The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate

(i) the radius of the solid sphere.

(ii) the number of cones recast. (Use π = 3.14).

(i) Surface area of a metallic sphere = 1256 cm².

Let the radius of this sphere be R.

$\therefore 4πR^2 = 1256 \\[1em] \Rightarrow R^2 = \dfrac{1256}{4π} \\[1em] \Rightarrow R^2 = \dfrac{1256}{4 \times 3.14} \\[1em] \Rightarrow R^2 = \dfrac{1256}{12.56} \\[1em] \Rightarrow R^2 = 100 \\[1em] \Rightarrow R = \sqrt{100} = 10 \text{ cm}.$

Hence, the radius of sphere = 10 cm.

(ii) Let the number of cones formed by recasting sphere be n.

Radius of cone (r) = 2.5 cm

Height of cone (h) = 8 cm.

Volume of sphere = n × Volume of each cone.

$\therefore \dfrac{4}{3}πR^3 = n \times \dfrac{1}{3}πr^2h$

Multiplying both sides by 3 and dividing by π.

$\Rightarrow 4R^3 = nr^2h \\[1em] \Rightarrow 4 \times 10^3 = n \times (2.5)^2 \times 8 \\[1em] \Rightarrow n = \dfrac{4 \times 1000}{6.25 \times 8} \\[1em] \Rightarrow n = \dfrac{4000}{50} \\[1em] \Rightarrow n = 80.$

Hence, the number of cones formed by recasting sphere are 80.