Question

A wire elongates by \[l\text{ }mm\] when a load W is changed from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation is
A) Zero
B) \[0.5l\]
C) \[l\]
D) \[2l\]

Answer 1

Hint: The Hooke’s law states that the force exerted by a compress spring by a certain length or distance along with the product of spring constant but we need to convert this statement according to the data given or to form the Young’s Modulus in the question i.e. the force in the Hooke’s law will be the ratio of the weight and the acceleration and the spring constant will be the ratio of original length to extended length. Now to find this extended length we will also need the elongated length whose formula is:
\[\Delta =\dfrac{WL}{E}\]
And the Young’s Modulus \[\left( E \right)\]:
\[E=\dfrac{WL}{AI}\]
where \[A\] is the area of the cross section of the rope, \[L\] is the original length, \[W\] is the weight of the wire and the elongated length is \[l\].

Complete step by step solution:
Now according to the Hooke’s law the modulus of elasticity of the wire in terms of weight, area and length is given as:
\[E=\dfrac{WL}{AI}\]
The load or the weight that is on the wire is given as \[W\]
The original length of the length without the extension is given as \[L\]
The area of the wire is given as \[A\]. Now placing all the variables in the formula, we get the Young’s Modulus as:
\[E=\dfrac{WL}{AI}\]
Now as we have the modulus let us find the elongation of the wires due to the weights added, the elongation value of the wire is:
\[\Delta =\dfrac{WL}{E}\]
Now to find the elongated length we need to find the product of length elongated and the elongation variable. Therefore, with tension on both sides the elongated length is:
\[=\dfrac{\dfrac{WL}{2}}{AE}\]
\[=\dfrac{WL}{2AE}\]
The length of the wire due to the tension \[W\] is taken as \[\dfrac{L}{2}\]. Therefore, replacing the values of \[\dfrac{WL}{AE}\] by \[\dfrac{l}{2}\], we get the total elongation as \[\dfrac{l}{2}\].
Now there are two weights are pulled from the pulley making the total elongation as \[\dfrac{l}{2}+\dfrac{l}{2}\] or \[l\text{ }mm\].
Therefore, elongation when two wires are stretched is \[l\text{ }mm\].

Note: Young’s modulus and modulus of elasticity are the same thing for those wondering if the both of them are different; both of them state the formula which shows the ratio between stress and strain. Although the Young’s Modulus has the variable \[E\] and modulus of elasticity as \[\lambda \]. Both of them are extended versions of the Hooke’s law or Hooke’s Law is the base for both the formulas. Hence both of them are correct if used.