Question

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?

Short answer

a) The resistance between the edges of the aluminum rod is$1.3\times {10}^{-3}\Omega$ .

b) The diameter of the copper rod is$4.6\times {10}^{-3}m$ .

Step-by-step solution

Identification of the given data

• 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,$\rho =2.75\times {10}^{-8}\Omega .m$

Understanding the concept of the resistance

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)

(a) Calculation of the resistance between the ends

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

(b) Calculation of the diameter of the copper rod

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$4.6\times {10}^{-3}m$ .