JEE (Main) · 2024
JEE (Main) 2024 — 1 February, 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 Calculus5
- Limits, Continuity and Differentiability3
- Sets, Relations and Functions2
7 easy · 21 medium · 2 hard
Physics
30 Q
- Electrostatics4
- Kinematics2
- Laws of Motion2
19 easy · 10 medium · 1 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- d- and f-Block Elements3
- Chemical Bonding and Molecular Structure2
20 easy · 10 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $\sqrt{x^2 - 25}$ needs $x^2 \ge 25$, i.e. $x \le -5$ or $x \ge 5$.
The denominator needs $x \ne \pm 2$ (already excluded).
The logarithm needs $x^2 + 2x - 15 = (x + 5)(x - 3) > 0$, i.e. $x < -5$ or $x > 3$.
Together: $x < -5$ or $x \ge 5$, so the domain is $(-\infty, -5) \cup [5, \infty)$.
Hence $\alpha = -5$, $\beta = 5$ and $\alpha^2 + \beta^3 = 25 + 125 = 150$.
Answer: (C). | ||||
| 2 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $R_1$: $1R_11$ would need $1 + 1 = 1$, which is false, so $R_1$ is not reflexive and not an equivalence relation.
$R_2$: $a + d = b + c \Leftrightarrow a - b = c - d$. So $(a, b)R_2(c, d)$ exactly when the pairs have the same value of $a - b$.
"Having the same value of $a - b$" is reflexive, symmetric and transitive, so $R_2$ is an equivalence relation.
Answer: (B). | ||||
| 3 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\left|z + \frac{1}{2}(3 + 4i)\right|$ is the distance of $z$ from the point $w = -\frac{3}{2} - 2i$, and $|w| = \frac{5}{2}$.
As printed ($|z| \ge 1$), $z = w$ itself is allowed since $|w| = \frac{5}{2} \ge 1$, so the minimum would be $0$, which is not an option.
The question evidently intends $|z| \le 1$ (the closed unit disc). The point of the disc nearest to $w$ lies on the segment from the origin to $w$, at distance $|w| - 1 = \frac{5}{2} - 1 = \frac{3}{2}$.
So the minimum value is $\frac{3}{2}$.
Answer: (A). | ||||
| 4 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha + \beta = -\frac{q}{p}$ and $\alpha\beta = -\frac{r}{p}$.
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{q}{r} = \frac{3}{4}$.
In the G.P. $p, q, r$ the common ratio is $\frac{r}{q} = \frac{4}{3}$, so $\frac{q}{p} = \frac{4}{3}$ and $\frac{r}{p} = \frac{16}{9}$.
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \frac{q^2}{p^2} + \frac{4r}{p} = \frac{16}{9} + \frac{64}{9} = \frac{80}{9}$.
Answer: (B). | ||||
| 5 | 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 For infinitely many solutions the coefficient determinant must vanish:
$\begin{vmatrix}1&2&3\\2&3&1\\4&3&\lambda\end{vmatrix} = 1(3\lambda - 3) - 2(2\lambda - 4) + 3(6 - 12) = -\lambda - 13 = 0$, so $\lambda = -13$.
The first two equations are independent, so the third must be a combination $a(\text{I}) + b(\text{II})$:
$x$: $a + 2b = 4$; $y$: $2a + 3b = 3$. Hence $b = 5$, $a = -6$; check $z$: $3a + b = -18 + 5 = -13 = \lambda$.
Then $\mu = 5a + 9b = -30 + 45 = 15$.
$\lambda + 2\mu = -13 + 30 = 17$.
Answer: (A). | ||||
| 6 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom{18}{r}\left(\frac{1}{3}x^{1/3}\right)^{18 - r}\left(\frac{1}{2}x^{-2/3}\right)^{r}$, whose coefficient is $\binom{18}{r}3^{-(18 - r)}2^{-r}$.
Seventh term ($r = 6$): $m = \binom{18}{6}3^{-12}2^{-6}$.
Thirteenth term ($r = 12$): $n = \binom{18}{12}3^{-6}2^{-12}$.
Since $\binom{18}{6} = \binom{18}{12}$, $\frac{n}{m} = \frac{3^{6}}{2^{6}} = \left(\frac{3}{2}\right)^6$.
$\left(\frac{n}{m}\right)^{1/3} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
Answer: (D). | ||||
| 7 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the first term be $a$ and common difference $d$.
$\frac{a + 9d}{a + 4d} = \frac{15}{7} \Rightarrow 7a + 63d = 15a + 60d \Rightarrow 8a = 3d$.
$S_{10} = 5(2a + 9d) = 390 \Rightarrow 2a + 9d = 78$. With $a = \frac{3d}{8}$: $\frac{3d}{4} + 9d = 78 \Rightarrow d = 8$, $a = 3$.
$S_{15} - S_5$ is the sum of the 6th to 15th terms: $\frac{10}{2}(T_6 + T_{15}) = 5[(3 + 40) + (3 + 112)] = 5 \times 158 = 790$.
Answer: (D). | ||||
| 8 | 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 $f$ is built from continuous functions, so it is continuous everywhere: $m = 0$.
Let $g(x) = 2x^2 + 5|x| - 3 = (2|x| - 1)(|x| + 3)$, which vanishes at $x = \pm\frac{1}{2}$ and changes sign there with non-zero slope, so $f = |g|$ has corners at $x = \pm\frac{1}{2}$.
Near $x = 0$, $g(0) = -3 < 0$, so $f = -g = -2x^2 - 5|x| + 3$, whose left and right derivatives at 0 are $5$ and $-5$: not differentiable at $0$.
So $n = 3$ and $m + n = 3$.
Answer: (C). | ||||
| 9 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f$ sends an odd number to an even one ($2x$) and an even number to an odd one ($x - 1$).
$f(y) = 21$ (odd): $y$ must be even with $y - 1 = 21$, so $y = 22$.
$f(z) = 22$ (even): $z$ must be odd with $2z = 22$, so $z = 11$.
$f(a) = 11$ (odd): $a$ must be even with $a - 1 = 11$, so $a = 12$.
As $x \to 12^-$: $\frac{|x|^3}{12} \to \frac{1728}{12} = 144$ and $\frac{x}{12} \to 1^-$, so $\left[\frac{x}{12}\right] = 0$.
The limit is $144 - 0 = 144$.
Answer: (B). | ||||
| 10 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.
$\int_0^{\pi/3}\cos^4 x\,dx = \frac{3}{8}\cdot\frac{\pi}{3} + \frac{1}{4}\sin\frac{2\pi}{3} + \frac{1}{32}\sin\frac{4\pi}{3} = \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64} = \frac{\pi}{8} + \frac{7\sqrt{3}}{64}$.
So $a = \frac{1}{8}$, $b = \frac{7}{64}$ and $9a + 8b = \frac{9}{8} + \frac{7}{8} = 2$.
Answer: (B). | ||||
| 11 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $g(x) = 2x^3 - 3x^2 - x + 1$. Then
$g(1 - x) = 2(1 - x)^3 - 3(1 - x)^2 - (1 - x) + 1 = -2x^3 + 3x^2 + x - 1 = -g(x)$.
Since the cube root is odd, $g(1 - x)^{1/3} = -g(x)^{1/3}$.
With $I = \int_0^1 g(x)^{1/3}dx$, the property $\int_0^1 h(x)dx = \int_0^1 h(1 - x)dx$ gives $I = -I$, so $I = 0$.
Answer: (B). | ||||
| 12 | 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 $f' - \alpha f = 3$ gives $f(x) = -\frac{3}{\alpha} + Ce^{\alpha x}$, and $f(0) = 2$ gives $C = 2 + \frac{3}{\alpha}$.
The limit condition is meant to fix $\alpha$: the exponential term must die out at the end where the limit is taken, leaving $-\frac{3}{\alpha} = 1$, i.e. $\alpha = -3$.
(With $\alpha = -3$ the term $e^{-3x}$ dies out as $x \to +\infty$, not as $x \to -\infty$; as printed, with $x \to -\infty$, no $\alpha$ satisfies all the conditions. The intended limit is $x \to \infty$.)
Then $C = 2 - 1 = 1$ and $f(x) = 1 + e^{-3x}$.
$f(-\log_e 2) = 1 + e^{3\log_e 2} = 1 + 8 = 9$.
Answer: (D). | ||||
| 13 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Take $P = (3\cos\theta, 2\sin\theta)$; then $Q = (3\cos\theta, 3\sin\theta)$ (same side of the $x$-axis).
$R$ divides $PQ$ internally in the ratio $4 : 3$: $R = \frac{3P + 4Q}{7} = \left(3\cos\theta, \frac{6\sin\theta + 12\sin\theta}{7}\right) = \left(3\cos\theta, \frac{18}{7}\sin\theta\right)$.
The locus is $\frac{x^2}{9} + \frac{y^2}{(18/7)^2} = 1$, an ellipse with $a = 3$, $b = \frac{18}{7}$.
$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{324/49}{9} = 1 - \frac{36}{49} = \frac{13}{49}$, so $e = \frac{\sqrt{13}}{7}$.
Answer: (A). | ||||
| 14 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections The origin lies on the circle (centre $C(0, 1)$, radius 1). For the midpoint $M(h, k)$ of a chord through $O$, $CM \perp OM$:
$(h, k)\cdot(h, k - 1) = 0 \Rightarrow h^2 + k^2 - k = 0$.
The locus is the circle $x^2 + y^2 - y = 0$, centre $\left(0, \frac{1}{2}\right)$, radius $\frac{1}{2}$.
Distance of the centre from $x + y = 1$: $\frac{|0 + \frac{1}{2} - 1|}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
$PQ = 2\sqrt{\frac{1}{4} - \frac{1}{8}} = 2\cdot\frac{1}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Answer: (C). | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line is $L(t) = (1 + 3t, -1 + 2t, 2 + t)$.
For the foot $F$: $(L - P)\cdot(3, 2, 1) = 0$, i.e. $3(3t - 2) + 2(2t - 5) + (t - 7) = 14t - 23 = 0$, so $t = \frac{23}{14}$.
$F = \left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right)$.
The image is $2F - P$, so $\alpha + \beta + \gamma = \frac{2(83 + 32 + 51)}{14} - (3 + 4 + 9) = \frac{332}{14} - 16$.
$14(\alpha + \beta + \gamma) = 332 - 224 = 108$.
Answer: (B). | ||||
| 16 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line is $(-3 + 8t, 4 + 2t, -1 + 2t)$. The foot $M$ of the perpendicular from $R$:
$(-4 + 8t, 2 + 2t, -4 + 2t)\cdot(8, 2, 2) = 72t - 36 = 0 \Rightarrow t = \frac{1}{2}$, so $M = (1, 5, 0)$.
$RM = \sqrt{0 + 9 + 9} = \sqrt{18} < 6$, so two such points exist, and $P$, $Q$ are symmetric about $M$: $P + Q = 2M$.
Centroid $= \frac{P + Q + R}{3} = \frac{(2, 10, 0) + (1, 2, 3)}{3} = (1, 4, 1)$.
$\alpha^2 + \beta^2 + \gamma^2 = 1 + 16 + 1 = 18$.
Answer: (A). | ||||
| 17 | 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 $\vec{AB} = (-3, 5, -2)$, $|\vec{AB}| = \sqrt{38}$; $\vec{AC} = (2, 3, 5)$, $|\vec{AC}| = \sqrt{38}$.
Since $AB = AC$, the bisector of $\angle A$ meets $BC$ at its midpoint: $D = \left(\frac{1}{2}, 7, \frac{7}{2}\right)$.
$\vec{AD} = \left(-\frac{1}{2}, 4, \frac{3}{2}\right)$ and $\vec{AD}\cdot\vec{AC} = -1 + 12 + \frac{15}{2} = \frac{37}{2}$.
Projection length $= \frac{37/2}{\sqrt{38}} = \frac{37}{2\sqrt{38}}$.
Answer: (B). | ||||
| 18 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $A$: Ajay appears, $V$: Vijay appears. $P(A) = 1 - \frac{2}{7} = \frac{5}{7}$ and $P(A \cap V) = \frac{1}{5}$.
$P(A \cap V') = P(A) - P(A \cap V) = \frac{5}{7} - \frac{1}{5} = \frac{25 - 7}{35} = \frac{18}{35}$.
Answer: (C). | ||||
| 19 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctMedium | |
A medium single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $\sum x_i = 10\cdot\frac{6}{5} = 12$, so $12 - 10\alpha = 2$ and $\alpha = 1$.
$\frac{\sum x_i^2}{10} = \frac{84}{25} + \frac{36}{25} = \frac{24}{5}$, so $\sum x_i^2 = 48$.
$\sum(x_i - \beta)^2 = 48 - 24\beta + 10\beta^2 = 40 \Rightarrow 5\beta^2 - 12\beta + 4 = 0 \Rightarrow \beta = 2$ or $\frac{2}{5}$.
$\beta$ is a positive integer, so $\beta = 2$ and $\frac{\beta}{\alpha} = 2$.
Answer: (B). | ||||
| 20 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions With $c = \cos x$: $4(1 - c^2) - 4c^3 + 9 - 4c = 0 \Rightarrow 4c^3 + 4c^2 + 4c = 13$.
For $c \in [-1, 1]$ the left side is increasing ($12c^2 + 8c + 4 > 0$) and its largest value, at $c = 1$, is $12 < 13$.
So the equation has no solution.
Answer: (A). | ||||
| 21 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $M^TM = 1$ says $M$ is a unit column vector. Then
$AM = M - 2M(M^TM) = -M$, so $\lambda = -1$ is possible (with $X = M$).
For a non-zero $X$ perpendicular to $M$ ($M^TX = 0$): $AX = X - 2M(M^TX) = X$, so $\lambda = 1$ is possible.
These are the only values: $A^2 = I - 4MM^T + 4M(M^TM)M^T = I$, so $AX = \lambda X$ gives $X = \lambda^2X$ and $\lambda^2 = 1$.
Sum of squares $= 1 + 1 = 2$.
Answer: 2. | ||||
| 22 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations With 20 lines in general position there would be $\binom{20}{2} = 190$ points.
The 10 parallel lines $L_1, L_3, \ldots, L_{19}$ meet nowhere among themselves: lose $\binom{10}{2} = 45$.
The 10 concurrent lines $L_2, L_4, \ldots, L_{20}$ meet in the single point $P$ instead of $\binom{10}{2} = 45$ points: lose $44$.
All other intersections (parallel line with concurrent line) can be made distinct.
Maximum $= 190 - 45 - 44 = 101$.
Answer: 101. | ||||
| 23 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the sides be $a, ar, ar^2$ with $a > 0$, $r > 1$, so $ar^2$ is the largest.
The triangle inequality needed is $a + ar > ar^2$, i.e. $r^2 - r - 1 < 0$, i.e. $r < \frac{1 + \sqrt{5}}{2} \approx 1.618$.
So $1 < r < 1.618$: $[r] = 1$ and $[-r] = -2$.
$3[r] + [-r] = 3 - 2 = 1$.
Answer: 1. | ||||
| 24 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $F'(x) = x f(x)$. Differentiating $F(x^2) = x^4 + x^5$: $2x\,F'(x^2) = 4x^3 + 5x^4$, i.e. $2x\cdot x^2 f(x^2) = 4x^3 + 5x^4$.
So $f(x^2) = 2 + \frac{5}{2}x$ for $x > 0$.
$\sum_{r=1}^{12} f(r^2) = 24 + \frac{5}{2}\cdot\frac{12 \cdot 13}{2} = 24 + 195 = 219$.
Answer: 219. | ||||
| 25 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $x^2 - \sqrt{x} = \sqrt{x}(x^{3/2} - 1) = \sqrt{x}(\sqrt{x} - 1)(x + \sqrt{x} + 1)$ and $x\sqrt{x} + x + \sqrt{x} = \sqrt{x}(x + \sqrt{x} + 1)$.
So the first term is $(\sqrt{x} + 1)(\sqrt{x} - 1) = x - 1$, with derivative 1.
The second term is $\frac{1}{15}(3\cos^5 x - 5\cos^3 x)$, with derivative $\frac{1}{15}(-15\cos^4 x\sin x + 15\cos^2 x\sin x) = \sin x\cos^2 x(1 - \cos^2 x) = \sin^3 x\cos^2 x$.
At $x = \frac{\pi}{6}$: $\sin^3 x\cos^2 x = \frac{1}{8}\cdot\frac{3}{4} = \frac{3}{32}$, so $y'\left(\frac{\pi}{6}\right) = 1 + \frac{3}{32} = \frac{35}{32}$.
$96 \cdot \frac{35}{32} = 105$.
Answer: 105. | ||||
| 26 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Write both curves as $x$ in terms of $y$: $x_1 = \frac{2y^2}{k}$ and $x_2 = y - \frac{k}{2}y^2$.
$x_2 - x_1 = y - cy^2$ with $c = \frac{k}{2} + \frac{2}{k} = \frac{k^2 + 4}{2k}$; the curves meet at $y = 0$ and $y = \frac{1}{c}$.
Area $= \left|\int_0^{1/c}(y - cy^2)\,dy\right| = \frac{1}{2c^2} - \frac{1}{3c^2} = \frac{1}{6c^2}$.
The area is largest when $|c| = \frac{k^2 + 4}{2|k|}$ is least. By A.M.–G.M., $\frac{|k|}{2} + \frac{2}{|k|} \ge 2$, with equality at $|k| = 2$.
So $k = 2$ or $k = -2$, and the sum of squares is $4 + 4 = 8$.
Answer: 8. | ||||
| 27 | 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 This is linear in $x$: $\frac{dx}{dy} - \frac{x}{y} = \frac{1 - y^2}{y}$, with integrating factor $\frac{1}{y}$.
$\frac{d}{dy}\left(\frac{x}{y}\right) = \frac{1 - y^2}{y^2} = \frac{1}{y^2} - 1$, so $\frac{x}{y} = -\frac{1}{y} - y + C$.
$x = 1$ at $y = 1$: $1 = -1 - 1 + C$, so $C = 3$ and $x = -1 - y^2 + 3y$.
At $y = 2$: $x = -1 - 4 + 6 = 1$, so $5x(2) = 5$.
Answer: 5. | ||||
| 28 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The chord $PQ$ is $y = (a - b)x + ab$. The area between it and $y = x^2$ from $-b$ to $a$:
$S_1 = \int_{-b}^{a}[(a - b)x + ab - x^2]\,dx = \int_{-b}^{a}(a - x)(x + b)\,dx = \frac{(a + b)^3}{6}$.
$S_2 = \frac{1}{2}|a\cdot b^2 - (-b)\cdot a^2| = \frac{1}{2}ab(a + b)$.
$\frac{S_1}{S_2} = \frac{(a + b)^2}{3ab} \ge \frac{4ab}{3ab} = \frac{4}{3}$, with equality at $a = b$.
So $\frac{m}{n} = \frac{4}{3}$ and $m + n = 7$.
Answer: 7. | ||||
| 29 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | NumericalHard | |
A hard numerical question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines $BC = 2\,AB\sin 60^\circ = AB\sqrt{3}$, so $AB = AC = 4$.
$B$ on the positive $x$-axis at distance 4 from $A(-1, 0)$: $B = (3, 0)$.
$C$ makes $120^\circ$ with $\vec{AB}$: $C = (-1 + 4\cos 120^\circ, \pm 4\sin 120^\circ) = (-3, \pm 2\sqrt{3})$.
Take $C = (-3, 2\sqrt{3})$: line $BC$ is $y = -\frac{x - 3}{\sqrt{3}}$. With $y = x + 3$: $\sqrt{3}x + 3\sqrt{3} = 3 - x$, so $\alpha = \frac{3(1 - \sqrt{3})}{1 + \sqrt{3}} = -3(2 - \sqrt{3})$ and $\beta = \alpha + 3 = 3(\sqrt{3} - 1)$.
$\beta^2 = 9(4 - 2\sqrt{3}) = 18(2 - \sqrt{3})$, so $\beta^4 = 324(2 - \sqrt{3})^2$, and $\alpha^2 = 9(2 - \sqrt{3})^2$: $\frac{\beta^4}{\alpha^2} = 36$.
Take $C = (-3, -2\sqrt{3})$: line $y = \frac{x - 3}{\sqrt{3}}$ gives $\alpha = -3(2 + \sqrt{3})$, $\beta = -3(1 + \sqrt{3})$, $\beta^2 = 18(2 + \sqrt{3})$, and again $\frac{\beta^4}{\alpha^2} = \frac{324(2 + \sqrt{3})^2}{9(2 + \sqrt{3})^2} = 36$.
Answer: 36. | ||||
| 30 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $(\vec{b} - \vec{c}) \times \vec{a} = \vec{0}$, so $\vec{c} = \vec{b} - \lambda\vec{a} = (-1 - \lambda, -8 - \lambda, 2 - \lambda)$.
The $\hat{i}$-component is 4, so $\lambda = -5$ and $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$.
With $\vec{d} = 3\hat{i} + 4\hat{j} + \hat{k}$: $\vec{c}\cdot\vec{d} = 12 - 12 + 7 = 7$, $|\vec{c}|^2 = 74$, $|\vec{d}|^2 = 26$.
$\cos^2\theta = \frac{49}{74 \cdot 26} = \frac{49}{1924}$, so $\tan^2\theta = \frac{1924 - 49}{49} = \frac{1875}{49} \approx 38.27$.
The greatest integer $\le \tan^2\theta$ is 38.
Answer: 38. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementSignificant figures | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on significant figures, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (A) 1001: all non-zero digits and the zeros between them count: 4 significant figures → (II).
(B) 010.1: the leading zero does not count; 1, 0, 1 do: 3 → (I).
(C) 100.100: in a number with a decimal point, the trailing zeros count: 1, 0, 0, 1, 0, 0 → 6 → (IV).
(D) 0.0010010: the zeros before the first 1 do not count; 1, 0, 0, 1, 0 do: 5 → (III).
So A→II, B→I, C→IV, D→III.
Answer: (B). | ||||
| 32 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Take north as positive. $v_A = 72$ km/h $= 20\ \mathrm{m\,s^{-1}}$ and $v_B = -108$ km/h $= -30\ \mathrm{m\,s^{-1}}$.
Velocity of B with respect to A: $v_B - v_A = -30 - 20 = -50\ \mathrm{m\,s^{-1}}$.
Velocity of the ground with respect to B: $0 - v_B = 30\ \mathrm{m\,s^{-1}}$.
Answer: (A). | ||||
| 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 The ball's momentum $0.12 \times 25 = 3\ \mathrm{kg\,m\,s^{-1}}$ is destroyed in $0.1$ s.
Average force $= \frac{\Delta p}{\Delta t} = \frac{3}{0.1} = 30$ N, and by Newton's third law the ball exerts the same magnitude on the hand.
Answer: (C). | ||||
| 34 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Net force $\vec{F}_1 + \vec{F}_2 = 8\hat{i} + 4\hat{j} + 4\hat{k}$.
$\vec{a} = \frac{\vec{F}}{m} = \frac{8\hat{i} + 4\hat{j} + 4\hat{k}}{4} = 2\hat{i} + \hat{j} + \hat{k}$.
Answer: (A). | ||||
| 35 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power On the smooth incline there is no friction, so no torque acts about the centre: the disc keeps spinning with its original angular speed, and its rotational kinetic energy does not change.
Only the translational kinetic energy $\frac{1}{2}Mv^2$ is turned into potential energy:
$\frac{1}{2}Mv^2 = Mgh \Rightarrow h = \frac{v^2}{2g}$.
Answer: (B). | ||||
| 36 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation The attraction supplies the centripetal force: $m\omega^2R = kR^{-3/2}$, so $\omega^2 \propto R^{-5/2}$.
$T = \frac{2\pi}{\omega}$ gives $T^2 \propto \frac{1}{\omega^2} \propto R^{5/2}$.
Answer: (C). | ||||
| 37 | Properties of Bulk MatterSurface tension | 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 surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Volume is conserved: $\frac{4}{3}\pi R^3 = 1000 \cdot \frac{4}{3}\pi r^3$, so $R = 10r$.
Surface energy is proportional to the total surface area:
before: $1000 \cdot 4\pi r^2$; after: $4\pi(10r)^2 = 100 \cdot 4\pi r^2$.
The surface energy becomes $\frac{100}{1000} = \frac{1}{10}$th of its value.
Answer: (B). | ||||
| 38 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics At constant pressure $W = nR\Delta T$ and $Q = nC_p\Delta T$ with $C_p = \frac{\gamma R}{\gamma - 1}$.
$Q = \frac{\gamma}{\gamma - 1}W = \frac{1.4}{0.4} \times 200 = 700$ J.
Answer: (B). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}} \propto \frac{1}{\sqrt{M}}$ at the same temperature.
$\frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{2}{32}} = \frac{1}{4}$, so $v_{O_2} = \frac{2}{4} = 0.5$ km/s.
Answer: (A). | ||||
| 40 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The inner cube $C_1$ encloses only $2Q$: $\phi_1 = \frac{2Q}{\varepsilon_0}$.
The outer cube $C_2$ encloses both charges, $2Q + 3Q = 5Q$: $\phi_2 = \frac{5Q}{\varepsilon_0}$.
$\phi_1 : \phi_2 = 2 : 5$.
Answer: (B). | ||||
| 41 | 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 Label the four corners of the bridge: left $L$, top $T$, right $R$, bottom $B$. The arms are $L$–$T$: $4\ \Omega$; $T$–$R$: $C_2 = 6\ \mu$F; $L$–$B$: $C_1 = 4\ \mu$F; $B$–$R$: $6\ \Omega$; and $G$ ($2\ \Omega$) between $T$ and $B$. The 6 V cell is across $L$ and $R$.
In the steady state no current flows through the capacitors, so the current path is $L \to 4\ \Omega \to T \to G \to B \to 6\ \Omega \to R$: total $12\ \Omega$, $I = \frac{6}{12} = 0.5$ A.
Taking $V_L = 6$ V and $V_R = 0$: $V_T = 6 - 0.5 \times 4 = 4$ V and $V_B = 4 - 0.5 \times 2 = 3$ V.
$C_1$ (between $L$ and $B$) has $6 - 3 = 3$ V: $Q_1 = 4 \times 3 = 12\ \mu$C.
$C_2$ (between $T$ and $R$) has $4 - 0 = 4$ V: $Q_2 = 6 \times 4 = 24\ \mu$C.
$\frac{Q_1}{Q_2} = \frac{12}{24} = \frac{1}{2}$.
Answer: (B). | ||||
| 42 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter If a fraction $f$ of the main current $I$ passes through the galvanometer, the shunt $S$ carries $(1 - f)I$ at the same potential difference:
$fI\,G = (1 - f)I\,S \Rightarrow S = \frac{f}{1 - f}G$, and the ammeter's resistance is $R_A = \frac{GS}{G + S} = fG$.
As printed ($f = 5\% = 0.05$): $S = \frac{G}{19}$ and $R_A = \frac{G}{20}$, which is not among the options.
The options are built for $f = 0.5\% = \frac{1}{200}$: then $S = \frac{G}{199}$ (option A, the shunt) and $R_A = \frac{G}{200}$.
So the intended resistance of the ammeter is $\frac{G}{200}$.
Answer: (B). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsTransformers | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on transformers, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Input power 4 kW; output power $= 0.8 \times 4000 = 3200$ W.
$I_s = \frac{P_{out}}{V_s} = \frac{3200}{240} = 13.33$ A.
Answer: (A). | ||||
| 44 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $\lambda = \frac{c}{f} = \frac{3 \times 10^8}{60 \times 10^6} = 5$ m.
Answer: (B). | ||||
| 45 | 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 The first minima of single-slit diffraction are at $a\sin\theta = \lambda$: $\sin\theta = \frac{2.0}{4.0} = \frac{1}{2}$, so $\theta = 30^\circ$ on each side.
The central maximum spreads from $-30^\circ$ to $+30^\circ$: angular spread $2\theta = 60^\circ$ (the screen distance is not needed).
Answer: (B). | ||||
| 46 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Energy of one photon: $E = h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 3.978 \times 10^{-19}$ J.
Photons per second: $n = \frac{P}{E} = \frac{2 \times 10^{-3}}{3.978 \times 10^{-19}} \approx 5.0 \times 10^{15}$.
Answer: (A). | ||||
| 47 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | StatementsMedium | |
A medium statements question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei (A) True: Bohr's quantisation condition is $L = n\hbar$.
(B) True: nuclear forces are short-ranged and fall off far faster than $\frac{1}{r^2}$.
(C) True: the nuclear force depends on the relative orientation of the nucleon spins.
(D) False: nuclear forces are (nearly) charge independent but they are non-central (they have a tensor part).
(E) Taken as true: the smaller the packing fraction, the more stable the nucleus, which is what 'inversely proportional' is meant to say.
So (A), (B), (C), (E) only.
Answer: (D). | ||||
| 48 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The threshold wavelength corresponds to the band gap: $E_g = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} = 3 \times 10^{-19}$ J.
In eV: $\frac{3 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.875 = \frac{15}{8}$ eV, so $X = 15$.
Answer: (C). | ||||
| 49 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | Single correctHard | |
A hard single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity The bridge arms are $AB = 0.8\ \mathrm{m\Omega}$, $BC = R$ (semiconductor), $AD = 1\ \mathrm{m\Omega}$, $DC = 3\ \mathrm{m\Omega}$.
Balance: $\frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}} \Rightarrow \frac{0.8}{R} = \frac{1}{3} \Rightarrow R = 2.4\ \mathrm{m\Omega}$.
In 10 s the arm is cooled by $2 \times 10 = 20^\circ$C, and its resistance changes from $3\ \mathrm{m\Omega}$ to $2.4\ \mathrm{m\Omega}$: a fractional change of magnitude $\frac{0.6}{3} = 0.2$.
$|\alpha| = \frac{0.2}{20} = 1 \times 10^{-2}\ ^\circ\mathrm{C}^{-1}$, and a semiconductor has a negative temperature coefficient, so $\alpha = -1 \times 10^{-2}\ ^\circ\mathrm{C}^{-1}$.
Answer: (A). | ||||
| 50 | Current ElectricityMetre bridge | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity Balance: $\frac{2}{X} = \frac{40}{60} \Rightarrow X = 3\ \Omega$.
Shunting with $2\ \Omega$: $X' = \frac{3 \times 2}{3 + 2} = 1.2\ \Omega$.
New balance length $l$: $\frac{2}{1.2} = \frac{l}{100 - l} \Rightarrow 3l = 5(100 - l) \Rightarrow l = 62.5$ cm.
The balance length changes by $62.5 - 40 = 22.5$ cm.
Answer: (B). | ||||
| 51 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $a = v\frac{dv}{dx} = 4\sqrt{x} \cdot \frac{4}{2\sqrt{x}} = 8\ \mathrm{m\,s^{-2}}$ (constant; equivalently $v^2 = 16x = 2ax$).
Answer: 8. | ||||
| 52 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The impulse $J = 0.2$ N s is taken perpendicular to the rod at $B$. Its angular impulse about the centre of mass equals the rod's angular momentum:
$J\cdot\frac{L}{2} = I\omega$ with $I = \frac{ML^2}{12} = \frac{2 \times 0.09}{12} = 0.015\ \mathrm{kg\,m^2}$.
$0.2 \times 0.15 = 0.015\,\omega \Rightarrow \omega = 2$ rad/s.
The rod then turns uniformly, so turning through $\frac{\pi}{2}$ takes $t = \frac{\pi/2}{2} = \frac{\pi}{4}$ s; $x = 4$.
Answer: 4. | ||||
| 53 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The upper wire holds both loads: $T_1 = (2 + 1)g = 30$ N. The lower wire holds only the 1 kg load: $T_2 = 10$ N.
For identical wires, strain $= \frac{T}{AY} \propto T$, so the ratio of strains is $\frac{30}{10} = 3$.
Answer: 3. | ||||
| 54 | 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 $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \propto \frac{1}{\sqrt{m}}$, so $\frac{f_1}{f_2} = \sqrt{\frac{9m}{m}} = 3$.
Answer: 3. | ||||
| 55 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The thread is tied to the wall. With the particle 10 cm from the wall on a 20 cm thread, the thread makes $\theta$ with the wall where $\sin\theta = \frac{10}{20}$, so $\theta = 30^\circ$.
Equilibrium: $T\sin\theta = qE$ and $T\cos\theta = mg$, so $qE = mg\tan 30^\circ$.
$q = \frac{mg\tan 30^\circ}{E} = \frac{0.002 \times 10}{\sqrt{3} \times 2 \times 10^4} = \frac{10^{-6}}{\sqrt{3}}$ C $= \frac{1}{\sqrt{3}}\ \mu$C.
So $x = 3$.
Answer: 3. | ||||
| 56 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Three branches join the same two rails: the 10 V cell with $R_1 = 4\ \Omega$; the capacitor $C = 10\ \mu$F with $R_2 = 5\ \Omega$; and $R_3 = 6\ \Omega$.
In the steady state no current flows in the capacitor branch, so the cell drives $I = \frac{10}{4 + 6} = 1$ A round the loop through $R_1$ and $R_3$.
The potential difference across $R_3$, $6 \times 1 = 6$ V, is also across the capacitor branch, and with no current in $R_2$ it all appears across $C$.
$Q = CV = 10 \times 6 = 60\ \mu$C.
Answer: 60. | ||||
| 57 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter At equilibrium $NBIA = k\theta$.
$NBIA = 100 \times 0.01 \times 10 \times 10^{-3} \times 2 \times 10^{-4} = 2 \times 10^{-6}$ N m.
$k = \frac{2 \times 10^{-6}}{0.05} = 4 \times 10^{-5}$ N m/rad, so $x = 4$.
Answer: 4. | ||||
| 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 $\omega = 2\pi \times 0.5 = \pi$ rad/s.
$\varepsilon_{max} = NBA\omega = 200 \times 0.01 \times 0.20 \times \pi = 0.4\pi = \frac{2\pi}{5}$ V, so $\beta = 5$.
Answer: 5. | ||||
| 59 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $I = I_{max}\cos^2\frac{\phi}{2}$. $I = \frac{I_{max}}{2}$ first when $\frac{\phi}{2} = \frac{\pi}{4}$, i.e. $\phi = \frac{\pi}{2}$, a path difference of $\frac{\lambda}{4}$.
$\frac{yd}{D} = \frac{\lambda}{4} \Rightarrow y = \frac{\lambda D}{4d} = \frac{5 \times 10^{-7} \times 1.0}{4 \times 10^{-3}} = 1.25 \times 10^{-4}$ m $= 125 \times 10^{-6}$ m.
Answer: 125. | ||||
| 60 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\nu \propto \frac{1}{n_1^2} - \frac{1}{n_2^2}$ for the same ion.
$2 \to 1$: $1 - \frac{1}{4} = \frac{3}{4}$; $3 \to 1$: $1 - \frac{1}{9} = \frac{8}{9}$.
$\nu_{31} = 3 \times 10^{15} \times \frac{8/9}{3/4} = 3 \times 10^{15} \times \frac{32}{27} = \frac{32}{9} \times 10^{15}$ Hz, so $x = 32$.
Answer: 32. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Radial nodes $= n - l - 1$. For 3p, $n = 3$ and $l = 1$: $3 - 1 - 1 = 1$.
Answer: (D). | ||||
| 62 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | StatementsEasy | |
A easy statements question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Statement I: a $\pi$ bonding MO concentrates electron density above and below the inter-nuclear axis (it has a nodal plane containing the axis itself). It does not have lower density there, so Statement I is false.
Statement II: the $\pi^*$ antibonding MO has, in addition, a nodal plane between the nuclei, so it has a node between them: true.
Answer: (D). | ||||
| 63 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure In o-nitrophenol the OH and NO2 groups are on adjacent ring carbons, so the O–H hydrogen bonds to an oxygen of the nitro group within the same molecule (a six-membered chelate ring).
$\mathrm{NH_3}$, $\mathrm{C_2H_5OH}$ and $\mathrm{H_2O}$ form only intermolecular hydrogen bonds.
Answer: (C). | ||||
| 64 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $W$ g per 100 mL is $10W$ g per litre, so the molar solubility is $s = \frac{10W}{M}$ mol/L.
$\mathrm{Ca_3(PO_4)_2 \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-}}$: $K_{sp} = (3s)^3(2s)^2 = 108s^5$.
$K_{sp} = 108 \times 10^5\left(\frac{W}{M}\right)^5 = 1.08 \times 10^7\left(\frac{W}{M}\right)^5 \approx 10^7\left(\frac{W}{M}\right)^5$.
Answer: (B). | ||||
| 65 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsEasy | |
A easy statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Statement I: the p-block contains both metals and non-metals, but every d-block element is a metal. So Statement I is false.
Statement II: non-metals hold their electrons more tightly, with higher ionisation enthalpies and electronegativities than metals: true.
Answer: (D). | ||||
| 66 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down group 15 the E–H bond becomes longer and weaker, so the hydrides give up hydrogen more easily and their reducing character increases: $\mathrm{NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3}$.
The strongest reducing agent is $\mathrm{BiH_3}$.
Answer: (B). | ||||
| 67 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: in group 14 the dioxides of carbon, silicon and germanium are acidic, while SnO, $\mathrm{SnO_2}$, PbO and $\mathrm{PbO_2}$ are amphoteric. True.
Statement II: carbon's allotropes arise from catenation and its ability to form $p\pi$–$p\pi$ multiple bonds (as in graphite). Carbon has no d orbitals, so $p\pi$–$d\pi$ bonding plays no part. False.
Answer: (C). | ||||
| 68 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The third ionisation removes an electron from the $M^{2+}$ ion. $\mathrm{Mn^{2+}}$ is $3d^5$, a stable half-filled configuration, so removing an electron from it is hardest.
(V$^{2+}$ $3d^3$, Cr$^{2+}$ $3d^4$, Fe$^{2+}$ $3d^6$ lose an electron more easily; the third ionisation enthalpies are about 2828, 2987, 3248 and 2957 kJ/mol for V, Cr, Mn, Fe.)
Answer: (C). | ||||
| 69 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Assertion and reasonMedium | |
A medium assertion and reason question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements (A) is true: $\mathrm{Cr^{2+}}$ ($d^4$) is readily oxidised to $\mathrm{Cr^{3+}}$, and $\mathrm{Mn^{3+}}$ ($d^4$) is readily reduced to $\mathrm{Mn^{2+}}$.
(R) is true, and it explains (A): $\mathrm{Cr^{3+}}$ is $d^3$ with a half-filled $t_{2g}$ level, and $\mathrm{Mn^{2+}}$ is $d^5$, half-filled. Each ion moves to the configuration with the extra stability of a half-filled set.
Answer: (A). | ||||
| 70 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Both contain $\mathrm{Co^{3+}}$ ($3d^6$).
$\mathrm{NH_3}$ is a strong-field ligand: the electrons pair up, the complex uses $d^2sp^3$ (inner orbital) hybridisation, and it is spin paired (diamagnetic).
$\mathrm{F^-}$ is a weak-field ligand: the electrons stay unpaired, the complex uses $sp^3d^2$ (outer orbital) hybridisation, and it is spin free (4 unpaired electrons).
So: spin paired, spin free. (Option C reverses inner/outer; option D is wrong about $\mathrm{[CoF_6]^{3-}}$.)
Answer: (A). | ||||
| 71 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In $\mathrm{KMnO_4}$ (Mn +7), $\mathrm{K_2Cr_2O_7}$ and $\mathrm{K_2CrO_4}$ (Cr +6) the metal has no d electrons ($d^0$), so their colour comes from ligand-to-metal charge transfer.
In $\mathrm{CuSO_4.5H_2O}$, $\mathrm{Cu^{2+}}$ is $3d^9$, and its blue colour is due to d–d transitions.
Answer: (D). | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Fusing the organic compound with sodium converts its nitrogen, sulphur, halogens and phosphorus into ionic forms (NaCN, $\mathrm{Na_2S}$, NaX, $\mathrm{Na_3PO_4}$) that are then tested in the sodium fusion extract.
So Lassaigne's test detects nitrogen, sulphur, phosphorus and halogens.
Answer: (D). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $-\mathrm{OR}$, $-\mathrm{OH}$ and $-\mathrm{NH_2}$ have a lone pair on the atom attached to the ring and donate electrons by resonance (+R).
$-\mathrm{COOH}$ has a C=O group in conjugation that withdraws electrons by resonance (−R).
Answer: (C). | ||||
| 74 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Electron-withdrawing groups (−R) are meta directing: $-\mathrm{NO_2}$, $-\mathrm{CHO}$, $-\mathrm{SO_3H}$, $-\mathrm{COR}$, $-\mathrm{COOH}$, $-\mathrm{COOR}$, $-\mathrm{CN}$.
$-\mathrm{NH_2}$, $-\mathrm{NHR}$, $-\mathrm{NHCOCH_3}$ and $-\mathrm{OCH_3}$ are ortho/para directing, so options (A), (C) and (D) each contain an o/p director.
Only (B) has all four meta directors.
Answer: (B). | ||||
| 75 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The first product is cis-pent-2-ene ($\mathrm{CH_3}$ and $\mathrm{C_2H_5}$ on the same side). Adding one mole of $\mathrm{H_2}$ on a palladium catalyst gives a cis alkene from an alkyne, so A is pent-2-yne, $\mathrm{CH_3C{\equiv}CC_2H_5}$. (n-Pentane is already saturated and cannot add $\mathrm{H_2}$.)
Reduction of but-2-yne with Na in liquid $\mathrm{NH_3}$ (Birch-type reduction) adds hydrogen anti, giving trans-but-2-ene: B.
Answer: (B). | ||||
| 76 | Haloalkanes and HaloarenesEnvironmental effects of polyhalogen compounds | Class 12: Haloalkanes and Haloarenes | Matching listEasy | |
A easy matching list question on environmental effects of polyhalogen compounds, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Carbon tetrachloride was used in fire extinguishers → (III).
Methylene chloride ($\mathrm{CH_2Cl_2}$) is a solvent used as a paint remover → (I).
DDT is a non-biodegradable insecticide → (IV).
Freons (chlorofluorocarbons) are refrigerants in refrigerators and air conditioners → (II).
So A→III, B→I, C→IV, D→II.
Answer: (D). | ||||
| 77 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Matching listEasy | |
A easy matching list question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (A) Heating phenol with zinc dust removes the OH group: benzene → (III).
(B) $\mathrm{CHCl_3}$/NaOH then acid is the Reimer–Tiemann reaction: salicylaldehyde → (I).
(C) $\mathrm{CO_2}$/NaOH then acid is Kolbe's reaction: salicylic acid → (II).
(D) Concentrated $\mathrm{HNO_3}$ gives 2,4,6-trinitrophenol, picric acid → (IV).
So A→III, B→I, C→II, D→IV.
Answer: (A). | ||||
| 78 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $\mathrm{C_2H_5Br}$ with alcoholic KOH undergoes elimination: A $= \mathrm{CH_2{=}CH_2}$.
Addition of $\mathrm{Br_2}$ in $\mathrm{CCl_4}$: B $= \mathrm{BrCH_2CH_2Br}$.
Excess KCN replaces both bromines: C $= \mathrm{NCCH_2CH_2CN}$.
Hydrolysis of both nitrile groups with excess $\mathrm{H_3O^+}$: D $= \mathrm{HOOCCH_2CH_2COOH}$, succinic acid.
Answer: (D). | ||||
| 79 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Butane (non-polar) < diethyl ether (weakly polar) < butanal (dipole–dipole) < butan-1-ol (intermolecular hydrogen bonding).
Butan-1-ol, $\mathrm{CH_3CH_2CH_2CH_2OH}$, boils highest (about 391 K).
Answer: (D). | ||||
| 80 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | StatementsMedium | |
A medium statements question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Statement I: in nickel dimethylglyoximate each DMG binds Ni through two N atoms, forming five-membered chelate rings (the six-membered rings are the O–H···O hydrogen-bonded rings, not chelate rings). So Statement I is false.
Statement II: Prussian blue is $\mathrm{Fe_4[Fe(CN)_6]_3}$, with $\mathrm{Fe^{3+}}$ outside and $\mathrm{Fe^{2+}}$ inside the complex ion: true.
Answer: (D). | ||||
| 81 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{C_xH_y + \left(x + \frac{y}{4}\right)O_2 \to xCO_2 + \frac{y}{2}H_2O}$.
By volume: 10 mL gives $10x = 40$ mL of $\mathrm{CO_2}$, so $x = 4$, and $5y = 50$ mL of water vapour, so $y = 10$.
The hydrocarbon is $\mathrm{C_4H_{10}}$: $4 + 10 = 14$ atoms.
Answer: 14. | ||||
| 82 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta G^\circ = -2.303RT\log K = -2.303 \times 8.314 \times 300 \times \log 10 = -5744$ J mol$^{-1}$ $= -5.744$ kJ mol$^{-1}$.
That is $-57.44 \times 10^{-1}$ kJ mol$^{-1}$; to the nearest integer, 57.
Answer: 57. | ||||
| 83 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = K_f\,m$: $24 = 1.86\,m$, so $m = \frac{24}{1.86}$ mol kg$^{-1}$.
Moles of glycol for 18.6 kg of water: $\frac{24}{1.86} \times 18.6 = 240$ mol.
Mass $= 240 \times 62 = 14880$ g $= 14.88$ kg; to the nearest integer, 15.
Answer: 15. | ||||
| 84 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\mathrm{2H_2O \to O_2 + 4H^+ + 4e^-}$: 2 mol of water give up 4 mol of electrons, so 1 mol needs 2 F.
$Q = 2 \times 96500 = 1.93 \times 10^5$ C; to the nearest integer, 2.
Answer: 2. | ||||
| 85 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Balanced: $\mathrm{2MnO_4^- + 6H^+ + 5H_2C_2O_4 \rightleftharpoons 2Mn^{2+} + 8H_2O + 10CO_2}$, so $n = 10$.
$E^\circ_{cell} = 1.51 - (-0.49) = 2.00$ V.
$\log K = \frac{nFE^\circ}{2.303RT} = \frac{10 \times 2.00}{0.05916} = 338.1$ at 298 K.
So $K \approx 10^{338}$ and $x = 338$.
Answer: 338. | ||||
| 86 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let $x$ atm of A decompose: total pressure $= (0.1 - x) + 2x + x = 0.1 + 2x = 0.28$, so $x = 0.09$ and $p_A = 0.01$ atm at 115 s.
$k = \frac{2.303}{t}\log\frac{p_0}{p_A} = \frac{2.303}{115}\log\frac{0.1}{0.01} = \frac{2.303}{115} = 2.0 \times 10^{-2}\ \mathrm{s^{-1}}$.
Answer: 2. | ||||
| 87 | 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 Moles of $\mathrm{H_2SO_4} = 0.010 \times 2 = 0.02$; it neutralises $2 \times 0.02 = 0.04$ mol of $\mathrm{NH_3}$.
Nitrogen $= 0.04 \times 14 = 0.56$ g in 1 g of compound, i.e. 56%.
Answer: 56. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques 2-Methylbutane, $\mathrm{(CH_3)_2CHCH_2CH_3}$, has four kinds of hydrogen:
1-chloro-2-methylbutane, $\mathrm{ClCH_2CH(CH_3)CH_2CH_3}$: C2 is chiral, 2 stereoisomers.
2-chloro-2-methylbutane: no stereocentre, 1.
2-chloro-3-methylbutane, $\mathrm{(CH_3)_2CHCHClCH_3}$: chiral, 2 stereoisomers.
1-chloro-3-methylbutane, $\mathrm{(CH_3)_2CHCH_2CH_2Cl}$: 1.
Total $= 2 + 1 + 2 + 1 = 6$.
Answer: 6. | ||||
| 89 | AminesClassification and identification of amines | Class 12: Amines | NumericalMedium | |
A medium numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hinsberg's reagent (benzenesulphonyl chloride) reacts with primary and secondary amines (N–H available on a basic amine nitrogen); tertiary amines, amides/imides, ureas, pyridine and diazonium salts do not react.
Benzenediazonium chloride: no. Dibenzamide (imide N–H, not an amine): no. Aniline (1°): yes.
N-Phenylpiperidine (3°): no. N-Methylbenzylamine (2°): yes. Trimethylamine (3°): no. Ethane-1,2-diamine (1°): yes.
Piperidine (2°): yes. Pyridine: no. Propan-1-amine (1°): yes. Urea (amide): no.
Count: aniline, N-methylbenzylamine, ethane-1,2-diamine, piperidine, propylamine $= 5$.
Answer: 5. | ||||
| 90 | BiomoleculesProteins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules A tripeptide is a sequence from the N-terminus to the C-terminus, so each ordering of the three different amino acids is a different tripeptide: $3! = 6$.
Answer: 6. | ||||
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