Question

A cylindrical tub of radius$16cm$ contains water to the depth of $30cm$. A spherical iron ball is dropped into the tub, as a result, the level of water rises by$9cm$. The radius of the ball is

• A. $10cm$
• B. $6cm$
• C. $12cm$
• D. None of the above

Answer 1

The correct option is C

$12cm$

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 $h=9cm$

A.T.Q We have,

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

$\begin{array}{rcl}\frac{\mathbf{4}}{\mathbf{3}}\mathit{\pi }{\mathit{R}}^{\mathbf{3}}& \mathbf{=}& \mathbf{}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{h}\mathbf{}\\ & \Rightarrow & \frac{4}{3}\pi R^3=\mathrm{\pi }\times {\left(16\right)}^{2}x9\\ & \Rightarrow & R^3=\frac{(27x{\left(16\right)}^2)}{4}\\ & \Rightarrow & R^3=1728\\ & \Rightarrow & R=\sqrt[3]{1728}\\ & \Rightarrow & R=12cm\end{array}$

Hence, the correct option is (C)