Question

A cylindrical disk of wood weighing 45.0 N and having a diameter of 30.0 cm floats on a cylinder of oil of density $\mathbf{0}\mathbf{.}\mathbf{850}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$(Fig. E12.21). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column? (b) Suppose now that someone puts a weight of 83.0 N on top of the wood, but no oil seeps around the edge of the wood. What is the change in pressure at (i) the bottom of the oil and (ii) halfway down in the oil?

Short answer

(a) The pressure on the top of the container is$636\mathrm{Pa}$

(b) (i) The gauge pressure at the bottom is $1174\mathrm{Pa}$.

(ii) The change in pressure at halfway down in the oil is $1174\mathrm{Pa}$.

Step-by-step solution

Identification of given information

The given data can be listed as,

• The weight of cylindrical disk is, $W=45N$.
• The diameter of disk is, $D=30cm\left(\frac{1m}{100cm}\right)=0.30m$.
• The height of cylinder containing oil is, role="math" localid="1656487801147" $h=75cm$.
• The density of oil is, $p=0.850g/{\mathrm{cm}}^3$.
• The weight put on the top of the wood is, $W==83\mathrm{N}$.

Definition of gauge pressure

There is a point in a fluid where the pressure is higher than the standard atmospheric pressure. In other words, the gauge pressure can be equal to the difference between standard atmospheric pressure and given absolute pressure.

Evaluation of pressure at the top of the oil container

The area of the lower part of the disk can be given by,

$$\begin{gathered} A=\pi r^2 \\ =\pi {\left(\frac{d}{2}\right)}^2 \end{gathered}$$

Here, $r$is the radius of the disk.

Substitute the value of rin the above equation.

$$\begin{gathered} A={\left(\frac{0.300m}{2}\right)}^2 \\ =\pi {\left(0.150m\right)}^2 \\ =0.0707m^2 \end{gathered}$$

At the top of the oil cylinder, the gauge pressure will produce a force that is equal to the weight of the wooden disk that can be given by,

$p-p_0=\frac{W}{A}$

Here,$p$is the absolute pressure of the container,$p_0$is the atmospheric pressure,$W$is the weight of the disk and$A$is the area of the disk.

Substitute all the values in the above equation.

$$\begin{gathered} p-p_0=\frac{45N}{0.0707m^2}\left(\frac{1Pa}{1N/m^2}\right) \\ =636Pa \end{gathered}$$

Thus, the pressure on the top of the container is$636Pa$

(b) i) Calculation of change in pressure at the bottom of the oil container

According to Pascal’s law, the increase in pressure is transferred to all points in the oil.When 83 N of weight has been put on wood, the pressure can be expressed as,

$∆p_{}=\frac{W}{A}$

Here, $W$is the weight on wood and $A$is the area of the disk.

Substitute values in the above equation.

$$\begin{gathered} ∆\mathrm{p}=\frac{83\mathrm{N}}{0.0707{\mathrm{m}}^2}\left(\frac{1\mathrm{Pa}}{1\mathrm{N}/{\mathrm{m}}^2}\right) \\ \approx 1174\mathrm{Pa} \end{gathered}$$

Thus, the gauge pressure at the bottom is $1174Pa$.

(b) ii) Evaluation of change in pressure at halfway down in the oil

As the pressure is equal at all the points in oil, the pressure at halfway down is equal to the pressure at the bottom, that is$1174\mathrm{Pa}$ .

Thus, the change in pressure at halfway down in the oil is $1174\mathrm{Pa}$.