In a metre bridge experiment , a null point is obtained at a length of 39.8 cm when a `2Omega` resistance is placed in the left gap and a `3Omega` resistance in the right gap . If the two resistance are interchanged , the null point is obtained at 60.8 cm . Calculated the end errors of the bridge .

Explanation

Suppose , resistance per unit lengh of the bridge wire `rhoOmega."cm"^(-1)` . End resistance of the left end of the bridge =resistance of `lamda_(1)` cm of the wire . End resistance of the right end of the bridge = resistance of `lamda_(2)` cm of hte wire . These two are the end errors of the bridge .
Therefore , if l be the position of the null point ,then according to the relation `(P)/(Q)=(R)/(S)` we have ,
`(R)/(S)=((l+lamda_(1))^(rho))/((100-l+lamda_(2))^(2))=(l+lamda_(1))/(100-l+lamda_(2))`
In the first case ,
`R=2Omega,S=3Omega and l = 39.8 "cm"`
`therefore(2)/(3)=(39.8+lamda_(1))/(60.2+lamda_(2))`
`"or, " 3lamda_(1)-2lamda_(2)=1` ...(1)
Again , in the second case ,
`R=3Omega,S=2Omega and l = 60.8 cm`
`therefore(3)/(2)=(60.8+lamda_(1))/(39.2+lamda_(2))` ...(2)
Solving (1) and (2) we get ,
`lamda_(1)=2.2 and lamda_(2)=2.8` So , the left end resistance and the right end resistanece of the bridge are equal to the resistances of 2.2 cm and 2.8 cm of the bridge wire repectively.