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 can be calculated using the bulk modulus \( B \), which is given by: \[ B = -\frac{V \Delta P}{\Delta V} \] where \( V \) is the volume, \( \Delta V \) is the change in volume, and \( \Delta P \) is the pressure change. The pressure at a depth \( h \) in a fluid is given by: \[ \Delta P = \rho g h \] where \( \rho \) is the density of water, \( g \) is the acceleration due to gravity, and \( h \) is the depth. Given: - Bulk modulus \( B = 2 \times 10^9 \, {N/m}^2 \), - Density of water \( \rho = 10^3 \, {kg/m}^3 \), - \( g = 10 \, {m/s}^2 \), - Depth \( h = 2500 \, {m} \). The fractional compression is calculated as: \[ \frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{\rho g h}{B} \] Substituting the values: \[ \frac{\Delta V}{V} = \frac{(10^3)(10)(2500)}{2 \times 10^9} = 1.5 \] Thus, the correct answer is 1.5.