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Chapter 7: Problem 16

A solid sphere of radius \(R\) made of a material of bulk modulus \(B\) is surrounded by a liquid in a cylindrical container. \(A\) massless piston of area \(A\) (the area of container is also \(A\) ) floats on the surface of the liquid. When a mass \(m\) is placed on the piston to compress the liquid, find the fractional change in radius of the sphere. (Given: \(\frac{M g}{A B}=0.3\) )

Short Answer

Expert verified
The fractional change in radius of the sphere is \(-0.1\), indicating a 10% decrease.

Step by step solution

01

Understand the Problem

We have a sphere submerged in a liquid, and a mass is added on top of a piston that compresses the liquid. We need to find how much the radius of the sphere changes when the liquid is compressed by this additional weight.
02

Relation of Pressure Increase

When the mass \(m\) is placed on the piston, it exerts an additional pressure \(P\) on the liquid. The pressure is given by \(P = \frac{mg}{A}\).
03

Bulk Modulus Definition

The bulk modulus \(B\) relates pressure change to volume change in the form of \(B = \frac{- ext{Pressure Change}}{ rac{ ext{Volume Change}}{ ext{Original Volume}}} \). The negative sign indicates that an increase in pressure results in a decrease in volume.
04

Calculate Volume Change

Rearranging the definition of bulk modulus gives us: \(\frac{\Delta V}{V} = -\frac{P}{B}\). Substituting \(P = \frac{mg}{A}\), we get \(\frac{\Delta V}{V} = -\frac{mg}{AB}\).
05

Relate Volume Change to Radius Change

For a sphere, the volume is related to the radius by \(V = \frac{4}{3}\pi R^3\). Using \(\frac{\Delta V}{V} = 3\frac{\Delta R}{R}\), we substitute \(\frac{\Delta V}{V}\) to find \(\frac{\Delta R}{R} = -\frac{1}{3}\frac{mg}{AB}\).
06

Use Given Values

We are given \(\frac{Mg}{AB} = 0.3\). Therefore, \(\frac{mg}{AB} = 0.3\), thus \(\frac{\Delta R}{R} = -\frac{1}{3} \times 0.3 = -0.1\).
07

Interpret the Result

The negative sign indicates the radius decreases. The fractional change in radius is \(-0.1\), meaning the radius decreases by 10%.

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Key Concepts

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

Pressure Change
In this exercise, the concept of pressure change is pivotal. When a mass is placed on the piston floating in the liquid, it introduces an additional pressure on the system. This pressure changes due to the added weight, which can be calculated using the formula \( P = \frac{mg}{A} \), where:
  • \( m \) is the mass placed on the piston,
  • \( g \) refers to the acceleration due to gravity, and
  • \( A \) is the area of the piston.
Pressure change is crucial because it initiates the sequence of reactions in the exercise as it compresses the liquid, affecting the submerged sphere inside the cylindrical container. Understanding how pressure is calculated and how it affects physical systems is critical for grasping the subsequent changes.
Volume Change
Volume change in this context arises from the compression of the liquid due to increased pressure. The concept is beautifully illustrated through the bulk modulus formula: \( B = \frac{-\text{Pressure Change}}{\frac{\text{Volume Change}}{\text{Original Volume}}} \). Bulk modulus \( B \) describes a material's resistance to uniform compression.
  • The negative sign indicates an inverse relationship between pressure and volume change.
  • Rearranging the bulk modulus formula gives \( \frac{\Delta V}{V} = -\frac{P}{B} \), showing that volume change is directly related to and proportionately dictated by pressure change, with moderation by the material's bulk modulus.
Therefore, when pressure increases, volume decreases, and this squeezed volume affects the sphere's radius.
Fractional Radius Change
The fractional radius change is derived from understanding how volume change translates to radius change. For a sphere, the fractional change in volume \( \frac{\Delta V}{V} \) is connected to the fractional change in radius \( \frac{\Delta R}{R} \) by the relationship \( \frac{\Delta V}{V} = 3 \frac{\Delta R}{R} \). This implies a change in volume results in a proportional change in radius.
  • Using \( \frac{\Delta V}{V} = -\frac{P}{B} \), the fractional change in radius becomes \( \frac{\Delta R}{R} = -\frac{1}{3} \frac{mg}{AB} \).
  • In this exercise, substituting the known value \( \frac{Mg}{AB} = 0.3 \), results in \( \frac{\Delta R}{R} = -0.1 \).
Thus, the radius decreases by 10% when the liquid's compression affects the sphere.
Solid Sphere
The solid sphere in this problem is central to understanding the effects of compression in a three-dimensional space. A solid sphere has the volume equation \( V = \frac{4}{3} \pi R^3 \), where \( R \) denotes the radius.
  • Its behavior under pressure not only showcases how solids react to external forces but also illustrates the principles of geometry when shapes experience physical changes.
  • The sphere's volume change, due to pressure, directly influences its radius, as derived from the bulk modulus principle.
  • Understanding the changes in the solid sphere illustrates broader physics principles applied in real-world scenarios such as engineering and design.
Solid spheres provide an excellent basis for studying compression and expansion in materials.
Cylindrical Container
A cylindrical container provides the perfect environment for observing how pressure transfers in contained systems. It allows us to visualize how pressure distributes uniformly over a surface area.
  • The container's role is to contain the liquid in which the sphere is submerged, ensuring all pressure changes impact the sphere.
  • The pressure from the piston, applied uniformly over the circular surface area \( A \), initiates the compression explored in the exercise.
  • This setup helps demonstrate the concept of pressure application in closed systems and its effects on submerged objects, like the change in the solid sphere's radius.
A cylindrical container's geometry allows straightforward analysis of pressure and volume changes, proving valuable in understanding fluid dynamics and statics.

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Most popular questions from this chapter

A small but heavy block of mass \(10 \mathrm{~kg}\) is attached to a wire \(0.3 \mathrm{~m}\) long. Its breaking stress is \(4.8 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}\). The area of the cross section of the wire is \(10^{-6} \mathrm{~m}^{2}\). The maximum angular velocity with which the block can be rotated in the horizontal circle is (1) \(4 \mathrm{rad} / \mathrm{s}\) (2) \(8 \mathrm{rad} / \mathrm{s}\) (3) \(10 \mathrm{rad} / \mathrm{s}\) (4) \(32 \mathrm{rad} / \mathrm{s}\)

Two wires \(A\) and \(B\) have the same cross section and are made of the same material, but the length of wire \(A\) is twice that of \(B\). Then, for a given load (1) the extension of \(A\) will be twice that of \(B\) (2) the extensions of \(A\) and \(B\) will be equal (3) the strain in \(A\) will be half that in \(B\) (4) the strains in \(A\) and \(B\) will be equal

A uniform plank is resting over a smooth horizontal floor and is pulled by applying a horizontal force at its one end. Which of the following statements are not correct? (1) Stress developed in plank material is maximum at the end at which force is applied and decrease linearly to zero at the other end. (2) A uniform tensile stress is developed in the plank material. (3) Since plank is pulled at one end only, plank starts to accelerate along direction of the force. Hence, no stress is developed in the plank material. (4) None of the above.

A ring of radius \(r\) (made of wire of density \(\rho\) ) placed on a smooth horizontal surface is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring as shown in figure. Determine the angular velocity in \(\mathrm{rad} / \mathrm{s}\) at which the ring breaks. The wire breaks at a tensile stress \(\sigma .\) (given \(\sigma / \rho=9 \mathrm{~N}-\mathrm{m} / \mathrm{kg}\) and \(r=1 \mathrm{~m}\) )

Choose the correct statements from the following: (1) Steel is more elastic than rubber. (2) The stretching of a coil spring is determined by the Young's modulus of the wire of the spring. (3) The frequency of a tuning fork is determined by the shear modulus of the material of the fork. (4) When a material is subjected to a tensile (stretching) stress the restoring forces are caused by interatomic attraction.
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