JEE Mains · Physics · STD 11- 8. mechanical properties of solids
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 \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) )
- A \(15\)
- B \(16\)
- C \(17\)
- D \(18\)
Answer & Solution
Correct Answer
(D) \(18\)
Step-by-step Solution
Detailed explanation
\(\beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}\) \(\Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}}\) \(\rho \mathrm{gh}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}}\)…
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