A right circular cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. The radius of the sphere is :

Question

Accepted Answer

To determine the radius of the sphere formed from the modelling clay originally shaped as a right circular cone, we need to use the principle of volume conservation. The volume of the cone is equal to the volume of the sphere since the clay is simply reshaped, not added to or removed.

1. **Volume of the Cone:**
   The formula for the volume of a cone is given by:
   $$
   V_{\text{cone}} = \frac{1}{3} \pi r^2 h
   $$
   where $ r $ is the radius of the base of the cone, and $ h $ is the height of the cone. For this cone, $ r = 6 $ cm and $ h = 24 $ cm. Plugging in these values:
   $$
   V_{\text{cone}} = \frac{1}{3} \pi (6)^2 (24) = \frac{1}{3} \pi \times 36 \times 24 = 288\pi \text{ cubic centimeters}
   $$

2. **Volume of the Sphere:**
   The formula for the volume of a sphere is:
   $$
   V_{\text{sphere}} = \frac{4}{3} \pi R^3
   $$
   where $ R $ is the radius of the sphere. Since the volumes are equal:
   $$
   \frac{4}{3} \pi R^3 = 288\pi
   $$
   Dividing both sides by $\pi$ and solving for $ R^3 $:
   $$
   \frac{4}{3} R^3 = 288
   $$
   $$
   R^3 = 288 \times \frac{3}{4} = 216
   $$
   Taking the cube root of both sides:
   $$
   R = \sqrt[3]{216} = 6 \text{ cm}
   $$

Thus, the radius of the sphere is 6 cm, which corresponds to Option 3.

**Explanation for Incorrect Options:**

- **Option 1 (2 cm):** If the radius were 2 cm, the volume of the sphere would be $\frac{4}{3} \pi (2)^3 = \frac{32}{3} \pi$, which is much less than the volume of the cone (288$\pi$).

- **Option 2 (3 cm):** If the radius were 3 cm, the volume of the sphere would be $\frac{4}{3} \pi (3)^3 = 36\pi$, which is still less than the volume of the cone.

- **Option 4 (1 cm):** If the radius were 1 cm, the volume of the sphere would be $\frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi$, which is far less than the volume of the cone.

Therefore, Option 3 is the only correct answer, as it satisfies the condition of equal volumes for the cone and the sphere.