Chapter 10, Problem 63GP

(a)

To determine

To calculate: The height of the bottle.

Answer to Problem 63GP

The height is $0.75\text{m}$ .

Explanation of Solution

Given:

Pressure is , $P_1=55\text{mm-Hg}$

Density of the fluid, $\rho =1.00{\text{g/cm}}^{\text{3}}={10}^3{\text{kg/m}}^{\text{3}}$ .

Formula used:

The pressure at the particular height is given as,

  $P=\rho gh$

Where, g is acceleration due to gravity, h is the height and $\rho$ is the density.

Calculation:

Now substitute the given values in the above formula and so, the height is,

  $\begin{array}{l}P=\rho gh\\ \Rightarrow h=\frac{P}{\rho g}=\frac{55\text{mm}-\text{Hg}}{\left({10}^3{\text{Kg/m}}^3\right)\times \left(9.8\right)}\\ =\frac{55\times 133\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}{\left({10}^3{\text{kg/m}}^{\text{3}}\right)\times \left(9.8\right)}=0.75\text{m}\end{array}$

Conclusion:

The height is $0.75\text{m}$ .

(b)

To determine

To calculate: The height of the bottle.

Answer to Problem 63GP

The height $h_1^{\text{'}}$ is $0.65\text{m}$ .

Explanation of Solution

Given:

The pressure is, $P_2=650{\text{mm-H}}_{\text{2}}\text{O}$

Density of the fluid, $\rho =1.00{\text{g/cm}}^{\text{3}}={10}^3{\text{kg/m}}^{\text{3}}$

Formula used:

The pressure at the particular height is given as,

  $P=\rho gh$

Where, $g$ is acceleration due to gravity, $h$ is the height and $\rho$ is the density.

Calculation:

Now substitute the given values in the above formula and so, the height is,

  $\begin{array}{l}P=\rho gh\\ \Rightarrow h=\frac{P}{\rho g}=\frac{650\times 9.81\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}{\left({10}^3{\text{kg/m}}^{\text{3}}\right)\times \left(9.8\right)}=0.65\text{m}\end{array}$

Conclusion:

The height $h_1^{\text{'}}$ is $0.65\text{m}$ .

(c)

To determine

To calculate: The height of the bottle.

Answer to Problem 63GP

The height where the bottle should be placed is 0.24 m.

Explanation of Solution

Given:

The height of the opening of the tank is $h_1$ and the height of the liquid surface is $h_2$ . Initially, the tank is at the rest. Assume, $v_2\approx 0$

Calculation:

For the case, where the pressure is $18\text{mm-Hg}$ above the atmospheric pressure,

  $P_3=18\text{mm-Hg}=1.8\text{cm-Hg}=1.8\text{m}\times 13.6\text{g}$

Now use this value of pressure in the given formula is,

  $\begin{array}{l}P=\rho gh\\ \Rightarrow h=\frac{P}{\rho g}=\frac{1.8\text{m}\times 13.6\times 9.8}{\left({10}^3{\text{kg/m}}^{\text{3}}\right)\times \left(9.8\right)}=24.48\times {10}^{-3}\text{cm}\\ =0.24\text{m}\end{array}$

Conclusion:

The height is 0.24 m.