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Figure shows a meter bridge,wire AC has uniform cross section. The length of wire AC is 100 cm X is a standard do resistor of 4 Omega and Y is a coil.When Y is immersed in melting ice, the null point is at 40 cm from point A. When the coil Y is heated to 100^@C, a 100 Omega resistor has to be connected in parallel with Y in order to keep the bridge balanced at the same point. Temperature coefficient of resistance of the coil is |
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Answer» `6.3xx10^(-4) K^(-1)` Since null point REMAINS unchanged `x/(R.)=40/60` `6=(100R_t)/(R_t+100) ,R_t = 6.38 Omega` `therefore alpha = (R_t- R_0)/(R_0 Deltat) = 6.3xx10^(-4) K^(-1)` |
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