JEE (Main) · 2022
JEE (Main) 2022 — 28 June, 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
- Three Dimensional Geometry3
- Binomial Theorem2
- Sets, Relations and Functions2
6 easy · 20 medium · 4 hard
Physics
30 Q
- Work, Energy and Power2
- Motion of System of Particles and Rigid Body2
- Electrostatics2
21 easy · 9 medium · 0 hard
Chemistry
30 Q
- p-Block Elements2
- Coordination Compounds2
- Organic Chemistry: Some Basic Principles and Techniques2
22 easy · 8 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 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctHard | |
A hard single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Write ${}^{31}C_{k-1} = {}^{31}C_{32-k}$. Then $\sum_{k=1}^{31} {}^{31}C_k\,{}^{31}C_{32-k}$ is the coefficient of $x^{32}$ in $(1+x)^{31}(1+x)^{31}$, i.e. ${}^{62}C_{32} = \frac{62!}{30!\,32!}$.
In the same way $\sum_{k=1}^{30} {}^{30}C_k\,{}^{30}C_{31-k} = {}^{60}C_{31} = \frac{60!}{29!\,31!}$.
In terms of $\frac{60!}{30!\,31!}$:
$\frac{62!}{30!\,32!} = \frac{60!}{30!\,31!}\cdot\frac{61 \cdot 62}{32}$ and $\frac{60!}{29!\,31!} = \frac{60!}{30!\,31!}\cdot 30$.
So $\alpha = \frac{61 \cdot 62}{32} - 30 = \frac{3782 - 960}{32} = \frac{2822}{32}$, and $16\alpha = 1411$.
Answer: (A). | ||||
| 2 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Look at what each piece produces.
Even $n$: $2n$ runs over $4, 8, 12, \ldots$ — all multiples of 4, each once.
$n = 4m - 1$: $n - 1 = 4m - 2$ runs over $2, 6, 10, \ldots$ — all numbers $\equiv 2 \pmod 4$, each once.
$n = 4m - 3$: $\frac{n+1}{2} = 2m - 1$ runs over $1, 3, 5, \ldots$ — all odd numbers, each once.
The three image sets are disjoint and together are all of $\mathbb{N}$; each piece is one-one. So $f$ is one-one and onto.
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 $\Delta = \begin{vmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & |\lambda| \end{vmatrix} = 2(|\lambda| + 1) - 3(|\lambda| - 1) - 1(-1 - 1) = 7 - |\lambda|$.
No solution needs $\Delta = 0$, so $|\lambda| = 7$.
Then the left side of the third equation is $-2(\text{first}) + 5(\text{second})$: $-2(2, 3, -1) + 5(1, 1, 1) = (1, -1, 7)$. Its right side must then be $-2(-2) + 5(4) = 24$ for the system to be consistent.
$\lambda = 7$: $4\lambda - 4 = 24$ — consistent (infinitely many solutions).
$\lambda = -7$: $4\lambda - 4 = -32 \ne 24$ — no solution.
Answer: (B). | ||||
| 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 For an $n \times n$ matrix, $\det(\text{adj}\,M) = (\det M)^{n-1}$ and $\det(kM) = k^n \det M$; here $n = 3$.
$\det(A^3) = 2^3 = 8$; $\det(\text{adj}(A^3)) = 8^2 = 64$; $\det(5\,\text{adj}(A^3)) = 5^3 \cdot 64 = 8000$.
$\det(\text{adj}(5\,\text{adj}(A^3))) = 8000^2 = 64 \times 10^6$.
$\det(2 \cdot \text{that}) = 2^3 \cdot 64 \times 10^6 = 512 \times 10^6$.
Answer: (A). | ||||
| 5 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations A multiple of 6 is even and has digit sum divisible by 3.
The six digits sum to 24, so the digit left out must be a multiple of 3: leave out 3 or 6.
Leave out 3 (digits 1, 2, 5, 6, 7): last digit 2 or 6, the rest in $4!$ ways: $2 \times 24 = 48$.
Leave out 6 (digits 1, 2, 3, 5, 7): last digit 2: $4! = 24$.
Total $48 + 24 = 72$.
Answer: (D). | ||||
| 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 Let $A_1 = a$, ratio $r > 1$.
$A_1A_3A_5A_7 = a^4r^{12} = \frac{1}{6^4}$, so $ar^3 = A_4 = \frac{1}{6}$.
Then $A_2 = \frac{7}{36} - \frac{1}{6} = \frac{1}{36}$, and $r^2 = \frac{A_4}{A_2} = 6$.
$A_6 + A_8 + A_{10} = A_2(r^4 + r^6 + r^8) = \frac{1}{36}(36 + 216 + 1296) = 1 + 6 + 36 = 43$.
Answer: (C). | ||||
| 7 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $g(x) = -8x^2 + 6x - 1 = -(4x - 1)(2x - 1)$: zero at $x = \frac{1}{4}, \frac{1}{2}$, largest value $g(\frac{3}{8}) = \frac{1}{8}$; $g(0) = -1$, $g(1) = -3$.
$g = -1$ at $x = 0$ and $x = \frac{3}{4}$; $g = -2$ where $8x^2 - 6x - 1 = 0$, i.e. $x = \frac{3 + \sqrt{17}}{8}$.
So $[g] = -1$ on $(0, \frac14)$; $0$ on $[\frac14, \frac12]$; $-1$ on $(\frac12, \frac34]$; $-2$ on $(\frac34, \frac{3 + \sqrt{17}}{8}]$; $-3$ on $(\frac{3 + \sqrt{17}}{8}, 1]$.
Integral $= -\frac14 - \frac14 - 2\left(\frac{3 + \sqrt{17}}{8} - \frac{6}{8}\right) - 3\left(1 - \frac{3 + \sqrt{17}}{8}\right)$
$= -\frac{4}{8} - \frac{2\sqrt{17} - 6}{8} - \frac{15 - 3\sqrt{17}}{8} = \frac{\sqrt{17} - 13}{8}$.
Answer: (C). | ||||
| 8 | 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 Simplify each piece. $x < 0$: $0 < e^x < 1$, so $f = 0$. $0 \le x < 1$: $[x - 1] = -1$, so $f = ae^x - 1$. $1 \le x < 2$: $f(1) = b$, and on $(1, 2)$ $-1 \le \sin \pi x < 0$, so $f = b - 1$. $x \ge 2$: $0 < e^{-x} < 1$, so $f = -c$.
At $x = 1$: $f(1) = b$ but $f(1^+) = b - 1$ — discontinuous for every $a, b, c$. So (A) is false.
At $x = 0$: continuous iff $a - 1 = 0$, i.e. $a = 1$. At $x = 2$: continuous iff $b - 1 = -c$, i.e. $b + c = 1$.
So $f$ can be discontinuous at exactly one point ($x = 1$), which makes (D) false, and that happens exactly when $a = 1$ and $b + c = 1$, giving $a + b + c = 2 \ne 1$. So (C) is true and (B) false.
Answer: (C). | ||||
| 9 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The region lies between the line $y = \sqrt{2}x$ (below) and the parabola branch $y = 2\sqrt{2}\sqrt{x}$ (above). They meet where $2x^2 = 8x$, i.e. $x = 4$; for $1 \le x \le 4$ the parabola is above the line.
Area $= \int_1^4 \left(2\sqrt{2}\sqrt{x} - \sqrt{2}x\right) dx = 2\sqrt{2}\cdot\frac{2}{3}\left(8 - 1\right) - \sqrt{2}\cdot\frac{16 - 1}{2} = \frac{28\sqrt{2}}{3} - \frac{15\sqrt{2}}{2} = \frac{11\sqrt{2}}{6}$.
Answer: (B). | ||||
| 10 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $y = vx$. The bracket becomes $\frac{1}{\sqrt{1 - v^2}} + e^v$ (for $x > 0$). Dividing by $x$:
$\left[\frac{1}{\sqrt{1 - v^2}} + e^v\right]\left(\frac{dy}{dx} - v\right) = 1$, and $\frac{dy}{dx} - v = x\frac{dv}{dx}$.
So $\left[\frac{1}{\sqrt{1 - v^2}} + e^v\right] dv = \frac{dx}{x}$, giving $\sin^{-1}v + e^v = \ln x + C$.
At $(1, 0)$: $0 + 1 = 0 + C$, so $C = 1$.
At $(2\alpha, \alpha)$: $v = \frac12$: $\frac{\pi}{6} + \sqrt{e} = \ln(2\alpha) + 1$, so $\alpha = \frac{1}{2}\exp\left(\frac{\pi}{6} + \sqrt{e} - 1\right)$.
Answer: (A). | ||||
| 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 Rewrite: $\frac{dy}{dx} + \frac{3x^2 - 1}{x - x^3}\,y = \frac{4x^2}{1 - x^2}$.
Since $\frac{d}{dx}(x - x^3) = 1 - 3x^2$, $\int \frac{3x^2 - 1}{x - x^3}dx = -\ln|x - x^3|$, and for $x > 1$ an integrating factor is $\frac{1}{x^3 - x}$.
$\frac{y}{x^3 - x} = \int \frac{4x^2}{1 - x^2}\cdot\frac{1}{x(x^2 - 1)}dx = -\int \frac{4x}{(x^2 - 1)^2}dx = \frac{2}{x^2 - 1} + C$.
So $y = 2x + C(x^3 - x)$. $y(2) = 4 + 6C = -2$ gives $C = -1$.
$y(3) = 6 - 24 = -18$.
Answer: (A). | ||||
| 12 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $f(x) = x^7 + 5x^3 + 3x + 1$. $f'(x) = 7x^6 + 15x^2 + 3 > 0$ for all $x$, so $f$ is strictly increasing and has at most one real zero.
It is a polynomial of odd degree, so $f \to -\infty$ and $+\infty$ at the two ends: exactly one real solution.
Answer: (B). | ||||
| 13 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $e^2 = 1 + \frac{b^2}{a^2} = \frac52$, so $b^2 = \frac32 a^2$.
Latus rectum $\frac{2b^2}{a} = 6\sqrt{2}$: $3a = 6\sqrt{2}$, so $a = 2\sqrt{2}$, $a^2 = 8$, $b^2 = 12$.
$y = mx + c$ touches the hyperbola when $c^2 = a^2m^2 - b^2 = 8 \cdot 4 - 12 = 20$.
Answer: (B). | ||||
| 14 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Centre $C(1, 2)$, radius $\sqrt{5}$.
Tangent at $O$ is perpendicular to $CO$, direction $(1, 2)$: $x + 2y = 0$.
$CP = (\sqrt{5}, 0)$ is horizontal, so the tangent at $P$ is the vertical line $x = 1 + \sqrt{5}$.
$Q = \left(1 + \sqrt{5}, -\frac{1 + \sqrt{5}}{2}\right)$.
$PQ$ is vertical with length $2 + \frac{1 + \sqrt{5}}{2} = \frac{5 + \sqrt{5}}{2}$, and $O$ is at horizontal distance $1 + \sqrt{5}$ from it.
Area $= \frac12(1 + \sqrt{5})\cdot\frac{5 + \sqrt{5}}{2} = \frac{10 + 6\sqrt{5}}{4} = \frac{5 + 3\sqrt{5}}{2}$.
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 Both planes pass through the origin, so the line does too. Its direction is $(-1, 2, -1) \times (3, -5, 2) = (-1, -1, -1)$, i.e. $(1, 1, 1)$.
Distance of $P$ from the line: $|OP|^2 = 14$; projection of $OP$ on $(1, 1, 1)$ is $\frac{1 - 2 + 3}{\sqrt{3}} = \frac{2}{\sqrt{3}}$; so $d^2 = 14 - \frac43 = \frac{38}{3}$.
$Q$ and $R$ are symmetric about the foot of this perpendicular: half of $QR = \sqrt{18 - \frac{38}{3}} = \frac{4}{\sqrt{3}}$, so $QR = \frac{8}{\sqrt{3}}$.
Area $= \frac12\cdot\frac{8}{\sqrt{3}}\cdot\sqrt{\frac{38}{3}} = \frac{4\sqrt{38}}{3}$.
Answer: (B). | ||||
| 16 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Planes through the intersection: $(5x + 8y + 13z - 29) + \lambda(8x - 7y + z - 20) = 0$.
Through $(2, 1, 3)$: $28 + \lambda(-8) = 0$, $\lambda = \frac72$. Normal $2(5, 8, 13) + 7(8, -7, 1) = (66, -33, 33) \parallel (2, -1, 1)$.
Through $(0, 1, 2)$: $5 + \lambda(-25) = 0$, $\lambda = \frac15$. Normal $5(5, 8, 13) + (8, -7, 1) = (33, 33, 66) \parallel (1, 1, 2)$.
$\cos\theta = \frac{2 - 1 + 2}{\sqrt{6}\sqrt{6}} = \frac12$, so $\theta = \frac{\pi}{3}$.
Answer: (A). | ||||
| 17 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $P$: $[\vec{r}\cdot(\hat{i} + 3\hat{j} - \hat{k}) - 6] + \mu[\vec{r}\cdot(-6\hat{i} + 5\hat{j} - \hat{k}) - 7] = 0$.
At $(2, 3, \frac12)$: $(2 + 9 - \frac12 - 6) + \mu(-12 + 15 - \frac12 - 7) = \frac92 - \frac92\mu = 0$, so $\mu = 1$.
$P$: $\vec{r}\cdot(-5\hat{i} + 8\hat{j} - 2\hat{k}) = 13$, so $\vec{a} = -5\hat{i} + 8\hat{j} - 2\hat{k}$, $d = 13$ (the ratio does not depend on scaling).
$\frac{|13\vec{a}|^2}{d^2} = \frac{169(25 + 64 + 4)}{169} = 93$.
Answer: (B). | ||||
| 18 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are 900 three-digit numbers. The hundreds digit has 5 odd and 4 even choices; the tens and units digits have 5 odd and 5 even each.
All three odd: $5 \cdot 5 \cdot 5 = 125$.
Exactly two odd: (H, T odd; U even) $125$; (H, U odd; T even) $125$; (T, U odd; H even) $4 \cdot 5 \cdot 5 = 100$.
Favourable $= 125 + 125 + 125 + 100 = 475$; probability $= \frac{475}{900} = \frac{19}{36}$.
Answer: (A). | ||||
| 19 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $CB = CQ = 80$ m. $PQ = 80\tan\frac{\pi}{8}$ and $AB = 80\tan\theta$.
$PQ = 2AB$ gives $\tan\theta = \frac12\tan\frac{\pi}{8} = \frac{\sqrt{2} - 1}{2}$.
$\tan^2\theta = \frac{(\sqrt{2} - 1)^2}{4} = \frac{3 - 2\sqrt{2}}{4}$.
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 $S_1 = (\sim p) \vee q \vee r = (\sim p) \vee (q \vee r)$, which is exactly $p \to (q \vee r) = S_2$.
So $S_1$ and $S_2$ always have the same truth value: (A), (B) and (D) are true, and (C) — $S_2$ false but $S_1$ true — is impossible.
Answer: (C). | ||||
| 21 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $R_1 - R_2$ holds the pairs $(p, p^n)$ with $n \ge 2$ and $p^n \le 50$.
$p = 2$: $4, 8, 16, 32$ (4 pairs). $p = 3$: $9, 27$ (2). $p = 5$: $25$ (1). $p = 7$: $49$ (1). Larger primes have $p^2 > 50$.
Total $4 + 2 + 1 + 1 = 8$.
Answer: 8. | ||||
| 22 | 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 Put $t = e^x > 0$ and divide by $t^2$: $\left(t^2 + \frac{1}{t^2}\right) + 4\left(t + \frac1t\right) - 58 = 0$.
With $u = t + \frac1t \ge 2$: $u^2 - 2 + 4u - 58 = 0$, i.e. $u^2 + 4u - 60 = 0$, $(u + 10)(u - 6) = 0$, so $u = 6$.
$t + \frac1t = 6$ gives $t = 3 \pm 2\sqrt{2}$, both positive, so two values of $x$.
Answer: 2. | ||||
| 23 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $\sum x = 15 \times 8 = 120$; $\sum x^2 = 15(3^2 + 8^2) = 1095$.
Corrected: $\sum x = 120 - 5 + 20 = 135$, mean $9$; $\sum x^2 = 1095 - 25 + 400 = 1470$.
Variance $= \frac{1470}{15} - 9^2 = 98 - 81 = 17$.
Answer: 17. | ||||
| 24 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}$ and $\vec{b}$ are not parallel, so $\vec{c} = \lambda\vec{a} + \mu\vec{b}$.
$\vec{a}\cdot\vec{a} = 14$, $\vec{a}\cdot\vec{b} = 12$, $\vec{b}\cdot\vec{b} = 19$.
$\vec{a}\cdot\vec{c} = 14\lambda + 12\mu = 5$ and $\vec{b}\cdot\vec{c} = 12\lambda + 19\mu = 0$.
From the second, $\lambda = -\frac{19\mu}{12}$; then $\frac{-266 + 144}{12}\mu = 5$, so $\mu = -\frac{30}{61}$ and $\lambda = \frac{95}{122}$.
$c_1 + c_2 + c_3 = 6\lambda + 7\mu = \frac{570}{122} - \frac{420}{122} = \frac{150}{122}$.
$122(c_1 + c_2 + c_3) = 150$.
Answer: 150. | ||||
| 25 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines The image of $P$ in the $x$-axis, $P'(2, -3)$, lies on the reflected ray $AQ$. Line $P'Q$: slope $\frac{7}{3}$; it meets $y = 0$ at $x = 2 + \frac{9}{7} = \frac{23}{7}$. So $A\left(\frac{23}{7}, 0\right)$.
$R$ divides $AQ$ in $2:1$: $R = \left(\frac{2 \cdot 5 + \frac{23}{7}}{3}, \frac{2 \cdot 4 + 0}{3}\right) = \left(\frac{31}{7}, \frac{8}{3}\right)$.
By the law of reflection the incident and reflected rays make equal angles with the normal at $A$, so the bisector of $\angle PAQ$ is the vertical line $x = \frac{23}{7}$.
The foot of the perpendicular from $R$ on it is $M\left(\frac{23}{7}, \frac{8}{3}\right)$.
$7\alpha + 3\beta = 23 + 8 = 31$.
Answer: 31. | ||||
| 26 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $P = (t, 2t^2 + t + 2)$; the slope of the tangent is $4t + 1$.
The normal passes through $Q$: $(4 - 2t^2 - t - 2)(4t + 1) = -(6 - t)$, which simplifies to $4t^3 + 3t^2 - 3t - 4 = 0$, i.e. $(t - 1)(4t^2 + 7t + 4) = 0$. The quadratic has no real root, so $t = 1$ and $P = (1, 5)$.
Area of $OPQ = \frac12|x_Py_Q - x_Qy_P| = \frac12|1 \cdot 4 - 6 \cdot 5| = 13$.
Answer: 13. | ||||
| 27 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalHard | |
A hard numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Write $A = \{b_1 < b_2 < \ldots < b_{20}\}$. The sums $b_1 + b_1 < b_1 + b_2 < b_2 + b_2 < b_2 + b_3 < \ldots < b_{20} + b_{20}$ are $2 \cdot 20 - 1 = 39$ distinct elements of $A + A$, so $|A + A| \ge 39$.
Equality means there are no other sums; in particular $b_{i-1} + b_{i+1}$ must equal $b_i + b_i$ (it lies strictly between $b_{i-1} + b_i$ and $b_i + b_{i+1}$), so $A$ is an arithmetic progression.
With 20 terms from 1 to 77: $d = \frac{76}{19} = 4$, so $A = \{1, 5, 9, \ldots, 77\}$.
$a_1 + \ldots + a_{18} = \frac{20(1 + 77)}{2} - 1 - 77 = 780 - 78 = 702$.
Answer: 702. | ||||
| 28 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = {}^{12}C_r\,(2x^3)^{12 - r}(3x^{-k})^r$; constant when $36 - 3r - kr = 0$, i.e. $r = \frac{36}{3 + k}$.
$3 + k$ must divide 36 with $k \ge 1$: $3 + k \in \{4, 6, 9, 12, 18, 36\}$, $r \in \{9, 6, 4, 3, 2, 1\}$.
The term is ${}^{12}C_r\,2^{12 - r}3^r$; its power of 2 is that of ${}^{12}C_r$ plus $12 - r$:
$r = 9$: $220 = 2^2 \cdot 55$, total $2^{5}$. $r = 6$: $924 = 2^2 \cdot 231$, total $2^8$. $r = 4$: $495$ odd, total $2^8$. $r = 3$: $220$, total $2^{11}$. $r = 2$: $66$, total $2^{11}$. $r = 1$: $12$, total $2^{13}$.
Only $r = 6$ ($k = 3$) and $r = 4$ ($k = 6$) work.
Answer: 2. | ||||
| 29 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $u = a - 3$, $v = b + 2$ (integers): we need $1 < u^2 + v^2 < 16$, i.e. $2 \le u^2 + v^2 \le 15$.
Points with $u^2 + v^2 \le 15$: $u = 0$: $|v| \le 3$, 7; $u = \pm 1$: $|v| \le 3$, 14; $u = \pm 2$: $|v| \le 3$, 14; $u = \pm 3$: $|v| \le 2$, 10. Total 45.
Remove those with $u^2 + v^2 \le 1$: $(0, 0), (\pm 1, 0), (0, \pm 1)$, 5 points.
$45 - 5 = 40$.
Answer: 40. | ||||
| 30 | CirclesRadius and centre | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Normals pass through the centre. From the two lines: $y = \sqrt{11} + 7\sqrt{7} - 2x$, and then $2\sqrt{11} + 14\sqrt{7} - 3x = 2\sqrt{11} + 6\sqrt{7}$, so $h = \frac{8\sqrt{7}}{3}$, $k = \sqrt{11} + \frac{5\sqrt{7}}{3}$.
Radius = distance of $(h, k)$ from $\sqrt{11}y - 3x - \frac{5\sqrt{77}}{3} - 11 = 0$:
$\sqrt{11}k - 3h - \frac{5\sqrt{77}}{3} - 11 = 11 + \frac{5\sqrt{77}}{3} - 8\sqrt{7} - \frac{5\sqrt{77}}{3} - 11 = -8\sqrt{7}$, so $r = \frac{8\sqrt{7}}{\sqrt{20}}$ and $r^2 = \frac{448}{20}$, $5r^2 = 112$.
$5h - 8k = \frac{40\sqrt{7}}{3} - 8\sqrt{11} - \frac{40\sqrt{7}}{3} = -8\sqrt{11}$, so $(5h - 8k)^2 = 704$.
$704 + 112 = 816$.
Answer: 816. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Assertion and reasonEasy | |
A easy assertion and reason question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $[P\,t] = [ML^{-1}T^{-2}][T] = [ML^{-1}T^{-1}]$, which is the dimension of the coefficient of viscosity: A is true.
Viscosity is defined by $F = \eta A\frac{dv}{dx}$, so $\eta = \frac{F}{A\,(dv/dx)}$. Force divided by velocity gradient alone has dimension $\frac{[MLT^{-2}]}{[T^{-1}]} = [MLT^{-1}]$, not that of $\eta$: R is false.
Answer: (C). | ||||
| 32 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $\frac{v^2}{r} = k^2rt^2$ gives $v = krt$, so the tangential acceleration is $\frac{dv}{dt} = kr$.
Only the tangential force $F_t = mkr$ does work (the centripetal force is perpendicular to the velocity).
$P = F_t v = mkr \cdot krt = mk^2r^2t$.
Answer: (C). | ||||
| 33 | KinematicsUniform circular motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $x = 4\cos\omega t$ and $y = 4\sin\omega t$, so $x^2 + y^2 = 16$: a circle of radius 4 m.
Answer: (A). | ||||
| 34 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Matching listEasy | |
A easy matching list question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Solid sphere about a tangent: $\frac25MR^2 + MR^2 = \frac75MR^2$ (II).
Hollow sphere about a tangent: $\frac23MR^2 + MR^2 = \frac53MR^2$ (I).
Ring about a diameter: half of $MR^2$, i.e. $\frac12MR^2$ (IV).
Disc about a diameter: half of $\frac12MR^2$, i.e. $\frac14MR^2$ (III).
Answer: (A). | ||||
| 35 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation By Kepler's third law $T^2 \propto r^3$: $\frac{r_A^3}{r_B^3} = \frac{T_A^2}{T_B^2} = 4$, so $r_A^3 = 4r_B^3$.
Answer: (C). | ||||
| 36 | Properties of Bulk MatterSurface tension | 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 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: $64r^3 = R^3$, so $r = \frac{R}{4}$ with $R = 1$ cm.
Increase in area $= 64 \cdot 4\pi r^2 - 4\pi R^2 = 4\pi R^2(4 - 1) = 12\pi R^2 = 12\pi \times 10^{-4}\ m^2$.
Gain in surface energy $= 0.075 \times 12\pi \times 10^{-4} \approx 2.8 \times 10^{-4}$ J.
Answer: (A). | ||||
| 37 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | StatementsEasy | |
A easy statements question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics In an adiabatic change $Q = 0$, so the work done by the gas is $W = -\Delta U = -\mu C_v(T_2 - T_1) = -\frac{\mu R(T_2 - T_1)}{\gamma - 1} = \frac{\mu R(T_2 - T_1)}{1 - \gamma}$: Statement I is true.
When work is done on the gas, $W < 0$, so $\Delta U = -W > 0$ and the temperature rises: Statement II is true.
Answer: (A). | ||||
| 38 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | StatementsMedium | |
A medium statements question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Statement I: the field at a point is now the external field plus the field of the charge. At a point where the positive charge's own field points along the external field, the resultant is larger. So the field may increase: true.
Statement II: in a non-uniform field the field at the two charges differs, so $q\vec{E}_1 - q\vec{E}_2 \ne 0$: the net force is not zero. True.
Answer: (A). | ||||
| 39 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Let $k = \frac{1}{4\pi\epsilon_0}$. A and C are each at distance $a$ from D, along perpendicular sides; B is at the diagonal distance $a\sqrt{2}$.
Fields of the charges at A and C: each $\frac{kq}{2a^2}$, at right angles, resultant $\frac{kq}{2a^2}\sqrt{2} = \frac{kq}{\sqrt{2}a^2}$, directed along BD (away from B).
Field of the charge at B: $\frac{kq}{2a^2}$, also along BD.
$E = \frac{kq}{a^2}\left(\frac{1}{\sqrt{2}} + \frac12\right)$.
Answer: (A). | ||||
| 40 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Apply Ampere's law on a circle of radius $r$ about the axis.
$r < R$: no current is enclosed, so $B = 0$.
$r > R$: the whole current $I$ is enclosed, $B \cdot 2\pi r = \mu_0 I$, so $B = \frac{\mu_0 I}{2\pi r}$, which jumps to $\frac{\mu_0 I}{2\pi R}$ at the surface and then falls as $\frac1r$.
That is graph (D).
Answer: (D). | ||||
| 41 | 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 Speed in the medium $v = \frac{E_0}{B_0} = \frac{2.25}{1.5 \times 10^{-8}} = 1.5 \times 10^8$ m/s.
The echo travels $2 \times 3$ km $= 6000$ m: $t = \frac{6000}{1.5 \times 10^8} = 4.0 \times 10^{-5}$ s.
Answer: (B). | ||||
| 42 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\mu = \frac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}} = \cot\frac{A}{2} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}$.
So $\sin\frac{A + \delta_m}{2} = \cos\frac{A}{2} = \sin\left(90^\circ - \frac{A}{2}\right)$, giving $\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}$ and $\delta_m = 180^\circ - 2A$.
Answer: (A). | ||||
| 43 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Resolving power of a telescope $= \frac{D}{1.22\lambda} = \frac{0.244}{1.22 \times 2440 \times 10^{-10}} = \frac{0.244}{2.977 \times 10^{-7}} \approx 8.2 \times 10^5$.
Answer: (C). | ||||
| 44 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Electron: $\lambda_e = \frac{h}{\sqrt{2mK}}$, so $K = \frac{h^2}{2m\lambda_e^2}$.
Photon of energy $K$: $\lambda_p = \frac{hc}{K} = \frac{2mc}{h}\lambda_e^2$.
So $\lambda_p \propto \lambda_e^2$.
Answer: (A). | ||||
| 45 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Energy conservation for a projectile striking a target at rest: kinetic energy of the products $= K_p + Q$.
The products must have positive kinetic energy (they also carry momentum), so the reaction can take place only if $K_p + Q > 0$. For an endothermic reaction ($Q < 0$) this says the projectile must bring more than $|Q|$.
Answer: (D). | ||||
| 46 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Diodes $D_1$ and $D_2$ have their cathodes towards the inputs, and X is pulled up to 10 V through R.
If either A or B is low (0), that diode conducts and holds X low; X is high only when both A and B are high. So X = A·B (an AND gate).
X drives the base of the npn transistor, whose collector is the output: high X saturates the transistor and makes Y low; low X cuts it off and Y is high (5 V). The transistor is a NOT gate.
$Y = \overline{AB}$ (NAND).
Answer: (C). | ||||
| 47 | Experimental SkillsSemiconductor experiments | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on semiconductor experiments, from Experimental Skills. NCERT: No matching NCERT chapter A diode has two terminals and conducts only when forward biased: the multimeter shows a low resistance one way and a very high resistance the other way. (Conduction both ways is a resistor; an initial deflection decaying to zero is a capacitor; three terminals is a transistor.)
Answer: (B). | ||||
| 48 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Assertion and reasonEasy | |
A easy assertion and reason question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits In an n-p-n transistor the main current is carried by electrons, in a p-n-p transistor by holes. Electrons have a higher mobility than holes, so for the same conditions the n-p-n transistor carries more current. Both are true and R explains A.
Answer: (A). | ||||
| 49 | Communication SystemsBandwidth of signals | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on bandwidth of signals, from Communication Systems. NCERT: No matching NCERT chapter Typical bandwidths: television signal about 6 MHz (IV); radio signal about 2 MHz (III); high-quality music about 20 kHz (II); human speech about 3 kHz (I).
A-IV, B-III, C-II, D-I.
Answer: (C). | ||||
| 50 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Beat frequency $= \frac{40}{12} = \frac{10}{3}$ Hz $= \frac{v}{4.08} - \frac{v}{4.16} = v\cdot\frac{0.08}{4.08 \times 4.16}$.
$v = \frac{10}{3}\cdot\frac{16.9728}{0.08} = \frac{10}{3} \times 212.16 = 707.2$ m/s.
Answer: (D). | ||||
| 51 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The bob falls through $h = l(1 - \cos 60^\circ) = 2.5 \times 0.5 = 1.25$ m to the lowest point, where its speed is greatest.
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 1.25} = 5$ m/s.
Answer: 5. | ||||
| 52 | Current ElectricityMetre bridge | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity With the end correction, the effective length on the side of A is $43 + 2 = 45$ cm, and on the other side $100 - 43 = 57$ cm.
$\frac{15}{R} = \frac{45}{57}$, so $R = \frac{15 \times 57}{45} = 19\ \Omega$.
Answer: 19. | ||||
| 53 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity No current in the 10 Ω arm means the bridge is balanced: $\frac{R}{4} = \frac{3}{6}$, so $R = 2\ \Omega$.
The 10 Ω arm can then be removed: $(2 + 4) = 6\ \Omega$ in parallel with $(3 + 6) = 9\ \Omega$ gives $\frac{54}{15} = 3.6\ \Omega$.
$I = \frac{36}{3.6} = 10$ A.
Answer: 10. | ||||
| 54 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\omega = 100$ rad/s: $X_L = \omega L = 10\ \Omega$, $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-4}} = 100\ \Omega$.
$Z = \sqrt{120^2 + (100 - 10)^2} = 150\ \Omega$. $V_{rms} = 20$ V, so $I_{rms} = \frac{20}{150} = \frac{2}{15}$ A.
Power in the resistor $= I_{rms}^2R = \frac{4}{225} \times 120 = \frac{32}{15}$ W.
Heat needed $= 2 \times 16 = 32$ J, so $t = \frac{32}{32/15} = 15$ s.
Answer: 15. | ||||
| 55 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\vec{L} = m\,\vec{r} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 0 \\ 0 & 3 & 1 \end{vmatrix} = -\hat{i} - 3\hat{j} + 9\hat{k}$.
$|\vec{L}| = \sqrt{1 + 9 + 81} = \sqrt{91}$.
Answer: 91. | ||||
| 56 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Momentum is conserved: $60v = (60 + 120) \times 2$, so $v = 6$ m/s.
Answer: 6. | ||||
| 57 | 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 The 2Mg pull on the 4M block exceeds the weight Mg of the hanging mass, so the block moves towards the pull and M rises.
For the system: $a = \frac{2Mg - Mg}{4M + M} = \frac{g}{5}$.
For the hanging mass (accelerating upwards): $T - Mg = M\frac{g}{5}$, so $T = \frac{6}{5}Mg$.
Answer: 6. | ||||
| 58 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | NumericalEasy | |
A easy numerical question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory A monoatomic gas has 3 degrees of freedom: $U = n\cdot\frac{3}{2}RT = 2 \times \frac32 \times 8.31 \times 300 = 7479$ J.
Answer: 7479. | ||||
| 59 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium 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 $m = 24 \times 1.67 \times 10^{-27} \approx 4.0 \times 10^{-26}$ kg; $K = 5 \times 10^3 \times 1.6 \times 10^{-19} = 8 \times 10^{-16}$ J.
$p = \sqrt{2mK} = \sqrt{2 \times 4.0 \times 10^{-26} \times 8 \times 10^{-16}} = 8.0 \times 10^{-21}$ kg m/s.
$r = \frac{p}{qB} = \frac{8.0 \times 10^{-21}}{1.6 \times 10^{-19} \times 0.5} = 0.10$ m $= 10$ cm.
Answer: 10. | ||||
| 60 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Total capacitance $C = 0.01 \times 100 = 1\ \mu F$. Impedance is least at resonance, $\omega L = \frac{1}{\omega C}$.
$\omega = 2\pi \times 500 = 1000\pi$, $\omega^2 = 10^6\pi^2 = 10^7$.
$L = \frac{1}{\omega^2 C} = \frac{1}{10^7 \times 10^{-6}} = 0.1$ H $= 100$ mH.
Answer: 100. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Solid StatePoint defects | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on point defects, from Solid State. NCERT: No matching NCERT chapter A vacancy defect removes particles from lattice sites without changing the volume, so it decreases the density. The other three statements are correct (Schottky: ion pairs missing, density falls; interstitial: extra particles, density rises; Frenkel: ions only shift to interstitial sites, density unchanged).
Answer: (D). | ||||
| 62 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter The zeta (electrokinetic) potential is the potential difference between the fixed layer and the diffuse layer of opposite charge around a colloidal particle; it arises from the charge on the particle's surface.
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 Period 4, group 16 is selenium. The element just above it is sulphur (period 3), with valence configuration $3s^2\,3p^4$.
Answer: (A). | ||||
| 64 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Assertion and reasonMedium | |
A medium assertion and reason question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter A: on the Ellingham diagram the $\Delta G^\circ$ lines for the formation of MgO and $Al_2O_3$ (per mole of $O_2$) cross at about 1350°C. Below it the MgO line lies lower, so Mg reduces $Al_2O_3$; above it the $Al_2O_3$ line lies lower, so Al reduces MgO. A is correct.
R: Mg melts at about 650°C and boils at about 1090°C; Al melts at about 660°C and boils at about 2470°C. R is correct as a fact.
R explains A: the MgO and $Al_2O_3$ lines run almost parallel while both metals are condensed; above the boiling point of Mg (about 1090°C, far below that of Al) MgO forms from Mg vapour, the entropy loss is much larger, and the Mg line becomes much steeper, so it crosses the Al line near 1350°C. The low boiling point of Mg is the reason for the crossover.
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 Dihydrogen reduces copper(II) oxide to the metal: $CuO + H_2 \rightarrow Cu + H_2O$.
Answer: (B). | ||||
| 66 | 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 Very pure nitrogen is made by heating barium azide: $Ba(N_3)_2 \rightarrow Ba + 3N_2$.
Barium nitrate gives $BaO$, $NO_2$ and $O_2$; sodium nitrate gives $NaNO_2$ and $O_2$; sodium nitrite does not give $N_2$ on heating.
Answer: (B). | ||||
| 67 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: the oxide in the higher oxidation state is the more acidic, so $E_2O_5$ is more acidic than $E_2O_3$. False.
Statement II: $N_2O_3$ and $P_2O_3$ are acidic, $As_2O_3$ and $Sb_2O_3$ amphoteric, $Bi_2O_3$ basic: acidic character decreases down the group. True.
Answer: (D). | ||||
| 68 | 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 $Eu^{2+}$ has the configuration $[Xe]4f^7$, a half-filled f subshell, which makes it the most stable divalent lanthanoid ion. ($Yb^{2+}$, $4f^{14}$, also exists but is a stronger reductant; $Sm^{2+}$ is strongly reducing; Ce prefers +3 and +4.)
Answer: (C). | ||||
| 69 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | StatementsMedium | |
A medium statements question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Statement I: $[Ni(CN)_4]^{2-}$ ($Ni^{2+}$, $3d^8$, strong-field $CN^-$) is indeed square planar, $dsp^2$, diamagnetic. But $[Ni(CO)_4]$ (Ni(0); the 4s electrons pair into 3d, giving $3d^{10}$) is tetrahedral, $sp^3$ and diamagnetic, not paramagnetic. So Statement I is false.
Statement II: $[NiCl_4]^{2-}$ has $Ni^{2+}$, $3d^8$, tetrahedral, two unpaired electrons (paramagnetic); $[Ni(CO)_4]$ has $3d^{10}$ and is diamagnetic. The d-configurations differ and only one is paramagnetic. False.
Answer: (B). | ||||
| 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 DDT and dieldrin (organochlorines) and organophosphates are insecticides/pesticides. Sodium arsenite is used as a herbicide (weed-killer), the class set apart from pesticides in this classification.
Answer: (D). | ||||
| 71 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Colourless spots on a TLC plate are seen under UV light, by placing the plate in a jar of iodine crystals, or by spraying a suitable reagent (e.g. ninhydrin for amino acids). A visualising agent is not added to the mobile phase.
Answer: (C). | ||||
| 72 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons A: cyclopentadienyl anion — planar, cyclic, fully conjugated, $6\pi$ electrons ($4n + 2$, $n = 1$): aromatic.
B: cycloheptatrienyl (tropylium) cation — $6\pi$ electrons over seven sp² carbons: aromatic.
C: as drawn, the two ring-fusion carbons carry hydrogens (sp³), so the $\pi$ system is not continuous round the rings (a dihydronaphthalene): not aromatic.
D: cycloheptatrienyl anion — $8\pi$ electrons ($4n$): not aromatic.
Answer: (B). | ||||
| 73 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons HBr adds to the C=C by an electrophilic mechanism. Protonation at the carbon next to $CH_2Br$ gives the benzylic carbocation $Ph-\overset{+}{C}H-CH_2-CH_2Br$, stabilised by resonance with the ring (protonation the other way would give a secondary cation destabilised by the nearby $-CH_2Br$).
$Br^-$ then attacks the benzylic carbon: $Ph-CHBr-CH_2-CH_2Br$.
Answer: (C). | ||||
| 74 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids This is the Cannizzaro reaction. $OD^-$ adds to one PhCHO; that intermediate transfers its own aldehydic H (as hydride) to the carbonyl carbon of the second PhCHO, becoming $PhCOO^-$.
The second molecule becomes $PhCH_2O^-$: both H on its carbon come from aldehyde C–H bonds, none from the solvent. It then takes $D^+$ from $D_2O$ at oxygen.
A $= PhCH_2OD$.
Answer: (A). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $S_N1$ needs a carbocation that can become planar. tert-Butyl, allyl and benzylic chlorides (A, B, D) give stabilised cations readily.
In C the Cl is on a bridgehead carbon of a rigid bicyclic system: the cation there cannot flatten (Bredt-type strain), so it does not form and the compound is inert to $S_N1$.
Answer: (C). | ||||
| 76 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines (1) $Br_2/H_2O$ brominates aniline at both ortho and the para positions: 2,4,6-tribromoaniline.
(2) $NaNO_2/HCl$ diazotises the $NH_2$ group: 2,4,6-tribromobenzenediazonium chloride.
(3) $H_3PO_2$ replaces $-N_2^+$ by H: 1,3,5-tribromobenzene.
Answer: (C). | ||||
| 77 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines A primary aliphatic amine forms an aliphatic diazonium salt, which is unstable even in the cold and on warming loses $N_2$; the carbocation is captured by water to give an alcohol: $RNH_2 + HNO_2 \rightarrow [RN_2^+] \rightarrow ROH + N_2$.
Answer: (B). | ||||
| 78 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Neoprene is polychloroprene, a homopolymer of chloroprene. Buna-S (butadiene-styrene) and PHBV (3-hydroxybutanoic acid + 3-hydroxypentanoic acid) are copolymers.
Answer: (B). | ||||
| 79 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The $\alpha$-helix is held by intramolecular hydrogen bonds between the C=O of one residue and the N–H of the residue four places along the chain.
Answer: (B). | ||||
| 80 | Practical Organic ChemistryDetection of elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on detection of elements, from Practical Organic Chemistry. NCERT: No matching NCERT chapter Sodium sulphide in the fusion extract reacts with sodium nitroprusside: $S^{2-} + [Fe(CN)_5NO]^{2-} \rightarrow [Fe(CN)_5NOS]^{4-}$, the purple complex $Na_4[Fe(CN)_5(NOS)]$.
Answer: (D). | ||||
| 81 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O$; $Cl_2 + 2KI \rightarrow 2KCl + I_2$; $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.
So 1 mol $MnO_2$ ≡ 1 mol $I_2$ ≡ 2 mol thiosulphate.
Thiosulphate $= 0.060 \times 0.1 = 6.0 \times 10^{-3}$ mol, so $MnO_2 = 3.0 \times 10^{-3}$ mol $= 3.0 \times 10^{-3} \times 87 = 0.261$ g.
Percentage $= \frac{0.261}{2.0} \times 100 = 13.05 \approx 13$.
Answer: 13. | ||||
| 82 | 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 $\lambda_{max} = \frac{hc}{\phi} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6.63 \times 10^{-19}} = 3 \times 10^{-7}$ m $= 300$ nm.
Answer: 300. | ||||
| 83 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure P has five bond pairs and no lone pair in $PF_5$ (trigonal bipyramidal): $sp^3d$, so $x = 3$, $y = 1$.
Answer: 1. | ||||
| 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 Expansion into vacuum is against zero external pressure, so $w = -p_{ext}\Delta V = 0$. For an ideal gas at constant temperature $\Delta U = 0$. By the first law $q = \Delta U - w = 0$.
Answer: 0. | ||||
| 85 | SolutionsRaoult's law | Class 12: Solutions | NumericalEasy | |
A easy numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions $p_A = 0.3 \times 50 = 15$ Torr, $p_B = 0.7 \times 100 = 70$ Torr, total $85$ Torr.
$y_B = \frac{70}{85} = \frac{14}{17}$.
Answer: 14. | ||||
| 86 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry $A_2X_3 \rightleftharpoons 2A^{3+} + 3X^{2-}$: $K_{sp} = (2s)^2(3s)^3 = 108s^5 = 1.1 \times 10^{-23}$, so $s^5 \approx 1.02 \times 10^{-25}$ and $s \approx 1.0 \times 10^{-5}$ mol $L^{-1} = 1.0 \times 10^{-2}$ mol $m^{-3}$.
The solution is so dilute that the molar conductivity is the limiting one: $\Lambda_m^\circ = \frac{\kappa}{c} = \frac{3 \times 10^{-5}}{1.0 \times 10^{-2}} = 3 \times 10^{-3}$ S $m^2$ $mol^{-1}$.
Answer: 3. | ||||
| 87 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$: each Cr goes from +6 to +3, so 6 mol of electrons, i.e. 6 F.
Answer: 6. | ||||
| 88 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics 67% completion leaves about one third: $t = \frac{2.303}{k}\log 3$; and $t_{1/2} = \frac{2.303}{k}\log 2$.
$\frac{t}{t_{1/2}} = \frac{\log 3}{\log 2} = \frac{0.4771}{0.3010} = 1.585 = 15.85 \times 10^{-1}$, so $x \approx 16$.
(Using exactly 33% remaining, $\log\frac{100}{33} = 0.4815$ gives 1.60, also 16.)
Answer: 16. | ||||
| 89 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Synergic bonding is the M←CO σ donation together with M→CO π back-donation found in metal carbonyls with the metal in a low oxidation state. All three species are metal carbonyls (metal in oxidation state 0), so all three show it.
Answer: 3. | ||||
| 90 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Molar mass of AgBr $= 188$. Mass of Br $= 0.40 \times \frac{80}{188} = 0.1702$ g.
Percentage $= \frac{0.1702}{0.5} \times 100 = 34.04 \approx 34$.
Answer: 34. | ||||
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