JEE (Main) · 2022
JEE (Main) 2022 — 29 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
- Complex Numbers2
- Matrices and Determinants2
- Sequences and Series2
8 easy · 21 medium · 1 hard
Physics
30 Q
- Electrostatics3
- Kinematics2
- Properties of Bulk Matter2
24 easy · 6 medium · 0 hard
Chemistry
30 Q
- Coordination Compounds3
- Some Basic Concepts of Chemistry2
- Chemical Bonding and Molecular Structure2
21 easy · 8 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $1 + x^2 + x^4 = 0$ is $1 + t + t^2 = 0$ with $t = x^2$, so $\alpha^2 = \omega$ or $\omega^2$, a non-real cube root of unity. Hence $\alpha^6 = 1$.
$1011 = 6 \cdot 168 + 3$, $2022 = 6 \cdot 337$, $3033 = 6 \cdot 505 + 3$, so the expression is $\alpha^3 + 1 - \alpha^3 = 1$.
Answer: (A). | ||||
| 2 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\arg(z-1) - \arg(z+1) = \frac{\pi}{4}$ means the segment from $-1$ to $1$ subtends the angle $\frac{\pi}{4}$ at $z$. For $z$ in the lower half-plane the difference is negative, so the locus is the arc above the real axis of the circle through $\pm 1$ for which the chord of length 2 subtends $\frac{\pi}{4}$: radius $R$ with $2 = 2R\sin\frac{\pi}{4}$, so $R = \sqrt{2}$, centre $(0, 1)$.
Circle: $x^2 + (y-1)^2 = 2$, i.e. $x^2 + y^2 - 2y - 1 = 0$; the arc has $0 < y \le 1 + \sqrt{2}$.
With $x^2 + y^2 = 9$: $9 - 2y - 1 = 0$, so $y = 4 > 1 + \sqrt{2}$. The two curves do not meet.
Answer: (C). | ||||
| 3 | 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 $adj A = \begin{pmatrix}2 & 1\\0 & 2\end{pmatrix}$, and by the binomial theorem (I commutes with everything) $B = (I - adj A)^5$.
$I - adj A = \begin{pmatrix}-1 & -1\\0 & -1\end{pmatrix} = -(I + N)$ with $N = \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}$, $N^2 = O$.
$B = -(I + N)^5 = -(I + 5N) = \begin{pmatrix}-1 & -5\\0 & -1\end{pmatrix}$. Sum of elements $= -7$.
Answer: (C). | ||||
| 4 | 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 The numerators 1, 5, 12, 22, 35, 51, 70 have differences 4, 7, 10, 13, 16, 19, so the $n$th is $\frac{n(3n-1)}{2}$. With $x = \frac{1}{6}$, $S = \sum_{n \ge 1} \frac{3n^2 - n}{2} x^{n-1}$.
Using $\sum n x^{n-1} = \frac{1}{(1-x)^2}$ and $\sum n^2 x^{n-1} = \frac{1+x}{(1-x)^3}$:
$\frac{1+x}{(1-x)^3} = \frac{7/6}{125/216} = \frac{252}{125}$, $\frac{1}{(1-x)^2} = \frac{36}{25} = \frac{180}{125}$.
$S = \frac{1}{2}\left(3 \cdot \frac{252}{125} - \frac{180}{125}\right) = \frac{1}{2} \cdot \frac{576}{125} = \frac{288}{125}$.
Answer: (C). | ||||
| 5 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $x^4 - 2x^3 + 2x - 1 = (x-1)^3(x+1)$, so the expression is $\frac{\sin^2(\pi x)}{(x-1)^2}$.
Put $x = 1 + h$: $\sin(\pi + \pi h) = -\sin(\pi h)$, so the limit is $\lim_{h \to 0}\frac{\sin^2(\pi h)}{h^2} = \pi^2$.
Answer: (D). | ||||
| 6 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f(3) = f(5) = 0$ and $f$ has exactly one critical point in $(3, 5)$, where $\frac{n_1}{x-3} + \frac{n_2}{x-5} = 0$, i.e. $\alpha = \frac{5n_1 + 3n_2}{n_1 + n_2}$.
On $(3, 5)$, $(x-3)^{n_1} > 0$ and $(x-5)^{n_2}$ has sign $(-1)^{n_2}$. If $n_2$ is even, $f > 0$ inside and vanishes at the ends, so $\alpha$ is a local maximum; if $n_2$ is odd, $f < 0$ inside and $\alpha$ is a local minimum.
(A) $n_2 = 4$: maximum, true. (B) $n_2 = 3$: minimum, true. (C) $n_2 = 5$: minimum, so the claimed maximum is NOT true. (D) $n_2 = 6$: maximum, true.
Answer: (C). | ||||
| 7 | 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 Let $a = \int_0^1 f(t)\,dt$, $b = \int_0^1 t f(t)\,dt$. Then $f(x) = (1 + a)x - b$.
$a = \int_0^1 ((1+a)t - b)\,dt = \frac{1+a}{2} - b$, so $a = 1 - 2b$.
$b = \int_0^1 ((1+a)t^2 - bt)\,dt = \frac{1+a}{3} - \frac{b}{2} = \frac{2 - 2b}{3} - \frac{b}{2}$, so $6b = 4 - 4b - 3b$, $b = \frac{4}{13}$, $a = \frac{5}{13}$.
$f(x) = \frac{18x - 4}{13}$. Then $f(6) = \frac{104}{13} = 8$, while $f(2) = \frac{32}{13}$, $f(1) = \frac{14}{13}$, $f(4) = \frac{68}{13}$.
Answer: (D). | ||||
| 8 | 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 left side is the area of the region $0 \le x \le 2$ between the semicircle $y = \sqrt{2x - x^2}$ (circle $(x-1)^2 + y^2 = 1$) and the parabola $y = \sqrt{2x}$ ($x = \frac{y^2}{2}$). Integrate along $y$ instead.
For $1 \le y \le 2$: $x$ runs from $\frac{y^2}{2}$ to 2, giving $\int_1^2 \left(2 - \frac{y^2}{2}\right)dy$.
For $0 \le y \le 1$: the points lie right of the parabola but outside the circle, i.e. $\frac{y^2}{2} \le x \le 1 - \sqrt{1-y^2}$ and $1 + \sqrt{1-y^2} \le x \le 2$. The first strip gives $\int_0^1\left(1 - \sqrt{1-y^2} - \frac{y^2}{2}\right)dy$, the second $\int_0^1 \left(1 - \sqrt{1 - y^2}\right)dy$.
So $I = \int_0^1 \left(1 - \sqrt{1 - y^2}\right)dy$.
Answer: (C). | ||||
| 9 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{1 + y^2} = -\frac{2e^x}{1 + e^{2x}}dx$, so $\tan^{-1} y = -2\tan^{-1}(e^x) + C$. With $y(0) = 0$: $C = \frac{\pi}{2}$.
From the equation at $x = 0$: $2y'(0) + 2 \cdot 1 \cdot 1 = 0$, so $y'(0) = -1$.
At $x = \log_e\sqrt{3}$, $e^x = \sqrt{3}$: $\tan^{-1} y = \frac{\pi}{2} - \frac{2\pi}{3} = -\frac{\pi}{6}$, $y = -\frac{1}{\sqrt{3}}$, $y^2 = \frac{1}{3}$.
$6\left(-1 + \frac{1}{3}\right) = -4$.
Answer: (C). | ||||
| 10 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A tangent of slope m makes $\frac{\pi}{4}$ with slope 3 when $\left|\frac{m - 3}{1 + 3m}\right| = 1$: $m = -2$ or $m = \frac{1}{2}$.
The point of contact of slope m is $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$: $A = \left(\frac{a}{4}, -a\right)$, $B = (4a, 4a)$; $S = (a, 0)$.
Slope SA $= \frac{-a}{a/4 - a} = \frac{4}{3}$, slope SB $= \frac{4a}{3a} = \frac{4}{3}$. So A, S, B are collinear for every $a > 0$ (the slopes multiply to $-1$, so AB is a focal chord).
Answer: (D). | ||||
| 11 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections The circle has centre $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and radius $R = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$.
As printed ($\angle BAC = \frac{\pi}{2}$), BC is a diameter, $BC = 2$, $AC = \sqrt{4 - 2} = \sqrt{2}$ and the area is 1, which is not among the options.
The options fit $\angle BAC = \frac{\pi}{3}$: then $BC = 2R\sin\frac{\pi}{3} = \sqrt{3}$, and $AB = 2R\sin C$ gives $\sin C = \frac{1}{\sqrt{2}}$, $C = \frac{\pi}{4}$ ($\frac{3\pi}{4}$ is too large beside $\frac{\pi}{3}$). So $B = \pi - \frac{\pi}{3} - \frac{\pi}{4} = \frac{5\pi}{12}$.
Area $= \frac{1}{2} \cdot AB \cdot BC \cdot \sin B = \frac{1}{2}\sqrt{2}\sqrt{3} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{6 + 2\sqrt{3}}{8} = \frac{3 + \sqrt{3}}{4}$.
Answer: (C). | ||||
| 12 | 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 The line passes through $(2, -1, -3)$ with direction $(3, -2, -1)$. The plane's normal is $(p, -q, 1)$.
Direction perpendicular to the normal: $3p + 2q - 1 = 0$. Point on the plane: $2p + q - 3 = 5$, so $q = 8 - 2p$.
$3p + 16 - 4p - 1 = 0$ gives $p = 15$, $q = -22$: the plane is $15x + 22y + z = 5$.
Distance from the origin $= \frac{5}{\sqrt{225 + 484 + 1}} = \frac{5}{\sqrt{710}} = \sqrt{\frac{25}{710}} = \sqrt{\frac{5}{142}}$.
Answer: (B). | ||||
| 13 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines The two sides meet at $A = (1, 1)$. Let B lie on $x - 2y + 1 = 0$ and C on $2x - y - 1 = 0$, with $H = \left(\frac{7}{3}, \frac{7}{3}\right)$.
BH ⊥ AC (slope 2), so BH has slope $-\frac{1}{2}$. With $B = (2t - 1, t)$: $\frac{7/3 - t}{10/3 - 2t} = -\frac{1}{2}$ gives $\frac{14}{3} - 2t = -\frac{10}{3} + 2t$, $t = 2$, $B = (3, 2)$.
CH ⊥ AB (slope $\frac{1}{2}$), so CH has slope $-2$. With $C = (s, 2s - 1)$: $\frac{10/3 - 2s}{7/3 - s} = -2$ gives $s = 2$, $C = (2, 3)$.
(Check: BC is $x + y = 5$, perpendicular to AH along $(1, 1)$.)
Centroid $= \left(\frac{1 + 3 + 2}{3}, \frac{1 + 2 + 3}{3}\right) = (2, 2)$; its distance from the origin is $2\sqrt{2}$.
Answer: (C). | ||||
| 14 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $1 + 4 + 2 - 16 = -9$, so $Q = (1, 2, 1) - \frac{2(-9)}{9}(1, 2, 2) = (3, 6, 5)$.
T contains $(0, 0, -1)$, the direction $(1, 1, 2)$ and the vector $(3, 6, 6)$ from $(0, 0, -1)$ to Q. Normal $= (1, 1, 2) \times (3, 6, 6) = (-6, 0, 3) \parallel (2, 0, -1)$.
T: $2x - z = 1$ (through $(0, 0, -1)$). Checking: (2, 1, 0) gives 4; (1, 2, 1) gives 1 ✓; (1, 2, 2) gives 0; (1, 3, 2) gives 0.
Answer: (B). | ||||
| 15 | Vector AlgebraAddition of vectors | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on addition of vectors, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{b} - \vec{a} = (1, \alpha - 4, 1)$ and $\vec{c} - \vec{a} = (2, -6, 2)$ are parallel only when $\alpha - 4 = -3$, i.e. $\alpha = 1$. So the smallest positive integer giving noncollinear points is $\alpha = 2$, $B = (2, 2, 4)$.
Midpoint of BC $= \left(\frac{5}{2}, 0, \frac{9}{2}\right)$. Median through A: $\left(\frac{3}{2}, -4, \frac{3}{2}\right)$, length $\sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \sqrt{\frac{82}{4}} = \frac{\sqrt{82}}{2}$.
Answer: (A). | ||||
| 16 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $\{x, y\} \times \{x, y\}$ has 4 pairs, so there are $2^4 = 16$ relations.
Symmetric: (x, y) and (y, x) together or not at all.
Without them, any subset of $\{(x, x), (y, y)\}$: 4 relations, all transitive.
With both (x, y) and (y, x), transitivity forces (x, x) and (y, y): only the full relation, 1 more.
Probability $= \frac{5}{16}$ (taking all 16 relations equally likely).
Answer: (A). | ||||
| 17 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctMedium | |
A medium single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean $= \frac{3 + 7 + 12 + 43}{5} = 13$.
$\sum x^2 = 9 + 49 + 144 + a^2 + (43 - a)^2 = 2a^2 - 86a + 2051$.
Variance $= \frac{2a^2 - 86a + 2051}{5} - 169 = \frac{2a^2 - 86a + 1206}{5}$.
Modulo 5 the numerator is $2a^2 - a + 1$, which for $a \equiv 0, 1, 2, 3, 4$ is $1, 2, 2, 1, 4$: never 0. So the variance is never an integer.
Answer: (A). | ||||
| 18 | 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 Let the tower have height h at horizontal distance d from the pole. From the pole's base, $h = d\tan 60° = \sqrt{3}d$.
From the tower's top the pole's base is depressed by 60°, so for the pole to subtend 30° its top is depressed by 30°: $\frac{h - 20}{d} = \tan 30° = \frac{1}{\sqrt{3}}$.
$\sqrt{3}d - 20 = \frac{d}{\sqrt{3}}$, so $\frac{2d}{\sqrt{3}} = 20$, $d = 10\sqrt{3}$ and $h = 30$ m.
Answer: (D). | ||||
| 19 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $\sim(A \Rightarrow B) \equiv A \wedge \sim B$. So the negation is $(p \vee q) \wedge \sim((\sim r) \vee p) = (p \vee q) \wedge r \wedge (\sim p)$.
With $\sim p$, $p \vee q$ reduces to q: $(\sim p) \wedge q \wedge r$.
Answer: (C). | ||||
| 20 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $9^n = (1 + 8)^n = 1 + 8n + \sum_{k=2}^{n}{}^nC_k 8^k$, so $\alpha = \sum_{k=2}^{n}{}^nC_k 8^{k-2}$. Likewise $\beta = \sum_{k=2}^{n}{}^nC_k 5^{k-2}$.
$\alpha - \beta = \sum_{k=2}^{n}{}^nC_k(8^{k-2} - 5^{k-2})$; the $k = 2$ term is 0, leaving ${}^nC_3(8 - 5) + {}^nC_4(8^2 - 5^2) + \ldots + {}^nC_n(8^{n-2} - 5^{n-2})$.
Answer: (C). | ||||
| 21 | 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 Cross both sides with $\vec{b}$: $\vec{b} \times \vec{a} + \vec{b} \times (\vec{b} \times \vec{c}) = \vec{0}$, i.e. $\vec{b} \times \vec{a} + (\vec{b} \cdot \vec{c})\vec{b} - |\vec{b}|^2\vec{c} = \vec{0}$.
$\vec{b} \times \vec{a} = (1, 1, 1) \times (1, -2, 3) = (5, -2, -3)$, so $3\vec{c} = 5(1, 1, 1) + (5, -2, -3) = (10, 3, 2)$.
$3(\vec{c} \cdot \vec{a}) = (10, 3, 2) \cdot (1, -2, 3) = 10 - 6 + 6 = 10$.
Answer: 10. | ||||
| 22 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} + \frac{2x}{x - 1}y = \frac{1}{(x-1)^2}$. Integrating factor $= e^{\int \frac{2x}{x-1}dx} = e^{2x + 2\ln(x-1)} = e^{2x}(x-1)^2$.
$\frac{d}{dx}\left[y\,e^{2x}(x-1)^2\right] = e^{2x}$, so $y\,e^{2x}(x-1)^2 = \frac{e^{2x}}{2} + C$.
At $x = 2$: $\frac{1 + e^4}{2e^4} \cdot e^4 = \frac{e^4}{2} + C$, so $C = \frac{1}{2}$.
At $x = 3$: $4e^6 y = \frac{e^6 + 1}{2}$, $y(3) = \frac{e^6 + 1}{8e^6}$. So $\alpha = 6$, $\beta = 8$, $\alpha + \beta = 14$.
Answer: 14. | ||||
| 23 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series First series: $3k$, $k = 1, \ldots, 78$ (up to 234). Second: $4j + 1$, $j = 1, \ldots, 59$ (5 to 237).
Common terms are multiples of 3 that are $\equiv 1 \pmod 4$: 9, 21, 33, … (common difference 12), up to 234: $9 + 12m \le 234$ gives $m \le 18$, so 19 terms, the last 225.
Sum $= \frac{19}{2}(9 + 225) = 19 \times 117 = 2223$.
Answer: 2223. | ||||
| 24 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin x = 1 - \sin^2 x$ gives $\sin^2 x + \sin x - 1 = 0$, $\sin x = \frac{\sqrt{5} - 1}{2} \approx 0.618$ (the other root is below $-1$).
With $\theta = \sin^{-1}(0.618) \approx 0.666$, the solutions are $\theta \approx 0.67$, $\pi - \theta \approx 2.48$, $2\pi + \theta \approx 6.95$, $3\pi - \theta \approx 8.76$; the next, $4\pi + \theta \approx 13.2$, exceeds 10.
Answer: 4. | ||||
| 25 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The ellipse lies inside the circle of radius a and contains the circle of radius b. So $\pi a^2 - \pi ab = 30\pi$ and $\pi ab - \pi b^2 = 18\pi$, i.e. $a(a - b) = 30$, $b(a - b) = 18$.
Subtracting: $(a - b)^2 = 12$.
Answer: 12. | ||||
| 26 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability The left side is $h'(x)$ with $h = f g'$.
f is even, so $f\left(\pm\frac{1}{4}\right) = f\left(\pm\frac{1}{2}\right) = 0$. g is even, so $g'$ is odd and $g'(0) = 0$. Hence h vanishes at least at $-\frac{1}{2}, -\frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}$.
By Rolle's theorem $h'$ has a zero in each of the four gaps between these five points, so at least 4 solutions. Nothing in the data forces any further zero of $g'$ or of $h'$ (for instance $f = (16x^2 - 1)(4x^2 - 1)\cdot k$ and $g = c(x^2 - \frac{9}{16})$ meet every condition with suitable constants), so the minimum is 4.
Answer: 4. | ||||
| 27 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{k+1} = {}^{15}C_k\,2^{15-k}(-1)^k x^{\frac{15 - 2k}{5}}$.
$x^{-1}$: $15 - 2k = -5$, $k = 10$: $m = {}^{15}C_{10}\,2^5 = {}^{15}C_5\,2^5$.
$x^{-3}$: $15 - 2k = -15$, $k = 15$: $n = -1$.
$mn^2 = {}^{15}C_5\,2^5$, so $r = 5$ (r = 10 would give ${}^{15}C_{5}\,2^{10}$, which is different).
Answer: 5. | ||||
| 28 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Let the last digit be $d \ne 0$. The first digit is a non-zero multiple of d ($\lfloor 9/d \rfloor$ ways); the second and third are multiples of d including 0 ($\lfloor 9/d \rfloor + 1$ ways each).
d = 1: $9 \cdot 10 \cdot 10 = 900$; d = 2: $4 \cdot 5 \cdot 5 = 100$; d = 3: $3 \cdot 4 \cdot 4 = 48$; d = 4: $2 \cdot 3 \cdot 3 = 18$; d = 5, 6, 7, 8, 9: $1 \cdot 2 \cdot 2 = 4$ each, 20.
Total $= 900 + 100 + 48 + 18 + 20 = 1086$.
Answer: 1086. | ||||
| 29 | 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 $M^2 = -\alpha^2 I$. With $t = -\alpha^2$, $N = (t + t^2 + \ldots + t^{49})I$ and $I - M^2 = (1 - t)I$.
$(1 - t)(t + \ldots + t^{49}) = t - t^{50} = -\alpha^2 - \alpha^{100}$.
So $\alpha^2 + \alpha^{100} = 2$, which (the left side increasing in $\alpha^2$) holds only for $\alpha^2 = 1$: $\alpha = 1$.
Answer: 1. | ||||
| 30 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Let $g(x) = ax + b$, $f(x) = px^2 + qx + r$.
$g(f(x)) = apx^2 + aqx + ar + b = 4x^2 + 6x + 1$: $ap = 4$, $aq = 6$, $ar + b = 1$.
Leading term of $f(g(x))$: $pa^2 = 8$, so $a = 2$, $p = 2$, $q = 3$.
$f(2x + b) = 2(2x + b)^2 + 3(2x + b) + r = 8x^2 + (8b + 6)x + 2b^2 + 3b + r = 8x^2 - 2x$: $b = -1$, $r = 1$ (and $ar + b = 1$ ✓).
$f(x) = 2x^2 + 3x + 1$, $g(x) = 2x - 1$: $f(2) + g(2) = 15 + 3 = 18$.
Answer: 18. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 From rest, $s = \frac{1}{2}at^2$: in time t it covers 10 m, in time 2t it covers $4 \times 10 = 40$ m. In the next t s it covers $40 - 10 = 30$ m.
Answer: (C). | ||||
| 32 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $\Delta D = D\alpha_L\Delta T$: $\Delta T = \frac{6.241 - 6.230}{6.230 \times 1.4 \times 10^{-5}} = \frac{0.011}{8.722 \times 10^{-5}} \approx 126$ K.
Required temperature $\approx 27 + 126 = 153$°C; the nearest option is 152.7°C.
Answer: (D). | ||||
| 33 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The components along the line of the charges cancel; the net force is along the bisector: $F = \frac{2kQq\,x}{\left(x^2 + \frac{d^2}{4}\right)^{3/2}}$.
$\frac{dF}{dx} = 0$ gives $x^2 + \frac{d^2}{4} = 3x^2$, so $x^2 = \frac{d^2}{8}$, $x = \frac{d}{2\sqrt{2}}$.
Answer: (D). | ||||
| 34 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Light is slower in B, so B is denser and the ray goes from denser to rarer. Critical angle: $\sin C = \frac{n_A}{n_B} = \frac{v_B}{v_A} = \frac{1.5}{2.0} = \frac{3}{4}$.
Total internal reflection needs $\theta > \sin^{-1}\left(\frac{3}{4}\right)$.
Answer: (D). | ||||
| 35 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (182, 74) $\xrightarrow{\alpha}$ (178, 72) $\xrightarrow{\beta^-}$ (178, 73) $\xrightarrow{\alpha}$ (174, 71) $\xrightarrow{\gamma}$ (174, 71).
Answer: (A). | ||||
| 36 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The light has angular frequencies $6 \times 10^{15}$ and $9 \times 10^{15}$ rad/s; the maximum kinetic energy comes from the higher one: $\nu = \frac{9 \times 10^{15}}{2\pi} = 1.432 \times 10^{15}$ Hz.
$h\nu = 4.14 \times 10^{-15} \times 1.432 \times 10^{15} = 5.93$ eV, so $K_{max} = 5.93 - 2.50 = 3.43 \approx 3.42$ eV.
Answer: (D). | ||||
| 37 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $q = q_0e^{-t/RC}$ and $U \propto q^2 \propto e^{-2t/RC}$.
$U = \frac{U_0}{2}$: $t_1 = \frac{RC\ln 2}{2}$. $q = \frac{q_0}{8}$: $t_2 = RC\ln 8 = 3RC\ln 2$.
$\frac{t_1}{t_2} = \frac{1}{6}$.
Answer: (D). | ||||
| 38 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics On a P–V diagram from the same initial state, the isobaric line stays at the initial pressure, the isotherm falls, and the adiabat falls fastest. The work is the area under each curve between $V_1$ and $V_2$, so $W_2 < W_1 < W_3$.
Answer: (D). | ||||
| 39 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter With equal currents in the same direction the fields cancel at the mid-point, so the currents are opposite and the fields add:
$B = 2 \cdot \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{\pi(0.04)} = 10^{-5} I$ T. Setting $10^{-5}I = 300 \times 10^{-6}$ gives $I = 30$ A.
Answer: (B). | ||||
| 40 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation $T \propto r^{3/2}$: $T' = 7 \times 3^{3/2} = 7 \times 5.196 \approx 36$ hours.
Answer: (B). | ||||
| 41 | Communication SystemsRange of a TV tower | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on range of a tv tower, from Communication Systems. NCERT: No matching NCERT chapter Range $d = \sqrt{2Rh}$, so $d \propto \sqrt{h}$. Doubling d needs $h' = 4h = 500$ m: an increase of $500 - 125 = 375$ m.
Answer: (C). | ||||
| 42 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\omega = \pi$ rad/s and $\omega = \sqrt{\frac{g}{L}}$, so $L = \frac{g}{\pi^2} = \frac{9.8}{9.87} \approx 0.993$ m $\approx 99.4$ cm.
Answer: (C). | ||||
| 43 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $n = \frac{16}{2} + \frac{128}{32} = 8 + 4 = 12$ mol. At STP, $V = 12 \times 22.4$ L $= 268.8$ L $\approx 2.7 \times 10^5\ cm^3 = 27 \times 10^4\ cm^3$.
Answer: (C). | ||||
| 44 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsEasy | |
A easy statements 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 Statement I: the electric force $q\vec{E}$ does work and changes the speed and kinetic energy; the magnetic force $q\vec{v} \times \vec{B}$ is perpendicular to $\vec{v}$ and does no work. Correct.
Statement II: a positive charge is accelerated along $\vec{E}$, not perpendicular to it, and the magnetic force is perpendicular to $\vec{B}$, not along it. Incorrect.
Answer: (C). | ||||
| 45 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Driving force $= 4 \times 10 = 40$ N; friction $= 0.02 \times 40 \times 10 = 8$ N; total mass 44 kg.
$a = \frac{40 - 8}{44} = \frac{8}{11}\ ms^{-2}$.
Answer: (D). | ||||
| 46 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The block falls from A to B through the height $y_0$ (A is at height y above the ground). Energy conservation: kinetic energy at B $= mgy_0$.
Answer: (D). | ||||
| 47 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion For the block: $Mg - N = Ma$ with $N = \frac{Mg}{4}$, so $a = \frac{3g}{4}$.
Answer: (C). | ||||
| 48 | 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 $E_x = -\frac{\partial V}{\partial x} = -6x = -6\ Vm^{-1}$ at x = 1: 6 $Vm^{-1}$ along the negative x-axis.
Answer: (D). | ||||
| 49 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Series: $I = \frac{2E}{2 + 2r}$. Parallel: $I = \frac{E}{2 + r/2}$.
Equal: $2\left(2 + \frac{r}{2}\right) = 2 + 2r$, i.e. $4 + r = 2 + 2r$, so $r = 2$ Ω.
Answer: (A). | ||||
| 50 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Maximum range $\frac{u^2}{g} = 100$ m. Thrown vertically with the same speed, it rises $\frac{u^2}{2g} = 50$ m.
Answer: (B). | ||||
| 51 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Observed reading $= 1.7 + 5 \times 0.01 = 1.75$ cm.
Corrected $= 1.75 - (-0.05) = 1.80$ cm $= 180 \times 10^{-2}$ cm.
Answer: 180. | ||||
| 52 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical 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 The ball enters the water at its terminal velocity: $v = \frac{2r^2(\rho - \sigma)g}{9\eta} = \frac{2 \times 10^{-8} \times (10^4 - 10^3) \times 10}{9 \times 10^{-5}} = \frac{1.8 \times 10^{-3}}{9 \times 10^{-5}} = 20\ ms^{-1}$.
Free fall: $v^2 = 2gh$, $h = \frac{400}{20} = 20$ m.
Answer: 20. | ||||
| 53 | 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 $\lambda = \frac{336}{400} = 0.84$ m $= 84$ cm, so $\frac{\lambda}{4} = 21$ cm.
First resonance: $l_1 + e = \frac{\lambda}{4}$, so the end correction $e = 21 - 20 = 1$ cm.
Third resonance: $l_3 + e = \frac{5\lambda}{4} = 105$ cm, so $l_3 = 104$ cm.
Answer: 104. | ||||
| 54 | 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 Voltmeter in parallel with 500 Ω: $\frac{500 \times 2000}{2500} = 400$ Ω. Total resistance $= 400 + 600 = 1000$ Ω, current $= \frac{20}{1000} = 0.02$ A.
Reading $= 0.02 \times 400 = 8$ V.
Answer: 8. | ||||
| 55 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - eV$, so $v^2 = u^2 - \frac{2eV}{m} = 36 \times 10^{10} - \frac{2 \times 1.6 \times 10^{-19} \times 0.4}{9 \times 10^{-31}} = 36 \times 10^{10} - 14.22 \times 10^{10} = 21.78 \times 10^{10}$.
$v = 4.667 \times 10^5 = \frac{14}{3} \times 10^5\ ms^{-1}$, so $x = 14$.
Answer: 14. | ||||
| 56 | Electromagnetic WavesDisplacement current | Class 12: Electromagnetic Waves | NumericalEasy | |
A easy numerical question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $I_d = C\frac{dV}{dt} = \frac{\varepsilon_0 A}{d}\frac{dV}{dt}$, so $d = \frac{8.85 \times 10^{-12} \times 40 \times 10^{-4} \times 10^6}{4.425 \times 10^{-6}} = \frac{3.54 \times 10^{-8}}{4.425 \times 10^{-6}} = 8 \times 10^{-3}$ m.
Answer: 8. | ||||
| 57 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $I_1 = \frac{ML^2}{3}$. Ring: $2\pi R = L$, $I_2 = \frac{MR^2}{2} = \frac{ML^2}{8\pi^2}$.
$\frac{I_1}{I_2} = \frac{8\pi^2}{3}$, so $x = 8$.
Answer: 8. | ||||
| 58 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $6.25\% = \frac{1}{16} = \left(\frac{1}{2}\right)^4$: four half-lives, $x = 4 \times 5 = 20$ years.
Answer: 20. | ||||
| 59 | 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 $\beta = \frac{\lambda D}{d}$, so $\Delta\beta = \frac{\lambda\,\Delta D}{d}$: $\lambda = \frac{3 \times 10^{-5} \times 10^{-3}}{5 \times 10^{-2}} = 6 \times 10^{-7}$ m $= 600$ nm.
Answer: 600. | ||||
| 60 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current In phase means resonance: $f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.5 \times 10^{-3} \times 200 \times 10^{-6}}} = \frac{1}{2\pi \times 3.162 \times 10^{-4}} \approx 503$ Hz $= 5.03 \times 10^2$ Hz.
To the nearest integer, 5.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\frac{0.02858 \times 0.112}{0.5702} = 0.005613\ldots$. In multiplication and division the result keeps as many significant figures as the least precise factor: 0.112 has three. So the answer is 0.00561.
Answer: (B). | ||||
| 62 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom For a 2s orbital $\Psi^2$ is largest at the nucleus, has one radial node ($n - l - 1 = 1$) where it is zero, then rises to a second, smaller maximum and decays. It can never be negative.
(A) has no node (1s); (C) goes negative, impossible for $\Psi^2$; (D) has two nodes (3s). Only (B) fits.
Answer: (B). | ||||
| 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 Each has 10 electrons ($CH_4$: 6 + 4; $NH_4^+$: 7 + 4 − 1; $BH_4^-$: 5 + 4 + 1), and each central atom has four bond pairs and no lone pair ($sp^3$): all are tetrahedral.
Answer: (B). | ||||
| 64 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Total moles at equilibrium $= \frac{PV}{RT} = \frac{6 \times 100}{0.082 \times 610} = \frac{600}{50.02} \approx 12$.
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$: moles $= 4 + (5 - x) + x + x = 9 + x = 12$, so $x = 3$: $PCl_5$ 2, $PCl_3$ 3, $Cl_2$ 3 mol.
Each mole contributes $\frac{6}{12} = 0.5$ atm: $K_p = \frac{1.5 \times 1.5}{1.0} = 2.25$ atm.
Answer: (A). | ||||
| 65 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctHard | |
A hard single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Concentration: 42.12 g per 100 mL $= 421.2$ g $L^{-1} = \frac{421.2}{58.5} = 7.2$ mol $L^{-1}$.
The coagulating value is the minimum concentration that coagulates the sol in 2 hours. Taking the time needed to be inversely proportional to the concentration, coagulation in 2 hours instead of 10 needs $7.2 \times \frac{10}{2} = 36$ mol $L^{-1}$.
Answer: (B). | ||||
| 66 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Assertion and reasonEasy | |
A easy assertion and reason question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Nitrogen ($2p^3$) has three singly occupied 2p orbitals; oxygen ($2p^4$) has one doubly occupied 2p orbital, whose paired electrons repel each other, so one is removed more easily. Hence $\Delta_iH_1$(O) < $\Delta_iH_1$(N): A is correct, and R (the extra repulsion among oxygen's four 2p electrons) is the reason.
Answer: (A). | ||||
| 67 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Siderite is $FeCO_3$, malachite $CuCO_3 \cdot Cu(OH)_2$, sphalerite (zinc blende) ZnS, calamine $ZnCO_3$: A-I, B-II, C-III, D-IV.
Answer: (A). | ||||
| 68 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | StatementsMedium | |
A medium statements question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds In $CuSO_4 \cdot 5H_2O$ the copper ion is surrounded by four water molecules in a plane and two sulphate oxygen atoms above and below: every donor is an oxygen atom, so Cu–O bonds are present (I correct) and no ligand binds through sulphur (II incorrect).
Answer: (C). | ||||
| 69 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Washing soda is $Na_2CO_3 \cdot 10H_2O$ (carbonate, $CO_3^{2-}$); baking soda is $NaHCO_3$ (hydrogencarbonate, $HCO_3^-$, not carbonate); caustic soda is NaOH. Only washing soda contains the carbonate anion.
Answer: (A). | ||||
| 70 | 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 Br has 7 valence electrons: 3 form bonds with F, the remaining 4 are two lone pairs. Five electron pairs ($sp^3d$, trigonal bipyramidal) with the lone pairs equatorial give a bent T-shape.
Answer: (C). | ||||
| 71 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Borax hydrolyses: $Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$. NaOH is a strong base and boric acid a very weak acid, so the solution is strongly alkaline. $B_2H_6$ gives boric acid (acidic); the borohydrides react with water releasing hydrogen to give metaborates, not a strongly basic solution of comparable kind.
Answer: (D). | ||||
| 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 In the atmosphere $SO_2$ is oxidised by oxygen and combines with water to form sulphuric acid: $2SO_2 + O_2 + 2H_2O \rightarrow 2H_2SO_4$. The other equations stop at $SO_3$ or use $H_2O_2$, not the complete atmospheric process.
Answer: (D). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques An allylic or dienylic cation spreads its charge over alternate carbons. It is most stabilised when the OCH3 sits on one of those charge-bearing carbons, so its lone pair can delocalise the charge onto oxygen ($+OCH_3$, an oxocarbenium form).
(A) and (B): the OCH3 is on a carbon that carries no positive charge in any resonance form, so it only withdraws inductively. (C): the charge reaches the OCH3-bearing carbon over an allyl system. (D): the charge reaches the OCH3-bearing carbon through a longer conjugated system, giving the most resonance forms. (D) is the most stable.
Answer: (D). | ||||
| 74 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $CH_3CH_2CH_2Cl$ with anhydrous $AlCl_3$ gives the primary n-propyl cation, which rearranges by a 1,2-hydride shift to the more stable secondary isopropyl cation $CH_3\overset{+}{C}HCH_3$. This attacks benzene, so the major product A is isopropylbenzene (cumene).
Answer: (C). | ||||
| 75 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes C: $\frac{0.522 \times 184}{12} = 8$; H: $\frac{0.049 \times 184}{1} = 9$; Br: $\frac{0.429 \times 184}{80} = 1$: $C_8H_9Br$.
A gives benzoic acid, so it has one side chain on the ring holding the Br; it is chiral and its Br is ionisable (benzylic, AgBr with alcoholic $AgNO_3$): $C_6H_5CHBrCH_3$.
B gives p-bromobenzoic acid: Br on the ring and an ethyl group para to it, 4-bromoethylbenzene (its aryl Br gives no precipitate).
Answer: (C). | ||||
| 76 | AminesBasicity of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines Aniline is a Lewis base and forms a salt with the Lewis acid $AlCl_3$: $C_6H_5\overset{+}{N}H_2-\overset{-}{A}lCl_3$. The nitrogen attached to the ring becomes positively charged, deactivating the ring, so Friedel–Crafts alkylation does not occur.
Answer: (D). | ||||
| 77 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Dacron (terylene) is formed by condensation of ethylene glycol (a diol) with terephthalic acid (a dicarboxylic acid); the repeating links are ester groups, which is exactly why it is a polyester. Both are correct and R explains A.
Answer: (A). | ||||
| 78 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Heating denatures a protein: hydrogen bonds and other weak interactions break, destroying the secondary, tertiary and quaternary structures. The primary structure (the amino-acid sequence held by peptide bonds) is unaffected.
Answer: (C). | ||||
| 79 | Chemistry in Everyday LifeAntibiotics and antiseptics | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on antibiotics and antiseptics, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter A mixture of chloroxylenol and terpineol is the well-known household antiseptic applied to wounds and skin.
Answer: (A). | ||||
| 80 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter $SO_3^{2-} + BaCl_2 \rightarrow BaSO_3$ (white precipitate). $BaSO_3 + 2HCl \rightarrow BaCl_2 + H_2O + SO_2$, and $SO_2$ has a characteristic pungent (burning sulphur) odour. Iodide, sulphide and nitrite give no white barium precipitate.
Answer: (B). | ||||
| 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 $n = \frac{0.90}{18} = 0.05$ mol; $P = \frac{32}{760} = 0.0421$ atm; T = 300 K.
$V = \frac{nRT}{P} = \frac{0.05 \times 0.082 \times 300}{0.0421} = \frac{1.23}{0.0421} = 29.2$ L $\approx 29$.
Answer: 29. | ||||
| 82 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $n = \frac{2.2}{44} = 0.05$ mol. At constant pressure, $q = nC\Delta T = 0.05 \times 100 \times (-40) = -200$ J.
Work done on the gas: $w = -P\Delta V = -1 \times (167.75 - 217.1)$ mL atm $= 49.35$ mL atm $= 0.04935 \times 101.3 \approx 5$ J.
$\Delta U = q + w = -200 + 5 = -195$ J, so $x = 195$.
Answer: 195. | ||||
| 83 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $K_b = \frac{4}{1.5}$ and $K_f = \frac{4}{4.5}$, so $\frac{K_b}{K_f} = \frac{4.5}{1.5} = 3$.
Answer: 3. | ||||
| 84 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $H_2 + Cu^{2+} \rightarrow 2H^+ + Cu$, $E^\ominus = 0.34$ V, n = 2.
$E = 0.34 - \frac{0.06}{2}\log\frac{[H^+]^2}{[Cu^{2+}]} = 0.34 - 0.03\log\frac{10^{-6}}{10^{-x}} = 0.34 - 0.03(x - 6)$.
$0.31 = 0.34 - 0.03(x - 6)$ gives $x - 6 = 1$, $x = 7$.
Answer: 7. | ||||
| 85 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Comparing with $k = Ae^{-E_a/RT}$: $\frac{E_a}{R} = 26000$ K, so $E_a = 26000 \times 8.314 = 216164$ J $mol^{-1} \approx 216$ kJ $mol^{-1}$.
Answer: 216. | ||||
| 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 Mn is +2 here: $3d^5$. $Br^-$ is a weak-field ligand, so the complex is high spin with 5 unpaired electrons.
$\mu = \sqrt{n(n+2)} = \sqrt{35} = 5.92 \approx 6$ B.M.
Answer: 6. | ||||
| 87 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Two chlorides are ionisable, so one is in the coordination sphere. Co(III) is six-coordinate: $[Co(NH_3)_5Cl]Cl_2$, so $x = 5$.
Answer: 5. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The alcohols $C_4H_{10}O$ are butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol. Only butan-2-ol, $CH_3CH(OH)CH_2CH_3$, has a carbon with four different groups.
Answer: 1. | ||||
| 89 | Alcohols, Phenols and EthersPrimary, secondary and tertiary alcohols | Class 12: Alcohols, Phenols and Ethers | NumericalMedium | |
A medium numerical question on primary, secondary and tertiary alcohols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (i) $K_2Cr_2O_7$ oxidises cyclohexanol to cyclohexanone. (ii), (iii) $C_6H_5MgBr$ then water give 1-phenylcyclohexan-1-ol. (iv) $H^+$, heat dehydrates it to 1-phenylcyclohexene (X), the double bond conjugated with the ring.
$sp^2$ carbons: 6 in the benzene ring + 2 of the ring C=C = 8.
Answer: 8. | ||||
| 90 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids With base cyclohexanone undergoes self aldol condensation; on heating the aldol dehydrates to 2-cyclohexylidenecyclohexan-1-one (P), an α,β-unsaturated ketone.
P has one C=C and one C=O: $2 + 2 = 4$ π electrons.
Answer: 4. | ||||
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