JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 26)
In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k$$\Omega $$ are used. The figure of merit of the galvanometer is 60 $$\mu A/$$division. In the absence of shunt resistance, the galvanometer produces a deflection of $$\theta $$ = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of $$\theta /2,$$ is closest to :
500 $$\Omega $$
220 $$\Omega $$
55 $$\Omega $$
110 $$\Omega $$
Explanation
Current required by unit deflection is 60 $$\mu$$A.
For, $$\theta$$ = 9 current is I = 9 $$\times$$ 60 $$\mu$$A
$$\Rightarrow$$ I = 540 $$\mu$$A = 540 $$\times$$ 10$$-$$6 A
Let G is resistance of galvanometer. Then,
$$540 \times {10^{ - 6}} = {6 \over {(11000 + G)}}$$
[11000 + G] 90 $$\times$$ 10$$-$$6 = 1
99000 + 9G = 105
9G = 100000 $$-$$ 99000
9G = 1000
$$G = {{1000} \over 9}\Omega $$
Also, in half deflection method,
$$G = {{RS} \over {R - S}} \Rightarrow {{1000} \over 9} = {{11000\,S} \over {11000 - S}}$$
$${1 \over 9} = {{11S} \over {11000 - S}} \Rightarrow 11000 - S = 99\,S$$
100 S = 11000 $$\Rightarrow$$ S = 110 $$\Omega$$
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