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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

Answer»

`6.3xx10^(-4) K^(-1)`
`4.3xx10^(-4) K^(-1)`
`8.3 xx 10^(-4) K^(-1)`
`2.3xx10^(-4) K^(-1)`

SOLUTION :`X/R_0=1/(100-l)=40/60 , R_0 = 6 Omega`
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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