Question

A wire elongates $l$ mm when a load W is hanged from it. 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 (in mm):
A. $l$
B. $2l$
C. Zero
D. $\dfrac{l}{2}$

Answer 1

Hint: When we apply force on an elastic material, it tends to increase in length. Upon removal of the force it regains its shape and size. Here the elastic material is the wire. So to solve the question we need to think in terms of Strain and apply $Y = \dfrac{{Stress}}{{Strain}}$.

Complete step by step answer:
According to Hooke’s Law, when we apply force (stress) on an elastic material it tends to differ in size and shape and upon removal of force it regains its shape and size, but only in the elastic limit. In this elastic limit, the stress is directly proportional to the strain. So,
Given when applied a force W, the elongation $\Delta l = l$mm.
Let the original length and area of cross section be L and A respectively. So, in the original scenario we have,
\[Y = \dfrac{{Stress}}{{Strain}} = \dfrac{{\dfrac{W}{A}}}{{\dfrac{l}{L}}}\]
Now the wire has been hung over a pulley. Here the tension remains the same on both sides of the pulley because the same load has been applied. Under normal conditions the Young’s modulus remains the same. Therefore,
\[\dfrac{{\dfrac{W}{A}}}{{\dfrac{l}{L}}} = \dfrac{{\dfrac{W}{A}}}{{\dfrac{{{l_1}}}{{L/2}}}} \Rightarrow {l_1} = \dfrac{l}{2}\]
Same goes for the other side ${l_2} = \dfrac{l}{2}$
Hence, the total elongation remains same i.e. ${l_{pulley}} = {l_1} + {l_2} = \dfrac{l}{2} + \dfrac{l}{2} = l$
So, the correct answer is “Option A”.

Note:
Without doing all the cumbersome process we can just apply a simple logic. When the wire goes around the pulley what is the tension in the wire? It is T=W. it is the same for the whole wire. Now in the original case the elongation was $l$mm when tension was W. Therefore the tension being the same, the elongation will also be the same i.e. $l$mm.