A metallic cube 30 cm side and weighing 450 N is lowered into a tank containing a two layer fluid of water and mercury. Determine the position of block at mercury-water interface at equilibrium condition. [ANNA UNIVERSITY ]

SOLUTION:

Given,

Side length of cube,`L=30 cm=0.3m`

Weight of metallic cube, `W=450 N`

Now, From the  principle of floatation,

Weight of cube = Buoyant force

=weight of water and mercury displaced by the block

or,`W=V_w*gamma_w+V_m*gamma_m`

or,`450=(0.3**0.3**h_1)**9810+13600**9.81**(0.3**0.3**h_2)`

or,`450=0.3**0.3**9.81**1000 (h_1+13.6h_2)`

or,`h_1+13.6h_2=0.5097`.... (i)

Again,from figure,

`h_1+h_2=0.3`.... (ii)

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SOLUTION:

Given,

Side length of cube,`L=30 cm=0.3m`

Weight of metallic cube, `W=450 N`

Now, From the  principle of floatation,

Weight of cube = Buoyant force

=weight of water and mercury displaced by the block

or,`W=V_w*gamma_w+V_m*gamma_m`

or,`450=(0.3**0.3**h_1)**9810+13600**9.81**(0.3**0.3**h_2)`

or,`450=0.3**0.3**9.81**1000 (h_1+13.6h_2)`

or,`h_1+13.6h_2=0.5097`.... (i)

Again,from figure,

`h_1+h_2=0.3`.... (ii)

Solving equations (i) and (ii), we get,

`h_1=0.283 m`

`h_2=0.01658 m`

Thus,the block is 0.01658 m below the interface.