Q 11-09-151JEE MainJEE Main 2018 (15 Apr, Shift 2)Medium
When an air bubble of radius $r$ rises from the bottom to the surface of a lake, its radius becomes $\dfrac{5r}{4}$. Taking the atmospheric pressure to be equal to $10\ \text{m}$ height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):
Answer: (D) $9.5\ \text{m}$
At constant temperature $pV$ is constant. Measuring pressure in metres of water:
$$(10 + h)\cdot\frac43\pi r^3 = 10\cdot\frac43\pi\left(\frac{5r}{4}\right)^3$$
$$10 + h = 10\times\frac{125}{64} = 19.53$$
$$h \approx 9.5\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics