Question

A cylindrical tub of radius$16cm$ contains water to a depth of $30cm$. A spherical iron ball is dropped into the tub and thus level of water is raised by$9cm$. What is the radius of the ball?

Answer 1

Finding the radius of the iron ball:

Let$R$ be the radius of the iron ball.

Given, Radius of Cylindrical tub, $r=16cm$

Depth of Cylindrical tub, $H=30cm$

Height of level of water raised of tub, $h=9cm$

According to the question, we have,

Volume of iron ball = Volume of water raised in the tub

$\begin{array}{rcl}\frac{4}{3}\pi R^3& =& \pi r^{2}h\\ & \Rightarrow & \frac{4}{3}\pi R^3=\mathrm{\pi }\times {\left(16\right)}^2\times 9\\ & \Rightarrow & R^3=\frac{(27\times {\left(16\right)}^2)}{4}\\ & \Rightarrow & R^3=1728\\ & \Rightarrow & R=\sqrt[3]{1728}\\ & \Rightarrow & R=12cm\end{array}$

Hence, the radius of the iron ball is $12cm$.