Part 1: When tension is applied to an aluminum rod (Y=70 PaY=70 Pa) of length 1

Question

Part 1: When tension is applied to an aluminum rod (Y=70 PaY=70 Pa) of length 1 m, it stretches by L. If the same tension is applied to another aluminum rod with the same cross-sectional area, but of length 2 m, by how much will it stretch?

a.) Less than L
b.) L
c.) More than L

Part 2: Now tension is applied to two aluminium rods of length 1 m1 m, one twice as thick as the other. If a compressive force F is applied to both wires, their lengths are reduced by Lthick rod and Lthin rod respectively. The ratio Lthick rod/Lthin rod is:

a.) =1

b.) <1

c.) >1

Explanation / Answer

1. stress = Y strain

(F / A) = Y (deltaL / L)

deltaL / L = constant (for this problem)

delta(L) / 1 = delta(L') / 2

delta(L') = 2 delta(L)

Ans: (C)

2. deltaL A = constant

L_thick (2A) = Lthin (A)

L_thick / L_thing = 1/2 < 1

Ans(b)

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