Question

A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same. Combination answers like 'f or s' are possible answers in some of the cases.

The new sphere has a density of ρ > ρ0 and a mass of m = m0.
The new sphere has a mass of m > m0 and a radius of r = r0.
The new sphere has a radius of r = r0 and a density of ρ > ρ0.

Answer 1

The correct response for each of the condition given in the questions are,

1. $d > d_0 , m=m_0$ ⇒ f or s
2. $m > m_0, r=r_0$⇒ r
3. $r=r_0,d > d_0$   ⇒ r

To find the answer, we have to know about the Archimedes principle.

How to solve the problem for different conditions?

• The Archimedes principle states that the upthrust F on a body is equal to the weight W of the displaced liquid.

• The sum of forces must be zero for the sphere to be in equilibrium.

                               $F-W=0\\F=W\\W=mg, where.\\m=density*volume=d*V\\V=\frac{4}{3} \pi r^3$

• Let's apply the idea of density to the body and water now. We are taking d instead of ρ.

                            $d_wVg = d_0(\frac{4}{3}\pi r^3 ) g \\d_wV = d_0(\frac{4}{3}\pi r^3 )$           (1)

• Let's examine each example for the initial condition with d₀, m₀, and r₀, where the height of the water is h.

Case 1:

• The new sphere's mass is m = m₀ .

• The new sphere's density, d > d₀.

• Here, the smaller, denser sphere with the same mass, If the sphere floats, the amount of water it displaces will be equal to its mass, which will be the same as the amount of water the original sphere displaces.
• Consequently, the water level is unchanged.

• But if the sphere descends, the water displaced is less than the sphere's mass, m = m0, and the level drops, f.

• Therefore, f or s is the appropriate response.

Case 2:

• The new sphere's radius, r = r₀, and mass, m > m₀.

• Consequently, the new sphere is denser than the old one.

• The right answer is r, because the mass of the water displaced where the sphere floats is m > m0, which is greater than the water displaced for the initial sphere.

Case 3:

• sphere with the same radius but a higher density.

• When the right side of equation (1) rises, the left side must also rise in order for the volume to rise and the height to rise as a result (r)

Thus, we can conclude that,

1. $d > d_0 , m=m_0$ ⇒ f or s
2. $m > m_0, r=r_0$⇒ r
3. $r=r_0,d > d_0$   ⇒ r

Learn more about Archimedes principle here:

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