Physics

As we know that root mean square speed is given as

$v=\sqrt[]{\frac{3RT}{M}},so$

$v_A>v_B>v_C\Rightarrow \frac{1}{v_A}<\frac{1}{v_B}<\frac{1}{v_C}as{m}_A<m_B<m_C$

$\overrightarrow{A}.\overrightarrow{B}=\left|\overrightarrow{A}*\overrightarrow{B}\right|\Rightarrow ABcos\theta =ABsin\theta \Rightarrow \theta =45°$             

$\left|\overrightarrow{A}-\overrightarrow{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }=\sqrt[]{A^2+B^2-2AB*\frac{1}{\sqrt[]{2}}}=\sqrt[]{A^2+B^2-\sqrt[]{2}AB}$                

As we know that radius of circular path in magnetic field is given as $r=\frac{mv}{qB}=\frac{\sqrt[]{2mK}}{qB},so$  

  $r=\frac{mv}{qB}=\frac{\sqrt[]{2mK}}{qB},so$

$\frac{r_d}{r_{\alpha }}=\sqrt[]{\frac{m_d}{m_{\alpha }}}*\frac{q_{\alpha }}{q_d}=\frac{1}{\sqrt[]{2}}*2=\sqrt[]{2}$               

Since process is isochoric, so

$Q=n{C}_v\Delta T=n\left(\frac{f}{2}R\right)\Delta T=4\left(\frac{5}{2}*R\right)\left(50\right)=500R$

Given ${}_{}{}^{}$ $\chi =499and{\mu }_0=4\pi *10^{-7}H/m$

As we know that

$$ ${\mu }_r=1+\chi =1+499=500$  Relative permeability

Absolute permeability = $\mu ={\mu }_r{\mu }_0=500*4\pi *10^{-7}=2\pi *10^{-4}H/m$

Let the true dip is $\delta$ $$ and apparent dip is $\delta \text{'}$ $$ , so

$tan\delta \text{'}=\frac{B_V}{B_{H}cos\theta }=\frac{tan\delta }{cos\theta }\Rightarrow tan\delta =tan\delta \text{'}cos\theta =tan45°cos30°=1*\frac{\sqrt[]{3}}{2}$

$\Rightarrow \delta =ta{n}^{-1}\left(\frac{\sqrt[]{3}}{2}\right)$

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

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