JEE (Main) · 2024
JEE (Main) 2024 — 4 April, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Matrices and Determinants4
- Limits, Continuity and Differentiability3
- Integral Calculus3
9 easy · 17 medium · 4 hard
Physics
30 Q
- Kinematics3
- Properties of Bulk Matter3
- Current Electricity3
19 easy · 9 medium · 2 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Chemical Bonding and Molecular Structure3
- Aldehydes, Ketones and Carboxylic Acids3
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 correctMedium | |
A medium 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 For $\sin^{-1}$ we need $-1 \le \frac{3x-22}{2x-19} \le 1$.
Upper bound: $\frac{3x-22}{2x-19} - 1 = \frac{x-3}{2x-19} \le 0$, so $3 \le x < \frac{19}{2}$.
Lower bound: $\frac{3x-22}{2x-19} + 1 = \frac{5x-41}{2x-19} \ge 0$, so $x \le \frac{41}{5}$ or $x > \frac{19}{2}$.
Together: $3 \le x \le \frac{41}{5}$.
For the logarithm: $\frac{3x^2-8x+5}{x^2-3x-10} = \frac{(3x-5)(x-1)}{(x-5)(x+2)} > 0$. The critical points are $-2, 1, \frac{5}{3}, 5$, and the sign is positive on $x < -2$, $1 < x < \frac{5}{3}$ and $x > 5$.
Intersecting with $[3, \frac{41}{5}]$ leaves $\left(5, \frac{41}{5}\right]$, so $\alpha = 5$, $\beta = \frac{41}{5}$.
$3\alpha + 10\beta = 15 + 82 = 97$.
Answer: (B). | ||||
| 2 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z = x + iy$. Then $(\bar{z})^2 = x^2 - y^2 - 2ixy$, and the equation gives
real part: $x^2 - y^2 + \sqrt{x^2+y^2} = 0$; imaginary part: $-2xy = 0$.
If $y = 0$: $x^2 + |x| = 0$, so $x = 0$ (the zero solution, excluded).
If $x = 0$: $-y^2 + |y| = 0$, so $|y| = 1$ for a non-zero solution: $z = i$ or $z = -i$.
Sum $\alpha = i + (-i) = 0$; product $\beta = i\cdot(-i) = 1$.
$4(\alpha^2 + \beta^2) = 4(0 + 1) = 4$.
Answer: (B). | ||||
| 3 | Quadratic EquationsForming equations from given roots | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on forming equations from given roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Product of roots: $2\cdot 6 = \frac{1}{a}$, so $a = \frac{1}{12}$. Sum of roots: $2 + 6 = -\frac{b}{a}$, so $b = -8a = -\frac{2}{3}$.
$2a + b = \frac{1}{6} - \frac{2}{3} = -\frac{1}{2}$, so the first new root is $-2$.
$6a + b = \frac{1}{2} - \frac{2}{3} = -\frac{1}{6}$, so the second new root is $-6$.
The equation with roots $-2$ and $-6$: $x^2 + 8x + 12 = 0$.
Answer: (A). | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A homogeneous system has a non-trivial solution exactly when its determinant is zero:
$\begin{vmatrix} 1 & \sqrt{2}\sin\alpha & \sqrt{2}\cos\alpha \\ 1 & \cos\alpha & \sin\alpha \\ 1 & \sin\alpha & -\cos\alpha \end{vmatrix} = 0$.
Expanding along the first row:
$1\cdot(-\cos^2\alpha - \sin^2\alpha) - \sqrt{2}\sin\alpha(-\cos\alpha - \sin\alpha) + \sqrt{2}\cos\alpha(\sin\alpha - \cos\alpha)$
$= -1 + \sqrt{2}(2\sin\alpha\cos\alpha + \sin^2\alpha - \cos^2\alpha) = -1 + \sqrt{2}(\sin 2\alpha - \cos 2\alpha)$.
So $\sin 2\alpha - \cos 2\alpha = \frac{1}{\sqrt{2}}$, i.e. $\sqrt{2}\sin\left(2\alpha - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, $\sin\left(2\alpha - \frac{\pi}{4}\right) = \frac{1}{2}$.
For $\alpha \in (0, \frac{\pi}{2})$, $2\alpha - \frac{\pi}{4} \in \left(-\frac{\pi}{4}, \frac{3\pi}{4}\right)$, so $2\alpha - \frac{\pi}{4} = \frac{\pi}{6}$ (the other value $\frac{5\pi}{6}$ is out of range).
$2\alpha = \frac{5\pi}{12}$, $\alpha = \frac{5\pi}{24}$.
Answer: (C). | ||||
| 5 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3\times 3$ matrix $M$, $\det(\operatorname{adj} M) = (\det M)^2$.
Let $B = 2A - A^T$. Then $A - 2A^T = -(2A^T - A) = -B^T$, so $\det(A - 2A^T) = (-1)^3\det B = -\det B$.
Hence $\det(\operatorname{adj}B\cdot\operatorname{adj}(-B^T)) = (\det B)^2(\det B)^2 = (\det B)^4 = 2^8$, so $|\det B| = 4$.
$B = 2A - A^T = \begin{bmatrix} 2 & 4 & 2\alpha \\ 2 & 0 & 2 \\ 0 & 2 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 2\alpha \\ 0 & 0 & 1 \\ -\alpha & 1 & 2 \end{bmatrix}$.
$\det B = 1(0 - 1) - 3(0 + \alpha) + 2\alpha(0 - 0) = -1 - 3\alpha$.
$|1 + 3\alpha| = 4$ with $\alpha > 0$ gives $\alpha = 1$.
$\det A = 1(0 - 1) - 2(2 - 0) + \alpha(1 - 0) = \alpha - 5 = -4$, so $(\det A)^2 = 16$.
Answer: (B). | ||||
| 6 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Left limit: $\frac{1-\cos 2x}{x^2} = \frac{2\sin^2 x}{x^2} \to 2$. So $\alpha = 2$.
Right limit: $\frac{\beta\sqrt{1-\cos x}}{x} = \frac{\beta\sqrt{2}\,|\sin(x/2)|}{x} \to \beta\sqrt{2}\cdot\frac{1}{2} = \frac{\beta}{\sqrt{2}}$ (for $x > 0$, $\sin(x/2) > 0$).
Continuity: $\frac{\beta}{\sqrt{2}} = 2$, so $\beta = 2\sqrt{2}$ and $\beta^2 = 8$.
$\alpha^2 + \beta^2 = 4 + 8 = 12$.
Answer: (C). | ||||
| 7 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Any 3 of the 18 points form a triangle unless all three lie on the same side.
Total choices: $^{18}C_3 = 816$.
Collinear choices: $^5C_3 + {}^6C_3 + {}^7C_3 = 10 + 20 + 35 = 65$.
Triangles: $816 - 65 = 751$.
Answer: (D). | ||||
| 8 | 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 General term: $T_{r+1} = {}^{15}C_r\, 2^{\frac{15-r}{5}}\, 5^{\frac{r}{3}}$, $r = 0, 1, \ldots, 15$.
It is rational when $\frac{15-r}{5}$ and $\frac{r}{3}$ are both integers, i.e. $r$ is a multiple of 5 and of 3: $r = 0$ or $r = 15$.
$r = 0$: $2^3 = 8$. $r = 15$: $5^5 = 3125$.
Sum $= 8 + 3125 = 3133$.
Answer: (B). | ||||
| 9 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the G.P. have ratio $r$: $p = 2r$, $q = 2r^2$. Let the A.P. have first term $a$ and common difference $d$.
$a + 6d = 2$, $a + 7d = p$, $a + 12d = q$. So $p - 2 = d$ and $q - 2 = 6d = 6(p-2)$.
$2r^2 - 2 = 6(2r - 2)$ gives $r^2 - 6r + 5 = 0$, so $r = 1$ or $r = 5$. Since $q \ne 2$, $r = 5$.
Then $p = 10$, $d = 8$, $a = 2 - 48 = -46$.
5th term of the G.P.: $2\cdot 5^4 = 1250$.
$-46 + (n-1)8 = 1250$ gives $n - 1 = 162$, $n = 163$.
Answer: (B). | ||||
| 10 | 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 $f(|x|)$: for $x \ne 0$, $|x| \in (0, 2]$ so $f(|x|) = |x| - 2$; at $x = 0$ it is $-2 = |0| - 2$ as well. So $f(|x|) = |x| - 2$ on $[-2, 2]$.
$\int_{-2}^{2}(|x| - 2)\,dx = 2\int_0^2 (x - 2)\,dx = 2(2 - 4) = -4$.
$|f(x)|$: equal to $2$ on $[-2, 0]$ and $2 - x$ on $(0, 2]$.
$\int_{-2}^{2}|f(x)|\,dx = 2\cdot 2 + \int_0^2 (2 - x)\,dx = 4 + 2 = 6$.
$\int_{-2}^{2} h(x)\,dx = -4 + 6 = 2$.
Answer: (B). | ||||
| 11 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy 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 $g$ is the inverse of $f$ ($f$ is strictly increasing since $f'(x) = 5x^4 + \frac{1}{2}e^{x/4} > 0$).
$f(0) = 0 + 2 = 2$, so $g(2) = 0$.
Differentiating $g(f(x)) = x$: $g'(f(x))\,f'(x) = 1$. At $x = 0$: $g'(2) = \frac{1}{f'(0)} = \frac{1}{1/2} = 2$.
$8g'(2) = 16$.
Answer: (D). | ||||
| 12 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The denominator $2x^2 + 3x + 8$ has discriminant $9 - 64 < 0$, so it is always positive.
Let $y = f(x)$: $2x^2(1 - y) - 3x(1 + y) + 8(1 - y) = 0$. For real $x$ (with $y \ne 1$; $y = 1$ occurs at $x = 0$):
$9(1+y)^2 - 64(1-y)^2 \ge 0$, i.e. $3(1+y) \ge 8|1-y|$ (both sides non-negative since $y > 0$).
$y \le 1$: $3 + 3y \ge 8 - 8y$ gives $y \ge \frac{5}{11}$. $y \ge 1$: $3 + 3y \ge 8y - 8$ gives $y \le \frac{11}{5}$.
So the range is $\left[\frac{5}{11}, \frac{11}{5}\right]$: minimum $\frac{5}{11}$, maximum $\frac{11}{5}$.
Sum $= \frac{25 + 121}{55} = \frac{146}{55}$, $\gcd(146, 55) = 1$.
$m + n = 146 + 55 = 201$.
Answer: (C). | ||||
| 13 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersections: $x(1 + 3x - 2x^2) = 1$, i.e. $2x^3 - 3x^2 - x + 1 = 0 = (2x - 1)(x^2 - x - 1)$.
So $x = \frac{1}{2}$ or $x = \frac{1 \pm \sqrt{5}}{2}$. For $x < 0$ the hyperbola branch is unbounded near $x = 0$, so the bounded region lies between $x = \frac{1}{2}$ and $x = \phi = \frac{1+\sqrt{5}}{2}$, where the parabola is above $y = \frac{1}{x}$ (at $x = 1$: $2 > 1$).
Area $= \int_{1/2}^{\phi}\left(1 + 3x - 2x^2 - \frac{1}{x}\right)dx = \left[x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln x\right]_{1/2}^{\phi}$.
Using $\phi^2 = \phi + 1$ and $\phi^3 = 2\phi + 1$:
at $\phi$: $\phi + \frac{3(\phi+1)}{2} - \frac{2(2\phi+1)}{3} - \ln\phi = \frac{7\phi}{6} + \frac{5}{6} - \ln\phi = \frac{17 + 7\sqrt{5}}{12} - \ln\phi$;
at $\frac{1}{2}$: $\frac{1}{2} + \frac{3}{8} - \frac{1}{12} + \ln 2 = \frac{19}{24} + \ln 2$.
Area $= \frac{34 + 14\sqrt{5} - 19}{24} - \ln(2\phi) = \frac{1}{24}(14\sqrt{5} + 15) - \log_e(1 + \sqrt{5})$.
$l = 14$, $m = 15$, $n = 1$: $l + m + n = 30$.
Answer: (A). | ||||
| 14 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $x^4 + 2x^3 + 3x^2 + 2x + 2 = (x^2 + 1)(x^2 + 2x + 2)$.
$\frac{2x^2 + 2x + 3}{(x^2+1)(x^2+2x+2)} = \frac{1}{x^2+1} + \frac{1}{x^2+2x+2}$ (check: $(x^2 + 2x + 2) + (x^2 + 1) = 2x^2 + 2x + 3$).
So $dy = \left(\frac{1}{x^2+1} + \frac{1}{(x+1)^2+1}\right)dx$ and $y = \tan^{-1}x + \tan^{-1}(x+1) + C$.
$y(-1) = -\frac{\pi}{4} + 0 + C = -\frac{\pi}{4}$ gives $C = 0$.
$y(0) = 0 + \tan^{-1}1 = \frac{\pi}{4}$.
Answer: (A). | ||||
| 15 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Centre $C = (5, 3)$, radius $r = \sqrt{25 + 9 - 30} = 2$.
The vertices are $C + \vec{v}_i$, where the four $\vec{v}_i$ have length 2 and, the square being symmetric about its centre, add to zero.
$\Sigma|C + \vec{v}_i|^2 = 4|C|^2 + 2C\cdot\Sigma\vec{v}_i + \Sigma|\vec{v}_i|^2 = 4(25 + 9) + 0 + 4\cdot 4 = 136 + 16 = 152$.
Check: with sides parallel to $y = x + 3$ the diagonals are horizontal and vertical, so the vertices are $(7, 3), (3, 3), (5, 5), (5, 1)$: $58 + 18 + 50 + 26 = 152$.
Answer: (B). | ||||
| 16 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines Sides: $AB$: $x - y + 4 = 0$; $BC$: $3x + 5y - 4 = 0$; $CA$: $x + y - 2 = 0$.
Inward shift by 1 unit (towards the opposite vertex):
$AB$: $C$ gives $3 + 1 + 4 > 0$, so the new line is $x - y + 4 - \sqrt{2} = 0$; distance from O: $\frac{4 - \sqrt{2}}{\sqrt{2}} = 2\sqrt{2} - 1 \approx 1.83$.
$BC$: $A$ gives $-3 + 15 - 4 > 0$, so the new line is $3x + 5y - 4 - \sqrt{34} = 0$; distance from O: $\frac{4 + \sqrt{34}}{\sqrt{34}} \approx 1.69$.
$CA$: $B$ gives $-2 + 2 - 2 < 0$, so the new line is $x + y - 2 + \sqrt{2} = 0$, i.e. $x + y - (2 - \sqrt{2}) = 0$; distance from O: $\frac{2 - \sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1 \approx 0.41$.
The nearest is $x + y - (2 - \sqrt{2}) = 0$.
Answer: (A). | ||||
| 17 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy 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 $\vec{PQ} = (4, -2, 4)$, $|\vec{PQ}| = 6$, unit vector $\frac{1}{3}(2, -1, 2)$.
Points at distance 9 from P: $P \pm 9\cdot\frac{1}{3}(2, -1, 2) = (1, -2, 3) \pm (6, -3, 6)$, i.e. $(7, -5, 9)$ or $(-5, 1, -3)$.
Squared distances from O: $49 + 25 + 81 = 155$ and $25 + 1 + 9 = 35$. The farther point is $(7, -5, 9)$.
$\alpha^2 + \beta^2 + \gamma^2 = 155$.
Answer: (D). | ||||
| 18 | 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 Let $\vec{C} = a\hat{i} + b\hat{j} + c\hat{k}$ with $a^2 + b^2 + c^2 = 1$.
$\frac{2a + 2b - c}{3} = \cos 60^\circ = \frac{1}{2}$, so $2a + 2b - c = \frac{3}{2}$.
$\frac{a - c}{\sqrt{2}} = \cos 45^\circ = \frac{1}{\sqrt{2}}$, so $c = a - 1$.
Then $a + 2b + 1 = \frac{3}{2}$, $b = \frac{1}{4} - \frac{a}{2}$.
$a^2 + \left(\frac{1}{4} - \frac{a}{2}\right)^2 + (a-1)^2 = 1$ gives $\frac{9}{4}a^2 - \frac{9}{4}a + \frac{1}{16} = 0$, i.e. $36a^2 - 36a + 1 = 0$, $a = \frac{1}{2} \pm \frac{\sqrt{2}}{3}$.
Taking $a = \frac{1}{2} + \frac{\sqrt{2}}{3}$: $b = -\frac{\sqrt{2}}{6} = -\frac{1}{3\sqrt{2}}$, $c = -\frac{1}{2} + \frac{\sqrt{2}}{3}$.
$\vec{C} + \left(-\frac{1}{2}, \frac{1}{3\sqrt{2}}, -\frac{\sqrt{2}}{3}\right) = \left(\frac{\sqrt{2}}{3}, 0, -\frac{1}{2}\right) = \frac{\sqrt{2}}{3}\hat{i} - \frac{1}{2}\hat{k}$.
(The other root $a = \frac{1}{2} - \frac{\sqrt{2}}{3}$ gives $-\frac{\sqrt{2}}{3}\hat{i} + \frac{\sqrt{2}}{3}\hat{j} - \left(\frac{1}{2} + \frac{2\sqrt{2}}{3}\right)\hat{k}$, which differs from option (B) in the sign of the $\hat{k}$ term, so it matches no option.)
Answer: (C). | ||||
| 19 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each urn is chosen with probability $\frac{1}{3}$ and each holds 12 balls.
$P(\text{black} \mid A) = \frac{5}{12}$, $P(\text{black} \mid B) = \frac{7}{12}$, $P(\text{black} \mid C) = \frac{6}{12}$.
By Bayes' theorem, $P(A \mid \text{black}) = \frac{\frac{1}{3}\cdot\frac{5}{12}}{\frac{1}{3}\left(\frac{5}{12} + \frac{7}{12} + \frac{6}{12}\right)} = \frac{5}{18}$.
Answer: (C). | ||||
| 20 | StatisticsMean deviation | Class 11: Statistics | Single correctEasy | |
A easy single correct question on mean deviation, from Statistics. NCERT: Class 11: Statistics Mean: $\frac{-3 + 4 + 7 - 6 + \alpha + \beta}{6} = 2$, so $\alpha + \beta = 10$.
Variance: $\frac{\Sigma x_i^2}{6} - 2^2 = 23$, so $\Sigma x_i^2 = 162$; $9 + 16 + 49 + 36 = 110$, so $\alpha^2 + \beta^2 = 52$.
$2\alpha\beta = 100 - 52 = 48$, $\alpha\beta = 24$: $\{\alpha, \beta\} = \{4, 6\}$.
Deviations from 2 of $-3, 4, 7, -6, 4, 6$: $5, 2, 5, 8, 2, 4$, total $26$.
Mean deviation $= \frac{26}{6} = \frac{13}{3}$.
Answer: (D). | ||||
| 21 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Students studying at least one subject: $220 - 10 = 210$.
$n(M \cup P \cup C) = n(M) + n(P) + n(C) - 30 - 50 - 40 + x$, where $x$ is the number studying all three.
So $x = 330 - (n(M) + n(P) + n(C))$.
The sum $n(M) + n(P) + n(C)$ lies between $125 + 85 + 75 = 285$ and $130 + 95 + 90 = 315$, so $15 \le x \le 45$.
But $x$ cannot exceed any pairwise intersection, and $n(P \cap C) = 30$, so $x \le 30$.
$x = 15$ (all three at their maxima) and $x = 30$ (sum 300, e.g. $M = 130, P = 90, C = 80$) both leave every region of the Venn diagram non-negative, so $m = 15$ and $n = 30$.
$m + n = 45$.
Answer: 45. | ||||
| 22 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The condition says every row of A has non-negative entries adding up to 3.
For a row $(a, b, c)$ with $a, b, c \ge 0$ and $a + b + c = 3$, its length is $\sqrt{a^2 + b^2 + c^2} \le a + b + c = 3$.
$|\det A|$ is the volume of the box spanned by the three rows, which is at most the product of their lengths (Hadamard's inequality): $|\det A| \le 3\cdot 3\cdot 3 = 27$.
$A = 3I$ satisfies the condition and has $\det A = 27$, so the maximum is 27.
Answer: 27. | ||||
| 23 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalHard | |
A hard numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $b = 1 + \sum_{n\ge 1}\frac{2^n}{n!} = \sum_{n\ge 0}\frac{2^n}{n!} = e^2$.
For $a$: $\frac{^nC_2}{(n+1)!} = \frac{n(n-1)}{2(n+1)!}$. Write $n(n-1) = (n+1)n - 2(n+1) + 2$, so
$\frac{n(n-1)}{(n+1)!} = \frac{1}{(n-1)!} - \frac{2}{n!} + \frac{2}{(n+1)!}$.
Summing over $n \ge 2$: $\sum\frac{1}{(n-1)!} = e - 1$, $\sum\frac{2}{n!} = 2(e - 2)$, $\sum\frac{2}{(n+1)!} = 2\left(e - \frac{5}{2}\right)$.
Total $= e - 1 - 2e + 4 + 2e - 5 = e - 2$, so $a = 1 + \frac{e - 2}{2} = \frac{e}{2}$.
$\frac{2b}{a^2} = \frac{2e^2}{e^2/4} = 8$.
Answer: 8. | ||||
| 24 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = 1$ both numerator ($6^{1/3} - 6^{1/3}$) and denominator ($\sqrt{5} - \sqrt{5}$) vanish, so use L'Hôpital's rule.
Numerator derivative at 1: $\frac{5}{3}(6)^{-2/3} - \frac{1}{3}(6)^{-2/3} = \frac{4}{3}\cdot 6^{-2/3}$.
Denominator derivative at 1: $\frac{2}{2}\cdot 5^{-1/2} - \frac{1}{2}\cdot 5^{-1/2} = \frac{1}{2\sqrt{5}}$.
Limit $= \frac{4}{3\cdot 6^{2/3}}\cdot 2\sqrt{5} = \frac{8\sqrt{5}}{3(6)^{2/3}} = \frac{8\sqrt{5}}{3(2\cdot 3)^{2/3}}$.
So $m = 8$, $n = 3$ (coprime), and $8m + 12n = 64 + 36 = 100$.
Answer: 100. | ||||
| 25 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Divide numerator and denominator by $\cos^2 x$: the integrand becomes $\frac{\tan^2 x}{\sec^2 x + \tan x}$. Put $t = \tan x$, $dx = \frac{dt}{1+t^2}$:
$I = \int_0^1\frac{t^2}{(t^2 + t + 1)(t^2 + 1)}dt$.
Partial fractions: $\frac{t^2}{(t^2+t+1)(t^2+1)} = \frac{t}{t^2+1} - \frac{t}{t^2+t+1}$ (check: $t(t^2+t+1) - t(t^2+1) = t^2$).
$\int_0^1\frac{t}{t^2+1}dt = \frac{1}{2}\ln 2$.
$\int_0^1\frac{t}{t^2+t+1}dt = \frac{1}{2}\int_0^1\frac{2t+1}{t^2+t+1}dt - \frac{1}{2}\int_0^1\frac{dt}{(t+\frac{1}{2})^2 + \frac{3}{4}} = \frac{1}{2}\ln 3 - \frac{1}{\sqrt{3}}\left[\tan^{-1}\frac{2t+1}{\sqrt{3}}\right]_0^1 = \frac{1}{2}\ln 3 - \frac{1}{\sqrt{3}}\left(\frac{\pi}{3} - \frac{\pi}{6}\right)$.
$I = \frac{1}{2}\ln 2 - \frac{1}{2}\ln 3 + \frac{\pi}{6\sqrt{3}} = \frac{1}{2}\log_e\left(\frac{2}{3}\right) + \frac{\pi}{6\sqrt{3}}$.
So $a = 2$, $b = 6$, $a + b = 8$.
Answer: 8. | ||||
| 26 | 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 $e^{-x}$: $\frac{d}{dx}(ye^{-x}) = e^{-x}(1 + 4\sin x)$.
$\int e^{-x}dx = -e^{-x}$ and $\int e^{-x}\sin x\,dx = -\frac{e^{-x}}{2}(\sin x + \cos x)$.
$ye^{-x} = -e^{-x} - 2e^{-x}(\sin x + \cos x) + C$, so $y = -1 - 2\sin x - 2\cos x + Ce^{x}$.
$y(\pi) = -1 - 0 + 2 + Ce^{\pi} = 1$ gives $C = 0$.
$y\left(\frac{\pi}{2}\right) = -1 - 2 - 0 = -3$, and $y\left(\frac{\pi}{2}\right) + 10 = 7$.
Answer: 7. | ||||
| 27 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalHard | |
A hard numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants By the Cayley–Hamilton theorem a $2\times 2$ matrix satisfies $A^2 - (\operatorname{tr}A)A + |A|I = O$, here $A^2 + 3A + 2I = O$.
Subtracting from $A^2 + xA + yI = O$: $(x - 3)A + (y - 2)I = O$.
If $x \ne 3$, A would be a scalar matrix $kI$; then $\operatorname{tr}A = 2k = -3$ and $|A| = k^2 = \frac{9}{4} \ne 2$, a contradiction. So $x = 3$, and then $y = 2$.
The equation is therefore satisfied by the single point $(3, 2)$. Infinitely many hyperbolas with transverse axis parallel to the x-axis pass through one point, so the data do not fix $e$ or $l$, and $e^4 + l^4$ cannot be determined from the question as printed.
Answer: not determined (question defective). | ||||
| 28 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections For $y^2 = 4ax$ with $a = 3$, a focal chord making angle $\theta$ with the axis has length $4a\csc^2\theta = \frac{12}{\sin^2\theta}$.
$\frac{12}{\sin^2\theta} = 15$ gives $\sin^2\theta = \frac{4}{5}$, $\cos^2\theta = \frac{1}{5}$, so the slope satisfies $\tan^2\theta = 4$, $m = \pm 2$.
The chord through the focus $(3, 0)$: $y = \pm 2(x - 3)$, i.e. $2x \mp y - 6 = 0$.
Distance from the origin: $p = \frac{6}{\sqrt{5}}$, $p^2 = \frac{36}{5}$.
$10p^2 = 72$.
Answer: 72. | ||||
| 29 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Points and directions: $\vec{a}_1 = (-2, -3, 5)$, $\vec{b}_1 = (2, 3, 4)$; $\vec{a}_2 = (3, 2, -4)$, $\vec{b}_2 = (1, -3, 2)$.
$\vec{b}_1\times\vec{b}_2 = (3\cdot 2 - 4(-3),\ 4\cdot 1 - 2\cdot 2,\ 2(-3) - 3\cdot 1) = (18, 0, -9)$, of length $9\sqrt{5}$.
$\vec{a}_2 - \vec{a}_1 = (5, 5, -9)$; $(\vec{a}_2 - \vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2) = 90 + 0 + 81 = 171$.
Shortest distance $= \frac{171}{9\sqrt{5}} = \frac{19}{\sqrt{5}} = \frac{38}{3\sqrt{5}}k$, so $k = \frac{3}{2}$.
On $[0, \frac{3}{2}]$: $[x^2] = 0$ on $[0, 1)$, $1$ on $[1, \sqrt{2})$, $2$ on $[\sqrt{2}, \frac{3}{2}]$ (since $\frac{3}{2} < \sqrt{3}$).
$\int_0^{3/2}[x^2]\,dx = (\sqrt{2} - 1) + 2\left(\frac{3}{2} - \sqrt{2}\right) = 2 - \sqrt{2}$.
So $\alpha - \sqrt{\alpha} = 2 - \sqrt{2}$, giving $\alpha = 2$, and $6\alpha^3 = 48$.
Answer: 48. | ||||
| 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 Area $= \frac{1}{2}|\vec{AB}\times\vec{AC}| = 15\sqrt{2}$, so $|\vec{AB}\times\vec{AC}|^2 = 1800$.
$\vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & -7\\ 6 & d & -2\end{vmatrix} = (7d - 4)\hat{i} - 40\hat{j} + (d - 12)\hat{k}$.
$(7d - 4)^2 + 1600 + (d - 12)^2 = 1800$ gives $50d^2 - 80d - 40 = 0$, i.e. $5d^2 - 8d - 4 = 0$, so $d = 2$ (as $d > 0$).
$\vec{BC} = \vec{AC} - \vec{AB} = 5\hat{i} + 0\hat{j} + 5\hat{k}$.
$|\vec{AB}|^2 = 1 + 4 + 49 = 54$, $|\vec{AC}|^2 = 36 + 4 + 4 = 44$, $|\vec{BC}|^2 = 25 + 25 = 50$.
The largest side squared is 54.
Answer: 54. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The argument of a sine or cosine must be dimensionless.
$\frac{2\pi nt}{\lambda}$ dimensionless: $nt$ has the dimension of $\lambda$, [L] — (A) is correct.
$\frac{2\pi x}{\lambda}$ dimensionless: $x$ is [L] — (B) is correct.
$n = \frac{[\mathrm{L}]}{[\mathrm{T}]} = [\mathrm{LT^{-1}}]$ — (C) is correct (n is a speed here).
$\frac{n}{\lambda} = \frac{[\mathrm{LT^{-1}}]}{[\mathrm{L}]} = [\mathrm{T^{-1}}]$, not [T] — (D) is the incorrect statement.
Answer: (D). | ||||
| 32 | 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 $v_x = \frac{dx}{dt} = 4$, $v_y = \frac{dy}{dt} = 3 + 16t$; $a_x = 0$, $a_y = 16$.
The acceleration $16\hat{j}$ is constant, so the motion is uniformly accelerated.
The initial velocity $(4, 3)$ is not along the acceleration, so the path bends: eliminating $t = \frac{x-2}{4}$ gives $y = \frac{3(x-2)}{4} + \frac{(x-2)^2}{2}$, a parabola.
Answer: (C). | ||||
| 33 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Let $F = kt$. While $F$ is less than the limiting static friction $\mu_s mg$, the block does not move, so $a = 0$.
After that, $a = \frac{kt - \mu_k mg}{m}$, which rises with $t$.
So the acceleration stays at zero for a while and only then grows. Of the four curves, only (A) shows a staying (nearly) zero at first and then rising; (B) and (C) have the block accelerating from the start, which ignores friction, and (D) is a constant acceleration, which needs a constant force.
Answer: (A). | ||||
| 34 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Distance in the $n$th second (uniform acceleration): $s_n = u + \frac{a}{2}(2n - 1)$.
$s_{n+2} - s_n = \frac{a}{2}\left[(2n + 3) - (2n - 1)\right] = 2a$.
$2a = 115.0 - 102.5 = 12.5$, so $a = 6.25$ m/s$^2$.
Answer: (B). | ||||
| 35 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Initial energy (at rest at height h): $mgh$. Energy at the top of the rebound: $mg\frac{h}{2}$.
Loss $= \frac{mgh - mgh/2}{mgh}\times 100\% = 50\%$.
Speed just before striking the ground: $\frac{1}{2}mv^2 = mgh$, so $v = \sqrt{2gh}$.
Answer: (B). | ||||
| 36 | GravitationNewton's law of gravitation | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation Radius of the arc: $\pi R = L$, so $R = \frac{L}{\pi}$; mass per length $\lambda = \frac{M}{L}$.
An element at angle $\theta$ has mass $\lambda R\,d\theta$ and pulls with $dF = \frac{Gm\lambda R\,d\theta}{R^2}$. Components along the arc's chord cancel; along the symmetry axis:
$F = \int_0^{\pi}\frac{Gm\lambda}{R}\sin\theta\,d\theta = \frac{2Gm\lambda}{R} = \frac{2GmM}{L}\cdot\frac{\pi}{L} = \frac{2GmM\pi}{L^2}$.
Answer: (C). | ||||
| 37 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | StatementsMedium | |
A medium statements question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Statement I: with the liquid at rest, Bernoulli's equation reduces to $P_1 + \rho g h_1 = P_2 + \rho g h_2$, i.e. $P_1 - P_2 = \rho g(h_2 - h_1)$ (h measured as height). Correct.
Statement II: in the venturi tube (horizontal), $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$, and the manometer gives $P_1 - P_2 = \rho g h$. Hence $2gh = v_2^2 - v_1^2$. At the narrow section $A_2 < A_1$, so $v_2 > v_1$ and $v_1^2 - v_2^2$ would be negative; the statement $2gh = v_1^2 - v_2^2$ is wrong.
Answer: (C). | ||||
| 38 | ThermodynamicsZeroth law and temperature | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on zeroth law and temperature, from Thermodynamics. NCERT: Class 11: Thermodynamics A change of temperature converts as $\Delta F = \frac{9}{5}\Delta C$ (the 32 offset cancels in a difference).
$\Delta F = \frac{9}{5}\times 40 = 72^\circ$F.
Answer: (A). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory For an ideal gas $P = \frac{\rho RT}{M}$, so at fixed density the P–T graph is a straight line of slope $\frac{\rho R}{M}$.
The steeper the line, the larger the density. The line marked $\rho_1$ is steepest and $\rho_3$ least steep, so $\rho_1 > \rho_2 > \rho_3$.
Of the options only $\rho_1 > \rho_2$ is true.
Answer: (D). | ||||
| 40 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Field of the line charge at distance r: $E = \frac{\lambda}{2\pi\varepsilon_0 r}$, directed outward, so the electron is pulled towards the wire.
This force provides the centripetal force: $\frac{mv^2}{r} = \frac{e\lambda}{2\pi\varepsilon_0 r}$.
$KE = \frac{1}{2}mv^2 = \frac{e\lambda}{4\pi\varepsilon_0}$, independent of r.
The graph is a horizontal line.
Answer: (D). | ||||
| 41 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Taking the resistance linear in temperature: $R_t = R_0 + \frac{R_{100} - R_0}{100}t$.
$R_{400} = 8 + \frac{10 - 8}{100}\times 400 = 8 + 8 = 16\ \Omega$.
Answer: (C). | ||||
| 42 | 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 Inside a long solenoid the magnetic field is uniform and along the axis.
The electron moves along the axis, so $\vec{v}\parallel\vec{B}$ and $\vec{F} = -e\,\vec{v}\times\vec{B} = 0$.
With no force it keeps moving with uniform velocity along the axis.
Answer: (D). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Current zero when voltage is maximum means current and voltage differ in phase by $90^\circ$.
A pure inductor (current lags by $90^\circ$) and a pure capacitor (current leads by $90^\circ$) do this. A combination of L and C alone has a purely reactive impedance, so the phase difference is also $90^\circ$ (lead or lag).
A pure resistor keeps current in phase with voltage, so C is excluded.
Answer: (C) A, B and D only. | ||||
| 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 The phase $\omega(t - z/c)$ shows the wave travels along $+z$.
$B_0 = \frac{E_0}{c} = \frac{40}{c}$, in phase with $\vec{E}$.
$\vec{B}$ must be perpendicular to both $\vec{E}$ ($\hat{i}$) and the direction of travel ($\hat{k}$), with $\vec{E}\times\vec{B}$ along $+\hat{k}$: $\hat{i}\times\hat{j} = \hat{k}$, so $\vec{B}$ is along $\hat{j}$.
$\vec{B} = \hat{j}\,\frac{40}{c}\cos\omega\left(t - \frac{z}{c}\right)$.
Answer: (C). | ||||
| 45 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For thin lenses in contact the powers add: $5P = 25$ D, so $P = 5$ D.
$f = \frac{1}{P} = \frac{1}{5}$ m $= 20$ cm.
Answer: (A). | ||||
| 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 The stopping potential depends only on the frequency (wavelength) of the light, not on its intensity: both curves must meet the V-axis at the same $-V_0$.
The saturation current is proportional to the intensity: with $I_2 > I_1$, the $I_2$ curve saturates higher.
Only graph (B) shows a common stopping potential with the $I_2$ curve above the $I_1$ curve.
Answer: (B). | ||||
| 47 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Reactants: ${}^1_0\mathrm{n} + {}^{235}_{92}\mathrm{U}$ give mass number $236$ and charge number $92$; both must be conserved.
(A) $153 + 99 + 3 = 255$: mass number not conserved.
(B) $144 + 89 + 3 = 236$ and $56 + 36 = 92$: both conserved.
(C) $140 + 94 + 3 = 237$ and $56 + 38 = 94$: neither conserved.
(D) $144 + 89 + 4 = 237$: mass number not conserved.
Answer: (B). | ||||
| 48 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The left terminal is at $-6$ V and the right at $-8$ V, so the left side is at the higher potential.
The diode in the 10 $\Omega$ branch points from left to right: it is forward biased and (taken as ideal) conducts.
The diode in the 5 $\Omega$ branch points from right to left: it is reverse biased and carries no current.
Net resistance $= 15\parallel 10 = \frac{15\times 10}{25} = 6\ \Omega$.
Answer: (D). | ||||
| 49 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Differentiating: $-\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2}$. For the maximum error the magnitudes add:
$\frac{\Delta f}{f^2} = \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}$, so $\Delta f = f^2\left[\frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}\right]$.
Answer: (D). | ||||
| 50 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity With external resistance R, the terminal voltage $V = \frac{ER}{R + r}$ is balanced against length $l$, so $V \propto l$.
$\frac{500}{400} = \frac{10/(10 + r)}{1/(1 + r)} = \frac{10(1 + r)}{10 + r}$.
$1.25(10 + r) = 10 + 10r$ gives $2.5 = 8.75r$, $r = \frac{2}{7} \approx 0.29\ \Omega \approx 0.3\ \Omega$.
Answer: (C). | ||||
| 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 Let the forces be $F$ and $3F$, with angle $\theta$; the resultant is $3F$.
$(3F)^2 = F^2 + (3F)^2 + 2\cdot F\cdot 3F\cos\theta$, so $F^2 + 6F^2\cos\theta = 0$, $\cos\theta = -\frac{1}{6}$.
So $\frac{1}{n} = -\frac{1}{6}$, $n = -6$, and $|n| = 6$.
Answer: 6. | ||||
| 52 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rolling without slipping, the kinetic energy is $\frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$, all of which turns into $mgh$:
$h = \frac{v^2}{2g}\left(1 + \frac{k^2}{R^2}\right)$.
Solid sphere: $\frac{k^2}{R^2} = \frac{2}{5}$, factor $\frac{7}{5}$. Hollow cylinder (thin-walled): $\frac{k^2}{R^2} = 1$, factor 2.
$\frac{h_1}{h_2} = \frac{7/5}{2} = \frac{7}{10}$, so $n = 7$.
Answer: 7. | ||||
| 53 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical 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 A soap bubble has two surfaces, so the work done is $W = T\cdot 2\cdot 4\pi(R_2^2 - R_1^2)$.
$R_1 = 3.5$ cm. $36960 = 40\times 8\times\frac{22}{7}(R_2^2 - 12.25)$.
$R_2^2 - 12.25 = \frac{36960\times 7}{320\times 22} = 36.75$, so $R_2^2 = 49$ and $R_2 = 7$ cm.
Answer: 7. | ||||
| 54 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Hooke's law: length $= L_0 + \frac{T}{k}$.
$a = L_0 + \frac{3}{k}$, $b = L_0 + \frac{2}{k}$, so $\frac{1}{k} = a - b$ and $L_0 = a - 3(a - b) = 3b - 2a$.
For length $3a - 2b$: $\frac{T}{k} = (3a - 2b) - (3b - 2a) = 5(a - b)$, so $T = 5$ N.
Answer: 5. | ||||
| 55 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Sheet: $E_s = \frac{\sigma_s}{2\varepsilon_0}$ at any distance.
Line charge: it lies in the plane $z = 4$ parallel to the y-axis, taken through $x = 0$ (the line $x = 0$, $z = 4$). Its distance from $(0, 0, 2)$ is 2 m, so $E_l = \frac{\lambda_e}{2\pi\varepsilon_0\cdot 2} = \frac{\lambda_e}{4\pi\varepsilon_0}$.
$\frac{E_s}{E_l} = \frac{\sigma_s/(2\varepsilon_0)}{\lambda_e/(4\pi\varepsilon_0)} = \frac{2\pi\sigma_s}{\lambda_e} = 2\pi\cdot 2 = 4\pi = \pi\sqrt{16}$.
So $n = 16$.
Answer: 16. | ||||
| 56 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalHard | |
A hard numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity From the figure the top face is a–b–c–d (a and c opposite corners), the bottom face is h–e–f–g, and the vertical edges are a–h, b–e, c–f, d–g. The battery (taken as ideal) holds $V_a = 6$ V, $V_c = 0$.
Symmetry: the plane through b, d, e, g is the perpendicular bisector of a and c, and swapping a with c (and h with f) reverses all potentials about 3 V. So $V_b = V_d = V_e = V_g = 3$ V and $V_h + V_f = 6$ V.
At node h (neighbours a, e, g, equal resistors): $V_h = \frac{V_a + V_e + V_g}{3} = \frac{6 + 3 + 3}{3} = 4$ V, so $V_f = 2$ V.
Check at e (neighbours b, h, f): $\frac{3 + 4 + 2}{3} = 3$ V, consistent.
$V_e - V_f = 3 - 2 = 1$ V.
Answer: 1. | ||||
| 57 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical 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 From the figure the loop spans $x = 2$ m to $2.5$ m and $y = 2$ m to $2.5$ m, with the current anticlockwise (right along the bottom edge, up the right edge, left along the top, down the left edge).
The field depends only on x, so the forces on the top and bottom edges (equal fields, opposite currents) cancel.
Left edge ($x = 2$): $B = 0.2(5) = 1.0$ T; current along $-\hat{j}$: $\vec{F} = IL(-\hat{j})\times B\hat{k} = -0.5\times 0.5\times 1.0\,\hat{i} = -0.25\hat{i}$ N.
Right edge ($x = 2.5$): $B = 0.2(6) = 1.2$ T; current along $+\hat{j}$: $\vec{F} = +0.5\times 0.5\times 1.2\,\hat{i} = 0.30\hat{i}$ N.
Net force $= 0.05$ N $= 50$ mN.
Answer: 50. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $i^2 = 36 + 12\sqrt{56}\sin(\ldots) + 56\sin^2(\ldots)$. Over a cycle the mean of $\sin$ is 0 and of $\sin^2$ is $\frac{1}{2}$.
$\langle i^2\rangle = 36 + \frac{56}{2} = 64$, so $i_{\text{rms}} = 8$ A.
Answer: 8. | ||||
| 59 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Bright fringes coincide where $n_1\lambda_1 = n_2\lambda_2$: $450n_1 = 650n_2$, i.e. $9n_1 = 13n_2$.
The smallest positive solution is $n_1 = 13$, $n_2 = 9$.
So the 9th order fringe of $\lambda_2$ is the first to overlap one of $\lambda_1$: $n = 9$.
Answer: 9. | ||||
| 60 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalHard | |
A hard numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Energy released: $E = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6\times\frac{5}{36} \approx 1.89$ eV.
The photon carries momentum $p = \frac{E}{c}$, and the atom recoils with the same momentum and kinetic energy $K = \frac{p^2}{2M} = \frac{E^2}{2Mc^2}$.
The photon's energy is lower by K, so $\frac{\Delta\lambda}{\lambda} = \frac{K}{E} = \frac{E}{2Mc^2}$.
$Mc^2 = 1.6\times 10^{-27}\times 9\times 10^{16} = 1.44\times 10^{-10}$ J $= 9\times 10^{8}$ eV.
$\frac{\Delta\lambda}{\lambda} = \frac{1.89}{1.8\times 10^{9}} \approx 1.05\times 10^{-9}$, i.e. a percentage change of about $1\times 10^{-7}$ %.
So $n = 7$.
Answer: 7. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | SolutionsConcentration terms | Class 12: Solutions | Single correctEasy | |
A easy single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Molar mass of NaCl $= 23 + 35.5 = 58.5$ g mol$^{-1}$, so $5.85$ g is $0.1$ mol.
Taking the volume of the solution as 500 mL $= 0.5$ L: $M = \frac{0.1}{0.5} = 0.2$ M.
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 CH$_4$ (tetrahedral) and PF$_5$ (trigonal bipyramidal) are symmetric: zero dipole moment.
NH$_3$ and NF$_3$ are both pyramidal with a lone pair on N. In NH$_3$ the N–H bond moments point towards N, the same way as the lone-pair moment, so they add ($\mu \approx 1.47$ D). In NF$_3$ the N–F bond moments point away from N and partly cancel the lone-pair moment ($\mu \approx 0.23$ D).
Maximum: NH$_3$.
Answer: (B). | ||||
| 63 | EquilibriumIonic equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium The weaker the acid, the stronger its conjugate base. Acid strengths: HCl > CH$_3$COOH > H$_2$O > ROH (alcohols are slightly weaker acids than water, because the alkyl group pushes electron density onto the oxygen).
So the basic strength runs the other way: RO$^-$ > OH$^-$ > CH$_3$COO$^-$ > Cl$^-$.
Answer: (B). | ||||
| 64 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry The calomel electrode is mercury in contact with sparingly soluble mercury(I) chloride (Hg$_2$Cl$_2$, calomel) in a solution of chloride ions: Hg | Hg$_2$Cl$_2$(s) | Cl$^-$(aq).
Its reaction is Hg$_2$Cl$_2$ + 2e$^-$ ⇌ 2Hg + 2Cl$^-$, so it is a metal – insoluble salt – anion electrode.
Answer: (C). | ||||
| 65 | ElectrochemistryNernst equation | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Electrode reaction: 2H$^+$ + 2e$^-$ → H$_2$, $E^\circ = 0$.
$E = 0 - \frac{0.059}{2}\log\frac{p_{\mathrm{H_2}}}{[\mathrm{H^+}]^2}$.
$E = 0$ requires $p_{\mathrm{H_2}} = [\mathrm{H^+}]^2$. In pure water at $25^\circ$C, $[\mathrm{H^+}] = 10^{-7}$ M, so $p_{\mathrm{H_2}} = 10^{-14}$ bar.
Answer: (B). | ||||
| 66 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across period 2 the first ionization enthalpy generally rises, with two exceptions:
B (2p$^1$) is below Be (2s$^2$), since the 2p electron is easier to remove than a 2s electron;
O (2p$^4$) is below N (half-filled 2p$^3$), since one 2p electron of O is paired.
Values (kJ mol$^{-1}$): B 801 < Be 899 < O 1314 < N 1402 < F 1681.
That is E < C < A < B < D.
Answer: (C). | ||||
| 67 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Group valency here means the valency equal to the number of valence electrons (3 for group 13, 4 for group 14, 5 for group 15, 6 for group 16, 7 for group 17).
B (3): BCl$_3$ — yes. Al (3): AlCl$_3$ — yes. C (4): CH$_4$ — yes. Si (4): SiCl$_4$ — yes. P (5): PCl$_5$ — yes. S (6): SF$_6$ — yes.
N (5): nitrogen has no d orbitals and its covalency cannot exceed 4, so it cannot show valency 5 — no.
O (6): oxygen's covalency stays at 2 (at most 3 or 4 in ions) — no.
F (7): fluorine is always univalent — no.
So 3 elements (N, O, F) cannot.
Answer: (B). | ||||
| 68 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Count σ bonds plus lone pairs (or a single unpaired electron) on the central atom:
NO$_3^-$: 3 σ bonds, no lone pair — sp$^2$.
BCl$_3$: 3 σ bonds, no lone pair — sp$^2$.
ClO$_2^-$: 2 σ bonds + 2 lone pairs on Cl — sp$^3$ (bent).
ClO$_3$ (neutral radical): 3 σ bonds + 1 unpaired electron on Cl — sp$^3$ (pyramidal).
Two species (ClO$_2^-$ and ClO$_3$) are sp$^3$.
Answer: (B). | ||||
| 69 | 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 Scandium ([Ar]3d$^1$4s$^2$) loses all three valence electrons to reach the noble-gas configuration and shows only +3.
Titanium shows +2, +3, +4; cobalt +2, +3 (and others); nickel +2 (and +3, +4 in some compounds).
Answer: (B). | ||||
| 70 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Spectrochemical series (increasing field strength): I$^-$ < Br$^-$ < S$^{2-}$ < SCN$^-$ < Cl$^-$ < F$^-$ < OH$^-$ < C$_2$O$_4^{2-}$ < H$_2$O < NCS$^-$ < EDTA$^{4-}$ < NH$_3$ < en < CN$^-$ < CO.
(A) CO > H$_2$O > F$^-$ > S$^{2-}$ follows the series.
(B) puts NCS$^-$ above CN$^-$ and CO — reversed. (C) puts OH$^-$ and F$^-$ above NH$_3$ and CN$^-$ — reversed. (D) puts S$^{2-}$ at the top — reversed.
Answer: (A). | ||||
| 71 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds H$_2$O is a weak-field ligand, so these octahedral complexes are high spin.
V$^{3+}$: d$^2$ — 2 unpaired (even).
Cr$^{2+}$: d$^4$, t$_{2g}^3$e$_g^1$ — 4 unpaired (even).
Fe$^{3+}$: d$^5$ — 5 unpaired (odd).
Ni$^{3+}$: d$^7$, t$_{2g}^5$e$_g^2$ — 3 unpaired (odd).
Cu$^{2+}$: d$^9$ — 1 unpaired (odd).
Two complexes have an even number of unpaired d electrons.
Answer: (B). | ||||
| 72 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter NH$_4$OH is a weak base: NH$_4$OH ⇌ NH$_4^+$ + OH$^-$. Adding NH$_4$Cl supplies the common ion NH$_4^+$ and suppresses this ionisation, lowering [OH$^-$].
The lowered [OH$^-$] is enough to exceed the very small solubility products of Fe(OH)$_3$, Al(OH)$_3$ and Cr(OH)$_3$, but not those of the hydroxides of later groups (Zn, Mn, Mg …), which stay in solution.
The purpose is to decrease the concentration of OH$^-$ ions (raising [NH$_4^+$] is the means, not the aim).
Answer: (A). | ||||
| 73 | 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 In Lassaigne's test nitrogen is detected as NaCN formed on fusion with sodium (then as Prussian blue). Forming CN$^-$ needs both carbon and nitrogen in the compound.
Urea (NH$_2$CONH$_2$), glycine (H$_2$NCH$_2$COOH) and phenylhydrazine (C$_6$H$_5$NHNH$_2$) contain carbon and give the test.
Hydrazine (NH$_2$NH$_2$) has no carbon, so no NaCN forms and the test fails.
Answer: (C). | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Matching listMedium | |
A medium matching list question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) The lone pair on the nitrogen of aniline is pushed into the ring (resonance structure with $^+$NH$_2$ and a negative charge in the ring): electrons released into the conjugated system, the +R effect → (IV).
(B) The π electrons of the C=C bond move towards the attacking H$^+$, which bonds to that carbon: transfer towards the reagent, the +E effect → (III).
(C) The π electrons move away from the carbon attacked by $^-$CN (the negative charge ends up on the other carbon): the −E effect → (I).
(D) The nitro group draws π electrons of the ring onto itself (resonance structure with a positive charge in the ring): the −R effect → (II).
(A) - (IV), (B) - (III), (C) - (I), (D) - (II).
Answer: (A). | ||||
| 75 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) True: the electromeric effect is the complete shift of a π pair at the demand of an attacking reagent, so it appears only when the reagent is present.
(B) True: it disappears when the reagent is removed — a temporary effect.
(C) True: when the two act in opposite directions, the electromeric effect predominates over the inductive effect.
(D) False: the effect is shown by the substrate, not by the reagent; and when H$^+$ attacks, the π electrons move towards the atom it bonds to, which is the positive (+E) electromeric effect.
Answer: (D). | ||||
| 76 | 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 (A) is 1-bromo-4-(2-bromoethyl)benzene. Alcoholic NaOH brings about dehydrohalogenation of the alkyl bromide side chain (the aryl C–Br bond is untouched): (B) is 4-bromostyrene, Br–C$_6$H$_4$–CH=CH$_2$.
HBr in ether (no peroxide) adds by Markovnikov's rule, H going to the terminal CH$_2$ and Br to the benzylic carbon (the more stable benzylic carbocation): (C) is 1-bromo-4-(1-bromoethyl)benzene, Br–C$_6$H$_4$–CHBr–CH$_3$.
(A) and (C) have the same formula C$_8$H$_8$Br$_2$ and the same carbon skeleton, and differ only in the position of the side-chain Br: position isomers.
Answer: (B). | ||||
| 77 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes To move the Br from C-1 to C-2, first eliminate HBr and then add HBr back by Markovnikov's rule.
X: concentrated alcoholic NaOH on heating dehydrohalogenates 1-bromopropane to propene (dilute aqueous NaOH would substitute, giving propan-1-ol).
Y: HBr (in acetic acid, no peroxide) adds to propene by Markovnikov's rule to give 2-bromopropane. Br$_2$/CHCl$_3$ would give 1,2-dibromopropane instead.
Answer: (B). | ||||
| 78 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | StatementsEasy | |
A easy statements question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Statement I: the base removes an acidic α-hydrogen to form the enolate, which adds to the carbonyl group of a second molecule. Correct.
Statement II: benzaldehyde has no α-hydrogen but can accept the enolate of ethanal; this cross-aldol (Claisen–Schmidt) condensation gives 3-hydroxy-3-phenylpropanal and then cinnamaldehyde. So a cross-aldol product does form. Incorrect.
Answer: (C). | ||||
| 79 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The starting compound is 3-acetylbenzaldehyde (a CHO group and, meta to it, a COCH$_3$ group).
Zn–Hg with concentrated HCl (Clemmensen reduction) reduces every aldehyde or ketone C=O group to CH$_2$: CHO → CH$_3$ and COCH$_3$ → CH$_2$CH$_3$. It does not stop at the alcohols.
Product: 1-ethyl-3-methylbenzene.
Answer: (A). | ||||
| 80 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules D-glucose in the Fischer projection has the OH groups on C-2, C-3, C-4, C-5 on the right, left, right, right.
L-glucose is its mirror image, so every OH changes side: left, right, left, left.
Option (C) shows exactly this (HO on the left at C-2, OH on the right at C-3, HO on the left at C-4 and C-5). The OH on C-5, the lowest stereocentre, on the left is what makes it an L-sugar.
Answer: (C). | ||||
| 81 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Bohr's condition $mvr = \frac{nh}{2\pi}$ gives $\lambda = \frac{h}{mv} = \frac{2\pi r_n}{n}$.
For hydrogen $r_n = n^2 a_0$, so $\lambda = 2\pi n a_0$.
$n = 4$: $\lambda = 8\pi a_0$.
Answer: 8. | ||||
| 82 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure From molecular orbital theory:
O$_2$ (16 electrons): $\pi^*2p_x^1\pi^*2p_y^1$ — 2 unpaired.
O$_2^-$ (17): $\pi^*$ holds 3 electrons — 1 unpaired.
NO (15): one electron in $\pi^*$ — 1 unpaired.
CN$^-$ (14, isoelectronic with N$_2$): 0 unpaired.
O$_2^{2-}$ (18): $\pi^*$ full — 0 unpaired.
Species with one unpaired electron: O$_2^-$ and NO, i.e. 2.
Answer: 2. | ||||
| 83 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics C$_2$H$_4$ + H$_2$ → C$_2$H$_6$.
Bonds broken: one C=C (615) + one H–H (435) = 1050 kJ (the four C–H bonds of ethylene survive into ethane).
Bonds formed: one C–C (347) + two new C–H (2 × 414 = 828) = 1175 kJ.
$\Delta H = 1050 - 1175 = -125$ kJ.
Answer: 125. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Molality: $m = \frac{\Delta T_b}{K_b} = \frac{2}{0.52} = 3.846$ mol kg$^{-1}$.
With the solute negligible beside the solvent, the mole fraction of solute is $x_2 \approx \frac{n_2}{n_1} = m\times\frac{18}{1000} = 3.846\times 0.018 = 0.0692$.
Raoult's law with $p^\circ = 760$ mm Hg (1 atm, as given): $\frac{p^\circ - p}{p^\circ} = 0.0692$, so $p = 760(1 - 0.0692) = 707.4$ mm Hg.
Nearest integer: 707.
Answer: 707. | ||||
| 85 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics With $k_i = A_i e^{-E_i/RT}$: $k = \frac{k_1k_2}{k_3} = \frac{A_1A_2}{A_3}e^{-(E_1 + E_2 - E_3)/RT}$.
So $E_a = E_1 + E_2 - E_3$: $400 = 300 + 200 - Ea_3$, giving $Ea_3 = 100$ kJ mol$^{-1}$.
Answer: 100. | ||||
| 86 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Fusion of MnO$_2$ with KOH in air: 2MnO$_2$ + 4KOH + O$_2$ → 2K$_2$MnO$_4$ + 2H$_2$O, so A is potassium manganate, K$_2$MnO$_4$ (Mn +6).
In neutral or acidic solution manganate disproportionates: 3MnO$_4^{2-}$ + 4H$^+$ → 2MnO$_4^-$ + MnO$_2$ + 2H$_2$O.
B and C are MnO$_4^-$ (Mn +7, 3d$^0$, $\mu = 0$) and MnO$_2$ (Mn +4, 3d$^3$, 3 unpaired, $\mu = \sqrt{3\times 5} = \sqrt{15} \approx 3.87$ BM).
Sum $\approx 3.87$ BM, nearest integer 4.
Answer: 4. | ||||
| 87 | 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 2MnO$_4^-$ + 5C$_2$O$_4^{2-}$ + 16H$^+$ → 2Mn$^{2+}$ + 10CO$_2$ + 8H$_2$O, so KMnO$_4$ : oxalic acid = 2 : 5.
Oxalic acid: $20\times 2 = 40$ mmol. KMnO$_4$ needed: $\frac{2}{5}\times 40 = 16$ mmol.
Molarity $= \frac{16}{2} = 8$ M.
Answer: 8. | ||||
| 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 The structural (chain) isomers of heptane:
1. heptane; 2. 2-methylhexane; 3. 3-methylhexane; 4. 2,2-dimethylpentane; 5. 2,3-dimethylpentane; 6. 2,4-dimethylpentane; 7. 3,3-dimethylpentane; 8. 3-ethylpentane; 9. 2,2,3-trimethylbutane.
Total 9.
Answer: 9. | ||||
| 89 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) Benzene + benzoyl chloride with anhydrous AlCl$_3$ is Friedel–Crafts acylation and gives benzophenone, C$_6$H$_5$COC$_6$H$_5$, not diphenylmethane. Incorrect.
(B) Benzoyl chloride with H$_2$/Pd–BaSO$_4$ (Rosenmund reduction) gives benzaldehyde, not benzoic acid. Incorrect.
(C) Benzene with CO and HCl over anhydrous AlCl$_3$/CuCl (Gattermann–Koch reaction) gives benzaldehyde. Correct.
(D) Benzamide on acid hydrolysis gives benzoic acid (and NH$_4^+$), not aniline (that needs Br$_2$/NaOH, Hofmann degradation). Incorrect.
One reaction is correct.
Answer: 1. | ||||
| 90 | AminesClassification and identification of amines | Class 12: Amines | NumericalEasy | |
A easy numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines C$_2$H$_5$NH$_2$ + HNO$_2$ → C$_2$H$_5$OH + N$_2$ + H$_2$O: one mole of N$_2$ per mole of ethylamine.
N$_2$: $\frac{2.24}{22.4} = 0.1$ mol, so ethylamine is 0.1 mol.
Molar mass of C$_2$H$_5$NH$_2$ $= 45$ g mol$^{-1}$: $X = 4.5$ g $= 45\times 10^{-1}$ g.
Answer: 45. | ||||
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