Two resistances of $400 Ω$ and $800 Ω$ are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance $10, 000 Ω$ is used to measure the potential difference across $400 Ω$ . The error in the measurement of potential difference in volts approximately is

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0.02

0.03

0.01

0.05

answer is D.

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Detailed Solution

Before connecting voltmeter potential difference across $400 Ω$ resistance is
$V_{i} = \frac{400}{(400 + 800)} \times 6 = 2 V$
After connecting voltmeter equivalent resistance between A and B
$= \frac{400 \times 10000}{(400 + 10000)} = 384.6 Ω$
Hence, potential difference measured by voltmeter
$V_{f} = \frac{384.6}{(384.6 + 800)} \times 6 = 1.95 V$
Error in measurement $= V_{i} - V_{f} = 2 - 1.95 = 0.05 V$