JEE Main 9 April 2024 Shift 2 question paper with solutions

Given the limit expression: $$\lim_{x \to 0} \frac{e - e^{\frac{1}{2x} \ln(1+2x)}}{x}$$ We can rewrite it as: $$= \lim_{x \to 0} \frac{e \left( e^{\frac{\ln(1+2x)}{2x} - 1} - 1 \right)}{x}$$ Simplifying further: $$= \lim_{x \to 0} (-e) \frac{\ln(1+2x) - 2x}{2x^2}$$ Evaluating the limit: $$= (-e) \times (-1) \frac{4}{2 \times 2} = e$$

Given $\($ $\frac{x-1}{2-1}$ = $\frac{y-2}{3-2}$ = $\frac{z-3}{5-3}$ $\)$. This implies $\($ $\frac{x-1}{1}$ = $\frac{y-2}{1}$ = $\frac{z-3}{2}$ = $\lambda$ $\)$. Point B is $\($ (1+$\lambda$, 2+$\lambda$, 3+2$\lambda$) $\)$. The direction ratios of AB are $\($ $\left$ $\)$. For point B $\($ $\left$( $\frac{5}{3}$, $\frac{8}{3}$, $\frac{13}{3}$ $\right$) $\)$, we have $\($ $\frac{3\lambda - 8}{3\lambda - 5}$ = $\frac{2}{1}$ $\)$. This implies $\($ 3$\lambda$ - 8 = 6$\lambda$ - 10 $\)$. Solving gives $\($ 3$\lambda$ = 2 $\)$, so $\($ $\lambda$ = $\frac{2}{3}$ $\)$. The distance AB is $\($ $\frac{\sqrt{36 + 9 + 36}}{3}$ = $\frac{9}{3}$ = 3 $\)$.

Given $\sqrt{1 - \left(y'(x)\right)^2} = y(x)$. $$1 - \left(\frac{dy}{dx}\right)^2 = y^2$$ $$\left(\frac{dy}{dx}\right)^2 = 1 - y^2$$ $$\frac{dy}{\sqrt{1-y^2}} = dx OR \frac{dy}{\sqrt{1-y^2}} = -dx$$ This implies $\sin^{-1} y = x + c$, $\sin^{-1} y = -x + c$. For $x = 0$, $y = 0$ implies $c = 0$. Thus, $\sin^{-1} y = x$, as $y \geq 0$. Therefore, $\sin x = y$. This implies $\frac{dy}{dx} = \cos x$. Then, $\frac{d^2 y}{dx^2} = -\sin x$. Thus, $-\sin x + \sin x + 1 = 1$.

Given $\($ $\frac{z - 2i}{z + 2i}$ + $\frac{\overline{z} + 2i}{\overline{z} - 2i}$ = 0 $\)$. $\($ zz - 2iz - 2iz + 4(-1) + z$\overline{z}$ + 2zi + 2z$\overline{i}$ + 4(-1) = 0 $\)$ $\($ $\Rightarrow$ 2|z|^2 = 8 $\Rightarrow$ |z| = 2 $\)$ $\($ |z - (6 + 8i)|_{maximum} = 10 + 2 = 12 $\)$

Given the equation of the ellipse $\frac{x^2}{18} + \frac{y^2}{6} = 1$. The line $y = x$ intersects the ellipse. Solving $\frac{x^2}{18} + \frac{3x^2}{18} = 1$ gives $4x^2 = 18$ which implies $x^2 = \frac{9}{2}$. The integral to find the area is $\int_{\frac{3}{\sqrt{2}}}^{\frac{3\sqrt{2}}{2}} \frac{\sqrt{18 - x^2}}{\sqrt{3}} \, dx$. This evaluates to: $$\frac{1}{\sqrt{3}} \left( x \frac{\sqrt{18 - x^2}}{2} + \frac{18}{2} \sin^{-1} \frac{x}{3\sqrt{2}} \right)_{\frac{3}{\sqrt{2}}}^{\frac{3\sqrt{2}}{2}}$$ Simplifying gives: $$= \frac{1}{\sqrt{3}} \left( 9 \times \frac{\pi}{2} - \frac{3}{2\sqrt{2}} \times \frac{3\sqrt{3}}{\sqrt{2}} - 9 \times \frac{\pi}{6} \right)$$ The required area is: $$= \frac{1}{2} \times \frac{9}{2} + \left( \frac{18\pi}{6} - \frac{9\sqrt{3}}{4} \right) \frac{1}{\sqrt{3}}$$ This simplifies to $= \sqrt{3} \pi$.

Given $e_1 = \sqrt{1 - \frac{75}{100}} = \frac{5}{10} = \frac{1}{2}$. $e_2 = 2$. $F_1(6, 1), F_2(-4, 1)$. $2ae_2 = 10 \Rightarrow a = \frac{5}{2} \Rightarrow 2a = 5$. $\Rightarrow \alpha = 5$. $4 = 1 + \frac{b^2}{a^2} \Rightarrow b^2 = 3a^2$. $b = \sqrt{3} \times \frac{5}{2}$. $\beta = 5\sqrt{3}$. $3\alpha^2 + 2\beta^2 = 3 \times 25 + 2 \times 25 \times 3 = 225$.

Given $M_{AB} = \frac{4}{5} \Rightarrow M_{DP} = \frac{5}{4}$. Equation of $PC$ is $y - 1 = \frac{5}{4}(x - 1)$ (1). $M_{AP} = \frac{2}{-2} = -1 \Rightarrow M_{BC} = +1$. Equation of $BC$ is $y - 3 = (x + 2)$ (2). On solving (1) and (2): $$x + 4 = \frac{5}{4}(x - 1) \Rightarrow 4x + 16 = 5x - 5 \Rightarrow \alpha = 21$$ $$\Rightarrow \beta = y = x + 5 = 26$$ $$\alpha + \beta = 47$$ Equation of $\perp$ bisector of $AP$: $$y - 0 = (x - 2) \ldots (3)$$ Equation of $\perp$ bisector of $AB$: $$y - 1 = \frac{5}{4}\left(x - \frac{1}{2}\right) \ldots (4)$$ On solving (3) $\&$ (4): $$(x - 3)4 = 5x - \frac{5}{2}$$ $$x = \frac{-19}{2} = h$$ $$y = \frac{-23}{2} = k$$ $$\Rightarrow 2\left(h + k\right) = -42$$

\begin{tabular}{|c|c|c|c|c|c|c|} \hline $x$ & $C$ & $2C$ & $3C$ & $4C$ & $5C$ & $6C$ \\ \hline $f$ & $2$ & $1$ & $1$ & $1$ & $1$ & $1$ \\ \hline \end{tabular} $$\bar{x} = \frac{(2+2+3+4+5+6)C}{7} = \frac{22C}{7}$$ $$\text{Var}(x) = \frac{c^2\left(2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2\right)}{7} - \left(\frac{22c}{7}\right)^2$$ $$= \frac{92c^2}{7} - c^2 \times \frac{484}{49}$$ $$= \frac{(644 - 484)c^2}{49} = \frac{160c^2}{49}$$ $$160 = \frac{160 \times c^2}{49} \Rightarrow c = 7$$

Given $f(x) = \frac{1}{2 + \sin 3x + \cos 3x}$. Consider the expressions: $$\left[ \frac{1}{2 + \sqrt{2}}, \frac{1}{2 - \sqrt{2}} \right]$$ Let $\frac{\alpha}{\beta} = \frac{a + b}{2\sqrt{ab}} = \frac{1}{2} \left( \sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} \right)$. Then, $$= \frac{1}{2} \left( \sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}} + \sqrt{\frac{2 + \sqrt{2}}{2 - \sqrt{2}}} \right)$$ $$= \frac{(2 - \sqrt{2}) + (2 + \sqrt{2})}{2 \times \sqrt{2}} = \sqrt{2}$$

Given $\bar{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\bar{a} = 2\hat{i} + \hat{j} - \hat{k}$. The cross product $\bar{a} \times \bar{r} = \bar{a} \times \bar{b}$ and $\bar{a} \cdot \bar{r} = 0$. This implies $\bar{a} \times (\bar{r} - \bar{b}) = 0$. Therefore, $\bar{a} = \lambda (\bar{r} - \bar{b})$. We have $\bar{a} \cdot \bar{a} = \lambda (\bar{a} \cdot \bar{r} - \bar{a} \cdot \bar{b})$. Solving $14 = -7\lambda$ gives $\lambda = -2$. Thus, $-\frac{\bar{a}}{2} = \bar{r} - \bar{b}$ implies $\bar{r} = \bar{b} - \frac{\bar{a}}{2}$. This results in $\bar{r} = \frac{2\bar{b} - \bar{a}}{2} = \frac{3\hat{i} + \hat{k}}{2}$. Statement (I) is incorrect. For the second part, $\cos 2A + \cos 2B + \cos 2C \geq -\frac{3}{2}$. Given $2A + 2B + 2C = 2\pi$, we have $\cos 2A + \cos 2B + \cos 2C = -1 - 4 \cos A \cdot \cos B \cdot \cos C$. This is greater than or equal to $-1 - 4 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = -\frac{3}{2}$. Statement (II) is correct.

The limit is given by $$ \lim_{x \to \frac{\pi}{2}} \frac{0 - \{ \sin(2x) + \cos(x) \} \cdot 3x^2}{2 \left( x - \frac{\pi}{2} \right)} $$ Simplifying, we have $$ = \lim_{x \to \frac{\pi}{2}} \frac{-\{ 2 \sin x \cos x + \cos x \} 3x^2}{2 \left( x - \frac{\pi}{2} \right)} $$ Using trigonometric identities, this becomes $$ = \lim_{x \to \frac{\pi}{2}} \left\{ \frac{2 \sin x \sin \left( \frac{\pi}{2} - x \right)}{2 \left( x - \frac{\pi}{2} \right)} + \frac{\sin \left( \frac{\pi}{2} - x \right)}{2 \left( \frac{\pi}{2} - x \right)} \right\} 3x^2 $$ Evaluating the limit, we get $$ = \left( 1(1) + \frac{1}{2} \right) 3 \left( \frac{\pi}{2} \right)^2 $$ Finally, the result is $$ = \frac{9\pi^2}{8} $$

$$T_{r+1} = {}^9C_r \left(x^{2/3}\right)^{9-r} \left(\frac{x^{-2/5}}{2}\right)^r$$ $$= {}^9C_r \left(\frac{1}{2}\right)^r x^{\left(6 - \frac{2r}{3} - \frac{2r}{5}\right)}$$ For coefficient of $x^{2/3}$, put $6 - \dfrac{2r}{3} - \dfrac{2r}{5} = \dfrac{2}{3}$ $\Rightarrow r = 5$ $\therefore$ Coefficient of $x^{2/3}$ is ${}^9C_5\left(\dfrac{1}{2}\right)^5$ For coefficient of $x^{-2/5}$, put $6 - \dfrac{2r}{3} - \dfrac{2r}{5} = -\dfrac{2}{5}$ $\Rightarrow r = 6$ $\therefore$ Coefficient of $x^{-2/5}$ is ${}^9C_6\left(\dfrac{1}{2}\right)^6$ $$\text{Sum} = {}^9C_5\left(\frac{1}{2}\right)^5 + {}^9C_6\left(\frac{1}{2}\right)^6 = \frac{21}{4}$$

Given $\mathrm{BCB}^{-1} = A$. Therefore, $\left( \mathrm{BCB}^{-1} \right) \left( \mathrm{BCB}^{-1} \right) = A \cdot A$. This implies $\mathrm{BCIBC}^{-1} = A^2$. Thus, $\mathrm{BC}^2 \mathrm{B}^{-1} = A^2$. Therefore, $\mathrm{B}^{-1} \left( \mathrm{BC}^2 \mathrm{B}^{-1} \right) \mathrm{B} = \mathrm{B}^{-1} (A \cdot A) \mathrm{B}$. From equation (1), $$C^2 = A^{-1} \cdot A \cdot B$$ $$C^2 = B$$ Also, $\mathrm{AB}^{-1} = A^{-1}$. Thus, $\mathrm{AB}^{-1} \cdot A = A^{-1} A = I$. Therefore, $A^{-1} \left( \mathrm{AB}^{-1} A \right) = A^{-1} I$. Thus, $\mathrm{B}^{-1} A = A^{-1}$. Now, the characteristic equation of $C^2$ is $$|C_2 - \lambda I| = 0$$ $$|B - \lambda I| = 0$$ $$\begin{vmatrix} 1 - \lambda & 3 \\ 1 & 5 - \lambda \end{vmatrix} = 0$$ This implies $(1 - \lambda)(5 - 1) - 3 = 0 \Rightarrow (\lambda^2 - 6\lambda + 5) - 3 = 0$. Thus, $\lambda^2 - 6\lambda + 2 = 0$. Therefore, $\beta^2 - 6\beta + 2I = 0$. Thus, $C^4 - 6C^2 + 2I = 0$. Let $\alpha = -6$ and $\beta = 2$. Therefore, $2\beta - \alpha = 4 + 6 = 10$.

Given $\ln(y) = 3 \sin^{-1} x$. $$\frac{1}{y} \cdot y' = 3 \left( \frac{1}{\sqrt{1-x^2}} \right)$$ $$\Rightarrow y' = \frac{3y}{\sqrt{1-x^2}} at x = \frac{1}{2}$$ $$\Rightarrow y' = 3e^{\frac{\pi}{6}} = 2 \sqrt{3} e^{\frac{\pi}{2}}$$ $$\Rightarrow y'' = 3 \left( \frac{\sqrt{1-x^2} y' - y \frac{1}{2\sqrt{1-x^2}} (-2x)}{(1-x^2)} \right)$$ $$\Rightarrow (1-x^2) y'' = 3 \left( 3y + \frac{xy}{\sqrt{1-x^2}} \right)$$ At $x = \frac{1}{2}$, $y = e^{3 \sin^{-1} \left( \frac{1}{2} \right)} = e^{3 \left( \frac{\pi}{6} \right)} = e^{\frac{\pi}{2}}$ $$\Rightarrow (1-x^2) y'' \big|_{at x = \frac{1}{2}} = 3 \left( 3e^{\frac{\pi}{2}} + \frac{1}{2} \left( e^{\frac{\pi}{2}} \right) \frac{\sqrt{3}}{2} \right)$$ $$= 3e^{\frac{\pi}{2}} \left( 3 + \frac{1}{\sqrt{3}} \right)$$ $$(1-x^2) y'' - xy' \big|_{at x = \frac{1}{2}}$$ $$= 3e^{\frac{\pi}{2}} \left( 3 + \frac{1}{\sqrt{3}} \right) - \frac{1}{2} \left( 2 \sqrt{3} e^{\frac{\pi}{2}} \right) = 9e^{\frac{\pi}{2}}$$

Given $$I = \int_{1/4}^{3/4} \cos \left( 2 \cot^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) \right) dx$$ This can be rewritten as $$\int_{1/4}^{3/4} \cos \left( 2 \left( \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right) \right) dx$$ Using the identity, we have $$\int_{1/4}^{3/4} \frac{1 - \tan^2 \left( \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right)}{1 + \tan^2 \left( \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right)} dx$$ This simplifies to $$\int_{1/4}^{3/4} \frac{1 - \left( \frac{1+x}{1-x} \right)}{1 + \left( \frac{1+x}{1-x} \right)} dx = \int_{1/4}^{3/4} \frac{-2x}{2} dx$$ Which further simplifies to $$\int_{1/4}^{3/4} (-x) dx = - \left( \frac{x^2}{2} \right) \bigg|_{1/4}^{3/4}$$ Evaluating the integral, we get $$= -\frac{1}{2} \left[ \frac{9}{16} - \frac{1}{16} \right]$$ Finally, $$= -\frac{1}{4}$$

Given $$\sum_{n=0}^{\infty} ar^n = 57$$ $$a + ar + ar^2 + \infty = 57$$ $$\frac{a}{1-r} = 57 \ldots (I)$$ $$\sum_{n=0}^{\infty} a^3 r^{3n} = 9747$$ $$a^3 + a^3 \cdot r^3 + a^3 \cdot r^6 + \ldots = 9746$$ $$\frac{a^3}{1-r^3} = 9746 \ldots (II)$$ From (I) and (II), $$\left(\frac{a}{1-r}\right)^3 = \frac{57^3}{9717} = 19$$ Thus, $$a = 19$$ Therefore, $$a + 18r = 19 + 18 \times \frac{2}{3} = 31$$ On solving, $$r = \frac{2}{3}$$ and $$r = \frac{3}{2}$$ (rejected) Thus, $$a = 19$$ Therefore, $$a + 18r = 19 + 18 \times \frac{2}{3} = 31$$

Favourable cases = $^6C_3$ Total outcomes = $6^3$ Probability of getting greater number than previous $$one = \frac{^6C_3}{r^3} = \frac{20}{216} = \frac{5}{54}$$

Given $$I = \int_{-1}^{2} \log_e \left( x + \sqrt{x^2 + 1} \right) dx$$ We have $$= x \log_e \left( x + \sqrt{x^2 + 1} \right) - \int_{-1}^{2} \left( \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{x + \sqrt{x^2 + 1}} \right) dx$$ This simplifies to $$= x \log_e \left( x + \sqrt{x^2 + 1} \right) - \int_{-1}^{2} \frac{x}{\sqrt{x^2 + 1}} dx$$ Further simplifying, we get $$= x \log_e \left( x + \sqrt{x^2 + 1} \right) - \sqrt{x^2 + 1} \bigg|_{-1}^{2}$$ Evaluating the limits, $$= (2 \log_e (2 + \sqrt{5}) - \sqrt{5})$$ $$- (- \log_e (-1 + \sqrt{2}) - \sqrt{2})$$ This results in $$= \log_e (2 + \sqrt{5})^2 - \sqrt{5} + \log_e (\sqrt{2} - 1) + \sqrt{2}$$ Simplifying further, $$= \log_e (2 + \sqrt{5})^2 - \sqrt{5} + \log_e (\sqrt{2} - 1) + \sqrt{2}$$ Finally, $$= \sqrt{2} - \sqrt{5} + \log_e \left( \frac{(2 + \sqrt{5})^2}{\sqrt{2} + 1} \right)$$ Which simplifies to $$= \sqrt{2} - \sqrt{5} + \log_e \left( \frac{9 + 4\sqrt{5}}{\sqrt{2} + 1} \right)$$

Given $x^2 - \sqrt{2}x - \sqrt{3} = 0 \langle \alpha, \beta \rangle$. $\alpha^{n+2} - \sqrt{2}\alpha^{n+1} - \sqrt{3}\alpha^n = 0$ and $\beta^{n+2} - \sqrt{2}\beta^{n+1} - \sqrt{3}\beta^n = 0$. Subtracting $$(\alpha^{n+2} - \beta^{n+2}) - \sqrt{2}(\alpha^{n+1} - \beta^{n+1}) - \sqrt{3}(\alpha^n - \beta^n) = 0$$ $$\Rightarrow P_{n+2} - \sqrt{2}P_{n+1} - \sqrt{3}P_n = 0$$ Put $n = 10$ $$P_{12} - \sqrt{2}P_{11} - \sqrt{3}P_{10} = 0$$ $n = 9$ $$P_{11} - \sqrt{2}P_{10} - \sqrt{3}P_9 = 0$$ $$11\left(\sqrt{3} \cdot P_{10} + \sqrt{2}P_{11} - P_{11}\right) - 10$$ $$= 0 - 10(-\sqrt{3}P_9) = 10\sqrt{3}P_9$$

Area of parallelogram $= \frac{1}{2} \left| \vec{d_1} \times \vec{d_2} \right|$. $$A = \frac{1}{2} \left| (\vec{a} + \vec{b}) \times (\vec{b} + \vec{c}) \right| = \frac{\sqrt{21}}{2}$$ So, $\vec{a} + \vec{b} = \hat{i} + \alpha \hat{j} + 2 \hat{k}$ and $\vec{b} + \vec{c} = -\hat{i} + \beta \hat{j}$. $$\begin{vmatrix} \vec{a} + \vec{b} \times \vec{b} + \vec{c} \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 2 \\ -1 & \beta & 0 \end{vmatrix}$$ $$= \hat{i}(-2\beta) - \hat{j}(2) + \hat{k}(\beta + \alpha)$$ $$\left| (\vec{a} + \vec{b}) \times (\vec{b} + \vec{c}) \right| = \sqrt{4\beta^2 + 4 + (\alpha + \beta)^2} = \sqrt{21}$$ $$4\beta^2 + 4 + \alpha^2 + \beta^2 + 2\alpha\beta = 21$$ $$\alpha^2 + 5\beta^2 - 12 = 17$$ $$\alpha^2 + 5\beta^2 = 29$$ and $\alpha\beta = -6$. Given $\alpha, \beta$ are integers, so, $\alpha = -3, \beta = 2$ or $\alpha = 3, \beta = -2$. $$(\alpha_1, \beta_1) = (-3, 2)$$ $$(\alpha_2, \beta_2) = (3, -2)$$ $$\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2 = 9 + 4 + 6 = 19$$

Given $T = S_1$. $x_1 + y_1 = x_1^2 + y_1^2$. $\alpha x_1 = x_1^2 + y_1^2$. $\alpha (2t^2) = 4t^4 + 16t^2$. $\alpha = 2t^2 + 8$. $\frac{\alpha-8}{20}=t^2$ Also, $4t^4 + 16t^2 - 4 < 0$. $t^2 = -2 + \sqrt{5}$. $\alpha = 4 + 2\sqrt{5}$. Therefore, $\alpha \in (8, 4 + 2\sqrt{5})$. Thus, $(2q - p)^2 = 80$.

Given $$f(x) = - (p^2 - 6p + 8) \cos 4n + 2(2 - p)n + 7$$ $$f^1(x) = +4 (p^2 - 6p + 8) \sin 4x + (4 - 2p) \neq 0$$ $$\sin 4x \neq \frac{2p - 4}{4(p - 4)(p - 2)}$$ $$\sin 4x \neq \frac{2(p - 2)}{4(p - 4)(p - 2)}$$ $$p \neq 2$$ $$\sin 4x \neq \frac{1}{2(p - 4)}$$ $$\Rightarrow \left| \frac{1}{2(p - 4)} \right| > 1$$ On solving we get $$\therefore p \in \left( \frac{7}{2}, \frac{9}{2} \right)$$ Hence $a = \frac{7}{2}$, $b = \frac{9}{2}$ $$\therefore 16ab = 252$$

\[ \frac{dy}{dx}-3y=\alpha \] If \[ \text{I.F.}=e^{\int -3\,dx}=e^{-3x} \] \[ y\,e^{-3x}=\int e^{-3x}\cdot \alpha\,dx \] \[ y\,e^{-3x}=\frac{\alpha e^{-3x}}{-3}+c \] \[ (*e^{3x}) \] \[ y=-\frac{\alpha}{3}+Ce^{3x} \] On substituting \(x=0,\ y=1\), \[ 1=-\frac{\alpha}{3}+C \] \[ C=1+\frac{\alpha}{3} \] We get \[ y=-\frac{\alpha}{3}+\left(1+\frac{\alpha}{3}\right)e^{3x} \] \[ y=7-6e^{3x} \] \[ 9f(-\log_e 3)=61 \]

Given $N = abc$. (i) All distinct digits $a + b + c = 14$ $a \geq 1$ $b, c \in \{0 to 9\}$ By hit and trial: 8 cases $(6, 5, 3)$ $(7, 6, 1)$ $(7, 5, 2)$ $(7, 4, 3)$ $(8, 6, 0)$ $(8, 5, 1)$ $(8, 4, 2)$ $(9, 4, 1)$ $(9, 3, 2)$ $(9, 5, 0)$ (ii) 2 same, 1 different $a = b; c$ $2a + c = 14$ By values: $$\begin{array}{c} (3, 8) \\ (4, 6) \\ (5, 4) \\ (6, 2) \\ (7, 0) \end{array} \frac{3!}{2!} \times 5 - 1$$ $= 14$ cases (iii) All same: $3a = 14$ $a = \frac{14}{3} \times rejected$ 0 cases Hence, Total cases: $8 \times 3! + 2 \times (4) + 14$ $= 48 + 22$ $= 70$

Given the equation $2x + 3y = 23$. For $x = 1$, $y = 7$. For $x = 4$, $y = 5$. For $x = 7$, $y = 3$. For $x = 10$, $y = 1$. Set $A = \{(1, 7), (4, 5), (7, 3), (10, 1)\}$ and $B = \{1, 4, 7, 10\}$. The number of one-one functions from $A$ to $B$ is equal to $4!$.

Given $m_{AB} = m_{AD}$, we have $$\frac{2}{t_1 + t_2} = \frac{2a (t_1 - t_3)}{at_1^2 - \alpha}$$ which implies $$at_1^2 - \alpha = a \{t_2^1 - t_1 t_3 + t_1 t_2 - t_2 t_3\}$$ leading to $$\alpha = a (t_1 t_3 + t_2 t_3 - t_1 t_2).$$ The length $AM = |2a (t_1 - t_3)|$, and $BN = |2a (t_2 - t_3)|$. For $CD = |at_3^2 - \alpha|$, we have $$CD = |at_3^2 - a (t_1 t_3 + t_2 t_3 - t_1 t_2)|$$ which simplifies to $$= a |t_3^2 - t_1 t_3 - t_2 t_3 + t_1 t_2|$$ and further to $$= a |t_3 (t_3 - t_1) - t_2 (t_3 - t_1)|$$ resulting in $$CD = a |(t_3 - t_2) (t_3 - t_1)|.$$ Therefore, $$\left(\frac{AM \cdot BN}{CD}\right)^2 = \left\{\frac{2a (t_1 - t_3) \cdot 2a (t_2 - t_3)}{a (t_3 - t_2) (t_3 - t_1)}\right\}^2.$$ Simplifying gives $$16a^2 = 16 \times \frac{9}{4} = 36.$$

Let M(3$\lambda$ + 1, 2$\lambda$, 4$\lambda$ + 2). $\overrightarrow{\mathrm{AM}}$ $\cdot$ $\overrightarrow{\mathbf{b}}$ = 0. $\Rightarrow$ 9$\lambda$ - 15 + 4$\lambda$ - 2 + 16$\lambda$ - 12 = 0. $\Rightarrow$ 29$\lambda$ = 29. $\Rightarrow$ $\lambda$ = 1. M(4, 2, 6), I = (2, 3, 7). Required Distance = $\sqrt{4 + 9 + 49}$ = $\sqrt{62}$. Ans. 62

Given $\($ $\left$( $\frac{1}{\alpha + 1}$ + $\frac{1}{\alpha + 2}$ + $\cdots$ + $\frac{1}{\alpha + 2012}$ $\right$) - $\left$$\{$ $\left$( $\frac{1}{1}$ - $\frac{1}{2}$ $\right$) + $\left$( $\frac{1}{3}$ - $\frac{1}{4}$ $\right$) + $\cdots$ + $\left$( $\frac{1}{2023}$ - $\frac{1}{2024}$ $\right$) $\right$$\}$ = $\frac{1}{2024}$ $\)$ $\($ $\Rightarrow$ $\left$( $\frac{1}{\alpha + 1}$ + $\frac{1}{\alpha + 2}$ + $\cdots$ + $\frac{1}{\alpha + 2012}$ $\right$) - $\left$$\{$ $\left$( $\frac{1}{1}$ - $\frac{1}{2}$ $\right$) + $\left$( $\frac{1}{1}$ + $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{4}$ $\right$) + $\cdots$ + $\frac{1}{2023}$ - $\frac{1}{2024}$ - 2 $\left$( $\frac{1}{2}$ + $\frac{1}{4}$ + $\cdots$ + $\frac{1}{2022}$ $\right$) $\right$$\}$ = $\frac{1}{2024}$ $\)$ $\($ $\Rightarrow$ $\left$( $\frac{1}{\alpha + 1}$ + $\frac{1}{\alpha + 2}$ + $\cdots$ + $\frac{1}{\alpha + 2012}$ $\right$) - $\left$( $\frac{1}{1}$ + $\frac{1}{2}$ + $\cdots$ + $\frac{1}{2023}$ $\right$) + $\frac{1}{2024}$ + $\left$( $\frac{1}{1}$ + $\frac{1}{2}$ + $\cdots$ + $\frac{1}{1011}$ $\right$) = $\frac{1}{2024}$ $\)$ $\($ $\Rightarrow$ $\frac{1}{\alpha + 1}$ + $\frac{1}{\alpha + 2}$ + $\cdots$ + $\frac{1}{\alpha + 2012}$ = $\frac{1}{1012}$ + $\frac{1}{1013}$ + $\cdots$ + $\frac{1}{2023}$ $\)$ $\($ $\Rightarrow$ $\alpha$ = 1011 $\)$

Given $2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}$. Therefore, $$\pi + \cos^{-1} x = \frac{2\pi}{5}$$ which implies $$\cos^{-1} x = \frac{-3\pi}{5}$$ which is not possible. Ans. 0

Given matrices $A = \begin{pmatrix} 2 & -5 \\ 3 & m \end{pmatrix}$ and $B = \begin{pmatrix} 20 \\ m \end{pmatrix}$, and vector $X = \begin{pmatrix} x \\ y \end{pmatrix}$. The equations are: $$2x - 5y = 20 \ldots (1)$$ $$3x + my = m \ldots (2)$$ From equation (2), solving for $y$: $$y = \frac{2m - 60}{2m + 15}$$ For $y < 0$, we have $m \in \left( -\frac{15}{2}, 30 \right)$. Solving for $x$: $$x = \frac{25}{2m + 15}$$ For $x < 0$, we have $m \in \left( -\frac{15}{2}, 0 \right)$. Thus, $m \in \left( -\frac{15}{2}, 0 \right)$. The determinant $|A| = 2m + 15$. Now, evaluate the integral: $$8 \int_{-\frac{15}{2}}^{0} (2m + 15) \, dm = 8 \left\{ m^2 + 15m \right\}_{-\frac{15}{2}}^{0}$$ This simplifies to: $$\Rightarrow 8 \left\{ 0 - \left( \frac{225}{4} - \frac{225}{2} \right) \right\}$$ Finally, we have: $$= 8 \times \frac{225}{4} = 450$$

By conservation of momentum $$p_i = p_f$$ $$0 = m_1 u_1 + m_2 u_2$$ $$\frac{u_1}{u_2} = -\left[ \frac{1}{2} \right] as \frac{m_1}{m_2} = \frac{2}{1}$$ move in opposite direction with speed ratio 1 : 2

Given $D = f_1 + f_2 = 25 \, \mathrm{cm}$. Paraxial parallel rays pass through focus and ray from focus of convex lens will become parallel.

K.E = $\frac{nf_1RT}{2}$ $T_i = -78^\circ \mathrm{C} \rightarrow 273 + [-78^\circ \mathrm{C}] = 195 \, \mathrm{K}$ K.E $\propto T$ To double the K.E energy temp also become double $T_f = 390 \, \mathrm{K}$ $T_f = 117^\circ \mathrm{C}$

Hydrogen will be in first excited state therefore it will emit one spectral line corresponding to transition between energy level 2 to 1.

Given $E_0 = B_0 C$. $$E_0 = 3 \times 10^8 \times (3.5 \times 10^{-7}) \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t)$$ $$E_0 = 105 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \mathrm{Vm^{-1}}$$ Data inconsistent while calculating speed of wave. You can challenge for data.

At $t = 10 \, sec$ complete loop is in magnetic field therefore no change in flux. $$e = \frac{d\phi}{dt} = 0$$ $e = 0$ for complete loop.

Theory Zener diode used as voltage regulator

Given the equations: $$P_1 - P_0 = \frac{4T}{r_1}$$ $$P_2 - P_0 = \frac{4T}{r_2}$$ We have: $$P_1 - P_0 = 3(P_2 - P_0)$$ Substituting the expressions: $$\frac{4T}{r_1} = 3 \cdot \frac{4T}{r_2}$$ Solving for $r_2$: $$r_2 = 3r_1$$ Now, considering the volumes: $$\frac{V_1}{V_2} = \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \frac{1}{27}$$

Given $\lambda = \frac{h}{\sqrt{2mE}}$ or $E = h\nu$. $$[ML^2 T^{-2}] = h [T^{-1}]$$ Therefore, $$h = [ML^2 T^{-1}]$$

Total energy = $\frac{-GMm}{2(2R)}$. If energy = $\frac{10^4 R}{6}$ is added then $$\frac{-GMm}{4R} + \frac{10^4 R}{6} = \frac{-GMm}{2r}$$ where r is new radius of revolving and $g = \frac{GM}{R^2}$. $$\frac{-mgR}{4} + \frac{10^4 R}{6} = \frac{-mgR^2}{2r} (m = 10^3 \, \mathrm{kg})$$ $$-10^3 \times 10 \times R + \frac{10^4 R}{6} = \frac{-10^3 \times 10 \times R^2}{2r}$$ $$-\frac{1}{4} + \frac{1}{6} = \frac{-R}{2r}$$ $$r = 6R$$

From symmetry we can remove two middle resistance. New circuit is

As per Gauss's theorem, $$\phi = \frac{q_{in}}{\varepsilon_0} = \frac{q + (-2q) + 5q}{\varepsilon_0}$$ $$\frac{4q}{\varepsilon_0}$$

In uniform magnetic field, $$R = \frac{mv}{qB} = \frac{\sqrt{2\, m\, (\mathrm{K.E})}}{qB}$$ Since same K.E $$R \propto \frac{\sqrt{m}}{q}$$ $$\therefore \frac{R_{deutron}}{R_{proton}} = \sqrt{\frac{m_d}{m_p}} \times \frac{q_p}{q_d}$$ $$= \sqrt{2} \times 1$$ $$\therefore \gamma_d : \gamma_p = \sqrt{2} : 1$$

Given $E_{photon} = (work function) + K \cdot E_{\max}$. Therefore, $4.13 = 3.13 + K \cdot E_{\max}$. Thus, $K_{\max} = 1 \, eV$.

In each fusion reaction, $^4_1 \mathrm{H}$ nucleus are used. Energy released per Nuclei of $^1_1 \mathrm{H} = \frac{26.7}{4} \mathrm{MeV}$. Therefore, energy released by $2 \mathrm{kg}$ hydrogen ($E_H$) is $$= \frac{2000}{1} \times N_A \times \frac{26.7}{4} \mathrm{MeV}$$ and energy released by $2 \mathrm{kg}$ Uranium ($E_V$) is $$= \frac{2000}{235} \times N_A \times 200 \mathrm{MeV}$$ So, $$\frac{E_H}{E_V} = 235 \times \frac{26.7}{4 \times 200} = 7.84$$ Therefore, approximately close to $7.62$.

Given $W_{AB} = \int \mathrm{PdV}$ (Assuming $T$ to be constant) $$= \int \frac{\mathrm{RTdV}}{V^3}$$ $$= \mathrm{RT} \int_2^4 V^{-3} \mathrm{dV}$$ $$= 8 \times 300 \times \left( -\frac{1}{2} \left[ \frac{1}{4^2} - \frac{1}{2^2} \right] \right)$$ $$= 225 \, \mathrm{J}$$ $W_{BC} = \mathrm{P} \int_4^2 \mathrm{dV} = 10(2 - 4) = -20 \, \mathrm{J}$ $W_{CA} = 0$ Therefore, $W_{cycle} = 205 \, \mathrm{J}$

Given $T_1 \sin 45^\circ = F$. $T_1 \cos 45^\circ = T_2 = 1 \times g$. Therefore, $\tan 45^\circ = \frac{F}{g}$. Thus, $F = 10 \, \mathrm{N}$.

Given $$V_T = \frac{2}{9} g \frac{R^2 [\rho_B - \rho_L]}{\eta}$$ We have $$\Rightarrow V_T = \frac{2}{9} \times \frac{10 \times (10^{-4})^2}{9.8 \times 10^{-6}} \left[10^5 - 10^3\right]$$ This simplifies to $$\Rightarrow V_T = 224.5$$ When the ball falls from height $h$, $$V = \sqrt{2gh}$$ Thus, $$h = \left(\frac{V^2}{2g}\right) = 2518 \, \mathrm{m}$$

For x, when 0 is A and 1 is B, the output E is 1. For y, when 1 is A and 0 is B, the output E is 1.

Magnetic moment of straight wire $= m x \ell = 44$. Magnetic moment of arc $$= m \times 2r$$ $$= m \times \frac{2\ell}{\pi}$$ $$= \frac{44 \times 2}{\pi} = \frac{88}{\pi} = 28$$

Given: $$m = 0.5 \, \mathrm{kg}$$ $$F = -50(x)$$ Using Newton's second law: $$ma = (-50x)$$ Substituting the value of $m$: $$0.5a = -50x$$ Solving for $a$: $$a = (-100x)$$ Given $W^2 = 100$, we find $w$: $$W^2 = 100 \implies (w = 10)$$ Now, calculating $T$: $$T = \frac{2\pi}{10} = \left(\frac{\pi}{5}\right) = \frac{22}{7 \times 15} = \left(\frac{22}{35}\right)$$ Equating: $$\frac{\pi}{35} = \frac{22}{35} \implies x = 22$$

Given $X_C = 4\sqrt{3} \, \Omega$ and $R = 4 \, \Omega$. The impedance $Z$ is given by $$Z = \sqrt{R^2 + X_L^2}$$ Substituting the values, $$Z = \sqrt{4^2 + (4\sqrt{3})^2} = 8 \, \Omega$$ The root mean square voltage $V_{rms}$ is $$V_{rms} = \frac{V}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \, V$$ The root mean square current $I_{rms}$ is $$I_{rms} = \frac{V_{rms}}{Z} = \frac{8}{8} = 1 \, A$$ The power dissipated is $$I_{rms}^2 \times R = 1 \times 4 = 4 \, W$$

Given $\vec{E} = 2x \hat{i}$. The flux $\phi$ is given by $\vec{E} \cdot \vec{A}$. For $\phi_{in}$, we have $-4 \times 4 = -16 \, \mathrm{Nm^2/c}$. For $\phi_{out}$, we have $8 \times 4 = 32 \, \mathrm{Nm^2/c}$. The net flux $d_{net}$ is $\phi_{in} + \phi_{out} = -16 + 32 = 16 \, \mathrm{Nm^2/c}$.

For slipping $a = g \sin \theta$ $$\ell = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2 \ell}{g \sin \theta}}$$ For rolling $$a' = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}} \left[ k = \frac{R}{\sqrt{2}} \right]$$ $$\implies a' = \frac{2 \, g \sin \theta}{3}$$ $$\ell = \frac{1}{2} a' (t')^2$$ $$\implies t' = \sqrt{\frac{6 \ell}{2 \, g \sin \theta}} = \sqrt{\frac{\alpha}{2}} \sqrt{\frac{2 \ell}{g \sin \theta}}$$ $$\implies \alpha = 3$$

Given $\($ Req = $\frac{10^4 R}{10^4 + R}$ $\)$ $\($ E = 4V, I = 2mA $\)$ $\($ I = $\frac{E}{Req}$ $\Rightarrow$ 2 $\times$ 10^{-3} = $\frac{4 \left(10^4 + R\right)}{10^4 R}$ $\)$ $\($ $\Rightarrow$ 20R = 40000 + 4R $\)$ $\($ 16R = 40000 $\)$ $\($ R = 2500 $\Omega$ $\)$

Given $B \cos \theta = \frac{B}{2}$. Therefore, $\theta = 60^\circ$. So, the angle between $\vec{A}$ and $\vec{B}$ is $90^\circ + 60^\circ = 150^\circ$.

$(\mu-1)t=n\lambda$ $(1.5-1)t=4\times500\times10^{-9}\ \mathrm{m}$ $t=4000\times10^{-9}\ \mathrm{m}$ $t=4\mu \mathrm{m}$

The work done is given by the integral of force with respect to displacement. $$W = \int_{x_1}^{x_2} F\,dx$$ Substituting the given force function, we have: $$W = \int_{2}^{4} (3x^2 + 2x - 5)\,dx$$ Integrating the function, we get: $$W = \left[x^3 + x^2 - 5x\right]_{2}^{4}$$ Evaluating the definite integral, we find: $$W = \left[60 - 2\right]\,\mathrm{J} = 58\,\mathrm{J}$$

Given the equation $R = R_0 (1 + \alpha \Delta T)$. Substitute the values: $$62 = 50 \left[1 + 2.4 \times 10^{-4} \Delta T \right]$$ Solve for $\Delta T$: $$\Delta T = 1000^\circ \mathrm{C}$$ Therefore, $$T - 27^\circ = 1000^\circ \mathrm{C}$$ So, $$T = 1027^\circ \mathrm{C}$$

The candela is the luminous intensity of a source that emits monochromatic radiation of frequency radiation of frequency $540 \times 10^{12} \, \mathrm{Hz}$ and has a radiant intensity in that direction of $\frac{1}{683} \, \mathrm{W/sr}$. It is unit of Candela.

CH_3-CH=CH-CH=O (III) Non Polar R.S. More No of covalent bond CH_3-$\overset{+}{CH}$-CH=CH$\overset{-}{O}$ (II) Having -ve charge on more electronegative atom CH_3-$\overset{-}{CH}$=CH-CH$\overset{+}{O}$ (I) Having -ve charge on less electronegative atom Stability order III > II > I

There are three stereo centers. So the number of stereoisomers is $2^3 = 8$.

Order of bond angle is $\mathrm{ClF_3} < \mathrm{PF_3} < \mathrm{BF_3}$

(A) $\mathrm{Br_2}$ water test is test of unsaturation in which reddish orange colour of bromine water disappears. (B) Alcohols given Red colour with ceric ammonium nitrate. (C) Phenol gives Violet colour with natural ferric chloride. (D) Aldehyde & Ketone give Yellow/Orange/Red Colour compounds with 2, 4-DNP i.e., 2, 4-Dinitrophenyl hydrazine.

A-IV, B-III, C-I, D-II

For the complex $\mathrm{K_2[Ni(CN)_4]}$, the oxidation state of nickel is $+2$. The electronic configuration of $\mathrm{Ni^{2+}}$ is $[\mathrm{Ar}]3d^8 4s^0$. Since $\mathrm{CN}$ is a strong field ligand, it causes pairing of electrons in the $3d$ orbitals. The pre-hybridization state of $\mathrm{Ni^{2+}}$ is shown with paired electrons in the $3d$ orbitals and empty $4s$ and $4p$ orbitals. The hybridization is $dsp^2$.

EDTA $^{4-}$ is a hexadentate ligand. $$[Ca(EDTA)]^{2-}$$ So the coordination environment is octahedral.

Glucose is soluble in water due to presence of alcohol functional group and extensive hydrogen bonding. Glucose exists in open chain as well as cyclic forms in its aqueous solution. Glucose has 6 carbon atoms so it is hexose and having an aldehyde functional group so it is aldose. Thus, aldohexose. Glucose is a monomer unit in sucrose with fructose.

The reaction starts with $\mathrm{CH_3 - CH_2 - CCl_2 - CH_3}$ treated with aqueous $\mathrm{KOH}$ to form $\mathrm{CH_3 - CH_2 - C(OH)_2 - CH_3}$. Upon dehydration, it forms $\mathrm{CH_3 - CH_2 - CO - CH_3}$. Further reaction with $\mathrm{KOI}$ leads to the formation of $\mathrm{CH_3 - CH_2 - COOK + CHI_3}$, where $\mathrm{CHI_3}$ is a yellow precipitate.

The equilibrium reaction is given by: $$\mathrm{AB_2}_{(s)} \rightleftharpoons \mathrm{A^{+2}}_{(aq)} + 2 \mathrm{B^-}_{(aq)}$$ The solubility product constant is: $$K_{sp} = [\mathrm{A^{+2}}][\mathrm{B^-}]^2$$ Substituting the given concentrations: $$= 1.2 \times 10^{-4} \times (2.4 \times 10^{-4})^2$$ Calculating the result: $$= 6.91 \times 10^{-12} \, \mathrm{M^3}$$

.

On moving down the group in transition elements, stability of higher oxidation state increases, due to increase in effective nuclear charge. $$\Rightarrow E^\circ_{\mathrm{Cu}^{+2}/\mathrm{Cu}} = 0.34 \, \mathrm{V}$$ $$\Rightarrow E^\circ_{\mathrm{H}^+/\mathrm{H}_2} = 0$$ SRP: $\mathrm{Cu}^{2+} > \mathrm{H}^+$ Cu can't liberate hydrogen gas from weak acid.

The carbon-carbon bonds in ethyne is stronger than that in ethene. (H - C $\equiv$ C - H) Ethyne is linear and carbon atoms are SP hybridised.

Option (A) $^7 \mathrm{N} : [\mathrm{He}] 2s^2 2p^3$ Option (B) $^{16} \mathrm{S} : [\mathrm{Ne}] 2s^2 3p^4$ Option ($C$) $^{35} \mathrm{Br} : [\mathrm{Ar}] 3d^{10} 4s^2 4p^5$ Option (D) $^{36} \mathrm{Kr} : [\mathrm{Ar}] 3d^{10} 4s^2 4p^6$

Melting point: $\mathrm{B} > \mathrm{Al} > \mathrm{Tl} > \mathrm{In} > \mathrm{Ga}$ Ionic radius $\left(\mathrm{M^{+3}/pm}\right)$: $\mathrm{Tl} > \mathrm{In} > \mathrm{Ga} > \mathrm{Al} > \mathrm{B}$ $\left(\Delta_{\mathrm{IEH}}\right)_1 \left[\frac{\mathrm{kJ}}{\mathrm{mol}}\right]$: $\mathrm{B} > \mathrm{Tl} > \mathrm{Al} \approx \mathrm{Ga} > \mathrm{In}$ Atomic radius (in pm): $\mathrm{Tl} > \mathrm{In} > \mathrm{Al} > \mathrm{Ga} > \mathrm{B}$

Apart from aldehyde, Formic acid $\mathrm{HCOOH}$ also gives silver mirror test with ammoniacal silver nitrate.

The reaction is given by: $$\mathrm{HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)}$$ The expression for $K_a$ is: $$K_a = \frac{\alpha^2 C}{1 - \alpha}$$ Rearranging gives: $$\alpha^2 C + K_a \alpha - K_a = 0$$ Substituting $\alpha = \frac{\lambda_m}{\lambda_m^\infty}$, we have: $$\left(\frac{\lambda_m}{\lambda_m^\infty}\right)^2 C + K_a \frac{\lambda_m}{\lambda_m^\infty} - K_a = 0$$ Simplifying, we get: $$\lambda_m^2 C + K_a \lambda_m \lambda_m^\infty - K_a (\lambda_m^\infty)^2 = 0$$

Einsteinium (atomic No = 99) : $[\mathrm{Rn}] 5f^{11} 6d^0 7s^2$

Fuming sulphuric acid is a mixture of conc. $\mathrm{H_2SO_4} + \mathrm{SO_3}$ or $\mathrm{H_2S_2O_7}$. So, Number of Oxygen atoms = 7

Given $X_{methylbenzene} = 0.5$. $Y_{methylbenzene} = \frac{P_{methylbenzene}}{P_{total}}$. $Y_{methylbenzene} = \frac{0.5 \times 24}{0.5 \times 80 + 0.5 \times 24}$. $= \frac{12}{40 + 12} = 0.23 = 23 \times 10^{-2}$.

If $M^{+2} = \mathrm{Co}^{+2}$ then $$\mathrm{Co}^{+2} + \mathrm{H_2 S} \rightarrow \mathrm{CoS} (Black) + byproduct$$ $$\mathrm{CoS} + aqua regia \rightarrow \mathrm{CoCl_2} + \mathrm{NOCl} + \mathrm{S} + \mathrm{H_2O}$$ $$\mathrm{CoCl_2} + \mathrm{KNO_2} \xrightarrow{\mathrm{CH_3COOH}} k_3 \left[ \mathrm{Co(NO_2)_6} \right] + byproduct$$ $[\mathrm{C}] = \mathrm{Co}^{+3} \rightarrow 3d^6 \Rightarrow \mathrm{NO_2}$ is SFL so $\mathrm{Co} +3$ has zero unpaired $e^-$. $$\Rightarrow n = 0 \Rightarrow u = \sqrt{n(n+2)} = 0 \mathrm{B.M}$$

The reaction is given by $\($ $\mathrm{A(g) \rightarrow 2 \, B(g) + C(g)}$ $\)$. The pressure at time $\($ t = 23 $\)$ is $\($ P_{23} = P_0 + 2x = 200 $\)$. The final pressure is $\($ P_$\infty$ = 3P_0 = 300 $\)$. The initial pressure is $\($ P_0 = 100 $\)$. The rate constant $\($ K $\)$ is given by $$ K = \frac{1}{t} \ln \frac{P_\infty - P_0}{P_\infty - P_t} $$ Substituting the values, we have $$ K = \frac{2.3}{23} \log \frac{300 - 100}{300 - 200} $$ Simplifying further, $$ K = \frac{2.3 \times 0.301}{23} = 0.0301 = 3.01 \times 10^{-2} \, \mathrm{sec^{-1}} $$

The formula for the retention factor $R_f$ is given by: $$R_f = \frac{Distance moved by substance from base line}{Distance moved by solvent from base line}$$ For substance A: $$(R_f)_A = \frac{4}{8}$$ For substance B: $$(R_f)_B = \frac{6}{8}$$ The ratio of $R_f$ values is: $$\frac{(R_f)_B}{(R_f)_A} = \frac{6}{8} \times \frac{8}{4}$$ Simplifying gives: $$(R_f)_B = 1.5 (R_f)_A$$ Given $x = 15$.

Given $m \Delta V \cdot \Delta x = \frac{h}{4 \pi}$. $$\Delta V = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^{-15} \times 4 \times 3.14}$$ $$= 57.97 \times 10^{9} \, \mathrm{m/sec}$$

Compounds which cannot undergo Friedel Crafts reaction are nitrobenzene, m-nitroaniline, and m-dinitrobenzene. These compounds contain electron-withdrawing groups that deactivate the benzene ring, making it less reactive towards electrophilic aromatic substitution reactions like Friedel Crafts.

For $\mathrm{O_2}(16\mathrm{e}): (\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 \left[ (\pi_{2p})^2 = (\pi_{2p})^2 \right], \left[ (\pi^*_{2p})^1 = (\mathrm{m}^*_{2p})^1 \right]$. Number of $\mathrm{e}^-$ present in $(\pi^*)$ of $\mathrm{O_2} = 2$. Number of $\mathrm{e}^-$ present in $(\pi^*)$ of $\mathrm{O_2}^+ = 1$. Number of $\mathrm{e}^-$ present in $(\pi^*)$ of $\mathrm{O_2}^- = 3$. So total $\mathrm{e}^-$ in $(\pi^*) = 2 + 1 + 3 = 6$.

At equilibrium $\Delta G_{PT} = 0$ $$\Delta H_{vap} = T \Delta S_{vap}$$ $$30 \times 1000 = T \times 75$$ $$T = 400 \, \mathrm{K}$$