Question

A beach ball is made of thin plastic. It has been inflated with air, but the plastic is not stretched. By swimming with fins on, you manage to take the ball from the surface of a pool to the bottom. Once the ball is completely submerged, what happens to the buoyant force exerted on the beach ball as you take it deeper? (a) It increases. (b) It remains constant. (c) It decreases. (d) It is impossible to determine.

Short answer

The buoyant force exerted on the beach ball as you take it deeper will decrease. Hence option (c) is the correct answer for this question.

Step-by-step solution

Step 1: Archimedes’ principle

When an object is partially or fully submerged in a fluid, the fluid exerts on the object an upward force called the buoyant force. According to Archimedes’s principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object:

$B={\rho }_{Fluid}g{V}_{disp}$

Where $V_{disp}$is the volume of fluid is displaced and${\rho }_{Fluid}$ is the density of the fluid.

Find what happens to the buoyant force exerted on the beach ball as you take it deeper

As it is motioned in the question that, ball is made up of plastic and can’t stretchable.

Therefore,BytheArchimedes’s principle,

$$\begin{gathered} B={\rho }_{Fluid}g{V}_{disp} \\ B\propto V_{disp}......\left(1\right) \end{gathered}$$

From equation (1) we can see that the buoyant force is directly proportional to Volume displaces and as the ball is compressed, the volume wills decreases and alternatively the buoyant force decreases. Hence option (c) is the correct answer for this question.