Question

A tall tube is evacuated, and its stopcock closed. The open end of the tube is immersed into a container of water (density 103 kg/m3) that is open to the atmosphere (pressure 105 N/m2). When the stopcock is opened, how far up the tube will the water rise?

Answer 1

Answer:

The water will rise to a height, H = 10.20 m

Given:

Atmospheric Pressure, P = $10^{5} N/m^{2}$

density of water, $\rho_{w} = 10^{3} kg/m^{3}$

Solution:

The water in the tube will rise to the point where the weight is balanced by the force as a result of the atmospheric pressure.

Therefore,

Pressure, P = $\rho_{w} gH$

where

g = acceleration due to gravity = 9.8 m/s$^{2}$

H = height of water in the tube

Now,

$H =\frac{P}{\rho_{w} g}$

Putting the values in the above formula:

$H =\frac{10^{5}}{10^{3}\times 9.8}$

H = 10.20 m

Answer 2

Answer:

The water will rise to a height of 10 meters.

Explanation:

• Density of water $d=10^{3}kg/m^{3}$.
• Atmospheric pressure $p=10^{5} N/m^{2}$.

The pressure exerted at a height h is given by :

$p=d\times g\times h$

g is acceleration due to gravity.

$h=\frac{P}{dg}$

$h=\frac{10^{5}}{10^{3} \times 9.8}$

h = 10 meters.

The water will rise to a height of 10 meters.

Learn more about the atmospheric pressure, refer:

• https://brainly.com/question/1524997