Question

Potential gradient along AB is

• A. 1/5 Vm${}^{-1}$
• B. 2/5 Vm${}^{-1}$
• C. 3/5 Vm${}^{-1}$
• D. 4/5 Vm${}^{-1}$

Answer 1

The correct option is B 2/5 Vm${}^{-1}$

Since 2 V is balanced across a length of 500 cm $=$ 5m, so the potential gradient $k=2/5V{m}^{-1}$

Reading of voltmeter $=$ potential difference across voltmeter

$=V_A-V_D=k\left(4.9\right)$

Reading of the voltmeter $=\frac{2}{5}\times 4.9=1.96V$

Terminal potential difference of the battery will also be 1.96 V. So

$1.96=2-i\times 10$ or $i=4\times {10}^{-3}A$

So, resistance of the voltmeter is$R_V=1.96/(4\times {10}^{-3})=490\omega$

$R_V=1.96/(4\times {10}^{-3})=490\omega$