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 determine the fractional compression \( \frac{\Delta V}{V} \) of water at the bottom of the ocean, we start by understanding the relationship given by the definition of the bulk modulus \( B \), which is:

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

Where:

• \( B = 2 \times 10^9 \, \text{N/m}^2 \) (bulk modulus of water)
• \( \Delta P \) is the change in pressure
• \( \frac{\Delta V}{V} \) is the fractional change in volume (compression)

We first need to determine the pressure increase \( \Delta P \) at the bottom of the ocean due to the water column. Using the hydrostatic pressure formula, we have:

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

Where:

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

Substitute the known values:

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

Using the bulk modulus equation, we solve for the fractional change in volume:

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

Substitute \( \Delta P \) and \( B \) values:

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

Expressing this in the given format \( \alpha \times 10^{-2} \), we find:

\[ \alpha = 2 \]

Verification:
The computed value of \( \alpha = 2 \) falls within the given range of 2,2, confirming the solution is correct.

Answer 2

The fractional compression \(\frac{\Delta V}{V}\) is given by:

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

The pressure \(\Delta P\) at the bottom of the ocean is:

\[ \Delta P = \rho gh = 1000 \times 10 \times 4000 = 4 \times 10^7 \, \text{Pa} \]

Thus,

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

Therefore, \(\alpha = 2\).