Question

long answer type question

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Answer 1

Here is your answer in the attachment given above

Answer 2

Answer:

given steel=6 x 10^11 N/m^2 area of cross

 section of wire =1  mm^2 ,g=10m/s^2

velocity at lowest position =√6g=√2*10=√20 m/s        ,∴g=10m/s^2

now force at lowest position F=6g+2*20/1=  2*10+40  = 20+40    = 60 N

Y=60/10^-6/ΔL/1  

2×10^11=60/ΔL*10^-6

60/2*×10^11*10^-6=ΔL

ΔL=3*10^-4 m

ΔL=0.9 mm

maximum extension in the length of wire is  0.3mm