Find the amount of water displaced by a solid spherical ball of diameter(i) $28 \mathrm{~cm}$(ii) $0.21 \mathrm{~m}$.

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Given:

Diameter of a solid spherical ball is

(i) $28 \mathrm{~cm}$
(ii) $0.21 \mathrm{~m}$.

To do:

We have to find the amount of water displaced by the solid spherical ball in each case.

Solution:

(i) Diameter of the  solid spherical ball $= 28\ cm$

Radius of the solid spherical ball $r = \frac{28}{2}\ cm$

$= 14\ cm$

Water displaced by the solid spherical ball $=$ Volume of the solid spherical ball

$=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times(14)^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14$

$=11498.66 \mathrm{~cm}^{3}$

(ii) Diameter of the  solid spherical ball $= 0.21\ m$

Radius of the solid spherical ball $r = \frac{0.21}{2}\ m$

$= 0.105\ m$

Water displaced by the solid spherical ball $=$ Volume of the solid spherical ball

$=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times 0.105 \times 0.105 \times 0.105$

$=0.004851 \mathrm{~m}^{3}$

Updated on 10-Oct-2022 13:46:39