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Modulus of rigidity of ideal liquids is [NCERT Exemplar] (a) infinity (b) zero (c) unity (d) some finite small non-zero constant value

Short Answer

Expert verified
(b) zero

Step by step solution

01

Understanding Modulus of Rigidity

The modulus of rigidity, also known as the shear modulus, measures a material's ability to withstand shear stress. It describes the material's response to shear (lateral) deformation.
02

Analyze Properties of Ideal Liquids

For ideal liquids, it is assumed they cannot resist shear forces. This means they do not maintain their shape when a shear force is applied and will start flowing as soon as any shear stress is applied.
03

Apply the Concept to Ideal Liquids

Since ideal liquids do not resist shear forces, they exhibit no shear deformation resistance. Hence, their modulus of rigidity is expected to be extremely low.
04

Conclusion Based on Ideal Liquid Characteristics

Because any shear stress applied to an ideal liquid results in flow without resistance, the modulus of rigidity for an ideal liquid is zero.

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Key Concepts

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

Shear Modulus
The shear modulus, interchangeably known as the modulus of rigidity, plays a crucial role in understanding the mechanical properties of materials. It is a measure of a material's ability to resist changes in shape when subjected to shear stress. In simpler terms, when forces are applied parallel to a material's surface, the shear modulus quantifies how much the material deforms.

Mathematically, the shear modulus \( G \) is defined as the ratio of shear stress \( \tau \) to the shear strain \( \gamma \):
\[ G = \frac{\tau}{\gamma} \]
  • Shear Stress (\( \tau \)): This is the force per unit area acting parallel to the material's surface.
  • Shear Strain (\( \gamma \)): This indicates the amount of deformation or angular change in the material.

Shear modulus is an important factor for engineers and scientists in determining and predicting how materials will react under different conditions, ensuring structural integrity and safety.
Ideal Liquids
Ideal liquids are theoretical constructs used to simplify the study of fluid dynamics. They are characterized by the absence of both viscosity and resistance to shear stress, meaning they can't sustain shear forces. As soon as a shear force is applied, an ideal liquid will start to flow without resistance. This is because ideal liquids have no internal friction; their particles can move past each other freely.

When considering the modulus of rigidity for ideal liquids:
  • They have no shear deformation resistance.
  • Their ability to resist shear is nonexistent, making their shear modulus zero.

This zero modulus of rigidity aligns with the definition that ideal liquids flow freely upon any shear stress application, emphasizing their theoretical perfection in fluid behavior.
Shear Stress
Shear stress is a critical concept in understanding how materials deform under applied forces. It represents the force acting parallel to the surface per unit area, causing the layers within an object or fluid to slide past each other.

In real-world applications:
  • Structural Materials: In materials like metals and plastics, shear stress analysis helps in determining how load and force distribute before causing yielding or failure.
  • Fluids: For fluids, shear stress impacts flow characteristics, viscosity, and the transition between fluid flow types (laminar to turbulent).

The measurement of shear stress is essential to ensure that materials and fluids will perform effectively and safely under expected use conditions, exemplifying the importance of fundamental physical concepts in practical scenarios.
Deformation Resistance
Deformation resistance is a material's ability to withstand alterations in shape under applied forces. This resistance is crucial for maintaining structural integrity and functionality in various fields, from construction to machinery.

Key points about deformation resistance include:
  • Shear Deformation Resistance: This is particularly relevant in assessing how a material can resist changes induced by shear stress.
  • Importance: It ensures that materials don't easily snap, bend, or reshape under regular use conditions, which is crucial for safety and longevity.

Understanding deformation resistance helps in selecting appropriate materials for specific applications, ensuring durability and performance standards are met across various industries.

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Most popular questions from this chapter

A steel cable with a radius of \(1.5 \mathrm{~cm}\) supports a chair lift at a ski area. If the maximum stress is not to exceed \(10^{8} \mathrm{~N} / \mathrm{m}^{2}\), what is the maximum load the cable can support? (a) \(7 \times 10^{5} \mathrm{~N}\) (b) \(7 \times 10^{6} \mathrm{~N}\) (c) \(7 \times 10^{4} \mathrm{~N}\) (d) \(9 \times 10^{5} \mathrm{~N}\)

A cube is compressed at \(0^{\circ} \mathrm{C}\) equally from all sides by an external pressure \(p\). By what amount should be temperature be raise to bring it back to the size it had before the external pressure was applied? (Given \(K\) is bulk modulus of elasticity of the material of the cube and \(\alpha\) is the coefficient of linear expansion.) (a) \(\frac{p}{K \alpha}\) (b) \(\frac{p}{3 K \alpha}\) (c) \(\frac{3 \pi \alpha}{p}\) (d) \(\frac{K}{3 p}\)

The following data were obtained when a wire was stretched within the elastic region Force applied to wire \(100 \mathrm{~N}\) Area of cross-section of wire \(10^{-6} \mathrm{~m}^{2}\) Extension of wire \(2 \times 10^{-9} \mathrm{~m}\) Which of the following deductions can be correctly made from this data? 1\. The value of Young's modulus is \(10^{11} \mathrm{Nm}^{-2}\) 2\. The strain is \(10^{-3}\) 3\. The energy stored in the wire when the load is applied is \(10 \mathrm{~J}\) (a) \(1,2,3\) are correct (b) 1,2 are correct (c) 1 only (d) 3 only

If the compressibility of water is \(\sigma\) per unit atmospheric pressure, then the decrease in volume (V) due to atmospheric pressure \(\boldsymbol{p}\) will be (a) \(\boldsymbol{\sigma} \boldsymbol{p} / \mathrm{V}\) (b) opV (c) \(\sigma / p V\) (d) \(\mathrm{oV} / \mathrm{P}\)

A metal wire of length \(L\), area of cross-section \(A\) and Young's modulus \(Y\) is stretched by a variable force \(F\) such that \(F\) is always slightly greater than the elastic forces of resistance in the wire. When the elongation of the wire is l (a) the work done by \(F\) is \(\frac{Y A l^{2}}{L}\) (b) the work done by \(F\) is \(\frac{Y A l^{2}}{2 L}\) (c) the elastic potential energy stored in the wire is \(\frac{Y A l^{2}}{2 L}\) (d) heat is produced during the elongation

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