JEE Advanced 28 August 2022 Paper 2 question paper with solutions

Given $\alpha \in \left(0, \frac{\pi}{4}\right)$, $\beta \in \left(-\frac{\pi}{4}, 0\right)$, it follows that $\alpha + \beta \in \left(\frac{\pi}{4}, \frac{\pi}{4}\right)$. $\sin(\alpha + \beta) = \frac{1}{3}$, $\cos(\alpha - \beta) = \frac{2}{3}$. $$\left( \frac{\sin \alpha}{\cos \beta} + \frac{\cos \alpha}{\sin \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\sin \beta}{\cos \alpha} \right)^2$$ $$= 4 \cos^2(\alpha - \beta) \left( \frac{1}{\sin 2\beta} + \frac{1}{\sin 2\alpha} \right)^2$$ $$= 4 \cos^2(\alpha - \beta) \left( \frac{2 \sin(\alpha + \beta) \cos(\alpha - \beta)}{\sin 2\alpha \sin 2\beta} \right) \cdots (1)$$ $$= \frac{16 \cos^4(\alpha - \beta) \sin^2(\alpha + \beta) \times 4}{(\cos 2(\alpha - \beta) - \cos 2(\alpha + \beta))^2}$$ $$= \frac{64 \cos^4(\alpha - \beta) \sin^2(\alpha + \beta)}{(2 \cos^2(\alpha - \beta) - 1 - 1 + 2 \sin^2(\alpha + \beta))^2}$$ $$= 64 \times \frac{16}{81} \times \frac{1}{9} \times \frac{1}{\left(2 \times \frac{4}{9} - 1 - 1 + \frac{2}{9}\right)^2}$$ $$= \frac{64 \times 16}{81 \times 9} \times \frac{81}{64} = \frac{16}{9}$$ $$\left[ \frac{16}{9} \right] = 1 Ans.$$

Given $f(x) = \log_2(x^3 + 1) = y$. $x^3 + 1 = 2^y \Rightarrow x = (2^y - 1)^{1/3} = f^{-1}(y)$. $f^{-1}(x) = (2^x - 1)^{1/3}$. $$= \int_1^2 \log_2(x^3 + 1) \, dx + \int_1^{\log_2 9} (2^x - 1)^{1/3} \, dx$$ $$= \int_1^2 f(x) \, dx + \int_1^{\log_2 9} f^{-1}(x) \, dx = 2 \log_2 9 - 1$$ $$= 8 < 9 < 2^{7/2} \Rightarrow 3 < \log_2 9 < \frac{7}{2}$$ $$= 5 < 2 \log_2 9 - 1 < 6$$ $$[2 \log_2 9 - 1] = 5$$

Given $x^{16(\log_5 x)^3 - 68 \log_5 x} = 5^{-16}$. Take log to the base 5 on both sides and put $\log_5 x = t$. $$16t^4 - 68t^2 + 16 = 0$$ $$\Rightarrow 4t^4 - 17t^2 + 4 = 0$$ $$\begin{cases} t_1 \\ t_2 \\ t_3 \\ t_4 \end{cases}$$ $t_1 + t_2 + t_3 + t_4 = 0$. $\log_5 x_1 + \log_5 x_2 + \log_5 x_3 + \log_5 x_4 = 0$. $x_1 x_2 x_3 x_4 = 1$

Given $$\beta = \lim_{x \to 0} \frac{e^{x^3} - (1 - x^3)^{1/3}}{x \sin^2 x} + \frac{\left((1 - x^2)^{1/2} - 1\right) \sin x}{x^2}$$ Use expansion $$\beta = \lim_{x \to 0} \frac{\left(1 + x^3\right) - \left(1 - \frac{x^3}{3}\right)}{x^3} + \lim_{x \to 0} \frac{\left(\left(1 - \frac{x^2}{2}\right) - 1\right) \sin x}{x^2}$$ $$\beta = \lim_{x \to 0} \frac{4x^3}{3x^3} + \lim_{x \to 0} \frac{-x^2}{2x^2}$$ $$\beta = \frac{4}{3} - \frac{1}{2} = \frac{5}{6}$$ $$6\beta = 5$$

Given $A = \begin{pmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{pmatrix}$ with $|A| = -1$. It follows that $|A^7 - (\beta - 1)A^6 - \beta A^5| = 0$. Thus, $|A|^5 |A^2 - (\beta - 1)A - \beta I| = 0$. Therefore, $|A|^5 (|A^2 - \beta A) + A - \beta I| = 0$. This implies $|A|^5 |A(A - \beta I) + I(A - \beta I)| = 0$. Hence, $|A|^5 |(A + I)(A - \beta I)| = 0$. Now, $A + I = \begin{pmatrix} \beta + 1 & 0 & 1 \\ 2 & 2 & -2 \\ 3 & 1 & -1 \end{pmatrix} \implies |A + I| = -4$, Here $|A| \neq 0$ and $|A + I| \neq 0$. For $A - \beta I = \begin{pmatrix} 0 & 0 & 1 \\ 2 & 1 - \beta & -2 \\ 3 & 1 & -2 - \beta \end{pmatrix}$. We have $|A - \beta I| = 2 - 3(1 - \beta) = 3\beta - 1 = 0 \implies \beta = \frac{1}{3}$. Thus, $9\beta = 3$.

Given $S_1 P - SP = 20$. $$\beta - \frac{\delta}{\sin \frac{\alpha}{2}} = 20$$ $$\beta^2 + \frac{\delta^2}{\sin^2 \frac{\alpha}{2}} - 400 = \frac{2 \beta \delta}{\sin \frac{\alpha}{2}}$$ $$\frac{1}{SP} = \frac{\sin \frac{\alpha}{2}}{\delta}$$ $$\cos \alpha = \frac{SP^2 + \beta^2 - 656}{\frac{2 \beta \delta}{\sin \frac{\alpha}{2}}}$$ $$\frac{2 \beta \delta}{\sin \frac{\alpha}{2}} - 256 = \cos \alpha$$ $$\frac{2 \beta \delta}{\sin \frac{\alpha}{2}} = 2 \beta \delta$$ $$\frac{\lambda - 128}{\lambda} = \cos \alpha$$ $$\lambda (1 - \cos \alpha) = 128$$ $$\frac{\beta \delta}{\sin \frac{\alpha}{2}} \cdot 2 \sin^2 \frac{\alpha}{2} = 128$$ $$\frac{\beta \delta}{9} \sin \frac{\alpha}{2} = \frac{64}{9} \implies \left\lfloor \frac{\beta \delta}{9} \sin \frac{\alpha}{2} \right\rfloor = 7$$ where $\lfloor . \rfloor$ denotes greatest integer function.

Given the equation $x^2 + \frac{5}{12} = \frac{2 - 8x}{3}$. Simplifying, we have: $$x^2 + \frac{8x}{3} + \frac{5}{12} - 2 = 0$$ Solving the quadratic equation: $$12x^2 + 32x - 19 = 0$$ $$12x^2 + 38x - 6x - 19 = 0$$ $$2x(6x + 19) - 1(6x + 19) = 0$$ $$(6x + 19)(2x - 1) = 0$$ Thus, $x = \frac{1}{2}$. Now, calculate $\alpha = 2A_1 + A_2$ where: $$\alpha = 2 \left( \int_0^{1/2} x^2 + \frac{5}{12} \, dx + \frac{1}{2} \times \frac{4}{3} \right)$$ Evaluating the integral: $$\Rightarrow \alpha = 2 \left[ \left( \frac{x^3}{3} + \frac{5x}{12} \right)_0^{1/2} + \frac{1}{12} \right]$$ Simplifying further: $$\Rightarrow \alpha = 2 \left[ \frac{1}{24} + \frac{5}{24} + \frac{1}{12} \right]$$ $$\Rightarrow \alpha = 2 \left[ \frac{1 + 5 + 2}{24} \right] \Rightarrow \alpha = 2 \times \frac{8}{24} \Rightarrow 9\alpha = 9 \times \frac{8}{12}$$ Thus, $\Rightarrow 9\alpha = 6$

∠PRQ = 70^$\circ$ - 40^$\circ$ = 30^$\circ$ ∠RQS = 70^$\circ$ - 15^$\circ$ = 55^$\circ$ ∠QSR = 180^$\circ$ - 55^$\circ$ - 70^$\circ$ = 55^$\circ$ ∴ QR = RS = 1 ∠QPR = 180^$\circ$ - 70^$\circ$ - 30^$\circ$ = 80^$\circ$ Apply sine-rule in $\triangle$ PRQ : $$\frac{\alpha}{\sin 30^\circ} = \frac{1}{\sin 80^\circ} \implies \alpha = \frac{1}{2 \sin 80^\circ} \ldots (1)$$ Apply sine-rule in $\triangle$ PRS $$\frac{\beta}{\sin 40^\circ} = \frac{1}{\sin \theta} \implies \beta \sin \theta = \sin 40^\circ \ldots (2)$$ $$4 \alpha \beta \sin \theta = \frac{4 \sin 40^\circ}{2 \sin 80^\circ} = \frac{4 \sin 40^\circ}{2 (2 \sin 40^\circ \cos 40^\circ)}$$ = $\sec$ 40^$\circ$ Now $\sec$ 30^$\circ$ < $\sec$ 40^$\circ$ < $\sec$ 45^$\circ$ $$\implies \frac{2}{\sqrt{3}} < \sec 40^\circ < \sqrt{2}$$

Given $\alpha = \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^4 + \left( \frac{1}{2} \right)^6 + \ldots$ $$\alpha = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}$$ Therefore, $g(x) = 2^{x/3} + 2^{1/3(1-x)}$. Thus, $g(x) = 2^{x/3} + \frac{2^{1/3}}{2^{x/3}}$. Where $g(0) = 1 + 2^{1/3}$ and $g(1) = 1 + 2^{1/3}$. Therefore, $g'(x) = \frac{1}{3} \left( 2^{x/3} - \frac{2^{1/3}}{2^{x/3}} \right) = 0$. $$\Rightarrow 2^{2x/3} = 2^{1/3} \Rightarrow x = \frac{1}{2} = critical point$$ Therefore, the graph of $g'(x)$ is: $$\begin{array}{c} graph of g'(x) = \begin{cases} + & for x > \frac{1}{2} \\ - & for x < \frac{1}{2} \end{cases} \end{array}$$ And $g\left( \frac{1}{2} \right) = 2^{7/6}$. Therefore, the graph of $g(x)$ in $[0, 1]$ is: $$\begin{array}{c} (0, 1 + 2^{1/3}) (1, 1 + 2^{1/3}) \\ \left( \frac{1}{2}, 2^{7/6} \right) \end{array}$$

Let $\left( z \right)^2 + \frac{1}{z^2} = m + in$, $m, n \in \mathbb{Z}$. $$\left( z \right)^2 + \frac{z^2}{|z|^4} = m + in$$ $$\Rightarrow \left( x^2 - y^2 \right) \left( 1 + \frac{1}{|z|^4} \right) = m \ldots (1)$$ $$\& -2xy \left( 1 + \frac{1}{|z|^4} \right) = n \ldots (2)$$ Equation $(1)^2 + (2)^2$ $$\left( 1 + \frac{1}{|z|^4} \right)^2 \left[ \left( x^2 + y^2 \right)^2 \right] = m^2 + n^2$$ $$\left( 1 + \frac{1}{|z|^4} \right)^2 \left( |z|^4 \right) = m^2 + n^2$$ $$\Rightarrow |z|^4 + \frac{1}{|z|^4} + 2 = m^2 + n^2$$ Now for option (A) $$|z|^4 = \frac{43 + 3\sqrt{205}}{2}$$ $$\Rightarrow m^2 + n^2 = 45$$ $$\Rightarrow m = \pm 6, \; n = \pm 3$$ Option (B) $$|z|^4 + \frac{1}{|z|^4} + 2 = \frac{7 + \sqrt{33}}{4} + \frac{7 - \sqrt{33}}{4} + 2 = \frac{7}{2} + 2 = \frac{11}{2}$$ Option (C) $$|z|^4 + \frac{1}{|z|^4} + 2 = \frac{9 + \sqrt{65}}{4} + \frac{9 - \sqrt{65}}{4} + 2 = \frac{18}{4} + 2 = \frac{9}{2} + 2 = \frac{13}{2}$$ Option (D) $$|z|^4 + \frac{1}{|z|^4} + 2 = \frac{7 + \sqrt{13}}{6} + \frac{7 - \sqrt{13}}{6} + 2 = \frac{14}{6} + 2 = \frac{7}{3} + 2 = \frac{13}{2}$$

Given $2(R + r) \sin \frac{\pi}{n} = 2r$. Therefore, $\($ $\frac{R + r}{r}$ = $\csc$ $\frac{\pi}{n}$ $\)$. (A) $n = 4$, $R + r = \sqrt{2} \, r$ (B) $n = 5$, $\frac{R + r}{r} = \csc \frac{\pi}{5} R$ (C) $n = 8$, $\frac{R + r}{r} = \csc \frac{\pi}{8} > \csc \frac{\pi}{4}$ $R + r > \sqrt{2} \, r$ (D) $n = 12$, $\frac{R + r}{r} = \csc \frac{\pi}{12} = \sqrt{2}(\sqrt{3} + 1)$ $R + r = \sqrt{2}(\sqrt{3} + 1)r$ $\sqrt{2}(\sqrt{3} + 1)r > R$

Given $\mathbf{a} = 3\hat{i} + \hat{j} - \hat{k}$, $\mathbf{b} = \hat{i} + b_2 \hat{j} + b_3 \hat{k}$, $\mathbf{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}$. $\[$ $\begin{pmatrix}$ 0 & -c_3 & c_2 $\\$ c_3 & 0 & -c_1 $\\$ -c_2 & c_1 & 0 $\end{pmatrix}$ $\begin{pmatrix}$ 1 $\\$ b_2 $\\$ b_3 $\end{pmatrix}$ = $\begin{pmatrix}$ 3 - c_1 $\\$ 1 - c_2 $\\$ -1 - c_3 $\end{pmatrix}$ $\]$ Multiply and compare: $b_2 c_3 - b_3 c_2 = c_1 - 3$ ...(1) $c_3 - b_3 c_1 = 1 - c_2$ ...(2) $c_2 - b_2 c_1 = 1 + c_3$ ...(3) $(1) \hat{i} - (2) \hat{j} + (3) \hat{k}$ $\[$ $\hat{i}$ (b_2 c_3 - b_3 c_2) - $\hat{j}$ (c_3 - b_3 c_1) + $\hat{k}$ (c_2 - b_2 c_1) $\]$ $\[$ = c_1 $\hat{i}$ + c_2 $\hat{j}$ + c_2 $\hat{k}$ - 3 $\hat{i}$ - $\hat{j}$ + $\hat{k}$ $\]$ $\[$ $\mathbf{b}$ $\times$ $\mathbf{c}$ = $\mathbf{c}$ - $\mathbf{a}$ $\]$ Take dot product with $\mathbf{b}$: $0 = \mathbf{c} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{b}$ $\mathbf{c} \cdot \mathbf{b} = 0$ $\mathbf{b} \perp \mathbf{c}$ $\mathbf{b} \hat{c} = 90^\circ$ Take dot product with $\mathbf{c}$: $0 = |\mathbf{c}|^2 - \mathbf{a} \cdot \mathbf{c}$ $\mathbf{a} \cdot \mathbf{c} = |\mathbf{c}|^2$ $\mathbf{a} \cdot \mathbf{c} \neq 0$ $\mathbf{b} \times \mathbf{c} = \mathbf{c} - \mathbf{a}$ Squaring: $|\mathbf{b}|^2 |\mathbf{c}|^2 = |\mathbf{c}|^2 + 11 - 2 |\mathbf{c}|^2$ $|\mathbf{b}|^2 |\mathbf{c}|^2 = 11 - |\mathbf{c}|^2$ $|\mathbf{c}|^2 (|\mathbf{b}|^2 + 1) = 11$ $|\mathbf{c}|^2 = \frac{11}{|\mathbf{b}|^2 + 1}$ $|\mathbf{c}| \leq \sqrt{11}$ Given $\mathbf{a} \cdot \mathbf{b} = 0$ $b_2 - b_3 = -3$ also $b_2^2 + b_3^2 - 2b_2 b_3 = 9$ $b_2 b_3 > 0$ $b_2^2 + b_3^2 = 9 + 2b_2 b_3$ $b_2^2 + b_3^2 = 9 + 2b_2 b_3 > 9$ $b_2^2 + b_3^2 > 9$ $|\mathbf{b}| = \sqrt{1 + b_2^2 + b_3^2}$ $|\mathbf{b}| > \sqrt{10}$

Given $\($ $\frac{dy}{dx}$ + 12y = $\cos$$\left$($\frac{\pi}{12}$x$\right$) $\)$ Linear D.E. I.F. = $\($ e^{$\int$ 12 $\,$ dx} = e^{12x} $\)$ Solution of DE $\($ y.e^{12x} = $\int$ e^{12x} $\cdot$ $\cos$$\left$($\frac{\pi}{12}$x$\right$) $\,$ dx $\)$ $\($ y.e^{12x} = $\frac{e^{12x}}{(12)^2 + \left(\frac{\pi}{12}\right)^2}$ $\left$( 12 $\cos$$\left$($\frac{\pi}{12}$x$\right$) + $\frac{\pi}{12}$ $\sin$$\left$($\frac{\pi}{12}$x$\right$) $\right$) + C $\)$ $\($ $\Rightarrow$ y = $\frac{(12)}{(12)^4 + \pi^2}$ $\left$( (12)^2 $\cos$$\left$($\frac{\pi x}{12}$$\right$) + $\pi$ $\sin$$\left$($\frac{\pi x}{12}$$\right$) $\right$) + $\frac{C}{e^{12x}}$ $\)$ Given $\($ y(0) = 0 $\)$ $\($ $\Rightarrow$ 0 = $\frac{12}{12^4 + \pi^2}$ (12^2 + 0) + C $\Rightarrow$ C = $\frac{-12^3}{12^4 + \pi^2}$ $\)$ $\($ $\therefore$ y = $\frac{12}{12^4 + \pi^2}$ $\left$( (12)^2 $\cos$$\left$($\frac{\pi x}{12}$$\right$) + $\pi$ $\sin$$\left$($\frac{\pi x}{12}$$\right$) - 12^2.e^{-12x} $\right$) $\)$ Now $\($ $\frac{dy}{dx}$ = $\frac{12}{12^4 + \pi^2}$ $\left$[ -12 $\pi$ $\sin$$\left$($\frac{\pi x}{12}$$\right$) + $\frac{\pi^2}{12}$ $\cos$$\left$($\frac{\pi x}{12}$$\right$) + 12^3 e^{-12x} $\right$] $\)$ $\($ $\left$( -$\sqrt{\frac{144 \pi^2 + \pi^4}{144}}$ = -12 $\pi$ $\sqrt{1 + \frac{\pi^2}{12^4}}$ $\right$) $\)$ $\($ $\Rightarrow$ $\frac{dy}{dx}$ > 0 $\)$ for $\($ x $\leq$ 0 $\)$ and may be negative/positive for $\($ x > 0 $\)$ So, $\($ f(x) $\)$ is neither increasing nor decreasing For some $\($ $\beta$ $\in$ $\mathbb{R}$, y = $\beta$ $\)$ intersects $\($ y = f(x) $\)$ at infinitely many points So option C is correct

Case-I: when exactly one box provides four balls (3R 1B or 2R 2B). Number of ways in this case $^5C_4 \left(^3C_1 \times ^2C_1\right)^3 \times 4$. Case-II: when exactly two boxes provide three balls (2R 1B or 1R 2B) each. Number of ways in this case $\left(^5C_3 - 1\right)^2 \left(^3C_1 \times ^2C_1\right)^2 \times 6$. Required number of ways = 21816. Language ambiguity: If we consider at least one red ball and exactly one blue ball, then required number of ways is 9504. None of the option is correct.

Given $$M = \begin{bmatrix} \frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2} \end{bmatrix}$$ We have $$M = \begin{bmatrix} \frac{3}{2} + 1 & \frac{3}{2} \\ -\frac{3}{2} & -\frac{3}{2} + 1 \end{bmatrix}$$ Thus, $$M = I + \frac{3}{2} \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$$ Let $$A = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$$ Then $$A^2 = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$ Therefore, $$M^{2022} = \left( I + \frac{3}{2} A \right)^{2022}$$ This simplifies to $$= I + 3033A$$ Thus, $$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 3033 \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$$ Finally, $$= \begin{bmatrix} 3034 & 3033 \\ -3033 & -3032 \end{bmatrix}$$

Box I: 8(R) 3(B) 5(G) Box II: 24(R) 9(B) 15(G) Box III: 1(B) 12(G) 3(y) Box IV: 10(G) 16(o) 6(w) A (one of the chosen balls is white) B (at least one of the chosen balls is green) $$\mathbb{P}\left( \frac{A}{B} \right) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$$ $$A \cap B \rightarrow (wG)$$ $$= \frac{\frac{5}{16} \times \frac{6}{32}}{\frac{5}{16} \times 1 + \frac{8}{16} \times \frac{15}{48} + \frac{3}{16} \times \frac{12}{16}}$$ $$= \frac{15}{156} = \frac{5}{52}$$

Given $$f(n) = n + \sum_{r=1}^{n} \frac{16r + (9 - 4r)n - 3n^2}{4rn + 3n^2}$$ Simplifying, $$f(n) = n + \sum_{r=1}^{n} \frac{(16r + 9n) - (4rn + 3n^2)}{4rn + 3n^2}$$ This becomes, $$f(n) = n + \left( \sum_{r=1}^{n} \frac{16r + 9n}{4rn + 3n^2} \right) - n$$ Taking the limit as $n$ approaches infinity, $$\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{16r + 9n}{4rn + 3n^2}$$ This is equal to, $$= \lim_{n \to \infty} \sum_{r=1}^{n} \frac{16\left(\frac{r}{n}\right) + 9}{4\left(\frac{r}{n}\right) + 3} \cdot \frac{1}{n}$$ Evaluating the integral, $$= \int_{0}^{1} \frac{16x + 9}{4x + 3} \, dx = \int_{0}^{1} 4 \, dx - \int_{0}^{1} \frac{3 \, dx}{4x + 3}$$ This simplifies to, $$= 4 - \frac{3}{4} \left( \ln|4x + 3| \right)_{0}^{1}$$ Finally, $$= 4 - \frac{3}{4} \cdot \frac{7}{3}$$

Torque about origin is zero. So angular momentum about origin remains conserved. $$ \begin{vmatrix} \mathrm{i} & \mathrm{j} & \mathrm{k} \\ \frac{1}{\sqrt{2}} & \sqrt{2} & 0 \\ -\sqrt{2} & \sqrt{2} & \frac{2}{\pi} \end{vmatrix} = \begin{vmatrix} \mathrm{i} & \mathrm{j} & \mathrm{k} \\ x & y & 0.5 \\ v_x & v_y & \frac{2}{\pi} \end{vmatrix} $$ $$ \hat{\mathrm{i}} \left[ \sqrt{2} \times \frac{2}{\pi} \right] - \hat{\mathrm{j}} \left[ \frac{\sqrt{2}}{\pi} \right] + \hat{\mathrm{k}} [1 + 2] = \hat{\mathrm{i}} \left[ y \times \frac{2}{\pi} - 0.5 v_y \right] - \hat{\mathrm{j}} \left[ x \times \frac{2}{\pi} - 0.5 v_x \right] + \mathrm{k} \left[ x v_y - y v_x \right] $$ $$x v_y - y v_x = 3$$

The reaction is $\mathrm{Th}_{90}^{230} \rightarrow \mathrm{Po}_{84}^{214} + n \alpha_2^4 + m \beta_{-1}^0$. $230 = 214 + 4n$ $n = \frac{16}{4} = 4$ $90 = 84 + n \times 2 - m \times 1$ $90 = 84 + 4 \times 2 - m \times 1$ $m = 92 - 90 = 2$ Hence $\frac{n}{m} = \frac{4}{2} = 2$ Ans.

For the balanced Wheatstone bridge $$\frac{R_1}{R_2} = \frac{\int_0^{0.5} \frac{\rho \, dx}{\pi r_x^2}}{\int_{0.5}^1 \frac{\rho \, dx}{\pi r_x^2}}$$ $$\frac{R_1}{R_2} = \frac{\left[ \frac{1}{r_x} \right]_0^{0.5}}{\left[ \frac{1}{r_x} \right]_{0.5}^1}$$ Therefore, $R_1 = 5R_2 = 5\, \Omega$

Given $\mathbf{B} = e^{\alpha} (m_e)^{\beta} h^{\gamma} k^{\delta}$. $$[\mathbf{B}] = \left[ e^{\alpha} \right] [m_e]^{\beta} [h]^{\gamma} [k]^{\delta}$$ $$[\mathbf{M}^{-1} \mathbf{T}^{-2} \mathbf{A}^{-1}] = [\mathbf{A} \mathbf{T}]^{\alpha} [\mathbf{m}]^{\beta} \left[ \mathbf{ML}^{2} \mathbf{T}^{-1} \right]^{\gamma} \left[ \mathbf{ML}^{3} \mathbf{A}^{-2} \mathbf{T}^{-4} \right]^{\delta}$$ $$\mathbf{M}^{-1} \mathbf{T}^{-2} \mathbf{A}^{-1} = \mathbf{m}^{\beta + \gamma + \delta} \mathbf{L}^{2\gamma + 3\delta} \mathbf{T}^{\alpha - \gamma - 4\delta} \mathbf{A}^{\alpha - 2\delta}$$ Compare: $\beta + \gamma + \delta = 1$; $2\gamma + 3\delta = 0$, $\alpha - \gamma - 4\delta = -2$, $\alpha - 2\delta = -1$. On solving $\alpha = 3$, $\beta = 2$, $\gamma = -3$, $\delta = 2$. $\alpha + \beta + \delta = 4$

Given $1 \sin 60^\circ = \frac{\sqrt{3}}{2} \sin \theta$. Therefore, $\theta_1 = 45^\circ$. $$\frac{\sqrt{3}}{2} \sin 45^\circ = \sqrt{3} \sin \theta_2$$ $$= \frac{\sqrt{3}}{2} \frac{1}{\sqrt{2}} = \sqrt{3} \sin \theta_2$$ $$= \theta_2 = 30^\circ$$ $h \tan 60^\circ + d \tan 45^\circ + d \tan 30^\circ$ $$\frac{1}{3} \sqrt{3} + \left(\frac{\sqrt{3} - 1}{2}\right) + \left(\frac{\sqrt{3} - 1}{2}\right) \frac{1}{\sqrt{3}}$$ $$\frac{2\sqrt{3} + 3\sqrt{3} - 3 + 3 - \sqrt{3}}{6}$$ $$\frac{4\sqrt{3}}{6}$$ Therefore, $n \cdot \frac{4\sqrt{3}}{6} = \frac{8}{\sqrt{3}}$. $$n = 4$$

From Gauss's law, $$\phi_{hemisphere} + \phi_{cone} = \frac{q}{\varepsilon_0}$$ Total flux produced from $q$ in $\alpha$ angle $$\phi = \frac{q}{2\varepsilon_0} [1 - \cos \alpha]$$ For hemisphere, $\alpha = \frac{\pi}{2}$ $$\phi_{hemisphere} = \frac{q}{2\varepsilon_0}$$ From equation (i) $$\frac{q}{2\varepsilon_0} + \phi_{cone} = \frac{q}{\varepsilon_0}$$ $$\phi_{cone} = \frac{q}{2\varepsilon_0}$$ $$\frac{4q}{6\varepsilon_0} = \frac{q}{2\varepsilon_0}$$ $n = 3$ Alternatively, $\phi \propto$ number of electric field lines passing through surface. $q$ is a point charge which has uniformly distributed electric field lines, thus half of electric field lines will pass through hemisphere and other half will pass through conical surface.

If $\ell > \ell_0$, then $k = k_1$. If $\ell < \ell_0$, then $k = k_2$. The time period of oscillation is given by $$T = \pi \sqrt{\frac{m}{k_1}} + \pi \sqrt{\frac{m}{k_2}}.$$ Substituting the values, we have $$T = \pi \sqrt{\frac{0.1}{0.009}} + \pi \sqrt{\frac{0.1}{0.016}}.$$ This simplifies to $$T = \frac{\pi}{0.3} + \frac{\pi}{0.4} \implies T = \frac{0.7}{0.12} \pi = 5.83\pi.$$ Therefore, $T \approx 6\pi$. So, $n = 6$.

Let $u = -30 \, \mathrm{cm}$ and $f = -10 \, \mathrm{cm}$. Then $v = \frac{fu}{u-f} = -15 \, \mathrm{cm}$. The lens formula is given by: $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ Differentiating with respect to time $t$: $$\frac{du}{dt} = -\frac{v^2}{u^2} \frac{dv}{dt}$$ The velocity of the image is: $$\vec{v}_m = -\left( \frac{v}{u} \right)^2 \vec{v}_o$$ Given $\vec{v}_i = \vec{0}$, we have: $$\vec{v}_i - \vec{v}_m = -\left( \frac{-15}{-30} \right)^2 (\vec{v}_{o/m})$$ Thus: $$\vec{v}_i - \vec{v}_m = -\frac{1}{4} \vec{v}_o + \frac{1}{4} \vec{v}_m$$ Given $\vec{v}_o = 15 \, \mathrm{cm/s} \, \hat{i}$ and $\vec{v}_i = 0 \, \mathrm{cm/s}$, we find: $$\frac{5}{4} \vec{v}_m = \frac{\vec{v}_o}{4}$$ Therefore: $$\vec{v}_m = \frac{\vec{v}_o}{4} = \frac{15 \, \mathrm{cm/s} \, \hat{i}}{5} = 3 \, \mathrm{m/s} \, \hat{i}$$ The magnitude of $\vec{v}_m$ is $3 \, \mathrm{cm/s} = 3$.

Given $q_1 = \int_0^r kr 4 \pi r'^2 \, dr' = \frac{4 \pi k}{4} = \pi k$. $q_2 = \int_1^r \frac{2k}{r} 4 \pi r'^2 \, dr' = \frac{8 \pi k \left( r^2 - 1^2 \right)}{2}$. $q_2 = 4 \pi k [r^2 - 1] = 4 \pi k r^2 - 4 \pi k$. $q_{net} = q_1 + q_2 = 4 \pi k r^2 - 3 \pi k$. $q_{net} = \pi k \left[ 4r^2 - 3 \right]$. (A) $E_{net} = 0 \Rightarrow q_{net} = 0 \Rightarrow r = \frac{\sqrt{3}}{2}$. (B) $V = \frac{k Q_{net}}{r} = \frac{1}{4 \pi \varepsilon_0} \frac{\pi k \left( 4r^2 - 3 \right)}{r}$. $V = \frac{k}{4 \varepsilon_0} \left[ 4r - \frac{3}{r} \right]$. $= \frac{k}{4 \varepsilon_0} \left[ 4 \times \frac{3}{2} - \frac{3 \times 2}{3} \right] = \frac{k}{\varepsilon_0}$. (C) $q_{net} = \pi k \left[ 4 \left( 2 \right)^2 - 3 \right] = 13 \pi k$. (D) $E_2 = \frac{kQ}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\pi k \left( 4r^2 - 3 \right)}{r^2}$. $= \frac{k}{4 \varepsilon_0} \left[ \frac{4 \left( \frac{5}{2} \right)^2 - 3}{\left( 5/2 \right)^2} \right]$. $= \frac{k}{25 \varepsilon_0} \left[ 25 - 3 \right] = \frac{22 \times k}{25 \varepsilon_0}$.

Case (i) When both switches are open equivalent resistance in circuit 1 $$R_{c_1} = \frac{16}{11} \, \Omega$$ Equivalent resistance in circuit 2 $$R_{c_2} = \frac{6}{11} \, \Omega$$ For voltage source $$P = \frac{V^2}{R}$$ $$P \propto \frac{1}{R}$$ Since $R_{c_1} > R_{c_2}$ $$\Rightarrow P_2 > P_1 (Option (A) correct)$$ For constant current source $$P = i^2 R$$ $$P \propto R$$ $$\Rightarrow P_1 > P_2 (Option (B) correct)$$ Case-II When switch is closed $$R'_{c_1} = \frac{5}{11} \, \Omega$$ $$R'_{c_2} = \frac{1}{2} \, \Omega$$ Since $R'_{c_1} P_1 (Option (C) correct)$$ And $R'_{c_1} > R'_{c_2}$ For current source $P \propto R$ $$Q_1 > Q_2 (Option (D) also correct)$$

The pressure of the gas is given by the equation $$P_{gas} = P_a + \frac{4S}{r}$$. For an adiabatic process, we have $$PV^\gamma = constant$$. Therefore, $$\left( P_{a_1} + \frac{4S}{r_1} \right) \left( \frac{4}{3} \pi r_1^3 \right)^{5/3} = \left( P_{a_2} + \frac{4S}{r_2} \right) \left( \frac{4}{3} \pi r_2^3 \right)^{5/3}$$. This implies $$\frac{r_1^3}{r_2^3} = \frac{\left( P_{a_2} + \frac{4S}{r_2} \right)}{\left( P_{a_1} + \frac{4S}{r_1} \right)}$$. For the process $$P^{1-\gamma} T^\gamma = constant$$, we have $$\left( P_{a_2} + \frac{4S}{r_2} \right)^{1-5/3} T_2^{5/3} = \left( P_{a_1} + \frac{4S}{r_1} \right)^{1-5/3} T_1^{5/3}$$. Therefore, $$\left( \frac{T_2}{T_1} \right)^{5/3} = \frac{\left( P_{a_1} + \frac{4S}{r_1} \right)}{\left( P_{a_2} + \frac{4S}{r_2} \right)}$$. Simplifying, $$\left( \frac{T_2}{T_1} \right)^{5/2} = \frac{\left( P_{a_2} + \frac{4S}{r_2} \right)}{\left( P_{a_1} + \frac{4S}{r_1} \right)}$$. Therefore, option (D) is correct.

The work-energy principle is given by $W_{el} + W_{ext} = k_f - k_i$. Substituting the values, we have $qv_i - qv_f + W_{ext} = k_f - k_i$. The expression becomes: $$\frac{q\sigma}{2\epsilon_0} \left[ \sqrt{R^2 + Z^2} - Z \right] - \frac{q\sigma R}{2\epsilon_0} + CZ = k_f - 0$$ Where $C = \frac{q\theta B}{2\epsilon_0}$. Substitute $\beta$ and $Z$, calculate kinetic energy at $z = 0$. If kinetic energy is positive, then the particle will reach the origin. If kinetic energy is negative, then the particle will not reach the origin.

Optical path difference $$\Delta x = n_1 (d \tan \theta) \sin \alpha - n_2 (d \tan \theta) \sin \theta$$ $$= (n_1 \sin \alpha - n_2 \sin \theta) d \tan \theta$$ $$= 0$$ $$\Rightarrow \Delta \phi = 0$$ Ans. (A,B)

For adiabatic process (A → B) $$P_A V_A^\gamma = P_B V_B^\gamma$$ $$10^5 \times (0.8)^{\frac{5}{3}} = 3 \times 10^5 \left(V_B\right)^{\frac{5}{3}}$$ $$\Rightarrow V_B = 0.8 \times \left(\frac{1}{3}\right)^{0.6} = 0.4$$ Work done in process A → B $$W_{AB} = \frac{P_A V_A - P_B V_B}{\gamma - 1}$$ $$\Rightarrow W_{AB} = \frac{10^5 \times 0.8 - 3 \times 10^5 \times 0.4}{\frac{5}{3} - 1}$$ $$\Rightarrow W_{AB} = -60 \, kJ \Rightarrow |W_{AB}| = 60 \, kJ$$ Work done in process B → C (Isothermal process) $$W_{BC} = nRT \ln \frac{V_C}{V_B} = P_B V_B \ln \frac{V_C}{V_B}$$ $$\Rightarrow W_{BC} = 3 \times 10^5 \times 0.4 \ln \frac{0.8}{0.4}$$ $$\Rightarrow W_{BC} = 84 \, kJ$$ Work done in process C → A $$W_{CA} = P \Delta V = 0 (\because \Delta V = 0)$$ So total work done in the process A→B→C $$W_{ABC} = W_{AB} + W_{BC} + W_{CA} = -60 + 84 + 0$$ $$W_{ABC} = 24 \, kJ$$ So correct options are (B, C, D)

Given $v = \omega (2R)$. $v = \omega_0 R$: no slipping. Therefore, $\omega_0 = 2\omega$. $$\vec{L} = m \vec{r} \times \vec{v}_c + I_c \omega_0$$ $$= M 2R v + \frac{1}{2} MR^2 \omega_0$$ $$= 4MR^2 \omega + \frac{1}{2} MR^2 (2\omega) = 5MR^2 \omega$$ Therefore, $n = 5$

Given $\($ $\frac{hc}{\lambda}$ = $\phi$ + 6 $\)$ $\ldots$ (i) $\($ $\frac{hc}{4\lambda}$ = $\phi$ + 0.6 $\)$ $\ldots$ (ii) $\($ $\frac{3hc}{4\lambda}$ = 5.4 $\,$ eV $\)$ $\therefore$ $\phi$ = 1.2 $\,$ eV $\)$ $\($ $\Rightarrow$ $\frac{3}{4}$ $\times$ $\frac{6.63 \times 10^{-24} \times 3 \times 10^{8}}{5.4 \times 1.6 \times 10^{-19}}$ = $\lambda$ = 1.72 $\times$ 10^{-7} $\,$ m $\)$

LC = $\frac{0.1}{100}$ = 0.001 $\,$ $\mathrm{mm}$ Zero error = 4 $\times$ 0.001 = 0.004 $\,$ $\mathrm{mm}$ Reading 1 = 0.5 $\times$ 4 + 20 $\times$ 0.001 - 0.004 = 2.16 $\,$ $\mathrm{mm}$ Reading 2 = 0.5 $\times$ 4 + 16 $\times$ 0.001 - 0.004 = 2.12 $\,$ $\mathrm{mm}$ Mean value = 2.14 $\,$ $\mathrm{mm}$ Mean absolute error = $\frac{0.02 + 0.02}{2}$ = 0.02 Diameter = 2.14 $\pm$ 0.02 Area = $\frac{\pi}{4}$ d^2

The magnetic field $\vec{B}$ is given by: $$\vec{B} = \frac{\mu_0 I}{4 \pi L} \sin 45^\circ (-\hat{k}) + \frac{\mu_0 I \pi}{4 \pi} \left( \frac{1}{L/2} \right) (-\hat{k}) + \frac{\mu_0 I}{4 \pi} \times \frac{\pi}{2} (-\hat{k})$$

The reaction of $\mathrm{H_2SO_4}$ is as follows: $$\mathrm{H_2SO_4} \rightleftharpoons \mathrm{HSO_4^-} + \mathrm{H^+}$$ Initially, the concentrations are $1 \, \mathrm{M}$ for $\mathrm{H_2SO_4}$ and $0 \, \mathrm{M}$ for $\mathrm{HSO_4^-}$ and $\mathrm{H^+}$. After dissociation, the concentrations are $1 \, \mathrm{M}$ for $\mathrm{HSO_4^-}$ and $\mathrm{H^+}$. For $\mathrm{Na_2SO_4}$: $$\mathrm{Na_2SO_4} \rightarrow 2\mathrm{Na^+} + \mathrm{SO_4^{2-}}$$ The initial concentration is $1.8 \times 10^{-2} \, \mathrm{M}$, leading to $3.6 \times 10^{-2} \, \mathrm{M}$ for $\mathrm{Na^+}$ and $1.8 \times 10^{-2} \, \mathrm{M}$ for $\mathrm{SO_4^{2-}}$. For $\mathrm{HSO_4^-}$: $$\mathrm{HSO_4^-} \rightleftharpoons \mathrm{H^+} + \mathrm{SO_4^{2-}}; K_{a_2} = 1.2 \times 10^{-2} \, \mathrm{M}$$ Initially, $1 \, \mathrm{M}$ for $\mathrm{HSO_4^-}$ and $\mathrm{H^+}$, and $1.8 \times 10^{-2} \, \mathrm{M}$ for $\mathrm{SO_4^{2-}}$. Since $Q_c > K_c$, it will move in the backward direction. The concentrations become $1 + x$ for $\mathrm{HSO_4^-}$, $1 - x$ for $\mathrm{H^+}$, and $1.8 \times 10^{-2} - x$ for $\mathrm{SO_4^{2-}}$. The equilibrium constant is: $$K_{a_2} = 1.2 \times 10^{-2} = \frac{(1-x)(1.8 \times 10^{-2} - x)}{(1+x)}$$ Since $x$ is very small, $(1 + x) \simeq 1$ and $(1 - x) \simeq 1$. Thus, $x = (1.8 \times 10^{-2} - 1.2 \times 10^{-2}) \, \mathrm{M}$. The concentration of $\mathrm{SO_4^{2-}}$ is: $$[\mathrm{SO_4^{2-}}] = (1.8 \times 10^{-2} - 0.6 \times 10^{-2}) \, \mathrm{M} = 1.2 \times 10^{-2} \, \mathrm{M}$$ For $\mathrm{PbSO_4}$: $$\mathrm{PbSO_4} \rightarrow \mathrm{Pb^{2+}} + \mathrm{SO_4^{2-}}$$ The concentration of $\mathrm{Pb^{2+}}$ is $s$ and $1.2 \times 10^{-2} \, \mathrm{M}$ for $\mathrm{SO_4^{2-}}$. The concentration of $\mathrm{PbSO_4}$ is $(s + 1.2 \times 10^{-2})$. The solubility product is: $$K_{sp} = s(s + 1.2 \times 10^{-2}) = 1.6 \times 10^{-8} (\mathrm{PbSO_4})$$ Here, $(s + 1.2 \times 10^{-2}) \simeq 1.2 \times 10^{-2}$ (since 's' is very small). Thus, $s(1.2 \times 10^{-2}) = 1.6 \times 10^{-8}$. Therefore, $s = \frac{1.6}{1.2 \times 10^{-6}} \mathrm{M} = X \times 10^{-Y} \mathrm{M}$. Thus, $Y = 6$.

0.1 mole ionic salt in 1.8 kg water at 35° C. Vapour pressure of solution = 59.724 $\mathrm{mm}$ $\mathrm{Hg}$. Vapour pressure of pure $\mathrm{H_2O}$ = 60.000 $\mathrm{mm}$ $\mathrm{Hg}$. Let the number of ions present per formula unit of the ionic salt be 'x'. $$\begin{array}{ccc} \mathrm{A_x} & \longrightarrow & \mathrm{xA} \\ (Salt) & & (Ions) \\ 0.1 & - & \\ 0.1 \left(1 - 0.9\right) & & \left(0.1 \times 0.9\right) x \end{array}$$ Total moles of non-volatile particles = 0.01 + 0.09 x in 1.8 kg water. Moles of water = $\frac{1.8 \times 10^3}{18}$ = 100 moles. Relative lowering of vapour pressure $\frac{P^o - P_s}{P^o}$ = Mole fraction of non-volatile particles. $$\frac{P^o - P_s}{P_s} = \frac{moles of non-volatile particles}{moles of water}$$ $$\frac{60.000 - 59.724}{59.724} = \frac{0.01 + 0.09x}{100}$$ $$(0.276) \times 100 = 0.59274 + (0.59274 \times 9)x$$ $$27.6 - 0.59274 = (0.59274 \times 9)x$$ $$\Rightarrow x \cong \frac{27}{0.6 \times 9} = 5$$

Given $\Lambda^\circ(U_m^p Y_p^\circ) = m \times \lambda^\circ_{Up} + p \times \lambda^\circ_{Ym} = 250$. $25m + 100p = 250$ $m + 4p = 10$ $\hspace{0.5cm}$ ......(1) $\Lambda^\circ(V_m^n X_n^\circ) = m \times \lambda^\circ_{Vn} + n \times \lambda^\circ_{Xm} = 440$ $100m + 80n = 440$ $5m + 4n = 22$ $\hspace{0.5cm}$ ......(2) From the extrapolation of curve $\Lambda^\circ(Z_m^n X_n) = 340$ $m \times \lambda^\circ_{Zp} + n \lambda^\circ_{Xm} = 340$ $50m + 80n = 340$ $5m + 8n = 34$ $\hspace{0.5cm}$ .......(3) (3) $-$ (2) $\Rightarrow$ $4n = 12 \Rightarrow n = 3$ Putting in (2) we get $m = 2$ Putting in (1) we get $p = 2$ $m + n + p = 2 + 3 + 2 = 7$

Xe + 2$\mathrm{O_2F_2}$ $\rightarrow$ $\mathrm{XeF_4}$ + 2$\mathrm{O_2}$ $$3\mathrm{XeF_4} + 6\mathrm{H_2O} \rightarrow 2\mathrm{Xe} + \mathrm{XeO_3} + \frac{3}{2}\mathrm{O_2} + 12\mathrm{HF}$$ Therefore, one mole of $\mathrm{XeF_4}$ gives 4 moles of $\mathrm{HF}$ on hydrolysis.

Given the reaction $\mathrm{AgNO_3} \rightarrow 2\mathrm{Ag} + 2\mathrm{NO_2} + \frac{1}{2}\mathrm{O_2}$. Both $\mathrm{NO_2}$ and $\mathrm{O_2}$ are paramagnetic. $\mathrm{NO_2}$ is an odd electron molecule with one unpaired electron. $\mathrm{O_2}$ has two unpaired electrons. Total number of antibonding electrons is 6.

The reaction sequence involves the following steps: 1. The starting compound is treated with sodium in liquid ammonia. 2. The resulting compound is then reacted with excess $\mathrm{Br_2}$. 3. Finally, the compound undergoes elimination with alcoholic KOH to form two isomers, cis and trans.

The given alkene undergoes ozonolysis with $\mathrm{O_3}$ and $\mathrm{Zn/H_2O}$ to form an aldehyde. The aldehyde is then oxidized by $\mathrm{KMnO_4}$ to form a carboxylic acid. This carboxylic acid undergoes Kolbe electrolysis in the presence of $\mathrm{NaOH}$ to form an alkane. Finally, the alkane is dehydrogenated using $\mathrm{Cr_2O_3}$ at $770 \, \mathrm{K}$ and $20 \, \mathrm{atm}$ to form benzene.

The compound (P) undergoes ozonolysis with $\mathrm{O_3/Zn/H_2O}$, resulting in the formation of several products. The products are analyzed for chirality. The first product is achiral. The second product is achiral. The third product is chiral. The fourth product is achiral. The fifth product is chiral. The sixth product is achiral.

For option (A) Let atomic mass of P be $M_P$ and atomic mass of Q be $M_Q$. Molar ratio of atoms P : Q in compound 3 is $$\frac{40}{M_P} : \frac{60}{M_Q} = 3 : 4$$ $$\frac{2M_Q}{3M_P} = \frac{3}{4} \Rightarrow 9M_P = 8M_Q$$ Molar ratio of atoms P : Q in compound 2 is $$\frac{44.4}{M_P} : \frac{55.6}{M_Q}$$ $$= 44.4 \, M_Q : 55.6 \, M_P$$ $$= 44.4 \, M_Q : 55.6 \times \frac{8M_Q}{9}$$ $$= 44.4 : 55.6 \times \frac{8}{9}$$ $$= 9 : 10$$ $\($$\Rightarrow$$\)$ Empirical formula of compound 2 is therefore $P_9Q_{10}$. Option (A) is incorrect. For option (B) Molar Ratio of atoms P : Q in compound 3 is $$\frac{40}{M_P} : \frac{60}{M_Q} = 3 : 2$$ $$\frac{2M_Q}{3M_P} = \frac{3}{2} \Rightarrow 9M_P = 4M_Q$$ If $M_P = 20$ $\($$\Rightarrow$$\)$ $$M_Q = \frac{9 \times 20}{4} = 45$$ Option (B) is correct. For option (C) Molar ratio of atoms P : Q in compound 2 is $$\frac{44.4}{M_P} : \frac{55.6}{M_Q} = 44.4M_Q : 55.6M_P = 1 : 1$$ $\($$\Rightarrow$$\)$ $$\frac{M_P}{M_Q} = \frac{44.4}{55.6}$$ Molar ratio of atoms P : Q in compound 1 is $$\frac{50}{M_P} : \frac{50}{M_Q} = M_Q : M_P$$ $$= 55.6 : 44.4$$ $$\simeq 5 : 4$$ Hence, empirical formula of compound 1 is $P_5Q_4$. Hence, option (C) is correct. For option (D) Molar ratio of atoms P : Q in compound 1 is $$\frac{50}{M_P} : \frac{50}{M_Q} = M_Q : M_P$$ $$= 35 : 70 = 1 : 2$$ Hence, empirical formula of compound 1 is $PQ_2$. Hence, option (D) is incorrect.

Given $\Delta G = \Delta H - T \Delta S$. $$\Delta G = \Delta H + T \left( \frac{d \Delta G}{dT} \right)_p$$ $$-nF \left( \frac{dE_{cell}}{dT} \right) = -\Delta S$$ $$\frac{dE_{cell}}{dT} = \frac{\Delta S}{nF} = \frac{R}{F} (given)$$ $$\Rightarrow \Delta S = nR$$ For the reaction, $M(g) + 2H^\oplus(aq) \longrightarrow H_2(g) + M^{2\oplus}(aq)$ $n = 2$ $$\Rightarrow \Delta S = 2R$$ Hence, option (A) is incorrect. For the reaction, $Pt_{(s)} | H_2(g), 1 bar| H^\oplus_{aq}(0.01M)|| H^\oplus_{aq}, 0.1M) | H_2(g, 1 bar)| Pt_{(s)}$ $$E_{cell} = E^\circ_{cell} - \frac{0.0591}{1} \log \frac{0.01}{0.1} = 0.0591 V$$ $E_{cell}$ is positive $\Rightarrow \Delta G 0$ ($\Delta H = 0$ for concentration cells) Hence, option (B) is correct. Racemization of an optically active compound is a spontaneous process. Here, $\Delta H = 0$ (similar type of bonds are present in enantiomers) $$\Rightarrow \Delta S > 0$$ Hence, option (C) is correct. $$\left[ Ni(H_2O)_6 \right]^{2+} + 3 en \rightarrow \left[ Ni(en)_3 \right]^{2+} + 6H_2O$$ is a spontaneous process More stable complex is formed $$\Rightarrow \Delta S > 0$$

(A) $2\mathrm{B} + 2\mathrm{NH}_3 \rightarrow 2\mathrm{BN} + 3\mathrm{H}_2$ Boron produced BN with ammonia but Boron is element not compound. So that this option not involve in answer. (B) $3\mathrm{B}_2\mathrm{H}_6 + 6\mathrm{NH}_3 \rightarrow 3[\mathrm{BH}_2(\mathrm{NH}_3)_2]^+[\mathrm{BH}_4]^-$ $$\xrightarrow{T = 200^\circ\mathrm{C}} 2\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6 + 12\mathrm{H}_2$$ $$\xrightarrow{T > 200^\circ\mathrm{C}} (\mathrm{BN})_x$$ (C) $\mathrm{B}_2\mathrm{O}_3(\ell) + 2\mathrm{NH}_3 \xrightarrow{1200^\circ\mathrm{C}} 2\mathrm{BN}_{(s)} + 3\mathrm{H}_2\mathrm{O}_{(g)}$ (D) $\mathrm{HBF}_4 + \mathrm{NH}_3 \rightarrow \mathrm{NH}_4[\mathrm{BF}_4]$

(A) $\mathrm{CaO} + \mathrm{SiO_2} \rightarrow \mathrm{CaSiO_3}$ (in the temperature range 900 – 1500 K) (B) In fusion zone molten iron becomes heavy by absorbing elemental impurities and produces Pig iron. (in the temperature range 900 – 1500 K) (C) $\mathrm{C} + \mathrm{CO_2} \rightarrow 2\mathrm{CO}$ (in the temperature range 900 – 1500 K) (D) Exhaust gases does not contain $\mathrm{NO_2}$.

The reaction starts with benzene and an acid chloride in the presence of $\mathrm{AlCl_3}$ to form compound (P), which is a carboxylic acid. Compound (P) is then reduced using $\mathrm{Zn/Hg}$ and $\mathrm{HCl}$ to form compound (Q), another carboxylic acid. Compound (Q) is treated with $\mathrm{SOCl_2}$ to form compound (R), an acid chloride. Compound (R) undergoes a Friedel-Crafts acylation with $\mathrm{AlCl_3}$ to form compound (S), which is a ketone. Finally, compound (R) reacts with $\mathrm{R_2Cd}$ to form the final product.

(a) The polymerisation of neoprene gives natural rubber. (b) is correct statement (c) is correct statement (d) Ethene at $350$-$570 \, \mathrm{K}$ temperature and $1000$-$2000 \, \mathrm{atm}$ pressure in the presence of a peroxide initiator yields low density polythene.

Atom 'X' occupies FCC lattice points as well as alternate tetrahedral voids of the same lattice. Therefore, $\frac{1}{4}$th distance of body diagonal is $\frac{\sqrt{3}a}{4} = 2r_x$. Thus, $a = \frac{8r_x}{\sqrt{3}}$. The number of atoms of X per unit cell is $4 + 4 = 8$ (FCC lattice points and alternate tetrahedral voids). The percentage packing efficiency is given by: $$Percentage packing efficiency = \frac{Volume occupied by X}{Volume of cubic unit cell} \times 100$$ $$= \frac{8 \times \frac{4}{3} \pi (r_x)^3}{a^3} \times 100$$ $$= \frac{8 \times \frac{4}{3} \pi (r_x)^3}{\left(\frac{8r_x}{\sqrt{3}}\right)^3} \times 100$$ $$= \left(8 \times \frac{4}{3} \times \pi \times \frac{1}{8^3} \times 3\sqrt{3}\right) \times 100$$ $$= \frac{\sqrt{3}\pi}{16} \times 100$$ $$= 34\%$$ Hence, option (B) is the most appropriate option.

The reaction is as follows: $$\mathrm{HClO_3 + HCl \rightarrow ClO_2 \,(Paramagnetic) + \frac{1}{2}Cl_2 + H_2O}$$ The subsequent reaction is: $$\mathrm{2ClO_2 + 2O_3 \rightarrow Cl_2O_6 + 2O_2}$$

The reaction is as follows: $$\mathrm{Pb(NO_3)_2 + 2NaCl \rightarrow PbCl_2 + 2NaNO_3}$$ In the presence of excess HCl, the following complex is formed: $$\mathrm{[PbCl_4]^{2-}}$$