Question

$\text{The fractional compression } \left( \frac{\Delta V}{V} \right) \text{ of water at the depth of } 2.5 \, \text{km below the sea level is } \_\_\_\_\_\_\_\_\_\_ \%. \text{ Given, the Bulk modulus of water } = 2 \times 10^9 \, \text{N m}^{-2}, \text{ density of water } = 10^3 \, \text{kg m}^{-3}, \text{ acceleration due to gravity } g = 10 \, \text{m s}^{-2}.$

• A. 1.25
• B. 1.0
• C. 1.75
• D. 1.5

Hint:

For problems involving pressure at depth, use the formula \( P = \rho g h \) and apply the bulk modulus to calculate fractional compression.

Answer 1

The Correct Option is A

To find the fractional compression of water at a depth of 2.5 km below sea level, we use the formula for fractional compression given by:

\(\frac{\Delta V}{V} = \frac{P}{K}\)

Where \(P\) is the pressure applied and \(K\) is the bulk modulus of the water.

The pressure \(P\) at a depth \(h\) is given by:

\(P = \rho \cdot g \cdot h\)

Given:

• \(h = 2.5 \, \text{km} = 2500 \, \text{m}\)
• \(\rho = 10^3 \, \text{kg m}^{-3}\)
• \(g = 10 \, \text{m s}^{-2}\)
• \(K = 2 \times 10^9 \, \text{N m}^{-2}\)

Substitute these values into the pressure formula:

\(P = 10^3 \cdot 10 \cdot 2500 = 2.5 \times 10^7 \, \text{N m}^{-2}\)

Now, substitute \(P\) and \(K\) into the fractional compression formula:

\(\frac{\Delta V}{V} = \frac{2.5 \times 10^7}{2 \times 10^9} = 0.0125\)

To express this as a percentage, multiply by 100:

\(0.0125 \times 100 = 1.25\%\)

Thus, the fractional compression of water at the given depth is 1.25%.