + 4 = 4x^2 + 4y^2 - 24y + 36$ $\Rightarrow 3x^2 + 3y^2 + 2x - 20y + 31 = 0 \Rightarrow x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0$. Centre $\left(-\frac{1}{3}, \frac{10}{3}\right)$; radius$^2 = \frac{1}{9} + \frac{100}{9} - \frac{31}{3} = \frac{101 - 93}{9} = \frac{8}{9}$, so radius $= \frac{2\sqrt{2}}{3}$. (A) TRUE. (B) FALSE (wrong centre). (C) FALSE ($\frac{\sqrt2}{3}$ is half the radius). (D) TRUE. Answer: (A), (D).
Q.8	Sets, Relations and FunctionsTypes of relations	Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions	NumericalMedium
A medium numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Reflexive: all $6$ pairs $(x, x)$ must be in $R$. That leaves $10 - 6 = 4$ more ordered pairs, all with $x \neq y$. Symmetric: such pairs come in twos, $(x, y)$ together with $(y, x)$. So the $4$ pairs are exactly $2$ unordered pairs $\{x, y\}$. There are $\binom{6}{2} = 15$ unordered pairs of distinct elements; choose $2$ of them: $\binom{15}{2} = 105$. Answer: $105$.
Q.9	Vector AlgebraAddition of vectors	Class 12: Vector Algebra	NumericalMedium
A medium numerical question on addition of vectors, from Vector Algebra. NCERT: Class 12: Vector Algebra Use position vectors $\vec p, \vec q, \vec r, \vec s$. The condition is $(\vec p - \vec s) + 5(\vec q - \vec s) + 6(\vec r - \vec s) = \vec 0$, so $\vec s = \frac{\vec p + 5\vec q + 6\vec r}{12}$. $\vec e = \frac{\vec p + \vec r}{2}$, $\vec f = \frac{\vec q + \vec r}{2}$. $\overrightarrow{EF} = \vec f - \vec e = \frac{\vec q - \vec p}{2}$, so $EF = \frac{1}{2}PQ$. $\overrightarrow{ES} = \vec s - \vec e = \frac{\vec p + 5\vec q + 6\vec r - 6\vec p - 6\vec r}{12} = \frac{5(\vec q - \vec p)}{12}$, so $ES = \frac{5}{12}PQ$. Ratio $= \frac{1/2}{5/12} = \frac{6}{5} = 1.2$. Answer: $1.20$.
Q.10	Permutations and CombinationsSimple applications	Class 11: Permutations and Combinations	NumericalMedium
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The first digit is $1$ or $2$; the other six are any of $0, 1, 2$. Let $A$ = "$0$ appears exactly twice", $B$ = "$1$ appears exactly twice". We want $\|A \cup B\| = \|A\| + \|B\| - \|A\cap B\|$. $\|A\|$: the two $0$s go in two of the last six places, $\binom{6}{2} = 15$ ways; the remaining $5$ places (first digit included) are each $1$ or $2$: $2^5 = 32$. $\|A\| = 480$. $\|B\|$: if the first digit is $1$, one more $1$ among the last six ($6$ ways), the other $5$ places from $\{0, 2\}$: $6 \cdot 32 = 192$. If the first digit is $2$, two $1$s among the last six ($15$ ways), the other $4$ places from $\{0, 2\}$: $15\cdot 16 = 240$. $\|B\| = 432$. $\|A\cap B\|$: digits are two $0$s, two $1$s, three $2$s. All arrangements: $\frac{7!}{2!\,2!\,3!} = 210$; those starting with $0$: $\frac{6!}{1!\,2!\,3!} = 60$. So $150$. $\|A\cup B\| = 480 + 432 - 150 = 762$. Answer: $762$.
Q.11	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 Expand near $0$: $\frac{1}{1-t^2} = 1 + t^2 + \dots$, so $\int_0^x \frac{dt}{1-t^2} = x + \frac{x^3}{3} + O(x^5)$. Also $x\cos x = x - \frac{x^3}{2} + O(x^5)$. The bracket is $\left(\frac{\alpha}{2} + \beta\right)x + \left(\frac{\alpha}{6} - \frac{\beta}{2}\right)x^3 + O(x^5)$. For the limit after dividing by $x^3$ to be finite, the $x$ term must vanish: $\frac{\alpha}{2} + \beta = 0$, so $\beta = -\frac{\alpha}{2}$. Then the limit is $\frac{\alpha}{6} - \frac{\beta}{2} = \frac{\alpha}{6} + \frac{\alpha}{4} = \frac{5\alpha}{12} = 2$, so $\alpha = \frac{24}{5} = 4.8$ and $\beta = -2.4$. $\alpha + \beta = 2.4$. Answer: $2.40$.
Q.12	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 Let $d$ be the common difference. Then $f(a_{i+1}) = f(a_i + d) = f(a_i)f(d)$, so $f(a_1), f(a_2), \dots$ is a geometric progression with ratio $r = f(d) > 0$. $f(a_{31}) = r^6 f(a_{25}) = 64 f(a_{25})$, so $r^6 = 64$ and, $r$ being positive, $r = 2$. Let $c = f(a_1)$. $\sum_{i=1}^{50} f(a_i) = c(2^{50} - 1) = c(2^{25} - 1)(2^{25} + 1) = 3(2^{25} + 1)$, so $c = \frac{3}{2^{25} - 1}$. $\sum_{i=6}^{30} f(a_i) = c(2^5 + 2^6 + \dots + 2^{29}) = c\cdot 2^5(2^{25} - 1) = 3 \cdot 32 = 96$. Answer: $96$.
Q.13	Differential EquationsSeparation of variables	Class 12: Differential Equations	NumericalMedium
A medium numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Each equation separates: $y = y(1)\exp\left(\int_1^x k(t)\,dt\right)$. Multiply the first two: $\frac{d}{dx}\ln(y_1y_2) = \sin^2 x + \cos^2 x = 1$, so $y_1y_2 = y_1(1)y_2(1)e^{x-1} = \frac{5}{3}e^{x-1}$. Third: $\frac{d}{dx}\ln y_3 = \frac{2}{x^3} - 1$, so $\ln y_3 = -\frac{1}{x^2} - x + C$. At $x = 1$: $\frac{3}{5e} = e^{-2 + C}$, so $e^{C} = \frac{3e}{5}$ and $y_3 = \frac{3}{5}e^{1 - x - 1/x^2}$. Product: $y_1y_2y_3 = \frac{5}{3}\cdot\frac{3}{5}\, e^{x - 1}\, e^{1 - x - 1/x^2} = e^{-1/x^2}$. Limit: $\frac{e^{-1/x^2} + 2x}{e^{3x}\sin x}$. As $x\to0^+$, $e^{3x}\to 1$, $\frac{2x}{\sin x}\to 2$, and $\frac{e^{-1/x^2}}{\sin x}\to 0$ (the exponential vanishes faster than any power of $x$). Answer: $2$.
Q.14	StatisticsMean deviation	Class 11: Statistics	Matching listMedium
A medium matching list question on mean deviation, from Statistics. NCERT: Class 11: Statistics Sum of frequencies: $5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19$, so $f_1 + f_2 = 7$. In increasing order: $4\,(5)$, $5\,(f_1)$, $6\,(1)$, $8\,(f_2)$, $9\,(2)$, $11\,(3)$, $12\,(1)$. With $19$ values the median is the $10$th. Cumulative counts: up to $5$ is $5 + f_1$, up to $6$ is $6 + f_1$. The $10$th value is $6$ only if $5 + f_1 < 10 \le 6 + f_1$, i.e. $f_1 = 4$; then $f_2 = 3$. (P) $7\cdot4 + 9\cdot3 = 55 \to (5)$. Mean: $(20 + 20 + 6 + 24 + 18 + 33 + 12)/19 = 133/19 = 7$. (Q) $\sum f\|x - 7\| = 5\cdot3 + 4\cdot2 + 1\cdot1 + 3\cdot1 + 2\cdot2 + 3\cdot4 + 1\cdot5 = 15 + 8 + 1 + 3 + 4 + 12 + 5 = 48$, so $19\alpha = 48 \to (3)$. (R) $\sum f\|x - 6\| = 5\cdot2 + 4\cdot1 + 0 + 3\cdot2 + 2\cdot3 + 3\cdot5 + 1\cdot6 = 10 + 4 + 6 + 6 + 15 + 6 = 47$, so $19\beta = 47 \to (2)$. (S) $\sum f(x - 7)^2 = 5\cdot9 + 4\cdot4 + 1\cdot1 + 3\cdot1 + 2\cdot4 + 3\cdot16 + 1\cdot25 = 45 + 16 + 1 + 3 + 8 + 48 + 25 = 146$, so $19\sigma^2 = 146 \to (1)$. P→5, Q→3, R→2, S→1. Answer: (C).
Q.15	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	Matching listHard
A hard matching list question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives (P) Let $p(x) = 10x^3 - 45x^2 + 60x + 35$. $p'(x) = 30(x - 1)(x - 2) \le 0$ on $[1, 2]$, so $p$ decreases from $p(1) = 60$ to $p(2) = 55$. $[p/n]$ is continuous on $[1, 2]$ only if $p/n$ never reaches an integer there (reaching an integer at an interior point or at either end makes the floor jump, since $p$ is strictly decreasing). So $[55/n, 60/n]$ must contain no integer. $n = 8$: $[6.875, 7.5]$ contains $7$; $n = 7$: $[7.86, 8.57]$ contains $8$; smaller $n$ give longer intervals, each containing an integer ($n = 6$: $10$; $n = 5$: $11, 12$; and so on). $n = 9$: $[6.11, 6.67]$, no integer. Minimum $n = 9 \to (2)$. (Q) $x^3 + 3x$ is strictly increasing ($3x^2 + 3 > 0$), so $g$ is increasing exactly when $2n^2 - 13n - 15 = (2n - 15)(n + 1) > 0$, i.e. $n > 7.5$. Minimum $n = 8 \to (1)$. (R) Near $x = 3$: $h(x) = (x - 3)^n\,(x + 3)^n(x^2 + 2x + 3)$ and the last two factors are positive. $h(3) = 0$, so $x = 3$ is a local minimum exactly when $(x - 3)^n \ge 0$ on both sides, i.e. $n$ even. Smallest even $n > 5$ is $6 \to (4)$. (S) $\cos\|u\| = \cos u$ is differentiable everywhere. $\sin\|x - k\|$ has left derivative $-\cos 0 = -1$ and right derivative $+1$ at $x = k$, so it has a corner at each $k = 0, 1, 2, 3, 4$; the other terms are smooth there, so the corners do not cancel. $5$ points $\to (3)$. P→2, Q→1, R→4, S→3. Answer: (B).
Q.16	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Matching listHard
A hard matching list question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Geometry first. $\vec w = \vec u\times\vec v$ is perpendicular to $\vec u$ and $\vec v$; $\vec u = \vec v\times\vec w$ is perpendicular to $\vec v$ and $\vec w$. So $\vec u, \vec v, \vec w$ are mutually perpendicular. Then $\|\vec w\| = \|\vec u\|\|\vec v\|$ and $\|\vec u\| = \|\vec v\|\|\vec w\|$. Multiply: $\|\vec u\|\|\vec w\| = \|\vec u\|\|\vec v\|^2\|\vec w\|$, so $\|\vec v\|^2 = 1$. (P)→(2). Also $\|\vec u\|^2 = \|\vec w\|^2 = 6$. $\vec u\cdot\vec w = 0$: $\alpha + \beta - 2\gamma = 0$. The three equations in $\alpha, \beta, \gamma$: adding $\alpha + \beta + \gamma$ to each and rearranging, they say $(t + 1)\alpha = (t + 1)\beta = (t + 1)\gamma = \alpha + \beta + \gamma$. Since $\vec u \neq \vec 0$, either $t = -1$ with $\alpha + \beta + \gamma = 0$, or $t \neq -1$ with $\alpha = \beta = \gamma$, and then $3\alpha = (t + 1)\alpha$ gives $t = 2$. Case $t = 2$, $\alpha = \beta = \gamma$: $\alpha + \beta - 2\gamma = 0$ holds, and $\|\vec u\|^2 = 3\alpha^2 = 6$ gives $\alpha^2 = 2$. Case $t = -1$: $\alpha + \beta + \gamma = 0$ and $\alpha + \beta - 2\gamma = 0$ give $\gamma = 0$, $\beta = -\alpha$; $\|\vec u\|^2 = 2\alpha^2 = 6$ gives $\alpha^2 = 3$. (Q) $\alpha = \sqrt3$ is the second case: $\gamma = 0$, $\gamma^2 = 0$ → (1). (R) $\beta + \gamma = -\sqrt3$, square $= 3$ → (4). (S) $\alpha = \sqrt2$ is the first case: $t = 2$, $t + 3 = 5$ → (5). P→2, Q→1, R→4, S→5. Answer: (A).

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	Oscillations and WavesSHM	Class 11: Oscillations; Class 11: Waves	Single correctMedium
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Let the line from the ring's centre to the disk's centre make a small angle $\theta$ with the vertical. The disk's centre moves on a circle of radius $R - r$, so its displacement along the path is $s = (R - r)\theta$ and its speed is $v = (R - r)\dot\theta$. Kinetic energy of a disk rolling without slipping: $\frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{1}{2}mr^2\left(\frac{v}{r}\right)^2 = \frac{1}{2}\cdot\frac{3}{2}m\,v^2$. Potential energy for small $\theta$: gravity $mg(R - r)(1 - \cos\theta) \approx \frac{1}{2}mg(R - r)\theta^2 = \frac{1}{2}\frac{mg}{R - r}s^2$; spring (stretched by $s$ along the periphery) $\frac{1}{2}ks^2$. So $E = \frac{1}{2}\cdot\frac{3m}{2}\dot s^2 + \frac{1}{2}\left(\frac{mg}{R - r} + k\right)s^2$, which is SHM with $\omega^2 = \frac{\frac{mg}{R-r} + k}{\frac{3m}{2}} = \frac{2}{3}\left(\frac{g}{R - r} + \frac{k}{m}\right)$. Answer: (A).
Q.2	Work, Energy and PowerCollisions	Class 11: Work, Energy and Power	Single correctMedium
A medium single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Work in the centre-of-mass frame. The CM moves at $V = \frac{2m\,u}{3m} = \frac{2u}{3}$. In that frame the heavy particle moves at $u - V = \frac{u}{3}$, and in an elastic collision it keeps that speed, only turning through some angle. Back in the lab frame its velocity is $\vec V + \vec{w}$, where $\|\vec V\| = \frac{2u}{3}$ and $\vec w$ (length $\frac{u}{3}$) can point any way. The tip of the lab velocity lies on a circle of radius $\frac{u}{3}$ centred at distance $\frac{2u}{3}$ from the origin. The largest angle is when the velocity is tangent to that circle: $\sin\theta_{max} = \frac{u/3}{2u/3} = \frac{1}{2} = \frac{m}{2m}$. So $\theta_{max} = \frac{\pi}{6}$. Answer: (D).
Q.3	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Single correctHard
A hard single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Let the side be $a$. Viewed from $+X$ (with $Y$ to the right and $Z$ up), a clockwise turn tips the loop from the $Z$ axis towards $+Y$. After time $t$ the loop makes angle $\omega t$ with the $XZ$ plane. For $0 < \omega t < \pi$ the loop lies in $y \ge 0$, where the field is. Only the part of the area facing $Z$ counts: the projection on the $XY$ plane is $a^2\sin\omega t$. So $\Phi = B_0\cos\omega t \cdot a^2\sin\omega t = \frac{B_0a^2}{2}\sin 2\omega t$, and $V = -\frac{d\Phi}{dt} = -B_0a^2\omega\cos 2\omega t$. This starts at its most negative value at $t = 0$, rises to a positive peak at $t = \frac{\pi}{2\omega}$ and falls back to the most negative value at $t = \frac{\pi}{\omega}$: one full cosine cycle. For $\pi < \omega t < 2\pi$ the loop is in $y < 0$, where there is no field: $\Phi = 0$, $V = 0$. The flux jumps to zero at $t = \frac{\pi}{\omega}$, so the graph drops straight to the axis there. From $2\pi/\omega$ the pattern repeats. That is graph (A). (D) has the right on/off pattern but a $-\sin 2\omega t$ shape; (B) and (C) have no zero stretch. Answer: (A).
Q.4	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 Read the scales from Fig. 1. The main scale has $10$ divisions in $1$ cm, so $1$ MSD $= 1$ mm. The zeros coincide (no zero error), and the $10$ vernier divisions span exactly $7$ main-scale divisions: $1$ VSD $= 0.7$ mm. Least count $= 1\text{ MSD} - 1\text{ VSD} = 0.3$ mm. Fig. 2: the vernier zero lies between the $1$ mm and $2$ mm marks, so the main-scale reading is $1$ mm. The $1$st vernier mark lines up with a main-scale mark (the $2$ mm mark). $D = 1\text{ mm} + 1 \times 0.3\text{ mm} = 1.3$ mm $= 0.13$ cm. (Check: the vernier zero sits at $2 - 0.7 = 1.3$ mm.) Answer: (C).
Q.5	Electromagnetic Induction and Alternating CurrentsFaraday's law	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Multiple correctMedium
A medium multiple correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current While the loop is entering, only its leading edge (length $L$) is in the field. EMF $= B_0Lv$, current $\frac{B_0Lv}{R}$, retarding force $\frac{B_0^2L^2v}{R}$. $M\frac{dv}{dt} = -\frac{B_0^2L^2}{R}v \Rightarrow \frac{dv}{dt} = -Kv \Rightarrow v = v_0e^{-Kt}$. Distance moved: $y(t) = \frac{v_0}{K}\left(1 - e^{-Kt}\right)$, which approaches $\frac{v_0}{K}$ but never reaches it. (A) $v_0 = 1.5KL$: the loop could go up to $1.5L > L$, so it gets fully inside (with speed $v_0 - KL > 0$; from $M\,dv = -\frac{B_0^2L^2}{R}dy$ the speed falls by $KL$ over a distance $L$). It does not stop. FALSE. (B) Once fully inside the uniform field the flux is constant: no EMF, no current, no force. TRUE. (C) $v_0 = \frac{KL}{10}$: $v = v_0e^{-Kt}$ only tends to zero; the loop never comes to rest at a finite time. FALSE. (D) $v_0 = 3KL$: $y = L$ when $1 - e^{-Kt} = \frac{KL}{v_0} = \frac{1}{3}$, i.e. $e^{-Kt} = \frac{2}{3}$, $t = \frac{1}{K}\ln\frac{3}{2}$. TRUE. Answer: (B), (D).
Q.6	Physical World and MeasurementSignificant figures	Class 11: Units and Measurement	Multiple correctEasy
A easy multiple correct question on significant figures, from Physical World and Measurement. NCERT: Class 11: Units and Measurement In cm: length $10.5$ (3 significant figures), breadth $0.05$ mm $= 0.005$ cm (1 significant figure), thickness $6.0\ \mu$m $= 6.0\times10^{-4}$ cm (2 significant figures). Volume $= 10.5 \times 0.005 \times 6.0\times10^{-4} = 3.15\times10^{-5}$ cm$^3$. A product keeps as many significant figures as the least precise factor: here $1$. So $V = 3\times10^{-5}$ cm$^3$. (A) $3.2\times10^{-5}$ has 2 significant figures, (B) $32.0\times10^{-6}$ has 3, (C) $3.0\times10^{-5}$ has 2: all too many. (D) is correct. Answer: (D).
Q.7	Oscillations and WavesWave motion	Class 11: Oscillations; Class 11: Waves	Multiple correctMedium
A medium multiple correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The tension $T$ is the same in all three strings, so the wave speed is $v = \sqrt{T/\mu}$ and, at fixed $\omega$, the wave number is $k \propto \sqrt{\mu}$: $k$ in $S_1$, $2k$ in $S_2$, $4k$ in $S_3$. At a joint where the wave passes into a denser string, the reflected wave is inverted (phase change $\pi$) and the transmitted wave is not. (A) Reflection at $P$ ($\mu \to 4\mu$, denser): it travels in $-x$ in $S_1$ with wave number $k$ and is inverted: $y = \alpha_1y_0\cos(\omega t + kx + \pi)$. TRUE. (B) Transmitted into $S_2$ the wave number is $2k$, not $k$: the form would be $\cos(\omega t - 2kx + \dots)$. FALSE. (C) Reflection at $Q$ happens in $S_2$: it must travel in $-x$ with wave number $2k$, i.e. $\cos(\omega t + 2kx + \dots)$. The given $\cos(\omega t - kx + \pi)$ travels in $+x$ with the wrong $k$. FALSE. (D) Transmitted into $S_3$: travels in $+x$, wave number $4k$, no phase reversal: $y = \alpha_4y_0\cos(\omega t - 4kx)$ (up to a constant phase from the path). TRUE. Answer: (A), (D).
Q.8	Laws of MotionNewton's laws	Class 11: Laws of Motion	NumericalEasy
A easy numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $\omega = \frac{2\pi}{T} = \frac{2\pi}{40\pi} = \frac{1}{20}$ s$^{-1}$. $a = \frac{d^2y}{dt^2} = -8\omega^2\sin\omega t$, so $\|a\|_{max} = 8\cdot\frac{1}{400} = 0.02$ m/s$^2$. Apparent weight $= m(g + a)$ ranges from $m(g - 0.02)$ to $m(g + 0.02)$. Maximum variation $= m \times 2 \times 0.02 = 50 \times 0.04 = 2$ N. Answer: $2$.
Q.9	Electromagnetic WavesProperties of electromagnetic waves	Class 12: Electromagnetic Waves	NumericalMedium
A medium numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Energy in the unit volume: $u = Nh\nu = 35\times10^7 \times 6\times10^{-34}\times10^{15} = 2.1\times10^{-10}$ J/m$^3$. Average energy density of a wave with field amplitude $B_0$ (electric and magnetic parts together): $u = \frac{B_0^2}{2\mu_0}$. $B_0^2 = 2\mu_0u = 2\times4\pi\times10^{-7}\times2.1\times10^{-10} = 16.8\pi\times10^{-17} = 16.8\times\frac{22}{7}\times10^{-17} = 52.8\times10^{-17}$. $B_0 = \sqrt{5.28\times10^{-16}} \approx 2.30\times10^{-8}$ T $= 22.98\times10^{-9}$ T. $\alpha \approx 22.98$ (the key accepts $21$ to $25$).
Q.10	Properties of Bulk MatterHeat transfer	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	NumericalMedium
A medium numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Between two parallel black plates the net flux is $\sigma(T_1^4 - T_2^4)$. So $W_0 = \sigma(T_P^4 - T_Q^4)$. With two plates (temperatures $T_1, T_2$) in between, in the steady state the same flux $W_S$ crosses each of the three gaps: $W_S = \sigma(T_P^4 - T_1^4) = \sigma(T_1^4 - T_2^4) = \sigma(T_2^4 - T_Q^4)$. Adding the three: $3W_S = \sigma(T_P^4 - T_Q^4) = W_0$. $\frac{W_0}{W_S} = 3$.
Q.11	OpticsPolarisation	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalHard
A hard numerical question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Set the sphere's centre $C$ at the origin and O at $(0, -R)$. The cavity (radius $\frac{R}{2}$) touches the sphere at O, so its centre is $(0, -\frac{R}{2})$ and its top is at $C$. At O the light reflects off the thin glass layer while travelling in the air of the cavity. Full polarization means Brewster's angle: $\tan i_B = \sqrt{3}$, $i_B = 60^\circ$ (the same layer seen from the glass side gives $30^\circ$, consistent with this). Trace the ray backwards from O. Inside the cavity it runs along a chord making $60^\circ$ with the vertical at O. A chord of a circle makes equal angles with the radii at its two ends, so it meets the cavity wall at B at $60^\circ$ to the normal; the chord length is $2\cdot\frac{R}{2}\cos60^\circ = \frac{R}{2}$, giving $B = \left(\frac{\sqrt3}{4}R, -\frac{3}{4}R\right)$. At B the ray crosses glass to air. Snell: $\sqrt3\sin i = 1\cdot\sin60^\circ$, so in the glass the ray makes $30^\circ$ with the normal at B. Working out the direction, the normal at B points along $(\sin60^\circ, -\cos60^\circ)$ and the ray in the glass turns out to be horizontal, at height $y = -\frac{3}{4}R$. This horizontal ray came from its reflection at the sphere's surface at A, where $y_A = -\frac{3}{4}R$, so $\|x_A\| = R\sqrt{1 - \frac{9}{16}} = \frac{\sqrt7}{4}R$. The angle of incidence at A is the angle between the horizontal ray and the radius $CA$: $\cos\theta = \frac{\|x_A\|}{R} = \frac{\sqrt7}{4}$, so $\sin\theta = \frac{3}{4} = 0.75$. (The other accepted answer, $0.5$, is what one gets by ignoring the cavity and treating the reflection at O as glass to air: Brewster angle $30^\circ$ in glass, and a straight chord from A to O makes the same angle at both ends, so $\theta = 30^\circ$.) Answer: $0.75$ (the key accepts $0.5$ or $0.75$).
Q.12	Physical World and MeasurementError analysis	Class 11: Units and Measurement	NumericalMedium
A medium numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $b = \frac{m\lambda D}{d} = \frac{3\times600\times10^{-9}\times1}{5\times10^{-3}} = 3.6\times10^{-4}$ m $= 360\ \mu$m. Taking the least counts as the errors, $\Delta D = 1$ cm $= 0.01$ m and $\Delta d = 1$ mm. Relative error: $\frac{\Delta b}{b} = \frac{\Delta D}{D} + \frac{\Delta d}{d} = \frac{0.01}{1} + \frac{1}{5} = 0.21$. $\Delta b = 0.21 \times 360 = 75.6\ \mu$m. (The key also accepts $94$ to $95$: that is the largest possible deviation worked out directly, $b_{max} - b = \frac{3\times600\times10^{-9}\times1.01}{4\times10^{-3}} - 360\ \mu\text{m} = 454.5 - 360 = 94.5\ \mu$m.) Answer: $75.6$.
Q.13	Dual Nature of Matter and Radiationde Broglie wavelength	Class 12: Dual Nature of Radiation and Matter	NumericalMedium
A medium numerical question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Bohr orbit: $r_n = \frac{n^2a_0}{Z}$, and the orbit holds $n$ whole wavelengths: $2\pi r_n = n\lambda$. For $n = 3$: $\lambda = \frac{2\pi r_3}{3} = \frac{2\pi}{3}\cdot\frac{9a_0}{Z} = \frac{6\pi a_0}{Z}$. Neutron with kinetic energy $k_BT$: $p = \sqrt{2m_Nk_BT}$, $\lambda = \frac{h}{\sqrt{2m_Nk_BT}}$. Equal wavelengths: $\frac{h^2}{2m_Nk_BT} = \frac{36\pi^2a_0^2}{Z^2}$, so $T = \frac{Z^2h^2}{72\pi^2a_0^2m_Nk_B}$. $\alpha = 72$.
Q.14	ElectrostaticsElectric dipole	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Matching listMedium
A medium matching list question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Each dipole is at distance $r$ from $X$, along the $x$ axis. Let $E_0 = \frac{p}{4\pi\epsilon_0r^3}$. A dipole pointing along $\pm\hat{j}$ has $X$ on its equatorial line: field $-E_0$ times its own direction. A dipole pointing along $\pm\hat{i}$ has $X$ on its axis: field $2E_0$ along its own direction. (P) both $+\hat j$: $-E_0\hat j - E_0\hat j = -2E_0\hat j = -\frac{p}{2\pi\epsilon_0r^3}\hat j$ → (2). (Q) $+\hat j$ and $-\hat j$: $-E_0\hat j + E_0\hat j = 0$ → (1). (R) left $+\hat j$: $-E_0\hat j$; right $+\hat i$ (axial): $2E_0\hat i$. Total $E_0(2\hat i - \hat j)$ → (4). (S) both $+\hat i$, both axial: $4E_0\hat i = \frac{p}{\pi\epsilon_0r^3}\hat i$ → (5). P→2, Q→1, R→4, S→5. Answer: (C).
Q.15	Electromagnetic Induction and Alternating CurrentsAC circuits	Class 12: Electromagnetic Induction; Class 12: Alternating Current	Matching listMedium
A medium matching list question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\omega = 400$ rad/s, $V_0 = 300$ V. For each load find $\|Z\|$, $i_0 = V_0/\|Z\|$ and the phase. (P) $R = 30\ \Omega$: $i = 10\sin(400t)$, in phase, peak $10$ at $400t = \frac{\pi}{2}$ → graph (3). (Q) $X_L = 400\times0.1 = 40\ \Omega$, $\|Z\| = \sqrt{30^2 + 40^2} = 50\ \Omega$, $i_0 = 6$ A, current lags by $\tan^{-1}\frac{4}{3} \approx 53^\circ$: $i = 6\sin(400t - 53^\circ)$, so $i(0) = -4.8$ and the peak comes after $\frac{\pi}{2}$ → graph (5). (R) $X_C = \frac{1}{400\times50\times10^{-6}} = 50\ \Omega$, $X_L = 400\times0.025 = 10\ \Omega$; net reactance $40\ \Omega$ capacitive. $\|Z\| = 50\ \Omega$, $i_0 = 6$ A, current leads by $53^\circ$: $i(0) = +4.8$, peak before $\frac{\pi}{2}$ → graph (2). (S) $X_C = 50\ \Omega$, $X_L = 400\times0.125 = 50\ \Omega$: resonance. $\|Z\| = R = 60\ \Omega$, $i = 5\sin(400t)$, in phase → graph (1). P→3, Q→5, R→2, S→1. Answer: (A).
Q.16	Atoms and NucleiNuclear binding energy	Class 12: Atoms; Class 12: Nuclei	Matching listMedium
A medium matching list question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (P) Hydrogen-like atoms: $E_n = -13.6\frac{Z^2}{n^2}$ eV, so transition energies go as $Z^2$ → (5). (Q) Moseley's law for characteristic ($K_\alpha$) x-rays: one electron screens the nucleus, $E \propto (Z - 1)^2$ → (1). (R) The Coulomb energy of $Z$ protons counts $\frac{Z(Z-1)}{2}$ proton pairs: $E_C \propto \frac{Z(Z-1)}{R} \propto \frac{Z(Z-1)}{A^{1/3}}$ → (2). (S) Binding energy per nucleon is nearly constant, about $8$ to $8.8$ MeV, for $A$ from about $30$ to $170$ → (4). (The continuous x-ray spectrum's maximum energy is set by the accelerating voltage, but (3) is not needed in the correct option.) P→5, Q→1, R→2, S→4. Answer: (C).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Single correctEasy
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Ammonium nitrite decomposes on gentle heating to nitrogen: $\mathrm{NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O}$. So X $= \mathrm{N_2}$. Ammonium nitrate at $200-250$ °C gives nitrous oxide: $\mathrm{NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O}$. So Y $= \mathrm{N_2O}$. In both, the $-3$ nitrogen of $\mathrm{NH_4^+}$ and the nitrogen of the anion meet in the middle (comproportionation). Answer: (A).
Q.2	Coordination CompoundsCrystal field theory	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All four are $\mathrm{Co^{3+}}$ ($d^6$) octahedral complexes, so the absorption band corresponds to the splitting $\Delta_o$, and $\lambda_{max} \propto \frac{1}{\Delta_o}$. Spectrochemical series: $\mathrm{CN^-} > \mathrm{NH_3} > \mathrm{H_2O} > \mathrm{Cl^-}$. Replacing one $\mathrm{NH_3}$ by a weaker ligand lowers $\Delta_o$. $\Delta_o$: $[\mathrm{Co(CN)_6}]^{3-} > [\mathrm{Co(NH_3)_6}]^{3+} > [\mathrm{Co(NH_3)_5(H_2O)}]^{3+} > [\mathrm{Co(NH_3)_5Cl}]^{2+}$. So $\lambda_{max}$ increases in exactly this order: $[\mathrm{Co(CN)_6}]^{3-} < [\mathrm{Co(NH_3)_6}]^{3+} < [\mathrm{Co(NH_3)_5(H_2O)}]^{3+} < [\mathrm{Co(NH_3)_5(Cl)}]^{2+}$. Answer: (A).
Q.3	d- and f-Block ElementsPotassium dichromate and permanganate	Class 12: The d- and f-Block Elements	Single correctEasy
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In neutral (or faintly alkaline) solution permanganate is reduced to $\mathrm{MnO_2}$ (a 3-electron change) and it oxidises iodide all the way to iodate (a 6-electron change): $\mathrm{2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-}$. (In acid, by contrast, iodide is oxidised only to $\mathrm{I_2}$.) Answer: (B) $\mathrm{IO_3^-}$.
Q.4	Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctHard
A hard single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Acidity is decided by how stable the carbanion left behind is. The strongest effect available here is aromaticity: a cyclopentadienyl anion ($6\pi$ electrons in a ring of five $sp^2$ carbons) is very stable. (B) The H is on a methyl group attached to the exocyclic carbon of a fulvene (ring with two C=C plus an exocyclic C=C). Removing it gives $\mathrm{^-CH_2{-}C(CH_3){=}C_{ring}}$, which in another resonance form is $\mathrm{CH_2{=}C(CH_3){-}}$ attached to a ring carrying the negative charge with its two C=C: an aromatic cyclopentadienide. Very stabilised, so this H is the most acidic. (A) and (C) The H is on an isopropyl group attached to a ring carbon of a methylene-fulvene. The anion is delocalised over the ring and the exocyclic $\mathrm{=CH_2}$, but in every resonance form one ring carbon stays double-bonded to an exocyclic carbon, so the ring never becomes a $6\pi$ aromatic anion. Only polyene (allyl-type) delocalisation. (D) The ring has an $sp^3$ $\mathrm{CH_2}$, so it cannot become a cyclopentadienyl anion at all; the anion is just a conjugated allyl/dienyl anion. Answer: (B).
Q.5	Chemical Bonding and Molecular StructureMolecular orbital theory	Class 11: Chemical Bonding and Molecular Structure	Multiple correctEasy
A easy multiple correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) $\mathrm{Ne_2}$ has 16 valence electrons: all bonding and antibonding $2s$ and $2p$ orbitals are filled, bond order $= \frac{8 - 8}{2} = 0$. Correct statement. (B) $\mathrm{F_2}$: $\dots\sigma_{2p}^2\,\pi_{2p}^4\,\pi^_{2p}{}^4$. The HOMO is $\pi^_{2p}$, a $\pi$-type orbital, not $\sigma$. INCORRECT statement. (C) Removing an electron from $\mathrm{O_2}$ takes it from an antibonding $\pi^*$ orbital: bond order rises from $2$ to $2.5$, so $\mathrm{O_2^+}$ has the LARGER bond energy. INCORRECT statement. (D) $\mathrm{Li_2}$ (bond order 1, large diffuse $2s$ orbitals, about $267$ pm) is longer than $\mathrm{B_2}$ (bond order 1, about $159$ pm). Correct statement. The incorrect statements are (B) and (C).
Q.6	d- and f-Block ElementsLanthanoids and actinoids	Class 12: The d- and f-Block Elements	Multiple correctMedium
A medium multiple correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements An ion is diamagnetic when it has no unpaired electrons: here, an empty or full $4f$ shell and no $5d$ electron. La $= [\mathrm{Xe}]5d^16s^2$, Ce $= [\mathrm{Xe}]4f^15d^16s^2$, Yb $= [\mathrm{Xe}]4f^{14}6s^2$, Lu $= [\mathrm{Xe}]4f^{14}5d^16s^2$. (A) $\mathrm{La^{3+}}$: $[\mathrm{Xe}]$; $\mathrm{Ce^{4+}}$: $[\mathrm{Xe}]$. Both diamagnetic. TRUE. (B) $\mathrm{Yb^{2+}}$: $[\mathrm{Xe}]4f^{14}$; $\mathrm{Lu^{3+}}$: $[\mathrm{Xe}]4f^{14}$. Both diamagnetic. TRUE. (C) $\mathrm{La^{2+}}$: $5d^1$ (one unpaired); $\mathrm{Ce^{3+}}$: $4f^1$. Paramagnetic. FALSE. (D) $\mathrm{Yb^{3+}}$: $4f^{13}$; $\mathrm{Lu^{2+}}$: $4f^{14}5d^1$. Paramagnetic. FALSE. Answer: (A), (B).
Q.7	Haloalkanes and HaloarenesNature of the C–X bond	Class 12: Haloalkanes and Haloarenes	Multiple correctMedium
A medium multiple correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes P = bromocyclohexane (HBr adds to cyclohexene). Finkelstein (NaI in acetone) swaps Br for I: Q = iodocyclohexane. Swarts (AgF, $\mathrm{Hg_2F_2}$ or $\mathrm{SbF_3}$) swaps Br for F: R = fluorocyclohexane. So X is Br in P, I in Q, F in R. (A) C–X bond length grows with the size of X: C–I > C–Br > C–F, i.e. Q > P > R. The option says Q > R > P. FALSE. (B) Bond enthalpy: C–F > C–Br > C–I, i.e. R > P > Q. TRUE. (C) $\mathrm{S_N2}$ reactivity follows leaving-group ability, I > Br > F: Q > P > R. The option says P > R > Q. FALSE. (D) Conjugate acids of the leaving groups: HF ($pK_a \approx 3.2$), HBr ($\approx -9$), HI ($\approx -10$). Order R > P > Q; the option says R > Q > P. FALSE. Answer: (B).
Q.8	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	NumericalEasy
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$: Cr goes from $+6$ to $+3$, $3$ electrons per Cr. $1$ mol $\mathrm{Cr^{3+}}$ needs $3$ F $= 3\times96500 = 289500$ C. Time $= 48.25\times60 = 2895$ s. $I = \frac{289500}{2895} = 100$ A.
Q.9	EquilibriumIonic equilibrium	Class 11: Equilibrium	NumericalMedium
A medium numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium The acid is so weak that $K_aC = 4\times10^{-14}$ is comparable with $K_w$, so water's own ionisation must be counted. Charge balance: $[\mathrm{H^+}] = [\mathrm{A^-}] + [\mathrm{OH^-}]$. The acid is hardly dissociated, so $[\mathrm{HA}] \approx C$ and $[\mathrm{A^-}] = \frac{K_aC}{[\mathrm{H^+}]}$; $[\mathrm{OH^-}] = \frac{K_w}{[\mathrm{H^+}]}$. $[\mathrm{H^+}]^2 = K_aC + K_w = 4\times10^{-14} + 1\times10^{-14} = 5\times10^{-14}$. $[\mathrm{H^+}] = \sqrt{5}\times10^{-7} = 2.236\times10^{-7}$ M. $X \approx 2.24$.
Q.10	States of MatterGas laws	No matching NCERT chapter	NumericalMedium
A medium numerical question on gas laws, from States of Matter. NCERT: No matching NCERT chapter $\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT$. Multiply by $V_m^2$ and expand: $PV_m^3 - (Pb + RT)V_m^2 + aV_m - ab = 0$. Coefficient of $V_m^2$: $-(Pb + RT) = -(300\times0.060 + 0.082\times300) = -(18 + 24.6) = -42.6$. Coefficient of $V_m$: $a = 6.0$. Ratio $= \frac{-42.6}{6.0} = -7.1$ mol dm$^{-3}$. (Dividing the whole cubic by $P$ first leaves the ratio unchanged.)
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 $144$ g of water $= \frac{144}{18} = 8$ mol. $\mathrm{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)}$: $8$ mol water gives $8$ mol $\mathrm{H_2}$ and $4$ mol $\mathrm{O_2}$, $12$ mol of gas. The liquid's own volume is negligible. Work done by the system at constant pressure: $P\Delta V = \Delta n_gRT = 12\times8.3\times300 = 29880$ J $= 29.88$ kJ. With the sign convention $w = -P\Delta V$ (work done on the system), $w = -29.88$ kJ. The key accepts either sign. Answer: $-29.88$ (or $29.88$).
Q.12	PolymersNylon, Teflon, Bakelite and PVC	No matching NCERT chapter	NumericalEasy
A easy numerical question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Nylon 6,6 is made from hexamethylenediamine, $\mathrm{H_2N(CH_2)_6NH_2}$, and adipic acid. Only the diamine (a primary amine) gives the carbylamine test, so X is hexamethylenediamine, with $2$ N atoms per molecule. Dumas method converts all the nitrogen into $\mathrm{N_2}$: $10$ mol X $\to 20$ mol N atoms $\to 10$ mol $\mathrm{N_2}$. Mass $= 10\times28 = 280$ g.
Q.13	Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro	Class 12: Aldehydes, Ketones and Carboxylic Acids	NumericalHard
A hard numerical question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids X is 2-(2-bromophenyl)-1,3-dioxolane: an aryl bromide whose aldehyde is protected as an acetal. Step 1, Na in dry ether (Wurtz–Fittig type coupling of two aryl bromides) and then $\mathrm{H_3O^+}$ (removes the acetal): P = biphenyl-2,2'-dicarbaldehyde, $\mathrm{C_{14}H_{10}O_2}$, $M = 168 + 10 + 32 = 210$, which matches the given molar mass. Two X make one P: $16$ mol X $\to 8$ mol P (100 %). Step 2, NaOH, heat, then $\mathrm{H_3O^+}$: intramolecular Cannizzaro, one CHO oxidised to COOH and the other reduced to $\mathrm{CH_2OH}$. Q = 2'-(hydroxymethyl)biphenyl-2-carboxylic acid; 50 %: $4$ mol. Step 3, NaOH/CaO, heat: decarboxylation of the acid (as its sodium salt). R = biphenyl-2-ylmethanol, $\mathrm{C_6H_5{-}C_6H_4{-}CH_2OH}$; 50 %: $2$ mol. Step 4, $\mathrm{PBr_3}$: T = 2-(bromomethyl)biphenyl; 50 %: $1$ mol. Step 5, T with R and NaH (Williamson ether synthesis: the alkoxide of R displaces Br from T): S = bis(biphenyl-2-ylmethyl) ether, $\mathrm{(C_6H_5C_6H_4CH_2)_2O}$ = $\mathrm{C_{26}H_{22}O}$; 50 % of $1$ mol: $0.5$ mol. $M(\mathrm{S}) = 26\times12 + 22\times1 + 16 = 312 + 22 + 16 = 350$. Mass of S $= 0.5\times350 = 175$ g.
Q.14	Principles of Qualitative AnalysisGroup analysis of cations	No matching NCERT chapter	Matching listEasy
A easy matching list question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter (P) $\mathrm{H_2S}$ in ammoniacal solution is the group IV reagent: it precipitates $\mathrm{MnS}$ (also ZnS, CoS, NiS) → (3). (Q) $\mathrm{(NH_4)_2CO_3}$ with $\mathrm{NH_4OH}$ is the group V reagent: $\mathrm{BaCO_3}$ → (4). (R) $\mathrm{NH_4OH}$ with $\mathrm{NH_4Cl}$ is the group III reagent: $\mathrm{Al(OH)_3}$ → (2). ($\mathrm{NH_4Cl}$ keeps $[\mathrm{OH^-}]$ low enough that $\mathrm{Mg(OH)_2}$ does not precipitate.) (S) $\mathrm{H_2S}$ in dilute HCl is the group II reagent: $\mathrm{CuS}$ → (1). $\mathrm{Mg^{2+}}$ (5) is in group VI and is not used. P→3, Q→4, R→2, S→1. Answer: (A).
Q.15	Organic Chemistry: Some Basic Principles and TechniquesTypes of organic reactions	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Matching listMedium
A medium matching list question on types of organic reactions, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Work out each product in List-II, then see which named reaction it starts. (1) Etard reaction: toluene with $\mathrm{CrO_2Cl_2}$, then hydrolysis, gives benzaldehyde. An aldehyde with no $\alpha$-H is the substrate for the Cannizzaro reaction → S. (2) Benzoic acid → benzoyl chloride → benzamide → (dehydration with $\mathrm{P_4O_{10}}$) benzonitrile. A nitrile is reduced to an aldehyde in the Stephen reaction → P. (3) Nitrobenzene → aniline (Fe/HCl) → benzenediazonium chloride ($\mathrm{NaNO_2}$/HCl at 273–278 K). A diazonium salt is the substrate for the Sandmeyer reaction → Q. (4) Toluene → benzal chloride ($\mathrm{Cl_2}/h\nu$) → benzaldehyde (water) → benzoic acid (Tollens') → benzoyl chloride ($\mathrm{SO_2Cl_2}$) → benzamide ($\mathrm{NH_3}$). A primary amide undergoes Hoffmann bromamide degradation → R. (5) Aniline → acetanilide → p-nitroacetanilide → p-nitroaniline; not needed for any entry. P→2, Q→3, R→4, S→1. Answer: (B).
Q.16	BiomoleculesProteins	Class 12: Biomolecules	Matching listMedium
A medium matching list question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules (P) A dipeptide methyl ester of 4-hydroxyphenylglycine and glycine. It has a free $\alpha$-$\mathrm{NH_2}$ (ninhydrin: purple) and a phenolic OH ($\mathrm{FeCl_3}$: violet) → (2). (Q) An N-acetyl dipeptide methyl ester (phenylglycine and valine). The amino group is acetylated, so it needs hydrolysis before ninhydrin reacts; after complete hydrolysis it gives free amino acids, which do. It has no phenol, so the phthalein dye test (which needs a phenol) is negative → (5). (R) Anilinium chloride. With benzenediazonium salt, aniline couples to give yellow diazoamino/aminoazobenzene dyes → (1). (S) 2,4,6-Trimethylphenylhydrazine. A hydrazine condenses with the carbonyl group of glucose to give the hydrazone → (3). (4) needs both N and S in the compound (thiocyanate gives the blood-red colour with $\mathrm{Fe^{3+}}$); none of the four contains sulphur. P→2, Q→5, R→1, S→3. Answer: (B).

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	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 Group the numerator: $3xe^x + 3x - \alpha e^x - \alpha x = 3x(e^x + 1) - \alpha(e^x + x)$, so $g(x) = x - \frac{\alpha}{3}\cdot\frac{e^x + x}{e^x + 1}$. At $x_0$, $e^{x_0} + x_0 = 0$, so $g(x_0) = x_0$ and $g(x_0) + e^{x_0} = x_0 + e^{x_0} = 0$. Hence $\lim_{x\to x_0}\frac{g(x) + e^{x_0}}{x - x_0} = \lim_{x\to x_0}\frac{g(x) - g(x_0)}{x - x_0} = g'(x_0)$. Let $h(x) = \frac{e^x + x}{e^x + 1}$. Then $h'(x) = \frac{(e^x + 1)(e^x + 1) - (e^x + x)e^x}{(e^x + 1)^2}$, and at $x_0$ the second term vanishes because $e^{x_0} + x_0 = 0$: $h'(x_0) = 1$. So $g'(x_0) = 1 - \frac{\alpha}{3}$. For $\alpha = 2$: $\left\|g'(x_0)\right\| = \frac{1}{3}$, so (A) and (B) are false. For $\alpha = 3$: $g'(x_0) = 0$, so (C) is true and (D) is false. Answer: (C).
Q.2	Integral CalculusArea under simple curves	Class 12: Integrals; Class 12: Application of Integrals	Single correctMedium
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The region lies above the hyperbola $y = \frac{1}{x}$ and below both lines $y = \frac{5x - 1}{4}$ and $y = \frac{17 - 4x}{4}$. Corners: the two lines meet where $5x - 1 = 17 - 4x$, i.e. $x = 2$, $y = \frac{9}{4}$; the first line meets the hyperbola where $\frac{1}{x} = \frac{5x - 1}{4}$, i.e. $5x^2 - x - 4 = 0$, so $x = 1$; the second line meets it where $\frac{1}{x} = \frac{17 - 4x}{4}$, i.e. $4x^2 - 17x + 4 = 0$, so $x = 4$ (the root $x = \frac{1}{4}$ lies outside the region, where the first line is below the hyperbola). So for $1 \le x \le 2$ the top is $\frac{5x - 1}{4}$, for $2 \le x \le 4$ it is $\frac{17 - 4x}{4}$, and the bottom is $\frac{1}{x}$ throughout. $\int_1^2 \frac{5x - 1}{4}dx = \frac{1}{4}\left[\frac{5x^2}{2} - x\right]_1^2 = \frac{1}{4}\left(8 - \frac{3}{2}\right) = \frac{13}{8}$. $\int_2^4 \frac{17 - 4x}{4}dx = \frac{1}{4}\left[17x - 2x^2\right]_2^4 = \frac{1}{4}(36 - 26) = \frac{5}{2} = \frac{20}{8}$. $\int_1^4 \frac{dx}{x} = \log_e 4$. Area $= \frac{13}{8} + \frac{20}{8} - \log_e 4 = \frac{33}{8} - \log_e 4$. Answer: (B).
Q.3	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctHard
A hard single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Write $t = \tan\theta$. Then $\frac{6t}{9 + t^2} = \frac{2(t/3)}{1 + (t/3)^2}$, so $\sin^{-1}\frac{6t}{9 + t^2} = 2\tan^{-1}\frac{t}{3}$ when $\|t\| \le 3$, $\pi - 2\tan^{-1}\frac{t}{3}$ when $t > 3$, and $-\pi - 2\tan^{-1}\frac{t}{3}$ when $t < -3$. Case $\|t\| \le 3$: the right side is $\tan^{-1}(2t) - \tan^{-1}\frac{t}{3}$, which lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. So $\theta$ must lie there too, and then $\theta = \tan^{-1}t$. The equation becomes $\tan^{-1}t + \tan^{-1}\frac{t}{3} = \tan^{-1}2t$. Taking tangents: $\frac{t + t/3}{1 - t^2/3} = 2t$, so $t = 0$ or $\frac{4}{3} = 2\left(1 - \frac{t^2}{3}\right)$, i.e. $t^2 = 1$. Check $t = 1$: $\tan^{-1}1 + \tan^{-1}\frac{1}{3} = \tan^{-1}\frac{4/3}{2/3} = \tan^{-1}2$, true (product $\frac{1}{3} < 1$, no extra $\pi$). By oddness $t = -1$ also works, and $t = 0$ works. So $\theta = 0, \frac{\pi}{4}, -\frac{\pi}{4}$. Case $t > 3$: the right side is $\tan^{-1}2t + \tan^{-1}\frac{t}{3} - \frac{\pi}{2} = \tan^{-1}2t - \tan^{-1}\frac{3}{t}$, which is below $\frac{\pi}{2}$; with $\theta = \tan^{-1}t$ the equation needs $\tan^{-1}t + \tan^{-1}\frac{3}{t} = \tan^{-1}2t$. But the two angles on the left have product of tangents $3 > 1$, so their sum exceeds $\frac{\pi}{2}$, while the right side is less than $\frac{\pi}{2}$. No solution. The case $t < -3$ is the mirror image: no solution. Total: 3 real solutions. Answer: (C).
Q.4	Conic SectionsHyperbola	Class 11: Conic Sections	Single correctMedium
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The second line is $4x + 3y = \frac{12}{\alpha}$. Multiplying the two equations eliminates $\alpha$: $(4x - 3y)(4x + 3y) = 144$, i.e. $16x^2 - 9y^2 = 144$, so $S$ is the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$. The tangent is parallel to $4x - \frac{3}{\sqrt 2}y = 0$, so its slope is $m = \frac{4\sqrt2}{3}$. Tangents of slope $m$ to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $y = mx \pm\sqrt{a^2m^2 - b^2} = mx \pm\sqrt{9\cdot\frac{32}{9} - 16} = mx \pm 4$. It meets the $y$-axis at $q > 0$, so $q = 4$ and $T: y = \frac{4\sqrt2}{3}x + 4$. At $y = 0$: $p = -\frac{3}{\sqrt2}$. $pq = -\frac{12}{\sqrt2} = -6\sqrt2$. Answer: (A).
Q.5	Matrices and DeterminantsAlgebra of matrices	Class 12: Matrices; Class 12: Determinants	Multiple correctMedium
A medium multiple correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let $R = \begin{pmatrix}a&b\\c&d\end{pmatrix}$. Since $P$ is diagonal, $RP = \begin{pmatrix}2a&3b\\2c&3d\end{pmatrix}$, so $QR = RP$ says $Q\begin{pmatrix}a\\c\end{pmatrix} = 2\begin{pmatrix}a\\c\end{pmatrix}$ and $Q\begin{pmatrix}b\\d\end{pmatrix} = 3\begin{pmatrix}b\\d\end{pmatrix}$. Both columns are non-zero, so $\det(Q - 2I) = 0$ and $\det(Q - 3I) = 0$: the characteristic polynomial of $Q$ is $(\lambda - 2)(\lambda - 3)$. Hence trace $x + 4 = 5$, so $x = 1$, and determinant $4x - yz = 6$, so $yz = -2$. (Such $R$ exists: e.g. from $(x - 2)a + yc = 0$, $a = yc \ne 0$.) (A) $\det(Q - 2I) = 0$: TRUE. (B) $\det(Q - 6I) = (2 - 6)(3 - 6) = 12$: TRUE. (C) $\det(Q - 3I) = 0$, not 15: FALSE. (D) $yz = -2$, not 2: FALSE. Answer: (A), (B).
Q.6	Conic SectionsParabola	Class 11: Conic Sections	Multiple correctHard
A hard multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Take the chord's ends as $(t_1^2, t_1)$ and $(t_2^2, t_2)$. Integrating along $y$, the area between the chord and $x = y^2$ is $\int_{t_1}^{t_2}(\text{line} - y^2)\,dy = \frac{\|t_2 - t_1\|^3}{6}$. Setting it to $\frac{4}{3}$: $\|t_2 - t_1\|^3 = 8$, so $\|t_2 - t_1\| = 2$. Mid-point: $k = \frac{t_1 + t_2}{2}$, $h = \frac{t_1^2 + t_2^2}{2} = \frac{(t_1 + t_2)^2 + (t_1 - t_2)^2}{4} = \frac{4k^2 + 4}{4} = k^2 + 1$. So $S$ is the parabola $x = y^2 + 1$. (A) $(4, \sqrt3)$: $3 + 1 = 4$. TRUE. (B) $(5, \sqrt2)$: $2 + 1 = 3 \ne 5$. FALSE. In the first quadrant the parabola $y^2 = x$ is $y = \sqrt x$ and $S$ is $y = \sqrt{x - 1}$, below it. So Area of $\mathcal R = \int_1^4\left(\sqrt x - \sqrt{x - 1}\right)dx = \frac{2}{3}\left(8 - 1\right) - \frac{2}{3}\cdot3^{3/2} = \frac{14}{3} - 2\sqrt3$. (C) TRUE, (D) FALSE. Answer: (A), (C).
Q.7	Conic SectionsEllipse	Class 11: Conic Sections	Multiple correctMedium
A medium multiple correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $C$ is the auxiliary circle of the ellipse, so $R$ and $S$ give the eccentric angles of $P$ and $Q$. $R = (3\cos\frac{\pi}{6}, 3\sin\frac{\pi}{6}) = \left(\frac{3\sqrt3}{2}, \frac{3}{2}\right)$ and $P = \left(\frac{3\sqrt3}{2}, 2\sin\frac{\pi}{6}\right) = \left(\frac{3\sqrt3}{2}, 1\right)$. $S = \left(\frac{3}{2}, \frac{3\sqrt3}{2}\right)$ and $Q = \left(\frac{3}{2}, \sqrt3\right)$. Slope of $PQ = \frac{\sqrt3 - 1}{\frac{3}{2} - \frac{3\sqrt3}{2}} = \frac{\sqrt3 - 1}{-\frac{3}{2}(\sqrt3 - 1)} = -\frac{2}{3}$, so the line is $2x + 3y = c$ with $c = 2\cdot\frac{3\sqrt3}{2} + 3\cdot1 = 3(1 + \sqrt3)$. (A) TRUE; (B) FALSE (slope $-2$ is wrong). (C) $\|N_2Q\| = \sqrt3$, $\|N_2S\| = \frac{3\sqrt3}{2}$: $3\sqrt3 = 2\cdot\frac{3\sqrt3}{2}$. TRUE. (D) $\|N_1P\| = 1$, $\|N_1R\| = \frac{3}{2}$: $9 \ne 4\cdot\frac{3}{2} = 6$. FALSE (in general the ratio is $b : a$, so $3\|N_1P\| = 2\|N_1R\|$). Answer: (A), (C).
Q.8	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	Multiple correctHard
A hard multiple correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Near $0$, with $u = \frac{\sin x}{x}$, $f(x) = \frac{6 + u}{2 + u} = 1 + \frac{4}{2 + u}$, and $u \to 1$, so $f$ is continuous with $f(0) = \frac{7}{3}$. For small $x \ne 0$, $u < 1$, so $\frac{4}{2 + u} > \frac{4}{3}$ and $f(x) > \frac{7}{3}$: $x = 0$ is a local minimum. (B) TRUE, (A) FALSE. For $x \ne 0$: $f'(x) = \frac{(6 + \cos x)(2x + \sin x) - (6x + \sin x)(2 + \cos x)}{(2x + \sin x)^2} = \frac{4(\sin x - x\cos x)}{(2x + \sin x)^2}$. For $x > 0$ the denominator is positive, so the sign of $f'$ is the sign of $h(x) = \sin x - x\cos x$, with $h'(x) = x\sin x$. The critical points are the roots of $\tan x = x$: one in each interval $\left(n\pi, n\pi + \frac{\pi}{2}\right)$, $n \ge 1$ (about $4.49, 7.73, 10.90, 14.07, 17.22, \ldots$). In $\left(n\pi, n\pi + \frac{\pi}{2}\right)$: for odd $n$, $\sin x < 0$, so $h$ is decreasing and changes from $+$ to $-$: a local maximum of $f$. For even $n$, $\sin x > 0$, $h$ changes from $-$ to $+$: a local minimum. (C) In $[\pi, 6\pi]$ the roots lie in $n = 1, \ldots, 5$ (the next is beyond $6\pi$). Maxima for $n = 1, 3, 5$: 3. At the end-points $f' \ne 0$, so they add nothing. TRUE. (D) In $[2\pi, 4\pi]$ the roots are for $n = 2$ (minimum, $\approx 7.73$) and $n = 3$ (maximum, $\approx 10.90$); the $n = 4$ root $\approx 14.07$ is past $4\pi \approx 12.57$. So 1 local minimum. TRUE. Answer: (B), (C), (D).
Q.9	Differential EquationsHomogeneous equations	Class 12: Differential Equations	NumericalMedium
A medium numerical question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Divide by $x^2$: $\frac{dy}{dx} = 1 + \left(\frac{y}{x}\right)^2 - \frac{y}{x}$, a homogeneous equation. Put $y = vx$: $v + x\frac{dv}{dx} = 1 + v^2 - v$, so $x\frac{dv}{dx} = (v - 1)^2$. $\int\frac{dv}{(v - 1)^2} = \int\frac{dx}{x}$ gives $\frac{1}{1 - v} = \ln x + C$. At $x = 1$, $v = 0$: $C = 1$. So $1 - v = \frac{1}{1 + \ln x}$, $v = \frac{\ln x}{1 + \ln x}$ and $y = \frac{x\ln x}{1 + \ln x}$ (defined for $x > \frac{1}{e}$). $y(e) = \frac{e}{2}$, $y(e^2) = \frac{2e^2}{3}$. $2\cdot\frac{e^2/4}{2e^2/3} = 2\cdot\frac{3}{8} = \frac{3}{4} = 0.75$. Answer: 0.75.
Q.10	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 $a_i = \binom{23}{i}\left(\frac{2}{5}\right)^i$, all positive. $\frac{a_{i+1}}{a_i} = \frac{23 - i}{i + 1}\cdot\frac{2}{5}$. This exceeds 1 when $2(23 - i) > 5(i + 1)$, i.e. $41 > 7i$, i.e. $i \le 5$. So $a_0 < a_1 < \cdots < a_6$ and $a_6 > a_7 > \cdots$ (at $i = 6$ the ratio is $\frac{17}{7}\cdot\frac{2}{5} = \frac{34}{35} < 1$). The largest is $a_6$, so $r = 6$. Answer: 6.
Q.11	ProbabilityBayes' theorem	Class 11: Probability; Class 12: Probability	NumericalEasy
A easy numerical question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(M_1) = \frac{2}{5}$, $P(M_2) = \frac{2}{5}$, $P(M_3) = \frac{1}{5}$; $P(D) = 0.2$; $P(D\mid M_1) = 0.15$. By Bayes: $P(M_2\mid D) = \frac{P(D\mid M_2)P(M_2)}{P(D)} = \frac{2}{5}$, so $P(D\mid M_2)\cdot0.4 = 0.08$ and $P(D\mid M_2) = 0.2$. Total probability: $0.2 = 0.4(0.15) + 0.4(0.2) + 0.2\,P(D\mid M_3) = 0.06 + 0.08 + 0.2\,P(D\mid M_3)$. So $P(D\mid M_3) = \frac{0.06}{0.2} = 0.3$. Answer: 0.3.
Q.12	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	NumericalMedium
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra First, $\vec x, \vec y, \vec z$ are not coplanar: $[\vec x\ \vec y\ \vec z] = \begin{vmatrix}1&2&3\\2&3&1\\3&1&2\end{vmatrix} = 5 - 2 - 21 = -18 \ne 0$. So $[\vec X\ \vec Y\ \vec Z] = D\cdot[\vec x\ \vec y\ \vec z]$, where $D$ is the determinant of the coefficients in the basis $\vec x, \vec y, \vec z$: $D = \begin{vmatrix}\alpha&\beta&-1\\-1&\alpha&\beta\\\beta&-1&\alpha\end{vmatrix}$, a circulant with entries $a = \alpha$, $b = \beta$, $c = -1$. $D = a^3 + b^3 + c^3 - 3abc = (a + b + c)\cdot\frac{1}{2}\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]$. Coplanar means $D = 0$. The bracket vanishes only if $a = b = c$, i.e. $\alpha = \beta = -1$, impossible for distinct positive numbers. So $a + b + c = 0$: $\alpha + \beta = 1$. $\alpha + \beta - 3 = -2$. Answer: $-2$.
Q.13	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 $\omega = e^{2\pi i/3}$, so $-\omega = e^{i(2\pi/3 - \pi)} = e^{-i\pi/3}$, a primitive sixth root of unity. Every block of six consecutive powers sums to zero. $2025 = 6\cdot337 + 3$, so the sum equals $z + z^2 + z^3$ with $z = e^{-i\pi/3}$. $z = \frac{1}{2} - \frac{\sqrt3}{2}i$, $z^2 = -\frac{1}{2} - \frac{\sqrt3}{2}i$, $z^3 = -1$. Sum $= -1 - \sqrt3\,i$, in the third quadrant, with principal argument $-\frac{2\pi}{3}$. $\frac{3\alpha}{\pi} = -2$. Answer: $-2$.
Q.14	Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	NumericalEasy
A easy numerical question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $g(x) = 2$ gives $1 + e^{-2x} = 2$, so $g^{-1}(2) = 0$. $(f\circ g^{-1})'(2) = \frac{f'(g^{-1}(2))}{g'(g^{-1}(2))} = \frac{f'(0)}{g'(0)}$. $f'(x) = \frac{2x + 2}{x^2 + 2x + 4}$, so $f'(0) = \frac{2}{4} = \frac{1}{2}$. $g'(x) = \frac{8e^{-2x}}{(1 + e^{-2x})^2}$, so $g'(0) = \frac{8}{4} = 2$. Derivative $= \frac{1/2}{2} = \frac{1}{4} = 0.25$. Answer: 0.25.
Q.15	TrigonometryTrigonometric identities	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	NumericalMedium
A medium numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\frac{1}{\sin a\,\sin(a + 1^\circ)} = \frac{\sin\left((a + 1^\circ) - a\right)}{\sin1^\circ\,\sin a\,\sin(a + 1^\circ)} = \frac{\cot a - \cot(a + 1^\circ)}{\sin1^\circ}$. So $\alpha\sin1^\circ = (\cot60^\circ + \cot62^\circ + \cdots + \cot118^\circ) - (\cot61^\circ + \cot63^\circ + \cdots + \cot119^\circ)$. Use $\cot(180^\circ - \theta) = -\cot\theta$. Even angles $60^\circ, 62^\circ, \ldots, 118^\circ$: $62^\circ$ cancels $118^\circ$, \ldots, $88^\circ$ cancels $92^\circ$, and $\cot90^\circ = 0$; only $\cot60^\circ = \frac{1}{\sqrt3}$ is left. Odd angles $61^\circ, \ldots, 119^\circ$: $61^\circ$ cancels $119^\circ$, \ldots, $89^\circ$ cancels $91^\circ$; the sum is 0. So $\alpha = \frac{1}{\sqrt3\sin1^\circ}$ and $\frac{\operatorname{cosec}1^\circ}{\alpha} = \sqrt3$. Its square is 3. Answer: 3.
Q.16	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	NumericalHard
A hard numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Substitute $x = \frac{1}{t}$: $dx = -\frac{dt}{t^2}$, $2x^2 - 3x + 2 = \frac{2 - 3t + 2t^2}{t^2}$, and the limits $\frac{1}{2} \to 2$ become $2 \to \frac{1}{2}$. So $\alpha = \int_{1/2}^{2}\frac{\tan^{-1}\frac{1}{t}}{2t^2 - 3t + 2}dt$. Adding the two forms, and using $\tan^{-1}x + \tan^{-1}\frac{1}{x} = \frac{\pi}{2}$ for $x > 0$: $2\alpha = \frac{\pi}{2}\int_{1/2}^{2}\frac{dx}{2x^2 - 3x + 2}$. $2x^2 - 3x + 2 = 2\left[\left(x - \frac{3}{4}\right)^2 + \frac{7}{16}\right]$, so $\int\frac{dx}{2x^2 - 3x + 2} = \frac{2}{\sqrt7}\tan^{-1}\frac{4x - 3}{\sqrt7}$. From $\frac{1}{2}$ to $2$: $\frac{2}{\sqrt7}\left[\tan^{-1}\frac{5}{\sqrt7} + \tan^{-1}\frac{1}{\sqrt7}\right]$. The product $\frac{5}{7} < 1$, so the sum is $\tan^{-1}\frac{6/\sqrt7}{1 - 5/7} = \tan^{-1}\frac{21}{\sqrt7} = \tan^{-1}(3\sqrt7)$. So $2\alpha = \frac{\pi}{2}\cdot\frac{2}{\sqrt7}\tan^{-1}(3\sqrt7)$, i.e. $\frac{2\alpha\sqrt7}{\pi} = \tan^{-1}(3\sqrt7)$. $\sqrt7\tan\left(\frac{2\alpha\sqrt7}{\pi}\right) = \sqrt7\cdot3\sqrt7 = 21$. Answer: 21.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	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 Work in units first. $S$ is volt per kelvin: $\mathrm{V\,K^{-1}}$. $\sigma$ is siemens per metre: $\mathrm{A\,V^{-1}\,m^{-1}}$. $\kappa$ is watt per metre per kelvin: $\mathrm{W\,m^{-1}\,K^{-1}}$. $Z = \frac{S^2\sigma}{\kappa} = \frac{\mathrm{V^2K^{-2}}\cdot\mathrm{A\,V^{-1}m^{-1}}}{\mathrm{W\,m^{-1}K^{-1}}} = \frac{\mathrm{V\,A}}{\mathrm{W}}\,\mathrm{K^{-1}} = \mathrm{K^{-1}}$, since $\mathrm{V\,A} = \mathrm{W}$. So $[Z] = [M^0L^0T^0I^0K^{-1}]$ (this is the thermoelectric figure of merit; $ZT$ is dimensionless). Answer: (B).
Q.2	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctHard
A hard single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The field is non-zero only in the dielectric, $\sqrt2R < r < 2R$: inside the inner conductor it is zero, and outside the grounded outer cylinder the total enclosed charge is zero. In the dielectric, $E = \frac{Q}{2\pi\epsilon_0\kappa\,\ell\,r}$, radial. For a radial field of this form, the flux through a strip of the plane of length $\ell$ that subtends an angle $d\phi$ at the axis is $E\cos\psi\,\ell\,ds = \frac{Q}{2\pi\epsilon_0\kappa}d\phi$ (since $\cos\psi\,ds = r\,d\phi$). On the plane, a point at distance $s$ from the foot of the perpendicular has $r = \sqrt{R^2 + s^2}$ and angle $\phi$ with $\cos\phi = \frac{R}{r}$. $r = \sqrt2R$ gives $\phi = 45^\circ$; $r = 2R$ gives $\phi = 60^\circ$. So on each side of the foot the field region subtends $15^\circ = \frac{\pi}{12}$, and $\frac{\pi}{6}$ in all. Flux $= \frac{Q}{2\pi\epsilon_0\kappa}\cdot\frac{\pi}{6} = \frac{Q}{12\epsilon_0\kappa} = \frac{Q}{60\epsilon_0}$. Answer: (C).
Q.3	Oscillations and WavesSimple pendulum and spring systems	Class 11: Oscillations; Class 11: Waves	Single correctEasy
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Turn the rod through a small angle $\theta$ about $O$. A spring attached at height $h$ is stretched or compressed by $h\theta$ and gives a restoring torque $kh\theta\cdot h = kh^2\theta$. The moment of inertia about $O$ is $I = \frac{ml^2}{3}$ in both cases, so $f \propto \sqrt{\sum kh^2}$. Fig. 1: $k\left(\frac{l}{2}\right)^2 + kl^2 = \frac{5}{4}kl^2$. Fig. 2: $2k\left(\frac{l}{2}\right)^2 = \frac{1}{2}kl^2$. $\frac{f_1}{f_2} = \sqrt{\frac{5/4}{1/2}} = \sqrt{\frac{5}{2}}$. Answer: (C).
Q.4	GravitationOrbital motion	Class 11: Gravitation	Single correctHard
A hard single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation The final key gives marks to all for this question. With the transfer slow and no mass leaving the pair, the orbital angular momentum is conserved. For a circular orbit, $L = \mu\sqrt{G(m_1 + m_2)r}$ with $\mu = \frac{m_1m_2}{m_1 + m_2}$; the total mass is constant, so $m_1^2m_2^2\,r$ is constant. Taking logarithmic derivatives with $\frac{dm_2}{dt} = -\gamma$ and $\frac{dm_1}{dt} = +\gamma$: $\frac{2}{m_1}\gamma - \frac{2}{m_2}\gamma + \frac{1}{r}\frac{dr}{dt} = 0$, so $\frac{1}{r}\frac{dr}{dt} = 2\gamma\left(\frac{1}{m_2} - \frac{1}{m_1}\right) \approx \frac{2\gamma}{m_2}$ for $m_1 \gg m_2$. The separation grows, so the rate is positive, while every option is negative. Option (B) has the right size but the wrong sign; that is why the question was given marks to all.
Q.5	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 whole conductor is at one potential, so find it at the centre, where every bit of induced charge is at distance $R$. (A) Neutral sphere: the induced charges sum to zero, so $V = \frac{kq}{d} = \frac{9\times10^9\times10^{-8}}{0.2} = 450$ V. TRUE. Grounded: $\frac{kq}{d} + \frac{kQ'}{R} = 0$, so $Q' = -\frac{qR}{d} = -\frac{10^{-8}\times0.1}{0.2} = -5\times10^{-9}$ C. (B) The sphere goes from 0 to $-5\times10^{-9}$ C, i.e. $5\times10^{-9}$ C of positive charge flows from the sphere to the ground (electrons flow up). TRUE. (C) After the ground is removed the charge is fixed at $-5\times10^{-9}$ C, whatever the point charge does later. TRUE. (D) With the point charge at 30 cm: $V = \frac{9\times10^9\times10^{-8}}{0.3} + \frac{9\times10^9\times(-5\times10^{-9})}{0.1} = 300 - 450 = -150$ V, not 300 V. FALSE. Answer: (A), (B), (C).
Q.6	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Multiple correctHard
A hard multiple correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Let the mirrors be a distance $D$ apart. Seen from either mirror, $S$ is behind half the slab: it lies $\frac{D}{2} - \frac{t}{2}$ of air plus $\frac{t}{2}$ of glass away, and its apparent distance is $\frac{D}{2} - \frac{t}{2} + \frac{t}{2n_0} = \frac{D}{2} - \frac{t}{2}\left(1 - \frac{1}{n_0}\right)$. The same shift applies to light coming back towards $S$ (paraxial rays, $f$ large). Way 1, each mirror images $S$ on itself: the apparent source must sit at the centre of curvature, $\frac{D}{2} - \frac{t}{2}\left(1 - \frac{1}{n_0}\right) = 2f$, so $D = 4f + \left(1 - \frac{1}{n_0}\right)t$. (A) is correct. Way 2, via both mirrors: put the apparent source at the focus of one mirror. Its light leaves that mirror parallel to the axis, crosses the slab undeviated (parallel rays normal to the slab keep their direction) and converges at the focus of the other mirror, which, by the same apparent-depth shift, is $S$. Condition $\frac{D}{2} - \frac{t}{2}\left(1 - \frac{1}{n_0}\right) = f$, so $D = 2f + \left(1 - \frac{1}{n_0}\right)t$. (B) is correct. (C) and (D) use $(n_0 - 1)t$, which would be the shift if the slab made the source look farther away; a slab makes it look nearer, so they are wrong. Answer: (A), (B).
Q.7	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 Each sheet gives $\frac{\sigma}{2\epsilon_0}$ pointing away from it. In a gap, the field (to the right) is $\frac{(\text{charge to the left}) - (\text{charge to the right})}{2\epsilon_0}$ per unit area. Configuration I ($+\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0$; total zero): regions 1, 3, 5 have $\frac{\sigma_0 - (-\sigma_0)}{2\epsilon_0} = \frac{\sigma_0}{\epsilon_0}$; regions 2 and 4 have zero. (A) Region 4 of I: zero. TRUE. (C) $\frac{\sigma_0}{\epsilon_0} = \frac{9\times10^{-6}}{9\times10^{-12}} = 10^6$ V/m over three gaps of $1\,\mu$m: $3$ V, not 5 V. FALSE. Configuration II ($+\frac{\sigma_0}{2}, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, -\frac{\sigma_0}{2}$; total zero): the charge to the left of the gaps is $\frac{\sigma_0}{2}, -\frac{\sigma_0}{2}, \frac{\sigma_0}{2}, -\frac{\sigma_0}{2}, \frac{\sigma_0}{2}$, so the fields are $+\frac{\sigma_0}{2\epsilon_0}, -\frac{\sigma_0}{2\epsilon_0}, +\frac{\sigma_0}{2\epsilon_0}, -\frac{\sigma_0}{2\epsilon_0}, +\frac{\sigma_0}{2\epsilon_0}$. (B) Region 3 of II: $\frac{\sigma_0}{2\epsilon_0}$, not $\frac{\sigma_0}{\epsilon_0}$. FALSE. (D) Potential difference across II: $\frac{\sigma_0}{2\epsilon_0}(1 - 1 + 1 - 1 + 1)\times10^{-6} = 0.5$ V, not zero. FALSE. Answer: (A).
Q.8	ThermodynamicsSecond law	Class 11: Thermodynamics	Multiple correctEasy
A easy multiple correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics (A) $W = \eta Q_H = 0.4\times150 = 60$ J. TRUE. (B) $\eta = 1 - \frac{T_C}{T_H}$: $T_C = 0.6\times1000 = 600$ K. TRUE. The heat pump is taken as ideal (Carnot), its coefficient of performance being heat delivered per unit work: $\frac{T_H}{T_H - T_C} = 10$. (C) $\frac{300}{300 - T_C} = 10$ gives $T_C = 270$ K. TRUE. (D) Heat delivered to its hot reservoir $= 10\times60 = 600$ J, not 540 J (540 J is the heat taken from the cold side). FALSE. Answer: (A), (B), (C).
Q.9	Magnetic Effects of Current and MagnetismMagnetic dipole	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	NumericalMedium
A medium numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter On a conductor the charge sits on the surface, so the charge is a spherical shell while the mass is a uniform solid sphere. Magnetic moment of a rotating charged shell: a ring at polar angle $\theta$ carries charge $\frac{Q}{2}\sin\theta\,d\theta$, current $\frac{\omega}{2\pi}$ times that, and area $\pi R^2\sin^2\theta$. So $\mu = \int_0^\pi\frac{Q\omega}{4\pi}\sin\theta\cdot\pi R^2\sin^2\theta\,d\theta = \frac{Q\omega R^2}{4}\cdot\frac{4}{3} = \frac{QR^2\omega}{3}$. Angular momentum of the solid sphere: $L = \frac{2}{5}MR^2\omega$. $\frac{\mu}{L} = \frac{Q/3}{2M/5} = \frac{5Q}{6M} = \frac{5}{3}\cdot\frac{Q}{2M}$. $\alpha = \frac{5}{3} \approx 1.67$. Answer: 1.67.
Q.10	Atoms and NucleiBohr model	Class 12: Atoms; Class 12: Nuclei	NumericalMedium
A medium numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Ionisation: $h\nu_1 = 13.6 + 10 = 23.6$ eV (the recoil of the proton is negligible). Positronium is a Bohr atom with reduced mass $\frac{m_e}{2}$, so its ground-state energy is half that of hydrogen: $-6.8$ eV. Energy balance for the capture: $10 = 5 + (-6.8) + h\nu_2$, so $h\nu_2 = 11.8$ eV. (Momentum checks: an electron of 10 eV and a 2-electron-mass atom of 5 eV have the same momentum, $p^2 = 2m_e\cdot10 = 2(2m_e)\cdot5$, so the photon's momentum is negligible, as it should be.) Difference $= 23.6 - 11.8 = 11.8$ eV. Answer: 11.8.
Q.11	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	NumericalMedium
A medium numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the diagram, $W \to X$ is the adiabatic expansion (volume up, temperature down) and $X \to Y$ the isobaric expansion (on a $V$-$T$ diagram an isobar is a line through the origin, $V \propto T$). Adiabatic, monatomic ($\gamma = \frac{5}{3}$): $TV^{2/3}$ is constant, so $T_X = T_W\left(\frac{64}{125}\right)^{2/3} = T_W\cdot\frac{16}{25}$, and $nRT_X = 0.64$ J. Isobaric: $V$ doubles from 125 to 250 cm$^3$, so $T_Y = 2T_X$ and $nR(T_Y - T_X) = 0.64$ J. Heat absorbed $= nC_p\Delta T = \frac{5}{2}nR\Delta T = \frac{5}{2}\times0.64 = 1.6$ J. Answer: 1.6.
Q.12	GravitationOrbital motion	Class 11: Gravitation	NumericalMedium
A medium numerical question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Kepler's third law: $T \propto r^{3/2}$. The geostationary satellite has $T_1 = 24$ h, so $T_2 = 24\left(\frac{r_2}{r_1}\right)^{3/2} = \frac{24}{1.21^{3/2}} = \frac{24}{1.331}$ h. The two move in opposite senses, so the angular speed of one relative to the other is $\omega_1 + \omega_2$: $\frac{1}{T_{rel}} = \frac{1}{24} + \frac{1.331}{24} = \frac{2.331}{24}$. So $T_{rel} = \frac{24}{2.331}$ h and $p = 2.331 \approx 2.33$. Answer: 2.33.
Q.13	Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation	Class 11: Kinetic Theory	NumericalEasy
A easy numerical question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The piston conducts heat, so both gases are at the same temperature, and both volumes are $\frac{AL}{2}$. By $PV = nRT$, $P_{left} = \frac{3}{2}P_{right}$. The spring is stretched from $\frac{2L}{5}$ to $\frac{L}{2}$, by $\frac{L}{10}$, so it pulls the piston to the left with force $\frac{kL}{10}$. Balance on the piston: $P_{left}A = P_{right}A + \frac{kL}{10}$, so $\frac{1}{2}P_{right}A = \frac{kL}{10}$ and $P_{right} = \frac{kL}{5A}$. $\alpha = 0.2$. Answer: 0.2.
Q.14	OpticsInterference and diffraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	NumericalMedium
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The two wedges together form a slab of uniform thickness $t$ and height $d + 2l = 4$ mm, the slanted interface running corner to corner; each slit is $l = 1$ mm from an end. Wedge $A$ is full thickness at its end and thins linearly to zero at the other end. At the slit near $A$'s thick end, $A$ is $\frac{3}{4}t = 9\,\mu$m thick and $B$ is $3\,\mu$m; at the other slit, $A$ is $3\,\mu$m and $B$ is $9\,\mu$m. Optical paths through the slab: $1.7(9) + 1.5(3) = 19.8\,\mu$m and $1.7(3) + 1.5(9) = 18.6\,\mu$m. Extra path at the first slit: $\Delta = 1.2\,\mu$m (equivalently $(1.7 - 1.5)\times6\,\mu$m). The central maximum moves towards that slit by $\frac{D\Delta}{d} = \frac{2\times1.2\times10^{-6}}{2\times10^{-3}} = 1.2\times10^{-3}$ m $= 1.2$ mm. Answer: 1.2.
Q.15	KinematicsProjectile motion	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	NumericalMedium
A medium numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The horizontal motion feels only the drag: $m\frac{dv_x}{dt} = -cv_x$, so $v_x = u\cos60^\circ\,e^{-t/\tau}$ with $\tau = \frac{m}{c} = \frac{0.2}{0.1} = 2$ s. $x(t) = u\cos60^\circ\,\tau\left(1 - e^{-t/\tau}\right)$. At $t = 2$ s: $x = 135\times2\times\left(1 - \frac{1}{e}\right) = 270\times\left(1 - \frac{1}{2.7}\right) = 270\times\frac{1.7}{2.7} = 170$ m. Answer: 170.
Q.16	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 The two pendulums swing as mirror images: they always move with equal speeds, both towards each other or both away, along the line joining them (which stays horizontal by symmetry). Their speed is greatest at the bottom: $v = \sqrt{2gL(1 - \cos\theta_0)} = \sqrt{2\times10\times8\times0.1} = 4$ m/s. The horizontal component $v\cos\theta$ is also greatest there. Approaching: $f_{max} = 660\cdot\frac{330 + 4}{330 - 4} = 660\cdot\frac{334}{326} \approx 676.2$ Hz. Receding: $f_{min} = 660\cdot\frac{330 - 4}{330 + 4} = 660\cdot\frac{326}{334} \approx 644.2$ Hz. Maximum variation $= f_{max} - f_{min} \approx 32.0$ Hz. Answer: 32.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
Q.1	Principles of Qualitative AnalysisAnalysis of anions	No matching NCERT chapter	Single correctEasy
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Sodium nitroprusside is $\mathrm{Na_2[Fe(CN)_5NO]}$, with the nitrosyl ligand present as $\mathrm{NO^+}$. Sulphide adds to the nitrosyl ligand, giving the violet complex: $\mathrm{[Fe(CN)_5NO]^{2-} + S^{2-} \to [Fe(CN)_5NOS]^{4-}}$. So the $\mathrm{NO^+}$ ligand becomes thionitrite, $\mathrm{NOS^-}$; the cyanide ligands are unchanged. Answer: (A).
Q.2	p-Block ElementsGroup 17 and 18 elements	No matching NCERT chapter	Single correctMedium
A medium single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Hydrolysis keeps the oxidation state of the central (less electronegative) halogen; the more electronegative halogen leaves as the halide. $\mathrm{ICl}$: I is $+1$; $\mathrm{ICl + H_2O \to HOI + HCl}$, giving $\mathrm{IO^-}$ (hypoiodite, $+1$). $\mathrm{ClF_3}$: Cl is $+3$; the oxoanion of Cl(+3) is chlorite, $\mathrm{ClO_2^-}$. $\mathrm{BrF_5}$: Br is $+5$; $\mathrm{BrF_5 + 3H_2O \to HBrO_3 + 5HF}$, giving bromate, $\mathrm{BrO_3^-}$. So the products are $\mathrm{IO^-}$, $\mathrm{ClO_2^-}$ and $\mathrm{BrO_3^-}$. Answer: (A).
Q.3	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctHard
A hard single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Count the $sp^2$ and $sp$ carbons of each product. P: 3-phenylpropanoic acid with $\mathrm{Br_2}$/red P then water (Hell-Volhard-Zelinsky) gives $\mathrm{PhCH_2CHBrCOOH}$. Unsaturated carbons: 6 (ring) + 1 (COOH) = 7. Q: benzaldehyde and acetaldehyde in aqueous NaOH (cross aldol, with dehydration to the conjugated product) give cinnamaldehyde $\mathrm{PhCH{=}CHCHO}$: 6 + 3 = 9. R: $\mathrm{NaNH_2}$ makes the acetylide, which the allyl bromide alkylates: $\mathrm{PhC{\equiv}C{-}CH_2CH{=}CH_2}$. $\mathrm{Hg^{2+}/H_3O^+}$ hydrates the triple bond to the ketone $\mathrm{PhCOCH_2CH_2CH{=}CH_2}$: 6 + 1 + 2 = 9. S: 2-methylindene. Ozonolysis ($\mathrm{O_3}$, Zn-$\mathrm{H_2O}$) opens the five-membered ring at the C=C: $o$-$\mathrm{OHC{-}C_6H_4{-}CH_2COCH_3}$. Two equivalents of $\mathrm{CH_3MgBr}$ turn the aldehyde into $\mathrm{{-}CH(OH)CH_3}$ and the ketone into $\mathrm{{-}CH_2C(OH)(CH_3)_2}$. Acid and heat dehydrate both alcohols: $o$-$\mathrm{CH_2{=}CH{-}C_6H_4{-}CH{=}C(CH_3)_2}$, a monocyclic diene: 6 + 2 + 2 = 10. S has the most unsaturated carbons. Answer: (D).
Q.4	Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) 2-Chloroethanol with NaCN gives $\mathrm{HOCH_2CH_2CN}$; hydrolysis gives 3-hydroxypropanoic acid, $\mathrm{HOCH_2CH_2COOH}$: one carboxyl group. (B) Bromine water oxidises only the aldehyde of glucose: gluconic acid, $\mathrm{HOCH_2(CHOH)_4COOH}$: one carboxyl group. (C) Ethanolic KOH eliminates HBr from bromocyclohexane to cyclohexene; hot acidic $\mathrm{KMnO_4}$ cleaves the C=C and oxidises both ends: adipic acid, $\mathrm{HOOC(CH_2)_4COOH}$, a dicarboxylic acid. (D) Chromic acid oxidises the primary alcohol of the hydroxy ketone to a carboxylic acid; the ketone stays. A keto acid with one carboxyl group. Answer: (C).
Q.5	Chemical Bonding and Molecular StructureHydrogen bonding	Class 11: Chemical Bonding and Molecular Structure	Multiple correctMedium
A medium multiple correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) Ion-ion energy varies as $\frac{1}{r}$, ion-dipole as $\frac{1}{r^2}$. The charge-charge energy falls off more slowly, not more rapidly. FALSE. (B) For freely rotating polar molecules the orientation-averaged dipole-dipole energy goes as $\frac{1}{r^6}$ (and as $\frac{1}{T}$), not $\frac{1}{r^3}$; $\frac{1}{r^3}$ is for fixed dipoles. FALSE. (C) The dipole-induced dipole energy is $-\frac{C\mu^2\alpha}{r^6}$; the induced dipole always follows the permanent one, so no thermal averaging enters and it does not depend on temperature. TRUE. (D) Nonpolar molecules attract through London dispersion forces between instantaneous and induced dipoles. TRUE. Answer: (C), (D).
Q.6	p-Block ElementsGroup 15 elements	No matching NCERT chapter	Multiple correctEasy
A easy multiple correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Count the hydrogens not on oxygen. (A) $\mathrm{H_3PO_4}$, $\mathrm{OP(OH)_3}$: all three H on O. No P–H. (B) $\mathrm{H_3PO_3}$, $\mathrm{HPO(OH)_2}$: one P–H (dibasic acid). Yes. (C) $\mathrm{H_4P_2O_7}$, $\mathrm{(HO)_2OP{-}O{-}PO(OH)_2}$: no P–H. (D) $\mathrm{H_3PO_2}$, $\mathrm{H_2PO(OH)}$: two P–H (monobasic acid). Yes. Answer: (B), (D).
Q.7	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 Hot acidic $\mathrm{KMnO_4}$ oxidises the saturated ring of tetralin at both benzylic positions: phthalic acid. With $\mathrm{NH_3}$ and heat, losing two waters, it gives phthalimide, X. Ethanolic KOH makes potassium phthalimide, and R–Br alkylates the nitrogen: N-alkylphthalimide, Y. NaOH hydrolysis (Gabriel synthesis) gives sodium phthalate (the aromatic compound) and the primary amine $\mathrm{RNH_2}$, Z. (A) Phthalimide and N-alkylphthalimide both contain oxygen. TRUE. (B) Y has no $\mathrm{NH_2}$ group, so it gives no isocyanide (carbylamine) test. FALSE. (C) Z is a primary amine and reacts with Hinsberg's reagent (benzenesulphonyl chloride). TRUE. (D) Z is an aliphatic primary amine: aryl halides do not undergo this substitution, so the Gabriel route does not give aromatic amines. FALSE. Answer: (A), (C).
Q.8	Alcohols, Phenols and EthersPreparation and properties	Class 12: Alcohols, Phenols and Ethers	Multiple correctHard
A hard multiple correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The lactone is 4-benzyltetrahydropyran-2-one: $\mathrm{O{=}C{-}CH_2{-}CH(CH_2Ph){-}CH_2{-}CH_2{-}O}$ in a ring. $\mathrm{LiAlH_4}$ reduces the ester to two primary alcohols: P $= \mathrm{HOCH_2CH_2{-}CH(CH_2Ph){-}CH_2CH_2OH}$, 3-benzylpentane-1,5-diol. (A) In P the carbon bearing the benzyl group carries two identical $\mathrm{{-}CH_2CH_2OH}$ arms, so it is not a stereocentre and P is achiral. FALSE. Chromic acid oxidises both primary alcohols: Q $= \mathrm{HOOC{-}CH_2{-}CH(CH_2Ph){-}CH_2{-}COOH}$. (C) Q is a carboxylic acid and liberates $\mathrm{CO_2}$ with aqueous $\mathrm{NaHCO_3}$. TRUE. Soda lime (NaOH and CaO, heat) removes both carboxyl groups: R $= \mathrm{CH_3CH(CH_2Ph)CH_3}$, isobutylbenzene. (D) R is an alkane side chain on benzene, not an alkyne. FALSE. Concentrated $\mathrm{H_2SO_4}$ at 443 K dehydrates the diol to an alkene (a diene). (B) S, having C=C, decolourises cold alkaline $\mathrm{KMnO_4}$ (Baeyer's test). TRUE. Answer: (B), (C).
Q.9	Solid StatePacking in fcc, bcc and hcp	No matching NCERT chapter	NumericalEasy
A easy numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter A ccp (fcc) unit cell holds $Z = 4$ atoms. $a = 400$ pm $= 4\times10^{-8}$ cm, so $a^3 = 64\times10^{-24} = 6.4\times10^{-23}$ cm$^3$. $\rho = \frac{ZM}{N_Aa^3} = \frac{4\times105.6}{6\times10^{23}\times6.4\times10^{-23}} = \frac{422.4}{38.4} = 11.0$ g cm$^{-3}$. Answer: 11.
Q.10	EquilibriumSolubility product	Class 11: Equilibrium	NumericalMedium
A medium numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Moles mixed: $\mathrm{Ba^{2+}}$ $200\times0.010 = 2$ mmol; $\mathrm{IO_3^-}$ $100\times0.10 = 10$ mmol; total volume 300 mL. $\mathrm{Ba(IO_3)_2}$ precipitates, using 2 mmol $\mathrm{Ba^{2+}}$ and 4 mmol $\mathrm{IO_3^-}$, leaving 6 mmol $\mathrm{IO_3^-}$ in 300 mL: $[\mathrm{IO_3^-}] = 0.020$ M. The solubility $s$ of the solid in this solution is the $[\mathrm{Ba^{2+}}]$ it supplies: $s(0.020)^2 = 1.58\times10^{-9}$ (the $2s$ added to iodate is negligible). $s = \frac{1.58\times10^{-9}}{4\times10^{-4}} = 3.95\times10^{-6}$ mol dm$^{-3}$. Answer: 3.95.
Q.11	Surface ChemistryFreundlich adsorption isotherm	No matching NCERT chapter	NumericalMedium
A medium numerical question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich: $\frac{x}{m} = KC^{1/n}$. Dividing the two data points: $\frac{10}{4} = \left(\frac{16}{10}\right)^{1/n}$, i.e. $2.5 = 1.6^{1/n}$. $\log2.5 = 1 - 2\log2 = 0.4$ and $\log1.6 = 4\log2 - 1 = 0.2$, so $\frac{1}{n} = 2$. $K = \frac{4}{10^2} = 0.04$. At $C = 20$: $\frac{x}{m} = 0.04\times400 = 16$ mg g$^{-1}$. Answer: 16.
Q.12	Chemical KineticsRate laws	Class 12: Chemical Kinetics	NumericalMedium
A medium numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let $[A]_0 = a$, so $[R]_0 = 10a$. At 40% completion (of A, the limiting reactant): $[A] = 0.6a$ and $[R] = 10a - 0.4a = 9.6a$. True rate: $k(0.6a)(9.6a)$. Pseudo-first-order rate: $k'[A] = k(10a)(0.6a)$. Relative error $= \frac{10 - 9.6}{9.6}\times100 = 4.17\%$ (taken against the true rate; against the assumed rate it is $4\%$, also inside the key's range). Answer: 4.17.
Q.13	SolutionsColligative properties	Class 12: Solutions	NumericalMedium
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Osmotic pressure $\pi = h\rho g = 0.0200\times1000\times10 = 200$ Pa. $\pi = \frac{c}{M}RT$ with $c = 2.00$ g dm$^{-3}$ $= 2000$ g m$^{-3}$: $M = \frac{cRT}{\pi} = \frac{2000\times8.3\times300}{200} = 24900$ g mol$^{-1} = 2.49\times10^4$ g mol$^{-1}$. Answer: 2.49.
Q.14	ElectrochemistryBatteries and fuel cells	Class 12: Electrochemistry	NumericalMedium
A medium numerical question on batteries and fuel cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\mathrm{C_4H_{10} + \frac{13}{2}O_2 \to 4CO_2 + 5H_2O}$. $\Delta G^o = 4(-394) + 5(-237) - (-18) = -1576 - 1185 + 18 = -2743$ kJ mol$^{-1}$. Electrons: $\frac{13}{2}$ mol $\mathrm{O_2}$ take $4$ electrons each, $n = 26$. $E^o = -\frac{\Delta G^o}{nF} = \frac{2743\times10^3}{26F} = \frac{105.5}{F}\times10^3$ V. Answer: 105.5.
Q.15	Coordination CompoundsCrystal field theory	Class 12: Coordination Compounds	NumericalEasy
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Both contain $\mathrm{Mn^{3+}}$, $d^4$, octahedral. $\mathrm{Br^-}$ is a weak-field ligand: high spin $t_{2g}^3e_g^1$, 4 unpaired electrons, $\mu = \sqrt{4\cdot6} = \sqrt{24} \approx 4.90$ B.M. $\mathrm{CN^-}$ is strong-field: low spin $t_{2g}^4$, 2 unpaired electrons, $\mu = \sqrt{2\cdot4} = \sqrt8 \approx 2.83$ B.M. Sum $\approx 7.73$ B.M. Answer: 7.73.
Q.16	BiomoleculesCarbohydrates	Class 12: Biomolecules	NumericalMedium
A medium numerical question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules A linear chain of 8 units has 7 glycosidic links; hydrolysis adds 7 water: mass of monosaccharides $= 1024 + 7\times18 = 1150$ g per mole. 2-Deoxyribose: $0.5826\times1150 = 670$ g, i.e. $\frac{670}{134} = 5$ units. The other 3 units weigh $1150 - 670 = 480$ g: with $r$ ribose and $g$ glucose, $r + g = 3$ and $150r + 180g = 480$. So $30g = 30$, $g = 1$, $r = 2$. Answer: 2.

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