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waterdemon (3810)

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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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