Question

A wire can bear a weight of 20kg before it breaks. If the wire is divided into two equal parts, then each part will support a maximum weight………

Answer 1

Hint: We have been provided with a wire which can bear weight of 20kg before it breaks. We know that weight is nothing but force under gravity i.e. F = mg. before braking of wire, wire has elastic energy and hence elongation occurs. So to calculate maximum weight after breaking of wire use the formula of young’s modulus for both conditions i.e. before breaking and after breaking. Take ratio and calculate maximum weight.

Formula used: Expression of young’s modulus is given by,
$\gamma =\dfrac{FL}{Al}$
Where, F = force acts on wire because of load.
L = length of wire
l = length of elongation
A = area of cross section.

Complete step by step answer:
Consider a wire of length L and radius r, which is suspended from a rigid support. Let the free and of wire be loaded by a weight F = mg. Let L + l be the length of wire, when elongation for given load is complete. Then
$\begin{align}
  & \text{young }\!\!'\!\!\text{ s modulus = }\dfrac{\text{longitudinal stress}}{\text{longitudinal strain}} \\
 & \gamma =\dfrac{FL}{Al} \\
 & F=\dfrac{\gamma Al}{L}.......\left( 1 \right) \\
\end{align}$
It is given that before the break wire can bear a weight of 20kg. We know that weight is nothing but mass or load under gravity and which can be written as, F =mg = 20kg.
Therefore, equation can be written as
${{F}_{initial}}=\dfrac{\gamma Al}{{{L}_{initial}}}......\left( 1 \right)$
When wire breaks, the wire of length L breaks into two equal parts. So equation (1) can be written as, $\begin{align}
  & {{F}_{final}}=\dfrac{\gamma Al}{{{L}^{1}}} \\
 & \because \text{ }{{\text{L}}^{1}}=\dfrac{L}{2} \\
 & \therefore \text{ }{{\text{F}}_{final}}=\dfrac{\gamma Al}{\dfrac{L}{2}}......\left( 3 \right) \\
\end{align}$
Take the ratio of (2) and (3) we get.
$\begin{align}
  & \dfrac{{{F}_{final}}}{{{F}_{initial}}}=\dfrac{\gamma Al}{\dfrac{L}{2}}\times \dfrac{L}{\gamma Al} \\
 & {{F}_{final}}=2{{F}_{initial}} \\
\end{align}$
We know that, $\begin{align}
  & {{F}_{initial}}=20kg \\
 & \\
\end{align}$
$\begin{align}
  & {{F}_{final}}=2\times 20kg \\
 & {{F}_{final}}=40kg \\
\end{align}$
Hence, after break each part will support a maximum weight of 40kg.

Note: When elongation or extension has been done in wire, at that time a small amount of work has been done. This can be expressed as work done $\dfrac{1}{2}\times \text{load}\times \text{extension}.$ Work done by stretching force is equal to energy gained by wire and this energy gained by wire is called strain energy.