Gauge pressure is the difference between absolute pressure and some reference pressure, most commonly atmospheric pressure. The increment in pressure caused by a static fluid is given by:

$\Delta P = \rho g d$ where $\rho$ is the density of the liquid, g is the accleration due to gravity and d is the depth.

Now, we see that $\Delta P$ is linearly proportional to d, and we can assume that $\rho$ remains constant, because liquids are usually not compressible.

Given that the greater depth is simply 3 times the smaller depth:

$d_2=3\cdot d_1\\9\,m= 3 \cdot 3\,m$

$\Delta P$ at $9\, m$ of depth will also be three times the gauge pressure at $3 \,m$ of depth.

We could also have calculated $\rho$ ny using:

$\Delta P = \rho \,g \,d\\\\\rho = \frac{\Delta P}{g \, d}\\\\\rho = \frac{\Delta P}{g \, d}= \frac{30 \,kPa}{9.8 \frac{m}{s^2} \, 3\,m}=1020.41 \frac{kg}{m^3}$

and used this result to calculate the gauge pressure. These are both similar methods that yield the same result

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