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Under the action of load F1, the length of a string is L1 and that under F2, is L2. the original length of the wire is(a) [L1F1 - L2F2] / [F1 + F2](b) [L1F2 - L2F1] / [F1 - F2](c) [L1F2 - L2F1] / [F2 - F1](d) [L1F2 - L2F1] / [F1 + F2]

sumit kumar , 11 Years ago
Grade Upto college level
anser 2 Answers
Saurabh Koranglekar

Last Activity: 4 Years ago

Dear student

Please find the link attached
https://www.askiitians.com/forums/Engineering-Entrance-Exams/under-the-action-of-load-f1-the-length-of-a-strin_77655.htm


Regards

Vikas TU

Last Activity: 4 Years ago

Let dL be the change in length

and L be the Original length

Therefore Y=(f1*L)/(A*dL) for L1

= (f1*L)/A*(L-L1)--------(1)

Y=(f2*L)/(A*dL) forL2

=(f2*L)/A(l-L2)-------(2)

Equating both 1&2

F1/(L-L1)=f2/(L-L2)

=>L=(f1L2-f2L1)/(f1-f2)

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