JEE (Main) · 2025
JEE (Main) 2025 — 3 April, Shift 2
Which chapter each question came from, 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.
Mathematics
25 Q
- Sets, Relations and Functions3
- Conic Sections3
- Three Dimensional Geometry2
8 easy · 16 medium · 1 hard
Physics
25 Q
- Optics4
- Properties of Bulk Matter3
- Physical World and Measurement2
17 easy · 8 medium · 0 hard
Chemistry
25 Q
- Some Basic Concepts of Chemistry3
- Thermodynamics3
- Organic Chemistry: Some Basic Principles and Techniques3
11 easy · 13 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions For each x there is exactly one y $= \max\{x, 1\}$:
$-2 \to 1$, $-1 \to 1$, $0 \to 1$, $1 \to 1$, $2 \to 2$, $3 \to 3$.
$R = \{(-2,1), (-1,1), (0,1), (1,1), (2,2), (3,3)\}$, so $l = 6$.
Reflexive: every $(a, a)$ must be in R. $(1,1), (2,2), (3,3)$ are there; $(-2,-2), (-1,-1), (0,0)$ must be added, so $m = 3$.
Symmetric: each pair needs its reverse. $(1,-2), (1,-1), (1,0)$ must be added; the diagonal pairs are their own reverses. So $n = 3$.
$l + m + n = 6 + 3 + 3 = 12$.
Answer: (B) $12$. | ||||
| 2 | Sets, Relations and FunctionsSets and their representation | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single 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 Inner logarithm: $x^2 - 9x + 18 > 0 \Rightarrow (x-3)(x-6) > 0 \Rightarrow x < 3$ or $x > 6$.
Outer logarithm: $1 - \log_4(x^2 - 9x + 18) > 0 \Rightarrow x^2 - 9x + 18 < 4 \Rightarrow x^2 - 9x + 14 < 0 \Rightarrow (x-2)(x-7) < 0 \Rightarrow 2 < x < 7$.
Both together: $(2, 3)\cup(6, 7)$.
$\alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18$.
Answer: (D) $18$. | ||||
| 3 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Put $x = 3$ (so $\frac{24}{x} = 8$): $f(3) + 3f(8) = 12$.
Put $x = 8$ (so $\frac{24}{x} = 3$): $f(8) + 3f(3) = 32$.
Adding: $4f(3) + 4f(8) = 44$, so $f(3) + f(8) = 11$.
Answer: (B) $11$. | ||||
| 4 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Write the equation as $(12 - k)x^2 + 2(12 - k)x - 2 = 0$. It is a quadratic only if $k \ne 12$.
Equal roots: discriminant zero, $4(12 - k)^2 + 8(12 - k) = 0 \Rightarrow 4(12 - k)(12 - k + 2) = 0$.
With $k \ne 12$: $14 - k = 0$, so $k = 14$.
The point is $(14, 7)$. Its distance from $3x + 4y + 5 = 0$:
$\frac{|3(14) + 4(7) + 5|}{\sqrt{9 + 16}} = \frac{75}{5} = 15$.
Answer: (C) $15$. | ||||
| 5 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations In an equilateral triangle the centroid is also the circumcentre, so the vertices are a rotation of one another by $\frac{2\pi}{3}$ about $z_0$:
$z_1 - z_0 = w$, $z_2 - z_0 = w\omega$, $z_3 - z_0 = w\omega^2$, where $\omega = e^{2\pi i/3}$.
$\sum(z_k - z_0)^2 = w^2(1 + \omega^2 + \omega^4) = w^2(1 + \omega^2 + \omega) = 0$.
Answer: (B) $0$. | ||||
| 6 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The numerator of the n-th term is the sum of the first n odd numbers, $n^2$. So the sum is $\sum_{n=1}^{\infty}\frac{n^2}{n!}$.
$\frac{n^2}{n!} = \frac{n}{(n-1)!} = \frac{(n-1) + 1}{(n-1)!}$.
$\sum_{n=1}^{\infty}\frac{n-1}{(n-1)!} = \sum_{m=1}^{\infty}\frac{1}{(m-1)!} = e$, and $\sum_{n=1}^{\infty}\frac{1}{(n-1)!} = e$.
Sum $= e + e = 2e$.
Answer: (B) $2e$. | ||||
| 7 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Any three of the 22 points make a triangle unless they are collinear.
$L_1$ carries O and the 12 points P: 13 collinear points. $L_2$ carries O and the 9 points Q: 10 collinear points.
Triangles $= \binom{22}{3} - \binom{13}{3} - \binom{10}{3} = 1540 - 286 - 120 = 1134$.
Answer: (C) $1134$. | ||||
| 8 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability The probabilities add to 1. With $t = \frac{1}{3}$, $\sum_{x=0}^{\infty}(x + 1)t^x = \frac{1}{(1 - t)^2} = \frac{9}{4}$.
So $k\cdot\frac{9}{4} = 1$, $k = \frac{4}{9}$.
$P(X \le 2) = \frac{4}{9}\left(1 + \frac{2}{3} + \frac{3}{9}\right) = \frac{4}{9}\cdot2 = \frac{8}{9}$.
$P(X \ge 3) = 1 - \frac{8}{9} = \frac{1}{9}$.
Answer: (D) $\frac{1}{9}$. | ||||
| 9 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctMedium | |
A medium single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean 5: $1 + 3 + a + 7 + b = 25 \Rightarrow a + b = 14$.
Variance 10: $\frac{1 + 9 + a^2 + 49 + b^2}{5} - 25 = 10 \Rightarrow a^2 + b^2 = 116$.
$2ab = 14^2 - 116 = 80$, $ab = 40$: a and b are the roots of $t^2 - 14t + 40 = 0$, i.e. 10 and 4. With $a > b$: $a = 10$, $b = 4$.
The observations are $1, 3, 10, 7, 4$, and $n + x_n$ gives $2, 5, 13, 11, 9$.
Mean $= \frac{40}{5} = 8$. Deviations $-6, -3, 5, 3, 1$; squares $36 + 9 + 25 + 9 + 1 = 80$.
Variance $= \frac{80}{5} = 16$.
Answer: (A) $16$. | ||||
| 10 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines Rearranged: $(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$, so every line passes through the meeting point of $x + 2y = 5$ and $3x + 7y = 17$.
From the first, $x = 5 - 2y$; then $15 - 6y + 7y = 17$, $y = 2$, $x = 1$. P $= (1, 2)$.
Of all lines through P, the one farthest from O is perpendicular to OP. OP has slope 2, so L has slope $-\frac{1}{2}$: $x + 2y = 5$ (the member $\lambda = 0$).
Distance of (3, 6) from $x + 2y - 5 = 0$: $d = \frac{|3 + 12 - 5|}{\sqrt{5}} = \frac{10}{\sqrt{5}}$, $d^2 = 20$.
Answer: (A) $20$. | ||||
| 11 | CirclesStandard and general form | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections The circle passes through the origin: $x^2 + y^2 + 2gx + 2fy = 0$.
(4, 6): $52 + 8g + 12f = 0 \Rightarrow 2g + 3f = -13$.
(−1, 5): $26 - 2g + 10f = 0 \Rightarrow -2g + 10f = -26$.
Adding: $13f = -39$, $f = -3$; then $g = -2$. Circle: $x^2 + y^2 - 4x - 6y = 0$, $r^2 = 4 + 9 = 13$.
(k, 3k): $10k^2 - 4k - 18k = 0 \Rightarrow k(10k - 22) = 0$. The points are distinct, so $k \ne 0$ and $k = \frac{11}{5}$.
$10k + r^2 = 22 + 13 = 35$.
Answer: (D) $35$. | ||||
| 12 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $ae = 2$ with $e = \frac{1}{2}$: $a = 4$, $b^2 = a^2(1 - e^2) = 12$, $b = 2\sqrt{3}$.
The smallest circle enclosing the ellipse is its auxiliary circle $x^2 + y^2 = 16$ (it must reach the vertices $(\pm4, 0)$).
E meets the negative y-axis at $(0, -2\sqrt{3})$, so QR lies on $y = -2\sqrt{3}$ and has length $2a = 8$.
The area $\frac{1}{2}\cdot8\cdot h$ is greatest when P on the circle is farthest from that line: P $= (0, 4)$, $h = 4 + 2\sqrt{3}$.
Maximum area $= 4(4 + 2\sqrt{3}) = 8(2 + \sqrt{3})$.
Answer: (A) $8(2 + \sqrt{3})$. | ||||
| 13 | 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 $-\frac{4}{1 + \sqrt{3}} = -\frac{4(\sqrt{3} - 1)}{2} = 2 - 2\sqrt{3}$.
With $s = \sin x$ and $\cos^2x = 1 - s^2$:
$(4 - \sqrt{3})s - 2\sqrt{3} + 2\sqrt{3}s^2 = 2 - 2\sqrt{3} \Rightarrow 2\sqrt{3}s^2 + (4 - \sqrt{3})s - 2 = 0$.
Discriminant: $(4 - \sqrt{3})^2 + 16\sqrt{3} = 19 + 8\sqrt{3} = (4 + \sqrt{3})^2$.
$s = \frac{-(4 - \sqrt{3}) \pm (4 + \sqrt{3})}{4\sqrt{3}}$: $s = \frac{1}{2}$ or $s = -\frac{2}{\sqrt{3}}$ (less than −1, rejected).
$\sin x = \frac{1}{2}$ in $\left[-2\pi, \frac{5\pi}{2}\right]$: $x = -\frac{11\pi}{6}, -\frac{7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}$ ($\frac{17\pi}{6}$ is past $\frac{15\pi}{6}$).
Five solutions.
Answer: (A) $5$. | ||||
| 14 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Let the angle with the x-axis be $\alpha$, so $\beta = \gamma = \frac{\alpha}{2}$, with $0 \le \alpha \le \pi$.
$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \Rightarrow \cos^2\alpha + 2\cos^2\frac{\alpha}{2} = 1 \Rightarrow \cos^2\alpha + 1 + \cos\alpha = 1$.
$\cos\alpha(\cos\alpha + 1) = 0$: $\alpha = \frac{\pi}{2}$ or $\alpha = \pi$.
So $\beta = \frac{\pi}{4}$ or $\beta = \frac{\pi}{2}$ (the second is the line along the negative x-axis, at right angles to the y- and z-axes; both satisfy the condition).
Sum $= \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$.
Answer: (C) $\frac{3\pi}{4}$. | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry From (7, 10, 11) move along the direction (2, 3, 6): $(7 + 2t, 10 + 3t, 11 + 6t)$.
The first line has $y = 4$ throughout, so $10 + 3t = 4$, $t = -2$, giving (3, 4, −1).
Check: $\frac{3 - 4}{1} = -1$ and $\frac{-1 - 2}{3} = -1$, so (3, 4, −1) is on the line.
Distance $= |t|\sqrt{4 + 9 + 36} = 2\times7 = 14$.
Answer: (B) $14$. | ||||
| 16 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Piece by piece:
$x < -2$: $|x + 2| - 2|x| = -(x + 2) + 2x = x - 2 < 0$, so $f(x) = 2 - x$, decreasing, reaching 4 at $x = -2$.
$-2 \le x < 0$: $(x + 2) + 2x = 3x + 2$, so $f(x) = |3x + 2|$: falls from 4 at $x = -2$ to 0 at $x = -\frac{2}{3}$, then rises to 2 at $x = 0$.
$x \ge 0$: $(x + 2) - 2x = 2 - x$, so $f(x) = |2 - x|$: falls from 2 at $x = 0$ to 0 at $x = 2$, then rises.
f keeps falling through $x = -2$, so that is no extremum. Local minima at $x = -\frac{2}{3}$ and $x = 2$ ($m = 2$); local maximum at $x = 0$ ($n = 1$).
$m + n = 3$.
Answer: (C) $3$. | ||||
| 17 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The circle is $x^2 + (y + 6)^2 = 1$: centre (0, −6), radius 1.
The shortest distance lies along the common normal: the normal to the parabola through the centre.
For $y^2 = 8x$ ($a = 2$), the normal at $(2t^2, 4t)$ is $y = -tx + 4t + 2t^3$. Through (0, −6): $2t^3 + 4t + 6 = 0 \Rightarrow t^3 + 2t + 3 = 0 \Rightarrow (t + 1)(t^2 - t + 3) = 0$, so $t = -1$ (the quadratic has no real root).
Foot on the parabola: (2, −4). Distance to the centre: $\sqrt{4 + 4} = 2\sqrt{2}$.
The parabola lies outside the circle, so the shortest distance is $2\sqrt{2} - 1$.
Answer: (A) $2\sqrt{2} - 1$. | ||||
| 18 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replace x by $\pi - x$ (the denominator is unchanged) and add the two forms:
$2I = \int_0^{\pi}\frac{8\pi\,dx}{4\cos^2x + \sin^2x}$, so $I = 4\pi\int_0^{\pi}\frac{dx}{4\cos^2x + \sin^2x} = 8\pi\int_0^{\pi/2}\frac{dx}{4\cos^2x + \sin^2x}$.
Divide by $\cos^2x$ and put $t = \tan x$: $\int_0^{\pi/2}\frac{\sec^2x\,dx}{4 + \tan^2x} = \int_0^{\infty}\frac{dt}{4 + t^2} = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4}$.
$I = 8\pi\cdot\frac{\pi}{4} = 2\pi^2$.
Answer: (C) $2\pi^2$. | ||||
| 19 | 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 $|x - y| \le y$ means $-y \le x - y \le y$, i.e. $x \ge 0$ and $y \ge \frac{x}{2}$.
So the region is $\frac{x}{2} \le y \le 4\sqrt{x}$. The curves meet where $\frac{x}{2} = 4\sqrt{x}$: $x = 0$ and $x = 64$.
Area $= \int_0^{64}\left(4\sqrt{x} - \frac{x}{2}\right)dx = \frac{8}{3}(64)^{3/2} - \frac{64^2}{4} = \frac{4096}{3} - 1024 = \frac{1024}{3}$.
Answer: (B) $\frac{1024}{3}$. | ||||
| 20 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $3\tan^2x + 3 = 3\sec^2x$, so the equation is $\frac{dy}{dx} + 3\sec^2x\,y = \sec^2x$.
Integrating factor $e^{3\tan x}$: $y\,e^{3\tan x} = \int\sec^2x\,e^{3\tan x}dx = \frac{1}{3}e^{3\tan x} + C$.
$y = \frac{1}{3} + Ce^{-3\tan x}$. $y(0) = \frac{1}{3} + C = \frac{1}{3} + e^3$, so $C = e^3$.
$y\left(\frac{\pi}{4}\right) = \frac{1}{3} + e^3e^{-3} = \frac{4}{3}$.
Answer: (D) $\frac{4}{3}$. | ||||
| 21 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Put $x = 1$: $\sum a_r = 3^{10}$. Put $x = -1$: $\sum(-1)^ra_r = 1$.
Odd coefficients: $a_1 + a_3 + \ldots + a_{19} = \frac{3^{10} - 1}{2} = \frac{59048}{2} = 29524$.
$a_2$: an $x^2$ comes from one factor's $x^2$ ($\binom{10}{1} = 10$ ways) or two factors' x ($\binom{10}{2} = 45$ ways), so $a_2 = 55$ and $11a_2 = 605$.
$29524 - 605 = 28919 = 121\times239$.
$k = 239$.
Answer: 239. | ||||
| 22 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalHard | |
A hard numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $|A| = \lambda(10 + 6) - 2(8 - 42) + 3(-4 - 35) = 16\lambda + 68 - 117 = 16\lambda - 49 = -1$, so $\lambda = 3$.
For an invertible 3 × 3 matrix M, $\mathrm{adj}\,M = |M|M^{-1}$.
$\mathrm{adj}(A^2) = |A|^2(A^2)^{-1} = A^{-2}$, so $A\,\mathrm{adj}(A^2) = A^{-1}$.
$\mathrm{adj}(A^{-1}) = |A^{-1}|A = -A$. So $B = (-A)^{-1} = -A^{-1}$.
$\lambda B + I = I - 3A^{-1} = A^{-1}(A - 3I)$, and $|\lambda B + I| = |A^{-1}|\,|A - 3I| = -|A - 3I|$.
$A - 3I = \begin{bmatrix}0&2&3\\4&2&6\\7&-1&-1\end{bmatrix}$: $|A - 3I| = 0 - 2(-4 - 42) + 3(-4 - 14) = 92 - 54 = 38$.
$|\lambda B + I| = -38$.
Answer: −38. | ||||
| 23 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The centre is the midpoint of the foci, (6, 2), and $c = 2$; the transverse axis is horizontal.
With the $x^2$ and $y^2$ coefficients 3 and −1, the equation is $3(x - 6)^2 - (y - 2)^2 = K$, i.e. $\frac{(x-6)^2}{K/3} - \frac{(y-2)^2}{K} = 1$.
$c^2 = a^2 + b^2 = \frac{K}{3} + K = \frac{4K}{3} = 4$, so $K = 3$.
Expand: $3x^2 - 36x + 108 - y^2 + 4y - 4 - 3 = 0 \Rightarrow 3x^2 - y^2 - 36x + 4y + 101 = 0$.
$\alpha = 36$, $\beta = 4$, $\gamma = 101$; $\alpha + \beta + \gamma = 141$.
Answer: 141. | ||||
| 24 | 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 $(\vec{b} - \vec{c})\times\vec{d} = \vec{0}$, so $\vec{d} = t(\vec{b} - \vec{c}) = t(\hat{i} - 2\hat{j} + \hat{k})$.
$\vec{a}\cdot\vec{d} = t(1 - 4 + 1) = -2t = 4 \Rightarrow t = -2$, $\vec{d} = -2\hat{i} + 4\hat{j} - 2\hat{k}$.
$|\vec{a}\times\vec{d}|^2 = |\vec{a}|^2|\vec{d}|^2 - (\vec{a}\cdot\vec{d})^2 = 6\times24 - 16 = 128$.
Answer: 128. | ||||
| 25 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability A $1^\infty$ form: $p = e^{L}$ with $L = \lim_{x\to0}\frac{1}{x^2}\left(\frac{\tan x}{x} - 1\right) = \lim_{x\to0}\frac{\tan x - x}{x^3}$.
$\tan x = x + \frac{x^3}{3} + \ldots$, so $L = \frac{1}{3}$ and $\log_e p = \frac{1}{3}$.
$96\log_e p = 32$.
Answer: 32. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Boltzmann constant: energy per kelvin, J K$^{-1}$ $= ML^2T^{-2}K^{-1}$ (III).
Coefficient of viscosity: from $F = \eta A\frac{dv}{dx}$, $\eta = \frac{F}{A\,(dv/dx)}$: $\frac{MLT^{-2}}{L^2T^{-1}} = ML^{-1}T^{-1}$ (IV).
Planck's constant: energy × time, J s $= ML^2T^{-1}$ (I).
Thermal conductivity: W m$^{-1}$ K$^{-1}$ $= \frac{ML^2T^{-3}}{L\,K} = MLT^{-3}K^{-1}$ (II).
A - III, B - IV, C - I, D - II.
Answer: (A). | ||||
| 27 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $R = \frac{u^2\sin2\theta}{g} = \frac{2u^2\sin\theta\cos\theta}{g}$ and $H = \frac{u^2\sin^2\theta}{2g}$.
$R = 3H$: $2\sin\theta\cos\theta = \frac{3}{2}\sin^2\theta \Rightarrow \tan\theta = \frac{4}{3}$, so $\sin\theta = \frac{4}{5}$, $\cos\theta = \frac{3}{5}$.
$R = \frac{2u^2}{g}\cdot\frac{4}{5}\cdot\frac{3}{5} = \frac{24u^2}{25g}$.
$n = 24$.
Answer: (D) $24$. | ||||
| 28 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $v = \frac{dx}{dt} = 2c_0t + 2c(t - 2)$, so the initial velocity is $v(0) = -4c$, not 4c.
$a = \frac{dv}{dt} = 2c_0 + 2c = 2(c + c_0)$, a constant.
Answer: (D) the acceleration of the particle is $2(c + c_0)$. | ||||
| 29 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Moving together, both blocks share one velocity. Momentum: $10\times3 = 15V$, $V = 2$ m/s.
Kinetic energy before: $\frac{1}{2}(10)(3^2) = 45$ J. After: $\frac{1}{2}(15)(2^2) = 30$ J.
The 15 J lost is stored in the spring (the table is frictionless): $\frac{1}{2}kx^2 = 15 \Rightarrow x^2 = \frac{30}{3000} = 0.01$, $x = 0.1$ m.
Answer: (D) 0.1 m. | ||||
| 30 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on 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 Connected, the levels become equal: $\frac{10 + 6}{2} = 8$ m in each.
A water column of height h has potential energy $\rho gA\frac{h^2}{2}$ (its centre of mass is at $\frac{h}{2}$). With $\rho gA = 10^3\times10\times2 = 2\times10^4$ N/m:
Before: $2\times10^4\left(\frac{100}{2} + \frac{36}{2}\right) = 2\times10^4\times68$ J.
After: $2\times10^4\left(\frac{64}{2} + \frac{64}{2}\right) = 2\times10^4\times64$ J.
Work done by gravity = loss of potential energy $= 2\times10^4\times4 = 8\times10^4$ J.
Answer: (C) $8\times10^4$ J. | ||||
| 31 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Work of the retarding force: $W = -\int_{0.1}^{1.9}kx\,dx = -\frac{10}{2}(1.9^2 - 0.1^2) = -5(3.61 - 0.01) = -18$ J.
Work-energy theorem: $\frac{1}{2}(1)v_f^2 = \frac{1}{2}(1)(10^2) - 18 = 32$, $v_f^2 = 64$, $v_f = 8$ m/s.
Answer: (B) 8 m/s. | ||||
| 32 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal velocity, weight = buoyancy + Stokes drag: $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$, so $\eta = \frac{2r^2(\rho - \sigma)g}{9v_t}$.
$r = 1.8\times10^{-3}$ m, $r^2 = 3.24\times10^{-6}$ m$^2$, $\rho - \sigma = 7825 - 925 = 6900$ kg m$^{-3}$.
$\eta = \frac{2\times3.24\times10^{-6}\times6900\times9.8}{9\times2.45\times10^{-2}} = \frac{0.4382}{0.2205} \approx 1.99$ Pa s.
Answer: (B) $1.99$. | ||||
| 33 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics The gas returns to its initial state, so $\Delta U = 0$ over the cycle and the net heat equals the net work.
Isothermal expansion $V_0 \to 4V_0$: $W_1 = P_0V_0\ln4 = 2\ln2\,P_0V_0$. The pressure falls to $\frac{P_0}{4}$.
Isobaric compression at $\frac{P_0}{4}$ from $4V_0$ to $V_0$: $W_2 = \frac{P_0}{4}(V_0 - 4V_0) = -0.75P_0V_0$.
Isochoric heating: $W_3 = 0$.
$Q = W_1 + W_2 = P_0V_0(2\ln2 - 0.75)$.
Answer: (A) $P_0V_0(2\ln2 - 0.75)$. | ||||
| 34 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory In a closed vessel the volume is fixed, so $P \propto T$ and $\frac{\Delta P}{P} = \frac{\Delta T}{T}$.
$0.004 = \frac{1}{T} \Rightarrow T = 250$ K.
Answer: (D) 250 K. | ||||
| 35 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Fundamental of a pipe closed at one end: $f = \frac{v}{4L}$.
$f_1 - f_2 = \frac{v}{4}\left(\frac{1}{1.0} - \frac{1}{1.2}\right) = \frac{v}{4}\cdot\frac{1}{6} = \frac{v}{24} = 15$.
$v = 360$ m/s.
Answer: (A) 360 m/s. | ||||
| 36 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy 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 $\tau = mB\sin60^\circ = 80\sqrt{3} \Rightarrow mB\cdot\frac{\sqrt{3}}{2} = 80\sqrt{3} \Rightarrow mB = 160$ J.
$U = -mB\cos60^\circ = -160\times\frac{1}{2} = -80$ J.
Answer: (B) −80 J. | ||||
| 37 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At its rated rms voltage the bulb draws its rated power: $I_{rms} = \frac{P}{V_{rms}} = \frac{100}{220} = 0.4545$ A.
$I_0 = \sqrt{2}\,I_{rms} = 1.414\times0.4545 \approx 0.64$ A.
Answer: (C) 0.64 A. | ||||
| 38 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charge on the first capacitor: $Q = 100\times60 = 6000$ pC. With the battery removed it is shared, not changed.
In parallel both capacitors are at 20 V: $(100 + C)\times20 = 6000 \Rightarrow 100 + C = 300$, $C = 200$ pF.
Answer: (B) 200. | ||||
| 39 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Input power $= VI = 100\times1 = 100$ W.
Loss $= (1 - 0.916)\times100 = 8.4$ W $= 8.4$ J/s.
With 1 cal $= 4.2$ J: $\frac{8.4}{4.2} = 2$ cal/s.
Answer: (D) 2. | ||||
| 40 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Assertion and reasonMedium | |
A medium assertion and reason question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter In a perpendicular field, $qvB = \frac{mv^2}{r}$, so $r = \frac{p}{qB}$: for equal momentum the radius depends only on the charge.
Assertion: $O^{2-}$ carries $2e$ and $H^+$ carries $e$, so $r_O = \frac{r_H}{2}$. The oxygen ion's path has the smaller radius of curvature, which is what the assertion states when "curvature" is read as radius of curvature. Taken that way, A is true.
Reason: a proton and an electron carry the same magnitude of charge, so with the same momentum they move on circles of the same radius (curving opposite ways). R is false.
Answer: (C) A is true but R is false. | ||||
| 41 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Amplitudes go as the square root of intensity: $a_1 : a_2 = 1 : 3$.
$\frac{I_{max}}{I_{min}} = \left(\frac{a_1 + a_2}{a_2 - a_1}\right)^2 = \left(\frac{4}{2}\right)^2 = 4$.
Answer: (A) 4:1. | ||||
| 42 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics In air: $\lambda_0 = \frac{c}{f} = \frac{3\times10^8}{5\times10^{14}} = 6\times10^{-7}$ m $= 600$ nm.
The frequency is unchanged on refraction and the speed falls by the factor $\mu$: $\lambda = \frac{\lambda_0}{\mu} = \frac{600}{2} = 300$ nm.
Answer: (A) 300 nm. | ||||
| 43 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The intensity from a slit is proportional to its width, so $I_1 : I_2 = 1 : 2$ and the amplitudes are $1 : \sqrt{2}$.
$\frac{I_{max}}{I_{min}} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$.
Answer: (B) $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$. | ||||
| 44 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Assertion and reasonEasy | |
A easy assertion and reason question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei A is true: Bohr's model gives correct energy levels only for one-electron systems (H, He$^+$, Li$^{2+}$, …).
R is true: the model treats a single electron in the Coulomb field of the nucleus and has no electron–electron repulsion.
That omission is exactly why it fails for many-electron atoms, so R explains A.
Answer: (A). | ||||
| 45 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A and B feed an OR gate; A is also taken straight to an AND gate, whose other input is the OR output.
$C = A\cdot(A + B) = A + AB = A$.
So C copies A: rows (0,0)→0, (0,1)→0, (1,0)→1, (1,1)→1. That is table (A).
(B) gives $C = AB$ (0 for A = 1, B = 0); (C) and (D) give $C = \overline{A + B}$.
Answer: (A). | ||||
| 46 | 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 For a product of powers the maximum relative errors add, each multiplied by its power:
$\frac{\Delta C}{C} = \frac{\Delta p}{p} + 2\frac{\Delta q}{q} + 3\frac{\Delta r}{r} + \frac{1}{2}\frac{\Delta s}{s} = 1 + 4 + 9 + 1 = 15\%$.
Answer: 15. | ||||
| 47 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy 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 A soap bubble has two surfaces: $\Delta P = \frac{4T}{R}$, so $\Delta P \propto \frac{1}{R}$.
$\Delta P_A = \frac{1}{2}\Delta P_B \Rightarrow R_A = 2R_B$.
$\frac{V_A}{V_B} = \left(\frac{R_A}{R_B}\right)^3 = 8$.
Answer: 8. | ||||
| 48 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Series (emfs aiding): $I_1 = \frac{1 + 2}{6 + 2 + 1} = \frac{1}{3}$ A.
Parallel: $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{0.5 + 2}{0.5 + 1} = \frac{5}{3}$ V, $r_{eq} = \frac{2\times1}{2 + 1} = \frac{2}{3}$ Ω.
$I_2 = \frac{5/3}{6 + 2/3} = \frac{5/3}{20/3} = \frac{1}{4}$ A.
$\frac{I_1}{I_2} = \frac{1/3}{1/4} = \frac{4}{3}$, so $x = 4$.
Answer: 4. | ||||
| 49 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Refraction at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$, with light travelling in the + direction from air into glass.
Convex surface towards the air: $R = +50$ cm, $v = +200$ cm, $\mu_1 = 1$, $\mu_2 = 1.5$.
$\frac{1.5}{200} - \frac{1}{u} = \frac{0.5}{50} \Rightarrow 0.0075 - \frac{1}{u} = 0.01 \Rightarrow \frac{1}{u} = -0.0025$, $u = -400$ cm.
The source is 400 cm $= 4$ m from the surface.
Answer: 4. | ||||
| 50 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The fourth excited state is $n = 5$: $E_5 = -\frac{13.6}{25} = -0.544$ eV.
After emitting 2.86 eV: $E_n = -0.544 - 2.86 = -3.404$ eV.
$-\frac{13.6}{n^2} = -3.4 \Rightarrow n^2 = 4$, $n = 2$.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2}$: one mole of Mg gives one mole of $\mathrm{H_2}$.
At STP one mole of gas occupies 22.4 L: $n(\mathrm{H_2}) = \frac{0.220}{22.4} = 9.82\times10^{-3}$ mol.
Mass of Mg $= 9.82\times10^{-3}\times24 = 0.2357$ g $= 235.7$ mg $\approx 236$ mg.
(235.7 g is a thousand times too much.)
Answer: (B) 236 mg. | ||||
| 52 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom Orbital angular momentum $L = \sqrt{l(l+1)}\frac{h}{2\pi}$.
2s: $l = 0$, so $L = 0$.
2p: $l = 1$, so $L = \sqrt{2}\frac{h}{2\pi}$.
($\sqrt{6}\frac{h}{2\pi}$ would be a d electron, $l = 2$.)
Answer: (A). | ||||
| 53 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | StatementsEasy | |
A easy statements question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Statement I: while ice and water coexist at constant pressure, heat supplied goes into melting (the enthalpy of fusion) and the temperature stays at the melting point until all the ice has melted. True.
Statement II: that heat is used to overcome the intermolecular attractions (hydrogen bonds) holding the ice lattice; the temperature, and so the average kinetic energy of the molecules, does not rise. True.
Answer: (A) Both Statement I and Statement II are true. | ||||
| 54 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctMedium | |
A medium single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry NaOH: $10\times2 = 20$ mmol; HCl: $20\times1 = 20$ mmol. They neutralise exactly, giving 20 mmol NaCl in 30 mL: a neutral NaCl solution.
10 mL of it carries only NaCl ($\frac{20}{3} \approx 6.7$ mmol), no acid or base.
The flask holds 2 mol HCl, made up to 100 mL $= 0.1$ L: $[\mathrm{HCl}] = \frac{2}{0.1} = 20$ M.
(The NaCl in it is only about 0.067 M, not 0.2 M.)
Answer: (A) 20 M HCl solution. | ||||
| 55 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry The strong acid reacts first: fast-moving $\mathrm{H^+}$ is replaced by slow $\mathrm{Na^+}$, so conductance falls (A to B). B, at 2.0 mL, is the end of HCl.
Then $\mathrm{CH_3COOH}$ (weakly ionised) is converted to $\mathrm{CH_3COO^-Na^+}$, so conductance rises slowly (B to C). C, at 5.0 mL, is the end of acetic acid. After C, excess $\mathrm{OH^-}$ makes it rise steeply.
HCl: $2.0\times0.1 = 0.2$ mmol in 40 mL, $\frac{0.2}{40} = 0.005$ M. True.
$\mathrm{CH_3COOH}$: $(5.0 - 2.0)\times0.1 = 0.3$ mmol, $\frac{0.3}{40} = 0.0075$ M, not 0.005 M.
HCl is neutralised first, not $\mathrm{CH_3COOH}$; C marks the end of the acetic acid, not of HCl.
Answer: (B). | ||||
| 56 | ElectrochemistryBatteries and fuel cells | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on batteries and fuel cells, from Electrochemistry. NCERT: Class 12: Electrochemistry In the methanol fuel cell, methanol is oxidised to $\mathrm{CO_2}$ at the anode and $\mathrm{O_2}$ is reduced to water at the cathode.
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \Rightarrow 1.21 = 1.229 - E^\circ_{CO_2/CH_3OH}$, so $E^\circ_{CO_2/CH_3OH} = 0.019$ V $= 19$ mV. (C) is true.
(A) False: oxygen is consumed at the cathode, not formed at the anode.
(B) False: methanol is oxidised, at the anode.
(D) False: a fuel cell is fed its reactants continuously.
Answer: (C). | ||||
| 57 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | StatementsMedium | |
A medium statements question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics A. False: the Arrhenius equation describes elementary and complex, homogeneous and heterogeneous reactions alike (with apparent A and $E_a$ for a complex one).
B. True: $k = Ae^{-E_a/RT}$ and the exponential has no units, so A has the units of k.
C. True: a smaller $E_a$ makes $e^{-E_a/RT}$ larger, so k is larger and the reaction faster.
D. False: A and $E_a$ are taken as constants, independent of temperature, over the range the equation is used.
E. False: A and $E_a$ are independent parameters; $E_a \gg RT$ is simply the usual condition, under which the temperature dependence of A is negligible.
Correct: B and C.
Answer: (B) B and C Only. | ||||
| 58 | PolymersNatural rubber and cellulose | No matching NCERT chapter | StatementsEasy | |
A easy statements question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Cellulose carries three free –OH groups on every glucose unit, so cotton hydrogen-bonds strongly with water, holds more of it and dries slowly. Statement I is true.
Nylon has far fewer hydrogen-bonding sites (only the amide groups, many of them bonded to each other between chains), so it binds less water and dries faster. Statement II, which says the opposite, is false.
Answer: (C) Statement I is true but Statement II is false. | ||||
| 59 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Matching listEasy | |
A easy matching list question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The period 7 p-block elements:
Mc, moscovium (Z = 115): group 15, a pnictogen.
Lv, livermorium (116): group 16, a chalcogen.
Ts, tennessine (117): group 17, a halogen.
Og, oganesson (118): group 18, a noble gas.
A-IV, B-III, C-I, D-II.
Answer: (B). | ||||
| 60 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter A. False: gallium is smaller than aluminium (poor shielding by its 3d electrons): radii B 85, Al 143, Ga 135, In 167, Tl 170 pm, so B < Ga < Al < In < Tl.
B. True on the values of the standard Group 13 table: B 2.0, Al 1.5, Ga 1.6, In 1.7, Tl 1.8.
C. False: density increases down the group (B 2.35, Al 2.70, Ga 5.90, In 7.31, Tl 11.85 g cm$^{-3}$), the reverse of the order given.
D. True: first ionisation enthalpies (kJ mol$^{-1}$) In 558 < Al 577 < Ga 579 < Tl 589 < B 801.
Correct: B and D.
Answer: (B) B and D Only. | ||||
| 61 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | StatementsEasy | |
A easy statements question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Down a transition group the higher oxidation states become more stable: Mo(VI) is more stable than Cr(VI). Statement II is false.
Because Cr(VI) is the less stable, it is readily reduced (to Cr(III)), so $\mathrm{CrO_3}$ is a strong oxidising agent while $\mathrm{MoO_3}$ is not. Statement I is true.
Answer: (C) Statement I is true but Statement II is false. | ||||
| 62 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | StatementsMedium | |
A medium statements question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A. $\mathrm{Mn^{3+}}$, $d^4$, with strong-field $\mathrm{CN^-}$: $t_{2g}^4$, two unpaired electrons. Paramagnetic.
B. $\mathrm{Co^{3+}}$, $d^6$, with $\mathrm{NH_3}$ (strong enough for $\mathrm{Co^{3+}}$): $t_{2g}^6$, all paired. Diamagnetic.
C. $\mathrm{Fe^{2+}}$, $d^6$, with $\mathrm{CN^-}$: $t_{2g}^6$. Diamagnetic.
D. $\mathrm{Co^{3+}}$, $d^6$, with weak-field $\mathrm{F^-}$ and $\mathrm{H_2O}$: high spin $t_{2g}^4e_g^2$, four unpaired electrons. Paramagnetic.
Diamagnetic: B and C.
Answer: (A) B and C Only. | ||||
| 63 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | StatementsMedium | |
A medium statements question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter A primary standard must be pure, stable in air, non-hygroscopic and of exact known composition.
A. $\mathrm{Na_2Cr_2O_7}$ is hygroscopic (the potassium salt is the one used as a primary standard). Not suitable.
B. Oxalic acid (as its dihydrate) is a standard primary standard for alkalis and permanganate. Suitable.
C. NaOH is hygroscopic and absorbs $\mathrm{CO_2}$ from air. Not suitable.
D. Ferrous sulphate loses water of crystallisation and is oxidised by air to Fe(III) (Mohr's salt is used instead). Not suitable.
E. Sodium tetraborate (borax) is a primary standard for acids. Suitable.
Not to be used: A, C and D.
Answer: (A) A, C and D Only. | ||||
| 64 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Pressure of dry nitrogen: $715 - 15 = 700$ mm Hg.
Volume at STP: $V_0 = 60\times\frac{700}{760}\times\frac{273}{300} = 50.29$ mL.
Mass of $\mathrm{N_2}$: $\frac{28}{22400}\times50.29 = 0.06286$ g.
% N $= \frac{0.06286}{0.4}\times100 = 15.71\%$.
Answer: (B) 15.71 %. | ||||
| 65 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsEasy | |
A easy statements question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Statement I: hyperconjugation (delocalisation of C–H σ electrons into an adjacent empty p orbital or π system) is present in the molecule's ground state all the time; it is a permanent effect. False.
Statement II: each alkyl group adds α C–H bonds that can donate into the empty p orbital of a carbocation, so more alkyl groups mean more hyperconjugation and a more stable cation ($3^\circ > 2^\circ > 1^\circ$). True.
Answer: (D) Statement I is false but Statement II is true. | ||||
| 66 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The carboxylic acid is the principal group: the parent is benzoic acid, with the $\mathrm{CO_2H}$ carbon as C-1.
Numbering towards the OH gives OH at C-2, Br at C-3 and $\mathrm{NO_2}$ at C-5 (locants {2, 3, 5}, the lowest set).
Prefixes go in alphabetical order: bromo, hydroxy, nitro.
3-Bromo-2-hydroxy-5-nitrobenzoic acid.
(A) and (B) are not alphabetical; (C) numbers wrongly (C-1 must carry the acid).
Answer: (D). | ||||
| 67 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Step 1: Br is an ortho/para director; sulfonation is reversible and bulky, so the para product dominates. A is 4-bromobenzenesulfonic acid.
Step 2: Br directs ortho/para (para is taken, so ortho to Br); $\mathrm{SO_3H}$ directs meta. The position ortho to Br is meta to $\mathrm{SO_3H}$: both groups direct there.
B is 3,4-dibromobenzenesulfonic acid, the second Br beside the first.
(C) puts the second Br beside $\mathrm{SO_3H}$ (ortho to it, against its meta direction); (A) and (B) start from the ortho isomer.
Answer: (D). | ||||
| 68 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Phenylglyoxal, $\mathrm{Ph{-}CO{-}CHO}$, has no α-hydrogen. With hot KOH it undergoes an intramolecular Cannizzaro reaction (a 1,2-hydride shift).
$\mathrm{OH^-}$ adds to the aldehyde carbon; the hydride on that carbon moves to the neighbouring keto carbon. The aldehyde end is oxidised to carboxylate and the keto carbon is reduced to a secondary alcohol.
P is potassium mandelate, $\mathrm{Ph{-}CH(OH){-}COO^-K^+}$.
Answer: (B). | ||||
| 69 | AminesDiazonium salts | Class 12: Amines | Single correctHard | |
A hard single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Target: $\mathrm{NH_2}$ at C-1 with Br at C-3, C-4 and C-5.
(C) 4-Nitroaniline with excess $\mathrm{Br_2}$: the strongly activating $\mathrm{NH_2}$ puts Br on both carbons ortho to it, giving 2,6-dibromo-4-nitroaniline. Diazotisation and CuBr (Sandmeyer) replace that $\mathrm{NH_2}$ by Br: 1,2,3-tribromo-5-nitrobenzene. Sn/HCl reduces $\mathrm{NO_2}$ to $\mathrm{NH_2}$: 3,4,5-tribromoaniline.
(A) gives 2,4,6-tribromoaniline. (B) gives 3-bromoaniline (one Br, meta to $\mathrm{NO_2}$). (D) gives bromobenzene, which does not react with $\mathrm{NH_3}$ under these conditions.
Answer: (C). | ||||
| 70 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | StatementsEasy | |
A easy statements question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Fat-soluble vitamins: A, D, E and K. Water-soluble: the B group and C.
Of those listed, E and K are fat soluble.
Answer: (C) C & E Only. | ||||
| 71 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics 1 → 2: constant volume (1000 cm³): $w = 0$.
2 → 3: expansion at constant 3.00 atm from 1.000 L to 2.000 L: $w = -P\Delta V = -3.00\times1.000 = -3.00$ L atm.
3 → 4: constant volume (2000 cm³): $w = 0$.
Total $w = -3.00$ L atm. Converting with $1$ L atm $= \frac{8.314}{0.082} = 101.4$ J (or 101.3 J): $w = -304$ J.
The blank asks for the magnitude after the printed minus sign: 304.
Answer: 304. | ||||
| 72 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat released $= C\Delta T = 5\times5 = 25$ kJ.
n-Octane $\mathrm{C_8H_{18}}$: $M = 96 + 18 = 114$ g mol$^{-1}$, so 1.14 g is 0.01 mol.
At constant volume: $|\Delta U| = \frac{25}{0.01} = 2500$ kJ mol$^{-1}$.
Answer: 2500. | ||||
| 73 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Enthalpies of atomisation (kJ mol$^{-1}$): Sc 326, Mn 281, Co 425, Cu 339. Cobalt is highest.
$\mathrm{Co^{2+}}$: $[\mathrm{Ar}]3d^7$, three unpaired electrons.
$\mu = \sqrt{n(n+2)} = \sqrt{15} = 3.87$ BM, which rounds to 4.
Answer: 4. | ||||
| 74 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{C_6H_5NO_2 \rightarrow C_6H_4(NO_2)_2}$, one mole to one mole.
m-Dinitrobenzene: $M = 72 + 4 + 28 + 64 = 168$ g mol$^{-1}$; $\frac{4.2}{168} = 0.025$ mol.
Nitrobenzene: $M = 72 + 5 + 14 + 32 = 123$ g mol$^{-1}$; $X = 0.025\times123 = 3.075$ g.
To the nearest integer: 3.
Answer: 3. | ||||
| 75 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{C_9H_{12}}$ is a benzene ring plus three more carbons, all saturated.
One substituent ($\mathrm{C_3H_7}$): propylbenzene and isopropylbenzene (cumene): 2.
Two substituents (methyl + ethyl): ortho, meta, para ethyltoluene: 3.
Three methyls: 1,2,3-, 1,2,4- and 1,3,5-trimethylbenzene: 3.
Total $= 2 + 3 + 3 = 8$.
Answer: 8. | ||||
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