Chapter 10, Problem 20P

(a)

To determine

The mass of the water in the tube on the basis of following information is to be found out.

Answer to Problem 20P

The mass of the water = $0.34Kg$

Explanation of Solution

Given:

The radius of the tube $r=.30cm$

Height of the tube $h=12m$

Formula used:

  $\begin{array}{l}\rho =m/V\\ Where\\ \rho =density\\ m=mass\\ V=volume\\ m=\rho V\\ =\rho \pi r^{2}h\\ \end{array}$

  $\pi r^{2}h$ = volume of the cylinder

Calculation:

  $\begin{array}{l}m=\rho V\\ =\rho \pi r^{2}h\end{array}$

  $\begin{array}{l}\rho =m/V\\ than\\ m=\rho V\\ =\rho \pi r^{2}h\\ $ =(1.00\times {10}^{3}kg/m^3)\pi {(0.30\times {10}^{-2}m)}^2(12m)$=0.3393kg\\ \approx .34kg\\ \end{array}$

Conclusion:

From the above discussion, it is concluded that the mass of the water tube is the product of the volume of the water tube and the density of the tube. And from the calculation the mass of the water tube is $.34kg$

(b)

To determine

To find out the value of the force exerted by the water in the barrel on the basis of information provided below.

Explanation of Solution

Given:

The radius of the barrel $=21cm$

Height of the tube $=12m$

Calculation:

In terms of pressure, force is the product of pressure and area.

  $\begin{array}{l}F=P_{gaug{e}_{}}A\\ =\rho gh\pi R^2\end{array}$

( $\pi R^2$ = area of the top of the barrel)

  $\begin{array}{l}$ =(1.00\times {10}^{3}kg/m^3)(9.8m/s^2)(12m)\pi {(0.21m)}^2$=1.6\times {10}^{4}N\end{array}$

Conclusion:

The net amount of force on the lid is the product of the gauge pressure of the water and the area of the lid. The value of force is found to be $1.6\times {10}^{4}N$ .