Q. When a large bubble rises from the bottom of a lake to the surface. Its radius doubles. If atmospheric pressure is equal to that of column of water height $H$, then the depth of lake is
A) $H$
B) $2H$
C) $7H$
D) $8H$
Explanation
(c) $P_{1}V_{1} = P_{2}V_{2} \Rightarrow \left( P_{0} + h\rho g \right) \times \frac{4}{3}\pi r^{3} = P_{0} \times \frac{4}{3}\pi(2r)^{3}$
Where, $h =$ depth of lake
$\Rightarrow h\rho g = 7P_{0} \Rightarrow h = 7 \times \frac{H\rho g}{\rho g} = 7H$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.