A coil of wire is kept in melting ice . Its resistance measured by a Wheatstone bridge is `5Omega` . If the coil is heated to `100^(@)C` and another wire of resistance `100Omega` is coonnected in parallel to it , the balanced condition of the brige remains unchaged . Determine the temperature coefficient of resistance of the wire of the coil .

Explanation

If the resistance of the coil at `100^(@)C` be rR then the equivalent resistance of the equivalent resistance of the parallel combination of r and `100Omrga ` is obviously `5Omega` .
`therefore" " (Rxx100)/(R+100)=5 " or, " 95R=500 " or, "500" or, "R=(100)/(19)Omega`
Again , if `alpha` be the temperture coefficient of resistance of the wire of coil , then
`R=R_(0)(1+alphat)`
`" or, "alpha=((R)/R_(0)-1)/(t)=((100)/(19xx5)-1)/(100)=(5)/(9500)`
`=5.26xx10^(-4@)C^(-1)`