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So here is the answer Well it was a difficult one and I tried putting it up for The GOIITians Here = Angle between the line (of R inthe fig. ) and the horizontal '' b ''First The pressure that will be acting on the potion aprt from the door will be dP = P ( a - Rsin )As the length of the door will be (a - R Sin )dFx = Pcg(a - R Sin )R.d * Sin![]() dFy = Pcg (a- Rsin )Rd * Cos![]() Knowing that T = F*R So the below relations are as per the horizontal and vertical components And is the density of the liquid that is inside the Cube.dTFx = cg ( a -Rin ) d .Sin . (R - RCos )dTFy = cg (a - Rsin )d .Cos .(RSin )Now resolving along the line of resultant "F" dT = dTFx + dTFy Where T is the torque = cg(a - Rsin )Rd .Sin. (R - Rcos ) + cg(a -Rsin )Rd .Cos .Rsin![]() = cgR2 [ (a-Rsin )Sin .(1-Cos )d + (a-Rsin )Cos .Sin .d ]= cgR2 [ (aSin - RSin2 )(1 - Cos )d + (aCos - R.Sin .cos )(Sin )d![]() = cgR2 [ aSin - aSin .Cos - RSin2 + RSin2 .Cos + aCos .Sin - RCos .Sin2 ] d![]() = cgR2 [ aSin - RSin2 ] d![]() = cgR2 [ -a [cos ] { from 0 to pi/2 } - R(1 - Cos2 / 2 ) ] d![]() = cgR2 [ a - R ( d / 2 - Cos2 / 2 .d ]= cgR2 [a - R /4 - (Sin2 /4).d![]() Sin 2 /4 = 0 as it is from 0 to /2= cgR2 [a - R /4 ]Therefore we will get Where F is the force to be applied on point T to avoid leakage of liquid F*R = T F*R = cgR2[a - R /4]F = cgR[a - R /4]Hence the answer plZ rate me for my efforts Cheers !!!!!!!!!!!!!!!!!!!!!!!!!!
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= Angle between the line (of R inthe fig. ) and the horizontal '' b ''
is the density of the liquid that is inside the Cube.
/4 - (Sin2








