JEE (Advanced) · 2008

JEE Advanced 2008, 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
45 Q
1. Integral Calculus7
2. Limits, Continuity and Differentiability5
3. Conic Sections4
9 easy · 29 medium · 7 hard
Physics
45 Q
1. Atoms and Nuclei6
2. Electrostatics6
3. Optics4
18 easy · 24 medium · 3 hard
Chemistry
45 Q
1. p-Block Elements5
2. Organic Chemistry: Some Basic Principles and Techniques4
3. Aldehydes, Ketones and Carboxylic Acids4
19 easy · 23 medium · 3 hard

Chapters

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

Integral Calculus7
Class 12: Integrals; Class 12: Application of Integrals
Limits, Continuity and Differentiability5
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Conic Sections4
Class 11: Conic Sections
Applications of Derivatives4
Class 12: Application of Derivatives
Circles4
Class 11: Conic Sections
Complex Numbers4
Class 11: Complex Numbers and Quadratic Equations
Vector Algebra3
Class 12: Vector Algebra
Three Dimensional Geometry3
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Straight Lines3
Class 11: Straight Lines
Probability2
Class 11: Probability; Class 12: Probability
Trigonometry1
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Matrices and Determinants1
Class 12: Matrices; Class 12: Determinants
Sequences and Series1
Class 11: Sequences and Series
Quadratic Equations1
Class 11: Complex Numbers and Quadratic Equations
Differential Equations1
Class 12: Differential Equations
Permutations and Combinations1
Class 11: Permutations and Combinations

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
1	Conic SectionsParabola	Class 11: Conic Sections	Single correctEasy
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Put $y^2 = 4x$ into the circle: $x^2 + 4x - 6x + 1 = 0$, that is $x^2 - 2x + 1 = (x-1)^2 = 0$. So $x = 1$ is a double root, and on the parabola $x = 1$ gives $y = \pm 2$: the common points are $(1, 2)$ and $(1, -2)$. A double root in $x$ means the curves meet tangentially there. Check at $(1, 2)$: for the parabola $2yy' = 4$, $y' = 1$; for the circle $2x + 2yy' - 6 = 0$, $y' = \frac{6-2}{4} = 1$. Same slope, so they touch; by symmetry about the x-axis they also touch at $(1, -2)$. The curves touch each other at exactly two points. Answer: (B).
2	TrigonometryInverse trigonometric functions and their properties	Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions	Single correctEasy
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta = \cot^{-1}x$, so $\cot\theta = x$ with $\theta$ in $(0, \frac{\pi}{2})$. Then $\cos\theta = \frac{x}{\sqrt{1+x^2}}$ and $\sin\theta = \frac{1}{\sqrt{1+x^2}}$. $x\cos\theta + \sin\theta = \frac{x^2 + 1}{\sqrt{1+x^2}} = \sqrt{1+x^2}$. Squaring: $1 + x^2$; subtracting 1: $x^2$; the square root is $x$ (as $x > 0$). The whole expression is $\sqrt{1+x^2}\cdot x = x\sqrt{1+x^2}$. Answer: (C).
3	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Single correctMedium
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra The volume is $\|[\hat{a}\;\hat{b}\;\hat{c}]\|$, and the square of a scalar triple product is the determinant of the dot products: $[\hat{a}\;\hat{b}\;\hat{c}]^2 = \begin{vmatrix}1 & \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & 1 & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} & 1\end{vmatrix}$. Expanding: $1\left(1 - \frac{1}{4}\right) - \frac{1}{2}\left(\frac{1}{2} - \frac{1}{4}\right) + \frac{1}{2}\left(\frac{1}{4} - \frac{1}{2}\right) = \frac{3}{4} - \frac{1}{8} - \frac{1}{8} = \frac{1}{2}$. Volume $= \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$. Answer: (A).
4	Conic SectionsStandard forms of conics	Class 11: Conic Sections	Single correctEasy
A easy single correct question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections $x^2 - 5xy + 6y^2 = (x - 2y)(x - 3y)$, so the second factor always gives the two lines $x = 2y$ and $x = 3y$. The first factor $ax^2 + by^2 + c = 0$: (A) $c = 0$, $a, b$ of the same sign: $ax^2 + by^2 = 0$ is only the point $(0, 0)$, not two lines. False. (B) $a = b$ and $c$ of opposite sign: $x^2 + y^2 = -\frac{c}{a} > 0$, a circle. Two lines and a circle. True. (C) $a, b$ of the same sign, $c$ opposite: $\frac{x^2}{-c/a} + \frac{y^2}{-c/b} = 1$ with both denominators positive, an ellipse, not a hyperbola. False. (D) The product gives one conic and two lines, never a circle and an ellipse together. False. Answer: (B).
5	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 Left of $x = 1$, $\|x - 1\| = 1 - x$, whose derivative is $-1$; so $p = -1$. Put $t = x - 1 \to 0^+$: $g = \frac{t^n}{m\log\cos t}$. Near $t = 0$, $\cos t \approx 1 - \frac{t^2}{2}$, so $\log\cos t \approx -\frac{t^2}{2}$ (that is, $\lim \frac{\log\cos t}{t^2} = -\frac{1}{2}$). Then $g \approx \frac{t^n}{-\frac{m t^2}{2}} = -\frac{2}{m}t^{n-2}$. For a finite non-zero limit we need $n = 2$, and then the limit is $-\frac{2}{m} = -1$, so $m = 2$. Answer: (C).
6	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	Single correctMedium
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives On $(-3, -1]$, $f = (2+x)^3$ is increasing, rising to $f(-1) = 1$. On $(-1, 2)$, $f = x^{2/3}$: as $x \to -1^+$, $f \to 1$; it decreases on $(-1, 0)$ to $f(0) = 0$ and increases on $(0, 2)$. At $x = -1$: $f$ rises up to the value 1 from the left and falls from 1 on the right, so $x = -1$ is a local maximum (f is continuous there). At $x = 0$: $f'$ does not exist, but $f(0) = 0 < f(x)$ nearby, a local minimum. There are no other turning points. Total: 2. Answer: (C).
7	CirclesIntersection of a line and a circle	Class 11: Conic Sections	Multiple correctHard
A hard multiple correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections PT and QR are two chords of the circumcircle meeting at S, so by intersecting chords $PS\cdot ST = QS\cdot SR$. AM-GM on $\frac{1}{PS}$ and $\frac{1}{ST}$: $\frac{1}{PS} + \frac{1}{ST} \ge \frac{2}{\sqrt{PS\cdot ST}} = \frac{2}{\sqrt{QS\cdot SR}}$, with equality only when $PS = ST$. AM-GM on $QS$ and $SR$: $\sqrt{QS\cdot SR} \le \frac{QS + SR}{2} = \frac{QR}{2}$, so $\frac{2}{\sqrt{QS\cdot SR}} \ge \frac{4}{QR}$, with equality only when $QS = SR$. Both equalities together would make S the midpoint of two different chords PT and QR. The chord bisected at a point S other than the centre is unique (it is perpendicular to the line from the centre to S), so that cannot happen: hence $\frac{1}{PS} + \frac{1}{ST} > \frac{4}{QR}$ strictly, and (D) is true. (B): the official key takes the first inequality as strict, the case $PS = ST$ being excluded with the centre (see note). Then $\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS\times SR}}$, true. (A) and (C) contradict these inequalities: false. Answer: (B), (D).
8	Conic SectionsParabola	Class 11: Conic Sections	Multiple correctMedium
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The ellipse is $\frac{x^2}{4} + y^2 = 1$: $a = 2$, $b = 1$, $e = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$. The foci are $(\pm\sqrt{3}, 0)$ and the semi-latus rectum is $\frac{b^2}{a} = \frac{1}{2}$, so the ends of the latera recta are $(\pm\sqrt{3}, \pm\frac{1}{2})$. The two with negative ordinates are $P(-\sqrt{3}, -\frac{1}{2})$ and $Q(\sqrt{3}, -\frac{1}{2})$ (see note). $PQ = 2\sqrt{3}$ is the latus rectum of the parabola, so $4A = 2\sqrt{3}$, $A = \frac{\sqrt{3}}{2}$. Its focus is the midpoint $(0, -\frac{1}{2})$ and its axis is the y-axis. Opening upward: vertex $(0, -\frac{1}{2} - \frac{\sqrt{3}}{2})$, $x^2 = 2\sqrt{3}\left(y + \frac{1}{2} + \frac{\sqrt{3}}{2}\right)$, so $x^2 - 2\sqrt{3}y = \sqrt{3} + 3$. This is (B). Opening downward: vertex $(0, -\frac{1}{2} + \frac{\sqrt{3}}{2})$, $x^2 = -2\sqrt{3}\left(y + \frac{1}{2} - \frac{\sqrt{3}}{2}\right)$, so $x^2 + 2\sqrt{3}y = 3 - \sqrt{3}$. This is (C). (A) and (D) have the wrong vertex for their direction of opening. Answer: (B), (C).
9	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	Multiple correctHard
A hard multiple correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Write each term as $\frac{1}{n}\cdot\frac{1}{1 + \frac{k}{n} + \left(\frac{k}{n}\right)^2} = \frac{1}{n}f\left(\frac{k}{n}\right)$ with $f(x) = \frac{1}{1 + x + x^2}$. On $[0, 1]$, $f$ is strictly decreasing. So $S_n$ (right end points, $k = 1 \ldots n$) is a lower Riemann sum and $T_n$ (left end points, $k = 0 \ldots n-1$) an upper sum: $S_n < \int_0^1 f(x)\,dx < T_n$. $\int_0^1 \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}} = \frac{2}{\sqrt{3}}\left[\tan^{-1}\frac{2x+1}{\sqrt{3}}\right]_0^1 = \frac{2}{\sqrt{3}}\left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \frac{\pi}{3\sqrt{3}}$. So $S_n < \frac{\pi}{3\sqrt{3}}$ (A true, B false) and $T_n > \frac{\pi}{3\sqrt{3}}$ (D true, C false). Answer: (A), (D).
10	Integral CalculusProperties of definite integrals	Class 12: Integrals; Class 12: Application of Integrals	Multiple correctMedium
A medium multiple correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Differentiate $f(x) = f(1-x)$: $f'(x) = -f'(1-x)$. (B) At $x = \frac{1}{2}$: $f'(\frac{1}{2}) = -f'(\frac{1}{2})$, so $f'(\frac{1}{2}) = 0$. True. (A) At $x = \frac{1}{4}$: $f'(\frac{3}{4}) = -f'(\frac{1}{4}) = 0$. So $f'$ vanishes at $\frac{1}{4}, \frac{1}{2}, \frac{3}{4}$; by Rolle's theorem on $f'$, $f''$ vanishes somewhere in $(\frac{1}{4}, \frac{1}{2})$ and in $(\frac{1}{2}, \frac{3}{4})$, at least twice on $[0, 1]$. True. (C) Let $h(x) = f(x + \frac{1}{2})$. Then $h(-x) = f(\frac{1}{2} - x) = f(1 - (\frac{1}{2} - x)) = f(\frac{1}{2} + x) = h(x)$, so $h$ is even and $h(x)\sin x$ is odd; its integral over $[-\frac{1}{2}, \frac{1}{2}]$ is 0. True. (D) In the right side put $t = 1 - u$: $\int_{1/2}^{1} f(1-t)e^{\sin\pi t}dt = \int_0^{1/2} f(u)e^{\sin(\pi - \pi u)}du = \int_0^{1/2} f(u)e^{\sin\pi u}du$, the left side. True. Answer: (A), (B), (C), (D).
11	Limits, Continuity and DifferentiabilityDerivatives up to order two	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Assertion and reasonMedium
A medium assertion and reason question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f'(x) = g'(x)\sin x + g(x)\cos x$, so $f'(0) = g(0)$: Statement-2 is true. $f''(x) = g''(x)\sin x + 2g'(x)\cos x - g(x)\sin x$, so $f''(0) = 2g'(0) = 0$. Statement-1: $g(x)\cot x - g(0)\,\mathrm{cosec}\,x = \frac{g(x)\cos x - g(0)}{\sin x}$. Since $g(x)\cos x = f'(x) - g'(x)\sin x$ and $g(0) = f'(0)$ (Statement-2), $\frac{g(x)\cos x - g(0)}{\sin x} = \frac{f'(x) - f'(0)}{\sin x} - g'(x) = \frac{f'(x) - f'(0)}{x}\cdot\frac{x}{\sin x} - g'(x) \to f''(0) - g'(0) = f''(0)$. (Directly, by L'Hospital: $\lim\frac{g'(x)\cos x - g(x)\sin x}{\cos x} = g'(0) = 0 = f''(0)$.) Statement-1 is true, and the step that turns the limit into $f''(0)$ is exactly $g(0) = f'(0)$, so Statement-2 explains it. Answer: (A).
12	Three Dimensional GeometryEquation of a line	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	Assertion and reasonMedium
A medium assertion and reason question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Normals: $\vec{n}_1 = (1, -1, 1)$, $\vec{n}_2 = (1, 1, -1)$, $\vec{n}_3 = (1, -3, 3)$. Directions of the lines: $L_1 \parallel \vec{n}_2\times\vec{n}_3 = (3 - 3,\; -1 - 3,\; -3 - 1) = (0, -4, -4)$, $L_2 \parallel \vec{n}_3\times\vec{n}_1 = (-3 + 3,\; 3 - 1,\; -1 + 3) = (0, 2, 2)$, $L_3 \parallel \vec{n}_1\times\vec{n}_2 = (1 - 1,\; 1 + 1,\; 1 + 1) = (0, 2, 2)$. All three are parallel to $(0, 1, 1)$: Statement-1 is false. Common point: $\vec{n}_3 = 2\vec{n}_1 - \vec{n}_2$ (check: $(2-1, -2-1, 2+1) = (1, -3, 3)$). If a point lay on $P_1$ and $P_2$, it would satisfy $2(x - y + z) - (x + y - z) = 2(1) - (-1) = 3$, i.e. $x - 3y + 3z = 3 \neq 2$. So no point is on all three: Statement-2 is true. Answer: (D).
13	Matrices and DeterminantsConsistency of linear systems	Class 12: Matrices; Class 12: Determinants	Assertion and reasonMedium
A medium assertion and reason question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Coefficient determinant: $D = \begin{vmatrix}1 & -2 & 3\\ -1 & 1 & -2\\ 1 & -3 & 4\end{vmatrix} = 1(4 - 6) + 2(-4 + 2) + 3(3 - 1) = -2 - 4 + 6 = 0$. Indeed row 3 $= 2\times$row 1 $+$ row 2 on the left sides; the right sides then need $2(-1) + k = 1$, i.e. $k = 3$. For $k \neq 3$ the equations are inconsistent: Statement-1 is true. Statement-2: $\begin{vmatrix}1 & 3 & -1\\ -1 & -2 & k\\ 1 & 4 & 1\end{vmatrix} = 1(-2 - 4k) - 3(-1 - k) - 1(-4 + 2) = -2 - 4k + 3 + 3k + 2 = 3 - k$, non-zero for $k \neq 3$: true. This determinant has the x-coefficients, the z-coefficients and the constants as its columns, so up to sign it is $D_y$ of Cramer's rule. With $D = 0$ and $D_y \neq 0$ the system has no solution; Statement-2 is the reason for Statement-1. Answer: (A).
14	ProbabilityProbability of an event	Class 11: Probability; Class 12: Probability	Assertion and reasonMedium
A medium assertion and reason question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are $2^4 = 16$ equally likely choices of $(a, b, c, d)$. A unique solution needs $ad - bc \neq 0$. $ad$ and $bc$ are each 0 or 1; $ad = 1$ in 1 way out of 4, $bc = 1$ in 1 way out of 4. $ad - bc \neq 0$ when ($ad = 1$, $bc = 0$): $1\times3 = 3$ ways, or ($ad = 0$, $bc = 1$): $3\times1 = 3$ ways. Total 6. Probability $= \frac{6}{16} = \frac{3}{8}$: Statement-1 is true. The system is homogeneous, so $x = y = 0$ always solves it: Statement-2 is true. But Statement-2 says nothing about uniqueness, so it does not explain Statement-1. Answer: (B).
15	CirclesStandard and general form	Class 11: Conic Sections	ComprehensionEasy
A easy comprehension question on standard and general form, from Circles. NCERT: Class 11: Conic Sections The centre lies on the normal to PQ at D, at distance 1 (the radius). The unit normal to $\sqrt{3}x + y - 6 = 0$ is $\frac{1}{2}(\sqrt{3}, 1)$. At the origin $\sqrt{3}x + y - 6 = -6 < 0$, so the origin's side is reached by moving from D along $-\frac{1}{2}(\sqrt{3}, 1)$: centre $= \left(\frac{3\sqrt{3}}{2} - \frac{\sqrt{3}}{2}, \frac{3}{2} - \frac{1}{2}\right) = (\sqrt{3}, 1)$. Circle: $(x - \sqrt{3})^2 + (y - 1)^2 = 1$. Answer: (D).
16	CirclesRadius and centre	Class 11: Conic Sections	ComprehensionMedium
A medium comprehension question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Centre $O' = (\sqrt{3}, 1)$, radius 1. In an equilateral triangle the three points of contact are $120^\circ$ apart on the incircle. $D - O' = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, at angle $30^\circ$. The other two are at $150^\circ$ and $270^\circ$: at $150^\circ$: $O' + \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) = \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$; at $270^\circ$: $O' + (0, -1) = (\sqrt{3}, 0)$. Only option (A) has these two points (E on QR and F on RP, consistent with Q.17: $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$ lies on $y = \sqrt{3}x$ and $(\sqrt{3}, 0)$ on $y = 0$). Answer: (A).
17	Straight LinesVarious forms of the equation of a line	Class 11: Straight Lines	ComprehensionMedium
A medium comprehension question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines Each side is the tangent to C at its point of contact, perpendicular to the radius there. At $E\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$ the radius direction is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, so the tangent is $-\frac{\sqrt{3}}{2}\left(x - \frac{\sqrt{3}}{2}\right) + \frac{1}{2}\left(y - \frac{3}{2}\right) = 0$, i.e. $-\sqrt{3}x + \frac{3}{2} + y - \frac{3}{2} = 0$: $y = \sqrt{3}x$ (side QR). At $F(\sqrt{3}, 0)$ the radius is vertical, so the tangent is $y = 0$ (side RP). Check: the distance from $(\sqrt{3}, 1)$ to $\sqrt{3}x - y = 0$ is $\frac{\|3 - 1\|}{2} = 1$; the two sides meet at the origin R, and the angle between them is $60^\circ$, as it must be. Answer: (D).
18	Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	ComprehensionMedium
A medium comprehension question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Differentiate $y^3 - 3y + x = 0$: $3y^2y' - 3y' + 1 = 0$, so $y' = -\frac{1}{3(y^2 - 1)}$. Differentiate again: $y'' = \frac{1}{3}\cdot\frac{2yy'}{(y^2 - 1)^2} = \frac{2y}{3(y^2-1)^2}\cdot\left(-\frac{1}{3(y^2-1)}\right) = -\frac{2y}{9(y^2 - 1)^3}$. At $y = 2\sqrt{2}$ (check: $16\sqrt{2} - 6\sqrt{2} - 10\sqrt{2} = 0$): $y^2 - 1 = 7$. $y'' = -\frac{4\sqrt{2}}{9\cdot 7^3} = -\frac{4\sqrt{2}}{7^3 3^2}$. Answer: (B).
19	Integral CalculusArea under simple curves	Class 12: Integrals; Class 12: Application of Integrals	ComprehensionMedium
A medium comprehension question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals For $x < -2$: $y^3 - 3y = -x > 2$, and $y^3 - 3y - 2 = (y - 2)(y + 1)^2 > 0$ needs $y > 2$. So $f(x) > 0$ there and the area is $\int_a^b f(x)\,dx$. By parts: $\int_a^b f(x)\,dx = \left[xf(x)\right]_a^b - \int_a^b xf'(x)\,dx$. From Q.18, $f'(x) = -\frac{1}{3((f(x))^2 - 1)}$, so Area $= bf(b) - af(a) + \int_a^b \frac{x}{3\left((f(x))^2 - 1\right)}dx$. Answer: (A).
20	Integral CalculusFundamental theorem of calculus	Class 12: Integrals; Class 12: Application of Integrals	ComprehensionMedium
A medium comprehension question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\int_{-1}^{1} g'(x)\,dx = g(1) - g(-1)$. If $y$ satisfies $y^3 - 3y + x = 0$, then $-y$ satisfies $(-y)^3 - 3(-y) + (-x) = 0$. So $x \mapsto -g(-x)$ is also a differentiable solution on $(-2, 2)$ with value 0 at 0; by uniqueness $-g(-x) = g(x)$, so $g$ is odd and $g(-1) = -g(1)$. Hence the integral is $g(1) + g(1) = 2g(1)$. Answer: (D).
21	Complex NumbersArgand diagram	Class 11: Complex Numbers and Quadratic Equations	ComprehensionMedium
A medium comprehension question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x + iy$: A is the half-plane $y \ge 1$; B is the circle with centre $(2, 1)$ and radius 3; and $(1 - i)(x + iy) = (x + y) + i(y - x)$, so C is the line $x + y = \sqrt{2}$. Distance from $(2, 1)$ to the line: $\frac{\|3 - \sqrt{2}\|}{\sqrt{2}} \approx 1.12 < 3$, so the line cuts the circle at two points. The foot of the perpendicular from the centre is $\left(\frac{1+\sqrt{2}}{2}, \frac{\sqrt{2}-1}{2}\right) \approx (1.21, 0.21)$, and the half-chord is $\sqrt{9 - \frac{(3-\sqrt{2})^2}{2}} = \sqrt{\frac{7 + 6\sqrt{2}}{2}} \approx 2.78$. Along the line, $y$ changes by $\pm\frac{2.78}{\sqrt{2}} \approx \pm1.97$, so the two points have $y \approx 2.18$ and $y \approx -1.76$. Only the first has $\mathrm{Im}\,z \ge 1$. One element. Answer: (B).
22	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	ComprehensionMedium
A medium comprehension question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations The points $-1 + i$ and $5 + i$, i.e. $(-1, 1)$ and $(5, 1)$, are at distance 3 on either side of the centre $(2, 1)$: they are the ends of a diameter of the circle B. z lies on B, so the angle subtended at z by this diameter is $90^\circ$, and by Pythagoras $\|z + 1 - i\|^2 + \|z - 5 - i\|^2 = (\text{diameter})^2 = 6^2 = 36$. 36 lies between 35 and 39. Answer: (C).
23	Complex NumbersModulus and argument	Class 11: Complex Numbers and Quadratic Equations	ComprehensionHard
A hard comprehension question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations From Q.21, z is the single point with $y = \frac{\sqrt{2}-1}{2} + \frac{1}{2}\sqrt{7 + 6\sqrt{2}} \approx 2.18$ and $x = \sqrt{2} - y \approx -0.76$, so $\|z\| \approx \sqrt{0.58 + 4.73} \approx 2.30$. w is inside the circle with centre $2 + i$ (modulus $\sqrt{5} \approx 2.24$) and radius 3. The origin is inside this disc, so $\|w\|$ takes every value in $[0, \sqrt{5} + 3) \approx [0, 5.24)$. Then $\|z\| - \|w\| + 3$ ranges over $(2.30 - 5.24 + 3,\; 2.30 + 3] \approx (0.07, 5.30]$. These values lie between $-3$ and 6, between $-6$ and 6, and between $-3$ and 9, but not between $-6$ and 3 (values above 3 occur). The official key accepts (B), (C) or (D). Answer: (B) (also (C) or (D)).

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
24	Physical World and MeasurementError analysis	Class 11: Units and Measurement	Single correctMedium
A medium single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $g = \frac{4\pi^2 L}{T^2}$ with $T = \frac{t}{n}$, so $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta t}{t}$ ($n$ is exact). With $\Delta L = 0.1$ cm and $\Delta t = 0.1$ s: I: $\frac{0.1}{64} + 2\cdot\frac{0.1}{128} = 0.00156 + 0.00156 = 0.0031$, $E_I \approx 0.31\%$. II: $\frac{0.1}{64} + 2\cdot\frac{0.1}{64} = 0.0047$, $E_{II} \approx 0.47\%$. III: $\frac{0.1}{20} + 2\cdot\frac{0.1}{36} = 0.0050 + 0.0056 = 0.0106$, $E_{III} \approx 1.06\%$. So $E_I$ is the minimum (not zero), $E_I \neq E_{II}$, and $E_{III}$, not $E_{II}$, is the maximum. Answer: (B).
25	Current ElectricityWheatstone bridge	Class 12: Current Electricity	Single correctMedium
A medium single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity All resistors are 1 Ω and $P = \frac{V^2}{R_{eq}} = \frac{9}{R_{eq}}$. R1: call the battery terminals T (top) and B (bottom) and the two junctions on the central column X (upper) and Y (lower). The resistors are T–X, X–Y, Y–B on the column, T–Y on the right and X–B on the left. This is a Wheatstone bridge (T–X–B and T–Y–B, each arm 1 Ω) with X–Y as the galvanometer arm; it is balanced, so no current flows in X–Y and $R_{eq} = (1 + 1)\parallel(1 + 1) = 1$ Ω. P1 = 9 W. R2: the top and bottom nodes are joined directly to the battery. Between them: the middle 1 Ω, the left pair in series (2 Ω) and the right pair in series (2 Ω), all in parallel. $\frac{1}{R_{eq}} = 1 + \frac{1}{2} + \frac{1}{2} = 2$, $R_{eq} = 0.5$ Ω. P2 = 18 W. R3: the diamond has two paths of 2 Ω each between top and bottom, giving 1 Ω, in series with the lower 1 Ω: $R_{eq} = 2$ Ω. P3 = 4.5 W. P2 > P1 > P3. Answer: (C).
26	Atoms and NucleiHydrogen spectrum	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (A) Moseley's law: $\sqrt{\nu} \propto (Z - b)$, so the characteristic frequency rises and the wavelength falls with Z. Correct statement. (B) The cut-off wavelength is $\lambda_{min} = \frac{hc}{eV}$, set only by the accelerating voltage, not by the target. This statement is WRONG. (C) More power (more electrons per second) gives more X-ray photons, so a higher intensity. Correct statement. (D) $\lambda_{min} = \frac{hc}{eV}$ depends on the electrons' energy $eV$. Correct statement. Answer: (B).
27	OpticsTotal internal reflection and prisms	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At minimum deviation the ray passes symmetrically, so the two refraction angles inside the prism are equal and add up to the prism angle: $r_1 = r_2 = \frac{A}{2} = 30^\circ$. This holds for any colour; only the angle of incidence (and the deviation) differ with the refractive index. Answer: (A).
28	Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation	Class 11: Kinetic Theory	Single correctEasy
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory From $PV = nRT$, $P \propto \frac{T}{V}$. Then $PT^2 = $ constant gives $\frac{T^3}{V} = $ constant, so $V = kT^3$. Coefficient of volume expansion: $\gamma = \frac{1}{V}\frac{dV}{dT} = \frac{3kT^2}{kT^3} = \frac{3}{T}$. Answer: (C).
29	GravitationOrbital motion	Class 11: Gravitation	Single correctEasy
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation For a circular orbit $\frac{V^2}{r} = g(r)$, the field at distance r. Inside ($r \le R$): only the mass within r pulls, $M(r) = \frac{4}{3}\pi r^3\rho_0$, so $g = \frac{GM(r)}{r^2} = \frac{4}{3}\pi G\rho_0 r$ and $V = \sqrt{gr} = r\sqrt{\frac{4}{3}\pi G\rho_0}$: V grows linearly from 0. Outside ($r > R$): $g = \frac{GM}{r^2}$, so $V = \sqrt{\frac{GM}{r}} \propto \frac{1}{\sqrt{r}}$, falling off. V rises linearly to a peak at R and then decreases: graph (C). Answer: (C).
30	Laws of MotionConservation of linear momentum	Class 11: Laws of Motion	Multiple correctMedium
A medium multiple correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion The total momentum before is $p\hat{i} - p\hat{i} = \vec{0}$, so after the collision $\vec{p}_1' + \vec{p}_2' = \vec{0}$, component by component. (A) The $\hat{k}$ component of the sum is $c_1 + 0 = c_1 \neq 0$. Not allowed. (B) Sum $= (c_1 + c_2)\hat{k}$, zero if $c_2 = -c_1$. Allowed. (C) Sum $= (a_1 + a_2)\hat{i} + (b_1 + b_2)\hat{j} + 0\hat{k}$, zero if $a_2 = -a_1$, $b_2 = -b_1$. Allowed. (D) The $\hat{j}$ component is $b_1 + b_1 = 2b_1 \neq 0$. Not allowed. Answer: (A), (D).
31	Atoms and NucleiNuclear fission and fusion	Class 12: Atoms; Class 12: Nuclei	Multiple correctMedium
A medium multiple correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei From the plot: B/A = 2 MeV for $A < 100$, 8 MeV for $100 < A < 200$, 4 MeV for $A > 200$. Energy is released when the total binding energy increases. (A) Two nuclei with $A < 50$ fuse to a nucleus with $A < 100$: B/A stays 2 MeV, total binding energy unchanged, no energy released. False. (B) Two nuclei with $51 < A < 100$ fuse to $102 < A < 200$: B/A rises from 2 to 8 MeV, energy released. True. (C) A nucleus with $100 < A < 200$ splits into halves with $50 < A < 100$: B/A falls from 8 to 2 MeV, energy absorbed. False. (D) A nucleus with $200 < A < 260$ splits into halves with $100 < A < 130$: B/A rises from 4 to 8 MeV, energy released. True. Answer: (B), (D).
32	Magnetic Effects of Current and MagnetismForce on moving charges and currents	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Multiple correctMedium
A medium multiple correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter In the field the particle moves on a circle of radius $r = \frac{mV}{qB}$, entering normal to the boundary, so the centre of the circle lies on the boundary line. (A), (B) It reaches Region III only if $r > \ell$, i.e. $V > \frac{q\ell B}{m}$. (A) true, (B) false. (C) If $r \le \ell$ it turns back along a semicircle of length $\pi r$, largest when $r = \ell$: $\pi\ell$. If $r > \ell$ it crosses on an arc $r\theta$ with $\sin\theta = \frac{\ell}{r}$; this is at most $\frac{\pi}{2}\ell$ (just above $r = \ell$) and less for faster particles. So the longest path, $\pi\ell$, is at $V = \frac{q\ell B}{m}$. True. (D) A returning particle always describes a semicircle, taking half a period: $t = \frac{\pi m}{qB}$, independent of V. True. Answer: (A), (C), (D).
33	OpticsInterference and diffraction	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Multiple correctMedium
A medium multiple correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Maxima occur at $d\sin\theta = n\lambda$, which needs $\|n\| \le \frac{d}{\lambda}$ with $\|\sin\theta\| < 1$ for a point on the screen. (A) $d = \lambda$: $n = 0$ gives the central maximum; $n = 1$ needs $\sin\theta = 1$ ($\theta = 90^\circ$), not on the screen. Only one maximum. True. (B) $\lambda < d < 2\lambda$: $n = \pm1$ gives $\sin\theta = \frac{\lambda}{d} < 1$, so there are more maxima besides the central one. True. With $I_1 = 4I_0$, $I_2 = I_0$: $I_{max} = (\sqrt{4I_0} + \sqrt{I_0})^2 = 9I_0$, $I_{min} = (2 - 1)^2I_0 = I_0$. (C) $I_1$ reduced to $I_0$: $I_{max} = 4I_0$, $I_{min} = 0$. Both decrease. False. (D) $I_2$ raised to $4I_0$: $I_{max} = 16I_0$ rises but $I_{min} = 0$ falls. False. Answer: (A), (B).
34	Current ElectricityMetre bridge	Class 12: Current Electricity	Assertion and reasonEasy
A easy assertion and reason question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity At balance, $\frac{X}{R} = \frac{l}{100 - l}$, where X is the unknown and R the standard resistance. Statement-2: a metal's resistance rises with temperature. True. Heating the (metal) unknown raises X. To keep the same null point the ratio $\frac{X}{R}$ must stay the same, so R must be increased, not decreased. Statement-1 is false. Answer: (D).
35	GravitationOrbital motion	Class 11: Gravitation	Assertion and reasonEasy
A easy assertion and reason question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation The space station and the astronaut both move under gravity alone: both are in free fall with the same acceleration $g$ at that height. So the floor need not push on the astronaut, the normal force is zero, and he feels weightless. Both statements are true, and the free fall of Statement-2 is the reason for the weightlessness in Statement-1. Answer: (A).
36	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	Assertion and reasonEasy
A easy assertion and reason question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rolling without slipping down an incline: $a = \frac{g\sin\theta}{1 + \frac{I}{MR^2}}$. Hollow cylinder: $I = MR^2$, $a = \frac{g\sin\theta}{2}$. Solid cylinder: $I = \frac{1}{2}MR^2$, $a = \frac{2}{3}g\sin\theta$. The solid cylinder has the larger acceleration and reaches the bottom first. Statement-1 is false. Statement-2: friction does no work in rolling without slipping, so each cylinder's total kinetic energy at the bottom equals $Mgh$; with equal masses and heights they are identical. True (only the split between translation and rotation differs). Answer: (D).
37	Properties of Bulk MatterFluid mechanics	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Assertion and reasonEasy
A easy assertion and reason question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Statement-2 is the equation of continuity, $Av = $ constant: true. Going up, the water slows down under gravity, so its cross-section A must grow: the stream spreads. Going down, it speeds up, so A shrinks: the stream narrows. Statement-1 is true, and it follows from Statement-2. Answer: (A).
38	Properties of Bulk MatterViscosity	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	ComprehensionEasy
A easy comprehension question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The buoyancy force is itself the net force of the liquid's pressure on the bubble's surface, so the "force due to the pressure of the liquid" is not an extra force besides the buoyancy. The other forces on a bubble moving through a real liquid are its weight (gravity) and the viscous drag opposing its motion. So, besides buoyancy: gravity and viscosity. Answer: (D).
39	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	ComprehensionMedium
A medium comprehension question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics No heat is exchanged, so the gas changes adiabatically: $P^{1-\gamma}T^\gamma = $ constant, i.e. $T \propto P^{\frac{\gamma - 1}{\gamma}} = P^{2/5}$ for $\gamma = \frac{5}{3}$. Pressure at the bottom: $P_0 + \rho_\ell gH$. At height y (depth $H - y$): $P_0 + \rho_\ell g(H - y)$. $T = T_0\left(\frac{P_0 + \rho_\ell g(H - y)}{P_0 + \rho_\ell gH}\right)^{2/5}$. Answer: (B).
40	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	ComprehensionMedium
A medium comprehension question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Buoyancy $= \rho_\ell Vg$, with the bubble's volume from $PV = nRT$: $V = \frac{nRT}{P}$. At height y, $P = P_0 + \rho_\ell g(H - y)$ and, from Q.39, $T = T_0\frac{P^{2/5}}{(P_0 + \rho_\ell gH)^{2/5}}$. $V = \frac{nRT_0}{(P_0 + \rho_\ell gH)^{2/5}}P^{2/5 - 1} = \frac{nRT_0}{(P_0 + \rho_\ell gH)^{2/5}P^{3/5}}$. Buoyancy $= \frac{\rho_\ell nRgT_0}{(P_0 + \rho_\ell gH)^{2/5}[P_0 + \rho_\ell g(H - y)]^{3/5}}$. Answer: (B).
41	Atoms and NucleiBohr model	Class 12: Atoms; Class 12: Nuclei	ComprehensionMedium
A medium comprehension question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Bohr energies: $E_n = -13.6\frac{Z^2}{n^2}$ eV. Excitation energy of H to $n = 2$: $13.6\left(1 - \frac{1}{4}\right) = 10.2$ eV. He$^+$ ($Z = 2$) in $n = 2$ has $E_2 = -\frac{54.4}{4} = -13.6$ eV. Adding 10.2 eV gives $-3.4$ eV. $-\frac{54.4}{n^2} = -3.4 \Rightarrow n^2 = 16$, $n = 4$. Answer: (C).
42	Atoms and NucleiHydrogen spectrum	Class 12: Atoms; Class 12: Nuclei	ComprehensionMedium
A medium comprehension question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei From $n = 4$ the He$^+$ ion can emit $4\to3$, $4\to2$, $4\to1$, $3\to2$, $3\to1$, $2\to1$. Energies ($54.4\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ eV): $4\to3$: $54.4\cdot\frac{7}{144} = 2.64$ eV; $3\to2$: $54.4\cdot\frac{5}{36} = 7.56$ eV; $4\to2$: 10.2 eV; the rest are larger. Only $4\to3$ lies in the visible (1.7 to 3.1 eV). $\frac{1}{\lambda} = RZ^2\left(\frac{1}{9} - \frac{1}{16}\right) = 4R\cdot\frac{7}{144} = \frac{7R}{36}$, so $\lambda = \frac{36}{7\times1.097\times10^7} \approx 4.7\times10^{-7}$ m. The nearest option is $4.8\times10^{-7}$ m. Answer: (C).
43	Atoms and NucleiBohr model	Class 12: Atoms; Class 12: Nuclei	ComprehensionEasy
A easy comprehension question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei In the Bohr model the kinetic energy equals $-E_n = 13.6\frac{Z^2}{n^2}$ eV. For $n = 2$: H ($Z = 1$) has 3.4 eV; He$^+$ ($Z = 2$) has 13.6 eV. Ratio $= \frac{1}{4}$. Answer: (A).
44	Work, Energy and PowerCollisions	Class 11: Work, Energy and Power	ComprehensionMedium
A medium comprehension question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power A to B: the incline is at $60^\circ$ to the horizontal with a horizontal run of $\sqrt{3}$ m, so the drop is $\sqrt{3}\tan60^\circ = 3$ m. Speed at B: $v = \sqrt{2gh} = \sqrt{60}$ m/s, directed along AB, $60^\circ$ below the horizontal. The second incline is $30^\circ$ below the horizontal, so the velocity makes $30^\circ$ with it. In a totally inelastic collision the component normal to the incline is destroyed and the component along it remains: $v' = \sqrt{60}\cos30^\circ = \sqrt{60}\cdot\frac{\sqrt{3}}{2} = \sqrt{45}$ m/s. Answer: (B).
45	Work, Energy and PowerConservation of energy	Class 11: Work, Energy and Power	ComprehensionMedium
A medium comprehension question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power B to C: horizontal run $3\sqrt{3}$ m at $30^\circ$, so the drop is $3\sqrt{3}\tan30^\circ = 3$ m. The surface is frictionless, so $v_C^2 = v_B'^2 + 2gh = 45 + 2\times10\times3 = 105$. $v_C = \sqrt{105}$ m/s. Answer: (B).
46	Work, Energy and PowerCollisions	Class 11: Work, Energy and Power	ComprehensionHard
A hard comprehension question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Just before B: $\vec{v} = \sqrt{60}(\cos60^\circ, -\sin60^\circ) = (\sqrt{15}, -\sqrt{45})$. Along the second incline: $\hat{t} = (\cos30^\circ, -\sin30^\circ)$; normal (out of the surface): $\hat{n} = (\sin30^\circ, \cos30^\circ)$. $v_t = \sqrt{15}\cdot\frac{\sqrt{3}}{2} + \sqrt{45}\cdot\frac{1}{2} = \sqrt{45}$; $v_n = \sqrt{15}\cdot\frac{1}{2} - \sqrt{45}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{15}}{2} - \frac{3\sqrt{15}}{2} = -\sqrt{15}$. An elastic collision reverses the normal component: after it, $v_n = +\sqrt{15}$, $v_t = \sqrt{45}$. Vertical component: $v_t(-\sin30^\circ) + v_n\cos30^\circ = -\frac{\sqrt{45}}{2} + \frac{\sqrt{15}\sqrt{3}}{2} = -\frac{\sqrt{45}}{2} + \frac{\sqrt{45}}{2} = 0$. (The block leaves B moving horizontally.) Answer: (C).

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
47	Organic Chemistry: Some Basic Principles and TechniquesElectronic effects	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctEasy
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Hyperconjugation is the delocalisation of the electrons of a C–H (or C–C) $\sigma$ bond on the carbon next to a carbocation, radical or double bond into the adjacent empty or partly filled p orbital (or $\pi$ system). The overlap is between a $\sigma$ bonding orbital and a p orbital: $\sigma - p$. Answer: (B).
48	Haloalkanes and HaloarenesSN1 and SN2 mechanisms	Class 12: Haloalkanes and Haloarenes	Single correctMedium
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate has two leaving groups: a secondary benzylic bromide (on a stereocentre) and an aryl fluoride. Aryl substitution (SNAr) needs an electron-withdrawing group ortho or para to the halogen. Here the NO2 group is para to the side chain and meta to F, so the ring fluorine is not activated and stays. PhS$^-$ is an excellent soft nucleophile and DMF is a polar aprotic solvent: the benzylic bromide undergoes SN2, with backside attack and inversion of configuration. Br on a wedge becomes SPh on a hashed bond. Product: (A). Answer: (A).
49	p-Block ElementsGroup 16 elements	No matching NCERT chapter	Single correctMedium
A medium single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Chlorine is a strong oxidant and takes thiosulphate (S average +2) all the way to sulphate (S +6), unlike iodine, which stops at tetrathionate: $\mathrm{Na_2S_2O_3 + 4Cl_2 + 5H_2O \rightarrow 2NaHSO_4 + 8HCl}$. The sodium-containing product is $\mathrm{NaHSO_4}$. (This is why thiosulphate is used as an "antichlor".) Answer: (B).
50	Isolation of MetalsOres and concentration	No matching NCERT chapter	Single correctEasy
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter In the cyanide process the metal is oxidised by air while cyanide complexes the ion: $\mathrm{4Ag + 8NaCN + O_2 + 2H_2O \rightarrow 4Na[Ag(CN)_2] + 4NaOH}$. Oxygen is the oxidising agent; without it silver does not dissolve. Answer: (B).
51	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	Single correctEasy
A easy single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics First order: $t_{1/2} = \frac{0.693}{k_1}$, so $k_1 = \frac{0.693}{40}$ s$^{-1}$. Zero order: $t_{1/2} = \frac{[A]_0}{2k_0}$, so $k_0 = \frac{1.386}{2\times20} = \frac{1.386}{40}$ mol dm$^{-3}$ s$^{-1}$. $\frac{k_1}{k_0} = \frac{0.693}{1.386} = 0.5$ mol$^{-1}$ dm$^3$. Answer: (A).
52	EquilibriumIonic equilibrium	Class 11: Equilibrium	Single correctHard
A hard single correct question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Base: $2.5\times0.4 = 1$ mmol. HCl needed: $\frac{1}{2/15} = 7.5$ mL. Total volume 10 mL, so at equivalence $[\mathrm{BH^+}] = \frac{1}{10} = 0.1$ M. $\mathrm{BH^+} \rightleftharpoons \mathrm{B + H^+}$ with $K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-12}} = 10^{-2}$. $K_a$ is large compared with $c$, so the usual approximation fails; solve exactly: $\frac{x^2}{0.1 - x} = 0.01$, $x^2 + 0.01x - 0.001 = 0$. $x = \frac{-0.01 + \sqrt{0.0001 + 0.004}}{2} = \frac{-0.01 + 0.0640}{2} = 0.0270$ M. $[\mathrm{H^+}] \approx 2.7\times10^{-2}$ M. (The approximate $\sqrt{K_ac} = 3.2\times10^{-2}$ M is option C, the trap.) Answer: (D).
53	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Multiple correctMedium
A medium multiple correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The compound is 2,3-dichlorobutane, $\mathrm{CH_3CHClCHClCH_3}$, drawn in a zig-zag with the H atoms hashed. C2: priorities Cl > C3 > CH3 > H; with H away from the viewer, Cl → C3 → CH3 runs clockwise: R. C3: Cl > C2 > CH3 > H, H hashed, Cl → C2 → CH3 also clockwise: R. So it is (2R,3R)-2,3-dichlorobutane. The two stereocentres have the same descriptor, so this is not the meso form: it is chiral and optically active. (A) true. A chiral molecule can have neither a plane nor a centre of symmetry. (B), (C) false. Rotating by $180^\circ$ about the axis through the midpoint of the C2–C3 bond (perpendicular to it) carries each CHCl(CH3) end into the other: it has a $C_2$ axis. (D) true. Answer: (A), (D).
54	Organic Chemistry: Some Basic Principles and TechniquesIsomerism	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Multiple correctMedium
A medium multiple correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques E is 3-methylpentan-2-one, $\mathrm{CH_3CH_2CH(CH_3)COCH_3}$. F and G are $\mathrm{CH_3CH_2C(CH_3){=}C(OH)CH_3}$, its enol, with OH and CH3 on the other end swapped between them. (A) Resonance structures differ only in the placing of electrons; E and its enols differ in the position of an H atom. False. (B) The ketone E and each enol interconvert by moving an H atom and a double bond: E, F and E, G are keto–enol tautomers. True. (C) F and G have the same connectivity and differ only in the arrangement about the C=C (each alkene carbon has two different groups): geometrical (E/Z) isomers. True. (D) Geometrical isomers are stereoisomers that are not mirror images: diastereomers. True. Answer: (B), (C), (D).
55	Principles of Qualitative AnalysisAnalysis of anions	No matching NCERT chapter	Multiple correctMedium
A medium multiple correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Boiling any ammonium salt with excess NaOH drives off $\mathrm{NH_3}$ (non-flammable): $\mathrm{NH_4^+ + OH^- \rightarrow NH_3 + H_2O}$. It stops when the ammonium ion is used up. Zn dust in alkali produces nascent hydrogen, which reduces nitrate and nitrite to ammonia, so $\mathrm{NH_3}$ is given off again: $\mathrm{NO_3^- + 4Zn + 7OH^- \rightarrow NH_3 + 4ZnO_2^{2-} + 2H_2O}$, and similarly for $\mathrm{NO_2^-}$. So the anion must be nitrate or nitrite: $\mathrm{NH_4NO_3}$ and $\mathrm{NH_4NO_2}$. Chloride and sulphate are not reduced to a gas. Answer: (A), (B).
56	ThermodynamicsFirst law and state functions	Class 11: Thermodynamics	Multiple correctMedium
A medium multiple correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT$. (A) For large $V_m$, $\frac{a}{V_m^2} \to 0$ and $b \ll V_m$, so it reduces to $PV_m = RT$. True. (B) At large pressures $V_m$ is small, and the $b$ term dominates: the gas deviates most. False. (C) $a$ and $b$ are constants characteristic of each gas, taken as independent of temperature. True. (D) $P = \frac{RT}{V_m - b} - \frac{a}{V_m^2}$: the attraction lowers P, but the excluded volume raises it, so P can be above or below the ideal value (it is above at high temperature or high pressure). Not true in general. The official key accepts A, C or A, C, D (the latter reading (D) as the effect of attraction alone). Answer: (A), (C).
57	HydrocarbonsElectrophilic substitution and directive influence	Class 11: Hydrocarbons	Assertion and reasonEasy
A easy assertion and reason question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Br is ortho/para directing (its lone pair stabilises the intermediates for ortho and para attack by resonance), and para attack is favoured sterically: 1,4-dibromobenzene is the major product. Statement-1 is true. The -I effect of Br governs the rate (the ring is deactivated), but the direction of attack is set by the +M effect. Statement-2 says the inductive effect dominates in directing the electrophile, which is false (if it did, Br would be meta directing). Answer: (C).
58	p-Block ElementsGroup 14 elements	No matching NCERT chapter	Assertion and reasonEasy
A easy assertion and reason question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down group 14 the ns$^2$ pair becomes reluctant to take part in bonding (inert pair effect), so the +2 state becomes more stable and the +4 state less stable for the heavier elements. $\mathrm{Pb^{4+}}$ readily goes to the stable $\mathrm{Pb^{2+}}$ and is a strong oxidant, while $\mathrm{Sn^{4+}}$ is the stable state of tin. Statement-1 is true. Statement-2 has it backwards: it is the lower (+2) state that is more stable for the heavier members. False. Answer: (C).
59	Chemical KineticsIntegrated rate laws and half-life	Class 12: Chemical Kinetics	Assertion and reasonMedium
A medium assertion and reason question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For light stable nuclei $N \approx Z$ (the $45^\circ$ line). For heavier ones the stable nuclei have more neutrons than protons ($N/Z$ up to about 1.5), so on a plot of Z against N the band bends below the $45^\circ$ line, towards the N (x) axis. Statement-1 is true. The reason is that the proton-proton Coulomb repulsion grows roughly as $Z^2$ while the short-range nuclear attraction does not; extra neutrons add attraction without repulsion. Statement-2 is true and explains Statement-1. Answer: (A).
60	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Assertion and reasonEasy
A easy assertion and reason question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics At equilibrium it is $\Delta_r G$ that is zero; the standard Gibbs energy is $\Delta_r G^\circ = -RT\ln K$, which is zero only when $K = 1$. Statement-1 is false. Statement-2 is the Gibbs energy criterion of spontaneity at constant T and P: true. Answer: (D).
61	Alcohols, Phenols and EthersPreparation and properties	Class 12: Alcohols, Phenols and Ethers	ComprehensionMedium
A medium comprehension question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Hex-3-ynal is $\mathrm{CH_3CH_2C{\equiv}CCH_2CHO}$. NaBH$_4$ reduces only the aldehyde (it does not touch the C≡C): hex-3-yn-1-ol, $\mathrm{CH_3CH_2C{\equiv}CCH_2CH_2OH}$. PBr$_3$ replaces OH by Br: $\mathbf{I}$ = 1-bromohex-3-yne, $\mathrm{CH_3CH_2C{\equiv}CCH_2CH_2Br}$, still with the triple bond (the acid chloride later in the scheme also has it). That is structure (D). Answer: (D).
62	Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids	Class 12: Aldehydes, Ketones and Carboxylic Acids	ComprehensionMedium
A medium comprehension question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids I, $\mathrm{CH_3CH_2C{\equiv}CCH_2CH_2Br}$, forms the Grignard reagent with Mg in ether; carboxylation with CO$_2$ and acid work-up adds one carbon as COOH: $\mathbf{J}$ = hept-4-ynoic acid, $\mathrm{CH_3CH_2C{\equiv}CCH_2CH_2COOH}$ (the triple bond is internal, so it has no acidic H to destroy the Grignard). J becomes the acid chloride shown in the scheme; the usual reagent is thionyl chloride, $\mathbf{K}$ = SOCl$_2$ ($\mathrm{RCOOH + SOCl_2 \rightarrow RCOCl + SO_2 + HCl}$). SO$_2$Cl$_2$ and CH$_3$SO$_2$Cl do not make acid chlorides this way. Answer: (A).
63	Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition	Class 12: Aldehydes, Ketones and Carboxylic Acids	ComprehensionMedium
A medium comprehension question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids H$_2$ over Pd/BaSO$_4$ poisoned with quinoline does two things here: the acid chloride is reduced to the aldehyde (Rosenmund reduction), $\mathrm{RCOCl \rightarrow RCHO}$; the internal alkyne is half-hydrogenated by syn addition on the poisoned catalyst to the cis alkene, which is not reduced further. $\mathbf{L}$ = (Z)-hept-4-enal, $\mathrm{CH_3CH_2CH{=}CHCH_2CH_2CHO}$ with a cis double bond: structure (C). Answer: (C).
64	p-Block ElementsGroup 15 elements	No matching NCERT chapter	ComprehensionEasy
A easy comprehension question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Phosphates are essential in humans (bones and teeth as calcium phosphate, ATP, DNA and RNA). False. (B), (C) Nitrates are highly soluble, so they are leached away by water and are found only in a few dry deposits; phosphates (apatite rocks) are far more abundant. So nitrates are the less abundant. (C) true, (B) false. (D) N in nitrate is already at its highest oxidation state, +5, so nitrate cannot be oxidised further. False. Answer: (C).
65	p-Block ElementsGroup 15 elements	No matching NCERT chapter	ComprehensionMedium
A medium comprehension question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter In $\mathrm{NH_3}$ (H–N–H about $107^\circ$) nitrogen is close to sp$^3$ hybridised, and its lone pair sits in a directional sp$^3$ orbital that overlaps well with an acceptor: a good $\sigma$ donor. In $\mathrm{PH_3}$ (H–P–H about $93^\circ$) the bonds use almost pure p orbitals, leaving the lone pair in an orbital of mostly s character, diffuse and nearly spherical, so it donates poorly. So $\mathrm{NH_3}$ is the better donor because its lone pair is in a more directional sp$^3$ orbital. Answer: (C).
66	p-Block ElementsGroup 15 elements	No matching NCERT chapter	ComprehensionEasy
A easy comprehension question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2}$. Phosphorus goes from 0 in $\mathrm{P_4}$ to $-3$ in $\mathrm{PH_3}$ (reduced) and to $+1$ in $\mathrm{NaH_2PO_2}$ (oxidised). The same element is both oxidised and reduced: disproportionation. Answer: (B).
67	SolutionsColligative properties	Class 12: Solutions	ComprehensionMedium
A medium comprehension question on colligative properties, from Solutions. NCERT: Class 12: Solutions In M ethanol (mole fraction 0.9) is the solvent and water the solute. Take 1 mol of mixture: 0.9 mol ethanol $= 0.9\times46 = 41.4$ g $= 0.0414$ kg, and 0.1 mol water. Molality $m = \frac{0.1}{0.0414} = 2.415$ mol kg$^{-1}$. $\Delta T_f = K_f^{ethanol}m = 2.0\times2.415 = 4.83$ K. Freezing point $= 155.7 - 4.83 = 150.9$ K. Answer: (D).
68	SolutionsRaoult's law	Class 12: Solutions	ComprehensionEasy
A easy comprehension question on raoult's law, from Solutions. NCERT: Class 12: Solutions The solute (water) is to be treated as non-volatile, so only the solvent ethanol contributes (Raoult's law): $P = x_{ethanol}P^\circ_{ethanol} = 0.9\times40 = 36.0$ mm Hg. Answer: (B).
69	SolutionsColligative properties	Class 12: Solutions	ComprehensionMedium
A medium comprehension question on colligative properties, from Solutions. NCERT: Class 12: Solutions Now water (mole fraction 0.9) is the solvent and ethanol (0.1) the solute, treated as non-volatile. In 1 mol of solution: 0.9 mol water $= 16.2$ g $= 0.0162$ kg, and 0.1 mol ethanol. Molality $m = \frac{0.1}{0.0162} = 6.17$ mol kg$^{-1}$. $\Delta T_b = K_b^{water}m = 0.52\times6.17 = 3.21$ K. Boiling point $= 373 + 3.2 = 376.2$ K. Answer: (B).

Mathematics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
1	Complex NumbersArgand diagram	Class 11: Complex Numbers and Quadratic Equations	Single correctEasy
A easy single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Start at $z_0 = 1 + 2i$, the point (1, 2). Moving horizontally away from the origin by 5 units increases the real part: $6 + 2i$. Moving vertically away from the origin by 3 units increases the imaginary part: $z_1 = 6 + 5i$. Moving $\sqrt{2}$ units along $\hat{i} + \hat{j}$ (unit vector $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$) adds 1 to each part: $7 + 6i$. Turning through $\frac{\pi}{2}$ anticlockwise on a circle centred at the origin is multiplication by $e^{i\pi/2} = i$: $z_2 = i(7 + 6i) = 7i + 6i^2 = -6 + 7i$. Answer: (D)
2	Applications of DerivativesIncreasing and decreasing functions	Class 12: Application of Derivatives	Single correctEasy
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Parity: $g(-u) = 2\tan^{-1}(e^{-u}) - \frac{\pi}{2}$. Since $e^{-u} = \frac{1}{e^u}$ and $e^u > 0$, $\tan^{-1}\frac{1}{e^u} = \frac{\pi}{2} - \tan^{-1}(e^u)$. So $g(-u) = \pi - 2\tan^{-1}(e^u) - \frac{\pi}{2} = \frac{\pi}{2} - 2\tan^{-1}(e^u) = -g(u)$. Hence $g$ is odd. Monotonicity: $g'(u) = \frac{2e^u}{1 + e^{2u}} > 0$ for every real $u$, so $g$ is strictly increasing on $(-\infty, \infty)$. Answer: (C)
3	Conic SectionsHyperbola	Class 11: Conic Sections	Single correctMedium
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Complete the squares: $(x - \sqrt{2})^2 - 2 - 2[(y + \sqrt{2})^2 - 2] = 6$, so $(x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 = 4$, that is $\frac{X^2}{4} - \frac{Y^2}{2} = 1$ with $X = x - \sqrt{2}$, $Y = y + \sqrt{2}$. So $a = 2$, $b^2 = 2$, and $e^2 = 1 + \frac{b^2}{a^2} = \frac{3}{2}$, $e = \sqrt{\frac{3}{2}}$. On one branch, take the vertex $A = (a, 0)$, the nearest focus $C = (ae, 0)$ and the end of the latus rectum through $C$, $B = \left(ae, \frac{b^2}{a}\right) = (ae, 1)$ (in the $X$, $Y$ frame; a shift of origin does not change areas). Triangle $ABC$ has a right angle at $C$, with $AC = ae - a = 2\sqrt{\frac{3}{2}} - 2$ and $BC = 1$. Area $= \frac{1}{2}\left(2\sqrt{\frac{3}{2}} - 2\right)(1) = \sqrt{\frac{3}{2}} - 1$. Answer: (B)
4	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 On $0 < x < \frac{\pi}{4}$, $\sin x > 0$, so the first curve lies above the second. The area is $\int_0^{\pi/4} \left(\sqrt{\frac{1 + \sin x}{\cos x}} - \sqrt{\frac{1 - \sin x}{\cos x}}\right) dx$. Put $t = \tan\frac{x}{2}$: $\sin x = \frac{2t}{1 + t^2}$, $\cos x = \frac{1 - t^2}{1 + t^2}$, $dx = \frac{2\,dt}{1 + t^2}$. Then $\frac{1 + \sin x}{\cos x} = \frac{(1 + t)^2}{1 - t^2} = \frac{1 + t}{1 - t}$ and $\frac{1 - \sin x}{\cos x} = \frac{1 - t}{1 + t}$. So the integrand is $\sqrt{\frac{1+t}{1-t}} - \sqrt{\frac{1-t}{1+t}} = \frac{(1 + t) - (1 - t)}{\sqrt{1 - t^2}} = \frac{2t}{\sqrt{1 - t^2}}$. The limits: $x = 0$ gives $t = 0$; $x = \frac{\pi}{4}$ gives $t = \tan\frac{\pi}{8} = \sqrt{2} - 1$. Area $= \int_0^{\sqrt{2}-1} \frac{2t}{\sqrt{1 - t^2}} \cdot \frac{2}{1 + t^2}\,dt = \int_0^{\sqrt{2}-1} \frac{4t}{(1 + t^2)\sqrt{1 - t^2}}\,dt$. Answer: (B)
5	Straight LinesCartesian coordinates and distance formula	Class 11: Straight Lines	Single correctHard
A hard single correct question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines Look at the family of points $X(t) = (\cos(\beta - \alpha + t), \sin(\beta - t))$. $t = 0$ gives $(\cos(\beta - \alpha), \sin\beta) = Q$. $t = \frac{\pi}{2}$ gives $(\cos(\beta - \alpha + \frac{\pi}{2}), \sin(\beta - \frac{\pi}{2})) = (-\sin(\beta - \alpha), -\cos\beta) = P$. $t = \theta$ gives $R$. Write $\sin(\beta - t) = \cos(t + \frac{\pi}{2} - \beta)$. Then $X(t) = (\cos(t + u), \cos(t + v))$ with $u = \beta - \alpha$, $v = \frac{\pi}{2} - \beta$. Such a curve is a straight segment only when $u - v$ is a multiple of $\pi$; otherwise it is an ellipse: eliminating $t$ gives $x^2 - 2xy\cos(u - v) + y^2 = \sin^2(u - v)$, a central conic which is an ellipse because $\cos^2(u - v) < 1$. Here $u - v = 2\beta - \alpha - \frac{\pi}{2}$, and with $0 < \alpha, \beta < \frac{\pi}{4}$ this lies strictly between $-\frac{3\pi}{4}$ and $0$, so it is never a multiple of $\pi$. The curve is a genuine ellipse, traced once as $t$ runs over a period of $2\pi$. $t = 0$, $\theta$ and $\frac{\pi}{2}$ are three different parameter values within one period (since $0 < \theta < \frac{\pi}{4}$), so $Q$, $R$, $P$ are three distinct points of an ellipse. A line meets an ellipse in at most two points, so no three distinct points of an ellipse are collinear. Hence $P$, $Q$, $R$ are non-collinear, and none of them lies on the segment joining the other two. Answer: (D)
6	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 $B$ have $k$ outcomes ($1 \le k \le 10$) and $A \cap B$ have $m$ outcomes. Independence: $P(A \cap B) = P(A)P(B)$, so $\frac{m}{10} = \frac{4}{10} \cdot \frac{k}{10}$, that is $m = \frac{2k}{5}$. $m$ must be a whole number, so $k$ must be a multiple of 5: $k = 5$ or $k = 10$. Both are possible: $k = 5$ needs $m = 2$ (two of $B$'s outcomes inside $A$), and $k = 10$ ($B$ the whole space) gives $m = 4$. Answer: (D)
7	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	Single correctMedium
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\|\vec{OP}\|^2 = \cos^2 t + \sin^2 t + 2\sin t\cos t\,(\hat{a}\cdot\hat{b}) = 1 + (\hat{a}\cdot\hat{b})\sin 2t$. The angle is acute, so $\hat{a}\cdot\hat{b} > 0$, and the maximum is at $\sin 2t = 1$, $t = \frac{\pi}{4}$: $M = (1 + \hat{a}\cdot\hat{b})^{\frac{1}{2}}$. At $t = \frac{\pi}{4}$, $\vec{OP} = \frac{\hat{a} + \hat{b}}{\sqrt{2}}$, which points along $\hat{a} + \hat{b}$, so $\hat{u} = \frac{\hat{a} + \hat{b}}{\|\hat{a} + \hat{b}\|}$. Answer: (A)
8	Integral CalculusIntegration by substitution	Class 12: Integrals; Class 12: Application of Integrals	Single correctMedium
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Multiply the numerator and denominator of $J$'s integrand by $e^{4x}$: $J = \int \frac{e^{3x}}{1 + e^{2x} + e^{4x}}\,dx$. So $J - I = \int \frac{e^{3x} - e^x}{e^{4x} + e^{2x} + 1}\,dx$. Put $u = e^x$, $du = e^x dx$: $J - I = \int \frac{u^2 - 1}{u^4 + u^2 + 1}\,du = \int \frac{1 - \frac{1}{u^2}}{u^2 + 1 + \frac{1}{u^2}}\,du$. Put $v = u + \frac{1}{u}$, $dv = \left(1 - \frac{1}{u^2}\right)du$, and $u^2 + \frac{1}{u^2} + 1 = v^2 - 1$: $J - I = \int \frac{dv}{v^2 - 1} = \frac{1}{2}\log\left\|\frac{v - 1}{v + 1}\right\| + C$. $\frac{v - 1}{v + 1} = \frac{u^2 - u + 1}{u^2 + u + 1}$, which is positive. Hence $J - I = \frac{1}{2}\log\left(\frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1}\right) + C$. Answer: (C)
9	Limits, Continuity and DifferentiabilityDerivatives up to order two	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	Single correctHard
A hard single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Take logs of $f(x + 1) = x f(x)$: $g(x + 1) = \log x + g(x)$. Differentiate twice: $g''(x + 1) = -\frac{1}{x^2} + g''(x)$, so $g''(x + 1) - g''(x) = -\frac{1}{x^2}$. Put $x = \frac{1}{2}, \frac{3}{2}, \ldots, N - \frac{1}{2}$ and add; the left side telescopes: $g''\left(N + \frac{1}{2}\right) - g''\left(\frac{1}{2}\right) = -\left[\frac{1}{(1/2)^2} + \frac{1}{(3/2)^2} + \cdots + \frac{1}{(N - 1/2)^2}\right] = -4\left\{1 + \frac{1}{9} + \frac{1}{25} + \cdots + \frac{1}{(2N - 1)^2}\right\}$. Answer: (A)
10	Sequences and SeriesGeometric progressions	Class 11: Sequences and Series	Assertion and reasonMedium
A medium assertion and reason question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let $a_k = ar^{k-1}$ with $a > 0$, $r > 0$ and $r \ne 1$ (the numbers are distinct). Then $b_1 = a$, $b_2 = a(1 + r)$, $b_3 = a(1 + r + r^2)$, $b_4 = a(1 + r + r^2 + r^3)$. A.P.? $b_2 - b_1 = ar$ and $b_3 - b_2 = ar^2$; equal only if $r = 1$. Not an A.P. G.P.? $b_2^2 = b_1 b_3$ needs $(1 + r)^2 = 1 + r + r^2$, i.e. $r = 0$. Not a G.P. So Statement-1 is true. H.P.? Then $\frac{2}{b_2} = \frac{1}{b_1} + \frac{1}{b_3}$: $\frac{2}{1 + r} = \frac{2 + r + r^2}{1 + r + r^2}$, i.e. $2 + 2r + 2r^2 = 2 + 3r + 2r^2 + r^3$, i.e. $r + r^3 = 0$, impossible for $r > 0$. So the $b$'s are not in H.P., and Statement-2 is false. Answer: (C)
11	Quadratic EquationsNature of roots	Class 11: Complex Numbers and Quadratic Equations	Assertion and reasonHard
A hard assertion and reason question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Statement-1. $p^2 - q$ and $b^2 - ac$ are a quarter of the discriminants of the two equations. If $\alpha$ were non-real, then (real coefficients) $\beta = \bar{\alpha}$, and the second equation, also with real coefficients, would have roots $\alpha$ and $\bar{\alpha}$; so $\frac{1}{\beta} = \bar{\alpha} = \beta$, giving $\beta^2 = 1$, which is excluded. So $\alpha$ is real, hence $\beta$ is real and $\frac{1}{\beta}$ is real: both equations have real roots, both discriminants are $\ge 0$, and the product is $\ge 0$. Statement-1 is true. Statement-2. If $b = pa$ and $c = qa$ (with $a \ne 0$), the second equation is $a(x^2 + 2px + q) = 0$, with the same roots $\{\alpha, \beta\}$. Then $\{\alpha, \frac{1}{\beta}\} = \{\alpha, \beta\}$ forces $\frac{1}{\beta} = \beta$ (or, if $\alpha = \beta$, $\frac{1}{\beta} = \alpha = \beta$), so $\beta^2 = 1$, excluded. Hence $b \ne pa$ or $c \ne qa$: Statement-2 is true. Statement-2 only says the two equations are not the same; it is not what makes the discriminants non-negative, which comes from the reality of the roots. So it does not explain Statement-1. Answer: (B)
12	CirclesIntersection of a line and a circle	Class 11: Conic Sections	Assertion and reasonMedium
A medium assertion and reason question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections $C$ has centre $(-3, 5)$ and radius $\sqrt{9 + 25 - 30} = 2$. $L_1$ is a chord when its distance from the centre is less than 2: $\frac{\|-6 + 15 + p - 3\|}{\sqrt{13}} = \frac{\|p + 6\|}{\sqrt{13}} < 2$, i.e. $-6 - 2\sqrt{13} < p < -6 + 2\sqrt{13}$. $L_2$ is a diameter only when it passes through the centre: $-6 + 15 + p + 3 = 0$, $p = -12$. Statement-1: $L_1$ is a chord for a whole interval of $p$, and $L_2$ is a diameter only for $p = -12$; so $L_2$ is not always a diameter. True. Statement-2: $L_1$ is a diameter when $p + 6 = 0$, $p = -6$. Then $L_2$ is $2x + 3y - 3 = 0$, at distance $\frac{\|-6 + 15 - 3\|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \approx 1.66 < 2$ from the centre, so $L_2$ is a chord. Statement-2 is false. Answer: (C)
13	Differential EquationsSeparation of variables	Class 12: Differential Equations	Assertion and reasonMedium
A medium assertion and reason question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate the variables: $\frac{dy}{y\sqrt{y^2 - 1}} = \frac{dx}{x\sqrt{x^2 - 1}}$, so $\sec^{-1}y = \sec^{-1}x + c$. At $x = 2$: $\sec^{-1}\frac{2}{\sqrt{3}} = \frac{\pi}{6}$ and $\sec^{-1}2 = \frac{\pi}{3}$, so $c = -\frac{\pi}{6}$. Hence $y = \sec\left(\sec^{-1}x - \frac{\pi}{6}\right)$: Statement-1 is true. Then $\frac{1}{y} = \cos\left(\sec^{-1}x - \frac{\pi}{6}\right)$. With $\theta = \sec^{-1}x$ ($x > 1$ near $x = 2$), $\cos\theta = \frac{1}{x}$, $\sin\theta = \sqrt{1 - \frac{1}{x^2}}$: $\frac{1}{y} = \frac{\sqrt{3}}{2}\cdot\frac{1}{x} + \frac{1}{2}\sqrt{1 - \frac{1}{x^2}}$. This is not $\frac{2\sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^2}}$ (at $x = 2$ that expression gives $\sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$, which happens to match, but the two differ elsewhere, e.g. at $x = 1$: $2\sqrt{3}$ against $\frac{\sqrt{3}}{2}$). Statement-2 is false. Answer: (C)
14	Limits, Continuity and DifferentiabilityDerivatives up to order two	Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability	ComprehensionMedium
A medium comprehension question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability With $D(x) = x^2 + ax + 1$ (positive for all $x$, since $a^2 < 4$), the quotient rule gives $f'(x) = \frac{(2x - a)(x^2 + ax + 1) - (x^2 - ax + 1)(2x + a)}{D^2} = \frac{2a(x^2 - 1)}{(x^2 + ax + 1)^2}$. So $f'(1) = f'(-1) = 0$, and $f'(1)f'(-1) = 0$, which is neither $(2 - a)^2$ nor $-(2 + a)^2$: (C) and (D) are false. Differentiate once more. At $x = \pm 1$ the factor $x^2 - 1$ vanishes, so only the derivative of that factor survives: $f''(x) = \frac{4ax}{D(x)^2}$ at $x = \pm 1$. $D(1) = 2 + a$, $D(-1) = 2 - a$, so $f''(1) = \frac{4a}{(2 + a)^2}$ and $f''(-1) = -\frac{4a}{(2 - a)^2}$. $(2 + a)^2 f''(1) + (2 - a)^2 f''(-1) = 4a - 4a = 0$. (A) is true; (B) gives $(2-a)^2\frac{4a}{(2+a)^2} + (2+a)^2\frac{4a}{(2-a)^2} \ne 0$. Answer: (A)
15	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	ComprehensionMedium
A medium comprehension question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives With $D(x) = x^2 + ax + 1$ (positive for all $x$, since $a^2 < 4$), the quotient rule gives $f'(x) = \frac{(2x - a)(x^2 + ax + 1) - (x^2 - ax + 1)(2x + a)}{D^2} = \frac{2a(x^2 - 1)}{(x^2 + ax + 1)^2}$. On $(-1, 1)$, $x^2 - 1 < 0$ and $a > 0$, so $f'(x) < 0$: $f$ is decreasing on $(-1, 1)$. At $x = 1$, $f'$ changes from negative (just left) to positive (just right, $x^2 > 1$), so $f$ has a local minimum at $x = 1$. (Equally, $f''(1) = \frac{4a}{(2 + a)^2} > 0$.) Answer: (A)
16	Integral CalculusFundamental theorem of calculus	Class 12: Integrals; Class 12: Application of Integrals	ComprehensionMedium
A medium comprehension question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals With $D(x) = x^2 + ax + 1$ (positive for all $x$, since $a^2 < 4$), the quotient rule gives $f'(x) = \frac{(2x - a)(x^2 + ax + 1) - (x^2 - ax + 1)(2x + a)}{D^2} = \frac{2a(x^2 - 1)}{(x^2 + ax + 1)^2}$. By the fundamental theorem and the chain rule, $g'(x) = \frac{f'(e^x)}{1 + e^{2x}}\cdot e^x = \frac{2a(e^{2x} - 1)e^x}{(1 + e^{2x})(e^{2x} + ae^x + 1)^2}$. Every factor except $e^{2x} - 1$ is positive. $e^{2x} - 1 < 0$ for $x < 0$ and $> 0$ for $x > 0$. So $g'(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$. Answer: (B)
17	Vector AlgebraScalar and vector products	Class 12: Vector Algebra	ComprehensionEasy
A easy comprehension question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Directions: $\vec{d_1} = (3, 1, 2)$, $\vec{d_2} = (1, 2, 3)$. $\vec{d_1}\times\vec{d_2} = \hat{i}(1\cdot3 - 2\cdot2) - \hat{j}(3\cdot3 - 2\cdot1) + \hat{k}(3\cdot2 - 1\cdot1) = -\hat{i} - 7\hat{j} + 5\hat{k}$, of length $\sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$. So a unit vector perpendicular to both is $\frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$. (A) and (D) are not perpendicular to $L_1$: e.g. $(-1, 7, 7)\cdot(3, 1, 2) = 18 \ne 0$. (C): $(-1, 7, 5)\cdot(3, 1, 2) = 14 \ne 0$. Answer: (B)
18	Three Dimensional GeometrySkew lines and shortest distance	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	ComprehensionEasy
A easy comprehension question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Directions: $\vec{d_1} = (3, 1, 2)$, $\vec{d_2} = (1, 2, 3)$. $\vec{d_1}\times\vec{d_2} = \hat{i}(1\cdot3 - 2\cdot2) - \hat{j}(3\cdot3 - 2\cdot1) + \hat{k}(3\cdot2 - 1\cdot1) = -\hat{i} - 7\hat{j} + 5\hat{k}$, of length $\sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$. Points on the lines: $A(-1, -2, -1)$ on $L_1$, $B(2, -2, 3)$ on $L_2$; $\vec{AB} = (3, 0, 4)$. Shortest distance $= \frac{\|\vec{AB}\cdot(\vec{d_1}\times\vec{d_2})\|}{\|\vec{d_1}\times\vec{d_2}\|} = \frac{\|-3 + 0 + 20\|}{5\sqrt{3}} = \frac{17}{5\sqrt{3}}$. Answer: (D)
19	Three Dimensional GeometryCoordinates of a point in space	Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry	ComprehensionEasy
A easy comprehension question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Directions: $\vec{d_1} = (3, 1, 2)$, $\vec{d_2} = (1, 2, 3)$. $\vec{d_1}\times\vec{d_2} = \hat{i}(1\cdot3 - 2\cdot2) - \hat{j}(3\cdot3 - 2\cdot1) + \hat{k}(3\cdot2 - 1\cdot1) = -\hat{i} - 7\hat{j} + 5\hat{k}$, of length $\sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$. The plane through $(-1, -2, -1)$ with normal $(-1, -7, 5)$: $-(x + 1) - 7(y + 2) + 5(z + 1) = 0$, i.e. $-x - 7y + 5z - 10 = 0$. Distance of $(1, 1, 1)$: $\frac{\|-1 - 7 + 5 - 10\|}{\sqrt{75}} = \frac{13}{\sqrt{75}}$. Answer: (C)
20	Straight LinesSlope, parallel and perpendicular lines	Class 11: Straight Lines	Matrix matchMedium
A medium matrix match question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines $L_1$ and $L_3$ meet where $x = 5 - 3y$ and $25 - 15y + 2y = 12$: $y = 1$, $x = 2$. (A) Concurrent: $L_2$ passes through $(2, 1)$: $6 - k - 1 = 0$, $k = 5$. A→s. (B) Slopes: $L_1$: $-\frac{1}{3}$; $L_3$: $-\frac{5}{2}$ (not parallel to each other); $L_2$: $\frac{3}{k}$. $L_2 \parallel L_1$: $\frac{3}{k} = -\frac{1}{3}$, $k = -9$. $L_2 \parallel L_3$: $\frac{3}{k} = -\frac{5}{2}$, $k = -\frac{6}{5}$. B→p, q. (C) The three lines form a triangle when no two are parallel and they are not concurrent: $k \ne -9, -\frac{6}{5}, 5$. Of the listed values only $k = \frac{5}{6}$. C→r. (D) They fail to form a triangle for $k = -9$, $-\frac{6}{5}$ (parallel pair) and $k = 5$ (concurrent). D→p, q, s. A→s; B→p, q; C→r; D→p, q, s.
21	Applications of DerivativesMaxima and minima of one variable	Class 12: Application of Derivatives	Matrix matchHard
A hard matrix match question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives (A) $\frac{x^2 + 2x + 4}{x + 2} = (x + 2) + \frac{4}{x + 2} - 2$. For $x + 2 > 0$, by AM ≥ GM, $(x + 2) + \frac{4}{x + 2} \ge 4$, with equality at $x = 0$; so the (local) minimum value is $4 - 2 = 2$. A→r. (B) Expanding, $(A + B)(A - B) = A^2 - AB + BA - B^2$ and $(A - B)(A + B) = A^2 + AB - BA - B^2$. Equality gives $AB = BA$. Then $(AB)^t = B^tA^t = (-B)A = -BA = -AB$. So $(-1)^k = -1$ and $k$ is odd: from the list, 1 and 3. B→q, s. (C) $\log_3 2$ lies in $(0, 1)$, so $a < 0$, and $3^{-a} = \frac{1}{3^a} = \frac{1}{\log_3 2} = \log_2 3$. $1 < 2^{-k + \log_2 3} < 2$ means $0 < -k + \log_2 3 < 1$, i.e. $\log_2 3 - 1 < k < \log_2 3$, i.e. $0.58 < k < 1.58$. So $k = 1$, which is less than 2 and less than 3. C→r, s. (D) $\sin\theta = \cos\varphi$ means $\cos\varphi = \cos\left(\frac{\pi}{2} - \theta\right)$, so $\varphi = 2n\pi \pm \left(\frac{\pi}{2} - \theta\right)$. With the plus sign: $\theta + \varphi - \frac{\pi}{2} = 2n\pi$. With the minus sign: $\theta - \varphi - \frac{\pi}{2} = -2n\pi$. So $\frac{1}{\pi}\left(\theta \pm \varphi - \frac{\pi}{2}\right)$ is an even integer: from the list, 0 and 2. D→p, r. A→r; B→q, s; C→r, s; D→p, r.
22	Permutations and CombinationsSimple applications	Class 11: Permutations and Combinations	Matrix matchMedium
A medium matrix match question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations ENDEANOEL has 9 letters: E three times, N twice, and D, A, O, L once each. (A) Glue ENDEA into one block. It uses E, N, D, E, A, leaving N, O, E, L: five distinct objects in all, arranged in $5!$ ways. A→p. (B) E at both ends; the middle seven letters are E, N, N, D, A, O, L: $\frac{7!}{2!} = 2520 = 21 \times 5!$. B→s. (C) D, L, N, N must all sit in the first four places: $\frac{4!}{2!} = 12$ ways; the last five places take E, E, E, A, O: $\frac{5!}{3!} = 20$ ways. Total $240 = 2 \times 5!$. C→q. (D) A, E, E, E, O fill the five odd places (1, 3, 5, 7, 9): $\frac{5!}{3!} = 20$; N, N, D, L fill the four even places: $\frac{4!}{2!} = 12$. Total $240 = 2 \times 5!$. D→q. A→p; B→s; C→q; D→q.

Physics questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
23	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance From the figure, AB passes through the centre O, so AB is a diameter ($AB = 2R$) and C lies on the negative x-axis at $(-R, 0)$. The angle in a semicircle is a right angle, so $\angle ACB = 90°$. With $\angle CAB = 60°$: $AC = AB\cos 60° = R$ and $CB = AB\sin 60° = \sqrt{3}R$. (C) Force between C and B: $\frac{1}{4\pi\varepsilon_0}\cdot\frac{(q/3)(2q/3)}{(\sqrt{3}R)^2} = \frac{2q^2/9}{12\pi\varepsilon_0 R^2} = \frac{q^2}{54\pi\varepsilon_0 R^2}$. True. (A) The equal charges at A and B are diametrically opposite, so their fields at O cancel. The field of $-\frac{2q}{3}$ at C points from O towards C, i.e. along the negative x-axis, with magnitude $\frac{2q/3}{4\pi\varepsilon_0 R^2} = \frac{q}{6\pi\varepsilon_0 R^2}$, not $\frac{q}{8\pi\varepsilon_0 R^2}$. False. (B) $U = \frac{1}{4\pi\varepsilon_0}\left[\frac{(q/3)^2}{2R} - \frac{2q^2/9}{R} - \frac{2q^2/9}{\sqrt{3}R}\right] \ne 0$ (the first term is $\frac{q^2}{18R}$, smaller than the negative terms). False. (D) $V_O = \frac{1}{4\pi\varepsilon_0 R}\left(\frac{q}{3} + \frac{q}{3} - \frac{2q}{3}\right) = 0$. False. Answer: (C)
24	Atoms and NucleiRadioactivity	Class 12: Atoms; Class 12: Nuclei	Single correctEasy
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Activity $A = \lambda N$. Let S2 have $N$ nuclei; S1 has $2N$. $\lambda_1 = \frac{5}{2N}$ and $\lambda_2 = \frac{10}{N}$ (in the same units), so $\frac{\lambda_1}{\lambda_2} = \frac{1}{4}$. Half-life is $\frac{\ln 2}{\lambda}$, so $T_1 = 4T_2$. Of the options only 20 years and 5 years fit. Answer: (A)
25	Oscillations and WavesWave motion	Class 11: Oscillations; Class 11: Waves	Single correctMedium
A medium single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v = 0.1$ m/s, $\lambda = 0.5$ m, $A = 0.1$ m. Angular frequency $\omega = \frac{2\pi v}{\lambda} = \frac{2\pi(0.1)}{0.5} = 0.4\pi$ rad/s. Each point of the string moves in SHM along y. At displacement $y = 0.05$ m its speed is $\omega\sqrt{A^2 - y^2} = 0.4\pi\sqrt{0.01 - 0.0025} = 0.4\pi\cdot\frac{\sqrt{3}}{20} = \frac{\sqrt{3}\pi}{50}$ m/s. Direction: for a wave moving in +x, the particle velocity is $v_y = -v\frac{\partial y}{\partial x}$. P is on the falling side just past a crest, where the slope is negative, so $v_y > 0$: P moves up (the crest behind it is arriving). Velocity $= \frac{\sqrt{3}\pi}{50}\hat{j}$ m/s. (The string's particles do not move along x, so (C) and (D) are out.) Answer: (A)
26	Oscillations and WavesSimple pendulum and spring systems	Class 11: Oscillations; Class 11: Waves	Single correctMedium
A medium single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The supports only press on the walls; they cannot pull on them. Displacing B by $x$ towards wall 1 compresses S1 (M1 is pushed against wall 1). S2 would be stretched, but M2 is free, so it is simply dragged along a distance $x$ from wall 2 (figure II) and S2 stays at its natural length. Stored energy: $\frac{1}{2}kx^2$. On release, S1 pushes B back. When B reaches the equilibrium position, S1 is at its natural length and M1 cannot hold it stretched (M1 just leaves wall 1). S2, still unstretched, has pushed M2 freely back; M2 reaches wall 2 exactly when B is at the equilibrium position. Beyond that point S2 is compressed against wall 2 while S1 exerts no force. All the energy goes into S2: $\frac{1}{2}kx^2 = \frac{1}{2}(4k)y^2$, so $y = \frac{x}{2}$ and $\frac{y}{x} = \frac{1}{2}$. Answer: (C)
27	Work, Energy and PowerMotion in a vertical circle	Class 11: Work, Energy and Power	Single correctMedium
A medium single correct question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Just reaching the top B on a string means the tension is just zero there: $v_B^2 = gL$. Energy from A to B: $V^2 = v_B^2 + 2g(2L) = 5gL$. At angle $\theta$ from the downward vertical the bob has risen $L(1 - \cos\theta)$: $v^2 = V^2 - 2gL(1 - \cos\theta)$. With $v = \frac{V}{2}$: $\frac{5gL}{4} = 5gL - 2gL(1 - \cos\theta)$, so $1 - \cos\theta = \frac{15}{8}$, $\cos\theta = -\frac{7}{8}$, $\theta \approx 151°$. $\cos\frac{3\pi}{4} = -0.707$ and $\cos\pi = -1$; $-0.875$ lies between them, so $\frac{3\pi}{4} < \theta < \pi$. Answer: (D)
28	Properties of Bulk MatterSurface tension	Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter	Single correctMedium
A medium single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The excess pressure inside a soap bubble is $\frac{4T}{R}$, where $R$ is its radius of curvature. At end 1 the bubble is a hemisphere of radius $r$, the smallest radius a bubble on a tube of radius $r$ can have. At end 2 the bubble is less than a hemisphere, so its radius of curvature is larger than $r$. So the pressure behind bubble 1 is higher: when the valve opens, air flows from end 1 to end 2. Bubble 1 loses air and its volume decreases (it flattens, its radius of curvature grows, and its excess pressure falls), while bubble 2 grows. Answer: (B)
29	Oscillations and WavesStanding waves and beats	Class 11: Oscillations; Class 11: Waves	Single correctEasy
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Third harmonic of a closed tube: $f = \frac{3v}{4L} = \frac{3 \times 340}{4 \times 0.75} = 340$ Hz. The string vibrates at 340 Hz. 4 beats: $n = 336$ or $344$ Hz. Raising the tension raises the string's frequency. The beats fall to 2, so the string has moved closer to the fork: the fork is above 340 Hz. $n = 344$ Hz. Answer: (A)
30	ElectrostaticsCapacitance	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Single correctMedium
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The plates are horizontal; the liquid fills a layer of thickness $h = \frac{d}{3} - Vt$ and air fills $d - h$. The two layers are capacitors in series. Per unit area, $C = \frac{\varepsilon_0}{(d - h) + \frac{h}{K}} = \frac{\varepsilon_0}{d - \frac{h}{2}}$. $d - \frac{h}{2} = d - \frac{d}{6} + \frac{Vt}{2} = \frac{5d + 3Vt}{6}$. So $C = \frac{6\varepsilon_0}{5d + 3Vt}$ and the time constant is $\tau = RC = \frac{6\varepsilon_0 R}{5d + 3Vt}$. Answer: (A)
31	OpticsTotal internal reflection and prisms	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Single correctEasy
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The boundaries are parallel, so $n\sin(\text{angle with the normal})$ is the same in every region (Snell's law applied at each boundary). The beam just misses entering Region IV when it grazes the III–IV boundary, i.e. its angle in Region IV would be 90°: $n_0\sin\theta = \frac{n_0}{8}\sin 90°$, so $\sin\theta = \frac{1}{8}$, $\theta = \sin^{-1}\frac{1}{8}$. (Region II and III have larger indices than $\frac{n_0}{8}$, so the beam does get through them at this angle.) Answer: (B)
32	KinematicsRelative velocity	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Assertion and reasonEasy
A easy assertion and reason question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Statement-2 is the definition of relative velocity: true. Statement-1 is the everyday observation: true. Every stationary object has the same velocity $-\vec{V_1}$ relative to the observer, nearby or distant. What differs is the angular velocity at which the line of sight turns, about $\frac{V_1}{\text{distance}}$: large for nearby objects, which therefore seem to rush backwards, and tiny for distant ones, which seem to stand still. Statement-2 gives the same relative velocity for near and far objects, so it does not explain why distant objects seem stationary. Answer: (B)
33	Laws of MotionFriction	Class 11: Laws of Motion	Assertion and reasonEasy
A easy assertion and reason question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Statement-1 (true): a pull is usually applied upwards at an angle, and its vertical component reduces the normal reaction, so friction $\mu N$ is smaller; a push at an angle has a downward component that increases $N$ and the friction. Statement-2 (true): the coefficient of friction depends on the nature of the surfaces. But Statement-1 is about the change in the normal reaction, not the nature of the surfaces, so Statement-2 does not explain it. Answer: (B)
34	ElectrostaticsElectric field and potential	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	Assertion and reasonEasy
A easy assertion and reason question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Statement-2 is the standard result for a charged spherical shell: true. Statement-1 is true: the earth is taken as zero potential. Why the earth serves as a fixed reference: by Statement-2, a charge $Q$ given to or taken from the earth changes its potential by $\frac{Q}{4\pi\varepsilon_0 R}$, and the earth's radius is so large (about $6.4 \times 10^6$ m) that this change is negligible. So its potential stays effectively constant whatever charge flows into it, and Statement-2 explains Statement-1. Answer: (A)
35	Magnetic Effects of Current and MagnetismMoving coil galvanometer	Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter	Assertion and reasonEasy
A easy assertion and reason question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Statement-1 is true: a soft-iron core concentrates the field (makes it radial and stronger), so the deflection per unit current, $\frac{NAB}{k}$, increases. Statement-2 is false: soft iron has high permeability and low retentivity and coercivity, so it is easily magnetized and easily demagnetized. Answer: (C)
36	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	ComprehensionEasy
A easy comprehension question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance By Gauss's law over a sphere of radius $R$, which encloses the whole nuclear charge $Ze$: $E(R)\cdot4\pi R^2 = \frac{Ze}{\varepsilon_0}$, so $E(R) = \frac{Ze}{4\pi\varepsilon_0 R^2}$. The total charge is fixed at $Ze$ however it is distributed inside, so $E(R)$ does not depend on $a$. Answer: (A)
37	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	ComprehensionMedium
A medium comprehension question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance For $a = 0$, the density falls linearly from $d$ at the centre to 0 at $R$: $\rho(r) = d\left(1 - \frac{r}{R}\right)$. Total charge: $Ze = \int_0^R \rho\,4\pi r^2\,dr = 4\pi d\left[\frac{r^3}{3} - \frac{r^4}{4R}\right]_0^R = 4\pi d\left(\frac{R^3}{3} - \frac{R^3}{4}\right) = \frac{\pi d R^3}{3}$. So $d = \frac{3Ze}{\pi R^3}$. Answer: (B)
38	ElectrostaticsGauss's law	Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance	ComprehensionMedium
A medium comprehension question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Inside, Gauss's law gives $E(r)\cdot4\pi r^2 = \frac{q(r)}{\varepsilon_0}$ with $q(r)$ the charge within radius $r$. $E \propto r$ means $q(r) \propto r^3$, i.e. the density is uniform throughout the nucleus. In the given profile the density is constant only on $0 \le r \le a$; for it to be constant all the way to $R$, the falling part must vanish: $a = R$. Answer: (C)
39	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	ComprehensionMedium
A medium comprehension question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take $x$ as the displacement of the CM from equilibrium. Both springs act on the axle, together $-2kx$. Friction $f$ (along x) acts at the contact point. Energy: $E = \frac{1}{2}Mv^2 + \frac{1}{2}\cdot\frac{MR^2}{2}\cdot\frac{v^2}{R^2} + \frac{1}{2}(2k)x^2 = \frac{3}{4}Mv^2 + kx^2$, conserved because static friction in rolling does no work. $\frac{dE}{dt} = 0$: $\frac{3}{2}Mva + 2kxv = 0$, so $a = -\frac{4k}{3M}x$. Net external force $= Ma = -\frac{4kx}{3}$. (It is the spring force $-2kx$ plus the friction $f = \frac{2kx}{3}$.) Answer: (D)
40	Oscillations and WavesSHM	Class 11: Oscillations; Class 11: Waves	ComprehensionMedium
A medium comprehension question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Take $x$ as the displacement of the CM from equilibrium. Both springs act on the axle, together $-2kx$. Friction $f$ (along x) acts at the contact point. Energy: $E = \frac{1}{2}Mv^2 + \frac{1}{2}\cdot\frac{MR^2}{2}\cdot\frac{v^2}{R^2} + \frac{1}{2}(2k)x^2 = \frac{3}{4}Mv^2 + kx^2$, conserved because static friction in rolling does no work. $\frac{dE}{dt} = 0$: $\frac{3}{2}Mva + 2kxv = 0$, so $a = -\frac{4k}{3M}x$. $a = -\omega^2 x$ with $\omega^2 = \frac{4k}{3M}$, so $\omega = \sqrt{\frac{4k}{3M}}$. Answer: (D)
41	Motion of System of Particles and Rigid BodyTorque and rotational dynamics	Class 11: Systems of Particles and Rotational Motion	ComprehensionHard
A hard comprehension question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take $x$ as the displacement of the CM from equilibrium. Both springs act on the axle, together $-2kx$. Friction $f$ (along x) acts at the contact point. Energy: $E = \frac{1}{2}Mv^2 + \frac{1}{2}\cdot\frac{MR^2}{2}\cdot\frac{v^2}{R^2} + \frac{1}{2}(2k)x^2 = \frac{3}{4}Mv^2 + kx^2$, conserved because static friction in rolling does no work. $\frac{dE}{dt} = 0$: $\frac{3}{2}Mva + 2kxv = 0$, so $a = -\frac{4k}{3M}x$. Friction needed: $f = Ma + 2kx = -\frac{4kx}{3} + 2kx = \frac{2kx}{3}$. It is largest at the extreme displacement, the amplitude $x_0 = \frac{V_0}{\omega}$ (the disk passes equilibrium with speed $V_0$). Rolling without slipping needs $f_{max} \le \mu Mg$: $\frac{2k}{3}\cdot\frac{V_0}{\omega} \le \mu Mg$. $V_0 \le \frac{3\mu Mg}{2k}\sqrt{\frac{4k}{3M}} = \mu g\cdot\frac{3M}{2k}\cdot2\sqrt{\frac{k}{3M}} = \mu g\sqrt{\frac{3M}{k}}$. Answer: (C)
42	KinematicsMotion graphs	Class 11: Motion in a Straight Line; Class 11: Motion in a Plane	Matrix matchMedium
A medium matrix match question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane (A) For small displacement $x$, the pendulum's potential energy is $U = \frac{mg}{2L}x^2$ above that at the mean position: a parabola. Graph s is its right half starting at the origin (displacement measured from the mean position, U = 0 there). Graph p is the same U-shaped curve when the displacement is measured from some other point and the energy from a level below the lowest point. A→p, s. (B) Zero acceleration: $x = x_0 + ut$, a straight line, through the origin (q) or with an intercept (r). Constant acceleration in the direction of motion: $x = ut + \frac{1}{2}at^2$, a rising curve (s). B→q, r, s. (C) $R = \frac{u^2\sin 2\theta}{g} \propto u^2$: a parabola from the origin, graph s. C→s. (D) $T^2 = \frac{4\pi^2}{g}L \propto L$: a straight line through the origin, graph q. D→q. A→p, s; B→q, r, s; C→s; D→q.
43	OpticsLenses and mirrors	Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics	Matrix matchMedium
A medium matrix match question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics (A) A concave mirror (reflecting face towards S). Object beyond F: real image, magnified when between F and 2F (real and magnified). Object at F: image at infinity. Object inside F: virtual, magnified image. A→p, q, r, s. (B) A convex mirror (bulging towards S). For every real object the image is virtual, erect and diminished, between the pole and the focus. B→q. (C) A convex (converging) lens: like the concave mirror, real (magnified between F and 2F), image at infinity (object at F), virtual magnified (object inside F). C→p, q, r, s. (D) A converging meniscus (concavo-convex lens, thicker in the middle): it is a converging lens, so again all four. D→p, q, r, s. A→p, q, r, s; B→q; C→p, q, r, s; D→p, q, r, s.
44	ThermodynamicsThermodynamic processes	Class 11: Thermodynamics	Matrix matchHard
A hard matrix match question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics For a polytropic process $PV^n = $ constant, $T \propto PV \propto V^{1-n}$ and the molar heat capacity is $C = C_V + \frac{R}{1 - n}$; for a monoatomic gas $C_V = \frac{3R}{2}$. (A) Free expansion into vacuum in an insulated vessel: $Q = 0$, $W = 0$, so $\Delta U = 0$ and the ideal gas's temperature stays the same. A→q (no heat in or out). (B) $n = 2$: $T \propto V^{-1}$, so doubling V halves T: temperature decreases (p). $C = \frac{3R}{2} + \frac{R}{1 - 2} = \frac{R}{2} > 0$, and $\Delta T < 0$, so $Q = nC\Delta T < 0$: the gas loses heat (r). B→p, r. (C) $n = \frac{4}{3}$: $T \propto V^{-1/3}$ decreases (p). $C = \frac{3R}{2} + \frac{R}{1 - 4/3} = \frac{3R}{2} - 3R = -\frac{3R}{2} < 0$; with $\Delta T < 0$, $Q > 0$: the gas gains heat (s). C→p, s. (D) From $V_1$ to $2V_1$ the pressure rises, so $PV$ and hence T rise (q). $\Delta U > 0$ and the gas does positive work, so $Q = \Delta U + W > 0$: it gains heat (s). D→q, s. A→q; B→p, r; C→p, s; D→q, s.

Chemistry questions, each with its chapter and our explanation
Q	Chapter and topic	NCERT	Kind and level
45	Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates	Class 11: Organic Chemistry – Some Basic Principles and Techniques	Single correctMedium
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Reading the structures: (I) $\mathrm{(CH_3)_2\overset{+}{C}{-}O{-}CH_2CH_3}$: a tertiary carbocation on a carbon bonded to oxygen. (II) $\mathrm{CH_3\overset{+}{C}H{-}CH_2{-}CH(CH_3)_2}$: a secondary alkyl carbocation. (III) $\mathrm{CH_3\overset{+}{C}H{-}O{-}CH(CH_3)_2}$: a secondary carbocation on a carbon bonded to oxygen. (IV) $\mathrm{\overset{+}{C}H_2{-}CH_2{-}CH(CH_3)_2}$: a primary alkyl carbocation. A lone pair on oxygen next to the cationic carbon gives strong resonance stabilisation ($\mathrm{C^+{-}\ddot{O}R \leftrightarrow C{=}O^+R}$, an oxocarbenium ion), far stronger than hyperconjugation. So I and III are the most stable, and of these the tertiary I (two extra methyls) is more stable than the secondary III. Among the plain alkyl cations, secondary II is more stable than primary IV. Order: I > III > II > IV. Answer: (D)
46	BiomoleculesCarbohydrates	Class 12: Biomolecules	Single correctMedium
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Cellulose is a linear chain of β-D-glucose units joined by β(1→4) glycosidic links. Each unit has three free OH groups, at C-2, C-3 and C-6. Excess acetic anhydride acetylates all three, giving cellulose triacetate; the glycosidic links are not changed. (A) shows β(1→4) links (each ring turned over relative to the next, as β-linked glucose drawn in Haworth form must be) with OAc at C-2, C-3 and C-6 of every unit: this is cellulose triacetate. (B) has only the C-6 groups acetylated; the C-2 and C-3 OH groups are still free (a monoacetate, not the triacetate). (C) has every ring in the same orientation with the links pointing down from C-1, i.e. α(1→4) links: the triacetate of amylose (starch), not cellulose. (D) also has α(1→4) links, and both secondary OAc groups drawn down (wrong configuration at C-3 for glucose). Answer: (A)
47	Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones	Class 12: Aldehydes, Ketones and Carboxylic Acids	Single correctMedium
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The starting acid is benzoylacetic acid, a β-keto acid; the labelled carbon is the CH$_2$ between the two carbonyl groups. Heating a β-keto acid removes $\mathrm{CO_2}$ (from the COOH carbon) through a cyclic transition state; the CH$_2$ becomes CH$_3$: E $= \mathrm{Ph{-}CO{-}\overset{}{C}H_3}$ (acetophenone labelled in the methyl group). Iodoform reaction ($\mathrm{I_2/NaOH}$): the methyl ketone's $\mathrm{CH_3}$ is converted to $\mathrm{CI_3}$, which is cleaved off as iodoform, and the rest becomes the carboxylate: F $= \mathrm{PhCOO^-Na^+}$ (unlabelled), G $= \mathrm{\overset{}{C}HI_3}$ (the label goes with the methyl carbon). (A) puts the label on the wrong carbon; (B) loses the label from iodoform; (D) gives $\mathrm{CH_3I}$, which is not formed. Answer: (C)
48	d- and f-Block ElementsTransition element properties	Class 12: The d- and f-Block Elements	Single correctEasy
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Colour in these compounds comes from d–d transitions, which need a partly filled d subshell. CuCl, $\mathrm{K_3[Cu(CN)_4]}$ and $\mathrm{[Cu(CH_3CN)_4]BF_4}$ all contain Cu(I), $3d^{10}$: no d–d transition, colourless. $\mathrm{CuF_2}$ contains Cu(II), $3d^9$: d–d transitions occur and it is coloured. Answer: (C)
49	Coordination CompoundsValence bond theory	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[Ni(CO)_4]}$: Ni(0), $3d^84s^2$. The strong-field CO pairs the electrons into $3d^{10}$; the empty 4s and 4p orbitals give $sp^3$, tetrahedral, diamagnetic. $\mathrm{[Ni(CN)_4]^{2-}}$: Ni(II), $3d^8$. Strong-field CN$^-$ pairs the eight d electrons into four orbitals, leaving one 3d empty: $dsp^2$, square planar, diamagnetic. Answer: (B)
50	Coordination CompoundsNomenclature	Class 12: Coordination Compounds	Single correctEasy
A easy single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The cation is named first, then the anion. Cation $\mathrm{[Ni(NH_3)_4]^{2+}}$: tetraamminenickel(II). Anion $\mathrm{[NiCl_4]^{2-}}$: a complex anion takes the ending -ate, tetrachloronickelate(II). The charges balance with Ni(II) in both. Name: tetraamminenickel(II) tetrachloronickelate(II). Answer: (C)
51	ElectrochemistryElectrochemical cells	Class 12: Electrochemistry	Single correctEasy
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry At the cathode $\mathrm{2H_2O + 2e^- \to H_2 + 2OH^-}$: 2 mol of electrons per mol of $\mathrm{H_2}$. 0.01 mol $\mathrm{H_2}$ needs 0.02 F $= 0.02 \times 96500 = 1930$ C. $t = \frac{Q}{I} = \frac{1930}{0.010} = 193000$ s $= 19.3 \times 10^4$ s. Answer: (B)
52	Surface ChemistryEmulsions, surfactants and micelles	No matching NCERT chapter	Single correctEasy
A easy single correct question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter The critical micelle concentration falls as the hydrophobic tail gets longer: a longer tail is more strongly driven out of water into the micelle core. The tails are C$_{16}$ in (A), C$_{12}$ in (B) and (D), and C$_8$ in (C) (counting the carboxylate carbon). Cetyltrimethylammonium bromide (A), with the longest tail, has the lowest CMC. Answer: (A)
53	EquilibriumSolubility product	Class 11: Equilibrium	Single correctEasy
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium MX: $K_{sp} = s^2$, $s = \sqrt{4.0 \times 10^{-8}} = 2 \times 10^{-4}$. $\mathrm{MX_2}$: $K_{sp} = s(2s)^2 = 4s^3$, $s^3 = 8 \times 10^{-15}$, $s = 2 \times 10^{-5}$. $\mathrm{M_3X}$: $K_{sp} = (3s)^3 s = 27s^4$, $s^4 = 1 \times 10^{-16}$, $s = 1 \times 10^{-4}$. Order: MX > $\mathrm{M_3X}$ > $\mathrm{MX_2}$. Answer: (D)
54	AminesDiazonium salts	Class 12: Amines	Assertion and reasonEasy
A easy assertion and reason question on diazonium salts, from Amines. NCERT: Class 12: Amines Aniline is diazotised to benzenediazonium chloride, which couples with β-naphthol (in alkali) to give 1-phenylazo-2-naphthol, an orange-red dye, not a dark blue precipitate. Statement-1 is false. The azo dye is coloured because the azo group links two aromatic systems into one extended conjugated system that absorbs visible light. Statement-2 is true. Answer: (D)
55	Coordination CompoundsValence bond theory	Class 12: Coordination Compounds	Assertion and reasonMedium
A medium assertion and reason question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds This is the brown-ring complex. NO is bound as $\mathrm{NO^+}$, so iron is Fe(I), $3d^7$ ($3d^64s^1$ rearranged into 3d with the weak-field water ligands), with three unpaired electrons; the measured moment is about 3.9 BM ($\sqrt{3 \times 5} = 3.87$). Statement-2 is true. A species with unpaired electrons is paramagnetic, so Statement-1 is true, and the three unpaired electrons are exactly why it is paramagnetic: Statement-2 explains Statement-1. Answer: (A)
56	Coordination CompoundsIsomerism in complexes	Class 12: Coordination Compounds	Assertion and reasonMedium
A medium assertion and reason question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[M(NH_3)_4Cl_2]}$ (octahedral) has cis and trans isomers. Each has a mirror plane (cis: the plane containing M and both Cl; trans: several), so neither is chiral: Statement-1 is true. Both have a proper axis of symmetry (cis: a $C_2$ bisecting the Cl–M–Cl angle; trans: a $C_4$ along Cl–M–Cl): Statement-2 is true. But an axis of symmetry does not make a molecule achiral (chiral molecules can have $C_2$ axes); optical inactivity comes from the mirror plane (an improper axis). So Statement-2 does not explain Statement-1. Answer: (B)
57	ThermodynamicsEntropy and Gibbs energy	Class 11: Thermodynamics	Assertion and reasonEasy
A easy assertion and reason question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Work can be turned completely into heat (friction, a resistor), but heat cannot be turned completely into work in a cycle. That is the asymmetry of Statement-1: true. Statement-2 is the Kelvin–Planck statement of the second law: true. It is exactly the reason for the asymmetry, so it explains Statement-1. Answer: (A)
58	Alcohols, Phenols and EthersPreparation and properties	Class 12: Alcohols, Phenols and Ethers	ComprehensionMedium
A medium comprehension question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Work back from the products. J + KOH → benzyl alcohol + L: a Cannizzaro reaction, so J is benzaldehyde, PhCHO, and L is potassium benzoate, $\mathrm{PhCOO^-K^+}$. K + KOH → only M, $\mathrm{Ph(CH_3)C{=}CH{-}COPh}$: the self aldol condensation of acetophenone (two $\mathrm{PhCOCH_3}$ units, losing water). So K is acetophenone, $\mathrm{PhCOCH_3}$. I is the alkene whose ozonolysis gives PhCHO and $\mathrm{PhCOCH_3}$: $\mathrm{PhCH{=}C(CH_3)Ph}$. H is the tertiary alcohol that dehydrates to I: $\mathrm{PhCH_2{-}C(OH)(CH_3){-}Ph}$ (dehydration towards the benzylic CH$_2$ gives the conjugated stilbene-type alkene). H, $\mathrm{PhCH_2C(OH)(CH_3)Ph}$, is made by adding benzylmagnesium bromide to acetophenone: $\mathrm{PhCOCH_3 + PhCH_2MgBr}$, then hydrolysis. (A) gives $\mathrm{Ph_2C(OH)CH_3}$, whose dehydration gives $\mathrm{Ph_2C{=}CH_2}$, not I. (C) and (D) start from an aldehyde and give secondary alcohols, not tertiary. Answer: (B)
59	HydrocarbonsAlkenes and alkynes	Class 11: Hydrocarbons	ComprehensionMedium
A medium comprehension question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Work back from the products. J + KOH → benzyl alcohol + L: a Cannizzaro reaction, so J is benzaldehyde, PhCHO, and L is potassium benzoate, $\mathrm{PhCOO^-K^+}$. K + KOH → only M, $\mathrm{Ph(CH_3)C{=}CH{-}COPh}$: the self aldol condensation of acetophenone (two $\mathrm{PhCOCH_3}$ units, losing water). So K is acetophenone, $\mathrm{PhCOCH_3}$. I is the alkene whose ozonolysis gives PhCHO and $\mathrm{PhCOCH_3}$: $\mathrm{PhCH{=}C(CH_3)Ph}$. H is the tertiary alcohol that dehydrates to I: $\mathrm{PhCH_2{-}C(OH)(CH_3){-}Ph}$ (dehydration towards the benzylic CH$_2$ gives the conjugated stilbene-type alkene). I is $\mathrm{PhCH{=}C(CH_3)Ph}$: one alkene carbon carries Ph and H, the other $\mathrm{CH_3}$ and Ph. Option (A) has exactly these groups (Ph and H on one carbon, $\mathrm{CH_3}$ and Ph on the other), and its ozonolysis gives PhCHO and $\mathrm{PhCOCH_3}$. (B) $\mathrm{CH_3CH{=}CPh_2}$ would give acetaldehyde and benzophenone. (C) $\mathrm{PhCH{=}C(CH_3)CH_2Ph}$ would give PhCHO and $\mathrm{PhCH_2COCH_3}$. (D) $\mathrm{(CH_3)(Ph)C{=}CH(CH_3)}$ would give acetophenone and acetaldehyde, and no benzyl alcohol could form from J. Answer: (A)
60	Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro	Class 12: Aldehydes, Ketones and Carboxylic Acids	ComprehensionMedium
A medium comprehension question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Work back from the products. J + KOH → benzyl alcohol + L: a Cannizzaro reaction, so J is benzaldehyde, PhCHO, and L is potassium benzoate, $\mathrm{PhCOO^-K^+}$. K + KOH → only M, $\mathrm{Ph(CH_3)C{=}CH{-}COPh}$: the self aldol condensation of acetophenone (two $\mathrm{PhCOCH_3}$ units, losing water). So K is acetophenone, $\mathrm{PhCOCH_3}$. I is the alkene whose ozonolysis gives PhCHO and $\mathrm{PhCOCH_3}$: $\mathrm{PhCH{=}C(CH_3)Ph}$. H is the tertiary alcohol that dehydrates to I: $\mathrm{PhCH_2{-}C(OH)(CH_3){-}Ph}$ (dehydration towards the benzylic CH$_2$ gives the conjugated stilbene-type alkene). So J = PhCHO, K = $\mathrm{PhCOCH_3}$, L = $\mathrm{PhCOO^-K^+}$. Answer: (D)
61	Solid StatePacking in fcc, bcc and hcp	No matching NCERT chapter	ComprehensionEasy
A easy comprehension question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter Corners: 12 atoms, each shared by 6 prisms: $12 \times \frac{1}{6} = 2$. Centres of the top and bottom hexagons: $2 \times \frac{1}{2} = 1$. The three atoms of the middle layer lie wholly inside: 3. Total $= 2 + 1 + 3 = 6$. Answer: (B)
62	Solid StatePacking in fcc, bcc and hcp	No matching NCERT chapter	ComprehensionMedium
A medium comprehension question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter Edge of the hexagon: neighbouring spheres touch, so $a = 2r$. Base area $= 6 \times \frac{\sqrt{3}}{4}a^2 = 6\sqrt{3}r^2$. Height: a middle-layer sphere sits on three touching spheres, forming a regular tetrahedron of edge $2r$, whose height is $2r\sqrt{\frac{2}{3}}$. The cell spans two such layers: $c = 4r\sqrt{\frac{2}{3}}$. Volume $= 6\sqrt{3}r^2 \times 4r\sqrt{\frac{2}{3}} = 24\sqrt{2}\,r^3$. Answer: (A)
63	Solid StatePacking in fcc, bcc and hcp	No matching NCERT chapter	ComprehensionEasy
A easy comprehension question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter Volume of the 6 spheres: $6 \times \frac{4}{3}\pi r^3 = 8\pi r^3$. Cell volume $24\sqrt{2}\,r^3$. Packing fraction $= \frac{8\pi}{24\sqrt{2}} = \frac{\pi}{3\sqrt{2}} \approx 0.74$, so 74% is filled and about 26% is empty. Answer: (D)
64	Principles Related to Practical ChemistryDetection of elements and functional groups	No matching NCERT chapter	Matrix matchHard
A hard matrix match question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter (p) Prussian blue (Lassaigne's test for N) needs both C and N in the compound, so that sodium fusion gives NaCN. (q) The neutral $\mathrm{FeCl_3}$ test is given by phenols. (r) A white precipitate with $\mathrm{AgNO_3}$ is AgCl, from chloride; bromide gives pale yellow AgBr and iodide yellow AgI. (s) Hydrazones form from compounds with a free $\mathrm{-NH{-}NH_2}$ group (hydrazine and its derivatives). (A) Hydrazine hydrochloride: no carbon, so no Prussian blue; chloride gives white AgCl (r); hydrazine condenses with aldehydes to hydrazones (s). A→r, s. (B) A phenolic amino acid hydroiodide: C and N present (p), phenol (q); iodide gives a yellow, not white, precipitate; no hydrazine group. B→p, q. (C) 4-Hydroxyanilinium chloride: C and N (p), phenol (q), chloride (r). C→p, q, r. (D) 2,4-Dinitrophenylhydrazinium bromide: C and N (p); bromide gives pale yellow AgBr; no phenol; 2,4-dinitrophenylhydrazine (Brady's reagent) forms 2,4-dinitrophenylhydrazones with aldehydes (s). D→p, s. A→r, s; B→p, q; C→p, q, r; D→p, s. The official key prints two accepted grids: rows A and D are accepted either as r and p alone or as r, s and p, s.
65	Isolation of MetalsRoasting and calcination	No matching NCERT chapter	Matrix matchMedium
A medium matrix match question on roasting and calcination, from Isolation of Metals. NCERT: No matching NCERT chapter (A) $\mathrm{2PbS + 3O_2 \to 2PbO + 2SO_2}$: heating a sulphide ore in air is roasting. A→p. (B) $\mathrm{CaCO_3 \to CaO + CO_2}$: heating a carbonate to drive off $\mathrm{CO_2}$ (no air needed) is calcination. B→q. (C) Zinc blende is roasted, $\mathrm{2ZnS + 3O_2 \to 2ZnO + 2SO_2}$, and the oxide reduced by carbon (coke), $\mathrm{ZnO + C \to Zn + CO}$. C→p, r. (D) $\mathrm{Cu_2S}$ is partly roasted to $\mathrm{Cu_2O}$, which then reacts with the remaining sulphide: $\mathrm{2Cu_2O + Cu_2S \to 6Cu + SO_2}$, self (auto) reduction. D→p, s. A→p; B→q; C→p, r; D→p, s.
66	Structure of AtomQuantum numbers	Class 11: Structure of Atom	Matrix matchHard
A hard matrix match question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom (A) The magnitude of the orbital angular momentum is $\sqrt{l(l+1)}\,\hbar$ (azimuthal, q), and its component along the field axis is $m_l\hbar$ (magnetic, r). A→q, r. (B) A one-electron wave function (spin-orbital) that obeys the Pauli principle is specified by all four quantum numbers $n$, $l$, $m_l$, $m_s$. B→p, q, r, s. (C) Size depends on $n$, shape on $l$, orientation on $m_l$. C→p, q, r. (D) The density at the nucleus is non-zero only for s orbitals ($l = 0$), and for them $\|\psi(0)\|^2 = \frac{Z^3}{\pi n^3 a_0^3}$ depends on $n$. So it depends on $n$ and $l$. D→p, q. A→q, r; B→p, q, r, s; C→p, q, r; D→p, q.

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