The total area of a solid metallic sphere is 1256 cm³. 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.

[Take π = 3.14]

(i) Given,

Total area of solid metallic sphere = 1256 cm³

Let R be the radius of metallic sphere.

By formula,

Total area of solid metallic sphere = 4πR²

∴ 4πR² = 1256

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

Hence, radius of sphere = 10 cm.

(ii) Given,

Radius of cone (r) = 2.5 cm

Height of cone (h) = 8 cm.

Let no. of cones formed be n.

Since, sphere is recasted into n cones.

∴ Volume of sphere = n × Volume of cones

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

Hence, no. of cones formed = 80.