Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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]

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]

Grade:Upto college level

2 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago
Vikas TU
14149 Points
one year 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)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free