JEE Advanced 18 May 2025 Paper 2 question paper with solutions

Given $e^{x_0} + x_0 = 0$. $$g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}$$ $$g(x) = x - \frac{\alpha (e^x + x)}{3(e^x + 1)}, g(x) + e^{x_0} = x + e^{x_0} - \frac{\alpha}{3} \left( \frac{e^x + x}{e^x + 1} \right)$$ $$g(x) + e^{x_0} = (x - x_0) - \frac{\alpha}{3} \left( \frac{e^x + x}{e^x + 1} \right)$$ As very clear $g(x_0) + x_0 \Rightarrow 0$ $$\lim_{x \to x_0} \frac{g(x) + e^{x_0}}{x - x_0} \frac{0 \uparrow}{0 \uparrow} = \lim_{x \to x_0} \left| g'(x_0) \right| = \left| g'(x_0) \right|$$ $$g'(x) = 1 - \frac{\alpha}{3} \left( \frac{(e^x + 1)(e^x + 1) - (e^x + x)e^x}{(e^x + 1)^2} \right)$$ $$g'(x_0) = 1 - \frac{\alpha}{3} \left( \frac{(e^x + 1)^2}{(e^x + 1)^2} \right) = 1 - \frac{\alpha}{3}$$ $$\Rightarrow \lim_{x \to x_0} \frac{g(x) + e^{x_0}}{x - x_0} = \left| g'(x_0) \right| = \left| 1 - \frac{\alpha}{3} \right|$$

Given the equations $5x - \frac{4}{x} = 1$ and $4x + \frac{4}{x} = 17$. Solving $5x - \frac{4}{x} = 1$ gives $x = 1$. Solving $4x + \frac{4}{x} = 17$ gives $x = 4$. For $5x - 4y = 1$ and $4x + 4y = 17$, solving gives $x = 2$, $y = \frac{9}{4}$. The area is calculated as follows: $$Area = \frac{1}{2} \left( 1 + \frac{9}{4} \right) + \frac{1}{2} \left( \frac{9}{4} + \frac{1}{4} \right) \cdot 2 - \int_{1}^{4} \frac{dx}{x}$$ This simplifies to: $$Area = \frac{13}{8} + \frac{20}{8} - \log_e 4 = \left( \frac{33}{8} - \log_e 4 \right)$$

Let $\alpha = \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right)$. $\newline$ $\theta = \tan^{-1}(2 \tan \theta) - \alpha$. $\newline$ $\theta + \alpha = \tan^{-1}(2 \tan \theta)$. $\newline$ $\($ $\tan$($\theta$ + $\alpha$) = 2 $\tan$ $\theta$ $\)$ $\newline$ $\($ $\frac{\tan \theta + \tan \alpha}{1 - \tan \alpha \tan \theta}$ = 2 $\tan$ $\theta$ $\)$ ....(1) $\newline$ $\($ $\sin$ 2 $\alpha$ = $\frac{6 \tan \theta}{9 + \tan^2 \theta}$ = $\frac{2 \tan \alpha}{1 + \tan^2 \alpha}$ $\)$ $\newline$ $\($ 3 $\tan$ $\theta$ + 3 $\tan$ $\theta$ $\tan$^2 $\alpha$ = 9 $\tan$ $\alpha$ + $\tan$ $\alpha$ $\tan$^2 $\theta$ $\)$ $\newline$ $\($ 3($\tan$ $\theta$ - 3 $\tan$ $\alpha$) = $\tan$ $\alpha$ ($\tan$ $\theta$ - 3 $\tan$ $\alpha$) $\)$ $\newline$ $\tan \theta = \frac{3}{\tan \alpha}$ or $\tan \theta = 3 \tan \alpha$. $\newline$ Case-I : $\tan$ $\theta$ = 3 $\tan$ $\alpha$ $\newline$ $\frac{\tan \theta + \frac{3}{\tan \theta}}{1 - \frac{\tan^2 \theta}{3}}$ = 2 $\tan$ $\theta$ $\newline$ $\tan$ $\theta$ = 0, $\frac{2}{3}$ = 1 - $\frac{\tan^2 \theta}{3}$ $\Rightarrow$ $\tan$ $\theta$ = 1, -1 $\newline$ $\tan$ $\theta$ = 0, -1, 1 $\Rightarrow$ $\theta$ = $\frac{\pi}{4}$, -$\frac{\pi}{4}$, 0 $\newline$ Case-II : $\tan$ $\theta$ = $\frac{3}{\tan \alpha}$ $\newline$ $\frac{\tan \theta + \frac{3}{\tan \theta}}{-2}$ = 2 $\tan$ $\theta$ $\newline$ $\tan$ $\theta$ + $\frac{3}{\tan \theta}$ = -4 $\tan$ $\theta$ $\Rightarrow$ $\frac{3}{\tan \theta}$ = -5 $\tan$ $\theta$ $\newline$ No Solution

Given the equations $4x - 3y = 12\alpha$ and $4x + 3y = \frac{12}{\alpha}$, we have: $$16x^2 - 9y^2 = 144$$ The equation of the curve is: $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$ For the tangent $T$: $y = mx \pm \sqrt{9m^2 - 16}$, where $$m = \frac{4\sqrt{2}}{3}$$ Thus, $$y = \frac{4\sqrt{2}x}{3} \pm \sqrt{32 - 16}$$ Simplifying gives: $$3y = 4\sqrt{2}x \pm 12$$ Since $q > 0$, we have: $$3y = 4\sqrt{2}x + 12$$ Therefore, $p = -\frac{3}{\sqrt{2}}$ and $q = 4$. Finally, $pq = -6\sqrt{2}$.

Given $QR = KP$. $$P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix}$$ $$\begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix} = \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$$ $$\begin{cases} xr_1 + yr_3 = 2r_1 \\ zr_1 + 4r_3 = 2r_3 \end{cases} \Rightarrow \begin{pmatrix} x - 2 \\ -y \end{pmatrix} = \begin{pmatrix} \frac{r_3}{r_1} \\ \frac{z}{-2} \end{pmatrix}$$ $$2x - 4 = yz$$ $$\begin{cases} xr_2 + yr_4 = 3r_2 \\ zr_2 + 4r_4 = 3r_4 \end{cases} \Rightarrow \begin{pmatrix} x - 3 \\ -y \end{pmatrix} = \begin{pmatrix} \frac{r_4}{r_2} \\ \frac{-z}{-z} \end{pmatrix}$$ $$x - 3 = yz$$ $$\Rightarrow 2x - 4 = x - 3 \Rightarrow x = 1 \& yz = -2$$ $$\Rightarrow Q - \lambda I = \begin{pmatrix} x - \lambda & y \\ z & 4 - \lambda \end{pmatrix}$$ $$|Q - \lambda I| = (x - \lambda)(4 - \lambda) - yz$$ $$= \lambda^2 - (x + 4)\lambda + 4x - yz$$ $$|Q - \lambda I| = \lambda^2 - 5\lambda + 6$$ Now verify the option

Given $T = S_1$. The equation is $ky - \left(\frac{x + h}{2}\right) = k^2 - h$. Rearranging gives $x - 2ky + 2k^2 - h = 0$. For $k^2 - h 0$. For the area, interchange $x$ and $y$: $y - 2kx + 2k^2 - h = 0$ and $y = x^2$. The equation becomes $x^2 - 2kx + (2k^2 - h) = 0$. Let the roots be $\alpha$ and $\beta$. Then $|\alpha - \beta| = \sqrt{4k^2 - 4(2k^2 - h)} = \sqrt{4h - 4k^2}$. The area $A = \int_{\alpha}^{\beta} ((2kx + h - 2k^2) - x^2) \, dx$. Solving gives $A = \frac{(4h - 4k^2)^{3/2}}{6} = \frac{4}{3}$. Given $(4h - 4k^2)^{3/2} = 8$, it implies $4h - 4k^2 = 4$. Therefore, $h - k^2 = 1$. The point $(4, \sqrt{3}) \in S$. The area $A = \int_{1}^{4} (\sqrt{x} - \sqrt{x-1}) \, dx = \frac{2}{3} \left( x^{3/2} - (x-1)^{3/2} \right) \bigg|_{1}^{4}$. Calculating gives $A = \frac{2}{3} (8 - 3\sqrt{3} - 1) = \frac{2}{3} (7 - 3\sqrt{3}) = \frac{14}{3} - 2\sqrt{3}$.

P is given by $P \equiv (3 \cos 30^\circ, 2 \sin 30^\circ) \equiv \left( \frac{3\sqrt{3}}{2}, 1 \right)$. Q is given by $Q \equiv (3 \cos 60^\circ, 2 \sin 60^\circ) \equiv \left( \frac{3}{2}, \sqrt{3} \right)$. R is $\left( \frac{3\sqrt{3}}{2}, \frac{3}{2} \right)$ and S is $\left( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right)$. The slope of $PQ$ is $m_{PQ} = \frac{\sqrt{3} - 1}{\frac{3}{2} - \frac{3\sqrt{3}}{2}} = -\frac{2}{3}$. The equation of line $PQ$ is $$y - \sqrt{3} = -\frac{2}{3} \left( x - \frac{3}{2} \right)$$ which simplifies to $$\Rightarrow 2x + 3y = 3(\sqrt{3} + 1)$$ option (A) is correct. Now, if $N_2 = (x_2, 0) = \left( \frac{3}{2}, 0 \right)$, $|N_2Q| = \sqrt{3}$ and $|N_2S| = \frac{3\sqrt{3}}{2}$ $\Rightarrow 3|N_2Q| = 2|N_2S|$ option (C) is correct. Now, if $N_1 = (x_1, 0) \Rightarrow N_1 = \left( \frac{3\sqrt{3}}{2}, 0 \right)$ $\Rightarrow |N_1P| = 1$, $|N_1R| = \frac{3}{2}$ option (D) is incorrect.

The limit of $f(x)$ as $x$ approaches $0^+$ is equal to the limit of $f(x)$ as $x$ approaches $0^-$, which is $3$. Since $3 > \frac{7}{3} > f(0)$, it implies $x = 0$ is a local minima. Therefore, option (B) is correct. Now, $$f(x) = \frac{6x + \sin x}{2x + \sin x} = 1 + \frac{4x}{2x + \sin x}$$ The derivative $f'(x)$ is given by: $$f'(x) = \frac{4\left[(2x + \sin x) \cdot 1 - x(2 + \cos x)\right]}{(2x + \sin x)^2} = \frac{4(\sin x - x \cos x)}{(2x + \sin x)^2}$$ This simplifies to: $$= \frac{4 \cos x (\tan x - x)}{(2x + \sin x)^2}$$

Put $y = vx$ which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Differential Equation (D.E.): $$x^2 \left( v + x \frac{dv}{dx} \right) + x^2 v = x^2 \left( 1 + v^2 \right)$$ This implies: $$v + x \frac{dv}{dx} - v = 1 + v^2$$ Thus: $$x \frac{dv}{dx} = 1 + v^2 - 2v$$ Integrating both sides: $$\int \frac{dv}{(v-1)^2} = \int \frac{dx}{x}$$ This gives: $$\frac{-1}{v-1} = \ln |x| + C$$ Rearranging: $$\frac{x}{x-y} = \ln |x| + C = \ln x + C (Since x > \frac{1}{e})$$ Given $y(1) = 0$, we find $C = 1$. So: $$\frac{x}{x-y} = \ln (ex)$$ Now $y = (e) = \frac{e}{2}$ and $y(e^2) = \frac{2e^2}{3}$. Therefore: $$\frac{2 \left( y(e) \right)^2}{y(e^2)} = \frac{2 \cdot \frac{e^2}{4}}{\frac{2e^2}{3}} = \frac{3}{4} = 0.75$$

For $x = 1$ $$\left(1 + \frac{2}{5}\right)^{23} = a_0 + a_1 + a_2 + \ldots + a_{23}$$ for numerically greatest term $$\frac{n+1}{1 + \left|\frac{a}{b}\right|} = \frac{23+1}{1 + \frac{5}{2}} = \frac{48}{7}$$ $$\Rightarrow \left\lfloor \frac{48}{7} \right\rfloor = 6 = m (where [.] greatest integer function)$$ so, $T_7$ is numerical greatest term. Hence $r = 6$

Now given probability $$P\left(Produced by M_2, defective\right) = \frac{40}{100} \times \frac{14-x}{40} \times \frac{40}{100} = \frac{2}{5}$$ $\($$\Rightarrow$ $\frac{14-x}{40}$ = $\frac{1}{5}$$\)$ $\($$\Rightarrow$ 14-x = 8$\)$ $\($$\Rightarrow$ x = 6$\)$ So, the required probability $$= \frac{6}{20} = 0.3$$

Given $[\bar{x} \; \bar{y} \; \bar{z}] = 0$. $$\begin{vmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{vmatrix} \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix} \neq 0 = 0$$ This implies: $$\Rightarrow (\alpha^3 + \beta^3 - 1) - (-\alpha \beta - \alpha \beta - \alpha \beta) = 0$$ $$\Rightarrow \alpha^3 + \beta^3 + 3 \alpha \beta = 1$$ $$\Rightarrow \alpha^3 + \beta^3 + (-1)^3 = 3(\alpha)(\beta)(-1)$$ $$\Rightarrow \alpha + \beta - 1 = 0$$ So, $\alpha + \beta - 3 = -2$.

Given $\alpha = \arg(-\omega + \omega^2 - \omega^3 + \ldots + (-\omega)^{2025})$. $$\alpha = \arg \left( \frac{-\omega ((-\omega)^{2025} - 1)}{-\omega - 1} \right)$$ $$\alpha = \arg \left( \frac{-\omega}{-\omega - 1} (-2) \right)$$ $$\alpha = \arg \left( \frac{-2\omega}{\omega + 1} \right)$$ $$\alpha = \arg \left( \frac{-2\omega}{-\omega^2} \right)$$ $$\alpha = \arg \left( \frac{2}{\omega} \right)$$ $$\alpha = \arg (2\omega^2)$$ $$\alpha = \frac{-2\pi}{3}$$ $$\frac{3\alpha}{\pi} = -2$$

Let $h(x) = f(g^{-1}(x))$ and $g(0) = 2$. $h'(x) = f'(g^{-1}(x))(g^{-1}(x))'$ $h'(2) = f'(g^{-1}(2))(g^{-1})'(2)$ $= f'(0)(g^{-1})'(2)$ Now $f(x) = \log_e(x^2 + 2x + 4)$ $$f'(x) = \frac{2x + 2}{x^2 + 2x + 4}$$ $f'(0) = \frac{1}{2}$ $g(x) = \frac{4}{1 + e^{-2x}}, \ g(0) = 2$ $g^{-1}(g(x)) = x$ $((g^{-1})'(g(x))) g'(x) = 1$ $$(g^{-1})'(2) = \frac{1}{g'(0)}$$ $= \frac{1}{2}$ So $h'(2) = \frac{1}{4} = 0.25$ $g(x) = \frac{4}{1 + e^{-2x}}$ $$g'(x) = \frac{8e^{-2x}}{(1 + e^{-2x})^2}$$ $$g'(0) = \frac{8}{4} = 2$$

Given $$\alpha = \sum_{r=30}^{59} \frac{1}{\sin(2r)^\circ \sin(2r+1)^\circ}$$ We have $$\frac{\alpha}{\csc 1^\circ} = \sum_{r=30}^{59} \frac{\sin 1^\circ}{\sin(2r)^\circ \sin(2r+1)^\circ}$$ This simplifies to $$= \sum_{r=30}^{59} (\cot(2r)^\circ - \cot(2r+1)^\circ)$$ Which equals $$= \cot 60^\circ - \cot 61^\circ + \cot 62^\circ - \cot 63^\circ$$ $$\vdots$$ $$+ \cot(118^\circ) - \cot(119^\circ)$$ Thus, $$\frac{\alpha}{\csc 1^\circ} = \cot 60^\circ = \frac{1}{\sqrt{3}}$$ Finally, $$\left( \frac{\csc 1^\circ}{\alpha} \right)^2 = 3$$

Given $$\alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} \, dx$$ ... (i) Let $$x = \frac{1}{t}$$ Then $$dx = -\frac{1}{t^2} \, dt$$ $$\alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} \left( \frac{1}{t} \right)}{\frac{2}{t^2} - \frac{3}{t} + 2} \left( -\frac{1}{t^2} \right) \, dt$$ $$\alpha = \int_{\frac{1}{2}}^{2} \frac{\cot^{-1} t}{2t^2 - 3t + 2} \, dt$$ ... (ii) Now by (i) + (ii) $$2\alpha = \int_{\frac{1}{2}}^{2} \frac{\pi}{2} \frac{1}{2x^2 - 3x + 2} \, dx$$ $$\alpha = \frac{\pi}{8} \int_{\frac{1}{2}}^{2} \frac{dx}{\left( x - \frac{3}{4} \right)^2 + \frac{7}{16}}$$ $$\alpha = \frac{\pi}{8 \times \sqrt{7}} \left[ \tan^{-1} \left( \frac{x - \frac{3}{4}}{\sqrt{\frac{7}{4}}} \right) \right]_{\frac{1}{2}}^{2}$$ $$\alpha = \frac{\pi}{2\sqrt{7}} \left[ \tan^{-1} \left( 4x - 3 \right) \sqrt{\frac{7}{4}} \right]_{\frac{1}{2}}^{2}$$ $$\alpha = \frac{\pi}{2\sqrt{7}} \left[ \tan^{-1} \frac{5}{\sqrt{7}} - \tan^{-1} \left( -\frac{1}{\sqrt{7}} \right) \right]$$ $$\alpha = \frac{\pi}{2\sqrt{7}} \tan^{-1} \left( \frac{\frac{5}{\sqrt{7}} + \frac{1}{\sqrt{7}}}{1 - \frac{5}{7}} \right)$$ $$\alpha = \frac{\pi}{2\sqrt{7}} \tan^{-1} (3\sqrt{7})$$ Now $$\sqrt{7} \tan \left( \frac{2\sqrt{7}\alpha}{\pi} \right)$$ $$\sqrt{7} \times \tan \left( \tan^{-1} (3\sqrt{7}) \right)$$ $$\sqrt{7} \times 3\sqrt{7}$$ $$= 21$$

S = emf per unit temperature difference σ = Electrical conductivity k = Thermal conductivity $[S] = [ML^2T^{-3}I^{-1}K^{-1}]$ $[\sigma] = [M^{-1}L^{-3}T^3I^2]$ $[K] = [M^1L^1T^{-3}K^{-1}]$ $$[Z] = \frac{S^2 \sigma}{K} = \frac{[M^1L^1T^{-3}K^{-2}]}{[M^1L^1T^{-3}K^{-1}]}$$ $[Z] = [K^{-1}]$

Here we are assuming that "r" is very large just for the sake of symmetry. Outside cylinder will have zero electric field inside, so the flux generated on the plate will be due to inner cylinder only in sections AB and CD, as section be will be at that place where electric field is zero. Flux through element will be $$d\phi = \mathbf{E} \cdot d\mathbf{S}$$ $$d\phi = \frac{2k\lambda}{r} \, dy \, \ell \, \cos \theta \ldots (1)$$ From figure we can say that $$\cos \theta = \frac{R}{r} \implies r = R \sec \theta$$ $$\tan \theta = \frac{y}{R} \implies y = R \tan \theta$$ $$\implies dy = R \sec^2 \theta \, d\theta$$ $$d\phi = \frac{2k\lambda}{R \sec \theta} \, R \sec^2 \theta \, \ell \, \cos \theta \, d\theta$$ $$d\phi = 2k\lambda \ell \, d\theta$$ $$\int_0^{\phi_{AB}} d\phi = 2k\lambda \ell \int_{\pi/4}^{\pi/3} d\theta$$ $$\phi_{AB} = 2k\lambda \ell \left[ \frac{\pi}{3} - \frac{\pi}{4} \right]$$ $$\phi_{AB} = 2kQ \left[ \frac{\pi}{12} \right]$$ $$\phi_{AB} = 2 \times \frac{1}{4\pi \epsilon_0 \epsilon_r} Q \left[ \frac{\pi}{12} \right]$$ $$\phi_{AB} = \frac{Q}{120 \epsilon_0}$$ $$\phi_{plate} = \phi_{AB} + \phi_{BC} + \phi_{CD} (\phi_{CD} = \phi_{AB})$$ $$\phi_{plate} = \frac{Q}{60 \epsilon_0}$$

For Fig. 1, the equation is given by: $$ \frac{ML^2}{3} \ddot{\theta} + K \frac{\ell}{2} \frac{\theta}{2} + K \ell \ddot{\ell} = 0 $$ Simplifying, we have: $$ \ddot{\theta} + \frac{15k}{4M} \theta = 0 $$ Thus, $$ \ddot{\theta} = -\left( \frac{15k}{4M} \right) \theta $$ The angular frequency is: $$ \omega_1 = \sqrt{\frac{15K}{4M}} $$ For Fig. 2, the equation is: $$ \frac{1}{3} ML^2 \ddot{\theta} + 2K \frac{L}{2} \theta \frac{L}{2} = 0 $$ Simplifying, we have: $$ \ddot{\theta} + \frac{3K}{2M} \theta = 0 $$ Thus, $$ \ddot{\theta} = -\frac{3K}{2M} \theta $$ The angular frequency is: $$ \omega_2 = \sqrt{\frac{3K}{2M}} $$ Finally, the ratio of the angular frequencies is: $$ \frac{\omega_1}{\omega_2} = \sqrt{\frac{15}{4} \times \frac{2}{3}} = \sqrt{\frac{5}{2}} $$

Given $m_2 \omega^2 r = \frac{G m_1 m_2}{r^2}$. $$\omega = \sqrt{\frac{G m_1}{r^3}}$$ $L = m_2 \omega r^2$ $$= m_2 \sqrt{\frac{G m_1}{r^3}} r^2$$ $L = m_2 \sqrt{G m_1} r = const.$ $\ln L = \ln m_2 + \ln G + \frac{1}{2} \ln m_1 + \frac{1}{2} \ln r$ $$0 = \frac{d m_2}{m_2} + \frac{1}{2} \frac{d m_1}{m_1} + \frac{1}{2} \frac{d r}{r}$$ $$\frac{d r}{r d t} = -\frac{2 d m_2}{m_2 d t} - \frac{d m_1}{m_1 d t} \approx -\frac{2 \gamma}{m_2}$$ Here $\left( \frac{d m}{d t} = \gamma \right)$

Before grounding $V_{sphere} = (V_c)_{net} = (V_c)_q + (V_c)_{ind}$ $$V_{sphere} = \frac{kq}{\ell} + 0 = \frac{9 \times 10^9 \times 10^{-8}}{0.2} = \frac{90}{0.2} = \frac{900}{2} = 450 \, volt$$ After grounding $$\frac{kQ}{\ell} + \frac{kq_s}{R} = V'_{sphere} = 0$$ $$q_s = -\frac{R}{\ell} q = -\frac{1}{2} \times 10^{-8} = -5 \times 10^{-9}$$ $q_s = -5 \times 10^{-9}$ Coulomb Charge flow from sphere to ground = $5 \times 10^{-9}$ Coulomb After grounding is removed $$(V_{sphere})_{final} = \frac{kq}{\ell'} + \frac{kq_s}{R}$$ $$= \frac{9 \times 10^9 \times 10^2 \times 10^{-8}}{30} - \frac{9 \times 10^9 \times 5 \times 10^{-9} \times 10^2}{10}$$ $$= \frac{9 \times 1000}{30} - 450 = 300 \, volt - 450 \, volt = -150 \, volt$$

Given the diagram, we start with the equation: $$\frac{S - t}{2} + \frac{t}{2n_0} = 2f$$ Solving for $S$, we have: $$S + t \left( \frac{1}{n_0} - 1 \right) = 4f$$ Therefore, $$S = 4f + \left( 1 - \frac{1}{n_0} \right) t$$ Also, we have: $$\frac{S - t}{2} + \frac{t}{2n_0} = f$$ Solving for $S$, we get: $$S = 2f + \left( 1 - \frac{1}{n_0} \right) t$$

Configuration I $$(E_4)_I = \frac{\sigma_0}{2 \epsilon_0} [1 - 1 + 1 - 1 - 1 + 1] = 0$$ $$(V_{First})_I = \frac{-\sigma_0}{2 \epsilon_0} [-1 + 2 - 3 + 4 - 5]d$$ $$= \frac{-\sigma_0}{2 \epsilon_0} [-3]d = \frac{\sigma_0 3d}{2 \epsilon_0}$$ $$(V_{Last})_I = \frac{-\sigma_0}{2 \epsilon_0} [1 - 2 + 3 - 4 + 5]$$ $$= \frac{\sigma_0}{2 \epsilon_0} [-3d]$$ $$(V_{First} - V_{Last})_I = \frac{3 \sigma_0 d}{\epsilon_0}$$ $$= \frac{3 \times 9 \times 10^{-6} \times 1 \times 10^{-6}}{9 \times 10^{-12}} = 3 volt$$ Configuration II $$(E_3)_{II} = \frac{\sigma_0}{2 \epsilon_0} \left[ \frac{1}{2} - 1 + 1 + 1 - 1 + \frac{1}{2} \right] = \frac{\sigma_0}{2 \epsilon_0}$$ $$= \frac{-\sigma_0}{2 \epsilon_0} \left[ -1 + 2 - 3 + 4 - \frac{5}{2} \right] d$$ $$= \frac{-\sigma_0}{2 \epsilon_0} [2 - 2.5]d = \frac{\sigma_0 d}{4 \epsilon_0}$$ $$(V_{Last})_{II} = \frac{-\sigma_0}{2 \epsilon_0} \left[ -1 + 2 + 3 - 4 + \frac{5}{2} \right] d$$ $$= \frac{-\sigma_0}{2 \epsilon_0} [6.5 - 6]d = \frac{-\sigma_0 d}{4 \epsilon_0}$$ $$(V_{First} - V_{Last})_{II} = \frac{\sigma_0 d}{2 \epsilon_0} \neq 0$$

Given $T_1 = 1000 \, \mathrm{K}$ and $T_3 = 300 \, \mathrm{K}$. $Q_1 = 150 \, \mathrm{J}$. The efficiency $\eta = 0.4$ and $(\mathrm{COP})_{\mathrm{HP}} = 10$. The work done $W = \eta \times Q_1 = 0.4 \times 150 = 60 \, \mathrm{J}$. The coefficient of performance is given by $$\mathrm{COP} = \frac{Dissered Effect}{Work input}.$$ Thus, $$10 = \frac{Q_3}{W}.$$ Solving for $Q_3$, we get $Q_3 = 600 \, \mathrm{J}$. Also, $$\eta = 0.4 = 1 - \frac{T_2}{T_1}.$$ Solving for $T_2$, we find $T_2 = 600 \, \mathrm{K}$. The coefficient of performance is also given by $$\mathrm{COP} = \frac{T_3}{T_3 - T_4} = 10.$$ Solving for $T_4$, we have $$\frac{300}{300 - T_4} = 10.$$ Therefore, $T_4 = 270 \, \mathrm{K}$.

Given the diagram, we have: $$dM = dI A$$ The area is given by: $$A = \pi r^2 = \pi (R \sin \theta)^2$$ The differential current is: $$dI = \frac{da}{T} = \frac{\sigma (2 \pi r)(R d\theta) \omega}{2 \pi}$$ Simplifying, we get: $$dI = \frac{\sigma 2 \pi R^2 \omega \sin \theta d\theta}{2 \pi}$$ The torque is: $$d\tau = \sigma R^2 \omega \sin \theta \, d\theta$$ Magnetic dipole moment: $$M = \int dM = \int_0^{\pi} \sigma R^2 \omega \pi R^2 \sin^3 \theta \, d\theta$$ Simplifying, we have: $$M = \sigma R^4 \omega \pi \int_0^{\pi} \sin^3 \theta \, d\theta \left( \because \int_0^{\pi} \sin^3 \theta \, d\theta = \frac{4}{3} \right)$$ Thus: $$M = \left( \frac{Q}{4 \pi R^2} \right) R^4 \omega \pi \left( \frac{4}{3} \right)$$ Magnetic dipole moment: $$M = \frac{Q R^2 \omega}{3}$$ Angular momentum: $$L = \left( \frac{2}{5} M R^2 \right) \omega$$ The ratio is: $$\frac{M}{L} = \frac{\frac{Q R^2 \omega}{3}}{\frac{2}{5} M R^2 \omega} = \frac{Q}{2M} \left( \frac{5}{3} \right)$$ Finally, we find: $$\alpha = \frac{5}{3} = 1.67$$

Given $h \nu_1 = 13.6 + 10 = 23.6 \, \mathrm{eV}$ ....(1) Energy of positronium in ground state $$-13.6 \frac{\mu}{m} \left( \frac{z}{n} \right)^2 \, \mathrm{eV}$$ $$= -13.6 \times \frac{1}{2} \, \mathrm{eV} = -6.8 \, \mathrm{eV}$$ So to make positronium $6.8 \, \mathrm{eV}$ must release $\&$ $5 \, \mathrm{eV}$ is the KE of COM. So total energy of photon released $(h \nu_2)$ will be: $$h \nu_2 = (10 - 5) + 6.8 = 11.8 \, \mathrm{eV} ...(2)$$ Therefore, difference in energy $= 23.6 - 11.8 = 11.8 \, \mathrm{eV}$

Given $nRT_w = P_w V_w = 1 \, \mathrm{J}$. $P_w = \frac{1}{64} \times 10^6 \, \mathrm{Pa}$. For WX process $$P_X V_X^\gamma = P_w V_w^\gamma$$ $$\Rightarrow P_X = P_w \left( \frac{V_w}{V_X} \right)^\gamma$$ Amount of heat absorbed in XY process $$Q = nC_P \Delta T = n \times \frac{5}{2} R \times [T_y - T_x] [For monoatomic gas C_P = \frac{5R}{2}]$$ $$Q = \frac{5}{2} [nRT_y - nRT_x]$$ $$= \frac{5}{2} [P_y V_y - P_X V_X]$$ $$= \frac{5}{2} P_X [V_y - V_X] [\because \; P_x = P_y ; Isobaric process]$$ $$= \frac{5}{2} \times P_w \times \left[ \frac{V_w}{V_X} \right]^\gamma [V_y - V_X]$$ Putting values: $$Q = 1.6 \, \mathrm{Joule}$$

Given $T \propto r^{3/2}$. $$\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$$ $$\frac{\omega_2}{\omega_1} = \left( \frac{r_1}{r_2} \right)^{3/2} = (1.21)^{3/2}$$ $$\omega_2 = \omega_1 (1.331) (i)$$ $$(\omega_2 + \omega_1) \, t_0 = 2\pi (ii)$$ $$t_0 = \frac{2\pi}{\omega_2 + \omega_1} = \frac{2\pi}{\left( 4 \over 3 + 1 \right) \omega_1} = \frac{6\pi}{7\omega_1}$$ $$t_0 = \frac{6\pi \left( T_{GSS} \right)}{2\pi \left( 7 \right)}$$ $$t_0 = \frac{3 \times 24 \, hours}{7} = \frac{24}{p} \, hours$$ $$p = \frac{7}{3} = 2.33$$

Extension in spring $x = 0.5 \, L - 0.4 \, L = 0.1 \, L$ FBD of piston $$kx + P_2 A = P_1 A$$ $$P_2 A = P_1 A - kx$$ $$P_2 = P_1 - \frac{kL}{A(10)} ....(i)$$ $P V = n_1 RT$ $P_2 V = n_2 RT$ $$\frac{P_1}{P_2} = \frac{n_1}{n_2} = \frac{3}{2}$$ $$P_1 = \frac{5}{2} P_2 ....(ii)$$ $$P_2 = \frac{3}{2} P_2 - \frac{kL}{10A}$$ $$\frac{P_2}{2} = \frac{kL}{10A}$$ $$P_2 = \frac{kL}{5A} = \frac{kL}{A} \alpha$$ $$\alpha = \frac{1}{5} = 0.2$$

Given $x + y = 12 \, \mu \mathrm{m}$. $$\frac{4}{12} = \frac{1}{x}$$ $x = 3 \, \mu \mathrm{m}$ $y = 6 \, \mu \mathrm{m}$ Therefore, $$\Delta = \frac{y d}{D} - (\mu_B - 1)x - (\mu_A - 1)y + (\mu_B - 1)y + (\mu_A - 1)x$$ $$-\frac{y d}{D} = -0.5 \times 3 - 0.7 \times 9 + 0.5 \times 9 + 0.7 \times 3$$ $$-\frac{y d}{D} = -0.5 \times 6 - 0.7 \times 6$$ $$\Rightarrow -\frac{y d}{D} = -1.2 \, \mu \mathrm{m}$$ $$\Rightarrow y = \frac{1.2 \times D}{d} = \frac{1.2 \times 2}{2 \times 10^{-3} \times 10^{-6}}$$ $= 1.2 \, \mathrm{mm}$

The net force is given by $\vec{F}_{net} = m \frac{d\vec{v}}{dt}$. The equation becomes $mg + \vec{F} = m \frac{d\vec{v}}{dt}$. Simplifying, we have $mg - C\vec{v} = m \frac{d\vec{v}}{dt}$. In the horizontal direction, $-Cv_x = m \frac{dv_x}{dt}$. Integrating, we have $$-\frac{C}{m} \int_0^t dt = \int_{v_{0x}}^{v_x} \frac{dv_x}{v_x}$$ which gives $$-\frac{t}{2} = \ln \frac{v_x}{v_{0x}}.$$ The velocity is given by $\frac{dx}{dt} = v_x = v_{0x} e^{-t/2}$. Integrating the position, $$\int_0^{S_x} dx = v_{0x} \int_0^t e^{-t/2} dt.$$ This results in $$S_x = 2v_{0x} \left(1 - e^{-t/2}\right).$$ At $t = 2$ sec, $$S_x = 2 \times 270 \times \cos 60^\circ \left[1 - \frac{1}{e}\right].$$ Simplifying, $$S_x = 270 \left[1 - \frac{1}{2.7}\right] = \frac{270}{2.7} \times (1.7) = 170 \, \mathrm{m}.$$ Therefore, $S_x = 170 \, \mathrm{m}$.

Given $\cos \theta_0 = 1 - \frac{\theta_0^2}{2} = 0.9$. $$\frac{\theta_0^2}{2} = 0.1 \Rightarrow \theta_0 = 10.2 = \frac{1}{\sqrt{5}}$$ For $f_{\max}$, $f_{\min}$: $$f_{\max} = \frac{v + v'}{v - v'} f$$ $$f_{\min} = \frac{v - v'}{v + v'} f$$ The difference $\Delta f_{\max}$ is given by: $$\Delta f_{\max} = f_{\max} - f_{\min} = \frac{v + v'}{v - v'} f - \frac{v - v'}{v + v'} f$$ Simplifying: $$= \frac{(v + v')^2 - (v - v')^2}{v^2 - v'^2} f$$ $$= \frac{4 v v'}{v^2 - v'^2} f .... (i)$$ Here, $v' = \ell \Omega_{\max}$ $$= \ell \theta_0 \cdot \omega (\omega = angular frequency)$$ $$= \ell \theta_0 \sqrt{\frac{g}{\ell}}$$ $$v' = \theta_0 \sqrt{g \ell}$$ $$v' = \frac{1}{\sqrt{5}} \sqrt{10 \times 8}$$ $$v' = 4$$ Put in equation (i): $$\Delta f_{\max} = \frac{4 \times 330 \times 4 \times 660}{330^2 - 4^2}$$ $$\approx \frac{16 \times 330 \times 660}{330} \approx 32$$

Solution: $\mathrm{Na_2[Fe(CN)_5(NO)]} + \mathrm{Na_2S} \rightarrow \mathrm{Na_4[Fe(CN)_5(NOS)]}$ Sodium nitroprusside purple solution

ICl + $\mathrm{H_2O}$ $\rightarrow$ $\mathrm{HIO}$ + $\mathrm{HCl}$ $\downarrow$ $\mathrm{IO}$^- $\mathrm{ClF_3}$ + $\mathrm{H_2O}$ $\rightarrow$ $\mathrm{HClO_2}$ + 3$\mathrm{HF}$ $\downarrow$ $\mathrm{ClO_2}$^- $\mathrm{BrF_5}$ + $\mathrm{H_2O}$ $\rightarrow$ $\mathrm{HBrO_3}$ + 5$\mathrm{HF}$ $\downarrow$ $\mathrm{BrO_3}$^-

The first reaction involves the Hell-Volhard-Zelinsky (H.V.Z.) reaction. The carboxylic acid is treated with $\mathrm{Br_2/Red\ phosphorus}$ followed by $\mathrm{H_2O}$ to yield $\mathrm{Ph-CH_2-CH(COOH)-Br}$. The second reaction is an aldol condensation. Benzaldehyde reacts with acetaldehyde in the presence of aqueous $\mathrm{NaOH}$ at $293\, \mathrm{K}$ to form $\mathrm{Ph-CH=CH-CHO}$. In the third reaction, phenylacetylene is treated with $\mathrm{NaNH_2}$ and $\mathrm{BrCH_2CH_2CH_2Br}$, followed by $\mathrm{Hg^{2+}, H_3O^+}$ to form $\mathrm{Ph-C\equiv C-CH_2-CH_2-CH=CH_2}$. The final reaction involves ozonolysis followed by Grignard reaction. The compound is treated with $\mathrm{O_3, Zn-H_2O}$, then $\mathrm{CH_3MgBr}$ (2 equivalents), and finally with $\mathrm{H^+, \Delta}$ to yield the final product.

(i) Ion – Ion → Interaction energy $\propto \frac{1}{r}$ Ion – dipole → Interaction energy $\propto \frac{1}{r^2}$ Ion – dipole Interaction energy approaches zero more rapidly as $r$ increases. (ii) Rotating Polar molecules → Interaction energy $\propto \frac{1}{r^6}$. (iii) Dipole – induced dipole forces are independent of temperature. (iv) Non-polar species show London dispersion forces.

The structures given are: (A) $\mathrm{H_3PO_4}$ (B) $\mathrm{H_3PO_3}$ (C) $\mathrm{H_4P_2O_7}$ (D) $\mathrm{H_3PO_2}$

The reaction starts with the compound containing two carboxylic acid groups. Upon treatment with $\mathrm{NH_3}$ and heat $\Delta$, two molecules of water $2\mathrm{H_2O}$ are removed, forming compound $X$ with two amide groups $\mathrm{C-NH_2}$. Next, compound $X$ undergoes a series of reactions: 1. Heating $\Delta$ 2. Treatment with alcoholic $\mathrm{KOH}$ 3. Reaction with $\mathrm{R-Br}$ These steps lead to the formation of compound $Y$, which contains an $\mathrm{R-N}$ group. Finally, compound $Y$ is treated with $\mathrm{NaOH}$, resulting in the formation of an aromatic compound with two carboxylate ions $\mathrm{COO^-}$ and $\mathrm{R-NH_2}$ as compound $Z$.

The reaction starts with the reduction of the ester using $\mathrm{LiAlH_4}$ to form a diol. The diol is then oxidized using $\mathrm{CrO_3}$ and $\mathrm{H_2SO_4}$ to form a diketone. The diketone undergoes dehydration in the presence of $\mathrm{H_2SO_4}$ at $443\, \mathrm{K}$ to form a conjugated diene. The diene undergoes an acid-catalyzed rearrangement to form the final product.

CCP: $\($(Z = 4, a = 400 $\,$ $\mathrm{pm}$, M = 105.6 $\,$ $\mathrm{g/mol}$)$\)$ Density = $\($$\frac{Z \times M}{a^3 \times N_A}$$\)$ $$= \frac{4 \times 105.6}{(400 \times 10^{-10})^3 \times 6 \times 10^{23}} = 11.00 \, \mathrm{g/cm^3}$$

The reaction is given by: $$\mathrm{Ba(NO_3)_2(aq) + 2NaIO_3(aq) \rightarrow Ba(IO_3)_2(s) + 2NaNO_3(aq)}$$ Initially, there are 2 mmol of $\mathrm{Ba(NO_3)_2}$ and 10 mmol of $\mathrm{NaIO_3}$. The limiting reagent (LR) is $\mathrm{Ba(NO_3)_2}$. After the reaction, 6 mmol of $\mathrm{NaIO_3}$ and 4 mmol of $\mathrm{Ba(NO_3)_2}$ are consumed. The concentration of $\mathrm{NaIO_3}$ is: $$[\mathrm{NaIO_3}] = \frac{6}{300} = 2 \times 10^{-2} \, \mathrm{M}$$ The equilibrium reaction is: $$\mathrm{Ba(IO_3)_2(s) \rightleftharpoons Ba^{2+} + 2IO_3^-}$$ Let $s$ be the solubility of $\mathrm{Ba(IO_3)_2}$, then: $$[\mathrm{IO_3^-}] = (2 \times 10^{-2} + 2s)$$ The solubility product $K_{sp}$ is given by: $$K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{IO_3^-}]^2$$ Substituting the values: $$1.58 \times 10^{-9} = [\mathrm{Ba^{2+}}] \times (2 \times 10^{-2})^2$$ Solving for $[\mathrm{Ba^{2+}}]$ gives: $$[\mathrm{Ba^{2+}}] = s = 3.95 \times 10^{-6} \, \mathrm{M}$$ Thus, $X = 3.95$.

Given $\($ $\frac{x}{m}$ = K $\times$ C^{1/n} $\)$. $\[$ $\log$ $\left$( $\frac{x}{m}$ $\right$) = $\log$ K + $\frac{1}{n}$ $\log$ C $\]$ $\[$ $\log$ 4 = $\log$ K + $\frac{1}{n}$ $\log$ 10 $\]$ $\[$ 0.6 = $\log$ K + $\frac{1}{n}$ ........ (1) $\]$ $\[$ $\log$ 10 = $\log$ K + $\frac{1}{n}$ $\log$ 16 $\]$ $\[$ 1 = $\log$ K + $\frac{1}{n}$ $\times$ 1.2 ........(2) $\]$ Equation (2) - equation (1) $\[$ 0.4 = $\frac{1}{n}$ $\times$ (0.2) $\Rightarrow$ n = 0.5 $\]$ and $\($ $\log$ K = -1.4 $\)$ $\[$ $\log$ $\frac{x}{m}$ = $\log$ K + $\frac{1}{n}$ $\times$ $\log$ C $\]$ $\[$ = -1.4 + 2 $\times$ $\log$ 20 $\]$ $\[$ = -1.4 + 2.6 = 1.2 $\]$ $\[$ $\frac{x}{m}$ = 10^{1.2} = 16 $\]$ $\($ $\log$ 2 = 0.3, 4 $\log$ 2 = 1.2, 16 = 10^{1.2} $\)$ OR $\($ $\frac{x}{m}$ = K $\times$ C^{1/n} $\)$ $\[$ 4 = K(10)^{1/n} .......... (1) $\]$ $\[$ 10 = K(16)^{1/n} .......... (2) $\]$ $\[$ X = K(20)^{1/n} .......... (3) $\]$ On solving equation (1) and (2) $\[$ $\frac{1}{n}$ = 2 $\]$ On solving equation (1) and (3) $\[$ $\frac{4}{X}$ = $\left$( $\frac{10}{20}$ $\right$)^2 $\]$ $\[$ X = 16 $\]$ On solving equation (2) and (3) $\[$ $\frac{10}{X}$ = $\left$( $\frac{16}{20}$ $\right$)^2 $\]$ $\[$ X = 15.625 $\]$

A + R $\rightarrow$ Product At time $t = 0$, $A_0$ and $10A_0$ are present. At time $t = t$, $0.6A_0$ and $9.6A_0$ are present. The rate is given by $Rate = k[A][R]$. Thus, $Rate_1 = k(0.6A_0) \times 9.6A_0$. A + R $\rightarrow$ Product At time $t = 0$, $A_0$ and $10A_0$ (excess) are present. At time $t = t$, $0.6A_0$ and $10A_0$ are present. The rate is given by $Rate = k'[A]$, where $k' = k[R]$. Thus, $Rate_2 = (k \times 10A_0) \times (0.6A_0)$. The percentage change in rate is given by $$100 \times \frac{\Delta Rate}{Rate_1} = \frac{(0.6 \times 10 - 0.6 \times 9.6)}{0.6 \times 9.6} \times 100 = 4.1666$$

Given $\pi = \rho gh = 10^3 \times 10 \times 2 \times 10^{-2} Pascal = 200 Pascal$. $\pi = CRT$ $$200 = \frac{2}{M} \times 1000 \times 8.3 \times 300$$ $$M = 24900 = 2.49 \times 10^4 g/mol$$ $X = 2.49$

The reaction is given by: $$\mathrm{C_4H_{10}(g) + \frac{13}{2}O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l)}$$ The change in Gibbs free energy is calculated as: $$\Delta_r G^\circ = 4 \Delta_f G^\circ_{\mathrm{CO_2}} + 5 \Delta_f G^\circ_{\mathrm{H_2O}} - \Delta_f G^\circ_{\mathrm{C_4H_{10}}}$$ Substituting the values: $$= 4 \times (-394) + 5 \times (-237) + 18$$ $$= -2743 \, \mathrm{kJ/mol}$$ Using the relation: $$\Delta_r G^\circ = -nFE^\circ$$ We have: $$-2743 \times 1000 = -26 \times F E^\circ$$ Solving for $E^\circ$: $$E^\circ = \frac{105.5}{F} \times 10^3 = 105.50$$

For $[\mathrm{MnBr}_6]^{3-}$, the spin configuration is shown as $\uparrow \uparrow \uparrow$. The magnetic moment $\mu = 4.89 \, \mathrm{B.M.}$ For $[\mathrm{Mn(CN)}_6]^{3-}$, the spin configuration is shown as $\uparrow \downarrow \uparrow \uparrow$. The magnetic moment $\mu = 2.84 \, \mathrm{B.M.}$ Sum of spin magnetic moments of both complexes is $7.70$ to $7.73 \, \mathrm{B.M.}$

Octasaccharide with molecular mass 1024 and 7 $\mathrm{H_2O}$ with molecular mass 126 react to form ribose, 2-deoxyribose, and glucose. The total mass is $1024 + 126 = 1150$. $$58.26 = \frac{134 \times n}{1150} \times 100$$ $$\frac{66.999}{100} = 134n$$ Solving for $n$, we get $n = 4.99 \approx 5$. There are 5 units of 2-deoxyribose. $$1150 = (5 \times 150) + (x \times 150) + (y \times 180)$$ $$1150 = \frac{750}{5 unit} + \frac{150x}{2 unit} + \frac{180y}{1 unit}$$ Solving for $n$, we find $n = 2.00$.