physics
posted by sand on .
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?

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 
Thanks a lot