JEE (Main) · 2025
JEE (Main) 2025 — 3 April, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
25 Q
- Conic Sections3
- Integral Calculus3
- Trigonometry2
4 easy · 19 medium · 2 hard
Physics
25 Q
- Thermodynamics3
- Optics3
- Magnetic Effects of Current and Magnetism3
11 easy · 13 medium · 1 hard
Chemistry
25 Q
- Organic Chemistry: Some Basic Principles and Techniques3
- Thermodynamics2
- Equilibrium2
10 easy · 15 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 | Single correctMedium | |
A medium single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions For each $x$, the condition is $-x^2 \le 2y \le 4 - x^2$ with $y \in A$.
$x = \pm 3$ ($x^2 = 9$): $-9 \le 2y \le -5$, so $y = -3$ only: 1 pair each, 2 in all.
$x = \pm 2$ ($x^2 = 4$): $-4 \le 2y \le 0$, so $y \in \{-2, -1, 0\}$: 3 each, 6 in all.
$x = \pm 1$ ($x^2 = 1$): $-1 \le 2y \le 3$, so $y \in \{0, 1\}$: 2 each, 4 in all.
$x = 0$: $0 \le 2y \le 4$, so $y \in \{0, 1, 2\}$: 3.
$l = 2 + 6 + 4 + 3 = 15$.
Reflexive pairs $(x, x)$ need $0 \le x^2 + 2x \le 4$: $(-3)$: $3$ ✓, $(-2)$: $0$ ✓, $(-1)$: $-1$ ✗, $(0)$: $0$ ✓, $(1)$: $3$ ✓, $(2)$: $8$ ✗, $(3)$: $15$ ✗.
So $(-1,-1), (2,2), (3,3)$ must be added: $m = 3$.
$l + m = 18$.
Answer: (C) 18. | ||||
| 2 | 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 Logarithm: $\frac{2x-3}{5+4x} > 0$, so $x < -\frac{5}{4}$ or $x > \frac{3}{2}$.
Inverse sine: $\left|\frac{4+3x}{2-x}\right| \le 1$ with $x \ne 2$, i.e. $(4+3x)^2 \le (2-x)^2$:
$9x^2 + 24x + 16 \le x^2 - 4x + 4 \Rightarrow 8x^2 + 28x + 12 \le 0 \Rightarrow (2x+1)(x+3) \le 0 \Rightarrow -3 \le x \le -\frac{1}{2}$.
Together: $-3 \le x < -\frac{5}{4}$, so $\alpha = -3$, $\beta = -\frac{5}{4}$.
$\alpha^2 + 4\beta = 9 - 5 = 4$.
Answer: (B) 4. | ||||
| 3 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Each root of $x^2 + \sqrt{3}x - 16 = 0$ satisfies $\alpha^{n+2} + \sqrt{3}\alpha^{n+1} - 16\alpha^n = 0$; adding the same for $\beta$:
$P_{n+2} + \sqrt{3}P_{n+1} = 16P_n$. With $n = 23$: $P_{25} + \sqrt{3}P_{24} = 16P_{23}$, so the first fraction is $\frac{16}{2} = 8$.
Likewise $Q_{n+2} + 3Q_{n+1} - Q_n = 0$, so $Q_{25} - Q_{23} = -3Q_{24}$ and the second fraction is $-3$.
Sum $= 8 - 3 = 5$.
Answer: (C) 5. | ||||
| 4 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z^2 + 3i = (2+3i)(z - 2 + i) = (2+3i)z + (2+3i)(-2+i) = (2+3i)z - 7 - 4i$.
So $z^2 - (2+3i)z + (7 + 7i) = 0$, with roots $z_1, z_2$: $z_1 + z_2 = 2 + 3i$, $z_1z_2 = 7 + 7i$.
(Neither root is $2 - i$: putting $z = 2 - i$ gives $3 - i \ne 0$.)
$z_1^2 + z_2^2 = (z_1+z_2)^2 - 2z_1z_2 = (4 + 12i - 9) - (14 + 14i) = -19 - 2i$.
Answer: (D) $-19 - 2i$. | ||||
| 5 | 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 For order $n = 3$: $|kM| = k^3|M|$ and $|\mathrm{adj}\,M| = |M|^2$.
$|2A| = 8 \cdot 5 = 40$, so $|\mathrm{adj}(2A)| = 40^2 = 1600$.
Let $B = 3A\,\mathrm{adj}(2A)$: $|B| = 27 \cdot |A| \cdot 1600 = 27 \cdot 5 \cdot 1600 = 216000 = 2^6 \cdot 3^3 \cdot 5^3$.
$|2\,\mathrm{adj}\,B| = 8|B|^2 = 2^3 \cdot 2^{12} \cdot 3^6 \cdot 5^6 = 2^{15} \cdot 3^6 \cdot 5^6$.
$\alpha + \beta + \gamma = 15 + 6 + 6 = 27$.
Answer: (C) 27. | ||||
| 6 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series In a G.P. $a_3a_5 = a_4^2 = 729$, so $a_4 = 27$ (positive terms).
With ratio $r$: $a_2 = \frac{27}{r^2}$, so $\frac{27}{r^2} + 27 = \frac{111}{4}$, giving $\frac{27}{r^2} = \frac{3}{4}$, $r^2 = 36$, $r = 6$ (increasing).
$a_1 = \frac{27}{6^3} = \frac{1}{8}$, and $a_1 + a_2 + a_3 = \frac{1}{8}(1 + 6 + 36) = \frac{43}{8}$.
$24 \cdot \frac{43}{8} = 129$.
Answer: (B) 129. | ||||
| 7 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Differences: $2, 8, 14, 20, 26, \ldots$, an A.P. with common difference 6. So the $n$th term is quadratic:
$T_n = 1 + \sum_{k=1}^{n-1}(6k - 4) = 1 + 3n(n-1) - 4(n-1) = 3n^2 - 7n + 5$ (check: $T_1 = 1$, $T_2 = 3$, $T_3 = 11$).
$S_{20} = 3\sum n^2 - 7\sum n + 5 \cdot 20 = 3 \cdot 2870 - 7 \cdot 210 + 100 = 8610 - 1470 + 100 = 7240$.
Answer: (C) 7240. | ||||
| 8 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Split: $\sum_{r=1}^{9} r\,{}^9C_r\,x^r + 3\sum_{r=1}^{9}{}^9C_r\,x^r$ with $x = \frac{1}{2}$.
$\sum r\,{}^9C_r x^r = 9x(1+x)^8 = \frac{9}{2}\left(\frac{3}{2}\right)^8 = 3\left(\frac{3}{2}\right)^9$.
$\sum_{r=1}^{9}{}^9C_r x^r = (1+x)^9 - 1 = \left(\frac{3}{2}\right)^9 - 1$.
Total $= 3\left(\frac{3}{2}\right)^9 + 3\left(\frac{3}{2}\right)^9 - 3 = 6\left(\frac{3}{2}\right)^9 - 3$.
$\alpha = 6$, $\beta = 3$, $(\alpha + \beta)^2 = 81$.
Answer: (D) 81. | ||||
| 9 | 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 The rational terms are those with an even power of $\sqrt{3}$, so their sum is $\frac{(2+\sqrt{3})^8 + (2-\sqrt{3})^8}{2}$.
Let $a_n = (2+\sqrt{3})^n + (2-\sqrt{3})^n$. The two numbers are roots of $t^2 - 4t + 1 = 0$, so $a_n = 4a_{n-1} - a_{n-2}$ with $a_0 = 2$, $a_1 = 4$:
$a_2 = 14$, $a_3 = 52$, $a_4 = 194$, $a_5 = 724$, $a_6 = 2702$, $a_7 = 10084$, $a_8 = 37634$.
Sum of rational terms $= \frac{37634}{2} = 18817$.
(Directly: $2^8 + 28 \cdot 2^6 \cdot 3 + 70 \cdot 2^4 \cdot 9 + 28 \cdot 2^2 \cdot 27 + 81 = 256 + 5376 + 10080 + 3024 + 81 = 18817$.)
Answer: (B) 18817. | ||||
| 10 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $g(x) = 3\tan x + 2x - \pi$. On each interval where $\tan x$ is continuous, $g'(x) = 3\sec^2 x + 2 > 0$, so $g$ is increasing and has at most one zero there.
The intervals are $[-2\pi, -\frac{3\pi}{2})$, $(-\frac{3\pi}{2}, -\frac{\pi}{2})$, $(-\frac{\pi}{2}, \frac{\pi}{2})$, $(\frac{\pi}{2}, \frac{3\pi}{2})$, $(\frac{3\pi}{2}, 2\pi]$.
On the three full branches $g$ runs from $-\infty$ to $+\infty$: one zero each.
On $[-2\pi, -\frac{3\pi}{2})$: $g(-2\pi) = -4\pi - \pi < 0$ and $g \to +\infty$ as $x \to -\frac{3\pi}{2}^-$: one zero.
On $(\frac{3\pi}{2}, 2\pi]$: $g \to -\infty$ at the left end and $g(2\pi) = 4\pi - \pi > 0$: one zero.
Total: 5 solutions.
Answer: (C) 5. | ||||
| 11 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The two parabolas are mirror images in the line $y = x$, so the smallest circle touching both has as diameter the shortest distance between them, and that is twice the shortest distance from $y = x^2 + 2$ to the line $y = x$.
The nearest point has its tangent parallel to $y = x$: $2x = 1$, $x = \frac{1}{2}$, $y = \frac{9}{4}$.
Its distance from $y = x$: $\frac{|\frac{1}{2} - \frac{9}{4}|}{\sqrt{2}} = \frac{7}{4\sqrt{2}}$.
Shortest distance between the parabolas $= \frac{7}{2\sqrt{2}}$; radius $= \frac{7}{4\sqrt{2}} = \frac{7\sqrt{2}}{8}$.
Answer: (D) $\frac{7\sqrt{2}}{8}$. | ||||
| 12 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $P$ is inside the ellipse ($\frac{5}{36} + \frac{5}{25} < 1$). Write the line as $x = \sqrt{5} + t\cos\theta$, $y = \sqrt{5} + t\sin\theta$ and substitute:
$\left(\frac{\cos^2\theta}{36} + \frac{\sin^2\theta}{25}\right)t^2 + (\ldots)t + \left(\frac{5}{36} + \frac{5}{25} - 1\right) = 0$.
The roots $t_1, t_2$ have opposite signs and $PA\cdot PB = |t_1t_2| = \frac{595/900}{\frac{\cos^2\theta}{36} + \frac{\sin^2\theta}{25}}$.
This is greatest when the denominator is least, i.e. $\sin\theta = 0$: the line $y = \sqrt{5}$.
Then $\frac{x^2}{36} = 1 - \frac{5}{25} = \frac{4}{5}$, $x = \pm\frac{12}{\sqrt{5}}$.
$PA = \frac{12}{\sqrt{5}} - \sqrt{5} = \frac{7}{\sqrt{5}}$, $PB = \frac{12}{\sqrt{5}} + \sqrt{5} = \frac{17}{\sqrt{5}}$.
$5(PA^2 + PB^2) = 49 + 289 = 338$.
Answer: (C) 338. | ||||
| 13 | 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 The line is $y = x$. With $L_1$: $3x + 6 = 0$, so $A = (-2, -2)$. With $L_2$: $6x = p$, so $B = \left(\frac{p}{6}, \frac{p}{6}\right)$.
$AB = \sqrt{2}\left|\frac{p}{6} + 2\right| = \frac{9}{\sqrt{2}}$ gives $\frac{p}{6} + 2 = \frac{9}{2}$, $p = 15$ (the other root is negative).
$L_2$: $4x + 2y - 15 = 0$. $AM$ is the distance of $A$ from $L_2$: $\frac{|-8 - 4 - 15|}{\sqrt{20}} = \frac{27}{2\sqrt{5}}$.
Triangle $AMB$ is right-angled at M: $BM^2 = AB^2 - AM^2 = \frac{81}{2} - \frac{729}{20} = \frac{81}{20}$, so $BM = \frac{9}{2\sqrt{5}}$.
$\frac{AM}{BM} = \frac{27}{9} = 3$.
Answer: (B) 3. | ||||
| 14 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy 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 Directions $\hat{k}$ and $\hat{j}$; $\hat{k} \times \hat{j} = -\hat{i}$.
Shortest distance $= |((\lambda - 1)\hat{i} + 3\hat{j} + 3\hat{k})\cdot(-\hat{i})| = |\lambda - 1| = 3$, so $\lambda = 4$ or $-2$: $\lambda_1 = 4$, $\lambda_2 = -2$.
$L_1$ is the line $x = 1$, $y = 2$. The distance of $(4, -2, 7)$ from it is $\sqrt{(4-1)^2 + (-2-2)^2} = 5$; its square is 25.
Answer: (A) 25. | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctHard | |
A hard single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Let $P = (4, 1, 0)$, $A = (1 + 2t, 2 + 3t, 3 + 4t)$ on $L_1$ and $B = (s + 6, s, 4 - s)$ on $L_2$.
P, A, B are collinear: $A - P = k(B - P)$:
$2t - 3 = k(s + 2)$, $3t + 1 = k(s - 1)$, $4t + 3 = k(4 - s)$.
Adding the first and third: $6t = 6k$, so $t = k$. Then the first gives $ks = -3$, and the second $3k + 1 = -3 - k$, so $k = -1$, $t = -1$, $s = 3$.
$A = (-1, -1, -1)$, $B = (9, 3, 1)$ (the third equation checks: $-1 = -1\cdot1$).
$\begin{vmatrix}1 & 0 & 1\\ -1 & -1 & -1\\ 9 & 3 & 1\end{vmatrix} = 1(-1 + 3) - 0 + 1(-3 + 9) = 2 + 6 = 8$.
Answer: (A) 8. | ||||
| 16 | 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 Expand along the first row:
$y = \sin x(28 - 27) - \cos x(27 - 27) + (\sin x + \cos x + 1)(27 - 28) = \sin x - \sin x - \cos x - 1 = -\cos x - 1$.
$\frac{d^2y}{dx^2} = \cos x$, so $\frac{d^2y}{dx^2} + y = -1$.
Answer: (B) $-1$. | ||||
| 17 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Left limit: $\lim_{x\to0^-}(1 + ax)^{1/x} = e^a$.
Right side: the numerator tends to 0, so for a finite limit the denominator must too: $c^{1/3} = 2$, $c = 8$.
Then $\lim_{x\to0^+}\frac{(4+x)^{1/2} - 4^{1/2}}{(8+x)^{1/3} - 8^{1/3}} = \frac{\frac{1}{2}\cdot4^{-1/2}}{\frac{1}{3}\cdot8^{-2/3}} = \frac{1/4}{1/12} = 3$.
Continuity: $e^a = 3 = 1 + b$, so $b = 2$.
$e^a bc = 3 \cdot 2 \cdot 8 = 48$.
Answer: (A) 48. | ||||
| 18 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $u = 3 - x^2$, $x\,dx = -\frac{du}{2}$, $x^2 = 3 - u$:
$f = -\frac{1}{2}\int(3 - u)\sqrt{u}\,du = -\frac{1}{2}\left(2u^{3/2} - \frac{2}{5}u^{5/2}\right) + C = -u^{3/2} + \frac{1}{5}u^{5/2} + C$.
At $x = \sqrt{2}$, $u = 1$: $f(\sqrt{2}) = -1 + \frac{1}{5} + C = -\frac{4}{5} + C$; $5f(\sqrt{2}) = -4 + 5C = -4$, so $C = 0$.
At $x = 1$, $u = 2$: $f(1) = -2\sqrt{2} + \frac{4\sqrt{2}}{5} = -\frac{6\sqrt{2}}{5}$.
Answer: (B) $-\frac{6\sqrt{2}}{5}$. | ||||
| 19 | 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 Domain: $\log_4\log_6(3 + 4x - x^2) > 0 \Rightarrow \log_6(3 + 4x - x^2) > 1 \Rightarrow 3 + 4x - x^2 > 6 \Rightarrow x^2 - 4x + 3 < 0 \Rightarrow 1 < x < 3$ (these also make the inner logarithms defined).
So $b - a = 2$.
$\int_0^2[x^2]\,dx = 0\cdot1 + 1(\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3}) = 5 - \sqrt{2} - \sqrt{3}$.
$p = 5$, $q = 2$, $r = 3$ (gcd 1): $p + q + r = 10$.
Answer: (C) 10. | ||||
| 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 Differentiate the integral equation: $g(x) = 1 - x\,g(x)$, so $g(x) = \frac{1}{1+x}$.
The equation becomes $\frac{dy}{dx} - y\tan x = 2\sec x$.
Integrating factor: $e^{-\int\tan x\,dx} = \cos x$. Then $\frac{d}{dx}(y\cos x) = 2\sec x\cos x = 2$, so $y\cos x = 2x + C$; $y(0) = 0$ gives $C = 0$.
$y\left(\frac{\pi}{3}\right) = \frac{2\pi/3}{1/2} = \frac{4\pi}{3}$.
Answer: (A) $\frac{4\pi}{3}$. | ||||
| 21 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations There are $9 \times 10^6$ seven-digit numbers. Fix the first six digits; of the ten choices of the last digit, exactly five make the digit sum even. So half of them qualify:
$\frac{9 \times 10^6}{2} = 4.5 \times 10^6 = 45 \times 10^5 = 9 \cdot 5 \cdot 10^5$.
So $m = 9$, $n = 5$ (no other $n$ from 1 to 9 gives a whole $m \le 9$: $n = 4$ needs $m = 112.5$, $n = 6$ needs $m = 0.75$).
$m + n = 14$.
Answer: 14. | ||||
| 22 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations There are $5! = 120$ words. $P(W_n) = 2^{n-1}P(W_1)$ and the probabilities add to 1: $P(W_1)(2^{120} - 1) = 1$, so $P(W_n) = \frac{2^{n-1}}{2^{120} - 1}$.
Rank of CDBEA: words beginning A or B: $2 \times 24 = 48$. Beginning CA or CB: $2 \times 6 = 12$ (60 so far). Beginning CDA: 2 (62). Then CDBAE is 63rd and CDBEA is 64th.
$P(\text{CDBEA}) = \frac{2^{63}}{2^{120} - 1}$, so $\alpha = 63$, $\beta = 120$.
$\alpha + \beta = 183$.
Answer: 183. | ||||
| 23 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections P is on H: $\frac{16}{a^2} - \frac{12}{b^2} = 1$.
Focal distances are $|ex - a|$ and $ex + a$, so their product is $e^2x^2 - a^2 = 16e^2 - a^2 = 32$. With $e^2 = 1 + \frac{b^2}{a^2}$: $\frac{16b^2}{a^2} - a^2 = 16$.
From the first equation $b^2 = \frac{12a^2}{16 - a^2}$, so $\frac{192}{16 - a^2} - a^2 = 16$, i.e. $192 = 256 - a^4$, $a^4 = 64$, $a^2 = 8$, and $b^2 = \frac{96}{8} = 12$.
$p = 2b$: $p^2 = 4b^2 = 48$. $q = \frac{2b^2}{a} = \frac{24}{2\sqrt{2}} = 6\sqrt{2}$: $q^2 = 72$.
$p^2 + q^2 = 120$.
Answer: 120. | ||||
| 24 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $(\vec{a} - \vec{b}) \times \hat{d} = \vec{0}$, so $\hat{d}$ is parallel to $\vec{a} - \vec{b} = -2\hat{i} - \hat{j} + 2\hat{k}$ (length 3): $\hat{d} = \pm\frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k})$.
$\vec{c} \perp \vec{a}$: $\lambda + \mu = 0$.
Take $\hat{d} = \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k})$: $\vec{c}\cdot\hat{d} = \frac{-\lambda + 2\mu}{3} = -\lambda = 1$, so $\lambda = -1$, $\mu = 1$.
$3\lambda\hat{d} + \mu\vec{c} = -(-2\hat{i} - \hat{j} + 2\hat{k}) + (-\hat{j} + \hat{k}) = 2\hat{i} - \hat{k}$; squared length 5.
(The other sign of $\hat{d}$ gives $\lambda = 1$, $\mu = -1$ and the same vector $2\hat{i} - \hat{k}$.)
Answer: 5. | ||||
| 25 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Compare the two pieces.
$x < 0$: $x|x-2| < 0 < |x|$, so $y = -x$.
$0 \le x \le 2$: $x(2 - x) \ge x \iff x \le 1$; so $y = 2x - x^2$ on $[0, 1]$ and $y = x$ on $[1, 2]$.
$x \ge 2$: $x(x - 2) \ge x \iff x \ge 3$; so $y = x$ on $[2, 3]$ and $y = x^2 - 2x$ on $[3, 4]$.
The curve is never below the $x$-axis. Area:
$\int_{-2}^{0}(-x)\,dx = 2$; $\int_0^1(2x - x^2)\,dx = \frac{2}{3}$; $\int_1^3 x\,dx = 4$; $\int_3^4(x^2 - 2x)\,dx = \left[\frac{x^3}{3} - x^2\right]_3^4 = \frac{16}{3} - 0 = \frac{16}{3}$.
Total $= 2 + \frac{2}{3} + 4 + \frac{16}{3} = 12$.
Answer: 12. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Physical World and MeasurementSignificant figures | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on significant figures, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Arithmetic sum: $435.42 + 226.3 + 0.125 = 661.845$ g.
In addition the result keeps as many decimal places as the least precise term. 226.3 g has one decimal place, so the sum is rounded to one decimal place: 661.8 g.
Answer: (D) 661.8 g. | ||||
| 27 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement A. From $F = \frac{Gm_1m_2}{r^2}$: $[G] = \frac{MLT^{-2}\cdot L^2}{M^2} = M^{-1}L^3T^{-2}$ → IV.
B. Potential energy is an energy: $ML^2T^{-2}$ → III.
C. Gravitational potential is energy per unit mass: $L^2T^{-2}$ → II.
D. Acceleration: $LT^{-2}$ → I.
A-IV, B-III, C-II, D-I.
Answer: (A). | ||||
| 28 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Take potential energy $mgh$ from the surface. Energy is conserved: $mgS = K + U$.
With $K = 3U$: $mgS = 4mgh$, so $h = \frac{S}{4}$.
$K = 3mg\frac{S}{4} = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{3gS}{2}}$.
Answer: (B) $\frac{S}{4}$, $\sqrt{\frac{3gS}{2}}$. | ||||
| 29 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane With speed $u$ at angle $\phi$ from the vertical, the vertical component is $u\cos\phi$, so
$h_m = \frac{u^2\cos^2\phi}{2g}$.
At $\phi = 0$ (straight up) $h_m$ is greatest, $\frac{u^2}{2g}$, and the curve starts flat there ($\frac{dh_m}{d\phi} = -\frac{u^2\sin2\phi}{2g} = 0$). It then falls steadily and reaches zero at $\phi = 90^\circ$ (horizontal projection).
Only the third graph starts at its maximum with a flat top at $\phi = 0$ and falls to zero at $90^\circ$. (The first falls steeply from the start and never reaches zero at $90^\circ$; the second rises; the fourth starts at zero.)
Answer: (C). | ||||
| 30 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | StatementsMedium | |
A medium statements question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane A. Phase against time is a straight line: a simple harmonic motion along a line has phase $\omega t + \phi_0$, linear in time (here with a negative initial phase). Possible.
B. Velocity against displacement is a closed circle (an ellipse in general units): this is the phase-space path of simple harmonic motion, $v^2 = \omega^2(A^2 - x^2)$. Possible.
C. Velocity against time is a circle: at one instant the particle would have two different velocities, which is impossible.
D. Total distance against time never decreases (it rises, at times more slowly): any real motion gives such a graph. Possible.
So A, B and D only.
Answer: (D). | ||||
| 31 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Let friction $f$ act forward at the contact point (its sign will come out). Then
translation: $F + f = ma$;
rotation about the centre: $(F - f)R = I\alpha = \frac{2}{5}mR^2\cdot\frac{a}{R}$, so $F - f = \frac{2}{5}ma$.
Adding: $2F = \frac{7}{5}ma$, $a = \frac{10F}{7m} = \frac{10 \times 49}{7 \times 20} = 3.5$ m/s$^2$.
Answer: (D) 3.5 m/s$^2$. | ||||
| 32 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Depth of the hole below the surface: $h = 1.6 - 0.9 = 0.7$ m.
The load adds a pressure $\Delta P = \frac{mg}{A} = \frac{50 \times 10}{0.5} = 1000$ Pa on the surface.
Bernoulli between the surface (speed negligible) and the hole: $\frac{1}{2}\rho v^2 = \rho gh + \Delta P$.
$v^2 = 2\left(10 \times 0.7 + \frac{1000}{1000}\right) = 2 \times 8 = 16$, so $v = 4$ m/s.
Answer: (C) 4 m/s. | ||||
| 33 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Ice is less dense than water, so on melting the volume of the system decreases. The atmosphere at constant pressure therefore does work $P|\Delta V|$ on the system: positive work is done on the ice-water system by the atmosphere (so (B) is wrong).
By the first law, $\Delta U = Q - P\Delta V$ with $Q$ = latent heat absorbed $> 0$ and $\Delta V < 0$, so $\Delta U > 0$: the internal energy increases ((C) and (D) are wrong).
Answer: (A). | ||||
| 34 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics For an adiabatic process $TV^{\gamma - 1}$ is constant:
$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 300 \times 4^{0.5} = 600$ K.
Change in temperature $= 600 - 300 = 300$ K.
Answer: (B) 300 K. | ||||
| 35 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctHard | |
A hard single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics The gas column has a fixed cross-section, so its volume is proportional to the piston's height. In an isothermal expansion from $L_0$ to $L_1$ the gas does work
$W_{gas} = nRT\ln\frac{V_1}{V_0} = nRT\ln\frac{L_1}{L_0}$.
As the piston rises by $L_1 - L_0$, its potential energy rises by $Mg(L_1 - L_0)$, and the spring (force $kx^3$) stores $\int kx^3\,dx = \frac{k}{4}x^4$.
The intended answer collects the three terms: $nRT\ln\left(\frac{L_1}{L_0}\right) + Mg(L_1 - L_0) + \frac{k}{4}(L_1^4 - L_0^4)$.
Of the options only (B) has the correct isothermal-work term $nRT\ln\frac{L_1}{L_0}$, the full $Mg(L_1 - L_0)$ and the quartic spring energy with coefficient $\frac{k}{4}$; (A) halves the gravity term and squares the ratio, (C) triples the gas term and uses a cubic spring energy, (D) has $\frac{3k}{4}$.
Answer: (B). | ||||
| 36 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | StatementsMedium | |
A medium statements question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Moving together, the blocks are a mass $M + m$ on a spring $k$:
A. $T = 2\pi\sqrt{\frac{M+m}{k}}$: correct.
B. $a = -\frac{kx}{M+m}$: correct.
C. Friction is the only horizontal force on the upper block, so $f = m|a| = \frac{mk|x|}{M+m}$. The statement has $\mu$ in place of $k$: incorrect.
D. No slipping needs $m|a|_{max} \le \mu mg$: $\frac{kA}{M+m} \le \mu g$, so the largest amplitude is $\frac{\mu(M+m)g}{k}$: correct.
E. Limiting friction on the upper block is $\mu mg$ (the normal force is $mg$), not $\mu(M+m)g$: incorrect.
A, B, D only.
Answer: (D). | ||||
| 37 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Resistance of the whole wire: $R = \frac{\rho L}{A} = \frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}} = 10\ \Omega$.
Between diametrically opposite points the two halves, $5\ \Omega$ each, are in parallel: $\frac{5 \times 5}{5 + 5} = 2.5\ \Omega$.
None of the printed options is $2.5\ \Omega$. The option one factor of ten away, 25 Ω, is what the same working gives with a resistivity of $2 \times 10^{-5}\ \Omega$ m (or an area of 0.5 mm$^2$), i.e. $R = 100\ \Omega$ and $\frac{R}{4} = 25\ \Omega$; it is marked as the intended answer.
Answer: (A) 25 Ω (see note). | ||||
| 38 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance First configuration: the two slabs are layers each of thickness $\frac{d}{2}$ across the whole plate area, i.e. two capacitors in series:
$C_1 = \frac{\varepsilon_0A}{\frac{d}{2}\left(\frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2}\right)} = \frac{2\varepsilon_0A}{d}\cdot\frac{\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2}$.
Second configuration: side by side, each filling the full gap over area $\frac{A}{2}$, i.e. in parallel:
$C_2 = \frac{\varepsilon_0A}{2d}(\varepsilon_1 + \varepsilon_2)$.
$\frac{C_1}{C_2} = \frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2}$.
Answer: (B). | ||||
| 39 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Inside a uniformly charged shell the field is zero, so the potential everywhere inside equals that on the surface: 120 V at the centre and 120 V at 5 cm.
Outside, $V = \frac{kQ}{r}$: $V(15\text{ cm}) = 120 \times \frac{10}{15} = 80$ V.
Answer: (C) 120 V, 120 V, 80 V. | ||||
| 40 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Treat the source as a point radiating equally in all directions, with the light falling normally on the surface.
Intensity at 2 m: $I = \frac{P}{4\pi r^2} = \frac{450}{16\pi} \approx 8.95$ W m$^{-2}$.
For a perfect reflector the pressure is $\frac{2I}{c} = \frac{2 \times 8.95}{3 \times 10^8} \approx 5.97 \times 10^{-8}$ Pa $\approx 6 \times 10^{-8}$ Pa.
Answer: (B) $6 \times 10^{-8}$ Pascals. | ||||
| 41 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | StatementsMedium | |
A medium statements question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At minimum deviation $i = e$ and $r_1 = r_2 = \frac{A}{2}$.
A. With $r_1 = r_2$ the ray inside crosses symmetrically, i.e. parallel to the base (of an isosceles prism): correct.
B. $\sin\frac{A + \delta_m}{2} = \mu\sin\frac{A}{2}$: a larger prism angle gives a larger minimum deviation: incorrect.
C. $i = e$: correct.
D. By reversibility of light, $i$ and $e$ can be interchanged with the same deviation, so every deviation other than the minimum occurs for two angles of incidence: correct.
E. The angle of refraction is half the prism angle, $r = \frac{A}{2}$, not double: incorrect.
A, C and D only.
Answer: (A). | ||||
| 42 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a biconvex lens $R_1 = +10$ cm, $R_2 = -15$ cm. Lens maker's formula:
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{10} + \frac{1}{15}\right) = (\mu - 1)\frac{1}{6}$.
$\frac{1}{12} = \frac{\mu - 1}{6}$, so $\mu - 1 = 0.5$, $\mu = 1.5$.
Answer: (D) 1.5. | ||||
| 43 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Threshold wavelength: $\lambda_0 = \frac{hc}{\phi} = \frac{1240\text{ eV nm}}{3\text{ eV}} \approx 413$ nm. Only light of shorter wavelength (higher photon energy) can eject electrons.
Of red, yellow, green and blue, blue has the shortest wavelength and highest photon energy, at the violet end of the visible range; red, yellow and green (above about 500 nm, photon energy below 2.5 eV) certainly cannot.
Answer: (D) Blue. | ||||
| 44 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Matching listMedium | |
A medium matching list question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei A. A neutron splits uranium-235 into two middle-sized nuclei: fission → III.
B. Hydrogen burning in oxygen involves only electrons rearranging: chemical reaction → I.
C. Two deuterons fuse to helium-3 and a neutron, releasing about 3.3 MeV: fusion with positive Q → II.
D. Binding energies: ${}^3$H about 8.48 MeV, ${}^1$H zero, two deuterons $2 \times 2.22 = 4.45$ MeV. Q = (binding after) − (binding before) $= 4.45 - 8.48 \approx -4.0$ MeV: negative → IV.
A-III, B-I, C-II, D-IV.
Answer: (A). | ||||
| 45 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The table gives $Y = 1$ exactly when $A = 1$: $Y = A$.
(A) $Y = \overline{A}\cdot(A + B) = \overline{A}B$: gives 1 only for $A = 0, B = 1$. No.
(B) $Y = (A + B)\cdot B = B$. No.
(C) $Y = A\cdot\overline{(A + B)} = 0$ always. No.
(D) $Y = A\cdot(A + B) = A$ (absorption law). Yes.
Answer: (D). | ||||
| 46 | GravitationNewton's law of gravitation | Class 11: Gravitation | NumericalMedium | |
A medium numerical question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation The only quantities the collision time can depend on are $G$, $m$ and $a$. Dimensionally ($[G] = M^{-1}L^3T^{-2}$) the only time is $\sqrt{\frac{a^3}{Gm}}$, so $T \propto \sqrt{\frac{a^3}{m}}$ (by symmetry the motion is the same shape in both cases, so the numerical factor is the same; treat the spheres as point masses).
New time: $T' = T\sqrt{\frac{(2a)^3/(2m)}{a^3/m}} = 4\sqrt{4} = 8$ s.
Answer: 8. | ||||
| 47 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy 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 $F = BIL\sin90^\circ = 0.15 \times 8 \times 0.040 = 0.048$ N $= 48$ mN.
Answer: 48. | ||||
| 48 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical 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 straight parts AB and CD lie on a line through O, so they give no field at O.
From the arrows, the current goes A→B, round the inner semicircle from B over to C, C→D, and back round the outer semicircle from D over to A. So it goes round the inner semicircle clockwise and the outer one anticlockwise: their fields at O are opposite.
Each semicircle gives $\frac{\mu_0I}{4R}$:
$B = \frac{\mu_0I}{4}\left(\frac{1}{R_2} - \frac{1}{R_1}\right) = \frac{4\pi \times 10^{-7} \times 12}{4}\left(\frac{1}{4\pi} - \frac{1}{6\pi}\right) = 12 \times 10^{-7}\left(\frac{1}{4} - \frac{1}{6}\right) = 1 \times 10^{-7}$ T.
$k = 1$.
Answer: 1. | ||||
| 49 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Ammeter (240 Ω) with its 10 Ω shunt: $\frac{240 \times 10}{250} = 9.6\ \Omega$.
Total resistance $= 150.4 + 9.6 = 160\ \Omega$; current from the 20 V cell $= \frac{20}{160} = 0.125$ A.
This divides inversely as the resistances; the ammeter carries $0.125 \times \frac{10}{250} = 0.005$ A $= 5$ mA.
Answer: 5. | ||||
| 50 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $I_{max} = (\sqrt{4I} + \sqrt{9I})^2 = 25I$, $I_{min} = (\sqrt{9I} - \sqrt{4I})^2 = I$.
Difference $= 24I$, so $x = 24$.
Answer: 24. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry For equal masses the number of atoms is $\frac{m}{M}N_A$, largest for the smallest molar mass.
Molar masses: Pr ≈ 141, Pt ≈ 195, Pb ≈ 207, Po ≈ 209 g mol$^{-1}$.
Praseodymium (Pr) has the smallest, so the most atoms.
Answer: (C) Pr. | ||||
| 52 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Quantum mechanics keeps discrete energy levels (A), stationary states that do not radiate (B), and emission of a photon on a downward transition (C).
It rejects a definite circular path for the electron: by the uncertainty principle the electron's position and momentum cannot both be exact, and it is described only by a probability distribution (an orbital), not an orbit.
Answer: (D). | ||||
| 53 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Solution 1 has 10 mol of x in about 10 L, i.e. about 1 mol per litre. One litre of it holds about 1 mol of x; adding 1 mol of x and 1 L of water makes about 2 mol in about 2 L, the same composition.
So the intensive properties (concentration, density, molar heat capacity) are unchanged.
Gibbs free energy is an extensive property: it depends on the amount of matter, and the amount in the system has changed, so it changes.
Answer: (A) Gibbs free energy. | ||||
| 54 | SolutionsColligative properties | Class 12: Solutions | Single correctEasy | |
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions Both solutes are non-electrolytes ($i = 1$). Total solute $= 4$ mol in 0.5 kg: molality $= 8$ mol kg$^{-1}$.
$\Delta T_b = K_bm = 0.52 \times 8 = 4.16$ K.
Boiling point $= 373.15 + 4.16 \approx 377.3$ K.
Answer: (C) 377.3 K. | ||||
| 55 | EquilibriumLe Chatelier's principle | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on le chatelier's principle, from Equilibrium. NCERT: Class 11: Equilibrium Adding an inert gas at constant pressure increases the volume. The partial pressures of the reacting gases fall, and the equilibrium shifts towards more gas moles (forward), but only partly.
At the new equilibrium $K_p = \frac{p_{\mathrm{PCl_3}}p_{\mathrm{Cl_2}}}{p_{\mathrm{PCl_5}}}$ is unchanged while the sum of the three partial pressures is less than before. If $p_{\mathrm{PCl_3}}$ and $p_{\mathrm{Cl_2}}$ did not fall, $p_{\mathrm{PCl_5}}$ would not fall either, and the sum could not decrease; so all three partial pressures, and hence all three concentrations ($c = \frac{p}{RT}$), decrease.
The moles of $\mathrm{PCl_3}$ and $\mathrm{Cl_2}$ increase, but their concentrations fall. So "$\mathrm{Cl_2}$ will decrease" is the true statement.
Answer: (B). | ||||
| 56 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Limiting molar conductivities at 298 K (S cm$^2$ mol$^{-1}$): $\mathrm{H^+}$ 349.6 (proton hopping), $\mathrm{Ca^{2+}}$ 119.0, $\mathrm{Mg^{2+}}$ 106.0, $\mathrm{K^+}$ 73.5, $\mathrm{Na^+}$ 50.1.
The doubly charged ions carry twice the charge; $\mathrm{Mg^{2+}}$ is smaller and more heavily hydrated than $\mathrm{Ca^{2+}}$, so it moves more slowly. $\mathrm{Na^+}$ is more hydrated than $\mathrm{K^+}$.
Order: $\mathrm{H^+ > Ca^{2+} > Mg^{2+} > K^+ > Na^+}$.
Answer: (D). | ||||
| 57 | EquilibriumChemical equilibrium | Class 11: Equilibrium | StatementsEasy | |
A easy statements question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium A catalyst lowers the activation energy of the forward and backward reactions equally, so equilibrium is reached faster but $K_c$ is unchanged: Statement I is true.
Since $K_c$ is unchanged, the equilibrium composition is unchanged too, whether the catalyst is homogeneous or heterogeneous: Statement II is false.
Answer: (C). | ||||
| 58 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics First order: $[A] = [A]_0\left(\frac{1}{2}\right)^{t/10}$, $[B] = [B]_0\left(\frac{1}{2}\right)^{t/40}$ ($t$ in minutes).
Equal when $8\left(\frac{1}{2}\right)^{t/10} = \left(\frac{1}{2}\right)^{t/40}$, i.e. $2^3 = 2^{t/10 - t/40} = 2^{3t/40}$.
$\frac{3t}{40} = 3$, $t = 40$ min.
(Check: after 40 min A has gone through 4 half-lives, $\frac{8}{16} = \frac{1}{2}$ of $[B]_0$; B through 1, also $\frac{1}{2}[B]_0$.)
Answer: (B) 40 min. | ||||
| 59 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A. $\mathrm{PF_5}$: 5 bond pairs, trigonal bipyramidal, $sp^3d$ → II.
B. $\mathrm{SF_6}$: 6 bond pairs, octahedral, $sp^3d^2$ → III.
C. $\mathrm{Ni(CO)_4}$: Ni(0), $3d^{10}$; CO pairs up nothing further, the 4s and 4p orbitals are used: tetrahedral $sp^3$ → IV.
D. $\mathrm{[PtCl_4]^{2-}}$: Pt(II), $5d^8$, square planar, $dsp^2$ → I.
A-II, B-III, C-IV, D-I.
Answer: (B). | ||||
| 60 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsMedium | |
A medium statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties A. False: for some atoms (noble gases, Be, Mg, N) adding an electron absorbs energy.
B. True: removing an electron always needs energy (ionisation enthalpy is positive).
C. True: B ($2s^22p^1$) loses a $2p$ electron that is shielded by the $2s$ pair, so $IE_1(\mathrm{B}) \approx 801 < IE_1(\mathrm{Be}) \approx 899$ kJ mol$^{-1}$.
D. False: electronegativity is not fixed; it depends on the hybridisation and the atoms attached, so C bonded to four Cl is more electronegative than C in $\mathrm{CH_4}$.
E. False: electropositive character increases down the group; Li is the least electropositive alkali metal.
B and C only.
Answer: (B). | ||||
| 61 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: The N–N single bond is indeed weaker than P–P (strong repulsion between the lone pairs on the small N atoms), but it is shorter (about 145 pm against about 221 pm for P–P). "Weaker and longer" is therefore false.
Statement II: Going down group 15 the +3 state becomes increasingly stable (inert pair effect); for As, Sb and Bi the +3 state is stable towards disproportionation. Only some intermediate states of N and P disproportionate (e.g. $\mathrm{3HNO_2 \rightarrow HNO_3 + 2NO + H_2O}$, $\mathrm{4H_3PO_3 \rightarrow 3H_3PO_4 + PH_3}$). As a general statement about group 15 it is false.
Both statements are false.
Answer: (B). | ||||
| 62 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | StatementsEasy | |
A easy statements question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mu = \sqrt{n(n+2)}$ B.M.; 4.9 B.M. means $n = 4$ unpaired electrons ($\sqrt{24} = 4.90$).
$\mathrm{Cr^{2+}}$ $3d^4$: 4 ✓. $\mathrm{Fe^{2+}}$ $3d^6$: 4 ✓. $\mathrm{Fe^{3+}}$ $3d^5$: 5 (5.92 B.M.). $\mathrm{Co^{2+}}$ $3d^7$: 3 (3.87 B.M.). $\mathrm{Mn^{3+}}$ $3d^4$: 4 ✓.
A, B and E only.
Answer: (A). | ||||
| 63 | 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 The energy absorbed matches the crystal field splitting $\Delta_o$, so the wavelength absorbed is larger when $\Delta_o$ is smaller.
Spectrochemical series: $\mathrm{Cl^- < H_2O < NH_3 < CN^-}$. All four are Co(III); replacing one $\mathrm{NH_3}$ by $\mathrm{H_2O}$ or $\mathrm{Cl^-}$ lowers $\Delta_o$, $\mathrm{Cl^-}$ most.
$\Delta_o$: C > B > A > D, so wavelength absorbed: D > A > B > C.
Answer: (D). | ||||
| 64 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsMedium | |
A medium statements question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A. Pentan-3-one ($\mathrm{C_2H_5COC_2H_5}$) and pentan-2-one ($\mathrm{CH_3COC_3H_7}$): same formula $\mathrm{C_5H_{10}O}$ and same functional group, with different alkyl groups on either side of it. These are metamers: correct.
B. Propyl cyanide ($\mathrm{C_3H_7CN}$) and propyl isocyanide ($\mathrm{C_3H_7NC}$): same formula, different functional groups: functional isomers: correct.
C. As drawn, the first is butan-1-ol ($\mathrm{C_4H_{10}O}$) and the second is pentan-2-ol ($\mathrm{CH_3CH_2CH_2CH(OH)CH_3}$, $\mathrm{C_5H_{12}O}$). Different molecular formulae, so they are not isomers at all: incorrect.
D. Butan-1-amine (a primary amine) and N-ethylpropan-1-amine (a secondary amine) belong to different classes, so they are not members of one homologous series: incorrect.
A & B only.
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 3-Methyl-6-oxoheptanal: $\mathrm{OHC{-}CH_2{-}CH(CH_3){-}CH_2{-}CH_2{-}CO{-}CH_3}$ (C1 to C7).
Ozonolysis of a ring alkene gives one open chain with a carbonyl at each end; joining the two carbonyl carbons (C1 and C6) back by a double bond rebuilds the ring: C1=C6–C5–C4–C3–C2–C1, a cyclohexene.
On that ring, C6 (the former ketone carbon) carries a methyl and is one alkene carbon; C1 (the former aldehyde) is the other, with an H. C3, bearing the second methyl, is three bonds from C6 either way round the ring, i.e. opposite to it.
So the alkene is 1,4-dimethylcyclohex-1-ene: a methyl on one alkene carbon and another at the para position. That is the third structure.
Answer: (C). | ||||
| 66 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines [A] gives benzenediazonium chloride with $\mathrm{NaNO_2}$/HCl at 273–278 K: diazotisation of a primary aromatic amine, so [A] is aniline.
Diazonium salt with KI: the diazo group is replaced by iodine, so [B] is iodobenzene.
Iodobenzene with sodium in dry ether (Fittig reaction): $\mathrm{2C_6H_5I + 2Na \rightarrow C_6H_5{-}C_6H_5 + 2NaI}$, so [C] is biphenyl.
Answer: (D). | ||||
| 67 | 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 Approximate $pK_a$ values:
(B) benzenesulfonic acid, a strong acid ($pK_a < 0$);
(C) butanoic acid, about 4.8;
(A) a phenol, about 9.5–10 (the meta-ethoxy group changes it little);
(D) the terminal alkyne C–H of ethyl propiolate: an sp carbon acid, far weaker even with the ester attached ($pK_a$ around 20 or more).
Least acidic: D.
Answer: (B), i.e. compound D. | ||||
| 68 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The iodoform test is given by $\mathrm{CH_3CO{-}}$ compounds (and ethanal) and by alcohols $\mathrm{CH_3CH(OH){-}}$ that are oxidised to them (and ethanol).
Positive: ethanol, isopropyl alcohol ($\mathrm{CH_3CH(OH)CH_3}$), bromoacetone ($\mathrm{CH_3COCH_2Br}$, still a methyl ketone), 2-butanol, 2-butanone, 2-pentanone.
Negative: butanal, 3-pentanone, pentanal, 3-pentanol (no $\mathrm{CH_3CO}$ or $\mathrm{CH_3CH(OH)}$ unit).
Number that cannot: 4.
Answer: (D) 4. | ||||
| 69 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Benzenediazonium chloride:
with ethanol it is reduced to benzene ($\mathrm{ArN_2Cl + CH_3CH_2OH \rightarrow ArH + N_2 + CH_3CHO + HCl}$), not converted to ethoxybenzene: (A) is NOT correct;
with $\mathrm{H_3PO_2}$/water it is also reduced to benzene: (B) correct;
with CuCN/KCN it gives benzonitrile (Sandmeyer): (C) correct;
with potassium iodide it gives iodobenzene: (D) correct.
Answer: (A). | ||||
| 70 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules D-Fructose in the Fischer projection (C1 at the top): C1 $\mathrm{CH_2OH}$, C2 C=O, C3 OH on the left, C4 OH on the right, C5 OH on the right, C6 $\mathrm{CH_2OH}$.
L-Fructose is its mirror image: every stereocentre is reversed, so C3 OH on the right, C4 OH on the left, C5 OH on the left (the OH on the lowest stereocentre, C5, on the left, as for every L-sugar).
The second structure has, below C=O, OH right, HO left, HO left: that is L-fructose.
(The first has left, right, left; the third right, right, left; the fourth left, left, left.)
Answer: (B). | ||||
| 71 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Formation: $\mathrm{2C(graphite) + 2H_2(g) \rightarrow C_2H_4(g)}$.
Energy in: sublime 2 C, $2 \times 710 = 1420$; break 2 H–H, $2 \times 436 = 872$. Total 2292 kJ.
Energy out: form 4 C–H, $4 \times 414 = 1656$; form 1 C=C, 611. Total 2267 kJ.
$\Delta H_f^\ominus = 2292 - 2267 = 25$ kJ mol$^{-1}$.
Answer: 25. | ||||
| 72 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Oxalic acid neutralised by KOH gives oxalate, which chelates iron(III): A is $\mathrm{K_3[Fe(C_2O_4)_3]}$, potassium trioxalatoferrate(III).
A tris-chelate octahedral complex $\mathrm{[M(AA)_3]}$ has no plane of symmetry; it exists as a non-superimposable pair of mirror images ($\Delta$ and $\Lambda$).
Number of optical isomers: 2.
Answer: 2. | ||||
| 73 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements This is the chromyl chloride test: a dichromate (A, e.g. $\mathrm{K_2Cr_2O_7}$) heated with NaCl and conc. $\mathrm{H_2SO_4}$ gives red vapours of $\mathrm{CrO_2Cl_2}$.
$\mathrm{CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O}$: B is sodium chromate.
Acidified: $\mathrm{2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O}$: C is the dichromate.
In $\mathrm{Cr_2O_7^{2-}}$ two $\mathrm{CrO_3}$ units share one bridging O; the other 6 oxygens are terminal.
Answer: 6. | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Mass of C $= 1.46 \times \frac{12}{44} = 0.3982$ g.
% C $= \frac{0.3982}{0.5} \times 100 = 79.6\% \approx 80$.
Answer: 80. | ||||
| 75 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium 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 X is piperazine, a six-membered ring of four $\mathrm{CH_2}$ and two NH: $\mathrm{C_4H_{10}N_2}$, molar mass $48 + 10 + 28 = 86$ g mol$^{-1}$.
Moles of X $= \frac{0.42}{86} = 4.884 \times 10^{-3}$; each molecule gives one $\mathrm{N_2}$.
Volume at STP $= 4.884 \times 10^{-3} \times 22400 = 109.4$ mL ≈ 109.
Answer: 109. | ||||
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