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. 4.5 cm
B. 6.0 cm
C. 7.5 cm
D. 9.0 cm

Answer

B

Explanation

Velocity of body before entering the surface of water `=sqrt(2 gh) =sqrt(2xx980xx0) cm//s`
This velocity of body is retarded due to upthrust of water experienced by body. The retardation is
`a=((V rho g - V sigma g)/(V rho)) =((rho - sigma)/(rho))g`
where `rho` = density of body and `sigma` = density of water
`:. a=(0.4-1)/(0.4)g =(-3)/(2)g`
If the body penetrates through distance h in water then using `u=sqrt(2xx980xx9) cm//s` ,
`a= -(3)/(2)xx980 cm//s`,
`v=0, S=h=?`
As, `v^(2)=u^(2) + 2 aS`
`:. 0=2xx980xx9+2 (-(3)/(2)xx980)h`
On solving, `h=6 cm`