A vessel is 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.

Given,

Radius of cone (R) = 5 cm

Height of cone (H) = 8 cm

By formula,

Volume of cone = $\dfrac{1}{3}πR^2H$

Substituting values we get :

$= \dfrac{1}{3} \times π \times 5^2 \times 8 \\[1em] = \dfrac{200π}{3} \text{ cm}^3.$

Volume of water in vessel = Volume of cone = $\dfrac{200π}{3}$ cm³.

Let no. of lead shots be n.

Given,

Radius of lead shots (r) = 0.5 cm

Since,

One-fourth of the water flows out after dropping lead shots in vessel.

∴ n × Volume of each lead shot = $\dfrac{1}{4} \times$ Volume of water

$\Rightarrow n \times \dfrac{4}{3}πr^3 = \dfrac{1}{4} \times \dfrac{200π}{3} \\[1em] \Rightarrow n \times \dfrac{4}{3}π \times (0.5)^3 = \dfrac{50π}{3} \\[1em] \Rightarrow n = \dfrac{50π \times 3}{4π \times 3 \times (0.5)^3} \\[1em] \Rightarrow n = \dfrac{50}{4 \times 0.125} \\[1em] \Rightarrow = \dfrac{50}{0.5} = 100.$

Hence, 100 lead shots were dropped in vessel.