A cube of mass `m =800 g` floats on the surface of water. Water wets it completely. The cube is `10 cm` on each edge. By what additional distance is it buoyed up or down by surface tension? Surface tension of water `=0.07Nm^(-1)`

Explanation

To solve the problem, we need to determine the additional distance by which the cube is buoyed up or down due to surface tension. We will follow these steps:

### Step 1: Understand the given data
- Mass of the cube, \( m = 800 \, \text{g} = 0.8 \, \text{kg} \)
- Edge length of the cube, \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Surface tension of water, \( T = 0.07 \, \text{N/m} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
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