A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ an

Q: Why does surface tension act upwards in this case?

The problem explicitly states that the surface tension applies an upward force. Physically, this happens when the liquid surface is depressed downwards by the droplet (like a heavy ball on a trampoline), so the 'elastic' surface tries to pull it back up to minimize its surface area.

Q: Why is the buoyant force based on $V/2$ and not $V$?

Archimedes' Principle states that the buoyant force equals the weight of the *displaced* fluid. Because the droplet is exactly half-immersed, it only pushes aside half of its own volume of liquid.

Q: How did we get $2\pi r$ for the surface tension force?

Surface tension ($T$) is defined as force per unit length. The force acts along the boundary where the liquid meets the solid. For a sphere submerged exactly to its middle, this boundary is the sphere's equator, which is a circle of circumference $2\pi r$.

Q: Why do we use two different densities, $d$ and $\rho$?

The density $d$ belongs to the material of the droplet itself, which determines its total weight. The density $\rho$ belongs to the liquid it is floating in, which determines the buoyant force pushing back up.

Q: What would happen to the surface tension force if the droplet was fully submerged?

If the droplet was fully submerged, there would be no liquid-air interface intersecting the droplet. Therefore, the surface tension force would become zero, and the equilibrium would only depend on weight and buoyancy.

Question

A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $\rho$ and surface tension $T$. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet)

Options:

(A) $r = \sqrt{\frac{3T}{(2d - \rho)g}}$

(B) $r = \sqrt{\frac{T}{(d - \rho)g}}$

(C) $r = \sqrt{\frac{T}{(d + \rho)g}}$

(D) $r = \sqrt{\frac{2T}{3(d + \rho)g}}$

Sigma Percentile · Accepted Answer

### The Physical Setup

Imagine a tiny spherical droplet resting peacefully on the surface of a liquid. It's not just sitting there; it's exactly half-immersed. This delicate balance is a beautiful interplay of three distinct physical forces. To unlock the secret of its radius, we must first understand the tug-of-war happening at the microscopic level.

When an object floats, it is in a state of mechanical equilibrium. This means the net force acting on it is zero. Let's break down the forces. Pulling downwards, we have the relentless force of gravity acting on the droplet's mass, which we call its weight ($W$). Pushing upwards, we have two heroes: the buoyant force ($F_B$) from the displaced liquid, and the surface tension ($F_T$) acting like an invisible elastic trampoline along the droplet's equator.

### Identifying the Forces

Let's translate these physical concepts into mathematical expressions. The weight of the droplet depends on its own density ($d$) and its total volume ($V$).

$$W = d \cdot V \cdot g$$

The buoyant force, according to Archimedes' Principle, is equal to the weight of the displaced liquid. Since the droplet is exactly half-immersed, it displaces half of its volume ($V/2$) of the liquid with density $\rho$.

$$F_B = \rho \cdot \frac{V}{2} \cdot g$$

Finally, the surface tension acts along the boundary where the liquid surface meets the droplet. For a half-immersed sphere, this boundary is the largest circle (the equator) of the sphere, which has a perimeter of $2\pi r$. The force is the surface tension ($T$) multiplied by this length.

$$F_T = T \cdot (2\pi r)$$

### The Master Equation

Since the droplet is in equilibrium, the total upward force must perfectly balance the downward force.

$$F_B + F_T = W$$

Substituting our raw expressions into this master equation, we get:

$$\rho \cdot \frac{V}{2} \cdot g + T \cdot 2\pi r = d \cdot V \cdot g$$

Now, we know the droplet is a perfect sphere, so its total volume is $V = \frac{4}{3}\pi r^3$. Let's plug this geometric truth into our physics equation.

$$\rho \cdot \left(\frac{2}{3}\pi r^3\right) \cdot g + T \cdot 2\pi r = d \cdot \left(\frac{4}{3}\pi r^3\right) \cdot g$$

### Algebraic Simplification

This equation looks a bit intimidating, but let's take a breath and simplify it. Notice that every term contains a $\pi$ and at least one $r$. We can divide the entire equation by $2\pi r$ to clean it up.

$$\rho \cdot \frac{1}{3} r^2 \cdot g + T = d \cdot \frac{2}{3} r^2 \cdot g$$

Now, let's group all the terms containing $r^2$ on one side of the equation to isolate our target variable.

$$T = d \cdot \frac{2}{3} r^2 \cdot g - \rho \cdot \frac{1}{3} r^2 \cdot g$$

Factoring out the common $r^2 g$ term, we get:

$$T = r^2 g \left(\frac{2d}{3} - \frac{\rho}{3}\right)$$

### The Final Radius

We are almost there! Let's combine the fractions inside the parenthesis.

$$T = r^2 g \left(\frac{2d - \rho}{3}\right)$$

Finally, we rearrange the equation to solve for $r^2$, and then take the square root to find the radius $r$.

$$r^2 = \frac{3T}{(2d - \rho)g}$$

**$$r = \sqrt{\frac{3T}{(2d - \rho)g}}$$**

And there we have it! A beautiful, elegant expression that perfectly describes the radius of our floating droplet, born from the harmony of gravity, buoyancy, and surface tension.

Animated Solution for Physics - Properties of Solids and Liquids: A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $ρ$ and surface tension $T$ . The radius of the droplet is (take note that the surface tension applies an upward force on the droplet)

Select Answer:

Visualized Solution

Visualizing the Setup

Equilibrium of Forces

Substituting Force Expressions

Volume of the Droplet

Algebraic Simplification

Final Radius Expression

The Way Forward

The Sigma Insight: Surface Tension and Capillary Action

The Physical Setup

Identifying the Forces

The Master Equation

Algebraic Simplification

The Final Radius

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