Two long, smooth, horizontal conducting rails a distance l apart are connected at one end by a capacitor of capacitance C, and a uniform vertical…

The motional emf $Blv$ is balanced entirely by the capacitor voltage because the resistance is negligible, so $q=CBlv$. The current is therefore $i=dq/dt=CBl\,a$, where $a$ is the rod's acceleration. The magnetic force on the rod opposes its motion with magnitude $Bil=CB^2l^2a$. Newton's second law gives $F-CB^2l^2a=ma$, so $a=\dfrac{F}{m+B^2l^2C}$, a constant. Then $v=at$ and $q=CBlv=\dfrac{CBlFt}{m+B^2l^2C}$. The capacitor effectively adds an inertia $B^2l^2C$ to the mass. Option A is what one obtains by ignoring the magnetic braking force, and option C has the wrong sign for the added inertia.