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Q.
A ball whose density is $0.5 \times 10^{3} kg / m ^{3}$ falls into water from a height of $10\, cm$. To what depth (in $cm$ ) does the ball sink?
Mechanical Properties of Fluids
Solution:
Velocity of ball before entering the water surface,
$v =\sqrt{2 gh }=\sqrt{2 g \times 10} cm / s$ ... (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,
$0-v^{2}=2 \times-g \times d$
$v^{2}=2 g d$
$\therefore 2 \times g \times 10=2 g d$ ....(Using (i)]
$\therefore d=10\, cm$