1.

The length of a metallic wire is L1. When the tension in the wire is T1; and length is L2, when the tension in the wire is T2. Find the original length of the wire.

Answer»

Let 'L' and 'a' be the original length and the area of cross-section of the wire respectively. If on applying a force F, extension produced is l, then

Y = (F/a)/(l/L) = (Fl)/(al)

In the first case, F = T1 and l = L1 - L

Therefore, Y = {T1L}/{a(L1 - L)}   ...(i)

In the second case, F = T2 and l = L2 - L

Therefore, Y = {T2L}/{a(L2 - L)}   ...(ii)

From the equations (i) and (ii), we have

{T1L}/{a(L1 - L)} = {T2L}/{a(L2 - L)}

or, T1(L2 - L) = T2(L1 - L)

or, L(T2 - T1) = T2L1 - T1L2

or, L = ({T2L1 - T1L2}/{T2 - T1})


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