Question

The fractional compression \( \frac{\Delta V}{V} \) of water at the depth of 2.5 km below the sea level is:

• A. 1.5
• B. 1.0
• C. 1.75
• D. 1.25

Hint:

To calculate fractional compression, use the formula \( \frac{\Delta V}{V} = \frac{\Delta P}{B} \), where \( \Delta P = \rho g h \) is the pressure change at a given depth.

Answer 1

The Correct Option is A

The fractional compression of water, defined as \( \frac{\Delta V}{V} \), is determined using the bulk modulus \( B \). The bulk modulus is expressed as: \[ B = -\frac{V \Delta P}{\Delta V} \] Here, \( V \) represents the volume, \( \Delta V \) denotes the change in volume, and \( \Delta P \) signifies the pressure change. The pressure at a depth \( h \) within a fluid is calculated using: \[ \Delta P = \rho g h \] In this formula, \( \rho \) is the fluid's density, \( g \) is the acceleration due to gravity, and \( h \) is the depth. The given parameters are: - Bulk modulus \( B = 2 \times 10^9 \, {N/m}^2 \) - Density of water \( \rho = 10^3 \, {kg/m}^3 \) - Acceleration due to gravity \( g = 10 \, {m/s}^2 \) - Depth \( h = 2500 \, {m} \) The fractional compression is computed using the following relation: \[ \frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{\rho g h}{B} \] Upon substituting the provided values: \[ \frac{\Delta V}{V} = \frac{(10^3)(10)(2500)}{2 \times 10^9} = 1.5 \] Consequently, the calculated fractional compression is 1.5.