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Hydrostatics air volume exam problem

  • Thread starter sovankc
  • Start date
9
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exam help!!

Hydrostatics
1. Homework Statement
A girl of mass 40 kg and volume of 40 liter wants to keep her head of volume 3 liter above water surface as she can't swim . how much air in Kg should she fill in her jacket??


2. Homework Equations



3. The Attempt at a Solution
mass of girl (m) = 40 kg
volume of girl (V) = 0.04 m^3(meter cube)
density of girl (D) = m / V = 1000 kg/m^3
volume of head (v') =0.003 m^3

let the mass of air in kg be x then

total mass of body = 40 + x
mass of water displaced = 40 + x
volume of water displaced = 1000(40 + x)
then
volume of immersion = V - v' = .04- .003 = .397

therefore from principle of flotation
volume of water displaced = volume of immersion of floating body
40000 + 1000x = .397
x=.397-40000 = -ve value





is my method correct??? and as " density of body = density of water " will require extra air to float.
 

Answers and Replies

Shooting Star
Homework Helper
1,975
4
It's not a good idea to double post (https://www.physicsforums.com/showthread.php?t=227700"). If you want people to take notice, bump it up once by replying to your own post and asking for help again.

volume of immersion = V - v' = .04- .003 = .397
0.037.

therefore from principle of flotation
volume of water displaced = volume of immersion of floating body
It should be:
weight of water displaced = weight of floating body.

is my method correct??? and as " density of body = density of water " will require extra air to float.
As pointed out. Also, the density of air is required to solve the problem.

Since the average density of the body is equal to the density of water, the head or any part will not stick out of the surface. The body can be at any depth. So, air is required to make any part of the body to be above the surface. In reality, however, the different parts of the body will have different densities, and the densest part, which happens to be the head, will sink down to achieve a stable equilibrium.
 
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