As shown in the figure, a meter bridge consisting of two resistances `P` and `Q` together in parallel with a meter-long wire `AB` of uniform cross-section. `C` is a movable contact that can slide along the wire `AB`. The resistors `P, Q` and resistances of segments `AC` and `CB` of the wire constitude the four arms of bridge. the length `AB = 100 cm, P = 4 Omega, Q` is a coil of wire. With `Q` immersed in melting ice the null point is found to be at a distance of `40 cm` from `A`. When the coil `Q` is heated to `100^@C`, a `100 Omega` resistor has to be connected in parallel with `Q` in order to keep the bridge balanced at teh same point. Calculate the temperature coefficient of resistance of the coil.
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Explanation

`R_0` : resistance of `Q` in melting ice `(0^@C)`
`(P)/(Q) = (40)/(60) rArr (4)/(R_0) = (2)/(3) rArr R_0 = 6 Omega`
`R_(100)` : resistance of `Q` at `100^@C`
`R_(100)` and `100 Omega` are in parallel,
Equivalent resistance `= (R_(100) xx 100)/(R_(100) + 100) = R'`
`(P)/(R') = (40)/(60) rArr (4)/(R') = (2)/(3) rArr R' = 6 Omega`
`R' = (100R_(100))/(100 + R_(100))`
`6 = (100 R_(100))/(100 + R_(100)) rArr R_(100) = (300)/(47) = 6.38 Omega`
`R_(100) = R_0[ 1 + alpha(100 - 0)]`
`6.38 = 6[ 1+ 100 alpha]`
`alpha = 6.3 xx 10^-4//.^@C`.