Question

27 spherical iron balls of radius 5 cm each are melted and recasted into a big sphere.The radius of the single big sphere(in cm) is______.

• A. 3 cm
• B. 9 cm
• C. 15 cm
• D. 5 cm

Answer 1

The correct option is C

15 cm

Given,

Radius of each small spherical ball,$r=5cm$

Let the radius of the big sphere be $Rcm$.

Volume of 27 small spherical balls $=27\times$ (Volume of each small spherical ball)

$=27\times \frac{4}{3}\pi r^3$

$=27\times \frac{4}{3}\times \pi \times 5^3$

Volume of big sphere $=\frac{4}{3}\pi R^3$

Since the volume of iron being melted and recast remains same, the total volume of 27 small spheres will be equal to the volume of the large single sphere.

$\therefore$ Volume of big sphere = Volume of 27 small spherical balls

$\Rightarrow \frac{4}{3}\times \pi R^3=27\times (\frac{4}{3}\times \pi r^3)$
$\Rightarrow R^3=27r^3$

$\Rightarrow R=3r=3\times 5=15cm$