Question

A rubber ball of mass $10g$ and volume $15c{m}^3$ is dipped in water to a depth of $10m$. Assuming density of water to be uniform throughout the depth, the time taken by it to reach the surface in seconds is (take $g=10m{s}^{-2})$

Answer 1

Density of ball ${\rho }_b=\frac{10}{15}=\frac{2}{3}gc{m}^{-3}$
At a depth $10m$
Net Force on the ball is $ma=F_B-mg={\rho }_{w}Vg-{\rho }_{b}Vg$
acceleration of ball is,
$a=g[\frac{{\rho }_w}{{\rho }_b}-1]=10[\frac{3}{2}-1]=5m{s}^{-2}$
Time taken to reach the surface is,
$t=\sqrt{\frac{2h}{a}}=\sqrt{\frac{2\times 10}{5}}=\sqrt{\frac{20}{5}}=2s$