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:

$\frac{21}{2}$

The correct answer is Option (1) → $\frac{21}{2}$

1. Volume of the Cone

The formula for the volume of a right circular cone ($V_c$) is:

$V_c = \frac{1}{3} \pi r_c^2 h$

Given:

• Height ($h$) = $42$ cm
• Radius ($r_c$) = $\frac{21}{2}$ cm

Substituting the values:

$V_c = \frac{1}{3} \pi \left( \frac{21}{2} \right)^2 \times 42$

$V_c = \frac{1}{3} \pi \times \frac{441}{4} \times 42$

$V_c = \pi \times \frac{441}{4} \times 14$

2. Volume of the Sphere

The formula for the volume of a sphere ($V_s$) with radius $R$ is:

$V_s = \frac{4}{3} \pi R^3$

3. Equating the Volumes

Since the cone is melted and recast into a sphere:

$V_s = V_c$

$\frac{4}{3} \pi R^3 = \frac{1}{3} \pi \left( \frac{21}{2} \right)^2 \times 42$

Cancel $\pi$ and $\frac{1}{3}$ from both sides:

$4R^3 = \left( \frac{21}{2} \right)^2 \times 42$

$4R^3 = \frac{441}{4} \times 42$

Multiply both sides by $\frac{1}{4}$ to solve for $R^3$:

$R^3 = \frac{441 \times 42}{16}$

$R^3 = \frac{21 \times 21 \times (21 \times 2)}{16}$

$R^3 = \frac{21^3 \times 2}{16}$

$R^3 = \frac{21^3}{8}$

4. Find the Radius ($R$)

Take the cube root of both sides:

$R = \sqrt[3]{\frac{21^3}{8}}$

$R = \frac{21}{2} \text{ cm}$

Final Answer: The radius of the sphere is $\frac{21}{2}$ cm.