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(D) Given: Resistance of galvanometer $G = 50 \Omega$,full-scale deflection current $I_g = 5 	imes 10^{-4} 	ext{ A}$,and target voltage $V = 3 	ext

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$5950$

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(D) Given: Resistance of galvanometer $G = 50 \Omega$,full-scale deflection current $I_g = 5 	imes 10^{-4} 	ext{ A}$,and target voltage $V = 3 	ext{ V}$.To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with it.The formula for the total resistance is $V = I_g(R + G)$.Rearranging for $R$: $R = \frac{V}{I_g} - G$.Substituting the values: $R = \frac{3}{5 	imes 10^{-4}} - 50$.$R = 0.6 	imes 10^4 - 50 = 6000 - 50 = 5950 \Omega$.Thus,the required resistance is $5950 \Omega$.

$A$ galvanometer of resistance $50 \Omega$ gives a full-scale deflection for a current of $5 \times 10^{-4} \text{ A}$. The resistance that should be connected in series with the galvanometer to read $3 \text{ V}$ is: (in $\Omega$)

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