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 \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(18\)
Step-by-step Solution
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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