JEE (Main) · 2023
JEE (Main) 2023 — 6 April, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Limits, Continuity and Differentiability3
- Integral Calculus3
- Three Dimensional Geometry3
8 easy · 20 medium · 2 hard
Physics
30 Q
- Electronic Devices2
- Properties of Bulk Matter2
- Atoms and Nuclei2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- Amines3
- Biomolecules2
- Redox Reactions2
16 easy · 12 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 FunctionsSets and their representation | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on sets and their representation, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Set A: $[x+3] + [x+4] = ([x] + 3) + ([x] + 4) = 2[x] + 7$. So $2[x] + 7 \le 3$ gives $[x] \le -2$, i.e. $x < -1$. Hence $A = (-\infty, -1)$.
Set B: $\sum_{r=1}^{\infty} \frac{3}{10^r} = 0.333\ldots = \frac{1}{3}$. The inequality becomes $3^x \cdot 3^{-(x-3)} < 3^{-3x}$, i.e. $3^3 < 3^{-3x}$, so $3 < -3x$, i.e. $x < -1$. Hence $B = (-\infty, -1)$.
So $A = B$.
Answer: (A). | ||||
| 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 $x^2 - 8x + 15 = (x-3)(x-5)$.
Case 1: $x \le 3$ or $x \ge 5$. Then $x^2 - 8x + 15 - 2x + 7 = 0$, i.e. $x^2 - 10x + 22 = 0$, $x = 5 \pm \sqrt{3}$. Now $5 + \sqrt{3} \approx 6.73 \ge 5$ is valid; $5 - \sqrt{3} \approx 3.27$ lies in $(3, 5)$, so it is rejected.
Case 2: $3 < x < 5$. Then $-(x^2 - 8x + 15) - 2x + 7 = 0$, i.e. $x^2 - 6x + 8 = 0$, $x = 2$ or $4$. Only $x = 4$ lies in $(3, 5)$.
Roots: $4$ and $5 + \sqrt{3}$. Sum $= 9 + \sqrt{3}$.
Answer: (B). | ||||
| 3 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Write $A = \begin{pmatrix} p & q \\ r & s \end{pmatrix}$ with all entries non-zero. Then
$A^2 = \begin{pmatrix} p^2 + qr & q(p+s) \\ r(p+s) & s^2 + qr \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.
Since $q \ne 0$, $q(p+s) = 0$ gives $p + s = 0$, so $a = p + s = 0$.
Then $s = -p$ and $p^2 + qr = 1$, so $b = |A| = ps - qr = -p^2 - qr = -(p^2 + qr) = -1$.
$3a^2 + 4b^2 = 0 + 4 = 4$.
Answer: (B). | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For infinitely many solutions the coefficient determinant must vanish:
$\begin{vmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{vmatrix} = 1(15 - 4) - 1(6 - 2) + a(4 - 5) = 7 - a = 0$, so $a = 7$.
The first equation must then be a combination of the other two: $\alpha(2, 5, 2) + \beta(1, 2, 3) = (1, 1, 7)$. From $2\alpha + \beta = 1$ and $5\alpha + 2\beta = 1$: $\alpha = -1$, $\beta = 3$; check the z-coefficient: $2(-1) + 3(3) = 7$. ✓
The right-hand sides must combine the same way: $b = 6\alpha + 3\beta = -6 + 9 = 3$.
$2a + 3b = 14 + 9 = 23$.
Answer: (B). | ||||
| 5 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $\frac{{}^{2n}C_3}{{}^{n}C_3} = \frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = \frac{4(2n-1)}{n-2} = 10$.
So $8n - 4 = 10n - 20$, giving $n = 8$.
$(n^2 + 3n) : (n^2 - 3n + 4) = (64 + 24) : (64 - 24 + 4) = 88 : 44 = 2 : 1$.
Answer: (C). | ||||
| 6 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Fifth term from the beginning: $T_5 = {}^nC_4 (2^{1/4})^{n-4} (3^{-1/4})^4$.
Fifth term from the end is the fifth term of $\left(3^{-1/4} + 2^{1/4}\right)^n$: ${}^nC_4 (3^{-1/4})^{n-4} (2^{1/4})^4$.
Ratio $= (2^{1/4})^{n-8} (3^{-1/4})^{8-n} = 2^{\frac{n-8}{4}} 3^{\frac{n-8}{4}} = 6^{\frac{n-8}{4}}$.
Setting this equal to $6^{1/2}$: $\frac{n-8}{4} = \frac{1}{2}$, so $n = 10$.
$T_3 = {}^{10}C_2 (2^{1/4})^8 (3^{-1/4})^2 = 45 \times 4 \times \frac{1}{\sqrt{3}} = \frac{180}{\sqrt{3}} = 60\sqrt{3}$.
Answer: (B). | ||||
| 7 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The differences $6, 8, 10, 12, \ldots$ form an AP, so $T_n$ is quadratic: $T_n = n^2 + 3n + 1$ (check: $T_1 = 5$, $T_2 = 11$, $T_3 = 19$).
$S_{20} = \sum n^2 + 3\sum n + 20 = \frac{20 \cdot 21 \cdot 41}{6} + 3 \cdot \frac{20 \cdot 21}{2} + 20 = 2870 + 630 + 20 = 3520$.
Answer: (D). | ||||
| 8 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Rationalise each term: $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$.
The sum telescopes to $\frac{\sqrt{a_n} - \sqrt{a_1}}{d}$.
The expression is $\sqrt{\frac{d}{n}} \cdot \frac{\sqrt{a_n} - \sqrt{a_1}}{d} = \frac{\sqrt{a_n} - \sqrt{a_1}}{\sqrt{d}\sqrt{n}}$.
With $a_n = a_1 + (n-1)d$, $\frac{\sqrt{a_n}}{\sqrt{n}} = \sqrt{\frac{a_1 + (n-1)d}{n}} \to \sqrt{d}$, while $\frac{\sqrt{a_1}}{\sqrt{n}} \to 0$.
Limit $= \frac{\sqrt{d}}{\sqrt{d}} = 1$.
Answer: (A). | ||||
| 9 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\frac{d}{dx}x^y = x^y\left(y'\ln x + \frac{y}{x}\right)$ and $\frac{d}{dx}y^x = y^x\left(\ln y + \frac{x}{y}y'\right)$.
Differentiating: $2x^y\left(y'\ln x + \frac{y}{x}\right) + 3y^x\left(\ln y + \frac{x}{y}y'\right) = 0$.
At $(2, 2)$, $x^y = y^x = 4$: $8(y'\ln 2 + 1) + 12(\ln 2 + y') = 0$.
$y'(8\ln 2 + 12) = -(8 + 12\ln 2)$, so $y' = -\frac{8 + 12\ln 2}{12 + 8\ln 2} = -\frac{2 + 3\ln 2}{3 + 2\ln 2} = -\left(\frac{2 + \log_e 8}{3 + \log_e 4}\right)$.
Answer: (D). | ||||
| 10 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replace x by $\frac{1}{x}$: $5f\left(\frac{1}{x}\right) + 4f(x) = x + 3$.
Multiply the given equation by 5 and this one by 4 and subtract:
$25f(x) - 16f(x) = \frac{5}{x} + 15 - 4x - 12$, so $f(x) = \frac{1}{9}\left(\frac{5}{x} - 4x + 3\right)$.
$18\int_1^2 f(x)\,dx = 2\int_1^2\left(\frac{5}{x} - 4x + 3\right)dx = 2\left[5\ln x - 2x^2 + 3x\right]_1^2 = 2(5\ln 2 - 6 + 3) = 10\log_e 2 - 6$.
Answer: (C). | ||||
| 11 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Note $\frac{d}{dx}(x\tan x + 1) = \tan x + x\sec^2 x$. Integrate by parts with $u = x^2$ and $dv = \frac{(x\tan x + 1)'}{(x\tan x + 1)^2}dx$, so $v = -\frac{1}{x\tan x + 1}$:
$I = -\frac{x^2}{x\tan x + 1} + \int \frac{2x}{x\tan x + 1}dx$.
$\frac{2x}{x\tan x + 1} = \frac{2x\cos x}{x\sin x + \cos x}$, and $\frac{d}{dx}(x\sin x + \cos x) = x\cos x$, so the integral is $2\ln|x\sin x + \cos x|$.
$I(x) = -\frac{x^2}{x\tan x + 1} + 2\ln(x\sin x + \cos x) + C$; $I(0) = 0$ gives $C = 0$.
At $x = \frac{\pi}{4}$: $x\tan x + 1 = \frac{\pi + 4}{4}$, so the first term is $-\frac{\pi^2}{16}\cdot\frac{4}{\pi + 4} = -\frac{\pi^2}{4(\pi + 4)}$.
$x\sin x + \cos x = \frac{1}{\sqrt{2}}\left(\frac{\pi}{4} + 1\right) = \frac{\pi + 4}{4\sqrt{2}}$, so $2\ln(\ldots) = \ln\frac{(\pi + 4)^2}{32}$.
$I\left(\frac{\pi}{4}\right) = \log_e\frac{(\pi + 4)^2}{32} - \frac{\pi^2}{4(\pi + 4)}$.
Answer: (A). | ||||
| 12 | 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 L meets the axes at $(5, 0)$ and $(0, 9)$. The points of trisection are $\left(\frac{10}{3}, 3\right)$ and $\left(\frac{5}{3}, 6\right)$.
Slopes from the origin: $m_1 = \frac{3}{10/3} = \frac{9}{10}$, $m_2 = \frac{6}{5/3} = \frac{18}{5}$; $m_1 + m_2 = \frac{9}{2}$.
Intersect $y = \frac{9}{2}x$ with $9x + 5y = 45$: $9x + \frac{45}{2}x = 45$, so $x = \frac{10}{7}$, $y = \frac{45}{7}$.
Check the options: $y - x = \frac{35}{7} = 5$. ✓ (The others give $\frac{25}{7}$, $\frac{15}{7}$, $\frac{105}{7}$.)
Answer: (C). | ||||
| 13 | 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 Plane through the line of intersection: $(2x - y + z - 3) + \lambda(4x - 3y + 5z + 9) = 0$, normal $(2 + 4\lambda, -1 - 3\lambda, 1 + 5\lambda)$.
Parallel to the line means the normal is perpendicular to $(-2, 4, 5)$:
$-2(2 + 4\lambda) + 4(-1 - 3\lambda) + 5(1 + 5\lambda) = -3 + 5\lambda = 0$, so $\lambda = \frac{3}{5}$.
Multiply by 5: $5(2x - y + z - 3) + 3(4x - 3y + 5z + 9) = 22x - 14y + 20z + 12 = 0$, i.e. $11x - 7y + 10z + 6 = 0$.
$a + b + c = 11 - 7 + 10 = 14$.
Answer: (C). | ||||
| 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 OP: through $(0,0,0)$ with direction $(3, 4, 5)$. The edges parallel to the z-axis that pass through neither O nor P are the lines through $(3, 0, 0)$ and $(0, 4, 0)$ with direction $(0, 0, 1)$.
$(3, 4, 5) \times (0, 0, 1) = (4, -3, 0)$, of length 5.
For the edge through $(3, 0, 0)$: distance $= \frac{|(3, 0, 0)\cdot(4, -3, 0)|}{5} = \frac{12}{5}$.
For the edge through $(0, 4, 0)$: $\frac{|(0, 4, 0)\cdot(4, -3, 0)|}{5} = \frac{12}{5}$ as well.
Answer: (A). | ||||
| 15 | 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 Take B as base: $\vec{BA} = (4, 3, 2\lambda - 3)$, $\vec{BC} = (-3, \lambda - 2, 1)$, $\vec{BD} = (-2, 3, 3)$.
Coplanar ⇔ $\begin{vmatrix} 4 & 3 & 2\lambda - 3 \\ -3 & \lambda - 2 & 1 \\ -2 & 3 & 3 \end{vmatrix} = 0$.
$= 4(3\lambda - 6 - 3) - 3(-9 + 2) + (2\lambda - 3)(-9 + 2\lambda - 4) = 12\lambda - 36 + 21 + (2\lambda - 3)(2\lambda - 13)$
$= 4\lambda^2 - 20\lambda + 24 = 4(\lambda - 2)(\lambda - 3)$.
So $S = \{2, 3\}$ and $\sum(\lambda + 2)^2 = 16 + 25 = 41$.
Answer: (C). | ||||
| 16 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{vmatrix} = (-6 + 8)\hat{i} - (3 - 2)\hat{j} + (4 - 2)\hat{k} = 2\hat{i} - \hat{j} + 2\hat{k}$.
$\vec{d} = t(2, -1, 2)$; $\vec{a}\cdot\vec{d} = t(4 - 3 + 8) = 9t = 18$, so $t = 2$ and $\vec{d} = (4, -2, 4)$, $|\vec{d}|^2 = 36$.
$|\vec{a} \times \vec{d}|^2 = |\vec{a}|^2|\vec{d}|^2 - (\vec{a}\cdot\vec{d})^2 = 29 \times 36 - 324 = 1044 - 324 = 720$.
Answer: (C). | ||||
| 17 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Combined variance of two groups:
$\sigma^2_{\text{total}} = \frac{n_1\sigma_1^2 + n_2\sigma_2^2}{n_1 + n_2} + \frac{n_1 n_2 (\bar{x}_1 - \bar{x}_2)^2}{(n_1 + n_2)^2}$.
$13 = \frac{15 \cdot 14 + 15\sigma^2}{30} + \frac{15 \cdot 15 \cdot 4}{900} = \frac{14 + \sigma^2}{2} + 1$.
So $14 + \sigma^2 = 24$ and $\sigma^2 = 10$.
Answer: (C). | ||||
| 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 A total of 5 comes from (1,4), (2,3), (3,2), (4,1): $p = \frac{4}{36} = \frac{1}{9}$, $q = \frac{8}{9}$.
$P(X \ge 4) = {}^5C_4 p^4 q + p^5 = \frac{5 \cdot 8 + 1}{9^5} = \frac{41}{3^{10}} = \frac{123}{3^{11}}$.
So $k = 123$.
Answer: (C). | ||||
| 19 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Depression of Q is $60^\circ$: $BQ = \frac{30}{\tan 60^\circ} = 10\sqrt{3}$ m.
C is on AB with $CB = PQ$, so C is level with P and $CP = BQ = 10\sqrt{3}$. Depression of P is $15^\circ$: $AC = CP\tan 15^\circ = 10\sqrt{3}(2 - \sqrt{3}) = 20\sqrt{3} - 30$.
$PQ = CB = 30 - AC = 60 - 20\sqrt{3}$.
BCPQ is a rectangle: area $= BQ \cdot PQ = 10\sqrt{3}(60 - 20\sqrt{3}) = 600\sqrt{3} - 600 = 600(\sqrt{3} - 1)$.
Answer: (B). | ||||
| 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 $(P \Rightarrow Q) \wedge (R \Rightarrow Q) \equiv (\sim P \vee Q) \wedge (\sim R \vee Q) \equiv (\sim P \wedge \sim R) \vee Q \equiv \sim(P \vee R) \vee Q \equiv (P \vee R) \Rightarrow Q$.
(Check (B): with P true, R false, Q false, $(P \wedge R) \Rightarrow Q$ is true but the given statement is false, so (B) is not equivalent.)
Answer: (D). | ||||
| 21 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Let $t = a - b$, an integer from $-9$ to $9$. We need $2t^2 + 3t \in \{0, 1, 2, 3, 4\}$.
$t = 0$: 0 ✓; $t = 1$: 5 ✗; $t = -1$: $-1$ ✗; $t = -2$: $8 - 6 = 2$ ✓; $t = 2$: 14 ✗; $t = -3$: 9 ✗; larger $|t|$ give larger values.
$t = 0$: $a = b$, 10 pairs. $t = -2$: $b = a + 2$ with $a = 1, \ldots, 8$: 8 pairs.
Total $= 18$.
Answer: 18. | ||||
| 22 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines The region lies above the line $x + y = 3$ and inside the circle $x^2 + y^2 = 9$, with $0 \le x \le 3$.
Above the line: $p + 1 > 3 - p$, so $p > 1$.
Inside the circle: $p^2 + (p + 1)^2 < 9$, i.e. $2p^2 + 2p - 8 < 0$, i.e. $p^2 + p - 4 < 0$, so $p < \frac{-1 + \sqrt{17}}{2}$ (about 1.56, within $0 \le x \le 3$).
So $a = 1$, $b = \frac{\sqrt{17} - 1}{2}$, and b is a root of $b^2 + b - 4 = 0$, i.e. $b^2 + b = 4$.
$b^2 + b - a^2 = 4 - 1 = 3$.
Answer: 3. | ||||
| 23 | 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 Each orange can go to any of the 3 children: $3^{20}$ ways. Remove those in which some child gets none (inclusion–exclusion):
$3^{20} - {}^3C_1 2^{20} + {}^3C_2 1^{20} = 3486784401 - 3 \times 1048576 + 3 = 3486784401 - 3145728 + 3 = 3483638676$.
Answer: 3483638676. | ||||
| 24 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = {}^{15}C_r (x^4)^{15 - r}\left(-\frac{1}{x^3}\right)^r = (-1)^r\,{}^{15}C_r\,x^{60 - 7r}$.
$60 - 7r = 18$ gives $r = 6$.
Coefficient $= (-1)^6\,{}^{15}C_6 = 5005$.
Answer: 5005. | ||||
| 25 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Since a is an integer, $f(x) = a + [13\sin x]$. On $(0, \pi)$, $13\sin x$ rises from 0 to 13 at $x = \frac{\pi}{2}$ and falls back to 0.
$[13\sin x]$ jumps (so f is discontinuous, hence not differentiable) wherever $13\sin x$ passes through an integer $k = 1, 2, \ldots, 12$: each such value is taken twice, giving $2 \times 12 = 24$ points.
At $x = \frac{\pi}{2}$, $13\sin x = 13$ exactly, while nearby it is just below 13: $f = a + 13$ there and $a + 12$ on either side, so f is discontinuous at $\frac{\pi}{2}$ too.
Total $= 24 + 1 = 25$.
Answer: 25. | ||||
| 26 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $x^2 \le 2y$: on or above the parabola $y = \frac{x^2}{2}$. $x^2 \ge 2y - y^2$: on or outside the circle $x^2 + (y - 1)^2 = 1$. And $y \le x$.
The parabola and the line $y = x$ meet at $x = 0$ and $x = 2$. Area between them: $\int_0^2\left(x - \frac{x^2}{2}\right)dx = 2 - \frac{4}{3} = \frac{2}{3}$.
Every point of the circle's disc has $x^2 \le 2y - y^2 \le 2y$, so the disc lies above the parabola. The line $y = x$ cuts the circle at $(0, 0)$ and $(1, 1)$; the part of the disc below the line is the minor segment cut off by this chord. Seen from the centre $(0, 1)$ the chord subtends $90^\circ$, so the segment's area is $\frac{\pi}{4} - \frac{1}{2}$.
Area of S $= \frac{2}{3} - \left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{7}{6} - \frac{\pi}{4}$.
Matching $\frac{n + 2}{n + 1} - \frac{\pi}{n - 1}$: $n = 5$ gives $\frac{7}{6} - \frac{\pi}{4}$.
Answer: 5. | ||||
| 27 | CirclesStandard and general form | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections The circle touching both axes in the first quadrant: $(x - r)^2 + (y - r)^2 = r^2$, with $A = (r, 0)$, $B = (0, r)$, and AB: $x + y = r$.
P above AB: $\alpha + \beta > r$, and $PQ = \frac{\alpha + \beta - r}{\sqrt{2}} = 11$, so $(\alpha + \beta - r)^2 = 242$.
P on the circle: $\alpha^2 + \beta^2 - 2r(\alpha + \beta) + r^2 = 0$. With $s = \alpha + \beta$, $\alpha^2 + \beta^2 = s^2 - 2\alpha\beta$, so $s^2 - 2rs + r^2 = 2\alpha\beta$, i.e. $(s - r)^2 = 2\alpha\beta$.
$\alpha\beta = \frac{242}{2} = 121$.
Answer: 121. | ||||
| 28 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Differentiate $x^2 + 2x - 4y + 9 = 0$: $2x + 2 - 4y' = 0$, so $y' = \frac{x + 1}{2} = 1$ at P. Tangent: $y - 3 = x - 1$, i.e. $y = x + 2$; it meets the y-axis at $A(0, 2)$.
Line through P parallel to $x - 3y = 6$: $x - 3y = 1 - 9 = -8$.
B is on this line and on $2x - 3y = 8$: subtracting, $x = 16$, then $3y = 24$, $y = 8$. Check: $8^2 = 64 = 4 \times 16$, so B(16, 8) is on the parabola. ✓
$(AB)^2 = 16^2 + (8 - 2)^2 = 256 + 36 = 292$.
Answer: 292. | ||||
| 29 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Divide by $x\cos x\,dx$: $\frac{dy}{dx} + \left(\tan x + \frac{1}{x}\right)y = \frac{\sec x}{x}$.
Integrating factor: $e^{\int(\tan x + 1/x)dx} = x\sec x$. Then $\frac{d}{dx}(xy\sec x) = \sec^2 x$, so $xy\sec x = \tan x + C$, i.e. $xy = \sin x + C\cos x$.
$\frac{\pi}{3}y\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} + \frac{C}{2} = \sqrt{3}$ gives $C = \sqrt{3}$: $xy = \sin x + \sqrt{3}\cos x$.
Let $u = xy$. Then $u' = y + xy'$ and $u'' = 2y' + xy''$, which is exactly the expression asked for.
$u'' = -\sin x - \sqrt{3}\cos x$; at $x = \frac{\pi}{6}$: $-\frac{1}{2} - \frac{3}{2} = -2$.
$|{-2}| = 2$.
Answer: 2. | ||||
| 30 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium 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 Image of P in $2x - y + z = 9$: $\frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z - 3}{1} = -2\cdot\frac{2 - 2 + 3 - 9}{6} = 2$, so $Q = (5, 0, 5)$.
$\vec{PQ} = (4, -2, 2)$, $\vec{PR} = (5, 8, 4)$.
$\vec{PQ} \times \vec{PR} = \left((-2)(4) - (2)(8),\ (2)(5) - (4)(4),\ (4)(8) - (-2)(5)\right) = (-24, -6, 42)$.
$|\vec{PQ} \times \vec{PR}|^2 = 576 + 36 + 1764 = 2376$.
(Area)$^2 = \frac{2376}{4} = 594$.
Answer: 594. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Communication SystemsRange of a TV tower | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on range of a tv tower, from Communication Systems. NCERT: No matching NCERT chapter The range of a tower of height h is $d = \sqrt{2Rh}$ (R the radius of the earth), so $d \propto \sqrt{h}$.
$\frac{d'}{d} = \sqrt{1.21} = 1.1$: the range increases by 10%.
Answer: (B). | ||||
| 32 | 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 The +5 V line feeds the top of the circuit, the "1" contacts of both switches and one end of the LED. The other end of the LED goes to the bottom wire, which is joined to the moving contacts of A and B and goes to ground through R.
If A or B (or both) is closed to "1", the bottom wire is connected straight to +5 V: both ends of the LED are at the same potential and it does not glow (Y = 0).
Only when both A and B are open (A = 0, B = 0) does current flow from +5 V through the LED and R to ground, so the LED glows (Y = 1).
Truth table: 00 → 1, 01 → 0, 10 → 0, 11 → 0, which is the NOR gate.
Answer: (B). | ||||
| 33 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal velocity the net force is zero: weight = buoyant force + viscous force.
Volume of the ball $V = \frac{M}{\rho}$, so the buoyant force is $V\rho_0 g = Mg\frac{\rho_0}{\rho}$.
$F = Mg - Mg\frac{\rho_0}{\rho} = Mg\left(1 - \frac{\rho_0}{\rho}\right)$.
Answer: (A). | ||||
| 34 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei From the figure: A is $4 \to 3$, B is $4 \to 2$, C is $3 \to 2$, D is $3 \to 1$.
$E_n = -\frac{13.6}{n^2}$ eV. Energies emitted: A: $0.85 - 1.51$, i.e. $0.66$ eV; B: $3.40 - 0.85 = 2.55$ eV; C: $3.40 - 1.51 = 1.89$ eV; D: $13.6 - 1.51 = 12.09$ eV.
The shortest wavelength corresponds to the largest energy ($\lambda = \frac{hc}{\Delta E}$): transition D.
Answer: (D). | ||||
| 35 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{\sqrt{2mK}}$, so compare $mK$.
Electron: $m_e \cdot 4K$. Proton: $m_p \cdot K$. Alpha: $4m_p \cdot 2K = 8m_pK$.
Since $m_e \ll m_p$: $m_\alpha K_\alpha > m_pK_p > m_eK_e$, hence $\lambda_\alpha < \lambda_p < \lambda_e$.
Answer: (B). | ||||
| 36 | 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 Snell's law: $\mu = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.
Frequency does not change on refraction: $\nu_2 = \nu_1$. Wavelength falls in proportion to the speed: $\lambda_2 = \frac{\lambda_1}{\mu} = \frac{\lambda_1}{\sqrt{2}}$.
Answer: (B). | ||||
| 37 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Average electric energy density: $\frac{1}{4}\varepsilon_0E_0^2$; average magnetic: $\frac{B_0^2}{4\mu_0}$.
With $E_0 = cB_0$ and $c^2 = \frac{1}{\mu_0\varepsilon_0}$: $\frac{1}{4}\varepsilon_0c^2B_0^2 = \frac{B_0^2}{4\mu_0}$.
The two are equal: ratio 1.
Answer: (A). | ||||
| 38 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | StatementsEasy | |
A easy statements question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current An emf is induced only when the flux $\Phi = BA\cos\theta$ through the coil changes.
A, B: translating the coil (at any speed) inside a uniform field leaves B, A and θ unchanged, so no emf.
C: rotating the coil changes θ, so the flux changes: emf induced.
D: changing the area changes the flux: emf induced.
Answer: (C) C and D only. | ||||
| 39 | 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 Let the speed be v and the radius r. Turning through $90^\circ$ takes time $t = \frac{\pi r/2}{v}$, and the displacement is the chord $r\sqrt{2}$.
Average velocity $= \frac{r\sqrt{2}}{\pi r/(2v)} = \frac{2\sqrt{2}v}{\pi}$.
Ratio $v : \frac{2\sqrt{2}v}{\pi} = \pi : 2\sqrt{2}$, so $x = 2$.
Answer: (C). | ||||
| 40 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Ampère's law on a circle of radius r about the axis.
Inside ($r < a$): enclosed current $I\frac{r^2}{a^2}$, so $B \cdot 2\pi r = \mu_0I\frac{r^2}{a^2}$, $B = \frac{\mu_0Ir}{2\pi a^2} \propto r$.
Outside ($r > a$): $B = \frac{\mu_0I}{2\pi r} \propto \frac{1}{r}$.
Answer: (C). | ||||
| 41 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits In a semiconductor the number of charge carriers grows roughly exponentially with temperature, so the resistivity falls with temperature, steeply at first and then more slowly (approximately $\rho \propto e^{E_g/2kT}$).
Graph (D), a curve falling steeply and levelling off, shows this. (A) is a straight-line fall, (B) and (C) show resistivity rising, as in a metal.
Answer: (D). | ||||
| 42 | 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 For a shell of charge Q and radius R: inside ($r \le R$) the field is zero, so the potential is constant, $V = \frac{kQ}{R}$; outside, $V = \frac{kQ}{r}$, which falls as $\frac{1}{r}$, a curve that flattens out.
Graph (B) shows a constant value up to R followed by a $\frac{1}{r}$ fall.
Answer: (B). | ||||
| 43 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves When m is displaced by x, one spring is stretched and the other compressed by x; both forces point back towards equilibrium: $F = -(K_1 + K_2)x$.
The springs act in parallel, $k_{eff} = K_1 + K_2$, and $T = 2\pi\sqrt{\frac{m}{K_1 + K_2}}$.
Answer: (A). | ||||
| 44 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics First law per unit time: $\frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt} = 1000 - 200 = 800$ W.
Answer: (D). | ||||
| 45 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Collision frequency $f = \frac{v_{avg}}{\lambda}$ with mean free path $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$, so $f = \sqrt{2}\pi d^2 n v_{avg} \propto n$ at a fixed temperature.
$\frac{f_1}{f_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}} = 0.25$.
Answer: (A). | ||||
| 46 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $R_p = \frac{R_1R_2}{R_1 + R_2} = \frac{150}{25} = 6\ \Omega$.
From $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$: $\frac{\Delta R_p}{R_p^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} = \frac{0.5}{100} + \frac{0.5}{225}$.
$\Delta R_p = 36\left(0.005 + 0.00222\right) = 0.26\ \Omega$.
Percentage error $= \frac{0.26}{6} \times 100 = 4.33\%$.
Answer: (A). | ||||
| 47 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Assertion and reasonEasy | |
A easy assertion and reason question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane On level ground, $R = \frac{u^2\sin 2\theta}{g}$ is greatest when $\sin 2\theta = 1$, i.e. $\theta = 45^\circ$.
So A is true, R is true, and R is exactly why A holds.
Answer: (A). | ||||
| 48 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion Let the extension be x. The spring force supplies the centripetal force at radius $0.2 + x$:
$kx = m\omega^2(0.2 + x)$: $7.5x = 0.1 \times 25 \times (0.2 + x) = 0.5 + 2.5x$, so $x = 0.1$ m.
Tension $= kx = 7.5 \times 0.1 = 0.75$ N.
Answer: (C). | ||||
| 49 | GravitationEscape velocity | Class 11: Gravitation | Assertion and reasonEasy | |
A easy assertion and reason question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation The moon's escape velocity (about 2.4 km/s) is much smaller than the earth's (about 11.2 km/s). Gas molecules, with their thermal speeds, escape the moon easily, so it has kept no atmosphere, while the earth holds its atmosphere.
A is true, R is true, and R explains A.
Answer: (A). | ||||
| 50 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g = \frac{GM}{R^2}$ and $M = \frac{4}{3}\pi R^3\rho$, so $g = \frac{4}{3}\pi G\rho R$.
Same density and double mass: $R^3$ doubles, $R' = 2^{1/3}R$, so $g' = 2^{1/3}g$.
Weight on the planet $= 2^{1/3}$ W.
Answer: (D). | ||||
| 51 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The wall receives (moving source) and re-emits; the driver hears the echo as a moving observer:
$f' = f\frac{v + u}{v - u} = f\frac{345}{315}$.
Change $= f' - f = f\frac{30}{315} = 40$, so $f = \frac{40 \times 315}{30} = 420$ Hz.
Answer: 420. | ||||
| 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 Sunlight at $30^\circ$ to the surface means an angle of incidence $i = 60^\circ$.
In water: $\sin r = \frac{\sin 60^\circ}{4/3} = \frac{3\sqrt{3}}{8}$, $\cos r = \frac{\sqrt{37}}{8}$, $\tan r = \frac{3\sqrt{3}}{\sqrt{37}} \approx 0.854$.
The shadow on the floor of the pool is made of two parts: the part of the pole above water (height h) throws the ray $h\tan 60^\circ = \sqrt{3}h$ out from the pole at the surface, and in 1.5 m of water the ray moves on by $1.5\tan r \approx 1.281$ m.
$\sqrt{3}h + 1.281 = 2.15$, so $h = \frac{0.869}{1.732} \approx 0.50$ m $= 50$ cm.
Answer: 50. | ||||
| 53 | 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 Bohr radius of the nth orbit: $r_n = r_0\frac{n^2}{Z}$. For $Li^{++}$, $Z = 3$, $n = 5$:
$r_5 = 0.51 \times \frac{25}{3} = 4.25$ Å $= 425 \times 10^{-12}$ m.
Answer: 425. | ||||
| 54 | Electromagnetic Induction and Alternating CurrentsTransformers | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on transformers, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current On the secondary side $P = \frac{V_s^2}{R_s}$, so $R_s = \frac{120^2}{60000} = \frac{14400}{60000} = 0.24\ \Omega = 240$ m$\Omega$.
Answer: 240. | ||||
| 55 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $R = \frac{\rho l}{A}$. New resistance $= \frac{\rho(1.2l)}{0.96A} = 1.25R$.
Percentage change $= 25\%$ (an increase).
Answer: 25. | ||||
| 56 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With a slab of thickness x and constant K: $C = \frac{\varepsilon_0A}{d - x + x/K} = \frac{\varepsilon_0A}{d - \frac{3x}{4}}$.
$x = \frac{d}{3}$: $C_1 = \frac{\varepsilon_0A}{3d/4} = \frac{4}{3}\frac{\varepsilon_0A}{d} = 2\ \mu F$, so $\frac{\varepsilon_0A}{d} = 1.5\ \mu F$.
$x = \frac{2d}{3}$: $C_2 = \frac{\varepsilon_0A}{d/2} = 2 \times 1.5 = 3\ \mu F$.
Answer: 3. | ||||
| 57 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Field of each loop at its centre: $B = \frac{\mu_0I}{2R} = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{0.4} = \sqrt{2}\pi \times 10^{-6}$ T.
The loops lie in perpendicular planes, so their fields at the common centre are perpendicular:
$B_{net} = \sqrt{2}B = 2\pi \times 10^{-6} = 6.28 \times 10^{-6}$ T $= 628 \times 10^{-8}$ T.
Answer: 628. | ||||
| 58 | 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 Strain $= \frac{F}{AY} = \frac{62.8 \times 10^3}{\pi(0.02)^2 \times 2 \times 10^{11}} = \frac{62800}{3.14 \times 4 \times 10^{-4} \times 2 \times 10^{11}} = \frac{62800}{2.512 \times 10^8} = 2.5 \times 10^{-4}$.
$= 25 \times 10^{-5}$.
Answer: 25. | ||||
| 59 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Retarding force $F = m(2x) = 0.01 \times 2x = 0.02x$ N.
Loss of kinetic energy = work done against it $= \int_0^x 0.02x\,dx = 0.01x^2 = \left(\frac{x}{10}\right)^2 = \left(\frac{10}{x}\right)^{-2}$ J.
So $n = 2$.
Answer: 2. | ||||
| 60 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Each centre is 0.2 m from the axis. By the parallel-axis theorem, each sphere gives $\frac{2}{5}MR^2 + Md^2 = \frac{2}{5}(2)(0.1)^2 + 2(0.2)^2 = 0.008 + 0.08 = 0.088$ kg m$^2$.
Total $= 2 \times 0.088 = 0.176$ kg m$^2$ $= 176 \times 10^{-3}$ kg m$^2$.
Answer: 176. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Solid StateClose packed structures | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on close packed structures, from Solid State. NCERT: No matching NCERT chapter In ccp there are 2 tetrahedral voids per atom. Take Y = 3 atoms: tetrahedral voids = 6, of which one third, 2, hold X.
X : Y = 2 : 3, so the formula is $X_2Y_3$.
Answer: (C). | ||||
| 62 | EquilibriumIonic equilibrium | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $A_2B_3 \rightleftharpoons 2A^{3+} + 3B^{2-}$. At equilibrium: $[A_2B_3] = c(1 - \alpha)$, $[A] = 2c\alpha$, $[B] = 3c\alpha$.
$K_{eq} = \frac{(2c\alpha)^2(3c\alpha)^3}{c(1 - \alpha)} = \frac{108c^5\alpha^5}{c(1 - \alpha)}$.
For a weak electrolyte $\alpha \ll 1$, so $1 - \alpha \approx 1$: $K_{eq} = 108c^4\alpha^5$ and $\alpha = \left(\frac{K_{eq}}{108c^4}\right)^{1/5}$.
Answer: (B). | ||||
| 63 | BiomoleculesProteins | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules A: sucrose is hydrolysed to glucose and fructose by invertase (III).
B: glucose is fermented to ethyl alcohol and $CO_2$ by zymase (I).
C: starch is broken down to maltose by diastase (IV).
D: proteins are hydrolysed to amino acids by pepsin (II).
A-III, B-I, C-IV, D-II.
Answer: (B). | ||||
| 64 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Approximate electron gain enthalpies (kJ mol$^{-1}$): Ne $+116$, Ar $+96$ (positive: noble gases resist an extra electron), F $-328$, Cl $-349$ (Cl is the most negative of all elements).
Differences: Ne–F 444, Ne–Cl 465, Ar–F 424, Ar–Cl 445.
The largest is between Ne (most positive) and Cl (most negative).
Answer: (B). | ||||
| 65 | Redox ReactionsOxidation number | Class 11: Redox Reactions | StatementsEasy | |
A easy statements question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions Au goes from +1 in $[Au(CN)_2]^-$ to 0 (reduced) and Zn from 0 to +2 (oxidised): a redox reaction (A true).
The more reactive zinc displaces gold from its complex: a (metal) displacement reaction (B true).
No single substance breaks into two (not decomposition) and two reactants do not join into one product (not combination).
Answer: (A) A and B only. | ||||
| 66 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Assertion and reasonMedium | |
A medium assertion and reason question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter A: removing the only electron of hydrogen leaves the bare nucleus, a proton, whose size is about $1.5 \times 10^{-3}$ pm, tiny beside ordinary atomic and ionic sizes (50 to 200 pm). True.
R: because it is so small, $H^+$ has an enormous charge density and does not exist free; it is always attached to other atoms or molecules (e.g. $H_3O^+$). True.
But R is a consequence of A, not the reason for it: the small size follows from losing the electron, not from H$^+$ being combined.
Answer: (B). | ||||
| 67 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry The energy of $M(s) \rightarrow M^+(aq)$ is built up through the cycle: sublimation $M(s) \rightarrow M(g)$, ionisation of the gaseous atom $M(g) \rightarrow M^+(g)$, and hydration $M^+(g) \rightarrow M^+(aq)$.
The electrode potential therefore depends on those three steps. "Ionisation of a solid metal atom" is not a step in the cycle (ionisation enthalpy is defined for a gaseous atom), so the potential does not depend on it.
Answer: (C). | ||||
| 68 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Gypsum ($CaSO_4\cdot 2H_2O$, about 2–3%) is added to clinker to slow down the setting of cement, so that it sets in a useful time. Limestone, clay and silica are raw materials of the cement itself.
Answer: (C). | ||||
| 69 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $N_2O_4$: two $NO_2$ units joined by one N–N bond (III).
$NO_2$: one N=O bond and one N–O bond (in resonance) (I).
$N_2O_5$: $O_2N–O–NO_2$, one N–O–N bridge (II).
$N_2O$: $N=N=O \leftrightarrow N\equiv N–O$, an N=N / N≡N bond (IV).
A-III, B-I, C-II, D-IV.
Answer: (B). | ||||
| 70 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The stable lanthanoid state is +3.
$Eu^{2+}$ ($4f^7$) readily gives up an electron to become $Eu^{3+}$: a strong reducing agent.
$Ce^{4+}$ ($4f^0$) readily takes an electron to become $Ce^{3+}$: a strong oxidising agent ($E^\circ \approx +1.74$ V).
Answer: (B). | ||||
| 71 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Photochemical smog forms when nitrogen oxides and hydrocarbons, emitted by vehicles and industry, react in sunlight to give ozone, PAN and other oxidants. These are most concentrated in industrial (and heavy-traffic) areas.
Answer: (B). | ||||
| 72 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Assertion and reasonMedium | |
A medium assertion and reason question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Both complexes contain $Fe^{3+}$, $3d^5$, so R is true.
$CN^-$ is a strong-field ligand: low spin $t_{2g}^5$, one unpaired electron, $\mu = \sqrt{1 \times 3} = 1.73 \approx 1.74$ BM.
$H_2O$ is weak-field: high spin $t_{2g}^3e_g^2$, five unpaired electrons, $\mu = \sqrt{5 \times 7} = 5.92$ BM. So A is true.
The difference comes from the ligands' field strength, not from the oxidation state, which is the same in both; R does not explain A.
Answer: (B). | ||||
| 73 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines P ($C_{14}H_{13}ON$) is neutral and is split by hot HCl into an acid and an amine: P is an amide.
Q, the residue that fizzes with $NaHCO_3$, is the carboxylic acid (insoluble in the acid medium). The amine stays in the filtrate as its salt; NaOH frees it as the oily liquid R.
R gives with Hinsberg's reagent a sulphonamide soluble in NaOH, so R is a primary amine.
(A) would give N-methylaniline, a secondary amine (its product is insoluble in NaOH). (B) is an amine–ketone, not hydrolysed to an acid. (C) is a primary amide giving ammonia, not an oily amine.
(D) $CH_3C_6H_4CONHC_6H_5$ ($C_{14}H_{13}NO$) gives p-toluic acid (Q) and aniline (R, primary, oily). ✓
Answer: (D). | ||||
| 74 | Practical Organic ChemistryDetection of elements | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on detection of elements, from Practical Organic Chemistry. NCERT: No matching NCERT chapter Lassaigne's tests:
Nitrogen: Prussian blue, $Fe_4[Fe(CN)_6]_3$ (III).
Sulphur: violet colour with sodium nitroprusside, $Na_2[Fe(CN)_5NO]$ (I).
Phosphorus: yellow precipitate with ammonium molybdate, $(NH_4)_2MoO_4$ (IV).
Halogen: precipitate with $AgNO_3$ (II).
A-III, B-I, C-IV, D-II.
Answer: (B). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes This is the Finkelstein reaction, an $S_N2$ substitution of a primary bromide in acetone.
(A) Acetic acid is protic and would solvate $I^-$ strongly, and NaBr would not precipitate from it; acetone is chosen for the opposite reasons. Not correct.
(B) NaBr is insoluble in acetone and precipitates, so $Br^-$ is removed and does not compete. Not correct.
(C) An $S_N2$ step forms no ions in the rate-determining step; it passes through a single transition state. Not correct.
(D) In the transition state $[I\cdots C\cdots Br]^-$ the negative charge is spread over two atoms, so it is less polar than the localised $I^-$ ion. Correct — this is why a polar aprotic solvent, which does not stabilise the localised anion strongly, favours the reaction.
Answer: (D). | ||||
| 76 | AminesClassification and identification of amines | Class 12: Amines | Single correctHard | |
A hard single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines $Br_2/NaOH$ on the amide $-CH_2CONH_2$ is the Hofmann bromamide degradation: it loses one carbon and gives $-CH_2NH_2$.
The new primary amine sits next to the ortho $-COOCH_3$ group; on heating it attacks the ester carbonyl and expels methanol, closing a five-membered lactam (ring atoms: two benzene carbons, $CH_2$, NH, C=O).
The product is isoindolin-1-one, structure (A).
Answer: (A). | ||||
| 77 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Matching listEasy | |
A easy matching list question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids HVZ: α-halogenation of acids with $Br_2$/red P, then water (III).
Iodoform: $I_2$ + NaOH (I).
Etard: chromyl chloride in $CS_2$, then hydrolysis (II).
Gattermann–Koch: CO + HCl with anhydrous $AlCl_3$ (IV).
A-III, B-I, C-II, D-IV.
Answer: (C). | ||||
| 78 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Acetanilide: the $-NHCOCH_3$ group is a moderately activating ortho/para director (the nitrogen lone pair is shared with the carbonyl). With $Br_2$ in acetic acid it is monobrominated, mainly at the less hindered para position: A = 4-bromoacetanilide. (Tribromination happens with aniline itself, not with the acetyl-protected amine.)
$LiAlH_4$ reduces an amide C=O to $CH_2$: $C_6H_5NHCOCH_3 \rightarrow C_6H_5NHCH_2CH_3$, so B = N-ethylaniline.
Answer: (A). | ||||
| 79 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Orlon (acrilan) is polyacrylonitrile, $-(CH_2-CH(CN))_n-$, made by addition polymerisation of acrylonitrile; it is used as a wool substitute.
Answer: (A). | ||||
| 80 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Vitamin A: xerophthalmia (III). Thiamine ($B_1$): beri-beri (I). Ascorbic acid (C): scurvy (IV). Riboflavin ($B_2$): cheilosis (II).
A-III, B-I, C-IV, D-II.
Answer: (B). | ||||
| 81 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $3BaCl_2 + 2Na_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6NaCl$.
5 mol $BaCl_2$ could give $\frac{5}{3} \approx 1.67$ mol; 2 mol $Na_3PO_4$ can give only $\frac{2}{2} = 1$ mol. $Na_3PO_4$ is limiting.
Maximum $Ba_3(PO_4)_2$ = 1 mol.
Answer: 1. | ||||
| 82 | 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 $\lambda = \frac{h}{\sqrt{2mK}} = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 4.5 \times 10^{-29}}} = \frac{6.6 \times 10^{-34}}{\sqrt{81 \times 10^{-60}}} = \frac{6.6 \times 10^{-34}}{9 \times 10^{-30}}$
$= 7.33 \times 10^{-5}$ m. Nearest integer: 7.
Answer: 7. | ||||
| 83 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $PF_5$: 5 bond pairs, trigonal bipyramidal.
$BrF_4^-$: 4 bond pairs + 2 lone pairs, square planar.
$IF_5$: 5 bp + 1 lp, square pyramidal ✓.
$BrF_5$: 5 bp + 1 lp, square pyramidal ✓.
$XeOF_4$: 5 bonded atoms + 1 lp on Xe, square pyramidal ✓.
$ICl_4^-$: 4 bp + 2 lp, square planar.
Count = 3.
Answer: 3. | ||||
| 84 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -54070 - 298 \times 10 = -57050$ J mol$^{-1}$.
$\Delta G^\circ = -2.303RT\log K$, so $\log K = \frac{57050}{5705} = 10$.
Answer: 10. | ||||
| 85 | SolutionsRaoult's law | Class 12: Solutions | NumericalMedium | |
A medium numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions Relative lowering of vapour pressure = mole fraction of solute: $\frac{\Delta p}{p^\circ} = \frac{n}{n + N} = 0.25$.
Water: $N = \frac{1000}{18} = 55.56$ mol. Then $n = 0.25(n + 55.56)$, $0.75n = 13.89$, $n = 18.52$ mol.
Urea, $M = 60$ g mol$^{-1}$: mass $= 18.52 \times 60 = 1111.1$ g. Nearest integer: 1111.
Answer: 1111. | ||||
| 86 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics At (a) G falls as the reaction proceeds (slope negative): the forward reaction is spontaneous.
At (b), the minimum, the slope is zero: equilibrium.
At (c) G rises with further reaction (slope positive): the forward reaction is non-spontaneous.
A: false (at b the system is at equilibrium, not spontaneous). B: true. C: true. D: false.
Number of true statements = 2.
Answer: 2. | ||||
| 87 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Taking the same pre-exponential factor for equal areas, $\frac{r_{Pt}}{r_{Ni}} = e^{(E_{Ni} - E_{Pt})/RT}$.
$\log\frac{r_{Pt}}{r_{Ni}} = \frac{E_{Ni} - E_{Pt}}{2.3RT} = \frac{41400 - 30000}{2.3 \times 8.3 \times 300} = \frac{11400}{5727} = 1.99$.
Nearest integer: 2.
Answer: 2. | ||||
| 88 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds An ambidentate ligand can bind through either of two different atoms. $SCN^-$ binds through S (thiocyanato) or N (isothiocyanato): ambidentate. Ethylenediamine binds through both N atoms at once: it is bidentate (chelating), not ambidentate.
The complex has four $SCN^-$ ligands, so 4.
Answer: 4. | ||||
| 89 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Free-radical substitution can replace any number of the six H atoms. Distinct products:
$C_2H_5Br$: 1. $C_2H_4Br_2$: 1,1- and 1,2-: 2. $C_2H_3Br_3$: 1,1,1- and 1,1,2-: 2. $C_2H_2Br_4$: 1,1,1,2- and 1,1,2,2-: 2. $C_2HBr_5$: 1. $C_2Br_6$: 1.
None of these has a stereocentre. Total $= 1 + 2 + 2 + 2 + 1 + 1 = 9$.
Answer: 9. | ||||
| 90 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions Ammonium phosphomolybdate is $(NH_4)_3PO_4\cdot 12MoO_3$, i.e. $(NH_4)_3[PMo_{12}O_{40}]$.
$3(+1) + (+5) + 12x + 40(-2) = 0$, so $12x = 72$ and $x = +6$ (as in $MoO_3$).
Answer: 6. | ||||
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