Chapter 18, Problem 22P

To determine

The ratio of the diameter of the longer wire to the diameter of the shorter wire.

Answer to Problem 22P

  $\frac{D_1}{D_2}=\sqrt{2}$

Explanation of Solution

Given info:

Two aluminium wires have the same resistance; $R_1=R_2$

One wire has twice the length of the other; $L_1=2L_2$

Formula used:

It is known that resistance:

  $R=\frac{\rho L}{A}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length of wire

  $A$ is the cross-sectional area of the wire.

The cross-sectional area:

  $A=\pi r^2$

Where, $\pi$ is constant and $r$ is a radius.

Formula of radius:

  $r=\frac{D}{2}$

Where, $D$ is a diameter.

Calculation:

By using the formula: $R=\frac{\rho L}{\pi r^2}$

Put the value of $r$ in above-mentioned formula,

  $R=\frac{\rho L}{\pi {\left(\frac{D}{2}\right)}^2}$

  $R=\frac{4\rho L}{\pi D^2}\mathrm{...}(i)$

From equation $\left(i\right)$ , values of $R_1$ and $R_2$

  $R_1=\frac{4{\rho }_1{L}_1}{\pi D_1^2}\mathrm{...}(ii)$

  $R_2=\frac{4{\rho }_2{L}_2}{\pi D_{{}^2}^2}\mathrm{...}(iii)$

According to the question, two aluminium wires have the same resistance. So, equations $\left(ii\right)$ and $\left(iii\right)$ are also equal.

Thus, $\frac{4{\rho }_1{L}_1}{\pi D_1^2}=\frac{4{\rho }_2{L}_2}{\pi D_{{}^2}^2}$

  $\frac{{\rho }_1{L}_1}{D_1^2}=\frac{{\rho }_2{L}_2}{D_{{}^2}^2}$

According to the question, two aluminium wires have the same resistance, they will have same resistivity; ${\rho }_1={\rho }_2$

Thus, $\frac{L_1}{D_1^2}=\frac{L_2}{D_{{}^2}^2}$

  $\frac{2L_2}{D_1^2}=\frac{L_2}{D_{{}^2}^2}\left[\because L_1=2L_2\right]$

  $\frac{2}{D_1^2}=\frac{1}{D_{{}^2}^2}\left[\because L_{2}willcancel\right]$

  $\frac{D_1^2}{D_{{}^2}^2}=2$

  ${\left(\frac{D_1^{}}{D_{{}^2}^{}}\right)}^2=2$

  $\frac{D_1^{}}{D_{{}^2}^{}}=\sqrt{2}$

Conclusion: The ratio of the diameter of the longer wire to the diameter of the shorter wire is $\frac{D_1^{}}{D_{{}^2}^{}}=\sqrt{2}$ .