What is the value of resistance to double the value of number of turns of coil for a galvanometer? Why?

(N/A) The torque on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin \theta$. The restoring torque is $\tau_r = k \phi$. In equilibrium,$N I A B = k \phi$,which implies $\phi = (\frac{NAB}{k}) I$. The current sensitivity is $S_i = \frac{\phi}{I} = \frac{NAB}{k}$. If the number of turns $N$ is doubled to $2N$,the length of the wire $l$ also doubles,as $l = N \times (2\pi r)$. Since resistance $R = \rho \frac{l}{A_{wire}}$,doubling the length $l$ results in the resistance $R$ becoming $2R$. Therefore,to double the number of turns,the resistance of the coil becomes $2$ times the original resistance.