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GUJCET · Physics · Current Electricity
Masses of three wires of copper are in the ratio of \(1: 3: 5\) and their lengths are in the ratio of \(5: 3: 1\). The ratio of their electrical resistance is _______.
- A \(5: 3: 1\)
- B \(\sqrt{125}: 15: 1\)
- C \(1: 15: 125\)
- D \(1: 3: 5\)
Answer & Solution
Correct Answer
(C) \(1: 15: 125\)
Step-by-step Solution
Detailed explanation
\(R =\frac{\rho l}{A}=\frac{\rho l^2}{A l}=\frac{\rho l^2}{V}\)
\({ll}\therefore R =\frac{\rho l^2}{m} d\left(\because \text { volume } V =\frac{m}{d}\right) \\ \therefore R \alpha \frac{l^2}{m} \\ \therefore R _1: R _2: R _3\)
\({l}=\frac{(1)^2}{5}: \frac{(3)^2}{3}: \frac{(5)^2}{1} \\ R_1: R_2: R_3=\frac{1}{5}: 3: 25 \\ R_1: R_2: R_3=1: 15: 125\)
\({ll}\therefore R =\frac{\rho l^2}{m} d\left(\because \text { volume } V =\frac{m}{d}\right) \\ \therefore R \alpha \frac{l^2}{m} \\ \therefore R _1: R _2: R _3\)
\({l}=\frac{(1)^2}{5}: \frac{(3)^2}{3}: \frac{(5)^2}{1} \\ R_1: R_2: R_3=\frac{1}{5}: 3: 25 \\ R_1: R_2: R_3=1: 15: 125\)
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