JEE (Main) · 2022
JEE (Main) 2022 — 26 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
- Integral Calculus3
- Sets, Relations and Functions2
- Matrices and Determinants2
7 easy · 21 medium · 2 hard
Physics
30 Q
- Kinematics3
- Electromagnetic Induction and Alternating Currents3
- Properties of Bulk Matter3
21 easy · 9 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques5
- p-Block Elements3
- Some Basic Concepts of Chemistry2
19 easy · 10 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f^2(x) = f\left(\frac{x-1}{x+1}\right) = \frac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1} = \frac{-2}{2x} = -\frac{1}{x}$.
$f^3(x) = f\left(-\frac{1}{x}\right) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1} = \frac{x+1}{1-x}$.
$f^4(x) = f^2(f^2(x)) = -\frac{1}{-1/x} = x$, so the compositions repeat with period 4.
Hence $f^6 = f^2$ and $f^7 = f^3$:
$f^6(6) = -\frac{1}{6}$, $f^7(7) = \frac{8}{-6} = -\frac{4}{3}$.
Sum $= -\frac{1}{6} - \frac{8}{6} = -\frac{3}{2}$.
Answer: (B). | ||||
| 2 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Set A: $|z+1| < |z-1|$ means z is nearer to $-1$ than to $1$, i.e. $\text{Re}(z) < 0$ (the open left half-plane).
Set B: $\arg\left(\frac{z-1}{z+1}\right) = \arg(z-1) - \arg(z+1)$ is the angle turned from the segment $z \to -1$ to the segment $z \to 1$. A positive value $\frac{2\pi}{3}$ puts z on the arc through $-1$ and $1$ lying above the real axis (for z below the axis this argument is negative).
The circle through $(\pm 1, 0)$ has centre $(0, k)$. On the y-axis the point of the arc is $z = iy$ with $\arg(iy-1) - \arg(iy+1) = \pi - 2\tan^{-1}y = \frac{2\pi}{3}$, so $y = \frac{1}{\sqrt{3}}$.
$1 + k^2 = \left(\frac{1}{\sqrt{3}} - k\right)^2 \Rightarrow k = -\frac{1}{\sqrt{3}}$; radius $= \sqrt{1 + \frac{1}{3}} = \frac{2}{\sqrt{3}}$.
So B is the arc of the circle centred at $\left(0, -\frac{1}{\sqrt{3}}\right)$ above the x-axis; intersecting with $\text{Re}(z) < 0$ leaves only the part in the second quadrant.
Answer: (B). | ||||
| 3 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3 \times 3$ matrix M, $|\text{adj}\,M| = |M|^2$ and $|kM| = k^3|M|$.
Left side: $|24A|^2 = (24^3|A|)^2 = 24^6|A|^2$.
$|\text{adj}(2A)| = |2A|^2 = 64|A|^2$, so $|3\,\text{adj}(2A)| = 27 \cdot 64|A|^2$ and the right side is $(27 \cdot 64)^2|A|^4 = 3^6 \cdot 2^{12}|A|^4$.
$2^{18}3^6|A|^2 = 3^6 2^{12}|A|^4 \Rightarrow |A|^2 = 2^6$.
Answer: (C). | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}3&-2&1\\5&-8&9\\2&1&a\end{vmatrix} = 3(-8a-9) + 2(5a-18) + 1(5+16) = -14a - 42$.
No solution needs $\Delta = 0$, so $a = -3$.
With $a = -3$ the coefficient rows are dependent: $\frac{3}{2}(3, -2, 1) - \frac{1}{2}(5, -8, 9) = (2, 1, -3)$.
The system is consistent only if the right sides obey the same relation: $\frac{3}{2}b - \frac{3}{2} = -1$, i.e. $b = \frac{1}{3}$.
So there is no solution when $a = -3$, $b \ne \frac{1}{3}$: among the options, $\left(-3, -\frac{1}{3}\right)$. (With $b = \frac{1}{3}$ there are infinitely many solutions.)
Answer: (C). | ||||
| 5 | 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 $2021 = 7 \times 288 + 5$, so $2021^{2023} \equiv 5^{2023} \equiv (-2)^{2023} \pmod 7$.
$(-2)^{2023} = -2 \cdot (-2)^{2022} = -2 \cdot 8^{674} = -2(7+1)^{674}$. By the binomial theorem $(7+1)^{674} = 7m + 1$, so $2021^{2023} = 7k - 2$.
Remainder $= 7 - 2 = 5$.
Answer: (C). | ||||
| 6 | 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 Put $\theta = \cos^{-1}x$, so $x = \cos\theta$ and $\theta \to \frac{\pi}{4}$.
$\frac{\sin\theta - \cos\theta}{1 - \frac{\sin\theta}{\cos\theta}} = \frac{(\sin\theta - \cos\theta)\cos\theta}{\cos\theta - \sin\theta} = -\cos\theta \to -\frac{1}{\sqrt{2}}$.
Answer: (D). | ||||
| 7 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Near $x = 0^-$: $f(x) = -(x+3) < 0$, so $(gof)(x) = f^2 + k_1 f$; near $0^+$: $f(x) = e^x \ge 1$, so $(gof)(x) = 4e^x + k_2$.
Derivatives at 0: left $= (2f(0^-) + k_1)f'(0^-) = (-6 + k_1)(-1) = 6 - k_1$; right $= 4e^0 = 4$. So $k_1 = 2$.
Continuity: $g(-3) = 9 - 3k_1 = 3$ must equal $4 + k_2$, so $k_2 = -1$.
$(gof)(-4)$: $f(-4) = -|-1| = -1$, $g(-1) = 1 - 2 = -1$.
$(gof)(4) = g(e^4) = 4e^4 - 1$.
Sum $= 4e^4 - 2 = 2(2e^4 - 1)$.
Answer: (D). | ||||
| 8 | 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 $3x - x^2 + 2 = 0$ at $x = \frac{3 \pm \sqrt{17}}{2}$; in $[-1, 2]$ only $r = \frac{3 - \sqrt{17}}{2} \approx -0.56$ lies.
On $[-1, r]$ the quadratic is negative: $f = x^2 - 3x - 2 - x = x^2 - 4x - 2$, decreasing there (vertex at 2); $f(-1) = 3$, $f(r) = -r = \frac{\sqrt{17} - 3}{2}$.
On $[r, 2]$: $f = -x^2 + 2x + 2$, with maximum 3 at $x = 1$, and $f(2) = 2$, $f(r) = \frac{\sqrt{17}-3}{2} \approx 0.56$.
Absolute maximum 3, absolute minimum $\frac{\sqrt{17} - 3}{2}$; sum $= \frac{\sqrt{17} + 3}{2}$.
Answer: (A). | ||||
| 9 | 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 (a, b) lies on the curve for every n, since $1^n + 1^n = 2$.
Differentiating: $\frac{n}{a}\left(\frac{x}{a}\right)^{n-1} + \frac{n}{b}\left(\frac{y}{b}\right)^{n-1}\frac{dy}{dx} = 0$; at (a, b): $\frac{1}{a} + \frac{1}{b}\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{b}{a}$.
Tangent: $y - b = -\frac{b}{a}(x - a)$, i.e. $\frac{x}{a} + \frac{y}{b} = 2$, for every natural n.
So $S = \mathbb{N}$.
Answer: (D). | ||||
| 10 | 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 $|x^2 - 9| \le 3$ when $6 \le x^2 \le 12$, i.e. $\sqrt{6} \le |x| \le 2\sqrt{3}$: there the curve lies below the line. By symmetry take $x > 0$ and double.
$\int_{\sqrt{6}}^{3} [3 - (9 - x^2)]dx = \left[\frac{x^3}{3} - 6x\right]_{\sqrt{6}}^{3} = -9 + 4\sqrt{6}$.
$\int_{3}^{2\sqrt{3}} [3 - (x^2 - 9)]dx = \left[12x - \frac{x^3}{3}\right]_{3}^{2\sqrt{3}} = 16\sqrt{3} - 27$.
Area $= 2(16\sqrt{3} + 4\sqrt{6} - 36) = 8(4\sqrt{3} + \sqrt{6} - 9)$.
Answer: (D). | ||||
| 11 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines The height of the triangle is the distance of R from the line: $h = \frac{|3 + 7 - 5|}{\sqrt{2}} = \frac{5}{\sqrt{2}}$.
For an equilateral triangle, side $= \frac{2h}{\sqrt{3}}$ and area $= \frac{h^2}{\sqrt{3}} = \frac{25/2}{\sqrt{3}} = \frac{25}{2\sqrt{3}}$.
Answer: (D). | ||||
| 12 | 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 centre lies on the perpendicular bisector of AB: midpoint $\left(\frac{5}{2}, \frac{3}{2}\right)$, AB along (1, 5), so $x + 5y = 10$.
Put $x = 10 - 5y$ in $(x-5)^2 + (y-1)^2 = \frac{13}{2}$: $25(1-y)^2 + (y-1)^2 = \frac{13}{2}$, so $(y-1)^2 = \frac{1}{4}$, $y = \frac{3}{2}$ or $\frac{1}{2}$.
$y = \frac{3}{2}$ gives the midpoint of AB (AB would be a diameter), so the centre is $\left(\frac{15}{2}, \frac{1}{2}\right)$.
$r^2 = \left(\frac{15}{2} - 2\right)^2 + \left(\frac{1}{2} + 1\right)^2 = \frac{121}{4} + \frac{9}{4} = \frac{65}{2}$.
Answer: (B). | ||||
| 13 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = 6x$ has $a = \frac{3}{2}$; take $P(at^2, 2at)$. Normal: $y = -tx + 2at + at^3$.
Through (5, −8): $-8 = -5t + 3t + \frac{3}{2}t^3$, i.e. $3t^3 - 4t + 16 = 0 = (t+2)(3t^2 - 6t + 8)$; the only real root is $t = -2$.
Tangent at P: $ty = x + at^2$. On the directrix $x = -a$: $y = \frac{a(t^2-1)}{t} = \frac{\frac{3}{2} \cdot 3}{-2} = -\frac{9}{4}$.
Answer: (B). | ||||
| 14 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $l_1$ has direction (3, −2, 0). $l_2$: $\frac{x-1}{1} = \frac{y + 3/2}{\alpha/2} = \frac{z+5}{2}$, direction $\left(1, \frac{\alpha}{2}, 2\right)$.
Perpendicular: $3 - \alpha = 0$, so $\alpha = 3$ and $l_2$ has direction $\left(1, \frac{3}{2}, 2\right) \parallel (2, 3, 4)$.
$l_3$: $\frac{x-1}{-3} = \frac{y - 1/2}{-2} = \frac{z}{4}$, direction (−3, −2, 4).
$\cos\theta = \frac{-6 - 6 + 16}{\sqrt{29}\sqrt{29}} = \frac{4}{29}$, so $\theta = \cos^{-1}\frac{4}{29} = \sec^{-1}\frac{29}{4}$.
Answer: (B). | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctHard | |
A hard 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 rotated plane contains the line of intersection: $2x + 3y + z + 20 + \lambda(x - 3y + 5z - 8) = 0$, and is perpendicular to the original plane:
$2(2+\lambda) + 3(3 - 3\lambda) + (1 + 5\lambda) = 14 - 2\lambda = 0$, so $\lambda = 7$.
Plane: $9x - 18y + 36z - 36 = 0$, i.e. $x - 2y + 4z - 4 = 0$.
For $P\left(2, -\frac{1}{2}, 2\right)$: $2 + 1 + 8 - 4 = 7$, and $|\vec{n}|^2 = 21$.
Image $= P - \frac{2 \cdot 7}{21}(1, -2, 4) = \left(\frac{4}{3}, \frac{5}{6}, -\frac{2}{3}\right)$, so $a : b : c = 8 : 5 : -4$.
Answer: (A). | ||||
| 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 $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c} = 3\vec{b} - \vec{c}$.
$\vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b}\cdot\vec{a})\vec{c} - (\vec{b}\cdot\vec{c})\vec{a} = \vec{c} - 2\vec{a}$.
$\vec{c} \times (\vec{b} \times \vec{a}) = (\vec{c}\cdot\vec{a})\vec{b} - (\vec{c}\cdot\vec{b})\vec{a} = 3\vec{b} - 2\vec{a}$.
The third vector is the sum of the first two, so the three are coplanar and their scalar triple product is 0.
Answer: (A). | ||||
| 17 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let p = P(head), q = 1 − p. $^5C_4p^4q = p^5 \Rightarrow 5q = p$, so $p = \frac{5}{6}$, $q = \frac{1}{6}$.
$P(X \le 2) = q^5 + 5pq^4 + 10p^2q^3 = \frac{1 + 25 + 250}{6^5} = \frac{276}{6^5} = \frac{46}{6^4}$.
Answer: (D). | ||||
| 18 | StatisticsMean deviation | Class 11: Statistics | Single correctEasy | |
A easy single correct question on mean deviation, from Statistics. NCERT: Class 11: Statistics Mean: $a + b + 23 = 30$, so $a + b = 7$.
Variance: $\frac{a^2 + b^2 + 64 + 25 + 100}{5} - 36 = 6.8$, so $a^2 + b^2 = 25$; hence {a, b} = {3, 4}.
Deviations from 6 of 3, 4, 8, 5, 10: 3, 2, 2, 1, 4; total 12, so $M = \frac{12}{5}$ and $25M = 60$.
Answer: (A). | ||||
| 19 | 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 $f'(x) = -\frac{2}{\sqrt{1-x^2}} - \frac{4}{1+x^2} - 6x - 2$.
On $(-1, 1)$: $\frac{2}{\sqrt{1-x^2}} \ge 2$, $\frac{4}{1+x^2} \ge 2$ and $-6x - 2 < 4$, so $f'(x) < 0$: f is decreasing.
$b = f(-1) = 2\pi + 4 \cdot \frac{3\pi}{4} - 3 + 2 + 10 = 5\pi + 9$; $a = f(1) = 0 + \pi - 3 - 2 + 10 = \pi + 5$.
$4a - b = 4\pi + 20 - 5\pi - 9 = 11 - \pi$.
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 Try the four choices.
$\nabla = \wedge$: $p \wedge q \Rightarrow ((p \Delta q) \wedge r)$ fails for p = q = T, r = F.
$\nabla = \vee$, $\Delta = \wedge$: $p \vee q \Rightarrow ((p \wedge q) \vee r)$ fails for p = T, q = F, r = F.
$\nabla = \vee$, $\Delta = \vee$: $p \vee q \Rightarrow (p \vee q \vee r)$ is a tautology.
So $\Delta = \nabla = \vee$ and $(p \nabla q)\Delta r = p \vee q \vee r$.
(A) $(p \vee r) \vee q$ is the same. (B) and (D) $(p \vee r) \wedge q$ and (C) $(p \wedge r) \vee q$ are false for p = T, q = F, r = F, where $p \vee q \vee r$ is true.
Answer: (A). | ||||
| 21 | 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 For the roots: $e_1 = \sum\alpha = 3$, $e_2 = \sum\alpha\beta = -2$, $e_3 = \sum\alpha\beta\gamma = -3$.
$p_1 = 3$, $p_2 = e_1^2 - 2e_2 = 9 + 4 = 13$.
Newton's identity: $p_3 = e_1p_2 - e_2p_1 + 3e_3 = 39 + 6 - 9 = 36$.
Answer: 36. | ||||
| 22 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations All groups: $^{10}C_3 \cdot {}^5C_3 = 120 \times 10 = 1200$.
Groups with both $B_1$ and $B_2$: $^8C_1 \cdot {}^5C_3 = 8 \times 10 = 80$.
Required $= 1200 - 80 = 1120$.
Answer: 1120. | ||||
| 23 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections A tangent to $y^2 = 4x$: $y = mx + \frac{1}{m}$. It touches $x^2 + y^2 = \frac{9}{4}$ when $\frac{|1/m|}{\sqrt{1+m^2}} = \frac{3}{2}$, i.e. $9m^4 + 9m^2 - 4 = 0$, so $m^2 = \frac{1}{3}$.
The two tangents $y = \pm\frac{1}{\sqrt{3}}x \pm \sqrt{3}$ meet on the x-axis at $x = -\frac{1}{m^2} = -3$: Q = (−3, 0), OQ = 3.
Ellipse: b = 3, a = 6. $e^2 = 1 - \frac{9}{36} = \frac{3}{4}$, $l = \frac{2b^2}{a} = 3$.
$\frac{l}{e^2} = \frac{3}{3/4} = 4$.
Answer: 4. | ||||
| 24 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $|x + k|$ is the distance of x from $-k$. The farthest of $-1, \ldots, -5$ is $-5$ when $x \ge -3$ and $-1$ when $x \le -3$.
So $f(x) = -(x+1)$ on $[-6, -3]$ and $x + 5$ on $[-3, 0]$.
$\int_{-6}^{-3} -(x+1)dx = \frac{5 + 2}{2} \cdot 3 = \frac{21}{2}$; $\int_{-3}^{0}(x+5)dx = \frac{2 + 5}{2} \cdot 3 = \frac{21}{2}$.
Total $= 21$.
Answer: 21. | ||||
| 25 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} = \frac{2x(x^2+4) + 6xy}{x^2+4}$, i.e. $\frac{dy}{dx} - \frac{6x}{x^2+4}y = 2x$.
Integrating factor: $e^{-3\ln(x^2+4)} = (x^2+4)^{-3}$.
$\frac{d}{dx}\left[\frac{y}{(x^2+4)^3}\right] = \frac{2x}{(x^2+4)^3}$, so $\frac{y}{(x^2+4)^3} = -\frac{1}{2(x^2+4)^2} + C$.
y(0) = 0: $C = \frac{1}{32}$.
At x = 2: $y = 512\left(\frac{1}{32} - \frac{1}{128}\right) = 512 \cdot \frac{3}{128} = 12$.
Answer: 12. | ||||
| 26 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin 10^\circ\sin 50^\circ\sin 70^\circ = \sin 10^\circ\sin(60^\circ - 10^\circ)\sin(60^\circ + 10^\circ) = \frac{1}{4}\sin 30^\circ = \frac{1}{8}$.
So the product is $\frac{1}{8}\sin 10^\circ\sin 20^\circ\sin 40^\circ$, and $\sin 20^\circ\sin 40^\circ = \frac{1}{2}\left(\cos 20^\circ - \frac{1}{2}\right)$.
Product $= \frac{1}{16}\sin 10^\circ\cos 20^\circ - \frac{1}{32}\sin 10^\circ$, with $\sin 10^\circ\cos 20^\circ = \frac{1}{2}(\sin 30^\circ - \sin 10^\circ) = \frac{1}{4} - \frac{1}{2}\sin 10^\circ$.
Product $= \frac{1}{64} - \frac{1}{32}\sin 10^\circ - \frac{1}{32}\sin 10^\circ = \frac{1}{64} - \frac{1}{16}\sin 10^\circ$.
So $\alpha = \frac{1}{64}$ and $16 + \alpha^{-1} = 80$.
Answer: 80. | ||||
| 27 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $A \cap B$ = even numbers from 2 to 200 divisible by neither 3 nor 5.
Sum of all even numbers to 200: $2 \cdot 5050 = 10100$.
Multiples of 6 (6 to 198, 33 terms): $6 \cdot \frac{33 \cdot 34}{2} = 3366$.
Multiples of 10 (10 to 200, 20 terms): $10 \cdot 210 = 2100$.
Multiples of 30 (30 to 180, 6 terms): $30 \cdot 21 = 630$.
Sum $= 10100 - 3366 - 2100 + 630 = 5264$.
Answer: 5264. | ||||
| 28 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $g(x) = \frac{3\pi x^2}{2} - x^3$ and $h(x) = \frac{\sin x}{1 + \cos^2 x}$; $h(\pi - x) = h(x)$.
$g(\pi - x) = \frac{\pi^3}{2} - \frac{3\pi x^2}{2} + x^3$, so $g(x) + g(\pi - x) = \frac{\pi^3}{2}$.
With $I = \int_0^\pi g h\,dx$: $2I = \frac{\pi^3}{2}\int_0^\pi \frac{\sin x}{1 + \cos^2 x}dx = \frac{\pi^3}{2}\left[-\tan^{-1}(\cos x)\right]_0^\pi = \frac{\pi^3}{2} \cdot \frac{\pi}{2}$.
$I = \frac{\pi^4}{8}$, and $\frac{48}{\pi^4} \cdot \frac{\pi^4}{8} = 6$.
Answer: 6. | ||||
| 29 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series For any i, j: $\min\{i, j\} + \max\{i, j\} = i + j$.
$A + B = \sum_{i=1}^{10}\sum_{j=1}^{10}(i + j) = 10\sum i + 10\sum j = 10 \cdot 55 + 10 \cdot 55 = 1100$.
Answer: 1100. | ||||
| 30 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{1}{1 + \sin 2x} = \frac{\sec^2 x}{(1 + \tan x)^2}$, so $y = -\frac{1}{1 + \tan x} + C$.
$y\left(\frac{\pi}{4}\right) = -\frac{1}{2} + C = \frac{1}{2}$ gives C = 1, and $y = \frac{\tan x}{1 + \tan x} = \frac{\sin x}{\sin x + \cos x}$.
Intersections with $y = \sqrt{2}\sin x$: $\sin x\left(\frac{1}{\sin x + \cos x} - \sqrt{2}\right) = 0$.
$\sin x = 0$: $x = \pi$.
$\sin x + \cos x = \frac{1}{\sqrt{2}}$: $\sin\left(x + \frac{\pi}{4}\right) = \frac{1}{2}$, so $x + \frac{\pi}{4} = \frac{5\pi}{6}$ or $\frac{13\pi}{6}$, i.e. $x = \frac{7\pi}{12}$ or $\frac{23\pi}{12}$.
All three lie in S. Sum $= \pi + \frac{7\pi}{12} + \frac{23\pi}{12} = \frac{42\pi}{12}$, so k = 42.
Answer: 42. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The argument of the logarithm is dimensionless, so $[\beta] = \frac{[kt]}{[x]}$. kt is an energy, $[ML^2T^{-2}]$, so $[\beta] = [MLT^{-2}]$.
P is dimensionless, so $[\alpha] = [\beta] = [MLT^{-2}]$.
Answer: (C). | ||||
| 32 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion The apparent weight is the normal reaction N. With a downward acceleration a: $mg - N = ma$, so $N = m(g - a) < mg$: the person feels lighter.
Upward acceleration gives $N = m(g + a) > mg$; uniform velocity gives $N = mg$.
Answer: (B). | ||||
| 33 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane At the highest point the velocity is zero, so the momentum mv is zero. Gravity still acts (force mg, acceleration g), and the potential energy is at its maximum.
Answer: (A). | ||||
| 34 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Let R be the vessel's radius. At angle $\alpha$ below the horizontal OP the ball has fallen $R\sin\alpha$, so $v^2 = 2gR\sin\alpha$.
Centripetal force: $\frac{mv^2}{R} = 2mg\sin\alpha$.
Radially: $N - mg\sin\alpha = \frac{mv^2}{R}$, so $N = 3mg\sin\alpha$.
$A = \frac{2mg\sin\alpha}{3mg\sin\alpha} = \frac{2}{3}$, a constant: a horizontal line.
Answer: (C). | ||||
| 35 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion No external torque acts about the axis, so angular momentum is conserved:
$MR^2 \cdot 2 = (MR^2 + 2mR^2)\omega$, so $\omega = \frac{2M}{M + 2m}$.
Answer: (C). | ||||
| 36 | 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 Inside a uniform earth, $g = \frac{GM}{R^3}r$: it rises linearly from 0 at the centre to its maximum at the surface.
Outside, $g = \frac{GM}{r^2}$: it falls as $\frac{1}{r^2}$.
Graph (A) shows a straight-line rise to r = R and a $\frac{1}{r^2}$-type fall. (D)'s rise is curved, not linear.
Answer: (A). | ||||
| 37 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{373} = \frac{100}{373} = 0.2681 = 26.81\%$.
Answer: (A). | ||||
| 38 | 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 In a lift accelerating upward, $g_{eff} = g + \frac{g}{6} = \frac{7g}{6}$.
$T' = 2\pi\sqrt{\frac{l}{g_{eff}}} = T\sqrt{\frac{g}{7g/6}} = \sqrt{\frac{6}{7}}\,T$.
Answer: (C). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The ordered kinetic energy of the gas becomes internal energy. Per mole: $\frac{1}{2}Mv^2 = C_V\Delta T$.
With $\gamma = 1.4$: $C_V = \frac{R}{\gamma - 1} = \frac{5R}{2}$.
$\Delta T = \frac{Mv^2}{2} \cdot \frac{2}{5R} = \frac{Mv^2}{5R}$.
Answer: (B). | ||||
| 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 Charge $C_1V$ is shared; at equilibrium both capacitors have the same potential difference:
$V' = \frac{C_1V}{C_1 + C_2}$.
Charge on $C_2$: $C_2V' = \frac{C_1C_2}{C_1 + C_2}V$.
Answer: (A). | ||||
| 41 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Assertion and reasonEasy | |
A easy assertion and reason question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance (A) is true: in a non-polar molecule the centres of positive and negative charge coincide in the absence of a field, so there is no permanent dipole moment.
(R) is false: in an external field the positive and negative charge centres are pulled apart, so they no longer coincide; this induced separation is what polarises the material.
Answer: (C). | ||||
| 42 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $|\varepsilon| = \frac{d\phi}{dt} = 15t^2 + 8t + 2 = 60 + 16 + 2 = 78$ V at t = 2 s.
$I = \frac{78}{5} = 15.6$ A.
Answer: (A). | ||||
| 43 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity At constant volume $A = \frac{V}{l}$, so $R = \frac{\rho l}{A} = \frac{\rho l^2}{V} \propto l^2$.
$\frac{\Delta R}{R} = 2\frac{\Delta l}{l} = 0.8\%$.
Answer: (C). | ||||
| 44 | 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 $r = \frac{mv}{qB}$; with the same v and B, $r \propto \frac{m}{q}$.
$\frac{r_\alpha}{r_p} = \frac{4m/2e}{m/e} = 2$, i.e. 2 : 1.
Answer: (C). | ||||
| 45 | 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 The wave travels along +z (phase $kz - \omega t$). $\vec{H} = \frac{1}{\eta_0}\hat{k} \times \vec{E}$ with $\eta_0 = \mu_0c = 4\pi \times 10^{-7} \times 3 \times 10^8 = 120\pi \approx 377\ \Omega$.
$\frac{301.6}{377} = 0.8$ and $\frac{452.4}{377} = 1.2$.
$\hat{k} \times \hat{a}_x = \hat{a}_y$ and $\hat{k} \times \hat{a}_y = -\hat{a}_x$, so
$\vec{H} = -0.8\sin(kz - \omega t)\hat{a}_y - 1.2\sin(kz - \omega t)\hat{a}_x$.
Answer: (C). | ||||
| 46 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\lambda = \frac{c}{f} = \frac{3 \times 10^8}{3 \times 10^9} = 0.1$ m, so the object is only 1 mm across, far smaller than the wavelength.
An obstacle much smaller than the wavelength cannot reflect or refract the wave (those need a size much larger than $\lambda$), and diffraction is marked when the size is comparable to $\lambda$. A particle much smaller than $\lambda$ scatters the wave in all directions.
Answer: (D). | ||||
| 47 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Equal wavelengths mean equal momenta: $p_e = p_{ph} = p$ (so (C) and (D) are wrong).
$E_e = \frac{p^2}{2m} = \frac{pv}{2}$ (non-relativistic electron), $E_{ph} = pc$.
$\frac{E_e}{E_{ph}} = \frac{v}{2c}$.
Answer: (B). | ||||
| 48 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Mass number falls by 238 − 206 = 32, all from alpha decays: $\frac{32}{4} = 8$ alpha particles.
8 alpha decays lower Z by 16, to 76; reaching 82 needs 6 $\beta^-$ decays (each raises Z by 1).
Answer: (D). | ||||
| 49 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Read from the graph: between 2 V and 2.1 V the current rises from 5 mA to 10 mA: $r_1 = \frac{0.1}{5 \times 10^{-3}} = 20\ \Omega$.
Between 4 V and 4.2 V it rises from 200 mA to 250 mA: $r_2 = \frac{0.2}{50 \times 10^{-3}} = 4\ \Omega$.
$r_1 : r_2 = 20 : 4 = 5 : 1$.
Answer: (B). | ||||
| 50 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter In amplitude modulation the amplitude of the high-frequency carrier wave is varied in accordance with the information (message, or modulating) signal; the result is the modulated wave.
Answer: (C). | ||||
| 51 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The bullet stays directly below the jet only if its horizontal velocity equals the jet's: $400\cos\theta = 200$, so $\cos\theta = \frac{1}{2}$ and $\theta = 60^\circ$.
Answer: 60. | ||||
| 52 | 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 Speed 10 m/s is reached after falling $d = \frac{v^2}{2g} = \frac{100}{20} = 5$ m.
Height above the ground $= 10 - 5 = 5$ m.
Answer: 5. | ||||
| 53 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium 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 From the graph a stress of 20 N/m² gives a strain of $1 \times 10^{-10}$, so $Y = \frac{20}{10^{-10}} = 2 \times 10^{11}\ N/m^2$.
Energy density $= \frac{1}{2}Y\varepsilon^2 = \frac{1}{2} \times 2 \times 10^{11} \times (5 \times 10^{-4})^2 = 2.5 \times 10^4\ J/m^3 = 25\ kJ/m^3$.
Answer: 25. | ||||
| 54 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The wire is stretched by the weight of the same hanging load, so $\Delta l = \frac{mgL}{AY} \propto g$.
$g_p = 10 \times \frac{6 \times 10^{-5}}{10^{-4}} = 6\ ms^{-2}$.
Answer: 6. | ||||
| 55 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Steady current: $I = \frac{20}{10} = 2$ A.
Average emf $= L\frac{\Delta I}{\Delta t} = 20 \times 10^{-3} \times \frac{2}{100 \times 10^{-6}} = 400$ V.
Answer: 400. | ||||
| 56 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At $M_2$ the angle of reflection is 30°, so the ray between the mirrors makes 60° with $M_2$. In the triangle formed with the mirrors (vertex angle 75°), it makes $180^\circ - 75^\circ - 60^\circ = 45^\circ$ with $M_1$, so the angle of incidence at $M_1$ is $\theta_1 = 45^\circ$.
Deviation at $M_1$: $180^\circ - 2 \times 45^\circ = 90^\circ$; at $M_2$: $180^\circ - 2 \times 30^\circ = 120^\circ$.
Total $= 210^\circ$ (in general $360^\circ - 2 \times 75^\circ$).
Answer: 210. | ||||
| 57 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter 1 MSD $= \frac{1}{20}$ cm $= 0.5$ mm. 10 VSD = 9 MSD, so 1 VSD = 0.9 MSD.
Vernier constant $= 1\text{ MSD} - 1\text{ VSD} = 0.1 \times 0.5$ mm $= 0.05$ mm $= 5 \times 10^{-2}$ mm.
Answer: 5. | ||||
| 58 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Read the battery symbol: the longer plate (positive terminal) is at the bottom, so current leaves the battery at the bottom-left, passes the 2 Ω resistor to the bottom-right node and must return to the top-left node.
Diodes (ideal): $D_2$ points from the top-left node towards the right, so it is reverse biased for this current and carries none. $D_1$ points from the right towards the top-left node: forward biased. $D_3$ points up the diagonal towards the top-left node: forward biased.
Path 1: bottom-right → 10 Ω → right side → 6 Ω → $D_1$ → top-left: 16 Ω.
Path 2: bottom-right → 6 Ω → $D_3$ → 10 Ω → top-left: 16 Ω.
These are in parallel: 8 Ω; with the 2 Ω in series, 10 Ω in all.
$I = \frac{10}{10} = 1$ A.
Answer: 1. | ||||
| 59 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current L and C are in parallel, and this pair is in series with the 55 Ω resistor.
At resonance of the parallel LC (ideal, no resistance), $X_L = X_C$ and the currents in L and C are equal and opposite, so the impedance of the pair is infinite.
No current flows from the source, so the current through 55 Ω is 0.
Answer: 0. | ||||
| 60 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Continuity: $av_1 = \frac{a}{2}v_2$, so $v_2 = 2v_1$. The wide end is h = 1 m above the narrow end.
Bernoulli: $P_1 + \frac{1}{2}\rho v_1^2 + \rho gh = P_2 + \frac{1}{2}\rho v_2^2$, so $P_1 - P_2 = \frac{3}{2}\rho v_1^2 - \rho gh$.
$4100 = 1200v_1^2 - 8000$, so $v_1^2 = \frac{12100}{1200} = \frac{121}{12}$ and $v_1 = \frac{11}{2\sqrt{3}} = \frac{\sqrt{363}}{6}$.
x = 363.
Answer: 363. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry 1 L of the acid weighs 1460 g, of which HCl is $0.35 \times 1460 = 511$ g.
Moles of HCl $= \frac{511}{36.5} = 14.0$, so the molarity is 14.0 M.
Answer: (C). | ||||
| 62 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Volume of the vessel $= \frac{135.0 - 40.0}{0.95} = 100$ mL $= 0.1$ L. Mass of gas $= 0.5$ g.
$n = \frac{PV}{RT} = \frac{0.82 \times 0.1}{0.082 \times 250} = 0.004$ mol.
$M = \frac{0.5}{0.004} = 125\ g\ mol^{-1}$.
Answer: (D). | ||||
| 63 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom $r_n = 0.529\frac{n^2}{Z}$ Å, so $r_n \propto n^2$: $\frac{r_4}{r_3} = \frac{16}{9}$.
Answer: (B). | ||||
| 64 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $O_2$ has 16 electrons and bond order 2. Each electron added goes into a $\pi^*$ orbital (lowering the bond order by $\frac{1}{2}$); removing one raises it by $\frac{1}{2}$.
$O_2^+$: 2.5, $O_2$: 2, $O_2^-$: 1.5, $O_2^{2-}$: 1.
Increasing order: $O_2^{2-} < O_2^- < O_2 < O_2^+$.
Answer: (A). | ||||
| 65 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry A reference electrode must keep a steady potential, so its emf should change as little as possible with temperature. Half cell C has the smallest temperature coefficient, $0.1 \times 10^{-4}$.
Answer: (C). | ||||
| 66 | 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 Down group 13 the inert pair effect grows: the $ns^2$ pair is less and less available for bonding, so the +1 state becomes steadily more stable (Tl⁺ is the most stable, Al⁺ the least).
Order: Al < Ga < In < Tl.
Answer: (A). | ||||
| 67 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | StatementsEasy | |
A easy statements question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Statement I is incorrect: an oxide with a lower (more negative) $\Delta G^\circ$ of formation is the more stable one.
Statement II is correct: a metal whose oxide line lies lower has a more negative $\Delta G^\circ$, so for its reaction with a higher-placed oxide the overall $\Delta G^\circ$ is negative and it reduces that oxide.
Answer: (D). | ||||
| 68 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Electrolysis of hydrogensulphate gives peroxodisulphate: $2HSO_4^- \rightarrow HO_3SOOSO_3H + 2e^-$; its hydrolysis gives back $2HSO_4^- + 2H^+$ and hydrogen peroxide, so A is $H_2O_2$.
$H_2O_2$ has an open-book structure: its dihedral angle is 111.5° in the gas phase but 90.2° in the solid at 110 K.
Answer: (C). | ||||
| 69 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Melting points: Be ≈ 1560 K, Mg ≈ 924 K, Ca ≈ 1124 K, Sr ≈ 1062 K. Magnesium is anomalously low (its metallic packing differs from the others).
Order: Be > Ca > Sr > Mg.
Answer: (D). | ||||
| 70 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter From $H_2S$ to $H_2Te$ the melting point rises with molar mass (stronger van der Waals forces): $H_2S$ (188 K) < $H_2Se$ (208 K) < $H_2Te$ (222 K). $H_2O$ (273 K) is highest because of hydrogen bonding.
Answer: (A). | ||||
| 71 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Only white phosphorus reacts with alkali as a disproportionation: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$. The oxoacid of the product, $H_3PO_2$, has two P–H bonds.
Red phosphorus does not react with alkali; $P_2O_3$ and $H_3PO_3$ are simply neutralised to phosphite, and $H_3PO_3$ has one P–H bond.
So no option gives an oxoacid of phosphorus with no P–H bond.
No option is correct as printed. | ||||
| 72 | Environmental ChemistryOzone depletion | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ozone depletion, from Environmental Chemistry. NCERT: No matching NCERT chapter In the Antarctic spring, polar stratospheric clouds provide surfaces on which chlorine nitrate is hydrolysed or reacts with HCl:
$ClONO_2 + H_2O \rightarrow HOCl + HNO_3$; $ClONO_2 + HCl \rightarrow Cl_2 + HNO_3$.
They facilitate the formation of HOCl (and $Cl_2$), which sunlight later splits into Cl radicals. $ClONO_2$ is consumed, not formed.
Answer: (B). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsEasy | |
A easy statements question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Statement I is correct: $Na + C + N + S \rightarrow NaSCN$, which gives a blood-red colour with $Fe^{3+}$.
Statement II is correct: with excess sodium, $NaSCN + 2Na \rightarrow NaCN + Na_2S$, so the usual cyanide and sulphide tests then work.
Answer: (A). | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $(C_7H_5O_2)_2$ is benzoyl peroxide, $C_6H_5CO-O-O-COC_6H_5$. Light breaks the weak O–O bond homolytically, giving two benzoyloxy radicals $C_6H_5COO^\bullet$; each loses $CO_2$ to give a phenyl radical.
X is the benzoyloxy radical, the structure with one unpaired electron on oxygen.
Answer: (D). | ||||
| 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 The reagent is the bis-Grignard $BrMg-CH_2-CH_2-MgBr$ and the ketone is pentane-2,4-dione, $CH_3COCH_2COCH_3$.
Each $CH_2-MgBr$ end adds to one C=O group. The two carbonyl carbons, the $CH_2$ between them and the two Grignard carbons close a five-membered ring, giving the bis-alkoxide A.
Hydrolysis (B) gives 1,3-dimethylcyclopentane-1,3-diol: each former carbonyl carbon carries OH and $CH_3$, separated by one ring $CH_2$. This is structure (A).
Answer: (A). | ||||
| 76 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Cyclohexane-1,3,5-trione can enolise all three C=O groups to give 1,3,5-trihydroxybenzene (phloroglucinol), an aromatic compound. Aromatic stabilisation makes the enol form overwhelmingly favoured, far more than for the 1,3-dione (a conjugated enol, but not aromatic), the 1,2-dione or the 1,4-dione.
Answer: (C). | ||||
| 77 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctHard | |
A hard single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques For the N lone pair to conjugate with the C=C, the enamine must lie flat with the pyrrolidine ring, so the groups near the nitrogen crowd each other.
(1) On the β-carbon, the OMe group should be trans to nitrogen, leaving only a small H next to the ring. This rules out (B) and (D), where OMe is cis to N.
(2) The α-Me group should point towards the unsubstituted ring $CH_2$, not towards the ring carbon carrying the COOH group. In (C) the Me is next to the COOH-bearing carbon; in (A) it is on the other side.
So (A) has the least steric strain.
Answer: (A). | ||||
| 78 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | StatementsEasy | |
A easy statements question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter (A) Buna-S (buta-1,3-diene + styrene) is a copolymer: correct.
(B) Nylon-6,6 is made by condensing hexamethylenediamine and adipic acid: correct.
(C) Nylon-6,6 is a fibre: correct.
(D) Terylene is a thermoplastic polyester fibre, not thermosetting: incorrect.
(E) Buna-N (buta-1,3-diene + acrylonitrile) is a copolymer, not a homopolymer: incorrect.
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 Histamine stimulates the secretion of pepsin and hydrochloric acid in the stomach. Cimetidine and Zantac (ranitidine) are antacids that block histamine's action on the receptors; antihistamines also oppose it.
Answer: (C). | ||||
| 80 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter In the ring test, $NO_3^-$ is reduced by $Fe^{2+}$ to NO, which binds to $[Fe(H_2O)_6]^{2+}$, giving the brown complex $[Fe(H_2O)_5(NO)]^{2+}$, i.e. $[Fe(H_2O)_5(NO)]SO_4$. This is nitrosoferrous sulphate (NO bound as a nitrosyl ligand), not a 'nitro' complex. It appears as a dark brown ring at the junction of the two layers, so (A) and (C) are true.
Heating a nitrate with conc. $H_2SO_4$ gives light brown fumes of $NO_2$, which deepen when copper turnings are added, so (D) is true.
The statement that is not true is (B).
Answer: (B). | ||||
| 81 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics A bomb calorimeter measures heat at constant volume: $\Delta U = -726$ kJ $mol^{-1}$.
$\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}$; T = 300 K.
$\Delta H = \Delta U + \Delta n_gRT = -726 - \frac{1}{2} \times 8.3 \times 10^{-3} \times 300 = -726 - 1.245 = -727.245$ kJ $mol^{-1}$.
x = 727 (nearest integer).
Answer: 727. | ||||
| 82 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions 0.5% by mass: 0.5 g KCl in 99.5 g water.
Molality $m = \frac{0.5/74.6}{0.0995} = \frac{0.006702}{0.0995} = 0.06736$ mol $kg^{-1}$.
$\Delta T_f = iK_fm$: $i = \frac{0.24}{1.80 \times 0.06736} = 1.979$.
For KCl, $i = 1 + \alpha$, so $\alpha = 0.979$, i.e. 97.9%, which rounds to 98.
Answer: 98. | ||||
| 83 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Acid: 5 mmol; NaOH added: 2.5 mmol. The mixture holds 2.5 mmol $CH_3COOH$ and 2.5 mmol $CH_3COO^-$ (half-neutralisation).
Henderson equation: $pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]} = 4.76 + \log 1 = 4.76 = 476 \times 10^{-2}$.
Answer: 476. | ||||
| 84 | 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 Half-lives independent of concentration mean first-order decay.
After t seconds: $[A] = [A]_0\left(\frac{1}{2}\right)^{t/100}$, $[B] = [B]_0\left(\frac{1}{2}\right)^{t/50}$, with $[A]_0 = [B]_0$.
$\frac{[A]}{[B]} = 2^{t/50 - t/100} = 2^{t/100} = 4$, so $\frac{t}{100} = 2$ and t = 200 s.
Answer: 200. | ||||
| 85 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter 2.0 g of $H_2$ is 1 mol. At 300 K and 1 bar its volume is $V = \frac{nRT}{P} = 0.083 \times 300 = 24.9$ L $= 24900$ mL.
Per gram of platinum: $\frac{24900}{2.5} = 9960$ mL.
Answer: 9960. | ||||
| 86 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Basic character falls as the oxidation state rises, so $V_2O_3$ (V +3) is the most basic ($V_2O_4$ is amphoteric, $V_2O_5$ amphoteric/acidic).
$V^{3+}$ is $3d^2$: n = 2 unpaired electrons. $\mu = \sqrt{n(n+2)} = \sqrt{8} = 2.83$ B.M., i.e. 3.
Answer: 3. | ||||
| 87 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds As octahedral complexes: $CoCl_3 \cdot 4NH_3 = [Co(NH_3)_4Cl_2]Cl$ (1 mol AgCl); $NiCl_2 \cdot 6H_2O = [Ni(H_2O)_6]Cl_2$ (2 mol AgCl); $PtCl_4 \cdot 2HCl = H_2[PtCl_6]$ (no AgCl).
So the complex is $[Ni(H_2O)_6]Cl_2$: $Ni^{2+}$ is $3d^8$ with 2 unpaired electrons in an octahedral field.
$\mu = \sqrt{2 \times 4} = 2.83$ B.M., i.e. 3.
Answer: 3. | ||||
| 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 Mass of C $= 0.20 \times \frac{12}{44} = 0.0545$ g.
% C $= \frac{0.0545}{0.30} \times 100 = 18.18\%$, i.e. 18.
Answer: 18. | ||||
| 89 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | NumericalEasy | |
A easy numerical question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers P is phenol. Dilute $HNO_3$ gives o-nitrophenol (A, intramolecular H-bond, steam volatile) and p-nitrophenol (B, intermolecular H-bonds).
Concentrated $HNO_3$ gives 2,4,6-trinitrophenol (picric acid), a yellow, strongly acidic solid: $C_6H_2(OH)(NO_2)_3$.
Oxygen atoms: 1 (OH) + 3 × 2 ($NO_2$) = 7.
Answer: 7. | ||||
| 90 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules The base found only in RNA is uracil; its nucleotide is uridine monophosphate, $C_9H_{13}N_2O_9P$.
Oxygens: uracil 2 (two C=O) + ribose part 4 (ring O, 2'-OH, 3'-OH, 5'-O) + phosphate 3 more (P=O and two P–OH) = 9.
Answer: 9. | ||||
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