JEE (Main) · 2025
JEE (Main) 2025 — 28 January, 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
22 Q
- Integral Calculus4
- Conic Sections3
- Sets, Relations and Functions2
5 easy · 17 medium · 0 hard
Physics
25 Q
- Optics3
- Physical World and Measurement2
- Properties of Bulk Matter2
18 easy · 6 medium · 1 hard
Chemistry
24 Q
- Thermodynamics2
- Solutions2
- Chemical Kinetics2
16 easy · 8 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 |
|---|---|---|---|---|
| 2 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Since both functions are onto, $A$ and $B$ are their ranges.
$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$, so on $[0, 3]$ $f$ rises to $x = 2$ and falls after it.
$f(0) = 7$, $f(2) = 16 - 60 + 72 + 7 = 35$, $f(3) = 54 - 135 + 108 + 7 = 34$. So $A = [7, 35]$.
For $x \ge 0$, $g(x) = 1 - \frac{1}{x^{2025} + 1}$ runs from $g(0) = 0$ upwards and never reaches 1, so $B = [0, 1)$.
Integers in $A$: $7, 8, \dots, 35$, that is 29 of them. Integers in $B$: only 0.
$n(S) = 29 + 1 = 30$.
Answer: (C) 30. | ||||
| 3 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium 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 The relation $f(x)f\left(\frac1x\right) = f(x) + f\left(\frac1x\right)$ for a polynomial gives $f(x) = 1 \pm x^n$; with degree 2, $f(x) = 1 \pm x^2$.
(Rewrite it as $(f(x) - 1)(f(1/x) - 1) = 1$; for $f(x) - 1 = ax^2 + bx + c$ the product with $a/x^2 + b/x + c$ must be the constant 1, which forces $b = c = 0$ and $a^2 = 1$.)
The codomain is $(-\infty, 1)$, so $f(x) < 1$ for every $x \ne 0$: $f(x) = 1 - x^2$.
$f(K) = -2K$: $1 - K^2 = -2K$, so $K^2 - 2K - 1 = 0$, $K = 1 \pm \sqrt2$.
Sum of squares $= (K_1 + K_2)^2 - 2K_1K_2 = 4 - 2(-1) = 6$.
Answer: (C) 6. | ||||
| 4 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Discriminant: $(3 - 2i)^2 + 4(2i - 2) = 9 - 12i - 4 + 8i - 8 = -3 - 4i$.
$(1 - 2i)^2 = 1 - 4i - 4 = -3 - 4i$, so $\sqrt{D} = \pm(1 - 2i)$.
Roots: $x = \frac{(3 - 2i) \pm (1 - 2i)}{2}$, that is $2 - 2i$ and $1$.
Check: sum $3 - 2i$, product $2 - 2i = -(2i - 2)$. Correct.
Take $\alpha = 2, \beta = -2$ and $\gamma = 1, \delta = 0$ (the order does not matter):
$\alpha\gamma + \beta\delta = 2 + 0 = 2$.
Answer: (B) 2. | ||||
| 5 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $P$ is a rotation matrix, so $P^TP = PP^T = I$.
$B^2 = PAP^TPAP^T = PA^2P^T$, and in the same way $B^{10} = PA^{10}P^T$.
$C = P^TB^{10}P = P^TPA^{10}P^TP = A^{10}$.
$A$ is upper triangular with diagonal $\frac1{\sqrt2}, 1$, so $A^{10}$ is upper triangular with diagonal $\left(\frac1{\sqrt2}\right)^{10} = \frac1{32}$ and $1$.
Trace of $C = \frac1{32} + 1 = \frac{33}{32}$; $m = 33$, $n = 32$, $m + n = 65$.
Answer: (D) 65. | ||||
| 6 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $a_k = \frac{(k+2)(k+3)}{4}$, so $\frac1{a_k} = \frac{4}{(k+2)(k+3)} = 4\left(\frac1{k+2} - \frac1{k+3}\right)$.
The sum telescopes: $S_n = 4\left(\frac13 - \frac1{n+3}\right)$.
$S_{2025} = 4\left(\frac13 - \frac1{2028}\right) = 4 \cdot \frac{676 - 1}{2028} = \frac{2700}{2028}$.
$507\,S_{2025} = \frac{507 \times 2700}{2028} = \frac{2700}{4} = 675$ (as $2028 = 4 \times 507$).
Answer: (C) 675. | ||||
| 7 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Coefficients of $T_r, T_{r+1}, T_{r+2}$ are $^{12}C_{r-1}, {}^{12}C_r, {}^{12}C_{r+1}$. For a G.P. the ratios of consecutive terms must be equal:
$\frac{^{12}C_r}{^{12}C_{r-1}} = \frac{13 - r}{r}$ and $\frac{^{12}C_{r+1}}{^{12}C_r} = \frac{12 - r}{r + 1}$.
$\frac{13 - r}{r} = \frac{12 - r}{r + 1} \Rightarrow (13 - r)(r + 1) = r(12 - r) \Rightarrow 13 + 12r - r^2 = 12r - r^2 \Rightarrow 13 = 0$, impossible. So $p = 0$.
General term of $\left(3^{1/4} + 4^{1/3}\right)^{12}$: $^{12}C_k\, 3^{(12-k)/4}\, 4^{k/3}$. It is rational when $4 \mid (12 - k)$ and $3 \mid k$, i.e. $k \equiv 0 \pmod{4}$ and $k \equiv 0 \pmod 3$, so $k = 0$ or $12$.
$k = 0$: $3^3 = 27$; $k = 12$: $4^4 = 256$. $q = 27 + 256 = 283$.
$p + q = 283$.
Answer: (A) 283. | ||||
| 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 Each bag is chosen with probability $\frac13$. $P(W \mid B_1) = \frac{6}{10}$, $P(W \mid B_2) = \frac{4}{10}$, $P(W \mid B_3) = \frac{5}{10}$.
By Bayes' theorem,
$P(B_2 \mid W) = \frac{\frac13 \cdot \frac{4}{10}}{\frac13\left(\frac{6}{10} + \frac{4}{10} + \frac{5}{10}\right)} = \frac{4}{15}$.
Answer: (A) $\frac{4}{15}$. | ||||
| 10 | Straight LinesSlope, parallel and perpendicular lines | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines The two equal sides meet at the vertex; the third side makes equal angles with both, so it is perpendicular to one of the two angle bisectors at the vertex.
Unit normals of the lines: $\vec n_1 = \frac{(-1, 2)}{\sqrt5}$ and $\vec n_2 = \frac{(1, 1)}{\sqrt2}$. The bisectors have normals $\vec n_1 \pm \vec n_2$, and a line perpendicular to a bisector runs along that normal, so its slope is (y-part)/(x-part) of $\vec n_1 \pm \vec n_2$.
Write $a = \sqrt2$, $b = \sqrt5$ (multiplying through by $\sqrt{10}$):
$\vec n_1 + \vec n_2 \propto (b - a,\ 2a + b)$, slope $m_1 = \frac{2a + b}{b - a}$;
$\vec n_1 - \vec n_2 \propto (-a - b,\ 2a - b)$, slope $m_2 = \frac{b - 2a}{a + b}$.
$m_1 + m_2 = \frac{(2a + b)(a + b) + (b - 2a)(b - a)}{b^2 - a^2} = \frac{(2a^2 + 3ab + b^2) + (b^2 - 3ab + 2a^2)}{3} = \frac{4a^2 + 2b^2}{3} = \frac{8 + 10}{3} = 6$.
Answer: (B) 6. | ||||
| 11 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Chord with a given midpoint: $T = S_1$.
$\frac{\sqrt2\,x}{9} + \frac{(4/3)\,y}{4} = \frac{2}{9} + \frac{16/9}{4} = \frac{2}{3}$, i.e. $\sqrt2\,x + 3y = 6$.
Substitute $y = \frac{6 - \sqrt2\,x}{3}$ into $4x^2 + 9y^2 = 36$: $4x^2 + (6 - \sqrt2\,x)^2 = 36 \Rightarrow 6x^2 - 12\sqrt2\,x = 0$, so $x = 0$ or $x = 2\sqrt2$.
End points: $(0, 2)$ and $\left(2\sqrt2, \frac23\right)$ (their midpoint is $\left(\sqrt2, \frac43\right)$, as required).
Length$^2 = 8 + \frac{16}{9} = \frac{88}{9}$, length $= \frac{2\sqrt{22}}{3}$. So $\alpha = 22$.
Answer: (C) 22. | ||||
| 12 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections From the circle $y^2 = 8x - x^2$; in $4x^2 - 9y^2 = 36$: $4x^2 - 72x + 9x^2 = 36 \Rightarrow 13x^2 - 72x - 36 = 0 \Rightarrow (x - 6)(13x + 6) = 0$.
$x = -\frac{6}{13}$ gives $y^2 < 0$; so $x = 6$, $y^2 = 48 - 36 = 12$: $A = (6, 2\sqrt3)$, $B = (6, -2\sqrt3)$.
Let $P = (h, k)$ with $2h - 3k + 4 = 0$. Centroid $(X, Y) = \left(\frac{h + 12}{3}, \frac{k}{3}\right)$, so $h = 3X - 12$, $k = 3Y$.
$2(3X - 12) - 9Y + 4 = 0 \Rightarrow 6X - 9Y = 20$.
Answer: (B) $6x - 9y = 20$. | ||||
| 13 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $A_r = \frac{\pi}{4} + (r - 1)\frac{\pi}{6}$ and $B_r = A_r + \frac{\pi}{6}$. Since $\sin(B_r - A_r) = \sin\frac{\pi}{6} = \frac12$,
$\frac{1}{\sin A_r \sin B_r} = \frac{1}{\sin\frac{\pi}{6}} \cdot \frac{\sin(B_r - A_r)}{\sin A_r \sin B_r} = 2(\cot A_r - \cot B_r)$.
The sum telescopes: $2\left(\cot\frac{\pi}{4} - \cot\left(\frac{\pi}{4} + \frac{13\pi}{6}\right)\right)$.
$\frac{\pi}{4} + \frac{13\pi}{6} = 2\pi + \frac{5\pi}{12}$, and $\cot\frac{5\pi}{12} = 2 - \sqrt3$.
Sum $= 2(1 - 2 + \sqrt3) = 2\sqrt3 - 2$, so $a = 2$, $b = -2$, $a^2 + b^2 = 8$.
Answer: (D) 8. | ||||
| 14 | Vector AlgebraAddition of vectors | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on addition of vectors, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\overrightarrow{OA}| = |\overrightarrow{OB}| = 2$, so the bisector is along $\overrightarrow{OA} + \overrightarrow{OB} = (1 + \sqrt3)(\hat{i} + \hat{j})$: the line $y = x$ (OA makes $30^\circ$ and OB $60^\circ$ with the x-axis; the bisector makes $45^\circ$).
Distance of $C(a, 1 - a)$ from $x - y = 0$: $\frac{|a - (1 - a)|}{\sqrt2} = \frac{|2a - 1|}{\sqrt2} = \frac{9}{\sqrt2}$.
$2a - 1 = \pm 9 \Rightarrow a = 5$ or $a = -4$. Sum $= 1$.
Answer: (C) 1. | ||||
| 15 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec a$ is the sum of its two components:
$\vec a = \frac{1}{11}\left[(48 - 4)\hat{i} + (16 - 5)\hat{j} + (-16 - 17)\hat{k}\right] = \frac{1}{11}(44\hat{i} + 11\hat{j} - 33\hat{k}) = 4\hat{i} + \hat{j} - 3\hat{k}$.
$\alpha^2 + \beta^2 + \gamma^2 = 16 + 1 + 9 = 26$.
Answer: (D) 26. | ||||
| 16 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium 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 Move from $P\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ along $(1, 4, 7)$: $\left(\frac{15}{7} + t, \frac{32}{7} + 4t, 7 + 7t\right)$.
A point of the line is $(3\lambda - 1, 5\lambda - 3, 7\lambda - 5)$.
z: $7 + 7t = 7\lambda - 5 \Rightarrow t = \lambda - \frac{12}{7}$.
x: $\frac{15}{7} + \lambda - \frac{12}{7} = 3\lambda - 1 \Rightarrow 2\lambda = \frac{10}{7}$, $\lambda = \frac57$, so $t = -1$.
y check: $\frac{32}{7} - 4 = \frac47$ and $5 \cdot \frac57 - 3 = \frac47$. Consistent.
Distance $= |t|\sqrt{1 + 16 + 49} = \sqrt{66}$; its square is 66.
Answer: (D) 66. | ||||
| 17 | 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 $F(2) = 2f(2) = 2$.
By parts: $\int_0^2 xF'(x)\,dx = [xF(x)]_0^2 - \int_0^2 F(x)\,dx = 4 - \int_0^2 F(x)\,dx = 6$, so $\int_0^2 F(x)\,dx = -2$.
By parts: $\int_0^2 x^2F''(x)\,dx = [x^2F'(x)]_0^2 - 2\int_0^2 xF'(x)\,dx = 4F'(2) - 12 = 40$, so $F'(2) = 13$.
$F'(2) + \int_0^2 F(x)\,dx = 13 - 2 = 11$.
Answer: (B) 11. | ||||
| 18 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $g'(x) = x f(x)$. Differentiate $g(x^3) = x^6 + x^7$: $3x^2 g'(x^3) = 6x^5 + 7x^6$.
$3x^2 \cdot x^3 f(x^3) = 6x^5 + 7x^6 \Rightarrow f(x^3) = \frac{6 + 7x}{3}$.
$\sum_{r=1}^{15} f(r^3) = \frac{1}{3}\left(6 \cdot 15 + 7 \cdot \frac{15 \cdot 16}{2}\right) = \frac{90 + 840}{3} = 310$.
Answer: (B) 310. | ||||
| 19 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $x = t^4$, $dx = 4t^3\,dt$:
$\int \frac{4t^3}{t(1 + t)}\,dt = 4\int \frac{t^2}{1 + t}\,dt = 4\int\left(t - 1 + \frac{1}{1 + t}\right)dt = 4\left(\frac{t^2}{2} - t + \ln(1 + t)\right) + C$.
$f(0) = C = -6$.
$f(1) = 4\left(\frac12 - 1 + \ln 2\right) - 6 = 4\ln 2 - 8 = 4(\log_e 2 - 2)$.
Answer: (C) $4(\log_e 2 - 2)$. | ||||
| 20 | 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 Intersections: $x = \frac{y^2}{2}$ in $x(1 + y^2) = 1$ gives $y^2(1 + y^2) = 2$, so $y^2 = 1$, $y = \pm1$, $x = \frac12$.
For $-1 \le y \le 1$ the curve $x = \frac{1}{1 + y^2}$ lies to the right of the parabola $x = \frac{y^2}{2}$ (at $y = 0$: $1 > 0$).
Area $= \int_{-1}^{1}\left(\frac{1}{1 + y^2} - \frac{y^2}{2}\right)dy = 2\left[\tan^{-1}y - \frac{y^3}{6}\right]_0^1 = 2\left(\frac{\pi}{4} - \frac16\right) = \frac{\pi}{2} - \frac13$.
Answer: (B) $\frac{\pi}{2} - \frac{1}{3}$. | ||||
| 21 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Angles in degrees: largest $219$, common difference 6, so the sum is $\frac{n}{2}\left[2 \cdot 219 - (n - 1)6\right] = n(222 - 3n)$.
Sum of interior angles $= (n - 2)180$.
$222n - 3n^2 = 180n - 360 \Rightarrow 3n^2 - 42n - 360 = 0 \Rightarrow n^2 - 14n - 120 = 0 \Rightarrow (n - 20)(n + 6) = 0$.
$n = 20$ (the smallest angle is then $219 - 114 = 105^\circ$).
Answer: 20. | ||||
| 23 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Reflection in $x + y + 4 = 0$ sends $(x, y)$ to $(-y - 4, -x - 4)$.
A point $(X, Y)$ is on the image when its reflection $(-Y - 4, -X - 4)$ is on $y^2 = 4x$: $(X + 4)^2 = -4(Y + 4)$.
With $Y = -5$: $(X + 4)^2 = 4$, $X = -2$ or $-6$. So $A = (-2, -5)$, $B = (-6, -5)$ and $d = 4$.
Focus $S = (1, 0)$ is at height 5 above the line $y = -5$: $a = \frac12 \cdot 4 \cdot 5 = 10$.
$a + d = 14$.
Answer: 14. | ||||
| 24 | 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 With $t = \tan\theta$: $\tan2\theta - \tan\theta = \frac{2t}{1 - t^2} - t = \frac{t + t^3}{1 - t^2}$.
So each term with $\theta = \frac{x}{2^{r+1}}$ is $\tan\frac{x}{2^r} - \tan\frac{x}{2^{r+1}}$, and the sum telescopes to $\tan x - \tan\frac{x}{2^{n+1}} \to \tan x$. Hence $f(x) = \tan x$.
$\frac{e^x - e^{\tan x}}{x - \tan x} = e^{\tan x}\cdot\frac{e^{x - \tan x} - 1}{x - \tan x}$. As $x \to 0$, $x - \tan x \to 0$ (and is non-zero near 0), so the fraction tends to 1 and $e^{\tan x} \to 1$.
Limit $= 1$.
Answer: 1. | ||||
| 25 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Let $s = \sin^{-1}\frac{x}{2}$, so $\frac{ds}{dx} = \frac{1}{\sqrt{4 - x^2}}$ and the equation becomes $\frac{dy}{ds} = (s^2 - y)s$, i.e. $\frac{dy}{ds} + sy = s^3$.
Integrating factor $e^{s^2/2}$: $y\,e^{s^2/2} = \int s^3e^{s^2/2}\,ds$. With $u = \frac{s^2}{2}$, $s^3\,ds = 2u\,du$, and $\int 2ue^u\,du = 2(u - 1)e^u$.
$y = s^2 - 2 + Ce^{-s^2/2}$.
At $x = 2$, $s = \frac{\pi}{2}$: $\frac{\pi^2}{4} - 2 + Ce^{-\pi^2/8} = \frac{\pi^2 - 8}{4}$, so $C = 0$.
$y = \left(\sin^{-1}\frac{x}{2}\right)^2 - 2$; $y(0) = -2$, $y^2(0) = 4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement P is on the perpendicular bisector (equatorial line) of the magnet: $B = \frac{\mu_0}{4\pi}\frac{M}{(d^2 + l^2)^{3/2}}$ with $M = m(2l)$, m the pole strength.
So $B \propto \frac{l}{(d^2 + l^2)^{3/2}}$, and for the largest relative error
$\frac{\Delta B}{B} = \frac{\Delta l}{l} + \frac32\cdot\frac{2d\,\Delta d + 2l\,\Delta l}{d^2 + l^2}$.
Both lengths carry a 1% error: $l = 10$, $\Delta l = 0.1$; $d = 10$, $\Delta d = 0.1$.
$\frac{\Delta B}{B} = 0.01 + \frac32\cdot\frac{2 + 2}{200} = 0.01 + 0.03 = 0.04 = 4\%$.
Answer: (B) or (D), depending on the reading noted. | ||||
| 27 | 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 (A) Angular impulse = change of angular momentum: $[ML^2T^{-1}]$ → (III).
(B) Latent heat = energy per unit mass: $[L^2T^{-2}] = [M^0L^2T^{-2}]$ → (I).
(C) Resistivity $\rho = \frac{RA}{l}$; $[R] = \frac{[V]}{[A]} = [ML^2T^{-3}A^{-2}]$, so $[\rho] = [ML^3T^{-3}A^{-2}]$ → (IV).
(D) EMF is work per unit charge: $\frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$ → (II).
Answer: (C) A→III, B→I, C→IV, D→II. | ||||
| 28 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Distance = area under the v–t graph from 0 to 4 s.
0 to 2 s: triangle, $\frac12 \times 2 \times 10 = 10$ m.
2 to 4 s: rectangle, $2 \times 10 = 20$ m.
Total $= 30$ m.
Answer: (D) 30 m. | ||||
| 29 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}}$. For the planet $M_p = \frac{M_e}{8}$, $R_p = \frac{R_e}{2}$:
$\frac{v_p}{v_e} = \sqrt{\frac{M_p}{M_e}\cdot\frac{R_e}{R_p}} = \sqrt{\frac18 \cdot 2} = \frac12$.
$v_p = 5.6$ km/s.
Answer: (B) 5.6. | ||||
| 30 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take moments about the edge at 40 cm.
Anticlockwise (left): $400 \times (40 - 10) = 12000$ g cm.
Clockwise (right): rod's weight at 50 cm, $250 \times 10 = 2500$, and the mass $m$ at 90 cm, $m \times 50$.
$2500 + 50m = 12000 \Rightarrow m = 190$ g.
Answer: (D) 190 g. | ||||
| 31 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The body is "placed" at P, so it starts from rest; the force is constant.
Average power = work / time: $W = \vec F\cdot\overrightarrow{PQ} = (2\hat i + 3\hat j)\cdot(3\hat i + 6\hat j) = 6 + 18 = 24$ J, so $P_{av} = \frac{24}{4} = 6$ W.
Acceleration $\vec a = \frac{\vec F}{m} = \frac{2\hat i + 3\hat j}{4}$; at $t = 4$ s, $\vec v = \vec a t = 2\hat i + 3\hat j$ m/s.
Instantaneous power $= \vec F\cdot\vec v = 4 + 9 = 13$ W.
Ratio $= 6 : 13$.
Answer: (D) 6 : 13. | ||||
| 32 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Let the upthrust be $U$ (unchanged when mass is released).
Before: $U - Mg = Ma \Rightarrow U = M(g + a)$.
After releasing $m$: $U - (M - m)g = (M - m)3a \Rightarrow M(g + a) = (M - m)(g + 3a)$.
$M - m = \frac{M(g + a)}{g + 3a} \Rightarrow m = M\cdot\frac{(g + 3a) - (g + a)}{g + 3a} = \frac{2Ma}{3a + g}$.
Answer: (B) $\frac{2Ma}{3a + g}$. | ||||
| 33 | Properties of Bulk MatterFluid mechanics | 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 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 Floating: weight = weight of water displaced. Displaced volume $= \frac{400\ \text{g}}{1\ \text{g cm}^{-3}} = 400$ cm$^3$.
Volume of the cube $= 10^3 = 1000$ cm$^3$, so $1000 - 400 = 600$ cm$^3$ is outside the water.
Answer: (B) 600 cm$^3$. | ||||
| 34 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Translational kinetic energy $= \frac32 nRT$, with $n = \frac{50}{44}$ mol and $T = 290$ K.
$\frac32 \times \frac{50}{44} \times 8.3 \times 290 \approx 4102.8$ J.
(With $R = 8.314$ J mol$^{-1}$ K$^{-1}$ it is about 4110 J; the nearest option, and the exact one for $R = 8.3$, is 4102.8 J.)
Answer: (A) 4102.8 J. | ||||
| 35 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Vapour density is proportional to molar mass, so $\frac{M_1}{M_2} = \frac{4}{25}$.
$v_{rms} = \sqrt{\frac{3RT}{M}}$, so at the same temperature $\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{25}{4}} = \frac52$.
Answer: (A) $\frac{5}{2}$. | ||||
| 36 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Assertion and reasonEasy | |
A easy assertion and reason question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $x = A\sin(\omega t + \phi)$, $p = m\omega A\cos(\omega t + \phi)$.
At $t = 0$: $x_0 = A\sin\phi$ and $p_0 = m\omega A\cos\phi$, so $A = \sqrt{x_0^2 + \frac{p_0^2}{m^2\omega^2}}$ and $\tan\phi = \frac{m\omega x_0}{p_0}$. (R) is true.
With amplitude and phase known, $x(t)$ and $p(t)$ are known for every $t$: (A) is true, and it is true precisely because of (R).
Answer: (A). | ||||
| 37 | 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 Field between the plates $E = \frac{V}{d} = \frac{20}{10^{-6}} = 2 \times 10^7$ V/m.
Energy density $u = \frac12\varepsilon_0E^2 = \frac12 \times 8.85 \times 10^{-12} \times 4 \times 10^{14} \approx 1.77 \times 10^3$ J/m$^3 \approx 1.8 \times 10^3$ J/m$^3$.
(The capacitance is not needed; air between the plates is assumed.)
Answer: (B) $1.8 \times 10^3$ J/m$^3$. | ||||
| 38 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter From the figure: the current comes in from infinity along the lower straight wire (a distance a below O) up to the point directly below O, goes clockwise round three quarters of a circle of radius a centred at O, and leaves along the upper straight wire, which starts at distance a to the right of O on the horizontal line through O.
Arc ($\frac34$ of a circle): $B_1 = \frac34\cdot\frac{\mu_0I}{2a} = \frac{3\mu_0I}{8a}$, into the page (clockwise current).
Upper wire: O lies on its line, so $B_2 = 0$.
Lower wire: semi-infinite, ending at the foot of the perpendicular from O, distance a: $B_3 = \frac{\mu_0I}{4\pi a}$; for current flowing in the $-x$ direction below O the field at O is also into the page.
$B = \frac{3\mu_0I}{8a} + \frac{\mu_0I}{4\pi a} = \frac{\mu_0}{4\pi}\frac{I}{a}\left[\frac{3\pi}{2} + 1\right]$.
Answer: (C). | ||||
| 39 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current For a disc rotating in a perpendicular field, the emf between centre and rim is $\varepsilon = \frac12B\omega r^2$.
$\varepsilon = \frac12 \times 0.4 \times 10\pi \times (0.2)^2 = 0.08\pi = 0.08 \times 3.14 = 0.2512$ V.
Answer: (A) 0.2512 V. | ||||
| 40 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves The phase $\omega(t - z/c)$ means the wave travels along $+\hat k$; $\vec E$ has the same phase, magnitude $E_0 = cB_0 = 30c$, and $\vec E \times \vec B$ must point along $+\hat k$, i.e. $\vec E = c\,\vec B \times \hat k$.
$\left(\frac{\sqrt3}{2}\hat i + \frac12\hat j\right) \times \hat k = \frac{\sqrt3}{2}(-\hat j) + \frac12\hat i = \frac12\hat i - \frac{\sqrt3}{2}\hat j$.
$\vec E = \left(\frac12\hat i - \frac{\sqrt3}{2}\hat j\right)30c\sin\left[\omega\left(t - \frac zc\right)\right]$.
Check: $\vec E\cdot\vec B = \frac{\sqrt3}{4} - \frac{\sqrt3}{4} = 0$.
Answer: (C). | ||||
| 41 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Light goes from A ($n_1 = 1.3$) into B ($n_2 = 1.4$); the surface is convex towards A, so its centre lies in B: $R > 0$ with the direction of light positive. $u = -13$ cm.
Magnification of a refracting surface: $m = \frac{n_1v}{n_2u} = -2 \Rightarrow v = -2\cdot\frac{1.4 \times (-13)}{1.3} = 28$ cm.
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$: $\frac{1.4}{28} + \frac{1.3}{13} = \frac{0.1}{R} \Rightarrow 0.05 + 0.1 = \frac{0.1}{R}$.
$R = \frac{0.1}{0.15} = \frac23$ cm.
Answer: (B) $\frac{2}{3}$ cm. | ||||
| 42 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $m = -\frac{v}{u} = -3 \Rightarrow v = 3u$: a real, inverted image on the object's side, three times as far.
$|v| - |u| = 2|u| = 20$ cm, so $u = -10$ cm, $v = -30$ cm.
$\frac1f = \frac1v + \frac1u = -\frac1{30} - \frac1{10} = -\frac{4}{30} \Rightarrow f = -7.5$ cm.
$|R| = 2|f| = 15$ cm.
Answer: (C) 15 cm. | ||||
| 43 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei In the Bohr model $v_n \propto \frac{1}{n}$ and $r_n \propto n^2$.
Frequency $f = \frac{v_n}{2\pi r_n} \propto \frac{1/n}{n^2} = \frac{1}{n^3}$.
Answer: (A) $\frac{1}{n^3}$. | ||||
| 44 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Reflection, refraction and diffraction are all explained by the wave theory.
In the Compton effect X-rays scattered by electrons come out with a longer wavelength, the shift depending on the scattering angle; this needs photons carrying momentum $\frac{h}{\lambda}$ colliding with electrons, which the wave theory cannot give.
Answer: (D) Compton effect. | ||||
| 45 | Electronic DevicesRectifiers | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on rectifiers, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The source $V = V_0\sin\omega t$ is the potential of its upper terminal relative to the lower one. A is joined to the upper terminal; the resistor R is across the two lines; the diode sits in the lower line, its arrow pointing away from the source towards the junction with R and B.
The diode conducts only when current can flow out of the source's lower terminal, through the diode, up through R and back to the upper terminal: that is, when the lower terminal is at the higher potential, i.e. during the negative half cycles ($V < 0$).
Then, with no drop across the diode, $V_{AB} = V_A - V_B = V$, which is negative. During positive half cycles the diode blocks, no current flows in R, and $V_{AB} = 0$ (B is then cut off from the source by the reverse-biased diode and sits at A's potential through R).
So $V_{AB}$ shows only the negative half cycles.
Answer: (C). | ||||
| 46 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $K = \frac{P}{\Delta V/V} \Rightarrow \Delta V = \frac{PV}{K}$.
$V = (0.1)^3 = 10^{-3}$ m$^3$: $\Delta V = \frac{7 \times 10^6 \times 10^{-3}}{1.4 \times 10^{11}} = 5 \times 10^{-8}$ m$^3$.
$1$ m$^3 = 10^9$ mm$^3$, so $\Delta V = 50$ mm$^3$.
Answer: 50. | ||||
| 47 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Potential energy $U = -pE\cos\theta$.
$W = U(180^\circ) - U(0^\circ) = pE - (-pE) = 2pE = 2 \times 6 \times 10^{-6} \times 10^6 = 12$ J.
Answer: 12. | ||||
| 48 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity With A and B at the same potential, no current flows in the 30 Ω resistor and the bridge is balanced:
$\frac{10}{20} = \frac{R}{40} \Rightarrow R = 20\ \Omega$.
The 30 Ω arm can then be removed: upper branch $10 + 20 = 30\ \Omega$, lower branch $20 + 40 = 60\ \Omega$, in parallel: $\frac{30 \times 60}{90} = 20\ \Omega$.
$I = \frac{40}{20} = 2$ A (ideal cell, key K closed).
Answer: 2. | ||||
| 49 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The rails form a V. A bar moving away from the vertex with speed v is at distance $vt$ from it at time t, and the length of bar between the rails is proportional to that distance: $l = kvt$.
Motional EMF $E = Blv = Bkv^2t \propto t$.
$n = 1$.
Answer: 1. | ||||
| 50 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalHard | |
A hard numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At normal incidence the condition for a transmission minimum changes by one order when the optical path $2\mu t$ changes by $\lambda$, so successive minima are separated by a thickness change
$\Delta t = \frac{\lambda}{2\mu} = \frac{560 \times 10^{-9}}{2 \times 1.4} = 2 \times 10^{-7}$ m.
Volume lost in 12 s: $\pi r^2\Delta t = \pi(1.8 \times 10^{-2})^2 \times 2 \times 10^{-7} = \pi \times 6.48 \times 10^{-11}$ m$^3$.
Rate $= \frac{\pi \times 6.48 \times 10^{-11}}{12} = \pi \times 5.4 \times 10^{-12} = 54\,\pi \times 10^{-13}$ m$^3$/s.
Answer: 54. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | StatementsEasy | |
A easy statements question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom In the hydrogen atom (one electron) the orbital energy depends only on n: all orbitals of a shell are degenerate.
(A) 1s (n=1) < 2p (n=2) < 3d (n=3) < 4s (n=4): correct.
(B) 1s < 2s = 2p < 3s = 3p: correct.
(C) puts 2s below 2p and 3s below 3p: not correct for hydrogen.
(D) puts 4s below 3d: not correct for hydrogen (3d, n = 3, is lower).
Not correct: (C) and (D).
Answer: (B) (C) and (D) only. | ||||
| 52 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Internal energy is a state function. In every case the gas starts at A and returns to A, so $\Delta U = 0$ for each cycle, whatever the path.
$\Delta U$(Case-I) $=$ $\Delta U$(Case-II) $=$ $\Delta U$(Case-III) $= 0$.
Answer: (D). | ||||
| 53 | SolutionsColligative properties | Class 12: Solutions | Single correctMedium | |
A medium single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions Read as intended, the outer solution has the same mole fractions of glucose and of water as the solution in the cell.
Equal mole fractions mean equal composition: the outer solution is also 0.9% (w/w) glucose. It is isotonic with the cell, so there is no net osmosis and the cell keeps its volume.
(Glucose neither dissociates nor associates appreciably in water, so no van't Hoff correction applies.)
Answer: (C). | ||||
| 54 | SolutionsConcentration terms | Class 12: Solutions | Single correctEasy | |
A easy single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Mass of solution containing 30 g HNO$_3$ $= \frac{30}{0.75} = 40$ g.
Volume $= \frac{40}{1.25} = 32$ mL.
Answer: (C) 32. | ||||
| 55 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Approximate solubility products at 298 K:
HgS $\sim 10^{-52}$ (one of the least soluble salts), PbS $\sim 10^{-28}$, AgBr $\sim 5 \times 10^{-13}$, $\mathrm{Ca(OH)_2}$ $\sim 5 \times 10^{-6}$ (sparingly soluble but far more soluble than the others).
Increasing order: HgS < PbS < AgBr < $\mathrm{Ca(OH)_2}$.
Answer: (B). | ||||
| 56 | 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: $\frac{dN}{dt} = -\lambda N$. Growth by the same law has the opposite sign: $\frac{dN}{dt} = kN$, so $N = N_0e^{kt}$.
$\frac{N}{N_0} = e^{kt}$ starts at 1 when $t = 0$ (not at zero) and rises more and more steeply with time: the exponential growth curve.
Answer: (B). | ||||
| 57 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For an elementary reaction the rate law follows the stoichiometry: rate $= k[\mathrm A][\mathrm B]$.
Reducing the volume to $\frac13$ triples each concentration, so the rate becomes $3 \times 3 = 9$ times.
Answer: (D) 9. | ||||
| 58 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | StatementsMedium | |
A medium statements question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter (A) Ammonium sulphide, $\mathrm{(NH_4)_2S}$, is colourless; the yellow reagent of qualitative analysis is yellow ammonium sulphide, the polysulphide $\mathrm{(NH_4)_2S_x}$.
(B) PbS is black.
(C) CuS is black.
(D) $\mathrm{As_2S_3}$ is yellow (group II precipitate).
(E) $\mathrm{As_2S_5}$ is yellow.
Yellow: (D) and (E).
Answer: (A) or (D), depending on the reading noted. | ||||
| 59 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | StatementsEasy | |
A easy statements question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Statement I: Newlands' Law of Octaves arranged the elements in increasing order of atomic WEIGHT (atomic numbers were unknown until Moseley, 1913). False.
Statement II: Lothar Meyer plotted physical properties such as atomic volume, melting point and boiling point against atomic WEIGHT and found a periodically repeated pattern. As written, "against their respective atomic numbers", it is false.
Answer: (B) Both Statement I and Statement II are false. | ||||
| 60 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Among the vanadium oxides, $\mathrm{V_2O_3}$ is basic, $\mathrm{V_2O_4}$ less basic, and $\mathrm{V_2O_5}$ is amphoteric (though mainly acidic).
With alkali: $\mathrm{V_2O_5 + 6OH^- \to 2VO_4^{3-} + 3H_2O}$, the vanadate ion.
In $\mathrm{VO_4^{3-}}$: $x + 4(-2) = -3 \Rightarrow x = +5$.
Answer: (C) +5. | ||||
| 61 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Matching listEasy | |
A easy matching list question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) $\mathrm{[CoF_6]^{3-}}$: $\mathrm{Co^{3+}}$, $3d^6$; F$^-$ is a weak-field ligand, high spin, outer orbital: $sp^3d^2$ → (III).
(B) $\mathrm{[NiCl_4]^{2-}}$: $\mathrm{Ni^{2+}}$, $3d^8$; Cl$^-$ weak field, tetrahedral: $sp^3$ → (II).
(C) $\mathrm{[Co(NH_3)_6]^{3+}}$: $\mathrm{Co^{3+}}$, $3d^6$; NH$_3$ pairs the electrons, inner orbital: $d^2sp^3$ → (I).
(D) $\mathrm{[Ni(CN)_4]^{2-}}$: $\mathrm{Ni^{2+}}$, $3d^8$; CN$^-$ strong field, square planar: $dsp^2$ → (IV).
Answer: (A) A→III, B→II, C→I, D→IV. | ||||
| 62 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A solid that passes directly into vapour on heating and back into solid on cooling, without melting, is purified by sublimation (camphor, naphthalene, iodine, anthracene are examples); non-volatile impurities are left behind.
Answer: (B) Sublimation. | ||||
| 63 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsEasy | |
A easy statements question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Statement I: the ring compound is methylcyclopentane, $\mathrm{C_6H_{12}}$; the open-chain compound is hex-3-ene, $\mathrm{C_6H_{12}}$. Same molecular formula, different structures: isomers (a cycloalkane and an alkene, ring–chain isomers). True.
Statement II: pentan-1-amine, $\mathrm{CH_3(CH_2)_4NH_2}$, $\mathrm{C_5H_{13}N}$, is a primary amine; N-ethylpropan-1-amine, $\mathrm{CH_3CH_2NHCH_2CH_2CH_3}$, $\mathrm{C_5H_{13}N}$, is a secondary amine. Same formula, different amino groups ($-\mathrm{NH_2}$ against $>\mathrm{NH}$): functional group isomers. True.
Answer: (A) Both Statement I and Statement II are true. | ||||
| 64 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Partial hydrogenation of propyne over a palladium catalyst stops at the alkene: [A] is propene, $\mathrm{CH_3-CH=CH_2}$.
Reductive ozonolysis cleaves the C=C: $\mathrm{CH_3CH=CH_2 \to CH_3CHO + HCHO}$.
[B] $\mathrm{CH_3CHO}$, [C] HCHO.
(Propane, as in option (D), has no C=C and could not undergo ozonolysis; option (C)'s ethanol is not an ozonolysis product.)
Answer: (A). | ||||
| 65 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate is $\mathrm{C_6H_5-CH(CH_3)-CHBr-CH_2-CHBr-CH_2-CH_3}$ (3,5-dibromo-2-phenylheptane).
Excess alcoholic KOH on heating removes two molecules of HBr (E2). Each double bond forms where it gives the most stable product:
the H on C2 (benzylic, tertiary) and the Br on C3 give C2=C3, conjugated with the ring and more substituted;
the Br on C5 and an H on C4 give C4=C5, which is conjugated with C2=C3.
Product: $\mathrm{C_6H_5-C(CH_3)=CH-CH=CH-CH_2CH_3}$, a conjugated diene also conjugated with the benzene ring: 2-phenylhepta-2,4-diene.
Answer: (D) 2-phenylhepta-2,4-diene. | ||||
| 66 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The starting compound is 1-(4-methylphenyl)prop-1-ene, $\mathrm{CH_3C_6H_4-CH=CH-CH_3}$.
HCl adds through the more stable carbocation, the benzylic one, so Cl goes to the carbon attached to the ring: (A) $\mathrm{CH_3C_6H_4-CHCl-CH_2CH_3}$.
AgCN is largely covalent; the nitrogen of the cyanide attacks, giving the isocyanide: (B) $\mathrm{CH_3C_6H_4-CH(NC)-CH_2CH_3}$.
Answer: (A). | ||||
| 67 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Hot $\mathrm{KMnO_4}$ oxidises a side chain to $-\mathrm{COOH}$ only if the carbon attached to the ring carries at least one hydrogen.
Toluene, ethylbenzene, isopropylbenzene (cumene), isobutylbenzene and propylbenzene have benzylic H: each gives benzoic acid (5).
tert-Butylbenzene and 2-phenylpropan-2-ol have no benzylic H: not oxidised.
4,4'-Dimethylbiphenyl is oxidised to biphenyl-4,4'-dicarboxylic acid, not benzoic acid.
Total $= 5$.
Answer: (C) 5. | ||||
| 68 | AminesClassification and identification of amines | Class 12: Amines | StatementsEasy | |
A easy statements question on classification and identification of amines, from Amines. NCERT: Class 12: Amines (A) False: primary aromatic amines give diazonium salts with $\mathrm{NaNO_2}$/HCl at 273–278 K (aliphatic ones give unstable diazonium salts that decompose).
(B) True: the carbylamine (isocyanide) test is given by both aliphatic and aromatic primary amines.
(C) False: secondary and tertiary amines do not give the carbylamine test.
(D) True: benzenesulfonyl chloride is Hinsberg's reagent.
(E) False: tertiary amines have no N–H and do not react with benzenesulfonyl chloride.
Answer: (C) (B) and (D) only. | ||||
| 69 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules (A) Sucrose: C1 of $\alpha$-D-glucose joined to C2 of $\beta$-D-fructose, $\alpha$1–$\beta$2 → (III).
(B) Maltose: two $\alpha$-D-glucose units, C1–C4: $\alpha$1–4 → (I).
(C) Lactose: C1 of $\beta$-D-galactose to C4 of glucose: $\beta$1–4 → (IV).
(D) Amylopectin: $\alpha$1–4 chains with $\alpha$1–6 branches → (II).
Answer: (A) A→III, B→I, C→IV, D→II. | ||||
| 70 | BiomoleculesCarbohydrates | Class 12: Biomolecules | StatementsEasy | |
A easy statements question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules (A) False: starch is a polymer of $\alpha$-D-glucose and gives glucose.
(B) True: sucrose (cane sugar) gives equal amounts of glucose and fructose (invert sugar).
(C) True: lactose (milk sugar) gives glucose and galactose.
(D) False: amylopectin gives only glucose.
(E) True: amylose gives only glucose.
Answer: (D) (B), (C) and (E) only. | ||||
| 71 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\mathrm{C_6H_6(l) + \frac{15}{2}O_2(g) \to 6CO_2(g) + 3H_2O(l)}$, $\Delta_cH = -3267.0$ kJ mol$^{-1}$.
$\Delta_cH = 6\Delta_fH(\mathrm{CO_2}) + 3\Delta_fH(\mathrm{H_2O}) - \Delta_fH(\mathrm{C_6H_6})$
$-3267.0 = 6(-393.5) + 3(-286.0) - \Delta_fH = -2361.0 - 858.0 - \Delta_fH$.
$\Delta_fH(\mathrm{C_6H_6}) = -3219.0 + 3267.0 = 48$ kJ mol$^{-1}$.
Answer: 48. | ||||
| 72 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry At the cathode $\mathrm{2H_2O + 2e^- \to H_2 + 2OH^-}$: one $\mathrm{OH^-}$ per electron.
pH 12 means $[\mathrm{OH^-}] = 10^{-2}$ M; in 0.600 L, $n(\mathrm{OH^-}) = 6 \times 10^{-3}$ mol, so $6 \times 10^{-3}$ mol of electrons passed (the initial solution taken as neutral, volume unchanged).
$Q = 6 \times 10^{-3} \times 96500 = 579$ C; $I = \frac{579}{300} = 1.93$ A $\approx 2$.
Answer: 2. | ||||
| 73 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{O_2}$: two unpaired electrons in $\pi^*$, paramagnetic.
$\mathrm{O_2^+}$: one unpaired electron, paramagnetic.
$\mathrm{O_2^-}$: one unpaired electron, paramagnetic.
NO: 15 electrons, odd, paramagnetic.
$\mathrm{NO_2}$: 23 electrons, odd, paramagnetic.
CO: 14 electrons, all paired, diamagnetic.
$\mathrm{K_2[NiCl_4]}$: $\mathrm{Ni^{2+}}$ $3d^8$, tetrahedral with weak-field Cl$^-$, two unpaired electrons, paramagnetic.
$\mathrm{[Co(NH_3)_6]Cl_3}$: $\mathrm{Co^{3+}}$ $3d^6$ low spin, diamagnetic.
$\mathrm{K_2[Ni(CN)_4]}$: $\mathrm{Ni^{2+}}$ $3d^8$ square planar, diamagnetic.
Paramagnetic: 6.
Answer: 6. | ||||
| 75 | 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 TiO ($\mathrm{Ti^{2+}}$) and VO ($\mathrm{V^{2+}}$) are reducing: $\mathrm{Ti^{2+}}$ and $\mathrm{V^{2+}}$ are readily oxidised. In $\mathrm{Mn_2O_3}$, $\mathrm{Mn^{3+}}$ ($3d^4$) is a strong oxidant, being reduced to the stable $\mathrm{Mn^{2+}}$ ($3d^5$, half-filled). So $\mathrm{Mn_2O_3}$ is the strongest oxidising agent.
$\mathrm{Mn^{3+}}$, $3d^4$ (high spin in the oxide): $n = 4$ unpaired electrons.
$\mu = \sqrt{n(n+2)} = \sqrt{24} \approx 4.90$ B.M., nearest integer 5.
Answer: 5. | ||||
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