Question

If the average depth of an ocean is 4000 m and the bulk modulus of water is \( 2 \times 10^9 \, \text{N/m}^2 \), then the fractional compression \( \frac{\Delta V}{V} \) of water at the bottom of the ocean is \( \alpha \times 10^{-2} \). The value of \( \alpha \) is ______ (Given \( g = 10 \, \text{m/s}^2 \), \( p = 1000 \, \text{kg/m}^3 \)).

Answer 1

Correct Answer: 2

To ascertain the fractional compression \( \frac{\Delta V}{V} \) of water at ocean depths, we apply the definition of the bulk modulus \( B \):

Bulk Modulus Relation:
\( B = -\frac{\Delta P}{\frac{\Delta V}{V}} \)

Where:

• \( B = 2 \times 10^9 \, \text{N/m}^2 \) denotes the bulk modulus of water.
• \( \Delta P \) represents the pressure increment.
• \( \frac{\Delta V}{V} \) signifies the fractional volume reduction.

The pressure increase \( \Delta P \) at the ocean floor is calculated using hydrostatic pressure:

Hydrostatic Pressure Formula:
\( \Delta P = \rho g h \)

Where:

• \( \rho = 1000 \, \text{kg/m}^3 \) is the density of water.
• \( g = 10 \, \text{m/s}^2 \) is the acceleration due to gravity.
• \( h = 4000 \, \text{m} \) is the ocean depth.

Substituting the given values yields:

\[ \Delta P = 1000 \times 10 \times 4000 = 4 \times 10^7 \, \text{N/m}^2 \]

Rearranging the bulk modulus equation to solve for fractional volume change:

\[ \frac{\Delta V}{V} = -\frac{\Delta P}{B} \]

Inserting the calculated \( \Delta P \) and provided \( B \) values:

\[ \frac{\Delta V}{V} = -\frac{4 \times 10^7}{2 \times 10^9} = -0.02 \]

Expressed in the format \( \alpha \times 10^{-2} \), this gives:

\[ \alpha = 2 \]

Confirmation:
The calculated \( \alpha = 2 \) aligns with the specified range of 2,2, validating the result.