A galvanometer shows a reading of 0.65mA. When a galvanometer is shunted with a $4 Ω$ resistance, the deflection is reduced to 0.13mA. If the resistance of $2 Ω$ is further connected parallel to $4 Ω$ resistance (i.e galvanometer is further shunted with a $2 Ω$ wire) the new reading will be (Assume main current remains the same).

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0.08mA

0.05mA

0.60mA

0.12mA

answer is B.

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Detailed Solution

$\begin{array}{l} I^{'} = \frac{4}{4 + G} I \Rightarrow \frac{4}{4 + G} = \frac{1}{5} (g i v e n) w h e r e G = 16 Ω \\ I^{''} = \frac{4 / 3}{\frac{4}{3} + 16} I \\ I^{''} = \frac{1}{13} I \\ o r \\ I'' = \frac{1}{13} (5 I') = \frac{5}{3} I^{'} \\ \frac{i}{5} G = 4 X \frac{4}{5} i \Rightarrow G = 16 W \\ I n \sec o n d c a s e (i - i') X 16 = \frac{4}{3} i' \\ R_{e q} = \frac{2 X 4}{2 + 4} = \frac{4}{3} \\ 4 i - 4 i' = \frac{i^{'}}{3} \Rightarrow 4 i = \frac{13 i}{3} \Rightarrow i' = \frac{12}{13} i \\ \Rightarrow i - i^{'} = \frac{1}{13} i = \frac{1}{13} X 0.65 m A = 0.05 m A \end{array}$