JEE Main 12 April 2023 Shift 1 question paper with solutions

We need to find the number of 5 digit numbers greater than 40000 and divisible by 5. Case I: Numbers start with 5. Here if the number starts with 5 it should end with 0 to be divisible by 5. $$4 \times 3 \times 2 = 24$$ Case II: Number starts with 7 and the last digit can have 0 or 5. $$4 \times 3 \times 2 \times 2 = 48$$ Case III: The number starts with 9 and the last digit should have 0 or 5. $$4 \times 3 \times 2 \times 2 = 48$$ Total numbers = 120. Hence this is the correct option.

To find the value of $\frac{\alpha^{23} + \beta^{23} + \alpha^{14} + \beta^{14}}{\alpha^{15} + \beta^{15} + \alpha^{10} + \beta^{10}}$, let $a_n = \alpha^n + \beta^n$. Hence, $$\frac{\alpha^{23} + \beta^{23} + \alpha^{14} + \beta^{14}}{\alpha^{15} + \beta^{15} + \alpha^{10} + \beta^{10}} = \frac{a_{23} + a_{14}}{a_{15} + a_{10}}.$$ Now, $x^2 + \sqrt{6}x + 3 = 0$ has roots $\alpha$ and $\beta$. So, $x = \frac{-\sqrt{6} \pm \sqrt{-6}}{2}$. $$\Rightarrow x = \sqrt{6} \left( \frac{-1 \pm i}{2} \right)$$ $$\Rightarrow x = \sqrt{3} \left( \frac{-1 \pm i}{\sqrt{2}} \right).$$ Hence, $\alpha = \sqrt{3} e^{i \frac{3\pi}{4}}$ and $\beta = \sqrt{3} e^{i \frac{5\pi}{4}}$. Now, solving $$\frac{\alpha^{23} + \beta^{23} + \alpha^{14} + \beta^{14}}{\alpha^{15} + \beta^{15} + \alpha^{10} + \beta^{10}}$$ $$= \left[ \left( \sqrt{3} \right)^{23} e^{i \frac{23 \times 3\pi}{4} + e^{i \frac{23 \times 5\pi}{4}}} + \left( \sqrt{3} \right)^{14} e^{i \frac{14 \times 3\pi}{4} + e^{i \frac{14 \times 5\pi}{4}}} \right]$$ $$= \left[ \left( \sqrt{3} \right)^{15} e^{i \frac{15 \times 3\pi}{4} + e^{i \frac{15 \times 5\pi}{4}}} + \left( \sqrt{3} \right)^{10} e^{i \frac{10 \times 3\pi}{4} + e^{i \frac{10 \times 5\pi}{4}}} \right]$$ $$= \left( \sqrt{3} \right)^9 \left[ e^{i \frac{23 \times 3\pi}{4} + e^{i \frac{23 \times 5\pi}{4}}} + e^{i \frac{14 \times 3\pi}{4} + e^{i \frac{14 \times 5\pi}{4}}} \right]$$ $$= \left( \sqrt{3} \right)^9 \left[ e^{i \frac{15 \times 3\pi}{4} + e^{i \frac{15 \times 5\pi}{4}}} + e^{i \frac{10 \times 3\pi}{4} + e^{i \frac{10 \times 5\pi}{4}}} \right]$$ $$= 9 \times \left( \sqrt{3} \right)^9 \left[ \frac{1+i}{\sqrt{2}} + 0 \right] \left[ \frac{1+i}{\sqrt{2}} \right]$$ $$= 9 \times \left( \sqrt{3} \right)^9 \left[ \frac{2i}{\sqrt{2}} \right]$$ $$= 9 \times \left( \sqrt{3} \right)^9 \left( \sqrt{3} \right)^4$$ $$= 81.$$

Given, $$S_n = \sum_{i=1}^{n} a_i = \frac{n^2 + 3n}{(n+1)(n+2)}$$ Now we know that, $$a_n = S_n - S_{n-1}$$ $$\Rightarrow \ a_n = \frac{n^2 + 3n}{(n+1)(n+2)} - \frac{(n-1)^2 + 3(n-1)}{n(n+1)}$$ $$\Rightarrow \ a_n = \frac{n}{(n+1)(n+2)}$$ Now solving, $$\sum_{k=1}^{10} \frac{1}{a_k} = \frac{1}{4} \sum_{k=1}^{10} k(k+1)(k+2)$$ $$\Rightarrow \ \sum_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} \sum_{k=1}^{10} k(k+1)(k+2)(k+3) - (k-1)k(k+1)(k+2)$$ $$\Rightarrow \ \sum_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} \left( [(1 \cdot 2 \cdot 3 \cdot 4) - (0 \cdot 1 \cdot 2 \cdot 3 \cdot 4)] \ldots + (10 \cdot 11 \cdot 12 \cdot 13 - 9 \cdot 10 \cdot 11 \cdot 12) \right)$$ $$\Rightarrow \ \sum_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} [10 \cdot 11 \cdot 12 \cdot 13 - 0]$$ $$\Rightarrow \ \sum_{k=1}^{10} \frac{1}{a_k} = \frac{1}{2} (5 \cdot 11 \cdot 3 \cdot 13)$$ $$\Rightarrow \ 28 \sum_{k=1}^{10} \frac{1}{a_k} = \frac{28 \times 5 \times 11 \times 3 \times 13}{2}$$ $$\Rightarrow \ 28 \sum_{k=1}^{10} \frac{1}{a_k} = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$$ Hence, there are six prime numbers in multiplications, So, $\ $m = 6

Given, the lines $L_1: \frac{x+5}{3} = \frac{y+4}{1} = \frac{z-2}{-2}$ and $L_2: 3x + 2y + z - 2 = 0 = x - 3y + 2z - 13$ be coplanar. Now direction ratio of line $L_2$ will be, $$\overrightarrow{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 1 & -3 & 2 \end{vmatrix} = \hat{i}(7) - \hat{j}(5) + \hat{k}(-11)$$ Let $R$ lies on $L_2$ and let $z = 0$. Therefore, $3x + 2y = 2$ and $x - 3y = 13$. Now solving above two equations we get, $$11y = -39 + 2 = -37 \Rightarrow y = \frac{-37}{11}$$ And $x = 13 + 3y \Rightarrow x = 13 + \frac{-37 \times 3}{11}$$ $$\Rightarrow x = \frac{143 - 111}{11} = \frac{32}{11}$$ Hence, the point $R $\left$( $\frac{32}{11}$, $\frac{-37}{11}$, 0 $\right$)$. Also $L_1, L_2$ and given point on line $L_1$ to $R$ are coplanar. So, by coplanar condition we get, $$\begin{vmatrix} -5 & \frac{32}{11} & -4 + \frac{37}{11} & \alpha \\ 3 & 1 & -2 & 0 \\ 7 & -5 & -11 \end{vmatrix} = 0$$ $$\Rightarrow \begin{vmatrix} -\frac{87}{11} & -\frac{7}{11} & \alpha \\ 3 & 1 & -2 \\ 7 & -5 & -11 \end{vmatrix} = 0$$ $$\Rightarrow \alpha = 7$$ Now let, $P(3k - 5, k - 4, -2k + 7)$ is nearest to point $(-4, -3, 2)$. So, by perpendicular condition we get, $$\overrightarrow{QP} \cdot \overrightarrow{n_1} = 0$$ $$\Rightarrow (3k - 1) \hat{i} + (k - 1) \hat{j} + (-2k + 5) \hat{k} \cdot (3 \hat{i} + \hat{j} - 2 \hat{k}) = 0$$ $$\Rightarrow 9k - 3 + k - 1 + 4k - 10 = 0$$ $$\Rightarrow 14k = 14 \Rightarrow k = 1$$ $$\Rightarrow P(-2, -3, 5) = (a, b, c)$$ Hence, $|a| + |b| + |c| = 2 + 3 + 5 = 10$

Given, $P \left( \frac{2\sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}} \right)$, $Q$, $R$ and $S$ be four points on the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ Now, $OP = r_1 = \sqrt{\left( \frac{2\sqrt{3}}{\sqrt{7}} \right)^2 + \left( \frac{6}{\sqrt{7}} \right)^2} = \sqrt{\frac{48}{7}}$ where $O$ is origin. Let $P$ be $(r_1 \cos \theta, r_1 \sin \theta)$ $P$ lies on ellipse, so we get, $$\frac{r_1^2 \cos^2 \theta}{4} + \frac{r_1^2 \sin^2 \theta}{9} = 1$$ $$\Rightarrow \frac{\cos^2 \theta}{4} + \frac{\sin^2 \theta}{9} = \frac{7}{48} \cdots (i)$$ Let $R$ be $(-r_2 \sin \theta, r_2 \cos \theta)$ as $PQ$ and $RS$ are perpendicular and pass through origin, So, $$\frac{r_2^2 \sin^2 \theta}{4} + \frac{r_2^2 \cos^2 \theta}{9} = 1$$ $$\Rightarrow \frac{\sin^2 \theta}{4} + \frac{\cos^2 \theta}{9} = \frac{1}{r_2^2} \cdots (ii)$$ Now adding equation $(i)$ and $(ii)$ we get, $$\frac{1}{r_2^2} = \frac{1}{4} + \frac{1}{9} - \frac{7}{48} = \frac{31}{144}$$ Now solving, $$\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4} \left( \frac{1}{OP^2} + \frac{1}{OR^2} \right)$$ $$\Rightarrow \frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4} \left( \frac{1}{r_1^2} + \frac{1}{r_2^2} \right)$$ $$\Rightarrow \frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4} \left( \frac{7}{48} + \frac{31}{144} \right) = \frac{13}{144} = \frac{p}{m}$$ Therefore, $p + m = 157$

Given that the vectors are coplanar. That means the determinant is equal to zero. $$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$$ $$\Rightarrow R_1 \rightarrow R_1 - R_2 and R_2 \rightarrow R_2 - R_3$$ $$\begin{vmatrix} a - 1 & 1 - b & 0 \\ 0 & b - 1 & 1 - c \\ 1 & 1 & c \end{vmatrix} = 0$$ Expand the determinant along $C_1$. $$\Rightarrow (a - 1) \left[c(b - 1) - (1 - c)\right] + 1[(1 - b)(1 - c)] = 0$$ $$\Rightarrow c(a - 1)(b - 1) - (a - 1)(1 - c) + (1 - b)(1 - c) = 0$$ $$\Rightarrow c(1 - a)(1 - b) + (1 - a)(1 - c) + (1 - a)(1 - b)(1 - c) = (1 - a)(1 - b)(1 - c)$$ $$\Rightarrow (1 - a)(1 - b) + (1 - a)(1 - c) + (1 - a)(1 - b)(1 - c) = (1 - a)(1 - b)(1 - c)$$ $$\Rightarrow \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$$ Hence this is the required option.

Given function is $f(x) = \left( \frac{\sqrt{3e}}{2 \sin x} \right) \sin^2 x$. For maxima or minima $f'(x) = 0$. $$f'(x) = f(x) \left[ 2 \sin x \cos x \times \ln \left( \frac{\sqrt{3e}}{2 \sin x} \right) + \sin^2 x \frac{2 \sin x}{\sqrt{3e}} \times \frac{\sqrt{3e}}{2} \left( -\frac{1}{\sin^2 x} \times \cos x \right) \right]$$ $$\Rightarrow f(x) \left[ \sin 2x \ln \left( \frac{\sqrt{3e}}{2 \sin x} \right) - \sin x \cos x \right] = 0$$ Now on equation we get, $\sin 2x = 0$ (not possible). So, $\ln \left( \frac{\sqrt{3e}}{2 \sin x} \right) = \frac{1}{2}$ $$\Rightarrow \frac{\sqrt{3} \times \sqrt{e}}{2 \sin x} = e^{\frac{1}{2}}$$ $$\Rightarrow \sin x = \frac{\sqrt{3}}{2}$$ $$\Rightarrow f_{\max} = \left( e \right)^{\frac{3}{8}} = \frac{e^{\frac{8}{8}}}{e} \Rightarrow k = e^{\frac{11}{8}}$$ $$\Rightarrow \left( \frac{k}{e} \right)^8 + \frac{k^8}{e^5} + k^8 = e^3 + e^6 + e^{11}$$

We have been given $f(x) = \sin^{-1} \left( \log_3 x \left( \frac{6 + 2 \log_3 x}{-5x} \right) \right)$. We know that the domain of $\sin^{-1} x$ is $-1 \leq x \leq 1$. Therefore, $\log_3 x \left( \frac{6 + 2 \log_3 x}{-5x} \right) \in [-1, 1]$, $3x > 0$, $3x \neq 1$, $\frac{6 + 2 \log_3 x}{-5x} > 0$. From (2), $x > 0$, from (3), $x \neq \frac{1}{3}$. From (4), $6 + 2 \log_3 x 0$) implies $\log_3 x 3x$ implies $\frac{5}{3} \leq 6 + 2 \log_3 x \frac{6}{23} \leq x \leq 3 > \frac{-15x^2 - 6}{2}$. Therefore, $1 > \frac{1}{3^{\frac{6}{23}}} \leq x \leq \frac{1}{3^{\frac{-15x^2 - 6}{2}}}$. Therefore, $3^{\frac{-23}{6}} \leq x < \frac{1}{27}$ ... (6), here $3^{\frac{-23}{6}}$ is a very small positive quantity. From (5) and (6), $3^{\frac{-23}{6}} < x < \frac{1}{27}$, $[x] = 0$. Therefore, $g(x) = x$. Hence, Range $g(x)$ is domain of $f(x)$. Therefore, $g(x) \in \left(3^{\frac{-23}{6}}, \frac{1}{27} \right) \equiv (\alpha, \beta)$. Therefore, $\alpha^2 + \frac{5}{3} = \left(3^{\frac{-23}{6}} \right)^2 + \frac{1}{27} \approx 135 \approx 135$. Note: This question was from a JEE Main April 2023 session.

Given, $$(1 + x^2) dy = y(x - y) dx$$ $\($$\Rightarrow$ $\frac{dy}{dx}$ = $\frac{x}{1 + x^2}$ y - $\frac{y^2}{1 + x^2}$ $\)$ $\($$\Rightarrow$ -$\frac{1}{y^2}$ $\frac{dy}{dx}$ + $\frac{x}{1 + x^2}$ = $\frac{1}{y}$ $\frac{1}{1 + x^2}$ $\)$ Now let $\($$\frac{1}{y}$ = t$\)$, we get, $\($-$\frac{1}{y^2}$ y' = $\frac{dt}{dx}$$\)$ So, the equation becomes, $\($$\frac{dt}{dx}$ + $\frac{x}{1 + x^2}$ t = $\frac{1}{1 + x^2}$$\)$ Now finding, IF $\($= e^{$\int$ $\frac{x}{1 + x^2}$ dx} = $\sqrt{1 + x^2}$$\)$ So, solution of the differential equation is given by, $$t \sqrt{1 + x^2} = \int \frac{\sqrt{1 + x^2}}{1 + x^2} dx$$ $\($$\Rightarrow$ $\frac{1}{y}$ $\sqrt{1 + x^2}$ = $\int$ $\frac{1}{\sqrt{1 + x^2}}$ dx$\)$ $\($$\Rightarrow$ $\frac{1}{y}$ $\sqrt{1 + x^2}$ = $\ln$ $\left$( x + $\sqrt{x^2 + 1}$ $\right$) + C$\)$ $\($$\therefore$ y(0) = 1 $\Rightarrow$ C = 1$\)$ $\($$\Rightarrow$ $\frac{1}{y}$ $\sqrt{1 + x^2}$ = $\ln$ $\left$( x + $\sqrt{x^2 + 1}$ $\right$) + 1$\)$ Now for $\($x = 2 $\sqrt{2}$, y = $\beta$$\)$ we get, $\($$\Rightarrow$ $\frac{3}{\beta}$ = $\ln$ |2 $\sqrt{2}$ + 3| + 1$\)$ $\($$\Rightarrow$ $\beta$ = $\frac{3}{1 + \ln |2 \sqrt{2} + 3|}$$\)$ $\($$\Rightarrow$ 3 $\beta$ - 1 = $\ln$ |2 $\sqrt{2}$ + 3|$\)$ $\($$\Rightarrow$ e^{3 $\beta$ - 1} = e^{$\ln$ $\left$( 3 + 2 $\sqrt{2}$ $\right$)}$\)$ $\($$\Rightarrow$ e^{3 $\beta$ - 1} = e $\left$( 3 + 2 $\sqrt{2}$ $\right$)$\)$

Given, $S_1 : (p \Rightarrow q) \land (p \land \sim q)$ $$\equiv (\sim p \lor q) \land (p \land \sim q) \{ as A \Rightarrow B \equiv \sim A \lor B \}$$ $$\equiv (\sim p \land (p \land \sim q)) \lor (q \land (p \land \sim q))$$ $$\equiv (f \land \sim q) \lor (f \land p)$$ $$\equiv f \lor f \equiv f$$ $S_2 : (p \land q) \lor ((\sim p) \land q) \lor (p \land (\sim q)) \lor ((\sim p) \land (\sim q))$ $$\equiv (p \land q) \lor (\sim p \land q) \lor (p \land \sim q) \lor (\sim p \land \sim q)$$ $$\equiv ((p \lor \sim p) \land q) \lor ((p \lor \sim p) \land \sim q)$$ $$\equiv (t \land q) \lor (t \land \sim q) \equiv q \lor \sim q \equiv t$$ Hence, both statement are true.

Given that $\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$. Also $\left( \vec{a} + \vec{b} + \vec{c} \right) \times \vec{c} = \vec{0}$. Therefore, $k \left( \vec{a} + \vec{b} \right) = \vec{c}$. $\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$. Now, take $\vec{a} \cdot \vec{c} = -17$. $$k \left( \lambda \hat{i} + \hat{j} - \hat{k} \right) \cdot \left( \lambda \hat{i} + \hat{j} - \hat{k} + 3\hat{i} - \hat{j} + 2\hat{k} \right) = -17$$ $$k \left( \lambda^2 + 3\lambda + 0 - 1 \right) = -17$$ $$k \left( \lambda^2 + 3\lambda - 1 \right) = -17 \cdots (1)$$ Similarly, on taking $\vec{b} \cdot \vec{c} = -20$, we get, $$k \left( 3\hat{i} - \hat{j} + 2\hat{k} \right) \cdot \left( \lambda \hat{i} + \hat{j} - \hat{k} + 3\hat{i} - \hat{j} + 2\hat{k} \right) = -20$$ $$k (3\lambda + 9 + 2) = -20$$ $$k (3\lambda + 11) = -20 \cdots (2)$$ Now on solving equation (1) and (2) we get, $$\Rightarrow 20\lambda^2 + 9\lambda - 207 = 0$$ $$\Rightarrow \lambda = 3, \frac{-69}{20}$$ For $\lambda = 3$, $k = -1$. $$\Rightarrow \vec{c} = -1 \left( \vec{a} + \vec{b} \right)$$ $$\Rightarrow \left( (\lambda + 3) \hat{i} + \hat{k} \right) = -6\hat{i} - \hat{k}$$ $$\Rightarrow \vec{c} \times \left( \lambda \hat{i} + \hat{j} + \hat{k} \right) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1 \end{vmatrix} = \hat{i} - 3\hat{j} + 6\hat{k}$$ $$\Rightarrow \left| \vec{c} \times \left( \lambda \hat{i} + \hat{j} + \hat{k} \right) \right|^2 = 46$$ Hence this is the correct option.

We know that $(1-x)^{100} = {}^{100}C_{0} - {}^{100}C_{1} \cdot x + {}^{100}C_{2} \cdot x^2 - \ldots - {}^{100}C_{49} \cdot x^{49} + \ldots + {}^{100}C_{100} \cdot x^{100}$. Now let, the sum of first fifty coefficients be $S$. $$S = {}^{100}C_{0} - {}^{100}C_{1} + {}^{100}C_{2} - \ldots - {}^{100}C_{49}$$ Let us substitute $x = 1$ in the expansion. $$(1-1)^{100} = {}^{100}C_{0} - {}^{100}C_{1} \cdot 1 + {}^{100}C_{2} \cdot 1^2 - \ldots - {}^{100}C_{49} \cdot 1^{49} + \ldots + {}^{100}C_{100} \cdot 1^{100}$$ $$0 = {}^{100}C_{0} - {}^{100}C_{1} + {}^{100}C_{2} - \ldots - {}^{100}C_{49} + {}^{100}C_{50} - {}^{100}C_{51} + \ldots + {}^{100}C_{100}$$ There are total of 101 terms and the middle term is $T_{51} = {}^{100}C_{50}$. As ${}^{n}C_{r} = {}^{n}C_{n-r}$ $$0 = {}^{100}C_{0} - {}^{100}C_{1} + {}^{100}C_{2} - \ldots - {}^{100}C_{49} + {}^{100}C_{50} - {}^{100}C_{49} + {}^{100}C_{48} - \ldots + {}^{100}C_{0}$$ $$0 = {}^{100}C_{50} + {}^{100}C_{0} - {}^{100}C_{1} + {}^{100}C_{2} - \ldots - {}^{100}C_{49} + {}^{100}C_{48} - \ldots + {}^{100}C_{0}$$ $$0 = {}^{100}C_{50} + 2 \left( {}^{100}C_{0} - {}^{100}C_{1} + {}^{100}C_{2} - \ldots - {}^{100}C_{49} \right)$$ $$-\frac{{}^{100}C_{50}}{2} = \left( {}^{100}C_{0} - {}^{100}C_{1} + {}^{100}C_{2} - \ldots - {}^{100}C_{49} \right)$$ $${}^{100}C_{0} - {}^{100}C_{1} + {}^{100}C_{2} - \ldots - {}^{100}C_{49} = -\frac{{}^{100}C_{50}}{2} = -\frac{1}{2} \cdot \frac{100 \cdot 99}{2} = -99\;{}^{99}C_{49}$$ Therefore, the required answer is $-99\;{}^{99}C_{49}$.

Given \(y=x^3\) ...(i) \[ \frac{dy}{dx}=3x^2 \] \[ \left(\frac{dy}{dx}\right)_{(-1,-1)}=3 \] Equation of tangent at \((-1,-1)\) \[ (y+1)=3(x+1) \] \[ y=3x+2 \] ...(ii) Solving (i) and (ii) \[ x^3=3x+2 \] \[ x^3-3x-2=0 \] \[ x^3-3x+2=0 \] \[ x=1,-1,2 \] Another point of intersection is \(Q(2,8)\) So, now plotting the diagram we get, Required area from the above diagram will be, \[ \int_{-1}^{1}(3x+2-x^3)\,dx \] \[ =\frac{3}{2}(4-1)+2(2-1)-\frac{1}{4}(16-1) \] \[ =\frac{27}{4} \] Hence this is the required option.

Given, $$A = \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix},$$ $$B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} = A \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}.$$ So, $B = MAN$. Now let $M = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}$ and $N = \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}$. Now solving, $$\begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$ Therefore, $MN = I = NM$. Now using, $B = MAN$ we get $$B^n = (MAN)^n = (MAN)(MAN) \ldots (MAN)$$ $$\Rightarrow B^n = MANMANMAN \ldots NMAN$$ $$\Rightarrow B^n = MAAAAA \ldots AN = MA^nN.$$ Now, using $A = \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix}$ $$\Rightarrow A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & \frac{1}{51} \\ 0 & 0 \end{bmatrix} = I + E.$$ Now finding, $E^2 = \begin{bmatrix} 0 & \frac{1}{51} \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & \frac{1}{51} \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$. $$\Rightarrow E^2 = 0,$$ so all higher power will also be zero, $$\therefore A^n = (I + E)^n = I + nE + \binom{n}{2}E^2 + \binom{n}{3}E^3 + \binom{n}{4}E^4 \ldots$$ $$\Rightarrow A^n = I + nE$$ $$\Rightarrow A^n = \begin{bmatrix} 1 & \frac{n}{51} \\ 0 & 1 \end{bmatrix}.$$ So, $$B^n = MA^nN = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & \frac{n}{51} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & \frac{n}{51} + 2 \\ -1 & -\frac{n}{51} - 1 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}.$$ $$\Rightarrow B^n = \begin{bmatrix} 1 & \frac{n}{51} + 2 \\ -1 & -\frac{n}{51} - 1 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{n}{51} - 1 & -\frac{n}{51} - 2 \\ \frac{n}{51} + 1 & \frac{n}{51} + 2 \end{bmatrix}.$$ $$\sum_{0}^{50} B^n = \begin{bmatrix} -50.5 & -50.5 \\ 50.5 & 50.5 \end{bmatrix} = \begin{bmatrix} 75 & 25 \\ -25 & 25 \end{bmatrix}.$$ Therefore, Sum = 100.

We know that if a plane passes through the line of intersection of planes $P_1$ and $P_2$ then the family of planes is given by $P_1 + \lambda P_2$ where $\lambda \in \mathbb{R}$. Equation of plane after rotation: $$(4x - y + z - 10) + \lambda (x + y - z - 4) = 0$$ $$\Rightarrow (4 + \lambda)x + y(\lambda - 1) + z(1 - \lambda) - 4\lambda - 10 = 0$$ Now angle between the new plane and $4x - y + z - 10$ is $90^\circ$. $$\Rightarrow \vec{n}_1 \cdot \vec{n}_2 = 0$$ $$\Rightarrow (4 + \lambda)4 + (\lambda - 1)(-1) + (1 - \lambda)1 = 0$$ $$\Rightarrow 16 + 4\lambda - \lambda + 1 + 1 - \lambda = 0$$ $$\Rightarrow 2\lambda = -18$$ $$\Rightarrow \lambda = -9$$ Therefore, the equation of the plane is: $$-5x - 10y + 10z + 26 = 0$$ Distance of plane from $(2, 3, -4)$ $$= \left| \frac{-10 - 30 - 40 + 26}{\sqrt{25 + 100 + 100}} \right| = \frac{54}{15} = \alpha$$ $$35\alpha = 35 \cdot \frac{54}{15} = 7 \times \frac{54}{3} = 7 \times 18 = 126$$ Hence this is the required option.

Given that $\frac{1}{n+1} {}^{n}C_{n} + \frac{1}{n} {}^{n}C_{n-1} + \ldots + \frac{1}{2} {}^{n}C_{1} + {}^{n}C_{0} = \frac{1023}{10}$. Therefore, $$\sum_{r=0}^{n} \frac{1}{r+1} {}^{n}C_{r} = \frac{1023}{10} \left(\because {}^{n+1}C_{r+1} = \frac{n+1}{r+1} {}^{n}C_{r}\right)$$ Thus, $$\sum_{r=0}^{n} \frac{1}{n+1} {}^{n+1}C_{r+1} = \frac{1023}{10}$$ Therefore, $$\frac{1}{n+1} \left[ {}^{n+1}C_{1} + {}^{n+1}C_{2} + \ldots + {}^{n+1}C_{n+1} \right] = \frac{1023}{10}$$ We know that ${}^{n}C_{0} + {}^{n}C_{1} + {}^{n}C_{2} + \ldots + {}^{n}C_{n} = 2^n$. Therefore, $$\frac{2^{n+1} - 1}{n+1} = \frac{1023}{10} = \frac{2^{10} - 1}{10}$$ Thus, $n + 1 = 10$. Therefore, $n = 9$. Hence this is the correct option.

Given, $z_0 = \frac{1+3i}{2}$, $z_1 = (1+i)$. So, $|z_1 - z_0| = \sqrt{\left(1 - \frac{1}{2}\right)^2 + \left(1 - \frac{3}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}$. And $z_2$ be outside circle, $|z_1 - z_0||z_2 - z_0| = 1$ $\Rightarrow \frac{1}{\sqrt{2}} |z_2 - z_0| = 1$ $\Rightarrow |z_2 - z_0| = \sqrt{2}$. Now given, $z_0$, $z_1$ $\&$ $z_2$ are collinear, So, by concept of rotation we get, $$\frac{z_2 - z_0}{z_1 - z_0} = \frac{|z_2 - z_0|}{|z_1 - z_0|} e^{i0}$$ $$\Rightarrow \frac{z_2 - z_0}{z_1 - z_0} = \frac{|z_2 - z_0|}{|z_1 - z_0|} (\pm 1)$$ $$\Rightarrow \frac{z_2 - z_0}{z_1 - z_0} = \pm 2$$ $$\Rightarrow z_2 = z_0 \pm 2(z_1 - z_0)$$ So, $z_2 = 2z_1 - z_0 = \frac{3}{2} + \frac{1}{2} i \Rightarrow |z_2| = \frac{5}{2}$ Or $z_2 = 3z_0 - 2z_1 = -1 + \frac{5}{2} i \Rightarrow |z_2|^2 = \frac{13}{2}$ Hence, the smaller value will be, $|z_2| = \frac{5}{2}$

The straight line is shown below with the intercepts. $$M \left( \frac{7}{2 \cos \theta}, \frac{7}{2 \sin \theta} \right) \equiv (h, k)$$ Therefore, $\cos \theta = \frac{7}{2h}$ and $\sin \theta = \frac{7}{2k}$. Thus, $\theta \in \left(0, \frac{\pi}{2}\right)$. $$\Rightarrow \frac{49}{4h^2} + \frac{49}{4k^2} = 1 \therefore x and y are positive$$ Therefore, Locus $= \frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{49}$. Now $\left( \alpha, \frac{7 \sqrt{3}}{3} \right)$ lies on curve. $$\Rightarrow \frac{1}{\alpha^2} + \frac{9}{147} = \frac{4}{49}$$ $$\frac{1}{\alpha^2} = \frac{1}{49}$$ Therefore, $\alpha = \pm 7$. Thus, $\alpha = 7$. Hence this is the required option.

Given that variance of $\alpha - \beta$ is $\frac{p}{q}$ $\alpha \in \{1, 2, 3, 4, 5, 6\}$ $\beta \in \{1, 2, 3, 4, 5, 6\}$ $(\alpha - \beta) = 0$ (6 cases) $(\alpha - \beta) = -1$ (5 cases) $(\alpha - \beta) = -2$ (4 cases) $(\alpha - \beta) = -3$ (3 cases) $(\alpha - \beta) = -4$ (2 cases) $(\alpha - \beta) = -5$ (1 case) $(\alpha - \beta) = 1$ (5 cases) $(\alpha - \beta) = 2$ (4 cases) $(\alpha - \beta) = 3$ (3 cases) $(\alpha - \beta) = 4$ (2 cases) $(\alpha - \beta) = 5$ (1 case) Mean $= 0$ We know that variance $= \frac{\sum_{i=1}^{n} f_i \left(x_i - \bar{x}\right)^2}{\sum_{i=1}^{n} f_i}$ Variance $= \sigma^2 = \frac{0^2 \times 6 + 2 \times 1^2 \times 5 + 2^2 \times 4 + 2 \times 3^2 \times 3 + 2 \times 4^2 \times 2 + 2 \times 5^2 \times 1}{36}$ $$= \frac{2}{36} \times (5 + 16 + 27 + 32 + 25) = \frac{36}{105} = \frac{35}{6}$$ Therefore, $p = 35$ Sum of divisors of $p = 1 + 5 + 7 + 35 = 48$ Option (3) is correct.

Given, $AB = 7$, $BC = 3$. $\Rightarrow c = 7$, $a = 3$. Apply cosine rule in the given equation $\cos A + 2 \cos B + \cos C = 2$, we get $$\Rightarrow \left( \frac{b^2 + c^2 - a^2}{2bc} \right) + 2 \left( \frac{c^2 + a^2 - b^2}{2ac} \right) + \left( \frac{a^2 + b^2 - c^2}{2ab} \right) = 2$$ $$\Rightarrow \left( \frac{b^2 + 49 - 9}{14b} \right) + \left( \frac{49 + 9 - b^2}{21} \right) + \left( \frac{9 + b^2 - 49}{6b} \right) = 2$$ $$\Rightarrow \left( \frac{b^2 + 40}{14b} \right) + \left( \frac{58 - b^2}{21} \right) + \left( \frac{b^2 - 40}{6b} \right) = 2$$ $$\Rightarrow b^3 - 5b^2 - 16b + 80 = 0$$ $$\Rightarrow (b - 4)(b + 4)(b - 5) = 0$$ $$\Rightarrow b = -4, 4, 5.$$ But $b \neq -4$. Also, if $b = 4$ triangle cannot be constructed. That means $b = 5$. $$\Rightarrow \cos A - \cos C = \left( \frac{b^2 + c^2 - a^2}{2bc} \right) - \left( \frac{a^2 + b^2 - c^2}{2ab} \right)$$ $$\Rightarrow \cos A - \cos C = \left( \frac{52 + 7^2 - 3^2}{2 \left( 5 \times 7 \right)} \right) - \left( \frac{3^2 + 5^2 - 7^2}{2 \left( 3 \times 5 \right)} \right)$$ $$\Rightarrow \cos A - \cos C = \left( \frac{65}{70} \right) - \left( \frac{-15}{30} \right) = \frac{195 + 105}{210} = \frac{300}{210}$$ $$\Rightarrow \cos A - \cos C = \frac{10}{7}$$

Given, a fair $n$ $(n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears, Now we know that, mean is given by, Mean $= \sum_{i=1}^{\infty} p_i x_i = 1 \cdot \frac{n-1}{n} + 2 \cdot \left( \frac{n-1}{n} \right) + 3 \cdot \frac{n-1}{n^2} + \cdots$ $$\Rightarrow \frac{n}{9} = \left( 1 - \frac{1}{n} \right) S \ldots \ldots \ldots (1)$$ Where, $S = 1 + \frac{2}{n} + \frac{3}{n^2} + \frac{4}{n^3} + \cdots \ldots \ldots (2)$ $$\frac{1}{n} S = \frac{1}{n} + \frac{2}{n^2} + \frac{3}{n^3} + \cdots \ldots \ldots (3)$$ Now subtracting equation $(2) - (3)$ we get, $$\left( 1 - \frac{1}{n} \right) S = 1 + \frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} + \cdots$$ $$\Rightarrow \left( 1 - \frac{1}{n} \right) S = \frac{1}{1 - \frac{1}{n}}$$ Now putting the value of $S$ in equation $(1)$ we get, $$\Rightarrow \frac{n}{9} = \left( 1 - \frac{1}{n} \right) \times \frac{1}{\left( 1 - \frac{1}{n} \right)^2} = \frac{n}{n-1}$$ $$\Rightarrow n = 10$$

Three numbers $a$, $b$, $c$ are in A.P. and they are used to make a 9-digit number using each digit thrice such that at least three consecutive digits are in A.P. So, $c$, $b$, $a$ will also be in A.P. Now, we have 7 positions to put $(a, b, c)$ or $(c, b, a)$ as $$\_\_\_\_\_\_\_$$ 7 positions. Number of such numbers is $$= \binom{2}{1} \cdot \binom{7}{1} \cdot \frac{6!}{2! \cdot 2! \cdot 2!}$$ $$= 1260$$

Given, $f(x) = |x| + \sqrt{x - [x]}$ $x - [x] \geq 0 \Rightarrow x \in \mathbb{R}$ Now using the property of greatest integer function and simplifying the function we get, $$f(x) = \begin{cases} 2 + \sqrt{x + 2}, & -2 < x < -1 \\ 1 + \sqrt{x + 1}, & -1 \leq x < 0 \\ \sqrt{x}, & 0 \leq x < 1 \end{cases}$$ Now plotting the diagram we get, Now from the diagram we can say that, $f(x)$ is discontinuous at two points $x = \{-1, 0\}$.

Given, the plane $x + 3y - 2z + 6 = 0$ meet the coordinate axes at the points $A$, $B$, $C$. So, $A(-6, 0, 0)$, $B(0, -2, 0)$, $C(0, 0, 3)$. And the orthocenter of the triangle $ABC$ is $\left( \alpha, \beta, \frac{6}{7} \right)$. We know that the circumcenter of a triangle is the point of intersection of perpendicular bisectors. So, plotting the diagram we get, Now from the diagram, we can see that $O$ is the midpoint of $BC \perp BC$. So, by the perpendicular vector formula, we get, $$(x - 0) \times 0 + (y + 1) \times 2 + \left( z - \frac{3}{2} \right) \times 3 = 0$$ $$\Rightarrow 4y + 6z - 5 = 0 \ldots (1)$$ Similarly, for side $AC$ we get, $$(x + 3) \times (-6) + (y - 0) \times 0 + \left( z - \frac{3}{2} \right) \times (-3) = 0$$ $$\Rightarrow 4x + 2z + 9 = 0 \ldots (2)$$ Similarly, for side $AB$ we get, $$(x + 3) \times (-6) + (y + 1) \times 2 + (z - 0) \times 0 = 0$$ $$\Rightarrow 3x - y + 8 = 0 \ldots (3)$$ Now from equation $(1)$, $(2)$, and $(3)$ we get, $$x = -\frac{9}{4}, y = \frac{5}{2}, z = \frac{3}{2}, z = x$$ Now using the relation between centroid, circumcenter, and orthocenter, we get, Now on comparing both sides we get, $$\alpha = \frac{15}{4}, \beta = \frac{9}{2}$$ $$98(\alpha + \beta)^2 = 98 \times 14^2 = 288$$

Given, $$I(x) = \int \frac{\sqrt{x+7}}{x} \, dx$$ Now let, $\frac{x+7}{x} = t^2 \Rightarrow \frac{-7}{x^2} dx = 2t \, dt$ $$\Rightarrow dx = \frac{-14t}{(t^2-1)^2} \, dt$$ So, $$I(x) = -14 \int \frac{t^2}{(t^2-1)^2} \, dt$$ $$\Rightarrow I(x) = -14 \int \frac{dt}{(t^2+1)(t^2-2)}$$ $$I(x) = \frac{-14}{2} \int \left[ \frac{1}{(t+1)^2} - \frac{1}{t^2-2} + \left( \frac{1}{t} - \frac{1}{t} \right)^2 \right] \, dt$$ $$\Rightarrow I(x) = -7 \left( \frac{1}{4} \ln \left| \frac{t+1}{t+1-2} \right| - \frac{1}{t} \right) + c$$ Now when $x = 9$, $t = \frac{4}{3}$ $$\Rightarrow I(9) = 12 + 7 \times \ln 7 = \frac{-7}{4} \ln \left( \frac{1}{7} \right)^2 + 7 \times \frac{12}{7} + c$$ $$\Rightarrow c = \frac{7}{2} \ln 7$$ Now when $x = 1$, $t = 2\sqrt{2}$ $$\Rightarrow I(1) = \frac{7}{4} \ln \left( \frac{2\sqrt{2}+1}{2\sqrt{2}-1} \right)^2 + 7 \times \frac{2\sqrt{2}}{7} + \frac{7}{2} \ln 7$$ $$\Rightarrow I(1) = \frac{7}{2} \ln \left( \frac{2\sqrt{2}+1}{7} \right)^2 + 2\sqrt{2} + \frac{7}{2} \ln 7$$ $$\Rightarrow I(1) = 7 \ln \left( 2\sqrt{2} + 1 \right) - \frac{7}{2} \ln 7 + 2\sqrt{2} + \frac{7}{2} \ln 7$$ $$\Rightarrow \alpha = 2\sqrt{2} \Rightarrow \alpha^4 = 64$$

Given that $D_k = \begin{vmatrix} 1 & 2k & 2k-1 \\ n & n^2 + n + 2 & n^2 \\ n & n^2 & n^2 + n + 2 \end{vmatrix}$ $$\Rightarrow \sum_{k=1}^{n} D_k = \begin{vmatrix} \sum 1 & 2 \sum k & 2 \sum k - \sum 1 \\ n & n^2 + n + 2 & n^2 \\ n & n^2 & n^2 + n + 2 \end{vmatrix}$$ $$\Rightarrow \sum_{k=1}^{n} D_k = \begin{vmatrix} n & n^2 + n + 2 & n^2 \\ n & n^2 + n & n^2 + n + 2 \\ n & n(n+1) & n^2 \end{vmatrix}$$ $$\Rightarrow \sum_{k=1}^{n} D_k = \begin{vmatrix} n & n^2 + n + 2 & n^2 \\ n & n^2 + n & n^2 + n + 2 \\ n & n^2 & n^2 + n + 2 \end{vmatrix}$$ Perform $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$: $$\Rightarrow \sum_{k=1}^{n} D_k = \begin{vmatrix} 0 & -2 & 0 \\ 0 & 2 & -n-2 \\ n & n^2 + n & n^2 + n + 2 \end{vmatrix}$$ Expand the determinant along $R_1$. $$\Rightarrow \sum_{k=1}^{n} D_k = 2((-n)(-n-2)) = 96$$ $$\Rightarrow n^2 + 2n = 48$$ $$\Rightarrow n = 6, \ -8$$ So, $n = 6$ Hence this is the required answer.

5 positive numbers $a_1, a_2, \ldots, a_5$ are in geometric progression. So, let the numbers be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$. According to question, $$\frac{\frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2}{5} = \frac{31}{10} \cdots (1)$$ And, $$\frac{r^2}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} = \frac{31}{40}$$ $$\Rightarrow \frac{1}{a} \left( \frac{5}{r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2}} \right) = \frac{31}{40} \cdots (2)$$ By (1) and (2), we get $a^2 = 4 \Rightarrow a = 2$. So, $$\left( r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2} \right) = \frac{31}{4}$$ $$\Rightarrow \left( r + \frac{1}{r} \right)^2 + r + \frac{1}{r} = \frac{27}{4} + 2$$ $$\Rightarrow t^2 + t = \frac{35}{4}$$ where, $t = r + \frac{1}{r}$. $$4t^2 + 4t - 35 = 0$$ $$\Rightarrow t = \frac{5}{2} \Rightarrow r = 2$$ So, numbers are $\frac{1}{2}, 1, 2, 4, 8$. Variance is $$= \frac{1}{4} \frac{1 + 1 + 4 + 16 + 64}{5} - \left( \frac{\frac{1}{2} + 1 + 2 + 4 + 8}{5} \right)^2$$ $$= \frac{186}{25} = \frac{m}{n}$$ $m + n = 211$. Hence this is the required option.

Given, set $A = \{1, 2, 3\}$. Now Cartesian product $A \times A = \{(1, 1), (2, 1), (1, 2), \ldots, (3, 3)\}$. Now, given the relation is reflexive, so, $(1, 1), (2, 2), (3, 3) \in R$. Also given $(1, 2), (2, 3) \in R$, $(1, 3)$ must $\in R$. Now finding, possible cases: Case-1: All of $(2, 1), (3, 2), (3, 1) \notin R \rightarrow 1$ relation. Case-2: Only one of $(2, 1), (3, 2), (3, 1) \in R \rightarrow 3$ relations. For example if relation is $\{(1, 1), (2, 2), (3, 3), (2, 1)\}$ then it is reflexive as well as transitive as $(2, 2), (2, 1) \rightarrow (2, 1)$ is present in relation. Note that exactly two of $(2, 1), (3, 2), (3, 1) \in R$ is not possible because if two of these $\in R$, third must $\in R$ to make relation transitive. Total number of relations $= 4$.

Given, $$\int_{-0.15}^{0.15} |100x^2 - 1| \, dx = \frac{k}{3000}$$ Now we know that, $$\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$$ if $f(x)$ is even. So, $$\int_{-0.15}^{0.15} |100x^2 - 1| \, dx = 2 \int_{0}^{0.15} |100x^2 - 1| \, dx$$ Now solving, $$2 \int_{0}^{0.15} |100x^2 - 1| \, dx = 2 \left( \int_{0}^{0.1} (100x^2 - 1) \, dx + \int_{0.1}^{0.15} 100x^2 - 1 \, dx \right)$$ $$\Rightarrow 2 \int_{0}^{0.15} |100x^2 - 1| \, dx = 2 \left( \left[ x - \frac{100x^3}{3} \right]_{0}^{0.1} + \left[ \frac{100x^3}{3} - x \right]_{0.1}^{0.15} \right)$$ $$\Rightarrow 2 \int_{0}^{0.15} |100x^2 - 1| \, dx = 2 \left( \left( \frac{1}{10} - \frac{1}{30} \right) + \left( \frac{100 \times 0.15^3}{3} - \frac{15}{100} \right) - \left( \frac{1}{30} - \frac{1}{10} \right) \right)$$ $$\Rightarrow 2 \int_{0}^{0.15} |100x^2 - 1| \, dx = 2 \left( \frac{2}{30} + \left( \frac{225}{3000} - \frac{15}{100} \right) + \frac{2}{30} \right)$$ $$\Rightarrow 2 \int_{0}^{0.15} |100x^2 - 1| \, dx = 2 \left( \frac{4}{30} + \left( -\frac{3}{40} \right) \right)$$ $$\Rightarrow 2 \int_{0}^{0.15} |100x^2 - 1| \, dx = \frac{575}{3000}$$ Now comparing with $$\int_{-0.15}^{0.15} |100x^2 - 1| \, dx = \frac{k}{3000}$$ we get $k = 575$

Assuming a circle of radius $r$ which touches both the axis is given by, $(x - r)^2 + (y - r)^2 = r^2$. Now given line $x + y = 2$ makes the intercept 2 the circle, so plotting the diagram we get, Now from diagram, $AC = BC = 1$, $OC = \sqrt{r^2 - 1}$. Now using the formula of perpendicular distance of point from the line we get, $$OC = \left| \frac{2r - 2}{\sqrt{2}} \right| = \sqrt{r^2 - 1}$$ $$\Rightarrow 2(r - 1)^2 = r^2 - 1$$ $$\Rightarrow r^2 - 4r + 3 = 0$$ $$\Rightarrow r_1 = 1, \ r_2 = 3$$ Hence, $r_1^2 + r_2^2 - r_1 r_2 = 1 + 9 - 3 = 7$

Let's assume that the original position of the bubble is $x$ from one side of the ice cube of side $a$. When the bubble is seen from one side, it can be written that $$\frac{x}{\mu} = 12 \ldots(1)$$ For the view from the other side, it can be written that $$\frac{a-x}{\mu} = 4 \ldots(2)$$ From equations (1) and (2), it is given that $$24 - 12\mu = 4\mu$$ $$\Rightarrow \mu = \frac{24}{16}$$ $$= 1.5$$

The formula to calculate the speed of a satellite while revolving around the Earth is given by $$v = \sqrt{\frac{GM_e}{R}} \ldots (1)$$ where, $v$ is the speed, $G$ is the Gravitational constant, $M_e$ is the mass of the Earth, $R$ is the radius of the orbit. As can be seen from the equation, speed is independent of mass of satellite. Hence, the speed of the satellites will be the same.

The carrier wave frequency is given by $$f_c = \frac{1000\pi}{2\pi}$$ $$= 500 \, Hz$$ The modulating wave frequency is given by $$f_m = \frac{4\pi}{2\pi}$$ $$= 2 \, Hz$$ Hence, the frequencies contained in the modulated signal are given by $$f = f_c - f_m, f_m, f_c + f_m$$ $$= 498 \, Hz, 500 \, Hz, 502 \, Hz$$

The change in collector current for the given transistor can be written as $$\Delta I_c = 16 \, \mathrm{mA} - 5 \, \mathrm{mA}$$ $$= 11 \, \mathrm{mA}$$ The change in base current for the given transistor can be written as $$\Delta I_b = 200 \, \mu\mathrm{A} - 100 \, \mu\mathrm{A}$$ $$= 100 \, \mu\mathrm{A}$$ The formula to calculate the current gain in CE configuration is given by $$\beta = \frac{\Delta I_c}{\Delta I_b} \ldots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the required quantity. $$\beta = \frac{\Delta I_c}{\Delta I_b}$$ $$= \frac{11 \, \mathrm{mA}}{100 \, \mu\mathrm{A}}$$ $$= 110$$

The formula to calculate the rms speed of the argon molecules is given by $$v_{Ar} = \sqrt{\frac{3RT}{M_{Ar}}} \cdots (1)$$ The formula to calculate the rms speed of chlorine molecule is given by $$v_{Cl} = \sqrt{\frac{3RT}{M_{Cl}}} \cdots (2)$$ Divide equation (1) by equation (2) and simplify to obtain the required ratio. $$\frac{v_{Ar}}{v_{Cl}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}} \cdots (3)$$ Substitute the values of the known parameters into equation (3) and solve to calculate the required value. $$\frac{v_{Ar}}{490 \, \mathrm{m \, s^{-1}}} = \sqrt{\frac{71}{40}}$$ $$\Rightarrow v_{Ar} = \sqrt{\frac{71}{40}} \times 490 \, \mathrm{m \, s^{-1}}$$ $$\approx 651.7 \, \mathrm{m \, s^{-1}}$$

The formula to calculate the de Broglie wavelength of the proton is given by $$\lambda_p = \frac{h}{\sqrt{2m_p q_p V}} \ldots (1)$$ The formula to calculate the de Broglie wavelength of the alpha particle is given by $$\lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V'}} \ldots (2)$$ Divide equation (1) by equation (2) and simplify to obtain the required ratio. $$\frac{\lambda_p}{\lambda_\alpha} = \frac{\sqrt{m_\alpha}}{\sqrt{m_p}} \times \sqrt{\frac{q_\alpha V'}{q_p V}} \ldots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required ratio. $$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4}{1}} \times \sqrt{\frac{2 \times 4}{1 \times 2}}$$ $$= 4 : 1$$

The properties of a diamagnetic material are as follows: 1. The susceptibility has a low negative value. 2. Susceptibility is independent of temperature. 3. The relative permeability is slightly less than one. 4. When placed in a non-uniform magnetic field they have a tendency to move away from the field. 5. When suspended freely in a uniform magnetic field, they set themselves perpendicular to the direction of the magnetic field. Hence, Statement I is false, but statement II is true.

The formula to calculate the resistance of the original wire is given by $$R = \rho \frac{L}{A} \ldots (1)$$ When the new wire is formed, its resistance is given by $$R' = \rho \frac{l}{4A} \ldots (2)$$ Divide equation (2) by equation (1) and simplify to obtain the expression for the new resistance. $$\frac{R'}{R} = \frac{\rho \frac{l}{4A}}{\rho \frac{L}{A}}$$ $$\Rightarrow R' = R \frac{l}{4l} \ldots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required new resistance. $$R' = \frac{l}{4l} \times 160 \, \Omega$$ $$= 10 \, \Omega$$

The formula to calculate the spring constant is $F = kx (1)$ The formula to calculate the angular speed is $\omega = \frac{2\pi}{T} (2)$ The formula to calculate the angular momentum is $L = mvr (3)$ The formula to calculate the moment of inertia is $I = MR^2 (4)$ Use the method of dimensional analysis to obtain the dimensions of the required quantities. From above relations, we can write $$[MLT^{-2}] = [k][L]$$ $$\Rightarrow [k] = [MT^{-2}]$$ $$[\omega] = \frac{1}{[T]}$$ $$= [T^{-1}]$$ $$[L] = [M][LT^{-1}][L]$$ $$= [ML^2T^{-1}]$$ $$[I] = [M][L^2]$$ $$= [ML^2]$$

When the force $F_1$ is removed, the particle starts accelerating because of the resultant force between $F_2$ and $F_3$. The resultant force can be calculated as follows: $$F = \sqrt{F_2^2 + F_3^2}$$ $$= 10 \, \mathrm{N}$$ Hence, the acceleration of the particle is given by $$a = \frac{10 \, \mathrm{N}}{5 \, \mathrm{kg}}$$ $$= 2 \, \mathrm{m \, s^{-2}}$$

The formula to calculate the rate of cooling of the object is given by $$\frac{dT}{dt} = K \left[ \frac{T_f + T_i}{2} - T_0 \right] \ldots (1)$$ For the first case, it can be written, using equation (1) that $$\frac{80 - 60}{5} = K \left[ \frac{80 + 60}{2} - 20 \right]$$ $$4 = 50K \ldots (2)$$ If $t$ is the required time for the second case, from equation (1), it can be written that $$\frac{60 - 40}{t} = K \left[ \frac{60 + 40}{2} - 20 \right]$$ $$\frac{20}{t} = 30K \ldots (3)$$ Divide equation (2) by equation (3) and solve to calculate the required time. $$\frac{4}{\frac{20}{t}} = \frac{50K}{30K}$$ $$\Rightarrow \frac{t}{5} = \frac{5}{3}$$ $$\Rightarrow t = \frac{25}{3} min = 500 s$$

The efficiency of Carnot engine can be calculated as follows: $$\eta_C = \left( 1 - \frac{273}{373} \right) \times 100\%$$ $$= 26.8\%$$ Hence, the efficiency of the engine is given by $$\eta_{engine} < \eta_{carnot}.$$

The work done in stopping the car and the truck can be written as $$W = Fx$$ $$= \Delta KE$$ As both the car and the truck are moving initially with the same kinetic energy, after being stopped, they will have the same change of kinetic energy. Also, they are stopped with the same applied force. Hence, from the above equation, it can be written that they have the same displacement, i.e., $x_1 = x_2$. Acceleration is a vector quantity. As the direction of motion of the car is different in the two given situations, the acceleration of the car will not be the same.

The formula to calculate the potential energy of the particle executing SHM is given by $$P = \frac{1}{2} m \omega^2 x^2 \ldots (1)$$ The formula to calculate the kinetic energy of the same particle is given by $$K = \frac{1}{2} m \omega^2 \left( A^2 - x^2 \right) \ldots (2)$$ Divide equation (1) by equation (2) to obtain the required ratio. $$\frac{P}{K} = \frac{\frac{1}{2} m \omega^2 x^2}{\frac{1}{2} m \omega^2 \left( A^2 - x^2 \right)}$$ $$= \frac{x^2}{A^2 - x^2} \ldots (3)$$ Substitute $\frac{A}{2}$ for $x$ into equation (3) to calculate the required ratio. $$\frac{P}{K} = \frac{\left( \frac{A}{2} \right)^2}{A^2 - \left( \frac{A}{2} \right)^2}$$ $$= \frac{1}{3}$$

The formula to calculate the velocity of the ball after 3 s is given by $$v_3 = u - 3g \ldots (1)$$ The formula to calculate the velocity of the ball after 5 s is given by $$v_5 = u - 5g \ldots (2)$$ Divide equation (1) by equation (2) to obtain the required ratio. $$\frac{v_3}{v_5} = \frac{u - 3g}{u - 5g} \ldots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required ratio. $$\frac{v_3}{v_5} = \frac{150 - 30}{150 - 50}$$ $$= \frac{120}{100}$$ $$= \frac{6}{5}$$ $$= \frac{5 + 1}{5}$$ Comparing the above equation with the given expression, it can be written that $x = 5$.

An electric dipole contains two equal and opposite charges. Thus, the net charge, as viewed from outside a surface enclosing the dipole, is zero. The net flux enclosed by any closed surface is, then, given by $$\phi = \frac{q_{enc}}{\varepsilon_0}$$ $$= 0$$

An optical telecommunication is communication at a distance using light to carry information. It can be performed visually or by using electronic devices. The wavelength of the microwaves are greater than that of the common optical light waves. Hence, the Assertion is not correct. But, because of the small wavelength, the electromagnetic waves are more energetic than the microwaves. Hence, the reason is true.

According to Bohr's postulates, if an electron absorbs a photon with energy equal to the difference between the energies of an excited state and the ground state, it will jump to a higher energy orbital. Considering that the energy of the colliding electron is $10.2 \, \mathrm{eV}$ or $12.09 \, \mathrm{eV}$, depending on how much energy is lost during the collision. $n = 3$ is the maximum excited state. Hence, there are spectral lines in a total of $$^3C_2 = \frac{3!}{2!1!}$$ $$= 3$$

The formula to calculate the escape velocity is given by $$v_e = \sqrt{\frac{2GM}{R}} \ldots(1)$$ For the Earth, it is given by $$v_e^E = \sqrt{\frac{2GM_E}{R_E}} \ldots(2)$$ and, for the planet, it is given by $$v_e^P = \sqrt{\frac{2GM_P}{R_P}} \ldots(3)$$ Divide equation (2) by equation (3) and simplify to obtain the required ratio. $$\frac{v_e^E}{v_e^P} = \sqrt{\frac{M_E}{M_P}} \times \sqrt{\frac{R_P}{R_E}} \ldots(4)$$ Substitute the values of the known parameters into equation (4) to obtain the required ratio. $$\frac{v_e^E}{v_e^P} = \sqrt{\frac{M_E}{16M_E}} \times \sqrt{\frac{4R_E}{R_E}}$$ $$= \frac{1}{2}$$ $$\Rightarrow v_e^P : v_e^E = 2 : 1$$

When an a.c. voltage is applied in a series LCR circuit, the current is increased first, attains a maximum and then decreases. The formula to calculate the power factor is given by $$\cos \phi = \frac{R}{\sqrt{R^2 + \left( X_L^2 - X_C^2 \right)}} \cdots (1)$$ At resonance, $X_L = X_C$. Hence, $$\cos \phi_{resonance} = 1$$ Thus, both the statements are true.

It is given, $f_5 = 405 \, \mathrm{Hz}$. As the first three resonance frequencies are in the ratio of $1 : 3 : 5$, therefore it must be a closed organ pipe. The frequency formula in a closed organ pipe is given by $f = \frac{nv}{4l}$. So, $$\frac{5v}{4l} = 405 \, \mathrm{Hz}$$ $$\Rightarrow \; l = \frac{405 \times 4}{5 \times 324} = 1 \, \mathrm{m}$$

The rotational kinetic energy is given by $K_{rot} = \frac{I \omega^2}{2} = \frac{1}{3} M R^2 \omega^2$. The total kinetic energy is given by $$K_{tot} = \frac{1}{2} m v^2 + \frac{1}{3} M R^2 \omega^2$$ $$= \frac{m R^2 \omega^2}{2} + \frac{1}{3} M R^2 \omega^2$$ $$= \frac{5}{6} M R^2 \omega^2$$ Thus, the ratio is $$\frac{K_{rot}}{K_{tot}} = \frac{\frac{1}{3} M R^2 \omega^2}{\frac{5}{6} M R^2 \omega^2} = \frac{2}{5}$$ Comparing with the given value, $$\frac{x}{5} = \frac{2}{5}$$ $$\Rightarrow x = 2$$

The formula for time period is given by $$T = 2\pi \sqrt{\frac{I}{MBH}}$$ Thus, $$T = 2\pi \sqrt{\frac{I}{MB \cos \theta}}$$ Let $$\theta_1 = 30^\circ$$ $$\theta_2 = 60^\circ$$ The ratio of the time periods is, $$\frac{T_1}{T_2} = \frac{\frac{60}{20}}{\frac{60}{30}} = \sqrt{\frac{B_2 \cos 60^\circ}{B_1 \cos 30^\circ}} = \sqrt{\frac{B_2}{B_1} \times \frac{1}{\frac{\sqrt{3}}{2}}}$$ Thus, $$\Rightarrow \frac{B_1}{B_2} = \frac{4}{\sqrt{243}}$$

The data given is $B = 0.4 \, \mathrm{T}$, $R = 2 \times 10^{-2} \, \mathrm{m}$. It is given that $\frac{dR}{dt} = 10^{-3} \, \mathrm{m \, s^{-1}}$. Area of the loop is $A = \pi R^2$. So, $\frac{dA}{dt} = 2 \pi R \frac{dR}{dt}$. The flux is, $$\phi = 2 \pi R \frac{dR}{dt} \times B$$ So, the magnitude of emf is $$\varepsilon = 2 \times \frac{22}{7} \times 2 \times 10^{-2} \times 10^{-3} \times 0.4$$ $$= 5.028 \times 10^{-5} \, \mathrm{V} = 50.28 \, \mu \mathrm{V} \approx 50 \, \mu \mathrm{V}$$

It is given, $v = 80 \, \mathrm{km \, h^{-1}}$ $m = 500 \, \mathrm{kg}$ $S = 4 \, \mathrm{km}$ $\mu = 0.04$ The power due to friction is $P_f = Fv$ where $F = \mu mg$ $$= \left( (0.04)(500 \times 9.8) \times 80 \times \frac{5}{18} \right) \, \mathrm{W}$$ Hence, the work done by the engine in overcoming the friction is $$W = P_f t = P_f \frac{d}{v}$$ $$\Rightarrow W = 20 \times 9.8 \times 80 \times \frac{5}{18} \times \frac{4 \times 10^3}{80 \times \frac{5}{18}} = 784 \, \mathrm{kJ}$$

The formula to calculate the Q-value of the given nuclear reaction can be written as $$Q = (m_U - m_{Th} - m_{He})c^2 \ldots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the required Q-value for the reaction. $$Q = (238.05060 \, \mathrm{u} - 234.04360 \, \mathrm{u} - 4.00260 \, \mathrm{u})c^2$$ $$= 0.0044c^2 \times 931.5 \, \mathrm{MeV}/c^2$$ $$\approx 4 \, \mathrm{MeV}$$

By Ohm's law, $V = IR$. Let the current and resistance at $0^\circ \mathrm{C}$ be $i_0$ and $R_0$ respectively. Let the current and resistance at $100^\circ \mathrm{C}$ be $i_{100}$ and $R_{100}$ respectively. By the given condition, $$i_0 R_0 = i_{100} R_{100}$$ $$\Rightarrow 2R_0 = 1.2R_0 \left(1 + 100\alpha\right)$$ $$\Rightarrow 1 + 100\alpha = \frac{5}{3}$$ $$\Rightarrow 50\alpha = \frac{1}{3}$$ Since the voltage remains constant, $$i_{50} R_{50} = i_0 R_0$$ $$\Rightarrow i_{50} = \frac{i_0 R_0}{R_{50}} = \frac{2 \times R_0}{R_0 \left(1 + 50\alpha\right)} = \frac{2}{1 + \frac{1}{3}} = 1.5 \, \mathrm{A}$$ $$\Rightarrow i_{50} = 15 \times 10^2 \, \mathrm{mA}$$

Let $L_1$ be the first lens and $L_2$ be the second lens. The first refraction is in $L_1$. Using lens formula, $$u_1 = \infty, f_1 = 20 \, \mathrm{cm},$$ $$\frac{1}{v_1} - 0 = \frac{1}{f_1}$$ $$\Rightarrow v_1 = 20 \, \mathrm{cm}$$ Hence, the distance between the second image and the $L_2$ would be $(60 - 20) = 40 \, \mathrm{cm}$. For the second lens $L_2$, $$u_2 = -40 \, \mathrm{cm}, f_2 = 20 \, \mathrm{cm},$$ $$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$$ $$\Rightarrow \frac{1}{v_2} + \frac{1}{40} = \frac{1}{20}$$ $$\Rightarrow \frac{1}{v_2} = \frac{1}{40}$$ $$\Rightarrow v_2 = 40 \, \mathrm{cm}$$ So, the distance of image from first lens is $$x = 40 \, \mathrm{cm} + 60 \, \mathrm{cm}$$ $$= 100 \, \mathrm{cm}$$

It is given that $A_1 = 10 \, \mathrm{cm}^2$, $A_2 = 5 \, \mathrm{cm}^2$, $\Delta P = 3 \, \mathrm{N} \, \mathrm{m}^{-2}$. By continuity equation, $$A_1 v_1 = A_2 v_2$$ $$\Rightarrow v_1 = \frac{A_2 v_2}{A_1} \cdots (i)$$ By Bernoulli's equation, $$\frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2 = \Delta P \cdots (ii)$$ Using equation (i) in (ii) $$\Delta P = \frac{\rho}{2} \left( v_2^2 - \left( \frac{A_2 v_2}{A_1} \right)^2 \right)$$ $$\Rightarrow \Delta P = \frac{\rho v_2^2}{2} \left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)$$ $$\Rightarrow 3 = \frac{1.25 \times 10^3 \times v_2^2}{2} \left( 1 - \frac{1}{4} \right)$$ $$\Rightarrow v_2^2 = \frac{6 \times 4}{1.25 \times 10^3 \times 3}$$ $$\Rightarrow v_2 = 8 \times 10^{-2} \, \mathrm{m} \, \mathrm{s}^{-1}$$ So, the discharge rate is, $$A_2 v_2 = x \times 10^{-5}$$ $$\Rightarrow x = \frac{5 \times 10^{-4} \times 8 \times 10^{-2}}{10^{-5}} = 4$$

The formula to calculate the potential $(V)$ of each tiny drop is given by $$V = k \frac{q}{r} \ldots (1)$$ When all the drops form the bigger drop, the radius of the new drop $(r')$ can be calculated as follows: Volume of big drop $= 64 \times$ volume of each tiny drop $$\Rightarrow \frac{4}{3} \pi r'^3 = 64 \times \frac{4}{3} \pi r^3$$ $$\Rightarrow r' = 4r$$ So, the potential of the new big drop $(V')$ can be written as $$V' = k \frac{64q}{r'}$$ $$= k \frac{64q}{4r}$$ $$= 16k \frac{q}{r} \ldots (2)$$ Divide equation (2) by equation (1) and simplify to obtain the potential of the new drop. $$\frac{V'}{V} = \frac{16k \frac{q}{r}}{k \frac{q}{r}}$$ $$\Rightarrow V' = 16V \ldots (3)$$ Substitute the value of the known parameter into equation (3) to calculate the required value of the potential for the new drop. $$V' = 16 \times 10 \, \mathrm{mV}$$ $$= 160 \, \mathrm{mV}$$

Aldehydes and ketones can react with Grignard reagents to form alcohols. Grignard reagents are organometallic compounds containing a carbon-magnesium bond (R-Mg-X). The reaction between ketone and a Grignard reagent involves the nucleophilic addition of the carbon atom of the Grignard reagent to the carbon-oxygen double bond of the ketone. This leads to the formation of a metal alkoxide intermediate. Subsequently, the metal alkoxide reacts with water (or a protic solvent) to yield the alcohol.

Critical temperature of a gas is the temperature at or above which vapor of the gas cannot be liquefied, no matter how much pressure is applied. The extent of adsorption is directly proportional to the critical temperature of the gas. Higher is the critical temperature of the gas, greater is the extent of adsorption. Order of $T_c$: $D > C > B > A$. Order of adsorption: $D > C > B > A$.

The distance between the nucleus and $5f$ orbital is more than the distance between the nucleus and $4f$ orbital. Hence the hold of nucleus on valence electron decreases in $5f$ orbital. $5f$ orbital not buried as $4f$ orbitals so electron present in $5f$ orbital experience less nuclear attraction than electron present in $4f$ orbital. Hence electrons of $5f$ orbital can take part in bonding to far great extent.

Nitrous acid is produced when sodium nitrate is reacted with acid. The nitrous acid can produce nitrosonium ion as electrophile. The tertiary amine given undergoes electrophilic substitution reaction with electrophile. The para substituted product is formed as major product. $$\mathrm{NaNO_2 + HX \rightarrow HNO_2 + NaX}$$ $$2 \mathrm{HNO_2 \rightarrow NO^+ + NO_2 + H_2O}$$ Therefore, Statement (2) is incorrect.

CFSE $=\left(-0.4\,n_{t_{2g}}+0.6\,n_{e_g}\right)\Delta_0$ $n_{t_{2g}}$ = Number of electrons in the $t_{2g}$ orbitals $n_{e_g}$ = Number of electrons in the $e_g$ orbitals Crystal Field Stabilisation Energy for the given complexes is as follows: (A) $[\mathrm{Cu(NH_3)_6}]^{2+}$ $\mathrm{Cu}^{2+}: 3d^9,\; t_{2g}^6\,e_g^3$ CFSE $=(-6\times0.4+3\times0.6)\Delta_0$ $=-0.6\Delta_0$ (B) $[\mathrm{Ti(H_2O)_6}]^{3+}$ $\mathrm{Ti}^{3+}: 3d^1,\; t_{2g}^1\,e_g^0$ CFSE $=-1\times0.4\Delta_0$ $=-0.4\Delta_0$ (C) $[\mathrm{Fe(CN)_6}]^{3-}$ $\mathrm{Fe}^{3+}: 3d^5,\; t_{2g}^5\,e_g^0$ CFSE $=-5\times0.4\Delta_0$ $=-2.0\Delta_0$ (D) $[\mathrm{NiF_6}]^{4-}$ $\mathrm{Ni}^{2+}: 3d^8,\; t_{2g}^6\,e_g^2$ CFSE $=(-6\times0.4+2\times0.6)\Delta_0$ $=-1.2\Delta_0$ So, the correct option is (B).

2-chloro-1,3-butadiene can undergo polymerization to form a polymer known as polychloroprene. Polychloroprene is a synthetic rubber commonly known as neoprene. Polyacrylonitrile (PAN) is an addition polymer. Addition polymers are formed through a polymerization process called addition polymerization, where monomers add to each other without the loss of any small molecules. Nylon-2-nylon-6 is a biodegradable polymer. Dacron is a type of polyester. Polyester is a polymer that belongs to the polyester family of polymers. So, A-II; B-I; C-IV; D-III

As we go down in a group, its size increases and also the volume of the atom increases as the number of electrons and protons increases. But the increase in mass is greater than that of the increase in volume, so the density, which is defined as mass per unit volume, will increase generally. Density of alkali metals increases down the group except potassium. Therefore, the correct order of density of given alkali metals is $\mathrm{K} < \mathrm{Na} < \mathrm{Rb} < \mathrm{Cs}$.

The oxidation state of $+5$ in pentahalides is more covalent as compared to the $+3$ oxidation state in trihalides. Due to the higher positive oxidation state of the central atom in pentahalide state, these atoms will have larger polarising power than the halogen atom attached to them since the polarising power is directly proportional to the charge. More is the polarisation, larger will be the covalent character of the bond. Hence, due to larger polarisation of bond in pentahalide state as compared to trihalide state, the $\mathrm{SbCl_5}$ is more covalent than $\mathrm{SbCl_3}$. Higher oxides of halogens are more stable than lower ones due to the formation of more number of bonds, which results in higher release of energy. So, both the statements I and II are correct.

Let the formula of metal chloride be $\mathrm{MCl}_x$. Mass of $100 \, \mathrm{mL}$ of vapours of $\mathrm{MCl}_x$ at STP $= 0.57 \, \mathrm{g}$. The mass of substance which occupies $22400 \, \mathrm{mL}$ of volume at STP is molar mass $(M)$. $$\frac{100 \times M}{22400} = 0.57$$ $$\therefore \text{Molar mass of } \mathrm{MCl}_x = 127.68 \, \mathrm{g \, mol^{-1}}$$ The mass percentage of chlorine $= \frac{\text{mass of chlorine}}{\text{mass of the compound}} \times 100$. $$\% \text{ of Cl} = \frac{35.5 \times x \times 100}{127.68} = 55$$ $$\Rightarrow x = 2$$ $$\therefore \text{Formula of metal chloride is } \mathrm{MCl}_2$$

For line II $\Delta S$ is more negative than line I. In the Ellingham diagram, a sharp change in slope of the line is observed for $\mathrm{Mg} - \mathrm{MgO}$ at $\sim 1120^\circ \mathrm{C}$ because that is the boiling point of magnesium. From diagram, $$\Delta S_I = (S_{solid}) - (S_{solid} + S_{gas})$$ $$\Delta S_{II} = (S_{liquid}) - (S_{solid} + S_{gas})$$ There is a large increase in entropy associated with the change of state of magnesium. So, both Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A).

Nitrogen oxides are produced by various natural and human activities, including combustion processes, industrial emissions, and vehicle exhaust. These are responsible for acid rains. Methane ($\mathrm{CH_4}$) is a potent greenhouse gas that contributes to global warming. While carbon dioxide ($\mathrm{CO_2}$) is the most well-known greenhouse gas. Carbon dioxide present in air, react with the rain water to form carbonic acid. The excessive use or runoff of phosphate fertilizers into bodies of water can contribute to eutrophication. Therefore, A-IV; B-III; C-II; D-I \begin{tabular}{|c|l|c|l|} \hline A & Nitrogen oxides in air & IV & Acid rain \\ \hline B & Methane in air & III & Global warming \\ \hline C & Carbon dioxide & II & pH of rain water becomes 5.6 \\ \hline D & Phosphate fertilisers in water & I & Eutrophication \\ \hline \end{tabular}

The lead storage battery is as shown in the diagram. Lead storage battery consists of lead anode and a grid of lead packed with lead oxide ($\mathrm{PbO_2}$) as cathode, a 38$\%$ solution of $\mathrm{H_2SO_4}$ is used as an electrolyte. During charging of battery $\mathrm{PbSO_4}$ deposited on cathode is converted in $\mathrm{PbO_2}$ and $\mathrm{PbSO_4}$ deposited on anode is converted into $\mathrm{Pb}$. The electrolyte used in battery is $\mathrm{H_2SO_4}$ which is about 38$\%$ by mass.

Ozone will cleave carbon-carbon double bonds to give ketones/carboxylic acids after oxidative workup. Here in this reaction cleavage occurs at unsaturation of hexene and acetic acid and butanoic acid are formed.

Aromatic compounds are cyclic, planar and must obey Huckel's rule i.e. molecule must have $4n + 2$ pi-electrons where $n$ is an integer (i.e. $n = 0, 1, 2, 3, 4 \ldots$ etc.). The compound $B$ follows the Huckel's rule. The acid-base reactions are fast reactions. Compound $(A)$, also called squaric acid, is a strong acid. So, the above reaction proceeds very fast. All $\mathrm{C} - \mathrm{C}$ bond lengths in $(B)$ are the same due to resonance. $(A)$ shows tautomerism.

The bond order of the acetylide ion $\mathrm{HC} \equiv \mathrm{C}^-$ is 3, indicating a triple bond between the carbon atoms. The acetylide ion consists of two carbon atoms with a triple bond between them and a negative charge. In the triple bond, there are six bonding electrons (two pi bonds and one sigma bond) and zero antibonding electrons. Among the given options $\mathrm{NO}^+$ will have the same bond order and magnetic property like acetylide ion.

When calcium chloride reacts with sodium carbonate, a double replacement reaction takes place. The products of this reaction are calcium carbonate and sodium chloride. $$\mathrm{CaCl_2 + Na_2CO_3 \rightarrow CaCO_3 \; (X) + \; 2 \, NaCl \; (Y)}$$ To obtain calcium chloride from calcium carbonate we need to react calcium carbonate with hydrochloric acid. $$\mathrm{CaCO_3 \; (X) + \; 2 \, HCl \; (Z) \rightarrow CaCl_2 + CO_2(g) + H_2O(l)}$$ Therefore, (X) $\mathrm{CaCO_3}$ (Y) $\mathrm{NaCl}$ (Z) $\mathrm{HCl}$

Boron is extremely hard as it exists as $\mathrm{B_{12}}$ icosahedral molecular units which accounts for its high lattice energy. So, statement-I is correct. Due to small atomic size boron forms strong covalent bonds with the neighbouring atoms. Thus, boron atoms are closely packed in its solid, thus boron has high melting and boiling points. So, statement-II is also correct.

Intramolecular Aldol condensations happen when a single molecule contains two aldehyde/ketone groups. When the alpha carbon of one group attacks the other, the molecule attacks itself forming a ring structure. The reaction is shown below.

$B_2H_6$, known as diborane, is an example of an electron-deficient hydride. HF, or hydrogen fluoride, is an example of an electron-rich hydride. Electron-rich hydrides are molecules that contain more electrons than what would be expected based on the normal valence electron count for the atoms involved. $CH_4$ is an electron precise hydride since there are exact number of electrons to form normal covalent bonds for completing octet. $MgH_2$ is a saline hydride, as this is the hydride that formed when hydrogen reacts with s-Block elements.

$\mathrm{SOCl_2}$ is a halogenating agent. The carboxylic acid group of reactant changes to acid chloride. The acid chloride formed reacts with primary amine to give secondary amide. The secondary amide undergo reduction with lithium aluminium hydride to give secondary amine.

The first law of thermodynamics for isothermal process $Q = -W$. The $W$ for isothermal reversible process is $$W = -nRT \ln \frac{V_2}{V_1}$$ $$\Delta S = \frac{Q}{T}$$ $$\Delta S_{surrounding} = -nR \ln \frac{V_2}{V_1}$$ $$= -1 \times 8.314 \times 2.303 \times \log \frac{3}{2}$$ $$= -3.37 \, \mathrm{JK^{-1}}$$ $$\simeq -3 \, \mathrm{JK^{-1}}$$ The given process is mentioned as isothermal reversible process hence the $\Delta S_{surrounding}$ is calculated.

RCHO + 2 $\left$[ $\mathrm{Ag}$ $\left$( $\mathrm{NH}$_3 $\right$)_2 $\right$] $\mathrm{OH}$ $\rightarrow$ $\mathrm{RCOONH}$_4 + 2 $\mathrm{Ag}$ + 3 $\mathrm{NH}$_3 + $\mathrm{H}$_2$\mathrm{O}$ The RCHO is cyclohexane carbaldehyde (C_6H_{11} $\mathrm{CHO}$). The molar mass is 112 $\mathrm{g}$. The number of moles of cyclohexane carbaldehyde = $\frac{131.8 \times 10^3}{112}$ = 1.176 $\times$ 10^3 mol 1 mole of aldehyde produces 3 moles of $\mathrm{NH}$_3 1.176 $\times$ 10^3 mol aldehyde will produce = 3 $\times$ 1.176 $\times$ 10^3 moles of $\mathrm{NH}$_3 Mass of $\mathrm{NH}$_3 produced = 3 $\times$ 1.176 $\times$ 10^3 $\times$ 17 = 59.97 $\times$ 10^3 g $\approx$ 60 kg

An oligopeptide is a short-chain peptide. The given oligopeptide is a tripeptide. The given Oligopeptide has the following structure In $sp^2$ hybridisation carbon can have one pi bond. The carbons marked with * are $sp^2$ carbons. It has 10 $sp^2$ hybridised C atoms.

The concentration of $\mathrm{H}^+$ or say HCl in both the given solution is $10^{-1}$ and $10^{-2}$ mole $\mathrm{L}^{-1}$ respectively. For the given amount of HCl the concentration depends over dilution hence $$M_1 V_1 = M_2 V_2$$ $$\frac{M_1}{M_2} = \frac{V_2}{V_1}$$ $$\frac{10^{-1}}{10^{-2}} = \frac{V_2}{1 \mathrm{L}}$$ $$V_2 = 10 \, \mathrm{L} = 10000 \, \mathrm{mL}$$ $$\Delta V = 10000 - 1000 \, \mathrm{mL} = 9000 \, \mathrm{mL}$$

In thin layer chromatography using hexane, the least polar compound will rise to maximum height and most polar compound will rise to minimum height. The retardation factor of each spot can be determined by dividing the distance that a particular compound has travelled by the distance between solvent front and initial spotting site. This parameter is dependent on the thin layer chromatographic plate and the solvent. Therefore, $R_f$ value for most polar compound $$= \frac{2}{8} = 0.25$$ $$= 25 \times 10^{-2}$$

One molal solution means one of solute present in one kg of solvent. $$\mathrm{MgCl_2} \rightleftharpoons \mathrm{Mg^{2+}} + 2 \mathrm{Cl^-}$$ $$\begin{array}{ccc} 1 & 0 & 0 \\ 1 - 0.8 & 0.8 & 1.6 \\ \end{array}$$ Hence, overall moles after dissociation $= 2.6$ The relative lowering of vapour pressure $$\frac{p^0 - p_s}{p^0} = \frac{moles of solute}{moles of solute + moles of solvent}$$ $$\frac{p^0 - p_s}{p^0} = \frac{2.6}{\frac{1000}{18} + 2.6}$$ For dilute solution $$\frac{p^0 - p_s}{p^0} = \frac{2.6}{\frac{1000}{18}}$$ $$p = 47.66 \approx 48 \, \mathrm{mm \, Hg}$$

The reaction of aldehydes and ketones with ammonia forms imine derivatives, also known as Schiff bases (compounds having a \(\mathrm{C=N}\) function). Imines can undergo nucleophilic addition reaction with KCN. Now, nitration takes place on benzene ring. Then amino group undergoes acylation reaction with acetic anhydride. Now, the carboxylic acid group undergoes esterification reaction with ethyl alcohol in acidic medium. Now, the nitro groups present on benzene undergo reduction to amines with \(\mathrm{H_2/Pd}\). Now, the amino groups undergo diazotization with nitrous acid. Then substitution of iodide takes place at diazonium groups upon reaction with \(\mathrm{NaI}\). \[ \therefore\ x = 15 \]

The root mean square velocity $V_{rms} = \sqrt{\frac{3RT}{M}}$. The most probable speed $V_{mp} = \sqrt{\frac{2RT}{M}}$. $T =$ Temperature in Kelvin. $M =$ Molar mass. $$\frac{V_{rms of X}}{V_{mp of Y}} = \sqrt{\frac{3T_X M_Y}{2T_Y M_X}}$$ $$1 = \sqrt{\frac{3 \times 600 \times M_Y}{2 \times 90 \times 40}} = \sqrt{\frac{M_Y}{4}}$$ $$M_Y = 4 \, gm \, mol^{-1}$$

On applying the law of mass action over the slowest step, $R = k[\mathrm{NOBr}_2][\mathrm{NO}]$ The slowest step is the rate-determining step. On applying the law of mass action on the equilibrium of Step I, $K = \frac{[\mathrm{NOBr}_2]}{[\mathrm{NO}][\mathrm{Br}_2]}$ $[\mathrm{NOBr}_2] = K[\mathrm{NO}][\mathrm{Br}_2]$ Hence, the overall rate equation will be $R = k[\mathrm{NO}][\mathrm{NOBr}_2]$ $= k[\mathrm{NO}]\left(K[\mathrm{NO}][\mathrm{Br}_2]\right)$ $= k'[\mathrm{NO}]^2[\mathrm{Br}_2]$ The overall order is equal to the sum of the powers of the concentration terms in the rate law. The overall order of the reaction is $3$.

The minimum amount of energy necessary to start the emission of electrons from a metal's surface is known as its work function. The work function is the minimum amount of energy required to remove an electron from a metal's surface to infinity. Energy of incident photon = $\frac{hc}{\lambda}$ $$= \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \, \mathrm{J}$$ $$= \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}} \, \mathrm{eV}$$ $$= 3.1 \, \mathrm{eV}$$ If radiation energy is greater than work function of metal, then metal show photo electric effect. Therefore, Li, Na and K will show photoelectric effect.