Question
Download Solution PDFThe ratio of Young's modulus of elasticity of two materials (E1 to E2) is 2.5. Find the ratio of the elongations in the two bars (δl1 to δl2) of these materials if they are of the same length and same area and subjected to the same force P.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
As per Hooke’s law,
∵ σ = ϵ × E
Given
Two bars of unit length, i.e L1 = L2 = L
Elongation ratio , i.e δL1: δL2 = ?
Both subjected to same tensile load, i.e P1 = P2 = P
Both have the same size, i.e A1 = A2 = A
Ratio of Young's modulus of elasticity (E1 : E2) = 2.5
Calculation
Elongation of the first bar,δL1 = \({PL\over AE_{1}}\)
Elongation of the second bar, δL2 = \({PL\over AE_{2}}\)
Taking ratio,
\({\delta_{L1} \over \delta_{L2}} = {{PL\over AE_{1}}\over {PL\over AE_{2}}}\)= \({E_{2}\over E_{1}} = {1 \over 2.5} \) = 0.4
Hence the ratio of the elongations in the two bars (δl1 :δl2) of these materials = 0.4
Last updated on May 28, 2025
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