In the given circuit, the galvanometer G shows zero deflection. The...

The given circuit is a Wheatstone bridge.

Since the galvanometer shows zero deflection, the Wheatstone bridge is in a balanced condition.

For a balanced Wheatstone bridge, the ratio of resistances in the adjacent arms is equal:

$\dfrac{R_{AB}}{R_{BC}} = \dfrac{R_{AD}}{R_{CD}}$

From the given circuit diagram:

$R_{AB} = 8\,\Omega$

$R_{BC} = 12\,\Omega$

$R_{CD} = 24\,\Omega$

The resistance in arm $AD$ is the series combination of $R$ and $6\,\Omega$:

$R_{AD} = R + 6$

Substituting these values into the balanced bridge condition:

$\dfrac{8}{12} = \dfrac{R + 6}{24}$

Simplifying the equation:

$\dfrac{2}{3} = \dfrac{R + 6}{24}$

$R + 6 = 24 \times \dfrac{2}{3}$

$R + 6 = 16$

$R = 10\,\Omega$

Answer: $10\,\Omega$