Answer:
Radius of aluminium sphere which has same mass as of sphere of iron with radius 14 m is 40.745 meters.
Step-by-step explanation:
Let the radius of aluminium sphere be
![r_(a)](https://img.qammunity.org/2020/formulas/chemistry/high-school/ufdqwy1hmam3r5599wh5tl18kaily4nbca.png)
From the relation between density, mass and volume we know that
![Mass=Density* volume...............(i)](https://img.qammunity.org/2020/formulas/physics/college/pn675ntmybj0y2ho4y86jy7rhbarchmbk7.png)
Applying equation 'i' separately to iron and aluminium sphere we get
![M_(a)=\rho _(a)* V_(a)](https://img.qammunity.org/2020/formulas/physics/college/nvzgjjvg8of95mptkev0cru244nz4mtr5a.png)
![M_(ir)=\rho _(ir)* V_(ir)](https://img.qammunity.org/2020/formulas/physics/college/kobbasoxh9g5z9b6daejlqecg6i5ftm2xw.png)
Equating the masses of iron and aluminium spheres we get
![M_(a)=M_(ir)](https://img.qammunity.org/2020/formulas/physics/college/8qv7y6jjpcm0pv70yjxl493pahfj2xsawf.png)
![\rho _(a)* (4\pi r_a^3)/(3)=\rho _(ir)* (4\pi r_(ir)^3)/(3)](https://img.qammunity.org/2020/formulas/physics/college/ro028k2ucpq76t94g4c2xexoow3unjlylm.png)
![\rho _(a)* r_(a)^3=\rho _(ir)* r_(ir)^(3)\\\\\therefore r_(a)=((\rho _(ir))/(\rho _(a)))^(1/3)\cdot r_(ir)\\\\r_(a)=(7.8610)/(2.701)* 14\\\\\therefore r_a=40.745m](https://img.qammunity.org/2020/formulas/physics/college/82bvtlr8art4yh0yoqpy443n9umu5qdnt6.png)