Question

The depth below the surface of the sea to which a rubber ball is taken so as to decrease its volume by 0.02% is _____ m. (Take density of sea water \( = 10^3 \, \text{kg m}^{-3} \), Bulk modulus of rubber \( = 9 \times 10^8 \, \text{N m}^{-2} \), and \( g = 10 \, \text{m s}^{-2} \))

Answer 1

Correct Answer: 18

To determine the depth at which a rubber ball's volume reduces by 0.02%, we utilize the bulk modulus \(B\):
\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
Here, \(\Delta P\) signifies the pressure change, and \(\frac{\Delta V}{V}\) represents the fractional volume change.
We are given \(\frac{\Delta V}{V} = 0.02\% = 0.0002\).
Rearranging the formula to solve for \(\Delta P\):
\[ \Delta P = -B \cdot \frac{\Delta V}{V} \]
Substituting \(B = 9 \times 10^8 \, \text{N m}^{-2}\):
\[ \Delta P = -9 \times 10^8 \times 0.0002 = -1.8 \times 10^5 \, \text{N m}^{-2} \]
The pressure difference \(\Delta P\) is also expressed as hydrostatic pressure:\[ \Delta P = \rho g h \]
where \(\rho = 10^3 \, \text{kg m}^{-3}\), \(g = 10 \, \text{m s}^{-2}\), and \(h\) is the depth.
Equating the two expressions for \(\Delta P\):
\[ \rho g h = 1.8 \times 10^5 \]
Inserting the known values:
\[ 10^3 \times 10 \times h = 1.8 \times 10^5 \]
\[ 10^4 h = 1.8 \times 10^5 \]
Solving for \(h\):
\[ h = \frac{1.8 \times 10^5}{10^4} = 18 \, \text{m} \]
This calculated depth of 18 m falls within the specified range (18, 18), validating the result. Consequently, the depth beneath the sea's surface is \(18 \, \text{m}\).