Question

A ductile metal wire has resistance R. Then the wire is stretched to three times its original length. The stretching shrinks the diameter of the wire, but does not affect its density or resistivity. What is the resistance of the stretched wire

Answer 1

Explanation:

Let us assume that the final resistance after stretching is $R_{f}$. Since, the volume of rod does not change and the initial and final volume will be the same as follows.

           V = $V_{f}$

        $LA = L_{f}A_{f}$

It is given that, $L_{f} = 3L$.

Therefore,      $LA = L_{f}A_{f}$

                      $LA = 3LA_{f}$

                   $A_{f} = \frac{A}{3}$

We know that relation between resistance and area is as follows.

          R = $\frac{\rho L}{A}$

    $R_{f} = \frac{\rho L_{f}}{A_{f}}$

             = $\frac{\rho (3L)}{\frac{A}{3}}$

             = 9 $(\frac{\rho L}{A})$  

             = 9R

Thus, we can conclude that the resistance of the stretched wire is 9R.

Answer 2

Answer:

Explanation:

initial length of the wire = l

initial area of crossetion of wire = A

initial resistance of the wire = R

Let the final resistance is r' area is A' and length is l' . Let  be the resistivity of the material of the wire.

l' = 3l

So, volume of the wire is constant.

l x A = l' x A'

l x A = 3l x A'

A' = A/3

The resistance of the wire is given by

$R=\rho \frac{l}{A}$    ... (1)

now the new resistance is R'

$R'=\rho \frac{l'}{A'}$

$R'=\rho \frac{3\times 3l}{A}$   .... (2)

divide equation (2) by equation (1)

R' = 9 R

Thus, the resistance of wire becomes 9 times.