Given the mass of the aluminium is \(\displaystyle{m}={115}\cdot{10}^{{-{3}}}\) kg

Density of the aluminium is \(\displaystyle\rho={2.70}\cdot{10}^{{3}}\ {k}\frac{{g}}{{m}^{{3}}}\)

The specidic resistance of the aluminium is \(\displaystyle\rho={2.82}\cdot{10}^{{-{8}}}\) Ohm/m

Now the volume of the block is \(\displaystyle{v}{o}{l}{u}{m}{e}{V}={\frac{{{m}}}{{\rho}}}={\frac{{{115}\times{10}^{{-{3}}}}}{{{2.70}\times{10}^{{3}}}}}\)

\(\displaystyle={4.26}\cdot{10}^{{-{5}}}{m}^{{3}}\)

a) Given the diameter is equal to its height \(d= h\)

We know the formula for the volume of the cylinder in terms of diameter is

\(\displaystyle{V}=\pi{r}^{{2}}{h}=\pi{\left({\frac{{{d}}}{{{2}}}}\right)}^{{2}}{d}={\frac{{\pi{d}^{{3}}}}{{{4}}}}\)

From this \(\displaystyle{d}={\left({\frac{{{4}{V}}}{{\pi}}}\right)}^{{{\frac{{{1}}}{{{3}}}}}}\)

\(\displaystyle={\left({\frac{{{4}{\left({4.26}\times{10}^{{-{5}}}\right)}}}{{{3.14}}}}\right)}^{{\frac{{{1}}}{{{3}}}}}\)

\(\displaystyle={0.03785}{m}\)

\(\displaystyle{R}={\frac{{\rho{l}}}{{{A}}}}={\frac{{\rho{d}}}{{\pi{\frac{{{d}^{{2}}}}{{{4}}}}}}}={\frac{{{4}\pi}}{{\pi{d}}}}\)

We know the formula for the resistance is

\(\displaystyle{\frac{{{4}{\left({2.82}\times{10}^{{-{8}}}\right)}}}{{{\left({3.14}\right)}{\left({0.03785}\right)}}}}={9.49}\times{10}^{{-{7}}}\) Ohms

b)In this case we have to calculate the resistance between theopposite faces of the aluminium cube.

We know the formula for the cube is \(\displaystyle{V}={L}^{{3}}\)

From this the length of the cube is \(\displaystyle{L}={V}^{{{\frac{{{1}}}{{{3}}}}}}={\left({4.26}\times{10}^{{-{5}}}\right)}^{{{\frac{{{1}}}{{{3}}}}}}\)

Now the resistance between the opposite faces of the cubeis

\(\displaystyle{R}={\frac{{\rho{L}}}{{{A}}}}={\frac{{\rho{L}}}{{{L}^{{2}}}}}={\frac{{\rho}}{{{L}}}}={\frac{{{2.82}\times{10}^{{-{8}}}}}{{{0.0349}}}}={8.07}\times{10}^{{-{7}}}\) Ohms