JEE (Main) · 2024
JEE (Main) 2024 — 30 January, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus5
- Sets, Relations and Functions2
- Matrices and Determinants2
11 easy · 14 medium · 5 hard
Physics
30 Q
- Laws of Motion3
- Kinematics2
- Properties of Bulk Matter2
24 easy · 6 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques5
- d- and f-Block Elements3
- Solutions2
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 FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium 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 Logarithm: $\frac{2x + 3}{4x^2 + x - 3} > 0$. Since $4x^2 + x - 3 = (4x - 3)(x + 1)$, the critical points are $-\frac{3}{2}, -1, \frac{3}{4}$, and the sign chart gives $x \in \left(-\frac{3}{2}, -1\right) \cup \left(\frac{3}{4}, \infty\right)$.
Inverse cosine: $-1 \le \frac{2x - 1}{x + 2} \le 1$.
For $x > -2$ multiply by $x + 2 > 0$: $-(x + 2) \le 2x - 1 \le x + 2$, i.e. $x \ge -\frac{1}{3}$ and $x \le 3$.
For $x < -2$ the inequalities reverse and would need $2x - 1 \ge x + 2$, i.e. $x \ge 3$: impossible.
So this part needs $x \in \left[-\frac{1}{3}, 3\right]$.
Intersection: $\left(\frac{3}{4}, 3\right]$, so $\alpha = \frac{3}{4}$, $\beta = 3$.
$5\beta - 4\alpha = 15 - 3 = 12$.
Answer: (D). | ||||
| 2 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Factor the cubic: $z^3 + 2z^2 + 2z + 1 = (z + 1)(z^2 + z + 1)$, so its roots are $-1, \omega, \omega^2$ (the non-real cube roots of unity).
$z = -1$: $(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \neq 0$. Not common.
$z = \omega$: $1985 = 3\cdot 661 + 2$ and $100 = 3\cdot 33 + 1$, so $\omega^{1985} + \omega^{100} + 1 = \omega^2 + \omega + 1 = 0$. Common.
$z = \omega^2$: $\omega^{3970} + \omega^{200} + 1$; $3970 = 3\cdot 1323 + 1$ and $200 = 3\cdot 66 + 2$, giving $\omega + \omega^2 + 1 = 0$. Common.
Number of common roots $= 2$.
Answer: (C). | ||||
| 3 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | StatementsHard | |
A hard statements question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let the common value be $k \neq 0$. Then no sine is zero and $x = \frac{k}{\sin\theta}$, $y = \frac{k}{\sin(\theta + 2\pi/3)}$, $z = \frac{k}{\sin(\theta + 4\pi/3)}$, all non-zero.
Write $a = \theta$, $b = \theta + \frac{2\pi}{3}$, $c = \theta + \frac{4\pi}{3}$.
Sum of pairwise products: $\sin a\sin b = \frac{1}{2}[\cos(a - b) - \cos(a + b)]$. Each difference $a - b$, $b - c$, $a - c$ has cosine $-\frac{1}{2}$, and $\cos(2\theta + \frac{2\pi}{3}) + \cos(2\theta + \frac{4\pi}{3}) + \cos(2\theta + 2\pi) = 0$. So $\sin a\sin b + \sin b\sin c + \sin c\sin a = \frac{1}{2}\left(-\frac{3}{2}\right) = -\frac{3}{4}$.
Product: $\sin\theta\sin(\theta + \frac{2\pi}{3})\sin(\theta + \frac{4\pi}{3}) = -\frac{1}{4}\sin 3\theta$ (non-zero, since none of the sines is zero).
Trace $(R) = k\left(\frac{1}{\sin a} + \frac{1}{\sin b} + \frac{1}{\sin c}\right) = k\cdot\frac{-3/4}{-\frac{1}{4}\sin 3\theta} = \frac{3k}{\sin 3\theta} \neq 0$. So (I) is false.
For a $3\times 3$ matrix, $\mathrm{adj}(\mathrm{adj}\,R) = (\det R)\,R$, so trace $(\mathrm{adj}(\mathrm{adj}\,R)) = xyz\cdot\mathrm{trace}(R)$. Here $xyz \neq 0$ and trace $(R) \neq 0$, so this trace is never 0; and $R$ always has three non-zero entries, not one. (II) is false as intended.
Answer: (D). | ||||
| 4 | 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 $D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{vmatrix} = (2\lambda - 3\lambda^2) - (\lambda - \lambda^2) + (3 - 2) = -2\lambda^2 + \lambda + 1 = -(2\lambda + 1)(\lambda - 1)$.
So the solution is unique exactly when $\lambda \neq 1$ and $\lambda \neq -\frac{1}{2}$.
$\lambda = 1$: subtracting the first equation, $y = 4$ and $2y = \mu - 5$; consistent (infinitely many) iff $\mu = 13$, inconsistent otherwise. So (B) and (D) are correct.
$\lambda = -\frac{1}{2}$: subtracting the first equation, $y - \frac{3}{4}z = 4$ and $2y - \frac{3}{2}z = \mu - 5$, i.e. $2\cdot 4 = \mu - 5$; consistent iff $\mu = 13$. So for $\lambda \neq 1$, $\mu = 13$ the system is always consistent: (A) is correct.
(C) fails: with $\lambda = -\frac{1}{2}$ and $\mu \neq 13$ the system has no solution at all.
Answer: (C). | ||||
| 5 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctHard | |
A hard single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem For four consecutive binomial coefficients $a_1, a_2, a_3, a_4$ (that is, ${}^nC_{r-1}, {}^nC_r, {}^nC_{r+1}, {}^nC_{r+2}$) the standard relation is
$\frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2a_2}{a_2 + a_3}$
(each fraction equals $\frac{r}{n+1}$, $\frac{r+2}{n+1}$ and $\frac{2(r+1)}{n+1}$ respectively).
Here $a_1 + a_2 = 2$ and $a_3 + a_4 = 2$, so
$\frac{2 - p}{2} + \frac{2 - \alpha}{2} = \frac{2p}{p + 2 - \alpha}$, i.e. $(4 - p - \alpha)(p + 2 - \alpha) = 4p$.
Expanding the left side: $4p + 8 - 4\alpha - p^2 - 2p + p\alpha - p\alpha - 2\alpha + \alpha^2 = -p^2 + \alpha^2 + 2p - 6\alpha + 8$.
Setting it equal to $4p$: $-p^2 + \alpha^2 - 2p - 6\alpha + 8 = 0$, so $p^2 - \alpha^2 + 6\alpha + 2p = 8$.
Answer: (C). | ||||
| 6 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series First GP: $ar^2 = b$, so $r^2 = \frac{b}{a}$ and $T_{11} = ar^{10} = a\left(\frac{b}{a}\right)^5$.
Second GP: $as^4 = b$, so $s^4 = \frac{b}{a}$ and $T_p = as^{p-1} = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$.
Equal, with $\frac{b}{a} \neq 1$: $\frac{p - 1}{4} = 5$, so $p = 21$.
Answer: (A). | ||||
| 7 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Put $x = y$: $f(1) = 1$.
Differentiate $f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$ with respect to $x$, holding $y$ fixed: $\frac{1}{y}f'\left(\frac{x}{y}\right) = \frac{f'(x)}{f(y)}$.
Now put $y = x$: $\frac{1}{x}f'(1) = \frac{f'(x)}{f(x)}$, so $xf'(x) = 2024 f(x)$.
(Indeed $f(x) = x^{2024}$ satisfies everything.)
$xf'(x) - 2024f(x) = 0$.
Answer: (A). | ||||
| 8 | 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 Continuity at $x = 1$: $1 + 3 + a = b + 2$. Differentiability at $x = 1$: $2(1) + 3 = b$, so $b = 5$ and $a = 3$.
$\int_{-2}^{1}(x^2 + 3x + 3)\,dx = \left[\frac{x^3}{3} + \frac{3x^2}{2} + 3x\right]_{-2}^{1} = \frac{29}{6} - \left(-\frac{8}{3}\right) = \frac{29}{6} + \frac{16}{6} = \frac{15}{2}$.
$\int_{1}^{2}(5x + 2)\,dx = \left[\frac{5x^2}{2} + 2x\right]_{1}^{2} = 14 - \frac{9}{2} = \frac{19}{2}$.
Total $= \frac{15}{2} + \frac{19}{2} = 17$.
Answer: (C). | ||||
| 9 | 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 $f'(x) = 2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2 = (x + 3)(x - 2)^2[2(x - 2) + 3(x + 3)] = 5(x + 3)(x - 2)^2(x + 1)$.
Critical points in $[-4, 4]$: $x = -3, -1, 2$.
Values: $f(-4) = 1\cdot(-6)^3 = -216$; $f(-3) = 0$; $f(-1) = 4\cdot(-27) = -108$; $f(2) = 0$; $f(4) = 49\cdot 8 = 392$.
$M = 392$, $m = -216$, $M - m = 608$.
Answer: (C). | ||||
| 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 $f(0) = a + b = -1$.
$f'(x) = 2ae^{2x} + be^x + c$, so $f'(\ln 2) = 8a + 2b + c = 21$.
$\int_0^{\ln 4}(ae^{2x} + be^x)\,dx = \frac{a}{2}(16 - 1) + b(4 - 1) = \frac{15a}{2} + 3b = \frac{39}{2}$, i.e. $5a + 2b = 13$.
With $b = -1 - a$: $5a - 2 - 2a = 13$, so $a = 5$, $b = -6$.
$c = 21 - 8a - 2b = 21 - 40 + 12 = -7$.
$|a + b + c| = |5 - 6 - 7| = 8$.
Answer: (A). | ||||
| 11 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Expand: $3(\sin\alpha\cos\beta + \cos\alpha\sin\beta) = 2(\sin\alpha\cos\beta - \cos\alpha\sin\beta)$.
So $\sin\alpha\cos\beta = -5\cos\alpha\sin\beta$; dividing by $\cos\alpha\cos\beta$: $\tan\alpha = -5\tan\beta$.
$k = -5$.
Answer: (C). | ||||
| 12 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f'(1) = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $f'(3) = \tan\frac{\pi}{4} = 1$.
With $u = f'(t)$, $du = f''(t)\,dt$: $\int_1^3((f')^2 + 1)f''\,dt = \left[\frac{u^3}{3} + u\right]_{1/\sqrt{3}}^{1}$.
At $u = 1$: $\frac{1}{3} + 1 = \frac{4}{3}$. At $u = \frac{1}{\sqrt{3}}$: $\frac{1}{9\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{10}{9\sqrt{3}}$.
Times 27: $36 - \frac{270}{9\sqrt{3}} = 36 - \frac{30}{\sqrt{3}} = 36 - 10\sqrt{3}$.
$\alpha = 36$, $\beta = -10$, $\alpha + \beta = 26$.
Answer: (A). | ||||
| 13 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f(f(x))$: $f(x)^4 = \frac{x^4}{1 + x^4}$, so $1 + f(x)^4 = \frac{1 + 2x^4}{1 + x^4}$ and $f(f(x)) = \frac{f(x)}{(1 + f(x)^4)^{1/4}} = \frac{x}{(1 + 2x^4)^{1/4}}$.
By the same step each further $f$ adds 1 to the coefficient: $g(x) = \frac{x}{(1 + 4x^4)^{1/4}}$.
$I = \int_0^{\sqrt{2\sqrt{5}}}\frac{x^3}{(1 + 4x^4)^{1/4}}\,dx$. Put $u = 1 + 4x^4$, $du = 16x^3\,dx$; at the upper limit $x^4 = 20$, $u = 81$.
$I = \frac{1}{16}\int_1^{81}u^{-1/4}\,du = \frac{1}{16}\cdot\frac{4}{3}\left[u^{3/4}\right]_1^{81} = \frac{1}{12}(27 - 1) = \frac{13}{6}$.
$18I = 39$.
Answer: (D). | ||||
| 14 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $A = (10, 0)$ and $B = \left(0, \frac{50}{7}\right)$.
$P = \frac{3A + 7B}{10} = \left(\frac{30}{10}, \frac{50}{10}\right) = (3, 5)$.
The perpendicular from $S = (ae, 0)$ to the $x$-axis is the line $x = ae$; it passes through $P$, so $ae = 3$.
Directrix $x = \frac{a}{e} = \frac{25}{3}$. Multiplying: $a^2 = 25$, $a = 5$, $e = \frac{3}{5}$.
$b^2 = a^2(1 - e^2) = 25\cdot\frac{16}{25} = 16$.
Latus rectum $= \frac{2b^2}{a} = \frac{32}{5}$.
Answer: (C). | ||||
| 15 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $a^2 = 9$, $b^2 = 4$, so $c^2 = 13$ and the foci are $(\pm\sqrt{13}, 0)$.
Area $= \frac{1}{2}\cdot 2\sqrt{13}\cdot y = \sqrt{13}\,y = 2\sqrt{13}$, so $y = 2$.
$\frac{x^2}{9} = 1 + \frac{4}{4} = 2$, so $x^2 = 18$.
$OP^2 = 18 + 4 = 22$.
Answer: (C). | ||||
| 16 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines Both lines have normal length $\sqrt{5}$, so the locus is $|x + 2y + 7| = |2x - y + 8|$, i.e. $(x + 2y + 7)^2 - (2x - y + 8)^2 = 0$ (the pair of angle bisectors).
$(x + 2y + 7)^2 = x^2 + 4y^2 + 4xy + 14x + 28y + 49$.
$(2x - y + 8)^2 = 4x^2 + y^2 - 4xy + 32x - 16y + 64$.
Difference: $-3x^2 + 3y^2 + 8xy - 18x + 44y - 15 = 0$. Divide by $-3$ to make the $x^2$ coefficient 1:
$x^2 - y^2 - \frac{8}{3}xy + 6x - \frac{44}{3}y + 5 = 0$.
So $h = -\frac{4}{3}$, $g = 3$, $f = -\frac{22}{3}$, $c = 5$.
$g + c + h - f = 3 + 5 - \frac{4}{3} + \frac{22}{3} = 14$.
Answer: (C). | ||||
| 17 | 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 $L_1 \perp L_2$: $(1)(3) + (-1)(1) + 2p = 0$, so $p = -1$.
Direction of $L_3$ is along $(\hat{i} - \hat{j} + 2\hat{k})\times(3\hat{i} + \hat{j} - \hat{k}) = \hat{i}(1 - 2) - \hat{j}(-1 - 6) + \hat{k}(1 + 3) = -\hat{i} + 7\hat{j} + 4\hat{k}$.
$L_3$ passes through the origin, so its points are $\delta(-1, 7, 4)$; with $\delta = 1$ we get $(-1, 7, 4)$.
Check the others: $(1, 7, -4)$ and $(1, -7, 4)$ are not multiples of $(-1, 7, 4)$.
Answer: (B). | ||||
| 18 | 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 $\left|(\vec{b}\times\vec{a}) - \vec{b}\right|^2 = |\vec{b}\times\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{b}\times\vec{a})\cdot\vec{b}$.
$\vec{b}\times\vec{a}$ is perpendicular to $\vec{b}$, so the dot product is 0.
Value $= 4 + 1 - 0 = 5$.
Answer: (D). | ||||
| 19 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\frac{\pi}{4} = |\vec{a}|\sqrt{6}\cdot\frac{1}{\sqrt{2}} = \sqrt{3}|\vec{a}| = 3\sqrt{2}$, so $|\vec{a}| = \sqrt{6}$.
$|\vec{a}|^2 = 1 + \alpha^2 + \beta^2 = 6$, so $\alpha^2 + \beta^2 = 5$.
$|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 36 - 18 = 18$.
Product $= 5\times 18 = 90$.
Answer: (B). | ||||
| 20 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(W\mid A) = \frac{3}{10}$, $P(W\mid B) = \frac{3}{5}$, and each bag has probability $\frac{1}{2}$.
$P(A\mid W) = \frac{\frac{1}{2}\cdot\frac{3}{10}}{\frac{1}{2}\cdot\frac{3}{10} + \frac{1}{2}\cdot\frac{6}{10}} = \frac{3}{3 + 6} = \frac{1}{3}$.
Answer: (B). | ||||
| 21 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions A symmetric relation on a 4-element set is fixed by choosing which of the 4 diagonal pairs $(a, a)$ and which of the $\binom{4}{2} = 6$ unordered pairs $\{a, b\}$ (each giving both $(a, b)$ and $(b, a)$) it contains: $2^{4 + 6} = 2^{10} = 1024$ relations.
Symmetric and reflexive: all 4 diagonal pairs forced, the 6 unordered pairs free: $2^6 = 64$.
Symmetric but not reflexive: $1024 - 64 = 960$.
Answer: 960. | ||||
| 22 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Either $x = 0$, or $x^2 + 3|x| + 5|x - 1| + 6|x - 2| = 0$.
Every term of the bracket is non-negative, so it is 0 only if $x = 0$, $x = 1$ and $x = 2$ at once, which is impossible. (In fact the bracket is at least $5|x-1| + 6|x-2| \ge 5 > 0$.)
So the only real solution is $x = 0$.
Answer: 1. | ||||
| 23 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Let $(a, b, c)$ be the numbers chosen from A, B, C: $a + b + c = 15$, $4 \le a \le 8$, $4 \le b, c \le 6$.
$a = 4$: $(b, c) = (5, 6), (6, 5)$: $\binom{8}{4}(6\cdot 1 + 1\cdot 6) = 70\cdot 12 = 840$.
$a = 5$: $(4, 6), (5, 5), (6, 4)$: $\binom{8}{5}(15 + 36 + 15) = 56\cdot 66 = 3696$.
$a = 6$: $(4, 5), (5, 4)$: $\binom{8}{6}(15\cdot 6 + 6\cdot 15) = 28\cdot 180 = 5040$.
$a = 7$: $(4, 4)$: $\binom{8}{7}\cdot 15\cdot 15 = 8\cdot 225 = 1800$.
($a = 8$ would need $b + c = 7$, impossible with both at least 4.)
Total $= 840 + 3696 + 5040 + 1800 = 11376$.
Check by complement: $\binom{20}{15} = 15504$. Fewer than 4 from A: only $a = 3, b = c = 6$: 56 ways. Fewer than 4 from B: $b = 3$: $20(15 + 48 + 28) = 1820$; $b = 2$: $15(6 + 8) = 210$; $b = 1$: $6$; total 2036, and the same for C. No two can fail together. $15504 - 56 - 2\cdot 2036 = 11376$.
Answer: 11376. | ||||
| 24 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Use $\frac{{}^nC_k}{k + 1} = \frac{{}^{n+1}C_{k+1}}{n + 1}$ and $\frac{{}^nC_{k+1}}{k + 2} = \frac{{}^{n+1}C_{k+2}}{n + 1}$.
$\alpha = \frac{1}{n + 1}\sum_k {}^{n+1}C_{k+1}\,{}^nC_{n-k} = \frac{{}^{2n+1}C_{n+1}}{n + 1}$ (Vandermonde: the lower indices add to $n + 1$).
$\beta = \frac{1}{n + 1}\sum_k {}^{n+1}C_{k+2}\,{}^nC_{n-k} = \frac{{}^{2n+1}C_{n+2}}{n + 1}$ (lower indices add to $n + 2$).
$\frac{\beta}{\alpha} = \frac{{}^{2n+1}C_{n+2}}{{}^{2n+1}C_{n+1}} = \frac{(2n + 1) - (n + 1)}{n + 2} = \frac{n}{n + 2}$.
$5\alpha = 6\beta$ gives $\frac{n}{n + 2} = \frac{5}{6}$, so $6n = 5n + 10$, $n = 10$.
Answer: 10. | ||||
| 25 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_k = \frac{k}{2}[6 + 4(k - 1)] = 2k^2 + k$.
$\sum_{k=1}^{n} S_k = \frac{2n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6}[2(2n + 1) + 3] = \frac{n(n+1)(4n + 5)}{6}$.
So the bracket equals $4n + 5$, and $40 < 4n + 5 < 42$ gives $35 < 4n < 37$, so $4n = 36$, $n = 9$.
Answer: 9. | ||||
| 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 The parabola has vertex $(1, 2)$ and opens to the right. The line $x = 2y - 4$ meets it where $(y - 2)^2 = 2y - 5$, i.e. $(y - 3)^2 = 0$: it touches the parabola at $(2, 3)$. The line meets the $y$-axis at $(0, 2)$.
The region (first quadrant, left of the parabola, below the line) is best taken in horizontal strips, $0 \le y \le 3$:
for $0 \le y \le 2$, $x$ runs from 0 to $(y - 2)^2 + 1$; for $2 \le y \le 3$, from the line $2y - 4$ to the parabola.
Area $= \int_0^2\left[(y - 2)^2 + 1\right]dy + \int_2^3\left[(y - 2)^2 + 1 - (2y - 4)\right]dy = \left(\frac{8}{3} + 2\right) + \int_2^3 (y - 3)^2\,dy = \frac{14}{3} + \frac{1}{3} = 5$.
Answer: 5. | ||||
| 27 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Write $p = Y'(x)$. The tangent meets the axes at $X = x - \frac{y}{p}$ and $Y = y - xp$.
Area $= \frac{1}{2}\left(x - \frac{y}{p}\right)(y - xp) = \frac{1}{2}\left(2xy - x^2p - \frac{y^2}{p}\right) = xy - \frac{x^2p}{2} - \frac{y^2}{2p}$.
Equating to $-\frac{y^2}{2p} + 1$: $xy - \frac{x^2p}{2} = 1$, so $p = \frac{2(xy - 1)}{x^2}$, i.e. $\frac{dy}{dx} - \frac{2}{x}y = -\frac{2}{x^2}$.
Integrating factor $x^{-2}$: $\frac{d}{dx}\left(\frac{y}{x^2}\right) = -\frac{2}{x^4}$, so $\frac{y}{x^2} = \frac{2}{3x^3} + C$.
$Y(1) = 1$: $1 = \frac{2}{3} + C$, $C = \frac{1}{3}$. So $y = \frac{2}{3x} + \frac{x^2}{3}$.
$Y(2) = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$, and $12Y(2) = 20$.
Answer: 20. | ||||
| 28 | CirclesRadius and centre | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Centres $O(0, 0)$ and $C(\alpha, 0)$, radii 5 and 4, and $P$ an intersection point. The angle $\theta$ at $P$ between $PO$ and $PC$ has $\sin\theta = \frac{\sqrt{63}}{8}$; taking the principal value, $\theta$ is acute and $\cos\theta = \frac{1}{8}$.
Cosine rule in triangle $OPC$: $\alpha^2 = 25 + 16 - 2\cdot 5\cdot 4\cdot\frac{1}{8} = 36$, so $\alpha = 6 \in (5, 9)$.
Half the common chord is the height of triangle $OPC$ from $P$: area $= \frac{1}{2}\cdot 5\cdot 4\sin\theta = \frac{5\sqrt{63}}{4}$, and height $= \frac{2\cdot\text{area}}{OC} = \frac{5\sqrt{63}}{2\cdot 6} = \frac{5\sqrt{63}}{12}$.
$\beta = \frac{5\sqrt{63}}{6}$, so $\alpha\beta = 5\sqrt{63}$ and $(\alpha\beta)^2 = 25\times 63 = 1575$.
Check with coordinates: subtracting the circles gives the chord $x = \frac{\alpha^2 + 9}{2\alpha} = \frac{45}{12} = \frac{15}{4}$, so half-chord $= \sqrt{25 - \frac{225}{16}} = \frac{\sqrt{175}}{4} = \frac{5\sqrt{7}}{4}$ and $\beta = \frac{5\sqrt{7}}{2} = \frac{5\sqrt{63}}{6}$.
Answer: 1575. | ||||
| 29 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $M = (1 + 3s, 2 + 2s, -1 - 2s)$ and $N = (-2 - 3t, 2 - 2t, 1 + 4t)$; let $Q = (-1, 2, 3)$.
$\vec{QM} = (2 + 3s, 2s, -4 - 2s)$ and $\vec{QN} = (-1 - 3t, -2t, -2 + 4t)$ must be parallel: $\vec{QM} = k\,\vec{QN}$.
Second components: $2s = -2kt$, so $k = -\frac{s}{t}$.
First: $2 + 3s = -\frac{s}{t}(-1 - 3t) = \frac{s}{t} + 3s$, so $s = 2t$ and $k = -2$.
Third: $-4 - 2s = -2(-2 + 4t)$, i.e. $-4 - 4t = 4 - 8t$, so $t = 2$, $s = 4$.
$M = (13, 10, -9)$, $\alpha + \beta + \gamma = 14$; $N = (-8, -2, 9)$, $a + b + c = -1$.
Check: $\vec{QM} = (14, 8, -12) = -2(-7, -4, 6) = -2\,\vec{QN}$.
Ratio $= \frac{14^2}{(-1)^2} = 196$.
Answer: 196. | ||||
| 30 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $N = 3 + 2 + 3 + 2 + 6 + 3 + 3 = 22$.
$\sum f_ix_i = 0 + 2 + 15 + 12 + 60 + 36 + 51 = 176$, so the mean is $\bar{x} = 8$.
$\sum f_i(x_i - 8)^2 = 3(64) + 2(49) + 3(9) + 2(4) + 6(4) + 3(16) + 3(81) = 192 + 98 + 27 + 8 + 24 + 48 + 243 = 640$.
$\sigma^2 = \frac{640}{22} = \frac{320}{11} \approx 29.09$.
Check: $\sum f_ix_i^2 = 0 + 2 + 75 + 72 + 600 + 432 + 867 = 2048$, and $\frac{2048}{22} - 64 = 29.09$.
Nearest integer: 29.
Answer: 29. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $[c] = LT^{-1}$, $[G] = M^{-1}L^3T^{-2}$, $[h] = ML^2T^{-1}$.
$[G^{-1/2}] = M^{1/2}L^{-3/2}T$ and $[h^{1/2}] = M^{1/2}LT^{-1/2}$; their product is $M L^{-1/2} T^{1/2}$.
For $m$ to have dimension $M$, $c^P$ must supply $L^{1/2}T^{-1/2}$: $L^PT^{-P} = L^{1/2}T^{-1/2}$, so $P = \frac{1}{2}$.
(This is the Planck mass, $\sqrt{hc/G}$.)
Answer: (D). | ||||
| 32 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane From the same height, equal times of flight need equal vertical components of the projection velocity. With angles measured from the vertical, the vertical components are $v_A\cos 45^\circ$ and $v_B\cos 60^\circ$:
$v_A\cdot\frac{1}{\sqrt{2}} = v_B\cdot\frac{1}{2} \Rightarrow \frac{v_A}{v_B} = \frac{1}{\sqrt{2}}$.
$v_A : v_B = 1 : \sqrt{2}$.
Answer: (B). | ||||
| 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 Whole system: $a = \frac{F}{m_A + m_B + m_C} = \frac{80}{5 + 3 + 2} = 8$ m s$^{-2}$.
$T_1$ alone pulls A: $T_1 = 5\times 8 = 40$ N.
$T_2$ pulls A and B together: $T_2 = (5 + 3)\times 8 = 64$ N.
(Check on C: $80 - T_2 = 2\times 8$, so $T_2 = 64$ N.)
Answer: (A). | ||||
| 34 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion On a slope of angle $\theta$ a block stays at rest if $\tan\theta \le \mu$.
Slope of the surface: $\tan\theta = \frac{dy}{dx} = \frac{x}{2}$. Limiting: $\frac{x}{2} = 0.5$, so $x = 1$ m.
Height $y = \frac{1^2}{4} = \frac{1}{4}$ m.
Answer: (A). | ||||
| 35 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Normal reaction $N = mg\cos 60^\circ = 1\times 10\times\frac{1}{2} = 5$ N (the 10 N push is parallel to the incline and does not change $N$).
Friction $f = \mu N = 0.1\times 5 = 0.5$ N.
Work against friction over 10 m $= 0.5\times 10 = 5$ J.
Answer: (B). | ||||
| 36 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}} \propto \sqrt{\frac{M}{R}}$.
$\frac{v_p}{v_e} = \sqrt{\frac{M/6}{R/3}} = \sqrt{\frac{1}{2}}$.
$v_p = \frac{11.2}{\sqrt{2}} \approx 7.9$ km/s.
Answer: (B). | ||||
| 37 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Heat supplied steadily to ice at $-10^\circ$C goes through four stages:
1. ice warms from $-10^\circ$C to $0^\circ$C — temperature rises;
2. ice melts at $0^\circ$C — temperature stays constant (latent heat of fusion);
3. water warms from $0^\circ$C to $100^\circ$C — temperature rises;
4. water boils at $100^\circ$C and turns to steam — temperature stays constant (latent heat of vaporisation; the longest flat part).
So the curve is rise, flat, rise, flat, ending at the boiling plateau, since the final state is steam at $100^\circ$C. Only graph (A) has this shape.
(B) has no boiling plateau; (C) and (D) show vertical jumps, i.e. a temperature change with no heat supplied, which is impossible.
Answer: (A). | ||||
| 38 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Both curves fall as the gas expands, and B is the steeper one. The adiabatic slope $\frac{dP}{dV} = -\gamma\frac{P}{V}$ is $\gamma$ times the isothermal slope $-\frac{P}{V}$, so B is adiabatic and A is isothermal.
Adiabatic: $PV^{\gamma} = k$; isothermal: $PV = k$ (equivalently $T = k$). Statement (B).
(A) is wrong (B is not isothermal); (D) puts the adiabatic relation on A and the isothermal one on B, the wrong way round.
Answer: (B). | ||||
| 39 | 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 $C_V$ of the mixture $= \frac{3\cdot\frac{3}{2}R + 2\cdot\frac{5}{2}R}{3 + 2} = \frac{4.5R + 5R}{5} = 1.9R$.
$C_P = C_V + R = 2.9R$.
$\gamma = \frac{2.9}{1.9} = \frac{29}{19} \approx 1.526 \approx 1.52$.
Check: $\frac{n_1 + n_2}{\gamma - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1} = \frac{3}{2/3} + \frac{2}{2/5} = 4.5 + 5 = 9.5$, so $\gamma - 1 = \frac{5}{9.5} = 0.526$.
Answer: (B). | ||||
| 40 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Field of an infinite line charge: $E = \frac{2k\lambda}{r}$, directed radially outward; the negative charge is pulled inward with force $\frac{2k\lambda q}{r}$.
This provides the centripetal force: $\frac{mv^2}{r} = \frac{2k\lambda q}{r}$, so $v = \sqrt{\frac{2k\lambda q}{m}}$ (independent of $r$).
$T = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{m}{2k\lambda q}}$.
Answer: (A). | ||||
| 41 | 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 Originally $W = \frac{V^2}{R}$.
Each half has resistance $\frac{R}{2}$; two of them in parallel give $\frac{R}{4}$.
New rate $= \frac{V^2}{R/4} = 4\frac{V^2}{R} = 4W$.
Answer: (C). | ||||
| 42 | Electromagnetic WavesDisplacement current | Class 12: Electromagnetic Waves | Matching listEasy | |
A easy matching list question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves A. Gauss's law for magnetism: the net magnetic flux through any closed surface is zero, $\oint\vec{B}\cdot\vec{da} = 0$ — II.
B. Faraday's law: the emf round a loop equals minus the rate of change of magnetic flux through it — III.
C. Ampere's law: $\oint\vec{B}\cdot\vec{dl} = \mu_0 I$ — IV.
D. Gauss's law of electrostatics: flux of $\vec{E}$ through a closed surface $= \frac{q_{\text{enc}}}{\varepsilon_0} = \frac{1}{\varepsilon_0}\int\rho\,dV$ — I.
A→II, B→III, C→IV, D→I.
Answer: (B). | ||||
| 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 In a resistor the current is in phase with the voltage: $i = i_0\sin 100\pi t$.
Half the peak: $\sin 100\pi t_1 = \frac{1}{2}$, so $100\pi t_1 = \frac{\pi}{6}$, $t_1 = \frac{1}{600}$ s.
Peak: $100\pi t_2 = \frac{\pi}{2}$, $t_2 = \frac{1}{200}$ s.
$\Delta t = \frac{1}{200} - \frac{1}{600} = \frac{1}{300}$ s $\approx 3.3$ ms.
Answer: (A). | ||||
| 44 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves For electromagnetic radiation completely absorbed, momentum delivered $p = \frac{U}{c}$.
$p = \frac{6.48\times 10^5}{3\times 10^8} = 2.16\times 10^{-3}$ kg m/s.
(For complete reflection it would be twice this.)
Answer: (A). | ||||
| 45 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics After the first polaroid unpolarised light is halved: $\frac{I_o}{2}$.
Malus's law at the second: $I = \frac{I_o}{2}\cos^2 45^\circ = \frac{I_o}{2}\cdot\frac{1}{2} = \frac{I_o}{4}$.
Answer: (C). | ||||
| 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 Einstein's equation: $E_k = h\nu - \phi_0$.
Plotted against $\nu$ this is a straight line of slope $h$ and intercept $-\phi_0$ (it cuts the $\nu$-axis at the threshold frequency).
So the slope $\tan\theta$ is Planck's constant. (The ratio $h/e$ would be the slope of stopping potential against $\nu$.)
Answer: (A). | ||||
| 47 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Energy released $Q = \Delta M c^2$ appears as kinetic energy of the three daughters.
Each daughter has mass $\approx\frac{M}{3}$ (the mass defect is small) and, being identical with zero total momentum, the same speed $v$.
$3\cdot\frac{1}{2}\cdot\frac{M}{3}v^2 = \Delta Mc^2 \Rightarrow \frac{1}{2}Mv^2 = \Delta Mc^2 \Rightarrow v = c\sqrt{\frac{2\Delta M}{M}}$.
(A) is an energy, not a speed; (D) has the wrong dimensions.
Answer: (B). | ||||
| 48 | 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 Magnetic moment of a circulating charge: $\mu = IA = \frac{ev}{2\pi r}\cdot\pi r^2 = \frac{evr}{2}$.
Equivalently $\mu = \frac{e}{2m}L$ with $L = \frac{nh}{2\pi}$, so $\mu_n = \frac{neh}{4\pi m} \propto n$.
(Check: $v \propto \frac{1}{n}$ and $r \propto n^2$, so $vr \propto n$.)
$x = 1$.
Answer: (B). | ||||
| 49 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Both diodes point along the current from the 15 V cell, so both are forward biased. Take the usual barrier drops: Ge 0.3 V, Si 0.7 V.
Voltage left for the resistors: $15 - 0.3 - 0.7 = 14$ V across $1.5 + 2.5 = 4$ k$\Omega$.
$I = \frac{14}{4000} = 3.5$ mA.
$V_L = 3.5\text{ mA}\times 2.5\text{ k}\Omega = 8.75$ V.
Answer: (B). | ||||
| 50 | 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 50 VSD = 49 MSD, so 1 VSD $= \frac{49}{50}$ MSD.
Vernier constant $= 1\text{ MSD} - 1\text{ VSD} = \frac{1}{50}$ MSD $= \frac{0.5}{50}$ mm $= 0.01$ mm.
Answer: (C). | ||||
| 51 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $|\vec{A}| = 5$ and $\hat{B} = \frac{4\hat{i} + 3\hat{j}}{5}$.
The vector is $5\hat{B} = 4\hat{i} + 3\hat{j}$: its $y$-component is 3, as stated, and its $x$-component is 4.
Answer: 4. | ||||
| 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 Angular momentum is conserved: $\omega = \frac{4\times 10 + 2\times 4}{4 + 2} = 8$ rad/s (taking both rotations in the same sense).
Initial KE $= \frac{1}{2}(4)(100) + \frac{1}{2}(2)(16) = 200 + 16 = 216$ J.
Final KE $= \frac{1}{2}(6)(64) = 192$ J.
Loss $= 24$ J.
Check: $\Delta K = \frac{1}{2}\frac{I_1I_2}{I_1 + I_2}(\omega_1 - \omega_2)^2 = \frac{1}{2}\cdot\frac{8}{6}\cdot 36 = 24$ J.
Answer: 24. | ||||
| 53 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Volume is conserved: $\frac{4}{3}\pi R^3 = 1000\cdot\frac{4}{3}\pi r^3$, so $R = 10r$.
$E_1 = 1000\cdot 4\pi r^2 T$ and $E_2 = 4\pi R^2 T = 100\cdot 4\pi r^2 T$.
$E_1 : E_2 = 1000 : 100 = 10 : 1$.
Answer: 10. | ||||
| 54 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For a point source the intensity falls as $\frac{1}{r^2}$. Taking the given $16\times 10^{-8}$ W m$^{-2}$ as the intensity at unit distance (1 m), $I(r) = \frac{16\times 10^{-8}}{r^2}$.
$I(2) = 4\times 10^{-8}$ W m$^{-2}$ and $I(4) = 1\times 10^{-8}$ W m$^{-2}$.
Difference $= 3\times 10^{-8}$ W m$^{-2}$.
Answer: 3. | ||||
| 55 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance For each sphere at equilibrium, $\tan\frac{\theta}{2} = \frac{F_e}{W_{\text{eff}}}$.
In air: $\tan\frac{\theta}{2} = \frac{F}{mg}$.
In the liquid the electric force becomes $\frac{F}{K}$ and the effective weight $mg\left(1 - \frac{\rho_l}{\rho}\right) = mg\left(1 - \frac{0.7}{1.4}\right) = \frac{mg}{2}$.
Same angle: $\frac{F}{mg} = \frac{F/K}{mg/2} = \frac{2F}{Kmg}$, so $K = 2$.
(The value of $\tan 37^\circ$ is not needed: the angle cancels.)
Answer: 2. | ||||
| 56 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity If the parallel combination of $100\ \Omega$ and the voltmeter has 1 V across it, the $200\ \Omega$ resistor has $4 - 1 = 3$ V.
Current $= \frac{3}{200} = 15$ mA.
Combination resistance $= \frac{1}{0.015} = \frac{200}{3}\ \Omega$.
$\frac{1}{R_V} = \frac{3}{200} - \frac{1}{100} = \frac{1}{200}$, so $R_V = 200\ \Omega$.
Answer: 200. | ||||
| 57 | 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 Each side is a finite wire at perpendicular distance $d = \frac{a}{2} = 0.5$ m from the centre, with ends at $45^\circ$:
$B_{\text{side}} = \frac{\mu_0 I}{4\pi d}(\sin 45^\circ + \sin 45^\circ) = 10^{-7}\cdot\frac{5}{0.5}\cdot\sqrt{2} = \sqrt{2}\times 10^{-6}$ T.
Four sides add: $B = 4\sqrt{2}\times 10^{-6} = 40\sqrt{2}\times 10^{-7}$ T.
(Formula check: $B = \frac{2\sqrt{2}\mu_0 I}{\pi a} = \frac{2\sqrt{2}\cdot 4\pi\times 10^{-7}\cdot 5}{\pi} = 40\sqrt{2}\times 10^{-7}$ T.)
$X = 40$.
Answer: 40. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsTransformers | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on transformers, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Input power $= V_pI_p = 2300\times 5 = 11500$ W.
Output power $= 0.9\times 11500 = 10350$ W.
Output current $= \frac{10350}{230} = 45$ A.
Answer: 45. | ||||
| 59 | Experimental SkillsOptics experiments | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on optics experiments, from Experimental Skills. NCERT: No matching NCERT chapter With distances measured from the foci, Newton's lens formula is $xy = f^2$: the data lie on a rectangular hyperbola.
On such a curve $x + y \ge 2\sqrt{xy} = 2f$, with equality only at $x = y = f$, so the straight line $x + y = 2f$ touches the curve at the point where $x = y$.
The plot shows the straight line from $(0, 40)$ to $(40, 0)$, i.e. $x + y = 40$, touching the curve at C $= (20, 20)$ cm, where $x = y$.
So $2f = 40$, $f = 20$ cm (and $f^2 = 20\times 20$).
Answer: 20. | ||||
| 60 | 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 At height $h = R$: $g' = \frac{g}{(1 + h/R)^2} = \frac{g}{4} = \frac{\pi^2}{4}$ m s$^{-2}$.
$T = 2\pi\sqrt{\frac{l}{g'}} = 2\pi\sqrt{\frac{4\times 4}{\pi^2}} = 2\pi\cdot\frac{4}{\pi} = 8$ s.
Answer: 8. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | SolutionsConcentration terms | Class 12: Solutions | Single correctEasy | |
A easy single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions The mole fraction of a component is its moles divided by the total moles of every component in the solution.
The solution contains A, B and C, so $x_C = \frac{n_C}{n_A + n_B + n_C}$.
Answer: (C). | ||||
| 62 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | StatementsMedium | |
A medium statements question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Statement I is false: $\mu(\mathrm{NH_3}) \approx 1.47$ D is much larger than $\mu(\mathrm{NF_3}) \approx 0.23$ D.
In $\mathrm{NH_3}$ the N–H bond dipoles point towards N (N is more electronegative), the same way as the lone-pair orbital dipole, so they add. In $\mathrm{NF_3}$ the N–F bond dipoles point away from N, opposite to the lone-pair dipole, so they largely cancel.
Statement II has the directions the wrong way round (it says opposite in $\mathrm{NH_3}$ and same in $\mathrm{NF_3}$), so it is false too.
Both statements are false.
Answer: (B). | ||||
| 63 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{PF_5}$ and $\mathrm{PCl_5}$: 5 bond pairs, no lone pair — trigonal bipyramidal.
$\mathrm{BrF_5}$: Br has 7 valence electrons, 5 bond pairs + 1 lone pair ($sp^3d^2$) — octahedral arrangement, square pyramidal shape.
$\mathrm{[Ni(CN)_4]^{2-}}$: $dsp^2$ — square planar.
Answer: (B). | ||||
| 64 | SolutionsColligative properties | Class 12: Solutions | Single correctMedium | |
A medium single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = iK_fm$. Compare at the same mass of solvent ($K_f$: water 1.86, benzene 5.12 K kg mol$^{-1}$).
(A) acetic acid in water: $\frac{180}{60} = 3$ mol, $i \approx 1$ (weak acid): $\propto 1.86\times 3 = 5.6$.
(B) glucose in water: $\frac{180}{180} = 1$ mol, $i = 1$: $\propto 1.86$.
(C) benzoic acid in benzene: $\frac{180}{122} = 1.48$ mol, dimerises, $i \approx \frac{1}{2}$: $\propto 5.12\times 0.74 = 3.8$.
(D) acetic acid in benzene: 3 mol, dimerises, $i \approx \frac{1}{2}$: $\propto 5.12\times 1.5 = 7.7$.
The largest depression is (D), about 7.7 K. But benzene freezes at 5.5 °C, so (D) freezes near −2.2 °C, while (A), acetic acid in water, freezes near −5.6 °C, the lowest freezing point. The stem asks for both at once, and each half picks a different option.
Answer: (D) or (A). | ||||
| 65 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry The higher the standard reduction potential, the more readily the species is reduced, i.e. the stronger it is as an oxidising agent.
$1.74 > 1.65 > 1.19$ V, so $\mathrm{BrO_4^- > IO_4^- > ClO_4^-}$.
Answer: (C). | ||||
| 66 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsEasy | |
A easy statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Statement I is false: across a period reactivity is highest at the two ends (alkali metals, which lose electrons easily, and halogens, which gain them easily) and lowest in the middle; group 18 noble gases are nearly unreactive.
Statement II is true: group 1 oxides ($\mathrm{Na_2O}$, $\mathrm{K_2O}$) are strongly basic, group 17 oxides ($\mathrm{Cl_2O_7}$ and others) are acidic.
Answer: (D). | ||||
| 67 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down the group the E–H bond gets longer and weaker.
A. True: thermal stability falls from $\mathrm{NH_3}$ to $\mathrm{BiH_3}$.
B. True: the weaker the E–H bond, the more easily H is given up, so reducing power rises from $\mathrm{NH_3}$ to $\mathrm{BiH_3}$.
C. False: it is the other way round — $\mathrm{BiH_3}$ is the strongest reducing agent and $\mathrm{NH_3}$ only a mild one.
D. False: basicity decreases down the group ($\mathrm{NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3}$), as the lone pair occupies a larger, more diffuse orbital.
A and B only.
Answer: (A). | ||||
| 68 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Acidic strength of the group 16 hydrides increases down the group: $\mathrm{H_2O < H_2S < H_2Se < H_2Te}$. So A is true.
The reason is the falling H–E bond dissociation enthalpy as the atom gets larger (Te–H is the weakest), so H$^+$ is released most easily from $\mathrm{H_2Te}$. R is true and explains A.
Answer: (A). | ||||
| 69 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Fusing $\mathrm{MnO_2}$ with KOH in the presence of an oxidant (air or $\mathrm{KNO_3}$) gives green potassium manganate: $\mathrm{2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O}$. So A is $\mathrm{MnO_4^{2-}}$.
Electrolytic oxidation in alkaline solution: $\mathrm{MnO_4^{2-} \rightarrow MnO_4^- + e^-}$, giving purple permanganate. So B is $\mathrm{MnO_4^-}$.
Answer: (C). | ||||
| 70 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Cr in $\mathrm{Cr_2O_7^{2-}}$ is +6 ($3d^0$) and Mn in $\mathrm{MnO_4^-}$ is +7 ($3d^0$). With no d electrons, d–d transitions are impossible.
The intense colours come from ligand-to-metal charge transfer (an electron moving from oxygen to the metal) in both.
Answer: (D). | ||||
| 71 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{Mn_2(CO)_{10}}$ is two $\mathrm{Mn(CO)_5}$ units joined by a Mn–Mn bond. Each Mn is bonded to five CO groups and to the other Mn: six positions in all, arranged octahedrally (the five CO make a square pyramid, and the Mn–Mn bond fills the sixth corner).
Answer: (A). | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Differential extraction separates a compound from a solution by shaking with a second, immiscible solvent in which it is more soluble; it relies on the difference in solubility in the two solvents.
Sublimation depends on volatility of a solid, distillation on boiling points, column chromatography on differential adsorption.
Answer: (C). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Alkyl groups stabilise a carbocation by +I effect and hyperconjugation; the more alkyl groups on the positive carbon, the more stable it is.
Tertiary (9 α-H) > secondary (6 α-H) > primary (3 α-H) > methyl (none):
$(\mathrm{CH_3})_3\mathrm{C}^+ > (\mathrm{CH_3})_2\mathrm{CH}^+ > \mathrm{CH_3CH_2}^+ > \mathrm{CH_3}^+$.
Answer: (C). | ||||
| 74 | 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 The nitrile carbon is part of the chain and is C1: $\mathrm{\overset{4}{C}H_3-\overset{3}{C}H(NH_2)-\overset{2}{C}H_2-\overset{1}{C}N}$.
Four carbons: butanenitrile; the amino group is on C3.
Name: 3-aminobutanenitrile.
Answer: (B). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | StatementsMedium | |
A medium statements question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Statement I: a secondary halide without bulky groups is open to backside attack, and a high concentration of a strong nucleophile favours the bimolecular path — $\mathrm{S_N2}$. True.
Statement II: in a large excess of ethanol (a weak nucleophile and a polar protic solvent) the reaction is solvolysis; the solvent stabilises the carbocation and there is no strong nucleophile, so the secondary halide reacts by $\mathrm{S_N1}$. True.
Answer: (A). | ||||
| 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 Reimer–Tiemann reaction: phenol with chloroform and aqueous NaOH (then acid) gives salicylaldehyde (2-hydroxybenzaldehyde). The electrophile is dichlorocarbene $:\mathrm{CCl_2}$ formed from $\mathrm{CHCl_3}$; the $\mathrm{-CHCl_2}$ group introduced ortho is hydrolysed to $\mathrm{-CHO}$.
With $\mathrm{CO_2}$/NaOH (Kolbe) the product is salicylic acid; with $\mathrm{CCl_4}$/NaOH it is also salicylic acid.
Answer: (C). | ||||
| 77 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The alkene is 1-methylcyclohexene (the C=C is in the ring at the carbon carrying $\mathrm{CH_3}$).
$\mathrm{H^+/H_2O}$: acid-catalysed hydration follows Markovnikov's rule through the tertiary carbocation, so OH goes to the carbon bearing $\mathrm{CH_3}$: A = 1-methylcyclohexanol.
$\mathrm{B_2H_6}$, then $\mathrm{H_2O_2}$/NaOH: hydroboration–oxidation adds H and OH anti-Markovnikov (syn addition), so OH goes to the neighbouring ring carbon: B = 2-methylcyclohexanol (trans to the methyl).
Answer: (A). | ||||
| 78 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids m-Chlorobenzaldehyde has no α-hydrogen, so with concentrated (50%) alkali it undergoes the Cannizzaro reaction: one molecule is oxidised to the carboxylate, another reduced to the alcohol.
$\mathrm{2\,ClC_6H_4CHO + KOH \rightarrow ClC_6H_4COO^-K^+ + ClC_6H_4CH_2OH}$.
The aryl C–Cl bond is not attacked by KOH under these conditions, so Cl stays meta in both products.
Option (A) is the benzoin product (needs cyanide); (B) and (D) wrongly replace Cl by OH.
Answer: (C). | ||||
| 79 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Benzene with conc. $\mathrm{HNO_3}$/conc. $\mathrm{H_2SO_4}$ at 323–333 K gives nitrobenzene; Sn/HCl reduces it to aniline. A = aniline.
Aniline with $\mathrm{NaNO_2}$/HCl at 273–278 K gives benzenediazonium chloride, which couples with phenol (in mildly alkaline medium) at the para position: B = p-hydroxyazobenzene, $\mathrm{HO-C_6H_4-N{=}N-C_6H_5}$ (an orange dye).
Para coupling is preferred over ortho for steric reasons, so (D) is not the product.
Answer: (C). | ||||
| 80 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Chromyl chloride is hydrolysed by alkali to sodium chromate: $\mathrm{CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O}$ (Cr stays +6; atoms balance). A = $\mathrm{Na_2CrO_4}$.
Acidified chromate with $\mathrm{H_2O_2}$ gives the blue peroxide $\mathrm{CrO_5}$ (Cr(VI) with two peroxo groups): $\mathrm{Na_2CrO_4 + 2HCl + 2H_2O_2 \rightarrow CrO_5 + 2NaCl + 3H_2O}$. B = $\mathrm{CrO_5}$.
Answer: (D). | ||||
| 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 Fifth excited state: $n = 6$; first excited state: $n = 2$.
Lines from all possible downward jumps between levels 6 and 2: $\frac{\Delta n(\Delta n + 1)}{2}$ with $\Delta n = 4$, i.e. $\frac{4\times 5}{2} = 10$.
(Count: from 6 — 4 lines, from 5 — 3, from 4 — 2, from 3 — 1; total 10.)
Answer: 10. | ||||
| 82 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Target = 3 × (second reaction) − (first reaction):
$\mathrm{3C + \frac{3}{2}O_2 \rightarrow 3CO}$, $\Delta H = -330$ kJ; reversing the first, $\mathrm{Fe_2O_3 \rightarrow 2Fe + \frac{3}{2}O_2}$, $\Delta H = +822$ kJ.
Adding: $\mathrm{3C + Fe_2O_3 \rightarrow 2Fe + 3CO}$, $\Delta H = -330 + 822 = +492$ kJ/mol.
(Same from enthalpies of formation: $3(-110) - (-822) = 492$.)
Answer: 492. | ||||
| 83 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Henderson equation: $\mathrm{pH = pK_a + \log\frac{[salt]}{[acid]}}$, so $\log\frac{[salt]}{[acid]} = 4.5 - 4.2 = 0.3 = \log 2$.
Both stock solutions are 1 M, so the ratio of moles equals the ratio of volumes: $V_{\text{salt}} = 2V_{\text{acid}}$.
$V_{\text{acid}} + 2V_{\text{acid}} = 300$ mL, so $V_{\text{acid}} = 100$ mL.
Answer: 100. | ||||
| 84 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions A species can disproportionate if its element is in an intermediate oxidation state with both higher and lower states available.
$\mathrm{H_2O_2}$ (O: −1 → 0 and −2): yes. $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$.
$\mathrm{ClO_3^-}$ (Cl: +5 → +7 and −1): yes. $\mathrm{4ClO_3^- \rightarrow 3ClO_4^- + Cl^-}$.
$\mathrm{P_4}$ (0 → −3 and +1): yes, with alkali gives $\mathrm{PH_3}$ and $\mathrm{H_2PO_2^-}$.
$\mathrm{Cl_2}$ (0 → −1 and +1 or +5): yes, with alkali.
$\mathrm{Cu^+}$ (+1 → 0 and +2): yes, in water.
$\mathrm{NO_2}$ (+4 → +3 and +5): yes, $\mathrm{2NO_2 + H_2O \rightarrow HNO_2 + HNO_3}$.
Ag (0, no lower state), $\mathrm{F_2}$ (no positive state), $\mathrm{K^+}$ (only 0 and +1): no.
Total = 6.
Answer: 6. | ||||
| 85 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate of disappearance of $\mathrm{N_2O_5}$ $= \frac{3 - 2.75}{30} = \frac{0.25}{30} = 8.33\times 10^{-3}$ mol L$^{-1}$ min$^{-1}$.
From the equation, $\frac{1}{4}\frac{d[\mathrm{NO_2}]}{dt} = -\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt}$, so $\frac{d[\mathrm{NO_2}]}{dt} = 2\times 8.33\times 10^{-3} = 16.67\times 10^{-3}$ mol L$^{-1}$ min$^{-1}$.
$x = 16.67$, nearest integer 17.
Answer: 17. | ||||
| 86 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Octahedral complexes of the types $\mathrm{[M(AA)_3]}$ and cis-$\mathrm{[M(AA)_2B_2]}$ have no plane of symmetry and are chiral; trans-$\mathrm{[M(AA)_2B_2]}$ has a mirror plane and is achiral.
cis-$\mathrm{[Cr(ox)_2Cl_2]^{3-}}$: optically active.
$\mathrm{[Co(en)_3]^{3+}}$: optically active.
cis-$\mathrm{[Pt(en)_2Cl_2]^{2+}}$ (Pt(IV), octahedral): optically active.
cis-$\mathrm{[Co(en)_2Cl_2]^+}$: optically active.
The two trans complexes: not optically active.
Total = 4.
Answer: 4. | ||||
| 87 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Reading the skeletal structure: $\mathrm{CHD{=}C(CH_3)-CH_2-C({=}CH-CH_3)-CH_2-C(CH_3){=}CHD}$ — two identical terminal $\mathrm{C{=}CHD}$ units joined through $\mathrm{CH_2}$ groups to a central carbon that carries an ethylidene ($\mathrm{{=}CH-CH_3}$) group.
Each terminal double bond is stereogenic: one carbon has H and D, the other $\mathrm{CH_3}$ and $\mathrm{CH_2R}$. Call their configurations $a$ and $b$ (E or Z).
The central double bond has H and $\mathrm{CH_3}$ on one carbon; on the other carbon are the two $\mathrm{CH_2}$ arms, which are identical in constitution and differ only if $a \neq b$.
If $a = b$ (EE or ZZ): the two arms are identical, the central bond is not stereogenic — 2 isomers.
If $a \neq b$ (EZ, the same as ZE): the arms differ, and the methyl of the ethylidene group can lie on the side of the E-arm or of the Z-arm — 2 isomers.
Total = 2 + 2 = 4.
Answer: 4. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Free-radical chlorination of $\mathrm{CH_3CHClCH_2CH_3}$ replaces one H on C1, C2, C3 or C4:
C1: 1,2-dichlorobutane, $\mathrm{ClCH_2CHClCH_2CH_3}$ — one stereocentre (C2): 2 optically active forms.
C2: 2,2-dichlorobutane — no stereocentre: 0.
C3: 2,3-dichlorobutane — two like stereocentres: (2R,3R) and (2S,3S) optically active, the meso form inactive: 2.
C4: 1,3-dichlorobutane, $\mathrm{CH_3CHClCH_2CH_2Cl}$ — one stereocentre (C2): 2.
Total optically active isomers $= 2 + 0 + 2 + 2 = 6$.
Answer: 6. | ||||
| 89 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules A. False — DNA, not RNA, is the reserve of genetic information.
B. True — DNA replicates itself during cell division.
C. False — proteins are synthesised by RNA (mRNA, tRNA, rRNA on ribosomes); DNA only carries the code.
D. True — the message for a particular protein is coded in DNA.
E. True — identical DNA copies pass to the daughter cells.
Correct statements: B, D, E = 3.
Answer: 3. | ||||
| 90 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Flame colours: $\mathrm{Sr^{2+}}$ crimson red, $\mathrm{Ba^{2+}}$ apple green, $\mathrm{Ca^{2+}}$ brick red, $\mathrm{Cu^{2+}}$ green / blue-green.
$\mathrm{Zn^{2+}}$, $\mathrm{Co^{2+}}$ and $\mathrm{Fe^{2+}}$ give no characteristic flame colour and are identified by other tests.
Total = 4.
Answer: 4. | ||||
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