The density of a metal at normal pressure is `rho`. Its density when it is subjected to an excess pressure p is `rho'` . If B is the bluk modulus of the metal, then find the ratio `rho'//rho`.

Options

A. `(1)/(1-(p)/(B))`
B. `1+(B)/(P)`
C. `(1)/(1-(B)/(P))`
D. `1+(p)/(B)`

Explanation

To find the ratio of the densities \(\frac{\rho'}{\rho}\) of a metal under normal pressure and under excess pressure \(p\), we can follow these steps:

### Step 1: Understand the relationship between density and volume
Density \(\rho\) is defined as mass per unit volume:
\[ \rho = \frac{m}{V} \]
where \(m\) is the mass and \(V\) is the volume.

### Step 2: Establish the relationship between densities
When the metal is subjected to an excess pressure \(p\), its density changes to \(\rho'\). The mass \(m\) remains constant, so we can express the new density as:
\[ \rho' = \frac{m}{V'} \]
where \(V'\) is the new volume after applying the pressure.

### Step 3: Relate the volumes
Since the mass is constant, we can write:
\[ \frac{\rho'}{\rho} = \frac{V}{V'} \]

### Step 4: Use the definition of bulk modulus
The bulk modulus \(B\) is defined as:
\[ B = -\frac{p}{\frac{dV}{V}} \]
where \(dV\) is the change in volume. Rearranging gives:
\[ \frac{dV}{V} = -\frac{p}{B} \]
This means that the new volume \(V'\) can be expressed as:
\[ V' = V(1 - \frac{p}{B}) \]

### Step 5: Substitute the new volume into the density ratio
Substituting \(V'\) into the density ratio gives:
\[ \frac{\rho'}{\rho} = \frac{V}{V(1 - \frac{p}{B})} = \frac{1}{1 - \frac{p}{B}} \]

### Step 6: Final expression for the density ratio
Thus, the ratio of the densities can be expressed as:
\[ \frac{\rho'}{\rho} = \frac{1}{1 - \frac{p}{B}} \]

### Conclusion
The final result for the ratio of densities under normal and excess pressure is:
\[ \frac{\rho'}{\rho} = \frac{1}{1 - \frac{p}{B}} \]

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