Question 148 of 192

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 length of the air column above mercury in the tube now ?
(Atmospheric pressure = $76cm$ of Hg)

A$16cm$B$22cm$C$38cm$D$6cm$

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Answer: (A) $16cm$

Using isothermal process conditions



$\Rightarrow {P}_{1}{V}_{1}={P}_{2}{V}_{2}\Rightarrow {P}_{1}A{\ell }_{1}={P}_{2}A{\ell }_{2}$

${8\text{p}}_{{ }_{0}}=\text{xp}\Rightarrow \text{p}=\dfrac{8{\text{p}}_{0}}{\text{x}}$ (here p0 = 76 cm of Hg)

$\dfrac{8{\text{p}}_{0}}{\text{x}}+{\text{ρ}}_{\text{m}}\text{g}\left(\text{54}-\text{x}\right)={\text{ρ}}_{\text{m}}\text{g}\text{(76)}$

$\dfrac{8}{\text{x}}+\dfrac{\text{54}-\text{x}}{\text{76}}=1$

x = 16 cm