JEE (Main) · 2024
JEE (Main) 2024 — 30 January, 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
- Limits, Continuity and Differentiability3
- Integral Calculus3
- Sets, Relations and Functions2
15 easy · 14 medium · 1 hard
Physics
30 Q
- Current Electricity3
- Electromagnetic Induction and Alternating Currents3
- Motion of System of Particles and Rigid Body2
22 easy · 7 medium · 1 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Aldehydes, Ketones and Carboxylic Acids3
- Chemical Bonding and Molecular Structure2
21 easy · 9 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $\cos^{-1}$ needs $-1 \le \frac{2 - |x|}{4} \le 1$, i.e. $-4 \le 2 - |x| \le 4$.
The right inequality gives $|x| \ge -2$ (always true); the left gives $|x| \le 6$, so $-6 \le x \le 6$.
$\{\log_e(3 - x)\}^{-1}$ needs $3 - x > 0$, i.e. $x < 3$, and $\log_e(3 - x) \ne 0$, i.e. $3 - x \ne 1$, $x \ne 2$.
Domain $= [-6, 3) - \{2\}$, so $\alpha = 6$, $\beta = 3$, $\gamma = 2$.
$\alpha + \beta + \gamma = 11$.
Answer: (C). | ||||
| 2 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations From $z^2 = -i\bar{z}$, take moduli: $|z|^2 = |\bar{z}| = |z|$. Since $z \ne 0$ ($xy \ne 0$), $|z| = 1$, so $|z^2| = 1$.
Check that such a $z$ exists: $(x + iy)^2 + i(x - iy) = (x^2 - y^2 + y) + i(2xy + x) = 0$.
Imaginary part: $x(2y + 1) = 0$ with $x \ne 0$ gives $y = -\frac{1}{2}$. Real part: $x^2 = y^2 - y = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.
Then $|z|^2 = \frac{3}{4} + \frac{1}{4} = 1$.
Answer: (B). | ||||
| 3 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Coefficient determinant:
$\Delta = \begin{vmatrix}1&1&1\\1&2&2\lambda\\1&3&4\lambda^2\end{vmatrix} = \begin{vmatrix}1&1&1\\0&1&2\lambda - 1\\0&2&4\lambda^2 - 1\end{vmatrix} = (4\lambda^2 - 1) - 2(2\lambda - 1) = (2\lambda - 1)^2$.
So for $\lambda \ne \frac{1}{2}$ the system has a unique solution for every $\mu$: (A) and (C) are true.
For $\lambda = \frac{1}{2}$ the equations are $x + y + z = 4\mu$, $x + 2y + z = 10\mu$, $x + 3y + z = \mu^2 + 15$. With $w = x + z$: the first two give $y = 6\mu$, $w = -2\mu$, and the third then needs $w + 3y = 16\mu = \mu^2 + 15$, i.e. $\mu^2 - 16\mu + 15 = 0$, $\mu = 1$ or $15$.
So for $\lambda = \frac{1}{2}$: infinitely many solutions when $\mu = 1$ or $\mu = 15$, none otherwise. (D) is true.
(B) says inconsistent whenever $\mu \ne 1$, but $\mu = 15$ gives infinitely many solutions. (B) is not correct.
Answer: (B). | ||||
| 4 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Write $c = \cos^4 x$, $s = \sin^4 x$, $S = \sin^2 2x$.
$R_2 \to R_2 - R_1$ gives $(3, 0, -3)$; $R_3 \to R_3 - R_1$ gives $(0, 3, -3)$.
$f(x) = \begin{vmatrix} 2c & 2s & 3 + S \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{vmatrix} = 2c(0 + 9) - 2s(-9 - 0) + (3 + S)(9) = 9(2c + 2s + 3 + S)$.
Now $2(\cos^4 x + \sin^4 x) = 2(1 - 2\sin^2 x\cos^2 x) = 2 - \sin^2 2x = 2 - S$.
So $f(x) = 9(2 - S + 3 + S) = 45$, a constant.
Hence $f'(x) = 0$ for all $x$, and $\frac{1}{5}f'(0) = 0$.
Answer: (A). | ||||
| 5 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are $11 \times 11 = 121$ equally likely ordered pairs.
$|x - y| > 5$ means $|x - y| \in \{6, 7, 8, 9, 10\}$.
Pairs with $x - y = d$ ($d \ge 1$): $11 - d$ of them. So $x - y = 6, 7, 8, 9, 10$ give $5 + 4 + 3 + 2 + 1 = 15$ pairs, and $y - x$ the same again: 30 pairs.
Probability $= \frac{30}{121}$.
Answer: (B). | ||||
| 6 | 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 $S_{20} = 10(2a + 19d) = 790 \Rightarrow 2a + 19d = 79$.
$S_{10} = 5(2a + 9d) = 145 \Rightarrow 2a + 9d = 29$.
Subtracting: $10d = 50$, $d = 5$, and $2a = 29 - 45 = -16$, $a = -8$.
$S_{15} = \frac{15}{2}(2a + 14d) = \frac{15}{2}(-16 + 70) = 405$.
$S_5 = \frac{5}{2}(2a + 4d) = \frac{5}{2}(-16 + 20) = 10$.
$S_{15} - S_5 = 395$.
Answer: (B). | ||||
| 7 | 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 Take $x > 0$. The base joins $(-x, y)$ and $(x, y)$, length $2x$; the height is $y$ (the distance of the origin from the line at height $y$).
Area $A(x) = \frac{1}{2}(2x)(y) = x(54 - 2x^2) = 54x - 2x^3$, with $0 < x < \sqrt{27}$ so that $y > 0$.
$A'(x) = 54 - 6x^2 = 0 \Rightarrow x = 3$; $A''(x) = -12x < 0$, a maximum.
$A(3) = 3(54 - 18) = 108$.
Answer: (B). | ||||
| 8 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Write the limit as $\lim\limits_{x \to 0} \frac{x^2}{e^{x^2} - 1}\cdot\frac{\int_0^x f(t)\,dt}{x}$.
The first factor tends to 1. For the second, L'Hospital's rule (or the fundamental theorem) gives $\lim\limits_{x \to 0} \frac{\int_0^x f(t)\,dt}{x} = f(0) = \frac{1}{2}$.
So $\alpha = \frac{1}{2}$ and $8\alpha^2 = 8\cdot\frac{1}{4} = 2$.
Answer: (B). | ||||
| 9 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f'(x) = \frac{1}{2}[g'(x) - g'(2 - x)]$.
$f'(1) = \frac{1}{2}[g'(1) - g'(1)] = 0$.
$f'\left(\frac{1}{2}\right) = \frac{1}{2}\left[g'\left(\frac{1}{2}\right) - g'\left(\frac{3}{2}\right)\right] = 0$ and $f'\left(\frac{3}{2}\right) = \frac{1}{2}\left[g'\left(\frac{3}{2}\right) - g'\left(\frac{1}{2}\right)\right] = 0$.
So $f'$ vanishes at $\frac{1}{2}$, $1$ and $\frac{3}{2}$. By Rolle's theorem applied to $f'$ on $\left[\frac{1}{2}, 1\right]$ and $\left[1, \frac{3}{2}\right]$, $f''$ vanishes at least once in each of $\left(\frac{1}{2}, 1\right)$ and $\left(1, \frac{3}{2}\right)$: at least two zeros in $(0, 2)$. (A) is true.
(B) and (C): $f''$ has at least one zero in $(0, 1)$, so (C) is false, and nothing forces exactly one, so (B) need not hold.
(D): $f'\left(\frac{3}{2}\right) + f'\left(\frac{1}{2}\right) = 0$, not 1.
Answer: (A). | ||||
| 10 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Divide numerator and denominator by $n^4$: the sum is $\frac{1}{n}\sum\limits_{k=1}^{n} \frac{1}{\left(1 + \frac{k^2}{n^2}\right)\left(1 + \frac{3k^2}{n^2}\right)}$, which tends to $\int_0^1 \frac{dx}{(1 + x^2)(1 + 3x^2)}$.
Partial fractions in $t = x^2$: $\frac{1}{(1 + t)(1 + 3t)} = \frac{-\frac{1}{2}}{1 + t} + \frac{\frac{3}{2}}{1 + 3t}$.
$\int_0^1 \frac{dx}{1 + x^2} = \frac{\pi}{4}$ and $\int_0^1 \frac{dx}{1 + 3x^2} = \frac{1}{\sqrt{3}}\tan^{-1}\sqrt{3} = \frac{\pi}{3\sqrt{3}}$.
Limit $= -\frac{\pi}{8} + \frac{3}{2}\cdot\frac{\pi}{3\sqrt{3}} = -\frac{\pi}{8} + \frac{\pi}{2\sqrt{3}} = \frac{\pi(4\sqrt{3} - 3)}{24}$.
Option (D): $\frac{13\pi}{8(4\sqrt{3} + 3)} = \frac{13\pi(4\sqrt{3} - 3)}{8(48 - 9)} = \frac{13\pi(4\sqrt{3} - 3)}{312} = \frac{\pi(4\sqrt{3} - 3)}{24}$. It matches.
Answer: (D). | ||||
| 11 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Work with $x$ as a function of $y$: parabola $x = \frac{y^2}{4} + 2$, line $x = \frac{y + 8}{2}$.
Intersection: $\frac{y^2}{4} + 2 = \frac{y + 8}{2} \Rightarrow y^2 - 2y - 8 = 0 \Rightarrow y = -2, 4$.
Between them the line lies to the right, so
Area $= \int_{-2}^{4}\left(\frac{y + 8}{2} - \frac{y^2}{4} - 2\right)dy = \int_{-2}^{4}\left(\frac{y}{2} + 2 - \frac{y^2}{4}\right)dy = \left[\frac{y^2}{4} + 2y - \frac{y^3}{12}\right]_{-2}^{4}$
$= \left(4 + 8 - \frac{16}{3}\right) - \left(1 - 4 + \frac{2}{3}\right) = \frac{20}{3} + \frac{7}{3} = 9$.
Answer: (D). | ||||
| 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 Multiply by $\cos x$: $dy = -\{2(1 - x)\sin x + x(2 - x)\cos x\}\,dx$.
Note $\frac{d}{dx}[x(2 - x)\sin x] = (2 - 2x)\sin x + x(2 - x)\cos x$, exactly the bracket.
So $y = -x(2 - x)\sin x + C$. $y(0) = 2$ gives $C = 2$.
$y(2) = -2(0)\sin 2 + 2 = 2$.
Answer: (B). | ||||
| 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 Turning clockwise by $15^\circ$ lowers the inclination from $30^\circ$ to $15^\circ$.
Slope $= \tan 15^\circ = 2 - \sqrt{3}$.
Line through $(9, 0)$: $y = (2 - \sqrt{3})(x - 9)$, i.e. $\frac{y}{2 - \sqrt{3}} = x - 9$.
So $x - \frac{y}{2 - \sqrt{3}} = 9$, which is $\frac{y}{\sqrt{3} - 2} + x = 9$.
Answer: (A). | ||||
| 14 | CirclesRadius and centre | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections First circle: centre $(-1, -2)$, radius $r$. Second: $(x - 2)^2 + (y - 2)^2 = 4$, centre $(2, 2)$, radius 2.
Distance between centres $d = \sqrt{3^2 + 4^2} = 5$.
Two distinct common points need $|r - 2| < d < r + 2$:
$r + 2 > 5 \Rightarrow r > 3$; $|r - 2| < 5 \Rightarrow r < 7$.
So $3 < r < 7$.
Answer: (C). | ||||
| 15 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Minor axis $= 2b$, distance between foci $= 2ae$. Given $2b = \frac{1}{2}(2ae)$, so $b = \frac{ae}{2}$.
With $b^2 = a^2(1 - e^2)$: $\frac{a^2e^2}{4} = a^2(1 - e^2) \Rightarrow \frac{5e^2}{4} = 1 \Rightarrow e^2 = \frac{4}{5}$.
$e = \frac{2}{\sqrt{5}}$.
Answer: (A). | ||||
| 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 A point of the line: $(5t - 3, 2t + 1, 3t - 4)$.
Vector from $P(1, 2, 3)$: $(5t - 4, 2t - 1, 3t - 7)$. It is perpendicular to $(5, 2, 3)$:
$5(5t - 4) + 2(2t - 1) + 3(3t - 7) = 38t - 43 = 0 \Rightarrow t = \frac{43}{38}$.
$\alpha + \beta + \gamma = (5t - 3) + (2t + 1) + (3t - 4) = 10t - 6 = \frac{430}{38} - 6$.
$19(\alpha + \beta + \gamma) = 215 - 114 = 101$.
Answer: (C). | ||||
| 17 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Area of a parallelogram $= \frac{1}{2}|\vec{AC}\times\vec{BD}|$.
$\vec{AC} = (-5, 1, -7)$, $\vec{BD} = (1, 2, 3)$.
$\vec{AC}\times\vec{BD} = (1\cdot 3 - (-7)\cdot 2,\ (-7)\cdot 1 - (-5)\cdot 3,\ (-5)\cdot 2 - 1\cdot 1) = (17, 8, -11)$.
$|\vec{AC}\times\vec{BD}|^2 = 289 + 64 + 121 = 474$.
Area $= \frac{1}{2}\sqrt{474}$.
Answer: (A). | ||||
| 18 | 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 $\vec{b}\cdot\vec{c} = 2\,\vec{b}\cdot(\vec{a}\times\vec{b}) - 3|\vec{b}|^2 = 0 - 48 = -48$.
$|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 16 - 4 = 12$.
Since $\vec{a}\times\vec{b} \perp \vec{b}$: $|\vec{c}|^2 = 4(12) + 9(16) = 192$, $|\vec{c}| = 8\sqrt{3}$.
$\cos\theta = \frac{-48}{4\cdot 8\sqrt{3}} = -\frac{3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$.
$\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.
Answer: (A). | ||||
| 19 | StatisticsMean, median and mode | Class 11: Statistics | Single correctEasy | |
A easy single correct question on mean, median and mode, from Statistics. NCERT: Class 11: Statistics $N = 3 + 9 + 10 + 8 + 6 = 36$, $\frac{N}{2} = 18$.
Cumulative frequencies: 3, 12, 22, 30, 36. The 18th value falls in the class 8 – 12.
$M = l + \frac{\frac{N}{2} - cf}{f}\times h = 8 + \frac{18 - 12}{10}\times 4 = 8 + 2.4 = 10.4$.
$20M = 208$.
Answer: (C). | ||||
| 20 | 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 $\sin 2x\cos x = 2\sin x\cos^2 x = 2\sin x - 2\sin^3 x$.
The equation becomes $2\sin^3 x + 2\sin x - 2\sin^3 x + 4\sin x - 4 = 0$, i.e. $6\sin x = 4$, $\sin x = \frac{2}{3}$.
Let $\theta = \sin^{-1}\frac{2}{3} \in \left(0, \frac{\pi}{2}\right)$. Solutions in $[0, \infty)$: $\theta,\ \pi - \theta,\ 2\pi + \theta,\ 3\pi - \theta, \ldots$
In $\left[0, \frac{5\pi}{2}\right]$ there are exactly three ($\theta$, $\pi - \theta$, $2\pi + \theta$); in $[0, 2\pi]$ only two, and in $[0, 3\pi]$ four. So $n = 5$.
$x^2 + 5x + 2 = 0$ has roots $\frac{-5 \pm \sqrt{17}}{2}$; both are negative ($\sqrt{17} < 5$), so they lie in $(-\infty, 0)$.
They are not integers, $\frac{-5 - \sqrt{17}}{2} \approx -4.56$ lies outside $\left(-\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2}\right)$, and neither is positive.
Answer: (C). | ||||
| 21 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations For each $a \in A$, $f(a)$ can be any subset of A that contains $a$: choose freely which of the other 6 elements to include, $2^6$ ways.
The 7 choices are independent, so the number of functions is $(2^6)^7 = 2^{42}$.
Writing $2^{42} = m^n$ with $m$ least ($m = 1$ is impossible): $m = 2$, $n = 42$.
$m + n = 44$.
Answer: 44. | ||||
| 22 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha + \beta = 70$, $\lambda = \alpha\beta$, with $\lambda$ divisible neither by 2 nor by 3.
So both roots are odd and neither is a multiple of 3. For a fixed sum, the product is smallest when the roots are farthest apart, so take the smallest allowed $\alpha$:
$\alpha = 1$: $\beta = 69 = 3\times 23$, not allowed. $\alpha = 3$: multiple of 3. $\alpha = 5$: $\beta = 65$, both fine.
So $\lambda = 5\times 65 = 325$, with $\{\alpha, \beta\} = \{5, 65\}$.
Value $= \frac{(\sqrt{4} + \sqrt{64})(325 + 35)}{60} = \frac{10\times 360}{60} = 60$.
Answer: 60. | ||||
| 23 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem General term: $T_{r+1} = \binom{824}{r}\,7^{\frac{824 - r}{2}}\,11^{\frac{r}{6}}$.
It is an integer when $\frac{r}{6}$ is an integer and $\frac{824 - r}{2}$ is an integer, i.e. $r$ is a multiple of 6 (then $r$ is even and so is $824 - r$).
$r = 0, 6, 12, \ldots, 822$ (since $822 = 6\times 137$ and $828 > 824$): $137 + 1 = 138$ values.
Answer: 138. | ||||
| 24 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The bases $1, 4, 8, 13, 19, 26, \ldots$ have differences $3, 4, 5, 6, 7, \ldots$, so
$T_n = 1 + \sum\limits_{k=1}^{n-1}(k + 2) = 1 + \frac{(n - 1)n}{2} + 2(n - 1) = \frac{n^2 + 3n - 2}{2}$ (check: $T_1 = 1$, $T_2 = 4$, $T_3 = 8$).
$4\alpha = \sum\limits_{n=1}^{10}(n^2 + 3n - 2)^2 = \sum(n^4 + 6n^3 + 5n^2 - 12n + 4)$.
$4\alpha - \beta = 6\sum n^3 + 5\sum n^2 - 12\sum n + 40$ with $\sum n^3 = 3025$, $\sum n^2 = 385$, $\sum n = 55$:
$= 18150 + 1925 - 660 + 40 = 19455$.
$55k + 40 = 19455 \Rightarrow k = \frac{19415}{55} = 353$.
Answer: 353. | ||||
| 25 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $h(x) = \frac{10x}{x + 1}$ increases from 0 at $x = 0$ to 9 at $x = 9$, so $\sqrt{h}$ runs from 0 to 3.
$\sqrt{h} < 1 \iff 10x < x + 1 \iff x < \frac{1}{9}$.
$\sqrt{h} < 2 \iff 10x < 4x + 4 \iff x < \frac{2}{3}$.
$\sqrt{h} < 3 \iff 10x < 9x + 9 \iff x < 9$.
So the integrand is 0 on $\left[0, \frac{1}{9}\right)$, 1 on $\left[\frac{1}{9}, \frac{2}{3}\right)$, and 2 on $\left[\frac{2}{3}, 9\right)$.
$\int_0^9 = 1\cdot\left(\frac{2}{3} - \frac{1}{9}\right) + 2\left(9 - \frac{2}{3}\right) = \frac{5}{9} + \frac{50}{3} = \frac{155}{9}$.
$9\times\frac{155}{9} = 155$.
Answer: 155. | ||||
| 26 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{\sqrt{3}(x^3 + 2)}{\sqrt{1 - x^2}}$.
Integrating factor: $e^{-\int\frac{x}{1 - x^2}dx} = e^{\frac{1}{2}\ln(1 - x^2)} = \sqrt{1 - x^2}$.
$\frac{d}{dx}\left(y\sqrt{1 - x^2}\right) = \sqrt{3}(x^3 + 2) \Rightarrow y\sqrt{1 - x^2} = \sqrt{3}\left(\frac{x^4}{4} + 2x\right) + C$.
$y(0) = 0$ gives $C = 0$.
At $x = \frac{1}{2}$: $y\cdot\frac{\sqrt{3}}{2} = \sqrt{3}\left(\frac{1}{64} + 1\right) = \frac{65\sqrt{3}}{64} \Rightarrow y = \frac{65}{32}$.
$m + n = 65 + 32 = 97$.
Answer: 97. | ||||
| 27 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ends of a latus rectum: $\left(c, \pm\frac{b^2}{a}\right)$ with $a = 3$, $c = \sqrt{9 + b^2}$.
It subtends $\frac{\pi}{3}$ at the centre, so each half makes $\frac{\pi}{6}$ with the axis: $\tan\frac{\pi}{6} = \frac{b^2/3}{c}$, i.e. $c = \frac{b^2}{\sqrt{3}}$.
Squaring: $9 + b^2 = \frac{b^4}{3} \Rightarrow b^4 - 3b^2 - 27 = 0 \Rightarrow b^2 = \frac{3 + \sqrt{9 + 108}}{2} = \frac{3 + 3\sqrt{13}}{2} = \frac{3}{2}(1 + \sqrt{13})$.
$l = 3$, $m = 2$, $n = 13$: $l^2 + m^2 + n^2 = 9 + 4 + 169 = 182$.
Answer: 182. | ||||
| 28 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry First pair. $x + 1 = 2y = -12z = t$: point $(-1, 0, 0)$, direction $\left(1, \frac{1}{2}, -\frac{1}{12}\right) \parallel (12, 6, -1)$.
$x = y + 2 = 6z - 6 = s$: point $(0, -2, 1)$, direction $\left(1, 1, \frac{1}{6}\right) \parallel (6, 6, 1)$.
$(12, 6, -1)\times(6, 6, 1) = (6 + 6, -6 - 12, 72 - 36) = (12, -18, 36) \parallel (2, -3, 6)$, of length 7.
Joining vector $(1, -2, 1)$; $(1, -2, 1)\cdot(2, -3, 6) = 2 + 6 + 6 = 14$. $d_1 = \frac{14}{7} = 2$.
Second pair: points $(1, -8, 4)$, $(1, 2, 6)$; directions $(2, -7, 5)$, $(2, 1, -3)$.
$(2, -7, 5)\times(2, 1, -3) = (21 - 5, 10 + 6, 2 + 14) = (16, 16, 16) \parallel (1, 1, 1)$, of length $\sqrt{3}$.
Joining vector $(0, 10, 2)$; dot product with $(1, 1, 1)$ is 12. $d_2 = \frac{12}{\sqrt{3}} = 4\sqrt{3}$.
$\frac{32\sqrt{3}\cdot 2}{4\sqrt{3}} = 16$.
Answer: 16. | ||||
| 29 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalHard | |
A hard numerical 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 Let $x = n(M\cap P\cap C)$ and write $a = n(M\cap P)$, $b = n(P\cap C)$, $c = n(M\cap C)$.
Inclusion–exclusion: $40 = 20 + 25 + 16 - (a + b + c) + x$, so $a + b + c = 21 + x$.
Each pairwise intersection contains the triple one: $a, b, c \ge x$, and $a \le 11$, $b, c \le 15$.
$x = 11$: then $a = 11$, and $b + c = 21$ with $b, c \ge 11$ is impossible ($22 > 21$). Larger $x$ fails because $x \le a \le 11$.
$x = 10$: take $a = 11$, $b = 10$, $c = 10$ (sum 31 = 21 + 10). Regions: only M $= 20 - 11 - 10 + 10 = 9$, only P $= 25 - 11 - 10 + 10 = 14$, only C $= 16 - 10 - 10 + 10 = 6$, M and P only $= 1$, P and C only $= 0$, M and C only $= 0$, all three $= 10$. Total $9 + 14 + 6 + 1 + 10 = 40$, all non-negative.
Maximum $= 10$.
Answer: 10. | ||||
| 30 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f$ is even, so it is enough to check $x = 2$.
Continuity: $4a + 2b = \frac{1}{2}$.
Derivatives: for $x > 2$, $f'(x) = -\frac{1}{x^2} \to -\frac{1}{4}$; for $x < 2$, $f'(x) = 2ax \to 4a$. So $4a = -\frac{1}{4}$, $a = -\frac{1}{16}$.
Then $2b = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$, $b = \frac{3}{8}$.
$a + b = -\frac{1}{16} + \frac{6}{16} = \frac{5}{16}$, and $48(a + b) = 15$.
Answer: 15. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (A) Viscosity: $F = \eta A\frac{dv}{dx}$, so $\eta = \frac{F}{A\,(dv/dx)} = \frac{\mathrm{MLT^{-2}}}{\mathrm{L^2}\cdot\mathrm{T^{-1}}} = [\mathrm{ML^{-1}T^{-1}}]$ → (III).
(B) Surface tension = force per unit length: $\frac{\mathrm{MLT^{-2}}}{\mathrm{L}} = [\mathrm{ML^0T^{-2}}]$ → (IV).
(C) Angular momentum $L = mvr$: $\mathrm{M}\cdot\mathrm{LT^{-1}}\cdot\mathrm{L} = [\mathrm{ML^2T^{-1}}]$ → (II).
(D) Rotational kinetic energy is an energy: $[\mathrm{ML^2T^{-2}}]$ → (I).
A→III, B→IV, C→II, D→I.
Answer: (B). | ||||
| 32 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion At the highest point the velocity is horizontal: $v = u\cos 30^\circ = \frac{\sqrt{3}}{2}u$.
The maximum height is $h = \frac{u^2\sin^2 30^\circ}{2g} = \frac{u^2}{8g}$.
The position vector from the point of projection has vertical part $h$, and the velocity is horizontal, so the moment arm of the momentum about that point is $h$:
$L = mvh = m\cdot\frac{\sqrt{3}}{2}u\cdot\frac{u^2}{8g} = \frac{\sqrt{3}}{16}\frac{mu^3}{g}$.
Answer: (C). | ||||
| 33 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Speed just before impact: $v_1 = \sqrt{2gh_1} = \sqrt{2\times 9.8\times 10} = 14$ m/s (downward).
Speed just after impact: $v_2 = \sqrt{2gh_2} = \sqrt{2\times 9.8\times 5} = \sqrt{98} \approx 9.90$ m/s (upward).
Impulse = change of momentum $= m(v_2 + v_1) = 0.1\times(9.90 + 14) \approx 2.39$ kg m s$^{-1}$, directed upward.
Answer: (A). | ||||
| 34 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion The string runs from the 2 kg block on the $30^\circ$ incline, over the fixed pulley at the top, down and round the movable pulley carrying the 4 kg block, and up to the ceiling.
Constraint: if the 4 kg block goes down by $x$, both strands beside the movable pulley lengthen by $x$, so the 2 kg block moves up the incline by $2x$. Hence $a_2 = 2a_4$.
Let $T$ be the tension.
2 kg block (up the incline): $T - 2g\sin 30^\circ = 2a_2$, i.e. $T - g = 2a_2$.
4 kg block with the pulley (downward): $4g - 2T = 4a_4 = 2a_2$.
From the first, $T = g + 2a_2$; substituting: $4g - 2g - 4a_2 = 2a_2$, so $6a_2 = 2g$ and $a_2 = \frac{g}{3}$.
Answer: (A). | ||||
| 35 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power A is 1 m above the lowest level and B is 0.5 m above it, so the particle falls through $1 - 0.5 = 0.5$ m from A to B.
It starts essentially from rest ("gently pushed") and the track is frictionless, so mechanical energy is conserved:
$\frac{1}{2}mv^2 = mg(0.5)$, $v = \sqrt{2\times 10\times 0.5} = \sqrt{10}$ m/s.
Answer: (B). | ||||
| 36 | GravitationGravitational potential energy | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation At distance $r$ from the centre: $V = -\frac{GM}{r}$ and $g = \frac{GM}{r^2}$, so $\frac{|V|}{g} = r$.
$r = \frac{5.12\times 10^7}{6.4} = 8\times 10^6$ m $= 8000$ km.
Height above the surface $= 8000 - 6400 = 1600$ km.
Answer: (A). | ||||
| 37 | 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 Young's modulus is a property of the material, not of the wire's dimensions. Changing the length or the cross-section changes how much the wire stretches under a load, but stress/strain stays the same.
So Young's modulus remains Y.
Answer: (C). | ||||
| 38 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctHard | |
A hard single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics On a $\log P$ – $\log V$ graph a process $PV^n = $ constant is a straight line of slope $-n$.
The intended reading: line A has slope of magnitude $\gamma$ ($\tan^{-1}\gamma$), the slope of an adiabatic process ($PV^\gamma = $ constant), and line B has slope of magnitude 1 ($45^\circ$), the slope of an isothermal process ($PV = $ constant).
Adiabatic: no heat is exchanged while the temperature changes, so $C_A = \frac{dQ}{n\,dT} = 0$.
Isothermal: heat is exchanged with no change of temperature, so $C_B = \infty$.
So $C_A = 0$ and $C_B = \infty$, which options (C) and (D) both state, in a different order.
Answer: (C) or (D). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$, so equal r.m.s. speeds need $\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$.
$T_{O_2} = 47 + 273 = 320$ K.
$T_{H_2} = 320\times\frac{2}{32} = 20$ K.
Answer: (A). | ||||
| 40 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance At a distance $r$ large compared with the dipole length, the potential of a dipole of moment $p$ is $V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}$.
So $V \propto \frac{1}{r^2}$ (the field falls as $\frac{1}{r^3}$).
Answer: (B). | ||||
| 41 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Hot resistance: $R = \frac{V}{I} = \frac{220}{2.75} = 80\,\Omega$.
$R = R_{27}(1 + \alpha\,\Delta T)$: $80 = 60(1 + 2\times 10^{-4}\,\Delta T)$, so $\Delta T = \frac{1/3}{2\times 10^{-4}} \approx 1667\,^\circ$C.
Temperature attained $= 27 + 1667 \approx 1694\,^\circ$C.
Answer: (C). | ||||
| 42 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Each loop gives at its centre $B_0 = \frac{\mu_0 I}{2a}$, along its own axis.
The loops are in perpendicular planes with a common centre O, so the two fields are perpendicular:
$B = \sqrt{B_0^2 + B_0^2} = \sqrt{2}\cdot\frac{\mu_0 I}{2a} = \frac{\mu_0 I}{\sqrt{2}\,a}$.
Answer: (A). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Power factor $\cos\phi = \frac{R}{\sqrt{R^2 + X_L^2}}$.
At $\omega = 1000$ rad/s: $\frac{1}{\sqrt{2}}$ means $X_L = \omega L = R$.
At $\omega = 2000$ rad/s the reactance doubles: $X_L = 2R$.
New power factor $= \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{1}{\sqrt{5}}$ (the amplitude of the emf does not matter).
Answer: (C). | ||||
| 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 The phase $(\omega t - kz)$ shows the wave travels along $+z$; $\vec{B}$ has the same phase.
The direction of travel is that of $\vec{E}\times\vec{B}$: with $\vec{E}$ along $\hat{i}$, $\hat{i}\times\hat{j} = \hat{k}$, so $\vec{B}$ is along $+\hat{j}$.
In free space $B_0 = \frac{E_0}{c}$.
$\vec{B} = \frac{E_0}{c}\cos(\omega t - kz)\,\hat{j}$.
Answer: (A). | ||||
| 45 | 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 In single-slit diffraction the minima on the focal plane are at $y_n = \frac{n\lambda f}{a}$. A secondary maximum lies between two successive minima, so its width is $\frac{\lambda f}{a}$.
$\frac{\lambda f}{a} = \frac{400\times 10^{-9}\times 1}{0.2\times 10^{-3}} = 2\times 10^{-3}$ m $= 2$ mm.
Answer: (A). | ||||
| 46 | 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 The threshold wavelength is where the photon energy equals the work function: $\lambda_0 = \frac{hc}{\phi}$.
$\lambda_0 = \frac{1240\ \text{eV nm}}{3.0\ \text{eV}} \approx 413$ nm (with $h = 6.63\times 10^{-34}$ J s: $\frac{6.63\times 10^{-34}\times 3\times 10^8}{3\times 1.6\times 10^{-19}} \approx 414$ nm).
Answer: (B). | ||||
| 47 | 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 In every Bohr orbit, $K = \frac{1}{2}\frac{ke^2}{r}$ and $U = -\frac{ke^2}{r}$, so $K = \frac{|U|}{2}$ whatever the level (the 5th excited state is $n = 6$, but $n$ does not matter).
$\frac{K}{|U|} = \frac{1}{2}$.
Answer: (B). | ||||
| 48 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Without the Zener the load would get $20\times\frac{500}{700} \approx 14.3$ V $> 10$ V, so the Zener breaks down and holds 10 V across itself and the $500\,\Omega$ load.
Current through $200\,\Omega$: $\frac{20 - 10}{200} = 50$ mA.
Load current: $\frac{10}{500} = 20$ mA.
Zener current $= 50 - 20 = 30$ mA.
Answer: (C). | ||||
| 49 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity The 4 V cell drives a single series chain: $3.3\,k\Omega$ and seven $100\,\Omega$ resistors, total $3300 + 700 = 4000\,\Omega$.
Current $I = \frac{4}{4000} = 1$ mA.
The tap is after the $3.3\,k\Omega$ and two $100\,\Omega$ resistors, so $V_0$ is taken across the remaining five $100\,\Omega$ resistors, $500\,\Omega$:
$V_0 = 1\ \text{mA}\times 500\,\Omega = 0.5$ V.
Answer: (C). | ||||
| 50 | Electromagnetic Induction and Alternating CurrentsTransformers | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on transformers, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Secondary voltage: $V_s = 220\times\frac{10}{100} = 22$ V.
This drives $15\,k\Omega$ and $7\,k\Omega$ in series ($22\,k\Omega$), and $V_0$ is taken across the $7\,k\Omega$:
$V_0 = 22\times\frac{7}{22} = 7$ V.
Answer: (B). | ||||
| 51 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Take the acceleration as uniform (the question implies it). Over 1 s the velocity rises by 50 m/s, so $a = 50$ m/s$^2$.
If $v$ is the velocity at time $t$: displacement in that second $= v(1) + \frac{1}{2}(50)(1)^2 = 125$, so $v = 100$ m/s.
The $(t+2)$th second runs from $t+1$ to $t+2$. At $t+1$ the velocity is $150$ m/s.
Distance in it $= 150\times 1 + \frac{1}{2}\times 50\times 1^2 = 175$ m.
Check: velocity at $t+2$ is 200 m/s; the average over the second is $\frac{150 + 200}{2} = 175$ m/s, i.e. 175 m.
Answer: 175. | ||||
| 52 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Moment of inertia of each disc: $I = \frac{1}{2}MR^2 = \frac{1}{2}\times 5\times 2^2 = 10$ kg m$^2$.
Initial kinetic energy: $\frac{1}{2}I\omega^2 = \frac{1}{2}\times 10\times 10^2 = 500$ J.
No external torque about the axis, so angular momentum is conserved: $I\omega = 2I\omega'$, $\omega' = 5$ rad/s.
Final kinetic energy: $\frac{1}{2}(2I)\omega'^2 = \frac{1}{2}\times 20\times 25 = 250$ J.
Energy dissipated $= 500 - 250 = 250$ J.
Check: for two equal bodies brought to a common speed half the energy is lost, $\frac{1}{2}\times 500 = 250$ J.
Answer: 250. | ||||
| 53 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical 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 P and Q lie on the smooth table, P joined to Q by wire A, Q joined by wire B over the pulley to the hanging block R.
Acceleration of the system: $a = \frac{m_R g}{m_P + m_Q + m_R} = \frac{30}{9} = \frac{10}{3}$ m/s$^2$.
Wire B pulls P and Q together: $T_B = (3 + 3)\times\frac{10}{3} = 20$ N.
(Check from R: $T_B = 3(g - a) = 3(10 - \frac{10}{3}) = 20$ N.)
Area $= 0.005$ cm$^2 = 5\times 10^{-7}$ m$^2$; stress $= \frac{20}{5\times 10^{-7}} = 4\times 10^7$ N m$^{-2}$.
Strain $= \frac{4\times 10^7}{2\times 10^{11}} = 2\times 10^{-4}$.
Answer: 2. | ||||
| 54 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Closed pipe, fundamental: $f = \frac{v}{4L}$, so $L = \frac{v}{4f}$.
Before: $L_1 = \frac{330}{4\times 30} = 2.75$ m. After: $L_2 = \frac{330}{4\times 110} = 0.75$ m.
The water fills $2.75 - 0.75 = 2$ m $= 200$ cm of the pipe.
Volume $= 200\times 2 = 400$ cm$^3$; with water's density 1 g/cm$^3$ the mass is 400 g.
Answer: 400. | ||||
| 55 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $E = \frac{1}{2}CV^2$. Join the capacitors in parallel, positive plate to positive plate.
Initial energy: $\frac{1}{2}CV^2 + \frac{1}{2}(2C)(2V)^2 = E + 8E = 9E$.
Common potential: $V' = \frac{CV + 2C\cdot 2V}{3C} = \frac{5V}{3}$.
Final energy: $\frac{1}{2}(3C)\left(\frac{5V}{3}\right)^2 = \frac{25}{6}CV^2 = \frac{25}{3}E$.
Loss $= 9E - \frac{25}{3}E = \frac{2}{3}E$.
Check with the formula $\Delta U = \frac{1}{2}\frac{C_1C_2}{C_1 + C_2}(V_1 - V_2)^2 = \frac{1}{2}\cdot\frac{2C}{3}\cdot V^2 = \frac{2}{3}E$.
$x = 2$.
Answer: 2. | ||||
| 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 In opposition the net emf is $8 - 2 = 6$ V and the total resistance is $2 + 4 = 6\,\Omega$ (no external resistance), so $I = 1$ A.
The current is driven by $E_1$, so it enters $E_2$ at its positive terminal: $E_2$ is being charged.
Terminal p.d. of $E_2$: $V_2 = E_2 + Ir_2 = 2 + 1\times 4 = 6$ V.
Check from $E_1$: its terminal p.d. is $8 - 1\times 2 = 6$ V, and the two cells share the same two terminals, as they must.
Answer: 6. | ||||
| 57 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The horizontal component points from south to north. A current from south-east to north-west makes $45^\circ$ with that direction.
$\frac{F}{L} = IB_H\sin 45^\circ = \sqrt{2}\times 3.5\times 10^{-5}\times\frac{1}{\sqrt{2}} = 3.5\times 10^{-5}$ N/m $= 35\times 10^{-6}$ N/m.
Answer: 35. | ||||
| 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 The blades turn in a horizontal plane, so only the vertical component of the earth's field cuts them:
$B_V = B\sin 30^\circ = 0.5\times\frac{1}{2} = 0.25$ G $= 2.5\times 10^{-5}$ T.
$\omega = 1200$ rpm $= \frac{1200\times 2\pi}{60} = 40\pi$ rad/s.
Emf between the centre and the tip of a rotating rod: $\varepsilon = \frac{1}{2}B_V\omega L^2 = \frac{1}{2}\times 2.5\times 10^{-5}\times 40\pi\times 0.8^2 = 32\pi\times 10^{-5}$ V.
The three blades are in parallel between hub and tips, so the emf is the same for one blade.
$N = 32$.
Answer: 32. | ||||
| 59 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a real image with magnification 2 the image distance is twice the object distance: $v = 2u$ (magnitudes), and object and image are on opposite sides, so $u + v = 45$ cm.
$3u = 45$: $u = 15$ cm, $v = 30$ cm.
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{30} + \frac{1}{15} = \frac{3}{30}$, $f = 10$ cm.
Check: $m = -\frac{v}{u}$ in sign convention, $u = -15$, $v = +30$: $\frac{1}{30} - \frac{1}{-15} = \frac{1}{10}$.
Answer: 10. | ||||
| 60 | 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 $E_n = -\frac{13.6}{n^2}$ eV: $n^2 = \frac{13.6}{0.85} = 16$, so $n = 4$.
From $n = 4$ the electron can reach the lower levels by the transitions 4→3, 4→2, 4→1, 3→2, 3→1, 2→1: $\binom{4}{2} = \frac{4\times 3}{2} = 6$.
Answer: 6. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | StatementsEasy | |
A easy statements question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Statement I is the definition of degenerate orbitals: true.
In the hydrogen atom (one electron) the energy depends only on the principal quantum number $n$, so 3s, 3p and 3d all have the same energy and are degenerate. (They split only in many-electron atoms, where energy depends on $n + l$.) Statement II is false.
Answer: (C). | ||||
| 62 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Count bond pairs and lone pairs on the central atom:
(A) $\mathrm{BrF_5}$: 5 bp + 1 lp (octahedral arrangement) → square pyramidal (IV).
(B) $\mathrm{H_2O}$: 2 bp + 2 lp (tetrahedral arrangement) → bent (III).
(C) $\mathrm{ClF_3}$: 3 bp + 2 lp (trigonal bipyramidal arrangement, lone pairs equatorial) → T-shape (I).
(D) $\mathrm{SF_4}$: 4 bp + 1 lp (trigonal bipyramidal arrangement) → see-saw (II).
A→IV, B→III, C→I, D→II.
Answer: (C). | ||||
| 63 | SolutionsColligative properties | Class 12: Solutions | Single correctEasy | |
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions Naphthalene is a non-volatile solute dissolved in the solvent benzene. A solute lowers the solvent's vapour pressure and hence its freezing point: $\Delta T_f = K_f\,m > 0$.
So the freezing point of benzene decreases.
Answer: (B). | ||||
| 64 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter In acidified aqueous solution aluminium chloride gives the hydrated ion $[\mathrm{Al(H_2O)_6}]^{3+}$: six water molecules round $\mathrm{Al^{3+}}$, using $sp^3d^2$ orbitals.
Six ligands → octahedral geometry.
Answer: (B). | ||||
| 65 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Assertion and reasonMedium | |
A medium assertion and reason question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) is true: from N to P the radius rises considerably (a new shell, no inner d electrons), but from As to Bi the rise is small because the filled 3d (and, for Bi, 4f) orbitals shield the nucleus poorly.
(R) is true as a general trend: radii increase down the group.
But (R) does not explain (A): (A) is about why the increase is small lower down the group, which comes from the poor shielding by d and f electrons, not from the general trend.
Answer: (B). | ||||
| 66 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Matching listEasy | |
A easy matching list question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Remove electrons from 4s first:
(A) Cr: $[\mathrm{Ar}]3d^5 4s^1$ → $\mathrm{Cr^{2+}}$: $3d^4$ (III).
(B) Mn: $[\mathrm{Ar}]3d^5 4s^2$ → $\mathrm{Mn^{+}}$: $3d^5 4s^1$ (IV).
(C) Ni: $[\mathrm{Ar}]3d^8 4s^2$ → $\mathrm{Ni^{2+}}$: $3d^8$ (I).
(D) V: $[\mathrm{Ar}]3d^3 4s^2$ → $\mathrm{V^{+}}$: $3d^3 4s^1$ (II).
A→III, B→IV, C→I, D→II.
Answer: (D). | ||||
| 67 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements An ion is diamagnetic when it has no unpaired electrons, i.e. $4f^0$ or $4f^{14}$.
$\mathrm{La^{3+}}$: $4f^0$; $\mathrm{Ce^{4+}}$: $4f^0$; $\mathrm{Lu^{3+}}$: $4f^{14}$ — diamagnetic.
$\mathrm{Nd^{3+}}$: $4f^3$; $\mathrm{Eu^{3+}}$: $4f^6$ — unpaired electrons, paramagnetic.
Only (B) has two diamagnetic ions.
Answer: (B). | ||||
| 68 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | StatementsMedium | |
A medium statements question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) True: ethane-1,2-diamine is a didentate ligand binding through both N atoms, forming a five-membered chelate ring.
(B) True: aluminium is extracted by electrolysis of alumina dissolved in molten cryolite.
(C) True: silver (like gold) is leached as $[\mathrm{Ag(CN)_2}]^-$ with cyanide.
(D) True as worded: Wilkinson's catalyst is $[\mathrm{RhCl(PPh_3)_3}]$, whose ligands are triphenylphosphine.
(E) False: the stability constants of the Ca and Mg EDTA complexes differ, and that difference is what lets the two ions be estimated separately in water hardness titrations.
Every option except (C) contains the false statement (E), so the answer is (A), (B), (C) only.
Answer: (C). | ||||
| 69 | 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 If the compound also contains N or S, the sodium fusion extract holds NaCN and $\mathrm{Na_2S}$. With silver nitrate these would give AgCN (white) and $\mathrm{Ag_2S}$ (black) and interfere with the silver halide test.
Boiling with dilute nitric acid decomposes them, driving off HCN and $\mathrm{H_2S}$:
$\mathrm{NaCN + HNO_3 \rightarrow NaNO_3 + HCN\uparrow}$, $\mathrm{Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S\uparrow}$.
Answer: (B). | ||||
| 70 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Pent-2-enal: a five-carbon chain with CHO as C-1 and a double bond between C-2 and C-3. 4-Methyl: a methyl group on C-4.
C-1 CHO, C-2=C-3, C-4 bearing $\mathrm{CH_3}$, C-5 $\mathrm{CH_3}$: $\mathrm{CH_3-CH(CH_3)-CH=CH-CHO}$.
That is option (B). The others are 3-methylpent-2-enal (A), 2-methylpent-2-enal (C) and 3-methylpent-4-enal (D).
Answer: (B). | ||||
| 71 | 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 Apply Hückel's rule (a cyclic, planar, fully conjugated system with $(4n + 2)\,\pi$ electrons is aromatic and specially stabilised; $4n$ is antiaromatic and destabilised):
(A) Cyclohexa-1,3-diene: not fully conjugated round the ring (two $sp^3$ carbons) — non-aromatic.
(B) Cyclopentadienyl cation: 4 $\pi$ electrons — antiaromatic.
(C) Drawn with a lone pair on the third carbon: the cyclopropenyl anion, 4 $\pi$ electrons — antiaromatic.
(D) Cyclopropenyl cation: 2 $\pi$ electrons ($n = 0$) — aromatic.
The aromatic cyclopropenyl cation is the most stable.
Answer: (D). | ||||
| 72 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Propyne is a terminal alkyne; its acidic $\equiv$C–H reacts with sodium to give sodium propynide: $\mathrm{A = CH_3C{\equiv}C^-Na^+}$.
The acetylide ion is a nucleophile and displaces bromide from a primary alkyl bromide. Product C is $\mathrm{CH_3-C{\equiv}C-CH_2-CH_2-CH_3}$ (hex-2-yne), so the three-carbon chain added came from $\mathrm{B = CH_3CH_2CH_2Br}$, and NaBr is the by-product.
Answer: (A). | ||||
| 73 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes A vinylic halide has the halogen on an $sp^2$ carbon of a C=C double bond (not part of an aromatic ring).
(A) X on the $\mathrm{CH_2}$ next to the ring double bond: allylic.
(B) X on the ring carbon next to the double bond: allylic.
(C) X on the =CH carbon of the exocyclic double bond: vinylic.
(D) X on a benzene ring: aryl halide.
Answer: (C). | ||||
| 74 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Assertion and reasonEasy | |
A easy assertion and reason question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (A) In $\mathrm{CH_2{=}CH-CH_2-Cl}$ the chlorine is on the carbon next to the C=C: it is allyl chloride. True.
(R) In an allylic halide the halogen is on an $sp^3$ carbon next to the double bond; a halogen on an $sp^2$ carbon describes a vinylic (or aryl) halide. False.
Answer: (C). | ||||
| 75 | 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 Both routes oxidise toluene's methyl group only as far as the aldehyde:
Route A: chromic oxide in acetic anhydride at 273–283 K converts $\mathrm{C_6H_5CH_3}$ to the benzylidene diacetate $\mathrm{C_6H_5CH(OCOCH_3)_2}$, which is hydrolysed by aqueous acid on warming to benzaldehyde. So A = $\mathrm{CrO_3}$.
Route B: chromyl chloride in $\mathrm{CS_2}$ forms a chromium complex (the Etard reaction), which is hydrolysed to benzaldehyde. So B = $\mathrm{CrO_2Cl_2}$.
Answer: (B). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Benzoyl chloride is hydrogenated over palladium on barium sulphate (a partly poisoned catalyst) and stops at benzaldehyde: $\mathrm{C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO + HCl}$.
This is the Rosenmund reduction. (Etard uses chromyl chloride on a methylarene; Stephen reduces a nitrile with $\mathrm{SnCl_2/HCl}$; Wolff-Kishner reduces C=O to $\mathrm{CH_2}$.)
Answer: (D). | ||||
| 77 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines (i) Bromobenzene with Mg in ether gives phenylmagnesium bromide; with $\mathrm{CO_2}$ and then acid it gives benzoic acid, $\mathrm{C_6H_5COOH}$.
(ii) With ammonia and heat the acid gives the ammonium salt, which loses water to give benzamide, $\mathrm{C_6H_5CONH_2}$.
(iii) $\mathrm{Br_2}$/NaOH on a primary amide is the Hofmann bromamide degradation: the amine with one carbon fewer, aniline $\mathrm{C_6H_5NH_2}$.
Answer: (D). | ||||
| 78 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Fehling's test needs a free (potential) aldehyde or $\alpha$-hydroxy ketone group: a reducing sugar.
Glucose, maltose and lactose are reducing sugars.
In sucrose the glycosidic link joins C-1 of glucose to C-2 of fructose, so neither reducing group is free: sucrose is non-reducing and gives no red-brown $\mathrm{Cu_2O}$.
Answer: (D). | ||||
| 79 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | StatementsEasy | |
A easy statements question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter A sulphide with dilute $\mathrm{H_2SO_4}$ gives $\mathrm{H_2S}$, which blackens lead acetate paper: $\mathrm{(CH_3COO)_2Pb + H_2S \rightarrow PbS\downarrow + 2CH_3COOH}$. This is the test for the sulphide ion: Statement I is true.
The black substance is lead sulphide, PbS, not lead sulphite ($\mathrm{PbSO_3}$, which is white). Statement II is false.
Answer: (C). | ||||
| 80 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines (A) Aniline is diazotised with nitrous acid, made from $\mathrm{NaNO_2 + HCl}$ at 0–5 °C, giving benzenediazonium chloride. ($\mathrm{HNO_3/H_2SO_4}$ would nitrate the ring, so option (C) is out.)
(B) The azo-dye test couples the diazonium salt with alkaline 2-naphthol ($\beta$-naphthol), giving the scarlet (orange-red) azo dye 1-phenylazo-2-naphthol. Coupling with aniline gives yellow p-aminoazobenzene, not a scarlet dye.
So A = $\mathrm{NaNO_2 + HCl}$, 0–5 °C and B = 2-naphthol.
Answer: (A). | ||||
| 81 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles needed $= M\times V = 0.35\times 0.250 = 0.0875$ mol.
Mass $= 0.0875\times 82.02 = 7.18$ g, which rounds to 7.
Answer: 7. | ||||
| 82 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure The number of molecular orbitals equals the number of atomic orbitals combined.
Each atom brings one 2s and three 2p orbitals, 4 in all; two atoms give 8 atomic orbitals, hence 8 molecular orbitals:
$\sigma 2s$, $\sigma^* 2s$, $\sigma 2p_z$, $\sigma^* 2p_z$, $\pi 2p_x$, $\pi 2p_y$, $\pi^* 2p_x$, $\pi^* 2p_y$.
Answer: 8. | ||||
| 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 Read the graph (V on the vertical axis, P on the horizontal): A (10 kPa, 10 dm$^3$), B (10 kPa, 30 dm$^3$), C (30 kPa, 10 dm$^3$). Note 1 kPa dm$^3$ = 1 J.
A→B: expansion at constant 10 kPa: work done by the gas $= 10\times 20 = 200$ J.
B→C: V falls from 30 to 10 dm$^3$ while P rises linearly from 10 to 30 kPa: work by the gas $= -(\text{average } P)\times 20 = -20\times 20 = -400$ J.
C→A: constant volume: no work.
Net work done by the gas $= 200 - 400 = -200$ J; check: the enclosed triangle has area $\frac{1}{2}\times 20\ \text{kPa}\times 20\ \text{dm}^3 = 200$ J.
In the chemistry convention ($w = -\int P\,dV$, work done on the system) the total work is $+200$ J.
Answer: 200. | ||||
| 84 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Precipitation starts when $[\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2 = K_{sp}$:
$[\mathrm{OH^-}]^2 = \frac{1\times 10^{-11}}{0.10} = 1\times 10^{-10}$, $[\mathrm{OH^-}] = 10^{-5}$ M.
pOH $= 5$, pH $= 14 - 5 = 9$ (at 25 °C).
Answer: 9. | ||||
| 85 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Mn goes from +7 to +4 (gains 3 e$^-$ each); I from −1 to 0.
Two $\mathrm{MnO_4^-}$ gain 6 electrons, so 6 $\mathrm{I^-}$ lose 6: $b = 6$, $x = 3$, $y = 2$.
Charge: left $-2 - 6 = -8$, so the right needs $z = 8$ hydroxide ions.
Oxygen: left $8 + c$, right $2\times 2 + 8 = 12$, so $c = 4$; hydrogen: $2c = 8 = z$. ✓
$2\mathrm{MnO_4^-} + 6\mathrm{I^-} + 4\mathrm{H_2O} \rightarrow 3\mathrm{I_2} + 2\mathrm{MnO_2} + 8\mathrm{OH^-}$.
Answer: 8. | ||||
| 86 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Area $= 0.05$ m$^2 = 500$ cm$^2$. Volume of silver $= 500\times 0.05 = 25$ cm$^3$.
Mass $= 25\times 7.9 = 197.5$ g; moles $= \frac{197.5}{108} = 1.829$.
Atoms $= 1.829\times 6.022\times 10^{23} = 11.01\times 10^{23}$, i.e. 11.
Answer: 11. | ||||
| 87 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For a first-order reaction rate $= k[\mathrm{A}]$, so the rates are in the ratio of the concentrations.
Over the 10 minutes between: $k = \frac{2.303}{10}\log\frac{0.04}{0.03} = \frac{2.303}{10}(2\log 2 - \log 3) = \frac{2.303}{10}(0.6020 - 0.4771) = \frac{2.303\times 0.1249}{10}$ min$^{-1}$.
$t_{1/2} = \frac{2.303\log 2}{k} = \frac{0.3010\times 10}{0.1249} = 24.1$ min, which rounds to 24.
(Check with natural logs: $10\ln 2/\ln\frac{4}{3} = 6.931/0.2877 = 24.09$.)
Answer: 24. | ||||
| 88 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | NumericalEasy | |
A easy numerical question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties IUPAC roots: un = 1, so "un-un-un-ium" (the printed spelling has an extra n) gives the atomic number 111 (roentgenium, Rg).
Element 111 sits in period 7 below Cu, Ag and Au, with configuration $[\mathrm{Rn}]5f^{14}6d^{10}7s^1$ (or $6d^9 7s^2$): group 11.
Answer: 11. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $R_f = \frac{\text{distance moved by the compound}}{\text{distance moved by the solvent front}} = \frac{3.5}{5} = 0.7 = 7\times 10^{-1}$.
Answer: 7. | ||||
| 90 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalEasy | |
A easy numerical question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Semicarbazide is $\mathrm{H_2N-NH-CO-NH_2}$. Its free $\mathrm{NH_2}$ (the one on the hydrazine N) condenses with the C=O of ethanal, losing water:
$\mathrm{CH_3CHO + H_2N-NH-CO-NH_2 \rightarrow CH_3CH{=}N-NH-CO-NH_2 + H_2O}$.
The semicarbazone keeps all three nitrogen atoms of semicarbazide.
Answer: 3. | ||||
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