First slide

Pressure in a fluid-mechanical properties of fluids

Question

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

Moderate

A

90
B

75
C

60
D

45

Solution

Let p₀ = atmospheric pressure, then
$\begin{array}{l} p_{1} V_{1} = p_{2} V_{2} \\ or p_{2} = p_{1} \frac{V_{1}}{V_{2}} \\ or p_{2} = p_{0} (\frac{40}{60}) = \frac{2}{3} p_{0} \end{array}$
Now, $p_{2} + (20 cm of Hg) = p_{0}$
$or \frac{2}{3} p_{0} + (20 cm of Hg) = p_{0}$
$or \frac{p_{0}}{3} = 20 cm of Hg$
$p_{0} = 60 cm of Hg$

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