A liquid is kept in a beaker of radius 4 cm. Consider a diameter of the beaker on the surface of the water. Find the force by which the surface on one side of the diameter pulls the surface on the other side. Surface tension of liquid `=0.075 Nm^(-1)`

Explanation

To solve the problem, we need to find the force exerted by the surface tension of the liquid across a diameter of the beaker. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Radius of the beaker, \( r = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Surface tension of the liquid, \( S = 0.075 \, \text{N/m} \)

### Step 2: Calculate the length of the diameter
The diameter \( L \) of the beaker can be calculated as:
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