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Bulk modulus of water is \(2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\). The change in pressure required to increase density of water by \(0.1 \%\) is (a) \(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) (b) \(2 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\) (c) \(2 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\) (d) \(2 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}\)

Short Answer

Expert verified
Change in pressure is \(2 \times 10^{2} \mathrm{~N/m^2}\), but no options match exactly.

Step by step solution

01

Understand the Bulk Modulus Formula

The bulk modulus (\(B\)) is a measure of a material's resistance to uniform compression and is defined by the formula \(B = - \frac{\Delta P}{\frac{\Delta V}{V}}\), where \(\Delta P\) is the change in pressure, \(\Delta V\) is the change in volume, and \(V\) is the original volume.
02

Relate Volume Change to Density Change

Since volume \(V\) and density \(\rho\) are inversely related (\(\rho = \frac{m}{V}\)), a change in density \(\Delta \rho\) is given by \(\Delta \rho = - \rho \frac{\Delta V}{V}\). Thus, \( \frac{\Delta V}{V} = - \frac{\Delta \rho}{\rho} \).
03

Substitute Into Bulk Modulus Equation

Substitute \( \frac{\Delta V}{V} = - \frac{\Delta \rho}{\rho} \) into the bulk modulus formula: \(B = - \frac{\Delta P}{- \frac{\Delta \rho}{\rho}} = \rho \frac{\Delta P}{\Delta \rho}\). Solve for \(\Delta P\): \(\Delta P = B \frac{\Delta \rho}{\rho}\).
04

Calculate Change in Pressure

Given that the change in density \(\Delta \rho / \rho = 0.1\% = 0.001\), and \(B = 2 \times 10^{5}\, \mathrm{N/m^2}\), calculate \(\Delta P = 2 \times 10^{5} \times 0.001 = 2 \times 10^{2} \).
05

Identify the Correct Answer

The change in pressure required is \(2 \times 10^{2}\, \mathrm{N/m^2}\), which matches none of the choices exactly. The closest value might have been a typographical error in the options provided. However, based on a literal calculation, none of the provided answer options (a to d) accurately reflects the calculated change in pressure.

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Key Concepts

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

Density Change
Understanding how density changes play a crucial role in bulk modulus calculations is key. Density is defined as mass per unit volume, represented as \( \rho = \frac{m}{V} \). As the volume of a material changes under pressure, its density will also change. When the volume decreases, the density increases, assuming the mass remains constant.

To relate this to bulk modulus, consider that the change in density, \( \Delta \rho \), can be described using the equation \( \Delta \rho = - \rho \frac{\Delta V}{V} \). This shows that a change in volume results in an inverse change in density. In the context of the given problem, if the density increases by \(0.1\%\), then we express this percentage change as \( \Delta \rho / \rho = 0.001 \). This directly impacts how we calculate pressure change via the bulk modulus equation.
Pressure Change
Pressure change is vital when considering how external forces affect a material's density and volume. The bulk modulus, \( B \), is defined in terms of pressure change using the formula \( B = - \frac{\Delta P}{\frac{\Delta V}{V}} \). Higher bulk modulus values indicate that more pressure change is needed to compress the material.

When seeking to find \( \Delta P \), we rearrange the equation after substituting \( \frac{\Delta V}{V} = - \frac{\Delta \rho}{\rho} \) from the density-volume relationship. Thus, the equation modifies to \( \Delta P = B \frac{\Delta \rho}{\rho} \). Using this, we can calculate the required change in pressure knowing the bulk modulus and the density change. For the given problem, this translates to \( \Delta P = 2 \times 10^{5} \, \mathrm{N/m^2} \times 0.001 = 2 \times 10^{2} \, \mathrm{N/m^2} \). Despite the options perhaps being inaccurate, this is a straightforward calculation method for understanding pressure effects.
Volume Compression
Volume compression is a concept tied closely to both density and pressure changes. In the context of materials' response to external pressures, volume changes accordingly – as pressure increases, volume typically decreases. Volume compression is quantitatively expressed in terms of fractional volume change \( \frac{\Delta V}{V} \), which directly influences the bulk modulus equation.

In a practical sense, comprehending how much volume change occurs under a specific pressure application helps predict material behavior. The expression \( \frac{\Delta V}{V} = - \frac{\Delta \rho}{\rho} \) allows us to convert density changes into volume changes without needing to measure them independently. This linkage is vital for calculating pressure changes in problems involving bulk modulus. By setting \( \Delta \rho / \rho = 0.001 \) in our calculations, we can find the necessary pressure change that causes a specific density increase, indicating volume compression within the material is directly related to these measurable parameters.

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Most popular questions from this chapter

A large number of liquid drops each of radius \(r\) coalesce to form a single drop of radius \(R\). The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big drop is (Given surface tension of liquid is \(T\), density of liquid is \(\rho\) ) (a) \(\sqrt{\frac{T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}\) (b) \(\sqrt{\frac{2 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}\) (c) \(\sqrt{\frac{4 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}\) (d) \(\sqrt{\frac{6 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}\)

A uniform solid cylinder made of steel, is compressed along the axis. Which of the following statements is correct? (a) Decrease in volume of the cylinder is independent of its area of cross- section but directly proportional to its length. (b) Decrease in volume of the cylinder is independent of its area of cross- section but inversely proportional to its length. (c) Decrease in volume of the cylinder is directly proportional to its cross- sectional area but is independent of its length (d) Decrease in volume of the cylinder is inversely proportional to its volume as well as length.

The height upto which water will rise in capillary tube will be (a) same at all temperatures (b) minimum when temperature of water is \(0^{\circ} \mathrm{C}\) (c) maximum when temperature of water is \(4^{\circ} \mathrm{C}\) (d) minimum when temperature of water is \(4^{\circ} \mathrm{C}\)

Water rises to a height of \(10 \mathrm{~cm}\) in a capillary tube and mercury falls to a depth of \(3.42 \mathrm{~cm}\) in the same capillary tube. If the density of mercury is \(13.6\) and angle of contact is \(135^{\circ}\), the ratio of surface tension for water and mercury is (Angle of contact for water and glass is \(8^{\circ}\) ) (a) \(1: 0.5\) (b) \(1: 3\) (c) \(1: 6.5\) (d) \(1.5: 1\)

Two parallel glass plates are dipped partly in the liquid of density \(d\) keeping them vertical. If the distance between the plates is \(x\), surface tension for liquids is \(T\) and angle of contact is \(\theta\), then rise of liquid between the plates due to capillary will be (a) \(\frac{T \cos \theta}{x d}\) (b) \(\frac{2 T \cos \theta}{x d g}\) (c) \(\frac{2 T}{x d g \cos \theta}\) (d) \(\frac{T \cos \theta}{x d g}\)

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