# 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

1. Radius $$r'$$ of the new sphere,
2. Ratio of $$S$$ and $$S’$$.

Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-8 | Question 9

## 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?

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’$$= 1:9$$

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