### Solution:

Concept: We are given the resistance of the lamp and the resistance of the conductor. They both are connected in series and so we can find the equivalent resistance using the series combination of the resistance. They are connected to an external source of EMF. The total current in the circuit may be determined by using Ohm's law to determine the individual voltage and current between the bulb and the conductor.Resistance of lamp R1=10Ω

Resistance of the conductor R2=2Ω

They both are connected in series, so equivalent resistance $R=10+2=12\Omega $

Now potential difference of the battery is 6 V. using Ohm’s law $V=\mathit{Ir}$

$i=\frac{V}{R}\mathrm{}\Rightarrow i=\frac{6}{12}\mathrm{}\therefore i=0.5A$

Since the current in a series combination is constant, the current flowing through the bulb and conductor is 0.5 A. We now use Ohm's law for lamps to account for any potential differences throughout the lamp:we apply Ohm’s law for lamp:

$V=\mathit{iR}\mathrm{}\Rightarrow V=0.5\times 10\mathrm{}\therefore V=5V$ As a result, the circuit has a current of 0.5 a and a potential difference of 5 V across the light.

Hence, the correct answer option (2) .