JEE (Main) · 2025
JEE (Main) 2025 — 4 April, 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
25 Q
- Conic Sections4
- Integral Calculus3
- Sets, Relations and Functions2
3 easy · 18 medium · 4 hard
Physics
25 Q
- Optics3
- Motion of System of Particles and Rigid Body2
- Properties of Bulk Matter2
20 easy · 5 medium · 0 hard
Chemistry
25 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Coordination Compounds3
- Solutions2
14 easy · 9 medium · 2 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsSets and their representation | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on sets and their representation, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Domain of f. Work from the outside in.
$\log_4(\cdot)$ needs $\log_3\log_7(8 - \log_2(x^2+4x+5)) > 0$, so $\log_7(8 - \log_2(x^2+4x+5)) > 1$, so $8 - \log_2(x^2+4x+5) > 7$, so $\log_2(x^2+4x+5) < 1$, i.e. $x^2 + 4x + 5 < 2$.
$x^2 + 4x + 3 < 0 \Rightarrow (x+3)(x+1) < 0 \Rightarrow -3 < x < -1$. (Here $x^2+4x+5 = (x+2)^2 + 1 > 0$ always, so the innermost logarithm is defined, and every intermediate logarithm is then positive.)
So $(\alpha, \beta) = (-3, -1)$.
Domain of g: $-1 \le \frac{7x+10}{x-2} \le 1$, $x \ne 2$.
$\frac{7x+10}{x-2} - 1 \le 0 \Rightarrow \frac{6x+12}{x-2} \le 0 \Rightarrow -2 \le x < 2$.
$\frac{7x+10}{x-2} + 1 \ge 0 \Rightarrow \frac{8x+8}{x-2} \ge 0 \Rightarrow x \le -1$ or $x > 2$.
Together: $[-2, -1]$, so $\gamma = -2$, $\delta = -1$.
$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 9 + 1 + 4 + 1 = 15$.
Answer: (C) $15$. | ||||
| 2 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions xRy means $y = 2x$ or $y = 2x - 1$, with y in A.
$x = -3$: $y = -6, -7$ (neither in A). $x = -2$: $y = -4, -5$ (neither). $x = -1$: $y = -2, -3$. $x = 0$: $y = 0, -1$. $x = 1$: $y = 2, 1$. $x = 2$: $y = 4$ (no), $3$. $x = 3$: $y = 6, 5$ (neither).
$R = \{(-1,-2), (-1,-3), (0,0), (0,-1), (1,2), (1,1), (2,3)\}$, so $l = 7$.
Reflexive: all seven pairs $(a, a)$ are needed; only $(0,0)$ and $(1,1)$ are present, so $m = 5$.
Symmetric: the five off-diagonal pairs $(-1,-2), (-1,-3), (0,-1), (1,2), (2,3)$ each need their reverse, none of which is in R, so $n = 5$.
$l + m + n = 7 + 5 + 5 = 17$.
Answer: (C) $17$. | ||||
| 3 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Multiply term by term.
$\sin^2\theta$: $(8+i)(1+8i) = 8 + 64i + i + 8i^2 = 65i$.
$\cos^2\theta$: $(7+4i)(4+7i) = 28 + 49i + 16i + 28i^2 = 65i$.
$\sin\theta\cos\theta$: $(8+i)(4+7i) + (7+4i)(1+8i) = (25 + 60i) + (-25 + 60i) = 120i$.
So $\omega_1\omega_2 = i(65 + 120\sin\theta\cos\theta) = i(65 + 60\sin2\theta)$.
Hence $\alpha = 0$, $\beta = 65 + 60\sin2\theta$, and $\alpha + \beta$ ranges over $[5, 125]$.
$p + q = 125 + 5 = 130$.
Answer: (A) $130$. | ||||
| 4 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Apply the relation repeatedly: $A^{50} = A^{48} + (A^2 - I) = A^{46} + 2(A^2 - I) = \ldots = A^2 + 24(A^2 - I) = 25A^2 - 24I$.
$A^2 = \begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix} = \begin{bmatrix}1&0&0\\1&1&0\\1&0&1\end{bmatrix}$.
$A^{50} = 25A^2 - 24I = \begin{bmatrix}1&0&0\\25&1&0\\25&0&1\end{bmatrix}$.
Sum of elements $= 1 + 1 + 1 + 25 + 25 = 53$.
Answer: (B) $53$. | ||||
| 5 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $r^4 + 3r^2 + 4 = (r^2 + 2)^2 - r^2 = (r^2 - r + 2)(r^2 + r + 2)$.
The two factors differ by $2r$, so
$T_r = \frac{4r}{(r^2 - r + 2)(r^2 + r + 2)} = 2\left[\frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2}\right]$.
Since $(r+1)^2 - (r+1) + 2 = r^2 + r + 2$, the sum telescopes:
$S_{20} = 2\left[\frac{1}{1 - 1 + 2} - \frac{1}{400 + 20 + 2}\right] = 2\left[\frac{1}{2} - \frac{1}{422}\right] = 1 - \frac{1}{211} = \frac{210}{211}$.
$m + n = 210 + 211 = 421$.
Answer: (B) $421$. | ||||
| 6 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Write A as $12 - d, 12, 12 + d$ (sum 36 gives middle term 12), and B as $12 - D, 12, 12 + D$.
$p = 12(144 - d^2)$, $q = 12(144 - D^2)$.
$\frac{p+q}{p-q} = \frac{288 - d^2 - D^2}{D^2 - d^2} = \frac{19}{5}$
$\Rightarrow 1440 - 5d^2 - 5D^2 = 19D^2 - 19d^2 \Rightarrow 24D^2 = 1440 + 14d^2$.
With $D = d + 3$: $24(d^2 + 6d + 9) = 1440 + 14d^2 \Rightarrow 10d^2 + 144d - 1224 = 0 \Rightarrow 5d^2 + 72d - 612 = 0$.
$d = \frac{-72 + \sqrt{5184 + 12240}}{10} = \frac{-72 + 132}{10} = 6$ (taking $d > 0$), so $D = 9$.
$p - q = 12(D^2 - d^2) = 12(81 - 36) = 540$.
Answer: (A) $540$. | ||||
| 7 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Use $r\,{}^nC_r = n\,{}^{n-1}C_{r-1}$ and $r^2 = r(r-1) + r$:
$\sum r^2\,{}^nC_r = n(n-1)2^{n-2} + n2^{n-1} = n(n+1)2^{n-2}$.
For $n = 15$: $15\cdot16\cdot2^{13} = 15\cdot2^{17} = 2^{17}\cdot3\cdot5$.
$m = 17$, $n = 1$, $k = 1$, so $m + n + k = 19$.
Answer: (B) $19$. | ||||
| 8 | StatisticsMean deviation | Class 11: Statistics | Single correctEasy | |
A easy single correct question on mean deviation, from Statistics. NCERT: Class 11: Statistics There are 8 observations.
Mean 4: $2 + 3 + 3 + 4 + 5 + 7 + a + b = 32 \Rightarrow a + b = 8$.
Variance 2: $\frac{\sum x^2}{8} - 16 = 2 \Rightarrow \sum x^2 = 144$. The known squares add to $4 + 9 + 9 + 16 + 25 + 49 = 112$, so $a^2 + b^2 = 32$.
Then $2ab = 64 - 32 = 32$, $ab = 16$, so $a = b = 4$.
Data: 2, 3, 3, 4, 4, 4, 5, 7. Mode $= 4$.
Mean deviation about 4: $\frac{2 + 1 + 1 + 0 + 0 + 0 + 1 + 3}{8} = \frac{8}{8} = 1$.
Answer: (D) $1$. | ||||
| 9 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections On $y = x$ in the first quadrant: vertex $V(1, 1)$ (distance $\sqrt{2}$), focus $S(2, 2)$ (distance $2\sqrt{2}$).
The directrix is perpendicular to the axis, as far behind V as S is in front: it passes through $(0, 0)$, so it is $x + y = 0$.
Parabola: $(x - 2)^2 + (y - 2)^2 = \frac{(x + y)^2}{2}$.
At $(1, k)$: $1 + (k - 2)^2 = \frac{(1 + k)^2}{2} \Rightarrow 2 + 2k^2 - 8k + 8 = k^2 + 2k + 1 \Rightarrow k^2 - 10k + 9 = 0$.
$k = 1$ or $k = 9$; of the options only 9 appears.
Answer: (D) $9$. | ||||
| 10 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Tangency of $y = mx + c$ to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: $c^2 = a^2m^2 + b^2 = 16 + 9 = 25$, so $p = \pm5$.
The point of contact is $\left(-\frac{a^2m}{c}, \frac{b^2}{c}\right)$: for $p = 5$, $A\left(-\frac{16}{5}, \frac{9}{5}\right)$; for $p = -5$, $B\left(\frac{16}{5}, -\frac{9}{5}\right)$.
$y = x$ meets E where $x^2\left(\frac{1}{16} + \frac{1}{9}\right) = 1$, $x = \pm\frac{12}{5}$: $C\left(\frac{12}{5}, \frac{12}{5}\right)$, $D\left(-\frac{12}{5}, -\frac{12}{5}\right)$.
CD is a diagonal with A and B on opposite sides of it. $CD = \frac{24\sqrt{2}}{5}$.
Distance of A from $x - y = 0$: $\frac{|-16/5 - 9/5|}{\sqrt{2}} = \frac{5}{\sqrt{2}}$; the same for B.
Area $= \frac{1}{2}\cdot CD\cdot\left(\frac{5}{\sqrt{2}} + \frac{5}{\sqrt{2}}\right) = \frac{1}{2}\cdot\frac{24\sqrt{2}}{5}\cdot\frac{10}{\sqrt{2}} = 24$.
Answer: (B) $24$. | ||||
| 11 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $F_1F_2$ is a diameter of C, so P on C gives $\angle F_1PF_2 = 90°$.
Let $PF_1 = r_1$, $PF_2 = r_2$: $r_1 + r_2 = 2a = 17$ and $r_1^2 + r_2^2 = (2c)^2 = 4c^2$.
Area $= \frac{1}{2}r_1r_2 = 30 \Rightarrow r_1r_2 = 60$.
$4c^2 = (r_1 + r_2)^2 - 2r_1r_2 = 289 - 120 = 169 \Rightarrow 2c = 13$.
(Check: $b^2 = a^2 - c^2 = 72.25 - 42.25 = 30$, and indeed the area equals $b^2$.)
Answer: (B) $13$. | ||||
| 12 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections For a point on the right branch the focal distances are $ex - a$ and $ex + a$; their sum is $2ex = 8e$.
$8e = 8\sqrt{\frac{5}{3}} \Rightarrow e^2 = \frac{5}{3}$, so $b^2 = a^2(e^2 - 1) = \frac{2a^2}{3}$.
P on H: $\frac{16}{a^2} - \frac{9}{b^2} = 1 \Rightarrow \frac{16}{a^2} - \frac{27}{2a^2} = 1 \Rightarrow \frac{5}{2a^2} = 1 \Rightarrow a^2 = \frac{5}{2}$, $b^2 = \frac{5}{3}$.
Latus rectum $l = \frac{2b^2}{a}$: $l^2 = \frac{4b^4}{a^2} = \frac{4\cdot\frac{25}{9}}{\frac{5}{2}} = \frac{40}{9}$, so $9l^2 = 40$.
Product of focal distances $m = (ex)^2 - a^2 = \frac{5}{3}\cdot16 - \frac{5}{2} = \frac{80}{3} - \frac{5}{2} = \frac{145}{6}$, so $6m = 145$.
$9l^2 + 6m = 40 + 145 = 185$.
Answer: (B) $185$. | ||||
| 13 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions The numerators 7, 19, 39, 67 have differences 12, 20, 28 (second difference 8), so the n-th is $4n^2 + 3$ and the n-th term is $\cot^{-1}\left(n^2 + \frac{3}{4}\right) = \tan^{-1}\frac{1}{n^2 + \frac{3}{4}}$.
$n^2 + \frac{3}{4} = 1 + \left(n + \frac{1}{2}\right)\left(n - \frac{1}{2}\right)$, so
$\tan^{-1}\frac{1}{1 + (n + \frac{1}{2})(n - \frac{1}{2})} = \tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(n - \frac{1}{2}\right)$.
The sum telescopes: $\lim_{N\to\infty}\tan^{-1}\left(N + \frac{1}{2}\right) - \tan^{-1}\frac{1}{2} = \frac{\pi}{2} - \tan^{-1}\frac{1}{2}$.
(Option (A) equals $\tan^{-1}\frac{1}{2}$, a different number.)
Answer: (C). | ||||
| 14 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Line 1 passes through $(-1, 0, 0)$ along $(3, 4, 5)$; line 2 through $(p, 2, 1)$ along $(2, 3, 4)$.
$\vec{n} = (3,4,5)\times(2,3,4) = (16 - 15, 10 - 12, 9 - 8) = (1, -2, 1)$, $|\vec{n}| = \sqrt{6}$.
Joining vector $(p + 1, 2, 1)$; its dot product with $\vec{n}$ is $p + 1 - 4 + 1 = p - 2$.
Shortest distance $\frac{|p - 2|}{\sqrt{6}} = \frac{1}{\sqrt{6}} \Rightarrow p = 1$ or $3$, so $a = 1$, $b = 3$.
Ellipse $\frac{x^2}{1} + \frac{y^2}{9} = 1$: the major axis is along y (semi-axis 3, minor semi-axis 1).
Latus rectum $= \frac{2(\text{minor})^2}{\text{major}} = \frac{2\cdot1}{3} = \frac{2}{3}$.
Answer: (B) $\frac{2}{3}$. | ||||
| 15 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $L_1$: $(7 + t, 5, 3 - t)$; $L_2$: $(1 + 3s, -3 + 4s, -7 + 5s)$. From $y$: $5 = -3 + 4s$, $s = 2$, giving $(7, 5, 3)$, which is $L_1$ at $t = 0$. So $A = (7, 5, 3)$.
Angle between the lines: $\cos\theta = \frac{(1,0,-1)\cdot(3,4,5)}{\sqrt{2}\cdot\sqrt{50}} = \frac{-2}{10}$, so $\sin^2\theta = 1 - \frac{1}{25} = \frac{24}{25}$ (the same for either choice of B and C on the lines).
Area $= \frac{1}{2}\cdot AB\cdot AC\cdot\sin\theta = \frac{15}{2}\sin\theta$.
Area$^2 = \frac{225}{4}\cdot\frac{24}{25} = 54$.
Answer: (A) $54$. | ||||
| 16 | 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 The limit has the form $1^\infty$: $\lim(f(2+x))^{3/x} = e^{\lim\frac{3(f(2+x) - 1)}{x}} = e^{3f'(2)} = e^{12}$, so $\alpha = 12$.
$y = 4x^3 - 4x^2 - 20x - 12 = 4(x^3 - x^2 - 5x - 3)$.
$x = -1$ is a root: $-1 - 1 + 5 - 3 = 0$. Dividing, $x^3 - x^2 - 5x - 3 = (x + 1)(x^2 - 2x - 3) = (x + 1)^2(x - 3)$.
The curve meets the x-axis at $x = -1$ (where it touches) and $x = 3$: two points.
Answer: (C) $2$. | ||||
| 17 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 18x^2 - 90ax + 108a^2 = 18(x - 2a)(x - 3a)$.
For $a > 0$, $f'$ changes from + to − at $2a$ (local maximum) and from − to + at $3a$ (local minimum): $x_1 = 2a$, $x_2 = 3a$.
$x_1x_2 = 6a^2 = 54 \Rightarrow a = 3$, so $x_1 = 6$, $x_2 = 9$.
$a + x_1 + x_2 = 3 + 6 + 9 = 18$.
Answer: (C) $18$. | ||||
| 18 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replace x by $\frac{1}{x}$: $f\left(\frac{1}{x}\right) + 2f(x) = \frac{1}{x^2} + 5$.
Twice this minus the given equation: $3f(x) = \frac{2}{x^2} + 10 - x^2 - 5$, so $f(x) = \frac{1}{3}\left(\frac{2}{x^2} + 5 - x^2\right)$.
$\alpha = \frac{1}{3}\left[-\frac{2}{x} + 5x - \frac{x^3}{3}\right]_1^2 = \frac{1}{3}\left(1 + 5 - \frac{7}{3}\right) = \frac{11}{9}$, so $9\alpha = 11$.
For g put $x = \frac{1}{2}$: $2g\left(\frac{1}{2}\right) - 3g\left(\frac{1}{2}\right) = \frac{1}{2}$, so $g\left(\frac{1}{2}\right) = -\frac{1}{2}$ and $g(x) = \frac{x}{2} - \frac{3}{4}$.
$\beta = \left[\frac{x^2}{4} - \frac{3x}{4}\right]_1^2 = \frac{3}{4} - \frac{3}{4} = 0$.
$9\alpha + \beta = 11$.
Answer: (D) $11$. | ||||
| 19 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Take $B = (2 + t^2, t)$, $t > 0$. The slope there is $\frac{dy}{dx} = \frac{1}{2y} = \frac{1}{2t}$.
The tangent passes through $(-2, 0)$: $0 - t = \frac{1}{2t}(-2 - 2 - t^2) \Rightarrow 2t^2 = 4 + t^2 \Rightarrow t = 2$. So $B = (6, 2)$.
Line AB: $y = \frac{x + 2}{4}$, i.e. $x = 4y - 2$. Parabola: $x = y^2 + 2$.
For $0 \le y \le 2$ the region lies between the line (left) and the parabola (right):
Area $= \int_0^2\left[(y^2 + 2) - (4y - 2)\right]dy = \int_0^2(y - 2)^2\,dy = \left[\frac{(y-2)^3}{3}\right]_0^2 = \frac{8}{3}$.
Answer: (D) $\frac{8}{3}$. | ||||
| 20 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Rewrite: $x^5\frac{dy}{dx} = 7x^4\cot y - e^x\,\mathrm{cosec}\,y$. Multiply by $\sin y$:
$x^5\sin y\frac{dy}{dx} = 7x^4\cos y - e^x$.
Put $u = \cos y$, $\frac{du}{dx} = -\sin y\frac{dy}{dx}$: $-x^5\frac{du}{dx} = 7x^4u - e^x$, i.e. $\frac{du}{dx} + \frac{7}{x}u = \frac{e^x}{x^5}$.
Integrating factor $x^7$: $\frac{d}{dx}(ux^7) = x^2e^x$, so $ux^7 = e^x(x^2 - 2x + 2) + C$.
At $x = 1$, $y = \frac{\pi}{2}$, $u = 0$: $0 = e + C$, $C = -e$.
At $x = 2$: $128\cos y = e^2(4 - 4 + 2) - e = 2e^2 - e$, so $\cos y = \frac{2e^2 - e}{128}$.
Answer: (B). | ||||
| 21 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha$ is a non-real cube root of unity ($\omega$ or $\omega^2$), so $\frac{1}{\alpha^k} = \alpha^{2k}$ and $\alpha^k + \alpha^{-k} = 2\cos\frac{2\pi k}{3}$.
This is $-1$ when 3 does not divide k, and $2$ when it does. Squared: 1 or 4.
Each block of three consecutive k contributes $1 + 1 + 4 = 6$.
$n = 9$: 18; $n = 10$: $18 + 1 = 19$; $n = 11$: $19 + 1 = 20$.
Answer: 11. | ||||
| 22 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalHard | |
A hard numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Write $m = 6a$, $n = 6b$ with $\gcd(a, b) = 1$ and $a < b$.
Two-digit: $10 \le 6a$ gives $a \ge 2$; $6b \le 99$ gives $b \le 16$. So count coprime pairs $2 \le a < b \le 16$.
For each b, the number of $a$ in $1 \le a < b$ coprime to b is $\varphi(b)$; excluding $a = 1$ leaves $\varphi(b) - 1$.
$\varphi(2), \ldots, \varphi(16) = 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8$, which add to 79.
Subtract one for each of the 15 values of b: $79 - 15 = 64$.
Answer: 64. | ||||
| 23 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Lost card a spade: prior $\frac{1}{4}$; then 12 spades remain and $P(\text{n spades}) = \frac{{}^{12}C_n}{{}^{51}C_n}$.
Lost card not a spade: prior $\frac{3}{4}$; 13 spades remain and $P = \frac{{}^{13}C_n}{{}^{51}C_n}$.
Bayes: $P(\text{lost spade}\mid\text{data}) = \frac{{}^{12}C_n}{{}^{12}C_n + 3\,{}^{13}C_n}$.
${}^{13}C_n = \frac{13}{13 - n}{}^{12}C_n$, so this is $\frac{1}{1 + \frac{39}{13 - n}} = \frac{13 - n}{52 - n}$.
$\frac{13 - n}{52 - n} = \frac{11}{50} \Rightarrow 650 - 50n = 572 - 11n \Rightarrow 39n = 78 \Rightarrow n = 2$.
Answer: 2. | ||||
| 24 | Vector AlgebraAddition of vectors | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on addition of vectors, from Vector Algebra. NCERT: Class 12: Vector Algebra The vectors close a triangle: $(2,-1,1) + (1,-3,-5) = (3,-4,-4)$.
Side lengths squared: $4 + 1 + 1 = 6$, $1 + 9 + 25 = 35$, $9 + 16 + 16 = 41$.
For any triangle, $GA^2 + GB^2 + GC^2 = \frac{1}{3}(a^2 + b^2 + c^2)$ (each median $m_a$ satisfies $4m_a^2 = 2b^2 + 2c^2 - a^2$ and $GA = \frac{2}{3}m_a$).
$GA^2 + GB^2 + GC^2 = \frac{6 + 35 + 41}{3} = \frac{82}{3}$.
$6\times\frac{82}{3} = 164$.
Answer: 164. | ||||
| 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 Let $t = \sqrt{1 + x^2} + x$; then $\frac{1}{t} = \sqrt{1 + x^2} - x$, so the integrand is $t^{10}\cdot t^9 = t^{19}$.
$\sqrt{1+x^2} = \frac{1}{2}\left(t + \frac{1}{t}\right)$, $x = \frac{1}{2}\left(t - \frac{1}{t}\right)$, $dx = \frac{1}{2}\left(1 + \frac{1}{t^2}\right)dt$.
$\int t^{19}\cdot\frac{1}{2}\left(1 + t^{-2}\right)dt = \frac{t^{20}}{40} + \frac{t^{18}}{36} + C$.
The given form: $n\sqrt{1+x^2} - x = \frac{1}{2}\left[(n - 1)t + \frac{n + 1}{t}\right]$, so
$\frac{1}{m}t^n\left(n\sqrt{1+x^2} - x\right) = \frac{(n - 1)t^{n+1} + (n + 1)t^{n-1}}{2m}$.
Matching powers: $n + 1 = 20$, $n = 19$. Coefficients: $\frac{18}{2m} = \frac{1}{40}$ and $\frac{20}{2m} = \frac{1}{36}$, both giving $m = 360$.
$m + n = 360 + 19 = 379$.
Answer: 379. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter One main scale division: $1\text{ MSD} = \frac{15}{300} = 0.05$ cm.
$25\text{ VSD} = 24\text{ MSD}$, so $1\text{ VSD} = \frac{24}{25}\text{ MSD}$.
Least count $= 1\text{ MSD} - 1\text{ VSD} = \frac{1}{25}\text{ MSD} = \frac{0.05}{25} = 0.002$ cm.
Answer: (B) 0.002. | ||||
| 27 | 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 Electric dipole moment $p = q\times d$: $[AT][L] = [LTA]$.
Magnetic dipole moment $m = I\times\text{area}$: $[AL^2]$.
$\frac{p}{m} = \frac{LTA}{AL^2} = [M^0L^{-1}TA^0]$.
So $P = 0$, $Q = -1$ (and $R = 1$, $S = 0$). This is the dimension of $\frac{1}{\text{speed}}$, as expected since $m = pc$ for matching dipoles.
Answer: (A) $0, -1$. | ||||
| 28 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion In rolling without slipping the centre moves at $v$ and $\omega R = v$.
Top point: $v + \omega R = 2v = 8$ m/s, so $v = 4$ m/s.
A rim point level with the centre has the translational velocity $v$ (horizontal) and the rotational velocity $\omega R = v$ (vertical), at right angles:
speed $= \sqrt{v^2 + v^2} = \sqrt{2}v = 4\sqrt{2}$ m/s.
(Equivalently, it is $\sqrt{2}R$ from the contact point, the instantaneous axis.)
Answer: (C) $4\sqrt{2}$ m/s. | ||||
| 29 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation The object is at $r = R + 3R = 4R$ from the earth's centre.
To escape, its total energy must reach zero: $\frac{1}{2}mv^2 - \frac{GMm}{4R} = 0$.
$v = \sqrt{\frac{2GM}{4R}} = \sqrt{\frac{GM}{2R}}$.
Answer: (D). | ||||
| 30 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | StatementsEasy | |
A easy statements question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane From the graph: O(0, 0), A(1, −5), B(2, 0), C(3, 5), D(5, 5), E(6, 10), F(7, 0), G(8, −5), H(9, 0), joined by straight lines.
(A) 0 to 3 s: $\frac{5 - 0}{3} = \frac{5}{3}$ m/s, not 10 m/s. False.
(B) 3 to 5 s: $x$ stays at 5 m, so the average velocity is 0. True.
(C) At $t = 2$ s the graph is on the straight segment A to C: slope $\frac{5 - (-5)}{3 - 1} = 5$ m/s. True.
(D) 5 to 7 s: $\frac{0 - 5}{2} = -2.5$ m/s; at 6.5 s (segment E to F) the slope is $\frac{0 - 10}{1} = -10$ m/s. Not equal. False.
(E) 0 to 9 s: $x$ returns to 0, so the average velocity is zero. True.
Answer: (D) (B), (C), (E) only. | ||||
| 31 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Temperature rise: $\Delta T = \frac{Q}{mC} = \frac{810}{0.1\times900} = 9$ K.
Areal expansion coefficient $\beta = 2\alpha = 6.2\times10^{-5}$ K$^{-1}$.
$A = 9\times4 = 36$ cm$^2 = 3.6\times10^{-3}$ m$^2$.
$\Delta A = \beta A\Delta T = 6.2\times10^{-5}\times3.6\times10^{-3}\times9 = 2.0\times10^{-6}$ m$^2$.
Answer: (A). | ||||
| 32 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Shear stress $= \frac{F}{A} = \frac{10^5}{\pi(0.04)^2} = \frac{10^5}{1.6\times10^{-3}\pi}$ N/m$^2$.
For a small displacement the shear strain is the angle: $\theta = \frac{F/A}{G} = \frac{10^5}{1.6\times10^{-3}\pi\times10^{10}} = \frac{1}{160\pi}$.
Answer: (D) $1/160\pi$. | ||||
| 33 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Matching listEasy | |
A easy matching list question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics First law: $\Delta Q = \Delta U + \Delta W$.
Isobaric: $\Delta W = P\Delta V$, so $\Delta Q = \Delta U + P\Delta V$ (IV).
Isochoric: $\Delta V = 0$, $\Delta W = 0$, so $\Delta Q = \Delta U$ (II).
Adiabatic: no heat exchanged, $\Delta Q = 0$ (III).
Isothermal (ideal gas): $\Delta U = 0$, so $\Delta Q = \Delta W$ (I).
Answer: (D) (A)-(IV), (B)-(II), (C)-(III), (D)-(I). | ||||
| 34 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\omega = 628$ rad/s, so $f = \frac{\omega}{2\pi} = \frac{628}{6.28} = 100$ Hz.
$\lambda = \frac{v}{f} = \frac{300}{100} = 3$ m.
Answer: (B) 3 m. | ||||
| 35 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With the dielectric, $C_1 = 4\times5 = 20$ $\mu$F.
$C_1$ and $C_2$ are in series: $\frac{20\times5}{20 + 5} = 4$ $\mu$F.
This branch is in parallel with $C_3$: $4 + 5 = 9$ $\mu$F.
Answer: (C) 9 $\mu$F. | ||||
| 36 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Each bulb's resistance: $R = \frac{V^2}{p}$.
In series across the same mains: total resistance $nR$, so the power is $\frac{V^2}{nR} = \frac{p}{n}$ (taking R independent of temperature).
Answer: (B) $\frac{p}{n}$. | ||||
| 37 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity (A) $R_1 \parallel (R_2 + R_3) \parallel R_4$: $\frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{10} = \frac{2}{5}$, $R = 2.5$ $\Omega$.
(B) $(R_1 + R_2 + R_4) \parallel R_3 = 20 \parallel 5 = \frac{100}{25} = 4$ $\Omega$.
(C) $(R_1 + R_2) \parallel (R_3 + R_4) = 10 \parallel 15 = \frac{150}{25} = 6$ $\Omega$.
(D) all in parallel: $\frac{1}{R} = \frac{3}{5} + \frac{1}{10} = \frac{7}{10}$, $R = \frac{10}{7}$ $\Omega$.
Answer: (C). | ||||
| 38 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Arc ABC is two quarter arcs, AB and BC, each giving a field of magnitude E at O.
By symmetry the field of a uniformly charged arc at its centre points along the arc's bisector (away from it, for positive charge). The bisectors of AB and BC are 90° apart.
Resultant: $\sqrt{E^2 + E^2} = \sqrt{2}E$.
Answer: (C) $\sqrt{2}$E. | ||||
| 39 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Convex mirror ($f = +30$ cm, $u = -30$ cm): $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{30} + \frac{1}{30}$, $v = 15$ cm behind the mirror.
This image is $30 + 15 = 45$ cm from the object.
A plane mirror forms its image as far behind it as the object is in front, so to put its image at the same place it must stand halfway: 22.5 cm from the object, facing it.
Distance between the mirrors $= 30 - 22.5 = 7.5$ cm.
Answer: (B) 7.5 cm. | ||||
| 40 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $P_1$ and $P_2$ are crossed (90°). Let $P_3$ make angle $\theta$ with $P_1$, hence $90° - \theta$ with $P_2$.
By Malus's law, $I = I_1\cos^2\theta\cos^2(90° - \theta) = \frac{I_1}{4}\sin^2 2\theta$, where $I_1$ is the intensity after $P_1$.
This is greatest when $2\theta = 90°$, $\theta = 45°$; the angle between $P_2$ and $P_3$ is then also $\frac{\pi}{4}$.
Answer: (D) $\frac{\pi}{4}$. | ||||
| 41 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | StatementsEasy | |
A easy statements question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Statement I: $E = h\nu$ gives h in J s $= ML^2T^{-1}$; angular momentum $mvr$ is also $ML^2T^{-1}$. Correct.
Statement II: Bohr's condition is $mvr = n\frac{h}{2\pi}$, an integral multiple of $\frac{h}{2\pi}$, not of h itself. Incorrect.
Answer: (C). | ||||
| 42 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei P only decays: its mass falls exponentially from the start.
Q is fed by P and itself decays: it starts at zero, rises while P's decay outpaces its own, peaks, then falls towards zero.
R is stable and only gains: it starts at zero and rises, levelling off as everything ends up as R.
Only the first figure shows all three: P falling, Q rising to a peak then falling, R rising to a plateau.
Answer: (A). | ||||
| 43 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | StatementsEasy | |
A easy statements question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits (A) In n-type material electrons are the majority carriers and holes the minority. True.
(B) n-type doping uses a pentavalent donor (P, As, Sb). True.
(C) In thermal equilibrium the mass-action law holds: $n_en_h = n_i^2$. So (C) is false.
(D) Since $n_en_h = n_i^2$, the sign $\ge$ is met only as an equality; the claim that doping raises the product above $n_i^2$ is false. In any case every option containing (D) also contains the false (C).
(E) Donors give electrons only; holes come from thermal breaking of bonds. True.
Answer: (A) (A), (B), (E) only. | ||||
| 44 | 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 Uniform velocity, so the forces balance. With F at 45° above the horizontal:
Normal reaction $N = mg - F\sin45°$; horizontal balance $F\cos45° = \mu N = \mu(mg - F\sin45°)$.
$\frac{F}{\sqrt{2}}(1 + \mu) = \mu mg \Rightarrow \frac{F}{\sqrt{2}} = \frac{0.25\times25\times9.8}{1.25} = 49$ N.
Work by F over 5 m: $W = F\cos45°\times5 = 49\times5 = 245$ J.
(With $g = 10$ m/s$^2$ it is 250 J; 245 J is the option.)
Answer: (A) 245 J. | ||||
| 45 | 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 Moles are conserved: $n = \frac{PV}{RT}$. Let the small vessel have volume V, the large 2V.
Before: $\frac{8\times2V}{1000R} + \frac{7\times V}{500R} = \frac{V}{R}(0.016 + 0.014) = \frac{0.030V}{R}$.
After, at a common pressure P and 600 K over $3V$: $\frac{P\times3V}{600R} = \frac{0.030V}{R} \Rightarrow P = 6$ kPa.
Answer: (A) 6. | ||||
| 46 | 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 Width of the central diffraction maximum: $\frac{2\lambda D}{a}$. Fringe width of the double-slit pattern: $\beta = \frac{\lambda D}{d}$.
20 maxima within it: $\frac{2\lambda D}{a} = 20\frac{\lambda D}{d} \Rightarrow a = \frac{d}{10} = \frac{1.5\text{ mm}}{10} = 0.15$ mm.
$0.15$ mm $= 0.015$ cm $= 15\times10^{-3}$ cm.
Answer: 15. | ||||
| 47 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalEasy | |
A easy numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $\frac{c}{v} = \sqrt{\mu_r\epsilon_r} = \sqrt{\frac{10}{\pi}\times\frac{1}{0.0885}} = \sqrt{\frac{10}{0.27803}} = \sqrt{35.97} \approx 6.0$.
Answer: 6. | ||||
| 48 | 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 $X_L = 2\pi fL = 2\pi\times50\times1 = 100\pi$ $\Omega$.
$Z = \sqrt{R^2 + X_L^2} = 100\pi\sqrt{2}$ $\Omega$.
The supply voltage is taken as rms: $I_{rms} = \frac{100\pi}{100\pi\sqrt{2}} = \frac{1}{\sqrt{2}}$ A.
Maximum (peak) current $I_0 = \sqrt{2}I_{rms} = 1$ A.
Answer: 1. | ||||
| 49 | 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 Fired perpendicular to B, the particle moves in a circle and first returns after one period:
$T = \frac{2\pi m}{qB} = \frac{2\times3.14\times16\times10^{-9}}{1.6\times10^{-6}\times6.28}$ s.
$\frac{2\times3.14}{6.28} = 1$, and $\frac{16\times10^{-9}}{1.6\times10^{-6}} = 10^{-2}$, so $T = 0.01$ s.
The stem does not ask for a rounded value, so the exact time is kept; an entry of 0, the nearest integer, is also accepted.
Answer: 0.01 | ||||
| 50 | 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 No external torque acts, and the intended reading is that the sphere's own angular momentum is conserved: $I_1\omega_1 = I_2\omega_2$.
Density stays uniform and the shape stays spherical, so $M \propto R^3$ and $I = \frac{2}{5}MR^2 \propto R^5$.
Radius halved: $I_2 = \frac{I_1}{2^5} = \frac{I_1}{32}$, so $\omega_2 = 32\omega_1$.
Answer: 32. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom Cr: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1$ (half-filled d, the exceptional configuration).
$l = 1$ (p electrons): $2p^6 + 3p^6 = 12$.
$l = 2$ (d electrons): $3d^5 = 5$.
Answer: (B) 12 and 5. | ||||
| 52 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics (A) The enthalpy of solution is −69.01 kJ mol$^{-1}$ with 10 mol water and −72.79 kJ mol$^{-1}$ with 40 mol: it depends on the amount of solvent. True.
(B) Dilution (b) − (a): $\Delta H = -72.79 - (-69.01) = -3.78$ kJ mol$^{-1}$. Heat is released; the value is −3.78, not +3.78 kJ mol$^{-1}$. False as written.
(C) Both values are negative: dissolving HCl gas here is exothermic. False.
(D) (a) and (b) start from HCl(g), not from the elements, so they are enthalpies of solution, not of formation. False.
Answer: (A). | ||||
| 53 | SolutionsColligative properties | Class 12: Solutions | StatementsMedium | |
A medium statements question on colligative properties, from Solutions. NCERT: Class 12: Solutions Statement I: $K_f = \frac{M_1RT_f^2}{\Delta H_{fus}}$ (with $M_1$ the solvent's molar mass in kg mol$^{-1}$). At the freezing point $\Delta S_{fus} = \frac{\Delta H_{fus}}{T_f}$, so $\frac{RT_f^2M_1}{\Delta H_{fus}} = \frac{M_1RT_f}{\Delta S_{fus}}$. Correct.
Statement II: $K_f$(benzene) $\approx 5.12$ K kg mol$^{-1}$, $K_f$(water) $= 1.86$ K kg mol$^{-1}$. Benzene's is larger, so the statement is incorrect.
Answer: (C). | ||||
| 54 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Zero order: $[A] = [A]_0 - kt$ and $t_{1/2} = \frac{[A]_0}{2k}$.
$1 = \frac{2.0}{2k} \Rightarrow k = 1.0$ mol L$^{-1}$ h$^{-1}$.
From 0.50 to 0.25 mol L$^{-1}$: $t = \frac{0.25}{1.0} = 0.25$ h $= 15$ min.
Answer: (C) 15 min. | ||||
| 55 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Arrhenius: $\log k = \log A - \frac{E_a}{2.303R}\cdot\frac{1}{T}$. The slope is $-\frac{E_a}{2.303R}$, so the steeper the line, the larger $E_a$.
Line 2 is nearly vertical (steepest), line 1 is moderately steep, line 3 is nearly flat.
$Ea_2 > Ea_1 > Ea_3$.
Answer: (A). | ||||
| 56 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | StatementsMedium | |
A medium statements question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure ClF$_3$ has 5 electron pairs on Cl (3 bond pairs, 2 lone pairs) in a trigonal bipyramid. The two lone pairs can be both equatorial (I), one axial and one equatorial (II), or both axial (III): these are the three possible arrangements, so Statement I is correct.
An axial position has three neighbours at 90°, an equatorial one only two. Lone pairs therefore go equatorial, where lp–lp and lp–bp repulsions at 90° are fewest. Structure I (the T-shape) is the most stable, not III. Statement II is incorrect.
Answer: (C). | ||||
| 57 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter First ionisation enthalpies (kJ mol$^{-1}$): B 801, Al 577, Ga 579, In 558, Tl 589.
Ga and Tl are raised by poor shielding of the d (and f) electrons, so the trend is irregular. Highest: B; lowest: In.
Answer: (B) B & In. | ||||
| 58 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: Across a period the nuclear charge rises and the size falls, so each group 14 element has a higher first ionisation enthalpy than its group 13 neighbour (C 1086 > B 801, Si 786 > Al 577, Ge 761 > Ga 579, Sn 708 > In 558, Pb 715 > Tl 589 kJ mol$^{-1}$). Correct.
Statement II: Group 14 elements, especially C (diamond/graphite) and Si with giant covalent networks, melt and boil far higher than B, Al, Ga, In and Tl. Gallium melts near 303 K. Incorrect.
Answer: (C). | ||||
| 59 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctHard | |
A hard single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All four contain $\mathrm{Fe^{3+}}$ ($3d^5$). The oxalate complex is a chelate (three five-membered rings), so it is the most stable.
Oxalate is a weak-field ligand: $\mathrm{Fe^{3+}}$ stays high spin, $t_{2g}^3e_g^2$. So $X = 3$.
$\mathrm{V_2O_3}$ (V in +3, a low oxidation state) is a basic oxide. (Vanadium oxides become more acidic as the oxidation state rises: $\mathrm{V_2O_3}$ basic, $\mathrm{V_2O_4}$ amphoteric, $\mathrm{V_2O_5}$ amphoteric to acidic.)
Answer: (C) Basic. | ||||
| 60 | 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 $\mathrm{[FeF_6]^{3-}}$: $\mathrm{Fe^{3+}}$, $3d^5$, F$^-$ weak field, high spin: 5 unpaired.
$\mathrm{[CoF_6]^{3-}}$: $\mathrm{Co^{3+}}$, $3d^6$, high spin: 4 unpaired.
$\mathrm{[Ni(CN)_4]^{2-}}$: $\mathrm{Ni^{2+}}$, $3d^8$, square planar ($dsp^2$) with strong-field CN$^-$: 0 unpaired.
$\mathrm{[Ni(CO)_4]}$: Ni(0), $3d^{10}$, tetrahedral: 0 unpaired.
Order: $5 > 4 > 0 = 0$.
Answer: (C). | ||||
| 61 | 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 Compare the energy to remove the next electron from each ion.
(A) $\mathrm{Cr^+}$ is $3d^5$ (stable half-filled), $\mathrm{Mn^+}$ is $3d^54s^1$ (loses an easy 4s electron): $\Delta_iH_2$ Mn 1509 < Cr 1592 kJ mol$^{-1}$. Correct relationship.
(B) $\mathrm{Mn^{2+}}$ is $3d^5$ (half-filled, hard to ionise), $\mathrm{Fe^{2+}}$ is $3d^6$ (loses the paired electron to reach $d^5$): $\Delta_iH_3$ Mn 3248 > Fe 2957 kJ mol$^{-1}$. So "$\mathrm{Mn^{2+} < Fe^{2+}}$" is incorrect.
(C) Removing an electron from the more highly charged $\mathrm{Fe^{3+}}$ costs more. Correct.
(D) Likewise $\mathrm{Mn^+} < \mathrm{Mn^{2+}}$. Correct.
Answer: (B). | ||||
| 62 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Matching listEasy | |
A easy matching list question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) Aniline is steam volatile and immiscible with water: steam distillation (IV).
(B) Glycerol decomposes at its normal boiling point, so it is recovered from spent lye by distillation under reduced pressure (III).
(C) Crude oil fractions have close, overlapping boiling points: fractional distillation (II).
(D) Chloroform (b.p. 334 K) and aniline (b.p. 457 K) differ widely: simple distillation (I).
Answer: (B) (A)-(IV), (B)-(III), (C)-(II), (D)-(I). | ||||
| 63 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) 1-methoxycyclohexyl cation vs 1-methylcyclohexyl cation. Both are tertiary, but the OMe oxygen donates a lone pair into the empty p orbital (+R), which outweighs hyperconjugation from Me. First more stable. Yes.
(B) A cyclohexyl carbanion carrying an o-nitrophenyl group vs the cyclohexyl carbocation carrying the same group. The negative charge is delocalised into the ring and onto the ortho NO$_2$ (−R, −I), stabilising the anion; the same electron-withdrawing group destabilises the cation. First more stable. Yes.
(C) Cyclohexylmethyl cation (primary, no resonance) vs cyclohex-1-enylmethyl cation (allylic, resonance stabilised). The second is more stable. No.
(D) tert-Butyl cation vs the cation with a methoxy group on the cationic carbon; +R donation from OMe makes the second more stable. No.
Pairs (A) and (B).
Answer: (C) (A) & (B) only. | ||||
| 64 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The principal group is OH (suffix -ol); the seven-carbon chain carries both multiple bonds.
OH is at C-4 from either end. The multiple bonds are at 1 and 6 from either end too.
On a tie, the double bond gets the lower number: number from the $\mathrm{CH_2{=}}$ end. C=C at 1, C≡C at 6, OH at 4.
Name: hept-1-en-6-yn-4-ol ("en" before "yn"). The "4-hydroxy…" names are wrong because OH must be the suffix.
Answer: (A). | ||||
| 65 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons HBr adds by way of the most stable carbocation (Markovnikov).
X has three C=C bonds: two disubstituted ones (in the left five-membered ring and in the six-membered ring) and one trisubstituted one in the lower five-membered ring, whose substituted end is a ring-fusion carbon.
Protonating the CH end of the trisubstituted alkene puts the positive charge on the ring-fusion carbon: a tertiary carbocation. Protonating either disubstituted alkene gives only secondary cations, none of them allylic.
Br$^-$ adds to the tertiary carbon, leaving the other two double bonds untouched: the first structure.
Answer: (A). | ||||
| 66 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | StatementsEasy | |
A easy statements question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Statement I: Aqueous KOH supplies OH$^-$ as a nucleophile; the alcohol forms by nucleophilic substitution, not elimination. Incorrect.
Statement II: In alcoholic KOH the alkoxide/hydroxide acts as a base, removing a β-hydrogen as Cl$^-$ leaves (dehydrohalogenation) to give an alkene. Correct.
Answer: (D). | ||||
| 67 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | StatementsEasy | |
A easy statements question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The yellow solid is iodoform, $\mathrm{CHI_3}$. It forms from a $\mathrm{CH_3CO{-}}$ group, or a $\mathrm{CH_3CH(OH){-}}$ group that the hypoiodite oxidises to one.
(A) Butan-2-ol has $\mathrm{CH_3CH(OH)-}$: yes.
(B) Propan-1-ol: no.
(C) Butan-2-one is a methyl ketone: yes.
(D) Acetic acid: the carboxyl carbon is not a ketone carbonyl, and the acid is simply neutralised: no.
(E) Propanal is not a methyl carbonyl: no.
Answer: (A) (A) and (C) Only. | ||||
| 68 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Work backwards from the clearest clues.
D, an inhaled anaesthetic made from an alcohol with concentrated sulphuric acid at about 140°C (413 K): diethyl ether, from ethanol. So C is ethanol, which is also blended into petrol to cut emissions.
B is reduced by diborane to ethanol and is a cooking component and food preservative: acetic acid (vinegar).
A is toxic and gives B with cyanide in aqueous acid: methanol ($\mathrm{CH_3OH \rightarrow CH_3CN \rightarrow CH_3COOH}$ on acidic hydrolysis).
Only option (D) fits: methanol; acetic acid; ethanol; diethyl ether.
Option (C) fails at once: diborane does not convert lactic acid into propanoic acid, and dipropyl ether is not an anaesthetic.
Answer: (D). | ||||
| 69 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctHard | |
A hard single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques All three are lactams; their basicity depends on how far the N lone pair is delocalised into C=O.
R is a bridged bicyclic lactam: the bridgehead N cannot lie in the plane of the C=O (the amide cannot become planar), so its lone pair is not delocalised. It behaves like an amine: most basic.
Q is an α,β-unsaturated lactam: the C=C is conjugated with the C=O and competes with N for it (cross-conjugation), so the N lone pair is less delocalised than in P.
P is an ordinary N-methyl lactam: the N lone pair is fully delocalised, so it is least basic.
R > Q > P.
Answer: (C). | ||||
| 70 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Nitrous acid replaces $\mathrm{-NH_2}$ by $\mathrm{-OH}$: $\mathrm{CH_3CH(NH_2)COOH \rightarrow CH_3CH(OH)COOH}$ (lactic acid). So y is alanine.
The cyclic molecule is the diketopiperazine with only $\mathrm{CH_2}$ groups in the ring, formed from two glycine molecules. So z is glycine.
The dipeptide contains alanine and glycine: alanine-glycine is the option.
Answer: (A). | ||||
| 71 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium pH 10 means pOH 4: $[\mathrm{OH^-}] = 10^{-4}$ M.
Each $\mathrm{Mg(OH)_2}$ gives two $\mathrm{OH^-}$: $[\mathrm{Mg(OH)_2}] = 5\times10^{-5}$ M.
In 1.0 L: $5\times10^{-5}\times58 = 2.9\times10^{-3}$ g $= 2.9$ mg $\approx 3$ mg.
Answer: 3. | ||||
| 72 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Kohlrausch's law: $\Lambda^\circ(\mathrm{NH_4OH}) = \Lambda^\circ(\mathrm{NH_4Cl}) - \lambda^\circ(\mathrm{Cl^-}) + \lambda^\circ(\mathrm{OH^-}) = 185 - 70 + 170 = 285$ S cm$^2$ mol$^{-1}$.
$\alpha = \frac{\Lambda_m}{\Lambda^\circ} = \frac{85.5}{285} = 0.3 = 3\times10^{-1}$.
Answer: 3. | ||||
| 73 | SolutionsConcentration terms | Class 12: Solutions | NumericalMedium | |
A medium numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 L of sea water: mass $= 1000\times2 = 2000$ g.
NaCl: $6\times58.5 = 351$ g, so the water (solvent) is $2000 - 351 = 1649$ g.
5.8 ppm by mass: $\mathrm{O_2} = \frac{5.8}{10^6}\times2000 = 1.16\times10^{-2}$ g $= \frac{0.0116}{32} = 3.625\times10^{-4}$ mol.
Molality $= \frac{3.625\times10^{-4}}{1.649} = 2.2\times10^{-4}$ m, so $x \approx 2$.
Answer: 2. | ||||
| 74 | 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 $\mathrm{CaCO_3 \rightarrow CaO + CO_2}$: 100 g of $\mathrm{CaCO_3}$ gives 56 g of CaO.
Pure $\mathrm{CaCO_3} = 0.75\times150 = 112.5$ kg.
CaO $= 112.5\times\frac{56}{100} = 63$ kg.
Answer: 63. | ||||
| 75 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{BrF_5}$: Br has 7 valence electrons, 5 used in Br–F bonds, leaving one lone pair (square pyramidal). So $x = 1$.
One $\mathrm{Cl^-}$ is ionisable, so the complex is $\mathrm{[MCl_3(NH_3)_3]Cl}$: octahedral ($sp^3d^2$), of the type $\mathrm{MA_3B_3}$.
$\mathrm{MA_3B_3}$ has two geometrical isomers, fac and mer.
Answer: 2. | ||||
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