JEE Main 8 April 2025 Shift 2 question paper with solutions

Given $\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$, $\vec{q} = 3\hat{i} + 4\hat{j} + 5\hat{k}$. $$\Rightarrow \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = -\hat{i} + 2\hat{j} - \hat{k}$$ $A \equiv (1, 2, 3)$, $B \equiv (\lambda, 4, 5)$ Shortest Distance = $$\frac{|\overrightarrow{AB} \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$$ $$\frac{1}{\sqrt{6}} = \left| \frac{((\lambda - 1)\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-\hat{i} + 2\hat{j} - \hat{k})}{\sqrt{6}} \right|$$ $$\Rightarrow |-\lambda + 1 + 4 - 2| = 1 \Rightarrow |\lambda - 3| = 1$$ $$\Rightarrow \lambda = 3 \pm 1 = 4, 2$$ Radius of circle passing through points $(0, 0), (4, 2) \& (2, 4)$ $$= \frac{abc}{4\Delta} = \frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 \end{vmatrix}} = \frac{20 \times 2\sqrt{2}}{2 \times 12}$$ $$= \frac{5\sqrt{2}}{3}$$

Given $x^2 + x + 1 = 0$. $\alpha$ is a root. Therefore, $\alpha^2 + \alpha + 1 = 0$. This implies $\alpha = \omega$ as $\omega^2$ [cube root of unity]. Also, $$\begin{bmatrix} 4 - a - 2b & 64 - a - 14b & 52 + 2a - 8b \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$$ Thus, $a + 2b = 4$ and $a + 14b = 64$. This implies $12b = 60 \Rightarrow b = 5$. Therefore, $a = -6$. Thus, $$\frac{4}{\alpha^4} + \frac{m}{\alpha^{-6}} + \frac{n}{\alpha^5} = 3$$ This implies $\frac{\omega}{1} + \frac{m}{1} + \frac{n}{\omega^2} = 3$. Therefore, $4\omega^2 + m + n\omega = 3$. This implies $$4 \left( -\frac{1}{2} - \frac{\sqrt{3}}{2} i \right) + m + n \left( -\frac{1}{2} + \frac{\sqrt{3}}{2} i \right) = 3$$ Thus, $-2 + m - \frac{n}{2} = 3$. (1) And, $$-\frac{4\sqrt{3}}{2} + \frac{n\sqrt{3}}{2} = 0$$ Therefore, $n = 4$ and $m = 7$. Thus, $m + n = 11$.

Given $f(x) = \frac{x}{3} + \frac{3}{x} + 3$, $x \neq 0$. Differentiating, we have: $$f'(x) = \frac{1}{3} - \frac{3}{x^2} = 0 \implies x = \pm 3$$ Thus, $$f'(x) = \frac{x^2 - 3}{3x^2}$$ $f'(x) > 0$ for $(-\infty, -3) \cup (3, \infty)$, indicating the function is increasing. $f'(x) < 0$ for $(-3, 0) \cup (0, 3)$, indicating the function is decreasing. The sum is given by: $$\sum_{i=1}^{5} \alpha_i^2 = (-3)^2 + (3)^2 + (-3)^2 + (0)^2 + (3)^2$$ $$= 36$$

Given $\mathrm{P}(A) = \frac{7}{10}$, $\mathrm{P}(B) = \frac{4}{10}$. $\mathrm{P}(A \cup \overline{B}) = \frac{5}{10}$. $\mathrm{P} \left( \frac{B}{A \cup \overline{B}} \right) = \frac{\mathrm{P}(B \cap (A \cup \overline{B}))}{\mathrm{P}(A \cup \overline{B})}$. $= \frac{\mathrm{P}((B \cap \overline{B}) \cup (B \cap A))}{\mathrm{P}(A \cup \overline{B})} = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(A \cup \overline{B})}$. $= \mathrm{P}(A) - \mathrm{P}(A \cap \overline{B})$. $= \mathrm{P}(A) + \mathrm{P}(\overline{B}) - \mathrm{P}(A \cap \overline{B}) = \frac{7}{10} + \left( 1 - \frac{4}{10} \right) - \frac{5}{10}$. $= \frac{2}{8} = \frac{1}{4}$.

Given $\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots \infty = \frac{\pi^4}{90} \ldots (i)$. $$\beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots$$ $$= \frac{1}{16} \left[ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots \right],$$ $$= \frac{1}{16} \times \frac{\pi^4}{90}$$ Using (ii) $\ldots$ (ii) $$\alpha = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots \infty$$ $$\left( \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \frac{1}{5^4} + \ldots \right)$$ $$- \left( \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots \right)$$ $$\alpha = \frac{\pi^4}{90} - \frac{1}{16} \times \frac{\pi^4}{90} [using (i) and (ii)]$$ $$\alpha = \frac{16 - 1}{16 \times 90} \times \pi^4 = \frac{15}{16 \times 90} \pi^4 = \frac{\pi^4}{96}$$ Therefore, $$\frac{\alpha}{\beta} = \frac{\frac{\pi^4}{96}}{\frac{\pi^4}{\frac{16 \times 90}{16 \times 90}}} = \frac{16 \times 90}{96} = 15$$

Given $|x - 2|^2 + 2|x - 2| - |x - 2| - 2 = 0$. This implies $ (|x - 2| + 2)(|x - 2| - 1) = 0$. Therefore, $|x - 2| = 1$. This gives $x = 2 \pm 1 = 3, 1$. The sum of the square of roots is $9 + 1 = 10$. Consider $x^2 - 2|x - 3| - 5 = 0$. Case-I: $x - 3 \geq 0$. This implies $x^2 - 2x + 1 = 0$. Therefore, $(x - 1)^2 = 0$. Thus, $x = 1$. But $x \geq 3$. Hence, $x \in \emptyset$. Case-II: $x - 3 0$ implies real and distinct roots. Let $f(x) = x^2 + 2x - 11$. Then $f(3) > 0$, $\frac{-p}{2a} = -1 < 3$. Thus, both roots $< 3$, both roots are acceptable. The sum of the square of roots is $(\alpha + \beta)^2 - 2\alpha\beta$. This equals $4 + 22 = 26$. Therefore, the final sum is $10 + 26 = 36$.

Slope of diagonal OB is $$\frac{\sqrt{3} + 1}{1 - \sqrt{3}}$$. Therefore, $$\alpha = \tan 105^\circ$$. Thus, $$\alpha = 60^\circ$$. Therefore, point A is at $$A (a \cos 60^\circ, a \sin 60^\circ)$$. Therefore, $$A \left( \frac{a}{2}, \frac{a \sqrt{3}}{2} \right)$$. A lies on the other diagonal. Therefore, $$\left( \frac{\sqrt{3} - 1}{2} \right) a - \left( \frac{\sqrt{3} + 1}{2} \right) \cdot \sqrt{3}a + 8\sqrt{3} = 0$$. Solving for $$a$$, $$a \left[ \frac{\sqrt{3} - 1 - 3 - \sqrt{3}}{2} \right] = -8\sqrt{3}$$. Therefore, $$a = 4\sqrt{3}$$. Thus, $$a^2 = 48$$.

Given $$I_1 = \int_{\frac{1}{2}}^1 2x f(2x(1-2x)) \, dx$$ Let $2x = t$, then $2 \, dx = dt$. Therefore, $$I_1 = \frac{1}{2} \int_{-1}^2 t f(t(1-t)) \, dt$$ Thus, $$2I_1 = \int_{-1}^2 (1-t) f(1-t)(1-(1-t)) \, dt$$ This implies, $$2I_1 = \int_{-1}^2 f \left( t(1-t) \right) \, dt - \int_{-1}^2 tf(t(1-t)) \, dt$$ Therefore, $$2I_1 = I_2 - 2I_1$$ Hence, $$4I_1 = I_2$$ So, $$I_2 = 4$$ Thus, $$I_1 = 4$$

Let vector $\vec{p}$ in plane of $\vec{a} \& \vec{b} = K(\vec{a} + \lambda \vec{b})$. $\vec{p} \perp \vec{a} = \vec{p} \cdot \vec{a} = 0$ $$\Rightarrow K(\vec{a} + \lambda \vec{b}) \cdot \vec{a} = 0$$ $$\Rightarrow \vec{a} \cdot \vec{a} + \lambda \vec{b} \cdot \vec{a} = 0$$ $$\Rightarrow 6 + \lambda (3) = 0$$ $$\Rightarrow \lambda = -2$$ $$\Rightarrow \vec{p} = (-3\hat{i} + 3\hat{k})$$ Unit vector $\rightarrow \pm \frac{(-\hat{i} + \hat{k})}{\sqrt{2}}$

Given the ellipse equation $E: \frac{x^2}{4/3} + \frac{y^2}{4/P} = 1$. The center of the circle is $(1, 2)$ with radius $r = \sqrt{1 + 4 + 11}$. Therefore, $r = 4$. Since $E$ passes through the center $(1, 2)$, we have $\frac{3}{4} + P = 1$. Thus, $P = \frac{1}{4}$, indicating a vertical ellipse. The eccentricity $e$ is calculated as follows: $$e = \sqrt{1 - \frac{4/3}{16}} = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}}.$$ Therefore, the focal distance of $C(h, k)$ is $b \pm ek$. Calculating the focal points: $$F_1 = 4 + \sqrt{\frac{11}{12}} \times 2,$$ $$F_2 = 4 - \sqrt{\frac{11}{12}} \times 2.$$ Thus, $F_1 F_2 = 16 - \frac{11}{3} = \frac{37}{3}$. Therefore, $6 F_1 F_2 - r = 74 - 4 = 70$.

Let, $I = \pi^2 \int_{-1}^{3/2} |x \sin \pi x| \, dx$. $$= \pi^2 \left\{ \int_{-1}^{1} x \sin \pi x \, dx - \int_{1}^{3/2} x \sin \pi x \, dx \right\}$$ $$= \pi^2 \left\{ 2 \int_{0}^{1} x \sin \pi x \, dx - \int_{-1}^{3/2} x \sin \pi x \, dx \right\}$$ Consider $$\int x \sin \pi x \, dx$$ $$= -x \cdot \frac{1}{\pi} \cos \pi x + \int 1 \cdot \frac{1}{\pi} \cos \pi x \, dx$$ $$= -\frac{x}{\pi} \cos \pi x + \frac{\sin \pi x}{\pi^2}$$ $$I = \pi^2 \left\{ 2 \left( -\frac{x}{\pi} \cos \pi x + \frac{\sin \pi x}{\pi^2} \right) \right\}_0^1 - \left( -\frac{x}{\pi} \cos \pi x + \frac{\sin \pi x}{\pi^2} \right)_1^{3/2}$$ $$= \pi^2 \left\{ \frac{2}{\pi} - \left( -\frac{1}{\pi^2} - \frac{1}{\pi} \right) \right\}$$ $$= \pi^2 \left\{ \frac{3}{\pi} + \frac{1}{\pi^2} \right\}$$ $$= 3\pi + 1$$

Given $\mathrm{m}_{PR} = 2 - \sqrt{3} = \tan 15^\circ$. Therefore, $\dfrac{\alpha}{2} = 15^\circ \Rightarrow \alpha = 30^\circ$. The equation of $PR$ is $y = \tan 15^\circ (x - a)$. Thus, $y = (2 - \sqrt{3})(x - a)$. The perpendicular distance from the origin is $\dfrac{1}{\sqrt{2}}$. \[ \left|\frac{\sqrt{3}a - 2a}{\sqrt{4 + 3 - 4\sqrt{3} + 1}}\right| = \frac{1}{\sqrt{2}} \] \[ |a|(2 - \sqrt{3}) = 1 \] \[ 2\sqrt{(2 - \sqrt{3})} = \sqrt{2} \] \[ |a| = \frac{\sqrt{2}}{2} = \sqrt{2(\sqrt{2} + \sqrt{3})} \] \[ a^2 = 2(2 + \sqrt{3}) \] \[ 3a^2 \tan^2\alpha - 2\sqrt{3} \] \[ 3 \times (4 + 2\sqrt{3}) \cdot \frac{1}{3} - 2\sqrt{3} = 4 \]

Given $^{12}C_3 - ^{5}C_3 = 210$.

Given $$1 + 10 \operatorname{Re} \left( \frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta} \right) = 0$$ Therefore, $$z + \bar{z} = 2 \operatorname{Re}(z)$$ $$\frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta} + \frac{2 \cos \theta - i \sin \theta}{\cos \theta + 3i \sin \theta} = 2 \times \left( \frac{-1}{10} \right)$$ $$\frac{(2 \cos^2 \theta - 3 \sin^2 \theta) + (2 \cos^2 \theta) - (3 \sin^2 \theta)}{\cos^2 \theta + 9 \sin^2 \theta} = \frac{-2}{10}$$ Thus, $$2 \cos^2 \theta - 3 \sin^2 \theta = \frac{-1}{10} \cos^2 \theta + 9 \sin^2 \theta$$ $$\Rightarrow 20 \cos^2 \theta - 30 \sin^2 \theta = -\cos^2 \theta - 9 \sin^2 \theta$$ $$21 \cos^2 \theta - 21 \sin^2 \theta = 0$$ Therefore, $$\cos 2\theta = 0$$ $$2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$$ Thus, $$\sum \theta^2 = \frac{\pi^2}{16} + \frac{9\pi^2}{16} + \frac{25\pi^2}{16} + \frac{49\pi^2}{16} = \frac{84\pi^2}{16} = \frac{21\pi^2}{4}$$

By given data $\overrightarrow{AB} + \overrightarrow{AC} = \overrightarrow{CB}$. Let pv of $\overrightarrow{A}$ are $\overrightarrow{O}$ then $\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A}$. i.e. pv of $\overrightarrow{B} = -2\hat{i} - \hat{j} + \hat{k}$. $\overrightarrow{CA} = \overrightarrow{A} - \overrightarrow{C}$. i.e. pv of $\overrightarrow{C} = -(\hat{i} - 3\hat{j} - 5\hat{k})$. Now pv of centroid $\overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3} = \frac{\overrightarrow{0} + (2, -1, 1) + (-1, 3, 5)}{3}$. $\overrightarrow{G} = \frac{1}{3}(\hat{i} + 2\hat{j} + 6\hat{k})$. Now $\overrightarrow{AG} = \frac{1}{3}(\hat{i} + 2\hat{j} + 6\hat{k})$. $\Rightarrow |\overrightarrow{AG}|^2 = \frac{1}{9} \times 41$. $\overrightarrow{BG} = \left(\frac{1}{3} - 2\right)\hat{i} + \left(\frac{2}{3} + 1\right)\hat{j} + (2 - 1)\hat{k}$. $\Rightarrow |\overrightarrow{BG}|^2 = \frac{59}{9}$. $\overrightarrow{CG} = \left(\frac{1}{3} + 1\right)\hat{i} + \left(\frac{2}{3} - 3\right)\hat{j} + (2 - 5)\hat{k}$. $\Rightarrow |\overrightarrow{CG}|^2 = \frac{146}{9}$. Now $6 \left[|\overrightarrow{AG}|^2 + |\overrightarrow{BG}|^2 + |\overrightarrow{CG}|^2\right] = 6 \times \left[\frac{41}{9} + \frac{59}{9} + \frac{146}{9}\right] = 6 \times \frac{246}{9} = 164$.

Given $T_r = \binom{1016}{r} \left( \frac{5}{2} \right)^{1016-r} \left( \frac{7}{8} \right)^r$. This implies $r = 0, 8, 16, 24, \ldots, 1016$. We have $1016 = 0 + (n-1)8$. Thus, $n - 1 = \frac{1016}{8} = 127$. So, $n = 128$.

Given $f(x) = x - 1$. $f(f(x)) = f(x) - 1 = x - 1 - 1 = x - 2$. $g(f(f(x))) = e^{x-2}$. Therefore, $\frac{dy}{dx} = e^{-2\sqrt{x}} \times e^{x-2} - \frac{1}{\sqrt{x}} y$. $\frac{dy}{dx} + \frac{1}{\sqrt{x}} y = e^{x-2\sqrt{x}-2}$ which is L.D.E. I.F. $= e^{\int \frac{dy}{\sqrt{x}}} = e^{2\sqrt{x}}$. Its solution is $$y \times e^{2\sqrt{x}} = \int e^{2\sqrt{x}} \times e^{x-2\sqrt{x}-2} dx + c$$ $$y \times e^{2\sqrt{x}} = \int e^{x-2} dx + c$$ $$y \times e^{2\sqrt{x}} = e^{x-2} + c$$ Given $x = 0$, $y = 0 \implies 0 = e^{-2} + c$; $c = -e^{-2}$. Therefore, $y \times e^{2\sqrt{x}} = e^{x-2} - e^{-2}$. When $x = 1$, $y \times e^2 = e^{-1} - e^{-2}$. $$y = \frac{e^{-1} - e^{-2}}{e^2} = \frac{\frac{1}{e} - \frac{1}{e^2}}{e^2} = \frac{e^2 - e}{e^5} = \frac{e-1}{e^4}$$ Option (1) is correct.

Given the expression: $$\cot^{-1}\left(\frac{|\sec 2|-1}{\tan 2}\right) - \cot^{-1}\left(\frac{|\sec \frac{1}{2}|+1}{\tan \frac{1}{2}}\right)$$ This simplifies to: $$= \cot^{-1}\left(\frac{-1 - \cos 2}{\sin 2}\right) - \cot^{-1}\left(\frac{1 + \cos \frac{1}{2}}{\sin \frac{1}{2}}\right)$$ Further simplification gives: $$= \pi - \cot^{-1}(\cot 1) - \cot^{-1}\left(\cot \frac{1}{4}\right)$$ This results in: $$= \pi - 1 - \frac{1}{4} = \pi - \frac{5}{4}$$

Given $\($ |A| = $\begin{vmatrix}$ 2 & 2+p & 2+p+q $\\$ 4 & 6+2p & 8+3p+2q $\\$ 6 & 12+3p & 20+6p+3q $\end{vmatrix}$ $\)$. Perform the row operations: $\($ C_3 $\rightarrow$ C_3 - C_2 - C_1 $\times$ $\frac{q}{2}$ $\)$. Then $\($ C_3 $\rightarrow$ C_2 - C_1 $\times$ $\left$( 1 + $\frac{p}{2}$ $\right$) $\)$. This results in: $$ |A| = \begin{vmatrix} 2 & 0 & 0 \\ 4 & 2 & 2+p \\ 6 & 6 & 8+3p \end{vmatrix} $$ Thus, $$ |A| = 2(16 + 6p - 12 - 6p) = 8 = 2^3 $$ Now, $$ |adj(adj(3A))| = |3A|^{(3-1)^2} = |3A|^4 $$ This implies, $$ = (3^3 |A|)^4 = (3^3 \times 2^3)^4 = 2^{12} \times 3^{12} $$ Therefore, $\($ m + n = 24 $\)$.

Given $\tan^{-1} x + \frac{1}{2} \left[ \ln(1+x) - \ln(1-x) \right] - 2x$. $$\lim_{x \to 0} \left( x - \frac{x^3}{3} + \frac{x^5}{5} \cdots \right) + \frac{1}{2} \left[ x - \frac{x^2}{2} + \frac{x^3}{3} \cdots - \left( -x - \frac{x^2}{2} - \frac{x^3}{3} \cdots \right) \right] - 2x$$ $$= \lim_{x \to 0} x^5$$ $$= 2x + \frac{2x^5}{5} \cdots - 2x = \frac{2}{5}$$ $$\lim_{x \to 1} x^{2(1-x)} = e^{\lim_{x \to 1} \left( 2(1-x) \right) (x-1)} = e^{-2}$$ Therefore, both statements are correct.

Given the equations $0 \leq 9x \leq y^2$ and $y \geq 3x - 6$. The required area $A$ is given by: $$A = \int_0^1 (-3\sqrt{x}) \, dx - \int_0^1 (3x - 6) \, dx$$ Calculating the integrals, we have: $$A = -3 \left( \frac{x^{3/2}}{3/2} \right) \bigg|_0^1 - \left( \frac{3x^2}{2} - 6x \right) \bigg|_0^1$$ Simplifying, we get: $$A = -2[1 - 0] \left[ \frac{3}{2} - 6 \right]$$ Further simplification gives: $$A = -2 - \frac{3}{2} + 6 = \frac{5}{2} Sq. unit$$ Thus, $6A = 6 \times \frac{5}{2} = 15$

Given $f(x) = \cos^{-1}\left(\frac{4x + 5}{3x - 7}\right)$. $$-1 \leq \frac{4x + 5}{3x - 7} \leq 1$$ $$\frac{4x + 5}{3x - 7} \geq -1$$ $$4x + 5 + 3x - 7 \geq 0$$ $$7x - 2 \geq 0$$ $$x \geq \frac{2}{7}$$ $$\frac{4x + 5}{3x - 7} \leq 1$$ $$4x + 5 \leq 3x - 7$$ $$x + 12 \leq 0$$ $$x \leq -12$$ Therefore, $x \in \left(-\infty, \frac{2}{7}\right] \cup \left(\frac{7}{3}, \infty\right)$ Domain of $f(x)$ is $\left[-12, \frac{2}{7}\right]$. Let $\alpha = -12$, $\beta = \frac{2}{7}$. $g(x) = \log_2(2 - 6 \log_{27}(2x + 5))$ Domain: $$2 - 6 \log_{27}(2x + 5) > 0$$ $$6 \log_{27}(2x + 5) 0 \Rightarrow x > -\frac{5}{2}$$ Domain is $x \in \left(-\frac{5}{2}, -1\right)$ Let $\gamma = \frac{-5}{2}$, $\delta = -1$ $$|7(\alpha + \beta) + 4(\gamma + \delta)| = \left|7\left(-12 + \frac{2}{7}\right) + 4\left(-\frac{5}{2} - 1\right)\right|$$ $$|-82 - 14| = 96$$

Given the lines: $$L_1: x + 2 = y - 1 = z = \ell$$ $$L_2: \frac{x}{5} = \frac{y - 3}{-1} = \frac{z - 2}{1} = m$$ $$L_3: \frac{x}{-3} = \frac{y}{5} = \frac{z - 1}{1} = n$$ Point of intersection of $L_1$ and $L_2$: $$\begin{cases} \ell - 2 = 5m + 3 \\ \ell + 1 = -m \\ \ell = m + 1 \end{cases}$$ Solving gives $\ell = 0$, $m = -1$. Therefore, $A(-2, 1, 0)$. Point of intersection of $L_2$ and $L_3$: $$\begin{cases} 5m + 3 = -3n \\ -m = 3n + 3 \\ m + 1 = n + 2 \end{cases}$$ Solving gives $m = 0$, $n = -1$. Therefore, $B(3, 0, 1)$. Point of intersection of $L_3$ and $L_4$: $$\begin{cases} -3n = \ell - 2 \\ 3n + 3 = \ell + 1 \\ n + 2 = \ell \end{cases}$$ Solving gives $\ell = 2$, $n = 0$. Therefore, $C(0, 3, 2)$. The area of triangle $\triangle ABC$ is given by: $$Ar(\triangle ABC) = \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -1 \\ -3 & 3 & 1 \end{vmatrix}$$ Calculating the determinant: $$A = \frac{1}{2} \left| \hat{i}(4) - \hat{j}(-8) + \hat{k}(-12) \right|$$ $$A = \frac{1}{2} \sqrt{16 + 64 + 144} = \sqrt{56}$$ Thus, $A^2 = 56$.

Given $ (1919)^{1919} = (1920 - 1)^{1919} $. $$ = \binom{1919}{0}(1920)^{1919} - \binom{1919}{1}(1920)^{1918} + \ldots $$ $$ + \binom{1919}{1918}(1920)^1 - \binom{1919}{1919} $$ $$ = 100\lambda + 1919 \times 1920 - 1 $$ $$ = 100\lambda + 3684480 - 1 $$ $$ = 100\lambda + \ldots 79 (last two digit) $$ Therefore, the number has last two digits 79. Thus, the product of the last two digits is 63.

The equation of the circle is $(x-a)^2 + (y-r)^2 = r^2$. Substituting $(4-a)^2 + (6-r)^2 = r^2$, we get $$16 + a^2 - 8a + 36 + r^2 - 12r = r^2$$ $$a^2 - 8a - 12r + 52 = 0$$ The tangent to the parabola at $(4, 6)$ is $6.4 = 9 \cdot \left(\frac{x+4}{2}\right)$ i.e. $3x - 4y + 12 = 0$. This is also tangent to the circle, so $CP = r$. Therefore, $$\frac{3a - 4r + 12}{5} = \pm r$$ $$3a + 12 = 4r \pm 5r$$ From equation (1), the equation of the circle is $(x-a)^2 + (y-r)^2 = r^2$. Substituting $P(4, 6)$, we get $a^2 - 8a - 12r + 52 = 0$. From equation (1), if $a + 4 = 3r$ then $a = +6$ (rejected). If $3a + 12 = -r$ then $a = -14$ and $r = 30$.

For a uniformly charged spherical shell: The electric field inside the shell is zero everywhere (from Gauss's law). When the electric field is zero, no work is done in moving a charge. Since work = force $\times$ displacement, and force = qE (where E = 0), work = 0. The potential inside is constant and equal to the value at the surface. Assertion A is correct: Since the electric field is zero everywhere inside the shell, no work is done moving a charge between any two points, regardless of path. Reason R is correct: The potential inside a uniformly charged spherical shell is indeed constant and equal to the value at the surface. Does R explain A? Yes, because: Constant potential means zero electric field (since E = -FW). Zero electric field means no force on a test charge. No force means no work done in moving the charge. Therefore, option (2) is correct: Both A and R are true and R is the correct explanation of A.

Given $L = 2 \pi R$. $$I = \frac{MR^2}{2} = \frac{\lambda \times L}{2} \times \left( \frac{L}{2\pi} \right)^2 = \frac{\lambda L^3}{8\pi^2}$$

Given $\frac{50 \, \mathrm{g}}{\pi r^2} = y \cdot \frac{\Delta \ell}{\ell}$. $$\frac{50 \times 3\pi}{\pi \times (3 \times 10^{-3})^2} = P \times 10^{11} \times \frac{0.1 \times 10^{-3}}{3}$$ Therefore, $$P = \frac{50 \times 3 \times 3}{3^2 \times 10^{-6} \times 10^{11} \times 0.1 \times 10^{-3}}$$ $$P = 5$$

Given $\sigma \propto \frac{1}{\mathrm{ROC}}$. For (1), $(\mathrm{ROC})_1 \sigma_3 > \sigma_2 = \sigma_4$.

Given $mgh = ms \Delta T$. $$\Delta T = \frac{gh}{s} = \frac{10 \times 200}{4200} = \frac{10}{21} \, \mathrm{K}$$

Given $\mu_1 = 1.3$, $R_1 = 30 \, \mathrm{cm}$, $R_2 = 20 \, \mathrm{cm}$, $\mu_2 = 1.5$. The formula for the focal length $f$ is given by: $$\frac{1}{f} = \left( \frac{1.3 - 1}{1} \right) \left( \frac{1}{\infty} - \frac{1}{-30} \right) + \left( \frac{1.5 - 1}{1} \right) \left( \frac{1}{-30} - \frac{1}{-30} \right)$$ Simplifying, we have: $$= \left( 0.3 \right) \left( 0 + \frac{1}{30} \right) + \left( 0.5 \right) \left( 0 \right)$$ $$= \frac{0.3}{30} + 0$$ $$= \frac{1}{100} + \frac{1}{120}$$ $$= \frac{6 + 5}{600}$$ $$= \frac{11}{600}$$ Thus, the focal length $f$ is: $$f = \frac{600}{11} \, \mathrm{cm}$$

Given $a = \sqrt{3} \, \mathrm{cm}$. The flux $\phi$ is given by $$\phi = \frac{q_{enc}}{\varepsilon_0} = \frac{\lambda \cdot \sqrt{3}a}{\varepsilon_0}$$ Substituting the values, $$= 2 \times 10^{-9} \times \sqrt{3} \times \sqrt{3} \times 10^{-2} \times 36\pi \times 10^9 \, \mathrm{Nm^2C^{-1}}$$ $$= 2.16\pi \, \mathrm{Nm^2C^{-1}}$$

Here $D_1$ is reverse biased and $D_2$ is forward biased. Therefore current flows through $D_2$ and $5 \, V$ drop on resistor. So, $V_{out} = 0$.

For a conducting sphere, $V = \frac{\sigma r}{\varepsilon_0}$. After contact, $V_1 = V_2$. $$\sigma_1 r_1 = \sigma_2 r_2$$ $$\sigma_1 = \frac{r_2}{r_1} \sigma_2$$ $$\sigma_1 = 3 \sigma_2$$

Fractional error = $2 \frac{\Delta X}{X} + \frac{3}{2} \frac{\Delta Y}{Y} + \frac{2}{5} \frac{\Delta Z}{Z}$. $$= 2(0.1) + \frac{3}{2} (0.2) + \frac{2}{5} (0.5)$$ $$= 0.2 + 0.3 + 0.2 = 0.7$$

Given $P_i V_i^\gamma = P_f V_f^\gamma$. $$\frac{P_f}{P_i} = \left(\frac{V_i}{V_f}\right)^\gamma = (8)^{5/3}$$ Therefore, $$\frac{P_f}{P_i} = 32$$

Equivalent focal length $$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$$ $$= \frac{1}{30} + \frac{1}{-20} = \frac{2 - 3}{60} = -\frac{1}{60}$$ $$f = -60 \, cm$$ Lens formula $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{v} - \frac{1}{-20} = \frac{1}{-60}$$ $$v = -15 \, cm$$

Given the equation for velocity, we have $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{T}{f \pi R^2}}$. The ratio of velocities is given by $\frac{v_2}{v_1} = \frac{R_1}{R_2} = 2$.

The magnetic field $\vec{B}$ is given by the sum of the magnetic fields due to each segment: $$\vec{B} = \vec{B}_{AB} + \vec{B}_{BC} + \vec{B}_{CD} + \vec{B}_{DA}$$ Calculating each component, we have: $$\vec{B} = \left[ \frac{-\mu_0 (2I/3)}{4\pi (a/2)} \sqrt{2} - \frac{\mu_0 (2I/3)}{4\pi (a/2)} \sqrt{2} \right.$$ $$+ \left. \frac{\mu_0 (I/3)}{4\pi (a/2)} \sqrt{2} + \frac{\mu_0 (I/3)}{4\pi (a/2)} \sqrt{2} \right] \hat{k}$$ Simplifying, we get: $$\vec{B} = \left[ \frac{-2\sqrt{2} \mu_0 I}{3\pi a} + \frac{\sqrt{2} \mu_0 I}{3\pi a} \right] \hat{k}$$ Finally, the magnetic field is: $$\vec{B} = \frac{-\sqrt{2} \mu_0 I}{3\pi a} \hat{k}$$

Given $\vec{a} = \frac{\mathbf{B}}{2} \hat{k} = 3 \hat{k}$, $t = \frac{5}{3} \, \mathrm{s}$. $\vec{u} = 3 \hat{i} + 4 \hat{j}$ $\vec{v} = \vec{u} + \vec{a} \, t = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$

Given, $(H_{max})_1 = 8 \times (H_{max})_2$. $$\frac{u^2 \sin^2 \theta_1}{2g} = 8 \times \frac{u^2 \sin^2 \theta_2}{2g}$$ Therefore, $\sin \theta_1 = 2\sqrt{2} \sin \theta_2$. $$T_1 = \frac{2u \sin \theta_1 / g}{2u \sin \theta_2 / g} = \frac{\sin \theta_1}{\sin \theta_2} = 2\sqrt{2}$$

Given the triangle, we calculate the side A using the cosine rule: $$A = \sqrt{2^2 + 4^2 + 2 \times 2 \times 4 \times \cos 120^\circ}$$ This simplifies to: $$= \sqrt{12} = 2\sqrt{3}$$ Next, we find $\tan \phi$: $$\tan \phi = \frac{2 \sin 120^\circ}{4 + 2 \cos 120^\circ} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$ Thus, $\phi$ is: $$\phi = \frac{\pi}{6}$$

Different colours will have different fringe width. Within a few fringes of red, there will be several fringes of violet. Also, there will be overlapping of colours.

The mass density ( $\rho$ ) of a nucleus is defined as the ratio of its mass to its volume: $\rho = Mass of nucleus / Volume of nucleus$ For a spherical nucleus with radius $R$, the volume is: $$Volume = \frac{4}{3} \pi R^3$$ The mass of the nucleus is related to the mass number $A$: $$Mass = A \times m_u$$ (where $m_u$ is the atomic mass unit, approximately $1.66 \times 10^{-27} kg$) Therefore: $$\rho = (A \times m_u) / \left[ \frac{4}{3} \pi R^3 \right]$$ Now, there's an important relationship between the mass number $A$ and nuclear radius $R$. Empirically, it has been found that: $$R = R_o \times A^{1/3}$$ Where $R_o$ is a constant approximately equal to $1.2 \times 10^{-15} m$ (1.2 fermi). Substituting this into the density equation: $$\rho = (A \times m_u) / \left[ \frac{4}{3} \pi (R_o \times A^{1/3})^3 \right]$$ $$\rho = (A \times m_u) / \left[ \frac{4}{3} \pi R_o^3 \times A \right]$$ $$\rho = m_u / \left[ \frac{4}{3} \pi R_o^3 \right]$$ This shows that the nuclear density is approximately constant for all nuclei, regardless of their mass number. This is a fundamental property of nuclear matter known as nuclear saturation density. The numerical value is: $$\rho = 2.3 \times 10^{17} kg/m^3$$

Given, Natural length of spring = 2 $\mathrm{m}$. Initial compression in spring (x_i) = 1 $\mathrm{m}$. Final compression in spring (x_f) = (2 - x) $\mathrm{m}$. Using energy conservation $$K_i + U_i = K_f + U_f$$ $$0 + \frac{1}{2} K x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} K x_f^2$$ $$\frac{1}{2} m v^2 = \frac{1}{2} K (x_i^2 - x_f^2)$$ $$\frac{1}{2} \times 2 \times v^2 = \frac{1}{2} \times 200 \times (1^2 - (2 - x)^2)$$ $$v^2 = 100 \left[ 1 - (2 - x)^2 \right]$$ $$v = 10 \left[ 1 - (2 - x)^2 \right]^{1/2}$$

Let the momentum of $e^-$ at any time $t$ is $p$ and its de-broglie wavelength is $\lambda$. Then, $p = \frac{h}{\lambda}$. $$\frac{dp}{dt} = -\frac{h}{\lambda^2} \frac{d\lambda}{dt}$$ $$ma = F = -\frac{h}{\lambda} \frac{d\lambda}{dt} [m = mass of e]$$ Where, -ve sign represents decrease in $\lambda$ with time. $$ma = -\frac{h}{(h/p)^2} \frac{d\lambda}{dt}$$ $$a = -\frac{p^2}{mh} \frac{d\lambda}{dt}$$ $$a = -\frac{h}{mv^2} \frac{d\lambda}{dt}$$ $$\frac{d\lambda}{dt} = -\frac{ah}{mv^2} \cdots (1)$$ Here, $a = \frac{qE}{m} = \frac{e}{m} \frac{\sigma}{2\varepsilon_0}$$ $$a = \frac{\sigma e}{2 m \varepsilon_0}$$ and $v = u + at$. $$v = at$$ Substituting values of $a$ \& $v$ in equation (1) $$\frac{d\lambda}{dt} = -\frac{2 h \varepsilon_0}{\sigma et^2}$$ $$\Rightarrow \frac{d\lambda}{dt} \propto \frac{1}{t^2}$$ $$\Rightarrow n = 2$$

Given, Initial pressure of liquid $(P_i) = 1 \, atm$ Final pressure of liquid $(P_f) = 5 \, atm$ Change in pressure $(\Delta P) = P_f - P_i = 4 \, atm = 4 \times 10^5 \, Pa$ Change in volume $(\Delta V) = -0.8 \, cm^3$ Bulk modulus $(B) = 2 \times 10^9 \, Pa$ Now, $B = -\frac{\Delta P}{(\Delta V / V)} \Rightarrow V = -B \left( \frac{\Delta V}{\Delta P} \right)$ $$\Rightarrow V = -2 \times 10^9 \times \left( \frac{-0.8 \times 10^{-6}}{4 \times 10^5} \right)$$ $$= 4 \times 10^{-3} \, m^3 = 4 \, litre$$

Given $$C_1 = \frac{5 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}} = 20 \, \mathrm{pF}$$ $$C_2 = \frac{3 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}} = 12 \, \mathrm{pF}$$ $$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{12 \times 20}{12 + 20} = 7.5 \, \mathrm{pF}$$ Finally equivalent capacitance $$(C_{eq})_{final} = 7.5 + 7.5 = 15 \, \mathrm{pF}$$

Given, $m = 1 \, \mathrm{kg}$. $\omega_i = 1800 \, \mathrm{rpm} = 1800 \times \frac{2\pi}{60} \, \mathrm{rad/sec} = 60\pi \, \mathrm{rad/sec}$ $\omega_f = 2100 \, \mathrm{rpm} = 2100 \times \frac{2\pi}{60} = 70\pi \, \mathrm{rad/sec}$ $\tau_{ext} = 25\pi \, \mathrm{Nm}$ $t = 40 \, \mathrm{sec}$ Using equation of motion $$\omega_f = \omega_i + \alpha t$$ $$70\pi = 60\pi + \alpha (40)$$ $$\alpha = \frac{\pi}{4} \, \mathrm{rad/sec^2}$$ Also, $\tau = I \alpha$ $$\tau = \frac{mR^2}{4} \alpha$$ $$25\pi = \frac{1 \times R^2}{4} \times \frac{\pi}{4}$$ $R = 20 \, \mathrm{m}$ Hence, diameter of disk $= 2R = 2 \times 20 = 40 \, \mathrm{m}$

Given, volume of block $= \left(10 \times 10^{-2}\right)^3 = 10^{-3} \, \mathrm{m^3}$. Let density of block $= \rho \, \mathrm{kg/m^3}$. Mass of block $= \rho \times 10^{-3} \, \mathrm{kg}$. Buoyant Force $(F_B) = 1000 \times \frac{10^{-3}}{2} \times 10 = 5 \, \mathrm{N}$. F.B.D. of blocks. Balancing torque about point $O$, we get $$mg(2 \times 10^{-2}) - F_B (2 \times 10^{-2}) = 0.2 \, g \left(25 \times 10^{-2}\right)$$ $$\rho \times 10^{-3} \times 10 \times 2 - 10 = 50$$ $$\rho = 3000 \, \mathrm{kg/m^3}$$ Hence, mass of block $= \rho \times 10^{-3}$ $$= 3000 \times 10^{-3} = 3 \, \mathrm{kg}$$

For 1st order reaction. When $C_t = C_0 / 4$, $t_1 = 2t_{50\%}$. When $C_t = C_0 / 8$, $t_2 = 3t_{50\%}$. So, $\frac{t_1}{t_2} = \frac{2}{3}$.

A benzamide shows positive Lassaigne's test. B benzoic acid gives effervescence with aq. $NaHCO_3$. C ester gives fruity smell.

Ethyl benzene reacts with $\mathrm{Br_2/Fe}$ to form the major product, brominated ethyl benzene. This compound then reacts with $\mathrm{Cl_2}$ under heat ($\Delta$) to form a chlorinated compound. Finally, treatment with alcoholic $\mathrm{KOH}$ leads to the formation of styrene (vinyl benzene) with a bromine substituent.

Homoleptic complex of type $[\mathrm{Ma}_6]$ (Where $a \Rightarrow$ monodentate ligand) cannot show geometrical as well as optical isomerism. Cis-platin and trans-platin has formula $[\mathrm{Pt(NH_3)_2Cl_2}]$ which is a heteroleptic complex of platinum.

Atomic no. 32 implies Ge. Atomic no. 35 implies Br. Atomic no. 87 implies Fr. Atomic no. 19 implies K. Lowest first ionization energy among the given elements will be of Fr [87]. Fr – [Rn] $7s^1$

Binary mixture of $\mathrm{C_6H_5OH}$ and $\mathrm{C_6H_5NH_2}$ shows negative deviation from Raoult's law. So vapour pressure of solution is less than V.P of pure $\mathrm{C_6H_5OH}$ and $\mathrm{C_6H_5NH_2}$. So B.P. of solution is greater than boiling point of pure $\mathrm{C_6H_5OH}$ and $\mathrm{C_6H_5NH_2}$. So shows maximum Boiling azeotrope.

Given $\Delta T_f = i k f m$. $$0.2 = i \times 1.8 \times 0.1$$ $$i = \frac{20}{18} = \frac{10}{9}$$ For $\mathrm{HA_{(aq)}} \rightleftharpoons \mathrm{H^+_{(aq)}} + \mathrm{A^-_{(aq)}}$ At $t = 0$, concentrations are $1$ and $0$. At $t = t_{eq}$, concentrations are $1 - \alpha$ and $\alpha$. Thus, $i = 1 + \alpha$. $$\frac{10}{9} = 1 + \alpha$$ $$\alpha = \frac{1}{9}$$ The equilibrium constant $K_{eq}$ is given by: $$K_{eq} = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} = \frac{C \alpha^2}{1 - \alpha}$$ Substituting the values: $$0.1 \left( \frac{1}{9} \right)^2$$ $$= \frac{1}{1 - \frac{1}{9}} = \frac{1}{\frac{8}{9}} = \frac{1}{720}$$ Thus, $$K_{eq} = 1.38 \times 10^{-3}$$

The compound shown is named 4-Ethylcyclopent-2-en-1-ol.

$Cu^{+} : [$\mathrm{Ar}$] 3d^{10}$, Spin only magnetic moment = 0 B.M. $Cu^{+2} : [$\mathrm{Ar}$] 3d^{9}$, Spin only magnetic moment = $\sqrt{3}$ B.M. $Cr^{+2} : [$\mathrm{Ar}$] 3d^{4}$, Spin only magnetic moment = $\sqrt{24}$ B.M. $Cr^{+3} : [$\mathrm{Ar}$] 3d^{3}$, Spin only magnetic moment = $\sqrt{15}$ B.M. Order of $\mu$ : $Cr^{+2} > Cr^{+3} > Cu^{+2} > Cu^{+}$

The reaction involves an elimination reaction between isobutyl sodium oxide and sec-butyl bromide. The major product formed is an alkene.

In the combustion of organic compound, all "C" in $\mathrm{CO_2}$ and all "H" in $\mathrm{H_2O}$ comes from organic compound $$\mathrm{C_xH_yO_z + O_2 \longrightarrow CO_2 + H_2O}$$ Weight of "C" in $\mathrm{CO_2} = \frac{12}{44} \times 0.307$$ $$= 0.0837 \, \mathrm{gm}$$ Weight of "H" in $$\mathrm{H_2O}$ = $\frac{2}{18}$ $\times$ 0.127 = 0.0141 $\,$ $\mathrm{g}$$$ % 'H' in compound $$= $\frac{0.0141}{0.21}$ $\times$ 100 = 6.719$\,$$\%$$$ $$= 6.72$\,$$\%$$$ Weight of "O" in compound $$= 0.210 - (0.0837 + 0.0141)$$ $$= 0.1122$$ % of "O" in compound $$= $\frac{0.1122}{0.21}$ $\times$ 100$$ $$= 53.41$\,$$\%$$$

Element with atomic number 9 is Fluorine. $$\mathrm{F(9) = 1s^2 \, 2s^2 \, 2p^5}$$ (A) 5 electrons can be up-spin $\left[ m_s = +\frac{1}{2} \right]$ and 4 electrons can be down spin $\left[ m_s = -\frac{1}{2} \right]$. (B) Unpaired electron can be in anyone of $p_x$, $p_y$ or $p_z$ orbital. (C) Last electron is in 2p subshell with $n = 2$, $\ell = 1$. (D) Angular node for s-orbital $= 0$ while of each p-orbital $= 1$. Sum of all angular node $= 3$.

In $[\mathrm{Co(NH_3)_6}]^{3+}$, $\mathrm{Co}^{+3}$: $[\mathrm{Ar}] 3d^6$, $\mathrm{NH_3}$ is S.F.L. Hybridisation state of $\mathrm{Co}^{3+}$ is $d^2sp^3$. In $\mathrm{SF_6}$, Hybridisation state of sulphur is $sp^3d^2$. In $[\mathrm{CrF_6}]^{3-}$, $\mathrm{Cr}^{+3}$: $[\mathrm{Ar}] 3d^3$. Hybridisation state of $\mathrm{Cr}^{3+}$ is $d^2sp^3$. $[\mathrm{CoF_6}]^{3-}$, $\mathrm{Co}^{+3}$: $[\mathrm{Ar}] 3d^6$, $\mathrm{F}^{-}$ is W.F.L. Hybridisation state of $\mathrm{Co}^{3+}$ is $sp^3d^2$. $[\mathrm{Mn(CN)_6}]^{3-}$, $\mathrm{Mn}^{+3}$: $[\mathrm{Ar}] 3d^4$, $\mathrm{CN}^{-}$ is S.F.L. Hybridisation state of $\mathrm{Mn}^{3+}$ is $d^2sp^3$. $[\mathrm{MnCl_6}]^{3-}$, $\mathrm{Mn}^{+3}$: $[\mathrm{Ar}] 3d^4$, $\mathrm{Cl}^{-}$ is W.F.L. Hybridisation state of $\mathrm{Cl}^{-}$ is $sp^3d^2$. Total number of $sp^3d^2$ hybridized molecules is 3.

Q3. (1) Carboxylic acid gives effervescence with sodium bicarbonate solution. (2) Phenolic-OH gives violet coloured complex with Neutral $\mathrm{FeCl_3}$. (3) Alcoholic-OH gives red colour with ceric ammonium nitrate. (4) When alkaline $\mathrm{KMnO_4}$ reacts with an unsaturated compound (alkene or alkyne) the purple colour of $\mathrm{KMnO_4}$ solution disappears, indicating positive test for unsaturation.

The reaction shown is an aldol condensation reaction. Initially, the enolate ion attacks the carbonyl carbon, forming a new carbon-carbon bond. This is followed by protonation to form a β-hydroxy ketone. Upon heating, dehydration occurs, resulting in the formation of an α,β-unsaturated ketone. The major product is the cyclopentanone derivative.

Acidic strength: $\mathrm{H_2Se} \mathrm{H_2Te}$ $[276 \, \mathrm{KJ/mol}]$ $[238 \, \mathrm{KJ/mol}]$

Given $\Delta H_{R.E} = \Delta H_f(exp) - \Delta H_f(Theo)$. $\Delta H_f(exp)$ for $X_2Y_{(g)} = 80 \, kJ/mole$. For $\Delta H_f(Theo)$, the reaction is $X_2(g) + \frac{1}{2} Y_2(g) \rightarrow X_2Y_{(g)}$ with $\Delta H_f = ?$. $$\Delta H_f(Theo) = \left( BE_{X \equiv X} + \frac{1}{2} BE_{Y \equiv Y} \right) - (BE_{X \equiv X} + BE_{X \equiv Y})$$ $$= \left( 940 + \frac{1}{2} \times 500 \right) - (410 + 602)$$ $$= 178 \, kJ/mole$$ $$\Delta H_{R.E} = 80 - 178$$ $$= -98 \, kJ/mol$$ $$|\Delta H_{R.E}| = 98$$

Given the total energy formula: $$E_T = -13.6 \frac{Z^2}{n^2} eV$$ For the energy of the H-atom, energy of the 1st Bohr orbit: $$E_1 = -13.6 eV \; [z = 1, n = 1]$$ For the Be$^{+3}$ ion, energy of the 1st E.S.: $$[z = 4, n = 2]$$ The ratio of energies is given by: $$\frac{E_H}{E_{Be^{+3}}} = \frac{Z_1^2}{n_1^2} \times \frac{n_2^2}{Z_2^2}$$ Substituting the values: $$\frac{E_H}{E_{Be^{+3}}} = \frac{1}{1} \times \frac{4}{16}$$ Therefore, $$E_{Be^{+3}} = -13.6 \times 4 = -54.4 eV$$ The magnitude of the energy is: $$|E_{Be^{+3}}| = 54.4 eV$$

The reaction is given by: $$\mathrm{NaI_{(aq)} + AgNO_3_{(aq)} \rightarrow AgI_{(s)} + NaNO_3_{(aq)}}$$ M, 20 ml excess 4.74 g Moles of $\mathrm{I^-}$ in $\mathrm{NaI}$ = Moles of $\mathrm{(I^-)}$ in $\mathrm{AgI} = \frac{4.74}{235}$ Moles of $\mathrm{NaI} = \frac{4.74}{235}$ Molarity $[\mathrm{NaI}] = \frac{4.74}{235 \times 0.02} = 1.008$

The reaction is given by $\mathrm{H_2O(g)} \rightleftharpoons \mathrm{H_2(g)} + \frac{1}{2} \mathrm{O_2(g)}$. At $t = 0$, we have 1 mole. At $t = t_{eq}$, the amounts are $1 - \alpha$, $\alpha$, and $\frac{\alpha}{2}$. The total number of moles $n_T = 1 + \frac{\alpha}{2} \simeq 1 (\alpha \ll 1)$. The equilibrium constant $k_P$ is given by: $$k_P = \frac{P_{\mathrm{H_2}} \cdot P_{\mathrm{O_2}}^{1/2}}{P_{\mathrm{H_2O}}} = \frac{(\alpha \cdot P) \left( \frac{\alpha}{2} P \right)^{1/2}}{(1 - \alpha) P}$$ Solving for $\alpha$, we have: $$8 \times 10^{-3} = \frac{\alpha^{3/2}}{\sqrt{2}}$$ $$\alpha^{3/2} = 8 \sqrt{2} \times 10^{-3}$$ $$\alpha^3 = 128 \times 10^{-6}$$ $$\alpha = \sqrt[3]{128} \times 10^{-2}$$ $$= 5.03 \times 10^{-2}$$

The reaction is given by: $$\mathrm{Cr_2O_7^{2-}}_{(aq)} + 14\mathrm{H^+}_{(aq)} + 6\mathrm{e^-} \rightarrow 2\mathrm{Cr^{3+}}_{(aq)} + 7\mathrm{H_2O}_{(\ell)}$$ The equation for the reduction potential is: $$E_R = E_R^0 - \frac{0.059}{6} \log \left( \frac{[\mathrm{Cr^{3+}}]^2}{[\mathrm{Cr_2O_7^{2-}}][\mathrm{H^+}]^{14}} \right)$$ Substituting the values, we have: $$0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[\mathrm{H^+}]^{14}} \right)$$ Simplifying gives: $$1.33 \times 6 = \frac{0.059}{[\mathrm{H^+}]^{14}} \log 10^{-6}$$ This results in: $$135.254 = -6 - 14 \log [\mathrm{H^+}]$$ Rearranging gives: $$141.254 = 14 \mathrm{pH}$$ Thus, the pH is: $$\mathrm{pH} = \frac{141.254}{14} = 10.08$$