Class 11th

In the frame of vehicle, vehicle is in equilibrium under the influence of pseudo force F_P

  $F_P=\frac{m{v}^2}{R}$            

N = mg cos 30° + F_P sin 30°

$\Rightarrow N=mgcos30°+\frac{m{v}^2}{R}sin30°.\left(i\right)$              

, and

$f_S=\frac{m{v}^2}{R}cos30°-mgsin30°$              

By doing (1) * cos 30° - (2) * sin 30°, we have

$\Rightarrow N=\frac{800*10}{0.87-0.2*0.5}=\frac{800*10}{0.77}=10.2*10^{3}kgm/s^2$

As we know that root mean square speed is given as

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

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

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

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

Since process is isochoric, so

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

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

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

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

2ⁿ = 4096 -> n = 12

¹²C₆ = 924 (the greatest coefficient)

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

$\Rightarrow \frac{6}{3}=\frac{2}{-t}\Rightarrow t=-1$

 ->R (-1,0)

    $P{R}^2+R{Q}^2=20+5=25$          

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

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