When a galvanometer is shunted with a 2\ resistance, the deflection...

Let the resistance of the galvanometer be $G$ and the main current be $I$.

When the galvanometer is shunted with $S_1 = 2\ \Omega$, the deflection is reduced to one-fourth. This means the current through the galvanometer is $I_{g1} = \dfrac{I}{4}$.

Using the formula for current division, $I_{g1} = \dfrac{S_1}{G + S_1} I$

$\dfrac{I}{4} = \dfrac{2}{G + 2} I \Rightarrow G + 2 = 8 \Rightarrow G = 6\ \Omega$

When the galvanometer is further shunted with a $1\ \Omega$ resistance, this new resistor is connected in parallel with the existing $2\ \Omega$ shunt. The new equivalent shunt resistance $S_2$ is:

$S_2 = \dfrac{2 \times 1}{2 + 1} = \dfrac{2}{3}\ \Omega$

The new current through the galvanometer $I_{g2}$ is:

$I_{g2} = \dfrac{S_2}{G + S_2} I = \dfrac{\dfrac{2}{3}}{6 + \dfrac{2}{3}} I = \dfrac{\dfrac{2}{3}}{\dfrac{20}{3}} I = \dfrac{I}{10}$

The further reduction in deflection is proportional to the decrease in the galvanometer current:

$\Delta I_g = I_{g1} - I_{g2} = \dfrac{I}{4} - \dfrac{I}{10} = \dfrac{5I - 2I}{20} = \dfrac{3I}{20}$

This reduction is given as a fraction $x$ of the deflection observed when shunted with $2\ \Omega$ ($I_{g1}$).

$x = \dfrac{\Delta I_g}{I_{g1}} = \dfrac{\dfrac{3I}{20}}{\dfrac{I}{4}} = \dfrac{3}{20} \times 4 = \dfrac{3}{5}$

Answer: $\dfrac{3}{5}$