Question

To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.2%? The bulk modulus of rubber is \( 9.8 \times 10^8 \, \text{N/m}^2 \) and the density of sea water is \( 10^3 \, \text{kg/m}^3 \) (g = 9.8 m/s\(^2\))

• A. 200 m
• B. 100 m
• C. 50 m
• D. 25 m

Hint:

The volume change due to pressure is related to the bulk modulus and the pressure change, which depends on depth and the density of the fluid.

Answer 1

The Correct Option is A

Step 1: Formula for Volume Change.
The change in volume due to pressure is related to the bulk modulus by the formula: \[ \Delta V = \frac{\Delta P}{B} V \] where \( B \) is the bulk modulus and \( \Delta P \) is the change in pressure. The pressure change at a depth \( h \) is \( \Delta P = \rho g h \). Substituting, we get: \[ \Delta V = \frac{\rho g h}{B} V \] Step 2: Solving for Depth.
We are given that the volume change is 0.2%, or \( \frac{\Delta V}{V} = 0.002 \). Solving for \( h \), we get: \[ h = \frac{B \times 0.002}{\rho g} \] Substituting the known values: \[ h = \frac{9.8 \times 10^8 \times 0.002}{10^3 \times 9.8} = 200 \, \text{m} \] Step 3: Final Answer.
Thus, the depth at which the volume will decrease by 0.2% is 200 m.