A wire has a resistance R If its length is doubled and radius is reduced to half then its resistance will become? (A) 2R (B) 4R(Cy8R D) 16R

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Answer 1 · 2018-06-13T23:00:01

$8R$ $"8R"$

Resistance of wire of circular cross section is given by

$R = \frac{ρL}{{πr}^{2}}$ $"R" = "ρL"/"πr"^2$

For the same type of wire

$R \propto \frac{L}{r^{2}}$ $"R" ∝ "L"/"r"^2$

From above

$\frac{R_{2}}{R_{1}} = \frac{L_{2}}{L_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2}$ $"R"_2/"R"_1 = "L"_2/"L"_1 × ("r"_1/"r"_2)^2$

$R_{2} = R_{1} \times \frac{L_{2}}{L_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2}$ $"R"_2 = "R"_1 × "L"_2/"L"_1 × ("r"_1/"r"_2)^2$

$color(white)("R"_2) = "R" × (2 cancel("L"_1))/(cancel("L"_1)) × ((cancel("r"_1))/(1/2 cancel("r"_1)))^2$

$color(white)("R"_2) = "R" × 2 × 4$

$color(white)("R"_2) = "8R"$