- #1
brotherbobby
- 699
- 163
- Homework Statement
- Two spheres of the same mass and radius, but one hollow and other solid, float in a vessel containing a certain amount of liquid as shown in the diagram below. Would the height of liquid be the same in the two occasions?
- Relevant Equations
- Law of flotation : Mass of the body = Mass of liquid displaced ##\left(m_B = \Delta m_L \right)##
Since the bodies have the same mass ##m_B##, the mass of liquid displaced is the same : ##\left(\Delta m_L \right)_1 = \left(\Delta m_L \right)_2 = m_B##.
Hence the volume of liquid displaced is the same : ##\left(\Delta V_L \right)_1 = \left(\Delta V_L \right)_2##.
The vessel is the same, with the same cross sectional area A.
Hence the heights of liquid displaced are the same : ##\left(\Delta h_L \right)_1 = \left(\Delta h_L \right)_2##.
The liquids had the same heights to begin with : ##\left(h_L \right)_1 = \left( h_L \right)_2##.
Hence the final height of the liquid is the same for the two cases :##\boxed{\left(h'_L \right)_1 = \left(h'_L \right)_2}##, which is my answer.This is my problem - given what I read from a textbook : (Knight - Physics for Scientists and Engineers).
The equation (14.14) above (in the text) written in my notation would read : ##\frac{\Delta V_L}{V_B} = \frac{\rho_B}{\rho_L}##. Note ##V_B## is the volume of the whole body.
The boxed warning in red clearly says that the equation above cannot be used if bodies have varying densities within them.
The hollow sphere has a non-uniform density - a very heavy metallic exterior and almost zero density hollow interior.
Yet, I used the same equation from the law of flotation for it.
Is that correct?