Doubt related to fluid mechanics question 
Question no. 281
My try:
The net force always passes through the centre of the cylinder, so net torque is always zero. But how can we use the fact that the net force is also zero. I am not getting any idea of calculating the force due to the liquids on the cylinder.
 A: The trick here is that for the cylinder to be in equilibrium,  the liquids on both sides of the cylinder are also in equilibrium. Construct vertical walls on the left and right sides of the left and right liquids respectively. 
Now, the force exerted by a wall on the corresponding liquid must be equal to the force exerted by the cylinder on the same liquid(but in opposite directions), to ensure the equilibrium of the liquid.  By Newton's third law, this is equal to the force exerted by the liquid in the cylinder. Thus we may conclude that the force exerted by the left liquid on the cylinder is equal to the force exerted by the left wall on the left liquid. The same applies to the right liquid and the right wall. 
Assume the width(going into the screen) of both walls to be w. The force by a wall on the corresponding liquid may be calculated by taking elemental areas (w)*(dy), multiplying it with the pressure at the depth of y and integrating it over the full height of the wall.
Let the densities be 2d and 3d.The force by the left wall on the left liquid is dgw*(h)^2. The force by the right wall on the right liquid is (3/2)dgw(R)^2
Equating them, h=R*sqrt(3/2)
Alternatively you could also go for the direct approach of taking elementary areas on the cylinder. But the calculation will be extremely tedious and time consuming. 
