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To pass 2% of the main current through the galvanometer of 98Ω the shunt resistance required is

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2Ω

49Ω

98Ω

50Ω

answer is C.

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Detailed Solution

$\begin{array}{l} S = \frac{G}{n - 1}; (∵ \frac{i_{g}}{i} = \frac{2}{100} = \frac{1}{50} = \frac{1}{n}) \\ S = \frac{98}{50 - 1} = \frac{98}{49} = 2 Ω \end{array}$

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