A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top which is open, is 2.5 cm. It is filled with water upto the rim. When lead shots, each of which is a sphere of radius 0.25 cm are dropped into the vessel, $\dfrac{2}{5}$ of the water flows out. Find the number of lead shots dropped into the vessel.

Radius of the top of the inverted cone, R = 2.5 cm

Height of the cone, H = 11 cm

Radius of lead shot, r = 0.25 cm

When lead shots are dropped into vessel, $\dfrac{2}{5}$ of water flows out.

∴ Volume of water flown out = $\dfrac{2}{5}$ Volume of cone.

$= \dfrac{2}{5} \times \dfrac{1}{3}π\text{R}^2 \text{H} \\[1em] = \dfrac{2}{15} \times π\times 2.5^2 \times 11 \\[1em] = \dfrac{2}{15} \times π\times 6.25 \times 11 \\[1em] = \dfrac{137.5}{15} π$

Let the number of spheres be n.

∴ Volume of water flown out = n × Volume of each lead shot

$\Rightarrow \dfrac{137.5}{15} π = \text{n} \times \dfrac{4}{3} \times π\text{r}^3 \\[1em] \text{Divide by π on both sides, we get:} \Rightarrow \dfrac{137.5}{15} = \text{n} \times \dfrac{4}{3} \times 0.25^3 \\[1em] \Rightarrow \dfrac{137.5}{15} = \text{n} \times \dfrac{4}{3} \times 0.015625 \\[1em] \Rightarrow \dfrac{137.5}{15} = \text{n} \times \dfrac{0.0625}{3} \\[1em] \Rightarrow \text{n} = \dfrac{137.5 \times 3}{0.0625 \times 15} \\[1em] \Rightarrow \text{n} = \dfrac{412.5}{0.9375} \\[1em] \Rightarrow \text{n} = 440.$

Hence, the number of lead shots are 440.