# Ex.13.3 Q2 Surface Areas and Volumes Solution - NCERT Maths Class 10

## Question

Metallic spheres of radii \(6 \,\rm{cm}\), \(8\, \rm{cm}\), and \(10\, \rm{cm}\), respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

## Text Solution

**What is known?**

Radii of three metallic spheres are \(6\, \rm{cm}\), \(8\, \rm{cm}\) and \(10 \,\rm{cm}\)

**What is unknown?**

The radius of the resulting sphere

**Reasoning:**

Draw a figure to visualize the shapes better

Since, three metallic spheres are melted and recast into a single solid sphere then the sphere formed by recasting these spheres will be same in volume as the sum of the volumes of these spheres.

Volume of the resulting sphere \(=\) Sum of the volumes of three spheres

We will find the volume of the sphere by using formula;

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

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

**Steps:**

Radius of 1^{st }sphere,\({r_1} = 6cm\)

Radius of 2^{nd }sphere,\({r_2} = 8cm\)

Radius of 3^{rd }sphere,\({r_3} = 10cm\)

Let the radius of the resulting sphere be *\(r\)*.

Volume of the resulting sphere \(=\) Sum of the volumes of three spheres

\[\begin{align}\frac{4}{3}\pi {r^3} &= \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 + \frac{4}{3}\pi r_3^3\\

\frac{4}{3}\pi {r^3} &= \frac{4}{3}\pi \left[ {r_1^3 + r_2^3 + r_3^3} \right]\\{r^3} &= \left[ {r_1^3 + r_2^3 + r_3^3} \right]\\{r^3} &= \left[ {{{\left( {6cm} \right)}^3} + {{\left( {8cm} \right)}^3} + {{\left( {10cm} \right)}^3}} \right]\\{r^3} &= \left[ {216c{m^3} + 512c{m^3} + 1000c{m^3}} \right]\\{r^3} &= 1728c{m^3}\\r &= 12cm\end{align}\]

Therefore, the radius of the sphere so formed will be \(12 \rm{cm.}\)