A measuring jar of internal diameter $10\ cm$ is partially filled with water. Four equal spherical balls of diameter $2\ cm$ each are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?

 Given:

A measuring jar of internal diameter $10\ cm$ is partially filled with water. Four equal spherical balls of diameter $2\ cm$ each are dropped in it and they sink down in water completely. 

To do:

We have to find the change in the level of water in the jar.

Solution:

Diameter of the measuring jar $= 10\ cm$

This implies,

Radius of the jar $(\mathrm{R})=\frac{10}{2}$

$=5 \mathrm{~cm}$

Diameter of the spherical ball $=2 \mathrm{~cm}$

Radius of the ball $(r)=\frac{2}{2}$

$=1 \mathrm{~cm}$

Therefore,

Volume of the ball $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi(1)^{3} \mathrm{~cm}^{3}$

Volume of $4 \mathrm{balls}=\frac{4}{3} \times \pi \times 4$

$=\frac{16}{3} \pi \mathrm{cm}^{2}$

Let the volume of the water raised be $h\ cm$

Therefore,

Volume of water in the jar $=\frac{16}{3} \pi \mathrm{cm}^{3}$

$\pi r^{2} h=\frac{16}{3} \pi$

$\pi(5)^{2} h=\frac{16}{3} \pi$

$25 h=\frac{16}{3}$

$h=\frac{16}{3} \times \frac{1}{25}$

$h=\frac{16}{75} \mathrm{~cm}$

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Updated on: 10-Oct-2022

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