JEE (Main) · 2022
JEE (Main) 2022 — 27 July, 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 Calculus4
- Sets, Relations and Functions2
- Matrices and Determinants2
2 easy · 20 medium · 8 hard
Physics
30 Q
- Kinematics3
- Current Electricity3
- Laws of Motion2
21 easy · 9 medium · 0 hard
Chemistry
30 Q
- Some Basic Concepts of Chemistry3
- Chemical Bonding and Molecular Structure2
- Classification of Elements and Periodicity2
22 easy · 8 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single 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 $\sin^{-1}[2x^2-3]$ needs $-1 \le [2x^2-3] \le 1$, i.e. $-1 \le 2x^2 - 3 < 2$, so $1 \le x^2 < \frac{5}{2}$: $x \in \left(-\sqrt{\frac{5}{2}}, -1\right] \cup \left[1, \sqrt{\frac{5}{2}}\right)$.
$\log_2(\log_{1/2} u)$ needs $\log_{1/2} u > 0$, i.e. $0 < u < 1$ with $u = x^2 - 5x + 5$.
$u < 1$: $x^2 - 5x + 4 < 0$, so $1 < x < 4$.
$u > 0$: $x < \frac{5-\sqrt{5}}{2} \approx 1.382$ or $x > \frac{5+\sqrt{5}}{2} \approx 3.618$.
Together: $x \in \left(1, \frac{5-\sqrt{5}}{2}\right) \cup \left(\frac{5+\sqrt{5}}{2}, 4\right)$.
Intersecting with $1 \le x < \sqrt{2.5} \approx 1.581$ leaves $\left(1, \frac{5-\sqrt{5}}{2}\right)$.
Answer: (C). | ||||
| 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 $\frac{1 - i\sin\alpha}{1 + 2i\sin\alpha} = \frac{(1 - i\sin\alpha)(1 - 2i\sin\alpha)}{1 + 4\sin^2\alpha} = \frac{1 - 2\sin^2\alpha - 3i\sin\alpha}{1 + 4\sin^2\alpha}$. Purely imaginary: $\sin^2\alpha = \frac{1}{2}$, so in $(\pi, 2\pi)$, $\alpha = \frac{5\pi}{4}$ or $\frac{7\pi}{4}$.
$\frac{1 + i\cos\beta}{1 - 2i\cos\beta} = \frac{(1 + i\cos\beta)(1 + 2i\cos\beta)}{1 + 4\cos^2\beta} = \frac{1 - 2\cos^2\beta + 3i\cos\beta}{1 + 4\cos^2\beta}$. Purely real: $\cos\beta = 0$, so $\beta = \frac{3\pi}{2}$.
$\cos 2\beta = \cos 3\pi = -1$; $\sin 2\alpha = \sin\frac{5\pi}{2} = 1$ or $\sin\frac{7\pi}{2} = -1$. So $Z = 1 - i$ or $-1 - i$.
For $Z = 1 - i$: $iZ = 1 + i$, $i\bar{Z} = i(1+i) = -1 + i$, $\frac{1}{-1+i} = \frac{-1-i}{2}$. Sum $= \frac{1}{2} + \frac{i}{2}$.
For $Z = -1 - i$: $iZ = 1 - i$, $i\bar{Z} = i(-1+i) = -1 - i$, $\frac{1}{-1-i} = \frac{-1+i}{2}$. Sum $= \frac{1}{2} - \frac{i}{2}$.
Total $= 1$.
Answer: (C). | ||||
| 3 | Quadratic EquationsForming equations from given roots | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on forming equations from given roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $t = \log_3 5$, so $\log_5 3 = \frac{1}{t}$.
$5^{\sqrt{\log_5 3}} = 3^{t \cdot t^{-1/2}} = 3^{\sqrt{t}} = 3^{\sqrt{\log_3 5}}$, so the first bracket is just 5.
$5^{(\log_5 3)^{2/3}} = 3^{t \cdot t^{-2/3}} = 3^{t^{1/3}} = 3^{(\log_3 5)^{1/3}}$, so the second bracket is $-1$.
The equation is (as intended, with the $x$ of the middle term) $x^2 - 5x - 3 = 0$: $\alpha + \beta = 5$, $\alpha\beta = -3$.
New roots: $\alpha + \frac{1}{\beta} = \frac{\alpha\beta + 1}{\beta} = -\frac{2}{\beta}$ and $\beta + \frac{1}{\alpha} = -\frac{2}{\alpha}$.
Sum $= -2\cdot\frac{\alpha+\beta}{\alpha\beta} = -2\cdot\frac{5}{-3} = \frac{10}{3}$; product $= \frac{4}{\alpha\beta} = -\frac{4}{3}$.
Equation: $x^2 - \frac{10}{3}x - \frac{4}{3} = 0$, i.e. $3x^2 - 10x - 4 = 0$.
Answer: (B). | ||||
| 4 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants By the Cayley–Hamilton theorem, $A^2 - (\operatorname{tr}A)A + (\det A)I = O$.
Subtracting the given relation: $(\gamma + \operatorname{tr}A)A + (18 - \det A)I = O$.
$A$ is not a multiple of $I$ (its off-diagonal entry is $-2$), so both coefficients vanish: $\gamma = -\operatorname{tr}A$ and $\det A = 18$.
(Check: $A^2 = \begin{pmatrix}16 - 2\alpha & -8 - 2\beta\\ 4\alpha + \alpha\beta & -2\alpha + \beta^2\end{pmatrix}$; the (1,2) entry of $A^2 + \gamma A + 18I$ gives $-8 - 2\beta - 2\gamma = 0$, $\gamma = -4 - \beta$, and the (1,1) entry then gives $16 - 2\alpha - 16 - 4\beta + 18 = 0$, i.e. $4\beta + 2\alpha = 18$, which is $\det A = 18$.)
Answer: (B). | ||||
| 5 | 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 The denominator tends to 0 as $x \to 0$, so for a finite limit the numerator must too: $(729p)^{1/7} = 3$, $729p = 2187$, $p = 3$.
Numerator: $(2187 + 3x)^{1/7} - 3 = 3\left[\left(1 + \frac{x}{729}\right)^{1/7} - 1\right] \approx \frac{3x}{7 \cdot 729}$.
Denominator: $(729 + qx)^{1/3} - 9 = 9\left[\left(1 + \frac{qx}{729}\right)^{1/3} - 1\right] \approx \frac{9qx}{3 \cdot 729} = \frac{3qx}{729}$.
So $f(0) = \lim_{x\to0} f(x) = \frac{3/7}{3q} = \frac{1}{7q}$.
With $p = 3$: $63q\,f(0) - p^2 = 9 - 9 = 0$. (The others give $2$, $-6$ and $-6$.)
Answer: (B). | ||||
| 6 | 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 Piece by piece:
$x < -1$: $f = 2 - x - (1 - x) - (x + 1) = -x$.
$-1 < x < 0$: $f = 2 - x - (1 - x) + (x + 1) = 2 + x$.
$0 < x < 1$: $f = 2 + x - (1 - x) + (x + 1) = 2 + 3x$.
$x > 1$: $f = 2 + x - (x - 1) + (x + 1) = 4 + x$.
(S1): $f'$ at the four points is $-1, 1, 3, 1$; the sum is 4, not 2. Wrong.
(S2): $\int_{-2}^{-1}(-x)\,dx = \frac{3}{2}$, $\int_{-1}^{0}(2 + x)\,dx = \frac{3}{2}$, $\int_0^1(2 + 3x)\,dx = \frac{7}{2}$, $\int_1^2(4 + x)\,dx = \frac{11}{2}$. Total $= 12$. Correct.
Answer: (D). | ||||
| 7 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series For an AP with an odd number of terms, the sum is the number of terms times the middle term: $S_{21} = \frac{21}{2}(a_1 + a_{21}) = 21\,a_{11}$.
Here $a_{11} = 10ar + 10 \cdot 10ar^2 = 10ar + 100ar^2$ and $S_{21} = \frac{21}{2}(2 \cdot 10ar + 20 \cdot 10ar^2) = 21(10ar + 100ar^2) = 21a_{11}$.
The G.P. data ($\frac{a}{1-r} = 5$, $1 - r^5 = \frac{98}{125}$) make $ar \ne 0$ and $r \ne 0.4$, so $22a_{11}$, $15a_{16}$ and $14a_{16}$ are not equal to it ($15a_{16} = 150ar + 2250ar^2$ equals $S_{21}$ only for $r = 0.4$).
Answer: (A). | ||||
| 8 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals For $0 \le y \le 4$ the region is the set of $x$ with $\frac{\sqrt{y}}{2} \le |x| \le 3\sqrt{y}$ (outside the narrow parabola $y = 4x^2$, inside the wide one $x^2 = 9y$).
Area $= 2\int_0^4\left(3\sqrt{y} - \frac{\sqrt{y}}{2}\right)dy = 5\int_0^4\sqrt{y}\,dy = 5\cdot\frac{2}{3}\cdot 8 = \frac{80}{3}$.
Answer: (D). | ||||
| 9 | 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 $2x^2 - 3x = x(2x - 3)$ is negative on $(0, \frac{3}{2})$ and positive on $(\frac{3}{2}, 2)$.
$\int_0^{3/2}(3x - 2x^2)\,dx = \frac{27}{8} - \frac{9}{4} = \frac{9}{8}$; $\int_{3/2}^2(2x^2 - 3x)\,dx = \frac{2}{3}\left(8 - \frac{27}{8}\right) - \frac{3}{2}\left(4 - \frac{9}{4}\right) = \frac{37}{12} - \frac{21}{8} = \frac{11}{24}$.
So $\int_0^2|2x^2 - 3x|\,dx = \frac{27}{24} + \frac{11}{24} = \frac{19}{12}$.
$\left[x - \frac{1}{2}\right]$ is $-1$ on $[0, \frac{1}{2})$, 0 on $[\frac{1}{2}, \frac{3}{2})$, 1 on $[\frac{3}{2}, 2]$: its integral is $-\frac{1}{2} + 0 + \frac{1}{2} = 0$.
Total $= \frac{19}{12}$.
Answer: (B). | ||||
| 10 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations From the figure the curve passes through (4, 2); $A_1$ is the area under the curve from $x = 4$ to $x$, and $A_2$ the area between the curve and the $y$-axis from $y = 2$ to $y$.
$\int_4^x y\,dx = 2\int_2^y x\,dy$. Differentiating with respect to $x$: $y = 2x\frac{dy}{dx}$, so $\frac{dy}{y} = \frac{dx}{2x}$, $y^2 = kx$; through (4, 2), $k = 1$: $y^2 = x$.
The line $2x - 12y = 15$ has slope $\frac{1}{6}$, so the normal has slope $-6$ and the tangent there slope $\frac{1}{6}$.
$\frac{dy}{dx} = \frac{1}{2y} = \frac{1}{6}$ gives $y = 3$, $x = 9$.
Normal: $y - 3 = -6(x - 9)$, i.e. $y = 57 - 6x$.
(6, 21): $57 - 36 = 21$, on it. (8, 9): $57 - 48 = 9$, on it. (12, $-15$): $57 - 72 = -15$, on it. (10, $-4$): $57 - 60 = -3 \ne -4$, not on it.
Answer: (C). | ||||
| 11 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines A is on AB and CA: $2x + y = 0$, $x - y = 3$ give A(1, $-2$). $PA^2 = 1 + 25 = 26$.
B on AB is $(t, -2t)$ with $(t - 2)^2 + (2t + 3)^2 = 26$: $5t^2 + 8t - 13 = 0$, $t = 1$ (A) or $t = -\frac{13}{5}$. B$\left(-\frac{13}{5}, \frac{26}{5}\right)$.
C on CA is $(s, s - 3)$ with $(s - 2)^2 + (s - 6)^2 = 26$: $s^2 - 8s + 7 = 0$, $s = 1$ (A) or 7. C(7, 4).
BC: $7 + 4p = 39$, $p = 8$ (check: $-\frac{13}{5} + 8\cdot\frac{26}{5} = 39$).
$AC^2 = 36 + 36 = 72 = 9p$: true. $AC^2 + p^2 = 72 + 64 = 136$: true.
Area $= \frac{1}{2}\left|1\left(\frac{26}{5} - 4\right) - \frac{13}{5}(4 + 2) + 7\left(-2 - \frac{26}{5}\right)\right| = \frac{1}{2}|1.2 - 15.6 - 50.4| = 32.4$.
So $32 < 32.4 < 36$ is true and $34 < \text{area} < 38$ is not.
Answer: (D). | ||||
| 12 | CirclesEquation from a diameter | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on equation from a diameter, from Circles. NCERT: Class 11: Conic Sections $C_1$: centre (2, 0), radius 2: $x^2 + y^2 - 4x = 0$. With $y = 2x$: $5x^2 - 4x = 0$, so A$\left(\frac{4}{5}, \frac{8}{5}\right)$.
OA is a diameter of $C_2$, so the tangent to $C_2$ at A is perpendicular to OA (slope 2): slope $-\frac{1}{2}$.
Tangent: $y - \frac{8}{5} = -\frac{1}{2}\left(x - \frac{4}{5}\right)$, i.e. $x + 2y = 4$. P(4, 0), Q(0, 2).
A divides QP: $QA = \sqrt{0.64 + 0.16} = \sqrt{0.8}$, $AP = \sqrt{10.24 + 2.56} = \sqrt{12.8}$; $\frac{QA}{AP} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Answer: (A). | ||||
| 13 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The distance from the focus to the tangent at the vertex is the focal length: $\frac{|a + a - a|}{\sqrt{2}} = \frac{|a|}{\sqrt{2}}$.
Latus rectum $= 4 \times \frac{|a|}{\sqrt{2}} = 2\sqrt{2}|a| = 16$, so $|a| = 4\sqrt{2}$.
Answer: (C). | ||||
| 14 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line is $M(-1 + 2t, 3 + 3t, 1 - t)$. $\vec{PM} = (2t - 1 - a, 3t - 1, -1 - t)$ is perpendicular to (2, 3, $-1$):
$2(2t - 1 - a) + 3(3t - 1) + (1 + t) = 0 \Rightarrow 14t = 2a + 4$, $t = \frac{a + 2}{7}$.
Then $49\,PM^2 = (5a + 3)^2 + (3a - 1)^2 + (a + 9)^2 = 35a^2 + 42a + 91 = 49 \times 24$, so $5a^2 + 6a - 155 = 0$, $(a - 5)(5a + 31) = 0$, $a = 5$.
$t = 1$, foot M(1, 6, 0); image Q $= 2M - P = (-3, 8, -2)$, $\sum\alpha_i = 3$.
$a + \sum\alpha_i = 5 + 3 = 8$.
Answer: (B). | ||||
| 15 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The line is along $(a, b, 0) \times (a, b, c) = (bc, -ac, 0) \parallel (b, -a, 0)$.
The plane $y - z + 2 = 0$ has normal (0, 1, $-1$). For a $30^\circ$ angle with the plane:
$\sin 30^\circ = \frac{|(b, -a, 0)\cdot(0, 1, -1)|}{\sqrt{a^2 + b^2}\sqrt{2}} = \frac{a}{\sqrt{2}\sqrt{a^2 + b^2}} = \frac{1}{2}$, so $a^2 = b^2$.
With $b = a$ the direction is $(1, -1, 0)$: direction cosines $\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0$.
(With $b = -a$ it would be $(1, 1, 0)$, option (A); see the note.)
Answer: (B). | ||||
| 16 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Mean $np$ and variance $npq$ are the roots of $t^2 - 24t + 128 = 0$: 16 and 8. The mean is the larger, so $np = 16$, $npq = 8$, $q = \frac{1}{2}$, $p = \frac{1}{2}$, $n = 32$.
$P(X > 29) = P(30) + P(31) + P(32) = \frac{\binom{32}{2} + \binom{32}{1} + 1}{2^{32}} = \frac{496 + 32 + 1}{2^{32}} = \frac{529}{2^{32}}$.
$k = 529$.
Answer: (B). | ||||
| 17 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let each prime (2, 3, 5) have probability $p$, each composite (4, 6) $c$, and 1 have $u$, with $3p = 6c = 2u = k$.
Then $p = \frac{k}{3}$, $c = \frac{k}{6}$, $u = \frac{k}{2}$, and $3p + 2c + u = k + \frac{k}{3} + \frac{k}{2} = \frac{11k}{6} = 1$, so $k = \frac{6}{11}$.
$P(1) = \frac{3}{11}$, $P(4) = \frac{1}{11}$; a perfect square (1 or 4) has probability $\frac{4}{11}$.
For two throws X is binomial with $n = 2$: mean $= 2 \times \frac{4}{11} = \frac{8}{11}$.
Answer: (D). | ||||
| 18 | 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 AQ $= 10$ (elevation $45^\circ$). B is at height $BR = \frac{d}{2}$ and $AR = \frac{\sqrt{3}d}{2}$, so $RQ = 10 - \frac{\sqrt{3}d}{2}$.
From B: $\tan 60^\circ = \frac{10 - d/2}{10 - \sqrt{3}d/2} = \sqrt{3}$, so $10 - \frac{d}{2} = 10\sqrt{3} - \frac{3d}{2}$, $d = 10(\sqrt{3} - 1)$.
Then $BR = 5(\sqrt{3} - 1)$ and $RQ = 10 - 5(3 - \sqrt{3}) = 5(\sqrt{3} - 1)$.
PQ and BR are the parallel sides: $\alpha = \frac{1}{2}(10 + 5\sqrt{3} - 5)\cdot 5(\sqrt{3} - 1) = \frac{25}{2}(\sqrt{3} + 1)(\sqrt{3} - 1) = 25$.
Answer: (A). | ||||
| 19 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctHard | |
A hard single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions With $B - A = \frac{\pi}{6}$: $\sec A\sec B = \frac{\sin(B - A)}{\sin\frac{\pi}{6}\cos A\cos B} = 2(\tan B - \tan A)$.
The sum telescopes: $2\left[\tan\left(\theta + \frac{9\pi}{6}\right) - \tan\theta\right] = 2(-\cot\theta - \tan\theta) = -\frac{2}{\sin\theta\cos\theta} = -\frac{4}{\sin 2\theta}$.
$-\frac{4}{\sin 2\theta} = -\frac{8}{\sqrt{3}}$ gives $\sin 2\theta = \frac{\sqrt{3}}{2}$, $\theta = \frac{\pi}{6}$ or $\frac{\pi}{3}$.
$\sum\theta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.
Answer: (C). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter An implication is false only when its antecedent is true and its consequent false.
$P \wedge \sim R$ true: P = T, R = F. Then $\sim R \wedge Q$ false needs Q = F.
(A) $P \vee Q$ = T, $\sim R$ = T: T. (B) $R \vee Q$ = F: T. (C) $\sim(P \vee Q)$ = F: T.
(D) $\sim(R \vee Q)$ = T and $\sim P$ = F: F.
Answer: (D). | ||||
| 21 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalHard | |
A hard numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3 \times 3$ matrix, $\det(\operatorname{adj}M) = (\det M)^2$, so four adjoints give $(\det A)^{16}$.
Adding row 1 to row 3 makes row 3 $(\alpha + \beta + \gamma)(1, 1, 1)$, so $\det A = (\alpha + \beta + \gamma)\det\begin{bmatrix}\alpha & \beta & \gamma\\ \alpha^2 & \beta^2 & \gamma^2\\ 1 & 1 & 1\end{bmatrix} = \pm(\alpha + \beta + \gamma)(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$.
The condition becomes $(\alpha + \beta + \gamma)^{16} = 2^{32} \cdot 3^{16} = 12^{16}$, so $\alpha + \beta + \gamma = 12$.
Sets of three distinct natural numbers with sum 12: {1,2,9}, {1,3,8}, {1,4,7}, {1,5,6}, {2,3,7}, {2,4,6}, {3,4,5}: 7 sets.
Each gives $3! = 6$ ordered triples: $7 \times 6 = 42$.
Answer: 42. | ||||
| 22 | 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 $x^2 - 10x + 9 \le 0 \Leftrightarrow 1 \le x \le 9$, so $A = \{1, \dots, 9\}$; $B = \{1, 4, 9, 16, 25, 36, \dots\}$.
Bound $(x - 3)^2 + 1$ and the number of squares not exceeding it:
$x = 1$: 5 → 2; $x = 2$: 2 → 1; $x = 3$: 1 → 1; $x = 4$: 2 → 1; $x = 5$: 5 → 2; $x = 6$: 10 → 3; $x = 7$: 17 → 4; $x = 8$: 26 → 5; $x = 9$: 37 → 6.
Number of functions $= 2 \cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 1440$.
Answer: 1440. | ||||
| 23 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom{n}{r}3^{n-r}(6x)^r$; at $x = \frac{3}{2}$, $6x = 9$ and $\frac{T_{r+1}}{T_r} = \frac{n - r + 1}{r}\cdot\frac{9}{3} = \frac{3(n - r + 1)}{r}$.
$T_9$ greatest: $\frac{T_9}{T_8} = \frac{3(n - 7)}{8} \ge 1 \Rightarrow n \ge 9.67$, and $\frac{T_{10}}{T_9} = \frac{3(n - 8)}{9} \le 1 \Rightarrow n \le 11$. Least $n_0 = 10$.
In $(3 + 6x)^{10}$ the coefficient of $x^r$ is $\binom{10}{r}3^{10-r}6^r = 3^{10}\binom{10}{r}2^r$.
$k = \frac{\binom{10}{6}2^6}{\binom{10}{3}2^3} = \frac{210 \times 8}{120} = 14$.
$k + n_0 = 14 + 10 = 24$.
Answer: 24. | ||||
| 24 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The $n$th numerator is $\sum_{j=1}^{n}\left[(2j)^3 - (2j - 1)^3\right] = \sum_{j=1}^{n}(12j^2 - 6j + 1)$
$= 2n(n + 1)(2n + 1) - 3n(n + 1) + n = n(4n^2 + 3n) = n^2(4n + 3)$.
The $n$th denominator is $n(4n + 3)$ (1×7, 2×11, 3×15, …, 15×63).
So the $n$th term is $n$, and the sum is $1 + 2 + \dots + 15 = 120$.
Answer: 120. | ||||
| 25 | Applications of DerivativesRate of change | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on rate of change, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives At depth $h$: radius $r = \frac{3h}{4}$, slant height $l = \frac{5h}{4}$.
$V = \frac{1}{3}\pi r^2h = \frac{3\pi h^3}{16}$, so $\frac{dV}{dt} = \frac{9\pi h^2}{16}\frac{dh}{dt} = 6$; at $h = 4$, $9\pi\frac{dh}{dt} = 6$, $\frac{dh}{dt} = \frac{2}{3\pi}$.
Wet curved area $S = \pi rl = \frac{15\pi h^2}{16}$, $\frac{dS}{dt} = \frac{15\pi h}{8}\frac{dh}{dt} = \frac{15\pi \cdot 4}{8}\cdot\frac{2}{3\pi} = 5$ m²/h.
Answer: 5. | ||||
| 26 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $(\alpha, \alpha)$: $2\alpha^2 - 3 + (-1)^5 = 0$, so $\alpha = \sqrt{2}$. Let $u = x^2 - y^2 - 1$ ($u = -1$ there).
First derivative: $2x + 2yy' + 5u^4(2x - 2yy') = 0$. At the point ($u^4 = 1$): $12x - 8yy' = 0$, $y' = \frac{3x}{2y} = \frac{3}{2}$.
Second derivative: $2 + 2y'^2 + 2yy'' + 20u^3u'(2x - 2yy') + 5u^4(2 - 2y'^2 - 2yy'') = 0$, with $u' = 2x - 2yy' = 2\sqrt{2} - 3\sqrt{2} = -\sqrt{2}$.
$20u^3(u')^2 = 20(-1)(2) = -40$. So $2 + \frac{9}{2} + 2\sqrt{2}y'' - 40 + 10 - \frac{45}{2} - 10\sqrt{2}y'' = 0$, i.e. $-46 - 8\sqrt{2}y'' = 0$, $y'' = -\frac{23}{4\sqrt{2}}$.
$3y' - y^3y'' = \frac{9}{2} - 2\sqrt{2}\left(-\frac{23}{4\sqrt{2}}\right) = \frac{9}{2} + \frac{23}{2} = 16$.
Answer: 16. | ||||
| 27 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $[x - k] = [x] - k$ is least for $k = 10$, so $f(x) = [x] - 10$.
On $[j, j + 1)$, $j = 0, \dots, 9$, $f = j - 10$, which runs through $-10, -9, \dots, -1$.
$\int f = -(1 + 2 + \dots + 10) = -55$, $\int f^2 = 1^2 + \dots + 10^2 = 385$, $\int|f| = 55$.
Sum $= -55 + 385 + 55 = 385$.
Answer: 385. | ||||
| 28 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Put $\lambda = \frac{\sqrt{3}\,t}{\sqrt{x}}$: then $\frac{\lambda^2x}{3} = t^2$, $d\lambda = \frac{\sqrt{3}}{\sqrt{x}}dt$, and $t$ runs from 0 to $\sqrt{x}$.
$f(x) = \frac{2}{\sqrt{x}}\int_0^{\sqrt{x}}f(t^2)\,dt$. With $s = \sqrt{x}$ and $h(s) = f(s^2)$: $s\,h(s) = 2\int_0^s h(t)\,dt$.
Differentiating: $h + sh' = 2h$, so $sh' = h$, $h = Cs$, i.e. $f(x) = C\sqrt{x}$.
$f(1) = \sqrt{3}$ gives $f(x) = \sqrt{3x}$. $\sqrt{3\alpha} = 6$ gives $\alpha = 12$.
Answer: 12. | ||||
| 29 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $y = mx + c$ touches the ellipse when $c^2 = 4m^2 + 9$ and the hyperbola when $c^2 = 42m^2 - 143$.
$4m^2 + 9 = 42m^2 - 143$ gives $m^2 = 4$, $c^2 = 25$.
Of $y = \pm2x \pm 5$, only $y = 2x + 5$ avoids the fourth quadrant ($x > 0 \Rightarrow y > 5$).
Point of contact with the ellipse: $\left(-\frac{a^2m}{c}, \frac{b^2}{c}\right) = \left(-\frac{8}{5}, \frac{9}{5}\right)$.
With the hyperbola: $\left(-\frac{a^2m}{c}, -\frac{b^2}{c}\right) = \left(-\frac{84}{5}, -\frac{143}{5}\right)$ (check: its tangent $-\frac{2x}{5} + \frac{y}{5} = 1$ is $y = 2x + 5$).
$|2x_1 + x_2| = \left|-\frac{16}{5} - \frac{84}{5}\right| = 20$.
Answer: 20. | ||||
| 30 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalHard | |
A hard numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Each vector is perpendicular to the other two, so the three are mutually perpendicular and
$|a||b| = 4|c|$, $|b||c| = 9|a|$, $|c||a| = \alpha|b|$.
Multiplying in pairs: $|b|^2 = 36$, $|a|^2 = 4\alpha$, $|c|^2 = 9\alpha$, so $|a| = 2\sqrt{\alpha}$, $|b| = 6$, $|c| = 3\sqrt{\alpha}$ and $|a| + |b| + |c| = 5\sqrt{\alpha} + 6$.
As printed the sum is $\frac{1}{36}$, which no $\alpha > 0$ can give, since $|b| = 6$ alone exceeds it (see the note).
Reading the sum as 36: $5\sqrt{\alpha} + 6 = 36$, $\sqrt{\alpha} = 6$, $\alpha = 36$.
Answer: 36. | ||||
| 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 The argument of the sine is dimensionless: $[\alpha] = \frac{[kt]}{[x]} = \frac{ML^2T^{-2}}{L} = MLT^{-2}$ ($kt$ is an energy).
$\frac{\alpha}{\beta}$ is an energy density, $ML^{-1}T^{-2}$, so $[\beta] = \frac{MLT^{-2}}{ML^{-1}T^{-2}} = L^2$, i.e. $[M^0L^2T^0]$.
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 Trajectory at $45^\circ$: $y = x - \frac{gx^2}{u^2}$. Through (20, 10): $10 = 20 - \frac{10 \times 400}{u^2}$, so $u^2 = 400$, $u = 20$ m/s.
$u_x = u_y = 10\sqrt{2}$ m/s; $T = \frac{2u_y}{g} = 2\sqrt{2}$ s, so $t = \frac{T}{\sqrt{2}} = 2$ s.
$v_y = 10\sqrt{2} - 20$, $v_x = 10\sqrt{2}$.
$\vec{p} = 10\left[10\sqrt{2}\,\hat{i} + (10\sqrt{2} - 20)\hat{j}\right] = 100\sqrt{2}\,\hat{i} + (100\sqrt{2} - 200)\hat{j}$.
Answer: (D). | ||||
| 33 | 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 At constant velocity the net force is zero, so the total contact force (normal reaction plus friction) balances the weight exactly.
Its magnitude is $\sqrt{(Mg\cos\theta)^2 + (Mg\sin\theta)^2} = Mg$.
Answer: (A). | ||||
| 34 | 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 A lift moving with uniform velocity is an inertial frame, so the acceleration along the incline is $g\sin\theta$.
$l = \frac{1}{2}g\sin\theta\,t^2$ gives $t \propto \frac{1}{\sqrt{\sin\theta}}$.
$t_2 = 2\sqrt{\frac{\sin 30^\circ}{\sin 45^\circ}} = 2\sqrt{\frac{0.5}{0.707}} = 2 \times 0.841 \approx 1.68$ s.
Answer: (C). | ||||
| 35 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Constant retardation $a$: $\left(\frac{u}{3}\right)^2 = u^2 - 2a(4)$, so $2a = \frac{8u^2}{9 \times 4} = \frac{2u^2}{9}$.
To stop: $u^2 = 2a(4 + x) = \frac{2u^2}{9}(4 + x)$, so $4 + x = 4.5$, $x = 0.5$ cm.
Answer: (C). | ||||
| 36 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Energy conservation with $v_e^2 = \frac{2GM}{R}$: $\frac{1}{2}m\lambda^2\frac{2GM}{R} - \frac{GMm}{R} = -\frac{GMm}{r}$.
$\frac{1}{r} = \frac{1 - \lambda^2}{R}$, so $r = \frac{R}{1 - \lambda^2}$ from the centre.
Answer: (B). | ||||
| 37 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The same tension F acts in both: $\Delta L = \frac{F}{A}\left(\frac{L_s}{Y_s} + \frac{L_c}{Y_c}\right)$.
$\frac{3.2}{2 \times 10^{11}} + \frac{4.4}{1.1 \times 10^{11}} = 1.6 \times 10^{-11} + 4 \times 10^{-11} = 5.6 \times 10^{-11}$ m³ N⁻¹.
$A = \frac{22}{7}(1.4 \times 10^{-3})^2 = 6.16 \times 10^{-6}$ m².
$F = \frac{1.4 \times 10^{-3} \times 6.16 \times 10^{-6}}{5.6 \times 10^{-11}} = 154$ N.
Answer: (D). | ||||
| 38 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Case 1: $\eta_1 = 1 - \frac{100}{300} = \frac{2}{3}$.
Case 2: the first engine takes $Q_1$ at 300 K and rejects $Q_2 = Q_1\frac{200}{300}$ at 200 K; the second takes this $Q_2$ and rejects $Q_3 = Q_2\frac{100}{200} = Q_1\frac{100}{300}$ at 100 K.
Combined efficiency $= 1 - \frac{Q_3}{Q_1} = 1 - \frac{100}{300} = \frac{2}{3}$: the same as case 1.
Answer: (A). | ||||
| 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 (A) False: n degrees of freedom are n independent ways of storing energy, not $n^2$.
(B) True: equipartition gives $\frac{1}{2}kT$ per molecule, $\frac{1}{2}RT$ per mole, per degree of freedom.
(C) False: a monatomic molecule has no rotational degree of freedom.
(D) True: $\text{CH}_4$ is a non-linear molecule: 3 translational + 3 rotational = 6 (vibration ignored).
Answer: (B). | ||||
| 40 | 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 With charges $q$ and $Q - q$ at a fixed distance, $F \propto q(Q - q)$, which is greatest at $q = \frac{Q}{2}$.
So 2 μC and 2 μC.
Answer: (B). | ||||
| 41 | Current ElectricityDrift velocity and mobility | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity $v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$.
(A) True: a hotter conductor has a shorter relaxation time $\tau$, so $v_d$ falls. (E) is therefore false.
(B) For a given applied potential difference, $v_d$ does not depend on the cross-section (only the current does); false in that sense.
(C) False: $v_d \propto V$.
(D) True: for a given V, $v_d \propto \frac{1}{L}$.
So (A) and (D).
Answer: (B). | ||||
| 42 | Magnetic Effects of Current and MagnetismEarth magnetism | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on earth magnetism, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter For an oscillation magnetometer the frequency $f \propto \sqrt{B_H}$, so $\frac{B_{H,P}}{B_{H,Q}} = \left(\frac{20}{10}\right)^2 = 4$.
$B_H = B\cos\delta$: $\frac{B_P\cos 30^\circ}{B_Q\cos 60^\circ} = 4$, so $B_P\frac{\sqrt{3}}{2} = 4B_Q\cdot\frac{1}{2}$ and $\frac{B_Q}{B_P} = \frac{\sqrt{3}}{4}$.
Answer: (A). | ||||
| 43 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Maximum kinetic energy $K = \frac{q^2B^2r^2}{2m} = \frac{(1.6 \times 10^{-19})^2(1)^2(0.6)^2}{2 \times 1.6 \times 10^{-27}} = 2.88 \times 10^{-12}$ J.
In eV: $\frac{2.88 \times 10^{-12}}{1.6 \times 10^{-19}} = 1.8 \times 10^7$ eV $= 18$ MeV.
Answer: (B). | ||||
| 44 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-8}}} = 10^4$ rad/s; 60% lower: $\omega = 0.4 \times 10^4 = 4000$ rad/s.
$X_L = 4000 \times 0.01 = 40\ \Omega$, $X_C = \frac{1}{4000 \times 10^{-6}} = 250\ \Omega$.
$Z = \sqrt{10^2 + 210^2} = \sqrt{44200} \approx 210.2\ \Omega$.
$I_m = \frac{50}{210.2} \approx 0.238$ A $= 238$ mA.
Answer: (C). | ||||
| 45 | 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 (A) False: the wave travels perpendicular to both fields, along $\vec{E} \times \vec{B}$.
(B) True: the average electric and magnetic energy densities are equal.
(C) False: the fields are perpendicular to each other.
(D) True.
(E) False: $\frac{B_0}{E_0} = \frac{1}{c}$; it is $\frac{E_0}{B_0}$ that equals $c$.
Answer: (B). | ||||
| 46 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Amplitudes are in the ratio 1 : 2, so $I_{max} \propto (1 + 2)^2 = 9$ and $I_{min} \propto (2 - 1)^2 = 1$.
$\frac{I_{max} + I_{min}}{I_{max} - I_{min}} = \frac{10}{8} = \frac{5}{4}$.
Matching $\frac{2\alpha + 1}{\beta + 3} = \frac{5}{4}$ term by term: $2\alpha + 1 = 5$, $\beta + 3 = 4$, so $\alpha = 2$, $\beta = 1$, $\frac{\alpha}{\beta} = 2$.
Answer: (B). | ||||
| 47 | 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 (A) True: $\frac{1}{2}mv_{max}^2 = h\nu - \phi$, so $v_{max}^2$ is linear in $\nu$.
(B) False: moving the source away lowers the intensity and so the saturation current.
(C) False: the maximum kinetic energy depends on frequency, not on power (intensity).
(D) False: immediate emission is exactly what the photon (particle) picture explains.
(E) True: the wave picture cannot explain a threshold.
Answer: (B). | ||||
| 48 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{6.4 \times 10^{-4}}{5 \times 10^{-6}} = 128 = 2^7$: seven half-lives, $7 \times 5 = 35$ days.
Answer: (D). | ||||
| 49 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{2\ \text{mA}}{5\ \mu\text{A}} = 400$.
Answer: (B). | ||||
| 50 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter The modulating signal has amplitude $A_m = 1$ V; the carrier amplitude is $A_c = 5$ V.
Modulation index $\mu = \frac{A_m}{A_c} = \frac{1}{5} = 0.2$.
Answer: (A). | ||||
| 51 | Experimental SkillsProperties of matter experiments | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on properties of matter experiments, from Experimental Skills. NCERT: No matching NCERT chapter $Y = \frac{FL}{A\,\Delta L}$, so $\frac{\Delta L}{F} = \frac{L}{AY}$: the graph's slope is $\frac{1}{AY}$.
From the graph, $\frac{\Delta L}{F}$ rises by $0.25 \times 10^{-5}$ per metre of length: $\frac{1}{AY} = 0.25 \times 10^{-5}$ N⁻¹.
$Y = \frac{1}{2 \times 10^{-6} \times 0.25 \times 10^{-5}} = 2 \times 10^{11}$ N m⁻².
Answer: 2. | ||||
| 52 | Current ElectricityMetre bridge | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity At balance $\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}} = \frac{l}{100 - l}$ for a uniform wire, whatever its radius.
Doubling the radius changes the resistance per unit length uniformly along the whole wire, so the ratio and the balancing length stay the same: 40 cm.
Answer: 40. | ||||
| 53 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The prisms are placed inverted with respect to each other (figure), so their mean deviations subtract.
$\delta = (n_1 - 1)A_1 - (n_2 - 1)A_2 = 0.5 \times 6^\circ - 0.55 \times 5^\circ = 3^\circ - 2.75^\circ = 0.25^\circ = \left(\frac{1}{4}\right)^\circ$.
$x = 4$.
Answer: 4. | ||||
| 54 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The loop's area vector is along $\hat{k}$ (area $\pi \times 1^2 = \pi$ m²), so only $B_z$ links it: $\Phi = 3t^2\pi$.
$|\varepsilon| = \frac{d\Phi}{dt} = 6t\pi = 12\pi$ V at $t = 2$ s.
$n = 12$.
Answer: 12. | ||||
| 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 In steady state no current flows in the capacitor branch, so the current is $I = \frac{E}{r + R} = \frac{10}{110} = \frac{1}{11}$ A through r and R.
The capacitor branch is in parallel with R, and R′ carries no current, so $V_C = IR = \frac{100}{11}$ V.
$Q = CV = 1.1 \times 10^{-6} \times \frac{100}{11} = 10 \times 10^{-6}$ C.
Answer: 10. | ||||
| 56 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The slab fills the gap over 1 cm × 4 cm; the rest (7 cm × 4 cm) is air. Two capacitors in parallel:
$C_1 = \frac{5\epsilon_0 (4 \times 10^{-4})}{4 \times 10^{-3}} = 0.5\epsilon_0$, $C_2 = \frac{\epsilon_0(28 \times 10^{-4})}{4 \times 10^{-3}} = 0.7\epsilon_0$.
$C = 1.2\epsilon_0$; with the battery connected, $U = \frac{1}{2}CV^2 = 0.6\epsilon_0 \times 400 = 240\epsilon_0$ J.
Answer: 240. | ||||
| 57 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Consecutive harmonics differ by the fundamental: $\frac{v}{2L} = 450 - 400 = 50$ Hz, so $v = 50 \times 0.6 = 30$ m/s.
$\mu = \frac{T}{v^2} = \frac{2700}{900} = 3$ kg/m.
Answer: 3. | ||||
| 58 | 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 Excess pressure in a soap bubble is $\frac{4T}{R}$. Inside the inner bubble: $P_0 + \frac{4T}{6} + \frac{4T}{3} = P_0 + \frac{4T}{2}$.
A single bubble with the same internal pressure has $\frac{4T}{r} = \frac{4T}{2}$, so $r = 2$ cm.
Answer: 2. | ||||
| 59 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The cylinder unwinds like a yo-yo. $mg - 2T = ma$ and $2TR = \frac{1}{2}mR^2\cdot\frac{a}{R}$, so $2T = \frac{ma}{2}$ and $a = \frac{2g}{3} = \frac{20}{3}$ m/s².
$v^2 = 2as$: $16 = 2 \times \frac{20}{3}s$, $s = 1.2$ m $= 120$ cm.
Answer: 120. | ||||
| 60 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Smooth planes assumed. AB is at $45^\circ$: length $10\sqrt{2}$ m, retardation $g\sin 45^\circ = 5\sqrt{2}$ m/s². The block just reaches B, so (reversing the motion) $10\sqrt{2} = \frac{1}{2}(5\sqrt{2})t_1^2$, $t_1 = 2$ s.
BC is at $30^\circ$: length $\frac{10}{\sin 30^\circ} = 20$ m, acceleration $g\sin 30^\circ = 5$ m/s², from rest: $20 = \frac{1}{2}(5)t_2^2$, $t_2 = 2\sqrt{2}$ s.
Total $= 2 + 2\sqrt{2} = 2(\sqrt{2} + 1)$ s, so $t = 2$.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom The orbitals are (A) 3s, (B) 4s, (C) 3p, (D) 3d, with $n + l$ = 3, 4, 4, 5.
Higher $n + l$ is higher in energy; for equal $n + l$ the larger $n$ is higher, so 4s > 3p.
Order: 3d > 4s > 3p > 3s, i.e. (D) > (B) > (C) > (A).
Answer: (A). | ||||
| 62 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) The subtractive combination $\psi_A - \psi_B$ is the anti-bonding orbital: III.
(B) $\mu = Q \times r$ is the dipole moment: I.
(C) $\frac{N_b - N_a}{2}$ is the bond order: IV.
(D) The additive combination $\psi_A + \psi_B$ is the bonding orbital: II.
Answer: (C). | ||||
| 63 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Adding acid to the base lowers the pH, so the curve falls: (B) and (D) rise and are out.
A weak base starts at a moderate pH (about 11), falls gently through the $\text{NH}_4^+/\text{NH}_4\text{OH}$ buffer region, and the pH drop at equivalence is short and not vertical, with the equivalence point below 7 (the salt $\text{NH}_4\text{Cl}$ is acidic). That is curve (A).
Curve (C), with a long vertical drop, is the strong base–strong acid shape.
Answer: (A). | ||||
| 64 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry KI is a strong electrolyte: its molar conductivity rises only slowly (linearly in $\sqrt{c}$) on dilution. Statement I is false.
Carbonic acid is a weak electrolyte: its degree of dissociation, and so its molar conductivity, rises steeply at high dilution. Statement II is false.
Answer: (B). | ||||
| 65 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter (A) is dialysis: dissolved ions and molecules diffuse through the parchment membrane while the colloid stays behind. True.
(R) states it the wrong way round: true-solution particles pass through parchment, colloidal particles do not. False.
Answer: (C). | ||||
| 66 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties The elements are Mg, Al, P and S.
Al (3p¹) is below Mg (filled 3s²): the 3p electron is easier to remove. S (3p⁴) is below P (half-filled 3p³): S loses a paired electron.
Values: Al 577 < Mg 738 < S 1000 < P 1012 kJ mol⁻¹, i.e. (B) < (A) < (D) < (C).
Answer: (B). | ||||
| 67 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Li, the smallest alkali-metal ion, has the largest hydration enthalpy in group 1.
Li shows a diagonal relationship with Mg.
Answer: (A). | ||||
| 68 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) True: boron's maximum covalency is 4 ($\text{BF}_4^-$), so $\text{BF}_6^{3-}$ does not form, while aluminium forms $\text{AlF}_6^{3-}$.
(R) True: boron is very small. Its small size (six fluoride ions cannot be packed round it) together with the lack of available d orbitals in the second period is why it cannot expand to six-coordination, so (R) is taken as explaining (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 In neutral or faintly alkaline solution permanganate is reduced to $\text{MnO}_2$ and oxidises thiosulphate to sulphate:
$8\text{MnO}_4^- + 3\text{S}_2\text{O}_3^{2-} + \text{H}_2\text{O} \rightarrow 8\text{MnO}_2 + 6\text{SO}_4^{2-} + 2\text{OH}^-$.
Answer: (D). | ||||
| 70 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A metal in a low oxidation state is electron-rich; ligands such as CO, which accept electron density into their empty $\pi^*$ orbitals (back-bonding), remove that excess charge and stabilise it, as in $\text{Ni(CO)}_4$ and $\text{Fe(CO)}_5$.
Answer: (A). | ||||
| 71 | Environmental ChemistryGreen chemistry | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on green chemistry, from Environmental Chemistry. NCERT: No matching NCERT chapter Statement I: fly ash and steel-plant slag are used by the cement industry. Correct.
Statement II: fuel made from plastic waste has a high octane rating and contains no lead. Correct.
Answer: (A). | ||||
| 72 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids NaOH removes an $\alpha$-hydrogen of the methyl ketone to give the enolate, which is alkylated by ethyl bromide ($S_N2$) at the $\alpha$-carbon:
$\text{R–CO–CH}_3 \rightarrow \text{R–CO–CH}_2^- \xrightarrow{\text{CH}_3\text{CH}_2\text{Br}} \text{R–CO–CH}_2\text{CH}_2\text{CH}_3$.
The carbonyl group stays; the product is the ketone with a propyl chain, (C).
Answer: (C). | ||||
| 73 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\text{Br}_2$ forms a bromonium ion with 2-methylbut-2-ene; methanol, present in excess, opens it at the more substituted (tertiary) carbon, which carries more positive charge. A is $(\text{CH}_3)_2\text{C}(\text{OCH}_3)-\text{CHBr}-\text{CH}_3$.
HI cleaves the ether: the protonated ether loses methanol to give the stable tertiary carbocation, which takes up iodide ($S_N1$). B is $(\text{CH}_3)_2\text{C(I)}-\text{CHBr}-\text{CH}_3$.
Answer: (B). | ||||
| 74 | 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 (A) Benzoyl chloride + $\text{H}_2$/Pd–BaSO₄ → benzaldehyde: Rosenmund reduction (IV).
(B) Nitrile + $\text{SnCl}_2$/HCl, then hydrolysis → aldehyde: Stephen reaction (III).
(C) Toluene + $\text{CrO}_2\text{Cl}_2$, then hydrolysis → benzaldehyde: Etard reaction (II).
(D) Benzene + CO, HCl with anhydrous $\text{AlCl}_3$ → benzaldehyde: Gattermann–Koch reaction (I).
Answer: (A). | ||||
| 75 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Neoprene is polychloroprene (II); Teflon is polytetrafluoroethene (III); Acrilan is polyacrylonitrile (I); natural rubber is cis-polyisoprene (IV).
Answer: (A). | ||||
| 76 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines A water-soluble, acidic salt containing N and Cl that gives a free amine (N, no Cl) with NaOH is an amine hydrochloride.
The amine diazotises with nitrous acid and couples with phenol to an orange azo dye, so it is a primary aromatic amine: aniline.
Anilinium chloride $\text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^-$ has $M = 93 + 36.5 = 129.5$, within $131 \pm 2$.
(B) is a neutral amine, not a salt, and its free base would still contain Cl; (A) and (C) are not aromatic amines and could not give a stable diazonium salt and azo dye.
Answer: (D). | ||||
| 77 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules (A) Prolonged heating with HI reduces glucose to n-hexane (IV).
(B) Bromine water oxidises the aldehyde group only: gluconic acid (I).
(C) Acetic anhydride acetylates the five OH groups: glucose pentaacetate (II).
(D) Nitric acid oxidises both ends: saccharic (glucaric) acid (III).
Answer: (A). | ||||
| 78 | Chemistry in Everyday LifeCleansing agents | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on cleansing agents, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Sodium rosinate is added to laundry soaps to make them lather well. Sodium stearate is the soap itself; sodium carbonate and trisodium phosphate are builders that act on hard water.
Answer: (C). | ||||
| 79 | 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 (A) Chloroform (b.p. 334 K) and aniline (b.p. 457 K) differ widely in boiling point: simple distillation (III).
(B) Benzoic acid and naphthalene are separated by crystallisation from a suitable solvent (IV).
(C) Aniline is steam volatile and immiscible with water: steam distillation (I).
(D) Naphthalene sublimes, sodium chloride does not: sublimation (II).
Answer: (D). | ||||
| 80 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter $4\text{Fe}^{3+} + 3[\text{Fe(CN)}_6]^{4-} \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3$, ferric ferrocyanide, Prussian blue.
Answer: (D). | ||||
| 81 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\text{H}_2\text{SO}_4$: $100 \times 0.1 \times 2 = 20$ meq. NaOH: $50 \times 0.1 = 5$ meq.
Acid left: 15 meq in 150 mL, so $N = \frac{15}{150} = 0.1 = 1 \times 10^{-1}$ N.
Answer: 1. | ||||
| 82 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry At high pressure the $a$ term is negligible: $P(V_m - b) = RT$, so $Z = \frac{PV_m}{RT} = 1 + \frac{Pb}{RT}$.
$Z = 2$ gives $b = \frac{RT}{P} = \frac{0.083 \times 298}{99} = 0.2498$ L mol⁻¹ $\approx 25 \times 10^{-2}$ L mol⁻¹.
Answer: 25. | ||||
| 83 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat released $= C\Delta T = 2.5 \times 0.45 = 1.125$ kJ.
Moles burnt $= \frac{1.125}{9} = 0.125$ mol; mass $= 0.125 \times 280 = 35$ g.
Answer: 35. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Relative lowering of vapour pressure $\frac{p^\circ - p}{p^\circ} = \frac{1}{2}$.
For a dilute solution $\frac{p^\circ - p}{p^\circ} \approx \frac{n_B}{n_A}$, with $n_A = \frac{100}{18} = 5.56$ mol of water.
$n_B = \frac{5.56}{2} = 2.78 \approx 3$.
Answer: 3. | ||||
| 85 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For a first-order reaction $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{70 \times 60\ \text{s}} = 1.65 \times 10^{-4}$ s⁻¹ $= 165 \times 10^{-6}$ s⁻¹.
Answer: 165. | ||||
| 86 | Isolation of MetalsOres and concentration | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Iron ores in the list: siderite ($\text{FeCO}_3$), haematite ($\text{Fe}_2\text{O}_3$), magnetite ($\text{Fe}_3\text{O}_4$) and limonite ($\text{Fe}_2\text{O}_3 \cdot 3\text{H}_2\text{O}$): 4.
The others are ores of Al (bauxite, kaolinite, cryolite), Cu (cuprite, malachite) and Zn (calamine, sphalerite).
Answer: 4. | ||||
| 87 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In basic medium: $2\text{MnO}_4^- + 3\text{H}_2\text{O}_2 \rightarrow 2\text{MnO}_2 + 3\text{O}_2 + 2\text{OH}^- + 2\text{H}_2\text{O}$.
Manganese in $\text{MnO}_2$ is +4.
Answer: 4. | ||||
| 88 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Planar: $\text{NO}_3^-$, $\text{BF}_3$, $\text{SO}_3$ (trigonal planar) and $\text{XeF}_4$ (square planar).
Non-planar: $\text{H}_2\text{O}_2$ (open-book), $\text{PCl}_3$ and $\text{XeO}_3$ (pyramidal), $\text{SF}_4$ (see-saw), $\text{PH}_4^+$ and $[\text{Al(OH)}_4]^-$ (tetrahedral): 6.
Answer: 6. | ||||
| 89 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Fehling's solution contains a Cu(II) tartrate complex. $\text{Cu}^{2+}$ is $d^9$ with one unpaired electron.
$\mu = \sqrt{n(n + 2)} = \sqrt{3} = 1.73$ B.M. $\approx 2$.
Answer: 2. | ||||
| 90 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Toluene $\text{C}_7\text{H}_8$ ($M = 92$) gives benzaldehyde $\text{C}_7\text{H}_6\text{O}$ ($M = 106$) mole for mole.
Moles of toluene $= \frac{5}{92}$; benzaldehyde $= \frac{5}{92} \times 106 \times 0.92 = 5.30$ g $= 530 \times 10^{-2}$ g.
Answer: 530. | ||||
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