# Ex.13.2 Q5 Surface Areas and Volumes Solution - NCERT Maths Class 10

## Question

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

## Text Solution

**What is known?**

An open vessel is in the form of an inverted cone with height \(=8\,\rm{cm}\) and radius of its top \(= 5\,\rm{cm}\), which is filled with water up to the drum.

Spherical lead shots each of radius \(=0.5\,\rm{cm},\) are dropped into the vessel until then one-fourth of the water in the vessel flows out.

**What is unknown?**

Number of lead shots dropped into the vessel.

**Reasoning:**

Draw a figure of the vessel and lead shot to visualize it

Since the water is filled up to the brim in the vessel

Volume of water in the vessel \(=\) volume of the conical vessel

Also, on dropping a certain number of lead shots (sphere) into the vessel one-fourth of the water flows out.

Volume of all lead shots dropped in the vessel \(\begin{align} = \frac{1}{4} \times \end{align}\)Volume of water in the vessel

Hence,

Number of lead shots \( \begin{align} = \frac{1}{4} \times\end{align} \)volume of the water in the vessel\( \div \) volume of each lead shot

We will find the volume of the water in the vessel and lead shot by using formulae;

Volume of the sphere\(\begin{align} = \frac{4}{3}\pi {r^3}\end{align}\)

where *\(r\)* is the radius of the hemisphere

Volume of the cone\( \begin{align}= \frac{1}{3}\pi {R^2}h\end{align}\)

where *\(R\)* and *\(h\)* are the radius and height of the cone respectively

**Steps:**

Height of the conical vessel,\(h = 8cm\)

Radius of the conical vessel,\(R = 5cm\)

Radius of the spherical lead shot,\(r = 0.5cm\)

Number of lead shots\( \begin{align}= \frac{1}{4} \times\end{align} \) volume of the water in the vessel \( \div \)volume of each lead shot

\[\begin{align}&= \left( {\frac{1}{4} \times \frac{1}{3}\pi {R^2}h} \right) \div \left( {\frac{4}{3}\pi {r^3}} \right)\\&= \frac{{\pi {R^2}h}}{{12}} \times \frac{3}{{4\pi {r^3}}}\\&= \frac{{{R^2}h}}{{16{r^3}}}\\&= \frac{{5cm \times 5cm \times 8cm}}{{16 \times 0.5cm \times 0.5cm \times 0.5cm}}\\ &= 100\end{align}\]