A right circular metal cone (solid) is 42 cm high and its radius is \(\frac{21}{2}\) cm. It is melted and recast into a sphere. Then the radius of the sphere will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\frac{21}{2}\)
When a solid is melted and recast, its volume does not change, so the volume of the sphere equals the volume of the cone.
\(\displaystyle \frac{4}{3}\pi R^3=\frac{1}{3}\pi r^2h\)
\(\displaystyle R^3=\frac{r^2h}{4}=\left(\frac{21}{2}\right)^2\times\frac{42}{4}\)
\(\displaystyle \frac{42}{4}=\frac{21}{2}\), so \(\displaystyle R^3=\left(\frac{21}{2}\right)^3\)
\(\displaystyle R=\frac{21}{2}\) cm