Question

Two spheres \(P\) and \(Q\) of equal radii have densities \(\rho_{1}\) and \(\rho_{2}\), respectively. The spheres are connected by a massless string and placed in liquids \(L_{1}\) and \(L_{2}\) of densities \(\sigma_{1}\) and \(\sigma_{2}\) and viscosities \(\eta_{1}\) and \(\eta_{2}\), respectively. They float in equilibrium with the sphere \(P\) in \(L_{1}\) and sphere \(Q\) in \(L_{2}\) and the string being taut (see figure). If sphere \(P\) alone in \(L_{2}\) has terminal velocity \(\vec{V}_{P}\) and \(Q\) alone in \(L_{1}\) has terminal velocity \(\vec{V}_{Q}\), then(A) \(\frac{\left|\vec{V}_{P}\right|}{\left|\vec{V}_{Q}\right|}=\frac{\eta_{1}}{\eta_{2}}\) (B) \(\frac{\left|\vec{V}_{P}\right|}{\left|\vec{V}_{Q}\right|}=\frac{\eta_{2}}{\eta_{1}}\) (C) \(\vec{V}_{P} \cdot \vec{V}_{Q}>0\) (D) \(\quad \vec{V}_{P} \cdot \vec{V}_{Q}<0\)

Short answer

The correct answer is (B) \(\frac{|\vec{V}_{P}|}{|\vec{V}_{Q}|} = \frac{\eta_{2}}{\eta_{1}}\).

Step-by-step solution

Understand terminal velocity

Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration. The terminal velocity, \(\vec{V}\), can be found by using the formula that equates the gravitational force with the force of resistance due to the medium, which is generally proportional to the velocity of the object and to the viscosity of the medium.

Analyze forces on spheres in equilibrium

For the spheres to float in equilibrium, the gravitational force must be balanced by the buoyant force from the fluid. Therefore, the weight of the sphere must be equal to the weight of the displaced liquid. The displaced volume is the same for both spheres since their radii are equal, and the densities of the spheres are given by \(\rho_{1}\) and \(\rho_{2}\), respectively.

Determine the terminal velocities \(\vec{V}_{P}\) and \(\vec{V}_{Q}\)

At terminal velocity, the force due to viscosity, which is proportional to the viscosity of the liquid and the terminal velocity, balances the net force (gravitational force minus buoyant force) on the sphere. Thus, the sphere P in liquid L2 would have a terminal velocity \(\vec{V}_{P}\) proportional to \(\eta_{2}\) and sphere Q in liquid L1 would have a terminal velocity \(\vec{V}_{Q}\) proportional to \(\eta_{1}\).

Compare the magnitudes of \(\vec{V}_{P}\) and \(\vec{V}_{Q}\)

Since the radii of the spheres and their respective buoyant forces in the liquids are equal (due to equal displaced volumes), the ratio of their respective terminal velocities will only depend on the ratio of the viscosities of the mediums L1 and L2, as the resistant forces are proportional to viscosities. Therefore, \(\frac{|\vec{V}_{P}|}{|\vec{V}_{Q}|} = \frac{\eta_{2}}{\eta_{1}}\).

Key concepts

Buoyant Force

Imagine submerging an object into a liquid; if it feels lighter than it does in air, that sensation is due to the buoyant force. This force is an upward push exerted by the fluid that opposes the weight of an immersed object. According to Archimedes' principle, the magnitude of this force equals the weight of the fluid displaced by the object.

In our exercise with spheres P and Q, the buoyant force plays a critical role in achieving equilibrium. Since the spheres have equal radii, they displace equal volumes of fluid when submerged. This implies that if the densities of the fluids \(L_{1}\) and \(L_{2}\) were the same, the buoyant force on both spheres would also be equal. However, since the densities of the liquids are different (\(\sigma_{1}\) and \(\sigma_{2}\)), the buoyant forces differ, affecting the spheres' behaviors and contributing to their terminal velocities.

Viscosity

Viscosity is a measure of a fluid's resistance to flow. You can think of it as the fluid's 'thickness'. For instance, honey has a higher viscosity than water because it flows more slowly. The internal friction of a fluid, which arises due to its molecular makeup, determines its viscosity.

In the context of our exercise involving terminal velocity, viscosity is directly proportional to the resistive force experienced by an object moving through the fluid. It means that a more viscous liquid will exert a larger force opposing the object's motion, leading to a slower terminal velocity. This is why sphere P has a terminal velocity \(\vec{V}_{P}\) proportional to the viscosity \(\eta_{2}\) in liquid \(L_{2}\), and the same applies to sphere Q in liquid \(L_{1}\) with viscosity \(\eta_{1}\) and terminal velocity \(\vec{V}_{Q}\). Understanding this relationship is crucial as it determines the rates at which the spheres settle in their respective fluids.

Gravitational Force

Gravitational force is the attraction between any two masses, which pulls objects toward each other. On Earth, this force causes objects to have weight and to fall downward when dropped. The force of gravity on an object near the Earth's surface is calculated by multiplying the object's mass by the acceleration due to gravity \(g\).

In the exercise, the gravitational force is significant because it influences how fast the spheres would fall in the absence of any other forces. When the spheres P and Q are in equilibrium, floating in the liquids \(L_{1}\) and \(L_{2}\), their weights are counteracted by the buoyant forces from the liquids. But as they move to reach terminal velocity, the balance between the gravitational pull downward and the upward buoyant force, combined with the resistive force due to viscosity, determines the spheres' final constant speed.