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NEET PHYSICSMechanical Properties of SolidsEasy

Question

The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length 100 cm100\text{ cm} to stretch it by 1 mm1\text{ mm} is: (given: Young's modulus of the wire Y=2.0×1011 N/m2Y = 2.0 \times 10^{11}\text{ N/m}^2)

A

1011 J/m310^{11}\text{ J/m}^3

B

1017 J/m310^{17}\text{ J/m}^3

C

107 J/m310^7\text{ J/m}^3

D

105 J/m310^5\text{ J/m}^3

Step-by-Step Solution

  1. Identify Given Values: Length of wire (LL) = 100 cm=1 m100\text{ cm} = 1\text{ m} Extension (ΔL\Delta L) = 1 mm=103 m1\text{ mm} = 10^{-3}\text{ m}
  • Young's Modulus (YY) = 2.0×1011 N/m22.0 \times 10^{11}\text{ N/m}^2
  1. Calculate Strain: Strain=ΔLL=1031=103\text{Strain} = \frac{\Delta L}{L} = \frac{10^{-3}}{1} = 10^{-3}
  2. Formula for Energy Density: The elastic potential energy per unit volume (uu) is given by: u=12×Stress×Strainu = \frac{1}{2} \times \text{Stress} \times \text{Strain} Using Young's Modulus (Y=Stress/StrainY = \text{Stress} / \text{Strain}), we can substitute Stress = Y×StrainY \times \text{Strain}: u=12×Y×(Strain)2u = \frac{1}{2} \times Y \times (\text{Strain})^2
  3. Calculation: u=12×(2.0×1011)×(103)2u = \frac{1}{2} \times (2.0 \times 10^{11}) \times (10^{-3})^2 u=1.0×1011×106u = 1.0 \times 10^{11} \times 10^{-6} u=105 J/m3u = 10^5\text{ J/m}^3

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Mechanical Properties of Solids. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMechanical Properties of Solidsamountelasticpotentialenergyvolume

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