JEE (Main) · 2022
JEE (Main) 2022 — 30 June, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Trigonometry3
- Integral Calculus3
- Matrices and Determinants2
10 easy · 17 medium · 3 hard
Physics
30 Q
- Kinematics3
- Properties of Bulk Matter2
- Thermodynamics2
23 easy · 7 medium · 0 hard
Chemistry
30 Q
- Aldehydes, Ketones and Carboxylic Acids3
- Coordination Compounds2
- Organic Chemistry: Some Basic Principles and Techniques2
23 easy · 6 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 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $S_1$: $x^2+3x+5 > 0$ for all x (discriminant $9-20<0$), and $-2+3x-x^2 = -(x-1)(x-2)$. So the condition is $\frac{x+2}{(x-1)(x-2)} \leq 0$.
Sign check: negative for $x<-2$, zero at $x=-2$, positive on $(-2,1)$, negative on $(1,2)$, positive for $x>2$. Hence $S_1 = (-\infty, -2] \cup (1, 2)$.
$S_2$: put $t = 3^x > 0$: $t^2 - 3t - 9t + 27 \leq 0$, i.e. $(t-3)(t-9) \leq 0$, so $3 \leq 3^x \leq 9$ and $x \in [1, 2]$.
$S_1 \cup S_2 = (-\infty, -2] \cup [1, 2]$.
Answer: (B). | ||||
| 2 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $(1+2i)^2(1-2i)^2 = 5^2 = 25$, so the numerator is $25(1+2i)^6$.
$(1+2i)^2 = -3+4i$, $(-3+4i)^2 = -7-24i$, $(1+2i)^6 = (-7-24i)(-3+4i) = 21 - 28i + 72i + 96 = 117 + 44i$.
Denominator: $(3+2i)(4+6i) = 12 + 18i + 8i - 12 = 26i$.
So the number is $\frac{25(117+44i)}{26i} = \frac{25(117+44i)(-i)}{26} = \frac{25(44 - 117i)}{26}$.
Real part $= \frac{1100}{26} = \frac{550}{13}$.
Answer: (D). | ||||
| 3 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations With roots p, q: $p+q = \frac{6-\alpha}{3}$, $pq = \frac{\alpha+3}{3}$.
$p^2+q^2 = \frac{(\alpha-6)^2}{9} - \frac{2(\alpha+3)}{3} = \frac{\alpha^2 - 18\alpha + 18}{9}$.
Real roots need $D = (\alpha-6)^2 - 12(\alpha+3) = \alpha^2 - 24\alpha \geq 0$, i.e. $\alpha \leq 0$ or $\alpha \geq 24$.
The parabola $\alpha^2 - 18\alpha + 18$ has its vertex at $\alpha = 9$, outside the allowed set; it decreases on $\alpha \leq 0$ and increases on $\alpha \geq 24$. Its least allowed values are at $\alpha = 0$ (value $\frac{18}{9} = 2$) and $\alpha = 24$ (value $\frac{162}{9} = 18$).
So the minimum is attained only at $\alpha = 0$ and S = {0} is a singleton. (Even if equal roots were excluded, the minimum over integers $\alpha \leq -1$ would be at $\alpha = -1$ alone: still a singleton.)
Answer: (B). | ||||
| 4 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $AB = \begin{bmatrix} 2+2+4\alpha & \alpha-4-5\alpha \\ 2\alpha-2-4 & \alpha^2+4+5 \end{bmatrix} = \begin{bmatrix} 4(1+\alpha) & -4(1+\alpha) \\ 2\alpha-6 & \alpha^2+9 \end{bmatrix}$.
$\det(AB) = 4(1+\alpha)(\alpha^2+9) + 4(1+\alpha)(2\alpha-6) = 4(1+\alpha)(\alpha^2+2\alpha+3)$.
The roots are $\alpha = -1$ and the two roots of $\alpha^2+2\alpha+3=0$ (sum $-2$). Sum of all values $= -3$, absolute value 3.
Answer: (A). | ||||
| 5 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For order 2, $\det(kM) = k^2 \det M$.
$\det(5(\det A)B) = \det(10B) = 100 \times 3 = 300$.
$\det(300A^2) = 300^2 (\det A)^2 = 90000 \times 4 = 360000 = 2^6 \cdot 3^2 \cdot 5^4$.
$a+b+c = 6+2+4 = 12$.
Answer: (B). | ||||
| 6 | 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 $T_{r+1} = \binom{10}{r} a^{10-r} b^r x^{\frac{10-r}{8} - \frac{r}{12}}$. The power is zero when $12(10-r) = 8r$, i.e. r = 6.
Constant term $= \binom{10}{6} a^4 b^6 = 210\, a^4 b^6$.
Let $u = \frac{1}{a^2}$, $v = \frac{1}{b^3}$, so $u + v = 4$ and $a^4 b^6 = \frac{1}{(uv)^2}$. By A.M. ≥ G.M., $uv \leq 4$, so $a^4b^6 \geq \frac{1}{16}$ (equality at u = v = 2).
Minimum constant term $= \frac{210}{16} = \frac{105}{8}$.
Answer: (C). | ||||
| 7 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{1}{1+2+\ldots+k} = \frac{2}{k(k+1)} = 2\left(\frac{1}{k} - \frac{1}{k+1}\right)$.
Summing k = 1 to 11 telescopes: $2\left(1 - \frac{1}{12}\right) = \frac{11}{6}$.
Answer: (B). | ||||
| 8 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The line is $x + y = 50 + \alpha = k$. Maximise $f(x) = x(k-x)^4$ for $0 < x < k$ (where $xy^4 > 0$).
$f'(x) = (k-x)^3[(k-x) - 4x] = 0$ gives $x = \frac{k}{5}$, $y = \frac{4k}{5}$ (f' changes from + to −: a maximum).
So $y = 4x$.
Answer: (A). | ||||
| 9 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 12x^2 - 22x + 8 = 2(2x-1)(3x-4)$, zero at $x = \frac{1}{2}$ and $x = \frac{4}{3}$.
$f' > 0$ for $x < \frac{1}{2}$, $f' < 0$ on $\left(\frac{1}{2}, \frac{4}{3}\right)$, $f' > 0$ for $x > \frac{4}{3}$.
So $x = \frac{1}{2}$ is a local maximum, $x = \frac{4}{3}$ a local minimum, and f decreases on $\left(\frac{1}{2}, \frac{4}{3}\right)$ (so (C) is false too).
Answer: (D). | ||||
| 10 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions For $x \in \left[0, \frac{\pi}{8}\right]$, $2x \in \left[0, \frac{\pi}{4}\right] \subset [-1, 1]$, so $\sin^{-1}2x + \cos^{-1}2x = \frac{\pi}{2}$.
$\sin 2x + \cos 2x = \sqrt{2}\sin\left(2x + \frac{\pi}{4}\right)$ with $2x + \frac{\pi}{4} \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right]$, so it ranges over $[1, \sqrt{2}]$.
$m = \frac{\pi}{2} + 1$, $M = \frac{\pi}{2} + \sqrt{2}$, $m + M = \pi + 1 + \sqrt{2}$.
Answer: (A). | ||||
| 11 | Integral CalculusPartial fractions | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on partial fractions, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Dividing by $n^2$: the sum is $\frac{1}{n}\sum_{r=1}^n \frac{r/n}{2(r/n)^2 - 7(r/n) + 6} \to \int_0^1 \frac{x\,dx}{2x^2 - 7x + 6}$.
$2x^2 - 7x + 6 = (2x-3)(x-2)$ and $\frac{x}{(2x-3)(x-2)} = \frac{-3}{2x-3} + \frac{2}{x-2}$.
$\int_0^1 = \left[-\frac{3}{2}\ln|2x-3| + 2\ln|x-2|\right]_0^1 = \left(0 + 0\right) - \left(-\frac{3}{2}\ln 3 + 2\ln 2\right) = \frac{3}{2}\ln 3 - 2\ln 2 = \log_e\frac{3\sqrt{3}}{4}$.
Answer: (B). | ||||
| 12 | 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 Cross-multiplying: $(ax + a)\,dx - cy\,dy - a\,dy - b(y\,dx + x\,dy) = 0$, and $y\,dx + x\,dy = d(xy)$.
Integrating: $\frac{a x^2}{2} + ax - \frac{c y^2}{2} - ay - bxy = K$.
For a circle the $xy$ term must vanish (b = 0) and the coefficients of $x^2$ and $y^2$ must be equal ($c = -a$, $a \neq 0$). Then $x^2 + y^2 + 2x - 2y = \frac{2K}{a}$, whose centre is $(-1, 1)$.
$\alpha + 2\beta = -1 + 2 = 1$.
Answer: (C). | ||||
| 13 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines Collinearity: $\begin{vmatrix} \alpha & -3 & 1 \\ 2 & 0 & 1 \\ 1 & \alpha & 1 \end{vmatrix} = \alpha(0 - \alpha) + 3(2 - 1) + (2\alpha - 0) = -\alpha^2 + 2\alpha + 3 = 0$, so $\alpha = -1, 3$.
The point is $(-1, 3)$ and the slope is $\tan\frac{\pi}{3} = \sqrt{3}$: $y - 3 = \sqrt{3}(x + 1)$, i.e. $\sqrt{3}x - y + \sqrt{3} + 3 = 0$.
Answer: (B). | ||||
| 14 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Write L as $mx - y + c = 0$ and let $s(P) = \frac{mx_P - y_P + c}{\sqrt{1+m^2}}$ be the signed distance of a centre P.
Each signed distance is $\pm r$; $C_1$ and $C_3$ on the same side means $s(0,0) = s(2,1)$: $c = 2m - 1 + c$, so $m = \frac{1}{2}$.
$C_2$ on the other side: $s(1,1) = -s(0,0)$: $m - 1 + c = -c$, so $c = \frac{1-m}{2} = \frac{1}{4}$.
$r = \frac{|c|}{\sqrt{1+m^2}} = \frac{1/4}{\sqrt{5}/2} = \frac{1}{2\sqrt{5}}$, $r^2 = \frac{1}{20}$.
$20(r^2 + c) = 20\left(\frac{1}{20} + \frac{1}{4}\right) = 1 + 5 = 6$.
Answer: (D). | ||||
| 15 | Conic SectionsStandard forms of conics | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections $y = e^x$ and $y = \log_e x$ are inverses, mirror images in $y = x$. The least value of $e^x - x$ is 1 (at x = 0), so the nearest point of $y = e^x$ to $y = x$ is at distance $\frac{1}{\sqrt{2}}$, and $a = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}$, $a^2 = 2$.
Ellipse $x^2 + 2y^2 = 50$: $\frac{x^2}{50} + \frac{y^2}{25} = 1$, $e_1^2 = 1 - \frac{25}{50} = \frac{1}{2}$.
Hyperbola $x^2 - 2y^2 = 5$: $\frac{x^2}{5} - \frac{y^2}{5/2} = 1$, $e_2^2 = 1 + \frac{1}{2} = \frac{3}{2}$.
$b^2 = \frac{e_1^2}{e_2^2} = \frac{1}{3}$, so $a^2 + \frac{1}{b^2} = 2 + 3 = 5$.
Answer: (D). | ||||
| 16 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy 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 has direction ratios $(-2, 2, 1)$ (since $\frac{2-x}{2} = \frac{x-2}{-2}$), of length 3.
Through (3, 2, −1): $(3 - 2t, 2 + 2t, -1 + t)$. In the plane: $3(3-2t) - (2+2t) + 4(-1+t) + 1 = 4 - 4t = 0$, so t = 1.
Distance $= |t| \times 3 = 3$.
Answer: (C). | ||||
| 17 | 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 $\vec{a} \times \vec{b} = (-2, 1, -3)$, $\vec{a} + \vec{b} = (1, 2, 0)$, $(\vec{a}+\vec{b}) \times (\vec{a}\times\vec{b}) = (-6, 3, 5)$, $\vec{a} - \vec{b} = (-3, 0, 2)$.
Let $\vec{c} = (x, y, z)$.
The k-component of $\vec{c} \times (\vec{a} - \vec{b})$ is $x \cdot 0 - y \cdot (-3) = 3y = 3$, so y = 1.
Coplanar: $\vec{c} \cdot (\vec{a}\times\vec{b}) = -2x + y - 3z = 0$, so $2x + 3z = 1$.
$-6x + 3y + 5z = -42$ gives $-6x + 5z = -45$.
Solving: from $x = \frac{1 - 3z}{2}$, $-3 + 9z + 5z = -45$, z = −3, x = 5.
$|\vec{c}|^2 = 25 + 1 + 9 = 35$.
Answer: (C). | ||||
| 18 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $q^5 = 5pq^4$ gives $q = 5p$, so $p = \frac{1}{6}$, $q = \frac{5}{6}$.
$\frac{P(X=2)}{P(X=3)} = \frac{\binom{5}{2}p^2q^3}{\binom{5}{3}p^3q^2} = \frac{q}{p} = 5$.
Answer: (D). | ||||
| 19 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta = \cos^{-1}\frac{1}{\sqrt{5}}$: $\sin\theta = \frac{2}{\sqrt{5}}$, $\sin 2\theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}$. So $\alpha = \tan\left(\frac{5\pi}{16} \cdot \frac{4}{5}\right) = \tan\frac{\pi}{4} = 1$.
$A = \sin^{-1}\frac{4}{5}$ and $B = \sec^{-1}\frac{5}{3}$ both have cosine $\frac{3}{5}$ and sine $\frac{4}{5}$. $\beta = \cos(A + B) = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25}$.
Sum $= \frac{18}{25}$, product $= -\frac{7}{25}$: $25x^2 - 18x - 7 = 0$.
Answer: (C). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter Let $A = p \wedge q$ and $B = (\sim p) \vee r$. The statement is $(A \rightarrow B) \vee (B \rightarrow A) = (\sim A \vee B) \vee (\sim B \vee A)$, which contains $A \vee \sim A$ and is always true: a tautology.
Answer: (A). | ||||
| 21 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations A multiple of 15 must end in 5 (no 0 is available) and have digit sum divisible by 3.
The seven digits sum to 28; leaving out digit d gives sum $28 - d$, divisible by 3 when $d \in \{1, 4, 7\}$ (5 is never left out).
For each choice, 5 is fixed last and the other five digits are arranged in $5! = 120$ ways.
Total $= 3 \times 120 = 360$.
Answer: 360. | ||||
| 22 | 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 Write $a_0 = ab$, $a_1 = (a+1)(2b)$, $a_2 = (a+2)(4b)$.
$f'(x) = 2a_0x + a_1$: $f'(0) = a_1 = 1$ and $f'(1) = 2a_0 + a_1 = 0$, so $a_0 = -\frac{1}{2}$.
$ab = -\frac{1}{2}$ and $2ab + 2b = 1$ give $-1 + 2b = 1$, b = 1, $a = -\frac{1}{2}$, so $a_2 = \frac{3}{2} \times 4 = 6$.
$f(x) = -\frac{x^2}{2} + x + 6$, $f(4) = -8 + 4 + 6 = 2$.
Answer: 2. | ||||
| 23 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $s = a + \sin\frac{x}{2}$ and $c = -b\cos x$. Expanding along the first row: $F = s(0 - s^2) - c(c^2 - 0) = -(s^3 + c^3)$.
For the limit to exist, F must vanish at 0 to third order. $F(0) = -(a^3 - b^3) = 0$ gives a = b.
Then $s^3 + c^3 = (s + c)(s^2 - sc + c^2)$ with $s + c = \sin\frac{x}{2} + a(1 - \cos x) = \frac{x}{2} + O(x^2)$ and $s^2 - sc + c^2 \to 3a^2$. So $F \approx -\frac{3a^2 x}{2}$, and $\frac{F}{x^3}$ has a limit only if a = 0 (hence b = 0).
With a = b = 0: $F(x) = -\sin^3\frac{x}{2}$ and $L = \lim \frac{-\sin^3(x/2)}{x^3} = -\frac{1}{8}$.
$-112L = 14$.
Answer: 14. | ||||
| 24 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The region is bounded by the lines $y = x + \alpha$, $y = -x - \alpha$, $y = 2 - x$ and $y = 2 + x$: two pairs of parallel lines meeting at right angles, so it is a rectangle (for $0 < \alpha < 2$).
Distance between $y = x + \alpha$ and $y = x + 2$: $\frac{2 - \alpha}{\sqrt{2}}$; between $y = -x - \alpha$ and $y = -x + 2$: $\frac{2 + \alpha}{\sqrt{2}}$.
Area $= \frac{(2-\alpha)(2+\alpha)}{2} = \frac{4 - \alpha^2}{2} = \frac{3}{2}$, so $\alpha = 1$.
Second region: $0 \leq y \leq x + 2$, $-1 \leq x \leq 1$ (here $x + 2 > 0$): area $= \int_{-1}^{1}(x + 2)\,dx = 4$.
Answer: 4. | ||||
| 25 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f'(1) = e \cdot \frac{1}{(1+2+2)^2} = \frac{e}{25}$.
For f(1) put $u = x^3$, $du = 3x^2dx$: $f(1) = \frac{1}{3}\int_0^1 \frac{e^u u^2}{(u^2 + 2u + 2)^2}du$.
With $Q = u^2 + 2u + 2$: $\frac{d}{du}\left(\frac{e^u}{Q}\right) = e^u\frac{Q - Q'}{Q^2} = e^u\frac{u^2}{Q^2}$.
So $f(1) = \frac{1}{3}\left[\frac{e^u}{Q}\right]_0^1 = \frac{1}{3}\left(\frac{e}{5} - \frac{1}{2}\right) = \frac{e}{15} - \frac{1}{6}$.
$f(1) + f'(1) = \frac{e}{15} + \frac{e}{25} - \frac{1}{6} = \frac{8e}{75} - \frac{1}{6}$, so $\alpha = \frac{8}{75}$ and $150\alpha = 16$.
Answer: 16. | ||||
| 26 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dN}{dt} = kN(100 - N)$. Separating: $\int\frac{dN}{N(100-N)} = \frac{1}{100}\ln\frac{N}{100 - N} = kt + C$, so $\frac{N}{100 - N} = Ae^{100kt}$.
t = 0: $\frac{2}{98} = \frac{1}{49} = A$. t = 4: $\frac{30}{70} = \frac{3}{7} = \frac{1}{49}e^{400k}$, so $e^{400k} = 21$.
t = 8: $\frac{N}{100 - N} = \frac{1}{49} \times 21^2 = 9$, so N = 90.
Answer: 90. | ||||
| 27 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Take $P(t_1^2, 2t_1)$, $Q(t_2^2, 2t_2)$ with $t_1t_2 = -1$ for a focal chord (a = 1). Its length is $\left(t_1 + \frac{1}{t_1}\right)^2 = 6.25$, so $\left|t_1 + \frac{1}{t_1}\right| = 2.5$ and $|t_1| = 2$ or $\frac{1}{2}$.
Area $= \frac{1}{2}|x_Py_Q - x_Qy_P| = \frac{1}{2}|2t_1^2t_2 - 2t_2^2t_1| = |t_1t_2||t_1 - t_2| = \left|t_1 + \frac{1}{t_1}\right| = 2.5$.
10 × area = 25.
Answer: 25. | ||||
| 28 | 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 $G = (g, g, g)$ with $g = \frac{2\alpha}{3}$. Points of the line: $D = (t, 0, 3 - t)$.
$GD^2 = (t - g)^2 + g^2 + (3 - t - g)^2$, least at $t = \frac{3}{2}$, where it equals $2\left(\frac{3}{2} - g\right)^2 + g^2 = 3g^2 - 6g + \frac{9}{2}$.
$3g^2 - 6g + \frac{9}{2} = \frac{57}{2}$ gives $g^2 - 2g - 8 = 0$, g = 4 (g > 0), so $\alpha = \frac{3g}{2} = 6$.
Answer: 6. | ||||
| 29 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability The probabilities add to 1 for every d; each must lie in [0, 1]: $d \leq 1$, $d \geq -\frac{1}{2}$, $d \leq \frac{1}{4}$, $d \geq -\frac{1}{3}$. So $-\frac{1}{3} \leq d \leq \frac{1}{4}$ and the minimum is $d = -\frac{1}{3}$.
Mean $= \frac{(1+2d) + 2(1-4d) + 3(1+3d)}{4} = \frac{6 + 3d}{4} = \frac{5}{4}$ at $d = -\frac{1}{3}$.
$60 \times \frac{5}{4} = 75$.
Answer: 75. | ||||
| 30 | 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 For $|t| \leq 1$ and odd $k \geq 3$, $t^k \leq t^2$ with equality only at t = 0 or 1. So $\sin^kx + \cos^kx \leq \sin^2x + \cos^2x = 1$, with equality only when $(\sin x, \cos x) = (0, 1)$ or $(1, 0)$, i.e. $x = 2n\pi$ or $x = 2n\pi + \frac{\pi}{2}$.
In $[0, 12\pi]$: $x = 0, 2\pi, \ldots, 12\pi$ (7 values) and $\frac{\pi}{2}, \ldots, 10\pi + \frac{\pi}{2}$ (6 values): $n(S_1) = 13$.
In $[0, 8\pi]$: 5 + 4 = 9 values: $n(S_2) = 9$.
$n(S_1) - n(S_2) = 4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 At release the truck (and ball) moves at $5 \times 20 = 100$ m/s along x. After release the ball keeps this horizontal velocity (no horizontal force) and gains $10 \times 1 = 10$ m/s downward.
$\vec{v} = 100\hat{i} - 10\hat{j}$ m/s.
Answer: (A). | ||||
| 32 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter 1 MSD $= \frac{10}{5} = 2$ mm. $(n+1)$ VSD $= n$ MSD, so 1 VSD $= \frac{n}{n+1}$ MSD.
Least count $= 1\text{ MSD} - 1\text{ VSD} = \frac{1}{n+1}$ MSD $= \frac{2}{n+1}$ mm.
Answer: (A). | ||||
| 33 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g = \frac{GM}{R^2} = \frac{4}{3}\pi G\rho R$, so $\frac{g_A}{g_B} = \frac{3\rho \times 2}{5\rho \times 3} = \frac{2}{5}$.
Answer: (D). | ||||
| 34 | 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 Equal maximum heights: $u_1^2\sin^2\theta_1 = u_2^2\sin^2 60°$, so $3\sin^2\theta_1 = 2 \times \frac{3}{4}$, $\sin^2\theta_1 = \frac{1}{2}$ and $\theta_1 = 45°$.
Answer: (C). | ||||
| 35 | 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 Rising steadily, the buoyant force balances the viscous drag (weight negligible): $\frac{4}{3}\pi r^3\sigma g = 6\pi\eta r v$, so $\eta = \frac{2r^2\sigma g}{9v}$.
Answer: (B). | ||||
| 36 | 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 Normal reaction N = 50 N, so limiting friction $= 0.5 \times 50 = 25$ N. When the block tends to move up, friction acts downward along with the weight.
Largest F without upward motion: $F = mg + f_{max} = 20 + 25 = 45$ N.
Answer: (D). | ||||
| 37 | 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 $K = \frac{p^2}{2m}$. With $p_B = 2p_A$: $\frac{K_A}{K_B} = \frac{p_A^2/5}{4p_A^2/8} = \frac{1/5}{1/2} = \frac{2}{5}$.
Answer: (B). | ||||
| 38 | ThermodynamicsZeroth law and temperature | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on zeroth law and temperature, from Thermodynamics. NCERT: Class 11: Thermodynamics At constant volume $P \propto T$: $T = 273 \times \frac{180}{100} = 491.4$ K ≈ 491 K.
Answer: (D). | ||||
| 39 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium 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 Turns per unit radial width: $\frac{n}{r_2 - r_1}$. A ring of radius r and width dr holds $dN = \frac{n\,dr}{r_2 - r_1}$ turns and gives $dB = \frac{\mu_0 I\,dN}{2r}$ at the centre.
$B = \frac{\mu_0 nI}{2(r_2 - r_1)}\int_{r_1}^{r_2}\frac{dr}{r} = \frac{\mu_0 nI}{2(r_2 - r_1)}\log_e\frac{r_2}{r_1}$.
Answer: (D). | ||||
| 40 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In Fig. 2 the dielectric fills half of the plate area across the full gap, so the capacitor is two capacitors in parallel, each with half the area: $C = \frac{C_o}{2} + \frac{\varepsilon_rC_o}{2} = \frac{C_o}{2}(1 + \varepsilon_r)$.
Answer: (A). | ||||
| 41 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $\gamma = \frac{5}{3}$. $P_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = 75 \times 8^{5/3} = 75 \times 32 = 2400$ kPa.
Work done on the gas $= \frac{P_2V_2 - P_1V_1}{\gamma - 1} = \frac{2400 \times 10^3 \times 150 \times 10^{-6} - 75 \times 10^3 \times 1200 \times 10^{-6}}{2/3} = \frac{360 - 90}{2/3} = 405$ J.
Answer: (B). | ||||
| 42 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $k = \frac{2\pi}{\lambda} = 0.5\pi\ cm^{-1}$ and $\omega = 2\pi\nu = 200\pi\ s^{-1}$, with the form $\sin(kx - \omega t)$.
(A) has $\omega = 100\pi$; (B) has $\omega = 100\pi$; (C) has $\omega = \frac{2\pi}{100}$; (D) has $k = 0.5\pi$, $\omega = 200\pi$: correct.
Answer: (D). | ||||
| 43 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium 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 At the dee radius R the proton's speed is $v = \omega R = 2\pi fR = 2\pi \times 10^7 \times 0.6 = 3.77 \times 10^7$ m/s.
$K = \frac{1}{2}mv^2 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (3.77 \times 10^7)^2 = 1.19 \times 10^{-12}$ J $= \frac{1.19 \times 10^{-12}}{1.6 \times 10^{-13}}$ MeV ≈ 7.4 MeV.
Answer: (A). | ||||
| 44 | 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 Using the given speed of light, $B_0 = \frac{E_0}{c} = \frac{300}{3 \times 10^8} = 1 \times 10^{-6}$ T.
Answer: (A). | ||||
| 45 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity For the same length, $R \propto \frac{1}{r^2}$. Each 3 mm wire: $14 \times \left(\frac{9}{3}\right)^2 = 126$ Ω. Seven in parallel: $\frac{126}{7} = 18$ Ω.
Answer: (B). | ||||
| 46 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $f = \frac{1}{2\pi\sqrt{LC}}$: LC becomes 4 times, so $f_2 = \frac{f_1}{2}$.
$Q = \frac{1}{R}\sqrt{\frac{L}{C}}$: L/C is unchanged, so $Q_2 = Q_1$.
Answer: (A). | ||||
| 47 | 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 $I_1 : I_2 = 9 : 16$, so amplitudes are $3 : 4$. $\frac{I_{max}}{I_{min}} = \frac{(4+3)^2}{(4-3)^2} = 49 : 1$.
Answer: (D). | ||||
| 48 | 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 $P = N\frac{hc}{\lambda}$, so at fixed power $N \propto \lambda$: $N = 9 \times 10^{20} \times \frac{800}{600} = 12 \times 10^{20}$ per second.
Answer: (A). | ||||
| 49 | 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 AM wave's amplitude varies between $A_c + A_m$ and $A_c - A_m$. Minimum $= 44 - 11 = 33$ V.
Answer: (A). | ||||
| 50 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $-13.6 + 12.09 = -1.51$ eV $= -\frac{13.6}{9}$ eV, so the electron goes to n = 3.
$\Delta L = (3 - 1)\frac{h}{2\pi} = \frac{2 \times 6.63 \times 10^{-34}}{2\pi} = 2.11 \times 10^{-34}$ Js.
Answer: (B). | ||||
| 51 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Path: O to P (200 m), the quarter arc P to Q ($\frac{\pi r}{2} = \frac{3.14 \times 200}{2} = 314$ m), Q to O (200 m). Total 714 m.
Time = 3 min 58 s = 238 s. Average speed $= \frac{714}{238} = 3$ m/s.
Answer: 3. | ||||
| 52 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves 72 km/h = 20 m/s. The wind carries the sound towards the employee, so the sound travels at $340 + 20 = 360$ m/s relative to the ground; the employee recedes from the stationary source at 20 m/s.
$f' = f\frac{(v + w) - v_o}{v + w} = 720 \times \frac{360 - 20}{360} = 680$ Hz.
Answer: 680. | ||||
| 53 | 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 O is the centre of the square, so each corner is at distance $\frac{1}{\sqrt{2}}$ m, $r^2 = \frac{1}{2}\ m^2$.
$I = (1 + 2 + 3 + 4) \times \frac{1}{2} = 5$ kg m².
Answer: 5. | ||||
| 54 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter For a drop $\Delta P = \frac{2T}{r}$: $T = \frac{500 \times 2 \times 10^{-3}}{2} = 0.5 = 500 \times 10^{-3}$ N/m.
Answer: 500. | ||||
| 55 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Volume conserved: $R = 2r$. Charge $Q = 8q$. Potential $V' = \frac{kQ}{R} = 4V$.
Energy $U = \frac{1}{2}QV$: $U' = \frac{1}{2}(8q)(4V) = 32 \times \frac{1}{2}qV = 32E$.
Answer: 32. | ||||
| 56 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity With $K_1$ open the balance length measures the emf: $E \propto 120$. With $K_1$ closed it measures the terminal voltage across 4 Ω: $V \propto 80$.
$r = R\left(\frac{l_1}{l_2} - 1\right) = 4\left(\frac{120}{80} - 1\right) = 2$ Ω.
Answer: 2. | ||||
| 57 | 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 Average power is greatest at resonance, $X_C = X_L = \frac{70}{11}$ Ω.
$C = \frac{1}{\omega X_C} = \frac{11}{2 \times \frac{22}{7} \times 50 \times 70} = \frac{11}{22000} = 5 \times 10^{-4}$ F $= 500$ μF.
Answer: 500. | ||||
| 58 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At minimum deviation $r = \frac{A}{2} = 30°$ and $e = i$. $\sin i = \mu\sin 30° = \frac{\sqrt{2}}{2}$, so $i = e = 45°$.
Answer: 45. | ||||
| 59 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Potential energy $U = 2E_n = -\frac{27.2}{n^2}$ eV. $-\frac{27.2}{n^2} = -1.08$ gives $n^2 \approx 25$, n = 5.
Photon energy $= E_5 - E_2 = 13.6\left(\frac{1}{4} - \frac{1}{25}\right) = 13.6 \times 0.21 = 2.856$ eV $= 285.6 \times 10^{-2}$ eV, so x ≈ 286.
(Using $E_5 = \frac{U}{2} = -0.54$ eV directly: $3.40 - 0.54 = 2.86$ eV, the same integer.)
Answer: 286. | ||||
| 60 | 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 Largest safe zener current $I_{max} = \frac{0.4}{10} = 0.04$ A. At the largest supply voltage, 15 V, the series resistor must drop 5 V at this current: $R_p = \frac{15 - 10}{0.04} = 125$ Ω.
Answer: 125. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter $\rho = \frac{ZM}{N_Aa^3}$ with Z = 2, $a^3 = (2 \times 10^{-8}\ cm)^3 = 8 \times 10^{-24}\ cm^3$.
$M = \frac{\rho N_Aa^3}{Z} = \frac{5 \times 6 \times 10^{23} \times 8 \times 10^{-24}}{2} = 12$ g/mol.
300 g is 25 mol, i.e. $25\ N_A$ atoms.
Answer: (D). | ||||
| 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 4p, n = 4, l = 1. Radial nodes $= n - l - 1 = 2$; total nodes $= n - 1 = 3$.
Answer: (A). | ||||
| 63 | SolutionsRaoult's law | Class 12: Solutions | Single correctMedium | |
A medium single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Henry's law: $p_{gas} = K_Hx_{gas} = K_H(1 - x_{H_2O})$. Plotted against $x_{H_2O}$, each line falls with slope $-K_H$.
The steepest line, p, belongs to the largest $K_H$ (A, 30.40 k bar); the flattest, s, to the smallest (C, $1.56 \times 10^{-5}$ k bar).
Answer: (A). | ||||
| 64 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium The second reaction is the first multiplied by $\frac{3}{4}$, so $K_2 = K_1^{3/4}$.
Answer: (C). | ||||
| 65 | ElectrochemistryNernst equation | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$: $[H^+]$ enters the Nernst equation, so the potential depends on pH. The other three half-reactions involve no $H^+$ or $OH^-$.
Answer: (B). | ||||
| 66 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Negative electron gain enthalpies (kJ/mol): O −141, S −200, Se −195, Te −190. Oxygen's is small because its compact 2p shell repels the incoming electron; from S downward the value falls with size.
So O < S > Se > Te.
Answer: (C). | ||||
| 67 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter For $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$, $\Delta S^\circ < 0$ (a gas is consumed), so $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ rises linearly with T.
At the boiling point of magnesium (1363 K; melting at 923 K changes it little) Mg becomes a gas, $\Delta S^\circ$ becomes much more negative, and the slope increases sharply: the Ellingham line bends upward, as in (A).
Answer: (A). | ||||
| 68 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter NaH is ionic (saline) → (ii). $SiH_4$ is a covalent molecular hydride → (iii). Vanadium forms a metallic, non-stoichiometric hydride (about $VH_{0.56}$) → (iv). Aluminium hydride is electron-deficient and acts as a Lewis acid → (i).
Answer: (A). | ||||
| 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 The superoxide ion $O_2^-$ has one unpaired electron, so superoxides are paramagnetic (and coloured). Oxides ($O^{2-}$) and peroxides ($O_2^{2-}$) have all electrons paired: diamagnetic and colourless (when pure).
Answer: (B). | ||||
| 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 Red phosphorus is made by heating white phosphorus at 573 K in an inert atmosphere. Red phosphorus heated in a sealed tube at 803 K gives α-black phosphorus (β-black phosphorus comes from white phosphorus at 473 K under high pressure).
A = white phosphorus, B = α-black phosphorus.
Answer: (D). | ||||
| 71 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A white precipitate with $BaCl_2$ needs free $SO_4^{2-}$; no precipitate with $AgNO_3$ needs no free halide. In $[Co(NH_3)_5Br]SO_4$ sulphate is the counter-ion and bromide is coordinated. (B) gives AgBr, not $BaSO_4$; (C) and (D) give silver halide precipitates and no sulphate.
Answer: (A). | ||||
| 72 | Environmental ChemistryGreen chemistry | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on green chemistry, from Environmental Chemistry. NCERT: No matching NCERT chapter Tetrachloroethene was the classic dry-cleaning solvent; it is being replaced by liquefied $CO_2$ with a suitable detergent, and $H_2O_2$ is used for bleaching clothes in the laundry, as the school text on green chemistry describes. Ethanol is not among them.
Answer: (C). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques On polar silica, a high $R_f$ means weak adsorption and good solubility in the mobile phase. A (0.72) is the least polar and elutes first: (A) and (C) are correct. C (0.20) is the most strongly adsorbed, so (D) is correct, and C is the least soluble in the mobile phase, so (B) is not correct.
Answer: (B). | ||||
| 74 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Red P and $Cl_2$ on a carboxylic acid with an α-hydrogen is the Hell–Volhard–Zelinsky reaction: the acid chloride formed is chlorinated at the α-carbon, and water then regenerates the acid. Cyclohexanecarboxylic acid gives 1-chlorocyclohexane-1-carboxylic acid.
Answer: (A). | ||||
| 75 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) Fittig reaction: aryl halide with sodium in dry ether couples to biphenyl: possible.
(B) The para nitro group activates the C–Cl bond to nucleophilic substitution; NaOH on heating and then acid gives 4-nitrophenol: possible.
(C) Tollens' reagent oxidises the aliphatic aldehyde group of phenylacetaldehyde to the carboxylate: possible.
(D) Aromatic aldehydes such as benzaldehyde do not reduce Fehling's/Benedict's $Cu^{2+}$: not possible.
Answer: (D). | ||||
| 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 The acid is $CH_3COCH_2CH_2COCH_2CH_2CO_2H$ (4,7-dioxooctanoic acid), a 1,4-diketone. Dilute KOH causes intramolecular aldol condensation giving a five-membered ring.
The enolate formed at $C_3$ (the $CH_2$ next to the $C_4$ ketone, carrying the $CH_2CO_2H$ arm) attacks the methyl ketone $C_7$; dehydration gives the conjugated, tetrasubstituted (most stable) enone: 3-methyl-2-(carboxymethyl)cyclopent-2-enone, after acidification with $H_3O^+$. This is the same ring closure that turns a 2,5-diketone with a long chain into a 2-alkyl-3-methylcyclopentenone.
The alternative closure through the terminal $CH_3$ enolate (option A) gives the less substituted enone.
Answer: (B). | ||||
| 77 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Ethanol reduces a benzenediazonium salt to benzene ($C_6H_5N_2^+Cl^- + CH_3CH_2OH \rightarrow C_6H_6 + N_2 + CH_3CHO + HCl$), so A is $C_6H_5N_2^+Cl^-$.
Benzene with oleum gives benzenesulphonic acid; fusion with NaOH gives sodium phenoxide, and HCl liberates phenol: B is phenol.
Answer: (B). | ||||
| 78 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Acrilan is polyacrylonitrile → (iii). Neoprene is polychloroprene → (i). Polystyrene → (ii). Buna-N is the butadiene–acrylonitrile copolymer → (iv).
Answer: (C). | ||||
| 79 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Complete hydrolysis of DNA gives phosphoric acid, the bases and 2-deoxy-D-ribose, a pentose.
Answer: (A). | ||||
| 80 | Principles Related to Practical ChemistryDetection of elements and functional groups | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Neutral $FeCl_3$ gives a violet/blue-green colour with phenols and a deep red colour (basic iron(III) acetate, brown-red precipitate on boiling) with acetate ion. It is not a test for alcohols or sulphide.
Answer: (B). | ||||
| 81 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $M(Cu_2O) = 143$, $M(Cu_2S) = 159$. Moles: $Cu_2O = \frac{2860}{143} = 20$, $Cu_2S = \frac{4770}{159} = 30$.
20 mol $Cu_2O$ needs only 10 mol $Cu_2S$, so $Cu_2O$ is limiting: Cu formed $= 3 \times 20 = 60$ mol $= 60 \times 63.5 = 3810$ g.
Answer: 3810. | ||||
| 82 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Polar: $NF_3$ (pyramidal), $CHCl_3$ (unsymmetrical tetrahedral), $H_2S$ (bent).
Non-polar by symmetry: $BF_3$ (trigonal planar), $BeF_2$ (linear), $SiF_4$ and $CCl_4$ (tetrahedral), $PF_5$ (trigonal bipyramidal).
Count = 3.
Answer: 3. | ||||
| 83 | 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 The curve is an isotherm at 300 K (reversible) from 6.0 bar to 3.0 bar.
$|W| = nRT\ln\frac{P_1}{P_2} = 1 \times 8.3 \times 300 \times 2.3 \times \log 2 = 2490 \times 2.3 \times 0.30 = 1718.1$ J ≈ 1718 J.
Answer: 1718. | ||||
| 84 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln\frac{A}{k} = \frac{E_a}{RT}$. $\frac{A}{k} = \frac{4 \times 10^{13}}{2 \times 10^{-6}} = 2 \times 10^{19}$, so $\ln\frac{A}{k} = 2.3(0.30 + 19) = 44.39$.
$\frac{E_a}{R} = \frac{8300}{8.3} = 1000$ K, so $T = \frac{1000}{44.39} = 22.53$ K $= 225.3 \times 10^{-1}$ K ≈ 225.
Answer: 225. | ||||
| 85 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Acetic acid removed $= 0.1 \times (0.3 - 0.125) = 0.0175$ mol $= 1.05$ g.
Per gram of charcoal: $\frac{1.05}{0.8} = 1.3125$ g $= 13125 \times 10^{-4}$ g.
Answer: 13125. | ||||
| 86 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter In the brown-ring complex NO is taken as nitrosonium, $NO^+$, with five neutral water ligands: $x + 1 = +2$, so iron is +1.
Answer: 1. | ||||
| 87 | 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 Fe is +3 ($3d^5$). $CN^-$ is a strong-field ligand, so the complex is low spin, $t_{2g}^5$, with one unpaired electron.
$\mu = \sqrt{1(1+2)} = 1.73$ B.M. ≈ 2.
Answer: 2. | ||||
| 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 Sulphur in 0.752 g $BaSO_4$ $= 0.752 \times \frac{32}{233} = 0.1033$ g.
Compound $= \frac{0.1033}{0.516} = 0.200$ g $= 2 \times 10^{-1}$ g.
Answer: 2. | ||||
| 89 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Moles of X $= \frac{10.0 \times 10^{-3}}{80} = 1.25 \times 10^{-4}$; moles of $H_2 = \frac{8.40}{22400} = 3.75 \times 10^{-4}$. So X has 3 C=C bonds.
$C_6H_8$ (M = 80) with three C=C and giving only HCHO and a dialdehyde on ozonolysis is hexa-1,3,5-triene, $CH_2=CH-CH=CH-CH=CH_2$.
Ozonolysis gives 2 HCHO + 2 OHC–CHO: 4 molecules.
Answer: 4. | ||||
| 90 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Veronal, Luminal and Seconal are barbiturates, which are used as tranquilizers (hypnotics/sedatives). Propanal is an aldehyde, not a drug.
Count = 3.
Answer: 3. | ||||
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