JEE (Main) · 2024
JEE (Main) 2024 — 5 April, 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
- Integral Calculus3
- Permutations and Combinations2
- Matrices and Determinants2
10 easy · 20 medium · 0 hard
Physics
30 Q
- Physical World and Measurement2
- Kinematics2
- Laws of Motion2
23 easy · 7 medium · 0 hard
Chemistry
30 Q
- Alcohols, Phenols and Ethers4
- Some Basic Concepts of Chemistry2
- Chemical Bonding and Molecular Structure2
21 easy · 8 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $f(1)$ and $f(3)$ are distinct elements of B with sum 14: the pairs are $\{2, 12\}$ and $\{4, 10\}$ ($7 + 7$ is not allowed, since f is one-one). Each pair can be assigned in 2 orders: $4$ ways.
The remaining three elements 7, 9, 11 of A go one-one into the remaining 5 elements of B: $5\times4\times3 = 60$ ways.
Total $= 4\times60 = 240$.
Answer: (C). | ||||
| 2 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | StatementsMedium | |
A medium statements question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Statement I: $\frac{z_1}{|z_1|}$ and $\frac{z_2}{|z_2|}$ are unit complex numbers, so by the triangle inequality $\left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| \le 1 + 1 = 2$. Multiplying by the positive number $|z_1| + |z_2|$ gives the inequality. Correct.
Statement II: let the common ratio be $k > 0$, so $a = k|y - z|$, $b = k|z - x|$, $c = k|x - y|$.
For any non-zero $w$, $\frac{|w|^2}{w} = \frac{w\bar{w}}{w} = \bar{w}$. Hence
$\frac{a^2}{y - z} + \frac{b^2}{z - x} + \frac{c^2}{x - y} = k^2\left(\overline{y - z} + \overline{z - x} + \overline{x - y}\right) = k^2\cdot\bar{0} = 0 \ne 1$. Incorrect.
Answer: (C). | ||||
| 3 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For order $n = 3$: $|kM| = k^3|M|$ and $|\mathrm{adj}\,M| = |M|^{2}$.
$|A^T| = |A| = 3$.
$|\mathrm{adj}(2A)| = |2A|^2 = (8\times3)^2 = 576$, so $|(\mathrm{adj}(2A))^{-1}| = \frac{1}{576}$.
$|\mathrm{adj}(4B)| = |4B|^2 = (64\times2)^2 = 16384$.
$|\mathrm{adj}(AB)| = |AB|^2 = 36$, so $|(\mathrm{adj}(AB))^{-1}| = \frac{1}{36}$.
The determinant of the product is the product of determinants:
$3\cdot3\cdot\frac{1}{576}\cdot16384\cdot\frac{1}{36}\cdot3\cdot3 = \frac{81\times16384}{576\times36} = \frac{81\times16384}{20736} = 64$.
Answer: (B). | ||||
| 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 For infinitely many solutions the coefficient determinant must vanish:
$\begin{vmatrix}11&1&\lambda\\2&3&5\\8&-19&-39\end{vmatrix} = 11(-117+95) - 1(-78-40) + \lambda(-38-24) = -242 + 118 - 62\lambda = -124 - 62\lambda = 0$, so $\lambda = -2$.
Then the third equation must be a combination $p(\text{first}) + q(\text{second})$:
x: $11p + 2q = 8$; y: $p + 3q = -19$. So $p = -19 - 3q$, $11(-19 - 3q) + 2q = 8$, $-31q = 217$, $q = -7$, $p = 2$.
z: $\lambda p + 5q = -4 - 35 = -39$, which agrees.
Constants: $\mu = -5p + 3q = -10 - 21 = -31$.
$\lambda^4 - \mu = 16 + 31 = 47$.
Answer: (B). | ||||
| 5 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | StatementsMedium | |
A medium statements question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = \cos x + 3 - \frac{2}{\pi}(2x + 1)$ and $f''(x) = -\sin x - \frac{4}{\pi} < 0$ on $\left(0, \frac{\pi}{2}\right)$.
So $f'$ is decreasing there: (II) is true.
Since $f'$ decreases, its least value on the interval is approached at $x = \frac{\pi}{2}$: $f'\left(\frac{\pi}{2}\right) = 0 + 3 - \frac{2}{\pi}(\pi + 1) = 1 - \frac{2}{\pi} > 0$.
Hence $f'(x) > 0$ throughout, and $f$ is increasing: (I) is true.
Answer: (D). | ||||
| 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 $\frac{1}{\sqrt{k} + \sqrt{k+1}} = \sqrt{k+1} - \sqrt{k}$, so the first sum telescopes: $m = \sqrt{100} - \sqrt{1} = 9$.
$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$, so $n = 1 - \frac{1}{100} = \frac{99}{100}$.
Check (A): $11\times9 - 100\times\frac{99}{100} = 99 - 99 = 0$. It lies on $11x - 100y = 0$.
(B) gives $88 - 99 \ne 0$; (C) gives $77 + 1 \ne 0$; (D) gives $88 + 101 \ne 0$.
Answer: (A). | ||||
| 7 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For the limit at 0 to be finite, the numerator must vanish to order $x^3$. Expand:
$\sin3x = 3x - \frac{9x^3}{2} + \ldots$, $\alpha\sin x = \alpha x - \frac{\alpha x^3}{6} + \ldots$, $-\beta\cos3x = -\beta + \frac{9\beta x^2}{2} - \ldots$
Constant term: $-\beta = 0$, so $\beta = 0$.
$x$ term: $3 + \alpha = 0$, so $\alpha = -3$.
$x^3$ term: $-\frac{9}{2} + \frac{3}{6} = -4$.
So $\lim_{x\to0} f(x) = -4$, and continuity gives $f(0) = -4$.
Answer: (D). | ||||
| 8 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy 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 $f(1) = 1 + 2 + 3 + 1 = 7$, so $g(7) = 1$.
Differentiating $g(f(x)) = x$: $g'(f(x))\,f'(x) = 1$. With $f'(x) = 5x^4 + 6x^2 + 3$, $f'(1) = 14$, so $g'(7) = \frac{1}{14}$.
$\frac{g(7)}{g'(7)} = 14$.
Answer: (C). | ||||
| 9 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let side AB (length 4) make an angle $\theta$ with side PQ. Each side of PQRS is made of the projections of two adjacent sides of ABCD:
$a = 4\cos\theta + 2\sin\theta$, $b = 4\sin\theta + 2\cos\theta$.
Area $= ab = 16\sin\theta\cos\theta + 8\cos^2\theta + 8\sin^2\theta + 4\sin\theta\cos\theta = 8 + 10\sin2\theta$.
This is greatest when $\sin2\theta = 1$, $\theta = \frac{\pi}{4}$: $a = b = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.
$(a + b)^2 = (6\sqrt{2})^2 = 72$.
Answer: (A). | ||||
| 10 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Split: $\frac{2y}{1 + \cos^2y}$ is odd and integrates to 0 over $[-\pi, \pi]$; $\frac{2y\sin y}{1 + \cos^2y}$ is even.
$I = 2\int_0^{\pi}\frac{2y\sin y}{1 + \cos^2y}dy = 4J$, where $J = \int_0^{\pi}\frac{y\sin y}{1 + \cos^2y}dy$.
Replacing $y$ by $\pi - y$: $J = \int_0^{\pi}\frac{(\pi - y)\sin y}{1 + \cos^2y}dy$, so $2J = \pi\int_0^{\pi}\frac{\sin y}{1 + \cos^2y}dy$.
With $t = \cos y$: $\int_0^{\pi}\frac{\sin y}{1 + \cos^2y}dy = \int_{-1}^{1}\frac{dt}{1 + t^2} = \frac{\pi}{2}$. So $J = \frac{\pi^2}{4}$.
$I = 4\cdot\frac{\pi^2}{4} = \pi^2$.
Answer: (A). | ||||
| 11 | 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 Write the numerator as a combination of the denominator and its derivative:
$136\sin x = A(3\sin x + 5\cos x) + B(3\cos x - 5\sin x)$.
$\sin x$: $3A - 5B = 136$; $\cos x$: $5A + 3B = 0$. So $B = -\frac{5A}{3}$, $3A + \frac{25A}{3} = 136$, $A = 12$, $B = -20$.
$\int\frac{136\sin x}{3\sin x + 5\cos x}dx = 12x - 20\ln|3\sin x + 5\cos x|$.
From 0 to $\frac{\pi}{4}$: $3\pi - 20\left[\ln\frac{8}{\sqrt{2}} - \ln5\right] = 3\pi - 20\ln(4\sqrt{2}) + 20\ln5 = 3\pi - 50\ln2 + 20\ln5$.
Answer: (D). | ||||
| 12 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor $e^{2x}$: $\frac{d}{dx}(ye^{2x}) = e^{2x}\sin2x$.
$\int e^{2x}\sin2x\,dx = \frac{e^{2x}(2\sin2x - 2\cos2x)}{8} = \frac{e^{2x}(\sin2x - \cos2x)}{4}$.
So $y = \frac{\sin2x - \cos2x}{4} + Ce^{-2x}$. $y(0) = -\frac{1}{4} + C = \frac{3}{4}$ gives $C = 1$.
At $x = \frac{\pi}{8}$: $\sin\frac{\pi}{4} = \cos\frac{\pi}{4}$, so $y = e^{-\pi/4}$.
Answer: (A). | ||||
| 13 | 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 The lines are $x = 3$ and $y = 2$. A circle of radius 1 touching both has centre $(3 \pm 1, 2 \pm 1)$. The one closest to the origin has centre $(2, 1)$.
Distance from $(5, 5)$ to the centre: $\sqrt{9 + 16} = 5$. The point is outside, so the shortest distance to the circle is $5 - 1 = 4$.
Answer: (C). | ||||
| 14 | 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 The triangle is right-angled and isosceles at O, with hypotenuse PQ on the line $3x + 4y = 12$.
The altitude from O to PQ is $p = \frac{12}{5}$. In a right isosceles triangle the altitude to the hypotenuse is half the hypotenuse: $PQ = 2p = \frac{24}{5}$.
By Pythagoras $OP^2 + OQ^2 = PQ^2$, so $l = 2PQ^2 = 2\cdot\frac{576}{25} = \frac{1152}{25} = 46.08$.
$[l] = 46$.
Answer: (D). | ||||
| 15 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $A = \left(\frac{k}{2}, 0\right)$, $B = \left(0, \frac{k}{3}\right)$. The circle on AB as diameter: $x\left(x - \frac{k}{2}\right) + y\left(y - \frac{k}{3}\right) = 0$, i.e. $x^2 + y^2 - \frac{k}{2}x - \frac{k}{3}y = 0$.
Comparing with $x^2 + y^2 - 3x - 2y = 0$: $k = 6$.
Ellipse $x^2 + 9y^2 = 36$: $\frac{x^2}{36} + \frac{y^2}{4} = 1$, $a = 6$, $b = 2$.
Latus rectum $= \frac{2b^2}{a} = \frac{8}{6} = \frac{4}{3}$, so $m = 4$, $n = 3$.
$2m + n = 11$.
Answer: (B). | ||||
| 16 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Put each line in standard form.
First: $\frac{x - 2}{-3} = \frac{y - \frac{2}{3}}{\frac{4\lambda + 1}{3}} = \frac{z - 4}{-1}$; direction $\left(-3, \frac{4\lambda + 1}{3}, -1\right)$.
Second: $\frac{x + 3}{3\mu} = \frac{y - \frac{1}{2}}{-3} = \frac{z - 5}{-7}$; direction $(3\mu, -3, -7)$.
Perpendicular: $-9\mu - (4\lambda + 1) + 7 = 0$, so $4\lambda + 9\mu = 6$.
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 General points: $(3s - 6, 2s, s - 1)$ and $(4t + 7, 3t + 9, 2t + 4)$.
From z: $s = 2t + 5$. From y: $2(2t + 5) = 3t + 9$, so $t = -1$, $s = 3$.
Check x: $3(3) - 6 = 3$ and $4(-1) + 7 = 3$. The point is $(3, 6, 2)$.
$d^2 = (7 - 3)^2 + (8 - 6)^2 + (9 - 2)^2 = 16 + 4 + 49 = 69$, so $d^2 + 6 = 75$.
Answer: (C). | ||||
| 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 The four points are coplanar: with $\vec{AB} = (4, 8, -8)$, $\vec{AC} = (2, 5, -12)$, $\vec{AD} = (-2, -3, -4)$, $\vec{AD}\times\vec{AC} = (56, -32, -4)$ and $\vec{AB}\cdot(56, -32, -4) = 224 - 256 + 32 = 0$.
For a plane quadrilateral, area $= \frac{1}{2}|\vec{AC}\times\vec{BD}|$.
$\vec{BD} = (-6, -11, 4)$. $\vec{AC}\times\vec{BD} = (5\cdot4 - (-12)(-11),\ (-12)(-6) - 2\cdot4,\ 2(-11) - 5(-6)) = (-112, 64, 8) = 8(-14, 8, 1)$.
$|\vec{AC}\times\vec{BD}| = 8\sqrt{196 + 64 + 1} = 8\sqrt{261} = 24\sqrt{29}$.
Area $= 12\sqrt{29}$.
Answer: (A). | ||||
| 19 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each of a, b, c is chosen from 8 values (repetition allowed): $8^3 = 512$ equally likely equations.
Repeated roots need $b^2 = 4ac$, so b is even and $ac = \frac{b^2}{4}$:
$b = 2$: $ac = 1$: (1, 1) — 1 way.
$b = 4$: $ac = 4$: (1, 4), (4, 1), (2, 2) — 3 ways.
$b = 6$: $ac = 9$: (3, 3) — 1 way.
$b = 8$: $ac = 16$: (2, 8), (8, 2), (4, 4) — 3 ways.
Favourable: 8. Probability $= \frac{8}{512} = \frac{1}{64}$.
Answer: (D). | ||||
| 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 Let $c = \cos\theta$; on $\left[0, \frac{\pi}{4}\right]$, $\sin\theta = \sqrt{1 - c^2}$.
$4c - 1 = 3\sqrt{1 - c^2}$; squaring: $16c^2 - 8c + 1 = 9 - 9c^2$, so $25c^2 - 8c - 8 = 0$.
$c = \frac{8 \pm \sqrt{64 + 800}}{50} = \frac{8 \pm 12\sqrt{6}}{50} = \frac{4 \pm 6\sqrt{6}}{25}$.
On $\left[0, \frac{\pi}{4}\right]$, $c \ge \frac{1}{\sqrt{2}}$, so $c = \frac{4 + 6\sqrt{6}}{25} \approx 0.748$ (and $4c - 1 > 0$, so the squaring added nothing).
Now $\frac{4}{3\sqrt{6} - 2} = \frac{4(3\sqrt{6} + 2)}{54 - 4} = \frac{12\sqrt{6} + 8}{50} = \frac{6\sqrt{6} + 4}{25}$, which matches.
Answer: (B). | ||||
| 21 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical 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 Write $a = n + f$ with $n = [a]$, $0 \le f < 1$. The right side is $3n + 2f$.
Case $a \ge \frac{1}{2}$: $2n + 2f - 1 = 3n + 2f$ gives $n = -1$, so $a < 0$: impossible.
Case $a < \frac{1}{2}$: $1 - 2n - 2f = 3n + 2f$ gives $f = \frac{1 - 5n}{4}$. For $0 \le f < 1$: $-\frac{3}{5} < n \le \frac{1}{5}$, so $n = 0$ and $f = \frac{1}{4}$: $a = \frac{1}{4}$ (check: $|{-\frac{1}{2}}| = \frac{1}{2} = 0 + \frac{1}{2}$).
$S = \left\{\frac{1}{4}\right\}$, so $72\sum a = 18$.
Answer: 18. | ||||
| 22 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $t = x + 1$. Then $|x||x + 2| = |(t - 1)(t + 1)| = |t^2 - 1|$, and the equation is $|t^2 - 1| - 5|t| - 1 = 0$, which is even in t.
For $t \ge 0$:
If $t \ge 1$: $t^2 - 1 = 5t + 1$, $t^2 - 5t - 2 = 0$, $t = \frac{5 + \sqrt{33}}{2}$ (the other root is negative).
If $0 \le t < 1$: $1 - t^2 = 5t + 1$, $t(t + 5) = 0$, $t = 0$.
By symmetry $t = -\frac{5 + \sqrt{33}}{2}$ also works. So $t \in \left\{0, \pm\frac{5 + \sqrt{33}}{2}\right\}$: three values, giving three distinct real x.
Answer: 3. | ||||
| 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 Count ordered $(x_1, x_2, x_3, x_4)$ with $1 \le x_i \le 6$ and sum 16. Put $y_i = x_i - 1 \in \{0, \ldots, 5\}$, sum 12.
Without the upper limit: $\binom{15}{3} = 455$.
Subtract those with some $y_i \ge 6$: $4\binom{9}{3} = 336$; add back those with two $y_i \ge 6$ (the rest summing to 0): $\binom{4}{2}\binom{3}{3} = 6$.
$455 - 336 + 6 = 125$.
Answer: 125. | ||||
| 24 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem General term of $\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$: $T_{r+1} = \binom{9}{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18 - 3r}$.
The constant term of the product takes $x^0$ (with 1), $x^{-1}$ (with 2x) and $x^{-3}$ (with $-3x^3$). $18 - 3r = -1$ has no integer solution.
$x^0$: $r = 6$: $\binom{9}{6}\left(\frac{3}{2}\right)^3\frac{1}{3^6} = 84\cdot\frac{27}{8}\cdot\frac{1}{729} = \frac{7}{18}$.
$x^{-3}$: $r = 7$: $\binom{9}{7}\left(\frac{3}{2}\right)^2\left(-\frac{1}{3^7}\right) = -36\cdot\frac{9}{4}\cdot\frac{1}{2187} = -\frac{1}{27}$; times $-3$ gives $\frac{1}{9}$.
$p = \frac{7}{18} + \frac{1}{9} = \frac{1}{2}$, so $108p = 54$.
Answer: 54. | ||||
| 25 | 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 With common difference d: $a_{2j-1}^2 - a_{2j}^2 = (a_{2j-1} - a_{2j})(a_{2j-1} + a_{2j}) = -d(a_{2j-1} + a_{2j})$.
So $A_k = -d\,S_{2k} = -dk(2a_1 + (2k - 1)d)$.
$A_3 = -153$: $d(2a_1 + 5d) = 51$. $A_5 = -435$: $d(2a_1 + 9d) = 87$.
Subtracting: $4d^2 = 36$, $d = \pm3$. With $d = 3$: $2a_1 + 15 = 17$, $a_1 = 1$. ($d = -3$ gives $a_1 = -1$, not positive.)
Check: $1 + 16 + 49 = 66$.
$a_{17} = 1 + 16\cdot3 = 49$; $A_7 = -3\cdot7(2 + 13\cdot3) = -861$.
$a_{17} - A_7 = 49 + 861 = 910$.
Answer: 910. | ||||
| 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 The limit is of the form $\frac{0}{0}$; by L'Hôpital's rule (differentiating in t): $2xf(x) - x^2f'(x) = 1$.
So $x^2f' - 2xf = -1$, i.e. $\frac{d}{dx}\left(\frac{f}{x^2}\right) = \frac{x^2f' - 2xf}{x^4} = -\frac{1}{x^4}$.
$\frac{f}{x^2} = \frac{1}{3x^3} + C$; $f(1) = 1$ gives $C = \frac{2}{3}$. So $f(x) = \frac{1}{3x} + \frac{2x^2}{3}$.
$f(2) = \frac{1}{6} + \frac{8}{3} = \frac{17}{6}$; $f(3) = \frac{1}{9} + 6 = \frac{55}{9}$.
$2f(2) + 3f(3) = \frac{17}{3} + \frac{55}{3} = 24$.
Answer: 24. | ||||
| 27 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersections: $x^2 - 5x = 7x - x^2$, $2x^2 - 12x = 0$, $x = 0, 6$. Between them $7x - x^2$ is the upper curve.
Area $= \int_0^6\left[(7x - x^2) - (x^2 - 5x)\right]dx = \int_0^6(12x - 2x^2)dx = 216 - 144 = 72$.
Answer: 72. | ||||
| 28 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = 12x$: $a = 3$, focus $(3, 0)$.
A focal chord of slope $m$ has length $l = 4a(1 + \cot^2\alpha) = \frac{12(1 + m^2)}{m^2}$, where $\tan\alpha = m$.
The chord $y = m(x - 3)$ is at distance $d = \frac{3|m|}{\sqrt{1 + m^2}}$ from the origin, so $d^2 = \frac{9m^2}{1 + m^2}$.
$ld^2 = 12\times9 = 108$, whatever m is.
Answer: 108. | ||||
| 29 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $(\vec{a} + 2\vec{b})\times\vec{c} + 3(\vec{a}\times\vec{c}) = \vec{0}$, so $(4\vec{a} + 2\vec{b})\times\vec{c} = \vec{0}$: $\vec{c} = \lambda(2\vec{a} + \vec{b}) = \lambda(4\hat{i} - 7\hat{j} + 15\hat{k})$.
$\vec{a}\cdot\vec{c} = \lambda(4 + 21 + 105) = 130\lambda = 130$, so $\lambda = 1$.
$\vec{b}\cdot\vec{c} = 8 + 7 + 15 = 30$.
Answer: 30. | ||||
| 30 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability The sample is drawn without replacement. $P(X = k) = \frac{\binom{3}{k}\binom{7}{5-k}}{\binom{10}{5}}$, with $\binom{10}{5} = 252$:
$P(0) = \frac{21}{252}$, $P(1) = \frac{105}{252}$, $P(2) = \frac{105}{252}$, $P(3) = \frac{21}{252}$.
$E(X) = \frac{105 + 210 + 63}{252} = \frac{378}{252} = \frac{3}{2}$.
$E(X^2) = \frac{105 + 420 + 189}{252} = \frac{714}{252} = \frac{17}{6}$.
$\sigma^2 = \frac{17}{6} - \frac{9}{4} = \frac{7}{12}$, so $96\sigma^2 = 56$.
Answer: 56. | ||||
| 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 Energy density: $[u] = \frac{\mathrm{ML^2T^{-2}}}{\mathrm{L^3}} = [\mathrm{ML^{-1}T^{-2}}]$.
From $F = \frac{Gm_1m_2}{r^2}$: $[G] = [\mathrm{M^{-1}L^3T^{-2}}]$.
$[uG] = [\mathrm{L^2T^{-4}}]$, so $[\sqrt{uG}] = [\mathrm{LT^{-2}}]$, an acceleration.
Force per unit mass is an acceleration: $[\mathrm{LT^{-2}}]$. Gravitational potential and energy per unit mass are $[\mathrm{L^2T^{-2}}]$; pressure gradient per unit mass is $\frac{\mathrm{ML^{-2}T^{-2}}}{\mathrm{M}} = [\mathrm{L^{-2}T^{-2}}]$.
Answer: (A). | ||||
| 32 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Resultant: $(2\vec{Q} + 2\vec{P}) + (2\vec{Q} - 2\vec{P}) = 4\vec{Q}$.
$4\vec{Q}$ points the same way as $\vec{Q}$, so the angle is $0^\circ$.
Answer: (D). | ||||
| 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 Take the block and cylinder together, mass $5 + 25 = 30$ kg, accelerating downward at $a = 0.1\ \mathrm{m\,s^{-2}}$.
Downward: $Mg - N = Ma$, so $N = M(g - a) = 30(9.8 - 0.1) = 291$ N.
By Newton's third law the system presses on the floor with the same force, 291 N.
Answer: (C). | ||||
| 34 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Gravity is a conservative force: the work done against it depends only on the change in height, not on the path.
Case 1 and Case 2 both raise the 50 kg body through $h = 20$ m, so each needs $mgh = 50\times10\times20 = 10^4$ J against gravity (taking $g = 10\ \mathrm{m\,s^{-2}}$).
Ratio $1 : 1$.
Answer: (D). | ||||
| 35 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Hollow sphere of radius R about a diameter: $I_1 = \frac{2}{3}MR^2$, so $k_1^2 = \frac{2}{3}R^2$.
Solid cylinder of radius R and length $L = 4R$: the axis AB is a diameter of its end face (perpendicular to the cylinder's axis, through the end).
About a parallel axis through the centre, $I_c = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right)$; shifting by $\frac{L}{2}$: $I_2 = M\left(\frac{R^2}{4} + \frac{L^2}{3}\right) = M\left(\frac{R^2}{4} + \frac{16R^2}{3}\right) = \frac{67}{12}MR^2$.
$\frac{k_1^2}{k_2^2} = \frac{2/3}{67/12} = \frac{8}{67}$, so $\frac{k_1}{k_2} = \sqrt{\frac{8}{67}}$ and $x = 67$.
Answer: (B). | ||||
| 36 | GravitationOrbital motion | Class 11: Gravitation | Matching listEasy | |
A easy matching list question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation For a planet in a circular orbit of radius $a$: $\frac{mv^2}{a} = \frac{GMm}{a^2}$.
(A) Kinetic energy $\frac{1}{2}mv^2 = \frac{GMm}{2a}$ → (II).
(B) Potential energy $-\frac{GMm}{a}$ → (I).
(C) Total energy $\frac{GMm}{2a} - \frac{GMm}{a} = -\frac{GMm}{2a}$ → (IV).
(D) To escape from the planet's surface, a unit mass needs energy equal to the magnitude of its potential energy there: $\frac{Gm}{r}$ → (III).
(A)-(II), (B)-(I), (C)-(IV), (D)-(III).
Answer: (B). | ||||
| 37 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | StatementsEasy | |
A easy statements 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 Capillary rise: $h = \frac{2T\cos\theta}{r\rho g}$.
Statement I: the liquid neither rises nor falls when $h = 0$, i.e. $\cos\theta = 0$, $\theta = 90^\circ$. A contact angle of $0^\circ$ gives the largest rise, not zero rise. False.
Statement II: the contact angle depends on the pair of materials — the solid and the liquid (and the surrounding medium). True.
Answer: (D). | ||||
| 38 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics In a cycle $\Delta U = 0$, so the heat absorbed equals the net work done, which is the area enclosed by the cycle on the P–V diagram.
The cycle is a circle spanning 60 to 340 on both axes: diameter 280 units, radius 140 units.
Area $= \pi(140)^2 = 61\,575$ (kPa)(cc). One kPa·cc is $10^3\ \mathrm{Pa}\times10^{-6}\ \mathrm{m^3} = 10^{-3}$ J, so the area is $61.6$ J.
The cycle runs clockwise with P on the vertical axis, so the net work is done by the system: heat absorbed $= +61.6$ J.
Answer: (C). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Collision frequency $Z = \frac{\bar{v}}{\lambda}$, with $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$.
In a closed chamber the number density $n$ is fixed, so $\lambda$ is unchanged, and $Z \propto \bar{v} \propto \sqrt{T}$.
$\frac{Z'}{Z} = \sqrt{\frac{400}{300}} = \frac{2}{\sqrt{3}}$, so $Z' = \frac{2}{\sqrt{3}}Z$.
Answer: (A). | ||||
| 40 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At height $h$: $g_h = \frac{GM}{(R+h)^2}$, so $T = 2\pi\sqrt{\frac{l}{g_h}} \propto (R + h)$.
At $h = R$: $T_1 \propto 2R$; at $h = 2R$: $T_2 \propto 3R$.
$\frac{T_1}{T_2} = \frac{2}{3}$, i.e. $3T_1 = 2T_2$.
Answer: (C). | ||||
| 41 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Both forces vary as $\frac{1}{r^2}$, so the ratio does not depend on the distance:
$\frac{F_e}{F_g} = \frac{ke^2}{Gm_em_p} = \frac{9\times10^9\times(1.6\times10^{-19})^2}{6.67\times10^{-11}\times9.1\times10^{-31}\times1.67\times10^{-27}} = \frac{2.30\times10^{-28}}{1.01\times10^{-67}} \approx 2.3\times10^{39}$.
Order $10^{39}$.
Answer: (B). | ||||
| 42 | 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 Name the junctions: P is where the top wire meets $R_2$, Q the top-right corner, S the bottom of $R_2$, T the bottom-right corner.
P and Q are joined by a plain wire, and so are S and T. $R_2$ runs from P to S, $R_3$ from Q to T, and $R_4$ (the diagonal) from S to Q. So all three are connected between the same two nodes (top and bottom): they are in parallel.
$\frac{1}{R_p} = \frac{1}{8} + \frac{1}{4} + \frac{1}{8} = \frac{1}{2}$, so $R_p = 2\,\Omega$.
$R_{eq} = R_1 + R_p = 10 + 2 = 12\,\Omega$, and $I = \frac{12}{12} = 1$ A.
Answer: (C). | ||||
| 43 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Take a circular Amperian loop coaxial with the cable: $B\cdot2\pi r = \mu_0 I_{enc}$.
Inside the inner conductor and between the conductors, $I_{enc} \ne 0$, so $B \ne 0$; inside the outer conductor only part of its return current is enclosed, so again $B \ne 0$.
Outside the cable, $I_{enc} = I - I = 0$, so $B = 0$.
Answer: (A). | ||||
| 44 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Let current $I$ flow in the large loop B (radius $b$). At its centre $B_0 = \frac{\mu_0 I}{2b}$; since $b \gg a$, this field is nearly uniform over the small loop A.
Flux through A: $\phi = \frac{\mu_0 I}{2b}\cdot\pi a^2$.
$M = \frac{\phi}{I} = \frac{\mu_0\pi a^2}{2b}$.
Answer: (D). | ||||
| 45 | Electromagnetic WavesDisplacement current | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves The displacement current between the plates equals the conduction current in the leads: $I_0 = \frac{V_0}{X_C} = V_0\,\omega C$.
$\omega = 2\pi\times4000 = 8000\pi\ \mathrm{rad\,s^{-1}}$.
$I_0 = 40\times8000\pi\times12\times10^{-6} = 3.84\pi \approx 12.06$ A $\approx 12$ A.
Answer: (B). | ||||
| 46 | OpticsWave optics | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on wave optics, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics A point source at the focus of a convex lens gives a parallel beam after the lens.
Rays are normal to the wavefront; parallel rays mean the wavefront is plane.
Answer: (D). | ||||
| 47 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | StatementsEasy | |
A easy statements question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation: $eV_0 = h\nu - \phi$, so $V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$. The graph of $V_0$ against $\nu$ is a straight line of slope $\frac{h}{e}$, the same for every material. Statement I is correct.
The line of $M_2$ meets the $\nu$-axis further right: $M_2$ has the larger threshold frequency, hence the larger work function. For the same incident frequency, $K_{max} = h\nu - \phi$ is smaller for $M_2$. Statement II is incorrect.
Answer: (C). | ||||
| 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 For circular motion under the Coulomb force: $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$, so $K = \frac{1}{2}mv^2 = \frac{kZe^2}{2r}$.
$U = -\frac{kZe^2}{r} = -2K$.
$E = K + U = -\frac{kZe^2}{2r} = \frac{U}{2}$, so $2E = U$.
Answer: (C). | ||||
| 49 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Read the gates from the figure.
Input A goes to both inputs of a NOR gate: output $\bar{A}$.
B and C go to a NAND gate: output $\overline{BC}$.
These two feed a NOR gate: $Y_1 = \overline{\bar{A} + \overline{BC}} = A\cdot BC$.
Input D goes to a gate with its inputs tied together, which acts as an inverter: $Y_2 = \bar{D}$.
The bulb (with R) is connected between $Y_1$ and $Y_2$; it glows only when the two ends are at different logic levels.
(A) $A=B=C=1$, $D=0$: $Y_1 = 1$, $Y_2 = 1$: no difference, OFF.
(B) $A=0$: $Y_1 = 0$; $D=1$: $Y_2 = 0$: OFF.
(C) $B=C=0$: $Y_1 = 0$; $D=0$: $Y_2 = 1$: a potential difference across the bulb, ON.
(D) $Y_1 = 0$, $Y_2 = 0$: OFF.
Answer: (C). | ||||
| 50 | Physical World and MeasurementSignificant figures | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on significant figures, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Mean $= \frac{4.62 + 4.632 + 4.6 + 4.64}{4} = \frac{18.492}{4} = 4.623$ s.
In a sum, the result keeps no more decimal places than the least precise term. The reading 4.6 s has only one decimal place, so the sum is known only to one decimal place, and so is the mean.
Rounded to one decimal place: 4.6 s.
Answer: (C). | ||||
| 51 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Starting from rest with uniform acceleration $a$, the distance covered in the $n$th second is $S_n = \frac{a}{2}(2n-1)$.
$\frac{S_{n-1}}{S_n} = \frac{2n-3}{2n-1}$. For $n = 10$: $\frac{17}{19} = 1 - \frac{2}{19}$.
So $x = 19$.
Answer: 19. | ||||
| 52 | Laws of MotionNewton's laws | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Take the three blocks with ropes 2 and 3 as one system of mass $4 + 6 + 10 = 20$ kg. The only external forces are $T_1$ upward and the weight $20g$ downward.
$T_1 - 20g = 20a$, so $T_1 = 20(10 + 2) = 240$ N.
Answer: 240. | ||||
| 53 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy 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 The stress is greatest at the top, where the wire carries its own whole weight: $\sigma = \frac{\rho A L g'}{A} = \rho L g'$.
At the breaking point: $L = \frac{\sigma_b}{\rho g'} = \frac{1.2\times10^8}{6\times10^4\times\frac{10}{3}} = \frac{1.2\times10^8}{2\times10^5} = 600$ m.
Answer: 600. | ||||
| 54 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Parallel: $C_p = 25 + 30 + 45 = 100\ \mu$F.
Series: $\frac{1}{C_s} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45} = \frac{18 + 15 + 10}{450} = \frac{43}{450}$, so $C_s = \frac{450}{43}\ \mu$F.
At the same voltage, energy $\propto C$: $\frac{E_s}{E} = \frac{450/43}{100} = \frac{9}{86}$.
So $x = 86$.
Answer: 86. | ||||
| 55 | 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 The field between the plates is proportional to the charge, which decays as $q = q_0e^{-t/RC}$.
$\frac{1}{3} = e^{-t/RC}$, so $t = RC\ln3$.
$R = \frac{t}{C\ln3} = \frac{6.6\times10^{-6}}{1.5\times10^{-6}\times1.1} = 4\ \Omega$.
Answer: 4. | ||||
| 56 | 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 Length of the wire: $R = \frac{\rho L}{A}$, so $L = \frac{RA}{\rho} = \frac{1\times10\times10^{-6}}{2\times10^{-6}} = 5$ m.
For the wire to hang in the air, the magnetic force (B perpendicular to the wire) balances the weight: $BIL = mg$.
$B = \frac{0.5\times10}{2\times5} = 0.5$ T $= 5\times10^{-1}$ T.
Answer: 5. | ||||
| 57 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current From $V = 50\sqrt{2}\sin100t$: $V_{rms} = 50$ V and $\omega = 100\ \mathrm{rad\,s^{-1}}$.
$X_L = \omega L = 100\ \Omega$; $X_C = \frac{1}{\omega C} = \frac{1}{100\times20\times10^{-6}} = 500\ \Omega$.
$Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{300^2 + 400^2} = 500\ \Omega$.
$I_{rms} = \frac{50}{500} = 0.1$ A, so $V_C = I_{rms}X_C = 0.1\times500 = 50$ V.
Answer: 50. | ||||
| 58 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Fringe width $\beta = \frac{\lambda D}{d} = \frac{5\times10^{-7}\times2}{3\times10^{-4}} = \frac{10}{3}\times10^{-3}$ m.
The third maximum is at $x = 3\beta = 10\times10^{-3}$ m $= 10$ mm.
Answer: 10. | ||||
| 59 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $3\,{}^4\mathrm{He} \rightarrow {}^{12}\mathrm{C}$. The atomic mass of $^{12}\mathrm{C}$ is exactly 12 u by definition of the unit (electron masses cancel when atomic masses are used on both sides).
Mass defect: $\Delta m = 3\times4.002603 - 12.000000 = 0.007809$ u.
Energy released: $0.007809\times931 = 7.270$ MeV $= 727.0\times10^{-2}$ MeV.
Answer: 727. | ||||
| 60 | Experimental SkillsElectricity experiments | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on electricity experiments, from Experimental Skills. NCERT: No matching NCERT chapter With the resistance box R in series with the galvanometer G and the cell (emf E, internal resistance neglected): $I = \frac{E}{R + G} = k\theta$, where $k$ is the figure of merit.
So $\frac{1}{\theta} = \frac{k}{E}R + \frac{kG}{E}$: a straight line in R.
From the graph, the points (2, 1), (4, 1.5), (6, 2) give slope $\frac{k}{E} = \frac{0.5}{2} = 0.25\ \mathrm{div^{-1}\,\Omega^{-1}}$, and the intercept is 0.5 (so $G = 2\,\Omega$).
$k = 0.25\times2 = 0.5$ A/division $= 5\times10^{-1}$ A/division.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryLaws of chemical combination | Class 11: Some Basic Concepts of Chemistry | StatementsEasy | |
A easy statements question on laws of chemical combination, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Dalton's theory says: matter consists of indivisible atoms; all atoms of a given element have identical properties, including identical mass; atoms of different elements differ in mass; compounds form when atoms of different elements combine in a fixed ratio; chemical reactions involve reorganisation of atoms.
(A), (C) and (E) are Dalton's postulates as stated, so they are correct.
(B) says atoms are divisible: contrary to Dalton, so incorrect.
(D) says atoms of one element have different properties including mass: contrary to Dalton, so incorrect.
The incorrect ones are (B) and (D) only.
Answer: (C). | ||||
| 62 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Ethylene is $\mathrm{H_2C{=}CH_2}$.
Four C–H single bonds: 4 $\sigma$.
The C=C double bond: 1 $\sigma$ + 1 $\pi$.
Total: 5 $\sigma$ and 1 $\pi$.
Answer: (B). | ||||
| 63 | EquilibriumLe Chatelier's principle | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on le chatelier's principle, from Equilibrium. NCERT: Class 11: Equilibrium $K_p = \frac{p_{\mathrm{CO_2}}^3}{p_{\mathrm{CO}}^3}$: the pure solid $\mathrm{Fe_2O_3}$ and the pure liquid Fe do not appear in the equilibrium expression.
Removing CO, or adding or removing $\mathrm{CO_2}$, changes the reaction quotient, so the equilibrium shifts.
Adding more solid $\mathrm{Fe_2O_3}$ does not change its activity (it stays 1), so the reaction quotient is unchanged and the equilibrium is not disturbed.
Answer: (D). | ||||
| 64 | ElectrochemistryBatteries and fuel cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on batteries and fuel cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Clocks commonly use the dry (Leclanché) cell. Its cathode reaction is
$\mathrm{MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3}$.
In $\mathrm{MnO_2}$ manganese is $+4$; in $\mathrm{MnO(OH)}$ it is $+3$. A cathode always carries out reduction.
So the cathode reaction is the reduction of Mn from $+4$ to $+3$.
Answer: (B). | ||||
| 65 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry A divalent cation $\mathrm{M^{2+}}$ and a divalent anion $\mathrm{X^{2-}}$ form the electrolyte MX, which gives one cation and one anion per formula unit.
By Kohlrausch's law, $\Lambda_m = \nu_+\lambda_+ + \nu_-\lambda_- = 1\times 57 + 1\times 73 = 130\ \mathrm{S\,cm^2\,mol^{-1}}$.
Answer: (A). | ||||
| 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 (A) Each species has 10 electrons: O (8+2), F (9+1), Na (11−1), Mg (12−2). True.
(B) Nuclear charges are +8, +9, +11, +12: different. False.
(C), (D) For isoelectronic species, the radius falls as nuclear charge rises: $\mathrm{O^{2-} > F^- > Na^+ > Mg^{2+}}$. So $\mathrm{O^{2-}}$ is the largest and $\mathrm{Mg^{2+}}$ the smallest. Both true.
Correct: (A), (C) and (D) only.
Answer: (B). | ||||
| 67 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: because of the inert pair effect, the $ns^2$ pair is held more and more tightly down group 13, so the $+1$ state becomes more stable (Tl(I) is more stable than Tl(III)). Correct.
Statement II: gallium's atomic radius (about 135 pm) is a little smaller than aluminium's (about 143 pm): the ten 3d electrons in Ga shield the nucleus poorly, so the extra nuclear charge pulls the shell in. Incorrect.
Answer: (C). | ||||
| 68 | 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 Boron's isotopes are $^{10}\mathrm{B}$ (about 20%) and $^{11}\mathrm{B}$ (about 80%). The more abundant is $^{11}\mathrm{B}$: neutrons $= 11 - 5 = 6$, so $x = 6$.
Amorphous boron burnt in air gives $\mathrm{B_2O_3}$ (with some BN); in both boron is $+3$, so $y = 3$.
$x + y = 9$.
Answer: (C). | ||||
| 69 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Manganese ($3d^5 4s^2$) can use all seven of its 4s and 3d electrons, so it shows oxidation states from $+2$ up to $+7$ (as in $\mathrm{KMnO_4}$): the highest state and the largest number of states in the first transition series.
Ti reaches only $+4$, Fe commonly $+2$, $+3$ (up to $+6$ rarely), Co $+2$, $+3$.
Answer: (C). | ||||
| 70 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Spectrochemical series (weak to strong field): $\mathrm{I^- < Br^- < SCN^- < Cl^- < S^{2-} < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < edta^{4-} < NH_3 < en < CN^- < CO}$.
(A) $\mathrm{Br^- < F^- < H_2O < NH_3}$ follows the series. Correct.
(B) puts $\mathrm{F^-}$ below $\mathrm{Br^-}$ and $\mathrm{I^-}$: wrong.
(C) puts $\mathrm{H_2O}$ below $\mathrm{OH^-}$ and $\mathrm{CN^-}$ below $\mathrm{NH_3}$: wrong.
(D) puts $\mathrm{Br^-}$ above $\mathrm{OH^-}$ and $\mathrm{Cl^-}$: wrong.
Answer: (A). | ||||
| 71 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Water is a weak-field ligand, so all four are high-spin octahedral complexes of $\mathrm{M^{2+}}$:
$\mathrm{Cr^{2+}}$ $d^4$: 4 unpaired; $\mathrm{Mn^{2+}}$ $d^5$: 5; $\mathrm{Fe^{2+}}$ $d^6$: 4; $\mathrm{Co^{2+}}$ $d^7$ ($t_{2g}^5 e_g^2$): 3.
Fewest unpaired electrons, so least paramagnetic: $\mathrm{[Co(H_2O)_6]^{2+}}$.
Answer: (D). | ||||
| 72 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Assertion and reasonMedium | |
A medium assertion and reason question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics (A): a strong acid and a strong base are completely ionised in dilute solution, so whichever pair is taken the only reaction is $\mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$, with $\Delta H \approx -57\ \mathrm{kJ\,mol^{-1}}$ (−57.1 kJ). True.
(R): it describes neutralisation as the combination of one mole of $\mathrm{H^+}$ with one mole of $\mathrm{OH^-}$ to give one mole of water. True.
Because the heat comes only from this one ionic reaction, which is the same for every strong acid–strong base pair, (R) is why the value in (A) is constant.
Answer: (A). | ||||
| 73 | Some Basic Concepts of ChemistryEmpirical and molecular formula | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Oxygen = 100 − 42.1 − 6.4 = 51.5%.
Atoms in 342 g: C $= \frac{0.421\times342}{12} = 12.0$; H $= \frac{0.064\times342}{1} = 21.9 \approx 22$; O $= \frac{0.515\times342}{16} = 11.0$.
Molecular formula $\mathrm{C_{12}H_{22}O_{11}}$ (check: $144 + 22 + 176 = 342$).
Answer: (A). | ||||
| 74 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | StatementsEasy | |
A easy statements question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Statement I: in the nitrating mixture, $\mathrm{H_2SO_4}$ protonates nitric acid, $\mathrm{HNO_3 + H_2SO_4 \rightarrow H_2\overset{+}{O}{-}NO_2 + HSO_4^-}$, and the protonated acid loses water: $\mathrm{H_2\overset{+}{O}{-}NO_2 \rightleftharpoons H_2O + NO_2^+}$. That is the step drawn. Correct.
Statement II: electrophilic substitution is promoted by a Lewis acid (for example $\mathrm{FeBr_3}$, anhydrous $\mathrm{AlCl_3}$), which generates or strengthens the electrophile; a Lewis base does not. Incorrect.
Answer: (C). | ||||
| 75 | 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 (B) pentane: non-polar, only dispersion forces; lowest (about 36 °C).
(A) ethyl propyl ether: weakly polar C–O–C, slightly heavier (M = 88); about 64 °C.
(C) pentan-3-one: strongly polar C=O, dipole–dipole attraction; about 102 °C.
(D) pentan-2-ol: hydrogen bonding; highest (about 119 °C).
Increasing order: (B) < (A) < (C) < (D).
Answer: (B). | ||||
| 76 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Assertion and reasonMedium | |
A medium assertion and reason question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A): in a cis alkene the two C–R bond dipoles point to the same side and add, giving a net dipole; in the trans isomer they point opposite ways and cancel (fully when the two ends are identical). So the cis form is more polar. True.
(R): trans-but-2-ene is symmetric, its two $\mathrm{CH_3{-}C}$ bond dipoles cancel, and its dipole moment is zero. True.
The zero (or smaller) dipole of the trans form, from cancelling bond dipoles, is exactly why the cis form is the more polar one, so (R) explains (A).
Answer: (A). | ||||
| 77 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | StatementsEasy | |
A easy statements question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Statement I: the –OH group activates the ring so strongly that phenol is brominated even without a Lewis acid; in $\mathrm{CHCl_3}$ or $\mathrm{CS_2}$ at low temperature it gives mono-bromophenols with no catalyst. False.
Statement II: in the bromination of less reactive rings (benzene), a Lewis acid such as $\mathrm{FeBr_3}$ polarises $\mathrm{Br_2}$ ($\mathrm{Br^{\delta+}{-}Br{-}FeBr_3^{\delta-}}$), producing the electrophile $\mathrm{Br^+}$. True.
Answer: (D). | ||||
| 78 | 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 Chlorobenzene with NaOH at 623 K and 300 atm (Dow process) gives sodium phenoxide, X.
Acidifying with HCl gives phenol, Y.
Phenol with concentrated nitric acid is nitrated at all three activated positions (2, 4 and 6), giving 2,4,6-trinitrophenol (picric acid), Z.
Answer: (D). | ||||
| 79 | 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 The ketone is butan-2-one, $\mathrm{CH_3COCH_2CH_3}$.
(i) Hydrazine forms the hydrazone $\mathrm{CH_3C(=NNH_2)CH_2CH_3}$.
(ii) Heating with KOH in ethylene glycol (Wolff–Kishner reduction) expels $\mathrm{N_2}$ and turns the C=O carbon into $\mathrm{CH_2}$.
A is butane, $\mathrm{CH_3CH_2CH_2CH_3}$.
Answer: (B). | ||||
| 80 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Ninhydrin reacts with free amino groups of amino acids and proteins to give a purple colour (Ruhemann's purple).
Egg albumin is a protein, so it gives a positive test. PVC, starch and cellulose have no amino groups.
Answer: (D). | ||||
| 81 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In Bohr's first orbit the total energy is $-R_H = -2.18\times10^{-18}$ J, and the kinetic energy equals its magnitude: $\frac{1}{2}mv^2 = 2.18\times10^{-18}$ J.
$v^2 = \frac{2\times2.18\times10^{-18}}{9.1\times10^{-31}} = 4.79\times10^{12}$, so $v = 2.19\times10^{6}\ \mathrm{m\,s^{-1}} = 21.9\times10^5\ \mathrm{m\,s^{-1}}$.
Nearest integer: 22.
Answer: 22. | ||||
| 82 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{NO_2^-}$ has $5 + 2\times6 + 1 = 18$ valence electrons.
Lewis structure: $\mathrm{O{=}\ddot{N}{-}O^-}$: nitrogen forms one double bond and one single bond and keeps one lone pair.
Electrons around nitrogen: $4$ (double bond) $+ 2$ (single bond) $+ 2$ (lone pair) $= 8$, an octet.
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 $\mathrm{C_6H_5COOH(s) + \frac{15}{2}O_2(g) \rightarrow 7CO_2(g) + 3H_2O(l)}$
$\Delta n_g = 7 - 7.5 = -\frac{1}{2}$.
$\Delta H = \Delta U + \Delta n_g RT = -321.30 + \left(-\frac{1}{2}\right)R(300) = -321.30 - 150R$.
So $x = 150$.
Answer: 150. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Effective concentration inside: glucose, $i = 1$: 0.2 M.
Outside: NaCl fully dissociated, $i = 2$: $2\times0.05 = 0.1$ M.
The osmotic pressure developed across the membrane depends on the difference: $\pi = \Delta(iC)RT = (0.2 - 0.1)\times0.083\times300 = 2.49$ bar $= 24.9\times10^{-1}$ bar.
Nearest integer: 25.
Answer: 25. | ||||
| 85 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let rate $= k[A]^a[B]^b$.
I and IV ([B] fixed): [A] rises 4 times, rate rises $\frac{2.40\times10^{-2}}{6.0\times10^{-3}} = 4$ times, so $a = 1$.
II and III ([A] fixed): [B] doubles, rate rises $\frac{2.88\times10^{-1}}{7.2\times10^{-2}} = 4$ times, so $b = 2$.
Check with I and II: $3^1\times2^2 = 12$ and $\frac{7.2\times10^{-2}}{6.0\times10^{-3}} = 12$.
Overall order $= 1 + 2 = 3$.
Answer: 3. | ||||
| 86 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mathrm{Ti^{2+}}$, $\mathrm{V^{2+}}$ and $\mathrm{Cr^{2+}}$ are strong reducing agents (they go readily to $+3$). $\mathrm{Co^{3+}}$ ($E^\circ_{\mathrm{Co^{3+}/Co^{2+}}} = +1.97$ V) is a strong oxidising agent in water.
$\mathrm{Co^{3+}}$ is $[\mathrm{Ar}]3d^6$: as a free (high-spin) ion it has 4 unpaired electrons.
$\mu = \sqrt{n(n+2)} = \sqrt{24} = 4.90$ BM, nearest integer 5.
Answer: 5. | ||||
| 87 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | NumericalHard | |
A hard numerical question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter In the borax bead test, the colour depends on the metal and on the part of the flame.
In the drawing, point A is at the tip of the flame (the outer, oxidising zone) and point B is lower down, inside the flame (the inner, reducing zone).
Hot bead colours in the reducing flame: copper red/opaque (colourless when hot), nickel grey, manganese colourless, iron bottle green. (In the oxidising flame, copper is green when hot, but that is point A.)
So the salt is the iron salt; in the reducing flame iron is held as Fe(II), $\mathrm{Fe(BO_2)_2}$: $\mathrm{Fe^{2+}}$ is $3d^6$ with 4 unpaired electrons.
$\mu = \sqrt{4\times6} = \sqrt{24} = 4.90$ BM, nearest integer 5.
Answer: 5. | ||||
| 88 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | NumericalEasy | |
A easy numerical question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Sandmeyer's reaction: an arenediazonium salt with CuCl/HCl or CuBr/HBr gives chlorobenzene or bromobenzene (also benzonitrile with CuCN).
Iodobenzene (IV) is made from the diazonium salt with KI, with no copper salt: not Sandmeyer's reaction. Fluorobenzene (I) is made by the Balz–Schiemann reaction (via the fluoroborate). Astatobenzene (V) is not made this way.
So II (chlorobenzene) and III (bromobenzene): 2.
Answer: 2. | ||||
| 89 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | NumericalMedium | |
A medium numerical question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Picric acid is prepared from phenol in two steps: phenol is first sulphonated with concentrated sulphuric acid to phenol-2,4-disulphonic acid (Product A), which concentrated nitric acid converts to 2,4,6-trinitrophenol (Product B), both $\mathrm{-SO_3H}$ groups being replaced by $\mathrm{-NO_2}$.
Product A, $\mathrm{C_6H_3(OH)(SO_3H)_2}$: $1 + 2\times3 = 7$ oxygen atoms.
Product B, $\mathrm{C_6H_2(OH)(NO_2)_3}$: $1 + 3\times2 = 7$ oxygen atoms.
Total: $7 + 7 = 14$.
Answer: 14. | ||||
| 90 | AminesClassification and identification of amines | Class 12: Amines | NumericalEasy | |
A easy numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Aniline with bromine water gives a white precipitate of 2,4,6-tribromoaniline, P: $\mathrm{C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr}$.
Moles of aniline $= \frac{9.3}{93} = 0.1$ mol, so 0.1 mol of P at best.
Molar mass of P $= 6(12) + 4(1) + 14 + 3(80) = 330$ g/mol, so the theoretical mass $= 33.0$ g.
Yield $= \frac{26.4}{33.0}\times100 = 80\%$.
Answer: 80. | ||||
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