When a resistance of $$5 \ \Omega$$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $$250$$ mA, however when $$1050 \ \Omega$$ resistance is connected with it in series, it gives full scale deflection for $$25$$ volt. The resistance of galvanometer is _____ $$\Omega$$.

We are given a moving coil galvanometer with two configurations: in one case it is used as an ammeter with a shunt resistance, and in the other as a voltmeter with a series resistance.

In the ammeter configuration a shunt resistance of $$S = 5 \; \Omega$$ is connected in parallel, and the full scale deflection corresponds to a total current of $$I = 250 \text{ mA} = 0.25 \text{ A}$$. In the voltmeter configuration a series resistance of $$R = 1050 \; \Omega$$ is connected, and the full scale deflection corresponds to a voltage of $$V = 25 \text{ V}$$.

Let the galvanometer resistance be $$G$$ and the full scale deflection current through the galvanometer be $$I_g$$.

From the voltmeter configuration we have the voltage across the galvanometer and series resistor equal to the applied voltage, so $$ I_g (G + 1050) = 25 $$ and hence $$ I_g = \frac{25}{G + 1050} \quad \cdots (1) $$.

From the ammeter (shunt) configuration the current through the shunt is $$(I - I_g)$$ and the voltage across the galvanometer equals the voltage across the shunt. Thus $$ I_g \cdot G = (I - I_g) \cdot S $$ which becomes $$ I_g \cdot G = (0.25 - I_g) \cdot 5 $$. This leads to $$ I_g (G + 5) = 1.25 $$ and therefore $$ I_g = \frac{1.25}{G + 5} \quad \cdots (2) $$.

Equating the expressions from (1) and (2) gives $$ \frac{25}{G + 1050} = \frac{1.25}{G + 5} $$. Multiplying through yields $$ 25(G + 5) = 1.25(G + 1050) $$, so $$ 25G + 125 = 1.25G + 1312.5 $$, leading to $$ 23.75G = 1187.5 $$. Solving yields $$ G = \frac{1187.5}{23.75} = 50 $$.

Therefore, the resistance of the galvanometer is 50 Ω.