Two resistances of 400 Ω and 800 Ω are connected in series with a 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

Before connecting voltmeter, potential difference across 400  Ω resistance isVi =  400400 +800 × 6=2VAfter connecting voltmeter, equivalent resistancebetween A and B 400 × 10000400 +10000= 384.6 ΩHence, potential difference measured by voltmeterVf =  384.6384.6 +800× 6=1.95 VError in measurement =Vi − Vf = 2−1.95=0.05  V