Chapter 10, Problem 16P

To determine

To Find: The density of the oil when water and oil are poured into U- shaped tube.

Answer to Problem 16P

The density of the oil is $\text{6}.\text{54}\times \text{1}{0}^{\text{2}}\text{kg}/{\text{m}}^{\text{3}}$ .

Explanation of Solution

Given:

Height of the water, $h_{\text{water}}=27.2\text{cm}$

Height of the oil, $h_{\text{oil}}=9.41\text{cm}$

Formula used:

If pressure is equal at point a & b then

  $\begin{array}{l}P{\text{}}_a=P_b\\ =P_0+{\rho }_{\text{oil}}g{h}_{\text{oil}}\\ =P_0+{\rho }_{\text{water}}g{h}_{\text{water}}\end{array}$

Here, $P_0$ is the atmospheric pressure,

  ${\rho }_{\text{oil}}$ is the density of the oil,

  ${\rho }_{\text{water}}$ is the density of the water,

  $h_{\text{oil}}$ is the height of the oil from the reference,

  $h_{\text{water}}$ is the height of the water from the reference.

From the above expression,

  $\begin{array}{l}\Rightarrow {\rho }_{\text{oil}}{h}_{\text{oil}}={\rho }_{\text{water}}{h}_{\text{water}}\\ {\rho }_{\text{oil}}=\frac{{\rho }_{\text{water}}{h}_{\text{water}}}{h_{\text{oil}}}\end{array}$

Calculation:

The pressure at point a and b is equal means the height is same at these two points in the same liquid of fluid. If these points are not equal (a & b), in that case the fluid would flow. Now calculate the pressure at point a and b. First, consider the atmospheric pressure at the top surface of each liquid.

  $\begin{array}{l}{\rho }_{\text{oil}}=\frac{{\rho }_{\text{water}}{h}_{\text{water}}}{h_{\text{oil}}}\\ =\frac{\left(1\times {10}^3{\text{kg/m}}^{\text{3}}\right)}{\left(0.272\text{m}\right)}\left(0.272\text{m}-0.0941\text{m}\right)\\ =6.54\times {10}^2{\text{kg/m}}^{\text{3}}\end{array}$

Conclusion:

The density of the oil is $6.54\times {10}^2{\text{kg/m}}^{\text{3}}$ .