The deflection in moving coil galvanometer falls from $$25$$ divisions to $$5$$ division when a shunt of $$24 \Omega$$ is applied. The resistance of galvanometer coil will be:

We need to find the resistance of a galvanometer coil, given that the deflection falls from 25 divisions to 5 divisions when a shunt of $$24 \, \Omega$$ is applied.

In a galvanometer, the deflection is directly proportional to the current passing through it:

$$\theta \propto I_g$$

So if the deflection drops from 25 to 5 divisions, the current through the galvanometer has reduced by a factor of 5:

$$\frac{I_g'}{I_g} = \frac{5}{25} = \frac{1}{5}$$

Without shunt: All the current flows through the galvanometer: $$I = I_g$$.

With shunt: The current divides between the galvanometer (resistance $$G$$) and the shunt (resistance $$S = 24 \, \Omega$$) in parallel. The total current $$I$$ remains the same (assuming the external circuit is unchanged), but now:

$$I_g' = I \times \frac{S}{G + S}$$

This is because in a parallel combination, the current divides inversely as the resistance.

$$\frac{I_g'}{I} = \frac{S}{G + S} = \frac{1}{5}$$

(Since $$I_g' = I_g/5$$ and $$I_g = I$$ without the shunt)

$$\frac{24}{G + 24} = \frac{1}{5}$$

$$5 \times 24 = G + 24$$

$$120 = G + 24$$

$$G = 96 \, \Omega$$

The correct answer is Option (2): $$96 \, \Omega$$.