Chapter 10, Problem 10P

To determine

To Find: The maximum depth up to which T can hide under the water without damaging his lungs

Answer to Problem 10P

  $1.15\text{ m}$

Explanation of Solution

Given info:

Pressure difference, $\Delta P=85\text{ mm - Hg}$

  $1\text{ mm - Hg = }133\text{ Pa}$

Density of water, $\rho =1000{\text{ kg / m}}^{\text{3}}$

Formula used:

Write the expression

  $\Delta P=\rho \times g\times \Delta h$

Write the expression of change in height.

  $\Delta h=\frac{\Delta P}{\rho \times g}$

Here, g is the acceleration due to gravity.

Calculation:

Substitute the given values, we get the difference in pressure

  $\Delta h=\frac{85\times 133}{1000\times 9.8}=1.15\text{ m}$

Conclusion:

Thus, maximum depth up to which T can hide under the water without damaging his lungs is $1.15\text{ m}\text{.}$ T can breathe up to $\text{1}\text{.15 m}$ before his lungs start to damage.