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Chapter 11: Problem 19

Determine the volume contraction of a solid copper cube, \(10 \mathrm{~cm}\) on an edge, when subjected to a hydraulic pressure of \(7 \times 10^{6} \mathrm{~Pa} . K\) for copper \(=140 \times 10^{9} \mathrm{~Pa}\). [NCERT] (a) \(5 \times 10^{-7} \mathrm{~m}^{3}\) (b) \(4 \times 10^{-6} \mathrm{~m}^{3}\) [c) \(5 \times 10^{-8} \mathrm{~m}^{3}\) (d) \(6 \times 10^{-4} \mathrm{~m}^{2}\)

Short Answer

Expert verified
The volume contraction is \( 5 \times 10^{-8} \mathrm{~m^3} \). Option (c).

Step by step solution

01

Understand the Problem

We need to find the volume contraction of a copper cube when it is subjected to a hydraulic pressure. The cube has an edge of length of 10 cm, and the bulk modulus \(K\) for copper is given as \(140 \times 10^9 \mathrm{~Pa}\). The hydraulic pressure applied is \(7 \times 10^6 \mathrm{~Pa}\).
02

Calculate Initial Volume

First, calculate the initial volume \( V_0 \) of the cube. The volume of a cube is given by \( V_0 = a^3 \), where \( a \) is the length of an edge.\[ V_0 = (10 \times 10^{-2} \mathrm{~m})^3 = 1 \times 10^{-3} \mathrm{~m^3} \]
03

Apply the Bulk Modulus Formula

The bulk modulus \( K \) is defined as \( K = - \frac{\Delta P}{\frac{\Delta V}{V_0}} \), where \( \Delta P \) is the change in pressure, and \( \Delta V \) is the change in volume. Rearranging for \( \Delta V \), \[ \Delta V = -\frac{\Delta P \times V_0}{K} \]
04

Substitute Values and Solve for Volume Contraction

Substitute the known values into the formula: \( \Delta P = 7 \times 10^6 \mathrm{~Pa} \), \( V_0 = 1 \times 10^{-3} \mathrm{~m^3} \), and \( K = 140 \times 10^9 \mathrm{~Pa} \).\[ \Delta V = -\frac{7 \times 10^6 \mathrm{~Pa} \times 1 \times 10^{-3} \mathrm{~m^3}}{140 \times 10^9 \mathrm{~Pa}} \]Compute \( \Delta V \):\[ \Delta V = -\frac{7 \times 10^{-3}}{140 \times 10^9} = -5 \times 10^{-8} \mathrm{~m^3} \]
05

Interpret Results

The negative sign indicates that the volume is contracting. Therefore, the volume contraction is \( 5 \times 10^{-8} \mathrm{~m^3} \).

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Key Concepts

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

Bulk Modulus
Understanding the bulk modulus is key to solving problems involving volume contraction due to pressure. Simply put, the bulk modulus is a measure of a material's resistance to uniform compression. Think of it as how much a material will "push back" when you try to squeeze it. In a formula, it is represented as \[ K = - \frac{\Delta P}{\frac{\Delta V}{V_0}} \]where
  • \( K \) is the bulk modulus,
  • \( \Delta P \) is the change in pressure applied to the material,
  • \( \Delta V \) is the change in volume,
  • \( V_0 \) is the original volume.
The negative sign indicates that the increase in pressure results in a decrease in volume鈥攈ence, contraction. For a copper cube, knowing its bulk modulus allows us to calculate how much it will shrink under a given hydraulic pressure. Remember, a high bulk modulus indicates a material is highly resistant to compression; for copper, this value is fairly high at \( 140 \times 10^9 \text{ Pa} \), showing it's not easily compressed.
Hydraulic Pressure
Hydraulic pressure is the force exerted by a fluid per unit area. This concept is crucial in understanding how pressure changes affect the volume of solids like a copper cube. When you apply hydraulic pressure to an object, you distribute that force evenly across its surface. Imagine filling a balloon with water; the pressure from the water pushes equally in all directions. Similarly, when hydraulic pressure is applied to a copper cube, it causes the cube to contract uniformly.In our exercise, the cube is subjected to a hydraulic pressure of \( 7 \times 10^6 \text{ Pa} \). This is a significant force, as it is seven million Pascals. But how does this relate to volume contraction? The pressure contributes to a change in volume, which we figure out using the bulk modulus formula. The higher the pressure, the more a material will try to compress, assuming the material's resistance to compression (expressed as bulk modulus) remains constant.
Copper Cube
Copper is well-known for its high thermal and electrical conductivity, but here we're interested in its mechanical properties, specifically its reaction to compression.When discussing volume contraction under pressure, a solid copper cube's initial properties are fundamental. In this exercise, we start with a cube where each side is 10 cm, leading to a volume of \( 1 \times 10^{-3} \text{ m}^3 \). Why a cube? A cube's geometry makes calculations straightforward. You use the formula for the volume of a cube \( V_0 = a^3 \) to find its original volume, where \( a \) = 10 cm (or 0.1 m).Once pressure is applied, copper, being relatively resistant to compression due to its high bulk modulus, undergoes only a small change in its volume. This change, or volume contraction, can then be calculated with our known values of pressure and bulk modulus. Thus, understanding the cube's dimensions and material properties allows for precise calculations of how much the volume will contract.

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Most popular questions from this chapter

The length of a wire is increased by \(1 \mathrm{~mm}\) on the application of a given load. In a wire of the same material, but of length and radius twice that of the first, on the application of the same load, extension is (a) \(0.25 \mathrm{~mm}\) (b) \(0.5 \mathrm{~mm}\) (c) \(2 \mathrm{~mm}\) (d) \(4 \mathrm{~mm}\)

Find the extension produced in a copper of length \(2 \mathrm{~m}\) and diameter \(3 \mathrm{~mm}\), when a force of \(30 \mathrm{~N}\) is applied. Young's modulus for copper \(=1.1 \times 10^{11} \mathrm{Nm}^{-2}\) (a) \(0.2 \mathrm{~mm}\) (b) \(0.04 \mathrm{~mm}\) (c) \(0.08 \mathrm{~mm}\) (d) \(0.68 \mathrm{~mm}\)

The Bulk Modulus for an incompressible liquid is [UP SEE 2008] (a) zero (b) unity (c) infinity (d) between 0 and 1

Two wires of the same material and length but diameters in the ratio \(1: 2\) are stretched by the same force. The potential energy per unit volume for the two wires when stretched will be in the ratio (a) \(16: 1\) (b) \(4: 1\) (c) \(2: 1\) (d) \(1: 1\)

A load of \(4.0 \mathrm{~kg}\) is suspended from a ceiling through a steel wire of length \(2.0 \mathrm{~m}\) and radius \(2.0 \mathrm{~mm}\). It is found that the length of the wire increases by \(0.031 \mathrm{~mm}\) as equilibrium is achieved. Taking, \(g=3.1 \pi \mathrm{ms}^{-2}\), the Young's modulus of steel is (a) \(2.0 \times 10^{8} \mathrm{Nm}^{-2}\) (b) \(2.0 \times 10^{9} \mathrm{Nm}^{-2}\) (c) \(2.0 \times 10^{11} \mathrm{Nm}^{-2}\) (d) \(2.0 \times 10^{13} \mathrm{Nm}^{-2}\)
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