Chapter 18, Problem 68GP

To determine

The new resistance.

Answer to Problem 68GP

The answer is $9.00\Omega$ .

Explanation of Solution

Given:

Length, $L=3.00$ times of its original.

Actual resistance is $1.00\Omega$

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.

Calculation:

A $1.00\Omega$ wire is stretched uniformly to $3.00$ times its original length. So,

  $L=3L_0$

Also, the area will change. It will reduce by one-third factor. Now, area is $A=\frac{1}{3}{A}_0$

So, by using the formula of resistance:

  $R=\rho \frac{L}{A}=\rho \frac{3L_0}{\frac{1}{3}{A}_0}=9\rho \frac{L_0}{A_0}=9R_0=9\times 1=9.00\Omega$

Conclusion:

The resistivity is $9.00\Omega$