Consider a ball of density `0.5 xx 10^3 kg//m^3` falling freely into the water from a height of 10 cm. Now, find the depth (in centimetres) to which the ball will sink:

Answer

A

Explanation

Velocity of ball before entering the water surface,
…………. (i), `v = sqrt(2gh) = sqrt(2g xx 10) cm//s`
When ball enters into water, due to upthrust of water the velocity of ball decreases.
`therefore` retardation, `a = ("apparent weight")/("mass of ball")`
` = (V(sigma - rho)g)/(V sigma) = (sigma - rho)/sigma.g`
`= (0.5-1)/0.5 xx g`
`=-g`
Let d be the depth upto which ball sinks then, using 3rd kinematical equation,
`0 -v^(2) = 2 xx -g xx d`
`v^(2) = 2gd`
`therefore 2 xx g xx 10 = 2gd`.........[using (i)]
`therefore d =10 cm`