Question

Suppose that the fraction of the body’s total volume that is above the water surface when a student floats in seawater (density 1030 kg/m3) is 5%.

A) Use this estimate and the student's weight of 700 N to calculate the total volume of the student's body.

B) What is the average density of the student's body? Express your answer with the appropriate units.

C) How does the average density from part B compare to the density of seawater? Express your answer as a ratio of the average density from part B to the density of seawater.

I got 950 kg/m^3 for density. The correct answer is supposed to be 980 kg/m^3. Please show me the error of my ways... Thank You!

Answer 1

(A) Suppose the total volume of student's body = V

5% of student's volume remains outside the sea-water.

So, total volume of student's body remains inside the water = (100-5)*V/100 = 0.95*V

Weight of the student, W = 700 N

So, mass of student, m = W/g = (700 / 9.81) kg = 71.36 kg

According to Archimedes Principle -

Buoyant force due to sea water = weight of the student

=> 0.95*V*density of sea water*g = 700 N

=> 0.95*V*1030*g = 700 N

=> V = 700 / (0.95*1030*9.81) = 0.0729 m^3 (Answer)

(B) Average density of student's body = m/V = 71.36 / 0.0729 = 978.56 kg/m^3

(C) Average density of student's body / density of sea water = 978.56 / 1030 = 0.95 (Answer)