Question

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

Answer 1

Step 1: Find the volume of the immersed spherical balls.

The volume of the immersed spherical balls $=4\times Volumeofoneball$ (Balls are identical i.e. radius are same)

Given that, the diameter of each ball $=2cm$

Thus, the radius of each ball $=\frac{2}{2}cm$ $(Radius=\frac{Diameter}{2})$

$=1cm$

The volume of a sphere of radius $r$ $=\frac{4}{3}{\mathrm{\pi r}}^3$

$$\begin{gathered} =\frac{4}{3}\mathrm{\pi }\times {\left(1\right)}^3 \\ =\frac{4}{3}\mathrm{\pi }{\mathrm{cm}}^3 \end{gathered}$$

The volume of $4$ spherical balls $=4\times \frac{4}{3}\mathrm{\pi }{\mathrm{cm}}^3$

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

Step 2: Find the volume of water risen in the jar.

Given that, the diameter of the jar $=10cm$

Thus, the radius of the jar $=\frac{10}{2}cm$ $(Radius=\frac{Diameter}{2})$

$=5cm$

After swing the ball in the water of the jar let the volume of water raised by $hcm$

The volume cylinder of height $h$ and radius $r$ $={\mathrm{\pi r}}^2\mathrm{h}$

The volume of water risen in the jar of radius $5cm$ and height $hcm$ $=\mathrm{\pi }{\left(5\right)}^2\mathrm{h}{\mathrm{cm}}^3$

$=25\mathrm{\pi h}{\mathrm{cm}}^3$

Step 3: Find the change in the level of water in the jar.

The volume of the water rise in the jar $=$ Volume of the immersed spherical balls

$\begin{array}{rcl}\therefore 25\pi h& =& \frac{16}{3}\pi \\ & \Rightarrow & h=\frac{16}{3\times 25}\\ & \Rightarrow & h=\frac{16}{75}cm\end{array}$

Hencee, the level of water is raised by $\frac{16}{75}cm$