Question:

Solution:

Radius of the sphere = 8 cm

Volume of the sphere $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times 8^{3}=2145.52 \mathrm{~cm}^{3}$

Volume of one lead ball $=\frac{4}{3} \times \frac{22}{7} \times 1^{3}=4.19 \mathrm{~cm}^{3}$
$\therefore$ Number of lead balls = $\frac{\text { volume of the sphere }}{\text { volume of one lead ball }}=\frac{2145.52}{4.19}=512.05 \approx 512$