Chapter 13, Problem 69A

To determine

Pressure exerted by two liquid at the bottom of the test tube.

Answer to Problem 69A

Two liquids exert $8.4\times {10}^2\text{}Pa$ pressure at the bottom of the test tube.

Explanation of Solution

Given:

Height of oil, $h_{oil}=2.5\text{cm}$

Density of oil, ${\rho }_{oil}\text{=0}\text{.81}{\text{g/cm}}^{\text{3}}$

Height of water, $h_{water}=\text{6}\text{.5}\text{cm}$

Density of water, ${\rho }_{water}\text{=1}{\text{.00×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}$

Acceleration due to gravity, $g=9.8\text{N/kg}$

Formula used:

To calculate the pressure in fluid, use formula:

  $P=\rho gh$

Where, $\rho$ is density of the fluid, $g$ is acceleration of gravity, and $h$ is depth of the fluid.

Calculation:

By using the formula, $P=\rho gh$ , calculate the pressure $P$ at the bottom of the test tube is.

  $P={\rho }_{oil}{h}_{oil}g+{\rho }_{water}{h}_{water}g$

Now, put the values and solve:

  $P=\text{(810}{\text{kg/m}}^{\text{3}}\text{)(0}\text{.025}\text{m)(}\text{9}\text{.8}\text{N/kg)+(1}{\text{.00×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\text{)(0}\text{.065}\text{m)×(9}\text{.8}\text{N/kg)}$

  $P=\text{8}{\text{.4×10}}^{\text{2}}\text{}\text{Pa}$

Conclusion:Pressure exerted by two liquids at the bottom of test tube is $8.4\times {10}^2\text{}Pa$