# Ex.13.8 Q9 Surface Areas and Volumes Solution - NCERT Maths Class 9

## Question

Twenty-seven solid iron spheres, each of radius \(r\) and surface area \(S\) are melted to form a sphere with surface area \(S’\). Find the

- Radius \(r'\) of the new sphere,
- Ratio of \(S\) and \(S’\).

## Text Solution

**Reasoning:**

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

**What is known?**

Number of spheres melted.

**What is unknown?**

Radius.

Ratio of \(S\) and \(S’\).

**Steps:**

Volume of a solid iron sphere \({=\frac{4}{3} \pi r^{3}}\)

Volume of \(27\) solid spheres \({=27\left(\frac{4}{3} \pi r^{3}\right)}={36 \pi r^{3}}\)

Volume of new sphere \({=36 \pi r^{3}}\)

Let the radius of the new sphere\({{= r' }}\)

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

According to the question,

\[\begin{align} \frac{4}{3} \pi r^{\prime 3}&= 36 \pi r^{3} \\ r^{3} &=\frac{\left(36 r^{3}\right)(3)}{4} \\ &=27 r^{3} \\ r^{\prime} &=\sqrt[3]{\left(27 r^{3}\right)} \\ r^{\prime} &=3 r \end{align}\]

Radius of the new sphere \( r=3r\)

\[\begin{align}{S}&={4 \pi r^{2}} \\ {s^{1}}&={4 \pi(3 r)^{2}} \\ {\frac{S}{s^{1}}}&={\frac{4 \pi r^{2}}{4 \pi(3 r)^{2}}} \\ {\frac{S}{s^{1}}}&={\frac{1}{3^{2}}}={\frac{1}{9}}\end{align}\]

So ratio of S:S’ is \(1:9\)

**Answer: **Radius r of new sphere = \(3r.\)

Ratio of S and S’\(= 1:9\)