Question

1.) The pressure of a given fluid doubles every (14 feet) below the surface. What is the density of the fluid? (Give your answer in N/m^3).

Answer 1

As given

Assume two sections 1 and 2

H2-H1= 14 ft =14*0.305m =4.27m

P2/P1 =2 as given double

Now

Pressure at 1 section (P1)= rho*g*H1

Pressure at 2 section(P2) = rho*g*H2

Substracts P2-P1= rho*g*(H2-H1)

Put value

P2-P1= rho*9.8*4.27

2P1-P1=P1=rho*41.846.......1.

So P2=rho*83.692........2.

P1 is at surface so it will be atomspheric pressure= 101325 N/m2

Put in 1

101325=rho*41.846

So rho=2421.37kg/m3

Rho=2421.37*9.8= 23729.5 N/m3

Rho is density

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