Question

A wire of resistance $R$ is stretched uniformly so that its length becomes $n$ times the original value. The new resistance of the wire is:

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

Answer 1

The correct option is A $n^{2}R$

Let $l$ be the original length of $A$.
Original area of cross-section, then
Original resistance $R=\frac{\rho l}{A}$
Now, length of the wire $=l^{\prime }=nl$
If $A^{\prime }$ be the new cross-sectional area, then $l^{\prime }{A}^{\prime }=lA$
$(\because$ volume of metal is a constant$)$.
$A^{\prime }=\frac{lA}{l^{\prime }}=\frac{lA}{nl}=\frac{A}{n}$
New resistance of the wire is
$R^{\prime }=\frac{\rho l^{\prime }}{Al}=\frac{\rho \left(nl\right)}{\frac{A}{n}}=\frac{\rho nl}{l}\times \frac{n}{A}=n^2\left(\frac{\rho l}{A}\right)=n^{2}R$