A galvanometer having coil resistance $$10 \Omega$$ shows a full scale deflection for a current of $$3$$ mA. For it to measure a current of $$8$$ A, the value of the shunt should be:

We need to find the shunt resistance to convert a galvanometer (coil resistance $$G = 10 \, \Omega$$, full-scale deflection current $$I_g = 3$$ mA) into an ammeter reading up to $$I = 8$$ A.

A shunt is a low resistance connected in parallel with the galvanometer. The voltage across both must be equal:

$$I_g \cdot G = (I - I_g) \cdot S$$

Solving for $$S$$:

$$S = \frac{I_g \cdot G}{I - I_g}$$

$$I_g = 3 \times 10^{-3}$$ A, $$G = 10 \, \Omega$$, $$I = 8$$ A:

$$S = \frac{3 \times 10^{-3} \times 10}{8 - 3 \times 10^{-3}} = \frac{0.03}{7.997}$$

$$S = \frac{0.03}{7.997} \approx 3.75 \times 10^{-3} \, \Omega$$

Note: Since $$I_g \ll I$$, we can approximate $$I - I_g \approx I = 8$$, giving $$S \approx 0.03/8 = 3.75 \times 10^{-3} \, \Omega$$, which confirms our answer.

The correct answer is Option (3): $$3.75 \times 10^{-3} \, \Omega$$.