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Question

Two rods each of length L2 and coefficient of linear expansion α2 each are connected freely to a third rod of length L1 and coefficient of expansion α1 to form an isosceles triangle. The arrangement is supported on a knife-edge at the midpoint of L1 which is horizontal. What relation must exist between L1 and L2 so that the apex of the isosceles triangle is to remain at a constant height from the knife edge as the temperature changes.

A
4L22α2=L21α1
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B
4L22α2=5L21α1
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C
2L22α2=3L21α1
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D
L22α2=3L21α1
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Solution

The correct option is A 4L22α2=L21α1
The rods AC and BC are of equal length, the length being L2.AB=L1. The rod AB is supported at its midpoints D so that the apex C is exactly above D and let CD = h.
For all temperature changes CD should remain a constant. Let a temperature increase of ΔT take place and AB extends to A1B1.

A1B1=L1+ΔL1Now A1C=B1C is the new length of rods AC and BCA1C=L2+ΔL2Let us consider the triangle ACD,CD2=h2=L22(L12)2(1)In the triangle A1CDCD2=h2=(L2+ΔL2)2(L1+ΔL12)2=L22+2L2ΔL2[(L12)2+2L12.ΔL12](2)Omitting the terms (ΔL1)2 and (ΔL2)2 because of (ΔL1) and (ΔL2) being very small.From eqn.(1) and (2)L22(L212)2=L22+2L2ΔL2)(L212)2L1ΔL12or 4L2ΔL2=L1ΔL1As ΔL1=L1α1ΔT and ΔL2=L2α2ΔTHence, 4L22α2=L21α1

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