Question

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?

Answer 1

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:

$u^{2} = v^{2} + 2gh$

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

u = $\sqrt{2gh}$                      (1)

Now, pressure is given by:

$\Delta p = \frac{1}{2}u^{2}\rho$        (2)

where,

$\Delta p$ = density of water = 1000 kg/$m^{3}$

g = 9.8 m/$s^{2}$

Using eqn (1) and (2):

$\Delta p = \frac{1}{2}(\sqrt{2gh})^{2}\rho$

⇒ $\Delta p = \frac{1}{2}\rho gh$

$\Delta p = \frac{1}{2}\times2\times 1000\times 9.8\times 50$ = $\Delta p$ =   490000 Pa

$p_{atm}$ = 101325 Pa

For absolute pressure, p:

p = $\Delta p$ + $p_{atm}$

p = 490000 - 101325

p = 388675 Pa

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