$\text{The fractional compression } \left( \frac{\Delta V}{V} \right) \text{ of water at the depth of } 2.5 \, \text{km below the sea level is } \_\_\_\_\_\_\_\_\_\_ \%. \text{ Given, the Bulk modulus of water } = 2 \times 10^9 \, \text{N m}^{-2}, \text{ density of water } = 10^3 \, \text{kg m}^{-3}, \text{ acceleration due to gravity } g = 10 \, \text{m s}^{-2}.$
$\text{The fractional compression } \left( \frac{\Delta V}{V} \right) \text{ of water at the depth of } 2.5 \, \text{km below the sea level is } \_\_\_\_\_\_\_\_\_\_ \%. \text{ Given, the Bulk modulus of water } = 2 \times 10^9 \, \text{N m}^{-2}, \text{ density of water } = 10^3 \, \text{kg m}^{-3}, \text{ acceleration due to gravity } g = 10 \, \text{m s}^{-2}.$
Show Hint
- 1.25
- 1.0
- 1.75
- 1.5
The Correct Option is A
Solution and Explanation
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%.
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