JEE (Main) · 2025
JEE (Main) 2025 — 2 April, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
25 Q
- Sets, Relations and Functions2
- Matrices and Determinants2
- Sequences and Series2
9 easy · 16 medium · 0 hard
Physics
25 Q
- Properties of Bulk Matter3
- Optics3
- Electrostatics3
22 easy · 3 medium · 0 hard
Chemistry
25 Q
- Solutions2
- Thermodynamics2
- Chemical Kinetics2
15 easy · 9 medium · 1 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 FunctionsSets and their representation | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on sets and their representation, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions First term: $10 + 3x - x^2 > 0 \Rightarrow x^2 - 3x - 10 < 0 \Rightarrow (x-5)(x+2) < 0 \Rightarrow -2 < x < 5$.
Second term: $x + |x| > 0$. For $x \le 0$, $x + |x| = 0$; for $x > 0$ it is $2x > 0$. So $x > 0$.
Domain: $(0, 5)$, so $a = 0$, $b = 5$.
$(1+a)^2 + b^2 = 1 + 25 = 26$.
Answer: (A) $26$. | ||||
| 2 | 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 Each pair needs $a_i = 2a_{i+1} + 1$ with every $a_i$ in A.
So the chain runs $a_{k+1} \to a_k = 2a_{k+1} + 1 \to \ldots$, each entry more than double the next. To make it as long as possible, start from the smallest last entry, $a_{k+1} = 1$:
$1, 3, 7, 15, 31, 63, 127, \ldots$
$127 > 100$, so the longest chain of entries in A is $63, 31, 15, 7, 3, 1$: six numbers, giving the five pairs $(63, 31), (31, 15), (15, 7), (7, 3), (3, 1)$.
Any other starting value $a_{k+1} \ge 2$ grows at least as fast, so no longer chain exists.
$k = 5$.
Answer: (A) $5$. | ||||
| 3 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The relation is $A^3 - 2A^2 - 4A + 4I = O$, so $A^3 = 2A^2 + 4A - 4I$.
$A^4 = A\cdot A^3 = 2A^3 + 4A^2 - 4A = 2(2A^2 + 4A - 4I) + 4A^2 - 4A = 8A^2 + 4A - 8I$.
$A^5 = A\cdot A^4 = 8A^3 + 4A^2 - 8A = 8(2A^2 + 4A - 4I) + 4A^2 - 8A = 20A^2 + 24A - 32I$.
So $\alpha = 20$, $\beta = 24$, $\gamma = -32$ and $\alpha + \beta + \gamma = 12$.
Answer: (B) $12$. | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The second equation is not a multiple of the first, so for infinitely many solutions the third equation must be a combination of the first two: (third) $= p\,$(first) $+ q\,$(second).
Coefficients of $x$: $2p + 3q = 4$. Constants: $5p + 7q = 9$.
Solving: $10p + 15q = 20$ and $10p + 14q = 18$ give $q = 2$, then $p = -1$.
Coefficients of $y$: $\lambda p + 2q = 5 \Rightarrow -\lambda + 4 = 5 \Rightarrow \lambda = -1$.
Coefficients of $z$: $3p - q = \mu \Rightarrow \mu = -3 - 2 = -5$.
Check: with $\lambda = -1$, $\mu = -5$ the determinant $\begin{vmatrix}2&-1&3\\3&2&-1\\4&5&-5\end{vmatrix} = 2(-10+5) + 1(-15+4) + 3(15-8) = -10 - 11 + 21 = 0$, as it must be.
$\lambda^2 + \mu^2 = 1 + 25 = 26$.
Answer: (C) $26$. | ||||
| 5 | 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 Let there be $2n$ terms, first term $a$, common difference $d$.
The odd-placed terms $a, a+2d, \ldots$ ($n$ of them) and the even-placed terms $a+d, a+3d, \ldots$ ($n$ of them) pair off, each even term exceeding its odd partner by $d$:
$30 - 24 = nd \Rightarrow nd = 6$.
Last minus first: $(2n-1)d = \frac{21}{2} \Rightarrow 2nd - d = \frac{21}{2} \Rightarrow 12 - d = \frac{21}{2} \Rightarrow d = \frac{3}{2}$, so $n = 4$: eight terms.
Sum of odd terms: $\frac{n}{2}[2a + (n-1)2d] = 4(a + 3d) = 4\left(a + \frac{9}{2}\right) = 24 \Rightarrow a = \frac{3}{2}$.
The terms are $\frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12$; the integers among them are $3, 6, 9, 12$.
Answer: (A) $4$. | ||||
| 6 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations From the figure the three rows hold 3, 3 and 2 boxes.
First choose which 5 of the 8 boxes get a letter, with every row used, then arrange the 5 distinct letters in them in $5!$ ways.
All choices of 5 boxes: $\binom{8}{5} = 56$.
Remove those leaving a row empty. Leaving the top row (3 boxes) empty: 5 boxes remain, $\binom{5}{5} = 1$. Leaving the middle row (3 boxes) empty: $1$. Leaving the bottom row (2 boxes) empty: $\binom{6}{5} = 6$. Two rows cannot both be empty, since at most 3 boxes would remain.
Good choices: $56 - 1 - 1 - 6 = 48$.
Ways $= 48 \times 5! = 48 \times 120 = 5760$.
Answer: (C) $5760$. | ||||
| 7 | 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 the bracket: $\frac{10^{r+1}-1}{10^r} = 10 - 10^{-r}$.
First part: $10\sum_{r=0}^{10}{}^{11}C_{r+1} = 10\sum_{s=1}^{11}{}^{11}C_s = 10(2^{11} - 1)$.
Second part: $\sum_{r=0}^{10}10^{-r}\,{}^{11}C_{r+1} = 10\sum_{s=1}^{11}{}^{11}C_s\left(\frac{1}{10}\right)^s = 10\left[\left(\frac{11}{10}\right)^{11} - 1\right]$.
Difference: $10\cdot2^{11} - 10 - \frac{11^{11}}{10^{10}} + 10 = \frac{10^{11}\cdot2^{11} - 11^{11}}{10^{10}} = \frac{20^{11} - 11^{11}}{10^{10}}$.
So $\alpha = 20$.
Answer: (C) $20$. | ||||
| 8 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean: $\frac{6+4+8+12+10+13+a+b}{8} = 9 \Rightarrow 53 + a + b = 72 \Rightarrow a + b = 19$.
Variance: $\frac{\sum x^2}{8} - 9^2 = 9.25 \Rightarrow \sum x^2 = 8 \times 90.25 = 722$.
Known squares: $36 + 16 + 64 + 144 + 100 + 169 = 529$, so $a^2 + b^2 = 193$.
$2ab = (a+b)^2 - (a^2+b^2) = 361 - 193 = 168 \Rightarrow ab = 84$.
$a + b + ab = 19 + 84 = 103$.
Answer: (B) $103$. | ||||
| 9 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each bag is chosen with probability $\frac{1}{3}$ and each holds 10 balls, so the $\frac{1}{3}\cdot\frac{1}{10}$ factors cancel in Bayes' theorem and the probabilities go by the ball counts.
Red: bags hold 3, 4, 5 red balls. $p = P(\text{I}\mid\text{Red}) = \frac{3}{3+4+5} = \frac{1}{4}$.
Green: bags hold 5, 3, 4 green balls. $q = P(\text{III}\mid\text{Green}) = \frac{4}{5+3+4} = \frac{1}{3}$.
$\frac{1}{p} + \frac{1}{q} = 4 + 3 = 7$.
Answer: (C) $7$. | ||||
| 10 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines In normal form the line is $x\cos45^\circ + y\sin45^\circ = p$, i.e. $x + y = p\sqrt{2}$. Its coefficients of $x$ and $y$ are equal, so in $x + by + c = 0$, $b = 1$.
The line $x + y + c = 0$ cuts the axes at $(-c, 0)$ and $(0, -c)$; the triangle's area is $\frac{1}{2}c^2 = 48$, so $c^2 = 96$.
$b^2 + c^2 = 1 + 96 = 97$.
Answer: (D) $97$. | ||||
| 11 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Here $4a = 16$, $a = 4$, focus $S = (4, 0)$, and points are $(4t^2, 8t)$.
P: $8t_1 = -4 \Rightarrow t_1 = -\frac{1}{2}$ (and $4t_1^2 = 1$ ✓).
For a focal chord $t_1t_2 = -1$, so $t_2 = 2$ and $Q = (16, 16)$.
Focal distances: $SP = x_P + a = 1 + 4 = 5$, $SQ = 16 + 4 = 20$.
S divides PQ in the ratio $5 : 20 = 1 : 4$, so $m^2 + n^2 = 1 + 16 = 17$.
Answer: (B) $17$. | ||||
| 12 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Minor axis $2b$, distance between foci $2ae$: $2b = \frac{1}{4}(2ae) \Rightarrow b = \frac{ae}{4}$.
With $b^2 = a^2(1 - e^2)$: $a^2(1 - e^2) = \frac{a^2e^2}{16} \Rightarrow 16 - 16e^2 = e^2 \Rightarrow e^2 = \frac{16}{17}$.
$e = \frac{4}{\sqrt{17}}$.
Answer: (B) $\frac{4}{\sqrt{17}}$. | ||||
| 13 | 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 Put $u = \text{cosec}\,\theta$: $\sqrt{3}u^2 - 2(\sqrt{3}-1)u - 4 = 0$.
Discriminant: $4(\sqrt{3}-1)^2 + 16\sqrt{3} = 4(4 - 2\sqrt{3}) + 16\sqrt{3} = 16 + 8\sqrt{3} = 4(\sqrt{3}+1)^2$.
$u = \frac{2(\sqrt{3}-1) \pm 2(\sqrt{3}+1)}{2\sqrt{3}}$, so $u = 2$ or $u = -\frac{2}{\sqrt{3}}$.
That is, $\sin\theta = \frac{1}{2}$ or $\sin\theta = -\frac{\sqrt{3}}{2}$.
In $\left[-\frac{7\pi}{6}, \frac{4\pi}{3}\right]$:
$\sin\theta = \frac{1}{2}$: $\theta = -\frac{7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}$ (3 values; $-\frac{11\pi}{6}$ is outside).
$\sin\theta = -\frac{\sqrt{3}}{2}$: $\theta = -\frac{2\pi}{3}, -\frac{\pi}{3}, \frac{4\pi}{3}$ (3 values; $\frac{5\pi}{3}$ is outside).
Both end points are solutions and are included. Total: 6.
Answer: (A) $6$. | ||||
| 14 | 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 Direction of AB: $B - A = (-1, -2, 2)$. A point of the line: $M = (4 - t, 7 - 2t, 1 + 2t)$.
M is the foot of the perpendicular from P when $(M - P)\cdot(-1, -2, 2) = 0$:
$(3 - t, 7 - 2t, -2 + 2t)\cdot(-1, -2, 2) = -3 + t - 14 + 4t - 4 + 4t = 9t - 21 = 0 \Rightarrow t = \frac{7}{3}$.
$M = \left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$.
The image is $Q = 2M - P = \left(\frac{7}{3}, \frac{14}{3}, \frac{25}{3}\right)$.
$\alpha + \beta + \gamma = \frac{46}{3}$.
Answer: (D) $\frac{46}{3}$. | ||||
| 15 | 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 $\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-3&2&4\\2&1&3\end{vmatrix} = (6-4)\hat{i} - (-9-8)\hat{j} + (-3-4)\hat{k} = 2\hat{i} + 17\hat{j} - 7\hat{k}$.
$|\vec{a}\times\vec{b}| = \sqrt{4 + 289 + 49} = \sqrt{342} = 3\sqrt{38}$.
Joining vector: $(7, 6, 2) - (5, 3, 4) = (2, 3, -2)$; its dot product with $\vec{a}\times\vec{b}$ is $4 + 51 + 14 = 69$.
Shortest distance $= \frac{69}{3\sqrt{38}} = \frac{23}{\sqrt{38}}$.
Answer: (D) $\frac{23}{\sqrt{38}}$. | ||||
| 16 | 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 $(\vec{a} - \vec{c})\times\vec{b} = \vec{a}\times\vec{b} - \vec{c}\times\vec{b} = \vec{a}\times\vec{b} + \vec{b}\times\vec{c} = \vec{a}\times\vec{b} + \vec{d}$.
$\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-3&1\\3&2&5\end{vmatrix} = (-15-2)\hat{i} - (10-3)\hat{j} + (4+9)\hat{k} = -17\hat{i} - 7\hat{j} + 13\hat{k}$.
$\vec{d} = (-18\hat{i} - 3\hat{j} + 12\hat{k}) - (-17\hat{i} - 7\hat{j} + 13\hat{k}) = -\hat{i} + 4\hat{j} - \hat{k}$.
$\vec{a}\cdot\vec{d} = -2 - 12 - 1 = -15$, so $|\vec{a}\cdot\vec{d}| = 15$.
(The condition $\vec{a}\cdot\vec{c} = 3$ is not needed for this.)
Answer: (C) $15$. | ||||
| 17 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Expand: $\cos 2x = 1 - 2x^2 + \frac{2}{3}x^4 - \ldots$, $\cos 4x = 1 - 8x^2 + \frac{32}{3}x^4 - \ldots$
Numerator $= (1 + a - b) - (2 + 8a)x^2 + \left(\frac{2}{3} + \frac{32a}{3}\right)x^4 + \ldots$
For a finite limit the $x^0$ and $x^2$ coefficients must vanish: $2 + 8a = 0 \Rightarrow a = -\frac{1}{4}$, and $b = 1 + a = \frac{3}{4}$.
$a + b = \frac{1}{2}$.
Answer: (A) $\frac{1}{2}$. | ||||
| 18 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Rationalise: $\frac{1}{\sqrt{3+x^2} + \sqrt{1+x^2}} = \frac{\sqrt{3+x^2} - \sqrt{1+x^2}}{2}$.
So $4\int_0^1(\ldots)dx = 2\int_0^1\sqrt{x^2+3}\,dx - 2\int_0^1\sqrt{x^2+1}\,dx$.
Use $\int\sqrt{x^2+k}\,dx = \frac{x}{2}\sqrt{x^2+k} + \frac{k}{2}\ln\left(x + \sqrt{x^2+k}\right)$.
$k = 3$: $\left[\frac{x}{2}\sqrt{x^2+3} + \frac{3}{2}\ln(x + \sqrt{x^2+3})\right]_0^1 = 1 + \frac{3}{2}\ln3 - \frac{3}{2}\ln\sqrt{3} = 1 + \frac{3}{4}\ln3$.
$k = 1$: $\frac{\sqrt{2}}{2} + \frac{1}{2}\ln(1 + \sqrt{2})$.
Twice the difference: $2 + \frac{3}{2}\ln3 - \sqrt{2} - \ln(1 + \sqrt{2})$.
Subtract $3\ln\sqrt{3} = \frac{3}{2}\ln3$: the value is $2 - \sqrt{2} - \log_e(1 + \sqrt{2})$.
Answer: (B). | ||||
| 19 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersection: $x^2 = 2(2x + 6) \Rightarrow x^2 - 4x - 12 = 0 \Rightarrow x = 6$ or $x = -2$. In the second quadrant $x = -2$, $y = 2$: $a = -2$, $b = 2$.
For $f(x) = \frac{9x^2}{1 + 5^x}$: $f(-x) = \frac{9x^2}{1 + 5^{-x}} = \frac{9x^2\,5^x}{5^x + 1}$, so $f(x) + f(-x) = 9x^2$.
Hence $I = \int_{-2}^{2}f(x)\,dx = \frac{1}{2}\int_{-2}^{2}9x^2\,dx = \int_0^2 9x^2\,dx = 3x^3\Big|_0^2 = 24$.
Answer: (C) $24$. | ||||
| 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 At $x = 1$: $0 = 5f(1) - 1 - 9 \Rightarrow f(1) = 2$.
Differentiate: $10f(x) = 5f(x) + 5xf'(x) - 5x^4 \Rightarrow xf'(x) - f(x) = x^4 \Rightarrow \frac{d}{dx}\left(\frac{f(x)}{x}\right) = x^2$.
So $\frac{f(x)}{x} = \frac{x^3}{3} + C$; at $x = 1$: $2 = \frac{1}{3} + C$, $C = \frac{5}{3}$.
$f(x) = \frac{x^4}{3} + \frac{5x}{3}$, and $f(3) = 27 + 5 = 32$.
Answer: (D) $32$. | ||||
| 21 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Both roots real and positive needs: discriminant $\ge 0$, sum $> 0$, product $> 0$.
Discriminant: $\frac{D}{4} = (a-3)^2 - 9(1-a) = a^2 + 3a \ge 0 \Rightarrow a \le -3$ or $a \ge 0$.
Product: $\frac{9}{1-a} > 0 \Rightarrow a < 1$.
Sum: $\frac{-2(a-3)}{1-a} > 0$; with $1 - a > 0$ this needs $a < 3$, already true.
Together: $a \in (-\infty, -3] \cup [0, 1)$. (At $a = 0$ the equation is $(x-3)^2 = 0$, a repeated positive root; at $a = -3$ it is $4x^2 - 12x + 9 = (2x-3)^2 = 0$.)
$\alpha = 3$, $\beta = 0$, $\gamma = 1$: $2\alpha + \beta + \gamma = 7$.
Answer: 7. | ||||
| 22 | 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 Factor: $1 + 4r^4 = (2r^2 + 1)^2 - (2r)^2 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$, and the two factors differ by $4r$.
So $\frac{4r}{1 + 4r^4} = \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1}$.
Since $2(r+1)^2 - 2(r+1) + 1 = 2r^2 + 2r + 1$, the sum telescopes:
$\sum_{r=1}^{10} = \frac{1}{2 - 2 + 1} - \frac{1}{200 + 20 + 1} = 1 - \frac{1}{221} = \frac{220}{221}$.
$m + n = 220 + 221 = 441$.
Answer: 441. | ||||
| 23 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta = \cos^{-1}\frac{x}{2}$, so $\cos\theta = \frac{x}{2}$ and $\sin\theta = \frac{\sqrt{4 - x^2}}{2}$ (as $\theta \in [0, \pi]$).
$y = \cos\frac{\pi}{3}\cos\theta - \sin\frac{\pi}{3}\sin\theta = \frac{x}{4} - \frac{\sqrt{3}\sqrt{4 - x^2}}{4}$.
$x - 4y = \sqrt{3}\sqrt{4 - x^2}$; squaring: $x^2 - 8xy + 16y^2 = 12 - 3x^2$.
$4x^2 - 8xy + 16y^2 = 12 \Rightarrow x^2 - 2xy + 4y^2 = 3 \Rightarrow (x - y)^2 + 3y^2 = 3$.
Answer: 3. | ||||
| 24 | Straight LinesCartesian coordinates and distance formula | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines Area of ABC: $\frac{1}{2}\left|(B - A)\times(C - A)\right| = \frac{1}{2}\left|(-3)(-1) - (3)(5)\right| = \frac{1}{2}\cdot12 = 6$.
In parallelogram AFDE, DF is parallel to AC and DE to AB. Let $BD : BC = t$. Then $AF = (1 - t)AB$ and $AE = t\,AC$, so
area$(AFDE) = AF\cdot AE\sin A = t(1 - t)\cdot AB\cdot AC\sin A = 2t(1 - t)\cdot[\triangle ABC]$.
$t(1 - t)$ is greatest, $\frac{1}{4}$, at $t = \frac{1}{2}$ (D the midpoint of BC), so the maximum area is $\frac{1}{2}[\triangle ABC] = 3$.
Answer: 3. | ||||
| 25 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor: $e^{\int2\sec^2x\,dx} = e^{2\tan x}$.
$y\,e^{2\tan x} = \int(2 + 3\tan x)\sec^2x\,e^{2\tan x}\,dx$. Put $t = \tan x$:
$\int(2 + 3t)e^{2t}dt = \frac{(2 + 3t)e^{2t}}{2} - \frac{3}{2}\int e^{2t}dt = \frac{(2 + 3t)e^{2t}}{2} - \frac{3e^{2t}}{4} + C$.
So $y = \frac{1}{4} + \frac{3}{2}\tan x + Ce^{-2\tan x}$.
$y(0) = \frac{1}{4} + C = \frac{5}{4} \Rightarrow C = 1$.
$y\left(\frac{\pi}{4}\right) = \frac{1}{4} + \frac{3}{2} + e^{-2} = \frac{7}{4} + e^{-2}$.
$12\left(y\left(\frac{\pi}{4}\right) - e^{-2}\right) = 12\cdot\frac{7}{4} = 21$.
Answer: 21. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Matching listEasy | |
A easy matching list question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Heat capacity $C = \frac{Q}{\Delta T}$: J K$^{-1}$ (II).
Specific heat capacity $c = \frac{Q}{m\Delta T}$: J kg$^{-1}$ K$^{-1}$ (III).
Latent heat $L = \frac{Q}{m}$: J kg$^{-1}$ (I).
Thermal conductivity, from $\frac{Q}{t} = kA\frac{\Delta T}{x}$: $k = \frac{Qx}{tA\Delta T}$, J m$^{-1}$ K$^{-1}$ s$^{-1}$ (IV).
A→II, B→III, C→I, D→IV.
Answer: (D). | ||||
| 27 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The force per unit length between two parallel currents is $\frac{F}{l} = \frac{\mu_0 I^2}{2\pi d}$, so $\mu_0 I^2$ has the dimensions of force.
Charge is current × time: $[q] = [I][T]$.
$[q\mu_0 I] = [\mu_0 I^2][T] = [\text{force}\times\text{time}] = [\text{momentum}] = MLT^{-1}$.
(In base units, $[\mu_0] = MLT^{-2}A^{-2}$, so $[q\mu_0 I] = (AT)(MLT^{-2}A^{-2})(A) = MLT^{-1}$.)
Answer: (D) $q\mu_0 I$. | ||||
| 28 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Note that r is the diameter, so the radius is $R = \frac{r}{2}$.
About a diameter, $I_d = \frac{1}{2}MR^2$. A tangent in the plane is parallel to a diameter at distance R, so by the parallel-axis theorem
$I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2 = \frac{3}{2}M\frac{r^2}{4} = \frac{3}{8}Mr^2$.
Answer: (C) $\frac{3}{8}Mr^2$. | ||||
| 29 | Laws of MotionEquilibrium of concurrent forces | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on equilibrium of concurrent forces, from Laws of Motion. NCERT: Class 11: Laws of Motion From the figure, $T_1$ makes $60^\circ$ and $T_2$ makes $30^\circ$ with the horizontal; the weight is $mg = 10$ N.
Horizontal: $T_1\cos60^\circ = T_2\cos30^\circ \Rightarrow \frac{T_1}{2} = \frac{\sqrt{3}T_2}{2} \Rightarrow T_1 = \sqrt{3}T_2$.
Vertical: $T_1\sin60^\circ + T_2\sin30^\circ = 10 \Rightarrow \sqrt{3}T_2\cdot\frac{\sqrt{3}}{2} + \frac{T_2}{2} = 2T_2 = 10$.
$T_2 = 5$ N, $T_1 = 5\sqrt{3}$ N.
Answer: (B) $5\sqrt{3}, 5$. | ||||
| 30 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane A and B are the ends of a diameter (figure), so each of A→B, B→A, A→B is half the circumference, $\pi r$.
Distance travelled along ABAB: $3\pi r$.
He starts at A and ends at B, so the displacement is the diameter AB $= 2r$.
Answer: (B) $3\pi r, 2r$. | ||||
| 31 | Properties of Bulk MatterSurface tension | 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 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 Volume is conserved: $\frac{4}{3}\pi R^3 = 2\cdot\frac{4}{3}\pi r^3 \Rightarrow R = 2^{1/3}r$.
Surface area before: $2\cdot4\pi r^2$; after: $4\pi R^2 = 4\pi r^2\cdot2^{2/3}$.
Energy released $= T\times(\text{decrease in area}) = 4\pi r^2T\left[2 - 2^{2/3}\right]$.
Answer: (B). | ||||
| 32 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | StatementsEasy | |
A easy statements question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics In an adiabatic process $Q = 0$, so by the first law the work and the change in internal energy are equal in magnitude ($W_{\text{on gas}} = \Delta U$). (B) is true.
$W = \frac{nR(T_1 - T_2)}{\gamma - 1}$ (by the gas), which is proportional to $(T_2 - T_1)$. (E) is true.
(A) false: the temperature changes, so U changes. (C) false: $TV^{\gamma-1}$ is constant, not $TV$. (D) false: $PV^\gamma$ is constant, not $PV$ (that is isothermal).
Answer: (A) (B), (E) only. | ||||
| 33 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The magnetic force $\vec{F} = -e\,\vec{v}\times\vec{B}$ is always perpendicular to the velocity, so it does no work: the speed stays $v_0$ (the electron moves on a circle in the xz-plane).
The momentum's magnitude $mv_0$ is unchanged, so $\lambda = \frac{h}{mv_0} = \lambda_0$ at every time.
Answer: (A) $\lambda_0$. | ||||
| 34 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Amplitude 2 cm. Wave number $k = \frac{2\pi}{\lambda} = \frac{2\pi}{7.5} \approx 0.838$ cm$^{-1}$.
Speed $v = \frac{1.2}{0.3} = 4$ cm/s, so $\omega = kv \approx 3.35$ rad/s.
A crest at $x = 0$ when $t = 0$ means $y(0, 0) = +2$ cm, which the cosine form gives.
$y = 2\cos(0.83x - 3.35t)$ cm, travelling along +x.
Answer: (D). | ||||
| 35 | 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 The speed of light in vacuum is $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$, so $\frac{1}{\mu_0\epsilon_0} = c^2$, whose dimension is $L^2/T^2$.
Answer: (D) $L^2/T^2$. | ||||
| 36 | 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 Focal length magnitude $f = \frac{R}{2} = 6$ cm; object distance $u = -18$ cm in both.
Convex mirror ($f = +6$): $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{6} + \frac{1}{18} = \frac{4}{18}$, $v = 4.5$ cm; $m = -\frac{v}{u} = \frac{4.5}{18} = \frac{1}{4}$.
Concave mirror ($f = -6$): $\frac{1}{v} = -\frac{1}{6} + \frac{1}{18} = -\frac{2}{18}$, $v = -9$ cm; $m = -\frac{v}{u} = -\frac{1}{2}$.
Ratio of sizes $= \frac{1/4}{1/2} = \frac{1}{2}$.
Answer: (B) $1/2$. | ||||
| 37 | 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 By the lens-maker's formula, $P = (\mu - 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$ for a bi-convex lens (magnitudes). Taking the same material, the power stays the same if $\frac{1}{R_1} + \frac{1}{R_2}$ stays the same.
Original: $6 + 6 = 12$ cm$^{-1}$.
(A) $6 + 9 = 15$; (B) $3 + 3 = 6$ (and $R_1 = R_2$); (C) $5 + 7 = 12$ ✓; (D) $3 + 7 = 10$.
Answer: (C) $\frac{1}{5}$ cm and $\frac{1}{7}$ cm. | ||||
| 38 | 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 Binding energy of each deuteron: $2\times1.1 = 2.2$ MeV; of two deuterons: $4.4$ MeV.
Binding energy of He-4: $4\times7.0 = 28.0$ MeV.
Energy released $= 28.0 - 4.4 = 23.6$ MeV.
Answer: (C) 23.6 MeV. | ||||
| 39 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Both inputs are 1.
NAND of the inputs: $X = \overline{1\cdot1} = 0$.
Each input passes through a NOT gate (giving 0, 0) into an OR gate: $Y = 0 + 0 = 0$.
P is the AND of X and Y: $P = 0\cdot0 = 0$.
X also passes through a NOT gate, giving $\bar{X} = 1$; this and Y go into the NOR gate: $Q = \overline{1 + 0} = 0$.
(The two wires that cross with a hop are not joined.)
Answer: (A) P = 0, Q = 0. | ||||
| 40 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei For a hydrogen-like ion, $r_n = \frac{n^2}{Z}a_0$.
$\mathrm{Li^{++}}$ has $Z = 3$; in the ground state $n = 1$: $r = \frac{a_0}{3}$, so $X = 3$.
Answer: (C) $3$. | ||||
| 41 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Assertion and reasonEasy | |
A easy assertion and reason question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In a polar dielectric each molecule has a permanent dipole moment, but with no external field thermal agitation orients these dipoles at random, so they cancel: the net dipole moment is zero.
So (A) is false and (R) is true (and (R) is in fact why the net moment is zero).
Answer: (D). | ||||
| 42 | 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 Between large parallel plates the field is uniform, $E = \frac{V}{d} = \frac{V}{0.10\ \text{m}}$, directed perpendicular to the plates (horizontal in the figure).
Only the displacement along the field changes the potential. In the right triangle ACB, $AC = 3$ cm (parallel to the plates) and $AB = 5$ cm, so $CB = \sqrt{5^2 - 3^2} = 4$ cm along the field.
$V_{AB} = E\times CB = \frac{V}{10\ \text{cm}}\times4\ \text{cm} = \frac{2}{5}V$.
Answer: (B) $\frac{2}{5}$V. | ||||
| 43 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Magnetic energy density in a medium of permeability $\mu = \mu_r\mu_0$ is $u = \frac{B^2}{2\mu}$.
With $\mu_r = 2$: $u = \frac{B^2}{4\mu_0}$.
Energy $= u\times\text{volume} = \frac{B^2}{4\mu_0}\cdot Al = \frac{B^2Al}{4\mu_0}$.
Answer: (C). | ||||
| 44 | 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 On the axis of a ring of radius R and charge Q: $E(z) = \frac{kQz}{(R^2 + z^2)^{3/2}}$.
$\frac{dE}{dz} = kQ\frac{(R^2 + z^2) - 3z^2}{(R^2 + z^2)^{5/2}} = 0 \Rightarrow R^2 = 2z^2 \Rightarrow z = \frac{R}{\sqrt{2}}$.
With $R = a\sqrt{2}$: $z = a$.
Answer: (C) $a$. | ||||
| 45 | 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 Voltage sensitivity $= \frac{\theta}{V} = \frac{NAB}{kR}$.
$M_1$: $\frac{15\times3.6\times10^{-3}\times0.25}{5} = \frac{0.0135}{5} = 2.7\times10^{-3}$ (per $k$).
$M_2$: $\frac{21\times1.8\times10^{-3}\times0.50}{7} = \frac{0.0189}{7} = 2.7\times10^{-3}$ (per $k$).
Ratio $= 1 : 1$.
Answer: (A) 1 : 1. | ||||
| 46 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The string leaves the rim tangentially, so the torque about the centre is $\tau = Fr = 10\times0.2 = 2$ N m.
$\tau = I\alpha \Rightarrow I = \frac{2}{2} = 1$ kg m$^2$.
(The value of g is not needed.)
Answer: 1. | ||||
| 47 | GravitationOrbital motion | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Orbit radius $r = 6.4\times10^6 + 0.27\times10^6 = 6.67\times10^6$ m.
For a circular orbit, $\frac{mv^2}{r} = \frac{GMm}{r^2}$, so $KE = \frac{GMm}{2r}$.
$\frac{GM}{r} = \frac{6.67\times10^{-11}\times6\times10^{24}}{6.67\times10^6} = 6\times10^7$ J/kg.
$KE = \frac{1}{2}\times1000\times6\times10^7 = 3\times10^{10}$ J.
Answer: 3. | ||||
| 48 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Within Hooke's law the extension is proportional to the tension: $l = l_0 + cT$.
$1.4 = l_0 + 5c$ and $1.56 = l_0 + 7c$ give $2c = 0.16$, $c = 0.08$ m/N.
$l_0 = 1.4 - 5\times0.08 = 1.0$ m.
Answer: 1. | ||||
| 49 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At minimum deviation the ray passes symmetrically: $r_1 = r_2 = \frac{A}{2} = 30^\circ$.
Snell's law at the first face: $\sin i = \mu\sin r = \sqrt{2}\times\frac{1}{2} = \frac{1}{\sqrt{2}}$.
$i = 45^\circ$.
Answer: 45. | ||||
| 50 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | NumericalMedium | |
A medium numerical question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory For a diatomic ideal gas (5 degrees of freedom), $U = \frac{5}{2}nRT = \frac{5}{2}PV$.
$V = 4\times4\times3 = 48$ m$^3$, $P = 1$ atm $= 1.013\times10^5$ Pa.
$U = 2.5\times1.013\times10^5\times48 = 1.2156\times10^7$ J $\approx 12.16\times10^6$ J.
To the nearest integer: 12. (With $P = 10^5$ Pa, $U = 12\times10^6$ J exactly.)
Answer: 12. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | SolutionsConcentration terms | Class 12: Solutions | Single correctMedium | |
A medium single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Molarity is moles of solute per litre of solution. The lid stops evaporation, and with the solubility unchanged the moles of dissolved NaCl stay the same, so M changes only through the volume of the solution: $M \propto \frac{1}{V}$.
Water (and a dilute aqueous solution) is densest near $4^\circ$C: from $1^\circ$C to about $4^\circ$C the volume shrinks, so M rises; above that the volume grows with temperature, so M falls steadily.
The plot that rises to a maximum at a few degrees and then falls is (A). (B) shows only a fall, (C) a dip then a rise, (D) only a rise.
Answer: (A). | ||||
| 52 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | StatementsEasy | |
A easy statements question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom (A) True: the azimuthal (subsidiary) quantum number $l$ fixes the orbital's shape (s, p, d, …).
(B) True: the drawing shows two lobes along the x-axis touching at the nucleus, the boundary surface of $2p_x$.
(C) False: the signs are the phase of the wave function, not electric charge.
(D) False: the nodal plane of $2p_x$ is the yz plane ($x = 0$). In the xy plane $\psi$ is non-zero except along the y-axis.
Answer: (A) (A) and (B) only. | ||||
| 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 Take the expansions and compressions between the same two states at constant temperature (ideal gas).
Reversible (infinite stages) expansion and compression trace the same isotherm, so $|w|$ is the same: $a = c = nRT\ln\frac{V_2}{V_1}$.
Single-stage expansion against the final (low) pressure does less work than the reversible path: $b < a$.
Single-stage compression against the final (high) pressure needs more work than the reversible path: $d > c$.
So $d > c = a > b$.
Answer: (A). | ||||
| 54 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctHard | |
A hard single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics For $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$: $\Delta H^\circ \approx -92$ kJ mol$^{-1}$ and $\Delta S^\circ \approx -198$ J K$^{-1}$ mol$^{-1}$, both nearly independent of T. So $\Delta H_R^\circ$ is a flat line near $-100$ and $\Delta S_R^\circ$ a flat line near $-200$.
$-\frac{\Delta H^\circ}{T} = \frac{92000}{T}$: about $+300$ J K$^{-1}$ mol$^{-1}$ near room temperature, falling as T rises.
$-\frac{\Delta G^\circ}{T} = -\frac{\Delta H^\circ}{T} + \Delta S^\circ$: about $300 - 200 = +100$ near room temperature, falling and becoming negative above about 465 K.
Graph (B) shows exactly this: $-\Delta H_R^\circ/T$ starting near 300, $-\Delta G_R^\circ/T$ starting near 100 and crossing zero, $\Delta H_R^\circ$ near $-100$, $\Delta S_R^\circ$ near $-200$. (A) swaps the two curves; (C) plots $-\Delta H^\circ$ and $\Delta S^\circ/T$; (D) has positive $\Delta H^\circ$ and $\Delta S^\circ$.
Answer: (B). | ||||
| 55 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Start with 1 mol A: at equilibrium $1 - \alpha$, $\alpha$, $\alpha$; total $1 + \alpha$.
$K_p = \frac{\left(\frac{\alpha}{1+\alpha}p\right)^2}{\frac{1-\alpha}{1+\alpha}p} = \frac{\alpha^2p}{1 - \alpha^2}$, so $\alpha = \sqrt{\frac{K_p}{K_p + p}}$.
As p increases, $\alpha$ decreases: (A) true, (B) false.
If $p \gg K_p$, $\alpha \to 0$, not 1: (C) false. If $K_p \gg p$, $\alpha \to 1$: (D) false.
Answer: (A). | ||||
| 56 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Read the profile off the data:
A → B is slow, so it has the largest activation energy: the first hump is the highest. It is endothermic, so B lies above A.
B → C and C → D are fast (small humps) and exothermic, so C lies below B and D below C.
Net heat is evolved, so D lies below A.
Profile (A) has all of these. (B) has the last hump highest and the intermediates rising; (C) has D above A (net endothermic) and C above B; (D) has the last hump highest.
Answer: (A). | ||||
| 57 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{PF_5}$: $sp^3d$, trigonal bipyramidal, axial and equatorial bonds differ, but no lone pair on P.
$\mathrm{XeF_2}$: $sp^3d$ with three lone pairs, but linear, both Xe–F bonds equal.
$\mathrm{SF_4}$: $sp^3d$ (4 bond pairs + 1 lone pair), see-saw; two axial and two equatorial S–F bonds of different lengths, and one lone pair on S. All three conditions hold.
$\mathrm{XeF_4}$: $sp^3d^2$, square planar.
Answer: (C) $\mathrm{SF_4}$. | ||||
| 58 | Classification of Elements and PeriodicityElectronegativity | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on electronegativity, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The elements are A = N, B = O, C = F, D = C (carbon), all in period 2.
Electronegativity rises across a period: C (2.5) < N (3.0) < O (3.5) < F (4.0).
So D < A < B < C.
Answer: (D). | ||||
| 59 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down group 16 the +4 state becomes more stable relative to +6 (inert pair effect), so the reducing power of the dioxides falls: $\mathrm{SO_2}$ is a reducing agent while $\mathrm{TeO_2}$ acts as an oxidising agent.
The acidic character of the hydrides increases down the group ($\mathrm{H_2O < H_2S < H_2Se < H_2Te}$) as the E–H bond weakens, so $\mathrm{TeH_2}$ is acidic.
Answer: (B) Oxidising and acidic. | ||||
| 60 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[Co(en)_3]^{3+}}$: $\mathrm{Co^{3+}}$, $d^6$; en is a strong-field ligand, low spin $t_{2g}^6e_g^0$, CFSE $= -2.4\Delta_o$ (with a large $\Delta_o$).
$\mathrm{[CoF_6]^{3-}}$: $d^6$, weak field, high spin $t_{2g}^4e_g^2$, CFSE $= -0.4\Delta_o$.
$\mathrm{[Mn(H_2O)_6]^{2+}}$: $d^5$ high spin $t_{2g}^3e_g^2$, CFSE $= 0$.
$\mathrm{[Zn(H_2O)_6]^{2+}}$: $d^{10}$, $t_{2g}^6e_g^4$, CFSE $= 0$.
Highest CFSE: $\mathrm{[Co(en)_3]^{3+}}$, $t_{2g}^6e_g^0$.
Answer: (D). | ||||
| 61 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Mn is in the +3 state: $\mathrm{Mn^{3+}}$, $3d^4$.
$\mathrm{Cl^-}$ is a weak-field ligand, so no pairing: the four 3d electrons stay unpaired and the six ligand pairs use 4s, 4p and two 4d orbitals, outer-orbital $sp^3d^2$.
Four unpaired electrons: paramagnetic.
Answer: (C). | ||||
| 62 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Sodium nitroprusside reacts with sulphide ion to give the purple/violet complex:
$\mathrm{S^{2-} + [Fe(CN)_5NO]^{2-} \rightarrow [Fe(CN)_5NOS]^{4-}}$, i.e. $\mathrm{Na_4[Fe(CN)_5NOS]}$ with the sodium ions.
So the test shows sulphide.
Answer: (C) Sulphide ion. | ||||
| 63 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Simple distillation separates liquids with widely different boiling points: chloroform (334 K) and aniline (457 K). A→III.
Fractional distillation separates liquids with close boiling points: petrol and diesel fractions of crude oil. B→I.
Distillation under reduced pressure is for liquids that decompose at their boiling point: glycerol from spent-lye in soap making. C→IV.
Steam distillation is for steam-volatile, water-immiscible liquids: aniline from water. D→II.
Answer: (B). | ||||
| 64 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Pressure of dry nitrogen: $715 - 15 = 700$ mm Hg.
Volume at STP: $V_0 = 60\times\frac{700}{760}\times\frac{273}{300} = 50.29$ mL.
Mass of nitrogen: $\frac{28}{22400}\times50.29 = 0.0629$ g.
Percentage: $\frac{0.0629}{0.5}\times100 = 12.57\%$.
Answer: (B) 12.57. | ||||
| 65 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Structure: $\mathrm{CH_2{=}CH{-}C(CH_3)_2{-}C{\equiv}C{-}CH_3}$ (C1 to C6, two methyls on C3).
$sp^2$: C1 and C2 (the double bond): 2.
$sp$: C4 and C5 (the triple bond): 2.
$sp^3$: C3, C6 and the two methyl carbons: 4.
Answer: (C) 4, 2, 2. | ||||
| 66 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | StatementsEasy | |
A easy statements question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Statement I: neopentane, $\mathrm{C(CH_3)_4}$, has twelve equivalent hydrogens (all on the four equivalent methyls), so it gives only one monosubstituted product. True.
Statement II: the nearly spherical, symmetrical neopentane molecule packs well in the solid, so it melts at about 257 K, far above n-pentane (about 143 K). True.
Answer: (A). | ||||
| 67 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Matching listEasy | |
A easy matching list question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (A) Two aryl halides coupled by sodium in dry ether to give biphenyl: Fittig reaction. A→III.
(B) Diazonium salt with copper powder and HCl giving ArCl: Gattermann reaction. B→IV.
(C) Halogen exchange with NaI in dry acetone: Finkelstein reaction. C→II.
(D) Tertiary alcohol with conc. HCl/anhydrous $\mathrm{ZnCl_2}$ (the Lucas reagent) giving the chloride: Lucas reaction. D→I.
Answer: (B). | ||||
| 68 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | StatementsMedium | |
A medium statements question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) Partial hydrolysis of a nitrile under mild acid conditions stops at the amide $\mathrm{RCONH_2}$. No.
(B) A Grignard reagent adds to $\mathrm{CO_2}$ and the salt is hydrolysed: $\mathrm{RCOOH}$. Yes.
(C) Stephen reduction ($\mathrm{SnCl_2/HCl}$, then hydrolysis) gives the aldehyde $\mathrm{RCHO}$. No.
(D) PCC oxidises a primary alcohol only as far as the aldehyde. No.
(E) Rosenmund reduction turns benzoyl chloride into benzaldehyde, and bromine water then oxidises the aldehyde to benzoic acid. Yes.
Answer: (C) B and E only. | ||||
| 69 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Nitrous acid diazotises sulphanilic acid: $\mathrm{HO_3S{-}C_6H_4{-}N_2^+}$. In acetic acid this couples with 1-naphthylamine at the 4-position to give the red azo dye
$\mathrm{HO_3S{-}C_6H_4{-}N{=}N{-}C_{10}H_6{-}NH_2}$ (4-[(4-amino-1-naphthyl)azo]benzenesulphonic acid).
Formula $\mathrm{C_{16}H_{13}N_3O_3S}$: $M = 192 + 13 + 42 + 48 + 32 = 327$ g mol$^{-1}$.
0.1 mol weighs 32.7 g $\approx 33$ g.
Answer: (B) 33. | ||||
| 70 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Each of the four different amino acids appears once, and a peptide chain has a direction (N-terminus to C-terminus), so every ordering is a different tetrapeptide.
Number of sequences $= 4! = 24$.
Answer: (C) 24. | ||||
| 71 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Read the half-life from the graph: the concentration falls from 40 g L$^{-1}$ (at 5 s) to 20 g L$^{-1}$ (at 15 s), and from 20 to 10 g L$^{-1}$ (at 25 s). Each halving takes 10 s, whatever the starting value, so the reaction is first order with $t_{1/2} = 10$ s.
$k = \frac{0.693}{10}$ s$^{-1}$ and $t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{t_{1/2}}{\log2}\log\frac{50}{2.5} = \frac{10\log20}{\log2}$.
$\log20 = 1 + 0.3010 = 1.3010$, so $t = \frac{10\times1.3010}{0.3010} = 43.2$ s.
Answer: 43. | ||||
| 72 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry 0.2% (w/v): 0.2 g in 100 mL, i.e. 2 g L$^{-1}$; $c = \frac{2}{40} = 0.05$ mol L$^{-1}$ $= 50$ mol m$^{-3}$.
Resistivity $\rho = 870.0$ m$\Omega$ m $= 0.870\ \Omega$ m, so $\kappa = \frac{1}{0.870} = 1.149$ S m$^{-1}$.
$\Lambda_m = \frac{\kappa}{c} = \frac{1.149}{50} = 2.299\times10^{-2}$ S m$^2$ mol$^{-1}$.
Convert: 1 S m$^2$ $= 10^3$ mS $\times 10^2$ dm$^2$ $= 10^5$ mS dm$^2$, so $\Lambda_m = 2299$ mS dm$^2$ mol$^{-1}$ $= 22.99\times10^2$.
Answer: 23. | ||||
| 73 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Delta T_b = K_b\frac{w/M}{W\text{ (kg)}}$, so $M = \frac{K_b\,w}{\Delta T_b\,W}$ (non-electrolytes assumed).
AB: $M_1 = \frac{0.5\times1}{2.7\times0.015} = \frac{1000}{81} = 12.35$ g mol$^{-1}$.
$\mathrm{AB_2}$: $M_2 = \frac{0.5\times1}{1.5\times0.015} = \frac{200}{9} = 22.22$ g mol$^{-1}$.
$A + B = 12.35$ and $A + 2B = 22.22$ give $B = 9.88$ and $A = 2M_1 - M_2 = \frac{2000 - 1800}{81} = 2.469$ amu.
$2.469 = 24.69\times10^{-1}$, nearest integer 25.
Answer: 25. | ||||
| 74 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Least enthalpy of atomisation among Ni, Zn, Mn, Cu: Zn (about 126 kJ mol$^{-1}$), since its filled $3d^{10}$ shell takes no part in metallic bonding. So M = Zn.
Diamagnetic complexes:
$\mathrm{K_2[NiCl_4]}$: $\mathrm{Ni^{2+}}$ $d^8$, tetrahedral with weak-field $\mathrm{Cl^-}$, two unpaired electrons: paramagnetic.
$\mathrm{[Zn(H_2O)_6]Cl_2}$: $\mathrm{Zn^{2+}}$ $d^{10}$: diamagnetic.
$\mathrm{K_3[Mn(CN)_6]}$: $\mathrm{Mn^{3+}}$ $d^4$, low spin $t_{2g}^4$, two unpaired: paramagnetic.
$\mathrm{[Cu(PPh_3)_3I]}$: $\mathrm{Cu^+}$ $d^{10}$: diamagnetic.
So $n = 2$ and the ion is $\mathrm{Zn^{2+}}$, $3d^{10}$, with no unpaired electrons: $\mu = \sqrt{0(0+2)} = 0$ BM.
Answer: 0. | ||||
| 75 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | NumericalEasy | |
A easy numerical question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes 2-Bromopentane $\mathrm{C_5H_{11}Br}$: $M = 60 + 11 + 80 = 151$ g mol$^{-1}$, so 151 g is 1 mol.
Alcoholic KOH eliminates HBr; the major product (Saytzeff) is pent-2-ene, P. At 80% yield: 0.8 mol.
$\mathrm{Br_2}$ adds across the double bond: Q is 2,3-dibromopentane, $\mathrm{C_5H_{10}Br_2}$, $M = 60 + 10 + 160 = 230$ g mol$^{-1}$. At 100% yield: 0.8 mol.
Mass of Q $= 0.8\times230 = 184$ g.
Answer: 184. | ||||
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