JEE (Main) · 2022
JEE (Main) 2022 — 25 July, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Conic Sections3
- Complex Numbers2
7 easy · 14 medium · 9 hard
Physics
30 Q
- Electromagnetic Induction and Alternating Currents3
- Physical World and Measurement2
- Kinematics2
25 easy · 5 medium · 0 hard
Chemistry
30 Q
- Some Basic Concepts of Chemistry2
- d- and f-Block Elements2
- p-Block Elements2
17 easy · 11 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 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Choose $f(1) = a$ and $f(2) = b$; then $f(3) = a + b$ must lie in $\{1, \ldots, 6\}$, so $a + b \le 6$ with $a, b \ge 1$.
For a sum $s = a + b$ there are $s - 1$ ordered pairs, so the number of pairs is $1 + 2 + 3 + 4 + 5 = 15$ (for $s = 2, \ldots, 6$).
$f(4)$ is free: 6 choices.
Total $15 \times 6 = 90$.
Answer: (B). | ||||
| 2 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Since $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$, the four roots are the fifth roots of unity other than 1, so each satisfies $x^5 = 1$.
$2021 = 5 \times 404 + 1$, so $\alpha^{2021} = \alpha$, and similarly for the others.
The sum is $\alpha + \beta + \gamma + \delta = -1$ (minus the coefficient of $x^3$).
Answer: (B). | ||||
| 3 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctHard | |
A hard single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $S_n$ is the circle with centre $C_1 = (3, -2)$ and radius $r_1 = \frac{n}{4}$; $T_n$ is the circle with centre $C_2 = (2, -3)$ and radius $r_2 = \frac{1}{n}$. The distance between the centres is $C_1C_2 = \sqrt{1 + 1} = \sqrt{2} \approx 1.414$.
The circles fail to meet when they lie outside each other, $r_1 + r_2 < \sqrt{2}$, or when one lies inside the other, $|r_1 - r_2| > \sqrt{2}$.
Outside each other: $\frac{n}{4} + \frac{1}{n} < \sqrt{2} \iff n^2 - 4\sqrt{2}\,n + 4 < 0 \iff 2\sqrt{2} - 2 < n < 2\sqrt{2} + 2$, i.e. $0.83 < n < 4.83$: $n = 1, 2, 3, 4$ (four values).
For $n = 5$: $r_1 + r_2 = 1.45 > 1.414$ and $r_1 - r_2 = 1.05 < 1.414$, so the circles meet; likewise for $n = 6$ ($1.667$ and $1.333$).
For $n \ge 7$, $r_1 - r_2 = \frac{n}{4} - \frac{1}{n} \ge 1.607 > \sqrt{2}$, so $T_n$ lies inside $S_n$ without touching it and the intersection is again empty.
So $S_n \cap T_n = \phi$ for $n = 1, 2, 3, 4$ and every $n \ge 7$: infinitely many values, none of the options.
No option is correct as printed. | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctHard | |
A hard single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The determinant of the coefficients is
$\Delta = 3\sin 3\theta(28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 21\sin 3\theta + 42\cos 2\theta - 42$.
$\Delta = 0 \iff \sin 3\theta = 2(1 - \cos 2\theta) = 4\sin^2\theta \iff 3\sin\theta - 4\sin^3\theta - 4\sin^2\theta = 0 \iff \sin\theta(2\sin\theta - 1)(2\sin\theta + 3) = 0$.
So $\sin\theta = 0$ or $\sin\theta = \frac{1}{2}$.
• $\sin\theta = 0$: $\sin 3\theta = 0$, $\cos 2\theta = 1$. The equations become $-y + z = 2$, $3x + 4y + 3z = 3$, $6x + 7y + 7z = 9$. The third minus twice the second gives $-y + z = 3$, contradicting the first: no solution. In $(0, 4\pi)$: $\theta = \pi, 2\pi, 3\pi$ (3 values).
• $\sin\theta = \frac{1}{2}$: $\sin 3\theta = 3 \cdot \frac{1}{2} - 4 \cdot \frac{1}{8} = 1$, $\cos 2\theta = 1 - 2 \cdot \frac{1}{4} = \frac{1}{2}$. The equations become $3x - y + z = 2$, $\frac{3}{2}x + 4y + 3z = 3$, $6x + 7y + 7z = 9$. The left side of the third is (first) + 2(second), but $2 + 2 \cdot 3 = 8 \ne 9$: no solution. In $(0, 4\pi)$: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$ (4 values).
Total $3 + 4 = 7$.
Answer: (B). | ||||
| 5 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For the limit to be finite the $n$ terms must cancel: $\sqrt{n^2 - n - 1} \approx n$, so $\alpha = -1$.
Then $\sqrt{n^2 - n - 1} - n = \frac{-n - 1}{\sqrt{n^2 - n - 1} + n} \to -\frac{1}{2}$, so the limit is $-\frac{1}{2} + \beta = 0$, giving $\beta = \frac{1}{2}$.
$8(\alpha + \beta) = 8\left(-\frac{1}{2}\right) = -4$.
Answer: (C). | ||||
| 6 | 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 $f'(x) = e^{(4x^3 - 12x^2 - 180x + 31)}\left[(2x - 2) + (x^2 - 2x + 7)(12x^2 - 24x - 180)\right]$.
$12x^2 - 24x - 180 = 12(x - 5)(x + 3)$, which is $\le 0$ on $[-3, 0]$; $x^2 - 2x + 7 > 0$; and $2x - 2 < 0$ on $[-3, 0]$.
So the bracket is negative throughout $[-3, 0]$: $f$ is strictly decreasing there, and its absolute maximum is at the left end, $x = -3$.
Answer: (B). | ||||
| 7 | 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 Q is $(0, 5)$ and $y'(0) = c = 3$.
Touching the $x$-axis at $(-2, 0)$: $y(-2) = 0$ and $y'(-2) = 0$.
$y(-2) = -8a + 4b - 6 + 5 = 0 \Rightarrow -8a + 4b = 1$.
$y'(-2) = 12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3$.
Adding: $4a = -2$, $a = -\frac{1}{2}$; then $4b = 1 - 4 = -3$, $b = -\frac{3}{4}$.
$y' = -\frac{3}{2}x^2 - \frac{3}{2}x + 3 = -\frac{3}{2}(x + 2)(x - 1)$, which changes from + to − at $x = 1$: local maximum there.
$y(1) = -\frac{1}{2} - \frac{3}{4} + 3 + 5 = \frac{27}{4}$.
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 The lines $y = x + 2$ and $y = 4 - 3x$ meet at $x = \frac{1}{2}$; for $x < \frac{1}{2}$ the smaller is $x + 2$, for $x > \frac{1}{2}$ it is $4 - 3x$.
$x^2 = x + 2$ at $x = -1, 2$; $x^2 = 4 - 3x$ at $x = 1, -4$. So the region runs from $x = -1$ to $x = 1$.
Area $= \int_{-1}^{1/2}(x + 2 - x^2)\,dx + \int_{1/2}^{1}(4 - 3x - x^2)\,dx$.
First: $\left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{1/2} = \frac{26}{24} - \left(-\frac{28}{24}\right) = \frac{9}{4}$.
Second: $\left[4x - \frac{3x^2}{2} - \frac{x^3}{3}\right]_{1/2}^{1} = \frac{13}{6} - \frac{19}{12} = \frac{7}{12}$.
Total $\frac{9}{4} + \frac{7}{12} = \frac{34}{12} = \frac{17}{6}$.
Answer: (B). | ||||
| 9 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $[k, k + 1)$ write $x = k + t$, $0 \le t < 1$: $\cos \pi x = (-1)^k \cos \pi t$.
If $k$ is odd, $f = t$ and the piece is $-\int_0^1 t\cos \pi t\,dt$; if $k$ is even, $f = 1 - t$ and the piece is $\int_0^1 (1 - t)\cos \pi t\,dt$.
$\int_0^1 \cos \pi t\,dt = 0$ and $\int_0^1 t\cos \pi t\,dt = \left[\frac{t\sin \pi t}{\pi}\right]_0^1 + \left[\frac{\cos \pi t}{\pi^2}\right]_0^1 = -\frac{2}{\pi^2}$.
So every unit interval contributes $\frac{2}{\pi^2}$, odd or even. There are 20 of them in $[-10, 10]$: the integral is $\frac{40}{\pi^2}$.
$\frac{\pi^2}{10} \cdot \frac{40}{\pi^2} = 4$.
Answer: (A). | ||||
| 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 Multiply numerator and denominator by $e^{2x}$: $\frac{dy}{dx} = \frac{2e^{4x} + 9e^{2x} - 6e^{x}}{2e^{2x} + 9} = e^{2x} - \frac{6e^x}{2e^{2x} + 9}$.
With $u = e^x$: $\int \frac{6e^x\,dx}{2e^{2x} + 9} = \int \frac{3\,du}{u^2 + \frac{9}{2}} = \sqrt{2}\tan^{-1}\frac{\sqrt{2}\,e^x}{3}$.
So $y = \frac{1}{2}e^{2x} - \sqrt{2}\tan^{-1}\frac{\sqrt{2}\,e^x}{3} + K$.
At $x = 0$: $\frac{1}{2} - \sqrt{2}\tan^{-1}\frac{\sqrt{2}}{3} + K = \frac{1}{2} + \frac{\pi}{2\sqrt{2}}$, so $K = \frac{\pi}{2\sqrt{2}} + \sqrt{2}\tan^{-1}\frac{\sqrt{2}}{3}$.
At $x = \alpha$, $y = \frac{1}{2}e^{2\alpha}$ gives $\sqrt{2}\tan^{-1}\frac{\sqrt{2}\,e^\alpha}{3} = K$, i.e. $\tan^{-1}\frac{\sqrt{2}\,e^\alpha}{3} = \frac{\pi}{4} + \tan^{-1}\frac{\sqrt{2}}{3}$.
Taking tangents: $\frac{\sqrt{2}\,e^\alpha}{3} = \frac{1 + \frac{\sqrt{2}}{3}}{1 - \frac{\sqrt{2}}{3}} = \frac{3 + \sqrt{2}}{3 - \sqrt{2}}$, so $e^\alpha = \frac{3}{\sqrt{2}}\left(\frac{3 + \sqrt{2}}{3 - \sqrt{2}}\right)$.
Answer: (B). | ||||
| 11 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $t = y^2$, $dt = 2y\,dy$: $(x - t)\,dx + \frac{1}{2}(5x + t)\,dt = 0$, so $\frac{dt}{dx} = \frac{2(t - x)}{5x + t}$, a homogeneous equation.
Put $t = vx$: $v + x\frac{dv}{dx} = \frac{2(v - 1)}{v + 5}$, so $x\frac{dv}{dx} = -\frac{v^2 + 3v + 2}{v + 5} = -\frac{(v + 1)(v + 2)}{v + 5}$.
Separating: $\int \frac{v + 5}{(v + 1)(v + 2)}\,dv = -\int \frac{dx}{x}$, with $\frac{v + 5}{(v + 1)(v + 2)} = \frac{4}{v + 1} - \frac{3}{v + 2}$.
$4\ln|v + 1| - 3\ln|v + 2| + \ln|x| = \text{const}$, i.e. $\frac{(v + 1)^4\,|x|}{|v + 2|^3} = C$.
With $v + 1 = \frac{t + x}{x}$ and $v + 2 = \frac{t + 2x}{x}$ the powers of $x$ cancel: $(t + x)^4 = C\,|(t + 2x)^3|$, that is $(y^2 + x)^4 = C\left|(y^2 + 2x)^3\right|$.
Answer: (A). | ||||
| 12 | Straight LinesCartesian coordinates and distance formula | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines Let $B = (t, t - 2)$. $AB^2 = (t - 4)^2 + (t - 5)^2 = \frac{29}{9}$, so $2t^2 - 18t + 41 = \frac{29}{9}$, i.e. $9t^2 - 81t + 170 = 0$, giving $t = \frac{81 \pm 21}{18} = \frac{17}{3}$ or $\frac{10}{3}$.
Slope of AB $= \frac{t - 5}{t - 4}$: for $t = \frac{17}{3}$ it is $\frac{2}{5}$; for $t = \frac{10}{3}$ it is $\frac{5}{2} > 1$.
So $B = \left(\frac{10}{3}, \frac{4}{3}\right)$, and $x + 2y = \frac{10}{3} + \frac{8}{3} = 6$.
(Checks of the others: $2x + y = 8$, $3x - 2y = \frac{22}{3}$, $2x - 3y = \frac{8}{3}$.)
Answer: (C). | ||||
| 13 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The circle touching the $x$-axis with centre $(\alpha, \beta)$, $\beta > 0$, has radius $\beta$. External contact with the circle of centre $(0, 1)$ and radius 1: $\sqrt{\alpha^2 + (\beta - 1)^2} = \beta + 1$, so $\alpha^2 = 4\beta$.
L is the parabola $x^2 = 4y$. It meets $y = 4$ at $x = \pm 4$.
Area $= \int_{-4}^{4}\left(4 - \frac{x^2}{4}\right)dx = 2\left[4x - \frac{x^3}{12}\right]_0^4 = 2\left(16 - \frac{16}{3}\right) = \frac{64}{3}$.
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 The two lines through the origin span a plane with normal $(2, 3, 5) \times (3, 7, 8) = (-11, -1, 5)$.
P is perpendicular to that plane, so $(-11, -1, 5)$ is parallel to P; P also contains the direction $(9, -1, -5)$.
Normal of P: $(9, -1, -5) \times (-11, -1, 5) = (-10, 10, -20) \parallel (1, -1, 2)$.
P passes through $(3, -4, 7)$: $x - y + 2z = 3 + 4 + 14 = 21$.
$d = \frac{|2 + 5 + 22 - 21|}{\sqrt{6}} = \frac{8}{\sqrt{6}}$, so $d^2 = \frac{64}{6} = \frac{32}{3}$.
Answer: (C). | ||||
| 15 | 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 Going round the triangle, $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
$|\vec{a}|^2 = |\vec{b} + \vec{c}|^2 = 12 + |\vec{c}|^2 + 2(12)$, so $72 = 36 + |\vec{c}|^2$ and $|\vec{c}| = 6$.
(S1): $(\vec{a} \times \vec{b}) + (\vec{c} \times \vec{b}) = (\vec{a} + \vec{c}) \times \vec{b} = -\vec{b} \times \vec{b} = \vec{0}$, so the left side is $0 - 6 = -6 \ne 6(2\sqrt{2} - 1)$. False.
(S2): $|\vec{c}|^2 = |\vec{a} + \vec{b}|^2 = 72 + 12 + 2\vec{a} \cdot \vec{b} = 36$, so $\vec{a} \cdot \vec{b} = -24$. The angle at C is between $\vec{CA} = \vec{b}$ and $\vec{CB} = -\vec{a}$: $\cos C = \frac{24}{6\sqrt{2} \cdot 2\sqrt{3}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$. True.
Answer: (C). | ||||
| 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 $np$, variance $npq$: $np + npq = 24$ and $np \cdot npq = 128$.
The two numbers are the roots of $t^2 - 24t + 128 = 0$: 16 and 8. Since the variance is smaller than the mean, $np = 16$ and $npq = 8$, so $q = \frac{1}{2}$, $p = \frac{1}{2}$, $n = 32$.
$P(X = 1) + P(X = 2) = \frac{\binom{32}{1} + \binom{32}{2}}{2^{32}} = \frac{32 + 496}{2^{32}} = \frac{528}{2^{32}} = \frac{33 \cdot 16}{2^{32}} = \frac{33}{2^{28}}$.
Answer: (C). | ||||
| 17 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability The quadratic is positive for all real $x$ iff its discriminant is negative: $\alpha^2 < 4\beta$.
Count over the 36 equally likely pairs:
$\alpha = 1$: $\beta > \frac{1}{4}$: 6. $\alpha = 2$: $\beta > 1$: 5. $\alpha = 3$: $\beta > \frac{9}{4}$: 4. $\alpha = 4$: $\beta > 4$: 2. $\alpha = 5$: $\beta > \frac{25}{4}$: 0. $\alpha = 6$: $\beta > 9$: 0.
Total 17, so the probability is $\frac{17}{36}$.
Answer: (A). | ||||
| 18 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions We need $\sin x \ge 0$ and $\sin x = |\cos x|$, i.e. $\tan x = \pm 1$ with $\sin x > 0$.
In one period $[0, 2\pi)$ this gives $x = \frac{\pi}{4}$ and $\frac{3\pi}{4}$: two solutions.
$[-4\pi, 4\pi]$ contains four full periods (the end points are not solutions), so there are $4 \times 2 = 8$ solutions.
Answer: (C). | ||||
| 19 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $AQ = \frac{QR}{\tan 60^\circ} = \frac{15}{\sqrt{3}} = 5\sqrt{3}$ m.
The elevation of the top P is $60^\circ + 15^\circ = 75^\circ$, so $PQ = AQ\tan 75^\circ = 5\sqrt{3}(2 + \sqrt{3}) = 10\sqrt{3} + 15 = 5(2\sqrt{3} + 3)$ m.
Answer: (A). | ||||
| 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 (A) $p$ false makes $(\sim p) \vee q$ true and $p$ false: the implication is false.
(B) $p$ true, $q$ false: $(\sim p) \vee q$ is false, so the implication is false.
(C) $p$ false, $q$ false: $(\sim p) \vee q$ is true and $q$ false: false.
(D) If $q$ is true then $(\sim p) \vee q$ is true; if $q$ is false the implication is true anyway. It is always true: a tautology.
Answer: (D). | ||||
| 21 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalHard | |
A hard numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Direct multiplication gives $A^2 = A$, so $A^n = A$ for every $n \ge 1$.
$B = A - I = \begin{pmatrix} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{pmatrix}$ and $B^2 = -B$, so $B^n = (-1)^{n-1}B$.
The condition becomes $A + \omega^n(-1)^{n-1}B = A + B$, i.e. $(-1)^{n-1}\omega^n = 1$, i.e. $(-\omega)^n = -1$.
$\omega = e^{2\pi i/3}$, so $-\omega = e^{-i\pi/3}$, a primitive sixth root of unity, and $(-\omega)^n = -1$ exactly when $n \equiv 3 \pmod 6$.
In $1, \ldots, 100$: $n = 3, 9, 15, \ldots, 99$, which is 17 values.
Answer: 17. | ||||
| 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 The letters in order are A, D, I, K, M, N, N (N twice).
Words beginning with A, D, I or K: each leaves 6 letters with N repeated, $\frac{6!}{2!} = 360$; total $4 \times 360 = 1440$.
Words beginning MA (nothing comes before A) and then D, I or K in third place: each leaves 4 letters with N twice, $\frac{4!}{2!} = 12$; total 36.
MAN then D or I in fourth place: $3! = 6$ each; total 12.
MANK then D in fifth place: $2! = 2$.
MANKI then D in sixth place: 1.
Then MANKIND itself: 1.
Serial number $1440 + 36 + 12 + 2 + 1 + 1 = 1492$.
Answer: 1492. | ||||
| 23 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom{15}{r}\left(t^2x^{1/5}\right)^{15 - r}\left(\frac{(1 - x)^{1/10}}{t}\right)^r$ has $t^{30 - 3r}$, so $r = 10$.
The term is $\binom{15}{10}x^{5/5}(1 - x)^{10/10} = 3003\,x(1 - x)$.
For $0 \le x \le 1$, $x(1 - x)$ is greatest at $x = \frac{1}{2}$, value $\frac{1}{4}$: $K = \frac{3003}{4}$.
$8K = 6006$.
Answer: 6006. | ||||
| 24 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\frac{1}{p} + \frac{1}{r} = \frac{p + r}{pr} = \frac{8a}{2a} = 4$ and $\frac{1}{q} + \frac{1}{s} = \frac{-12b}{6b} = -2$.
Let the A.P. be $A, A + d, A + 2d, A + 3d$. Then $2A + 2d = 4$ and $2A + 4d = -2$, so $d = -3$, $A = 5$.
$\frac{1}{p} = 5$, $\frac{1}{q} = 2$, $\frac{1}{r} = -1$, $\frac{1}{s} = -4$.
$pr = 2a$: $\frac{1}{5} \cdot (-1) = 2a$, $a = -\frac{1}{10}$. $qs = 6b$: $\frac{1}{2} \cdot \left(-\frac{1}{4}\right) = 6b$, $b = -\frac{1}{48}$.
(Check: $p + r = -\frac{4}{5} = 8a$ and $q + s = \frac{1}{4} = -12b$.)
$a^{-1} - b^{-1} = -10 + 48 = 38$.
Answer: 38. | ||||
| 25 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $a_n = 2n - 1$, and $b_n = a_1 + a_2 + \cdots + a_n = 1 + 3 + \cdots + (2n - 1) = n^2$.
$\sum_{n=1}^{15}(2n - 1)n^2 = 2\sum n^3 - \sum n^2 = 2\left(\frac{15 \cdot 16}{2}\right)^2 - \frac{15 \cdot 16 \cdot 31}{6} = 28800 - 1240 = 27560$.
Answer: 27560. | ||||
| 26 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $g(x) = 4x^2 - 8x + 5 = 4(x - 1)^2 + 1 > 0$, so $|g| = g$.
$8x^2 - 6x + 1 = (4x - 1)(2x - 1) < 0$ exactly on $\left(\frac{1}{4}, \frac{1}{2}\right)$. Outside it $f = g$, a polynomial.
On $\left(\frac{1}{4}, \frac{1}{2}\right)$, $g$ decreases from $g\left(\frac{1}{4}\right) = 3.25$ to $g\left(\frac{1}{2}\right) = 2$, so $[g]$ is 3 and then 2, jumping where $g = 3$: $4x^2 - 8x + 2 = 0$, $x = 1 - \frac{1}{\sqrt{2}} \approx 0.293$, inside the interval.
• $x = \frac{1}{4}$: $f\left(\frac{1}{4}\right) = 3.25$ but $f \to 3$ from the right: discontinuous.
• $x = 1 - \frac{1}{\sqrt{2}}$: $[g]$ jumps from 3 to 2: discontinuous.
• $x = \frac{1}{2}$: $f$ is continuous (both sides give 2), but the left derivative is 0 (constant 2) and the right derivative is $g'\left(\frac{1}{2}\right) = -4$.
So $f$ is not differentiable at 3 points.
Answer: 3. | ||||
| 27 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Left: $(nk + 1) + \cdots + (nk + n) = n^2k + \frac{n(n + 1)}{2}$, and $\frac{(n + 1)^{k-1}}{n^{k+1}} \cdot n^2\left(k + \frac{n + 1}{2n}\right) \to k + \frac{1}{2}$.
Right: $\frac{1}{n^{k+1}}\sum_{r=1}^{n} r^k = \frac{1}{n}\sum\left(\frac{r}{n}\right)^k \to \int_0^1 x^k\,dx = \frac{1}{k + 1}$.
So $\frac{2k + 1}{2} = \frac{33}{k + 1}$, i.e. $2k^2 + 3k - 65 = 0$, $(k - 5)(2k + 13) = 0$; the integral value is $k = 5$.
Answer: 5. | ||||
| 28 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The centre is $(1, -f)$ and lies on both diameters: $2p + f = 1$ and $2 - pf = 4p$.
With $f = 1 - 2p$: $2 - p + 2p^2 = 4p$, so $2p^2 - 5p + 2 = 0$, $p = 2$ or $\frac{1}{2}$, giving $f = -3$ or $0$.
The radius is $\sqrt{1 + f^2 - 1} = |f|$; $f = 0$ gives a point circle, so $f = -3$ and the centre is $(1, 3)$.
The hyperbola is $\frac{x^2}{1} - \frac{y^2}{3} = 1$; its tangents of slope $m$ are $y = mx \pm \sqrt{m^2 - 3}$.
Through $(1, 3)$: $(3 - m)^2 = m^2 - 3$, so $9 - 6m = -3$, $m = 2$ (positive, and $m^2 > 3$).
Answer: 2. | ||||
| 29 | Conic SectionsParabola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The point is on the parabola ($75 \cdot \frac{64}{25} = 192 = 64 \cdot 3$). Differentiating, $150x = 320y'$, so at $x = \frac{8}{5}$ the tangent slope is $\frac{3}{4}$ and the normal has direction $(3, -4)$.
A circle touching the parabola there has its centre on the normal: $\left(\frac{8}{5} + 3u, \frac{6}{5} - 4u\right)$ with radius $5|u|$.
Touching the $y$-axis: $\left|\frac{8}{5} + 3u\right| = 5|u|$.
$u > 0$: $\frac{8}{5} + 3u = 5u$, $u = \frac{4}{5}$, radius 4.
$u < 0$: $\frac{8}{5} + 3u = -5u$, $u = -\frac{1}{5}$, radius 1.
(The remaining sign choices give no solution.) The diameters are 8 and 2; their sum is 10.
Answer: 10. | ||||
| 30 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalHard | |
A hard 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 The line of shortest distance is perpendicular to both lines: direction $(0, 1, 1) \times (2, 2, 1) = (-1, 2, -2)$, of length 3.
With the normal $(a, -1, -1)$ of P: $\frac{|(-1, 2, -2) \cdot (a, -1, -1)|}{3\sqrt{a^2 + 2}} = \frac{a}{3\sqrt{a^2 + 2}}$.
This is the sine of the angle with the plane. It can never equal $\sin\left(\cos^{-1}\sqrt{\frac{2}{27}}\right) = \frac{5}{3\sqrt{3}}$, since $\frac{a}{\sqrt{a^2 + 2}} < 1 < \frac{5}{\sqrt{3}}$. The given value $\sqrt{\frac{2}{27}}$ must be read as this sine (the cosine of the angle with the normal), as the setter evidently intended:
$\frac{a}{3\sqrt{a^2 + 2}} = \frac{\sqrt{2}}{3\sqrt{3}} \Rightarrow 3a^2 = 2a^2 + 4 \Rightarrow a = 2$.
P: $2x - y - z = 0$. For $(1, 1, -5)$: $2 - 1 + 5 = 6$, so the image is $(1, 1, -5) - \frac{2 \cdot 6}{6}(2, -1, -1) = (-3, 3, -3)$.
$\alpha + \beta - \gamma = -3 + 3 + 3 = 3$.
Answer: 3. | ||||
| 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 From $F = \eta A\frac{dv}{dx}$: $\eta = \frac{F}{A \cdot (dv/dx)}$.
Force is momentum per time, $[P\,T^{-1}]$, and the velocity gradient has dimension $[T^{-1}]$.
So $[\eta] = \frac{[P\,T^{-1}]}{[A][T^{-1}]} = [P\,A^{-1}\,T^0]$.
(Check: $\frac{MLT^{-1}}{L^2} = ML^{-1}T^{-1}$, the usual dimension of viscosity.)
Answer: (A). | ||||
| 32 | 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 $\vec{D} = \varepsilon_0\vec{E}$, and for a charged sheet or conductor $D = \sigma$: both are charge per area, C m⁻², $[AT\,L^{-2}]$.
(B) A current (A) and an electric field (V m⁻¹) differ. (C) Current density is A m⁻², surface charge density A s m⁻². (D) Potential is energy per charge.
Answer: (A). | ||||
| 33 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The arc AB subtends $135^\circ = \frac{3\pi}{4}$ at the centre, so $r\cdot\frac{3\pi}{4} = 60$ and $r = \frac{80}{\pi} \approx 25.5$ m.
Displacement $AB = \sqrt{r^2 + r^2 - 2r^2\cos 135^\circ} = r\sqrt{2 + 1.4} = 25.5 \times 1.844 \approx 47$ m.
Answer: (B). | ||||
| 34 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power By the work-energy theorem $W = \frac{1}{2}m(v_2^2 - v_1^2)$.
$v(0) = 4$ m/s, $v(2) = 16$ m/s.
$W = \frac{1}{2}(0.5)(256 - 16) = 60$ J.
Answer: (B). | ||||
| 35 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rolling without slipping from height h: $Mgh = \frac{1}{2}Mv^2\left(1 + \frac{k^2}{R^2}\right)$, so $v^2 = \frac{2gh}{1 + k^2/R^2}$.
Cylinder: $\frac{k^2}{R^2} = \frac{1}{2}$, $v^2 = \frac{4gh}{3}$. Sphere: $\frac{k^2}{R^2} = \frac{2}{5}$, $v^2 = \frac{10gh}{7}$.
$\frac{v_c}{v_s} = \sqrt{\frac{4/3}{10/7}} = \sqrt{\frac{14}{15}}$.
Answer: (D). | ||||
| 36 | GravitationNewton's law of gravitation | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation P is 13 m from B along the perpendicular bisector, so $PA = PC = 13\sqrt{2}$ m and PA, PC make $45^\circ$ with PB.
Force from B: $\frac{G(100)(100)}{13^2} = \frac{10^4G}{169}$, along PB.
Forces from A and C: each $\frac{10^4G}{338}$; their components across PB cancel and along PB they add to $2 \cdot \frac{10^4G}{338}\cos 45^\circ = \frac{10^4G}{169\sqrt{2}}$.
$F = \frac{10^4G}{169}\left(1 + \frac{1}{\sqrt{2}}\right) = 59.2G \times 1.707 \approx 101G \approx 100G$.
Answer: (B). | ||||
| 37 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Isothermal doubling of volume: $P_1 = \frac{2 \times 10^7}{2} = 10^7$ Pa.
Adiabatic doubling: $P_2 = P_1\left(\frac{1}{2}\right)^{1.5} = \frac{10^7}{2\sqrt{2}} = 3.536 \times 10^6$ Pa.
Answer: (B). | ||||
| 38 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The average kinetic energy of a molecule is $\frac{3}{2}kT$: it depends on temperature alone.
(A) True: lower temperature, lower average kinetic energy.
(B) False: at constant temperature it does not change with pressure.
(C) False: volume by itself does not fix it.
(D) Intended as "at constant volume": then $P \propto T$ and the pressure rises with temperature. True as intended.
(E) False: at constant pressure the volume increases with temperature.
Only (A) and (D) can be correct, and (A) and (D) only is an option.
Answer: (A). | ||||
| 39 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves A mass between two springs fixed to opposite walls feels both springs together: effective constants add.
(A): mass $3m$, constant $k + k = 2k$: $T_1 = 2\pi\sqrt{\frac{3m}{2k}}$.
(B): mass $m$, constant $k + 2k = 3k$: $T_2 = 2\pi\sqrt{\frac{m}{3k}}$.
$\frac{T_1}{T_2} = \sqrt{\frac{3m}{2k} \cdot \frac{3k}{m}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.
Answer: (A). | ||||
| 40 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $V = \frac{Q}{C}$ is proportional to Q: a straight line through the origin.
At $Q = 5$ C: $V = \frac{5}{2 \times 10^{-6}} = 2.5 \times 10^6$ V.
That is the straight line ending at 2.5 (×10⁶ V) at Q = 5 C.
Answer: (A). | ||||
| 41 | 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 $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, so with equal K, $q \propto \frac{\sqrt{m}}{r}$.
$\frac{q_1}{q_2} = \sqrt{\frac{9}{4}} \cdot \frac{5}{6} = \frac{3}{2} \cdot \frac{5}{6} = \frac{5}{4}$.
Answer: (B). | ||||
| 42 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $f_0 = \frac{1}{2\pi\sqrt{LC}}$ depends only on L and C, not on R or on the source frequency.
A capacitor added in series lowers the total capacitance ($\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$), which raises $f_0$.
Answer: (C). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Let current I flow in the large loop. Each side, at distance $\frac{L}{2}$ from O and seen at $45^\circ$ each way, gives $\frac{\mu_0 I}{4\pi(L/2)}(\sin 45^\circ + \sin 45^\circ) = \frac{\sqrt{2}\mu_0 I}{2\pi L}$. Four sides: $B = \frac{2\sqrt{2}\mu_0 I}{\pi L}$.
Since $l \ll L$ the field is nearly uniform over the small loop: $\phi = Bl^2 = \frac{2\sqrt{2}\mu_0 I l^2}{\pi L}$.
$M = \frac{\phi}{I} = \frac{2\sqrt{2}\mu_0 l^2}{\pi L}$.
Answer: (C). | ||||
| 44 | 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 $I_{rms} = \frac{V_{rms}}{X_C} = V_{rms}\,\omega C$.
$C = \frac{6.9 \times 10^{-6}}{230 \times 600} = \frac{6.9 \times 10^{-6}}{1.38 \times 10^5} = 5 \times 10^{-11}$ F $= 50$ pF.
Answer: (B). | ||||
| 45 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics In the primary rainbow, light is refracted, reflected once inside the drop and refracted out; red emerges at about $42^\circ$ and violet at about $40^\circ$ to the antisolar direction, so red is on the outside (top) and violet on the inside (bottom). (A) is correct and (B) wrong.
(C) Two internal reflections produce the secondary rainbow, not the primary.
(D) The secondary rainbow, with one more reflection, is the fainter one.
Answer: (A). | ||||
| 46 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For thickness d: $t_1 = \frac{d}{v_A}$, $t_2 = \frac{d}{v_B}$, and $\frac{v_A}{v_B} = \frac{\mu_B}{\mu_A} = 2$, so $v_B = \frac{v_A}{2}$.
$t_2 - t_1 = \frac{2d}{v_A} - \frac{d}{v_A} = \frac{d}{v_A} = 5 \times 10^{-10}$ s.
$d = 5 \times 10^{-10}\,v_A$ m.
Answer: (A). | ||||
| 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 Photon energies: $\frac{1230}{800} = 1.5375$ eV and $\frac{1230}{500} = 2.46$ eV.
$K_1 = 1.5375 - \phi$ and $K_2 = 2.46 - \phi = 2K_1 = 3.075 - 2\phi$.
So $\phi = 3.075 - 2.46 = 0.615$ eV.
Answer: (C). | ||||
| 48 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Bohr's quantisation condition: $mvr = \frac{nh}{2\pi}$.
So the linear momentum is $p = mv = \frac{nh}{2\pi r}$.
Answer: (A). | ||||
| 49 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The orbiting electron is a current loop: $I = \frac{ev}{2\pi r}$, area $\pi r^2$, so $\mu = \frac{evr}{2} = \frac{e}{2m}(mvr) = \frac{eL}{2m}$.
The electron's charge is negative, so the magnetic moment points opposite to the angular momentum: $\vec{\mu}_L = -\frac{e\vec{L}}{2m}$.
Answer: (B). | ||||
| 50 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Each diode has its cathode at the input (A or B) and its anode at the common point Y, which is pulled up to 5 V through R.
If either input is at 0 V, that diode is forward biased and holds Y near 0 V: Y = 0.
Only when both inputs are at 5 V are both diodes without forward bias, and Y stays at 5 V: Y = 1.
This is an AND gate: Y = 1 only for A = B = 1.
Answer: (A). | ||||
| 51 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane With the same braking deceleration a, the stopping distance is $s = \frac{u^2}{2a} \propto u^2$.
One third the speed gives one ninth the distance: $\frac{27}{9} = 3$ m.
Answer: 3. | ||||
| 52 | Laws of MotionEquilibrium of concurrent forces | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on equilibrium of concurrent forces, from Laws of Motion. NCERT: Class 11: Laws of Motion From the figure: 2 N acts up and to the left at $45^\circ$ to the vertical, 1 N up and to the right at $45^\circ$ to the vertical, $F_1$ horizontally to the right and $F_2$ vertically down.
Horizontal: $F_1 + 1\sin 45^\circ - 2\sin 45^\circ = 0$, so $F_1 = \frac{1}{\sqrt{2}}$ N.
Vertical: $2\cos 45^\circ + 1\cos 45^\circ - F_2 = 0$, so $F_2 = \frac{3}{\sqrt{2}}$ N.
$F_1 : F_2 = 1 : 3$, so $x = 3$.
Answer: 3. | ||||
| 53 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $\Delta L = \frac{FL}{\pi r^2 Y}$.
For the second wire: $\Delta L' = \frac{(4F)(4L)}{\pi(4r)^2 Y} = \frac{16FL}{16\pi r^2 Y} = \Delta L = 5$ cm.
Answer: 5. | ||||
| 54 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter For the difference to stay constant the two expansions must be equal: $L_b\alpha_b\Delta T = L_i\alpha_i\Delta T$.
$L_i = \frac{40 \times 1.8}{1.2} = 60$ cm (and indeed $60 - 40 = 20$ cm).
Answer: 60. | ||||
| 55 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Observer speed $18$ km/h $= 5$ m/s; the source and the hill are at rest.
Direct sound: the observer moves away from the source behind him: $f_1 = 640 \times \frac{320 - 5}{320} = 630$ Hz.
Reflected sound: the hill reflects 640 Hz, and the observer moves towards it: $f_2 = 640 \times \frac{320 + 5}{320} = 650$ Hz.
Beat frequency $= 650 - 630 = 20$ Hz.
Answer: 20. | ||||
| 56 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The number of lines per unit area is the field strength at the surface.
$\rho = 2\ \mu$C cm⁻³ $= 2 \times 10^{-6} \times 10^{6} = 2$ C m⁻³.
By Gauss's law $E = \frac{\rho R}{3\varepsilon_0} = \frac{2 \times 6}{3 \times 8.85 \times 10^{-12}} = 4.52 \times 10^{11} = 45.2 \times 10^{10}$ N C⁻¹.
To the nearest integer, 45.
Answer: 45. | ||||
| 57 | 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 Each branch is a cell (2 V, 4 V or 6 V, negative plate at the grounded wire) in series with 1 kΩ, all joined at the output node. No current is drawn from $V_o$.
Kirchhoff's current law at the node: $\frac{2 - V_o}{1} + \frac{4 - V_o}{1} + \frac{6 - V_o}{1} = 0$.
$V_o = \frac{2 + 4 + 6}{3} = 4$ V.
Answer: 4. | ||||
| 58 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity One wire: $R_1 = \frac{\rho l}{\pi d^2/4}$; eight in parallel: $R = \frac{R_1}{8} = \frac{4\rho l}{8\pi d^2}$.
Single wire of length $2l$, diameter D: $\frac{4\rho(2l)}{\pi D^2} = \frac{4\rho l}{8\pi d^2}$, so $D^2 = 16d^2$ and $D = 4d$.
Answer: 4. | ||||
| 59 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits $E_g = \frac{hc}{\lambda} = \frac{12400\ \text{eV Å}}{4000\ \text{Å}} = 3.1$ eV, which rounds to 3.
Answer: 3. | ||||
| 60 | Communication SystemsRange of a TV tower | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on range of a tv tower, from Communication Systems. NCERT: No matching NCERT chapter Area to be covered $= \frac{6.03 \times 10^5}{100} = 6030$ km².
Range $d = \sqrt{2Rh}$, so the area is $\pi d^2 = 2\pi Rh$.
$h = \frac{6030}{2\pi \times 6400}$ km $= 0.150$ km $= 150$ m.
Answer: 150. | ||||
| 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 One mole of $\mathrm{SO_2Cl_2}$ gives 1 mol $\mathrm{H_2SO_4}$ (needs 2 mol NaOH) and 2 mol HCl (need 2 mol NaOH): 4 mol NaOH in all.
Moles of $\mathrm{SO_2Cl_2} = \frac{16}{4} = 4$.
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 $l$ can take the values $0, 1, \ldots, n - 1$; for $n = 3$ the largest is $l = 2$. So $l = 3$ with $n = 3$ is not allowed.
The other sets satisfy $|m_l| \le l < n$ and $s = \pm\frac{1}{2}$.
Answer: (C). | ||||
| 63 | Solutionsvan't Hoff factor | Class 12: Solutions | Single correctMedium | |
A medium single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions 0.5 mL of formic acid per litre is $0.5 \times 1.05 = 0.525$ g, i.e. $\frac{0.525}{46} = 0.0114$ mol per litre. In so dilute a solution this is also the molality, about 0.0114 mol kg⁻¹.
$\Delta T_f = i\,K_f\,m$: $0.0405 = i \times 1.86 \times 0.0114$, so $i = \frac{0.0405}{0.0212} \approx 1.9$.
Answer: (C). | ||||
| 64 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{NH_4OH}$: $20 \times 0.1 = 2$ mmol; HCl: $40 \times 0.05 = 2$ mmol. They neutralise exactly, leaving 2 mmol $\mathrm{NH_4Cl}$ in 60 mL: $c = \frac{1}{30}$ M.
For the salt of a weak base and strong acid: $\mathrm{pH} = 7 - \frac{1}{2}\mathrm{p}K_b - \frac{1}{2}\log c = 7 - 2.5 - \frac{1}{2}\log\frac{1}{30} = 4.5 + \frac{1}{2}(1.48) = 5.24$.
Nearest: 5.2.
Answer: (C). | ||||
| 65 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter (A) Haber process: iron oxide promoted with $\mathrm{K_2O}$ and $\mathrm{Al_2O_3}$ → (III).
(B) Methanation, CO + 3H₂ → CH₄ + H₂O: nickel → (IV).
(C) CO + H₂ → HCHO: copper → (I).
(D) Methanol synthesis: $\mathrm{Cu/ZnO - Cr_2O_3}$ → (II).
A→III, B→IV, C→I, D→II.
Answer: (C). | ||||
| 66 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Radon has 86 electrons; adding $14 + 1 + 2 = 17$ gives $Z = 103$.
IUPAC roots: 1 = un, 0 = nil, 3 = tri, so 103 is Unniltrium (Unt), i.e. lawrencium.
Answer: (D). | ||||
| 67 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In copper extraction the iron in the copper pyrites ore is converted to FeO during roasting and smelting; silica is added as flux and removes it as slag: $\mathrm{FeO + SiO_2 \rightarrow FeSiO_3}$.
CaO is the flux (forming $\mathrm{CaSiO_3}$ slag) in the extraction of iron, not copper; $\mathrm{Al_2O_3}$, ZnO and NiO are not the slag-forming oxides of this process.
Answer: (D). | ||||
| 68 | 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 acid, $\mathrm{H_2O_2}$ reduces permanganate to $\mathrm{Mn^{2+}}$:
$\mathrm{2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O}$.
Answer: (A). | ||||
| 69 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Densities (g cm⁻³): Li 0.53, Na 0.97, K 0.86, Rb 1.53, Cs 1.90.
Density generally rises down the group, but potassium is lighter than sodium (its atomic volume grows more than its mass).
Order: Li < K < Na < Rb < Cs.
Answer: (A). | ||||
| 70 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{2BF_3 + 6NaH \xrightarrow{450\ K} B_2H_6 + 6NaF}$: A is diborane.
Diborane is cleaved by the Lewis base trimethylamine: $\mathrm{B_2H_6 + 2NMe_3 \rightarrow 2BH_3 \cdot NMe_3}$.
In the adduct boron has four bonds (three H and the N donor) and no lone pair: $sp^3$, tetrahedral.
Answer: (B). | ||||
| 71 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter With excess fluorine bromine forms $\mathrm{BrF_5}$: $\mathrm{Br_2 + 5F_2 \rightarrow 2BrF_5}$.
Bromine is in the +5 state, the oxidation state of the halogen in a halate ($\mathrm{BrO_3^-}$), as against +1 in a hypohalite, +3 in a halite and +7 in a perhalate.
So the product corresponds to a halate.
Answer: (B). | ||||
| 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 Photochemical (oxidising) smog forms from nitrogen oxides and hydrocarbons in sunlight; it contains NO, $\mathrm{NO_2}$, ozone, PAN, and aldehydes such as HCHO and acrolein.
$\mathrm{SO_2}$ is characteristic of classical (reducing) smog, not photochemical smog.
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 A is $\mathrm{CH_2{=}C(CH_3)CH_2C(CH_3)_3}$; both reagents cleave the C=C bond, giving the same ketone $\mathrm{CH_3COCH_2C(CH_3)_3}$ from the substituted end.
The $\mathrm{CH_2{=}}$ end gives HCHO with ozonolysis ($\mathrm{O_3}$ then $\mathrm{H_2O/Zn}$), but is oxidised further to $\mathrm{CO_2}$ and water by hot acidic $\mathrm{KMnO_4}$.
'a' oxidises to formic acid, the acid of ants: so a = HCHO and X is $\mathrm{O_3, H_2O/Zn}$. Y is $\mathrm{KMnO_4/H^+}$ (b = $\mathrm{CO_2}$).
Cold dilute $\mathrm{KMnO_4}$ (Baeyer's reagent) would give a diol, not a ketone.
Answer: (D). | ||||
| 74 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The reactant is 3-methylpentan-2-ol. Assigning priorities from the drawing: at C-2, OH > C-3 > CH₃ > H with OH on a wedge, giving R; at C-3, C-2 > ethyl > CH₃ > H with ethyl on a wedge, giving S. The reactant is (2R,3S).
(i) Tosyl chloride in pyridine converts OH into OTs without touching the C–O bond: configuration retained.
(ii) Cyanide displaces tosylate by $S_N2$ in DMF (polar aprotic): inversion at C-2. C-3 is untouched.
At C-2 the priorities stay in the same order (CN > C-3 > CH₃ > H), so inversion turns R into S; C-3 keeps S. The product is (2S,3S)-2-cyano-3-methylpentane (2,3-dimethylpentanenitrile).
Checking the drawings: (A) is (2R,3S), retention; (B) is (2S,3S); (C) is (2S,3R); (D) is (2S,3R), the mirror image of (A).
Answer: (B). | ||||
| 75 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) Rosenmund reduction: acid chloride + H₂ over Pd–BaSO₄ gives the aldehyde. Correct.
(B) DIBAL-H reduces a nitrile to an imine, hydrolysed to the aldehyde. Correct.
(C) DIBAL-H ($\mathrm{AlH(i\text{-}Bu)_2}$) reduces an ester to the aldehyde. Correct.
(D) Acidified sodium dichromate is a strong oxidant: a primary alcohol is oxidised past the aldehyde to the carboxylic acid (4-methylbenzoic acid). The pairing shown is not correct.
Answer: (D). | ||||
| 76 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines $\mathrm{C_8H_5NO_2}$ that gives a primary amine via ethanolic KOH, an alkyl halide and alkaline hydrolysis is phthalimide: this is the Gabriel synthesis.
Phthalimide comes from phthalic acid: with $\mathrm{NH_3}$ it forms ammonium phthalate (B), which on strong heating loses water to give phthalimide (C).
So A is phthalic acid (benzene-1,2-dicarboxylic acid).
Answer: (C). | ||||
| 77 | 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 Melamine–formaldehyde polymer is made by condensing melamine, 2,4,6-triamino-1,3,5-triazine (a six-membered ring of alternating C and N carrying three $\mathrm{NH_2}$ groups), with formaldehyde.
Answer: (A). | ||||
| 78 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Denaturation (by heat, pH change etc.) breaks hydrogen bonds and other weak interactions, unfolding the secondary, tertiary and quaternary structure.
The peptide bonds are not broken, so the primary structure (the amino-acid sequence) remains intact.
Answer: (A). | ||||
| 79 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Drugs that bind to a receptor site and block it, so that the natural messenger cannot act, are antagonists. Agonists mimic the natural messenger and switch the receptor on.
Answer: (B). | ||||
| 80 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Statement I: heating glycerol with $\mathrm{KHSO_4}$ removes two molecules of water, giving acrolein $\mathrm{CH_2{=}CH{-}CHO}$. Correct.
Statement II: acrolein is used to test for glycerol, but its odour is sharp, pungent and irritating, not fruity. Incorrect.
Answer: (C). | ||||
| 81 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure By molecular orbital theory:
$\mathrm{N_2}$ (14 e⁻): all paired, diamagnetic.
$\mathrm{N_2^+}$ (13), $\mathrm{N_2^-}$ (15): one unpaired electron, paramagnetic.
$\mathrm{N_2^{2-}}$ (16, like $\mathrm{O_2}$): two unpaired electrons in $\pi^*$, paramagnetic.
$\mathrm{O_2}$ (16): two unpaired, paramagnetic. $\mathrm{O_2^+}$ (15), $\mathrm{O_2^-}$ (17): one unpaired, paramagnetic.
$\mathrm{O_2^{2-}}$ (18): $\pi^*$ full, diamagnetic.
Diamagnetic: $\mathrm{N_2}$ and $\mathrm{O_2^{2-}}$, i.e. 2.
Answer: 2. | ||||
| 82 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\mathrm{3C(graphite) + 4H_2(g) \rightarrow C_3H_8(g)}$.
By Hess's law, $\Delta_fH = 3(-393.5) + 4(-285.8) - (-2220.0) = -1180.5 - 1143.2 + 2220.0 = -103.7$ kJ mol⁻¹.
Magnitude 103.7, nearest integer 104.
Answer: 104. | ||||
| 83 | 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 Pressure of the dry gas $= 4 - 0.4 = 3.6$ atm. Doubling the volume at constant temperature halves it (Boyle's law): 1.8 atm.
The water vapour stays saturated at the same temperature (liquid water is present), so its pressure is still 0.4 atm.
New pressure $= 1.8 + 0.4 = 2.2$ atm $= 22 \times 10^{-1}$ atm.
Answer: 22. | ||||
| 84 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry $E^\circ_{cell} = 0.008 - (-0.763) = 0.771$ V.
Nernst: $E = E^\circ - \frac{0.06}{n}\log Q$, so $0.801 = 0.771 - \frac{0.06}{n}(-2) = 0.771 + \frac{0.12}{n}$.
$\frac{0.12}{n} = 0.03$, $n = 4$.
Answer: 4. | ||||
| 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 $4.0$ min $= 240$ s: the half-life is the same at both initial pressures.
Since $t_{1/2} \propto P_0^{1-n}$, a half-life independent of the initial pressure means $1 - n = 0$: first order.
Answer: 1. | ||||
| 86 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All four are Co(III), $d^6$. The shortest wavelength absorbed means the largest splitting $\Delta_o$, which comes from the strongest-field ligand: $\mathrm{CN^-}$, so $\mathrm{[Co(CN)_6]^{3-}}$.
Co(III) with $\mathrm{CN^-}$ is low spin, $t_{2g}^6e_g^0$: no unpaired electrons.
Spin-only moment $= 0$ B.M.
Answer: 0. | ||||
| 87 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mathrm{Ti^{2+}}$, $\mathrm{V^{2+}}$ and $\mathrm{Cr^{2+}}$ are strong reducing agents (negative $E^\circ$ for $\mathrm{M^{3+}/M^{2+}}$) and can reduce $\mathrm{H^+}$ to $\mathrm{H_2}$. $\mathrm{Co^{3+}}$ is a strong oxidising agent ($E^\circ_{\mathrm{Co^{3+}/Co^{2+}}} = +1.97$ V) and cannot.
Gaseous $\mathrm{Co^{3+}}$: $3d^6$, with four unpaired electrons (free ion, Hund's rule).
$\mu = \sqrt{4(4 + 2)} = \sqrt{24} = 4.90$ B.M., nearest integer 5.
Answer: 5. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy 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 $\mathrm{H_2SO_4}$ used $= 2.5 \times 2 = 5$ mmol, which neutralises $2 \times 5 = 10$ mmol $\mathrm{NH_3}$.
Nitrogen $= 10 \times 14 = 140$ mg $= 0.14$ g.
$\%\mathrm{N} = \frac{0.14}{0.25} \times 100 = 56$.
Answer: 56. | ||||
| 89 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{C_4H_5N}$ has $\frac{2(4) + 2 + 1 - 5}{2} = 3$ degrees of unsaturation; acyclic, so three π bonds.
The nitriles of this formula, $\mathrm{CH_2{=}CH{-}CH_2{-}CN}$ (allyl cyanide), $\mathrm{CH_3{-}CH{=}CH{-}CN}$ and $\mathrm{CH_2{=}C(CH_3){-}CN}$, each have one $sp^3$ carbon (the $\mathrm{CH_2}$ or $\mathrm{CH_3}$); the other three carbons are $sp^2$ or $sp$.
No structure can have two $sp^3$ carbons: the other two carbons and N would then have to carry three π bonds, which is impossible.
Answer: 1. | ||||
| 90 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalHard | |
A hard numerical question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The reactant is ethyl 1-methyl-2-(2-oxopropyl)cyclopentane-1-carboxylate. The ester carbon is attached to a quaternary carbon, so the ester has no α-hydrogen; ethoxide forms the enolate of the ketone.
The enolate of the terminal $\mathrm{CH_3}$ of the ketone attacks the ester carbonyl intramolecularly (Claisen/Dieckmann-type condensation), expelling ethoxide and closing a six-membered ring (ester C, C-1, C-2, $\mathrm{CH_2}$, C=O, $\mathrm{CH_2}$). Attack by the internal $\mathrm{CH_2}$ would give a strained four-membered ring.
A is a bicyclic 1,3-diketone: a cyclohexane-1,3-dione fused to the cyclopentane, with the methyl group at the ring junction.
Chiral carbons in A: the ring-junction carbon bearing $\mathrm{CH_3}$ (bonded to $\mathrm{CH_3}$, C=O, the other junction carbon and a ring $\mathrm{CH_2}$) and the other junction carbon (bonded to H and three different carbons). The $\mathrm{CH_2}$ groups and carbonyl carbons are not chiral.
Number of chiral carbons $= 2$.
Answer: 2. | ||||
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