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The scale of galvanometer of resistance 100Ω contains 25 divisions. It gives a deflection of one division on passing a current of 4×10^-4 ampere. The resistance, in ohms, to be added to it, so that it may become a voltmeter of range 2.5 volt, is

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150

100

250

300

answer is B.

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Detailed Solution

$\begin{array}{l} (∵ V = 2 .5 v, G = 100 Ω, i_{g} = 4 \times 10^{- 4} \times 25 A) \\ V = i_{g} (G + R), 2 .5 = 25 \times 10^{- 4} (100 + R) \\ ∴ R = 150 Ω \end{array}$

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