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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

A bar of mass m and length l is hanging from point A as shown in...

A bar of mass mm and length ll is hanging from point AA as shown in the figure. If the Young's modulus of elasticity of the bar is YY and area of cross-section of the wire is AA, then the increase in its length due to its own weight will be

A bar of mass m and length l is hanging from point A as shown in the figure. If the Young's modulus of elasticity of

Options

  1. AmgL2AY\dfrac{mgL}{2AY}
  2. BmgA2LY\dfrac{mgA}{2LY}
  3. Cmg2LAY\dfrac{mg}{2LAY}
  4. D2LYmgA\dfrac{2LY}{mgA}

Correct answer

A. mgL2AY\dfrac{mgL}{2AY}

Step-by-step solution

A bar of mass m and length l is hanging from point A as shown in the figure. If the Young's modulus of elasticity of

Consider a small section dxdx of the bar at a distance xx from B. The weight of the bar for a length xx is, W=(mgL)x.W = \left(\dfrac{mg}{L}\right)x .

Elongation in xx will be

de=(WAY)dx=(mgLAY)dx⋅xde = \left(\dfrac{W}{AY}\right)dx = \left(\dfrac{mg}{LAY}\right)dx ⋅ x .

Total elongation of the bar can be obtained by integrating this expression for x=0tox=L.x = 0tox = L .

∴\Deltal=∫x=0x=Lde=(mgLAY)∫0Lx⋅dx∴ \Deltal = \mathop{∫}_{x = 0}^{x = L}de = \left(\dfrac{mg}{LAY}\right)\mathop{∫}_{0}^{L}x ⋅ dx

⇒\Deltal=mgL2AY⇒ \Deltal = \dfrac{mgL}{2AY}
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