# A vessel in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one- fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer :

Height of the conical vessel, h = 8 cm.

Its radius r = 5 cm

Volume of cone = Volume of water in cone $$= \ \frac{1}{3} \pi r^2 h$$

$$= \ \frac{1}{3} \ × \ \frac{22}{7} \ × \ 5 \ × \ 5 \ × \ 8 \ = \ \frac{4400}{21}$$ cm3

Volume of water flows out = Volume of lead shots

$$= \ \frac{1}{4}$$ of the volume of water in the cone

$$= \ \frac{1}{4} \ × \ \frac{4400}{21} \ = \ \frac{1100}{21}$$ cm3

Radius of the lead shot = 0.5 cm

Volume of one spherical lead shot $$= \ \frac{4}{3} \pi r^3 \ = \ \frac{4}{3} \ × \ \frac{22}{7} \ × \ ( \frac{5}{10})^3$$

$$= \ \frac{11}{21}$$ cm3

$$\therefore$$ Number of lead shots dropped into the vessel

$$= \frac{ \ Volume \ of \ water \ flows \ out}{Volume \ of \ one \ lead \ shot} \ = \frac{ \frac{1100}{21}}{ \frac{11}{21}}$$

$$= \ \frac{1100}{21} \ × \ \frac{21}{11} \ = \ 100$$