Question

If aqua man can withstand a pressure of 2.0 x 10^8 Pa, what must be his depth limit?

Answer 1

Answer:

Approximately $20,\!000\; {\rm m}$, assuming that $g = 9.81\; {\rm N\cdot kg^{-1}}$ and that density of the liquid is uniformly $\rho = 1.00 \times 10^{3}\; {\rm kg\cdot m^{-3}}$.

Explanation:

At a depth of $h$ in a liquid of uniform density $\rho$, the pressure from the liquid would be:

$P = \rho\, g\, h$.

In this question, assuming that $\rho = 1.00\times 10^{3}\; {\rm kg\cdot m^{-3}}$ and $g = 9.81\; {\rm N\cdot kg^{-1}}$, the goal is to find depth $h$ given pressure $P$. Rearrange the equation to obtain the value of depth:

$\begin{aligned}h &= \frac{P}{\rho\, g} \\ &= \frac{2.0 \times 10^{8}\; {\rm Pa}}{(1.00 \times 10^{3}\; {\rm kg\cdot m^{-3}})\, (9.81\; {\rm N\cdot kg^{-1}})} \\ &\approx 20,\!000\; {\rm m}\end{aligned}$.

In other words, the maximum possible depth under these assumptions would be approximately $20,\!000\; {\rm m}$.