A cube with a mass `= 20 g` wettable water floats on the surface of water. Each face of the cube is `alpha = 3 cm` long. Surface tension of water is `70 dyn//cm`. The distance of the lower face of the cube from the surface of water is (`g= 980 cm s^(-12)`)

Options

A. `2.3cm`
B. `4.6cm`
C. `9.7cm`
D. `12.7cm`

Explanation

To solve the problem of finding the distance of the lower face of the cube from the surface of water, we can follow these steps:

### Step 1: Understand the Problem
We have a cube of mass \( m = 20 \, \text{g} \) and side length \( a = 3 \, \text{cm} \) that is floating on the surface of water. The surface tension of water is given as \( T = 70 \, \text{dyn/cm} \). We need to find the distance \( x \) of the lower face of the cube from the surface of water.

### Step 2: Write Down the Forces Acting on the Cube
The forces acting on the cube are:
1. The weight of the cube \( W = mg \).
2. The buoyant force \( F_b \) acting upwards.
3. The surface tension force \( F_T \) acting downwards.

### Step 3: Calculate the Weight of the Cube
The weight of the cube can be calculated using the formula:
\[ W = mg \]
Given \( m = 20 \, \text{g} \) and \( g = 980 \, \text{cm/s}^2 \):
\[ W = 20 \, \text{g} \times 980 \, \text{cm/s}^2 = 19600 \, \text{dyn} \]

### Step 4: Calculate the Buoyant Force
The buoyant force is given by Archimedes' principle:
\[ F_b = \rho \cdot V_{\text{submerged}} \cdot g \]
Where \( V_{\text{submerged}} \) is the volume of the submerged part of the cube. If \( x \) is the height submerged, then:
\[ V_{\text{submerged}} = a^2 \cdot x = (3 \, \text{cm})^2 \cdot x = 9x \, \text{cm}^3 \]
Assuming the density of water \( \rho = 1 \, \text{g/cm}^3 \):
\[ F_b = 1 \, \text{g/cm}^3 \cdot 9x \, \text{cm}^3 \cdot 980 \, \text{cm/s}^2 = 8820x \, \text{dyn} \]

### Step 5: Calculate the Surface Tension Force
The surface tension force can be calculated as:
\[ F_T = T \cdot L \]
Where \( L \) is the length of the perimeter of the cube in contact with the water. For a cube, \( L = 4a \):
\[ L = 4 \cdot 3 \, \text{cm} = 12 \, \text{cm} \]
Thus,
\[ F_T = 70 \, \text{dyn/cm} \cdot 12 \, \text{cm} = 840 \, \text{dyn} \]

### Step 6: Apply the Equilibrium Condition
For the cube to float in equilibrium:
\[ F_b = W + F_T \]
Substituting the values we calculated:
\[ 8820x = 19600 + 840 \]
\[ 8820x = 20440 \]
\[ x = \frac{20440}{8820} \approx 2.317 \, \text{cm} \]

### Step 7: Conclusion
The distance of the lower face of the cube from the surface of water is approximately \( 2.317 \, \text{cm} \).

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