Oil having a density of 930 kg/m3 floats on water. Arectangular block of wood 4.00 cm high and with a density of 960kg/m3 floats partly in the oil and partly in the water. The
oil completely covers the block. How far below the interfacebetween the two liquids is the bottom of the block?

x=1.7 cm

Explanation:

We are given following data for block of wood in oil and water alloy:  

p_water = 1000 kg/m^3

    p_oil = 930 kg/m^3

p_wood = 960 kg/m^3

           h = 4 cm

Oil is above the water because the density of oil is lower than density of  

water. Density of wood is higher than oil density, but lower than water  

density. it means that it is positioned somewhere between the oil and water.  

Equilibrium equation:  

p_wood*g*h-p_oil*g*(h-x)-p_water*g*x=0

Solving it for x to see how far below the interface between the two liquids  

is the bottom of the block:  

x=(p_wood-p_oil)/(p_water-p_oil)*h

x=1.7 cm

b

explanation:

bc its just b

answer; the correct answer is technician a; ///100%good luck