Interpret the Wheatstone bridge

Let the four arms be:

• $AB = 10\,\Omega$
• $BC = 15\,\Omega$
• $AD = 10\,\Omega$
• $DC = 15\,\Omega$

and the galvanometer of resistance $R_g = 15\,\Omega$ is connected across $BD$.

A potential difference of $V_{AC} = 10\,\text{V}$ is applied across $A$ and $C$.

Check if the bridge is balanced

For a Wheatstone bridge, balance condition is $\frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}}$

Substituting: $\frac{10}{15} = \frac{10}{15}$

This is true, so the bridge is balanced.

Consequence of balance

In a balanced Wheatstone bridge, the potentials at points $B$ and $D$ are equal. Therefore, $V_B = V_D$

Hence the potential difference across the galvanometer is $V_{BD} = 0$

Current through galvanometer

Using Ohm’s law, $I_g = \frac{V_{BD}}{R_g} = \frac{0}{15} = 0$

So the current through the galvanometer is $\boxed{0\,\text{A}}$

Compare with given options

None of the listed options matches $0$.

Comparison with stored correct answer

Stored correct answer is B: $4.87\,\text{mA}$.

But for the given resistances, the bridge is exactly balanced, so galvanometer current must be zero. Therefore the stored answer appears inconsistent with the stated data.