A galvanometer of resistance \(25 \Omega\) is connected to a battery of \(2 \mathrm{~V}\) along with a resistance in series.
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A galvanometer of resistance \(25 \Omega\) is connected to a battery of \(2 \mathrm{~V}\) along with a resistance in series. When the value of this resistance is \(3000 \Omega\), a full scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 20 units, the resistance in series will be

(a) \(4513 \Omega\)

(b) \(5413 \Omega\)

(c) \(2000 \Omega\)

(d) \(6000 \Omega\)

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The correct option of this question will be (a).

Solution —

Current through galyanometer,

$$I_{k}=\frac{V}{R+G}=\frac{2}{3000+25}=k \times 30$$

where $k$ is figure of merit of galvanometer. The current corresponding 1020 units deflection of galvanometer:

$$I=\frac{I_{g}}{30} \times 20=\frac{2}{3} I_{g}=\frac{2}{3} \times \frac{2}{3025} \mathrm{~A}$$

If $R^{\prime}$ is the resistance to be used in series of galvanometer, then

$$I=\frac{V}{R^{\prime}+G} \text { or } \frac{2}{3} \times \frac{2}{3025}=\frac{2}{R^{\prime}+25}$$

On solving, we get

$$R^{\prime}=4512.5 \Omega \approx 4513 \Omega$$

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