Question

A solid metallic right circular cylinder and a solid metallic right circular cone are given. The cylinder and cone both have same height $h$ and same base radii $r$. The two solids are melted together and recast into a solid cylinder of radius $\frac{1}{2}r$. Then the height of the cylinder is $\frac{16}{3}h$.

• A. True
• B. False

Answer 1

The correct option is A True

Volume of the resulting cylinder $=$ Volume of Cylinder $+$ Volume of Cone which are melted.

Let the height of the resulting cylinder be $H$.

Volume of a cone $=\frac{1}{3}\pi r^{2}h$, where $r$ is the radius of the base of the cone and $h$ is the height.

Volume of a cylinder of radius $R$ and height $h$, $=\pi R^{2}h$

So, $\pi {\left(\frac{r}{2}\right)}^{2}H=\pi r^{2}h+\frac{1}{3}\pi r^{2}h$
$\Rightarrow \frac{H}{4}=\frac{4}{3}h$
$\Rightarrow H=\frac{16}{3}h$
Hence, the height of the resulting cylinder is $\frac{16}{3}h$.