At a certain depth in an incompressible liquid, the absolute pressure is p. At twice this depth, will the absolute pressure be equal to 2p, greater than 2p, or less than 2p? Justify your answer.

The given data can be written below,

The entire pressure experienced by a liquid is referred to as absolute pressure. It is computed by beginning with zero pressure and adjusting for temperature and air pressure.

As the gauge pressure falls below the standard atmospheric pressure, it alters and becomes negative.

The absolute pressure at a depth of h is expressed as,

$$\begin{gathered} p=p_0+\rho gh \\ \rho gh=p-p_0 \end{gathered}$$

Here, $p$is pressure in the fluid at uniform density and depth h, $p_0$is the pressure at the surface of the fluid, $\rho$is the uniform density of the fluid, $g$is the acceleration due to gravity (g > 0), and $h$is depth below the surface.

At twice depth, the new pressure becomes,

$$\begin{gathered} p_1=p_0+\rho g\left(2h\right) \\ =p_0+2\left(p-p_0\right) \\ =2p-p_0 \\ <2p \end{gathered}$$

Here, the first term of the equation stays the same, but the second term doubles itself. This means that the final pressure is less than two times the original absolute pressure.

Thus,the absolute pressure will be less than 2p.