The surface area of a solid metallic sphere is 2464 cm². It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate :

(i) the radius of the sphere.

(ii) the number of cones recast.

(i) Given,

Surface area of solid metallic sphere = 2464

Let radius of solid metallic sphere be R cm.

By formula,

Surface area of solid metallic sphere = 4πR²

∴ 4πR² = 2464

⇒ $4 \times \dfrac{22}{7} \times R^2 = 2464$

⇒ $R^2 = \dfrac{2464 \times 7}{22 \times 4}$

⇒ $R^2 = \dfrac{17248}{88}$

⇒ R² = 196

⇒ R = $\sqrt{196}$

⇒ R = 14 cm.

Hence, the radius of sphere = 14 cm.

(ii) Given,

Radius of cones (r) = 3.5 cm

Height of cones (h) = 7 cm.

Let no. of cones formed be n.

Since,

Sphere is melted and recasted into cones.

∴ Volume of sphere = n × Volume of cone

$\Rightarrow \dfrac{4}{3}πR^3 = \dfrac{1}{3}πr^2h \\[1em] \Rightarrow 4R^3 = n \times r^2h \\[1em] \Rightarrow n = \dfrac{4R^3}{r^2h} \\[1em] \Rightarrow n = \dfrac{4 \times (14)^3}{(3.5)^2 \times 7} \\[1em] \Rightarrow n = \dfrac{4 \times 2744}{12.25 \times 7} \\[1em] \Rightarrow n = \dfrac{10976}{85.75} \\[1em] \Rightarrow n = 128.$

Hence, no. of cones formed = 128.