JEE (Main) · 2022
JEE (Main) 2022 — 29 July, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Sequences and Series3
- Integral Calculus3
- Sets, Relations and Functions2
9 easy · 19 medium · 2 hard
Physics
30 Q
- Kinematics4
- Properties of Bulk Matter2
- Electrostatics2
27 easy · 3 medium · 0 hard
Chemistry
30 Q
- d- and f-Block Elements3
- Amines3
- Chemical Bonding and Molecular Structure2
18 easy · 12 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 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions The condition is only on b: b must be a product of two primes $p, q \geq 3$ (not necessarily distinct) with $b \leq 60$.
With 3: $9, 15, 21, 33, 39, 51, 57$; with 5 (and $q \geq 5$): $25, 35, 55$; with 7: $49$.
So b takes 11 values, while a can be any of the 60 elements.
$n(R) = 60 \times 11 = 660$.
Answer: (B). | ||||
| 2 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z^5 + \overline{z}^5 = 2\,\mathrm{Re}(z^5)$.
$z^2 = 4 - 9 + 12i = -5 + 12i$, $z^4 = 25 - 144 - 120i = -119 - 120i$.
$z^5 = (-119 - 120i)(2 + 3i) = -238 - 357i - 240i + 360 = 122 - 597i$.
So $z^5 + \overline{z}^5 = 2 \times 122 = 244$.
Answer: (A). | ||||
| 3 | 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 If A were invertible, $AB = 0$ would give $B = A^{-1}0 = 0$, contrary to $B \neq 0$; so $|A| = 0$. Likewise if B were invertible, $A = 0$; so $|B| = 0$ and (C) is false.
Since $|A| = 0$, the homogeneous system $AX = 0$ has non-trivial solutions, hence infinitely many: (B) true, (A) false.
$|\mathrm{adj}\,A| = |A|^2 = 0$, so (D) is false.
Answer: (B). | ||||
| 4 | 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 The numbers $20 - a, 40 - a, \ldots, 200 - a$ are in A.P. with common difference 20, so each term is $\frac{1}{20}\left(\frac{1}{20k - a} - \frac{1}{20k + 20 - a}\right)$.
The sum telescopes: $\frac{1}{20}\left(\frac{1}{20 - a} - \frac{1}{200 - a}\right) = \frac{1}{20}\cdot\frac{180}{(20-a)(200-a)} = \frac{9}{(20-a)(200-a)}$.
So $(20 - a)(200 - a) = 9 \times 256 = 2304$, i.e. $a^2 - 220a + 1696 = 0$.
$a = \frac{220 \pm \sqrt{48400 - 6784}}{2} = \frac{220 \pm 204}{2} = 212$ or $8$.
Maximum value: 212.
Answer: (C). | ||||
| 5 | 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 The denominator behaves as $x^3$. Expanding the numerator:
$(\alpha + \beta) + (\alpha - \beta + \gamma)x + \frac{\alpha + \beta}{2}x^2 + \frac{\alpha - \beta - \gamma}{6}x^3 + \ldots$
For a finite limit: $\alpha + \beta = 0$ and $\alpha - \beta + \gamma = 0$; then the limit is $\frac{\alpha - \beta - \gamma}{6} = \frac{2}{3}$, so $\alpha - \beta - \gamma = 4$.
With $\beta = -\alpha$: $2\alpha + \gamma = 0$, $2\alpha - \gamma = 4$, giving $\alpha = 1$, $\beta = -1$, $\gamma = -2$.
(A) $1 + 1 + 4 = 6$ ✓. (B) $-1 + 2 - 2 + 1 = 0$ ✓. (C) $\alpha\beta^2 + \beta\gamma^2 + \gamma\alpha^2 + 3 = 1 - 4 - 2 + 3 = -2 \neq 0$ ✗. (D) $1 - 1 + 4 = 4$ ✓.
Answer: (C). | ||||
| 6 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t = \tan\frac{x}{2}$: $\sin x = \frac{2t}{1+t^2}$, $\cos x = \frac{1-t^2}{1+t^2}$, $dx = \frac{2\,dt}{1+t^2}$; the limits become 0 and 1.
The denominator becomes $\frac{3(1+t^2) + 4t + 1 - t^2}{1+t^2} = \frac{2t^2 + 4t + 4}{1+t^2}$.
Integral $= \int_0^1 \frac{2\,dt}{2t^2 + 4t + 4} = \int_0^1 \frac{dt}{(t+1)^2 + 1} = \left[\tan^{-1}(t+1)\right]_0^1 = \tan^{-1}2 - \frac{\pi}{4}$.
Answer: (B). | ||||
| 7 | 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 $\frac{dy}{dx} + y = \frac{1}{1 + e^{2x}}$; the integrating factor is $e^x$:
$\frac{d}{dx}(e^x y) = \frac{e^x}{1 + e^{2x}}$, so $e^x y = \tan^{-1}(e^x) + C$.
At $x = 0$: $\frac{\pi}{2} = \frac{\pi}{4} + C$, so $C = \frac{\pi}{4}$.
As $x \to \infty$, $\tan^{-1}(e^x) \to \frac{\pi}{2}$, so $e^x y \to \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.
Answer: (B). | ||||
| 8 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections A line through the intersection: $bx + 10y - 8 + \lambda(2x - 3y) = 0$. Through $(1, 1)$: $b + 2 - \lambda = 0$, so $\lambda = b + 2$.
L: $(3b + 4)x + (4 - 3b)y - 8 = 0$.
It touches $x^2 + y^2 = \frac{16}{17}$: $\frac{8}{\sqrt{(3b+4)^2 + (4-3b)^2}} = \frac{4}{\sqrt{17}}$, so $18b^2 + 32 = 68$ and $b^2 = 2$.
The ellipse $\frac{x^2}{5} + \frac{y^2}{2} = 1$ has $e = \sqrt{1 - \frac{2}{5}} = \sqrt{\frac{3}{5}}$.
Answer: (B). | ||||
| 9 | 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 minus the foot is $\left(1, \frac{1}{2}, \frac{3}{2}\right) \parallel (2, 1, 3)$, the normal; so $m = 1$, $n = 3$ (check: $-4 + \frac{7}{2} + \frac{9}{2} = 4$ ✓). P: $2x + y + 3z = 4$.
The line from A with direction $(3, -1, -4)$: $(-1 + 3t, 4 - t, 3 - 4t)$.
On P: $2(-1 + 3t) + (4 - t) + 3(3 - 4t) = 4 \Rightarrow 11 - 7t = 4 \Rightarrow t = 1$, the point $(2, 3, -1)$.
Distance $= \sqrt{9 + 1 + 16} = \sqrt{26}$.
Answer: (B). | ||||
| 10 | 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 $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c} = \vec{b} + \lambda\vec{c}$.
Since $\vec{b}$ and $\vec{c}$ are non-parallel, compare coefficients: $\vec{a}\cdot\vec{c} = 1$ and $\lambda = -\vec{a}\cdot\vec{b} = -(3 + 2 + 0) = -5$.
Answer: (A). | ||||
| 11 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let O be the foot of the tower and h its height. OA points north and AB west, so $\angle OAB = 90^\circ$ and $OA = \sqrt{15^2 - 9^2} = 12$.
At B: $\cos\theta = \frac{3}{\sqrt{13}}$ gives $\tan\theta = \frac{2}{3}$, so $h = 15 \times \frac{2}{3} = 10$.
$\cot\alpha = \frac{OA}{h} = \frac{12}{10} = \frac{6}{5}$.
Answer: (A). | ||||
| 12 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $(p \wedge q) \Rightarrow (p \wedge r) \equiv \sim p \vee \sim q \vee (p \wedge r)$.
Since $\sim p \vee (p \wedge r) \equiv \sim p \vee r$, this is $\sim p \vee \sim q \vee r \equiv \sim(p \wedge q) \vee r \equiv (p \wedge q) \Rightarrow r$.
(Check of the others: with p false, (A) fails for q true, r any, while the given statement is true.)
Answer: (D). | ||||
| 13 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines A and C lie on $x = a$, so the perpendicular bisector of AC is $y = \frac{3 + b}{2}$; it passes through P: $b = -1$. Then $ab > 0$ needs $a < 0$.
$PA = PB$: $(a - 1)^2 + 4 = 4 + 16$, so $a - 1 = \pm 4$ and $a = -3$.
A$(-3, 3)$, B$(-1, 5)$, C$(-3, -1)$.
AP: slope $\frac{1 - 3}{1 + 3} = -\frac{1}{2}$, line $x + 2y = 3$. BC: slope $\frac{5 + 1}{-1 + 3} = 3$, line $y = 3x + 8$.
Solving: $x + 6x + 16 = 3 \Rightarrow x = -\frac{13}{7}$, $y = \frac{17}{7}$.
$k_1 + k_2 = \frac{4}{7}$.
Answer: (B). | ||||
| 14 | 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 $\hat{a}\cdot\hat{b} = \frac{1}{\sqrt{2}}$, $|\hat{a} \times \hat{b}| = \frac{1}{\sqrt{2}}$, and $\hat{a} \times \hat{b}$ is perpendicular to both.
Let $\vec{u} = \hat{a} + \hat{b}$, $\vec{v} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})$.
$\vec{u}\cdot\vec{v} = 1 + 2\cdot\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 2 = 3 + \frac{3}{\sqrt{2}}$.
$|\vec{u}|^2 = 2 + \sqrt{2}$; $|\vec{v}|^2 = 1 + 4 + 4\cdot\frac{1}{\sqrt{2}} + 4\cdot\frac{1}{2} = 7 + 2\sqrt{2}$.
$\cos^2\theta = \frac{\left(3 + \frac{3}{\sqrt{2}}\right)^2}{(2 + \sqrt{2})(7 + 2\sqrt{2})} = \frac{\frac{27}{2} + 9\sqrt{2}}{18 + 11\sqrt{2}} = \frac{(27 + 18\sqrt{2})(18 - 11\sqrt{2})}{2(324 - 242)} = \frac{90 + 27\sqrt{2}}{164}$.
$164\cos^2\theta = 90 + 27\sqrt{2}$.
Answer: (A). | ||||
| 15 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals In $f\left(\frac{1}{\alpha}\right) = \int_1^{1/\alpha} \frac{\log_{10} t}{1 + t}dt$ put $t = \frac{1}{u}$: $f\left(\frac{1}{\alpha}\right) = \int_1^{\alpha} \frac{-\log_{10} u}{1 + \frac{1}{u}}\cdot\left(-\frac{du}{u^2}\right) = \int_1^{\alpha} \frac{\log_{10} u}{u(1 + u)}du$.
Adding: $f(\alpha) + f\left(\frac{1}{\alpha}\right) = \int_1^{\alpha} \log_{10} t\left(\frac{1}{1+t} + \frac{1}{t(1+t)}\right)dt = \int_1^{\alpha} \frac{\ln t}{t\ln 10}dt = \frac{(\ln\alpha)^2}{2\ln 10}$.
With $\alpha = e^3$: $\frac{9}{2\log_e 10}$.
Answer: (D). | ||||
| 16 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $y = x - 1$ meets $x^2 + y^2 = 5$ (with $y \geq 0$) at $(2, 1)$; $y = 1 - x$ meets it at $(-1, 2)$.
Area $= \int_{-1}^{2}\sqrt{5 - x^2}\,dx - \int_{-1}^{2}|x - 1|\,dx$.
Second integral: $\int_{-1}^{1}(1 - x)dx + \int_1^2 (x - 1)dx = 2 + \frac{1}{2} = \frac{5}{2}$.
First: $\left[\frac{x}{2}\sqrt{5 - x^2} + \frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}\right]_{-1}^{2} = (1 + 1) + \frac{5}{2}\left(\sin^{-1}\frac{2}{\sqrt{5}} + \sin^{-1}\frac{1}{\sqrt{5}}\right) = 2 + \frac{5}{2}\cdot\frac{\pi}{2}$ (the two angles are complementary).
Area $= 2 + \frac{5\pi}{4} - \frac{5}{2} = \frac{5\pi}{4} - \frac{1}{2}$.
Answer: (D). | ||||
| 17 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections L passes through the focus $(1, 0)$ of P, so $c = -m$.
Tangency to $\frac{x^2}{4} - \frac{y^2}{4} = 1$: $c^2 = 4m^2 - 4$, so $m^2 = 4m^2 - 4$ and $m^2 = \frac{4}{3}$.
H has $e = \sqrt{2}$, so $F = (2\sqrt{2}, 0)$.
For M, N: $x = \frac{y}{m} + 1$ in $y^2 = 4x$ gives $y^2 - \frac{4}{m}y - 4 = 0$, so $|y_M - y_N| = \sqrt{\frac{16}{m^2} + 16} = 4\sqrt{\frac{3}{4} + 1} = 2\sqrt{7}$.
O and F are on the x-axis with M and N on opposite sides, so the area is $\frac{1}{2}\cdot OF\cdot|y_M - y_N| = \frac{1}{2}\cdot 2\sqrt{2}\cdot 2\sqrt{7} = 2\sqrt{14}$.
Answer: (B). | ||||
| 18 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $|x - 1|\sin|x - 1| = (x - 1)\sin(x - 1)$ for all x (both factors change sign together), and $\cos|x - 2| = \cos(x - 2)$; so the first term is differentiable everywhere.
Second term: $(x - 3)|(x - 1)(x - 4)|$. A factor $|x - a|$ with a simple zero makes the product non-differentiable at a unless the remaining factor vanishes there.
At $x = 1$: the rest is $(x - 3)|x - 4| = -6 \neq 0$; at $x = 4$: $(x - 3)|x - 1| = 3 \neq 0$. Both points are non-differentiable.
Answer: (B). | ||||
| 19 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability $2022 = 2 \times 3 \times 337$ (337 is prime).
The numbers in S coprime to 2022 number $\phi(2022) = 2022\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{337}\right) = 1 \times 2 \times 336 = 672$.
Probability $= \frac{672}{2022} = \frac{112}{337}$.
Answer: (D). | ||||
| 20 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctHard | |
A hard single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $g(x) = (x^2 - 2)^3 + 4$, so $f = 3^g$, $f' = f\ln 3\,g'$, $f'' = f\ln 3\left(\ln 3\,g'^2 + g''\right)$.
$g' = 6x(x^2 - 2)^2$, $g'' = 6(x^2 - 2)(5x^2 - 2)$.
P: $f'$ has the sign of $g' = 6x(x^2 - 2)^2$, which goes from − to + at $x = 0$: local minimum. True.
Q: at $x = \sqrt{2}$, $g' = g'' = 0$ so $f'' = 0$. Near $\sqrt{2}$, $g'^2$ is of order $(x - \sqrt{2})^4$ while $g''$ is of order $(x - \sqrt{2})$ and changes sign (negative just left, positive just right), so $f''$ changes sign: a point of inflection. True.
R: for $x > \sqrt{2}$, $g'' > 0$ and $g'^2 \geq 0$, so $f'' > 0$ and $f'$ is increasing. True.
Answer: (D). | ||||
| 21 | 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 Let $c = \cos 2\theta$: $\cos^2\theta = \frac{1 + c}{2}$, $\sin^2\theta = \frac{1 - c}{2}$.
$\frac{7(1 + c) - 3(1 - c)}{2} - 2c^2 = 2 \Rightarrow 2 + 5c - 2c^2 = 2 \Rightarrow c(5 - 2c) = 0$, so $\cos 2\theta = 0$.
In $(0, 2\pi)$: $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ — four equations.
For each, $\tan^2\theta = \cot^2\theta = 1$, so the sum of roots is $2(1 + 1) = 4$.
Total: $4 \times 4 = 16$.
Answer: 16. | ||||
| 22 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $\overline{x^2} = \sigma^2 + \bar{x}^2 = 9 + 225 = 234$.
Mean of $(x_i + \alpha)^2$ $= \overline{x^2} + 2\alpha\bar{x} + \alpha^2 = 234 + 30\alpha + \alpha^2 = 178$.
$\alpha^2 + 30\alpha + 56 = 0 \Rightarrow \alpha = -2$ or $-28$.
Maximum $\alpha = -2$; its square is 4.
Answer: 4. | ||||
| 23 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $3a + 4a - 14b = 0 \Rightarrow a = 2b$.
$ab - 4a^2 + 14 = 0 \Rightarrow 2b^2 - 16b^2 + 14 = 0 \Rightarrow b^2 = 1$, $a^2 = 4$.
The line: $\frac{x + 1}{5} = \frac{y - 2}{3} = \frac{z}{1} = t$, i.e. $(5t - 1, 3t + 2, t)$.
In the plane: $5t - 1 - 3t - 2 + t = 0 \Rightarrow t = 1$, the point $(4, 5, 1)$.
$\alpha + \beta + \gamma = 10$.
Answer: 10. | ||||
| 24 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series With $a_r = a + (r - 1)d$: $\sum \frac{a}{2^r} = a$ and $\sum_{r \geq 1} \frac{r - 1}{2^r} = \sum \frac{r}{2^r} - \sum \frac{1}{2^r} = 2 - 1 = 1$.
So the sum is $a + d = a_2 = 4$, and $4a_2 = 16$.
Answer: 16. | ||||
| 25 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_5 = {}^nC_4\, 2^{\frac{n-4}{4}}\, 3^{-1}$; the fifth from the end is ${}^nC_4\, 2^{1}\, 3^{-\frac{n-4}{4}}$.
Ratio: $2^{\frac{n-4}{4} - 1}\cdot 3^{\frac{n-4}{4} - 1} = 6^{\frac{n-8}{4}} = 6^{\frac{1}{4}}$, so $n = 9$.
$T_6 = {}^9C_5\,(2^{1/4})^4\,(3^{-1/4})^5 = 126 \times 2 \times \frac{1}{3\cdot 3^{1/4}} = \frac{84}{\sqrt[4]{3}}$.
$\alpha = 84$.
Answer: 84. | ||||
| 26 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The sum is the number of 1s among the 9 entries; it must be 2, 3, 5 or 7.
Count: ${}^9C_2 + {}^9C_3 + {}^9C_5 + {}^9C_7 = 36 + 84 + 126 + 36 = 282$.
Answer: 282. | ||||
| 27 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Take $p!$, $(p+1)!$, $(p+2)!$ out of rows 1, 2, 3:
$\Delta = p!(p+1)!(p+2)!\begin{vmatrix} 1 & p+1 & (p+1)(p+2) \\ 1 & p+2 & (p+2)(p+3) \\ 1 & p+3 & (p+3)(p+4) \end{vmatrix}$.
$R_2 - R_1$ and $R_3 - R_2$ give rows $(0, 1, 2(p+2))$ and $(0, 1, 2(p+3))$, so the determinant is $2(p+3) - 2(p+2) = 2$.
$\Delta = 2\,p!\,(p+1)!\,(p+2)!$.
Since p and $p + 2$ are both prime, $p \geq 3$ is odd and $2p > p + 2$: each of $p!, (p+1)!, (p+2)!$ contains the prime p exactly once, so $\alpha = 3$. The prime $p + 2$ appears only in $(p+2)!$, once, so $\beta = 1$.
$\alpha + \beta = 4$.
Answer: 4. | ||||
| 28 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right]$.
Sum over $n = 2$ to 100: $\frac{1}{2}\left[\frac{1}{6} - \frac{1}{101 \times 102}\right] = \frac{1}{2}\cdot\frac{1717 - 1}{101 \times 102} = \frac{1716}{2 \times 101 \times 102}$.
So $k = \frac{1716}{204}$ and $34k = \frac{1716}{6} = 286$.
Answer: 286. | ||||
| 29 | Sets, Relations and FunctionsSets and their representation | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on sets and their representation, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions A is the set of numbers $4x + 6y + 9z$ with non-negative integers $x, y, z$, not all zero.
Even numbers: $4x + 6y$ gives every even number $\geq 4$, so every even element of T (10 to 1000) is in A.
Odd numbers need z odd, so they are $9 + (\text{0 or an even number} \geq 4) + \ldots$: 9, 13, 15, 17, … — every odd number $\geq 13$ and 9 itself. 11 is not reachable ($11 - 9 = 2$ is not a sum of 4s and 6s).
So $T - A = \{11\}$ and the sum is 11.
Answer: 11. | ||||
| 30 | CirclesStandard and general form | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections $c_1$ has centre $(1, 3)$ and radius$^2$ $10 - \alpha$.
The image of $(h, k)$ in $y = x + 1$ is $(k - 1, h + 1)$, so the centre of $c_2$ is $(2, 2)$; hence $g = f = -2$.
$c_2$: $x^2 + y^2 - 4x - 4y + \frac{38}{5} = 0$, so $r^2 = 4 + 4 - \frac{38}{5} = \frac{2}{5}$.
Reflection keeps the radius: $10 - \alpha = \frac{2}{5}$, $\alpha = \frac{48}{5}$.
$\alpha + 6r^2 = \frac{48}{5} + \frac{12}{5} = 12$.
Answer: 12. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 $[\rho r^3] = [M]$ and $[S] = [MT^{-2}]$, so $[S^{3/2}] = [M^{3/2}T^{-3}]$.
$\left[\frac{\rho r^3}{S^{3/2}}\right] = [M^{-1/2}T^3]$, and its square root is $[M^{-1/4}T^{3/2}]$ — not a time.
So the formula as given is not dimensionally correct: (A) is false. (R) states exactly this: true. (The correct form is $T = K\sqrt{\rho r^3/S}$.)
Answer: (D). | ||||
| 32 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $u^2 = 2gh$. At height $\frac{h}{3}$: $\frac{h}{3} = ut - \frac{1}{2}gt^2$, so $t = \frac{u \pm \sqrt{u^2 - \frac{2gh}{3}}}{g} = \frac{\sqrt{2gh}\left(1 \pm \sqrt{\frac{2}{3}}\right)}{g}$.
Ratio $\frac{t_1}{t_2} = \frac{1 - \sqrt{2/3}}{1 + \sqrt{2/3}} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$.
Answer: (B). | ||||
| 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 $\sqrt{x} = t - 4$, so $x = (t - 4)^2$ and $\frac{dx}{dt} = 2(t - 4)$.
At $t = 4$: $\frac{dx}{dt} = 0$.
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 The wall's normal reaction is the only horizontal force and supplies the centripetal force: $N = \frac{mv^2}{R}$.
N grows as $v^2$: a parabola through the origin, curving upward — graph (A).
Answer: (A). | ||||
| 35 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane At the highest point only the horizontal velocity $u\cos 60^\circ = \frac{u}{2}$ remains.
KE $= \frac{1}{2}m\frac{u^2}{4} = \frac{E}{4}$.
Answer: (C). | ||||
| 36 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\vec{r}_{cm} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{4} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4} = -2\hat{i} - \hat{j} + \hat{k}$, of magnitude $\sqrt{6}$.
Magnitudes of the options: (A) $\sqrt{6}$, (B) $\sqrt{14}$, (C) $\sqrt{8}$, (D) 3.
Answer: (A). | ||||
| 37 | Properties of Bulk MatterSurface tension | 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 surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Water does not wet an oily or greasy surface: the angle of contact between water and oil/grease is obtuse, so water forms drops on the stain instead of spreading over it and cannot get between the grease and the fibres. That is why water alone does not remove such stains (detergents are added to lower the angle of contact).
So (A) is true, (R) is true and explains it.
Answer: (A). | ||||
| 38 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Young's modulus is a property of the material, not of the wire's dimensions. Changing length and radius changes the extension for a given load, not Y.
Answer: (A). | ||||
| 39 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The vehicle accelerates at $g\sin\alpha$ down the incline. In its frame the effective gravity is $\vec{g} - \vec{a}$, whose component along the incline cancels, leaving $g\cos\alpha$ perpendicular to the incline.
$T = 2\pi\sqrt{\frac{L}{g\cos\alpha}}$.
Answer: (A). | ||||
| 40 | 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 Charge within radius r: $q = \int_0^r \rho_0\left(\frac{3}{4} - \frac{x}{R}\right)4\pi x^2\,dx = 4\pi\rho_0\left(\frac{r^3}{4} - \frac{r^4}{4R}\right) = \pi\rho_0 r^3\left(1 - \frac{r}{R}\right)$.
Gauss's law: $E\cdot 4\pi r^2 = \frac{q}{\varepsilon_0}$, so $E = \frac{\rho_0 r}{4\varepsilon_0}\left(1 - \frac{r}{R}\right)$.
Answer: (C). | ||||
| 41 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In electrostatic equilibrium the field inside a conductor is zero, so the whole conductor including its surface is one equipotential: Statement I is correct.
The surface is an equipotential, so the field just outside has no tangential component and is normal to the surface at every point: Statement II is correct.
Answer: (A). | ||||
| 42 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Each wire has length l and area A. In series: $\frac{2l}{\sigma A} = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A}$.
$\frac{2}{\sigma} = \frac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2}$, so $\sigma = \frac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}$.
Answer: (B). | ||||
| 43 | 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 $X_L = \omega L = 100\pi\cdot\frac{\sqrt{2}}{\pi} = 100\sqrt{2}\ \Omega$.
$E_{rms} = \frac{440}{\sqrt{2}}$ V, so the ammeter reads $I_{rms} = \frac{440/\sqrt{2}}{100\sqrt{2}} = 2.2$ A.
Answer: (C). | ||||
| 44 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\tau = \frac{L}{R} = 10$ ms and $I_0 = \frac{6}{100} = 0.06$ A; $i = I_0(1 - e^{-t/\tau})$.
(a) $i = \frac{I_0}{2}$ when $t = \tau\ln 2 = 6.93 \approx 7$ ms.
(b) At 15 ms: $i = 0.06(1 - e^{-3/2}) = 0.06 \times 0.75 = 0.045$ A; $U = \frac{1}{2}Li^2 = \frac{1}{2}(0.045)^2 \approx 1.0 \times 10^{-3}$ J = 1 mJ.
Answer: (C). | ||||
| 45 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves UV kills germs — water purification (ii). X-rays — diagnosis in medicine (i). Microwaves — communication and radar (iii). Infrared penetrates haze — visibility in fog (iv).
Answer: (B). | ||||
| 46 | 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 $E = \frac{hc}{\lambda} - \phi$ and $2E = \frac{hc}{\lambda'} - \phi$.
Subtracting: $\frac{hc}{\lambda'} = \frac{hc}{\lambda} + E = \frac{hc + E\lambda}{\lambda}$, so $\lambda' = \frac{hc\lambda}{E\lambda + hc}$.
Answer: (B). | ||||
| 47 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $E_n = -\frac{13.6}{n^2}$ eV.
(i) $2 \to 1$: $13.6\left(1 - \frac{1}{4}\right) = 13.6 \times \frac{3}{4}$.
(ii) $\infty \to 1$: $13.6$.
Ratio $= \frac{3}{4}$, i.e. 3 : 4.
Answer: (A). | ||||
| 48 | 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 The amplitude varies by 8 V between maximum and minimum: $A_{max} - A_{min} = 8$ V with $A_{max} = 9$ V, so $A_{min} = 1$ V.
$\mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}} = \frac{8}{10} = 0.8$.
Answer: (A). | ||||
| 49 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter 1 MSD $= \frac{1}{20}$ cm $= 0.05$ cm. 25 VSD = 24 MSD, so 1 VSD $= \frac{24}{25}$ MSD.
Least count $= 1\text{ MSD} - 1\text{ VSD} = \frac{1}{25}\times 0.05 = 0.002$ cm. (The total of 50 vernier divisions does not change the least count.)
Answer: (C). | ||||
| 50 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{0.5}{50} = 0.01$ mm.
Observed reading $= 2.5 + 45 \times 0.01 = 2.95$ mm.
A negative zero error of 0.03 mm is added back: diameter $= 2.95 + 0.03 = 2.98$ mm.
Answer: (C). | ||||
| 51 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Maximum range: $\theta = 45^\circ$, $R = \frac{u^2}{g}$.
Second: $\frac{u^2\sin 2\theta'}{g} = \frac{u^2}{2g}$, so $\sin 2\theta' = \frac{1}{2}$ and $2\theta' = 30^\circ$ or $150^\circ$: $\theta' = 15^\circ$ (or the complementary $75^\circ$).
Answer: 15. | ||||
| 52 | GravitationAcceleration due to gravity | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation For $h \ll R_e$: $g_h = g\left(1 - \frac{2h}{R_e}\right)$ and $g_d = g\left(1 - \frac{d}{R_e}\right)$.
Equal when $d = 2h$, so $\alpha = 2$.
Answer: 2. | ||||
| 53 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $TV^{\gamma - 1}$ is constant, and $V \propto \frac{1}{d}$, so $T \propto d^{\gamma - 1}$.
$\frac{T_2}{T_1} = 32^{2/5} = 4$.
Answer: 4. | ||||
| 54 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | NumericalEasy | |
A easy numerical question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $C_V = \frac{1\cdot\frac{3}{2}R + 3\cdot\frac{5}{2}R}{1 + 3} = \frac{9R}{4}$.
$\frac{\alpha^2}{4} = \frac{9}{4}$, so $\alpha = 3$.
Answer: 3. | ||||
| 55 | 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 Name the junctions from left to right 1, 2, 3, 4, with the 9 $\Omega$ resistors between 1–2, 2–3 and 3–4. The upper wire joins 1 to 3 and the lower wire joins 2 to 4.
So there are only two distinct nodes, X (1 and 3) and Y (2 and 4), and each resistor runs between X and Y: the three are in parallel, $R = 3\ \Omega$. The battery is across X and Y.
$I = \frac{6}{3} = 2$ A.
Answer: 2. | ||||
| 56 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B = \frac{\mu_0 N I}{2r}$, so $I = \frac{2rB}{\mu_0 N} = \frac{2 \times 0.05 \times 37.68 \times 10^{-4}}{4 \times 3.14 \times 10^{-7} \times 100} = \frac{3.768 \times 10^{-4}}{1.256 \times 10^{-4}} = 3$ A.
Answer: 3. | ||||
| 57 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At A (phase 0): $(\sqrt{4I} + \sqrt{9I})^2 = 25I$. At B (phase $\pi$): $(\sqrt{9I} - \sqrt{4I})^2 = I$.
Difference $= 24I$.
Answer: 24. | ||||
| 58 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $2\pi r = 3.14$ m, so $r = 0.5$ m and $A = \pi r^2 = 3.14 \times 0.25 = 0.785$ m$^2$.
$m = IA = 14 \times 0.785 = 10.99 \approx 11$ A m$^2$.
Answer: 11. | ||||
| 59 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $|\vec{P}| = \sqrt{48 + 27 + 25} = 10$. The normal is the z-axis, so $\cos i = \frac{5}{10}$ and $i = 60^\circ$.
Snell: $\sqrt{2}\sin 60^\circ = \sqrt{3}\sin r \Rightarrow \sin r = \frac{\sqrt{2}\cdot\frac{\sqrt{3}}{2}}{\sqrt{3}} = \frac{1}{\sqrt{2}}$, $r = 45^\circ$.
$i - r = 15^\circ$.
Answer: 15. | ||||
| 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 $E = \frac{V}{d} = \frac{0.6}{6 \times 10^{-6}} = 1 \times 10^5$ N/C.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | 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 NO has 11 valence electrons: an odd-electron molecule. In $\mathrm{H_2SO_4}$ sulphur forms six bonds (two S=O, two S–OH), 12 electrons around S: an expanded octet.
$\mathrm{BCl_3}$ is electron-deficient (6 electrons on B), not odd-electron; $\mathrm{SF_6}$ is expanded but the pair in (C) has no odd-electron molecule.
Answer: (B). | ||||
| 62 | 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 $n(\mathrm{N_2}) = \frac{20}{28} = 0.714$ mol; $n(\mathrm{H_2}) = \frac{5}{2} = 2.5$ mol.
0.714 mol $\mathrm{N_2}$ needs $3 \times 0.714 = 2.14$ mol $\mathrm{H_2}$, less than 2.5: $\mathrm{N_2}$ is limiting.
$n(\mathrm{NH_3}) = 2 \times 0.714 \approx 1.42$ mol.
Answer: (C). | ||||
| 63 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Egg albumin is a protein whose large molecules are strongly solvated by water: stirring it into water (here a dilute salt solution) gives a colloidal solution directly, which is a lyophilic sol. A 5% NaCl solution does not salt the protein out, and a lyophilic sol is not easily coagulated by a small amount of electrolyte.
Answer: (A). | ||||
| 64 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Al ($3s^2 3p^1$) loses a 3p electron, which is shielded by the 3s pair and easier to remove than Mg's 3s electron; so $\Delta_i H_1$(Al) is less than Mg's 737 but more than Na's 496. Of the options only 577 kJ mol$^{-1}$ fits (the known value).
Answer: (C). | ||||
| 65 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Gangue (matrix) is the unwanted earthy and siliceous material — sand, clay, rock — that accompanies the ore and is removed during concentration.
Answer: (A). | ||||
| 66 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Zinc is amphoteric: $\mathrm{Zn + 2OH^- + 2H_2O \rightarrow [Zn(OH)_4]^{2-} + H_2}$. In excess aqueous alkali the zinc is present as the tetrahydroxidozincate(II) ion.
Answer: (C). | ||||
| 67 | 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 Lithium, like magnesium (diagonal relationship), forms a nitrate that decomposes to the oxide: $\mathrm{4LiNO_3 \rightarrow 2Li_2O + 4NO_2 + O_2}$.
Other alkali-metal nitrates give the nitrite: $\mathrm{2NaNO_3 \rightarrow 2NaNO_2 + O_2}$.
Answer: (C). | ||||
| 68 | 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 $\mathrm{SCl_2}$: S has 6 valence electrons, 2 bond pairs, so 2 lone pairs.
$\mathrm{O_3}$: the central O forms one double and one single (coordinate) bond and keeps 1 lone pair.
$\mathrm{ClF_3}$: 7 electrons, 3 bonds, 2 lone pairs (T-shape).
$\mathrm{SF_6}$: 6 bonds, 0 lone pairs.
Answer: (B). | ||||
| 69 | 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 Colour needs d–d transitions, so a partly filled d subshell.
$\mathrm{Sc^{3+}}$ is $3d^0$ and $\mathrm{Zn^{2+}}$ is $3d^{10}$: both colourless.
$\mathrm{Cu^{2+}}$ ($d^9$), $\mathrm{V^{2+}}$ ($d^3$), $\mathrm{Ti^{3+}}$ ($d^1$), $\mathrm{Mn^{2+}}$ ($d^5$, pale pink) are coloured.
Answer: (A). | ||||
| 70 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In neutral or faintly alkaline medium permanganate is reduced to $\mathrm{MnO_2}$:
$\mathrm{8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-}$.
Mn goes from +7 to +4: a change of 3.
Answer: (D). | ||||
| 71 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Sodium chlorate ($\mathrm{NaClO_3}$) and sodium arsenite are herbicides (weedicides). Aldrin and dieldrin are organochlorine insecticides, so every other pair contains an insecticide.
Answer: (D). | ||||
| 72 | AminesBasicity of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines (A) triethylamine and (B) diethylamine are ordinary aliphatic amines; in water their basicity is limited by the bulky ethyl groups hindering solvation of the cation (and in triethylamine, approach to N).
(C) is an enamine (N-methyl-2-pyrroline): the N lone pair is delocalised into the C=C bond, lowering its basicity.
(D) quinuclidine (1-azabicyclo[2.2.2]octane): the alkyl groups are tied back in a rigid cage, so the lone pair is fully exposed and the conjugate acid is well solvated; it is the strongest base of the four (conjugate-acid $pK_a \approx 11.3$, against about 10.8 for triethylamine and 11.0 for diethylamine).
Answer: (D). | ||||
| 73 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Each pair shows the two Kekulé structures of a methylbenzene. Ozonolysis cleaves each C=C and each fragment joined by a single bond becomes a 1,2-dicarbonyl.
Toluene, m-xylene and p-xylene: both Kekulé structures give the same set (for m- and p-xylene: two methylglyoxal + one glyoxal; for toluene: one methylglyoxal + two glyoxal).
o-Xylene: when the two methyl-bearing carbons are joined by a single bond, one structure gives dimethylglyoxal + 2 glyoxal; the other, with the double bond between them, gives 2 methylglyoxal + glyoxal. Different products.
Answer: (C). | ||||
| 74 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Cyanide is ambident. AgCN is largely covalent (Ag–C bond), so the alkyl halide is attacked through nitrogen: ethyl isocyanide, A = $\mathrm{CH_3CH_2NC}$.
NaCN is ionic; the free $\mathrm{CN^-}$ attacks through the more nucleophilic carbon: ethyl cyanide, B = $\mathrm{CH_3CH_2CN}$.
Answer: (C). | ||||
| 75 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $\mathrm{Br_2}$ (with the strongly activating OH) brominates both positions ortho to OH: A = 3,5-dibromo-4-hydroxybenzaldehyde.
$\mathrm{NH_2OH}$ converts CHO to the oxime CH=NOH: B.
$\mathrm{P_2O_5}$, heat, dehydrates the aldoxime to the nitrile: C = 3,5-dibromo-4-hydroxybenzonitrile, with OH, both Br and CN.
Answer: (D). | ||||
| 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 The last step ($\mathrm{Br_2}$/NaOH on $\mathrm{C_8H_8ClNO}$) is the Hofmann bromamide degradation: an amide $\mathrm{ArCONH_2}$ gives $\mathrm{ArNH_2}$, the NH2 taking the place of the CONH2.
The product is 4-chloro-3-methylaniline, so the amide is 4-chloro-3-methylbenzamide (C8H8ClNO ✓), formed from the acid chloride with $\mathrm{NH_3}$.
A = 4-chloro-3-methylbenzoyl chloride, $\mathrm{C_8H_6Cl_2O}$: COCl with Cl para and CH3 meta to it — structure (C).
Answer: (C). | ||||
| 77 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Benzenediazonium chloride couples with 2-naphthol (β-naphthol) in alkaline solution to give 1-phenylazo-2-naphthol, an orange-red dye; this is the standard test for a diazonium salt / primary aromatic amine.
Answer: (A). | ||||
| 78 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids DIBAL-H, $\mathrm{AlH(i\text{-}Bu)_2}$, then water, reduces a nitrile to the aldehyde: A = 4-aminobenzaldehyde.
With $\mathrm{CH_3CHO}$ in dilute NaOH and heat, the enolate of acetaldehyde adds to ArCHO (cross aldol, Claisen–Schmidt) and the aldol dehydrates: B = $\mathrm{H_2N{-}C_6H_4{-}CH{=}CH{-}CHO}$ (4-aminocinnamaldehyde).
Answer: (B). | ||||
| 79 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Amytal (amobarbital) is a barbiturate, a hypnotic/sedative. Seldane (terfenadine) is an antihistamine, aspartame an artificial sweetener, prontosil an antibacterial sulpha drug.
Answer: (B). | ||||
| 80 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Soluble in NaOH but not in $\mathrm{NaHCO_3}$: an acid weaker than carbonic acid — a phenol, not a carboxylic acid. A violet colour with neutral $\mathrm{FeCl_3}$ is the test for a phenolic OH.
Cyclohexanol, benzyl alcohol and cyclohexylmethanol are alcohols, not acidic enough to dissolve in NaOH and give no $\mathrm{FeCl_3}$ colour. X is phenol.
Answer: (B). | ||||
| 81 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry $\kappa = \frac{G^*}{R}$: $\kappa_1 = \frac{129}{100} = 1.29$ S m$^{-1}$, $\kappa_2 = \frac{129}{50} = 2.58$ S m$^{-1}$.
74.5 ppm = 74.5 mg per litre (dilute) = $1 \times 10^{-3}$ mol L$^{-1}$; 149 ppm = $2 \times 10^{-3}$ mol L$^{-1}$.
$\Lambda = \frac{\kappa}{c}$, so $\frac{\Lambda_1}{\Lambda_2} = \frac{\kappa_1}{\kappa_2}\cdot\frac{c_2}{c_1} = \frac{1}{2} \times 2 = 1 = 1000 \times 10^{-3}$.
Answer: 1000. | ||||
| 82 | Solid StateIonic radii and radius ratio | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on ionic radii and radius ratio, from Solid State. NCERT: No matching NCERT chapter ccp anions with cations in all octahedral voids is the rock-salt structure; along the cube edge cation and anion touch: $a = 2(r^+ + r^-) = 2(102 + 181) = 566$ pm.
Answer: 566. | ||||
| 83 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $\Delta x\,m\,\Delta v \geq \frac{h}{4\pi}$ with $\Delta x = 2a_0 = 105.8$ pm.
$\Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.1416 \times 9.1 \times 10^{-31} \times 105.8 \times 10^{-12}} = \frac{6.63 \times 10^{-34}}{1.2099 \times 10^{-39}} = 5.48 \times 10^5$ m s$^{-1}$ = 548 km s$^{-1}$.
Answer: 548. | ||||
| 84 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics $n(\mathrm{H^+}) = 0.12$ mol, $n(\mathrm{OH^-}) = 0.04$ mol: 0.04 mol is neutralised.
Heat $= 0.04 \times 57000 = 2280$ J.
The 1000 mL of solution is taken as 1000 g of water: $\Delta T = \frac{2280}{1000 \times 4.2} = 0.543$ °C $= 54.3 \times 10^{-2}$ °C.
Answer: 54. | ||||
| 85 | SolutionsConcentration terms | Class 12: Solutions | NumericalMedium | |
A medium numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Henry's law: $x_{\mathrm{O_2}} = \frac{p}{K_H} = \frac{0.920}{46820} = 1.965 \times 10^{-5}$.
1 L of water is $\frac{1000}{18} = 55.6$ mol; since the solubility is tiny, $n_{\mathrm{O_2}} \approx x \cdot n_{\mathrm{water}} = 1.965 \times 10^{-5} \times 55.6 = 1.09 \times 10^{-3}$ mol = 1.09 mmol.
Answer: 1. | ||||
| 86 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $K_{sp} = s^2$, so $s = \sqrt{8 \times 10^{-28}} = 2\sqrt{2} \times 10^{-14} = 2.82 \times 10^{-14} = 282 \times 10^{-16}$ mol L$^{-1}$.
Answer: 282. | ||||
| 87 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate $= k[X]$. From I: $k = \frac{2 \times 10^{-3}}{0.1} = 0.02$ min$^{-1}$.
II: $4 \times 10^{-3} = 0.02L$, so $L = 0.2$.
III: rate $= 0.02 \times 0.4 = 8 \times 10^{-3}$, so $M = 8$.
$\frac{M}{L} = \frac{8}{0.2} = 40$.
Answer: 40. | ||||
| 88 | BiomoleculesProteins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules A linear chain of 4 amino acids has 3 peptide bonds: $4 - 3 = 1$.
Answer: 1. | ||||
| 89 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{CH_3C{\equiv}CH + 2Br_2 \rightarrow CH_3CBr_2CHBr_2}$; molar mass of the product $= 3(12) + 4(1) + 4(80) = 360$ g mol$^{-1}$.
1 g $\mathrm{Br_2}$ = $\frac{1}{160}$ mol, giving at most $\frac{1}{320}$ mol of product = $\frac{360}{320} = 1.125$ g.
At 27% yield: $0.27 \times 1.125 = 0.304$ g $= 3.04 \times 10^{-1}$ g.
Answer: 3. | ||||
| 90 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{Fe^{3+}}$ is $3d^5$. An inner-orbital ($d^2sp^3$) complex is low spin: $t_{2g}^5 e_g^0$.
CFSE $= 5 \times (-0.4\Delta_o) = -2.0\Delta_o$ (pairing energy ignored).
Answer: 2. | ||||
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