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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) $6000 \Omega$

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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