A galvanometer of resistance $50 Ω$ is connected to a battery of 3 V along with a resistance of $2950 Ω$ in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

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$6050 Ω$

$4450 Ω$

$5050 Ω$

$5550 Ω$

answer is D.

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Detailed Solution

Current through the galvanometer,

$I = \frac{3}{(50 + 2950)} = 10^{- 3} A$

Current for 30 divisions $= 10^{- 3} A$

Current for 20 divisions = $\frac{10^{- 3}}{30} \times 20 = \frac{2}{3} \times 10^{- 3} A$

For the same deflection to obtain for 20 division, let resistance added be R.

$\begin{array}{l} ∴ \frac{2}{3} \times 10^{- 3} = \frac{3}{(50 + R)} \\ or R = 4450 Ω \end{array}$

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