An open glass tube is immersed in mercury in such a way that a length of 8cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46cm. What will be the length of the air column above mercury in the tube now? (in cm) (atmospheric pressure =76cm of Hg )

Let the length of air column be 'x'

Since the amount of air trapped in the top of glass tube is same, the final pressure inside of the glass tube finally will be  $\mathrm{P}=\frac{8.{\mathrm{P}}_{\mathrm{o}}}{\mathrm{x}}$, here P_o is atmospheric pressure.

 And considering any two points on the surface of the mercury. The pressures will be equal.

$\frac{8{\mathrm{P}}_{\mathrm{o}}}{\mathrm{x}}+\mathrm{\rho g}\left(54-\mathrm{x}\right)={\mathrm{P}}_{\mathrm{o}}$...(1)

Since ${\mathrm{P}}_{\mathrm{o}}=\mathrm{\rho g}\left(76\right)$

On substituting in (1)

$\frac{8\mathrm{\rho g}\left(76\right)}{\mathrm{x}}+\mathrm{\rho g}\left(54-\mathrm{x}\right)=\mathrm{\rho g}\left(76\right)$

on solving this we get x=16cm