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To find the resistance or a galvanometer by the half deflection method the following circuit is used with resistances R1=9970 Ω, R2 = 30 Ωand R3 = 0. The deflection in the galvanometer is d. With R3 =107 Ω the deflection changed to d2 The galvanometer resistance is approximately

This is answer

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Answer plz

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bhbnnnnn
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I'm just giving Hint.
Apply Kirchoffs second law in the loop containing R2 ,G,and R3 taking two cases.(First R2=0 and Second R2=107) Two equation will come and solve it simultaneously.(Take total current as same in both the cases). 
You will get the answer as G=77ohm.
Note that current in second case in the galvanometer will be half of current in galvanometer in first case.
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Yes this ans is right

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At initial condition, R3 = 0

Equivalent resistance of R2 & R3 = 30  Ω

When R3 =107 Ω & R2 = 30 Ω, then equivalent resistance should be 30/2

= 15 Ω

this will be when equivalent resistance , R(g) , R3 will be parallel to R2, we get resistance of 15 Ω

Let Rg - R3 = R(eq)

1/R2 + 1/R(eq) = 1/30 + 1/30 = 1/15

R(eq) = 30 Ω

∴ R3- Rg = 30

∴ 107 - Rg = 30

∴ Rg = 77 Ω
 
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