Fluids at rest exert a normal force to the walls of the container or to the sruface of the Body immersed in the fluid. The pressure exerted by this force at a point inside the liqid is the sum of atmospheric pressure and a factor which depends on the density of the liquid, the acceleration due to gravity and the height of the liquid, above that point. The upthrust acting on a body immersed in a stationary liquid is the net force acting on the body in the upward direction. A number of phenomenon of liquids in motion can be explain by Bernoulli's theorem which relates the pressure, flow speed and height for flow of an ideal incompressible fluid.
A container of large uniform corss sectional area. A resting on a horizontal surface holds two immiscible, non viscous and incompressile liquids of densities `d` and `2d`, each of height `H//2` as shown in the figure. The lower density liquid is open to the atmosphere having pressure `P_(0)`.
Situation I:
A homogeneous solid cylinder of length `L(LltH//2)`. cross sectional area `A//5` is immersed such that it floats with its axis vertical at liquid -liquid interface with lenght `L//4` in the denser liquid.
The density of the solid is

Options

A. `(5d)/4`
B. `d/4`
C. `(2d)/4`
D. `(3d)/4`

Answer

A

Explanation

Since the body is in equlibrium the net fore acting should be zero. i.e., the vector sum of forces should be zero.

The forces on the body are mg downwards, buoyant force due to denser liquid upwards and buoyant force due to lighter liquid in upward direction.
`F_(mg)=("Density" xx"volume")g=Dxx(AL)/5g`............i
`F_("buoyant")=d xx A/5 xx (3L)/4g+2d xx A/5xxL/4g`
`=((5d)/4xxA/5xxL)g=(AdL)/4g`...........ii
From eqn i and ii we have `D xx (AL)/5g=(AdL)/4g`
`impliesD=(5d)/4`