Question

A 1.7 cm thick bar of soap is floating in water, with 1.1 cm of the bar underwater. Bath oil with a density of 890.0 kg/m{eq}^3 {/eq} is added and floats on top of the water. How high on the side of the bar will the oil reach when the soap is floating in only the oil?

Answer 1

The height of the oil on the side of the bar when the soap is floating in only the oil is 1.236 cm

Step-by-step 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.