JEE (Main) · 2022
JEE (Main) 2022 — 25 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
- Integral Calculus3
- Applications of Derivatives3
- Conic Sections3
4 easy · 23 medium · 3 hard
Physics
30 Q
- Kinematics3
- Laws of Motion3
- Magnetic Effects of Current and Magnetism3
26 easy · 4 medium · 0 hard
Chemistry
30 Q
- Biomolecules3
- Equilibrium2
- Electrochemistry2
17 easy · 11 medium · 2 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $|x+1| < 2 \Rightarrow A = (-3, 1)$; $|x-1| \geq 2 \Rightarrow B = (-\infty, -1] \cup [3, \infty)$.
$A - B = A \cap (-1, 3) = (-1, 1)$: true.
$B - A = (-\infty, -3] \cup [3, \infty) = \mathbb{R} - (-3, 3)$, not $\mathbb{R} - (-3, 1)$ (the points of $[1, 3)$ are not in $B$): false.
$A \cap B = (-3, -1]$: true.
$A \cup B = (-\infty, 1) \cup [3, \infty) = \mathbb{R} - [1, 3)$: true.
Answer: (B). | ||||
| 2 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations A repeated root needs discriminant zero: $4b^2 = 60a$, so $a = \frac{b^2}{15}$, and the root is $\alpha = \frac{b}{a} = \frac{15}{b}$.
$\alpha$ is a root of $x^2 - 2bx + 21 = 0$: $\frac{225}{b^2} - 30 + 21 = 0 \Rightarrow b^2 = 25$.
Then $\alpha + \beta = 2b$, $\alpha\beta = 21$: $\alpha^2 + \beta^2 = 4b^2 - 42 = 100 - 42 = 58$.
Answer: (B). | ||||
| 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 Taking conjugates of $\bar{z}_1 = i\bar{z}_2$: $z_1 = -iz_2$, so $\arg z_1 = \arg z_2 - \frac{\pi}{2}$.
$\arg\left(\frac{z_1}{\bar{z}_2}\right) = \arg z_1 + \arg z_2 = \pi$.
Hence $2\arg z_2 - \frac{\pi}{2} = \pi$, giving $\arg z_2 = \frac{3\pi}{4}$ and $\arg z_1 = \frac{\pi}{4}$.
Options (A) and (B) give values of $\arg z_2$ that do not occur; $\arg z_1 = \frac{\pi}{4}$.
Answer: (C). | ||||
| 4 | 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 $\Delta = \begin{vmatrix}-k&3&-14\\-15&4&-k\\-4&1&3\end{vmatrix} = -k(12+k) - 3(-45-4k) - 14(-15+16) = 121 - k^2$.
For $k \neq \pm 11$, $\Delta \neq 0$ and the system has a unique solution.
$k = 11$: from the third equation $y = 4 + 4x - 3z$; the first gives $x - 23z = 13$ and the second $x - 23z = -13$: inconsistent.
$k = -11$: the first gives $23(x - z) = 13$ and the second $x - z = -13$: inconsistent.
So the system is consistent exactly for $k \in \mathbb{R} - \{-11, 11\}$.
Answer: (D). | ||||
| 5 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $s = \sin x \to 1$. Rationalising, the bracket is $\frac{(2s^2+3s+4) - (s^2+6s+2)}{\sqrt{2s^2+3s+4} + \sqrt{s^2+6s+2}} = \frac{(s-1)(s-2)}{\sqrt{2s^2+3s+4} + \sqrt{s^2+6s+2}}$, whose denominator $\to 3 + 3 = 6$.
$\tan^2 x = \frac{s^2}{(1-s)(1+s)}$.
The product is $\frac{-s^2(s-2)}{(1+s)\left(\sqrt{2s^2+3s+4} + \sqrt{s^2+6s+2}\right)} \to \frac{-1 \cdot (-1)}{2 \cdot 6} = \frac{1}{12}$.
Answer: (A). | ||||
| 6 | 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 Write both as $x$ in terms of $y$: $x = \frac{y^2+1}{2}$ and $x = \frac{y^2+3}{4}$. They meet where $2y^2 + 2 = y^2 + 3$, i.e. $y = \pm 1$.
For $|y| < 1$ the second curve lies to the right, so
Area $= \int_{-1}^{1}\left(\frac{y^2+3}{4} - \frac{y^2+1}{2}\right)dy = \frac{1}{4}\int_{-1}^{1}(1 - y^2)\,dy = \frac{1}{4}\cdot\frac{4}{3} = \frac{1}{3}$.
Answer: (A). | ||||
| 7 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The terms form a G.P. with first term $(5+x)^{500}$, ratio $\frac{x}{5+x}$ and 501 terms:
sum $= \frac{(5+x)^{501} - x^{501}}{(5+x) - x} = \frac{(5+x)^{501} - x^{501}}{5}$.
The coefficient of $x^{101}$ is $\frac{1}{5}\,{}^{501}C_{101}\,5^{400} = {}^{501}C_{101}\,5^{399}$.
Answer: (A). | ||||
| 8 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let $S = 1 + 2\cdot 3 + 3\cdot 3^2 + \ldots + 10\cdot 3^9$. Then $3S = 1\cdot 3 + 2\cdot 3^2 + \ldots + 9\cdot 3^9 + 10\cdot 3^{10}$.
Subtracting, $-2S = 1 + 3 + 3^2 + \ldots + 3^9 - 10\cdot 3^{10} = \frac{3^{10} - 1}{2} - 10\cdot 3^{10}$.
So $S = \frac{20\cdot 3^{10} - 3^{10} + 1}{4} = \frac{19\cdot 3^{10} + 1}{4}$.
Answer: (B). | ||||
| 9 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $P$: $(x + 3y - z - 5) + \lambda(2x - y + z - 3) = 0$. Through $(2, 1, -2)$: $2 + \lambda(-2) = 0$, so $\lambda = 1$ and $P$: $3x + 2y - 8 = 0$.
Let $f(x, y, z) = 3x + 2y - 8$:
$X(1, -2, 4)$: $f = -9 < 0$; $Y(5, -1, 2)$: $f = 5 > 0$;
$X + Y = (6, -3, 6)$: $f = 4 > 0$; $Y - X = (4, 1, -2)$: $f = 6 > 0$; $X - Y = (-4, -1, 2)$: $f = -22 < 0$.
(A) $X$ and $X+Y$ have opposite signs: false. (B) $Y$ and $Y-X$ are on the same side: false. (C) $X$ and $Y$ have opposite signs: true. (D) $X+Y$ and $X-Y$ have opposite signs: false.
Answer: (C). | ||||
| 10 | CirclesStandard and general form | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Let the centre be $(h, k)$. Touching the $y$-axis, the radius is $|h|$; touching $x + y = 0$, the radius is $\frac{|h + k|}{\sqrt{2}}$.
$2h^2 = (h + k)^2 \Rightarrow h^2 - k^2 - 2hk = 0$.
Locus: $x^2 - y^2 = 2xy$.
Answer: (D). | ||||
| 11 | Applications of DerivativesRate of change | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on rate of change, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The vessel has radius 7 and height 35, so for the water $h = 5r$ and the slant height is $l = \sqrt{r^2 + 25r^2} = \sqrt{26}\,r$.
Volume $V = \frac{1}{3}\pi r^2 (5r) = \frac{5}{3}\pi r^3$, so $\frac{dV}{dt} = 5\pi r^2\frac{dr}{dt} = 1$.
Wet surface $S = \pi r l = \sqrt{26}\,\pi r^2$, so $\frac{dS}{dt} = 2\sqrt{26}\,\pi r\frac{dr}{dt} = \frac{2\sqrt{26}}{5r}$.
At $h = 10$, $r = 2$: $\frac{dS}{dt} = \frac{2\sqrt{26}}{10} = \frac{\sqrt{26}}{5}$.
Answer: (C). | ||||
| 12 | 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 Using $\cos^2 A - \cos^2 B = -\sin(A+B)\sin(A-B)$ with $A = (n+1)x$, $B = nx$:
$b_{n+1} - b_n = \int_0^{\pi/2} \frac{-\sin(2n+1)x\,\sin x}{\sin x}dx = -\int_0^{\pi/2}\sin(2n+1)x\,dx = -\frac{1 - \cos\frac{(2n+1)\pi}{2}}{2n+1} = -\frac{1}{2n+1}$.
So $b_3 - b_2 = -\frac{1}{5}$, $b_4 - b_3 = -\frac{1}{7}$, $b_5 - b_4 = -\frac{1}{9}$.
These are not in A.P. or G.P. ((A) and (C) fail); their reciprocals $-5, -7, -9$ are in A.P. with common difference $-2$, so (B) fails and (D) holds.
Answer: (D). | ||||
| 13 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $y = vx$: $2x^2\left(v + x\frac{dv}{dx}\right) - 2x^2 v + 3v^2x^2 = 0 \Rightarrow 2x\frac{dv}{dx} = -3v^2$.
$\int\frac{2\,dv}{v^2} = -3\int\frac{dx}{x} \Rightarrow -\frac{2}{v} = -3\ln x - C$, i.e. $\frac{2x}{y} = 3\ln x + C$.
$y(e) = \frac{e}{3}$: $6 = 3 + C$, so $C = 3$.
At $x = 1$: $\frac{2}{y} = 3$, $y(1) = \frac{2}{3}$.
Answer: (B). | ||||
| 14 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $\frac{dx}{dt} = 12(1 + \cos 2t) = 24\cos^2 t$, $\frac{dy}{dt} = 24(1 + \sin t)\cos t$.
$\frac{dy}{dx} = \frac{1 + \sin t}{\cos t} = \tan\left(\frac{\pi}{4} + \frac{t}{2}\right) = \tan\frac{\pi}{3}$, so $\frac{t}{2} = \frac{\pi}{12}$, $t = \frac{\pi}{6}$.
$y_0 = 12\left(1 + \frac{1}{2}\right)^2 = 27$.
Answer: (C). | ||||
| 15 | 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 $2\sin 12^\circ - \sin 72^\circ = \sin 12^\circ - (\sin 72^\circ - \sin 12^\circ) = \sin 12^\circ - 2\cos 42^\circ\sin 30^\circ = \sin 12^\circ - \sin 48^\circ$.
$\sin 12^\circ - \sin 48^\circ = -2\cos 30^\circ\sin 18^\circ = -\sqrt{3}\cdot\frac{\sqrt{5} - 1}{4} = \frac{\sqrt{3}\left(1 - \sqrt{5}\right)}{4}$.
Answer: (D). | ||||
| 16 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(2) = \frac{1}{2}$, $P(4) = \frac{1}{4}$, $P(8) = \frac{1}{8}$, $P(16) = \frac{1}{16}$, and each face marked 32 has probability $\frac{1}{32}$, so $P(32) = \frac{1}{16}$ (the total is 1).
Three numbers from $\{2, 4, 8, 16, 32\}$ adding to 48: $16 + 16 + 16$ and $32 + 8 + 8$ (in 3 orders).
$P = \left(\frac{1}{16}\right)^3 + 3\cdot\frac{1}{16}\cdot\left(\frac{1}{8}\right)^2 = \frac{1}{2^{12}} + \frac{3}{2^{10}} = \frac{1 + 12}{2^{12}} = \frac{13}{2^{12}}$.
Answer: (D). | ||||
| 17 | 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 $(\sim p) \vee q \equiv p \Rightarrow q$, and $(\sim q) \wedge p \equiv \sim(p \Rightarrow q)$.
So the expression is $\sim(p \Rightarrow q) \Rightarrow (p \Rightarrow q) \equiv (p \Rightarrow q) \vee (p \Rightarrow q) \equiv p \Rightarrow q$.
Its negation is $\sim(p \Rightarrow q)$.
Answer: (C). | ||||
| 18 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $x - x^2 = 4 + kx \Rightarrow x^2 + (k-1)x + 4 = 0$; tangency needs $(k-1)^2 = 16$, and $k > 0$ gives $k = 5$.
Then $x^2 + 4x + 4 = 0$, $x = -2$, $y = -6$: $P(-2, -6)$.
$y = \frac{1}{4} - \left(x - \frac{1}{2}\right)^2$, so $V\left(\frac{1}{2}, \frac{1}{4}\right)$.
Slope of $PV = \frac{\frac{1}{4} + 6}{\frac{1}{2} + 2} = \frac{25/4}{5/2} = \frac{5}{2}$.
Answer: (C). | ||||
| 19 | 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 $\cos\frac{15\pi}{4} = \cos\left(4\pi - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, so the argument is $\frac{\frac{1}{\sqrt{2}} - 1}{\frac{1}{\sqrt{2}}} = 1 - \sqrt{2} = -\tan\frac{\pi}{8}$.
$\tan^{-1}(1 - \sqrt{2}) = -\frac{\pi}{8}$.
Answer: (B). | ||||
| 20 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $x^2 + 2(x+1)^2 = 4 \Rightarrow 3x^2 + 4x - 2 = 0$: $x_1 + x_2 = -\frac{4}{3}$, $x_1x_2 = -\frac{2}{3}$.
$(x_1 - x_2)^2 = \frac{16}{9} + \frac{8}{3} = \frac{40}{9}$; on a line of slope 1, $PQ^2 = 2(x_1 - x_2)^2 = \frac{80}{9}$.
$r^2 = \frac{PQ^2}{4} = \frac{20}{9}$, so $(3r)^2 = 20$.
Answer: (A). | ||||
| 21 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A^2 = \begin{pmatrix}2&-2\\1&-1\end{pmatrix} = A$ and $B^2 = \begin{pmatrix}-1&2\\-1&2\end{pmatrix} = B$, so $A^n = A$ and $B^m = B$ for all $n, m \geq 1$.
$nA + mB = \begin{pmatrix}2n - m & -2n + 2m\\ n - m & -n + 2m\end{pmatrix} = I$ needs $n = m$ and $2n - m = 1$, so $n = m = 1$ (then $-n + 2m = 1$ also holds).
Only $(1, 1)$ works.
Answer: 1. | ||||
| 22 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f(g(x)) = \left[2g(x)^2 + 1\right]$.
On $(-1, 0)$: $g = 2x - 3$ runs through $(-5, -3)$, so $g^2$ decreases strictly through $(9, 25)$ and $2g^2 + 1$ through $(19, 51)$. It crosses each integer $20, 21, \ldots, 50$ once: 31 discontinuities.
On $(0, 1)$: $g = 2x + 3 \in (3, 5)$, $2g^2 + 1$ increases through $(19, 51)$: again 31 discontinuities.
At $x = 0$: as $x \to 0^-$, $2g^2 + 1 \to 19^+$, so $fog \to 19$; $fog(0) = [19] = 19$ and to the right $2g^2 + 1 \to 19^+$ as well. So $fog$ is continuous at 0.
Total $= 31 + 31 = 62$.
Answer: 62. | ||||
| 23 | Integral CalculusPartial fractions | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on partial fractions, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{1}{(x^2-1)(x^2-4)} = \frac{1}{3}\left(\frac{1}{x^2 - 4} - \frac{1}{x^2 - 1}\right)$.
$12\int\frac{dx}{(x^2-1)(x^2-4)} = 4\left[\frac{1}{4}\ln\frac{x-2}{x+2} - \frac{1}{2}\ln\frac{x-1}{x+1}\right] = \ln\frac{(x-2)(x+1)^2}{(x+2)(x-1)^2} = F(x)$.
$F(3) = \ln\frac{1\cdot 16}{5\cdot 4} = \ln\frac{4}{5}$, so $F(b) = \ln\frac{49}{40} + \ln\frac{4}{5} = \ln\frac{49}{50}$.
$\frac{(b-2)(b+1)^2}{(b+2)(b-1)^2} = \frac{49}{50}$; $b = 6$ gives $\frac{4\cdot 49}{8\cdot 25} = \frac{49}{50}$. (The left side increases with $b$ for $b > 3$, rising from $\frac{4}{5}$ towards 1, so the root is unique.)
Answer: 6. | ||||
| 24 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = {}^{10}C_r\,2^{10-r}3^r x^{30 - 4r}$. The power $30 - 4r$ is always even, and positive for $r \leq 7$.
Sum of all coefficients $= (2 + 3)^{10} = 5^{10}$. Subtract the terms $r = 8, 9, 10$:
$r = 8$: $45\cdot 4\cdot 3^8 = 60\cdot 3^9$; $r = 9$: $10\cdot 2\cdot 3^9 = 20\cdot 3^9$; $r = 10$: $3^{10} = 3\cdot 3^9$.
Sum $= 5^{10} - 83\cdot 3^9$, so $\beta = 83$.
Answer: 83. | ||||
| 25 | StatisticsMean deviation | Class 11: Statistics | NumericalMedium | |
A medium numerical question on mean deviation, from Statistics. NCERT: Class 11: Statistics Mean $= \frac{n+1}{2}$. With $n = 2k + 1$, the deviations are $k, k-1, \ldots, 1, 0, 1, \ldots, k$, summing to $k(k+1) = \frac{(n-1)(n+1)}{4}$.
M.D. $= \frac{(n-1)(n+1)}{4n} = \frac{5(n+1)}{n} \Rightarrow n - 1 = 20$, $n = 21$.
Answer: 21. | ||||
| 26 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}\times\vec{b}$ is perpendicular to $\vec{b}$: $13 - 1 - 4\lambda = 0$, $\lambda = 3$, $\vec{b} = (1, 1, 3)$, $|\vec{b}|^2 = 11$.
$\vec{b}\times(\vec{a}\times\vec{b}) = 11\vec{a} - (\vec{a}\cdot\vec{b})\vec{b} = 11\vec{a} + 21\vec{b}$.
$\vec{b}\times(13, -1, -4) = (-1, 43, -14)$, so $11\vec{a} = (-1, 43, -14) - 21(1, 1, 3) = (-22, 22, -77)$ and $\vec{a} = (-2, 2, -7)$.
(Check: $\vec{a}\cdot\vec{b} = -21$, $\vec{a}\times\vec{b} = (13, -1, -4)$.)
$\vec{b} - \vec{a} = (3, -1, 10)$, dotted with $(0, -1, 1)$: $1 + 10 = 11$.
$\vec{b} + \vec{a} = (-1, 3, -4)$, dotted with $(1, 0, -1)$: $-1 + 4 = 3$.
Total $= 14$.
Answer: 14. | ||||
| 27 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Three-digit numbers: 900. All digits different: $9\cdot 9\cdot 8 = 648$. All three digits the same: 9.
Exactly one digit appearing twice: $900 - 648 - 9 = 243$.
Answer: 243. | ||||
| 28 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalHard | |
A hard numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $(x-1)(x^2 - 2x - 3) = (x+1)(x-1)(x-3) = x^3 - 3x^2 - x + 3$, positive on $(0, 1)$ and $(3, 4)$, negative on $(1, 3)$.
On $(0, 1)$ and $(3, 4)$: $f = x^3 - 3x^2$, $f' = 3x(x-2)$: negative on $(0, 1)$, positive on $(3, 4)$.
On $(1, 3)$: $f = -x^3 + 3x^2 + 2x - 6$, $f' = -3x^2 + 6x + 2$, which is $5$ at $x = 1$, $-7$ at $x = 3$, and vanishes once at $x = 1 + \sqrt{\frac{5}{3}} \approx 2.29$ (a local maximum).
At $x = 1$, $f$ changes from decreasing to increasing: local minimum. At $x = 3$, from decreasing to increasing: local minimum.
$m = 2$, $M = 1$, $m + M = 3$.
Answer: 3. | ||||
| 29 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $b^2 = a^2(e^2 - 1) = \frac{9}{16}a^2$. The point lies on the curve: $\frac{64}{5a^2} - \frac{144}{25}\cdot\frac{16}{9a^2} = \frac{320 - 256}{25a^2} = 1$, so $a^2 = \frac{64}{25}$, $b^2 = \frac{36}{25}$.
Normal at $(x_1, y_1)$: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$, i.e. $\frac{8\sqrt{5}}{25}x + \frac{15}{25}y = 4$, or $8\sqrt{5}x + 15y = 100$.
$\beta = 15$, $\lambda = 100$, $\lambda - \beta = 85$.
Answer: 85. | ||||
| 30 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $l_1$ passes through $\left(\frac{1}{8}, 0, 0\right)$ and $\left(0, \frac{1}{4\sqrt{2}}, 0\right)$: direction $\left(-\frac{1}{8}, \frac{1}{4\sqrt{2}}, 0\right) \parallel (-1, \sqrt{2}, 0)$.
$l_2$ passes through $\left(-\frac{1}{8}, 0, 0\right)$ and $\left(0, 0, -\frac{1}{6\sqrt{3}}\right)$: direction $\left(\frac{1}{8}, 0, -\frac{1}{6\sqrt{3}}\right) \parallel (3\sqrt{3}, 0, -4)$.
$\vec{n} = (-1, \sqrt{2}, 0)\times(3\sqrt{3}, 0, -4) = (-4\sqrt{2}, -4, -3\sqrt{6})$, $|\vec{n}|^2 = 32 + 16 + 54 = 102$.
Joining vector $\left(-\frac{1}{4}, 0, 0\right)$; its dot with $\vec{n}$ is $\sqrt{2}$.
$d = \frac{\sqrt{2}}{\sqrt{102}}$, $d^{-2} = \frac{102}{2} = 51$.
Answer: 51. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Assertion and reasonEasy | |
A easy assertion and reason question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Equal ranges with the same speed need complementary angles $\theta$ and $90^\circ - \theta$, so $h_1 = \frac{u^2\sin^2\theta}{2g}$ and $h_2 = \frac{u^2\cos^2\theta}{2g}$: R is true.
Then $\sqrt{h_1h_2} = \frac{u^2\sin\theta\cos\theta}{2g}$ and $4\sqrt{h_1h_2} = \frac{2u^2\sin\theta\cos\theta}{g} = \frac{u^2\sin 2\theta}{g} = R$: A is true, and it follows directly from R.
Answer: (A). | ||||
| 32 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $v_P = \frac{dX_P}{dt} = \alpha + 2\beta t$, $v_Q = f - 2t$.
$\alpha + 2\beta t = f - 2t \Rightarrow t = \frac{f - \alpha}{2(1 + \beta)}$.
Answer: (D). | ||||
| 33 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion Static friction supplies the centripetal force: $m\omega^2R \leq \mu mg$, so $R \leq \frac{\mu g}{\omega^2}$.
Answer: (B). | ||||
| 34 | 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 $6a^2 = 24 \Rightarrow a = 2$ m, $V = 8\ m^3$. $\gamma = 3\alpha = 1.5 \times 10^{-3}\ ^\circ C^{-1}$.
$\Delta V = V\gamma\Delta T = 8 \times 1.5 \times 10^{-3} \times 10 = 0.12\ m^3 = 1.2 \times 10^5\ cm^3$.
Answer: (B). | ||||
| 35 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Heat given by the copper cooling to 0°C: $Q = 5000 \times 0.39 \times 500 = 9.75 \times 10^5$ J.
Ice melted: $m = \frac{9.75 \times 10^5}{335} \approx 2910$ g $\approx 2.9$ kg.
Answer: (C). | ||||
| 36 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $C_V = \frac{f}{2}R$ and $C_P = C_V + R = \left(\frac{f}{2} + 1\right)R$, so $\frac{C_P}{C_V} = 1 + \frac{2}{f}$.
Answer: (B). | ||||
| 37 | KinematicsUniform circular motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane In uniform circular motion the acceleration has magnitude $\frac{v^2}{R}$ and points from $P$ towards the centre, i.e. along $-(\cos\theta\,\hat{i} + \sin\theta\,\hat{j})$.
$\vec{a} = -\frac{v^2}{R}\cos\theta\,\hat{i} - \frac{v^2}{R}\sin\theta\,\hat{j}$.
Answer: (C). | ||||
| 38 | 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 A metal slab of thickness $t$ reduces the effective gap to $d - t = \frac{d}{2}$.
$C' = \frac{\varepsilon_0A}{d/2} = 2C$, so $C' : C = 2 : 1$.
Answer: (A). | ||||
| 39 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity $I = \frac{2E}{R + r_1 + r_2}$. The terminal voltage of the second cell is $E - Ir_2 = 0$:
$E(R + r_1 + r_2) = 2Er_2 \Rightarrow R = r_2 - r_1$.
Answer: (A). | ||||
| 40 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsEasy | |
A easy statements 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 Statement I: for a paramagnet $\chi = \frac{C}{T}$ (Curie's law), and for a ferromagnet above its Curie temperature $\chi = \frac{C}{T - T_C}$; both increase as the temperature falls. True.
Statement II: an applied field changes the orbital motion of the electrons so that the induced moments oppose the field; this is diamagnetism. True.
Answer: (A). | ||||
| 41 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B = \mu_0nI$. With $n \to \frac{n}{2}$ and $I \to 2I$: $B' = \mu_0\cdot\frac{n}{2}\cdot 2I = B$.
Answer: (A). | ||||
| 42 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $X_L = \omega L = 3000 \times 0.01 = 30\ \Omega$; $X_C = \frac{1}{\omega C} = \frac{1}{3000 \times 25 \times 10^{-6}} = 13.33\ \Omega$.
$\tan\Phi = \frac{X_L - X_C}{R} = \frac{16.67}{100} \approx 0.17$, so $\Phi = \tan^{-1}(0.17)$.
Answer: (A). | ||||
| 43 | 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 $v = \frac{c}{\sqrt{\mu_r\varepsilon_r}} \Rightarrow \varepsilon_r = \left(\frac{c}{v}\right)^2 = \left(\frac{3 \times 10^8}{2 \times 10^8}\right)^2 = 2.25$.
Answer: (A). | ||||
| 44 | 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 $\sqrt{4} : 1 = 2 : 1$, so $I_{max} \propto (2+1)^2 = 9$ and $I_{min} \propto (2-1)^2 = 1$.
$\frac{I_{max} + I_{min}}{I_{max} - I_{min}} = \frac{10}{8} = \frac{5}{4}$, so $x = 4$.
Answer: (B). | ||||
| 45 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Light from the polaroid is plane polarised. At Brewster's angle the component of the electric field in the plane of incidence is not reflected at all, so light polarised in that plane is transmitted completely into the prism and there is no reflected ray.
(A) describes the geometry at Brewster's angle when a reflected ray exists, but here there is none; (B) and (C) do not happen.
Answer: (D). | ||||
| 46 | 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 For the same kinetic energy $K$, $\lambda = \frac{h}{\sqrt{2mK}} \propto \frac{1}{\sqrt{m}}$.
$m_e < m_p < m_n < m_\alpha$ (the neutron is slightly heavier than the proton), so $\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e$.
Answer: (B). | ||||
| 47 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $R = R_0A^{1/3}$, so $\ln\left(\frac{R}{R_0}\right) = \frac{1}{3}\ln A$: a straight line through the origin with slope $\frac{1}{3}$.
That is the straight line of option (B); (A), (C) and (D) are curves.
Answer: (B). | ||||
| 48 | 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 Each input goes into a NOR gate with both its inputs joined, which acts as a NOT gate: the outputs are $\bar{A}$ and $\bar{B}$.
These feed a NOR gate: $Y = \overline{\bar{A} + \bar{B}} = A\cdot B$ (De Morgan).
The circuit is an AND gate.
Answer: (A). | ||||
| 49 | Communication SystemsElements of a communication system | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on elements of a communication system, from Communication Systems. NCERT: No matching NCERT chapter Facsimile (fax) transmits a static document image: A–I.
A guided medium is a physical channel such as an optical fibre: B–IV.
Frequency modulation is used in local (FM) broadcast radio: C–II.
A digital signal has discrete levels, a rectangular wave: D–III.
Answer: (B). | ||||
| 50 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The figure of merit is the current per division, so the galvanometer current is $I_g = nK$.
With the shunt in parallel, $I_gG = (I - I_g)S \Rightarrow I = I_g\frac{G + S}{S} = \frac{nK(G + S)}{S}$.
Answer: (D). | ||||
| 51 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalEasy | |
A easy numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\frac{\Delta z}{z} = 3\frac{\Delta x}{x} + \frac{1}{2}\frac{\Delta y}{y} = 3(4\%) + \frac{1}{2}(12\%) = 12\% + 6\% = 18\%$.
Answer: 18. | ||||
| 52 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion On a level curve $v_{max} = \sqrt{\mu gr}$, so $v \propto \sqrt{r}$.
$v' = 30\sqrt{\frac{48}{75}} = 30 \times \frac{4}{5} = 24$ m/s.
Answer: 24. | ||||
| 53 | Laws of MotionEquilibrium of concurrent forces | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on equilibrium of concurrent forces, from Laws of Motion. NCERT: Class 11: Laws of Motion On the smooth $60^\circ$ incline, balance along the incline: $F\cos 60^\circ = mg\sin 60^\circ$.
$F = mg\tan 60^\circ = 0.2 \times 10 \times \sqrt{3} = 2\sqrt{3} = \sqrt{12}$ N, so $x = 12$.
Answer: 12. | ||||
| 54 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion With radius $2R$: $I_1 = \frac{2}{5}M(2R)^2 = \frac{8}{5}MR^2$, $I_2 = \frac{1}{2}M(2R)^2 = 2MR^2$, $I_3 = \frac{1}{4}M(2R)^2 = MR^2$, $I_4 = \frac{1}{2}M(2R)^2 = 2MR^2$.
$2(I_2 + I_3) + I_4 = 6MR^2 + 2MR^2 = 8MR^2 = x\cdot\frac{8}{5}MR^2 \Rightarrow x = 5$.
Answer: 5. | ||||
| 55 | GravitationOrbital motion | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Orbital speed $v = \sqrt{\frac{GM}{r}} \propto \frac{1}{\sqrt{r}}$.
$\frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}} = \sqrt{\frac{800}{3200}} = \frac{1}{2}$, so $x = 2$.
Answer: 2. | ||||
| 56 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | NumericalEasy | |
A easy numerical question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory At constant volume $P \propto T$, so $\frac{\Delta P}{P} = \frac{\Delta T}{T}$: $0.004 = \frac{1}{T}$, $T = 250$ K.
Answer: 250. | ||||
| 57 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Volume is conserved: $R = 27^{1/3}r = 3r$; charge $Q = 27q$.
$V = \frac{kQ}{R} = \frac{27kq}{3r} = 9\times 22 = 198$ V.
Answer: 198. | ||||
| 58 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Stretching keeps the volume fixed, so the area halves when the length doubles: $R = \frac{\rho l}{A}$ becomes $\frac{\rho(2l)}{A/2} = 4R$.
Increase $= \frac{4R - R}{R} \times 100\% = 300\%$.
Answer: 300. | ||||
| 59 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $X_L = 2\pi fL = 2\pi \times 50 \times 0.1 = 31.42\ \Omega$; $X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50 \times 10^{-4}} = 31.83\ \Omega$.
$Z = \sqrt{10^2 + (0.41)^2} \approx 10.01\ \Omega$, so $I = \frac{220}{10.01} \approx 22$ A.
Answer: 22. | ||||
| 60 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits From the graph, $I_C$ rises by 5 mA for every 100 µA rise in $I_B$: $\beta = \frac{5 \times 10^{-3}}{100 \times 10^{-6}} = 50$.
Voltage gain $= \beta\frac{R_{out}}{R_{in}} = 50 \times \frac{60}{200} = 15$.
Answer: 15. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | 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 The minimum photon energy is the work function: $E_0 = h\nu_0 = 6.6 \times 10^{-34} \times 1.3 \times 10^{15} = 8.58 \times 10^{-19}$ J.
Answer: (C). | ||||
| 62 | ThermodynamicsHess law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics By Hess's law, $\Delta H = 3\Delta_cH(C_2H_2) - \Delta_cH(C_6H_6) = 3(-1300) - (-3268) = -3900 + 3268 = -632$ kJ $mol^{-1}$.
Answer: (C). | ||||
| 63 | Solutionsvan't Hoff factor | Class 12: Solutions | Single correctMedium | |
A medium single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Molality $m = \frac{0.7/93}{0.042} = 0.1792$ mol kg⁻¹, so $K_fm = 1.86 \times 0.1792 = 0.3333$ K.
$i = \frac{\Delta T_f}{K_fm} = \frac{0.2}{0.3333} = 0.6$.
For dimerisation $2A \rightleftharpoons A_2$, $i = 1 - \frac{\alpha}{2}$: $\alpha = 2(1 - 0.6) = 0.8$, i.e. 80%.
Answer: (D). | ||||
| 64 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $Bi_2S_3 \rightleftharpoons 2Bi^{3+} + 3S^{2-}$: $K_{sp} = (2s)^2(3s)^3 = 108s^5$.
$s^5 = \frac{1.08 \times 10^{-73}}{108} = 10^{-75}$, so $s = 1.0 \times 10^{-15}$ mol $L^{-1}$.
Answer: (A). | ||||
| 65 | BiomoleculesProteins | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Zymase comes from yeast (A–II); diastase from malt (B–III); urease from soyabean (C–IV); pepsin is secreted in the stomach (D–I).
Answer: (B). | ||||
| 66 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Electron gain enthalpies (kJ mol⁻¹, negative values): Cl $\approx -349$, F $\approx -328$, Te $\approx -190$, Po $\approx -174$.
Chlorine's is the most negative (fluorine's small size causes electron–electron repulsion), and down group 16 the values become less negative.
Arranged as signed values: Cl < F < Te < Po.
Answer: (D). | ||||
| 67 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | StatementsMedium | |
A medium statements question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In the electrolytic refining of copper, the impure (blister) copper is the anode and a thin strip of pure copper the cathode, in acidified copper sulphate: Statement II is true.
As the blister copper anode dissolves, the less reactive precious metals in it (silver, gold, platinum) do not dissolve and settle below it as anode mud; so blister copper yields (deposits) the precious metals: Statement I is true.
Answer: (A). | ||||
| 68 | EquilibriumIonic equilibrium | Class 11: Equilibrium | Assertion and reasonEasy | |
A easy assertion and reason question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Water's amphoteric nature is explained by the Brønsted–Lowry concept (it can donate or accept a proton), not the Lewis concept: A is false.
$H_2O + NH_3 \rightleftharpoons OH^- + NH_4^+$ (water donates a proton, an acid); $H_2O + H_2S \rightleftharpoons H_3O^+ + HS^-$ (water accepts a proton, a base): R is true.
Answer: (D). | ||||
| 69 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Standard reduction potentials (V): $Cl_2/Cl^-$ +1.36, $Ag^+/Ag$ +0.80, $I_2/I^-$ +0.54, $Na^+/Na$ −2.71, $Li^+/Li$ −3.04.
Order: A > C > B > D > E.
Answer: (A). | ||||
| 70 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $2Pb(NO_3)_2 \xrightarrow{673\ K} 2PbO + 4NO_2 + O_2$, so A is $NO_2$.
$2NO_2 \rightarrow N_2O_4$ (B). In $N_2O_4$ the two N atoms are bonded directly to each other ($O_2N–NO_2$); all four O atoms are terminal.
Bridging oxygen atoms: 0.
Answer: (A). | ||||
| 71 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Free ions: $V^{2+}$ $3d^3$ (3 unpaired), $Ni^{2+}$ $3d^8$ (2), $Cr^{2+}$ $3d^4$ (4), $Fe^{2+}$ $3d^6$ (4).
Fewest unpaired electrons: $Ni^{2+}$, $\mu = \sqrt{2 \times 4} = 2.83$ BM.
Answer: (B). | ||||
| 72 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Clean water has a BOD below 5 ppm, while highly polluted water may have a BOD of 17 ppm or more: A is correct.
BOD measures the oxygen needed by bacteria to break down only the biodegradable organic matter; the oxygen needed to oxidise both biodegradable and non-biodegradable matter is the COD: R is not correct.
Answer: (C). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Assertion and reasonEasy | |
A easy assertion and reason question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Benzoic acid and naphthalene both dissolve in benzene, so benzene cannot separate them: A is false.
Benzoic acid dissolves in hot water while naphthalene does not, so the mixture is separated by crystallisation from hot water: R is true.
Answer: (D). | ||||
| 74 | Practical Organic ChemistryDetection of elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on detection of elements, from Practical Organic Chemistry. NCERT: No matching NCERT chapter If N or S is present, the extract contains $NaCN$ or $Na_2S$, which would give precipitates with $AgNO_3$ and interfere. Boiling with concentrated $HNO_3$ decomposes them: $NaCN + HNO_3 \rightarrow NaNO_3 + HCN\uparrow$, $Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S\uparrow$.
Answer: (B). | ||||
| 75 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 3,4-Dihydro-2H-pyran is an enol ether. $Br_2$ adds to the C=C to give a bromonium ion; the ring oxygen stabilises positive charge on the carbon next to it (oxocarbenium character), so methanol attacks that carbon (C-2), from the face opposite the bromine.
Product: 3-bromo-2-methoxytetrahydropyran with Br and $OCH_3$ trans (anti addition), with $OCH_3$ on the carbon next to the ring O.
(A) has $OCH_3$ on C-2 and Br on C-3, trans (one wedge, one hash); (B) is the cis isomer; (C) and (D) put Br on C-2.
Answer: (A). | ||||
| 76 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The product is triphenylmethanol, $(C_6H_5)_3COH$. An ester reacts with two equivalents of Grignard reagent: methyl benzoate supplies one phenyl and the carbinol carbon, so the Grignard reagent must be $C_6H_5MgBr$.
$2C_6H_5Br \xrightarrow{2Mg,\ THF} 2C_6H_5MgBr$; $C_6H_5COOCH_3 + 2C_6H_5MgBr \rightarrow (C_6H_5)_3COMgBr \xrightarrow{H_3O^+} (C_6H_5)_3COH$.
A is bromobenzene.
Answer: (B). | ||||
| 77 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) $C_6H_5CHO + CH_3MgBr$ gives 1-phenylethanol; $Na_2Cr_2O_7/H^+$ oxidises it to acetophenone.
(B) $CH_3CHO + C_6H_5MgBr$ gives the same 1-phenylethanol; PCC oxidises it to acetophenone.
(C) An ester with two equivalents of $CH_3MgBr$: the acetophenone formed first reacts again, giving the tertiary alcohol 2-phenylpropan-2-ol, not acetophenone.
(D) $CH_3MgBr + CdCl_2$ gives dimethylcadmium, which converts benzoyl chloride into acetophenone and does not attack the ketone further.
Answer: (C). | ||||
| 78 | 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 Acid protonates the C=C of dihydropyran at C-3, giving a carbocation at C-2 stabilised by the ring oxygen (an oxocarbenium ion). 3-Methylbutan-1-ol adds to that carbon.
The product is the tetrahydropyranyl (THP) ether, 2-(3-methylbutoxy)tetrahydropyran, with the alkoxy group on the carbon next to the ring O.
(A) has it on C-3; (B) and (C) are ring-opened products that are not formed.
Answer: (D). | ||||
| 79 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids An enamine forms when the carbinolamine (or iminium ion) from a secondary amine loses a proton from an α-carbon, so the ketone needs an α-hydrogen.
Pentan-3-one, butan-2-one and cyclohexanone all have α-hydrogens. Di-tert-butyl ketone, $(CH_3)_3C–CO–C(CH_3)_3$, has none (and is also very hindered), so it cannot give an enamine.
Answer: (C). | ||||
| 80 | Chemistry in Everyday LifeAntibiotics and antiseptics | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on antibiotics and antiseptics, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Dettol is a mixture of chloroxylenol and terpineol. Chloroxylenol has a benzene ring ($6\pi$ electrons): A. α-Terpineol has a single ring C=C ($2\pi$ electrons): B.
Answer: (B). | ||||
| 81 | BiomoleculesProteins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The protein must contain at least one glycine unit, so $0.30\%$ of its minimum molar mass is 75 g:
$M_{min} = \frac{75 \times 100}{0.30} = 25000$ g mol⁻¹ $= 25 \times 10^3$ g mol⁻¹.
Answer: 25. | ||||
| 82 | 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 A rigid tank keeps the volume constant, so $\frac{P_1}{T_1} = \frac{P_2}{T_2}$:
$P_2 = 30 \times \frac{318}{300} = 31.8$ atm $\approx 32$ atm.
Answer: 32. | ||||
| 83 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $BeF_2$ (linear), $BF_3$ (trigonal planar) and $CCl_4$ (tetrahedral) are symmetric: zero dipole moment.
$H_2O$ (bent), $NH_3$ (pyramidal) and HCl (polar diatomic) have non-zero dipole moments.
Answer: 3. | ||||
| 84 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For order $n$, $t_{1/2} \propto P_0^{1-n}$.
$\frac{340}{170} = \left(\frac{55.5}{27.8}\right)^{1-n}$, i.e. $2 = 2^{1-n}$ (to a close approximation), so $1 - n = 1$ and $n = 0$.
Half-life proportional to the initial pressure is the mark of a zero-order reaction.
Answer: 0. | ||||
| 85 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $Fe^{3+} + 3e^- \rightarrow Fe$. Moles of Fe $= \frac{0.3482}{56} = 6.218 \times 10^{-3}$.
Charge $Q = 3 \times 6.218 \times 10^{-3} \times 96500 = 1800$ C.
$t = \frac{Q}{I} = \frac{1800}{1.5} = 1200$ s $= 20$ min.
Answer: 20. | ||||
| 86 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $PCl_3$ is hydrolysed stepwise, each step replacing Cl by OH: $PCl_3 \rightarrow P(OH)Cl_2$ (A) $\rightarrow$ … and complete hydrolysis gives phosphorous acid, $PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$.
B is taken as $H_3PO_3$, which exists as $HP(O)(OH)_2$: one H is bonded directly to P and is not ionisable; the two O–H hydrogens are.
Ionisable protons: 2.
Answer: 2. | ||||
| 87 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalHard | |
A hard numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $K_3[Fe(CN)_6]$: $Fe^{3+}$ ($3d^5$) with the strong-field ligand $CN^-$ is low spin, $d^2sp^3$ (inner orbital), with 1 unpaired electron.
$[Co(NH_3)_6]Cl_3$: $Co^{3+}$ ($3d^6$) with $NH_3$ is low spin, $d^2sp^3$ (inner orbital), with 0 unpaired electrons.
The hydrated iron(III) chloride has weak-field water ligands (high spin), not an inner-orbital complex.
Of the two inner-orbital complexes, $CN^-$ stands higher than $NH_3$ in the spectrochemical series, so $[Fe(CN)_6]^{3-}$ has the larger splitting and absorbs at the shorter wavelength.
$\mu = \sqrt{n(n+2)} = \sqrt{3} = 1.73$ BM $\approx 2$.
Answer: 2. | ||||
| 88 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Novolac is the linear phenol–formaldehyde polymer. Each repeating unit comes from one phenol and one formaldehyde with loss of water: $-[C_6H_3(OH)-CH_2]-$, mass $94 + 30 - 18 = 106$ g mol⁻¹.
Number of units $= \frac{963}{106} \approx 9.08 \approx 9$.
Answer: 9. | ||||
| 89 | BiomoleculesProteins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The Biuret test (violet colour with alkaline $CuSO_4$) needs at least two peptide (–CO–NH–) bonds.
Glycine: no peptide bond. Glycylalanine (a dipeptide): one peptide bond only, negative. A tripeptide: two peptide bonds, positive. Biuret itself, $H_2N-CO-NH-CO-NH_2$, gives the test (it is named after it): positive.
Answer: 2. | ||||
| 90 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry At neutralisation, equivalents of acid = equivalents of base: $10 \times 0.1 \times n = 30 \times 0.05 \times 2$, so $n = 3$.
Answer: 3. | ||||
Sit this paper free.
Timed like the exam or chapter by chapter, with a solution for each question. Run by a non-profit trust.