Question

aluminium wire of resistance R is pulled so that its length becomes 4 times its original length find the new resistance of the wire in terms of its original resistance

Answer 1

the resistance will decrease 1/4th of the original resistance

Answer 2

The new resistance of the aluminum wire after pulling it four times its original length is 16 times of old resistance of the wire.

Explanation:

$R=\frac{\rho L}{A} \rightarrow(1)$

Where R is the resistance, L is the length of the wire and A is the cross section area of the wire

So according to the given question, the aluminum wire is pulled to increase its length 4 times its original length.

So let the new length be,

$L^{\prime}=4 L \rightarrow(2)$

As when the wire is pulled, the length will increase but the volume of the wire will be constant. So there will be reduction in the wire’s cross sectional area when it is pulled. The new cross sectional area can be obtained using below equations.

$\text{Volume of wire}=\pi r^{2} L \rightarrow(3)$

Where $\pi r^2$ is the wire’s cross sectional area and L is the wire’s original length. As the wire’s volume tends to remain constant.

After pulling the wire to 4 times its original length, the new cross sectional area A’ is

$\text{Volume of original wire} = \text{Volume of pulled wire}$

$A \times L=A^{\prime} \times 4 L$

$A^{\prime}=\frac{A}{4} \rightarrow(4)$

There is no change in the resistivity of the wire of the wire even after pulling the wire, so it is denoted as ρ.

So the new resistance,

$R^{\prime}=\frac{\rho L^{\prime}}{A^{\prime}} \rightarrow(5)$

Substituting the equation (2) & (4) in equation (5), we get

$R^{\prime}=\frac{4 \rho L}{A} \times 4$

$R^{\prime}=\frac{16 \rho L}{A} \rightarrow(6)$

Substituting equation (1) in (6), we get

$\text{New resistance},R^{\prime}=16 R$