Q. When an air bubble of radius $'r'$ rises from the bottom to the surface of a lake, its radius becomes $5r/4$ (the pressure of the atmosphere is equal to the $10\ m$ height of water column). If the temperature is constant and the surface tension is neglected, the depth of the lake is
A) $3.53\ m$
B) $6.53\ m$
C) $9.53\ m$
D) $12.53\ m$
Explanation
(c) According to Boyle’s law
$\left( P_{1}V_{1} \right)_{At\ top\ of\ the\ lake} = \left( P_{2}V_{2} \right)_{At\ the\ bottom\ of\ the\ lake}$
$\Rightarrow P_{1}V_{1} = \left( P_{1} + h \right)V_{2} \Rightarrow 10 \times \frac{4}{3}\pi\left( \frac{5r}{4} \right)^{3}$
$\Rightarrow (10 + h) \times \frac{4}{3}\pi r^{3} \Rightarrow h = \frac{610}{64} = 9.53m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.