If \(\mathrm{x}\), longitudinal strain is produced in a wire of young's modulus \(\mathrm{y}\) then energy stored in the material of the wire per unit volume is...... (A) \(\mathrm{yx}^{2}\) (B) \(2 \mathrm{yx}^{2}\) (C) \((1 / 2) \mathrm{y}^{2} \mathrm{x}\) (D) \((1 / 2) \mathrm{yx}^{2}\)

Short Answer

Expert verified
The energy stored in the material of the wire per unit volume is \(\frac{1}{2} Yx^2\). Therefore, the correct answer is (D) \(\frac{1}{2} \mathrm{yx}^{2}\).

Step by step solution

01

Recall the formula for elastic potential energy

The elastic potential energy, U, stored in a wire when it is subjected to longitudinal strain is given by the formula: \(U = \frac{1}{2} \cdot F \cdot \Delta L\) Where F is the force exerted on the wire and ΔL is the change in length of the wire.
02

Express force in terms of stress and area

Stress is the force applied per unit area. We can express the force acting on the wire (F) as the product of stress and the wire's cross-sectional area (A): \(F = \sigma \cdot A\)
03

Relate stress with Young's modulus and strain

Young's modulus (Y) is defined as the ratio of stress to strain: \(\sigma = Y \cdot x\) Where x is the longitudinal strain.
04

Substitute stress in the Force equation

Substitute the expression for stress in terms of Young's modulus and strain into the force equation: \(F = Y \cdot x \cdot A\)
05

Substitute force and change in length in the elastic potential energy formula

Substitute the expression for force and change in length (ΔL) in terms of strain (x) and original length (L) into the elastic potential energy formula: \(U = \frac{1}{2} \cdot (Y \cdot x \cdot A) \cdot (x \cdot L)\)
06

Simplify the elastic potential energy expression

Simplify the expression for elastic potential energy: \(U = \frac{1}{2} \cdot Y \cdot x^{2} \cdot A \cdot L\)
07

Divide the elastic potential energy by volume

Divide the elastic potential energy by the volume of the wire (V = A ⋅ L) to find the energy stored per unit volume: \(\frac{U}{V} = \frac{1}{2} \cdot Y \cdot x^{2}\) So, the energy stored in the material of the wire per unit volume is (1/2)yx². Therefore, the correct answer is (D) \((1 / 2) \mathrm{yx}^{2}\).

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