A wire of length L is fixed at one end. It is elongates by I when a load W is hanged from other end. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be ......

Options

A. `l/2`
B. l
C. 2l
D. 4l

Answer

B

Explanation

`Y=( F//A)/(l //L) `
`therefore l = (WL )/(AY) " "[because F =W]`
Now wire goes over a pulley and weight W is suspended then length in both part increase suppose to l.
`therefore . = (WL//2)/( AY) = 1/2 (WL)/(AY) = l/2`
`therefore` Total extension in wire `= l ^(1) + l ^(1) = (l)/(2) + l/2=l`