Question

An open glass tube is immersed in mercury in such a way that a length of $8\,cm$ 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 $46\,cm.$ Then, the length of the air column above mercury in the tube will be $\ldots \ldots cm$ . (Atmospheric pressure $=76\,cm$ of $Hg$ )

Answer 1

In this question, the system is accelerating horizontally, i.e. no component of acceleration in vertical direction. Hence, the pressure in the vertical direction will remain unaffected.
i.e. $p=p_{0}+\rho gh$
Again, we have to use the concept that the pressure in the same level will be same.

For air trapped in tube, $p_{1}V_{1}=p_{2}V_{2}$
$p_{1}=p_{atm}=\rho g76$
$\because v_{1}=A\cdot 8$ ( $A=$ area of cross-section)
$\Rightarrow p_{2}=p_{atm}-\rho g\left(54 - x\right)=\rho g\left(22 + x\right)$
$\because V_{2}=A\cdot x$
or $\rho g76\times 8A=\rho g\left(22 + x\right)Ax$
$x^{2}+22x-78\times 8=0\Rightarrow x=16\,cm$