Fluid MechanicsHard
Question
A cubical block of wood of edge 10cm and mass 0.92 kg floats on a tank of water with oil of rel. density 0.6. Thickness of oil is 4cm above water. When the block attains equilibrium with four of its sides edges vertical :
Options
A.1 cm of it will be above the free surface of oil.
B.5 cm of it will be under water.
C.2 cm of it will be above the common surface of oil and water.
D.8 cm of it will be under water.
Solution
Let completely submerged in water, then
Fb = 1000 > mg(920) So, not possible
Let complete in oil
Fb = (0.6) (4) (1000 + (1) (6) (100) = 840
Fb < mg So, not possible
So, let ′X′ part in oil and remaining in water
920 = [(1) (10 − x) + (0.6) (x)] 100
9.2 = 10 − x + 0.6 x
0.4 x = 0.8
x = 2 cm.
Fb = 1000 > mg(920) So, not possible
Let complete in oil
Fb = (0.6) (4) (1000 + (1) (6) (100) = 840
Fb < mg So, not possible
So, let ′X′ part in oil and remaining in water
920 = [(1) (10 − x) + (0.6) (x)] 100
9.2 = 10 − x + 0.6 x
0.4 x = 0.8
x = 2 cm.
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