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Q.

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 (in Ω)

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answer is 4450.

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Detailed Solution

Total initial resistance $= G + R = 50 + 2950 = 3000 Ω$
Current, $i = \frac{3 V}{3000 W} = 1 \times 10^{- 3} A = 1 mA$
If the deflection has to be reduced to 20 divisions, then Current
Let x be the effective resistance of the circuit, then
$\begin{array}{l} 3 V = 3000 Ω \times 1 mA = xΩ \times \frac{2}{3} mA \\ x = 3000 \times \frac{3}{2} = 4500 Ω \end{array}$
Resistance to be added $= (4500 - 50) = 4450 Ω$

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