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

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0.08mA

0.60mA

0.05mA

0.12mA

answer is D.

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Detailed Solution

according to shunt formula $S = \frac{G}{n - 1} [h e r e n = \frac{0.65}{0.13} = 5]$

$4 = \frac{G}{4} \Rightarrow G = 16$

next time $s = \frac{4 \times 2}{4 + 2} = \frac{4}{3}$

again S= $\frac{4}{3} = \frac{16}{n - 1} \Rightarrow \Rightarrow$ n= 13

$i = \frac{I}{13} = \frac{0.65}{13} = 0.05$

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