Question

A beaker is filled with two non-mixing liquids. The lower liquid has density twice that of the upper one. A cylinder of height h floats with 1/4th of its height submerged in the lower liquid and half of its height submerged in the upper liquid. Another beaker is filled with the denser of the two liquids alone. If the same cylinder is kept in the second beaker, the height of the submerged position would be

Answer 1

Given:

Liquids = 2

Density = Twice than upper

Height = h =1/4 submerged in lower liquid

Height = x = 1/2 submerged in upper liquid

To find:

Height of the submerged position would be

Solution:

Let the mass of cylinder be = m,

Area of the cross section be = a,

Density of floating liquid be = d

Thus,

mg = h/4a2dg + h/2adg = hadg -- 1

Similarly, in second case when it is submerged -

mg = xa2dg -- 2

From 1 and 2

xa2dg = hadg

x = h/2

Answer: Height of the submerged position will be h/2