Question

The resistance of a wire of length 250 m is 1 ohm if the resistivity of the material of wire is 1.6 into 10 to the power of minus 8 ohm metre find the area of cross section of the wire how much does the resistance change if the diameter is doubled

Answer 1

$l = 250 \: m \\ \\ \rho = 1.6 \times {10}^{ - 8} \ ohm \: m\\ \\ R = 1\ ohm\\ \\ A = ?\\ \\ R = \frac{ \rho l}{A}\\ \\ \Rightarrow A = \frac{ \rho l}{R}\\ \\ \Rightarrow A = \frac{ 1.6 \times 10^{-8} \times 250}{1}\\ \\ \Rightarrow A = 4 \times 10^{-6} \ m^2 = 4\ mm^2$

$\text{If diameter is doubled,} \\ \\ R = \frac{ \rho l}{A}\\ \\ \Rightarrow R \: \: \alpha \: \frac{1}{A} \\ \\ \Rightarrow R \: \: \alpha \: \frac{1}{ {d}^{2} } \\ \\ \text{If diameter is doubled,} \\ \text{ resistance will become one - fourth.}$

Answer 2

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$l = 250 \: m \\ \\ \rho = 1.6 \times {10}^{ - 8} \ ohm \: m\\ \\ R = 1\ ohm\\ \\ A = ?\\ \\ R = \frac{ \rho l}{A}\\ \\ \Rightarrow A = \frac{ \rho l}{R}\\ \\ \Rightarrow A = \frac{ 1.6 \times 10^{-8} \times 250}{1}\\ \\ \Rightarrow A = 4 \times 10^{-6} \ m^2 = 4\ mm^2$

$\text{If diameter is doubled,} \\ \\ R = \frac{ \rho l}{A}\\ \\ \Rightarrow R \: \: \alpha \: \frac{1}{A} \\ \\ \Rightarrow R \: \: \alpha \: \frac{1}{ {d}^{2} } \\ \\ \text{If diameter is doubled,} \\ \text{ resistance will become one - fourth.}$

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