Q. Radius of an air bubble at the bottom of the lake is $r$ and it becomes $2r$when the air bubble rises to the top surface of the lake. If $\rho$ cm of water be the atmospheric pressure, then the depth of lake is
A) $2p$
B) $8p$
C) $4p$
D) $7p$
Explanation
(d) Let depth of lake is $x$ cm.
$\therefore\ p_{1}V_{1} = p_{2}V_{2}$
$(pdg + xdg)\left( \frac{4}{3}\pi r^{3} \right) = pdg\left\lbrack \frac{4}{3}\pi(2r)^{3} \right\rbrack$
$(p + x)r^{3} = p\left( 8r^{3} \right)$
$x = 8p - p$
$x = 7p$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.