Question

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 $46cm.$ Then, the length of the air column above mercury in the tube will be $\dots \dots cm$ . (Atmospheric pressure $=76cm$ of $Hg$ )

Answer 1

Answer: 16

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 $\rho g76\times 8A=\rho g\left(22+x\right)Ax$
$x^2+22x-78\times 8=0\Rightarrow x=16cm$