If only 1/50 of the main current is to be passed through a galvanometer of resistance G, then resistance of the shunt will be

The shunt is a resistor that is connected in parallel with the galvanometer to divert most of the current away from the galvanometer. Let R be the resistance of the shunt. Then, the current through the galvanometer is given by

$I_g=\frac{1}{50}{I}_m$

where $I_m$​ is the main current. The current through the shunt is,

$I_s=\frac{49}{50}{I}_m$

By Kirchhoff's current law, the total current through the circuit is

$I_m=I_g+I_s=\frac{1}{50}{I}_m+\frac{49}{50}{I}_m=I_m$

Therefore, the resistance of the shunt is given by

$R=\frac{V}{I_s}=\frac{V}{{\displaystyle \frac{49}{50}}{I}_m}=\frac{50V}{50I_m}$

where V is the voltage across the shunt. Since the voltage across the galvanometer and the shunt is the same, we have

$V=I_{g}G=\frac{1}{50}{I}_{m}G$

Substituting this into the expression for R, we get

$R=\frac{50C}{49I_m}=\frac{50}{49}\times \frac{1}{50}G=\frac{G}{49}$

Hence the correct answer is $\frac{G}{49}.$