The gage pressure in a liquid at a depth of 3 m is read to...

Given that the gage pressure at depth of 3 m is 42 kPa. To determine the gage pressure at depth 9 m we can consider that the pressure of the first case ${\displaystyle P_1}$ is the gage pressure at depth of ${\displaystyle h_1=3m}$ and similarly we can define ${\displaystyle P_2}$ as the gage pressure at depth of ${\displaystyle h_2=9m}$.
Then we could obtain ${\displaystyle P_1\text{and}{P}_2}$ as:
${\displaystyle P_1=\rho Lg{h}_1}$ (1)
${\displaystyle P_2=\rho Lg{h}_2}$ (2)
Dividing equ. (2) by equ. (1)
${\displaystyle \frac{P_2}{P_1}=\frac{\rho Lg{h}_2}{\rho Lg{h}_1}}$
${\displaystyle \therefore P_2=P_1\frac{h_2}{h_1}}$
${\displaystyle =42\left(kPa\right)\cdot \frac{9}{3}}$
${\displaystyle =126kPa}$

Given,
Depth ${\displaystyle 1=h_1=3m}$
Gage pressure at depth ${\displaystyle 1=P_{g1}=42kPa}$
Depth ${\displaystyle 2=h_2=9m}$
Total pressure in liquid is given by,
${\displaystyle P=P_a+\rho gh}$
Where,
${\displaystyle P_a=}$ Atmoshpheric pressure
${\displaystyle \rho =}$ Density of liquid
${\displaystyle g=}$ Gravitational acceleration
${\displaystyle h=}$ depth
Gage pressure is given by,
${\displaystyle P_g=P-P_a=\rho gh}$
Step 2
Gage pressure at depth 1 is given by,
${\displaystyle P_{g1}=\rho g{h}_1(eq.i)}$
Gage pressure at depth 2 is given by,
${\displaystyle P_{g2}=\rho g{h}_2(eq.ii)}$
Taking ration of equaion (i) and (ii),
${\displaystyle \therefore \frac{P_{g1}}{P_{g2}}=\frac{\rho g{h}_1}{\rho g{h}_2}}$
${\displaystyle \therefore \frac{P_{gq}}{P_{g2}}=\frac{h_1}{h_2}}$
${\displaystyle \therefore \frac{3}{9}}$
${\displaystyle \therefore P_{g2}=42\times \frac{9}{3}}$
${\displaystyle \therefore P_{g2}=126kPa}$
Answer:
Gage pressure at depth of ${\displaystyle 9m=126kPa}$