When a galvanometer is shunted with a $4\,\Omega$ resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with $2\,\Omega$ wire, the further reduction in the deflection will be (the main current remains same)
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When only $4 \Omega$ resistance is shunted $\left(i_{g}\right)_{1}=i / 5$
$G \times 1 / 5=4 \times(4 / 5)=16 \Omega$
$\left(i_{g}\right)_{2}=\frac{4 / 3}{16+(4 / 3)} i=\frac{1}{13} i$
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