
We are given a galvanometer connected in the first circuit in the above figure, where E=6V, R=11kΩ and the deflection θ=9 divisions. The galvanometer has a resistance of G. The current in the circuit,
I=R+GE...(1)
Also, since the figure of merit is 60μA/division
I=60×9μA=540μA...(2)
Substituting equation (2) in (1),
540×10−6=11,000+G6⇒G=540×10−66−11,000⇒G=111Ω...(3)
In the second circuit, it is given that the deflection is half that of the previous circuit. So, current flowing through the galvanometer half the first circuit. Hence,
2I=R+G+SGSE×S+GS⇒2I=R(S+G)+GSES
⇒S=E−2(R+G)IRG×2I
Substituting the values from equations (1),(2) and (3),
S=6−(26)11×103×111×2540×10−6=110Ω