$F \Delta t=M V$

$i=\frac{E_{\text {induced }}}{R}$

$h=\frac{v^2}{2 g}$

The horizontanl component of magnetic field due to solenoid will exert force on ring in vertical direction.

$F=B_H i(2 \pi r)$

$F \Delta t=M V$

$i=\frac{(\Delta \phi / \Delta t)}{\left(\rho \frac{(2 \pi r)}{A}\right)}$

$B_H i(2 \pi r) \Delta t=M V$

$V=\frac{B_H \Delta \phi A}{\rho M}=\frac{K}{\rho M}$

$h=\frac{V^2}{2 g}=\frac{K^2}{\rho^2 M^2}$

$h_A > h_B$

$\frac{K^2}{\rho_A^2 M_A^2}>\frac{K^2}{\rho_B^2 M_B^2}$

$\Rightarrow \rho_B M_B>\rho_A M_A$ to Using this we get