Discussion

Explanation:

Volume of iron sheet = 882 × 109 × 0.5 cm³ 

Let r be the radius of sphere which is equal to side of cube 

Volume of sphere = $\frac{4}{3}{\mathrm{\pi r}}^3$ = $\frac{4}{3}\times \frac{22}{7}\times {\mathrm{r}}^3$ = $\frac{88}{21}{r}^3$

Volume of cube = (side)³ = r³

Total volume of iron after reformation = $\frac{88}{21}{r}^3$ + r³ = $\frac{109}{21}{r}^3$

Volume of iron before reformation and after reformation should be same 

⇒ 221 × 218 × 1 = $\frac{109}{21}{r}^3$

⇒ r³ = 21 × 221 × 2 

⇒ r³ = 21 × 441 

⇒ r = 21

Hence, 21.

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