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(C) $1$. Voltmeter conversion: $V = I_g(R_g + R_v) \implies 0.1 = 2 	imes 10^{-6} (10 + R_v) \implies 50000 = 10 + R_v \implies R_v = 49990 \Omega \a

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$2, 3$

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(C) $1$. Voltmeter conversion: $V = I_g(R_g + R_v) \implies 0.1 = 2 	imes 10^{-6} (10 + R_v) \implies 50000 = 10 + R_v \implies R_v = 49990 \Omega \approx 50 	ext{ k} \Omega$. Statement $(2)$ is correct.$2$. Ammeter conversion: $I_g R_g = (I - I_g) R_s \implies 2 	imes 10^{-6} 	imes 10 = (10^{-3} - 2 	imes 10^{-6}) R_s \implies 2 	imes 10^{-5} = 0.998 	imes 10^{-3} R_s \implies R_s \approx 0.02004 \Omega$. The total resistance of the ammeter is $R_A = \frac{R_g R_s}{R_g + R_s} \approx R_s \approx 0.02 \Omega$. Statement $(3)$ is correct.$3$. Measured resistance: The voltmeter is in parallel with $R = 1000 \Omega$. The equivalent resistance is $R_{eq} = \frac{R 	imes R_v}{R + R_v} = \frac{1000 	imes 50000}{1000 + 50000} = \frac{50000}{51} \approx 980.39 \Omega$. Statement $(1)$ is incorrect.$4$. Internal resistance: Adding internal resistance $r = 5 \Omega$ in series with the circuit reduces the total current $x$ flowing through the ammeter,but the measured value of $R$ is determined by the ratio of voltage across the voltmeter to the current through the resistor $R$. Since the voltmeter is in parallel with $R$,the measured resistance remains $R_{eq} \approx 980.39 \Omega$,which is independent of the internal resistance of the cell. Statement $(4)$ is incorrect.Thus,statements $(2)$ and $(3)$ are correct.

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