Water from old faithful geyser shoots upward to a height of about 50 m once every 90 minutes. what must be the pressure in the hot springs below the ground in order for the water to go that high?

Solution:

Given:

height of geyser, h = 50 m

speed of water at ground can be given by third eqn of motion with v = 0 m/s:

putting v = 0 m/s in the above eqn, we get:

u =                      (1)

Now, pressure is given by:

       (2)

where,

= density of water = 1000 kg/

g = 9.8 m/

Using eqn (1) and (2):

⇒

= =   490000 Pa

= 101325 Pa

For absolute pressure, p:

p = +

p = 490000 - 101325

p = 388675 Pa

Therefore, pressure in hot springs for the water to attain the height of 50m is 388675 Pa

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explanation: