Question

A block of wood floats partially submerged in water. Oil is poured on top of the water until the block of wood is totally submerged, partially in water and partially in oil as shown in the figure. A clear cup is partially filled with water, and has oil floating on top of the water, almost filling the vessel. Completely submerged in the fluids is a block. The block floats so that part of it is in the water and part of it is in the oil. If 86% of the wood is submerged in water before the oil is added, determine the percent submerged when oil with a density of 853 kg/m3 covers the block. Be sure to account for the buoyant force of air before the oil is added. (Use 1000 kg/m3 as the density of water and 1.29 kg/m3 as the density of air.)

I cant get the picture of the figure on here. But, There are no numbers or key info to be used from the figure

Answer 1

Density of oil o = 853 kg/m³

density of water _w = 1000 kg/m³

density of air a = 1.29kg/m³

density of block = 86% of density of water = 860 kg/m³

Let V1 and V2 be the volumes of the block in oi and in water .

so , volume pf block V = V1 + V2

Weight of block = up thrust on the block due to two liquids,

*V*g = V1*o*g + V2*_w*g

*V*g = (V - V2)*o*g + V2*_w*g

*V = (V - V2)*o +  V2*_w

V2 = ( - o/_w-o)*V

Therefore, the fraction submerged when oil covers the block is given by

V2/V =  ( - o/_w-o)

V2/V = (860-853/1000-853)

V2/V = 0.0476

V2/V = 4.67 %