A body is floating in water with 3/5th of its volume below the water surface. What will be density of the body?

Answer

To solve the problem of finding the density of a body floating in water with 3/5th of its volume submerged, we can follow these steps:

### Step 1: Understand the Problem
We have a body floating in water, and we know that 3/5 of its volume is below the water surface. We need to find the density of the body.

### Step 2: Use the Principle of Buoyancy
According to Archimedes' principle, the weight of the water displaced by the submerged part of the body is equal to the weight of the body itself when it is floating.

### Step 3: Define Variables
Let:
- \( V \) = Total volume of the body
- \( \rho_b \) = Density of the body
- \( \rho_w \) = Density of water (approximately \( 1000 \, \text{kg/m}^3 \))
- The volume of the submerged part of the body = \( \frac{3}{5} V \)

### Step 4: Write the Equation for Buoyancy
The weight of the body can be expressed as:
\[ \text{Weight of the body} = \rho_b \cdot V \cdot g \]
The weight of the water displaced is:
\[ \text{Weight of the displaced water} = \rho_w \cdot \left(\frac{3}{5} V\right) \cdot g \]
Setting these two weights equal gives us:
\[ \rho_b \cdot V \cdot g = \rho_w \cdot \left(\frac{3}{5} V\right) \cdot g \]

### Step 5: Simplify the Equation
We can cancel \( g \) and \( V \) (assuming \( V \neq 0 \)):
\[ \rho_b = \rho_w \cdot \frac{3}{5} \]

### Step 6: Substitute the Value of Water Density
Substituting \( \rho_w = 1000 \, \text{kg/m}^3 \):
\[ \rho_b = 1000 \cdot \frac{3}{5} \]

### Step 7: Calculate the Density of the Body
\[ \rho_b = 1000 \cdot 0.6 = 600 \, \text{kg/m}^3 \]

### Final Answer
The density of the body is \( 600 \, \text{kg/m}^3 \).

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