Question

The length of a metal wire is \[{l_1}\] ​ when the tension in it is \[{T_1}\] and is \[{l_2}\] when the tension is \[{T_2}\] ​. The natural length of wire is
(a)\[\dfrac{{{l_1} + {l_2}}}{2}\]
(b)$\sqrt {{l_1}{l_2}} $
(c) $\dfrac{{{l_1}{T_2} - {l_2}{T_1}}}{{{T_2} - {T_1}}}$
(d) $\dfrac{{{l_1}{T_2} + {l_2}{T_1}}}{{{T_2} + {T_1}}}$

Answer 1

Hint: Length, change in length and tension appear together in the expression for Young’s Modulus of elasticity. Y does not change for a given material.

Formula Used:
1. Young’s Modulus of elasticity: $Y = \dfrac{{Stress}}{{Strain}}$ ……(1)
2. Stress is defined as: $Stress = \dfrac{F}{A}$ ……(2)
Where,
F is force acted on the object
A is the area on which the force acts.
3. Strain is defined as: $Strain = \dfrac{{\Delta l}}{{{l_0}}}$ ……(3)
Where,
$\Delta l$ is the change in length of the object when force is applied on it
${l_0}$ is the original length of the object or natural length.
4. Change in length $\Delta l = {l_f} - {l_i}$ ……(4)
Where,
${l_i}$ is the initial length of the object
${l_f}$ is the final length of the object

Complete step by step answer:
Given:
1. Length of string in 1^st case: \[{l_1}\]
2. Tension on string in 1^st case: \[{T_1}\]
3. Length of string in 2^nd case: \[{l_2}\]
4. Tension on string in 2^nd case: \[{T_2}\]

To find: The natural length of the string.

Step 1:
Use eq (1), (2) and (3) to find the expression for Y in terms of length, tension and area.
\[Y = \dfrac{F}{A} \times \dfrac{{{l_0}}}{{\Delta l}}\] ……(5)
Let the natural length of the wire be ${l_0}$ and area be A.
Consider case 1.
Find $\Delta l$ using eq (4):
$\Delta l = {l_1} - {l_0}$
Force acting on the wire is tension. So, $F = {T_1}$
Use eq (5) to find Y:
\[Y = \dfrac{{{T_1}}}{A} \times \dfrac{{{l_0}}}{{{l_1} - {l_0}}}\] ……(6)

Step 2:
Consider case 2.
Find $\Delta l$ using eq (4):
$\Delta l = {l_2} - {l_0}$
Force acting on the wire is tension. So, $F = {T_2}$
Use eq (5) to find Y:
\[Y = \dfrac{{{T_2}}}{A} \times \dfrac{{{l_0}}}{{{l_2} - {l_0}}}\] ……(7)

Step 3:
Now, Young’s modulus for a material remains the same. As the wire is the same, its Y will not change. So, we can equate eq (6) and (7):
\[\dfrac{{{T_1}}}{A} \times \dfrac{{{l_0}}}{{{l_1} - {l_0}}} = \dfrac{{{T_2}}}{A} \times \dfrac{{{l_0}}}{{{l_2} - {l_0}}}\]
Simplifying:
\[
  \dfrac{{{T_1}}}{{{l_1} - {l_0}}} = \dfrac{{{T_2}}}{{{l_2} - {l_0}}} \\
  {T_1}({l_2} - {l_0}) = {T_2}({l_1} - {l_0}) \\
  {T_1}{l_2} - {T_1}{l_0} = {T_2}{l_1} - {T_2}{l_0} \\
  {l_0} = \dfrac{{{l_1}{T_2} - {l_2}{T_1}}}{{{T_2} - {T_1}}} \\
 \]

Final Answer
The natural length of wire is: (c) $\dfrac{{{l_1}{T_2} - {l_2}{T_1}}}{{{T_2} - {T_1}}}$

Note: Young’s modulus of a material is an intrinsic property of the material. It is the measure of tensile strength of a material. For a particular material it is isotropic, but differs material to material.