JEE Main 27 August 2021 Shift 2 question paper with solutions

Given $n = 2(\ell + m)$. $\ell m + n(\ell + m) = 0$ $\ell m + 2(\ell + m)^2 = 0$ $2\ell^2 + 2m^2 + 5m\ell = 0$ $2\left(\frac{\ell}{m}\right)^2 + 2 + 5\left(\frac{\ell}{m}\right) = 0$ $2t^2 + 5t + 2 = 0$ $(t + 2)(2t + 1) = 0$ $\Rightarrow t = -2; -\frac{1}{2}$ (i) $\frac{\ell}{m} = -2$ $\frac{n}{m} = -2$ $(-2\, m, \, m, \, -2\, m)$ $(-2, \, 1, \, -2)$ (ii) $\frac{\ell}{m} = -\frac{1}{2}$ $n = -2\ell$ $(\ell, \, -2\ell, \, -2\ell)$ $(1, \, -2, \, -2)$ $\cos \theta = \frac{-2 - 2 + 4}{\sqrt{9} \sqrt{9}} = 0 \Rightarrow \theta = \frac{\pi}{2}$

Given the determinant equation: $$\begin{vmatrix} [x+1] & [x+2] & [x+3] \\ [x] & [x+3] & [x+3] \\ [x] & [x+2] & [x+4] \end{vmatrix} = 192$$ Perform the row operations: $$R_1 \rightarrow R_1 - R_3 \& R_2 \rightarrow R_2 - R_3$$ This gives: $$\begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ [x] & [x] + 2 & [x] + 4 \end{vmatrix} = 192$$ Expanding the determinant, we have: $$2[x] + 6 + [x] = 192 \Rightarrow [x] = 62$$

Let $g(x)=\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$ $g(x)\in[1,\sqrt{2}]$ for $x\in\left[0,\frac{\pi}{2}\right]$ $f(x)=\tan^{-1}(\sin x+\cos x)\in\left[\frac{\pi}{4},\tan^{-1}\sqrt{2}\right]$ $\tan\left(\tan^{-1}\sqrt{2}-\frac{\pi}{4}\right) =\frac{\sqrt{2}-1}{1+\sqrt{2}} \times\frac{\sqrt{2}-1}{\sqrt{2}-1} =3-2\sqrt{2}$

C-I '0' Head TTT $$\left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^3 = \frac{1}{64}$$ C-II '1' head HTT $$\left( \frac{3}{8} \right) \left( \frac{3}{8} \right) = \frac{9}{64}$$ C-III '2' Head HHT $$\left( \frac{3}{8} \right) \left( \frac{3}{8} \right) = \frac{9}{64}$$ C-IV '3' Heads HHH $$\left( \frac{1}{8} \right) \left( \frac{1}{8} \right) = \frac{1}{64}$$ Total probability = $\frac{5}{16}$ .

Given $\($ $\alpha$ . $\mathrm{R}$ = $\frac{|3(2) + 4(-3) - 5|}{5}$ = $\frac{11}{5}$ $\)$ $\($ (x - h)^2 = $\frac{11}{5}$ (y - k) $\)$ differentiate with respect to $\($ x' $\)$: $\($ 2(x - h) = $\frac{11}{5}$ $\frac{dy}{dx}$ $\)$ again differentiate $\($ 2 = $\frac{11}{5}$ $\frac{d^2y}{dx^2}$ $\)$ $\($ $\frac{11}{dx^2}$ $\frac{d^2y}{dx^2}$ = 10. $\)$

Locus is directrix of parabola $$x - 3 + 4 = 0 \Rightarrow x + 1 = 0$$

Equation of planes are $\vec{r}\cdot(\hat{i}+\hat{j}+\hat{k})-1=0$ $\Rightarrow x+y+z-1=0$ and $\vec{r}\cdot(2\hat{i}+3\hat{j}-\hat{k})+4=0$ $\Rightarrow 2x+3y-z+4=0$ Equation of planes through line of intersection of these planes is $(x+y+z-1)+\lambda(2x+3y-z+4)=0$ $\Rightarrow (1+2\lambda)x+(1+3\lambda)y+(1-\lambda)z-1+4\lambda=0$ But this plane is parallel to $x$-axis whose direction ratios are $(1,0,0)$ $\therefore (1+2\lambda)1+(1+3\lambda)0+(1-\lambda)0=0$ $\lambda=-\dfrac{1}{2}$ $\therefore$ Required plane is $0x+\left(1-\dfrac{3}{2}\right)y+\left(1+\dfrac{1}{2}\right)z-1+4\left(-\dfrac{1}{2}\right)=0$ $\Rightarrow -\dfrac{y}{2}+\dfrac{3}{2}z-3=0$ $\Rightarrow y-3z+6=0$ $\Rightarrow \vec{r}\cdot(\hat{j}-3\hat{k})+6=0$

(2x - 10y^3) $\,$ dy + y $\,$ dx = 0 $\Rightarrow$ $\frac{dx}{dy}$ + $\left$( $\frac{2}{y}$ $\right$) x = 10y^2 I. $\ $F. = e^{$\int$ $\frac{2}{y}$ $\,$ dy} = e^{2 $\ln$(y)} = y^2 Solution of D.E. is $\therefore$ $\ $x $\cdot$ y = $\int$ (10y^2) $\,$ y^2 $\cdot$ dy $$xy^2 = \frac{10y^5}{5} + C \Rightarrow xy^2 = 2y^5 + C$$ It passes through (0, 1) $\rightarrow$ 0 = 2 + C $\Rightarrow$ C = -2 $\therefore$ Curve is xy^2 = 2y^5 - 2 Now, it passes through (2, $\beta$) $$2\beta^2 = 2\beta^5 - 2 \Rightarrow \beta^5 - \beta^2 - 1 = 0$$ $\therefore$ $\beta$ is root of an equation y^5 - y^2 - 1 = 0 Ans.

Given the determinant equation: $$\left| \begin{array}{ccc} \frac{1}{2} & a & 0 \\ b & 2b+1 & 1 \\ 0 & b & 1 \end{array} \right| = 1$$ Expanding the determinant, we have: $$\begin{vmatrix} a & 0 & 1 \\ b & 2b+1 & 1 \\ 0 & b & 1 \end{vmatrix} = \pm 2$$ This simplifies to: $$a(2b + 1 - b) - 0 + 1(b^2 - 0) = \pm 2$$ Thus, $$a = \frac{\pm 2 - b^2}{b+1}$$ Therefore, $$a = \frac{2 - b^2}{b+1} and a = \frac{-2 - b^2}{b+1}$$ The sum of possible values of 'a' is: $$= \frac{-2b^2}{a+1} Ans.$$

Given the determinant $$D = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 2 & 5 \\ 9 & 4 & 28 + [\lambda] \end{vmatrix}$$ we have: $$= -24 - [\lambda] + 15 = -[\lambda] - 9$$ If $[\lambda] + 9 \neq 0$ then there is a unique solution. If $[\lambda] + 9 = 0$ then $D_1 = D_2 = D_3 = 0$, so there are infinite solutions. Hence $\lambda$ can be any real number.

Given equation becomes $$a^2 - (k+1)ab + kb^2 = 0$$ $$a(a-kb) - b(a-kb) = 0$$ $$(a-kb)(a-b) = 0 \Rightarrow a = kb or a = b (reject) $$ Therefore, $a = kb$ Thus, $$3x^2 + 4x + 3 = k \left(3x^2 + 4x + 2\right)$$ $$\Rightarrow 3(k-1)x^2 + 4(k-1)x + (2k-3) = 0$$ For real roots $$D \geq 0$$ $$\Rightarrow 16(k-1)^2 - 4(3(k-1))(2k-3) \geq 0$$ $$\Rightarrow 4(k-1)\{4(k-1) - 3(2k-3)\} \geq 0$$ $$\Rightarrow 4(k-1)\{-2k+5\} \geq 0$$ $$\Rightarrow -4(k-1)\{2k-5\} \geq 0$$ $$\Rightarrow (k-1)(2k-5) \leq 0$$ Therefore, $k \in \left[1, \frac{5}{2}\right]$ Thus, $k \neq 1$ Therefore, $k \in \left(1, \frac{5}{2}\right]$ Ans.

The volume $V$ is given by $V = \ell \cdot b \cdot h = (a - 2x)(b - 2x)x$. Thus, $V(x) = (2x - a)(2x - b)x$. Expanding, $V(x) = 4x^3 - 2(a + b)x^2 + abx$. Differentiating with respect to $x$, we have: $$\frac{d}{dx}V(x) = 12x^2 - 4(a + b)x + ab.$$ Setting the derivative to zero for critical points: $$\frac{d}{dx}(V(x)) = 0 \Rightarrow 12x^2 - 4(a + b)x + ab = 0.$$ This is a quadratic equation in $x$: $$x = \frac{4(a + b) \pm \sqrt{16(a + b)^2 - 48ab}}{2(12)}.$$ Simplifying, we get: $$x = \frac{(a + b) \pm \sqrt{a^2 + b^2 - ab}}{6}.$$ Let $x = \alpha = \frac{6}{(a+b)+\sqrt{a^2+b^2-ab}}$ and $x = \beta = \frac{(a+b)-\sqrt{a^2+b^2-ab}}{6}$. Now, $12(x - \alpha)(x - \beta) = 0$. The sign chart shows that $x = \beta$ is a maximum. Therefore, $x = \beta = \frac{a+b-\sqrt{a^2+b^2-ab}}{b}$.

Given $(p \land q) \Rightarrow ((r \land q) \land p)$ $$\sim (p \land q) \lor ((r \land q) \land p)$$ $$\sim (p \land q) \lor ((r \land q) \land (p \land q))$$ $$\Rightarrow [\sim (p \land q) \lor (p \land q)] \land (\sim (p \land q) \lor (r \land p))$$ $$\Rightarrow t \land [\sim (p \land q) \lor (r \land p)]$$ $$\Rightarrow \sim (p \land q) \lor (r \land p)$$ $$\Rightarrow (p \land q) \Rightarrow (r \land p)$$ Aliter: given statement says "if p and q both happen then p and q and r will happen" it Simply implies "If p and q both happen then 'r' too will happen " i.e. "if p and q both happen then r and p too will happen i.e. $(p \land q) \Rightarrow (r \land p)$

The equations of the circles are given by: $$(x - 2)^2 + y^2 \leq 4$$ $$x^2 + y^2 \leq 4$$ The number of points common in $C_1$ and $C_2$ is 5. The points are $(0, 0)$, $(1, 0)$, $(2, 0)$, $(1, 1)$, $(1, -1)$. Similarly, in $C_2$ and $C_3$ the number of common points is 5. The number of relations is $2^{5 \times 5} = 2^{25}$.

Given $y = 3 \Rightarrow x = 2$. Point is $(2, 3)$. Differentiate with respect to $x$: $$2(y - 2)y' = 1$$ which implies $$y' = \frac{1}{2(y-2)}$$ Therefore, $$y'(2,3) = \frac{1}{2}$$ Thus, $$\frac{y-3}{x-2} = \frac{1}{2} \Rightarrow x - 2y + 4 = 0$$ The area is given by $$\int_{0}^{3} \left( (y-2)^2 + 1 - (2y-4) \right) \, dy$$ which equals $9$ square units.

Given $$y(x) = \cot^{-1} \left[ \frac{\cos \frac{x}{2} + \sin \frac{x}{2} - \sin \frac{x}{2} - \cos \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2} - \sin \frac{x}{2} + \cos \frac{x}{2}} \right]$$ Simplifying, $$y(x) = \cot^{-1} \left( \tan \frac{x}{2} \right) = \frac{\pi}{2} - \frac{x}{2}$$ Differentiating, $$y'(x) = -\frac{1}{2}$$

Given $\tan \theta = \frac{1}{2}$. $\tan(\theta + \alpha) = \frac{x}{b}$, $\tan \alpha = \frac{x}{a+b}$. Therefore, $$\frac{1}{2} + \frac{x}{a+b}$$ implies $$\frac{\frac{1}{2} + \frac{x}{a+b}}{1 - \frac{1}{2} \cdot \frac{x}{a+b}} = \frac{x}{b}.$$ This leads to $$x^2 - 2ax + ab + b^2 = 0.$$

Given $$y = \left(1 - \frac{1}{2}\right)x^2 + \left(1 - \frac{1}{3}\right)x^3 + \ldots$$ This can be rewritten as $$= \left(x^2 + x^3 + x^4 + \ldots\right) - \left(\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots\right)$$ Simplifying further, we have $$= \frac{x^2}{1-x} + x - \left(x + \frac{x^2}{2} + \frac{x^2}{3} + \ldots\right)$$ This simplifies to $$= \frac{x}{1-x} + \ln(1-x)$$ Substituting $x = \frac{1}{2}$, we get $$y = 1 - \ln 2$$ Thus, $$e^{1+y} = e^{1+1-\ln 2}$$ This equals $$= e^{2-(\ln 2)} = \frac{e^2}{2}$$

Given $$I = \int_0^1 \frac{\sqrt{x}}{(1+x)(1+3x)(3+x)} \, dx$$ Let $x = t^2 \Rightarrow dx = 2t \cdot dt$ $$I = \int_0^1 \frac{t(2t)}{(t^2+1)(1+3t^2)(3+t^2)} \, dt$$ $$I = \int_0^1 \frac{(3t^2+1) - (t^2+1)}{(3t^2+1)(t^2+1)(3+t^2)} \, dt$$ $$I = \int_0^1 \frac{dt}{(t^2+1)(3+t^2)} - \int_0^1 \frac{dt}{(1+3t^2)(3+t^2)}$$ $$= \frac{1}{2} \int_0^1 \frac{dt}{t^2+1} - \frac{1}{2} \int_0^1 \frac{dt}{t^2+3} + \frac{1}{8} \int_0^1 \frac{dt}{t^2+3} - \frac{3}{8} \int_0^1 \frac{dt}{1+3t^2}$$ $$= \frac{1}{2} \int_0^1 \frac{dt}{t^2+1} - \frac{3}{8} \int_0^1 \frac{dt}{t^2+3} - \frac{3}{8} \int_0^1 \frac{dt}{1+3t^2}$$ $$= \frac{1}{2} \left( \tan^{-1}(t) \right)_0^1 - \frac{3}{8\sqrt{3}} \left( \tan^{-1} \left( \frac{t}{\sqrt{3}} \right) \right)_0^1$$ $$- \frac{3}{8\sqrt{3}} \left( \tan^{-1}(\sqrt{3}t) \right)_0^1$$ $$= \frac{1}{2} \left( \frac{\pi}{4} \right) - \frac{\sqrt{3}}{8} \left( \frac{\pi}{6} \right) - \frac{\sqrt{3}}{8} \left( \frac{\pi}{3} \right)$$ $$= \frac{\pi}{8} - \frac{\sqrt{3}}{16} \pi$$ $$= \frac{\pi}{8} \left( 1 - \frac{\sqrt{3}}{2} \right)$$

Given $\($ $\lim$_{x $\to$ $\infty$} $\left$( $\sqrt{x^2 - x + 1}$ - ax $\right$) = b $\)$ ($\($ $\infty$ - $\infty$ $\)$). This implies $\($ a > 0 $\)$. Now, $\($ $\lim$_{x $\to$ $\infty$} $\frac{x^2 - x + 1 - a^2 x^2}{\sqrt{x^2 - x + 1} + ax}$ = b $\)$. $\($ $\Rightarrow$ $\lim$_{x $\to$ $\infty$} $\frac{(1-a^2)x^2 - x + 1}{\sqrt{x^2 - x + 1} + ax}$ = b $\)$. $\($ $\Rightarrow$ $\lim$_{x $\to$ $\infty$} $\frac{(1-a^2)x^2 - x + 1}{x \left( \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + a \right)}$ = b $\)$. $\($ $\Rightarrow$ 1 - a^2 = 0 $\Rightarrow$ a = 1 $\)$. Now, $\($ $\lim$_{x $\to$ $\infty$} $\frac{-x + 1}{x \left( \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + a \right)}$ = b $\)$. $\($ $\Rightarrow$ $\frac{-1}{1 + a}$ = b $\Rightarrow$ b = -$\frac{1}{2}$ $\)$. Thus, $\($(a, b) = $\left$(1, -$\frac{1}{2}$$\right$)$\)$.

Given equation $$\sin^4 \theta + \cos^4 \theta - \sin \theta \cos \theta = 0$$ This implies $$1 - \sin^2 \theta \cos^2 \theta - \sin \theta \cos \theta = 0$$ which simplifies to $$2 - (\sin 2\theta)^2 - \sin 2\theta = 0$$ Rearranging gives $$(\sin 2\theta)^2 + (\sin 2\theta) - 2 = 0$$ Factoring, we have $$(\sin 2\theta + 2)(\sin 2\theta - 1) = 0$$ Thus, $$\sin 2\theta = 1 or \sin 2\theta = -2 (not possible)$$ Therefore, $$2\theta = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \frac{13\pi}{2}$$ This gives $$\theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}$$ Summing these, we find $$S = \frac{\pi}{4} + \frac{5\pi}{4} + \frac{9\pi}{4} + \frac{13\pi}{4} = 7\pi$$ Finally, $$\frac{8S}{\pi} = \frac{8 \times 7\pi}{\pi} = 56.00$$

Since $R(3, 5, \gamma)$ lies on the plane $2x - y + z + 3 = 0$. Therefore, $6 - 5 + \gamma + 3 = 0$ implies $\gamma = -4$. Now, direction ratios of line $QS$ are $2, -1, 1$. The equation of line $QS$ is $$\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = \lambda (say)$$ implies $F(2\lambda + 1, -\lambda + 3, \lambda + 4)$. F lies in the plane $$2(2\lambda + 1) - (-\lambda + 3) + (\lambda + 4) + 3 = 0$$ $$\Rightarrow 4\lambda + 2 + \lambda - 3 + \lambda + 7 = 0$$ $$\Rightarrow 6\lambda + 6 = 0 \Rightarrow \lambda = -1$$ implies $F(-1, 4, 3)$. Since, $F$ is the midpoint of $QS$. Therefore, coordinates of $S$ are $(-3, 5, 2)$. So, $SR = \sqrt{36 + 0 + 36} = \sqrt{72}$. $SR^2 = 72$.

Given $\sum \mathrm{P}(X) = 1 \Rightarrow k + 2k + 2k + 3k + k = 1$. Therefore, $k = \frac{1}{9}$. Now, $p = \mathrm{P}\left(\frac{kX<4}{X<3}\right) = \frac{\mathrm{P}(X=2)}{\mathrm{P}(X<3)} = \frac{2k}{k + 9k} = \frac{2k}{9k} = \frac{2}{3}$. Thus, $p = \frac{2}{3}$. Now, $5p = \lambda k$. Therefore, $(5)\left(\frac{2}{3}\right) = \lambda (1/9)$. Thus, $\lambda = 30$.

Given $|z - 3| = \mathrm{Re}(z)$. Let $Z = x = iy$. $$\Rightarrow (x - 3)^2 + y^2 = x^2$$ $$\Rightarrow x^2 + 9 - 6x + y^2 = x^2$$ $$\Rightarrow y^2 = 6x - 9$$ $$\Rightarrow y^2 = 6 \left( x - \frac{3}{2} \right)$$ $z_1$ and $z_2$ lie on the parabola mentioned in eq.(1). $$\arg(z_1 - z_2) = \frac{\pi}{4}$$ $$\Rightarrow Slope of PQ = 1.$$ Let $P \left( \frac{3}{2} + \frac{3}{2} t_1^2, 3t_1 \right)$ and $Q \left( \frac{3}{2} + \frac{3}{2} t_2^2, 3t_2 \right)$. Slope of $PQ = \frac{3(t_2 - t_1)}{\frac{3}{2} (t_2^2 - t_1^2)} = 1$ $$\Rightarrow \frac{2}{t_2 + t_1} = 1$$ $$\Rightarrow t_2 + t_1 = 2$$ $$\mathrm{Im}(z_1 + z_2) = 3t_1 + 3t_2 = 3(t_1 + t_2) = 3(2)$$ Ans. 6.00 Aliter: Let $z_1 = x_1 + iy_1; z_2 = x_2 + iy_2$. $z_1 - z_2 = (x_1 - x_2) + i (y_1 - y_2)$ $$\therefore \arg(z_1 - z_2) = \frac{\pi}{4} \Rightarrow \tan^{-1} \left( \frac{y_1 - y_2}{x_1 - x_2} \right) = \frac{\pi}{4}$$ $$y_1 - y_2 = x_1 - x_2 (1)$$ $$|z_1 - 3| = \mathrm{Re}(z_1) \Rightarrow (x_1 - 3)^2 + y_1^2 = x_1^2 (2)$$ $$|z_2 - 3| = \mathrm{Re}(z_2) \Rightarrow (x_2 - 3)^2 + y_2^2 = x_2^2 (3)$$ Sub (2) & (3) $$(x_1 - 3)^2 - (x_2 - 3)^2 + y_1^2 - y_2^2 = x_1^2 - x_2^2$$ $$(x_1 - x_2)(x_1 + x_2 - 6) + (y_1 - y_2)(y_1 + y_2) = (x_1 - x_2)(x_1 + x_2)$$ $$x_1 + x_2 - 6 + y_1 + y_2 = x_1 + x_2 \Rightarrow y_1 + y_2 = 6.$$

3n type $\rightarrow 3, 6, 9 = P$ 3n - 1 type $\rightarrow 2, 5 = Q$ 3n - 2 type $\rightarrow 1, 4 = R$ Number of subset of $S$ containing one element which are not divisible by 3 $= \binom{2}{1} + \binom{2}{1} = 4$ Number of subset of $S$ containing two numbers whose sum is not divisible by 3 $$= \binom{3}{1} \times \binom{2}{1} + \binom{3}{1} \times \binom{2}{1} + \binom{2}{2} + \binom{2}{2} = 14$$ Number of subsets containing 3 elements whose sum is not divisible by 3 $$= \binom{3}{2} \times \binom{4}{1} + \left( \binom{2}{2} \times \binom{2}{1} \right) 2 + \binom{3}{1} \left( \binom{2}{2} + \binom{2}{2} \right) = 22$$ Number of subsets containing 4 elements whose sum is not divisible by 3 $$= \binom{3}{3} \times \binom{4}{1} + \binom{3}{2} \left( \binom{2}{2} + \binom{2}{2} \right) + \left( \binom{3}{1} \binom{2}{1} \times \binom{2}{1} \right) 2$$ $$= 4 + 6 + 12 = 22.$$ Number of subsets of $S$ containing 5 elements whose sum is not divisible by 3. $$= \binom{3}{3} \left( \binom{2}{2} + \binom{2}{2} \right) + \left( \binom{3}{2} \binom{2}{1} \times \binom{2}{2} \right) \times 2 = 2 + 12 = 14$$ Number of subsets of $S$ containing 6 elements whose sum is not divisible by 3 $= 4$ $\Rightarrow$ Total subsets of Set $A$ whose sum of digits is not divisible by 3 $= 4 + 14 + 22 + 22 + 14 + 4 = 80.$

Since, point $A(\sec \theta, 2 \tan \theta)$ lies on the hyperbola $$2x^2 - y^2 = 2$$ Therefore, $2 \sec^2 \theta - 4 \tan^2 \theta = 2$ $$\Rightarrow 2 + 2 \tan^2 \theta - 4 \tan^2 \theta = 2$$ $$\Rightarrow \tan \theta = 0 \Rightarrow \theta = 0$$ Similarly, for point $B$, we will get $\phi = 0$. But according to question $\theta + \phi = \frac{\pi}{2}$ which is not possible. Hence it must be a 'BONUS'.

Slope of line joining centres of circles $= \frac{4}{3} = \tan \theta$. $$\Rightarrow \cos \theta = \frac{3}{5}, \sin \theta = \frac{4}{5}$$ Now using parametric form $$\frac{x-1}{\cos \theta} = \frac{y-2}{\sin \theta} = \pm 5$$ $\oplus$ $(x, y) = (1 + 5 \cos \theta, 2 + 5 \sin \theta)$ $(\alpha, \beta) = (4, 6)$ $\Theta(x, y) = (\gamma, \delta) = (1 - 5 \cos \theta, 2 - 5 \sin \theta)$ $(\gamma, s) = (-2, -2)$ $$\Rightarrow |(\alpha + \beta)(\gamma + \delta)| = |10x - 4| = 40$$

Given $$3(1+6)^{22} + 2 \cdot (1+9)^{22} - 44 = (3+2-44) = 18 \cdot I$$ $$= -39 + 18 \cdot I$$ $$= (54 - 39) + 18(I - 3)$$ $$= 15 + 18I_1$$ Therefore, the remainder is 15.

Given $\sigma_b^2 = 2$ (variance of boys) $n_1 = no. of boys$. $\bar{x}_b = 12$ and $n_2 = no. of girls$. $\sigma_g^2 = 2$. $\bar{x}_g = \frac{50 \times 15 - 12 \times \sigma_b}{30} = \frac{750 - 12 \times 20}{30} = 17 = \mu$. Variance of combined series: $$\sigma^2 = \frac{n_1 \sigma_b^2 + n_2 \sigma_g^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)^2} \left( \bar{x}_b - \bar{x}_g \right)^2$$ $$\sigma^2 = \frac{20 \times 2 + 30 \times 2}{20 + 30} + \frac{20 \times 30}{(20 + 30)^2} (12 - 17)^2$$ $$\sigma^2 = 8.$$ Therefore, $\mu + \sigma^2 = 17 + 8 = 25$.

Given $$\int \frac{2e^x}{4e^x + 7e^{-x}} \, dx + 3 \int \frac{e^{-x}}{4e^x + 7e^{-x}} \, dx$$ Let $4e^x + 7 = T$ $$8e^x \, dx = dT$$ $$2e^x \, dx = \frac{dT}{4}$$ Let $4 + 7e^{-2x} = t$ $$-14e^{-2x} \, dx = dt$$ $$e^{-2x} \, dx = -\frac{dt}{14}$$ $$\int \frac{dT}{4T} - \frac{3}{14} \int \frac{dt}{t}$$ $$= \frac{1}{4} \log T - \frac{3}{14} \log t + C$$ $$= \frac{1}{4} \log (4e^x + 7) - \frac{3}{14} \log (4 + 7e^{-2x}) + C$$ $$= \frac{1}{14} \left[ \frac{1}{2} \log (4e^x + 7e^{-x}) + \frac{13}{2} x \right] + C$$ $$u = \frac{13}{2}, \; v = \frac{1}{2} \Rightarrow u + v = 7$$ Aliter: $$2e^x + 3e^{-x} = A (4e^x + 7e^{-x}) + B (4e^x - 7e^{-x}) + \lambda$$ $$2 = 4A + 4B; 3 = 7A - 7B; \; \lambda = 0$$ $$A + B = \frac{1}{2}$$ $$A - B = \frac{3}{7}$$ $$A = \frac{1}{2} \left( \frac{1}{2} + \frac{3}{7} \right) = \frac{7 + 6}{28} = \frac{13}{28}$$ $$B = A - \frac{3}{7} = \frac{13}{28} - \frac{3}{7} = \frac{13 - 12}{28} = \frac{1}{28}$$ $$\int \frac{5}{28} \, dx + \frac{1}{28} \int \frac{4e^x - 7e^{-x}}{4e^x + 7e^{-x}} \, dx$$ $$\frac{13}{28} x + \frac{1}{28} \ln |4e^x + 7e^{-x}| + C$$ $$u = \frac{13}{2}, \; v = \frac{1}{2}$$ $$\Rightarrow u + v = 7$$

The equations for the focal lengths are given by: $$\frac{1}{f_1} = (\mu_1 - 1) \left(\frac{1}{R}\right)$$ $$\frac{1}{f_2} = (\mu_2 - 1) \left(-\frac{1}{R}\right)$$ Adding these equations, we have: $$\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{f_{eq}} = \frac{(\mu_1 - 1) - (\mu_2 - 1)}{R}$$ Simplifying, we get: $$\frac{1}{f_{eq}} = \frac{(\mu_1 - \mu_2)}{R}$$ Therefore, the equivalent focal length is: $$\frac{R}{f_{eq}} = (\mu_1 - \mu_2)$$

$(m_{1}g-2T)=m_{1}a \qquad (1)$ $T-m_{2}g=m_{2}(2a)$ $2T-2m_{2}g=4m_{2}a \qquad (2)$ From (1) and (2), $m_{1}g-2m_{2}g=(m_{1}+4m_{2})a$ $a=\frac{(8-4)g}{(8+8)}$ $=\frac{4}{16}g$ $=\frac{g}{4}$ $a=\frac{10}{4}\ \mathrm{m/s^{2}}$ $S=\frac{1}{2}at^{2}$ $0.2=\frac{1}{2}\times\frac{10}{4}\times t^{2}$ $\frac{0.2\times2\times4}{10}=t^{2}$ $t=0.4\ \mathrm{s}$

Given $\alpha = \frac{I_C}{I_E}$, $\beta = \frac{I_C}{I_B}$; $I_E = I_C + I_B$. $$\alpha = \frac{I_C}{I_C + I_B} = \frac{I_C / I_B}{I_C / I_B + 1} = \frac{\beta}{\beta + 1}$$ $$1 + \frac{1}{\beta} = \frac{1}{\alpha}$$ $$\frac{1}{\beta} = \frac{1 - \alpha}{\alpha}$$ $$\beta = \frac{\alpha}{1 - \alpha}$$

Given $H = \frac{1}{2} g t^2$. $$\frac{9.8 \times 2}{9.8} = t^2$$ $t = \sqrt{2} sec$ $\Delta t$: time interval between drops $h = \frac{1}{2} g (\sqrt{2} - \Delta t)^2$ $0 = \frac{1}{2} g (\sqrt{2} - 2 \Delta t)^2$ $\Delta t = \frac{1}{\sqrt{2}}$ $h = \frac{1}{2} g \left( \sqrt{2} - \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} \times 9.8 \times \frac{1}{2} = \frac{9.8}{4} = 2.45 \, m$ $H - h = 9.8 - 2.45 = 7.35 \, m$

From conservation of angular momentum we get $$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega$$ $$\omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$$ $$k_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2$$ $$k_f = \frac{1}{2} (I_1 + I_2) \omega^2$$ $$k_i - k_f = \frac{1}{2} \left[ I_1 \omega_1^2 + I_2 \omega_2^2 - \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{I_1 + I_2} \right]$$ Solving above we get $$k_i - k_f = \frac{1}{2} \left( \frac{I_1 I_2}{I_1 + I_2} \right) (\omega_1 - \omega_2)^2$$

$(V_D - V) C_2 + (V_D - 0) C_3 = 0$ $(V_D - V) 6 + (V_D - 0) 12 = 0$ $V_D - V + 2V_D = 0$ $$V_D = \frac{V}{3}$$ $q_2 = (V - V_D) C_2$ $$= \left( V - \frac{V}{3} \right) (6 \,\mu F)$$ $$q_2 = (4V)\,\mu F$$ $q_3 = (V_D - 0) C_3$ $$= \frac{V}{3} \times 12\,\mu F$$ $$= 4V\,\mu F$$ $q_1 = (V - 0) C_1$ $$= V (2\,\mu F)$$ $$q_1 : q_2 : q_3 = 2 : 4 : 4$$ $$q_1 : q_2 : q_3 = 1 : 2 : 2$$

Given $R = 75 \times 10^2 \pm 5\%$ of $7500$. Calculating, $R = (7500 \pm 375) \, \Omega$.

Given $h$: height of antenna and $\lambda$: wavelength of signal. $h h$ $\lambda > 400 \, \mathrm{m}$

The root mean square velocity $V_{rms}$ is given by $$V_{rms} = \sqrt{\frac{3kT}{M}}$$ The ratio of $V_{rms}$ for $\mathrm{O_2}$ and $\mathrm{H_2}$ is $$\left( V_{rms} \right)_{\mathrm{O_2}} / \left( V_{rms} \right)_{\mathrm{H_2}} = \sqrt{\frac{M_{\mathrm{H_2}}}{M_{\mathrm{O_2}}}} = \sqrt{\frac{2}{32}}$$ The velocity of $\mathrm{H_2}$ is $$\left( V_{mm} \right)_{\mathrm{H_2}} = 4 \times \left( V_{mms} \right)_{\mathrm{O_2}}$$ Calculating gives $$= 4 \times 160$$ $$= 640 \, \mathrm{m/s}$$

Given the formula for emf as $emf = BLV$. Substituting the values, we have: $$emf = 1 \cdot (2R) \cdot 1$$ $$= 2 \, \mathrm{V}$$

The gravitational potential at point A is given by: $$V_A = \left[ -\frac{GM_1}{r} - \frac{GM_2}{R} \right]$$ Substituting the given values: $$= \left[ -\frac{50}{25}G - \frac{100}{50}G \right]$$ Simplifying, we get: $$= -4G$$

Given $$I_{max} = \frac{50}{2} = 25 \, mA$$ $$\mathbf{R} = \frac{V}{I} = \frac{50 \, mV}{25 \, mA} = 2 \, \Omega$$

$K E_{\max}=\dfrac{hc}{\lambda_i}-\phi$ or $eV_0=\dfrac{hc}{\lambda_i}-\phi$ when $\lambda_i=670.5\ \mathrm{nm};\qquad V_0=0.48$ when $\lambda_i=474.6\ \mathrm{nm};\qquad V_0=?$ $e(0.48)=\dfrac{1240}{670.5}-\phi$ So, $e(V_0)=\dfrac{1240}{474.6}-\phi$

SI unit of Rydberg const. = $\mathrm{m^{-1}}$ SI unit of Planck's const. = $\mathrm{kg \, m^2 \, s^{-1}}$ SI unit of Magnetic field energy density = $\mathrm{kg \, m^{-1} \, s^{-2}}$ SI unit of coeff. of viscosity = $\mathrm{kg \, m^{-1} \, s^{-1}}$

Density $= \left[ F^a L^b T^c \right]$ $\left[ ML^{-3} \right] = \left[ M^a L^a T^{-2a} L^b T^c \right]$ $\left[ M^1 L^{-3} \right] = \left[ M^a L^{a+b} T^{-2a+c} \right]$ $a = 1; \quad a + b = -3; \quad -2a + c = 0$ $1 + b = -3 \quad c = 2a$ $b = -4 \quad c = 2$ So, density $= \left[ F^1 L^{-4} T^2 \right]$

When $x < a$: $$B_1(2 \pi x) = \mu_0 \left( \frac{i_0}{\pi a^2} \right) \pi x^2$$ $$B(2 \pi x) = \frac{\mu_0 i_0 a^2}{a^2}$$ $$B_1 = \frac{\mu_0 i_0 x}{2 \pi a^2}$$ When $a < x < b$: $$B_2(2 \pi x) = \mu_0 i_0$$ $$B_2 = \frac{\mu_0 i_0}{2 \pi x}$$ $$\frac{B_1}{B_2} = \frac{\mu_0 i_0 \frac{x}{2 \pi a^2}}{\frac{\mu_0 i_0}{2 \pi x}} = \frac{x^2}{a^2}$$

Change in P.E. = Heat energy $$mgh = mS \Delta T$$ $$\Delta T = \frac{gh}{S}$$ $$= \frac{10 \times 63}{4200 \, \mathrm{J/kg}^\circ \mathrm{C}}$$ $$= 0.147^\circ \mathrm{C}$$

Given $$H = \frac{U^2 \sin^2 \theta}{2g}$$ Substituting the values, $$= \frac{(25)^2 \cdot (\sin 45)^2}{2 \times 10}$$ $$= 15.625 \, \mathrm{m}$$ For time, $$T = \frac{U \sin \theta}{g}$$ Substituting the values, $$= \frac{25 \times \sin 45^\circ}{10}$$ $$= 2.5 \times 0.7$$ $$= 1.77 \, \mathrm{s}$$

Given $$I_{max} = \left( \sqrt{I_1} + \sqrt{I_2} \right)^2$$ which simplifies to $$= 4I_0$$

Given $\varepsilon = \frac{2k\lambda}{R} \sin\left(\frac{\theta}{2}\right) \left(-\hat{i}\right)$. $\lambda = \left(\frac{-Q}{R\theta}\right) = \left(\frac{-Q}{R \cdot \frac{2\pi}{3}}\right)$. $\lambda = \frac{-3Q}{2\pi R}$. $\varepsilon = \frac{2k}{R} \cdot \frac{-3Q}{2\pi R} \cdot \sin(60^\circ) \left(-\hat{i}\right)$. $\varepsilon = \frac{3\sqrt{3}Q}{8\pi^2 \varepsilon_0 R^2} \left(+\hat{i}\right)$.

Given $$y_1 = 10 \sin \left( 3\pi t + \frac{\pi}{3} \right) \Rightarrow Amplitude = 10$$ $$y_2 = 5 (\sin 3\pi t + \sqrt{3} \cos 3\pi t)$$ $$y_2 = 10 \left( \frac{1}{2} \sin 3\pi t + \frac{\sqrt{3}}{2} \cos 3\pi t \right)$$ $$y_2 = 10 \left( \cos \frac{\pi}{3} \sin 3\pi t + \sin \frac{\pi}{3} \cos 3\pi t \right)$$ $$y_2 = 10 \sin \left( 3\pi t + \frac{\pi}{3} \right) \Rightarrow Amplitude = 10$$ So ratio of amplitudes = $\frac{10}{10}$ = 1

Number of different wavelengths is given by the formula: $$\frac{n(n-1)}{2}$$ Substituting the values, we have: $$\frac{6 \times (6 - 1)}{2} = \frac{6 \times 5}{2} = 15$$

When unregulated voltage is 14 V voltage across zener diode must be 10 V So potential difference across resistor $\Delta V_{R_s} = 4$ V and $P_{\text{zener}} = 2$ W $VI = 2$ $I = \dfrac{2}{10} = 0.2\,\text{A}$ $\Delta V_{R_s} = IR_s$ $4 = 0.2R_s \Rightarrow R_s = \dfrac{40}{2} = 20\,\Omega$

$B.S_1=\frac{T_{\max}}{8\times10^{-7}}$ $\Rightarrow T_{1\max}=8\times1.25\times100$ $=1000\,\mathrm{N}$ $B.S_2=\frac{T_{2\max}}{4\times10^{-7}}$ $\Rightarrow T_{2\max}=4\times1.25\times100$ $=500\,\mathrm{N}$ $m=\frac{500-100}{10}$ $=40\,\mathrm{kg}$

Given $V' = \sqrt{5gR} = \sqrt{5 \times 10 \times 0.5}$. $V' = 5 \, \mathrm{m/s}$. $m_1 \, V = m_2 \times 5 - m_1 \times 100$. $$\frac{10}{1000} \times V = 5 - \frac{10}{1000} \times 100$$ $V = 400 \, \mathrm{m/s}$.

Given $\cos \phi = \frac{R}{\sqrt{R^2 + 3R^2}}$ and $\cos \phi' = \frac{R}{\sqrt{R^2 + R^2}}$. This simplifies to $\cos \phi = \frac{1}{\sqrt{10}} = \frac{1}{\sqrt{10}}$ and $\cos \phi' = \frac{1}{\sqrt{2}}$. Therefore, $\frac{\cos \phi'}{\cos \phi} = \frac{\sqrt{10}}{\sqrt{2}} = \frac{\sqrt{5}}{1}$. Thus, $x = 1$.

Given $$R_{eq open} = \frac{3R}{2}$$ $$R_{eq closed} = 2 \times \frac{R \times 2R}{3R} = \frac{4R}{3}$$ Therefore, $$R_{eq open} \times R_{eq closed} = \frac{3R}{2} \times \frac{3}{4R} = \frac{9}{8}$$ Thus, $$x = 9$$

Given $|\mathbf{B}| = \frac{|\mathbf{E}|}{C} = \frac{6}{3 \times 10^8}$. This equals $2 \times 10^{-8} \, \mathrm{T}$. Therefore, $x = 2$.

Given $\frac{\lambda}{4} = \ell \Rightarrow \lambda = 4\ell$. The frequency $f = \frac{V}{\lambda} = \frac{V}{4\ell}$. Therefore, $$250 = \frac{340}{4\ell}$$ which implies $$\ell = \frac{34}{4 \times 25} = 0.34 \, \mathrm{m}$$ Thus, $\ell = 34 \, \mathrm{cm}$.

(1) Standard enthalpy of formation for alkali metal bromides becomes more negative on descending down the group. (2) In case of CsI, lattice energy is less, but $\mathrm{Cs^+}$ is having less hydration enthalpy due to which it is less soluble in water. (3) For alkali metal fluorides, the solubility in water increases from lithium to caesium. LiF is least soluble in water. (4) Standard enthalpy of formation for LiF is most negative among alkali metal fluorides.

The reaction is given by: $$\mathrm{Cr^{3+} }+ NaOH \xrightarrow{dil.} Cr_2O_3 \cdot (H_2O)_n (precipitate)$$

Statement 1 is true. But it consumes 3 moles of Grignard Reagent (G R). So statement 2 is false.

Ozone in the stratosphere is a product of UV radiations acting on dioxygen ($\text{O}_2$) molecules. $\text{O}_2(\text{g}) \xrightarrow{\text{UV}} \text{O(g)} + \text{O(g)}$ $\text{O(g)} + \text{O}_2(\text{g}) \xrightleftharpoons{\text{UV}} \text{O}_3(\text{g})$

Q5 (1) (a) Shows intra molecular H-bonding (b) Shows significant intermolecular H-bonding (c) It do not show intermolecular H-bonding due to steric hindrance.

Histamine stimulates the secretion of $\mathrm{HCl}$. Histamine structure

1. $\mathrm{PbO_2} + 2\mathrm{H_2O} \rightarrow \mathrm{Pb(OH)_4}$ 2. $\mathrm{Na_2O_2} + 2\mathrm{H_2O} \rightarrow 2\mathrm{NaOH} + \mathrm{H_2O_2}$ this reaction is possible at room temperature 3. $\mathrm{SnO_2} + 2\mathrm{H_2O} \rightarrow \mathrm{Sn(OH)_4}$ 4. Acidified $\mathrm{BaO_2.8H_2O}$ gives $\mathrm{H_2O_2}$ after evaporation.

All gives propanoic acid as product but option 4 gives butanoic as product. $$\mathrm{CH_3CH_2CH_2Br} \xrightarrow[Dry ether]{Mg} \mathrm{CH_3CH_2CH_2MgBr}$$ $$\mathrm{CH_3CH_2CH_2MgBr} \xrightarrow{CO_2} \mathrm{CH_3CH_2CH_2C(=O)OMgBr}$$ $$\mathrm{CH_3CH_2CH_2C(=O)OMgBr} \xrightarrow{H_3O^+} \mathrm{CH_3CH_2CH_2C(=O)OH}$$ Butanoic acid

The correct order of ionic radii is $\mathrm{P^{3-} > S^{2-} > Cl^{-} > K^{+} > Ca^{2+}}$. All the given species are isoelectronic species. In isoelectronic species, size increases with increase of negative charge and size decreases with increase in positive charge.

The reaction begins with the addition of 2 equivalents of $\mathrm{CH_3MgBr}$ to the nitrile group $\mathrm{N\equiv C}$, forming a Grignard reagent intermediate. This intermediate undergoes a reaction with the ketone group, resulting in the formation of a tertiary alcohol. Upon treatment with $\mathrm{H_3O^+}$, the intermediate is protonated, leading to the formation of a stable alcohol. Finally, dehydration occurs in the presence of $\mathrm{H_2SO_4}$, resulting in the formation of the final product with a double bond.

The reaction involves an elimination process (E2) where the base $\mathrm{CH_3O^-}$ in $\mathrm{CH_3OH}$ abstracts a proton from the $eta$-carbon. This leads to the formation of a double bond, resulting in the major product (A) which is an alkene: $$CH_3-CH_2-C \equiv CH_2$$ This is the major product due to the stability of the formed alkene.

$\mathrm{O_2F_2}$ oxidises plutonium to $\mathrm{PuF_6}$, and the reaction is used for removing plutonium as $\mathrm{PuF_6}$ from spent nuclear fuel.

In the lyophilic colloids, the colloidal particles are extensively solvated.

The reaction involves the deprotonation of phenol by a base to form the phenoxide ion. This ion then undergoes an $S_N^2$ reaction with the alkyl bromide to form the ether product.

The reaction is given by: $$2\mathrm{KMnO_4} \xrightarrow{\Delta \, 200^\circ \mathrm{C}} \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2}$$ In $\mathrm{K_2MnO_4}$, manganese oxidation state is $+6$ and hence it has one unpaired $e^-$.

In Seliwanoff's reagent, Cu is not present. In Barfoed, Biuret and in Benedict reagent Cu is present.

Sucrose is formed by $\alpha - \mathrm{D}(+)$ Glucose $+ \beta - \mathrm{D}(-)$ Fructose. We obtain these monomers on hydrolysis.

Name of ore/mineral (a) Calamine $\mathrm{ZnCO_3}$ (b) Malachite $\mathrm{CuCO_3 \cdot Cu(OH)_2}$ (c) Siderite $\mathrm{FeCO_3}$ (d) Sphalerite $\mathrm{ZnS}$

When red phosphorus is heated in a sealed tube at $803 \, \mathrm{K}$, $\alpha$-black phosphorus is formed.

The reaction starts with the reduction of the nitro group to an amine using $\mathrm{H_2/Pd}$ in $\mathrm{C_2H_5OH}$, forming compound (A). Compound (A) is then reacted with acetic anhydride in a nucleophilic acyl substitution reaction to form an amide. The final product is compound (B), which is the major product.

Given $K_{700} = 6.36 \times 10^{-3} \, \mathrm{s^{-1}}$, $K_{600} = x \times 10^{-6} \, \mathrm{s^{-1}}$, $E_a = 209 \, \mathrm{kJ/mol}$. Applying: $$\log \left( \frac{K_{T_2}}{K_{T_1}} \right) = \frac{-E_a}{2.303R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$$ $$\log \left( \frac{K_{700}}{K_{600}} \right) = \frac{-E_a}{2.303R} \left( \frac{1}{700} - \frac{1}{600} \right)$$ $$\log \left( \frac{6.36 \times 10^{-3}}{K_{600}} \right) = \frac{+209 \times 1000}{2.303 \times 8.31} \left( \frac{100}{700 \times 600} \right)$$ $$\log(6.36 \times 10^{-3}) - \log K_{600} = 2.6$$ $$\Rightarrow \log K_{600} = -2.19 - 2.6 = -4.79$$ $$\Rightarrow K_{600} = 10^{-4.79} = 1.62 \times 10^{-5}$$ $$= 16.2 \times 10^{-6}$$ $$= x \times 10^{-6}$$ $$\Rightarrow x = 16$$

The number of optical isomers for $[\mathrm{Cr}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$ is two.

Applying; $(n_I + n_{II})_{initial} = (n_I + n_{II})_{final}$ Assuming the system attains a final temperature of $T$ (such that $300 < T < 60$) $$\Rightarrow \left( Heat lost by N_2 of container I \right) = \left( Heat gained by N_2 of container II \right)$$ $$\Rightarrow n_I C_m (300 - T) = n_{II} C_m (T - 60)$$ $$\Rightarrow \left( \frac{2.8}{28} \right) (300 - T) = \left( \frac{0.2}{28} \right) (T - 60)$$ $$\Rightarrow 14(300 - T) = T - 60$$ $$\Rightarrow \frac{(14 \times 300 + 60)}{15} = T$$ $$\Rightarrow T = 284 \, \mathrm{K} (final temperature)$$ If the final pressure $= P$ $$\Rightarrow (n_I + n_{II})_{final} = \left( \frac{3.0}{28} \right)$$ $$\Rightarrow \frac{P}{RT} (V_I + V_{II}) = \frac{3.0 \, \mathrm{gm}}{28 \, \mathrm{gm/mol}}$$ $$P = \left( \frac{3}{28} \, \mathrm{mol} \right) \times 8.31 \, \frac{\mathrm{J}}{\mathrm{mol} \, \mathrm{K}} \times \frac{284 \, \mathrm{K}}{3 \times 10^{-3} \, \mathrm{m}^3} \times 10^{-5} \, \frac{\mathrm{bar}}{\mathrm{Pa}}$$ $$\Rightarrow 0.84287 \, \mathrm{bar}$$ $$\Rightarrow 84.28 \times 10^{-2} \, \mathrm{bar}$$ $$\Rightarrow 84$$

The reaction is given by $$\mathrm{C_3H_8_{(g)} + 5O_2_{(g)} \rightarrow 3CO_2_{(g)} + 4H_2O_{(\ell)}}$$ At $t = 0$, the moles are $2.27$ mole and $31.25$ mol. At $t = \infty$, the moles are $0$, $19.9$ mol, $6.81$ mol, and $9.08$ mol. The mole fraction of $\mathrm{CO_2}$ in the final reaction mixture (heterogeneous) is calculated as $$X_{\mathrm{CO_2}} = \frac{6.81}{19.9 + 6.81 + 9.08}$$ $$= 0.1902 = 19.02 \times 10^{-2}$$ Thus, the answer is $19$.

Given the molality, $$molality = \frac{\left( \frac{40}{180} \right) mol}{0.2 Kg} = \left( \frac{10}{9} \right) molal$$ implies $$\Delta T_f = T_f - T'_f = 1.86 \times \frac{10}{9}$$ implies $$T'_f = 273.15 - 1.86 \times \frac{10}{9}$$ $$= 271.08 K$$ $$\simeq 271 K (nearest-integer)$$

Given $K = \frac{1}{R} \times \ell / A = \left( \left( \frac{1}{1500} \right) \times 1.14 \right) \, \mathrm{S \, cm^{-1}}$. Therefore, $\wedge_m = 1000 \times \frac{\left( \frac{1.14}{1500} \right)}{0.001} \, \mathrm{S \, cm^2 \, mol^{-1}} = 760 \, \mathrm{S \, cm^2 \, mol^{-1}}$. Thus, $\Rightarrow 760$.

Energy emitted in 0.1 sec. $$= 0.1 \, sec. \times 10^{-3} \, \frac{\mathrm{J}}{\mathrm{s}}$$ $$= 10^{-4} \, \mathrm{J}$$ If 'n' photons of $\lambda = 1000 \, \mathrm{nm}$ are emitted, then; $$10^{-4} = n \times \frac{hc}{\lambda}$$ $$\Rightarrow 10^{-4} = \frac{n \times 6.63 \times 10^{-34} \times 3 \times 10^8}{1000 \times 10^{-9}}$$ $$\Rightarrow n = 5.02 \times 10^{14} = 50.2 \times 10^{13}$$ $$\Rightarrow 50 \; (nearest integer)$$

Moles of $\mathrm{NH_4HS}$ initially taken $= \frac{5.1 \, \mathrm{g}}{51 \, \mathrm{g/mol}} = 0.1 \, \mathrm{mol}$. Volume of vessel $= 2 \, \ell$. $$\mathrm{NH_4HS_{(s)}} \rightleftharpoons \mathrm{NH_3}_{(g)} + \mathrm{H_2S}_{(g)}$$ At $t = 0$, $0.1 \, \mathrm{mol}$. At $t = \infty$, $0.1(1 - 0.2)$, $0.1 \times 0.2$, $0.1 \times 0.2$. Partial pressure of each component: $$P = \frac{nRT}{V} = \frac{0.1 \times 0.2 \times 0.082 \times 300}{2} = 0.246 \, \mathrm{atm}$$ $$\therefore \, K_P = P_{\mathrm{NH_3}} \times P_{\mathrm{H_2S}} = (0.246)^2 = 0.060516$$ $$= 6.05 \times 10^{-2}$$ $$\Rightarrow 6$$

The first structure is $\mathrm{SO_3}$, which is trigonal planar. The second structure is $\mathrm{NO_3^-}$, which is also trigonal planar. The third structure is $\mathrm{CO_3^{2-}}$, which is trigonal planar. The fourth structure is $\mathrm{PCl_3}$, which is pyramidal. Hence non-pyramidal species are $\mathrm{SO_3}$, $\mathrm{NO_3^-}$ and $\mathrm{CO_3^{2-}}$.

Given $$T_{min} = \left( \frac{\Delta^0 H}{\Delta^0 S} \right)$$ $$\Delta^0 H_{rxn} = \left[ \Delta^0_f H(\mathrm{Fe}) + \Delta^0_f H(\mathrm{CO}) \right] - \left[ \Delta^0_f H(\mathrm{FeO}) + \Delta^0_f H(\mathrm{C(graphite)}) \right]$$ $$= [0 - 110.5] - [-266.3 + 0]$$ $$= 155.8 \, \mathrm{kJ/mol}$$ $$\Delta^0 S_{rxn} = \left[ \Delta^0 S(\mathrm{Fe}) + \Delta^0 S(\mathrm{CO}) \right] - \left[ \Delta^0 S(\mathrm{FeO}) + \Delta^0 S(\mathrm{C(graphite)}) \right]$$ $$= [27.28 + 197.6] - [57.49 + 5.74]$$ $$= 161.65 \, \mathrm{J/mol \cdot K}$$ $$T_{min} = \frac{155.8 \times 10^3 \, \mathrm{J/mol}}{161.65 \, \mathrm{J/mol \cdot K}} = 963.8 \, \mathrm{K}$$ $$\simeq 964 \, \mathrm{K} \text{ (nearest integer)}$$