A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal end containing air at the same pressure P. When the tube is held at an angle of `60^@` with the vertical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure P in centimeters of mercury. (The temperature of the system is kept at `30^@C`).

Options

A. `75.4`
B. `45.8`
C. `67.5`
D. `89.3`

Answer

A

Explanation

Let a be the area of cross-section of the tube. When the tube is horizontal, the 5 cm column of Hg is in the middle, so length of air column on either side at pressure `P=(46+44.5)/(2)=45.25 cm`
When the rube is held at `60^(@)` with the vertical, the lengths of air columns at the bottom and the top are `44.5` cm and 46 cm respectively. If `P_(1) and P_(2)` are their pressures. then
`P_(1)-P_(2)5 cos60^(@)=5xx1/2=5/2` cm of Hg
Using Boyle's law for constant temperature,
`PV =P_(1)V_(1)=P_(2)V_(2)`
`PxxAxx45.25=P_(1)xxAxx44.5=P_(2)xxA46`
`therefore(Pxx45.25)/(44.5)-(Pxx45.25)/(46)=5/2`
`or P(5xx44.5xx46)/(2xx45.25xx1.5)=75.4` cm of Hg