JEE (Main) · 2022
JEE (Main) 2022 — 28 June, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Sets, Relations and Functions2
- Sequences and Series2
9 easy · 20 medium · 1 hard
Physics
30 Q
- Physical World and Measurement2
- Kinematics2
- Work, Energy and Power2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- Some Basic Concepts of Chemistry2
- Classification of Elements and Periodicity2
- p-Block Elements2
21 easy · 7 medium · 2 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Both relations are reflexive ($|a-a| = 0$) and symmetric ($|a-b| = |b-a|$).
$R_1$ is not transitive: $(1, 14) \in R_1$ and $(14, 27) \in R_1$, but $|1 - 27| = 26 > 13$, so $(1, 27) \notin R_1$.
$R_2$ is not transitive: $(1, 2) \in R_2$ and $(2, 14) \in R_2$, but $|1 - 14| = 13$, so $(1, 14) \notin R_2$.
Neither is an equivalence relation.
Answer: (B). | ||||
| 2 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Write $f(x) = a(x + 1)(x - r)$ with $a \ne 0$.
$f(-2) = a(-1)(-2 - r) = a(2 + r)$ and $f(3) = a(4)(3 - r) = a(12 - 4r)$.
$f(-2) + f(3) = a(14 - 3r) = 0$, so $r = \frac{14}{3}$.
Sum of roots $= -1 + \frac{14}{3} = \frac{11}{3}$.
Answer: (A). | ||||
| 3 | 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 Fix $c_2 \in \{4, \dots, 7\}$ and $c_3 \in \{2, \dots, 6\}$. Then $c_1 + c_4 = 30 - c_2 - c_3$ with $c_1, c_4 \ge 0$, which has $31 - c_2 - c_3$ solutions.
Total $= \sum_{c_2=4}^{7}\sum_{c_3=2}^{6}(31 - c_2 - c_3) = 20 \times 31 - 5(4+5+6+7) - 4(2+3+4+5+6) = 620 - 110 - 80 = 430$.
Answer: (D). | ||||
| 4 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem General term of the bracket: $T_{r+1} = \binom{11}{r}\left(\frac{5}{2}\right)^{11-r}\left(-\frac{1}{5}\right)^r x^{33 - 5r}$.
We need $x^0$ (with 1), $x^{-2}$ (with $-x^2$) and $x^{-3}$ (with $3x^3$).
$33 - 5r = 0$ and $33 - 5r = -3$ have no integer solution; $33 - 5r = -2$ gives $r = 7$.
Term independent of $x$ $= -\binom{11}{7}\left(\frac{5}{2}\right)^4\left(-\frac{1}{5}\right)^7 = 330 \cdot \frac{625}{16} \cdot \frac{1}{78125} = \frac{330}{2000} = \frac{33}{200}$.
Answer: (B). | ||||
| 5 | Sequences and SeriesArithmetic and geometric means | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic and geometric means, from Sequences and Series. NCERT: Class 11: Sequences and Series Common difference $d = \frac{100 - a}{n + 1}$. First mean $a + d$, last mean $100 - d$.
$7(a + d) = 100 - d$ gives $7a + 8d = 100$.
With $a = 33 - n$: $d = \frac{67 + n}{n + 1}$, so $7(33 - n)(n + 1) + 8(67 + n) = 100(n + 1)$.
This simplifies to $7n^2 - 132n - 667 = 0$, so $n = \frac{132 \pm 190}{14}$, and the positive root is $n = 23$.
(Check: $a = 10$, $d = 3.75$, first mean 13.75, last mean 96.25, ratio $1 : 7$.)
Answer: (C). | ||||
| 6 | 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 For $x < 0$: $g(x) = e^x - x$ is decreasing with $g(x) > 1$ and $g(x) \to 1$ as $x \to 0^-$. So $f(g(x)) = g(x) - 1 \to 0$.
At $x = 0$: $g(0) = 0$, $f(0) = 1$. So $fog$ is discontinuous at $x = 0$.
For $0 < x < 2$: $g(x) = (x-1)^2 - 1 \in [-1, 0)$, so $f(g(x)) = [g(x)] = -1$ (constant, including at $x = 1$ where $g = -1$).
At $x = 2$: $g(2) = 0$, $f = 1$; for $x > 2$, $g(x) > 0$ and $f(g(x)) = |1 - g(x)| \to 1$. The left-hand limit is $-1$, so $fog$ is discontinuous at $x = 2$.
Elsewhere $fog$ is continuous. Exactly two points.
Answer: (B). | ||||
| 7 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The integrand is $\frac{d}{dt}(f(t)\sec t)$, so $g(x) = f\left(\frac{\pi}{4}\right)\sec\frac{\pi}{4} - f(x)\sec x = 2 - \frac{f(x)}{\cos x}$.
As $x \to \frac{\pi}{2}^-$, $\frac{f(x)}{\cos x}$ is $\frac{0}{0}$; by L'Hospital's rule it tends to $\frac{f'(\pi/2)}{-\sin(\pi/2)} = -1$.
So $\lim g(x) = 2 - (-1) = 3$.
Answer: (B). | ||||
| 8 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replacing $x$ by $x + k$: $f(x + k) + f(x + 2k) = n$, so $f(x + 2k) = f(x)$: $f$ has period $2k$.
Over one period: $\int_0^{2k} f = \int_0^k (f(x) + f(x + k))dx = nk$.
$I_1$ covers $\frac{4nk}{2k} = 2n$ periods: $I_1 = 2n \cdot nk = 2n^2k$.
$I_2$ covers $\frac{4k}{2k} = 2$ periods: $I_2 = 2nk$.
$I_1 + nI_2 = 2n^2k + 2n^2k = 4n^2k$.
Answer: (C). | ||||
| 9 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals For $x \le -1$: $y = \frac{7}{2} + 2x$, zero at $x = -\frac{7}{4}$.
For $-1 \le x \le \frac{1}{2}$: $y = \frac{3}{2}$.
For $x \ge \frac{1}{2}$: $y = \frac{5}{2} - 2x$, zero at $x = \frac{5}{4}$.
The region is a trapezium with parallel sides $\frac{3}{2}$ (from $-1$ to $\frac{1}{2}$) and $3$ (from $-\frac{7}{4}$ to $\frac{5}{4}$), height $\frac{3}{2}$.
Area $= \frac{1}{2}\left(\frac{3}{2} + 3\right)\cdot\frac{3}{2} = \frac{27}{8}$.
Answer: (C). | ||||
| 10 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Put $x = vy^2$, so $dx = y^2dv + 2vy\,dy$.
$2ye^v(y^2dv + 2vy\,dy) + (y^2 - 4vy^2e^v)dy = 0 \Rightarrow 2y^3e^v dv + y^2dy = 0 \Rightarrow 2e^v dv = -\frac{dy}{y}$.
Integrating: $2e^v = -\ln y + C$. At $y = 1$, $x = 0$ so $v = 0$: $C = 2$.
At $y = e$: $2e^v = 1$, $v = -\ln 2$, so $x = vy^2 = -e^2\log_e 2$.
Answer: (D). | ||||
| 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 $\frac{dy}{dx} + 2\tan x\cdot y = 2\sin x$. Integrating factor $e^{\int 2\tan x\,dx} = \sec^2 x$.
$\frac{d}{dx}(y\sec^2 x) = 2\sin x\sec^2 x = 2\sec x\tan x$, so $y\sec^2 x = 2\sec x + C$.
At $\left(\frac{\pi}{4}, 0\right)$: $0 = 2\sqrt{2} + C$, $C = -2\sqrt{2}$.
$y = 2\cos x - 2\sqrt{2}\cos^2 x$.
$\int_0^{\pi/2} y\,dx = 2 - 2\sqrt{2}\cdot\frac{\pi}{4} = 2 - \frac{\pi}{\sqrt{2}}$.
Answer: (B). | ||||
| 12 | Straight LinesSection formula | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on section formula, from Straight Lines. NCERT: Class 11: Straight Lines Let $A = \left(a, \frac{12 + 4a}{3}\right)$ on $L_2$ and $B = \left(b, \frac{10 - 2b}{5}\right)$ on $L_1$. $AP : PB = 1 : 3$ gives $P = \frac{3A + B}{4}$.
$x$: $3a + b = 8$. $y$: $(12 + 4a) + \frac{10 - 2b}{5} = 12$, i.e. $10a - b = -5$.
So $a = \frac{3}{13}$, $b = \frac{95}{13}$: $A = \left(\frac{3}{13}, \frac{56}{13}\right)$, $B = \left(\frac{95}{13}, -\frac{12}{13}\right)$.
$L_1 \cap L_2$: $C = \left(-\frac{15}{13}, \frac{32}{13}\right)$.
$\vec{CA} = \frac{1}{13}(18, 24)$, $\vec{CB} = \frac{1}{13}(110, -44)$.
Area $= \frac{1}{2}\cdot\frac{|18(-44) - 24(110)|}{169} = \frac{3432}{338} = \frac{132}{13}$.
Answer: (B). | ||||
| 13 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $e^2 = \frac{a^2 + b^2}{a^2}$, $l = \frac{2b^2}{a}$; $(e')^2 = \frac{a^2 + b^2}{b^2}$, $l' = \frac{2a^2}{b}$.
First condition: $\frac{a^2 + b^2}{a^2} = \frac{11b^2}{7a}$, so $7(a^2 + b^2) = 11ab^2$.
Second: $\frac{a^2 + b^2}{b^2} = \frac{11a^2}{4b}$, so $4(a^2 + b^2) = 11a^2b$.
Dividing: $\frac{7}{4} = \frac{b}{a}$, $b = \frac{7a}{4}$. Then $4a^2\cdot\frac{65}{16} = \frac{77a^3}{4}$ gives $a = \frac{65}{77}$, $b = \frac{65}{44}$.
$77a + 44b = 65 + 65 = 130$.
Answer: (D). | ||||
| 14 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a} \times \vec{b} = (2 + \alpha)\hat{i} - (\alpha - 2)\hat{j} + (\alpha^2 + 4)\hat{k}$.
$|\vec{a}\times\vec{b}|^2 = (\alpha + 2)^2 + (\alpha - 2)^2 + (\alpha^2 + 4)^2 = (\alpha^2 + 4)(\alpha^2 + 6)$.
Setting this equal to $15(\alpha^2 + 4)$: $\alpha^2 = 9$.
$|\vec{a}|^2 = \alpha^2 + 5 = 14$, $|\vec{b}|^2 = \alpha^2 + 5 = 14$, $\vec{a}\cdot\vec{b} = -2\alpha + 2\alpha - 1 = -1$.
$2(14) + (-1)(14) = 14$.
Answer: (D). | ||||
| 15 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The distance from the vertex to the directrix is $a = \frac{|4(2) - 3(-1) - 21|}{5} = \frac{10}{5} = 2$.
Latus rectum $= 4a = 8$.
Answer: (B). | ||||
| 16 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The normal is perpendicular to both given normals: $(2, 1, -5) \times (3, 5, -7) = (18, -1, 7)$.
Plane: $18x - y + 7z = 36 - 3 - 35 = -2$. With $d > 0$: $-18x + y - 7z = 2$, and $\gcd(18, 1, 7, 2) = 1$.
$a + 7b + c + 20d = -18 + 7 - 7 + 40 = 22$.
Answer: (D). | ||||
| 17 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are $5 \cdot 4 \cdot 3 \cdot 2 = 120$ one-one functions. We need distinct values with $f(a) + 2f(b) = f(c) + f(d)$.
The right side is at most $5 + 4 = 9$, so $f(a) + 2f(b) \le 9$ with $f(a) \ne f(b)$:
$f(b) = 1$: $f(a) + 2 = f(c) + f(d)$ with $c, d$ values distinct from 1 and $f(a)$: $f(a) = 3$ gives 5 = 2 + 3 (uses 3), no; $f(a) = 4$ gives 6 = 2 + 4 (uses 4), no; $f(a) = 5$ gives 7 = 2 + 5 or 3 + 4: $\{3, 4\}$ works, 2 ordered ways; $f(a) = 2$ gives 4 = 1 + 3 (uses 1), no.
$f(b) = 2$: $f(a) + 4 \le 9$; $f(a) = 1$: 5 = 1 + 4 or 2 + 3, both use 1 or 2, no; $f(a) = 3$: 7 = 2 + 5 or 3 + 4, no; $f(a) = 4$: 8 = 3 + 5, works, 2 ways; $f(a) = 5$: 9 = 4 + 5, no.
$f(b) = 3$: $f(a) \le 3$; $f(a) = 1$: 7 = 2 + 5, works (3 + 4 uses 3), 2 ways; $f(a) = 2$: 8 = 3 + 5, uses 3, no.
$f(b) = 4$: $f(a) = 1$: 9 = 4 + 5, uses 4, no.
Favourable: 6. Probability $= \frac{6}{120} = \frac{1}{20}$.
Answer: (D). | ||||
| 18 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\frac{1}{r^2 + 3r + 3} = \frac{(r + 2) - (r + 1)}{1 + (r + 1)(r + 2)}$, so each term is $\tan^{-1}(r + 2) - \tan^{-1}(r + 1)$.
The sum telescopes to $\tan^{-1}(n + 2) - \tan^{-1}2 \to \frac{\pi}{2} - \tan^{-1}2 = \cot^{-1}2$.
$6\tan(\cot^{-1}2) = 6\cdot\frac{1}{2} = 3$.
Answer: (C). | ||||
| 19 | 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 $\vec{a} = (x, y, z)$. $\vec{a} \times (2, 0, 1) = (y, 2z - x, -2y)$.
So $y = 2$ (consistent with $-2y = -4$) and $2z - x = -13$.
Perpendicularity: $3x + \frac{1}{2}y + 2z = 0$, i.e. $3x + 2z = -1$.
Subtracting: $4x = 12$, $x = 3$, $z = -5$. $\vec{a} = (3, 2, -5)$.
Projection on $(2, 2, 1)$: $\frac{6 + 4 - 5}{3} = \frac{5}{3}$.
Answer: (C). | ||||
| 20 | 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 $\tan\alpha = 1$ with $\alpha$ in the third quadrant: $\alpha = \frac{5\pi}{4}$.
$\cos\beta = -\frac{3}{5}$, $\beta$ in the second quadrant: $\sin\beta = \frac{4}{5}$, $\tan\beta = -\frac{4}{3}$.
$\tan(\alpha + \beta) = \frac{1 - \frac{4}{3}}{1 + \frac{4}{3}} = -\frac{1}{7}$.
$\beta = \pi - \cos^{-1}\frac{3}{5} \approx 2.214$, so $\alpha + \beta \approx 3.927 + 2.214 = 6.141$, between $\frac{3\pi}{2}$ and $2\pi$: the fourth quadrant.
Answer: (A). | ||||
| 21 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of $L$: $M = (6 + 3t, 1 + 2t, 2 + 3t)$. $\vec{PM} = (5 + 3t, -1 + 2t, -1 + 3t)$ must be perpendicular to $(3, 2, 3)$:
$15 + 9t - 2 + 4t - 3 + 9t = 10 + 22t = 0$, $t = -\frac{5}{11}$. Foot $M = \left(\frac{51}{11}, \frac{1}{11}, \frac{7}{11}\right)$.
$Q = 2M - P = \left(\frac{91}{11}, -\frac{20}{11}, -\frac{19}{11}\right)$.
$R = \frac{3P + Q}{4} = \frac{1}{44}(124, 46, 80)$, so $\alpha + \beta + \gamma = \frac{250}{44}$.
$22(\alpha + \beta + \gamma) = 125$.
Answer: 125. | ||||
| 22 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Variance 20 for 7 students means $\sum(x_i - 62)^2 = 7 \times 20 = 140$.
A failing student has $x_i < 50$, so $(x_i - 62)^2 > 144 > 140$, which alone exceeds the whole sum.
Hence no student can fail.
Answer: 0. | ||||
| 23 | CirclesRadius and centre | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections The first circle has centre $C_1 = (\sqrt{2}, 3\sqrt{2})$ and radius$^2$ $= 2 + 18 - 14 = 6$.
The second circle's centre is $C_2 = (2\sqrt{2}, 2\sqrt{2})$; $C_1C_2^2 = 2 + 2 = 4$.
The diameter is a chord of the second circle with midpoint $C_1$, so it is perpendicular to $C_2C_1$ and its half-length is the first circle's radius $\sqrt{6}$:
$r^2 = 6 + C_1C_2^2 = 6 + 4 = 10$.
Answer: 10. | ||||
| 24 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability The numerator vanishes at $x = 1$. Since $\sin u = u + O(u^3)$ and $3x^2 - 4x + 1 = (3x - 1)(x - 1)$:
numerator $= (x - 1)[(3x - 1) - (x + 1)] + O((x-1)^3) = 2(x - 1)^2 + O((x-1)^3)$.
For a finite non-zero limit the denominator must have a double root at 1: $2 - 7 + a + b = 0$ and $6 - 14 + a = 0$, so $a = 8$, $b = -3$.
Then $2x^3 - 7x^2 + 8x - 3 = (x - 1)^2(2x - 3)$ and the limit is $\frac{2}{2 - 3} = -2$, as required.
$a - b = 11$.
Answer: 11. | ||||
| 25 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_n = \frac{n^2}{1 - \frac{1}{(n+1)^2}} = \frac{n^2(n + 1)^2}{n(n + 2)} = \frac{n(n + 1)^2}{n + 2} = n^2 + 1 - \frac{2}{n + 2}$.
So $S_n + \frac{2}{n+1} - n - 1 = n^2 - n + \frac{2}{n + 1} - \frac{2}{n + 2}$.
$\sum_{n=1}^{50}(n^2 - n) = 42925 - 1275 = 41650$; the telescoping part is $2\left(\frac{1}{2} - \frac{1}{52}\right) = 1 - \frac{1}{26}$.
Total $= \frac{1}{26} + 41650 + 1 - \frac{1}{26} = 41651$.
Answer: 41651. | ||||
| 26 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalEasy | |
A easy numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Infinitely many solutions: $\frac{\alpha}{2} = \frac{5}{-3} = \frac{\beta + 1}{\gamma + 5}$.
$\alpha = -\frac{10}{3}$, so $9\alpha = -30$.
$3(\beta + 1) = -5(\gamma + 5)$, so $3\beta + 5\gamma = -28$.
$|9\alpha + 3\beta + 5\gamma| = |-30 - 28| = 58$.
Answer: 58. | ||||
| 27 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A^2 = \begin{pmatrix} i & 1 + i \\ 1 - i & -i \end{pmatrix}$, $A^3 = \begin{pmatrix} 0 & i \\ 1 & 1 - i \end{pmatrix}$, $A^4 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$.
$A^2$, $A^3$ and $A^4$ all differ from $A$, and the powers repeat with period 4, so $A^n = A$ exactly when $n \equiv 1 \pmod 4$.
$n = 1, 5, 9, \dots, 97$: 25 values.
Answer: 25. | ||||
| 28 | 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 $z + \bar{z} = (1 + i)z^2$. With $z = x + iy$: $(1 + i)z^2 = 2x$, so $z^2 = x(1 - i)$.
$x^2 - y^2 = x$ and $2xy = -x$.
If $x = 0$: $y = 0$, $z = 0$.
If $y = -\frac{1}{2}$: $x^2 - x - \frac{1}{4} = 0$, $x = \frac{1 \pm \sqrt{2}}{2}$, and $|z|^2 = x^2 + \frac{1}{4} = x + \frac{1}{2}$.
Sum of $|z|^2$: $0 + \left(\frac{1 + \sqrt{2}}{2} + \frac{1}{2}\right) + \left(\frac{1 - \sqrt{2}}{2} + \frac{1}{2}\right) = 2$.
Answer: 2. | ||||
| 29 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalHard | |
A hard numerical 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 Symmetry and $f(a, b) \ge a$ give $f(a, b) \ge \max(a, b)$; $f$ is fixed by its values on unordered pairs.
Pairs with maximum 4 ((1,4), (2,4), (3,4), (4,4)) must take 4.
Pairs with maximum 3 ((1,3), (2,3), (3,3)): each 3 or 4, 8 ways.
Pairs with maximum 2 ((1,2), (2,2)): each 2, 3 or 4.
(1,1): any of 1–4.
Onto: 1 can only come from (1,1), so $f(1,1) = 1$. 2 can only come from the max-2 pairs: $3^2 - 2^2 = 5$ ways with at least one 2.
Of these $5 \times 8 = 40$, remove those with no 3 anywhere: max-2 pairs from $\{2, 4\}$ with at least one 2 (3 ways) and max-3 pairs all 4 (1 way): 3.
Count $= 40 - 3 = 37$.
Answer: 37. | ||||
| 30 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter Try $p = T$, $r = F$, $s = F$ (any $q$):
$p \vee r \vee s$: T (p). $p \vee r \vee \sim s$: T. $p \vee \sim q \vee s$: T. $\sim p \vee \sim r \vee s$: T ($\sim r$). $\sim p \vee \sim r \vee \sim s$: T. $\sim p \vee q \vee \sim s$: T ($\sim s$). $q \vee r \vee \sim s$: T ($\sim s$). $q \vee \sim r \vee \sim s$: T. $\sim p \vee \sim q \vee \sim s$: T ($\sim s$).
All nine are true at once, so the maximum is 9.
Answer: 9. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Time $= \frac{v}{a}$ and length $= \frac{v^2}{a}$, so the same relations hold between the two systems:
$T_2 = \frac{v_2}{a_2} = \frac{n/m^2}{1/(mn)}\cdot\frac{v_1}{a_1} = \frac{n^2}{m}T_1$.
$L_2 = \frac{v_2^2}{a_2} = \frac{n^2/m^4}{1/(mn)}\cdot\frac{v_1^2}{a_1} = \frac{n^3}{m^3}L_1$.
Answer: (A). | ||||
| 32 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\omega = \int(6t^2 - 2t)dt = 2t^3 - t^2 + 10$ (using $\omega(0) = 10$).
$\theta = \int\omega\,dt = \frac{t^4}{2} - \frac{t^3}{3} + 10t + 4$ (using $\theta(0) = 4$).
Answer: (B). | ||||
| 33 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Work done by the retarding force: $W = -\int_{0.5}^{1.5}12x\,dx = -6(1.5^2 - 0.5^2) = -12$ J.
Initial KE $= \frac{1}{2}(2)(4^2) = 16$ J, so the final KE $= 4$ J and $v = \sqrt{\frac{2 \times 4}{2}} = 2.0\ ms^{-1}$.
Answer: (C). | ||||
| 34 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The top of the ladder is $\sqrt{34 - 9} = 5$ m above the floor. Weight $W = 100$ N acts at the middle, 1.5 m horizontally from the foot.
The wall is frictionless, so $F_w$ is horizontal. Torques about the foot: $F_w \times 5 = 100 \times 1.5$, $F_w = 30$ N.
The floor supplies the vertical 100 N and a horizontal friction of 30 N, so $F_f = \sqrt{100^2 + 30^2} = 10\sqrt{109}$ N.
$\frac{F_w}{F_f} = \frac{30}{10\sqrt{109}} = \frac{3}{\sqrt{109}}$.
Answer: (C). | ||||
| 35 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Mass flow $= \frac{9 \times 10^4}{3600} = 25$ kg/s. Power of falling water $= 25 \times 10 \times 40 = 10^4$ W.
Electrical power $= 0.5 \times 10^4 = 5000$ W, enough for $\frac{5000}{100} = 50$ lamps.
Answer: (B). | ||||
| 36 | GravitationNewton's law of gravitation | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation The masses become $\frac{2m}{3}$ and $\frac{4m}{3}$, whose product is $\frac{8}{9}m^2$. At the same separation the force becomes $\frac{8}{9}F$.
Answer: (C). | ||||
| 37 | 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 Density of water $= 10^6\ g\ m^{-3} = 10^3\ kg\ m^{-3}$.
$v_t = \frac{2r^2\rho g}{9\eta} = \frac{2 \times 10^{-12} \times 10^3 \times 10}{9 \times 1.8 \times 10^{-5}} = \frac{2 \times 10^{-8}}{1.62 \times 10^{-4}} = 123.4 \times 10^{-6}\ ms^{-1}$.
Answer: (D). | ||||
| 38 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Over the whole cycle $\Delta U = 0$, so the net work by the gas equals the net heat absorbed: $W_{net} = 40 + 0 - 60 = -20$ J.
AB is vertical on the P–V diagram (constant volume): $W_{AB} = 0$. During BC, $W_{BC} = -50$ J (work done on the gas).
$W_{CA} = -20 - 0 - (-50) = 30$ J; positive, as it should be, since C→A is an expansion at constant pressure.
Answer: (B). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$. $T$ doubles and $M$ halves (32 to 16), so $\frac{T}{M}$ becomes 4 times and $v_{rms}$ doubles.
Answer: (B). | ||||
| 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 At the midpoint both fields point from A towards B and add: $E = 2\cdot\frac{kq}{(d/2)^2} = \frac{8kq}{d^2}$.
$d^2 = \frac{8 \times 9 \times 10^9 \times 8 \times 10^{-6}}{6.4 \times 10^4} = 9$, so $d = 3.0$ m.
Answer: (B). | ||||
| 41 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity With $R = R_0(1 + \alpha t)$ ($R_0$ at 0°C): $\frac{3}{2} = \frac{1 + 30\alpha}{1 + 10\alpha}$, so $3 + 30\alpha = 2 + 60\alpha$ and $\alpha = \frac{1}{30} \approx 0.033\ °C^{-1}$.
Answer: (A). | ||||
| 42 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B = \mu_0(1 + \chi)H$ against $B_0 = \mu_0H$, so $\frac{B - B_0}{B_0} = \chi = 1.2 \times 10^{-5}$.
Answer: (A). | ||||
| 43 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $\frac{F}{l} = \frac{\mu_0 x^2}{2\pi d} = \frac{2 \times 10^{-7}x^2}{0.2} = 10^{-6}x^2 = 2 \times 10^{-6}$, so $x = \sqrt{2} \approx 1.4$.
Answer: (C). | ||||
| 44 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The induced emf is $\varepsilon \propto N$. The wire length is proportional to $N$ (same coil radius) and its resistance $R = \frac{\rho L}{\pi r^2} \propto \frac{N}{r^2}$.
$P = \frac{\varepsilon^2}{R} \propto \frac{N^2 r^2}{N} = Nr^2$.
With $N \to \frac{N}{2}$ and $r \to 2r$: $P \to \frac{1}{2} \times 4 = 2$ times. The power is doubled.
Answer: (D). | ||||
| 45 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $k = \frac{2\pi}{\lambda} = \frac{2\pi}{8 \times 10^{-3}} = \frac{\pi}{4} \times 10^3\ m^{-1}$ and the wave speed is $3 \times 10^8\ ms^{-1}$.
$E_0 = 60\ Vm^{-1}$, so $B_0 = \frac{E_0}{c} = 2 \times 10^{-7}$ T.
Answer: (B). | ||||
| 46 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The plate adds a path difference $(\mu - 1)t = 0.5x\lambda$. The intensity there is unchanged if this is a whole number of wavelengths: $0.5x = n$, $x = 2n$. Of the options only $x = 2$ works.
Answer: (B). | ||||
| 47 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $K_1 = \frac{hc}{3\lambda_2} - \phi$ and $\frac{K_2}{3} = \frac{hc}{3\lambda_2} - \frac{\phi}{3}$.
Since $\phi > 0$, $K_1 < \frac{K_2}{3}$.
Answer: (B). | ||||
| 48 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | StatementsEasy | |
A easy statements question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (A) False: radioactive decay does not depend on physical or chemical conditions.
(B) True: $N = N_0e^{-\lambda t}$.
(C) True in magnitude: $\ln N = \ln N_0 - \lambda t$, slope $-\lambda = -\frac{1}{\tau}$.
(D) False: $\lambda T_{1/2} = \ln 2$, a constant.
Answer: (C). | ||||
| 49 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $D_1$ and $D_2$ are connected in parallel with opposite polarity across the output. When $V_{in} > 0.6$ V, one diode conducts and holds $V_o$ at $+0.6$ V; when $V_{in} < -0.6$ V the other conducts and holds $V_o$ at $-0.6$ V. Between these, both are off and $V_o = V_{in}$.
The input peak exceeds 0.6 V, so the output is the sine wave clipped at $\pm 0.6$ V: waveform (D).
Answer: (D). | ||||
| 50 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter The modulating frequency is $f_m = 10^4$ Hz = 10 kHz. The sidebands are at $f_c \pm f_m$, so the bandwidth is $2f_m = 20$ kHz.
Answer: (C). | ||||
| 51 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalEasy | |
A easy numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Mean $= \frac{1.22 + 1.23 + 1.19 + 1.20}{4} = 1.21$ mm.
Absolute errors: 0.01, 0.02, 0.02, 0.01; mean absolute error $= 0.015$ mm.
Percentage error $= \frac{0.015}{1.21} \times 100 = \frac{150}{121}\%$, so $x = 150$.
Answer: 150. | ||||
| 52 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits With the zener regulating, $V_L = 8$ V and the current through R is $\frac{10 - 8}{100} = 20$ mA.
$I_L = 20 - I_Z$ with $0 \le I_Z \le 10$ mA, so $10 \le I_L \le 20$ mA.
$R_{L,max} = \frac{8}{10\ mA} = 800\ \Omega$, $R_{L,min} = \frac{8}{20\ mA} = 400\ \Omega$. Ratio $= 2$.
Answer: 2. | ||||
| 53 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Angular fringe width $= \frac{\lambda}{d}$; in the medium the wavelength becomes $\frac{\lambda}{\mu}$.
New angular width $= \frac{0.35°}{7/5} = 0.25° = \frac{1}{4}$ degree, so $\alpha = 4$.
Answer: 4. | ||||
| 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 $V_L = V_C$, so the circuit is at resonance and the whole 220 V is across R: $I = \frac{220}{5} = 44$ A, $V_R = 220$ V.
$V_L = 2V_R$ gives $X_L = 2R = 10\ \Omega$.
$L = \frac{X_L}{2\pi f} = \frac{10}{100\pi} = \frac{1}{10\pi}$ H.
As printed (mH) this is $\frac{100}{\pi}$ mH, i.e. $K = 0.01$; with the unit read as henry, evidently what the setter meant, $\frac{1}{K\pi}$ H gives $K = 10$.
Answer: 10 or 0.01. | ||||
| 55 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity From the battery: one resistor in series, then two branches; each branch has a resistor followed by two branches; each of these has a resistor followed by two resistors in parallel to the common point that returns to the battery.
Innermost pair: $\frac{1}{2}\ \Omega$; with its series resistor $1.5\ \Omega$; two in parallel: $0.75\ \Omega$; with the next series resistor $1.75\ \Omega$; two in parallel: $0.875\ \Omega$; with the first resistor $R = 1.875 = \frac{15}{8}\ \Omega$.
$I = \frac{3}{15/8} = \frac{8}{5}$ A, so $a = 8$.
Answer: 8. | ||||
| 56 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Initial charge $Q = 5 \times 30 = 150\ \mu C$. After sharing, the common potential is $V = \frac{150}{5 + 10} = 10$ V.
Charge on $C_2 = 10 \times 10 = 100\ \mu C$.
Answer: 100. | ||||
| 57 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\lambda = \frac{340}{340} = 1$ m $= 100$ cm. A tube closed at one end resonates at lengths $\frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, \dots = 25, 75, 125$ cm.
So 125 cm is the $\frac{5\lambda}{4}$ resonance. Pouring water shortens the column; the next resonance is at 75 cm, which needs $125 - 75 = 50$ cm of water.
Answer: 50. | ||||
| 58 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Continuity: $v_2 = 2v_1$. Bernoulli: $\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}(750)(3v_1^2) = 1125v_1^2 = 4500$, so $v_1 = 2\ ms^{-1}$.
Rate of flow $= A_1v_1 = 1.2 \times 10^{-2} \times 2 = 24 \times 10^{-3}\ m^3s^{-1}$.
Answer: 24. | ||||
| 59 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Block: $mg - T = ma$. Disc: $TR = \frac{1}{2}MR^2\cdot\frac{a}{R}$, so $T = \frac{Ma}{2} = 2a$.
$20 - 2a = 2a$, $a = 5\ ms^{-2}$, $T = 10$ N.
Answer: 10. | ||||
| 60 | 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 For equal distances, $v_{avg} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} = \frac{3}{\frac{1}{11} + \frac{1}{22} + \frac{1}{33}} = \frac{3 \times 66}{6 + 3 + 2} = 18\ ms^{-1}$.
Answer: 18. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryEmpirical and molecular formula | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles per 100 g: C $= \frac{74}{12} = 6.17$, H $= \frac{8.7}{1} = 8.7$, N $= \frac{17.3}{14} = 1.236$.
Dividing by 1.236: C : H : N $= 5 : 7 : 1$, empirical formula $C_5H_7N$ (mass 81).
$\frac{162}{81} = 2$, so the molecular formula is $C_{10}H_{14}N_2$.
Answer: (D). | ||||
| 62 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | StatementsEasy | |
A easy statements question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom (A) True.
(B) False: $l$ runs from 0 to $n - 1$.
(C) True: $m_l = -l, \dots, +l$.
(D) True: $m_s = \pm\frac{1}{2}$.
(E) False: $l = 5$ has $2l + 1 = 11$ orbitals.
Answer: (C). | ||||
| 63 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $SF_4$ has 4 bond pairs and 1 lone pair (trigonal bipyramidal arrangement, see-saw shape). The lone pair takes an equatorial position, where it is at 90° to only the two axial bond pairs (an axial lone pair would have three such 90° repulsions).
Answer: (A). | ||||
| 64 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium $K_a = \frac{[H^+][A^-]}{[HA]}$, so $\frac{[A^-]}{[HA]} = \frac{K_a}{[H^+]} = \frac{1.3 \times 10^{-5}}{10^{-4}} = 0.13$.
Answer: (B). | ||||
| 65 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | MatchingEasy | |
A easy matching question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Metal sulphide sols such as CdS are negatively charged: A→II.
Starch is a macromolecular colloid: B→III.
Hydrated ferric oxide sol is positively charged: C→I.
Cheese is a gel (liquid dispersed in solid): D→IV.
Answer: (C). | ||||
| 66 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | MatchingEasy | |
A easy matching question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties $Cl_2O_7$ (anhydride of $HClO_4$) is acidic: A→IV. $Na_2O$ is basic: B→II. $Al_2O_3$ is amphoteric: C→I. $N_2O$ is neutral: D→III.
Answer: (B). | ||||
| 67 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In copper smelting the iron present as FeO is the impurity (gangue); silica is the flux, and the fusible $FeSiO_3$ formed is the slag.
Answer: (D). | ||||
| 68 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Isotopes have the same number of protons (same atomic number) and the same electronic configuration; they differ in the number of neutrons and hence in atomic mass.
Answer: (D). | ||||
| 69 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties $SO_3$ and $SiO_2$ are acidic, $Al_2O_3$ is amphoteric, and CaO (an alkaline earth metal oxide) is basic.
Answer: (C). | ||||
| 70 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $N_2O$ is linear N–N–O; $N_2O_3$ is $O=N-NO_2$; $N_2O_4$ is $O_2N-NO_2$: all have an N–N bond.
$N_2O_5$ is $O_2N-O-NO_2$, with no N–N bond.
Three compounds.
Answer: (C). | ||||
| 71 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Thiosulphuric acid $H_2S_2O_3$ has a central S (+6) bonded to a terminal S (−2): average +2, two different states.
In $H_2S_2O_6$ both S are +5; in $H_2S_2O_7$ and $H_2S_2O_8$ both are +6.
Answer: (A). | ||||
| 72 | 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 Photochemical smog forms in warm, dry, sunny climates from the action of sunlight on nitrogen oxides and (unsaturated) hydrocarbons; it is oxidising. Smoke + fog + $SO_2$ is classical (reducing) smog, typical of cool humid climates.
Answer: (D). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The compound is $O_2N-CH(CH_3)-CH=C(CH_3)-CO-CH_2-CHO$.
The aldehyde is the principal group, so C1 is the CHO carbon: C2 = $CH_2$, C3 = C=O (oxo), C4 = $C(CH_3)$, C5 = CH, C6 = $CH(NO_2)$, C7 = $CH_3$; the double bond is C4=C5.
Name: 4-methyl-6-nitro-3-oxohept-4-enal.
Answer: (C). | ||||
| 74 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctHard | |
A hard single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The alcohol is 2,2,5-trimethylcyclohexan-1-ol. $H^+$ protonates OH and water leaves, giving a secondary carbocation at C1.
A 1,2-methyl shift from the adjacent quaternary C2 gives a more stable tertiary carbocation at C2 (C1 now carries a Me).
Loss of the H on C1 gives the most substituted (tetrasubstituted) alkene C1=C2, each carrying a Me, with the original Me on the ring: 1,2,4-trimethylcyclohex-1-ene, structure (C).
Answer: (C). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes 4-Bromophenylacetic acid is $4\text{-}Br\text{-}C_6H_4\text{-}CH_2COOH$: one carbon more than a benzylic $CH_2$.
4-Bromotoluene: (i) $Cl_2/\Delta$ chlorinates the side chain, giving $BrC_6H_4CH_2Cl$; (ii) $CN^-$ substitutes, giving $BrC_6H_4CH_2CN$; (iii) acid hydrolysis gives $BrC_6H_4CH_2COOH$.
Answer: (C). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Crossed aldol: the α-H of isobutyraldehyde adds to HCHO, giving A = $HOCH_2-C(CH_3)_2-CHO$.
KCN forms the cyanohydrin B = $HOCH_2-C(CH_3)_2-CH(OH)-CN$.
Hydrolysis gives $HOCH_2-C(CH_3)_2-CH(OH)-COOH$, a γ-hydroxy acid that lactonises to the stable five-membered γ-lactone (pantolactone): ring O–$CH_2$–$C(CH_3)_2$–CH(OH)–C(=O). The OH is on the carbon next to the C=O and the gem-dimethyl carbon is next to it: structure (C).
Answer: (C). | ||||
| 77 | AminesBasicity of amines | Class 12: Amines | StatementsMedium | |
A medium statements question on basicity of amines, from Amines. NCERT: Class 12: Amines In strongly acidic medium aniline is largely protonated to the anilinium ion, which is meta-directing; nitration gives p-nitroaniline (about 51 %) and m-nitroaniline (about 47 %) with little ortho. (B) is true, (A) false.
In the nitrating mixture $H_2SO_4$ protonates $HNO_3$ to give $NO_2^+$: $H_2SO_4$ acts as the acid and $HNO_3$ as a base. (D) is true, (C) false.
Answer: (C). | ||||
| 78 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter (A) is true: natural rubber is linear cis-1,4-polyisoprene and is elastic.
(R) is false: the coiled chains are held together by weak van der Waals forces, not strong polar interactions; that is why rubber stretches.
Answer: (C). | ||||
| 79 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Hydrolysis of sucrose gives two isomers, glucose and fructose ($C_6H_{12}O_6$). Glucose (A) is oxidised by $HNO_3$ to saccharic acid; fructose (B) is laevorotatory. Maltose gives only glucose, lactose gives glucose and galactose (neither laevorotatory), and starch gives only glucose.
Answer: (B). | ||||
| 80 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Tagamet is the antacid cimetidine: a 5-methylimidazole joined through $-CH_2-S-CH_2CH_2-NH-$ to a cyanoguanidine group, structure (C). (A) is histamine, (B) is ranitidine, (D) is the antihistamine terfenadine.
Answer: (C). | ||||
| 81 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $n = \frac{PV}{RT} = \frac{1.5 \times 416}{0.083 \times 300} = \frac{624}{24.9} = 25.06$ mol.
$M = \frac{100}{25.06} = 3.99 \approx 4\ g\ mol^{-1}$.
Answer: 4. | ||||
| 82 | 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 $Mg(s) + \frac{1}{2}O_2(g) \to MgO(s)$: $\Delta n_g = -\frac{1}{2}$.
$\Delta U = \Delta H - \Delta n_g RT = -601.70 + \frac{1}{2} \times 8.3 \times 300 \times 10^{-3} = -601.70 + 1.245 = -600.455$ kJ.
Magnitude $\approx 600$ kJ.
Answer: 600. | ||||
| 83 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $C = \frac{\pi}{RT} = \frac{5.03 \times 10^{-3}}{0.083 \times 300} = 2.02 \times 10^{-4}$ M; in 0.5 L, $n = 1.01 \times 10^{-4}$ mol.
Molar mass of the protein $= \frac{2.5}{1.01 \times 10^{-4}} \approx 24752\ g\ mol^{-1}$.
With glycine's molar mass 75: number of units $= \frac{24752}{75} \approx 330$.
Answer: 330. | ||||
| 84 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Add free energies, not potentials: $Sn^{4+} + 2e^- \to Sn^{2+}$ is ($Sn^{4+} \to Sn$) minus ($Sn^{2+} \to Sn$).
$-2FE^o = -4F(0.010) - [-2F(-0.140)]$, so $2E^o = 0.040 + 0.280 = 0.320$, $E^o = 0.16$ V $= 16 \times 10^{-2}$ V.
Answer: 16. | ||||
| 85 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\frac{A}{A_0} = \left(\frac{1}{2}\right)^{83/200}$, so $\log\frac{A_0}{A} = \frac{83}{200} \times 0.301 = 0.125$.
$\frac{A_0}{A} = $ antilog $0.125 = 1.333$, so $\frac{A}{A_0} = 0.75$: 75 % remains.
Answer: 75. | ||||
| 86 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $CN^-$ is a strong-field ligand.
$[Fe(CN)_6]^{4-}$: $Fe^{2+}$ $d^6$ low spin, $t_{2g}^6$: diamagnetic.
$[Fe(CN)_6]^{3-}$: $Fe^{3+}$ $d^5$ low spin, one unpaired electron: paramagnetic.
$[Ti(CN)_6]^{3-}$: $Ti^{3+}$ $d^1$: paramagnetic.
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$ $d^8$ square planar: diamagnetic.
$[Co(CN)_6]^{3-}$: $Co^{3+}$ $d^6$ low spin: diamagnetic.
Two paramagnetic complexes.
Answer: 2. | ||||
| 87 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (a) $[Co(NH_3)_4Cl_2]Cl$: octahedral $MA_4B_2$, shows cis-trans isomerism.
(b) $[Co(NH_3)_5Cl]Cl_2$: $MA_5B$, no geometrical isomers.
(c) $[Co(NH_3)_6]Cl_3$: $MA_6$, none.
(d) $[Co(NH_3)_5Cl](NO_3)_2$ (or with $NO_3^-$ in the sphere): $MA_5B$ either way, none.
One complex.
Answer: 1. | ||||
| 88 | 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 Mass of C $= 0.793 \times \frac{12}{44} = 0.2163$ g; mass of H $= 0.442 \times \frac{2}{18} = 0.0491$ g.
Mass of O $= 0.492 - 0.2163 - 0.0491 = 0.2266$ g.
% O $= \frac{0.2266}{0.492} \times 100 = 46.06 \approx 46$.
Answer: 46. | ||||
| 89 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $Br_2$ with light ($h\nu$) brings about free-radical substitution. The most reactive C–H bonds are the allylic ones: the $CH_2$ of the ethyl group is allylic (next to the $C=CH_2$) and secondary, so it forms the most stable radical.
Monobromination there gives $CH_3-CHBr-C(=CH_2)-Ar$ (Ar = benzofuran-4-yl). The aromatic benzofuran ring is not attacked under these conditions.
The product contains one bromine atom.
Answer: 1. | ||||
| 90 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter $MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O$.
Moles of $Fe^{2+}$ $= 20 \times 0.05 = 1$ mmol, needing $0.2$ mmol $KMnO_4$, i.e. $\frac{0.2}{0.01} = 20$ mL.
Left in the burette: $50 - 20 = 30$ mL.
Answer: 30. | ||||
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