Answer
See answers.
Work Step by Step
a. The mass of water in the tube equals the density of the water multiplied by the tube’s volume.
$$m=\rho V = \rho \pi r^2 h=(1000\;kg/m^3)\pi(0.0030\;m)^2(12\;m)=0.34\;kg$$
b. The net force exerted on the lid is the pressure multiplied by the area of the lid. Use equation 10–3a.
$$F=PA = (\rho gh)(\pi R^2)=(1000\;kg/m^3)(9.80\;m/s^2)(12\;m)\pi(0.21\;m)^2=1.6\times 10^4 \;N$$
