/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A body weighs \(5 \mathrm{~N}\) ... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 26

A body weighs \(5 \mathrm{~N}\) in air and \(2 \mathrm{~N}\) when immersed in a liquid. The buoyant force is: (a) \(2 \mathrm{~N}\) (b) \(3 \mathrm{~N}\) (c) \(5 \mathrm{~N}\) (d) \(7 \mathrm{~N}\)

Short Answer

Expert verified
The buoyant force is \(3 \mathrm{~N}\).

Step by step solution

01

Understanding the Context

When an object is placed in a fluid, it experiences an upward buoyant force. This force is equal to the weight of the fluid displaced by the object.
02

Identify the Given Values

From the problem, the weight of the body in air is given as \(5 \mathrm{~N}\), and the weight of the body when immersed in the liquid is \(2 \mathrm{~N}\).
03

Apply the Concept of Buoyant Force

The buoyant force can be calculated as the difference between the weight of the object in air and its weight when submerged in the liquid. This is because the apparent loss in weight of the body when submerged equals the buoyant force.
04

Calculate the Buoyant Force

Using the given values, the buoyant force \(F_b\) is:\[ F_b = ext{Weight in air} - ext{Weight in liquid} = 5 \mathrm{~N} - 2 \mathrm{~N} = 3 \mathrm{~N} \]
05

Select the Correct Answer

From the calculation, the buoyant force is \(3 \mathrm{~N}\). Therefore, the correct answer is (b) \(3 \mathrm{~N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Archimedes' principle
Archimedes' principle is a fundamental concept in fluid mechanics. It states that any object wholly or partially immersed in a fluid experiences an upward buoyant force. The magnitude of this force is equal to the weight of the fluid that the object displaces. This principle helps us understand why objects float or sink when placed in liquids.
For example, when you place a block of wood in water, it displaces a certain amount of water equal to its own volume. The water pushes back with a force equal to the weight of the water displaced, keeping the block afloat.
A key takeaway is that the buoyant force does not depend on the objects' weight but on the fluid it displaces. Thus, Archimedes' principle provides the fundamental explanation for buoyancy.
apparent weight
Apparent weight refers to the perceived weight of an object when it is immersed in a fluid. When an object is submerged, it seems to weigh less than its actual weight in the air. This is because the buoyant force acts in the opposite direction of gravity, effectively reducing the object's weight.
To calculate the apparent weight of an object, you subtract the buoyant force from its actual weight in the air. In the provided exercise, a body weighing 5 N in air weighs only 2 N in the liquid, which means it experiences a buoyant force of 3 N. Consequently, its apparent weight is the difference, resulting in 2 N.
Understanding apparent weight is crucial in scenarios ranging from engineering applications to recreational activities such as swimming. It helps predict how an object will behave when submerged.
weight in fluid
Weight in fluid is a crucial parameter when studying objects in different environments. It defines how much an object weighs when it is submerged in a fluid, which is different from its true weight in air.
This altered weight results from the buoyant force that acts upward against the force of gravity, making objects seem lighter. The weight in a fluid can be computed by adjusting for the buoyant force, which creates a net change in the object's perceived weight.
In practical terms, understanding the concept of weight in fluid is fundamental for designing ships, submarines, and even explaining the working of hot air balloons. It is the reason why ships made of steel can float, and hot air balloons lift off the ground.

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Most popular questions from this chapter

There is a wide tank of cross-section area \(A\) contain a liquid to a height \(H\) has a small orifice at its base of area \(' a^{\prime}(a<1)\) (a) \(t=\left(\frac{A}{a}\right)\left[\sqrt{\frac{1}{g}}\left(\sqrt{H}-\sqrt{\frac{H}{\eta}}\right)\right]\) (b) \(t=\left(\frac{a}{A}\right)\left[\sqrt{\frac{2}{g}}\left(\sqrt{H}-\sqrt{\frac{H}{\eta}}\right)\right]\) (c) \(t=\left(\frac{a}{A}\right)\left[\sqrt{\frac{1}{g}}\left(\sqrt{H}-\sqrt{\frac{H}{\eta}}\right)\right]\) (d) \(t=\left(\frac{A}{a}\right)\left[\frac{1}{\sqrt{2 g}}\left(\sqrt{H}-\sqrt{\frac{H}{\eta}}\right)\right]\)

In air, a metallic sphere with an internal cavity weighs \(40 \mathrm{~g}\) and in water it weighs \(20 \mathrm{~g}\). What is the volume of cavity if the density of material with cavity be \(8 \mathrm{~g} / \mathrm{cm}^{3}\) ? (a) zero (b) \(15 \mathrm{~cm}^{3}\) (c) \(5 \mathrm{~cm}^{3}\) (d) \(20 \mathrm{~cm}^{3}\)

In English, the phrase 'tip of the iceberg' is used to mean a small visible fraction of something that is mostly hidden. For a real iceberg, what is this fraction if the density of sea water is \(1.03 \mathrm{~g} / \mathrm{cc}\) and that of ice is \(0.92 \mathrm{~g} / \mathrm{cc} ?\) (a) \(0.106\) (b) \(10.6\) (c) \(0.901\) (d) \(0.801\)

A rectangular plate \(2 \mathrm{~m} \times 3 \mathrm{~m}\) is immersed in water in such a way that its greatest and least depth are \(6 \mathrm{~m}\) and \(4 \mathrm{~m}\) respectively from the water surface. The total thrust on the plate is:(a) \(294 \times 10^{3} \mathrm{~N}\) (b) \(294 \mathrm{~N}\) (c) \(100 \times 10^{5} \mathrm{~N}\) (d) \(400 \times 10^{3} \mathrm{~N}\)

Which of the following is/are correct about pressure? (a) Pressure at a point acts equally in all directions (b) Liquid at rest exerts lateral pressure which decreases with depth (c) Pressure acts normally on any area whatever orientation the area may be held (d) Both (a) and (c) are correct
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