The length of the wire is 36 m.
Step-by-step explanation:
Given, Diameter of sphere = 6 cm
We know that, radius can be found by taking the half in the diameter value. So,
![\text { sphere radius, } R=(D)/(2)=(6)/(2)=3 \mathrm{cm}=3 * 10^(-2) \mathrm{m}](https://img.qammunity.org/2020/formulas/physics/middle-school/jgkcdwc7sga0rhfg49ikg5iqbui2egjgwi.png)
Similarly,
![\text { wire radius, } r=(0.2)/(2)=0.1 \mathrm{mm}=1 * 10^(-3) \mathrm{m}](https://img.qammunity.org/2020/formulas/physics/middle-school/v6gni7kn0gv2dgruw34sykolmjdd04mzi6.png)
We know the below formulas,
![\text {volume of sphere}=(4)/(3) * \pi * R^(3)](https://img.qammunity.org/2020/formulas/physics/middle-school/nnab6xyfe2gbtdgxg0q37gsc55z3q4258t.png)
![\text {volume of wire}=\pi * r^(2) * l](https://img.qammunity.org/2020/formulas/physics/middle-school/d85us2xfoy570hguxryuonrswyiske6gsu.png)
When equating both the equations, we can find length of wire as below, where
![\pi=(22)/(7)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nl8n6jnmuehjrwyj380ezu975a168s2ley.png)
![(4)/(3) * \pi * R^(3)=\pi * r^(2) * l](https://img.qammunity.org/2020/formulas/physics/middle-school/iyiudbrijre4hlz30jqeovh8h54p4rfr0c.png)
![(4)/(3) * (22)/(7) *\left(3 * 10^(-2)\right)^(3)=(22)/(7) *\left(1 * 10^(-3)\right)^(2) * l](https://img.qammunity.org/2020/formulas/physics/middle-school/phpzm9rzkofhovct6em7uco6q9cm8gwjaq.png)
The
value gets cancelled as common on both sides, we get
![(4)/(3) * 27 * 10^(-6)=10^(-6) * l](https://img.qammunity.org/2020/formulas/physics/middle-school/mj5onpoit5n0ylnw68aumvejawnd9toz5k.png)
The
value gets cancelled as common on both sides, we get
![l=4 * 9=36 m](https://img.qammunity.org/2020/formulas/physics/middle-school/f7gd9qthpbs03uw51ruclvqah4rmgl5ylt.png)