JEE (Main) · 2022
JEE (Main) 2022 — 26 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
- Conic Sections4
- Limits, Continuity and Differentiability3
- Integral Calculus3
9 easy · 18 medium · 3 hard
Physics
30 Q
- Laws of Motion3
- Electrostatics3
- Properties of Bulk Matter2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- Aldehydes, Ketones and Carboxylic Acids3
- Biomolecules2
- Amines2
21 easy · 9 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $fog(x) = \frac{x^2}{x^2-1} - 1 = \frac{1}{x^2-1}$, $x \in \mathbb{R} - \{1,-1\}$.
It is even, e.g. $fog(2) = fog(-2) = \frac{1}{3}$, so it is not one-one.
$x^2 - 1$ takes the values $[-1, 0) \cup (0, \infty)$, so $\frac{1}{x^2-1}$ takes $(-\infty, -1] \cup (0, \infty)$; values such as $0$ or $-\frac{1}{2}$ are never taken, so it is not onto $\mathbb{R}$.
Answer: (D). | ||||
| 2 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For infinitely many solutions the coefficient determinant must vanish:
$\begin{vmatrix}\alpha&1&1\\1&2&3\\1&3&5\end{vmatrix} = \alpha(10-9) - 1(5-3) + 1(3-2) = \alpha - 1 = 0$, so $\alpha = 1$.
Then the left side of the third equation is $2(x+2y+3z) - (x+y+z)$, so consistency needs $\beta = 2(4) - 5 = 3$. With $\alpha = 1, \beta = 3$ the third equation is a combination of the first two, leaving a line of solutions.
Answer: (C). | ||||
| 3 | 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 For odd $n$ the term is $\frac{1}{2^n}$; for even $n$ it is $\frac{1}{4^n}$.
$\sum_{n \text{ odd}} \frac{1}{2^n} = \frac{1/2}{1 - 1/4} = \frac{2}{3}$, $\sum_{n \text{ even}} \frac{1}{4^n} = \frac{1/16}{1 - 1/16} = \frac{1}{15}$.
$A = \frac{2}{3} + \frac{1}{15} = \frac{11}{15}$, $B = -\frac{2}{3} + \frac{1}{15} = -\frac{9}{15}$.
$\frac{A}{B} = -\frac{11}{9}$.
Answer: (C). | ||||
| 4 | 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 $\cos(\sin x) - \cos x = 2\sin\left(\frac{x + \sin x}{2}\right)\sin\left(\frac{x - \sin x}{2}\right)$.
As $x \to 0$: $\frac{x + \sin x}{2} \approx x$ and $\frac{x - \sin x}{2} \approx \frac{x^3}{12}$ (since $x - \sin x = \frac{x^3}{6} - \dots$).
So the numerator $\approx 2 \cdot x \cdot \frac{x^3}{12} = \frac{x^4}{6}$, and the limit is $\frac{1}{6}$.
Answer: (C). | ||||
| 5 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $x\sin x \ge 0$ on $[0, \pi]$ and $\le 0$ on $[\pi, 2\pi]$. Hence $f(x) = 1$ on $[0, \pi]$ and $f(x) = 1 + x\sin x$ on $[\pi, 2\pi]$.
Both pieces are continuous and agree at $x = \pi$, so $f$ is continuous everywhere: $n = 0$.
At $x = \pi$: left derivative $0$; right derivative $\sin x + x\cos x$ at $\pi$ $= -\pi$. Not differentiable there.
Inside $(0, \pi)$ and $(\pi, 2\pi)$ $f$ is smooth (near $0$ and $2\pi$ only one formula applies), so $m = 1$.
Answer: (B). | ||||
| 6 | 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 Surface areas: cuboid $2(8x^2 + 10x^2 + 20x^2) = 76x^2$; closed hemisphere $2\pi r^2 + \pi r^2 = 3\pi r^2$. So $76x^2 + 3\pi r^2 = k$.
Volume $V = 40x^3 + \frac{2}{3}\pi r^3$.
From the constraint, $152x\,dx + 6\pi r\,dr = 0$, so $\frac{dr}{dx} = -\frac{152x}{6\pi r}$.
$\frac{dV}{dx} = 120x^2 + 2\pi r^2\frac{dr}{dx} = 120x^2 - \frac{152}{3}xr = 0 \Rightarrow 120x = \frac{152}{3}r \Rightarrow \frac{x}{r} = \frac{152}{360} = \frac{19}{45}$.
This is the only interior stationary point, and it is the ratio the question asks for.
Answer: (B). | ||||
| 7 | 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 Intersection: $8x = 48 - 16x \Rightarrow x = 2$, $y = \pm 4$.
In terms of $y$, the region lies between $x = \frac{y^2}{8}$ (left) and $x = 3 - \frac{y^2}{16}$ (right):
Area $= \int_{-4}^{4}\left(3 - \frac{y^2}{16} - \frac{y^2}{8}\right)dy = \int_{-4}^{4}\left(3 - \frac{3y^2}{16}\right)dy = 2\left[3y - \frac{y^3}{16}\right]_0^4 = 2(12 - 4) = 16$.
Answer: (C). | ||||
| 8 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{1}{x}\sqrt{\frac{1-x}{1+x}} = \frac{1-x}{x\sqrt{1-x^2}} = \frac{1}{x\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}}$.
For $0 < x \le 1$: $\int \frac{dx}{x\sqrt{1-x^2}} = \log_e\left(\frac{x}{1 + \sqrt{1-x^2}}\right)$ (put $x = \sin\theta$: $\int \csc\theta\,d\theta = \log\tan\frac{\theta}{2}$).
So $g(x) = \log_e\left(\frac{x}{1+\sqrt{1-x^2}}\right) - \sin^{-1}x + C$; $g(1) = 0 - \frac{\pi}{2} + C = 0 \Rightarrow C = \frac{\pi}{2}$.
$g\left(\frac{1}{2}\right) = \log_e\left(\frac{1/2}{1 + \sqrt{3}/2}\right) - \frac{\pi}{6} + \frac{\pi}{2} = \log_e(2 - \sqrt{3}) + \frac{\pi}{3}$.
Since $\frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{2} = 2 - \sqrt{3}$, this is option (A).
Answer: (A). | ||||
| 9 | 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 Multiplying by $x$: $x^2y' + 2xy = x^2e^x$, i.e. $(x^2y)' = x^2e^x$.
$x^2y = e^x(x^2 - 2x + 2) + C$; $y(1) = 0 \Rightarrow 0 = e + C$, $C = -e$.
$z(x) = x^2y - e^x = e^x(x^2 - 2x + 1) - e = e^x(x-1)^2 - e$.
$z'(x) = e^x\left[(x-1)^2 + 2(x-1)\right] = e^x(x-1)(x+1)$: $z'$ changes from $+$ to $-$ at $x = -1$, a local maximum.
$z(-1) = 4e^{-1} - e = \frac{4}{e} - e$.
Answer: (D). | ||||
| 10 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Let $u = e^x(x^2 - 2x)$. Then $u' = e^x(x^2 - 2x + 2x - 2) = e^x(x^2 - 2)$.
The equation reads $y' + u'y = u\,u'$ (since $(x^2-2x)e^x \cdot (x^2-2)e^x = u\,u'$).
Integrating factor $e^{u}$: $(ye^u)' = u\,u'e^u$, so $ye^u = (u - 1)e^u + C$, i.e. $y = u - 1 + Ce^{-u}$.
At $x = 0$: $u = 0$, so $0 = -1 + C$, $C = 1$.
At $x = 2$: $u = e^2(4 - 4) = 0$, so $y(2) = 0 - 1 + 1 = 0$.
Answer: (C). | ||||
| 11 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections A tangent of slope $m$ to the ellipse: $y = mx \pm\sqrt{16m^2 + 9}$; to the circle: $y = mx \pm \sqrt{12(1 + m^2)}$.
For a common tangent, $16m^2 + 9 = 12 + 12m^2 \Rightarrow 4m^2 = 3 \Rightarrow 12m^2 = 9$.
Answer: (B). | ||||
| 12 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections If $(h, k)$ is the mid point, the point on the ellipse is $(2h - 4, 2k - 3)$:
$(2h-4)^2 + 2(2k-3)^2 = 4 \Rightarrow (h-2)^2 + 2\left(k - \frac{3}{2}\right)^2 = 1 \Rightarrow \frac{(h-2)^2}{1} + \frac{(k - 3/2)^2}{1/2} = 1$.
$e^2 = 1 - \frac{1/2}{1} = \frac{1}{2}$, so $e = \frac{1}{\sqrt{2}}$ (the same shape as the original ellipse, scaled by $\frac{1}{2}$).
Answer: (C). | ||||
| 13 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The point lies on the hyperbola: $\frac{64}{a^2} - \frac{27}{9} = 1 \Rightarrow a^2 = 16$.
Normal at $(x_1, y_1)$: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2 \Rightarrow \frac{16x}{8} + \frac{9y}{3\sqrt{3}} = 25 \Rightarrow 2x + \sqrt{3}y = 25$.
Check: $(-1, 9\sqrt{3})$ gives $-2 + 27 = 25$. ✓ (The others give 24, 24 and 16.)
Answer: (C). | ||||
| 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 A plane through the line of intersection: $(2x + y - 5z) + \lambda(3x - y + 4z - 7) = 0$, normal $(2 + 3\lambda, 1 - \lambda, -5 + 4\lambda)$.
After a rotation of $\frac{\pi}{2}$ it is perpendicular to the first plane:
$2(2 + 3\lambda) + (1 - \lambda) - 5(-5 + 4\lambda) = 30 - 15\lambda = 0 \Rightarrow \lambda = 2$.
Plane: $8x - y + 3z - 14 = 0$. Check $(1, 0, 2)$: $8 + 6 - 14 = 0$. ✓
Answer: (C). | ||||
| 15 | 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 Directions $\vec{d_1} = (0, 3, -1)$, $\vec{d_2} = (2, 0, -3)$; $\vec{d_1} \times \vec{d_2} = (-9, -2, -6)$, so the normal is $(9, 2, 6)$.
Coplanarity: $\left[(\alpha - 1, 0, -1)\right] \cdot (9, 2, 6) = 9(\alpha - 1) - 6 = 0 \Rightarrow \alpha = \frac{5}{3}$.
Plane through $(1, -1, 1)$: $9x + 2y + 6z = 9 - 2 + 6 = 13$.
Distance of $\left(\frac{5}{3}, 0, 0\right)$: $\frac{|15 - 13|}{\sqrt{81 + 4 + 36}} = \frac{2}{11}$.
Answer: (B). | ||||
| 16 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Let $\vec{v} = p\vec{a} + q\vec{b} = (p + 2q,\ p - 3q,\ 2p + q)$.
Projection on $\vec{c}$: $\frac{(p + 2q) - (p - 3q) + (2p + q)}{\sqrt{3}} = \frac{2p + 6q}{\sqrt{3}} = \frac{2}{\sqrt{3}} \Rightarrow p + 3q = 1$.
$\vec{v}\cdot\hat{j} = p - 3q = 7$. So $p = 4$, $q = -1$ and $\vec{v} = (2, 7, 7)$.
$\vec{v}\cdot(\hat{i} + \hat{k}) = 2 + 7 = 9$.
Answer: (D). | ||||
| 17 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctMedium | |
A medium single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Wrong sum $= 50 \times 15 = 750$; correct sum $= 50 \times 16 = 800$. With $a$ the wrong value and $b$ the correct one: $b - a = 50$, $a + b = 70$, so $a = 10$, $b = 60$.
Wrong $\sum x^2 = 50(2^2 + 15^2) = 11450$. Correct $\sum x^2 = 11450 - 10^2 + 60^2 = 14950$.
Correct variance $= \frac{14950}{50} - 16^2 = 299 - 256 = 43$.
Answer: (C). | ||||
| 18 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin\theta\sin(60^\circ - \theta)\sin(60^\circ + \theta) = \frac{1}{4}\sin 3\theta$. With $\theta = 20^\circ$: $\sin 20^\circ\sin 40^\circ\sin 80^\circ = \frac{1}{4}\sin 60^\circ = \frac{\sqrt{3}}{8}$.
$16 \times \frac{\sqrt{3}}{8} = 2\sqrt{3}$.
Answer: (B). | ||||
| 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 With $\theta = \tan^{-1}\frac{4}{3}$ (first quadrant): $\cos\theta = \frac{3}{5}$, $\sin\theta = \frac{4}{5}$.
$\frac{3}{10}\cdot\frac{3}{5} + \frac{2}{5}\cdot\frac{4}{5} = \frac{9}{50} + \frac{16}{50} = \frac{1}{2}$.
$\cos^{-1}\frac{1}{2} = \frac{\pi}{3}$.
Answer: (C). | ||||
| 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 The statement is $(r \vee \sim p) \Rightarrow ((p \wedge q) \vee r)$.
$r = p$: antecedent $p \vee \sim p$ is always true, consequent $(p\wedge q) \vee p = p$; false when $p$ is false. Not a tautology.
$r = q$: with $p, q$ both false, antecedent true, consequent $q$ false. Not a tautology.
$r = \sim p$: antecedent $\sim p$; whenever it is true, the consequent $(p \wedge q) \vee \sim p$ is true. A tautology.
$r = \sim q$: with $p$ false, $q$ true, antecedent true, consequent $F \vee F$ false. Not a tautology.
Answer: (C). | ||||
| 21 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $y = 0$: $f(x) = 2^xf(0) + f(x) \Rightarrow f(0) = 0$.
Swapping $x$ and $y$: $2^xf(y) + 4^yf(x) = 2^yf(x) + 4^xf(y)$, so $f(x)(4^y - 2^y) = f(y)(4^x - 2^x)$, hence $f(x) = c(4^x - 2^x)$.
$f(2) = 12c = 3 \Rightarrow c = \frac{1}{4}$.
$f'(x) = c\ln 2\,(2\cdot 4^x - 2^x)$, so $\frac{f'(4)}{f'(2)} = \frac{512 - 16}{32 - 4} = \frac{496}{28}$.
$14 \cdot \frac{496}{28} = 248$.
Answer: 248. | ||||
| 22 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $s = pq$. $p^2 + q^2 = 9 - 2s$ and $p^4 + q^4 = (9 - 2s)^2 - 2s^2 = 81 - 36s + 2s^2 = 369$.
$s^2 - 18s - 144 = 0 \Rightarrow s = 24$ or $s = -6$.
$p, q$ are real roots of $t^2 - 3t + s = 0$, which needs $9 - 4s \ge 0$; so $s = -6$.
$\left(\frac{1}{p} + \frac{1}{q}\right)^{-2} = \left(\frac{pq}{p+q}\right)^2 = \left(\frac{-6}{3}\right)^2 = 4$.
Answer: 4. | ||||
| 23 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z$ is a non-real cube root of unity ($\omega$ or $\omega^2$), so $z^3 = 1$.
$\left(z^n + \frac{(-1)^n}{z^n}\right)^2 = z^{2n} + z^{-2n} + 2(-1)^n$.
Over $n = 1$ to $15$ (five full periods of 3), $\sum z^{2n} = 0$ and $\sum z^{-2n} = 0$.
$\sum_{n=1}^{15} 2(-1)^n = 2(-1) = -2$ (eight odd and seven even $n$).
So the sum is $-2$ and its modulus is $2$.
Answer: 2. | ||||
| 24 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $X^3 = O$, and $Y\,Z = \alpha^3 I$ (the $X$ and $X^2$ terms cancel), so $Y^{-1} = \frac{1}{\alpha^3}Z = \frac{1}{\alpha}I - \frac{\beta}{\alpha^2}X + \frac{\beta^2 - \alpha\gamma}{\alpha^3}X^2$.
The given $Y^{-1} = \frac{1}{5}I - \frac{2}{5}X + \frac{1}{5}X^2$. Comparing:
$\frac{1}{\alpha} = \frac{1}{5} \Rightarrow \alpha = 5$; $\frac{\beta}{25} = \frac{2}{5} \Rightarrow \beta = 10$; $\frac{100 - 5\gamma}{125} = \frac{1}{5} \Rightarrow \gamma = 15$.
$(\alpha - \beta + \gamma)^2 = (5 - 10 + 15)^2 = 100$.
Answer: 100. | ||||
| 25 | 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 $\gcd(N, 36) = 2$ means $N = 2m$ with $m$ odd and $3 \nmid m$.
$100 \le N \le 999 \Rightarrow 50 \le m \le 499$. Odd $m$ in this range: $51, 53, \dots, 499$, i.e. $225$.
Odd multiples of 3 among them: $51, 57, \dots, 495$, i.e. $\frac{495 - 51}{6} + 1 = 75$.
Count $= 225 - 75 = 150$.
Answer: 150. | ||||
| 26 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Using ${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$ repeatedly (start with ${}^{40}C_0 = {}^{41}C_0$):
${}^{41}C_0 + {}^{41}C_1 = {}^{42}C_1$, ${}^{42}C_1 + {}^{42}C_2 = {}^{43}C_2$, …, so the sum is ${}^{61}C_{20}$.
${}^{61}C_{20} = \frac{61!}{20!\,41!} = \frac{61}{41}\cdot\frac{60!}{20!\,40!} = \frac{61}{41}\,{}^{60}C_{20}$.
$m = 61$, $n = 41$ (coprime), $m + n = 102$.
Answer: 102. | ||||
| 27 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let $a_n = ar^{n-1}$. From $3ar + ar^2 = 2ar^3$: $2r^2 - r - 3 = 0$, so $r = \frac{3}{2}$ or $r = -1$.
$r = -1$: $-a - a = 2a + 1$ gives $a = -\frac{1}{4} < 0$, rejected.
$r = \frac{3}{2}$: $a\left(\frac{3}{2} + \frac{27}{8} - \frac{9}{2}\right) = 1 \Rightarrow \frac{3a}{8} = 1 \Rightarrow a = \frac{8}{3}$.
$a_2 + a_4 + 2a_5 = a\left(\frac{3}{2} + \frac{27}{8} + \frac{81}{8}\right) = \frac{8}{3}\cdot\frac{120}{8} = 40$.
Answer: 40. | ||||
| 28 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Divide numerator and denominator by $x^2$: the integrand becomes $\frac{\left(\frac{2}{x^2} - 1\right)dx}{\left(x + \frac{2}{x}\right)\sqrt{x^2 + \frac{4}{x^2}}}$.
Put $t = x + \frac{2}{x}$: $dt = \left(1 - \frac{2}{x^2}\right)dx$ and $x^2 + \frac{4}{x^2} = t^2 - 4$. As $x$ goes from $0$ to $\sqrt{2}$, $t$ goes from $\infty$ to $2\sqrt{2}$.
$I = \int_{\infty}^{2\sqrt{2}}\frac{-dt}{t\sqrt{t^2 - 4}} = \int_{2\sqrt{2}}^{\infty}\frac{dt}{t\sqrt{t^2 - 4}} = \frac{1}{2}\left[\sec^{-1}\frac{t}{2}\right]_{2\sqrt{2}}^{\infty} = \frac{1}{2}\left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \frac{\pi}{8}$.
$\frac{24}{\pi}\cdot\frac{\pi}{8} = 3$.
Answer: 3. | ||||
| 29 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The point is the foot $(h, k)$ of the perpendicular from the origin to a tangent. $L_1$ is perpendicular to the direction $(h, k)$ through $(h, k)$: $hx + ky = h^2 + k^2$, i.e. $y = -\frac{h}{k}x + \frac{h^2 + k^2}{k}$.
Tangency to $\frac{x^2}{16} - \frac{y^2}{4} = 1$ ($c^2 = a^2m^2 - b^2$): $\frac{(h^2 + k^2)^2}{k^2} = 16\frac{h^2}{k^2} - 4$, so $(h^2 + k^2)^2 = 16h^2 - 4k^2$.
$\alpha = 16$, $\beta = -4$, $\alpha + \beta = 12$.
Answer: 12. | ||||
| 30 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are $2^6 = 64$ such numbers.
Divisibility by 7: every such number equals $111111 + 7\sum 10^i$ (sum over the places holding 8), and $111111 = 7 \times 15873$, so all 64 are multiples of 7.
Divisibility by 3: with $k$ eights the digit sum is $8k + (6 - k) = 7k + 6$, a multiple of 3 iff $3 \mid k$, i.e. $k = 0, 3, 6$: ${}^6C_0 + {}^6C_3 + {}^6C_6 = 22$ numbers.
$p = \frac{22}{64}$, $96p = 33$.
Answer: 33. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $M = \frac{\Phi}{I}$, and flux $\Phi$ has the dimensions of energy per current ($U = \frac{1}{2}MI^2$ gives $M = \frac{2U}{I^2}$).
$[M] = \frac{[ML^2T^{-2}]}{[A^2]} = [ML^2T^{-2}A^{-2}]$.
Answer: (C). | ||||
| 32 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Take downward accelerations as positive. Let $P_2$ and $P_3$ be the second and third (movable) pulleys.
Fixed pulley (string length constant): $a_1 + a_{P_2} = 0$, so $a_{P_2} = -a_1$.
Pulley $P_2$: relative to it the two ends move oppositely, $(a_2 - a_{P_2}) + (a_{P_3} - a_{P_2}) = 0$, so $a_2 + a_{P_3} = 2a_{P_2} = -2a_1$.
Pulley $P_3$: $a_3 + a_4 = 2a_{P_3} = 2(-2a_1 - a_2)$.
Hence $4a_1 + 2a_2 + a_3 + a_4 = 0$.
Answer: (A). | ||||
| 33 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Work done is the signed area under the $F$–$x$ graph.
(a) The line rises from $-F$ to $F$, crossing zero at $x_0$ midway: the negative and positive triangles cancel, $W_1 = 0$.
(b) The same, followed by a rectangle of height $F$ from $x_1$ to $x_2$: $W_2 > 0$.
(c) As (b) plus a further positive triangle after $x_2$: $W_3 > W_2$.
(d) A small positive triangle up to $x_0$, then a negative triangle and a negative rectangle of height $F$ up to $x_2$, and only a small positive triangle after $x_2$: $W_4 < 0$.
So $W_3 > W_2 > W_1 > W_4$.
Answer: (A). | ||||
| 34 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For rolling without slipping, $\omega = \frac{v}{R}$ and $I = \frac{2}{5}MR^2$:
$K_{rot} = \frac{1}{2}\cdot\frac{2}{5}MR^2\cdot\frac{v^2}{R^2} = \frac{1}{5}Mv^2$, $K_{total} = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$.
Ratio $= \frac{1/5}{7/10} = \frac{2}{7}$.
Answer: (B). | ||||
| 35 | GravitationAcceleration due to gravity | Class 11: Gravitation | Assertion and reasonMedium | |
A medium assertion and reason question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation Because of the earth's rotation, the effective $g' = g - \omega^2R\cos^2\lambda$ changes in magnitude with latitude $\lambda$ (and the earth is not a perfect sphere either); its direction also deviates slightly from the centre at intermediate latitudes. So A is false.
At the equator the centrifugal term $m\omega^2R$ points radially outward, along the same line as gravity, so the effective $g$ there points towards the centre. R is true.
Answer: (D). | ||||
| 36 | Properties of Bulk MatterViscosity | 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 viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Reynolds number is the ratio of inertial to viscous forces: $R_e = \frac{\rho v d}{\eta}$, which is dimensionless.
Answer: (C). | ||||
| 37 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The average (translational) kinetic energy per molecule is $\frac{3}{2}kT$, which depends only on temperature. Both gases are at the same temperature, so the ratio is $1 : 1$, whatever the mass ratio.
Answer: (D). | ||||
| 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 The three capacitors are in series across 13 V: $\frac{1}{C} = \frac{1}{10} + \frac{1}{15} + \frac{1}{20} = \frac{6 + 4 + 3}{60} = \frac{13}{60}$, so $C = \frac{60}{13}\,\mu F$.
Every capacitor in series carries the same charge: $Q = CV = \frac{60}{13} \times 13 = 60\,\mu C$.
Answer: (A). | ||||
| 39 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The slab fills half the gap (thickness $\frac{d}{2}$) over the whole plate area, so the capacitor is an air part and a dielectric part in series:
$C = \frac{\varepsilon_0 A}{\frac{d}{2} + \frac{d}{2K}} = \frac{\varepsilon_0 A}{d}\cdot\frac{2K}{K + 1} = 4 \times \frac{6}{4} = 6\,\mu F$.
Answer: (C). | ||||
| 40 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Volume is conserved: $\frac{4}{3}\pi R^3 = 64\cdot\frac{4}{3}\pi r^3 \Rightarrow R = 4r$. Charge $Q = 64q$.
$\frac{\sigma_{big}}{\sigma_{small}} = \frac{64q/(4\pi \cdot 16r^2)}{q/(4\pi r^2)} = 4$.
Answer: (B). | ||||
| 41 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Work along the outer path from A, collapsing as we go.
A to the top-left node: two 5 Ω in series ($10$ Ω) in parallel with the direct 10 Ω: $5$ Ω.
Add the next 5 Ω: $10$ Ω, in parallel with the 10 Ω from A to the next node: $5$ Ω.
Add the next 5 Ω: $10$ Ω, in parallel with the next 10 Ω from A: $5$ Ω.
Add the last 5 Ω to reach B: $10$ Ω, in parallel with the 10 Ω from A to B: $5$ Ω.
Answer: (C). | ||||
| 42 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $W = mB(\cos\theta_1 - \cos\theta_2) = mB(1 - \cos 60^\circ) = 2.0 \times 10^5 \times 14 \times 10^{-5} \times \frac{1}{2} = 14$ J.
Answer: (A). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current In the figure the current enters the first coil at its left end and leaves at its right end, then (along the lower wire) enters the second coil at its right end and leaves at its left end. The two coils are wound alike, so the current circulates in opposite senses in them and their fluxes oppose.
For each coil the induced emf is then $L_1\frac{dI}{dt} - M\frac{dI}{dt}$ and $L_2\frac{dI}{dt} - M\frac{dI}{dt}$, so $L_{eq} = L_1 + L_2 - 2M$.
Answer: (D). | ||||
| 44 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The plane of rotation (vertical, east-west) is perpendicular to the horizontal (north-south) component of the earth's field, so that component is cut.
$\varepsilon = \frac{1}{2}B\omega l^2 = \frac{1}{2} \times 0.2 \times 10^{-4} \times 5 \times 1^2 = 5 \times 10^{-5}$ V $= 50\,\mu V$.
Answer: (B). | ||||
| 45 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves In order of increasing wavelength the spectrum runs: gamma rays ($< 10^{-11}$ m), X-rays ($10^{-11}$–$10^{-8}$ m), ultraviolet, visible ($4$–$7 \times 10^{-7}$ m), infrared, microwaves ($10^{-3}$–$0.1$ m), radio.
So $\lambda_{gamma} < \lambda_{X} < \lambda_{visible} < \lambda_{microwave}$.
Answer: (B). | ||||
| 46 | OpticsWave optics | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on wave optics, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For $v \ll c$, the Doppler shift of light is $\frac{\Delta\lambda}{\lambda} = \frac{v}{c}$ (red shift: the galaxy recedes).
$v = \frac{0.7}{670} \times 3 \times 10^8 \approx 3.13 \times 10^5\ ms^{-1}$.
Answer: (C). | ||||
| 47 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Momentum from the circular path: $p = eBr = 1.6 \times 10^{-19} \times 2 \times 10^{-3} \times 2 \times 10^{-3} = 6.4 \times 10^{-25}$ kg m/s.
$K = \frac{p^2}{2m} = \frac{(6.4 \times 10^{-25})^2}{2 \times 9.1 \times 10^{-31}} \approx 2.25 \times 10^{-19}$ J $\approx 1.41$ eV.
Photon energy: $\frac{1240}{450} \approx 2.76$ eV.
$\phi = 2.76 - 1.41 \approx 1.35$ eV, i.e. approximately 1.36 eV.
Answer: (A). | ||||
| 48 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Decay constants add: $\lambda = \lambda_1 + \lambda_2$, so $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2} = \frac{1}{3} + \frac{1}{4.5} = \frac{5}{9}$.
$T = 1.8$ hours.
Answer: (D). | ||||
| 49 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits In an oscillator a portion of the output power is fed back to the input in phase (positive feedback), so the oscillation sustains itself with no external input. Feedback is through an LC (tank) or RC network, not an LR one, and the base-collector junction stays reverse biased in the active region.
Answer: (B). | ||||
| 50 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter Carrier amplitude $A_c = 40$, message amplitude $A_m = 20$, modulation index $\mu = \frac{20}{40} = 0.5$.
The AM wave has components at $\omega_c$ (amplitude $A_c$) and $\omega_c \pm \omega_m$ (each amplitude $\frac{\mu A_c}{2}$).
The lowest-frequency component is the lower side band $\omega_c - \omega_m$, of amplitude $\frac{0.5 \times 40}{2} = 10$.
Answer: (D). | ||||
| 51 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Heights: $y_1 = 50t - 5t^2$, $y_2 = 50(t - 2) - 5(t - 2)^2$ for $t \ge 2$.
$y_1 = y_2$: $0 = -100 - 5(-4t + 4) = -120 + 20t \Rightarrow t = 6$ s (both at height 120 m).
Answer: 6. | ||||
| 52 | Laws of MotionNewton's laws | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Impulse = change in momentum $= m(v - (-v)) = 2mv = 2 \times 0.4 \times 15 = 12$ Ns.
Answer: 12. | ||||
| 53 | Laws of MotionNewton's laws | Class 11: Laws of Motion | NumericalMedium | |
A medium numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion When the 6th ball has just left the table, balls 1–6 hang and balls 7–10 are on the smooth table.
Acceleration of the system: $a = \frac{6mg}{10m} = 0.6g = 6\ ms^{-2}$ (taking $g = 10\ ms^{-2}$).
The string between the 7th and 8th balls pulls balls 8, 9 and 10: $T = 3ma = 3 \times 2 \times 6 = 36$ N.
Answer: 36. | ||||
| 54 | 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 | NumericalEasy | |
A easy numerical 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 needed per minute: $2000 \times 4.2 \times 40 = 3.36 \times 10^5$ J.
Fuel per minute: $\frac{3.36 \times 10^5}{8 \times 10^3} = 42$ g.
Answer: 42. | ||||
| 55 | ThermodynamicsSecond law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Efficiency $\eta = 1 - \frac{180}{300} = 0.4$. No engine beats a Carnot engine between the same reservoirs: $1 - \frac{324}{T_H} \ge 0.4 \Rightarrow T_H \ge \frac{324}{0.6} = 540$ K.
Answer: 540. | ||||
| 56 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $f_{20} = f_1 + 19 \times 4 = f_1 + 76$ and $f_{20} = 2f_1$, so $f_1 = 76$ Hz and $f_{20} = 152$ Hz.
Answer: 152. | ||||
| 57 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical 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 $F = \frac{\mu_0 I_1I_2 L}{2\pi d} \Rightarrow d = \frac{2 \times 10^{-7} \times 5 \times 5 \times 0.1}{10^{-5}} = 0.05$ m $= 5$ cm.
Answer: 5. | ||||
| 58 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Light emerges only within the critical angle: $\sin C = \frac{3}{4}$, $\tan C = \frac{3}{\sqrt{7}}$.
Radius of the bright circle: $r = h\tan C = \sqrt{7} \times \frac{3}{\sqrt{7}} = 3$ m. Area $= 9\pi\ m^2$, so $x = 9$.
Answer: 9. | ||||
| 59 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter 1 MSD $= \frac{1}{40}$ cm $= 0.025$ cm; 1 VSD $= \frac{49}{50}$ MSD.
Least count $= 1\text{ MSD} - 1\text{ VSD} = \frac{1}{50} \times 0.025$ cm $= 5 \times 10^{-4}$ cm $= 5 \times 10^{-6}$ m.
Answer: 5. | ||||
| 60 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon energy: $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6630 \times 10^{-10}} = 3.0 \times 10^{-19}$ J.
$K_{max} = eV_0 = 1.6 \times 10^{-19} \times 0.42 = 0.672 \times 10^{-19}$ J.
$h\nu_0 = 3.0 \times 10^{-19} - 0.672 \times 10^{-19} = 2.328 \times 10^{-19}$ J, so $\nu_0 = \frac{2.328 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 3.51 \times 10^{14} = 35.1 \times 10^{13}$ /s.
Answer: 35. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom For 4d: $n = 4$, $l = 2$. Angular nodes $= l = 2$; radial nodes $= n - l - 1 = 1$.
Answer: (A). | ||||
| 62 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Invertase hydrolyses cane sugar (sucrose) into glucose and fructose: A-IV.
Zymase ferments glucose into ethanol and $CO_2$: B-III.
Diastase converts starch into maltose: C-I.
Maltase hydrolyses maltose into glucose: D-II.
Answer: (C). | ||||
| 63 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Metalloids lie along the boundary between metals and non-metals: B, Si, Ge, As, Sb, Te (and Po). Sc is a transition metal, and Pb and Bi are metals.
Answer: (D). | ||||
| 64 | Isolation of MetalsFroth flotation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on froth flotation, from Isolation of Metals. NCERT: No matching NCERT chapter A depressant (e.g. NaCN in an ore of ZnS and PbS) selectively stops one sulphide from forming a froth, so the two can be separated. Froth stabilisers (cresols, aniline) stabilise the froth, and collectors (pine oil, xanthates) make the mineral non-wettable.
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 On boiling: $Ca(HCO_3)_2 \rightarrow CaCO_3\downarrow + H_2O + CO_2$ and $Mg(HCO_3)_2 \rightarrow Mg(OH)_2\downarrow + 2CO_2$.
Magnesium is precipitated as the hydroxide because $Mg(OH)_2$ is less soluble than $MgCO_3$.
Answer: (A). | ||||
| 66 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Li gives a crimson-red flame, Na golden yellow and Rb red-violet. Be (like Mg) has electrons held too tightly, being small, for the flame to excite them, so it imparts no colour.
Answer: (D). | ||||
| 67 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $NO_2$ has $5 + 2 \times 6 = 17$ valence electrons, an odd number; the unpaired electron sits on nitrogen (which is why $NO_2$ dimerises to $N_2O_4$). $N_2O$, $N_2O_3$ and $N_2O_5$ all have even electron counts.
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), Mn in $MnO_4^{2-}$ (+6) is both oxidised to $MnO_4^-$ (+7) and reduced to $MnO_2$ (+4): one species is oxidised and reduced at once, a disproportionation.
(B) is only a reduction half-reaction; (C) and (D) are ordinary redox reactions between two different species.
Answer: (A). | ||||
| 69 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Ce ($4f^15d^16s^2$) readily forms $Ce^{4+}$, which has the noble-gas configuration $4f^0$. La ($4f^0$ as $La^{3+}$), Gd ($4f^7$, half filled as $Gd^{3+}$) and Lu ($4f^{14}$ as $Lu^{3+}$) are all especially stable in +3.
Answer: (A). | ||||
| 70 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Clean water has a BOD below 5 ppm, while highly polluted water has a BOD of 17 ppm or more. So B (18 ppm) and C (21 ppm) are highly polluted.
Answer: (C). | ||||
| 71 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes For the same attacking atom, the anion is a stronger nucleophile than its neutral conjugate acid: $OH^- > H_2O$, $RO^- > ROH$, $NH_2^- > NH_3$. Along a period nucleophilicity follows basicity: $OH^- > F^-$.
Only (D) is correct.
Answer: (D). | ||||
| 72 | 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 Chromic oxide in acetic anhydride converts the methyl group to a benzylidene diacetate, $C_6H_5CH(OCOCH_3)_2$, which protects the aldehyde from further oxidation; acid hydrolysis then gives benzaldehyde.
$CrO_3$/acetic acid and $KMnO_4$ oxidise toluene all the way to benzoic acid. CO/HCl with anhydrous $AlCl_3$ (Gattermann–Koch) formylates the ring, giving p-tolualdehyde from toluene, not benzaldehyde.
Answer: (B). | ||||
| 73 | 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 The alkene is 3,3-dimethylbut-1-ene, $(CH_3)_3C-CH=CH_2$.
Oxymercuration ($Hg(OAc)_2$, $H_2O$) adds OH by Markovnikov's rule through a bridged mercurinium ion, so no free carbocation forms and no methyl shift occurs; $NaBH_4$ then replaces HgOAc by H.
Product: $(CH_3)_3C-CH(OH)-CH_3$, 3,3-dimethylbutan-2-ol. (Acid-catalysed hydration would instead rearrange to 2,3-dimethylbutan-2-ol, option C.)
Answer: (A). | ||||
| 74 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons In electrophilic halogenation the strongest activating group decides the position.
(C) o-Cresol: $-OH$ (at C-2, with $CH_3$ at C-1) is far more activating than $-CH_3$ and directs to its ortho and para positions, C-3 and C-5. Both are meta to the methyl group.
(A), (D): the $-CHO$ / $-NO_2$ groups are deactivating, so the activating $CH_3$ wins and the halogen goes ortho/para to it.
(B): $CH_3$ (ortho/para) and $NO_2$ (meta to itself) both point to C-4 and C-6, which are para and ortho to $CH_3$.
Answer: (C). | ||||
| 75 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Passing the vapour of the acid (with formic acid) over MnO at about 573 K gives the aldehyde in one step: $C_6H_5COOH + HCOOH \xrightarrow{MnO,\ 573\ K} C_6H_5CHO + CO_2 + H_2O$.
$LiAlH_4$ reduces the acid right down to benzyl alcohol, $NaBH_4$ does not reduce carboxylic acids, and $KMnO_4$ is an oxidant.
Answer: (C). | ||||
| 76 | 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 HCN adds to the carbonyl group of butan-2-one: $CH_3CH_2C(OH)(CN)CH_3$ (cyanohydrin).
Hot 95% $H_2SO_4$ hydrolyses $-CN$ to $-COOH$ and dehydrates the tertiary alcohol, forming the more substituted, conjugated double bond: $CH_3-CH=C(CH_3)-COOH$ (2-methylbut-2-enoic acid).
Answer: (A). | ||||
| 77 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hinsberg's reagent (benzenesulphonyl chloride or p-toluenesulphonyl chloride) gives with a primary amine a sulphonamide that still has an acidic N–H and dissolves in alkali; with a secondary amine the sulphonamide has no N–H and is insoluble in alkali; tertiary amines do not react.
So statement C is wrong.
Answer: (C). | ||||
| 78 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Sn/HCl reduces nitrobenzene to aniline (A). $NaNO_2$/HCl at 0–5 °C gives benzenediazonium chloride (B). In NaOH, β-naphthol couples at C-1, next to its OH group, giving the orange-red dye 1-phenylazo-2-naphthol (C).
Answer: (C). | ||||
| 79 | 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 (A) sodium dodecylbenzenesulphonate is an anionic detergent; (C) cetyltrimethylammonium bromide is a cationic detergent; (D) a polyethylene glycol stearate is a non-ionic detergent.
(B) sodium stearate is a soap, the sodium salt of a fatty acid, not a synthetic detergent.
Answer: (B). | ||||
| 80 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Water-soluble vitamins (B group, C) are generally excreted in urine and must be supplied regularly; vitamin $B_{12}$ is the exception, being stored in the liver and not readily excreted.
Answer: (D). | ||||
| 81 | 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 $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$.
$CH_4 = \frac{100}{16} = 6.25$ mol; $O_2 = \frac{208}{32} = 6.5$ mol. Burning 6.25 mol $CH_4$ needs 12.5 mol $O_2$, so oxygen is limiting.
$CO_2 = \frac{6.5}{2} = 3.25$ mol $= 3.25 \times 44 = 143$ g.
Answer: 143. | ||||
| 82 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter Per unit cell: A (ccp) $= 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$; B in all octahedral voids $= 12 \times \frac{1}{4} + 1 = 4$.
The atoms at face centres are A atoms. Removing the two on opposite faces removes $2 \times \frac{1}{2} = 1$ A: A $= 3$, B $= 4$, i.e. $A_3B_4$.
$x = 3$.
Answer: 3. | ||||
| 83 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Lone pairs on the central atom: $SF_4$ 1 (see-saw), $XeF_4$ 2 (square planar), $CF_4$ 0, $H_2O$ 2 (bent).
So two species: $XeF_4$ and $H_2O$.
Answer: 2. | ||||
| 84 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics For $H_2O(l) \rightarrow H_2O(g)$, $\Delta n_g = 1$ per mole, so $\Delta U = \Delta H - \Delta n_gRT = 41.1 - \frac{8.31 \times 373}{1000} = 41.1 - 3.10 = 38.0$ kJ mol$^{-1}$.
Answer: 38. | ||||
| 85 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $n = \frac{2.0}{60000} = 3.33 \times 10^{-5}$ mol; taking the solution volume as 0.200 L, $C = 1.667 \times 10^{-4}$ mol L$^{-1}$.
$\pi = CRT = 1.667 \times 10^{-4} \times 0.083 \times 300 = 4.15 \times 10^{-3}$ bar $= 415$ Pa.
Answer: 415. | ||||
| 86 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $2HI \rightleftharpoons H_2 + I_2$. From 1 mol HI with 40% decomposed: HI 0.6, $H_2$ 0.2, $I_2$ 0.2 ($\Delta n_g = 0$, so the pressure does not matter).
$K = \frac{0.2 \times 0.2}{0.6^2} = \frac{1}{9}$.
$\Delta G^\ominus = -RT\ln K = 8.31 \times 300 \times 2.3 \times \log 9 = 2493 \times 2.3 \times 0.954 = 5470$ J mol$^{-1}$.
Answer: 5470. | ||||
| 87 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cu is oxidised (anode), $Sn^{2+}$ reduced (cathode): $E^\ominus_{cell} = -0.14 - 0.34 = -0.48$ V, $n = 2$.
$E = E^\ominus - \frac{0.059}{2}\log\frac{[Cu^{2+}]}{[Sn^{2+}]} = -0.48 - 0.0295\log 10 = -0.5095$ V.
$\Delta G = -nFE = -2 \times 96500 \times (-0.5095) = 98334$ J $\approx 98.3$ kJ $= 983 \times 10^{-1}$ kJ mol$^{-1}$.
Answer: 983. | ||||
| 88 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\frac{k_{cat}}{k_{uncat}} = e^{\Delta E_a/RT}$ with the same $A$. $x = \frac{10000}{8.31 \times 300} = 4.01 \approx 4$.
Answer: 4. | ||||
| 89 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Excess $NH_3$ with air oxidises cobalt(II) to $[Co(NH_3)_6]^{3+}$. Co(III) is $3d^6$; with strong-field $NH_3$ it is low spin, $t_{2g}^6e_g^0$, which is diamagnetic.
Electrons in $t_{2g}$: 6.
Answer: 6. | ||||
| 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 $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$. Water $= \frac{81}{18} = 4.5$ mol, so methane $= 2.25$ mol $= 225 \times 10^{-2}$ mol.
Answer: 225. | ||||
Sit this paper free.
Timed like the exam or chapter by chapter, with a solution for each question. Run by a non-profit trust.