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Radius of an air bubble at the bottom of the lake is r and· it becomes 2 r when the air bubble rises to the top surface of the lake. If P cm of water be the atmospheric pressure, then the depth of the lake is:

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8 P

4 P

7 P

2 P

answer is D.

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Detailed Solution

(PV)_bottom=(PV)_surface
$(P_{0} + {hρ}_{w} g) \frac{4}{3} {πr}^{3} = P_{0} \times \frac{4}{3} π {(2 r)}^{3} {hρ}_{w} g = 7 P_{0} = 7 [P \times ρ_{w} g] \Rightarrow h = 7 P$

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