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A vessel is in the form of an inverted cone. Its height is $8 \operatorname{cm}$ and the radius of its top, which is open, is $5 \operatorname{cm}$. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.5 \operatorname{cm}$, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
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Sol. Radius of cone $= 5 \operatorname{cm}$, height of cone $= 8 \operatorname{cm}$
Volume of water in the cone $= \frac{1}{3} \pi \times (5)^2 \times 8$
$= \frac{200\pi}{3} \operatorname{cm}^3$
Volume of water flows out $= \frac{1}{4}$ (Volume of water in the cone)
$= \frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3} \operatorname{cm}^3$
Radius of sphere (lead shot) $= 0.5 = \frac{1}{2} \operatorname{cm}$
Volume of one lead shot $= \frac{4}{3} \pi \times (\frac{1}{2})^3$
$= \frac{\pi}{6} \operatorname{cm}^3$
Number of lead shots $= \frac{\frac{50\pi}{3}}{\frac{\pi}{6}} = 100$
Volume of water in the cone $= \frac{1}{3} \pi \times (5)^2 \times 8$
$= \frac{200\pi}{3} \operatorname{cm}^3$
Volume of water flows out $= \frac{1}{4}$ (Volume of water in the cone)
$= \frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3} \operatorname{cm}^3$
Radius of sphere (lead shot) $= 0.5 = \frac{1}{2} \operatorname{cm}$
Volume of one lead shot $= \frac{4}{3} \pi \times (\frac{1}{2})^3$
$= \frac{\pi}{6} \operatorname{cm}^3$
Number of lead shots $= \frac{\frac{50\pi}{3}}{\frac{\pi}{6}} = 100$