Question

(a) A 115-g mass of aluminum is formed into a right circular cylinder, shaped so that its diameter equals its height.Calculate the resistance between

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asked 2021-05-08
(a) A 115-g mass of aluminum is formed into a right circular cylinder, shaped so that its diameter equals its height.Calculate the resistance between the top and bottom faces of thecylinder at 20 degrees C. (b) Calculate the resistance betweenopposite faces if the same mass of aluminum is formed into acube.

Answers (1)

2021-05-10

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

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