Question

A cone of radius 3 m and height 12 m is recasted into following shapes. Pair the radius in each case.

• A. $3\times \sqrt[3]{32}m$
  
• B. $2\sqrt{3}m$
• C. $3m$

Answer 1

When a solid is melted and moulded to form another solid shape, the volume of the solid formed will be equal to the volume of the solid melted.

Given: Radius of cone, r = 3 m
Height of the cone, h = 12 m

Let the radius of sphere be R.
Volume of the cone = Volume of sphere
$\therefore$$\frac{1}{3}\pi r^{2}h=\frac{4}{3}\pi R^3{R}^3=\frac{r^{2}h}{4}{R}^3=\frac{3\times 3\times 12}{4}R=3\text{m}$
$\therefore$ Radius of the sphere will be 3 m.

Let the radius of hemisphere be R.
Volume of the cone = Volume of hemisphere
$\therefore$$\frac{1}{3}\pi r^{2}h=\frac{2}{3}\pi R^3{R}^3=\frac{r^{2}h}{2}{R}^3=\frac{3\times 3\times 12}{2}R=3\sqrt[3]{32}\text{m}$
$\therefore$ Radius of hemisphere will be $3\sqrt[3]{32}\text{m}$

Let the radius of cylinder be R.
Height of cylider, $H=3\text{m}$
Volume of the cone = Volume of cylinder
$\therefore$$\frac{1}{3}\pi r^{2}h=\pi R^{2}H{R}^2=\frac{r^{2}h}{3\times 3}{R}^2=\frac{3\times 3\times 12}{3\times 3}=12R=2\sqrt{3}\text{m}$
$\therefore$ Radius of cylinder will be $2\sqrt{3}\text{m}.$