Q 11-09-143JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
A submarine experiences a pressure of $5.05\times10^6\ \text{Pa}$ at a depth of $d_1$ in a sea. When it goes further to a depth of $d_2$, it experiences a pressure of $8.08\times10^6\ \text{Pa}$. Then $d_2 - d_1$ is approximately (density of water $= 10^3\ \text{kg/m}^3$ and acceleration due to gravity $= 10\ \text{m s}^{-2}$):
Answer: (C) $300\ \text{m}$
$$\Delta P = \rho g\,(d_2 - d_1) \;\Rightarrow\; d_2 - d_1 = \frac{3.03\times10^6}{10^3\times10} \approx 300\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics