Question

:
A cube made of material having a density of
0.9x1000kg/m floats between water and a
liquid of density 0.7x1000 kg/m
which is immiscible with water. What part
of the cube is immersed in water -
​

Answer 1

Answer:

$b = \frac{2}{3}a$

Explanation:

since the cube floats between water and other material we can say that

weight of object = buoyancy force by water + buoyancy force by liquid

buoyancy force = weight of liquid displaced

weight =  volume * density * g

volume = Area * Height

⇒ Weight = Area * Height * density * g

we have to find the height in water

let cube is having side a and b is the length up-to which it is immersed in water.

so (a-b) is the length up to witch the cube is immersed in water.

buoyancy force by water

=  Area * Height * density * g

$= a^{2} * b * density_{water} * g \\= bga^{2}d_{water}$

buoyancy force by water

=  Area * Height * density * g

$= a^{2} * (a-b) * density_{liquid} * g \\= (a-b)ga^{2}d_{liquid}$

buoyancy force by water + buoyancy force by liquid

$= [bga^{2}d_{water}] + [a^{2} * g * (a-b) * density_{liquid}]$

$= a^{2}g[bd_{water} + (a-b)d_{liquid}]$

weight of object

Weight = Area * Height * density * g

$= a^{2} * a * density_{cube} * g \\= ga^{3}d_{cube}$

weight of object = buoyancy force by water + buoyancy force by liquid

$ga^{3}d_{cube}$ $= a^{2}g[bd_{water} + (a-b)d_{liquid}]$

$ad_{cube} = bd{water} + (a-b)d_{liquid}\\900a = 1000b + (a-b)700\\200a = 300b\\b = \frac{2}{3}a$