JEE Advanced 28 August 2022 Paper 1 question paper with solutions

Given $\cos^{-1} \left( \frac{\sqrt{2}}{2 + \pi^2} \right) = \tan^{-1} \frac{\pi}{\sqrt{2}}$. Using the triangle, $\theta = \sin^{-1} \left( \frac{\sqrt{2\sqrt{2}\pi}}{2 + \pi^2} \right) = \sin^{-1} \left( \frac{2 \times \frac{\pi}{\sqrt{2}}}{1 + \left( \frac{\pi}{\sqrt{2}} \right)^2} \right) = \pi - 2 \tan^{-1} \left( \frac{\pi}{\sqrt{2}} \right)$. As $\sin^{-1} \left( \frac{2x}{1 + x^2} \right) = \pi - 2 \tan^{-1} x, x \geq 1$ and $\tan^{-1} \frac{\sqrt{2}}{\pi} = \cot^{-1} \left( \frac{\pi}{\sqrt{2}} \right)$. Therefore, the expression is $$ \frac{3}{2} \left( \tan^{-1} \frac{\pi}{\sqrt{2}} \right) + \frac{1}{4} \left( \pi - 2 \tan^{-1} \frac{\pi}{\sqrt{2}} \right) + \cot^{-1} \left( \frac{\pi}{\sqrt{2}} \right) $$ $$ = \left( \frac{3}{2} - \frac{2}{4} \right) \tan^{-1} \frac{\pi}{\sqrt{2}} + \frac{\pi}{4} + \cot^{-1} \frac{\pi}{\sqrt{2}} $$ $$ = \left( \tan^{-1} \frac{\pi}{\sqrt{2}} + \cot^{-1} \frac{\pi}{\sqrt{2}} \right) + \frac{\pi}{4} $$ $$ = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} $$ $$ = 2.35 or 2.36 $$

Given $\lim_{x \to \alpha^+} \frac{2 \ln \left( \sqrt{x} - \sqrt{\alpha} \right)}{\ln \left( e^{\sqrt{x}} - e^{\sqrt{\alpha}} \right)} \left( \frac{0}{0} form \right)$. Using L'Hopital's rule, $$= 2 \lim_{x \to \alpha^+} \frac{\left( \frac{1}{\sqrt{x} - \sqrt{\alpha}} \right) \cdot \frac{1}{2\sqrt{x}}}{\left( \frac{1}{e^{\sqrt{x}} - e^{\sqrt{\alpha}}} \right) \cdot e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}}$$ $$= \frac{2}{e^{\sqrt{\alpha}}} \lim_{x \to \alpha^+} \frac{\left( e^{\sqrt{x}} - e^{\sqrt{\alpha}} \right)}{\left( \sqrt{x} - \sqrt{\alpha} \right)} \left( \frac{0}{0} \right)$$ $$= \frac{2}{e^{\sqrt{\alpha}}} \lim_{x \to \alpha^+} \frac{\left( e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} - 0 \right)}{\left( \frac{1}{2\sqrt{x}} - 0 \right)} = 2$$ So, $\lim_{x \to \alpha^+} f\left( g(x) \right) = \lim_{x \to \alpha^+} f(2)$ $$= f(2) = \sin \frac{\pi}{6} = \frac{1}{2}$$ $$= 0.50$$

Given $n(U) = 900$. Let $A \equiv$ Fever, $B \equiv$ Cough, $C \equiv$ Breathing problem. Therefore, $n(A) = 190$, $n(B) = 220$, $n(C) = 220$. $n(A \cup B) = 330$, $n(B \cup C) = 350$, $n(A \cup C) = 340$, $n(A \cap B \cap C) = 30$. Now $n(A \cup B) = n(A) + n(B) - n(A \cap B)$. Therefore, $330 = 190 + 220 - n(A \cap B)$. Therefore, $n(A \cap B) = 80$. Similarly, $350 = 220 + 220 - n(B \cap C)$. Therefore, $n(B \cap C) = 90$. And $340 = 190 + 220 - n(A \cap C)$. Therefore, $n(A \cap C) = 70$. Therefore, $n(A \cup B \cup C) = (190 + 220 + 220) - (80 + 90 + 70) + 30 = 660 - 240 = 420$. Therefore, number of person without any symptom $= n(U) - n(A \cup B \cup C) = 900 - 420 = 480$. Now, number of person suffering from exactly one symptom $= (n(A) + n(B) + n(C)) - 2(n(A \cap B) + n(B \cap C) + n(C \cap A)) + 3n(A \cap B \cap C) = (190 + 220 + 220) - 2(80 + 90 + 70) + 3(30) = 630 - 480 + 90 = 240$. Therefore, number of person suffering from at most one symptom $= 480 + 240 = 720$. Therefore, probability $= \frac{720}{900} = \frac{8}{10} = \frac{4}{5} = 0.80$.

Given that $z \neq \overline{z}$. Let $\alpha = \frac{2 + 3z + 4z^2}{2 - 3z + 4z^2} = \frac{\left(2 - 3z + 4z^2\right) + 6z}{2 - 3z + 4z^2}$. Therefore, $\alpha = 1 + \frac{6z}{2 - 3z + 4z^2}$. If $\alpha$ is a real number, then $\alpha = \overline{\alpha}$. Therefore, $\[$ $\frac{z}{2 - 3z + 4z^2}$ = $\frac{\overline{z}}{2 - 3\overline{z} + 4z^2}$ $\]$ Thus, $\[$ 2(z - $\overline{z}$) = 4z$\overline{z}$(z - $\overline{z}$) $\]$ This implies $\[$ (z - $\overline{z}$)(2 - 4z$\overline{z}$) = 0 $\]$ As $z \neq \overline{z}$ (Given), $\[$ $\Rightarrow$ z$\overline{z}$ = $\frac{2}{4}$ = $\frac{1}{2}$ $\]$ $\[$ $\Rightarrow$ |z|^2 = 0.50 $\]$

Given $\overline{z} - z^2 = i(\overline{z} + z)$. $$\Rightarrow (1-i)\overline{z} = (1+i)z^2$$ $$\Rightarrow \frac{(1-i)}{(1+i)} \overline{z} = z^2$$ $$\Rightarrow \left(\frac{-2i}{2}\right) \overline{z} = z^2$$ Thus, $z^2 = -i \overline{z}$. Let $z = x + iy$, Therefore, $(x^2 - y^2) + i(2xy) = -i(x - iy)$ so, $x^2 - y^2 + y = 0 \ldots(1)$ and $(2y + 1)x = 0 \ldots(2)$ $$\Rightarrow x = 0 or y = -\frac{1}{2}$$ Case I: When $x = 0$ Therefore, $(1) \Rightarrow y(1-y) = 0 \Rightarrow y = 0, 1$ Thus, $(0,0), (0,1)$ Case II: When $y = -\frac{1}{2}$ Therefore, $(1) \Rightarrow x^2 - \frac{1}{4} = 0 \Rightarrow x^2 = \frac{3}{4} \Rightarrow x = \pm \frac{\sqrt{3}}{2}$ Thus, $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right), \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ Therefore, the number of distinct $z$ is equal to 4.

Given $$A_{51} - A_{50} = 1000 \Rightarrow \ell_{51} w_{51} - \ell_{50} w_{50} = 1000$$ $$\Rightarrow (\ell_1 + 50d_1)(w_1 + 50d_2) - (\ell_1 + 49d_1)(w_1 + 49d_2) = 1000$$ $$\Rightarrow (\ell_1 d_2 + w_1 d_1) = 10 \ldots(1)$$ (As $d_1 d_2 = 10$) Therefore, $$A_{100} - A_{90} = \ell_{100} w_{100} - \ell_{90} w_{90}$$ $$= (\ell_1 + 99d_1)(w_1 + 99d_2) - (\ell_1 + 89d_1)(w_1 + 89d_2)$$ $$= 10(\ell_1 d_2 + w_1 d_1) + (99^2 - 89^2) d_1 d_2$$ $$= 10(10) + (99 - 89)(99 + 89)(10)$$ (As, $d_1 d_2 = 10$) $$= 100 (1 + 188) = 100 (189)$$ $$= 18900$$

The solution involves calculating the number of 4-digit integers in the range [2022, 4482]. (1) For the number starting with 2, the digits can be 0, 2, and any of 2, 3, 4, 6, 7. This gives: $$5$$ (2) For the number starting with 2, the digits can be 0, and any of 3, 4, 6, 7. This gives: $$4 \times 6 = 24$$ (3) For the number starting with 2, the digits can be 2, 3, 4, 6, 7. This gives: $$5 \times 6 \times 6 = 180$$ (4) For the number starting with 3, the digits can be any number. This gives: $$6 \times 6 \times 6 = 216$$ (5) For the number starting with 4, the digits can be 0, 2, 3, 4. This gives: $$4 \times 6 \times 6 = 144$$ Number of 4 digit integers in [2022, 4482] is: $$5 + 24 + 180 + 216 + 144 = 569$$

Given $4 - \sqrt{10} = 0.83$ or $0.84$. $C_1 \left( \frac{1}{2}, \frac{3}{2} \right)$ and $r_1 = \frac{\sqrt{10}}{2}$. $C_2 = (r, r)$. Therefore, circle $C_2$ touches $C_1$ internally. $$\Rightarrow C_1 C_2 = \left| r - \frac{\sqrt{10}}{2} \right|$$ $$\Rightarrow \left( r - \frac{1}{2} \right)^2 + \left( r - \frac{3}{2} \right)^2 = \left( r - \frac{\sqrt{10}}{2} \right)^2$$ $$r^2 - 4r + \sqrt{10}r = 0$$ $r = 0$ (reject) or $r = 4 - \sqrt{10}$

Given $$\int_1^e \frac{(\log_e x)^{1/2}}{x \left( a - (\log_e x)^{3/2} \right)^2} = 1$$. Let $$a - (\log_e x)^{3/2} = t$$. Then $$\frac{(\log_e x)^{1/2}}{x} \, dx = -\frac{2}{3} \, dt$$. Thus, $$= \frac{2}{3} \int_a^{a-1} \frac{-dt}{t^2} = \frac{2}{3} \left( \frac{1}{t} \right)_{a}^{a-1} = 1$$. This simplifies to $$\frac{2}{3a(a-1)} = 1$$. Solving the quadratic equation, $$3a^2 - 3a - 2 = 0$$. The solution is $$a = \frac{3 \pm \sqrt{33}}{6}$$.

Given $a_1 = 7$, $d = 8$. $T_{n+1} - T_n = a_n \forall n \geq 1$ $S_n = T_1 + T_2 + T_3 + \ldots + T_{n-1} + T_n$ $S_n = T_1 + T_2 + T_3 + \ldots + T_{n-1} + T_n$ On subtraction: $T_n = T_1 + a_1 + a_2 + \ldots + a_{n-1}$ $T_n = 3 + (n - 1)(4n - 1)$ $T_n = 4n^2 - 5n + 4$ $$\sum_{k=1}^{n} T_k = 4 \sum n^2 - 5 \sum n + 4n$$ $T_{20} = 1504$ $T_{30} = 3454$ $$\sum_{k=1}^{30} T_k = 35615$$ $$\sum_{k=1}^{20} T_k = 10510$$

The line of intersection is $\frac{x}{0} = \frac{y - 4}{-4} = \frac{z}{5}$. (1) Any skew line with the line of intersection of given planes can be an edge of the tetrahedron. (2) Any intersecting line with the line of intersection of given planes must lie either in plane $P_1$ or $P_2$ and can be an edge of the tetrahedron.

Given $\vec{r} = \hat{k} + t(-\hat{i} + \hat{j}) + p(-\hat{i} + \hat{k})$ and $\vec{n} = \hat{i} + \hat{j} + \hat{k}$. This implies $x + y + z = 1$. Consider points $Q(10, 15, 20)$ and $S(\alpha, \beta, \gamma)$. We have: $$\frac{\alpha - 10}{1} = \frac{\beta - 15}{1} = \frac{\gamma - 20}{1} = -2 \left( \frac{10 + 15 + 20 - 1}{1 + 1 + 1} \right)$$ $$= -\frac{88}{3}$$ Thus, $ (\alpha, \beta, \gamma) = \left( -\frac{58}{3}, -\frac{43}{3}, -\frac{28}{3} \right) $. Therefore, A, B, C are correct options.

Let the equation of the tangent with slope $m$ be $$y = mx + \frac{1}{m}$$ The tangent passes through $(-2, 1)$, so $$1 = -2m + \frac{1}{m}$$ This implies $m = -1$ or $m = \frac{1}{2}$. Points are given by $\($ $\left$( $\frac{a}{m^2}$, $\frac{2a}{m}$ $\right$) $\)$. So, one point will be $(1, -2)$ and $(4, 4)$. Let $P_1(4, 4)$ and $P_2(1, -2)$. For $P_1S$: $4x - 3y - 4 = 0$. For $P_2S$: $x - 1 = 0$. Calculate $PQ_1$: $$PQ_1 = \frac{|4(-2) - 3(1) - 4|}{5} = 3$$ $SP = \sqrt{10}$; $PQ_2 = 3$; $SQ_1 = 1 = SQ_2$. Comparing areas: $$\frac{1}{2} \left( \frac{Q_1Q_2}{2} \right) \times \sqrt{10} = \frac{1}{2} \times 3 \times 1$$ This implies $$Q_1Q_2 = \frac{2 \times 3}{\sqrt{10}} = \frac{3 \sqrt{10}}{5}$$

Given $\($ f($\theta$) = $\frac{1}{2}$ $\begin{vmatrix}$ 1 & $\sin$ $\theta$ & 1 $\\$ -$\sin$ $\theta$ & 1 & $\sin$ $\theta$ $\\$ -1 & -$\sin$ $\theta$ & 1 $\end{vmatrix}$ + $\begin{vmatrix}$ $\sin$ $\pi$ & $\cos$ $\left$( $\theta$ + $\frac{\pi}{4}$ $\right$) & $\tan$ $\left$( $\theta$ - $\frac{\pi}{4}$ $\right$) $\\$ $\sin$ $\left$( $\theta$ - $\frac{\pi}{4}$ $\right$) & -$\cos$ $\frac{\pi}{2}$ & $\log$_e $\left$( $\frac{4}{\pi}$ $\right$) $\\$ $\cot$ $\left$( $\theta$ + $\frac{\pi}{4}$ $\right$) & $\log$_e $\frac{\pi}{4}$ & $\tan$ $\pi$ $\end{vmatrix}$ $\)$ $\($ f($\theta$) = $\frac{1}{2}$ $\begin{vmatrix}$ 2 & $\sin$ $\theta$ & 1 $\\$ 0 & 1 & $\sin$ $\theta$ $\\$ 0 & -$\sin$ $\theta$ & 1 $\end{vmatrix}$ + $\begin{vmatrix}$ 0 & -$\sin$ $\left$( $\theta$ - $\frac{\pi}{4}$ $\right$) & $\tan$ $\left$( $\theta$ - $\frac{\pi}{4}$ $\right$) $\\$ $\sin$ $\left$( $\theta$ - $\frac{\pi}{4}$ $\right$) & 0 & $\log$_e $\left$( $\frac{4}{\pi}$ $\right$) $\\$ -$\tan$ $\left$( $\theta$ - $\frac{\pi}{4}$ $\right$) & -$\log$_e $\left$( $\frac{4}{\pi}$ $\right$) & 0 $\end{vmatrix}$ $\)$ $\($ f($\theta$) = (1 + $\sin$^2 $\theta$) + 0 $\)$ (skew symmetric) $\($ g($\theta$) = $\sqrt{f(\theta) - 1 + \sqrt{f\left( \frac{\pi}{2} - \theta \right) - 1}}$ $\)$ $\($ = |$\sin$ $\theta$| + |$\cos$ $\theta$| for $\theta$ $\in$ $\left$[ 0, $\frac{\pi}{2}$ $\right$] $\)$ $\($ g($\theta$) $\in$ $\left$[ 1, $\sqrt{2}$ $\right$] $\)$ Again let $\($ P(x) = k(x - $\sqrt{2}$)(x - 1) $\)$ $\($ 2 - $\sqrt{2}$ = k(2 - $\sqrt{2}$)(2 - 1) $\)$ $\($ $\Rightarrow$ k = 1 (P(2) = 2 - $\sqrt{2}$ given) $\)$ $\($ $\therefore$ P(x) = (x - $\sqrt{2}$)(x - 1) $\)$ For option (A) $\($ P$\left$( $\frac{3 + \sqrt{2}}{4}$ $\right$) 0 $\)$ correct Option (D) $\($ P$\left$( $\frac{5 - \sqrt{2}}{4}$ $\right$) > 0 $\)$ incorrect

For (I), given $x \in \left[ -\frac{2\pi}{3}, \frac{2\pi}{3} \right]$, $\cos x + \sin x = 1$. $$\cos x + \sin x = 1$$ $$\Rightarrow \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}}$$ $$\Rightarrow \cos \left( x - \frac{\pi}{4} \right) = \cos \frac{\pi}{4}$$ $$\Rightarrow x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4} ; \; n \in \mathbb{Z}$$ $$\Rightarrow x = 2n\pi ; \; x = 2n\pi + \frac{\pi}{2} ; \; n \in \mathbb{Z}$$ $$\Rightarrow x \in \left( 0, \frac{\pi}{2} \right)$$ in given range has two solutions. For (II), given $x \in \left[ -\frac{5\pi}{18}, \frac{5\pi}{18} \right]$, $\sqrt{3} \tan 3x = 1$. $$\sqrt{3} \tan 3x = 1 \Rightarrow \tan 3x = \frac{1}{\sqrt{3}} \Rightarrow 3x = n\pi + \frac{\pi}{6}$$ $$\Rightarrow x = (6n+1) \frac{\pi}{18} ; \; n \in \mathbb{Z}$$ $$\Rightarrow x \in \left( \frac{\pi}{18}, -\frac{5\pi}{18} \right)$$ in given range has two solutions. For (III), given $x \in \left[ -\frac{6\pi}{5}, \frac{\pi}{5} \right]$, $2 \cos(2x) = \sqrt{3}$. $$2 \cos 2x = \sqrt{3}$$ $$\Rightarrow \cos 2x = \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6}$$ $$\Rightarrow 2x = 2n\pi \pm \frac{\pi}{6} ; \; n \in \mathbb{Z}$$ $$\Rightarrow x = n\pi \pm \frac{\pi}{12} ; \; n \in \mathbb{Z}$$ $$x \in \left\{ \frac{\pi}{12}, \pi \pm \frac{\pi}{12}, -\pi \pm \frac{\pi}{12} \right\}$$ Six solutions in given range. For (IV), given $x \in \left[ -\frac{7\pi}{4}, \frac{7\pi}{4} \right]$, $\sin x - \cos x = 1$. $$\cos x - \sin x = -1$$ $$\Rightarrow \cos \left( x + \frac{\pi}{4} \right) = \frac{-1}{\sqrt{2}} = \cos \frac{3\pi}{4}$$ $$\Rightarrow x + \frac{\pi}{4} = 2n\pi \pm \frac{3\pi}{4} ; \; n \in \mathbb{Z}$$ $$\Rightarrow x = 2n\pi + \frac{\pi}{2} \; or \; x = 2n\pi - \pi ; \; n \in \mathbb{Z}$$ $$\Rightarrow x \in \left( \frac{\pi}{2}, -\frac{3\pi}{2}, -\pi, -\pi \right)$$ four solutions in given range.

P(draw in 1 round) = $\frac{6}{36}$ = $\frac{1}{6}$ P(win in 1 round) = $\frac{1}{2}$ $\left$( 1 - $\frac{1}{6}$ $\right$) = $\frac{5}{12}$ P(loss in 1 round) = $\frac{5}{12}$ P(X_2 > Y_2) = P(10,0) + P(7,2) = $\frac{5}{12}$ $\times$ $\frac{5}{12}$ + $\frac{5}{12}$ $\times$ $\frac{1}{6}$ $\times$ 2 = $\frac{45}{144}$ = $\frac{5}{16}$ P(X_2 = Y_2) = P(5,5) + P(4,4) = $\frac{5}{12}$ $\times$ $\frac{5}{12}$ $\times$ 2 + $\frac{1}{6}$ $\times$ $\frac{1}{6}$ = $\frac{25 + 2}{72}$ = $\frac{3}{8}$ P(X_3 = Y_3) = P(6,6) + P(7,7) = $\frac{1}{6 \times 6 \times 6}$ + $\frac{5}{12}$ $\times$ $\frac{1}{6}$ $\times$ $\frac{5}{12}$ $\times$ 6 = $\frac{2}{432}$ + $\frac{75}{432}$ = $\frac{77}{432}$ P(X_3 > Y_3) = $\frac{1}{2}$ $\left$( 1 - $\frac{77}{432}$ $\right$) = $\frac{355}{864}$

If $\frac{q}{r} = 10 \Rightarrow A = D \Rightarrow D_x = D_y = D_z = 0$. So, there are infinitely many solutions. Look of infinitely many solutions can be given as $$x + y + z = 1$$ $$10x + 10y + 1000z = 0 \Rightarrow x + 10y + 100z = 0$$ Let $z = \lambda$. Then $x + y = 1 - \lambda$. And $x + 10y = -100\lambda$. $$\Rightarrow x = \frac{10}{9} + 10\lambda; y = \frac{-1}{9} - 11\lambda$$ $$i.e., (x, y, z) \equiv \left( \frac{10}{9} + 10\lambda, \frac{-1}{9} - 11\lambda, \lambda \right)$$ $$Q \left( \frac{10}{9}, \frac{-1}{9}, 0 \right) valid for \lambda = 0$$ $$P \left( 0, \frac{10}{9}, \frac{-1}{9} \right) not valid for any \lambda.$$ (I) $\rightarrow Q,R,T$ (II) If $\frac{p}{r} \neq 100$, then $D_y \neq 0$. So no solution. (II) $\rightarrow (S)$ (III) If $\frac{p}{q} \neq 10$, then $D_z \neq 0$ so, no solution. (III) $\rightarrow (S)$ (IV) If $\frac{p}{q} = 10 \Rightarrow D_z = 0 \Rightarrow D_x = D_y = 0$ So infinitely many solution. (IV) $\rightarrow Q,R,T$

Let $F(2\cos \phi, 2\sin \phi)$ and $E(2\cos \phi, \sqrt{3} \sin \phi)$. EG: $\frac{x}{2} \cos \phi + \frac{y}{\sqrt{3}} \sin \phi = 1$. Therefore, $G\left(\frac{2}{\cos \phi}, 0\right)$ and $\alpha = 2\cos \phi$. $ar(\Delta FGH) = \frac{1}{2} HG \times FH$. $$= \frac{1}{2} \left(\frac{2}{\cos \phi} - 2\cos \phi\right) \times 2\sin \phi$$ $f(\phi) = 2\tan \phi \sin^2 \phi$. Therefore, (I) $f\left(\frac{\pi}{4}\right) = 1$ (II) $f\left(\frac{\pi}{3}\right) = \frac{3\sqrt{3}}{2}$ (III) $f\left(\frac{\pi}{6}\right) = \frac{1}{2\sqrt{3}}$ (IV) $f\left(\frac{\pi}{12}\right) = 2(2 - \sqrt{3})\left(\frac{\sqrt{3} - 1}{2\sqrt{2}}\right)^2 = (4 - 2\sqrt{3})\frac{(\sqrt{3} - 1)^2}{8} = \frac{(\sqrt{3} - 1)^4}{8}$. Therefore, (I) $\to$ (Q); (II) $\to$ (T); (III) $\to$ (S); (IV) $\to$ (P).

Given $R_A = R_B = R$. $M_B = 2M_A$. Calculation of escape velocity for A: Radius of remaining star $= \frac{R_A}{2}$. Mass of remaining star $= \rho_A \frac{4}{3} \pi \frac{R_A^3}{8} = \frac{M_A}{8}$. $$\frac{-GM_{A/B}}{R_{A/2}} + \frac{1}{2} m v_A^2 = 0 \implies v_A = \sqrt{\frac{2GM_{A/B}}{R_{A/2}}} = \sqrt{\frac{GM_A}{2R}}$$ Calculation of escape velocity for B: Mass collected over B $= \frac{7}{8} M_A$. Let the radius of B becomes $r$. $$\therefore \frac{4}{3} \pi (r^3 - R_B^3) \rho_A = \frac{7}{8} \rho_A \frac{4}{3} \pi R_A^3 \implies r^3 = \frac{7}{8} R_A^3 + R_B^3 = \frac{(15)^{1/3} R}{2}$$ $$\therefore \frac{V_B^2}{V_A^2} = \frac{23 GM_A}{8 \times 15^{1/3} \frac{R}{2}} = \frac{23 GM_A}{4 \times 15^{1/3} R}$$ $$\therefore V_B = \frac{\sqrt{23 GM_A}}{2 \sqrt{15^{1/3}} R}$$ $$\therefore \frac{V_B}{V_A} = \sqrt{\frac{23}{15^{1/3}}} = \sqrt{\frac{10 \times 2.30}{15^{1/3}}}$$ $n = 2.30$

The reaction is given by $^{16}_{7} \mathrm{N} + ^{4}_{2} \mathrm{He} \rightarrow ^{1}_{1} \mathrm{He} + ^{19}_{8} \mathrm{O}$. The atomic masses are $16.006$, $4.003$, $1.008$, and $19.003$. For maximum loss of kinetic energy, $4v_0 = 1v_1 + 19v_2 = 20v_2$. Therefore, $v_0 = \frac{v_2}{5}$. The energy required is calculated as $(1.008 + 19.003 - 16.006 - 4.003) \times 930 = 1.86$. $$\frac{1}{2} 4v_0^2 - \frac{1}{2} 20v^2 = 1.86$$ $$\frac{1}{2} 4v_0^2 - 10 \frac{v_0^2}{25} 20v^2 = 1.86$$ $$2v_0^2 - \frac{2}{5} v_0^2 = 1.86$$ $$\frac{8}{5} v_0^2 = 1.86$$ Solving for $v_0^2$, we get: $$v_0^2 = \frac{1.86 \times 5}{8}$$ The kinetic energy is given by: $$\mathrm{KE} = \frac{1}{2} 4v_0^2 = 2v_0^2 = \frac{18.6 \times 5}{4}$$ The final result is $= 2.325$.

Potential difference across the terminals of $C_3$ is $2\, \mathrm{V}$. Therefore, $Q_3 = CV = (4\mu)(2) = 8\mu \mathrm{C}$.

Given $\frac{h_i}{h_0} = -\frac{30\sqrt{3}}{13}$, $\frac{v}{u} = \frac{13}{\sqrt{3}}$. Therefore, $\frac{v}{-\frac{43}{3}} = \frac{430}{13}$ cm. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, we have $\frac{1}{v} = \frac{1}{10} - \frac{3}{40}$, which gives $v = 40$ cm. The distance $x = 40 - \frac{430}{13} = \frac{90}{13}$ cm. For the angle, $\tan \alpha = \frac{30\sqrt{3}}{\frac{13}{90}} = \frac{1}{\sqrt{3}}$, which implies $\alpha = 30^\circ = \frac{\pi}{6}$. Therefore, $N = 6$.

At $t = 0$, $\omega = 0$. At $t = \sqrt{\pi}$, $\omega = \alpha t = \frac{2}{3} \sqrt{\pi}$, $v = \omega R = \frac{2}{3} \sqrt{\pi}$. $$\theta = \frac{1}{2} \alpha t^2$$ $$\theta = \frac{1}{2} \times \frac{2}{3} \times \pi = \frac{\pi}{3}$$ $\theta = 60^\circ$ $v_y = v \sin 60 = \frac{\sqrt{3}}{2} V$ $$h = \frac{u_y^2}{2g} = \frac{3}{4} \frac{v^2}{2g}$$ $$h = \frac{3}{4} \frac{4}{9} \pi \frac{1}{2g}$$ $$h = \frac{3\pi}{9 \times 2g} = \frac{\pi}{6g}$$ Maximum height from plane, $H = \frac{R}{2} + h$ $$H = \frac{1}{2} + \frac{\pi}{6 \times 10}$$ $$x = \frac{\pi}{6} ; x = 0.52$$

Solid sphere 1 kg, 1 m $5 + 1 - 1 - f = 1a$ $5 - f = a$ About COM $f \cdot 1 - 2(1(0.5)) = \frac{2}{5} Mr^2 \alpha$ $$\Rightarrow f - 1 = \frac{2}{5} a \Rightarrow f = 1 + \frac{2}{5} a$$ $$5 - a = 1 + \frac{2}{5} a$$ $$\Rightarrow 4 = \frac{7a}{5} \Rightarrow a = \frac{20}{7} = 2.86 \, \mathrm{m/s^2}$$

Maximum energy will be $$\frac{q_0^2}{2C} = \frac{1}{2} LI_0^2$$ $$\frac{q_0^2}{CL} = I_0^2$$ $$I_0 = \frac{q_0}{\sqrt{LC}}$$ $$I_0 = \frac{CV}{\sqrt{LC}}$$ $$I_0 = \sqrt{\frac{C}{L}} \times V$$ $$V = emf = \left| \frac{AdB}{dt} \right|$$ $$I_0 = \sqrt{\frac{10^{-3}}{0.1}} \times 0.04$$ $$V = (1 \times 0.04)$$ Maximum current $I_0 = 0.004 = 4 mA$

The initial range is given by $$d = \frac{v^2 \sin 2\theta}{g}$$ The maximum height is $$H_{max} = \frac{v^2 \sin^2 \theta}{2g}$$ The effective height equation is $$\frac{1}{2} g_{eff} t^2 = H_{max} \implies t^2 = \frac{2H_{max}}{g_{eff}}$$ Thus, $$t = \sqrt{\frac{v^2 \sin^2 \theta \times 0.81}{g^2}} = \frac{0.9v \sin \theta}{g}$$ The time squared is $$t^2 = \frac{2 \times v^2 \sin^2 \theta}{2g} \left( \frac{g}{0.81} \right)$$ The new range $d'$ is $$d' = \frac{d}{2} + d_1$$ where $$d_1 = v \cos \theta \cdot t$$ Substituting, $$= \frac{v^2 \sin^2 \theta \cos \theta \times 0.9}{g}$$ Thus, $$d' = \frac{v^2 \sin 2\theta}{2g} + \frac{v^2 \sin 2\theta \times 0.9}{2g}$$ Simplifying, $$= \frac{v^2 \sin 2\theta}{g} \left( \frac{1.0}{2} \right) = 0.95d$$ Therefore, $n = 0.95$

For figure (a), $$E_0 = \frac{V}{d} ; C = \frac{K \varepsilon_0 A}{d}$$ For figure (b), $$C' = \frac{\varepsilon_0 A}{2d - d + d/k} ;$$ $$C' = \frac{K \varepsilon_0 A}{(K+1)d} ; C' = \frac{C}{K+1}$$

From KCL $i_1 + i_2 + i_3 = 0$ $$\frac{18 - V_0}{3/2} + \frac{12 - V_0}{1/2} + \frac{0 - V_0}{3/2} = 0$$ $$\Rightarrow 18 - V_0 + 36 - 3V_0 - V_0 = 0$$ $$\Rightarrow 54 = 5V_0$$ $$2 \left( \frac{54}{5} - v' \right) + \frac{18 - v'}{1} = 0$$ $$\Rightarrow \frac{108}{5} + 18 = 3v'$$ $$\Rightarrow v' = \frac{198}{5 \times 3} = \frac{66}{5} \, \mathrm{V}$$ $$I_{R_1} = \frac{36}{5} = 7.2 \, \mathrm{A}$$ $$I_{R_2} = \frac{6}{5} = 1.2 \, \mathrm{A}$$ $$I_{R_3} = \frac{24}{5} = 4.8 \, \mathrm{A}$$ $$I_{R_5} = \frac{12}{5} = 2.4 \, \mathrm{A}$$

Given $A_2 = 0.4 \, \mathrm{m^2}$, $A_1 = 0.1 \, \mathrm{m^2}$, $P_1 = 0.2 \, \mathrm{kg/m^3}$, $P_1 = 600 \, \mathrm{Pa}$, $T_1 = 300 \, \mathrm{K}$. The rate of change of mass is given by $\frac{dm}{dt} = \rho_1 A_1 v_1 = 0.8 \, \mathrm{kg/s}$. Solving for $v_1$: $$v_1 = \frac{0.8}{0.2 \times 0.1} = 40 \, \mathrm{m/s}$$ Given $g = 10 \, \mathrm{m/s^2}$ and $\gamma = 2$. Gas undergoes adiabatic expansion, $$P^{1-\gamma} T^\gamma = Constant$$ $$\frac{P_2}{P_1} = \left( \frac{T_1}{T_2} \right)^{\frac{\gamma}{1-\gamma}}$$ $$P_2 = \left( \frac{300}{150} \right)^{2-1} \times 600$$ $$P_2 = \frac{600}{4} = 150 \, \mathrm{Pa}$$ Now $\rho = \frac{PM}{RT} \Rightarrow \rho \propto \frac{P}{T}$ $$\frac{\rho_1}{\rho_2} = \left( \frac{P_1}{P_2} \right) \left( \frac{T_1}{T_2} \right) = \left( \frac{150}{600} \right) \left( \frac{300}{150} \right) = \frac{1}{2}$$ $$\rho_2 = \frac{\rho_1}{2} = 0.1 \, \mathrm{kg/m^3}$$ Now $\rho_2 A_2 v_2 = 0.8 \Rightarrow v_2 = \frac{0.8}{0.1 \times 0.4} = 20 \, \mathrm{m/s}$ Now work done on gas $= \Delta K + \Delta U + (Internal energy)$ $$P_1 A_1 \Delta x_1 - P_2 A_2 \Delta x_2 = \frac{1}{2} \Delta m v_2^2 - \frac{1}{2} \Delta m v_1^2 + \Delta m gh + \frac{f}{2} (P_2 \Delta V_2 - P_1 \Delta V_1)$$ $$\Rightarrow 2P_1 \frac{\Delta V_1}{\Delta m} - 2P_2 \frac{\Delta V_2}{\Delta m} = \frac{V_2^2 - V_1^2}{2} + gh$$ $$\Rightarrow 2 \times 600 \frac{20}{0.2} - 2 \times 150 \frac{40}{0.1} = \frac{20^2 - 40^2}{2} + 10h$$ $$h = 360 \, \mathrm{m}$$

Ray will come out after one reflection for $\theta = 30^\circ$ and $0 < \ell < L$. For $\theta = 60^\circ$ and $\ell = \frac{L}{2}$, ray will come out after two reflections. For $\ell = \frac{L}{3}$ and $\theta = 60^\circ$, ray will come out after five reflections. For $\theta = 60^\circ$ and $0 < \ell < \frac{L}{2}$, ray will come out after five reflections.

Due to symmetry $\vec{E}_0 = 0$. $$E_{net} = \frac{kq}{(2d)^2} \times 2 = \frac{2q \times 4}{4 \pi \varepsilon_0 \cdot 4 \cdot 3a^2}$$ $$= \frac{q}{6 \pi \varepsilon_0 a^2}$$ $$v = \frac{7kq}{2d} = \frac{7q}{4 \pi \varepsilon_0 \cdot \sqrt{3}a} = \frac{7q}{4 \sqrt{3} \pi \varepsilon_0 q}$$ $$v = \frac{2kq}{2d} = \frac{2q}{4 \pi \varepsilon_0 \cdot \sqrt{3}a} = \frac{q}{2 \sqrt{3} \pi \varepsilon_0 q}$$

Binding energy of proton and neutron due to nuclear force is the same. So difference in binding energy is only due to electrostatic P.E. and it is positive. $$E_0^p - E_0^n = electrostatic P.E.$$ $$= Z \times P.E. of one proton$$ $$= Z \times \frac{1}{4 \pi \varepsilon_0} \frac{(Z-1)e^2}{R}$$ Where $R = R_0 A^{1/3}$ $$= \frac{1}{4 \pi \varepsilon_0} \frac{Z(Z-1)e^2}{R_0 A^{1/3}}$$

Solution. (I) $\mathbf{B} = \frac{\mu_0 m I}{\sqrt{2}} \left( \sin \omega t \, \mathbf{\hat{j}} + \cos \omega t \, \mathbf{\hat{k}} \right)$ $\phi = \mathbf{B} \cdot \mathbf{A} = \frac{\mu_0 m I}{\sqrt{2}} \cos (\omega t) \cdot A$ $\varepsilon = \frac{d \phi}{dt} = \frac{\mu_0 m I \omega A}{\sqrt{2}} \sin (\omega t)$ $i = \frac{\varepsilon}{R} = \frac{\mu_0 m I \omega A}{\sqrt{2} R} \sin (\omega t)$ $\mathbf{\bar{M}} = i \mathbf{\bar{A}} = i A \left( \mathbf{\hat{k}} \right) = \frac{\mu_0 m I \omega A^2}{\sqrt{2} R} \sin (\omega t) \left( \mathbf{\hat{k}} \right)$ $\mathbf{\tau} = \mathbf{\bar{M}} \times \mathbf{B} = \frac{\mu_0 m^2 I^2 \omega A^2}{\sqrt{2} R} \sin^2 (\omega t) \left( -\mathbf{\hat{i}} \right)$ $= -\left( \frac{\alpha}{4} \right) \mathbf{\hat{i}}$ (II) $\mathbf{B} = \frac{\mu_0 m I}{\sqrt{2}} \left( \sin \omega t \, \mathbf{\hat{i}} + \cos \omega t \, \mathbf{\hat{j}} \right)$ $\phi = 0, \varepsilon = 0, i = 0, t = 0$ (III) $\mathbf{B} = \frac{\mu_0 m I}{\sqrt{2}} \left( \sin \omega t \, \mathbf{\hat{i}} + \cos \omega t \, \mathbf{\hat{k}} \right)$ $\phi = \mathbf{B} \cdot \mathbf{A} = \frac{\mu_0 m I}{\sqrt{2}} \cdot \cos (\omega t) \cdot A$ $\varepsilon = -\frac{d \phi}{dt} = \frac{\mu_0 m I \omega A}{\sqrt{2}} \sin (\omega t)$ $i = \frac{\varepsilon}{R} = \frac{\mu_0 m I \omega A}{\sqrt{2} R} \sin (\omega t)$ $\mathbf{\bar{M}} = i \mathbf{\bar{A}} = i A \left( \mathbf{\hat{k}} \right) = \frac{\mu_0 m I \omega A^2}{\sqrt{2} R} \sin (\omega t) \left( \mathbf{\hat{k}} \right)$ $\mathbf{\tau} = \mathbf{\bar{M}} \times \mathbf{B} = \frac{\mu_0 m^2 I^2 \omega A^2}{2 R} \sin^2 (\omega t) \left( +\mathbf{\hat{j}} \right)$ $= \frac{\alpha}{4} \mathbf{\hat{j}}$ (IV) $\mathbf{B} = \frac{\mu_0 m I}{\sqrt{2}} \left( \cos \omega t \, \mathbf{\hat{j}} + \sin \omega t \, \mathbf{\hat{k}} \right)$ $\phi = \mathbf{B} \cdot \mathbf{A} = \frac{\mu_0 m I}{\sqrt{2}} \cdot \sin (\omega t) \cdot A$ $\varepsilon = -\frac{d \phi}{dt} = \frac{\mu_0 m I \omega A}{\sqrt{2}} \cos (\omega t)$ $i = \frac{\varepsilon}{R} = \frac{\mu_0 m I \omega A}{\sqrt{2} R} \cos (\omega t)$ $\mathbf{\bar{M}} = i \mathbf{\bar{A}} = i A \left( \mathbf{\hat{k}} \right) = -\frac{\mu_0 m I \omega A^2}{\sqrt{2} R} \cos (\omega t) \left( \mathbf{\hat{k}} \right)$ $\mathbf{\tau} = \mathbf{\bar{M}} \times \mathbf{B} = -\frac{\mu_0 m^2 I^2 \omega A^2}{2 R} \cos^2 (\omega t) \left( -\mathbf{\hat{i}} \right)$ $= \alpha \cdot \cos^2 \left( \frac{\pi}{6} \right) \mathbf{\hat{i}}$ $= \frac{3 \alpha}{4} \mathbf{\hat{i}}$ Answer. (C) I-Q, II-P, III-S, IV-R

Given $v_{BA}^2 = v_A^2 + v_B^2 - 2v_{AB} \cos \theta$. As $\omega_A = \omega_B$, $\theta = 90^\circ$ remains constant. Also, $v_A = v_B = 1 \, \mathrm{m/s}$. So, $v_{BA} = \sqrt{2} \, \mathrm{m/s}$. (II) $\vec{u}_A = \frac{5\pi}{2} \hat{i} + \frac{5\pi}{2} \hat{j}$ $\vec{v}_A = \frac{5\pi}{2} \hat{i} + \left( \frac{5\pi}{2} - 10 \cdot \frac{\pi}{3} \right) \hat{j}$ $$= \frac{5\pi}{2} \hat{i} - \frac{5\pi}{2} \hat{j}$$ $\vec{u}_B = -\frac{5\pi}{2} \hat{i} + \frac{5\pi}{2} \hat{j}$ $\vec{u}_B = -\frac{5\pi}{2} \hat{i} - \left( \frac{5\pi}{6} + 1 \right) \hat{j}$ $\vec{v}_{BA} = -5\pi \hat{i} - \hat{j}$ $v_{BA} = \sqrt{25\pi^2 + 1}$ (III) $x_A = \sin t$ $v_A = \cos t = \frac{1}{2} \, \mathrm{m/s}$ $x_B = \cos t$ $v_B = -\sin t = -\frac{\sqrt{3}}{2} \, \mathrm{m/s}$ $v_{BA} = -\frac{\sqrt{3}}{2} \cdot \frac{1}{2}$ (IV) $\vec{v}_A$ and $\vec{v}_B$ are always perpendicular. So, $|\vec{v}_{BA}| = \sqrt{v_A^2 + v_B^2} = \sqrt{10} \, \mathrm{m/s}$

Sol. (I) $\Delta U = \Delta Q - \Delta W$ $$= \left\{ (10^{-3} \times 2250) - \frac{10^5 (10^{-3} - 10^{-6})}{10^3} \right\} kJ$$ $$= (2.25 - 0.0999) kJ$$ $$= (2.1501) kJ$$ (II) $\Delta U = nC_V \Delta T$ $$= \frac{5}{2} nR \Delta T$$ $$= \frac{5}{2} \cdot (0.2) (8) (1500 - 500) J$$ $$= 4 kJ$$ (III) $P_1 V_1^{\frac{5}{3}} = P_2 V_2^{\frac{5}{3}}$ $$\Rightarrow 2 \left( \frac{1}{3} \right)^{\frac{5}{3}} = P_2 \left( \frac{1}{24} \right)^{\frac{5}{3}}$$ $$\Rightarrow P_2 = 64 kPa$$ $$\Delta U = nC_V \Delta T = \frac{3}{2} (P_2 V_2 - P_1 V_1)$$ $$= \frac{3}{2} \left( 64 \times \frac{1}{24} - 2 \times \frac{1}{3} \right) kJ$$ $$= 3 kJ$$ (IV) $\Delta U = nC_V \Delta T$ $$= n \cdot \frac{7}{2} R \Delta T$$ $$= \frac{7}{9} \Delta Q$$ $$= 7 kJ$$

Solution. (I) $v_1 = \frac{uf}{u+f}$ $$= \frac{(-20)(10)}{(-20)+(10)} = +20$$ $u_2 = +15$ $$v_2 = \frac{(15)(15)}{(15)+(15)} = +7.5$$ (II) $v_1 = +20$ $u_2 = +15$ $$v_2 = \frac{(15)(-10)}{(15)+(-10)} = -30$$ (III) $v_1 = +20$ $u_2 = +15$ $$v_2 = \frac{(15)(-20)}{(15)+(-20)} = 60$$ (IV) $v_1 = \frac{(-20)(-20)}{(-20)+(-20)} = -10$ $u_2 = -15$ $$v_2 = \frac{(-15)(10)}{(-15)+(10)} = 30$$

Given $Q_{rxn} = C \Delta T$. $|\Delta U| \times 2 = 20 \times 14.8$. $|\Delta U| = 148 \, kJ/mol$. $\Delta U = -148 \, kJ/mol$. $Hg(g) + \frac{1}{2} O_2(g) \rightarrow HgO(s) : \Delta U = -148 \, kJ/mol$. $\Delta H = \Delta U + \Delta n_g \, RT$. $$= -148 - \frac{3}{2} \times \frac{8.3}{1000} \times 298 = -151.7101$$ $Hg(l) + \frac{1}{2} O_2(g) \rightarrow HgO(s)$. $\Delta H = -151.7101 + 61.32 = -90.39 \, kJ/mol$. Ans. 90.39

For the required reaction $\Delta G^\circ = \Delta G^\circ_1 + \Delta G^\circ_2 + \Delta G^\circ_3$. $$\Rightarrow 7 \times \mathrm{E} = 1.68 \times 3 + 1.21 \times 2 + (-1.03) \times 2$$ $$\mathrm{E} = \frac{5.4}{7} = 0.7714$$ Ans. = 0.77

The reaction is $\mathrm{H_2CO_3 + NaOH \rightarrow NaHCO_3 + H_2O}$. Milli moles: $\begin{array}{ccc} \mathrm{H_2CO_3} & \mathrm{NaOH} & \mathrm{NaHCO_3} \\ 10 & 10 & - \\ \end{array}$ At end: $\begin{array}{ccc} 0 & 0 & 10 + 10 = 20 \\ \end{array}$ Final mixture has 20 milli moles $\mathrm{NaHCO_3}$ and 10 milli moles $\mathrm{Na_2CO_3}$. $$\mathrm{pH} = \mathrm{pK_{a_2}} + \log \left( \frac{Salt}{Acid} \right)$$ $$\mathrm{pH} = \mathrm{pK_{a_2}} + \log \left( \frac{10}{20} \right) [Buffer : \mathrm{Na_2CO_3 + NaHCO_3}]$$ $$= 10.32 - \log 2 = 10.02$$

2Cu($\mathrm{NO}$_3)_2 + 5$\mathrm{KI}$ $\longrightarrow$ $\mathrm{Cu}$_2$\mathrm{I}$_2 + $\mathrm{KI}$_3 + 4$\mathrm{KNO}$_3 0.02 $\hspace{2cm}$ 0.01 $\mathrm{KI}$_3 + $\mathrm{H}$_2$\mathrm{S}$ $\longrightarrow$ $\mathrm{S}$ $\downarrow$ + $\mathrm{KI}$ + 2$\mathrm{HI}$ 0.01 $\hspace{2cm}$ 0.01 n_S = 0.01 mole weight of sulphur = 32 $\times$ 0.01 = 0.32 gm

Mole of $P_4 = \frac{1.24}{31 \times 4} = 0.01$ $P_4 + 3\mathrm{NaOH} + 3\mathrm{H_2O} \longrightarrow \mathrm{PH_3} + 3\mathrm{NaH_2PO_2}$ $0.01$ mole $ 0.01$ mole $2\mathrm{PH_3} + 3\mathrm{CuSO_4} \rightarrow \mathrm{Cu_3P_2} + 3\mathrm{H_2SO_4}$ $0.01 \frac{3}{2} \times 0.01$ $$= \frac{0.03}{2}$$ moles $W_{\mathrm{CuSO_4}} = \frac{0.03}{2} \times 159 = 2.385$ gm Ans = 2.38 or 2.39

The molecular weight (M.W.) of the compound is 250 $\mathrm{g/mol}$. For 1 $\mathrm{g}$ of R, the number of moles is given by $$\frac{1}{250} moles.$$ The number of Br atoms corresponds to $$\frac{2}{250} moles.$$ Therefore, the moles of AgBr formed are $$\frac{2}{250} moles.$$ The mass of AgBr is calculated as $$\frac{2}{250} \times (108 + 80) = 1.504.$$

Starting with $3 \mathrm{HC} \equiv \mathrm{CH}$ and $15$ moles, using iron at red hot conditions, we obtain $4$ moles of benzene with an $80\%$ yield. Next, reacting with $\mathrm{H_3C(CH_2)_3Cl}$ and $\mathrm{AlCl_3}$, we get $2$ moles of isopropylbenzene with a $50\%$ yield. Then, using $1. \mathrm{O_2}$ and $2. \mathrm{H_3O^+}$, we convert it to $1$ mole of phenol with a $50\%$ yield. Finally, reacting with $\mathrm{CH_3COCl}$ and pyridine, we obtain $1$ mole of acetophenone with a $100\%$ yield, resulting in $\mathrm{C_8H_8O_2}$ with a mass of $136 \, \mathrm{g}$.

(A) $p_z$ $p_z$ $\rightarrow$ $\sigma$ molecular orbital. Total 2 nodal plane. (B) $p_z$ $p_z$ $\rightarrow$ $\sigma^*$ molecular orbital. Zero node in the $xz$ plane. (C) $p_z$ $p_z$ $\rightarrow$ $\pi$-molecular orbital. Zero node in the plane which is perpendicular to the molecular axis and goes through the center of the molecule. (D) $p_z$ $p_z$ $\rightarrow$ $\pi^*$-molecular orbital. One node in $xy$ plane containing the molecular axis.

Chemisorption is unimolecular layered. Enthalpy of physisorption is much less in magnitude. Chemisorption of gases on solids is exothermic. As physisorption is exothermic so lowering temperature favours it.

(A) Electrochemical extraction of Aluminum from bauxite done below $2500^\circ \mathrm{C}$. (B) $2\mathrm{Na[Al(OH)_4]_{aq}} + 2\mathrm{CO_2}_{(g)} \rightarrow \mathrm{Al_2O_3} \cdot 3\mathrm{H_2O}_{(s)} \downarrow + 2\mathrm{NaHCO_3}_{(aq)}$ The sodium aluminate present in solution is neutralised by passing $\mathrm{CO_2}$ gas and hydrated $\mathrm{Al_2O_3}$ is precipitated. (C) $\mathrm{Al_2O_3}_{(s)} + 2\mathrm{NaOH}_{(aq)} + 3\mathrm{H_2O}_{(l)} \rightarrow 2\mathrm{Na[Al(OH)_4]_{aq}}$ Concentration of bauxite is carried out by heating the powdered ore with hot concentrated solution of $\mathrm{NaOH}$. (D) In metallurgy of aluminum, $\mathrm{Al_2O_3}$ is mixed with $\mathrm{Na_3AlF_6}$.

The reaction is given by: $$3\mathrm{PbS} + 8\mathrm{HNO_3} \rightarrow 3\mathrm{Pb(NO_3)_2} + 2\mathrm{NO} + 4\mathrm{H_2O} + \mathrm{S}$$ NO is a neutral oxide, paramagnetic, has linear geometry, and is a colourless gas.

The reaction starts with propanoic acid, $\mathrm{CH_3CH_2COOH}$, which is treated with red phosphorus and bromine to form 2-bromopropanoic acid, $\mathrm{BrCH(CH_3)COOH}$. This compound is more acidic than propanoic acid. Next, 2-bromopropanoic acid is treated with $\mathrm{NH_4OH}$ followed by $\mathrm{H^+}$ to form alanine, $\mathrm{NH_2CH(CH_3)COOH}$. Alanine is then reacted with $\mathrm{N^\circ K^+}$ to form a compound $Q$. Compound $Q$ is treated with $\mathrm{NaOH}$ and $\mathrm{H_3O^+}$ to form a mixture of alanine and phthalic acid. Finally, alanine is treated with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ to form lactic acid, $\mathrm{OHCH(CH_3)COOH}$, and nitrogen gas.

The compound (P) is reduced using $\mathrm{H_2/Pd}$ or $\mathrm{Sn/HCl}$ to form compound (Q). Compound (Q) is then treated with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ or $\mathrm{HNO_2}$ to form compound (R). Compound (R) undergoes a reaction to form compound (S) with $\mathrm{N_2^+Cl^-}$ and water, resulting in compound (T). Compound (S) can also undergo a reaction with $\mathrm{H_3PO_2}$ or ethanol, followed by $\mathrm{KMnO_4}$ and $\mathrm{KOH}$ with heat, to form a carboxylic acid.

(I) rate = $\frac{k[x]}{x_s + [x]}$ = $\frac{k}{\frac{x_s}{[x]} + 1}$. If [x] $\to$ $\infty$ $\Rightarrow$ rate $\to$ k $\Rightarrow$ order = 0 $\Rightarrow$ (I) - (R), (P). (II) [x] > x_s $\Rightarrow$ rate = k $\Rightarrow$ order = 0 $\Rightarrow$ (III) - (P), (S). (IV) rate = $\frac{k[x]^2}{x_s + [x]}$. [x] >> x_s $\Rightarrow$ rate = k[x] $\Rightarrow$ (IV) - (Q), (T).

Solution: (P) $\mathrm{Mg(HCO_3)_2} + 2\mathrm{Ca(OH)_2} \rightarrow \mathrm{Mg(OH)_2} + 2\mathrm{CaCO_3} + 2\mathrm{H_2O}$ (Q) $\mathrm{BaO_2} + \mathrm{H_2SO_4} \rightarrow \mathrm{H_2O_2} + \mathrm{BaSO_4}$ (R) $\mathrm{Ca(OH)_2} + \mathrm{MgCl_2} \rightarrow \mathrm{Mg(OH)_2} + \mathrm{CaCl_2}$ (S) $\mathrm{BaO_2} + 2\mathrm{HCl} \rightarrow \mathrm{BaCl_2} + \mathrm{H_2O_2}$ (T) $\mathrm{Ca(HCO_3)_2} + \mathrm{Ca(OH)_2} \rightarrow 2\mathrm{CaCO_3} + 2\mathrm{H_2O}$

(1) $[Cr(CN)_6]^{4-}$ $Cr^{+2} = [Ar] 3d^4 4s^0$; low spin complex $$\begin{array}{c} \uparrow\downarrow \\ \uparrow\downarrow \\ \uparrow\downarrow \\ \uparrow \end{array} \Delta_0 > P e_g^0 t_{2g}^4$$ P, R, T (2) $[RuCl_6]^{2-}$ $Ru^{+4} = [Kr] 4d^4 5s^0$; low spin complex $$\begin{array}{c} \uparrow\downarrow \\ \uparrow\downarrow \\ \uparrow\downarrow \\ \uparrow \end{array} \Delta_0 > P e_g^0 t_{2g}^4$$ P, R, S, T (3) $[Cr(H_2O)_6]^{2+}$ $Cr^{+2} = [Ar] 3d^4 4s^0$; high spin complex $$\begin{array}{c} \uparrow \\ \uparrow \\ \uparrow \\ \uparrow \end{array} \Delta_0 < P e_g^1 t_{2g}^3$$ Q, T (4) $[Fe(H_2O)_6]^{2+}$ $Fe^{+2} = [Ar] 3d^6$; High spin complex $$\begin{array}{c} \uparrow\downarrow \\ \uparrow \\ \uparrow \\ \uparrow \end{array} \Delta_0 < P e_g^2 t_{2g}^4$$ P, Q

Aniline: Blue colour in Lassign test due to presence of N. o-Cresol: Violet colour with $\mathrm{FeCl_3}$ due to presence of phenolic OH. Cystein: It gives blood red colour with $\mathrm{NaSCN}$. Caprolactam: Blue colour in Lassign test due to presence of N.