JEE (Main) · 2025
JEE (Main) 2025 — 22 January, Shift 1
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
- Integral Calculus4
- Sets, Relations and Functions2
- Sequences and Series2
6 easy · 18 medium · 1 hard
Physics
21 Q
- Properties of Bulk Matter4
- Optics4
- Current Electricity3
11 easy · 8 medium · 2 hard
Chemistry
21 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Electrochemistry2
- Chemical Kinetics2
15 easy · 6 medium · 0 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 FunctionsSets and their representation | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium 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 Every element of $B$ is a fraction $\frac{m}{n}$ in lowest terms with $1 \le m < n \le 10$. Two different such pairs give different numbers (a fraction in lowest terms is unique), so $n(B)$ is the number of pairs.
For a fixed denominator $n$, the numerators $m < n$ with $\gcd(m, n) = 1$ number $\varphi(n)$ (Euler's totient).
$n = 2$: 1; $n = 3$: 2; $n = 4$: 2; $n = 5$: 4; $n = 6$: 2; $n = 7$: 6; $n = 8$: 4; $n = 9$: 6; $n = 10$: 4.
(For $n = 1$ there is no $m < 1$.)
Total: $1 + 2 + 2 + 4 + 2 + 6 + 4 + 6 + 4 = 31$.
Answer: (B) $31$. | ||||
| 2 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions An equivalence relation on a set is the same thing as a partition of the set into classes. (Every equivalence relation on a non-empty set contains the pairs $(a, a)$, so all of them are non-empty.)
Partitions of $\{1, 2, 3\}$:
$\{1\},\{2\},\{3\}$ — one way;
one pair and one single: $\{1,2\},\{3\}$; $\{1,3\},\{2\}$; $\{2,3\},\{1\}$ — three ways;
$\{1, 2, 3\}$ — one way.
Total $1 + 3 + 1 = 5$.
Answer: (B) $5$. | ||||
| 3 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Take natural logarithms of both sides (both are positive): $5(\ln x)^2 + 3 = 8\ln x$.
Put $t = \ln x$: $5t^2 - 8t + 3 = 0$, i.e. $(5t - 3)(t - 1) = 0$, so $t = 1$ or $t = \frac{3}{5}$. Both give a valid $x > 0$: $x = e$ and $x = e^{3/5}$.
Product of the solutions $= e^{t_1}\cdot e^{t_2} = e^{t_1 + t_2} = e^{8/5}$ (sum of roots $= \frac{8}{5}$).
Answer: (D) $e^{8/5}$. | ||||
| 4 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations On $|z| = 1$: $z_1 = e^{-i\pi/4}$, $z_2 = 1$, $z_3 = e^{i\pi/4}$, and $\bar z = e^{-i\theta}$ for $z = e^{i\theta}$.
$z_1\bar z_2 = e^{-i\pi/4}$, $z_2\bar z_3 = e^{-i\pi/4}$, $z_3\bar z_1 = e^{i\pi/4}e^{i\pi/4} = e^{i\pi/2} = i$.
Sum $= 2e^{-i\pi/4} + i = 2\left(\frac{1}{\sqrt2} - \frac{i}{\sqrt2}\right) + i = \sqrt2 + i(1 - \sqrt2)$.
$|\,\cdot\,|^2 = 2 + (1 - \sqrt2)^2 = 2 + 3 - 2\sqrt2 = 5 - 2\sqrt2$.
So $\alpha = 5$, $\beta = -2$ and $\alpha^2 + \beta^2 = 25 + 4 = 29$.
Answer: (B) $29$. | ||||
| 5 | 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 Write $S_n = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$. Then
$T_n = S_n - S_{n-1} = \frac{(2n-1)(2n+1)(2n+3)}{64}\left[(2n+5) - (2n-3)\right] = \frac{(2n-1)(2n+1)(2n+3)}{8}$.
So $\frac{1}{T_r} = \frac{8}{(2r-1)(2r+1)(2r+3)}$.
Telescoping: $\frac{1}{(2r-1)(2r+1)(2r+3)} = \frac{1}{4}\left[\frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)}\right]$ (the difference of the two fractions is $\frac{(2r+3) - (2r-1)}{(2r-1)(2r+1)(2r+3)} = \frac{4}{(2r-1)(2r+1)(2r+3)}$).
Hence $\sum_{r=1}^{n}\frac{1}{T_r} = 8\cdot\frac{1}{4}\left[\frac{1}{1\cdot 3} - \frac{1}{(2n+1)(2n+3)}\right] \to 2\cdot\frac{1}{3} = \frac{2}{3}$ as $n \to \infty$.
(Strictly, the given formula is not $0$ at $n = 0$, so $T_1$ would come out as $\frac{105}{64}$; the question intends the general term above for every $r$.)
Answer: (C) $\frac{2}{3}$. | ||||
| 6 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series In a G.P., $a_1a_5 = a_2a_4$ (both equal $a_1^2r^4$). So $a_2a_4 = 28$ and $a_2 + a_4 = 29$.
$a_2, a_4$ are the roots of $t^2 - 29t + 28 = 0$, i.e. $1$ and $28$. The terms increase, so $a_2 = 1$, $a_4 = 28$.
Then $r^2 = \frac{a_4}{a_2} = 28$ and $a_6 = a_4r^2 = 28 \times 28 = 784$.
Answer: (C) $784$. | ||||
| 7 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Once the five letters are chosen, the alphabetical arrangement is fixed, so we only count choices.
The middle (third) letter is M, so two letters come before M and two after it.
Letters before M: A to L, 12 letters, choose 2: $\binom{12}{2} = 66$.
Letters after M: N to Z, 13 letters, choose 2: $\binom{13}{2} = 78$.
Total $= 66 \times 78 = 5148$.
Answer: (D) $5148$. | ||||
| 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 Take a fair coin: the 8 outcomes are equally likely. X counts the places where H is immediately followed by T.
HHH: 0, HHT: 1, HTH: 1, HTT: 1, THH: 0, THT: 1, TTH: 0, TTT: 0.
So $P(X = 1) = \frac{4}{8} = \frac{1}{2}$, $P(X = 0) = \frac{1}{2}$ (X cannot be 2 in three tosses).
$\mu = E(X) = \frac{1}{2}$, $E(X^2) = \frac{1}{2}$, $\sigma^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.
$64(\mu + \sigma^2) = 64 \times \frac{3}{4} = 48$.
Answer: (B) $48$. | ||||
| 9 | 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 $B_1$, $B_2$ be \"first ball black\" and \"second ball black\".
$P(B_1 \cap B_2) = \frac{6}{10}\cdot\frac{5}{9} = \frac{1}{3}$.
$P(B_2) = \frac{6}{10}\cdot\frac{5}{9} + \frac{4}{10}\cdot\frac{6}{9} = \frac{30 + 24}{90} = \frac{3}{5}$ (by symmetry it equals $P(B_1)$).
$P(B_1 \mid B_2) = \frac{1/3}{3/5} = \frac{5}{9}$.
So $m = 5$, $n = 9$ and $m + n = 14$.
Answer: (D) $14$. | ||||
| 10 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines Reflection in a line maps the centroid of a triangle to the centroid of the image triangle, so reflect the centroid.
Centroid of the given triangle: $\left(\frac{1 + 3 + 2}{3}, \frac{3 + 1 + 4}{3}\right) = \left(2, \frac{8}{3}\right)$.
Image of $(x_0, y_0)$ in $ax + by + c = 0$: $(x_0, y_0) - \frac{2(ax_0 + by_0 + c)}{a^2 + b^2}(a, b)$.
Here $a = 1$, $b = 2$, $c = -2$: $ax_0 + by_0 + c = 2 + \frac{16}{3} - 2 = \frac{16}{3}$, and $\frac{2\cdot 16/3}{5} = \frac{32}{15}$.
$\alpha = 2 - \frac{32}{15} = -\frac{2}{15}$, $\beta = \frac{8}{3} - \frac{64}{15} = \frac{40 - 64}{15} = -\frac{24}{15}$.
$15(\alpha - \beta) = -2 + 24 = 22$.
Answer: (C) $22$. | ||||
| 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 C lies in the second quadrant and touches both axes with radius 2, so its centre is $(-2, 2)$.
Distance between the centres: $d = \sqrt{(2 + 2)^2 + (5 - 2)^2} = 5$.
Two circles of radii $r$ and $2$ cut at exactly two points when $|r - 2| < d < r + 2$:
$r + 2 > 5 \Rightarrow r > 3$, and $|r - 2| < 5 \Rightarrow r < 7$.
So $r \in (3, 7)$: $\alpha = 3$, $\beta = 7$, and $3\beta - 2\alpha = 21 - 6 = 15$.
Answer: (D) $15$. | ||||
| 12 | CirclesStandard and general form | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections The parabola meets the y-axis at $(0, -3)$ and the x-axis at the roots $x_1, x_2$ of $x^2 + px - 3 = 0$ (real, since the product of roots is $-3 < 0$).
The centre of a circle through $(x_1, 0)$ and $(x_2, 0)$ lies on the perpendicular bisector $x = \frac{x_1 + x_2}{2} = -\frac{p}{2}$. The centre is $(-1, -1)$, so $-\frac{p}{2} = -1$ and $p = 2$.
Then $x^2 + 2x - 3 = (x + 3)(x - 1)$: the points are $(1, 0)$, $(-3, 0)$, $(0, -3)$.
Check the centre's y-coordinate: equal distances from $(-1, k)$ to $(1, 0)$ and $(0, -3)$: $4 + k^2 = 1 + (k + 3)^2 \Rightarrow 6k = -6 \Rightarrow k = -1$. It matches.
Area of the triangle: base from $-3$ to $1$ on the x-axis is $4$, height $3$: $\frac{1}{2}\cdot 4\cdot 3 = 6$.
Answer: (C) $6$. | ||||
| 13 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The foci lie on the line $x = 1$: the transverse axis is vertical, the centre is the midpoint $(1, 1)$ and $2c = 26$, so $c = 13$.
The point $(1, 6)$ is on the transverse axis, so it is a vertex: $a = 6 - 1 = 5$.
(Check: $|PF_1 - PF_2| = |8 - 18| = 10 = 2a$.)
$b^2 = c^2 - a^2 = 169 - 25 = 144$.
Length of the latus rectum $= \frac{2b^2}{a} = \frac{288}{5}$.
Answer: (B) $\frac{288}{5}$. | ||||
| 14 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions For $|x| \ge 1$, $\sec^{-1}x + \mathrm{cosec}^{-1}x = \frac{\pi}{2}$. Let $\theta = \sec^{-1}x \in [0, \pi]$, $\theta \ne \frac{\pi}{2}$.
$(\sec^{-1}x)^2 + (\mathrm{cosec}^{-1}x)^2 = \theta^2 + \left(\frac{\pi}{2} - \theta\right)^2 = 2\left(\theta - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$.
Minimum at $\theta = \frac{\pi}{4}$ ($x = \sqrt2$): $\frac{\pi^2}{8}$.
Maximum at the end of $[0, \pi]$ farthest from $\frac{\pi}{4}$, i.e. $\theta = \pi$ ($x = -1$): $2\cdot\frac{9\pi^2}{16} + \frac{\pi^2}{8} = \frac{5\pi^2}{4}$.
Sum $= \frac{\pi^2}{8} + \frac{5\pi^2}{4} = \frac{11\pi^2}{8}$; times 16 gives $22\pi^2$.
Answer: (B) $22\pi^2$. | ||||
| 15 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctHard | |
A hard single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry General points: $P = (1 + 2s, 2 + 3s, 3 + 4s)$ on $L_1$, $Q = (2 + 3t, 4 + 4t, 5 + 5t)$ on $L_2$.
$\vec{PQ} = (1 + 3t - 2s,\ 2 + 4t - 3s,\ 2 + 5t - 4s)$ must be perpendicular to both directions.
With $(2, 3, 4)$: $16 + 38t - 29s = 0$. With $(3, 4, 5)$: $21 + 50t - 38s = 0$.
From the first, $29s = 16 + 38t$; substituting, $29(21 + 50t) = 38(16 + 38t)$, so $609 + 1450t = 608 + 1444t$, giving $t = -\frac{1}{6}$ and $s = \frac{1}{3}$.
$P = \left(\frac{5}{3}, 3, \frac{13}{3}\right)$, $Q = \left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)$, and $\vec{PQ} = \left(-\frac16, \frac13, -\frac16\right) \parallel (-1, 2, -1)$ (which is also $(2,3,4)\times(3,4,5)$).
Line of shortest distance: $\left(\frac{5}{3}, 3, \frac{13}{3}\right) + \lambda(-1, 2, -1)$.
$\lambda = -3$ gives $\left(\frac{5}{3} + 3, 3 - 6, \frac{13}{3} + 3\right) = \left(\frac{14}{3}, -3, \frac{22}{3}\right)$, option (D).
The others fail: for (A) $x$ needs $\lambda = \frac{10}{3}$ but then $y = \frac{29}{3} \ne -7$; for (B) $\lambda = -1$ gives $y = 1$; for (C) $\lambda = -\frac13$ gives $y = \frac73$.
Answer: (D). | ||||
| 16 | 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 From $f(x + y) = f(x)f(y)$: $f'(x) = \lim_{h\to0}\frac{f(x)f(h) - f(x)}{h} = f(x)f'(0)$ (with $f(0) = 1$; $f \equiv 0$ is ruled out since $f'(0) = 4a \ne 0$). So $f(x) = e^{kx}$ with $k = f'(0) = 4a$.
Put it in $f'' - 3af' - f = 0$: $k^2 - 3ak - 1 = 0$, i.e. $16a^2 - 12a^2 - 1 = 0$, so $a^2 = \frac{1}{4}$ and $a = \frac{1}{2}$.
Then $f(x) = e^{2x}$ and $f(ax) = e^{x}$.
Area $= \int_0^2 e^x\,dx = e^2 - 1$.
Answer: (B) $e^2 - 1$. | ||||
| 17 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Put $x = y = 0$: $f(0) = 2f(0)f'(0)$, so $f'(0) = \frac{1}{2}$.
Put $y = 0$: $f(x) = f(x)f'(0) + f'(x)f(0) = \frac{1}{2}f(x) + f'(x)$, so $f'(x) = \frac{1}{2}f(x)$.
With $f(0) = 1$: $f(x) = e^{x/2}$. (Check: $f(x)f'(y) + f'(x)f(y) = \frac12 e^{(x+y)/2} + \frac12 e^{(x+y)/2} = f(x+y)$.)
$\log_e f(n) = \frac{n}{2}$, so $\sum_{n=1}^{100}\frac{n}{2} = \frac{1}{2}\cdot\frac{100\cdot101}{2} = 2525$.
Answer: (C) $2525$. | ||||
| 18 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Group the terms: $f(x) = 7\tan^6x(\tan^2x + 1) - 3\tan^2x(\tan^2x + 1) = (7\tan^6x - 3\tan^2x)\sec^2x$.
With $t = \tan x$: $\int f(x)\,dx = \tan^7x - \tan^3x$. Call this $g(x)$.
$I_1 = g(\pi/4) - g(0) = (1 - 1) - 0 = 0$.
$I_2$ by parts: $I_2 = [x\,g(x)]_0^{\pi/4} - \int_0^{\pi/4} g(x)\,dx = \frac{\pi}{4}\cdot 0 - \int_0^{\pi/4}(\tan^7x - \tan^3x)\,dx$.
$\tan^7x - \tan^3x = \tan^3x(\tan^2x - 1)(\tan^2x + 1) = (\tan^5x - \tan^3x)\sec^2x$, so
$\int_0^{\pi/4}(\tan^7x - \tan^3x)\,dx = \left[\frac{t^6}{6} - \frac{t^4}{4}\right]_0^1 = -\frac{1}{12}$, and $I_2 = \frac{1}{12}$.
$7I_1 + 12I_2 = 0 + 1 = 1$.
Answer: (C) $1$. | ||||
| 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 The circle has centre $(2\sqrt3, 0)$ and radius $2\sqrt3$; it passes through the origin. Its equation is $x^2 - 4\sqrt3x + y^2 = 0$.
Intersections with $y^2 = 2\sqrt3x$: $x^2 - 2\sqrt3x = 0$, so $x = 0$ or $x = 2\sqrt3$ (where $y = \pm 2\sqrt3$).
For $0 < x < 2\sqrt3$ the circle's upper arc $y = \sqrt{4\sqrt3x - x^2}$ lies above the parabola's $y = \sqrt{2\sqrt3x}$, since $4\sqrt3x - x^2 > 2\sqrt3x \iff x < 2\sqrt3$. For $x > 2\sqrt3$ the circle lies inside the parabola. So the region is between the two curves for $0 \le x \le 2\sqrt3$, above and below the axis.
Area $= 2\left[\int_0^{2\sqrt3}\sqrt{12 - (x - 2\sqrt3)^2}\,dx - \int_0^{2\sqrt3}\sqrt{2\sqrt3x}\,dx\right]$.
The first integral is a quarter of the circle: $\frac{1}{4}\pi(2\sqrt3)^2 = 3\pi$.
The second: $\sqrt{2\sqrt3}\cdot\frac{2}{3}(2\sqrt3)^{3/2} = \frac{2}{3}(2\sqrt3)^2 = 8$.
Area $= 2(3\pi - 8) = 6\pi - 16$.
Answer: (D) $6\pi - 16$. | ||||
| 20 | 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 Treat $x$ as the dependent variable: $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$, linear in $x$.
Integrating factor $e^{\int y^{-2}dy} = e^{-1/y}$.
$x\,e^{-1/y} = \int \frac{1}{y^3}e^{-1/y}\,dy$. Put $u = -\frac{1}{y}$, $du = \frac{dy}{y^2}$, $\frac{1}{y} = -u$: $\int -u\,e^u\,du = (1 - u)e^u = \left(1 + \frac{1}{y}\right)e^{-1/y}$.
So $x = 1 + \frac{1}{y} + Ce^{1/y}$.
$x(1) = 1$: $1 = 2 + Ce$, so $C = -\frac{1}{e}$.
$x\left(\frac12\right) = 1 + 2 - \frac{1}{e}e^{2} = 3 - e$.
Answer: (B) $3 - e$. | ||||
| 21 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For order 3: $\mathrm{adj}(kA) = k^2\,\mathrm{adj}A$, $\mathrm{adj}(\mathrm{adj}A) = |A|A$, $\det(kA) = k^3\det A$.
$\mathrm{adj}(3A) = 9\,\mathrm{adj}A$, so $-6\,\mathrm{adj}(3A) = -54\,\mathrm{adj}A$.
$\mathrm{adj}(-54\,\mathrm{adj}A) = (-54)^2\,\mathrm{adj}(\mathrm{adj}A) = 2916\cdot(-2)A = -5832A$.
$3\times(-5832A) = -17496A$, and $17496 = 2^3\cdot 3^7$.
$\det(-17496A) = (-17496)^3\cdot(-2) = 2\cdot(2^3\cdot3^7)^3 = 2^{10}\cdot 3^{21}$.
So $m + n = 10$ and $mn = 21$: $m = 7$, $n = 3$ (as $m > n$).
$4m + 2n = 28 + 6 = 34$.
Answer: 34. | ||||
| 22 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Use $\frac{1}{k+1}\binom{11}{k} = \frac{1}{12}\binom{12}{k+1}$ (since $\binom{12}{k+1} = \frac{12}{k+1}\binom{11}{k}$).
With $k = 2r + 1$: $\frac{{}^{11}C_{2r+1}}{2r+2} = \frac{1}{12}\,{}^{12}C_{2r+2}$.
$\sum_{r=0}^{5}{}^{12}C_{2r+2} = {}^{12}C_2 + {}^{12}C_4 + \cdots + {}^{12}C_{12}$ = (sum of all even-index coefficients) $- {}^{12}C_0 = 2^{11} - 1 = 2047$.
So the sum is $\frac{2047}{12}$. $2047 = 23\times 89$ shares no factor with 12, so $m = 2047$, $n = 12$.
$m - n = 2035$.
Answer: 2035. | ||||
| 23 | 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 $|\vec a| = 3$ and $\vec b\cdot\vec a = \lambda + 8$, so $\vec c = \frac{\vec b\cdot\vec a}{|\vec a|^2}\vec a = \frac{\lambda + 8}{9}\vec a$.
$\vec a + \vec c = \left(1 + \frac{\lambda + 8}{9}\right)\vec a$, so $|\vec a + \vec c| = \frac{\lambda + 17}{9}\cdot 3 = 7$, giving $\lambda + 17 = 21$, $\lambda = 4$.
Then $\vec c = \frac{12}{9}\vec a = \frac{4}{3}\vec a$ and $\vec b = 4\hat i + 4\hat k$.
Area $= |\vec b\times\vec c| = \frac{4}{3}|\vec b\times\vec a|$.
$\vec b\times\vec a = (0\cdot2 - 4\cdot2)\hat i - (4\cdot2 - 4\cdot1)\hat j + (4\cdot2 - 0\cdot1)\hat k = -8\hat i - 4\hat j + 8\hat k$, of length $\sqrt{64 + 16 + 64} = 12$.
Area $= \frac{4}{3}\times12 = 16$.
Answer: 16. | ||||
| 24 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Points: $L_1$: $(1 + 3s, 1 - s, -1)$; $L_2$: $(2 + 2t, 0, -4 + \alpha t)$.
Intersection: $1 - s = 0 \Rightarrow s = 1$; $1 + 3s = 2 + 2t \Rightarrow t = 1$; $-1 = -4 + \alpha \Rightarrow \alpha = 3$. So $B = (4, 0, -1)$.
$L_2$ has direction $(2, 0, 3)$. Foot $P = (2 + 2t, 0, -4 + 3t)$ from $A(1, 1, -1)$: $\vec{AP} = (1 + 2t, -1, -3 + 3t)$ with $\vec{AP}\cdot(2, 0, 3) = 13t - 7 = 0$, so $t = \frac{7}{13}$.
$P = \left(\frac{40}{13}, 0, -\frac{31}{13}\right)$, $\vec{PB} = \left(\frac{12}{13}, 0, \frac{18}{13}\right)$, $(PB)^2 = \frac{144 + 324}{169} = \frac{468}{169}$.
$26\,\alpha(PB)^2 = 26\cdot3\cdot\frac{468}{169} = 6\cdot36 = 216$.
Answer: 216. | ||||
| 25 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Continuity at $x = 1$: $-3a - 2 = a^2 + b$. Equal derivatives at $x = 1$: $-6a = b$.
So $-3a - 2 = a^2 - 6a$, i.e. $a^2 - 3a + 2 = 0$, $a = 1$ or $2$; with $a > 1$, $a = 2$ and $b = -12$.
$f(x) = -6x^2 - 2$ for $x < 1$ and $f(x) = 4 - 12x$ for $x \ge 1$; its largest value is $-2$ at $x = 0$.
$f(x) = -20$: on the left, $6x^2 = 18$, $x = -\sqrt3$; on the right, $12x = 24$, $x = 2$.
Area $= \int_{-\sqrt3}^{1}(18 - 6x^2)\,dx + \int_1^2(24 - 12x)\,dx$.
First: $[18x - 2x^3]_{-\sqrt3}^{1} = 16 - (-18\sqrt3 + 6\sqrt3) = 16 + 12\sqrt3$.
Second: $[24x - 6x^2]_1^2 = 24 - 18 = 6$.
Area $= 22 + 12\sqrt3$, so $\alpha + \beta = 34$.
Answer: 34. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | 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 For a long straight wire $B = \frac{\mu_0 I}{2\pi r}$, so $\frac{B}{\mu_0} = \frac{I}{2\pi r}$ has the dimensions of current divided by length.
$\left[\frac{B}{\mu_0}\right] = \frac{[A]}{[L]} = L^{-1}A$.
(Directly: $[B] = MT^{-2}A^{-1}$ and $[\mu_0] = MLT^{-2}A^{-2}$; the ratio is $L^{-1}A$.)
Answer: (D) $L^{-1}A$. | ||||
| 27 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | StatementsEasy | |
A easy statements question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter In a vernier callipers, $N$ vernier divisions span $(N - 1)$ main scale divisions, so $1\ \text{VSD} = \frac{N-1}{N}\ \text{MSD}$, which is smaller than $1$ MSD. Statement I is true.
The vernier constant (least count) is $1\ \text{MSD} - 1\ \text{VSD} = \frac{1\ \text{MSD}}{N}$: one main scale division DIVIDED by the number of vernier divisions, not multiplied. Statement II is false.
Answer: (C). | ||||
| 28 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The string just becomes slack at the top D, so the tension there is zero: $\frac{mv_D^2}{l} = mg$, $v_D^2 = gl$.
Energy from A to D (rise $2l$): $v_0^2 = v_D^2 + 4gl = 5gl$.
Heights above A (from the figure, $\angle AOB = 60^\circ$ and $\angle DOC = 60^\circ$):
B: $l - l\cos60^\circ = \frac{l}{2}$; C: $l + l\cos60^\circ = \frac{3l}{2}$.
$KE_B = \frac{1}{2}m(5gl - gl) = \frac{1}{2}m(4gl)$, $KE_C = \frac{1}{2}m(5gl - 3gl) = \frac{1}{2}m(2gl)$.
$KE_B : KE_C = 4 : 2 = 2$.
Answer: (B) $2$. | ||||
| 29 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The hole has radius $\frac{R}{2}$ and, as drawn, its centre is $\frac{R}{2}$ from O (it touches the rim and passes through O).
Mass removed (area ratio $\frac{1}{4}$): $m = \frac{M}{4}$.
Its moment of inertia about the axis through O (parallel axes): $\frac{1}{2}m\left(\frac{R}{2}\right)^2 + m\left(\frac{R}{2}\right)^2 = \frac{3}{8}m R^2 = \frac{3}{32}MR^2$.
Remaining: $\frac{1}{2}MR^2 - \frac{3}{32}MR^2 = \frac{13}{32}MR^2$.
Answer: (A) $\frac{13}{32}MR^2$. | ||||
| 30 | GravitationNewton's law of gravitation | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation $F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$.
As drawn, the cavity touches the surface on the side of P, so its centre is $R - \frac{R}{3} = \frac{2R}{3}$ from O and $2R - \frac{2R}{3} = \frac{4R}{3}$ from P.
Mass removed: $M\left(\frac{1}{3}\right)^3 = \frac{M}{27}$. Its pull on m: $\frac{G(M/27)m}{(4R/3)^2} = \frac{GMm}{27}\cdot\frac{9}{16R^2} = \frac{GMm}{48R^2}$.
By superposition, $F_2 = \frac{GMm}{4R^2} - \frac{GMm}{48R^2} = \frac{11\,GMm}{48R^2}$.
$F_1 : F_2 = \frac{12}{48} : \frac{11}{48} = 12 : 11$.
Answer: (B) $12 : 11$. | ||||
| 31 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Same material, so the same emissivity. By the Stefan–Boltzmann law the power radiated is $P = e\sigma A T^4 \propto r^2T^4$.
$\frac{P_{large}}{P_{small}} = \left(\frac{0.8}{0.2}\right)^2\left(\frac{400}{800}\right)^4 = 16\times\frac{1}{16} = 1$.
So the larger body radiates $E$ as well.
Answer: (D) $E$. | ||||
| 32 | 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 | Single correctEasy | |
A easy single correct 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 Add the heat for each stage, with $m = 10^{-3}$ kg:
Ice $-10^\circ$C to $0^\circ$C: $10^{-3}\times2100\times10 = 21$ J.
Melting: $10^{-3}\times3.35\times10^5 = 335$ J.
Water $0^\circ$C to $100^\circ$C: $10^{-3}\times4180\times100 = 418$ J.
Boiling: $10^{-3}\times2.25\times10^6 = 2250$ J.
Steam $100^\circ$C to $110^\circ$C: $10^{-3}\times1920\times10 = 19.2$ J.
Total $= 21 + 335 + 418 + 2250 + 19.2 = 3043.2$ J $\approx 3043$ J.
Answer: (A) 3043 J. | ||||
| 33 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Speed of sound $v = \sqrt{B/\rho}$; same $B$, so $\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} = 4$.
Closed tube, 9th harmonic: $f = \frac{9v_1}{4L_1}$. Open tube, 4th harmonic: $f = \frac{4v_2}{2L_2} = \frac{2v_2}{L_2}$.
Equal: $\frac{9\cdot4v_2}{4\times10} = \frac{2v_2}{L_2}$, so $\frac{9}{10} = \frac{2}{L_2}$ and $L_2 = \frac{20}{9}$ cm.
Answer: (D) $\frac{20}{9}$ cm. | ||||
| 34 | 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 The supply stays connected, so V = 100 V throughout.
Before: $Q_1 = CV = 40\times10^{-6}\times100 = 4$ mC, $U_1 = \frac{1}{2}CV^2 = 0.2$ J.
After filling with $K = 2$: $C' = 80\,\mu$F, $Q_2 = 8$ mC, $U_2 = \frac{1}{2}C'V^2 = 0.4$ J.
Extra charge $= 4$ mC; change in energy $= 0.2$ J (an increase).
Answer: (A) 4 mC and 0.2 J. | ||||
| 35 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The horizontal velocity is unchanged ($10^6$ m/s), so the time in the field is $t = \frac{0.1}{10^6} = 10^{-7}$ s.
Field $E = 9.1$ V/cm $= 910$ V/m. Acceleration $a = \frac{eE}{m} = \frac{1.6\times10^{-19}\times910}{9.1\times10^{-31}} = 1.6\times10^{14}$ m/s$^2$.
Vertical velocity $v_y = at = 1.6\times10^{14}\times10^{-7} = 1.6\times10^{7} = 16\times10^6$ m/s.
Answer: (D). | ||||
| 36 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity A standard resistor must keep its value as the temperature changes, and a wire-wound one needs a fairly high resistivity so that a reasonable length gives the resistance. Alloys such as manganin and constantan are used: high resistivity and a very small temperature coefficient.
That is a nearly flat line well above zero: option (C).
(A) rises sharply (large positive coefficient), (B) falls steeply like a semiconductor, (D) falls steadily: none is stable with temperature.
Answer: (C). | ||||
| 38 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | StatementsMedium | |
A medium statements question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity For cells $(E_1, r_1)$ and $(E_2, r_2)$ in parallel:
$E_{eq} = \frac{E_1r_2 + E_2r_1}{r_1 + r_2}$ and $r_{eq} = \frac{r_1r_2}{r_1 + r_2}$.
$E_{eq}$ is a weighted average of $E_1$ and $E_2$, so it lies between them (it equals both when they are equal). It is not less than the smaller emf. Statement-I is false.
$r_{eq} = \frac{r_1r_2}{r_1 + r_2} < r_1$ and $< r_2$ for positive resistances. Statement-II is true.
Answer: (D). | ||||
| 39 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity The slider divides $R_p$ into two halves of $0.5\,\Omega$ each. As drawn, $R_e$ joins the left end of the wire to the slider, so it is in parallel with the left half.
Parallel part: $\frac{0.5\times2}{0.5 + 2} = 0.4\,\Omega$. In series with the right half: $0.4 + 0.5 = 0.9\,\Omega$.
Current from the 0.9 V cell: $I = \frac{0.9}{0.9} = 1.0$ A.
Answer: (C) 1.0 A. | ||||
| 40 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Assertion and reasonEasy | |
A easy assertion and reason question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Fringe width $\beta = \frac{\lambda D}{d}$. In a medium of refractive index $\mu > 1$ the wavelength is $\lambda' = \frac{v}{f} = \frac{\lambda}{\mu}$, so $\beta' = \frac{\beta}{\mu}$: the fringes come closer. (A) is true.
In a denser medium light slows down ($v = c/\mu$) while the frequency, fixed by the source, stays the same. (R) is true.
The slower speed at the same frequency is exactly why the wavelength, and hence the fringe width, decreases. So (R) explains (A).
Answer: (A). | ||||
| 41 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The silvered lens acts as a mirror: light passes the lens, reflects at the polished face, and passes the lens again.
Lens: $\frac{1}{f_L} = (\mu - 1)\left(\frac{1}{R} + \frac{1}{R}\right) = \frac{2(\mu - 1)}{R}$. Polished convex face, seen from inside, is a concave mirror of focal length $\frac{R}{2}$, power $\frac{2}{R}$.
Equivalent mirror power: $P = 2P_L + P_M = \frac{4(\mu - 1)}{R} + \frac{2}{R} = \frac{2(2\mu - 1)}{R}$, so $f_{eq} = \frac{R}{2(2\mu - 1)}$ (concave).
The image falls on the object when the object is at the centre of curvature of the equivalent mirror, $2f_{eq} = \frac{R}{2\mu - 1}$.
Answer: (B) $R/(2\mu - 1)$. | ||||
| 42 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctHard | |
A hard single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For thin lenses in contact the power is the sum of the powers of all refracting surfaces, $P = \sum\frac{n_2 - n_1}{R}$. The top of the upper water and the bottom of the lower water are flat and add nothing; only the two curved glass faces count.
Take light travelling downward. Both glass faces are bowl-shaped, so each centre of curvature is above the face, on the side the light comes from: $R = -|R|$ for both.
Upper face, water to glass: $\frac{\frac32 - \frac43}{-|R_1|} = -\frac{1}{6|R_1|}$.
Lower face, glass to water: $\frac{\frac43 - \frac32}{-|R_2|} = +\frac{1}{6|R_2|}$.
Total: $P = -\frac{1}{6}\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)$ (in units of inverse length). The same result follows for light travelling upward.
Answer: (B). | ||||
| 44 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Bohr radii go as $n^2$: $\frac{8.48\times10^{-10}}{5.3\times10^{-11}} = 16 = n^2$, so the excited orbit is $n = 4$ (the third excited state, as stated).
Bohr's condition $mvr = \frac{nh}{2\pi}$ means $n\lambda = 2\pi r$, so $\lambda = \frac{2\pi r}{n} \propto \frac{n^2}{n} = n$.
$\frac{\lambda_4}{\lambda_1} = 4$; ground-to-excited is $\frac{1}{4}$. The options carry only the size of the ratio, 4.
Answer: (B) $4$. | ||||
| 45 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A diode is forward biased when its anode (the triangle side) is at a higher potential than its cathode (the bar side).
(A) anode $-10$ V, cathode $0$ V: anode lower, reverse biased.
(B) anode $-10$ V, cathode $-15$ V: anode higher, forward biased.
(C) anode $4$ V, cathode $2$ V: forward biased.
(D) the diode points left: anode at $-10$ V (right), cathode at $-5$ V (left): reverse biased.
(E) the diode points left: anode at $2$ V, cathode earthed at $0$ V: forward biased.
Forward biased: (B), (C) and (E).
Answer: (C). | ||||
| 48 | 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 Excess pressure inside a soap bubble is $\frac{4T}{r}$. The common film separates the two bubbles, so its own excess pressure is the difference:
$\frac{4T}{R} = \frac{4T}{r_1} - \frac{4T}{r_2}$, giving $R = \frac{r_1r_2}{r_2 - r_1}$.
$R = \frac{2\times4}{4 - 2} = 4$ cm (curving into the larger bubble).
Answer: 4. | ||||
| 49 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Conductors 1 and 2 are in parallel between $100^\circ$C and $\theta$; conductor 3 carries the same heat current from $\theta$ to $0^\circ$C. Let each length be $L$ and the area of 1 and 2 be $A$ (so 3 has $2A$).
$\frac{(k_1 + k_2)A(100 - \theta)}{L} = \frac{k_3(2A)(\theta - 0)}{L}$.
$180(100 - \theta) = 270\,\theta$, so $18000 = 450\,\theta$ and $\theta = 40$.
Answer: 40. | ||||
| 50 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalHard | |
A hard numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Convex mirror: $f = +\frac{R}{2} = 1$ m. Object distance $u = -24$ m, speed $|\dot u| = 90$ km/h $= 25$ m/s, constant.
Mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ gives $v = \frac{uf}{u - f}$.
Differentiate: $\dot v = -\frac{f^2}{(u - f)^2}\dot u$, and again (with $\ddot u = 0$): $\ddot v = \frac{2f^2}{(u - f)^3}\dot u^2$.
At $u = -24$: $u - f = -25$, $(u - f)^3 = -15625$.
$|\ddot v| = \frac{2\times1\times625}{15625} = 0.08$ m/s$^2$.
$100a = 8$.
Answer: 8. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 53 | SolutionsColligative properties | Class 12: Solutions | Single correctEasy | |
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions Elevation of boiling point $\Delta T_b = iK_bm$ grows with the number of solute particles, $i \times$ concentration (dilute solutions, molarity $\approx$ molality).
Urea does not ionise, $i = 1$: (ii) $1\times10^{-4}$.
NaCl gives two ions, $i = 2$: (i) $2\times10^{-4}$, (iii) $2\times10^{-3}$, (iv) $2\times10^{-2}$.
Increasing boiling point: (ii) < (i) < (iii) < (iv).
Answer: (D). | ||||
| 54 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{CO_2(g) + C(s) \rightleftharpoons 2CO(g)}$. Graphite is a solid and does not enter $K_p$.
If $x$ atm of $\mathrm{CO_2}$ reacts: $p_{\mathrm{CO_2}} = 0.5 - x$, $p_{\mathrm{CO}} = 2x$, total $0.5 + x = 0.8$, so $x = 0.3$.
$p_{\mathrm{CO_2}} = 0.2$ atm, $p_{\mathrm{CO}} = 0.6$ atm.
$K_p = \frac{(0.6)^2}{0.2} = 1.8$ atm.
Answer: (A) 1.8 atm. | ||||
| 55 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Charge passed: $Q = It = 2\times30\times60 = 3600$ C.
$\mathrm{Al^{3+} + 3e^- \rightarrow Al}$: 3 faradays deposit 27 g.
Mass $= \frac{Q}{3F}\times27 = \frac{3600\times27}{3\times96500} = \frac{97200}{289500} = 0.3358$ g $\approx 0.336$ g.
Answer: (D) 0.336 g. | ||||
| 56 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Peroxodisulphuric acid is made by electrolysing cold, concentrated sulphuric acid at a high current density. At the anode the hydrogensulphate ions are oxidised: $\mathrm{2HSO_4^- \rightarrow H_2S_2O_8 + 2e^-}$.
In dilute sulphuric acid or in sodium sulphate solutions, water is oxidised instead and oxygen is evolved at the anode.
Answer: (B). | ||||
| 57 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Radioactive decay is first order: $t_{1/2} = \frac{\ln 2}{\lambda}$, so (A) is true.
It is a nuclear process, unaffected by temperature, pressure or chemical state, so the decay constant does not depend on temperature: (B) is true and (D) is NOT true.
After three half-lives $\left(\frac12\right)^3 = \frac18$ remains: (C) is true.
Answer: (D). | ||||
| 58 | Classification of Elements and PeriodicityElectronegativity | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on electronegativity, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Pauling values: Mg 1.31, Be 1.57, Al 1.61, B 2.04, Si 1.90, C 2.55, S 2.58, N 3.04, Cl 3.16, O 3.44, F 3.98.
(A) Mg < Be < B < N: correct.
(B) Al < Mg is wrong: across period 3 electronegativity rises from Mg (1.31) to Al (1.61). So (B) is the incorrect order.
(C) Al < Si < C < N: correct.
(D) S < Cl < O < F: correct.
Answer: (B). | ||||
| 59 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Matching listEasy | |
A easy matching list question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties (A) $\mathrm{Al^{3+}, Mg^{2+}, Na^+, F^-}$ are isoelectronic (10 electrons); the radius grows as the nuclear charge falls: ionic radii → (IV).
(B) B < C < O < N is the ionisation enthalpy order of period 2 (N is above O because of its half-filled 2p shell) → (I).
(C) B < Al < Mg < K: metallic character grows down a group and to the left of a period → (II).
(D) Si < P < S < Cl: electronegativity grows across period 3 → (III).
A→IV, B→I, C→II, D→III.
Answer: (A). | ||||
| 60 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Ground configurations: Sm $[\mathrm{Xe}]4f^66s^2$, Eu $[\mathrm{Xe}]4f^76s^2$, Gd $[\mathrm{Xe}]4f^75d^16s^2$, Tb $[\mathrm{Xe}]4f^96s^2$.
$\mathrm{Eu^{2+}}$: $4f^7$. $\mathrm{Gd^{3+}}$: $4f^7$. $\mathrm{Eu^{3+}}$: $4f^6$. $\mathrm{Tb^{3+}}$: $4f^8$. $\mathrm{Sm^{2+}}$: $4f^6$.
So (A) and (B).
Answer: (A). | ||||
| 61 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[NiCl_4]^{2-}}$: Ni is +2, $3d^8$. $\mathrm{Cl^-}$ is a weak-field ligand and does not pair the electrons, so two unpaired electrons remain (paramagnetic) and the hybridisation is $sp^3$: tetrahedral.
$\mathrm{[Ni(CO)_4]}$: Ni is 0, $3d^84s^2$. The strong-field CO pushes the 4s electrons into 3d, giving $3d^{10}$ with no unpaired electron (diamagnetic) and $sp^3$ hybridisation: tetrahedral.
Answer: (B). | ||||
| 63 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium 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 compound is $\mathrm{CH_3-CH(COOH)-CH_2-CH_2-CH(COOCH_3)-CH_3}$: 9 carbon atoms in all.
The main chain runs from one carboxyl carbon through the four chain carbons to the far $\mathrm{CH_3}$: six carbons, a hexane-derived chain with the other carboxyl group and a methyl as substituents.
Count carbons in each option:
(A) hexanoate 6 + two methyls 2 + carboxy 1 + ester methyl 1 = 10: wrong compound.
(D) hexanoic acid 6 + two methyls 2 + methoxycarbonyl 2 = 10: wrong compound.
(B) treats a hexane with both groups as prefixes; it ignores that an acid or ester must be the principal characteristic group named as suffix.
(C) Methyl 5-carboxy-2-methylhexanoate: the ester carbon is C1, the methyl is on C2 and the $\mathrm{-COOH}$ on C5 of the hexanoate chain. This describes the compound and is in the proper form.
(Strictly, IUPAC ranks the acid above the ester, which would give 5-(methoxycarbonyl)-2-methylhexanoic acid; that name is not among the options.)
Answer: (C). | ||||
| 64 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Pent-3-en-2-ol, $\mathrm{CH_3-CH(OH)-CH=CH-CH_3}$.
C2 carries four different groups ($\mathrm{CH_3}$, OH, H, $\mathrm{CH=CHCH_3}$): one stereocentre, R or S.
The C3=C4 double bond has two different groups on each carbon: E or Z.
The two are independent: $2\times2 = 4$ stereoisomers.
Answer: (A) $4$. | ||||
| 65 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) Propene, $\mathrm{CH_2=CH-CH_3}$, has two H atoms on one doubly bonded carbon: no geometrical isomers. Incorrect.
(B) Correct: that is the definition of trans.
(C) Correct: in trans-but-2-ene the bond dipoles cancel (nearly zero); cis has a net dipole.
(D) 2-methylbut-2-ene, $\mathrm{(CH_3)_2C=CH-CH_3}$, has two methyls on one carbon: no geometrical isomers. Incorrect.
(E) Trans isomers pack better in the solid and usually melt higher than cis. Incorrect.
Incorrect statements: (A), (D) and (E).
Answer: (D). | ||||
| 66 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | StatementsEasy | |
A easy statements question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Propyne $\mathrm{CH_3-C\equiv C-H}$ has one acidic (terminal) hydrogen.
Statement I: $\mathrm{2CH_3C\equiv CH + 2Na \rightarrow 2CH_3C\equiv CNa + H_2}$: 1 mol propyne gives $\frac{1}{2}$ mol $\mathrm{H_2}$. Correct.
Statement II: 4 g of propyne ($M = 40$ g mol$^{-1}$) is 0.1 mol, and $\mathrm{CH_3C\equiv CH + NaNH_2 \rightarrow CH_3C\equiv CNa + NH_3}$ gives 0.1 mol $\mathrm{NH_3}$ = 2240 mL at STP, not 224 mL. Incorrect.
Answer: (C). | ||||
| 67 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | StatementsMedium | |
A medium statements question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Statement I: ionisation of $\mathrm{CH_3OCH_2Cl}$ gives $\mathrm{CH_3O-\overset{+}{C}H_2 \leftrightarrow CH_3\overset{+}{O}=CH_2}$, a carbocation strongly stabilised by the lone pair on oxygen. So this primary halide reacts by $\mathrm{S_N1}$. Correct.
Statement II: neopentyl chloride, $\mathrm{(CH_3)_3C-CH_2Cl}$, is primary, but the bulky tert-butyl group next to the reacting carbon blocks the back-side attack that $\mathrm{S_N2}$ needs. It reacts very slowly by $\mathrm{S_N2}$. Correct.
Answer: (A). | ||||
| 68 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines (i) $\mathrm{-NO_2}$ is a meta director: bromination gives m-bromonitrobenzene.
(ii) Sn/HCl reduces $\mathrm{-NO_2}$ to $\mathrm{-NH_2}$: m-bromoaniline.
(iii) $\mathrm{NaNO_2}$/HCl at 273 K diazotises it: m-bromobenzenediazonium chloride.
(iv) Ethanol reduces the diazonium group to H and is itself oxidised to ethanal: $\mathrm{ArN_2^+Cl^- + CH_3CH_2OH \rightarrow ArH + N_2 + CH_3CHO + HCl}$.
Products: bromobenzene (A) and $\mathrm{CH_3CHO}$ (B).
Answer: (B). | ||||
| 69 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Ascorbic acid is vitamin C, a water-soluble vitamin.
Aspartic acid is an amino acid, adipic acid (hexanedioic acid) is a monomer of nylon-6,6, and saccharic (glucaric) acid is an oxidation product of glucose. None of them is a vitamin.
Answer: (B) Ascorbic acid. | ||||
| 70 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Fehling's solution is reduced by aliphatic aldehydes and by $\alpha$-hydroxy ketones such as fructose, but not by aromatic aldehydes or by ordinary ketones.
(A) Benzaldehyde: aromatic aldehyde, negative.
(B) Acetophenone: ketone, negative.
(C) $\mathrm{HOCH_2-CO-(CHOH)_3-CH_2OH}$ is fructose, an $\alpha$-hydroxy ketone: positive.
(D) Ethanal: aliphatic aldehyde, positive.
(E) Phenylacetaldehyde, $\mathrm{C_6H_5CH_2CHO}$: the CHO is on a $\mathrm{CH_2}$, not on the ring, so it behaves as an aliphatic aldehyde: positive.
Positive: (C), (D) and (E).
Answer: (A). | ||||
| 71 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry HCl used: $0.1\times40 = 4$ mmol. $\mathrm{Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O}$, so the unreacted $\mathrm{Ca(OH)_2}$ was 2 mmol.
The $\mathrm{CO_2}$ used up exactly half the $\mathrm{Ca(OH)_2}$, so the other half, also 2 mmol, reacted: $\mathrm{CO_2 + Ca(OH)_2 \rightarrow CaCO_3 + H_2O}$ needs 2 mmol $\mathrm{CO_2}$.
At 1 atm and 273 K one mole of gas occupies 22.4 L, so $x = 2\times10^{-3}\times22400 = 44.8$ cm$^3 \approx 45$ cm$^3$.
Answer: 45. | ||||
| 72 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Linear: $\mathrm{BeCl_2}$ (2 bond pairs), $\mathrm{CO_2}$, $\mathrm{N_3^-}$ and $\mathrm{NO_2^+}$ (16-electron triatomics, $sp$ centre), $\mathrm{XeF_2}$ and $\mathrm{I_3^-}$ (2 bond pairs and 3 lone pairs on the centre, lone pairs equatorial).
Bent: $\mathrm{SO_2}$, $\mathrm{NO_2}$, $\mathrm{O_3}$ (a lone pair or odd electron on the centre) and $\mathrm{F_2O}$ (two lone pairs on O).
Linear count: 6.
Answer: 6. | ||||
| 73 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Arrhenius: $k = Ae^{-E_a/RT}$, so $\log_{10}k = \log_{10}A - \frac{E_a}{2.303RT}$.
$\frac{E_a}{2.303RT} = \frac{191480}{2.303\times8.314\times1000} = \frac{191480}{19147} \approx 10.0$.
$\log_{10}k = 20 - 10 = 10$, so $k = 10^{10}$ s$^{-1}$ (frequency factor taken in s$^{-1}$).
First order: $t_{1/2} = \frac{0.693}{k} = 6.93\times10^{-11}$ s $= 69.3$ ps $\approx 69$ ps.
(Without rounding $2.303$, $\frac{E_a}{RT} = 23.031$ and $t_{1/2} = 69.7$ ps, which rounds to 70.)
Answer: 69. | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Molar mass of AgCl $= 108 + 35.5 = 143.5$ g mol$^{-1}$, so 143.5 mg of AgCl contains 35.5 mg of chlorine.
Percentage of Cl $= \frac{35.5}{180}\times100 = 19.72\%$, which as an integer is $20$.
Answer: 20. | ||||
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