Question

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 t₁=0°C and t₂=100°C are 50 cm and 60 cm respectively. If the standard resistance taken out is S=4Ω in both trials, the temperature coefficient of the coil is

• A. 0.05°C^-1
• B. 0.02°C^-1
• C. 0.005°C^-1
• D. 2.0°C^-1

Answer 1

The Correct Option is C

Given: - The balance points at temperatures \( t_1 = 0^\circ C \) and \( t_2 = 100^\circ C \) are 50 cm and 60 cm, respectively. - Standard resistance taken out is \( S = 4 \, \Omega \) in both trials. Using the formula for the temperature coefficient of resistance: \[ \alpha = \frac{R_2 - R_1}{R_1 (t_2 - t_1)} \] Where: - \( R_1 = \frac{50}{100} \times S \) and \( R_2 = \frac{60}{100} \times S \). Now, calculating: \[ R_1 = \frac{50}{100} \times 4 = 2 \, \Omega \] \[ R_2 = \frac{60}{100} \times 4 = 2.4 \, \Omega \] Now, substituting into the formula: \[ \alpha = \frac{2.4 - 2}{2 \times 100} = \frac{0.4}{200} = 0.005 \, \Omega/^\circ C \] Thus, the temperature coefficient of resistance is \( 0.005^\circ C^{-1} \).

Answer 2

The temperature coefficient of resistance \( \alpha \) is given by: \[ \alpha = \frac{R_2 - R_1}{R_1 \cdot (T_2 - T_1)} \] Where: \( R_1 \) and \( R_2 \) are the resistances at temperatures \( t_1 = 0^\circ C \) and \( t_2 = 100^\circ C \), respectively. The resistance at \( t_1 = 0^\circ C \) is given by: \[ R_1 = \frac{S \cdot l_1}{100 - l_1} = \frac{4 \times 50}{100 - 50} = 4 \, \Omega \] The resistance at \( t_2 = 100^\circ C \) is given by: \[ R_2 = \frac{S \cdot l_2}{100 - l_2} = \frac{4 \times 60}{100 - 60} = 4.8 \, \Omega \] Thus, the temperature coefficient of resistance is: \[ \alpha = \frac{4.8 - 4}{4 \times (100 - 0)} = 0.005^\circ C^{-1} \]