A metallic sphere 1 dm in diameter is beaten into

Radius of metallic sphere $r=\frac{10}{2} \mathrm{~cm}$

Thickness of circular sheet

$h=1 \mathrm{~mm}$

$=\frac{1}{10} \mathrm{~cm}$

Let r1 be the radius of sheet.

Therefore,

Volume of circular sheet = volume of metallic sphere

$r_{1}^{2} \times \frac{1}{10}=\frac{4}{3} \times(5)^{3}$

$r_{1}^{2}=\frac{4 \times 125 \times 10}{3}$

$=\frac{5000}{3}$

$r_{1}=\sqrt{\frac{5000}{3}}$

$r_{1}=40.8 \mathrm{~cm}$

Hence, the radius of circular sheet = 40.8 cm