JEE (Main) · 2025
JEE (Main) 2025 — 7 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
- Conic Sections3
- Sets, Relations and Functions2
- Sequences and Series2
9 easy · 14 medium · 2 hard
Physics
25 Q
- Electrostatics4
- Optics3
- Physical World and Measurement2
19 easy · 6 medium · 0 hard
Chemistry
25 Q
- Coordination Compounds3
- Thermodynamics2
- Solutions2
13 easy · 12 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and 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 $A$ is the rectangle $-3 \le \alpha \le 5$, $-1 \le \beta \le 11$.
$B$ is the region inside the ellipse $\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \le 1$: centre $(2, 6)$, semi-axes 3 (along $\alpha$) and 4 (along $\beta$), so every point of $B$ has $-1 \le \alpha \le 5$ and $2 \le \beta \le 10$.
Both ranges lie inside those of the rectangle, so every point of $B$ is in $A$: $B \subset A$. The corner $(-3, -1)$ of $A$ is not in $B$, so $A \not\subset B$, and $A \cup B = A$, which is not the set in option (C).
Answer: (B) $B \subset A$. | ||||
| 2 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Put $y = f(x)$: $y(x^2 - 3x + 2) = 5 - x$, that is $yx^2 + (1 - 3y)x + (2y - 5) = 0$.
For a real $x$ (when $y \ne 0$) the discriminant must be non-negative:
$(1 - 3y)^2 - 4y(2y - 5) = y^2 + 14y + 1 \ge 0$.
The roots of $y^2 + 14y + 1 = 0$ are $y = -7 \pm 4\sqrt3$, so $y \le -7 - 4\sqrt3$ or $y \ge -7 + 4\sqrt3$. ($y = 0$ is taken at $x = 5$, and the excluded $x = 1, 2$ give no value of $y$, so nothing else is lost.)
So $\alpha = -7 - 4\sqrt3$, $\beta = -7 + 4\sqrt3$ and
$\alpha^2 + \beta^2 = 2(49 + 48) = 194$.
Answer: (D) $194$. | ||||
| 3 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Split at $x = 2$ and $x = 3$.
$x \ge 3$: $x(x - 2) + 3(x - 3) + 1 = x^2 + x - 8 = 0$, $x = \frac{-1 \pm \sqrt{33}}{2}$; the larger is about $2.37 < 3$, so no root here.
$2 \le x < 3$: $x(x - 2) + 3(3 - x) + 1 = x^2 - 5x + 10 = 0$; discriminant $25 - 40 < 0$, no root.
$x < 2$: $x(2 - x) + 3(3 - x) + 1 = -x^2 - x + 10 = 0$, so $x^2 + x - 10 = 0$, $x = \frac{-1 \pm \sqrt{41}}{2}$: $2.70$ (not below 2, rejected) and $-3.70$ (accepted).
Exactly one real root.
Answer: (A) $1$. | ||||
| 4 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\frac{\bar z - 1}{2\bar z - i}$ is the conjugate of $w = \frac{z - 1}{2z + i}$, and a number and its conjugate have the same real part. So the condition is $2\,\mathrm{Re}(w) = 2$, $\mathrm{Re}(w) = 1$.
Write $z = x + iy$. $w = \frac{(z - 1)(2\bar z - i)}{|2z + i|^2}$, and the numerator is $2|z|^2 - iz - 2\bar z + i$, whose real part is $2(x^2 + y^2) + y - 2x$ (since $\mathrm{Re}(-iz) = y$).
$|2z + i|^2 = 4x^2 + (2y + 1)^2$.
$\mathrm{Re}(w) = 1$: $2x^2 + 2y^2 - 2x + y = 4x^2 + 4y^2 + 4y + 1$, so $x^2 + y^2 + x + \frac{3}{2}y + \frac{1}{2} = 0$.
Centre $(a, b) = \left(-\frac12, -\frac34\right)$, $r^2 = \frac14 + \frac{9}{16} - \frac12 = \frac{5}{16}$.
$\frac{15ab}{r^2} = \frac{15\cdot\frac38}{\frac{5}{16}} = 18$.
Answer: (C) $18$. | ||||
| 5 | 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 For infinitely many solutions the coefficient determinant is zero:
$\begin{vmatrix}1&5&-1\\4&3&-3\\24&1&\lambda\end{vmatrix} = (3\lambda + 3) - 5(4\lambda + 72) - (4 - 72) = -17\lambda - 289 = 0$, so $\lambda = -17$.
The third equation must then be $p(\text{first}) + q(\text{second})$: $p + 4q = 24$, $5p + 3q = 1$ give $q = 7$, $p = -4$ (check: $z$-coefficient $4 - 21 = -17$), so $\mu = -4 + 49 = 45$.
Solve the first two: from the first, $x = 1 - 5y + z$; into the second, $-17y + z = 3$. So $z = 3 + 17y$, $x = 4 + 12y$, with $y$ free.
$x + y + z = 7 + 30y$, and $7 \le 7 + 30y \le 77$ gives $0 \le y \le \frac73$.
$x, y, z$ integers: $12y$ and $17y$ integers force $y$ to be an integer (as $\gcd(12, 17) = 1$), so $y = 0, 1, 2$.
Three solutions.
Answer: (A) $3$. | ||||
| 6 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_7 = 7a_4 = 7$, so $a_4 = 1$; with $a_6 = 7$, $2d = 6$, $d = 3$ and $a_1 = 1 - 9 = -8$.
$S_n = \frac{n}{2}\left[-16 + 3(n - 1)\right] = \frac{n(3n - 19)}{2} = 700$, so $3n^2 - 19n - 1400 = 0$.
$n = \frac{19 + \sqrt{361 + 16800}}{6} = \frac{19 + 131}{6} = 25$.
$a_{25} = -8 + 24\times3 = 64$.
Answer: (B) $64$. | ||||
| 7 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $ar(1 + r^2 + r^4) = 21$ and $ar^7(1 + r^2 + r^4) = 15309$. Dividing, $r^6 = 729$, so $r = 3$ (positive terms).
Then $3a(1 + 9 + 81) = 273a = 21$, $a = \frac{1}{13}$.
$S_9 = \frac{a(r^9 - 1)}{r - 1} = \frac{19682}{13\times2} = 757$.
Answer: (C) $757$. | ||||
| 8 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations No three vertices of a regular polygon are collinear, so $p = {}^nC_3$ and $q = {}^nC_4$.
$p + q = {}^nC_3 + {}^nC_4 = {}^{n+1}C_4 = 126 = {}^9C_4$, so $n = 8$.
Ellipse $\frac{x^2}{16} + \frac{y^2}{8} = 1$: $e^2 = 1 - \frac{8}{16} = \frac12$, $e = \frac{1}{\sqrt2}$.
Answer: (B) $\frac{1}{\sqrt2}$. | ||||
| 9 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $P(X = 2) = P(X = 3) = q$, with $2p + 2q = 1$.
$E(X) = p + 2q + 3q = p + 5q$ and $E(X^2) = p + 4q + 9q = p + 13q$.
$p + 13q = 2p + 10q$ gives $p = 3q$; then $8q = 1$, $q = \frac18$, $p = \frac38$.
$8p - 1 = 3 - 1 = 2$.
Answer: (B) $2$. | ||||
| 10 | 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 $P(\text{unbiased}) = \frac{19}{20}$, $P(H\mid\text{unbiased}) = \frac12$; $P(\text{two-headed}) = \frac{1}{20}$, $P(H\mid\text{two-headed}) = 1$.
By Bayes' theorem, $P(\text{unbiased}\mid H) = \frac{\frac{19}{40}}{\frac{19}{40} + \frac{2}{40}} = \frac{19}{21}$.
$m = 19$, $n = 21$: $n^2 - m^2 = 441 - 361 = 80$.
Answer: (D) $80$. | ||||
| 11 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines The first two lines meet at $x + 1 = 4x - 8$, $x = 3$: vertex $A(3, 4)$.
The altitude from $A$ passes through the orthocentre $H(3, -1)$, so it is the line $x = 3$; the opposite side $y = mx + c$ is perpendicular to it, so $m = 0$ and that side is $y = c$.
Its ends: $B(c - 1, c)$ on $y = x + 1$ and $C\left(\frac{c + 8}{4}, c\right)$ on $y = 4x - 8$.
The altitude from $B$ is perpendicular to $y = 4x - 8$, slope $-\frac14$, and passes through $H$:
$\frac{-1 - c}{3 - (c - 1)} = -\frac14 \Rightarrow -4 - 4c = -4 + c \Rightarrow c = 0$.
Check: $C = (2, 0)$, and $HC$ has slope $\frac{0 + 1}{2 - 3} = -1$, perpendicular to $y = x + 1$.
$m - c = 0$.
Answer: (A) $0$. | ||||
| 12 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $f(t) = \left(t + \frac12\right)^2 + \frac{11}{12} - \frac14$, minimum $\frac{2}{3}$, so $e = \frac23$.
With the major axis along $x$: $b^2 = a^2(1 - e^2) = \frac{5a^2}{9}$, and the latus rectum $\frac{2b^2}{a} = \frac{10a}{9} = 10$, so $a = 9$.
$a^2 = 81$, $b^2 = 45$, $a^2 + b^2 = 126$.
Answer: (D) $126$. | ||||
| 13 | Conic SectionsStandard forms of conics | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections As $b < 5$, the first ellipse has its major axis along $y$: $e_1^2 = 1 - \frac{b^2}{25}$. For the hyperbola $e_2^2 = 1 + \frac{b^2}{16}$.
$e_1^2e_2^2 = 1$: $(25 - b^2)(16 + b^2) = 400 \Rightarrow 9b^2 - b^4 = 0 \Rightarrow b^2 = 9$.
Then $e_1 = \frac45$ and the ellipse's foci are $(0, \pm 5e_1) = (0, \pm4)$; $e_2 = \frac54$ and the hyperbola's foci are $(\pm 4e_2, 0) = (\pm5, 0)$.
The ellipse through $(\pm5, 0)$ and $(0, \pm4)$ is $\frac{x^2}{25} + \frac{y^2}{16} = 1$, with $e^2 = 1 - \frac{16}{25} = \frac{9}{25}$, $e = \frac35$.
Answer: (C) $\frac35$. | ||||
| 14 | 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 $\cos2\theta\cos\frac\theta2 = \frac12\left[\cos\frac{5\theta}{2} + \cos\frac{3\theta}{2}\right]$, and $2\cos^3u = \frac12(\cos3u + 3\cos u)$ with $u = \frac{5\theta}{2}$.
The equation becomes $\frac32\cos\frac{5\theta}{2} + \frac12\cos\frac{3\theta}{2} = \frac12\cos\frac{15\theta}{2} + \frac32\cos\frac{5\theta}{2}$, that is $\cos\frac{15\theta}{2} = \cos\frac{3\theta}{2}$.
So $\frac{15\theta}{2} = 2k\pi \pm \frac{3\theta}{2}$: $6\theta = 2k\pi$, $\theta = \frac{k\pi}{3}$, or $9\theta = 2k\pi$, $\theta = \frac{2k\pi}{9}$.
In $\left[-\frac\pi2, \frac\pi2\right]$: $\theta = 0, \pm\frac\pi3$ from the first family and $\theta = 0, \pm\frac{2\pi}{9}, \pm\frac{4\pi}{9}$ from the second.
Distinct values: $0, \pm\frac{2\pi}{9}, \pm\frac{\pi}{3}, \pm\frac{4\pi}{9}$, which is 7.
Answer: (C) $7$. | ||||
| 15 | 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 Let $P = |\vec a + \vec b|$, $M = |\vec a - \vec b|$. From $\frac{P + M}{P - M} = \sqrt2 + 1$ (componendo–dividendo): $\frac{P}{M} = \frac{\sqrt2 + 2}{\sqrt2} = 1 + \sqrt2$, so $\frac{P^2}{M^2} = 3 + 2\sqrt2$.
As $|\vec a| = |\vec b|$, $P^2 + M^2 = 2|\vec a|^2 + 2|\vec b|^2 = 4|\vec a|^2$.
$\frac{P^2}{|\vec a|^2} = \frac{4P^2}{P^2 + M^2} = \frac{4(3 + 2\sqrt2)}{4 + 2\sqrt2} = \frac{2(3 + 2\sqrt2)(2 - \sqrt2)}{2} = 6 - 3\sqrt2 + 4\sqrt2 - 4 = 2 + \sqrt2$.
Answer: (A) $2 + \sqrt2$. | ||||
| 16 | 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 The point $\left(0, -\frac12, 0\right)$ lies on the line: $\frac{0 - 1}{-2} = \frac12$, so $\frac{-\frac12 + 4}{d} = \frac12$ gives $d = 7$, and $\frac{0 - c}{-4} = \frac12$ gives $c = 2$.
The direction $(-2, 7, -4)$ is perpendicular to both given directions:
$-2 + 7a - 4b = 0$ and $2b + 7a - 20 = 0$.
Subtracting, $-2 - 4b - 2b + 20 = 0$, so $b = 3$, and then $7a = 14$, $a = 2$.
$a + b + c + d = 2 + 3 + 2 + 7 = 14$.
Answer: (D) $14$. | ||||
| 17 | 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 Both lines have direction $(1, 1, 1)$.
On $L_1$: $Q = (1 + t, 2 + t, t)$, and $\vec{PQ}\cdot(1, 1, 1) = (t - 4) + (t + 1) + (t + 3) = 3t = 0$, so $Q = (1, 2, 0)$.
On $L_2$: $R = (2 + s, s, 1 + s)$, and $(s - 3) + (s - 1) + (s + 4) = 3s = 0$, so $R = (2, 0, 1)$.
$\vec{PQ} = (-4, 1, 3)$, $\vec{PR} = (-3, -1, 4)$, $\vec{PQ}\times\vec{PR} = (1\cdot4 - 3\cdot(-1),\ 3\cdot(-3) - (-4)\cdot4,\ (-4)(-1) - 1\cdot(-3)) = (7, 7, 7)$.
$A = \frac12\cdot7\sqrt3$, so $4A^2 = 49\times3 = 147$.
Answer: (C) $147$. | ||||
| 18 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The limit exists and equals 5 only if $f$ has no constant or $x$ term and the $x^2$ coefficient is 5: $f(x) = ax^4 + bx^3 + 5x^2$.
$f'(x) = x(4ax^2 + 3bx + 10)$, and $f'(4) = f'(5) = 0$, so $4ax^2 + 3bx + 10 = 4a(x - 4)(x - 5) = 4a(x^2 - 9x + 20)$.
Constant terms: $80a = 10$, $a = \frac18$; $x$ terms: $3b = -36a = -\frac92$, $b = -\frac32$.
$f(2) = \frac18\cdot16 - \frac32\cdot8 + 5\cdot4 = 2 - 12 + 20 = 10$.
Answer: (B) $10$. | ||||
| 19 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Divide by $x^2 + 1$: $y' - \frac{2x}{x^2 + 1}y = (x^2 + 1)\cos x$. Integrating factor $e^{-\ln(x^2 + 1)} = \frac{1}{x^2 + 1}$.
$\frac{d}{dx}\left(\frac{y}{x^2 + 1}\right) = \cos x$, so $\frac{y}{x^2 + 1} = \sin x + C$, and $y(0) = 1$ gives $C = 1$.
$y = (x^2 + 1)(1 + \sin x)$.
$\int_{-3}^{3} y\,dx = \int_{-3}^{3}(x^2 + 1)\,dx + \int_{-3}^{3}(x^2 + 1)\sin x\,dx = 2(9 + 3) + 0 = 24$ (the second integrand is odd).
Answer: (B) $24$. | ||||
| 20 | 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 $x + 7 = 11 - 3x$ at $x = 1$; for $x \le 1$ the minimum is $x + 7$, for $x \ge 1$ it is $11 - 3x$.
$1 + x^2 = x + 7$ at $x = -2, 3$; $1 + x^2 = 11 - 3x$ at $x = 2, -5$. So the region runs from $x = -2$ to $x = 2$.
$A = \int_{-2}^{1}(x + 6 - x^2)\,dx + \int_{1}^{2}(10 - 3x - x^2)\,dx$.
First: $\left[\frac{x^2}{2} + 6x - \frac{x^3}{3}\right]_{-2}^{1} = \frac{37}{6} + \frac{22}{3} = \frac{81}{6}$.
Second: $\left[10x - \frac{3x^2}{2} - \frac{x^3}{3}\right]_{1}^{2} = \frac{34}{3} - \frac{49}{6} = \frac{19}{6}$.
$A = \frac{100}{6} = \frac{50}{3}$, so $3A = 50$.
Answer: (A) $50$. | ||||
| 21 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The term with ${}^{20}C_j$ is $(-1)^j(j - 2)(j - 3)\,{}^{20}C_j$ for $j = 4, 5, \ldots, 20$: at $j = 4$ the weight is $2\times1$, at $j = 5$ it is $3\times2$, at $j = 20$ it is $18\times17$, as printed.
Over the full range $j = 0$ to $20$: $(j - 2)(j - 3)$ is a quadratic in $j$, and for $n = 20$
$\sum(-1)^j{}^{n}C_j = 0$, $\sum(-1)^j j\,{}^{n}C_j = 0$, $\sum(-1)^j j(j - 1)\,{}^{n}C_j = 0$
(these are the values at $x = 1$ of $(1 - x)^n$ and its first two derivatives, up to sign). So the full sum is 0.
The terms $j = 0$ to $3$ are: $j = 0$: $(-2)(-3) = 6$; $j = 1$: $-(-1)(-2)\cdot20 = -40$; $j = 2, 3$: $0$. Together $-34$.
Required sum $= 0 - (-34) = 34$.
Answer: 34. | ||||
| 22 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Focus $(-ae, 0) = (-5, 0)$ and directrix $x = -\frac ae = -\frac95$: $ae = 5$, $\frac ae = \frac95$, so $a^2 = 9$, $a = 3$, $e = \frac53$, and $b^2 = a^2(e^2 - 1) = 16$.
The point lies on $\frac{x^2}{9} - \frac{y^2}{16} = 1$: $\frac{\alpha^2}{9} = 1 + \frac{20}{16} = \frac94$, $|\alpha| = \frac92$.
Focal distances are $e|\alpha| \pm a = \frac{15}{2} \pm 3 = \frac{21}{2}, \frac92$.
$p = \frac{189}{4}$, $4p = 189$.
Answer: 189. | ||||
| 23 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $u = t + 2 \to 1^+$ and $h = u - 1 \to 0^+$. Then $u^{1/k} - 1 = \frac{h}{k} + O(h^2)$.
Dividing the equation by $h$, the coefficients tend to $\frac17$, $\frac16$, $\frac{1}{21}$, so the roots tend to the roots of $\frac{x^2}{7} + \frac{x}{6} + \frac{1}{21} = 0$, i.e. $6x^2 + 7x + 2 = 0$: $x = -\frac12, -\frac23$.
$a + b = -\frac76$ (only the sum matters), and $72(a + b)^2 = 72\cdot\frac{49}{36} = 98$.
Answer: 98. | ||||
| 24 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals For $x > 0$: $\sqrt[23]{3x^{-24} + x^{-26}} = \sqrt[23]{x^{-23}\left(3x^{-1} + x^{-3}\right)} = \frac{1}{x}\left(3x^{-1} + x^{-3}\right)^{1/23}$.
So the integrand is $\left(\frac{1}{x^2} + \frac{1}{x^4}\right)\left(\frac{3}{x} + \frac{1}{x^3}\right)^{1/23}$.
Put $u = \frac3x + \frac{1}{x^3}$: $du = -3\left(\frac{1}{x^2} + \frac{1}{x^4}\right)dx$.
$\int = -\frac13\int u^{1/23}\,du = -\frac13\cdot\frac{23}{24}u^{24/23} + C = -\frac{23}{3\cdot24}\left(3x^{-1} + x^{-3}\right)^{\frac{24}{23}} + C$.
Matching: $\alpha = 23$, $\beta = -1$, $\gamma = -3$, and $\alpha + \beta + \gamma = 19$.
Answer: 19. | ||||
| 25 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f(0) = \lim_{x\to0}f(x)$. Expand to order $x^3$:
$\tan x = x + \frac{x^3}{3} + \ldots$, so $\tan(\tan x) = \tan x + \frac{\tan^3x}{3} + \ldots = x + \frac{2x^3}{3} + \ldots$
$\sin x = x - \frac{x^3}{6} + \ldots$, so $\sin(\sin x) = \sin x - \frac{\sin^3x}{6} + \ldots = x - \frac{x^3}{3} + \ldots$
Numerator $= x^3 + O(x^5)$; denominator $\tan x - \sin x = \frac{x^3}{2} + O(x^5)$.
$f(0) = \frac{1}{1/2} = 2$.
Answer: 2. | ||||
| 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 $\mu_0\epsilon_0 = \frac{1}{c^2}$, so $\sqrt{\frac{\mu_0}{\epsilon_0}} = \sqrt{\mu_0^2c^2} = \mu_0c$.
$\mu_0$ has the units of inductance per length (H m$^{-1}$), so $\mu_0c$ has the units of H s$^{-1}$. From $V = L\frac{dI}{dt}$, H s$^{-1}$ = V A$^{-1}$ = $\Omega$.
(Indeed $\sqrt{\mu_0/\epsilon_0} \approx 377\ \Omega$ is the impedance of free space.)
Answer: (B) Resistance. | ||||
| 27 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Mass density = mass/volume: $[ML^{-3}T^0]$ → (IV).
Impulse = force × time = change of momentum: $[MLT^{-1}]$ → (II).
Power = work/time: $[ML^2T^{-2}]/[T] = [ML^2T^{-3}]$ → (I).
Moment of inertia = mass × (distance)$^2$: $[ML^2T^0]$ → (III).
Answer: (D) A→IV, B→II, C→I, D→III. | ||||
| 28 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $a = v\frac{dv}{dx} = 4\sqrt{x}\cdot\frac{4}{2\sqrt{x}} = 8$ m s$^{-2}$, a constant.
$F = ma = 0.5\times8 = 4$ N.
Answer: (B) 4 N. | ||||
| 29 | 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 The object keeps the helicopter's horizontal speed $360$ km/h $= 100$ m/s.
In 20 s it moves $100\times20 = 2000$ m $= 2$ km horizontally, and it falls the 2 km altitude (check: $\frac12\times10\times20^2 = 2000$ m).
Displacement of O from the release point $= \sqrt{2^2 + 2^2} = 2\sqrt2$ km.
Answer: (B) $2\sqrt2$ km. | ||||
| 30 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power A potential energy can be defined only for a conservative force, one whose work between two points does not depend on the path.
Coulomb's force, gravitational force and the (elastic) restoring force are conservative.
Friction always opposes the motion, so its work depends on the path length and is not recovered: it is non-conservative and has no potential energy.
Answer: (D) Frictional force. | ||||
| 31 | GravitationKepler's laws | Class 11: Gravitation | Assertion and reasonEasy | |
A easy assertion and reason question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation (A) is Kepler's second law: $\frac{dA}{dt}$ is constant. True.
(R): a central force acts along the radius vector, so its torque about the centre is $\vec r\times\vec F = 0$ and the angular momentum $\vec L$ is conserved. True.
The areal velocity is $\frac{dA}{dt} = \frac{L}{2m}$, so it is constant because $L$ is constant: (R) explains (A).
Answer: (A). | ||||
| 32 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on 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 Vertical rise: $h = \frac{2T\cos\theta}{r\rho g} = \frac{2\times0.07\times1}{10^{-4}\times1000\times9.8} = \frac{1}{7}$ m $= 14.29$ cm.
In a tube inclined at $30^\circ$ to the vertical the water still rises to the same vertical height, so the length of water in the tube is
$l = \frac{h}{\cos30^\circ} = \frac{14.29}{0.866} \approx 16.5$ cm.
The nearest option is $\frac{82}{5} = 16.4$ cm.
Answer: (B) $\frac{82}{5}$. | ||||
| 33 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The average translational kinetic energy per molecule of an ideal gas is $\frac32kT$, which depends only on the temperature, not on the molar mass. Both gases are monatomic, so this is all their kinetic energy.
At the same 300 K the ratio is $1 : 1$ (the molar masses change the speeds, not the energies).
Answer: (A) 1 : 1. | ||||
| 34 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Matching listEasy | |
A easy matching list question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Isothermal (ideal gas): temperature constant, so $\Delta U = 0$ → (IV).
Adiabatic: no heat exchanged, $\Delta Q = 0$ → (II).
Isobaric: $\Delta W = p\Delta V \ne 0$ and the temperature changes, so $\Delta U \ne 0$ → (III).
Isochoric: no volume change, $\Delta W = 0$ → (I).
Answer: (D) A→IV, B→II, C→III, D→I. | ||||
| 35 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Particle velocity $\frac{\partial y}{\partial t} = 10\pi\cos(20\pi x + 10\pi t)$, so the speed is $10\pi|\cos(20\pi x + 10\pi t)|$.
Two points have the same speed at every instant when their phases differ by $\pi$ (the cosines are then equal and opposite). The smallest such separation is half a wavelength.
$k = 20\pi$ m$^{-1}$, $\lambda = \frac{2\pi}{k} = 0.1$ m $= 10$ cm, so $\frac{\lambda}{2} = 5$ cm.
Answer: (B) 5.0 cm. | ||||
| 36 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Assertion and reasonEasy | |
A easy assertion and reason 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 (A): no isolated magnetic pole has ever been found; poles always occur in N–S pairs. True.
(R): magnetic field lines do not start or end anywhere; they run continuously (through a magnet too) and close on themselves. True.
Gauss's law for magnetism, $\oint\vec B\cdot d\vec A = 0$, ties the two together: since there is no magnetic charge for lines to start or end on, they must form closed loops, so the net flux through any closed surface is zero. (R) is taken as the explanation of (A).
Answer: (A). | ||||
| 37 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Dipole moment $p = qd = 2\times10^{-6}\times0.5\times10^{-6} = 10^{-12}$ C m.
Field between the plates $E = \frac{V}{d} = \frac{5}{0.5\times10^{-3}} = 10^4$ V/m.
$\tau = pE\sin30^\circ = 10^{-12}\times10^4\times\frac12 = 5\times10^{-9}$ N m.
Answer: (B) $5\times10^{-9}$ Nm. | ||||
| 38 | 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 (R): in electrostatics the field inside a charge-free cavity in a conductor is zero, whatever the charges and fields outside (electrostatic shielding). True.
(A): the metal body of an aircraft is such a closed conducting shell; the charge of a lightning strike stays on and flows over the outer surface, and the field inside the cabin stays (nearly) zero, so the people inside are protected. True, and it is so because of (R).
Answer: (A). | ||||
| 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 The intensity of an electromagnetic wave is $I = \frac12\epsilon_0cE_0^2$, where $E_0$ is the amplitude of the electric field.
So $\sqrt{\frac{2I}{\epsilon_0c}} = E_0$, an electric field, with unit N C$^{-1}$ (= V m$^{-1}$).
Answer: (B) NC$^{-1}$. | ||||
| 40 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics A magnification of $+\frac14$ (erect, diminished) is given by a convex mirror. With the object at distance $x$ in front, $m = -\frac vu = \frac14$ puts the virtual image $\frac x4$ behind the mirror.
Object–image distance: $x + \frac x4 = 40$, so $x = 32$ cm and the image is 8 cm behind.
$\frac1f = \frac1v + \frac1u = \frac18 - \frac{1}{32} = \frac{3}{32}$, $f = \frac{32}{3} \approx 10.7$ cm.
(Reading the magnification as $-\frac14$ for a concave mirror gives $x - \frac x4 = 40$, $x = \frac{160}{3}$, $v = \frac{40}{3}$ cm, and again $|f| = \frac{32}{3} \approx 10.7$ cm.)
Answer: (B) 10.7 cm. | ||||
| 41 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Assertion and reasonEasy | |
A easy assertion and reason question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics (A): glass ($n \approx 1.5$) is optically denser than air ($n \approx 1.0003$). True.
(R): optical density (refractive index) is not proportional to mass density. For example, turpentine has a higher refractive index than water although it is less dense, and refractive index is not doubled when mass density doubles. False.
Answer: (C) (A) is correct but (R) is not correct. | ||||
| 42 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At the side face: $\sin\theta = 1.25\sin r$.
The side and top faces are perpendicular, so the ray meets the top face at an angle of incidence $90^\circ - r$.
Total internal reflection there needs $\sin(90^\circ - r) \ge \sin\theta_C = \frac{1}{1.25} = 0.8$, i.e. $\cos r \ge 0.8$, $\sin r \le 0.6$.
So $\sin\theta = 1.25\sin r \le 0.75$: the largest $\theta$ is $\sin^{-1}\left(\frac34\right)$ (at that angle the ray just grazes the top face, as drawn).
Answer: (C) $\sin^{-1}(3/4)$. | ||||
| 43 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation, with work function $\phi = \frac{hc}{\lambda_o}$: $K_{max} = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_o} = hc\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_o}\right)$.
$\lambda_e = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mhc\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_o}\right)}} = \sqrt{\frac{h^2}{2mhc\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_o}\right)}} = \sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_o}\right)}}$.
Option (C) lacks the square root over $h$ and so has the wrong dimensions.
Answer: (D). | ||||
| 44 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Assertion and reasonEasy | |
A easy assertion and reason question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (R): $R = R_0A^{1/3}$ with $R_0 \approx 1.2$ fm. True.
Then the nuclear density is $\rho = \frac{Am_u}{\frac43\pi R_0^3A} = \frac{3m_u}{4\pi R_0^3}$, independent of $A$: copper and carbon nuclei have (nearly) the same density, about $2.3\times10^{17}$ kg m$^{-3}$.
So (A) is false.
Answer: (D) (A) is not correct but (R) is correct. | ||||
| 45 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits From the figure: A and B go to an AND gate; B and $\bar A$ (A through a NOT gate) go to an OR gate; the two outputs go to a NAND gate.
$Y = \overline{(A\cdot B)\cdot(B + \bar A)}$.
$Y = 0$ only when both $A\cdot B = 1$ and $B + \bar A = 1$. $A\cdot B = 1$ needs $A = B = 1$, and then $B + \bar A = 1$ as well.
Check the others: $(0,0)$: $AB = 0$, $Y = 1$; $(0,1)$: $AB = 0$, $Y = 1$; $(1,0)$: $AB = 0$, $Y = 1$.
Answer: (D) A = 1 and B = 1. | ||||
| 46 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The small disc touches the rim, so its centre O' is $R - \frac R3 = \frac{2R}{3}$ from O. Its mass (same surface density) is $M\cdot\frac{(R/3)^2}{R^2} = \frac M9$.
About the axis through O perpendicular to the plane:
whole disc: $\frac12MR^2$;
small disc (parallel axes): $\frac12\cdot\frac M9\cdot\frac{R^2}{9} + \frac M9\cdot\frac{4R^2}{9} = \frac{MR^2}{162} + \frac{8MR^2}{162} = \frac{9MR^2}{162}$.
Remaining: $\frac{81 - 9}{162}MR^2 = \frac{72}{162}MR^2 = \frac49MR^2$.
So $x = 9$.
Answer: 9. | ||||
| 47 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Thermal resistance $R = \frac{L}{K\pi r^2}$, so $\frac{R_A}{R_B} = \frac{L_A}{L_B}\cdot\frac{K_B}{K_A}\cdot\left(\frac{r_B}{r_A}\right)^2 = \frac12\times2\times\frac14 = \frac14$.
In the steady state the same heat current flows through both: $\frac{400 - T}{R_A} = \frac{T - 200}{R_B}$, so $4(400 - T) = T - 200$, $5T = 1800$, $T = 360$ K.
Answer: 360. | ||||
| 48 | 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 $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + 50^2} = 50\sqrt2\ \Omega$.
$I_{rms} = \frac{10}{50\sqrt2} = \frac{1}{5\sqrt2}$ A.
$P = I_{rms}^2R = \frac{1}{50}\times50 = 1$ W.
Answer: 1. | ||||
| 49 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With free charge $Q$ on the plates and a slab filling the gap, the induced (bound) charge on each face of the slab is $Q' = Q\left(1 - \frac1K\right)$.
$4\times10^{-6} = 5\times10^{-6}\left(1 - \frac1K\right) \Rightarrow 1 - \frac1K = \frac45 \Rightarrow K = 5$.
Answer: 5. | ||||
| 50 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance A surface parallel to the $x$-$z$ plane has its area vector along $\hat j$, so only $E_y = 4\times10^3$ N/C contributes to the flux.
$\Phi = E_yA \Rightarrow A = \frac{6.0}{4\times10^3} = 1.5\times10^{-3}$ m$^2 = 15$ cm$^2$.
Answer: 15. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | StatementsMedium | |
A medium statements question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom A half-filled subshell (such as $d^5$ in Cr or Mn$^{2+}$) is extra stable because:
(A) its electrons are distributed symmetrically over the degenerate orbitals. True.
(D) the number of ways of exchanging parallel-spin electrons among the degenerate orbitals is largest, so the exchange energy (a stabilising term) is large. True.
(B) electrons of the same spin in different orbitals keep apart, so the coulombic repulsion energy is smaller. True.
(E) electrons in the same subshell shield one another relatively poorly, so each feels a larger effective nuclear charge. True.
(C) is false: the parallel-spin electrons are in degenerate orbitals (same energy) of one subshell; that is exactly where exchange energy arises.
Correct: (A), (B), (D) and (E).
Answer: (C). | ||||
| 52 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics (A) False: the standard state fixes the pressure (1 bar), not the temperature; the temperature must be stated (298 K is merely the usual one).
(B) False for the same reason: a pure gas at 1 bar is in its standard state at any stated temperature, not only at 273 K.
(C) False: O(g) is not the reference form of oxygen; forming it from $\mathrm{O_2}$(g) needs half the bond dissociation energy, so $\Delta_fH^\circ$ of O(g) is about +249 kJ mol$^{-1}$.
(D) True: $\mathrm{O_2}$(g) is the reference state of oxygen, and the standard enthalpy of formation of an element in its reference state is zero at every temperature, including 500 K.
Answer: (D). | ||||
| 53 | ThermodynamicsHess law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Dissolving can be split into two steps (Hess's law):
1. Break the lattice: $\mathrm{KCl(s) \to K^+(g) + Cl^-(g)}$. The lattice energy $-z$ is for forming the lattice from the gaseous ions, so this step needs $+z$.
2. Hydrate the ions: $\mathrm{K^+(g) + Cl^-(g) \to K^+(aq) + Cl^-(aq)}$, $\Delta H = -x - y$.
$\Delta_{sol}H = z - x - y = z - (x + y)$ kJ/mol.
Answer: (D) $z - (x + y)$. | ||||
| 54 | SolutionsRaoult's law | Class 12: Solutions | Matching listEasy | |
A easy matching list question on raoult's law, from Solutions. NCERT: Class 12: Solutions (A) Chloroform and acetone hydrogen-bond to each other (C–H···O=C), stronger than the like attractions: negative deviation from Raoult's law, so a maximum boiling azeotrope → (III).
(B) Ethanol and water: adding ethanol breaks some of water's hydrogen bonds, positive deviation, so a minimum boiling azeotrope (95% ethanol) → (I).
(C) Benzene and toluene form a nearly ideal solution: $\Delta V_{mix} = 0$ (and $\Delta H_{mix} = 0$) → (IV).
(D) Acetic acid in benzene forms hydrogen-bonded dimers → (II).
Answer: (C) A→III, B→I, C→IV, D→II. | ||||
| 55 | SolutionsRaoult's law | Class 12: Solutions | Single correctEasy | |
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions By Raoult's and Dalton's laws, $y_A = \frac{x_Ap_A^\circ}{P}$ and $y_B = \frac{x_Bp_B^\circ}{P}$, so
$\frac{y_A}{y_B} = \frac{x_A}{x_B}\cdot\frac{350}{750} < \frac{x_A}{x_B}$.
The vapour is richer in the more volatile B. So (A) holds and (B), (D) do not.
(C): since $x_A + x_B = 1 = y_A + y_B$, $x_B - y_B = -(x_A - y_A)$, and $x_A - y_A > 0$; so $x_A - y_A > x_B - y_B$ and (C) is false.
Answer: (A). | ||||
| 56 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | StatementsMedium | |
A medium statements question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry At the cathode, the ion with the highest (most positive) reduction potential is reduced first, at the lowest applied voltage.
Statement I: $0.80 > 0.79 > 0.24$ V, so as the voltage is raised Ag deposits first, then Hg, then Cu. Correct.
Statement II: $\mathrm{Mg^{2+}}$ ($-2.37$ V) is far harder to reduce than water, so Mg is indeed not deposited; but the cathode then gives hydrogen ($\mathrm{2H_2O + 2e^- \to H_2 + 2OH^-}$), not oxygen. Oxygen is evolved at the anode. Incorrect.
Answer: (C) Statement I is correct but Statement II is incorrect. | ||||
| 57 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let the initial pressure of A be $p_0$. If A falls by $x$: $p_A = p_0 - x$, and $p_t = (p_0 - x) + x + x = p_0 + x$.
At completion $p_\infty = 2p_0$, so $p_0 = \frac{p_\infty}{2}$ and $x = p_t - p_0$.
$p_A = p_0 - x = 2p_0 - p_t = p_\infty - p_t$.
First order: $k = \frac1t\ln\frac{p_0}{p_A} = \frac1t\ln\frac{p_\infty}{2(p_\infty - p_t)}$.
Answer: (B). | ||||
| 58 | 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 $\mathrm{SO_2}$: S has 2 σ-bonds and 1 lone pair, steric number 3 → sp$^2$ (bent).
$\mathrm{NO_2^-}$: N has 2 σ-bonds and 1 lone pair, steric number 3 → sp$^2$ (bent).
$\mathrm{N_3^-}$: $[\mathrm{N{=}N{=}N}]^-$, the central N has 2 σ-bonds and no lone pair, steric number 2 → sp (linear).
Answer: (C) sp$^2$, sp$^2$ and sp. | ||||
| 59 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across a period atomic radius falls; down a group it rises.
(A) At and Cs are in period 6, Cs at the left: $r_{At} < r_{Cs}$. Correct trend.
(B) Mg and Al are in period 3 with Al to the right of Mg, so Al is smaller (Mg about 160 pm, Al about 143 pm). $r_{Mg} < r_{Al}$ is the incorrect trend.
(C) Rb above Cs in group 1: $r_{Rb} < r_{Cs}$. Correct.
(D) Br (period 4, group 17) is smaller than K (period 4, group 1). Correct.
Answer: (B). | ||||
| 60 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down group 13 the inert pair effect makes the +1 state more stable than +3 for thallium.
(A) True: $\mathrm{Tl^{3+}}$ readily takes 2 electrons to become $\mathrm{Tl^+}$, so it is a strong oxidising agent.
(B) True: $E^\circ(\mathrm{Al^{3+}/Al}) = -1.66$ V, so $\mathrm{Al^{3+}}$ is hard to reduce.
(C) False: $\mathrm{Tl^{3+}}$ is not stable in solution; it is reduced to $\mathrm{Tl^+}$.
(D) True, by the inert pair effect.
(E) True: +3 is the stable state for Al and +1 for Tl.
Correct: (A), (B), (D) and (E).
Answer: (C). | ||||
| 61 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Oxides of a metal become more acidic as its oxidation state rises.
$\mathrm{V_2O_3}$ (V +3) is basic; $\mathrm{VO_2}$ (V +4) is amphoteric; $\mathrm{V_2O_5}$ (V +5) is amphoteric, though mainly acidic; $\mathrm{CrO_3}$ (Cr +6) and $\mathrm{Mn_2O_7}$ (Mn +7) are acidic.
So the acidic oxides are $\mathrm{CrO_3}$ and $\mathrm{Mn_2O_7}$: X = 2.
In $\mathrm{[Co(en)_3]_2(SO_4)_3}$, three sulfates (charge −6) balance two complex ions, so each is $\mathrm{[Co(en)_3]^{3+}}$ and Co is +3. Primary valency (oxidation state) Y = 3.
X + Y = 5.
Answer: (C) $5$. | ||||
| 62 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[Fe(CN)_6]^{3-}}$: Fe$^{3+}$, $d^5$, strong-field CN$^-$, low spin $t_{2g}^5$: 1 unpaired.
$\mathrm{[FeF_6]^{3-}}$: Fe$^{3+}$, $d^5$, weak-field F$^-$, high spin $t_{2g}^3e_g^2$: 5 unpaired.
$\mathrm{[CoF_6]^{3-}}$: Co$^{3+}$, $d^6$, weak field, high spin $t_{2g}^4e_g^2$: 4 unpaired.
$\mathrm{[Mn(CN)_6]^{3-}}$: Mn$^{3+}$, $d^4$, strong field, low spin $t_{2g}^4$: 2 unpaired.
So 1, 5, 4, 2.
Answer: (D). | ||||
| 63 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | Matching listMedium | |
A medium matching list question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Primary valency = oxidation state of the metal; secondary valency = coordination number.
(A) $\mathrm{[Co(en)_2Cl_2]^+}$: Co +3; two bidentate en and two Cl give CN 6 → 3, 6 → (I).
(B) $\mathrm{[Pt(NH_3)_2Cl(NO_2)]}$: neutral, so Pt +2; four donor atoms, CN 4 → 2, 4 → (IV).
(D) $\mathrm{[Mg(EDTA)]^{2-}}$: Mg +2; EDTA$^{4-}$ is hexadentate, CN 6 → 2, 6 → (III).
(C) $\mathrm{Hg[Co(SCN)_4]}$: CN 4. The only option with A→I, B→IV and D→III is the one with C→II (3, 4).
Answer: (C) A→I, B→IV, C→II, D→III. | ||||
| 64 | 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 NaOH converts benzoic acid into sodium benzoate, an ionic salt that dissolves in the aqueous layer: $\mathrm{C_6H_5COOH + NaOH \to C_6H_5COO^-Na^+ + H_2O}$.
Chlorobenzene is neutral and aniline is a base; neither reacts with NaOH, so both stay in the organic (ethyl acetate) layer.
Answer: (C) chlorobenzene and aniline. | ||||
| 65 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | StatementsMedium | |
A medium statements question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Statement I: in the cis isomers the two C–X bond dipoles add. C–Cl is more polar than C–Br (Cl is more electronegative), so cis-1,2-dichloroethene (about 1.9 D) is more polar than cis-1,2-dibromoethene (about 1.35 D). Correct.
Statement II: in trans-1,2-dibromoethene the two C–Br dipoles point in opposite directions and cancel, so its dipole moment is (nearly) zero; it has weaker dipole–dipole attraction and boils lower than the cis isomer (about 108 °C against 112 °C). The first part is right, but it is less polar, not more. Incorrect.
Answer: (C) Statement I is correct but Statement II is incorrect. | ||||
| 66 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The compound is $\mathrm{CH_3{-}CH{=}CH{-}CH(CH_3){-}CH{=}CH{-}CH(CH_3){-}CH{=}CH{-}CH_3}$ with two stereocentres.
Complete ozonolysis cleaves all three C=C bonds, each carbon of a double bond becoming a C=O:
the two ends give $\mathrm{CH_3CHO}$ (twice), and each $\mathrm{{=}CH{-}CH(CH_3){-}CH{=}}$ unit gives $\mathrm{OHC{-}CH(CH_3){-}CHO}$ (2-methylpropanedial, twice).
In $\mathrm{OHC{-}CH(CH_3){-}CHO}$ the central carbon carries two identical CHO groups, so it is not a stereocentre; ethanal is achiral too.
The stereocentres are destroyed: none of the products is optically active.
Answer: (A) $0$. | ||||
| 67 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Matching listEasy | |
A easy matching list question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Each conversion replaces Cl on the ring by OH. Plain chlorobenzene resists nucleophilic substitution (partial double-bond character of C–Cl); nitro groups at the ortho and para positions withdraw electrons and stabilise the intermediate, so each extra nitro group makes the reaction easier.
(A) Chlorobenzene → phenol needs NaOH at 623 K and 300 atm, then acid → (IV).
(B) 4-Nitrochlorobenzene → 4-nitrophenol: NaOH at 443 K, then acid → (III).
(C) 2,4-Dinitrochlorobenzene → 2,4-dinitrophenol: NaOH at 368 K, then acid → (II).
(D) 2,4,6-Trinitrochlorobenzene → picric acid with warm water alone → (I).
Answer: (D) A→IV, B→III, C→II, D→I. | ||||
| 68 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A positive 2,4-DNP test shows a carbonyl group; a negative Tollens test rules out an aldehyde. So P is a ketone, $\mathrm{C_6H_{12}O}$, and it must have a stereocentre.
(A) 4-methylpentan-2-one: the CH carries two CH$_3$ groups, no stereocentre. Not optically active.
(B) 3-methylpentan-2-one: C-3 carries H, CH$_3$, C$_2$H$_5$ and COCH$_3$, four different groups, so it is chiral. A ketone, Tollens negative. Fits.
(C) 3-methylpentanal is chiral but an aldehyde: it would give a positive Tollens test.
(D) hexan-2-one has no stereocentre.
Answer: (B). | ||||
| 69 | AminesBasicity of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines Aliphatic amines are much stronger bases than aromatic amines, whose nitrogen lone pair is delocalised into the ring.
In water, $\mathrm{(CH_3)_2NH}$ (p$K_b$ 3.27) is more basic than $\mathrm{CH_3NH_2}$ (3.38): E > D.
Among the anilines: the $-$OMe group at the para position donates electrons by resonance and raises the electron density on N, so 4-methoxyaniline (B) is more basic than aniline (A); the para $-$NO$_2$ group withdraws electrons strongly, so 4-nitroaniline (C) is the weakest.
Order: E > D > B > A > C.
Answer: (A). | ||||
| 70 | BiomoleculesProteins | Class 12: Biomolecules | StatementsMedium | |
A medium statements question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Statement I: oligopeptides (a few amino acid units) do give few $\alpha$-amino acids on hydrolysis, but proteins also give $\alpha$-amino acids, not $\beta$-amino acids. Incorrect.
Statement II: denaturation (by heat, or by a change in pH such as added acid) disturbs the secondary and tertiary structure. In the standard description it is the water-soluble globular proteins that coagulate into an insoluble, fibre-like form (as egg white does on boiling). Fibrous proteins are already insoluble in water, so the statement as worded is incorrect.
Answer: (B) Both Statement I and Statement II are incorrect. | ||||
| 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 Butane: $\frac{174000}{58} = 3000$ mol. Oxygen: $\frac{320000}{32} = 10000$ mol.
3000 mol of butane would need $6.5\times3000 = 19500$ mol $\mathrm{O_2}$, so oxygen is the limiting reagent.
Water formed $= 10000\times\frac{5}{6.5} = 7692.3$ mol $= 7692.3\times18 = 138461.5$ g.
With density 1 g mL$^{-1}$ that is 138461.5 mL $= 138.46$ L.
Answer: 138. | ||||
| 72 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Basic buffer: $\mathrm{pOH} = \mathrm{p}K_b + \log\frac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]}$.
Before: $\frac{0.10}{0.10} = 1$, pOH $= 4.745$, pH $= 9.255$.
HCl converts 0.05 mol $\mathrm{NH_3}$ into $\mathrm{NH_4^+}$: $\mathrm{NH_3}$ 0.05 mol, $\mathrm{NH_4^+}$ 0.15 mol.
After: pOH $= 4.745 + \log3 = 5.222$, pH $= 8.778$.
Change in pH $= 0.477 = 47.7\times10^{-2}$, which to the nearest integer is 48.
Answer: 48. | ||||
| 73 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Unpaired electrons:
$\mathrm{[Co(NH_3)_6]^{3+}}$: Co$^{3+}$ $d^6$, low spin $t_{2g}^6$: 0 (diamagnetic).
$\mathrm{[Co(C_2O_4)_3]^{3-}}$: Co$^{3+}$ $d^6$, low spin: 0 (diamagnetic).
$\mathrm{[MnCl_6]^{3-}}$: Mn$^{3+}$ $d^4$, weak field, high spin: 4.
$\mathrm{[Mn(CN)_6]^{3-}}$: Mn$^{3+}$ $d^4$, low spin $t_{2g}^4$: 2.
$\mathrm{[CoF_6]^{3-}}$: Co$^{3+}$ $d^6$, high spin: 4.
$\mathrm{[Fe(CN)_6]^{3-}}$: Fe$^{3+}$ $d^5$, low spin: 1.
$\mathrm{[FeF_6]^{3-}}$: Fe$^{3+}$ $d^5$, high spin: 5.
The only paramagnetic species sharing a count are $\mathrm{[MnCl_6]^{3-}}$ and $\mathrm{[CoF_6]^{3-}}$ (4 each): 2 species.
Answer: 2. | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Pressure of dry $\mathrm{N_2}$ $= 715 - 15 = 700$ mm Hg.
Volume at STP: $V_0 = 50\times\frac{700}{760}\times\frac{273}{300} = 41.91$ mL.
Mass of nitrogen $= \frac{28}{22400}\times41.91 = 0.05239$ g.
%N $= \frac{0.05239}{0.292}\times100 = 17.9\% \approx 18$.
(With $n = \frac{PV}{RT} = \frac{(700/760)\times0.050}{0.0821\times300} = 1.870\times10^{-3}$ mol, mass $= 0.0524$ g, the same 17.9%.)
Answer: 18. | ||||
| 75 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons A: $\mathrm{PCl_5}$ turns the C=O of acetophenone into a gem-dichloride, $\mathrm{Ph{-}CCl_2{-}CH_3}$.
B: $\mathrm{NaNH_2}$ removes two HCl to give phenylacetylene, and the third equivalent removes the acidic terminal H: $\mathrm{Ph{-}C{\equiv}C^-\,Na^+}$.
C: acidifying gives $\mathrm{Ph{-}C{\equiv}CH}$.
D: hydroboration–oxidation of a terminal alkyne adds water anti-Markovnikov; the enol tautomerises to the aldehyde $\mathrm{Ph{-}CH_2{-}CHO}$ (phenylacetaldehyde).
sp$^2$ carbons in D: the 6 ring carbons and the CHO carbon $= 7$ (the $\mathrm{CH_2}$ is sp$^3$).
Answer: 7. | ||||
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