Question

Depth of a river is \(100 \, {m}\). Magnitude of compressibility of the water is \(0.5 \times 10^{-9} \, {N}^{-1}{m}^2\). The fractional compression in water at the bottom of the river is (Acceleration due to gravity = \(10 \, {m/s}^2\))

• A. \(0.9 \times 10^3\)
• B. \(0.5 \times 10^{-3}\)
• C. \(2 \times 10^{-3}\)
• D. \(1.3 \times 10^{-2}\)

Hint:

Remember, compressibility is the reciprocal of the bulk modulus, which relates stress to the relative change in volume.

Answer 1

The Correct Option is B

To calculate the fractional compression in water, use the bulk modulus formula related to compressibility: \[ \text{Bulk Modulus} = \frac{1}{\text{Compressibility}} \] \[ \text{Pressure} = \text{Depth} \times \text{Density of water} \times \text{Gravity} \] \[ = 100 \, \text{m} \times 1000 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 = 10^6 \, \text{Pa} \] The fractional change in volume (or compression) is given by: \[ \frac{\Delta V}{V} = \text{Pressure} \times \text{Compressibility} \] \[ = 10^6 \, \text{Pa} \times 0.5 \times 10^{-9} \, \text{N}^{-1}\text{m}^2 \] \[ = 0.5 \times 10^{-3} \]