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 Jul 1, 2025
-> SSC JE notification 2025 for Civil Engineering has been released on June 30.
-> Candidates can fill the SSC JE CE application from June 30 to July 21.
-> SSC JE Civil Engineering written exam (CBT-1) will be conducted on 21st to 31st October.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.
-> The Staff Selection Commission conducts the SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.