JEE (Main) · 2022
JEE (Main) 2022 — 26 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
- Limits, Continuity and Differentiability2
- Matrices and Determinants2
6 easy · 20 medium · 4 hard
Physics
30 Q
- Laws of Motion2
- Electromagnetic Waves2
- Atoms and Nuclei2
20 easy · 9 medium · 1 hard
Chemistry
30 Q
- Hydrocarbons4
- Alcohols, Phenols and Ethers2
- Amines2
22 easy · 7 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $g(x) = f(x) - \frac{x}{2}$. Then $g(3x) - g(x) = f(3x) - f(x) - \frac{3x}{2} + \frac{x}{2} = x - x = 0$, so $g(x) = g\left(\frac{x}{3}\right) = g\left(\frac{x}{9}\right) = \dots = g\left(\frac{x}{3^n}\right)$.
Letting $n \to \infty$ and using continuity, $g(x) = g(0)$: $g$ is a constant $c$, and $f(x) = \frac{x}{2} + c$.
$f(8) = 4 + c = 7$ gives $c = 3$, so $f(14) = 7 + 3 = 10$.
Answer: (B). | ||||
| 2 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations OB is the hypotenuse, so the right angle is at A and $AB = OA$: $\vec{AB}$ is $\vec{AO}$ turned through $90^\circ$, i.e. $z_2 - z_1 = \pm i z_1$.
$z_2 = (1+2i)(1+i) = -1 + 3i$ or $z_2 = (1+2i)(1-i) = 3 + i$. Since $\mathrm{Re}(z_2) < 0$, $z_2 = -1 + 3i$.
(A) $z_2$ lies in the second quadrant with $\tan = -3$: $\arg z_2 = \pi - \tan^{-1}3$. True.
(B) $z_1 - 2z_2 = 1 + 2i + 2 - 6i = 3 - 4i$, fourth quadrant: $\arg = -\tan^{-1}\frac{4}{3}$. True.
(C) $|z_2| = \sqrt{1 + 9} = \sqrt{10}$. True.
(D) $2z_1 - z_2 = 2 + 4i + 1 - 3i = 3 + i$, $|3 + i| = \sqrt{10} \ne 5$. Not true.
Answer: (D). | ||||
| 3 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}8&1&4\\1&1&1\\\lambda&-3&0\end{vmatrix} = 8(0+3) - 1(0-\lambda) + 4(-3-\lambda) = 12 - 3\lambda$. Infinitely many solutions need $\Delta = 0$: $\lambda = 4$.
From the second equation $z = -x - y$; the first becomes $8x + y - 4x - 4y = -2$, i.e. $4x - 3y = -2$. The third is $4x - 3y = \mu$, so the system is consistent (with infinitely many solutions) only for $\mu = -2$.
Distance of $(4, -2, -\frac{1}{2})$ from $8x + y + 4z + 2 = 0$: $\frac{|32 - 2 - 2 + 2|}{\sqrt{64 + 1 + 16}} = \frac{30}{9} = \frac{10}{3}$.
Answer: (D). | ||||
| 4 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $(A + I)(\mathrm{Adj}A + I) = A\,\mathrm{Adj}A + A + \mathrm{Adj}A + I = (\det A)I + A + \mathrm{Adj}A + I = A + \mathrm{Adj}A$, since $\det A = -1$.
For $A = \begin{pmatrix}a&b\\c&d\end{pmatrix}$, $\mathrm{Adj}A = \begin{pmatrix}d&-b\\-c&a\end{pmatrix}$, so $A + \mathrm{Adj}A = (a + d)I$.
$\det((a+d)I) = (a+d)^2 = 4$, so the trace $a + d = \pm 2$. Of the options, 2.
Answer: (B). | ||||
| 5 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals For $1 \le y \le 3$, $y^a > 0$, so the area is $\int_1^3 y^a\,dy = \frac{3^{a+1} - 1}{a + 1}$.
For $a = 5$: $\frac{729 - 1}{6} = \frac{728}{6} = \frac{364}{3}$. (For $a = 3$ it is 20, and it grows with a.)
Answer: (B). | ||||
| 6 | 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 Product of all terms $= 2^{1 + 2 + \dots + 60} \cdot 4^{1 + 2 + \dots + n} = 2^{1830 + n(n+1)}$.
GM $= 2^{\frac{1830 + n^2 + n}{60 + n}} = 2^{\frac{225}{8}}$, so $8(1830 + n^2 + n) = 225(60 + n)$, i.e. $8n^2 - 217n + 1140 = 0$.
$n = \frac{217 \pm \sqrt{47089 - 36480}}{16} = \frac{217 \pm 103}{16}$; the natural value is $n = 20$.
$\sum_{k=1}^{20} k(20 - k) = 20 \cdot \frac{20 \cdot 21}{2} - \frac{20 \cdot 21 \cdot 41}{6} = 4200 - 2870 = 1330$.
Answer: (C). | ||||
| 7 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Numerator: $\log_e[(1 + x^2)^2 - x^2] = \log_e(1 + x^2 + x^4)$. Denominator: $\sec x - \cos x = \frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x}$.
$\lim_{x \to 0} \frac{\log_e(1 + x^2 + x^4)}{x^2 + x^4} \cdot \frac{x^2 + x^4}{x^2} \cdot \frac{x^2}{\sin^2 x} \cdot \cos x = 1 \cdot 1 \cdot 1 \cdot 1 = 1$.
Continuity needs $k = 1$.
Answer: (A). | ||||
| 8 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Continuity of f at 0: $a = |0 - 4| = 4$. Continuity of g at 0: $0 + 1 = 16 + b$, so $b = -15$.
$(gof)(2) = g(|2 - 4|) = g(2) = (2 - 4)^2 - 15 = -11$.
$(fog)(-2) = f(-2 + 1) = f(-1) = -1 + 4 = 3$.
Sum $= -11 + 3 = -8$.
Answer: (D). | ||||
| 9 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives For $x \le 1$: $f'(x) = 3x^2 - 2x + 10 > 0$ (discriminant $4 - 120 < 0$), so f increases up to $f(1) = 1 - 1 + 10 - 7 = 3$.
For $x > 1$: f decreases, with $\lim_{x \to 1^+} f(x) = -2 + \log_2(b^2 - 4)$.
f has its maximum at $x = 1$ when $-2 + \log_2(b^2 - 4) \le 3$, i.e. $0 < b^2 - 4 \le 32$, i.e. $4 < b^2 \le 36$.
So $b \in [-6, -2) \cup (2, 6]$.
Answer: (C). | ||||
| 10 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $a = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} \frac{2}{1 + (k/n)^2} = 2\int_0^1 \frac{dx}{1 + x^2} = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}$, so $\frac{a}{2} = \frac{\pi}{4}$.
For $x \in (0, 1)$, $f(x) = \sqrt{\tan^2\frac{x}{2}} = \tan\frac{x}{2}$ and $f'(x) = \frac{1}{2}\sec^2\frac{x}{2}$.
$f\left(\frac{\pi}{4}\right) = \tan\frac{\pi}{8} = \sqrt{2} - 1$; $f'\left(\frac{\pi}{4}\right) = \frac{1}{2}\left(1 + (\sqrt{2} - 1)^2\right) = \frac{1}{2}(4 - 2\sqrt{2}) = 2 - \sqrt{2} = \sqrt{2}(\sqrt{2} - 1)$.
So $f'\left(\frac{a}{2}\right) = \sqrt{2}f\left(\frac{a}{2}\right)$.
Answer: (C). | ||||
| 11 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor $e^{\int 2\tan x\,dx} = \sec^2 x$.
$y\sec^2 x = \int \sin x\sec^2 x\,dx = \int \sec x\tan x\,dx = \sec x + C$, so $y = \cos x + C\cos^2 x$.
$y\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{C}{4} = 0$ gives $C = -2$: $y = \cos x - 2\cos^2 x$.
With $t = \cos x \in (0, 1)$, $t - 2t^2$ is greatest at $t = \frac{1}{4}$: $\frac{1}{4} - \frac{1}{8} = \frac{1}{8}$.
Answer: (A). | ||||
| 12 | 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 $(x - 1)^2 + (y - 2)^2 + (x + 2)^2 + (y - 1)^2 = 14$ gives $2x^2 + 2y^2 + 2x - 6y - 4 = 0$, i.e. $x^2 + y^2 + x - 3y - 2 = 0$.
$y = 0$: $x^2 + x - 2 = 0$, $x = 1, -2$: A, B are 3 apart, on either side of O.
$x = 0$: $y^2 - 3y - 2 = 0$, $y = \frac{3 \pm \sqrt{17}}{2}$: C, D are $\sqrt{17}$ apart, on either side of O.
ACBD has perpendicular diagonals AB and CD, so its area is $\frac{1}{2} \cdot 3 \cdot \sqrt{17} = \frac{3\sqrt{17}}{2}$.
Answer: (B). | ||||
| 13 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The line has slope $-1$, so the tangent has slope 1. For $y^2 = 4ax$ with $a = 6$, the point of contact for slope m is $\left(\frac{a}{m^2}, \frac{2a}{m}\right) = (6, 12)$: $\alpha = 6$, $\beta = 12$.
Hyperbola $\frac{x^2}{36} - \frac{y^2}{144} = 1$ at $(10, 16)$ (check: $\frac{100}{36} - \frac{256}{144} = 1$).
Normal: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$: $\frac{36x}{10} + \frac{144y}{16} = 180$, i.e. $2x + 5y = 100$.
$(25, 10)$: 100; $(20, 12)$: 100; $(30, 8)$: 100; $(15, 13)$: $30 + 65 = 95 \ne 100$.
Answer: (D). | ||||
| 14 | 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 Direction of the given line: $(1, 1, -1) \times (1, -2, 3) = (3 - 2, -1 - 3, -2 - 1) = (1, -4, -3)$.
From $(1, 2, 4)$ to $(1, -2, 5)$: $\vec{v} = (0, -4, 1)$, $|\vec{v}|^2 = 17$; $\vec{v} \cdot \vec{d} = 16 - 3 = 13$; $|\vec{d}|^2 = 26$.
Perpendicular distance$^2 = 17 - \frac{13^2}{26} = 17 - \frac{13}{2} = \frac{21}{2}$, so the length is $\sqrt{\frac{21}{2}}$.
Answer: (A). | ||||
| 15 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a} \times \vec{b} = (1\cdot(-\alpha) - (-1)\cdot 1)\hat{i} - (\alpha(-\alpha) - (-1)\cdot 2)\hat{j} + (\alpha - 2)\hat{k} = (1 - \alpha)\hat{i} + (\alpha^2 - 2)\hat{j} + (\alpha - 2)\hat{k}$.
Projection on $\frac{1}{3}(-1, 2, -2)$: $\frac{-(1 - \alpha) + 2(\alpha^2 - 2) - 2(\alpha - 2)}{3} = \frac{2\alpha^2 - \alpha - 1}{3} = 30$.
$2\alpha^2 - \alpha - 91 = 0$, $\alpha = \frac{1 \pm 27}{4}$; with $\alpha > 0$, $\alpha = 7$.
Answer: (D). | ||||
| 16 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $np = \alpha$, $npq = \frac{\alpha}{3}$, so $q = \frac{1}{3}$, $p = \frac{2}{3}$.
$P(X = 1) = n \cdot \frac{2}{3} \cdot \left(\frac{1}{3}\right)^{n-1} = \frac{2n}{3^n} = \frac{4}{243}$, i.e. $\frac{n}{3^n} = \frac{2}{243} = \frac{6}{729}$: $n = 6$.
$P(4) + P(5) = \binom{6}{4}\frac{2^4}{3^6} + \binom{6}{5}\frac{2^5}{3^6} = \frac{240 + 192}{729} = \frac{432}{729} = \frac{16}{27}$.
Answer: (C). | ||||
| 17 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each probability lies in $[0, 1]$: $-\frac{2}{3} \le p \le \frac{4}{3}$; $-6 \le p \le 2$; $-1 \le p \le 1$.
Mutually exclusive, so the sum is at most 1: $\frac{4(2 + 3p) + 3(2 - p) + 12(1 - p)}{24} = \frac{26 - 3p}{24} \le 1$, i.e. $p \ge \frac{2}{3}$.
Together: $\frac{2}{3} \le p \le 1$, so $p_1 + p_2 = 1 + \frac{2}{3} = \frac{5}{3}$.
Answer: (B). | ||||
| 18 | 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 $u = 8^{2\sin^2\theta}$; then $8^{2\cos^2\theta} = \frac{64}{u}$, and $u + \frac{64}{u} = 16$ gives $(u - 8)^2 = 0$, $u = 8$: $\sin^2\theta = \frac{1}{2}$.
$S = \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\right\}$, $n(S) = 4$.
$\sec A\,\mathrm{cosec}\,A = \frac{2}{\sin 2A}$ with $2A = \frac{\pi}{2} + 4\theta$: the term is $\frac{2}{\cos 4\theta}$. For each $\theta \in S$, $4\theta$ is an odd multiple of $\pi$, so $\cos 4\theta = -1$ and the term is $-2$.
Total $= 4 + 4(-2) = -4$.
Answer: (C). | ||||
| 19 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $2\tan^{-1}\frac{1}{5} = \tan^{-1}\frac{2/5}{1 - 1/25} = \tan^{-1}\frac{5}{12}$; $2\tan^{-1}\frac{1}{8} = \tan^{-1}\frac{1/4}{1 - 1/64} = \tan^{-1}\frac{16}{63}$.
$\sec^{-1}\frac{\sqrt{5}}{2}$: $\cos = \frac{2}{\sqrt{5}}$, so it is $\tan^{-1}\frac{1}{2}$.
$\tan^{-1}\frac{5}{12} + \tan^{-1}\frac{16}{63} = \tan^{-1}\frac{\frac{315 + 192}{756}}{1 - \frac{80}{756}} = \tan^{-1}\frac{507}{676} = \tan^{-1}\frac{3}{4}$.
$\tan\left(\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{2}\right) = \frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{8}} = \frac{5/4}{5/8} = 2$.
Answer: (B). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $\sim(p \Leftrightarrow \sim q) \equiv (p \Leftrightarrow q)$, and $(p \Leftrightarrow q) \wedge q$ is true only when q is true and then p is true: it is $p \wedge q$.
It is neither a tautology nor a contradiction.
$(p \Rightarrow q) \wedge q \equiv q$ (with q true, $p \Rightarrow q$ is true), which is not $p \wedge q$.
$(p \Rightarrow q) \wedge p \equiv p \wedge q$ (modus ponens: p true forces q true). So the statement is equivalent to $(p \Rightarrow q) \wedge p$.
Answer: (D). | ||||
| 21 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations The equation is $(px - q)^2 + (qx - r)^2 = 0$. A real root x needs $px = q$ and $qx = r$: p, q, r are in G.P. with common ratio x.
The roots of $x^2 + 2x - 8 = 0$ are 2 and $-4$. A positive ratio would give p, q, r one sign, so the ratio is $x = -4$.
$\frac{q^2 + r^2}{p^2} = x^2 + x^4 = 16 + 256 = 272$.
Answer: 272. | ||||
| 22 | 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 $36 = 2^2 \cdot 3^2$; no digit is 0, and the usable digits are 1, 2, 3, 4, 6, 9. The digit multisets and their arrangements:
$\{9, 4, 1, 1, 1\}$: $\frac{5!}{3!} = 20$
$\{9, 2, 2, 1, 1\}$: $\frac{5!}{2!2!} = 30$
$\{6, 6, 1, 1, 1\}$: $\frac{5!}{2!3!} = 10$
$\{6, 3, 2, 1, 1\}$: $\frac{5!}{2!} = 60$
$\{4, 3, 3, 1, 1\}$: $\frac{5!}{2!2!} = 30$
$\{3, 3, 2, 2, 1\}$: $\frac{5!}{2!2!} = 30$
Total $= 20 + 30 + 10 + 60 + 30 + 30 = 180$.
Answer: 180. | ||||
| 23 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The sets have 1, 3, 5, … elements; the first 10 sets hold $1 + 3 + \dots + 19 = 100$ multiples, and the 11th holds the next 21: $3 \times 101, \dots, 3 \times 121$.
Sum $= 3(101 + 102 + \dots + 121) = 3 \cdot \frac{21(101 + 121)}{2} = 3 \cdot 2331 = 6993$.
Answer: 6993. | ||||
| 24 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Expanded: $P(x) = x^8 - x^7 - x^6 + x^5 + 3x^4 - 4x^3 - 2x^2 + 4x - 1$.
$P(1) = 0$; dividing by $(x - 1)$ gives $x^7 - x^5 + 3x^3 - x^2 - 3x + 1$, which again vanishes at 1; dividing gives $x^6 + x^5 + 3x^2 + 2x - 1$, which vanishes at $-1$; dividing gives $x^5 + 3x - 1$.
So $P(x) = (x - 1)^2(x + 1)(x^5 + 3x - 1)$.
$x^5 + 3x - 1$ is strictly increasing (derivative $5x^4 + 3 > 0$), so it has exactly one real root, which lies in $(0, 1)$ (values $-1$ and 3 at the ends) and so is neither 1 nor $-1$.
Distinct real roots: 1, $-1$ and that root, 3 in all.
Answer: 3. | ||||
| 25 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Coefficient of x: $p - q = -3$.
Coefficient of $x^2$: $\binom{p}{2} - pq + \binom{q}{2} = \frac{(p - q)^2 - (p + q)}{2} = \frac{9 - (p + q)}{2} = -5$, so $p + q = 19$.
$p = 8$, $q = 11$ (both $\le 15$).
Coefficient of $x^3$: $\binom{8}{3} - \binom{8}{2}\cdot 11 + 8\binom{11}{2} - \binom{11}{3} = 56 - 308 + 440 - 165 = 23$.
Answer: 23. | ||||
| 26 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $I_m = \int_0^1 (1 - x^n)^m dx$. By parts (with $x$ as the second factor): $I_m = [x(1 - x^n)^m]_0^1 + mn\int_0^1 x^n(1 - x^n)^{m-1}dx = mn(I_{m-1} - I_m)$.
So $I_m = \frac{mn}{mn + 1}I_{m-1}$, and $I_{2n+1} = \frac{n(2n+1)}{n(2n+1) + 1}I_{2n}$.
The condition becomes $n(2n+1)I_{2n} = 1177 \cdot \frac{n(2n+1)}{n(2n+1) + 1}I_{2n}$, i.e. $n(2n + 1) + 1 = 1177$.
$2n^2 + n - 1176 = 0$, $n = \frac{-1 + \sqrt{1 + 9408}}{4} = \frac{-1 + 97}{4} = 24$.
Answer: 24. | ||||
| 27 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\int_3^x y\,dt = \frac{y^3}{x^3}$. Differentiating: $y = \frac{3y^2y'}{x^3} - \frac{3y^3}{x^4}$; dividing by $\frac{y}{x^4}$: $x^4 = 3xyy' - 3y^2$.
Put $y = vx$, $y' = v + xv'$: $x^4 = 3x^2v(v + xv') - 3v^2x^2 = 3x^3vv'$, so $3v\,dv = x\,dx$ and $\frac{3}{2}v^2 = \frac{x^2}{2} + C$.
Through $(3, 3)$, $v = 1$: $\frac{3}{2} = \frac{9}{2} + C$, $C = -3$. So $3\frac{y^2}{x^2} = x^2 - 6$, i.e. $3y^2 = x^4 - 6x^2$.
At $y = 6\sqrt{10}$: $1080 = \alpha^4 - 6\alpha^2$, $\alpha^2 = \frac{6 + \sqrt{36 + 4320}}{2} = \frac{6 + 66}{2} = 36$, so $\alpha = 6$ (first quadrant).
Answer: 6. | ||||
| 28 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | NumericalHard | |
A hard numerical question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines A is on AB and CA: $2x + y = 0$, $x - y = 3$ give $A = (1, -2)$.
AH $\perp$ BC: $\vec{AH} = (1, a + 2)$ is parallel to BC's normal $(1, p)$, so $p = a + 2$.
C on $x - y = 3$: $C = (c, c - 3)$. CH $\perp$ AB (direction $(1, -2)$): $(2 - c) - 2(a - c + 3) = 0$, so $c = 2a + 4$, $C = (2a + 4, 2a + 1)$.
C on BC: $2a + 4 + (a + 2)(2a + 1) = 15a$, i.e. $2a^2 - 8a + 6 = 0$, $a = 1$ or 3. With $-\frac{1}{2} < a < 2$, $a = 1$ and $p = 3$.
Check with B: BH $\perp$ CA gives $B = (-2 - a, 4 + 2a) = (-3, 6)$, and $-3 + 3 \cdot 6 = 15 = 15a$. ✓
Answer: 3. | ||||
| 29 | Conic SectionsParabola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $f'(x) = 4x - \frac{1}{x}$ changes sign at $x = \frac{1}{2}$: $a = \frac{1}{2}$. The parabola is $y^2 = 2x$, and the points are $(4, 3)$ and $(-2, 0)$.
A tangent of slope m: $y = mx + \frac{1}{2m}$. Through $(4, 3)$: $3 = 4m + \frac{1}{2m}$, $8m^2 - 6m + 1 = 0$, $m = \frac{1}{2}$ or $\frac{1}{4}$.
Through $(-2, 0)$ would need $-2m + \frac{1}{2m} = 0$, $m = \pm\frac{1}{2}$; so $m = \frac{1}{4}$.
Point of contact $\left(\frac{1/2}{m^2}, \frac{1}{m}\right) = (8, 4)$. Normal (slope $-4$): $y - 4 = -4(x - 8)$, i.e. $4x + y = 36$, i.e. $\frac{x}{9} + \frac{y}{36} = 1$.
$\alpha + \beta = 9 + 36 = 45$.
Answer: 45. | ||||
| 30 | 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 point on the line: $(-1 + 2t, -2 + 3t, 1 + 2t)$. Its distance$^2$ from P: $(2t - 5)^2 + (3t - 4)^2 + (2t - 6)^2 = 17t^2 - 68t + 77 = 26$, so $t^2 - 4t + 3 = 0$, $t = 1, 3$.
$Q = (1, 1, 3)$, $R = (5, 7, 7)$; $QR = |(4, 6, 4)| = \sqrt{68}$.
The foot of the perpendicular from P is at $t = 2$: $(3, 4, 5)$; height $= |(1, -2, 2)| = 3$.
Area $= \frac{1}{2}\sqrt{68} \cdot 3$, and its square $= \frac{68 \cdot 9}{4} = 153$.
Answer: 153. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctHard | |
A hard single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Let M accelerate at a (in the direction of F). The pulley is fixed to M, and $m_2$ hangs from it, so $m_2$ shares M's horizontal acceleration a and rises at $2\ ms^{-2}$ relative to it.
$m_2$ vertically: $T - m_2g = m_2 \cdot 2$, so $T = 20 \times 12 = 240$ N.
As $m_2$ rises, $m_1$ slides back (away from the pulley) at $2\ ms^{-2}$ relative to M, so its ground acceleration is $a - 2$; the only horizontal force on it is T: $m_1(a - 2) = 240$, $a - 2 = 24$, $a = 26\ ms^{-2}$.
For the whole system the only external horizontal force is F: $F = Ma + m_2a + m_1(a - 2) = 100(26) + 20(26) + 10(24) = 2600 + 520 + 240 = 3360$ N.
Answer: (A). | ||||
| 32 | 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 $\lambda = \frac{c}{f}$: at 6 MHz, $\lambda = \frac{3 \times 10^8}{6 \times 10^6} = 50$ m; at 10 MHz, $\lambda = 30$ m.
Wavelength bandwidth $= 50 - 30 = 20$ m.
Answer: (B). | ||||
| 33 | 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 $R = R_0e^{-\lambda t}$, so $\lambda = \frac{2.303}{t}\log_{10}\frac{R_0}{R} = \frac{2.303}{10}\log_{10}\frac{4250}{2250}$.
$\frac{4250}{2250} \approx 1.88$, so $\lambda \approx 0.2303 \times 0.274 \approx 0.063\ min^{-1}$.
Answer: (C). | ||||
| 34 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Energy of a photon: $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{900 \times 10^{-9}} \approx 2.21 \times 10^{-19}$ J.
Power through $1\ cm^2$: $100 \times 10^{-4} = 10^{-2}$ W.
Photons per second $= \frac{10^{-2}}{2.21 \times 10^{-19}} \approx 4.5 \times 10^{16}$.
Answer: (B). | ||||
| 35 | 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 $\beta = \frac{\lambda D}{d}$, and in water the wavelength becomes $\frac{\lambda}{\mu}$.
$\beta' = \frac{\beta}{\mu} = \frac{12}{4/3} = 9$ mm.
Answer: (B). | ||||
| 36 | 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 Wave speed $\frac{\omega}{k} = \frac{1.5 \times 10^{11}}{0.5 \times 10^3} = 3 \times 10^8$ m/s; $E_0 = cB_0 = 3 \times 10^8 \times 2 \times 10^{-8} = 6\ Vm^{-1}$.
The phase $kx + \omega t$ means the wave travels along $-x$. $\vec{E}$ is perpendicular to both $\vec{B}$ ($\hat{j}$) and the direction of travel ($\hat{i}$), so it lies along the z-axis (with $\hat{E} \times \hat{B}$ along $-\hat{i}$, $\vec{E}$ is along $+\hat{k}$ when $\vec{B}$ is along $+\hat{j}$).
Answer: (C). | ||||
| 37 | 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 LR circuit: $P_1 = \frac{R}{\sqrt{R^2 + X_L^2}} = \frac{R}{\sqrt{2}R} = \frac{1}{\sqrt{2}}$.
With $X_C = X_L$ the circuit is at resonance: $P_2 = \frac{R}{R} = 1$.
$\frac{P_1}{P_2} = \frac{1}{\sqrt{2}}$.
Answer: (B). | ||||
| 38 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The magnetic force $q\vec{v} \times \vec{B}$ is perpendicular to $\vec{B}$, so $\vec{a} \cdot \vec{B} = 0$: $2\alpha - 12 = 0$, $\alpha = 6$.
Answer: (B). | ||||
| 39 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter At the centre of a coil, $B = \frac{\mu_0NI}{2r}$. With equal I and r, $\frac{B_X}{B_Y} = \frac{200}{400} = \frac{1}{2}$.
Answer: (B). | ||||
| 40 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity Name the nodes: L (left, + terminal), X (between the lower-left 4 Ω and the 2 Ω), Y (between the 2 Ω and the upper-right 4 Ω) and R (right, − terminal).
Resistors: L–X 4 Ω, L–Y 4 Ω (top), X–Y 2 Ω, Y–R 4 Ω, X–R 4 Ω (bottom).
This is a Wheatstone bridge with $\frac{4}{4} = \frac{4}{4}$: it is balanced, X and Y are at the same potential and the 2 Ω carries no current.
Equivalent resistance $= (4 + 4) \parallel (4 + 4) = 4$ Ω, so $I = \frac{40}{4} = 10$ A.
Answer: (A). | ||||
| 41 | 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 In parallel, $C = 1 + 2 + 4 + 3 = 10\ \mu F$; $Q = CV = 10\ \mu F \times 20$ V $= 200\ \mu C$.
Answer: (A). | ||||
| 42 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v^2 = \omega^2(A^2 - x^2)$, i.e. $\frac{x^2}{A^2} + \frac{v^2}{\omega^2A^2} = 1$: an ellipse in the x–v plane (a circle only in the special case $\omega = 1$ in the chosen units).
Answer: (B). | ||||
| 43 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Internal energy depends only on temperature: $\Delta U = nC_V\Delta T = 7 \times \frac{3}{2} \times 8.3 \times 40 = 3486$ J.
Answer: (B). | ||||
| 44 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Constant entropy means adiabatic: $PV^\gamma$ is constant with $\gamma = \frac{5}{3}$.
$P' = P\left(\frac{V}{V/8}\right)^{5/3} = P \cdot 8^{5/3} = 32P$.
Answer: (C). | ||||
| 45 | Properties of Bulk MatterSurface tension | 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 surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Volume is conserved: $729r^3 = R^3$, so $r = \frac{R}{9}$.
Increase in area $= 4\pi(729r^2 - R^2) = 4\pi R^2(9 - 1) = 32\pi\ cm^2 = 100.48\ cm^2 = 1.0048 \times 10^{-2}\ m^2$.
$S = 75$ dyne/cm $= 0.075$ N/m, so $\Delta E = 0.075 \times 1.0048 \times 10^{-2} \approx 7.5 \times 10^{-4}$ J.
Answer: (C). | ||||
| 46 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation For $h \ll R$, $g' = g\left(1 - \frac{2h}{R}\right)$, so the fractional decrease is $\frac{2h}{R} = \frac{64}{6400} = 0.01$, i.e. 1 %.
Answer: (A). | ||||
| 47 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The total momentum is zero, so at maximum elongation both blocks are momentarily at rest and all the kinetic energy is in the spring.
$2 \times \frac{1}{2}(0.25)v^2 = \frac{1}{2}(2)x^2$, so $0.25v^2 = x^2$ and $x = \frac{v}{2}$.
Answer: (B). | ||||
| 48 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Going down with $a = 4$: $T = m(g - a) = 50 \times 6 = 300$ N < 350 N, so the rope holds.
Going up with $a = 5$: $T = m(g + a) = 50 \times 15 = 750$ N > 350 N, so the rope breaks.
Answer: (C). | ||||
| 49 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Time to the top $= \frac{u\sin\theta}{g}$. Equal times: $u_1\sin 30^\circ = u_2\sin 45^\circ$, so $\frac{u_1}{u_2} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.
Answer: (C). | ||||
| 50 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{0.5}{50} = 0.01$ mm; diameter $d = 1.5 + 7 \times 0.01 = 1.57$ mm $= 0.157$ cm.
Curved surface area $= \pi dL = 3.14 \times 0.157 \times 6.8 \approx 3.35\ cm^2$.
The length 6.8 cm has two significant figures, so the area is $3.4\ cm^2$.
Answer: (B). | ||||
| 51 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Trajectory: $y = x\tan\theta - \frac{gx^2}{2u_x^2}$. Here $y = 5x - 5x^2$ with $u_x = 1$ m/s (and indeed $\frac{g}{2u_x^2} = 5$).
$\tan\theta = \frac{u_y}{u_x} = 5$, so $u_y = 5$ m/s: the y component is $5\hat{j}$.
Answer: 5. | ||||
| 52 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Moment of inertia about the axis: $I = \frac{1}{2}(1)R^2 + (1)R^2 = \frac{3}{2}R^2$.
The body falls through 2R (the disc's centre does not move): $mg(2R) = \frac{1}{2}I\omega^2$, i.e. $20R = \frac{3}{4}R^2\omega^2$.
$\omega^2 = \frac{80}{3R}$, $\omega = \sqrt{\frac{16 \times 5}{3R}} = 4\sqrt{\frac{5}{3R}}$, so $x = 5$.
Answer: 5. | ||||
| 53 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalMedium | |
A medium numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $Y = \frac{4MgL}{\pi d^2\,\Delta l} = \frac{4 \times 1 \times 10 \times 1}{\pi (0.4 \times 10^{-3})^2(0.4 \times 10^{-3})} = \frac{40}{\pi \times 6.4 \times 10^{-11}} \approx 1.99 \times 10^{11}\ Nm^{-2}$.
$\frac{\Delta Y}{Y} = \frac{\Delta(\Delta l)}{\Delta l} + 2\frac{\Delta d}{d} = \frac{0.02}{0.4} + 2 \times \frac{0.01}{0.4} = 0.05 + 0.05 = 0.1$.
$\Delta Y = 0.1 \times 1.99 \times 10^{11} \approx 1.99 \times 10^{10} \approx 2 \times 10^{10}\ Nm^{-2}$, so $x = 2$.
Answer: 2. | ||||
| 54 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Approaching: $100 = f\frac{v}{v - v_s}$; receding: $50 = f\frac{v}{v + v_s}$.
Adding the reciprocals: $\frac{1}{100} + \frac{1}{50} = \frac{(v - v_s) + (v + v_s)}{fv} = \frac{2}{f}$, so $f = \frac{2}{0.03} = \frac{200}{3}$ Hz.
An observer moving with the car hears the true frequency $f = \frac{200}{3}$ Hz, so $x = 200$.
Answer: 200. | ||||
| 55 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The two layers are capacitors in series: $C_1 = \frac{3\varepsilon_0A}{0.5\text{ mm}}$, $C_2 = \frac{4\varepsilon_0A}{1\text{ mm}}$, so $C_1 : C_2 = 6 : 4$.
In series the voltages divide inversely as C: $V_1 = 100 \times \frac{4}{10} = 40$ V across layer 1 ($\varepsilon_{r1} = 3$, next to the plate on the battery's positive side) and $V_2 = 60$ V across layer 2.
Taking the plate on the negative terminal as 0 V, the foil is at $V_2 = 60$ V.
Answer: 60. | ||||
| 56 | Current ElectricityMetre bridge | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity First balance: $\frac{P}{Q} = \frac{40}{60}$, so $Q = \frac{3}{2} \times 4 = 6$ Ω.
With x in series with P: $\frac{4 + x}{6} = \frac{80}{20} = 4$, so $4 + x = 24$, $x = 20$ Ω.
Answer: 20. | ||||
| 57 | 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 At very high frequency, $X_C = \frac{1}{\omega C} \to 0$ (capacitors act as short circuits) and $X_L = \omega L \to \infty$ (inductors act as open circuits).
First block: the 8 Ω–20 mH branch is open; the 0.5 μF–1 Ω branch leaves 1 Ω.
The vertical 0.5 μF capacitors short the top and bottom wires together at both junctions.
Middle block: 4 Ω (top) parallel with 4 Ω (bottom) $= 2$ Ω.
Last block: the 50 mH–8 Ω branch is open; the 0.8 μF–2 Ω branch leaves 2 Ω.
Total $= 1 + 2 + 2 = 5$ Ω, so $I = \frac{220}{5} = 44$ A.
Answer: 44. | ||||
| 58 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. At B, $\frac{1}{u} = -0.10\ cm^{-1}$ and $\frac{1}{v} = 0$, so $\frac{1}{f} = 0.10\ cm^{-1}$, $f = 10$ cm (A gives the same).
Lens maker's formula for an equiconvex lens: $\frac{1}{f} = (1.5 - 1)\frac{2}{R} = \frac{1}{R}$, so $R = f = 10$ cm.
Answer: 10. | ||||
| 59 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Lyman first line ($2 \to 1$): $\frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right)$, $\lambda = \frac{4}{3R}$.
Paschen third line ($6 \to 3$): $\frac{1}{\lambda_P} = R\left(\frac{1}{9} - \frac{1}{36}\right) = \frac{R}{12}$, $\lambda_P = \frac{12}{R}$.
Balmer second line ($4 \to 2$): $\frac{1}{\lambda_B} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$, $\lambda_B = \frac{16}{3R}$.
$\lambda_P - \lambda_B = \frac{36 - 16}{3R} = \frac{20}{3R} = 5 \cdot \frac{4}{3R} = 5\lambda$, so $a = 5$.
Answer: 5. | ||||
| 60 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The load sits across the zener, so it is always at 60 V: $I_L = \frac{60}{10\text{ k}\Omega} = 6$ mA.
The zener current is largest at the largest input, 120 V: $I = \frac{120 - 60}{4\text{ k}\Omega} = 15$ mA.
$I_z = I - I_L = 15 - 6 = 9$ mA.
Answer: 9. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) $BrF_5$: 5 bond pairs + 1 lone pair, square pyramidal (II).
(B) $[CrF_6]^{3-}$: six ligands, $d^2sp^3$, octahedral (IV).
(C) $O_3$: 2 bonding domains + 1 lone pair on the central O, bent (I).
(D) $PCl_5$: 5 bond pairs, trigonal bipyramidal (III).
Answer: (C). | ||||
| 62 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter (A) Contact process for $SO_3$: $V_2O_5$ (III).
(B) Ostwald process (oxidation of $NH_3$ to NO): Pt–Rh gauze (II).
(C) Haber process: iron (I).
(D) Hydrogenation of vegetable oils: finely divided nickel (IV).
Answer: (B). | ||||
| 63 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsEasy | |
A easy statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Statement I: the covalent radius is half the internuclear distance in $Cl_2$; it is the atomic radius of chlorine in the covalent bond, not double it. Incorrect.
Statement II: adding electrons increases electron–electron repulsion while the nuclear charge stays the same, so an anion is always larger than its parent atom. Correct.
Answer: (D). | ||||
| 64 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In liquation the impure metal is heated on a sloping hearth: the easily fusible (low melting) metal melts and flows away from the higher-melting impurities. It is used for metals such as tin, lead and bismuth.
Answer: (A). | ||||
| 65 | 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 $H_2O_2$ decomposes slowly ($2H_2O_2 \rightarrow 2H_2O + O_2$), faster in light and in the presence of traces of metal. It is stored in wax-lined plastic or coloured bottles, and a little urea is added as a stabiliser (negative catalyst).
Answer: (A). | ||||
| 66 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Beryllium hydride is made by reducing beryllium chloride with lithium aluminium hydride:
$2BeCl_2 + LiAlH_4 \rightarrow 2BeH_2 + LiCl + AlCl_3$
The products are $AlCl_3$ (A), $BeH_2$ (B) and LiCl (D).
Answer: (B). | ||||
| 67 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter $3B_2H_6 + 6NH_3 \rightarrow 3[BH_2(NH_3)_2]^+[BH_4]^-$, which on heating gives $2B_3N_3H_6 + 12H_2$.
X = $B_2H_6$, Y = $NH_3$.
Answer: (B). | ||||
| 68 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions In a disproportionation one element is both oxidised and reduced.
(A) O goes from −1 to −2 and 0. Disproportionation.
(B) N goes from +4 to +5 ($HNO_3$) and +3 ($HNO_2$). Disproportionation.
(C) Mn goes only from +7 to +4: a reduction half-reaction, not a disproportionation.
(D) Mn goes from +6 to +7 and +4. Disproportionation.
Answer: (C). | ||||
| 69 | 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 $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$.
Mn goes from +7 (purple $MnO_4^-$) to +2 (nearly colourless $Mn^{2+}$): a change of 5.
Answer: (A). | ||||
| 70 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons A chlorine radical abstracts a hydrogen atom from methane (the propagation step of free-radical chlorination):
$\dot{C}l + CH_4 \rightarrow \dot{C}H_3 + HCl$.
Answer: (C). | ||||
| 71 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Neither p-nitrophenol (intermolecular hydrogen bonding) nor picric acid is steam-volatile, and neither is separated by sublimation.
The two differ in polarity, so a chromatographic method is needed. For a small quantity such as 100 mg, preparative thin layer chromatography is the suitable scale; a 2–5 ft long silica gel column is meant for much larger quantities.
Answer: (D). | ||||
| 72 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers In water (polar), phenol partly ionises to the strongly activating phenoxide ion and $Br_2$ is polarised, so 2,4,6-tribromophenol precipitates.
In chloroform (non-polar, low temperature) there is no ionisation and less polarisation of $Br_2$, so mainly monobromophenols (o- and p-) form.
The difference comes from the polarity of the solvent.
Answer: (B). | ||||
| 73 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A) Cyclopentadienyl anion: cyclic, planar, fully conjugated, $6\pi$ electrons ($4n + 2$, n = 1). Aromatic.
(B) Cyclopropenyl cation: $2\pi$ electrons ($n = 0$). Aromatic.
(C) [10]Annulene has $10\pi$ electrons, but the two inward-pointing hydrogens clash, so the ring cannot be planar and the conjugation is broken. Not aromatic.
(D) Naphthalene: planar, $10\pi$ electrons. Aromatic.
Answer: (C). | ||||
| 74 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Tollens' reagent, $[Ag(NH_3)_2]^+OH^-$, oxidises only the aldehyde: the CHO becomes COOH (as its salt, acidified on work-up); the ring ketone is untouched. A = 4-oxocyclohexanecarboxylic acid.
$NaBH_4$ reduces the ketone to a secondary alcohol but does not reduce the carboxylic acid. B = 4-hydroxycyclohexanecarboxylic acid.
Answer: (C). | ||||
| 75 | 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 alcohol is $(Ph)_2C(OH)CH_3$, a tertiary alcohol. One mole of PhMgBr adds one phenyl group to a ketone: $PhCOCH_3 + PhMgBr$, then $H_3O^+$, gives $Ph_2C(OH)CH_3$.
Nitriles (A, B) give ketones after hydrolysis, not alcohols. The ester (C) would need two moles of PhMgBr and would give $Ph_2C(OH)CH_3$ only after losing phenoxide, using a second mole.
Answer: (D). | ||||
| 76 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Na/liq. $NH_3$ with ethanol is a Birch reduction: the ring becomes a non-conjugated cyclohexa-1,4-diene.
An electron-donating group ($OCH_3$) ends up on a carbon that stays part of a double bond; an electron-withdrawing group ($NO_2$) ends up on a carbon that is reduced (sp³, carrying the added H), along with the carbon para to it.
So the product has $OCH_3$ on a ring double bond and $NO_2$ on the adjacent sp³ carbon with H: option (A).
Answer: (A). | ||||
| 77 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines An arenediazonium ion is stabilised by delocalisation of the positive charge into the ring; electron-donating groups at the para position help and electron-withdrawing groups hurt.
(A) p-$OCH_3$ (+R, donating) is the most stable; (C) unsubstituted comes next; (D) p-CN and (B) p-$NO_2$ withdraw electrons, $NO_2$ more strongly.
Order: (A) > (C) > (D) > (B).
Answer: (B). | ||||
| 78 | Chemistry in Everyday LifeCleansing agents | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on cleansing agents, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Stearic acid esterifies with polyethylene glycol: $CH_3(CH_2)_{16}COO(CH_2CH_2O)_nCH_2CH_2OH$. The molecule carries no ionic group: a non-ionic detergent.
Answer: (D). | ||||
| 79 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules A sugar is reducing when it has a free hemiacetal (or hemiketal) group, which opens to a free aldehyde or ketone.
(A) The anomeric carbon carries a free OH and the ring oxygen: a hemiacetal. Reducing.
(B) The anomeric OH is replaced by $OCH_3$: an acetal (glycoside). Non-reducing.
(C) The two rings are joined through both anomeric carbons, so neither has a free hemiacetal. Non-reducing.
(D) There is no free aldehyde or hemiacetal group. Non-reducing.
Answer: (A). | ||||
| 80 | AminesBasicity of amines | Class 12: Amines | Assertion and reasonMedium | |
A medium assertion and reason question on basicity of amines, from Amines. NCERT: Class 12: Amines Aniline, a Lewis base, forms a salt with the Lewis acid $AlCl_3$. The nitrogen then carries a positive charge and strongly deactivates the ring, so aniline does not undergo Friedel–Crafts alkylation at all.
(A) is true: no o- or p-methylaniline is formed.
(R) is false: the reaction does not go on to give m-methylaniline either.
Answer: (C). | ||||
| 81 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions 2 L of dilute aqueous solution ≈ 2000 g. 48 ppm means 48 g of Mg per $10^6$ g: mass of Mg $= 48 \times 10^{-6} \times 2000 = 0.096$ g.
Moles $= \frac{0.096}{24} = 0.004$; atoms $= 0.004 \times 6.02 \times 10^{23} = 2.408 \times 10^{21} = 24.08 \times 10^{20}$.
Nearest integer: 24.
Answer: 24. | ||||
| 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 In 100 g: $H_2$ 40 g = 20 mol; $O_2$ 60 g = 1.875 mol.
Mole fraction of $H_2 = \frac{20}{21.875} = 0.914$; $p_{H_2} = 0.914 \times 2.2 = 2.01$ bar.
Nearest integer: 2.
Answer: 2. | ||||
| 83 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $\lambda = \frac{h}{mv}$; equal wavelengths need $m_ev_e = m_nv_n$.
$x = \frac{v_e}{v_n} = \frac{m_n}{m_e} = \frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}} = 1758.2$.
Nearest integer: 1758.
Answer: 1758. | ||||
| 84 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat released $= C\Delta T = 20.0 \times 2 = 40$ kJ, for $\frac{2.4}{12} = 0.2$ mol of carbon.
$\Delta U = -\frac{40}{0.2} = -200$ kJ mol⁻¹ (bomb: constant volume).
$C(s) + O_2(g) \rightarrow CO_2(g)$ has $\Delta n_g = 0$, so $\Delta H = \Delta U = -200$ kJ mol⁻¹: x = 200.
Answer: 200. | ||||
| 85 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Initial moles $= 0.8 \times 0.5 = 0.4$ mol. Evaporated: $\frac{11.5}{63} = 0.1825$ mol. Remaining: 0.2175 mol in 0.4 L.
Molarity $= \frac{0.2175}{0.4} = 0.544$ M $= 54.4 \times 10^{-2}$ M.
Nearest integer: 54.
Answer: 54. | ||||
| 86 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium The second reaction is the first reversed and halved: $K' = \left(\frac{1}{K}\right)^{1/2} = \left(\frac{1}{2 \times 10^{15}}\right)^{1/2} = (5 \times 10^{-16})^{1/2} = 2.236 \times 10^{-8}$.
Nearest integer: 2.
Answer: 2. | ||||
| 87 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry In $Fe_3O_4$ the average oxidation state of iron is $+\frac{8}{3}$ ($FeO \cdot Fe_2O_3$: one $Fe^{2+}$ and two $Fe^{3+}$).
Reducing 3 mol of Fe to the metal takes $2 + 3 + 3 = 8$ F, so one mole takes $\frac{8}{3} = 2.67$ F.
Nearest integer: 3.
Answer: 3. | ||||
| 88 | 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 $t_{1/2} \propto [A]_0^{1-n}$: $\frac{100}{50} = \left(\frac{0.5}{1.0}\right)^{1-n}$, i.e. $2 = 2^{n-1}$, so $n = 2$.
Answer: 2. | ||||
| 89 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[Co(H_2O)_6]^{2+}$: $Co^{2+}$ is $3d^7$; water is a weak-field ligand, so high spin $t_{2g}^5e_g^2$ with 3 unpaired electrons: $\mu = \sqrt{15} = 3.87$ BM.
$[Cr(H_2O)_6]^{3+}$: $Cr^{3+}$ is $3d^3$, $t_{2g}^3$, 3 unpaired electrons: $\mu = 3.87$ BM.
Difference $= 0$.
Answer: 0. | ||||
| 90 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons In sunlight (UV), benzene adds three molecules of chlorine by a free-radical addition: $C_6H_6 + 3Cl_2 \rightarrow C_6H_6Cl_6$ (benzene hexachloride, 1,2,3,4,5,6-hexachlorocyclohexane).
X has 6 hydrogens.
Answer: 6. | ||||
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