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

• A. 38 cm
• B. 6 cm
• C. 16 cm
• D. 22 cm

Answer 1

Answer: (C)

16 cm

Using isothermal process conditions

We have $P_1{V}_1=P_2{V}_2$

Thus

$8A{P}_0=PAx$(A is the cross-sectional area of the tube and P is the pressure in the tube when it is raised)

Thus $P=\frac{8P_o}{x}$

Thus equating the pressure at points 1 and 2 we get

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

or

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

or

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

Thus $x=16$ cm