Question

water is flowing steadily through a horizontal tube of non uniform cross section.if the pressure of water is 4*10^4N/mm^2 at a point where cross se ction is 0.02 m^2 velocity of flow is 2m/s what is pressure at a poin . where crossection reduces to 0.01m^2

Answer 1

Answer:

As we know 

A1V1=A2V2

Where A1=0.02 m2

V1=2 m/s

A2=0.01 m2

⟹0.02×2=0.01×V2

⟹V2=4 m/s

Now using Bernoulli's theorem  

ρP1+21V12=ρP2+21v22

⟹P2=P1+2ρ1[V12−V12]

⟹P2=4×10

Explanation:

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