A ball whose density is $0.4 \times 10^3 \, kg/m^3$ falls into water from a height of $9\,cm$. To what depth does the ball sink?

Updated On: Jun 18, 2024

9 cm
6 cm
4.5 cm
2.25 cm

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The Correct Option is B

Solution and Explanation

The velocity of ball before entering the water surface
$v =\overline{2 gh }=\overline{2 g } X$
When ball enters into water, due to upthrust of water,
the velocity of ball decreases (or retarded)
The retardationm
$a =\frac{\text { apparent weight }}{\text { mass of ball }} $
$=\frac{ v (\rho-\sigma) g }{ V \rho}=\frac{(\rho-\sigma}{=} \frac{(\rho-\sigma) g }{\rho}$
$=\left(\frac{(0.4-1)}{0.4}\right) g =-\overline{2} g$
If $h$ be the depth upto which ball $\sin x$, then
$0- v ^{2}=2 \times h \left(\frac{-3}{2} g \right) X$
$\Rightarrow 2 g \times 9=3 \,gh $
$\therefore h =6\, cm$

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