JEE Advanced 26 May 2024 Paper 2 question paper with solutions

JEE Advanced 26 May 2024 Paper 2: all 51 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

Maths

Question 1

Maths · Inverse Trigonometric Functions · Single correct

Considering only the principal values of the inverse trigonometric functions, the value of $$\tan \left( \sin^{-1} \left( \frac{3}{5} \right) - 2 \cos^{-1} \left( \frac{2}{\sqrt{5}} \right) \right)$$ is

$\frac{7}{24}$
$\frac{-7}{24}$
$\frac{-5}{24}$
$\frac{5}{24}$

Answer: (b)

Solution

Let $\sin^{-1} \frac{3}{5} = \alpha$, $2 \cos^{-1} \frac{2}{\sqrt{5}} = \beta$. Therefore, $\cos \frac{\beta}{2} = \frac{2}{\sqrt{5}}$. Since $\sin \alpha = \frac{3}{5}$, it follows that $\tan \alpha = \frac{3}{4}$. For $\tan \beta$, we have: $$\tan \beta = \frac{2 \tan \frac{\beta}{2}}{1 - \tan^2 \frac{\beta}{2}} = \frac{2 \times \frac{2}{2}}{1 - \frac{1}{4}} = \frac{4}{3}.$$ Therefore, $\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{\frac{3}{4} - \frac{4}{3}}{1 + \frac{3}{4} \times \frac{4}{3}} = -\frac{7}{24}$.

Question 2

Maths · Applications of Integrals · Single correct

Let $S=\{(x,y)\in\mathbb{R}\times\mathbb{R}: x\ge0,\ y\ge0,$ $y^2\le4x,\ y^2\le12-2x,$ $3y+\sqrt{8}\,x\le5\sqrt{8}\}$. If the area of the region $S$ is $\alpha\sqrt{2},$ then $\alpha$ is equal to

$\frac{17}{2}$
$\frac{17}{3}$
$\frac{17}{4}$
$\frac{17}{5}$

Answer: (b)

Solution

Given $y^2 = 4x$, $y^2 = 12 - 2x \implies x = 2$, $y = \sqrt{8}$. $$A = \int_{0}^{2} 2 \sqrt{x} \, dx + \frac{1}{2} \times 3 \times \sqrt{8}$$ $$= \left[ 2 \times \frac{2}{3} x^{\frac{3}{2}} \right]_{0}^{2} + 3 \sqrt{2} = \frac{4}{3} \times 2 \sqrt{2} + 3 \sqrt{2} = \frac{17}{3} \sqrt{2}$$ Therefore, $A = \alpha \sqrt{2} \implies \alpha = \frac{17}{3}$ Option (B) is correct.

Question 3

Maths · Limits and Derivatives · Single correct

Let $k \in \mathbb{R}$. If $\lim_{x \to 0^+} \left( \sin(\sin kx) + \cos x + x \right)^{\frac{2}{x}} = e^6$, then the value of $k$ is

Answer: (b)

Solution

Given $I = \lim_{x \to 0^+} \left( \sin(\sin kx) + \cos x + x \right)^{\frac{2}{x}} = e^6$. Therefore, $\ln I = \lim_{x \to 0^+} \frac{2}{x} \left( \sin(\sin kx) + \cos x + x - 1 \right)$. This implies $\ln I = \lim_{x \to 0^+} 2 \left( \frac{\sin(\sin kx)}{\sin kx} \cdot \frac{\sin kx}{kx} \cdot \frac{kx}{x} + 1 - \frac{(1 - \cos x)}{x^2} \cdot x \right)$. Thus, $\ln I = 2(k + 1)$ which implies $I = e^{2(k + 1)} = e^6$. From $k + 1 = 3$, we find $k = 2$.

Question 4

Maths · Continuity and Differentiability · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined by $$f(x) = \begin{cases} x^2 \sin \left( \frac{\pi}{x^2} \right), & if x \neq 0, \\ 0, & if x = 0. \end{cases}$$ Then which of the following statements is TRUE?

$f(x) = 0$ has infinitely many solutions in the interval $\left[ \frac{1}{10^{10}}, \infty \right)$.
$f(x) = 0$ has no solutions in the interval $\left[ \frac{1}{\pi}, \infty \right)$.
The set of solutions of $f(x) = 0$ in the interval $\left( 0, \frac{1}{10^{10}} \right)$ is finite.
$f(x) = 0$ has more than 25 solutions in the interval $\left( \frac{1}{\pi}, \frac{1}{2} \right)$.

Answer: (d)

Solution

Given $$f(x) = \begin{cases} x^2 \sin\left(\frac{\pi}{x^2}\right), & if x \neq 0, \\ 0, & if x = 0. \end{cases}$$ For $f(x) = 0$, we have $\sin\left(\frac{\pi}{x^2}\right) = 0$. This implies $$\frac{\pi}{x^2} = n\pi$$ which gives $$x^2 = \frac{1}{n}$$ and thus $$x = \frac{1}{\sqrt{n}}$$ For $$\frac{1}{\sqrt{n}} \in \left[\frac{1}{10^{10}}, \infty\right)$$ we have $$\sqrt{n} \in (0, 10^{10}]$$ which implies $$n \in (0, (10^{10})^2]$$ This gives finite values of $n$. For $$\frac{1}{\sqrt{n}} \in \left[\frac{1}{\pi}, \infty\right)$$ we have $$\sqrt{n} \in (0, \pi]$$ which implies $$n \in (0, \pi^2]$$ This gives $n = 1, 2, 3, \ldots, 9$. For $$\sqrt{n} \in (10^{10}, \infty)$$ $n$ is infinite. If $$x \in \left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)$$ then $$\sqrt{n} \in (\pi, \pi^2)$$ which implies $$n \in (\pi^2, \pi^4)$$ This gives $n \in (9.8, 97.2, \ldots)$ Thus, there are more than 25 solutions.

Question 5

Maths · Limits and Derivatives · Multiple correct

Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $$\lim_{x \to \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin\left(\frac{1}{x^2}\right)}{x^\alpha (\log_e (1+x))^\beta} = 0$$ Then which of the following is (are) correct?

$(-1, 3) \in S$
$(-1, 1) \in S$
$(1, -1) \in S$
$(1, -2) \in S$

Answer: (b), (c)

Solution

Given $$$ $\lim$_{x $\to$ $\infty$} $\frac{\sin(x^2) \sin\left(\frac{1}{x^2}\right)(\ln x)^\alpha}{x^{\alpha \beta} (\ln(1+x))^\beta}$ = 0 $$$. $\[$ $\lim$_{x $\to$ $\infty$} $\frac{(\sin x^2) \sin\left(\frac{1}{x^2}\right) \frac{1}{x^2}}{\frac{1}{x^2} x^{\alpha \beta} (\ln(1+x))^\beta}$ = 0 $\]$ It is possible if $$$ $\alpha$ $\beta$ + 2 > 0 $$$. $$$ $\alpha$ $\beta$ > -2 $$$ (A) $$$ $\alpha$ $\beta$ = -3 $$$ (B) $$$ $\alpha$ $\beta$ = -1 $$$ (C) $$$ $\alpha$ $\beta$ = -1 $$$ (D) $$$ $\alpha$ $\beta$ = -2 $$$

Question 6

Maths · Three Dimensional Geometry · Multiple correct

A straight line drawn from the point $P(1,3,2)$, parallel to the line $\frac{x-2}{1} = \frac{y-4}{2} = \frac{z-6}{1}$, intersects the plane $L_1: x-y+3z=6$ at the point $Q$. Another straight line which passes through $Q$ and is perpendicular to the plane $L_1$ intersects the plane $L_2: 2x-y+z=-4$ at the point $R$. Then which of the following statements is(are) TRUE?

The length of the line segment $PQ$ is $\sqrt{6}$
The coordinates of $R$ are $(1,6,3)$
The centroid of the triangle $PQR$ is $\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)$
The perimeter of the triangle $PQR$ is $\sqrt{2} + \sqrt{6} + \sqrt{11}$

Answer: (a), (c)

Solution

Equation of line parallel to $$$ $\frac{x-2}{1}$ = $\frac{y-4}{2}$ = $\frac{z-6}{1}$ $$$ through $$$ P(1,3,2) $$$ is $$$ $\frac{x-1}{1}$ = $\frac{y-3}{2}$ = $\frac{z-2}{1}$ = $\lambda$ $$$ (let). Now, putting any point $$$ ($\lambda$ + 1, 2$\lambda$ + 3, $\lambda$ + 2) $$$ in $$$ L_1 $$$, $\[$ $\lambda$ = 1 $\]$ $$$ $\Rightarrow$ $$$ Point $$$ Q(2,5,3) $$$. Equation of line through $$$ Q(2,5,3) $$$ perpendicular to $$$ L_1 $$$ is $\[$ $\frac{x-2}{1}$ = $\frac{y-5}{-1}$ = $\frac{z-3}{3}$ = $\mu$ (Let) $\]$ Putting any point $$$ ($\mu$ + 2, -$\mu$ + 5, 3$\mu$ + 3) $$$ in $$$ L_2 $$$, $\[$ $\mu$ = -1 $\]$ $$$ $\Rightarrow$ $$$ Point $$$ R(1, 6, 0) $$$. $$$ (A) $$$ $$$ PQ = $\sqrt{1 + 4 + 1}$ = $\sqrt{6}$ $$$ $$$ (B) $$$ $$$ R(1, 6, 0) $$$ $$$ (C) $$$ Centroid $$$ $\left$( $\frac{4}{3}$, $\frac{14}{3}$, $\frac{5}{3}$ $\right$) $$$ $$$ (D) $$$ $$$ PQ + QR + PR = $\sqrt{6}$ + $\sqrt{11}$ + $\sqrt{13}$ $$$

Question 7

Maths · Conic Sections · Multiple correct

Let $A_1$, $B_1$, $C_1$ be three points in the $xy$-plane. Suppose that the lines $A_1C_1$ and $B_1C_1$ are tangents to the curve $y^2 = 8x$ at $A_1$ and $B_1$, respectively. If $O = (0,0)$ and $C_1 = (-4,0)$, then which of the following statements is (are) TRUE?

The length of the line segment $OA_1$ is $4\sqrt{3}$
The length of the line segment $A_1B_1$ is $16$
The orthocentre of the triangle $A_1B_1C_1$ is $(0, 0)$
The orthocentre of the triangle $A_1B_1C_1$ is $(1, 0)$

Answer: (a), (c)

Solution

Let $A_1 = (2t_1^2, 4t_1)$ and $B_1 = (2t_2^2, 4t_2)$. $C \equiv (-4, 0) \equiv (2t_1t_2, 2(t_1 + t_2))$. Therefore, $t_2 = -t_1$ and $t_1(-t_1) = -2$. Thus, $t_1 = \sqrt{2}$, $t_2 = -\sqrt{2}$. $A_1 \equiv (4, 4\sqrt{2})$, $B_1 \equiv (4, -4\sqrt{2})$. Therefore, $OA_1 = \sqrt{4^2 + (4\sqrt{2})^2} = 4\sqrt{3}$. $A_1B_1 = 8\sqrt{2}$. Altitude $C_1M : y = 0$ $\ldots$ (i). Altitude $B_1N : \sqrt{2}x + y = 0$ $\ldots$ (ii). Therefore, Orthocentre = $(0, 0)$.

Question 8

Maths · Relations and Functions · Numerical

Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$, and $g : \mathbb{R} \to (0, \infty)$ be a function such that $g(x+y) = g(x) g(y)$ for all $x, y \in \mathbb{R}$. If $f\left( -\frac{3}{5} \right) = 12$ and $g\left( -\frac{1}{3} \right) = 2$, then the value of $\left( f\left( \frac{1}{4} \right) + g(-2) - 8 \right) g(0)$ is

Answer: 51

Solution

Given $f(x + y) = f(x) + f(y)$, we have $f(x) = kx$. $$f\left(\frac{-3}{5}\right) = 12 \Rightarrow k = -20$$ Thus, $f(x) = -20x$. For $g(x + y) = g(x)g(y)$, we have $g(x) = a^x$. $$g\left(\frac{-1}{3}\right) = 2 \Rightarrow a = \frac{1}{8}$$ Thus, $g(x) = \left(\frac{1}{8}\right)^x$. $$\left(f\left(\frac{1}{4}\right) + g(-2) - 8\right)g(0) = (-5 - 64) \times 1 = 51$$

Question 9

Maths · Probability · Numerical

A bag contains $N$ balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For $i = 1, 2, 3$, let $W_i$, $G_i$, and $B_i$ denote the events that the ball drawn in the $i^{th}$ draw is a white ball, green ball, and blue ball, respectively. If the probability $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$ and the conditional probability $P(B_3 \mid W_1 \cap G_2) = \frac{2}{9}$, then $N$ equals .

Answer: 11

Solution

Given $N$ balls $= 3W + 6G + (N - 9)B$. $$P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$$ $$\Rightarrow \frac{3}{N} \times \frac{6}{N-1} \times \frac{N-9}{N-2} = \frac{2}{5N}$$ $$\Rightarrow N^2 - 48N + 407 = 0$$ $$\Rightarrow N = 11 or 37$$ $$P(B_3 \mid W_1 \cap G_2) = \frac{2}{9}$$ $$\Rightarrow \frac{P(W_1 \cap G_2 \cap B_3)}{P(W_1 \cap G_2)} = \frac{2}{9}$$ $$\Rightarrow \frac{\frac{2}{5N}}{\frac{3}{N} \times \frac{6}{N-1}} = \frac{2}{9}$$ $$\Rightarrow \frac{N-1}{45} = \frac{2}{9}$$ $$\Rightarrow N = 11$$

Question 10

Maths · Relations and Functions · Numerical

Let the function $f : \mathbb{R} \to \mathbb{R}$ be defined by $$f(x) = \frac{\sin x \left(x^{2023} + 2024x + 2025\right)}{e^{\pi x} \left(x^2 - x + 3\right)} + \frac{2 \left(x^{2023} + 2024x + 2025\right)}{e^{\pi x} \left(x^2 - x + 3\right)}.$$ Then the number of solutions of $f(x) = 0$ in $\mathbb{R}$ is ______.

Answer: 1

Solution

Given $f(x) = 0$. $$\Rightarrow \frac{x^{2023} + 2024x + 2025}{(x^2 - x + 3)} \left[ \frac{\sin x + 2}{e^{\pi x}} \right] = 0$$ $$\Rightarrow x^{2023} + 2024x + 2025 = 0$$ Let $g(x) = x^{2023} + 2024x + 2025$. $$g'(x) = 2023x^{2022} + 2024 > 0 \forall x \in \mathbb{R}$$ Therefore, $f(x) = 0$ has only one solution.

Question 11

Maths · Vector Algebra · Fill in the blank

Let $\vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{q} = \hat{i} - \hat{j} + \hat{k}$. If for some real numbers $\alpha$, $\beta$ and $\gamma$, we have $15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha(2\vec{p} + \vec{q}) + \beta(\vec{p} - 2\vec{q}) + \gamma(\vec{p} \times \vec{q})$, then the value of $\gamma$ is ____.

Answer: 2

Solution

Given $2\vec{p} + \vec{q} = 5\hat{i} + \hat{j} + 7\hat{k}$. $\vec{p} - 2\vec{q} = 0\hat{i} + 3\hat{j} + \hat{k}$. The cross product $\vec{p} \times \vec{q}$ is given by: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(4) - \hat{j}(-1) + \hat{k}(-3)$$ This simplifies to $4\hat{i} + \hat{j} - 3\hat{k}$. Now, $15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha(5\hat{i} + \hat{j} + 7\hat{k}) + \beta(3\hat{j} + \hat{k}) + \gamma(4\hat{i} + \hat{j} - 3\hat{k})$. Therefore, $$15 = 5\alpha + 4\gamma$$ $$10 = \alpha + 3\beta + \gamma$$ $$6 = 7\alpha + \beta - 3\gamma$$ Solving these equations gives $\alpha = \frac{7}{5}$, $\beta = \frac{11}{5}$, $\gamma = 2$. Thus, $\gamma = 2$.

Question 12

Maths · Conic Sections · Numerical

A normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point $(0, -\alpha)$ to the parabola $x^2 = -4ay$, where $a > 0$. Let $L$ be the line passing through $(0, -\alpha)$ and parallel to the directrix of the parabola. Suppose that $L$ intersects the parabola at two points $A$ and $B$. Let $r$ denote the length of the latus rectum and $s$ denote the square of the length of the line segment $AB$. If $r : s = 1 : 16$, then the value of $24a$ is

Answer: 12

Solution

Given $x^2 = -4ay$. Equation of normal $$y = mx - 2a - \frac{a}{m^2}$$ $$-\alpha = -2a - \frac{a}{1} = -8a$$ $$\Rightarrow \alpha = 8a$$ Equation of required line $$y = -\alpha$$ $$\Rightarrow y = -8a$$, solving with $x^2 = -4ay$ $$\Rightarrow x^2 = 32a^2$$ $$\Rightarrow x = \pm 4\sqrt{2a}$$ $$= \pm \frac{\alpha}{\sqrt{2}}$$ $$A\left(\frac{\alpha}{\sqrt{2}}, -\alpha\right), B\left(-\frac{\alpha}{\sqrt{2}}, -\alpha\right) \Rightarrow AB = \sqrt{2\alpha}$$ $$\Rightarrow \frac{r}{s} = \frac{4a}{2\alpha^2} = \frac{1}{16} \Rightarrow \frac{4a}{2 \times 64a^2} = \frac{1}{16}$$ $$\Rightarrow a = \frac{1}{2}$$ $$\Rightarrow 24a = 12$$

Question 13

Maths · Applications of Integrals · Fill in the blank

Let the function $f:[1,\infty)\to\mathbb{R}$ be defined by $f(t)=(-1)^{n+1}2,$ if $t=2n-1,\ n\in\mathbb{N}$ $f(t)=\dfrac{2n+1-t}{2}f(2n-1)$ $+\dfrac{t-(2n-1)}{2}f(2n+1),$ if $2n-1<t<2n+1,\ n\in\mathbb{N}$ Define $g(x)=\int_1^x f(t)\,dt,\ x\in(1,\infty)$. Let $\alpha$ denote the number of solutions of $g(x)=0$ in $(1,8]$ and $\beta=\lim\limits_{x\to1^+}\dfrac{g(x)}{x-1}$. Then the value of $\alpha+\beta$ is equal to \_\_\_\_\_.

Answer: 5

Solution

Given $$f(t) = \frac{(2n+1) - t}{2}(-1)^{n+1}2 + \frac{t - (2n-1)}{2}(-1)^{n+2}2, t \in (2n-1, 2n+1)$$ $$\Rightarrow f(t) = 2(-1)^{n+1}(2n-t), t \in (2n-1, 2n+1)$$ $$\Rightarrow g(x) = \int_1^x f(t) dt, x \in (1, 8]$$ $$= \begin{cases} \int_1^x 2(2-t) dt, & 1 < x \leq 3, \ n = 1 \\ \int_1^3 2(2-t) dt + \int_3^x (2t-8) dt, & 3 < x \leq 5, \ n = 2 \\ \int_1^3 2(2-t) dt + \int_3^5 (2t-8) dt + \int_5^x 2(6-t) dt, & 5 < x \leq 7, \ n = 3 \\ \int_1^3 2(2-t) dt + \int_3^5 (2t-8) dt + \int_5^7 2(6-t) dt + \int_7^x (2t-16) dt, & x \in (7, 8], \ n = 4 \end{cases}$$ $$= \begin{cases} -x^2 + 4x - 3, & 1 < x \leq 3, \\ x^2 - 8x + 15, & 3 < x \leq 5 \\ -x^2 + 12x - 35, & 5 < x \leq 7 \\ x^2 - 16x + 63, & 7 < x \leq 8 \end{cases}$$ $$\Rightarrow g(x) = 0 \Rightarrow x = 3, 5, 7 \Rightarrow \alpha = 3$$ $$\beta = \lim_{x \to 1^+} \frac{g(x)}{x-1} = \lim_{x \to 1^+} \frac{-(x-1)(x-3)}{x-1} = 2$$ $$\Rightarrow \alpha + \beta = 5$$

Question 14

Maths · Permutations and Combinations · Fill in the blank

Let $S=\{1,2,3,4,5,6\}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties: i. $R$ has exactly $6$ elements. ii. For each $(a,b)\in R$, we have $|a-b|\geq2$. Let $Y=\{R\in X:\text{the range of }R\text{ has exactly one element}\}$ and $Z=\{R\in X:R\text{ is a function from }S\text{ to }S\}$. Let $n(A)$ denote the number of elements in a set $A$. (There are two questions based on PARAGRAPH "I", the question given below is one of them) If $n(X) = \binom{m}{6}$, then the value of $m$ is _______.

Answer: 20

Solution

Let $S = \{1, 2, 3, 4, 5, 6\}$ and $R : S \to S$. The number of elements in $R = 6$ and for each $(a, b) \in R$, $|a - b| \geq 2$. Let $X \to$ set of all relation $R : S \to S$. If $a = 1$, $b = 3, 4, 5, 6$ then there are 4 possibilities. If $a = 2$, $b = 4, 5, 6$ then there are 3 possibilities. If $a = 3$, $b = 1, 5, 6$ then there are 3 possibilities. If $a = 4$, $b = 1, 2, 6$ then there are 3 possibilities. If $a = 5$, $b = 1, 2, 3$ then there are 3 possibilities. If $a = 6$, $b = 1, 2, 3, 4$ then there are 4 possibilities. The total number of ordered pairs $(a, b)$ such that $|a - b| \geq 2$ is 20. Therefore, $n(X) =$ number of elements in $X = \binom{20}{6}$. Thus, $m = 20$. Let $S = \{1, 2, 3, 4, 5, 6\}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties: i. $R$ has exactly 6 elements. ii. For each $(a, b) \in R$, we have $|a - b| \geq 2$. Let $Y = \{R \in X : The range of R has exactly one element\}$ and $Z = \{R \in X : R is a function from S to S\}$. Let $n(A)$ denote the number of elements in a set $A$.

Question 15

Maths · Basics Of Mathematics · Fill in the blank

Let $S=\{1,2,3,4,5,6\}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties: i. $R$ has exactly $6$ elements. ii. For each $(a,b)\in R$, we have $|a-b|\geq2$. Let $Y=\{R\in X:\text{the range of }R\text{ has exactly one element}\}$ and $Z=\{R\in X:R\text{ is a function from }S\text{ to }S\}$. Let $n(A)$ denote the number of elements in a set $A$. (There are two questions based on PARAGRAPH "I", the question given below is one of them) If the value of $n(Y) + n(Z)$ is $k^2$, then $|k|$ is ____________.

Answer: 36

Solution

Given $S = \{1, 2, 3, 4, 5, 6\}$ and $R : S \to S$. Number of elements in $R = 6$ and for each $(a, b) \in R; \ |a - b| \geq 2$. Let $X$ be the set of all relations $R : S \to S$. The total number of ordered pairs $(a, b)$ such that $|a - b| \geq 2$ is $20$. Therefore, $n(X) = number of elements in X = \binom{20}{6}$. Thus, $m = 20$. Let $Y = \{R \in X : The range of R has exactly one element\}$. From above, if the range of $R$ has exactly one element, then the maximum number of elements in $R$ will be $4$. Therefore, $n(Y) = 0$. Let $Z = \{R \in X : R is a function from S to S\}$. Then, $$n(Z) = \binom{4}{1} \times \binom{3}{1} \times \binom{3}{1} \times \binom{3}{1} \times \binom{3}{1} \times \binom{4}{1}$$ $$= (36)^2$$ $$n(Y) + n(Z) = 0 + (36)^2 = k^2$$ $$\Rightarrow |k| = 36$$ PARAGRAPH II Let $f : \left[0, \frac{\pi}{2}\right] \to [0, 1]$ be the function defined by $f(x) = \sin^2 2x$ and let $g : \left[0, \frac{\pi}{2}\right] \to [0, \infty)$ be the function defined by $$g(x) = \sqrt{\frac{\pi x}{2} - x^2}.$$ (There are two questions based on PARAGRAPH "II", the question given below is one of them)

Question 16

Maths · Integrals · Fill in the blank

Let $f:\left[0,\frac{\pi}{2}\right]\to[0,1]$ be the function defined by $f(x)=\sin^2x$ and let $g:\left[0,\frac{\pi}{2}\right]\to[0,\infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$. (There are two questions based on PARAGRAPH "II", the question given below is one of them) The value of $2 \left\{ \int_{0}^{\frac{\pi}{2}} f(x)g(x) \, dx - \int_{0}^{\frac{\pi}{2}} g(x) \, dx \right\}$ is _______.

Answer: 0

Question 17

Maths · Applications of Integrals · Fill in the blank

Let $f:\left[0,\frac{\pi}{2}\right]\to[0,1]$ be the function defined by $f(x)=\sin^2x$ and let $g:\left[0,\frac{\pi}{2}\right]\to[0,\infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$. (There are two questions based on PARAGRAPH "II", the question given below is one of them) The value of $\frac{16}{3\pi} \int_{0}^{\frac{\pi}{2}} f(x)g(x) \, dx$ is ______.

Answer: 0.25

Solution

According to Q.16, $$2 \int_0^{\frac{\pi}{2}} f(x)g(x)dx = \pi \int_0^{\frac{\pi}{2}} g(x)dx = I_1 (let)$$ Now, $$I_1 = \int_0^{\frac{\pi}{2}} g(x)dx = \int_0^{\frac{\pi}{2}} \sqrt{\frac{\pi}{2} - x^2} \, dx$$ $$I_1 = \int_0^{\frac{\pi}{2}} \left( \sqrt{\left( \frac{\pi}{4} \right)^2 - \left( \frac{\pi}{4} - x \right)^2} \right) dx$$ Put $$\frac{\pi}{4} - x = t$$ $$\Rightarrow dx = -dt$$ $$I_1 = -\int_{\frac{\pi}{4}}^{-\frac{\pi}{4}} \sqrt{\left( \frac{\pi}{4} \right)^2 - t^2} \, dt$$ $$I_1 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{\left( \frac{\pi}{4} \right)^2 - t^2} \, dt$$ $$I_1 = 2 \int_0^{\frac{\pi}{4}} \sqrt{\left( \frac{\pi}{4} \right)^2 - t^2} \, dt = 2 \left[ \frac{t}{2} \sqrt{\left( \frac{\pi}{4} \right)^2 - t^2} - \frac{\pi^2}{32} \sin^{-1} \left( \frac{4t}{\pi} \right) \right]_0^{\frac{\pi}{4}}$$ $$I_1 = \frac{\pi^3}{32}$$ Now, $$I = \frac{8}{3\pi} I_1$$ $$I = \frac{1}{4} = 0.25$$

Physics

Question 18

Physics · Electromagnetic Induction · Single correct

A region in the form of an equilateral triangle (in $x-y$ plane) of height $L$ has a uniform magnetic field $\mathbf{B}$ pointing in the $+z$-direction. A conducting loop $PQR$, in the form of an equilateral triangle of the same height $L$, is placed in the $x-y$ plane with its vertex $P$ at $x=0$ in the orientation shown in the figure. At $t=0$, the loop starts entering the region of the magnetic field with a uniform velocity $\mathbf{v}$ along the $+x$-direction. The plane of the loop and its orientation remain unchanged throughout its motion. Which of the following graph best depicts the variation of the induced emf $(E)$ in the loop as a function of the distance $(x)$ starting from $x=0$?

Answer: (a)

Solution

For $x < L$: Area $= \frac{x}{2} \times \frac{x}{2} \tan 30 \times 4 \times \frac{1}{2} = \frac{1}{2} x^2 \tan 30$. $\phi' = B_0 \times \tan 30 V \propto x$. For $x \geq L$: Area $= A_0 - A_1 - A_2 - A_3$. $= A_0 - 2A_1 - (x - L)(x - L) \tan 30$. $= A_0 - (x - L)^2 \tan 30 - \{L \tan 30 - (x - L) \tan 30\}^2 \frac{1}{2} \times \frac{1}{2} \times \tan 60 \times 2$. $= A_0 - (x - L)^2 \tan 30 - \tan 30 \cdot (2L - x)^2 \frac{1}{2}$. $\varepsilon' = -2(x - L) \tan 30 V - \tan 30 \cdot 2(2L - x)(-1)V$. $= (4L - x - 2x + 2L) \tan 30 V$. $= (4L - 3x)V$. $0$ at $x = \frac{4L}{3}$. From 1 and 2, $1.33 < 1.5$.

Question 19

Physics · Gravitation · Single correct

A particle of mass $m$ is under the influence of the gravitational field of a body of mass $M (>> m)$. The particle is moving in a circular orbit of radius $r_0$ with time period $T_0$ around the mass $M$. Then, the particle is subjected to an additional central force, corresponding to the potential energy $V_c(r) = m \alpha / r^3$, where $\alpha$ is a positive constant of suitable dimensions and $r$ is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius $r_0$ in the combined gravitational potential due to $M$ and $V_c(r)$, but with a new time period $T_1$, then $\left( T_1^2 - T_0^2 \right) / T_1^2$ is given by [ $G$ is the gravitational constant. ]

$\frac{3 \alpha}{G M r_0^2}$
$\frac{\alpha}{2 G M r_0^2}$
$\frac{\alpha}{G M r_0^2}$
$\frac{2 \alpha}{G M r_0^2}$

Answer: (a)

Solution

Given $$$ $\frac{Gmm}{r_0^2}$ - $\frac{3 \alpha m}{r_0^4}$ = $\frac{mv^2}{r_0}$ $$$. $\[$ T = $\frac{2 \pi r_0}{\sqrt{\frac{Gmr_0^2 - 3 \alpha}{r_0^3}}}$ $\]$ $\[$ T_0^2 = $\frac{4 \pi^2}{Gm}$ r_0^3 $\]$ $\[$ $\frac{T^3 - T_0^2}{T_1^2}$ = 1 - $\frac{T_0^2}{T_1^2}$ $\]$ $\[$ = 1 - $\frac{4 \pi^2 r_0^3}{Gm}$ $\frac{Gmr_0^2 - 3 \alpha}{4 \pi^2 r_0^2}$ $\frac{1}{r_0^3}$ $\]$ $\[$ = 1 - 1 + $\frac{3 \alpha}{Gmr_0^2}$ $\]$ $\[$ = $\frac{3 \alpha}{GMr_0}$ $\]$

Question 20

Physics · Dual Nature of Radiation and Matter · Single correct

A metal target with atomic number $Z = 46$ is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio $r$ of the wavelengths of the $K_{\alpha}$-line and the cut-off is found to be $r = 2$. If the same electron beam bombards another metal target with $Z = 41$, the value of $r$ will be

2.53
1.27
2.24
1.58

Answer: (a)

Solution

Given $$$ $\frac{1}{\lambda_\alpha}$ = $\frac{3}{4}$ R(Z-1)^2 p $$$. $$$ $\lambda$_{cut} = $\frac{hc}{eV}$ $$$ Therefore, Ratio $$$ $\propto$ $\frac{1}{(Z-1)^2}$ $$$ for the same beam. $\[$ $\frac{Z}{x}$ = $\frac{40^2}{45^2}$ $\Rightarrow$ x = $\frac{45^2}{40^2}$ $\cdot$ 2 $\approx$ 2.53 $\]$

Question 21

Physics · Moving Charges and Magnetism · Single correct

A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass $m$ and radius $r$ and it is in a uniform vertical magnetic field $B_0$, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity $g$, on two conducting supports at $P$ and $Q$. When a current $I$ is passed through the loop, the loop turns about the line $PQ$ by an angle $\theta$ given by

$\tan \theta = \pi r I B_0 / (mg)$
$\tan \theta = 2 \pi r I B_0 / (mg)$
$\tan \theta = \pi r I B_0 / (2mg)$
$\tan \theta = mg / (\pi r I B_0)$

Answer: (a)

Solution

Now for equilibrium, $\tau = mg \cdot r \sin \theta$. $l \pi r^2 B_0 \cos \theta = mg r \sin \theta$. Therefore, $$\tan \theta = \frac{l \pi r B_0}{mg}.$$

Question 22

Physics · Electric Charges and Fields · Multiple correct

A small electric dipole $\vec{p}_0$, having a moment of inertia $I$ about its center, is kept at a distance $r$ from the center of a spherical shell of radius $R$. The surface charge density $\sigma$ is uniformly distributed on the spherical shell. The dipole is initially oriented at a small angle $\theta$ as shown in the figure. While staying at a distance $r$, the dipole is free to rotate about its center. If released from rest, then which of the following statement(s) is (are) correct? $[\varepsilon_0$ is the permittivity of free space.]

The dipole will undergo small oscillations at any finite value of $r$.
The dipole will undergo small oscillations at any finite value of $r > R$.
The dipole will undergo small oscillations with an angular frequency of $\sqrt{\frac{2 \sigma p_0}{\varepsilon_0 I}}$ at $r = 2R$.
The dipole will undergo small oscillations with an angular frequency of $\sqrt{\frac{\sigma p_0}{100 \varepsilon_0 I}}$ at $r = 10R$.

Answer: (b), (d)

Solution

The torque is given by $\tau = |\vec{p} \times \vec{E}|$. We have $I \alpha = p_0 E \sin \theta$. The angular acceleration $\alpha$ is given by $$\alpha = \frac{\rho_0}{I} \left( \frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi R^2}{r^2} \right).$$ Simplifying, we get $$\alpha = \left( \frac{\rho_0 \sigma R^2}{I \varepsilon_0 r^2} \right) \cdot \theta.$$ Therefore, $$\omega = \sqrt{\frac{\rho_0 \sigma R^2}{I \varepsilon_0 r^2}}.$$ For $r = 2R$, $$\omega = \frac{\rho_0 \sigma}{4 I \varepsilon_0}$$ (C is incorrect). Also, for $r = 10R$, $$\omega = \frac{\rho_0 \sigma}{4 I (100)}$$ (D is correct). It will oscillate for any finite value of $r > R$. (B is correct)

Question 23

Physics · Mechanical Properties of Fluids · Multiple correct

A table tennis ball has radius $(3/2) \times 10^{-2} \, \mathrm{m}$ and mass $(22/7) \times 10^{-3} \, \mathrm{kg}$. It is slowly pushed down into a swimming pool to a depth of $d = 0.7 \, \mathrm{m}$ below the water surface and then released from rest. It emerges from the water surface at speed $v$, without getting wet, and rises up to a height $H$. Which of the following option(s) is (are) correct? [Given: $\pi = 22/7$, $g = 10 \, \mathrm{m/s^2}$, density of water $= 1 \times 10^3 \, \mathrm{kg/m^3}$, viscosity of water $= 1 \times 10^{-3} \, \mathrm{Pa\cdot s}$.]

The work done in pushing the ball to the depth $d$ is $0.077 \, \mathrm{J}$.
If we neglect the viscous force in water, then the speed $v = 7 \, \mathrm{m/s}$.
If we neglect the viscous force in water, then the height $H = 1.4 \, \mathrm{m}$.
The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500/9$.

Answer: (a), (b), (d)

Solution

Work done in pushing the ball $$W = (\rho v g) d - (\sigma v g) d$$ Where $\rho \to$ Density of water, $\sigma \to$ Density of ball $$\Rightarrow W = \frac{4}{3} \pi R^3 \times 10 \times 0.7 \left[ 1000 - \frac{3}{4} \times 10^3 \frac{1}{R^3} \right]$$ $$W = 0.077 \, \mathrm{J}$$ [A is correct] When ball is released at bottom, same work (i.e. 0.077 J) is done on ball. $$\therefore \frac{1}{2} mv^2 = 0.077$$ $$v = \sqrt{\frac{0.077 \times 2}{\frac{22}{7} \times 10^{-3}}}$$ $$= 7 \, \mathrm{m/s}$$ [B is correct] also, $$H = \frac{v^2}{2g} = \frac{7 \times 7}{2 \times 10} = 2.45 \, \mathrm{m}$$ [C is incorrect] Net force $F_{net} = \nu \sigma g - \nu \sigma g = 0.11 \, \mathrm{N}$ Also, viscous force is maximum when $v = 7 \, \mathrm{m/s}$. $$\therefore (F_v)_{max} = 6 \pi \eta v$$ $$= 6 \times \frac{22}{7} \times 10^{-3} \left( \frac{3}{2} \times 10^{-2} \right) \times 7$$ $$= 18 \times 11 \times 10^{-5} \, \mathrm{N}$$ Now, $$\frac{F_{net}}{(F_v)_{max}} = \frac{500}{9}$$ [D is correct]

Question 24

Physics · Moving Charges and Magnetism · Multiple correct

A positive, singly ionized atom of mass number $A_M$ is accelerated from rest by the voltage $192 \, \mathrm{V}$. Thereafter, it enters a rectangular region of width $w$ with magnetic field $\mathbf{B}_0 = 0.1 \hat{k} \, \mathrm{Tesla}$, as shown in the figure. The ion finally hits a detector at the distance $x$ below its starting trajectory. [Given: Mass of neutron/proton = $(5/3) \times 10^{-27} \, \mathrm{kg}$, charge of the electron = $1.6 \times 10^{-19} \, \mathrm{C}$.]

The value of $x$ for $H^+$ ion is $4 \, \mathrm{cm}$.
The value of $x$ for an ion with $A_M = 144$ is $48 \, \mathrm{cm}$.
For detecting ions with $1 \leq A_M \leq 196$, the minimum height $(x_1 - x_0)$ of the detector is $55 \, \mathrm{cm}$.
The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M = 196$ is $56 \, \mathrm{cm}$.

Answer: (a), (b)

Solution

Solution: $x = 2R$ $$= \frac{2mv}{qB}$$ $$= \frac{2\sqrt{2m(e \Delta V)}}{qB}$$ For $\mathrm{H}^+$ ion $x = 3.91 \, \mathrm{cm}$ $\approx 4 \, \mathrm{cm}$ (A is correct) For $m = 144 \, (m_p)$ $$= 12 \left( x_{H^+} \right)$$ $$= 48 \, \mathrm{cm}$$ (B is correct) For $1 \leq A_M \leq 196$ $$\Rightarrow (x_1 - x_0)_{\min} = 2R_{196} - 2R_1$$ $$= (14 \times 4) - 4$$ $$= 52 \, \mathrm{cm}$$ (C is incorrect) For $A_M = 196$ $$w_{\min} = R_{196} = 28 \, \mathrm{cm}$$ (D is incorrect)

Question 25

Physics · Physical World, Units and Measurements · Numerical

The dimensions of a cone are measured using a scale with a least count of 2 $\mathrm{mm}$. The diameter of the base and the height are both measured to be 20.0 $\mathrm{cm}$. The maximum percentage error in the determination of the volume is

Answer: 3

Solution

Given $V = \frac{1}{3} \pi R^2 H$. Therefore, $$\frac{dV}{V} = 2 \cdot \frac{dR}{R} + \frac{dH}{H}$$ Therefore, the percentage error in measuring volume is $$= \left[ 2 \times \frac{0.2}{20} + \frac{0.2}{20} \right] \times 100$$ $$= 3$$

Question 26

Physics · Motion in a Plane · Fill in the blank

A ball is thrown from the location $(x_0, y_0) = (0,0)$ of a horizontal playground with an initial speed $\upsilon_0$ at an angle $\theta_0$ from the $+x$-direction. The ball is to be hit by a stone, which is thrown at the same time from the location $(x_1, y_1) = (L, 0)$. The stone is thrown at an angle $(180^\circ - \theta_1)$ from the $+x$-direction with a suitable initial speed. For a fixed $\upsilon_0$, when $(\theta_0, \theta_1) = (45^\circ, 45^\circ)$, the stone hits the ball after time $T_1$, and when $(\theta_0, \theta_1) = (60^\circ, 30^\circ)$, it hits the ball after time $T_2$. In such a case, $(T_1/T_2)^2$ is ________.

Answer: 2

Solution

Let $B$: Ball $S$: Stone $v_0$: Initial speed of stone. Since relative acceleration $= 0$ Path seen would be straight line To meet, $v_0 \sin \theta_0 = v \sin \theta_1$ And $$\Delta t = \frac{L}{v_0 \cos \theta_1 + v_0 \cos \theta_0}$$ Case I: $v_0 = v_0 \Rightarrow \Delta t_1 = T_1 = \frac{L}{v_0 \left[ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right]} = \frac{L}{\sqrt{2} v_0}$ Case II: $\sqrt{3} v_0 = v_0 \Rightarrow \Delta t_2 = T_2 = \frac{L}{\sqrt{3} v_0 \cdot \frac{\sqrt{3}}{2} + v_0 \cdot \frac{\sqrt{3}}{2}} = \frac{L}{2 v_0}$ $$\Rightarrow \left( \frac{T_1}{T_2} \right)^2 = (\sqrt{2})^2 = 2$$

Question 27

Physics · Electric Charges and Fields · Numerical

A charge is kept at the central point $P$ of a cylindrical region. The two edges subtend a half-angle $\theta$ at $P$, as shown in the figure. When $\theta = 30^\circ$, then the electric flux through the curved surface of the cylinder is $\Phi$. If $\theta = 60^\circ$, then the electric flux through the curved surface becomes $\Phi / \sqrt{n}$, where the value of $n$ is ______.

Answer: 3

Solution

For any $\theta$, let us first find the flux inside a cone of half angle $\theta$. We know that for such a cone, solid angle subtended at centre is $\Omega = 2\pi [1 - \cos \theta]$. Therefore, flux through 1 cone $= \phi_0 = \frac{\Omega}{4\pi} \cdot \frac{Q}{\varepsilon_0} = \frac{Q}{2\varepsilon_0} [1 - \cos \theta]$. Flux through curved surface $= \frac{Q}{\varepsilon_0} - 2\phi_0 = \frac{Q}{\varepsilon_0} - \frac{Q}{\varepsilon_0} [1 - \cos \theta] = \frac{Q}{\varepsilon_0} \cos \theta$. Therefore, $\phi = \frac{Q}{\varepsilon_0} \cdot \frac{\sqrt{3}}{2}$. And $\frac{\phi}{\sqrt{n}} = \frac{Q}{\varepsilon_0} \cdot \frac{1}{2}$. Therefore, $\sqrt{n} = \sqrt{3}$, so $n = 3$.

Question 28

Physics · Ray Optics and Optical Instruments · Fill in the blank

Two equilateral-triangular prisms $P_1$ and $P_2$ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism $P_1$ at an angle of incidence $\theta$ such that the outgoing ray undergoes minimum deviation in prism $P_2$. If the respective refractive indices of $P_1$ and $P_2$ are $\sqrt{\frac{3}{2}}$ and $\sqrt{3}$, $\theta = \sin^{-1}\left[ \sqrt{\frac{3}{2}} \sin\left( \frac{\pi}{\beta} \right) \right]$, where the value of $\beta$ is $\_$$\_$$\_$$\_$$\_$.

Answer: 12

Solution

By using optical reversibility principle. For prism $P_2$: Minimum deviation. $1 \times \sin \theta_1 = \sqrt{3} \sin r$, $r_1 = r_2 = \frac{A}{2}$. $\sin \theta_1 = \sqrt{3} \times \frac{1}{2}$, $r_1 = r_2 = 30^\circ$. $\Rightarrow \; i = e = 60^\circ$. For prism $P_1$: Incident angle will be $60^\circ$. $1 \times \sin 60^\circ = \frac{\sqrt{3}}{\sqrt{2}} \sin r_1$. $$\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{2}} \sin r_1$$. $r_1 + r_2 = 60^\circ$. $\sin r_1 = \frac{1}{\sqrt{2}}$. $$r_1 = 45^\circ$$. $r_2 = 15^\circ$. $$\frac{\sqrt{3}}{\sqrt{2}} \sin(45^\circ) = 1 \times \sin \theta$$. $$15^\circ - \frac{\pi \times 15}{180} \; \mathrm{rad} = \frac{\pi}{12} \; \mathrm{rad}$$. $$\theta = \sin^{-1} \left[ \frac{\sqrt{3}}{\sqrt{2}} \sin \left( \frac{\pi}{12} \right) \right]$$. $\beta = 12$

Question 29

Physics · Electrostatic Potential and Capacitance · Numerical

An infinitely long thin wire, having a uniform charge density per unit length of 5 $\mathrm{nC/m}$, is passing through a spherical shell of radius 1 $\mathrm{m}$, as shown in the figure. A 10 $\mathrm{nC}$ charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points $P$ and $R$, in Volt, is _______. [Given: In SI units $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9$, $\ln 2 = 0.7$. Ignore the area pierced by the wire.]

Answer: 171

Solution

The electric field due to a line charge is given by $$E_{Line charge} = \frac{\lambda}{2 \pi \varepsilon_0 \, r}$$ Therefore, $$\Delta V_{Line charge} = \int_{0.5}^{2} \frac{\lambda}{2 \pi \varepsilon_0 \, r} \, dr = \frac{\lambda}{2 \pi \varepsilon_0} \ln 4 (i)$$ The potential difference due to the sphere is $$\Delta V_{Sphere} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R} - \frac{1}{4 \pi \varepsilon_0} \frac{Q}{2R}$$ which simplifies to $$= \frac{1}{4 \pi \varepsilon_0} \frac{Q}{2} (ii)$$ Thus, the net potential difference is $$\Delta V_{Net} = \frac{\lambda}{2 \pi \varepsilon_0} \ln 4 + \frac{1}{4 \pi \varepsilon_0} \frac{Q}{2}$$ This results in a potential difference of 171 Volts.

Question 30

Physics · Mechanical Properties of Fluids · Numerical

A spherical soap bubble inside an air chamber at pressure $P_0 = 10^5 \, \mathrm{Pa}$ has a certain radius so that the excess pressure inside the bubble is $\Delta P = 144 \, \mathrm{Pa}$. Now, the chamber pressure is reduced to $\frac{8P_0}{27}$ so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure $\Delta P$ in both the cases to be much smaller than the chamber pressure. The new excess pressure $\Delta P$ in Pa is

Answer: 96

Solution

Since the situation follows isothermal condition. $P_1 V_1 = P_2 V_2$ $$V_1 = \frac{4}{3} \pi R_1^3, \ V_2 = \frac{4}{3} \pi R_2^3$$ $$P_1 = P_0 + \Delta P_1, \ \Delta P_1 = \frac{4T}{R_1}$$ and $$P_2 = \frac{8P_0}{27} + \Delta P_2, \ \Delta P_2 = \frac{4T}{R_2}$$ So for isothermal condition $$\left( P_0 + \Delta P_1 \right) \times \frac{4}{3} \pi R_1^3 = \left( \frac{8P_0}{27} + \Delta P_2 \right) \times \frac{4}{3} \pi R_2^3$$ here $P_0 = 10^5 \, \mathrm{Pa}$ $\Delta P_1 = 144 \, \mathrm{Pa}$ and $\Delta P_1 \ll P_0$ So $$\left( P_0 + \Delta P_1 \right) \left( \frac{4T}{\Delta P_1} \right)^3 = \left( \frac{8P_0}{27} + \Delta P_2 \right) \left( \frac{4T}{\Delta P_2} \right)^3$$ $$\frac{P_0}{(\Delta P_1)^3} \frac{8P_0}{27} \times \frac{1}{(\Delta P_2)^3}$$ $$\Delta P_2 = \frac{2}{3} \Delta P_1 = \frac{2}{3} \times (144 \, \mathrm{Pa})$$ $$\Delta P_2 = 96 \, \mathrm{Pa}$$

Question 31

Physics · Wave Optics · Numerical

In a Young's double slit experiment, each of the two slits $A$ and $B$, as shown in the figure, are oscillating about their fixed center and with a mean separation of $0.8\,\mathrm{mm}$. The distance between the slits at time $t$ is given by $d=(0.8+0.04\sin\omega t)\,\mathrm{mm}$, where $\omega=0.08\,\mathrm{rad\,s^{-1}}$. The distance of the screen from the slits is $1\,\mathrm{m}$ and the wavelength of the light used to illuminate the slits is $6000\,\mathrm{\AA}$. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point $O$. The $8^{th}$ bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer ($\mu$ m), is

Answer: 601.5

Solution

As central bright fringe position is not changing, the two slits are oscillating with a phase difference of $\pi$. For 8th bright fringe $$y = \frac{8 \lambda D}{(0.8 + 0.04 \sin \omega t)} \times 10^3$$ $$= \frac{8 \times 6000 \times 10^{-10} \times 10^3}{(0.8 + 0.04 \sin \omega t)}$$ $$y = \frac{48 \times 10^{-4}}{(0.8 + 0.04 \sin \omega t)}$$ $d$ varies from 0.84 mm to 0.76 mm $$\Delta y = 6.015 \times 10^{-4}$$ $$= 601.50 \, \mu \mathrm{m}$$

Question 32

Physics · Wave Optics · Numerical

In a Young's double slit experiment, each of the two slits $A$ and $B$, as shown in the figure, are oscillating about their fixed center and with a mean separation of $0.8\,\mathrm{mm}$. The distance between the slits at time $t$ is given by $d=(0.8+0.04\sin\omega t)\,\mathrm{mm}$, where $\omega=0.08\,\mathrm{rad\,s^{-1}}$. The distance of the screen from the slits is $1\,\mathrm{m}$ and the wavelength of the light used to illuminate the slits is $6000\,\mathrm{\AA}$. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point $O$. The maximum speed in $\mu \, \mathrm{m/s}$ at which the $8^{th}$ bright fringe will move is .

Answer: 24

Solution

Finding speed $$\frac{\delta y}{\delta t} = \frac{\delta}{\delta t} \left( \frac{8 \lambda D}{d} \right)$$ $$= -\frac{8 \lambda D}{d^2} \frac{\delta d}{(\delta t)}$$ $$v = -\frac{8 \lambda D}{d^2} (0.04 \omega \cos \omega t) \times 10^{-3}$$ $$v_{\max} = \frac{8 \lambda D}{d^2} \times 4 \omega \times 10^{-5}$$ $$= \frac{8 \times 6 \times 10^{-7} \times 1 \times 4 \times 8 \times 10^{-7}}{64 \times 10^{-8}}$$ $$= 24 \times 10^{-6}$$ $$= 24 \, \mu \mathrm{m/s}$$

Question 33

Physics · System of Particles and Rotational Motion · Numerical

Two particles, $1$ and $2$, each of mass $m$, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at $x_0$, are oscillating with amplitude $a$ and angular frequency $\omega$. Thus, their positions at time $t$ are given by $x_1(t)=(x_0+d)+a\sin\omega t$ and $x_2(t)=(x_0-d)-a\sin\omega t$, respectively, where $d>2a$. Particle $3$ of mass $m$ moves towards this system with speed $v_0=a\omega/2$, and undergoes instantaneous elastic collision with particle $2$, at time $t_0$. Finally, particles $1$ and $2$ acquire a center of mass speed $v_{cm}$ and oscillate with amplitude $b$ and the same angular frequency $\omega$. If the collision occurs at time $t_0 = 0$, the value of $v_{cm}/(a \omega)$ will be

Answer: 0.75

Solution

At $t = 0$, 2 is at mean position. Therefore, $u_2 = a \omega$ towards left after collision, velocity will exchange. Thus, $v_2 = \frac{a \omega}{2}$ towards right. $u_1 = a \omega$ towards right. Therefore, $v_{cm} = \frac{3 a \omega}{4}$. $$\frac{v_{cm}}{a \omega} = \frac{3}{4} = 0.75$$ At $t = \frac{\pi}{2 \omega}$, $u_2 = 0$. After collision, $v_2 = \frac{a \omega}{2}$ towards right. $u_1 = 0$. Therefore, $v_{cm} = \frac{a \omega}{4}$ towards right. With respect to the centre of mass: $$v = \omega \sqrt{A^2 - x^2}$$ $$\frac{a \omega}{4} = \omega \sqrt{A^2 - a^2}$$ $$\frac{a^2}{16} + a^2 = A^2$$ $$\frac{17}{16} a^2 = A^2 = b^2$$ Therefore, $b^2 = \frac{17}{16} a^2$. $$\frac{4b^2}{a^2} = \frac{17}{4}$$ $$= 4.25$$

Question 34

Physics · Work, Energy and Power · Numerical

Answer: 4.25

Solution

At $t = 0$, 2 is at mean position. Therefore, $u_2 = a \omega$ towards left after collision, velocity will exchange. Thus, $v_2 = \frac{a \omega}{2}$ towards right. $u_1 = a \omega$ towards right. Therefore, $v_{cm} = \frac{3a \omega}{4}$. $$\frac{v_{cm}}{a \omega} = \frac{3}{4} = 0.75$$ At $t = \frac{\pi}{2 \omega}$, $u_2 = 0$. After collision, $v_2 = \frac{a \omega}{2}$ towards right. $u_1 = 0$. Therefore, $v_{cm} = \frac{a \omega}{4}$ towards right with respect to the center of mass. $$v = \omega \sqrt{A^2 - x^2}$$ $$\frac{a \omega}{4} = \omega \sqrt{A^2 - a^2}$$ $$\frac{a^2}{16} + a^2 = A^2$$ $$\frac{17}{16} a^2 = A^2 = b^2$$ Therefore, $b^2 = \frac{17}{16} a^2$. $$\frac{4b^2}{a^2} = \frac{17}{4}$$ $$= 4.25$$

Chemistry

Question 35

Chemistry · Structure of Atom · Single correct

According to Bohr's model, the highest kinetic energy is associated with the electron in the

First orbit of H atom
First orbit of He$^+$
Second orbit of He$^+$
Second orbit of Li$^{2+}$

Answer: (b)

Solution

K.E. of electron in $n^{th}$ Bohr's orbit, $$K.E. = 13.6 \frac{Z^2}{n^2} eV/atom$$ For $n = 1$ (H-atom), $K.E. \propto \frac{1^2}{1^2} = 1$. For $n = 1$ (He$^+$ ion), $K.E. \propto \frac{2^2}{1^2} = 4$. For $n = 2$ (He$^+$ ion), $K.E. \propto \frac{2^2}{2^2} = 1$. For $n = 2$ (Li$^{2+}$ ion), $K.E. \propto \frac{3^2}{2^2} = \frac{9}{4}$. Highest for $\rightarrow n = 1$ of He$^+$ ion.

Question 36

Chemistry · Redox Reactions · Single correct

In a metal deficient oxide sample, $\mathrm{M}_x\mathrm{Y}_2\mathrm{O}_4$ (M and Y are metals), M is present in both $+2$ and $+3$ oxidation states and Y is in $+3$ oxidation state. If the fraction of $\mathrm{M}^{2+}$ ions present in M is $\frac{1}{3}$, the value of X is _____.

0.25
0.33
0.67
0.75

Answer: (a)

Solution

Given $\mathrm{M}_x\mathrm{Y}_2\mathrm{O}_4$. $\mathrm{M}^{+2} = \frac{X}{3}$, $\mathrm{M}^{+3} = \frac{2X}{3}$. So, total of O.N. of all atoms $$\frac{2X}{3} + 3\left(\frac{2X}{3}\right) + 2(+3) + 4(-2) = 0$$ $$\frac{2X}{3} + 2X + 6 - 8 = 0$$ $$\frac{8X}{3} = 2$$ $$X = \frac{6}{8} = \frac{3}{4} = 0.75$$

Question 37

Chemistry · Hydrocarbons · Single correct

In the following reaction sequence, the major product Q is

Answer: (d)

Solution

L-Glucose $\mathrm{C_6H_{12}O_6}$ is treated with HI and heat to form $\mathrm{C_6H_{14}}$ (n-Hexane). This is then reacted with $\mathrm{Cr_2O_3}$ at $775 \, \mathrm{K}$ and $10{-}20 \, \mathrm{atm}$ to form benzene. Finally, benzene is chlorinated with excess $\mathrm{Cl_2}$ under UV light to form benzene hexachloride (BHC).

Question 38

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct

The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are

$[\mathrm{PF}_4]^+[\mathrm{PF}_6]^-$ and $[\mathrm{PCl}_4]^+[\mathrm{PF}_6]^-$
$[\mathrm{PCl}_4]^+[\mathrm{PCl}_4\mathrm{F}_2]^-$ and $[\mathrm{PCl}_4]^+[\mathrm{PF}_6]^-$
$\mathrm{PF}_3$ and $\mathrm{PCl}_3$
$\mathrm{PF}_5$ and $\mathrm{PCl}_3$

Answer: (b)

Solution

If $\mathrm{PCl_5}$ is fluorinated in a polar solvent, ionic isomers are formed. e.g.: $[\mathrm{PCl_4}]^+[\mathrm{PCl_4F_2}]^-$ (colourless crystals) and $[\mathrm{PCl_4}]^+[\mathrm{PF_6}]^-$ (white crystals).

Question 39

Chemistry · Electrochemistry · Multiple correct

An aqueous solution of hydrazine ($\mathrm{N_2H_4}$) is electrochemically oxidized by $\mathrm{O_2}$, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is $\mathrm{N_2(g)}$. Choose the correct statement(s) about the above process

$\mathrm{OH^-}$ ions react with $\mathrm{N_2H_4}$ at the anode to form $\mathrm{N_2(g)}$ and water, releasing 4 electrons to the anode.
At the cathode, $\mathrm{N_2H_4}$ breaks to $\mathrm{N_2(g)}$ and nascent hydrogen released at the electrode reacts with oxygen to form water.
At the cathode, molecular oxygen gets converted to $\mathrm{OH^-}$.
Oxides of nitrogen are major by-products of the electrochemical process.

Answer: (a), (c)

Solution

At anode: $\mathrm{N_2H_4} + 4\mathrm{OH}^- \rightarrow \mathrm{N_2} + 4\mathrm{H_2O} + 4e^-$. At cathode: $\mathrm{O_2} + 2\mathrm{H_2O} + 4e^- \rightarrow 4\mathrm{OH}^-$. Complete reaction: $\mathrm{N_2H_4} + \mathrm{O_2} \rightarrow \mathrm{N_2} + 2\mathrm{H_2O}$. Statements (A) and (C) are correct.

Question 40

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Multiple correct

The option(s) with correct sequence of reagents for the conversion of $P$ to $Q$ is(are)

i) Lindlar's catalyst, $H_2$; ii) $SnCl_2/HCl$; iii) $NaBH_4$; iv) $H_3O^+$
i) Lindlar's catalyst, $H_2$; ii) $H_3O^+$; iii) $SnCl_2/HCl$; iv) $NaBH_4$
i) $NaBH_4$; ii) $SnCl_2/HCl$; iii) $H_3O^+$; iv) Lindlar's catalyst, $H_2$
i) Lindlar's catalyst, $H_2$; ii) $NaBH_4$; iii) $SnCl_2/HCl$; iv) $H_3O^+$

Answer: (c), (d)

Solution

The reaction sequence involves the following steps: 1. The compound (P) is reduced using $\mathrm{NaBH_4}$ to form an intermediate. 2. This intermediate undergoes a reaction with $\mathrm{SnCl_2}$ and $\mathrm{HCl}$ to form another compound. 3. The resulting compound is then treated with $\mathrm{H_3O^+}$ to form a carboxylic acid and an aldehyde. 4. Finally, the compound is subjected to hydrogenation using Lindlar's catalyst and $\mathrm{H_2}$ to form the final product (Q).

Question 41

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Multiple correct

The compound(s) having peroxide linkage is(are)

$\mathrm{H_2S_2O_7}$
$\mathrm{H_2S_2O_8}$
$\mathrm{H_2S_2O_5}$
$\mathrm{H_2SO_5}$

Answer: (b), (d)

Solution

The solution is $\mathrm{H_2S_2O_7}$. The structures for the given compounds are shown. $\mathrm{H_2S_2O_7}$ is the correct structure.

Question 42

Chemistry · Surface Chemistry · Numerical

To form a complete monolayer of acetic acid on $1\,\mathrm{g}$ of charcoal, $100\,\mathrm{mL}$ of $0.5\,\mathrm{M}$ acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, $40\,\mathrm{mL}$ of $1\,\mathrm{M}$ NaOH solution was required. If each molecule of acetic acid occupies $P\times10^{-23}\,\mathrm{m^2}$ surface area on charcoal, the value of $P$ is $\underline{\hspace{2cm}}$ $\text{[Use given data:}$ $\text{Surface area of charcoal }=1.5\times10^2\,\mathrm{m^2\,g^{-1}};$ $\text{Avogadro's number }(N_A)=6.0\times10^{23}\,\mathrm{mol^{-1}}\text{]}$

Answer: 2500

Solution

Number of moles of unadsorbed CH$_3$COOH = $$$ $\frac{40 \times 1}{1000}$ = 4 $\times$ 10^{-2} $$$ mol Number of moles of adsorbed CH$_3$COOH = $$$ $\frac{100 \times 0.5}{1000}$ - 4 $\times$ 10^{-2} $$$ $$$ = 10^{-2} $$$ mol Surface area occupied by one molecule of CH$_3$COOH = $$$ $\frac{1.5 \times 10^2}{10^{-2} \times 6 \times 10^{23}}$ = $\frac{150 \times 10^2 \times 10^{-23}}{6}$ $$$ $$$ = 2500 $\times$ 10^{-23} $$$ m$^2$ Therefore, as per question P = 2500

Question 43

Chemistry · Solutions · Fill in the blank

Vessel-1 contains $w_2$ g of a non-volatile solute $X$ dissolved in $w_1$ g of water. Vessel-2 contains $w_2$ g of another non-volatile solute $Y$ dissolved in $w_1$ g of water. Both the vessels are at the same temperature and pressure. The molar mass of $X$ is 80$\%$ of that of $Y$. The van’t Hoff factor for $X$ is 1.2 times of that of $Y$ for their respective concentrations. The elevation of boiling point for solution in Vessel-1 is _____ $\%$ of the solution in Vessel-2.

Answer: 150

Solution

Sol. Vessel-I $$ (\Delta T_B)_I = i_X \frac{w_2}{M_X} \cdot \frac{1}{w_1} \times 1000 \times K_b $$ $M_X =$ Molar mass of 'X' Vessel-II $$ (\Delta T_B)_{II} = i_Y \frac{w_2}{M_Y} \cdot \frac{1}{w_1} \times 1000 \times K_B $$ $M_Y =$ Molar mass of 'Y' $$ \frac{(\Delta T_b)_I}{(\Delta T_b)_{II}} \times 100 = \frac{i_X}{i_Y} \cdot \frac{M_Y}{M_X} \times 100 $$ $$ = 1.2 \times \frac{100}{80} \times 100 $$ $$ = 150\% $$

Question 44

Chemistry · Biomolecules · Numerical

For a double strand DNA, one strand is given below: The amount of energy required to split the double strand DNA into two single strands is _____ kcal mol$^{-1}$. [Given: Average energy per H-bond for A-T base pair = 1.0 kcal mol$^{-1}$, G-C base pair = 1.5 kcal mol$^{-1}$, and A-U base pair = 1.25 kcal mol$^{-1}$. Ignore electrostatic repulsion between the phosphate groups.]

Answer: 41

Solution

Total energy = [BE H-bond A – T × No. of A = T pair × 2] + [BE H-bond G – C × No. of G ≡ C pair × 3] = [1 × 7 × 2] + [1.5 × 6 × 3] = 14 + 27 = 41 kcal

Question 45

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into Pb-206 while the rest of it remains undisintegrated. When the age of the sample is $P \times 10^8$ years, the ratio of mass of Pb-206 to that of U-238 in the sample is found to be 7. The value of $P$ is ______. [Given: Half-life of U-238 is $4.5 \times 10^9$ years; $\log_e 2 = 0.693$]

Answer: 143

Solution

Life of sample $\rightarrow t$ years $[A]_0 \propto$ Initial mole of U-238 $[A]_t \propto$ Final mole of U-238 $$\frac{[A]_0}{[A]_t} = \frac{\frac{1}{238} + \frac{7}{206}}{\frac{1}{238}}$$ $$= \frac{0.0042 + 0.0340}{0.0042}$$ $$= 9.1$$ $$= \frac{2.303 \log 2 \times t}{4.5 \times 10^9} = 2.303 \log 9.1$$ $$t = 14.27 \times 10^9 years$$ $$= 142.7 \times 10^9 years$$ $P = 142.7$ $P \simeq 143$

Question 46

Chemistry · Chemical Bonding and Molecular Structure · Numerical

Among $[\mathrm{Co(CN)}_4]^{4-}$, $[\mathrm{Co(CO)}_3(\mathrm{NO})]$, $\mathrm{XeF}_4$, $[\mathrm{PCl}_4]^+$, $[\mathrm{PdCl}_4]^{2-}$, $[\mathrm{ICl}_4]^-$, $[\mathrm{Cu(CN)}_4]^{3-}$ and $\mathrm{P}_4$ the total number of species with tetrahedral geometry is .

Answer: 3

Solution

[$\mathrm{Co(CN)_4}$]^{4-} $\rightarrow$ $\mathrm{Co}$^{0} = 3d^74s^2 Due to SFL, $\mathrm{CN}$^- pairing and transference of electron takes place and hybridisation is $dsp^2$ Geometry $\Rightarrow$ Square planer [$\mathrm{Co(CO)_3NO}$] $\mathrm{Co}$^{-1} $\rightarrow$ 3d^{10} due to SFL CO and NO $sp^3$ hybridisation Geometry = Tetrahedral $\mathrm{XeF_4}$ $\Rightarrow$ 4bp + 2lp $\Rightarrow$ sp^3d^2 Square planer $\mathrm{PCl_4}$^+ $\Rightarrow$ 4pb + 0lp $sp^3 \Rightarrow$ tetrahedral [$\mathrm{PdCl_4}$]^{2-} $\rightarrow$ $\mathrm{Pd}$^{2+}, $\mathrm{Cl}$^- behaves as SFL $\mathrm{Pd}$^{2+} $\rightarrow$ 4d^8 $\Rightarrow$ dsp^2 $\Rightarrow$ square planer $\mathrm{ICl_4}$^- $\Rightarrow$ 4bp + 2lp $sp^3d^2$ square planer [$\mathrm{Cu(CN)_4}$]^{3-} $\rightarrow$ $\mathrm{Cu}$^{+1} $\rightarrow$ 3d^{10} $\Rightarrow$ sp^3 Tetrahedral

Question 47

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

An organic compound $P$ having molecular formula $C_6H_6O_3$ gives ferric chloride test and does not have intramolecular hydrogen bond. The compound $P$ reacts with 3 equivalents of $\mathrm{NH_2OH}$ to produce oxime $Q$. Treatment of $P$ with excess methyl iodide in the presence of $\mathrm{KOH}$ produces compound $R$ as the major product. Reaction of $R$ with excess iso-butylmagnesium bromide followed by treatment with $\mathrm{H_3O^+}$ gives compound $S$ as the major product. The total number of methyl ($-\mathrm{CH_3}$) group(s) in compound $S$ is .

Answer: 12

Solution

The reaction starts with compound (P), which is converted to a compound with three carbonyl groups. This compound reacts with 3 equivalents of $\mathrm{NH_2OH}$ to form compound (Q), which has three $\mathrm{NOH}$ groups. Next, the compound with carbonyl groups reacts with excess $\mathrm{CH_3I}$ in the presence of $\mathrm{KOH}$ to form compound (R), which is a methylated derivative. Compound (R) then undergoes a reaction with excess $\mathrm{CH_3CH_2CH_2MgBr}$ followed by hydrolysis with $\mathrm{H_3O^+}$ to form the final product. The final product contains 12 $\mathrm{CH_3}$ groups.

Question 48

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Fill in the blank

An organic compound $P$ with molecular formula $\mathrm{C_9H_{18}O_2}$ decolorizes bromine water and also shows positive iodoform test. $P$ on ozonolysis followed by treatment with $\mathrm{H_2O_2}$ gives $Q$ and $R$. While compound $Q$ shows positive iodoform test, compound $R$ does not give positive iodoform test. Both $Q$ and $R$ on oxidation with pyridinium chlorochromate (PCC) followed by heating give $S$ and $T$, respectively. Both $S$ and $T$ show positive iodoform test. Complete copolymerization of $500$ moles of $Q$ and $500$ moles of $R$ gives one mole of a single acyclic copolymer $U$. [Given, atomic mass: H=1, C=12, O=16] Sum of number of oxygen atoms in S and T is _____.

Answer: 2

Solution

An organic compound P with molecular formula $C_9H_{18}O_2$ decolorizes bromine water and also shows positive iodoform test. P on ozonolysis followed by treatment with $H_2O_2$ gives Q and R. While compound Q shows positive iodoform test, compound R does not give positive iodoform test. Q and R on oxidation with pyridinium chlorochromate (PCC) followed by heating give S and T, respectively. Both S and T show positive iodoform test. Complete copolymerization of 500 moles of Q and 500 moles of R gives one mole of a single acyclic copolymer U. [Given, atomic mass: $H = 1$, $C = 12$, $O = 16$]

Question 49

Chemistry · Some Basic Concepts of Chemistry · Fill in the blank

Answer: 103018

Solution

Mol. wt. of polymer = (106 $\times$ 500) + (118 $\times$ 500) - 18 $\times$ 499 = 53000 + 59000 - 8982 = 103018 $\,$ $\mathrm{g}$ When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex $P$ is formed. In a strong acidic medium, the equilibrium shifts completely towards $P$. Addition of zinc chloride to $P$ in a slightly acidic medium results in a sparingly soluble complex $Q$.

Question 50

Chemistry · Some Basic Concepts of Chemistry · Fill in the blank

When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex $P$ is formed. In a strong acidic medium, the equilibrium shifts completely towards $P$. Addition of zinc chloride to $P$ in a slightly acidic medium results in a sparingly soluble complex $Q$. The number of moles of potassium iodide required to produce two moles of P is _______.

Answer: 2

Solution

From this equation we need 2 mol of KI $$2\mathrm{KI} + 2\mathrm{K_3[Fe(CN)_6]} \rightarrow \mathrm{I_2} + 2\mathrm{K_4[Fe(CN)_6]}$$ $$2\mathrm{K_4[Fe(CN)_6]} + 3\mathrm{ZnCl_2} \rightarrow \mathrm{K_2Zn_3[Fe(CN)_6]_2} + 6\mathrm{KCl}$$ When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex $P$ is formed. In a strong acidic medium, the equilibrium shifts completely towards $P$. Addition of zinc chloride to $P$ in a slightly acidic medium results in a sparingly soluble complex $Q$.

Question 51

Chemistry · Co-ordination Compounds · Fill in the blank

When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex $P$ is formed. In a strong acidic medium, the equilibrium shifts completely towards $P$. Addition of zinc chloride to $P$ in a slightly acidic medium results in a sparingly soluble complex $Q$. The number of zinc ions present in the molecular formula of Q is _____.

Answer: 3

Solution