A uniform cube of side 10 cm weighs 880gxxg. It is floating in saline water (specific gravity of saline water = 1.1). What will be the thrust on each face of the cube?

A uniform cube of side 10 cm weighs 880gxxg. It is floating in saline

1.	A uniform cube of side 10 cm weighs 880gxxg. It is floating in saline water (specific gravity of saline water = 1.1). What will be the thrust on each face of the cube?
Answer» Solution :Density of the material of the cube = `880/10^(3)=0.88g*cm^(-3)`. Since the density is less than the density of saline WATER, some part of the cube remains above water. Let the lower SURFACE of the cube be at a depth of x cm from the upper surface of saline water. So, according to the CONDITION of floatation, weight of the cube = weight of the saline water DISPLACED by the cube or, `880xxg=x xx10xx10xx1.1xxgor,x=8cm` So, the area of the part of a side of the cube that remains immersed in the saline water, `A=8xx10=80cm^(2)` and the average depth that of the surface = `(0+x)/2=x/2`. Hence, the thrust on each lateral surface of the cube `=x/2xx1.1xxgxxA=8/2xx1.1xx981xx80` `=3.45xx10^(5)DYN`.