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JEE Mains · Physics · STD 11- Mechanical properties of solids
  • JEE Mains 2024
  • Hard
  • Numerical

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by \(0.02 \%\) is ______ \(\mathrm{m}\). (Take density of sea water \(=10^3 \mathrm{kgm}^{-3}\), Bulk modulus of rubber \(=9 \times 10^8 \mathrm{Nm}^{-2}\), and \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) )

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Answer & Solution

Correct Answer

(D) \(18\)

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Detailed explanation

\(\beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}\)

\(\Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}}\)

\(\rho \mathrm{gh}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}}\)

\(10^3 \times 10 \times \mathrm{h}=-9 \times 10^8 \times\left(-\frac{0.02}{100}\right)\)

\(\Rightarrow \mathrm{h}=18 \mathrm{~m}\)

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