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Chapter 10: Q29P (page 260)

Because gasoline is less dense than water, drums containing gasoline will float in water. Suppose a 210-L steel drum is completely full of gasoline. What can the total volume of steel be used in making the drum if the gasoline-filled drum is to float in freshwater?

Short Answer

Expert verified

The total volume of steel used to make the drum if the gasoline-filled drum is to float in freshwater is 9.9 x 10-3m3.

Step by step solution

01

Understanding the principle of buoyancy

Buoyancy is an upward force exerted by the fluid in the upward direction that opposes the weight of a fully or partially submerged body. It is expressed in Newtons.

02

Identification of the given data

The volume of the gasoline in the steel drum is Vgasoline = 210L.

03

Determination of the total volume of steel

As the steel drum is filled with gasoline, the buoyancy force (FB) will be equal to the sum of the weight of the steel drum (Wsteel) and weight of gasoline (Wgasoline) filled inside it.


Here, msteeland mgasoline is the mass of the steel drum and gasoline respectively. And we know that the mass and volume ratio equals density. Thus, mass can be written as the product of volume and density.

Here, Vgasolineand Vsteel are the volume of gasoline and steel drum respectively. Ρgasoline, ρsteel and ρwater are the densities of gasoline, steel and water respectively.

Rearranging,

04

Calculations to determine the desired result

Substitute the standard values of densities, that are, Ρgasoline= 680 kg/m3, ρsteel= 7800 kg/m3and ρwater= 1000 kg/m3, in the above formula.

Hence, the total volume of steel used to make the drum if the gasoline-filled drum is to float in freshwater is 9.9 x 10-3m3.

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Most popular questions from this chapter

(II) Poiseuille’s equation does not hold if the flow velocity is high enough that turbulence sets in. The onset of turbulence occurs when the Reynolds number, \(Re\) , exceeds approximately 2000. \(Re\) is defined as

\({\mathop{\rm Re}\nolimits} = \frac{{2\overline v r\rho }}{\eta }\)

where \(\overline v \) is the average speed of the fluid, \(\rho \) is its density, \(\eta \) is its viscosity, and \(r\) is the radius of the tube in which the fluid is flowing. (a) Determine if blood flow through the aorta is laminar or turbulent when the average speed of blood in the aorta \(\left( {{\bf{r = 0}}{\bf{.80}}\;{\bf{cm}}} \right)\) during the resting part of the heart’s cycle is about \({\bf{35}}\;{\bf{cm/s}}\). (b) During exercise, the blood-flow speed approximately doubles. Calculate the Reynolds number in this case, and determine if the flow is laminar or turbulent.

Why do airplanes normally take off into the wind?

(a) Show that the flow speed measured by a venturi meter (see Fig. 10-29) is given by the relation

\({{\bf{v}}_{\bf{1}}}{\bf{ = }}{{\bf{A}}_{\bf{2}}}\sqrt {\frac{{{\bf{2}}\left( {{{\bf{P}}_{\bf{1}}}{\bf{ - }}{{\bf{P}}_{\bf{2}}}} \right)}}{{{\bf{\rho }}\left( {{\bf{A}}_{\bf{1}}^{\bf{2}}{\bf{ - A}}_{\bf{2}}^{\bf{2}}} \right)}}} \).

(b) A venturi meter is measuring the flow of water; it has a main diameter of \({\bf{3}}{\bf{.5\;cm}}\) tapering down to a throat diameter of \({\bf{1}}{\bf{.0\;cm}}\). If the pressure difference is measured to be \({\bf{18\;mm - Hg}}\), what is the speed of the water entering the venturi throat?

Question: A simple model (Fig. 10–56) considers a continent as a block (density\( \approx {\bf{2800}}\;{\bf{kg/}}{{\bf{m}}^{\bf{3}}}\)) floating in the mantle rock around it (density\( \approx {\bf{3300}}\;{\bf{kg/}}{{\bf{m}}^{\bf{3}}}\)). Assuming the continent is 35 km thick (the average thickness of the Earth’s continental crust), estimate the height of the continent above the surrounding mantle rock.

Figure 10-56

Explain how the tube in Fig. 10–45, known as a siphon, can transfer liquid from one container to a lower one even though the liquid must flow uphill for part of its journey. (Note that the tube must be filled with liquid to start with.)
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