(a) 198 g
When the rock is submerged into the water, there are two forces acting on the rock:
- its weight, equal to [tex]W=mg[/tex] (m=mass, g=acceleration of gravity), downward
- the buoyant force, equal to [tex]B=m_w g[/tex] ([tex]m_w[/tex]=mass of water displaced), upward
So the resultant force, which is the apparent weight of the rock (W'), is
[tex]W'=W-B[/tex]
which can be rewritten as
[tex]m'g = mg-m_w g[/tex]
where m' is the apparent mass of the rock. Using:
m = 540 g
m' = 342 g
we find the mass of water displaced
[tex]m_w = m-m'=540 g-342 g=198 g[/tex]
(b) [tex]1.98\cdot 10^{-4} m^3[/tex]
If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.
The volume of water displaced is given by
[tex]V_w = \frac{m_w}{\rho_w}[/tex]
where
[tex]m_w = 198 g = 0.198 kg[/tex] is the mass of the water displaced
[tex]\rho_w = 1000 kg/m^3[/tex] is the density of the water
Substituting,
[tex]V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3[/tex]
And so this is also the volume of the rock.
(c) [tex]2727 kg/m^3[/tex]
The average density of the rock is given by
[tex]\rho = \frac{m}{V}[/tex]
where
m = 540 g = 0.540 kg is the mass of the rock
[tex]V=1.98\cdot 10^{-4} m^3[/tex] is its volume
Substituting into the equation, we find
[tex]\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3[/tex]
