A solid piece of iron in the form a cuboid of dimensions (49 cm × 33 cm × 24 cm) is moulded into a solid sphere.

Solution:

(b) 21 cm

Volume of the cuboid $=(l \times b \times h)=49 \times 33 \times 24 \mathrm{~cm}^{3}$

Let the radius of the sphere be $r \mathrm{~cm}$.

 

Volume of the sphere $=\frac{4}{3} \pi r^{3}$

The volume of the sphere and the cuboid are the same.
Therefore,

$\frac{4}{3} \pi r^{3}=49 \times 33 \times 24$

$\Rightarrow \frac{4}{3} \times \frac{22}{7} \times r^{3}=49 \times 33 \times 24$

$\Rightarrow r^{3}=49 \times 33 \times 24 \times \frac{21}{88}$

$\Rightarrow r^{3}=21 \times 21 \times 21$

$\Rightarrow r^{3}=(21)^{3}$

 

$\Rightarrow r=21 \mathrm{~cm}$

Hence, the radius of the sphere is 21 cm.