Answer to Question #113106 in Classical Mechanics for Caylin

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, pressure in excess of atmospheric pressure is 50kPa. How tall is the tube?

Answer 1

As per the question,

Diameter of the tube $(d)=3.0 cm=3\times 10^{-2}m$

so radius $(r)=\dfrac{3\times 10^{-2}m}{2}$

mercury pressure at the bottom of the tube (P)=?

The atmospheric pressure$(\rho g h)=50kPa$

Pressure at the bottom of the tube $P=P_{atm}+\rho gh$

We know that,

Density of mercury $(\rho)= 13546kg/m^3$

gravitational acceleration $(g)=9.8 m/sec^2$

Bottom pressure in excess of the atmospheric pressure

$50kPa = P - P_{atm}$

$\Rightarrow \ \rho g h=50\times 10^3Pa$

$\Rightarrow h=\dfrac{50\times 10^3}{ 9.8\times 13546}=0.37m =37cm$

$\Rightarrow h=37cm$