Question

The volume of an air bubble become three times as it rises from the bottom of a lake to its surface . Assuming atmospheric pressure to be 75 cm of Hg an density of water to be 1/10 of the density of mercury , the depth of the lake :-

Answer 1

Answer on Question #74100, Physics / Mechanics | Relativity

The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be $75~\mathrm{cm}$ of $\mathrm{Hg}$ and density of water to be 1/10 of the density of mercury, the depth of the lake.

Solution:

Pressure at the bottom of lake,

$$P _ {1} = \text{Atmospheric pressure} + \text{pressure due to water column in the lake}$$ $$P _ {1} = 75 \times \rho_ {Hg} \times g + h \times \rho_ {w} \times g$$

Volume of bubble at the bottom, $V_{1} = V$

Volume of bubble at the surface, $V_{2} = 2V$

Pressure at the surface

$$P _ {2} = \text{Atmospheric pressure}$$ $$P _ {2} = 75 \times \rho_ {Hg} \times g$$

From Boyle's law,

$$P _ {1} V _ {1} = P _ {2} V _ {2}$$ $$(75 \times \rho_ {Hg} \times g + h \times \rho_ {w} \times g) \times V = 75 \times \rho_ {Hg} \times g \times 2V$$ $$h = 75 \times \left(\rho_ {Hg} / \rho_ {w}\right)$$ $$h = 75~\mathrm{cm} \times \left(\frac{13.56~g/\mathrm{cm}^3}{13.56~g/\mathrm{cm}^3 \times 1/10}\right)$$ $$h = 75~\mathrm{cm} \times 10 = 750~\mathrm{cm} = 7.5~m$$

Answer: 7.5 m

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