Question

A mild steel wire of length 2l meter cross-sectional area A m2 is fixed horizontally between two pillars. A small mass m kg is suspended from the mid point of the wire. If extension in wire are within elastic limit. Then depression at the mid point of the wire will be​

Answer 1

Explanation:

Given A mild steel wire of length 2l meter cross-sectional area A m^2 is fixed horizontally between two pillars. A small mass m kg is suspended from the mid point of the wire. If extension in wire are within elastic limit. Then depression at the mid point of the wire will be

• Now let the vertical displacement m be very small compared to L.SO change in length will be the difference of total length and the initial length 2L
• Therefore increase in length of the wire will be ΔL = 2(L^2 + m^2)^1/2 – 2L
•                                           = 2L (1 + m^2/L^2)^1/2 – 2L
• By applying Binomial theorem we get
•         So ΔL = 2L [ (1 + m^2/L^2)^1/2 – 1]
•                    = 2L[1 + ½ m^2/L^2 – 1]
•                   = m^2 / L  (because m < L)
•  Strain is the ratio of change in length to the original length.
•  So strain = ΔL / 2L
•                    = m^2/L / 2L
•                    = m^2 / 2L^2

Reference link will be

https://brainly.in/question/1457824