JEE (Main) · 2023
JEE (Main) 2023 — 13 April, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus4
- Matrices and Determinants3
- Three Dimensional Geometry3
13 easy · 16 medium · 1 hard
Physics
30 Q
- Properties of Bulk Matter3
- Motion of System of Particles and Rigid Body2
- Work, Energy and Power2
20 easy · 10 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Chemical Bonding and Molecular Structure2
- p-Block Elements2
16 easy · 11 medium · 3 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Replace $x$ by $\frac{1}{x}$: $3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$.
Multiply the given equation by 3 and this one by 2 and subtract: $9f(x) - 4f(x) = \frac{3}{x} - 30 - 2x + 20$, so $f(x) = \frac{1}{5}\left(\frac{3}{x} - 2x - 10\right)$.
$f(3) = \frac{1}{5}(1 - 6 - 10) = -3$.
$f'(x) = \frac{1}{5}\left(-\frac{3}{x^2} - 2\right)$, so $f'\left(\frac{1}{4}\right) = \frac{1}{5}(-48 - 2) = -10$.
$\left|f(3) + f'\left(\frac{1}{4}\right)\right| = |-3 - 10| = 13$.
Answer: (D). | ||||
| 2 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $g(x) = x|x - 1| + |x + 2|$; the equation is $g(x) = -a$.
For $x < -2$: $g(x) = x(1 - x) - (x + 2) = -x^2 - 2$, with $g'(x) = -2x > 0$.
For $-2 \le x < 1$: $g(x) = x(1 - x) + x + 2 = -x^2 + 2x + 2$, with $g'(x) = 2 - 2x > 0$.
For $x \ge 1$: $g(x) = x^2 - x + x + 2 = x^2 + 2$, with $g'(x) = 2x > 0$.
The pieces join continuously ($g(-2) = -6$, $g(1) = 3$), so $g$ is continuous and strictly increasing, with $g \to -\infty$ as $x \to -\infty$ and $g \to \infty$ as $x \to \infty$.
Hence $g(x) = -a$ has exactly one real root for every real $a$: the set is $(-\infty, \infty)$.
Answer: (C). | ||||
| 3 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3 \times 3$ matrix, $|\operatorname{adj} A| = |A|^2 = 4$.
$|B| = 1(8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(\alpha - 2\alpha) = -\alpha^2 + 6\alpha - 4$.
$-\alpha^2 + 6\alpha - 4 = 4 \Rightarrow \alpha^2 - 6\alpha + 8 = 0 \Rightarrow \alpha = 2$ or $4$; since $\alpha > 2$, $\alpha = 4$.
The product is $\alpha^2\, u B u^T$ with $u = \begin{bmatrix}1 & -2 & 1\end{bmatrix}$ and $B = \begin{bmatrix}1 & 3 & 4\\1 & 2 & 3\\4 & 4 & 4\end{bmatrix}$.
$B u^T = \begin{bmatrix}1 - 6 + 4\\1 - 4 + 3\\4 - 8 + 4\end{bmatrix} = \begin{bmatrix}-1\\0\\0\end{bmatrix}$, so $u B u^T = -1$.
Value $= 16 \times (-1) = -16$.
Answer: (B). | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}2 & 4 & 2a\\1 & 2 & 3\\2 & -5 & 2\end{vmatrix} = 2(4 + 15) - 4(2 - 6) + 2a(-5 - 4) = 54 - 18a$.
For $a \ne 3$ (for example $a = 6$ or $a = 8$), $\Delta \ne 0$ and the system has a unique solution whatever $b$ is: (A) and (B) are correct.
For $a = 3$ the first equation is $2x + 4y + 6z = b$, i.e. $2(x + 2y + 3z) = b$. With the second equation this needs $b = 8$.
If $b = 8$ the first equation repeats the second, and the remaining two independent equations give infinitely many solutions: (C) is correct.
If $b = 6$ the first two equations contradict each other ($x + 2y + 3z$ cannot be both 3 and 4): no solution, so (D) is NOT correct.
Answer: (D). | ||||
| 5 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A symmetric $3 \times 3$ matrix is fixed by its 3 diagonal entries and the 3 entries above the diagonal (those below are equal to them).
Each of these 6 entries can be any of the 10 digits: $10^6$ matrices.
Answer: (C). | ||||
| 6 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series For the $i$-th A.P., $a_i = i$ and $d_i = 2i - 1$, so $s_i = \frac{12}{2}(2a_i + 11d_i) = 12a_i + 66d_i$.
$\sum a_i = 1 + 2 + \cdots + 10 = 55$ and $\sum d_i = 1 + 3 + \cdots + 19 = 100$.
$\sum_{i=1}^{10} s_i = 12(55) + 66(100) = 660 + 6600 = 7260$.
Answer: (A). | ||||
| 7 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals (S1): $2 + 4 + \cdots + 2n = n(n + 1)$, so the limit is $\lim \frac{n(n+1)}{n^2} = 1$. True.
(S2): as a Riemann sum, $\frac{1}{n^{16}}\sum_{r=1}^{n} r^{15} = \frac{1}{n}\sum_{r=1}^{n}\left(\frac{r}{n}\right)^{15} \to \int_0^1 x^{15}\,dx = \frac{1}{16}$. True.
Both are true.
Answer: (C). | ||||
| 8 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The plane is $(x + 2y + az - 2) + \lambda(x - y + z - 3) = 0$, and it must be a multiple $k$ of $5x - 11y + bz - (6a - 1) = 0$:
$1 + \lambda = 5k$, $2 - \lambda = -11k$, $a + \lambda = bk$, $-2 - 3\lambda = -(6a - 1)k$.
Adding the first two: $3 = -6k$, $k = -\frac{1}{2}$, so $\lambda = -\frac{7}{2}$.
Constant term: $-2 + \frac{21}{2} = \frac{17}{2} = \frac{6a - 1}{2}$, so $a = 3$.
Then $3 - \frac{7}{2} = -\frac{b}{2}$ gives $b = 1$. The plane is $5x - 11y + z = 17$.
Distance from $(3, -c, c)$: $\frac{|15 + 11c + c - 17|}{\sqrt{25 + 121 + 1}} = \frac{|12c - 2|}{7\sqrt{3}} = \frac{2}{\sqrt{3}}$, so $|12c - 2| = 14$: $c = \frac{4}{3}$ or $c = -1$. As $c \in \mathbb{Z}$, $c = -1$.
$\frac{a + b}{c} = \frac{4}{-1} = -4$.
Answer: (A). | ||||
| 9 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $(\vec{d} - \vec{c}) \times \vec{b} = \vec{0}$, so $\vec{d} = \vec{c} + t\vec{b}$.
$\vec{d} \cdot \vec{a} = \vec{c} \cdot \vec{a} + t\,\vec{b} \cdot \vec{a} = (2 - 4 + 8) + t(3 - 8 + 14) = 6 + 9t = 24$, so $t = 2$.
$\vec{d} = (2, -1, 4) + 2(3, -2, 7) = (8, -5, 18)$.
$|\vec{d}|^2 = 64 + 25 + 324 = 413$.
Answer: (D). | ||||
| 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 Let $f(x) = x - \sin 2x + \frac{1}{3}\sin 3x$. Then $f'(x) = 1 - 2\cos 2x + \cos 3x$.
Write $c = \cos x$: $\cos 2x = 2c^2 - 1$, $\cos 3x = 4c^3 - 3c$, so $f'(x) = 4c^3 - 4c^2 - 3c + 3 = (c - 1)(4c^2 - 3)$.
On $(0, \pi)$: $c - 1 < 0$, and $4c^2 - 3 > 0$ for $x < \frac{\pi}{6}$ or $x > \frac{5\pi}{6}$, $< 0$ in between. So $f$ decreases on $\left[0, \frac{\pi}{6}\right]$, increases on $\left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$ and decreases on $\left[\frac{5\pi}{6}, \pi\right]$.
The maximum is therefore at $x = 0$ or $x = \frac{5\pi}{6}$. $f(0) = 0$, while
$f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} - \sin\frac{5\pi}{3} + \frac{1}{3}\sin\frac{5\pi}{2} = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 2 + 3\sqrt{3}}{6} \approx 3.82$.
Answer: (B). | ||||
| 11 | Integral CalculusPartial fractions | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on partial fractions, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t = e^x$, $dx = \frac{dt}{t}$: the integral is $\int_1^{\infty} \frac{6\,dt}{t(t + 1)(t + 2)(t + 3)}$.
Partial fractions: $\frac{6}{t(t+1)(t+2)(t+3)} = \frac{1}{t} - \frac{3}{t + 1} + \frac{3}{t + 2} - \frac{1}{t + 3}$.
An antiderivative is $\ln\frac{t(t + 2)^3}{(t + 1)^3(t + 3)}$, which tends to $\ln 1 = 0$ as $t \to \infty$ and equals $\ln\frac{1 \cdot 27}{8 \cdot 4} = \ln\frac{27}{32}$ at $t = 1$.
Integral $= 0 - \ln\frac{27}{32} = \log_e\frac{32}{27}$.
Answer: (D). | ||||
| 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 $\sin x = \cos x$ at $x = -\frac{3\pi}{4}$ and $x = \frac{\pi}{4}$. So $f = \sin x$ on $\left[-\pi, -\frac{3\pi}{4}\right]$ and $\left[\frac{\pi}{4}, \pi\right]$, and $f = \cos x$ on $\left[-\frac{3\pi}{4}, \frac{\pi}{4}\right]$.
Below the axis: $\int_{-\pi}^{-3\pi/4} |\sin x|\,dx = 1 - \frac{1}{\sqrt{2}}$ and $\int_{-3\pi/4}^{-\pi/2} |\cos x|\,dx = 1 - \frac{1}{\sqrt{2}}$.
Above the axis: $\int_{-\pi/2}^{\pi/4} \cos x\,dx = \frac{1}{\sqrt{2}} + 1$ and $\int_{\pi/4}^{\pi} \sin x\,dx = 1 + \frac{1}{\sqrt{2}}$.
Total area $= 2\left(1 - \frac{1}{\sqrt{2}}\right) + 2\left(1 + \frac{1}{\sqrt{2}}\right) = 4$.
Answer: (B). | ||||
| 13 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $4^{2022} = 16^{1011} = (15 + 1)^{1011} = 15m + 1$ for an integer $m$ (every other term of the binomial expansion is a multiple of 15).
So $\frac{4^{2022}}{15} = m + \frac{1}{15}$, and the fractional part is $\frac{1}{15}$.
Answer: (A). | ||||
| 14 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{y + 7} = dx$ gives $y + 7 = Ce^x$.
$y_1(0) = 0$: $C = 7$, $y_1 = 7e^x - 7$. $y_2(0) = 1$: $C = 8$, $y_2 = 8e^x - 7$.
$y_2 - y_1 = e^x > 0$ for every $x$, so the curves never meet.
Answer: (A). | ||||
| 15 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Tangent at $(3\sqrt{3}, 1)$: $\frac{3\sqrt{3}\,x}{36} + \frac{y}{4} = 1$; at $x = 0$, $y = 4$, so $A = (0, 4)$. Its slope is $-\frac{1}{\sqrt{3}}$.
Normal: slope $\sqrt{3}$, $y - 1 = \sqrt{3}(x - 3\sqrt{3})$; at $x = 0$, $y = -8$, so $B = (0, -8)$.
Circle on $AB$ as diameter: centre $(0, -2)$, radius 6: $x^2 + (y + 2)^2 = 36$.
$PQ$ is the chord of contact of $(\alpha, \beta)$: $\alpha x + (\beta + 2)(y + 2) = 36$. For this to be the line $x = 2\sqrt{5}$: $\beta + 2 = 0$ and $2\sqrt{5}\,\alpha = 36$.
So $\beta = -2$, $\alpha = \frac{18}{\sqrt{5}}$, and $\alpha^2 - \beta^2 = \frac{324}{5} - 4 = \frac{304}{5}$.
Answer: (C). | ||||
| 16 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Here $a = 9$; points are $(9t^2, 18t)$. A focal chord inclined at $\theta$ has length $\frac{4a}{\sin^2\theta} = \frac{36}{\sin^2\theta} = 100$, so $\sin\theta = \frac{3}{5}$ and the slope is $\tan\theta = \frac{3}{4}$.
For a focal chord $t_1 t_2 = -1$ and the slope is $\frac{2}{t_1 + t_2} = \frac{3}{4}$, so $t_1 + t_2 = \frac{8}{3}$: $t^2 - \frac{8}{3}t - 1 = 0$, $t = 3$ or $-\frac{1}{3}$.
$P$ (positive ordinate) $= (81, 54)$, $Q = (1, -6)$ (check: $PQ = \sqrt{80^2 + 60^2} = 100$).
$M = \frac{1 \cdot P + 3 \cdot Q}{4} = \left(\frac{81 + 3}{4}, \frac{54 - 18}{4}\right) = (21, 9)$.
Line through $M$ perpendicular to $PQ$ (slope $-\frac{4}{3}$): $4x + 3y = 111$.
$(3, 33)$: $12 + 99 = 111$ ✓; $(-6, 45)$: $-24 + 135 = 111$ ✓; $(6, 29)$: $24 + 87 = 111$ ✓; $(-3, 43)$: $-12 + 129 = 117$ ✗.
Answer: (C). | ||||
| 17 | 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 The line of shortest distance is perpendicular to both lines, so it is parallel to $(2\hat{i} + \hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) = \hat{i} - \hat{j} - 2\hat{k}$.
Line through $(-1, 2, 3)$ in this direction: $(-1 + t, 2 - t, 3 - 2t)$.
It meets $x - 2y + 3z = 10$ when $(-1 + t) - 2(2 - t) + 3(3 - 2t) = 10$, i.e. $4 - 3t = 10$, $t = -2$.
Distance $= |t|\sqrt{1 + 1 + 4} = 2\sqrt{6}$.
Answer: (B). | ||||
| 18 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(H) = \frac{3}{4}$, $P(T) = \frac{1}{4}$.
$X = 1$: H, probability $\frac{3}{4}$. $X = 2$: TH, probability $\frac{3}{16}$. $X = 3$: TT followed by anything (TTH or TTT), probability $\frac{1}{16}$.
$E(X) = 1 \cdot \frac{3}{4} + 2 \cdot \frac{3}{16} + 3 \cdot \frac{1}{16} = \frac{12 + 6 + 3}{16} = \frac{21}{16}$.
Answer: (A). | ||||
| 19 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Put $t = g(x) = \sqrt{x} + 1$, so $\sqrt{x} = t - 1$ and $x = (t - 1)^2$.
$f(t) = (t - 1)^2 + 3 - (t - 1) = t^2 - 3t + 5$.
$f(0) = 5$.
Answer: (D). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $A \wedge (B \vee C)$ true means $A$ is true, and then $A \vee B$ is true. So $(A \wedge (B \vee C)) \Rightarrow (A \vee B)$ is a tautology $T$.
The statement is $T \Rightarrow A$, which is equivalent to $A$.
Its negation is equivalent to $\sim A$.
Answer: (C). | ||||
| 21 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x + iy$: $w = (x^2 + y^2 + k_1 x - k_2 y + \lambda) + i(k_1 y + k_2 x + \lambda)$.
$\operatorname{Re}(w) = 0$: $x^2 + y^2 + k_1 x - k_2 y + \lambda = 0$, centre $\left(-\frac{k_1}{2}, \frac{k_2}{2}\right)$.
A circle of radius 1 in the first quadrant touching the $y$-axis and $y = 1$ has centre $(1, 2)$. So $k_1 = -2$, $k_2 = 4$, and $1 + 4 - \lambda = 1$ gives $\lambda = 4$.
$\operatorname{Im}(w) = 0$: $-2y + 4x + 4 = 0$, i.e. $2x - y + 2 = 0$.
Distance of $(1, 2)$ from this line: $\frac{|2 - 2 + 2|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$.
$AB^2 = 4\left(1 - \frac{4}{5}\right) = \frac{4}{5}$, so $30(AB)^2 = 24$.
Answer: 24. | ||||
| 22 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Write each digit as $d_i = 1 + e_i$ with $0 \le e_i \le 3$. Then $e_1 + \cdots + e_7 = 12 - 7 = 5$.
Without the upper bound: $\binom{5 + 6}{6} = 462$ solutions.
Remove those with some $e_i \ge 4$ (only one can be): choose $i$ (7 ways), then the rest sum to 1: $\binom{1 + 6}{6} = 7$ ways, giving $49$.
Count $= 462 - 49 = 413$.
Answer: 413. | ||||
| 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 $T_{r+1} = \binom{n}{r}(-6)^r x^{\frac{n - r}{2} - \frac{3r}{2}} = \binom{n}{r}(-6)^r x^{\frac{n - 4r}{2}}$.
A constant term needs $n = 4r$, so $n \in \{4, 8, 12\}$.
Sum of all coefficients $= (1 - 6)^n = (-5)^n$, so $(-5)^n - \alpha = 649$.
$n = 4$: $\alpha = \binom{4}{1}(-6) = -24$ and $625 + 24 = 649$ ✓. ($n = 8, 12$ give sums far from 649.)
Coefficient of $x^{-4}$: $\frac{4 - 4r}{2} = -4 \Rightarrow r = 3$: $\binom{4}{3}(-6)^3 = -864$.
$\lambda = \frac{-864}{-24} = 36$.
Answer: 36. | ||||
| 24 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Group the 20 terms in 10 pairs: the $k$-th pair is $2(2k)^2 - (2k + 1)^2 = 8k^2 - 4k^2 - 4k - 1 = 4k^2 - 4k - 1$.
Sum $= 4\sum_{k=1}^{10} k^2 - 4\sum_{k=1}^{10} k - 10 = 4(385) - 4(55) - 10 = 1540 - 220 - 10 = 1310$.
Answer: 1310. | ||||
| 25 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $C_1 = 1 - \int_0^1 x\,dx = \frac{1}{2}$, $S_1(x) = \frac{x}{2} + \int_0^x t\,dt = \frac{x}{2} + \frac{x^2}{2}$.
$C_2 = 1 - \int_0^1\left(\frac{x}{2} + \frac{x^2}{2}\right)dx = 1 - \frac{1}{4} - \frac{1}{6} = \frac{7}{12}$.
$S_2(x) = \frac{7x}{12} + 2\int_0^x\left(\frac{t}{2} + \frac{t^2}{2}\right)dt = \frac{7x}{12} + \frac{x^2}{2} + \frac{x^3}{3}$.
$S_2(3) = \frac{7}{4} + \frac{9}{2} + 9 = \frac{61}{4}$.
$C_3 = 1 - \int_0^1 S_2(x)\,dx = 1 - \left(\frac{7}{24} + \frac{1}{6} + \frac{1}{12}\right) = 1 - \frac{13}{24} = \frac{11}{24}$, so $6C_3 = \frac{11}{4}$.
$S_2(3) + 6C_3 = \frac{61}{4} + \frac{11}{4} = 18$.
Answer: 18. | ||||
| 26 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections For $\frac{y^2}{25} - \frac{x^2}{16} = 1$, the line $y = mx + c$ is a tangent when $c^2 = 25 - 16m^2$.
Through P(4, 1): $(1 - 4m)^2 = 25 - 16m^2 \Rightarrow 32m^2 - 8m - 24 = 0 \Rightarrow 4m^2 - m - 3 = 0$, so $m = 1$ or $-\frac{3}{4}$. Thus $|m_1|, |m_2| = 1, \frac{3}{4}$.
Slope 1: $c^2 = 9$; the $x$-intercept $-c$ is positive for $c = -3$: $y = x - 3$, $\alpha = 3$.
Slope $\frac{3}{4}$: $c^2 = 25 - 9 = 16$; the $x$-intercept $-\frac{4c}{3}$ is positive for $c = -4$: $y = \frac{3}{4}x - 4$, $\beta = \frac{16}{3}$.
They meet where $x - 3 = \frac{3}{4}x - 4$: $x = -4$, $y = -7$. So Q $= (-4, -7)$.
$(PQ)^2 = 8^2 + 8^2 = 128$ and $\alpha\beta = 16$, so $\frac{(PQ)^2}{\alpha\beta} = 8$.
Answer: 8. | ||||
| 27 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical 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 At the given point $x - 2y + z - 2 = \frac{5}{3} - \frac{10}{3} + \frac{8}{3} - 2 = -1$.
Image: $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right) - \frac{2(-1)}{1 + 4 + 1}(1, -2, 1) = \left(\frac{5}{3} + \frac{1}{3}, \frac{5}{3} - \frac{2}{3}, \frac{8}{3} + \frac{1}{3}\right) = (2, 1, 3)$.
$PQ^2 = 4^2 + 3^2 + (\alpha - 3)^2 = 169$, so $(\alpha - 3)^2 = 144$, $\alpha = 15$ or $-9$. With $\alpha > 0$, $\alpha = 15$.
Answer: 15. | ||||
| 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 $|\vec{a}|^2 = 11$, $|\vec{c}|^2 = 22$.
$|\vec{b} \times \vec{c}|^2 = |\vec{b}|^2|\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2$: $11 = 50 \times 22 - (\vec{b} \cdot \vec{c})^2$, so $(\vec{b} \cdot \vec{c})^2 = 1089$, $\vec{b} \cdot \vec{c} = \pm 33$.
$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2\,\vec{b} \cdot \vec{c} = 72 \pm 66$.
$\left|72 - |\vec{b} + \vec{c}|^2\right| = 66$ either way.
Answer: 66. | ||||
| 29 | StatisticsMean deviation | Class 11: Statistics | NumericalEasy | |
A easy numerical question on mean deviation, from Statistics. NCERT: Class 11: Statistics $\sum f = 64 + \alpha$, $\sum fx = 4 + 72 + 140 + 7\alpha + 72 = 288 + 7\alpha$.
Mean 5: $288 + 7\alpha = 5(64 + \alpha)$, so $\alpha = 16$ and $N = 80$.
$\sum f|x - 5| = 4(4) + 24(2) + 28(0) + 16(2) + 8(4) = 128$, so $m = \frac{128}{80} = 1.6$.
$\sum f(x - 5)^2 = 4(16) + 24(4) + 0 + 16(4) + 8(16) = 352$, so $\sigma^2 = \frac{352}{80} = 4.4$.
$\frac{3\alpha}{m + \sigma^2} = \frac{48}{6} = 8$.
Answer: 8. | ||||
| 30 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin^{-1}\frac{u}{\sqrt{u^2 + 1}} = \tan^{-1}u$ for every real $u$. With $u = x + 1$ ($x^2 + 2x + 2 = (x+1)^2 + 1$) and $u = x$, the equation is $\tan^{-1}(x + 1) - \tan^{-1}x = \frac{\pi}{4}$.
The left side lies in $\left(0, \frac{\pi}{2}\right)$, so taking tangents: $\frac{1}{1 + x(x + 1)} = 1$, i.e. $x^2 + x = 0$: $S = \{0, -1\}$.
For both, $x^2 + x + 5 = 5$: $\sin\frac{5\pi}{2} - \cos 5\pi = 1 - (-1) = 2$.
Sum over $S$: $2 + 2 = 4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 $K = \frac{1}{2}mv^2 = \frac{1}{2}(5)(20)^2 = 1000$ J.
$\frac{\Delta K}{K} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v} = \frac{0.5}{5} + 2 \times \frac{0.4}{20} = 0.1 + 0.04 = 0.14$.
$\Delta K = 0.14 \times 1000 = 140$ J, so $K = (1000 \pm 140)$ J.
Answer: (B). | ||||
| 32 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Speeds: $108$ km/h $= 30$ m/s, $72$ km/h $= 20$ m/s.
To cross the tunnel a train covers the tunnel length plus its own length.
$t_A = \frac{L + l}{30} = \frac{61l}{30}$, $t_B = \frac{L + 4l}{20} = \frac{64l}{20} = \frac{96l}{30}$.
$t_B - t_A = \frac{35l}{30} = 35$ s, so $l = 30$ m and $L = 60l = 1800$ m.
Answer: (A). | ||||
| 33 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion In half a rotation without slipping the centre moves forward by half the circumference, $\pi R$.
Point A goes from the top of the disc to the bottom: its vertical displacement is $2R$ and its horizontal displacement equals that of the centre, $\pi R$.
Displacement $= \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$.
Answer: (D). | ||||
| 34 | 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 $K = \frac{p^2}{2m}$, so for the same $p$, $m \propto \frac{1}{K}$.
$m_1 : m_2 = K_2 : K_1 = 9 : 16$.
Answer: (B). | ||||
| 35 | 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 The impulse on the bullet is its change of momentum: $0.01 \times 600 = 6$ N s.
By Newton's third law the gun receives an equal and opposite impulse, of magnitude 6 N s (the barrel length and the gun's mass are not needed).
Answer: (C). | ||||
| 36 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}} \propto \sqrt{\frac{M}{R}}$.
$v = 11.2\sqrt{\frac{9}{4}} = 11.2 \times 1.5 = 16.8$ km/s.
Answer: (C). | ||||
| 37 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single 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 Continuity: $A_A v_A = A_B v_B$ with $A_A = 150$ mm$^2$, $A_B = 25$ mm$^2$, $v_B = 0.6$ m/s, so $v_A = 0.1$ m/s.
Bernoulli on the horizontal axis: $P_A - P_B = \frac{1}{2}\rho(v_B^2 - v_A^2) = 500(0.36 - 0.01) = 175$ Pa.
Answer: (B). | ||||
| 38 | Properties of Bulk MatterElasticity | 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 elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $B = -V\frac{dP}{dV} = -V(-3aV^{-4}) = 3aV^{-3} = 3P$.
Answer: (A). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$ and $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$, so $\frac{v_{rms}}{v_{avg}} = \sqrt{\frac{3\pi}{8}}$.
$1 + \frac{5}{x} = \frac{3\pi}{8} = \frac{3 \times 22}{8 \times 7} = \frac{33}{28} = 1 + \frac{5}{28}$, so $x = 28$.
Answer: (D). | ||||
| 40 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Total energy minus potential energy is the kinetic energy: $K = \frac{1}{2}k(A^2 - x^2)$.
It is largest at the mean position $x = 0$, zero at $x = \pm A$, symmetric about $x = 0$, and curves downward (an inverted parabola).
Only graph (A) has this shape: (B) is made of straight lines, (C) is the potential energy, and (D) is not symmetric.
Answer: (A). | ||||
| 41 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $P_1 = \frac{300 \times g \times 100}{300} = 100g$ W, $P_2 = \frac{50 \times g \times 100}{120} = \frac{125g}{3}$ W.
$\frac{P_1}{P_2} = \frac{300}{125} = \frac{12}{5}$.
$\frac{3\sqrt{x}}{\sqrt{x} + 1} = \frac{12}{5} \Rightarrow 15\sqrt{x} = 12\sqrt{x} + 12 \Rightarrow \sqrt{x} = 4$, $x = 16$.
Answer: (C). | ||||
| 42 | Electromagnetic WavesDisplacement current | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $\oint \vec{E} \cdot \vec{dl} = -\frac{\partial \phi_B}{\partial t}$ (Faraday's law) is the form that describes time-varying fields: a changing magnetic flux produces a non-conservative electric field.
$\oint \vec{E} \cdot \vec{dl} = 0$ holds only for static fields; $\oint \vec{B} \cdot \vec{dl} = \mu_0 I$ without the displacement-current term is the static (magnetostatic) form; Gauss's law $\oint \vec{D} \cdot \vec{dA} = Q$ holds in both cases.
The equation meant is Faraday's law.
Answer: (B). | ||||
| 43 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | StatementsMedium | |
A medium statements question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves (A) A permanent magnet gives a steady field. Not a source.
(B) An electric field changing linearly with time gives a constant displacement current, which produces a steady (time-independent) magnetic field. Not a source.
(C) A direct current gives a steady field. Not a source.
(D) A decelerating (accelerated) charge radiates: its magnetic field changes with time. A source.
(E) An antenna fed with a digital signal carries a current that switches between levels; at each switch the field changes, so in principle it too gives a time-varying field, but no option contains (D) and (E) together.
Of the options, only "(D) only" is consistent: (B) and (C) are wrong in every option that contains them.
Answer: (B). | ||||
| 44 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For layers viewed normally from above, the apparent depth is the sum of each real thickness divided by its refractive index.
$d_{app} = \frac{d/2}{n_1} + \frac{d/2}{n_2} = \frac{d(n_1 + n_2)}{2n_1 n_2}$.
Answer: (D). | ||||
| 45 | 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 $\lambda_0 = \frac{hc}{\phi}$: $\lambda_A = \frac{1242}{9} = 138$ nm, $\lambda_B = \frac{1242}{4.5} = 276$ nm.
Difference $= 276 - 138 = 138$ nm.
Answer: (D). | ||||
| 46 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Mass defect: $\Delta m = 238.05079 - 234.04363 - 4.00260 = 0.00456$ u.
$Q = 0.00456 \times 931.5 \approx 4.25$ MeV.
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 Input A passes through a NOR gate with its inputs joined, which gives $\bar{A}$; input B passes through an OR gate with its inputs joined, which gives $B$. Both feed a NAND gate:
$Y = \overline{\bar{A} \cdot B} = A + \bar{B}$.
Reading the inputs from the waveforms:
before $t_1$: A = 0, B = 0, Y = 1; $t_1$–$t_2$: A = 0, B = 1, Y = 0; $t_2$–$t_3$: A = 1, B = 1, Y = 1; $t_3$–$t_4$: A = 0, B = 0, Y = 1; $t_4$–$t_5$: A = 1, B = 1, Y = 1; $t_5$–$t_6$: A = 1, B = 0, Y = 1; after $t_6$: A = 0, B = 0, Y = 1.
So Y is 1 throughout except between $t_1$ and $t_2$, where it is 0: graph (C).
Answer: (C). | ||||
| 48 | Communication SystemsLayers of the atmosphere | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on layers of the atmosphere, from Communication Systems. NCERT: No matching NCERT chapter Troposphere: up to about 10 km (C – I).
D layer of the ionosphere: about 65–75 km (B – IV).
E layer: about 100 km (D – III).
$F_1$ layer: about 170–190 km (A – II).
A – II, B – IV, C – I, D – III.
Answer: (C). | ||||
| 49 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $p = q \cdot 2a = 0.01 \times 0.4 \times 10^{-3} = 4 \times 10^{-6}$ C m.
$E = 10$ dyne/C $= 10 \times 10^{-5}$ N/C $= 10^{-4}$ N/C.
$\tau = pE\sin 30° = 4 \times 10^{-6} \times 10^{-4} \times \frac{1}{2} = 2 \times 10^{-10}$ N m.
Answer: (B). | ||||
| 50 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity The same current $I$ enters each combination, so $P = I^2 R_{eq}$ and the order of power is the order of $R_{eq}$.
(A) two in parallel, then one in series: $\frac{R}{2} + R = \frac{3R}{2}$.
(B) two in series, parallel with one: $\frac{2R \cdot R}{3R} = \frac{2R}{3}$.
(C) three in parallel: $\frac{R}{3}$.
(D) three in series: $3R$.
$\frac{R}{3} < \frac{2R}{3} < \frac{3R}{2} < 3R$, so $P_C < P_B < P_A < P_D$.
Answer: (B). | ||||
| 51 | 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 In Bohr's model $r \propto \frac{n^2}{Z}$.
$r_1 \propto \frac{2^2}{2} = 2$ ($He^+$, $Z = 2$), $r_2 \propto \frac{4^2}{4} = 4$ ($Be^{3+}$, $Z = 4$).
$\frac{r_2}{r_1} = 2$, so $x = 2$.
Answer: 2. | ||||
| 52 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Let the bird be at height $y$ above the water and the fish at depth $h$. Seen from the water, the bird appears at height $\mu y$, so the apparent distance between them is $s = \mu y + h$.
$\frac{ds}{dt} = \mu\frac{dy}{dt} + \frac{dh}{dt}$: in magnitude, $12 = \frac{4}{3}v_b + 8$.
$v_b = 3$ m s$^{-1}$.
Answer: 3. | ||||
| 53 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Rate of storing energy in the inductor: $P = Li\frac{di}{dt} = i(E - iR)$.
This is maximum when $\frac{dP}{di} = E - 2iR = 0$, i.e. $i = \frac{E}{2R}$, giving $P_{max} = \frac{E^2}{4R} = \frac{E^2}{100}$ for $R = 25\ \Omega$.
$\frac{E^2}{100} = \frac{E^2}{2 \times 50}$, so $a = 2$, $b = 50$ and $\frac{b}{a} = 25$.
Answer: 25. | ||||
| 54 | 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 Between the line and the sheet, the field of the positive line points away from the line (towards the sheet) and that of the positive sheet points away from the sheet (towards the line): they are opposite.
At distance $r$ from the line: $E = \frac{\lambda}{2\pi\varepsilon_0 r} - \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}\left(\frac{1}{\pi r} - \frac{1}{2}\right)$ with $\lambda = 2\sigma$.
P, $r = \frac{3}{\pi}$: $\frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$, so $E_P = \frac{\sigma}{6\varepsilon_0}$.
Q, $r = \frac{4}{\pi}$: $\frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$, so $E_Q = \frac{\sigma}{4\varepsilon_0}$.
$\frac{E_P}{E_Q} = \frac{4}{6}$, so $a = 6$.
Answer: 6. | ||||
| 55 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $U = \frac{1}{2}\frac{YA}{L}x^2$.
$80 = \frac{1}{2} \times \frac{2 \times 10^{11} \times A}{20} \times (0.02)^2 = 2 \times 10^{6} A$.
$A = 4 \times 10^{-5}$ m$^2 = 40$ mm$^2$.
Answer: 40. | ||||
| 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 About its own axis, $L = I\omega = \frac{2}{5}MR^2\omega$.
Total kinetic energy of rolling: $E = \frac{1}{2}Mv^2 + \frac{1}{2}\cdot\frac{2}{5}MR^2\omega^2 = \frac{7}{10}MR^2\omega^2$.
$\frac{L}{E} = \frac{2/5}{(7/10)\omega} = \frac{4}{7\omega} = \frac{\pi}{22}$, so $\omega = \frac{88}{7\pi}$.
With $\pi = \frac{22}{7}$, $\omega = 4$ rad/s (with $\pi = 3.1416$ it is 4.001, still 4).
Answer: 4. | ||||
| 57 | 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 $y = A\cos 30° = \frac{\sqrt{3}}{2}A$, so $A^2 - y^2 = \frac{A^2}{4}$.
$K = \frac{1}{2}k(A^2 - y^2) = \frac{kA^2}{8} = \frac{k(0.4)^2}{8} = 0.02k$.
$0.02k = 200 \Rightarrow k = 10^4$ N m$^{-1}$, so $x = 4$.
Answer: 4. | ||||
| 58 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Let $G$ be the resistance and $I_g$ the full-scale current of the galvanometer.
With the 5 Ω shunt: $I_g G = (0.25 - I_g) \times 5$, so $I_g(G + 5) = 1.25$.
With 1050 Ω in series: $I_g(G + 1050) = 25$.
Dividing: $\frac{G + 1050}{G + 5} = 20 \Rightarrow G + 1050 = 20G + 100 \Rightarrow G = 50$ Ω.
Answer: 50. | ||||
| 59 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits From the graph, $I_C$ rises by 10 mA for every 100 μA rise of $I_B$: current gain $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{10 \times 10^{-3}}{100 \times 10^{-6}} = 100$.
Voltage gain $= \beta\frac{R_C}{R_B} = 100 \times \frac{1}{10} = 10$.
Power gain $= \beta \times$ voltage gain $= \beta^2\frac{R_C}{R_B} = 100 \times 10 = 10^3$, so $x = 3$.
Answer: 3. | ||||
| 60 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $P_1 = \frac{V_0^2}{R}$.
Each half has resistance $\frac{R}{2}$, so each dissipates $\frac{2V_0^2}{R}$ and together $P_2 = \frac{4V_0^2}{R}$.
$P_2 : P_1 = 4 : 1 = \sqrt{16} : 1$, so $x = 16$.
Answer: 16. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In Bohr's model $E_n = \frac{E_1}{n^2}$ for hydrogen.
$E_3 = \frac{E_1}{9}$: one ninth of the first-orbit value ($-2.42 \times 10^{-19}$ J).
Answer: (B). | ||||
| 62 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure NO (15 electrons) has one unpaired electron in $\pi^*2p$: bond order 2.5, paramagnetic. Removing it gives $NO^+$ (14 electrons, isoelectronic with $N_2$): bond order 3, diamagnetic. Both conditions hold.
$O_2 \to O_2^+$: bond order 2 → 2.5, but $O_2^+$ is still paramagnetic.
$O_2 \to O_2^{2-}$: becomes diamagnetic, but bond order falls from 2 to 1.
$N_2 \to N_2^+$: bond order falls from 3 to 2.5 and the ion becomes paramagnetic.
Answer: (A). | ||||
| 63 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter A lyophilic sol acts as a protective colloid: its particles form a layer (film) around the lyophobic particles and protect them from coagulation by electrolytes.
Answer: (C). | ||||
| 64 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsMedium | |
A medium statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties A: Not correct. Cl has the more negative electron gain enthalpy; in the small 2p shell of F the added electron suffers strong repulsion.
B: Not correct as a general statement. Ionization enthalpy usually decreases down a group, but not throughout: in group 13, Ga has a higher first ionization enthalpy than Al, and Tl than In (poor shielding by d and f electrons); in group 14, Pb is higher than Sn.
C: Correct. Electronegativity is not a fixed property of an element; it varies with the atom it is bonded to (and with its hybridisation and oxidation state).
D: Not correct. $Al_2O_3$ is amphoteric, but NO is a neutral oxide.
Not correct: A, B and D.
Answer: (A). | ||||
| 65 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Titanium (and zirconium) is purified by the van Arkel method through the volatile iodide. Nickel is purified by the Mond process through $Ni(CO)_4$. Copper is refined electrolytically.
Liquation is used for low-melting metals such as tin; zinc is refined by distillation (it is volatile). So "Zinc – Liquation" is the mismatch.
Answer: (C). | ||||
| 66 | HydrogenHydrogen and its compounds | No matching NCERT chapter | StatementsEasy | |
A easy statements question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Statement I: the synthetic resin (ion-exchange) method is more efficient than the permutit (zeolite) process, which softens water by exchanging $Ca^{2+}$/$Mg^{2+}$ for $Na^+$. Statement I is incorrect.
Statement II: in the synthetic resin method the cation exchanger replaces cations by $H^+$ and the anion exchanger replaces anions by $OH^-$, which combine to water; it is the permutit process that leaves soluble sodium salts in the water. Statement II is incorrect.
Answer: (B). | ||||
| 67 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter $Be(OH)_2$ is amphoteric; with the strong base $Sr(OH)_2$ it acts as an acid: $Be(OH)_2 + Sr(OH)_2 \to Sr[Be(OH)_4]$.
The salt contains both Sr and Be (A true); the reaction is an acid–base neutralisation (B true); Be is in the anion $[Be(OH)_4]^{2-}$, tetrahedrally coordinated (D true).
"Be is present in the cationic part" is incorrect.
Answer: (C). | ||||
| 68 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter In $ClF_5$, Cl has 5 bond pairs and 1 lone pair: $sp^3d^2$, octahedral electron geometry, square pyramidal shape.
$ClF_5$ boils at about $-13\,°C$, so at room temperature it is a colourless gas.
Answer: (C). | ||||
| 69 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Borazine $B_3N_3H_6$ ("inorganic benzene") is a planar six-membered ring of alternate B and N (cyclic), with the nitrogen lone pairs delocalised into the empty boron p orbitals, and it is hydrolysed by water (to boric acid, ammonia and hydrogen).
Banana (three-centre two-electron) bonds are found in diborane, not borazine.
Answer: (B). | ||||
| 70 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The third ionization removes an electron from the $M^{2+}$ ion. $Eu^{2+}$ is $4f^7$ (half-filled) and $Yb^{2+}$ is $4f^{14}$ (completely filled); removing an electron from these stable configurations needs a lot of energy.
So Eu and Yb both have high third ionization energies (which is also why both show a stable +2 state).
Answer: (B). | ||||
| 71 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | StatementsEasy | |
A easy statements question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A: Chlorophyll contains Mg, not Co (Co is in vitamin $B_{12}$). Mismatched.
B: Hardness of water is estimated by titration with EDTA. Correct.
C: In photography the undeveloped AgBr is dissolved as $[Ag(S_2O_3)_2]^{3-}$; $[Ag(CN)_2]^-$ is used in silver plating and extraction. Mismatched.
D: Wilkinson's catalyst is $[(Ph_3P)_3RhCl]$. Correct.
E: D-penicillamine is a chelating ligand used to remove excess copper or lead. Correct.
Mismatched: A and C.
Answer: (B). | ||||
| 72 | Environmental ChemistryOzone depletion | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ozone depletion, from Environmental Chemistry. NCERT: No matching NCERT chapter UV radiation breaks CFCs to give chlorine radicals: $CF_2Cl_2 \xrightarrow{h\nu} {}^{\bullet}CF_2Cl + Cl^{\bullet}$.
$Cl^{\bullet} + O_3 \to ClO^{\bullet} + O_2$ and $ClO^{\bullet} + O \to Cl^{\bullet} + O_2$: the chlorine radical is regenerated and destroys ozone catalytically.
Answer: (C). | ||||
| 73 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure In (D) the central C=C bond polarises so that the three-membered ring carries a positive charge (cyclopropenium, $2\pi$ electrons, aromatic) and the five-membered ring a negative charge (cyclopentadienide, $6\pi$ electrons, aromatic). Both charge-separated rings are aromatic, so this dipolar form contributes heavily and the molecule has a large dipole moment.
(A) and (C) are symmetric hydrocarbons whose two halves are identical: no net dipole. (B) cyclopropanone has a C=O dipole, but much smaller than that of the aromatic zwitterion-like (D).
Answer: (D). | ||||
| 74 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons n-Alkanes heated with anhydrous $AlCl_3$ and HCl isomerise to branched alkanes (through carbocation rearrangements).
n-Hexane gives mainly 2-methylpentane, $CH_3CH(CH_3)CH_2CH_2CH_3$, with some 3-methylpentane.
Answer: (D). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes 2-Methylpropyl (isobutyl) bromide, $(CH_3)_2CHCH_2Br$, is a primary halide.
With the strong nucleophile $C_2H_5O^-$ it reacts by $S_N2$ without rearrangement: A = isobutyl ethyl ether, $(CH_3)_2CHCH_2OC_2H_5$.
With the weak nucleophile $C_2H_5OH$ (solvolysis) the reaction goes through a carbocation ($S_N1$); the primary cation rearranges by a hydride shift to the tertiary $(CH_3)_3C^+$, which ethanol captures: B = tert-butyl ethyl ether.
Answer: (A). | ||||
| 76 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctHard | |
A hard single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Protonation and loss of water give a secondary carbocation on the carbon joining the cyclobutyl ring (A) and the cyclopentyl ring (B).
1. Ring A (4-membered, strained) expands: a ring bond migrates to the cationic carbon, giving a five-membered ring with the positive charge on a ring carbon; a hydride shift then makes the cation tertiary at the carbon carrying ring B.
2. Ring B (5-membered) next to this cation expands to a six-membered ring, giving a spiro cation.
3. The former ring A (now five-membered) is next to the new cation and expands once more, giving a fused bicyclic (decalin-type) tertiary cation in which both rings are six-membered; loss of a proton gives the alkene.
Six-membered rings are strain-free, so expansion stops there: both rings finally become six-membered.
Answer: (A). | ||||
| 77 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines The starting compound is N-methylpyrrolidin-2-one, a cyclic amide (lactam).
Hot aqueous NaOH hydrolyses the amide bond, opening the ring to the sodium salt of 4-(methylamino)butanoic acid; acidification gives $MeHN-CH_2CH_2CH_2-COOH$.
Answer: (A). | ||||
| 78 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctHard | |
A hard single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques HCl protonates the amine (it becomes $-NH_3^+$ and takes no further part) and the tertiary benzylic OH; loss of water gives a tertiary benzylic carbocation next to the spiro carbon of the cyclopentane ring.
A bond of the cyclopentane ring migrates to the cationic carbon (ring expansion), turning the five-membered ring into a six-membered ring fused to the tetralin: the cationic carbon becomes a quaternary ring-junction carbon carrying the $CH_2CH_2NH_2$ chain, and the positive charge moves to the former spiro carbon (now the other ring-junction carbon); chloride traps it.
KOH then eliminates HCl towards the neighbouring $CH_2$ of the middle ring, giving the alkene with the double bond in the middle ring, and frees the amine.
This is structure (D).
Answer: (D). | ||||
| 79 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Nylon 6 is made from caprolactam (a→III). Vulcanised rubber is cross-linked by sulphur bridges (b→II). cis-1,4-Polyisoprene is natural rubber (c→I). Polychloroprene is neoprene (d→IV).
Answer: (B). | ||||
| 80 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules HCN adds to the CHO group of D-glyceraldehyde from both faces, creating a new stereocentre: two diastereomeric cyanohydrins.
Hydrolysis gives the two 2,3,4-trihydroxybutanoic acids, and $HNO_3$ oxidises the terminal $CH_2OH$ to COOH: both become 2,3-dihydroxybutanedioic (tartaric) acids.
One has the two OH groups on the same side in the Fischer projection: meso-tartaric acid (optically inactive). The other is the chiral D-tartaric acid (optically active).
Answer: (B). | ||||
| 81 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions $2KMnO_4 + 10FeSO_4(NH_4)_2SO_4 \cdot 6H_2O + 8H_2SO_4 \to K_2SO_4 + 2MnSO_4 + 5Fe_2(SO_4)_3 + 10(NH_4)_2SO_4 + 68H_2O$.
Each $MnO_4^-$ takes 5 electrons, so 2 $KMnO_4$ oxidise 10 Mohr's salt units. The redox step itself gives $8H_2O$ ($2MnO_4^- + 16H^+ + 10Fe^{2+} \to 2Mn^{2+} + 10Fe^{3+} + 8H_2O$), and the 10 hexahydrate units release $60H_2O$.
Total water molecules: $8 + 60 = 68$.
Answer: 68. | ||||
| 82 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $PV = ZnRT$, so for the same amount of gas $\frac{P_1V_1}{Z_1T_1} = \frac{P_2V_2}{Z_2T_2}$.
$V_2 = V_1 \cdot \frac{P_1}{P_2} \cdot \frac{Z_2}{Z_1} \cdot \frac{T_2}{T_1} = 0.15 \times \frac{100}{300} \times \frac{1.4}{1.07} \times \frac{300}{500} = 0.03 \times 1.3084 = 0.039252$ dm$^3$.
$= 392.5 \times 10^{-4}$ dm$^3$, which rounds to 393.
Answer: 393. | ||||
| 83 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Let the bond enthalpies be $x$ ($A_2$), $0.5x$ ($B_2$) and $x$ (AB).
$\Delta H_f^0 = -200$ kJ mol$^{-1}$ is per mole of AB, so for $A_2 + B_2 \to 2AB$, $\Delta H = 2 \times (-200) = -400$ kJ.
$\Delta H = $ (bonds broken) $-$ (bonds formed) $= x + 0.5x - 2x = -0.5x = -400$, so $x = 800$ kJ mol$^{-1}$.
Answer: 800. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Isotonic solutions at the same temperature have the same molar concentration: $\frac{12/M}{1\ \text{L}} = 0.05$, so $M = 240$ g mol$^{-1}$.
(With empirical formula $CH_2O$ of mass 30, the molecular formula is $C_8H_{16}O_8$.)
Answer: 240. | ||||
| 85 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium After mixing equal volumes the concentrations halve: $[Ba^{2+}] = 0.025$ M, $[F^-] = 0.010$ M.
Ionic product $= 0.025 \times (0.010)^2 = 2.5 \times 10^{-6}$.
Ratio $= \frac{2.5 \times 10^{-6}}{0.5 \times 10^{-6}} = 5$.
Answer: 5. | ||||
| 86 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Volume of nickel $= 100$ cm$^2 \times 10^{-4}$ cm $= 0.01$ cm$^3$; mass $= 0.01 \times 10 = 0.1$ g; moles $= \frac{0.1}{60} = \frac{1}{600}$.
$Ni^{2+} + 2e^- \to Ni$: charge $= 2 \times \frac{1}{600} \times 96500 = 321.67$ C.
$x = \frac{321.67}{2} = 160.8$ s, which rounds to 161.
Answer: 161. | ||||
| 87 | 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 87.5% completion leaves $\frac{1}{8} = \left(\frac{1}{2}\right)^3$ of the reactant: three half-lives for a first-order reaction.
$t_{87.5} = 3t_{50}$, so $x = 3$.
Answer: 3. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Mass of C $= 0.220 \times \frac{12}{44} = 0.060$ g. Since C is 24%, the mass of compound $= \frac{0.060}{0.24} = 0.25$ g.
Mass of H $= 0.126 \times \frac{2}{18} = 0.014$ g.
% H $= \frac{0.014}{0.25} \times 100 = 5.6 = 56 \times 10^{-1}$.
Answer: 56. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A is 2,2,4-trimethylpentan-3-ol, $(CH_3)_3C-CH(OH)-CH(CH_3)_2$. With $H^+$ and heat it loses water to give a secondary carbocation at C-3, which rearranges to tertiary carbocations in two ways:
1. 1,2-Hydride shift from C-4: $(CH_3)_3C-CH_2-\overset{+}{C}(CH_3)_2$. Loss of a proton gives $(CH_3)_3C-CH=C(CH_3)_2$ (2,4,4-trimethylpent-2-ene) and $(CH_3)_3C-CH_2-C(CH_3)=CH_2$ (2,4,4-trimethylpent-1-ene).
2. 1,2-Methyl shift from C-2: $(CH_3)_2\overset{+}{C}-CH(CH_3)-CH(CH_3)_2$. Loss of a proton gives $(CH_3)_2C=C(CH_3)-CH(CH_3)_2$ (2,3,4-trimethylpent-2-ene) and $CH_2=C(CH_3)-CH(CH_3)-CH(CH_3)_2$ (2,3,4-trimethylpent-1-ene).
None of these alkenes shows geometrical isomerism. Total products from the tertiary carbocations: 4.
Answer: 4. | ||||
| 90 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Standardisation: $35.5 - 25.5 = 10.0$ mL of $Ca(OH)_2$ neutralise $20 \times 0.5 = 10$ mmol HCl. $Ca(OH)_2$ gives 2 $OH^-$, so it holds 5 mmol in 10 mL: 0.5 M.
20 mL of this $Ca(OH)_2$ = 10 mmol, which neutralises 10 mmol $H_2SO_4$ (1 : 1).
$[H_2SO_4] = \frac{10}{10} = 1$ M.
Answer: 1. | ||||
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