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:

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.

To do:

We have to find the number of lead shots dropped in the vessel.

Solution:

Height of the cone $=8 \mathrm{~cm}$

The radius of the cone $=5 \mathrm{~cm}$

This implies,

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

$=\frac{1}{3} \pi(5)^{2} \times 8$

$=\frac{200}{3} \pi \mathrm{cm}^{3}$

The radius of one spherical lead shot $=0.5 \mathrm{~cm}$

Therefore,

Volume of one spherical lead shot $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi(0.5)^{3}$

$=\frac{4 \times 0.125}{3} \pi \mathrm{cm}^{3}$

$=\frac{0.5}{3} \pi \mathrm{cm}^{3}$
When spherical lead is dropped in the vessel, one-fourth of the water flows out

Let the number of lead shots be $n$.

Volume of $n$ spherical shots $=\frac{1}{4}$ volume of conical vessel

$n(\frac{0.5}{3} \pi)=\frac{1}{4}(\frac{200}{3} \pi)$

$n(0.5)=50$

$n=\frac{50 \times 10}{5}$

$n=100$

The number of lead shots dropped in the vessel is 100.

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Updated on: 10-Oct-2022

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