Find the P.D in the attached file.Please solve this problem as soon as possible.

The answer is (d) 12/11.

The technique is simple.

First calculate the total capacitance of the circuit which comes out to be 16/11 microfarad(uF). Now applying q=cv find the total charge flowing here it comes 192/11. Now divide the circuit into 2 components in series out of which one is 2 uF capcitor and the other one will be the combined of other all (=16/3 uF).again apply v=q/C and find potential drop across the combined one (=36/11).now this same potential drop is across both 4uF capacitor and {combination of 2uF and 4 uF capacitor} (because in parallel potential remains constant). Now find the Q flowing across combination of 2uF and 4uF capacitor by Q=Cv(=48/11) and NOW FINALLY You can find the potential drop across a and b by applying V=Q/C(C=4uF and hence answer is 12/11V).

REMEMBER THAT IN THESE TYPE OF QUESTIONS TO FIRST REDUCE THE CIRCUIT DIAGRAM IN STEPS AND REACH THE END TO FIND EITHER CHARGE OR POTENTIAL AS REQUIRED THROUGH NET CAPCITANCE AND GO REVERSE BACK TO THE FIRST  DIAGRAM SOLVING THE VALUES OF CHARGE AND POTENTIAL IN EACH CAPACITOR.