Question

A 3.0-cm-diameter tube is held upright and filled to the top with mercury. The mercury pressure at the bottom of the tube— the pressure in excess of atmospheric pressure—is 50 kPa. How tall is the tube?

Short answer

The height of tube is 0.38 m

Step-by-step solution

Given information

Diameter of tube =3.0-cm = 0.03 m

so radius is 0.015m

It is filled to the top with mercury.

The mercury pressure at the bottom of the tube—
the pressure in excess of atmospheric pressure—is 50 kPa.

Explanation

The pressure exerted by the mercury is given as

P = hρg ...............................(1)

p = 50 x 10³ Pa

h = depth of liquid

ρ= density of liquid

g = gravitational acceleration

Calculate hρg , consider g = 9.8 m/s² and density of mercury is 13.6 x10³kg/m³

hρg= 50 x 10³ Pa = h x 13.6 x10³kg/m³ x 9.8 m/s²

$$\begin{gathered} h=\frac{50\times {10}^{3}Pa}{(13.6\times {10}^{3}kg/m^3)\times (9.8m/s^2)} \\ h=0.38m \end{gathered}$$

Height of tube = 0.38 m.