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What you need first, is to find the acceleration of the system.

yeah i have tried tat ..bt the ans is not right could u please give the eqs for few steps Vincent

What is your expression for the acceleration?

Using free body diagram we can find acceleration ,a=gsin(theta)-xgcos(theta)
Let the angle made by the surface with the incline be @. Now refer to the diagram in the file attached. We have considered the a water column of length L and area of cross section A.
Let the two ends of the cross section be at a depth of h1 &h2 as shown.
Pressure on the area at depth h1=p1=P+h1*d*gcos(theta) , where P is the atmospheric pressure
Similarly pressure at depth h2=p2=P+h2*d*gcos(theta)
Force at area A at depth h1=p1*A
Force at area A at depth h2=p2*A
Net force on the water column=F2-F1=(h2-h1)*d*gcos(theta)
F2=F1=m*a=L*A*d*(gsin(theta)-xgcos(theta))
Hence (h2-h1)*d*gcos(theta) = L*A*d*(gsin(theta)-xgcos(theta))
Rearranging and cancelling the like terms,
(h2-h1)/L=tan(theta)-x
Using trigonometry,
tan@=tan(theta)-x
****NOTE THAT tan(theta)>x [OTHERWISE CYLINDER WILL NOT SLIDE]. tan@ IS POSITIVE AND @ IS ACUTE ANGLE ******

@ Vincent a=gsintheta -ugcostheta

@ diwakar ..that seems to be right ...bt i got the answer from other method..

tan inverse (sin(theta) - xcos(theta))/cos(theta)