The sensitivity of a galvanometer that measures current is decreased by 1/40 times by using shunt resistance of 10Ω. Then, the value of the resistance of the galvanometer is

Given,

$\frac{S_1}{S_2}=40$

Sensitivity of the galvanometer is given by $S=\frac{\Phi }{I}$, where $\Phi$ is the magnetic flux inside the coil and I is the current flowing through the coil.

Also,

$\varphi \propto V$ here V is the voltage across coil,

$\varphi =kV$

Thus, $S_1=\frac{\varphi }{I_1}=\frac{kV}{I_1}=k{R}_G$

$S_2=\frac{\varphi }{I_2}=\frac{kV}{I_2}=k\left(R_G\right|\left|R_{sh}\right)$

Thus, $\frac{R_G}{R_G\left|\right|R_{sh}}=40$

$R_G=R_{sh}\times \left(40-1\right)=390\Omega$

Hence the correct answer is $390\Omega .$