An aluminum rod with a square cross section is 1.3 mlong and 5.2 mmon edge. (a) What is the resistance between its ends? (b)What must be the diameter of a cylindrical copper rod of length 1.3 mif its resistance is to be the same as that of the aluminum rod?

• Length of the aluminum rod is,$L=L_{AI}=1.3m$ .
• Breadth of the rod is,$b_{AI}=side=5.2mm=5.2\times {10}^{-3}m$ .
• Length of the cylindrical copper rod is,$L_{Cu}=1.3m$ .
• Resistance of aluminum rod is the resistance of the copper rod.
• Resistivity of the aluminum rod is,$\rho =2.75\times {10}^{-8}\Omega .m$

In this problem, by using the relation between resistance and resistivity, we can find the resistance of the aluminum rod and the diameter of the cylindrical copper rod.

Formulae:

Area of the square is,

A = side x side … (i)

Area of the circle in terms of diameter is,

$A=\mathrm{\pi }\frac{{\mathrm{d}}^2}{4}$ … (ii)

The resistance of the material is,

$R=\rho \frac{L}{A}$ … (iii)

Area of cross section of the rod is the area of the square.

Thus, the value of the area of the cross-section can be given using equation (i) as follows:

$$\begin{gathered} A=5.2\times {10}^{-3}m\times 5.2\times {10}^{-3}m \\ =2.7\times {10}^{-5}{m}^2 \end{gathered}$$

Now, the resistance between the aluminum ends can be calculated using equation (iii) as follows:

$$\begin{gathered} R=2.75\times {10}^{-8}\Omega .m\times \frac{1.3m}{2.7\times {10}^{-5}{m}^2} \\ =1.3\times {10}^{-3}\Omega \end{gathered}$$

Hence, the value of the resistance is$1.3\times {10}^{-3}\Omega$ .

Resistance of copper rod is the resistance of the Aluminum rod

Thus,

$1.3\times {10}^{-3}\Omega$

Resistivity of copper is,$\rho =1.69\times {10}^{-8}\Omega .m$ .

Now, the area of the copper rod can be calculated using equation (iii) and the above values as follows:

$A=\rho \frac{L_{Cu}}{R}$

Substitute the values in the above equation.

$$\begin{gathered} A=1.69\times {10}^{-8}\Omega .m\times \frac{1.3m}{1.3\times {10}^{-3}\Omega } \\ =1.69\times {10}^{-5}{m}^2 \end{gathered}$$

Now, the value of the diameter of the copper rod can be calculated using the above area in equation (ii) as follows:

$d=2\times \sqrt{\frac{A}{\mathrm{\pi }}}$

Substitute the values in the above equation.

$$\begin{gathered} d=2\times \sqrt{\frac{1.69\times {10}^{-5}{m}^2}{\mathrm{\pi }}} \\ =4.6\times {10}^{-3}m \end{gathered}$$

Hence, the value of the diameter of the rod is$4.6\times {10}^{-3}m$ .