The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.
<h3>
What mass of lead should be placed on the cube?</h3>
Given: Side of the cube (a) = 20cm
The density of the cube (ρc) = ![$$650 kg/m^3](https://tex.z-dn.net/?f=%24%24650%20kg%2Fm%5E3)
a) Applying the force balance, the buoyant force must be equal to the weight of the cube
ρcgV = ρg × (Ax)
Substituting the values in the above equation, we get
![(650*(0.2 m)^3)=1000*(0.2 m)^2*x](https://tex.z-dn.net/?f=%28650%2A%280.2%20m%29%5E3%29%3D1000%2A%280.2%20m%29%5E2%2Ax)
x = 0.13
where x is the height of the cube in the water
is the area of the cross-section
ρ is the density of the water
V is the volume of the cube
Now, the height above the surface of the water would be
h = a − x
Substituting the values, then we get
h = 0.2 − 0.13
h = 0.07 m
b) The mass added is "m" so the complete cube is submerged in the water, therefore
ρcgV + mg = ρg × (V)
![$(650*(0.2 m)^3)+m=1000*(0.2 m)^3](https://tex.z-dn.net/?f=%24%28650%2A%280.2%20m%29%5E3%29%2Bm%3D1000%2A%280.2%20m%29%5E3)
m = 2.8 kg
The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.
To learn more about buoyant force refer to:
brainly.com/question/11884584
#SPJ4