Correct Answer - A
Charge on capacitor when it is connected with battery `q = CV`
when it is connected with another uncharged capacitor.
Common potential difference across capacitors
`V_(C) = (q_(1)+q_(2))/(C_(1)+C_(2)) = (q + 0)/(C+C) = (q)/(2C) rArr V_(C) = (V)/(2)`
Initial energy, `U_(1) = (1)/(2)CV^(2)`
Final energy, `U_(f) = (1)/(2) [2C] [(V)/(2)]^(2) = (1)/(2) U_(1)`
Hence total electrostatic energy of resulting system is decreased by a factor of `2`.