Correct Answer - B
The discharging current in the circuit is
`i=i_(0)e^((-t)/(CR))` Here, `i_(0)=` initial current `=(V)/(R)`
Here, V is the potential with which capacitor was charged since, V and R for both te capacitors are same, initial discharging current will be sae but non-zero further `tau_(c)=CR`
`C_(1)ltC_(2)` or `tau_(C_(1))lttau_(C_(2))`
or `C_(1)` loses its `50%` of initial charge sooner than `C_(2)`