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Mathematics

Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.

Mensuration

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Answer

Given,

Radius of first solid sphere (r1) = 2 cm

Radius of second solid sphere (r2) = 4 cm

Height of cone (h) = 8 cm

Let radius of cone be r cm.

Since, two spheres are melted and recasted into cone.

∴ Volume of 1st sphere + Volume of 2nd sphere = Volume of cone

⇒43π(r1)3+43π(r2)3=13πr2h⇒4(r1)3+4(r2)3=r2h⇒r2=4r13+4r23h⇒r2=4×23+4×438⇒r2=4×8+4×648⇒r2=32+2568⇒r2=2888⇒r2=36⇒r=36⇒r=6 cm.\Rightarrow \dfrac{4}{3}π(r1)^3 + \dfrac{4}{3}π(r2)^3 = \dfrac{1}{3}πr^2h \\[1em] \Rightarrow 4(r1)^3 + 4(r2)^3 = r^2h \\[1em] \Rightarrow r^2 = \dfrac{4r1^3 + 4r2^3}{h} \\[1em] \Rightarrow r^2 = \dfrac{4 \times 2^3 + 4 \times 4^3}{8} \\[1em] \Rightarrow r^2 = \dfrac{4 \times 8 + 4 \times 64}{8} \\[1em] \Rightarrow r^2 = \dfrac{32 + 256}{8} \\[1em] \Rightarrow r^2 = \dfrac{288}{8} \\[1em] \Rightarrow r^2 = 36 \\[1em] \Rightarrow r = \sqrt{36} \\[1em] \Rightarrow r = 6 \text{ cm}.

Hence, the radius of cone formed = 6 cm.

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