Question

A cable breaks if stretched by more than 2 mm. It is cut into 2 equal parts. How much either part can be stretched without breaking?

Answer 1

Hint: The parameters stretch and the length are related by Young’s modulus. So, using Young’s modulus formula we will begin with the formation of the expression that relates the stretch and the length of the cable before and after cutting the cable into equal parts.
Formula used:
\[Y=\dfrac{FL}{A\Delta L}\]

Complete answer:
The parameters stretch and the length are related by Young’s modulus. The formula for computing the same is given as follows.
\[Y=\dfrac{FL}{A\Delta L}\]
Where \[F\] is the force applied, \[L\] is the original length, \[A\] is the area and \[\Delta L\] is the change in the length.
From the given information, we have the data as follows.
The length of the cable before cutting into 2 equal parts, is, \[L=2mm\].
Now consider the before and after situation of the cable in terms of Young’s modulus.
The original length of the cable.
\[{{Y}_{1}}=\dfrac{{{F}_{1}}{{L}_{1}}}{{{A}_{1}}\Delta {{L}_{1}}}\]
When the cable is cut into 2 equal parts.
\[{{Y}_{2}}=\dfrac{{{F}_{2}}{{L}_{2}}}{{{A}_{2}}\Delta {{L}_{2}}}\]
As the stretch per unit length remains constant, so,
\[{{Y}_{1}}={{Y}_{2}}\]
\[\Rightarrow \dfrac{{{F}_{1}}{{L}_{1}}}{{{A}_{1}}\Delta {{L}_{1}}}=\dfrac{{{F}_{2}}{{L}_{2}}}{{{A}_{2}}\Delta {{L}_{2}}}\]
The force applied to stretch the cable and the area of the cable remains constant, so,
\[\Rightarrow \dfrac{{{L}_{1}}}{\Delta {{L}_{1}}}=\dfrac{{{L}_{2}}}{\Delta {{L}_{2}}}\]
Substitute the value of the length in the above equation.
\[\begin{align}
  & \Rightarrow \dfrac{{{L}_{1}}}{\Delta {{L}_{1}}}=\dfrac{{{L}_{2}}}{\Delta {{L}_{2}}} \\
 & \Delta {{L}_{2}}=\dfrac{{{L}_{2}}}{{{L}_{1}}}\Delta {{L}_{1}} \\
 & \Rightarrow \Delta {{L}_{2}}=\dfrac{{}^{L}/{}_{2}}{L}\times \Delta {{L}_{1}} \\
 & \therefore \Delta {{L}_{2}}=\dfrac{1}{2}\times \Delta {{L}_{1}} \\
\end{align}\]
Now substitute the value of the length of the cable in the above equation.
\[\begin{align}
  & \Delta {{L}_{2}}=\dfrac{1}{2}\times 2 \\
 & \Rightarrow \Delta {{L}_{2}}=1mm \\
\end{align}\]
\[\therefore \]By 1 mm either part of the cable can be stretched without breaking.

Note:
The concept that the amount of stretch per unit length remains constant should be known to solve this problem. Even the force applied and the area remains constant. As Young’s modulus formula relates the stretch and the length of a material, so, using the same, this problem is solved.