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

To solve for the depth to which the rubber ball is taken, given that its volume decreases by 0.5%, we will use the relationship between pressure change and volume change in the context of bulk modulus. The equation for bulk modulus \( B \) is:

\( B = -\frac{\Delta P}{\frac{\Delta V}{V}} \)

Where:

• \(\Delta P\) is the change in pressure.
• \(\Delta V/V\) is the fractional change in volume.
• Given \( B = 9.8 \times 10^8 \, \text{N/m}^2 \) and \(\frac{\Delta V}{V} = -0.005\), since the volume decreases.

The pressure change \(\Delta P\) due to depth in a fluid is calculated as:

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

Where:

• \(\rho = 10^3 \, \text{kg/m}^3\) is the density of sea water.
• \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity.
• \(h\) is the depth.

Equating the two expressions for pressure change, we have:

\(B = \rho g h / 0.005\)

Rearranging for \(h\), we get:

\(h = \frac{B \times 0.005}{\rho g}\)

Plugging in the values:

\(h = \frac{(9.8 \times 10^8 \, \text{N/m}^2) \times 0.005}{10^3 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2}\)

Simplifying gives:

\(h = \frac{4.9 \times 10^6}{9.8 \times 10^3}\)

\(h = \frac{4.9}{9.8} \times 1000\)

\(h = 500\, \text{m}\)

The depth \(h = 500 \, \text{m}\) matches the expected range of 500 to 500, confirming its accuracy.