Question Video: Relation of Object Buoyancy to Density of Fluid | Nagwa Question Video: Relation of Object Buoyancy to Density of Fluid | Nagwa

# Question Video: Relation of Object Buoyancy to Density of Fluid

A woman floats in fresh water of density 1.00 × 10³ kg/m³. 4.00% of her volume is above the water’s surface. What is the density of the woman? What percent of the woman’s volume would be above water if she floated in sea water of density 1.03 × 10³ kg/m³?

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### Video Transcript

A woman floats in fresh water of density 1.00 times 10 to the third kilograms per cubic meter. 4.00 percent of her volume is above the water’s surface. What is the density of the woman? What percent of the woman’s volume would be above water if she floated in sea water of density 1.03 times 10 to the third kilograms per meter cubed?

In part one, we want to solve for the density of the woman. We can call that 𝜌 sub 𝑤. And in part two, we want to solve for the percent of the woman’s volume that would be above water if she was in sea water. We’ll call that percent capital 𝑃. We’re told the density of fresh water as well as salt water. And we can name those densities 𝜌 sub 𝑓 and 𝜌 sub 𝑠, respectively.

In the initial example, where the woman is in fresh water, we’re told that 4.00 percent of her volume is above that water when she floats. This means that, compared to the density of fresh water, the density of the woman, 𝜌 sub 𝑤, is four percent less. Written as an equation, we can say that 𝜌 sub 𝑤 equals 𝜌 sub 𝑓 minus four percent of 𝜌 sub 𝑓, 0.04 times that density. This is equal to 0.96 times 𝜌 sub 𝑓, or 0.96 times 1.00 times 10 to the third kilograms per cubic meter. This product equals 960 kilograms per cubic meter. That’s the density of the woman.

In part two, we imagine that the woman moves from fresh water to salt water. And now we want to know what percentage of her volume is above that level of the water. We’ll figure this out using the density of salt water, 𝜌 sub 𝑠, and the density of the woman from part one, 𝜌 sub 𝑤. Here’s one way to think of this. Imagine that 𝜌 sub 𝑤 and 𝜌 sub 𝑠 were the same, so that this fraction is equal to one. In that case, 100 percent of the woman would be submerged and zero percent would be above water.

So, to solve for 𝑃, the percent above water, we’ll want to subtract this ratio from 100 percent. And then, more than that, we’ll want to turn this ratio from a decimal into a percentage value. And we’ll do that by multiplying it by 100. Now, we have an expression, which when we calculate it, we’ll solve for the percentage of the woman’s volume that is above the level of the saltwater.

When we plug in the densities for the woman and for salt water and calculate this expression, we find it’s equal, to three significant figures, to 6.80 percent. That’s the percentage of the woman’s volume that would be above water level if she was floating in sea water.

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