Question

The length of wire is $2.01m$ when $5kg$ hanging from it is and $2.02m$ when $10kg$ hanging, then natural length of wire is
A) $2m$
B) $1.95m$
C) $1.9m$
D) $2.005m$

Answer 1

In this question, we have to find original length by comparing because of the change in length i.e. longitudinal strain. So, the material of the wire has an elastic modulus known as Young’s modulus of elasticity. It remains constant for the material of the wire.

Formula used:
When we apply a force $F$ to the cross section area $A$ of wire then a stress is produced in the wire by which a strain is also produced.
When the stress and strain are produced in a longitudinal way, it is called longitudinal stress and longitudinal strain respectively.
According to the Hooke’s law- The longitudinal stress produced in any wire is directly proportional to strain produced in it.
Longitudinal stress $\propto$ Longitudinal strain
Or $\dfrac{{stress(longitudinal)}}{{strain(longitudinal)}} = Y$
Here, $Y$ is the Young’s modulus of elasticity which is constant for the material of the wire.

Complete step by step solution:
When a body of some mass is hung with any wire it’s length increases. Hence a strain will produce in the wire. And the mass which is hung with wire acts like an agent of force (gravitational force). If we consider the per unit area, this force will be equal to the stress exerted by the wire. So we can easily apply these conditions in this question.
According to the equation, we have-
The changed line ${L_1} = 2.01m$
With the hanging mass ${m_1} = 5kg$
And the changed length ${L_2} = 2.02m$
With the mass ${m_2} = 10kg$
Let the original length of the wire is $L$. Now, by using Young’s modulus-
$Y = \dfrac{{stress(longitudinal)}}{{strain(longitudinal)}}$
Here, we have-
Longitudinal stress$= \dfrac{F}{A}$
And the longitudinal strain $= \dfrac{{\Delta L}}{L}$
Where $\Delta L$ is the change in the length.
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$
Now, we know that $F = mg$.
So, $Y = \dfrac{{mgL}}{{A\Delta L}}$...................(i)
Now, we have a wire which has changes in length ${L_1}$and ${L_2}$respectively for different masses.
So, using equation (i), we get-
For first mass, ${Y_1} = \dfrac{{{M_1}gL}}{{A\Delta {L_1}}}$..................(ii)
And for second mass, ${Y_2} = \dfrac{{{M_2}gL}}{{A\Delta {L_2}}}$..............(iii)
But we know that the Young’s modulus remains constant, so,

$$\Rightarrow {Y_1} = {Y_2} \\\ \Rightarrow \dfrac{{{M_1}gL}}{{A\Delta {L_1}}} = \dfrac{{{M_2}gL}}{{A\Delta {L_2}}}$$

$\Rightarrow \dfrac{{{M_1}}}{{\Delta {L_1}}} = \dfrac{{{M_2}}}{{\Delta {L_2}}}$
Now, putting $\Delta {L_1} = {L_1} - L$ and $\Delta {L_2} = {L_2} - L$. So, we get-
$\Rightarrow \dfrac{{{M_1}}}{{{L_1} - L}} = \dfrac{{{M_2}}}{{{L_2} - L}}$
Substituting the values of ${M_1},{M_2},{L_1}$ and ${L_2}$, we get-
$\Rightarrow \dfrac{5}{{(2.01 - L)}} = \dfrac{{10}}{{(2.02 - L)}} \\\ \Rightarrow 5 \times (2.02 - L) = 10 \times (2.01 - L)$
$\Rightarrow 10.10 - 5L = 20.1 - 10L \\\ \Rightarrow 10L - 5L = 2.1 - 10.10 \\\ \Rightarrow 5L = 10 \\\ \Rightarrow L = 2m$
Hence, the original length of the wire is $2m$.

Therefore, option A is correct.

Note: The most important thing to remember is that the Young’s modulus of any material remains constant and the cross section area is the same as we add extra weight to the original length not in strained wire.