Q. A glass tube 80 cm long and open at both ends is half immersed in mercury. Then the top of the tube is closed and it is taken out of the mercury. A column of mercury 20 cm long then remains in the tube. The atmospheric pressure (in cm of Hg) is
A) 90
B) 75
C) 60
D) 45
Explanation
(c) Let $p_{0} =$ atmospheric pressure. Then
$p_{1}V_{1} = p_{2}V_{2}$ or $p_{2} = p_{1}\frac{V_{1}}{V_{2}}$
or $p_{2} = p_{0}\ \left( \frac{40}{60} \right) = \frac{2}{3}\ p_{0}$
$p_{2} + (20\ cm\ of\ Hg) = p_{0}$
or $\frac{2}{3}p_{0} + (20\ cm\ of\ Hg) = p_{0}$
or $\frac{p_{0}}{3} = 20\ cm\ of\ Hg$
$p_{0} = 60\ cm\ of\ Hg$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.