The seave of a galvanometer of resistance 100 Ω contains 25 divisions. It gives a deflection of 1 division on passing a current of 4 x 10-4 A. The resistance in ohm to be added to it, so that it may become a voltmeter of range 2 .5 V is

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100

150

250

300

answer is B.

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Detailed Solution

For full scale deflection , ig = n k∴ ig=25×4×10−4=10−2ANow R=Vig−G=2⋅510−2−100=250−100=150Ω

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