Question

A wire of resistance R is elongated n-fold to make a new uniform wire. The resistance of new wire

• A. $nR$
• B. $n^{2}R$
• C. $2nR$
• D. $2n^{2}R$

Answer 1

Answer: (B)

$n^{2}R$

We know that resistance $R=\rho \frac{l}{A}$, where $\rho$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the cross-sectional area of the wire.

The wire is elongated keeping the volume fixed.

$\text{Old Volume}=\text{New volume}$

$\Rightarrow l\times A=l^{\prime }\times A^{\prime }$

$\Rightarrow l\times A=nl\times A^{\prime }$

$\Rightarrow A^{\prime }=\frac{A}{n}$

Thus, the area is decreased by n times.

So, the altered resistance would be $R^{\prime }=\rho \frac{nl}{\frac{A}{n}}=n^2\rho \frac{l}{A}=n^{2}R$