JEE (Main) · 2022
JEE (Main) 2022 — 27 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
- Integral Calculus3
- Conic Sections3
- Sets, Relations and Functions2
11 easy · 16 medium · 3 hard
Physics
30 Q
- Properties of Bulk Matter3
- Current Electricity3
- Kinematics2
20 easy · 10 medium · 0 hard
Chemistry
30 Q
- Solutions2
- p-Block Elements2
- d- and f-Block Elements2
23 easy · 7 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $R_1$: reflexive ($a^2 \ge 0$) and symmetric ($ab = ba$), but not transitive: $1 \cdot 0 \ge 0$ and $0 \cdot (-1) \ge 0$, yet $1 \cdot (-1) < 0$. So $R_1$ is not an equivalence relation.
$R_2$: reflexive and transitive, but not symmetric ($2 \ge 1$ while $1 \not\ge 2$). So $R_2$ is not one either.
Answer: (D). | ||||
| 2 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $(f+g)(a) = f(a) + a + 1$.
$(f+g)(4) = 2 + 5 = 7$ and $(f+g)(5) = 1 + 6 = 7$, so $f+g$ is not one-one.
For $a \ge 2$, $f(a) \ge 1$, so $(f+g)(a) \ge 4$: the values 1, 2, 3 are never taken, so $f+g$ is not onto.
Answer: (D). | ||||
| 3 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations The sum of squared distances from the points $0$, $3$ and $6i$ is least at their centroid: $z_0 = \frac{0 + 3 + 6i}{3} = 1 + 2i$.
$v_0 = |1+2i|^2 + |-2+2i|^2 + |1-4i|^2 = 5 + 8 + 17 = 30$.
$z_0^2 = -3 + 4i$, so $2z_0^2 = -6 + 8i$. $\bar{z}_0 = 1 - 2i$, $\bar{z}_0^2 = -3 - 4i$, $\bar{z}_0^3 = (-3-4i)(1-2i) = -11 + 2i$.
$2z_0^2 - \bar{z}_0^3 + 3 = -6 + 8i + 11 - 2i + 3 = 8 + 6i$, whose modulus squared is 100.
Total $= 100 + 30^2 = 1000$.
Answer: (A). | ||||
| 4 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants tr $A = -4$, $\det A = -5 + 4 = -1$, so by Cayley–Hamilton $A^2 + 4A - I = O$, i.e. $A^2 = -4A + I$.
$\alpha A^2 + \beta A = (\beta - 4\alpha)A + \alpha I = 2I$ gives $\alpha = 2$ and $\beta = 4\alpha = 8$.
$\alpha + \beta = 10$.
Answer: (D). | ||||
| 5 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $2021 = 7 \cdot 288 + 5 \equiv -2$ and $2022 \equiv -1 \pmod 7$.
$(-2)^{2022} = 2^{2022} = 8^{674} = (7+1)^{674} \equiv 1$, and $(-1)^{2021} = -1$.
Sum $\equiv 1 - 1 = 0$.
Answer: (A). | ||||
| 6 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{S_5}{S_9} = \frac{5(a + 2d)}{9(a + 4d)} = \frac{5}{17}$, so $17a + 34d = 9a + 36d$, i.e. $d = 4a$.
$a_{15} = a + 14d = 57a$, and $110 < 57a < 120$ with $a$ natural gives $a = 2$, $d = 8$.
$S_{10} = 5(2 \cdot 2 + 9 \cdot 8) = 5 \times 76 = 380$.
Answer: (B). | ||||
| 7 | 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 As $x \to -1^-$: $[x] = -2$, $\sin(-\pi) = 0$, and $2 - x \to 3^+$ so $[2-x] = 3$: the limit is 3.
As $x \to -1^+$: $[x] = -1$, $\sin\left(-\frac{\pi}{2}\right) = -1$, and $2 - x \to 3^-$ so $[2-x] = 2$: the limit is $2 - a$.
Equal limits give $a = -1$.
$\int_0^4 -\sin\frac{\pi[x]}{2}\,dx = -(0 + 1 + 0 - 1) = 0$ (unit intervals with $[x] = 0, 1, 2, 3$).
$\int_0^4 [2 - x]\,dx = 1 + 0 + (-1) + (-2) = -2$ (on $(0,1), (1,2), (2,3), (3,4)$).
So $\int_0^4 f(x)\,dx = -2$.
Answer: (B). | ||||
| 8 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $g(x) = \frac{8\sin x - \sin 2x}{x} = \frac{2\sin x\,(4 - \cos x)}{x}$. On $\left[\frac{\pi}{4}, \frac{\pi}{3}\right]$ it decreases ($g(\pi/4) \approx 5.93$, $g(\pi/3) \approx 5.79$), and the interval has length $\frac{\pi}{12}$.
So $\frac{\pi}{12}g\left(\frac{\pi}{3}\right) < I < \frac{\pi}{12}g\left(\frac{\pi}{4}\right)$, i.e. $\frac{7\sqrt{3}}{8} < I < \frac{4\sqrt{2} - 1}{3}$, i.e. $1.516 < I < 1.553$. A numerical evaluation gives $I \approx 1.536$.
This lies in none of the four printed ranges: (C) ends at $\frac{\sqrt{2}}{3}\pi \approx 1.481$ and (A) begins at $\frac{\pi}{2} \approx 1.571$.
No option is correct as printed. | ||||
| 9 | 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 The circle has centre $(-2\sqrt{3}, 0)$ and radius 4. Substituting $y^2 = 8x + 4$: $x^2 + (8 + 4\sqrt{3})x = 0$, so $x = 0$ ($y = \pm 2$); the other root is negative enough to make $y^2 < 0$.
The smaller region lies between the parabola $x = \frac{y^2 - 4}{8}$ on the left and the arc $x = -2\sqrt{3} + \sqrt{16 - y^2}$ on the right, for $-2 \le y \le 2$:
$\int_{-2}^{2}\sqrt{16 - y^2}\,dy = 2\left[\frac{y}{2}\sqrt{16 - y^2} + 8\sin^{-1}\frac{y}{4}\right]_0^2 = 4\sqrt{3} + \frac{8\pi}{3}$,
$\int_{-2}^{2}(-2\sqrt{3})\,dy = -8\sqrt{3}$, and $-\int_{-2}^{2}\frac{y^2 - 4}{8}\,dy = -\frac{1}{8}\left(\frac{16}{3} - 16\right) = \frac{4}{3}$.
Area $= \frac{4}{3} - 4\sqrt{3} + \frac{8\pi}{3} = \frac{1}{3}\left(4 - 12\sqrt{3} + 8\pi\right) \approx 2.78$, clearly the smaller of the two parts.
Answer: (C). | ||||
| 10 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $y' - y = x$ has integrating factor $e^{-x}$: $ye^{-x} = -(x+1)e^{-x} + C$, so $y = Ce^x - x - 1$.
$y_1(0) = 0$ gives $C = 1$; $y_2(0) = 1$ gives $C = 2$.
$y_1 = y_2$ would need $e^x = 2e^x$, which never holds: no point of intersection.
Answer: (A). | ||||
| 11 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The centre is $(5, 7)$. With $P = (2t^2, 4t)$, the tangent is $ty = x + 2t^2$.
Through $(5, 7)$: $7t = 5 + 2t^2$, i.e. $2t^2 - 7t + 5 = 0$, so $t = 1$ or $t = \frac{5}{2}$.
Points: $(2, 4)$ and $\left(\frac{25}{2}, 10\right)$. $A = 2 \cdot \frac{25}{2} = 25$, $B = 4 \cdot 10 = 40$.
$A + B = 65$.
Answer: (D). | ||||
| 12 | 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 Reading the cross product as $-\hat{i} + 9\hat{j} + 12\hat{k}$ (see the note):
$\vec{a}\times\vec{b} = (4 + 5\beta)\hat{i} + (3\beta - 4\alpha)\hat{j} + (-5\alpha - 3)\hat{k}$.
$4 + 5\beta = -1$ gives $\beta = -1$; $-5\alpha - 3 = 12$ gives $\alpha = -3$; then $3\beta - 4\alpha = -3 + 12 = 9$, as required.
$\vec{a} = (-3, 1, -1)$, $\vec{b} - 2\vec{a} = (9, -7, 6)$, $\vec{b} + \vec{a} = (0, -4, 3)$ with $|\vec{b} + \vec{a}| = 5$.
Projection $= \frac{0 + 28 + 18}{5} = \frac{46}{5}$.
Answer: (D). | ||||
| 13 | 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 $\vec{a}\times\vec{b} = (-2 - 5\beta)\hat{i} + (5\alpha - 4)\hat{j} + (2\beta + \alpha)\hat{k}$.
For $\vec{v} = (v_1, v_2, v_3)$, $\vec{v}\times\hat{i} = (0, v_3, -v_2)$, so $(\vec{v}\times\hat{i})\cdot\hat{k} = -v_2 = 4 - 5\alpha = \frac{23}{2}$, giving $\alpha = -\frac{3}{2}$.
$\vec{b}\times\hat{j} = -2\hat{i} + \alpha\hat{k}$, so $\left|\vec{b}\times 2\hat{j}\right| = 2\sqrt{4 + \frac{9}{4}} = 2 \cdot \frac{5}{2} = 5$ (independent of $\beta$).
Answer: (B). | ||||
| 14 | 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 There are 90000 five-digit numbers.
Multiples of 7: from $10003 = 7 \cdot 1429$ to $99995 = 7 \cdot 14285$, i.e. $14285 - 1429 + 1 = 12857$.
Multiples of 35: from $10010 = 35 \cdot 286$ to $99995 = 35 \cdot 2857$, i.e. $2857 - 286 + 1 = 2572$.
Favourable: $12857 - 2572 = 10285$, so $p = \frac{10285}{90000}$ and $9p = 1.0285$.
Answer: (C). | ||||
| 15 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $A$ be the foot of the tower, $AP = x$ and $t = \tan\alpha$.
From $P$: $\tan 2\alpha = \frac{h}{x}$, so $x = \frac{h(1 - t^2)}{2t}$.
After moving $d$ away from the tower: $\tan\alpha = \frac{2h}{x + d}$, so $x + d = \frac{2h}{t}$.
$\frac{h(1 - t^2)}{2t} + \sqrt{7}h = \frac{2h}{t}$ gives $1 - t^2 + 2\sqrt{7}t = 4$, i.e. $t^2 - 2\sqrt{7}t + 3 = 0$, so $t = \sqrt{7} \pm 2$.
Since $2\alpha < 90^\circ$, $t < 1$: $\tan\alpha = \sqrt{7} - 2$.
Answer: (C). | ||||
| 16 | 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 When $p$ is false, both sides of $\Leftrightarrow$ are false, so the statement is true, matching $\sim p$ = T for any $r$.
When $p$ is true, the statement reduces to $r \Leftrightarrow \sim q$, and it must be false (to match $\sim p$ = F). That requires $r$ to have the truth value of $q$ in every case, i.e. $r = q$.
Check of the others: $r = \sim q$ makes it true when $p$ is true; $r = p$ or $\sim p$ fails for one value of $q$.
Answer: (C). | ||||
| 17 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium 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 Perpendicular normals: $(2, k, -5)\cdot(3k, -k, 1) = 6k - k^2 - 5 = 0$, so $k = 1$ or $5$; $k < 3$ gives $k = 1$.
$P$: $(2x + y - 5z - 1) + \lambda(3x - y + z - 5) = 0$. Its $x$-intercept is $\frac{1 + 5\lambda}{2 + 3\lambda} = 1$, so $\lambda = \frac{1}{2}$.
$P$: $\frac{7}{2}x + \frac{1}{2}y - \frac{9}{2}z - \frac{7}{2} = 0$, i.e. $7x + y - 9z = 7$.
Its $y$-intercept is 7.
Answer: (D). | ||||
| 18 | Straight LinesSection formula | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on section formula, from Straight Lines. NCERT: Class 11: Straight Lines Triangles $APB$ and $ABC$ share the height from $A$, so $BP : BC = 4 : 7$.
$P = B + \frac{4}{7}(C - B) = \left(-4 + \frac{8}{7},\ 3 - \frac{32}{7}\right) = \left(-\frac{20}{7}, -\frac{11}{7}\right)$.
$AP$ has slope $\frac{1 + 11/7}{1 + 20/7} = \frac{2}{3}$: $y - 1 = \frac{2}{3}(x - 1)$ meets the $x$-axis at $x = -\frac{1}{2}$.
$AC$ has slope 2: $y - 1 = 2(x - 1)$ meets the $x$-axis at $x = \frac{1}{2}$.
The triangle with vertices $(1, 1)$, $\left(-\frac{1}{2}, 0\right)$, $\left(\frac{1}{2}, 0\right)$ has base 1 and height 1: area $\frac{1}{2}$.
Answer: (C). | ||||
| 19 | CirclesStandard and general form | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Through $(6, 1)$: $36 + 1 - 12g + 6 - 19c = 0$, i.e. $12g + 19c = 43$.
Centre $(g, -3)$ on the line: $g + 6c = 8$.
Then $12(8 - 6c) + 19c = 43$ gives $c = 1$, $g = 2$: the circle is $x^2 + y^2 - 4x + 6y - 19 = 0$.
On $y = 0$: $x^2 - 4x - 19 = 0$, whose roots differ by $\sqrt{16 + 76} = 2\sqrt{23}$.
Answer: (D). | ||||
| 20 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $\int_0^4 (5 - |t - 3|)\,dt = \int_0^3 (2 + t)\,dt + \int_3^4 (8 - t)\,dt = \frac{21}{2} + \frac{9}{2} = 15$. Continuity: $16 + 4b = 15$, so $b = -\frac{1}{4}$.
For $x < 4$: $f'(x) = 2x - \frac{1}{4}$; for $x > 4$: $f'(x) = 5 - |x - 3| = 8 - x$.
(A) $f'(4^-) = \frac{31}{4}$, $f'(4^+) = 4$: not differentiable — true.
(B) $f'(3) + f'(5) = \frac{23}{4} + 3 = \frac{35}{4}$ — true.
(C) $f' < 0$ for $x < \frac{1}{8}$ and for $x > 8$, so $f$ decreases there — NOT true.
(D) $f'$ changes from negative to positive at $\frac{1}{8}$: local minimum — true.
Answer: (C). | ||||
| 21 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalHard | |
A hard numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\cos(\sin^{-1}x) = \sqrt{1 - x^2}$, and $\cot\left(\tan^{-1}\sqrt{1 - x^2}\right) = \frac{1}{\sqrt{1 - x^2}}$.
So the equation is $\cos\left(\sin^{-1}\frac{x}{\sqrt{1 - x^2}}\right) = \sqrt{1 - \frac{x^2}{1 - x^2}} = \sqrt{\frac{1 - 2x^2}{1 - x^2}} = k$ (defined since $|x| < \frac{1}{\sqrt{2}}$).
$x^2 = \frac{1 - k^2}{2 - k^2}$, so the two solutions are $\alpha = -\beta$: $\frac{\alpha}{\beta} = -1$ and $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{2(2 - k^2)}{1 - k^2}$.
Product of the roots of $x^2 - bx - 5 = 0$: $-\frac{2(2 - k^2)}{1 - k^2} = -5$, so $4 - 2k^2 = 5 - 5k^2$, $k^2 = \frac{1}{3}$, and the other root is 5.
Sum: $b = 5 + (-1) = 4$. Then $\frac{b}{k^2} = 12$.
Answer: 12. | ||||
| 22 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Wrong sums: $\sum x = 150$, and $\frac{\sum x^2}{10} - 225 = 15$ gives $\sum x^2 = 2400$.
Corrected: $\sum x = 140$ (mean 14), $\sum x^2 = 2400 - 625 + 225 = 2000$.
Variance $= 200 - 196 = 4$, so the standard deviation is 2.
Answer: 2. | ||||
| 23 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A plane through the second line is $(4ax - y + 5z - 7a) + \mu(2x - 5y - z - 3) = 0$. It contains the first line if it contains $(4, -1, 0)$ and is parallel to $(1, -2, 1)$:
$9a + 1 + 10\mu = 0$ and $4a + 7 + 11\mu = 0$. Eliminating $\mu$: $11(9a + 1) = 10(4a + 7)$, so $a = 1$, $\mu = -1$. (The plane $2x - 5y - z = 3$ alone does not contain $(4, -1, 0)$.)
The plane is $2x + 4y + 6z - 4 = 0$, i.e. $x + 2y + 3z = 2$.
A point of the given line is $(3 + 7s, 2 - s, 3 - 4s)$: $3 + 7s + 4 - 2s + 9 - 12s = 2$ gives $s = 2$, so $P = (17, 0, -5)$.
$\alpha + \beta + \gamma = 12$.
Answer: 12. | ||||
| 24 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $H$: $\frac{y^2}{64} - \frac{x^2}{49} = 1$, vertices $(0, \pm 8)$, transverse axis along the $y$-axis, $e_H^2 = 1 + \frac{49}{64} = \frac{113}{64}$.
So the major axis of $E$ is along the $y$-axis and $b = 8$. $e_E = \frac{1}{2e_H} = \frac{4}{\sqrt{113}}$, so $1 - \frac{a^2}{64} = \frac{16}{113}$ and $a^2 = \frac{64 \cdot 97}{113}$.
$l = \frac{2a^2}{b} = \frac{2 \cdot 64 \cdot 97}{8 \cdot 113} = \frac{1552}{113}$, so $113l = 1552$.
Answer: 1552. | ||||
| 25 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $u = x^2$, $du = 2x\,dx$: $\sin 2u\,\ln(\tan u)\,dy + \left(2y - 2\sqrt{2}\sin\left(u - \frac{\pi}{4}\right)\right)du = 0$.
Divide by $\sin 2u$ and note $\frac{d}{du}\ln\tan u = \frac{2}{\sin 2u}$, and $2\sqrt{2}\sin\left(u - \frac{\pi}{4}\right) = 2(\sin u - \cos u)$:
$\frac{d}{du}\left(y\ln\tan u\right) = \frac{\sin u - \cos u}{\sin u\cos u} = \sec u - \csc u$.
$y\ln\tan u = \ln(\sec u + \tan u) - \ln\tan\frac{u}{2} + C$.
At $u = \frac{\pi}{6}$, $y = 1$: $-\frac{1}{2}\ln 3 = \frac{1}{2}\ln 3 - \ln(2 - \sqrt{3}) + C$, so $C = -\ln 3 + \ln(2 - \sqrt{3})$.
At $u = \frac{\pi}{3}$: $\frac{y}{2}\ln 3 = \ln(2 + \sqrt{3}) + \frac{1}{2}\ln 3 - \ln 3 + \ln(2 - \sqrt{3}) = -\frac{1}{2}\ln 3$, so $y = -1$.
$\left|y\left(\sqrt{\pi/3}\right)\right| = 1$.
Answer: 1. | ||||
| 26 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Differentiating: $5y^4y' - 9y - 9xy' + 2 = 0$, so $y' = \frac{9y - 2}{5y^4 - 9x}$.
Horizontal tangent: $y = \frac{2}{9}$; then the curve gives $\left(\frac{2}{9}\right)^5 - 2x + 2x = 0$, impossible. $M = 0$.
Vertical tangent: $9x = 5y^4$; on the curve $y^5 - 5y^5 + \frac{10}{9}y^4 = 0$, so $y = 0$ (point $(0, 0)$) or $y = \frac{5}{18}$ (point $\left(\frac{5}{9}\left(\frac{5}{18}\right)^4, \frac{5}{18}\right)$). At both, $9y - 2 \ne 0$, so the tangent is vertical. $N = 2$.
$M + N = 2$.
Answer: 2. | ||||
| 27 | 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 $f(x) \ge -\frac{9}{8}$, so the condition is $2n^2 - n - 801 \le 0$, i.e. $\frac{1 - \sqrt{6409}}{4} \le n \le \frac{1 + \sqrt{6409}}{4}$, about $-19.8 \le n \le 20.3$.
So $S = \{-19, -18, \ldots, 20\}$, 40 integers.
$\sum n = 20$, $\sum n^2 = 2 \cdot \frac{19 \cdot 20 \cdot 39}{6} + 400 = 5340$.
$\sum f(n) = 2(5340) - 20 - 40 = 10620$.
Answer: 10620. | ||||
| 28 | 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 tr$(A^{\mathrm{T}}A) = \sum_{i,j} a_{ij}^2$, which here is the number of non-zero entries.
Six of the nine entries are non-zero: choose them in $\binom{9}{6} = 84$ ways, each $\pm 1$ in $2^6 = 64$ ways.
$84 \times 64 = 5376$.
Answer: 5376. | ||||
| 29 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $(x + 1)^2 + 4(y + 1)^2 = \lambda + 5$, i.e. $\frac{(x+1)^2}{\lambda + 5} + \frac{(y+1)^2}{(\lambda + 5)/4} = 1$.
Latus rectum $= \frac{2b^2}{a} = \frac{(\lambda + 5)/2}{\sqrt{\lambda + 5}} = \frac{\sqrt{\lambda + 5}}{2} = 4$, so $\lambda + 5 = 64$, $\lambda = 59$.
Major axis $l = 2\sqrt{64} = 16$. $\lambda + l = 75$.
Answer: 75. | ||||
| 30 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x + iy$: $x^2 - y^2 + x = 0$ and $2xy - y = 0$.
$y = 0$: $x = 0$ or $-1$. $x = \frac{1}{2}$: $y^2 = \frac{3}{4}$, $y = \pm\frac{\sqrt{3}}{2}$.
$S = \left\{0, -1, \frac{1}{2} \pm \frac{\sqrt{3}}{2}i\right\}$, and the sum of $\mathrm{Re} + \mathrm{Im}$ is $0 - 1 + \frac{1}{2} + \frac{1}{2} + 0 = 0$.
Answer: 0. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Torque has dimensions $[ML^2T^{-2}]$, so its fractional error is $\frac{\Delta M}{M} + 2\frac{\Delta L}{L} + 2\frac{\Delta T}{T} = 5\% + 10\% + 10\% = 25\%$.
Answer: (B). | ||||
| 32 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Take upward as positive. The bullet starts with $v = -100$ m/s and accelerates downward at $10\,\mathrm{m/s^2}$: $v = -100 - 10t$, reaching $-200$ m/s at $t = 10$ s.
It then stops at once and $v = 0$ from 10 s to 20 s.
This is graph (A): a straight line from $-100$ to $-200$ m/s, a sudden jump to zero at 10 s, then zero.
Answer: (A). | ||||
| 33 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The force needed to give the falling sand the belt's velocity is $F = v\frac{dm}{dt} = 5 \times 0.5 = 2.5$ N.
Power $P = Fv = v^2\frac{dm}{dt} = 25 \times 0.5 = 12.5$ W.
Answer: (D). | ||||
| 34 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Relative to the belt the bag starts with speed 2 m/s and friction decelerates it at $\mu g = 4\,\mathrm{m/s^2}$.
Slipping distance on the belt $= \frac{2^2}{2 \times 4} = 0.5$ m.
(Relative to the ground the bag also moves $\frac{2^2}{2 \times 4} = 0.5$ m while it speeds up to 2 m/s.)
Answer: (B). | ||||
| 35 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power A column of height $h$ has potential energy $\rho A g\frac{h^2}{2}$. The common final level is 125 cm.
Initial: $\rho A g\frac{1^2 + 1.5^2}{2} = \rho A g(1.625)$. Final: $\rho A g \cdot 2\cdot\frac{1.25^2}{2} = \rho A g(1.5625)$.
Work by gravity = loss of potential energy $= \rho A g(0.0625) = 10^3 \times 16\times 10^{-4} \times 10 \times 0.0625 = 1$ J.
Answer: (B). | ||||
| 36 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation For a circular orbit $E = -\frac{GMm}{2r}$, so $\frac{E_A}{E_B} = \frac{m_A}{m_B}\cdot\frac{r_B}{r_A} = \frac{4}{3}\cdot\frac{4}{3} = \frac{16}{9}$.
Answer: (B). | ||||
| 37 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Thermal resistance $R = \frac{L}{KA}$: $\frac{R_1}{R_2} = \frac{L_1}{L_2}\cdot\frac{K_2}{K_1}\cdot\frac{A_2}{A_1} = 2 \times 9 \times \frac{1}{2} = 9$.
The same heat current flows through both: $\frac{450 - T}{R_1} = \frac{T - 0}{R_2}$, so $450 - T = 9T$ and $T = 45^\circ$C.
Answer: (C). | ||||
| 38 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter A. Newton's law of cooling: for a small difference the rate is proportional to the difference, so doubling it doubles the rate — correct.
B. Stefan's law: $\left(\frac{293}{283}\right)^4 \approx 1.149$, i.e. 1 : 1.15 — correct.
C. $\eta = 1 - \frac{100}{400} = 75\%$ — correct.
D. Quadrupling the difference quadruples the rate, not doubles it — wrong.
Answer: (A). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$ depends only on the gas and its temperature: the same in both (A true, D false).
$p = \frac{NkT}{V}$ with equal $V$ and $T$: $p_1 : p_2 = 1 : 4$ (B true, C false).
Answer: (C). | ||||
| 40 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Displace $q_0$ by $x$ towards one charge. Net force $F = \frac{Qq_0}{4\pi\varepsilon_0}\left[\frac{1}{(a + x)^2} - \frac{1}{(a - x)^2}\right] \approx -\frac{Qq_0}{4\pi\varepsilon_0}\cdot\frac{4x}{a^3} = -\frac{Qq_0}{\pi\varepsilon_0 a^3}x$.
$\omega^2 = \frac{Qq_0}{\pi\varepsilon_0 m a^3}$, so $T = 2\pi\sqrt{\frac{\pi\varepsilon_0 m a^3}{Qq_0}} = \sqrt{\frac{4\pi^3\varepsilon_0 m a^3}{q_0 Q}}$.
Answer: (A). | ||||
| 41 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Current $I = \frac{2E}{R + r_1 + r_2}$. Terminal voltage of the first source: $E - Ir_1 = 0$, so $I = \frac{E}{r_1}$.
$2Er_1 = E(R + r_1 + r_2)$ gives $R = r_1 - r_2$.
Answer: (A). | ||||
| 42 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $T = 2\pi\sqrt{\frac{I}{MB_H}}$, so $M \propto \frac{I}{T^2}$.
$\frac{M_1}{M_2} = \frac{I_1}{I_2}\cdot\frac{T_2^2}{T_1^2} = \frac{3}{2}\cdot\frac{16}{9} = \frac{8}{3}$.
Answer: (B). | ||||
| 43 | Magnetic Effects of Current and MagnetismEarth magnetism | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on earth magnetism, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter In a vertical plane at angle $\beta$ to the meridian the horizontal component is $B_H\cos\beta$, so the apparent dip satisfies $\tan\delta' = \frac{\tan\delta}{\cos\beta}$.
$\tan\delta = \tan 60^\circ\cos 45^\circ = \frac{\sqrt{3}}{\sqrt{2}}$, so $\delta = \tan^{-1}\sqrt{\frac{3}{2}}$.
Answer: (A). | ||||
| 44 | 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 DC: $I^2R = 16 \times 3 = 48$ W. AC: $I_{rms}^2R = \frac{16}{2} \times 2 = 16$ W.
Ratio $48 : 16 = 3 : 1$.
Answer: (B). | ||||
| 45 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $B = \frac{E}{c}$ (along $z$), and the charge moves along $y$, perpendicular to $\vec{B}$.
$\frac{F_E}{F_B} = \frac{qE}{qvB} = \frac{c}{v} = \frac{3\times 10^8}{3 \times 10^7} = 10$.
Answer: (C). | ||||
| 46 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Resolving power of a microscope $= \frac{2n\sin\theta}{1.22\lambda}$, proportional to the refractive index $n$ of the medium.
Going from $n = 1$ to $n = 2$ doubles it.
Answer: (B). | ||||
| 47 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Force on the electron $= (-e)(-E_0\hat{i}) = eE_0\hat{i}$, along its motion, so $v = v_0 + \frac{eE_0}{m}t$.
$\lambda = \frac{h}{mv} = \frac{h}{mv_0\left(1 + \frac{eE_0t}{mv_0}\right)} = \frac{\lambda_0}{1 + \frac{eE_0t}{mv_0}}$.
Answer: (D). | ||||
| 48 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{1}{16} = \left(\frac{1}{2}\right)^4$: four half-lives in 30 years, so $T_{1/2} = 7.5$ years.
Answer: (C). | ||||
| 49 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Reading the waveforms interval by interval: Y is high only in the two intervals where A and B are both high (the first and third pulses of A); during the second pulse of A, B is low and Y is low; Y is low whenever either input is low.
That is the truth table of an AND gate.
Answer: (A). | ||||
| 50 | Communication SystemsRange of a TV tower | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on range of a tv tower, from Communication Systems. NCERT: No matching NCERT chapter Range $d = \sqrt{2Rh}$, so $h \propto d^2$. Tripling the range needs $9 \times 100 = 900$ m.
Answer: (D). | ||||
| 51 | Current ElectricityMetre bridge | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity At balance $\frac{S}{R} = \frac{l}{100 - l} = \frac{30}{70}$.
$S = 5600 \times \frac{3}{7} = 2400\,\Omega$.
Answer: 2400. | ||||
| 52 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= 1\text{ mm} - \frac{9}{10}\text{ mm} = 0.1$ mm $= 0.01$ cm.
Zero error: Vernier zero to the right, 4th division coinciding: $+4 \times 0.01 = +0.04$ cm.
Reading: $4.1 + 6 \times 0.01 = 4.16$ cm. Corrected diameter $= 4.16 - 0.04 = 4.12$ cm $= 412 \times 10^{-2}$ cm.
Answer: 412. | ||||
| 53 | 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 $I_R = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi = 5I + 4I\cos\phi$.
At A: $5I$. At B: $5I + 4I \cdot \frac{1}{2} = 7I$. Difference $2I$, so $x = 2$.
Answer: 2. | ||||
| 54 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Lamp: $R = \frac{100^2}{50} = 200\,\Omega$, rated current $\frac{50}{100} = 0.5$ A. For the lamp to work at its rating, $Z = \frac{200}{0.5} = 400\,\Omega$.
$X_C = \sqrt{400^2 - 200^2} = 200\sqrt{3}\,\Omega$.
$C = \frac{1}{2\pi f X_C} = \frac{1}{100\pi \cdot 200\sqrt{3}}$ F $= \frac{10^6}{2\times 10^4\sqrt{3}\,\pi}\,\mu$F $= \frac{50}{\pi\sqrt{3}}\,\mu$F.
So $x = 3$.
Answer: 3. | ||||
| 55 | Current ElectricityDrift velocity and mobility | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity Field in the wire $E = \rho J = \frac{\rho I}{A} = \frac{1.7\times 10^{-8} \times 1}{2\times 10^{-6}} = 8.5 \times 10^{-3}$ V/m.
Force on an electron $= eE = 1.6\times 10^{-19} \times 8.5 \times 10^{-3} = 1.36 \times 10^{-21}$ N $= 136 \times 10^{-23}$ N.
Answer: 136. | ||||
| 56 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Gauss's law on a coaxial cylinder of radius $x$ and length $L$: $E \cdot 2\pi xL = \frac{\rho\pi x^2L}{\varepsilon_0}$, so $E = \frac{\rho x}{2\varepsilon_0}$.
With $x = \frac{2\varepsilon_0}{\rho}$: $E = 1$ V m⁻¹.
Answer: 1. | ||||
| 57 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At the mean position momentum is conserved: $v' = \frac{Mv}{M + m}$.
$A = \frac{v_{max}}{\omega}$ with $\omega = \sqrt{k/\text{mass}}$: $A_1 = v\sqrt{\frac{M}{k}}$ and $A_2 = \frac{Mv}{M + m}\sqrt{\frac{M + m}{k}}$.
$\frac{A_1}{A_2} = \sqrt{\frac{M + m}{M}} = \sqrt{\frac{1.024}{0.9}} = \frac{32}{30} = \frac{16}{15}$.
$\frac{\alpha}{\alpha - 1} = \frac{16}{15}$ gives $\alpha = 16$.
Answer: 16. | ||||
| 58 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The force acts on the narrow face of area $0.60 \times 0.15 = 0.09\,\mathrm{m^2}$: shear stress $= \frac{18\times 10^4}{0.09} = 2 \times 10^6$ Pa.
Shear strain $= \frac{2\times 10^6}{25\times 10^9} = 8 \times 10^{-5}$.
The upper edge is 60 cm above the riveted lower edge: $\Delta x = 0.60 \times 8\times 10^{-5} = 4.8 \times 10^{-5}$ m $= 48\,\mu$m.
Answer: 48. | ||||
| 59 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\alpha = \frac{Fr}{I} = \frac{1.5(12t - 3t^2)}{4.5} = 4t - t^2$.
Starting from rest, $\omega = 2t^2 - \frac{t^3}{3}$, which returns to zero (and then reverses) at $t = 6$ s.
$\theta = \int_0^6\left(2t^2 - \frac{t^3}{3}\right)dt = 144 - 108 = 36$ rad.
Rotations $= \frac{36}{2\pi} = \frac{18}{\pi}$, so $K = 18$.
Answer: 18. | ||||
| 60 | 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 Time of flight $T = \frac{2u_y}{g}$, so equal times mean equal vertical components of launch velocity.
Maximum height $H = \frac{u_y^2}{2g}$ is then the same for both, independent of mass: ratio $1 : 1$, so $x = 1$.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | SolutionsConcentration terms | Class 12: Solutions | Single correctMedium | |
A medium single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Carbon $= 0.108 \times 250 = 27$ g. Glucose $C_6H_{12}O_6$ (180 g mol⁻¹) is $\frac{72}{180} = 40\%$ carbon, so glucose $= \frac{27}{0.4} = 67.5$ g $= 0.375$ mol.
Water $= 250 - 67.5 = 182.5$ g.
Molality $= \frac{0.375}{0.1825} = 2.05 \approx 2.06$ m.
Answer: (B). | ||||
| 62 | Solid StateClassification of solids | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on classification of solids, from Solid State. NCERT: No matching NCERT chapter Statement I: $O_2$ (two unpaired $\pi^*$ electrons), $Cu^{2+}$ ($3d^9$) and $Fe^{3+}$ ($3d^5$) are paramagnetic — weakly attracted and magnetized along the field. Correct.
Statement II: NaCl and $H_2O$ have all electrons paired: diamagnetic, weakly magnetized opposite to the field. Correct.
Answer: (A). | ||||
| 63 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom As the nuclear charge rises, an electron in a given orbital is held more strongly and its energy falls (becomes more negative): R is true.
Li ($Z = 3$) therefore has a lower-energy 2s orbital than H ($Z = 1$), so the 2s of H is higher in energy: A is true, and R is exactly the reason.
Answer: (A). | ||||
| 64 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Easily liquefiable gases — those with HIGHER critical temperatures — are adsorbed more by charcoal (stronger van der Waals forces).
$SO_2$ (critical temperature about 430 K) is adsorbed far more than $CH_4$ (about 190 K): A is correct; R states the opposite and is wrong.
Answer: (C). | ||||
| 65 | SolutionsColligative properties | Class 12: Solutions | Single correctEasy | |
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions Equal elevation of boiling point (same solvent, non-electrolytes) means equal molality.
Taking the same amount of solvent (dilute solutions): $\frac{2}{M_A} = \frac{8}{M_B}$, so $M_B = 4M_A$.
(Strictly, per 98 g and 92 g of water, $M_B \approx 4.26M_A$; (B) is the intended relation.)
Answer: (B). | ||||
| 66 | 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 (A) IE falls down group 1: K < Na < Li — correct.
(B) Rn, below Xe, has a lower IE — correct.
(C) Rb (37, ~403 kJ mol⁻¹) < Sr (38, ~549 kJ mol⁻¹) — correct.
(D) Zn (30) has a stable filled $3d^{10}4s^2$ shell, IE ~906 kJ mol⁻¹, while Ga ($4p^1$) is ~579 kJ mol⁻¹: Ga is LOWER — the statement is incorrect.
Answer: (D). | ||||
| 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 Liquation (Sn), electrolysis (Cu, Zn) and distillation (Zn, Hg) are refining methods.
Calcination is a step in converting the concentrated ore to oxide, and leaching is a method of concentrating the ore; neither refines a metal.
Answer: (A). | ||||
| 68 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Statement I: $H_2O_2$ oxidises in acid (e.g. $2Fe^{2+} \to 2Fe^{3+}$) and in base (e.g. $2Fe^{2+} \to 2Fe^{3+}$, $Mn^{2+} \to Mn^{4+}$). True.
Statement II: density of $H_2O_2$ at 298 K is about 1.44 g cm⁻³, higher than that of $D_2O$ (about 1.11 g cm⁻³). False.
Answer: (C). | ||||
| 69 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Statement I: $BeCl_2$ and $AlCl_3$ are Cl-bridged and soluble in organic solvents, but they are electron-deficient and act as Lewis ACIDS, not bases. False.
Statement II: $Be(OH)_2 + 2OH^- \to [Be(OH)_4]^{2-}$ and $Al(OH)_3 + OH^- \to [Al(OH)_4]^-$ — both amphoteric. True.
Answer: (D). | ||||
| 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 Pyrophosphorous acid $H_4P_2O_5$ (5 O), hypophosphoric acid $H_4P_2O_6$ (6 O), phosphoric acid $H_3PO_4$ (4 O), pyrophosphoric acid $H_4P_2O_7$ (7 O).
Answer: (D). | ||||
| 71 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Statement I: $Fe^{3+}$ and acidified dichromate do oxidise $I^-$ to $I_2$, but neutral (or faintly alkaline) $KMnO_4$ oxidises $I^-$ to iodate, $IO_3^-$, not $I_2$. False.
Statement II: manganate $MnO_4^{2-}$ ($d^1$) is paramagnetic, but its Mn–O $\pi$ bonding is $p\pi - d\pi$, not $p\pi - p\pi$. False.
Answer: (B). | ||||
| 72 | 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 $Mn_2O_7$ is $O_3Mn{-}O{-}MnO_3$: two tetrahedra sharing one bridging oxygen. Each Mn has three terminal Mn=O bonds: $2 \times 3 = 6$.
Answer: (C). | ||||
| 73 | 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 Excess sulphate: laxative effect (III). Excess nitrate: methemoglobinemia, "blue baby" syndrome (I). Lead above 50 ppb: damage to kidneys, liver and reproductive system (IV). Fluoride above 2 ppm: brown mottling of teeth (II).
Answer: (B). | ||||
| 74 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons R is correct: an aromatic system must be cyclic, planar, fully conjugated and have $(4n + 2)\,\pi$ electrons.
[6] Annulene (benzene) is aromatic. [8] Annulene (cyclooctatetraene) is tub-shaped and non-aromatic.
[10] Annulene, although it has 10 $\pi$ electrons, cannot be planar: the inward hydrogens (shown) clash, or the all-cis form has too much angle strain. It is non-aromatic, so A is wrong in calling it aromatic.
Answer: (D). | ||||
| 75 | 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 HCl on heating replaces the benzylic OH (a reactive, alcoholic OH) by Cl: A is 4-(chloromethyl)phenol. The phenolic OH, attached to the ring, is not replaced.
NaI (in acetone) exchanges the benzylic Cl for I by $S_N2$ (Finkelstein): B is 4-(iodomethyl)phenol, $HOC_6H_4CH_2I$.
Answer: (A). | ||||
| 76 | 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 Phenol–formaldehyde (linear) resin: Novolac (II). Butadiene–styrene copolymer: Buna-S (III). Glycol + phthalic acid: Glyptal (I). Glycol + terephthalic acid: Dacron (IV).
Answer: (B). | ||||
| 77 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Furfural is the dehydration product of a pentose (hexoses give hydroxymethylfurfural). With resorcinol (Seliwanoff's test) ketoses dehydrate fast and colour quickly; aldoses dehydrate slowly and colour only after some time.
A slow-reacting pentose giving furfural is an aldopentose. Tetroses cannot form furfural, and oxalic acid is not a sugar.
Answer: (A). | ||||
| 78 | 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 A is morphine: narcotic analgesic (III).
B is chloroxylenol (4-chloro-3,5-dimethylphenol): antiseptic (IV).
C is phenelzine ($C_6H_5CH_2CH_2NHNH_2$): anti-depressant (I).
D is saccharin: about 550 times sweeter than cane sugar (II).
Answer: (C). | ||||
| 79 | 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 % Br $= \frac{80}{188} \times \frac{0.36}{0.45} \times 100 = 0.4255 \times 0.8 \times 100 = 34.04\%$.
Answer: (A). | ||||
| 80 | 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 Benzenesulphonyl chloride is Hinsberg's reagent (III). The Hoffmann bromamide degradation goes through an isocyanate (IV). The carbylamine reaction is a test for primary amines (I). Hoffmann elimination gives the less substituted alkene: anti-Saytzeff (II).
Answer: (C). | ||||
| 81 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \to 2Cr^{3+} + 6Fe^{3+} + 7H_2O$.
Dichromate $= 20 \times 0.02 = 0.4$ mmol, so $Fe^{2+} = 2.4$ mmol in 10 mL: $0.24$ M $= 24 \times 10^{-2}$ M.
Answer: 24. | ||||
| 82 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Runs 1 and 2 keep $p_{H_2}$ fixed. Halving $p_{NO}$ ($40.0 \to 20.1$) cuts the rate from 0.135 to 0.033, a factor of about 4.1.
$\left(\frac{40.0}{20.1}\right)^n = \frac{0.135}{0.033}$, i.e. $2^n \approx 4$, so $n = 2$.
(Runs 3 and 4 show first order in $H_2$.)
Answer: 2. | ||||
| 83 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Odd-electron species are paramagnetic: $KO_2$ (superoxide $O_2^-$, one unpaired electron), $NO_2$ (17 valence electrons), $ClO_2$ (19), NO (11).
$Na_2O$ ($O^{2-}$), $N_2O$, $SO_2$ and $Cl_2O$ have all electrons paired.
Answer: 4. | ||||
| 84 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $C_v = C_p - R = 20.785 - 8.314 = 12.471$ J K⁻¹ mol⁻¹.
$\Delta U = nC_v\Delta T$: $n = \frac{5000}{12.471 \times 200} = 2.005 \approx 2$.
Answer: 2. | ||||
| 85 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Bond orders: $CN^-$ (14 electrons) 3; $NO^+$ (14) 3; $O_2$ (16) 2; $O_2^+$ (15) 2.5; $O_2^{2+}$ (14) 3.
Three species — $CN^-$, $NO^+$, $O_2^{2+}$ — share the bond order 3.
Answer: 3. | ||||
| 86 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Solubility $= 2.34 \times 10^{-2}$ g L⁻¹ $= \frac{2.34\times 10^{-2}}{78} = 3 \times 10^{-4}$ mol L⁻¹.
$K_{sp} = s(2s)^2 = 4s^3 = 4 \times 27 \times 10^{-12} = 1.08 \times 10^{-10} = 0.0108 \times 10^{-8}$.
In the printed units (× 10⁻⁸) the answer is 0.0108 (0 to the nearest integer); if the scale was meant as × 10⁻¹⁰ it is 1.08 (1 to the nearest integer).
Answer: 0.0108. | ||||
| 87 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A 1:1 electrolyte means one ionisable chloride: $[Co(NH_3)_4Cl_2]Cl$.
In Werner's theory the primary (ionisable) valency is the oxidation state of the metal: Co is +3 here, so the primary valency is 3 (secondary valency, the coordination number, is 6).
Answer: 3. | ||||
| 88 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions $2MnO_4^- + 5H_2C_2O_4 + 6H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$.
Carbon goes from +3 in oxalic acid to +4 in $CO_2$: a change of 1 for each carbon atom.
Answer: 1. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Optical purity $= \frac{\text{observed rotation}}{\text{rotation of pure enantiomer}} \times 100 = \frac{12.6}{30} \times 100 = 42\%$.
Answer: 42. | ||||
| 90 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | NumericalEasy | |
A easy numerical question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Overall yield is the product of the step yields: $0.60 \times 0.50 = 0.30$, i.e. 30%.
Answer: 30. | ||||
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