Question

a spherical ball of density p=0.70kg/l and has a radius r=10cm.if the ball is placed on the surface of a water and released .how much the ball become submerged?​

Answer 1

$Given:-$

• Density of a spherical ball p = 0.70kg/l
• Radius = 10cm
• Ball placed on the surface of a water released.

$To\:Find:-$

• Volume of water displaced

$Solution:-$

Volume of sphere :

$\implies\:\: V = \dfrac{4}{3} \pi r^{3}$

$\implies\:\:V = \dfrac{4}{3}\pi (10cm)^{3} = \dfrac{4000\pi}{3}cm^{3}$

$\implies\:\:V = \dfrac{4000\pi}{3}cm^{3}\bigg(\dfrac{1L}{1000cm^{3}}\bigg) = \dfrac{4\pi}{3}L$

Now, multiplying this value by density

$m = \bigg( \dfrac{4\pi}{3}L\bigg) \bigg( 0.70\dfrac{kg}{L}\bigg) = 2.93 kg$

Calculating the volume of water displaced :

$\implies \:\:V = \dfrac{m}{\rho}$

$\implies \:\:V = \dfrac{ 0.9\overline{3}kg}{\bigg(1.0\frac{kg}{L}\bigg)}$

$\implies \:\:V = 2.93L$