Question

When a rubber ball is taken to a depth of \(\dots\dots\dots\) m in deep sea, its volume decreases by \( 0.5% \).
(The bulk modulus of rubber \(= 9.8 \times 10^8 \, Nm^{-2} \)
Density of sea water \(= 10^3 \, kg m^{-3} \)
\( g = 9.8 \, m/s^2 \))

Hint:

Remember to convert percentage volume changes into decimals (\(0.5% \to 0.005\)) before using them in physical formulas.

Answer 1

Correct Answer: 500

Step 1: Understanding the Concept:
The Bulk Modulus \( B \) relates the change in pressure \( \Delta P \) to the fractional change in volume \( \Delta V / V \). At a depth \( h \) in the sea, the increase in pressure is due to the weight of the water column.
Step 2: Key Formula or Approach:
1. Bulk Modulus: \( B = \frac{\Delta P}{\Delta V / V} \implies \Delta P = B \cdot \frac{\Delta V}{V} \).
2. Hydrostatic Pressure: \( \Delta P = \rho g h \).
Step 3: Detailed Explanation:
Given:
- \( \frac{\Delta V}{V} = 0.5% = 0.005 \).
- \( B = 9.8 \times 10^8 \, Nm^{-2} \).
1. Calculate required pressure change:
\[ \Delta P = B \times \frac{\Delta V}{V} = (9.8 \times 10^8) \times 0.005 \]
\[ \Delta P = 0.049 \times 10^8 = 4.9 \times 10^6 \, Pa \]
2. Calculate depth h:
\[ \rho g h = 4.9 \times 10^6 \]
\[ (10^3) \times (9.8) \times h = 4.9 \times 10^6 \]
\[ 9.8 \times 10^3 \times h = 4.9 \times 10^6 \]
\[ h = \frac{4.9 \times 10^6}{9.8 \times 10^3} = \frac{1}{2} \times 10^3 = 500 \, m \]
Step 4: Final Answer:
The ball is taken to a depth of 500 m.