Two resistances of $400 \Omega$ and $800 \Omega$ are connected in series with $6 \mathrm{~V}$ battery of negligible internal resistance. A voltmeter of resistance $10000 \Omega$ is used to measure the potential difference across $400 \Omega$. The error in the measurement of potential difference in volts approximately is :

• A

  (a) 0.01
• B

  0.02
• C

  0.03
• Correct answer
  D

  0.05

Solution

$R_1=400 \Omega, R_2=800 \Omega$

$P D$ across $400 \Omega$ resistance (when voltmeter is not connected) $V_1=\frac{6}{(400+800)} \times 400$ $=\frac{6 \times 400}{1200}=2 \mathrm{~V}$ when voltmeter is connected Total resistance of the circuit, $R=\left(\frac{10000 \times 400}{10000+400}\right)+800$ $=\frac{10000 \times 400}{10400}+800$ $=\frac{40000}{104}+800$ $=\frac{40000+83200}{104}$ $=\frac{123200}{104}=\frac{30800}{26}=\frac{15400}{13}$ New $P D$ across $400 \Omega$ resistance, $V_2=6-800\left(\frac{6}{15400 / 13}\right)$ $=6-\frac{800 \times 13 \times 6}{15400}$ $=6-4.052=1.95 \mathrm{~V}$ $\therefore \quad$ Error $=V_1-V_2$ $=2-1.95$ $=0.05$

Hence, the correct answer is option: 'D'.