The volume of an air bubble becomes three times as it rises from the bottom of a take to its surface. Assuming atmospheric pressure to be $75 c m$ of $H g$ and the density of water to be $\frac{1}{10}$ of the density of mercury, the depth of the take is :

5 m

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10 m

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15 m

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20 m

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Solution

The correct option is C $15 m$
Since, the temperature of the surroundings remains constant, we can safely assume that the process is isothermal.
Therefore, applying Boyle's Law
$P_{1} V_{1} = P_{2} V_{2}$

Let 1 denote the surface of water and 2 denote the depth

$P_{1} = 75 c m H g$

$P_{2} = 75 + h / 10$ where h is the depth of the lake

$V_{1} = 3 V_{2}$

Substituting the values,

$75 \times 3 = 75 + \frac{h}{10}$

Solving, h = 1500 cm or 15 m

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