JEE (Main) · 2023
JEE (Main) 2023 — 6 April, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Three Dimensional Geometry3
- Conic Sections2
11 easy · 18 medium · 1 hard
Physics
30 Q
- Properties of Bulk Matter3
- Kinematics2
- Gravitation2
19 easy · 10 medium · 1 hard
Chemistry
30 Q
- Classification of Elements and Periodicity2
- Electrochemistry2
- Coordination Compounds2
18 easy · 9 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 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | StatementsMedium | |
A medium statements question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Domain: we need $\lceil x \rceil - x > 0$. For an integer $x$, $\lceil x \rceil = x$ and the expression is 0; for a non-integer $x$, $0 < \lceil x \rceil - x < 1$. So $A = \mathbb{R} - \mathbb{Z}$.
Range: $\lceil x \rceil - x$ takes every value in $(0, 1)$, so $\sqrt{\lceil x \rceil - x} \in (0, 1)$ and $f(x) \in (1, \infty)$. So $B = (1, \infty)$.
(S1): $A \cap B$ = numbers greater than 1 that are not integers $= (1, \infty) - \mathbb{N}$. True.
(S2): $A \cup B$ contains, for example, $-0.5 \in A$, which is not in $(1, \infty)$. False.
Only (S1) is true.
Answer: (A). | ||||
| 2 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Both languages: $75 + 40 - 100 = 15$. Only English $\alpha = 75 - 15 = 60$; only Hindi $\beta = 40 - 15 = 25$.
The ellipse: $\frac{x^2}{(\alpha/5)^2} + \frac{y^2}{(\beta/5)^2} = 1$, i.e. $\frac{x^2}{144} + \frac{y^2}{25} = 1$.
$e = \sqrt{1 - \frac{25}{144}} = \frac{\sqrt{119}}{12}$.
Answer: (B). | ||||
| 3 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $z = x + iy$ and $u = \mathrm{Re}(z^2) = x^2 - y^2$. The conditions are
$au + bx = a$ and $bu + ax = b$.
Subtracting: $(a - b)(u - x) = a - b$, so $u - x = 1$. Adding: $(a + b)(u + x) = a + b$, so (for $a + b \ne 0$) $u + x = 1$.
Hence $x = 0$ and $u = 1$, i.e. $-y^2 = 1$, which has no real $y$. So X is empty: 0 elements.
Answer: (A). | ||||
| 4 | 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 Reduce powers with $P^2 = I - P$:
$P^3 = P - P^2 = 2P - I$, $P^4 = 2P^2 - P = 2I - 3P$, $P^5 = 5P - 3I$, $P^6 = 5I - 8P$, $P^7 = 13P - 8I$, $P^8 = 13I - 21P$.
Adding the two given equations: $2P^\alpha = (\gamma + \delta)I - 42P$, so $P^\alpha = \frac{\gamma + \delta}{2}I - 21P$, giving $\alpha = 8$ and $\gamma + \delta = 26$.
Subtracting: $2P^\beta = (\gamma - \delta)I - 16P$, so $P^\beta = \frac{\gamma - \delta}{2}I - 8P$, giving $\beta = 6$ and $\gamma - \delta = 10$.
So $\gamma = 18$, $\delta = 8$, and $\alpha + \beta + \gamma - \delta = 8 + 6 + 18 - 8 = 24$.
Answer: (B). | ||||
| 5 | 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 $\Delta = \begin{vmatrix}1&1&1\\1&2&\alpha\\1&3&5\end{vmatrix} = (10 - 3\alpha) - (5 - \alpha) + (3 - 2) = 6 - 2\alpha$.
For $\alpha \ne 3$ the system has a unique solution, so (A) is true.
For $\alpha = 3$: $R_2 - R_1$ gives $y + 2z = 4$, and $R_3 - R_1$ gives $2y + 4z = \beta - 6$. These agree only if $\beta - 6 = 8$, i.e. $\beta = 14$ (infinitely many solutions, (C) true); for $\beta \ne 14$ there is no solution ((B) true).
So (D) is false: with $\alpha = 3$, $\beta \ne 14$ there is no solution, not a unique one.
Answer: (D). | ||||
| 6 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | StatementsMedium | |
A medium statements question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem (S1): $a^n - b^n$ is divisible by $a - b$. Here $a - b = 2023 - 1999 = 24$, a multiple of 8. True.
(S2): $13(13)^n = 13^{n+1} = (1 + 12)^{n+1} = 1 + 12(n + 1) + 144k$ for some integer $k$ (every later term has $12^2$).
So $13^{n+1} - 11n - 13 = 1 + 12n + 12 + 144k - 11n - 13 = n + 144k$.
This is divisible by 144 exactly when $144 \mid n$, which holds for infinitely many $n$. True.
Answer: (D). | ||||
| 7 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem In $\left(ax^2 + \frac{1}{2bx}\right)^{11}$: $T_{r+1} = \binom{11}{r}(ax^2)^{11-r}\left(\frac{1}{2bx}\right)^r$, power of $x$ is $22 - 3r = 7$, so $r = 5$. Coefficient $= \binom{11}{5}\frac{a^6}{32b^5}$.
In $\left(ax - \frac{1}{3bx^2}\right)^{11}$: $T_{r+1} = \binom{11}{r}(ax)^{11-r}\left(-\frac{1}{3bx^2}\right)^r$, power $11 - 3r = -7$, so $r = 6$. Coefficient $= \binom{11}{6}\frac{a^5}{729b^6}$.
Since $\binom{11}{5} = \binom{11}{6}$: $\frac{a^6}{32b^5} = \frac{a^5}{729b^6}$, so $729ab = 32$.
Answer: (D). | ||||
| 8 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Alphabetical order of the letters: B, C, I, L, P, U.
Words before PUBLIC:
starting with B, C, I or L: $4 \times 5! = 480$;
starting P, second letter B, C, I or L: $4 \times 4! = 96$;
PU, third letter B (smallest left), so none; PUB, fourth letter from {C, I, L} smaller than L: C or I, $2 \times 2! = 4$;
PUBL, fifth letter from {C, I} smaller than I: C, $1 \times 1! = 1$; then PUBLIC itself.
Serial number $= 480 + 96 + 4 + 1 + 1 = 582$.
Answer: (D). | ||||
| 9 | 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 For every $k \ge 1$, $1 < 2^{\frac{1}{2k+1}} < 2^{\frac{1}{2}}$, so each factor satisfies $0 < 2^{\frac{1}{2}} - 2^{\frac{1}{2k+1}} < \sqrt{2} - 1$.
Hence $0 < P_n < (\sqrt{2} - 1)^n$, and $(\sqrt{2} - 1)^n \to 0$ because $0 < \sqrt{2} - 1 < 1$.
By the squeeze theorem the limit is 0.
Answer: (B). | ||||
| 10 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Pair the terms: $(2k-1)^2 - (2k)^2 = -(4k - 1)$ for $k = 1, \ldots, 1011$.
Sum of pairs $= -\sum_{k=1}^{1011}(4k - 1) = -(2 \cdot 1011 \cdot 1012 - 1011) = -1011 \cdot 2023$.
Adding $2023^2$: total $= 2023(2023 - 1011) = 2023 \cdot 1012$.
So $m^2 n = 2023 = 7 \cdot 17^2$, giving $m = 17$, $n = 7$ (gcd 1).
$m^2 - n^2 = 289 - 49 = 240$.
Answer: (A). | ||||
| 11 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $I = \int_0^\pi f(x)\sin x\, dx$. Replacing $x$ by $\pi - x$: $I = \int_0^\pi f(\pi - x)\sin x\, dx$.
Adding: $2I = \int_0^\pi \pi^2 \sin x\, dx = 2\pi^2$, so $I = \pi^2$.
Answer: (D). | ||||
| 12 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Let $t = x\log_e x$; then $\frac{dt}{dy} = (1 + \log_e x)\frac{dx}{dy}$, and the equation becomes $\frac{dt}{dy} - t = e^y$.
Integrating factor $e^{-y}$: $\frac{d}{dy}(te^{-y}) = 1$, so $te^{-y} = y + C$.
At $(1, 0)$: $t = 0$, so $C = 0$. The curve is $x\log_e x = ye^y$.
At $(\alpha, 2)$: $\alpha\log_e\alpha = 2e^2$, so $\alpha^\alpha = e^{\alpha\log_e\alpha} = e^{2e^2}$.
Answer: (D). | ||||
| 13 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Centre $C\left(1, -\frac{1}{2}\right)$, radius $r$ with $r^2 = 1 + \frac{1}{4} + 5 = \frac{25}{4}$, so $r = \frac{5}{2}$.
$CR^2 = \left(\frac{5}{4}\right)^2 + \left(\frac{5}{2}\right)^2 = \frac{125}{16}$; tangent length $L = \sqrt{\frac{125}{16} - \frac{100}{16}} = \frac{5}{4}$.
Area of the triangle formed by the two tangents and the chord of contact $= \frac{rL^3}{r^2 + L^2} = \frac{\frac{5}{2}\cdot\frac{125}{64}}{\frac{125}{16}} = \frac{5}{8}$.
Answer: (C). | ||||
| 14 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $y = 3 - 2x$ for $x < 1$, $y = 1$ for $1 \le x \le 2$, $y = 2x - 3$ for $x > 2$. It meets $y = 3$ at $x = 0$ and $x = 3$.
The region is a trapezium with parallel sides 3 (at $y = 3$, from 0 to 3) and 1 (at $y = 1$, from 1 to 2), height 2.
Area $= \frac{1}{2}(3 + 1)\times 2 = 4$.
Answer: (B). | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry P: $(x + y + z - 6) + \lambda(2x + 3y + 4z + 5) = 0$.
Through $(0, 2, -2)$: $-6 + \lambda(6 - 8 + 5) = 0$, so $\lambda = 2$, and P is $5x + 7y + 9z + 4 = 0$.
Distance of $(12, 12, 18)$: $\frac{|60 + 84 + 162 + 4|}{\sqrt{25 + 49 + 81}} = \frac{310}{\sqrt{155}}$.
Square $= \frac{310^2}{155} = 620$.
Answer: (C). | ||||
| 16 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the given line: $(1 + 2t, 2 + 3t, 3 + 4t)$. Direction of L from $A(0, 1, 2)$: $(1 + 2t, 1 + 3t, 1 + 4t)$.
Parallel to the plane: $2(1 + 2t) + (1 + 3t) - 3(1 + 4t) = -5t = 0$, so $t = 0$ and L has direction $(1, 1, 1)$.
$\vec{AP} = (1, -10, 0)$, $|\vec{AP}|^2 = 101$; its component along $(1, 1, 1)$ is $\frac{-9}{\sqrt{3}}$, squared 27.
Distance$^2 = 101 - 27 = 74$, distance $= \sqrt{74}$.
Answer: (C). | ||||
| 17 | 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 With $A(1, -2, 3)$: $\vec{AB} = (1, -1, 1)$, $\vec{AC} = (\alpha, 2, -1)$, $\vec{AD} = (8, \alpha - 6, 3)$.
Coplanar: $\begin{vmatrix}1&-1&1\\\alpha&2&-1\\8&\alpha-6&3\end{vmatrix} = 0$.
$= 1(6 + \alpha - 6) + 1(3\alpha + 8) + (\alpha^2 - 6\alpha - 16) = \alpha^2 - 2\alpha - 8 = 0$.
Roots $\alpha = 4, -2$; their sum is 2.
Answer: (B). | ||||
| 18 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $[\vec{a}\ \ \vec{b} + \vec{c}\ \ \vec{a} + 2\vec{b} + 3\vec{c}] = [\vec{a}\ \ \vec{b} + \vec{c}\ \ 2\vec{b} + 3\vec{c}]$ (the $\vec{a}$ part repeats a vector).
$= [\vec{a}\ \vec{b}\ 3\vec{c}] + [\vec{a}\ \vec{c}\ 2\vec{b}] = 3[\vec{a}\ \vec{b}\ \vec{c}] - 2[\vec{a}\ \vec{b}\ \vec{c}] = [\vec{a}\ \vec{b}\ \vec{c}]$.
The volume is V.
Answer: (A). | ||||
| 19 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Favourable outcomes: $6 \times 5 \times 4 = 120$ out of $6^3 = 216$.
Probability $= \frac{120}{216} = \frac{5}{9}$, so $p = 5$, $q = 9$, $q - p = 4$.
Answer: (A). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | StatementsEasy | |
A easy statements question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter (S1): $p \Rightarrow q \equiv \sim p \vee q$, and $(\sim p) \wedge q$ already implies $\sim p \vee q$. So the statement is $\sim p \vee q$, which is false for $p$ true, $q$ false. Not a tautology.
(S2): $(q \Rightarrow p) \Rightarrow X \equiv \sim(q \Rightarrow p) \vee X = (q \wedge \sim p) \vee ((\sim p) \wedge q) = (\sim p) \wedge q$. This is true for $p$ false, $q$ true. Not a contradiction.
Neither is true.
Answer: (D). | ||||
| 21 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $M = \frac{\alpha + \beta}{2}$ and $d = |\alpha - \beta|$. By the parallelogram (Apollonius) identity,
$|z - \alpha|^2 + |z - \beta|^2 = 2|z - M|^2 + \frac{d^2}{2}$.
So the locus is $|z - M|^2 = \lambda - \frac{d^2}{4}$, a circle of radius$^2$ $\lambda - \frac{d^2}{4}$.
Setting this equal to $\lambda - 1$: $d^2 = 4$, $d = 2$.
Answer: 2. | ||||
| 22 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Distinct vowels in UNIVERSE: U, I, E. Distinct consonants: N, V, R, S.
Choose 2 vowels: $\binom{3}{2} = 3$; choose 2 consonants: $\binom{4}{2} = 6$; arrange the 4 distinct letters: $4! = 24$.
Number of words $= 3 \times 6 \times 24 = 432$.
Answer: 432. | ||||
| 23 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Divide by $20^{19}$ and put $x = \frac{21}{20}$: $k = 1 + 2x + 3x^2 + \ldots + 20x^{19} = S$.
Arithmetico-geometric sum: $S - xS = 1 + x + \ldots + x^{19} - 20x^{20} = \frac{x^{20} - 1}{x - 1} - 20x^{20}$.
With $x - 1 = \frac{1}{20}$: $\frac{x^{20} - 1}{x - 1} = 20x^{20} - 20$, so $(1 - x)S = -20$, and $S = \frac{-20}{-1/20} = 400$.
Answer: 400. | ||||
| 24 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $y' = 5x^4 - 60x^2 + 50 = 5(x^4 - 12x^2 + 10)$, zero at $x^2 = 6 \pm \sqrt{26}$, i.e. $x \approx \pm 3.33$ and $x \approx \pm 0.95$.
Values at the critical points: $y(-3.33) \approx 164.6 > 0$, $y(-0.95) \approx -29.1 < 0$, $y(0.95) \approx 33.1 > 0$, $y(3.33) \approx -160.6 < 0$.
With $y \to -\infty$ as $x \to -\infty$ and $y \to +\infty$ as $x \to +\infty$, the signs alternate $-, +, -, +, -, +$ across the six monotonic pieces, so there is exactly one root in each of the five gaps.
The curve crosses the x-axis at 5 points.
Answer: 5. | ||||
| 25 | 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 Tangent at $(x, y)$ meets the y-axis at $c = y - xy'$. The condition $xc = 3$ gives $xy - x^2y' = 3$, i.e. $y' - \frac{y}{x} = -\frac{3}{x^2}$.
Integrating factor $\frac{1}{x}$: $\frac{d}{dx}\left(\frac{y}{x}\right) = -\frac{3}{x^3}$, so $\frac{y}{x} = \frac{3}{2x^2} + C$, $y = \frac{3}{2x} + Cx$.
Through $P\left(1, \frac{3}{2}\right)$: $C = 0$, so $y = \frac{3}{2x}$. At Q: $\frac{1}{2} = \frac{3}{2a}$, $a = 3$.
$PQ^2 = (3 - 1)^2 + \left(\frac{1}{2} - \frac{3}{2}\right)^2 = 4 + 1 = 5$.
Answer: 5. | ||||
| 26 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f(f(x)) = \frac{x}{(1 + 2x^n)^{1/n}}$, and by induction $f^n(x) = \frac{x}{(1 + nx^n)^{1/n}}$.
$I_n = \int_0^1 \frac{x^{n-1}}{(1 + nx^n)^{1/n}}dx$. Put $u = 1 + nx^n$, $du = n^2x^{n-1}dx$:
$I_n = \frac{1}{n^2}\int_1^{1+n}u^{-1/n}du = \frac{1}{n^2}\cdot\frac{(1 + n)^{1 - 1/n} - 1}{1 - 1/n} = \frac{(1 + n)^{1 - 1/n} - 1}{n(n - 1)}$.
Since $(1 + n)^{1/n} \to 1$, the numerator behaves like $n$, so $I_n \approx \frac{n}{n(n - 1)} \to 0$.
Answer: 0. | ||||
| 27 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Hyperbola $\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$ is rectangular, $e = \sqrt{2}$; its foci are at $(\pm 1, 0)$.
Ellipse: $e = \frac{1}{\sqrt{2}}$, so $b^2 = a^2(1 - e^2) = \frac{a^2}{2}$.
An ellipse and a hyperbola cut at right angles exactly when they are confocal: $a^2 - b^2 = \frac{1}{2} + \frac{1}{2} = 1$.
So $\frac{a^2}{2} = 1$, $a^2 = 2$, $b^2 = 1$.
Latus rectum $= \frac{2b^2}{a} = \frac{2}{\sqrt{2}} = \sqrt{2}$; its square is 2.
Answer: 2. | ||||
| 28 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry First line: $(1 + 2s, 2 + 3s, 3 + \alpha s)$ (note $\frac{2 - y}{-3} = \frac{y - 2}{3}$). Second: $(4 + 5t, 1 + 2t, \beta t)$.
$x$: $1 + 2s = 4 + 5t$; $y$: $2 + 3s = 1 + 2t$. Solving: $s = t = -1$.
$z$: $3 - \alpha = -\beta$, so $\beta = \alpha - 3$.
$8\alpha\beta = 8(\alpha^2 - 3\alpha) = 8\left(\alpha - \frac{3}{2}\right)^2 - 18$, minimum $-18$ at $\alpha = \frac{3}{2}$.
Its magnitude is 18.
Answer: 18. | ||||
| 29 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $N = 40 + \alpha + \beta$, $\sum f_ix_i = 360 + 6\alpha + 12\beta$.
Mean 9: $360 + 6\alpha + 12\beta = 9(40 + \alpha + \beta)$, so $3\beta = 3\alpha$, $\alpha = \beta$.
Variance about the mean: $\sum f_i(x_i - 9)^2 = 4\cdot 49 + 4\cdot 25 + 9\alpha + 15 + 8 + 9\beta + 4\cdot 25 + 5\cdot 49 = 664 + 18\alpha$.
$\frac{664 + 18\alpha}{40 + 2\alpha} = 15.08$ gives $664 + 18\alpha = 603.2 + 30.16\alpha$, so $\alpha = 5 = \beta$.
$\alpha^2 + \beta^2 - \alpha\beta = 25 + 25 - 25 = 25$.
Answer: 25. | ||||
| 30 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\tan\theta + \cot\theta = \frac{2}{\sin 2\theta}$.
$\tan 9^\circ + \tan 81^\circ = \tan 9^\circ + \cot 9^\circ = \frac{2}{\sin 18^\circ}$; $\tan 27^\circ + \tan 63^\circ = \frac{2}{\sin 54^\circ}$.
With $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$ and $\sin 54^\circ = \frac{\sqrt{5} + 1}{4}$:
value $= \frac{8}{\sqrt{5} - 1} - \frac{8}{\sqrt{5} + 1} = \frac{8 \cdot 2}{4} = 4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 Object distance $u = 100 - 80 = 20$ cm, image distance $v = 180 - 100 = 80$ cm.
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ (magnitudes), so $f = \frac{uv}{u + v} = \frac{1600}{100} = 16$ cm.
Each distance is the difference of two scale readings, each uncertain by the least count 0.2 cm, so $\Delta u = \Delta v = 0.4$ cm.
Differentiating $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$: $\frac{\Delta f}{f^2} = \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2} = \frac{0.4}{400} + \frac{0.4}{6400} = \frac{0.4 \times 17}{6400}$.
$\Delta f = 256 \times \frac{6.8}{6400} = 0.272$ cm, and $\frac{\Delta f}{f}\times 100 = \frac{0.272}{16}\times 100 = 1.70\%$.
Taking each distance as uncertain by one least count instead, $\Delta u = \Delta v = 0.2$ cm, gives exactly half: $0.85\%$, option (B).
Answer: (D) or (B). | ||||
| 32 | KinematicsUniform circular motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Let the radius be R. The arc AB subtends $120^\circ = \frac{2\pi}{3}$ at the centre.
Time from A to B $= \frac{\text{arc}}{\text{speed}} = \frac{(2\pi/3)R}{\pi} = \frac{2R}{3}$.
Displacement = chord $AB = 2R\sin 60^\circ = \sqrt{3}R$.
Average velocity $= \frac{\sqrt{3}R}{2R/3} = \frac{3\sqrt{3}}{2} = 1.5\sqrt{3}$ m/s (independent of R).
Answer: (C). | ||||
| 33 | 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 The acceleration is uniform, so the first equation of motion applies: $v = u + at$.
Here $u = 10.0\ \mathrm{ms^{-1}}$, $v = 60.0\ \mathrm{ms^{-1}}$ and $a = 2.0\ \mathrm{ms^{-2}}$.
$60 = 10 + 2t$, so $2t = 50$ and $t = 25$ s.
(Check: the velocity rises by 2 m/s every second, and it must rise by 50 m/s.)
Answer: (B). | ||||
| 34 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion $\omega = \frac{2\pi}{T} = \frac{2 \times 3.14}{3.14} = 2$ rad/s.
Centrifugal force (in the rotating frame) $= m\omega^2 r = 5 \times 4 \times 2 = 40$ N.
Answer: (A). | ||||
| 35 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation (A) True: the earth's acceleration towards the body is $\frac{F}{M_e}$, negligible because $M_e$ is enormous.
(B) True: in a circular orbit $v = \sqrt{\frac{GM}{r}}$ is constant.
(C) True: gravity is conservative, so total mechanical energy is conserved.
(D) False: by Kepler's second law (conservation of angular momentum) the planet moves faster near the sun and slower far from it.
Answer: (D). | ||||
| 36 | 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 $F \propto \frac{1}{(R + h)^2}$ with $h = \frac{R}{4}$: $F = 100 \times \left(\frac{R}{5R/4}\right)^2 = 100 \times \frac{16}{25} = 64$ N.
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 | Assertion and reasonEasy | |
A easy assertion and reason 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 R is the statement of Pascal's law: a pressure change in an enclosed incompressible fluid is transmitted undiminished throughout. True.
Squeezing one end of the tube raises the pressure there; this increase is transmitted through the paste to the open end, pushing it out. So A is an application of Pascal's principle, and R explains it.
Answer: (A). | ||||
| 38 | Properties of Bulk MatterHeat transfer | 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 heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Newton's law of cooling: the excess over the surroundings falls by the same factor in equal times.
First 7 min: $\frac{40 - 10}{60 - 10} = \frac{3}{5}$. Next 7 min: $T - 10 = 30 \times \frac{3}{5} = 18$, so $T = 28$°C.
(The average-temperature form gives the same: $\frac{20}{7} = k(50 - 10)$ and $\frac{40 - T}{7} = k\left(\frac{40 + T}{2} - 10\right)$ give $T = 28$°C.)
Answer: (A). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T}$. Temperature rises by a factor 4, so $v_{rms}$ doubles: $2v_0$.
Answer: (B). | ||||
| 40 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v = \sqrt{\frac{\gamma RT}{M}}$. Both gases are diatomic (same $\gamma$), so $\frac{v_H}{v_O} = \sqrt{\frac{M_O}{M_H}} = \sqrt{\frac{32}{2}} = 4$.
Ratio 4 : 1.
Answer: (D). | ||||
| 41 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter $\mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}}$: $0.6 = \frac{A_{max} - 3}{A_{max} + 3}$.
$0.6A_{max} + 1.8 = A_{max} - 3$, so $0.4A_{max} = 4.8$ and $A_{max} = 12$ V.
Answer: (C). | ||||
| 42 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Assertion and reasonMedium | |
A medium assertion and reason question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A: In forward bias the barrier is lowered, majority carriers diffuse across in large numbers, and the diffusion current far exceeds the (small, minority-carrier) drift current. True.
R: The diffusion current is carried by holes moving p → n and electrons moving n → p; both give a conventional current from the p-side to the n-side. So the diffusion current is not from n to p. False.
Answer: (C). | ||||
| 43 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Force $F = -\frac{dU}{dr} = -m\omega^2r$ (towards the origin). For a circular orbit: $\frac{mv^2}{r} = m\omega^2r$, so $v = \omega r$.
Bohr quantization: $mvr = \frac{nh}{2\pi}$, i.e. $m\omega r^2 = \frac{nh}{2\pi}$.
So $r^2 \propto n$ and $r \propto \sqrt{n}$.
Answer: (C). | ||||
| 44 | 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 $eV_0 = h\nu - \phi$, so $V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$. The slope $\frac{h}{e}$ is the same for every metal; only the intercept depends on the work function.
Ratio of slopes = 1.
Answer: (B). | ||||
| 45 | OpticsWave optics | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Assertion and reasonMedium | |
A medium assertion and reason question on wave optics, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics In a medium of index $\mu$ the wavelength is $\frac{\lambda}{\mu}$, so a thickness $t$ holds $\frac{\mu t}{\lambda}$ wavelengths. R is true.
Two waves crossing equal thicknesses of media with indices $\mu_1$ and $\mu_2$ acquire a phase difference $\Delta\phi = \frac{2\pi}{\lambda}(\mu_1 - \mu_2)t \ne 0$, so their phase difference changes. A is true, and it happens precisely because the wavelengths differ in the two media: R explains A.
Answer: (A). | ||||
| 46 | 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 Electric energy density $u_E = \frac{1}{2}\epsilon_0E^2$; magnetic energy density $u_B = \frac{B^2}{2\mu_0}$. (In an EM wave the two are equal, since $B = \frac{E}{c}$ and $c^2 = \frac{1}{\mu_0\epsilon_0}$.)
Answer: (D). | ||||
| 47 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $X_C = \frac{1}{\omega C} = \frac{1}{120\pi \times 150 \times 10^{-6}} = \frac{1}{0.018\pi} \approx 17.7\ \Omega$.
$I_0 = \frac{E_0}{X_C} = 36 \times 0.018\pi \approx 2.04$ A $\approx 2$ A.
Answer: (D). | ||||
| 48 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter To verify Ohm's law we need the voltage across $R_T$ and the current through it.
Voltmeter: a galvanometer with a large resistance in series ($G_1$ with $R_1 = 10\,M\Omega$), connected in parallel with $R_T$, so it draws almost no current.
Ammeter: a galvanometer with a very small shunt in parallel ($G_2$ with $R_2 = 0.001\,\Omega$), connected in series with $R_T$, so it adds almost no resistance.
Only circuit (D) has both arranged this way.
(A) and (C): $G_1$ shunted by $0.001\,\Omega$ placed across $R_T$ short-circuits $R_T$; in (A) the $10\,M\Omega$ in the main line also chokes the current.
(B): $G_1$ with $10\,M\Omega$ in parallel is still a low-resistance path across $R_T$, not a voltmeter, and $G_2$ in series carries the full current without a shunt.
Answer: (D). | ||||
| 49 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Let V be the potential of the junction. Current in = current out:
$\frac{30 - V}{10} = \frac{V - 12}{20} + \frac{V - 2}{30}$.
Multiplying by 60: $180 - 6V = 3V - 36 + 2V - 4$, so $11V = 220$ and $V = 20$ V.
Current through 20 $\Omega$ $= \frac{20 - 12}{20} = 0.4$ A.
Answer: (C). | ||||
| 50 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctHard | |
A hard single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Masses: $m$ (charge $+q$) and $2m$ (charge $-q$), separated by $l$.
Restoring torque for a small angle $\theta$: $\tau = pE\sin\theta \approx qlE\theta$.
Taking the rotation about the midpoint of the dipole: $I = m\left(\frac{l}{2}\right)^2 + 2m\left(\frac{l}{2}\right)^2 = \frac{3ml^2}{4}$.
$\omega = \sqrt{\frac{qlE}{I}} = \sqrt{\frac{4qE}{3ml}}$.
Answer: (A). | ||||
| 51 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $\frac{v_1}{v_2} = 4$, so $\frac{K_2}{K_1} = \frac{1}{16}$.
Fractional loss $= 1 - \frac{1}{16} = \frac{15}{16}$, i.e. $93.75\%$.
$\frac{x}{4} = 93.75$, so $x = 375$.
Answer: 375. | ||||
| 52 | 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 Ring about its axis: $I = MR_r^2$, so $k = R_r$. Solid sphere about a diameter: $I = \frac{2}{5}MR_s^2$, so $k = \sqrt{\frac{2}{5}}R_s$.
Equal $k$: $\frac{R_r}{R_s} = \sqrt{\frac{2}{5}}$, so $x = 5$.
Answer: 5. | ||||
| 53 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Area $= \pi r^2 = \frac{22}{7}\times(7\times 10^{-3})^2 = 154 \times 10^{-6}\ \mathrm{m^2}$.
Tension $= \text{stress} \times \text{area} = 7\times 10^5 \times 154\times 10^{-6} = 107.8$ N $= mg$.
$m = \frac{107.8}{9.8} = 11$ kg.
Answer: 11. | ||||
| 54 | 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 Field at the centre of the large coil ($R = 10$ m, $N_2 = 200$) carrying current $I$: $B = \frac{\mu_0N_2I}{2R}$, nearly uniform over the small coil ($r = 0.01$ m, $N_1 = 10$).
Flux through the small coil: $\Phi = N_1B\pi r^2$, so $M = \frac{\mu_0N_1N_2\pi r^2}{2R}$.
$M = \frac{4\pi\times 10^{-7}\times 10\times 200\times\pi\times 10^{-4}}{20} = \frac{4\pi^2\times 2000\times 10^{-11}}{20} = 4\times 10\times 100\times 10^{-11} = 4\times 10^{-8}$ H.
Answer: 4. | ||||
| 55 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Bright fringes coincide where $n_1\lambda_1 = n_2\lambda_2$: $7000n_1 = 5500n_2$, i.e. $14n_1 = 11n_2$; least $n_1 = 11$, $n_2 = 14$.
$y = \frac{n_1\lambda_1D}{d} = \frac{11\times 7000\times 10^{-10}\times 1.5}{2.5\times 10^{-3}} = 4.62\times 10^{-3}$ m $= 462\times 10^{-5}$ m.
Answer: 462. | ||||
| 56 | 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 Total energy of the electron in orbit $= -\frac{ke^2}{2r}$, so the energy to separate it is $\frac{ke^2}{2r} = 12.8$ eV.
$r = \frac{ke^2}{2\times 12.8\times 1.6\times 10^{-19}} = \frac{9\times 10^9\times 1.6\times 10^{-19}}{25.6} = \frac{9}{16}\times 10^{-10}$ m.
$x = 16$.
Answer: 16. | ||||
| 57 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $\frac{1}{2}mv^2 = 2\times 1.6\times 10^{-19}$ J gives $v = \sqrt{\frac{6.4\times 10^{-19}}{1.6\times 10^{-27}}} = 2\times 10^4$ m/s.
Component along B: $v\cos 60^\circ = 10^4$ m/s.
Period $T = \frac{2\pi m}{qB} = \frac{2\pi\times 1.6\times 10^{-27}}{1.6\times 10^{-19}\times\frac{\pi}{2}\times 10^{-3}} = 4\times 10^{-5}$ s.
Pitch $= 10^4\times 4\times 10^{-5} = 0.4$ m $= 40$ cm.
Answer: 40. | ||||
| 58 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity The 3 V cell drives current through 2 $\Omega$ in series with (5 $\Omega$ ∥ voltmeter). The parallel part has 2 V, so the 2 $\Omega$ has 1 V and the current is $I = 0.5$ A.
Resistance of the parallel part $= \frac{2}{0.5} = 4\ \Omega$: $\frac{1}{4} = \frac{1}{5} + \frac{1}{R_V}$, so $R_V = 20\ \Omega$.
Answer: 20. | ||||
| 59 | 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 From the figure: the outer plates are 5 mm apart; the two inner plates are joined by a wire and are 1 mm apart, leaving gaps $a$ and $b$ with $a + c + b = d$, so $a + b = 4$ mm.
The arrangement is two capacitors in series, $\frac{\epsilon_0A}{a}$ and $\frac{\epsilon_0A}{b}$:
$C = \frac{\epsilon_0A}{a + b} = \frac{\epsilon_0\times 200\times 10^{-4}}{4\times 10^{-3}} = 5\epsilon_0$ F.
$x = 5$.
Answer: 5. | ||||
| 60 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Maximum speed at the lowest point: $v = A\omega = A\sqrt{\frac{g}{L}} = 0.1\sqrt{10}$ m/s, so $v^2 = 0.1$ m²/s².
$T_{max} = mg + \frac{mv^2}{L} = 0.25\times 10 + 0.25\times\frac{0.1}{1} = 2.5 + 0.025 = 2.525$ N.
$\frac{x}{40} = 2.525$, so $x = 101$.
Answer: 101. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{Ba(OH)_2 + 2HBr \to BaBr_2 + 2H_2O}$.
Moles of $\mathrm{OH^-}$ $= 2 \times 0.01 \times 10.0 = 0.2$ mmol, so 0.2 mmol HBr is needed.
$V = \frac{0.2}{0.02} = 10.0$ mL.
Answer: (D). | ||||
| 62 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Bohr's condition $mvr = \frac{nh}{2\pi}$ gives $2\pi r_n = n\lambda$.
$r_3 = 9a_0$, so $\lambda = \frac{2\pi\times 9a_0}{3} = 6\pi a_0$.
Answer: (B). | ||||
| 63 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Each peak of the energy profile is an activated complex (transition state): there are 3.
Each valley between two peaks is an intermediate: there are 2.
The rate-determining step is the one with the highest energy barrier, step II.
Answer: (A). | ||||
| 64 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Pure boiling water is at 100°C.
Li melts at about 181°C: solid. Br boils at about 59°C: a gas at 100°C. Cs melts at about 28°C but reacts violently with water, so it does not remain.
Ga melts at about 30°C and boils at about 2400°C, and does not react with water: it stays liquid.
Answer: (B). | ||||
| 65 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Electrolysis of brine (aqueous NaCl): at the anode $\mathrm{2Cl^- \to Cl_2 + 2e^-}$; at the cathode $\mathrm{2H_2O + 2e^- \to H_2 + 2OH^-}$; $\mathrm{Na^+}$ and $\mathrm{OH^-}$ remain as NaOH.
HCl is not a product.
Answer: (C). | ||||
| 66 | 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 Solid $\mathrm{BeCl_2}$ is a chain polymer with chloride bridges. In the vapour it is a chloro-bridged dimer, which breaks into linear monomers at very high temperature (about 1200 K).
Polymeric, Dimeric, Monomeric.
Answer: (B). | ||||
| 67 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Hydration enthalpy grows with charge density. For ions of the same charge, the smallest ion, $\mathrm{Be^{2+}}$, has the highest; it falls down the group: $\mathrm{Be^{2+} > Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}}$.
Answer: (A). | ||||
| 68 | 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 Metallic character, and with it the basic character of the oxides, increases down group 13: $\mathrm{B_2O_3}$ is acidic, $\mathrm{Al_2O_3}$ and $\mathrm{Ga_2O_3}$ are amphoteric, and the oxides of In and Tl are basic.
The most basic is $\mathrm{Tl_2O_3}$.
Answer: (D). | ||||
| 69 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements A change of 3 means Mn goes from +7 to +4, i.e. $\mathrm{MnO_4^-}$ is reduced to $\mathrm{MnO_2}$. This happens in neutral or faintly alkaline solution:
$\mathrm{8MnO_4^- + 3S_2O_3^{2-} + H_2O \to 8MnO_2 + 6SO_4^{2-} + 2OH^-}$.
In acidic medium permanganate is reduced to $\mathrm{Mn^{2+}}$ (a change of 5), so every option containing 'acidic' is wrong.
Answer: (A). | ||||
| 70 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Assertion and reasonMedium | |
A medium assertion and reason question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A: CO is a neutral ligand, so Ni in $\mathrm{Ni(CO)_4}$ and Fe in $\mathrm{Fe(CO)_5}$ are in the zero oxidation state. True.
R: Low oxidation states are stabilised by ligands that are $\pi$-acceptors (they take back electron density from the metal into empty $\pi^*$ orbitals, synergic bonding), not $\pi$-donors. False.
Answer: (C). | ||||
| 71 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter DDT and aldrin are both organochlorine insecticides (pesticides).
(A) Tetrachloroethene is a dry-cleaning solvent. (B) PAN (peroxyacetyl nitrate) is a constituent of photochemical smog. (D) Sodium chlorate and sodium arsenite are herbicides (weedicides), classed separately from the pesticides.
Answer: (C). | ||||
| 72 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The complex ion is an anion, so the metal name ends in -ate: cobaltate. Co: $x + 3(-2) = -3$, so $x = +3$.
Three oxalato ligands: 'trioxalato' in the school textbook convention (as in potassium trioxalatochromate(III) for $\mathrm{K_3[Cr(C_2O_4)_3]}$), and 'tris(oxalato)' in the current IUPAC recommendations. Name: potassium trioxalatocobaltate(III), equally written potassium tris(oxalato)cobaltate(III).
(A) and (C) lack the -ate ending required for an anionic complex.
Answer: (B) or (D). | ||||
| 73 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter $\mathrm{Pb^{2+}}$ is confirmed by precipitates: yellow lead chromate with $\mathrm{K_2CrO_4}$, yellow lead iodide with KI, and white lead sulphate with dilute $\mathrm{H_2SO_4}$.
Lead nitrate is soluble; it forms no precipitate and is no test.
Answer: (D). | ||||
| 74 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Nessler's reagent is an alkaline solution of potassium tetraiodomercurate(II), $\mathrm{K_2[HgI_4]}$ in KOH. It contains K, Hg, I (and O, H), but no nitrogen; it is used to detect $\mathrm{NH_4^+}$ / $\mathrm{NH_3}$.
Answer: (C). | ||||
| 75 | 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 In acid the secondary alcohol is protonated and loses water, giving a secondary carbocation next to the tert-butyl group.
A 1,2-methyl shift turns it into the more stable tertiary carbocation, $\mathrm{-CH(CH_3)-\overset{+}{C}(CH_3)_2}$.
The ring double bond attacks this cation intramolecularly. Counting ring C-1 (bearing the chain), $\mathrm{CH_2}$, $\mathrm{CH_2}$, $\mathrm{CH(CH_3)}$, $\mathrm{C(CH_3)_2}$ and ring C-2, the new ring is six-membered (favoured), and the positive charge moves to ring C-1, a tertiary centre.
Loss of a proton gives the tetrasubstituted double bond at the ring fusion: an octahydronaphthalene with the two original ring methyls, a gem-dimethyl next to the fusion and a methyl beside it, which is option (D).
Without the methyl shift, cyclisation would give the five-membered ring with a tert-butyl group, option (C), from a less stable secondary cation; (A) and (B) are ethers that would need the OH to stay in the product.
Answer: (D). | ||||
| 76 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsMedium | |
A medium statements question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In the column, band c is nearest the outlet and a is highest. The least strongly adsorbed (least polar) compound moves fastest, so c is least polar and elutes first; a is most polar, most strongly adsorbed, and stays longest.
A: false (c is the least polar). B: false (a is the most polar). C: false (b comes out after c and before a). D: true.
Incorrect statements: A, B and C.
Answer: (B). | ||||
| 77 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids MeMgBr with CuI forms a methyl cuprate, which adds 1,4 (conjugate addition) to the enone: the methyl goes to C-3 (the $\beta$-carbon, which already has a methyl), giving the enolate with negative charge at C-2.
n-PrI then alkylates the enolate at C-2.
Product: 3,3-dimethyl-2-propylcyclohexan-1-one, option (C).
(A) is the 1,2-addition product of a Grignard without copper; (B) and (D) put the groups on the wrong carbons.
Answer: (C). | ||||
| 78 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Electron-withdrawing groups stabilise the phenoxide ion and raise acidity; electron-donating groups lower it.
At the meta position, $\mathrm{-NO_2}$ withdraws strongly by its −I effect, more than −Cl; $\mathrm{-CH_3}$ donates.
Order: 3-nitrophenol (pKa about 8.4) > 3-chlorophenol (about 9.1) > phenol (10.0) > 3-methylphenol (about 10.1).
Answer: (A). | ||||
| 79 | BiomoleculesProteins | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules One-letter codes: arginine R, aspartic acid D, asparagine N, alanine A.
(A) – IV, (B) – I, (C) – II, (D) – III.
Answer: (A). | ||||
| 80 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | StatementsEasy | |
A easy statements question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Statement I: morphine is a narcotic analgesic, but in medicinal doses narcotics relieve pain and produce sleep. The words 'without producing sleep' make it false.
Statement II: morphine and its derivatives (codeine, heroin) come from the opium poppy. True.
Answer: (D). | ||||
| 81 | Solid StateSeven crystal systems | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on seven crystal systems, from Solid State. NCERT: No matching NCERT chapter Body-centred lattices occur in the cubic, tetragonal and orthorhombic systems only (monoclinic has end-centred; hexagonal, rhombohedral and triclinic only primitive).
Answer: 3. | ||||
| 82 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{XeF_4}$: 4 bond pairs + 2 lone pairs, square planar. Yes.
$\mathrm{SF_4}$: 4 bp + 1 lp, see-saw. $\mathrm{SiF_4}$, $\mathrm{BF_4^-}$: tetrahedral.
$\mathrm{BrF_4^-}$: 4 bp + 2 lp, square planar. Yes.
$\mathrm{[Cu(NH_3)_4]^{2+}}$: $d^9$, $dsp^2$, square planar. Yes.
$\mathrm{[FeCl_4]^{2-}}$: weak-field $\mathrm{Cl^-}$, $sp^3$, tetrahedral.
$\mathrm{[PtCl_4]^{2-}}$: $5d^8$, square planar. Yes.
Count: 4.
Answer: 4. | ||||
| 83 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\mathrm{2C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \to C_2H_5OH(l)}$.
$\Delta H_f = 2(-393.5) + 3(-241.8) - (-1234.7) = -787.0 - 725.4 + 1234.7 = -277.7\ \mathrm{kJ\ mol^{-1}}$.
To the nearest integer: −278.
Answer: 278. | ||||
| 84 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Isotonic means equal $iC$ (complete dissociation).
A: $2\times 1 = 2$ and $1\times 2 = 2$. B: $3\times 1 = 3$ and $2\times 1.5 = 3$. C: $4\times 1.5 = 6$ and $3\times 2 = 6$. D: $2\times 2.5 = 5$ and $5\times 1 = 5$.
All 4 pairs are isotonic.
Answer: 4. | ||||
| 85 | EquilibriumLe Chatelier's principle | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on le chatelier's principle, from Equilibrium. NCERT: Class 11: Equilibrium Check: $K_c = \frac{0.40}{0.2\times 0.1} = 20$.
Taking the volume as 1 L, adding 0.2 mol $\mathrm{Cl_2}$ makes $[\mathrm{Cl_2}] = 0.3$. Let $x$ react:
$\frac{0.40 + x}{(0.2 - x)(0.3 - x)} = 20 \Rightarrow 0.40 + x = 1.2 - 10x + 20x^2 \Rightarrow 20x^2 - 11x + 0.8 = 0$.
$x = \frac{11 - \sqrt{57}}{40} = \frac{11 - 7.55}{40} = 0.0863$ (the other root exceeds 0.2).
$[\mathrm{PCl_5}] = 0.4863\ \mathrm{mol\ L^{-1}} = 48.6\times 10^{-2}$, i.e. 49 to the nearest integer.
Answer: 49. | ||||
| 86 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\mathrm{NO_3^-}$ ($E^\theta = 0.97$ V) can oxidise any metal whose reduction potential is lower, since then $E^\theta_{cell} > 0$.
V (−1.19), Fe (−0.04) and Ag (0.80) qualify; Au (1.40) does not.
Answer: 3. | ||||
| 87 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Gem stones: solid in solid. Paints: solid in liquid (sol). Smoke: solid in gas. Cheese: liquid in solid (gel). Milk: liquid in liquid (emulsion). Hair cream: liquid in liquid (emulsion). Insecticide sprays: liquid in gas. Froth and soap lather: gas in liquid (foam).
Liquid dispersion medium: paints, milk, hair cream, froth, soap lather = 5.
Answer: 5. | ||||
| 88 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure In ice each O is tetrahedrally surrounded by four other O atoms: it donates two hydrogen bonds through its H atoms and accepts two through its lone pairs.
Answer: 4. | ||||
| 89 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A positive iodoform test needs $\mathrm{CH_3CO-}$ or $\mathrm{CH_3CH(OH)-}$.
(a) $\mathrm{C_6H_5CH_2COCH_2CH_3}$: no methyl ketone. No.
(b) $\mathrm{(CH_3)_2C(OH)CH_2CH_3}$: tertiary alcohol. No.
(c) $\mathrm{CH_3CH(OH)CH(CH_3)_2}$: yes.
(d) $\mathrm{C_6H_5CH(OH)CH_3}$: yes.
(e) $\mathrm{CH_3COC(CH_3)_3}$: yes.
(f) $\mathrm{C_6H_5CH_2CH(OH)CH_3}$: yes.
Count: 4.
Answer: 4. | ||||
| 90 | AminesClassification and identification of amines | Class 12: Amines | NumericalHard | |
A hard numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Gabriel synthesis gives primary amines from primary alkyl (or benzyl) halides by $S_N2$ on the phthalimide anion; it cannot make arylamines (NH2 on the ring), since aryl halides do not undergo this substitution.
$\mathrm{C_8H_{11}N}$ has 4 degrees of unsaturation: a benzene ring plus two more carbons. Primary amines with NH2 on a $\mathrm{CH_2}$:
$\mathrm{C_6H_5CH_2CH_2NH_2}$, and o-, m- and p-$\mathrm{CH_3C_6H_4CH_2NH_2}$.
That is 4.
Answer: 4. | ||||
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