JEE (Main) · 2024
JEE (Main) 2024 — 6 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
- Sets, Relations and Functions3
- Integral Calculus3
- Differential Equations3
13 easy · 15 medium · 2 hard
Physics
30 Q
- Work, Energy and Power2
- Kinematics2
- Properties of Bulk Matter2
25 easy · 5 medium · 0 hard
Chemistry
30 Q
- Chemical Bonding and Molecular Structure4
- d- and f-Block Elements3
- Organic Chemistry: Some Basic Principles and Techniques3
19 easy · 9 medium · 2 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 $R_1$: $2x - 3y = 2$ gives $x = \frac{2 + 3y}{2}$, so $y$ must be even.
$y = 2, 4, 6, 8, 10, 12$ give $x = 4, 7, 10, 13, 16, 19$ (at $y = 14$, $x = 22 > 20$). So $R_1$ has 6 pairs: $(4,2), (7,4), (10,6), (13,8), (16,10), (19,12)$.
No pair has $x = y$, and no reversed pair $(y, x)$ is already in $R_1$ (e.g. $(2,4)$: $4 - 12 \ne 2$). So all 6 reverses must be added: $M = 6$.
$R_2$: $4y = 5x$, so $x$ is a multiple of 4: $(4,5), (8,10), (12,15), (16,20)$ — 4 pairs, none of whose reverses is in $R_2$. So $N = 4$.
$M + N = 10$.
Answer: (B). | ||||
| 2 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions The denominator $x^2 - 4x + 9 = (x - 2)^2 + 5 > 0$, so $f$ is defined on all of $\mathbb{R}$.
Not one-one: $x^2 + 2x - 15 = (x + 5)(x - 3)$, so $f(-5) = f(3) = 0$.
Range: $y = f(x)$ gives $(y - 1)x^2 - (4y + 2)x + (9y + 15) = 0$. For real $x$ (when $y \ne 1$) the discriminant must be $\ge 0$:
$(4y + 2)^2 - 4(y - 1)(9y + 15) = -20y^2 - 8y + 64 \ge 0$, i.e. $5y^2 + 2y - 16 \le 0$.
So the range is a bounded interval, not $\mathbb{R}$: $f$ is not onto.
Answer: (D) neither one-one nor onto. | ||||
| 3 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Total triangles: $\binom{8}{3} = 56$.
Triangles with exactly two sides of the octagon (three consecutive vertices): 8.
Triangles with exactly one side of the octagon: choose the side (8 ways), then the third vertex must not be next to either end: $8 - 4 = 4$ ways, giving $8 \times 4 = 32$.
Triangles with no side of the octagon: $56 - 8 - 32 = 16$.
(Equivalently $\frac{n(n - 4)(n - 5)}{6} = \frac{8 \cdot 4 \cdot 3}{6} = 16$.)
Answer: (A). | ||||
| 4 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $p = t^2 - 5t + 6$. Then $\alpha + \beta = p$, $\alpha\beta = 1$, and each root satisfies $x^2 = px - 1$.
Multiplying by $x^{2023}$ and adding for both roots: $a_{2025} = p\,a_{2024} - a_{2023}$, so $\frac{a_{2023} + a_{2025}}{a_{2024}} = p$.
$p = t^2 - 5t + 6 = \left(t - \frac{5}{2}\right)^2 - \frac{1}{4}$, whose minimum is $-\frac{1}{4}$ at $t = \frac{5}{2}$.
At $p = -\frac{1}{4}$ the discriminant $p^2 - 4 \ne 0$, so the roots are distinct (complex conjugates), as required.
Answer: (A). | ||||
| 5 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Split the first column: $(r, 2r, 3r - 2) = r(1, 2, 3) + (0, 0, -2)$.
The determinant with first column $r(1, 2, 3)$ is 0, since that column is $r$ times the second column $(1, 2, 3)$.
So $A_r = \begin{vmatrix} 0 & 1 & \frac{n^2}{2} + \alpha \\ 0 & 2 & n^2 - \beta \\ -2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix} = -2\left[(n^2 - \beta) - 2\left(\frac{n^2}{2} + \alpha\right)\right] = -2(-\beta - 2\alpha) = 4\alpha + 2\beta$,
independent of $r$. Hence $2A_{10} - A_8 = 2(4\alpha + 2\beta) - (4\alpha + 2\beta) = 4\alpha + 2\beta$.
Answer: (D). | ||||
| 6 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctHard | |
A hard single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability The intended reading: put $x = \frac{1}{a}$ and write the expression as $a^2 f\left(\frac{1}{a}\right)$:
$\frac{a(a + 1)}{2} = a^2\cdot\frac{1 + x}{2}$, $a^2 = a^2\cdot 1$, $-2\log_e a = 2\log_e x = a^2\cdot 2x^2\log_e x$.
So the expression is $a^2 f\left(\frac{1}{a}\right)$ with $f(x) = \frac{1 + x}{2}\tan^{-1}x + 1 + 2x^2\log_e x$, and by the given property its limit is $f'(1)$.
$f'(x) = \frac{1}{2}\tan^{-1}x + \frac{1 + x}{2(1 + x^2)} + 4x\log_e x + 2x$.
$f'(1) = \frac{1}{2}\cdot\frac{\pi}{4} + \frac{2}{4} + 0 + 2 = \frac{5}{2} + \frac{\pi}{8}$.
(Taken literally, the limit does not exist: $\frac{a(a + 1)}{2}\tan^{-1}\frac{1}{a} \approx \frac{a + 1}{2}$ and $a^2 - 2\log_e a$ both grow without bound. See the note.)
Answer: (A). | ||||
| 7 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions There are $700 - 100 + 1 = 601$ integers in $[100, 700]$.
Multiples of 3: $102, 105, \ldots, 699$: $\frac{699 - 102}{3} + 1 = 200$.
Multiples of 4: $100, 104, \ldots, 700$: $\frac{700 - 100}{4} + 1 = 151$.
Multiples of 12: $108, 120, \ldots, 696$: $\frac{696 - 108}{12} + 1 = 50$.
Multiples of 3 or 4: $200 + 151 - 50 = 301$.
Neither: $601 - 301 = 300$.
Answer: (C). | ||||
| 8 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium 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 For $x \ne 0$: $f'(x) = 3x^2\sin\frac{1}{x} - x\cos\frac{1}{x}$; and $f'(0) = \lim_{h \to 0} h^2\sin\frac{1}{h} = 0$.
$f''(0) = \lim_{h \to 0}\frac{f'(h)}{h} = \lim_{h \to 0}\left(3h\sin\frac{1}{h} - \cos\frac{1}{h}\right)$, which does not exist. So (A) and (C) are false.
For $x \ne 0$: $f''(x) = 6x\sin\frac{1}{x} - 3\cos\frac{1}{x} - \cos\frac{1}{x} - \frac{1}{x}\sin\frac{1}{x} = 6x\sin\frac{1}{x} - 4\cos\frac{1}{x} - \frac{1}{x}\sin\frac{1}{x}$.
At $x = \frac{2}{\pi}$: $\frac{1}{x} = \frac{\pi}{2}$, $\sin = 1$, $\cos = 0$, so $f''\left(\frac{2}{\pi}\right) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24 - \pi^2}{2\pi}$.
Answer: (B). | ||||
| 9 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f(x) = e^{x\log_e x}$, so $f'(x) = x^x(\log_e x + 1)$.
$f'(x) > 0$ for $x > \frac{1}{e}$ and $f'(x) < 0$ for $0 < x < \frac{1}{e}$; $f'$ vanishes only at the single point $x = \frac{1}{e}$.
So $f$ is strictly increasing on $\left[\frac{1}{e}, \infty\right)$.
Answer: (A). | ||||
| 10 | 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 line is $y + 9 = m(x - 4)$.
$x$-intercept: $y = 0$ gives $x = 4 + \frac{9}{m}$; $y$-intercept: $x = 0$ gives $y = -9 - 4m$.
With $m > 0$: $OA + OB = \left(4 + \frac{9}{m}\right) + (9 + 4m) = 13 + 4m + \frac{9}{m}$.
By AM–GM, $4m + \frac{9}{m} \ge 2\sqrt{36} = 12$, with equality at $m = \frac{3}{2}$.
Minimum $= 13 + 12 = 25$.
Answer: (B). | ||||
| 11 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Divide numerator and denominator by $\cos^6 x$: the integrand becomes $\frac{\tan^2 x\sec^2 x}{(1 + \tan^3 x)^2}$.
Put $t = \tan x$, $dt = \sec^2 x\,dx$; the limits become 0 to 1:
$\int_0^1\frac{t^2}{(1 + t^3)^2}\,dt = \left[-\frac{1}{3(1 + t^3)}\right]_0^1 = -\frac{1}{6} + \frac{1}{3} = \frac{1}{6}$.
Answer: (B). | ||||
| 12 | 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 Corners: $y = 3x$ and $2y = 27 - 3x$ meet at $(3, 9)$. $y = 3x$ and $y = 3x - x^{3/2}$ meet at $(0, 0)$. $2y = 27 - 3x$ and $y = 3x - x^{3/2}$: $9x - 2x^{3/2} = 27$, satisfied by $x = 9$, giving $(9, 0)$.
The curve $y = 3x - x^{3/2}$ lies below $y = 3x$ and is $\ge 0$ on $[0, 9]$, so the region is bounded above by $y = 3x$ (for $0 \le x \le 3$) and by the line (for $3 \le x \le 9$), and below by the curve.
$A = \int_0^3 x^{3/2}\,dx + \int_3^9\left(\frac{27 - 3x}{2} - 3x + x^{3/2}\right)dx$.
First: $\frac{2}{5}\cdot 3^{5/2} = \frac{18\sqrt{3}}{5}$.
Second: $\frac{27}{2}\cdot 6 - \frac{9}{4}(81 - 9) + \frac{2}{5}\left(243 - 9\sqrt{3}\right) = 81 - 162 + \frac{486}{5} - \frac{18\sqrt{3}}{5}$.
$A = -81 + \frac{486}{5} = \frac{81}{5} = 16.2$, so $10A = 162$.
Answer: (B). | ||||
| 13 | 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 $\frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1}x}}{1 + x^2}$. Integrating factor $e^{\tan^{-1}x}$.
$y\,e^{\tan^{-1}x} = \int\frac{e^{2\tan^{-1}x}}{1 + x^2}\,dx = \frac{1}{2}e^{2\tan^{-1}x} + C$, so $y = \frac{1}{2}e^{\tan^{-1}x} + Ce^{-\tan^{-1}x}$.
$y(1) = 0$: $\frac{1}{2}e^{\pi/4} + Ce^{-\pi/4} = 0$, $C = -\frac{1}{2}e^{\pi/2}$.
$y(0) = \frac{1}{2} + C = \frac{1}{2}\left(1 - e^{\pi/2}\right)$.
Answer: (C). | ||||
| 14 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Divide by $2x$: $\log_e x\,\frac{dy}{dx} + \frac{y}{x} = \frac{3\log_e x}{2x^2}$. The left side is $\frac{d}{dx}(y\log_e x)$.
$y\log_e x = \frac{3}{2}\int\frac{\log_e x}{x^2}\,dx = \frac{3}{2}\left(-\frac{\log_e x}{x} - \frac{1}{x}\right) + C$.
At $x = e^{-1}$, $y = 0$: $0 = \frac{3}{2}(e - e) + C$, so $C = 0$.
At $x = e$: $y\cdot 1 = \frac{3}{2}\left(-\frac{1}{e} - \frac{1}{e}\right) = -\frac{3}{e}$.
Answer: (B). | ||||
| 15 | CirclesRadius and centre | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Inradius of an equilateral triangle of side 12: $r = \frac{12}{2\sqrt{3}} = 2\sqrt{3}$.
A square inscribed in the circle has diagonal $2r = 4\sqrt{3}$, so side $\frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$.
$m = 24$, $n = 8\sqrt{6}$, $n^2 = 384$.
$m + n^2 = 408$.
Answer: (A). | ||||
| 16 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections A circle touching both lines $y = \sqrt{3}|x|$ inside the angle between them has its centre on the $y$-axis, $(0, k)$, with radius $r = \frac{k}{2}$ (distance from $(0, k)$ to $\sqrt{3}x - y = 0$).
The lines meet the parabola at $(\pm\sqrt{3}, 3)$.
Intended solution: take the circle that touches the parabola at its vertex $(0, 6)$: $6 - k = \frac{k}{2}$, so $k = 4$, $r = 2$, and $C: x^2 + (y - 4)^2 = 4$.
It touches $y = \sqrt{3}x$: $x^2 + (\sqrt{3}x - 4)^2 = 4$ gives $(x - \sqrt{3})^2 = 0$, tangency at $(\sqrt{3}, 3)$; similarly at $(-\sqrt{3}, 3)$.
Check the options: $(2, 4)$: $4 + 0 = 4$ — on $C$. $(1, 2)$: $1 + 4 = 5$; $(2, 2)$: $4 + 4 = 8$; $(1, 1)$: $1 + 9 = 10$ — not on $C$.
Answer: (A). | ||||
| 17 | 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 Points $(3, -15, 9)$ and $(-1, 1, 9)$; directions $(2, -7, 5)$ and $(2, 1, -3)$.
$(2, -7, 5)\times(2, 1, -3) = (21 - 5,\ 10 + 6,\ 2 + 14) = (16, 16, 16) \parallel (1, 1, 1)$.
Joining vector $(-4, 16, 0)$; its dot product with $(1, 1, 1)$ is 12.
Shortest distance $= \frac{12}{\sqrt{3}} = 4\sqrt{3}$.
Answer: (C). | ||||
| 18 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Given: $\sum x = 20 \times 10 = 200$ and $\sum x^2 = 20(2^2 + 10^2) = 2080$.
Corrected: $\sum x = 200 - 8 + 12 = 204$, $\sum x^2 = 2080 - 64 + 144 = 2160$.
Mean $= 10.2$; variance $= \frac{2160}{20} - 10.2^2 = 108 - 104.04 = 3.96$.
Standard deviation $= \sqrt{3.96}$.
Answer: (B). | ||||
| 19 | 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 The four points are coplanar: $\vec{AB} = \left(-\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right)$, $\vec{AC} = (-1, 1, 2)$, $\vec{AD} = \left(\frac{1}{3}, -\frac{1}{3}, \frac{2}{3}\right)$; a normal to $AB, AC$ is $(-1,1,1)\times(-1,1,2) = (1, 1, 0)$, and $\vec{AD}\cdot(1, 1, 0) = 0$.
For a plane quadrilateral, area $= \frac{1}{2}|\vec{AC}\times\vec{BD}|$.
$\vec{BD} = \left(\frac{5}{3}, -\frac{5}{3}, -\frac{2}{3}\right)$.
$\vec{AC}\times\vec{BD} = \left(-\frac{2}{3} + \frac{10}{3},\ \frac{10}{3} - \frac{2}{3},\ \frac{5}{3} - \frac{5}{3}\right) = \left(\frac{8}{3}, \frac{8}{3}, 0\right)$, of length $\frac{8\sqrt{2}}{3}$.
Area $= \frac{4\sqrt{2}}{3}$.
Answer: (D). | ||||
| 20 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(S) = 0.6 \times 0.8 + 0.4 \times 0.9 = 0.48 + 0.36 = 0.84$.
$p = P(B \mid S) = \frac{0.36}{0.84} = \frac{3}{7}$.
$126p = 54$.
Answer: (D). | ||||
| 21 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $P(x) = 4x^4 + 8x^3 - 17x^2 - 12x + 9 = 4\prod(x - x_k)$.
Since $4 + x_k^2 = (2i - x_k)(-2i - x_k)$, $\prod(4 + x_k^2) = \frac{P(2i)}{4}\cdot\frac{P(-2i)}{4} = \frac{|P(2i)|^2}{16}$.
$P(2i) = 4(16) + 8(-8i) - 17(-4) - 12(2i) + 9 = 141 - 88i$.
$|P(2i)|^2 = 141^2 + 88^2 = 19881 + 7744 = 27625$.
$\frac{27625}{16} = \frac{125}{16}m$ gives $m = \frac{27625}{125} = 221$.
Answer: 221. | ||||
| 22 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A non-trivial solution $(x, y, z)$ exists only if the determinant of the coefficients is zero:
$\begin{vmatrix} \alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma \end{vmatrix} = \alpha(\beta\gamma - 6) - (\gamma - 6) + 2(2 - 2\beta) = 0$.
$\alpha\beta\gamma - 6\alpha - \gamma + 6 + 4 - 4\beta = 0$, and with $\alpha\beta\gamma = 45$: $6\alpha + 4\beta + \gamma = 55$.
Answer: 55. | ||||
| 23 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_2 = n x^{n-1}y = 135$, $T_3 = \binom{n}{2}x^{n-2}y^2 = 30$, $T_4 = \binom{n}{3}x^{n-3}y^3 = \frac{10}{3}$.
$\frac{T_3}{T_2} = \frac{n - 1}{2}\cdot\frac{y}{x} = \frac{2}{9}$ and $\frac{T_4}{T_3} = \frac{n - 2}{3}\cdot\frac{y}{x} = \frac{1}{9}$.
Dividing: $\frac{3(n - 1)}{2(n - 2)} = 2$, so $3n - 3 = 4n - 8$, $n = 5$. Then $\frac{y}{x} = \frac{1}{9}$.
$T_2 = 5x^4\cdot\frac{x}{9} = 135$ gives $x^5 = 243$, $x = 3$, $y = \frac{1}{3}$.
$6\left(125 + 9 + \frac{1}{3}\right) = 750 + 54 + 2 = 806$.
Answer: 806. | ||||
| 24 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Divide the recurrence by $3^r$: $\frac{T_r}{3^r} = \frac{T_{r-1}}{3^{r-1}} + 2^r$, with $\frac{T_1}{3} = 2$.
So $\frac{T_r}{3^r} = 2 + 2^2 + \cdots + 2^r = 2^{r+1} - 2$, i.e. $T_r = 2\cdot 6^r - 2\cdot 3^r$.
$S_n = 2\cdot\frac{6(6^n - 1)}{5} - 2\cdot\frac{3(3^n - 1)}{2} = \frac{12\cdot 6^n - 12 - 15\cdot 3^n + 15}{5} = \frac{3}{5}\left(4\cdot 6^n - 5\cdot 3^n + 1\right)$.
Comparing: $n^2 - 12n + 39 = 3$, i.e. $(n - 6)^2 = 0$, $n = 6$.
Answer: 6. | ||||
| 25 | 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 Let $I_k = \int_0^1(1 - x^7)^k\,dx$. By parts with $u = (1 - x^7)^{k+1}$, $dv = dx$:
$I_{k+1} = \left[x(1 - x^7)^{k+1}\right]_0^1 + 7(k + 1)\int_0^1 x^7(1 - x^7)^k\,dx = 7(k + 1)(I_k - I_{k+1})$.
So $(7k + 8)I_{k+1} = 7(k + 1)I_k$ and $r_k = \frac{I_k}{I_{k+1}} = \frac{7k + 8}{7k + 7}$.
$r_k - 1 = \frac{1}{7(k + 1)}$, so $\frac{1}{7(r_k - 1)} = k + 1$.
$\sum_{k=1}^{10}(k + 1) = 55 + 10 = 65$.
Answer: 65. | ||||
| 26 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A line through $P$: $y = mx + 1$ (a vertical line through $P$ meets this vertical-axis parabola only once without touching it).
Substituting: $9x^2 + (12 + 18m)x + 4 = 0$. Tangency: $(12 + 18m)^2 = 144$, so $m = 0$ or $m = -\frac{4}{3}$.
$L_1$ has direction $(1, 0)$, $L_2$ has unit direction $\left(\frac{3}{5}, -\frac{4}{5}\right)$.
Since $PQ = PR$, $QR$ is perpendicular to a bisector of the angle at $P$.
Bisector directions: $(1, 0) + \left(\frac{3}{5}, -\frac{4}{5}\right) \parallel (2, -1)$ (slope $-\frac{1}{2}$) and $(1, 0) - \left(\frac{3}{5}, -\frac{4}{5}\right) \parallel (1, 2)$ (slope 2).
So $QR$ has slope 2 or $-\frac{1}{2}$: $m_1^2 + m_2^2 = 4 + \frac{1}{4} = \frac{17}{4}$.
$16\cdot\frac{17}{4} = 68$.
Answer: 68. | ||||
| 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 $\frac{dy}{dx} = \frac{1}{2}\cdot\frac{y + 5}{x - 3}$, so $\frac{2\,dy}{y + 5} = \frac{dx}{x - 3}$ and $(y + 5)^2 = \lambda(x - 3)$.
Through $(4, -2)$: $9 = \lambda$. $C: (y + 5)^2 = 9(x - 3)$, a parabola with vertex $(3, -5)$ and $4a = 9$, $a = \frac{9}{4}$.
$(7, 1)$ is on $C$: $36 = 9 \times 4$.
Focal distance $= (x - 3) + a = 4 + \frac{9}{4} = \frac{25}{4}$.
$12d = 75$.
Answer: 75. | ||||
| 28 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Direction of the line: $(-8, 12, -16) \parallel (2, -3, 4)$. A point on it: $Q = (2 + 2t, -5 - 3t, 11 + 4t)$.
$\vec{RQ} = (1 + 2t, -12 - 3t, 5 + 4t)$ must be perpendicular to $(2, -3, 4)$:
$2 + 4t + 36 + 9t + 20 + 16t = 58 + 29t = 0$, so $t = -2$ and $Q = (-2, 1, 3)$.
$\vec{PQ} = (-12, 3, 4)$, $PQ = \sqrt{144 + 9 + 16} = 13$.
Answer: 13. | ||||
| 29 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}\times\vec{b} = (15 - 16,\ 12 + 10,\ 8 + 9) = (-1, 22, 17)$.
So $(\vec{a} + \vec{b})\times\vec{c} = (1, 8, 13) - (-1, 22, 17) = (2, -14, -4)$, with $\vec{a} + \vec{b} = (5, 1, -1)$.
With $\vec{c} = (x, y, z)$: $(5, 1, -1)\times(x, y, z) = (y + z,\ -x - 5z,\ 5y - x) = (2, -14, -4)$.
So $y + z = 2$ and $x = 5y + 4$ (the middle equation then holds too).
$\vec{a}\cdot\vec{c} = 2x - 3y + 4z = 2(5y + 4) - 3y + 4(2 - y) = 3y + 16 = 13$, so $y = -1$, $z = 3$, $x = -1$.
$\vec{b}\cdot\vec{c} = -3 - 4 - 15 = -22$, and $24 - (-22) = 46$.
Answer: 46. | ||||
| 30 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions For positive arguments $\cot^{-1}k = \tan^{-1}\frac{1}{k}$.
$\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{7/12}{11/12} = \tan^{-1}\frac{7}{11}$.
$\tan^{-1}\frac{7}{11} + \tan^{-1}\frac{1}{5} = \tan^{-1}\frac{46/55}{48/55} = \tan^{-1}\frac{23}{24}$.
$\tan^{-1}\frac{1}{n} = \frac{\pi}{4} - \tan^{-1}\frac{23}{24} = \tan^{-1}\frac{1 - 23/24}{1 + 23/24} = \tan^{-1}\frac{1}{47}$.
$n = 47$.
Answer: 47. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 Kinetic energy in terms of momentum: $K = \frac{p^2}{2m}$.
With the same $p$, $K$ is largest for the smallest mass, $\frac{m}{2}$, i.e. particle $A$.
Answer: (A). | ||||
| 32 | 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 an Atwood machine $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$.
Let $r = \frac{m_1}{m_2}$: $\frac{1 - r}{1 + r} = \frac{1}{\sqrt{2}}$, so $\sqrt{2} - \sqrt{2}r = 1 + r$ and $r = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$.
Answer: (A). | ||||
| 33 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation The body must gain potential energy from $-\frac{GMm}{R_E}$ to 0, so it needs kinetic energy $\frac{GMm}{R_E}$.
With $GM = gR_E^2$: $K = mgR_E$.
Answer: (D). | ||||
| 34 | 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 Distance while accelerating: $\frac{0 + 80}{2}t = 40t$. Distance at constant speed: $80 \times 3t = 240t$.
Average speed $= \frac{40t + 240t}{4t} = 70$ km/h.
Answer: (A). | ||||
| 35 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal velocity the net force is zero: weight = buoyancy + viscous force.
Volume $V = \frac{m}{\rho}$, buoyancy $= V\rho_0 g = mg\frac{\rho_0}{\rho}$.
Viscous force $= mg - mg\frac{\rho_0}{\rho} = mg\left(1 - \frac{\rho_0}{\rho}\right)$.
Answer: (A). | ||||
| 36 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics First law for one mole at constant pressure: $C_P\,dT = C_V\,dT + P\,dV$.
From $PV^2 = RT$ at constant $P$: $2PV\,dV = R\,dT$, so $P\,dV = \frac{R}{2V}dT$.
$C_P = C_V + \frac{R}{2V}$.
Answer: (A). | ||||
| 37 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory In a mixture both gases are at the same temperature, and $v_{rms} = \sqrt{\frac{3RT}{M}} \propto \frac{1}{\sqrt{M}}$.
$\frac{v_{He}}{v_{O_2}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2\sqrt{2}$.
Answer: (A). | ||||
| 38 | 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 By Gauss's law, just outside the shell $E = \frac{Q}{4\pi\epsilon_0 R^2} = \frac{\sigma\cdot 4\pi R^2}{4\pi\epsilon_0 R^2} = \frac{\sigma}{\epsilon_0}$; inside it is zero.
The field at the surface, as the question intends (just outside), is $\frac{\sigma}{\epsilon_0}$.
Answer: (A). | ||||
| 39 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $v = \frac{c}{\sqrt{\mu_r\epsilon_r}}$, so $\mu_r\epsilon_r = \left(\frac{3\times 10^8}{1.5\times 10^8}\right)^2 = 4$.
$\epsilon_r = \frac{4}{2} = 2$.
Answer: (B). | ||||
| 40 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity Arm $AB$: two 12 $\Omega$ resistors in series (24 $\Omega$) in parallel with 24 $\Omega$: $R_{AB} = 12\ \Omega$.
Arm $BC$: 1 $\Omega \parallel$ 1 $\Omega$ $= 0.5\ \Omega$.
Arm $AD$: $x$ in series with 12 $\Omega \parallel$ 12 $\Omega$ $= x + 6$.
Arm $DC$: 0.5 $\Omega$.
Balanced bridge: $\frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}}$, so $\frac{12}{0.5} = \frac{x + 6}{0.5}$, giving $x = 6\ \Omega$.
Answer: (B). | ||||
| 41 | 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 Least count $= \frac{1\text{ mm}}{100} = 0.01$ mm.
Diameter $= 1 + 42\times 0.01 = 1.42$ mm $= \frac{x}{50}$, so $x = 71$.
Answer: (C). | ||||
| 42 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Torque = force × distance: $[ML^2T^{-2}]$ → IV.
Magnetic field from $F = qvB$: $[MLT^{-2}]/([AT][LT^{-1}]) = [MT^{-2}A^{-1}]$ → III.
Magnetic moment = current × area: $[L^2A]$ → II.
Permeability from $B = \frac{\mu_0 I}{2\pi r}$: $[MT^{-2}A^{-1}][L]/[A] = [MLT^{-2}A^{-2}]$ → I.
A-IV, B-III, C-II, D-I.
Answer: (C). | ||||
| 43 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Biot–Savart law; the element along $x$ and the point on the $y$-axis are perpendicular ($\sin 90^\circ = 1$):
$\Delta B = \frac{\mu_0}{4\pi}\frac{I\,\Delta x}{r^2} = 10^{-7}\times\frac{10\times 0.01}{0.25} = 4\times 10^{-8}$ T.
Answer: (A). | ||||
| 44 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | StatementsEasy | |
A easy statements question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Statement I: $I = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}$ is largest when $X_L = X_C$, i.e. at resonance. True.
Statement II: for the same resistance $R$, $Z = \sqrt{R^2 + (X_L - X_C)^2} \ge R$, so the current $\frac{V}{R}$ in the purely resistive circuit is never less than $\frac{V}{Z}$. True.
Answer: (A). | ||||
| 45 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $T = 2\pi\sqrt{\frac{m}{k}}$ gives $k = \frac{4\pi^2 m}{T^2}$.
$\frac{\Delta k}{k} = \frac{\Delta m}{m} - 2\frac{\Delta T}{T} = (-1\%) - 2(+2\%) = -5\%$.
(The maximum-error rule gives the same: $1\% + 2\times 2\% = 5\%$.)
Answer: (B) 5%. | ||||
| 46 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | StatementsEasy | |
A easy statements question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Reflection, diffraction, interference and polarisation are all explained by the wave theory of light.
The photoelectric effect (instant emission, threshold frequency, kinetic energy independent of intensity) needs the photon picture and is not explained by the wave nature.
Answer: (C) C only. | ||||
| 47 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $h\nu = \phi + eV_0$, so $\phi = 2.48 - 0.5 = 1.98$ eV.
Answer: (A). | ||||
| 48 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Shortest wavelength = series limit ($n_2 \to \infty$).
Balmer: $\frac{1}{\lambda_B} = R\cdot\frac{1}{4}$; Lyman: $\frac{1}{\lambda_L} = R$.
$\frac{\lambda_B}{\lambda_L} = 4$, i.e. $4 : 1$.
Answer: (D). | ||||
| 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 From the circuit: $A$ goes through a NOT gate to one input of the OR gate; $A$ and $B$ go to an AND gate whose output is the other OR input. So $Y = \bar{A} + AB$.
$A = 0$: $\bar{A} = 1$, $Y = 1$ for $B = 0, 1$.
$A = 1, B = 0$: $Y = 0 + 0 = 0$. $A = 1, B = 1$: $Y = 0 + 1 = 1$.
Y column: 1, 1, 0, 1 — the second table.
Answer: (B). | ||||
| 50 | 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 Fractional loss $= 1 - \left(\frac{40}{100}\right)^2 = 1 - 0.16 = 0.84$, i.e. 84%.
Answer: (B). | ||||
| 51 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\mu = \frac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$ with $\delta_m = A$: $\mu = \frac{\sin A}{\sin\frac{A}{2}} = 2\cos\frac{A}{2}$.
$2\cos\frac{A}{2} = \sqrt{3}$, so $\frac{A}{2} = 30^\circ$ and $A = 60^\circ$.
Answer: 60. | ||||
| 52 | 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.
$v_{max} = A\omega = 0.06\times 2 = 0.12$ m/s $= 12$ cm/s.
Answer: 12. | ||||
| 53 | 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 $\vec{A}\cdot(\vec{B}\times\vec{C}) = \begin{vmatrix} -x & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix}$
$= -x(12 + 3) + 6(-3 + 24) - 2(1 + 32) = -15x + 126 - 66 = 60 - 15x$.
Setting it to zero: $x = 4$.
Answer: 4. | ||||
| 54 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Volume is conserved: $\frac{4}{3}\pi R^3 = 1000\cdot\frac{4}{3}\pi r^3$, so $R = 10r$.
Surface energy $\propto$ total area: $\frac{1000\cdot 4\pi r^2}{4\pi R^2} = \frac{1000r^2}{100r^2} = 10 = \frac{10}{x}$.
$x = 1$.
Answer: 1. | ||||
| 55 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current With dc the inductor shows only its resistance: $R = \frac{100}{5} = 20\ \Omega$.
With ac: $Z = \sqrt{20^2 + (20\sqrt{3})^2} = 40\ \Omega$; peak current $= \frac{200}{40} = 5$ A, rms current $\frac{5}{\sqrt{2}}$ A.
Power $= I_{rms}^2 R = \frac{25}{2}\times 20 = 250$ W.
Answer: 250. | ||||
| 56 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion With no external torque, $L = I\omega$ is conserved; $I = \frac{2}{5}MR^2 \propto R^2$ and $\omega \propto \frac{1}{T}$, so $T \propto R^2$.
$T' = 24\times\left(\frac{3}{4}\right)^2 = 24\times\frac{9}{16} = 13.5$ h $= 13$ hours 30 minutes.
Answer: 13. | ||||
| 57 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $M = NIA = 200\times 100\times 10^{-6}\times 2.5\times 10^{-4} = 5\times 10^{-6}$ A m$^2$.
$W = MB(\cos 0^\circ - \cos 90^\circ) = 5\times 10^{-6}\times 1 = 5\times 10^{-6}$ J $= 5\ \mu$J.
Answer: 5. | ||||
| 58 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Volume is unchanged: area becomes $\frac{1}{4}$, so length becomes 4 times.
$R = \frac{\rho l}{A}$: $R' = \rho\frac{4l}{A/4} = 16R$.
Answer: 16. | ||||
| 59 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Each sheet gives a field of magnitude $\frac{|\sigma_i|}{2\epsilon_0}$, away from a positive sheet and towards a negative one. $P$ lies between $x = a$ and $x = 3a$.
Sheet $-\sigma$ at $x = -a$ (left of $P$): field towards it, along $-x$: $\frac{\sigma}{2\epsilon_0}$.
Sheet $-2\sigma$ at $x = a$ (left of $P$): along $-x$: $\frac{2\sigma}{2\epsilon_0}$.
Sheet $+\sigma$ at $x = 3a$ (right of $P$): away from it, along $-x$: $\frac{\sigma}{2\epsilon_0}$.
Total: $\frac{\sigma}{2\epsilon_0}(1 + 2 + 1) = \frac{2\sigma}{\epsilon_0}$ along $-x$. So $x = 2$.
Answer: 2. | ||||
| 60 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $r_n = n^2a_0$: $n^2 = \frac{8.48}{0.529} \approx 16$, so $n = 4$.
$E_n = \frac{E}{n^2} = \frac{E}{16}$, so $x = 16$.
Answer: 16. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | SolutionsConcentration terms | Class 12: Solutions | Single correctEasy | |
A easy single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions 3 molal: 3 mol NaOH (120 g) in 1000 g water, so the solution weighs 1120 g.
Its volume $= \frac{1120\text{ g}}{1.12\text{ g mL}^{-1}} = 1000$ mL $= 1$ L.
Molarity $= \frac{3}{1} = 3.0$ M, so $x = 3.0$.
Answer: (C). | ||||
| 62 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{SO_2Cl_2}$: S has four bond pairs and no lone pair — tetrahedral (III).
NO: 11 valence electrons, one unpaired electron in $\pi^*$ — paramagnetic (I).
$\mathrm{NO_2^-}$: 18 valence electrons, all paired — diamagnetic (II).
$\mathrm{I_3^-}$: central I has two bond pairs and three lone pairs (sp$^3$d) — linear (IV).
A-III, B-I, C-II, D-IV.
Answer: (B). | ||||
| 63 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Equal numbers of molecules at 1 atm total: each partial pressure is $\frac{1}{3}$ atm.
$K_p = \frac{p_{HI}^2}{p_{H_2}\,p_{I_2}} = \frac{(1/3)^2}{(1/3)(1/3)} = 1$.
($\Delta n_g = 0$, so R and T do not enter.) $1 = x\times 10^{-1}$ gives $x = 10$.
Answer: (B). | ||||
| 64 | Solid StateClassification of solids | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on classification of solids, from Solid State. NCERT: No matching NCERT chapter Germanium and silicon are the standard elemental semiconductors, and copper oxide ($\mathrm{Cu_2O}$) is a well-known semiconducting oxide.
Graphite has delocalised $\pi$ electrons across its layers and conducts like a metal (a conductor, not a semiconductor).
Answer: (B) Graphite. | ||||
| 65 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry At infinite dilution a weak electrolyte is already completely dissociated, and its molar conductivity has reached its limiting value $\Lambda_m^\circ$.
Adding more water cannot raise it further, so it remains the same; and the conductance of such a dilute solution is so small that it cannot be measured accurately.
Answer: (C). | ||||
| 66 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | StatementsMedium | |
A medium statements question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Electron affinity is negative when energy must be supplied to add the electron (positive electron gain enthalpy).
Be ($2s^2$, filled subshell) and N ($2p^3$, half-filled subshell) resist an extra electron: endothermic.
$\mathrm{O \to O^{2-}}$: the second electron is added against the repulsion of $\mathrm{O^-}$, and the overall process is endothermic.
Na ($3s^1$) and Al ($3p^1$) release energy on gaining an electron: positive electron affinity.
So A, B and C.
Answer: (A). | ||||
| 67 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: gallium melts at about 303 K and stays liquid up to about 2500 K, so it is used in high-temperature thermometers. True.
Statement II: below 303 K gallium is solid, so a gallium thermometer cannot read 256 K. False.
Answer: (C). | ||||
| 68 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{SF_4}$: 4 bond pairs + 1 lone pair — see-saw (III).
$\mathrm{BrF_3}$: 3 bond pairs + 2 lone pairs — bent T-shape (IV).
$\mathrm{BrO_3^-}$: 3 bond pairs + 1 lone pair — pyramidal (II).
$\mathrm{NH_4^+}$: 4 bond pairs — tetrahedral (I).
A-III, B-IV, C-II, D-I.
Answer: (C). | ||||
| 69 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The lanthanoids are Ce (58) to Lu (71): Eu (63), Er (68), Tb (65), Yb (70) and Lu (71) are all lanthanoids.
Cm (curium, 96) is an actinoid.
So only 1 element is not a lanthanoid.
Answer: (B). | ||||
| 70 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{sp^3}$ — tetrahedral (III); $\mathrm{dsp^2}$ — square planar (IV); $\mathrm{sp^3d}$ — trigonal bipyramidal (I); $\mathrm{sp^3d^2}$ — octahedral (II).
A-III, B-IV, C-I, D-II.
Answer: (A). | ||||
| 71 | 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 Wavenumber absorbed $\propto$ crystal field splitting $\Delta$.
B, $\mathrm{[Co(CN)_6]^{3-}}$: six strong-field $\mathrm{CN^-}$ ligands — largest $\Delta$.
C, $\mathrm{[Co(NH_3)_5(H_2O)]^{3+}}$ vs A, $\mathrm{[CoCl(NH_3)_5]^{2+}}$: same $\mathrm{Co^{3+}}$, and $\mathrm{H_2O}$ is a stronger ligand than $\mathrm{Cl^-}$, so C > A.
D, $\mathrm{[Cu(H_2O)_4]^{2+}}$: a $\mathrm{M^{2+}}$ ion with weak-field water ligands, absorbing in the red — smallest.
D < A < C < B.
Answer: (A). | ||||
| 72 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Group reagents of cation analysis:
$\mathrm{NH_4Cl + NH_4OH}$ — group III, precipitates $\mathrm{Al(OH)_3}$: A-III.
$\mathrm{NH_4OH}$ + carbonate — group V, precipitates $\mathrm{SrCO_3}$: B-IV.
$\mathrm{NH_4OH + NH_4Cl + H_2S}$ — group IV, precipitates MnS: C-I.
Dilute HCl — group I, precipitates $\mathrm{PbCl_2}$: D-II.
A-III, B-IV, C-I, D-II.
Answer: (B). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsMedium | |
A medium statements question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A: glycerol decomposes at its boiling point, so it is distilled under reduced pressure. Correct.
B: steam distillation suits aniline because it is steam-volatile and immiscible with water; the reason given ('miscible') is wrong. Incorrect.
C: ethanol and water form an azeotrope (95.6% ethanol), so ordinary fractional distillation cannot separate them; azeotropic distillation (adding a third component) is used. Correct.
D: the mixed melting point test — a sample mixed with the pure compound melts sharply at the same temperature only if it is the same pure compound. Correct.
So A, C, D.
Answer: (C). | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques X is $\mathrm{C_6H_5{-}NH{-}CO{-}C_6H_{11}}$ (N-phenylcyclohexanecarboxamide), $\mathrm{C_{13}H_{17}NO}$.
Metamers have the same molecular formula and the same functional group, with different groups on either side of it.
(A) $\mathrm{C_6H_{11}{-}NH{-}CO{-}C_6H_5}$ (N-cyclohexylbenzamide): $\mathrm{C_{13}H_{17}NO}$, an amide with the phenyl and cyclohexyl groups exchanged — a metamer.
(B) has two cyclohexyl groups ($\mathrm{C_{13}H_{23}NO}$) and (C) two phenyl groups ($\mathrm{C_{13}H_{11}NO}$): different formulae.
(D) is an aldehyde and a secondary amine — a functional isomer, not a metamer.
Answer: (A). | ||||
| 75 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A sulphonic acid is $\mathrm{R{-}SO_3H}$, i.e. $\mathrm{R{-}S(=O)_2{-}OH}$: the functional group is $\mathrm{-SO_3H}$.
Option (D), $\mathrm{-S(=O){-}OH}$, is the sulphinic acid group ($\mathrm{-SO_2H}$).
Answer: (C). | ||||
| 76 | Haloalkanes and HaloarenesEnvironmental effects of polyhalogen compounds | Class 12: Haloalkanes and Haloarenes | Matching listEasy | |
A easy matching list question on environmental effects of polyhalogen compounds, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Iodoform — antiseptic (III); carbon tetrachloride — fire extinguisher (I); CFCs (freons) — refrigerants (IV); DDT — insecticide (II).
A-III, B-I, C-IV, D-II.
Answer: (C). | ||||
| 77 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | StatementsEasy | |
A easy statements question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Statement I: picric acid is 2,4,6-trinitrophenol; 2,4,6-trinitrotoluene is TNT. Incorrect.
Statement II: phenol is first sulphonated to phenol-2,4-disulphonic acid, which conc. $\mathrm{HNO_3}$ converts to 2,4,6-trinitrophenol (the sulphonic groups are replaced by nitro groups). Correct.
Answer: (D). | ||||
| 78 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Phenol with $\mathrm{CHCl_3}$ and aqueous NaOH: the phenoxide ion attacks dichlorocarbene ($:\mathrm{CCl_2}$) at the ortho position, giving sodium 2-(dichloromethyl)phenoxide.
This intermediate is hydrolysed by the alkali ($\mathrm{-CHCl_2 \to -CHO}$) and acidified to give salicylaldehyde.
The intermediate is the phenoxide carrying $\mathrm{-CHCl_2}$ ortho, option (D).
Answer: (D). | ||||
| 79 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Alkyl groups push electron density onto the carbonyl carbon (+I) and crowd the approach of the nucleophile.
Formaldehyde, HCHO, has no alkyl group: its carbonyl carbon is the most electrophilic and least hindered.
Answer: (A). | ||||
| 80 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules The four DNA bases are adenine, guanine, cytosine and thymine.
(A) 6-aminopurine — adenine, correct.
(B) a 2-amino-6-oxopurine carrying a $\mathrm{CH_3}$ at C-8 — guanine has H at C-8, so this structure is incorrect.
(C) 4-amino-2-oxopyrimidine — cytosine, correct.
(D) 5-methyl-2,4-dioxopyrimidine — thymine, correct.
Answer: (B). | ||||
| 81 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalHard | |
A hard numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Frequency of the electron wave $= \frac{v}{\lambda}$ with $\lambda = \frac{h}{mv}$, so $f = \frac{mv^2}{h} = \frac{2K}{h}$.
In the first Bohr orbit the kinetic energy equals $|E_1| = R_H = 2.18\times 10^{-18}$ J.
$f = \frac{2\times 2.18\times 10^{-18}}{6.6\times 10^{-34}} = 6.606\times 10^{15}$ Hz $= 660.6\times 10^{13}$ Hz.
(Equivalently $f = \frac{v}{2\pi r}$, the number of revolutions per second, since the orbit holds exactly one wavelength.)
Nearest integer: 661.
Answer: 661. | ||||
| 82 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Hydrogen bonding needs H bonded to N, O or F.
$\mathrm{CH_3OH}$, $\mathrm{H_2O}$, HF, $\mathrm{NH_3}$: intermolecular hydrogen bonding — 4.
o-Nitrophenol: intramolecular hydrogen bond between the OH and the adjacent $\mathrm{NO_2}$ — 1.
$\mathrm{C_2H_6}$ and $\mathrm{C_6H_6}$: no H on N, O or F — none.
Total 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 Irreversible adiabatic expansion: $q = 0$, $\Delta U = w = -P_{ext}(V_2 - V_1)$.
$V_1 = \frac{nR\cdot 298}{5}$ and $V_2 = 2V_1$, so $w = -1\times V_1 = -\frac{298nR}{5}$.
$n\cdot\frac{5}{2}R(T_2 - 298) = -\frac{298}{5}nR$, so $T_2 - 298 = -\frac{2\times 298}{25} = -23.84$.
$T_2 = 274.16$ K $\approx 274$ K.
Answer: 274. | ||||
| 84 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Degree of dissociation: $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.2\times 10^{-5}}{0.03}} = \sqrt{4\times 10^{-4}} = 0.02$.
van't Hoff factor $i = 1 + \alpha = 1.02$.
$\pi = iCRT = 1.02\times 0.03\times 0.083\times 300 = 1.02\times 0.747 = 0.762$ bar $= 76.2\times 10^{-2}$ bar.
Nearest integer: 76.
Answer: 76. | ||||
| 85 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics First order: $t = \frac{2.303}{k}\log\frac{a}{a - x}$.
99.9%: $\log 1000 = 3$; 90%: $\log 10 = 1$.
Ratio $= 3$.
Answer: 3. | ||||
| 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 CrO is basic, $\mathrm{Cr_2O_3}$ amphoteric, $\mathrm{CrO_3}$ acidic.
CrO: $\mathrm{Cr^{2+}}$, $3d^4$, $n = 4$: $\mu = \sqrt{24} = 4.899$ BM.
$\mathrm{Cr_2O_3}$: $\mathrm{Cr^{3+}}$, $3d^3$, $n = 3$: $\mu = \sqrt{15} = 3.873$ BM.
Sum $= 8.772$ BM $= 877.2\times 10^{-2}$ BM.
Nearest integer: 877.
Answer: 877. | ||||
| 87 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In $\mathrm{KMnO_4}$, Mn is +7 ($3d^0$): $\mu = 0$.
In acidic medium permanganate is reduced to $\mathrm{Mn^{2+}}$ ($3d^5$, 5 unpaired electrons): $\mu = \sqrt{35} = 5.92$ BM.
Difference $= 5.92 \approx 6$ BM.
Answer: 6. | ||||
| 88 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Na in liquid $\mathrm{NH_3}$ (Birch-type reduction) converts but-2-yne to trans-but-2-ene.
Cold dilute $\mathrm{KMnO_4}$ (Baeyer's reagent, 273 K) hydroxylates the double bond: butane-2,3-diol, $\mathrm{CH_3CH(OH)CH(OH)CH_3}$.
P has 2 oxygen atoms.
Answer: 2. | ||||
| 89 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | NumericalHard | |
A hard numerical question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Both routes begin with $\mathrm{Br_2}$ adding to cyclohexene: trans-1,2-dibromocyclohexane.
Route to B: alc. KOH (3 eq.) removes two HBr by double dehydrohalogenation, giving cyclohexa-1,3-diene — two C=C, 4 $\pi$ electrons.
Route to A: one equivalent of sodium propargyl oxide, $\mathrm{HC{\equiv}C{-}CH_2O^-Na^+}$, replaces one bromine (Williamson ether synthesis), giving 1-bromo-2-(prop-2-ynyloxy)cyclohexane — one C≡C, 4 $\pi$ electrons.
Total $= 4 + 4 = 8$.
Answer: 8. | ||||
| 90 | AminesDiazonium salts | Class 12: Amines | NumericalEasy | |
A easy numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines Aniline, $\mathrm{C_6H_5NH_2}$, M = 93: 9.3 g = 0.1 mol.
Diazotisation gives benzenediazonium chloride, which couples with phenol to give the orange dye p-hydroxyazobenzene, $\mathrm{C_6H_5{-}N{=}N{-}C_6H_4{-}OH}$ ($\mathrm{C_{12}H_{10}N_2O}$, M = 198).
1 mol aniline gives 1 mol dye: $0.1\times 198 = 19.8$ g.
Nearest integer: 20.
Answer: 20. | ||||
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