A sphere of density `rho` is half submerged in a liquid of density `sigma` and surface tension T. The sphere remains in equilibrium. Find radius of the sphere (assume the force due to surface tension acts tangentially to surface of sphere)

Options

A. `r=sqrt((T)/((d-rho)g))`
B. `r=sqrt((T)/((d+rho)g))`
C. `r=sqrt((2T)/(3(d+rho)g))`
D. `r=sqrt((3T)/((2d-rho)g))`a

Answer

To solve the problem of finding the radius of a sphere that is half submerged in a liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Sphere**:
- The sphere experiences three main forces:
- Weight (W) acting downwards.
- Buoyant force (F_B) acting upwards.
- Surface tension force (F_T) acting upwards.

2. **Write the Equation for Equilibrium**:
- Since the sphere is in equilibrium, the net force acting on it must be zero. Therefore, we can write:
\[ W - F_B - F_T = 0 \]
- Rearranging gives us:
\[ W = F_B + F_T \]

3. **Calculate the Weight of the Sphere**:
- The weight of the sphere can be expressed as:
\[ W = mg = \rho V g \]
- Here, \(V\) is the volume of the sphere, which is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Thus, the weight becomes:
\[ W = \rho \left(\frac{4}{3} \pi r^3\right) g \]

4. **Calculate the Buoyant Force**:
- The buoyant force acting on the submerged part of the sphere (half submerged) is:
\[ F_B = \sigma \left(\frac{V}{2}\right) g = \sigma \left(\frac{1}{2} \cdot \frac{4}{3} \pi r^3\right) g = \frac{2}{3} \sigma \pi r^3 g \]

5. **Calculate the Surface Tension Force**:
- The surface tension force acts along the circumference of the sphere at the water's surface. The length of the contact line is \(2\pi r\), so:
\[ F_T = T \cdot (2\pi r) \]

6. **Substitute Forces into the Equilibrium Equation**:
- Now substituting the expressions for \(W\), \(F_B\), and \(F_T\) into the equilibrium equation:
\[ \rho \left(\frac{4}{3} \pi r^3\right) g = \frac{2}{3} \sigma \pi r^3 g + T (2\pi r) \]

7. **Simplify the Equation**:
- Dividing through by \(\pi\) (assuming \(\pi \neq 0\)):
\[ \rho \left(\frac{4}{3} r^3\right) g = \frac{2}{3} \sigma r^3 g + 2Tr \]
- Rearranging gives:
\[ \frac{4}{3} \rho r^3 g - \frac{2}{3} \sigma r^3 g = 2Tr \]

8. **Factor Out Common Terms**:
- Factor out \(r^3 g\):
\[ \left(\frac{4}{3} \rho - \frac{2}{3} \sigma\right) r^3 g = 2Tr \]

9. **Solve for \(r^2\)**:
- Dividing both sides by \(r\) (assuming \(r \neq 0\)):
\[ \left(\frac{4}{3} \rho - \frac{2}{3} \sigma\right) r^2 g = 2T \]
- Rearranging gives:
\[ r^2 = \frac{2T}{g \left(\frac{4}{3} \rho - \frac{2}{3} \sigma\right)} \]

10. **Final Expression for Radius**:
- Taking the square root of both sides:
\[ r = \sqrt{\frac{2T}{g \left(\frac{4}{3} \rho - \frac{2}{3} \sigma\right)}} \]