A10-μf capacitor in an lc circuit made entirely of superconducting materials ( r = 0 ω ) is charged to 100 μc. then a superconducting switch is closed. at t = 0 s, plate 1 is positively charged and plate 2 is negatively charged. at a later time, vab = +10v . at that time, vdc is

Vdc=10V

Explanation:

in a closed loop consisting of a super charged capacitor and an inductor, the super charge capacitor acts as a supply when the loop is closed, at t=0, the emf stored in the capacitor is 10V (q/c); and at that same time Vl= voltage across the inductor or loop too would be 10V,

if the loop remains closed for  a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.

option number c is the right answer.