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KCET · 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 resistances is

  1. A \(1: 3: 5\)
  2. B \(5: 3: 1\)
  3. C \(1: 15: 125\)
  4. D \(125: 15: 1\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(125: 15: 1\)

Step-by-step Solution

Detailed explanation

Given, \(\quad m_{1}: m_{2}: m_{3}=1: 3: 5\)
and \(\quad I_{1}: I_{2}: I_{3}=5: 3: 1\)
We know that, electrical resistance,
\(R=\rho \frac{l}{A}\)
and \(d=\frac{m}{V}=\frac{m}{A l}\)
or \(A=\frac{m}{d l}\)
\(\Rightarrow A \propto \frac{m}{l}\)
(\(\because\) density is same for the three wires)
\(\therefore R_{1}: R_{2}: R_{3}=\frac{l_{1}}{A_{1}}: \frac{I_{2}}{A_{2}}: \frac{I_{3}}{A_{3}}\)
or \(R_{1}: R_{2}: R_{3}=\frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{3}^{2}}{m_{3}}\)
\(=\frac{25}{1}: \frac{9}{3}: \frac{1}{5}\)
\(\Rightarrow R_{1}: R_{2}: R_{3}=125: 15: 1\)