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An air bubble of volume $15 {cm}^{3}$ is formed at a depth of 50 m in a lake. If the temperature of the bubble, while rising remains constant, then the volume of the bubble at the surface will be (consider $g = 10 {ms}^{- 2}$ and atmospheric pressure = 10 metre of water)

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$80 {cm}^{3}$

$40 {cm}^{3}$

$100 {cm}^{3}$

$90 {cm}^{3}$

answer is B.

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Detailed Solution

$P_{1} V_{1} = P_{2} V_{2}$

$(10 + 50) (15) = 10$ $V_{2}$

$V_{2}$ = $90 {cm}^{3}$

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