JEE (Main) · 2025
JEE (Main) 2025 — 23 January, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
23 Q
- Sets, Relations and Functions2
- Matrices and Determinants2
- Conic Sections2
9 easy · 13 medium · 1 hard
Physics
23 Q
- Properties of Bulk Matter3
- Optics3
- Physical World and Measurement2
21 easy · 2 medium · 0 hard
Chemistry
23 Q
- Solutions2
- Equilibrium2
- d- and f-Block Elements2
12 easy · 9 medium · 2 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 FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions On the line $x = 0$: $A$ needs $|y| \ge 3$ and $B$ needs $|y| \le 3$, so $y = \pm3$: the points $(0, 3)$ and $(0, -3)$.
On the line $y = 0$: likewise $x = \pm3$: the points $(3, 0)$ and $(-3, 0)$.
So $C = \{(0, 3), (0, -3), (3, 0), (-3, 0)\}$, and each point has $|x + y| = 3$.
$\sum_{(x, y)\in C}|x + y| = 4\times3 = 12$.
Answer: (D) $12$. | ||||
| 2 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Statement I: $b_1 = b_1$ (reflexive); $b_1 = b_2 \Rightarrow b_2 = b_1$ (symmetric); $b_1 = b_2,\ b_2 = b_3 \Rightarrow b_1 = b_3$ (transitive). R is an equivalence relation: true.
Statement II: $(x, y)\,R\,(a, b)$ means $y = b$, so $S$ is the horizontal line $y = b$, parallel to the x-axis, never to $y = x$, whatever $(a, b)$ is: false.
Answer: (C). | ||||
| 3 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Write $z = e^{i\theta}$, $0 \le \theta < 2\pi$. Then $\frac{z}{\bar{z}} = e^{2i\theta}$ and $\frac{\bar{z}}{z} = e^{-2i\theta}$, so the sum is $2\cos2\theta$.
$|2\cos2\theta| = 1 \Rightarrow \cos2\theta = \pm\frac{1}{2}$.
As $\theta$ runs over $[0, 2\pi)$, $2\theta$ runs over $[0, 4\pi)$; in each turn of $2\pi$ the equations $\cos\phi = \frac12$ and $\cos\phi = -\frac12$ have two solutions each, so 4 per turn, 8 in all.
$\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{11\pi}{6}$.
Answer: (C) $8$. | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}1&1&1\\1&2&5\\1&5&\lambda\end{vmatrix} = (2\lambda - 25) - (\lambda - 5) + (5 - 2) = \lambda - 17$.
For $\lambda \ne 17$ the solution is unique, so no solution needs $\lambda = 17$.
With $\lambda = 17$ the third left side is a combination of the first two: $(1, 5, 17) = -3(1, 1, 1) + 4(1, 2, 5)$.
So the system is consistent only if $\mu = -3(6) + 4(9) = 18$; for $\mu \ne 18$ it has no solution.
Answer: (A) $\lambda = 17,\ \mu \ne 18$. | ||||
| 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 Let $c_1, c_2, c_3$ be the columns of $A$. The first condition gives $c_2 = (0, 0, 1)^T$.
Second: $4c_1 + c_2 + 3c_3 = (0, 1, 0)^T$, so $4c_1 + 3c_3 = (0, 1, -1)^T$.
Third: $2c_1 + c_2 + 2c_3 = (1, 0, 0)^T$, so $2c_1 + 2c_3 = (1, 0, -1)^T$, i.e. $4c_1 + 4c_3 = (2, 0, -2)^T$.
Subtracting: $c_3 = (2, 0, -2)^T - (0, 1, -1)^T = (2, -1, -1)^T$.
$a_{23}$ is the second entry of the third column: $a_{23} = -1$.
Answer: (C) $-1$. | ||||
| 6 | 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 $(1 + x)^p(1 - x)^q = \left(1 + px + \binom{p}{2}x^2 + \cdots\right)\left(1 - qx + \binom{q}{2}x^2 - \cdots\right)$.
Coefficient of $x$: $p - q = 1$.
Coefficient of $x^2$: $\binom{p}{2} + \binom{q}{2} - pq = \frac{(p - q)^2 - (p + q)}{2} = -2$, so $1 - (p + q) = -4$, $p + q = 5$.
$p = 3$, $q = 2$, and $p^2 + q^2 = 9 + 4 = 13$.
Answer: (B) $13$. | ||||
| 7 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Number of ways to choose 2 squares: $\binom{16}{2} = 120$.
Pairs sharing a side: in each of the 4 rows there are 3 horizontally adjacent pairs, and in each of the 4 columns 3 vertically adjacent pairs: $12 + 12 = 24$.
Pairs with no common side: $120 - 24 = 96$.
Probability $= \frac{96}{120} = \frac{4}{5}$.
Answer: (D) $\frac{4}{5}$. | ||||
| 8 | Straight LinesSection formula | Class 11: Straight Lines | Single correctHard | |
A hard single correct question on section formula, from Straight Lines. NCERT: Class 11: Straight Lines Let $A = (a, a + 2)$ and $B = (b, -2)$. $AB = 8$: $(a - b)^2 + (a + 4)^2 = 64$.
P divides AB in $2 : 1$: $P = \frac{A + 2B}{3} = \left(\frac{a + 2b}{3}, \frac{a - 2}{3}\right)$.
So $a = 3y + 2$ and $b = \frac{3x - a}{2} = \frac{3x - 3y - 2}{2}$, giving $a - b = \frac{9y - 3x + 6}{2}$ and $a + 4 = 3y + 6$.
Then $\frac{(9y - 3x + 6)^2}{4} + (3y + 6)^2 = 64$, i.e. $9(3y - x + 2)^2 + 36(y + 2)^2 = 256$.
Expanding: $9x^2 + 81y^2 - 54xy - 36x + 108y + 36 + 36y^2 + 144y + 144 = 256$,
$9x^2 + 117y^2 - 54xy - 36x + 252y - 76 = 0$, i.e. $9(x^2 + 13y^2 - 6xy - 4x + 28y) - 76 = 0$.
So $\alpha = 13$, $\beta = -6$, $\gamma = -4$, and $\alpha - \beta - \gamma = 13 + 6 + 4 = 23$.
Answer: (C) $23$. | ||||
| 9 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Chord with a given mid-point: $T = S_1$: $\frac{x}{4} + \frac{y}{4} = \frac{1}{4} + \frac{1}{8}$, i.e. $x + y = \frac{3}{2}$.
Substitute $y = \frac32 - x$: $\frac{x^2}{4} + \frac{(3/2 - x)^2}{2} = 1 \Rightarrow 6x^2 - 12x + 1 = 0$.
$|x_1 - x_2| = \frac{\sqrt{144 - 24}}{6} = \frac{\sqrt{120}}{6}$. On a line of slope $-1$, length $= \sqrt2\,|x_1 - x_2| = \frac{\sqrt{240}}{6} = \frac{4\sqrt{15}}{6} = \frac{2}{3}\sqrt{15}$.
Answer: (D) $\frac{2}{3}\sqrt{15}$. | ||||
| 10 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The normal at $(t^2, 2t)$ is $y = -tx + 2t + t^3$; through $(a, 0)$: $a = 2 + t^2$ (for $a > 2$ there is a $t \ne 0$; for $a \le 2$ the nearest point is the vertex and the distance is $a \le 2$, not 4).
Distance$^2$ $= (t^2 - a)^2 + 4t^2 = 4 + 4t^2 = 16 \Rightarrow t^2 = 3$, $a = 5$.
The circle passes through $(5, 0)$ and the focus $(1, 0)$ with its centre on the x-axis: centre $(3, 0)$, radius 2.
$(x - 3)^2 + y^2 = 4$, i.e. $x^2 + y^2 - 6x + 5 = 0$.
Answer: (C). | ||||
| 12 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Through $(1, 4, 0)$ draw a line parallel to $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$: $(1 + t, 4 + 2t, 3t)$.
It meets $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ when $\frac{t - 1}{2} = \frac{2t - 2}{3} = \frac{3t - 3}{4}$, which holds for $t = 1$: the point $(2, 6, 3)$.
Distance from $(1, 4, 0)$: $\sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.
Answer: (A) $\sqrt{14}$. | ||||
| 13 | 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 $\vec{a}_1 = (2, 1, -3)$, $\vec{b}_1 = (1, 2, -3)$; $\vec{a}_2 = (-1, -3, -5)$, $\vec{b}_2 = (2, 4, -5)$.
$\vec{b}_1\times\vec{b}_2 = (2\cdot(-5) - (-3)\cdot4,\ (-3)\cdot2 - 1\cdot(-5),\ 1\cdot4 - 2\cdot2) = (2, -1, 0)$, of length $\sqrt5$.
$\vec{a}_2 - \vec{a}_1 = (-3, -4, -2)$; its dot product with $(2, -1, 0)$ is $-6 + 4 = -2$.
Shortest distance $= \frac{2}{\sqrt5}$, its square $\frac{4}{5}$: $m + n = 9$.
Answer: (B) $9$. | ||||
| 14 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\frac{2x^2 - 3x + 5}{3x^2 + 5x + 4} \to \frac{2}{3}$.
$\frac{(3x - 1)^{x/2}}{(3x + 2)^{x/2}} = \left(1 - \frac{3}{3x + 2}\right)^{x/2} \to e^{\lim \left(-\frac{3}{3x + 2}\cdot\frac{x}{2}\right)} = e^{-1/2}$.
Limit $= \frac{2}{3}e^{-1/2} = \frac{2}{3\sqrt e}$.
Answer: (B). | ||||
| 15 | Applications of DerivativesRate of change | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on rate of change, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let the ball have radius $r$ and the layer thickness $h$. Ice-cream volume $V = \frac43\pi(r + h)^3 - \frac43\pi r^3$, so $\frac{dV}{dt} = 4\pi(r + h)^2\frac{dh}{dt}$.
$81 = 4\pi(r + 1)^2\cdot\frac{1}{4\pi} = (r + 1)^2 \Rightarrow r + 1 = 9$, $r = 8$ cm.
Surface area of the ball $= 4\pi(8)^2 = 256\pi$ cm$^2$.
Answer: (C) $256\pi$. | ||||
| 16 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals For $x \le 0$ the upper curve is $a + e^{-x} - e^{-x} = a > 0$; for $x \ge 0$ it is $a + e^x - e^{-x} > 0$. So the area is $\int_{-1}^{1}(a + e^{|x|} - e^{-x})\,dx$.
$\int_{-1}^1 a\,dx = 2a$; $\int_{-1}^1 e^{|x|}dx = 2(e - 1)$; $\int_{-1}^1 e^{-x}dx = e - \frac1e$.
Area $= 2a + 2e - 2 - e + \frac1e = 2a - 2 + e + \frac1e$.
Given area $= e + 8 + \frac1e$, so $2a - 2 = 8$, $a = 5$.
Answer: (A) $5$. | ||||
| 17 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals By parts repeatedly: $\int x^3\sin x\,dx = -x^3\cos x + 3x^2\sin x + 6x\cos x - 6\sin x + C$, so $g(x) = -x^3\cos x + 3x^2\sin x + 6x\cos x - 6\sin x$.
$g\left(\frac\pi2\right) = 3\cdot\frac{\pi^2}{4} - 6$ and $g'\left(\frac\pi2\right) = \left(\frac\pi2\right)^3\sin\frac\pi2 = \frac{\pi^3}{8}$.
$8\left(\frac{3\pi^2}{4} - 6 + \frac{\pi^3}{8}\right) = \pi^3 + 6\pi^2 - 48$: $\alpha = 1$, $\beta = 6$, $\gamma = -48$.
$\alpha + \beta - \gamma = 1 + 6 + 48 = 55$.
Answer: (A) $55$. | ||||
| 19 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $x = vy$: $\frac{dx}{dy} = v + y\frac{dv}{dy}$, so $x - y\frac{dx}{dy} = -y^2\frac{dv}{dy}$.
The equation becomes $y = -y^2\frac{dv}{dy}\sin v$, i.e. $\sin v\,dv = -\frac{dy}{y}$.
Integrating: $\cos v = \ln y + c$. At $y = 1$, $v = \frac{\pi}{2}$: $0 = 0 + c$, so $\cos\frac{x}{y} = \ln y$.
At $y = 2$: $\cos\frac{x(2)}{2} = \ln2$, and $\cos x(2) = 2\cos^2\frac{x(2)}{2} - 1 = 2(\log_e2)^2 - 1$.
Answer: (B). | ||||
| 20 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\overrightarrow{OA} = \frac{1}{r + 1}(5r - 1,\ 5r - 1,\ 10r + 2)$.
$\overrightarrow{OQ}\cdot\overrightarrow{OA} = \frac{5(5r - 1) + 5(5r - 1) + 10(10r + 2)}{r + 1} = \frac{150r + 10}{r + 1}$.
With $u = 5r - 1$, $w = 10r + 2$: $(-1, -1, 2)\times(u, u, w) = (-w - 2u,\ 2u + w,\ 0)$, and $2u + w = 20r$, so $\left|\overrightarrow{OP}\times\overrightarrow{OA}\right|^2 = \frac{800r^2}{(r + 1)^2}$.
$\frac{150r + 10}{r + 1} - \frac{160r^2}{(r + 1)^2} = 10 \Rightarrow (150r + 10)(r + 1) - 160r^2 = 10(r + 1)^2$.
$-10r^2 + 160r + 10 = 10r^2 + 20r + 10 \Rightarrow 20r^2 = 140r \Rightarrow r = 7$ (as $r > 0$).
Answer: (C) $7$. | ||||
| 21 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_{11} = 11a_6 = 88 \Rightarrow a_6 = 8$. Then $a_{10} = 8 + 4\cdot\frac32 = 14$ and $a_{11} = 15.5$.
Sum of roots $\frac{p}{3} = 29.5 \Rightarrow p = 88.5$; product $\frac{q}{3} = 14\times15.5 = 217 \Rightarrow q = 651$.
$q - 2p = 651 - 177 = 474$.
Answer: 474. | ||||
| 22 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Since $\alpha^2 = a\alpha + b$ and $\beta^2 = a\beta + b$: $P_{n+1} = aP_n + bP_{n-1}$.
$P_5 = aP_4 + bP_3$: $11 = -3a - 5b$. $P_6 = aP_5 + bP_4$: $45 = 11a - 3b$.
Solving: $a = 3$, $b = -4$; the equation is $x^2 - 3x + 4 = 0$, $\alpha + \beta = 3$, $\alpha\beta = 4$.
$\alpha^2 + \beta^2 = 9 - 8 = 1$; $\alpha^4 + \beta^4 = 1^2 - 2\cdot16 = -31$.
(Check: $\alpha - \beta = -\sqrt7\,i$ and $P_3 = (\alpha - \beta)(\alpha^2 + \alpha\beta + \beta^2) = -5\sqrt7\,i$.)
$|\alpha^4 + \beta^4| = 31$.
Answer: 31. | ||||
| 23 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations All boys together: treat them as one block with the 4 girls: $5!$ arrangements of the units, $5!$ inside the block: $120\times120 = 14400$.
No two boys together: arrange the girls ($4!$), then put the 5 boys in the 5 gaps, one each ($5!$): $24\times120 = 2880$.
The two cases cannot both happen, so the total is $14400 + 2880 = 17280$.
Answer: 17280. | ||||
| 24 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics An A.P. with $a = 8$, $d = 13$: $320 = 8 + 13(n - 1) \Rightarrow n = 25$.
Variance of $n$ terms of an A.P. $= \frac{d^2(n^2 - 1)}{12} = \frac{169\times624}{12} = 169\times52 = 8788$.
Answer: 8788. | ||||
| 25 | CirclesStandard and general form | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections $y^2 = 4(x + 4)$ has focus $(-3, 0)$; C: $(x + 3)^2 + y^2 = 25$.
Intersection: $y = 3x$, $x(1 + 3\lambda) = 4$, so $x = 4k$, $y = 12k$ with $k = \frac{1}{1 + 3\lambda}$.
$(4k + 3)^2 + 144k^2 = 25 \Rightarrow 160k^2 + 24k - 16 = 0 \Rightarrow 20k^2 + 3k - 2 = 0 \Rightarrow k = \frac14$ or $k = -\frac25$.
$1 + 3\lambda = 4 \Rightarrow \lambda = 1$; $1 + 3\lambda = -\frac52 \Rightarrow \lambda = -\frac76$. So $\lambda_1 = -\frac76$, $\lambda_2 = 1$.
$12\lambda_1 + 29\lambda_2 = -14 + 29 = 15$.
Answer: 15. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (A) $\mu_0$: from $F = \frac{\mu_0 I_1I_2 l}{2\pi d}$, $[\mu_0] = \frac{MLT^{-2}}{A^2} = [MLT^{-2}A^{-2}]$ → (III).
(B) $B$: from $F = qvB = ILB$, $[B] = \frac{MLT^{-2}}{AL} = [MT^{-2}A^{-1}]$ → (II).
(C) Magnetic moment $m = IA$: $[L^2A]$ → (IV).
(D) Torsional constant $C = \frac{\tau}{\theta}$: torque per radian, $[ML^2T^{-2}]$ → (I).
Answer: (A) A→III, B→II, C→IV, D→I. | ||||
| 27 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\ln E = 3\ln\alpha - \beta t$, so $\frac{\Delta E}{E} = 3\frac{\Delta\alpha}{\alpha} + \beta\,\Delta t = 3\frac{\Delta\alpha}{\alpha} + \beta t\frac{\Delta t}{t}$ (errors added for the maximum).
$= 3(1.2\%) + (0.3\times5)(1.6\%) = 3.6\% + 2.4\% = 6\%$.
Answer: (A) 6%. | ||||
| 28 | 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 At the highest point only the horizontal component $u\cos60^\circ = \frac{u}{2}$ remains.
$KE_{top} = \frac12m\frac{u^2}{4} = \frac{KE}{4}$.
Answer: (D) $\frac{KE}{4}$. | ||||
| 29 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy 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 $\omega = \frac{d\theta}{dt} = 10t - 8$, $\alpha = 10\ \mathrm{rad\,s^{-2}}$. At $t = 2$ s, $\omega = 12\ \mathrm{rad\,s^{-1}}$.
About the perpendicular axis through the centre, $I = \frac12MR^2$, so $\tau = I\alpha = 5MR^2$.
$P = \tau\omega = 5MR^2\times12 = 60MR^2$ (W).
Answer: (B) $60MR^2$. | ||||
| 30 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Kepler's third law: $T \propto r^{3/2}$.
$T_s = 27\times\left(\frac19\right)^{3/2} = \frac{27}{27} = 1$ day.
Answer: (D) 1 day. | ||||
| 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 Hooke's law: tension $= kx$, so $x_1 = \frac{5}{k}$ and $x_2 = \frac{7}{k}$.
$5x_1 - 2x_2 = \frac{25 - 14}{k} = \frac{11}{k}$, which needs a tension of 11 N.
Answer: (B) 11 N. | ||||
| 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 correctEasy | |
A easy single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Bernoulli along the horizontal pipe: $P_1 = P_2 + \frac12\rho v^2$, so $v = \sqrt{\frac{2(P_1 - P_2)}{\rho}} \propto \sqrt{P_1 - P_2}$.
Answer: (B). | ||||
| 33 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics From the diagram: A $(2V_0, P_0)$ → B $(3V_0, P_0)$ → C $(3V_0, 2P_0)$ → D $(V_0, 2P_0)$.
A→B, isobaric expansion: $W = P_0(3V_0 - 2V_0) = P_0V_0$.
B→C, isochoric: $W = 0$.
C→D, isobaric compression: $W = 2P_0(V_0 - 3V_0) = -4P_0V_0$.
Total: $P_0V_0 - 4P_0V_0 = -3P_0V_0$.
Answer: (D) $-3P_0V_0$. | ||||
| 34 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $dS = \frac{dQ}{T} = \frac{mc\,dT}{T}$, so $\Delta S = mc\ln\frac{T_2}{T_1}$.
With $c = 1$ this is $m\ln\frac{T_2}{T_1}$; heating raises the entropy, so the sign is positive.
Answer: (A) $m\ln\left(\frac{T_2}{T_1}\right)$. | ||||
| 37 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $S = \frac{I_gG}{I - I_g} = \frac{0.02\times30}{3 - 0.02} = \frac{0.6}{2.98} = \frac{30}{149}\ \Omega$.
Answer: (A) 149. | ||||
| 38 | 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 Separation $r = 14$ cm $= 0.14$ m; $\frac{1}{4\pi\epsilon_0} \approx 9\times10^9$.
$U = \frac{9\times10^9\times7\times10^{-6}\times(-4\times10^{-6})}{0.14} = \frac{-0.252}{0.14} = -1.8$ J.
Answer: (C) −1.8 J. | ||||
| 39 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $B = \frac{E}{c} = \frac{9.3}{3\times10^8} = 3.1\times10^{-8}$ T.
$\vec{E}\times\vec{B}$ points along $+x$; with $\vec{E}$ along $+y$, $\hat{y}\times\hat{z} = \hat{x}$, so $\vec{B}$ is along $+z$.
Answer: (C) $B_z = 3.1\times10^{-8}$ T. | ||||
| 40 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\mu = \frac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$ with $\delta_m = A$: $\mu = \frac{\sin A}{\sin\frac A2} = 2\cos\frac A2$.
$2\cos\frac A2 = \sqrt3 \Rightarrow \frac A2 = 30^\circ$, $A = 60^\circ$.
Answer: (C) 60°. | ||||
| 41 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Field amplitudes are in the ratio of the widths: $E_1 : E_2 = 1 : x$.
$\frac{I_{max}}{I_{min}} = \left(\frac{x + 1}{x - 1}\right)^2 = \frac94 \Rightarrow \frac{x + 1}{x - 1} = \frac32 \Rightarrow 2x + 2 = 3x - 3 \Rightarrow x = 5$.
(The other root, $x = \frac15$, is not among the options.)
Answer: (B) 5. | ||||
| 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 A mirror's focal length is $f = \frac{R}{2}$, fixed by its geometry alone; reflection does not depend on the surrounding medium.
So the focal length stays $f$.
Answer: (C) $f$. | ||||
| 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 $\frac{hc}{\lambda} = \phi + eV_0 = 2.14 + 2 = 4.14$ eV.
$\lambda = \frac{1242}{4.14} = 300$ nm.
Answer: (B) 300 nm. | ||||
| 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 (A): for mass numbers about 30 to 170 the binding energy per nucleon stays close to 8 MeV, nearly constant: true.
(R): the nuclear force is short range (a few fm); its saturation, each nucleon bonding only with its near neighbours, is exactly why the binding energy per nucleon is nearly constant: false.
Answer: (C). | ||||
| 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 In the figure the longer plate of the cell is on the left, so current leaves the cell to the left, goes up to the left junction and passes through both branches from left to right.
Both diodes point left to right, so both are forward biased and (ideal) conduct with no drop: the two 20 $\Omega$ resistors are in parallel, 10 $\Omega$.
$I = \frac{5}{10} = 0.5$ A.
Answer: (B) 0.5 A. | ||||
| 46 | GravitationOrbital motion | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Orbit radius $r = R + \frac R3 = \frac{4R}{3}$; orbital speed $v = \sqrt{\frac{GM}{r}}$.
$L = mvr = m\sqrt{GMr} = \frac M2\sqrt{GM\cdot\frac{4R}{3}} = M\sqrt{\frac{GMR}{3}}$.
$x = 3$.
Answer: 3. | ||||
| 47 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter An air bubble in a liquid has one surface: excess pressure $\frac{2T}{r} = \frac{2\times0.095}{10^{-3}} = 190$ Pa.
Liquid pressure at the depth: $\rho gh = 10^3\times10\times0.2 = 2000$ Pa.
$P_{in} - P_{atm} = 2000 + 190 = 2190\ \mathrm{N/m^2}$.
Answer: 2190. | ||||
| 48 | Electromagnetic WavesDisplacement current | Class 12: Electromagnetic Waves | NumericalEasy | |
A easy numerical question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $i_d = \epsilon_0\frac{d\Phi_E}{dt} = C\frac{dV}{dt}$.
$\frac{dV}{dt} = \frac{0.25\times10^{-3}}{2.5\times10^{-6}} = 100\ \mathrm{V\,s^{-1}}$.
Answer: 100. | ||||
| 49 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Maximum current at resonance: $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1\times25\times10^{-9}}} = \frac{1}{\sqrt{2.5\times10^{-9}}} = \frac{1}{5\times10^{-5}} = 2\times10^4\ \mathrm{rad\,s^{-1}}$.
Answer: 2. | ||||
| 50 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance At steady state no current flows into the capacitor; the 10 $\Omega$ and 15 $\Omega$ resistors carry $I = \frac{5}{25} = 0.2$ A in series.
The 8 $\mu$F capacitor is connected across the 10 $\Omega$ resistor, so its voltage is $0.2\times10 = 2$ V.
$Q = CV = 8\times2 = 16\ \mu$C.
Answer: 16. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 52 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta G = \Delta H - T\Delta S$.
(A) $\Delta H > 0$, $\Delta S < 0$: $\Delta G > 0$ at every T, non-spontaneous: correct.
(B) $\Delta H > 0$, $\Delta S > 0$: spontaneous only at high T; at low T $\Delta G > 0$: incorrect.
(C) $\Delta H < 0$, $\Delta S < 0$: spontaneous at low T (non-spontaneous at high T): incorrect.
(D) $\Delta H < 0$, $\Delta S > 0$: $\Delta G < 0$ at every T, spontaneous: correct.
Incorrect: (B) and (C).
Answer: (C). | ||||
| 53 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\Delta G^\circ = -nFE^\circ_{cell}$ with $E^\circ_{cell} = E^\circ_{right} - E^\circ_{left}$; for a fixed $n = 2$ (per mole of the reaction as balanced with a divalent ion) the largest positive $E^\circ_{cell}$ gives the most negative $\Delta G^\circ$.
(A) $-2.37 - 0.80 = -3.17$ V; (B) $-2.37 - (-0.76) = -1.61$ V: both give $\Delta G^\circ > 0$.
(C) $0.80 - 0.34 = 0.46$ V; (D) $0.80 - (-0.76) = 1.56$ V.
(D): $\Delta G^\circ = -2F(1.56)$, the most negative.
Answer: (D). | ||||
| 54 | SolutionsRaoult's law | Class 12: Solutions | Single correctEasy | |
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Relative lowering: $\frac{\Delta p}{p^\circ} = x_{solute}$: $\frac{10}{p^\circ} = 0.2 \Rightarrow p^\circ = 50$ mm Hg.
For $\Delta p = 20$ mm: $x_{solute} = \frac{20}{50} = 0.4$, so $x_{solvent} = 0.6$.
Answer: (B) 0.6. | ||||
| 55 | SolutionsRaoult's law | Class 12: Solutions | Single correctMedium | |
A medium single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Raoult: $P = x_1P_1^0 + (1 - x_1)P_2^0$ and $y_1 = \frac{x_1P_1^0}{P}$.
$\frac{1}{y_1} = \frac{P_2^0 + x_1(P_1^0 - P_2^0)}{x_1P_1^0} = \frac{P_2^0}{P_1^0}\cdot\frac{1}{x_1} + \frac{P_1^0 - P_2^0}{P_1^0}$.
Solving for $\frac{1}{x_1}$: $\frac{1}{x_1} = \frac{P_1^0}{P_2^0}\cdot\frac{1}{y_1} + \frac{P_2^0 - P_1^0}{P_2^0}$.
Plotted as $\frac{1}{x_1}$ (vertical) against $\frac{1}{y_1}$: slope $\frac{P_1^0}{P_2^0}$, intercept $\frac{P_2^0 - P_1^0}{P_2^0}$.
Answer: (B). | ||||
| 56 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Starting with 1 mol: $\mathrm{X_2Y}$ $1 - x$, $\mathrm{X_2}$ $x$, $\mathrm{Y_2}$ $\frac{x}{2}$; total $1 + \frac{x}{2}$, total pressure $p$.
$K_p = \frac{\left(\frac{x}{1 + x/2}p\right)\left(\frac{x/2}{1 + x/2}p\right)^{1/2}}{\frac{1 - x}{1 + x/2}p}$.
For $x \ll 1$: $K_p \approx x\left(\frac{x}{2}\right)^{1/2}p^{1/2} = \frac{x^{3/2}p^{1/2}}{\sqrt2}$.
$x^{3/2} = \frac{\sqrt2K_p}{\sqrt p} \Rightarrow x^3 = \frac{2K_p^2}{p} \Rightarrow x = \sqrt[3]{\frac{2K_p^2}{p}}$.
Answer: (B). | ||||
| 57 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For a zero order reaction, rate $= k$, constant, and $[R] = [R]_0 - kt$: concentration falls linearly with time.
(A) a falling rate and (C) a decaying rate are not zero order; (B) a constant concentration means no reaction.
(D), concentration falling in a straight line, is zero order.
Answer: (D). | ||||
| 58 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Ionisation of water is endothermic, so $K_w$ increases on heating; $[\mathrm{H^+}] = [\mathrm{OH^-}] = \sqrt{K_w}$ both increase (the water stays neutral).
With larger $[\mathrm{H^+}]$, pH $= -\log[\mathrm{H^+}]$ falls below 7.
Answer: (B) Decrease. | ||||
| 60 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter The numbers are C (6), Si (14), Sn (50) and Pb (82).
Melting points: C (diamond/graphite) above 3500 °C, Si about 1410 °C, Pb about 327 °C, Sn about 232 °C.
Tin has the lowest melting point in the group: atomic number 50.
Answer: (D) 50. | ||||
| 61 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In alkali dichromate turns to chromate: $\mathrm{K_2Cr_2O_7 + 2KOH \rightarrow 2K_2CrO_4 + H_2O}$, [A] $= \mathrm{K_2CrO_4}$.
In acid chromate turns back to dichromate: $\mathrm{2K_2CrO_4 + H_2SO_4 \rightarrow K_2Cr_2O_7 + K_2SO_4 + H_2O}$, [B] $= \mathrm{K_2Cr_2O_7}$.
Answer: (C). | ||||
| 62 | 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 Bronze is copper with tin (IV); brass is copper with zinc (III); UK "silver" coins are cupro-nickel, copper with nickel (I); stainless steel is iron with chromium, nickel and carbon (II).
A→IV, B→III, C→I, D→II.
Answer: (B). | ||||
| 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 (A) $\mathrm{[FeO_4]^{2-}}$: Fe(+6), $3d^2$.
(B) $\mathrm{[Mn(CN)_6]^{3-}}$: Mn(+3), $3d^4$.
(C) $\mathrm{[Fe(CN)_6]^{3-}}$: Fe(+3), $3d^5$.
(D) $\mathrm{Cr_2(CH_3COO)_4(H_2O)_2}$, chromium(II) acetate: Cr(+2), $3d^4$.
(E) $\mathrm{[NiF_6]^{2-}}$: Ni(+4), $3d^6$.
$d^4$: (B) and (D).
Answer: (A). | ||||
| 64 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter In group analysis lead is precipitated as the black sulphide PbS (A), which dissolves in hot dilute $\mathrm{HNO_3}$ to give $\mathrm{Pb(NO_3)_2}$.
With $\mathrm{H_2SO_4}$ the white precipitate $\mathrm{PbSO_4}$ (B) forms.
$\mathrm{PbSO_4}$ dissolves in ammonium acetate (as lead acetate); acidified with acetic acid and treated with $\mathrm{K_2CrO_4}$ it gives yellow $\mathrm{PbCrO_4}$ (C).
$\mathrm{Pb(SO_4)_2}$ is not formed and $\mathrm{Pb(CH_3COO)_2}$ is soluble, not a yellow precipitate, which rules out (B) and (C). Between (A) and (D), which differ only in A: lead reaches the nitric-acid step of group analysis as the group II sulphide PbS, dissolved in hot dilute $\mathrm{HNO_3}$.
Answer: (D) PbS, PbSO₄, PbCrO₄. | ||||
| 65 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Hydration $\mathrm{R_2C{=}O + H_2O \rightleftharpoons R_2C(OH)_2}$ is favoured by small groups (less crowding in the $sp^3$ hydrate) and by electron-withdrawing groups (they destabilise the carbonyl).
Statement I: formaldehyde has only H atoms on the carbonyl carbon; its hydration constant is about $2.3\times10^3$ (2280): true.
Statement II: in chloral the strongly electron-withdrawing $\mathrm{CCl_3}$ group ($-I$ effect) makes the carbonyl carbon very electrophilic; its hydration constant is about $2\times10^3$, chloral hydrate being a stable compound: true.
Answer: (A). | ||||
| 66 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons All four are tetrahydroindenes with a methyl group and two C=C bonds, one of them the ring-fusion bond C3a=C7a. Ozonolysis cuts both double bonds; since the fusion bond is in both rings, each compound gives one open-chain tetracarbonyl compound.
(D) C1=C2 with the methyl on C1: C1 becomes $\mathrm{CH_3CO-}$, C2 becomes CHO; C7a and C3a become ketones. Chain: $\mathrm{CH_3CO{-}CO{-}(CH_2)_4{-}CO{-}CH_2{-}CHO}$ = (II).
(B) C2=C3 with the methyl on C2: C3 becomes CHO next to the C3a ketone, C2 becomes $\mathrm{CH_3CO-}$ beyond C1. Chain: $\mathrm{OHC{-}CO{-}(CH_2)_4{-}CO{-}CH_2{-}COCH_3}$ = (IV).
(A) the second double bond C6=C7 in the six-membered ring, the methyl on C2 of the saturated five-membered ring: $\mathrm{OHC{-}CO{-}CH_2{-}CH(CH_3){-}CH_2{-}CO{-}CH_2CH_2{-}CHO}$ = (III).
(C) the second double bond C4=C5, the methyl on C6: $\mathrm{OHC{-}CO{-}(CH_2)_3{-}CO{-}CH_2{-}CH(CH_3){-}CHO}$ = (I).
A→III, B→IV, C→I, D→II.
Answer: (D). | ||||
| 67 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Solvolysis goes by $S_N1$; its rate follows the stability of the carbocation formed.
(D) 2-bromo-3-methylpentane gives a secondary alkyl cation: slowest.
(A) a 1-bromo-1-methylcyclohexane (with 4,4-dimethyl) gives a tertiary alkyl cation, stabilised only by hyperconjugation and +I effect.
(B) 1-bromo-4-methyltetralin gives a secondary benzylic cation, delocalised into the benzene ring: more stable than the tertiary alkyl cation.
(C) bromodiphenylmethane gives the benzhydryl cation, delocalised into two rings: fastest.
Ascending order: (D) < (A) < (B) < (C).
Answer: (D). | ||||
| 68 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Chlorobenzene with NaOH at 623 K and 300 atm (Dow process) gives sodium phenoxide; acidification gives phenol, [A].
Phenol is oxidised by chromic acid ($\mathrm{Na_2Cr_2O_7/H_2SO_4}$) to p-benzoquinone, [B].
Answer: (D). | ||||
| 69 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Statement I: more carbon atoms means larger molecules and stronger van der Waals forces, so boiling points rise: true.
Statement II: the O–H group forms intermolecular hydrogen bonds, which ethers and haloalkanes of similar mass cannot; alcohols and phenols boil higher: true.
Answer: (A). | ||||
| 70 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The secondary structure of a protein is the shape of the polypeptide chain held by hydrogen bonds between N–H and C=O groups of the backbone: the $\alpha$-helix and the $\beta$-pleated sheet.
Answer: (B) secondary structure. | ||||
| 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 $\mathrm{4Al + 3O_2 \rightarrow 2Al_2O_3}$. Al: $\frac{81}{27} = 3$ mol; $\mathrm{O_2}$: $\frac{128}{32} = 4$ mol.
3 mol Al needs $\frac34\times3 = 2.25$ mol $\mathrm{O_2}$, less than 4: Al is limiting.
$\mathrm{Al_2O_3}$ formed: $\frac{3}{2} = 1.5$ mol $\times102$ g mol$^{-1}$ $= 153$ g.
Answer: 153. | ||||
| 72 | Chemical Bonding and Molecular StructureIonic bonding and lattice enthalpy | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on ionic bonding and lattice enthalpy, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Born–Haber cycle: $\Delta H_f = \Delta H_{sub} + \Delta H_i + \frac12\Delta H_{bond} + \Delta H_{eg} - \Delta H_{lattice}$ (the lattice enthalpy is given for breaking the lattice, so it enters with a minus sign).
$-400 = 100 + 500 + \frac12\Delta H_{bond} - 300 - 800 \Rightarrow \frac12\Delta H_{bond} = 100$.
$\Delta H_{bond} = 200$ kJ mol$^{-1}$.
Answer: 200. | ||||
| 73 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{H_2O}$: $\frac{0.9}{18} = 0.05$ mol, holding $0.1$ mol H atoms from 0.01 mol of X: 10 H atoms per molecule, 10 g of H per mole.
10% of the molar mass is hydrogen: $M = \frac{10}{0.10} = 100$ g mol$^{-1}$.
Answer: 100. | ||||
| 74 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons X takes up 2 mol $\mathrm{H_2}$, so it has two C=C bonds; oxidative cleavage gives one fragment per alkene end.
Acetone comes from $\mathrm{(CH_3)_2C{=}}$; acetic acid from $\mathrm{{=}CH{-}CH_3}$; 4-oxopentanoic acid from $\mathrm{CH_3{-}C({=}){-}CH_2CH_2{-}CH{=}}$ (its ketone carbon and its acid carbon were the two alkene carbons).
X: $\mathrm{(CH_3)_2C{=}C(CH_3){-}CH_2CH_2{-}CH{=}CH{-}CH_3}$, $\mathrm{C_{10}H_{18}}$.
$\sigma$ bonds: 9 C–C (an open chain of 10 carbons with branches has 9 C–C links, each double bond counting one $\sigma$) + 18 C–H = 27.
Answer: 27. | ||||
| 75 | AminesDiazonium salts | Class 12: Amines | NumericalMedium | |
A medium numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines 4-Ethoxyaniline with $\mathrm{NaNO_2/HCl}$ at 0–5 °C gives the diazonium salt A, $\mathrm{EtO{-}C_6H_4{-}N_2^+Cl^-}$.
Coupling with phenol in NaOH (para position), then dilute HCl: B $= \mathrm{EtO{-}C_6H_4{-}N{=}N{-}C_6H_4{-}OH}$, $\mathrm{C_{14}H_{14}N_2O_2}$.
NaOH then $\mathrm{CH_3CH_2Br}$ (Williamson) ethylates the phenol: C $= \mathrm{EtO{-}C_6H_4{-}N{=}N{-}C_6H_4{-}OEt}$, $\mathrm{C_{16}H_{18}N_2O_2}$.
Only the two ethyl groups have $sp^3$ carbons: $2\times2 = 4$; the 12 ring carbons are $sp^2$.
Answer: 4. | ||||
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