JEE (Main) · 2025
JEE (Main) 2025 — 8 April, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
25 Q
- Integral Calculus3
- Sets, Relations and Functions2
- Matrices and Determinants2
8 easy · 15 medium · 2 hard
Physics
24 Q
- Electrostatics4
- Motion of System of Particles and Rigid Body3
- Oscillations and Waves3
18 easy · 6 medium · 0 hard
Chemistry
25 Q
- Coordination Compounds3
- Organic Chemistry: Some Basic Principles and Techniques3
- Structure of Atom2
13 easy · 12 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 FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | StatementsEasy | |
A easy statements question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Count the pairs.
$\max\{x, y\} = 3$: both entries in $\{0, 1, 2, 3\}$ and not both in $\{0, 1, 2\}$: $4^2 - 3^2 = 7$ pairs.
$\max\{x, y\} = 4$: both entries in $\{0, \dots, 4\}$ and not both in $\{0, \dots, 3\}$: $5^2 - 4^2 = 9$ pairs.
So $R$ has $7 + 9 = 16$ elements, not 18: $(S_1)$ is false.
Symmetric: $\max\{x, y\} = \max\{y, x\}$, so $(x, y) \in R \Rightarrow (y, x) \in R$. Yes.
Reflexive: $(0, 0) \notin R$ since $\max = 0$. No.
Transitive: $(0, 3) \in R$ and $(3, 0) \in R$ but $(0, 0) \notin R$. No.
So $(S_2)$ is true.
Answer: (D) only $(S_2)$ is true. | ||||
| 2 | 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 First equation: put $t = |x - 2| \ge 0$: $t^2 + t - 2 = 0$, so $(t + 2)(t - 1) = 0$ and $t = 1$ (as $t \ge 0$). Then $x = 3$ or $x = 1$; squares $9 + 1 = 10$.
Second equation:
For $x \ge 3$: $x^2 - 2x + 6 - 5 = 0$, i.e. $(x - 1)^2 = 0$, $x = 1$, which is not $\ge 3$: no root.
For $x < 3$: $x^2 + 2(x - 3) - 5 = 0$, i.e. $x^2 + 2x - 11 = 0$, $x = -1 \pm 2\sqrt3$. Both ($\approx 2.46$ and $\approx -4.46$) are less than 3, so both are roots.
Sum of their squares $= (x_1 + x_2)^2 - 2x_1x_2 = 4 + 22 = 26$.
Total: $10 + 26 = 36$.
Answer: (D) $36$. | ||||
| 3 | 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 Multiply numerator and denominator by the conjugate $\cos\theta + 3i\sin\theta$:
$\frac{(2\cos\theta + i\sin\theta)(\cos\theta + 3i\sin\theta)}{\cos^2\theta + 9\sin^2\theta}$, whose real part is $\frac{2\cos^2\theta - 3\sin^2\theta}{\cos^2\theta + 9\sin^2\theta}$.
The condition becomes $\cos^2\theta + 9\sin^2\theta + 20\cos^2\theta - 30\sin^2\theta = 0$, i.e. $21\cos^2\theta = 21\sin^2\theta$, so $\tan^2\theta = 1$.
In $[0, 2\pi]$: $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
$\sum\theta^2 = \frac{\pi^2}{16}(1 + 9 + 25 + 49) = \frac{84}{16}\pi^2 = \frac{21}{4}\pi^2$.
Answer: (D) $\frac{21}{4}\pi^2$. | ||||
| 4 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$:
$\det A = \begin{vmatrix} 2 & 2 + p & 2 + p + q \\ 0 & 2 & 4 + p \\ 0 & 6 & 14 + 3p \end{vmatrix} = 2[2(14 + 3p) - 6(4 + p)] = 2 \cdot 4 = 8$.
$\det(3A) = 3^3 \cdot 8 = 2^3 \cdot 3^3$.
For an $n \times n$ matrix $B$, $\det(\mathrm{adj}\,B) = (\det B)^{n-1}$, so $\det(\mathrm{adj}(\mathrm{adj}\,B)) = (\det B)^{(n-1)^2} = (\det B)^4$ for $n = 3$.
$\det(\mathrm{adj}(\mathrm{adj}(3A))) = (2^3 3^3)^4 = 2^{12} \cdot 3^{12}$, so $m + n = 24$.
Answer: (C) $24$. | ||||
| 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 The product gives three equations:
$4 - a - 2b = 0$, $64 - a - 14b = 0$, $52 + 2a - 8b = 0$.
Subtracting the first from the second: $60 - 12b = 0$, so $b = 5$ and $a = 4 - 10 = -6$. The third checks: $52 - 12 - 40 = 0$.
$\alpha$ is a non-real cube root of unity ($\alpha^3 = 1$, $\alpha^2 + \alpha + 1 = 0$), and $m, n$ are taken real.
$\frac{4}{\alpha^4} = \frac{4}{\alpha} = 4\alpha^2$, $\frac{m}{\alpha^{-6}} = m\alpha^6 = m$, $\frac{n}{\alpha^5} = \frac{n}{\alpha^2} = n\alpha$.
So $4\alpha^2 + m + n\alpha = 3$. With $\alpha^2 = -1 - \alpha$: $(m - 7) + (n - 4)\alpha = 0$.
Since $\alpha$ is not real, $m = 7$ and $n = 4$, so $m + n = 11$.
Answer: (B) $11$. | ||||
| 6 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let $S = \sum \frac{1}{k^4} = \frac{\pi^4}{90}$.
The even terms: $\beta = \sum \frac{1}{(2k)^4} = \frac{1}{16}S$.
The odd terms: $\alpha = S - \beta = \frac{15}{16}S$.
$\frac{\alpha}{\beta} = 15$.
Answer: (B) $15$. | ||||
| 7 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Any 3 of the 12 points form a triangle unless all three come from the 5 collinear points.
Number $= \binom{12}{3} - \binom{5}{3} = 220 - 10 = 210$.
Answer: (A) $210$. | ||||
| 8 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom{1016}{r}\, 5^{\frac{1016 - r}{2}}\, 7^{\frac{r}{8}}$.
It is an integer when $8 \mid r$ and $2 \mid (1016 - r)$. Since 1016 is even, the second condition follows from the first.
$r = 0, 8, 16, \dots, 1016$: that is $\frac{1016}{8} + 1 = 128$ values.
Answer: (C) $128$. | ||||
| 9 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(A \cap B) = P(A) - P(A \cap \overline{B}) = 0.7 - 0.5 = 0.2$.
$P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B}) = 0.7 + 0.6 - 0.5 = 0.8$.
$B \cap (A \cup \overline{B}) = A \cap B$, so
$P(B \mid A \cup \overline{B}) = \frac{0.2}{0.8} = \frac14$.
Answer: (B) $\frac14$. | ||||
| 10 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines After the rotation the line makes the angle $\alpha - \frac{\alpha}{2} = \frac{\alpha}{2}$ with the $x$-axis, so $\tan\frac{\alpha}{2} = 2 - \sqrt3$, giving $\frac{\alpha}{2} = 15^\circ$ and $\alpha = 30^\circ$, $\tan^2\alpha = \frac13$.
New line: $y = (2 - \sqrt3)(x - a)$, i.e. $(2 - \sqrt3)x - y - (2 - \sqrt3)a = 0$.
Distance from the origin: $\frac{(2 - \sqrt3)|a|}{\sqrt{1 + (2 - \sqrt3)^2}}$. Here $1 + (2 - \sqrt3)^2 = 8 - 4\sqrt3 = 4(2 - \sqrt3)$, so the distance is $\frac{|a|\sqrt{2 - \sqrt3}}{2}$.
Setting it equal to $\frac{1}{\sqrt2}$: $a^2(2 - \sqrt3) = 2$, so $a^2 = \frac{2}{2 - \sqrt3} = 2(2 + \sqrt3) = 4 + 2\sqrt3$.
$3a^2\tan^2\alpha - 2\sqrt3 = a^2 - 2\sqrt3 = 4$.
Answer: (D) $4$. | ||||
| 11 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines The first diagonal passes through the origin O, so it is OB; the other diagonal AC is $(\sqrt3 - 1)x - (\sqrt3 + 1)y + 8\sqrt3 = 0$.
The diagonals of a square bisect each other at right angles, so the distance from O to AC is half a diagonal, $\frac{a\sqrt2}{2} = \frac{a}{\sqrt2}$.
Distance $= \frac{8\sqrt3}{\sqrt{(\sqrt3 - 1)^2 + (\sqrt3 + 1)^2}} = \frac{8\sqrt3}{\sqrt8} = 2\sqrt6$.
$\frac{a}{\sqrt2} = 2\sqrt6 \Rightarrow a = 4\sqrt3$, $a^2 = 48$.
Answer: (C) $48$. | ||||
| 12 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Circle: centre $C = (1, 2)$, $r = \sqrt{1 + 4 + 11} = 4$.
Through $C$: $3 + 4p = 4$, so $p = \frac14$ and the ellipse is $\frac{x^2}{4/3} + \frac{y^2}{16} = 1$, with major axis along $y$ (semi-major axis 4).
$e^2 = 1 - \frac{4/3}{16} = \frac{11}{12}$.
For a point $(x, y)$ on it the focal distances are $4 \pm ey$; at $C$, $y = 2$: $f_1f_2 = 16 - 4e^2 = 16 - \frac{11}{3} = \frac{37}{3}$.
$6f_1f_2 - r = 74 - 4 = 70$.
Answer: (B) $70$. | ||||
| 13 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctHard | |
A hard single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions The angles are in radians.
First term: $2 \in (\frac{\pi}{2}, \pi)$, so $\sec 2 < 0$ and $\sqrt{1 + \tan^2 2} = |\sec 2| = -\sec 2$.
$\frac{-\sec 2 - 1}{\tan 2} = \frac{-1 - \cos 2}{\sin 2} = -\frac{2\cos^2 1}{2\sin 1\cos 1} = -\cot 1$.
$\cot^{-1}(-\cot 1) = \pi - \cot^{-1}(\cot 1) = \pi - 1$ (as $1 \in (0, \pi)$).
Second term: $\frac12 \in (0, \frac{\pi}{2})$, so $\sqrt{1 + \tan^2\frac12} = \sec\frac12$ and
$\frac{\sec\frac12 + 1}{\tan\frac12} = \frac{1 + \cos\frac12}{\sin\frac12} = \cot\frac14$, so $\cot^{-1}(\cot\frac14) = \frac14$.
Value $= \pi - 1 - \frac14 = \pi - \frac54$.
Answer: (A) $\pi - \frac54$. | ||||
| 14 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra A vector in the plane of $\vec{a}, \vec{b}$ perpendicular to $\vec{a}$ is the part of $\vec{b}$ perpendicular to $\vec{a}$:
$\vec{b} - \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\vec{a}$, with $\vec{a}\cdot\vec{b} = 2 + 2 - 1 = 3$, $|\vec{a}|^2 = 6$.
$\vec{b} - \frac12\vec{a} = \frac32\hat{i} + 0\hat{j} - \frac32\hat{k} \parallel \hat{i} - \hat{k}$.
So $\hat{c} = \pm\frac{1}{\sqrt2}(\hat{i} - \hat{k})$; option (C) is $\frac{1}{\sqrt2}(-\hat{i} + \hat{k})$.
Check against the normal of the plane, $\vec{a}\times\vec{b} = (-3, 3, -3)$: $(-1, 0, 1)$ is perpendicular to both $\vec{a}$ and this normal, so it lies in the plane and is perpendicular to $\vec{a}$. The vectors in (A), (B) and (D) are perpendicular to $\vec{a}$ but have dot products $-9$, $9$ and $9$ with the normal, so they are not in the plane.
Answer: (C). | ||||
| 15 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $\vec{d_1} = (2, 3, 4)$, $\vec{d_2} = (3, 4, 5)$, $\vec{d_1}\times\vec{d_2} = (15 - 16, 12 - 10, 8 - 9) = (-1, 2, -1)$, of length $\sqrt6$.
Joining vector: $(\lambda - 1, 2, 2)$.
SD $= \frac{|-(\lambda - 1) + 4 - 2|}{\sqrt6} = \frac{|3 - \lambda|}{\sqrt6} = \frac{1}{\sqrt6}$, so $\lambda = 2$ or $4$.
Circle through $(0, 0)$, $(2, 4)$, $(4, 2)$: $x^2 + y^2 + 2gx + 2fy = 0$.
$20 + 4g + 8f = 0$ and $20 + 8g + 4f = 0$ give $g = f = -\frac53$.
$r = \sqrt{g^2 + f^2} = \frac{5\sqrt2}{3}$.
Answer: (D) $\frac{5\sqrt2}{3}$. | ||||
| 16 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | StatementsMedium | |
A medium statements question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Statement I: $\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$ and $\log_e\sqrt{\frac{1 + x}{1 - x}} = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$
Their sum minus $2x$ is $\frac{2x^5}{5} + O(x^7)$, so the limit is $\frac25$. True.
Statement II: a $1^\infty$ form; the limit is $e^{L}$ with $L = \lim_{x\to1}\frac{2}{1 - x}(x - 1) = -2$. So the limit is $e^{-2} = \frac{1}{e^2}$. True.
Answer: (A) Both statements are true. | ||||
| 17 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = \frac13 - \frac{3}{x^2} = \frac{x^2 - 9}{3x^2}$.
$f' > 0$ for $|x| > 3$: increasing on $(-\infty, -3) \cup (3, \infty)$, so $\alpha_1 = -3$, $\alpha_2 = 3$.
$f' < 0$ for $0 < |x| < 3$: decreasing on $(-3, 0) \cup (0, 3)$, so $\alpha_3 = -3$, $\alpha_4 = 0$, $\alpha_5 = 3$.
$\sum\alpha_i^2 = 9 + 9 + 9 + 0 + 9 = 36$.
Answer: (B) $36$. | ||||
| 18 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals In $I_1$ put $t = 2x$: $I_1 = \int_{-1}^{2} t\, f\big(t(1 - t)\big)\,\frac{dt}{2}$.
Let $J = \int_{-1}^{2} t\, f(t(1 - t))\,dt$. Replace $t$ by $(-1 + 2) - t = 1 - t$: $t(1 - t)$ is unchanged, so $J = \int_{-1}^{2}(1 - t) f(t(1 - t))\,dt$.
Adding: $2J = \int_{-1}^{2} f(t(1 - t))\,dt = I_2$, so $J = \frac{I_2}{2}$ and $I_1 = \frac{J}{2} = \frac{I_2}{4}$.
$\frac{I_2}{I_1} = 4$ (the positivity of $f$ makes $I_2 \ne 0$).
Answer: (D) $4$. | ||||
| 19 | 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 By parts, an antiderivative of $\pi^2x\sin\pi x$ is $F(x) = -\pi x\cos\pi x + \sin\pi x$ (check: $F' = -\pi\cos\pi x + \pi^2 x\sin\pi x + \pi\cos\pi x$).
$x\sin\pi x$ is even and $\ge 0$ on $[-1, 1]$; on $[1, \frac32]$ it is negative ($x > 0$, $\sin\pi x < 0$).
$\int_{-1}^{1} = 2[F(1) - F(0)] = 2(\pi - 0) = 2\pi$.
$\int_{1}^{3/2}|\dots| = -[F(\tfrac32) - F(1)] = -[(0 - 1) - \pi] = 1 + \pi$.
Total $= 3\pi + 1$.
Answer: (A) $1 + 3\pi$. | ||||
| 20 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $f(f(x)) = x - 2$, so $g(f(f(x))) = e^{x - 2}$ and
$\frac{dy}{dx} + \frac{y}{\sqrt{x}} = e^{x - 2 - 2\sqrt{x}}$.
Integrating factor: $e^{\int x^{-1/2}dx} = e^{2\sqrt{x}}$.
$\frac{d}{dx}\left(y e^{2\sqrt{x}}\right) = e^{x - 2}$, so $y e^{2\sqrt{x}} = e^{x - 2} + C$.
$y(0) = 0$: $0 = e^{-2} + C$, $C = -e^{-2}$.
At $x = 1$: $y(1)e^2 = e^{-1} - e^{-2}$, so $y(1) = \frac{e^{-1} - e^{-2}}{e^2} = \frac{e - 1}{e^4}$.
Answer: (A) $\frac{e - 1}{e^4}$. | ||||
| 21 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Domain of $f$: $-1 \le \frac{4x + 5}{3x - 7} \le 1$.
$\frac{4x + 5}{3x - 7} \le 1 \iff \frac{x + 12}{3x - 7} \le 0 \iff -12 \le x < \frac73$.
$\frac{4x + 5}{3x - 7} \ge -1 \iff \frac{7x - 2}{3x - 7} \ge 0 \iff x \le \frac27$ or $x > \frac73$.
Together: $[-12, \frac27]$, so $\alpha + \beta = -12 + \frac27 = -\frac{82}{7}$ and $7(\alpha + \beta) = -82$.
Domain of $g$: $2x + 5 > 0$ and $2 - 6\log_{27}(2x + 5) > 0$, i.e. $\log_{27}(2x + 5) < \frac13$, $2x + 5 < 3$, $x < -1$. With $x > -\frac52$: $(\gamma, \delta) = (-\frac52, -1)$, $4(\gamma + \delta) = -14$.
$|-82 - 14| = 96$.
Answer: 96. | ||||
| 22 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The last two digits are $1919^{1919} \bmod 100 = 19^{1919} \bmod 100$.
$19^{1919} = 19\cdot(361)^{959} = 19(1 + 360)^{959}$. Modulo 100, $(1 + 360)^{959} \equiv 1 + 959\cdot360 \pmod{100}$ (higher terms contain $360^2$, a multiple of 100).
$959 \cdot 360 = 345240 \equiv 40$, so $(361)^{959} \equiv 41$ and $19^{1919} \equiv 19 \cdot 41 = 779 \equiv 79 \pmod{100}$.
Check another way: $19^2 \equiv 61$, $19^4 \equiv 21$, $19^5 \equiv 99 \equiv -1$, so $19^{10} \equiv 1$ and $19^{1919} \equiv 19^9 = 19^5\cdot19^4 \equiv -21 \equiv 79$.
Last two digits 7 and 9; product $= 63$.
Answer: 63. | ||||
| 23 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent to $y^2 = 9x$ at $(4, 6)$: $6y = \frac92(x + 4)$, i.e. $3x - 4y + 12 = 0$.
The circle touches the parabola at $(4, 6)$, so its centre lies on the normal there, which has direction $(3, -4)$: centre $= (4 + 3t, 6 - 4t)$ and $r = 5|t|$.
It touches the $x$-axis at $(a, 0)$, so the centre is $(a, \pm r)$ with $a = 4 + 3t < 0$, so $t < -\frac43$; then $6 - 4t > 0$, so the centre is above the axis: $6 - 4t = r = -5t$, giving $t = -6$.
$r = 30$, centre $(-14, 30)$, $a = -14 < 0$. ✓
Answer: 30. | ||||
| 24 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $L_1: (-2 + t, 1 + t, t)$; $L_2: (3 + 5s, -s, 1 + s)$; $L_3: (-3u, 3 + 3u, 2 + u)$.
$L_1 \cap L_2$: $t = 1 + s$ and $1 + t = -s$ give $s = -1$, $t = 0$; $x$: $-2 = 3 - 5$ ✓. Point $P(-2, 1, 0)$.
$L_1 \cap L_3$: $t = 2 + u$ and $1 + t = 3 + 3u$ give $u = 0$, $t = 2$; $x$: $0 = 0$ ✓. Point $Q(0, 3, 2)$.
$L_2 \cap L_3$: $1 + s = 2 + u$ and $-s = 3 + 3u$ give $u = -1$, $s = 0$; $x$: $3 = 3$ ✓. Point $R(3, 0, 1)$.
$\vec{PQ} = (2, 2, 2)$, $\vec{PR} = (5, -1, 1)$, $\vec{PQ}\times\vec{PR} = (4, 8, -12)$, of squared length $16 + 64 + 144 = 224$.
$A = \frac12\sqrt{224}$, $A^2 = \frac{224}{4} = 56$.
Answer: 56. | ||||
| 25 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $0 \le 9x \le y^2$ means $x \ge 0$ and $y^2 \ge 9x$, i.e. $y \ge 3\sqrt{x}$ or $y \le -3\sqrt{x}$ (outside the parabola $y^2 = 9x$, right of the $y$-axis).
Upper part ($y \ge 3\sqrt{x}$, $y \ge 3x - 6$): it runs up without limit, so it is not bounded.
Lower part ($3x - 6 \le y \le -3\sqrt{x}$): bounded, from $x = 0$ to where $3x - 6 = -3\sqrt{x}$, i.e. $\sqrt{x} = 1$, $x = 1$.
$A = \int_0^1\left(-3\sqrt{x} - (3x - 6)\right)dx = \left[-2x^{3/2} - \frac{3x^2}{2} + 6x\right]_0^1 = -2 - \frac32 + 6 = \frac52$.
$6A = 15$.
Answer: 15. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement For $Q = X^aY^bZ^c$ the maximum fractional error is $|a|\frac{\Delta X}{X} + |b|\frac{\Delta Y}{Y} + |c|\frac{\Delta Z}{Z}$.
$= 2(0.1) + \frac32(0.2) + \frac25(0.5) = 0.2 + 0.3 + 0.2 = 0.7$.
Answer: (C) 0.7. | ||||
| 27 | 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 Acceleration $= \frac{6}{2} = 3$ m s$^{-2}$ along $+z$; the $x$ and $y$ components are unchanged.
$v_z = 3 \times \frac53 = 5$ m s$^{-1}$.
$\vec{v} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ m s$^{-1}$.
Answer: (C). | ||||
| 28 | 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 Mass $M = \lambda L$; $2\pi R = L$, so $R = \frac{L}{2\pi}$.
About a diameter, a ring has $I = \frac12MR^2 = \frac12\lambda L\cdot\frac{L^2}{4\pi^2} = \frac{\lambda L^3}{8\pi^2}$.
Answer: (B) $\frac{\lambda L^3}{8\pi^2}$. | ||||
| 29 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The block performs SHM about the natural length position ($x = 2$ m) with amplitude $A = 2 - 1 = 1$ m and $\omega = \sqrt{k/m} = \sqrt{200/2} = 10$ rad/s.
At distance $x$ from the wall, the displacement from equilibrium is $2 - x$, so
$v = \omega\sqrt{A^2 - (2 - x)^2} = 10\left[1 - (2 - x)^2\right]^{1/2}$ m/s.
(Energy check: $\frac12 kA^2 = \frac12 k(2 - x)^2 + \frac12 mv^2$ gives $v^2 = 100[1 - (2 - x)^2]$.)
Answer: (C). | ||||
| 30 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $H = \frac{u^2\sin^2\theta}{2g}$ and $T = \frac{2u\sin\theta}{g}$, so with the same $u$, $T \propto \sin\theta \propto \sqrt{H}$.
$\frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}} = \sqrt8 = 2\sqrt2$.
Answer: (C) $2\sqrt2 : 1$. | ||||
| 31 | Properties of Bulk MatterElasticity | 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 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 $Y = \frac{FL}{A\,\Delta L} = \frac{(50 \times 3\pi)(3)}{\pi(3 \times 10^{-3})^2(0.1 \times 10^{-3})}$
$= \frac{450\pi}{9\pi \times 10^{-10}} = 5 \times 10^{11}$ N m$^{-2}$.
So P = 5.
Answer: (A) 5. | ||||
| 32 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter All the potential energy of a mass $m$ of water becomes heat in that same water: $mgh = mc\,\Delta T$.
$\Delta T = \frac{gh}{c} = \frac{10 \times 200}{4200} = 0.476 \approx 0.48$ K.
Answer: (B) 0.48 K. | ||||
| 33 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics A sudden compression in an insulated container is adiabatic: $PV^\gamma$ = constant.
$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma = 8^{5/3} = (2^3)^{5/3} = 2^5 = 32$.
Answer: (A) 32. | ||||
| 34 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v = \sqrt{\frac{T}{\mu}}$ with $\mu = \rho\pi r^2$ (same material, so same $\rho$). With the same $T$, $v \propto \frac1r$.
$\frac{v_2}{v_1} = \frac{R}{R/2} = 2$.
Answer: (B) 2. | ||||
| 35 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Assertion and reasonEasy | |
A easy assertion and reason question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Inside a uniformly charged shell the field is zero, so the potential is the same everywhere inside and equal to its value on the surface, $\frac{kQ}{R}$. R is true.
Work done by an external agent in moving a test charge $q$ slowly between two points is $q(V_B - V_A) = 0$, whatever the path (the field is conservative). A is true.
A follows directly from R.
Answer: (A). | ||||
| 37 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | StatementsMedium | |
A medium statements question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance On a charged conductor the surface charge density is larger where the surface is more sharply curved (smaller radius of curvature).
From the figure: the top point (σ1) is a sharp tip, the sharpest; the bottom point (σ3) is more sharply curved than the circle drawn dotted inside; the side points σ2 and σ4 lie where the outline follows the dotted circle, and the shape is symmetric left to right.
So $\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$.
A: $\sigma_1 > \sigma_3$ and $\sigma_2 = \sigma_4$. True.
B: $\sigma_1 > \sigma_2$ and $\sigma_3 > \sigma_4$. True.
C: $\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$. True.
D: reverses the order. False.
E: the density is not uniform on an irregular conductor. False.
Answer: (C) A, B and C Only. | ||||
| 38 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In contact the spheres come to the same potential: $\frac{kq_1}{R} = \frac{kq_2}{3R}$, so $\frac{q_1}{q_2} = \frac13$ (treating the spheres as far apart after separation, as usual).
$\frac{\sigma_1}{\sigma_2} = \frac{q_1/4\pi R^2}{q_2/4\pi(3R)^2} = \frac13 \times 9 = 3$.
(Equivalently, equal potential means $\sigma r$ is the same, so $\sigma \propto \frac1r$.)
Answer: (C) 3. | ||||
| 39 | 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 The two paths are in parallel: $i_1$ (through ABC, resistance $r$) $= \frac{2I}{3}$ and $i_2$ (through ADC, resistance $2r$) $= \frac{I}{3}$.
Field at the centre from one side of length $a$ carrying $i$, at perpendicular distance $\frac{a}{2}$, with the ends subtending $45^\circ$ on each side:
$B = \frac{\mu_0 i}{4\pi(a/2)}(\sin45^\circ + \sin45^\circ) = \frac{\sqrt2\mu_0 i}{2\pi a}$.
Current A→B→C goes one way round the centre and A→D→C the other, so their fields are opposite:
$B_{net} = 2\cdot\frac{\sqrt2\mu_0}{2\pi a}(i_1 - i_2) = \frac{\sqrt2\mu_0}{\pi a}\cdot\frac{I}{3} = \frac{\sqrt2\mu_0 I}{3\pi a}$.
(The straight leads point along the diagonal AC through the centre, so they give no field there.)
Answer: (A). | ||||
| 40 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi} = \sqrt{16 + 4 + 2(4)(2)\cos120^\circ} = \sqrt{20 - 8} = 2\sqrt3$.
$\tan\delta = \frac{A_2\sin\phi}{A_1 + A_2\cos\phi} = \frac{2 \cdot \frac{\sqrt3}{2}}{4 - 1} = \frac{1}{\sqrt3}$, so $\delta = \frac{\pi}{6}$.
Answer: (A) $\left[2\sqrt3, \frac{\pi}{6}\right]$. | ||||
| 41 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Take the concave surface to have radius 30 cm and the convex surface 20 cm (the order in which they are named).
Glass meniscus: $\frac{1}{f_g} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{30}\right) = 0.5 \times \frac{1}{60} = \frac{1}{120}$ (converging, since the convex face is more curved).
Liquid filling the concave face: a plano-convex lens with radius 30 cm: $\frac{1}{f_l} = (1.3 - 1)\frac{1}{30} = \frac{1}{100}$.
Thin lenses in contact: $\frac1F = \frac{1}{120} + \frac{1}{100} = \frac{5 + 6}{600} = \frac{11}{600}$, so $F = \frac{600}{11}$ cm.
Answer: (A) $\frac{600}{11}$ cm. | ||||
| 42 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics One slit now passes only red light and the other only green light. Waves of different frequencies have no constant phase relation, so they are not coherent and cannot form a steady interference pattern. Each colour comes from only one slit, so neither colour interferes with itself either.
The screen shows only the overlapping single-slit illumination (red and green), with no interference fringes.
Answer: (C). | ||||
| 43 | 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 $\frac1F = \frac{1}{30} - \frac{1}{20} = -\frac{1}{60}$, so $F = -60$ cm (diverging).
$u = -20$ cm: $\frac1v = \frac1F + \frac1u = -\frac{1}{60} - \frac{1}{20} = -\frac{4}{60}$, so $v = -15$ cm.
The image is virtual, 15 cm from the lens on the object's side.
Answer: (C) 15. | ||||
| 44 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $R = R_0A^{1/3}$, so the volume $\frac43\pi R^3 = \frac43\pi R_0^3 A$ is proportional to A, as is the mass ($\approx A \times$ nucleon mass).
Density $= \frac{A m_N}{\frac43\pi R_0^3 A} = \frac{3m_N}{4\pi R_0^3}$, the same for all nuclei.
Answer: (D) Independent of A. | ||||
| 45 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Both diodes have their cathodes at the inputs (D1 at +5 V, D2 at ground) and their anodes at the output node, which is pulled up to +5 V through the resistor (a diode AND gate).
D2: the resistor pulls its anode toward +5 V while its cathode is at 0 V, so it is forward biased and conducts; an ideal diode then holds the output at 0 V.
D1: anode at 0 V, cathode at +5 V, so it is reverse biased and carries no current.
Current flows from the +5 V supply through the resistor and D2 to ground, and $V_{out} = 0$ V (AND of 1 and 0 is 0).
Answer: (B) 0 V. | ||||
| 46 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The wedge is 25 cm from the 200 g mass, so $27 - 25 = 2$ cm from the cube.
Half the cube is in water: displaced volume $= \frac12(10)^3 = 500$ cm$^3$, buoyant force = weight of 500 g.
Moments about the wedge (the rod's own weight is not given, so the rod is taken as light):
$(M - 500)\,g \times 2 = 200\,g \times 25$, so $M - 500 = 2500$ and $M = 3000$ g = 3 kg.
Answer: 3. | ||||
| 47 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\Delta\omega = (2100 - 1800)\frac{2\pi}{60} = 10\pi$ rad/s.
Angular impulse: $\tau t = I\Delta\omega$, so $I = \frac{25\pi \times 40}{10\pi} = 100$ kg m$^2$.
For a thin disk about a diameter, $I = \frac14MR^2$: $\frac14(1)R^2 = 100$, $R = 20$ m.
Diameter $= 40$ m.
Answer: 40. | ||||
| 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 $B = \frac{\Delta P}{\Delta V/V}$, so $V = \frac{B\,\Delta V}{\Delta P}$.
$\Delta P = 4$ atm $= 4 \times 10^5$ Pa, $\Delta V = 0.8 \times 10^{-6}$ m$^3$.
$V = \frac{2 \times 10^9 \times 0.8 \times 10^{-6}}{4 \times 10^5} = 4 \times 10^{-3}$ m$^3$ = 4 litre.
Answer: 4. | ||||
| 49 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The two slabs, each of thickness $\frac d2$, are stacked between the plates, so they act as capacitors in series:
$C = \frac{\epsilon_0 A}{\frac{d/2}{5} + \frac{d/2}{3}} = \frac{\epsilon_0A}{d}\cdot\frac{1}{\frac1{10} + \frac16} = \frac{15}{4}\cdot\frac{\epsilon_0A}{d}$.
$\frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}} = 2 \times 10^{-12}$ F = 2 pF.
$C = 7.5$ pF; in parallel with 7.5 pF the total is $15$ pF.
Answer: 15. | ||||
| 50 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The sheet's field $\frac{\sigma}{2\epsilon_0}$ is uniform, so the electron (repelled by the negative sheet) has a constant acceleration $a = \frac{e\sigma}{2\epsilon_0 m}$ and, starting from rest, $v = at$.
$\lambda = \frac{h}{mv} = \frac{h}{mat}$, so $\frac{d\lambda}{dt} = -\frac{h}{ma}\cdot\frac{1}{t^2}$.
The rate varies as $t^{-2}$: $n = 2$.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | StatementsMedium | |
A medium statements question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom Fluorine (Z = 9): $1s^2\,2s^2\,2p^5$.
A: four orbitals are filled (two electrons of opposite spin each) and one 2p orbital has a single electron, so the spins can be 5 of $+\frac12$ and 4 of $-\frac12$. True.
B: the single unpaired 2p electron may be in any one of the three degenerate 2p orbitals; nothing makes it $2p_z$. Not a correct statement.
C: the last electron enters a 2p orbital: $n = 2$, $l = 1$. True.
D: s orbitals have no angular node and each p orbital has one; the occupied orbitals 1s, 2s, $2p_x$, $2p_y$, $2p_z$ have $0 + 0 + 1 + 1 + 1 = 3$ angular nodes, not 1. False.
Correct: A and C.
Answer: (C) A and C Only. | ||||
| 52 | SolutionsRaoult's law | Class 12: Solutions | Single correctEasy | |
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions A minimum boiling azeotrope forms when the mixture shows positive deviation from Raoult's law (the A–B attractions are weaker than A–A and B–B).
Water + ethyl methyl ketone, carbon disulphide + acetone, and methanol + chloroform all show positive deviation (the hydrogen bonding or dipolar attractions of one component are broken on mixing) and form minimum boiling azeotropes.
Phenol + aniline form strong O–H···N hydrogen bonds between the two components: negative deviation, so they give a maximum boiling azeotrope.
Answer: (D) $\mathrm{C_6H_5OH + C_6H_5NH_2}$. | ||||
| 53 | Solutionsvan't Hoff factor | Class 12: Solutions | Single correctMedium | |
A medium single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = iK_fm$: $0.20 = i \times 1.8 \times 0.1$, so $i = \frac{10}{9}$.
For HA → H⁺ + A⁻, $i = 1 + \alpha$, so $\alpha = \frac19$.
$K_a = \frac{c\alpha^2}{1 - \alpha} = \frac{0.1 \times \frac{1}{81}}{\frac89} = \frac{0.1}{72} = 1.39 \times 10^{-3}$.
Answer: (B) $1.38 \times 10^{-3}$. | ||||
| 54 | 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 For a first order reaction $t = \frac1k\ln\frac{[A]_0}{[A]}$.
$t_1 = \frac{\ln4}{k} = \frac{2\ln2}{k}$ and $t_2 = \frac{\ln8}{k} = \frac{3\ln2}{k}$ (two and three half-lives).
$\frac{t_1}{t_2} = \frac23$.
Answer: (D) $\frac23$. | ||||
| 55 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[Co(NH_3)_6]^{3+}}$: Co³⁺ $d^6$, strong field, electrons paired in $t_{2g}$: inner orbital $d^2sp^3$.
$\mathrm{SF_6}$: six bond pairs on S using 3s, 3p and two 3d: $sp^3d^2$. ✓
$\mathrm{[CrF_6]^{3-}}$: Cr³⁺ $d^3$ leaves two 3d orbitals empty: $d^2sp^3$.
$\mathrm{[CoF_6]^{3-}}$: Co³⁺ $d^6$ with weak-field F⁻, high spin (4 unpaired): outer orbital $sp^3d^2$. ✓
$\mathrm{[Mn(CN)_6]^{3-}}$: Mn³⁺ $d^4$, strong field, low spin: $d^2sp^3$.
$\mathrm{[MnCl_6]^{3-}}$: Mn³⁺ $d^4$ with weak-field Cl⁻, high spin (4 unpaired): $sp^3d^2$. ✓
Number = 3.
Answer: (D) 3. | ||||
| 56 | 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 19 is K, 87 is Fr (both alkali metals, $ns^1$), 35 is Br and 32 is Ge (p-block, much higher ionisation enthalpies).
Down group 1 the ionisation enthalpy falls as the outer electron is farther from the nucleus: Fr (about 393 kJ mol⁻¹) is below K (419 kJ mol⁻¹).
Lowest: atomic number 87.
Answer: (B) 87. | ||||
| 57 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down group 16 the central atom grows, the E–H bond becomes longer and weaker, and the bond dissociation enthalpy falls: H–Se is stronger than H–Te. Statement II is true.
Because the H–Te bond breaks more easily, $\mathrm{H_2Te}$ is the stronger acid (acidity: $\mathrm{H_2O < H_2S < H_2Se < H_2Te}$). Statement I is false.
Answer: (D). | ||||
| 58 | 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 Spin-only moment $\mu = \sqrt{n(n + 2)}$ BM, $n$ = unpaired electrons:
Cr²⁺: $3d^4$, $n = 4$, 4.90 BM.
Cr³⁺: $3d^3$, $n = 3$, 3.87 BM.
Cu²⁺: $3d^9$, $n = 1$, 1.73 BM.
Cu⁺: $3d^{10}$, $n = 0$, 0 BM.
Order: Cr²⁺ > Cr³⁺ > Cu²⁺ > Cu⁺.
Answer: (C). | ||||
| 59 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | StatementsEasy | |
A easy statements question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Statement I: a homoleptic octahedral complex with monodentate ligands is $\mathrm{MA_6}$; all six positions are equivalent, so there are no geometrical or optical isomers. True.
Statement II: cisplatin and transplatin are $\mathrm{[Pt(NH_3)_2Cl_2]}$; they are heteroleptic, but the metal is platinum, not palladium. False.
Answer: (C). | ||||
| 60 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Matching listMedium | |
A medium matching list question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A. $\mathrm{[Ni(CO)_4]}$: Ni(0), $3d^{10}$, $sp^3$: tetrahedral, 0 BM → III.
B. $\mathrm{[Ni(CN)_4]^{2-}}$: Ni²⁺ $d^8$ with strong-field CN⁻ pairs up, $dsp^2$: square planar, 0 BM → II.
C. $\mathrm{[NiCl_4]^{2-}}$: Ni²⁺ $d^8$ with weak-field Cl⁻, $sp^3$, 2 unpaired: tetrahedral, $\sqrt8 \approx 2.83$ BM → I.
D. $\mathrm{[MnBr_4]^{2-}}$: Mn²⁺ $d^5$, 5 unpaired, $sp^3$: tetrahedral, $\sqrt{35} \approx 5.92$ BM → IV.
A-III, B-II, C-I, D-IV.
Answer: (B). | ||||
| 61 | Practical Organic ChemistryDetection of functional groups | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on detection of functional groups, from Practical Organic Chemistry. NCERT: No matching NCERT chapter A. Sodium bicarbonate gives brisk effervescence of CO₂ with a carboxylic acid → II.
B. Neutral ferric chloride gives a violet (or other) colour with a phenol → III.
C. Ceric ammonium nitrate turns from yellow to red with an alcohol → IV.
D. Cold dilute alkaline KMnO₄ (Baeyer's reagent) is decolourised by a C=C double bond → I.
A-II, B-III, C-IV, D-I.
Answer: (A). | ||||
| 62 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques % H $= \frac{2}{18} \times \frac{0.127}{0.210} \times 100 = 6.72$.
% C $= \frac{12}{44} \times \frac{0.307}{0.210} \times 100 = 39.87$.
% O $= 100 - 6.72 - 39.87 = 53.41$.
Hydrogen 6.72 %, oxygen 53.41 %.
Answer: (D). | ||||
| 63 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The ring is a cyclopentene carrying OH and an ethyl group; the double bond lies between the two ring carbons that join the C–OH and the C–ethyl carbons.
OH is the principal group, so it is named as the suffix "-ol" and its carbon is C1. Numbering towards the double bond gives it the locants 2,3, and the ethyl carbon is then C4.
Name: 4-Ethylcyclopent-2-en-1-ol.
(A) and (B) wrongly name OH as a "hydroxy" prefix; (D) gives OH a higher locant than the substituent.
Answer: (C). | ||||
| 64 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Matching listEasy | |
A easy matching list question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A. A carbocation has an $sp^2$ carbon with an empty p orbital → III.
B. A carbon free radical has an $sp^2$ (planar) or $sp^3$ (pyramidal) carbon with one unpaired electron → IV.
C. A nucleophile donates an electron pair → I.
D. An electrophile accepts an electron pair → II.
A-III, B-IV, C-I, D-II.
Answer: (D). | ||||
| 65 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (i) Alcoholic KOH removes one HBr: 1-bromocyclooctene.
(ii) NaNH₂ removes the second HBr: cyclooctyne (the smallest isolable cycloalkyne).
(iii) Hg²⁺/H⁺ hydrates the triple bond (Markovnikov, through the enol): cyclooctanone.
(iv) Zn–Hg/HCl (Clemmensen reduction) turns C=O into CH₂: cyclooctane.
P is cyclooctane.
Answer: (C). | ||||
| 66 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Target: 4-bromostyrene (Br on the ring para to a CH=CH₂ group).
Br₂/Fe brominates the ring; the ethyl group is ortho/para directing and the para product dominates: 4-bromoethylbenzene.
Cl₂ with heat (free radical) chlorinates the side chain at the benzylic carbon: 1-(4-bromophenyl)-1-chloroethane.
Alcoholic KOH eliminates HCl to give the C=C: 4-bromostyrene.
The others fail: (A) puts Cl on the ring and no halogen on the side chain; (C) aqueous KOH substitutes (gives an alcohol) rather than eliminates; (D) puts both halogens on the ring.
Answer: (B). | ||||
| 67 | Alcohols, Phenols and EthersWilliamson synthesis | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on williamson synthesis, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Williamson synthesis works when the alkyl halide is methyl or primary (SN2); with a secondary or tertiary halide the strongly basic alkoxide mainly causes elimination.
(A) cyclohexoxide + CH₃Br: methyl halide, SN2: gives the ether.
(B) isobutoxide + sec-butyl bromide: a secondary halide with a strong, bulky base: E2 elimination (butenes) is the major path, so the ether is not the major product.
(C) phenoxide + n-propyl bromide: primary halide: gives the ether.
(D) tert-butoxide + ethyl bromide: primary halide; the bulky alkoxide is only the nucleophile: gives the ether.
Answer: (B). | ||||
| 68 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The compound is 6-oxoheptanal, $\mathrm{CH_3CO(CH_2)_4CHO}$.
The aldehyde carbonyl is the more electrophilic, so the enolate attacks the CHO carbon. The enolate formed at the CH₂ next to the ketone (C5) attacking the CHO carbon (C1) closes a five-membered ring (C1–C5), the most favoured ring size.
Dehydration of the β-hydroxy ketone puts the double bond in conjugation with the ketone: 1-acetylcyclopentene (1-(cyclopent-1-en-1-yl)ethan-1-one).
The alternative five-membered closure (the aldehyde's α-CH₂ attacking the ketone) would give option (D), but attack on the less electrophilic ketone carbon is the minor path.
Answer: (C). | ||||
| 69 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A contains N and has molar mass 121. Benzamide, $\mathrm{C_6H_5CONH_2}$: $7(12) + 7(1) + 14 + 16 = 121$. ✓
NaOH then H₃O⁺ hydrolyses the amide (NH₃ is lost) to benzoic acid, B, which liberates CO₂ with NaHCO₃ (effervescence). ✓
EtOH with H₂SO₄ and heat esterifies B to ethyl benzoate, C, which has a fruity smell. ✓
Option (B)'s A, $\mathrm{C_6H_5CH_2NHNH_2}$, has molar mass 122 and would not give benzoic acid by hydrolysis; in (C) and (D) aminobenzaldehyde (M = 121) would not be converted by NaOH/H₃O⁺ into the products shown.
Answer: (A). | ||||
| 70 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The compound is valine, $\mathrm{H_2N{-}CH(CH(CH_3)_2){-}COOH}$; in water near neutral pH it is a zwitterion.
At pH 2 (strongly acidic, below its isoelectric point ≈ 6) both groups are protonated: $\mathrm{H_3N^+{-}CH(R){-}COOH}$, the cation (A).
At pH 10 (basic, above the isoelectric point) both groups are deprotonated: $\mathrm{H_2N{-}CH(R){-}COO^-}$, the anion (B).
Answer: (B). | ||||
| 71 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry NaI + AgNO₃ → AgI + NaNO₃ (1 : 1).
Molar mass of AgI $= 108 + 127 = 235$ g mol⁻¹; moles of AgI $= \frac{4.74}{235} = 0.02017$ mol = moles of NaI.
Molarity $= \frac{0.02017}{0.020} = 1.009$ M ≈ 1.
Answer: 1. | ||||
| 72 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom For a hydrogen-like ion $E_n = -13.6\frac{Z^2}{n^2}$ eV.
Be³⁺: $Z = 4$; first excited state $n = 2$: $E = -13.6 \times \frac{16}{4} = -54.4$ eV.
Magnitude 54.4 eV ≈ 54.
Answer: 54. | ||||
| 73 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Only the structure X=X=Y has all its bond energies given (the X–Y single bond of the other is not), so the enthalpy of formation is calculated for it:
Bonds broken: X≡X 940 + ½(Y=Y) 250 = 1190 kJ mol⁻¹.
Bonds formed in X=X=Y: X=X 410 + X=Y 602 = 1012 kJ mol⁻¹.
Calculated $\Delta_fH = 1190 - 1012 = 178$ kJ mol⁻¹.
The real molecule is more stable than this one structure by the resonance energy:
Resonance energy $= 178 - 80 = 98$ kJ mol⁻¹.
Answer: 98. | ||||
| 74 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Start with 1 mol H₂O: at equilibrium H₂O $1 - \alpha$, H₂ $\alpha$, O₂ $\frac{\alpha}{2}$; total $1 + \frac{\alpha}{2} \approx 1$, so at $P = 1$ bar the partial pressures are about $1$, $\alpha$, $\frac{\alpha}{2}$.
$K_p = \frac{\alpha\left(\frac{\alpha}{2}\right)^{1/2}}{1} = \frac{\alpha^{3/2}}{\sqrt2} = 8.0 \times 10^{-3}$.
$\alpha^3 = 2 \times 64 \times 10^{-6} = 1.28 \times 10^{-4}$, so $\alpha = 0.0504 = 5.04 \times 10^{-2}$.
(Check: $\alpha = 0.05$ gives $\frac{0.05^{1.5}}{\sqrt2} = 7.9 \times 10^{-3}$.)
Nearest integer: 5.
Answer: 5. | ||||
| 75 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry $E = E^\circ - \frac{0.059}{6}\log\frac{[\mathrm{Cr^{3+}}]^2}{[\mathrm{Cr_2O_7^{2-}}][\mathrm{H^+}]^{14}}$.
$\log\frac{[\mathrm{Cr^{3+}}]^2}{[\mathrm{Cr_2O_7^{2-}}][\mathrm{H^+}]^{14}} = -6 + 14\,\mathrm{pH}$.
$E = 0$: $1.33 = \frac{0.059}{6}(-6 + 14\,\mathrm{pH})$, so $-6 + 14\,\mathrm{pH} = \frac{1.33 \times 6}{0.059} = 135.25$.
$14\,\mathrm{pH} = 141.25$, pH $= 10.09 \approx 10$.
Answer: 10. | ||||
Sit this paper free.
Timed like the exam or chapter by chapter, with a solution for each question. Run by a non-profit trust.