JEE Main 6 April 2023 Shift 1 question paper with solutions

Given: $$5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3 \ldots (1)$$ Replace $x \rightarrow \frac{1}{x}$ $$5f\left(\frac{1}{x}\right) + 4f(x) = x + 3 \ldots (2)$$ Eliminating $f\left(\frac{1}{x}\right)$ from (1) and (2), we get $$9f(x) = \frac{5}{x} + 15 - 4x - 12$$ $$\Rightarrow 9f(x) = \frac{5}{x} - 4x + 3$$ $$\Rightarrow 18f(x) = \frac{10}{x} - 8x + 6$$ $$\Rightarrow 18 \int_1^2 f(x) dx = \int_1^2 \left(\frac{10}{x} - 8x + 6\right) dx$$ $$\Rightarrow 18 \int_1^2 f(x) dx = \left[10 \log x - 4x^2 + 6x\right]_1^2$$ $$\Rightarrow 18 \int_1^2 f(x) dx = [(10 \log 2 - 4) - (10 \log 1 + 2)]$$ $$\Rightarrow 18 \int_1^2 f(x) dx = 10 \log 2 - 6$$ Hence this is the correct option.

Given that, a pair of dice is thrown 5 times. Required Sample space is $\($ $\{$(1,1),(1,2),(1,3),$\ldots$,(6,6)$\}$ $\)$. Total number of observations = 36. Now, a sum of 5 is observed in $\($ $\{$(1,4),(4,1),(2,3),(3,2)$\}$ $\)$. Number of favourable outcomes = 4. $\($ P(success) = p = $\frac{4}{36}$ = $\frac{1}{9}$ $\)$ $\($ P(failure) = q = 1 - $\frac{1}{9}$ = $\frac{8}{9}$ $\)$ We know that for a binomial distribution, $\($ B(n,p) $\)\[$ P(X = k) = \binom{n}{k} p^k q^{n-k} $\]$ where \( k \geq 0 \)and \(p + q = 1 \). Therefore, the required probability $\[ $\Rightarrow P(X \geq 4) = P(X = 4) + P(X = 5) $\]$ $\[$ = $\binom{5}{4}$ $\left$( $\frac{1}{9}$ $\right$)^4 $\left$( $\frac{8}{9}$ $\right$) + $\binom{5}{5}$ $\left$( $\frac{1}{9}$ \right$)^5 $\]$ $\[$ = 5 \cdot \frac{8}{9^5} + \frac{1}{9^5} = \frac{41}{9^5} $\]$ $\[$ = \frac{41 \times 3}{3^{10} \times 3} = \frac{123}{3^{11}} $\]$ But given that \( P(X \geq 4) = \frac{k}{3^{11}} \) $\[$ \Rightarrow \frac{k}{3^{11}} = \frac{123}{3^{11}} \] = \[k = 123 ] Hence, this is the correct option.

We have been given that ${}^{2n}C_{3} : {}^{n}C_{3} = 10$. We know that ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$. $$\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 10$$ $$\Rightarrow \frac{(2n)!}{(2n-3)!} \times \frac{(n-3)!}{n!} = 10$$ $$\Rightarrow \frac{(2n)(2n-1)(2n-2)}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 10$$ $$\Rightarrow \frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 10$$ $$\Rightarrow \frac{(4)(2n-1)}{(n-2)} = 10$$ $$\Rightarrow 8n - 4 = 10n - 20$$ $$\Rightarrow n = 8$$ Hence, the value of $\frac{n^2 + 3n}{n^2 - 3n + 4}$ is $\frac{8^2 + 3(8)}{8^2 - 3(8) + 4} = \frac{88}{44} = 2$. Therefore, the value of $\frac{n^2 + 3n}{n^2 - 3n + 4}$ is $2 : 1$.

Given expansion is $\left( \sqrt[4]{2} + \frac{1}{\sqrt[4]{3}} \right)^n$. We know that general term of an expansion $(x + a)^n$ is $T_{r+1} = \binom{n}{r} x^{n-r} a^r$. (5th term from beginning) $$T_{4+1} = \binom{n}{4} \left( \sqrt[4]{2} \right)^{n-4} \cdot \left( \frac{1}{\sqrt[4]{3}} \right)^4$$ (5th term from end) We know that $r$th term from the end in a binomial expansion is $T_{n-r+2}$ from the beginning. $$T'_5 = T_{n-4+1} = \binom{n}{4} \left( \frac{1}{\sqrt[4]{3}} \right)^{n-4} \left( \sqrt[4]{2} \right)^4$$ It is also given that the ratio of the fifth term from the beginning to the fifth term from the end is $\sqrt{6} : 1$. Now, $$\frac{T_5}{T'_5} = \frac{\sqrt{6}}{1}$$ $$\Rightarrow \frac{2^{\frac{n-8}{4}} \cdot 3^{-\frac{n-8}{4}}}{3^{-1}} = \sqrt{6}$$ $$\therefore \left( \sqrt{6} \right)^{\frac{n-8}{2}} = \left( \sqrt{6} \right)$$ $$\therefore n = 10$$ $$\therefore T_3 = \binom{10}{2} \left( \sqrt[4]{2} \right)^8 \left( \frac{1}{\sqrt[4]{3}} \right)^2 = \frac{45 \times 4}{\sqrt{3}}$$ $$= 60 \sqrt{3}$$ Hence, this is the correct option.

Given that, \[ \vec{a}=2\hat{i}+3\hat{j}+4\hat{k},\qquad \vec{b}=\hat{i}-2\hat{j}-2\hat{k} \] and \[ \vec{c}=-\hat{i}+4\hat{j}+3\hat{k}. \] Let us find $\vec{b}$ $\times$ $\vec{c}$ $_$ $\Rightarrow$ $\vec{b}$ $\times$ $\vec{c}$ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -2 & -2\\ -1 & 4 & 3 \end{vmatrix} \] \[ =-(6+8)\hat{i}-(3-2)\hat{j}+(4-2)\hat{k} \] \[ \Rightarrow \vec{b}\times\vec{c} = -14\hat{i}-\hat{j}+2\hat{k} \] Since $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, $\vec{d}$ = $\lambda(2\hat{i}-\hat{j}+2\hat{k})$ Also, $\vec{a}\cdot\vec{d}=18$ \[ \Rightarrow \lambda(2\hat{i}+3\hat{j}+4\hat{k}) \cdot (2\hat{i}-\hat{j}+2\hat{k}) =18 \] \[ \Rightarrow 9\lambda=18 \] \[ \Rightarrow \lambda=2 \] Now let us apply Lagrange's identity which is as follows, $|\vec{a}\times\vec{d}|^2$ = $|\vec{a}|^2\,|\vec{d}|^2$ - $(\vec{a}\cdot\vec{d})^2$ \[ = |2\hat{i}+3\hat{j}+4\hat{k}|^2 \,|2(2\hat{i}-\hat{j}+2\hat{k})|^2 -(18)^2 \] \[ =720 \]

Given, the straight lines $l_1$ and $l_2$ pass through the origin and trisect the line segment of the line $L : 9x + 5y = 45$ between the axes. And $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$. Now on plotting the diagram we get, given equation of line, $L : 9x + 5y = 45$ $$\frac{x}{5} + \frac{y}{9} = 1$$ Now using the section formula between point $A(5, 0)$ and $B(0, 9)$ we get the value of point $C$ and $D$ $$C \equiv \left( \frac{10}{3}, 3 \right) and D \equiv \left( \frac{5}{3}, 6 \right)$$ Now finding the slope $m_1$ and $m_2$ we get, $$m_1 = \frac{3 - 0}{10/3 - 5} = \frac{9}{10} \& m_2 = \frac{6 \times 3}{5} = \frac{18}{5}$$ So, equation of line $y = (m_1 + m_2)x$ will be, $$y = \left( \frac{9}{10} + \frac{36}{10} \right)x = \frac{9}{2}x$$ So, intersection point with $L$ will be, $$7y = 45 \Rightarrow y = \frac{45}{7}, x = \frac{10}{7}$$ Hence, $y - x = \frac{45 - 10}{7} = 5$

Given that from the top $A$ of a vertical wall $AB$ of height $30\,\mathrm{m}$, the angles of depression of the top $P$ and bottom $Q$ of a vertical tower $PQ$ are $15^\circ$ and $60^\circ$ respectively, $B$ and $Q$ are on the same horizontal level and $C$ is a point on $AB$ such that $CB=PQ$. The required diagram will be From the diagram, \[ \Rightarrow \tan 60^\circ = \frac{AB}{BQ} \] \[ \Rightarrow \frac{30}{x} = \sqrt{3} \] \[ \Rightarrow x = 10\sqrt{3}\ \mathrm{m} \] Also, \[ \tan 15^\circ = \frac{AC}{CP} \] \[ \Rightarrow \frac{30-h}{x} = 2-\sqrt{3} \] \[ \Rightarrow 30-h = 10\sqrt{3}(2-\sqrt{3}) \] \[ \Rightarrow 30-h = 20\sqrt{3}-30 \] \[ \Rightarrow h = 60-20\sqrt{3} \] Now, area of the quadrilateral \(BCPQ\) is \[ BC \times BQ \] \[ \therefore \text{Area} = hx \] \[ = (60-20\sqrt{3})\,10\sqrt{3} \] \[ = 200\sqrt{3}(3-\sqrt{3}) \] \[ = 600(\sqrt{3}-1)\ \mathrm{m^2} \] Hence, this is the correct option.

Let the given series be $S_n$, then we can write as $S_n=5+11+19+29+41+\cdots+T_n$. $S_n=5+11+19+29+41+\cdots+T_{n-1}+T_n$. Subtracting the above equations, we get $0=5+6+8+10+\cdots-T_n$. $\Rightarrow\ T_n=5+\left[\frac{n-1}{2}\left(12+(n-2)\cdot 2\right)\right]$ $\Rightarrow\ T_n=5+(n-1)(n+4)$ $\Rightarrow\ T_n=n^2+3n+1$ So, $S_{20}=\sum_{n=1}^{20}T_n$ $=\sum_{n=1}^{20}(n^2+3n+1)$ $\Rightarrow\ S_{20} =\frac{20\times21\times41}{6} +\frac{3\times20\times21}{2} +20$ $\Rightarrow\ S_{20}=3520$.

Given that, for set 1: Mean = 12 and Variance = 14. For set 2: Mean = 14 and Variance = $\sigma$^2. Now, for set 1 we have, $$\frac{\sum_{i=1}^{15} x_i}{15} = 12$$ $$\Rightarrow \sum_{i=1}^{15} x_i = 15 \times 12$$ Also, $$\frac{\sum_{i=1}^{15} x_i^2}{15} - \left( \frac{\sum_{i=1}^{15} x_i}{15} \right)^2 = 14$$ $$\Rightarrow \frac{\sum_{i=1}^{15} x_i^2}{15} - 12^2 = 14$$ $$\Rightarrow \sum_{i=1}^{15} x_i^2 = (14 + 144) \times 15$$ Now for set 2: $$\frac{\sum_{i=1}^{15} y_i}{15} = 14$$ and $$\sum_{i=1}^{15} y_i = 15 \times 14$$ Also, $$\frac{\sum_{i=1}^{15} y_i^2}{15} - \left( \frac{\sum_{i=1}^{15} y_i}{15} \right)^2 = \sigma^2$$ $$\Rightarrow \frac{\sum_{i=1}^{15} y_i^2}{15} - 14^2 = \sigma^2$$ $$\Rightarrow \sum_{i=1}^{15} y_i^2 = (\sigma^2 + 196) \times 15$$ Now, the combined variance will be $$13 = \frac{\left( \sum_{i=1}^{15} x_i^2 \right) + \left( \sum_{i=1}^{15} y_i^2 \right)}{15 + 15} - \left( \frac{\sum_{i=1}^{15} x_i + \sum_{i=1}^{15} y_i}{15 + 15} \right)^2$$ $$\Rightarrow 13 = \frac{(14 + 144) \times 15 + (\sigma^2 + 196) \times 15}{30} - 13^2$$ $$\Rightarrow \sigma^2 = 10$$ Hence, $\($ $\sigma^2$ = 10 $\)$.

Let, $$A = \begin{bmatrix} p & q \\ r & s \end{bmatrix}$$ $$A^2 = \begin{bmatrix} p & q \\ r & s \end{bmatrix} \begin{bmatrix} p & q \\ r & s \end{bmatrix} = \begin{bmatrix} p^2 + qr & pq + qs \\ rp + rs & qr + s^2 \end{bmatrix}$$ Now given, $A^2 = I$ $$\Rightarrow \begin{bmatrix} p^2 + qr & pq + qs \\ rp + rs & qr + s^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ Now on comparing both sides we get, $$\Rightarrow p^2 + qr = 1, \; q(p + s) = 0$$ And $r(p + s) = 0, \; qr + s^2 = 1$ Now on solving above relation we get, $q \neq 0 \Rightarrow p + s = 0 \Rightarrow a = 0$ And $b = |A| = ps - qr = -p^2 - qr = -1 (\because s = -p)$ $$\therefore 3a^2 + 4b^2 = 4$$

Given that $I(x) = \int x^2 \left( \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} \right) dx$. Apply Integration by parts. $$\int f(x)g(x)dx = f(x)\int g(x)dx - \int (f'(x) \int g(x)dx)dx$$ $$= x^2 \int \left( \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} \right) dx - \int \left( \frac{d x^2}{dx} \left( \int \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} \right) dx \right) dx$$ Let $x \tan x + 1 = p$ $$\Rightarrow (x \sec^2 x + \tan x) dp = dx$$ $$= x^2 \int \frac{dp}{p^2} - \int \left( 2x \int \frac{dp}{p^2} \right) dx$$ $$= \frac{-x^2}{x \tan x + 1} + \int \frac{2x}{x \tan x + 1} dx$$ Let $I_1 = 2 \int \frac{x}{x \tan x + 1} dx$$ $$= 2 \int \frac{x \cos x}{\sin x + \cos x} dx$$ Let $x $\sin$ x + $\cos$ x = t$ $$\Rightarrow (x \cos x \sin x - \sin x) dx = dt$$ $$\Rightarrow (x \cos x) dx = dt$$ $$= 2 \int \frac{dt}{t} = 2 \log t + c$$ $$= 2 \log |x \sin x + \cos x| + c$$ $$\therefore \int \frac{x (x \sec^2 x + \tan x)}{(x \tan x + 1)^2} dx$$ $$= \frac{-x^2}{x \tan x + 1} + 2 \log \left| x \sin x + \cos x \right| + c$$ But $I(0) = 0$ $$\Rightarrow c = 0$$ Also, $$I\left( \frac{\pi}{4} \right) = - \left( \frac{\pi}{4} \right)^2 \frac{1}{\frac{\pi}{4} \times 1 + 1} + 2 \log \left| \frac{1}{\sqrt{2}} \left( \frac{\pi}{4} \right) + \frac{1}{\sqrt{2}} \right|$$ $$= \log_e \left( \frac{\pi + 4}{32} \right) - \frac{\pi^2}{4(\pi + 4)}$$ Hence, this is the correct option.

Given, the equation of the plane passing through the line of intersection of the planes $2x - y + z = 3$, $4x - 3y + 5z + 9 = 0$ and parallel to the line $\($ $\frac{x+1}{-2}$ = $\frac{y+3}{4}$ = $\frac{z-2}{5}$ $\)$ is $ax + by + cz + 6 = 0$. Now let, $P_1 = 2x - y + z = 3$ and $P_2 = 4x - 3y + 5z + 9 = 0$. Now using family of planes we get, $$P_1 + \lambda P_2 = 0$$ $$\Rightarrow (2x - y + z - 3) + \lambda (4x - 3y + 5z + 9) = 0$$ $$\Rightarrow (2 + 4\lambda)x - (1 + 3\lambda)y + (1 + 5\lambda)z - (3 - 9\lambda) = 0$$ Now let, $$P_3 = (2 + 4\lambda)x - (1 + 3\lambda)y + (1 + 5\lambda)z - (3 - 9\lambda) = 0$$ Now given $P_3$ is parallel to line, $\($ $\frac{x+1}{-2}$ = $\frac{y+3}{4}$ = $\frac{z-2}{5}$ $\)$. So, using perpendicular condition as normal vector of plane will be perpendicular to given line we get, $$-2(2 + 4\lambda) - 4(1 + 3\lambda) + 5(1 + 5\lambda) = 0$$ $$\Rightarrow -3 + 5\lambda = 0$$ $$\Rightarrow \lambda = \frac{3}{5}$$ Hence, equation of plane $P_3$ will be, $$P_3 : \frac{22x}{5} - \frac{14y}{5} + \frac{20z}{5} + \frac{12}{5} = 0$$ $$\Rightarrow P_3 = 11x - 7y + 10z + 6 = 0$$ Now on comparing with $ax + by + cz + 6 = 0$, we get, $a = 11$, $b = -7$ and $c = 10$. Therefore, $a + b + c = 11 - 7 + 10 = 14$.

Given that $(P \Rightarrow Q) \land (R \Rightarrow Q)$. We know that $P \Rightarrow Q \equiv \sim P \lor Q$. $$\Rightarrow (\sim P \lor Q) \land (\sim R \lor Q)$$ $$\Rightarrow (\sim P \land \sim R) \lor Q$$ $$\Rightarrow \sim (P \lor R) \lor Q$$ $$\Rightarrow (P \lor R) \Rightarrow Q$$ Hence this is the correct option.

Given, $$|x^2 - 8x + 15| - 2x + 7 = 0$$ Now taking Case I when, $$x^2 - 8x + 15 \geq 0 \Rightarrow (x - 3)(x - 5) \geq 0$$ $$\Rightarrow x \leq 3 OR x \geq 5$$ $$|x^2 - 8x + 15| - 2x + 7 = 0$$ $$\Rightarrow x^2 - 8x + 15 - 2x + 7 = 0$$ $$\Rightarrow x^2 - 10x + 22 = 0$$ $$\Rightarrow x = 5 + \sqrt{3}, 5 - \sqrt{3} (rejected)$$ So, $\alpha = 5 + \sqrt{3}$ Now taking Case II when, $$|x^2 - 8x + 15| - 2x + 7 = 0$$ $$\Rightarrow x^2 - 8x + 15 < 0 \Rightarrow 3 < x < 5$$ $$\Rightarrow x^2 - 8x + 15 + 2x - 7 = 0$$ $$\Rightarrow x^2 - 6x + 8 = 0 \Rightarrow x = 4, 2 (rejected)$$ So, $\gamma = 4$ Hence, $\alpha + \gamma = 9 + \sqrt{3}$

Given, $a_1, a_2, a_3, \ldots, a_n$ are terms of an A.P. So common difference will be, $d = a_2 - a_1 = a_3 - a_2 = \ldots = a_n - a_{n-1}$. Now solving, $$\lim_{n \to \infty} \sqrt{\frac{d}{n} \left( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} \right)}$$ $$= \lim_{n \to \infty} \sqrt{\frac{d}{n} \left( \frac{\sqrt{a_2} - \sqrt{a_1}}{a_2 - a_1} + \frac{\sqrt{a_3} - \sqrt{a_2}}{a_3 - a_2} + \cdots + \frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{a_n - a_{n-1}} \right)}$$ $$= \lim_{n \to \infty} \sqrt{\frac{d}{n} \times \frac{1}{d} \left( \sqrt{a_n} - \sqrt{a_1} \right)}$$ Now using the formula $a_n = a_1 + (n-1)d$ we get, $$= \lim_{n \to \infty} \frac{1}{\sqrt{d}} \left( \frac{\sqrt{a_1 + (n-1)d} - \sqrt{a_1}}{\sqrt{n}} \right)$$ $$= \lim_{n \to \infty} \frac{1}{\sqrt{d}} \left( \sqrt{\frac{n \left( \frac{a_1}{n} + d - \frac{d}{n} \right) - \sqrt{\frac{a_1}{n}}}{\sqrt{n}}} \right)$$ $$= \frac{1}{\sqrt{d}} \left( \sqrt{0 + d - 0 - \sqrt{0}} \right)$$ $$= \frac{1}{\sqrt{d}} \times \sqrt{d} = 1.$$

Given system of equations are $$x + y + az = b \ldots (i)$$ $$2x + 5y + 2z = 6 \ldots (ii)$$ $$x + 2y + 3z = 3 \ldots (iii)$$ It is given that the system of equations have infinite solutions. $$\Rightarrow \Delta = 0$$ $$\begin{vmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{vmatrix} = 0$$ $$\Rightarrow 1(15 - 4) - 1(6 - 2) + a(4 - 5) = 0$$ $$\Rightarrow a = 7$$ Also, we have $$\Rightarrow \Delta_1 = 0$$ $$\begin{vmatrix} b & 1 & a \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{vmatrix} = 0$$ $$\Rightarrow b(15 - 4) - 1(18 - 6) + 7(12 - 15) = 0$$ $$\Rightarrow b = 3$$ Therefore, $a = 7$, $b = 3$. Thus, $2a + 3b = 14 + 9 = 23$. Hence, this is the correct option.

Given, $2xy + 3y^x = 20$ Now differentiating both sides w.r.t $x$, we get $$2xy \left( y' \ln x + \frac{y}{x} \right) + 3y^x \left( \ln y + \frac{x}{y} \cdot y' \right) = 0 \{here \ln = \log_e\}$$ Putting $x = 2$ and $y = 2$, we get $$\Rightarrow 8(\ln 2 \cdot y' + 1) + 12(\ln y + y') = 0$$ $$\Rightarrow 8(\ln 2 \cdot y' + 1) + 12(\ln y + y') = 0$$ $$\Rightarrow y' = -\left( \frac{3 \ln 2 + 2}{2 \ln 2 + 3} \right)$$ $$\Rightarrow y' = -\left( \frac{2 + \ln 8}{3 + \ln 4} \right)$$

Given, one vertex of a rectangular parallelepiped is at the origin $O$ and the lengths of its edges along $x$, $y$ and $z$ axes are $3$, $4$ and $5$ units respectively, and $P$ be the vertex $(3, 4, 5)$. Now plotting the diagram we get, Now equation of line $OP$: $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$ And equation of line $AB$: $\frac{x - 0}{3} = \frac{y - 0}{0} = \frac{z}{1}$ Now finding, $\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 0 & 0 & 1 \end{vmatrix}$ $$\Rightarrow \vec{n_1} \times \vec{n_2} = \hat{i}(4) - \hat{j}(3) + \hat{k}(0)$$ $$\Rightarrow \vec{n_1} \times \vec{n_2} = 4\hat{i} - 3\hat{j}$$ Now we know that, Shortest Distance, $S. D = \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{n_1} \times \vec{n_2})}{|\vec{n_1} \times \vec{n_2}|}$ $$\Rightarrow S. D = \frac{(3\hat{i} + 0\hat{j} + 0\hat{k} - 0\hat{i} + 0\hat{j} + 0\hat{k}) \cdot (4\hat{i} - 3\hat{j})}{5}$$ $$\Rightarrow S. D = \frac{(3\hat{i}) \cdot (4\hat{i} - 3\hat{j})}{5}$$ $$\Rightarrow S. D = \frac{12}{5}$$

Given that: $A(5, 5, 2\lambda)$, $B(1, 2, 3)$, $C(-2, \lambda, 4)$ and $D(-1, 5, 6)$. Now, $$\overrightarrow{AB} \equiv \begin{pmatrix} -4, -3, 3 - 2\lambda \end{pmatrix}$$ $$\overrightarrow{AC} \equiv \begin{pmatrix} -7, \lambda - 5, 4 - 2\lambda \end{pmatrix}$$ $$\overrightarrow{AD} \equiv \begin{pmatrix} -6, 0, 6 - 2\lambda \end{pmatrix}$$ Given that $A, B, C$ and $D$ are coplanar. Therefore, $$\begin{vmatrix} \overrightarrow{AB} & \overrightarrow{AC} & \overrightarrow{AD} \end{vmatrix} = 0$$ $$\begin{vmatrix} -4 & -3 & 3 - 2\lambda \\ -7 & \lambda - 5 & 4 - 2\lambda \\ -6 & 0 & 6 - 2\lambda \end{vmatrix} = 0$$ Expand the determinant along the first row. $$\Rightarrow -4[(\lambda - 5)(6 - 2\lambda) - 0] - (-3)[-7(6 - 2\lambda) - (-6)(4 - 2\lambda)] + (3 - 2\lambda)[0 - (-6)(\lambda - 5)]$$ On solving the above equation we get, Therefore, $\lambda = 3, \lambda = 2$ We have two values of $\lambda$. $$\Rightarrow \sum_{\lambda \in S} (\lambda + 2)^2 = (3 + 2)^2 + (2 + 2)^2$$ $$= 41$$ Hence, $\sum_{\lambda \in S} (\lambda + 2)^2 = 41$

Given, $$A = \left\{ x \in \mathbb{R} : [x+3] + [x+4] \leq 3 \right\} and B = \left\{ x \in \mathbb{R} : 3^x \left( \sum_{r=1}^{\infty} \frac{3}{10^r} \right)^{x-3} < 3^{-3x} \right\},$$ Now solving $A$ we get, $$[x+3] + [x+4] \leq 3$$ $$\Rightarrow [x] + 3 + [x] + 4 \leq 3$$ $$\Rightarrow 2[x] \leq -4$$ $$\Rightarrow [x] \leq -2$$ $$\Rightarrow x < -1 \Rightarrow A = (-\infty, -1)$$ Now solving $B$ we get, $$3^x \left( \sum_{r=1}^{\infty} \frac{3}{10^r} \right)^{x-3} < 3^{-3x}$$ Now using infinite G.P. formula $S_\infty = \frac{a}{1-r}$, we get, $$3^x \left( \frac{\frac{3}{10}}{1 - \frac{1}{10}} \right)^{x-3} < 3^{-3x}$$ $$\Rightarrow 3^x \left( \frac{1}{3} \right)^{x-3} < 3^{-3x}$$ $$\Rightarrow 3^{x-x+3x} < 1$$ $$\Rightarrow 3^{3(x+1)} < 1$$ $$\Rightarrow 3(x+1) < 0$$ $$\Rightarrow x < -1 \Rightarrow B = (-\infty, -1)$$ $$\Rightarrow A = B$$

Given, $$f(x) = [a + 13 \sin x], \ x \in (0, \pi)$$ $$\Rightarrow f(x) = [13 \sin x] + a, as a \in Integer$$ Now let $g(x) = 13 \sin x$, now plotting the diagram of $y = 13 \sin x$ we get, Now from above diagram we can see that, $$f(x) = [13 \sin x]$$ is not differentiable at intersection points of $y = k$, $k \in [1, 13]$ and $k \in \mathbb{Z}$ $$\Rightarrow Points of non-differentiability = 12 \times 2 + 1 = 25$$

Let the equation of circle touching both the coordinate axis be, $$(x - r)^2 + (y - r)^2 = r^2$$ which passes through $(\alpha, \beta)$. So, $$(\alpha - r)^2 + (\beta - r)^2 = r^2 \ldots (1)$$ Also equation of line $AB$: $x + y = r$. Let $Q(h, k)$. Now using the formula of foot of perpendicular, $$\frac{h - \alpha}{1} = \frac{k - \beta}{1} = \frac{- (\alpha + \beta - r)}{2}$$ $$\Rightarrow \left(h, k\right) = \left(\alpha - \frac{\alpha - \beta + r}{2}, \beta - \frac{\alpha - \beta + r}{2}\right)$$ $$\Rightarrow (h, k) = \left(\frac{\alpha - \beta + r}{2}, \frac{- \alpha + \beta + r}{2}\right)$$ Now using the distance formula we get, $$(PQ)^2 = \left(\frac{- \alpha - \beta + r}{2}\right)^2 + \left(\frac{- \alpha - \beta + r}{2}\right)^2 = 121$$ $$\Rightarrow (r - (\alpha + \beta))^2 = 242$$ $$\Rightarrow r^2 + (\alpha + \beta)^2 - 2r(\alpha + \beta) = 242$$ $$\Rightarrow \alpha^2 + \beta^2 + 2\alpha\beta + r^2 - 2r\alpha - 2r\beta = 242$$ $$\Rightarrow 2\alpha\beta = 242 \{from equation (1), (\alpha - r)^2 + (\beta - r)^2 = r^2\}$$ $$\Rightarrow \alpha\beta = 121$$

Given, we have to distribute 20 identical oranges to 3 children. Let children A get $O_A$ oranges, children B gets $O_B$ oranges and children C gets $O_C$ oranges. So, $O_A + O_B + O_C = 20$. Now this can be distributed using the multinomial theorem. So, the number of ways will be, coefficient of $x^{20}$ in $\left( x + x^2 + \ldots + x^{18} \right)^3$. Coefficient of $x^{20}$ in $x^3 \left( 1 + x + x^2 + \ldots x^{17} \right)^3$. Coefficient of $x^{17}$ in $\left( \frac{1-x^{18}}{1-x} \right)^3$. Coefficient of $x^{17}$ in $(1-x)^{-3}$. $$\binom{19}{2} = 171$$ As we know that expansion of $(1-x)^{-3}$ is $1 + 3\binom{1}{1}x + 4\binom{2}{2}x^2 + 5\binom{3}{3}x^3 + \ldots + 19\binom{17}{17}x^{17} \ldots \infty$. Note: This is bonus question as here 20 oranges are considered identical but in question it is considered distinct.

Given, $S = \{(x, y) : 2y - y^2 \leq x^2 \leq 2y, x \geq y\}$. Now plotting the diagram, we get. Now from above diagram, given region are, $2y - y^2 \leq x^2$, $x^2 \leq 2y$ and $x \geq y$. Now the intersecting point of $x^2 + y^2 - 2y = 0$ and $x = y$ will be $(1, 1)$ and $x^2 = 2y$ and $x = y$ will be $(2, 2)$. So, required area will be, $$A = \int_0^1 \left(1 + \sqrt{1 - x^2} - \frac{x^2}{2}\right) + \int_1^2 \left(x - \frac{x^2}{2}\right) dx$$ $$\Rightarrow A = \left(x - \frac{x}{2} \sqrt{1 - x^2} - \frac{1}{2}\right) + \frac{1}{2} x \sin^{-1} x - \frac{x^3}{6} \bigg|_0^1 + \left(\frac{x^2}{2} - \frac{x^3}{6}\right) \bigg|_1^2$$ $$\Rightarrow A = \frac{7}{6} - \frac{\pi}{4} = \frac{n+2}{n+1} - \frac{\pi}{x-1}$$ So, on comparing we get, $n = 5$

Given, the point $(p, p+1)$ lies inside the region $E = \{(x, y): 3-x \leq y \leq \sqrt{9-x^2}, 0 \leq x \leq 3\}$, and the set of all values of $p$ is the interval $(a, b)$. Now by observation $(p, p+1)$ lies on $y-x=1$. Also point $(p, p+1)$ lies on $x+y=3$. So, solving $y-x=1$ and $x+y=3$ we get, $P(1, 2)$. Again given point $(p, p+1)$ lies on $x^2+y^2=9$. So, solving $y-x=1$ and $x^2+y^2=9$, we get $$x^2+(1+x)^2=9$$ $$\Rightarrow 2x^2+2x-8=0$$ $$\Rightarrow x^2+x-4=0 \ldots (i)$$ $$\Rightarrow x=\frac{-1 \pm \sqrt{1+4 \cdot 4}}{2} = \frac{\sqrt{17}-1}{2}$$ Hence, $p \in \left(1, \frac{\sqrt{17}-1}{2}\right)$. Now on comparing with $p \in (a, b)$ we get, $$a=1, \ b=\frac{\sqrt{17}-1}{2} \ [using (i)]$$ Hence, the value of $b^2-b-4=1-3=3$.

Given, $$(x \cos x) dy + (xy \sin x + y \cos x - 1) dx = 0$$ $$\Rightarrow (x \cos x) \frac{dy}{dx} + (xy \sin x + y \cos x) = 1$$ $$\Rightarrow \frac{dy}{dx} + y \left( \frac{x \sin x + \cos x}{x \cos x} \right) = \frac{1}{x \cos x}$$ Which is a linear differential equation of form $\frac{dy}{dx} + P(x)y = Q(x)$. Now we know that integrating factor will be $e^{\int P dx}$. $$\Rightarrow IF = e^{\int \left( \tan x + \frac{1}{x} \right) dx} = e^{\ln |\sec x| + \ln x} = x \cdot \sec x$$ Hence, solution of linear differential equation is given by, $$y \cdot x \cdot \sec x = \int \frac{x \sec x}{x \cos x} dx$$ $$\Rightarrow xy \sec x = \tan x + C$$ Now using $\frac{\pi}{3} y \left( \frac{\pi}{3} \right) = \sqrt{3}$ we get, $$\Rightarrow \frac{\pi}{3} y \left( \frac{\pi}{3} \right) \sec \frac{\pi}{3} = \tan \frac{\pi}{3} + C$$ $$\Rightarrow C = \sqrt{3}$$ Hence, $xy \sec x = \tan x + \sqrt{3}$ $$\Rightarrow y(x) = 2 \sin \left( x + \frac{\pi}{3} \right)$$ Now differentiating both side we get, $$\Rightarrow xy'(x) + y(x) = 2 \cos \left( x + \frac{\pi}{3} \right)$$ Now again differentiating we get, $$\Rightarrow xy''(x) + 2y'(x) = -2 \sin \left( x + \frac{\pi}{3} \right)$$ Thus $\frac{\pi}{6} y'' \left( \frac{\pi}{6} \right) + 2y' \left( \frac{\pi}{6} \right) = -2$ Hence $\left| \frac{\pi}{6} y'' \left( \frac{\pi}{6} \right) + 2y' \left( \frac{\pi}{6} \right) \right| = 2$

Given, binomial expression $\left( x^4 - \frac{1}{x^3} \right)^{15}$. Now general term is given by, $$T_{r+1} = \binom{15}{r} (x^4)^{15-r} \left( -\frac{1}{x^3} \right)^r$$ $$\Rightarrow T_{r+1} = \binom{15}{r} x^{60-7r} (-1)^r$$ For coefficient of $x^{18}$, equate $60 - 7r = 18$ $$\Rightarrow r = 6$$ Coefficient of $x^{18} = (-1)^6 \cdot \binom{15}{6} = 5005$$

Given, $$A = \{1, 2, 3, \ldots, 10\} B = \{0, 1, 2, 4\}$$ $$(a, b) \in A \times A such that $$ $$2(a-b)^2 + 3(a-b) - k = 0$$ where $k \in \{0, 1, 2, 3, 4\}$ Now finding discriminant $D = 9 - 4 \times 2(-k)$ And $9 - 4 \times 2(-k)$ a perfect square for any possible $(a, b)$ as $(a-b)$ will be an integer, So, $9 + 8k$ is a perfect square $$\Rightarrow k = 0 or k = 2$$ Now for $k = 0$, $$2(a-b)^2 + 3(a-b) = 0$$ $$\Rightarrow (a-b) \left[ 2(a-b) + 3 \right] = 0$$ $$\Rightarrow a-b = 0 \Rightarrow (a, b) \in \{(1, 1), (2, 2) \ldots (10, 10)\}$$ $$\Rightarrow Total 10 elements belonging to R.$$ And, $a-b = \frac{3}{2}$ is not possible Now for $k = 2$, $$2(a-b)^2 + 3(a-b) - 2 = 0$$ $$\Rightarrow a-b = -2 or a-b = \frac{1}{2} (not possible)$$ Now for $a-b = -2$ possible pair will be, $$\Rightarrow (a, b) \in \{(1, 3), (2, 4), \ldots (8, 10)\}$$ $$\Rightarrow 8 elements belonging to R$$ Total number of elements will be $= 18$

Given, the image of the point $P(1, 2, 3)$ in the plane $2x - y + z = 9$ be $Q$. The coordinates of the point $R$ are $(6, 10, 7)$. Now plotting the diagram we get, $R$ lies on the plane as $(6, 10, 7)$ satisfy the plane equation $2x - y + z = 9$. Now length of $PR = \sqrt{5^2 + 8^2 + 4^2} = \sqrt{105}$. Now finding the angle between $PR$ and $PM$ (which is normal vector to plane) we get, $$\cos \theta = \frac{\overrightarrow{PR} \cdot \overrightarrow{PM}}{|\overrightarrow{PR}| \cdot |\overrightarrow{PM}|} = \frac{(5\hat{i} + 8\hat{j} + 4\hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k})}{\sqrt{105} \times \sqrt{6}}$$ $$\Rightarrow \cos \theta = \frac{6}{\sqrt{630}}$$ Now $Area(\triangle PQR) = 2 area(\triangle PMR)$ $$\Rightarrow \triangle PQR = 2 \cdot \frac{1}{2} \cdot PR \sin \theta \cdot PR \cos \theta$$ $$\Rightarrow \triangle PQR = 2 \cdot \frac{1}{2} \left( PR \right)^2 \sin \theta \cos \theta$$ $$\Rightarrow \triangle PQR = 105 \cdot \frac{6}{\sqrt{630}} \cdot \frac{\sqrt{594}}{\sqrt{630}}$$ $$\Rightarrow \triangle PQR = \sqrt{594}$$ Hence, area of the square will be 594.

Given, the tangent to the curve $x^2 + 2x - 4y + 9 = 0$ at the point $P(1, 3)$ on it meets the $y$-axis at $A$. So, the tangent to the circle at point $P(1, 3)$ is given by, $$x + (x + 1) - 2(y + 3) + 9 = 0$$ $$\Rightarrow 2x - 2y = -4$$ So, on the $y$-axis, $A(0, 2)$. And the line passing through $P$ and parallel to the line $x - 3y = 6$ meets the parabola $y^2 = 4x$ at $B$. So, the equation of the line through $P$ will be $y - 3 = \frac{1}{3}(x - 1)$. $$\Rightarrow 3y = x + 8$$ Now finding the intersection of $y^2 = 4x$ and $3y = x + 8$, we get, $B = (4, 4)$ and $(16, 8)$. Also, $B$ lies on the line $2x - 3y = 8$, so $(4, 4)$ will not satisfy. Hence, $B = (16, 8)$. $$(AB)^2 = (0 - 16)^2 + (2 - 8)^2 = 256 + 36 = 292$$

The formula to calculate the average electric energy density is given by $$U_E = \frac{1}{2} \varepsilon_0 E^2 \cdots (1)$$ and, the formula to calculate average magnetic energy density is given by $$U_B = \frac{1}{2\mu_0} B^2 \cdots (2)$$ For any particular electromagnetic wave, both magnetic and electric field are equally involved in contributing to energy density. Hence, the ratio of average electric energy density to magnetic energy density is 1.

Let's assume that when the path is short, the state is called state 1 and when it is open, the state is called state 0. Depending upon whether the path is short or not in the circuit, the following table can be obtained: The above table is similar to the truth table corresponding to NOR gate. Hence, the circuit indicates a NOR logic gate.

The buoyant force is given by $$B = V \rho_0 g$$ where $V$ is the volume of the ball. The weight of the ball $$W = Mg = V \rho g,$$ is acting in the opposite direction of the buoyant force. Hence, at terminal velocity $$Mg - B = F_v$$ $$\Rightarrow B + F_v = Mg$$ $$\Rightarrow F_v = V \rho g - V \rho_0 g$$ $$\Rightarrow F_v = \rho V g \left( 1 - \frac{\rho_0}{\rho} \right)$$ $$\Rightarrow F_v = Mg \left( 1 - \frac{\rho_0}{\rho} \right)$$

The given data is $n_1 = 3 \times 10^{19}$ $n_2 = 12 \times 10^{19}$ The mean time between two collisions has the meaning of a period $\tau$, since it indicates the repetitive time intervals in which on average collisions take place. The reciprocal of the time $\tau$ can be understood as the collision frequency $f$, which indicates the number of collisions per unit time. The collision frequency is also directly proportional to the number of molecules per unit volume. $$f \propto \frac{1}{\tau}$$ $$\Rightarrow f \propto n$$ $$\Rightarrow \frac{f_1}{f_2} = \frac{n_1}{n_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}} = 0.25$$

According to the first law of thermodynamics, $Q = W + U$. Upon differentiation, we get $$\frac{dQ}{dt} = \frac{dW}{dt} + \frac{dU}{dt}$$ $$\Rightarrow \frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt}$$ Given: $\frac{dQ}{dt} = 1000 \, \mathrm{W}$, $\frac{dW}{dt} = 200 \, \mathrm{W}$. Therefore, $$\frac{dU}{dt} = 1000 \, \mathrm{W} - 200 \, \mathrm{W} = 800 \, \mathrm{W}$$

In circular motion with constant speed, magnitude of instantaneous velocity, $v_{ins} = \omega r$. Time taken to turn by $90^\circ$ will be $= \frac{\pi}{2 \omega} = \frac{\pi}{2 \omega}$. Displacement will be equal to the length of $AB = \sqrt{r^2 + r^2} = r \sqrt{2}$. Average velocity $$v_{avg} = \frac{displacement}{time}$$ $$= \frac{r \sqrt{2} \times 2 \omega}{\pi} = \frac{2 \sqrt{2}}{\pi} \omega r$$ Hence, required ratio $$\frac{v_{ins}}{v_{avg}} = \frac{\omega r \pi}{2 \sqrt{2} \omega r} = \frac{\pi}{2 \sqrt{2}}$$ Therefore, $x = 2$.

Centripetal force will be provided by the spring force. Let the elongation in the spring be $x$, then we can write $$kx = m\omega^2 (r + x)$$ $$7.5 \, x = 2.5 (0.2 + x)$$ $$x = \frac{0.5}{5} = 0.1$$ Therefore, required tension will be $T = kx = 0.75 \, \mathrm{N}$.

By using Snell's law, $n_1 \sin \theta_1 = n_2 \sin \theta_2$, the refractive index is $$\mu = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2}$$ Also using the relation $n = \frac{c}{v}$ where $n$ is the refractive index, $c$ is the velocity of light in air and $v$ is the velocity in the medium $$\Rightarrow \frac{c_{air}}{v} = \sqrt{2} = \frac{\lambda_1}{\lambda_2}$$ The velocities are given by $$v = \nu \lambda_2$$ $$c = \nu \lambda_1$$ $$\Rightarrow \lambda_1 = \sqrt{2} \lambda_2 and \nu_1 = \nu_2.$$ The frequency of light does not change when the medium changes.

Horizontal range of a projectile from a horizontal plane is given by, $R = \frac{u^2 \sin 2\theta}{g}$. Clearly for maximum range, $\sin 2\theta = 1$. Therefore, $\theta = 45^\circ$.

For a parallel combination the equivalent resistance is $$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \cdots (i)$$ A general expression for error can be written as $x \pm \Delta x$. Differentiating equation (i) $$\frac{dR}{R^2} = \frac{dR_1}{R_1^2} + \frac{dR_2}{R_2^2} \cdots (ii)$$ The equivalent resistance with $R_1 = 10 \, \Omega$ and $R_2 = 15 \, \Omega$ is $$\frac{1}{R} = \frac{1}{10} + \frac{1}{15}$$ $$\Rightarrow R = \frac{150}{25} = 6 \, \Omega$$ Substituting the above value in equation (ii) $$dR = 36 \left( \frac{0.5}{100} + \frac{0.5}{15 \times 15} \right) = 0.26$$ $$\Rightarrow R = 6 \pm 0.26$$ Hence, the percentage error is $$\frac{dR}{R} \times 100 = \frac{0.26}{6} \times 100 = 4.33\%$$

The formula for acceleration due to gravity is $g = \frac{GM}{R^2}$. Let the density of earth and planet be $\rho$, and the radius of the planet be $R'$. Let $R$ be the radius of the earth and $g'$ be the acceleration due to gravity of planet. So, the masses are $$M = \frac{4}{3} \pi R^3 \rho \cdots (i)$$ $$M' = \frac{4}{3} \pi (R')^3 \rho \cdots (ii)$$ So, $$\frac{g'}{g} = \frac{\frac{GM'}{R'^2}}{\frac{GM}{R^2}} = \frac{M'}{M} \left( \frac{R}{R'} \right)^2 \cdots (iii)$$ From equation (i) and (ii) we get $R' = 2^{\frac{1}{3}} R$. Substituting the value in equation (iii), $$\frac{g'}{g} = (2^{1}) 2^{-\frac{2}{3}} = 2^{\frac{1}{3}} g$$ Weight is given by $W' = mg' = 2^{\frac{1}{3}} W$

The formula to calculate the escape velocity is given by $$v_e = \sqrt{2gR} \cdots (1)$$ As both the acceleration and radius of the Moon is much less than that of the Earth, the value of the escape velocity in Moon is much less than that on the Earth. The average velocity of gas molecules remains higher than the escape velocity due to the lower escape velocity on the moon's surface, making the moon unable to maintain an atmosphere.

The potential is constant inside the sphere as the electric field is zero inside the sphere. So, the value is given by $$V = \frac{Q}{4\pi \varepsilon_0 R} for r R$$

When a modest amount of energy is applied to semiconductors, electrons quickly travel to the conduction band and are then free to move across the crystal's lattice. If thermal energy is provided, the conduction of electrons will increase. Hence, as the temperature is increased, the resistivity of the semiconductor is decreased according to the following diagram:

The formula to calculate the de Broglie wavelength of a particle is given by $$ \lambda = \frac{h}{\sqrt{2mK}} \ldots (1) $$ Now, the masses of the given particles are interrelated by the following relations: $$ m_e \approx \frac{m_p}{1837} $$ $$ m_\alpha = 4m_p $$ The de Broglie wavelength for electron is given by $$ \lambda_e = \frac{h}{\sqrt{2m_e \,(4K)}} $$ $$ = \frac{h}{\sqrt{\frac{8}{1837} m_p K}} \ldots (2) $$ The de Broglie wavelength for proton is given by $$ \lambda_p = \frac{h}{\sqrt{2m_p \,(2K)}} $$ $$ = \frac{h}{\sqrt{4m_p K}} \ldots (3) $$ And, the de Broglie wavelength of alpha particle is given by $$ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha K}} $$ $$ = \frac{h}{\sqrt{2\,(4m_p)\,K}} $$ $$ = \frac{h}{\sqrt{8m_p K}} \ldots (4) $$ From equation (2), (3) and (4), it can be concluded that $\lambda_e > \lambda_p > \lambda_\alpha$.

The range of a TV tower antenna is given by $$d = \sqrt{2hR}$$ New height is given as $$h' = 1.21h$$ Let the new range be $$d' = \sqrt{2h'R}$$ $$\Rightarrow d' = \sqrt{1.21} \sqrt{2hR} = 1.1d$$ So, the increase in the range is $$= \frac{1.1d - d}{d} \times 100 = 10\%$$

Wavelength of photons released during de-excitation is given by, $$\frac{1}{\lambda} = R \left[ \frac{1}{(n_1)^2} - \frac{1}{(n_2)^2} \right].$$ Clearly, shortest wavelength will correspond to highest energy difference in de-excitation, which in the given figure will correspond to transition from $n = 3$ to $n = 1$.

For the given combination, if the block is displaced towards right, elongation in one spring will be equal to the compression in other and the direction of force due to both spring will be the same and restoring in nature. Hence, both springs are effectively in parallel combination. Therefore, $$K_{eff} = K_1 + K_2$$ and $$T = 2\pi \sqrt{\frac{m}{K_{eff}}} = 2\pi \sqrt{\frac{m}{K_1 + K_2}}.$$

Induced emf is given by Faraday's law $$|\varepsilon| = \frac{d\phi}{dt}$$ where $\phi$ is the magnetic flux. The rate of change of flux through a circuit is defined as emf. The formula for flux is $\phi = \vec{B} \cdot \vec{A}$. In the statements $A$ and $B$ given in the question, the magnetic flux in the coil in uniform magnetic field does not change. So, no emf is induced. But in statement $C$, the angle between the area vector and the magnetic field continuously changes hence, an emf is generated. In statement $D$, the area is changing in unit time so magnetic flux will change and an emf is generated.

The formula to calculate the magnetic field inside the conducting wire is given by $$B_{in} = \frac{\mu_0 J r}{2} \cdots (1)$$ The formula to calculate the magnetic field outside the conducting wire is given by $$B_{out} = \frac{\mu_0 J a^2}{2r} \cdots (2)$$ From equation (1), it can be concluded that the magnetic field is directly proportional to the radial distance inside the wire, whereas from equation (2), it can be concluded that the magnetic field is inversely proportional to the radial distance outside the wire.

Let angle $BAC$ be $\theta$. Applying Snell's law at $E$, we get $1 \times \sin 60^\circ = \frac{4}{3} \times \sin \theta$. Therefore, $$\sin \theta = \frac{3 \sqrt{3}}{8}$$ $$\Rightarrow \tan \theta = \frac{\sin \theta}{\sqrt{1 - (\sin \theta)^2}} = \frac{3 \sqrt{3}}{\sqrt{37}}$$ Now, $BC = 1.5 \times \tan \theta = 1.28$. Then, $AB = 2.15 - BC = 0.86 \, \mathrm{m}$. Required value, $DF = AB \times \tan 30^\circ = 0.5015 \, \mathrm{m} \approx 50 \, \mathrm{cm}$.

The given data is $l' = 1.2l$ $A' = 0.96A$ Using the formula $R = \frac{\rho l}{A}$, the new resistance is $$R' = \frac{1.2 \rho l}{0.96 A}$$ $$\Rightarrow R' = 1.25R$$ $$\Rightarrow \Delta R = 1.25R - R = 0.25R$$ Hence, the percentage increase is $$= \frac{\Delta R}{R} \times 100$$ $$= \frac{0.25R}{R} \times 100 = 25\%$$

Acceleration can be written as, $a = v \frac{dv}{dx}$. Given: $a = -2x$ (-ve sign is due to retardation). Therefore, $$v \frac{dv}{dx} = -2x$$ $$\Rightarrow v dv = (-2x) dx$$ Integrating both sides, we get $$\frac{1}{2} \left( v_f^2 - v_i^2 \right) = -2 \left( \frac{x^2}{2} \right) = x^2$$ $$\Rightarrow \frac{1}{2} m \left( v_f^2 - v_i^2 \right) = mx^2$$ $$\Rightarrow |\Delta KE| = 0.01 x^2 = \left( \frac{10}{x} \right)^{-2}$$ Hence, $n = 2$.

The formula to calculate the magnetic field ($B$) at the centre of a current carrying circular loop is given by $$B = \frac{\mu_0 i}{2R} \cdots (1)$$ where, $\mu_0$ is the permeability of free space, $i$ is the current and $R$ is the radius of the loop. $B_V$, $B_H$ are the magnetic field due to the vertical and the horizontal loop, which by magnitude are equal. The resultant magnetic field ($B_{net}$) at the centre is, then, given by $$B_{net} = \sqrt{B^2 + B^2} = \sqrt{2}B$$ $$= \frac{\sqrt{2} \times 4 \pi \times 10^{-7} \times \sqrt{2}}{2 \times 0.2}$$ $$= 628 \times 10^{-8} \, \mathrm{T}$$

The formula to calculate the frequency $(f')$ as heard by an observer with respect to the frequency $(f)$ produced from a source is given by $$f' = \frac{v + v_o}{v + v_s} f \ldots (1)$$ where, $v$ is the speed of sound in air, $v_o$ is the speed of the observer and $v_s$ is the speed of the source. In the first situation, the source is the moving car and the observer is the wall. Hence, the frequency $(f_w)$ as heard on the wall can be written as $$f_w = \frac{v - 0}{v - v_c} f$$ $$= \frac{v}{v - v_c} f \ldots (2)$$ where, $v_c$ is the speed of the car. In the second situation, when the sound reflects back from the wall, the source of sound is the wall and the observer is the car. Hence, the frequency $(f_c)$ of sound as heard by the car after the reflection is given by $$f_c = \frac{v + v_c}{v - 0} f_w$$ $$= \frac{v + v_c}{v} f_w \ldots (3)$$ From equation (2) and (3), it follows that $$f_c = \frac{v + v_c}{v} \times \frac{v}{v - v_c} f$$ $$= \frac{v + v_c}{v - v_c} f \ldots (4)$$ It is given in the problem that $$f_c - f = 40 \ldots (5)$$ Substitute the expression from equation (4) into equation (5) and solve to calculate the frequency of the horn. $$\frac{v + v_c}{v - v_c} f - f = 40$$ $$\Rightarrow \frac{2v_c}{v - v_c} f = 40$$ $$\Rightarrow f = 40 \times \frac{v - v_c}{2v_c} = 40 \times \frac{330 - 15}{2 \times 15} \, \mathrm{m} \, \mathrm{s}^{-1} = 420 \, \mathrm{m} \, \mathrm{s}^{-1}$$

The formula to calculate the radius of Lithium-ion is given by $$R_{Li} = R_{H} \times \frac{n^2}{Z} \cdots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the required radius of the ion. $$R_{Li} = \frac{0.51 \times 25}{3} \times 10^{-10} \, m$$ $$= 425 \times 10^{-12} \, m$$

Young's modulus is given as $$Y = \frac{Stress}{Strain} = \frac{\frac{F}{A}}{\frac{\Delta L}{L}}$$ The given data is Stress = $62.8 \, kN$, $r = 20 \times 10^{-3} \, m$, $Y = 2 \times 10^{11} \, N m^{-2}$. Thus, $$Strain = \frac{F}{A \times Y}$$ $$\Rightarrow \frac{\Delta L}{L} = \frac{62.8 \times 10^3 \, N}{\pi \times (20 \times 10^{-3} \, m)^2 \times (2 \times 10^{11} \, N m^{-2})} = 25 \times 10^{-5} \, N$$

The given data is $P = 60 \times 10^3 \, \mathrm{W}$ $V_s = 120 \, \mathrm{V}$ A transformer has a primary coil and a secondary coil. The formula for the power delivered is given by $$P = \frac{V_s^2}{R_s}$$ $$\Rightarrow 60 \times 10^3 = \frac{120 \times 120}{R_s}$$ $$\Rightarrow R_s = \frac{120 \times 120}{60 \times 10^3} = 240 \, \mathrm{m\Omega}$$

The moment of inertia of a sphere is given by the formula $I = \frac{2}{5} MR^2$. Since there are two masses, the total moment of inertia about the axis perpendicular to the rod passing through the centre is $$I = 2 \times \left[ \frac{2}{5} MR^2 + M \left( \frac{l}{2} + r \right)^2 \right] \cdots (i)$$ The given data is $M = 2 \, \mathrm{kg}$, $l = (0.4 - 0.2) \, \mathrm{m} = 0.2 \, \mathrm{m}$, $R = 0.1 \, \mathrm{m}$. Substituting the values in equation (i) $$I = \left( \frac{4}{5} \left( 2 \times 0.1^2 \right) + 2 \times 2 \left( \frac{0.2}{2} + 0.1 \right)^2 \right) \, \mathrm{kg} \, \mathrm{m}^2$$ $$\Rightarrow I = \left( \left( \frac{4}{5} \times 0.02 \right) + (4 \times 0.04) \right) \, \mathrm{kg} \, \mathrm{m}^2$$ $$\Rightarrow I = (0.016 + 0.16) \, \mathrm{kg} \, \mathrm{m}^2 = 0.176 \, \mathrm{kg} \, \mathrm{m}^2$$

The formula to calculate the capacitance of a parallel plate capacitor having thickness $d$ and with a dielectric of thickness $x$ between the plates is given by $$C = \frac{\varepsilon_0 A}{(d-x) + \frac{x}{K}} \ldots (1)$$ Substitute the values of the parameters for the first case into equation (1) to obtain the capacitance. $$C_1 = \frac{\varepsilon_0 A}{\frac{2d}{3} + \frac{d}{3 \cdot 4}}$$ $$= \frac{4 \varepsilon_0 A}{3d} \ldots (2)$$ Similarly, substitute the parameters into equation (1) for the second case to obtain the capacitance $$C_2 = \frac{\varepsilon_0 A}{\frac{2d}{3} + \frac{d}{3 \cdot 4}}$$ $$= \frac{4 \varepsilon_0 A}{2d}$$ $$= \frac{2 \varepsilon_0 A}{d} \ldots (3)$$ Divide equation (3) by equation (2) to obtain the capacitance in the second case. $$\frac{C_2}{C_1} = \frac{\frac{2 \varepsilon_0 A}{d}}{\frac{4 \varepsilon_0 A}{3d}}$$ $$= \frac{3}{2}$$ $$C_2 = \frac{3}{2} C_1 \ldots (4)$$ Substitute the value of the capacitance in the first case into equation (4) to obtain the value of the capacitance in the second case. $$C_2 = \frac{3}{2} \times 2 \, \mu F$$ $$= 3 \, \mu F$$

Atoms of Element Y are present in cubic close packing. In cubic close packing, the number of atoms per unit cell is 4. So, the effective number of element Y per unit cell is 4. Element X is present at one third of the tetrahedral voids. Number of tetrahedral voids = Twice the number of atoms in unit cell. Number of atoms of Element X are present per unit cell = $\($ $\frac{1}{3}$ $\times$ 8 = $\frac{8}{3}$ $\)$. X : Y $\($ $\frac{8}{3}$ $\)$ : 4 $\($ 2 $\)$ : 3. Formula of the compound = $\($ $\mathrm{X_2Y_3}$ $\)$.

The nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide (NaCN) soluble in water. This is converted into sodium ferrocyanide by the addition of sufficient quantity of ferrous sulphate. $$6 \, \mathrm{NaCN} + \mathrm{FeSO_4} \rightarrow \mathrm{Na_4[Fe(CN)_6]} + \mathrm{Na_2 \, SO_4}$$ Ferric ions generated during the process react with ferrocyanide to form Prussian blue precipitate of ferric ferrocyanide. $$\mathrm{Na_4} \left[ \begin{array}{c} \mathrm{Fe} \left( \mathrm{CN} \right) \end{array} \right]_6 + \mathrm{Fe^{3+}} \rightarrow \mathrm{Fe_4[Fe(CN)_6]_3}$$ Sulphur present in organic compounds are detected by Lassaigne’s test. Here, a small piece of Na metal is heated in a fusion tube with the organic compound. The principle is that, in doing so, Na converts all the elements present into ionic form. $$2 \mathrm{Na} + \mathrm{S} \rightarrow \mathrm{Na_2 S}$$ Test of sulphur: The extract is treated with sodium nitroprusside. The appearance of violet colour indicates the presence of sulphur. The following reaction occurs: $$\mathrm{S^{2-}} + \left[ \mathrm{Fe(CN)_5 \, NO} \right]^{2-} \rightarrow \left[ \mathrm{Fe(CN)_5 \, NOS} \right]^{4-}$$ The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus. The halogen present in the compound converted into sodium halide, and it will be precipitated with silver nitrate.

The stability of $\mathrm{M}^+$ ions in aqueous medium depends on three factors: (i) Enthalpy of atomisation (ii) First ionisation enthalpies of isolated gaseous atom (iii) Hydration enthalpy. An element in $\mathrm{M}^+$ state in aqueous medium is stabler if the electrode potential ($\mathrm{M}^+/\mathrm{M}$) value of more negative. Down the group, the tendency to form $\mathrm{M}^+$ ion increases. In this process, ionisation of solid metal is not involved in this process.

The addition polymerisation of acrylonitrile in presence of a peroxide catalyst leads to the formation of polyacrylonitrile. The commercial name of polyacrylonitrile is orlon (acrilan). That is, it is used as a substitute for wool in making commercial fibres as orlon or acrilan.

Electron gain enthalpy is defined as the amount of energy released when an electron is added to an isolated gaseous atom. During the addition of an electron, energy can either be released or absorbed. Electron gain enthalpy decreases down the group because as we move down the group, the size of an element increases, and the effective nuclear charge decreases. Due to this, the incoming electron will experience less attraction. Thus, electron gain enthalpy decreases as we move down the group. $\Delta H_{eg}$ for chlorine $= -349 \, \mathrm{kJ \, mole^{-1}}$ $\Delta H_{eg}$ for Neon $= +116 \, \mathrm{kJ \, mole^{-1}}$ $\Delta H_{eg}$ for Argon $= +96 \, \mathrm{kJ \, mol^{-1}}$ $\Delta H_{eg}$ for fluorine $= -333 \, \mathrm{kJ \, mol^{-1}}$

The Zymase enzyme converts glucose and fructose into ethanol and carbon dioxide. And the Invertase enzyme is used to convert sucrose or cane sugar into glucose and fructose. The enzyme diastase converts starch to maltose. The enzyme pepsin converts proteins to amino acids.

Photochemical smog is a mixture of pollutants that are formed when nitrogen oxides and volatile organic compounds (VOCs) react to sunlight, creating a brown haze above cities. It tends to occur more often in summer, because that is when we have the most sunlight. Photochemical smog formation will be more at industrial areas.

Cement or Portland cement mainly contains oxides of calcium, silicon, aluminium, and iron. When water is mixed with cement, the oxides get hydrated, and cement sets extremely fast. To control the setting time of cement, gypsum is added to it, which increases the setting time of cement.

Loss of an electron from a hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to its small size it cannot exist free. Loss of the electron from hydrogen atom results in nucleus size ($\mathrm{H}^+$) of $1.5 \times 10^{-3}$. A proton is a positively charged subatomic particle that is present in the nucleus of an atom, and it can exist in various forms, including as a part of an atomic nucleus, as a free particle, or as a constituent of a larger particle such as a molecule. $\mathrm{H}^+$ always exists in combined form. Both A and R are correct but there is no relation between two statements.

Q is giving effervescence with sodium bicarbonate, it means Q contain carboxylic acid functional group. The compound R is reacting with Hinsberg's reagent and giving solid which soluble in sodium hydroxide means R contain primary amine functional group. Hence, The compound P can be a secondary amide. The overall reaction is shown below.

(A) Hell-Volhard-Zelinsky reaction: Carboxylic acid react with chlorine or bromine in the presence of phosphorus to give $\alpha$-halogenated carboxylic acids. The halogenation continues till all the hydrogen atom in the $\alpha$-position are replaced by halogen atoms. This reaction is called HVZ (Hell-Volhard-Zelinsky) reaction. $$CH_3COOH \xrightarrow{Cl_2,\, P_4} ClCH_2COOH \xrightarrow{Cl_2,\, P_4} Cl_2CHCOOH \xrightarrow{Cl_2,\, P_4} Cl_3CCOOH$$ (B) Iodoform reaction: The Iodoform reaction is a chemical reaction where a iodoform ($CHI_3$) is produced by the full iodonation of a methyl ketone ($RCOCH_3$, where R can be either a hydrogen atom, an alkyl or an aryl group) in the presence of a base. $$2NaOH + I_2 \rightarrow NaOI + NaI$$ $$CH_3COCH_3 + 3NaOI \rightarrow CH_3COONa + CHI_3 + 2NaOH$$ (C) Etard reaction: In Etard reaction, chromyl chloride oxidises methyl group to a chromium complex, which on hydrolysis gives corresponding Benzaldehyde. The equation for the reaction is: $$C_6H_5 - CH_3 + CrO_2Cl_2 \xrightarrow{CS_2} C_6H_5 - CH(O CrOHCl_2) \xrightarrow{H_3O^+} C_6H_5 - CHO$$ (D) Gatterman-Koch reaction: When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, it gives Benzaldehyde or substituted Benzaldehyde. The equation for the reaction is: $$C_6H_5 - CH_3 \xrightarrow{CO + HCl} C_6H_5 - CHO$$ $$Anh.AlCl_3 / CuCl$$

Acetanilide undergo electrophilic bromination with bromine in acetic acid. p-bromoacetanilide is the major product in this reaction. Acetanilide undergo reduction reaction with lithium aluminium hydride.

$[\mathrm{Fe(H_2O)_6}]^{3+}$ has a magnetic moment of $5.92\,\mathrm{BM}$ because $\mathrm{H_2O}$ is a weak ligand and electron pairing does not occur. It involves $sp^3d^2$ hybridisation with five unpaired electrons and forms an outer-orbital complex. The magnetic moment can be calculated using the formula $\mu_m=\sqrt{n(n+2)}\,\mathrm{BM}$ where $\mu_m$ is the magnetic moment and $n$ is the number of unpaired electrons. $\mu_m=1.74\,\mathrm{BM},\quad n=1$ $\mu_m=5.92\,\mathrm{BM},\quad n=5$ In the $[\mathrm{Fe(CN)_6}]^{3-}$ complex, $\mathrm{CN^-}$ is a stronger ligand than $\mathrm{H_2O}$ according to the spectrochemical series. Hence, four electrons pair up. It involves $d^2sp^3$ hybridisation with one unpaired electron and forms an inner-orbital complex. Therefore, it has a magnetic moment of only $1.74\,\mathrm{BM}$. In both complexes, the oxidation state of iron is $+3$.

Vitamin A Deficiency: Xerophthalmia: With this condition, the eyes may become very dry and crusted, which may damage the cornea and retina. Thiamine Deficiency: Beriberi is a disease in which the body does not have enough thiamine (vitamin B1). Scurvy is a disease caused by a serious vitamin C (Ascorbic acid) deficiency. Riboflavin deficiency is often associated with cheilosis (chapping and fissuring of the lips). Riboflavin deficiency can be associated with various developmental abnormalities, including cleft lip.

Gold is then extracted from this complex by displacement technique by using a more electropositive zinc metal. In this method, zinc (Zn) acts as a reducing agent and it reduces $\mathrm{Au}^{+}$ to Au and itself gets oxidized from Zn to $\mathrm{Zn}^{+2}$ ions which combine with $\mathrm{CN}^{-}$ ions and form soluble complex, sodium tetracyanozincate (II). $$2[\mathrm{Au(CN)_2}]^{-} \,(\mathrm{aq}) + \mathrm{Zn} \,(\mathrm{s}) \rightarrow 2\, \mathrm{Au} \,(\mathrm{s}) + [\mathrm{Zn(CN)_4}]^{2-} \,(\mathrm{aq})$$

The coplanarity of $\mathrm{N_2O_4}$ results from a delicate balance of the repulsive forces, which favour the skew structure, and of the effects of bonding which favour the planar structure. $\mathrm{NO_2}$ is a bent molecule. The $\mathrm{N_2O_5}$ molecule is a trigonal planar molecule. $\mathrm{N_2O}$ has a linear, unsymmetrical structure that may be thought of as a hybrid of two resonance forms.

$E_{\mathrm{Ce}^{4+}/\mathrm{Ce}^{3+}} = +1.74 \, \mathrm{V}$ Lanthanoids have $+3$ oxidation state as stable oxidation state. Hence, in their $+4$ oxidation state, they act as oxidising agents and $+2$ oxidation state, they act as reducing agents.

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine. This reaction involves the degradation of amide and is popularly known as Hoffmann bromamide degradation reaction. In the given molecule, amide part convert to amine with bromine in sodium hydroxide. Now, nucleophilic acyl substitution reaction takes place to get the final product as shown below.

Given that concentration of $\mathrm{A_2B_3}$ is $C$ and degree of dissociation is $\alpha$. The equilibrium reaction of weak electrolyte $\mathrm{A_2B_3}$ is $$\mathrm{A_2B_3} \rightleftharpoons 2 \mathrm{A^{3+}} + 3 \mathrm{B^{2-}}$$ $$C \rightarrow C(1-\alpha) 2C\alpha 3C\alpha$$ Now, the equilibrium constant is the ratio of the concentrations of reactants and products raised to the stoichiometric coefficients. $$K_{eq} = \frac{(2C\alpha)^2 (3C\alpha)^3}{C(1-\alpha)} = 108C^4\alpha^5$$ $$\frac{K_{eq}}{108C^4} = \alpha^5 \Rightarrow \alpha = \left(\frac{K_{eq}}{108C^4}\right)^{\frac{1}{5}}$$

As mentioned the reaction follows $S_N2$ reaction mechanism. The Finkelstein reaction classically involves the conversion of Alkyl bromides into Alkyl iodides by the treatment with a solution of Sodium iodide in Acetone. NaBr is not readily soluble in acetone and hence reaction shifts in forward reaction. For $S_N2$ reactions, transition state formed is less polar than the localised anion.

Given: Kinetic energy (K.E.) = 4.55 $\times$ 10^{-29} $\,$ $\mathrm{J}$ Mass of electron $$ (m_e) = 9.1 \times 10^{-31} \, \mathrm{kg} $$ $$ h = 6.6 \times 10^{-34} \, \mathrm{J \cdot sec} $$ De Broglie wavelength $\lambda_d = \frac{h}{mv} = \frac{h}{\sqrt{2m \mathrm{KE}}}$ Here $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, $m$ is the mass of the particle, $v$ is its velocity, and $p$ is the momentum of the particle, which is equal to $mv$. $$ \lambda = \frac{h}{\sqrt{2m \mathrm{KE}}} $$ $$ = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 4.5 \times 10^{-29}}} = \frac{6.6 \times 10^{-34}}{9 \times 10^{-30}} = \frac{6.6}{9} \times 10^{-4} $$ $$ = \frac{66}{9} \times 10^{-5} $$ $$ = 7.33 \times 10^{-5} $$

A substitution reaction is a chemical reaction during which one functional group in a chemical compound is replaced by another functional group. The number of different products possible when ethane is allowed to react with bromine in sunlight are 9. $CH_3-CH_3+Br_2\xrightarrow{\text{sunlight}}CH_3-CH_2Br+CH_2Br-CH_2Br+CH_2Br-CHBr_2+CHBr_2-CHBr_2+CBr_3-CBr_3+CH_3-CHBr_2+CH_3-CBr_3+CH$ 6 Bromine atoms: 1 product possible 5 Bromine atoms: 1 product possible 4 Bromine atoms: 2 products possible 3 Bromine atoms: 2 products possible 2 Bromine atoms: 2 products possible 1 Bromine atom: 1 product possible Total 9 products possible

For spontaneous process, $\Delta G = -\, \mathrm{ve}$, at equilibrium $\Delta G = 0$ and for non-spontaneous process $\Delta G = +\, \mathrm{ve}$. The relation between the change in Gibbs reaction energy and Gibbs energy can be defined as the slope of the Gibbs energy plotted against the extent of reaction at constant pressure and temperature. At point a: Slope $= -\, \mathrm{ve}$ $$\Delta G = -\, \mathrm{ve}$$ Hence, reaction is spontaneous at point a. At point b: Slope $= 0$ $$\Delta G = 0$$ Hence, it is the equilibrium condition. At point c: Slope $= +\, \mathrm{ve}$ $$\Delta G = +\, \mathrm{ve}$$ Hence, reaction is non-spontaneous at point c.

Relative lowering of vapour pressure equals mole fraction of solute. $$\frac{Lowering of vapour pressure}{Vapour pressure of solution} = \frac{moles of solute}{moles of solvent}$$ $$\frac{25}{100}$$ $$\frac{25}{75} = \frac{moles of urea}{\left(\frac{1000}{18}\right)}$$ $$\frac{1}{3} \times \frac{1000}{18} = moles of urea$$ Mass of urea equals $$\frac{1}{3} \times \frac{1000}{18} \times 60$$ $$= 1111.11 \, gm$$

We can use the relationship between the equilibrium constant (K) and the standard Gibbs free energy change ($\Delta G^\circ$) to calculate the value of $\log K$ at $298 \, \mathrm{K}$. Given, $\Delta H^\circ = -54.07 \, \mathrm{kJ \, mol^{-1}}$ $\Delta S^\circ = 10 \, \mathrm{J \, K^{-1} \, mol^{-1}}$ We know, $$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$ $$= -54.07 - \frac{298(10)}{1000}$$ $$= -57.05 \, \mathrm{kJ/mole}$$ $$\Delta G^\circ = -2.303 \, RT \, \log K_{eq}$$ $$-57.05 \times 1000 = -2.303 \times 8.314 \times 298 \, \log K_{eq}$$ $$-57.05 \times 1000 = -5705 \, \log e_{eq}$$ $$10 = \log K_{eq}$$

Among the given, $\mathrm{IF_5}$, $\mathrm{BrF_5}$, and $\mathrm{XeOF_4}$ have square pyramidal structure. VSEPR theory helps us to predict the shape of molecules from the number of electron pairs and lone pairs of electrons that surrounds the central atoms.

Ambidentate ligand has two or more donor atoms, but during complex formation, only one donor atom is attached to central metal. SCN$^{-}$ ion has donor atoms N and S. Examples of ambidentate ligands include $\mathrm{NO_2^{-}}$ and $\mathrm{SO_3^{2-}}$. So, 4 SCN$^{-}$ ligands are ambidentate. $$[\mathrm{M(en)(SCN)_4}]$$ $$\mathrm{S = C = N^-}$$

According to Arrhenius's equation, $$k = A e^{-E_a/RT}$$ $$\log k = \log A - \frac{E_a}{2.3 \, RT}$$ $E_{a1}$ = For the adsorption of hydrogen on platinum, the activation energy $E_{a2}$ = for the adsorption of hydrogen on nickel, the activation energy $$\log \left( \frac{k_2}{k_1} \right) = \frac{E_{a2} - E_{a1}}{2.3 \, RT} = \frac{11.4 \times 1000}{2.3 \times 8.3 \times 300}$$ $$= 1.990 \approx 2$$

The balanced chemical equation for the reaction between $\mathrm{BaCl_2}$ and $\mathrm{Na_3 PO_4}$ is: $$3 \mathrm{BaCl_2} + 2 \mathrm{Na_3 PO_4} \rightarrow \mathrm{Ba_3(PO_4)_2} + 6 \mathrm{NaCl}$$ 5 moles of barium chloride will require $\frac{5 \times 2}{3} = 3.3$ moles of sodium phosphate. Here, sodium phosphate is the limiting reagent. 2 moles of sodium phosphate produce 1 mole of barium phosphate. Thus, 2 moles of sodium phosphate will produce 1 mole of barium phosphate. Hence, the maximum no. of moles of $\mathrm{Ba_3(PO_4)_2}$ that can be formed is one mole.

Phosphomolybdate is an anion containing both phosphorus and molybdenum, with the chemical formula $\mathrm{(NH_4)_3 PO_4 \cdot 12MoO_3}$. The ammonium ion has a total charge of $+3$, and the phosphate ion has a total charge of $-3$. Therefore, the $12 \ \mathrm{MoO_3}$ units in the compound must have a total charge of $+3 - (-3) = +6$. Each $\mathrm{MoO_3}$ unit contains one Mo atom and three O atoms. The total oxidation state of the three O atoms is $-6$, since each O atom has an oxidation state of $-2$. So, $$\mathrm{Mo} + (-6) = 0$$ $$\mathrm{Mo} = +6$$