The density `rho` of water of buk modulus B at a depth y then ocean is related to the density at surface `rho_(0)` by the relation

Options

A. `rho = rho_(0)[1-(rho_(0)gy)/(B)]`
B. `rho = rho_(0)[1+(rho_(0)gy)/(B)]`
C. `rho = rho_(0)[1+ (B)/(rho_(0)hgy)]`
D. `rho = rho_(0)[1-(B)/(rho_(0)gy)]`

Answer

To find the relationship between the density of water at a depth \( y \) (denoted as \( \rho \)) and the density at the surface (denoted as \( \rho_0 \)), we can use the concept of bulk modulus and hydrostatic pressure.

### Step-by-Step Solution:

1. **Understanding Bulk Modulus**:
The bulk modulus \( B \) is defined as:
\[ B = -\frac{\Delta P}{\Delta V / V} \]
where \( \Delta P \) is the change in pressure, \( \Delta V \) is the change in volume, and \( V \) is the original volume of the fluid.

2. **Hydrostatic Pressure at Depth**:
The pressure at a depth \( y \) in a fluid is given by:
\[ P = P_0 + \rho_0 g y \]
where \( P_0 \) is the atmospheric pressure at the surface, \( \rho_0 \) is the density of water at the surface, \( g \) is the acceleration due to gravity, and \( y \) is the depth.

3. **Change in Pressure**:
The change in pressure \( \Delta P \) when moving from the surface to depth \( y \) is:
\[ \Delta P = \rho_0 g y \]

4. **Volume Change**:
The change in volume \( \Delta V \) can be expressed in terms of the bulk modulus:
\[ \Delta V = -\frac{\Delta P \cdot V}{B} \]

5. **Density Relation**:
The density \( \rho \) at depth \( y \) can be expressed as:
\[ \rho = \frac{m}{V + \Delta V} \]
where \( m \) is the mass of the water, which remains constant.

Substituting \( \Delta V \):
\[ \rho = \frac{m}{V - \frac{\Delta P \cdot V}{B}} = \frac{m}{V \left(1 - \frac{\Delta P}{B}\right)} \]

6. **Expressing Density in Terms of Surface Density**:
Since \( \rho_0 = \frac{m}{V} \), we can substitute this into the equation:
\[ \rho = \frac{\rho_0 V}{V \left(1 - \frac{\Delta P}{B}\right)} = \frac{\rho_0}{1 - \frac{\Delta P}{B}} \]

7. **Final Expression**:
Substituting \( \Delta P = \rho_0 g y \):
\[ \rho = \frac{\rho_0}{1 - \frac{\rho_0 g y}{B}} \]

### Final Relationship:
\[ \rho = \frac{\rho_0}{1 - \frac{\rho_0 g y}{B}} \]