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

Options

A. 9 cm
B. 6 cm
C. 4.5 cm
D. 2.25 cm

Answer

B

Explanation

The velocity of ball before entering the water surface.
`v=sqrt(2gh)=sqrt(2g xx 9)`
The retardation in ball due to upthrust of water is given by
`a=("Apparent weight")/("Mass of ball")`
`a=(V(p-sigma)g)/(V_(p))=((p-sigma)g)/(p)`
`=((0.4-1)/(0.4))g=-(3)/(2)g`
Let h be the depth upto which ball sink, then
`0-v^(2)=2xx((-3)/(2)g)xx h`
`rArr 2g xx 9=3gh`
`therefore h=6 cm`