JEE (Advanced) · 2024
JEE Advanced 2024, question by question
Which chapter each question came from, how the paper was spread, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level. One year is not a forecast.
Mathematics
34 Q
- Sets, Relations and Functions5
- Integral Calculus4
- Probability3
9 easy · 21 medium · 4 hard
Physics
34 Q
- Optics6
- Electrostatics5
- Oscillations and Waves4
6 easy · 24 medium · 4 hard
Chemistry
34 Q
- Coordination Compounds6
- Aldehydes, Ketones and Carboxylic Acids4
- p-Block Elements3
14 easy · 14 medium · 6 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 11: Oscillations; Class 11: Waves
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 11: Units and Measurement
Class 12: Atoms; Class 12: Nuclei
Class 11: Systems of Particles and Rotational Motion
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 11: Thermodynamics
Class 11: Gravitation
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Class 11: Work, Energy and Power
Class 12: Coordination Compounds
Class 12: Aldehydes, Ketones and Carboxylic Acids
No matching NCERT chapter
Class 12: Biomolecules
Class 11: Structure of Atom
Class 11: Chemical Bonding and Molecular Structure
Class 12: Chemical Kinetics
Class 12: Electrochemistry
Class 11: Hydrocarbons
Class 12: Alcohols, Phenols and Ethers
Class 11: Some Basic Concepts of Chemistry
Class 11: Thermodynamics
No matching NCERT chapter
No matching NCERT chapter
No matching NCERT chapter
Class 12: Solutions
No matching NCERT chapter
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Probability; Class 12: Probability
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 11: Conic Sections
Class 12: Matrices; Class 12: Determinants
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 12: Vector Algebra
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 12: Differential Equations
Class 11: Sequences and Series
Class 11: Complex Numbers and Quadratic Equations
Class 11: Permutations and Combinations
Class 11: Conic Sections
Class 12: Application of Derivatives
03
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations The limit is of the form $\frac{0}{0}$ as $t \to x$. Apply L'Hospital's rule in $t$ (with $x$ fixed):
$\lim_{t\to x} \frac{10t^9 f(x) - x^{10} f'(t)}{9t^8} = \frac{10x^9 f(x) - x^{10} f'(x)}{9x^8} = 1$.
So $10x f(x) - x^2 f'(x) = 9$, i.e. $f'(x) - \frac{10}{x} f(x) = -\frac{9}{x^2}$, a linear equation.
Integrating factor $e^{-\int \frac{10}{x}dx} = x^{-10}$: $\frac{d}{dx}\left(x^{-10} f\right) = -9x^{-12}$, so $x^{-10} f = \frac{9}{11}x^{-11} + C$.
Hence $f(x) = \frac{9}{11x} + Cx^{10}$. With $f(1) = 2$: $\frac{9}{11} + C = 2$, so $C = \frac{13}{11}$.
$f(x) = \frac{9}{11x} + \frac{13}{11}x^{10}$.
Answer: (B). | ||||
| Q.2 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $K$ = the student knows the answer, $G$ = he guesses, $C$ = the answer is correct. Put $P(K) = k$, $P(G) = 1 - k$.
$P(C \mid K) = 1$ and $P(C \mid G) = \frac{1}{2}$, so $P(C) = k + \frac{1-k}{2}$.
Bayes: $P(G \mid C) = \frac{\frac{1}{2}(1-k)}{k + \frac{1}{2}(1-k)} = \frac{1}{6}$.
So $3(1-k) = k + \frac{1}{2}(1-k)$, i.e. $\frac{5}{2}(1-k) = k$, giving $5 - 5k = 2k$ and $k = \frac{5}{7}$.
Answer: (C). | ||||
| Q.3 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Regroup the expression by $\sin 6x$ and $\cos 6x$:
$\left(\sin 6x\cos\frac{11x}{2} - \cos 6x \sin\frac{11x}{2}\right) + \left(\cos 6x\cos\frac{11x}{2} + \sin 6x\sin\frac{11x}{2}\right)$
$= \sin\left(6x - \frac{11x}{2}\right) + \cos\left(6x - \frac{11x}{2}\right) = \sin\frac{x}{2} + \cos\frac{x}{2}$.
In the second quadrant with $\cot x = -\frac{5}{\sqrt{11}}$: the hypotenuse is $\sqrt{25 + 11} = 6$, so $\sin x = \frac{\sqrt{11}}{6}$.
$\left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2 = 1 + \sin x = \frac{6 + \sqrt{11}}{6}$, and since $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ both terms are positive.
So the value is $\sqrt{\frac{6+\sqrt{11}}{6}} = \sqrt{\frac{12 + 2\sqrt{11}}{12}} = \frac{\sqrt{(\sqrt{11}+1)^2}}{2\sqrt{3}} = \frac{\sqrt{11}+1}{2\sqrt{3}}$.
Answer: (B). | ||||
| Q.4 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The ends of the minor axis are $(0, \pm 2)$, with tangents $y = \pm 2$. $S$ is in the first quadrant, so the tangent through it is $y = 2$, giving $q = 2$.
$O = (0,0)$, $R = (3, 0)$. Area of $\Delta ORT = \frac{1}{2}\cdot 3 \cdot |y_T| = \frac{3}{2}$, so $|y_T| = 1$; in the fourth quadrant $y_T = -1$.
Then $\frac{x_T^2}{9} = 1 - \frac{1}{4}$, so $x_T = \frac{3\sqrt{3}}{2}$.
Tangent at $T$: $\frac{x \cdot \frac{3\sqrt3}{2}}{9} + \frac{y \cdot (-1)}{4} = 1$, i.e. $\frac{\sqrt{3}}{6}x - \frac{y}{4} = 1$.
$S$ lies on it with $y = 2$: $\frac{\sqrt3}{6}p = \frac{3}{2}$, so $p = \frac{9}{\sqrt3} = 3\sqrt{3}$.
Check: $\frac{27}{9} + \frac{4}{4} = 4 > 1$, so $S$ is outside the ellipse.
Answer: (A). | ||||
| Q.5 | Sets, Relations and FunctionsSets and their representation | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Multiple correctMedium | |
A medium multiple correct question on sets and their representation, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions (A) TRUE. Every integer $a$ is $a + 0\sqrt2 \in S$. $S$ is closed under multiplication: $(a + b\sqrt2)(c + d\sqrt2) = (ac + 2bd) + (ad + bc)\sqrt2$. Since $-1 + \sqrt2$ and $1 + \sqrt2$ are in $S$, all their powers are too, so $\mathbb{Z} \cup T_1 \cup T_2 \subset S$.
(B) FALSE. $0 < \sqrt2 - 1 < \frac{1}{2}$, so $(\sqrt2 - 1)^n$ is positive and tends to $0$; for large $n$ it lies in $\left(0, \frac{1}{2024}\right)$; for instance $(\sqrt2 - 1)^9 \approx 0.00036 < \frac{1}{2024} \approx 0.00049$. So the intersection is not empty.
(C) TRUE. $1 + \sqrt2 > 2$, so $(1+\sqrt2)^n > 2^n > 2024$ for $n \ge 11$.
(D) TRUE. $\cos\theta + i\sin\theta$ has modulus 1, so it is an integer only when it equals $\pm 1$, i.e. when $\theta = \pi(a + b\sqrt2)$ is an integer multiple of $\pi$, i.e. $a + b\sqrt2 \in \mathbb{Z}$. As $\sqrt2$ is irrational, this happens exactly when $b = 0$; and when $b = 0$ the value is $(-1)^a \in \mathbb{Z}$.
Answer: (A), (C), (D). | ||||
| Q.6 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants First describe $S$. Taking $y = 0$ gives $a > 0$. For $y \ne 0$ put $t = \frac{x}{y}$: $at^2 + 2bt + c > 0$ for all real $t$ needs $a > 0$ and discriminant $4b^2 - 4ac < 0$. So $(a,b,c) \in S \iff a > 0,\ ac - b^2 > 0$ (and then $c > 0$ too).
(A) FALSE. $ac = 12$ but $b^2 = \frac{49}{4} = 12.25 > 12$.
(B) TRUE. $3 \cdot \frac{1}{12} - b^2 > 0$ gives $b^2 < \frac{1}{4}$, so $|2b| < 1$.
(C) TRUE. The determinant of the system is $ac - b^2 > 0$, not zero, so the solution is unique.
(D) TRUE. The determinant is $(a+1)(c+1) - b^2 = (ac - b^2) + a + c + 1 > 0$, so the homogeneous system has only the zero solution, which is unique.
Answer: (B), (C), (D). | ||||
| Q.7 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctHard | |
A hard multiple correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry For $X = (x, y, z)$: $|X - P|^2 - |X - Q|^2 = 2X\cdot(Q - P) + |P|^2 - |Q|^2$, with $Q - P = (3, 0, 4)$, $|P|^2 = 14$, $|Q|^2 = 69$.
$S$: $2(3x + 4z) - 55 = 50$, i.e. the plane $3x + 4z = 52.5$.
$T$: $|Y - Q|^2 - |Y - P|^2 = -2Y\cdot(Q-P) + 55 = 50$, i.e. the plane $3x + 4z = 2.5$.
These are parallel planes with normal $(3, 0, 4)$, a distance $\frac{52.5 - 2.5}{5} = 10$ apart.
(A) TRUE. $S$ is a whole plane, which contains triangles of every area, e.g. of area 1.
(B) TRUE. $T$ is a plane, so the segment joining any two of its points lies in it.
(C) TRUE. A rectangle with two vertices on each plane must have them adjacent (if opposite, the common midpoint of the diagonals would lie on both planes). So one side $AB$ lies in $S$, the opposite side $DC$ in $T$, and the sides $AD$, $BC$ cross the gap, each of length $w \ge 10$. With perimeter 48 the other side is $l = 24 - w$. Taking $AD$ along the normal, $w = 10$ and $l = 14$, and the rectangle can be placed anywhere and turned about the normal: infinitely many.
(D) TRUE. A square of perimeter 48 has side 12. Let $AD$ (length 12) cross the gap obliquely: normal component 10, component along the planes $\sqrt{144 - 100} = \sqrt{44}$. Choose $AB$ (length 12) in $S$ perpendicular to that in-plane component; then $AB \perp AD$, $B$ lies in $S$, and $C = B + \overrightarrow{AD}$ lies in $T$. This is a square.
Answer: (A), (B), (C), (D). | ||||
| Q.8 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $a^2 = 18$, so $\log_a 18 = 2$ and $\log_a 18^{5/4} = \frac{5}{4}\cdot 2 = \frac{5}{2}$. Hence $3x + 2y = \frac{5}{2}$.
$1080 = 216 \cdot 5 = 6^3 \cdot 5$, so $\sqrt{1080} = 6^{3/2}\,5^{1/2}$. And $b = 5^{-1/6}\,6^{-1/2}$, so $b^{-3} = 5^{1/2}\,6^{3/2} = \sqrt{1080}$. Hence $\log_b\sqrt{1080} = -3$ and $2x - y = -3$.
From the second, $y = 2x + 3$. Then $3x + 4x + 6 = \frac{5}{2}$, so $x = -\frac{1}{2}$ and $y = 2$.
$4x + 5y = -2 + 10 = 8$.
Answer: 8. | ||||
| Q.9 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $4x^3 + 3ax^2 + 2bx = x(4x^2 + 3ax + 2b)$. As $i\sqrt3 \ne 0$, it is a root of $4x^2 + 3ax + 2b = 0$; with real coefficients, $-i\sqrt3$ is the other root.
Sum of roots $0 = -\frac{3a}{4}$, so $a = 0$. Product $3 = \frac{2b}{4}$, so $b = 6$.
$f(1) = 1 + 0 + 6 + c = -9$, so $c = -16$.
$f(x) = x^4 + 6x^2 - 16 = (x^2 + 8)(x^2 - 2)$, with roots $\pm 2\sqrt2\,i$ and $\pm\sqrt2$.
Sum of $|\alpha|^2$: $8 + 8 + 2 + 2 = 20$.
Answer: 20. | ||||
| Q.10 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Expand along the first row: $|A| = 0 - 1\cdot(e - d) + c\,(b - a) = (d - e) + c(b - a)$.
Each of $d - e$ and $b - a$ takes the value $-1$ in 1 way, $0$ in 2 ways and $1$ in 1 way (out of 4).
Case $c = 0$: need $d - e = \pm1$: 2 choices of $(d, e)$, and $a, b$ free (4): 8 matrices.
Case $c = 1$: need $(d - e) + (b - a) = \pm 1$, i.e. one of them is $0$ and the other is $\pm1$: $2\cdot2 + 2\cdot2 = 8$ matrices.
Total $8 + 8 = 16$.
Answer: 16. | ||||
| Q.11 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Without restriction: $\binom{9}{2}\binom{7}{3} = 36 \cdot 35 = 1260$.
$s_1 \in X$: the other member of $X$ from 8, then $Y$ from 7: $8 \cdot \binom{7}{3} = 280$.
$s_2 \in Y$: two more for $Y$ from 8, then $X$ from 6: $\binom{8}{2}\binom{6}{2} = 28 \cdot 15 = 420$.
Both: $X = \{s_1, \text{one of 7}\}$, $Y = \{s_2, \text{two of the remaining 6}\}$: $7 \cdot 15 = 105$.
Inclusion-exclusion: $1260 - 280 - 420 + 105 = 665$.
Answer: 665. | ||||
| Q.12 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Write $p = \frac{\alpha-1}{\alpha}$, $q = \frac{\beta - 1}{\beta}$. The scalar triple product is
$\begin{vmatrix} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & \frac{1}{2}\end{vmatrix} = p\left(\frac{q}{2} - 1\right) - \left(\frac{1}{2} - 1\right) + (1 - q) = \frac{pq}{2} - p - q + \frac{3}{2} = 0$.
So $pq - 2p - 2q + 3 = 0$, i.e. $(p - 2)(q - 2) = 1$.
$p - 2 = -\frac{\alpha + 1}{\alpha}$ and $q - 2 = -\frac{\beta+1}{\beta}$, so $(\alpha+1)(\beta+1) = \alpha\beta$, giving $\alpha + \beta = -1$.
On the plane: $3\alpha + 3\beta - 2 + l = 0$, so $-3 - 2 + l = 0$ and $l = 5$.
Answer: 5. | ||||
| Q.13 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $P(X = x) = mx + c$ for $x = 0, \ldots, 4$.
Total probability: $10m + 5c = 1$. Mean: $\sum x(mx + c) = 30m + 10c = \frac{5}{2}$.
From the first, $10c = 2 - 20m$; then $30m + 2 - 20m = \frac{5}{2}$, so $m = \frac{1}{20}$ and $c = \frac{1}{10}$.
Probabilities: $0.10, 0.15, 0.20, 0.25, 0.30$ (sum 1).
$E(X^2) = 0 + 0.15 + 4(0.20) + 9(0.25) + 16(0.30) = 0.15 + 0.8 + 2.25 + 4.8 = 8$.
$\alpha = 8 - \left(\frac{5}{2}\right)^2 = \frac{7}{4}$, so $24\alpha = 42$.
Answer: 42. | ||||
| Q.14 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | Matching listHard | |
A hard matching list question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\alpha + \beta = -1$, so $1 + \alpha + \beta = 0$, and $\alpha, \beta = \frac{-1 \pm \sqrt5}{2}$ are irrational.
Three entries of $T$ sum to 0 only when they are $1, \alpha, \beta$ once each: a sum $p\cdot1 + q\alpha + r\beta$ with $p + q + r = 3$ equals $(p - r) + (q - r)\alpha$ (using $\beta = -1 - \alpha$), which is 0 only if $p = q = r = 1$.
(P) Every row and every column is a permutation of $(1, \alpha, \beta)$: a Latin square of order 3. There are $3! \cdot 2 = 12$. (P) → (2).
(Q) Columns are permutations of $T$; symmetry makes rows the same, so these are symmetric Latin squares. For order 3 the diagonal must hold all three symbols (each off-diagonal symbol appears in pairs, so each symbol appears an odd number of times, i.e. once, on the diagonal). Choose the diagonal: $3! = 6$ ways; then $a_{12}$ must be the symbol not in positions $(1,1)$, $(2,2)$, and so on — everything is forced. 6 matrices. (Q) → (4).
(R) $M = \begin{pmatrix}0 & a_{12} & a_{13}\\ -a_{12} & 0 & a_{23}\\ -a_{13} & -a_{23} & 0\end{pmatrix}$ with all off-diagonal entries non-zero. A $3\times3$ skew-symmetric matrix has determinant 0 and here rank 2; its null space is spanned by $w = (a_{23}, -a_{13}, a_{12})$. Since $M^T = -M$, the system $Mv = b$ is consistent exactly when $b \cdot w = 0$: $a_{12}a_{23} + 0 - a_{23}a_{12} = 0$. Consistent with a one-dimensional family of solutions: infinitely many. (R) → (3).
(S) $R_i = 0$ for all $i$ means $M\begin{pmatrix}1\\1\\1\end{pmatrix} = 0$, so $M$ is singular: $|\det M| = 0$. (S) → (5).
So (P)→(2), (Q)→(4), (R)→(3), (S)→(5).
Answer: (C). | ||||
| Q.15 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Matching listEasy | |
A easy matching list question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Tangency: distance of $(0, \alpha)$ from $2x - y = 0$ is $\frac{\alpha}{\sqrt5} = r$, so $\alpha = r\sqrt5$.
$\alpha + r = r(\sqrt5 + 1) = 5 + \sqrt5 = \sqrt5(\sqrt5 + 1)$, so $r = \sqrt5$ and $\alpha = 5$.
$A_1$ is the foot of the perpendicular from $(0, 5)$ to $2x - y = 0$: $(0,5) - \frac{2\cdot0 - 5}{5}(2, -1) = (0,5) + (2, -1) = (2, 4)$.
$B_1 = 2(0, 5) - (2, 4) = (-2, 6)$.
(P)→(4), (Q)→(2), (R)→(5), (S)→(3).
Answer: (C). | ||||
| Q.16 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Matching listMedium | |
A medium matching list question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Points: $L_1$: $(-11 + t, -21 + 2t, -29 + 3t)$; $L_2$: $(-16 + 3s, -11 + 2s, -4 + \gamma s)$.
$x$: $t = 3s - 5$. $y$: $-21 + 2(3s - 5) = -11 + 2s$, so $4s = 20$, $s = 5$, $t = 10$.
Point: $(-1, -1, 1)$. $z$: $-29 + 30 = 1 = -4 + 5\gamma$, so $\gamma = 1$. (P)→(3); $\overrightarrow{OR_1} = -\hat i - \hat j + \hat k$, (R)→(1).
Normal: $(1, 2, 3) \times (3, 2, 1) = (2 - 6,\ 9 - 1,\ 2 - 6) = (-4, 8, -4) \parallel (1, -2, 1)$, so $\hat n = \frac{1}{\sqrt6}(1, -2, 1)$. (Q)→(4).
$\overrightarrow{OR_1}\cdot\hat n = \frac{-1 + 2 + 1}{\sqrt6} = \frac{2}{\sqrt6} = \sqrt{\frac{2}{3}}$. (S)→(5).
Answer: (C). | ||||
| Q.17 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Matching listHard | |
A hard matching list question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions (P) $h(x) = g(x) + g\left(\frac12 - x\right)$. For $0 \le x \le \frac12$, also $0 \le \frac12 - x \le \frac12$, so $h = (1 - 2x) + (1 - 1 + 2x) = 1$. For $x < 0$ or $x > \frac12$ both terms are 0. Range $\{0, 1\}$: (P)→(5).
(Q) $h = f$. For $x \ne 0$ it is differentiable; at 0, $\frac{f(x) - 0}{x} = |x|\sin\frac1x \to 0$, so $f'(0) = 0$. Differentiable on $\mathbb{R}$. It is even ($f(-x) = f(x)$), so not one-one, and it is bounded below (for $|x| \ge 1$, $x^2\sin\frac{1}{x} > 0$; for $|x| < 1$, $|f| < 1$), so not onto. (Q)→(3).
(R) $h(x) = x - g(x)$: $h = x$ for $x < 0$ and $x > \frac12$, and $h = 3x - 1$ on $\left[0, \frac12\right]$. The pieces cover $(-\infty, 0)$, $\left[-1, \frac12\right]$ and $\left(\frac12, \infty\right)$, so $h$ is onto. It is not one-one ($h(-\frac12) = h(\frac16) = -\frac12$) and is discontinuous at 0 ($h(0^-) = 0$, $h(0) = -1$), so not differentiable. (R)→(2).
(S) $h = g$: range $[0, 1]$ (from $1 - 2x$ on $\left[0, \frac12\right]$, and 0 elsewhere). (S)→(4).
So (P)→(5), (Q)→(3), (R)→(2), (S)→(4).
Answer: (C). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\frac{e^2}{4\pi\varepsilon_0 r}$ is an energy, so $\frac{e^2}{\varepsilon_0}$ has dimensions of energy × length. $hc$ also has dimensions of energy × length ($E = \frac{hc}{\lambda}$).
So $\frac{e^2}{\varepsilon_0 h c}$ is dimensionless (it is $4\pi$ times the fine-structure constant), and so is any power of it: $\left(\frac{e^2}{\varepsilon_0 hc}\right)^n = e^{2n}\varepsilon_0^{-n}h^{-n}c^{-n}$.
Check by M, L, T, A: $[e] = AT$, $[\varepsilon_0] = M^{-1}L^{-3}T^4A^2$, $[h] = ML^2T^{-1}$, $[c] = LT^{-1}$. Then $[e^2/(\varepsilon_0 hc)] = A^2T^2 \cdot ML^3T^{-4}A^{-2}\cdot M^{-1}L^{-2}T \cdot L^{-1}T = M^0L^0T^0A^0$.
$(\alpha, \beta, \gamma, \delta) = (2n, -n, -n, -n)$.
Answer: (A). | ||||
| Q.2 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The field is azimuthal, $B = \frac{\mu_0 I}{2\pi r}$, so along any path $\vec B\cdot\vec{dl} = \frac{\mu_0 I}{2\pi r}\,(r\,d\phi) = \frac{\mu_0 I}{2\pi}d\phi$. The integral depends only on the change in the azimuthal angle $\phi$.
Start $(-\sqrt3a, a)$: $\phi = 150^\circ$. End $(a, a)$: $\phi = 45^\circ$. The straight segment (at $y = a$) does not cross the wire, and $|\Delta\phi| = 105^\circ = \frac{7\pi}{12}$.
$\left|\int\vec B\cdot\vec{dl}\right| = \frac{\mu_0 I}{2\pi}\cdot\frac{7\pi}{12} = \frac{7\mu_0 I}{24}$.
Answer: (A). | ||||
| Q.3 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Equilibrium is diametrically opposite the fixed bead. Let the moving bead be displaced by a small angle $\phi$ (at the centre) from that point. The chord between the beads is $d = 2R\cos\frac{\phi}{2}$.
Potential energy $U = \frac{kq^2}{2R\cos(\phi/2)}$ with $k = \frac{1}{4\pi\varepsilon_0}$. For small $\phi$, $\frac{1}{\cos(\phi/2)} \approx 1 + \frac{\phi^2}{8}$, so $U \approx \text{const} + \frac{kq^2}{16R}\phi^2$.
Kinetic energy $\frac{1}{2}mR^2\dot\phi^2$. Comparing with $\frac{1}{2}I\dot\phi^2 + \frac{1}{2}K\phi^2$: $K = \frac{kq^2}{8R}$, $I = mR^2$.
$\omega^2 = \frac{K}{I} = \frac{kq^2}{8mR^3} = \frac{q^2}{32\pi\varepsilon_0 R^3 m}$.
Answer: (B). | ||||
| Q.4 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $F = -20(x - 0.5)$: SHM about $x_0 = 0.5$ m with $\omega^2 = \frac{20}{5} = 4$, $\omega = 2$ rad/s.
Starting from rest at $x = 1$ m, the amplitude is $0.5$ m: $x = 0.5 + 0.5\cos 2t$, $v = -\sin 2t$ (m/s).
At $t = \frac{\pi}{4}$: $\cos\frac{\pi}{2} = 0$, so $x = 0.5$ m; $v = -1$ m/s, momentum $p = 5 \times (-1) = -5$ kg m/s.
Answer: (C). | ||||
| Q.5 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Multiple correctMedium | |
A medium multiple correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Circular orbit: $\frac{mv^2}{r} = kr$, so $v = r\sqrt{\frac{k}{m}}$.
$L = mvr = r^2\sqrt{mk} = n\hbar$, so $r^2 = \frac{n\hbar}{\sqrt{mk}} = n\hbar\sqrt{\frac{1}{mk}}$. (A) TRUE.
$v^2 = \frac{k}{m}r^2 = \frac{k}{m}\cdot\frac{n\hbar}{\sqrt{mk}} = n\hbar\sqrt{\frac{k}{m^3}}$. (B) TRUE.
$\frac{L}{mr^2} = \frac{v}{r} = \sqrt{\frac{k}{m}}$. (C) TRUE.
$E = \frac{1}{2}mv^2 + \frac{1}{2}kr^2 = \frac12kr^2 + \frac12kr^2 = kr^2 = n\hbar\sqrt{\frac{k}{m}}$, not $\frac{n\hbar}{2}\sqrt{\frac{k}{m}}$. (D) FALSE.
Answer: (A), (B), (C). | ||||
| Q.6 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Multiple correctMedium | |
A medium multiple correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Wave speeds: $v_1 = \sqrt{T/\mu}$ on PO, $v_2 = \sqrt{T/4\mu} = \frac{v_1}{2}$ on OQ. Note $\nu_0 = \frac{v_1}{2L}$.
Node at O: each part is a string fixed at both ends. PO: $f = p\frac{v_1}{2L} = p\nu_0$. OQ: $f = q\frac{v_2}{2(2L)} = q\frac{v_1}{8L} = \frac{q}{4}\nu_0$. Smallest common value: $p = 1$, $q = 4$, $f = \nu_0$. (A) TRUE.
At that frequency PO has nodes only at P and O; OQ vibrates in its 4th harmonic with 5 nodes including O and Q. Distinct nodes: P, O, three inside OQ, and Q: 6. (C) TRUE.
Antinode at O: each part is fixed at one end and free (antinode) at O. PO: $L = (2p - 1)\frac{\lambda_1}{4}$, $f = (2p-1)\frac{\nu_0}{2}$. OQ: $2L = (2q - 1)\frac{\lambda_2}{4}$, $f = (2q - 1)\frac{\nu_0}{8}$. Equal frequencies need $4(2p - 1) = 2q - 1$: even = odd, impossible. So no such mode exists: (D) TRUE and (B) FALSE.
Answer: (A), (C), (D). | ||||
| Q.7 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The image falls back on the object when the rays strike the plane mirror STU normally, i.e. travel parallel to the axis inside the glass. So O must be at the focus of the system above the mirror (taken as thin, paraxial rays).
Above the convex glass is a thin layer of liquid with a flat top (QPR) and a concave lower face of radius 9 cm: a plano-concave liquid lens, power $-\frac{n - 1}{9}$ (cm$^{-1}$). The glass is plano-convex, power $\frac{1.6 - 1}{9}$. Together (in contact): $P = \frac{1.6 - n}{9}$, the same as a single refraction at the liquid–glass surface.
So $h = \frac{1}{P} = \frac{9}{1.6 - n}$ cm.
(A) $n = 1.42$: $h = \frac{9}{0.18} = 50$ cm. TRUE.
(B) $n = 1.35$: $h = \frac{9}{0.25} = 36$ cm. TRUE.
(C) $n = 1.45$: $h = \frac{9}{0.15} = 60$ cm, not 65. FALSE.
(D) $n = 1.48$: $h = \frac{9}{0.12} = 75$ cm, not 85. FALSE.
Answer: (A), (B). | ||||
| Q.8 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $-73\ ^\circ$C $= 200$ K and $27\ ^\circ$C $= 300$ K.
$Q = m\int_{200}^{300} kT\,dT = 1 \cdot \frac{k}{2}\left(300^2 - 200^2\right) = \frac{k}{2}(90000 - 40000) = 25000\,k$.
Answer: 25000. | ||||
| Q.9 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion No external torque acts about the large disc's axis, and everything starts at rest, so the total angular momentum about that axis stays zero.
Let the large disc turn at $\Omega$ (opposite to the small disc). The small disc's centre is carried round a circle of radius $R$, contributing $MR^2\Omega$; its spin contributes $\frac{1}{2}M\left(\frac{R}{2}\right)^2\omega = \frac{MR^2}{8}\omega$ (with $\omega$ its angular speed in the ground frame); the large disc contributes $\frac{1}{2}MR^2\Omega$.
$\frac{MR^2}{8}\omega = \left(\frac{1}{2}MR^2 + MR^2\right)\Omega = \frac{3}{2}MR^2\Omega$, so $\Omega = \frac{\omega}{12}$.
Answer: 12. | ||||
| Q.10 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Unpolarized light through the first polaroid: $\frac{I_B}{2}$. Through the second (Malus, $45^\circ$): $\frac{I_B}{2}\cos^2 45^\circ = \frac{I_B}{4}$.
New ratio $r = \frac{I_A}{I_B/4} = 4 \times 2 = 8$.
Answer: 8. | ||||
| Q.11 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Let $c$ be the speed of sound (still air).
Case 1 (approaching): $288 = 240\cdot\frac{c + v}{c - v}$, so $\frac{c + v}{c - v} = 1.2$.
Case 2 (receding): $n = 240\cdot\frac{c - v}{c + v} = \frac{240}{1.2} = 200$.
Answer: 200. | ||||
| Q.12 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Let $A$ be the tank's cross-section, $a$ the hole's, and $y$ the water height at a moment.
Tank 1: efflux speed $\sqrt{2gy}$ (Torricelli), so $A\frac{dy}{dt} = -a\sqrt{2gy}$ and $t_1 = \frac{A}{a\sqrt{2g}}\int_0^h\frac{dy}{\sqrt y} = \frac{A}{a\sqrt{2g}}\cdot 2\sqrt h$.
Tank 2: the pipe (same radius, full of water) delivers at the ground, a drop of $y + H$ below the free surface, so by Bernoulli the outflow speed is $\sqrt{2g(y + H)}$. Then $t_2 = \frac{A}{a\sqrt{2g}}\int_0^h\frac{dy}{\sqrt{y + H}} = \frac{A}{a\sqrt{2g}}\cdot2\left(\sqrt{h + H} - \sqrt H\right)$.
With $H = \frac{16}{9}h$: $\sqrt{h + H} - \sqrt{H} = \frac{5}{3}\sqrt h - \frac{4}{3}\sqrt h = \frac{1}{3}\sqrt h$.
$\frac{t_1}{t_2} = \frac{\sqrt h}{\frac13\sqrt h} = 3$.
Answer: 3. | ||||
| Q.13 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalHard | |
A hard numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For the rod and string to turn together about O, the rod must start rotating about O as a rigid extension of the string: its centre (distance $\frac{3L}{2}$ from O) moves at $v = \frac{3L}{2}\omega$ and it spins at the same $\omega$. The string can only pull along its length, so it gives no impulse perpendicular to the rod.
Linear impulse (perpendicular to the rod): $P = mv = m\cdot\frac{3L}{2}\omega$.
Angular impulse about the centre of mass: $Px = I_{cm}\omega = \frac{mL^2}{12}\omega$.
Divide: $x = \frac{L^2/12}{3L/2} = \frac{L}{18}$, on the side of the centre away from O (as in the figure).
Answer: 18. | ||||
| Q.14 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Matching listMedium | |
A medium matching list question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the diagram: J $(P_0, T_0)$, K $(P_0, 3T_0)$, L $(2P_0, 3T_0)$, M $(2P_0, T_0)$.
JK isobaric heating: $W = \mathcal R\Delta T = 2\mathcal RT_0$, $\Delta U = \frac32\mathcal R(2T_0) = 3\mathcal RT_0$.
KL isothermal at $3T_0$, pressure doubles so volume halves: $W = 3\mathcal RT_0\ln\frac12 = -3\mathcal RT_0\ln2$; $\Delta U = 0$, so heat $= W = -3\mathcal RT_0\ln 2$.
LM isobaric cooling: $W = \mathcal R(T_0 - 3T_0) = -2\mathcal RT_0$.
MJ isothermal at $T_0$, pressure halves: $W = \mathcal RT_0\ln 2$; $\Delta U = 0$.
Net work: $2\mathcal RT_0 - 3\mathcal RT_0\ln2 - 2\mathcal RT_0 + \mathcal RT_0\ln 2 = -2\mathcal RT_0\ln 2$.
(P)→(4), (Q)→(3), (R)→(5), (S)→(2).
Answer: (B). | ||||
| Q.15 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Matching listMedium | |
A medium matching list question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Each gap (1–2, 2–3, 3–4) is a capacitor $C_0$.
(P) Three gaps in series: $\frac{C_0}{3}$. (P)→(3).
(Q) Shorting $S_2$–$S_3$ removes the middle gap; gaps 1–2 and 3–4 are in series: $\frac{C_0}{2}$. (Q)→(2).
(R) Terminals $S_1$ and $S_3$; node N $= S_2 + S_4$. Gap 1–2 is $C_0$ between $S_1$ and N. Gaps 2–3 and 3–4 both join N to $S_3$: $2C_0$ in parallel. In series: $\frac{C_0\cdot 2C_0}{3C_0} = \frac{2C_0}{3}$. (R)→(4).
(S) Terminal A $= S_1 + S_3$, terminal B $= S_2 + S_4$. Every gap (1–2, 2–3, 3–4) lies between A and B: three $C_0$ in parallel, $3C_0$. (S)→(1).
Answer: (C). | ||||
| Q.16 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Matching listMedium | |
A medium matching list question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Deviation with one internal reflection: refraction in $(\theta_0 - \phi_0)$, reflection $(180^\circ - 2\phi_0)$, refraction out $(\theta_0 - \phi_0)$, so $\alpha = 180^\circ + 2\theta_0 - 4\phi_0$.
$\alpha = 180^\circ$ needs $\theta_0 = 2\phi_0$. One solution is always $\theta_0 = \phi_0 = 0$ (normal incidence, straight back). Otherwise $\sin2\phi_0 = n\sin\phi_0$ gives $\cos\phi_0 = \frac{n}{2}$.
(P) $n = 2$: $\cos\phi_0 = 1$, only $\phi_0 = 0$, so $\theta_0 = 0^\circ$ only. (P)→(5).
(Q), (R) $n = \sqrt3$: $\cos\phi_0 = \frac{\sqrt3}{2}$, $\phi_0 = 30^\circ$, $\theta_0 = 60^\circ$; together with the normal ray: $\theta_0 \in \{60^\circ, 0^\circ\}$, $\phi_0 \in \{30^\circ, 0^\circ\}$. (Q)→(2), (R)→(1).
(S) $n = \sqrt2$, $\theta_0 = 45^\circ$: $\sin\phi_0 = \frac{1}{2}$, $\phi_0 = 30^\circ$; $\alpha = 180^\circ + 90^\circ - 120^\circ = 150^\circ$. (S)→(4).
Answer: (A). | ||||
| Q.17 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Matching listMedium | |
A medium matching list question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current From the figure: with K1 closed (K2 open) the battery (20 V), $R_0 = 5\ \Omega$ and $L = 25$ mH form a series loop. With K2 closed and K1 open, $C_0 = 10\ \mu$F is connected across $L$ alone.
(P) Just after K1 closes, the inductor keeps the current at its earlier value, zero: $I_1 = 0$. (P)→(1).
(Q) In steady state the inductor is a plain wire: $I_2 = \frac{20}{5} = 4$ A. (Q)→(3).
(R) LC circuit: $\omega_0 = \frac{1}{\sqrt{LC_0}} = \frac{1}{\sqrt{25\times10^{-3}\times10\times10^{-6}}} = \frac{1}{5\times10^{-4}} = 2000$ rad/s $= 2$ krad/s. (R)→(2).
(S) Energy: $\frac12 LI_2^2 = \frac12 C_0V_0^2$, so $V_0 = I_2\sqrt{\frac{L}{C_0}} = 4\sqrt{\frac{25\times10^{-3}}{10\times10^{-6}}} = 4 \times 50 = 200$ V. (S)→(5).
Answer: (A). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Same vessel and temperature, so partial pressure $\propto$ moles. X gives 2 atm; Y adds $6 - 2 = 4$ atm, so $n_Y = 2n_X$.
Molar masses: $M_X = \frac{10}{n_X}$, $M_Y = \frac{80}{2n_X} = \frac{40}{n_X}$, so $M_Y = 4M_X$.
$u_{rms} = \sqrt{\frac{3RT}{M}}$, so $\frac{u_X}{u_Y} = \sqrt{\frac{M_Y}{M_X}} = \sqrt4 = 2$, i.e. $2 : 1$.
Answer: (D). | ||||
| Q.2 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Nitrous acid (N in $+3$) is unstable in water and disproportionates to nitric acid ($+5$) and nitric oxide ($+2$):
$\mathrm{3HNO_2 \rightarrow HNO_3 + 2NO + H_2O}$, i.e. in water $\mathrm{3HNO_2 \rightarrow H_3O^+ + NO_3^- + 2NO}$.
Check: one N goes up by 2, two N go down by 1 each. $\mathrm{NO_2}$ ($+4$) and $\mathrm{N_2O}$ ($+1$) are not the products, and $\mathrm{NO^-}$ is not formed.
Answer: (A). | ||||
| Q.3 | BiomoleculesProteins | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules \"Aspartyl phenylalanine methyl ester\" (Asp-Phe-OMe): the $\alpha$-carboxyl of aspartic acid forms the peptide (amide) bond with the amino group of phenylalanine; the side-chain $\mathrm{CH_2COOH}$ of aspartic acid stays free; the carboxyl of phenylalanine is a methyl ester; both residues keep their natural configuration, as drawn in the given structures.
(A) joins aspartic acid through its side-chain carboxyl (a $\beta$-peptide). Wrong.
(B) $\mathrm{HOOC{-}CH_2{-}CH(NH_2){-}CO{-}NH{-}CH(CH_2Ph){-}COOCH_3}$, with the configurations of the given amino acids: this is aspartame.
(C) is Phe-Asp: phenylalanine supplies the free amine, the wrong order.
(D) has the methyl ester on aspartic acid's side chain and a free acid on phenylalanine.
Answer: (B). | ||||
| Q.4 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) $\mathrm{[Ni(CO)_4]}$ is tetrahedral: no geometrical isomers. Fails Set-I.
(B) $\mathrm{[Co(NH_3)_6][Cr(CN)_6]}$ and $\mathrm{[Cr(NH_3)_6][Co(CN)_6]}$ exchange ligands between the metals: coordination isomers, not ionization isomers. Fails Set-II.
(C) $\mathrm{[Co(NH_3)_3(NO_2)_3]}$ ($\mathrm{MA_3B_3}$) has fac and mer isomers; $\mathrm{[Co(en)_2Cl_2]}$ has cis and trans. $\mathrm{[Co(NH_3)_5Cl]SO_4}$ and $\mathrm{[Co(NH_3)_5(SO_4)]Cl}$ give different ions in solution ($\mathrm{SO_4^{2-}}$ vs $\mathrm{Cl^-}$): ionization isomers. Both sets satisfied.
(D) $\mathrm{[Cr(NH_3)_5Cl]Cl_2}$ ($\mathrm{MA_5B}$) has no geometrical isomers, and the Set-II pair are hydrate isomers.
Answer: (C). | ||||
| Q.5 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | Multiple correctEasy | |
A easy multiple correct question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom (A) TRUE. Position and momentum cannot both be known exactly, so a definite trajectory (orbit) cannot be specified.
(B) TRUE. A bound electron has negative energy; an electron infinitely far away (at rest) has zero energy. So the 2s electron is lower in energy.
(C) TRUE. $E_n = -\frac{13.6Z^2}{n^2}$ eV is most negative for $n = 1$, the most stable orbit.
(D) FALSE. In Bohr's model $v_n \propto \frac{Z}{n}$: the speed decreases as $n$ increases.
Answer: (A), (B), (C). | ||||
| Q.6 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Multiple correctMedium | |
A medium multiple correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Cumene process: cumene hydroperoxide splits into phenol and acetone, so P $= \mathrm{CH_3COCH_3}$.
Chlorination of the methyl ketone (3 $\mathrm{Cl_2}$) replaces one methyl's H's: Q $= \mathrm{CH_3COCCl_3}$ (1,1,1-trichloroacetone).
Haloform cleavage with $\mathrm{Ca(OH)_2}$: R $= \mathrm{CHCl_3}$ and S $= \mathrm{(CH_3COO)_2Ca}$.
(A) TRUE. $\mathrm{CHCl_3}$ with KOH gives $\mathrm{^-CCl_3}$, which adds to acetone; acidification gives $\mathrm{(CH_3)_2C(OH)CCl_3}$ (chloretone), the structure shown.
(B) TRUE. Chloroform is oxidised by air in light: $\mathrm{2CHCl_3 + O_2 \rightarrow 2COCl_2 + 2HCl}$ (phosgene).
(C) FALSE. Q with aqueous NaOH undergoes the haloform reaction: $\mathrm{CHCl_3 + CH_3COONa}$. $\mathrm{Cl_3CCH_2OH}$ and $\mathrm{Cl_3CCOONa}$ come from the Cannizzaro reaction of chloral, not from Q.
(D) TRUE. Dry distillation of calcium acetate: $\mathrm{(CH_3COO)_2Ca \xrightarrow{\Delta} CH_3COCH_3 + CaCO_3}$.
Answer: (A), (B), (D). | ||||
| Q.7 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Multiple correctMedium | |
A medium multiple correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Follow the octet rule (every atom other than H has 8 electrons; H its duplet): $\mathrm{CO_2}$, $\mathrm{C_2H_4}$, $\mathrm{HCl}$, $\mathrm{O_3}$.
Do not: $\mathrm{NO}$ and $\mathrm{NO_2}$ (odd-electron molecules, N has 7 electrons), $\mathrm{BCl_3}$ (B has only 6), $\mathrm{H_2SO_4}$ (S has an expanded octet, 12 electrons in its usual structure).
(A) $\mathrm{CO_2}$, $\mathrm{C_2H_4}$, $\mathrm{HCl}$ follow; NO does not: three. TRUE.
(B) $\mathrm{O_3}$, $\mathrm{HCl}$ follow; $\mathrm{NO_2}$, $\mathrm{H_2SO_4}$ do not: two. FALSE.
(C) none of $\mathrm{BCl_3}$, NO, $\mathrm{NO_2}$, $\mathrm{H_2SO_4}$: zero. FALSE.
(D) $\mathrm{CO_2}$, $\mathrm{O_3}$, $\mathrm{C_2H_4}$ follow; $\mathrm{BCl_3}$ does not: three. TRUE.
Answer: (A), (D). | ||||
| Q.8 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics From the diagram: X (335 K, 10 L) → Y (335 K, 20 L) is isothermal, Y → Z (415 K, 20 L) is isochoric.
Enthalpy of an ideal gas depends only on temperature, and it is a state function: $\Delta H = nC_{P,m}\Delta T$ from X to Z.
$C_{P,m} = C_{V,m} + R = 12 + 8.3 = 20.3$ J K$^{-1}$ mol$^{-1}$; $\Delta T = 415 - 335 = 80$ K.
$\Delta H = 5 \times 20.3 \times 80 = 8120$ J.
Answer: 8120. | ||||
| Q.9 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The slow step decides the rate: rate $= k_2[\mathrm{N_2O_2}][\mathrm{H_2}]$.
From the fast pre-equilibrium: $[\mathrm{N_2O_2}] = \frac{k_1}{k_{-1}}[\mathrm{NO}]^2$.
Rate $= \frac{k_1k_2}{k_{-1}}[\mathrm{NO}]^2[\mathrm{H_2}]$: order $2 + 1 = 3$.
Answer: 3. | ||||
| Q.10 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Three crossed aldol additions of HCHO on the $\alpha$-carbon of acetaldehyde give $\mathrm{(HOCH_2)_3C{-}CHO}$; a crossed Cannizzaro with HCHO then reduces it to pentaerythritol and oxidises HCHO to formate.
P $= \mathrm{C(CH_2OH)_4}$ (no Tollens' test); Q $= \mathrm{HCOONa}$, which on acidification gives formic acid (positive Tollens' test).
P with excess cyclohexanone and acid: each pair of $\mathrm{CH_2OH}$ groups forms a cyclic ketal (1,3-dioxane ring) with one cyclohexanone. R is the dispiro bis-ketal: a central C bearing four $\mathrm{CH_2O}$ groups, each pair of O atoms bound to the spiro carbon of a cyclohexane ring.
$\mathrm{CH_2}$ groups: 4 (from pentaerythritol) $+ 2 \times 5$ (each cyclohexane ring: 5 $\mathrm{CH_2}$ and the spiro C) $= 14$. Oxygen atoms: 4.
Sum $= 14 + 4 = 18$.
Answer: 18. | ||||
| Q.11 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Isoelectronic species have the same total number of electrons. CO has $6 + 8 = 14$ electrons.
$\mathrm{Ni(CO)_4}$: $28 + 4 \times 14 = 84$.
$\mathrm{V(CO)_6}$: $23 + 84 = 107$. $\mathrm{Cr(CO)_5}$: $24 + 70 = 94$. $\mathrm{Cu(CO)_3}$: $29 + 42 = 71$. $\mathrm{Mn(CO)_5}$: $25 + 70 = 95$. $\mathrm{Fe(CO)_5}$: $26 + 70 = 96$.
$\mathrm{[Co(CO)_3]^{3-}}$: $27 + 42 + 3 = 72$. $\mathrm{[Cr(CO)_4]^{4-}}$: $24 + 56 + 4 = 84$. $\mathrm{Ir(CO)_3}$: $77 + 42 = 119$.
Only $\mathrm{[Cr(CO)_4]^{4-}}$ has 84 electrons (it has the same tetrahedral 4-carbonyl framework as $\mathrm{Ni(CO)_4}$): 1 species.
Counted as valence electrons on the metal (the 18-electron count of $\mathrm{Ni(CO)_4}$: $10 + 4\times2$): $\mathrm{Fe(CO)_5}$ $8 + 10 = 18$, $\mathrm{[Co(CO)_3]^{3-}}$ $9 + 3 + 6 = 18$, $\mathrm{[Cr(CO)_4]^{4-}}$ $6 + 4 + 8 = 18$; the others have 15–17. That gives 3.
Answer: 1 or 3. | ||||
| Q.12 | Chemistry in Everyday LifeCleansing agents | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on cleansing agents, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Start: $\mathrm{HC{\equiv}C{-}(CH_2)_{15}{-}CO_2Et}$ (18 carbons in the acid chain).
(i) $\mathrm{Hg^{2+}/H_3O^+}$: Markovnikov hydration of the terminal alkyne gives the methyl ketone $\mathrm{CH_3CO(CH_2)_{15}CO_2Et}$.
(ii) Clemmensen reduction ($\mathrm{Zn{-}Hg/HCl}$): $\mathrm{C{=}O \rightarrow CH_2}$, giving $\mathrm{CH_3(CH_2)_{16}CO_2Et}$.
(iii) $\mathrm{H_3O^+}$, $\Delta$: ester hydrolysis. P $= \mathrm{CH_3(CH_2)_{16}COOH}$, stearic acid ($\mathrm{C_{17}H_{35}COOH}$).
Q = glyceryl tristearate. Saponification gives 3 mol sodium stearate per mol Q; with $\mathrm{CaCl_2}$: R $= \mathrm{(C_{17}H_{35}COO)_2Ca}$, so 1 mol Q gives $\frac{3}{2}$ mol R.
Molar mass of R: $\mathrm{C_{17}H_{35}COO}$ $= 18(12) + 35(1) + 2(16) = 283$; R $= 2(283) + 40 = 606$ g/mol.
Mass $= 1.5 \times 606 = 909$ g.
Answer: 909. | ||||
| Q.13 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{Mn^{3+}}$ is $d^4$: 4 (high spin) or 2 (low spin) unpaired electrons, paramagnetic either way — $\mathrm{[Mn(NH_3)_6]^{3+}}$ and $\mathrm{[MnCl_6]^{3-}}$ are paramagnetic.
$\mathrm{Fe^{3+}}$ is $d^5$: 5 or 1 unpaired — $\mathrm{[FeF_6]^{3-}}$ and $\mathrm{[Fe(NH_3)_6]^{3+}}$ are paramagnetic.
$\mathrm{Co^{3+}}$ is $d^6$: with weak-field $\mathrm{F^-}$, high spin $t_{2g}^4e_g^2$ (4 unpaired), paramagnetic; with strong-field en, low spin $t_{2g}^6$ (0 unpaired), diamagnetic.
Only $\mathrm{[Co(en)_3]^{3+}}$ is diamagnetic.
Answer: 1. | ||||
| Q.14 | ElectrochemistryConductance | Class 12: Electrochemistry | Matching listMedium | |
A medium matching list question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Dilution is negligible, so watch which ions are replaced by which.
(P) $\mathrm{KCl + AgNO_3}$: $\mathrm{Cl^-}$ (7.6) is precipitated as AgCl and replaced by $\mathrm{NO_3^-}$ (7.2): a slight fall. After the end point excess $\mathrm{Ag^+ + NO_3^-}$ raises the conductance: graph (3).
(Q) $\mathrm{AgNO_3 + KCl}$: $\mathrm{Ag^+}$ (6.2) is replaced by $\mathrm{K^+}$ (7.4): a slight rise. After the end point excess $\mathrm{K^+ + Cl^-}$ (15.0) gives a steeper rise: graph (4).
(R) NaOH + HCl: $\mathrm{OH^-}$ (19.9) is replaced by $\mathrm{Cl^-}$ (7.6): a steep fall. After the end point excess $\mathrm{H^+ + Cl^-}$ (42.6) gives a steep rise: graph (2).
(S) NaOH + $\mathrm{CH_3COOH}$: $\mathrm{OH^-}$ (19.9) is replaced by $\mathrm{CH_3COO^-}$ (4.1): a fall. After the end point the weak acid, its ionisation further suppressed by the acetate present, adds almost no ions: nearly flat. Graph (5).
P→3, Q→4, R→2, S→5.
Answer: (C). | ||||
| Q.15 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Xe has 8 valence electrons; each F uses 1, each O (double bond) uses 2.
(P) $\mathrm{XeF_2}$: 2 bond pairs + 3 lone pairs = 5 electron pairs, trigonal bipyramidal arrangement (linear shape). (5).
(Q) $\mathrm{XeF_4}$: 4 bp + 2 lp = 6, octahedral (square planar shape). (3).
(R) $\mathrm{XeO_3}$: 3 $\sigma$ bp + 1 lp = 4, tetrahedral (pyramidal shape). (2).
(S) $\mathrm{XeO_3F_2}$: 5 $\sigma$ bp, 0 lp, trigonal bipyramidal. (4).
P→5, Q→3, R→2, S→4.
Answer: (B). | ||||
| Q.16 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Matching listHard | |
A hard matching list question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (P) Cyclohexene: ozonolysis gives hexanedial $\mathrm{OHC(CH_2)_4CHO}$; base and heat give the intramolecular aldol condensation product cyclopent-1-ene-1-carbaldehyde. The CHO is protected as the ethylene acetal. Hydroboration–oxidation puts OH on the less substituted alkene carbon (the ring CH next to the acetal carbon). $\mathrm{H_3O^+}$ frees the aldehyde: 2-hydroxycyclopentane-1-carbaldehyde; $\mathrm{NaBH_4}$ reduces it to 2-(hydroxymethyl)cyclopentan-1-ol. (P)→(3).
(Q) 1-Methylcyclohexene: ozonolysis gives 6-oxoheptanal $\mathrm{CH_3CO(CH_2)_4CHO}$. In the aldol the more reactive aldehyde is the acceptor, attacked by the ketone's ring-side $\alpha$-carbon (five-membered ring); dehydration gives 1-acetylcyclopentene. The ketone is protected as a ketal; hydroboration–oxidation puts OH on the ring CH next to it; deprotection gives 2-acetylcyclopentanol; $\mathrm{NaBH_4}$ gives 2-(1-hydroxyethyl)cyclopentan-1-ol. (Q)→(5).
(R) 2-Methylcyclopent-2-en-1-one: ketal protection, then oxymercuration–demercuration adds OH Markovnikov, to the methyl-bearing carbon. Deprotection gives 2-hydroxy-2-methylcyclopentanone; $\mathrm{NaBH_4}$ gives 1-methylcyclopentane-1,2-diol. (R)→(4).
(S) Same ketal, but hydroboration–oxidation adds OH anti-Markovnikov, to C-3 (the CH), with H on the methyl carbon. Deprotection gives 3-hydroxy-2-methylcyclopentanone; $\mathrm{NaBH_4}$ gives 2-methylcyclopentane-1,3-diol. (S)→(1).
P→3, Q→5, R→4, S→1.
Answer: (A). | ||||
| Q.17 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Matching listMedium | |
A medium matching list question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (P) Benzenesulfonic acid, fused with NaOH then acidified, gives phenol; concentrated $\mathrm{HNO_3}$ nitrates it fully to picric acid, 2,4,6-trinitrophenol. (P)→(3).
(Q) Nitrobenzene → m-dinitrobenzene (nitration, meta) → m-phenylenediamine (Sn/HCl) → bis-diazonium salt → resorcinol (warm water) → nitration of the very activated ring at 2, 4, 6: 2,4,6-trinitroresorcinol. (Q)→(5).
(R) Resorcinol: conc. $\mathrm{H_2SO_4}$ sulfonates the open activated positions 4 and 6, blocking them; nitration then goes to C-2 (between the two OH); $\mathrm{H_3O^+}$, $\Delta$ removes the sulfonic groups: 2-nitroresorcinol. (R)→(4).
(S) Toluene → benzoic acid ($\mathrm{KMnO_4}$) → 3,5-dinitrobenzoic acid (nitration with heat, meta to COOH) → 3,5-dinitrobenzamide ($\mathrm{SOCl_2}$, $\mathrm{NH_3}$) → 3,5-dinitroaniline (Hofmann, $\mathrm{Br_2}$/NaOH) → diazonium → 3,5-dinitrophenol. (S)→(1).
P→3, Q→5, R→4, S→1.
Answer: (C). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | 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 $a = \sin^{-1}\frac{3}{5}$, an angle in $(0, \frac{\pi}{2})$ with $\cos a = \frac{4}{5}$, so $\tan a = \frac{3}{4}$.
Let $b = \cos^{-1}\frac{2}{\sqrt5}$, in $(0, \frac{\pi}{2})$ with $\sin b = \frac{1}{\sqrt5}$, so $\tan b = \frac{1}{2}$.
$\tan 2b = \frac{2\cdot\frac12}{1 - \frac14} = \frac{4}{3}$.
$\tan(a - 2b) = \frac{\frac34 - \frac43}{1 + \frac34\cdot\frac43} = \frac{-\frac{7}{12}}{2} = -\frac{7}{24}$.
Answer: (B). | ||||
| Q.2 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The parabolas $y^2 = 4x$ and $y^2 = 12 - 2x$ meet where $4x = 12 - 2x$, i.e. at $x = 2$, $y = 2\sqrt2$ (taking $y \ge 0$).
The line $3y + 2\sqrt2\,x = 10\sqrt2$ also passes through $(2, 2\sqrt2)$: $6\sqrt2 + 4\sqrt2 = 10\sqrt2$. Writing it as $x = 5 - \frac{3y}{2\sqrt2}$:
- with $y^2 = 4x$: $y^2 + 3\sqrt2\,y - 20 = 0$, roots $y = 2\sqrt2$ and $-5\sqrt2$, so for $y \ge 0$ they meet only at $(2, 2\sqrt2)$;
- with $y^2 = 12 - 2x$: $y^2 - \frac{3}{\sqrt2}y - 2 = 0$, roots $y = 2\sqrt2$ and $-\frac{1}{\sqrt2}$, again only that point.
Cut the region in horizontal strips. For $0 \le y \le 2\sqrt2$ the point must have $x \ge \frac{y^2}{4}$, $x \le 6 - \frac{y^2}{2}$ and $x \le 5 - \frac{3y}{2\sqrt2}$. Since $\left(6 - \frac{y^2}{2}\right) - \left(5 - \frac{3y}{2\sqrt2}\right) = 1 + \frac{3y}{2\sqrt2} - \frac{y^2}{2} \ge 0$ on this range (it is $1$ at $y = 0$ and $0$ at $y = 2\sqrt2$), the line is the binding right boundary. For $y > 2\sqrt2$ we would need $\frac{y^2}{4} \le 6 - \frac{y^2}{2}$, impossible, so the region stops at $y = 2\sqrt2$.
Area $= \int_0^{2\sqrt2}\left(5 - \frac{3y}{2\sqrt2} - \frac{y^2}{4}\right)dy = 10\sqrt2 - \frac{3}{2\sqrt2}\cdot 4 - \frac{(2\sqrt2)^3}{12} = 10\sqrt2 - 3\sqrt2 - \frac{4\sqrt2}{3} = \frac{17\sqrt2}{3}$.
So $\alpha = \frac{17}{3}$.
Answer: (B). | ||||
| Q.3 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability As $x \to 0^+$ the base $\to 0 + 1 + 0 = 1$ and the exponent $\to \infty$: a $1^\infty$ form, so
$\lim (\text{base})^{2/x} = \exp\left(\lim_{x\to0^+} \frac{2}{x}\left(\sin(\sin kx) + \cos x + x - 1\right)\right)$.
Near $0$: $\sin(\sin kx) \approx kx$, $\cos x - 1 \approx -\frac{x^2}{2}$. So $\frac{2}{x}(kx - \frac{x^2}{2} + x) \to 2(k + 1)$.
$2(k + 1) = 6$ gives $k = 2$.
Answer: (B). | ||||
| Q.4 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions For $x \ne 0$, $f(x) = 0$ exactly when $\sin\frac{\pi}{x^2} = 0$, i.e. $\frac{\pi}{x^2} = n\pi$ with $n$ a positive integer, so $x = \pm\frac{1}{\sqrt n}$, $n = 1, 2, 3, \dots$ (and $x = 0$ itself).
(A) In $\left[\frac{1}{10^{10}}, \infty\right)$ we need $\frac{1}{\sqrt n} \ge 10^{-10}$, i.e. $n \le 10^{20}$: finitely many. False.
(B) In $\left[\frac{1}{\pi}, \infty\right)$ we need $n \le \pi^2 \approx 9.87$: $n = 1, \dots, 9$ give 9 solutions. False.
(C) In $\left(0, \frac{1}{10^{10}}\right)$ every $n > 10^{20}$ works: infinitely many. False.
(D) In $\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)$ we need $\pi < \sqrt n < \pi^2$, i.e. $\pi^2 < n < \pi^4$, so $9.87 < n < 97.4$: $n = 10, 11, \dots, 97$, which is 88 solutions, more than 25. True.
Answer: (D). | ||||
| Q.5 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctHard | |
A hard multiple correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability As $x \to \infty$: $\sin\frac{1}{x^2} \sim \frac{1}{x^2}$ and $\log_e(1+x) \sim \log_e x$. So the expression behaves like
$\sin(x^2)\cdot\frac{(\log_e x)^{\alpha - \beta}}{x^{\alpha\beta + 2}}$.
$\sin(x^2)$ keeps oscillating between $-1$ and $1$, so the limit is $0$ exactly when $\frac{(\log_e x)^{\alpha-\beta}}{x^{\alpha\beta+2}} \to 0$: that needs $\alpha\beta + 2 > 0$, or $\alpha\beta + 2 = 0$ with $\alpha - \beta < 0$.
(A) $(-1, 3)$: $\alpha\beta + 2 = -1 < 0$, the factor $x\,(\log x)^{-4} \to \infty$ and the product keeps oscillating unboundedly. Not in $S$.
(B) $(-1, 1)$: $\alpha\beta + 2 = 1 > 0$: $\frac{(\log x)^{-2}}{x} \to 0$. In $S$.
(C) $(1, -1)$: $\alpha\beta + 2 = 1 > 0$: $\frac{(\log x)^{2}}{x} \to 0$. In $S$.
(D) $(1, -2)$: $\alpha\beta + 2 = 0$ and $\alpha - \beta = 3 > 0$: the factor $(\log x)^3 \to \infty$. Not in $S$.
Answer: (B), (C). | ||||
| Q.6 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctMedium | |
A medium multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Line through $P$ with direction $(1, 2, 1)$: $(1 + t, 3 + 2t, 2 + t)$. In $L_1$: $(1+t) - (3+2t) + 3(2+t) = 4 + 2t = 6$, so $t = 1$ and $Q = (2, 5, 3)$.
Line through $Q$ along the normal $(1, -1, 3)$ of $L_1$: $(2 + s, 5 - s, 3 + 3s)$. In $L_2$: $2(2+s) - (5-s) + (3+3s) = 2 + 6s = -4$, so $s = -1$ and $R = (1, 6, 0)$.
(A) $PQ = |(1, 2, 1)| = \sqrt6$. True.
(B) $R = (1, 6, 0)$, not $(1, 6, 3)$. False.
(C) Centroid $= \left(\frac{1+2+1}{3}, \frac{3+5+6}{3}, \frac{2+3+0}{3}\right) = \left(\frac43, \frac{14}{3}, \frac53\right)$. True.
(D) $QR = |(-1, 1, -3)| = \sqrt{11}$, $PR = |(0, 3, -2)| = \sqrt{13}$, so the perimeter is $\sqrt6 + \sqrt{11} + \sqrt{13}$, not $\sqrt2 + \sqrt6 + \sqrt{11}$. False.
Answer: (A), (C). | ||||
| Q.7 | Conic SectionsParabola | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections For $y^2 = 8x$, $a = 2$; a point is $(2t^2, 4t)$ and the tangent there is $ty = x + 2t^2$.
Through $C_1 = (-4, 0)$: $0 = -4 + 2t^2$, so $t = \pm\sqrt2$. The points of contact are $A_1 = (4, 4\sqrt2)$, $B_1 = (4, -4\sqrt2)$ (or the other way round).
(A) $OA_1 = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt3$. True.
(B) $A_1B_1 = 8\sqrt2$, not 16. False.
(C), (D) The triangle is symmetric about the $x$-axis, so its orthocentre is some $H = (h, 0)$. $A_1H$ must be perpendicular to $B_1C_1$, whose direction is $C_1 - B_1 = (-8, 4\sqrt2)$: $(h - 4, -4\sqrt2)\cdot(-8, 4\sqrt2) = -8h + 32 - 32 = -8h = 0$, so $h = 0$. The orthocentre is $(0, 0)$: (C) true, (D) false.
Answer: (A), (C). | ||||
| Q.8 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Additivity gives $f(nx) = nf(x)$ for integers $n$, hence $f(q) = q\,f(1)$ for every rational $q$.
$f(-\frac35) = -\frac35 f(1) = 12$, so $f(1) = -20$ and $f(\frac14) = -5$.
Likewise $g > 0$ and $g(x+y) = g(x)g(y)$ give $g(0) = 1$ and $g(q) = g(1)^q$ for rational $q$.
$g(-\frac13) = g(1)^{-1/3} = 2$, so $g(1) = \frac18$ and $g(-2) = 8^2 = 64$.
Value $= (-5 + 64 - 8)\cdot 1 = 51$.
Answer: 51. | ||||
| Q.9 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are $N - 9$ blue balls. After a white and a green ball are drawn, $N - 2$ balls remain, $N - 9$ of them blue:
$P(B_3 \mid W_1 \cap G_2) = \frac{N - 9}{N - 2} = \frac29$, so $9N - 81 = 2N - 4$, $N = 11$.
Check: $P(W_1 \cap G_2 \cap B_3) = \frac{3}{11}\cdot\frac{6}{10}\cdot\frac{2}{9} = \frac{36}{990} = \frac{2}{55} = \frac{2}{5\cdot 11}$. Consistent.
Answer: 11. | ||||
| Q.10 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | NumericalEasy | |
A easy numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Factor: $f(x) = \frac{(\sin x + 2)\left(x^{2023} + 2024x + 2025\right)}{e^{\pi x}(x^2 - x + 3)}$.
$\sin x + 2 \ge 1 > 0$, $e^{\pi x} > 0$, and $x^2 - x + 3 > 0$ always (discriminant $1 - 12 < 0$). So $f(x) = 0$ exactly when $h(x) = x^{2023} + 2024x + 2025 = 0$.
$h'(x) = 2023x^{2022} + 2024 > 0$, so $h$ is strictly increasing; it has odd degree, so it takes every real value exactly once. Hence exactly one root (it is $x = -1$: $-1 - 2024 + 2025 = 0$).
Answer: 1. | ||||
| Q.11 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec p\times\vec q = \begin{vmatrix}\hat i & \hat j & \hat k\\ 2 & 1 & 3\\ 1 & -1 & 1\end{vmatrix} = (1 + 3)\hat i - (2 - 3)\hat j + (-2 - 1)\hat k = 4\hat i + \hat j - 3\hat k$.
Take the dot product of both sides with $\vec p\times\vec q$, which is perpendicular to $\vec p$ and $\vec q$, so the $\alpha$ and $\beta$ terms vanish:
$(15, 10, 6)\cdot(4, 1, -3) = 60 + 10 - 18 = 52 = \gamma\,|\vec p\times\vec q|^2 = 26\gamma$.
So $\gamma = 2$.
Answer: 2. | ||||
| Q.12 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Points of $x^2 = -4ay$: $(2at, -at^2)$. Differentiating, $\frac{dy}{dx} = -\frac{x}{2a} = -t$, so the normal there has slope $\frac{1}{t}$.
Slope $\frac{1}{\sqrt6}$ gives $t = \sqrt6$, foot $(2\sqrt6\,a, -6a)$. Normal: $y + 6a = \frac{1}{\sqrt6}(x - 2\sqrt6\,a)$. At $x = 0$: $y = -6a - 2a = -8a$, so $\alpha = 8a$.
The directrix $y = a$ is horizontal, so $L$ is $y = -8a$. On the parabola $x^2 = 32a^2$, $x = \pm 4\sqrt2\,a$, so $AB = 8\sqrt2\,a$ and $s = 128a^2$.
$r = 4a$. $\frac{r}{s} = \frac{4a}{128a^2} = \frac{1}{32a} = \frac{1}{16}$ gives $a = \frac12$.
$24a = 12$.
Answer: 12. | ||||
| Q.13 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals At the odd integers $f(1) = 2$, $f(3) = -2$, $f(5) = 2$, $f(7) = -2$, $f(9) = 2$, and between consecutive odd integers $f$ is the straight line joining these values. So $f$ is a zigzag between $+2$ and $-2$, crossing $0$ at $t = 2, 4, 6, 8$.
On each $[2n - 1, 2n + 1]$ the graph is a line from $\pm2$ to $\mp2$, whose integral is $0$. Hence $g(3) = g(5) = g(7) = 0$.
In between $g$ never vanishes: on $(1, 3)$, $g$ rises while $f > 0$ (to $t = 2$) and falls back to $0$ only at $3$, so $g > 0$; on $(3, 5)$, $g < 0$; on $(5, 7)$, $g > 0$; on $(7, 8]$, $g < 0$ (it would return to $0$ only at $9$).
So $\alpha = 3$ (the zeros $3, 5, 7$).
$\beta = \lim_{x\to1^+}\frac{g(x) - g(1)}{x - 1} = g'(1^+) = f(1) = 2$ ($f$ is continuous).
$\alpha + \beta = 5$.
Answer: 5. | ||||
| Q.14 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions A relation from $S$ to $S$ is a subset of $S\times S$. Count the ordered pairs allowed by (ii).
$S\times S$ has $36$ pairs. Those with $|a - b| \le 1$: the $6$ pairs with $a = b$ and the $2\times5 = 10$ pairs with $|a - b| = 1$, so $16$. Allowed pairs: $36 - 16 = 20$.
$X$ is the set of all 6-element subsets of these 20 pairs: $n(X) = {}^{20}C_6$.
So $m = 20$.
Answer: 20. | ||||
| Q.15 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $Y$: all 6 pairs would share the same second element $b$. The first elements $a$ with $|a - b| \ge 2$ number: $b = 1$: $\{3,4,5,6\}$, 4; $b = 2$: $\{4,5,6\}$, 3; $b = 3$: $\{1,5,6\}$, 3; $b = 4$: $\{1,2,6\}$, 3; $b = 5$: $\{1,2,3\}$, 3; $b = 6$: $\{1,2,3,4\}$, 4. At most 4 pairs, never 6, so $n(Y) = 0$.
$Z$: a function from $S$ to $S$ has exactly 6 pairs, one for each $a$, so we only need each image to satisfy $|a - f(a)| \ge 2$. Choices: $a = 1$: 4 ($3,4,5,6$); $a = 2$: 3 ($4,5,6$); $a = 3$: 3 ($1,5,6$); $a = 4$: 3 ($1,2,6$); $a = 5$: 3 ($1,2,3$); $a = 6$: 4 ($1,2,3,4$).
$n(Z) = 4\cdot3\cdot3\cdot3\cdot3\cdot4 = 1296 = 36^2$.
$n(Y) + n(Z) = 1296 = k^2$, so $|k| = 36$.
Answer: 36. | ||||
| Q.16 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $2f(x) - 1 = 2\sin^2 x - 1 = -\cos 2x$, so the value is $-\int_0^{\pi/2}\cos 2x\; g(x)\,dx$.
Replace $x$ by $\frac{\pi}{2} - x$: $g\left(\frac\pi2 - x\right) = \sqrt{\frac{\pi}{2}\left(\frac\pi2 - x\right) - \left(\frac\pi2 - x\right)^2} = \sqrt{\frac{\pi x}{2} - x^2} = g(x)$, while $\cos(\pi - 2x) = -\cos 2x$.
So the integral $I = \int_0^{\pi/2}\cos 2x\,g(x)\,dx$ equals $-I$, hence $I = 0$.
Answer: 0. | ||||
| Q.17 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $I = \int_0^{\pi/2}\sin^2 x\,g(x)\,dx$. With $x \to \frac\pi2 - x$ (and $g$ symmetric, as $g(\frac\pi2 - x) = g(x)$): $I = \int_0^{\pi/2}\cos^2 x\,g(x)\,dx$. Adding, $2I = \int_0^{\pi/2} g(x)\,dx$.
$y = \sqrt{\frac{\pi x}{2} - x^2}$ means $\left(x - \frac\pi4\right)^2 + y^2 = \frac{\pi^2}{16}$, $y \ge 0$: the upper half of a circle of radius $\frac\pi4$ centred at $(\frac\pi4, 0)$, running exactly from $x = 0$ to $x = \frac\pi2$.
So $\int_0^{\pi/2} g\,dx = \frac12\pi\left(\frac\pi4\right)^2 = \frac{\pi^3}{32}$, and $I = \frac{\pi^3}{64}$.
$\frac{16}{\pi^3}\cdot\frac{\pi^3}{64} = \frac14 = 0.25$.
Answer: 0.25. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctHard | |
A hard single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Both triangles are symmetric about the same axis (the $x$-axis through P). The loop points its vertex P along $+x$; the field region points its vertex back towards P. At distance $x'$ from P's starting point, the field region is $\frac{2x'}{\sqrt3}$ wide; when the loop's vertex has advanced to $x$, the loop at $x'$ is $\frac{2(x - x')}{\sqrt3}$ wide. The overlap at each $x'$ is the narrower of the two. Write $c = \frac{2}{\sqrt3}$.
$0 \le x \le L$: the overlap is two small triangles meeting at $x' = \frac{x}{2}$, area $A = c\left(\frac{x}{2}\right)^2 = \frac{c x^2}{4}$. So $\frac{dA}{dx} = \frac{cx}{2}$: the emf magnitude $B v\frac{dA}{dx}$ grows linearly from $0$ to $\frac{cBvL}{2}$ at $x = L$.
$L \le x \le 2L$: the overlap runs from $x' = x - L$ (the loop's base) to $x' = L$ (the field's base): $A = c\left[\frac{x^2}{4} - (x - L)^2\right]$, so $\frac{dA}{dx} = c\left(2L - \frac{3x}{2}\right)$. This falls linearly from $\frac{cL}{2}$ at $x = L$, through $0$ at $x = \frac{4L}{3}$ (between $L$ and $\frac{3L}{2}$), to $-cL$ at $x = 2L$: the opposite sign and twice the magnitude reached at $x = L$.
Beyond $2L$ the loop is clear of the field, so the emf drops at once to $0$.
With $E = -\frac{d\Phi}{dt}$ the graph starts at $0$, goes linearly to a negative extreme at $x = L$, rises linearly, crosses zero just before $\frac{3L}{2}$, reaches a positive value twice the size of that extreme at $x = 2L$, and then falls straight to zero. That is graph (A); (B) has the wrong final height, (C) and (D) the wrong shape.
Answer: (A). | ||||
| Q.2 | GravitationOrbital motion | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation The extra force is $F = -\frac{dV_c}{dr} = \frac{3m\alpha}{r^4}$, outwards.
Without it: $m\omega_0^2 r_0 = \frac{GMm}{r_0^2}$, so $\omega_0^2 = \frac{GM}{r_0^3}$.
With it, the net inward force is $\frac{GMm}{r_0^2} - \frac{3m\alpha}{r_0^4} = m\omega_1^2 r_0$, so $\omega_1^2 = \frac{GM}{r_0^3} - \frac{3\alpha}{r_0^5}$.
$\frac{T_1^2 - T_0^2}{T_1^2} = 1 - \frac{T_0^2}{T_1^2} = 1 - \frac{\omega_1^2}{\omega_0^2} = \frac{3\alpha/r_0^5}{GM/r_0^3} = \frac{3\alpha}{GMr_0^2}$.
Answer: (A). | ||||
| Q.3 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The cut-off wavelength $\lambda_{min} = \frac{hc}{eV}$ depends only on the accelerating voltage, so it is the same for both targets.
By Moseley's law the $K_\alpha$ line has $\frac{1}{\lambda_{K\alpha}} \propto (Z - 1)^2$, so $\lambda_{K\alpha} \propto \frac{1}{(Z-1)^2}$.
$r' = r\cdot\frac{(46 - 1)^2}{(41 - 1)^2} = 2\cdot\frac{2025}{1600} = 2.53$.
Answer: (A). | ||||
| Q.4 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The loop hangs in a vertical plane containing PQ, touching PQ at its top. Its magnetic moment $\mu = I\pi r^2$ is normal to the loop, so initially horizontal and perpendicular to the vertical field: the magnetic torque about PQ is $\mu B_0$.
After turning by $\theta$ about PQ, the normal makes angle $90^\circ - \theta$ with $\vec B_0$, so the magnetic torque is $\mu B_0\cos\theta$.
The weight acts at the loop's centre, a distance $r$ below PQ; turned by $\theta$, its torque about PQ is $mg\,r\sin\theta$, opposing the turn.
Equilibrium: $I\pi r^2 B_0\cos\theta = mgr\sin\theta$, so $\tan\theta = \frac{\pi r I B_0}{mg}$.
Answer: (A). | ||||
| Q.5 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Shell charge $Q = 4\pi R^2\sigma$. Inside ($r < R$) the field is zero; outside it is radial, $E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{\sigma R^2}{\varepsilon_0 r^2}$.
A dipole at a small angle $\theta$ to the field feels a restoring torque $p_0E\sin\theta \approx p_0E\theta$, so it performs angular SHM with $\omega = \sqrt{\frac{p_0E}{I}}$.
(A) For $r < R$ there is no field and no torque, so no oscillation. False.
(B) For any finite $r > R$ the field is non-zero and the dipole, nearly along the outward field, oscillates. True.
(C) At $r = 2R$: $E = \frac{\sigma}{4\varepsilon_0}$, $\omega = \sqrt{\frac{\sigma p_0}{4\varepsilon_0 I}}$, not $\sqrt{\frac{2\sigma p_0}{\varepsilon_0 I}}$. False.
(D) At $r = 10R$: $E = \frac{\sigma}{100\varepsilon_0}$, $\omega = \sqrt{\frac{\sigma p_0}{100\varepsilon_0 I}}$. True.
Answer: (B), (D). | ||||
| Q.6 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctMedium | |
A medium multiple correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Volume $V = \frac43\pi r^3 = \frac43\cdot\frac{22}{7}\cdot\frac{27}{8}\times10^{-6} = \frac{99}{7}\times10^{-6}$ m$^3$.
Buoyancy $= \rho Vg = 10^3\cdot\frac{99}{7}\times10^{-6}\cdot10 = \frac{0.99}{7}$ N; weight $= \frac{22}{7}\times10^{-3}\cdot10 = \frac{0.22}{7}$ N. Net upward force (excluding viscosity) $F = \frac{0.77}{7} = 0.11$ N.
(A) Pushing slowly against $F$ through $0.7$ m: $W = 0.11\times0.7 = 0.077$ J. True.
(B) Without viscosity this work returns as kinetic energy at the surface: $\frac12mv^2 = 0.077$, $v^2 = \frac{0.154}{(22/7)\times10^{-3}} = 49$, $v = 7$ m/s. True.
(C) $H = \frac{v^2}{2g} = \frac{49}{20} = 2.45$ m, not $1.4$ m. False.
(D) Stokes' drag $6\pi\eta r v$ is greatest at the greatest speed, about $7$ m/s (the terminal speed $\frac{0.11}{6\pi\eta r} \approx 390$ m/s is never approached, so viscosity changes $v$ very little): $6\cdot\frac{22}{7}\cdot10^{-3}\cdot1.5\times10^{-2}\cdot7 = 1.98\times10^{-3}$ N. Ratio $\frac{0.11}{1.98\times10^{-3}} = \frac{500}{9}$. True.
Answer: (A), (B), (D). | ||||
| Q.7 | 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 The ion enters the field perpendicular to its edge, turns through a semicircle and comes back out at a distance $x = 2R$ below its line of entry, where $R = \frac{mv}{qB} = \frac{\sqrt{2mqV}}{qB}$.
For $A_M = 1$: $2mqV = 2\cdot\frac53\times10^{-27}\cdot1.6\times10^{-19}\cdot192 = 1024\times10^{-46}$, so $mv = 3.2\times10^{-22}$; $qB = 1.6\times10^{-20}$; $R = 0.02$ m $= 2$ cm. In general $R = 2\sqrt{A_M}$ cm.
(A) $H^+$: $x = 2R = 4$ cm. True.
(B) $A_M = 144$: $R = 24$ cm, $x = 48$ cm. True.
(C) $A_M$ from 1 to 196: $x$ from $4$ cm to $2\cdot2\sqrt{196} = 56$ cm, so the detector must span $56 - 4 = 52$ cm, not 55 cm. False.
(D) The semicircle reaches a depth $R$ into the field, so $w \ge R = 2\sqrt{196} = 28$ cm, not 56 cm. False.
(Direction check: $q\vec v\times\vec B$ with $\vec v$ along $+y$ and $\vec B$ along $+z$ points along $+x$, downwards in the figure.)
Answer: (A), (B). | ||||
| Q.8 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalEasy | |
A easy numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $V = \frac13\pi\left(\frac{D}{2}\right)^2h = \frac{\pi}{12}D^2h$, so $\frac{\Delta V}{V} = 2\frac{\Delta D}{D} + \frac{\Delta h}{h}$.
Taking the least count as the error: $\Delta D = \Delta h = 0.2$ cm, $\frac{0.2}{20.0} = 1\%$.
Maximum percentage error $= 2(1\%) + 1\% = 3\%$.
Answer: 3. | ||||
| Q.9 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Both move under the same gravity, so relative to the ball the stone moves in a straight line at constant velocity. To hit, the relative velocity must point along the line joining them, which is horizontal: the vertical components must be equal, $u\sin\theta_1 = v_0\sin\theta_0$ ($u$ the stone's speed). They then close the gap $L$ at the rate $v_0\cos\theta_0 + u\cos\theta_1$, so $T = \frac{L}{v_0\cos\theta_0 + u\cos\theta_1}$.
$(45^\circ, 45^\circ)$: $u = v_0$, $T_1 = \frac{L}{2v_0\cos45^\circ} = \frac{L}{\sqrt2\,v_0}$.
$(60^\circ, 30^\circ)$: $u\cdot\frac12 = v_0\frac{\sqrt3}{2}$, $u = \sqrt3\,v_0$; closing speed $\frac{v_0}{2} + \sqrt3v_0\cdot\frac{\sqrt3}{2} = 2v_0$, so $T_2 = \frac{L}{2v_0}$.
$\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{2}{\sqrt2}\right)^2 = 2$.
Answer: 2. | ||||
| Q.10 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance A flat end seen from P under a half-angle $\theta$ subtends the solid angle $2\pi(1 - \cos\theta)$, so it receives $\frac{q}{4\pi\varepsilon_0}\cdot2\pi(1 - \cos\theta) = \frac{q}{2\varepsilon_0}(1 - \cos\theta)$.
The curved surface gets the rest: $\frac{q}{\varepsilon_0} - 2\cdot\frac{q}{2\varepsilon_0}(1 - \cos\theta) = \frac{q}{\varepsilon_0}\cos\theta$.
$\theta = 30^\circ$: $\Phi = \frac{\sqrt3}{2}\frac{q}{\varepsilon_0}$. $\theta = 60^\circ$: $\frac12\frac{q}{\varepsilon_0} = \frac{\Phi}{\sqrt3}$.
So $n = 3$.
Answer: 3. | ||||
| Q.11 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Minimum deviation in P$_2$ (angle $60^\circ$): the ray inside makes $30^\circ$ with each face normal, so the angle of incidence on P$_2$ is given by $\sin i = \sqrt3\sin30^\circ = \frac{\sqrt3}{2}$, $i = 60^\circ$.
The facing faces of P$_1$ and P$_2$ are parallel, so the ray leaves P$_1$ at $60^\circ$ to that face's normal. Inside P$_1$: $\sqrt{\frac32}\sin r_2 = \sin60^\circ = \frac{\sqrt3}{2}$, so $\sin r_2 = \frac{1}{\sqrt2}$, $r_2 = 45^\circ$.
In P$_1$, $r_1 + r_2 = 60^\circ$, so $r_1 = 15^\circ = \frac{\pi}{12}$.
At entry: $\sin\theta = \sqrt{\frac32}\sin\frac{\pi}{12}$, so $\beta = 12$.
Answer: 12. | ||||
| Q.12 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance From the figure, the wire runs along a diameter; P is $0.5$ m from it (inside the shell) and R is $2$ m from it on the same perpendicular line through the centre (outside the shell).
Shell ($Q = 10$ nC, radius 1 m): $V_P = \frac{kQ}{1} = 90$ V (inside), $V_R = \frac{kQ}{2} = 45$ V. Difference $45$ V.
Wire: $V_P - V_R = 2k\lambda\ln\frac{r_R}{r_P} = 2\cdot9\times10^9\cdot5\times10^{-9}\cdot\ln4 = 90\times1.4 = 126$ V.
Both make P higher, so $|V_P - V_R| = 45 + 126 = 171$ V.
Answer: 171. | ||||
| Q.13 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Excess pressure of a soap bubble: $\Delta P = \frac{4T}{r}$, so $\Delta P \propto \frac1r$.
The air inside is at $P_{out} + \Delta P \approx P_{out}$ (as $\Delta P \ll P_{out}$). Isothermal: $P_0\cdot\frac43\pi r_1^3 = \frac{8P_0}{27}\cdot\frac43\pi r_2^3$, so $r_2^3 = \frac{27}{8}r_1^3$ and $r_2 = \frac32r_1$.
$\Delta P_2 = 144\cdot\frac{r_1}{r_2} = 144\cdot\frac23 = 96$ Pa.
Answer: 96. | ||||
| Q.14 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The 8th bright fringe is at $y = \frac{8\lambda D}{d} = \frac{8\cdot6\times10^{-7}\cdot1}{d} = \frac{4.8\times10^{-6}\ \text{m}^2}{d}$.
$d$ swings between $0.76$ mm and $0.84$ mm, so $y$ swings between $\frac{4.8\times10^{-6}}{0.76\times10^{-3}} = 6.3158$ mm and $\frac{4.8\times10^{-6}}{0.84\times10^{-3}} = 5.7143$ mm.
Separation $= 4.8\left(\frac{1}{0.76} - \frac{1}{0.84}\right)$ mm $= 0.6015$ mm $= 601.50\ \mu$m.
(The small-change estimate $\Delta y \approx y\frac{\Delta d}{d} = 6\text{ mm}\times\frac{0.08}{0.8} = 600\ \mu$m is close.)
Answer: 601.50 (about 600). | ||||
| Q.15 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalHard | |
A hard numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $y = \frac{8\lambda D}{d}$, so $\left|\frac{dy}{dt}\right| = \frac{8\lambda D}{d^2}\left|\frac{dd}{dt}\right|$, with $\frac{dd}{dt} = 0.04\times10^{-3}\cdot0.08\cos\omega t = 3.2\times10^{-6}\cos\omega t$ m/s.
The slits move fastest when $\cos\omega t = \pm1$, i.e. at $d = 0.8$ mm:
$\left|\frac{dy}{dt}\right| = \frac{4.8\times10^{-6}\cdot3.2\times10^{-6}}{(0.8\times10^{-3})^2} = \frac{15.36\times10^{-12}}{6.4\times10^{-7}} = 2.4\times10^{-5}$ m/s $= 24\ \mu$m/s.
(Strictly, $\frac{\cos\omega t}{d^2}$ peaks slightly before $d$ reaches its mean, at $\sin\omega t \approx -0.1$, giving $24.12\ \mu$m/s; the difference is under 0.5 %.)
Answer: 24. | ||||
| Q.16 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Velocities: $\dot x_1 = a\omega\cos\omega t$, $\dot x_2 = -a\omega\cos\omega t$. At $t_0 = 0$: $v_1 = a\omega$, $v_2 = -a\omega$.
Particle 3 moves at $+\frac{a\omega}{2}$. An instantaneous elastic collision of equal masses exchanges their velocities: particle 2 now has $+\frac{a\omega}{2}$ (particle 3 bounces back at $-a\omega$); particle 1 is unaffected during the instant.
$v_{cm} = \frac{v_1 + v_2}{2} = \frac{a\omega + \frac{a\omega}{2}}{2} = \frac{3a\omega}{4}$.
$\frac{v_{cm}}{a\omega} = 0.75$.
Answer: 0.75. | ||||
| Q.17 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalHard | |
A hard numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At $t_0 = \frac{\pi}{2\omega}$: $\cos\omega t_0 = 0$, so both particles are momentarily at rest, at their extreme positions (each displaced $a$ from its mean position about the centre of mass).
Equal-mass elastic collision: particle 2 takes particle 3's velocity $\frac{a\omega}{2}$; particle 1 stays at rest. So $v_{cm} = \frac{a\omega}{4}$.
In the centre-of-mass frame particle 2 moves at $\frac{a\omega}{2} - \frac{a\omega}{4} = \frac{a\omega}{4}$ and particle 1 at $-\frac{a\omega}{4}$, each still displaced $a$ from its mean position (the collision is instantaneous).
Each particle performs SHM with the same $\omega$ about its mean position, so its new amplitude is $b = \sqrt{a^2 + \left(\frac{v}{\omega}\right)^2} = \sqrt{a^2 + \frac{a^2}{16}}$, $b^2 = \frac{17}{16}a^2$.
$\frac{4b^2}{a^2} = \frac{17}{4} = 4.25$.
Answer: 4.25. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In Bohr's model the kinetic energy equals $-E_n = 13.6\,\frac{Z^2}{n^2}$ eV.
(A) H, $n = 1$: $13.6$ eV. (B) $\mathrm{He^+}$, $n = 1$: $13.6\times4 = 54.4$ eV. (C) $\mathrm{He^+}$, $n = 2$: $13.6$ eV. (D) $\mathrm{Li^{2+}}$, $n = 2$: $13.6\times\frac94 = 30.6$ eV.
The largest is the first orbit of $\mathrm{He^+}$.
Answer: (B). | ||||
| Q.2 | Solid StatePoint defects | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on point defects, from Solid State. NCERT: No matching NCERT chapter Charge balance: $\mathrm{Y_2}$ gives $+6$ and $\mathrm{O_4}$ gives $-8$, so the M ions together carry $+2$.
With $\frac13$ of M as $\mathrm{M^{2+}}$ and $\frac23$ as $\mathrm{M^{3+}}$, the average charge per M is $\frac13\cdot2 + \frac23\cdot3 = \frac83$.
$X\cdot\frac83 = 2$, so $X = \frac34 = 0.75$.
Answer: (D). | ||||
| Q.3 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Heating glucose (L or D, it does not matter) with HI reduces it completely to n-hexane, $\mathrm{CH_3(CH_2)_4CH_3}$.
n-Hexane over $\mathrm{Cr_2O_3}$ at 775 K and 10–20 atm undergoes aromatisation (cyclisation and dehydrogenation) to benzene: P is benzene.
Benzene with excess $\mathrm{Cl_2}$ under UV light adds chlorine across all three double bonds (free-radical addition, not substitution), giving 1,2,3,4,5,6-hexachlorocyclohexane (benzene hexachloride).
That is the saturated ring with six Cl, structure (D); (A) is hexachlorobenzene, a substitution product formed with a Lewis acid, not under UV.
Answer: (D). | ||||
| Q.4 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctHard | |
A hard single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter In a polar solvent phosphorus pentachloride does not stay as trigonal-bipyramidal $\mathrm{PCl_5}$ molecules; it ionises, and the stable cation is tetrahedral $[\mathrm{PCl_4}]^+$.
Fluorination replaces Cl by F in the anionic part: the products are salts of $[\mathrm{PCl_4}]^+$ with the partly fluorinated anion $[\mathrm{PCl_4F_2}]^-$ and with the fully fluorinated octahedral $[\mathrm{PF_6}]^-$.
(A) needs $[\mathrm{PF_4}]^+$, which is not the cation formed. (C) and (D) are neutral molecules, which is what the reaction gives in a non-polar medium or the gas phase, not in a polar solvent.
Answer: (B). | ||||
| Q.5 | ElectrochemistryBatteries and fuel cells | Class 12: Electrochemistry | Multiple correctMedium | |
A medium multiple correct question on batteries and fuel cells, from Electrochemistry. NCERT: Class 12: Electrochemistry This is a hydrazine fuel cell (alkaline).
Anode (oxidation): $\mathrm{N_2H_4 + 4OH^- \rightarrow N_2 + 4H_2O + 4e^-}$.
Cathode (reduction): $\mathrm{O_2 + 2H_2O + 4e^- \rightarrow 4OH^-}$.
Overall: $\mathrm{N_2H_4 + O_2 \rightarrow N_2 + 2H_2O}$.
(A) Exactly the anode reaction, releasing 4 electrons. True.
(B) Hydrazine is oxidised at the anode, not the cathode, and no nascent hydrogen is involved. False.
(C) At the cathode $\mathrm{O_2}$ is reduced to $\mathrm{OH^-}$. True.
(D) The products are $\mathrm{N_2}$ and water; oxides of nitrogen are not formed. False.
Answer: (A), (C). | ||||
| Q.6 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Multiple correctHard | |
A hard multiple correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Changes needed from P to Q: both $\mathrm{C\equiv C}$ to cis $\mathrm{C=C}$ (Lindlar, $\mathrm{H_2}$); ring ketone to $\mathrm{CH(OH)}$ ($\mathrm{NaBH_4}$); $\mathrm{CN}$ to $\mathrm{CHO}$ (Stephen reduction: $\mathrm{SnCl_2/HCl}$ gives the aldimine salt, which water/acid hydrolyses); ethyl ester to $\mathrm{CO_2H}$ and acetate to $\mathrm{OH}$ (acid hydrolysis, $\mathrm{H_3O^+}$).
The order matters: $\mathrm{NaBH_4}$ must act before any $\mathrm{CHO}$ (or the imine that leads to it) is present, or that group is reduced too.
(A) $\mathrm{NaBH_4}$ comes after $\mathrm{SnCl_2/HCl}$, so it meets the imine (iminium) salt and reduces it towards the amine; no $\mathrm{CHO}$ would form. Incorrect.
(B) After $\mathrm{H_3O^+}$ and the Stephen step the aldehyde is formed, and the final $\mathrm{NaBH_4}$ reduces it to $\mathrm{CH_2OH}$. Incorrect.
(C) $\mathrm{NaBH_4}$ first reduces only the ketone (ester, acetate and nitrile untouched); $\mathrm{SnCl_2/HCl}$ then $\mathrm{H_3O^+}$ give $\mathrm{CHO}$, $\mathrm{CO_2H}$ and the free $\mathrm{OH}$; finally Lindlar's catalyst reduces the alkynes to cis alkenes without touching the aldehyde. Correct.
(D) Lindlar first (cis alkenes), $\mathrm{NaBH_4}$ (ketone to alcohol), then $\mathrm{SnCl_2/HCl}$ and $\mathrm{H_3O^+}$ (imine to $\mathrm{CHO}$, ester and acetate hydrolysed). Correct.
Answer: (C), (D). | ||||
| Q.7 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) $\mathrm{H_2S_2O_7}$, pyrosulphuric acid: two $\mathrm{SO_2(OH)}$ units joined by $\mathrm{S{-}O{-}S}$. No peroxide.
(B) $\mathrm{H_2S_2O_8}$, peroxodisulphuric acid: $\mathrm{HO_3S{-}O{-}O{-}SO_3H}$. Peroxide.
(C) $\mathrm{H_2S_2O_5}$, disulphurous acid: has an $\mathrm{S{-}S}$ bond, no $\mathrm{O{-}O}$. No peroxide.
(D) $\mathrm{H_2SO_5}$, peroxomonosulphuric (Caro's) acid: $\mathrm{HO{-}O{-}SO_2{-}OH}$. Peroxide.
Answer: (B), (D). | ||||
| Q.8 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Acetic acid taken: $0.100\times0.5 = 0.050$ mol. Unadsorbed (equal to NaOH used): $0.040\times1 = 0.040$ mol. Adsorbed: $0.010$ mol $= 0.010\times6.0\times10^{23} = 6.0\times10^{21}$ molecules.
These cover $150$ m$^2$, so each occupies $\frac{150}{6.0\times10^{21}} = 2.5\times10^{-20}$ m$^2 = 2500\times10^{-23}$ m$^2$.
$P = 2500$.
Answer: 2500. | ||||
| Q.9 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\Delta T_b = iK_b m = iK_b\frac{w_2/M}{w_1/1000}$. Same $w_1$, $w_2$ and solvent, so $\Delta T_b \propto \frac{i}{M}$.
$\frac{\Delta T_{b,1}}{\Delta T_{b,2}} = \frac{i_X}{i_Y}\cdot\frac{M_Y}{M_X} = 1.2\cdot\frac{1}{0.8} = 1.5$, i.e. $150\%$.
Answer: 150. | ||||
| Q.10 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules The strand has 13 bases: A at positions 1, 5, 9, 10 (4); T at 3, 8, 12 (3); G at 2, 7, 11 (3); C at 4, 6, 13 (3).
In the double strand each A or T forms an A–T pair and each G or C a G–C pair: $4 + 3 = 7$ A–T pairs and $3 + 3 = 6$ G–C pairs. (A–U pairs occur only in RNA, so that value is not used.)
A–T pairs have 2 hydrogen bonds, G–C pairs 3.
Energy $= 7\times2\times1.0 + 6\times3\times1.5 = 14 + 27 = 41$ kcal mol$^{-1}$.
Answer: 41. | ||||
| Q.11 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Mass ratio 7 means a ratio of atoms $\frac{N_{Pb}}{N_U} = 7\times\frac{238}{206} = 8.087$.
Every Pb atom came from a U atom, so $\frac{N_0}{N} = 1 + 8.087 = 9.087$.
$t = \frac{t_{1/2}}{\ln2}\ln\frac{N_0}{N} = \frac{4.5\times10^9}{0.693}\times\ln 9.087 = 6.494\times10^9\times2.2069 = 1.433\times10^{10}$ years $= 143.3\times10^8$ years.
The answer is an integer, so $P = 143$.
Answer: 143. | ||||
| Q.12 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[\mathrm{Co(CN)_4}]^{4-}$: cobalt in the 0 state, $sp^3$ with four $\mathrm{CN^-}$: tetrahedral.
$[\mathrm{Co(CO)_3(NO)}]$: with linear $\mathrm{NO^+}$, cobalt is $-1$ ($d^{10}$), $sp^3$: tetrahedral.
$\mathrm{XeF_4}$: 4 bond pairs + 2 lone pairs: square planar.
$[\mathrm{PCl_4}]^+$: 4 bond pairs, $sp^3$: tetrahedral.
$[\mathrm{PdCl_4}]^{2-}$: $\mathrm{Pd^{2+}}$ ($4d^8$), $dsp^2$: square planar.
$[\mathrm{ICl_4}]^-$: 4 bond pairs + 2 lone pairs: square planar.
$[\mathrm{Cu(CN)_4}]^{3-}$: $\mathrm{Cu^+}$ ($d^{10}$), $sp^3$: tetrahedral.
$\mathrm{P_4}$: four P atoms at the corners of a tetrahedron: tetrahedral.
Tetrahedral: $[\mathrm{Co(CN)_4}]^{4-}$, $[\mathrm{Co(CO)_3(NO)}]$, $[\mathrm{PCl_4}]^+$, $[\mathrm{Cu(CN)_4}]^{3-}$, $\mathrm{P_4}$, so 5.
Answer: 5. | ||||
| Q.13 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | NumericalHard | |
A hard numerical question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers $\mathrm{C_6H_6O_3}$, a phenol (ferric chloride test) with no intramolecular hydrogen bond (so no two OH groups ortho): benzene-1,3,5-triol (phloroglucinol).
It reacts in its keto form, cyclohexane-1,3,5-trione, so it takes 3 equivalents of $\mathrm{NH_2OH}$ to give the trioxime Q.
With excess $\mathrm{CH_3I}$/KOH the very acidic $\mathrm{CH_2}$ groups between the carbonyls are methylated completely (C-methylation): R is 2,2,4,4,6,6-hexamethylcyclohexane-1,3,5-trione, with 6 $\mathrm{CH_3}$.
Excess iso-butylmagnesium bromide adds to all three $\mathrm{C=O}$; after $\mathrm{H_3O^+}$, S is the triol carrying three $\mathrm{-CH_2CH(CH_3)_2}$ groups, each with 2 $\mathrm{CH_3}$.
Methyl groups in S: $6 + 3\times2 = 12$.
Answer: 12. | ||||
| Q.14 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalHard | |
A hard numerical question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $\mathrm{C_9H_{18}O_2}$ has one degree of unsaturation: the $\mathrm{C=C}$ (it decolorises bromine water). The two O are OH groups. Oxidative ozonolysis turns each alkene carbon into $\mathrm{COOH}$, so Q and R are hydroxy acids.
PCC turns each OH into $\mathrm{C=O}$; heating then decarboxylates only a $\beta$-keto acid, giving a methyl ketone (iodoform positive). So in Q and R the OH is on the $\beta$ carbon.
Q is iodoform positive: $\mathrm{CH_3CH(OH)CH_2COOH}$ (3-hydroxybutanoic acid). PCC gives $\mathrm{CH_3COCH_2COOH}$, which on heating loses $\mathrm{CO_2}$: S = acetone, $\mathrm{CH_3COCH_3}$.
R has the other 5 carbons and is iodoform negative: $\mathrm{CH_3CH_2CH(OH)CH_2COOH}$ (3-hydroxypentanoic acid). PCC gives $\mathrm{CH_3CH_2COCH_2COOH}$, heating gives T = butanone, $\mathrm{CH_3CH_2COCH_3}$.
So P is $\mathrm{CH_3CH(OH)CH_2CH{=}CHCH_2CH(OH)CH_2CH_3}$ (non-4-ene-2,7-diol), $\mathrm{C_9H_{18}O_2}$, iodoform positive through its $\mathrm{CH_3CH(OH)}$ end.
S (acetone) has 1 O and T (butanone) has 1 O: the sum is 2.
Answer: 2. | ||||
| Q.15 | PolymersBiodegradable polymers | No matching NCERT chapter | NumericalHard | |
A hard numerical question on biodegradable polymers, from Polymers. NCERT: No matching NCERT chapter $\mathrm{C_9H_{18}O_2}$ has one degree of unsaturation: the $\mathrm{C=C}$ (it decolorises bromine water). The two O are OH groups. Oxidative ozonolysis turns each alkene carbon into $\mathrm{COOH}$, so Q and R are hydroxy acids.
PCC turns each OH into $\mathrm{C=O}$; heating then decarboxylates only a $\beta$-keto acid, giving a methyl ketone (iodoform positive). So in Q and R the OH is on the $\beta$ carbon.
Q is iodoform positive: $\mathrm{CH_3CH(OH)CH_2COOH}$ (3-hydroxybutanoic acid). PCC gives $\mathrm{CH_3COCH_2COOH}$, which on heating loses $\mathrm{CO_2}$: S = acetone, $\mathrm{CH_3COCH_3}$.
R has the other 5 carbons and is iodoform negative: $\mathrm{CH_3CH_2CH(OH)CH_2COOH}$ (3-hydroxypentanoic acid). PCC gives $\mathrm{CH_3CH_2COCH_2COOH}$, heating gives T = butanone, $\mathrm{CH_3CH_2COCH_3}$.
So P is $\mathrm{CH_3CH(OH)CH_2CH{=}CHCH_2CH(OH)CH_2CH_3}$ (non-4-ene-2,7-diol), $\mathrm{C_9H_{18}O_2}$, iodoform positive through its $\mathrm{CH_3CH(OH)}$ end.
Q ($\mathrm{C_4H_8O_3}$, $M = 104$) and R ($\mathrm{C_5H_{10}O_3}$, $M = 118$) are hydroxy acids, which condense into a polyester (the biodegradable copolymer PHBV). One acyclic chain of 1000 monomer units has 999 ester links, each releasing one $\mathrm{H_2O}$.
$M(U) = 500\times104 + 500\times118 - 999\times18 = 52000 + 59000 - 17982 = 93018$.
Answer: 93018. | ||||
| Q.16 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Iodide reduces ferricyanide to ferrocyanide and is itself oxidised to iodine:
$\mathrm{2K_3[Fe(CN)_6] + 2KI \rightleftharpoons 2K_4[Fe(CN)_6] + I_2}$.
(In strong acid the ferrocyanide is protonated, which pulls the equilibrium completely to the right.) P is potassium ferrocyanide, $\mathrm{K_4[Fe(CN)_6]}$.
Each mole of P needs one electron, i.e. one $\mathrm{I^-}$: 2 mol of P need 2 mol of KI.
Answer: 2. | ||||
| Q.17 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds P is $\mathrm{K_4[Fe(CN)_6]}$. With zinc chloride in slightly acidic solution it gives the white, sparingly soluble potassium zinc ferrocyanide:
$\mathrm{2K_4[Fe(CN)_6] + 3ZnCl_2 \rightarrow K_2Zn_3[Fe(CN)_6]_2 + 6KCl}$.
Q $= \mathrm{K_2Zn_3[Fe(CN)_6]_2}$, which contains 3 zinc ions.
Answer: 3. | ||||
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