Physics

Since binary mass system performs circular motion about is common centre of mass, so

$m_A{\omega }_A^2{r}_A=\frac{G{m}_B{m}_A}{{\left(r_A+r_B\right)}^2}=\frac{G{m}_B{m}_A}{r^2}$

$m_A{\omega }_A^2*\frac{m_B}{\left(m_A+m_B\right)}r=\frac{G{m}_B{m}_A}{r^2}$

${\omega }_A=\sqrt[]{\frac{G\left(m_A+m_B\right)}{r^3}}$

Similarly we can show that

${\omega }_B=\sqrt[]{\frac{G\left(m_A+m_B\right)}{r^3}}$

Hence their angular velocity will be same, time period will be same, i.e. TA = TB

Time period of Satellite = T = $\frac{2\pi R}{v}={\left(\frac{4{\pi }^2{R}^3}{GM}\right)}^{\frac{1}{2}}$  

$\Rightarrow T\alpha R^{\frac{3}{2}}\Rightarrow \frac{T_2}{T_1}=\frac{{\left(1.02R\right)}^{\frac{3}{2}}}{R^{\frac{3}{2}}}={\left(1.02\right)}^{\frac{3}{2}}={\left(1+0.02\right)}^{\frac{3}{2}}$     

The percentage difference in the time periods of the two satellites = $\frac{\Delta T}{T_1}$ * 100

= (0.03) * 100 = 3%

As we know that ${\upsilon }^2={\omega }^2\left(A^2-x^2\right)$ ${}^{}{}^{}{}^{}{}^{}$ for SHM, so

${\upsilon }_1^2={\omega }^2\left(A^2-x_1^2\right).......\left(i\right),and{\upsilon }_2^2={\omega }^2\left(A^2-x_2^2\right).........\left(ii\right)$

Subtracting equation (ii) from equation (i), we have

${\upsilon }_1^2-{\upsilon }_2^2={\omega }^2\left(x_2^2-x_1^2\right)\Rightarrow \omega =\frac{2\pi }{T}=\sqrt[]{\frac{{\upsilon }_1^2-{\upsilon }_2^2}{x_2^2-x_1^2}}\Rightarrow T=2\pi \sqrt[]{\frac{x_2^2-x_1^2}{{\upsilon }_1^2-{\upsilon }_2^2}}$

According to Doppler's effect in light, we can write

$\frac{v}{c}=\frac{\Delta \lambda }{\lambda }\Rightarrow \frac{v}{c}=\frac{5896-5890}{5890}=\frac{6}{5890}=\frac{3}{2945}\Rightarrow v=\frac{3}{2945}*3*10^5\approx 305.6km/s$

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

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