Q 11-08-046JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
The depth below the surface of sea to which a rubber ball must be 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 $= 9\times10^8\ \text{N m}^{-2}$, and $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 18
$$\Delta P = B\frac{\Delta V}{V} = 9\times10^8\times2\times10^{-4} = 1.8\times10^5\ \text{Pa}$$
$$h = \frac{\Delta P}{\rho g} = \frac{1.8\times10^5}{10^4} = 18\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics