# Absolute pressure. depth

At what depth beneath the surface of a lake is the absolute pressure 6 times the atmospheric pressure of 1.01 x 105 Pa that acts on the lake's surface?

pressure = density * g * h

I thought to multiply 6 * 1.10 X 10^5 pa. Then divide that answer by 1000 kg/m3 (density of water, right?) * 9.8 m/s/s.

This was marked wrong my webassign.
Can someone help me figure out what I did wrongly?

Thank you.

SpaceTiger
Staff Emeritus
Gold Member
cavery4 said:
At what depth beneath the surface of a lake is the absolute pressure 6 times the atmospheric pressure of 1.01 x 105 Pa that acts on the lake's surface?

You were almost right, but you calculated the depth at which the change in pressure was six times the surface pressure. This makes the total pressure at that depth slightly more than six times the surface value.

$$h\ =\ depth$$

$$P-P_s=\Delta P=\rho gh$$

$$\frac{P+\Delta P}{P}=6=1+\frac{\rho gh}{P}$$

$$h=\frac{5P}{\rho g}$$

Just a little smaller than your value.