Physics

Assume the length of train be I and its acceleration be a.

$v^2=u^2+2al\Rightarrow al=\frac{v^2-u^2}{2}$

Velocity when middle point crosses the post,

$T=2\pi \sqrt[]{\frac{r^3}{G{M}_e}}$

$T_B-T_A=\frac{2\pi }{\sqrt[]{G{M}_e}}\left(r_B^{3/2}-r_A^{3/2}\right)=\frac{2*3.14}{\sqrt[]{6.67*10^{-11}*6*10^{24}}}\left[{\left(8*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(7*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$\lambda =\frac{h}{p}or\lambda =\frac{h}{mv}$

${\lambda }_p={\lambda }_{\alpha }\left[Given\right]$

$\frac{h}{m_p{V}_p}=\frac{h}{m_{\alpha }{v}_{\alpha }}\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{m_{\alpha }}{m_p}$

$\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{4m}{m}=\frac{4}{1}$ or 4 : 1

Kindly consider the following figure

Amplitude is proportional to the slit – width, thus,

^{$\frac{A_1}{A_2}=3$}

^{$\frac{l_{max}}{l_{min}}=\frac{{\left(A_1+A_2\right)}^2}{{\left(A_1-A_2\right)}^2}=\frac{{\left(\frac{A_1}{A_2}+1\right)}^2}{{\left(\frac{A_1}{A_2}-1\right)}^2}=\frac{{\left(3+1\right)}^2}{{\left(3-1\right)}^2}=4.$}

If $\mu$  is Poisson's ratio,

Y = 3K (1 - 2 $\mu$ )      ……… (1)

and Y = 2 $n\left(1+\mu \right)$  ……… (2)

With the help of equations (1) and (2), we can write

  $\frac{3}{Y}=\frac{1}{\eta }+\frac{1}{3k}\Rightarrow K=\frac{\eta Y}{9\eta -3Y}$

$h{f}_1=R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=R*\frac{8}{9}$

$h{f}_2=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R*\frac{3}{4}$

$\frac{f_1}{f_2}=\frac{8/9}{3/4}=\frac{32}{27}$

f2 = f1 * $\left(\frac{27}{32}\right)$

$=\left(2.92*10^{15}\right)*\left(\frac{27}{32}\right)$

$=2.46*10^{15}Hz$

Kindly consider the following figure

For mirror

msx (16 – 10) = msy (20 – 16)

$\Rightarrow \frac{s_x}{s_y}=\frac{2}{3}$

$m{s}_y\left(26-20\right)=m{s}_z\left(30-26\right)$

$\Rightarrow \frac{s_y}{s_z}=\frac{2}{3}$