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Question

Two rods of length L2 and coefficient of linear expansion α2 are connected freely to a third rod of length L1 of coefficient of linear expansion α1 to form an isosceles triangle. The arrangement is supported on the knife edge at the midpoint of rod of length L1, which is horizontal. The apex of the isosceles triangle remains at a constant distance from the knife edge, if L1L2=n(α2α1)12. Then value of n is

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Solution

The apex of the isosceles triangle to remain at a constant distance from the knife edge. Thus, DC should remain constant before and after heating.

Before expansion: In triangle ADC
(DC)2=L22(L12)2 (i)
After expansion:
(DC)2=[L2(1+α2t)]2[L12(1+α1t)]2 (ii)
Equating Eqs. (i) and (ii), we get
L22(L12)2=[L2(1+α2t)]2[L12(1+α1t)]2
L22L214=L22+L22×2α2×tL214L214×2α1×t
(Neglecting higher terms)
L22(2α2)t=L214(2α1t)L1L2=2α2α1
Why this question?

Application of binomial theorem (1+x)n=1+nx when x0 in physics.

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