Question

the water tank on the roof of a building has its water level at a height of 10 m above a water tap on the ground floor . calculate (1) the hydrostatic pressure at the water tap , (2) total pressure at a point inside the pipe at the level of the tap, and (3) the pressure with wh
ich water rushes out of the tap . take atmospheric pressure = 10^5 pa​

Answer 1

Answer:

pressure

Explanation:

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Answer 2

1.) 2*10^5 pa   2.) 2*10^5 pa   3.) 10^5 pa

Solution:

Let P0 be the atmospheric pressure at the top of the tank

P1 be the pressure at the tap level

P2 be the pressure in pipe of the tap

P3 be the pressure outside near the tap

Now for calculating P1 we can use, (Assuming density of water ρ=10^3 kg/m^3, g=10m/s^2)

P1-P0 = ρ*g*h

P1=P0+ρ*g*h

P1= 10^5+10^3*10*10

P1= 10^5+10^5

P1=2*10^5 pa

Since P2 is at the same horizontal level as P1 within the same fluid, P2=P1=2*10^5 pa

[can be proven by using the above formula with h=0m]

and Since P3 is in air, the pressure difference for a 10m depth change is negligible compared to the atmospheric pressure[ΔP=122.5pa], So we can safely assuem that

P3=P0=10^5pa

Therefore, the pressure with which the water rushes out is given by

P2-P3=2*10^5-10^5

ΔP= 10^5 pa