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(D) Let $R_1 = 400 \Omega$ and $R_2 = 800 \Omega$.$1$. Potential difference $(V_1)$ across $400 \Omega$ resistor without the voltmeter:$V_1 = \frac{R_

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$(d)$ $0.05$

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(D) Let $R_1 = 400 \Omega$ and $R_2 = 800 \Omega$.$1$. Potential difference $(V_1)$ across $400 \Omega$ resistor without the voltmeter:$V_1 = \frac{R_1}{R_1 + R_2} 	imes V = \frac{400}{400 + 800} 	imes 6 = \frac{400}{1200} 	imes 6 = 2 	ext{ V}$.$2$. Potential difference $(V_2)$ across $400 \Omega$ resistor with the voltmeter (resistance $R_v = 10000 \Omega$):The equivalent resistance of the parallel combination of $400 \Omega$ and $10000 \Omega$ is:$R_p = \frac{400 	imes 10000}{400 + 10000} = \frac{4000000}{10400} = \frac{40000}{104} \approx 384.62 \Omega$.The total resistance of the circuit is $R_{eq} = R_p + R_2 = 384.62 + 800 = 1184.62 \Omega$.The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{6}{1184.62} \approx 0.005065 	ext{ A}$.The potential difference measured by the voltmeter is $V_2 = I 	imes R_p = 0.005065 	imes 384.62 \approx 1.948 	ext{ V}$.$3$. The error in measurement is:$	ext{Error} = V_1 - V_2 = 2 - 1.948 = 0.052 	ext{ V}$.Rounding to the nearest given option,the error is approximately $0.05 	ext{ V}$.

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