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If \(S\) is stress and \(Y\) is Young's modulus of material of a wire, then the energy stored in the wire per unit volume is (A) \(2 S^{2} Y\) (B) \(S^{2} / 2 Y\) (C) \(2 Y / S^{2}\) (D) \(S / 2 Y\)

Short Answer

Expert verified
The short answer is: (B) \(S^{2} / 2 Y\)

Step by step solution

01

Write the expression for Strain energy density

The given formula for strain energy density (U) is: \(U\) = (1/2) × stress × strain
02

Re-write the strain from the given Young's modulus formula

Using the relationship (Y = S /strain), we rewrite the strain in terms of stress and Young's modulus: strain = S / Y
03

Substitute the value of the strain obtained in Step 2 into the formula in Step 1

Now we substitute the strain value (S/Y) into the strain energy density formula: \(U\) = (1/2) × stress × (S/Y)
04

Simplify the expression

We can now simplify the expression for strain energy density: \(U\) = (1/2) × S × (S/Y) \(U\) = \(S^{2}\) / (2Y)
05

Match the expression with the given options

Comparing the derived expression, \(U\) = \(S^{2}\) / (2Y), with the given options, we find that it matches with the option (B). So the correct answer to this exercise is (B) \(S^{2} / 2 Y\).

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Key Concepts

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

Stress in Physics
Understanding stress is a fundamental concept in physics, especially when it comes to the fields of material science and mechanical engineering. Stress is a measure of the internal force an object experiences per unit area when a load is applied. It's essentially the force causing a deformable object to deform. The formula for stress (\textbf{S}) is given by the ratio of force (\textbf{F}) to the area (\textbf{A}) over which the force is distributed:
\[ S = \frac{F}{A} \]
Various forms of stress include tensile stress, compressive stress, and shear stress, each relating to different types of loads and deformations. Understanding how materials respond to stress is crucial for designing structures and machinery that can withstand everyday use without failure.
The exercise provided asks us to relate stress to strain energy density, showcasing the critical role stress plays in determining how much energy is stored within a material when it is deformed.
Young's Modulus
Young's Modulus, represented as (\textbf{Y}), is a fundamental mechanical property that is a measure of a material's stiffness. It is a ratio of stress to strain and is used to predict the elongation or compression of an object when a force is applied to it. The relationship is expressed as:
\[ Y = \frac{S}{\text{strain}} \]
In the context of the given exercise, Young's Modulus plays a pivotal role in determining the strain energy density of a material. Generally, materials with a high Young's Modulus are more rigid and less deformable under load, which implies that they would store a different amount of strain energy compared to materials that are more flexible. Young's Modulus is not only a crucial concept for physics students but also for engineers, who use this property to choose the right materials for their applications.
Material Deformation Physics

Understanding Material Deformation

Material deformation physics focuses on how materials deform under various stresses. When a force is applied to an object, it can undergo either elastic or plastic deformation. Elastic deformation is reversible, meaning the material will return to its original shape after the stress is removed. Plastic deformation, on the other hand, leads to permanent shape changes.
The aforementioned exercise leverages the concept of elastic deformation, given that strain energy is stored and potentially recoverable. Strain energy is a form of potential energy that arises due to the deformation of an object. This energy is directly related to both the stress applied to the material and the resultant strain—a consequence of how well the material can resist deformation, which in turn depends on its Young's Modulus.
Physics for JEE MAIN
Aspiring students preparing for the Joint Entrance Examination (JEE) Main, which is a gateway for engineering courses in India, must grasp concepts like stress, Young's Modulus, and material deformation to excel.
Problems such as the one presented in the exercise are common in the JEE Main Physics syllabus because they integrate concepts of mechanics and material properties—critical for understanding the fundamental principles of physical forces and responses in materials. Such problems require not only conceptual understanding but also application skills. Through practicing these types of questions, students hone their analytical and problem-solving skills, which are vital for success in the examination and subsequent engineering studies. Additionally, recognizing the interrelation between these concepts can often be the key to figuring out complex problems and is an indicative approach of JEE Main’s testing pattern.

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Most popular questions from this chapter

An iron tyre is to be fitted onto a wooden wheel \(1.0 \mathrm{~m}\) in diameter. The diameter of the tyre is \(6 \mathrm{~mm}\) smaller than that of wheel. The tyre should be heated so that its temperature increases by a minimum of (coefficient of volumetric expansion of iron is \(3.6 \times 10^{-5} /{ }^{\circ} \mathrm{C}\) ) (A) \(167^{\circ} \mathrm{C}\) (B) \(334^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(1000^{\circ} \mathrm{C}\)

A soap bubble in vacuum has a radius of \(3 \mathrm{~cm}\) and another soap bubble in vacuum has a radius of \(4 \mathrm{~cm}\). If two bubbles coalesce under isothermal conditions, then the radius of the new bubble is (A) \(2.3 \mathrm{~cm}\) (B) \(4.5 \mathrm{~cm}\) (C) \(5 \mathrm{~cm}\) (D) \(7 \mathrm{~cm}\)

As the temperature is increased, the period of pendulum, (A) Increases as its effective length increases even though its centre of mass still remains at the centre of the bob. (B) Decreases as its effective length increases even through its centre of mass still remains at the centre of the bob. (C) Increases as its effective length increases due to shifting to centre of mass below the centre of the bob. (D) Decreases as its effective length remains same but the centre of mass shifts above the centre of the bob.

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly, (A) its speed of rotation increases. (B) its speed of rotation decreases. (C) its speed of rotation remains same. (D) its speed in increases because its moment of inertia increases.

An aluminium sphere is dipped into water. Which of the following is true? (A) Buoyancy will be less in water at \(0^{\circ} \mathrm{C}\) than that in water at \(4^{\circ} \mathrm{C}\) (B) Buoyancy will be more in water at \(0^{\circ} \mathrm{C}\) than that in water at \(4^{\circ} \mathrm{C}\) (C) Buoyancy in water at \(0^{\circ} \mathrm{C}\) will be same as that in water at \(4^{\circ} \mathrm{C}\) (D) Buoyancy may be more or less in water at \(4^{\circ} \mathrm{C}\) depending on the radius of the sphere

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