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Find the amount of water displaced by a solid spherical ball of diameter
(i) $ 28 \mathrm{~cm} $
(ii) $ 0.21 \mathrm{~m} $.
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
$=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \times \frac{22}{7} \times 0.105 \times 0.105 \times 0.105$
$=0.004851 \mathrm{~m}^{3}$