A 8 cm side cube weighing 4N is immersed in a liquid of relative density 0.8 contained in a rectangular tank of cross sectional area 12 cm ** 12 cm. If the tank contains liquid to a height of 6.4 cm before the immersion, determine the levels of the bottom of the cube and the liquid surface.

SOLUTION:

Given,

Side length of cube,`L=8 cm=0.08m`

Weight of  cube, `W=4 N`

Relative density of liquid ,`S=0.8`

Let,

`h_1` be the height to which the bottom of the cube falls below original liquid level

`h_2` be the height of rise of liquid above the original liquid surface

Thus, `h_1 + h_2` will be the depth of submergence of the cube. 

Now, from figure,

Volume L = volume M�?� (i)

Where,

Volume L = `8**8**h_1`

And volume .... Show More

SOLUTION:

Given,

Side length of cube,`L=8 cm=0.08m`

Weight of  cube, `W=4 N`

Relative density of liquid ,`S=0.8`

Let,

`h_1` be the height to which the bottom of the cube falls below original liquid level

`h_2` be the height of rise of liquid above the original liquid surface

Thus, `h_1 + h_2` will be the depth of submergence of the cube. 

Now, from figure,

Volume L = volume M�?� (i)

Where,

Volume L = `8**8**h_1`

And volume M = `12**12**h_2-8**8**h_2` (see figure)

Thus from equation (i),

or,`8**8**h_1=12**12**h_2-8**8**h_2`

or,`h_1=1.25h_2`... (ii)

Also, From the law of floatation,

Weight of cube = Buoyant force 

=weight of water displaced by immersed portion

or,`4=0.8**9.81**1000**8**8 (h_1+h_2)**10^(-6)`

or,`h_1+h_2=7.9638`... (iii)

Now, from equations (ii) and (iii),

`h_1=4.424 cm`

`h_2=3.5395 cm`

Thus the level of cube above BC is

`=6.4-4.424`

`=1.976 cm`

Also the level of liquid surface above plane BC is,

`=6.4+3.5395`

`=9.94 cm`