JEE (Main) · 2025
JEE (Main) 2025 — 2 April, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
25 Q
- Conic Sections3
- Matrices and Determinants3
- Vector Algebra2
5 easy · 18 medium · 2 hard
Physics
25 Q
- Optics4
- Motion of System of Particles and Rigid Body3
- Electrostatics3
16 easy · 9 medium · 0 hard
Chemistry
25 Q
- Some Basic Concepts of Chemistry3
- Solutions2
- Equilibrium2
11 easy · 11 medium · 3 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 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Focus $S = (1, 0)$. Take $P = (t^2, 2t)$ with $t > 0$ (first quadrant).
Slope of PS: $\frac{2t}{t^2 - 1} = \tan 60^\circ = \sqrt{3}$, so $\sqrt{3}t^2 - 2t - \sqrt{3} = 0$, giving $t = \frac{2 \pm 4}{2\sqrt{3}}$; the positive root is $t = \sqrt{3}$.
So $P = (3, 2\sqrt{3})$.
The circle on PS as diameter has centre $\left(\frac{3+1}{2}, \frac{2\sqrt{3}}{2}\right) = (2, \sqrt{3})$ and radius $\frac{PS}{2} = \frac{x_P + 1}{2} = 2$.
The centre is at distance 2 from the $y$-axis, equal to the radius, so the circle touches the $y$-axis at $(0, \sqrt{3})$: $\alpha = \sqrt{3}$.
$5\alpha^2 = 15$.
Answer: (A) $15$. | ||||
| 2 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Cross-multiply: $2 + k^2z = kz(k + \bar z) = k^2z + kz\bar z = k^2z + k|z|^2 = k^2z + k$.
So $k = 2$, and the point is $k + ik^2 = 2 + 4i$.
The circle has centre $1 + 2i$ and radius 1. Distance of $2 + 4i$ from the centre: $|1 + 2i| = \sqrt{5}$.
The point is outside the circle, so its maximum distance from a point of the circle is $\sqrt{5} + 1$.
Answer: (C) $\sqrt{5} + 1$. | ||||
| 3 | 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 A quadratic in $c = \cos\theta$: $2\sqrt{2}c^2 + (2 - \sqrt{6})c - \sqrt{3} = 0$.
Discriminant: $(2 - \sqrt{6})^2 + 4\cdot2\sqrt{2}\cdot\sqrt{3} = 10 - 4\sqrt{6} + 8\sqrt{6} = 10 + 4\sqrt{6} = (2 + \sqrt{6})^2$.
$c = \frac{(\sqrt{6} - 2) \pm (2 + \sqrt{6})}{4\sqrt{2}}$, so $c = \frac{2\sqrt{6}}{4\sqrt{2}} = \frac{\sqrt{3}}{2}$ or $c = \frac{-4}{4\sqrt{2}} = -\frac{1}{\sqrt{2}}$.
$\cos\theta = \frac{\sqrt{3}}{2}$ in $[-2\pi, 2\pi]$: $\theta = \pm\frac{\pi}{6}, \pm\frac{11\pi}{6}$, 4 solutions.
$\cos\theta = -\frac{1}{\sqrt{2}}$: $\theta = \pm\frac{3\pi}{4}, \pm\frac{5\pi}{4}$, 4 solutions.
Total 8.
Answer: (A) $8$. | ||||
| 4 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $a^2 = 18$, $b^2 = 9$, $e^2 = 1 - \frac{9}{18} = \frac{1}{2}$.
For $P = (x, y)$ on the ellipse the focal distances are $a + ex$ and $a - ex$, so $SP\cdot S'P = a^2 - e^2x^2 = 18 - \frac{x^2}{2}$, with $0 \le x^2 \le 18$.
Maximum (at $x = 0$): 18. Minimum (at $x^2 = 18$): $18 - 9 = 9$.
Sum $= 27$.
Answer: (D) $27$. | ||||
| 5 | 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 Put A at the origin and $\vec{b} = p\hat i$, $\vec{c} = q\hat j$, $\vec{d} = r\hat k$ along the three perpendicular edges.
Then $\frac{pq}{2} = 5$, $\frac{qr}{2} = 6$, $\frac{rp}{2} = 7$.
Area of BCD $= \frac{1}{2}|(\vec c - \vec b)\times(\vec d - \vec b)| = \frac{1}{2}|qr\,\hat i + rp\,\hat j + pq\,\hat k|$.
So $(\text{area})^2 = \left(\frac{qr}{2}\right)^2 + \left(\frac{rp}{2}\right)^2 + \left(\frac{pq}{2}\right)^2 = 36 + 49 + 25 = 110$.
Area $= \sqrt{110}$.
Answer: (C) $\sqrt{110}$. | ||||
| 6 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The power of the prime 3 in $50!$ is $\left\lfloor\frac{50}{3}\right\rfloor + \left\lfloor\frac{50}{9}\right\rfloor + \left\lfloor\frac{50}{27}\right\rfloor + \left\lfloor\frac{50}{81}\right\rfloor = 16 + 5 + 1 + 0 = 22$.
So the largest $n$ is 22.
Answer: (B) $22$. | ||||
| 7 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability The denominator tends to 0, so for a finite limit the numerator must too: $\gamma - 1 = 0$, $\gamma = 1$.
The numerator is then $x^2\sin\alpha x = \alpha x^3 + O(x^5)$.
Denominator: $\sin 2x - \beta x = (2 - \beta)x - \frac{8x^3}{6} + \ldots$. If $\beta \ne 2$ the limit would be 0, so $\beta = 2$ and the denominator is $-\frac{4}{3}x^3 + \ldots$.
Limit $= \frac{\alpha}{-4/3} = -\frac{3\alpha}{4} = 3$, so $\alpha = -4$.
$\beta + \gamma - \alpha = 2 + 1 + 4 = 7$.
Answer: (D) $7$. | ||||
| 8 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 6x^2 - 18ax + 12a^2 = 6(x - a)(x - 2a)$.
With $a > 0$, $f'$ changes from $+$ to $-$ at $x = a$ (local maximum) and from $-$ to $+$ at $x = 2a$ (local minimum). So $p = a$, $q = 2a$.
$p^2 = q$: $a^2 = 2a$, $a = 2$.
$f(x) = 2x^3 - 18x^2 + 48x + 1$, $f(3) = 54 - 162 + 144 + 1 = 37$.
Answer: (A) $37$. | ||||
| 9 | 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 The sequence has five 1s, three 2s and so two 0s. The number of arrangements of these ten symbols is
$\frac{10!}{5!\,3!\,2!} = \frac{3628800}{120\cdot6\cdot2} = 2520$.
Answer: (D) $2520$. | ||||
| 10 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Focus: $ae = \sqrt{10}$. Directrix: $\frac{a}{e} = \frac{9}{\sqrt{10}}$.
Multiplying, $a^2 = 9$; dividing, $e^2 = \frac{10}{9}$.
$b^2 = a^2(e^2 - 1) = 9\cdot\frac{1}{9} = 1$, and $l = \frac{2b^2}{a} = \frac{2}{3}$.
$9(e^2 + l) = 9\left(\frac{10}{9} + \frac{2}{3}\right) = 10 + 6 = 16$.
Answer: (B) $16$. | ||||
| 11 | 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 $\vec a = x\hat i + y\hat j + z\hat k$. The first two vectors have magnitude 3, $\hat k$ has magnitude 1, so the projections are $\frac{2x - y + 2z}{3}$, $\frac{x + 2y - 2z}{3}$ and $z$.
Equal projections: $2x - y + 2z = 3z$ and $x + 2y - 2z = 3z$, i.e. $2x - y - z = 0$ and $x + 2y - 5z = 0$.
From the first, $y = 2x - z$; then $x + 4x - 2z - 5z = 0$, so $x = \frac{7z}{5}$ and $y = \frac{9z}{5}$.
$\vec a \parallel 7\hat i + 9\hat j + 5\hat k$, of magnitude $\sqrt{49 + 81 + 25} = \sqrt{155}$.
Unit vector: $\pm\frac{1}{\sqrt{155}}(7\hat i + 9\hat j + 5\hat k)$; the $+$ sign is offered.
Answer: (D). | ||||
| 12 | 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 Reflexive? $(f, f) \in R$ needs $f(0) = f(1)$, false for e.g. $f(x) = x$. Not reflexive.
Symmetric? If $f(0) = g(1)$ and $f(1) = g(0)$, then $g(0) = f(1)$ and $g(1) = f(0)$, so $(g, f) \in R$. Symmetric.
Transitive? From $(f, g), (g, h) \in R$: $f(0) = g(1) = h(0)$ and $f(1) = g(0) = h(1)$. For $(f, h) \in R$ we need $f(0) = h(1)$, i.e. $f(0) = f(1)$, which need not hold. E.g. $f(x) = x$, $g(x) = 1 - x$, $h = f$: $(f, g), (g, f) \in R$ but $(f, f) \notin R$. Not transitive.
So R is symmetric but neither reflexive nor transitive.
Answer: (B). | ||||
| 13 | 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 line passes through $A(-3, 1, -4)$ with direction $\vec d = (5, 2, 3)$, $|\vec d| = \sqrt{38}$.
$\vec{AP} = (3, 1, 7)$, and $\vec{AP}\times\vec d = (1\cdot3 - 7\cdot2,\ 7\cdot5 - 3\cdot3,\ 3\cdot2 - 1\cdot5) = (-11, 26, 1)$, of magnitude $\sqrt{121 + 676 + 1} = \sqrt{798}$.
Distance of P from the line: $h = \sqrt{\frac{798}{38}} = \sqrt{21}$.
Area $= \frac{1}{2}\cdot QR\cdot h = \frac{5\sqrt{21}}{2} = \frac{m}{n}$, so $2m = 5\sqrt{21}\,n$, i.e. $2m - 5\sqrt{21}\,n = 0$.
Answer: (D). | ||||
| 14 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctHard | |
A hard single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Collect the terms: $f(2x + 2y)(\sin x\cos y - \cos x\sin y) = f(2x - 2y)(\sin x\cos y + \cos x\sin y)$, i.e.
$f(2x + 2y)\sin(x - y) = f(2x - 2y)\sin(x + y)$.
Put $u = 2x + 2y$, $v = 2x - 2y$ (any reals): $f(u)\sin\frac{v}{2} = f(v)\sin\frac{u}{2}$.
So $\frac{f(u)}{\sin(u/2)}$ is the same constant $k$ for all $u$: $f(x) = k\sin\frac{x}{2}$.
$f'(x) = \frac{k}{2}\cos\frac{x}{2}$, $f'(0) = \frac{k}{2} = \frac{1}{2}$, so $k = 1$.
$f''(x) = -\frac{1}{4}\sin\frac{x}{2}$, $f''\left(\frac{5\pi}{3}\right) = -\frac{1}{4}\sin\frac{5\pi}{6} = -\frac{1}{8}$.
$24f''\left(\frac{5\pi}{3}\right) = -3$.
Answer: (B) $-3$. | ||||
| 15 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A = (A + I) - I = \begin{bmatrix}0 & a & 1\\2 & 0 & 0\\a & 1 & 1\end{bmatrix}$.
Expanding along the second row: $\det A = -2\begin{vmatrix}a & 1\\1 & 1\end{vmatrix} = -2(a - 1) = -4$, so $a = 3$.
Then $\det(4\,\mathrm{adj}(2A)) = 4^3\det(\mathrm{adj}(2A)) = 64\,(\det 2A)^2$ (for order 3, $\det(\mathrm{adj}B) = (\det B)^2$).
$\det(2A) = 8\cdot(-4) = -32$, so the value is $64\cdot1024 = 2^{16} = 2^{16}3^0$.
$m = 16$, $n = 0$, $m + n = 16$.
Answer: (C) $16$. | ||||
| 16 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem With $t = x^{1/3}$: $x + 1 = t^3 + 1 = (t + 1)(t^2 - t + 1)$, so the first fraction is $x^{1/3} + 1$.
With $s = x^{1/2}$: $\frac{x - 1}{x - x^{1/2}} = \frac{(s - 1)(s + 1)}{s(s - 1)} = 1 + x^{-1/2}$.
The bracket is $x^{1/3} - x^{-1/2}$.
General term of $(x^{1/3} - x^{-1/2})^{10}$: $T_{r+1} = \binom{10}{r}(-1)^r x^{\frac{10 - r}{3} - \frac{r}{2}}$.
Independent of $x$: $\frac{10 - r}{3} = \frac{r}{2}$, $20 - 2r = 3r$, $r = 4$.
$T_5 = \binom{10}{4} = 210$.
Answer: (B) $210$. | ||||
| 17 | Quadratic EquationsForming equations from given roots | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on forming equations from given roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $s = \alpha + \beta = P_1 = 1$ and $p = \alpha\beta$. Since $\alpha, \beta$ are roots of $x^2 - sx + p = 0$, $P_{n} = sP_{n-1} - pP_{n-2}$.
$P_{10} = P_9 - pP_8$: $123 = 76 - 47p$, so $p = -1$.
For the roots $\frac{1}{\alpha}, \frac{1}{\beta}$: sum $= \frac{s}{p} = -1$, product $= \frac{1}{p} = -1$.
Equation: $x^2 - (-1)x + (-1) = 0$, i.e. $x^2 + x - 1 = 0$.
Answer: (B). | ||||
| 18 | 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 $\det A = \alpha\beta + 6 = 0$ and $\alpha + \beta = 1$: $\alpha^2 - \alpha - 6 = 0$, $\alpha = 3$ (as $\alpha > 0$), $\beta = -2$.
$A = \begin{bmatrix}3 & -1\\6 & -2\end{bmatrix}$ has trace 1 and determinant 0, so by Cayley–Hamilton $A^2 = A$, and $A^k = A$ for all $k \ge 1$.
$(I + A)^8 = \sum_{k=0}^{8}\binom{8}{k}A^k = I + (2^8 - 1)A = I + 255A$.
$= \begin{bmatrix}1 + 765 & -255\\1530 & 1 - 510\end{bmatrix} = \begin{bmatrix}766 & -255\\1530 & -509\end{bmatrix}$.
Answer: (C). | ||||
| 19 | 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 Let the common difference be $d$. $a_1, a_3, \ldots, a_{23}$ are 12 terms of an A.P. with difference $2d$:
$\sum = 12a_1 + \frac{12\cdot11}{2}\cdot2d = 12a_1 + 132d = -\frac{72}{5}a_1$.
So $132d = -\frac{132}{5}a_1$, $d = -\frac{a_1}{5}$.
$\sum_{k=1}^{n}a_k = \frac{n}{2}\left(2a_1 + (n - 1)d\right) = \frac{n}{2}a_1\left(2 - \frac{n - 1}{5}\right) = 0$ gives $n - 1 = 10$, $n = 11$.
Answer: (B) $11$. | ||||
| 20 | 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 Infinitely many solutions need $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$.
$\Delta_z = \begin{vmatrix}3 & 1 & 3\\2 & \alpha & -3\\1 & 2 & 4\end{vmatrix} = 3(4\alpha + 6) - (8 + 3) + 3(4 - \alpha) = 9\alpha + 19 = 0$, so $\alpha = -\frac{19}{9}$.
$\Delta = \begin{vmatrix}3 & 1 & \beta\\2 & \alpha & -1\\1 & 2 & 1\end{vmatrix} = 3(\alpha + 2) - 3 + \beta(4 - \alpha) = 3\alpha + 3 + \beta(4 - \alpha) = 0$.
With $\alpha = -\frac{19}{9}$: $-\frac{10}{3} + \frac{55}{9}\beta = 0$, so $\beta = \frac{6}{11}$.
(Check: $\Delta_x = 3(\alpha + 2) - 1 - \beta(6 + 4\alpha) = -\frac{1}{3} - 1 + \frac{6}{11}\cdot\frac{22}{9} = 0$.)
$22\beta - 9\alpha = 12 + 19 = 31$.
Answer: (D) $31$. | ||||
| 21 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $|4 - x^2| \le x^2$: for $x^2 \le 4$ this needs $x^2 \ge 2$; for $x^2 > 4$ it always holds. $|4 - x^2| \le y \le 4$ needs $x^2 \le 8$. So $\sqrt{2} \le x \le 2\sqrt{2}$.
On $[\sqrt{2}, 2]$: $4 - x^2 \le y \le x^2$. Area $= \int_{\sqrt{2}}^{2}(2x^2 - 4)\,dx = \left[\frac{2x^3}{3} - 4x\right]_{\sqrt{2}}^{2} = -\frac{8}{3} + \frac{8\sqrt{2}}{3}$.
On $[2, 2\sqrt{2}]$: $x^2 - 4 \le y \le 4$. Area $= \int_{2}^{2\sqrt{2}}(8 - x^2)\,dx = \left[8x - \frac{x^3}{3}\right]_{2}^{2\sqrt{2}} = \frac{32\sqrt{2}}{3} - \frac{40}{3}$.
Total $= \frac{40\sqrt{2}}{3} - 16 = \frac{80\sqrt{2}}{6} - 16$.
So $\alpha = 6$, $\beta = 16$, $\alpha + \beta = 22$.
Answer: 22. | ||||
| 22 | CirclesStandard and general form | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections The lines meet at $(3, 0)$; their angle bisectors are $y = 0$ and $x = 3$.
At $(-9, 4)$ both $x + y - 3$ and $x - y - 3$ are negative, the region $x < 3 - |y|$, which contains the bisector ray $y = 0$, $x < 3$. So the centre is $(h, 0)$ with $h < 3$, and the radius is its distance from either line: $r = \frac{3 - h}{\sqrt{2}}$.
Through $(-9, 4)$: $(h + 9)^2 + 16 = \frac{(3 - h)^2}{2}$, i.e. $h^2 + 42h + 185 = 0$, $h = -5$ or $h = -37$.
$r^2 = \frac{(3 - h)^2}{2}$: $\frac{64}{2} = 32$ and $\frac{1600}{2} = 800$.
Difference $= 800 - 32 = 768$.
Answer: 768. | ||||
| 23 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The integrand is $[e^{1 - x}]$, a decreasing function equal to $e$ at $x = 0$.
$[e^{1-x}] = 2$ while $e^{1-x} \ge 2$, i.e. $0 \le x \le 1 - \ln 2$.
$[e^{1-x}] = 1$ while $1 \le e^{1-x} < 2$, i.e. $1 - \ln 2 < x \le 1$.
$[e^{1-x}] = 0$ for $x > 1$ (up to $e^3$).
Integral $= 2(1 - \ln 2) + 1\cdot\ln 2 = 2 - \ln 2$.
So $\alpha = 2$ and $\alpha^3 = 8$.
Answer: 8. | ||||
| 24 | 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 $f'' = f$ has the general solution $f(x) = Ae^{x} + Be^{-x}$.
$f(0) = A + B = 0$ and $f'(0) = A - B = 3$, so $A = \frac{3}{2}$, $B = -\frac{3}{2}$: $f(x) = \frac{3}{2}(e^x - e^{-x})$, which is odd and has $f'(x) = \frac{3}{2}(e^x + e^{-x}) > 0$, as required.
$f(\ln 3) = \frac{3}{2}\left(3 - \frac{1}{3}\right) = 4$.
$9f(\ln 3) = 36$.
Answer: 36. | ||||
| 25 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | NumericalHard | |
A hard numerical question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Count the 3-element subsets that form a G.P. Written in increasing order, the terms are $a, ar, ar^2$ with $r = \frac{p}{q} > 1$ in lowest terms. Since $ar^2 = \frac{ap^2}{q^2}$ is an integer and $\gcd(p, q) = 1$, $a = kq^2$, and the terms are $kq^2, kpq, kp^2$ with $kp^2 \le 40$.
For each coprime pair $p > q \ge 1$ there are $\left\lfloor\frac{40}{p^2}\right\rfloor$ choices of $k$:
$p = 2$ ($q = 1$): 10.
$p = 3$ ($q = 1, 2$): $4 + 4 = 8$.
$p = 4$ ($q = 1, 3$): $2 + 2 = 4$.
$p = 5$ ($q = 1, 2, 3, 4$): $1 \times 4 = 4$.
$p = 6$ ($q = 1, 5$): $1 + 1 = 2$.
$p \ge 7$: $p^2 > 40$, none.
Favourable: $10 + 8 + 4 + 4 + 2 = 28$. Total: $\binom{40}{3} = 9880$.
Probability $= \frac{28}{9880} = \frac{7}{2470}$, so $m + n = 2477$.
Answer: 2477. | ||||
| 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 Only like quantities can be added, so $\frac{a}{V^2}$ has the dimensions of pressure: $[a] = [P][V^2]$.
Likewise $b$ is subtracted from $V$, so $[b] = [V]$.
$[ab^{-2}] = \frac{[P][V^2]}{[V^2]} = [P] = ML^{-1}T^{-2}$.
Energy density (energy per volume) has $\frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$, the same.
(Compressibility is $M^{-1}LT^2$, Planck's constant $ML^2T^{-1}$, strain is dimensionless.)
Answer: (B) Energy density. | ||||
| 27 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (A) Viscosity: $F = \eta A\frac{dv}{dx}$, so $[\eta] = \frac{MLT^{-2}}{L^2\cdot T^{-1}} = ML^{-1}T^{-1}$ → (IV).
(B) Intensity is power per area: $\frac{ML^2T^{-3}}{L^2} = ML^0T^{-3}$ → (I).
(C) Pressure gradient is pressure per length: $\frac{ML^{-1}T^{-2}}{L} = ML^{-2}T^{-2}$ → (II).
(D) Compressibility is the reciprocal of bulk modulus (a pressure): $M^{-1}LT^2$ → (III).
A→IV, B→I, C→II, D→III.
Answer: (D). | ||||
| 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 The spokes are massless, so all the mass is in the rim: a ring, $I = MR^2 = 10\times(0.1)^2 = 0.1$ kg m$^2$.
Work done by the pull while 1 m of cord unwinds: $W = F\,s = 20\times1 = 20$ J. There is no friction, so all of it becomes rotational kinetic energy:
$\frac{1}{2}I\omega^2 = 20 \Rightarrow \omega^2 = \frac{40}{0.1} = 400$, $\omega = 20$ rad/s.
Answer: (B) 20 rad/s. | ||||
| 29 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Only the component of the boat's velocity relative to the water that is across the river carries it across; the river's flow is along the bank.
Across-river component: $27\sin150^\circ = 27\times\frac{1}{2} = 13.5$ km h$^{-1}$ $= 13.5\times\frac{5}{18} = 3.75$ m/s.
Time $= 30$ s, so width $= 3.75\times30 = 112.5$ m.
Answer: (A) 112.5 m. | ||||
| 30 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The pivot at O can supply any force, so the lamina stays at rest when the net torque about O is zero.
Take O as origin, OA along $+x$, OC along $+y$, side $a = 0.1$ m; anticlockwise torque $\tau = xF_y - yF_x$ is positive.
At C $(0, a)$: 10 N up has no arm (it acts along OC), $\tau = 0$. F to the left: $\tau = -a(-F) = +aF$.
At B $(a, a)$: 10 N up gives $+10a$; 10 N to the right gives $-10a$. Together 0.
At A $(a, 0)$: 10 N to the right acts along OA, $\tau = 0$; 10 N down gives $-10a$.
Net: $aF - 10a = 0$, so $F = 10$ N.
Answer: (C) 10 N. | ||||
| 31 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics In an adiabatic process no heat is exchanged: $dQ = 0$.
Molar heat capacity $C = \frac{dQ}{n\,dT} = 0$ while the temperature changes, so it is zero. (It is infinite for an isothermal process.)
(A) is wrong: from $dQ = dU + dW = 0$, the work done by the gas equals the decrease of internal energy, not the increase.
(B) is wrong: for an ideal gas the internal energy rises with temperature.
Answer: (C). | ||||
| 32 | 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 $\alpha = \frac{ML^2}{12}$.
Each half has mass $\frac{M}{2}$ and length $\frac{L}{2}$. In the cross the two halves cross at their midpoints, so the axis passes through the centre of each and is normal to it:
$I_{half} = \frac{1}{12}\cdot\frac{M}{2}\cdot\left(\frac{L}{2}\right)^2 = \frac{ML^2}{96} = \frac{\alpha}{8}$.
Cross: $2\times\frac{\alpha}{8} = \frac{\alpha}{4}$.
Answer: (D) $\alpha/4$. | ||||
| 33 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Both motions have $\omega = 5$ rad/s, so their sum is an SHM of the same $\omega$ with amplitude
$A^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\frac{\pi}{3} = 7 + 28 + 2\cdot\sqrt7\cdot2\sqrt7\cdot\frac{1}{2} = 35 + 14 = 49$, $A = 7$ cm.
Maximum acceleration $= \omega^2A = 25\times7 = 175$ cm/s$^2$ $= 175\times10^{-2}$ m s$^{-2}$.
$x = 175$.
Answer: (D) $175$. | ||||
| 34 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The sheet's field is uniform, horizontal and of size $E = \frac{\sigma}{2\epsilon_0}$.
The bob is held by tension, weight $mg$ (down) and the electric force $qE$ (horizontal, away from the sheet). With the string at $45^\circ$ to the vertical sheet, $\tan45^\circ = \frac{qE}{mg}$, so $qE = mg$.
$m = 100$ mg $= 10^{-4}$ kg, $mg = 10^{-3}$ N, $q = 10^{-5}$ C, so $E = \frac{10^{-3}}{10^{-5}} = 100$ N/C.
$\sigma = 2\epsilon_0E = 2\times8.85\times10^{-12}\times100 = 1.77\times10^{-9}$ C/m$^2$ $= 1.77$ nC/m$^2$.
Answer: (A). | ||||
| 35 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Charge stored: $Q = 5800$ mAh $= 5.8\ \text{A}\times3600\ \text{s} = 20880$ C.
Energy $= VQ = 4.2\times20880 = 87696$ J $\approx 87.7$ kJ.
Answer: (B) 87.7 kJ. | ||||
| 36 | 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 $B = \mu_0(H + M) = \mu_0(1 + \chi)H$, and $B = \mu H$.
So $\mu = \mu_0(1 + \chi)$, i.e. $\mu_r = \frac{\mu}{\mu_0} = 1 + \chi$, giving $\chi = \frac{\mu}{\mu_0} - 1$.
Answer: (C). | ||||
| 37 | 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 Each large plate acts as a sheet of charge whose field is $\frac{|\sigma|}{2\epsilon_0}$, the same at every distance.
Between the plates: the $+\sigma$ plate pushes $+q$ away from it (towards the $-2\sigma$ plate) with field $\frac{\sigma}{2\epsilon_0}$; the $-2\sigma$ plate pulls it towards itself, the same direction, with $\frac{2\sigma}{2\epsilon_0}$.
$E = \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$, the same anywhere between the plates.
(For conducting plates the charges rearrange on the faces, but the field between them is still $\frac{\sigma_1 - \sigma_2}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$.)
$F = qE = \frac{3\sigma q}{2\epsilon_0}$, towards the $-2\sigma$ plate.
Answer: (A). | ||||
| 38 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B_1 = \frac{\mu_0I}{2R}$ and $B_2 = \frac{\mu_0IR^2}{2(R^2 + x^2)^{3/2}}$, so $\frac{B_2}{B_1} = \frac{R^3}{(R^2 + x^2)^{3/2}}$.
With $x = 3k$, $R = 4k$: $R^2 + x^2 = 25k^2$, $(R^2 + x^2)^{3/2} = 125k^3$.
$\frac{B_2}{B_1} = \frac{64k^3}{125k^3} = \frac{64}{125}$.
Answer: (D) 64 : 125. | ||||
| 39 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Between two like charges the fields oppose. At $(x, 0, 0)$ with $0 < x < d$:
$\frac{kq}{x^2} = \frac{k\cdot9q}{(d - x)^2} \Rightarrow d - x = 3x \Rightarrow x = \frac{d}{4}$.
Answer: (B) $(d/4, 0, 0)$. | ||||
| 40 | OpticsWave optics | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on wave optics, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics A wave travels along the normal to its wave fronts. The normal to the planes $x + y + z = $ constant is $\hat{i} + \hat{j} + \hat{k}$, unit vector $\frac{1}{\sqrt3}(\hat{i} + \hat{j} + \hat{k})$.
Its direction cosine with the $x$-axis is $\frac{1}{\sqrt3}$, so the angle is $\cos^{-1}\left(\frac{1}{\sqrt3}\right)$.
Answer: (D). | ||||
| 41 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics From the figure: A is on the axis 30 cm from the lens; B is 1 cm nearer the lens and 2 cm above the axis. $f = +20$ cm. Use $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, i.e. $v = \frac{uf}{u + f}$.
A: $u = -30$ cm, $v = \frac{(-30)(20)}{-10} = 60$ cm, on the axis.
B: $u = -29$ cm, $v = \frac{(-29)(20)}{-9} = \frac{580}{9} \approx 64.44$ cm; magnification $m = \frac{v}{u} = -\frac{20}{9}$, so its height is $2\times\left(-\frac{20}{9}\right) = -\frac{40}{9} \approx -4.44$ cm.
The image runs from $(60, 0)$ to $(64.44, -4.44)$: it moves $\frac{40}{9}$ cm along the axis and $\frac{40}{9}$ cm below it.
$\tan\theta = \frac{-40/9}{40/9} = -1$, so the image makes $-45^\circ$ with the principal axis (below it).
(Small-object check: transverse magnification $-2$, longitudinal magnification $m^2 = 4$, so 1 cm along becomes 4 cm and 2 cm up becomes 4 cm down.)
Answer: (D) $-45^\circ$. | ||||
| 42 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Light goes from O (in medium 1) to the right. The surface is convex towards O with C on the far side, so $R = +0.4$ m; $u = -0.2$ m, $n_1 = 1$, $n_2 = 1.5$.
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$: $\frac{1.5}{v} + \frac{1}{0.2} = \frac{0.5}{0.4}$, so $\frac{1.5}{v} = 1.25 - 5 = -3.75$ and $v = -0.4$ m.
The image is virtual, 0.4 m to the left of the surface (on the same side as O).
Answer: (A). | ||||
| 43 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | StatementsEasy | |
A easy statements question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei For a hydrogen-like ion $E_n = -13.6\frac{Z^2}{n^2}$ eV.
H ground state ($Z = 1$, $n = 1$): $-13.6$ eV.
(A) He$^+$ first excited state ($Z = 2$, $n = 2$): $-13.6\cdot\frac{4}{4} = -13.6$ eV. True.
(B) Li$^{++}$ second excited state ($Z = 3$, $n = 3$): $-13.6\cdot\frac{9}{9} = -13.6$ eV. True.
(C) He$^+$ ground state: $-13.6\times4 = -54.4$ eV. False.
(D) He$^+$ ($n = 2$) is $-13.6$ eV but Li$^{++}$ ground state is $-13.6\times9 = -122.4$ eV. False.
Answer: (A) (A), (B) only. | ||||
| 44 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation: $K_{max} = \frac{hc}{\lambda} - \phi$, so the electron's momentum is $p = \sqrt{2mK_{max}}$.
In a field perpendicular to its velocity it moves on a circle of radius $r = \frac{p}{eB} = \frac{\sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)}}{eB}$.
Its velocity at A is normal to the plate, so the centre of the circle lies on the plate; the electron returns to the plate after a semicircle, at the other end of a diameter.
$AB = 2r = \frac{2\sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)}}{eB} = \frac{\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)}}{eB}$.
Answer: (A). | ||||
| 45 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The load is in parallel with the zener, so it sits at 5 V. The series resistor drops $25 - 5 = 20$ V and carries $I = \frac{20}{400} = 0.05$ A $= 50$ mA.
$I = I_Z + I_L = 4I_L + I_L = 5I_L$, so $I_L = 10$ mA (and $I_Z = 40$ mA).
$R_L = \frac{5}{0.01} = 500\ \Omega$.
Answer: (C). | ||||
| 46 | 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 Poisson's ratio $= \frac{\text{transverse strain}}{\text{longitudinal strain}}$, so longitudinal strain $\varepsilon = \frac{10^{-3}}{0.2} = 5\times10^{-3}$.
Energy density $u = \frac{1}{2}\times\text{stress}\times\text{strain} = \frac{1}{2}Y\varepsilon^2 = \frac{1}{2}\times2\times10^{11}\times25\times10^{-6} = 2.5\times10^6$ J m$^{-3}$ $= 25\times10^5$.
Answer: 25. | ||||
| 47 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Both sides are filled to the top, so at the window (depth 3 m) the atmospheric pressure cancels and the net pressure on the door is the difference of the liquid pressures:
$\Delta P = (\rho_2 - \rho_1)gh = (1500 - 1000)\times10\times3 = 1.5\times10^4$ Pa.
The window is small, so the pressure over it is taken as that at its depth: net force $= \Delta P\cdot A = 1.5\times10^4\times100\times10^{-4} = 150$ N, pushing from the denser liquid towards the water.
A force of 150 N must be applied to hold the door shut.
Answer: 150. | ||||
| 48 | 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 With $f$ degrees of freedom, $\gamma = 1 + \frac{2}{f}$.
Gas A: $f = 3$, $\gamma_A = \frac{5}{3}$.
Gas B: a vibrational mode carries two quadratic terms (kinetic and potential energy), so $f = 3 + 3 + 2 = 8$ and $\gamma_B = 1 + \frac{2}{8} = \frac{5}{4}$.
$\frac{\gamma_A}{\gamma_B} = \frac{5/3}{5/4} = \frac{4}{3} = 1 + \frac{1}{3}$, so $n = 3$.
(Counting the vibrational mode as one degree of freedom would give $\gamma_B = \frac{9}{7}$ and $\frac{\gamma_A}{\gamma_B} = \frac{35}{27}$, which is not of the form $1 + \frac{1}{n}$ with whole $n$.)
Answer: 3. | ||||
| 49 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Total distance $= x + \frac{3}{2}x = \frac{5}{2}x$. Total time $= \frac{x}{5} + \frac{3x}{2v_2}$.
$\frac{\frac{5}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}} = \frac{50}{7} \Rightarrow \frac{1}{5} + \frac{3}{2v_2} = \frac{5}{2}\cdot\frac{7}{50} = \frac{7}{20}$.
$\frac{3}{2v_2} = \frac{7}{20} - \frac{4}{20} = \frac{3}{20}$, so $v_2 = 10$ m/s.
Answer: 10. | ||||
| 50 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Minima of a single slit of width $a$: $a\sin\theta = m\lambda$.
The second minimum on one side is at $\theta_2$ and the third on the other side at $\theta_3$; their separation is $\theta_2 + \theta_3 = 30^\circ = \frac{\pi}{6}$.
With $\sin\theta \approx \theta$: $\frac{2\lambda}{a} + \frac{3\lambda}{a} = \frac{\pi}{6}$, so $a = \frac{30\lambda}{\pi} = \frac{30\times628\times10^{-9}}{3.14} = 6\times10^{-6}$ m $= 6\ \mu$m.
(Solving $\sin^{-1}\frac{2\lambda}{a} + \sin^{-1}\frac{3\lambda}{a} = \frac{\pi}{6}$ without the approximation gives $a \approx 6.08\ \mu$m, which also rounds to 6.)
Answer: 6. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles of HCl $= 0.250\ \text{L}\times0.76\ \text{M} = 0.19$ mol. Moles of $\mathrm{CaCO_3} = \frac{1000}{100} = 10$ mol, far more than the $0.095$ mol the acid can use, so HCl is the limiting reagent.
2 mol HCl give 1 mol $\mathrm{CaCl_2}$: $n(\mathrm{CaCl_2}) = \frac{0.19}{2} = 0.095$ mol.
Molar mass of $\mathrm{CaCl_2} = 40 + 71 = 111$ g mol$^{-1}$; mass $= 0.095\times111 = 10.545$ g.
Answer: (D) 10.545 g. | ||||
| 52 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In Bohr's model $r_n = 0.529\,n^2$ Å for hydrogen, so $\frac{r_{n_1}}{r_{n_2}} = \frac{n_1^2}{n_2^2}$.
(A) $\frac{16}{4} = 4$: correct.
(B) $\frac{36}{16} = 2.25$, not 3: incorrect.
(C) $\frac{64}{16} = 4$: correct.
(D) $\frac{9}{1} = 9$: correct.
Answer: (B). | ||||
| 53 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics This is Joule's free-expansion experiment. Only vessel A holds gas; B is evacuated, so the pressure in B before the stopcock is opened is zero, and (D) is true.
The gas expands against zero external pressure, so $dw = -p_{ext}\,dV = 0$; (A) is false.
The water's temperature does not change, so no heat passes between the gas and the water: $dq = 0$; (B) is false.
By the first law $dU = dq + dw = 0$; (C) is false. (For an ideal gas this is also why the temperature stays the same.)
Answer: (D). | ||||
| 54 | SolutionsRaoult's law | Class 12: Solutions | Single correctHard | |
A hard single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Henry's law: $p = K_H\,x$. A larger $K_H$ means a smaller mole fraction dissolved at the same pressure, i.e. a less soluble gas.
Order of $K_H$: of these gases helium is the least soluble in water, so it has the largest $K_H$ (about 14.5 GPa at 25 °C), then nitrogen (about 8.6 GPa), while methane and oxygen dissolve better (about 4 GPa). So He > N$_2$ > CH$_4$, and O$_2$ lies below N$_2$; plots (A) and (C), which put O$_2$ on top and He at the bottom, have the order reversed.
Temperature: gases become less soluble as water warms, so $K_H$ rises at first; for these gases it passes through a maximum (solubility through a minimum) and then falls at higher temperature. That is the shape of (B), not the straight lines of (C) or the minimum of (D).
Answer: (B). | ||||
| 55 | SolutionsRaoult's law | Class 12: Solutions | Single correctEasy | |
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions $x_A = \frac{1}{4}$, $x_B = \frac{3}{4}$. Raoult's law (ideal solution): $p = x_Ap_A^\circ + x_Bp_B^\circ$.
$500 = \frac{1}{4}(200) + \frac{3}{4}p_B^\circ \Rightarrow \frac{3}{4}p_B^\circ = 450 \Rightarrow p_B^\circ = 600$ mm Hg.
A has the lower vapour pressure (200 < 600 mm Hg), so A is the less volatile component.
Answer: (C) 600 mm Hg, A. | ||||
| 56 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{AB_2}$ gives $\mathrm{A^{2+}}$ and XY gives $\mathrm{Y^-}$. Mixing equal volumes halves each concentration. A precipitate forms when the ionic product $Q = [\mathrm{A^{2+}}][\mathrm{Y^-}]^2$ exceeds $K_{sp} = 5.2\times10^{-7}$.
(A) $[\mathrm{A^{2+}}] = 10^{-2}$, $[\mathrm{Y^-}] = 10^{-2}$: $Q = 10^{-2}\times10^{-4} = 1\times10^{-6} > 5.2\times10^{-7}$. Precipitate.
(B) $0.75\times10^{-4}\times(0.75\times10^{-3})^2 \approx 4.2\times10^{-11}$. No.
(C) $1.0\times10^{-4}\times(0.4\times10^{-3})^2 = 1.6\times10^{-11}$. No.
(D) $1.8\times10^{-3}\times(2.5\times10^{-4})^2 \approx 1.1\times10^{-10}$. No.
Answer: (A). | ||||
| 57 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Lone pairs on the central atom and polarity:
$\mathrm{SO_2}$: 1 lone pair, bent, polar. $\mathrm{NF_3}$, $\mathrm{NH_3}$: 1 lone pair, pyramidal, polar. $\mathrm{SF_4}$: 1 lone pair, see-saw, polar.
$\mathrm{XeF_2}$: 3 lone pairs, but linear, so its dipole moment is zero; it is excluded.
$\mathrm{ClF_3}$: 3 bond pairs + 2 lone pairs, T-shaped, polar.
So the polar molecule with the most lone pairs is $\mathrm{ClF_3}$: 5 electron pairs, $sp^3d$.
Answer: (B) $sp^3d$. | ||||
| 58 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | StatementsEasy | |
A easy statements question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties From F to I: covalent radius (64, 99, 114, 133 pm) and ionic radius of $\mathrm{X^-}$ (133, 184, 196, 220 pm) rise steadily; first ionisation energy (1681, 1256, 1142, 1008 kJ mol$^{-1}$) falls steadily. These are regular.
Electron affinity is irregular: Cl has a larger electron affinity than F, because the incoming electron in fluorine's small, crowded 2p subshell meets strong repulsion (order Cl > F > Br > I).
Answer: (B) B only. | ||||
| 59 | 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 Statement I: the metallic radius of Al is 143 pm and that of Ga is 135 pm. Ga is smaller because its ten 3d electrons shield the nucleus poorly, so the greater effective nuclear charge pulls its electrons in. Statement I is incorrect.
Statement II: the ionic radii are $\mathrm{Al^{3+}}$ 53.5 pm and $\mathrm{Ga^{3+}}$ 62 pm, so $\mathrm{Al^{3+}}$ is smaller. Statement II is correct.
Answer: (D). | ||||
| 60 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | StatementsEasy | |
A easy statements question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Statement I: if the splitting $\Delta_o$ is smaller than the pairing energy P, electrons occupy the $e_g$ level before pairing (high spin); if $\Delta_o > P$ they pair in $t_{2g}$ first (low spin). Correct.
Statement II: $\Delta_t \approx \frac{4}{9}\Delta_o$, which is almost always smaller than P, so tetrahedral complexes are nearly always high spin; low-spin ones are rare. Correct.
Answer: (A). | ||||
| 61 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A monoatomic p-block element is a noble gas; the rarest non-radioactive one (radon is radioactive) is xenon, which also has the lowest ionisation enthalpy of the three. X = F (most electronegative) and Y = O (second). The molecule is $\mathrm{XeOF_4}$.
Xe has 8 valence electrons: four Xe–F bonds and one Xe=O bond use 6, leaving 1 lone pair. Six electron domains ($sp^3d^2$, octahedral arrangement) with one lone pair give a square pyramidal shape.
Answer: (C) Square pyramidal. | ||||
| 62 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | StatementsMedium | |
A medium statements question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter (A) $\mathrm{Cu^{2+}}$ with potassium ferrocyanide in acetic acid gives chocolate-brown $\mathrm{Cu_2[Fe(CN)_6]}$. Correct.
(B) $\mathrm{Fe^{3+}}$ gives Prussian blue, $\mathrm{Fe_4[Fe(CN)_6]_3}$. Correct.
(C) $\mathrm{Zn^{2+}}$ gives a white or bluish-white precipitate of zinc ferrocyanide; this is the confirmatory test for zinc. Correct.
(D) $\mathrm{Mg^{2+}}$ gives no blue precipitate with ferrocyanide (magnesium is confirmed with disodium hydrogen phosphate or magneson). Incorrect.
(E) $\mathrm{Ba^{2+}}$ is not identified by a ferrocyanide precipitate (it is confirmed as yellow $\mathrm{BaCrO_4}$ or by its green flame). Incorrect.
Answer: (A) A, B and C only. | ||||
| 63 | Some Basic Concepts of ChemistryEmpirical and molecular formula | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Mass of C $= 1.46\times\frac{12}{44} = 0.398$ g, i.e. $0.0332$ mol.
Mass of H $= 0.567\times\frac{2}{18} = 0.063$ g, i.e. $0.063$ mol.
Mass of O (by difference) $= 1.0 - 0.398 - 0.063 = 0.539$ g, i.e. $0.0337$ mol.
C : H : O $= 0.0332 : 0.063 : 0.0337 \approx 1 : 2 : 1$, so the empirical formula is $\mathrm{CH_2O}$.
Empirical formula mass $= 12 + 2 + 16 = 30$ g.
Answer: (B) 30. | ||||
| 64 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques X is $\mathrm{H{-}C{\equiv}C{-}CH_2{-}CH(CH_3){-}CH_3}$; I, II, III, IV label the carbons whose C–H bond is broken.
I: the alkynyl radical $\mathrm{HC{\equiv}C^\bullet}$ type, with the odd electron in an sp orbital, held close to the nucleus and not delocalised: the least stable.
II: a propargylic radical, $\mathrm{HC{\equiv}C{-}\dot{C}H{-}}$, stabilised by resonance with the triple bond ($\mathrm{HC^\bullet{=}C{=}CH{-}}$): the most stable.
III: a tertiary alkyl radical, stabilised only by hyperconjugation, less than the resonance-stabilised II.
IV: a primary alkyl radical, less stable than III but still more stable than the sp radical I.
Most stable II, least stable I.
Answer: (A) II, I. | ||||
| 65 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Hückel's rule: a planar, cyclic, fully conjugated system with $(4n + 2)$ π electrons is aromatic.
(a) Cyclopentadienyl anion: 6 π electrons. Aromatic.
(b) Cyclopentadienyl cation: 4 π electrons. Antiaromatic, not aromatic.
(c) Cyclobutadiene dication: 2 π electrons ($n = 0$). Aromatic.
(d) Cyclobutadiene dianion: 6 π electrons. Aromatic.
(e) Cycloheptatrienyl (tropylium) cation: 6 π electrons. Aromatic.
(f) Cyclooctatetraene: 8 π electrons; it is tub-shaped, not planar. Not aromatic.
(g) Cyclobutadiene: 4 π electrons. Antiaromatic, not aromatic.
(h) Cyclopropenyl cation: 2 π electrons. Aromatic.
Aromatic: a, c, d, e, h; not aromatic: b, f, g.
Answer: (B). | ||||
| 66 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons [A]: the only chiral $\mathrm{C_4H_9Br}$ is 2-bromobutane, $\mathrm{CH_3CHBrCH_2CH_3}$.
[B]: alcoholic KOH eliminates HBr; by Saytzeff's rule the major alkene is but-2-ene, $\mathrm{CH_3CH{=}CHCH_3}$.
[C]: $\mathrm{Br_2}$ adds to give 2,3-dibromobutane.
[D]: alcoholic $\mathrm{NaNH_2}$ removes two HBr: but-2-yne, $\mathrm{CH_3C{\equiv}CCH_3}$ (a gas at the reaction temperature; it boils near 27 °C).
[E]: with $\mathrm{HgSO_4}$/dilute $\mathrm{H_2SO_4}$ one mole of water (18 g) adds: the enol $\mathrm{CH_3C(OH){=}CHCH_3}$ tautomerises to $\mathrm{CH_3COCH_2CH_3}$, butan-2-one.
Answer: (A) Butan-2-one. | ||||
| 67 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Acid derivatives are hydrolysed by nucleophilic acyl substitution (addition of water to C=O, then loss of the leaving group). The rate rises with the electrophilic character of the carbonyl carbon and with the ease of losing the leaving group.
Leaving group: $\mathrm{Cl^-}$ (best) > carboxylate $\mathrm{CH_3COO^-}$ > ethoxide > $\mathrm{NH_2^-}$ (worst). Resonance donation to C=O, which lowers reactivity, grows in the same direction: Cl < OCOR < OR < NH$_2$.
Order: acid chloride (p) > anhydride (q) > ester (r) > amide (s).
Answer: (D) p > q > r > s. | ||||
| 68 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | StatementsEasy | |
A easy statements question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Vanillin is 4-hydroxy-3-methoxybenzaldehyde.
Statement I: its phenolic OH is acidic and reacts with NaOH to give the phenoxide; its –CHO group is oxidised by Tollens' reagent (silver mirror). Correct.
Statement II: aldol condensation needs an α-hydrogen. The CHO carbon of vanillin is bonded directly to the ring and has no α-H, so it cannot undergo self aldol condensation. Incorrect.
Answer: (C). | ||||
| 69 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines In water the basicity of an amine depends on the +I effect of the alkyl groups, hydrogen-bond solvation of its conjugate acid, and steric crowding.
For ethyl groups the result is $\mathrm{(C_2H_5)_2NH}$ (p$K_b$ 3.0) > $\mathrm{(C_2H_5)_3N}$ (3.25) > $\mathrm{C_2H_5NH_2}$ (3.29) > $\mathrm{NH_3}$ (4.75).
Hydrazine is weaker than ammonia (p$K_b$ about 6): the electron-withdrawing –NH$_2$ on nitrogen lowers its basicity.
Order: $\mathrm{H_2N{-}NH_2 < NH_3 < C_2H_5NH_2 < (C_2H_5)_3N < (C_2H_5)_2NH}$.
(Option C is the order for methyl amines, where the tertiary amine falls below the primary.)
Answer: (A). | ||||
| 70 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules (A) False: isoleucine and threonine have two chiral centres.
(B) False: glycine ($\mathrm{H_2NCH_2COOH}$) has no chiral centre and is optically inactive.
(C) True: the free –SH groups of two cysteine molecules are easily oxidised to a disulfide bond (–S–S–), giving the dimer cystine.
(D) False: aspartic acid also has a –COOH group in its side chain.
Answer: (C). | ||||
| 71 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalHard | |
A hard numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Anode (left): $\mathrm{QH_2 \rightarrow Q + 2H^+ + 2e^-}$. Its reduction potential from the Nernst equation:
$E_{\mathrm{Q/QH_2}} = E^\circ - \frac{0.06}{2}\log\frac{1}{[\mathrm{H^+}]^2} = 0.7 - 0.06\,\text{pH}$.
Cathode: $E_{\mathrm{Ag^+/Ag}} = 0.8$ V (with $[\mathrm{Ag^+}] = 1$ M).
$E_{cell} = 0.8 - (0.7 - 0.06\,\text{pH}) = 0.1 + 0.06\,\text{pH} = 0.4$, so pH $= 5$.
The $\mathrm{NH_4X}$ solution is acidic by hydrolysis of $\mathrm{NH_4^+}$ (salt of a weak base and a strong acid): pH $= 7 - \frac{1}{2}\text{p}K_b - \frac{1}{2}\log C$.
$5 = 7 - \frac{1}{2}\text{p}K_b - \frac{1}{2}\log(0.01) = 7 - \frac{1}{2}\text{p}K_b + 1$, so $\text{p}K_b = 6$.
Answer: 6. | ||||
| 72 | 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 The half-life is directly proportional to $[A]_0$ (a straight line through the origin), which is true of a zero-order reaction: $t_{1/2} = \frac{[A]_0}{2k}$.
Slope $= \frac{1}{2k} = 76.92$ min L mol$^{-1}$, so $k = \frac{1}{153.84} = 6.5\times10^{-3}$ mol L$^{-1}$ min$^{-1}$.
Zero order: $[A] = [A]_0 - kt = 2.5 - 6.5\times10^{-3}\times10 = 2.5 - 0.065 = 2.435$ mol L$^{-1}$ $= 2435\times10^{-3}$ mol L$^{-1}$.
Answer: 2435. | ||||
| 73 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalHard | |
A hard numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Take 5 bar as the total pressure at equilibrium. Total moles then $= \frac{pV}{RT} = \frac{5\times2}{0.08\times500} = 0.25$ mol.
At equilibrium: $\mathrm{CH_3OH} = 0.04$ mol, CO $= 0.1 - 0.04 = 0.06$ mol, so $\mathrm{H_2} = 0.25 - 0.04 - 0.06 = 0.15$ mol.
Each partial pressure is $n\frac{RT}{V} = 20n$ bar: $p_{\mathrm{CH_3OH}} = 0.8$, $p_{\mathrm{CO}} = 1.2$, $p_{\mathrm{H_2}} = 3.0$ bar.
$K_p^\theta = \frac{p_{\mathrm{CH_3OH}}}{p_{\mathrm{CO}}\,p_{\mathrm{H_2}}^2} = \frac{0.8}{1.2\times9} = 0.0741 = 74.1\times10^{-3}$ (standard pressure 1 bar).
Nearest integer: 74.
Answer: 74. | ||||
| 74 | 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 $E^\circ(\mathrm{M^{3+}/M^{2+}})$: Cr $-0.41$ V, Fe $+0.77$ V, Mn $+1.57$ V, Co $+1.97$ V. The highest is cobalt.
In $\mathrm{[Co(CN)_6]^{4-}}$: $x + 6(-1) = -4$, so Co is +2, $d^7$.
$\mathrm{CN^-}$ is a strong-field ligand, so the octahedral complex is low spin: $t_{2g}^6e_g^1$.
Electrons in $e_g$: 1.
Answer: 1. | ||||
| 75 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Count the atoms of P (a 5-iodouracil base joined through N to a 2-fluoro sugar ring).
Base: ring C4, N2, two C=O oxygens; H on C6 and on N3; I on C5.
Sugar: ring C1'–C4' and ring O; C5' as $\mathrm{CH_2OH}$; H on C1', C2', C3', C4'; F on C2'; OH on C3'.
C: 4 + 5 = 9. H: 2 (base) + 4 (ring CH) + 2 ($\mathrm{CH_2}$) + 2 (two OH) = 10. N: 2. O: 2 + 1 + 2 = 5. F: 1. I: 1.
Formula $\mathrm{C_9H_{10}FIN_2O_5}$: $M = 108 + 10 + 19 + 127 + 28 + 80 = 372$ g mol$^{-1}$.
0.1 mol weighs 37.2 g $= 372\times10^{-1}$ g.
Answer: 372. | ||||
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