# A U-Tube of base length "l" filled with same volume of two liquids of densities t & 2t is moving with an acceleration "a" on the horizontal?

plane. If the height difference between the two surfaces(open atmosphere) becomes zero, then the height his given by -

(A) (a/2g)l

(B) (3a/2g)l

(C) (a/g)l

(D) (2a/3g)l Relevance

The pressures in the liquids at the bottom of the LHS and RHS columns are Po+2tgh and Po+tgh respectively, where Po is atmospheric pressure. (The pressure must be greater at the LHS, because a net force to the right is needed to accelerate the liquid in the horizontal part of the tube.)

The masses of liquids in the LH and RH parts of the horizontal section are 2t(L/2)A and t(L/2)A, where A is the cross-section area of the tube. (It is not stated, but I presume that the interface between the two liquids is exactly at the midpoint of the horizontal section.)

The difference in pressure between the LH and RH ends of the horizontal section is tgh. This must be sufficient to cause the acceleration of the mass of liquid in between. Apply F=ma :

tghA = [2t(L/2)A + t(L/2)A]a

gh = 3(L/2)a

h = (3a/2g)L.