Chapter 13, Problem 94A

To determine

To calculate: The pressure inside the water.

Answer to Problem 94A

The required depth is $1.35\text{m}$ .

Explanation of Solution

Given:

The depth of mercury beaker is $10\text{cm}$ .

Formula used:

The pressure exerted by the fluid is given by,

  $P=\rho hg$  ...... (1)

Here, $\rho$ is the density of the fluid, $h$ is the height of the column of fluid and $g$ is the acceleration due to gravity.

Calculation:

The density of the water is $1\times {10}^3{\text{kg/m}}^{\text{3}}$ .

The density of the water is $13.55\times {10}^3{\text{kg/m}}^{\text{3}}$ .

Substitute the value of $13.55\times {10}^3{\text{kg/m}}^{\text{3}}$ for $\rho$ , $10.0\times {10}^{-2}\text{m}$ for $h$ and $9.8\text{m}\cdot {\text{s}}^{\text{-2}}$ for $g$ in equation (1),

  $\begin{array}{l}P=\left(1.35\times {10}^3\right)\times \left(10.0\times {10}^{-2}\right)\times \left(9.8{\text{m/s}}^{\text{2}}\right)\\ P=1.33\times {10}^4\text{Pa}\end{array}$

Equation (1) can be rearranged,

  $h=\frac{P}{\rho g}$  ...... (2)

Now, it is required to calculate the depth of the water container that has a pressure of $1.33\times {10}^4\text{Pa}$ . The density of the water is $1.0\times {10}^3{\text{kg/m}}^{\text{3}}$ .

Substitute $1.33\times {10}^4\text{Pa}$ for $P$ , $1.0\times {10}^3{\text{kg/m}}^{\text{3}}$ for $\rho$ and $9.8{\text{m/s}}^{\text{2}}$ in equation (2),

  $\begin{array}{l}h=\frac{1.33\times {10}^4\text{Pa}}{\left(1.0\times {10}^3{\text{kg/m}}^{\text{3}}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ h=1.35\text{m}\end{array}$

Thus, the water container should be $1.35\text{m}$ deep.

Conclusion:

Hence, the required depth is $1.35\text{m}$ .