JEE (Main) · 2025
JEE (Main) 2025 — 24 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
21 Q
- Conic Sections3
- Sets, Relations and Functions2
- Matrices and Determinants2
5 easy · 16 medium · 0 hard
Physics
23 Q
- Magnetic Effects of Current and Magnetism4
- Thermodynamics3
- Properties of Bulk Matter2
18 easy · 5 medium · 0 hard
Chemistry
22 Q
- Thermodynamics2
- Chemical Kinetics2
- Chemical Bonding and Molecular Structure2
19 easy · 3 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 FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy 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 Divide numerator and denominator by $2^{-x}$: $f(x) = \frac{4^x - 1}{4^x + 1} = 1 - \frac{2}{4^x + 1}$.
As $x$ increases, $4^x + 1$ increases, so $f$ is strictly increasing: $f$ is one-one.
As $x \to -\infty$, $f(x) \to -1$; as $x \to \infty$, $f(x) \to 1$. The range is $(-1, 1)$.
The codomain is $(-\infty, 1)$, and values such as $-2$ are never taken, so $f$ is not onto.
Answer: (A) One-one but not onto. | ||||
| 2 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Set A: $\log_{(2/\pi)}|\sin x \cos x| = 2$, so $|\sin x\cos x| = \frac{4}{\pi^2}$, i.e. $|\sin 2x| = \frac{8}{\pi^2} \approx 0.81$.
For $x \in (0, \pi)$, $2x \in (0, 2\pi)$, and $|\sin 2x| = k$ with $0 < k < 1$ has four solutions there (two where $\sin 2x = k$, two where $\sin 2x = -k$). None is $\frac{\pi}{2}$. So $n(A) = 4$.
Set B: put $t = \sqrt{x} \geq 0$: $t^2 - 4t - 3|t - 2| + 6 = 0$.
If $t \geq 2$: $t^2 - 7t + 12 = 0$, $t = 3$ or $4$, so $x = 9, 16$.
If $0 \leq t < 2$: $t^2 - 4t + 3t - 6 + 6 = 0$, $t^2 - t = 0$, $t = 0$ or $1$, so $x = 0, 1$.
So $B = \{0, 1, 9, 16\}$, $n(B) = 4$.
The elements of A lie in $(0, \pi)$ and none of them is $1$ (at $x = 1$, $|\sin 2| \approx 0.909 \neq 0.81$), so A and B are disjoint.
$n(A \cup B) = 4 + 4 = 8$.
Answer: (D) $8$. | ||||
| 3 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $|x - 3| \leq |x + 2|$ exactly when $x \geq \frac{1}{2}$.
Case $x \geq \frac{1}{2}$: the right side is $|x - 3|$. For $\frac{1}{2} \leq x \leq 3$ it is at most $2.5$, while $x^2 + 3x + 2 \geq 3.75$; for $x > 3$, $x - 3 < x^2 + 3x + 2$. No solution.
Case $x < \frac{1}{2}$: the right side is $|x + 2|$.
For $-2 \leq x < \frac{1}{2}$: $x^2 + 3x + 2 = x + 2 \Rightarrow x(x + 2) = 0$, so $x = 0$ or $x = -2$; both lie in the range.
For $x < -2$: $x^2 + 3x + 2 = -x - 2 \Rightarrow (x + 2)^2 = 0$, $x = -2$, not in this range.
So there are two real solutions, $x = 0$ and $x = -2$.
Answer: (C) $2$. | ||||
| 4 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants As $x \to 0$, $\frac{\sin x}{x} \to 1$ and the determinant is continuous in its entries, so
$\lim_{x\to0} f(x) = \begin{vmatrix} a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 \end{vmatrix}$.
Subtract row 1 from rows 2 and 3: rows become $(a+1, 1, b)$, $(-1, 1, 0)$, $(-1, 0, 1)$.
Expanding: $(a+1)(1 - 0) - 1(-1 - 0) + b(0 + 1) = a + b + 2$.
So $\lambda = 2$, $\mu = 1$, $\nu = 1$ and $(\lambda + \mu + \nu)^2 = 4^2 = 16$.
Answer: (B) $16$. | ||||
| 5 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The first two equations are independent (their $x$ and $z$ coefficients are not proportional), so for infinitely many solutions the third equation must be a combination $p(\text{eq. 1}) + q(\text{eq. 2})$ of them.
Coefficients of $x$: $p + 2q = 14$; constants: $2p + 5q = 33$. Hence $q = 5$, $p = 4$.
Coefficients of $y$: $2p + \lambda q = 3 \Rightarrow 8 + 5\lambda = 3 \Rightarrow \lambda = -1$.
Coefficients of $z$: $-3p + 5q = \mu \Rightarrow \mu = -12 + 25 = 13$.
$\lambda + \mu = -1 + 13 = 12$.
Answer: (C) $12$. | ||||
| 6 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Split the series: $5 + \sum_{n\geq1}\frac{5}{7^n} + \alpha\sum_{n\geq1}\frac{n}{7^n}$.
$\sum_{n\geq1}\frac{5}{7^n} = 5\cdot\frac{1/7}{1 - 1/7} = \frac{5}{6}$.
$\sum_{n\geq1} n r^n = \frac{r}{(1-r)^2}$ with $r = \frac{1}{7}$: $\frac{1/7}{36/49} = \frac{7}{36}$.
So $7 = 5 + \frac{5}{6} + \frac{7\alpha}{36}$, giving $\frac{7\alpha}{36} = \frac{7}{6}$ and $\alpha = 6$.
Answer: (A) $6$. | ||||
| 9 | 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 Let $m = 2n - 1$. The $(r+1)^{\text{th}}$ term has coefficient $\binom{m}{r}$, so $A = \binom{m}{29}$ and $B = \binom{m}{11}$.
$2\binom{m}{29} = 5\binom{m}{11}$. Since $\binom{m}{29} = \binom{m}{m-29}$, try $m - 29 = 12$, i.e. $m = 41$:
$\binom{41}{12} = \binom{41}{11}\cdot\frac{41 - 11}{12} = \frac{30}{12}\binom{41}{11} = \frac{5}{2}\binom{41}{11}$, which is exactly $2A = 5B$.
The other options do not work: for $m = 39$, $\binom{39}{29} = \binom{39}{10} < \binom{39}{11}$, and for $m = 43$, $\binom{43}{29} = \binom{43}{14} = \frac{32\cdot31\cdot30}{14\cdot13\cdot12}\binom{43}{11} > \frac{5}{2}\binom{43}{11}$.
$2n - 1 = 41 \Rightarrow n = 21$.
Answer: (C) $21$. | ||||
| 10 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Let $b$ boys come from A; then $5 - b$ girls come from A, $4 - b$ boys and $b - 1$ girls from B. A has only 3 girls, so $b \geq 2$; and $b \leq 4$.
$b = 2$: $\binom{7}{2}\binom{3}{3}\binom{6}{2}\binom{5}{1} = 21\cdot1\cdot15\cdot5 = 1575$.
$b = 3$: $\binom{7}{3}\binom{3}{2}\binom{6}{1}\binom{5}{2} = 35\cdot3\cdot6\cdot10 = 6300$.
$b = 4$: $\binom{7}{4}\binom{3}{1}\binom{6}{0}\binom{5}{3} = 35\cdot3\cdot1\cdot10 = 1050$.
Total $= 1575 + 6300 + 1050 = 8925$.
Answer: (C) $8925$. | ||||
| 11 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines Vertices: $x + y = 11$ and $x + 2y = 16$ meet at $(6, 5)$; $x + y = 11$ and $2x + 3y = 29$ at $(4, 7)$; $x + 2y = 16$ and $2x + 3y = 29$ at $(10, 3)$.
The vertical line $x = \frac{11}{2}$ lies between $x = 4$ and $x = 6$, so it cuts the sides from $(4, 7)$: the side $x + y = 11$ at $y = \frac{11}{2}$ and the side $2x + 3y = 29$ at $y = \frac{29 - 11}{3} = 6$.
So $\frac{11}{2} \leq \alpha \leq 6$, and the product of the extreme values is $\frac{11}{2}\cdot6 = 33$.
Answer: (B) $33$. | ||||
| 12 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The axis is perpendicular to the directrix, direction $(1, 2)$. The foot of the perpendicular from V to $x + 2y = 0$ is $V - t(1, 2)$ with $t = \frac{3/2 + 6}{5} = \frac{3}{2}$: the point $(0, 0)$.
The vertex is midway between this foot and the focus, so the focus is $S = V + \frac{3}{2}(1, 2) = (3, 6)$.
Parabola: $(x - 3)^2 + (y - 6)^2 = \frac{(x + 2y)^2}{5}$.
$5x^2 + 5y^2 - 30x - 60y + 225 = x^2 + 4xy + 4y^2 \Rightarrow 4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0$.
So $\alpha = 4$, $\beta = 1$, $\gamma = 4$ and $\alpha + \beta + \gamma = 9$.
Answer: (D) $9$. | ||||
| 13 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The chord with mid-point $(x_1, y_1)$ is $T = S_1$: $\frac{xx_1}{25} + \frac{yy_1}{16} = \frac{x_1^2}{25} + \frac{y_1^2}{16}$.
With $(3, 1)$: $\frac{3x}{25} + \frac{y}{16} = \frac{9}{25} + \frac{1}{16}$. Multiply by 400: $48x + 25y = 144 + 25 = 169$.
Answer: (D) $48x + 25y = 169$. | ||||
| 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 $\beta + \frac{1 + \beta^2}{\alpha - \beta} = \frac{1 + \alpha\beta}{\alpha - \beta} > 0$, so the first term is $\tan^{-1}\frac{\alpha - \beta}{1 + \alpha\beta} = \tan^{-1}\alpha - \tan^{-1}\beta$ (all positive).
Likewise the second term is $\tan^{-1}\beta - \tan^{-1}\gamma$.
The third argument is $\frac{1 + \gamma\alpha}{\gamma - \alpha} < 0$. For $y < 0$, $\cot^{-1}y = \pi - \cot^{-1}|y|$, so the third term is $\pi - (\tan^{-1}\alpha - \tan^{-1}\gamma)$.
Sum: $(\tan^{-1}\alpha - \tan^{-1}\beta) + (\tan^{-1}\beta - \tan^{-1}\gamma) + \pi - \tan^{-1}\alpha + \tan^{-1}\gamma = \pi$.
Answer: (D) $\pi$. | ||||
| 16 | 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 Centroid: $G = \frac{1}{3}\left[(4 - 5 + 2)\vec{p} + (1 + 1 - 1)\vec{q} + (-3 + 2 + 2)\vec{r}\right] = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$.
On the Euler line, G divides HO in the ratio $2 : 1$: $G = \frac{H + 2O}{3}$, so $O = \frac{3G - H}{2}$.
$O = \frac{1}{2}\left[(\vec{p} + \vec{q} + \vec{r}) - \frac{\vec{p} + \vec{q} + \vec{r}}{4}\right] = \frac{3}{8}(\vec{p} + \vec{q} + \vec{r})$.
So $\alpha = \beta = \gamma = \frac{3}{8}$ and $\alpha + 2\beta + 5\gamma = 8\cdot\frac{3}{8} = 3$.
Answer: (B) $3$. | ||||
| 17 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $|x - 2|$ is continuous everywhere, so $f$ is discontinuous exactly where $[x]$ jumps: at $x = -1, 0, 1, 2$ in $(-2, 3)$. So $m = 4$.
A function is not differentiable where it is not continuous, so these four points count for $n$. Elsewhere $[x]$ is constant and $|x - 2|$ has its only corner at $x = 2$, which is already counted. So $n = 4$.
$m + n = 8$.
Answer: (C) $8$. | ||||
| 18 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = \frac{2}{x - 2} - 2x + a$. The interval of increase ends at $x = 3$, so $f'(3) = 0$: $2 - 6 + a = 0$, $a = 4$.
Check: $f'(x) = \frac{2}{x - 2} - 2(x - 2) = \frac{2[1 - (x - 2)^2]}{x - 2} > 0$ exactly for $2 < x < 3$.
Then $g(x) = (x - 1)^3(x - 2)^2$ and $g'(x) = (x - 1)^2(x - 2)[3(x - 2) + 2(x - 1)] = (x - 1)^2(x - 2)(5x - 8)$.
$g' < 0$ when $(x - 2)(5x - 8) < 0$, i.e. $\frac{8}{5} < x < 2$. So $b = \frac{8}{5}$, $c = 2$.
$100(a + b - c) = 100(4 + 1.6 - 2) = 360$.
Answer: (C) $360$. | ||||
| 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 For $x \geq 0$, $|e^x - 1| = e^x - 1 < e^x$: the curves never meet, so no bounded region lies on that side.
For $x < 0$, $|e^x - 1| = 1 - e^x$. It meets $y = e^x$ where $e^x = \frac{1}{2}$, $x = -\log_e 2$. Between there and the y-axis, $e^x$ lies above $1 - e^x$.
Area $= \int_{-\ln 2}^{0}\left[e^x - (1 - e^x)\right]dx = \left[2e^x - x\right]_{-\ln 2}^{0} = 2 - (1 + \ln 2) = 1 - \log_e 2$.
Answer: (C) $1 - \log_e 2$. | ||||
| 20 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $x^2f' - 2xf = 3$. Divide by $x^4$: $\frac{d}{dx}\left(\frac{f}{x^2}\right) = \frac{3}{x^4}$.
Integrate: $\frac{f(x)}{x^2} = -\frac{1}{x^3} + C$. With $f(1) = 4$: $C = 5$.
$f(x) = 5x^2 - \frac{1}{x}$, so $f(2) = 20 - \frac{1}{2} = \frac{39}{2}$ and $2f(2) = 39$.
Answer: (A) $39$. | ||||
| 22 | 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 A point of L is $(1 + 2t, -1 + 3t, 4t)$. The foot F of the perpendicular from Q satisfies
$(2t - 6, 3t + 1, 4t - 5)\cdot(2, 3, 4) = 29t - 29 = 0$, so $t = 1$ and $F = (3, 2, 4)$.
The image is $P = 2F - Q = (-1, 6, 3)$.
R has $x = 5$, so $t = 2$ and $R = (5, 5, 8)$.
PQ is perpendicular to L and bisected at F, and R lies on L, so RF is the height of the triangle on base PQ.
$QF = |(4, -4, 1)| = \sqrt{33}$, so $PQ = 2\sqrt{33}$; $RF = |(2, 3, 4)| = \sqrt{29}$.
Area $= \frac{1}{2}\cdot2\sqrt{33}\cdot\sqrt{29} = \sqrt{957}$, and its square is $957$.
Answer: 957. | ||||
| 23 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Write $2x^2 + 5x + 9 = 2(x^2 + x + 1) + \frac{3}{2}(2x + 1) + \frac{11}{2}$. Let $u = x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}$ and $L = \log_e\left|x + \frac{1}{2} + \sqrt{u}\right|$.
$\int 2\sqrt{u}\,dx = 2\left[\frac{x + \frac{1}{2}}{2}\sqrt{u} + \frac{3}{8}L\right] = \left(x + \frac{1}{2}\right)\sqrt{u} + \frac{3}{4}L$.
$\int\frac{\frac{3}{2}(2x + 1)}{\sqrt{u}}dx = 3\sqrt{u}$; $\int\frac{11/2}{\sqrt{u}}dx = \frac{11}{2}L$.
Total: $x\sqrt{u} + \left(\frac{1}{2} + 3\right)\sqrt{u} + \left(\frac{3}{4} + \frac{11}{2}\right)L = x\sqrt{u} + \frac{7}{2}\sqrt{u} + \frac{25}{4}L$.
$\alpha = \frac{7}{2}$, $\beta = \frac{25}{4}$, $\alpha + 2\beta = \frac{7}{2} + \frac{25}{2} = 16$.
Answer: 16. | ||||
| 24 | 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 Divide by $2\cos x$: $\frac{dy}{dx} + 2\tan x\cdot y = \sin x$. Integrating factor $e^{\int2\tan x\,dx} = \sec^2x$.
$\frac{d}{dx}(y\sec^2x) = \sin x\sec^2x = \sec x\tan x$, so $y\sec^2x = \sec x + C$.
$y\left(\frac{\pi}{3}\right) = 0$: $0 = 2 + C$, $C = -2$. Hence $y = \cos x - 2\cos^2x$.
$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} - 1$; $y' = -\sin x + 4\sin x\cos x$, so $y'\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} + 2$.
Sum $= 1$.
Answer: 1. | ||||
| 25 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $H_1$: $e_1^2 = 1 + \frac{b^2}{a^2} = \frac{5}{2}$, so $b^2 = \frac{3}{2}a^2$. Latus rectum $\frac{2b^2}{a} = 3a = 15\sqrt{2}$, so $a = 5\sqrt{2}$ and the transverse axis is $2a = 10\sqrt{2}$.
$H_2$ has its transverse axis along the y-axis, of length $2B$: $10\sqrt{2}\cdot2B = 100\sqrt{10}$ gives $B = 5\sqrt{5}$.
Its latus rectum is $\frac{2A^2}{B} = 12\sqrt{5}$, so $A^2 = 6\sqrt{5}\cdot5\sqrt{5} = 150$.
$e_2^2 = 1 + \frac{A^2}{B^2} = 1 + \frac{150}{125} = \frac{11}{5}$, and $25e_2^2 = 55$.
Answer: 55. | ||||
| 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 Every term of the equation must have the dimensions of $E^2$.
$pc$ has the dimensions of energy ($[MLT^{-1}][LT^{-1}] = [ML^2T^{-2}]$), so $p^2c^2$ is an energy squared; $pc^2$ is not.
$mc^2$ is an energy, so $m^2c^4$ is an energy squared; $m^2c^2$ is not.
Only $E^2 = p^2c^2 + m^2c^4$ is consistent, and it is the relativistic energy–momentum relation.
Answer: (C) $E^2 = p^2c^2 + m^2c^4$. | ||||
| 27 | ThermodynamicsZeroth law and temperature | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on zeroth law and temperature, from Thermodynamics. NCERT: Class 11: Thermodynamics $\frac{C}{5} = \frac{F - 32}{9}$, so $C = \frac{5}{9}(F - 32)$: a straight line of positive slope $\frac{5}{9}$ on axes with F across and C up.
At $F = 0$, $C = -\frac{160}{9}$ (negative C intercept); at $C = 0$, $F = 32$ (positive F intercept).
Only graph (B) rises and cuts the C axis below the origin and the F axis to the right of it.
Answer: (B). | ||||
| 29 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\vec{v} = \frac{d\vec{r}}{dt} = 10t\hat{i} - 5\hat{j}$; at $t = 2$ s, $\vec{v} = 20\hat{i} - 5\hat{j}$ m/s.
$|\vec{v}| = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17}$ m/s.
The velocity points into the fourth quadrant. Its angle $\theta$ with the $-$ve Y axis has $\tan\theta = \frac{20}{5} = 4$ (with the $+$ve X axis it would be $\tan^{-1}\frac{1}{4}$).
Answer: (A). | ||||
| 30 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy 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 For rolling without slipping down an incline, $a = \frac{g\sin\theta}{1 + \frac{k^2}{R^2}}$.
Solid sphere: $\frac{k^2}{R^2} = \frac{2}{5}$, $a_1 = \frac{5}{7}g\sin\theta$. Hollow sphere: $\frac{k^2}{R^2} = \frac{2}{3}$, $a_2 = \frac{3}{5}g\sin\theta$.
$a_1 > a_2$, and for the same length $t = \sqrt{\frac{2s}{a}}$, so the solid sphere takes less time: $t_1 < t_2$.
Answer: (A) $t_1 < t_2$. | ||||
| 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 Newton's law of cooling (average form): $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$.
First 4 min: $\frac{16}{4} = k(32 - 16)$, so $k = \frac{1}{4}$ per minute.
Next 4 min: $\frac{24 - T}{4} = \frac{1}{4}\left(\frac{24 + T}{2} - 16\right)$, so $48 - 2T = T - 8$, $T = \frac{56}{3}\ ^\circ$C.
(The exact exponential law gives the same: the excess over the air falls from 24 to 8, a factor $\frac{1}{3}$ in 4 min, so it becomes $\frac{8}{3}$ and $T = 16 + \frac{8}{3} = \frac{56}{3}\ ^\circ$C.)
Answer: (D) $\frac{56}{3}\ ^\circ$C. | ||||
| 32 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics In a cycle $\Delta U = 0$, so the heat exchanged equals the work, the area enclosed on the P–V diagram.
The path A → B → C is the upper half of a circle centred at (300 cc, 300 kPa) with semi-axes 100 kPa and 100 cc; C → A is the horizontal diameter. The enclosed region is a half ellipse.
Area $= \frac{1}{2}\pi ab = \frac{1}{2}\pi(100 \times 10^3\ \text{Pa})(100 \times 10^{-6}\ \text{m}^3) = \frac{1}{2}\pi(10\ \text{J}) = 5\pi$ J.
Answer: (B) $5\pi$. | ||||
| 33 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Assertion and reasonEasy | |
A easy assertion and reason question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics (A): in an adiabatic compression $Q = 0$ and the work done on the gas is positive, so $\Delta U > 0$ and the temperature rises ($TV^{\gamma-1}$ = constant, so halving V raises T). (A) is false.
(R): in free expansion the gas expands into vacuum suddenly; no heat is exchanged (adiabatic) and the process cannot be reversed (irreversible). (R) is true.
Answer: (D) (A) is false but (R) is true. | ||||
| 34 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $x = \sin^2\frac{t}{2} = \frac{1 - \cos t}{2}$: simple harmonic motion about $x = \frac{1}{2}$ with amplitude $\frac{1}{2}$ m and $\omega = 1$ rad/s, between $x = 0$ and $x = 1$.
$K = \frac{1}{2}m\omega^2\left[A^2 - \left(x - \frac{1}{2}\right)^2\right] = \frac{1}{2}m\left[\frac{1}{4} - \left(x - \frac{1}{2}\right)^2\right] = \frac{1}{2}m(x - x^2)$.
This is an inverted parabola, zero at $x = 0$ and $x = 1$ and greatest at $x = \frac{1}{2}$: graph (A).
Answer: (A). | ||||
| 35 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Assertion and reasonEasy | |
A easy assertion and reason question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The magnetic force is $\vec{F} = q\vec{v}\times\vec{B}$. If $\vec{B}$ is along $\vec{v}$, $\vec{v}\times\vec{B} = 0$, so there is no force and the electron keeps moving with constant velocity in a straight line.
So (A) can hold in a uniform field, and (R) states exactly the condition that makes it hold: both are true and (R) explains (A).
Answer: (A). | ||||
| 36 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Ampere's law with a circle of radius r:
Inside ($r < a$): enclosed current $I\frac{r^2}{a^2}$, so $B = \frac{\mu_0Ir}{2\pi a^2}$, growing linearly from zero.
Outside ($r \geq a$): $B = \frac{\mu_0I}{2\pi r}$, falling as $\frac{1}{r}$. The two agree at $r = a$, where B is greatest.
Answer: (B). | ||||
| 37 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Identical spheres in contact share the charge equally: each has $2 \times 10^{-8}$ C.
$F = \frac{kq^2}{r^2} \Rightarrow r^2 = \frac{9\times10^9 \times (2\times10^{-8})^2}{9\times10^{-3}} = \frac{3.6\times10^{-6}}{9\times10^{-3}} = 4\times10^{-4}\ \text{m}^2$.
$r = 2\times10^{-2}$ m $= 2$ cm.
Answer: (B) 2 cm. | ||||
| 38 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Configuration (1): four pairs at distance $a$ (sides) and two at $a\sqrt{2}$ (diagonals):
$U_1 = \frac{Kq_0^2}{a}\left(4 + \frac{2}{\sqrt{2}}\right) = \frac{Kq_0^2}{a}(4 + \sqrt{2})$.
Configuration (2): the mid-points form a square of side $\frac{a}{\sqrt{2}}$ whose diagonals are $a$: four pairs at $\frac{a}{\sqrt{2}}$ and two at $a$:
$U_2 = \frac{Kq_0^2}{a}(4\sqrt{2} + 2)$.
$U_2 - U_1 = \frac{Kq_0^2}{a}(4\sqrt{2} + 2 - 4 - \sqrt{2}) = \frac{Kq_0^2}{a}(3\sqrt{2} - 2)$.
Answer: (B). | ||||
| 39 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The rotating ring of charges is a current loop. In one period $T = \frac{2\pi}{\omega}$ all N charges pass any cross-section of the ring once, so the (time-averaged) current is $I = \frac{Nq}{T} = \frac{Nq\omega}{2\pi}$.
Loop A goes round the ring at one place, so the ring's current passes once through it: $I_A = \frac{Nq\omega}{2\pi}$.
Loop B is large and, as the figure shows, the ring crosses the surface it bounds twice, once in each direction (equivalently, it encloses the whole circle), so the net current through it is $I_B = 0$.
$I_A - I_B = \frac{N}{2\pi}q\omega$.
Answer: (D). | ||||
| 40 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves In order of increasing wavelength the spectrum runs: X-rays (about $10^{-10}$ m), ultraviolet ($10^{-8}$ m), infrared ($10^{-5}$ m), microwaves ($10^{-2}$ m).
So $\lambda_4 < \lambda_2 < \lambda_3 < \lambda_1$.
Answer: (D). | ||||
| 41 | 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 The landscape is $20$ km $\times$ $20$ km and its image is $2$ cm $\times$ $2$ cm, so the linear magnification is $m = \frac{2\times10^{-2}}{2\times10^{4}} = 10^{-6}$.
The object is very far away, so the image forms at the focal plane: $v \approx f$ and $m = \frac{f}{u}$.
$f = m\,u = 10^{-6} \times 18\times10^{3}$ m $= 1.8\times10^{-2}$ m $= 1.8$ cm.
Answer: (C) 1.8 cm. | ||||
| 43 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics After $P_1$ and $P_2$ each beam has intensity $\frac{I_0}{2}$, polarised at right angles to the other, so they cannot interfere yet.
$P_3$ at $45^\circ$ to both passes $\frac{I_0}{2}\cos^245^\circ = \frac{I_0}{4}$ of each, now with the same polarisation.
Path difference $\frac{\lambda}{3}$ gives phase difference $\phi = \frac{2\pi}{3}$, $\cos\phi = -\frac{1}{2}$.
$I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi = \frac{I_0}{4} + \frac{I_0}{4} - \frac{I_0}{4} = \frac{I_0}{4}$.
Answer: (D) $\frac{I_0}{4}$. | ||||
| 44 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | StatementsEasy | |
A easy statements question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation: $eV_0 = h\nu - \phi_0$, so $V_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}$.
(A) True: a straight line.
(B) False: the slope is $\frac{h}{e}$, not $\frac{\phi_0}{h}$.
(C) True: the slope gives h (as $h = e \times$ slope).
(D) False: finding h from the slope needs the electron's charge e.
(E) True: the intercept on the $V_0$ axis is $-\frac{\phi_0}{e}$, so $\phi_0 = e \times$ (intercept) without h (equally, the threshold frequency read off the $\nu$ axis).
Answer: (A) (A), (C) and (E) only. | ||||
| 45 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | StatementsEasy | |
A easy statements question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The first gate is an OR gate: its output is $X + Y$. It feeds both inputs of the second gate (drawn with its inputs joined), which then acts as a NOT gate.
Output $= \overline{X + Y}$ (NOR). It is 1 only for $X = Y = 0$ and 0 for the other three cases.
Low output for (B), (C) and (D).
Answer: (C) (B), (C) and (D) only. | ||||
| 46 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion Conical pendulum: the horizontal component of tension gives the centripetal force, $T\sin\theta = M\omega^2R$, with $R = L\sin\theta$.
So $T = M\omega^2L$.
$\omega = 2\pi f = 2\pi \times \frac{3}{\pi} = 6$ rad/s, so $T = 36\,ML$.
Answer: 36. | ||||
| 47 | GravitationAcceleration due to gravity | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g = \frac{GM}{R^2}$. With the same M and radius $\frac{R}{3}$: $g' = \frac{GM}{(R/3)^2} = 9\frac{GM}{R^2} = 9g$.
Answer: 9. | ||||
| 48 | 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 $\Delta p = B\frac{|\Delta V|}{V} = 2.15\times10^9 \times 0.002 = 4.3\times10^6$ N m$^{-2}$ $= 43\times10^5$ N m$^{-2}$.
Answer: 43. | ||||
| 49 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Period of circular motion in a field B: $T = \frac{2\pi m}{eB}$, so $B = \frac{2\pi m}{eT}$.
Inside a long solenoid $B = \mu_0nI$. Hence $n = \frac{2\pi m}{eT\mu_0I} = \frac{2\pi \times 9\times10^{-31}}{1.6\times10^{-19}\times75\times10^{-9}\times4\pi\times10^{-7}\times1.5}$.
$n = \frac{18\times10^{-31}}{2\times1.6\times75\times1.5\times10^{-35}} = \frac{18\times10^{-31}}{360\times10^{-35}} = 250$ turns per metre.
Answer: 250. | ||||
| 50 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Power $P = N\frac{hc}{\lambda}$, so $\frac{P_1}{P_2} = \frac{N_1/\lambda_1}{N_2/\lambda_2} = 2$.
$N_2 = \frac{N_1\lambda_2}{2\lambda_1} = \frac{2\times10^{15}\times300}{2\times600} = 5\times10^{14}$ per second.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Some Basic Concepts of ChemistryEmpirical and molecular formula | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles in 100 g: C $\frac{54.2}{12} = 4.52$; H $\frac{9.2}{1} = 9.2$; O $\frac{36.6}{16} = 2.29$.
Divide by 2.29: C 1.97, H 4.02, O 1, i.e. $\mathrm{C_2H_4O}$ (empirical mass 44).
$n = \frac{132}{44} = 3$, so the molecular formula is $\mathrm{C_6H_{12}O_3}$.
Answer: (A) $\mathrm{C_6H_{12}O_3}$. | ||||
| 52 | 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 the principal quantum number n: $E_n = -\frac{13.6}{n^2}$ eV.
The n = 3 orbitals ($3p_x$, $3d_{x^2-y^2}$, $3d_{z^2}$) are equal in energy and lower than the n = 4 orbitals (4s, $4p_z$).
Answer: (C) (B), (C) and (D) only. | ||||
| 53 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics The temperature rise is (heat released) / (heat capacity of the mixture), i.e. proportional to the moles neutralised divided by the total volume, with the enthalpy of neutralisation per mole.
(A) 20 mmol of a strong acid–strong base in 70 mL: $\frac{20}{70} = 0.29$ mmol/mL.
(B) 30 mmol in 60 mL (0.50 mmol/mL), but acetic acid is weak: part of the heat goes into its ionisation, so less heat per mole.
(C) 25 mmol in 70 mL (0.36 mmol/mL), weak acid again.
(D) 30 mmol of strong acid–strong base in 60 mL: 0.50 mmol/mL with the full $-57$ kJ mol$^{-1}$.
The largest rise is for (D).
Answer: (D) 30 mL HCl and 30 mL NaOH. | ||||
| 54 | ThermodynamicsHess law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics As enthalpy changes: (1) $\Delta H_1 = -2x$; (2) $\Delta H_2 = -y$.
$\mathrm{S} + \mathrm{O_2} \to \mathrm{SO_2}$ is (1) $-$ (2): $\Delta H_f(\mathrm{SO_2}) = -2x - (-y) = y - 2x$ kcal.
Answer: (A) $y - 2x$ kcal. | ||||
| 55 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Starting from $\mathrm{H_2}$ and $\mathrm{I_2}$, both reactants are used up equally and $\mathrm{HI}$ forms from zero, until all three concentrations become constant at equilibrium.
Graph (B) shows exactly this: $\mathrm{H_2}$ and $\mathrm{I_2}$ fall and level off, HI rises from zero and levels off.
(A) has HI falling and $\mathrm{H_2}$ rising together with $\mathrm{I_2}$ falling, (C) has $\mathrm{I_2}$ rising while $\mathrm{H_2}$ falls, and (D) has all three rising: none is consistent with the stoichiometry.
Answer: (B). | ||||
| 57 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | StatementsEasy | |
A easy statements question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Statement I: for a first order reaction $t_{1/2} = \frac{0.693}{k}$, independent of $[\mathrm{R}]_0$, so the plot is a horizontal line. True.
Statement II: $k = \frac{2.303}{t}\log\frac{[\mathrm{R}]_0}{[\mathrm{R}]}$, so $\log\frac{[\mathrm{R}]}{[\mathrm{R}]_0} = -\frac{k}{2.303}t$: a straight line through the origin with slope $-\frac{k}{2.303}$, going down. The graph drawn rises with slope $+\frac{k}{2.303}$, so it is not valid. False.
Answer: (C) Statement I is true but Statement II is false. | ||||
| 58 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | StatementsEasy | |
A easy statements question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Statement I: both O–O bonds in ozone are equal (128 pm), longer than O=O (121 pm) and shorter than O–O (148 pm). True.
Statement II: the intermediate, equal bond lengths are due to resonance (each bond has bond order 1.5), not solely to lone pair–lone pair repulsion. Statement II also misstates the fact (\"smaller than that of a double bond\"). False.
Answer: (C) Statement I is true but Statement II is false. | ||||
| 59 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter NiS and CoS (black, group IV) do not dissolve in dilute HCl and are dissolved in aqua regia.
$\mathrm{Co^{2+}}$ with $\mathrm{KNO_2}$ in acetic acid gives the yellow precipitate of potassium cobaltinitrite, $\mathrm{K_3[Co(NO_2)_6]}$.
$\mathrm{Ni^{2+}}$ gives no such precipitate (it is identified by the red dimethylglyoxime complex); MnS and ZnS dissolve in dilute HCl and need no aqua regia.
So A is CoS.
Answer: (B) CoS. | ||||
| 60 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[Ni(H_2O)_6]^{2+}}$ is green. Replacing water by the stronger-field ligand en step by step raises $\Delta_o$, so the absorbed light moves to shorter wavelength and the colour seen changes:
$\mathrm{[Ni(H_2O)_4(en)]^{2+}}$ pale blue $\to$ $\mathrm{[Ni(H_2O)_2(en)_2]^{2+}}$ blue $\to$ $\mathrm{[Ni(en)_3]^{2+}}$ violet.
Answer: (A) Green $\to$ Pale Blue $\to$ Blue $\to$ Violet. | ||||
| 61 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds For a $d^4$ ion, the fourth electron enters $e_g$ (giving $t_{2g}^3e_g^1$) only when the splitting $\Delta_o$ is smaller than the pairing energy, which is the case with weak field ligands.
With four unpaired electrons the complex is high spin. (Strong field ligands would give $t_{2g}^4e_g^0$, low spin.)
Answer: (A) weak field ligand, high spin complex. | ||||
| 62 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Matching listEasy | |
A easy matching list question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mu = \sqrt{n(n + 2)}$ B.M. for n unpaired electrons.
$\mathrm{Ti^{3+}}$ $3d^1$: n = 1, $\mu = 1.73$ → (III).
$\mathrm{V^{2+}}$ $3d^3$: n = 3, $\mu = 3.87$ → (I).
$\mathrm{Ni^{2+}}$ $3d^8$: n = 2, $\mu = 2.84$ → (IV).
$\mathrm{Sc^{3+}}$ $3d^0$: n = 0, $\mu = 0$ → (II).
Answer: (C) A→III, B→I, C→IV, D→II. | ||||
| 64 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The ionisation energies rise steadily for the first three electrons and then jump about sevenfold from 3658 to 25024 kJ/mol: the fourth electron is taken from a filled inner shell.
So the element has three valence electrons ($ns^2np^1$): Group 13 (the values are those of boron).
Answer: (A) Group 13. | ||||
| 65 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure The structure is $\mathrm{CH_3{-}CO{-}CH{=}CH{-}CH_2{-}C{\equiv}C{-}CH_2{-}CH{=}CH{-}C{\equiv}N}$.
sp carbons: the two carbons of $\mathrm{C{\equiv}C}$ and the carbon of $\mathrm{C{\equiv}N}$: 3.
$\mathrm{sp^2}$ carbons: the carbonyl carbon and the four carbons of the two C=C double bonds: 5.
(The $\mathrm{CH_3}$ and the two $\mathrm{CH_2}$ carbons are $\mathrm{sp^3}$.)
Answer: (A) 3 and 5. | ||||
| 66 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | StatementsEasy | |
A easy statements question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A) True: both donate electrons to the ring by resonance (+R) and activate it.
(B) False: $-\mathrm{OH}$ is ortho/para directing.
(C) True: $-\mathrm{CN}$ and $-\mathrm{SO_3H}$ withdraw electrons and direct meta.
(D) True: activating groups increase electron density mainly at the ortho and para positions.
(E) False: halogens are deactivating (−I outweighs +R), though ortho/para directing.
Answer: (B) (A), (C) and (D) only. | ||||
| 67 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The aryl C–I and C–Br bonds have partial double-bond character and are not substituted by AgCN under these conditions; only the benzylic C–Cl reacts.
AgCN is largely covalent, so the cyanide attacks through nitrogen (the carbon end is bound to silver), and the product is the isocyanide $\mathrm{ArCH_2{-}N{\equiv}C}$.
Answer: (B) the benzyl isocyanide with I and Br unchanged. | ||||
| 68 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Matching listEasy | |
A easy matching list question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) $\mathrm{RCN} \xrightarrow{\text{SnCl}_2/\text{HCl};\ \text{H}_3\text{O}^+} \mathrm{RCHO}$: Stephen reaction → (IV).
(B) $\mathrm{C_6H_5COCl} \xrightarrow{\text{H}_2,\ \text{Pd-BaSO}_4} \mathrm{C_6H_5CHO}$: Rosenmund reduction → (III).
(C) Toluene $\xrightarrow{\text{CrO}_2\text{Cl}_2,\ \text{CS}_2;\ \text{H}_3\text{O}^+}$ benzaldehyde: Etard reaction → (I).
(D) Benzene $\xrightarrow{\text{CO, HCl; anhyd. AlCl}_3/\text{CuCl}}$ benzaldehyde: Gattermann–Koch reaction → (II).
Answer: (A) A→IV, B→III, C→I, D→II. | ||||
| 69 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Direct bromination of aniline gives 2,4,6-tribromoaniline, so the ring must be moderated and the para position blocked.
$\mathrm{H_2SO_4}$ (with heating) sulphonates aniline at the para position (sulphanilic acid), blocking it.
$\mathrm{Ac_2O}$ acetylates the $-\mathrm{NH_2}$, moderating its activation so that only one bromine enters.
$\mathrm{Br_2}$ then substitutes ortho to the acetamido group (para is blocked).
Heating with water/acid ($\mathrm{H_2O}$, $\Delta$) removes the $-\mathrm{SO_3H}$ group (desulphonation) and hydrolyses the amide; NaOH frees the amine: o-bromoaniline.
(A) gives mainly p-bromoaniline (para is free); (C) leaves the $-\mathrm{SO_3H}$ group on the ring (no desulphonation step); (D) brominates the free amine at all three positions.
Answer: (B). | ||||
| 71 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $i = \frac{\text{normal molar mass}}{\text{observed molar mass}} = \frac{164}{65.6} = 2.5$.
$\mathrm{MX_2 \to M^{2+} + 2X^-}$ gives 3 particles: $i = 1 + 2\alpha$, so $\alpha = \frac{2.5 - 1}{2} = 0.75$.
Degree of ionisation $= 75\%$.
Answer: 75. | ||||
| 72 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics With $k_i = A_ie^{-E_i/RT}$: $k = \sqrt{\frac{A_1A_3}{A_2}}\,e^{-(E_1 + E_3 - E_2)/2RT}$.
$E_a = \frac{E_1 + E_3 - E_2}{2} = \frac{60 + 10 - 30}{2} = 20$ kJ mol$^{-1}$.
Answer: 20. | ||||
| 73 | 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 AgBr $= 188$ g mol$^{-1}$; mass of Br in 0.15 g AgBr $= \frac{80}{188}\times0.15 = 0.06383$ g.
% Br $= \frac{0.06383}{0.25}\times100 = 25.53\%$ $= 255.3\times10^{-1}\%$.
To the nearest integer: 255.
Answer: 255. | ||||
| 74 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Carbon: $0.90\times80 = 72$ g, i.e. 6 C atoms; hydrogen: $8$ g, 8 H atoms. X is $\mathrm{C_6H_8}$.
Degree of unsaturation $= \frac{2C + 2 - H}{2} = \frac{12 + 2 - 8}{2} = 3$.
Answer: 3. | ||||
| 75 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques 5-Phenylpent-4-en-2-ol is $\mathrm{CH_3{-}CH(OH){-}CH_2{-}CH{=}CH{-}C_6H_5}$.
C-2 is a stereocentre (H, OH, $\mathrm{CH_3}$, $\mathrm{CH_2CH{=}CHPh}$): 2 configurations.
The C4=C5 double bond has different groups on each carbon: E and Z.
Stereoisomers $= 2\times2 = 4$.
Answer: 4. | ||||
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