First slide

Fluid statics

Question

An inverted (bell) lying at the bottom of a lake 47.6 m deep has 50 cm³ of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be atmospheric pressure = 70 cm of Hg and density of Hg= 13.6 g/cm³):

Easy

A

350 cm³
B

300 cm³
C

250 cm³
D

22 cm³

Solution

$P_{1} V_{1} = P_{2} V_{2} (P_{o} + {hρ}_{w} g) V_{1} = P_{o} V_{2} V_{2} = (1 + \frac{{hρ}_{w} g}{P_{o}}) V_{o} = (1 + \frac{47.6 \times 10^{2} \times 1 \times 1000}{70 \times 13.6 \times 1000}) 50 = 300 {cm}^{3}$

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