25 मिमी व्यास और 250 मिमी लंबाई के पीतल की छड़ के यंग मापांक ज्ञात करें, जिसे 50 kN के तन्य भार के अधीनकृत किया जाता है, जब रॉड का विस्तार 0.3 मिमी के बराबर होता है 

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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
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20 Questions 20 Marks 16 Mins

Detailed Solution

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Concept:

Elongation of a circular bar:

Extension of the rod under the axial pull,

 \(\delta l = \frac{P.L}{(\frac{\pi}{4}).(d^2)E}\)

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

\(0.3 = \frac{50000 \times 250}{(\frac{\pi}{4}).(25^2)E}\)

⇒ E = 84882.64 MPa

⇒ E = 84.883 GPa or 84.883 GN/m2

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