91Ó°ÊÓ

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, $W==83\mathrm{N}$.

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.

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

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

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

Substitute the value of rin the above equation.

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

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$636Pa$

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.

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

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

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 $1174\mathrm{Pa}$.