Question

The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter, the change in the resistance of the wire will be

• A. 200%
• B. 100%
• C. 50%
• D. 300%

Answer 1

Answer: (D)

Let the length of the wire be l, radius of the wire be r
$∴$ Resistance $R = \rho \frac{l ⁡}{\pi r ⁡^{2}} \rho =$ resistivity of the wire
Now $l $ is increased by 100%
$∴ l ^{' ⁡} = l ⁡ + \frac{1 0 0}{1 0 0} l ⁡ = 2 l ⁡$
As length is increased, its radius is going to be decreased in such a way that the volume of the cylinder remains constant.
$\pi r ^{2} \times l ⁡ = \pi r ⁡^{' ⁡}^{2} \times l ⁡^{' ⁡} ⇒ r ⁡^{' ⁡ 2} = \frac{r ⁡^{2} \times l ⁡}{l ⁡^{' ⁡}} = \frac{r ⁡^{2} \times l ⁡}{2 l ⁡} = \frac{r ⁡^{2}}{2}$
$∴$ The new resistance $R ^{' ⁡ 2} = \rho \frac{l ⁡^{' ⁡}}{\pi r ⁡^{' ⁡ 2}} = \rho \frac{2 l ⁡}{\pi \times \frac{r ⁡^{2}}{2}} = 4 R ⁡$
$∴$ Change in resistance $= R ^{' ⁡} - R ⁡ = 3 R ⁡$
$∴$ % change $= \frac{3 R }{R ⁡} \times 1 0 0 \% ⁡ = 3 0 0 \%. ⁡$