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A galvanometer of resistance 50 Omega is connected to a battery of 3 V along with a resistance of 2950 Omega in series shows full-scale deflection of 30 divisions. The additional series resistance required to reduce the deflection to 20 divisions is |
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Answer» `1500 Omega` `I=(3)/((50+2950))` `I=10^(-3)A` Current for 30 DIVISION `= 10^(-3)A` Current for 20 division `= (10^(-3))/(30)xx20-(2)/(3)xx10^(-3)A` Let the SERIES resistance = R `therefore (2)/(3)xx 10^(-3)=(3)/((50+R))` `R = 4450 Omega` Option (c ) is nearest option in the given options. 
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