Question

A cylindrical disk of wood weighing 𝟓𝟎.𝟎 [N] and having a diameter of 𝟐𝟓.𝟎 [cm] floats on a liquid of density 𝝆=𝟎.𝟕𝟓𝟎 [g/cm3]. The cylinder of liquid is 𝟔𝟎.𝟎 [cm] high and has a diameter that is the same as that of the wood. What is the gauge pressure halfway down the liquid?

Answer 1

The gauge pressure is the pressure ignoring the atmospheric pressure. So, the wooden cylinder creates a pressure of

$$p_1=\frac FA=\frac{4F}{\pi d^2}.$$

The pressure of liquid at the depth of 30 cm is

$$p_2=\rho gh.$$

The gauge pressure:

$$P=p_1+p_2=\rho gh+\frac{4F}{\pi d^2}=3223.6\text{ Pa}.$$