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) =
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
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)
m = 2.8 kg
The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.
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