Question

A tall tube is evacuated and its stopcock is closed. The open end of the tube is immersed in a container of water (density 10^3 kg/m^3) that is open to the atmosphere (pressure 10^5 N/m^2). When the stopcock is opened, how far up the tube will the water rise?

Answer 1

Answer:

10.19 m

Explanation:

Water will rise to equalize the pressure inside and outside the tube.

The equation of pressure is given by

$p=\rho gh$

Where,

p = Pressure of air = 10⁵ N/m²

ρ = Density of water = 10³ kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Height the water will rise

$h=\frac{p}{gh}\\\Rightarrow h=\frac{10^5}{9.81\times 10^3}\\\Rightarrow h=10.19\ m$

∴ The water will rise by 10.19 m.