JEE (Main) · 2024
JEE (Main) 2024 — 4 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
30 Q
- Integral Calculus4
- Sets, Relations and Functions2
- Matrices and Determinants2
10 easy · 19 medium · 1 hard
Physics
30 Q
- Kinematics2
- Gravitation2
- Properties of Bulk Matter2
24 easy · 6 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Chemical Bonding and Molecular Structure3
- Some Basic Concepts of Chemistry2
18 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 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: $(x, y)\,R\,(x, y)$ since $x \le x$. True.
Not symmetric: $(1, 1)\,R\,(2, 2)$ because $1 \le 2$, but $(2, 2)\,R\,(1, 1)$ fails since $2 \not\le 1$ and $2 \not\le 1$. So (I) is true.
Transitive? Take $(2, 4)\,R\,(3, 1)$ (as $2 \le 3$) and $(3, 1)\,R\,(1, 2)$ (as $1 \le 2$). But $(2, 4)\,R\,(1, 2)$ needs $2 \le 1$ or $4 \le 2$, both false. So R is not transitive and (II) is false.
Only (I) is correct.
Answer: (A). | ||||
| 2 | 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 Write $z = x + iy$. Then $z + \bar{z} = 2x$ and $i(z - \bar{z}) = i(2iy) = -2y$, so the second condition is $2x - 2y \le 2$, i.e. $y \ge x - 1$.
$|z - 1| \le 2$ is the disc with centre $(1, 0)$ and radius 2, and $\mathrm{Im}(z) \ge 0$ keeps its upper half.
The line $y = x - 1$ passes through the centre $(1, 0)$ at $45^\circ$. Measured from the centre, $y \ge 0$ keeps the directions $0^\circ$ to $180^\circ$, and $y \ge x - 1$ keeps the directions $45^\circ$ to $225^\circ$.
Their common part is the sector from $45^\circ$ to $180^\circ$, an angle of $135^\circ = \frac{3\pi}{4}$.
Area $= \frac{1}{2}r^2\theta = \frac{1}{2}\cdot 4\cdot\frac{3\pi}{4} = \frac{3\pi}{2}$.
Answer: (A). | ||||
| 3 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\mathrm{adj}\,A = \begin{bmatrix}1&-2\\0&1\end{bmatrix}$. By induction, $(\mathrm{adj}\,A)^k = \begin{bmatrix}1&-2k\\0&1\end{bmatrix}$.
$B = \sum_{k=0}^{10}\begin{bmatrix}1&-2k\\0&1\end{bmatrix} = \begin{bmatrix}11&-2(55)\\0&11\end{bmatrix} = \begin{bmatrix}11&-110\\0&11\end{bmatrix}$.
Sum of the elements $= 11 - 110 + 0 + 11 = -88$.
Answer: (D). | ||||
| 4 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $2\,{}^nC_5 = {}^nC_4 + {}^nC_6$. Dividing by ${}^nC_5$: $2 = \frac{5}{n-4} + \frac{n-5}{6}$.
Multiply by $6(n-4)$: $12(n-4) = 30 + (n-5)(n-4)$, i.e. $12n - 48 = 30 + n^2 - 9n + 20$, so $n^2 - 21n + 98 = 0$.
$(n - 7)(n - 14) = 0$, so $n = 7$ or $14$. The maximum is 14.
Answer: (B). | ||||
| 5 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series With $n = 100$, the numerator is $\sum k(k+1)^2 = \sum (k^3 + 2k^2 + k)$ and the denominator is $\sum k^2(k+1) = \sum (k^3 + k^2)$.
Using $\sum k = \frac{n(n+1)}{2}$, $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$, $\sum k^3 = \frac{n^2(n+1)^2}{4}$:
Numerator $= \frac{n(n+1)}{12}\left[3n(n+1) + 4(2n+1) + 6\right] = \frac{n(n+1)(3n^2 + 11n + 10)}{12} = \frac{n(n+1)(n+2)(3n+5)}{12}$.
Denominator $= \frac{n(n+1)}{12}\left[3n(n+1) + 2(2n+1)\right] = \frac{n(n+1)(3n^2 + 7n + 2)}{12} = \frac{n(n+1)(n+2)(3n+1)}{12}$.
Ratio $= \frac{3n+5}{3n+1} = \frac{305}{301}$.
Answer: (B). | ||||
| 6 | Sequences and SeriesArithmetic and geometric means | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic and geometric means, from Sequences and Series. NCERT: Class 11: Sequences and Series The mean of an A.P. of three terms is the middle term, so $b = 8$ and $a + c = 16$.
G.P.: $b^2 = (a+1)(c+3)$, so $64 = (a+1)(19-a)$, i.e. $a^2 - 18a + 45 = 0$, $a = 15$ or $3$.
Since $a > 10$, $a = 15$ and $c = 1$.
Cube of the geometric mean $= abc = 15\times 8\times 1 = 120$.
Answer: (C). | ||||
| 7 | 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 domain is $[2, 4]$. Put $x - 2 = 2\sin^2\theta$, so $4 - x = 2\cos^2\theta$, with $\theta \in [0, \frac{\pi}{2}]$.
$f = \sqrt{2}\,(3\sin\theta + \cos\theta)$.
Maximum: $3\sin\theta + \cos\theta \le \sqrt{10}$, reached inside the range, so $\beta = \sqrt{20}$ and $\beta^2 = 20$.
Minimum: $3\sin\theta + \cos\theta$ rises then falls on $[0, \frac{\pi}{2}]$, so its least value is at an end: $1$ at $\theta = 0$, $3$ at $\theta = \frac{\pi}{2}$. Hence $\alpha = \sqrt{2}$, $\alpha^2 = 2$.
$\alpha^2 + 2\beta^2 = 2 + 40 = 42$.
Answer: (D). | ||||
| 8 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Numerator: $72^x - 9^x - 8^x + 1 = (9^x - 1)(8^x - 1) \approx (x\ln 9)(x\ln 8) = 6x^2\ln 2\ln 3$.
Denominator: $\sqrt{2} - \sqrt{1 + \cos x} = \frac{1 - \cos x}{\sqrt{2} + \sqrt{1 + \cos x}} \approx \frac{x^2/2}{2\sqrt{2}} = \frac{x^2}{4\sqrt{2}}$.
$\lim_{x\to 0} f(x) = 6\ln 2\ln 3\cdot 4\sqrt{2} = 24\sqrt{2}\,\ln 2\ln 3$.
Continuity gives $a = 24\sqrt{2}$, so $a^2 = 1152$.
Answer: (C). | ||||
| 9 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The limit is of the form $\frac{0}{0}$; by L'Hospital's rule it equals $\lim \frac{f'(x)}{3x^2} = \lim\frac{x + \sin(1 - e^x)}{3x^2}$.
Near 0: $1 - e^x = -x - \frac{x^2}{2} + O(x^3)$, and $\sin u = u + O(u^3)$, so $\sin(1 - e^x) = -x - \frac{x^2}{2} + O(x^3)$.
Then $x + \sin(1 - e^x) = -\frac{x^2}{2} + O(x^3)$, and the limit is $\frac{-1/2}{3} = -\frac{1}{6}$.
Answer: (D). | ||||
| 10 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replace $x$ by $-x$: $I = \int_{-1}^{1}\frac{3^x\cos\alpha x}{1 + 3^x}dx$. Adding the two forms, $2I = \int_{-1}^{1}\cos\alpha x\,dx = \frac{2\sin\alpha}{\alpha}$, so $I = \frac{\sin\alpha}{\alpha}$.
$\frac{\sin\alpha}{\alpha} = \frac{2}{\pi}$ holds for $\alpha = \frac{\pi}{2}$.
Check the others: $\frac{\pi}{4}$ gives $\frac{2\sqrt{2}}{\pi}$, $\frac{\pi}{3}$ gives $\frac{3\sqrt{3}}{2\pi}$, $\frac{\pi}{6}$ gives $\frac{3}{\pi}$.
Answer: (C). | ||||
| 11 | 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 In terms of $y$: the region lies between $x = \frac{y^2}{2}$ (on the left) and $x = \frac{y+1}{4}$ (on the right).
They meet where $\frac{y^2}{2} = \frac{y+1}{4}$, i.e. $2y^2 - y - 1 = 0$, $y = -\frac{1}{2}$ or $1$.
Area $= \int_{-1/2}^{1}\left(\frac{y+1}{4} - \frac{y^2}{2}\right)dy = \left[\frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6}\right]_{-1/2}^{1}$
$= \left(\frac{1}{8} + \frac{1}{4} - \frac{1}{6}\right) - \left(\frac{1}{32} - \frac{1}{8} + \frac{1}{48}\right) = \frac{5}{24} + \frac{7}{96} = \frac{27}{96} = \frac{9}{32}$.
Answer: (A). | ||||
| 12 | 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 $2x^3y + 8xy = 2xy(x^2 + 4)$, so the equation is $\frac{dy}{dx} + \frac{2x}{x^2 + 4}y = \frac{2}{(x^2 + 4)^2}$.
Integrating factor: $e^{\int\frac{2x}{x^2+4}dx} = x^2 + 4$.
$y(x^2 + 4) = \int\frac{2}{x^2 + 4}dx = \tan^{-1}\frac{x}{2} + C$; $y(0) = 0$ gives $C = 0$.
$y(2) = \frac{\tan^{-1}1}{8} = \frac{\pi}{32}$.
Answer: (A). | ||||
| 13 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections PQ lies on $x + y = 2$, at distance $\frac{2}{\sqrt{2}} = \sqrt{2}$ from the centre.
A chord of length 2 has half-length 1, so its distance from the centre is $\sqrt{10 - 1} = 3$. MN has slope $-1$, so it lies on $x + y = \pm 3\sqrt{2}$.
Both chords are perpendicular to the same direction $(1, 1)$. If MN is on the same side as PQ, the distance between them is $3 - \sqrt{2}$; on the other side it is $3 + \sqrt{2}$.
Only $3 - \sqrt{2}$ is among the options.
Answer: (B). | ||||
| 14 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Let H be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with foci $(\pm c, 0)$.
The nearest points of H to the origin are its vertices, so $C_1$ has radius $a$: $\pi a^2 = 36\pi$, $a = 6$.
$C_2$ is centred at the focus $(c, 0)$ and passes through the nearer vertex $(a, 0)$, so its radius is $c - a$: $\pi(c - a)^2 = 4\pi$, $c = 8$.
$b^2 = c^2 - a^2 = 64 - 36 = 28$.
Latus rectum $= \frac{2b^2}{a} = \frac{56}{6} = \frac{28}{3}$.
Answer: (A). | ||||
| 15 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Chord with a given midpoint: $T = S_1$. For $y^2 = 12x$ and the point $(4, 1)$: $y\cdot 1 - 6(x + 4) = 1^2 - 12\cdot 4$.
$y - 6x - 24 = -47$, so the line is $y = 6x - 23$.
At $x = \frac{1}{2}$: $y = 3 - 23 = -20$. The point $\left(\frac{1}{2}, -20\right)$ lies on it. (The others give $y = -5$ at $x = 3$, $-11$ at $x = 2$, $-14$ at $x = \frac{3}{2}$.)
Answer: (D). | ||||
| 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 Points of the lines: $(2 + t, 4 + 5t, 2 + t)$ and $(3 + 2s, 2 + 3s, 3 + 2s)$.
$x$ (and $z$): $2 + t = 3 + 2s$; $y$: $4 + 5t = 2 + 3s$. Then $t = 1 + 2s$, $9 + 10s = 2 + 3s$, $s = -1$, $t = -1$. So $P = (1, -1, 1)$.
$4x = 2y = z = \lambda$ gives $(x, y, z) = \lambda\left(\frac{1}{4}, \frac{1}{2}, 1\right)$: the line through the origin with direction $\vec{d} = (1, 2, 4)$.
Distance $= \frac{|\vec{OP}\times\vec{d}|}{|\vec{d}|}$. $\vec{OP}\times\vec{d} = (1, -1, 1)\times(1, 2, 4) = (-6, -3, 3)$, of length $\sqrt{54}$; $|\vec{d}| = \sqrt{21}$.
Distance $= \sqrt{\frac{54}{21}} = \sqrt{\frac{18}{7}} = \frac{3\sqrt{14}}{7}$.
Answer: (A). | ||||
| 17 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{b} + \vec{c} = (x + 2)\hat{i} + 6\hat{j} - 2\hat{k}$, and $\vec{a}\cdot(\vec{b} + \vec{c}) = x + 6$.
$\vec{a}\cdot\vec{d} = 1$ means $x + 6 = |\vec{b} + \vec{c}| = \sqrt{(x + 2)^2 + 40}$. Squaring: $x^2 + 12x + 36 = x^2 + 4x + 44$, so $x = 1$ (and $x + 6 > 0$ as needed).
$(\vec{a}\times\vec{b})\cdot\vec{c} = \begin{vmatrix}1&1&1\\2&4&-5\\1&2&3\end{vmatrix} = 1(12 + 10) - 1(6 + 5) + 1(4 - 4) = 11$.
Answer: (D). | ||||
| 18 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $(\vec{a} + \vec{b})\cdot(\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 = 0$, so $1 + \lambda^2 + 9 = 14$ and $\lambda = 2$.
$\vec{a}\cdot\vec{b} = 3 - 2 - 6 = -5$, and $|\vec{a}||\vec{b}| = 14$, so $\cos\theta = -\frac{5}{14}$.
$(14\cos\theta)^2 = 25$.
Answer: (A). | ||||
| 19 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Total probability: $a + 2a + (a + b) + 2b + 3b = 4a + 6b = 1$.
Mean: $0 + 4a + 4(a + b) + 12b + 24b = 8a + 40b = \frac{46}{9}$.
From the first, $8a = 2 - 12b$; so $2 + 28b = \frac{46}{9}$, $b = \frac{1}{9}$, and $a = \frac{1 - 6/9}{4} = \frac{1}{12}$.
$E(X^2) = 4(2a) + 16(a + b) + 36(2b) + 64(3b) = 24a + 280b = 2 + \frac{280}{9} = \frac{298}{9}$.
Variance $= \frac{298}{9} - \left(\frac{46}{9}\right)^2 = \frac{2682 - 2116}{81} = \frac{566}{81}$.
Answer: (B). | ||||
| 20 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $A = \cos^{-1}x$ and $B = \sin^{-1}y$, so $x = \cos A$, $y = \sin B$ and $\alpha = A - B$.
$\sin\alpha = \sin A\cos B - \cos A\sin B$, so
$x^2 + y^2 + 2xy\sin\alpha = \cos^2A + \sin^2B - 2\cos^2A\sin^2B + 2\sin A\cos A\sin B\cos B$.
Since $\cos^2A + \sin^2B - 2\cos^2A\sin^2B = \cos^2A\cos^2B + \sin^2A\sin^2B$, the expression is $(\cos A\cos B + \sin A\sin B)^2 = \cos^2(A - B) = \cos^2\alpha$.
Its least value is 0, at $\alpha = \frac{\pi}{2}$ (for example $x = 0$, $y = 0$).
Answer: (C). | ||||
| 21 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Suppose $g(x) = \frac{Ax}{\sqrt{1 + Bx^2}}$. Then
$f(g(x)) = \frac{2g}{\sqrt{1 + 9g^2}} = \frac{2Ax/\sqrt{1 + Bx^2}}{\sqrt{\frac{1 + Bx^2 + 9A^2x^2}{1 + Bx^2}}} = \frac{2Ax}{\sqrt{1 + (B + 9A^2)x^2}}$.
So each further $f$ doubles $A$ and adds $9A^2$ to $B$. Starting from $A_1 = 2$, $B_1 = 9$: $A_k = 2^k$ and
$B_{10} = 9(1 + 4 + 4^2 + \cdots + 4^9) = 9\cdot\frac{4^{10} - 1}{3} = 3(4^{10} - 1)$.
$9\alpha = B_{10}$ gives $\alpha = \frac{4^{10} - 1}{3}$, so $3\alpha + 1 = 4^{10}$ and $\sqrt{3\alpha + 1} = 2^{10} = 1024$.
Answer: 1024. | ||||
| 22 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $t = \sin^2 2\theta \in [0, 1]$. Then $\sin^4\theta + \cos^4\theta = 1 - \frac{t}{2}$ and $\sin^6\theta + \cos^6\theta = 1 - \frac{3t}{4}$.
Real roots need $\sin^2 2\theta \ge 4\left(1 - \frac{t}{2}\right)\left(1 - \frac{3t}{4}\right)$, i.e. $t \ge 4 - 5t + \frac{3t^2}{2}$, i.e. $3t^2 - 12t + 8 \le 0$.
The roots of $3t^2 - 12t + 8 = 0$ are $t = 2 \pm \frac{2}{\sqrt{3}}$, so $t \in \left[2 - \frac{2}{\sqrt{3}}, 2 + \frac{2}{\sqrt{3}}\right]$. With $t \le 1$: $S = \left[2 - \frac{2}{\sqrt{3}}, 1\right]$.
$\alpha = 2 - \frac{2}{\sqrt{3}}$, $\beta = 1$: $3\left(\frac{4}{3} + 0\right) = 4$.
Answer: 4. | ||||
| 23 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let $A = \begin{bmatrix}a&b\\b&d\end{bmatrix}$. Then $a + b = 3$, $b + d = 7$ and $ad - b^2 = 1$.
$(3 - b)(7 - b) - b^2 = 21 - 10b = 1$, so $b = 2$, $a = 1$, $d = 5$: $A = \begin{bmatrix}1&2\\2&5\end{bmatrix}$.
By the Cayley–Hamilton theorem, $A^2 - (\mathrm{tr}\,A)A + (\det A)I = 0$: $A^2 - 6A + I = 0$. Multiplying by $A^{-1}$: $A^{-1} = 6I - A$.
$\alpha = -1$, $\beta = 6$, $\alpha + \beta = 5$.
Answer: 5. | ||||
| 24 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Let $k$ men (and $4 - k$ women) come from Group A. Then Group B must give $4 - k$ men and $k$ women.
Ways $= \sum_{k=0}^{4}\left[{}^4C_k\,{}^5C_{4-k}\right]\left[{}^5C_{4-k}\,{}^4C_k\right] = \sum_{k=0}^{4}\left({}^4C_k\,{}^5C_{4-k}\right)^2$.
$k = 0$: $(1\cdot 5)^2 = 25$; $k = 1$: $(4\cdot 10)^2 = 1600$; $k = 2$: $(6\cdot 10)^2 = 3600$; $k = 3$: $(4\cdot 5)^2 = 400$; $k = 4$: $(1\cdot 1)^2 = 1$.
Total $= 25 + 1600 + 3600 + 400 + 1 = 5626$.
Answer: 5626. | ||||
| 25 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | NumericalHard | |
A hard numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Note that $\frac{d}{dx}(ff') = f'^2 + ff''$ and $\frac{d^2}{dx^2}(ff') = 2f'f'' + f'f'' + ff''' = 3f'f'' + ff'''$. So we need the zeros of $(ff')''$.
Zeros of $f$: $x = 0$, and by the intermediate value theorem one each in $(1, 2)$, $(2, 3)$, $(3, 4)$ (the sign changes there). So $f$ has at least 4 zeros.
By Rolle's theorem $f'$ has a zero between each pair of consecutive zeros of $f$: at least 3 more zeros. Taking the zeros of $f$ as consecutive, these lie strictly between them, so they are different points. So $ff'$ has at least $4 + 3 = 7$ zeros.
Rolle again: $(ff')'$ has at least 6 zeros and $(ff')''$ at least 5.
So $3f'f'' + ff'''$ has at least 5 zeros.
Answer: 5. | ||||
| 26 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Reduction formula (from integration by parts): $\int\mathrm{cosec}^n x\,dx = -\frac{\mathrm{cosec}^{n-2}x\cot x}{n - 1} + \frac{n - 2}{n - 1}\int\mathrm{cosec}^{n-2}x\,dx$.
$n = 5$: $\int\mathrm{cosec}^5x\,dx = -\frac{1}{4}\mathrm{cosec}^3x\cot x + \frac{3}{4}\int\mathrm{cosec}^3x\,dx$.
$n = 3$: $\int\mathrm{cosec}^3x\,dx = -\frac{1}{2}\mathrm{cosec}\,x\cot x + \frac{1}{2}\log_e\left|\tan\frac{x}{2}\right|$.
Together: $-\frac{1}{4}\cot x\,\mathrm{cosec}\,x\left(\mathrm{cosec}^2x + \frac{3}{2}\right) + \frac{3}{8}\log_e\left|\tan\frac{x}{2}\right| + C$.
$\alpha = -\frac{1}{4}$, $\beta = \frac{3}{8}$, $8(\alpha + \beta) = -2 + 3 = 1$.
Answer: 1. | ||||
| 27 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Put $u = x + y + 2$: $\frac{du}{dx} = 1 + \frac{dy}{dx} = 1 + u^2$, so $\tan^{-1}u = x + C$. At $x = 0$, $u = 0$, so $C = 0$ and $u = \tan x$.
$y = \tan x - x - 2$, and $y' = \sec^2x - 1 \ge 0$, so $y$ increases on $\left[0, \frac{\pi}{3}\right]$.
$\beta = y(0) = -2$ and $\alpha = y\left(\frac{\pi}{3}\right) = \sqrt{3} - \frac{\pi}{3} - 2$.
$3\alpha + \pi = 3\sqrt{3} - 6$, $(3\sqrt{3} - 6)^2 = 27 - 36\sqrt{3} + 36 = 63 - 36\sqrt{3}$; adding $\beta^2 = 4$: $67 - 36\sqrt{3}$.
$\gamma = 67$, $\delta = -36$, $\gamma + \delta = 31$.
Answer: 31. | ||||
| 28 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines The angles are $A = 120^\circ$, $B = 30^\circ$, so $C = 30^\circ$: the triangle is isosceles and the foot M of the perpendicular from A to BC is the midpoint of BC.
For the line $x - y + 4 = 0$: $AM = \frac{|1 - 2 + 4|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$, and $M = (1, 2) - \frac{3}{2}(1, -1) = \left(-\frac{1}{2}, \frac{7}{2}\right)$.
$BM = AM\cot 30^\circ = \frac{3\sqrt{3}}{\sqrt{2}}$. Along the line the unit direction is $\frac{(1, 1)}{\sqrt{2}}$, so B and C are $M \pm \frac{3\sqrt{3}}{2}(1, 1)$.
Their x-coordinates are $-\frac{1}{2} \pm \frac{3\sqrt{3}}{2}$, so $\alpha^2 + \gamma^2 = 2\cdot\frac{1}{4} + 2\cdot\frac{27}{4} = 14$.
Answer: 14. | ||||
| 29 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry L: $(1, 2, 1) + t(1, -1, -2)$.
Foot F of the perpendicular from A: $\vec{PA} = (1, 0, 1)$, $t = \frac{(1, 0, 1)\cdot(1, -1, -2)}{6} = -\frac{1}{6}$, so $F = \left(\frac{5}{6}, \frac{13}{6}, \frac{4}{3}\right)$.
Image $= 2F - A = \left(-\frac{1}{3}, \frac{7}{3}, \frac{2}{3}\right)$.
$\alpha + \beta + 6\gamma = -\frac{1}{3} + \frac{7}{3} + 4 = 6$.
Answer: 6. | ||||
| 30 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $x + y = 10$, so $|x - y| = |2x - 10| \le 2$ means $x = 4, 5$ or $6$. With the matches independent, $x$ is binomial with $n = 10$, $p = \frac{1}{3}$.
$p = \frac{{}^{10}C_4\,2^6 + {}^{10}C_5\,2^5 + {}^{10}C_6\,2^4}{3^{10}} = \frac{210\cdot 64 + 252\cdot 32 + 210\cdot 16}{3^{10}} = \frac{13440 + 8064 + 3360}{3^{10}} = \frac{24864}{3^{10}}$.
$3^9p = \frac{24864}{3} = 8288$.
Answer: 8288. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Experimental SkillsSemiconductor experiments | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on semiconductor experiments, from Experimental Skills. NCERT: No matching NCERT chapter The dynamic resistance $r_d = \frac{\Delta V}{\Delta I}$ is measured on the forward characteristic, so the diode must be forward biased: its p side (the arrow's tail) connected towards the positive terminal of the cell, with a series resistor to limit the current.
(A) The long (positive) plate of the cell is grounded, so the top wire is at $-5$ V: the diode's anode is at the lower potential. Reverse biased.
(B) The long plate is at the top, so current flows from $+5$ V through R into the anode of $D_2$ and out to ground. Forward biased, with a current-limiting resistor. Correct.
(C) The source is drawn with two equal plates, so no polarity (it reads as a capacitor); it cannot set a steady forward bias.
(D) The cell is drawn with plates of nearly equal length, so the drawing sets no clear bias; it is not the standard forward-bias arrangement with the diode's p side towards the positive terminal.
Answer: (B). | ||||
| 32 | 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 $[G] = [\mathrm{M^{-1}L^3T^{-2}}]$, so $[GM] = [\mathrm{L^3T^{-2}}]$.
(A) $\frac{r^3}{GM}$ has dimensions $\frac{\mathrm{L^3}}{\mathrm{L^3T^{-2}}} = [\mathrm{T^2}]$, matching $T^2$.
(B) $\frac{r^2}{GM}$ gives $[\mathrm{L^{-1}T^2}]$; (C) $r^3$ gives $[\mathrm{L^3}]$; (D) $\frac{r}{GM^2}$ gives $[\mathrm{M^{-1}L^{-2}T^2}]$. None is $[\mathrm{T^2}]$.
Answer: (A). | ||||
| 33 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane In the figure P is at the top and S at the left end of the horizontal diameter, so OP and OS are perpendicular radii of 2 km.
Displacement is the straight line PS, whatever the path: $PS = \sqrt{2^2 + 2^2} = \sqrt{8}$ km.
Answer: (C). | ||||
| 34 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Sliding just begins when the angle of the incline equals the angle of repose: $\mu_s = \tan\theta = \tan 45^\circ = 1$ (independent of the mass).
Answer: (A). | ||||
| 35 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At height $2R$ the distance from the centre is $3R$, so $g' = g\left(\frac{R}{3R}\right)^2 = \frac{g}{9}$.
Force $= mg' = 90\times\frac{10}{9} = 100$ N.
Answer: (D). | ||||
| 36 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power From the figure, A is on the circle with its radius at $45^\circ$ above the horizontal, so A is $R\sin 45^\circ$ above the centre; B is the lowest point, $R$ below the centre.
Height fallen: $h = R + R\sin 45^\circ = 14 + \frac{14}{\sqrt{2}} = 14 + 7(1.4) = 23.8$ m.
With no friction: $v = \sqrt{2gh} = \sqrt{2\times 10\times 23.8} = \sqrt{476} \approx 21.8$ m/s, which is option (A), 21.9 m/s (the nearest).
Answer: (A). | ||||
| 37 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation For an orbit of radius $r = R + h$: $\frac{GMm}{r^2} = m\left(\frac{2\pi}{T}\right)^2r$, so $r^3 = \frac{GMT^2}{4\pi^2}$.
With $GM = gR^2$: $R + h = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3}$, so $h = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R$.
Answer: (D). | ||||
| 38 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | StatementsEasy | |
A easy statements 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 Statement I: the contact angle depends on the pair of materials (the solid and the liquid). True.
Statement II: capillary rise $h = \frac{2T\cos\theta}{r\rho g}$ is inversely proportional to the inner radius $r$. False.
Answer: (C). | ||||
| 39 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $TV^{\gamma - 1} = $ constant, so $T_2 = T\left(\frac{1}{2}\right)^{1/2} = \frac{T}{\sqrt{2}}$.
Work by the gas: $W = \frac{\mu R(T_1 - T_2)}{\gamma - 1} = \frac{RT\left(1 - \frac{1}{\sqrt{2}}\right)}{1/2} = 2RT\left(1 - \frac{1}{\sqrt{2}}\right) = RT\left[2 - \sqrt{2}\right]$.
Answer: (A). | ||||
| 40 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Every molecule has 3 translational degrees of freedom. $\mathrm{CH_4}$ is a non-linear (tetrahedral) polyatomic molecule, so it can rotate about all three axes: 3 rotational degrees of freedom.
Answer: (C). | ||||
| 41 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Put an identical cube on the other side of that face: the two cubes together enclose $q$, with total flux $\frac{q}{\epsilon_0}$. By symmetry each cube gets half.
Flux through the cube $= \frac{q}{2\epsilon_0}$.
Answer: (A). | ||||
| 42 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Resistance of the bulb: $R = \frac{V^2}{P} = \frac{200^2}{50} = 800\ \Omega$ (taken as constant).
At 100 V: $P = \frac{100^2}{800} = 12.5$ W.
Answer: (A). | ||||
| 43 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Potential energy $U = -mB\cos\theta$. Most stable: $\theta = 0$, $U = -mB$; most unstable: $\theta = 180^\circ$, $U = +mB$.
$W = 2mB = 2\times 0.5\times 8\times 10^{-2} = 8\times 10^{-2}$ J.
Answer: (C). | ||||
| 44 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Matching listEasy | |
A easy matching list question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Read the phasor diagrams: I has $I$ at $90^\circ$ ahead of $V$; II has $I$ and $V$ along the same line; III has $V$ ahead of $I$ by an angle $\theta$; IV has $V$ at $90^\circ$ ahead of $I$.
A. Purely capacitive: current leads voltage by $90^\circ$ → I.
B. Purely inductive: voltage leads current by $90^\circ$ → IV.
C. Series LCR at resonance: $X_L = X_C$, current in phase with voltage → II.
D. Series LCR in general: a phase angle $\theta$ between $V$ and $I$ → III.
A-I, B-IV, C-II, D-III.
Answer: (B). | ||||
| 45 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves In the electromagnetic spectrum, wavelength increases from gamma rays to X-rays to ultraviolet, visible, infrared and then microwaves.
So $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$.
Answer: (D). | ||||
| 46 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The intensity from a slit is proportional to its width, so $I_1 : I_2 = 4 : 1$ and the amplitudes are in the ratio $2 : 1$.
$\frac{I_{max}}{I_{min}} = \left(\frac{a_1 + a_2}{a_1 - a_2}\right)^2 = \left(\frac{2 + 1}{2 - 1}\right)^2 = 9$.
Answer: (A). | ||||
| 47 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Assertion and reasonEasy | |
A easy assertion and reason question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter A: the number of photons depends on the intensity (power) of the light. For a given intensity it is $\frac{P}{h\nu}$, which actually falls as the frequency rises. A is not correct.
R: Einstein's equation $K_{max} = h\nu - \phi$: the maximum kinetic energy rises with frequency. R is correct.
Answer: (D). | ||||
| 48 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Bohr's quantum condition: $L = \frac{nh}{2\pi}$. For $n = 4$: $L = \frac{4h}{2\pi} = 2\frac{h}{\pi}$.
Answer: (A). | ||||
| 49 | 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 Each input passes through a NOT gate, giving $\bar{A}$ and $\bar{B}$, which feed an AND gate. The solid dot at the AND gate's output is a junction, not an inverting bubble (the inverting bubbles in the figure are hollow).
$Y = \bar{A}\cdot\bar{B} = \overline{A + B}$ (De Morgan), which is a NOR gate.
Answer: (D). | ||||
| 50 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For a spring–block oscillator the total energy is $E = \frac{1}{2}kA^2$ (equal to $\frac{1}{2}m\omega^2A^2$ with $\omega^2 = \frac{k}{m}$).
It depends on the spring constant and the amplitude only, so doubling the mass at the same amplitude leaves it $E$.
Answer: (D). | ||||
| 51 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Mass of the products: $139.9054 + 93.9063 + 1.0086 = 234.8203$ u.
Mass defect: $\Delta m = 235.0439 - 234.8203 = 0.2236$ u.
$Q = 0.2236\times 931 = 208.17$ MeV, which rounds to 208.
Answer: 208. | ||||
| 52 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Critical angle: $\sin C = \frac{1}{\sqrt{2}}$, $C = 45^\circ$; so the angle of incidence is $i = 45^\circ$.
Refraction into the slab: $\sin r = \frac{\sin 45^\circ}{\sqrt{2}} = \frac{1}{2}$, $r = 30^\circ$.
Lateral displacement: $d = \frac{t\sin(i - r)}{\cos r} = \frac{4\sqrt{3}\times\sin 15^\circ}{\cos 30^\circ} = \frac{4\sqrt{3}\times 0.25}{\sqrt{3}/2} = 2$ cm.
Answer: 2. | ||||
| 53 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Each half of the rod (length $\frac{l}{2} = 0.3$ m) rotating about the centre develops an emf $\frac{1}{2}B\omega\left(\frac{l}{2}\right)^2 = \frac{1}{2}(0.5)(20)(0.09) = 0.45$ V between the centre and its end.
Both ends are therefore at the same potential relative to the centre (each 0.45 V from it, in the same sense), so the potential difference between the two ends is 0.
Answer: 0. | ||||
| 54 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter From the figure both wires carry current in the same direction (upward): $I$ on the left, $2I$ on the right, $2r$ apart. A is $r$ to the left of the first wire and C is $r$ to the right of the second.
At A (both wires on the same side, same current direction) the fields add: $B_A = \frac{\mu_0}{2\pi}\left(\frac{I}{r} + \frac{2I}{3r}\right) = \frac{\mu_0 I}{2\pi r}\cdot\frac{5}{3}$.
At C the fields also add: $B_C = \frac{\mu_0}{2\pi}\left(\frac{2I}{r} + \frac{I}{3r}\right) = \frac{\mu_0 I}{2\pi r}\cdot\frac{7}{3}$.
$\frac{B_A}{B_C} = \frac{5}{7}$, so $x = 5$.
Answer: 5. | ||||
| 55 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Same material and same mass means the same volume $V = Al$, so $l = \frac{V}{A}$ and $R = \frac{\rho l}{A} = \frac{\rho V}{A^2} \propto \frac{1}{r^4}$.
$\frac{R_A}{R_B} = \left(\frac{r_B}{r_A}\right)^4 = 2^4 = 16$, so $R_A = 16\times 2 = 32\ \Omega$.
Answer: 32. | ||||
| 56 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\omega = \frac{2\pi}{T} = \frac{2\times 3.14}{3.14} = 2$ rad/s.
$v = \frac{dx}{dt} = 10\omega\cos\left(\omega t + \frac{\pi}{3}\right)$; at $t = 0$: $v = 10\times 2\times\cos\frac{\pi}{3} = 10$ m/s.
Answer: 10. | ||||
| 57 | 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 The tube is open at the top, so the pressure at the bottom is the atmospheric pressure transmitted by the mercury plus the mercury column's own pressure:
$P = P_0 + \rho gh = 10^5 + 1.36\times 10^4\times 10\times 0.30 = 10^5 + 40800 = 140800$ Pa.
Area $= \pi r^2 = \frac{22}{7}\times(0.02)^2 = 1.257\times 10^{-3}$ m$^2$.
$F = PA = 140800\times\frac{22}{7}\times 4\times 10^{-4} = 177.0$ N.
Answer: 177. | ||||
| 58 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For the centre of mass to stay put, $m_1\Delta x_1 = m_2\Delta x_2$ (both moving towards it).
$3\times 2 = 2\times\Delta x_2$, so $\Delta x_2 = 3$ cm.
Answer: 3. | ||||
| 59 | 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 $u = 72$ km/h $= 20$ m/s, $v = 0$, $t = 4$ s.
With uniform retardation, distance $= \frac{u + v}{2}t = \frac{20 + 0}{2}\times 4 = 40$ m.
Answer: 40. | ||||
| 60 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Initial energy: $U_i = \frac{1}{2}CV^2 = \frac{1}{2}\times 12.5\times 10^{-12}\times 144 = 900\times 10^{-12}$ J.
With the battery disconnected the charge stays the same, and the capacitance becomes $6C$, so $U_f = \frac{Q^2}{2(6C)} = \frac{U_i}{6} = 150\times 10^{-12}$ J.
Change $= 900 - 150 = 750\times 10^{-12}$ J (a decrease).
Answer: 750. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryLaws of chemical combination | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on laws of chemical combination, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Dalton's theory says matter is made of indivisible atoms (A), that atoms of one element are identical in mass and properties (B), and that chemical reactions rearrange atoms (D).
It says compounds form when atoms of different elements combine in a fixed ratio, not in any ratio. So (C) is the incorrect statement.
Answer: (C). | ||||
| 62 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | StatementsEasy | |
A easy statements question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A: true — a hydrogen bond forms when H is bonded to a highly electronegative atom (F, O, N).
B: false — $o$-nitrophenol has intramolecular hydrogen bonding (between the OH and the neighbouring $\mathrm{NO_2}$).
C: false — HF has intermolecular hydrogen bonding (the zig-zag chains).
D: true — hydrogen bonding is strongest in the solid state and weakest in the gaseous state.
E: true — it strongly affects structure and properties such as boiling point and solubility.
A, D, E only.
Answer: (D). | ||||
| 63 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium The required reaction is the given one reversed and doubled, so $K_c' = \left(\frac{1}{K_c}\right)^2$.
$\frac{1}{4.9\times 10^{-2}} = 20.41$; $K_c' = (20.41)^2 = 416.5 \approx 416$.
Answer: (B). | ||||
| 64 | ElectrochemistryBatteries and fuel cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on batteries and fuel cells, from Electrochemistry. NCERT: Class 12: Electrochemistry A: true — $\mathrm{H_2}$–$\mathrm{O_2}$ fuel cells powered the Apollo space programme.
B: false — the efficiency of a fuel cell is about 70%, not 40% (that is closer to a thermal power plant).
C: false — the electrodes use finely divided platinum or palladium as catalyst, not aluminium.
D: true — the only product is water, so it is pollution-free.
E: true — it converts the energy of a redox reaction directly into electrical energy, as a galvanic cell does.
A, D, E only.
Answer: (C). | ||||
| 65 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Kohlrausch: $\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$. The slope $A$ has the units of $\frac{\Lambda_m}{\sqrt{c}}$:
$\frac{\mathrm{S\,cm^2\,mol^{-1}}}{(\mathrm{mol\,L^{-1}})^{1/2}} = \mathrm{S\,cm^2\,mol^{-3/2}\,L^{1/2}}$.
Answer: (D). | ||||
| 66 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsMedium | |
A medium statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties First ionization enthalpies (kJ mol$^{-1}$): Na 496, Li 520, Cl 1251, F 1681. So Na < Li < Cl < F — Statement I is true.
Negative electron gain enthalpies (kJ mol$^{-1}$): Na 53, Li 60, F 328, Cl 349. Cl is above F because the small 2p shell of F repels the incoming electron. So Na < Li < F < Cl — Statement II is true.
Answer: (A). | ||||
| 67 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties First ionization enthalpies (kJ mol$^{-1}$): B 801, Al 577, Ga 579, In 558, Tl 589.
Ga is slightly above Al because the poorly shielding 3d electrons raise its effective nuclear charge; Tl is higher still because of the poor shielding by the 4f and 5d electrons.
So Tl > Ga > Al is correct; B > Al > Ga fails at Al > Ga, and the others are reversed.
Answer: (C). | ||||
| 68 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{S_2O_3^{2-}}$: central S has four bonded atoms (three O, one S) and no lone pair — tetrahedral.
$\mathrm{SO_4^{2-}}$: tetrahedral.
$\mathrm{SO_3^{2-}}$: S has three bonded O and one lone pair — trigonal pyramidal.
$\mathrm{S_2O_7^{2-}}$: each S is bonded to four O with no lone pair — tetrahedral around each S.
Only one species is pyramidal.
Answer: (A). | ||||
| 69 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mu = \sqrt{n(n+2)}$: $n = 3$ gives $\sqrt{15} = 3.87$ BM, so the ion has 3 unpaired electrons.
Among the options: $\mathrm{Mn^{2+}}$ (25) is $d^5$, 5 unpaired; $\mathrm{V^{2+}}$ (23) is $d^3$, 3 unpaired; $\mathrm{Ti^{2+}}$ (22) is $d^2$, 2 unpaired; $\mathrm{Fe^{2+}}$ (26) is $d^6$, 4 unpaired.
The metal is vanadium, $Z = 23$.
Answer: (B). | ||||
| 70 | 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 Co is in the +3 state: $\mathrm{Co^{3+}}$ is $3d^6$.
With the high charge of $\mathrm{Co^{3+}}$, even water gives a crystal field splitting larger than the pairing energy, so the complex is low spin: $t_{2g}^6e_g^0$ (it is diamagnetic).
Unpaired d-electrons: 0.
Answer: (A). | ||||
| 71 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{Fe^{3+}}$ is $d^5$. No electron in $e_g$ means low spin $t_{2g}^5$, which needs strong-field ligands in an octahedral complex: coordination number 6, so $x + y = 6$.
Charge check: $+3 + 0\cdot x - y = -1$ gives $y = 4$, so $x = 2$: $\mathrm{[Fe(NH_3)_2(CN)_4]^-}$.
Answer: (D). | ||||
| 72 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter The greenish yellow gas is chlorine. A chloride heated with $\mathrm{MnO_2}$ and conc. $\mathrm{H_2SO_4}$ gives it:
$\mathrm{MnO_2 + 2Cl^- + 4H^+ \rightarrow Mn^{2+} + Cl_2 + 2H_2O}$.
NaBr would give reddish-brown $\mathrm{Br_2}$, $\mathrm{CaI_2}$ violet $\mathrm{I_2}$ vapour; $\mathrm{KNO_3}$ gives no halogen. So A is $\mathrm{NH_4Cl}$.
Answer: (B). | ||||
| 73 | 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 In adsorption chromatography (column or thin-layer) the stationary phase is an adsorbent, usually silica gel or alumina.
Quick lime is not used. So the most appropriate answer is A and B only.
Answer: (C). | ||||
| 74 | 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 d, the cyclopentadienyl anion, is aromatic ($6\pi$ electrons, planar, cyclic): the most stable.
a, the cyclopropenyl anion, has $4\pi$ electrons in a planar ring: antiaromatic, the least stable.
b and c are ordinary (non-aromatic) cycloalkyl carbanions and lie in between.
Only option (A) puts d first and a last: d > c > b > a.
Answer: (A). | ||||
| 75 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Step A cleaves the ring C=C of 1-methylcyclohexene into a ketone and an aldehyde: reductive ozonolysis ($\mathrm{O_3}$, then $\mathrm{Zn/H_2O}$) gives 6-oxoheptanal, $\mathrm{CH_3CO(CH_2)_4CHO}$.
Step B turns the methyl ketone $\mathrm{-COCH_3}$ into the sodium carboxylate $\mathrm{-COO^-Na^+}$ while the aldehyde stays: the haloform (iodoform) reaction, $\mathrm{NaOH/I_2}$, which removes $\mathrm{CH_3}$ as $\mathrm{CHI_3}$.
$\mathrm{KMnO_4}$ would also oxidise the aldehyde, and $\mathrm{H_2O/H^+}$ only hydrates the alkene.
Answer: (B). | ||||
| 76 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes In 3,5-dibromocyclopentene both C–Br carbons are allylic, so both bromines are easily replaced by nucleophilic substitution.
Dimethylamine substitutes each Br at the carbon that carried it, the double bond staying where it was.
The product is 3,5-bis(dimethylamino)cyclopentene, with the two $\mathrm{NMe_2}$ groups on the carbons flanking the double bond: option (A).
Answer: (A). | ||||
| 77 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate is 2-bromo-3-methyl-1-phenylbutane, $\mathrm{PhCH_2-CHBr-CH(CH_3)_2}$. Alcoholic KOH on heating causes E2 elimination.
A β-hydrogen can be lost from the benzylic $\mathrm{CH_2}$ or from the $\mathrm{CH(CH_3)_2}$.
Loss from the benzylic carbon gives $\mathrm{PhCH=CH-CH(CH_3)_2}$, whose double bond is conjugated with the benzene ring; this conjugation makes it the more stable alkene and the major product (it outweighs the extra alkyl substitution of $\mathrm{PhCH_2CH=C(CH_3)_2}$).
Answer: (A). | ||||
| 78 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Benzene-1,2-diol is catechol; benzene-1,3-diol is resorcinol; benzene-1,4-diol is quinol (hydroquinone); o-cresol is 2-methylphenol.
Answer: (C). | ||||
| 79 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers NaOH in ethanol eliminates HBr from 1-bromopropane: A is propene, $\mathrm{CH_3CH=CH_2}$.
B: acid-catalysed hydration follows Markovnikov's rule, giving 2-propanol.
C: hydroboration–oxidation adds water anti-Markovnikov, giving 1-propanol.
Answer: (A). | ||||
| 80 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules A. Glucose and galactose differ only in the configuration at C-4: epimers → IV.
B. $\alpha$- and $\beta$-glucose differ only at the anomeric carbon C-1: anomers → III.
C. Glucose (an aldohexose) and fructose (a ketohexose) share $\mathrm{C_6H_{12}O_6}$ with different functional groups: functional isomers → I.
D. Ribose ($\mathrm{C_5H_{10}O_5}$) and glucose ($\mathrm{C_6H_{12}O_6}$) are aldoses differing by one $\mathrm{CHOH}$ unit: homologous → II.
A-IV, B-III, C-I, D-II.
Answer: (C). | ||||
| 81 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom For $n = 4$, $l = 0, 1, 2, 3$ (4s, 4p, 4d, 4f). Each subshell has exactly one orbital with $m_l = 0$.
Number of orbitals $= 4$.
Answer: 4. | ||||
| 82 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Zero dipole moment (symmetric): $\mathrm{BeCl_2}$ (linear), $\mathrm{BCl_3}$ (trigonal planar), $\mathrm{XeF_4}$ (square planar), $\mathrm{CCl_4}$ (tetrahedral), $\mathrm{CO_2}$ (linear), $\mathrm{H_2}$.
Non-zero: $\mathrm{NF_3}$ (pyramidal), $\mathrm{H_2O}$ and $\mathrm{H_2S}$ (bent), HBr, HCl.
Count $= 5$.
Answer: 5. | ||||
| 83 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Work done on the gas against the constant external pressure: $w = -P_{ext}\Delta V = -5(20 - 60) = +200$ L atm.
Isothermal ideal gas: $\Delta U = 0$, so $q = -w = -200$ L atm (heat is given out).
The blank follows the printed minus sign, so it is 200.
Answer: 200. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Moles: water $= \frac{2700}{18} = 150$; acetic acid $= \frac{2700}{60} = 45$. Water is in excess (and the stem says acetic acid behaves as a solute in water), so water is the solvent.
Molality of acetic acid $= \frac{45}{2.7} = 16.67$ mol kg$^{-1}$.
$\Delta T_f = K_f m = 1.86\times 16.67 = 31.0$ K, so the freezing point is $-31$ °C.
Answer: 31. | ||||
| 85 | 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 A and B start equal and are used 1 : 1, so $[A] = [B]$ throughout and rate $= k[A]$: first order in A.
$t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{0.046}\log 10 = 50.07$ s, which rounds to 50.
Answer: 50. | ||||
| 86 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Vanadium has the highest enthalpy of atomisation in the first transition series; its oxides are $\mathrm{V_2O_3}$, $\mathrm{V_2O_4}$ and $\mathrm{V_2O_5}$.
Their character runs from basic $\mathrm{V_2O_3}$ through less basic $\mathrm{V_2O_4}$ to amphoteric $\mathrm{V_2O_5}$.
In $\mathrm{V_2O_5}$ vanadium is +5, $3d^0$: no unpaired electrons, so the spin-only moment is 0 BM.
Answer: 0. | ||||
| 87 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{C_6H_5NH_2 \rightarrow C_6H_5NHCOCH_3}$, one mole to one mole. Molar masses: aniline 93, acetanilide 135 g mol$^{-1}$.
Moles of aniline $= \frac{6.55}{93} = 0.07043$; mass of acetanilide $= 0.07043\times 135 = 9.51$ g $= 95.1\times 10^{-1}$ g.
Nearest integer: 95.
Answer: 95. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Structure: $\mathrm{HOOC-CO-CH_2-C\equiv C-CH_3}$ ($\mathrm{C_6H_6O_3}$).
σ bonds: C–C 5, C–H 5 (2 on $\mathrm{CH_2}$, 3 on $\mathrm{CH_3}$), two C=O give 2, C–OH 1, O–H 1: total 14.
π bonds: two C=O give 2, C≡C gives 2: total 4.
σ + π $= 18$.
Answer: 18. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Vanillin is 4-hydroxy-3-methoxybenzaldehyde, $\mathrm{C_8H_8O_3}$: 3 oxygen atoms (OH, $\mathrm{OCH_3}$, CHO).
π electrons: the benzene ring has 3 π bonds (6 electrons), the C=O has 1 (2 electrons): 8.
Sum $= 3 + 8 = 11$.
Answer: 11. | ||||
| 90 | AminesClassification and identification of amines | Class 12: Amines | NumericalEasy | |
A easy numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines KOH forms potassium phthalimide; benzyl chloride then alkylates the nitrogen (first step of the Gabriel synthesis): P is N-benzylphthalimide.
π bonds: the phthalimide benzene ring 3, the two C=O 2, the benzyl ring 3: total 8.
Answer: 8. | ||||
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