First slide

Pressure in a fluid-mechanical properties of fluids

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)

Moderate

A

350 cm³
B

300 cm³
C

250 cm³
D

22 cm³

Solution

According to Boyle's law, pressure and volume are inversely proportional to each other i.e. $P \propto \frac{1}{V}$
$\begin{array}{l} \Rightarrow P_{1} V_{1} = P_{2} V_{2} \\ \Rightarrow (P_{0} + h ρ_{w} g) V_{1} = P_{0} V_{2} \\ \Rightarrow V_{2} = (1 + \frac{h ρ_{w} g}{P_{0}}) V_{1} \\ \Rightarrow V_{2} = (1 + \frac{47.6 \times 10^{2} \times 1 \times 1000}{70 \times 13.6 \times 1000}) V_{1} \\ \Rightarrow V_{2} = (1 + 5) 50 {cm}^{3} = 300 {cm}^{3} . \\ [As P_{2} = P_{0} = 70 cm of Hg = 70 \times 13.6 \times 1000] \end{array}$

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