JEE (Main) · 2024
JEE (Main) 2024 — 1 February, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Complex Numbers2
- Matrices and Determinants2
9 easy · 16 medium · 5 hard
Physics
30 Q
- Physical World and Measurement2
- Kinematics2
- Laws of Motion2
22 easy · 6 medium · 2 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Chemical Bonding and Molecular Structure3
- Redox Reactions3
22 easy · 8 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 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Write $z = x + iy$. Then $z + \bar{z} = 2x$ and $z - \bar{z} = 2iy$, so $-i(z - \bar{z}) = 2y$.
The second condition is the line $2(\sqrt{2} - 1)x + 2y = 2\sqrt{2}$, i.e. $y = \sqrt{2} - (\sqrt{2} - 1)x$.
The first condition is the circle $(x - 1)^2 + y^2 = 1$.
Substituting: $(x - 1)^2 + (\sqrt{2} - (\sqrt{2} - 1)x)^2 = 1$, which simplifies to $(2 - \sqrt{2})x^2 - (3 - \sqrt{2})x + 1 = 0$.
$x = 1$ is a root; the product of the roots is $\frac{1}{2 - \sqrt{2}} = 1 + \frac{1}{\sqrt{2}}$, so the other root is $x = 1 + \frac{1}{\sqrt{2}}$.
So $S$ has two points: $1 + i$ (with $|z|^2 = 2$) and $\left(1 + \frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}}i$ (with $|z|^2 = 1 + \sqrt{2} + \frac{1}{2} + \frac{1}{2} = 2 + \sqrt{2}$).
Hence $z_1 = \left(1 + \frac{1}{\sqrt{2}}\right) + \frac{i}{\sqrt{2}}$ and $z_2 = 1 + i$.
$\sqrt{2}z_1 = (\sqrt{2} + 1) + i$, so $\sqrt{2}z_1 - z_2 = \sqrt{2}$ and $|\sqrt{2}z_1 - z_2|^2 = 2$.
Answer: (B). | ||||
| 2 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Since $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 1$, put $t = (\sqrt{3} + \sqrt{2})^x > 0$; then $(\sqrt{3} - \sqrt{2})^x = \frac{1}{t}$.
$t + \frac{1}{t} = 10 \Rightarrow t^2 - 10t + 1 = 0 \Rightarrow t = 5 \pm 2\sqrt{6}$, both positive.
$(\sqrt{3} + \sqrt{2})^2 = 5 + 2\sqrt{6}$, so $t = 5 + 2\sqrt{6}$ gives $x = 2$ and $t = 5 - 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^{-2}$ gives $x = -2$.
S has 2 elements.
Answer: (C). | ||||
| 3 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\det A = 2 + 1 = 3$, $\det B = 1$, $\det A^T = \det A = 3$.
$\det C = \det A \cdot \det B \cdot \det A^T = 9$.
$\det X = \det A^T \cdot (\det C)^2 \cdot \det A = 3 \times 81 \times 3 = 729$.
Answer: (B). | ||||
| 4 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The offices are indistinguishable, so we count the ways of splitting 5 distinct people into at most 4 non-empty groups (empty offices are not told apart).
Into 1 group: 1 way. Into 2 groups: $\frac{2^5 - 2}{2} = 15$. Into 3 groups: sizes (3,1,1) give $\binom{5}{3} = 10$ and sizes (2,2,1) give $\frac{1}{2}\binom{5}{2}\binom{3}{2} = 15$, total 25. Into 4 groups: sizes (2,1,1,1) give $\binom{5}{2} = 10$.
$n = 1 + 15 + 25 + 10 = 51$.
Answer: (A). | ||||
| 5 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For infinitely many solutions every one of $D, D_x, D_y, D_z$ must vanish.
$D_z = \begin{vmatrix}2 & 3 & 5\\1 & \alpha & -4\\3 & -1 & 7\end{vmatrix} = 2(7\alpha - 4) - 3(7 + 12) + 5(-1 - 3\alpha) = -\alpha - 70 = 0$, so $\alpha = -70$.
$D = \begin{vmatrix}2 & 3 & -1\\1 & \alpha & 3\\3 & -1 & \beta\end{vmatrix} = 2(\alpha\beta + 3) - 3(\beta - 9) - (-1 - 3\alpha) = 2\alpha\beta - 3\beta + 3\alpha + 34 = 0$.
With $\alpha = -70$: $-140\beta - 3\beta - 210 + 34 = 0$, so $\beta = -\frac{176}{143} = -\frac{16}{13}$.
Check: $D_x = 5(\alpha\beta + 3) + 3(4\beta + 21) - (4 - 7\alpha) = 5\alpha\beta + 12\beta + 7\alpha + 74 = \frac{5600 - 192 - 6370 + 962}{13} = 0$.
$13\alpha\beta = 13 \times (-70) \times \left(-\frac{16}{13}\right) = 1120$.
Answer: (B). | ||||
| 6 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the common difference be $d$: $a = 3 + d$, $b = 3 + 2d$, $c = 3 + 3d$.
The G.P. is $3,\ 2 + d,\ 4 + 2d,\ 12 + 3d$.
$(2 + d)^2 = 3(4 + 2d) \Rightarrow d^2 - 2d - 8 = 0 \Rightarrow d = 4$ or $d = -2$.
$d = -2$ gives the terms $3, 0, 0, 6$, not a G.P. With $d = 4$: $3, 6, 12, 24$, a G.P. with ratio 2.
So $a = 7$, $b = 11$, $c = 15$ and their arithmetic mean is $\frac{33}{3} = 11$.
Answer: (A). | ||||
| 7 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Replace $x$ by $\frac{1}{x}$: $5f\left(\frac{1}{x}\right) + 4f(x) = \frac{1}{x^2} - 2$.
Multiply the given equation by 5 and this one by 4 and subtract: $9f(x) = 5x^2 - 10 - \frac{4}{x^2} + 8 = 5x^2 - 2 - \frac{4}{x^2}$.
So $y = 9x^2f(x) = 5x^4 - 2x^2 - 4$ (for $x \ne 0$).
$y' = 20x^3 - 4x = 4x(5x^2 - 1) > 0$ when $-\frac{1}{\sqrt{5}} < x < 0$ or $x > \frac{1}{\sqrt{5}}$.
Answer: (C). | ||||
| 8 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Continuity at 0: $f(0) = 2$, so $\lim_{x \to 0^-}\frac{a - b\cos 2x}{x^2}$ must be finite, which needs $a = b$. Then $\frac{a(1 - \cos 2x)}{x^2} = \frac{2a\sin^2 x}{x^2} \to 2a$, so $2a = 2$, $a = b = 1$.
Continuity at 1: $1 + c + 2 = 2(1) + 1 = 3$, so $c = 0$.
Differentiability at 0: for $x < 0$, $f(x) = \frac{2\sin^2 x}{x^2} = 2 - \frac{2x^2}{3} + \ldots$, so the left derivative is 0; the right derivative of $x^2 + 2$ is 0. Differentiable.
At 1: left derivative $2x = 2$, right derivative 2. Differentiable.
Elsewhere each piece is differentiable, so $m = 0$.
$m + a + b + c = 0 + 1 + 1 + 0 = 2$.
Answer: (B). | ||||
| 9 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The first curve is $y = \frac{16}{x + 4}$. Meeting $y = 6 - x$: $(6 - x)(x + 4) = 16 \Rightarrow x^2 - 2x - 8 = 0 \Rightarrow x = -2, 4$.
Between them the line lies above the curve, so
Area $= \int_{-2}^{4}\left(6 - x - \frac{16}{x + 4}\right)dx = \left[6x - \frac{x^2}{2}\right]_{-2}^{4} - 16\log_e\frac{8}{2}$
$= (24 - 8) - (-12 - 2) - 16\log_e 4 = 30 - 32\log_e 2$.
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 Replacing $x$ by $\frac{\pi}{4} - x$ swaps $\sin 2x$ and $\cos 2x$, so the denominator is unchanged and
$2I = \frac{\pi}{4}\int_0^{\pi/4}\frac{dx}{\sin^4 2x + \cos^4 2x}$. With $t = 2x$: $I = \frac{\pi}{16}\int_0^{\pi/2}\frac{dt}{\sin^4 t + \cos^4 t}$.
Divide numerator and denominator by $\cos^4 t$ and put $u = \tan t$: $\int_0^{\pi/2}\frac{\sec^4 t\,dt}{\tan^4 t + 1} = \int_0^{\infty}\frac{1 + u^2}{1 + u^4}du$.
$\int_0^{\infty}\frac{1 + u^{-2}}{(u - u^{-1})^2 + 2}du = \left[\frac{1}{\sqrt{2}}\tan^{-1}\frac{u - u^{-1}}{\sqrt{2}}\right]_0^{\infty} = \frac{1}{\sqrt{2}}\left(\frac{\pi}{2} + \frac{\pi}{2}\right) = \frac{\pi}{\sqrt{2}}$.
$I = \frac{\pi}{16}\cdot\frac{\pi}{\sqrt{2}} = \frac{\sqrt{2}\pi^2}{32}$.
Answer: (B). | ||||
| 11 | 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 Put $u = x + y$: $\frac{du}{dx} = 1 + \frac{dy}{dx} = 2xu^3 - xu$, with $u(0) = 1$.
This is a Bernoulli equation. Put $w = u^{-2}$: $\frac{dw}{dx} = -2u^{-3}\frac{du}{dx} = -4x + 2xw$, i.e. $\frac{dw}{dx} - 2xw = -4x$.
Integrating factor $e^{-x^2}$: $\frac{d}{dx}(we^{-x^2}) = -4xe^{-x^2}$, so $we^{-x^2} = 2e^{-x^2} + C$ and $w = 2 + Ce^{x^2}$.
At $x = 0$: $w = 1$, so $C = -1$ and $\frac{1}{u^2} = 2 - e^{x^2}$.
At $x = \frac{1}{\sqrt{2}}$: $u^2 = \left(\frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right)\right)^2 = \frac{1}{2 - \sqrt{e}}$.
Answer: (A). | ||||
| 12 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Hyperbola: $\frac{x^2}{5} - \frac{y^2}{5\sin^2\theta} = 1$, so $e_H^2 = 1 + \sin^2\theta$.
Ellipse: $\frac{x^2}{5\sin^2\theta} + \frac{y^2}{5} = 1$, major axis along $y$, so $e_E^2 = 1 - \sin^2\theta = \cos^2\theta$.
$e_H^2 = 7e_E^2$: $1 + \sin^2\theta = 7 - 7\sin^2\theta \Rightarrow \sin^2\theta = \frac{3}{4}$, so $\theta = \frac{\pi}{3}$.
Answer: (B). | ||||
| 13 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections For the ellipse $e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2}$, so $\frac{b^2}{a^2} = \frac{1}{2}$ (the latus rectum $\frac{2b^2}{a} = a = \sqrt{14}$ fixes the size but is not needed).
For the hyperbola $e^2 = 1 + \frac{b^2}{a^2} = \frac{3}{2}$.
Answer: (A). | ||||
| 14 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions For $x \ge 1$: $f(x) = \log_e x \ge 0$, so $g(f(x)) = \log_e x$, taking every value in $[0, \infty)$.
For $0 < x < 1$: $f(x) = \log_e x < 0$, so $g(f(x)) = e^{\log_e x} = x$, values in $(0, 1)$.
For $x \le 0$: $f(x) = e^{-x} \ge 1$, so $g(f(x)) = e^{-x}$, values in $[1, \infty)$.
The range is $[0, \infty) \ne \mathbb{R}$: not onto.
$g(f(e^2)) = 2 = g(f(-\log_e 2))$: not one-one.
Answer: (D). | ||||
| 15 | CirclesStandard and general form | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections C: centre $(0, 0)$, $r_1 = 2$. C': centre $(2\lambda, 0)$, $r_2 = \sqrt{4\lambda^2 - 9}$ (needs $|\lambda| > \frac{3}{2}$). Distance of centres $d = 2|\lambda|$.
Two distinct points need $|r_1 - r_2| < d < r_1 + r_2$.
Since $r_2 < 2|\lambda|$ and $2 < 2|\lambda|$, the left inequality always holds. The right one: $2|\lambda| - 2 < \sqrt{4\lambda^2 - 9} \Rightarrow 4\lambda^2 - 8|\lambda| + 4 < 4\lambda^2 - 9 \Rightarrow |\lambda| > \frac{13}{8}$.
So the set is $\mathbb{R} - \left[-\frac{13}{8}, \frac{13}{8}\right]$: $a = -\frac{13}{8}$, $b = \frac{13}{8}$.
The point is $(-13 + 12, 26 - 20) = (-1, 6)$, and $6(1) + 36 = 42$: it lies on $6x^2 + y^2 = 42$. (The others give $-1$, $-143$ and $114$ on their left sides.)
Answer: (B). | ||||
| 16 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy 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 Points $A(\lambda, 2, 1)$, $B(\sqrt{3}, 1, 2)$; directions $\vec{b}_1 = (-2, 1, 1)$, $\vec{b}_2 = (1, -2, 1)$.
$\vec{b}_1 \times \vec{b}_2 = (1 + 2,\ 1 + 2,\ 4 - 1) = (3, 3, 3)$, of length $3\sqrt{3}$.
$\vec{AB} = (\sqrt{3} - \lambda, -1, 1)$, and $\vec{AB}\cdot(3, 3, 3) = 3(\sqrt{3} - \lambda)$.
Shortest distance $= \frac{|\sqrt{3} - \lambda|}{\sqrt{3}} = 1 \Rightarrow \lambda = 0$ or $\lambda = 2\sqrt{3}$.
Sum $= 2\sqrt{3}$.
Answer: (D). | ||||
| 17 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Take every number $w = 0, 1, \ldots, 8$ of white balls as equally likely beforehand.
The draw (2 white, 2 black) is possible only for $w = 2, \ldots, 6$, with likelihood $\frac{\binom{w}{2}\binom{8 - w}{2}}{\binom{8}{4}}$.
Numerators: $w = 2$: $1 \times 15 = 15$; $w = 3$: $3 \times 10 = 30$; $w = 4$: $6 \times 6 = 36$; $w = 5$: 30; $w = 6$: 15. Total 126.
By Bayes' theorem, $P(w = 4 \mid \text{draw}) = \frac{36}{126} = \frac{2}{7}$.
Answer: (D). | ||||
| 18 | StatisticsMean deviation | Class 11: Statistics | Single correctMedium | |
A medium single correct question on mean deviation, from Statistics. NCERT: Class 11: Statistics Three of the known values (190, 210, 230) exceed 170, so for 170 to be the 4th of 7 values, $a, b \le 170$.
Mean deviation about the median: $45 + 0 + 60 + 20 + 40 + (170 - a) + (170 - b) = 205$, so $a + b = 300$.
Mean $= \frac{170 + 125 + 230 + 190 + 210 + 300}{7} = \frac{1225}{7} = 175$.
Deviations from 175: $5, 50, 55, 15, 35$, and $(175 - a) + (175 - b) = 350 - 300 = 50$ (both $a, b \le 170$). Total $= 210$.
Mean deviation about the mean $= \frac{210}{7} = 30$.
Answer: (B). | ||||
| 19 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\-5 & 1 & -3\\1 & 2 & -4\end{vmatrix} = (-4 + 6)\hat{i} - (20 + 3)\hat{j} + (-10 - 1)\hat{k} = 2\hat{i} - 23\hat{j} - 11\hat{k}$.
For any $\vec{v} = (v_1, v_2, v_3)$: $\vec{v} \times \hat{i} = (0, v_3, -v_2)$.
$(2, -23, -11) \times \hat{i} = (0, -11, 23)$; $\times \hat{i}$ again: $(0, 23, 11)$; $\times \hat{i}$ again: $\vec{c} = (0, 11, -23)$.
$\vec{c}\cdot(-1, 1, 1) = 11 - 23 = -12$.
Answer: (B). | ||||
| 20 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $s = \sqrt{x^2 + x + 1}$. $\tan A + \tan B = \frac{1 + x}{\sqrt{x}\,s}$ and $\tan A\tan B = \frac{1}{s^2}$.
$\tan(A + B) = \frac{(1 + x)/(\sqrt{x}\,s)}{(s^2 - 1)/s^2} = \frac{(1 + x)s}{\sqrt{x}\,x(x + 1)} = \frac{s}{x\sqrt{x}}$.
$\tan C = \sqrt{\frac{1 + x + x^2}{x^3}} = \frac{s}{x\sqrt{x}}$, so $\tan(A + B) = \tan C$.
As $\tan A\tan B < 1$, $A + B < \frac{\pi}{2}$, and $C$ is also in $(0, \frac{\pi}{2})$; hence $A + B = C$.
Answer: (A). | ||||
| 21 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $R_1$: $a$ divides $b$. $R_2$: $b$ divides $a$. A pair of $R_1$ is also in $R_2$ exactly when $a \mid b$ and $b \mid a$, i.e. $a = b$.
Number of pairs with $a \mid b$: $\sum_{a=1}^{20}\left\lfloor\frac{20}{a}\right\rfloor = 20 + 10 + 6 + 5 + 4 + 3 + 2 + 2 + 2 + 2 + 10 \times 1 = 66$.
Removing the 20 pairs $(a, a)$: $|R_1 - R_2| = 66 - 20 = 46$.
Answer: 46. | ||||
| 22 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | NumericalHard | |
A hard numerical question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations P is the disc with centre $C(-2, 3)$ and radius 1. With $z = x + iy$, $z(1 + i) + \bar{z}(1 - i) = 2(x - y)$, so Q is the half-plane $x - y \le -4$.
C satisfies $-5 \le -4$, and its distance from $x - y + 4 = 0$ is $\frac{1}{\sqrt{2}} < 1$, so $P \cap Q$ is the disc minus a cap.
We measure distance from $A(3, -2)$. $\vec{CA} = (5, -5)$ is along $(1, -1)$, perpendicular to the chord line.
Farthest point: $C$ moved 1 unit away from A, $z_1 = \left(-2 - \frac{1}{\sqrt{2}}\right) + \left(3 + \frac{1}{\sqrt{2}}\right)i$, which lies in Q ($x - y = -5 - \sqrt{2}$).
$|z_1|^2 = \left(2 + \frac{1}{\sqrt{2}}\right)^2 + \left(3 + \frac{1}{\sqrt{2}}\right)^2 = 14 + 5\sqrt{2}$.
Nearest point: the disc point nearest A is cut off (it has $x - y = -5 + \sqrt{2} > -4$), so the minimum is on the chord, at the foot of the perpendicular from A to $x - y = -4$, which is $(-1.5, 2.5)$, inside the disc. So $z_2 = -\frac{3}{2} + \frac{5}{2}i$, $|z_2|^2 = \frac{17}{2}$.
$|z_1|^2 + 2|z_2|^2 = 31 + 5\sqrt{2}$: $\alpha = 31$, $\beta = 5$, $\alpha + \beta = 36$.
Answer: 36. | ||||
| 23 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem A general term is $\binom{6}{i}x^{-i}\cdot\binom{7}{j}x^{2j}\cdot\binom{8}{k}(-1)^kx^{3k}$; we need $-i + 2j + 3k = 30$ with $0 \le i \le 6$, $0 \le j \le 7$, $0 \le k \le 8$.
$k = 6$ ($2j - i = 12$): $(i, j) = (0, 6), (2, 7)$: $\binom{8}{6}\left[\binom{6}{0}\binom{7}{6} + \binom{6}{2}\binom{7}{7}\right] = 28(7 + 15) = 616$.
$k = 7$ ($2j - i = 9$): $(1, 5), (3, 6), (5, 7)$: $-8\left[6 \cdot 21 + 20 \cdot 7 + 6 \cdot 1\right] = -8 \times 272 = -2176$.
$k = 8$ ($2j - i = 6$): $(0, 3), (2, 4), (4, 5), (6, 6)$: $1\left[35 + 15 \cdot 35 + 15 \cdot 21 + 1 \cdot 7\right] = 882$.
($k \le 5$ would need $2j - i \ge 15$, impossible as $2j \le 14$.)
$\alpha = 616 - 2176 + 882 = -678$, so $|\alpha| = 678$.
Answer: 678. | ||||
| 24 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations For each $z = 0, 1, \ldots, 14$, the number of $(x, y)$ with $x + 2y = 42 - 3z$ is $\left\lfloor\frac{42 - 3z}{2}\right\rfloor + 1$.
These are $22, 20, 19, 17, 16, 14, 13, 11, 10, 8, 7, 5, 4, 2, 1$ for $z = 0, \ldots, 14$.
Sum $= 169$.
Answer: 169. | ||||
| 25 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The first A.P. is the numbers $\equiv 3 \pmod 4$ from 3 to 403; the second is the numbers $\equiv 2 \pmod 3$ from 2 to 404.
Common terms are $\equiv 11 \pmod{12}$: $11, 23, 35, \ldots$ up to 403, the last being $11 + 12 \times 32 = 395$.
That is 33 terms, with sum $\frac{33}{2}(11 + 395) = 33 \times 203 = 6699$.
Answer: 6699. | ||||
| 26 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Right of 0: $\{x\} = x \to 0^+$. Write $f = \frac{\cos^{-1}(1 - x^2)}{x}\cdot\frac{\sin^{-1}(1 - x)}{1 - x^2}$.
With $\cos^{-1}(1 - x^2) = \theta$: $x^2 = 1 - \cos\theta = 2\sin^2\frac{\theta}{2}$, so $\frac{\theta}{x} \to \sqrt{2}$. The second factor $\to \frac{\pi}{2}$. So $R = \frac{\pi}{\sqrt{2}}$.
Left of 0: $\{x\} = 1 + x$; put $h = -x \to 0^+$, $\{x\} = 1 - h$.
$1 - \{x\}^2 = 2h - h^2 \to 0$, so $\cos^{-1}(1 - \{x\}^2) \to \frac{\pi}{2}$; $\sin^{-1}(1 - \{x\}) = \sin^{-1}h \approx h$; the denominator $\{x\}(1 - \{x\}^2) = (1 - h)(2h - h^2) \approx 2h$.
So $L = \frac{\pi}{2}\cdot\frac{h}{2h} = \frac{\pi}{4}$.
$\frac{32}{\pi^2}\left(\frac{\pi^2}{16} + \frac{\pi^2}{2}\right) = 2 + 16 = 18$.
Answer: 18. | ||||
| 27 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Add the integrand at $x$ and $-x$: $\frac{1}{1 + e^{s}} + \frac{1}{1 + e^{-s}} = 1$, so
$I = \int_0^{\pi/2}\frac{8\sqrt{2}\cos x}{1 + \sin^4 x}dx = 8\sqrt{2}\int_0^1\frac{dt}{1 + t^4}$ ($t = \sin x$).
$\frac{1}{1 + t^4} = \frac{1}{2}\cdot\frac{1 + t^2}{1 + t^4} + \frac{1}{2}\cdot\frac{1 - t^2}{1 + t^4}$.
$\int_0^1\frac{1 + t^{-2}}{(t - t^{-1})^2 + 2}dt = \left[\frac{1}{\sqrt{2}}\tan^{-1}\frac{t - t^{-1}}{\sqrt{2}}\right]_0^1 = \frac{\pi}{2\sqrt{2}}$.
$\frac{1 - t^2}{1 + t^4} = -\frac{1 - t^{-2}}{(t + t^{-1})^2 - 2}$. With $u = t + t^{-1}$ (running from $\infty$ down to 2 as $t$ goes from 0 to 1): $\int_0^1\frac{1 - t^2}{1 + t^4}dt = \int_2^\infty\frac{du}{u^2 - 2} = \frac{1}{2\sqrt{2}}\log_e\frac{2 + \sqrt{2}}{2 - \sqrt{2}} = \frac{1}{2\sqrt{2}}\log_e(3 + 2\sqrt{2})$.
So $\int_0^1\frac{dt}{1 + t^4} = \frac{\pi}{4\sqrt{2}} + \frac{\log_e(3 + 2\sqrt{2})}{4\sqrt{2}}$ (about 0.867).
$I = 8\sqrt{2}\cdot\frac{\pi + \log_e(3 + 2\sqrt{2})}{4\sqrt{2}} = 2\pi + 2\log_e(3 + 2\sqrt{2})$.
$\alpha = \beta = 2$, $\alpha^2 + \beta^2 = 8$.
Answer: 8. | ||||
| 28 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalEasy | |
A easy numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dx}{dt} - \frac{2x}{t + 1} = (t + 1)^3$. Integrating factor $\frac{1}{(t + 1)^2}$:
$\frac{d}{dt}\left(\frac{x}{(t + 1)^2}\right) = t + 1 \Rightarrow \frac{x}{(t + 1)^2} = \frac{(t + 1)^2}{2} + C$.
$x(0) = 2$: $2 = \frac{1}{2} + C$, $C = \frac{3}{2}$.
$x(1) = 4\left(2 + \frac{3}{2}\right) = 14$.
Answer: 14. | ||||
| 29 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections $x^2 = 2y$ in $x^2 + y^2 = 3$: $y^2 + 2y - 3 = 0$, $y = 1$ (as $y \ge 0$), $x = \sqrt{2}$. So $P = (\sqrt{2}, 1)$ and $\alpha = 2 + 1 = 3$.
A centre $(0, k)$ is at distance $\frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3}$ from L, so $k = 9$ or $-3$: $Q_1 = (0, 9)$, $Q_2 = (0, -3)$.
Area of $PQ_1Q_2 = \frac{1}{2}\times 12 \times \sqrt{2} = 6\sqrt{2}$; its square is 72.
Answer: 72. | ||||
| 30 | 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 $P = (1 + \lambda, 2 - \lambda, 3 + \lambda)$, $Q = (4 + \mu, 5 + \mu, 6 - \mu)$, $\vec{PQ} = (3 + \mu - \lambda,\ 3 + \mu + \lambda,\ 3 - \mu - \lambda)$.
$\vec{PQ}\cdot(1, -1, 1) = 3 - \mu - 3\lambda = 0$ and $\vec{PQ}\cdot(1, 1, -1) = 3 + 3\mu + \lambda = 0$.
Solving: $\lambda = \frac{3}{2}$, $\mu = -\frac{3}{2}$. So $P = \left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right)$, $Q = \left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right)$.
$\alpha + \beta + \gamma = \frac{1}{2}\left(\frac{15}{2} + \frac{27}{2}\right) = \frac{21}{2}$, so $2(\alpha + \beta + \gamma) = 21$.
Answer: 21. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Angular impulse = torque × time = change in angular momentum.
$[\tau t] = [\mathrm{M L^2 T^{-2}}][\mathrm{T}] = [\mathrm{M L^2 T^{-1}}]$.
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 The speed is $v = \frac{2\pi R}{T}$.
Maximum height $H = \frac{v^2\sin^2\theta}{2g} = 4R \Rightarrow \sin^2\theta = \frac{8gR}{v^2} = \frac{8gRT^2}{4\pi^2R^2} = \frac{2gT^2}{\pi^2R}$.
$\theta = \sin^{-1}\left[\frac{2gT^2}{\pi^2R}\right]^{1/2}$.
Answer: (B). | ||||
| 33 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Friction on the 20 kg trolley: $f_k = \mu mg = 0.04 \times 20 \times 10 = 8$ N.
The hanging 6 kg block pulls with its weight, 60 N.
For the whole system: $a = \frac{60 - 8}{20 + 6} = \frac{52}{26} = 2\ \mathrm{m\,s^{-2}}$.
Answer: (D). | ||||
| 34 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion For a ball whirled in a horizontal circle on a string of length $l$ (the string making angle $\theta$ with the vertical), the horizontal component gives $T\sin\theta = m\omega^2(l\sin\theta)$, so $T = m\omega^2 l$.
$400 = 0.5 \times \omega^2 \times 0.5 \Rightarrow \omega^2 = 1600 \Rightarrow \omega = 40$ rad/s.
Answer: (B). | ||||
| 35 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Momentum is conserved in the (perfectly inelastic) collision: $0.01 \times 200 = 1.01\,v \Rightarrow v = \frac{2}{1.01} \approx 1.98$ m/s.
Then energy is conserved during the swing: $h = \frac{v^2}{2g} = \frac{(1.98)^2}{20} \approx 0.196$ m $\approx 0.20$ m.
Answer: (A). | ||||
| 36 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At height $2R$ the distance from the centre is $3R$, so $g' = \frac{g}{9} = \frac{\pi^2}{9}$.
A second's pendulum has period 2 s: $2 = 2\pi\sqrt{\frac{L}{g'}} \Rightarrow L = \frac{g'}{\pi^2} = \frac{1}{9}$ m.
Answer: (D). | ||||
| 37 | 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 As temperature rises, the interatomic forces weaken (the atoms vibrate more and their mean separation grows), so a smaller stress produces the same strain.
Young's modulus therefore decreases with rise in temperature.
Answer: (A). | ||||
| 38 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics For a polytropic process $PV^n = K$ with $n = \frac{3}{2}$:
$W = \int_{V_1}^{V_2} KV^{-n}dV = \frac{K\left(V_2^{1 - n} - V_1^{1 - n}\right)}{1 - n} = \frac{P_2V_2 - P_1V_1}{1 - n} = \frac{P_1V_1 - P_2V_2}{n - 1}$.
With $n - 1 = \frac{1}{2}$: $W = 2(P_1V_1 - P_2V_2)$ (work done by the gas).
Answer: (C). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $C_V$ is $\frac{3}{2}R$ for the monoatomic gas and $\frac{5}{2}R$ for the diatomic gas.
$C_{V,\text{mix}} = \frac{2 \times \frac{3}{2}R + 6 \times \frac{5}{2}R}{2 + 6} = \frac{3R + 15R}{8} = \frac{9}{4}R$.
Answer: (B). | ||||
| 40 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charges $CV$ and $2CV$; joining like plates gives common potential $V' = \frac{CV + 2CV}{2C} = \frac{3V}{2}$.
Initial energy $= \frac{1}{2}CV^2 + \frac{1}{2}C(2V)^2 = \frac{5}{2}CV^2$. Final energy $= \frac{1}{2}(2C)\left(\frac{3V}{2}\right)^2 = \frac{9}{4}CV^2$.
Decrease $= \frac{5}{2}CV^2 - \frac{9}{4}CV^2 = \frac{1}{4}CV^2$.
Answer: (B). | ||||
| 41 | 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 The circuit is a single loop of 8 identical cells (emf 5 V, internal resistance $0.2\ \Omega$), all connected in the same sense; the ideal voltmeter draws no current.
Current $I = \frac{8 \times 5}{8 \times 0.2} = \frac{40}{1.6} = 25$ A.
The voltmeter reads the terminal voltage of one cell: $V = E - Ir = 5 - 25 \times 0.2 = 0$ V.
Answer: (C). | ||||
| 42 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Full-scale current through galvanometer and series resistor: $I_g(G + R) = 100$ V.
$R = \frac{100}{5 \times 10^{-3}} - 50 = 20000 - 50 = 19950\ \Omega$.
Answer: (A). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Resonance frequency $\omega_0 = \frac{1}{\sqrt{LC}}$ stays the same if $LC$ stays the same: $L' \times 4C = LC$, so $L' = \frac{L}{4}$.
The inductance is reduced by $L - \frac{L}{4} = \frac{3}{4}L$.
Answer: (D). | ||||
| 44 | Electromagnetic WavesDisplacement current | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $I_{rms} = \frac{V_{rms}}{X_C} = V_{rms}\,\omega C = 230 \times 300 \times 200 \times 10^{-12} = 1.38 \times 10^{-5}$ A $= 13.8\ \mu$A.
The displacement current between the plates equals the conduction current in the wires, so it is also $13.8\ \mu$A.
Answer: (A). | ||||
| 45 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Width of the central maximum $= \frac{2\lambda f}{a} = \frac{2 \times 6 \times 10^{-7} \times 0.2}{1 \times 10^{-5}} = 2.4 \times 10^{-2}$ m $= 24$ mm.
Answer: (B). | ||||
| 46 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{mv} \Rightarrow v = \frac{h}{m\lambda}$.
$\frac{v_p}{v_\alpha} = \frac{m_\alpha\lambda_\alpha}{m_p\lambda_p} = \frac{4m \times 2\lambda}{m \times \lambda} = 8$.
Ratio $8 : 1$.
Answer: (C). | ||||
| 47 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Balmer lines end on $n = 2$, so the atom must first be excited to at least $n = 3$.
Energy needed: $13.6\left(1 - \frac{1}{9}\right) = 12.09 \approx 12.1$ eV.
Answer: (C). | ||||
| 48 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctHard | |
A hard single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Maximum Zener current: $I_{Z,\max} = \frac{P}{V_Z} = \frac{10\ \text{mW}}{5\ \text{V}} = 2$ mA. Load current: $I_L = \frac{5\ \text{V}}{1\ \text{k}\Omega} = 5$ mA. Voltage across $R_s$: $8 - 5 = 3$ V.
The source current is $I_s = I_L + I_Z$. The Zener stays within its rating only if $I_s \le 7$ mA, i.e. $R_s \ge \frac{3}{7}$ k$\Omega \approx 429\ \Omega$, and it stays in breakdown only if $\frac{8 \times 1}{R_s + 1} \ge 5$ (in k$\Omega$), i.e. $R_s \le 600\ \Omega$.
None of 10 k$\Omega$, 1 k$\Omega$, 10 $\Omega$, 5 k$\Omega$ lies in 429–600 $\Omega$: with 1 k$\Omega$ the load voltage is only 4 V (no regulation), and 10 $\Omega$ would burn out the Zener.
No option is correct as printed. | ||||
| 49 | 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 $11$ VSD $= 10$ MSD, so $1$ VSD $= \frac{10}{11}$ MSD.
Least count $= 1$ MSD $- 1$ VSD $= \frac{1}{11}$ MSD $= \frac{5}{11}$ units.
Answer: (B). | ||||
| 50 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\rho = \frac{R\pi r^2}{l}$, so $\frac{\Delta\rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$.
$= \frac{10}{100} + 2 \times \frac{0.05}{0.35} + \frac{0.2}{15} = 0.1 + 0.2857 + 0.0133 = 0.399$.
Percentage error $\approx 39.9\%$.
Answer: (A). | ||||
| 51 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $v = \frac{dx}{dt} = -9t^2 + 36t + 16$ and $a = \frac{dv}{dt} = -18t + 36$.
$a = 0$ at $t = 2$ s, when $v = -36 + 72 + 16 = 52$ m/s.
Answer: 52. | ||||
| 52 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The three equal masses are at $(0, 0)$, $(4, 0)$ and $(0, 4)$.
Centre of mass $= \left(\frac{4}{3}, \frac{4}{3}\right)$, at distance $\frac{4\sqrt{2}}{3}$ from the origin.
So $x = 3$.
Answer: 3. | ||||
| 53 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Speeds: $180$ km/h $= 50$ m/s (below), $252$ km/h $= 70$ m/s (above).
By Bernoulli's principle, $\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}(4900 - 2500) = 1200$ Pa.
Lift on both wings $= 1200 \times 2 \times 40 = 96000$ N $= mg$, so $m = 9600$ kg.
Answer: 9600. | ||||
| 54 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For the same wire and length, the frequency is proportional to $\sqrt{T}$.
At 6 N the wire's frequency equals the fork's, $f$. At 54 N it is $f\sqrt{\frac{54}{6}} = 3f$.
Beats: $3f - f = 12 \Rightarrow f = 6$ Hz.
Answer: 6. | ||||
| 55 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In equilibrium $\tan\frac{\theta}{2} = \frac{F}{W}$, with $F$ the electric force and $W$ the (effective) weight.
In water the electric force becomes $\frac{F}{K}$ and the effective weight $W\left(1 - \frac{\rho_w}{\rho}\right) = W\left(1 - \frac{1}{1.5}\right) = \frac{W}{3}$.
Same angle: $\frac{F/K}{W/3} = \frac{F}{W} \Rightarrow K = 3$.
Answer: 3. | ||||
| 56 | Current ElectricityDrift velocity and mobility | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity $q = \int_1^2 (3t^2 + 4t^3)dt = \left[t^3 + t^4\right]_1^2 = (8 + 16) - (1 + 1) = 22$ C.
Answer: 22. | ||||
| 57 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Each side is $a = \frac{4\pi}{6} = \frac{2\pi}{3}$ m; its distance from the centre is $d = \frac{a}{2}\cot 30^\circ = \frac{\sqrt{3}}{2}a = \frac{\pi}{\sqrt{3}}$ m. Each side subtends $30^\circ$ on either side of the perpendicular.
Field of one side: $\frac{\mu_0I}{4\pi d}(\sin 30^\circ + \sin 30^\circ) = \frac{\mu_0I}{4\pi d}$.
Total: $B = 6 \times 10^{-7} \times \frac{I}{d} = 6 \times 10^{-7} \times \frac{4\pi\sqrt{3}}{\pi/\sqrt{3}} = 6 \times 10^{-7} \times 12 = 72 \times 10^{-7}$ T.
Answer: 72. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalHard | |
A hard numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Area $A = 12 \times 5 = 60\ \mathrm{cm^2} = 6 \times 10^{-3}\ \mathrm{m^2}$.
The field decreases along $+x$ at $10^{-3}$ T/cm $= 0.1$ T/m, and the loop moves along $+x$ at $0.05$ m/s, so the field it sees falls at $0.1 \times 0.05 = 5 \times 10^{-3}$ T/s from the motion, plus $10^{-3}$ T/s from the change in time.
Both reduce the flux: $\left|\frac{d\Phi}{dt}\right| = A(5 \times 10^{-3} + 10^{-3}) = 6 \times 10^{-3} \times 6 \times 10^{-3} = 3.6 \times 10^{-5}$ V.
Power $= \frac{\varepsilon^2}{R} = \frac{(3.6 \times 10^{-5})^2}{6 \times 10^{-3}} = \frac{1.296 \times 10^{-9}}{6 \times 10^{-3}} = 2.16 \times 10^{-7}$ W $= 216 \times 10^{-9}$ W.
Answer: 216. | ||||
| 59 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The virtual image is on the same side as the object: $u = -x$, $v = -3x$, and the separation is $2x = 20$ cm, so $x = 10$ cm.
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = -\frac{1}{30} + \frac{1}{10} = \frac{2}{30}$, so $f = 15$ cm.
Answer: 15. | ||||
| 60 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $R \propto A^{1/3}$, so $A = 64\left(\frac{4}{4.8}\right)^3 = 64 \times \frac{125}{216} = \frac{1000}{27}$.
So $x = 27$.
Answer: 27. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom de Broglie: $\lambda = \frac{h}{p}$, so $\lambda$ is inversely proportional to $p$.
A plot of $\lambda$ against $p$ is a rectangular hyperbola (option A). Against $\frac{1}{p}$ it would be a straight line through the origin, not a curve (so C is wrong), and $\lambda$ does not rise with $p$ (B and D are wrong).
Answer: (A). | ||||
| 62 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Ionic character grows with the electronegativity difference (Pauling values) of the bonded atoms:
N–N: 0; Cl–F: $3.98 - 3.16 = 0.82$; S–O: $3.44 - 2.58 = 0.86$; K–O: $3.44 - 0.82 = 2.62$; Li–F: $3.98 - 0.98 = 3.00$.
Order: $\mathrm{N_2 < ClF_3 < SO_2 < K_2O < LiF}$.
Answer: (B). | ||||
| 63 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $q = 0$. Free expansion (against zero external pressure): $w = 0$.
So $\Delta U = q + w = 0$, and for an ideal gas $U$ depends only on $T$, so $\Delta T = 0$.
Answer: (C). | ||||
| 64 | Solutionsvan't Hoff factor | Class 12: Solutions | Single correctMedium | |
A medium single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions In a real NaCl solution the ions attract each other (ion pairing / interionic attraction), so $i$ is somewhat below 2, and this effect weakens on dilution.
As the solution gets more dilute, $i$ approaches its limiting value 2. So the most dilute solution C has the largest $i$: $i_A < i_B < i_C$.
Answer: (A). | ||||
| 65 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties $\mathrm{F^-}$, Ne and $\mathrm{Na^+}$ all have 10 electrons in the same shells ($n = 2$ outermost), so neither $n$ nor the electron count differs.
What differs is the nuclear charge (9, 10, 11): the larger it is, the stronger the pull on the same electron cloud, and the smaller the species: $\mathrm{F^- > Ne > Na^+}$.
Answer: (B). | ||||
| 66 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | Assertion and reasonEasy | |
A easy assertion and reason question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) is correct: $\mathrm{NH_3}$ (b.p. about 240 K) boils higher than $\mathrm{PH_3}$ (about 185 K).
(R) has it the wrong way round: $\mathrm{NH_3}$ molecules are associated by hydrogen bonding (N is small and highly electronegative), while $\mathrm{PH_3}$ molecules are held only by van der Waals forces. So (R) is not correct.
Answer: (C). | ||||
| 67 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions In acid, dichromate is reduced to $\mathrm{Cr^{3+}}$: each Cr goes from +6 to +3, so 2 Cr take $6e^-$.
The 7 O atoms become 7 $\mathrm{H_2O}$, needing $14\mathrm{H^+}$:
$\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$ (charge: $-2 + 14 - 6 = +6$ on both sides).
X = 14, Y = 6, Z = 7, A = $\mathrm{Cr^{3+}}$.
Answer: (C). | ||||
| 68 | Redox ReactionsOxidation number | Class 11: Redox Reactions | StatementsMedium | |
A medium statements question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions Disproportionation: one species in one oxidation state is both oxidised and reduced.
(A) Cu(+1) gives Cu(+2) and Cu(0): disproportionation.
(B) Mn(+6) in $\mathrm{MnO_4^{2-}}$ gives Mn(+7) in $\mathrm{MnO_4^-}$ and Mn(+4) in $\mathrm{MnO_2}$: disproportionation.
(C) Mn(+7) is reduced (to +6 and +4) while oxygen (−2) is oxidised to $\mathrm{O_2}$: two different elements change, so it is an ordinary (intramolecular) redox decomposition, not disproportionation.
(D) Mn(+7) and Mn(+2) both become Mn(+4): this is comproportionation, the reverse.
So (A) and (B).
Answer: (A). | ||||
| 69 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | StatementsEasy | |
A easy statements question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Statement I: $\mathrm{[Ni(H_2O)_6]^{2+}}$ (octahedral $d^8$, two unpaired electrons, weak-field ligand) shows d–d absorption in the visible region and is green. Correct.
Statement II: $\mathrm{CN^-}$ is a strong-field ligand; $\mathrm{[Ni(CN)_4]^{2-}}$ is square planar, low-spin $d^8$ with all electrons paired and a large splitting, so it absorbs little visible light and is described as colourless. Correct.
Answer: (A). | ||||
| 70 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A homoleptic complex has only one kind of ligand around the metal.
The first three carry both $\mathrm{NH_3}$ and $\mathrm{Cl^-}$ (heteroleptic); $\mathrm{[Ni(CN)_4]^{2-}}$ has only $\mathrm{CN^-}$.
Answer: (D). | ||||
| 71 | 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 Kjeldahl's method the compound is digested with conc. $\mathrm{H_2SO_4}$ to convert its nitrogen into ammonium sulphate. $\mathrm{CuSO_4}$ is added as a catalyst to speed up the digestion ($\mathrm{K_2SO_4}$ raises the boiling point).
Answer: (C). | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsEasy | |
A easy statements question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Aniline is the common name of $\mathrm{C_6H_5NH_2}$, and aminobenzene (IUPAC: benzenamine) names the same compound.
So Statement I is correct and Statement II is incorrect.
Answer: (C). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsEasy | |
A easy statements question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In heterolytic cleavage the shared pair stays with one atom, giving ions (carbocations, carbanions); reactions that go through such ions are called ionic (polar) reactions.
Homolytic cleavage gives free radicals (primary, secondary, …), which belong to free-radical reactions, not ionic ones.
So only (B).
Answer: (B). | ||||
| 74 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Assertion and reasonEasy | |
A easy assertion and reason question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (A) is correct: KCN is ionic and gives free $\mathrm{CN^-}$, which attacks through the more nucleophilic carbon, giving alkyl cyanides (R–CN); AgCN is mainly covalent, so the nitrogen lone pair attacks, giving isocyanides (R–NC).
(R) is wrong: KCN is ionic but AgCN is predominantly covalent. That difference is exactly the reason for (A).
Answer: (C). | ||||
| 75 | 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 Side-chain chlorination of toluene with $\mathrm{Cl_2}/h\nu$ (controlled) gives benzal chloride, $\mathrm{C_6H_5CHCl_2}$ (A).
Hydrolysis of the gem-dichloride with water at 373 K gives the gem-diol $\mathrm{C_6H_5CH(OH)_2}$, which loses water to give benzaldehyde, $\mathrm{C_6H_5CHO}$ (B).
Answer: (C). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Matching listMedium | |
A medium matching list question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) Ester to aldehyde: partial reduction with DIBAL-H, then water (IV).
(B) Ketone C=O to $\mathrm{CH_2}$: Clemmensen reduction, Zn(Hg) and conc. HCl (II).
(C) Aldehyde to a secondary alcohol with one more carbon: $\mathrm{CH_3MgBr}$, then water (I).
(D) Ketone reduced to an alcohol while the ester survives: $\mathrm{NaBH_4}$ (III), which does not reduce esters.
A→IV, B→II, C→I, D→III.
Answer: (B). | ||||
| 77 | AminesClassification and identification of amines | Class 12: Amines | StatementsEasy | |
A easy statements question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Statement I: the lone pair on N is delocalised into the ring (+R effect), raising electron density at the ortho and para positions; $\mathrm{-NH_2}$ is a strong activating, o/p-directing group. Correct.
Statement II: aniline (a Lewis base) forms a salt with the Lewis acid $\mathrm{AlCl_3}$; the N then carries a positive charge and strongly deactivates the ring, so Friedel–Crafts alkylation and acylation do not occur. Correct.
Answer: (A). | ||||
| 78 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The ring most rich in electrons reacts fastest with an electrophile.
$\mathrm{-OH}$ is a strongly activating group (+R effect of the O lone pair outweighs its −I effect); $\mathrm{-CH_3}$ activates only weakly (+I, hyperconjugation); $\mathrm{-Cl}$ deactivates (−I outweighs +R).
Reactivity: phenol > toluene > benzene > chlorobenzene.
Answer: (C). | ||||
| 79 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Bases pair A with T and G with C.
A T G C T T C A pairs with T A C G A A G T.
Answer: (A). | ||||
| 80 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | StatementsEasy | |
A easy statements question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Statement I: potassium hydrogen phthalate is pure, stable, non-hygroscopic and of high molar mass, so it is a standard primary standard for NaOH. Correct.
Statement II: it is a weak acid titrated with a strong base, so the end point is in the basic range (pH about 8–10), where phenolphthalein changes colour. Correct.
Answer: (A). | ||||
| 81 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles divided by coefficients: $\mathrm{PbCl_2}$: $\frac{72}{3} = 24$; $\mathrm{(NH_4)_3PO_4}$: $\frac{50}{2} = 25$.
$\mathrm{PbCl_2}$ is limiting; 3 mmol of it gives 1 mmol of $\mathrm{Pb_3(PO_4)_2}$, so 72 mmol gives 24 mmol.
Answer: 24. | ||||
| 82 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{PF_5}$, $\mathrm{PCl_5}$: 5 bond pairs, no lone pair, $sp^3d$: trigonal bipyramidal.
$\mathrm{Fe(CO)_5}$: $dsp^3$, trigonal bipyramidal.
$\mathrm{BrF_5}$: 5 bp + 1 lp: square pyramidal. $\mathrm{[PtCl_4]^{2-}}$: $dsp^2$, square planar. $\mathrm{BF_3}$: trigonal planar.
Count = 3.
Answer: 3. | ||||
| 83 | EquilibriumIonic equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium For a salt of a weak acid and a weak base, $\mathrm{pH} = 7 + \frac{1}{2}(\mathrm{p}K_a - \mathrm{p}K_b)$ (at 298 K, $\mathrm{p}K_w = 14$).
Here $K_a = K_b$, so $\mathrm{p}K_a = \mathrm{p}K_b$ and $\mathrm{pH} = 7$.
Answer: 7. | ||||
| 84 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions In $\mathrm{A_2B}$ the atom B has oxidation number −2 (A is +1). A lowest oxidation number of −2 means the atom gains 2 electrons to complete its octet, so it has $8 - 2 = 6$ valence electrons (a group 16 element, like O or S).
Answer: 6. | ||||
| 85 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Nernst equation: $E = E^\circ - \frac{0.06}{2}\log\frac{P_{\mathrm{H_2}}}{[\mathrm{H^+}]^2} = 0 - 0.03\log\frac{2}{1} = -0.03 \times 0.3 = -0.009$ V.
That is $-0.9 \times 10^{-2}$ V; the blank takes the magnitude, 0.9, which to the nearest integer is 1.
Answer: 1. | ||||
| 86 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Radioactive decay is first order. $\frac{1}{8} = \left(\frac{1}{2}\right)^3$, so 3 half-lives have passed.
Age $= 3 \times 5730 = 17190$ years.
Answer: 17190. | ||||
| 87 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Amphoteric: $\mathrm{Al_2O_3}$, $\mathrm{SnO_2}$, $\mathrm{PbO_2}$ (they dissolve in both acids and alkalis).
Acidic: $\mathrm{Cl_2O_7}$, $\mathrm{SiO_2}$, $\mathrm{N_2O_5}$. Neutral: CO, $\mathrm{N_2O}$, NO.
Count = 3.
Answer: 3. | ||||
| 88 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The groups, in order: the ester group attached through its carbonyl carbon ($\mathrm{-COOCH_3}$), the acetamido group $\mathrm{-NHCOCH_3}$, the methylamino group $\mathrm{-NHCH_3}$, the cyano group $\mathrm{-C{\equiv}N}$, and the methoxy group $\mathrm{-OCH_3}$.
$\mathrm{-COOCH_3}$ and $\mathrm{-CN}$ withdraw electrons from the ring (−R, −I): deactivating.
$\mathrm{-NHCOCH_3}$ (moderately), $\mathrm{-NHCH_3}$ and $\mathrm{-OCH_3}$ (strongly) donate their lone pair to the ring (+R): activating.
Deactivating groups = 2.
Answer: 2. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In $\mathrm{CH_3CHClCH_2CH_3}$ carbon 2 carries four different groups (H, Cl, $\mathrm{CH_3}$, $\mathrm{C_2H_5}$): one chiral centre.
Number of optical isomers $= 2^1 = 2$ (the d and l enantiomers).
Answer: 2. | ||||
| 90 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter White: $\mathrm{SrSO_4}$, $\mathrm{Mg(NH_4)PO_4}$, $\mathrm{Mn(OH)_2}$ (white when freshly precipitated, browning in air), $\mathrm{PbSO_4}$, $\mathrm{CaC_2O_4}$.
Coloured: $\mathrm{BaCrO_4}$ and $\mathrm{PbCrO_4}$ yellow, AgBr pale yellow, $\mathrm{PbI_2}$ yellow, basic iron(III) acetate reddish brown.
Count = 5.
Answer: 5. | ||||
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