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?

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.

answer:

a

explanation: