JEE Main 11 April 2023 Shift 1 question paper with solutions

Let $I = \int_{-\log_e 2}^{\log_e 2} e^x \left[ \log_e \left( e^x + \sqrt{1 + e^{2x}} \right) \right] dx$. Let us substitute $e^x = t$. \[ \Rightarrow e^x dx = dt \] Lower limit $= e^{-\log_e (2)} = \frac{1}{2}$ Upper limit $= e^{\log_e (2)} = 2$ \[ \Rightarrow I = \int_{\frac{1}{2}}^{2} 1 \times \log_e \left[ t + \sqrt{1 + t^2} \right] dt \] Apply integration by parts. \[ \int u(t) v(t) dt = u(t) \int v(t) dt - \int (u'(t) \int v(t) dt) dt \] Take $u(t) = \log_e \left[ t + \sqrt{1 + t^2} \right]$, $v(t) = 1$ \[ \Rightarrow u'(t) = \frac{1}{t + \sqrt{1 + t^2}} \left( 1 + \frac{2t}{2\sqrt{1 + t^2}} \right) = t \] \[ = \left[ t \ln \left( \sqrt{t^2 + 1 + x} \right) \right]_{\frac{1}{2}}^{2} - \int_{\frac{1}{2}}^{2} \frac{t}{\sqrt{t^2 + 1}} dt \] \[ = \left[ t \ln \sqrt{t^2 + 1 + t - \sqrt{t^2 + 1}} \right]_{\frac{1}{2}}^{2} \] \[ = \left[ 2 \ln \left( \sqrt{5} + 2 \right) - \sqrt{5} \right] - \left[ \frac{1}{2} \ln \left( \sqrt{5} + 2 + \frac{1}{2} \right) - \sqrt{\frac{5}{2}} \right] \] \[ = \ln \left( \frac{\sqrt{2} (2 + \sqrt{5})^2}{1 + \sqrt{5}} \right) - \frac{\sqrt{5}}{2} \] Hence this is the correct option.

Given the equation of the plane that contains the point $(-2, 3, 5)$ and is perpendicular to each of the planes $2x + 4y + 5z = 8$ and $3x - 2y + 3z = 5$ is $$\alpha x + \beta y + \gamma z + 97 = 0$$ Now satisfying the point $(-2, 3, 5)$ in plane $P : \alpha x + \beta y + \gamma z + 97 = 0$ we get, $$-2\alpha + 3\beta + 5\gamma + 97 = 0 \ldots (i)$$ And using perpendicular condition $2\alpha + 4\beta + 5\gamma = 0$ we get, $$2\alpha + 4\beta + 5\gamma = 0 \ldots (ii)$$ Similarly, $3\alpha - 2\beta + 3\gamma = 0 \ldots (iii)$ Now solving equation $(i)$, $(ii)$ $\&$ $(iii)$ we get, $$\alpha = 22, \beta = 9 \& \gamma = -16$$ Hence, $\alpha + \beta + \gamma = 22 + 9 - 16 = 15$

Given, R be a rectangle given by the lines $x = 0$, $x = 2$, $y = 0$ and $y = 5$. And point $A(\alpha, 0)$ and $B(0, \beta)$, $\alpha \in [0, 2]$ and $\beta \in [0, 5]$. Now plotting the diagram we get, Now given that line segment $AB$ divides the ratio of area in $4 : 1$, we get $$\frac{10 - \frac{1}{2} \alpha \beta}{\frac{1}{2} \alpha \beta} = \frac{4}{1}$$ $$\Rightarrow 20 - \alpha \beta = 4 \alpha \beta$$ $$\Rightarrow \alpha \beta = 4 \ldots \ldots (1)$$ Now using midpoint formula we get, $$h = \frac{\alpha}{2} \& \beta = \frac{k}{2}$$ Now using equation (1) we get, $$\Rightarrow 4hk = 4$$ $$\Rightarrow xy = 1$$ which is a equation of rectangular hyperbola.

Let the elements in set $A$ be $\{x_1, x_2, x_3, x_4, x_5\}$. Now given mean is $= 5$. Now subtracting 3 from each term we get, new mean as $(x_1 - 3, x_2 - 3, \ldots, x_5 - 3) = 5 - 3 = 2$. So, sum of elements will be $2 \times 5 = 10$. Also given variance, $Var(X) = 12$. Now we know that subtracting 3 from each term will not change the variance, so, $Var(x_1 - 3, x_2 - 3, \ldots, x_5 - 3) = 12$. $$\Rightarrow \frac{\sum (x_i - 3)^2}{5} = 12$$ $$\Rightarrow \sum (x_i - 3)^2 = 80$$ Now let elements in set $B$ be $\{y_1, y_2, \ldots, y_5\}$. Given mean $(y_1, y_2, \ldots, y_5) = 8$. Now adding each element by 2 we get, new mean $(y_1 + 2, y_2 + 2, \ldots, y_5 + 2) = 10$. So, sum of elements will be $10 \times 5 = 50$. Also given $Var(y_1, y_2, \ldots, y_5) = 20$. Similarly new variance, $Var(y_1 + 2, y_2 + 2, \ldots, y_5 + 2) = 20$. $$\Rightarrow \frac{\sum (y_i + 2)^2}{5} = 20$$ $$\Rightarrow \sum (y_i + 2)^2 = 120 \times 5$$ Now finding, combined mean we get, $$\frac{\sum_{i=1}^{5} (x_i - 3) + \sum (y_i + 2)}{10} = \frac{10 + 50}{10} = 6$$ And combined variance is $$\frac{\sum (x_i - 3)^2 + \sum (y_i + 2)^2}{10} - 6^2$$ $$= \frac{80 + 120 \times 5}{10} - 36 = 32$$ Hence, the sum of combined mean and variance will be $32 + 6 = 38$.

Given, $$f(x) = [x^2 - x] + |-x + [x]|$$ $$\Rightarrow f(x) = [x^2 - x] + |x - [x]|, \{ as |A| = |-A| \}$$ $$\Rightarrow f(x) = [x^2 - x] + \{x\}, \{ as [x] + \{x\} = x \}$$ $$\Rightarrow f(x) = [x^2 - x] + \{x\} (\because \{x\} \geq 0)$$ Now at $x = 0$, $f(0) = 0$ And $f(0^+) = -1$, $\{ as x^2 - x < 0 for x \to 0^+ \}$ So, function is discontinuous at $x = 0$ Now at $x = 1$, $f(1) = 0$, $$f(1^+) = 0 + 0 = 0 and f(1^-) = -1 + 1 = 0$$ Hence, the function is continuous at $x = 1$.

Given, $$x + y + z = 15$$ Now we know that, Non-negative integral solution of equation $a + b + c = n$ is given by $\binom{n+3-1}{3-1}$ where $a = b = c$ and $a = b \neq c$ are also possibilities. So by above formula we get, Total number of non-negative solution will be, $$\binom{15+3-1}{3-1} = \binom{17}{2}$$ Now solving if any of these 2 are equal So, the equation will become $x + 2y = 15$ Now finding possible cases we get, $$y = 0 x = 15$$ $$y = 1 x = 13$$ $$y = 2 x = 11$$ $$y = 3 x = 9$$ $$y = 4 x = 7$$ $$y = 5 x = 5 \rightarrow x = y = z = 5$$ $$y = 6 x = 3$$ $$y = 7 x = 1$$ So, total possibilities where $x$, $y$ and $z$ are distinct will be, $$= \binom{17}{2} - \binom{3}{2} \times 8 + 2$$ {Note adding 2 because the cases $x = y = z$ is subtracted three times} $$= 136 - 24 + 2 = 114$$

Given, $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$, with $|0|a_i| < 1$, $i = 1, 2, 3$. Let us assume that $|a_1| \leq |a_2| \leq |a_3|$. Now we know that, $$|\vec{a}|^2 = |\vec{a_1}|^2 + |\vec{a_2}|^2 + |\vec{a_3}|^2$$ And $$|\vec{a_3}|^2 \leq |\vec{a_1}|^2 + |\vec{a_2}|^2 + |\vec{a_3}|^2$$ So, combining both above equations we get, $$\Rightarrow |\vec{a}|^2 \geq |\vec{a_3}|^2$$ $$\Rightarrow |\vec{a}| \geq |\vec{a_3}| = \max \left\{ |\vec{a_1}|, |\vec{a_2}|, |\vec{a_3}| \right\}$$ Hence, (A) is true. Now again solving, $$|\vec{a}|^2 = |\vec{a_1}|^2 + |\vec{a_2}|^2 + |\vec{a_3}|^2$$ And $$|\vec{a_1}|^2 + |\vec{a_2}|^2 + |\vec{a_3}|^2 \leq |\vec{a_3}|^2 + |\vec{a_3}|^2$$ $$\Rightarrow |\vec{a_1}|^2 + |\vec{a_2}|^2 + |\vec{a_3}|^2 \leq 3 |\vec{a_3}|^2$$ $$\Rightarrow |\vec{a}|^2 \leq 3 |\vec{a_3}|^2$$ $$\Rightarrow |\vec{a}| \leq \sqrt{3} |\vec{a_3}| = \max \left\{ |\vec{a_1}|, |\vec{a_2}|, |\vec{a_3}| \right\}$$ Hence, (B) is also true.

Given, two complex numbers $w_1$ and $w_2$. $w_1 = 3 + 5i$ and $w_2 = 5 + 4i$ are both rotated by $90^\circ$ with respect to origin anticlockwise and clockwise respectively. So by concept of rotation we get, $$w_3 = iw_1 = i(3 + 5i) = -5 + 3i$$ and $$w_4 = -iw_2 = -i(5 + 4i) = 4 - 5i$$ Now principal argument of $w_3 - w_4 = -9 + 8i$ will be $$\pi - \tan^{-1} \frac{8}{9}$$ {as complex number is in second quadrant so principal argument is given by $\pi - \tan^{-1} \left| \frac{y}{x} \right|$}

Let $A$, $B$ and $C$ denote the set of men who received medals in event $A$, event $B$ and event $C$ respectively. Then, $n(A) = 48$, $n(B) = 25$, $n(C) = 18$, $n(A \cup B \cup C) = 60$ and $n(A \cap B \cap C) = 5$. Now we know that, $$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)$$ $$\Rightarrow 60 = 48 + 25 + 18 - n(A \cap B) - n(A \cap C) - n(B \cap C) + 5$$ $$\Rightarrow n(A \cap B) + n(A \cap C) + n(A \cap C) = 48 + 25 + 18 + 5 - 60 = 36$$ Therefore, the number of people who received medals in exactly two of the three sports will be $36 - 3(n(A \cap B \cap C)) = 36 - 15 = 21$.

Given, $$S = \{ M = [a_{ij}], \ a_{ij} \in \{0, 1, 2\}, \ \{1 \leq i, j \leq 2\} \}$$ So, let $$M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ where $$a, b, c, d \in \{0, 1, 2\}$$ So, total sample space will be $$3^4 = 81$$ Now we will find $$P(A)$$ or possibilities for $$|M| = 0$$ So, for $$|M| = 0$$ cases will be, 1. $$\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$ or $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$ or $$\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \rightarrow$$ Total matrix = 3 2. Two 1's and Two 0's $$\rightarrow$$ Total matrix = 4 3. Two 2's and Two 0's $$\rightarrow$$ Total matrix = 4 4. Two 1's and Two 2's $$\rightarrow$$ Total matrix = 4 5. One 1 and three 0's $$\rightarrow$$ Total matrix = 4 6. One 2 and Three 0's $$\rightarrow$$ Total matrix = 4 7. One 1 and one 2 and two 0's $$\rightarrow$$ Total matrix = 8 So, total cases for which $$|M| = 0$$ is 31, So probability will be $$P(\overline{A}) = \frac{31}{81}$$ Hence, $$P(A) = 1 - \frac{31}{81} = \frac{50}{81}$$

Given, $E_k : kx^2 + k^2y^2 = 1$ $$\Rightarrow E_k : \left( \frac{x}{\frac{1}{\sqrt{k}}} \right)^2 + \left( \frac{y}{\frac{1}{k}} \right)^2 = 1$$ Now equation of the chord joining the points $\left( \frac{1}{\sqrt{k}}, 0 \right)$ and $\left( 0, \frac{1}{k} \right)$ will be, $$L_k : \frac{x}{\left( \frac{1}{\sqrt{k}} \right)} + \frac{y}{\left( \frac{1}{k} \right)} = 1$$ $$\Rightarrow \sqrt{k}x + ky - 1 = 0$$ Now $r_k =$ Perpendicular distance of $L_k$ from $(0, 0)$ we get, $$r_k = \left| \frac{-1}{\sqrt{k} + k^2} \right|$$ $$\Rightarrow r_k^2 = \frac{1}{k + k^2}$$ Now putting the value of $r_k^2$ in $\sum_{k=1}^{20} \frac{1}{r_k^2}$ we get, $$\sum_{k=1}^{20} \frac{1}{r_k^2} = \sum_{k=1}^{20} k + k^2 = \frac{20 \times 21}{2} + 20 \times 21 \times 41$$ $$= 210 + 2870 = 3080$$

We have been given that $\log \left( z + \frac{7}{z-3} \right)^2 \geq 0$. We need to find the domain. For $\log_{g(x)}(f(x))$ is defined when $f(x) > 0$ and $g(x) > 0$, $g(x) \neq 1$. $\Rightarrow x + \frac{7}{z} > 0$ $\Rightarrow x > -\frac{7}{z}$ Also, $x + \frac{7}{z} \neq 1$ $\Rightarrow x \neq -\frac{7}{z} + 1$ And $z - 3 \neq 0$ $\Rightarrow x \neq 7$ and $x \neq \frac{3}{2}$ Domain : $\left( -\frac{7}{2}, \infty \right) - \left\{ -\frac{5}{2}, 0, \frac{3}{2} \right\}$ Now let us take different cases. Case I: $0 0$ $\ldots$ (2) And $\frac{x-7}{z-3} - 1 \leq 0$ $\Rightarrow \frac{x-7}{z-3} - 1 \leq 0$ $\Rightarrow \frac{2x-3}{3} \leq 0$ $\Rightarrow \frac{2x-3}{3} \leq 0$ $\Rightarrow \frac{2x-3}{3} \leq 0$ $\ldots$ (3) From (1), (2), (3) we can tell that there is no intersection, no solution Case II: $\Rightarrow x + \frac{7}{z} > 1$ $\Rightarrow x > -\frac{5}{2}$ $\Rightarrow \left( \frac{x-7}{z-3} \right) \geq 1$ $\Rightarrow \frac{x-7}{z-3} \geq 1$ $\Rightarrow \frac{x-7}{z-3} < -1$ $\Rightarrow x \in \left( -\frac{4}{3}, \frac{3}{2} \right) \cup \left( \frac{3}{2}, \frac{10}{3} \right)$ Hence there are total 6 integers $\{-2, -1, 0, 1, 2, 3\}$. Hence this is the required option.

Given, $$x^2 + (y - 2)^2 \leq 4 and x^2 \geq 2y$$ Now on solving, $$x^2 + (y - 2)^2 = 4 and x^2 = 2y$$ we get intersecting point of the curves as $$(\pm 2, 2),$$ Now plotting the diagram we get, So, from above diagram area of required region will be, $$= 2 \left[ \frac{1}{4} \left( \pi \times 2^2 \right) - \int_0^2 \sqrt{2} \cdot \sqrt{y} dy \right]$$ $$= 2 \left[ \pi - \left. \frac{\sqrt{2} y^{\frac{3}{2}}}{\frac{3}{2}} \right|_0^2 \right]$$ $$= 2 \left[ \pi - \frac{2\sqrt{2}}{3} \cdot 2\sqrt{2} \right]$$ $$= 2 \left[ \pi - \frac{8}{3} \right]$$ $$= 2\pi - \frac{16}{3}$$

Given, domain and range of function, $f : [2, 4] \rightarrow \mathbb{R}$. $$(x \log_e x) f'(x) + (\log_e x) f(x) + f(x) \geq 1, \; x \in [2, 4]$$ $$\Rightarrow (x \log_e x) \frac{d}{dx} f(x) + (\log_e x) f(x) \frac{d}{dx} (x) + x f(x) \frac{d}{dx} (\log_e x) \geq 1$$ $$\Rightarrow \frac{d}{dx} [x \ln x f(x)] \geq 1$$ $$\Rightarrow \frac{d}{dx} [x \ln x f(x)] - \frac{d}{dx} (x) \geq 0$$ $$\Rightarrow \frac{d}{dx} [x \ln x f(x) - x] \geq 0$$ Hence, $h(x) = x(\ln x) f(x) - x$ is an increasing function, Therefore, $h(x) \geq h(2), \; x \in [2, 4]$ $$\Rightarrow x \ln x \times f(x) - x \geq 2 \ln 2 \times f(2) - 2$$ $$\Rightarrow x \ln x f(x) - x \geq \ln 2 - 2$$ Similarly, $h(x) \leq h(4)$ $$\Rightarrow x \ln x f(x) - x \leq \ln 4 - 4$$ So, $$\frac{\ln 2 - 2}{x \ln x} + \frac{1}{\ln x} \leq f(x) \leq \frac{\ln 4 - 4}{x \ln x} + \frac{1}{\ln x}$$ Now for $x \in [2, 4]$, $$\frac{\ln 4 - 4}{x \ln x} + \frac{1}{\ln x} \leq \frac{\ln 4 - 4}{2 \ln 2} + \frac{1}{\ln 2} = 1 - \frac{1}{\ln 2} \frac{1}{8}$$ Hence, $$f(x) \geq \frac{1}{8}$$ Hence both A & B are correct. Note this question was bonus in Jee Main 2023 April session, as LMVT on $f(x) \cdot x \ln x$ can't be satisfied. Hence, no such $f(x)$ exist.

Given, $$(1 - x^2 y^2) dx = y dx + x dy$$ which implies $$dx = \frac{y dx + x dy}{1 - (xy)^2}$$ $$\Rightarrow dx = \frac{d(yx)}{1 - (xy)^2}$$ $$\Rightarrow dx = \frac{1}{2} \left( \frac{d(xy)}{1 - xy} + \frac{d(xy)}{1 + xy} \right)$$ Now integrating both sides we get, $$\Rightarrow 2x + c = \ln \left| \frac{1 + xy}{1 - xy} \right|$$ $$\Rightarrow \left| \frac{xy + 1}{xy - 1} \right| = e^c e^{2x}$$ Now given, $y(1) = 2$ so putting the value in the above equation we get, $$3 = e^c e^2$$ $$\Rightarrow e^c = \frac{3}{e^2}$$ Hence, the equation becomes $$\left| \frac{xy + 1}{xy - 1} \right| = 3e^{x - 2}$$ Now finding $y(2)$ so putting $x = 2$ in the above equation we get, $$\left| \frac{2y + 1}{2y - 1} \right| = 3e^2$$ $$\Rightarrow 2y + 1 = 2 \cdot 3e^2 y - 3e^2$$ $$\Rightarrow 1 + 3e^2 = 2y(3e^2 - 1)$$ $$\Rightarrow y(2) = \frac{1 + 3e^2}{2(3e^2 - 1)}$$

Given, A be a 2 $\times$ 2 matrix with real entries such that $A' = \alpha A + I$, Now let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ Now putting the value in $A' = \alpha A + I$ we get, $$\Rightarrow \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} \alpha a + 1 & \alpha b \\ \alpha c & \alpha d + 1 \end{bmatrix}$$ Now equating both sides we get, $a = \alpha a + 1 \Rightarrow a = \frac{1}{1-\alpha} \ldots \ldots \ldots (i)$ $b = \alpha c \ldots \ldots (ii)$ and $c = \alpha b \ldots \ldots \ldots (iii)$ Now from equation $(ii)$ and $(iii)$ we get, $c = 0$ or $\alpha = \pm 1$ (not possible as mentioned in question) $\therefore c = 0$ $\Rightarrow c = 0, \ b = 0$ Also $d = \alpha d + 1 \Rightarrow d = \frac{1}{1-\alpha}$ Also given, $|A^2 - A| = 4$ $$\Rightarrow |A||A - I| = 4$$ $$\Rightarrow \left( \frac{1}{1-\alpha} \right)^2 \left( \frac{1}{1-\alpha} - 1 \right)^2 = 4$$ Now let $\frac{1}{1-\alpha} = t$ So, $t^2(t-1)^2 = 4$ $$\Rightarrow t(t-1) = \pm 2$$ $$\Rightarrow t = 2 or t = -1$$ $$\Rightarrow \alpha = \frac{1}{2}, 2$$ So, sum of all possible value will be $\frac{1}{2} + 2 = \frac{5}{2}$

Given plane is $2x + y - 3z = 6$. Since, $(\alpha, \beta, \gamma)$ is the image point of $(2, 3, 5)$ in the given plane, so $$\frac{\alpha - 2}{2} = \frac{\beta - 3}{1} = \frac{\gamma - 5}{-3} = \frac{-2 \left(4 + 3 - 15 - 6\right)}{4 + 1 + 9} = 2$$ $$\Rightarrow \frac{\alpha - 2}{2} = 2; \frac{\beta - 3}{1} = 2; \frac{\gamma - 5}{-3} = 2$$ $$\therefore \alpha = 6, \beta = 5, \gamma = -1$$ Hence, $\alpha + \beta + \gamma = 6 + 5 - 1 = 10$ Hence this is the required option.

Given, $\vec{a}$ be a non-zero vector parallel to the line of intersection of the two planes described by $\hat{i} + \hat{j}, \hat{i} + \hat{k}$. So finding the normal vector to the above plane, we get, $$\vec{n}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix} = \hat{i} - \hat{j} - \hat{k}$$ And normal vector to planes $\hat{i} - \hat{j}, \hat{j} - \hat{k}$ will be, $$\vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i} + \hat{j} + \hat{k}$$ Now $\vec{a} = \lambda (\vec{n}_1 \times \vec{n}_2)$ So, direction ratio of $\vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{vmatrix} = -2\hat{j} + 2\hat{k}$ Hence, direction ratio of $\vec{a} = (0, -2, 2) = (0, -1, 1)$ Now given, $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$ And $\vec{a} = \lambda (-\hat{j} + \hat{k})$ So, $\vec{a} \cdot \vec{b} = 6 = \lambda \left(2 + 1\right)$ $$\Rightarrow \lambda = 2$$ $$\therefore \vec{a} = -2\hat{j} + 2\hat{k}$$ Now we know that, $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$ $$\Rightarrow 6 = 2\sqrt{2} \times 3 \cos \theta$$ $$\Rightarrow \cos \theta = \frac{1}{\sqrt{2}}$$ $$\Rightarrow \theta = \frac{\pi}{4}$$ Now finding, $\left| \vec{a} \times \vec{b} \right| = \left| \vec{a} \right| \left| \vec{b} \right| \sin \theta = 2\sqrt{2} \times 3 \times \frac{1}{\sqrt{2}} = 6$ Hence, ordered pair $\left( \theta, \left| \vec{a} \times \vec{b} \right| \right) = \left( \frac{\pi}{4}, 6 \right)$

Given, $$3 \cos^4 \theta - 5 \cos^2 \theta - 2 \sin^6 \theta + 2 = 0$$ $\($ $\Rightarrow$ $\cos$^2 $\theta$ $\left$[ 3 $\cos$^2 $\theta$ - 5 $\right$] - 2 $\sin$^6 $\theta$ + 2 = 0 $\)$ $\($ $\Rightarrow$ (1 - $\sin$^2 $\theta$) (3 - 3 $\sin$^2 $\theta$ - 5) - 2 $\sin$^6 $\theta$ + 2 = 0 $\)$ $\($ $\Rightarrow$ ($\sin$^2 $\theta$ - 1) (3 $\sin$^2 $\theta$ + 2) - 2 $\sin$^6 $\theta$ + 2 = 0 $\)$ Let $\($ $\sin$^2 $\theta$ = t $\)$ $\($ $\Rightarrow$ (t - 1) (3t + 2) - 2t^3 + 2 = 0 $\)$ $\($ $\Rightarrow$ (t - 1) (3t + 2) - 2(t^3 - 1) = 0 $\)$ We know that $\($ a^3 - b^3 = (a - b) (a^2 + ab + b^2) $\)$ $\($ $\Rightarrow$ (t - 1) (3t + 2) - 2(t - 1) (t^2 + t + 1) = 0 $\)$ $\($ $\Rightarrow$ (t - 1) $\left$[ 3t + 2 - 2(t^2 + t + 1) $\right$] = 0 $\)$ $\($ (t - 1) $\left$[ 2t^2 - t $\right$] = 0 $\)$ $\($ $\Rightarrow$ t = 0, 1, $\frac{1}{2}$ $\)$ For $\($ $\sin$^2 $\theta$ = 0, $\theta$ = 0, $\pi$, 2$\pi$ $\)$ $\($ $\sin$^2 $\theta$ = 1, $\theta$ = $\frac{\pi}{2}$, $\frac{3\pi}{2}$ $\)$ $\($ $\sin$^2 $\theta$ = $\frac{1}{2}$, $\theta$ = $\frac{\pi}{6}$, $\frac{5\pi}{6}$, $\frac{7\pi}{6}$, $\frac{11\pi}{6}$ $\)$ Therefore, total solution = 9 Hence this is the required option.

Given, Mean of $x_1,x_2,\ldots,x_{100}$ is $200$. So, the sum of observations will be, $\sum x_i=100\times200$ Now using the sum of A.P. formula in above equation as all terms are in arithmetic progression we get, $\frac{100}{2}(x_1+x_{100})=100\times200$ $\Rightarrow 50(2+x_{100})=100\times200$ $\Rightarrow x_{100}=398$ $\Rightarrow x_1+99d=398$ $\Rightarrow d=4$ Now, $x_i=2+(i-1)4=4i-2$ So, $y_i=i(x_i-i)=3i^2-2i$ Now finding mean we get, $\bar y=\frac{1}{100}\sum y_i$ $\Rightarrow \bar y=\frac{1}{100}\sum(3i^2-2i)$ $\Rightarrow \bar y=\frac{1}{100}\left[3\times\frac{100\times101\times201}{6}-2\times\frac{100\times101}{2}\right]$ $\Rightarrow \bar y=\frac{1}{100}\left[3\times338350-10100\right]$ $\Rightarrow \bar y=\frac{1}{100}(1004950)$ $\Rightarrow \bar y=10049.5$

We know that. Binomial expansion of $(2 + x)^9$ is given by, $$(2 + x)^9 = {}^{9}C_{0} x^0 2^9 + {}^{9}C_{1} x^1 2^8 + \ldots + {}^{9}C_{9} x^9 2^0$$ Now put $x = 1$ we get, $$(2 + 1)^9 = {}^{9}C_{0} 2^9 + {}^{9}C_{1} 2^8 + {}^{9}C_{2} 2^7 + \ldots + {}^{9}C_{9} 2^0$$ $$\Rightarrow 3^9 = {}^{9}C_{0} 2^9 + {}^{9}C_{1} 2^8 + {}^{9}C_{2} 2^7 + \ldots + {}^{9}C_{9} 2^0$$ So, sum of coefficient of $x$, $x^2$, $x^3$, $\ldots$, $x^7$ will be, $$\Rightarrow {}^{9}C_{1} 2^8 + {}^{9}C_{2} 2^7 + \ldots + {}^{9}C_{7} 2^2 = 3^9 - {}^{9}C_{0} 2^9 - {}^{9}C_{8} 2^1 - {}^{9}C_{9} 2^0$$ $$\Rightarrow {}^{9}C_{1} 2^8 + {}^{9}C_{2} 2^7 + \ldots + {}^{9}C_{7} 2^2 = 3^9 - 2^9 - 18 - 1$$ $$\Rightarrow {}^{9}C_{1} 2^8 + {}^{9}C_{2} 2^7 + \ldots + {}^{9}C_{7} 2^2 = 19683 - 512 - 19$$ $$\Rightarrow {}^{9}C_{1} 2^8 + {}^{9}C_{2} 2^7 + \ldots + {}^{9}C_{7} 2^2 = 19152$$ Now mean is given by $\frac{19152}{7} = 2736$

Given, $$S = 109 + \frac{108}{5} + \frac{107}{5^2} + \cdots + \frac{2}{5^{107}} + \frac{1}{5^{108}} \cdots (1)$$ $$\frac{S}{5} = \frac{109}{5} + \frac{108}{5^2} + \cdots + \cdots + \frac{2}{5^{108}} + \frac{1}{5^{109}} \cdots (2)$$ Now subtracting above equations we get, $$\frac{-4S}{5} = -109 + \left( \frac{1}{5} + \frac{1}{5^2} + \cdots + \frac{1}{5^{108}} + \frac{1}{5^{109}} \right)$$ $$\Rightarrow \frac{-4S}{5} = -109 + \frac{1}{5} \left( \frac{1 - \left( \frac{1}{5} \right)^{109}}{1 - \frac{1}{5}} \right)$$ $$\Rightarrow \frac{4S}{5} = -109 + \frac{1}{4} \left( 1 - \frac{1}{5^{109}} \right)$$ $$\Rightarrow 16S = -436 + \left( 1 - \frac{1}{5^{109}} \right)$$ $$\Rightarrow 16S = -436 \times 5 + (5 - 5^{-108})$$ $$\Rightarrow 16S - (25)^{-54} = 2175$$

Given, $$\alpha(m, n) = \int_0^2 t^m (1 + 3t)^n dt$$ Now using integration by parts we get, $$\Rightarrow \alpha(m, n) = (1 + 3t)^n \cdot \frac{t^{m+1}}{m+1} \bigg|_0^2 - \int_0^2 n (1 + 3t)^{n-1} \times 3 \cdot \frac{t^{m+1}}{m+1} dt$$ $$\Rightarrow (m + 1) \alpha(m, n) = (1 + 3t)^n \cdot \left( \frac{t^{m+1}}{m+1} \right) \bigg|_0^2 - 3n(m + 1) \int_0^2 (1 + 3t)^{n-1} \times t^{m+1} dt$$ $$\Rightarrow (m + 1) \alpha(m, n) = (1 + 3 \times 2)^n \cdot \left( 2^{m+1} \right) - 3n(m + 1) \alpha(m + 1, n - 1)$$ $$\Rightarrow (m + 1) \alpha(m, n) = 7^n \cdot 2^{m+1} - 3n \alpha(m + 1, n - 1)$$ $$\Rightarrow (m + 1) \alpha(m, n) + 3n \alpha(m + 1, n - 1) = 7^n \cdot 2^{m+1}$$ Now put $m = 10$, $n = 6$ in above equation we get, $$11 \alpha(10, 6) + 18 \alpha(11, 5) = 7^6 \cdot 2^{11} = 32 \times (14)^6$$ Hence, on comparing with $11 \alpha(10, 6) + 18 \alpha(11, 5) = p \times (14)^6$ we get, $$\Rightarrow p = 32$$

5 boys with allotted roll numbers and seat numbers are seated in such a way that no one sits on the allotted seats. The number of ways of such seating arrangement is equal to the derangement of 5 boys i.e., $$= 5! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right)$$ $$= 5! - 5! + 60 - 20 + 5 - 1$$ $$= 44$$

Given, a line $L$ passes through the origin and be perpendicular to the lines $L_1 : \vec{r} = \left( \hat{i} - 11 \hat{j} - 7 \hat{k} \right) + \lambda \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec{r} = \left( -\hat{i} + \hat{k} \right) + \mu \left( 2 \hat{i} + 2 \hat{j} + \hat{k} \right)$, $\mu \in \mathbb{R}$. So, direction ratio of line will be, $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i} (4) - \hat{j} (-5) + \hat{k} (-2) = -4 \hat{i} + 5 \hat{j} - 2 \hat{k}$$ Hence, the equation of line $L$ which passes through origin will be, $$L : \vec{r} = \sigma \left( -4 \hat{i} + 5 \hat{j} - 2 \hat{k} \right)$$ Now finding the intersection point $P$ we get, $$1 + \lambda = -4 \sigma \ldots (1)$$ $$-11 + 2 \lambda = 5 \sigma \ldots (2)$$ $$-7 + 3 \lambda = -2 \sigma \ldots (3)$$ So, from equation (1) & (3) we get, $1 + \lambda = -14 + 6 \lambda \Rightarrow \lambda = 3$, $\sigma = -1$. Hence, point $P = (4, -5, 2)$. Now finding the point $Q$ which is foot of perpendicular from point $P(4, -5, 2)$ on line $L_2 : \left( -\hat{i} + \hat{k} \right) + \mu \left( 2 \hat{i} + 2 \hat{j} + \hat{k} \right)$ we get, Now any point on line $L_2$ will be $Q \left( (2 \mu - 1) \hat{i} + 2 \mu \hat{j} + (\mu + 1) \hat{k} \right)$. Now direction ratio of $PQ$ will be $(-5 + 2 \mu) \hat{i} + (2 \mu + 5) \hat{j} + (\mu - 1) \hat{k} = 0$. Now using the perpendicular condition we get, $$2(-5 + 2 \mu) + 2(2 \mu + 5) + 1(\mu - 1) = 0$$ $$\Rightarrow \mu = \frac{1}{9}$$ $$P (4, -5, 2)$$ $$M(-1 + 2 \mu, 2 \mu, 1 + \mu)$$ Hence, $\alpha + \beta + \gamma = 2 \mu - 1 + 2 \mu + \mu + 1 = 5 \mu = \frac{5}{9}$ $$\Rightarrow 9(\alpha + \beta + \gamma) = 5$$

Given expansion is $\left(3^{\frac{1}{2}} + 5^{\frac{1}{4}}\right)^{680}$. The general term in the expansion $(x + a)^n$ is $T_{r+1} = {}^{n}C_{r} \cdot x^{n-r} \cdot a^r$. Therefore, $$T_{r+1} = {}^{680}C_{r} \left(3^{\frac{1}{2}}\right)^{680-r} \left(5^{\frac{1}{4}}\right)^r, 0 \leq r \leq 680$$ $$= {}^{680}C_{r} (3)^{\frac{680-r}{2}} (5)^{\frac{r}{4}}.$$ For $(3)^{\frac{680-r}{2}} (5)^{\frac{r}{4}}$ to be rational, $r$ should be a multiple of 4. Therefore, $$r = 0, 4, 8, 12, \ldots, 680$$ $$a_n = a + (n - 1)d$$ $$680 = 0 + (n - 1)4$$ $$n = 171.$$ That means $r$ can take 171 values. Hence, the required answer is 171.

Given, Expression $(p \lor q) \land (p \lor r) \Rightarrow (q \lor r)$. Now by truth table we get, $$ \begin{array}{|c|c|c|c|c|c|c|} \hline p & q & r & p \lor q & p \lor r & (p \lor q) \land (p \lor r) & (p \lor q) \land (p \lor r) \Rightarrow (q \lor r) \\ \hline T & T & T & T & T & T & T \\ T & T & F & T & T & T & T \\ T & F & T & T & T & T & T \\ T & F & F & T & T & T & F \\ F & T & T & T & T & T & T \\ F & T & F & T & F & F & T \\ F & F & T & F & T & F & T \\ F & F & F & F & F & F & T \\ \hline \end{array} $$ Now from above truth table we can say that there are total 7 triplet are possible.

Given, equation of hyperbola, $$H_n : \frac{x^2}{1+n} - \frac{y^2}{3+n} = 1, \; n \in \mathbb{N}$$ Now using the eccentricity formula we get, $$e^2 = 1 + \frac{3+n}{1+n}$$ $$\Rightarrow e^2 = \frac{2n+4}{n+1} = \frac{2(n+2)}{n+1}$$ Now check when $(n+1) = 9, 25, 49, \ldots$ is a perfect square and then at the same time $2(n+2)$ should also be a perfect square. So checking for $n = 8 \rightarrow e^2 = \frac{20}{9}$ $$n = 24 \rightarrow e^2 = \frac{52}{25}$$ $$n = 48 \rightarrow e^2 = \frac{100}{49} \Rightarrow e = \frac{10}{7}$$ Hence, for $n = 48$ both $n+1$ and $2(n+2)$ are perfect squares. Now we know that, length of latus rectum is given by $l = \frac{2b^2}{a} = \frac{2(n+3)}{n+1} = \frac{2 \times 51}{7}$ Hence, $21l = \frac{42 \times 51}{7} = 306$

Since, $a$ and $b$ are roots of $x^2 - 7x - 1 = 0$, $$a^2 - 7a - 1 = 0$$ $$\Rightarrow a^2 - 1 = 7a$$ $$\Rightarrow a^4 + 1 - 2a^2 = 49a^2$$ $$\Rightarrow a^4 + 1 = 51a^2 \ldots(1)$$ Similarly, $$b^4 + 1 = 51b^2 \ldots(2)$$ Now, $$\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}$$ $$= \frac{a^{19}(a^4 + 1) + b^{17}(b^4 + 1)}{a^{19} + b^{19}}$$ $$= \frac{a^{19} + b^{19}}{a^{19} + b^{19}}$$ $$= \frac{51a^{19} + 51b^{19}}{a^{19} + b^{19}}$$ $$= 51$$

Given, $$A = \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}$$ $$\Rightarrow A^2 = \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix} = \begin{bmatrix} a + 2 & 2c & 3 \\ a + 3c & 2a \\ ca & 1 & 2 + 3c \end{bmatrix}$$ Now finding, $$A^3 = \begin{bmatrix} a + 2 & 2c & 3 \\ a + 3c & 2a \\ ca & 1 & 2 + 3c \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix} = \begin{bmatrix} 2ca + 3 & a + 2 + 3c & 2a + 4 + 6c \\ a^2 + 3ca + 2a & 3 + 2ac & 6 + 3a + 9c \\ a + 2 + 3c & ca + 2c + 3c^2 & 2ca + 3 \end{bmatrix}$$ Now equating $A^3 = A$ and comparing both sides we get, $$2ca + 3 = 0 \Rightarrow c = \frac{-3}{2a}$$ And $a + 2 + 3c = 1$ $$\Rightarrow a + 2 + 3 \left( \frac{-3}{2a} \right) = 1$$ $$\Rightarrow a + 1 - \frac{9}{2a} = 0$$ $$\Rightarrow 2a^2 + 2a - 9 = 0$$ $$\Rightarrow a = \frac{-2 \pm \sqrt{4 + 4 \times 9 \times 2}}{4} = \frac{-2 \pm \sqrt{76}}{4} \approx \frac{6.7}{4} \approx 1.4$$ Hence, $a \in (1, 2]$ So, on comparing with $a \in (n - 1, n]$ we get $n = 2$

Given that, $\vec{E} = 20 \sin \omega \left( t - \frac{x}{c} \right) \hat{j} \, \mathrm{N \, C^{-1}}$. The average energy density of an electromagnetic wave is $\frac{1}{2} \varepsilon_0 E_0^2$. The total energy stored is $$E_{total} = \frac{1}{2} \varepsilon_0 E_0^2 \times volume$$ $$\Rightarrow E_{total} = \frac{1}{2} \times 8.85 \times 10^{-12} \times \left( 20 \right)^2 \times 5 \times 10^{-4}$$ $$= 8.85 \times 10^{-13} \, \mathrm{J}$$

The area under the velocity time graph gives the displacement. The area for $t = 0 \, \mathrm{s}$ to $t = 25 \, \mathrm{s}$ is $$(Area) = \left( \frac{1}{2} \times 5 \times 10 \right) + (5 \times 10) + \left( \frac{1}{2} \times 30 \times 5 \right) + \left( \frac{1}{2} \times 20 \times 5 \right) - \left( \frac{1}{2} \times 20 \times 5 \right)$$ $$\Rightarrow Area = 25 + 50 + 75 + 50 - 50 = 150 \, \mathrm{m}$$ Hence, the displacement = $150 \, \mathrm{m}$ The distance is defined as the path taken by the object to travel from initial to final position. Also note that displacement is a vector quantity, but distance is a scalar quantity. Distance $$= \left( \frac{1}{2} \times 5 \times 10 \right) + (5 \times 10) + \left( \frac{1}{2} \times 30 \times 5 \right) + \left( \frac{1}{2} \times 20 \times 5 \right) + \left( \frac{1}{2} \times 20 \times 5 \right)$$ $$= (200 + 50) = 250 \, \mathrm{m}$$ Therefore, required ratio $$\frac{distance}{displacement} = \frac{250}{150} = \left( \frac{5}{3} \right)$$

Acceleration due to gravity is written as $$g = \frac{GM}{r^2}$$ $$= \frac{G \times \frac{4}{3} \times \pi \times r^3 \times \rho}{r^2}$$ $$= \left( \frac{4 \pi G}{3} \right) \rho r$$ Given: $r_A = R$ $r_B = 4R$ $\rho_A = \rho$ $\rho_B = \frac{\rho}{3}$ Hence, the ratio is $$\frac{g_A}{g_B} = \frac{R \times \rho}{4R \times \frac{\rho}{3}} = \frac{3}{4}.$$

The coin will slip when $\mu mg = m \omega^2 r$. $$\Rightarrow \mu g = \omega^2 \times 1 \, \mathrm{cm} \cdots (i)$$ Now, $\omega' = \frac{\omega}{2}$ $$\mu g = \left( \frac{\omega}{2} \right)^2 R \cdots (ii)$$ Hence we can write from (i) and (ii) $$\omega^2 \times 1 \, \mathrm{cm} = \frac{\omega^2}{4} R$$ $$\Rightarrow R = 4 \, \mathrm{cm}$$

All the gates used in the given configuration are NOR gates. Thus, the output can be found as follows: $$Y = \overline{\overline{a + b}}$$ $$= \overline{\overline{ab}}$$ $$= ab$$ The output of the given configuration can also be verified by the following truth table: The calculation and the truth table clearly shows that the configuration represents gate.

The charge on the plates of the first capacitor when connected against the potential difference $V$ is given by $Q = CV = 2V$. When both the capacitors are connected, from the conservation of charge, it can be written that $$2V = 2V' + 2V'$$ $$\Rightarrow V' = \left( \frac{1}{2} V \right)$$ where, $V'$ is the new potential difference across each capacitor. The formula to calculate the energy stored in the first capacitor is given by $$E_1 = \frac{1}{2} \times C \times V^2$$ $$= \frac{1}{2} \times 2 \times V^2$$ $$= V^2 \ldots (1)$$ For the second case, the energy stored in the combination of capacitors is given by $$E_2 = \frac{1}{2} CV'^2$$ $$= \frac{1}{2} \times 2 \times \frac{V^2}{4} \times 2$$ $$= \left( \frac{V^2}{2} \right) \ldots (2)$$ Divide equation (2) by equation (1) to obtain the required ratio of the stored energy. $$\frac{E_2}{E_1} = \frac{\frac{V^2}{2}}{V^2}$$ $$= \left( \frac{1}{2} \right)$$

Let the resistance of the filaments be $R$ and the voltage be $V$. For the series combination the power is $$P_1 = \left( \frac{V^2}{2R} \right)$$ For parallel combination, $$P_2 = \frac{V^2}{\left( \frac{R}{2} \right)} = \left( \frac{2V^2}{R} \right)$$ Hence, the ratio is $$\Rightarrow \frac{P_2}{P_1} = \frac{4}{1}$$

The formula to calculate the range of the antenna is given by $$\gamma = \sqrt{2 R h} \ldots (1)$$ Substitute the values of the known parameters into equation (1) and solve to obtain the required height of the antenna. $$4 \times 10^3 = \sqrt{2} \times 6400000 \times h$$ $$\Rightarrow 16 \times 10^6 = 2 \times 64 \times 10^5 \times h$$ $$\Rightarrow h = \left( \frac{160}{2 \times 64} \right) \mathrm{m} = \left( \frac{10}{8} \right) \mathrm{m}$$ $$= \frac{1000}{8} \times 10^{-2} \mathrm{m}$$ $$= 125 \times 10^{-2} \mathrm{m}$$

The formula to calculate the inductive reactance is $X_L = \omega L \ldots (1)$ The formula to calculate capacitive reactance is $X_C = \left( \frac{1}{\omega C} \right) \ldots (2)$ As can be seen from equation (1), the inductive reactance varies directly with the frequency and equation (2) indicates that capacitive reactance varies inversely with the frequency. Hence, curve $A$ represents the capacitive reactance and curve $B$ represents the inductive reactance.

The work to be done in the process is given by $$dW = PdV$$ $$= 1 \times 10^5 \, \mathrm{Pa} \times (1.671 - 0.001) \, \mathrm{m}^3$$ $$= 1.670 \times 10^5 \, \mathrm{J}$$ The change in heat energy during the vaporisation process can be calculated as follows- $$\Delta Q_{supplied} = 2257 \times 1 \times 10^3 \, \mathrm{J}$$ $$= 22.57 \times 10^5 \, \mathrm{J}$$ Hence, the change in internal energy in the process is given by $$\Delta U = \Delta Q_{supplied} - \Delta W$$ $$= (22.57 - 1.67) \times 10^5 \, \mathrm{J}$$ $$= 20.9 \times 10^5 \, \mathrm{J}$$ $$= 2090 \, \mathrm{kJ}$$

The critical angle and the refractive index of a media are related by the formula: $$\sin \theta_c = \left( \frac{1}{\mu} \right) \ldots (1)$$ Substitute the value of the critical angle into equation (1) and solve to calculate the refractive index of the medium. $$\sin 45^\circ = \frac{1}{\mu}$$ $$\Rightarrow \frac{1}{\sqrt{2}} = \frac{1}{\mu}$$ $$\Rightarrow \mu = \sqrt{2}$$ The refractive index of the medium can also be written as $$\mu = \frac{c}{v} \ldots (2)$$ Substitute the values of the known parameters into equation (2) and solve to calculate the required speed of light within the medium. $$\sqrt{2} = \frac{3 \times 10^8 \, \mathrm{m \, s^{-1}}}{v}$$ $$\Rightarrow v = \frac{3 \times 10^8 \, \mathrm{m \, s^{-1}}}{\sqrt{2}}$$ $$= 2.12 \times 10^8 \, \mathrm{m \, s^{-1}}$$

Let the threshold frequency be $\lambda_0$. By the equation of photoelectric effect, for wavelength $\lambda$, $$eV_0 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \cdots (i)$$ For the wavelength $2\lambda$, $$\frac{eV_0}{4} = hc \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) \cdots (ii)$$ Dividing (i) by (ii) $$4 = \frac{hc \left( \frac{\lambda_0 - \lambda}{\lambda_0 \lambda} \right)}{hc \left( \frac{\lambda_0 - 2\lambda}{2\lambda \lambda_0} \right)}$$ $$\Rightarrow 4 = \frac{2(\lambda_0 - \lambda)}{\lambda_0 - 2\lambda}$$ $$\Rightarrow 2\lambda_0 = 6\lambda$$ $$\Rightarrow \lambda_0 = 3\lambda$$

It is given that $\chi_m = 2 \times 10^{-2}$. The permeability is $\mu_r = 1 + \chi_m = 1 + 0.02 = 1.02$. The magnetic field is $B = \mu_r B_0 = 1.02 B_0$. Hence, percentage increase in magnetic field is $$\frac{B - B_0}{B_0} \times 100\% = \frac{B_0 (1.02 - 1)}{B_0} \times 100\% = 2\%$$

The current sensitivity is given as $$S_i = \frac{NBA}{C}.$$ The voltage sensitivity is $$S_v = \left( \frac{NAB}{CR} \right),$$ where $R$ is the galvanometer resistance. Since the resistance is constant, $$\Delta S_i = (\Delta S_v)$$ Hence, the percentage change in voltage sensitivity is, $$(\Delta S_v) = +25\%. $$

The formula to calculate the kinetic energy of a particle executing SHM is given by $$K = \frac{1}{2} m \omega^2 \left(A^2 - x^2\right) \ldots(1)$$ Substitute 0 for $x$ into equation (1) to obtain the kinetic energy at the mean position. $$K = \frac{1}{2} m \omega^2 \left(A^2 - 0^2\right)$$ $$= \frac{1}{2} m \omega^2 A^2$$ Substitute $A$ for $x$ into equation (1) to obtain the kinetic energy at the end points. $$K = \frac{1}{2} m \omega^2 \left(A^2 - A^2\right)$$ $$= 0$$ Hence, the variation can be shown as follows:

The relation between two temperature scales can be written as follows $$\frac{65 - x}{65 - (-15)} = \frac{212 - F}{180} \ldots (1)$$ Substitute the values of the known parameters into equation (1) and solve to calculate the required temperature in Fahrenheit scale. $$\frac{65 + 95}{80} = \frac{212 - F}{180}$$ $$\Rightarrow 2 \times 180 = 212 - F$$ $$\Rightarrow F = 212 - 360$$ $$= -148^\circ \mathrm{F}$$

Astronomical unit denotes roughly the distance from Earth to the Sun. Parsec is a unit of length used to measure the large distances to astronomical objects outside the Solar System. A light-year is another large unit of length used to express astronomical distances. The relation between these units are as follows: $$1 Parsec = 2 \times 10^5 Au$$ $$1 Au = 1.58 \times 10^{-5} ly$$ Thus, $1 Au < 1 ly < 1 Parsec$. Hence, Statements I is true but Statements II is false.

The RMS speed is given $v_{rms} = \sqrt{\frac{3RT}{M}}$. The temperature and pressure is same for all the gases. So the speed is inversely proportional to the square root of the molar mass of the gases. $$v_{rms}(mono) = \sqrt{\frac{3RT}{4 \times 10^{-3}}}$$ $$v_{rms}(dia) = \sqrt{\frac{3RT}{71 \times 10^{-3}}}$$ $$v_{rms}(poly) = \sqrt{\frac{3RT}{352 \times 10^{-3}}}$$ $$v_{rms} \propto \frac{1}{\sqrt{M}}$$ Hence, $$v_{rms}(poly) < v_{rms}(dia) < v_{rms}(mono).$$

The formula to calculate the net momentum of the gun and bullet system can be written as $F = nmv \ldots (1)$ Substitute the values of the known parameters into equation (1) and solve to calculate the required number of bullets. $$125 = n \times \frac{10}{1000} \times \left(250\right)$$ $$\Rightarrow n = 50$$

It is given that $T_{\frac{1}{2}}(A) = T_{av}(B)$. The half life of a substance is given by $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$. So, $$\frac{\ln 2}{\lambda_A} = \frac{1}{\lambda_B}$$ $$\Rightarrow \lambda_A = \lambda_B \ln 2$$

Let the higher state be $n$. The number of wavelengths emitted is given by, $$\frac{n(n-1)}{2} = 6$$ $$\Rightarrow n^2 - n - 12 = 0$$ $$\Rightarrow (n-4)(n+3) = 0$$ $$\Rightarrow n = 4$$ For hydrogen atom, $$h\nu = 13.6 \left( \frac{1}{1} - \frac{1}{16} \right) = \left( 13.6 \times \frac{15}{16} \right) \, \mathrm{eV}$$ $$\Rightarrow \nu = \frac{E}{h}$$ $$= \frac{13.6 \times 15 \times 1.6 \times 10^{-19}}{16 \times 6.62 \times 10^{-34}}$$ $$= 3 \times 10^{15} \, \mathrm{Hz}$$

The power of a lens is given by $$P_1 = \frac{1}{f}$$ Using Lens Maker's formula, $$P_1 = \left(1.8 - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ and $$P_2 = \frac{1}{f_{immersed}} = \left(\frac{1.8}{1.5} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ Hence, the ratio is, $$\frac{P_1}{P_2} = \frac{\left(1.8 - 1\right)}{\left(\frac{1.8}{1.5} - 1\right)} = \frac{0.8 \times 1.5}{0.3} = 4$$

The equation of the wave is $Y = 10^{-2} \sin 2\pi \left(160t - 0.5x + \frac{\pi}{4}\right)$. The velocity of a wave is given by $v = \frac{\omega}{k}$ (i). A wave equation is given by $Y = A \sin(\omega t - kx + \phi)$. Using equation (i), $$v = \left(\frac{2\pi \times 160}{2\pi \times 0.5}\right) = 320 \, \mathrm{m} \, \mathrm{s}^{-1}$$ $$\Rightarrow v = 320 \times \frac{18}{5}$$ $$\Rightarrow v = 1152 \, \mathrm{km} \, \mathrm{h}^{-1}$$

The formula to calculate the work done by a variable force is given by $$W = \int_{x_1}^{x_2} F dx \ldots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the required work done. $$W = \int_{0}^{4} \left( 2 + 3x \right) dx$$ $$= \left[ 2x + \frac{3}{2} x^2 \right]_{0}^{4}$$ $$= [8 + 24] \, \mathrm{J}$$ $$= 32 \, \mathrm{J}$$

For the condition of equilibrium, the Coulomb force is equal to $mg \sin \theta$. Hence, $$\frac{kq_0^2}{4h^2} = \left( mg \sin \theta \right)$$ $$\Rightarrow \frac{kq_0^2}{4h^2} = (20 \times 10^{-3}) \times 10 \times \frac{1}{2}$$ $$\Rightarrow \frac{9 \times 10^9 \times 4 \times 10^{-12}}{4h^2} = 20 \times 10^{-3} \times 10 \times \frac{1}{2}$$ $$\Rightarrow h = \left( \sqrt{\frac{9}{100}} \right) \mathrm{m} = 0.3 \, \mathrm{m} = 300 \times 10^{-3} \, \mathrm{m}$$ $$h^2 = \left( \frac{9}{100} \right) \Rightarrow h = \left( \frac{3}{10} \right) \mathrm{m} = 0.3 \, \mathrm{m}$$ $$= 300 \times 10^{-3} \, \mathrm{m}$$

The angular momentum about the diameter is $L_{diameter} = \frac{2}{5} MR^2 \omega$. The moment of inertia of a sphere is $I_{CM} = \frac{2}{5} MR^2$. The moment of inertia about the tangent is $$I_{tangent} = I_{CM} + MR^2 = \frac{2}{5} MR^2 + MR^2 = \frac{7}{5} MR^2.$$ It is given that $\omega = 10 \, rad \, s^{-1}$. Using the given relation between the moment of inertia and the angular momentum, $$\frac{7}{5} MR^2 = x \times 10^{-2} \times \frac{2}{5} MR^2 \omega$$ $$\Rightarrow x = \frac{3.5 \times 10^2}{\omega} = 3.5 \times 10 = 35$$

Young's modulus is given by $Y = \frac{F}{A} \frac{\Delta l}{l}$. $$\Rightarrow \frac{F}{A} = Y \left( \frac{\Delta l}{l} \right)$$ $$\Rightarrow \Delta l = \frac{Fl}{AY}$$ $$\Rightarrow l + \Delta l = l_1$$ $$\Rightarrow \left( \frac{FL}{AY} + L \right) = L \left( \frac{F_1}{AY} + 1 \right) = l_1 \cdots (i)$$ Similarly, $$l_2 = \left( \frac{F_2}{AY} + 1 \right) L$$ $$\Rightarrow 10l_2 = 11l_1$$ It is given that $10l_1 = 11l_2$ $$\Rightarrow 10 \times L \left[ \frac{F_2}{AY} + 1 \right] = 11L \left[ \frac{F_1}{AY} + 1 \right]$$ $$\Rightarrow \frac{10F_2}{AY} + 10 = \frac{11F_1}{AY} + 11$$ Substituting $F_1 = 100 \, \mathrm{N}$, $F_2 = 120 \, \mathrm{N}$ $$\Rightarrow \frac{1200}{AY} - \frac{1100}{AY} = 1$$ $$\Rightarrow \frac{100}{AY} = 1$$ $$\Rightarrow \frac{1}{AY} = \frac{1}{100}$$ Using equation (i), $$L = \frac{l_1}{\left( \frac{F_1}{AY} + 1 \right)} = \frac{l_1}{\frac{100}{100} + 1} = \frac{l_1}{2}$$

From the graph, it can be seen that the value of the magnetic field at $t = 2 \, \mathrm{s}$ is $4 \, \mathrm{mT}$, whereas the magnetic field at $t = 4 \, \mathrm{s}$ is $8 \, \mathrm{mT}$. The formula to calculate the magnitude of the induced emf is given by $$\varepsilon_{ind} = \frac{d\phi}{dt}$$ $$= \frac{d}{dt}(BA)$$ $$= A \frac{dB}{dt} \cdots (1)$$ $\frac{dB}{dt}$ is the slope of the given graph. Hence, the value of the induced emf can be calculated as follows: $$\varepsilon_{ind} = 4 \, \mathrm{m^2} \times \left( \frac{8 \, \mathrm{mT} - 4 \, \mathrm{mT}}{4 \, \mathrm{s} - 2 \, \mathrm{s}} \right)$$ $$= 8 \, \mathrm{mV}$$

The time taken to reach maximum height is $T' = \frac{u \sin \theta}{g}$. The total time of flight is $T = (3 + 5) \, \mathrm{s} = 8 \, \mathrm{s}$. Hence, $$T' = \left( \frac{T}{2} \right) = \left( \frac{3+5}{2} \right) = \left( \frac{u \sin \theta}{g} \right)$$ $$\Rightarrow 4 = \left( u \right) \times \frac{1}{2} \frac{1}{10}$$ $$\Rightarrow u = 80 \, \mathrm{m} \, \mathrm{s}^{-1}$$

Let the current flowing through the circuit be $I \, \mathrm{A}$ in anti-clockwise direction. The net resistance is $R = 4.5 \, \Omega + 2 \, \Omega + 1.5 \, \Omega = 8 \, \Omega$. Hence, $$I = \frac{V}{R} = \frac{4 \, \mathrm{V}}{8 \, \Omega} = 0.5 \, \mathrm{A}$$ The current flowing through the $3 \, \Omega$ resistor is $$\frac{6}{6+3} \times 0.5 = \frac{1}{3} \, \mathrm{A}$$ So, $x = 1$

The tetrose is giving Schiff's test means it has aldehyde group. The compound X on reduction with $\mathrm{NaBH_4}$ gives a chiral compound so the compound can be identified as erythrose. The compound can give triacetate on acetylation.

High-density polyethylene (HDPE) is commonly prepared in the presence of a Ziegler-Natta catalyst. The information provided describes the preparation and characteristics of a high-density polyethylene (HDPE) polymer. HDPE is a widely used thermoplastic polymer known for its high strength, durability, and excellent chemical resistance. The presence of trimethylammoniumpropyl (TMP) and titanium tetrachloride ($\mathrm{TiCl_4}$) indicates that the polymerization process is likely carried out using a Ziegler-Natta catalyst system. This catalyst system enables the production of high-density polyethylene by controlling the polymerization conditions.

The brown ring test is a typical nitrate test that involves adding iron(II) sulphate to a nitrate solution, then slowly adding strong sulfuric acid until the acid forms a layer below the aqueous solution. The existence of the nitrate ion will be shown by the formation of a brown ring at the junction of the two layers. Dark brown ring is formed in the confirmatory test of $\mathrm{NO_3^-}$ ions. Deep red precipitate with $\mathrm{FeCl_3}$ is formed in the presence of $\mathrm{CH_3COO^-}$ ions.

[$\mathrm{CoCl(NH_3)_5}$]^{2+} absorbs with higher wavelength of light with respect to [$\mathrm{Co(NH_3)_5(H_2O)}$]^{3+} because $\mathrm{Cl}$^- is weaker ligand than $\mathrm{H_2O}$. The crystal field stabilisation is higher for [$\mathrm{Co(NH_3)_5(H_2O)}$]^{3+}, hence we can say that wavelength of light absorbed depends upon the oxidation state of the metal ion.

To determine the ion(s) present in very small quantity in the solution, we need to consider the chemical reaction that occurs when silver nitrate reacts with potassium iodide. The reaction between $\mathrm{AgNO_3}$ and KI is a double replacement reaction and can be represented as follows: $$\mathrm{AgNO_3} + \mathrm{KI} \rightarrow \mathrm{AgI} + \mathrm{KNO_3}$$ Millimoles of $\mathrm{AgNO_3} = 25$ Millimoles of KI $= 25 \times 1.05 = 26.25$ Therefore, KI is in excess and $\mathrm{AgI}$ forms negatively charged colloid. (Some $\mathrm{Ag^+}$ remains in solution) Ions $\mathrm{Ag^+}$ and $\mathrm{F^-}$ are therefore, present in very small quantity.

Oxidation of this substituted ketone will give a ketone and a carboxylic acid. This reaction takes place by the cleavage of carbon atoms near the carbonyl carbon and the nearby carbons. This cleavage can be from both the sides of carbonyl carbon thus producing two different carboxylic acids and ketones. The ketone formed converts to hydrocarbon in the presence of hydrazine in alkaline medium. The overall reaction is shown below.

Ligand which has two different donor atoms and either of the two donor atoms is attached to the metal during complex formation is called ambidentate ligand. Among the given ligands $\mathrm{NCS}^- (\mathrm{SCN}^-)$ and $\mathrm{NO}_2^- (\mathrm{ONO}^-)$ are ambidentate ligands. $\mathrm{C_2O_4^{2-}}$ (bidentate), ethylenediamine (bidentate) and $\mathrm{H_2O}$ (monodentate) ligands are not ambidentate.

The $S_{N}1$ reactions take place via formation of carbocation. The speed of the reaction depends on the stability of carbocation. Electron releasing groups support the $S_{N}1$ reactions. The $S_{N}2$ reactions take place through the formation of transition state. The transition can be stabilized by electron withdrawing groups.

The first ionisation energy varies predictably across the periodic table. As we move from left to right across a period, the ionisation energy of elements increases. This is due to the decrease in the size of atoms across a period. The valence electrons get closer to the nucleus of an atom as we move from left to right due to increased nuclear charge. The force of attraction between the nucleus and the electrons increases and hence more energy is required to remove an electron from the valence shell. Ionisation energy of boron being unexpectedly less than that for beryllium due to the 2s orbital being totally filled in beryllium, whereas boron has one electron in a 2p orbital as well, and the 2s orbital is shielded much more than the 2p orbital. Ionisation energy is more for nitrogen than oxygen because nitrogen is more stable due to its half-filled electronic configuration. $$\mathrm{Li} < \mathrm{B} < \mathrm{Be} < \mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}$$

The molecular geometry of $\mathrm{H_3O^+}$ is pyramidal. It consists of three hydrogen atoms bonded to the central oxygen atom, with one lone pair of electrons on the oxygen atom. The acetylide ion consists of two carbon atoms bonded together by a triple bond ($\mathrm{C \equiv C}$) with a lone pair of electrons on the terminal carbon atom. So it is linear. The $\mathrm{NH_4^+}$ ion consists of a central nitrogen atom bonded to four hydrogen atoms. It is tetrahedral. The $\mathrm{ClO_2^-}$ ion is bent in shape due to its molecular geometry. It consists of one central chlorine atom bonded to two oxygen atoms, with an additional lone pair of electrons on the central chlorine atom.

Gallous tetrachloro aluminate ($\mathrm{GaAlCl_4}$) exists as $\mathrm{Ga^+ \ AlCl_4^-}$. Due to the presence of d-orbital in the case of Ga, the shielding effect of the d-orbital electrons is very poor so the size of the Ga atom is smaller than expected and so it can attract the shared pairs of electrons more effectively. Hence, the electronegativity of Ga is more than Al.

The copper(I) sulphide produced is converted to copper with a final blast of air. The end product of this is called blister copper - a porous brittle form of copper, about 98 - 99.5$\%$ pure. The reactions given are carried out in the production of blister copper using self reduction.

Lithium can indeed participate in thermonuclear reactions, specifically in certain fusion reactions. The movement of $\mathrm{K^+}$ ions (potassium ions) is indeed involved in the functioning of the sodium-potassium pump. KOH (potassium hydroxide) can act as an absorbent of carbon dioxide. This property is primarily due to the alkaline nature of KOH. KCl (potassium chloride) is commonly used as a fertilizer in agriculture due to its high potassium content. So option A is correct.

The process of extraction of different compounds absorbed on the column by means of a suitable solvent (eluent) is called 'elution'. Silica gel and solvent methylene chloride both are polar. The polar solvent and silica gel bind the polar compound, so it moves very slowly, whereas the non-polar compound moves faster on the silica gel. Correct order of elution $\rightarrow$ A $>$ C $>$ B

The type of the complex $[\mathrm{MA_3B_3}]$ exhibit the geometrical isomerism fac-mer isomerism. When three ligands and the metal are in one plane, the isomer is said to be meridional, or mer. $[\mathrm{Co(NH_3)_3(NO_2)_3}]$ can show facial and meridional isomerism.

In steam-methane reforming, methane reacts with steam under $3-25 \, \mathrm{bar}$ pressure in the presence of a nickel catalyst to produce hydrogen, carbon monoxide, and a relatively small amount of carbon dioxide. Steam reforming is endothermic, that is, heat must be supplied to the process for the reaction to proceed. $$\mathrm{NaNO_2 + NH_4Cl \xrightarrow{\Delta} N_2 + NaCl + H_2O}$$ as $$\mathrm{NH_4NO_2 \xrightarrow{\Delta} N_2 + H_2O}$$ Hence, both the statements are correct.

Drinking water has a BOD level of $1 - 2$ ppm. When the BOD value of water is in the range $3 - 5$ ppm, the water is moderately clean. Polluted water has a BOD value in the range of $6 - 9$ ppm. Healthy water should generally have dissolved oxygen concentrations above $6.5 - 8$ mg/L. Drinking water with concentrations of nitrate (measured as nitrate-nitrogen) below $10$ mg of nitrate per liter of water (mg/L) is considered safe. Permissible limit of Zn in potable water is $5$ ppm.

The more nucleophilic group will make ring more prone to aromatic substitution reaction. Benzene becomes more reactive towards Electrophilic Aromatic Substitution when there is more electron density. In structure (a) $-$ CH$_2$ group shows $+H$ effect, while in (b) $-$ OH group shows $+R$ effect. In (c) $+R$ effect is shown by functional group and in (d) both $-$ OH and $-$ O show $+R$ effect. So correct increasing order is: a < b < c < d.

[$\mathrm{Fe}$_3($\mathrm{OH}$)_2($\mathrm{OAc}$)_6]$\mathrm{Cl}$ is a coordination complex containing iron ions coordinated with hydroxide and acetate ligands. When dissolved in water, this complex can undergo hydrolysis and dissociation. So, [$\mathrm{Fe}$_3($\mathrm{OH}$)_2($\mathrm{OAc}$)_6]$\mathrm{Cl}$ dissolves in water. Rest of the complexes form precipitate.

Benzotriazole can be prepared by treating o-phenylenediamine with nitrous acid (liberated during the reaction between sodium nitrite and acetic acid) to form mono-diazonium salt that follows spontaneous intramolecular cyclization reaction to produce benzotriazole.

The reaction is given by: $$\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$$ Initial moles are 1 for both $\mathrm{CO}$ and $\mathrm{H_2O}$. At equilibrium, the moles are $1 - 0.4$ for both $\mathrm{CO}$ and $\mathrm{H_2O}$, and $0.4$ for both $\mathrm{CO_2}$ and $\mathrm{H_2}$. The molarity is calculated as follows: $$0.6/10, \; 0.6/10, \; 0.4/10, \; 0.4/10$$ Now, the equilibrium constant $K_C$ is given by: $$K_C = \frac{[\mathrm{CO_2}] [\mathrm{H_2}]}{[\mathrm{CO}] [\mathrm{H_2O}]}$$ Substituting the values, we get: $$K_C = \frac{0.04 \times 0.04}{0.06 \times 0.06} = 0.44$$ Therefore, $K_C \times 10^2 = 44$

In case of $\mathrm{Cr^{3+}}$ ion, the valence shell electronic configuration is same whether the ligand is strong field or weak field. The spin only magnetic moment, $\mu_{eff} = \sqrt{n(n+2)}$ BM. The valence shell electronic configuration of $\mathrm{Cr^{3+}}$ is $3d^3$, hence, it has three unpaired electrons. $$\mu_{eff} of [\mathrm{Cr(CN)_6}]^{3-} = \sqrt{15} B.M.$$ $$\mu_{eff} of [\mathrm{Cr(H_2O)_6}]^{3+} = \sqrt{15} B.M.$$ Ratio $= 1$

The density of crystal can be calculated as follows, $$\rho = \frac{Z \times M}{N_A \times a^3}$$ $\rho$ = density of the unit cell $Z$ = Effective atomic number $M$ = Molar mass $N_A$ = Avogadro's number $a$ = Edge length of the unit cell $$3 = \frac{Z \times 12}{6.02 \times 10^{23} \times \left(3 \right)^3 \times 10^{-24}}$$ $$Z = \frac{6.02 \times 10^{-1} \times 27}{4} = 4$$

Acid-base reactions are faster than nucleophilic addition reaction. So, in the first step acid-base reaction takes place between alcoholic functional group and Grignard reagent. Now, the second mole of Grignard reagent to react with aldehyde functional group. Hence, at the end of the reaction a diol is produced on acidification.

In the first step of the reaction, protonation of alcohol takes place and carbocation is formed after removal of water. The carbocation formed undergoes rearrangement to get stability. In the given, methyl shift takes place to convert secondary carbocation to tertiary carbocation. The alpha hydrogen to the positive charged carbon are involved in hyperconjugation. So, number of hyperconjugation structures in most stable carbocation $$= 6 + 1 = 7$$

Molten iron can be prepared by alumino thermite process as shown below. $$\mathrm{Fe_2 O_3 + 2Al \rightarrow 2Fe + Al_2 O_3}$$ The iron oxide and aluminium metal ratio is 1 : 2 in the above reaction. Now, the Enthalpy of the reaction is equal to the difference of enthalpy of formation of products to the enthalpy of formation of reactants. $$\Delta H^0_{reaction} = -1700 + (840)$$ Ratio of $\mathrm{Fe_2 O_3}$ and $\mathrm{Al}$ $$= 160 : 54$$ $$= 2.96 in 214 gm mixture$$ Therefore, $$\frac{\Delta H^0}{1 gm mixture} = \frac{-860}{214} = -4 \, kJ/gram$$

Mass of solution $= 200 + 500 = 700 \, \mathrm{g}$ $25\%$ of sugar solution means $100 \, \mathrm{g}$ solution contain $25 \, \mathrm{g}$ sugar $40\%$ of sugar solution means $100 \, \mathrm{g}$ solution contain $40 \, \mathrm{g}$ sugar Mass of sugar $= 0.25 \times 200 + 0.40 \times 500$ $$= 50 + 200 = 250 \, \mathrm{g}$$ Mass $\%$ of resulting solution $$= \frac{250}{700} \times 100 = 36\%$$

$\mathrm{KClO_3 + 6FeSO_4 + 3H_2SO_4 \rightarrow KCl + 3Fe_2(SO_4)_3 + 3H_2O}$ From the above equation, rate of decomposition of $\mathrm{FeSO_4}$ and rate of production of $\mathrm{Fe_2(SO_4)_3}$ can be written as follows, $\frac{1}{6}\times \text{Rate of decomposition}_{\mathrm{FeSO_4}}=\frac{1}{3}\times \text{Rate of production}_{\mathrm{Fe_2(SO_4)_3}}$ Rate of decomposition of $\mathrm{FeSO_4}$ $\frac{\text{Change in concentration}}{\text{time in seconds}}=\frac{(10-8.8)}{30\times60}$ $=\frac{1.2}{1800}\ \mathrm{mol}\ \mathrm{L}^{-1}\ \mathrm{s}^{-1}$ Rate of production of $\mathrm{Fe_2(SO_4)_3}$ $=\frac{0.6}{1800}=3.33\times10^{-4}$ $=333.3\times10^{-6}\ \mathrm{mol}\ \mathrm{s}^{-1}$

Osmotic pressure $\pi = iCRT$ where $i =$ Van't Hoff Factor, $C =$ Molarity, $R =$ Same gas constant, $T =$ Temperature. For isotonic solutions, osmotic pressures are equal. Hence, we have $$(0.004) \times i = 0.01$$ $$\Rightarrow i = \frac{0.01}{0.004}$$ $$= \frac{10}{4} = \frac{5}{2} = 1 + 2\alpha$$ $$2\alpha = 1.5$$ $$\alpha = 0.75$$ Therefore, $\% \alpha = 75$.

The relation between Gibbs Free energy and emf of the cell can be related as follows, $$\Delta G = -nF E_{cell}$$ The electrode reactions can be written as follows, $$\mathrm{Pb^{2+} + 2e^- \rightarrow Pb} \Delta G_1 = -2F E^\circ_{\mathrm{Pb^{2+}/Pb}}$$ $$\mathrm{Pb^{4+} + 4e^- \rightarrow Pb} \Delta G_2 = -4F E^\circ_{\mathrm{Pb^{4+}/Pb}}$$ $$\mathrm{Pb^{4+} + 2e^- \rightarrow Pb^{2+}} \Delta G_3 = -2F E^\circ_{\mathrm{Pb^{4+}/Pb^{2+}}}$$ Now, $$\Delta G_3 = \Delta G_2 - \Delta G_1$$ $$-2F E^\circ_{\mathrm{Pb^{4+}/Pb^{2+}}} = -4F E^\circ_{\mathrm{Pb^{4+}/Pb}} + 2F E^\circ_{\mathrm{Pb^{2+}/Pb}}$$ $$\Rightarrow 4E^\circ_{\mathrm{Pb^{4+}/Pb}} = 2E^\circ_{\mathrm{Pb^{2+}/Pb}} + 2E^\circ_{\mathrm{Pb^{4+}/Pb^{2+}}}$$ $$\Rightarrow 4n = 2m + 2E^\circ_{\mathrm{Pb^{4+}/Pb^{2+}}}$$ $$\Rightarrow E^\circ_{\mathrm{Pb^{4+}/Pb^{2+}}} = 2n - m$$ $$\Rightarrow E^\circ_{\mathrm{Pb^{2+}/Pb^{4+}}} = m - 2n$$ Hence, the value of $x = 2$