Question

The mass M was initially hanging from a free end of a rod of length L and area of cross-section A. Extension caused by this mass is ∆L. Another rod (R) of same material and same length L but area of cross section A/2 is carrying a load of 3*M. Find length extension experienced by rod R in terms of ∆L​

Answer 1

Answer:

in 1st case-

longitudinal stress= weight of load /area of cross section

weight= mass×acceleration due to gravity

= M×g

∴stress=( M×g)/A

longitudinal strain= change in length/original length

∴ strain= ΔL/L

stress/strain=MgL/AΔL

in 2nd case-

weight= 3*M×g

∴stress= (3*M×g)/(A/2)

strain= ΔL'/L

stress/strain=6MgL/AΔL'

By applying Hooke's law

Stress/strain= constant for same material

∴MgL/AΔL=6MgL/AΔL'

⇒1/ΔL=6/ΔL'

∴ΔL'=6ΔL

Ans:- length extension experienced by rod R is 6ΔL