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Chapter 7: Problem 36

A ball falling in a lake of depth \(200 \mathrm{~m}\) shows a decrease of \(0.1 \%\) in its volume at the bottom. The bulk modulus of the elasticity of the material of the ball is (take \(g=10 \mathrm{~ms}^{-2}\) ) (1) \(10^{9} \mathrm{~N} / \mathrm{m}^{2}\) (2) \(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) (3) \(3 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) (4) \(4 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\)

Short Answer

Expert verified
The bulk modulus of the material is \(2 \times 10^9 \text{ N/m}^2\), option (2).

Step by step solution

01

Understanding Bulk Modulus

The bulk modulus \( K \) is a measure of a substance's resistance to uniform compression. It is defined as \( K = - \frac{\Delta P}{\frac{\Delta V}{V}} \), where \( \Delta P \) is the change in pressure, \( \Delta V \) is the change in volume, and \( V \) is the initial volume.
02

Calculate Pressure at Depth

The pressure at the bottom of the lake can be calculated using the formula for pressure due to a column of fluid: \( P = \rho gh \). Assuming the density of the water \( \rho = 1000 \text{ kg/m}^3 \), and depth \( h = 200 \text{ m} \), and \( g = 10 \text{ m/s}^2 \), the pressure is \( P = 1000 \times 10 \times 200 = 2 \times 10^6 \text{ N/m}^2 \).
03

Calculate Volume Strain

The problem states the volume decreases by \(0.1\%\) at the bottom of the lake. Hence, \( \frac{\Delta V}{V} = \frac{0.1}{100} = 0.001 \).
04

Calculate Bulk Modulus

Using the relation \( K = - \frac{\Delta P}{\frac{\Delta V}{V}} \) and substituting the values \( \Delta P = 2 \times 10^6 \text{ N/m}^2 \) and \( \frac{\Delta V}{V} = 0.001 \), we have: \[ K = - \frac{2 \times 10^6}{0.001} = -2 \times 10^9 \text{ N/m}^2 \]. The negative sign indicates compression, which is often dropped when reporting bulk modulus, so \( K = 2 \times 10^9 \text{ N/m}^2 \).

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Key Concepts

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

Elasticity
Elasticity refers to the ability of a material to return to its original shape and size after removing the forces causing deformation. It is a fundamental property that dictates how substances deform under stress.
Each material has its own elasticity, which determines how it will respond to applied forces. For instance:
  • Metals tend to have high elasticity, meaning they can undergo significant shape changes but return to their original form.
  • Rubber is very elastic, capable of stretching and then springing back to its starting size.
Elasticity is quantified in different ways, depending on the type of deformation:
  • Tensile elasticity, described by Young鈥檚 Modulus, relates to changes in length.
  • Bulk elasticity, quantified by Bulk Modulus, pertains to changes in volume, crucial for understanding how materials behave when submerged in fluids.
Volume Compression
Volume compression occurs when a material's volume decreases under pressure. It is an essential concept in understanding how objects behave in environments with high pressure, like deep underwater.
For the ball sinking to the lake's bottom, the change in its volume tells us how compressible the material is. In physics, this is a common scenario where volume compression is analyzed using:
  • Volume Strain: The fractional change in volume, expressed as \( \frac{\Delta V}{V} \), showing the percent decrease or increase in volume.
  • Bulk Modulus: Reflects a material鈥檚 resistance to changing volume, with higher values indicating less compressibility.
Volume compression provides insights into material integrity under pressure, vital for applications in engineering and material science.
Pressure in Fluids
In the realm of physics, pressure in fluids is a compelling topic, especially in contexts involving depth, such as lakes or oceans. Understanding fluid pressure is crucial because it helps explain how forces exert influence over submerged objects.
Here are some key points:
  • Hydrostatic Pressure: Increases with depth and is calculated using the equation: \( P = \rho gh \), where \( P \) is pressure, \( \rho \) is fluid density, \( g \) is gravitational acceleration, and \( h \) is depth.
  • Applications: This principle explains phenomena such as buoyancy, where submerged objects experience an upward force.
  • Implications for Material Design: Adjusting object characteristics and choosing appropriate materials can mitigate adverse effects from pressure changes.
In our example, understanding pressure variations can help in designing robust materials for underwater equipment.
JEE Advanced Physics
For students preparing for JEE Advanced Physics, grasping concepts like elasticity, volume compression, and pressure in fluids is indispensable. These topics not only appear in exams but also form the backbone of understanding physical phenomena in engineering and science.
To tackle JEE Advanced questions effectively:
  • Focus on fundamental formulas and their applications, like calculating pressure using \( P = \rho gh \).
  • Understand the interplay between physical concepts, such as how elasticity relates to volume changes.
  • Practice problem-solving to develop insight into applying these principles in complex scenarios.
For instance, questions often combine principles, requiring students to use multiple steps to arrive at a solution, similar to the exercise about the ball in the lake.

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Most popular questions from this chapter

A long wire hangs vertically with its upper end clamped. A torque of \(8 \mathrm{Nm}\) applied to the free end twists it through \(45^{\circ}\). The potential energy of the twisted wire is (1) \(\pi \mathrm{J}\) (2) \(\frac{\pi}{2} \mathrm{~J}\) (3) \(\frac{\pi}{4} \mathrm{~J}\) (4) \(\frac{\pi}{8} \mathrm{~J}\)

A piece of copper wire has twice the radius of a piece of steel wire. Young's modulus for steel is twice that of the copper. One end of the copper wire is joined to one end of the steel wire so that both can be subjected to the same longitudinal force. By what fraction of its length will the steel have stretched when the length of the copper has increased by \(1 \%\) ? (1) \(1 \%\) (2) \(2 \%\) (3) \(2.5 \%\) (4) \(3 \%\)

A metal wire length \(L\), cross-sectional area \(A\) and Young's modulus \(Y\) is stretched by a variable force \(F . F\) is varying in such a way that \(F\) is always slightly greater than the elastic forces of resistance in the wire. When the elongation in the wire is \(l\), up to this instant (1) the work done by \(F\) is \(\frac{Y A l^{2}}{2 L}\) (2) the work done by \(F\) is \(\frac{Y A l^{2}}{L}\) (3) the elastic potential energy stored in wire is \(\frac{Y A l^{2}}{2 L}\) (4) no energy is lost during elongation

A nylon rope \(2 \mathrm{~cm}\) in diameter has a breaking strength of \(1.5 \times 10^{5} \mathrm{~N}\). The breaking strength of a similar rope \(1 \mathrm{~cm}\) in diameter is (1) \(0.375 \times 10^{5} \mathrm{~N}\) (2) \(2 \times 10^{5} \mathrm{~N}\) (3) \(6 \times 10^{5} \mathrm{~N}\) (4) \(9 \times 10^{4} \mathrm{~N}\)

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
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