A galvanometer with 50 divisions on the scale has a resistance of 25Ω. A current of $2 \times 10^{- 4} A$ gives a deflection of one scale division. To convert it into a voltmeter reading upto 25 volt, the additional series resistance required is of

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$\frac{25}{99} Ω$

25Ω

25 x 99Ω

25 x100Ω

answer is B.

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Detailed Solution

$\begin{array}{l} V = i_{g} (G + R) (\begin{matrix} ∵ i_{g} = 50 \times 2 \times 10^{- 4} = 10^{- 2} A \\ V = 25 v, G = 25 Ω \end{matrix}) \\ ∴ R = 25 \times 99 Ω \end{array}$

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