JEE (Main) · 2022
JEE (Main) 2022 — 26 July, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Conic Sections3
- Matrices and Determinants2
12 easy · 13 medium · 5 hard
Physics
30 Q
- Electrostatics3
- Properties of Bulk Matter3
- Kinematics2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- Some Basic Concepts of Chemistry3
- p-Block Elements3
- Chemical Bonding and Molecular Structure2
23 easy · 6 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 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations The equation is $x^2 + (3 - a)x + (1 - 2a) = 0$, so $\alpha + \beta = a - 3$ and $\alpha\beta = 1 - 2a$.
$\alpha^2 + \beta^2 = (a - 3)^2 - 2(1 - 2a) = a^2 - 2a + 7 = (a - 1)^2 + 6$.
The discriminant $(3 - a)^2 - 4(1 - 2a) = a^2 + 2a + 5 = (a + 1)^2 + 4 > 0$, so the roots are real for every $a$.
The minimum is 6, at $a = 1$.
Answer: (C). | ||||
| 2 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|z - i| = |z + 5i|$ puts z on the perpendicular bisector of $i$ and $-5i$, the line $y = -2$.
$|z| = 2$ then gives $x^2 + 4 = 4$, so $x = 0$: $z = -2i$, i.e. $(x, y) = (0, -2)$.
Check: (A) $0 - 4 - 4 \ne 0$; (B) $0 - 2 - 4 \ne 0$; (C) $0 - 4 + 4 = 0$; (D) $0 + 2 + 3 \ne 0$.
Answer: (C). | ||||
| 3 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants With A the column of ones, $A'BA$ is the sum of all the entries of B.
$81 - 100 + 121 + 144 + 169 - 196 - 225 + 256 + 289 = 539$.
Answer: (D). | ||||
| 4 | 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 Summing over all pairs, $\sum_{i,j} {}^{n}C_i\,{}^{n}C_j = \left(\sum_i {}^{n}C_i\right)^2 = 2^{2n}$.
The diagonal terms $i = j$ give $\sum_i ({}^{n}C_i)^2 = {}^{2n}C_n$.
So the sum over $i \ne j$ is $2^{2n} - {}^{2n}C_n$.
Answer: (A). | ||||
| 5 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctHard | |
A hard single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The circle has centre $C(1, -1)$ and radius 1, and lies wholly below the parabola. The shortest distance is along the common normal: the normal to $y = x^2$ at $Q(t, t^2)$ must pass through C, and P is where CQ meets the circle.
Normal at Q has direction $(-2t, 1)$, i.e. $(1 - t) + 2t(-1 - t^2) = 0$: $2t^3 + 3t - 1 = 0$.
$g(t) = 2t^3 + 3t - 1$ is increasing; $g(0.3) = -0.046$, $g(0.32) = 0.026$, so $t \approx 0.313$ and $Q \approx (0.313, 0.098)$.
$\vec{CQ} \approx (-0.687, 1.098)$, of length $\approx 1.295$; the unit vector is $\approx (-0.530, 0.848)$.
$P = C + $ unit vector $\approx (0.470, -0.152)$, so the abscissa of P is about 0.47, in $\left(\frac{1}{4}, \frac{1}{2}\right)$.
Answer: (C). | ||||
| 6 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctHard | |
A hard single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f_a'(x) = \frac{2}{1 + 4x^2} - 3a \ge 0$ on the interval. The least value of $\frac{2}{1 + 4x^2}$ there is approached at $x = \pm\frac{\pi}{6}$: $\frac{2}{1 + \pi^2/9} = \frac{18}{9 + \pi^2}$.
So $3a \le \frac{18}{9 + \pi^2}$ and $\bar{a} = \frac{6}{9 + \pi^2}$.
$3\bar{a}\cdot\frac{\pi}{8} = \frac{18\pi}{8(9 + \pi^2)} = \frac{9\pi}{4(9 + \pi^2)}$.
Then $f_{\bar{a}}\left(\frac{\pi}{8}\right) = \tan^{-1}\frac{\pi}{4} + 7 - \frac{9\pi}{4(9 + \pi^2)}$.
With $\tan^{-1}\frac{\pi}{4} \approx 0.6658$ and $\frac{9\pi}{4(9 + \pi^2)} \approx 0.3746$, $f_{\bar{a}}\left(\frac{\pi}{8}\right) \approx 7.291$.
The options are about 7.625 (A), 7.899 (B), 4.608 (C) and 7.215 (D). (A) has the right last term but takes $\tan^{-1}\frac{\pi}{4}$ as 1, which is false.
No option is correct as printed. | ||||
| 7 | 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 The denominator behaves like $3\alpha x^2$. The numerator is $\alpha x - 3x - \frac{9x^2}{2} - \ldots$; for a finite limit the $x$ term must vanish, so $\alpha = 3$.
Then $\beta = \lim \frac{-\frac{9}{2}x^2}{9x^2} = -\frac{1}{2}$.
$\alpha + \beta = 3 - \frac{1}{2} = \frac{5}{2}$.
Answer: (C). | ||||
| 8 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\log_{\cos x}\mathrm{cosec}\,x = \frac{\ln \mathrm{cosec}\,x}{\ln\cos x} = -\frac{\ln\sin x}{\ln\cos x}$.
Its derivative is $-\frac{\cot x\,\ln\cos x + \tan x\,\ln\sin x}{(\ln\cos x)^2}$.
At $x = \frac{\pi}{4}$: $\ln\sin x = \ln\cos x = -\frac{1}{2}\ln 2$ and $\tan x = \cot x = 1$, so the derivative is $-\frac{2(-\frac{1}{2}\ln 2)}{\frac{1}{4}(\ln 2)^2} = \frac{4}{\ln 2}$.
Multiplying by $\log_e 2$ gives 4.
Answer: (D). | ||||
| 9 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $(|\sin x| + |\cos x|)^2 = 1 + |\sin 2x|$.
$|\sin 2x|$ has period $\frac{\pi}{2}$ and $\int_0^{\pi/2}|\sin 2x|\,dx = 1$; $[0, 20\pi]$ holds 40 periods.
The integral is $20\pi + 40 = 20(\pi + 2)$.
Answer: (D). | ||||
| 10 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor: $e^{\int \frac{x}{x^2 - 1}dx} = \sqrt{1 - x^2}$ on $(-1, 1)$.
$\frac{d}{dx}\left(y\sqrt{1 - x^2}\right) = x^4 + 2x$, so $y\sqrt{1 - x^2} = \frac{x^5}{5} + x^2 + C$; through the origin, $C = 0$.
$f(x) = \frac{x^5/5 + x^2}{\sqrt{1 - x^2}}$. Over the symmetric interval the odd part integrates to zero.
$\int_{-\sqrt{3}/2}^{\sqrt{3}/2}\frac{x^2}{\sqrt{1 - x^2}}dx = 2\int_0^{\pi/3}\sin^2\theta\,d\theta = \left[\theta - \frac{\sin 2\theta}{2}\right]_0^{\pi/3} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$.
Answer: (B). | ||||
| 11 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The ellipse is $\frac{x^2}{5/2} + \frac{y^2}{5/3} = 1$; a tangent of slope m is $y = mx \pm\sqrt{\frac{5}{2}m^2 + \frac{5}{3}}$.
Through $(1, 3)$: $(3 - m)^2 = \frac{5}{2}m^2 + \frac{5}{3}$, i.e. $9m^2 + 36m - 44 = 0$.
$m_1 + m_2 = -4$, $m_1m_2 = -\frac{44}{9}$; $|m_1 - m_2| = \sqrt{16 + \frac{176}{9}} = \frac{8\sqrt{5}}{3}$.
$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| = \frac{8\sqrt{5}/3}{35/9} = \frac{24\sqrt{5}}{35} = \frac{24}{7\sqrt{5}}$.
Answer: (B). | ||||
| 12 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A tangent to $y = x^2$ with slope m is $y = mx - \frac{m^2}{4}$.
It touches $y = -(x - 2)^2$ when $x^2 + (m - 4)x + 4 - \frac{m^2}{4} = 0$ has equal roots: $(m - 4)^2 - 4\left(4 - \frac{m^2}{4}\right) = 2m^2 - 8m = 0$, so $m = 0$ or 4.
$m = 4$ gives $y = 4x - 4 = 4(x - 1)$.
Answer: (B). | ||||
| 13 | CirclesEquation from a diameter | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on equation from a diameter, from Circles. NCERT: Class 11: Conic Sections The circle on PQ as diameter is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$, i.e. $(x^2 - 4x - 6) + (y^2 + 2y - 7) = 0$.
$x^2 + y^2 - 4x + 2y - 13 = 0$: $a = -2$, $b = 1$, $c = -13$.
$a + b - c = -2 + 1 + 13 = 12$.
Answer: (A). | ||||
| 14 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $\frac{x^2}{6/k} - \frac{y^2}{6} = 1$: $a^2 = \frac{6}{k}$, $b^2 = 6$, $e^2 = 1 + \frac{b^2}{a^2} = 1 + k$.
Directrix $x = \frac{a}{e} = 1$ gives $a^2 = e^2$: $\frac{6}{k} = 1 + k$, so $k^2 + k - 6 = 0$ and $k = 2$ (k > 0).
The hyperbola is $2x^2 - y^2 = 6$. Check: (A) $40 - 36 = 4$; (B) $10 - 9 = 1$; (C) $10 - 4 = 6$; (D) $40 - 54 = -14$.
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 Normal to the first plane: $\hat{i} \times (\hat{i} + \hat{j}) = \hat{k}$.
Normal to the second: $(\hat{i} - \hat{j}) \times (\hat{i} + \hat{k}) = -\hat{i} - \hat{j} + \hat{k}$.
The line of intersection is along $\hat{k} \times (-\hat{i} - \hat{j} + \hat{k}) = \hat{i} - \hat{j}$.
$\cos\theta = \pm\frac{(1)(1) + (-1)(-2)}{\sqrt{2}\cdot 3} = \pm\frac{1}{\sqrt{2}}$, so the angles are $\frac{\pi}{4}$ and $\frac{3\pi}{4}$; the obtuse one is $\frac{3\pi}{4}$.
Answer: (A). | ||||
| 16 | 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 $\frac{\alpha}{\alpha + \beta} = \frac{\sin^{-1}x}{\sin^{-1}x + \cos^{-1}x} = \frac{2}{\pi}\sin^{-1}x$, so $\frac{2\pi\alpha}{\alpha + \beta} = 4\sin^{-1}x = 4\theta$ with $\sin\theta = x$.
$\sin 4\theta = 2\sin 2\theta\cos 2\theta = 2\cdot 2x\sqrt{1 - x^2}\cdot(1 - 2x^2) = 4x\sqrt{1 - x^2}(1 - 2x^2)$.
Answer: (B). | ||||
| 17 | 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 $q \Rightarrow p \equiv (\sim q) \vee p$.
If p is true, the right side is true and the biconditional is true. If p is false, the right side is $\sim q$, and $p \Leftrightarrow \sim q$ is true exactly when $\sim q$ is false, i.e. q true.
So $p \Leftrightarrow (q \Rightarrow p) \equiv p \vee q$, and its negation is $(\sim p) \wedge (\sim q)$.
Answer: (D). | ||||
| 18 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $np = 4$, $npq = \frac{4}{3}$, so $q = \frac{1}{3}$, $p = \frac{2}{3}$, $n = 6$.
$P(X \le 2) = \frac{1}{3^6}\left(1 + 6\cdot 2 + 15\cdot 4\right) = \frac{73}{729}$.
$54 \times \frac{73}{729} = \frac{146}{27}$.
Answer: (B). | ||||
| 19 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Write the integrand as $\frac{\sqrt{3}}{2}\left(1 - \frac{1}{\sqrt{3}}\right)\frac{\cos x - \sin x}{\frac{\sqrt{3}}{2} + \sin 2x} = \frac{\sqrt{3} - 1}{2}\cdot\frac{\sqrt{2}\cos\left(x + \frac{\pi}{4}\right)}{\frac{\sqrt{3}}{2} + \sin 2x}$.
Differentiate option (A). Since $\frac{d}{dx}\ln\tan\left(\frac{x}{2} + c\right) = \frac{1}{\sin(x + 2c)}$, its derivative is $\frac{1}{2}\left[\frac{1}{\sin(x + \frac{\pi}{6})} - \frac{1}{\sin(x + \frac{\pi}{3})}\right] = \frac{1}{2}\cdot\frac{\sin(x + \frac{\pi}{3}) - \sin(x + \frac{\pi}{6})}{\sin(x + \frac{\pi}{6})\sin(x + \frac{\pi}{3})}$.
Numerator: $2\cos\left(x + \frac{\pi}{4}\right)\sin\frac{\pi}{12}$. Denominator: $\frac{1}{2}\left[\cos\frac{\pi}{6} - \cos\left(2x + \frac{\pi}{2}\right)\right] = \frac{1}{2}\left(\frac{\sqrt{3}}{2} + \sin 2x\right)$.
So the derivative is $\frac{2\sin\frac{\pi}{12}\cos(x + \frac{\pi}{4})}{\frac{\sqrt{3}}{2} + \sin 2x}$, and $2\sin\frac{\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{2} = \frac{\sqrt{2}(\sqrt{3} - 1)}{2}$: exactly the integrand.
(C) is $-2$ times (A) up to a constant, and (B), (D) give different denominators.
Answer: (A). | ||||
| 20 | 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 $|x^2 - 1| = 1$ at $x = 0$ and $x = \pm\sqrt{2}$; for $|x| \le \sqrt{2}$ the curve is below $y = 1$.
Area $= 2\left[\int_0^1 \left(1 - (1 - x^2)\right)dx + \int_1^{\sqrt{2}}\left(1 - (x^2 - 1)\right)dx\right] = 2\left[\frac{1}{3} + \left(2x - \frac{x^3}{3}\right)_1^{\sqrt{2}}\right]$.
$\left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) - \frac{5}{3} = \frac{4\sqrt{2} - 5}{3}$, so the area is $2\cdot\frac{4\sqrt{2} - 4}{3} = \frac{8}{3}(\sqrt{2} - 1)$.
Answer: (D). | ||||
| 21 | Sets, Relations and FunctionsSets and their representation | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical question on sets and their representation, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $A \cap B = \{3, 6, 7\}$. A subset of A misses B exactly when it is a subset of $\{1, 2, 4, 5\}$: $2^4 = 16$ such subsets.
Required number $= 2^7 - 2^4 = 128 - 16 = 112$.
Answer: 112. | ||||
| 22 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Both lines pass through $(1, 1, 0)$. Normal: $(1, a, -1) \times (-1, 1, -a) = (1 - a^2,\ 1 + a,\ 1 + a) = (1 + a)(1 - a, 1, 1)$ (for $a = -1$ the lines coincide in direction and give no plane).
Plane: $(1 - a)(x - 1) + (y - 1) + z = 0$. Distance of $(2, 1, 4)$: $\frac{|(1 - a) + 4|}{\sqrt{(1 - a)^2 + 2}} = \sqrt{3}$.
$(5 - a)^2 = 3(a^2 - 2a + 3)$, i.e. $a^2 + 2a - 8 = 0$, so $a = 2$ or $-4$.
The largest value is 2.
Answer: 2. | ||||
| 23 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Four-digit numbers with first digit 1 or 2; divisible by 4 means the last two digits form a multiple of 4.
Possible endings from distinct digits of 1–6: 12, 16, 24, 32, 36, 52, 56, 64.
First digit 1: endings without 1 are 24, 32, 36, 52, 56, 64 (6); the middle digit has 3 choices: 18.
First digit 2: endings without 2 are 16, 36, 56, 64 (4); 3 choices each: 12.
Total $18 + 12 = 30$.
Answer: 30. | ||||
| 24 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $k^4 + k^2 + 1 = (k^2 - k + 1)(k^2 + k + 1)$, so $\frac{k}{k^4 + k^2 + 1} = \frac{1}{2}\left[\frac{1}{k^2 - k + 1} - \frac{1}{k^2 + k + 1}\right]$.
Since $k^2 + k + 1 = (k + 1)^2 - (k + 1) + 1$, the sum telescopes: $\frac{1}{2}\left[1 - \frac{1}{111}\right] = \frac{55}{111}$.
$\gcd(55, 111) = 1$, so $m + n = 166$.
Answer: 166. | ||||
| 25 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions First equation: $2\sin^2\theta - (1 - 2\sin^2\theta) = 0$, so $\sin^2\theta = \frac{1}{4}$, $\sin\theta = \pm\frac{1}{2}$.
Second: $2 - 2\sin^2\theta + 3\sin\theta = 0$, i.e. $2\sin^2\theta - 3\sin\theta - 2 = 0$, so $\sin\theta = -\frac{1}{2}$ (2 is impossible).
Common: $\sin\theta = -\frac{1}{2}$, $\theta = \frac{7\pi}{6}, \frac{11\pi}{6}$ in $[0, 2\pi]$. Sum $= 3\pi$.
Answer: 3. | ||||
| 26 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $\sum x = 40 \times 30 = 1200$; $\sum x^2 = 40(5^2 + 30^2) = 37000$.
Omitting 12 and 10: $\sum x = 1178$, $\sum x^2 = 37000 - 144 - 100 = 36756$, $n = 38$; mean $= 31$.
$38\sigma^2 = \sum x^2 - 38\bar{x}^2 = 36756 - 38 \times 961 = 36756 - 36518 = 238$.
Answer: 238. | ||||
| 27 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalHard | |
A hard numerical question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A plane through L is $lx - y + 3(1 - l)z - 1 + \lambda(x + 2y - z - 2) = 0$: coefficients $(l + \lambda,\ 2\lambda - 1,\ 3 - 3l - \lambda)$, constant $1 + 2\lambda$.
It is $3x - 8y + 7z = 4$, so $\frac{2\lambda - 1}{-8} = \frac{1 + 2\lambda}{4}$, giving $\lambda = -\frac{1}{6}$ and common ratio $\frac{1}{6}$; then $l - \frac{1}{6} = \frac{1}{2}$, $l = \frac{2}{3}$ (the z-coefficient $3 - 2 + \frac{1}{6} = \frac{7}{6}$ checks).
L is along $(2, -3, 3) \times (1, 2, -1) = (-3, 5, 7)$ (normals scaled), with $|d|^2 = 83$.
$\cos^2\theta = \frac{25}{83}$, so $415\cos^2\theta = 125$.
Answer: 125. | ||||
| 28 | 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 Integrating factor $e^{-x}$: $(ye^{-x})' = 2e^{-x} - e^{-2x}$, so $ye^{-x} = -2e^{-x} + \frac{1}{2}e^{-2x} + C$ and $y = -2 + \frac{1}{2}e^{-x} + Ce^{x}$.
A finite limit at infinity needs $C = 0$: $y = -2 + \frac{1}{2}e^{-x}$.
$y(0) = -\frac{3}{2}$, $y'(0) = -\frac{1}{2}$. Tangent: $y = -\frac{1}{2}x - \frac{3}{2}$, so $b = -\frac{3}{2}$ and $a = -3$.
$a - 4b = -3 + 6 = 3$.
Answer: 3. | ||||
| 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 $199 = 100 + (n - 1)d$, so $(n - 1)d = 99$ with $3 \le n \le 33$, i.e. $n - 1 \in [2, 32]$ dividing 99.
Divisors of 99 are 1, 3, 9, 11, 33, 99; those in $[2, 32]$ are 3, 9, 11, giving $d = 33, 11, 9$.
Sum $= 33 + 11 + 9 = 53$.
Answer: 53. | ||||
| 30 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalHard | |
A hard numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A = A^{-1}$ means $A^2 = I$: $a^2 + bc = 1$, $d^2 + bc = 1$, $b(a + d) = 0$, $c(a + d) = 0$.
If $a + d \ne 0$: then $b = c = 0$, $a^2 = d^2 = 1$, $a = d$: $A = I$ or $-I$, 2 matrices.
If $d = -a$: both a and $-a$ lie in $\{-1, \ldots, 10\}$, so $a \in \{-1, 0, 1\}$, and $bc = 1 - a^2$.
$a = 0$: $bc = 1$, $(b, c) = (1, 1)$ or $(-1, -1)$: 2.
$a = \pm 1$: $bc = 0$; with 12 values available, $12 + 12 - 1 = 23$ pairs each: 46.
Total $2 + 2 + 46 = 50$.
Answer: 50. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $R = \frac{u^2\sin 2\theta}{g}$, so $R_1 : R_2 = \sin 90^\circ : \sin 60^\circ = 1 : \frac{\sqrt{3}}{2} = 2 : \sqrt{3}$.
Answer: (C). | ||||
| 32 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= 1\text{ MSD} - 1\text{ VSD} = 1 - 0.9 = 0.1$ mm.
Observed reading $= 30 + 6 \times 0.1 = 30.6$ mm.
Zero error: the Vernier zero is to the left of the main-scale zero, a negative error. One Vernier division is 0.9 mm; if the Vernier zero sits $x$ to the left and the 4th Vernier division meets a main-scale mark, $-x + 4 \times 0.9 = 3$ mm, so $x = 0.6$ mm and the zero error is $-0.6$ mm.
Corrected reading $= 30.6 - (-0.6) = 31.2$ mm $= 3.12$ cm, which is not an option. (C) 3.10 cm follows only from taking the error as $-4 \times 0.1 = -0.4$ mm, which is not how a negative zero error is read.
No option is correct as printed. | ||||
| 33 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Change of momentum $= 2mv = 2 \times 0.15 \times 12 = 3.6$ kg m s⁻¹.
$\Delta t = \frac{\Delta p}{F} = \frac{3.6}{100} = 0.036$ s.
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 $p = \sqrt{2mK}$; with equal K, $p_1 : p_2 = \sqrt{8} : \sqrt{2} = 2 : 1$.
Answer: (B). | ||||
| 35 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Connected, the spheres reach a common potential $V = \frac{kQ}{R}$; the surface field is $E = \frac{kQ}{R^2} = \frac{V}{R}$.
$E_A : E_B = \frac{1}{R_A} : \frac{1}{R_B} = 10 : 5 = 2 : 1$ (treating each sphere as isolated, as is usual).
Answer: (B). | ||||
| 36 | 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 speed is $v = \frac{\omega}{k} = \frac{1000\pi \times 4 \times 10^8}{1000\pi \times 5} = 8 \times 10^7$ m s⁻¹ (the wave is in a medium).
$E_0 = vB_0 = 8 \times 10^7 \times 5 \times 10^{-6} = 400 = 4 \times 10^2$ V m⁻¹.
Answer: (D). | ||||
| 37 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\sin C = \frac{n_2}{n_1} = \frac{v_1}{v_2} = \frac{1.5}{2.0} = \frac{3}{4}$.
$\cos C = \frac{\sqrt{7}}{4}$, so $\tan C = \frac{3}{\sqrt{7}}$ and $C = \tan^{-1}\frac{3}{\sqrt{7}}$.
Answer: (A). | ||||
| 38 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v^2 = \frac{1}{9}\cdot 2gR$. Energy conservation: $\frac{1}{2}v^2 = gR - \frac{gR^2}{R + h}$, giving $h = \frac{Rv^2}{2gR - v^2}$.
$h = \frac{R \cdot \frac{2gR}{9}}{2gR - \frac{2gR}{9}} = \frac{R}{8} = 800$ km.
Answer: (A). | ||||
| 39 | 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 $\mu = \frac{V_{max} - V_{min}}{V_{max} + V_{min}} = \frac{60 - 20}{60 + 20} = 0.5$, i.e. 50%.
Answer: (B). | ||||
| 40 | 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 Momentum is conserved from rest, so the two parts have equal and opposite momenta.
$\lambda = \frac{h}{p}$ is the same for both: 1 : 1.
Answer: (C). | ||||
| 41 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Area of the walls $= 2(0.6 \times 0.5 + 0.5 \times 0.2 + 0.6 \times 0.2) = 1.04$ m².
Heat current $= \frac{kA\Delta T}{d} = \frac{0.05 \times 1.04 \times 40}{0.01} = 208$ W.
Rate of melting $= \frac{208}{3.4 \times 10^5} = 6.1 \times 10^{-4} = 61 \times 10^{-5}$ kg s⁻¹.
Answer: (B). | ||||
| 42 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $C_V = \frac{n}{2}R$, $C_P = \left(\frac{n}{2} + 1\right)R$, so $\frac{C_V}{C_P} = \frac{n}{n + 2}$.
Answer: (A). | ||||
| 43 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Wave velocity $\frac{\omega}{k}$; maximum particle velocity $A\omega = 2\omega$.
$\frac{\omega}{k} = 2\omega$ gives $k = \frac{1}{2}$ cm⁻¹, so $\lambda = \frac{2\pi}{k} = 4\pi$ cm.
Answer: (A). | ||||
| 44 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity The network is a Wheatstone bridge: 3 Ω and 3 Ω in the upper arm, 6 Ω and 6 Ω in the lower, 5 Ω between the midpoints.
$\frac{3}{3} = \frac{6}{6}$, so the bridge is balanced and the 5 Ω carries no current.
Upper path 6 Ω in parallel with lower path 12 Ω: $\frac{6 \times 12}{18} = 4$ Ω. With the 2 Ω in series, total 6 Ω.
$I = \frac{6}{6} = 1$ A.
Answer: (A). | ||||
| 45 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With K closed both capacitors are in parallel across V: $E_1 = 2 \times \frac{1}{2}CV^2 = CV^2$.
With K open, the first capacitor stays across the source; the second is isolated with charge $Q = CV$.
After the dielectric: first, $C' = 5C$ at voltage V, energy $\frac{1}{2}(5C)V^2 = 2.5\,CV^2$; second, charge CV on 5C, energy $\frac{(CV)^2}{2 \times 5C} = 0.1\,CV^2$.
$E_2 = 2.6\,CV^2$, so $\frac{E_1}{E_2} = \frac{1}{2.6} = \frac{5}{13}$.
Answer: (C). | ||||
| 46 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium 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 From the figure, the inner loop carries I anticlockwise (moment along $+\hat{k}$) and the outer loop clockwise (moment along $-\hat{k}$), viewing the X-Y plane with z out of the page.
$\vec{M} = I\pi(r_1^2 - r_2^2)\hat{k} = 7\pi(0.09 - 0.25)\hat{k} = -1.12\pi\,\hat{k} \approx -3.52\,\hat{k}$ A m².
This is approximately $-\frac{7}{2}\hat{k}$ A m².
Answer: (B). | ||||
| 47 | 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 $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{2.56 \times 10^{14}} = 1.6 \times 10^7$ m s⁻¹.
Undeflected: $E = vB = 1.6 \times 10^7 \times 12 \times 10^{-3} = 1.92 \times 10^5$ V m⁻¹ $= 192$ kV m⁻¹.
Answer: (A). | ||||
| 48 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $a = \frac{(M_2 - M_1)g}{M_1 + M_2}$.
$M_2 = 2M_1$: $a_1 = \frac{g}{3}$. $M_2 = 3M_1$: $a_2 = \frac{2g}{4} = \frac{g}{2}$.
$\frac{a_1}{a_2} = \frac{2}{3}$.
Answer: (B). | ||||
| 49 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $R = R_0A^{1/3}$, so the volume is proportional to A, and so is the mass.
Density $\propto \frac{A}{A}$ is independent of A: ratio 1 : 1.
Answer: (C). | ||||
| 50 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The rope's breaking stress is fixed, so the area must grow in proportion to the load.
$A = 2.5 \times 10^{-4} \times \frac{25}{10} = 6.25 \times 10^{-4}$ m².
Answer: (A). | ||||
| 51 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Component $= \frac{\vec{A}\cdot\vec{B}}{|\vec{B}|} = \frac{2 + 6 - 2}{\sqrt{1 + 4 + 4}} = \frac{6}{3} = 2$ m.
Answer: 2. | ||||
| 52 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For a thin rod about a perpendicular axis through its centre, $I = \frac{ML^2}{12}$, so $k = \frac{L}{\sqrt{12}} = \frac{10\sqrt{3}}{2\sqrt{3}} = 5$ m.
Answer: 5. | ||||
| 53 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The figure marks the angle of incidence inside the glass at face AC as 60° (measured from the normal to AC).
Grazing emergence along AC: $\mu\sin 60^\circ = n\sin 90^\circ$, so $n = 1.5 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4} = \frac{\sqrt{27}}{4}$.
$x = 27$.
Answer: 27. | ||||
| 54 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Binding energy of Y: $4 \times 7.6 = 30.4$ MeV; of two X: $2 \times 2 \times 1.1 = 4.4$ MeV.
Energy released $= 30.4 - 4.4 = 26$ MeV.
Answer: 26. | ||||
| 55 | 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 Elongation under its own weight: $\Delta l = \frac{MgL}{2AY} = \frac{20 \times 10 \times 20}{2 \times 0.4 \times 2 \times 10^{11}} = \frac{4000}{1.6 \times 10^{11}} = 2.5 \times 10^{-8}$ m.
$= 25 \times 10^{-9}$ m, so $x = 25$.
Answer: 25. | ||||
| 56 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits From the graph, $I_C$ rises by 5 mA for every 100 μA of $I_B$: $\beta = \frac{5 \times 10^{-3}}{100 \times 10^{-6}} = 50$.
Voltage gain $= \beta\frac{R_L}{R_i} = 50 \times \frac{2}{0.5} = 200$.
Answer: 200. | ||||
| 57 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Force from B: $\frac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.16 \times 10^{-6}}{(0.03)^2} = 8$ N, along AB.
Force from C: $\frac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.3 \times 10^{-6}}{(0.03)^2} = 15$ N, along AC.
AB ⟂ AC, so the resultant is $\sqrt{8^2 + 15^2} = 17$ N.
Answer: 17. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $|e| = \left|\frac{d\phi}{dt}\right| = \frac{4t}{3}$; the flux is zero at $t = 3$ s.
$H = \int_0^3 \frac{e^2}{R}dt = \int_0^3 \frac{16t^2}{9 \times 8}dt = \frac{2}{9}\cdot\frac{t^3}{3}\Big|_0^3 = \frac{2}{9} \times 9 = 2$ J.
Answer: 2. | ||||
| 59 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Potential gradient $= \frac{20\text{ mV}}{60\text{ cm}}$, so the whole 300 cm wire has $100$ mV $= 0.1$ V across it.
$\frac{4R}{780 + R} = 0.1$: $4R = 78 + 0.1R$, $R = 20$ Ω (taking the standard cell to have no internal resistance).
Answer: 20. | ||||
| 60 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Figure (a): springs in series, $k_a = \frac{k \cdot 2k}{3k} = \frac{2k}{3}$. Figure (b): in parallel, $k_b = 3k$.
$\frac{T_b}{T_a} = \sqrt{\frac{k_a}{k_b}} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}$, so $T_b = \sqrt{2}$ s and $x = 2$.
Answer: 2. | ||||
| 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 Mass of Fe $= 3.3 \times 0.0034 = 0.01122$ g $= \frac{0.01122}{56} = 2.0 \times 10^{-4}$ mol.
Atoms $= 2.0 \times 10^{-4} \times 6.022 \times 10^{23} = 1.21 \times 10^{20}$.
Answer: (C). | ||||
| 62 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure By Fajans' rules, with the same cation, the larger and more polarisable the anion, the greater the covalent character: $\mathrm{F^- < Cl^- < Br^- < I^-}$.
Order: $\mathrm{CaF_2 < CaCl_2 < CaBr_2 < CaI_2}$, i.e. A < B < C < D.
Answer: (B). | ||||
| 63 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium pOH $= 14 - 8.26 = 5.74$. Henderson equation: $\mathrm{pOH} = \mathrm{p}K_b + \log\frac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]}$, so $\log\frac{[\mathrm{NH_4^+}]}{0.2} = 1$.
$[\mathrm{NH_4^+}] = 2$ M: 2 mol in one litre, $2 \times 53.5 = 107.0$ g.
Answer: (C). | ||||
| 64 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics A half-life independent of concentration means first order: $t = \frac{t_{1/2}}{\log 2}\log\frac{[A]_0}{[A]}$.
80% decomposed leaves 20%: $\log 5 = 1 - 0.30 = 0.70$.
$t = \frac{200 \times 0.70}{0.30} = 466.7 \approx 467$ s.
Answer: (C). | ||||
| 65 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter A: the finest gold sol is red; as the particle size grows it looks purple, then blue and finally golden. True.
R: the colour of a colloidal solution depends on the wavelength of light scattered by the dispersed particles, which in turn depends on their size and nature. True, and it is why the colour of the gold sol changes with particle size.
Answer: (A). | ||||
| 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 Gallium melts at about 303 K (it melts in the hand), unusually low, and sits in Group 13 next to germanium and below/beside the metalloids.
Al and In melt far higher (933 K, 430 K), and Se is a non-metal.
Answer: (B). | ||||
| 67 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Lead (galena, PbS) and zinc (zinc blende, ZnS) are extracted from sulfide ores by roasting and reduction.
Aluminium has no sulfide ore at all; it is extracted from bauxite, $\mathrm{Al_2O_3 \cdot xH_2O}$, by electrolysis.
Iron is extracted from its oxide ores (haematite, magnetite), not from iron pyrites.
Taking the intended answer as the metal whose extraction never involves a sulfide ore: aluminium.
Answer: (A). | ||||
| 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, permanganate oxidises hydrogen peroxide to oxygen and is itself reduced to $\mathrm{Mn^{2+}}$:
$\mathrm{2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2}$.
Answer: (D). | ||||
| 69 | Chemical Bonding and Molecular StructureIonic bonding and lattice enthalpy | Class 11: Chemical Bonding and Molecular Structure | Assertion and reasonEasy | |
A easy assertion and reason question on ionic bonding and lattice enthalpy, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A: LiF is only sparingly soluble, because its lattice enthalpy (small $\mathrm{Li^+}$ with small $\mathrm{F^-}$) is very high. True.
R: $\mathrm{Li^+}$ is the smallest ion of the group, and so has the highest hydration enthalpy, not the least. False.
Answer: (C). | ||||
| 70 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Boric acid is a weak monobasic acid. It is not a protonic acid: as a Lewis acid it takes $\mathrm{OH^-}$ from water and releases $\mathrm{H^+}$:
$\mathrm{B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^- + H_3O^+}$.
Both are correct, and this way of ionising is why it is weak.
Answer: (A). | ||||
| 71 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) Cu(I), $3d^{10}$: no unpaired electron, diamagnetic.
(B) Cu(II), $3d^9$: one unpaired electron.
(C) Fe(I), $3d^7$: odd electron count, paramagnetic.
(D) Fe(II) with weak-field $\mathrm{Cl^-}$, high-spin $3d^6$: four unpaired electrons.
Answer: (A). | ||||
| 72 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Microorganisms (pathogens) come from domestic sewage: A-II.
Plant nutrients (nitrates, phosphates) from chemical fertilizers: B-III.
Toxic heavy metals from chemical factories: C-IV.
Sediment from erosion by strip mining: D-I.
Answer: (A). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Priority: $\mathrm{-COOH > -SO_3H > -COOCOR > -COOR > -COX > -CONH_2 > -CN > -CHO > >C{=}O > -OH > -NH_2}$.
(A) puts $\mathrm{-CONH_2}$ above $\mathrm{-COCl}$; (C) puts $\mathrm{-NH_2}$ above $\mathrm{>C{=}O}$; (D) puts $\mathrm{-CONH_2}$ above $\mathrm{-COCl}$. Only (B) is in the right order.
Answer: (B). | ||||
| 74 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Benzenoid compounds contain benzene rings. (A) is built of fused benzene rings, (C) is naphthalene and (D) is aniline.
(B) is cyclooctatetraene, an eight-membered ring with no benzene ring (and itself non-aromatic, 8 π electrons, tub-shaped).
Answer: (B). | ||||
| 75 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Carbolic acid is phenol. Warming benzenediazonium chloride with water hydrolyses it to phenol: $\mathrm{C_6H_5N_2^+Cl^- + H_2O \rightarrow C_6H_5OH + N_2 + HCl}$.
Cumene gives phenol only after oxidation to cumene hydroperoxide; benzal chloride hydrolyses to benzaldehyde; a ketal hydrolyses to a ketone and the diol.
Answer: (B). | ||||
| 76 | 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 DIBAL-H followed by water reduces an ester to an aldehyde and a nitrile (through the imine) to an aldehyde; the ring C=C is untouched.
Both $\mathrm{EtOOC{-}CH_2{-}}$ and $\mathrm{{-}CH_2CH_2CN}$ therefore become aldehydes: $\mathrm{OHC{-}CH_2{-}}$(ring)$\mathrm{{-}CH_2CH_2CHO}$.
Answer: (A). | ||||
| 77 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Start: 4-nitroaniline; target: 4-iodophenol.
$\mathrm{HNO_2}$ diazotises the $\mathrm{NH_2}$; KI replaces $\mathrm{N_2^+}$ by I, giving 4-iodonitrobenzene.
$\mathrm{Fe/H^+}$ reduces $\mathrm{NO_2}$ to $\mathrm{NH_2}$ (4-iodoaniline); $\mathrm{HNO_2}$ diazotises it again; warm water converts the diazonium salt to the phenol.
Order: $\mathrm{HNO_2}$, KI, $\mathrm{Fe/H^+}$, $\mathrm{HNO_2}$, $\mathrm{H_2O}$/warm.
Answer: (B). | ||||
| 78 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Vulcanisation is heating natural rubber, cis-polyisoprene, with sulphur (and an additive), which forms sulphur cross-links between chains.
Of the options, isoprene and sulphur describes it.
Answer: (C). | ||||
| 79 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Glycogen, the storage polysaccharide of animals, is called animal starch; its structure resembles amylopectin but is more highly branched.
Answer: (C). | ||||
| 80 | EquilibriumIonic equilibrium | Class 11: Equilibrium | Assertion and reasonEasy | |
A easy assertion and reason question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium A: phenolphthalein is colourless in acid and pink in base. True.
R: it is a weak acid, but it does dissociate in basic medium; the pink colour comes from its anion. False.
Answer: (C). | ||||
| 81 | 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 $n = \frac{PV}{RT} = \frac{6 \times 10^5 \times 0.0125}{8.3 \times 300} = \frac{7500}{2490} = 3.01$ mol.
With m g of He: $\frac{10 - m}{2} + \frac{m}{4} = 3.01$, so $20 - m = 12.05$ and $m = 7.95$ g.
Nearest integer: 8.
Answer: 8. | ||||
| 82 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom Neutrons $= 48 - 22 = 26$; electrons $= 22 + 3 = 25$.
$a = \frac{26 - 25}{25} \times 100 = 4$.
Answer: 4. | ||||
| 83 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta n_g = 2 - 1 = 1$.
$\Delta H = \Delta U + \Delta n_gRT = -59.6 + \frac{8.314 \times 300}{1000} = -59.6 + 2.49 = -57.1$ kJ mol⁻¹.
Nearest integer: 57.
Answer: 57. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $K_b = \frac{3}{1} = 3$ K kg mol⁻¹ and $K_f = \frac{6}{2} = 3$ K kg mol⁻¹ (the same solute, same van't Hoff factor).
$K_b : K_f = 1 : 1$, so $X = 1$.
Answer: 1. | ||||
| 85 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter From the equations, $2\mathrm{Cu^{2+}} \equiv \mathrm{I_2} \equiv 2\mathrm{S_2O_3^{2-}}$, so moles of $\mathrm{Cu^{2+}}$ = moles of thiosulphate.
Thiosulphate $= 20 \times 0.02 = 0.4$ mmol, so $[\mathrm{Cu^{2+}}] = \frac{0.4}{10} = 0.04 = 4 \times 10^{-2}$ M.
Answer: 4. | ||||
| 86 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{3C_2H_5OH + PCl_3 \rightarrow 3C_2H_5Cl + H_3PO_3}$, so A is phosphorous acid.
Phosphorous acid with $\mathrm{PCl_3}$ gives pyrophosphorous acid: $\mathrm{5H_3PO_3 + PCl_3 \rightarrow 3H_4P_2O_5 + 3HCl}$.
In $\mathrm{H_4P_2O_5}$, $\mathrm{HO{-}P(H)(=O){-}O{-}P(H)(=O){-}OH}$, two H are on P (non-ionisable) and two on O (ionisable).
Answer: 2. | ||||
| 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 The highest oxidation state, Mn(IV) in $\mathrm{MnF_4}$, is the strongest oxidising agent of the three (MnF4 readily loses fluorine to give $\mathrm{MnF_3}$).
Mn(IV) is $3d^3$: three unpaired electrons, $\mu = \sqrt{3 \times 5} = \sqrt{15} = 3.87$ B.M.
Nearest integer: 4.
Answer: 4. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Chlorine can replace H at five kinds of position.
On the $\mathrm{CH_3}$: (chloromethyl)cyclohexane, achiral: 1.
At C-1: 1-chloro-1-methylcyclohexane, achiral: 1.
At C-2: two stereocentres, cis and trans each a pair of enantiomers: 4.
At C-3: likewise two stereocentres: 4.
At C-4: cis and trans, both achiral (a plane of symmetry): 2.
Total $1 + 1 + 4 + 4 + 2 = 12$.
Answer: 12. | ||||
| 89 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{C_2H_5MgBr + CH_3OH \rightarrow C_2H_6 + Mg(OCH_3)Br}$: the gas is ethane.
$\frac{2.24}{22400} = 1 \times 10^{-4}$ mol of ethane $= 1 \times 10^{-4} \times 30$ g $= 3$ mg.
Answer: 3. | ||||
| 90 | Chemistry in Everyday LifeAntibiotics and antiseptics | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on antibiotics and antiseptics, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Ofloxacin is a broad-spectrum antibiotic.
Penicillin G has a narrow spectrum; terpineol is a constituent of antiseptic preparations, not an antibiotic; Salvarsan is an arsenic drug for syphilis, not broad spectrum.
Count: 1.
Answer: 1. | ||||
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