KCET · Physics · Current Electricity
Masses of three wires of copper are in the ration \( 1: 3: 5 \) and their lengths are in the ratio \( 5: 3 \) : 1. The ratio of their electrical resistance are
- A 1:3:5
- B 5:3:1
- C 1:15:125
- D 125:15:1
Answer & Solution
Correct Answer
(D) 125:15:1
Step-by-step Solution
Detailed explanation
\( (C) \)
Resistance \( \mathrm{R}=\frac{\delta l}{A} \)
\(A=\frac{\text { volume }}{l}=\frac{\text { mass }}{l \times \text { Density }}=\frac{m}{l d} \)
\(R=\frac{s l}{\frac{m}{l d}}=\frac{s d l^{2}}{m} \)
\(R \propto \frac{l^{2}}{m}\)
\(R_{1}: R_{2}: R_{3}=\frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{3}^{2}}{m_{3}} \)
\(=\frac{5 l^{2}}{m}: \frac{3 l^{2}}{3 m}: \frac{l^{2}}{5 m} \)
\(=25: 3: \frac{1}{5} \ldots\left(\frac{l}{m} \text { is a constant }\right)=125: 15: 1\)
Resistance \( \mathrm{R}=\frac{\delta l}{A} \)
\(A=\frac{\text { volume }}{l}=\frac{\text { mass }}{l \times \text { Density }}=\frac{m}{l d} \)
\(R=\frac{s l}{\frac{m}{l d}}=\frac{s d l^{2}}{m} \)
\(R \propto \frac{l^{2}}{m}\)
\(R_{1}: R_{2}: R_{3}=\frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{3}^{2}}{m_{3}} \)
\(=\frac{5 l^{2}}{m}: \frac{3 l^{2}}{3 m}: \frac{l^{2}}{5 m} \)
\(=25: 3: \frac{1}{5} \ldots\left(\frac{l}{m} \text { is a constant }\right)=125: 15: 1\)
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