The length of a given cylindrical wire is increased by 150%. Due to the consequent decrease in diameter the change in the resistance of the wire will be

200 %

525 %

300 %

400 %

Solution

525 %

If suppose initial length, l₁ = 100,

then the length of the cylindrical wire increased by 150%

l₂ = l₁ + 150

l₂ = 100 + 150

l₂ = 250

∴ $\frac{R_{1}}{R_{2}} = {(\frac{l_{1}}{l_{2}})}^{2}$

= ${(\frac{100}{250})}^{2}$

⇒ $\frac{R_{1}}{R_{2}} = \frac{4}{25}$

⇒ R₂ = $\frac{25}{4}$ R₁

⇒ $\frac{∆ R}{R} \times 100 = \frac{R_{2} - R_{1}}{R_{1}} \times 100$

= $\frac{\frac{25}{4} R_{1} - R_{2}}{R_{1}} \times 100$

⇒ $\frac{∆ R}{R}$ × 100 = 525 %

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