A galvanometer, whose resistance is 50 ohm, has 25 division in it. When a current of $4 \times 10^{- 4} A$ passes through it, is its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range $2.5 V,$ it should be connected to a resistance of:

6250 o h m

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250 o h m

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200 o h m

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6200 o h m

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Solution

The correct option is C: $6200 o h m$
Given Data:
Resistance of galvanometer, $R_{g} = 50 o h m$
Current passing through the galvanometer $I_{g} =$ $4 \times 10^{- 4} A$
Suppose we connect a residstance of $R o h m$ in series
So, $2.5 = (R + R_{g}) \times$ $4 \times 10^{- 4} A$
$2.5 = (R + 50) \times$ $4 \times 10^{- 4} A$
$R + 50 = \frac{2.5}{4 \times 10^{- 4}} = 6200 o h m$

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