The surface area of a solid metallic sphere is 616 cm2. It is melted and recast into solid right circular cones of diameter 3.5 cm and height 3.5 cm. Find the:
(a) radius of the sphere.
(b) number of cones formed.
$\left(\pi = \frac{22}{7}\right)$
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(a) Given,
Surface area of the solid metallic sphere = 616 cm2
Let the radius of the sphere be $r$ cm.
By formula,
Surface area of sphere = $4\pi r^2$
Equating this to the given surface area and substituting $\pi = \frac{22}{7}$, we get :
$$\begin{aligned}\Rightarrow 4 \times \frac{22}{7} \times r^2 = 616 \\[1em] \Rightarrow \frac{88}{7} \times r^2 = 616 \\[1em] \Rightarrow r^2 = \frac{616 \times 7}{88} \\[1em] \Rightarrow r^2 = 49 \\[1em] \Rightarrow r = 7 \text{ cm}.\end{aligned}$$
Hence, the radius of the sphere = 7 cm.
(b) We know that,
When a solid is melted and recast into a number of smaller solids, the material only changes its shape and no material is lost.
∴ Volume of the sphere = Number of cones × Volume of one cone
$\Rightarrow \text{Number of cones} = \frac{\text{Volume of sphere}}{\text{Volume of one cone}}$
So, we first find the volume of the sphere and the volume of one cone.
Volume of the sphere :
From part (a), radius of the sphere ($r$) = 7 cm.
By formula,
Volume of sphere = $\frac{4}{3} \pi r^3$
Substituting values we get :
$$\begin{aligned}= \frac{4}{3} \times \frac{22}{7} \times 7^3 \\[1em] = \frac{4}{3} \times \frac{22}{7} \times 343 \\[1em] = \frac{4}{3} \times 22 \times 49 \\[1em] = \frac{4312}{3} \text{ cm}^3.\end{aligned}$$
Volume of one cone :
Given,
Diameter of cone = 3.5 cm
$\Rightarrow \text{Radius of cone (r)} = \frac{\text{Diameter}}{2} = \frac{3.5}{2} = \frac{7}{4} \text{ cm}$
Height of cone ($h$) = 3.5 cm = $\frac{7}{2}$ cm
By formula,
Volume of cone = $\frac{1}{3} \pi r^2 h$
Substituting values we get :
$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times \Big(\frac{7}{4}\Big)^2 \times \frac{7}{2} \\[1em] = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{16} \times \frac{7}{2} \\[1em] = \frac{22 \times 49 \times 7}{3 \times 7 \times 16 \times 2} \\[1em] = \frac{539}{48} \text{ cm}^3.\end{aligned}$$
Number of cones :
$$\begin{aligned}\text{Number of cones} = \frac{\text{Volume of sphere}}{\text{Volume of one cone}} \\[1em] = \frac{\frac{4312}{3}}{\frac{539}{48}} \\[1em] = \frac{4312}{3} \times \frac{48}{539} \\[1em] = \frac{4312}{539} \times \frac{48}{3} \\[1em] = 8 \times 16 \\[1em] = 128.\end{aligned}$$
Hence, the number of cones formed = 128.