JEE (Main) · 2022
JEE (Main) 2022 — 27 June, 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
- Matrices and Determinants3
- Integral Calculus3
- Sequences and Series2
10 easy · 17 medium · 3 hard
Physics
30 Q
- Physical World and Measurement3
- Electrostatics3
- Laws of Motion2
26 easy · 3 medium · 1 hard
Chemistry
30 Q
- Some Basic Concepts of Chemistry2
- Chemical Bonding and Molecular Structure2
- p-Block Elements2
23 easy · 7 medium · 0 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 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|z - (4 + 3i)| = 2$ is the circle with centre $(4, 3)$ and radius 2.
$|z| + |z - 4| = 6$ is the ellipse with foci $(0, 0)$ and $(4, 0)$ and $2a = 6$: centre $(2, 0)$, $a = 3$, $c = 2$, $b^2 = 9 - 4 = 5$, so $\frac{(x - 2)^2}{9} + \frac{y^2}{5} = 1$.
Test points of the circle against the ellipse, using $E(x, y) = \frac{(x - 2)^2}{9} + \frac{y^2}{5}$:
• the lowest point of the circle, $(4, 1)$: $E = \frac{4}{9} + \frac{1}{5} = 0.64 < 1$, inside the ellipse;
• the centre's level points $(2, 3)$ and $(6, 3)$, and the top $(4, 5)$: $E = \frac{9}{5}$, $\frac{16}{9} + \frac{9}{5}$, $\frac{4}{9} + 5$, all $> 1$, outside.
So the circle dips into the ellipse only around its lowest point: one arc of the circle lies inside, the rest outside. The circle enters and leaves the ellipse once each, crossing it at exactly 2 points (a check of $E$ all round the circle confirms only two sign changes).
Answer: (C). | ||||
| 2 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Expand along the first row:
$f(x) = a(a \cdot a + ax) + 1 \cdot (a \cdot ax + ax^2) = a^3 + a^2x + a^2x + ax^2 = a^3 + 2a^2x + ax^2$.
$f'(x) = 2a^2 + 2ax$, so $f'(10) = 2a^2 + 20a$ and $f'(5) = 2a^2 + 10a$.
$2f'(10) - f'(5) + 100 = 4a^2 + 40a - 2a^2 - 10a + 100 = 2a^2 + 30a + 100 = 0$, i.e. $a^2 + 15a + 50 = 0$, so $a = -5$ or $a = -10$.
Sum of squares $= 25 + 100 = 125$.
Answer: (C). | ||||
| 3 | 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 A homogeneous system has more than one solution only when its determinant is zero:
$4i\,\bar{a} - (1 + i) \cdot 8\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 0$.
$(1 + i)\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -\frac{1 + \sqrt{3}}{2} + \frac{\sqrt{3} - 1}{2}i$, so the second term is $-4(1 + \sqrt{3}) + 4(\sqrt{3} - 1)i$.
With $\bar{a} = \alpha + i\beta$: $4i\bar{a} = -4\beta + 4\alpha i$.
Equating real and imaginary parts: $-4\beta = -4(1 + \sqrt{3})$, so $\beta = 1 + \sqrt{3}$; $4\alpha = 4(\sqrt{3} - 1)$, so $\alpha = \sqrt{3} - 1$.
$\frac{\alpha}{\beta} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.
Answer: (B). | ||||
| 4 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For an $n \times n$ matrix M, $|adj\,M| = |M|^{n-1}$; here $n = 3$.
$AB = I$ gives $|B| = 8$.
$|2A| = 2^3 \cdot \frac{1}{8} = 1$, so $|adj(2A)| = 1^2 = 1$.
$|B\,adj(2A)| = 8 \times 1 = 8$, so $|adj(B\,adj(2A))| = 8^2 = 64$.
Answer: (C). | ||||
| 5 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The numerators $2, 6, 12, 20, 30, \ldots$ are $n(n + 1)$, so $S = \sum_{n \ge 1} n(n + 1)x^{n-1}$ with $x = \frac{1}{7}$.
Differentiating $\sum_{n \ge 0} x^{n+1} = \frac{x}{1 - x}$ twice: $\sum n(n + 1)x^{n-1} = \frac{2}{(1 - x)^3}$.
$S = \frac{2}{(6/7)^3} = \frac{2 \cdot 343}{216} = \frac{343}{108}$, so $4S = \frac{343}{27} = \left(\frac{7}{3}\right)^3$.
Answer: (C). | ||||
| 6 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $a_{10} = a_1 + 9d = 3$ gives $d = \frac{1}{9}$, so $a_4 = 2 + \frac{3}{9} = \frac{7}{3}$.
$b_1 = \frac{1}{a_1} = \frac{1}{2}$ and $b_{10} = \frac{1}{a_{10}} = \frac{1}{3}$, so the common difference of the b's is $\frac{1/3 - 1/2}{9} = -\frac{1}{54}$ and $b_4 = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$.
$a_4b_4 = \frac{7}{3} \cdot \frac{4}{9} = \frac{28}{27}$.
Answer: (D). | ||||
| 7 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives By the fundamental theorem of calculus,
$f'(x) = 2x \cdot \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} = \frac{2x(x - 1)(x + 1)(x - 2)(x + 2)}{2 + e^{x^2}}$.
The denominator is positive, so the sign of $f'$ is that of $x(x^2 - 1)(x^2 - 4)$, which changes at $-2, -1, 0, 1, 2$:
$x < -2$: −; $(-2, -1)$: +; $(-1, 0)$: −; $(0, 1)$: +; $(1, 2)$: −; $x > 2$: +.
From − to + (local minimum) at $x = -2, 0, 2$; from + to − (local maximum) at $x = -1, 1$.
So $m = 2$, $n = 3$.
Answer: (B). | ||||
| 8 | 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 Differentiate both sides with respect to x:
$-\cos^2 x\, f(\cos x) \cdot (-\sin x) = 3\sin^2 x \cos x - \sin x$,
so $f(\cos x) = \frac{3\sin x \cos x - 1}{\cos^2 x}$.
Put $t = \cos x$, $\sin x = \sqrt{1 - t^2}$: $f(t) = \frac{3\sqrt{1 - t^2}}{t} - \frac{1}{t^2}$.
$f'(t) = 3 \cdot \frac{-\frac{t^2}{\sqrt{1 - t^2}} - \sqrt{1 - t^2}}{t^2} + \frac{2}{t^3} = -\frac{3}{t^2\sqrt{1 - t^2}} + \frac{2}{t^3}$.
At $t = \frac{1}{\sqrt{3}}$: $t^2 = \frac{1}{3}$, $\sqrt{1 - t^2} = \sqrt{\frac{2}{3}}$, so $f'(t) = -\frac{9}{\sqrt{2/3}} + 6\sqrt{3} = -\frac{9\sqrt{3}}{\sqrt{2}} + 6\sqrt{3}$.
$\frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) = 6 - \frac{9}{\sqrt{2}}$.
Answer: (B). | ||||
| 9 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals For $\frac{1}{k + 1} < x \le \frac{1}{k}$, $\left[\frac{1}{x}\right] = k$. So
$I = \sum_{k=1}^{\infty} 7^{-k}\left(\frac{1}{k} - \frac{1}{k + 1}\right)$.
$\sum_{k \ge 1} \frac{7^{-k}}{k} = -\log_e\left(1 - \frac{1}{7}\right) = \log_e\frac{7}{6}$.
$\sum_{k \ge 1} \frac{7^{-k}}{k + 1} = 7\sum_{k \ge 1} \frac{7^{-(k+1)}}{k + 1} = 7\left(\log_e\frac{7}{6} - \frac{1}{7}\right) = 7\log_e\frac{7}{6} - 1$.
$I = \log_e\frac{7}{6} - 7\log_e\frac{7}{6} + 1 = 1 - 6\log_e\frac{7}{6} = 1 + 6\log_e\left(\frac{6}{7}\right)$.
Answer: (A). | ||||
| 10 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Treat x as a function of y: $\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1} y}{1 + y^2}$, linear in x.
Integrating factor $e^{\tan^{-1} y}$. With $t = \tan^{-1} y$:
$x e^{t} = \int t e^{t}\,dt = (t - 1)e^{t} + C$.
At $(1, 0)$, $t = 0$: $1 = -1 + C$, so $C = 2$ and $x = t - 1 + 2e^{-t}$.
For $y = \tan 1$, $t = 1$: $x = 0 + \frac{2}{e} = \frac{2}{e}$.
Answer: (B). | ||||
| 11 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The axis is perpendicular to the directrix through the vertex. The foot of the perpendicular from $V(5, 4)$ to $3x + y - 29 = 0$: $3(5) + 4 - 29 = -10$, so the foot is $(5, 4) + \frac{10}{10}(3, 1) = (8, 5)$.
The vertex is midway between the focus and this foot, so the focus is $S = (2 \cdot 5 - 8, 2 \cdot 4 - 5) = (2, 3)$.
Parabola: $(x - 2)^2 + (y - 3)^2 = \frac{(3x + y - 29)^2}{10}$.
$10x^2 + 10y^2 - 40x - 60y + 130 = 9x^2 + y^2 + 6xy - 174x - 58y + 841$
$x^2 + 9y^2 - 6xy + 134x - 2y - 711 = 0$.
So $a = 9$, $b = -6$, $c = 134$, $d = -2$, $k = -711$, and $a + b + c + d + k = -576$.
Answer: (D). | ||||
| 12 | 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 $x^2 + y^2 - 3x + 2y + \frac{k}{4} = 0$: centre $\left(\frac{3}{2}, -1\right)$, $r^2 = \frac{9}{4} + 1 - \frac{k}{4} = \frac{13 - k}{4}$.
Inside the fourth quadrant: the distances of the centre from the axes are 1 (from the x-axis) and $\frac{3}{2}$ (from the y-axis), so $0 < r < 1$: $0 < 13 - k < 4$, i.e. $9 < k < 13$.
The point $\left(1, -\frac{1}{3}\right)$ on or inside: $4 + \frac{4}{9} - 12 - \frac{8}{3} + k \le 0$, i.e. $k \le 12 - 4 - \frac{4}{9} + \frac{8}{3} = \frac{92}{9}$.
Together: $k \in \left(9, \frac{92}{9}\right]$.
Answer: (D). | ||||
| 13 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy 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 line is $(-2 + 4t, 1 + 2t, -1 + 3t)$. Its join with $(1, 2, 4)$ is $(-3 + 4t, -1 + 2t, -5 + 3t)$, perpendicular to $(4, 2, 3)$:
$4(-3 + 4t) + 2(-1 + 2t) + 3(-5 + 3t) = 29t - 29 = 0$, so $t = 1$ and $P = (2, 3, 2)$.
Distance from the plane: $\frac{|3(2) + 4(3) + 12(2) + 23|}{\sqrt{9 + 16 + 144}} = \frac{65}{13} = 5$.
Answer: (A). | ||||
| 14 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Points $A(3, 2, 1)$, $B(-3, 6, 5)$; directions $\vec{d_1} = (2, 3, -1)$, $\vec{d_2} = (2, 1, 3)$.
$\vec{d_1} \times \vec{d_2} = (3 \cdot 3 - (-1)(1),\ (-1)(2) - 2 \cdot 3,\ 2 \cdot 1 - 3 \cdot 2) = (10, -8, -4)$, of length $\sqrt{180} = 6\sqrt{5}$.
$\vec{AB} = (-6, 4, 4)$; $\vec{AB} \cdot (10, -8, -4) = -60 - 32 - 16 = -108$.
Shortest distance $= \frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}$.
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 A parallelogram with diagonals $\vec{a}$, $\vec{b}$ has area $\frac{1}{2}|\vec{a} \times \vec{b}|$, so $|\vec{a} \times \vec{b}| = 4\sqrt{2}$.
$|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ with an acute angle θ gives $\cos\theta = \sin\theta$, $\theta = \frac{\pi}{4}$.
$|\vec{a}||\vec{b}|\sin\frac{\pi}{4} = 4\sqrt{2}$ with $|\vec{a}| = 1$ gives $|\vec{b}| = 8$.
$\vec{a} \times \vec{b}$ is perpendicular to $\vec{b}$, so $\vec{b} \cdot \vec{c} = -2|\vec{b}|^2 = -128$ and
$|\vec{c}|^2 = 8|\vec{a} \times \vec{b}|^2 + 4|\vec{b}|^2 = 8(32) + 4(64) = 512$, $|\vec{c}| = 16\sqrt{2}$.
$\cos\phi = \frac{-128}{8 \cdot 16\sqrt{2}} = -\frac{1}{\sqrt{2}}$, so $\phi = \frac{3\pi}{4}$.
Answer: (D). | ||||
| 16 | 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 Mean: $36 + x + y = 48$, so $x + y = 12$.
Variance: $\frac{\sum x_i^2}{8} - 36 = \frac{9}{4}$, so $\sum x_i^2 = 306$. The known squares sum to $16 + 25 + 36 + 36 + 49 + 64 = 226$, so $x^2 + y^2 = 80$.
Then $2xy = 144 - 80 = 64$, $xy = 32$: x and y are 4 and 8, and $x < y$ gives $x = 4$, $y = 8$.
$x^4 + y^2 = 256 + 64 = 320$.
Answer: (B). | ||||
| 17 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability $x + 2y = 6$ meets the y-axis at $(0, 3)$; $5x - 6y = 30$ meets it at $(0, -5)$; the two lines meet where $5(6 - 2y) - 6y = 30$, i.e. $y = 0$, $x = 6$.
The region is the triangle $(0, 3), (0, -5), (6, 0)$, of area $\frac{1}{2} \cdot 8 \cdot 6 = 24$.
The part with $y \ge 1$ is the triangle $(0, 3), (0, 1), (4, 1)$ (on $x + 2y = 6$, $y = 1$ gives $x = 4$), of area $\frac{1}{2} \cdot 2 \cdot 4 = 4$.
The point is taken uniformly over the region, so $P(y < 1) = \frac{24 - 4}{24} = \frac{5}{6}$.
Answer: (B). | ||||
| 18 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\frac{1}{1 + n(n + 1)} = \frac{(n + 1) - n}{1 + n(n + 1)}$, so $\tan^{-1}\frac{1}{1 + n + n^2} = \tan^{-1}(n + 1) - \tan^{-1}n$.
The sum telescopes to $\tan^{-1}51 - \tan^{-1}1 = \tan^{-1}\frac{51 - 1}{1 + 51} = \tan^{-1}\frac{25}{26}$.
Its cotangent is $\frac{26}{25}$.
Answer: (A). | ||||
| 19 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\cos 72^\circ = \sin 18^\circ = \frac{\sqrt{5} - 1}{4}$, so $\sin^2 36^\circ = \frac{1 - \cos 72^\circ}{2} = \frac{5 - \sqrt{5}}{8}$.
Let $s = \alpha^2$: $8s - 5 = -\sqrt{5}$, so $(8s - 5)^2 = 5$, i.e. $64s^2 - 80s + 20 = 0$, or $16s^2 - 20s + 5 = 0$.
So α satisfies $16x^4 - 20x^2 + 5 = 0$.
Answer: (C). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter (A) contains $\sim q \wedge q$: always false.
(B) contains $p \wedge \sim p$: always false.
(C) $p \vee \sim p$ is always true, and anything $\vee$ a true statement is true: a tautology.
(D) is $r \wedge \sim r$ with $r = p \wedge q$: always false.
Answer: (C). | ||||
| 21 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions f sends $1, 2, 3, 4, 5 \to 2, 4, 6, 8, 10$ and $6, 7, 8, 9, 10 \to 1, 3, 5, 7, 9$: a bijection, so $g(n) = f^{-1}(f(g(n)))$.
$f^{-1}$: $2 \to 1$, $4 \to 2$, $6 \to 3$, $8 \to 4$, $10 \to 5$, $1 \to 6$, $3 \to 7$, $5 \to 8$, $7 \to 9$, $9 \to 10$.
$g(1) = f^{-1}(2) = 1$, $g(2) = f^{-1}(1) = 6$, $g(3) = f^{-1}(4) = 2$, $g(4) = f^{-1}(3) = 7$, $g(5) = f^{-1}(6) = 3$, $g(10) = f^{-1}(9) = 10$.
$g(10)(1 + 6 + 2 + 7 + 3) = 10 \times 19 = 190$.
Answer: 190. | ||||
| 22 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha + \beta = 4\lambda$ and $\alpha + \gamma = 3\sqrt{2} + 2\sqrt{3}$. Adding and using $\beta + \gamma = 3\sqrt{2}$: $2\alpha = 4\lambda + 2\sqrt{3}$, so $\alpha = 2\lambda + \sqrt{3}$.
α is a root of the first equation: $(2\lambda + \sqrt{3})^2 - 4\lambda(2\lambda + \sqrt{3}) + 5 = -4\lambda^2 + 8 = 0$, so $\lambda = \sqrt{2}$ ($\lambda > 0$).
$\alpha = 2\sqrt{2} + \sqrt{3}$, $\beta = 4\sqrt{2} - \alpha = 2\sqrt{2} - \sqrt{3}$ (check $\alpha\beta = 8 - 3 = 5$), $\gamma = 3\sqrt{2} - \beta = \sqrt{2} + \sqrt{3}$ (check $\alpha\gamma = 7 + 3\sqrt{6} = 7 + 3\lambda\sqrt{3}$).
$\alpha + 2\beta + \gamma = (2\sqrt{2} + \sqrt{3}) + (4\sqrt{2} - 2\sqrt{3}) + (\sqrt{2} + \sqrt{3}) = 7\sqrt{2}$, whose square is 98.
Answer: 98. | ||||
| 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 p is 3, 5 or 7. Count ordered 4-tuples of entries from 0 to 5 with the given sum: the coefficient of $x^p$ in $\left(\frac{1 - x^6}{1 - x}\right)^4$.
Sum 3: $\binom{6}{3} = 20$ (no entry can exceed 5).
Sum 5: $\binom{8}{3} = 56$.
Sum 7: $\binom{10}{3} - 4\binom{4}{3} = 120 - 16 = 104$ (removing the tuples with an entry of 6 or 7).
Total $20 + 56 + 104 = 180$.
Answer: 180. | ||||
| 24 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom{7}{r}2^r x^{n(7 - r) - 5r}$, with coefficients for $r = 0, \ldots, 7$: 1, 14, 84, 280, 560, 672, 448, 128.
The power $7n - r(n + 5)$ decreases with r when $n > -5$, so the positive powers are those for $r = 0, 1, \ldots, k$. The partial sums are 1, 15, 99, 379, 939, …: 939 needs exactly $r = 0$ to 4.
(For $n \le -5$ the positive powers would be a final block of terms; the sums 128, 576, 1248, … never give 939.)
So the power is positive at $r = 4$ and not positive at $r = 5$:
$3n - 20 > 0 \Rightarrow n \ge 7$; $2n - 25 \le 0 \Rightarrow n \le 12$.
$n = 7, 8, 9, 10, 11, 12$, whose sum is 57.
Answer: 57. | ||||
| 25 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability As $x \to 0^-$: $[x] = -1$, $\{x\} = x + 1 \to 1$, $[1 + x] = 0$.
$2[x] + \{x\} \to -2 + 1 = -1$, so the numerator tends to $\alpha^{-1} - 1 - 1$ and the denominator to $-1$.
Left-hand limit $= 0 + \frac{\frac{1}{\alpha} - 2}{-1} = 2 - \frac{1}{\alpha}$.
$2 - \frac{1}{\alpha} = \alpha - \frac{4}{3}$ gives $3\alpha^2 - 10\alpha + 3 = 0$, so $\alpha = 3$ or $\frac{1}{3}$.
The integral value is 3.
Answer: 3. | ||||
| 26 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $y = x^{x^2}$, $\ln y = x^2 \ln x$, so $y' = y(2x\ln x + x)$.
$y'' = y(2x\ln x + x)^2 + y(2\ln x + 3)$.
At $x = 1$: $y = 1$, $y' = 1$, $y'' = 1 + 3 = 4$.
$\frac{d^2x}{dy^2} = -\frac{y''}{(y')^3} = -4$, so $\frac{d^2x}{dy^2} + 20 = 16$.
Answer: 16. | ||||
| 27 | 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 The whole astroid $x^{2/3} + y^{2/3} \le 1$ has area $\frac{3\pi}{8}$ (with $x = \cos^3\theta$, $y = \sin^3\theta$, the first-quadrant area is $\int_0^{\pi/2} 3\sin^4\theta\cos^2\theta\,d\theta = \frac{3\pi}{32}$).
The region is the whole first-quadrant part, $\frac{3\pi}{32}$, together with the second-quadrant part where $y \ge -x = |x|$.
The astroid is symmetric about the line $y = -x$, which halves its second-quadrant part, so that piece has area $\frac{3\pi}{64}$.
$A = \frac{3\pi}{32} + \frac{3\pi}{64} = \frac{9\pi}{64}$, and $\frac{256A}{\pi} = 36$.
Answer: 36. | ||||
| 28 | 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 $\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{x^3 + 2}{\sqrt{1 - x^2}}$.
Integrating factor $e^{\int -\frac{x}{1 - x^2}dx} = e^{\frac{1}{2}\ln(1 - x^2)} = \sqrt{1 - x^2}$.
$\frac{d}{dx}\left(y\sqrt{1 - x^2}\right) = x^3 + 2$, so $y\sqrt{1 - x^2} = \frac{x^4}{4} + 2x + C$, and $y(0) = 0$ gives $C = 0$.
$k = \int_{-1/2}^{1/2}\left(\frac{x^4}{4} + 2x\right)dx = 2 \cdot \frac{1}{4} \cdot \frac{(1/2)^5}{5} = \frac{1}{320}$ (the odd part $2x$ integrates to 0).
$k^{-1} = 320$.
Answer: 320. | ||||
| 29 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections $L_1 \perp L_2$ (since $4 \cdot 3 + 3 \cdot (-4) = 0$), as it must be: the radius PQ is perpendicular to the tangent.
Q: $4x + 3y = -2$, $3x - 4y = 11$ give $x = 1$, $y = -2$.
P is on $L_1$ at distance 5 from Q; the unit direction of $L_1$ is $\frac{(3, -4)}{5}$, so $P = (1, -2) \pm (3, -4) = (4, -6)$ or $(-2, 2)$.
The circle lies below the x-axis, so $P = (4, -6)$ (its top point $y = -1$ is below the axis).
Distance from $5x - 12y + 51 = 0$: $\frac{|20 + 72 + 51|}{13} = \frac{143}{13} = 11$.
Answer: 11. | ||||
| 30 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(A) = \frac{\text{sum of the indices in A}}{36}$, and the indices 1 to 8 add to 36.
$P(A) \ge \frac{4}{5}$ means the index sum is at least $28.8$, i.e. at least 29; equivalently the complement's index sum is at most 7.
Count subsets of $\{1, \ldots, 8\}$ with sum at most 7 (distinct parts):
sum 0: 1 (empty); 1: {1}; 2: {2}; 3: {3}, {1,2}; 4: {4}, {1,3}; 5: {5}, {1,4}, {2,3}; 6: {6}, {1,5}, {2,4}, {1,2,3}; 7: {7}, {1,6}, {2,5}, {3,4}, {1,2,4}.
Total $1 + 1 + 1 + 2 + 2 + 3 + 4 + 5 = 19$.
Answer: 19. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Pascal is pressure: $[ML^{-1}T^{-2}]$. Multiplying by a second: $[ML^{-1}T^{-1}]$ (the dimensions of the coefficient of viscosity).
Answer: (A). | ||||
| 32 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement 2000 seconds of arc $= \frac{2000}{3600}$ degree $= \frac{2000}{3600} \times \frac{\pi}{180}$ rad $\approx 9.70 \times 10^{-3}$ rad.
Diameter $= $ distance $\times$ angle $= 1.5 \times 10^{11} \times 9.70 \times 10^{-3} \approx 1.45 \times 10^{9}$ m.
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 Fall of 4.9 m from rest: $t = \sqrt{\frac{2 \times 4.9}{9.8}} = 1$ s, and $v = 9.8$ m/s.
It then sinks for $4 - 1 = 3$ s at 9.8 m/s: depth $= 29.4$ m.
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 With elongation x, the object moves in a circle of radius $l_0 + x$ and the spring force supplies the centripetal force:
$kx = m\omega^2(l_0 + x)$, so $x(k - m\omega^2) = m\omega^2 l_0$ and $x = \frac{m\omega^2 l_0}{k - m\omega^2}$.
Answer: (C). | ||||
| 35 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At the lowest point the velocity u is horizontal. With the string horizontal the stone has risen L, and its velocity v is vertical, with $v^2 = u^2 - 2gL$.
The two velocities are perpendicular, so $|\Delta\vec{v}| = \sqrt{u^2 + v^2} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$.
$x = 2$.
Answer: (B). | ||||
| 36 | GravitationGravitational potential energy | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation Sum over every pair:
• four sides, separation d: $-4\frac{Gm^2}{d}$;
• two diagonals, separation $\sqrt{2}d$: $-2\frac{Gm^2}{\sqrt{2}d} = -\sqrt{2}\frac{Gm^2}{d}$;
• M with each corner, separation $\frac{d}{\sqrt{2}}$: $-4\frac{\sqrt{2}GMm}{d}$.
Total $= -\frac{Gm}{d}\left[(4 + \sqrt{2})m + 4\sqrt{2}M\right]$.
Answer: (A). | ||||
| 37 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory For a fixed amount of gas at constant pressure, $V = \frac{nR}{P}T$: the V–T line has slope $\frac{nR}{P}$.
The $P_2$ line is steeper, so $P_2$ is the smaller pressure: $P_1 > P_2$.
Answer: (A). | ||||
| 38 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | StatementsEasy | |
A easy statements question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory A: false. Molecular motion does not stop at 0 °C (273 K); kinetic energy is proportional to absolute temperature.
B: true. $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$ falls as the number density n rises.
C: true. $n = \frac{P}{k_BT}$, so $\lambda = \frac{k_BT}{\sqrt{2}\pi d^2 P}$ grows with T at constant P.
D: false. The energy per degree of freedom is $\frac{1}{2}k_BT$; $\frac{3}{2}k_BT$ is the whole energy of a monoatomic molecule.
Answer: (B). | ||||
| 39 | Properties of Bulk MatterThermal expansion and calorimetry | 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 thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Heat needed per kg: $c\Delta T + L = 125 \times 200 + 2.5 \times 10^4 = 5 \times 10^4$ J.
$0.4 \times \frac{1}{2}v^2 = 5 \times 10^4$, so $v^2 = 2.5 \times 10^5$ and $v = 500$ m/s.
Answer: (B). | ||||
| 40 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v = \frac{dx}{dt} = \pi\cos\pi\left(t + \frac{1}{3}\right)$.
At $t = 1$: $\cos\frac{4\pi}{3} = -\frac{1}{2}$, so the speed is $\frac{\pi}{2} = 1.57$ m/s $= 157$ cm/s.
Answer: (B). | ||||
| 41 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The charge sits at the centre of the flat face, taken as just inside the closed surface, so the total flux out of the closed hemisphere is $\frac{q}{\epsilon_0}$ (Gauss's law).
The field of a point charge is radial and spreads equally in all directions. Seen from a point just above the centre of the flat face, the flat face fills one half of all directions (a solid angle of $2\pi$) and the curved surface the other half.
So half the flux leaves through the flat face: $\frac{q}{2\epsilon_0}$.
Answer: (B). | ||||
| 42 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Each ball is repelled by the other two with forces of equal size F (same charges, same separation 1 m), at $60^\circ$ to each other.
Resultant $= 2F\cos 30^\circ = \sqrt{3}F$.
Ratio $= \sqrt{3} : 1$.
Answer: (D). | ||||
| 43 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct 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 From the figure both currents flow along +x; $S_1$ (4 A) is above P at 4 cm and $S_2$ (2 A) is below P at 6 cm.
At P (below $S_1$): $B_1 = \frac{\mu_0 \cdot 4}{2\pi(0.04)} = 2 \times 10^{-5}$ T along $-\hat{k}$.
At P (above $S_2$): $B_2 = \frac{\mu_0 \cdot 2}{2\pi(0.06)} = \frac{2}{3} \times 10^{-5}$ T along $+\hat{k}$.
Net $\vec{B} = -\frac{4}{3} \times 10^{-5}\hat{k}$ T.
$\vec{F} = q\vec{v} \times \vec{B} = 3\pi\left(-\frac{4}{3} \times 10^{-5}\right)(2\hat{i} + 3\hat{j}) \times \hat{k} = -4\pi \times 10^{-5}(-2\hat{j} + 3\hat{i}) = 4\pi \times 10^{-5}(-3\hat{i} + 2\hat{j})$ N.
So $x = 3$.
Answer: (C). | ||||
| 44 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement RC and $\frac{L}{R}$ are the time constants of RC and LR circuits, and $\sqrt{LC} = \frac{1}{\omega}$ for an LC circuit: all times.
$\frac{L}{C} = \frac{L}{R} \cdot \frac{1}{RC} \cdot R^2$ has the dimensions of $R^2$ (resistance squared), not time.
Answer: (D). | ||||
| 45 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | StatementsEasy | |
A easy statements question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Statement I is the basis of electromagnetic waves (Maxwell's displacement current and Faraday's law): true.
Statement II: $\frac{1}{\sqrt{\mu_0\epsilon_0}}$ is the speed in vacuum; in a medium it is $\frac{1}{\sqrt{\mu\epsilon}}$: false.
Answer: (C). | ||||
| 46 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Cutting along the principal axis leaves each half with both curved faces unchanged, so the half $L_1$ keeps power P.
Cutting the other half perpendicular to the axis splits it into two plano-convex pieces $L_2$ and $L_3$, each with one curved face: $P = P_2 + P_3$ with equal faces, so each has $\frac{P}{2}$.
So (B), (C) and (D) are correct, and the incorrect one is 'Power of $L_1 = \frac{P}{2}$'.
Answer: (A). | ||||
| 47 | 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 Frequency is fixed by the source and does not change on refraction. In a denser medium the speed falls, and $\lambda = \frac{v}{f}$ falls with it.
Answer: (C). | ||||
| 48 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | StatementsEasy | |
A easy statements question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Statement I: a jump from a lower to a higher orbit absorbs energy, it does not emit radiation, and $E_1 - E_2$ would be negative: false.
Statement II: a jump from $E_2$ down to $E_1$ emits a photon of $hf = E_2 - E_1$, Bohr's frequency condition: true.
Answer: (D). | ||||
| 49 | 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 As a switch the transistor is either fully off (cut-off, no collector current: switch open) or fully on (saturation: switch closed). The active region is used for amplification.
Answer: (D). | ||||
| 50 | Communication SystemsNeed for modulation | No matching NCERT chapter | StatementsEasy | |
A easy statements question on need for modulation, from Communication Systems. NCERT: No matching NCERT chapter (a) True: an antenna needs a size of about $\frac{\lambda}{4}$, impractically large for long wavelengths.
(b) False: the power radiated by an antenna varies as $\left(\frac{l}{\lambda}\right)^2$, so a long-wavelength baseband signal radiates little power.
(c) True: sending every baseband signal directly would mix signals from different transmitters.
(d) True: this is modulation, superimposing the signal on a high-frequency carrier.
Answer: (C). | ||||
| 51 | Laws of MotionEquilibrium of concurrent forces | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on equilibrium of concurrent forces, from Laws of Motion. NCERT: Class 11: Laws of Motion Take the rope as massless. At the midpoint, the upper half balances the 30 N horizontal force and the 100 N weight carried by the lower half.
$T\sin\theta = 30$ and $T\cos\theta = 100$, so $\tan\theta = 0.3 = 3 \times 10^{-1}$.
$x = 3$.
Answer: 3. | ||||
| 52 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalHard | |
A hard numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The wheel rolls without slipping from P down PQ, through a height h. The string stays taut, so the 3 kg block moves up PR through the same length; both inclines make angle α with the ground, so the block rises h. Its speed equals the speed v of the wheel's centre.
Energy: $12gh - 3gh = \frac{1}{2}(12)v^2 + \frac{1}{2}I\omega^2 + \frac{1}{2}(3)v^2$.
Taking the wheel as a ring ($I = mR^2$, so $\frac{1}{2}I\omega^2 = \frac{1}{2}(12)v^2$):
$9gh = 6v^2 + 6v^2 + 1.5v^2 = 13.5v^2$, so $v^2 = \frac{2}{3}gh$ and $v = \frac{1}{2}\sqrt{\frac{8}{3}gh}$.
So $x = \frac{8}{3} \approx 2.67$, which rounds to 3. (A uniform disc, $I = \frac{1}{2}mR^2$, gives $10.5v^2 = 9gh$, $x = \frac{24}{7} \approx 3.43$: also 3.)
Answer: 3. | ||||
| 53 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Isobaric: $W = nR\Delta T = 400$ J and $Q = nC_p\Delta T = \frac{\gamma}{\gamma - 1}nR\Delta T = \frac{1.4}{0.4} \times 400 = 1400$ J.
Answer: 1400. | ||||
| 54 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves From the extreme position, $x = A\cos\omega t$. $x = \frac{A}{2}$ when $\omega t = \frac{\pi}{3}$, i.e. $t = \frac{T}{6} = \frac{6}{6} = 1$ s.
Answer: 1. | ||||
| 55 | 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 The upper conductor has three steps, each b below the last; the lower conductor three steps, each b above the last; the innermost facing steps are b apart.
So the facing pairs of plates are b apart (inner), $b + 2b = 3b$ (middle) and $3b + 2b = 5b$ (outer).
These three capacitors share the same two conductors, so they are in parallel:
$C = \epsilon_0 A\left(\frac{1}{b} + \frac{1}{3b} + \frac{1}{5b}\right) = \frac{23}{15}\frac{\epsilon_0 A}{b}$.
$x = 23$.
Answer: 23. | ||||
| 56 | Current ElectricityDrift velocity and mobility | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity $I = J \cdot \pi\left(r^2 - \frac{r^2}{4}\right) = 10^6 \times \pi \times \frac{3}{4} \times 16 \times 10^{-6} = 12\pi$ A.
$x = 12$.
Answer: 12. | ||||
| 57 | 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 Three resistors ma in parallel: $\frac{ma}{3}$. Two resistors $\frac{a}{m}$ in parallel: $\frac{a}{2m}$. These are in series:
$R = \frac{ma}{3} + \frac{a}{2m}$.
$\frac{dR}{dm} = \frac{a}{3} - \frac{a}{2m^2} = 0$ gives $m^2 = \frac{3}{2}$ (a minimum, since $R \to \infty$ as $m \to 0$ or $\infty$).
$m = \sqrt{\frac{3}{2}}$, so $x = 3$.
Answer: 3. | ||||
| 58 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$. Same K, same charge, $m_d = 2m_p$:
$\frac{r_d}{r_p} = \sqrt{\frac{m_d}{m_p}} = \sqrt{2}$.
$x = 2$.
Answer: 2. | ||||
| 59 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The rod lies north–south and moves east. The horizontal (northward) component is along the rod, so $\vec{v} \times \vec{B}_H$ is vertical, across the rod: no emf from it.
The vertical component $B_V = B_H\tan 45^\circ = 4 \times 10^{-3}$ T gives $\vec{v} \times \vec{B}_V$ along the rod:
$\varepsilon = B_V v l = 4 \times 10^{-3} \times 20 \times 0.2 = 1.6 \times 10^{-2}$ V $= 16$ mV.
Answer: 16. | ||||
| 60 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $D_1$ and $D_2$ point in opposite directions, so whichever way the 1 V cell drives current, exactly one branch (a 60 Ω resistor and a forward-biased diode) conducts and the other is reverse biased.
$I = \frac{1 - 0.6}{60 + 40} = \frac{0.4}{100}$ A $= 4$ mA.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry For an ideal gas $pV = \frac{w}{M}RT$, so $p = \frac{dRT}{M}$: at a fixed temperature p is proportional to d, a straight line through the origin with slope $\frac{RT}{M}$.
The higher the temperature, the steeper the line, so $T_3$ (the highest) has the steepest line, then $T_2$, then $T_1$. That is plot (B).
Answer: (B). | ||||
| 62 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure P in $PCl_5$ is $sp^3d$ hybridised and the molecule is trigonal bipyramidal, with the three equatorial bonds in one plane: (A), (B), (D) are correct.
The two axial bonds suffer more repulsion from the bond pairs (three at $90^\circ$) and are longer and weaker than the equatorial bonds. (C) is incorrect.
Answer: (C). | ||||
| 63 | Isolation of MetalsOres and concentration | No matching NCERT chapter | StatementsEasy | |
A easy statements question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Statement I: leaching needs air (oxygen) as the oxidant: $4Au + 8CN^- + 2H_2O + O_2 \to 4[Au(CN)_2]^- + 4OH^-$, and the complex is of Au(I). Incorrect.
Statement II: $2[Au(CN)_2]^- + Zn \to [Zn(CN)_4]^{2-} + 2Au$: zinc goes from 0 to +2, so it is oxidised. Correct.
Answer: (D). | ||||
| 64 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $CH_4$ forms no hydrogen bonds (C–H is nearly non-polar and C has no lone pair).
HCN forms only weak C–H···N hydrogen bonds (the H on the sp carbon is slightly acidic).
$NH_3$ forms N–H···N hydrogen bonds, stronger than those of HCN (and weaker than those of water, which rules out (A)).
Order: $CH_4 < HCN < NH_3$.
Answer: (C). | ||||
| 65 | 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 All five ions have 10 electrons. In an isoelectronic series the radius falls as the nuclear charge rises: $N^{3-}$ (Z = 7) $> O^{2-}$ (8) $> F^-$ (9) $> Na^+$ (11) $> Mg^{2+}$ (12).
Increasing order: $Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$.
Answer: (A). | ||||
| 66 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $NH_4Cl(aq) + NaNO_2(aq) \to N_2(g) + 2H_2O(l) + NaCl(aq)$ (via ammonium nitrite, which decomposes on warming): the laboratory preparation of nitrogen.
Answer: (B). | ||||
| 67 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Fluorine forms only one oxoacid, HOF (hypofluorous acid): A is correct.
Fluorine is the smallest and most electronegative halogen (and has no d orbitals for higher oxidation states): R is correct.
Its small size and high electronegativity are exactly why it forms only this one oxoacid, so R explains A.
Answer: (A). | ||||
| 68 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $E^\circ(M^{2+}/M)$: Cr −0.91 V, Fe −0.44 V, Cu +0.34 V, Zn −0.76 V.
Copper alone has a positive value (its high enthalpy of atomisation and ionisation are not compensated by hydration), so it is the highest.
Answer: (C). | ||||
| 69 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Lanthanide ions $Ln^{n+}$ have the configuration $[Xe]4f^{Z - 54 - n}$.
$Eu^{2+}$: $4f^7$; $Tm^{2+}$: $4f^{13}$; $Sm^{2+}$: $4f^6$; $Tm^{3+}$: $4f^{12}$; $Tb^{4+}$: $4f^7$; $Yb^{2+}$: $4f^{14}$; $Dy^{3+}$: $4f^9$; $Yb^{3+}$: $4f^{13}$.
Half-filled ($4f^7$) and completely filled ($4f^{14}$) together: $Tb^{4+}$ and $Yb^{2+}$.
Answer: (C). | ||||
| 70 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A: $Fe^{3+}$, $d^5$, weak-field $F^-$, high spin: 5 unpaired electrons.
B: $Fe^{3+}$, $d^5$, strong-field $CN^-$, low spin ($t_{2g}^5$): 1 unpaired.
C: $Mn^{3+}$, $d^4$, high spin: 4 unpaired.
D: $Mn^{3+}$, $d^4$, strong-field $CN^-$, low spin ($t_{2g}^4$): 2 unpaired.
$\mu = \sqrt{n(n + 2)}$ BM rises with n: B (1) < D (2) < C (4) < A (5).
Answer: (B). | ||||
| 71 | Environmental ChemistryOzone depletion | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ozone depletion, from Environmental Chemistry. NCERT: No matching NCERT chapter $ClONO_2 + H_2O \to HOCl + HNO_3$, so A = HOCl and B = $HNO_3$.
$ClONO_2 + HCl \to Cl_2 + HNO_3$, so C = $Cl_2$ (B again $HNO_3$).
Answer: (A). | ||||
| 72 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Apply Hückel's rule to each cyclic, planar, fully conjugated species:
(A) cyclopropenyl cation: 2 π electrons ($4n + 2$, $n = 0$): aromatic, strongly stabilised.
(B) cyclopropenyl anion: 4 π electrons: antiaromatic, very unstable.
(C) cyclopentadienyl cation: 4 π electrons: antiaromatic.
(D) cyclohexa-1,3-diene: the $CH_2$ groups break the cyclic conjugation: non-aromatic.
The aromatic cyclopropenyl cation is the most stable.
Answer: (A). | ||||
| 73 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (i) n-BuLi removes the acidic terminal alkyne hydrogen of hex-1-yne ($n\text{-Bu}-C \equiv CH$); the acetylide is alkylated by $n\text{-}C_5H_{11}Cl$, giving $n\text{-Bu}-C \equiv C-C_5H_{11}$ (undec-5-yne).
(ii) Lindlar's catalyst adds $H_2$ syn and stops at the alkene: the cis (Z) alkene, with n-Bu and $C_5H_{11}$ on the same side.
That is structure (C). (A) and (B) are trisubstituted alkenes, not formed here; (D) is the trans alkene (from Na/liquid $NH_3$).
Answer: (C). | ||||
| 74 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (a) $Br_2$/Fe brominates the ring; the ethyl group is ortho/para directing and the para product is major: 1-bromo-4-ethylbenzene.
(b) $Cl_2$ with heat substitutes at the benzylic carbon (free-radical chlorination): 1-bromo-4-(1-chloroethyl)benzene.
(c) Alcoholic KOH eliminates HCl (dehydrohalogenation; the aryl C–Br is untouched): 4-bromostyrene, $Br-C_6H_4-CH=CH_2$.
Answer: (D). | ||||
| 75 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Matching listEasy | |
A easy matching list question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers A: phenol → salicylaldehyde (ortho-CHO): Reimer–Tiemann reaction, $CHCl_3$/NaOH (IV).
B: phenol → benzene: distillation with zinc dust (III).
C: phenol → benzoquinone: oxidation with $Na_2Cr_2O_7/H_2SO_4$ (II).
D: phenol → p-bromophenol: $Br_2$ in a non-polar solvent such as $CS_2$ (monobromination) (I).
A-IV, B-III, C-II, D-I.
Answer: (A). | ||||
| 76 | 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 Number the ring with COOH at C-1. The six acids have the two $NH_2$ groups at 2,3; 2,4; 2,5; 2,6; 3,4; 3,5. Removing the COOH leaves a diaminobenzene:
• 2,3 and 3,4 → 1,2-diaminobenzene (o-phenylenediamine): two acids, product B.
• 2,4, 2,6 and 3,5 → 1,3-diaminobenzene (m-phenylenediamine): three acids, product A.
• 2,5 → 1,4-diaminobenzene (p-phenylenediamine): one acid, product C.
Of the three isomers the para one, being the most symmetrical, packs best and melts highest: about 142 °C (the meta isomer about 63 °C, the ortho about 104 °C).
Answer: (D). | ||||
| 77 | 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 Buna-N (nitrile rubber) is the copolymer of 1,3-butadiene and acrylonitrile; 'N' stands for the nitrile group. (B) describes Buna-S; (A) describes polybutadiene; it is synthetic, not natural.
Answer: (C). | ||||
| 78 | BiomoleculesCarbohydrates | Class 12: Biomolecules | StatementsEasy | |
A easy statements question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Maltose is two α-D-glucose units joined by a C1–C4 glycosidic link; the second unit's C1 is free (a hemiacetal), so maltose is a reducing sugar. Statement I is true and Statement II is false.
Answer: (C). | ||||
| 79 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Antipyretics reduce fever (III); analgesics reduce pain (I); tranquilizers reduce stress and anxiety (II); antacids reduce excess stomach acidity (IV).
Answer: (A). | ||||
| 80 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter $CO_3^{2-}$ gives $CO_2$: brisk effervescence, lime water milky (IV).
$S^{2-}$ gives $H_2S$: lead acetate paper black (PbS) (I).
$SO_3^{2-}$ gives $SO_2$: acidified dichromate turns green ($Cr^{3+}$) (II).
$NO_2^-$ gives brown $NO_2$ fumes, which liberate iodine from KI: starch blue (III).
A-IV, B-I, C-II, D-III.
Answer: (D). | ||||
| 81 | Some Basic Concepts of ChemistryEmpirical and molecular formula | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles: H $\frac{7.5}{1} = 7.5$; O $\frac{60}{16} = 3.75$; C $\frac{48.5}{12} \approx 4.04$.
Dividing by 3.75: C : H : O $\approx 1.08 : 2 : 1$, i.e. the empirical formula $CH_2O$.
$CH_3COOH$ ($C_2H_4O_2$) and HCHO ($CH_2O$) have this empirical formula; $CH_3OOCH_3$ ($C_2H_6O_2$, 1 : 3 : 1) and $CH_3CHO$ ($C_2H_4O$, 2 : 4 : 1) do not.
So 2 formulae agree.
Answer: 2. | ||||
| 82 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom l can be 0 to $n - 1$, and $m_l$ from $-l$ to $+l$.
A: $l = 3$ is not allowed for $n = 3$. Wrong.
B: $n = 3$, $l = 2$, $m_l = -2$: allowed.
C: $n = 2$, $l = 1$, $m_l = +1$: allowed.
D: $l = 2$ is not allowed for $n = 2$. Wrong.
Two sets are correct.
Answer: 2. | ||||
| 83 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter $BeO + 2NH_3 + 4HF \to (NH_4)_2[BeF_4] + H_2O$, so A is ammonium tetrafluoroberyllate; on heating, $(NH_4)_2[BeF_4] \to BeF_2 + 2NH_4F$ (B is $BeF_2$).
In $[BeF_4]^{2-}$: $x + 4(-1) = -2$, so $x = +2$.
Answer: 2. | ||||
| 84 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $|W| = 2.303\,nRT\log\frac{V_2}{V_1} = 2.303 \times 5 \times 8.3 \times 300 \times 0.3010$.
$5 \times 8.3 \times 300 = 12450$; $12450 \times 2.303 = 28672.35$; $\times 0.3010 = 8630.4$ J.
To the nearest integer, 8630 J.
Answer: 8630. | ||||
| 85 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Delta T_b = 373.535 - 373.15 = 0.385$ K.
$M = \frac{K_b \times w \times 1000}{\Delta T_b \times W} = \frac{0.52 \times 2.5 \times 1000}{0.385 \times 75} = \frac{1300}{28.875} = 45.02$ g mol⁻¹.
Answer: 45. | ||||
| 86 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium NaOH is fully dissociated: $[OH^-] = 10^{-3}$ M, $pOH = 3$, $pH = 14 - 3 = 11$ (at 25 °C).
Answer: 11. | ||||
| 87 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $\frac{1}{2}H_2(g) + Ag^+(aq) \to H^+(aq) + Ag(s)$, with $n = 1$.
$\Delta G^\circ = -nFE^\circ_{cell} = -1 \times 96500 \times 0.5332 = -51\,454$ J mol⁻¹ $\approx -51$ kJ mol⁻¹.
Answer: −51. | ||||
| 88 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ with $T_1 = 300$ K, $T_2 = 309$ K.
$\ln 2 = 2.3 \times 0.30 = 0.69$; $\frac{1}{300} - \frac{1}{309} = \frac{9}{92\,700}$.
$E_a = \frac{0.69 \times 8.3 \times 92\,700}{9} = 0.69 \times 8.3 \times 10\,300 = 58\,988$ J mol⁻¹ $\approx 59$ kJ mol⁻¹.
Answer: 59. | ||||
| 89 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich isotherm: $\log\frac{x}{m} = \log k + \frac{1}{n}\log p$.
From the graph, slope $\frac{1}{n} = 1$ and intercept $\log k = 0.602$, so $k = 4$.
$\frac{x}{m} = kp = 4 \times 0.03 = 0.12$ g $= 12 \times 10^{-2}$ g.
Answer: 12. | ||||
| 90 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Molar mass of AgCl $= 108 + 35.5 = 143.5$.
% Cl $= \frac{35.5}{143.5} \times \frac{0.40}{0.25} \times 100 = 0.2474 \times 160 = 39.58\%$, which rounds to 40.
Answer: 40. | ||||
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