(b) [image:3_image175.emf]
Since deflection in galvanometer is zero so current will flow as shown in the above diagram
Current $I = \frac{V_{A}}{R_{1} + R} = \frac{12}{500 + 100} = \frac{12}{600}$
So $V_{B} = IR$
$= \frac{12}{600} \times 100 = 2V$