Question

The resistance of a wire, of length 'l', is RΩ. The wire is stretched to double its length keeping the volume constant. The resistance of the wire will become?

• A. 4 R Ω
• B. 2 R Ω
• C. $\frac{R}{2}$ Ω
• D. $\frac{R}{4}$ Ω

Answer 1

The correct option is A

4 R Ω

Here Volume = $length\times Area$ is constant. So if the wire is stretched to increase its length, its area will decrease proportionally i.e., if length is doubled then its area will be halved.
when length gets doubled l = 2l
area a = a/2
Therefore according to the formula, $R=\rho \frac{l}{a}$
putting l = 2l and a = a/2, new resistance R' will be
$R^{\prime }=\rho \frac{2l}{a/2}$
$R^{\prime }=\rho \frac{l2\times 2}{a}$
$R^{\prime }=4\left(\rho \frac{l}{a}\right)$
R' = 4(R)
which will make the new resistance R' four times the initial value R.