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Why does an ocean liner float?

(a) It is made of steel, which floats.

(b) It's very big size changes the way water supports it.

(c) It is held up in the water by large Styrofoam compartments.

(d) The average density of the ocean liner is less than that of seawater.

(e) Remember the Titanic—ocean liners do not float.

Short Answer

Expert verified

The correct option is (d) the average density of the ocean liner is less than that of seawater.

Step by step solution

01

Understanding the average density

The average density can be described as the ratio of the total mass of an object/region and the total volume of the object/region. The average density of the object helps in the analysis of the buoyancy force that acts on the object.

02

Explanation of the relation between buoyancy force and weight force

Whenever an object (ocean liner) floats in seawater, it means the buoyancy force would be more than the weight force of the object (ocean liner), whereas if the buoyancy force is less than the weight force of floating object (ocean liner), then the object will sink not float.

The direction of buoyancy force works on the object in the vertically upward direction, whereas the weight force acts on the object in the vertically downward direction. The weight force depends on the object's mass density, whereas the buoyancy force depends on the mass density of the liquid (seawater).

The mass density of steel is more than the mass density of seawater, so the steel cannot float; it will sink.

Thus, option (a) is not correct.

03

Explanation of the incorrect options

An ocean liner that has big size does not change the way water supports it because its size does not make the ocean liner supported differently by water.

There are no Styrofoam compartments on an ocean liner, so the ocean liner cannot be held up in the water.

In the context of Titanic, the statement “Ocean liners do not float” is sarcastic and not absolutely right.

Thus, options (b), (c) and (e) are not correct, and the correct choice is (d).

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Most popular questions from this chapter

In Fig. 10-54, take into account the speed of the top surface of the tank and show that the speed of fluid leaving an opening near the bottom is \({{\bf{v}}_{\bf{1}}}{\bf{ = }}\sqrt {\frac{{{\bf{2gh}}}}{{\left( {{\bf{1 - A}}_{\bf{1}}^{\bf{2}}{\bf{/A}}_{\bf{2}}^{\bf{2}}} \right)}}} \),

where \({\bf{h = }}{{\bf{y}}_{\bf{2}}} - {{\bf{y}}_{\bf{1}}}\), and \({{\bf{A}}_{\bf{1}}}\) and \({{\bf{A}}_{\bf{2}}}\) are the areas of the opening and of the top surface, respectively. Assume \({{\bf{A}}_{\bf{1}}}{\bf{ < < }}{{\bf{A}}_{\bf{2}}}\) so that the flow remains nearly steady and laminar.

Figure 10-54

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