JEE (Main) · 2023
JEE (Main) 2023 — 10 April, 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
- Trigonometry3
- Integral Calculus3
- Three Dimensional Geometry3
7 easy · 21 medium · 2 hard
Physics
30 Q
- Electrostatics2
- Oscillations and Waves2
- Properties of Bulk Matter2
25 easy · 5 medium · 0 hard
Chemistry
30 Q
- Some Basic Concepts of Chemistry2
- Classification of Elements and Periodicity2
- Coordination Compounds2
20 easy · 9 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 FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy 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 An element of R is a choice of $(a_1, b_1)$ and $(a_2, b_2)$ from $A \times B$ with $a_1 \mid b_2$ and $a_2 \mid b_1$. The two conditions involve the separate pairs $(a_1, b_2)$ and $(a_2, b_1)$.
Pairs $(a, b) \in A \times B$ with $a \mid b$: $2 \mid 8$, $2 \mid 12$, $3 \mid 9$, $3 \mid 12$, $4 \mid 8$, $4 \mid 12$, i.e. 6 pairs.
So $(a_1, b_2)$ can be chosen in 6 ways and, independently, $(a_2, b_1)$ in 6 ways.
Number of elements $= 6 \times 6 = 36$.
Answer: (D). | ||||
| 2 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Put $u = 9^{\tan^2 x}$. Then $\frac{9}{u} + u = 10$, so $u^2 - 10u + 9 = 0$, $u = 1$ or $u = 9$, i.e. $\tan^2 x = 0$ or $\tan^2 x = 1$.
In $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$: $x = 0$ or $x = \pm\frac{\pi}{4}$. So $S = \left\{0, \frac{\pi}{4}, -\frac{\pi}{4}\right\}$.
$\beta = \tan^2 0 + 2\tan^2\frac{\pi}{12} = 2(2 - \sqrt{3})^2 = 2(7 - 4\sqrt{3}) = 14 - 8\sqrt{3}$.
$\frac{1}{6}(\beta - 14)^2 = \frac{1}{6}(64 \cdot 3) = 32$.
Answer: (C). | ||||
| 3 | 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 With $z = x + iy$: $2z - 3i = 2x + i(2y - 3)$ and $4z + 2i = 4x + i(4y + 2)$.
The quotient is real when the imaginary part of $(2z - 3i)\overline{(4z + 2i)}$ is zero:
$(2y - 3)(4x) - (2x)(4y + 2) = 8xy - 12x - 8xy - 4x = -16x = 0$, so $x = 0$; and the denominator must not vanish, so $z \ne -\frac{i}{2}$, i.e. $y \ne -\frac{1}{2}$.
So $S = \{iy : y \ne -\frac{1}{2}\}$.
(A) $x = 0$: true. (B) $y \ne -\frac{1}{2}$: true. (C) with $x = 0$, $y + y^2 \ne -\frac{1}{4}$ means $\left(y + \frac{1}{2}\right)^2 \ne 0$, i.e. $y \ne -\frac{1}{2}$: true. (D) $(0, -\frac{1}{2})$ is exactly the excluded point: NOT correct.
Answer: (D). | ||||
| 4 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Take $5!$, $6!$, $7!$ out of the three rows of the bracketed matrix M:
$\det M = 5!\,6!\,7!\begin{vmatrix}1 & 6 & 42\\1 & 7 & 56\\1 & 8 & 72\end{vmatrix} = 5!\,6!\,7!\begin{vmatrix}1 & 6 & 42\\0 & 1 & 14\\0 & 2 & 30\end{vmatrix} = 5!\,6!\,7! \cdot (30 - 28) = 2 \cdot 5!\,6!\,7!$.
The factor $\frac{1}{5!\,6!\,7!}$ is meant to cancel this, giving $|A| = 2$ (see note).
For a $3 \times 3$ matrix, $|2A| = 2^3 |A| = 16$, and $|\operatorname{adj}(\operatorname{adj} B)| = |B|^{(n-1)^2} = |B|^4$.
So $|\operatorname{adj}(\operatorname{adj}(2A))| = 16^4 = 2^{16}$.
Answer: (B). | ||||
| 5 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations With at most 3 in each of 3 cars and 8 persons, the only split is $3, 3, 2$.
Choose which car carries 2 persons: 3 ways. Then distribute the persons: $\frac{8!}{3!\,3!\,2!} = 560$ ways.
Total $= 3 \times 560 = 1680$.
Answer: (C). | ||||
| 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 $(1 + x)^p(1 - x)^q = \left(1 + px + \binom{p}{2}x^2 + \ldots\right)\left(1 - qx + \binom{q}{2}x^2 - \ldots\right)$.
Coefficient of $x$: $p - q = 4$.
Coefficient of $x^2$: $\binom{p}{2} + \binom{q}{2} - pq = \frac{(p - q)^2 - (p + q)}{2} = -5$, so $16 - (p + q) = -10$, $p + q = 26$.
Hence $p = 15$, $q = 11$ and $2p + 3q = 30 + 33 = 63$.
Answer: (B). | ||||
| 7 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Modulo 3: $22 \equiv 1$, so $22^{2022} \equiv 1$; $2022 \equiv 0$, so $2022^{22} \equiv 0$. Hence $\alpha = 1$.
Modulo 7: $22 = 21 + 1 \equiv 1$, so $22^{2022} \equiv 1$; $2022 = 7 \cdot 289 - 1 \equiv -1$, so $2022^{22} \equiv (-1)^{22} = 1$. Hence $\beta = 2$.
$\alpha^2 + \beta^2 = 1 + 4 = 5$.
Answer: (A). | ||||
| 8 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The differences $7, 10, 13, 16, \ldots$ form an AP with common difference 3, so the $n$th term is quadratic:
$T_n = 4 + \sum_{k=1}^{n-1}(3k + 4) = 4 + \frac{3n(n-1)}{2} + 4(n - 1) = \frac{3n^2 + 5n}{2}$ (check: $T_1 = 4$, $T_2 = 11$, $T_3 = 21$).
$S_{29} - S_9 = \sum_{n=10}^{29} T_n = \frac{3}{2}\sum_{n=10}^{29} n^2 + \frac{5}{2}\sum_{n=10}^{29} n$.
$\sum_{n=10}^{29} n^2 = \frac{29 \cdot 30 \cdot 59}{6} - \frac{9 \cdot 10 \cdot 19}{6} = 8555 - 285 = 8270$; $\sum_{n=10}^{29} n = 435 - 45 = 390$.
$S_{29} - S_9 = 12405 + 975 = 13380$, and $\frac{13380}{60} = 223$.
Answer: (B). | ||||
| 9 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $g'(x) = f'(x) - f'(1 - x)$ and $g''(x) = f''(x) + f''(1 - x) > 0$, so $g'$ is strictly increasing on $(0, 1)$.
$g'\left(\frac{1}{2}\right) = f'\left(\frac{1}{2}\right) - f'\left(\frac{1}{2}\right) = 0$, so $g' < 0$ on $\left(0, \frac{1}{2}\right)$ and $g' > 0$ on $\left(\frac{1}{2}, 1\right)$. Hence $\alpha = \frac{1}{2}$.
The sum is $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3$. Since $2 \cdot 3 > 1$, $\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1}\frac{2 + 3}{1 - 6} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Total $= \frac{\pi}{4} + \frac{3\pi}{4} = \pi$.
Answer: (D). | ||||
| 10 | Integral CalculusIntegral as an antiderivative | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{d}{dx}\left(\frac{x}{e}\right)^{2x} = \left(\frac{x}{e}\right)^{2x}\frac{d}{dx}[2x(\ln x - 1)] = \left(\frac{x}{e}\right)^{2x}(2\ln x - 2 + 2) = 2\ln x\left(\frac{x}{e}\right)^{2x}$.
Likewise $\frac{d}{dx}\left(\frac{e}{x}\right)^{2x} = \left(\frac{e}{x}\right)^{2x}\frac{d}{dx}[2x(1 - \ln x)] = -2\ln x\left(\frac{e}{x}\right)^{2x}$.
So the integral is $\frac{1}{2}\left(\frac{x}{e}\right)^{2x} - \frac{1}{2}\left(\frac{e}{x}\right)^{2x} + C$: $\alpha = \beta = \gamma = \delta = 2$.
$\alpha + 2\beta + 3\gamma - 4\delta = 2 + 4 + 6 - 8 = 4$.
Answer: (D). | ||||
| 11 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Differentiate both sides with respect to $t$ (Leibniz rule): $\left(f(t^2) + t^4\right)\cdot 2t = 4t^2$, so $f(t^2) = 2t - t^4$ for $t > 0$.
For $t^2 = \frac{\pi^2}{4}$, $t = \frac{\pi}{2}$: $f\left(\frac{\pi^2}{4}\right) = \pi - \frac{\pi^4}{16} = \pi\left(1 - \frac{\pi^3}{16}\right)$.
Answer: (D). | ||||
| 12 | CirclesStandard and general form | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Let $B = (4\cos\theta, 4\sin\theta)$. P divides AB in the ratio $3:2$, so $P = \frac{2A + 3B}{5} = \left(\frac{2 + 12\cos\theta}{5}, \frac{4 + 12\sin\theta}{5}\right)$.
The locus of P is the circle $\left(x - \frac{2}{5}\right)^2 + \left(y - \frac{4}{5}\right)^2 = \left(\frac{12}{5}\right)^2$, with centre $C = \left(\frac{2}{5}, \frac{4}{5}\right)$.
$AC = \sqrt{\left(1 - \frac{2}{5}\right)^2 + \left(2 - \frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \frac{3\sqrt{5}}{5}$.
Answer: (A). | ||||
| 13 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The ellipse is $\frac{x^2}{19} + \frac{y^2}{15} = 1$ (major axis along the x-axis, minor axis along the y-axis); the circle is $x^2 + y^2 = 16$.
A line $y = mx + c$ touches the ellipse when $c^2 = 19m^2 + 15$ and touches the circle when $c^2 = 16(1 + m^2)$.
So $19m^2 + 15 = 16m^2 + 16$, $m^2 = \frac{1}{3}$, $m = \pm\frac{1}{\sqrt{3}}$: the tangents make $\frac{\pi}{6}$ with the x-axis (major axis), hence $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$ with the minor axis.
Answer: (A). | ||||
| 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 $\vec{AB} = (0, 2, 1)$, $\vec{AC} = (-1, 3, 1)$. Normal $\vec{AB} \times \vec{AC} = (2 - 3, -1 - 0, 0 + 2) = (-1, -1, 2)$, or $(1, 1, -2)$.
Plane: $(x - 1) + (y - 2) - 2z = 0$, i.e. $x + y - 2z = 3$.
For P: $1 + 2 - 12 - 3 = -12$. The image is $Q = P - \frac{2(-12)}{1 + 1 + 4}(1, 1, -2) = (1, 2, 6) + 4(1, 1, -2) = (5, 6, -2)$.
$\alpha^2 + \beta^2 + \gamma^2 = 25 + 36 + 4 = 65$.
Answer: (B). | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A general point of the first line is $(t, 6 - 2t, 5t - 8)$.
Meeting the second line: $\frac{t - 5}{4} = \frac{-1 - 2t}{3} = 5t - 6$. From the first and third, $t - 5 = 20t - 24$, $t = 1$; check: $-1 = -1 = -1$. So $A = (1, 4, -3)$.
Meeting the third line: $\frac{t + 3}{6} = \frac{2t - 3}{3} = 5t - 14$. From the first and third, $t + 3 = 30t - 84$, $t = 3$; check: $1 = 1 = 1$. So $B = (3, 0, 7)$.
Mid-point $M = (2, 2, 2)$. Distance from $2x - 2y + z - 14 = 0$: $\frac{|4 - 4 + 2 - 14|}{\sqrt{4 + 4 + 1}} = \frac{12}{3} = 4$.
Answer: (C). | ||||
| 16 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\2 & 7 & -1\\3 & 0 & 5\end{vmatrix} = 35\hat{i} - 13\hat{j} - 21\hat{k}$.
$\vec{d} = \lambda(35, -13, -21)$ and $\vec{c}\cdot\vec{d} = \lambda(35 + 13 - 42) = 6\lambda = 12$, so $\lambda = 2$ and $\vec{d} = (70, -26, -42)$.
$\vec{c} \times \vec{d} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & -1 & 2\\70 & -26 & -42\end{vmatrix} = (42 + 52)\hat{i} - (-42 - 140)\hat{j} + (-26 + 70)\hat{k} = (94, 182, 44)$.
$(-1, 1, -1)\cdot(94, 182, 44) = -94 + 182 - 44 = 44$.
Answer: (C). | ||||
| 17 | Vector AlgebraAddition of vectors | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on addition of vectors, from Vector Algebra. NCERT: Class 12: Vector Algebra Take the circumcentre P as the origin and let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of A, B, C, so $|\vec{a}| = |\vec{b}| = |\vec{c}|$.
Let $\vec{h} = \vec{a} + \vec{b} + \vec{c}$. Then $(\vec{h} - \vec{a})\cdot(\vec{b} - \vec{c}) = (\vec{b} + \vec{c})\cdot(\vec{b} - \vec{c}) = |\vec{b}|^2 - |\vec{c}|^2 = 0$, so the point $\vec{h}$ lies on the altitude from A; similarly on the other altitudes. Hence $\vec{h}$ is the orthocentre Q.
$\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \vec{a} + \vec{b} + \vec{c} = \overrightarrow{PQ}$.
Answer: (C). | ||||
| 18 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability On each roll P(odd) = P(even) = $\frac{1}{2}$, so P(odd exactly $r$ times) $= \binom{n}{r}\frac{1}{2^n}$.
$\binom{n}{7} = \binom{n}{9}$ gives $n = 16$.
P(even exactly twice) $= \binom{16}{2}\frac{1}{2^{16}} = \frac{120}{2^{16}} = \frac{60}{2^{15}}$, so $k = 60$.
Answer: (C). | ||||
| 19 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $\sum f_i = (k + 2) + 2k + 3k^2 - 1 + (k - 3) = 3k^2 + 4k - 2 = 62$, so $3k^2 + 4k - 64 = 0$, $(k - 4)(3k + 16) = 0$, $k = 4$.
Frequencies: 6, 8, 15, 15, 17, 1 (total 62).
$\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2$, so $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N}$.
$\sum f_i x_i^2 = 0 + 8 + 60 + 135 + 272 + 25 = 500$, so $\mu^2 + \sigma^2 = \frac{500}{62} \approx 8.06$.
$[\mu^2 + \sigma^2] = 8$.
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 By De Morgan's law, $\sim[p \vee (\sim(p \wedge q))] \equiv (\sim p) \wedge (p \wedge q)$, which is a contradiction (it needs both $p$ and $\sim p$).
(C) $(p \wedge q) \wedge (\sim p)$ is the same statement.
The others are not contradictions: (A) $\equiv (\sim p) \wedge q$, true for $p$ false, $q$ true; (B) is true for $p$, $q$ both false; (D) is true for $p$ false.
Answer: (C). | ||||
| 21 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sec^{-1} t$ is defined for $|t| \ge 1$. So $\left|\frac{2x}{5x + 3}\right| \ge 1$ with $x \ne -\frac{3}{5}$:
$4x^2 \ge (5x + 3)^2 \iff 21x^2 + 30x + 9 \le 0 \iff 7x^2 + 10x + 3 \le 0 \iff (7x + 3)(x + 1) \le 0 \iff -1 \le x \le -\frac{3}{7}$.
Removing $x = -\frac{3}{5}$: domain $= \left[-1, -\frac{3}{5}\right) \cup \left(-\frac{3}{5}, -\frac{3}{7}\right]$.
$\alpha = -1$, $\beta = \gamma = -\frac{3}{5}$, $\delta = -\frac{3}{7}$.
$3\alpha + 10(\beta + \gamma) + 21\delta = -3 - 12 - 9 = -24$, and its absolute value is 24.
Answer: 24. | ||||
| 22 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalHard | |
A hard numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants No solution requires $\Delta = \begin{vmatrix}6\lambda & -3 & 3\\2 & 6\lambda & 4\\3 & 2 & 3\lambda\end{vmatrix} = 0$:
$\Delta = 6\lambda(18\lambda^2 - 8) + 3(6\lambda - 12) + 3(4 - 18\lambda) = 108\lambda^3 - 84\lambda - 24 = 12(9\lambda^3 - 7\lambda - 2) = 12(\lambda - 1)(3\lambda + 1)(3\lambda + 2)$.
So $\lambda \in \left\{1, -\frac{1}{3}, -\frac{2}{3}\right\}$. In each case the coefficient rows have rank 2; write row 3 as $a\cdot$row 1 $+ b\cdot$row 2 and compare the right sides.
$\lambda = 1$: $a = \frac{1}{3}$, $b = \frac{1}{2}$; right side $\frac{4}{3} + \frac{1}{2} = \frac{11}{6} \ne 1$: no solution.
$\lambda = -\frac{1}{3}$: rows $(-2, -3, 3)$, $(2, -2, 4)$, $(3, 2, -1)$; $a = -1$, $b = \frac{1}{2}$; right side $-\frac{4}{9} + \frac{1}{2} = \frac{1}{18} \ne -\frac{1}{3}$: no solution.
$\lambda = -\frac{2}{3}$: rows $(-4, -3, 3)$, $(2, -4, 4)$, $(3, 2, -2)$; $a = -\frac{8}{11}$, $b = \frac{1}{22}$; right side $-\frac{8}{11}\cdot\frac{16}{9} + \frac{1}{22} = -\frac{247}{198} \ne -\frac{2}{3}$: no solution.
So $S = \left\{1, -\frac{1}{3}, -\frac{2}{3}\right\}$, $\sum|\lambda| = 2$ and $12 \times 2 = 24$.
Answer: 24. | ||||
| 23 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Number of arrangements of 2, 1, 2, 3: $\frac{4!}{2!} = 12$.
In a given place, 1 appears $\frac{3!}{2!} = 3$ times, 3 appears 3 times and 2 appears $3! = 6$ times.
Sum of the digits in each place $= 1 \cdot 3 + 3 \cdot 3 + 2 \cdot 6 = 24$.
Total $= 24 \times 1111 = 26664$.
Answer: 26664. | ||||
| 24 | 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 the terms as $a$, $2(a + d)$, $4(a + 2d)$, $8(a + 3d)$, $16(a + 4d)$.
Third term: $4(a + 2d) = 2$, so $a + 2d = \frac{1}{2}$.
Sum: $a + 2a + 2d + 2 + 8a + 24d + 16a + 64d = 27a + 90d + 2 = \frac{49}{2}$, so $27a + 90d = \frac{45}{2}$.
With $a = \frac{1}{2} - 2d$: $\frac{27}{2} + 36d = \frac{45}{2}$, $d = \frac{1}{4}$, $a = 0$.
Terms: $0, \frac{1}{2}, 2, 6, 16$ (sum $\frac{49}{2}$). So $a_4 = 16(a + 4d) = 16$.
Answer: 16. | ||||
| 25 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $f(x) = ax^2 + bx + c$. Conditions: $f(-1) = a - b + c = 0$; $f(1) = a + b + c = 1$; $f'(1) = 2a + b = 1$.
So $b = \frac{1}{2}$, $a = \frac{1}{4}$, $c = \frac{1}{4}$: $f(x) = \frac{(x + 1)^2}{4}$.
$(\alpha, \alpha + 1)$ on the curve: $\frac{(\alpha + 1)^2}{4} = \alpha + 1$; in the first quadrant $\alpha + 1 \ne 0$, so $\alpha = 3$ and the point is $(3, 4)$.
$f'(3) = \frac{3 + 1}{2} = 2$, so the normal has slope $-\frac{1}{2}$: $y - 4 = -\frac{1}{2}(x - 3)$. At $y = 0$, $x = 3 + 8 = 11$.
Answer: 11. | ||||
| 26 | 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 We need $|x^2 - 2| \le x$. For $x \ge \sqrt{2}$: $x^2 - x - 2 \le 0$ gives $x \le 2$. For $0 \le x < \sqrt{2}$: $x^2 + x - 2 \ge 0$ gives $x \ge 1$. So $1 \le x \le 2$.
$A = \int_1^2 x\,dx - \int_1^{\sqrt{2}}(2 - x^2)\,dx - \int_{\sqrt{2}}^2(x^2 - 2)\,dx$.
$\int_1^2 x\,dx = \frac{3}{2}$; $\int_1^{\sqrt{2}}(2 - x^2)\,dx = \frac{4\sqrt{2}}{3} - \frac{5}{3}$; $\int_{\sqrt{2}}^2(x^2 - 2)\,dx = -\frac{4}{3} + \frac{4\sqrt{2}}{3}$.
$A = \frac{3}{2} - \frac{8\sqrt{2}}{3} + 3 = \frac{9}{2} - \frac{8\sqrt{2}}{3}$.
$6A + 16\sqrt{2} = 27 - 16\sqrt{2} + 16\sqrt{2} = 27$.
Answer: 27. | ||||
| 27 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Tangent at $P(x, y)$: it meets the x-axis at $A = \left(x - \frac{y}{y'}, 0\right)$ and the y-axis at $B = (0, y - xy')$.
$AP : PB = 1 : k$ means $P = \frac{kA + B}{k + 1}$. Comparing x-coordinates: $x = \frac{k}{k + 1}\left(x - \frac{y}{y'}\right)$, so $-\frac{y}{y'} = \frac{x}{k}$, i.e. $\frac{dy}{dx} = -\frac{ky}{x}$.
Hence $y = Cx^{-k}$. Through $(1, 1)$: $C = 1$. Through $\left(\frac{1}{10}, 100\right)$: $10^k = 100$, $k = 2$.
The equation becomes $e^{y'} = 2x + 1$, i.e. $y' = \ln(2x + 1)$, with $y(0) = 2$.
$y = \frac{(2x + 1)\ln(2x + 1) - (2x + 1)}{2} + C_1$; $y(0) = -\frac{1}{2} + C_1 = 2$, $C_1 = \frac{5}{2}$.
$y(1) = \frac{3\ln 3 - 3}{2} + \frac{5}{2} = \frac{3}{2}\ln 3 + 1$, so $4y(1) - 5\ln 3 = 4 + 6\ln 3 - 5\ln 3 = 4 + \ln 3 \approx 5.10$.
Rounded to the nearest integer: 5.
Answer: 5. | ||||
| 28 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines The two sides meet at $(-1, 7)$, which is not on $3x + 7y = 23$ ($-3 + 49 = 46$), so it is the vertex B.
The diagonal AC meets the sides at $2x - 3y = -23$, $3x + 7y = 23$: $(-4, 5)$, and $5x + 4y = 23$, $3x + 7y = 23$: $(3, 2)$. These are A and C.
Mid-point of AC: $\left(-\frac{1}{2}, \frac{7}{2}\right)$. Diagonal BD passes through $(-1, 7)$ and this point: slope $\frac{7/2 - 7}{1/2} = -7$, so BD is $7x + y = 0$.
Distance of A from BD (either labelling of A gives the same value, since BD bisects AC): $d = \frac{|7(-4) + 5|}{\sqrt{50}} = \frac{23}{\sqrt{50}}$.
$50d^2 = 529$.
Answer: 529. | ||||
| 29 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $4 - 3 + 2 + 3 = 6$, so $N = (4, 3, 1) - \frac{6}{6}(1, -1, 2) = (3, 4, -1)$ and $AN = \sqrt{6}$.
B on P: $5 - \alpha + 2\beta + 3 = 0$, so $\alpha = 8 + 2\beta$.
AN is perpendicular to the plane, so the angle at N is a right angle: area $= \frac{1}{2}\cdot AN \cdot NB = 3\sqrt{2}$ gives $NB = 2\sqrt{3}$, $NB^2 = 12$.
$NB^2 = 2^2 + (\alpha - 4)^2 + (\beta + 1)^2 = 4 + (4 + 2\beta)^2 + (\beta + 1)^2 = 12$, so $5\beta^2 + 18\beta + 9 = 0$, $(\beta + 3)(5\beta + 3) = 0$.
Integer solution: $\beta = -3$, $\alpha = 2$.
$\alpha^2 + \beta^2 + \alpha\beta = 4 + 9 - 6 = 7$.
Answer: 7. | ||||
| 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 From the figure, ABDC is a rectangle (right angles at A, B and D), E lies on BD produced, $\theta_1 = \angle CBA$ and $\theta_2 = \angle ECD$.
Let $AB = CD = a$ and $AC = BD = h$. Then $\tan\theta_1 = \frac{h}{a}$, $BE = \frac{4a}{\sqrt{3}}$, $ED = \frac{4a}{\sqrt{3}} - h$ and $\tan\theta_2 = \frac{ED}{CD} = \frac{4}{\sqrt{3}} - \tan\theta_1$.
Since $\theta_2 = \frac{\pi}{2} - \theta_1$, $\tan\theta_2 = \cot\theta_1$. With $t = \tan\theta_1$: $t + \frac{1}{t} = \frac{4}{\sqrt{3}}$, $t = \sqrt{3}$ or $\frac{1}{\sqrt{3}}$.
$\frac{\theta_2}{\theta_1}$ is largest for $\theta_1 = \frac{\pi}{6}$, $\theta_2 = \frac{\pi}{3}$.
Area of $\Delta CAB = \frac{1}{2}ah = \frac{a^2}{2\sqrt{3}} = 2\sqrt{3} - 3$, so $a^2 = 12 - 6\sqrt{3} = (3 - \sqrt{3})^2$, $a = 3 - \sqrt{3}$.
In right triangle CDE with $\angle C = \frac{\pi}{3}$: $CE = 2a$, $ED = \sqrt{3}a$. Perimeter $= a(3 + \sqrt{3}) = (3 - \sqrt{3})(3 + \sqrt{3}) = 6$.
Answer: 6. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 $C_1 = \frac{\varepsilon_0 A}{d}$. A metal sheet of thickness $t$ has no field inside it, so the effective gap becomes $d - t$:
$C_2 = \frac{\varepsilon_0 A}{d - \frac{2d}{3}} = \frac{3\varepsilon_0 A}{d} = 3C_1$.
$\frac{C_2}{C_1} = 3:1$.
Answer: (C). | ||||
| 32 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $y = \sin\omega t + \cos\omega t = \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin\omega t + \frac{1}{\sqrt{2}}\cos\omega t\right) = \sqrt{2}\sin\left(\omega t + \frac{\pi}{4}\right)$.
Amplitude $= \sqrt{1^2 + 1^2} = \sqrt{2}$.
Answer: (C). | ||||
| 33 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Internal energy $U = \frac{f}{2}nRT$.
Oxygen (diatomic, no vibration): $f = 5$, $U_1 = \frac{5}{2}\cdot 2RT = 5RT$.
Neon (monatomic): $f = 3$, $U_2 = \frac{3}{2}\cdot 4RT = 6RT$.
Total $= 11RT$.
Answer: (C). | ||||
| 34 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $Q = 0$, so $\Delta U = Q - W_{\text{by gas}} = W_{\text{on gas}}$.
In compression work is done on the gas, so $\Delta U > 0$ and the temperature rises.
(A) true; (B) true; (D) true (no heat exchanged). (C) "no change in internal energy" is NOT true.
Answer: (C). | ||||
| 35 | Properties of Bulk MatterElasticity | 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 elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $\Delta L = \frac{FL}{AY}$. With the same F and L, $\Delta L \propto \frac{1}{AY}$.
$\frac{\Delta L_A}{\Delta L_B} = \frac{A_B Y_B}{A_A Y_A} = \frac{3 \times 4}{1 \times 1} = 12$.
Ratio $= 12:1$.
Answer: (C). | ||||
| 36 | GravitationAcceleration due to gravity | Class 11: Gravitation | StatementsEasy | |
A easy statements question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation Because of rotation, at latitude $\lambda$: $g' = g - \omega^2 R\cos^2\lambda$.
Statement I is true: rotation does change g (except at the poles).
At the equator ($\lambda = 0$) the reduction $\omega^2 R$ is the largest; at the poles ($\lambda = 90^\circ$) it is zero. So the effect is maximum at the equator and minimum at the poles, the reverse of Statement II, which is false.
Answer: (C). | ||||
| 37 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Orbit radius $r = 2R$. $T = 2\pi\sqrt{\frac{r^3}{GM}}$ with $GM = gR^2$:
$T = 2\pi\sqrt{\frac{8R^3}{gR^2}} = 2\pi\sqrt{\frac{8R}{\pi^2}} = 2\sqrt{8R} = \sqrt{32R}$.
Answer: (D). | ||||
| 38 | 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 $H = \frac{u^2\sin^2\theta}{2g}$, so $\frac{H_1}{H_2} = \frac{\sin^2 30^\circ}{\sin^2 60^\circ} = \frac{1/4}{3/4} = \frac{1}{3}$.
Answer: (A). | ||||
| 39 | 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 Least count $= 0.1$ mm $= 0.01$ cm.
Zero error: the vernier zero is to the right of the main-scale zero, so the error is positive: $+6 \times 0.01 = +0.06$ cm.
Observed reading $= 3.2 + 4 \times 0.01 = 3.24$ cm.
Corrected diameter $= 3.24 - 0.06 = 3.18$ cm.
Answer: (D). | ||||
| 40 | 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 Total displacement $= 2x$; total time $= \frac{x}{v_1} + \frac{x}{v_2}$.
$v = \frac{2x}{\frac{x}{v_1} + \frac{x}{v_2}}$, so $\frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}$.
Answer: (D). | ||||
| 41 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Assertion and reasonEasy | |
A easy assertion and reason question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion When the current is switched off the driving torque stops, but the blades keep turning because a body in motion tends to stay in motion (inertia of motion, here rotational inertia); friction slowly brings them to rest.
So A is correct, R is correct, and R explains A.
Answer: (A). | ||||
| 42 | 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 An AM wave has side bands at $f_c \pm f_m$, so its bandwidth is $2f_m = 2 \times 3$ kHz $= 6$ kHz.
Answer: (D). | ||||
| 43 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits From the figure: a 5 V cell in series with 25 $\Omega$ feeds three parallel branches; $I_1$ is the current from the cell, $I_2$ flows in the branch of $D_3$, $I_3$ in the branch of $D_2$ and $I_4$ in the branch of $D_1$, each branch having a 125 $\Omega$ resistor.
$D_1$ and $D_3$ point along the current (forward biased); $D_2$ is reversed, so $I_3 = 0$.
Each conducting branch: $25 + 125 = 150\ \Omega$; two in parallel: $75\ \Omega$; with the series 25 $\Omega$: $100\ \Omega$.
$I_1 = \frac{5}{100} = 0.05$ A, and $I_2 = I_4 = 0.025$ A.
So $\frac{I_1}{I_2} = 2$; (A), (B) and (C) fail ($I_3 = 0$).
Answer: (D). | ||||
| 44 | 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 When $\frac{7}{8}$ has disintegrated, $\frac{1}{8} = \left(\frac{1}{2}\right)^3$ remains: three half-lives.
Time $= 3T$.
Answer: (C). | ||||
| 45 | 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 From the graph the stopping potential is zero at the threshold frequency $\nu_0 = 5 \times 10^{14}$ Hz.
$\phi = h\nu_0 = 6.63 \times 10^{-34} \times 5 \times 10^{14} = 3.32 \times 10^{-19}$ J $= \frac{3.32 \times 10^{-19}}{1.6 \times 10^{-19}}$ eV $\approx 2.07$ eV.
Answer: (C). | ||||
| 46 | Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The falling magnet induces eddy currents in the copper tube; by Lenz's law they oppose its motion with a force that grows with its speed.
The speed rises until this magnetic drag balances the weight, after which the magnet falls with an almost constant (terminal) speed.
Answer: (C). | ||||
| 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 For two waves of equal intensity $I_0$: $I = 4I_0\cos^2\frac{\phi}{2}$.
$I_P \propto \cos^2\frac{\pi}{6} = \frac{3}{4}$; $I_Q \propto \cos^2\frac{\pi}{4} = \frac{1}{2}$.
$I_P : I_Q = \frac{3}{4} : \frac{1}{2} = 3:2$.
Answer: (A). | ||||
| 48 | 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 $E_0 = cB_0 = 3 \times 10^8 \times 6.0 \times 10^{-7} = 180$ V m$^{-1}$.
Answer: (B). | ||||
| 49 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsEasy | |
A easy statements question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Diamagnetic materials have small negative susceptibility; the limit $\chi = -1$ is reached by a superconductor (perfect diamagnet). So $-1 \le \chi < 0$: Statement I is true.
Their induced moment opposes the field, so they are pushed from stronger to weaker regions of a non-uniform field: Statement II is true.
Answer: (A). | ||||
| 50 | Current ElectricityDrift velocity and mobility | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity The field points from higher to lower potential; electrons, being negative, are pushed the other way, so their drift is from lower to higher potential: (A) is wrong.
Between collisions with the ions the electrons accelerate under the field and their velocity keeps changing in direction and size, so their paths are curved, not straight lines with uniform velocity: (B) and (C) are wrong.
They move along curved paths with a net drift from lower to higher potential.
Answer: (D). | ||||
| 51 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\omega = \frac{2\pi}{T} = \frac{2 \times 3.14}{3.14} = 2$ rad s$^{-1}$, so $k = m\omega^2 = 5 \times 4 = 20$ N m$^{-1}$.
Maximum force $= kA = 20 \times 1 = 20$ N.
Answer: 20. | ||||
| 52 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Equation of continuity: $A_1v_1 = A_2v_2$.
$A_1 = 2.0$ cm$^2 = 200$ mm$^2$, $A_2 = 10$ mm$^2$.
$v_2 = \frac{200 \times 4}{10} = 80$ cm s$^{-1}$.
Answer: 80. | ||||
| 53 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Position of the point of application (origin) relative to $(2, -3, 0)$: $\vec{r} = (0 - 2)\hat{i} + (0 + 3)\hat{j} = -2\hat{i} + 3\hat{j}$.
$\vec{\tau} = \vec{r} \times \vec{F} = (-2\hat{i} + 3\hat{j}) \times (-P\hat{k}) = 2P(\hat{i} \times \hat{k}) - 3P(\hat{j} \times \hat{k}) = -2P\hat{j} - 3P\hat{i}$.
So $a = -3$, $b = -2$, $\frac{a}{b} = \frac{3}{2} = \frac{x}{2}$, $x = 3$.
Answer: 3. | ||||
| 54 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At uniform speed the motor's force balances weight and friction: $F = mg + f = 1400 \times 10 + 2000 = 16000$ N.
$P = Fv = 16000 \times 3 = 48000$ W $= 48$ kW.
Answer: 48. | ||||
| 55 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Field of a long line charge (Gauss's law): $E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r}$.
Centripetal force: $\frac{mv^2}{r} = eE = \frac{2ke\lambda}{r}$, so $v^2 = \frac{2ke\lambda}{m}$, independent of r.
$v^2 = \frac{2 \times 9 \times 10^9 \times 1.6 \times 10^{-19} \times 2 \times 10^{-8}}{9 \times 10^{-31}} = 6.4 \times 10^{13}$, $v = 8 \times 10^6$ m s$^{-1}$.
Answer: 8. | ||||
| 56 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The lowest wavelength of a series is its series limit ($n_2 \to \infty$).
Lyman: $\frac{1}{\lambda_L} = R\left(\frac{1}{1^2}\right) = R$, so $\frac{1}{R} = 917$ Å.
Balmer: $\frac{1}{\lambda_B} = R\left(\frac{1}{2^2}\right) = \frac{R}{4}$, so $\lambda_B = 4 \times 917 = 3668$ Å.
Answer: 3668. | ||||
| 57 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens $L_1$ ($f = 24$ cm), $u = -6$ cm: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{24} - \frac{1}{6} = -\frac{1}{8}$, $v = -8$ cm: a virtual image 8 cm to the left of $L_1$.
For $L_2$ ($f = 9$ cm) this image is $8 + 10 = 18$ cm to the left: $u = -18$ cm, $\frac{1}{v} = \frac{1}{9} - \frac{1}{18} = \frac{1}{18}$, $v = 18$ cm to the right of $L_2$.
Distance from the object: $6 + 10 + 18 = 34$ cm.
Answer: 34. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Inside the solenoid $B = \mu_0 n I_0\sin\omega t$ with $n = 50$ cm$^{-1} = 5000$ m$^{-1}$; the loop's area $A = (0.02)^2 = 4 \times 10^{-4}$ m$^2$.
$\Phi = \mu_0 n I_0 A\sin\omega t$, so the emf amplitude is $\mu_0 n I_0\omega A$:
$4\pi \times 10^{-7} \times 5000 \times 2.5 \times 700 \times 4 \times 10^{-4} = 14\pi \times 10^{-4} = 14 \times \frac{22}{7} \times 10^{-4} = 44 \times 10^{-4}$ V.
$x = 44$.
Answer: 44. | ||||
| 59 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The straight parts lie along the line through O, so they give no field at O ($d\vec{l} \parallel \vec{r}$).
Semicircular arc: $B = \frac{\mu_0 I}{4r} = \frac{4\pi \times 10^{-7} \times 14}{4 \times 0.022} = \frac{\pi \times 14 \times 10^{-7}}{0.022}$.
With $\pi = \frac{22}{7}$: $B = \frac{44 \times 10^{-7}}{0.022} = 2 \times 10^{-4}$ T.
Answer: 2. | ||||
| 60 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity The rectangular faces are the 1 cm $\times$ 100 cm faces; opposite ones are 1 cm apart.
$R = \frac{\rho l}{A} = \frac{3 \times 10^{-7} \times 0.01}{0.01 \times 1} = 3 \times 10^{-7}\ \Omega$.
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Matching listEasy | |
A easy matching list question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry A: 16 g of $CH_4$ is 1 mol; each molecule has $6 + 4 = 10$ electrons, so $10 \times 6.02 \times 10^{23} = 60.2 \times 10^{23}$ electrons: II.
B: 1 g of $H_2$ is 0.5 mol; at STP (1 bar, 273 K, 22.7 L mol$^{-1}$) it occupies $0.5 \times 22.7 \approx 11.4$ L: IV.
C: 1 mol of $N_2$ weighs 28 g: I.
D: 0.5 mol of $SO_2$ weighs $0.5 \times 64 = 32$ g: III.
A-II, B-IV, C-I, D-III.
Answer: (B). | ||||
| 62 | 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 fcc: spheres touch along the face diagonal, $\sqrt{2}a = 4r$, so $a = 2\sqrt{2}r$.
bcc: spheres touch along the body diagonal, $\sqrt{3}a = 4r$.
Answer: (A). | ||||
| 63 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Blood is a colloidal sol whose particles carry negative charge. By the Hardy–Schulze rule the coagulating power of an ion of opposite charge rises steeply with its charge, so the triply charged $Fe^{3+}$ ions neutralise and coagulate the blood colloid, stopping bleeding.
Answer: (C). | ||||
| 64 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Metallic character decreases across a period and increases down a group.
K (group 1, period 4) is more metallic than Ca (group 2, period 4); Ca is more metallic than Be (group 2, period 2).
K > Ca > Be.
Answer: (A). | ||||
| 65 | 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 A reducing agent can reduce an oxide at a temperature where its own oxidation line lies below the oxide's line (the combined $\Delta G^\circ$ is negative).
At 600 °C, from the figure: $2Fe + O_2 \to 2FeO$ is at about $-280$ kJ mol$^{-1}$; $C + O_2 \to CO_2$ is at about $-320$; $2CO + O_2 \to 2CO_2$ is at about $-320$; $2Zn + O_2 \to 2ZnO$ is at about $-540$.
(A) The C line is below the FeO line: C can reduce FeO. True.
(B) The C line is above the ZnO line: C cannot reduce ZnO at 600 °C. False.
(C) The CO line is above the ZnO line: False.
(D) The CO line is below the FeO line, so CO can reduce FeO. False.
Answer: (A). | ||||
| 66 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Protium, deuterium and tritium have masses about 1, 2 and 3 u: the relative mass difference (doubling, tripling) is the largest among all isotopes of any element.
Physical properties such as density, boiling point and rate of diffusion depend on mass, so they differ markedly among the hydrogen isotopes (their chemical properties are nearly the same). So A is true, R is true, and R explains A.
Answer: (A). | ||||
| 67 | 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 sodium carbonate decahydrate, $Na_2CO_3 \cdot 10H_2O$: 10 water molecules.
Soda ash is anhydrous sodium carbonate, $Na_2CO_3$: 0 water molecules.
Answer: (C). | ||||
| 68 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Assertion and reasonMedium | |
A medium assertion and reason question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Forming $Mg^+$ needs only the first ionisation enthalpy (about 738 kJ mol$^{-1}$); forming $Mg^{2+}$ needs the first plus the second (about 1451 kJ mol$^{-1}$), in all about 2189 kJ mol$^{-1}$, so A is true.
R is a true statement: $Mg^{2+}$ is smaller and more highly charged than $Mg^+$.
The second electron is pulled out of a positive, contracted species against a larger effective nuclear charge, which is the size-and-charge effect R describes, so R is taken as the explanation of A.
Answer: (A). | ||||
| 69 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A. $Fe^{3+}$, $d^5$ with strong-field $CN^-$: low spin $t_{2g}^5$: 1 unpaired.
B. $Fe^{3+}$, $d^5$ with weak-field $F^-$: high spin: 5 unpaired.
C. $Co^{3+}$, $d^6$ with $F^-$: high spin $t_{2g}^4e_g^2$: 4 unpaired.
D. $Cr^{3+}$, $d^3$: $t_{2g}^3$: 3 unpaired.
E. $Ni(0)$ in $Ni(CO)_4$: $3d^{10}$: 0 unpaired.
Order: E (0) < A (1) < D (3) < C (4) < B (5).
Answer: (C). | ||||
| 70 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Matching listMedium | |
A medium matching list question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds List II gives the crystal field stabilisation energy in units of $\Delta_0$: each $t_{2g}$ electron contributes $-0.4\Delta_0$ and each $e_g$ electron $+0.6\Delta_0$. Water is a weak-field ligand, so the complexes are high spin.
A. $Ti^{2+}$, $d^2$: $t_{2g}^2$: $-0.8\Delta_0$: IV.
B. $V^{2+}$, $d^3$: $t_{2g}^3$: $-1.2\Delta_0$: I.
C. $Mn^{3+}$, $d^4$: $t_{2g}^3e_g^1$: $-1.2 + 0.6 = -0.6\Delta_0$: II.
D. $Fe^{3+}$, $d^5$: $t_{2g}^3e_g^2$: $-1.2 + 1.2 = 0$: III.
A-IV, B-I, C-II, D-III.
Answer: (A). | ||||
| 71 | 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 Respiration is part of the natural carbon–oxygen cycle: the $CO_2$ it releases is taken up again by photosynthesis, which returns the $O_2$, so it does not upset the balance.
Burning coal and petroleum adds extra $CO_2$ and uses $O_2$, and deforestation removes the plants that absorb $CO_2$: these disturb the balance.
Answer: (D). | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In the Carius method for sulphur, the sulphur of the compound is oxidised to sulphuric acid (Y), which with $BaCl_2$ gives a precipitate of $BaSO_4$ (Z), weighed to estimate sulphur.
So X must contain sulphur. Methionine ($CH_3SCH_2CH_2CH(NH_2)COOH$) does; a nucleotide contains phosphorus, cytosine only C, H, N, O, and chloroxylenol chlorine.
Answer: (B). | ||||
| 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 Hydride affinity is the tendency of a carbocation to take up $H^-$; it is larger the less stable the cation.
From the figure: A is a tertiary allylic cation ($CH_2=CH-\overset{+}{C}(CH_3)_2$), B is the triphenylmethyl cation, C is the tert-butyl cation and D is the tricyclopropylmethyl cation.
Stability: D (three cyclopropyl groups, very strong σ-conjugation with the bent bonds) > B (charge spread over three rings) > A (tertiary plus allylic resonance) > C (tertiary, hyperconjugation only).
Hydride affinity is the reverse: C > A > B > D.
Answer: (A). | ||||
| 74 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The substrate is $Ar-CH_2CH_2CH_2-CHCl-CH_3$, where the ring carries $OCH_3$ meta and $NO_2$ para to the side chain.
Anhydrous $AlCl_3$ makes the secondary carbocation $Ar-(CH_2)_3-\overset{+}{C}H-CH_3$, which attacks a ring carbon ortho to the chain (intramolecular Friedel–Crafts alkylation), closing a six-membered ring: a tetralin with the $CH_3$ on the new benzylic carbon.
The two ortho positions: one lies between the chain and $OCH_3$ (ortho to $OCH_3$, crowded); the other is para to $OCH_3$. Both are meta to $NO_2$. Attack para to the strongly activating $OCH_3$, at the uncrowded position, is preferred.
Product: 1-methyl-tetralin with $OCH_3$ and $NO_2$ on the far side of the aromatic ring (positions 6 and 7), as drawn in option (A). (B) is the crowded ortho product, (C) an intermolecular dimer and (D) puts the $CH_3$ on a non-benzylic carbon.
Answer: (A). | ||||
| 75 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (A) Hydroboration–oxidation of an alkene gives an alcohol (anti-Markovnikov).
(B) A ketone with RMgBr, then hydrolysis, gives a tertiary alcohol.
(C) Ozonolysis of an alkene cleaves the double bond and gives aldehydes and/or ketones, not alcohols: incorrect method.
(D) An alkyl halide with aqueous NaOH gives an alcohol by substitution.
Answer: (C). | ||||
| 76 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers C is phenol, D is 4-methoxyphenol and E is 4-nitrophenol.
Phenols are more acidic than alcohols, because the phenoxide ion is resonance-stabilised.
Among the phenols, $-NO_2$ (electron-withdrawing, −R and −I) at the para position stabilises the phenoxide and raises acidity; $-OCH_3$ (+R) at the para position destabilises it and lowers acidity: E > C > D.
Among the alcohols, the electron-releasing methyl groups of tert-butanol destabilise its alkoxide (and hinder its solvation), so $CH_3OH$ > $(CH_3)_3COH$: A > B.
Order: E > C > D > A > B.
Answer: (A). | ||||
| 77 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The substrate (figure) is $H_2NCO-CH(CH_3)-C(=CH_2)-CO-CH_2CH_3$: an amide, a C=C and a ketone.
$LiAlH_4$ then $H_3O^+$: the amide $-CONH_2$ is reduced to $-CH_2NH_2$ (not to a carbinolamine), the ketone $C=O$ to $CH-OH$, and the isolated $C=C$ is left untouched (hydride adds to the carbonyl carbon, 1,2-reduction).
X $= H_2N-CH_2-CH(CH_3)-C(=CH_2)-CH(OH)-CH_2CH_3$, as drawn in option (B).
(A) and (D) wrongly keep an OH on the carbon next to N; (C) and (D) wrongly reduce the C=C.
Answer: (B). | ||||
| 78 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Buna-S is the copolymer of 1,3-butadiene ($CH_2=CH-CH=CH_2$) and styrene ($C_6H_5CH=CH_2$).
In the chain, butadiene enters by 1,4-addition as $-CH_2-CH=CH-CH_2-$ and styrene as $-CH(C_6H_5)-CH_2-$.
Repeat unit: $-[CH_2-CH=CH-CH_2-CH(C_6H_5)-CH_2]_n-$.
Answer: (D). | ||||
| 79 | Chemistry in Everyday LifeCleansing agents | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on cleansing agents, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Soap is made by saponification: a fat (triglyceride, an ester) is heated with aqueous NaOH, which hydrolyses it to glycerol and the sodium salts of the fatty acids (soap). This is alkaline hydrolysis of an ester.
Answer: (B). | ||||
| 80 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Assertion and reasonEasy | |
A easy assertion and reason question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Hydrated oxalic acid is $H_2C_2O_4 \cdot 2H_2O$: molar mass $= 90 + 36 = 126$ g mol$^{-1}$, so R is true.
Moles $= \frac{3.1500}{126} = 0.025$ mol in 0.2500 L: molarity $= \frac{0.025}{0.25} = 0.1$ M, so A is true.
The calculation in A rests on the molar mass in R, so R explains A.
Answer: (A). | ||||
| 81 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom For a hydrogen-like ion, $\Delta E = R_H Z^2\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$ with $Z = 3$ for $Li^{2+}$.
$1.47 \times 10^{-17} = 2.18 \times 10^{-18} \times 9 \times \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$, so $\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{1.47 \times 10^{-17}}{1.962 \times 10^{-17}} \approx 0.75 = \frac{3}{4}$.
$n = 1$: $1 - \frac{1}{4} = \frac{3}{4}$. ✓
Answer: 1. | ||||
| 82 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Counting all lone pairs in each Lewis structure:
$H_2O$: 2 on O, total 2 ✓.
$N_2$ ($:N\equiv N:$): one on each N, total 2 ✓.
$CO$ ($:C\equiv O:$): one on C, one on O, total 2 ✓.
$XeF_4$: 2 on Xe + 3 on each F = 14.
$NH_3$: 1.
$NO$: 1 on N (plus the odd electron) and 2 on O = 3.
$CO_2$: 2 on each O = 4.
$F_2$: 3 on each F = 6.
Molecules with exactly two lone pairs: $H_2O$, $N_2$, CO, i.e. 3.
Answer: 3. | ||||
| 83 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics A and B break bonds (atomisation, bond dissociation): endothermic.
C is vaporisation: endothermic.
D is combustion of carbon: exothermic.
E: dissolving $NH_4Cl$ in water absorbs heat (the solution cools): endothermic.
Endothermic: A, B, C, E, i.e. 4.
Answer: 4. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\pi V = \frac{w}{M}RT$, so $M = \frac{wRT}{\pi V}$.
$\pi = 1.29$ mbar $= 1.29 \times 10^{-3}$ bar, $V = 0.300$ L.
$M = \frac{0.63 \times 0.083 \times 300}{1.29 \times 10^{-3} \times 0.300} = \frac{15.687}{3.87 \times 10^{-4}} \approx 40534.9$ g mol$^{-1}$.
Nearest integer: 40535.
Answer: 40535. | ||||
| 85 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics At constant T and V, partial pressures are proportional to amounts. If A falls by $p$: A $= 450 - p$, B $= 2p$, C $= p$.
Total: $450 - p + 2p + p = 450 + 2p = 720$, so $p = 135$ mm Hg.
Fraction decomposed $= \frac{135}{450} = 0.3 = 3 \times 10^{-1}$.
Answer: 3. | ||||
| 86 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry $\Lambda_m = \frac{1000\kappa}{C} = \frac{1000 \times 5 \times 10^{-5}}{0.0025} = 20$ S cm$^2$ mol$^{-1}$.
$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{20}{400} = 0.05$.
$K_a = \frac{C\alpha^2}{1 - \alpha} = \frac{0.0025 \times 0.0025}{0.95} = 6.58 \times 10^{-6} = 65.8 \times 10^{-7}$.
Nearest integer: 66.
Answer: 66. | ||||
| 87 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics A: for zero order, $t_{1/2} = \frac{[A]_0}{2k}$; as the concentration falls each successive half-life is shorter. Correct.
B: a reactant whose order is zero does not affect the rate. Correct.
C: order can be fractional, but molecularity is always a whole number. Incorrect.
D: zero order $k$: mol L$^{-1}$ s$^{-1}$; second order $k$: mol$^{-1}$ L s$^{-1}$. Correct.
Incorrect statements: only C, i.e. 1.
Answer: 1. | ||||
| 88 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Complete hydrolysis (disproportionation): $6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$.
The oxidised product is $XeO_3$, Xe in +6; in $XeF_4$ Xe is +4.
Difference $= 6 - 4 = 2$.
Answer: 2. | ||||
| 89 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Spin-only moment $\mu = \sqrt{n(n + 2)}$ BM. $\sqrt{4 \times 6} = \sqrt{24} = 4.90$, so $n = 4$.
Answer: 4. | ||||
| 90 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In neutral or faintly alkaline medium permanganate is reduced to manganese dioxide:
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$; Mn goes from +7 to +4, a gain of 3 electrons.
Answer: 3. | ||||
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