Question

Calculate the (a) liquid pressure, and (b) total pressure
on a scubadiver 6 m below the surface of a lake?
Given that atmospheric pressure, Po = 1.01 105 Pa,
and density of lake water = 10kg m-3. You may take
g= 10 m s-​

Answer 1

Explanation:

we know,pressure=h,r,g

where h=depth of liquid,r=density g=acceleration due to gravity

a)h=6m

r=10kgm^-3

g=10ms-

p=6×10×10=600Pa

b)total pressure=Po + hrg(p)

Po=1.01105Pa(atmospheric pressure)

tp=1.01105+600=601.01105Pa