KCET · Physics · Current Electricity
In an experiment to determine the temperature coefficient of resistance of a conductor, a coil of wire \(X\) is immersed in a liquid. It is heated by an external agent. A meter bridge set up is used to determine resistance of the coil \(X\) at different temperatures. The balancing points measured at temperatures \(t_1=0^{\circ} \mathrm{C}\) and \(t_2=100^{\circ} \mathrm{C}\) are 50 cm and 60 cm respectively. If the standard resistance taken out is \(S=4 \Omega\) in both trials, the temperature coefficient of the coil is
- A \(0.05^{\circ} \mathrm{C}^{-1}\)
- B \(0.02^{\circ} \mathrm{C}^{-1}\)
- C \(0.005^{\circ} \mathrm{C}^{-1}\)
- D \(2.0^{\circ} \mathrm{C}^{-1}\)
Answer & Solution
Correct Answer
(C) \(0.005^{\circ} \mathrm{C}^{-1}\)
Step-by-step Solution
Detailed explanation
Standard resistance, \(S=4 \Omega\)
At \(t_1=0^{\circ} \mathrm{C}, l_1=50 \mathrm{~cm}\)
\(\therefore R_1=\frac{l_1}{100-1} \times S=\frac{50}{100-50} \times 4=4 \Omega\)
At \(t_2=100^{\circ}, l_2=60 \mathrm{~cm}\)
\(\therefore R_2=\frac{l_2}{100-l_2} \times S=\frac{60}{100-60} \times 4=6 \Omega\)
\(\therefore\) Temperature coefficient of the coil,
\(\alpha=\frac{R_2-R_1}{R_1 t_2-R_2 t_1}=\frac{6-4}{4 \times 100-6 \times 0}=\frac{1}{200}\)
\(=0.005^{\circ} \mathrm{C}^{-1}\)
At \(t_1=0^{\circ} \mathrm{C}, l_1=50 \mathrm{~cm}\)
\(\therefore R_1=\frac{l_1}{100-1} \times S=\frac{50}{100-50} \times 4=4 \Omega\)
At \(t_2=100^{\circ}, l_2=60 \mathrm{~cm}\)
\(\therefore R_2=\frac{l_2}{100-l_2} \times S=\frac{60}{100-60} \times 4=6 \Omega\)
\(\therefore\) Temperature coefficient of the coil,
\(\alpha=\frac{R_2-R_1}{R_1 t_2-R_2 t_1}=\frac{6-4}{4 \times 100-6 \times 0}=\frac{1}{200}\)
\(=0.005^{\circ} \mathrm{C}^{-1}\)
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