Find Young's modulus of a brass rod of diameter 25 mm and length 250 mm which is subjected to a tensile load of 50 kN, when the extension of the rod is equal to 0.3 mm. 

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BDL MT Mechanical 16 April 2022 Official Paper
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  1. 12.775 GN/m2
  2. 6.945 GN/m2
  3. 23.112 GN/m2
  4. 84.883 GN/m2

Answer (Detailed Solution Below)

Option 4 : 84.883 GN/m2
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ST 1: BDL MT Thermodynamics
20 Qs. 20 Marks 16 Mins

Detailed Solution

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Concept:

Elongation of a circular bar:

Extension of the rod under the axial pull,

 

Where, P= load applied, L = length of the bar, E =  Modulus of elasticity, d = diameter of the bar

Calculation:

Given,

L = 250 mm

δL = 0.3 mm, P = 50 kN, d = 25 mm

⇒ E = 84882.64 MPa

⇒ E = 84.883 GPa or 84.883 GN/m2

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