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A hollow speherical silver ball has an external diameter 4 cm and internal diameter 2 cm thick. Then the volume of the silver used in the ball is :
External radius, $$R = 2$$ cm and Internal radius, $$r = 1$$ cm
=> Volume of external sphere = $$\frac{4}{3} \pi R^3$$
= $$\frac{4}{3}\pi (2)^3 = \frac{32 \pi}{3}$$
Volume of internal sphere = $$\frac{4}{3} \pi r^3$$
= $$\frac{4}{3}\pi (1)^3 = \frac{4 \pi}{3}$$
$$\therefore$$ Volume of silver = $$\frac{32 \pi}{3} - \frac{4 \pi}{3}$$
= $$\frac{28 \pi}{3} cm^3$$
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