Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length $100\text{ cm}$ to stretch it by $1\text{ mm}$ is: (given: Young's modulus of the wire $Y = 2.0 \times 10^{11}\text{ N/m}^2$ )

$10^{11}\text{ J/m}^3$

$10^{17}\text{ J/m}^3$

$10^7\text{ J/m}^3$

$10^5\text{ J/m}^3$

Step-by-Step Solution

Identify Given Values: Length of wire ( $L$ ) = $100\text{ cm} = 1\text{ m}$ Extension ( $\Delta L$ ) = $1\text{ mm} = 10^{-3}\text{ m}$

Young's Modulus ( $Y$ ) = $2.0 \times 10^{11}\text{ N/m}^2$

Calculate Strain: $\text{Strain} = \frac{\Delta L}{L} = \frac{10^{-3}}{1} = 10^{-3}$
Formula for Energy Density: The elastic potential energy per unit volume ( $u$ ) is given by: $u = \frac{1}{2} \times \text{Stress} \times \text{Strain}$ Using Young's Modulus ( $Y = \text{Stress} / \text{Strain}$ ), we can substitute Stress = $Y \times \text{Strain}$ : $u = \frac{1}{2} \times Y \times (\text{Strain})^2$
Calculation: $u = \frac{1}{2} \times (2.0 \times 10^{11}) \times (10^{-3})^2$ $u = 1.0 \times 10^{11} \times 10^{-6}$ $u = 10^5\text{ J/m}^3$

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