Answer:
a) s,f,r b) r c) f
Explanation:
To determine what happens with the sphere we use Newton's second law with the Archimedes principle that states that the thrust (B) on a body is equal to the weight of the liquid dislodged
For the sphere to be in equilibrium the sum of forces is zero
B - W = 0
B = W = mg
Now let's use the concept of density for the body and water
Solid sphere
ρ = m / V
V = 4/3 π r³
m = ρ₀ (4/3 π r³)
W = ρ₀ (4/3 π r³) g
Water (a)
ρ = mₐ / Vₐ
mₐ = ρ Vₐ
B = ρ Vₐ g
Let's replace and simplify
ρ Vₐ g = ρ₀ (4/3 π r³) g
ρ Vf = ρ₀ (4/3 π r³) (1)
For the initial condition with rho, mo and ro the height of the water is H, let's analyze each case
a) We have the same mass, but less radius, as density is mass over volume density increases
r <ro V <V₀ ⇒ ρ₁> ρ₀
When analyzing the equation (1) on the right side, this case is the most complicated because I can make the relationship between the density of the sphere and its volume change even when the mass is constant
Assume the three possibilities
- The product of (ρ₁ V) that does not matter in that case the left side does not change and the mark remains the same (s)
- The product (ρ₁ V) increases the left side must increase so the mark goes up (r)
- The product (ρ₁ V) decreases the left side should go down, so the low mark (f)
b) sphere the same radius, but the density increases.
In this case the right side of the equation (1) increases, therefore the left side must increase so that the volume must increase and consequently increase the height (r)
c) you have the same radius, but the mass decreases
r = r₀ V = V₀ m <m₀ ρ₁ <ρ₀
The right side of the equation decreases, because the density decreases, the left side must decrease, for this the volume must decrease, lowering the height (f)
