JEE (Main) · 2024
JEE (Main) 2024 — 9 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
30 Q
- Integral Calculus3
- Trigonometry2
- Matrices and Determinants2
10 easy · 18 medium · 2 hard
Physics
30 Q
- Magnetic Effects of Current and Magnetism3
- Kinematics2
- Motion of System of Particles and Rigid Body2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques3
- Chemical Bonding and Molecular Structure2
- Solutions2
22 easy · 7 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin^{-1}$ needs $-1 \le \frac{x - 1}{2x + 3} \le 1$ with $x \ne -\frac{3}{2}$, i.e. $|x - 1| \le |2x + 3|$.
Squaring: $(x - 1)^2 \le (2x + 3)^2 \Rightarrow 0 \le 3x^2 + 14x + 8 = (3x + 2)(x + 4)$.
So $x \le -4$ or $x \ge -\frac{2}{3}$; the excluded point $-\frac{3}{2}$ already lies in the gap.
Domain $= \mathbb{R} - \left(-4, -\frac{2}{3}\right)$, so $\alpha = -4$, $\beta = -\frac{2}{3}$.
$12\alpha\beta = 12\times(-4)\times\left(-\frac{2}{3}\right) = 32$.
Answer: (B). | ||||
| 2 | Quadratic EquationsForming equations from given roots | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on forming equations from given roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha + \beta = -2\sqrt{2}$, $\alpha\beta = -1$.
$\alpha^2 + \beta^2 = 8 + 2 = 10$, $\alpha^2\beta^2 = 1$.
$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 = 100 - 2 = 98$.
$\alpha^6 + \beta^6 = (\alpha^2 + \beta^2)^3 - 3\alpha^2\beta^2(\alpha^2 + \beta^2) = 1000 - 30 = 970$, so $\frac{1}{10}(\alpha^6 + \beta^6) = 97$.
Sum of the new roots $= 195$, product $= 98\times 97 = 9506$.
Equation: $x^2 - 195x + 9506 = 0$.
Answer: (C). | ||||
| 3 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For infinitely many solutions the coefficient determinant must vanish:
$\begin{vmatrix}3&5&\lambda\\7&11&-9\\97&155&-189\end{vmatrix} = 3(-2079 + 1395) - 5(-1323 + 873) + \lambda(1085 - 1067) = -2052 + 2250 + 18\lambda = 198 + 18\lambda$.
So $\lambda = -11$.
The third equation must then be a combination $a(\text{first}) + b(\text{second})$ of the other two: $3a + 7b = 97$, $5a + 11b = 155$ give $b = 10$, $a = 9$.
Check on $z$: $9\lambda + 10(-9) = -99 - 90 = -189$. Correct.
Right-hand side: $\mu = 9\times 3 + 10\times 2 = 47$.
$\mu + 2\lambda = 47 - 22 = 25$.
Answer: (C). | ||||
| 4 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctHard | |
A hard single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The general term is $x^k(1 + x)^{100 - k}$, $k = 2, 3, \ldots, 54$. Its coefficient of $x^{70}$ is ${}^{100 - k}C_{70 - k} = {}^{100 - k}C_{30}$.
With $n = 100 - k$ running from 46 to 98, the coefficient is $\sum_{n = 46}^{98} {}^{n}C_{30}$.
By the hockey-stick identity $\sum_{n = 30}^{N} {}^{n}C_{30} = {}^{N + 1}C_{31}$:
$\sum_{n = 46}^{98} {}^{n}C_{30} = {}^{99}C_{31} - {}^{46}C_{31}$.
So $p = 31$ or $68$ and $q = 31$ or $15$; the possible values of $p + q$ are $62, 46, 99, 83$.
Of the options only 83 ($p = 68$, $q = 15$) occurs.
Answer: (B). | ||||
| 5 | 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 The terms $1, 1 + d, \ldots, 1 + 10d$ are in A.P., so each term telescopes:
$\frac{1}{(1 + kd)(1 + (k + 1)d)} = \frac{1}{d}\left(\frac{1}{1 + kd} - \frac{1}{1 + (k + 1)d}\right)$.
Sum $= \frac{1}{d}\left(1 - \frac{1}{1 + 10d}\right) = \frac{10}{1 + 10d}$.
$\frac{10}{1 + 10d} = 5 \Rightarrow 1 + 10d = 2 \Rightarrow d = \frac{1}{10}$.
$50d = 5$.
Answer: (A). | ||||
| 6 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy 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 $f(1) = a + b + c + 41 = 40 \Rightarrow a + b + c = -1$.
$f'(x) = 3ax^2 + 2bx + c$: $3a + 2b + c = 2$.
$f''(x) = 6ax + 2b$: $6a + 2b = 4$, i.e. $3a + b = 2$.
Subtracting the first from the second: $2a + b = 3$. With $3a + b = 2$: $a = -1$, $b = 5$, then $c = -1 - a - b = -5$.
$a^2 + b^2 + c^2 = 1 + 25 + 25 = 51$.
Answer: (A). | ||||
| 7 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines Let L be $\frac{x}{a} + \frac{y}{b} = 1$ with $a, b > 0$. Through (3, 5): $\frac{3}{a} + \frac{5}{b} = 1$.
By A.M. $\ge$ G.M.: $1 = \frac{3}{a} + \frac{5}{b} \ge 2\sqrt{\frac{15}{ab}}$, so $ab \ge 60$.
Area $= \frac{1}{2}ab \ge 30$, with equality when $\frac{3}{a} = \frac{5}{b} = \frac{1}{2}$, i.e. $a = 6$, $b = 10$.
Minimum area $= 30$.
Answer: (D). | ||||
| 8 | Integral CalculusIntegral as an antiderivative | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{2 - \tan x}{3 + \tan x} = \frac{2\cos x - \sin x}{3\cos x + \sin x}$.
Write the numerator as $A(\text{denominator}) + B(\text{its derivative})$: $2\cos x - \sin x = A(3\cos x + \sin x) + B(\cos x - 3\sin x)$.
$\cos x$: $3A + B = 2$; $\sin x$: $A - 3B = -1$. So $A = B = \frac{1}{2}$.
Integral $= \frac{1}{2}x + \frac{1}{2}\log_e|3\cos x + \sin x| + C = \frac{1}{2}\left(x + \log_e|\sin x + 3\cos x|\right) + C$.
So $\alpha = 1$ and $\frac{\gamma}{\beta} = 3$ (any common multiple of $\beta, \gamma$ only changes the constant).
$\alpha + \frac{\gamma}{\beta} = 4$.
Answer: (C). | ||||
| 9 | 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 Intersection: $x^2 + 4x - 5 = 0 \Rightarrow x = 1$, $y = \pm 2$.
The part inside the parabola ($y^2 \le 4x$) and inside the circle is symmetric about the x-axis; its upper half is bounded by $y = 2\sqrt{x}$ for $0 \le x \le 1$ and by $y = \sqrt{5 - x^2}$ for $1 \le x \le \sqrt{5}$.
$\int_0^1 2\sqrt{x}\,dx = \frac{4}{3}$.
$\int_1^{\sqrt{5}} \sqrt{5 - x^2}\,dx = \left[\frac{x}{2}\sqrt{5 - x^2} + \frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}\right]_1^{\sqrt{5}} = \frac{5\pi}{4} - 1 - \frac{5}{2}\sin^{-1}\frac{1}{\sqrt{5}}$.
Area $= 2\left[\frac{4}{3} - 1 + \frac{5\pi}{4} - \frac{5}{2}\sin^{-1}\frac{1}{\sqrt{5}}\right] = \frac{2}{3} + \frac{5\pi}{2} - 5\sin^{-1}\frac{1}{\sqrt{5}}$.
Since $\sin^{-1}\frac{1}{\sqrt{5}} = \frac{\pi}{2} - \sin^{-1}\frac{2}{\sqrt{5}}$, area $= \frac{2}{3} + 5\sin^{-1}\left(\frac{2}{\sqrt{5}}\right) \approx 6.2$.
The circle has area $5\pi \approx 15.7$, so the other part is about 9.5 and this one is the smaller.
Answer: (D). | ||||
| 10 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} = \frac{x^2 + y^2}{5xy}$, homogeneous. Put $y = vx$: $v + x\frac{dv}{dx} = \frac{1 + v^2}{5v}$, so $x\frac{dv}{dx} = \frac{1 - 4v^2}{5v}$.
Separate: $\frac{5v\,dv}{1 - 4v^2} = \frac{dx}{x} \Rightarrow -\frac{5}{8}\ln|1 - 4v^2| = \ln|x| + c$.
So $|1 - 4v^2|^{-5} = Kx^8$, i.e. $\frac{x^{10}}{|x^2 - 4y^2|^5} = Kx^8$, or $|x^2 - 4y^2|^5 = \frac{x^2}{K}$.
$y(1) = 0$: $1 = \frac{1}{K}$, so $K = 1$.
Solution: $|x^2 - 4y^2|^5 = x^2$.
Answer: (D). | ||||
| 11 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $(5 - 2y)\frac{dy}{dx} = 3 \Rightarrow (5 - 2y)\,dy = 3\,dx \Rightarrow 5y - y^2 = 3x + c$.
At (0, 1): $5 - 1 = c = 4$. So $y^2 - 5y + 3x + 4 = 0$.
Complete the square: $\left(y - \frac{5}{2}\right)^2 = \frac{25}{4} - 4 - 3x = \frac{9}{4} - 3x = -3\left(x - \frac{3}{4}\right)$.
A parabola with vertex $\left(\frac{3}{4}, \frac{5}{2}\right)$.
$2x + 3y = \frac{3}{2} + \frac{15}{2} = 9$, so the vertex lies on $2x + 3y = 9$.
Answer: (A). | ||||
| 12 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $g(10) = \frac{10}{1} = 10$, so $a = f(10) = 109$. $f(3) = 18$, so $b = g(18) = \frac{18}{9} = 2$.
Ellipse $\frac{x^2}{109} + \frac{y^2}{2} = 1$: major axis along x, semi-axes $\sqrt{109}$ and $\sqrt{2}$.
$e^2 = 1 - \frac{2}{109} = \frac{107}{109}$.
Latus rectum $l = \frac{2\times 2}{\sqrt{109}} = \frac{4}{\sqrt{109}}$, $l^2 = \frac{16}{109}$.
$8e^2 + l^2 = \frac{856 + 16}{109} = \frac{872}{109} = 8$.
Answer: (C). | ||||
| 13 | CirclesStandard and general form | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Solve by Cramer: $\Delta = 15c^2 - 5(2 + c) = 5(3c^2 - c - 2) = 5(3c + 2)(c - 1)$.
$x_c = \frac{5c^2 - 5}{\Delta} = \frac{5(c - 1)(c + 1)}{5(3c + 2)(c - 1)} = \frac{c + 1}{3c + 2} \to \frac{2}{5}$.
$y_c = \frac{3 - (2 + c)}{\Delta} = \frac{1 - c}{5(3c + 2)(c - 1)} = -\frac{1}{5(3c + 2)} \to -\frac{1}{25}$.
Centre $\left(\frac{2}{5}, -\frac{1}{25}\right)$: the circle is $x^2 + y^2 - \frac{4}{5}x + \frac{2}{25}y + c_0 = 0$.
Through (2, 0): $4 - \frac{8}{5} + c_0 = 0 \Rightarrow c_0 = -\frac{12}{5}$.
Multiply by 25: $25x^2 + 25y^2 - 20x + 2y - 60 = 0$.
Answer: (A). | ||||
| 14 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines The reflected ray, produced backwards, passes through the image of P in the x-axis, $P'(1, -2)$.
Line $P'R$: slope $\frac{3 + 2}{4 - 1} = \frac{5}{3}$; it meets $y = 0$ where $x = 1 + \frac{3}{5}\cdot 2 = \frac{11}{5}$. So $Q = \left(\frac{11}{5}, 0\right)$.
In parallelogram PQRS the diagonals bisect each other: $S = P + R - Q = \left(1 + 4 - \frac{11}{5},\ 2 + 3 - 0\right) = \left(\frac{14}{5}, 5\right)$.
$hk^2 = \frac{14}{5}\times 25 = 70$.
Answer: (B). | ||||
| 15 | 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 A point of the first line: $P = (2 + t, -t, 1 + t)$. A point of the second: $Q = \left(-1 + \frac{s}{2}, 1 + \frac{s}{2}, s - 1\right)$.
L has direction (3, 1, 2), so $P - Q = \left(3 + t - \frac{s}{2},\ -1 - t - \frac{s}{2},\ 2 + t - s\right) = \mu(3, 1, 2)$.
From the second component $\mu = -1 - t - \frac{s}{2}$. The first gives $4t + s + 6 = 0$; the third gives $3t + 4 = 0$.
So $t = -\frac{4}{3}$, $s = -\frac{2}{3}$, and $P = \left(\frac{2}{3}, \frac{4}{3}, -\frac{1}{3}\right)$.
L: $\left(\frac{2}{3} + 3\lambda, \frac{4}{3} + \lambda, -\frac{1}{3} + 2\lambda\right)$. For $x = -\frac{1}{3}$, $\lambda = -\frac{1}{3}$, giving $\left(-\frac{1}{3}, 1, -1\right)$.
Answer: (A). | ||||
| 16 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $\vec{b}_1 = (4, -11, 5)$, $\vec{b}_2 = (3, -6, 1)$; $\vec{b}_1\times\vec{b}_2 = (-11 + 30,\ 15 - 4,\ -24 + 33) = (19, 11, 9)$, with $|\vec{b}_1\times\vec{b}_2| = \sqrt{361 + 121 + 81} = \sqrt{563}$.
Points: (3, −7, 1) and (5, 9, −2); difference (2, 16, −3).
$(2, 16, -3)\cdot(19, 11, 9) = 38 + 176 - 27 = 187$.
Shortest distance $= \frac{187}{\sqrt{563}}$.
Answer: (D). | ||||
| 17 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\vec{OA}\times\vec{OC}| = |2\vec{a}\times 3\vec{b}| = 6|\vec{a}\times\vec{b}| = 15$, so $|\vec{a}\times\vec{b}| = \frac{5}{2}$.
Area of quadrilateral OABC $= \frac{1}{2}|\vec{OB}\times\vec{AC}|$, with diagonals $\vec{OB} = 6\vec{a} + 5\vec{b}$ and $\vec{AC} = 3\vec{b} - 2\vec{a}$.
$(6\vec{a} + 5\vec{b})\times(3\vec{b} - 2\vec{a}) = 18\,\vec{a}\times\vec{b} - 10\,\vec{b}\times\vec{a} = 28\,\vec{a}\times\vec{b}$.
Area $= \frac{1}{2}\times 28\times\frac{5}{2} = 35$.
Answer: (B). | ||||
| 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 Area $= \frac{1}{2}|\vec{a}\times\vec{b}| = 5\sqrt{6}$, so $|\vec{a}\times\vec{b}|^2 = 600$.
$\vec{a}\times\vec{b} = (16 - 6,\ 10 - 4\alpha,\ 3\alpha - 20) = (10, 10 - 4\alpha, 3\alpha - 20)$.
$100 + (10 - 4\alpha)^2 + (3\alpha - 20)^2 = 25\alpha^2 - 200\alpha + 600 = 600 \Rightarrow \alpha(\alpha - 8) = 0$, so $\alpha = 8$.
$\vec{c} = \vec{a} - \vec{b} = (3, 1, -2)$, $|\vec{c}|^2 = 9 + 1 + 4 = 14$.
Answer: (C). | ||||
| 19 | StatisticsMean deviation | Class 11: Statistics | Single correctMedium | |
A medium single correct question on mean deviation, from Statistics. NCERT: Class 11: Statistics $5 + 8 + 5 + 12 + x + y = 40 \Rightarrow x + y = 10$.
Median: the 20th and 21st values. Cumulative frequencies 5, 13, 18, 30, so both are 18 and the median is 18.
$\sum f|x_i - 18| = 5(3) + 8(2) + 5(1) + 12(0) + x(1) + y(2) = 36 + x + 2y$.
Mean deviation $= \frac{36 + x + 2y}{40} = 1.25 \Rightarrow x + 2y = 14$.
With $x + y = 10$: $y = 4$, $x = 6$.
$4x + 5y = 24 + 20 = 44$.
Answer: (B). | ||||
| 20 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions The angles 60 are in degrees ($60^\circ = \frac{\pi}{3}$). Identity: $\cos\theta\cos(60^\circ - \theta)\cos(60^\circ + \theta) = \frac{1}{4}\cos 3\theta$.
The condition becomes $\frac{1}{4}|\cos 3\theta| \le \frac{1}{8}$, i.e. $|\cos 3\theta| \le \frac{1}{2}$.
So the largest value $\cos 3\theta$ can take is $\frac{1}{2}$.
$\cos 3\theta = \frac{1}{2}$ with $3\theta \in [0, 6\pi]$: $3\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3}$.
Sum of these $3\theta$ values $= \frac{54\pi}{3} = 18\pi$, so the sum of the $\theta$ values is $6\pi$.
Answer: (A). | ||||
| 21 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions An element of R is a pair $((a_1, b_1), (a_2, b_2))$ with $a_1 + a_2 = b_1 + b_2 = s$. Count ordered pairs by their sum.
From A: sum 4: 1, 5: 2, 6: 1, 8: 2, 9: 4, 10: 2, 12: 1, 13: 2, 14: 1.
From B: sum 8: 1, 9: 2, 10: 3, 11: 2, 12: 3, 13: 2, 14: 2, 16: 1.
Common sums: $s = 8$: $2\times 1 = 2$; $s = 9$: $4\times 2 = 8$; $s = 10$: $2\times 3 = 6$; $s = 12$: $1\times 3 = 3$; $s = 13$: $2\times 2 = 4$; $s = 14$: $1\times 2 = 2$.
Total $= 2 + 8 + 6 + 3 + 4 + 2 = 25$.
Answer: 25. | ||||
| 22 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|z - 1| \le 1$ with integer $a, b$: $(a - 1)^2 + b^2 \le 1$ gives $z = 1, 0, 2, 1 + i, 1 - i$.
$|z - 5| \le |z - 5i|$: $z$ is at least as close to (5, 0) as to (0, 5), i.e. on or below the bisector $y = x$: $b \le a$.
All five points satisfy $b \le a$ ($0 \le 0$ and $1 \le 1$ included).
Sum of $|z|^2$: $1 + 0 + 4 + 2 + 2 = 9$.
Answer: 9. | ||||
| 23 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For order 3: $\det(kM) = k^3\det M$ and $\det(\mathrm{adj}\,M) = (\det M)^2$. Let $d = \det A$.
$\det(dA) = d^4$; $\det\mathrm{adj}(dA) = d^8$; $\det(2\,\mathrm{adj}(dA)) = 8d^8$; $\det\mathrm{adj}(\ldots) = 64d^{16}$; $\det(3\ldots) = 27\times 64\,d^{16} = 3^3\cdot 2^6\cdot d^{16}$.
$3^3\cdot 2^6\cdot d^{16} = 3^{-13}\cdot 2^{-10} \Rightarrow d^{16} = 3^{-16}\cdot 2^{-16}$, so $d^2 = \frac{1}{36}$.
$\det(3\,\mathrm{adj}(2A)) = 27(\det 2A)^2 = 27(8d)^2 = 27\times 64\times\frac{1}{36} = 48 = 2^4\cdot 3^1$.
$m = 4$, $n = 1$: $|3m + 2n| = 14$.
Answer: 14. | ||||
| 24 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Left limit: put $x = \frac{\pi}{2} - h$. $\tan 8x = \tan(4\pi - 8h) = -\tan 8h \approx -8h$ and $\tan 7x = \tan\left(\frac{7\pi}{2} - 7h\right) = \cot 7h \approx \frac{1}{7h}$.
So $\frac{\tan 8x}{\tan 7x} \approx -56h^2 \to 0$ and the left limit is $\left(\frac{8}{7}\right)^0 = 1$.
Right limit: put $x = \frac{\pi}{2} + h$. $|\cot x| = \tan h \to 0$ and $|\tan x| = \cot h$, so $(1 + \tan h)^{\frac{b}{a}\cot h} \to e^{b/a}$.
Continuity: $1 = a - 8 = e^{b/a}$, so $a = 9$ and $b = 0$.
$a^2 + b^2 = 81$.
Answer: 81. | ||||
| 25 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $428 = 420 + 8 = 21\times 20 + 8$, so $428^{2024} \equiv 8^{2024} \pmod{21}$.
$8^2 = 64 = 63 + 1 \equiv 1 \pmod{21}$.
$8^{2024} = (8^2)^{1012} \equiv 1 \pmod{21}$. (Equivalently, $(63 + 1)^{1012}$ expanded binomially leaves remainder 1.)
Answer: 1. | ||||
| 26 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series From $f(m + n) = f(m) + f(n)$ and $f(1) = 1$: $f(n) = n$ for every natural $n$ (induction: $f(n + 1) = f(n) + 1$).
$\sum_{k=1}^{2022}(\lambda + k) = 2022\lambda + \frac{2022\times 2023}{2} \le 2022^2$.
Divide by 2022: $\lambda + \frac{2023}{2} \le 2022 \Rightarrow \lambda \le 1010.5$.
Largest natural $\lambda = 1010$.
Answer: 1010. | ||||
| 27 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $g(x) = 1 + \lambda^2x - x^3$: $g'(x) = \lambda^2 - 3x^2 = 0$ at $x = \pm\frac{\lambda}{\sqrt{3}}$; $g''(x) = -6x > 0$ at $x = -\frac{\lambda}{\sqrt{3}}$, the point of local minimum.
$x^2 + x + 2 > 0$ always (discriminant $-7$), so the inequality needs $x^2 + 5x + 6 = (x + 2)(x + 3) < 0$, i.e. $-3 < x < -2$.
$-3 < -\frac{\lambda}{\sqrt{3}} < -2 \Rightarrow 2\sqrt{3} < \lambda < 3\sqrt{3}$.
$\alpha^2 + \beta^2 = 12 + 27 = 39$.
Answer: 39. | ||||
| 28 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The $r$-th pair ($r = 1, \ldots, n$) is $\frac{n}{\sqrt{n^4 + r^4}} - \frac{2nr^2}{(n^2 + r^2)\sqrt{n^4 + r^4}} = \frac{n(n^2 - r^2)}{(n^2 + r^2)\sqrt{n^4 + r^4}}$.
With $t = \frac{r}{n}$ this is $\frac{1}{n}\cdot\frac{1 - t^2}{(1 + t^2)\sqrt{1 + t^4}}$, so the limit is $\int_0^1 \frac{1 - t^2}{(1 + t^2)\sqrt{1 + t^4}}\,dt$.
Divide numerator and denominator by $t^2$: $\frac{\frac{1}{t^2} - 1}{\left(t + \frac{1}{t}\right)\sqrt{t^2 + \frac{1}{t^2}}}$. Put $u = t + \frac{1}{t}$, $du = \left(1 - \frac{1}{t^2}\right)dt$, $t^2 + \frac{1}{t^2} = u^2 - 2$.
As $t$ goes from 0 to 1, $u$ goes from $\infty$ to 2: integral $= \int_2^{\infty}\frac{du}{u\sqrt{u^2 - 2}} = \frac{1}{\sqrt{2}}\left[\sec^{-1}\frac{u}{\sqrt{2}}\right]_2^{\infty} = \frac{1}{\sqrt{2}}\left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \frac{\pi}{4\sqrt{2}}$.
So $k = 4\sqrt{2}$ and $k^2 = 32$.
Answer: 32. | ||||
| 29 | 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 circle passes through the origin, so its radius is $r = \sqrt{h^2 + k^2}$ (distance from its centre to O); also $h = \frac{1}{2}$.
Both (0, 0) and (1, 0) lie inside $x^2 + y^2 = 9$, so the circle touches it internally: distance between centres $= 3 - r$.
The distance between centres is $\sqrt{h^2 + k^2} = r$, so $r = 3 - r$, $r = \frac{3}{2}$.
$h^2 + k^2 = r^2 = \frac{9}{4}$ for every such centre, and $4(h^2 + k^2) = 9$.
Answer: 9. | ||||
| 30 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are $4^3 = 64$ equally likely outcomes. Real roots need $b^2 \ge 4ac$.
$b = 1$: none. $b = 2$: $ac \le 1$: (1, 1), 1 case.
$b = 3$: $ac \le 2$: (1, 1), (1, 2), (2, 1), 3 cases.
$b = 4$: $ac \le 4$: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (3, 1), (4, 1), 8 cases.
Favourable $= 12$; probability $= \frac{12}{64} = \frac{3}{16}$.
$m + n = 19$.
Answer: 19. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Latent heat $L = \frac{Q}{m}$ is energy per unit mass (J/kg).
$[L] = \frac{[ML^2T^{-2}]}{[M]} = [M^0L^2T^{-2}]$.
Answer: (B). | ||||
| 32 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Let the total distance be $2d$. First half: time $t_1 = \frac{d}{6}$.
Second half: two equal intervals $t$ each, $9t + 15t = d \Rightarrow t = \frac{d}{24}$, total $\frac{d}{12}$.
Total time $= \frac{d}{6} + \frac{d}{12} = \frac{d}{4}$.
Average speed $= \frac{2d}{d/4} = 8$ m/s.
Answer: (B). | ||||
| 33 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy 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 Take the bar uniform, of length $l$, its weight acting at the middle, and the shoulder pushing vertically up with force $N$ at the far end.
Torques about the end on the ground: $N\,(l\cos\theta) = W\left(\frac{l}{2}\cos\theta\right)$.
$N = \frac{W}{2}$, whatever the angle.
Answer: (D). | ||||
| 34 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion For the Atwood machine $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$ (taking $m_2$ as the heavier block).
$\frac{m_2 - m_1}{m_2 + m_1} = \frac{1}{8}$. With $r = \frac{m_2}{m_1}$: $8(r - 1) = r + 1 \Rightarrow r = \frac{9}{7}$.
Answer: (B). | ||||
| 35 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At $x = 0$, $v = 0$; at $x = d$, $v = \alpha\sqrt{d}$.
By the work-energy theorem, $W = \Delta K = \frac{1}{2}m\alpha^2 d - 0 = \frac{m\alpha^2 d}{2}$.
Answer: (B). | ||||
| 36 | GravitationOrbital motion | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation $h = 318.5$ km $= \frac{R_e}{20}$, so the orbit radius is $r = \frac{21}{20}R_e$.
On the surface (at rest): $E_1 = -\frac{GM_em}{R_e}$.
In the orbit: $E_2 = -\frac{GM_em}{2r} = -\frac{10\,GM_em}{21R_e}$.
$\Delta E = \frac{GM_em}{R_e}\left(1 - \frac{10}{21}\right) = \frac{11\,GM_em}{21R_e}$, so $x = 11$.
Answer: (C). | ||||
| 37 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on 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 "Just floats": fully immersed, weight equals the buoyant force on the whole outer volume.
Weight $= \sigma\rho_w g\cdot\frac{\pi}{6}(D^3 - d^3)$; buoyancy $= \rho_w g\cdot\frac{\pi}{6}D^3$.
$\sigma(D^3 - d^3) = D^3 \Rightarrow D^3(\sigma - 1) = \sigma d^3$.
$\frac{D}{d} = \left(\frac{\sigma}{\sigma - 1}\right)^{1/3}$.
Answer: (B). | ||||
| 38 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $TV^{\gamma - 1}$ is constant, so $T_2 = T\left(\frac{1}{2}\right)^{1/2} = \frac{T}{\sqrt{2}}$.
$W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{R\left(T - \frac{T}{\sqrt{2}}\right)}{1/2} = 2RT\left(1 - \frac{1}{\sqrt{2}}\right) = RT\left[2 - \sqrt{2}\right]$.
Answer: (A). | ||||
| 39 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $P_1V_1^\gamma = P_2V_2^\gamma \Rightarrow \frac{P_1}{P_2} = \left(\frac{V_2}{V_1}\right)^{\gamma} = \left(\frac{4}{5}\right)^{3/2} = \frac{8}{5\sqrt{5}}$.
Answer: (B). | ||||
| 40 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Each stair faces the flat plate over area $\frac{A}{3}$, at separations $d$, $2d$ and $3d$. The three parts are in parallel.
$C = \frac{\epsilon_0 A}{3}\left(\frac{1}{d} + \frac{1}{2d} + \frac{1}{3d}\right) = \frac{\epsilon_0 A}{3d}\cdot\frac{11}{6} = \frac{11\epsilon_0 A}{18d}$.
Answer: (A). | ||||
| 41 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Read the drawing: from A the 6 Ω resistor runs along the top to the top middle junction T; from B the 8 Ω resistor runs down the left side to the left middle junction L.
A vertical wire joins T to the bottom middle junction M, and a horizontal wire joins L to the right middle junction R; where they cross the hump shows they are not connected. So T and M are one node, and L and R are one node.
Between these two nodes there are three branches: 10 Ω + 5 Ω (top right), 7 Ω + 8 Ω (bottom right) and 4 Ω + 11 Ω (bottom left), each 15 Ω.
Three 15 Ω in parallel give 5 Ω.
$R_{AB} = 6 + 5 + 8 = 19\,\Omega$.
Answer: (C). | ||||
| 42 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsMedium | |
A medium statements question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Statement I: for time-varying currents the forces between current elements need not be equal and opposite; momentum conservation (and so the third law) holds only when the momentum of the electromagnetic field is included. True.
Statement II: Ampere's circuital law is not new in content; it can be derived from the Biot-Savart law, and both relate the magnetic field to its source current. So it does depend on the Biot-Savart law. False.
Answer: (C). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The dielectric raises the capacitance $C$, so the capacitive reactance $X_C = \frac{1}{\omega C}$ falls.
The impedance $\sqrt{R^2 + X_C^2}$ falls, the current rises, and the bulb glows brighter.
Answer: (B). | ||||
| 44 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $\vec{E}$ along $y$ and propagation along $x$: $\vec{B}$ must be along $z$ ($\hat{j}\times\hat{k} = \hat{i}$).
$B_0 = \frac{E_0}{c} = \frac{60}{3\times 10^8} = 2\times 10^{-7}$ T.
$k = \frac{2\pi}{\lambda} = \frac{2\pi}{4\times 10^{-3}} = \frac{\pi}{2}\times 10^3$ m$^{-1}$, and $\omega = kc$.
$B_z = 2\times 10^{-7}\sin\left[\frac{\pi}{2}\times 10^3(x - 3\times 10^8 t)\right]\hat{k}$ T.
Answer: (D). | ||||
| 45 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | StatementsEasy | |
A easy statements question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Statement I: for a concave lens of focal length $-F$ and an object at $u = -2F$: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = -\frac{1}{F} - \frac{1}{2F}$, so $v = -\frac{2F}{3}$. The image is virtual, on the same side as the object, not at the centre of curvature on the other side. False.
Statement II: for a real object a concave lens always gives a virtual, erect, diminished image. True.
Answer: (D). | ||||
| 46 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{\sqrt{2mE}}$: at equal kinetic energy, $\lambda \propto \frac{1}{\sqrt{m}}$.
$m_e < m_p < m_\alpha$, so $\lambda_e > \lambda_p > \lambda_\alpha$, i.e. $\lambda_\alpha < \lambda_p < \lambda_e$.
Answer: (C). | ||||
| 47 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $E = mc^2 = 10^{-3}\times(3\times 10^8)^2 = 9\times 10^{13}$ J.
In MeV: $\frac{9\times 10^{13}}{1.6\times 10^{-13}} \approx 5.6\times 10^{26}$ MeV.
Answer: (D). | ||||
| 48 | 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 $\lambda = \frac{hc}{E_g} = \frac{1240\text{ eV nm}}{1.42\text{ eV}} \approx 873$ nm, closest to 875 nm.
Answer: (D). | ||||
| 49 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter $(n + 1)$ VSD $= n$ MSD $= nm$, so 1 VSD $= \frac{nm}{n + 1}$.
Least count $= 1$ MSD $- 1$ VSD $= m - \frac{nm}{n + 1} = \frac{m}{n + 1}$.
Answer: (B). | ||||
| 50 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter A shunt $S$ in parallel: $I_gG = (I - I_g)S$.
$S = \frac{20\times 10^{-6}\times 200}{20\times 10^{-3} - 20\times 10^{-6}} = \frac{4\times 10^{-3}}{19.98\times 10^{-3}} \approx 0.20\,\Omega$.
Answer: (C). | ||||
| 51 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Square: $a^2 + b^2 + 2ab\cos\theta = 2(a^2 + b^2 - 2ab\cos\theta)$, so $a^2 + b^2 = 6ab\cos\theta = \frac{10}{3}ab$.
With $a = nb$: $n^2 + 1 = \frac{10}{3}n \Rightarrow 3n^2 - 10n + 3 = 0 \Rightarrow n = 3$ or $\frac{1}{3}$.
The integer value is $n = 3$.
Answer: 3. | ||||
| 52 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Torque $\tau = FR = 40\times 0.1 = 4$ N m.
$\alpha = \frac{\tau}{I} = \frac{4}{0.40} = 10$ rad/s$^2$.
$\omega = \alpha t = 10\times 10 = 100$ rad/s.
Answer: 100. | ||||
| 53 | 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 The tension in the wire is 200 N (one pull balances the other).
$\Delta l = \frac{Fl}{AY} = \frac{200\times 2}{2\times 10^{-4}\times 10^{11}} = 2\times 10^{-5}$ m $= 20\ \mu$m.
Answer: 20. | ||||
| 54 | 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 $|a| = \omega^2|x| \Rightarrow \omega^2 = \frac{16}{4} = 4$.
$v^2 = \omega^2(A^2 - x^2) \Rightarrow 4 = 4(A^2 - 16) \Rightarrow A^2 = 17$.
$A = \sqrt{17}$ m, so $x = 17$.
Answer: 17. | ||||
| 55 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Every element of the half ring is at distance $R$ from the centre, so $V = \frac{kQ}{R} = \frac{k\lambda\pi R}{R} = k\lambda\pi$.
$V = 9\times 10^9\times 4\times 10^{-9}\times\pi = 36\pi$ V.
Answer: 36. | ||||
| 56 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Read the drawing: A–B is 4 Ω, A–D is 4 Ω, B–C is 2 Ω, C–D is 2 Ω; the branch A–C holds the 10 V cell in series with 1 Ω (current $I_1$); B–D holds the 5 V cell E.
Use superposition. With only the 5 V cell acting, the network is balanced: A sits midway between B and D (4 Ω and 4 Ω) and so does C (2 Ω and 2 Ω), so A and C are at the same potential and no current flows in the A–C branch. The 5 V cell contributes nothing to the 1 Ω current.
With only the 10 V cell acting (5 V cell replaced by a wire, joining B and D): from A, the two 4 Ω resistors in parallel give 2 Ω to the B–D node, and from there the two 2 Ω resistors in parallel give 1 Ω to C.
Current in the 1 Ω branch $= \frac{10}{1 + 2 + 1} = 2.5$ A.
So $I_1 = 2.5$ A $= \frac{25}{10}$ A, and $n = 25$.
Answer: 25. | ||||
| 57 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The two edges parallel to the x-axis feel equal and opposite forces (the same field pattern along each, opposite currents), so they cancel.
The two edges parallel to the y-axis, at $x_1$ and $x_1 + 2$, feel opposite forces of size $ILB$, with different fields.
Net force $= IL\,[B(x_1 + 2) - B(x_1)] = IL\cdot B_0\cdot 4\cdot 2 = 2\times 2\times 5\times 8 = 160$ N.
Answer: 160. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current DC: $R = \frac{20}{5} = 4\,\Omega$. AC: $Z = \frac{20}{4} = 5\,\Omega$.
$X_L = \sqrt{Z^2 - R^2} = 3\,\Omega$.
$L = \frac{X_L}{2\pi f} = \frac{3}{2\times 3\times 50} = 0.01$ H $= 10$ mH.
Answer: 10. | ||||
| 59 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $I = I_{max}\cos^2\frac{\phi}{2} = \frac{I_{max}}{4} \Rightarrow \cos\frac{\phi}{2} = \frac{1}{2}$, smallest $\phi = \frac{2\pi}{3}$.
Path difference $= \frac{\lambda}{3}$, so $y = \frac{\lambda D}{3d} = \frac{600\times 10^{-9}\times 1}{3\times 10^{-3}} = 2\times 10^{-4}$ m $= 200\ \mu$m.
Answer: 200. | ||||
| 60 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Mass defect per reaction: $3\times 4.0026 - 12 = 0.0078$ u.
$Q = 0.0078\times 931.5$ MeV $\approx 7.27$ MeV $\approx 7.27\times 1.6\times 10^{-13}$ J $\approx 1.164\times 10^{-12}$ J.
Reactions per second $= \frac{5.808\times 10^{30}}{1.164\times 10^{-12}} \approx 4.99\times 10^{42} \approx 5\times 10^{42}$ s$^{-1}$.
(The mass of the star is not needed.)
$n = 5$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom Use the $(n + l)$ rule; for equal $(n + l)$ the lower $n$ has the lower energy.
(A) 4p: 5; (B) 4d: 6; (C) 3p: 4; (D) 3d: 5; (E) 4s: 4.
(C) 3p < (E) 4s (both 4, lower $n$ first) < (D) 3d < (A) 4p (both 5) < (B) 4d.
Answer: (B). | ||||
| 62 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $NH_2^-$: 2 bond pairs + 2 lone pairs, $sp^3$. $H_2O$: $sp^3$.
$BF_3$: 3 bond pairs, $sp^2$. $NO_2^-$: 2 bond pairs + 1 lone pair, $sp^2$. ($NO_2$ is also $sp^2$, but it is paired with $H_2O$.)
Only $BF_3$ and $NO_2^-$ are both $sp^2$.
Answer: (B). | ||||
| 63 | SolutionsColligative properties | Class 12: Solutions | Single correctMedium | |
A medium single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions The membrane passes only the solvent, so the ions never meet and no green $Cu_2Cr_2O_7$ forms on either side.
Effective particle concentrations: $CuSO_4$: $0.05\times 2 = 0.10$ M; $K_2Cr_2O_7$: $0.01\times 3 = 0.03$ M.
Water moves by osmosis from the lower concentration (side X, $K_2Cr_2O_7$) to the higher (side Y, $CuSO_4$).
So the $CuSO_4$ solution is diluted (its molarity falls) and the $K_2Cr_2O_7$ solution becomes more concentrated.
Answer: (D). | ||||
| 64 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry A strong electrolyte shows a small, nearly linear fall of $\Lambda_m$ with $\sqrt{C}$ (Debye–Hückel–Onsager): curve B.
A weak electrolyte shows $\Lambda_m$ rising steeply as $C \to 0$ because its degree of dissociation increases sharply on dilution: curve A.
A is weak, B is strong.
Answer: (A). | ||||
| 65 | Redox ReactionsOxidation number | Class 11: Redox Reactions | StatementsEasy | |
A easy statements question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions Statement I: oxidation state is the charge an atom would carry if the bonding electrons were assigned on the basis of electronegativity, not electron gain enthalpy. Incorrect.
Statement II: second-period elements (C, N, O) are small, so their p orbitals overlap well sideways and they readily form $p\pi - p\pi$ multiple bonds; heavier elements seldom do. Correct.
Answer: (D). | ||||
| 66 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A): both allotropes of sulphur are made of puckered $S_8$ rings; oxygen is the diatomic $O_2$. Correct.
(R): oxygen, small and highly electronegative, forms strong $p\pi - p\pi$ double bonds (O=O, C=O, N=O); sulphur, larger, does not, and so prefers S–S single bonds in rings and chains. Correct.
That is exactly why oxygen is $O_2$ and sulphur is $S_8$, so (R) explains (A).
Answer: (A). | ||||
| 67 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Hydroxyapatite is $3Ca_3(PO_4)_2\cdot Ca(OH)_2$; fluoride replaces the hydroxide, giving fluoroapatite $3Ca_3(PO_4)_2\cdot CaF_2$.
Answer: (C). | ||||
| 68 | 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 Although $3d^{10}$ looks more stable, in aqueous solution $Cu^{2+}$ is the more stable ion: its much more negative hydration enthalpy more than makes up for the second ionisation enthalpy of copper.
Accordingly $Cu^+(aq)$ disproportionates: $2Cu^+ \to Cu^{2+} + Cu$.
Answer: (A). | ||||
| 69 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Assertion and reasonEasy | |
A easy assertion and reason question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[Co(en)_2Cl_2]^+$ is octahedral (coordination number 6), so (R) is correct.
Its geometrical isomers are only two, cis and trans. (Three is the total number of stereoisomers: trans plus the two optical forms of cis.) So (A) is not correct.
Answer: (D). | ||||
| 70 | 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 The choice of method (crystallisation, distillation, sublimation, extraction, chromatography) depends on the nature of the compound and on the impurities present in it.
Answer: (C). | ||||
| 71 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (I) is the neutral structure with every atom having an octet: most stable.
(II) separates charge with the negative charge on the more electronegative oxygen and the positive on carbon: next.
(III) puts the positive charge on oxygen and the negative on carbon, against electronegativity: least stable.
(I) > (II) > (III).
Answer: (D). | ||||
| 72 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Benzene + succinic anhydride / $AlCl_3$ (Friedel–Crafts acylation): A $= C_6H_5COCH_2CH_2COOH$ (4-oxo-4-phenylbutanoic acid).
Zn–Hg/HCl (Clemmensen) reduces the ketone C=O to $CH_2$: B $= C_6H_5CH_2CH_2CH_2COOH$ (4-phenylbutanoic acid).
Conc. $H_2SO_4$: intramolecular Friedel–Crafts acylation closes a six-membered ring with the carbonyl next to the benzene ring: C = 1-tetralone ($\alpha$-tetralone).
Answer: (A). | ||||
| 73 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Assertion and reasonMedium | |
A medium assertion and reason question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (A): benzyl bromide undergoes $S_N2$ much faster than ethyl bromide. Correct.
(R): in the $S_N2$ transition state the central carbon is $sp^2$-like with a p orbital partly bonded to both the nucleophile and the leaving group; for the benzylic carbon this p orbital overlaps with the ring π system, which lowers the energy of the transition state. Correct, and it is the reason for (A).
Answer: (A). | ||||
| 74 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The intended route works through the more reactive C–Br bond first.
Na/dry ether (Wurtz coupling at C–Br): two molecules of 1-bromo-3-chlorocyclobutane join, A = 3,3'-dichloro-1,1'-bi(cyclobutyl), Cl–C$_4$H$_6$–C$_4$H$_6$–Cl.
(i) Mg/ether turns each C–Cl into C–MgCl; (ii) $D_2O$ replaces each MgCl by D: B = D–C$_4$H$_6$–C$_4$H$_6$–D.
$CoF_2$ exchanges Cl for F (Swarts-type halogen exchange): C = F–C$_4$H$_6$–C$_4$H$_6$–F.
This matches option (A).
Answer: (A). | ||||
| 75 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Benzene with CO and HCl in the presence of anhydrous $AlCl_3$/CuCl is the Gattermann–Koch reaction: formyl substitution gives benzaldehyde, $C_6H_5CHO$.
Answer: (C). | ||||
| 76 | 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 Acid-catalysed hydration follows Markovnikov's rule through the more stable secondary carbocation: A = propan-2-ol, $CH_3CH(OH)CH_3$.
Hydroboration–oxidation adds water anti-Markovnikov: B = propan-1-ol, $CH_3CH_2CH_2OH$.
Answer: (B). | ||||
| 77 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The compounds: (A) phenol, (B) 4-nitrophenol, (C) 3-methoxyphenol, (D) 3-nitrophenol, (E) 4-methoxyphenol.
Electron-withdrawing $NO_2$ stabilises the phenoxide and lowers $pK_a$, most strongly from the para position (−M and −I): (B) < (D).
Electron-donating $OCH_3$ (+M at para) destabilises the phenoxide and raises $pK_a$: (E) is the weakest acid.
The intended order: (B) < (D) < (A) < (C) < (E) (measured values about 7.1, 8.4, 10.0, 9.6, 10.2).
Of the options, (C) is the one that places the nitrophenols first and 4-methoxyphenol last.
Answer: (C). | ||||
| 78 | AminesBasicity of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines Piperidine: $sp^3$ nitrogen, a localised lone pair, a secondary aliphatic amine ($pK_b \approx 2.9$): strongest base.
Pyridine: $sp^2$ nitrogen; its lone pair is not part of the aromatic sextet but has more s character, so it is held more tightly ($pK_b \approx 8.8$).
Pyrrole: the nitrogen lone pair is part of the aromatic sextet and is not available ($pK_b \approx 13.6$): weakest.
Piperidine > Pyridine > Pyrrole.
Answer: (A). | ||||
| 79 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $3PbS + 8HNO_3(\text{dil}) \to 3Pb(NO_3)_2 + 2NO + 3S + 4H_2O$.
Lead nitrate, nitric oxide (NO) and sulphur are formed; nitrous oxide ($N_2O$) is not.
Answer: (C). | ||||
| 80 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | StatementsEasy | |
A easy statements question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter A primary standard is pure and available dry (A true), stable in air (B true), readily soluble (D true), and reacts instantly and stoichiometrically, but it must be non-hygroscopic, so (C) is incorrect.
$KMnO_4$ (not obtainable pure, decomposes in light) and NaOH (hygroscopic, absorbs $CO_2$) are secondary standards, so (E) is incorrect.
Answer: (D). | ||||
| 81 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 L of solution: mass $= 1250$ g; it holds 0.2 mol $CuSO_4$ ($M = 159.5$ g/mol), i.e. $0.2\times 159.5 = 31.9$ g.
Mass of water $= 1250 - 31.9 = 1218.1$ g.
Molality $= \frac{0.2}{1.2181} = 0.1642$ m $= 164\times 10^{-3}$ m.
Answer: 164. | ||||
| 82 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure By molecular orbital filling:
$N_2$ (14 e): 0 unpaired. $O_2$ (16 e): 2 unpaired. $C_2^-$ (13 e): one electron in $\sigma 2p_z$, 1 unpaired.
$O_2^-$ (17 e): 1 unpaired. $O_2^{2-}$ (18 e): 0. $H_2^+$ (1 e): 1 unpaired.
$CN^-$ (14 e, like $N_2$): 0. $He_2^+$ (3 e, $\sigma 1s^2\sigma^*1s^1$): 1 unpaired.
One unpaired electron: $C_2^-, O_2^-, H_2^+, He_2^+$, four species.
Answer: 4. | ||||
| 83 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics $CuSO_4(s) + aq \to CuSO_4(aq)$, $\Delta H_1 = -70$ kJ.
$CuSO_4\cdot 5H_2O(s) + aq \to CuSO_4(aq)$, $\Delta H_2 = +12$ kJ.
Hydration $CuSO_4 + 5H_2O \to CuSO_4\cdot 5H_2O$: by Hess's law $\Delta H = \Delta H_1 - \Delta H_2 = -70 - 12 = -82$ kJ.
So $x = 82$.
Answer: 82. | ||||
| 84 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Both are strong acids, so the heat released is proportional to the moles of $H^+$ neutralised.
Equal volumes of 1 M HCl and 1 M $H_2SO_4$ supply $H^+$ in the ratio 1 : 2, and the NaOH is in excess.
$\frac{y}{x} = 2$.
Answer: 2. | ||||
| 85 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Dichromate ($E^\circ = 1.33$ V) oxidises a metal when the metal's reduction potential is lower, so that $E^\circ_{cell} = 1.33 - E^\circ_{metal} > 0$.
Fe ($-0.04$ V), Ni ($-0.25$ V) and Ag ($0.80$ V): yes. Au ($1.40$ V): no.
Number of metals $= 3$.
Answer: 3. | ||||
| 86 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Statement I: doubling both, $r' = k(2[A])^2(2[B]) = 8r$, so $x = 8$.
Statement II: concentration falls linearly with time from $[R]_0$ with slope $-k$: $[R] = [R]_0 - kt$, a zero order reaction, so $y = 0$.
$x + y = 8$.
Answer: 8. | ||||
| 87 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements An $Ln^{3+}$ ion is colourless when it has no f–f transition: $4f^0$ or $4f^{14}$.
$La^{3+}$ ($4f^0$) and $Lu^{3+}$ ($4f^{14}$) are colourless.
$Eu^{3+}$ ($4f^6$, pale pink), $Nd^{3+}$ ($4f^3$, lilac) and $Sm^{3+}$ ($4f^5$, yellow) are coloured.
Answer: 2. | ||||
| 88 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds An ambidentate ligand can bind through either of two different donor atoms, one at a time.
$NO_2^-$ (N or O), $SCN^-$ (S or N), $CN^-$ (C or N): ambidentate.
$C_2O_4^{2-}$ is didentate (both O at once); $NH_3$, $H_2O$ are monodentate; $SO_4^{2-}$ binds through O only.
Answer: 3. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques All three effects need a polar group (inductive), a conjugated π system with a group that donates or withdraws by resonance (mesomeric), and C–H bonds on a carbon next to a π system (hyperconjugation).
Anisole: −I and +M from $OCH_3$, but its methyl is on oxygen, not next to the π system: no hyperconjugation. Not counted.
$(CH_3)_2CH-CH=CH-CO-CH_3$: C=C conjugated with C=O (mesomeric), polar C=O (inductive), α C–H of the isopropyl and acetyl methyl (hyperconjugation). Counted.
Benzene: no substituent, no inductive effect. Not counted. Chlorocyclohexane: inductive only. Not counted.
2-Nitrocumene: $NO_2$ (−I, −M) and the isopropyl C–H next to the ring (hyperconjugation). Counted.
3-(2-Nitrophenyl)but-1-ene: $NO_2$ (−I, −M) and the benzylic C–H (hyperconjugation). Counted.
m-Xylene: methyl groups give +I and hyperconjugation, but there is no mesomeric group. Not counted.
1-Acetyl-2-methylcyclohexene: C=C conjugated with C=O (mesomeric, inductive) and the methyl and ring $CH_2$ next to the C=C (hyperconjugation). Counted.
Total $= 4$.
Answer: 4. | ||||
| 90 | BiomoleculesProteins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Essential amino acids (the textbook list of ten): valine, leucine, isoleucine, arginine, lysine, threonine, methionine, phenylalanine, tryptophan, histidine.
From the list: arginine, phenylalanine, histidine, valine are essential; aspartic acid, cysteine and proline are not.
Answer: 4. | ||||
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