In the left gap of a meter bridge there is a coil of copper and in the right gap there is fixed resistance . If the coil of copper is dipped in ice the balance point is obtained at 41.2 cm of the bridge wire . Next the coil is taken of form ice and placed in a container of hoy water . Now the balcnce point is shifted by a distance 8.1 cm towards right . What is the temperature of hot water ? (Tempertaure cofficient of resistance of copper =`42.5xx10^(-4@)C^(-1)` . )

Explanation

Suppose , fixed resistance = R , temperature coefficient of resistance of copper `=alpha` , resistance of the coil at `0^(@)C=R_(0)` , position of the balance point `=l_(0)` , resistance of the coil at `t^(@)C=R_(t)` and position of the balance point = l .
`therefore" " R_(t)=R_(0)(1+alphat)` according to the principle of the metre brige ,
In case of `0^(@)C,(R_(0))/(R)=(L_(0))/(100-l_(0))" " ..(1)`
In casr of `t^(@)C,(R_(0)(1+alphat))/(R)=(l)/(100l)" " ...(2)`
Dividing (2) by (1) we have ,
`l+alphat=(l)/(l_(0))xx(100-l_(0))/(100-l)`
`=(79.3)/(41.3)xx(100-41.2)/(100-49.3)[l=41.2+8.1=49.3]`
`=(49.3)/(41.2)xx(58.8)/(50.7)=1.388`
`therefore" "t=(1.388-1)/alpha=(0.388)/(42.5xx10^(-4))=91.3^(@)C`