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 ......

A wire of length L is fixed at one end. It is elongates by I when a lo

1.	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 ......
Answer» `l/2` l 2l 4l Solution :`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`