# Ex.13.8 Q2 Surface Areas and Volumes Solution - NCERT Maths Class 9

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## Question

Find the amount of water displaced by a solid spherical ball of diameter.

(i) $$28\, \rm{cm}$$

(ii) $$0.21 \,\rm{m}$$

Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-8 | Question 2

## Text Solution

Reasoning:

The amount of water displaced by a solid spherical ball is nothing but its volume.

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

What is known?

Diameter of the sphere.

What is unknown?

Amount of water displaced by the solid.

Steps:

1. Diameter

Radius \begin{align}\,r = \frac{{28}}{2} = 14\,\,cm \end{align}

Amount of water displaced by the solid sphere $$=$$ Volume of the sphere.

\begin{align}&= \frac{4}{3}\pi {r^3} \\&= \frac{4}{3} \times \frac{{22}}{7} \times {(14)^3}\\ &= 11498.66\,\,\,c{m^3} \end{align}

1. Diameter

Radius\begin{align}\,\,\,r = \frac{{0.21}}{2} = 0.105\,\,m \end{align}

Amount of water displaced by the solid sphere $$=$$ Volume of the sphere.

\begin{align}&= \frac{4}{3}\pi {r^3} \\&= \frac{4}{3} \times \frac{{22}}{7} \times {(0.105)^3}\\ &= .004851\,\,{m^3} \end{align}

1. Amount of water displaced\begin{align}\,\,\, = 11498.66\,\,c{m^3} \end{align}
2. Amount of water displaced \begin{align}\,\, = .004851\,\,{m^3} \end{align}