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 \( g = 10 \, \text{m s}^{-2} \))
Correct Answer: 18
Approach Solution - 1
To find the depth at which the volume of a rubber ball decreases by 0.02%, we use the formula for bulk modulus \(B\):
\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
where \(\Delta P\) is the change in pressure, and \(\frac{\Delta V}{V}\) is the fractional change in volume.
Given \(\frac{\Delta V}{V} = 0.02\% = \frac{0.02}{100} = 0.0002\).
Rearranging the formula gives:
\[ \Delta P = -B \cdot \frac{\Delta V}{V} \]
Substitute \(B = 9 \times 10^8 \, \text{N m}^{-2}\):
\[ \Delta P = -9 \times 10^8 \times 0.0002 = -1.8 \times 10^5 \, \text{N m}^{-2} \]
The pressure difference \(\Delta P\) is also given by the 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 expressions for \(\Delta P\):
\[ \rho g h = 1.8 \times 10^5 \]
Substituting the known values:
\[ 10^3 \times 10 \times h = 1.8 \times 10^5 \]
\[ 10^4 h = 1.8 \times 10^5 \]
\[ h = \frac{1.8 \times 10^5}{10^4} = 18 \, \text{m} \]
This computed depth of 18 m is within the provided range (18, 18), confirming the solution. Therefore, the depth below the surface of the sea is \(18 \, \text{m}\).
Approach Solution -2
Given:
\[ \beta = -\frac{\Delta P}{\Delta V / V} \] \[ \Delta P = -\beta \frac{\Delta V}{V} \]Pressure difference due to sea water:
\[ \rho g h = -\beta \frac{\Delta V}{V} \]Substituting the given values:
\[ 10^3 \times 10 \times h = -9 \times 10^8 \times \left(-\frac{0.02}{100}\right) \]Simplifying:
\[ h = 18 \, \text{m} \]Top JEE Main Physics Questions
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