A capacitor of capacitance  C=3 μF is first charged by connecting it across a  10V battery by closing key K1. It is then allowed to get discharged through  2 Ω and  4 Ω resistor by closing the key K2 and opening key K1 .The total energy dissipated in the  2 Ω resistor is equal to

Before K1 opened the energy stored over capacity  C1=12CV2=12×3×10−6×102=150 μJNow this energy will be dissipated  E=i2Rt in the two resistors in  the ratio of R1:R2 as they are in series (current flowing through the resistors is same) when switch K2 is closed.∴150 μJ is lost at the ratio  2:4=1:2In 2 Ω , 13rd of 150 μJ = 50 μJ and In 4 Ω ,23rd of 150 μJ=100 μJ will be lost.Therefore, the correct answer is ( B )