The length of a give cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be……………..%.

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alok kumar singh | Contributor-Level 10

2 months ago

volume of wire will remains constant

$π r_{1}^{2} l_{1} = π r_{2}^{2} l_{2}$

$\Rightarrow l_{2} = 2 l_{1} (G i v e n)$

$r_{2} = \frac{r_{1}}{\sqrt[]{2}}$

$R_{2} = \frac{ρ l_{2}}{π r_{2}^{2}} = \frac{ρ 2 l_{1}}{π {(\frac{r_{1}}{\sqrt[]{2}})}^{2}}$

$= \frac{R_{2} - R_{1}}{R_{1}} \times 100 = \frac{4 R_{1} - R_{1}}{R_{1}} \times 100 = \frac{3 R_{1}}{R_{1}} \times 100 = 300 %$

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The length of a give cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be……………..%.

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