Q 11-12-023NEETJEE MainMedium
An air bubble released at the bottom of a lake has three times its original volume when it reaches the surface. Assuming the temperature is the same throughout, the depth of the lake is about (atmospheric pressure $= 10^5$ Pa, $g = 10$ m s$^{-2}$)
Answer: (A) $20$ m
Boyle's law: $p_{\text{bottom}}V = p_0(3V)$, so $p_{\text{bottom}} = 3 \times 10^5$ Pa.
$\rho gh = 2 \times 10^5$ Pa, so $h = \dfrac{2 \times 10^5}{1000 \times 10} = 20$ m.
Solution by Sreeraj P, M.Sc Physics