Given densities

The ball has density $\rho$ such that $\rho_{oil} < \rho < \rho_{water}.$

Also, oil and water are immiscible, so they form two layers:

• oil on top (less dense)
• water at bottom (more dense)

What happens if the ball is fully in oil?

Buoyant force in oil would be $F_{b,oil} = \rho_{oil} V g.$ Weight of the ball is $W = \rho V g.$

Since $\rho > \rho_{oil}$, $W > F_{b,oil}.$

So the ball cannot float entirely in oil; it would sink through the oil.

What happens if the ball is fully in water?

Buoyant force in water would be $F_{b,water} = \rho_{water} V g.$

Since $\rho < \rho_{water}$, $F_{b,water} > W.$

So the ball cannot remain fully submerged in water; it would rise up.

Therefore, equilibrium position

Since the ball sinks in oil but rises in water, its equilibrium must be at the interface of oil and water, with part of its volume in oil and part in water.

Let volume in oil be $V_o$ and volume in water be $V_w$. Then $V_o + V_w = V.$

For equilibrium, $\rho_{oil} V_o g + \rho_{water} V_w g = \rho V g.$

This is possible because $\rho$ lies between $\rho_{oil}$ and $\rho_{water}$.

Physical interpretation

The ball will be partly immersed in oil and partly in water, crossing the boundary between the two liquids.

Correct option

Hence the correct diagram is the one showing the ball at the oil-water interface.

Therefore, the correct answer is: $\boxed{B}$