# Ex.13.3 Q1 Surface Areas and Volumes Solution - NCERT Maths Class 10

## Question

A metallic sphere of radius \(4.2\rm \,{cm}\) is melted and recast into the shape of a cylinder of radius \(6\rm \,{cm}\). Find the height of the cylinder.

## Text Solution

**What is known?**

Radius of the metallic sphere \(4.2 \,\rm{cm}\) and radius of the cylinder \(6 \,\rm{cm}\)

**What is unknown?**

The height of the cylinder.

**Reasoning:**

Draw a figure to visualize the shapes better

Since, a metallic sphere is melted and recast into the shape of a cylinder then their volume must be same.

Volume of the sphere \(=\) Volume of the cylinder

We will find the volume of the sphere and cylinder by using formulae;

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

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

Volume of the cylinder \( = \pi {r^2}h\)

where *\(r\)* and *\(h\)* are radius and height of the cylinder respectively

**Steps:**

Radius of the hemisphere, \({r_1} = 4.2 \rm cm\)

Radius of the cylinder, \({r_2} = 6 \rm cm\)

Let the height of the cylinder be *\(h\)*.

Volume of sphere \(=\) Volume of cylinder

\[\begin{align}\frac{4}{3}\pi r_1^3 &= \,\pi r_2^2h\\\frac{4}{3}r_1^3 &= \,r_2^2h\end{align}\]

\[\begin{align}h &= \frac{{4r_1^3}}{{3\,r_2^2}}\\&= \frac{\begin{pmatrix} 4 \times 4.2 \rm cm \times \\ 4.2 \rm cm \times 4.2cm\end{pmatrix} }{{3 \times 6 \rm cm \times 6cm}}\\&= 2.74 \rm cm\end{align}\]

Hence, the height of the cylinder so formed will be \(2.74\rm{ cm.}\)