In a galvanometer, a current of 1μA produces a deflection of 20 divisions. It has a resistance of 10 Ω. If the galvanometer has 50 divisions on its scale and a shunt of 2.5 Ω is connected across the galvanometer, the maximum current that the Galvanometer can measure now is

The maximum current that the galvanometer can measure with the shunt of 2.5Ω connected across it can be calculated as follows:
The resistance of the galvanometer is given as 10Ω and a current of 1μA produces a deflection of 20 divisions. Therefore, the full-scale deflection current of the galvanometer is:

$I_{fs}=\frac{20}{1\mu A}=20\times {10}^{6}A$

When the shunt of 2.5Ω is connected across the galvanometer, the total resistance of the circuit becomes:

$R_{total}=\frac{R_g{R}_s}{R_g+R_s}=\frac{10\times 2.5}{10+2.5}=2\Omega$

The maximum current that the galvanometer can measure with the shunt connected is given by:

$I_{max}=\frac{I_{fs}}{1+{\displaystyle \frac{R_g}{R_s}}}=\frac{20\times {10}^6}{1+{\displaystyle \frac{10}{2.5}}}=12.5\mu A$

Hence the correct answer is $12.5\mu A.$