Answer :
The distance from horizontal top surface of the cube to water level is 0.125 m. The mass of lead required to be placed on the cube is 2,600 kg.
(a) Edge dimension = 20 cm
Density = d - 650 KG/m³
Volume = V
Mass = m
Let's use Archimedes' law, which establishes that the pressure of the water is equal to the weight of the dislodged liquid.
Let's write the forces in the equilibrium condition;
= B - W = 0
= B = W
Where,
B = Thrust = ρ (liquid) X g X V(water)
Body weight = mg
Density = ρ = m/V
Substituting the values,
= W = ρ(body) X g X V(body)
Volume of body = L³
Volume of water = yL²
Substituting the values,
= ρ(liquid) X g and L² = ρ(body) X g X L³
= ρ(liquid) X y = ρ(body) X L
= y = L X ρ(body) /ρ(liquid)
= y = 650/ 100 X 20
= y = 0.325 m
This is the distance from the bottom of the cube, the distance from the top of the cube;
= y ’= 0.2 - y
= y ' = 0.2 - 0.325
= y ' = 0.125 m
b) In this case the equilibrium equation is;
B - W(body) - W(lead) = 0
B = W(body) + W(lead)
ρ(liquid) X g X V(liquid) = ρ(body) X g X V(body) + m(lead) X g = 0
ρ(liquid) X L³ = ρ(body) X L³ + m(lead) = 0
m(lead) = L3 (ρ(liquid) - ρ(body))
Let's calculate;
= m(lead) = 0.20 3 (1000 - 651)
= m(lead) = 2,600 kg
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