Question

Oil having a density of 923 kg/m^3 floats on water. A rectangular block of wood 4.81 cm high and with a density of 961 kg/m^3 floats partly in the oil and partly in the water. The oil completely covers the block.
(a) How far below the interface between the two liquids is the bottom of the block?

Answer 1

Answer:

x = 2.37 cm

Explanation:

given,

density of oil = 923 kg/m³

height of the block = 4.81 cm

density of the block = 961 Kg/m³

density of water = 1000 Kg/m³

Let x be the depth of the block in the water.

 now,

Weight of block will be acting downward.

Buoyancy force will be acting on the block because of the oil and water.

At equilibrium position both weight and the buoyant force will balance.

 $\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx= 0$

 $961\times 0.0481 - 923\times 0.0481 = -923\times x +1000\times x$

 $1.8278 = 77 x$

   x = 0.0237 m

  x = 2.37 cm

Hence, the block is 2.37 cm in the water.