The diameter of a metallic sphere …

We have,
Diameter of the sphere = 6 cm
{tex}\therefore{/tex} Radius of the sphere = {tex}\frac { 6 } { 2 } \mathrm { cm } = 3 \mathrm { cm }{/tex}
{tex}\Rightarrow{/tex}  Volume of the sphere ={tex}= \frac { 4 } { 3 } \times \pi \times 3 ^ { 3 } \mathrm { cm } ^ { 3 } = 36 \pi \mathrm { cm } ^ { 3 }{/tex} {tex}\left[ \text { Using } V = \frac { 4 } { 3 } \pi r ^ { 3 } \right]{/tex}
Let the radius of cross-section of wire be r cm. It is given that the length of the cylindrical shaped wire is 36 m.
{tex}\therefore{/tex} Volume of the wire = {tex}\left( \pi r ^ { 2 } \times 3600 \right) \mathrm { cm } ^ { 3 }{/tex}
Since metallic sphere is converted into cylindrical shaped wire. Therefore, Volume of the wire = Volume of the sphere
{tex}\Rightarrow \quad \pi r ^ { 2 } \times 3600 = 36 \pi{/tex}
{tex}\Rightarrow \quad r ^ { 2 } = \frac { 36 \pi } { 3600 \pi } = \frac { 1 } { 100 }{/tex}
{tex}\Rightarrow \quad r = \frac { 1 } { 10 } \mathrm { cm } = 1 \mathrm { mm }{/tex}