Question

If an iron ball is melted and recasted into 3 small spherical balls of radius 3 cm, 4 cm and 5 cm, then the radius of the larger iron ball is .

• A. 6 cm
• B. 12 cm
• C. 18 cm
• D. 24 cm

Answer 1

The correct option is A 6 cm

Let the radius of the larger iron ball be R cm.
Given:
Radius of first iron ball = $r_1=3cm$
Radius of second iron ball = $r_2=4cm$
Radius of third iron ball = $r_3=5cm$

We know that, when we combine many solids to form one solid, the volume remains same.

$\Rightarrow$ Volume of the larger iron ball = Sum of the volume of three small iron balls

$\Rightarrow \frac{4}{3}\pi R^3=\frac{4}{3}\pi {r_1}^3+\frac{4}{3}\pi {r_2}^3+\frac{4}{3}\pi {r_3}^3$
$\Rightarrow R^3={r_1}^3+{r_2}^3+{r_3}^3$
$\Rightarrow R^3=3^3+4^3+5^3$
$\Rightarrow R^3=27+64+125$
$\Rightarrow R^3=216$
$\Rightarrow R^3=6^{3}c{m}^3$
$\Rightarrow R=6cm$

Hence, the radius of the larger iron ball is 6 cm.