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When a weight W is hung from one end of ...

When a weight `W` is hung from one end of a wire of length `L` (other end being fixed), the length of the wire increases by `l`. If the same wire is passed over a pulley and two weights `W` each are hung at the two ends,what will be the total elongation in the wire?

A

`l`

B

`2l`

C

`3l`

D

`(l)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
In case (a), let `Y` be Young's modulus of the material of the wire. IF a is its area of cross section, then we can write
`F = (F//a)/(l//L) = (WL)/(al) ( :' F = W)`
Incrases in length of wire `l = (WL)/(aY)`
In case (b), we can treat either segment of length `L//2` in Fig. as if one of its ends is fixed while the other end is attached to the load `W` beacuse the point `M` remains stationary, i.e.m fixed all the time due to symmeterical load. Thus, when the wire is passed over the pulley let `'l'` be the incrases in the length of each segement.
Since each segment is of length `L//2` we have `Y = (W(L//2))/(al')`
Increases in length of wire of one side of pulley
`I' = (1)/(2) (WL)/(aY) = (1)/(2)`
Therefore, increases in length of wire on one side of pulley
`l' = (1)/(2) (WL)/(aY) = (1)/(2)`
Therefore, increases in length of both the segments of the wire `l' + l' = l//2 = l`
So the increases in length remains the same.
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