Questions
If a rod of length , very small area of cross-section A, Young’s modulus of elasticity Y is acted upon by two parallel forces 3F and F respectively (as shown) and placed on a smooth horizontal plane. If the elastic limit is not crossed and then to study the change in length of rod and it’s elastic potential energy (U) the rod is segmented into four equal parts where magnitude of change in lengths are and elastic potential energy stored in each segment are respectively as shown then which is/are correct?
detailed solution
Correct option is A
In steady state, the F.B.D. of the rod can be drawn by dividing it into four equal parts as shown. We have YΔll=stressaverage⇒Δl=lY.FavgA=lY.F1+F22A For any rod . If Now, Δl1compression=Favg(length1)AY=F+02l4AY=Fl8AY;Δl2elongation=Fl8AY acceleration of the system=a=3F+F4m=FmFrom FBD of all blocks we get N1,N2 and N3 are the normal reactions betweeen 1st and 2nd blocks, 2nd and 3rd blocks, and 3rd and 4th blocks on solving we get N1=0, N2=F and N3=2FΔl3( elongation )=2F+F2l4AY=3Fl8AY,Δl4( elongation )=2F+3F2l4AY=5Fl8AYΔl=8FI8AY=FlAY ΔU∝Strain 2⇒ΔU∝(Δl)2Talk to our academic expert!
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A composite rod consists of a steel rod of length 25 cm and area 2A and a copper rod of length 50 cm and area A. The composite rod is subjected to an axial load F. If the Young's moduli of steel and copper are in the ratio 2:1, then
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