JEE (Main) · 2024
JEE (Main) 2024 — 8 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
- Applications of Derivatives3
- Integral Calculus3
- Complex Numbers2
7 easy · 21 medium · 2 hard
Physics
30 Q
- Physical World and Measurement2
- Kinematics2
- Laws of Motion2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- Coordination Compounds3
- Organic Chemistry: Some Basic Principles and Techniques3
- Chemical Bonding and Molecular Structure2
18 easy · 10 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 FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11$, so $A = \{2, 3, 5, 7, 11\}$.
$x = 2$: $\left[\frac{8}{5}\right] = 1$, $\log_2 5 \approx 2.32$, $f = 2$.
$x = 3$: $\left[\frac{27}{5}\right] = 5$, $\log_2 14 \approx 3.81$, $f = 3$.
$x = 5$: $\left[\frac{125}{5}\right] = 25$, $\log_2 50 \approx 5.64$, $f = 5$.
$x = 7$: $\left[\frac{343}{5}\right] = 68$, $\log_2 117 \approx 6.87$, $f = 6$.
$x = 11$: $\left[\frac{1331}{5}\right] = 266$, $\log_2 387 \approx 8.60$, $f = 8$.
The range of $f$ is $\{2, 3, 5, 6, 8\}$: 5 elements.
One-to-one functions from a 5-element set to a 5-element set: $5! = 120$.
Answer: (C). | ||||
| 2 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $w = z + 2$, so $|w| = 1$ and $\frac{1}{w} = \bar{w}$.
$\frac{z+1}{z+2} = \frac{w-1}{w} = 1 - \frac{1}{w} = 1 - \bar{w}$.
$\mathrm{Im}(1 - \bar{w}) = -\mathrm{Im}(\bar{w}) = \mathrm{Im}(w) = \frac{1}{5}$.
Then $(\mathrm{Re}\,w)^2 = 1 - \frac{1}{25} = \frac{24}{25}$, so $|\mathrm{Re}\,w| = \frac{2\sqrt{6}}{5}$.
$\mathrm{Re}(\overline{z+2}) = \mathrm{Re}(\bar{w}) = \mathrm{Re}(w)$, so the required value is $\frac{2\sqrt{6}}{5}$.
Answer: (A). | ||||
| 3 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $t = 8^x > 0$: $t^2 - 16t + 48 = 0$, so $(t - 4)(t - 12) = 0$ and $t = 4$ or $t = 12$.
The solutions are $x_1 = \log_8 4$ and $x_2 = \log_8 12$.
$x_1 + x_2 = \log_8 48 = \log_8(8\cdot 6) = 1 + \log_8 6$.
Answer: (D). | ||||
| 4 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctHard | |
A hard single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The given relation is $A^3 - 4A^2 + A + 21I = O$. Matching it with the characteristic equation $\lambda^3 - (\mathrm{tr}\,A)\lambda^2 + (\text{sum of principal minors})\lambda - \det A = 0$ (which $A$ satisfies):
$\mathrm{tr}\,A = 2 + 3 + b = 4$, so $b = -1$.
Sum of principal $2\times 2$ minors: $(6 - a) + 2b + (3b - 5) = 1 - a + 5b = 1$, so $a = 5b = -5$.
Check: $\det A = 2(3b - 5) - a(b) = 2(-8) - (-5)(-1) = -16 - 5 = -21$, as required ($-\det A = 21$).
Direct multiplication with $a = -5$, $b = -1$ confirms $A^3 - 4A^2 + A + 21I = O$.
$2a + 3b = -10 - 3 = -13$.
Answer: (B). | ||||
| 5 | Straight LinesSection formula | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on section formula, from Straight Lines. NCERT: Class 11: Straight Lines Take $B = (b, 14 - 4b)$ on AB and $C = \left(c, \frac{3c - 5}{2}\right)$ on AC.
The point dividing BC in the ratio $2:1$ is $\frac{B + 2C}{3} = \left(2, -\frac{4}{3}\right)$:
$b + 2c = 6$ and $(14 - 4b) + (3c - 5) = -4$, i.e. $-4b + 3c = -13$.
With $b = 6 - 2c$: $-24 + 8c + 3c = -13$, so $c = 1$ and $b = 4$.
$B = (4, -2)$, $C = (1, -1)$. Slope of BC $= \frac{-1 + 2}{1 - 4} = -\frac{1}{3}$.
BC: $y + 1 = -\frac{1}{3}(x - 1)$, i.e. $x + 3y + 2 = 0$.
Answer: (A). | ||||
| 6 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $a = 42 - 5b \geq 1$ gives $b = 1, 2, \ldots, 8$, so $m = 8$.
$1! = 1$: $i^1 = i$. $2! = 2$: $i^2 = -1$. $3! = 6$: $i^6 = -1$. For $n \geq 4$, $n!$ is a multiple of 4, so $i^{n!} = 1$.
Sum $= (1 - i) + 2 + 2 + 0 + 0 + 0 + 0 + 0 = 5 - i$, so $x = 5$, $y = -1$.
$m + x + y = 8 + 5 - 1 = 12$.
Answer: (A). | ||||
| 7 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | StatementsEasy | |
A easy statements question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = -\sin x - 1 \leq 0$ for all $x$, and it is zero only at isolated points, so $f$ is decreasing on the whole real line. (S2) is false.
$f(0) = 2 > 0$ and $f(\pi) = -1 - \pi + 1 = -\pi < 0$; a strictly decreasing continuous function crosses zero exactly once in $[0, \pi]$. (S1) is true.
Answer: (B). | ||||
| 8 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = \frac{2}{3}(x - 2)^{-1/3}(2x + 1) + 2(x - 2)^{2/3} = \frac{(2)(2x + 1) + 6(x - 2)}{3(x - 2)^{1/3}} = \frac{10(x - 1)}{3(x - 2)^{1/3}}$.
$f'(x) = 0$ at $x = 1$, and $f'(x)$ does not exist at $x = 2$ (where $f$ is defined).
Critical points: $x = 1$ and $x = 2$, i.e. 2.
Answer: (C). | ||||
| 9 | 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 $f'(x) = (12\cos^2 x + 6\sqrt{3}\cos x)(-\sin x) = -6\sin x\cos x\,(2\cos x + \sqrt{3})$.
Zeros in $(0, 2\pi)$: $\cos x = 0$ ($\frac{\pi}{2}, \frac{3\pi}{2}$), $\cos x = -\frac{\sqrt{3}}{2}$ ($\frac{5\pi}{6}, \frac{7\pi}{6}$), $\sin x = 0$ ($\pi$).
Sign of $f'$: $(0, \frac{\pi}{2})$ −; $(\frac{\pi}{2}, \frac{5\pi}{6})$ +; $(\frac{5\pi}{6}, \pi)$ −; $(\pi, \frac{7\pi}{6})$ +; $(\frac{7\pi}{6}, \frac{3\pi}{2})$ −; $(\frac{3\pi}{2}, 2\pi)$ +.
$f'$ changes from + to − at $x = \frac{5\pi}{6}$ and $x = \frac{7\pi}{6}$: two local maxima (the minima are at $\frac{\pi}{2}, \pi, \frac{3\pi}{2}$).
Answer: (B). | ||||
| 10 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t = 1 - \cot x$, $dt = \mathrm{cosec}^2 x\,dx$: $I = \int \frac{6}{t^2}\,dt = -\frac{6}{1 - \cot x} + C = \frac{6\sin x}{\cos x - \sin x} + C$.
$I(0) = 0 + C = 3$, so $C = 3$.
$I\left(\frac{\pi}{12}\right) = \frac{6\tan 15^\circ}{1 - \tan 15^\circ} + 3$ with $\tan 15^\circ = 2 - \sqrt{3}$:
$\frac{2 - \sqrt{3}}{\sqrt{3} - 1} = \frac{(2 - \sqrt{3})(\sqrt{3} + 1)}{2} = \frac{\sqrt{3} - 1}{2}$.
$I\left(\frac{\pi}{12}\right) = 3(\sqrt{3} - 1) + 3 = 3\sqrt{3}$.
Answer: (B). | ||||
| 11 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals By parts, $I_n = \left[x(1 - x^k)^n\right]_0^1 + \int_0^1 x\cdot nk\,x^{k-1}(1 - x^k)^{n-1}\,dx = nk\int_0^1 x^k(1 - x^k)^{n-1}\,dx$.
Writing $x^k = 1 - (1 - x^k)$: $I_n = nk(I_{n-1} - I_n)$, so $I_n(1 + nk) = nk\,I_{n-1}$.
For $n = 21$: $\frac{I_{20}}{I_{21}} = \frac{1 + 21k}{21k} = \frac{148}{147}$.
$147 + 147\cdot 21k = 148\cdot 21k$, so $21k = 147$ and $k = 7$.
Answer: (D). | ||||
| 12 | Differential EquationsOrder and degree | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on order and degree, from Differential Equations. NCERT: Class 12: Differential Equations $\int_0^a f(x)\,dx = e^{-a} + 4a^2 + a - 1$; differentiating in $a$: $f(a) = -e^{-a} + 8a + 1$.
$y = c_1 f(x) + c_2$ gives $y' = c_1 f'(x) = c_1(e^{-x} + 8)$ and $y'' = c_1 f''(x) = -c_1 e^{-x}$.
Eliminating $c_1$: $\frac{y''}{y'} = \frac{-e^{-x}}{e^{-x} + 8} = \frac{-1}{1 + 8e^x}$.
So $(8e^x + 1)\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$.
Answer: (C). | ||||
| 13 | 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 Separating: $\frac{dy}{1 + y^2} = -\frac{e^{\tan x}\sec^2 x}{1 + e^{2\tan x}}\,dx$.
With $u = e^{\tan x}$, $du = e^{\tan x}\sec^2 x\,dx$: $\tan^{-1} y = -\tan^{-1}(e^{\tan x}) + C$.
At $x = 0$, $y = 1$: $\frac{\pi}{4} = -\frac{\pi}{4} + C$, so $C = \frac{\pi}{2}$.
At $x = \frac{\pi}{4}$: $\tan^{-1} y = \frac{\pi}{2} - \tan^{-1} e = \tan^{-1}\frac{1}{e}$, so $y = \frac{1}{e}$.
Answer: (B). | ||||
| 14 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium 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 $L_1$: point $(2, 1, 3)$, direction $\vec{b}_1 = (1, -3, 4)$. $L_2$: point $(2, 3, 5)$, direction $\vec{b}_2 = (2, 3, 1)$.
$\vec{b}_1 \times \vec{b}_2 = (-3 - 12,\ 8 - 1,\ 3 + 6) = (-15, 7, 9)$, $|\vec{b}_1 \times \vec{b}_2| = \sqrt{225 + 49 + 81} = \sqrt{355}$.
Joining vector $(0, 2, 2)$; $(0, 2, 2)\cdot(-15, 7, 9) = 14 + 18 = 32$.
Shortest distance $= \frac{32}{\sqrt{355}}$; $\gcd(32, 355) = 1$, so $m + n = 32 + 355 = 387$.
Answer: (D). | ||||
| 15 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections $(6, 6) = \frac{1\cdot(\alpha, \beta) + 2\cdot\left(8, \frac{15}{2}\right)}{3}$, so $(\alpha, \beta) = (18 - 16, 18 - 15) = (2, 3)$ and $\alpha + \beta = 5$.
The circles touch at $(6, 6)$, so $r_1 = \sqrt{(6 - 2)^2 + (6 - 3)^2} = 5$ and $r_2 = \sqrt{(8 - 6)^2 + \left(\frac{15}{2} - 6\right)^2} = \frac{5}{2}$ (indeed $r_1 : r_2 = 2 : 1$).
$4(r_1^2 + r_2^2) = 4\left(25 + \frac{25}{4}\right) = 125$.
$(\alpha + \beta) + 4(r_1^2 + r_2^2) = 5 + 125 = 130$.
Answer: (C). | ||||
| 16 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The transverse axis is the $y$-axis: $e^2 = 1 + \frac{a^2}{b^2} = 3$, so $a^2 = 2b^2$; latus rectum $\frac{2a^2}{b} = 4b = 4\sqrt{3}$, so $b^2 = 3$, $a^2 = 6$.
$(\alpha, 6)$ on $H$: $\frac{36}{3} - \frac{\alpha^2}{6} = 1$, so $\alpha^2 = 66$.
Foci $(0, \pm be) = (0, \pm 3)$. Distances: $\sqrt{66 + 9} = 5\sqrt{3}$ and $\sqrt{66 + 81} = 7\sqrt{3}$.
$\beta = 5\sqrt{3}\cdot 7\sqrt{3} = 105$, and $\alpha^2 + \beta = 66 + 105 = 171$.
Answer: (C). | ||||
| 17 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $z = \gamma\cos\phi$ and $OQ = \gamma\sin\phi$, so $x = \gamma\sin\phi\cos\theta$, $y = \gamma\sin\phi\sin\theta$.
Distance from the $x$-axis $= \sqrt{y^2 + z^2} = \sqrt{\gamma^2 - x^2} = \gamma\sqrt{1 - \sin^2\phi\cos^2\theta}$.
Answer: (A). | ||||
| 18 | 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 Obtuse angle needs $\vec{a}\cdot\vec{b} < 0$: $\alpha t^2 - 12 + 6\alpha t < 0$ for all $t$.
$\alpha = 0$: $-12 < 0$ for all $t$, so $\alpha = 0$ works.
$\alpha \neq 0$: need $\alpha < 0$ and discriminant $36\alpha^2 + 48\alpha < 0$, i.e. $-\frac{4}{3} < \alpha < 0$.
(The vectors are never antiparallel for these $\alpha$: that would need $-2k = 6$, $k = -3$, and then $\alpha = -3$.)
The set is $\left(-\frac{4}{3}, 0\right]$.
Answer: (C). | ||||
| 19 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability The pairs are $(x, 24 - x)$, $x = 1, 2, \ldots, 23$: 23 equally likely ordered pairs.
The greatest product is $12\cdot 12 = 144$; we need $x(24 - x) \geq 108$, i.e. $x^2 - 24x + 108 \leq 0$, $(x - 6)(x - 18) \leq 0$, $6 \leq x \leq 18$: 13 values.
Probability $= \frac{13}{23}$, so $n - m = 23 - 13 = 10$.
Answer: (D). | ||||
| 20 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions In the third quadrant $\cos x = -\frac{4}{5}$ and $\tan x = \frac{3}{4}$.
$80\left(\frac{9}{16} + \frac{4}{5}\right) = 45 + 64 = 109$.
Answer: (A). | ||||
| 21 | 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 Let $s = \sin^2\theta \in [0, 1]$: $f = \frac{s^2 + 3(1 - s)}{s^2 + 1 - s} = 1 + \frac{2(1 - s)}{s^2 - s + 1}$.
With $u = 1 - s \in [0, 1]$, $s^2 - s + 1 = u^2 - u + 1$ and $g(u) = \frac{u}{u^2 - u + 1}$ has $g'(u) = \frac{1 - u^2}{(u^2 - u + 1)^2} \geq 0$ on $[0, 1]$.
So $g$ runs from $g(0) = 0$ to $g(1) = 1$, and $f$ runs over $[1, 3]$: $\alpha = 1$, $\beta = 3$.
Sum of the G.P. $= \frac{64}{1 - \frac{1}{3}} = 96$.
Answer: 96. | ||||
| 22 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\mathrm{tr}\,A = 3$, $\det A = 3$, so the eigenvalues satisfy $\lambda^2 - 3\lambda + 3 = 0$: $\lambda = \frac{3 \pm i\sqrt{3}}{2} = \sqrt{3}\,e^{\pm i\pi/6}$.
$\mathrm{tr}(A^{13}) = \lambda_1^{13} + \lambda_2^{13} = 2(\sqrt{3})^{13}\cos\frac{13\pi}{6} = 2\cdot 3^{13/2}\cdot\frac{\sqrt{3}}{2} = 3^7$.
(Equivalently, $A^2 = 3A - 3I$ gives $t_{k+1} = 3t_k - 3t_{k-1}$ for $t_k = \mathrm{tr}\,A^k$, reaching $t_{13} = 2187$.)
$n = 7$.
Answer: 7. | ||||
| 23 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Total 3-digit numbers without repetition: $5\cdot 4\cdot 3 = 60$.
Divisible by 3 when the digit sum is: $\{2, 3, 4\}$ (9), $\{2, 3, 7\}$ (12), $\{3, 4, 5\}$ (12), $\{3, 5, 7\}$ (15) — the other six triples give 10, 11, 13, 14, 14, 16.
These 4 triples give $4\cdot 3! = 24$ numbers.
Not divisible by 3: $60 - 24 = 36$.
Answer: 36. | ||||
| 24 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $\sum {}^nC_r = 2^n$, $\sum r\,{}^nC_r = n2^{n-1}$, $\sum r^2\,{}^nC_r = n(n + 1)2^{n-2}$.
$\alpha = 4n(n + 1)2^{n-2} + 2n2^{n-1} + 2^n = 2^n\left[n(n + 1) + n + 1\right] = 2^n(n + 1)^2$.
$\sum \frac{{}^nC_r}{r + 1} = \frac{2^{n+1} - 1}{n + 1}$, so $\beta = \frac{2^{n+1}}{n + 1}$.
$\frac{2\alpha}{\beta} = \frac{2^{n+1}(n + 1)^2(n + 1)}{2^{n+1}} = (n + 1)^3$.
$140 < (n + 1)^3 < 281$ gives $(n + 1)^3 = 216$, $n = 5$.
Answer: 5. | ||||
| 25 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The first $k$ rows hold $1 + 2 + \cdots + k = \frac{k(k + 1)}{2}$ numbers, so row $k$ ends with $\frac{k(k + 1)}{2}$.
$\frac{102\cdot 103}{2} = 5253 < 5310 \leq 5356 = \frac{103\cdot 104}{2}$.
So 5310 is in row 103.
Answer: 103. | ||||
| 26 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $P(x) = \prod_{k=1}^{10}(\cos kx)^{1/k}$. Then $\ln P = \sum_{k=1}^{10}\frac{1}{k}\ln\cos kx$.
For small $x$, $\ln\cos kx \approx -\frac{k^2x^2}{2}$, so $\ln P \approx -\frac{x^2}{2}\sum_{k=1}^{10} k = -\frac{55x^2}{2}$.
Then $1 - P \approx \frac{55x^2}{2}$, and the limit is $2\cdot\frac{55}{2} = 55$.
Answer: 55. | ||||
| 27 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\sin x = \cos x$ at $x = -\frac{3\pi}{4}$ and $\frac{\pi}{4}$; on $\left(-\frac{3\pi}{4}, \frac{\pi}{4}\right)$ the minimum is $\sin x$, elsewhere $\cos x$.
$\int_{-\pi}^{-3\pi/4}|\cos x|\,dx = \frac{\sqrt{2}}{2}$.
$\int_{-3\pi/4}^{0}|\sin x|\,dx = 1 + \frac{\sqrt{2}}{2}$, $\int_0^{\pi/4}\sin x\,dx = 1 - \frac{\sqrt{2}}{2}$.
$\int_{\pi/4}^{\pi/2}\cos x\,dx = 1 - \frac{\sqrt{2}}{2}$, $\int_{\pi/2}^{\pi}|\cos x|\,dx = 1$.
$A = \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} + 1 = 4$, so $A^2 = 16$.
Answer: 16. | ||||
| 28 | 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 $(\vec{r} - \vec{b} - \vec{c})\times\vec{a} = \vec{0}$, so $\vec{r} = \vec{b} + \vec{c} + t\vec{a}$.
$\vec{r}\cdot(\vec{b} - \vec{c}) = |\vec{b}|^2 - |\vec{c}|^2 + t\,\vec{a}\cdot(\vec{b} - \vec{c}) = 0$.
$|\vec{b}|^2 = 227$, $|\vec{c}|^2 = 294$; $\vec{b} - \vec{c} = (-14, 9, -14)$, $\vec{a}\cdot(\vec{b} - \vec{c}) = -126 - 117 - 350 = -593$.
$-67 - 593t = 0$, so $t = -\frac{67}{593}$ and $593\vec{r} + 67\vec{a} = 593(\vec{b} + \vec{c})$.
$\frac{|593\vec{r} + 67\vec{a}|^2}{593^2} = |\vec{b} + \vec{c}|^2 = |(20, 5, -12)|^2 = 400 + 25 + 144 = 569$.
Answer: 569. | ||||
| 29 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | NumericalHard | |
A hard numerical question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines The centroid divides the segment from circumcentre to orthocentre in the ratio $1:2$: $G = \frac{2(3, 4) + (-6, -8)}{3} = (0, 0)$. So the first triangle's orthocentre is $H = (0, 0)$.
Vertex of $2x + 3y = 1$ and $x + 2y = 1$: $(-1, 1)$. The altitude from it passes through $H$ along $(1, -1)$ and is perpendicular to $ax + by = 1$, so $(a, b) \parallel (1, -1)$: $b = -a$.
The vertex on $2x + 3y = 1$ and $ax + by = 1$ lies on the altitude through $H$ perpendicular to $x + 2y = 1$, i.e. on $(s, 2s)$: $2s + 6s = 1$, $s = \frac{1}{8}$, vertex $\left(\frac{1}{8}, \frac{1}{4}\right)$.
On $ax - ay = 1$: $\frac{a}{8} - \frac{a}{4} = 1$, so $a = -8$, $b = 8$.
Check: the third vertex $\left(\frac{1}{4}, \frac{3}{8}\right)$ of $x + 2y = 1$ and $-8x + 8y = 1$ lies on $(2u, 3u)$, the altitude through $H$ perpendicular to $2x + 3y = 1$.
$|a - b| = 16$.
Answer: 16. | ||||
| 30 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each drawn ball is blue with probability $\frac{5}{9}$, so by linearity $\overline{X} = 3\cdot\frac{5}{9} = \frac{5}{3}$ and $\overline{Y} = 3 - \overline{X} = \frac{4}{3}$.
(From the distribution: $P(X = 0, 1, 2, 3) = \frac{4}{84}, \frac{30}{84}, \frac{40}{84}, \frac{10}{84}$, mean $\frac{30 + 80 + 30}{84} = \frac{5}{3}$.)
$7\cdot\frac{5}{3} + 4\cdot\frac{4}{3} = \frac{35 + 16}{3} = 17$.
Answer: 17. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementSignificant figures | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on significant figures, from Physical World and Measurement. NCERT: Class 11: Units and Measurement A number written as $a \times 10^b$ (with $1 \leq a < 10$) is rounded to a power of ten: if $a \leq 5$ it is taken as $10^b$, and if $a > 5$ as $10^{b+1}$.
So the exponent $b$ is the order of magnitude when $a \leq 5$.
Answer: (C). | ||||
| 32 | KinematicsUniform circular motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane In 30 minutes the second hand makes 30 revolutions: distance $= 30\times 2\pi\times 0.75 = 45\pi$ m.
The minute hand makes half a revolution: distance $= \pi\times 0.60 = 0.6\pi$ m.
$x = 45\pi - 0.6\pi = 44.4\pi = 44.4\times 3.14 \approx 139.4$ m.
Answer: (B). | ||||
| 33 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Momentum is conserved from zero: $m_A v_A = m_B v_B = p$.
$K = \frac{1}{2}mv^2 = \frac{1}{2}pv$, so $\frac{K_B}{K_A} = \frac{v_B}{v_A}$.
Answer: (A). | ||||
| 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 $F = \frac{\Delta p}{\Delta t} = \frac{0.15\times 20}{0.1} = 30$ N.
Answer: (C). | ||||
| 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 $p = \sqrt{2mK}$, and $K$ is the same, so $p \propto \sqrt{m}$.
$\sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6} = 1 : \sqrt{3} : \sqrt{4} = 1 : \sqrt{3} : 2$.
Answer: (A). | ||||
| 36 | Properties of Bulk MatterFluid mechanics | 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 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 Along a streamline of an ideal fluid, pressure energy, potential energy and kinetic energy per unit volume add to a constant: $P + \rho gh + \frac{1}{2}\rho v^2 = \text{constant}$.
Answer: (C). | ||||
| 37 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Points $a$ and $b$ lie on the hotter isotherm $T_1$, and $d$ and $c$ on the colder isotherm $T_2$; $a \to d$ and $b \to c$ are adiabatic.
On an adiabat $TV^{\gamma - 1}$ is constant: $T_1V_a^{\gamma-1} = T_2V_d^{\gamma-1}$ and $T_1V_b^{\gamma-1} = T_2V_c^{\gamma-1}$.
So $\left(\frac{V_a}{V_d}\right)^{\gamma-1} = \frac{T_2}{T_1} = \left(\frac{V_b}{V_c}\right)^{\gamma-1}$, giving $\frac{V_a}{V_d} = \frac{V_b}{V_c}$.
Answer: (A). | ||||
| 38 | GravitationKepler's laws | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation In a circular orbit $v = \sqrt{\frac{GM}{r}}$, so $L = mvr = m\sqrt{GMr}$ and $r = \frac{L^2}{m^2GM}$.
$\frac{r_1}{r_2} = \left(\frac{L}{3L}\right)^2\left(\frac{m_2}{m_1}\right)^2 = \frac{1}{9}\left(\frac{m_2}{m_1}\right)^2$.
$T \propto r^{3/2}$: $\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2} = \frac{1}{27}\left(\frac{m_2}{m_1}\right)^3$.
Answer: (A). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Molar heat capacity at constant volume: monoatomic $C_V = \frac{3}{2}R$ (3 degrees of freedom), rigid diatomic $C_V = \frac{5}{2}R$ (5 degrees of freedom).
Ratio $= \frac{3/2}{5/2} = \frac{3}{5}$.
Answer: (B). | ||||
| 40 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity The two $4\ \Omega$ resistors are in parallel: $2\ \Omega$.
Current $I = \frac{3}{2 + 1} = 1$ A.
Terminal p.d. $= E - Ir = 3 - 1\times 1 = 2$ V.
Answer: (A). | ||||
| 41 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsEasy | |
A easy statements question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter A: true — a paramagnetic sample aligns along the external field.
B: false — paramagnetic substances are only weakly attracted (strong attraction is ferromagnetic).
C: true — the susceptibility is small and positive.
D: false — they move from weaker to stronger field regions (diamagnetic substances move from strong to weak).
Answer: (B). | ||||
| 42 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves For a perfectly absorbing surface, radiation pressure $= \frac{I}{c}$, so $F = \frac{IA}{c}$.
$I = 360$ W/cm$^2$ $= 3.6\times 10^6$ W/m$^2$.
$A = \frac{Fc}{I} = \frac{2.4\times 10^{-4}\times 3\times 10^8}{3.6\times 10^6} = 0.02$ m$^2$ (the duration does not matter for an average force).
Answer: (C). | ||||
| 43 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Total internal reflection occurs going from the denser (first) medium to the rarer (second): $\sin C = \frac{n_2}{n_1}$.
$\frac{n_1}{n_2} = \frac{1}{\sin 45^\circ} = \sqrt{2}$, so $n_1 : n_2 = \sqrt{2} : 1$.
Answer: (C). | ||||
| 44 | 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 Equal $\lambda = \frac{h}{p}$ means equal momenta. $K = \frac{p^2}{2m}$, so $K_p : K_e = m_e : m_p = 1 : 1836$.
Answer: (A). | ||||
| 45 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Connected conductors reach the same potential: $\frac{kq_1}{a} = \frac{kq_2}{b}$.
$\frac{q_1}{q_2} = \frac{a}{b}$.
Answer: (A). | ||||
| 46 | 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 The mass defect is $\Delta m = \frac{E}{c^2} = \frac{18\times 10^8}{9\times 10^{16}} = 2\times 10^{-8}$ kg $= 2\times 10^{-5}$ g $= 20\ \mu$g.
Answer: (B). | ||||
| 47 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Upper OR gate: inputs $A$ and $\overline{B}$, output $A + \overline{B}$.
Lower AND gate: inputs $B$ (tapped from the B line) and $\overline{A}$ (A through a NOT gate), output $\overline{A}B$.
Final AND gate: $Y = (A + \overline{B})\,\overline{A}B = A\overline{A}B + \overline{B}\,\overline{A}B = 0 + 0 = 0$.
Answer: (A). | ||||
| 48 | 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 $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l} = \frac{5}{500} + \frac{0.02}{2} = 0.01 + 0.01 = 0.02$.
Percentage error $= 2\%$.
Answer: (C). | ||||
| 49 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= 1$ MSD $- 1$ VSD $= 1 - \frac{9}{10} = 0.1$ mm.
Diameter $= 20 + 2\times 0.1 = 20.2$ mm $= 2.02$ cm, radius $1.01$ cm.
Volume $= \frac{4}{3}\pi(1.01)^3 \approx 4.316$ cm$^3$.
Density $= \frac{8.635}{4.316} \approx 2.0$ g/cm$^3$.
Answer: (A). | ||||
| 50 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At resonance $X_L = X_C$, so the impedance is $Z = R$ and $I_0 = \frac{V_0}{R}$. The resonant frequency $\frac{1}{2\pi\sqrt{LC}}$ does not depend on $R$.
Halving $R$ doubles the current amplitude.
Answer: (B). | ||||
| 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 Take $\overrightarrow{OP}$ along $+x$ and $\overrightarrow{OQ}$ along $+y$; $\overrightarrow{OR}$ is $45^\circ$ below $\overrightarrow{OP}$.
Sum: $x$-component $A + \frac{A}{\sqrt{2}}$, $y$-component $A - \frac{A}{\sqrt{2}}$.
$R^2 = A^2\left[\left(1 + \frac{1}{\sqrt{2}}\right)^2 + \left(1 - \frac{1}{\sqrt{2}}\right)^2\right] = A^2\cdot 2\left(1 + \frac{1}{2}\right) = 3A^2$.
$R = A\sqrt{3}$, so $x = 3$.
Answer: 3. | ||||
| 52 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The plate is the rectangle from $(0, 0)$ to $(3, 2)$ (area 6, centre $(1.5, 1)$) with the square from $(1, 1)$ to $(2, 2)$ removed (area 1, centre $(1.5, 1.5)$).
$x_{cm} = \frac{6(1.5) - 1(1.5)}{5} = 1.5$, $y_{cm} = \frac{6(1) - 1(1.5)}{5} = 0.9$.
$\frac{x_{cm}}{y_{cm}} = \frac{1.5}{0.9} = \frac{5}{3} = \frac{15}{9}$, so $n = 15$.
Answer: 15. | ||||
| 53 | 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 A soap bubble has two surfaces: excess pressure $= \frac{4T}{r}$.
$\frac{4T}{r} = h\rho g = 0.04\times 10^{-2}\times 8\times 10^3\times 10 = 32$ Pa.
$r = \frac{4\times 0.28}{32} = 0.035$ m $= 3.5$ cm, so the diameter is 7 cm.
Answer: 7. | ||||
| 54 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Closed pipe: only odd harmonics, so the seventh overtone is the 15th harmonic, $\frac{15v}{4L}$.
Open pipe: the seventh overtone is the 8th harmonic, $\frac{8v}{2L} = \frac{16v}{4L}$.
Ratio $= \frac{15}{16} = \frac{a - 1}{a}$, so $a = 16$.
Answer: 16. | ||||
| 55 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $\phi = \vec{E}\cdot\hat{n}\,A = \frac{(2)(2) + (6)(1) + (8)(1)}{6}\times 4 = \frac{18}{6}\times 4 = 12$ V m.
Answer: 12. | ||||
| 56 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $R_t = R_0(1 + \alpha t)$: $10.2 = 10(1 + 100\alpha)$ gives $\alpha = 2\times 10^{-4}$ per °C.
$10.95 = 10(1 + 2\times 10^{-4}\,t)$ gives $t = \frac{0.095}{2\times 10^{-4}} = 475$ °C.
In kelvin: $475 + 273 = 748$ K (748.15 K with 273.15, which rounds to 748).
Answer: 748. | ||||
| 57 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy 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 $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2\times 5\times 1.6\times 10^{-19}}{9\times 10^{-31}}} = \sqrt{\frac{16}{9}\times 10^{12}} = \frac{4}{3}\times 10^6$ m/s.
Undeflected motion needs $eE = evB$: $E = vB = \frac{4}{3}\times 10^6\times 3\times 10^{-6} = 4$ N/C.
Answer: 4. | ||||
| 58 | Atoms and NucleiRutherford model and alpha scattering | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on rutherford model and alpha scattering, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei At closest approach the kinetic energy has become potential energy: $\frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0}\frac{(2e)(Ze)}{r_0}$.
$\frac{1}{2}mv^2 = \frac{9\times 10^9\times 2\times 80\times(1.6\times 10^{-19})^2}{4.5\times 10^{-14}} = 8.192\times 10^{-13}$ J.
$v^2 = \frac{2\times 8.192\times 10^{-13}}{6.72\times 10^{-27}} \approx 2.438\times 10^{14}$, so $v \approx 1.561\times 10^7$ m/s $= 156.1\times 10^5$ m/s.
Answer: 156. | ||||
| 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 Single-slit minima: $a\sin\theta = n\lambda$. For $n = 2$: $\theta \approx \frac{2\lambda}{a} = \frac{2\times 600\times 10^{-9}}{0.4\times 10^{-3}} = 3\times 10^{-3}$ rad.
Answer: 3. | ||||
| 60 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Side $l = \sqrt{3.6\times 10^{-3}} = 0.06$ m. The loop leaves the field in 1.0 s, so $v = 0.06$ m/s.
emf $= NBlv = 10\times 0.5\times 0.06\times 0.06 = 0.018$ V; current $I = \frac{0.018}{100} = 1.8\times 10^{-4}$ A.
Work = heat produced $= \frac{\varepsilon^2}{R}t = \frac{(0.018)^2}{100}\times 1 = 3.24\times 10^{-6}$ J.
To the nearest integer: 3.
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O}$.
900 g of glucose is $\frac{900}{180} = 5$ mol, needing $5\times 6 = 30$ mol $\mathrm{O_2}$.
Mass of $\mathrm{O_2} = 30\times 32 = 960$ g.
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 (A) $\mathrm{NH_3}$: 3 bond pairs + 1 lone pair → trigonal pyramidal (III).
(B) $\mathrm{BrF_5}$: 5 bond pairs + 1 lone pair → square pyramid (I).
(C) $\mathrm{PCl_5}$: 5 bond pairs → trigonal bipyramidal (IV).
(D) $\mathrm{CH_4}$: 4 bond pairs → tetrahedral (II).
A-III, B-I, C-IV, D-II.
Answer: (C). | ||||
| 63 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Adding the three reactions gives $\mathrm{X \rightleftharpoons W}$, so the constants multiply: $K = K_1K_2K_3 = 1.0\times 2.0\times 4.0 = 8.0$.
Answer: (B). | ||||
| 64 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions With iodine, S goes from +2 (in $\mathrm{S_2O_3^{2-}}$) only to +2.5 (in $\mathrm{S_4O_6^{2-}}$); with bromine it goes all the way to +6 (in $\mathrm{SO_4^{2-}}$).
Thiosulphate is oxidised in both reactions, and both halogens are reduced; bromine oxidises it further because bromine is the stronger oxidising agent.
Answer: (A). | ||||
| 65 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Disproportionation needs the element to go both up and down in oxidation state from 0.
Fluorine, the most electronegative element, shows no positive oxidation state, so $\mathrm{F_2}$ cannot disproportionate.
$\mathrm{Cl_2}$, $\mathrm{Br_2}$ and $\mathrm{I_2}$ can (e.g. $\mathrm{3X_2 + 6OH^- \to 5X^- + XO_3^- + 3H_2O}$).
Answer: (C). | ||||
| 66 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down group 13 the $ns^2$ pair is increasingly reluctant to take part in bonding (inert pair effect), so the +1 state grows more stable: Ga < In < Tl. A is true.
R is true and is exactly the reason for A.
Answer: (A). | ||||
| 67 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | StatementsMedium | |
A medium statements question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Statement I: both have a lone pair on the donor atom and act as ligands. Correct.
Statement II: $\mathrm{N(CH_3)_3}$ is only a σ-donor (N has no low-lying vacant d orbitals), while $\mathrm{P(CH_3)_3}$ is a σ-donor and also a π-acceptor through back-bonding. The bonding is not the same. Incorrect.
Answer: (C). | ||||
| 68 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Matching listEasy | |
A easy matching list question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties (A) Cl and S have the most negative electron gain enthalpies in groups 17 and 16 (more than F and O, whose small size crowds the added electron) → (IV).
(B) Ge and As are metalloids → (III).
(C) Fr and Ra are the largest atoms of groups 1 and 2 → (II).
(D) F and O are the most electronegative of their groups → (I).
A-IV, B-III, C-II, D-I.
Answer: (D). | ||||
| 69 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | StatementsEasy | |
A easy statements question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The catalysis runs in two steps:
$\mathrm{2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2}$ (Fe(III) oxidises iodide — A),
$\mathrm{2Fe^{2+} + S_2O_8^{2-} \to 2Fe^{3+} + 2SO_4^{2-}}$ (Fe(II) reduces persulphate — D).
Persulphate cannot be oxidised further, and iodide cannot be reduced, so B and C are wrong.
Answer: (C). | ||||
| 70 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Two $\mathrm{Cl^-}$ are outside the coordination sphere, so one Cl is a ligand. Octahedral: 6 ligands = 1 Cl + 5 $\mathrm{NH_3}$, so $n = 5$: $\mathrm{[Co(NH_3)_5Cl]Cl_2}$.
Oxidation state: $x - 1 = +2$, so $x = +3$.
$x + n = 8$.
Answer: (D). | ||||
| 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 $\mathrm{Fe^{2+}}$, $d^6$, high spin with water: $t_{2g}^4e_g^2$ → 4 (even).
$\mathrm{Co^{2+}}$, $d^7$, high spin: $t_{2g}^5e_g^2$ → 5 (odd).
$\mathrm{Co^{3+}}$, $d^6$, low spin (large $\Delta_o$ for a +3 ion): $t_{2g}^6$ → 6 (even). (Even high spin it would be $t_{2g}^4$, still even.)
$\mathrm{Cu^{2+}}$, $d^9$: $t_{2g}^6e_g^3$ → 6 (even).
$\mathrm{Cr^{2+}}$, $d^4$, high spin: $t_{2g}^3e_g^1$ → 3 (odd).
Three complexes.
Answer: (C). | ||||
| 72 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter (A) Borax bead: the salt is fused with borax, giving the metal metaborate, which is reduced in the flame → (III).
(B) Charcoal cavity: the salt is heated with sodium carbonate on charcoal: carbonate → oxide → metal → (IV).
(C) Cobalt nitrate test: the oxide residue is heated with cobalt nitrate to give a coloured $\mathrm{CoO\cdot MO}$ → (I).
(D) Flame test: the salt is moistened with conc. HCl to form the volatile chloride, whose metal ion colours the flame → (II).
A-III, B-IV, C-I, D-II.
Answer: (D). | ||||
| 73 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter (A) Ferric ferrocyanide is Prussian blue → (III).
(B) The nitroprusside–sulphide complex (test for sulphur) is violet → (I).
(C) Thiocyanatoiron(III) is blood red → (II).
(D) Ammonium phosphomolybdate is a yellow precipitate → (IV).
A-III, B-I, C-II, D-IV.
Answer: (B). | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsMedium | |
A medium statements question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Compound A is a benzene ring with Cl and two $\mathrm{NO_2}$ groups, one ortho and one para to Cl. Numbering from Cl gives the locant set $\{1, 2, 4\}$; numbering from a nitro group gives $\{1, 3, 4\}$. The lowest set wins: 1-chloro-2,4-dinitrobenzene. Statement I is incorrect.
Compound B is aniline ($\mathrm{NH_2}$ at C-1) with $\mathrm{CH_3}$ at C-2 and $\mathrm{C_2H_5}$ at C-4; prefixes in alphabetical order: 4-ethyl-2-methylaniline. Statement II is correct.
Answer: (D). | ||||
| 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 The compound is $\mathrm{CH_3{-}CH(CH_3){-}CH_2{-}CH(CH_3){-}CH_3}$ (2,4-dimethylpentane).
A 2° carbon is bonded to two other carbons. The $\mathrm{CH_3}$ groups are 1°, the two CH carbons are 3° (three carbon neighbours each), and only the central $\mathrm{CH_2}$ is 2°.
Answer: (D). | ||||
| 76 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Aromatic: cyclic, planar, fully conjugated, with $(4n + 2)$ π electrons.
A: as drawn, the bicyclic ring has only four C=C bonds — there is no double bond at the ring fusion — so it has $\mathrm{sp^3}$ carbons, is not fully conjugated, and neither ring is a benzene ring. Not aromatic.
B: styrene (vinylbenzene) has a benzene ring. Aromatic.
C: [10]annulene has 10 π electrons, but the inward-pointing hydrogens prevent a planar ring. Not aromatic.
D: [14]annulene, 7 C=C, 14 π electrons ($n = 3$), planar enough to be aromatic.
Aromatic: B and D.
Answer: (D). | ||||
| 77 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $\mathrm{S_N2}$ is fastest for the least hindered carbon bearing the halogen.
(A) 1-bromo-1-methylcyclobutane: tertiary.
(B) a tertiary bromide on a carbon carrying a cyclobutyl group: tertiary.
(C) (bromomethyl)cyclobutane: primary — least hindered.
(D) (1-bromoethyl)cyclobutane: secondary.
Answer: (C). | ||||
| 78 | 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 1-Methylcyclohexan-1-ol is a tertiary alcohol.
With conc. $\mathrm{H_2SO_4}$ and heat it dehydrates by E1; the more substituted (Saytzeff) alkene, 1-methylcyclohexene (endocyclic C=C), is the major product A.
With $\mathrm{CH_3COCl}$ in pyridine the OH is acetylated: B is the ester 1-methylcyclohexyl acetate ($-\mathrm{OCOCH_3}$ on the ring carbon carrying $\mathrm{CH_3}$).
Answer: (B). | ||||
| 79 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $\mathrm{Br_2}$/red P followed by water is the Hell–Volhard–Zelinsky reaction: a carboxylic acid with an α-hydrogen is brominated at the α-carbon, and the acyl bromide formed is hydrolysed back to the acid.
The α-carbon of cyclopentanecarboxylic acid is the ring carbon carrying COOH, so P is 1-bromocyclopentane-1-carboxylic acid.
Answer: (A). | ||||
| 80 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules The Fischer projection (CHO; OH right, left, right, right; $\mathrm{CH_2OH}$) is D-glucose.
(A) C-2 to C-5 are asymmetric: 4. True.
(B) Bromine water is a mild oxidant: it oxidises only the CHO group, giving gluconic acid, a monocarboxylic acid. (A dicarboxylic acid, saccharic acid, needs nitric acid.) False — this is the incorrect statement.
(C) Glucose does not give Schiff's test, as the free CHO is present only in small amounts in the open-chain form. True.
(D) The open chain is in equilibrium with the α- and β-pyranose forms. True.
Answer: (B). | ||||
| 81 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom The marked 1.5 pm runs from a crest to the next point where the wave crosses the axis: a quarter of a wavelength.
$\lambda = 4\times 1.5 = 6$ pm $= 6\times 10^{-12}$ m.
$\nu = \frac{c}{\lambda} = \frac{3\times 10^8}{6\times 10^{-12}} = 5\times 10^{19}$ Hz.
Answer: 5. | ||||
| 82 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Odd-electron molecules: $\mathrm{NO_2}$, $\mathrm{ClO_2}$.
Incomplete octet: $\mathrm{BF_3}$ (6 electrons on B), $\mathrm{BeF_2}$ (4 on Be).
Expanded octet: $\mathrm{H_2SO_4}$ (12 on S), $\mathrm{PCl_5}$ (10 on P).
$\mathrm{CO_2}$, $\mathrm{CH_4}$, $\mathrm{SiF_4}$, $\mathrm{C_2H_6}$, $\mathrm{CHCl_3}$, $\mathrm{CBr_4}$ obey the octet rule.
Exceptions: 6.
Answer: 6. | ||||
| 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 At A the gas occupies 10 L; moving the piston to B adds 90 L, so the final volume is 100 L.
Reversible isothermal work (by the gas): $w = nRT\ln\frac{V_2}{V_1} = 1\times 0.08206\times 291.15\times\ln 10$.
$= 23.892\times 2.3026 \approx 55.01$ L atm.
To the nearest integer, $x = 55$ (the work done on the gas is $-55$ L atm).
Answer: 55. | ||||
| 84 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Molality $= \frac{10/200}{0.1} = 0.5$ mol kg$^{-1}$.
$\Delta T_b = iK_bm$: $0.52 = i\times 0.52\times 0.5$, so $i = 2$.
For $\mathrm{AB_2 \to A^{2+} + 2B^-}$: $i = 1 + 2\alpha = 2$, so $\alpha = 0.5 = 5\times 10^{-1}$.
Answer: 5. | ||||
| 85 | 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 $t_{1/4} = 2t_{1/2}$ means the half-life does not depend on concentration: first order in A.
[B] falling linearly with time means a constant rate in B: zero order in B.
Overall order $= 1 + 0 = 1$.
Answer: 1. | ||||
| 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 Metallic radii (pm): Sc 164, Ti 147, V 135, Cr 129, Mn 137, Zn 137. The least is Cr.
In $\mathrm{CrO_4^{2-}}$, Cr is +6: $3d^0$, no unpaired electrons.
Spin-only moment $= \sqrt{n(n + 2)} = 0$ BM.
Answer: 0. | ||||
| 87 | AminesDiazonium salts | Class 12: Amines | NumericalEasy | |
A easy numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines $\mathrm{C_6H_5N_2^+Cl^- + C_6H_5NH_2 \to}$ p-aminoazobenzene (aniline yellow, $\mathrm{C_{12}H_{11}N_3}$) + HCl.
Aniline: $\frac{279}{93} = 3$ mol, giving 3 mol of aniline yellow.
Molar mass of $\mathrm{C_{12}H_{11}N_3} = 144 + 11 + 42 = 197$ g mol$^{-1}$.
Mass $= 3\times 197 = 591$ g.
Answer: 591. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The compound is a perhydroanthracene (three linearly fused cyclohexane rings) with Br on one ring-fusion carbon and $\mathrm{CH_3}$ on the end ring next to the other fusion.
Stereocentres: the fusion carbon carrying Br; the other fusion carbon of that pair (H, the C–Br carbon, and two different ring $\mathrm{CH_2}$ paths); both fusion carbons of the second pair; and the carbon carrying $\mathrm{CH_3}$. Each has four different groups: 5 stereocentres.
The molecule has no symmetry that could make any isomer meso, so the number of optical isomers is $2^5 = 32$.
Answer: 32. | ||||
| 89 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{KMnO_4}$–KOH, heat, oxidises the whole side chain of ethylbenzene to COOH (after acidification): A is benzoic acid.
COOH is meta-directing, so nitration gives B = 3-nitrobenzoic acid.
π bonds: 3 in the ring, 1 in C=O, 1 in $\mathrm{NO_2}$ = 5.
Answer: 5. | ||||
| 90 | AminesClassification and identification of amines | Class 12: Amines | NumericalMedium | |
A medium numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines With Hinsberg's reagent ($\mathrm{C_6H_5SO_2Cl}$) a primary amine gives an N-alkyl/aryl sulphonamide with an acidic N–H, which dissolves in NaOH. Secondary amines give NaOH-insoluble sulphonamides, and tertiary amines do not react.
Primary amines in the list: aniline; the compound $\mathrm{C_6H_5NH{-}C_6H_6{-}NH_2}$ (it carries a free $\mathrm{NH_2}$); o-anisidine (2-methoxyaniline); propan-1-amine; cyclohexylamine — 5.
Not counted: benzamide, semicarbazide and urea (amides, not amines); N-methylaniline (secondary); triethylamine and N-ethyl-N-methylcyclohexylmethanamine (tertiary).
Answer: 5. | ||||
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