A solid piece of iron of dimensions 49 × 33 × 24 cm is moulded into a sphere. The radius of the sphere is · ABC26GN5011

Volume of the solid $=(49 \times 33 \times 24) \mathrm{cm}^{3}$. Let the radius of the sphere be $r$. $$\begin{array}{r} \text { Then, } \frac{4}{3} \pi r^{3}=(49 \times 33 \times 24) \Leftrightarrow r^{3}=\left(\frac{49 \times 33 \times 24 \times 3 \times 7}{4 \times 22}\right) \\ =(21)^{3} \Leftrightarrow r=21 . \end{array}$$