1.

In the circuit shown R_(1)=R_(2)=10Omega and resitance per unit length of wire PQ=1Omega//cm andlength PQ=10Omegacm. If R_(2) is made 20Omega then to get zero deflection in galvanometer S is midpoint of wire PQ

Answer»

The jockey at `P` can be moved towards right `2 cm`
The jockey at `Q` can be moved towards LEFT `2cm`
The jockey at `S` can be moved towards left a distance `5//3m`
The jockey at all position fixed and `R_(1)` should be made `20Omega`.

Solution :Equivalent DIAGRAM is shown is `P` is moved `2cm`right them `R_(1)=12,R_(3)=3(R_(1))/(R_(3))=(R_(2))/(R_(4))` (Hence wheat STONE will be BALANCED) if `s` is moved left `5/3` cm then `R_(3)=10/3` and `R_(4)=20/3` hence `(R_(1))/(R_(3))=(R_(2))/(R_(4))`
(hence Wheatstone will be balanced)


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