JEE (Main) · 2022
JEE (Main) 2022 — 24 June, 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
- Applications of Derivatives4
- Matrices and Determinants3
- Trigonometry3
4 easy · 22 medium · 4 hard
Physics
30 Q
- Properties of Bulk Matter2
- Magnetic Effects of Current and Magnetism2
- Current Electricity2
21 easy · 9 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques3
- p-Block Elements2
- Hydrocarbons2
19 easy · 9 medium · 2 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 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations A is the closed ring between the circles of radii 1 and 2 about $P(1, 1)$.
B is the set of points of the circle $|z - (1 - i)| = 1$ (centre $Q(1, -1)$, radius 1) that lie in A.
$PQ = 2$, so the distance from P of a point on the circle about Q runs from $2 - 1 = 1$ to $2 + 1 = 3$. Every point of that circle whose distance from P lies between 1 and 2 belongs to B: this is a whole arc of the circle (from the point $(1, 0)$, at distance 1, out to the two points at distance 2).
An arc has infinitely many points, so B is an infinite set.
Answer: (D). | ||||
| 2 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $3^{2022} = 9^{1011} = (10 - 1)^{1011}$. Expanding by the binomial theorem, every term except the last is a multiple of 10, and the last is $(-1)^{1011} = -1$.
So $3^{2022} = 10k - 1 = 5(2k - 1) + 4$: the remainder is 4.
Answer: (D). | ||||
| 3 | Applications of DerivativesRate of change | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on rate of change, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $S = 4\pi r^2$ and $\frac{dS}{dt} = c$ (constant), so $S$, and hence $r^2$, is a linear function of t: $r^2 = 9 + kt$.
At $t = 5$: $49 = 9 + 5k$, so $k = 8$.
At $t = 9$: $r^2 = 9 + 72 = 81$, so $r = 9$.
Answer: (A). | ||||
| 4 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let E be the event \"1 red and 1 black\".
From bag A (6 balls): $P(E|A) = \frac{3 \times 1}{^6C_2} = \frac{3}{15} = \frac{1}{5}$.
From bag B ($n + 5$ balls): $P(E|B) = \frac{2 \times 3}{^{n+5}C_2} = \frac{12}{(n+5)(n+4)}$.
By Bayes' theorem (each bag chosen with probability $\frac{1}{2}$):
$\frac{\frac{1}{5}}{\frac{1}{5} + P(E|B)} = \frac{6}{11}$, so $\frac{11}{5} = \frac{6}{5} + 6P(E|B)$, giving $P(E|B) = \frac{1}{6}$.
Then $(n+5)(n+4) = 72 = 9 \times 8$, so $n = 4$.
Answer: (C). | ||||
| 5 | CirclesStandard and general form | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections The slope of $y = x^2$ at (2, 4) is 4. The circle touches the parabola there, so its centre $(h, k)$ lies on the normal at (2, 4): $k - 4 = -\frac{1}{4}(h - 2)$.
The centre is equidistant from (2, 4) and (0, 6): $(h-2)^2 + (k-4)^2 = h^2 + (k-6)^2$, which simplifies to $-4h + 4k = 16$, i.e. $k = h + 4$.
Then $h = -\frac{1}{4}(h - 2)$, so $h = \frac{2}{5}$, $k = \frac{22}{5}$.
Hence $A = -2h = -\frac{4}{5}$, $B = -2k = -\frac{44}{5}$.
Through (0, 6): $36 + 6B + C = 0$, so $C = -36 + \frac{264}{5} = \frac{84}{5}$.
$A + C = -\frac{4}{5} + \frac{84}{5} = 16$.
Answer: (A). | ||||
| 6 | 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 $\Delta = \begin{vmatrix}1&1&1\\ \alpha&2\alpha&3\\ 1&3\alpha&5\end{vmatrix} = (10\alpha - 9\alpha) - (5\alpha - 3) + (3\alpha^2 - 2\alpha) = 3\alpha^2 - 6\alpha + 3 = 3(\alpha - 1)^2$.
For $\alpha \ne 1$ the system has a unique solution. For $\alpha = 1$:
$x + y + z = 1$, $x + 2y + 3z = -1$, $x + 3y + 5z = 4$.
Subtracting successive equations: $y + 2z = -2$ and $y + 2z = 5$, a contradiction. So the system is inconsistent only for $\alpha = 1$: one value.
Answer: (B). | ||||
| 7 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha + \beta = -\frac{\lambda}{3}$, $\alpha\beta = -\frac{1}{3}$.
$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = 9\left(\frac{\lambda^2}{9} + \frac{2}{3}\right) = \lambda^2 + 6 = 15$, so $\lambda^2 = 9$.
$\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = -\frac{\lambda^3}{27} - \frac{\lambda}{3} = -\frac{\lambda(\lambda^2 + 9)}{27} = -\frac{2\lambda}{3}$.
$6(\alpha^3 + \beta^3)^2 = 6 \cdot \frac{4\lambda^2}{9} = 6 \cdot 4 = 24$.
Answer: (B). | ||||
| 8 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $t = \tan^{-1}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$; then $\cot^{-1}x = \frac{\pi}{2} - t$.
$t^3 + \left(\frac{\pi}{2} - t\right)^3 = \frac{\pi^3}{8} - \frac{3\pi}{2}t\left(\frac{\pi}{2} - t\right) = \frac{3\pi}{2}t^2 - \frac{3\pi^2}{4}t + \frac{\pi^3}{8}$.
This quadratic in t has its minimum at $t = \frac{\pi}{4}$: $\frac{3\pi^3}{32} - \frac{3\pi^3}{16} + \frac{\pi^3}{8} = \frac{\pi^3}{32}$.
As $t \to -\frac{\pi}{2}$ it tends to $\frac{3\pi^3}{8} + \frac{3\pi^3}{8} + \frac{\pi^3}{8} = \frac{7\pi^3}{8}$ (not attained); as $t \to \frac{\pi}{2}$ it tends to $\frac{\pi^3}{8}$, which is smaller.
So the expression takes exactly the values in $\left[\frac{\pi^3}{32}, \frac{7\pi^3}{8}\right)$, and $k \in \left[\frac{1}{32}, \frac{7}{8}\right)$.
Answer: (A). | ||||
| 9 | 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 $\det A = 1(1 - 0) - 0 + a(0 + a) = 1 + a^2$.
For a $3 \times 3$ matrix, $\det(\text{adj } A) = (\det A)^2 = (1 + a^2)^2$.
With $a = \sqrt{n}$, $n = 1, 3, \ldots, 49$: $a^2 = n$, so the sum is $\sum (1 + n)^2 = 2^2 + 4^2 + \cdots + 50^2 = 4\sum_{k=1}^{25} k^2 = 4 \cdot \frac{25 \cdot 26 \cdot 51}{6} = 22100$.
$100\lambda = 22100$, so $\lambda = 221$.
Answer: (B). | ||||
| 10 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = \frac{4}{x-1} - 4x + 4 = \frac{4}{x-1} - 4(x-1) = \frac{4(1 - (x-1)^2)}{x-1} = \frac{4x(2-x)}{x-1}$.
(A) $f' > 0$ on (1, 2) and $f' < 0$ on $(2, \infty)$: correct.
(B) $f(2) = 0 - 8 + 8 + 5 = 5$ is the maximum, and $f \to -\infty$ as $x \to 1^+$ and as $x \to \infty$. So $f(x) = -1$ has one root in (1, 2) and one in $(2, \infty)$: correct.
(C) $f''(x) = -\frac{4}{(x-1)^2} - 4$, so $f''(2) = -8$. $f'(e) = \frac{4}{e-1} - 4(e-1) \approx 2.33 - 6.87 = -4.55$. Then $f'(e) - f''(2) \approx -4.55 + 8 = 3.45 > 0$. The statement is NOT correct.
(D) $f(e) = 4\ln(e-1) - 2e^2 + 4e + 5 \approx 2.17 - 14.78 + 10.87 + 5 = 3.26 > 0$ and $f(e+1) = 4 - 2(e+1)^2 + 4(e+1) + 5 \approx 4 - 27.66 + 14.87 + 5 = -3.79 < 0$, so a root lies in $(e, e+1)$: correct.
Answer: (C). | ||||
| 11 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The tangent at $(x_1, y_1)$ has slope $3x_1^2 + 6x_1$. It passes through the origin, so $y_1 = x_1(3x_1^2 + 6x_1)$:
$x_1^3 + 3x_1^2 + 5 = 3x_1^3 + 6x_1^2$, i.e. $2x_1^3 + 3x_1^2 - 5 = 0$, i.e. $(x_1 - 1)(2x_1^2 + 5x_1 + 5) = 0$. The quadratic has no real root, so $x_1 = 1$, $y_1 = 9$.
Check (1, 9): (A) $1 + 1 = 2$ ✓; (B) $9 - 1 = 8$ ✓; (C) $4 + 5 = 9$ ✓; (D) $\frac{1}{3} - 81 \ne 2$.
Answer: (D). | ||||
| 12 | 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 $2x^2 + 3x - 2 = (2x - 1)(x + 2)$ is negative on $\left[0, \frac{1}{2}\right)$ and positive on $\left(\frac{1}{2}, 1\right]$. Write $\sin x\cos x = \frac{1}{2}\sin 2x$.
On $\left[0, \frac{1}{2}\right]$: $f = -2x^2 - 3x + 2 + \frac{1}{2}\sin 2x$, $f' = -4x - 3 + \cos 2x < 0$: decreasing.
On $\left[\frac{1}{2}, 1\right]$: $f = 2x^2 + 3x - 2 + \frac{1}{2}\sin 2x$, $f' = 4x + 3 + \cos 2x > 0$: increasing.
Minimum: $f\left(\frac{1}{2}\right) = \sin\frac{1}{2}\cos\frac{1}{2} = \frac{1}{2}\sin 1$.
Maximum: larger of $f(0) = 2$ and $f(1) = 3 + \sin 1\cos 1 = 3 + \frac{1}{2}\sin 2$; it is $3 + \frac{1}{2}\sin 2$.
Sum $= 3 + \frac{1}{2}(\sin 1 + \sin 2) = 3 + \frac{1}{2}(\sin 1 + 2\sin 1\cos 1) = 3 + \frac{1}{2}(1 + 2\cos 1)\sin 1$.
Answer: (B). | ||||
| 13 | 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 Let $a_1 = a$. Then $\sum_{i=1}^{n} a_i = \frac{n}{2}(2a + n - 1) = 192$.
The even-placed terms $a + 1, a + 3, \ldots, a + n - 1$ are $\frac{n}{2}$ terms with sum $\frac{n}{4}(2a + n) = 120$.
So $n(2a + n - 1) = 384$ and $n(2a + n) = 480$. Subtracting: $n = 96$.
Answer: (B). | ||||
| 14 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dx}{dy} - \frac{2}{y}x = y^2(y + 1)e^y$, a linear equation in x with integrating factor $e^{-\int \frac{2}{y}dy} = \frac{1}{y^2}$.
$\frac{d}{dy}\left(\frac{x}{y^2}\right) = (y + 1)e^y$, so $\frac{x}{y^2} = ye^y + C$.
$x(1) = 0$: $0 = e + C$, $C = -e$. Hence $x = y^2(ye^y - e)$.
$x(e) = e^2(e \cdot e^e - e) = e^3(e^e - 1)$.
Answer: (A). | ||||
| 15 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The hyperbola is $\frac{x^2}{b^2/a^2} - \frac{y^2}{b^2} = 1$. The line is $y = \frac{\lambda}{2}x - \frac{\mu}{2}$.
The line $y = mx + c$ touches $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ when $c^2 = A^2m^2 - B^2$:
$\frac{\mu^2}{4} = \frac{b^2}{a^2}\cdot\frac{\lambda^2}{4} - b^2$. Dividing by $\frac{b^2}{4}$: $\frac{\mu^2}{b^2} = \frac{\lambda^2}{a^2} - 4$.
So $\left(\frac{\lambda}{a}\right)^2 - \left(\frac{\mu}{b}\right)^2 = 4$.
Answer: (D). | ||||
| 16 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{c}$ and $\vec{c} \times \hat{a}$ are perpendicular, so $|\hat{b}|^2 = |\vec{c}|^2 + 4|\vec{c} \times \hat{a}|^2 = |\vec{c}|^2\left(1 + 4\sin^2\frac{\pi}{12}\right) = 1$.
$\sin^2 15^\circ = \frac{1 - \cos 30^\circ}{2} = \frac{2 - \sqrt{3}}{4}$, so $1 + 4\sin^2 15^\circ = 3 - \sqrt{3}$.
$|\vec{c}|^2 = \frac{1}{3 - \sqrt{3}} = \frac{3 + \sqrt{3}}{6}$.
$|6\vec{c}|^2 = 36 \cdot \frac{3 + \sqrt{3}}{6} = 6(3 + \sqrt{3})$.
Answer: (C). | ||||
| 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 $3q^{33} = 33pq^{32}$ gives $q = 11p$, so $p = \frac{1}{12}$, $q = \frac{11}{12}$ and $\frac{q}{p} = 11$.
Since $^{33}C_{15} = {}^{33}C_{18}$: $\frac{P(X=15)}{P(X=18)} = \frac{p^{15}q^{18}}{p^{18}q^{15}} = \left(\frac{q}{p}\right)^3 = 1331$.
Since $^{33}C_{16} = {}^{33}C_{17}$: $\frac{P(X=16)}{P(X=17)} = \frac{q}{p} = 11$.
Difference: $1331 - 11 = 1320$.
Answer: (A). | ||||
| 18 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Numerator: $x^2 - 9 \ne 0$ and $\left|\frac{(x-2)(x-3)}{(x-3)(x+3)}\right| \le 1$. For $x \ne 3$ this is $\left|\frac{x-2}{x+3}\right| \le 1$, i.e. $(x-2)^2 \le (x+3)^2$, i.e. $10x \ge -5$, i.e. $x \ge -\frac{1}{2}$.
Denominator: $x^2 - 3x + 2 > 0$, i.e. $x < 1$ or $x > 2$; and $\log_e(x^2 - 3x + 2) \ne 0$, i.e. $x^2 - 3x + 1 \ne 0$, i.e. $x \ne \frac{3 \pm \sqrt{5}}{2}$ (both of which lie in the set found).
Together: $\left[-\frac{1}{2}, 1\right) \cup (2, \infty) - \left\{\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}\right\}$ (and, strictly, $x \ne 3$, where the fraction is $\frac{0}{0}$).
Answer: (D). | ||||
| 19 | 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 $\tan\theta(\sin\theta + 1) = 2\sin\theta\cos\theta$. Multiplying by $\cos\theta \ne 0$: $\sin\theta\left[(\sin\theta + 1) - 2\cos^2\theta\right] = 0$.
Either $\sin\theta = 0$: $\theta = -\pi, 0, \pi$.
Or $\sin\theta + 1 = 2 - 2\sin^2\theta$, i.e. $2\sin^2\theta + \sin\theta - 1 = 0$, i.e. $\sin\theta = \frac{1}{2}$ or $-1$; $\sin\theta = -1$ gives $\theta = -\frac{\pi}{2}$, excluded. $\sin\theta = \frac{1}{2}$: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$.
So $n(S) = 5$ and $T = 1 + 1 + 1 + \cos\frac{\pi}{3} + \cos\frac{5\pi}{3} = 3 + \frac{1}{2} + \frac{1}{2} = 4$.
$T + n(S) = 9$.
Answer: (B). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $\wedge$: with p, q both true, $p \wedge q$ is T but $(p \wedge \sim q) \vee (\sim p \wedge q)$ is F. Not a tautology.
$\vee$: $(p \vee \sim q) \vee (\sim p \vee q)$ contains $p \vee \sim p$, so it is always T; the implication is a tautology.
$\Rightarrow$: $(p \Rightarrow \sim q) \vee (\sim p \Rightarrow q) = (\sim p \vee \sim q) \vee (p \vee q)$, always T; a tautology.
$\Leftrightarrow$: $p \Leftrightarrow \sim q$ and $\sim p \Leftrightarrow q$ are both $\sim(p \Leftrightarrow q)$, so the statement is $(p \Leftrightarrow q) \Rightarrow \sim(p \Leftrightarrow q)$, false when $p \Leftrightarrow q$ is true. Not a tautology.
Two choices: $\vee$ and $\Rightarrow$.
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 | NumericalHard | |
A hard 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 $f(b) = 2f(a) + 3f(c) + f(d)$ with four distinct values from 0 to 10, so $2f(a) + 3f(c) + f(d) \le 10$. Count by $f(c)$.
$f(c) = 0$: $f(b) = 2f(a) + f(d) \le 10$, $f(a), f(d) \ge 1$, distinct from each other and from $f(b)$ (automatic since $f(b) > f(a), f(d)$).
• $f(a) = 1$: $f(d) = 2, \ldots, 8$: 7.
• $f(a) = 2$: $f(d) = 1, 3, 4, 5, 6$: 5.
• $f(a) = 3$: $f(d) = 1, 2, 4$: 3.
• $f(a) = 4$: $f(d) = 1, 2$: 2.
Subtotal 17.
$f(c) = 1$: $f(b) = 2f(a) + f(d) + 3 \le 10$, so $2f(a) + f(d) \le 7$, with $f(a), f(d) \ne 1$ and distinct.
• $f(a) = 0$: $f(d) = 2, \ldots, 7$: 6.
• $f(a) = 2$: $f(d) = 0, 3$: 2.
• $f(a) = 3$: $f(d) = 0$: 1.
Subtotal 9.
$f(c) = 2$: $2f(a) + f(d) \le 4$, values $\ne 2$, distinct.
• $f(a) = 0$: $f(d) = 1, 3, 4$: 3.
• $f(a) = 1$: $f(d) = 0$: 1.
Subtotal 4.
$f(c) = 3$: $2f(a) + f(d) \le 1$: $f(a) = 0, f(d) = 1$: 1.
$f(c) \ge 4$ is impossible with the other values distinct (it would need $2f(a) + f(d) \le -2$).
In every case $f(b)$ differs from $f(c)$ (it is larger whenever $f(a)$ or $f(d)$ is positive, which always holds). Total $17 + 9 + 4 + 1 = 31$.
Answer: 31. | ||||
| 22 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Let c questions be right and w wrong, $c + w \le 5$: $3c - 2w = 5$.
c must be odd: $c = 1$ gives $w = -1$ (impossible); $c = 3$ gives $w = 2$ ($c + w = 5$ ✓); $c = 5$ gives $w = 5$ ($c + w = 10$, impossible).
So 3 right and 2 wrong: choose the 3 right questions in $^5C_3 = 10$ ways; each wrong question can be answered wrongly in 2 ways: $2^2 = 4$.
Total $10 \times 4 = 40$.
Answer: 40. | ||||
| 23 | Matrices and DeterminantsArea of triangles using determinants | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on area of triangles using determinants, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $B = \left(-\frac{3}{\sqrt{a}}, \sqrt{a}\right)$, $C = \left(-\frac{3}{\sqrt{a}}, -\sqrt{a}\right)$.
Area of $\triangle ACD = \frac{1}{2}|\det(A - C, D - C)|$ with $A - C = \left(\frac{6}{\sqrt{a}}, 2\sqrt{a}\right)$, $D - C = \left(3\cos\theta + \frac{3}{\sqrt{a}}, a\sin\theta + \sqrt{a}\right)$:
$\det = \frac{6}{\sqrt{a}}(a\sin\theta + \sqrt{a}) - 2\sqrt{a}\left(3\cos\theta + \frac{3}{\sqrt{a}}\right) = 6\sqrt{a}(\sin\theta - \cos\theta)$.
Area $= 3\sqrt{a}\,|\sin\theta - \cos\theta|$. In the fourth quadrant ($\theta \in (-\frac{\pi}{2}, 0)$), $|\sin\theta - \cos\theta| = \sqrt{2}\,|\sin(\theta - \frac{\pi}{4})|$ reaches $\sqrt{2}$ at $\theta = -\frac{\pi}{4}$.
Maximum area $3\sqrt{2a} = 12$, so $2a = 16$, $a = 8$.
Answer: 8. | ||||
| 24 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $A = (7 + 3s, 1 - s, -2 + s)$, $B = (2t, 7 + 3t, t)$.
$\vec{AB} = (2t - 3s - 7,\ 3t + s + 6,\ t - s + 2)$ must be $k(1, -4, 2)$.
From the first and third components: $2(2t - 3s - 7) = t - s + 2$, i.e. $3t - 5s = 16$.
From the first and second: $3t + s + 6 = -4(2t - 3s - 7)$, i.e. $11t - 11s = 22$, i.e. $t = s + 2$.
Then $3s + 6 - 5s = 16$: $s = -5$, $t = -3$.
$A = (-8, 6, -7)$, $B = (-6, -2, -3)$, $\vec{AB} = (2, -8, 4)$ (indeed $2(1, -4, 2)$).
$AB^2 = 4 + 64 + 16 = 84$.
Answer: 84. | ||||
| 25 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability On $x < -1$ and $x > 1$ the pieces are continuous.
On $(-1, 1)$: $4x^2 - 1$ runs over $[-1, 3)$. $[4x^2 - 1]$ jumps where $4x^2 - 1 = 0, 1, 2$, i.e. $x^2 = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$: $x = \pm\frac{1}{2}, \pm\frac{1}{\sqrt{2}}, \pm\frac{\sqrt{3}}{2}$ — 6 points. At $x = 0$, $4x^2 - 1 = -1$ is its least value, so the floor stays $-1$ near 0: continuous.
At $x = -1$: $f(-1) = |2 + 3 - 7| = 2$; right-hand limit $[4x^2 - 1] \to [3^-] = 2$. Continuous.
At $x = 1$: $f(1) = 2 + 1 = 3$; left-hand limit $[3^-] = 2$. Discontinuous.
Total $6 + 1 = 7$.
Answer: 7. | ||||
| 26 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $\alpha = \int_{-\pi/2}^{\pi/2} f(t)dt$, $\beta = \int_{-\pi/2}^{\pi/2} tf(t)dt$. Then $f(\theta) = (1 + \alpha)\sin\theta + \beta\cos\theta$.
$\alpha = \int_{-\pi/2}^{\pi/2}[(1+\alpha)\sin t + \beta\cos t]dt = 0 + 2\beta$.
$\beta = \int_{-\pi/2}^{\pi/2}[(1+\alpha)t\sin t + \beta t\cos t]dt = (1+\alpha)\cdot 2\int_0^{\pi/2} t\sin t\,dt + 0 = 2(1 + \alpha)$, since $\int_0^{\pi/2} t\sin t\,dt = [-t\cos t + \sin t]_0^{\pi/2} = 1$ and $t\cos t$ is odd.
So $\beta = 2 + 4\beta$, $\beta = -\frac{2}{3}$, $\alpha = -\frac{4}{3}$, and $f(\theta) = -\frac{1}{3}\sin\theta - \frac{2}{3}\cos\theta$.
$\int_0^{\pi/2} f = -\frac{1}{3} - \frac{2}{3} = -1$; its absolute value is 1.
Answer: 1. | ||||
| 27 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $g(x) = \frac{9 - x^2}{5 - x}$. $g'(x) = \frac{x^2 - 10x + 9}{(5 - x)^2} = \frac{(x-1)(x-9)}{(5-x)^2}$: g increases on [0, 1] and decreases on [1, 2].
$g(0) = \frac{9}{5}$, $g(1) = 2$, $g(2) = \frac{5}{3}$. So $\alpha = 2$, $\beta = \frac{5}{3}$, and the limits are $\beta - \frac{8}{3} = -1$ to $2\alpha - 1 = 3$.
$g(x) - x = \frac{9 - 5x}{5 - x}$, which (for $x < 5$) is $\ge 0$ exactly when $x \le \frac{9}{5}$.
So the integral is $\int_{-1}^{9/5} g\,dx + \int_{9/5}^{3} x\,dx$, with $g(x) = x + 5 + \frac{16}{x - 5}$:
$\int_{-1}^{9/5} g = \left[\frac{x^2}{2} + 5x + 16\ln|x - 5|\right]_{-1}^{9/5} = \frac{28}{25} + 14 + 16\ln\frac{16/5}{6} = \frac{28}{25} + 14 + 16\ln\frac{8}{15}$.
$\int_{9/5}^{3} x\,dx = \frac{1}{2}\left(9 - \frac{81}{25}\right) = \frac{72}{25}$.
Total $= 18 + 16\ln\frac{8}{15}$: $\alpha_1 = 18$, $\alpha_2 = 16$, $\alpha_1 + \alpha_2 = 34$.
Answer: 34. | ||||
| 28 | Conic SectionsParabola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A tangent to $y^2 = 4x$ is $y = mx + \frac{1}{m}$. Through $(\alpha, \beta)$: $\alpha m^2 - \beta m + 1 = 0$, with roots $m$ and $4m$.
$5m = \frac{\beta}{\alpha}$ and $4m^2 = \frac{1}{\alpha}$. Eliminating m: $\frac{4\beta^2}{25\alpha^2} = \frac{1}{\alpha}$, so $4\beta^2 = 25\alpha$.
On the ellipse: $25\alpha^2 + 4\beta^2 = 1$, so $25\alpha^2 + 25\alpha = 1$, i.e. $\alpha^2 + \alpha = \frac{1}{25}$.
$(10\alpha + 5)^2 = 100(\alpha^2 + \alpha) + 25 = 4 + 25 = 29$.
$16\beta^2 = 100\alpha$, so $(16\beta^2 + 50)^2 = 10^4\left(\alpha^2 + \alpha + \frac{1}{4}\right) = 10^4\left(\frac{1}{25} + \frac{1}{4}\right) = 400 + 2500 = 2900$.
Sum $= 29 + 2900 = 2929$.
Answer: 2929. | ||||
| 29 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The curves meet at (0, 0) and (1, 1); S lies between $y = \sqrt{x}$ (above) and $y = x^3$ (below) for $0 \le x \le 1$, with area $\int_0^1(\sqrt{x} - x^3)dx = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$.
For $x \ge 0$, $y = 2|x| = 2x$ meets $y = \sqrt{x}$ at $x = \frac{1}{4}$ and lies above $x^3$ on [0, 1]. It cuts off the part between $\sqrt{x}$ and $2x$ for $0 \le x \le \frac{1}{4}$:
$R_1 = \int_0^{1/4}(\sqrt{x} - 2x)dx = \frac{2}{3}\cdot\frac{1}{8} - \frac{1}{16} = \frac{1}{48}$.
$R_2 = \frac{5}{12} - \frac{1}{48} = \frac{19}{48}$.
$\frac{R_2}{R_1} = 19$.
Answer: 19. | ||||
| 30 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $\vec{b_1} \times \vec{b_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-a&0\\1&-1&1\end{vmatrix} = -a\hat{i} - \hat{j} + (a - 1)\hat{k}$.
$\vec{a_2} - \vec{a_1} = \hat{i} - \hat{j} - \hat{k}$; dot product with the cross product: $-a + 1 - a + 1 = 2 - 2a$.
$|\vec{b_1} \times \vec{b_2}|^2 = a^2 + 1 + (a-1)^2 = 2(a^2 - a + 1)$.
$\frac{4(1 - a)^2}{2(a^2 - a + 1)} = \frac{2}{3}$ gives $6(a^2 - 2a + 1) = 2(a^2 - a + 1)$, i.e. $2a^2 - 5a + 2 = 0$, so $a = 2$ or $\frac{1}{2}$.
The integral value is 2.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 $B = \frac{\Delta P}{\Delta V / V}$, so $\Delta P = B \cdot \frac{\Delta V}{V} = 3 \times 10^{10} \times 0.02 = 6 \times 10^8\ Nm^{-2}$.
Answer: (C). | ||||
| 32 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Assertion and reasonEasy | |
A easy assertion and reason question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The magnetic force $q\vec{v} \times \vec{B}$ is always perpendicular to $\vec{v}$, so it does no work: the kinetic energy, and hence the speed, cannot change. (R) is true and is exactly why (A) is true.
Answer: (A). | ||||
| 33 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Two identical cells in parallel act as one cell of emf 1.5 V and internal resistance $\frac{r}{2}$. The external resistance is $20 \parallel 20 = 10$ Ω.
The voltmeter reads the terminal voltage: $V = \frac{E R}{R + r/2}$, so $1.2 = \frac{1.5 \times 10}{10 + r/2}$, giving $10 + \frac{r}{2} = 12.5$ and $r = 5$ Ω.
Answer: (C). | ||||
| 34 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (A) Both are reciprocal lengths, $[L^{-1}]$.
(B) Both are force per area, $[ML^{-1}T^{-2}]$.
(C) Both are measured in $A\,m^{-1}$, $[AL^{-1}]$.
(D) Specific heat capacity is $J\,kg^{-1}K^{-1}$, $[L^2T^{-2}K^{-1}]$; latent heat is $J\,kg^{-1}$, $[L^2T^{-2}]$: different.
Answer: (D). | ||||
| 35 | 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 The velocity becomes horizontal at the top, after $t = \frac{u\sin\theta}{g}$, so $u\sin\theta = gt$.
The time of flight is 2t, so $R = u\cos\theta \cdot 2t$, i.e. $u\cos\theta = \frac{R}{2t}$.
$\tan\theta = \frac{gt}{R/2t} = \frac{2gt^2}{R} = \frac{20t^2}{R}$, so $\theta = \tan^{-1}\left(\frac{20t^2}{R}\right) = \cot^{-1}\left(\frac{R}{20t^2}\right)$.
Answer: (D). | ||||
| 36 | 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 Retardation $a = \mu g = 0.5 \times 9.8 = 4.9\ ms^{-2}$.
$s = \frac{v^2}{2a} = \frac{9.8^2}{2 \times 4.9} = 9.8$ m.
Answer: (B). | ||||
| 37 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion The tension supplies the centripetal force (gravity neglected): $T = m\omega^2 L$, so $80 = 0.1 \times \omega^2 \times 2$, $\omega^2 = 400$, $\omega = 20$ rad/s.
In rev/min: $\frac{20 \times 60}{2\pi} = \frac{600}{\pi}$. So $K = 600$.
Answer: (C). | ||||
| 38 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $qE = mg$: $q = \frac{10^{-4} \times 9.8}{4.9 \times 10^5} = 2.0 \times 10^{-9}$ C.
Answer: (B). | ||||
| 39 | 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 $W = \int_1^2 4x\,dx + \int_2^3 3y^2\,dy = 2(4 - 1) + (27 - 8) = 6 + 19 = 25$ J.
By the work-energy theorem the kinetic energy changes by 25 J.
Answer: (C). | ||||
| 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 $g_h = \frac{gR^2}{(R + h)^2} = \frac{g}{3}$, so $R + h = \sqrt{3}R$ and $h = 0.732 \times 6400 \approx 4685$ km.
Answer: (B). | ||||
| 41 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Take $i = i_0\cos\omega t$ (maximum at $t = 0$). It equals $\frac{i_0}{\sqrt{2}}$ when $\omega t = \frac{\pi}{4}$, i.e. after $t = \frac{T}{8}$.
$T = \frac{1}{50}$ s $= 20$ ms, so $t = 2.5$ ms.
Answer: (A). | ||||
| 42 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The phase difference is $2\pi \times 1.5 = 3\pi$, an odd multiple of $\pi$: the waves are in opposite phase.
Resultant amplitude $= 5 - 3 = 2$ cm.
Answer: (A). | ||||
| 43 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Speed in the medium: $v = \frac{c}{\sqrt{\mu_r\varepsilon_r}} = \frac{3 \times 10^8}{\sqrt{1.61 \times 6.44}} = \frac{3 \times 10^8}{3.22} \approx 9.32 \times 10^7$ m/s.
$B = \mu H = \mu_r\mu_0 H = 1.61 \times 4\pi \times 10^{-7} \times 4.5 \times 10^{-2} \approx 9.10 \times 10^{-8}$ T.
$E = vB \approx 9.32 \times 10^7 \times 9.10 \times 10^{-8} \approx 8.48\ Vm^{-1}$.
Answer: (C). | ||||
| 44 | Atoms and NucleiRutherford model and alpha scattering | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on rutherford model and alpha scattering, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (A) It is the other way round: in Thomson's model the electrons sit in stable equilibrium in the ground state, while in Rutherford's model they revolve and always feel a net (centripetal) force. False.
(B) Reversed: Thomson's atom has a nearly continuous mass distribution; Rutherford's has its mass concentrated in a tiny nucleus. False.
(C) An orbiting (accelerated) electron radiates energy classically and spirals into the nucleus: Rutherford's classical atom is unstable. True.
(D) In both models the positive part carries most of the mass. False.
Answer: (C). | ||||
| 45 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $Q = (105 + 115) \times 6.4 - 220 \times 5.6 = 220 \times 0.8 = 176$ MeV.
Answer: (D). | ||||
| 46 | Communication SystemsSize of antenna | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on size of antenna, from Communication Systems. NCERT: No matching NCERT chapter The transmitted (modulated) signal is at the carrier frequency, $\lambda = \frac{3 \times 10^8}{3.5 \times 10^9} \approx 0.0857$ m $= 85.7$ mm.
The minimum antenna size is $\frac{\lambda}{4} \approx 21.4$ mm.
Answer: (C). | ||||
| 47 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\eta = 1 - \frac{T_2}{T_1}$ with $T_2 = 300$ K: $0.25 = 1 - \frac{300}{T_1}$, $T_1 = 400$ K.
Efficiency doubled to 0.5: $0.5 = 1 - \frac{300}{T_1'}$, $T_1' = 600$ K.
The source temperature must increase by 200 K, i.e. 200°C.
Answer: (B). | ||||
| 48 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $Q_{max} = CV_{max} = \frac{K\varepsilon_0 A}{d} \cdot E_{max}d = K\varepsilon_0 A E_{max}$.
$\varepsilon_0 A E_{max} = \frac{30\pi \times 10^{-4} \times 3.6 \times 10^7}{4\pi \times 9 \times 10^9} = \frac{1.08 \times 10^5}{3.6 \times 10^{10}} = 3 \times 10^{-6}$ C.
$K = \frac{7 \times 10^{-6}}{3 \times 10^{-6}} \approx 2.33$.
Answer: (D). | ||||
| 49 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B = \frac{\mu_0 I}{2r}$ at the centre and $B_x = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$ on the axis.
$\frac{B_x}{B} = \frac{r^3}{(r^2 + r^2/4)^{3/2}} = \left(\frac{4}{5}\right)^{3/2} = \left(\frac{2}{\sqrt{5}}\right)^3$.
Answer: (C). | ||||
| 50 | Properties of Bulk MatterHeat transfer | 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 heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter In the steady state the same heat current flows through both blocks:
$\frac{K_1 A(100 - 80)}{16} = \frac{K A(80 - 0)}{8}$, so $K_1 \cdot \frac{20}{16} = 10K$ and $K_1 = 8K$.
Answer: (B). | ||||
| 51 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | NumericalMedium | |
A medium numerical question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $n = \frac{56}{28} = 2$ mol; $T_1 = 400$ K.
Molecular speed $\propto \sqrt{T}$, so doubling it needs $T_2 = 4T_1 = 1600$ K, $\Delta T = 1200$ K.
In a closed vessel the volume is constant, and for diatomic $N_2$, $C_V = \frac{5}{2}R = 5$ cal mol⁻¹K⁻¹.
$Q = nC_V\Delta T = 2 \times 5 \times 1200 = 12000$ cal $= 12$ kcal.
Answer: 12. | ||||
| 52 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For each glass lens: $\frac{1}{15} = (1.5 - 1)\frac{2}{R}$, so $R = 15$ cm (equal radii).
The liquid between them forms a biconcave lens with radii 15 cm: $\frac{1}{f_l} = (1.25 - 1)\left(-\frac{2}{15}\right) = -\frac{1}{30}$.
$\frac{1}{F} = \frac{1}{15} + \frac{1}{15} - \frac{1}{30} = \frac{3}{30} = \frac{1}{10}$, so $F = 10$ cm.
Answer: 10. | ||||
| 53 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Current gain $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{1.5\ mA}{10\ \mu A} = 150$.
Input resistance $r_i = \frac{\Delta V_{BE}}{\Delta I_B} = \frac{10\ mV}{10\ \mu A} = 1$ kΩ.
Voltage gain $A_V = \beta\frac{R_L}{r_i} = 150 \times \frac{5}{1} = 750$.
Answer: 750. | ||||
| 54 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current In a pure inductor the current lags the voltage by $\frac{\pi}{2}$, so the current is at its peak when the voltage is zero.
$X_L = 2\pi fL = 2\pi \times 50 \times 0.2 = 20\pi$ Ω. With 220 V as the rms value, $V_0 = 220\sqrt{2}$ V.
$I_0 = \frac{220\sqrt{2}}{20\pi} = \frac{11\sqrt{2}}{\pi} = \frac{\sqrt{242}}{\pi}$ A.
So $a = 242$.
Answer: 242. | ||||
| 55 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The first (secondary) maximum of single-slit diffraction lies at $y \approx \frac{3\lambda D}{2a}$.
$\Delta y = \frac{3D}{2a}\Delta\lambda = \frac{3 \times 2.0}{2 \times 0.5 \times 10^{-3}} \times 5 \times 10^{-9} = 6000 \times 5 \times 10^{-9} = 3 \times 10^{-5}$ m.
Answer: 3. | ||||
| 56 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\frac{1}{2}mv^2 = h\nu - h\nu_0$.
At $2\nu_0$: $\frac{1}{2}mv_1^2 = h\nu_0$. At $5\nu_0$: $\frac{1}{2}mv_2^2 = 4h\nu_0$.
$\frac{v_2^2}{v_1^2} = 4$, so $v_2 = 2v_1$ and $x = 2$.
Answer: 2. | ||||
| 57 | 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 Taking downward positive with height h: $h = -ut_1 + \frac{1}{2}gt_1^2$ and $h = ut_2 + \frac{1}{2}gt_2^2$.
Eliminating u: multiply the first by $t_2$, the second by $t_1$ and add: $h(t_1 + t_2) = \frac{1}{2}gt_1t_2(t_1 + t_2)$, so $h = \frac{1}{2}gt_1t_2$.
For the ball dropped from rest, $h = \frac{1}{2}gt_3^2$, so $t_3 = \sqrt{t_1t_2} = \sqrt{6 \times 1.5} = 3$ s.
Answer: 3. | ||||
| 58 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The ball falls h onto the platform and then moves down a further $\frac{h}{2}$ before coming to rest. Taking the platform and spring as massless, the loss of gravitational potential energy equals the energy stored in the spring:
$mg\left(h + \frac{h}{2}\right) = \frac{1}{2}k\left(\frac{h}{2}\right)^2$
$0.1 \times 10 \times 0.15 = \frac{1}{2}k(0.05)^2$, so $0.15 = 0.00125k$ and $k = 120\ Nm^{-1}$.
Answer: 120. | ||||
| 59 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity In a potentiometer the balancing length is proportional to the emf: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.
$\frac{3}{2} = \frac{75}{l_2}$, so $l_2 = 50$ cm.
Difference $= 75 - 50 = 25$ cm.
Answer: 25. | ||||
| 60 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The scale's weight acts at the 50 cm mark (it balances at its centre). Taking moments about the new knife edge at 40 cm:
$20\ g \times (40 - 10) = M \times (50 - 40)$, so $M = 60$ g $= 6 \times 10^{-2}$ kg.
Answer: 6. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $C_{15}H_{30} + \frac{45}{2}O_2 \rightarrow 15CO_2 + 15H_2O$.
1 L of fuel weighs 756 g; molar mass $= 15 \times 12 + 30 = 210$ g mol⁻¹, so 3.6 mol.
$O_2$: $3.6 \times 22.5 \times 32 = 2592$ g. $CO_2$: $3.6 \times 15 \times 44 = 2376$ g.
Answer: (C). | ||||
| 62 | 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 Degenerate orbitals have the same energy: the same n and l (the same subshell), differing only in $m_l$.
(A) 3p and 3d: different l. Not degenerate.
(B) Both 3d ($m_l = -2$ and $-1$): degenerate.
(C) 4d and 3d: different n. Not degenerate.
Answer: (B). | ||||
| 63 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) $[PtCl_4]^{2-}$: Pt(II), $5d^8$, square planar, $dsp^2$ (III).
(B) $BrF_5$: 5 bond pairs + 1 lone pair = 6 pairs, $sp^3d^2$ (IV).
(C) $PCl_5$: 5 bond pairs, $sp^3d$ (I).
(D) $[Co(NH_3)_6]^{3+}$: Co(III), $3d^6$ low spin with strong-field $NH_3$, inner-orbital $d^2sp^3$ (II).
Answer: (B). | ||||
| 64 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Starting from 1 mol A: at equilibrium A $1 - \alpha$, B $\alpha$, C $\frac{\alpha}{2}$; total $1 + \frac{\alpha}{2} = \frac{2 + \alpha}{2}$.
Partial pressures: $p_A = \frac{2(1-\alpha)}{2+\alpha}p$, $p_B = \frac{2\alpha}{2+\alpha}p$, $p_C = \frac{\alpha}{2+\alpha}p$.
$K = \frac{p_B\,p_C^{1/2}}{p_A} = \frac{\frac{2\alpha p}{2+\alpha}\left(\frac{\alpha p}{2+\alpha}\right)^{1/2}}{\frac{2(1-\alpha)p}{2+\alpha}} = \frac{\alpha^{3/2}p^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}$.
Answer: (B). | ||||
| 65 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | StatementsEasy | |
A easy statements question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter Statement I is correct: oil-in-water emulsions are unstable and can separate into two layers on standing.
Statement II is incorrect: an emulsion is stabilised by adding an emulsifying agent (soap, protein, gum). Adding electrolyte breaks emulsions (demulsification) rather than stabilising them.
Answer: (C). | ||||
| 66 | 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 $Na_2O$ is basic; $As_2O_3$ is amphoteric; $N_2O$ and NO are neutral; $Cl_2O_7$ is strongly acidic.
Only one amphoteric oxide.
Answer: (B). | ||||
| 67 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Sphalerite (zinc blende) is ZnS (IV); calamine is $ZnCO_3$ (III); galena is PbS (II); siderite is $FeCO_3$ (I).
Answer: (A). | ||||
| 68 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter The largest single use of dihydrogen is the synthesis of ammonia (Haber process, $N_2 + 3H_2 \rightarrow 2NH_3$), which is then used for nitric acid and nitrogenous fertilisers. So it goes mostly into compounds of nitrogen.
Answer: (B). | ||||
| 69 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | StatementsMedium | |
A medium statements question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter (A) Correct: by the diagonal relationship, LiCl and $MgCl_2$ (largely covalent) are both soluble in ethanol.
(B) Incorrect: Li and Mg do not form superoxides; their oxides $Li_2O$ and MgO do not combine with excess oxygen to give superoxides.
(C) Correct: LiF (very high lattice enthalpy) is much less soluble than the other alkali metal fluorides.
(D) Incorrect: $Li_2O$, like LiF, is comparatively much less soluble than the other alkali metal oxides.
Answer: (A). | ||||
| 70 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) False: the four terminal B–H bonds (2c–2e) differ from the four bridging B–H–B bonds.
(B) False: there are two 3-centre-2-electron (banana) bonds.
(C) True: diborane is electron-deficient and accepts electron pairs (e.g. with $NH_3$, CO).
(D) True: $4BF_3 + 3NaBH_4 \rightarrow 3NaBF_4 + 2B_2H_6$ uses both.
(E) False: the terminal H atoms lie in one plane and the bridging H atoms above and below it.
So (C) and (D) only.
Answer: (C). | ||||
| 71 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Nitrogen forms its strongest bonds with the small, highly electronegative fluorine atom. $NF_3$ is thermodynamically stable, while $NCl_3$, $NBr_3$ and $NI_3$ are unstable (explosive), since the N–X bond weakens as X grows.
Answer: (A). | ||||
| 72 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Tooth enamel is hydroxyapatite $3Ca_3(PO_4)_2 \cdot Ca(OH)_2$, which fluoride in water converts to the harder fluorapatite $3Ca_3(PO_4)_2 \cdot CaF_2$.
It contains $Ca^{2+}$, $F^-$ and phosphorus in the +5 state (phosphate). Phosphorus in the +3 state is not present.
Answer: (B). | ||||
| 73 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The products are styrenes, so $C_8H_{10}$ is ethylbenzene.
A: nitration; the ethyl group is ortho/para directing and the bulky group favours para: 4-nitroethylbenzene.
B: $Br_2$ with heat brominates the benzylic position (radical substitution): $O_2N-C_6H_4-CHBr-CH_3$.
C: alcoholic KOH eliminates HBr: $O_2N-C_6H_4-CH=CH_2$ (4-nitrostyrene).
Answer: (B). | ||||
| 74 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | StatementsEasy | |
A easy statements question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Statement I is correct: an acid with an even number of carbon atoms packs better in the crystal, so its melting point is higher than those of the odd-carbon acids on either side of it (the alternation of melting points).
Statement II is correct: the first few acids are miscible with water, and the solubility falls as the hydrocarbon part grows with molar mass.
Answer: (A). | ||||
| 75 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In a conjugated diketone the two C=O groups are joined through a system of alternating single and double bonds.
(A), (B) and (D): the carbonyl groups are separated by $sp^3$ $CH_2$ groups, so they are not conjugated with each other.
(C) 1,4-Benzoquinone: O=C–C=C–C=O runs continuously round the ring, so the two carbonyls are conjugated through the ring C=C bonds.
Answer: (C). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (i) NaCN substitutes the benzylic bromide: $Ar-CH_2-CN$ (Ar = 4-methoxyphenyl).
(ii) $OH^-$ removes a proton from the $CH_2$, which is acidic because it is benzylic and α to the nitrile: $Ar-\bar{C}H-CN$.
(iii) The carbanion adds to the C=O of cyclohexanone (nucleophilic addition); after protonation: $Ar-CH(CN)-C(OH)<$ (a 1-hydroxycyclohexyl group).
(iv) $H_2$/Ni reduces $-C\equiv N$ to $-CH_2NH_2$: $Ar-CH(CH_2NH_2)-C(OH)$(cyclohexane), i.e. 1-[2-amino-1-(4-methoxyphenyl)ethyl]cyclohexan-1-ol.
This is structure (D).
Answer: (D). | ||||
| 77 | PolymersBiodegradable polymers | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biodegradable polymers, from Polymers. NCERT: No matching NCERT chapter PHBV, poly-β-hydroxybutyrate-co-β-hydroxyvalerate, is a copolymer of 3-hydroxybutanoic and 3-hydroxypentanoic acids joined by ester links: a biodegradable polyester.
Butadiene-styrene (Buna-S) and neoprene are addition polymers; melamine polymer is a melamine-formaldehyde condensation polymer, not an ester.
Answer: (D). | ||||
| 78 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Y gives gluconic acid with bromine water, so Y is glucose and X is a glucose polymer. Starch, amylose and amylopectin have α-glycosidic links; cellulose is a linear chain of β-D-glucose units joined only by β(1→4) glycosidic linkages.
Answer: (B). | ||||
| 79 | Chemistry in Everyday LifeAntibiotics and antiseptics | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on antibiotics and antiseptics, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Penicillin G acts mainly against Gram-positive bacteria: it is a narrow spectrum antibiotic.
Ampicillin (a modified penicillin) and ofloxacin are broad spectrum, and the syllabus text counts vancomycin among the important broad spectrum antibiotics.
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 In ammoniacal (alkaline) solution, $Ni^{2+}$ gives a bright red (rosy red) precipitate of nickel dimethylglyoximate with dimethylglyoxime. Nessler's reagent is the test for ammonium ion (brown precipitate).
Answer: (A). | ||||
| 81 | Solid StateClose packed structures | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on close packed structures, from Solid State. NCERT: No matching NCERT chapter For N atoms of X in hcp there are 2N tetrahedral voids. Y occupies $\frac{2}{3} \times 2N = \frac{4N}{3}$ of them.
X : Y $= N : \frac{4N}{3} = 3 : 4$, so the formula is $X_3Y_4$.
Percentage of X (by number of atoms) $= \frac{3}{7} \times 100 = 42.86 \approx 43$.
Answer: 43. | ||||
| 82 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Start with 1 mol $O_3$; half dissociates: $O_3$ 0.5 mol, $O_2$ $\frac{3}{2} \times 0.5 = 0.75$ mol; total 1.25 mol at 1 atm.
$p_{O_3} = 0.4$ atm, $p_{O_2} = 0.6$ atm.
$K_p = \frac{(0.6)^3}{(0.4)^2} = \frac{0.216}{0.16} = 1.35$.
$\Delta G^\circ = -RT\ln K_p = -8.3 \times 300 \times 0.3 = -747\ J\ mol^{-1}$.
Answer: 747. | ||||
| 83 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\pi = CRT$: $C = \frac{7.47}{0.083 \times 300} = \frac{7.47}{24.9} = 0.3$ mol L⁻¹.
In grams: $0.3 \times 180 = 54\ g\ L^{-1}$.
Answer: 54. | ||||
| 84 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $H_2 + Cu^{2+} \rightarrow 2H^+ + Cu$, $n = 2$, $E^\circ_{cell} = 0.34$ V (taking $p_{H_2} = 1$ bar).
$E = E^\circ - \frac{0.06}{2}\log\frac{[H^+]^2}{[Cu^{2+}]}$
$0.576 = 0.34 - 0.03\log\frac{[H^+]^2}{0.01}$, so $\log\frac{[H^+]^2}{0.01} = -7.87$.
$2\log[H^+] + 2 = -7.87$, $\log[H^+] = -4.93$, pH $= 4.93 \approx 5$.
Answer: 5. | ||||
| 85 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln k = \ln A - \frac{E_a}{RT} = \ln A - \frac{E_a}{1000R}\cdot\frac{10^3}{T}$.
So the slope $= -\frac{E_a}{1000R} = -18.5$, giving $E_a = 18.5 \times 1000 \times 8.31 = 153735\ J\ mol^{-1} \approx 153.7\ kJ\ mol^{-1}$.
To the nearest integer: 154.
Answer: 154. | ||||
| 86 | 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 chromate, $CrO_4^{2-}$: $x - 8 = -2$, $x = +6$.
In dichromate, $Cr_2O_7^{2-}$: $2x - 14 = -2$, $x = +6$.
Difference $= 0$.
Answer: 0. | ||||
| 87 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds In $Co_2(CO)_8$ the two Co atoms are joined by one Co–Co bond and by two bridging CO groups; each Co also carries three terminal CO groups.
X = 1, Y = 6, X + Y = 7.
Answer: 7. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Kjeldahl's method.
Acid taken: $50.0 \times 0.5 = 25.0$ meq. Acid left over (neutralised by NaOH): $30.0 \times 0.25 = 7.5$ meq.
Acid used by $NH_3$: $25.0 - 7.5 = 17.5$ meq, so $NH_3 = 17.5$ mmol and N $= 17.5 \times 14 = 245$ mg $= 0.245$ g.
% N $= \frac{0.245}{0.166} \times 100 = 147.6 \approx 148$.
Answer: 148.
Read literally, the acid consumed by the ammonia is 25 − 7.5 = 17.5 meq, which gives 1.4 × 17.5 / 0.166 = 147.6 % nitrogen, impossible. The data are evidently meant so that the 7.5 meq is the acid used by the ammonia: % N = 1.4 × 7.5 / 0.166 = 63.25, so 63.
Answer: 63. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesTypes of organic reactions | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on types of organic reactions, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The electron-poor carbon atoms that a nucleophile can attack are:
1. The carbonyl carbon of the ketone ($C^{\delta+}=O^{\delta-}$).
2. The β-carbon of the C=C conjugated with the C=O (the α,β-unsaturated ketone), which is electron-poor by resonance ($-CH=CH-C=O \leftrightarrow -\overset{+}{C}H-CH=C-O^-$).
3. The nitrile carbon of $-C\equiv N$.
The naphthalene ring carbons and the $CH_2$ groups are not electrophilic centres.
Answer: 3. | ||||
| 90 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalHard | |
A hard numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $(CH_3)_2C=CH-CH_2-CH_2-CH=C(CH_3)_2$. Protonation of one C=C at its CH end gives a tertiary carbocation, $(CH_3)_2\overset{+}{C}-CH_2-CH_2-CH_2-CH=C(CH_3)_2$.
The other C=C attacks this cation intramolecularly; bonding through its CH carbon closes a five-membered ring and leaves a tertiary cation outside the ring: a 2,2-dimethylcyclopentyl ring carrying $-\overset{+}{C}(CH_3)_2$.
Loss of a proton gives the most substituted (tetrasubstituted) alkene: 1,1-dimethyl-2-(propan-2-ylidene)cyclopentane.
The product has one C=C, so 2 $sp^2$ carbon atoms (the other carbons are all $sp^3$).
Answer: 2. | ||||
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