JEE (Main) · 2022
JEE (Main) 2022 — 24 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
- Conic Sections3
12 easy · 17 medium · 1 hard
Physics
30 Q
- Electrostatics3
- Work, Energy and Power2
- Thermodynamics2
23 easy · 7 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques3
- Isolation of Metals2
- s-Block Elements2
23 easy · 6 medium · 1 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 | 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 $x * 1 = x^2 + 1$, so $(x * 1) * 1 = (x^2 + 1)^2 + 1$.
$1 * 1 = 1 + 1 = 2$, so $x * (1 * 1) = x * 2 = x^2 + 8$.
Equating: $x^4 + 2x^2 + 2 = x^2 + 8$, i.e. $x^4 + x^2 - 6 = 0$, $(x^2 + 3)(x^2 - 2) = 0$, so $x^2 = 2$ and $x^4 = 4$.
Then $\frac{x^4 + x^2 - 2}{x^4 + x^2 + 2} = \frac{4 + 2 - 2}{4 + 2 + 2} = \frac{1}{2}$, and $2\sin^{-1}\frac{1}{2} = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3}$.
Answer: (B). | ||||
| 2 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $e^{2x} = 4$ gives $x = \log_e 2$.
$6e^{2x} - 5e^x + 1 = 0$ is $(2e^x - 1)(3e^x - 1) = 0$, so $e^x = \frac{1}{2}$ or $\frac{1}{3}$, i.e. $x = -\log_e 2$ or $x = -\log_e 3$.
Sum of the real roots $= \log_e 2 - \log_e 2 - \log_e 3 = -\log_e 3$.
Answer: (B). | ||||
| 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 Determinant of the coefficients: $\begin{vmatrix}1&1&\alpha\\3&1&1\\1&0&2\end{vmatrix} = 1(2) - 1(6 - 1) + \alpha(0 - 1) = -3 - \alpha$, so a unique solution needs $\alpha \ne -3$.
From the third equation $x = 1 - 2z$; the second gives $3 - 6z + y + z = 4$, so $y = 1 + 5z$; the first gives $1 - 2z + 1 + 5z + \alpha z = 2$, i.e. $(3 + \alpha)z = 0$, so $z = 0$.
Hence $(x^*, y^*, z^*) = (1, 1, 0)$.
The points $(\alpha, 1)$, $(1, \alpha)$, $(1, -1)$ are collinear:
$\begin{vmatrix}\alpha&1&1\\1&\alpha&1\\1&-1&1\end{vmatrix} = \alpha(\alpha + 1) - 1(0) + 1(-1 - \alpha) = \alpha^2 - 1 = 0$, so $\alpha = \pm 1$ (both $\ne -3$).
Sum of absolute values $= 1 + 1 = 2$.
Answer: (C). | ||||
| 4 | Sequences and SeriesRelation between A.M. and G.M. | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on relation between a.m. and g.m., from Sequences and Series. NCERT: Class 11: Sequences and Series Write $3x + 2y = x + x + x + y + y$ and apply A.M. ≥ G.M. to these five positive terms:
$\frac{3x + 2y}{5} \ge (x^3y^2)^{1/5} = (2^{15})^{1/5} = 2^3 = 8$.
So $3x + 2y \ge 40$, with equality when $x = y = 8$ (and indeed $8^3 \cdot 8^2 = 2^{15}$).
Answer: (D). | ||||
| 5 | 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 On $(-2, -1)$: $[x] = -2$, so $f(x) = \frac{\sin(x + 2)}{x + 2}$.
On $|x| < 1$: $[|x|] = 0$, so $f(x) = \max\{2x, 0\}$, which is $0$ for $-1 < x \le 0$ and $2x$ for $0 \le x < 1$.
Elsewhere ($x \le -2$, $x = -1$, $x \ge 1$): $f(x) = 1$.
At $x = -2$: with $h = x + 2 \to 0^+$, $\frac{\sin h}{h} \to 1 = f(-2)$, so $f$ is continuous; $\frac{\sin h}{h} = 1 - \frac{h^2}{6} + \ldots$ has right derivative $0$, and the left derivative is $0$, so $f$ is differentiable there.
At $x = -1$: left limit $\sin 1$, $f(-1) = 1$, right limit $0$: not continuous, not differentiable.
At $x = 0$: $\max\{2x, 0\}$ is continuous but has a corner (slopes $0$ and $2$): not differentiable.
At $x = 1$: left limit $2$, $f(1) = 1$: not continuous, not differentiable.
So $m = 2$ (at $-1, 1$) and $n = 3$ (at $-1, 0, 1$).
Answer: (C). | ||||
| 6 | 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 Let $I$ be the integral. Replacing $x$ by $-x$: $I = \int_{-\pi/2}^{\pi/2} \frac{e^x\,dx}{(1 + e^x)(\sin^6 x + \cos^6 x)}$. Adding,
$2I = \int_{-\pi/2}^{\pi/2} \frac{dx}{\sin^6 x + \cos^6 x} = 2\int_0^{\pi/2} \frac{dx}{\sin^6 x + \cos^6 x}$, so $I = \int_0^{\pi/2} \frac{dx}{\sin^6 x + \cos^6 x}$.
Divide numerator and denominator by $\cos^6 x$ and put $t = \tan x$: $I = \int_0^{\infty} \frac{(1 + t^2)^2}{1 + t^6}\,dt = \int_0^{\infty} \frac{1 + t^2}{t^4 - t^2 + 1}\,dt$.
Dividing by $t^2$: $\int_0^{\infty} \frac{1 + \frac{1}{t^2}}{\left(t - \frac{1}{t}\right)^2 + 1}\,dt$; with $u = t - \frac{1}{t}$ running from $-\infty$ to $\infty$, $I = \int_{-\infty}^{\infty} \frac{du}{u^2 + 1} = \pi$.
Answer: (C). | ||||
| 7 | 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 The $r$-th term is $\frac{n^2}{(n^2 + r^2)(n + r)} = \frac{1}{n} \cdot \frac{1}{\left(1 + \frac{r^2}{n^2}\right)\left(1 + \frac{r}{n}\right)}$, $r = 1, \ldots, n$.
So the limit is $\int_0^1 \frac{dx}{(1 + x^2)(1 + x)}$.
Partial fractions: $\frac{1}{(1 + x)(1 + x^2)} = \frac{1/2}{1 + x} + \frac{-\frac{1}{2}x + \frac{1}{2}}{1 + x^2}$.
Integral $= \frac{1}{2}\log_e 2 - \frac{1}{4}\log_e 2 + \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8} + \frac{1}{4}\log_e 2$.
Answer: (A). | ||||
| 8 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Let $P = (x, y)$. The tangent $Y - y = y'(X - x)$ meets the $x$-axis at $Q = \left(x - \frac{y}{y'}, 0\right)$.
The midpoint of $PQ$ lies on the $y$-axis: $x + x - \frac{y}{y'} = 0$, so $\frac{dy}{dx} = \frac{y}{2x}$.
Separating, $\frac{2\,dy}{y} = \frac{dx}{x}$, so $y^2 = cx$. Through $(3, 3)$: $c = 3$, so $C$ is $y^2 = 3x$.
This parabola has $4a = 3$: latus rectum $3$ and focus $\left(\frac{3}{4}, 0\right)$.
Answer: (A). | ||||
| 9 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Take the vertex $A(-a, 0)$ and the side parallel to the $y$-axis joining $(a\cos\theta, \pm 2\sin\theta)$.
Area $= \frac{1}{2}(4\sin\theta)(a\cos\theta + a) = 2a\sin\theta(1 + \cos\theta)$.
$\frac{d}{d\theta}[\sin\theta + \sin\theta\cos\theta] = \cos\theta + \cos 2\theta = 0$ gives $2\cos^2\theta + \cos\theta - 1 = 0$, $\cos\theta = \frac{1}{2}$, $\theta = \frac{\pi}{3}$ (a maximum).
Maximum area $= 2a \cdot \frac{\sqrt{3}}{2} \cdot \frac{3}{2} = \frac{3\sqrt{3}}{2}a = 6\sqrt{3}$, so $a = 4$.
$e = \sqrt{1 - \frac{4}{16}} = \frac{\sqrt{3}}{2}$.
Answer: (A). | ||||
| 10 | Matrices and DeterminantsArea of triangles using determinants | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on area of triangles using determinants, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Area $= \frac{1}{2}\left|1(0 - \alpha) + \alpha(\alpha - \alpha) + 0(\alpha - 0)\right| = \frac{|\alpha|}{2} = 4$, so $\alpha = \pm 8$ and $\alpha^2 = 64$.
The points $(\alpha, -\alpha)$ and $(-\alpha, \alpha)$ both lie on the line $y = -x$, so $(\alpha^2, \beta)$ must too: $\beta = -\alpha^2 = -64$.
Answer: (C). | ||||
| 11 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $f(x) = x^7 - 7x - 2$. $f'(x) = 7x^6 - 7 = 0$ at $x = \pm 1$, so $f$ increases on $(-\infty, -1)$, decreases on $(-1, 1)$ and increases on $(1, \infty)$.
$f(-1) = -1 + 7 - 2 = 4 > 0$ (local maximum) and $f(1) = 1 - 7 - 2 = -8 < 0$ (local minimum).
With $f \to -\infty$ as $x \to -\infty$ and $f \to \infty$ as $x \to \infty$, the graph crosses the $x$-axis once in each of the three intervals: 3 distinct real roots.
Answer: (D). | ||||
| 12 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $k + 2k + 4k + 6k + 8k = 21k = 1$.
$P(X \le 2) = k + 2k + 4k = 7k$. The event $\{1 < X < 4\} \cap \{X \le 2\}$ is $\{X = 2\}$, with probability $4k$.
$P(1 < X < 4 \mid X \le 2) = \frac{4k}{7k} = \frac{4}{7}$.
Answer: (A). | ||||
| 13 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\cos(A + B)\cos(A - B) = \cos^2 A - \sin^2 B$ with $A = \frac{\pi}{3}$, $B = x$: the left side is $\frac{1}{4} - \sin^2 x$.
So $1 - 4\sin^2 x = \cos^2 2x$. With $c = \cos 2x$, $4\sin^2 x = 2(1 - c)$: $2c - 1 = c^2$, i.e. $(c - 1)^2 = 0$, so $\cos 2x = 1$.
Then $x = k\pi$, and in $[-3\pi, 3\pi]$: $k = -3, -2, \ldots, 3$, i.e. 7 solutions.
Answer: (D). | ||||
| 14 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Points $(1, 2, 3)$ and $(2, 4, 5)$, joining vector $(1, 2, 2)$; directions $\vec{d_1} = (2, 3, \lambda)$, $\vec{d_2} = (1, 4, 5)$.
$\vec{d_1} \times \vec{d_2} = (15 - 4\lambda, \lambda - 10, 5)$.
$(1, 2, 2) \cdot (\vec{d_1} \times \vec{d_2}) = 15 - 4\lambda + 2\lambda - 20 + 10 = 5 - 2\lambda$.
$|\vec{d_1} \times \vec{d_2}|^2 = (15 - 4\lambda)^2 + (\lambda - 10)^2 + 25 = 17\lambda^2 - 140\lambda + 350$.
$\frac{(5 - 2\lambda)^2}{17\lambda^2 - 140\lambda + 350} = \frac{1}{3}$ gives $12\lambda^2 - 60\lambda + 75 = 17\lambda^2 - 140\lambda + 350$, i.e. $\lambda^2 - 16\lambda + 55 = 0$, $\lambda = 5$ or $11$.
Sum $= 16$.
Answer: (A). | ||||
| 15 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry P is the perpendicular bisector plane of the segment joining the two points, so its normal is along $(2 - (-4), -2 - 2, 3 - 1) = (6, -4, 2)$, i.e. $(3, -2, 1)$.
The other plane has normal $(2, 1, 3)$.
$\cos\theta = \frac{|3 \cdot 2 - 2 \cdot 1 + 1 \cdot 3|}{\sqrt{14}\sqrt{14}} = \frac{7}{14} = \frac{1}{2}$, so $\theta = \frac{\pi}{3}$.
Answer: (C). | ||||
| 16 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | StatementsMedium | |
A medium statements question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\hat{a} + \hat{b}$ is perpendicular to $\hat{a} \times \hat{b}$, so $|\hat{a} + \hat{b}|^2 + 4|\hat{a} \times \hat{b}|^2 = 4$:
$2 + 2\cos\theta + 4\sin^2\theta = 4$, i.e. $2\cos^2\theta - \cos\theta - 1 = 0$, so $\cos\theta = -\frac{1}{2}$ (since $\theta \ne 0$), $\theta = \frac{2\pi}{3}$.
(S1): $2|\hat{a} \times \hat{b}| = 2\sin\frac{2\pi}{3} = \sqrt{3}$ and $|\hat{a} - \hat{b}| = \sqrt{2 - 2\cos\theta} = \sqrt{3}$: true.
(S2): $|\hat{a} + \hat{b}| = \sqrt{2 + 2\cos\theta} = 1$ and $\hat{a} \cdot (\hat{a} + \hat{b}) = 1 + \cos\theta = \frac{1}{2}$, so the projection is $\frac{1/2}{1} = \frac{1}{2}$: true.
Answer: (C). | ||||
| 17 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability With $t = x^3$: $\sec t - \tan t = \frac{1 - \sin t}{\cos t} = \frac{\left(\cos\frac{t}{2} - \sin\frac{t}{2}\right)^2}{\cos^2\frac{t}{2} - \sin^2\frac{t}{2}} = \frac{1 - \tan\frac{t}{2}}{1 + \tan\frac{t}{2}} = \tan\left(\frac{\pi}{4} - \frac{t}{2}\right)$.
For $\frac{\pi}{2} < t < \frac{3\pi}{2}$, $\frac{\pi}{4} - \frac{t}{2} \in \left(-\frac{\pi}{2}, 0\right)$, inside the principal range, so $y = \frac{\pi}{4} - \frac{x^3}{2}$.
$y' = -\frac{3x^2}{2}$, $y'' = -3x$.
Check (B): $x^2(-3x) - 6\left(\frac{\pi}{4} - \frac{x^3}{2}\right) + \frac{3\pi}{2} = -3x^3 - \frac{3\pi}{2} + 3x^3 + \frac{3\pi}{2} = 0$.
(A) gives $-6x^2$, (C) gives $\frac{3\pi}{2}$, (D) gives $3x^2$: none is zero.
Answer: (B). | ||||
| 18 | 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 The statement is $(A \wedge C) \to B$.
$\sim(p \to q) \equiv p \wedge (\sim q)$, so its negation is $(A \wedge C) \wedge (\sim B)$, i.e. $(\sim B) \wedge (A \wedge C)$.
Answer: (B). | ||||
| 19 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Slope of the tangent $= -\frac{xy - x^2y^2 - 1}{x^2}$, so $x^2\frac{dy}{dx} = 1 + x^2y^2 - xy$.
Put $v = xy$: $\frac{dv}{dx} = y + x\frac{dy}{dx} = y + \frac{1 + v^2 - v}{x} = \frac{v + 1 + v^2 - v}{x} = \frac{1 + v^2}{x}$.
Separating: $\tan^{-1} v = \log_e x + C$. At $(1, 1)$: $v = 1$, so $C = \frac{\pi}{4}$.
At $x = e$: $\tan^{-1}(e\,y(e)) = 1 + \frac{\pi}{4}$, so $e \cdot y(e) = \tan\left(1 + \frac{\pi}{4}\right) = \frac{1 + \tan(1)}{1 - \tan(1)}$.
Answer: (D). | ||||
| 20 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'_\lambda(x) = 12\lambda x^2 - 72\lambda x + 36 \ge 0$ for all $x$. For $\lambda > 0$ the discriminant must be $\le 0$: $72^2\lambda^2 - 4(12\lambda)(36) \le 0$, i.e. $5184\lambda^2 \le 1728\lambda$, $\lambda \le \frac{1}{3}$. (For $\lambda < 0$ the quadratic is eventually negative.) So $\lambda^* = \frac{1}{3}$.
$f(x) = \frac{4}{3}x^3 - 12x^2 + 36x + 48$. In $f(1) + f(-1)$ the odd powers cancel: $2(-12 + 48) = 72$.
Answer: (D). | ||||
| 21 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x + iy$: $z(4 + 3i) + \bar{z}(4 - 3i) = 2\,\mathrm{Re}[(x + iy)(4 + 3i)] = 2(4x - 3y)$, so the second condition is $4x - 3y \le 12$.
$S$ is the part of the disc with centre $(3, 0)$, radius 1, on the side $4x - 3y \le 12$ (the line passes through the centre).
The point of the full disc nearest to $(0, 4)$ lies on the segment to the centre: the distance from $(0, 4)$ to $(3, 0)$ is 5, so the point is $(3, 0) + \frac{1}{5}(-3, 4) = \left(\frac{12}{5}, \frac{4}{5}\right)$.
It satisfies $4\cdot\frac{12}{5} - 3\cdot\frac{4}{5} = \frac{36}{5} \le 12$, so it is in $S$ and is the closest point.
$25(\alpha + \beta) = 25 \cdot \frac{16}{5} = 80$.
Answer: 80. | ||||
| 22 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $n(n + 1)$ is always even, and $n = 1$ requires $A^2 = I$.
$A^2 = \begin{pmatrix}1 & -a + ab\\0 & b^2\end{pmatrix} = I$ needs $b^2 = 1$, so $b = 1$; then $-a + a = 0$ for every $a$.
With $A^2 = I$, every even power is $I$, so such $A$ lie in every $T_n$; and any $A$ in all $T_n$ must have $A^2 = I$.
So the intersection is $\{A : b = 1, a \in \{1, \ldots, 100\}\}$: 100 elements.
Answer: 100. | ||||
| 23 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalHard | |
A hard numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The digits are used once each; their sum is 31. Let $S_o$ be the sum of the 4 digits in odd places and $S_e$ of the 3 in even places: $S_o + S_e = 31$ and $11 \mid S_o - S_e$.
$S_o - S_e$ is odd and lies between $-21$ and $31$, so it is $\pm 11$ (33 is impossible).
$S_o - S_e = 11$: $S_e = 10$. Three of the digits summing to 10: $\{1, 2, 7\}$, $\{1, 4, 5\}$, $\{2, 3, 5\}$: 3 ways.
$S_o - S_e = -11$: $S_o = 10$. Four digits summing to 10: only $\{1, 2, 3, 4\}$ (then $S_e = 5 + 7 + 9 = 21$): 1 way.
Each choice gives $4! \times 3! = 144$ arrangements.
Total $= 4 \times 144 = 576$.
Answer: 576. | ||||
| 24 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $HCF(\alpha, 24) = 1$ means $\alpha$ is divisible by neither 2 nor 3.
Sum of $1$ to $100$: $5050$. Multiples of 2: $2(1 + \ldots + 50) = 2550$. Multiples of 3: $3(1 + \ldots + 33) = 1683$. Multiples of 6: $6(1 + \ldots + 16) = 816$.
Sum of numbers divisible by 2 or 3 $= 2550 + 1683 - 816 = 3417$.
Required sum $= 5050 - 3417 = 1633$.
Answer: 1633. | ||||
| 25 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The sum is $\frac{3^{2022} - 1}{2}$.
$3^{2022} = 9^{1011} = (10 - 1)^{1011}$. By the binomial theorem, modulo 100 only the last two terms matter: $(10 - 1)^{1011} \equiv 1011 \cdot 10 \cdot (-1)^{1010} + (-1)^{1011} = 10110 - 1 \equiv 10 - 1 = 9 \pmod{100}$.
So $3^{2022} - 1 = 100k + 8$, and $\frac{3^{2022} - 1}{2} = 50k + 4$.
Remainder $= 4$.
Answer: 4. | ||||
| 26 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersection: $\frac{y^2}{2} = 4 - y$, i.e. $y^2 + 2y - 8 = 0$, $y = 2$ or $y = -4$.
Between them the line ($x = 4 - y$) is to the right of the parabola ($x = \frac{y^2}{2}$):
Area $= \int_{-4}^{2}\left(4 - y - \frac{y^2}{2}\right)dy = \left[4y - \frac{y^2}{2} - \frac{y^3}{6}\right]_{-4}^{2} = \left(8 - 2 - \frac{4}{3}\right) - \left(-16 - 8 + \frac{32}{3}\right) = \frac{14}{3} + \frac{40}{3} = 18$.
Answer: 18. | ||||
| 27 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Touching the $x$-axis at $(1, 0)$ with $k > 0$: centre $(1, r)$, so $h = 1$, $k = r$.
Distance of the centre from $x + y = 0$: $\frac{1 + r}{\sqrt{2}}$. Half-chord 1: $r^2 - \frac{(1 + r)^2}{2} = 1$, i.e. $r^2 - 2r - 3 = 0$, $r = 3$.
$h + k + r = 1 + 3 + 3 = 7$.
Answer: 7. | ||||
| 28 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $i$ of the first 4 and $j$ of the other 6 be right, $i + j = 8$: $(i, j) = (4, 4), (3, 5), (2, 6)$. The guesses are taken as independent.
$P = \sum \binom{4}{i}\frac{3^i}{4^4} \cdot \binom{6}{j}\frac{3^{6-j}}{4^6}$, so $4^{10}P = \sum \binom{4}{i}3^i\binom{6}{j}3^{6-j}$:
$(4, 4)$: $1 \cdot 81 \cdot 15 \cdot 9 = 10935$
$(3, 5)$: $4 \cdot 27 \cdot 6 \cdot 3 = 1944$
$(2, 6)$: $6 \cdot 9 \cdot 1 \cdot 1 = 54$
Total $= 12933 = 27 \times 479$.
So $k = 479$.
Answer: 479. | ||||
| 29 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $E$: $\frac{x^2}{4} + \frac{y^2}{3} = 1$, latus rectum $\frac{2 \cdot 3}{2} = 3$, $e_E^2 = 1 - \frac{3}{4} = \frac{1}{4}$.
$H$: latus rectum $\frac{2 \cdot 1}{a} = 3$, so $a = \frac{2}{3}$ and $e_H^2 = 1 + \frac{1}{a^2} = 1 + \frac{9}{4} = \frac{13}{4}$.
$12\left(\frac{13}{4} + \frac{1}{4}\right) = 12 \cdot \frac{14}{4} = 42$.
Answer: 42. | ||||
| 30 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The axis of $P_1$ has direction $(4 - 3, 4 - 2) = (1, 2)$, and the vertex is the midpoint of the focus and the foot of the directrix on the axis: that foot is $2(3, 2) - (4, 4) = (2, 0)$.
The directrix is perpendicular to the axis through $(2, 0)$: $x + 2y = 2$.
Its image in $x + 2y = 6$ is the parallel line on the other side at the same distance: $x + 2y = 6 + (6 - 2) = 10$. This is the directrix of $P_2$.
Answer: 10. | ||||
| 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 Velocity gradient $\frac{dv}{dx}$: $\frac{[LT^{-1}]}{[L]} = [T^{-1}]$. Decay constant $\lambda$ (from $N = N_0e^{-\lambda t}$): $[T^{-1}]$. Same.
Wien's constant $[LK]$ and Stefan's constant $[MT^{-3}K^{-4}]$ differ; angular frequency $[T^{-1}]$ and angular momentum $[ML^2T^{-1}]$ differ; wave number $[L^{-1}]$ and Avogadro number $[\text{mol}^{-1}]$ differ.
Answer: (A). | ||||
| 32 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Kepler's third law: $T^2 \propto R^3$, so $T' = T\left(\frac{3R}{R}\right)^{3/2} = 3\sqrt{3}$ years (with $T = 1$ year).
Answer: (D). | ||||
| 33 | 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 With the string at angle $\theta$ from the lowest position, the radial equation is $T - mg\cos\theta = \frac{mv^2}{r}$, so $T = \frac{mv^2}{r} + mg\cos\theta$.
With $v$ constant, $T$ is least when $\cos\theta = -1$, i.e. at the highest point, where $T = \frac{mv^2}{r} - mg$.
Answer: (B). | ||||
| 34 | 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 In limiting equilibrium the Coulomb repulsion equals the limiting friction: $\frac{kq^2}{L^2} = \mu mg$.
$kq^2 = 9 \times 10^9 \times (2 \times 10^{-7})^2 = 3.6 \times 10^{-4}$ N m$^2$; $\mu mg = 0.25 \times 0.01 \times 10 = 0.025$ N.
$L^2 = \frac{3.6 \times 10^{-4}}{0.025} = 0.0144$ m$^2$, $L = 0.12$ m $= 12$ cm.
Answer: (A). | ||||
| 35 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $T_1 = 1000$ K, $T_2 = 400$ K, efficiency $\eta = 1 - \frac{400}{1000} = 0.6$.
$W = \eta Q_1 = 0.6 \times 5000$ kcal $= 3000$ kcal $= 3000 \times 4200$ J $= 12.6 \times 10^6$ J.
Answer: (C). | ||||
| 36 | 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 $v_{max} = A\omega$, so $A_1\omega_1 = A_2\omega_2$ and $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$.
$\omega_1 = \sqrt{\frac{2k}{0.05}} = \sqrt{40k}$, $\omega_2 = \sqrt{\frac{9k}{0.1}} = \sqrt{90k}$.
$\frac{A_1}{A_2} = \sqrt{\frac{90}{40}} = \frac{3}{2}$.
Answer: (B). | ||||
| 37 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity (A) $\frac{2 \cdot 6}{8} + 4 = 5.5$ Ω. (B) $\frac{2 \cdot 4}{6} + 6 = \frac{4}{3} + 6 = \frac{22}{3}$ Ω. (C) $\frac{8 \cdot 4}{12} = \frac{8}{3}$ Ω. (D) $\frac{10 \cdot 2}{12} = \frac{5}{3}$ Ω.
Only (B) gives $\frac{22}{3}$ Ω.
Answer: (B). | ||||
| 38 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter An electromagnet must be strongly magnetised by a small current (high permeability) and lose its magnetism when the current is switched off (low retentivity, low coercivity). Soft iron has high permeability and low retentivity.
Answer: (C). | ||||
| 39 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy 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 $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, so at equal $K$, $r \propto \frac{\sqrt{m}}{q}$.
Proton $(m, e)$: $1$; deuteron $(2m, e)$: $\sqrt{2}$; α-particle $(4m, 2e)$: $\frac{2}{2} = 1$.
Ratio $1 : \sqrt{2} : 1$.
Answer: (D). | ||||
| 40 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | StatementsEasy | |
A easy statements question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Statement I: in a series LC (or LCR) circuit at resonance $X_L = X_C$, so the net reactance is zero although both elements are present: true.
Statement II: $P_{avg} = V_{rms}I_{rms}\cos\phi$; for a purely inductive or purely capacitive circuit $\phi = 90°$ and the average power is zero: false.
Answer: (C). | ||||
| 41 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At equilibrium $F = -\frac{dU}{dr} = 0$: $\frac{dU}{dr} = -\frac{10A}{r^{11}} + \frac{5B}{r^6} = 0$, so $r^5 = \frac{2A}{B}$ and $r = \left(\frac{2A}{B}\right)^{1/5}$.
Answer: (C). | ||||
| 42 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Going up, gravity and drag both oppose motion: $a_1 = g + \frac{f}{m} = 10 + 2 = 12$ m s$^{-2}$. Coming down, drag opposes gravity: $a_2 = 10 - 2 = 8$ m s$^{-2}$.
The same height is covered from rest (descent) and to rest (ascent): $h = \frac{1}{2}a_1t_1^2 = \frac{1}{2}a_2t_2^2$.
$\frac{t_1}{t_2} = \sqrt{\frac{a_2}{a_1}} = \sqrt{\frac{8}{12}} = \frac{\sqrt{2}}{\sqrt{3}}$.
Answer: (B). | ||||
| 43 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\theta = \frac{1}{2}\alpha t^2$: in the first second $5 = \frac{1}{2}\alpha$, so $\alpha = 10$ rad s$^{-2}$.
In 2 s: $\theta = \frac{1}{2}(10)(4) = 20$ rad. In the second second: $20 - 5 = 15$ rad.
Answer: (B). | ||||
| 44 | 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 Kinetic energy of the hammer $= \frac{1}{2}(1.5)(60)^2 = 2700$ J; one fourth, $675$ J, heats the nail.
$m c\,\Delta T = 675$ with $mc = 100 \times 0.42 = 42$ J/°C: $\Delta T = \frac{675}{42} \approx 16.07$ °C.
Answer: (C). | ||||
| 45 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $U = \frac{q^2}{2C}$, so $\frac{(q + 2)^2}{q^2} = 1.44$, $\frac{q + 2}{q} = 1.2$, $q = 10$ C.
Answer: (A). | ||||
| 46 | 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 Gauss's law for a coaxial cylinder of radius $r > R$, length $l$: $E(2\pi rl) = \frac{\rho\pi R^2l}{\varepsilon_0}$, so $E = \frac{\rho R^2}{2\varepsilon_0 r}$.
The electric force provides the centripetal force (the particle's charge is opposite to ρ): $qE = \frac{mv^2}{r}$, so $mv^2 = \frac{q\rho R^2}{2\varepsilon_0}$.
$K = \frac{1}{2}mv^2 = \frac{\rho qR^2}{4\varepsilon_0}$ (in magnitude).
Answer: (A). | ||||
| 47 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Radiated power $= 0.035 \times 200 = 7$ W. Intensity at 4 m: $I = \frac{7}{4\pi(4)^2} = \frac{7}{64\pi} \approx 3.48 \times 10^{-2}$ W m$^{-2}$.
$I = \frac{cB_0^2}{2\mu_0}$, so $B_0^2 = \frac{2\mu_0 I}{c} = \frac{2 \times 4\pi \times 10^{-7} \times 3.48 \times 10^{-2}}{3 \times 10^8} \approx 2.92 \times 10^{-16}$ T$^2$.
$B_0 \approx 1.71 \times 10^{-8}$ T.
Answer: (B). | ||||
| 48 | 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 $K_{max} = h\nu - \phi$: $3.8 - 0.6 = 3.2$ eV and $1.4 - 0.6 = 0.8$ eV, ratio $4 : 1$.
$v_{max} \propto \sqrt{K_{max}}$: ratio $2 : 1$.
Answer: (B). | ||||
| 49 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Amplitudes are in the ratio $\sqrt{9} : \sqrt{4} = 3 : 2$.
$\frac{I_{max}}{I_{min}} = \frac{(3 + 2)^2}{(3 - 2)^2} = \frac{25}{1}$.
Answer: (D). | ||||
| 50 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei In the $n$-th orbit $K = \frac{13.6}{n^2}$ eV, $P = -2K = -\frac{27.2}{n^2}$ eV, $E = -K = -\frac{13.6}{n^2}$ eV.
As $n$ increases, $K$ decreases while $P$ and $E$ (negative quantities) increase towards zero.
Answer: (B). | ||||
| 51 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane At 45° the initial components are equal: $u_x = u_y = a$.
After 2 s: $v_x = a$, $v_y = a - 20$, and $a^2 + (a - 20)^2 = 400$, i.e. $2a^2 - 40a = 0$, so $a = 20$ m s$^{-1}$ (at that instant the body is at the top, moving horizontally).
$H = \frac{u_y^2}{2g} = \frac{400}{20} = 20$ m.
Answer: 20. | ||||
| 52 | Communication SystemsSize of antenna | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on size of antenna, from Communication Systems. NCERT: No matching NCERT chapter An antenna must be at least about a quarter of the wavelength in its medium: $l = \frac{\lambda}{4}$, so the largest wavelength it can radiate is $\lambda = 4 \times 5.0$ mm $= 20$ mm.
Speed in the medium: $v = \frac{c}{\sqrt{\varepsilon_r\mu_r}} = \frac{3 \times 10^8}{\sqrt{6.25}} = 1.2 \times 10^8$ m s$^{-1}$.
Minimum frequency $f = \frac{v}{\lambda} = \frac{1.2 \times 10^8}{0.02} = 6 \times 10^9$ Hz $= 6$ GHz.
Answer: 6. | ||||
| 53 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Current in the primary circuit (battery taken as ideal): $I = \frac{25}{20 + 30} = 0.5$ A.
Potential drop across the wire $= 0.5 \times 20 = 10$ V over 10 m, so the gradient is 1 V/m.
$E = 1 \times 2.5 = 2.5$ V $= \frac{25}{10}$, so $x = 25$.
Answer: 25. | ||||
| 54 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The amplitude at position $x$ is $|10\cos\pi x|$.
At $x = \frac{4}{3}$: $\left|10\cos\frac{4\pi}{3}\right| = 10 \times \frac{1}{2} = 5$ cm.
Answer: 5. | ||||
| 55 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A 10 V source feeds, through 800 Ω, a 5 V Zener diode (reverse biased) in parallel with $R_L = 1$ kΩ.
Without the Zener, the voltage across $R_L$ would be $10 \times \frac{1000}{1800} \approx 5.56$ V $> 5$ V, so the Zener is in breakdown and holds the load at 5 V.
$I_L = \frac{5}{1000}$ A $= 5$ mA.
Answer: 5. | ||||
| 56 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Equal masses with atomic weights $1 : 2$ give initial numbers of atoms $N_A : N_B = 2 : 1$.
In 16 s, A goes through 4 half-lives and B through 2: $N_A = \frac{2}{16} = \frac{1}{8}$ and $N_B = \frac{1}{4}$ (in the same units).
$\frac{N_A}{N_B} = \frac{1/8}{1/4} = \frac{1}{2} = \frac{50}{100}$, so $x = 50$.
Answer: 50. | ||||
| 57 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\sin r = \frac{\sin 60°}{\sqrt{3}} = \frac{1}{2}$, so $r = 30°$.
Lateral shift $d = \frac{t\sin(i - r)}{\cos r} = \frac{t\sin 30°}{\cos 30°} = \frac{t}{\sqrt{3}}$.
$\frac{t}{\sqrt{3}} = 4\sqrt{3}$, so $t = 12$ cm.
Answer: 12. | ||||
| 58 | 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 $\varepsilon_{max} = NBA\omega = 1000 \times 0.07 \times 1 \times 2\pi \approx 439.8$ V, i.e. 440 V to the nearest integer.
Answer: 440. | ||||
| 59 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics First law: $\Delta U = Q - W = \frac{3Q}{4}$.
For $n$ moles, $\Delta U = nC_V\Delta T = n\frac{3R}{2}\Delta T$ and $Q = nC\Delta T$.
$n\frac{3R}{2}\Delta T = \frac{3}{4}nC\Delta T$, so $C = 2R$.
Answer: 2. | ||||
| 60 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter From the graph, $\Delta T$ falls from 60 °C at $t = 0$ to 40 °C at $t = 6$ min and to 20 °C at $t = t_2$. The surroundings are at $80 - 60 = 20$ °C.
Newton's law of cooling in its usual average form, $\frac{\Delta\theta}{t} = k(\bar{\theta} - \theta_s)$:
first interval: $\frac{60 - 40}{6} = k \cdot \frac{60 + 40}{2}$, so $k = \frac{1}{15}$ min$^{-1}$;
second interval (from 40 to 20): $\frac{40 - 20}{t} = \frac{1}{15} \cdot \frac{40 + 20}{2} = 2$, so $t = 10$ min.
$t_2 = 6 + 10 = 16$ min.
(The exact exponential law gives $t_2 = 6\frac{\ln 3}{\ln 1.5} \approx 16.3$ min, also 16 to the nearest integer.)
Answer: 16. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Carbon: $\frac{12}{44} \times 330 = 90$ g, i.e. $\frac{90}{120} \times 100 = 75\%$.
Hydrogen: $\frac{2}{18} \times 270 = 30$ g, i.e. $\frac{30}{120} \times 100 = 25\%$ (and $90 + 30 = 120$ g, as it must be).
Answer: (D). | ||||
| 62 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $E = N_A\frac{hc}{\lambda} = 6.02 \times 10^{23} \times \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 6.02 \times 10^{23} \times 6.63 \times 10^{-19}$ J $\approx 3.99 \times 10^5$ J mol$^{-1}$ $= 399$ kJ mol$^{-1}$.
Answer: (C). | ||||
| 63 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Electron counts: $\text{C}_2^{2-}$ 14 (like $\text{N}_2$), bond order 3; $\text{N}_2^{2-}$ 16 (like $\text{O}_2$), bond order 2; $\text{O}_2^{2-}$ 18, bond order 1.
So $\text{O}_2^{2-} < \text{N}_2^{2-} < \text{C}_2^{2-}$.
Answer: (B). | ||||
| 64 | ThermodynamicsHess law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics $2\text{C(graphite)} + 3\text{H}_2\text{(g)} \to \text{C}_2\text{H}_6\text{(g)}$.
By Hess's law, $\Delta_f H = 2\Delta_c H(\text{C}) + 3\Delta_c H(\text{H}_2) - \Delta_c H(\text{C}_2\text{H}_6) = 2(-394) + 3(-286) - (-1560) = -788 - 858 + 1560 = -86$ kJ mol$^{-1}$.
Answer: (C). | ||||
| 65 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $t_{90\%} = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303}{k}$ and $t_{1/2} = \frac{2.303 \log 2}{k}$.
$x = \frac{t_{90\%}}{t_{1/2}} = \frac{1}{\log 2} = \frac{1}{0.3010} \approx 3.32$.
Answer: (C). | ||||
| 66 | 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 Melting points: Hg about $-39$ °C (liquid at room temperature), Ga about 30 °C, Cs about 28 °C, Ag about 962 °C. Silver, with strong metallic bonding involving its d electrons, melts highest.
Answer: (B). | ||||
| 67 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In the Hall–Héroult process purified alumina dissolved in molten cryolite is electrolysed with carbon electrodes; the oxygen set free burns the carbon anode, so the overall reaction is $2\text{Al}_2\text{O}_3 + 3\text{C} \to 4\text{Al} + 3\text{CO}_2$.
(A) is the thermite (aluminothermic) reduction, (C) reduction of iron oxide in a blast furnace, (D) recovery of gold in the cyanide process.
Answer: (B). | ||||
| 68 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter NaOH is made by electrolysing brine (chlor-alkali process): $2\text{NaCl} + 2\text{H}_2\text{O} \to 2\text{NaOH} + \text{Cl}_2 + \text{H}_2$. Hydrogen is set free at the cathode as a byproduct.
Sodium metal (electrolysis of fused NaCl) gives chlorine, not hydrogen; NaCl and $\text{Na}_2\text{CO}_3$ (Solvay process) give none.
Answer: (A). | ||||
| 69 | 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 Baking soda ($\text{NaHCO}_3$) releases $\text{CO}_2$ when heated or acidified, which smothers a fire; it is used in soda-acid and dry-chemical fire extinguishers.
Answer: (A). | ||||
| 70 | 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 To form five bonds the central atom must expand its octet. Phosphorus (n = 3) has vacant 3d orbitals; nitrogen (n = 2) has no d orbitals in its valence shell, so its covalency is limited to 4 and $\text{NCl}_5$ cannot form.
Answer: (B). | ||||
| 71 | 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 All four have the same ligand and the same charge (+3) on the metal. $\Delta_0$ rises sharply down a group (3d < 4d < 5d), because the larger 4d and 5d orbitals interact more strongly with the ligands.
Os is a 5d metal, Mo 4d, Cr and Fe 3d, so $[\text{Os(H}_2\text{O)}_6]^{3+}$ has the largest $\Delta_0$.
Answer: (D). | ||||
| 72 | 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 $\text{CH}_4$, $\text{O}_3$ and water vapour absorb infrared radiation and are greenhouse gases. $\text{N}_2$, a homonuclear diatomic molecule, has no changing dipole moment, does not absorb IR, and does not cause the greenhouse effect.
Answer: (D). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctHard | |
A hard single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A is the cyclopentyl cation (a plain secondary carbocation).
B is a tetrahydrofuran-2-yl cation: the positive carbon is next to the ring oxygen, whose lone pair delocalises the charge ($\text{C}^+\text{–O} \leftrightarrow \text{C=O}^+$, an oxocarbenium ion). This +R effect makes B the most stable.
C is a tetrahydrofuran-3-yl cation: the oxygen is one carbon away, so it cannot donate by resonance and only withdraws by its −I effect. C is the least stable.
So the true order is B > A > C.
None of the four options states this.
No option is correct as printed. | ||||
| 74 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | StatementsEasy | |
A easy statements question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Statement I: the π bond (sideways overlap) is weaker than a σ bond, so alkenes are less stable and more reactive than alkanes: correct.
Statement II: the C=C bond enthalpy (about 681 kJ mol$^{-1}$) is greater than that of the C–C single bond (about 348 kJ mol$^{-1}$): correct.
Answer: (A). | ||||
| 75 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers A is 1-methyl-4-methylenecyclohexane; B is 4-methylcyclohexane-1-carbaldehyde, so the $=\text{CH}_2$ carbon must become $\text{–CHO}$ without losing a carbon.
Hydroboration–oxidation adds water anti-Markovnikov: $\text{ring–CH}_2\text{OH}$ (a primary alcohol). PCC then oxidises it to the aldehyde and stops there: B.
(A) PCC does not attack the alkene; (B) ozonolysis gives 4-methylcyclohexanone and HCHO; (D) HBr adds Markovnikov, giving a tertiary bromide and then a tertiary alcohol that $\text{K}_2\text{Cr}_2\text{O}_7$ does not oxidise to an aldehyde.
Answer: (C). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Hex-4-en-2-ol: $\text{CH}_3\text{CH(OH)CH}_2\text{CH=CHCH}_3$.
PCC oxidises the secondary alcohol: A = hex-4-en-2-one, $\text{CH}_3\text{COCH}_2\text{CH=CHCH}_3$.
NaOI (iodoform reaction of the methyl ketone): B = sodium pent-3-enoate, $\text{CH}_3\text{CH=CHCH}_2\text{COONa}$ (+ $\text{CHI}_3$).
Soda lime decarboxylates it: C = $\text{CH}_3\text{CH=CHCH}_3$, 2-butene.
Answer: (C). | ||||
| 77 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines One carbon must be added. $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{SOCl}_2} \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \xrightarrow{\text{KCN}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} \xrightarrow{\text{H}_2/\text{Ni or Na(Hg)/C}_2\text{H}_5\text{OH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2$.
(B) and (D) add no carbon; (C) and (D) with $\text{CH}_3\text{NH}_2$ would give N-methyl (secondary) amines.
Answer: (A). | ||||
| 78 | 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 Nylon 6,6 (hexamethylenediamine + adipic acid), Dacron (ethylene glycol + terephthalic acid) and silicones (from hydrolysis and condensation of alkylchlorosilanes) are condensation polymers.
Buna-N is an addition copolymer of 1,3-butadiene and acrylonitrile.
Answer: (C). | ||||
| 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 The molecule has a 5-methylimidazole ring joined through $\text{–CH}_2\text{SCH}_2\text{CH}_2\text{NH–}$ to an N-cyano-N′-methylguanidine group. This is cimetidine, an antihistamine ($\text{H}_2$-receptor blocker) used as an antacid.
Ranitidine has a furan ring and a nitroethenediamine group instead; Seldane (terfenadine) and codeine are quite different.
Answer: (C). | ||||
| 80 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter A green flame with a blue centre is characteristic of copper(II). Strontium gives crimson red, calcium brick red, and barium an apple (yellowish) green without the blue centre.
Answer: (A). | ||||
| 81 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry At the same $P$ and $V$, $nT$ is the same for both gases: $n_A \times 300 = \frac{0.2}{2.0} \times 200 = 20$, so $n_A = \frac{1}{15}$ mol.
$M_A = \frac{3.0}{1/15} = 45$ g mol$^{-1}$.
Answer: 45. | ||||
| 82 | SolutionsConcentration terms | Class 12: Solutions | NumericalMedium | |
A medium numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Henry's law: $x_{\text{CO}_2} = \frac{p}{K_H} = \frac{0.835}{1670} = 5.0 \times 10^{-4}$.
1 L of water is about 1000 g, i.e. $\frac{1000}{18} \approx 55.5$ mol. Since the mole fraction is small, $n_{\text{CO}_2} \approx x_{\text{CO}_2} \times n_{\text{water}} = 5.0 \times 10^{-4} \times 55.5 = 2.775 \times 10^{-2}$ mol.
Mass $= 2.775 \times 10^{-2} \times 44 = 1.221$ g $= 1221 \times 10^{-3}$ g (taking O = 16).
Answer: 1221. | ||||
| 83 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Total moles at equilibrium: $n = \frac{PV}{RT} = \frac{2.46 \times 200}{0.082 \times 600} = \frac{492}{49.2} = 10$.
If $y$ mol of $\text{PCl}_5$ dissociate: $(5 - y) + y + y + 2 = 10$, so $y = 3$: $\text{PCl}_5$ 2 mol, $\text{PCl}_3$ 3 mol, $\text{Cl}_2$ 3 mol.
Each mole contributes a partial pressure $\frac{RT}{V} = \frac{0.082 \times 600}{200} = 0.246$ atm.
$K_p = \frac{(3 \times 0.246)^2}{2 \times 0.246} = \frac{9 \times 0.246}{2} = 1.107$ atm $= 1107 \times 10^{-3}$.
Answer: 1107. | ||||
| 84 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell constant $G^* = \kappa R = 0.152 \times 10^{-3} \times 1750 = 0.266$ cm$^{-1}$ $= 266 \times 10^{-3}$ cm$^{-1}$.
Answer: 266. | ||||
| 85 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Acetic acid adsorbed $= 0.200 \text{ L} \times (0.2 - 0.1)$ M $= 0.02$ mol $= 0.02 \times 60 = 1.2$ g.
Per gram of charcoal: $\frac{1.2}{0.6} = 2$ g.
Answer: 2. | ||||
| 86 | Isolation of MetalsOres and concentration | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Baryte is $\text{BaSO}_4$ (a sulphate). Galena $\text{PbS}$, zinc blende $\text{ZnS}$ and copper pyrites $\text{CuFeS}_2$ are sulphides.
Number of sulphide minerals: 3.
Answer: 3. | ||||
| 87 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In acid, manganate disproportionates: $3\text{MnO}_4^{2-} + 4\text{H}^+ \to 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}$.
Mn(VI) goes to Mn(VII) in $\text{MnO}_4^-$ and Mn(IV) in $\text{MnO}_2$. Difference $= 7 - 4 = 3$.
Answer: 3. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques 22.400 mL of $\text{N}_2$ at STP is $\frac{22.4}{22400} = 10^{-3}$ mol $= 0.028$ g.
% N $= \frac{0.028}{0.2} \times 100 = 14$.
Answer: 14. | ||||
| 89 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | NumericalMedium | |
A medium numerical question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The reactant is 4-chloro-3-methylpent-1-ene, $\text{CH}_3\text{–CHCl–CH(CH}_3\text{)–CH=CH}_2$, a secondary chloride.
In aqueous NaOH it reacts by $S_N1$: the secondary carbocation undergoes a 1,2-hydride shift from C-3 to give a tertiary, allylic cation $\text{CH}_3\text{CH}_2\text{–C}^+\text{(CH}_3\text{)–CH=CH}_2 \leftrightarrow \text{CH}_3\text{CH}_2\text{C(CH}_3\text{)=CH–CH}_2^+$, which water captures to give 3-methylpent-1-en-3-ol or 3-methylpent-2-en-1-ol (without rearrangement, 3-methylpent-4-en-2-ol).
Whichever alcohol forms, P keeps exactly one C=C double bond, so it has 2 π electrons.
Answer: 2. | ||||
| 90 | BiomoleculesProteins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The name lists five amino acid residues: Ala–Gly–Leu–Ala–Val. A chain of $n$ residues has $n - 1$ peptide (–CO–NH–) linkages: $5 - 1 = 4$.
Answer: 4. | ||||
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