Question

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 ends containing air at the same pressure P${}_o$. When the tube is held at an angle 60${}^o$ with the vertical, the lengths of the air column above and below the mercury are 46 and 44.5 cm respectively. The pressure P${}_o$ in cm of Hg is: (The temperature of the system is left at 30 K.)

• A. $80cmofHg$
• B. $75.4cmofHg$
• C. $70cmofHg$
• D. $90cmofHg$

Answer 1

The correct option is B $75.4cmofHg$

Since the total length of the tube remains same before and after tilting,

$2l+5=46+5+44.5$

When the tube is tilted,

$(P_{bottom}=P_{top}+5\cos {60}^{\circ })cmofHg$

Also PV=constant holds for each air column.

Therefore, $P_{bottom}=\frac{l}{l_{bottom}}{P}_0$

And $P_{top}=\frac{l}{l_{top}}{P}_0$

Solving above four equations gives $P_0=75.4cmofHg$