Two long horizontal smooth conducting rails, 0.4 m apart, lie in a uniform vertical magnetic field B=0.5 T and are joined at one end by an ideal…

With negligible resistance the capacitor voltage equals the motional emf, $q=CBLv$. Hence $I=\dfrac{dq}{dt}=CBL\,a$. Newton's law gives $F-BIL=ma$, so $a=\dfrac{F}{m+B^2L^2C}=\dfrac{0.8}{0.2+0.25\times0.16\times5}=2\ \mathrm{m\,s^{-2}}$ constant, and $I=5\times0.5\times0.4\times2=2\ \mathrm{A}$ constant; (A) and (B) are true. At $t=3$ s, $v=6\ \mathrm{m\,s^{-1}}$ and $q=It=6\ \mathrm{C}$. The rod's kinetic energy is $\tfrac12(0.2)(6)^2=3.6\ \mathrm{J}$, so (D) is true. The capacitor energy is $q^2/2C=36/10=3.6\ \mathrm{J}$, not $7.2\ \mathrm{J}$, so (C) is false. The work done by $F$ is $F\cdot\tfrac12at^2=7.2\ \mathrm{J}$, which splits equally between kinetic and capacitor energy. Note that the capacitor behaves like an extra inertia $B^2L^2C=0.2\ \mathrm{kg}$ added to the rod, which is why the acceleration is smaller than $F/m=4\ \mathrm{m\,s^{-2}}$.