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 vetical 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).

A thin tube of uniform cross-section is sealed at both ends. It lies h

1.	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 vetical 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).
Answer» <P>`75.4` `45.8` `67.5` `89.3` Solution :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