Question

Metallic spheres of radii 6 cm,8 cm and 10 cm respectively are melted and recasted into a single solid sphere.Taking $\pi$=3.1,find the surface area of solid sphere formed.

• A. 14 cm
• B. 12 cm
• C. 18 cm
• D. 23 cm

Answer 1

The correct option is B

12 cm

Let r1, r2 ,r3 be the radius of the given 3 spheres & R be the radius of a single solid sphere.

Given :

r1= 6cm, r2= 8cm, r3= 10 cm

Volume of first metallic sphere (V1)= $\frac{4}{3}\pi r{1}^3=\frac{4}{3}\pi 6^3$

Volume of second metallic sphere (V2)=$\frac{4}{3}\pi r{2}^3=\frac{4}{3}\pi 8^3$

Volume of third metallic sphere (V3) =$\frac{4}{3}\pi r{3}^3=\frac{4}{3}\pi {10}^3$

Volume of single solid sphere(V)= $\frac{4}{3}\pi R^3$

Volume of 3 metallic spheres= volume of single solid sphere

V1+V2+V3 = V

$\frac{4}{3}\pi$ (6³+8³+10³) = $\frac{4}{3}\pi$R³

216+ 512+ 1000 = R³

1728= R³

(12×12×12) = R³

12³= R³

R= 12

Hence, the radius of the resulting sphere = 12 cm

Surface area of sphere =$4\pi R^2$ =$4\pi \times {12}^2=1785.6c{m}^2$