Physics

According to ideal gas law we know that

PV = nRT

According to question, we can write

Assuming Tension in metal wire suspended from roof varies linearly and 0 and T0 (developed due its own weight) are tensions are ends of wire of length $\mathcal{l}.$ $$ So

$T_1\alpha \left({\mathcal{l}}_1-\mathcal{l}\right),and{T}_2\alpha \left({\mathcal{l}}_2-\mathcal{l}\right)$

$\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}\Rightarrow \mathcal{l}=\frac{T_2{\mathcal{l}}_1-T_1{\mathcal{l}}_2}{T_2-T_1}$

As we know that direction of propagation $\left(\widehat{p}\right)$ $\stackrel{}{}$ of wave is given as

$\widehat{p}=\widehat{E}*\widehat{B}=\widehat{i}*\widehat{k}=\left(-\widehat{j}\right)$

$K_T=\frac{m{v}^2}{2}and{K}_R=\frac{l{\omega }^2}{2}=\frac{l{v}^2}{2R^2}$

According to question, we can write

$K_R=\frac{1}{2}{K}_T\Rightarrow \frac{l{v}^2}{2R^2}=\frac{1}{2}*\frac{m{v}^2}{2}\Rightarrow l=\frac{m{R}^2}{2}\Rightarrow Solidcylinder$

$R=-\frac{dN}{dt}=\lambda N=\lambda N_0{e}^{-\lambda t}\Rightarrow \mathcal{l}n\left(R\right)=\mathcal{l}n\left(\lambda N_0\right)-\lambda t,so$

$\lambda =tan\theta =\frac{6}{40}=\frac{3}{20}{s}^{-1}\Rightarrow$ Slope of graph

$\Rightarrow t_{\frac{1}{2}}=\frac{0.693}{\lambda }=\frac{0.693*20}{3}=4.62s$

$\gamma =1+\frac{2}{f}\Rightarrow \frac{2}{f}=\gamma -1\Rightarrow f=\frac{2}{\gamma -1}$

Let the distance travelled is x, so

$$ $\left|{\overrightarrow{v}}_{Eg}\right|=\frac{x}{t_2}\Rightarrow$ speed of escalator for ground

${\stackrel{}{}}_{}$ $\left|{\overrightarrow{v}}_{BE}\right|=\frac{x}{t_1}\Rightarrow$ speed of boy concerning the escalator

${\stackrel{}{}}_{}$ $\left|{\overrightarrow{v}}_{Bg}\right|=\frac{x}{t_1}+\frac{x}{t_2}\Rightarrow$ speed of boy concerning ground

The time taken by him to walk up the moving escalator = t =  $\frac{x}{\left|{\overrightarrow{v}}_{Bg}\right|}$ $\frac{}{{\stackrel{}{}}_{}}$

$\Rightarrow t=\frac{x}{\frac{x}{t_1}+\frac{x}{t_2}}=\frac{t_1{t}_2}{t_1+t_2}$

$\left|\overrightarrow{P}\right|=\left|\overrightarrow{Q}\right|=x$

let the angle between $\overrightarrow{P}and\overrightarrow{Q}$ $\stackrel{}{}\stackrel{}{}$ is , so according to question, we can write $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\text{exception 25:}\text{exception 25:}$

^{$\left|\overrightarrow{P}+\overrightarrow{Q}\right|=n*\left|\overrightarrow{P}-\overrightarrow{Q}\right|\Rightarrow \sqrt[]{P^2+Q^2+2PQcos\theta }=n*\sqrt[]{P^2+Q^2-2PQcos\theta }$}

$\Rightarrow \sqrt[]{x^2+x^2+2x^{2}cos\theta }=n*\sqrt[]{x^2+x^2-2x^{2}cos\theta }$

$\Rightarrow \theta =co{s}^{-1}\left(\frac{n^2-1}{n^2+1}\right)$

Total surface energy before coalesce = $U_i=2*4\pi R^2*\sigma =8\pi R^2\sigma ..........\left(i\right)$ ${}_{}{}^{}{}^{}$

Let new radius becomes r, so according to conservation energy we can write

$2*\frac{4\pi R^3}{3}=\frac{4\pi r^3}{3}\Rightarrow r=2^{\frac{1}{3}}R$

Total surface energy after coalesce = $U_f=4\pi r^2*\sigma =2^{\frac{2}{3}}*4\pi R^2\sigma ..........\left(ii\right)$ ${}_{}{}^{}{}^{\frac{}{}}{}^{}$

$\Rightarrow \frac{U_i}{U_f}=\frac{8\pi R^2\sigma }{2^{\frac{2}{3}}*4\pi R^2\sigma }=2^{\frac{1}{3}}:1$