When a galvanometer is shunted with a 4Ω resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with is further shunted with a 2Ω wire (2Ωand4Ω resistors are connected in parallel), the further reduction in the deflection will be(the main current remains the same).
(813) of the deflection when shunted with 4Ω only
Case I : Rg×15=(I−15)×4⇒Rg=16Ω
Case II : 16l1=4×26(l−l1)⇒l1=l13
So decrease in current to previous current =l5−l13l5