A lead brick with volume of 7*10^-4m^3 is floating in bath of liquid mercury.
a) What fraction of the lead brick’s volume is above the surface of mercury?
A student uses a stick to push the lead brick below the mercury surface so that it is completely submerged. What force is required to hold the lead brick below the mercury surface?
physics - drwls, Monday, December 26, 2011 at 12:09pm
You need to know the specific gravity (or density) of both lead and mercury to answer this question.
Lead has a specific gravity of 11.35, so it is 11.35 times as dense as water. It's density in SI units is
rho1 = 11.35*10^3 kg/m^3
Mercury has a specific gravity of 13.56 and density
rho2 = 13.56*10^3 kg/m^3
The buoyancy force is (rho2)*g*V'
where V' is the displaced volume of mercury. The brick's weight is W = (rho1*g*V)
where V is the volume of the brick.
Setting them equal,
V' = (rho1/rho2)*V = 0.837
The fraction above the surface must be 0.163 or 16.3%
To hold it all beneath the surface, the force required is (rho2 - rho1)*g*V =
2.21*10^3*9.8*7*10^-4 = 15.2 N
physics - sand, Monday, December 26, 2011 at 6:53pm
Thanks a lot