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Question

In an experiment to determine the resistance of a galvanometer by half deflection method, the circuit shown is used. In one set of readings, if R = 10Ω and S =4Ω, then the resistance of the galvanometer is
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A
203Ω
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B
403Ω
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C
503Ω
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D
103Ω
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Solution

The correct option is A 203Ω
When K1 is closed, the current through the galvanometer is IG=ER+G.......(1)
If θ is the deflection shown in galvanometer due to flow of current IG, then IGθ or IG=kθ where k is known as figure of merit.
so, using (1) ER+G=kθ.....(2)
When K2 is also closed, the current through galvanometer is IG=SG+SI where I= current flow in the main circuit.
Now the value of S is adjusted such that the deflection θ will be half i.e θ2
Thus, IG=kθ2
or SG+SI=kθ2........(3)
Total resistance of the circuit is Rt=R+SGS+G
Thus, I=ERt=ER+SGS+G....(4)
(3)/(2)SI(R+G)(G+S)E=12
now substituting the value of I then after solving we get the resistance of galvanometer is G=RSRS
or G=10×4104=203Ω

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