Answer:
The height of the oil on the side of the bar when the soap is floating in only the oil is 1.236 cm
Explanation:
The water level on the bar soap = 1.1 m mark
Therefore, the proportion of the bar soap that is under the water is given by the relation;
Volume of bar soap = LW1.7
Volume under water = LW1.1
Volume floating = LW0.6
The relative density of the bar soap = Density of bar soap/(Density of water)
= m/LW1.7/(m/LW1.1) = 1.1/1.7
Given that the oil density = 890 kg/m³
Relative density of the oil to water = Density of the oil/(Density of water)
Relative density of the oil to water = 890/1000 = 0.89
Therefore, relative density of the bar soap to the relative density of the oil = (1.1/1.7)/0.89
Relative density of the bar soap to the oil = (1.1/0.89/1.7) = 1.236/1.7
Given that the relative density of the bar soap to the oil = Density of bar soap/(Density of oil) = m/LW1.7/(m/LWX) = X/1.7 = 1.236/1.7
Where:
X = The height of the oil on the side of the bar when the soap is floating in only the oil
Therefore;
X = 1.236 cm.
