Question

A ball whose density is $0.5\times 10^{3}kg/m^3$ falls into water from a height of $10cm$. To what depth (in $cm$ ) does the ball sink?

Answer 1

Answer: 10

Velocity of ball before entering the water surface,
$v=\sqrt{2gh}=\sqrt{2g\times 10}cm/s$ ... (i)
When ball enters into water, due to upthrust of water the velocity of ball decreases.
$\therefore$ retardation, $a=\frac{\text{ apparent weight }}{\text{ mass of ball }}$
$=\frac{V(\sigma -\rho )g}{V\sigma }=\left(\frac{\sigma -\rho }{\sigma }\right)g$
$=\left(\frac{0.5-1}{0.5}\right)\times g$
$=-g$
Let $d$ be the depth upto which ball sinks then, using III ${}^{rd}$, kinematical equation,
$0-v^2=2\times -g\times d$
$v^2=2gd$
$\therefore 2\times g\times 10=2gd$ ....(Using (i)]
$\therefore d=10cm$