Question

A nylon rope of $2cm$ in diameter has a breaking strength of $1.5\times {{10}^{5}}N$. Find the breaking strength of rope having $1cm$ diameter?
$\begin{align}
  & \text{A}\text{. }0.375\times {{10}^{5}}N \\
 & \text{B}\text{. 2}\times {{10}^{5}}N \\
 & \text{C}\text{. 6}\times {{10}^{5}}N \\
 & \text{D}\text{. 9}\times {{10}^{4}}N \\
\end{align}$

Answer 1

Hint: Breaking strength of a rope or a wire is directly proportional to the cross-sectional of rope. A more thick wire can withstand larger force before breaking as compared to a thin wire of the same material. We will use the relation of breaking strength and radius of rope to calculate the required breaking strength.

Formula used:
$F\propto {{r}^{2}}$
Where,
$F$ is the breaking force
$r$ is the radius of the rope

Complete step by step solution:
The breaking strength of a material is defined as the maximum amount of tensile stress that the material can withstand before its failure, such as breaking or permanent deformation. It is expressed as the minimum tensile stress, that is, force per unit area needed to split the material apart.
Breaking Strength is expressed as the tensile or compressive load required to fracture or to cause the rope to fail in handling that load and results in breaking itself. Breaking strength of a rope is directly proportional to its cross-sectional area. A more thick wire can withstand larger force before breaking as compared to a thin wire of the same material.
We are given that a nylon rope of $2cm$ in diameter has a breaking strength of $1.5\times {{10}^{5}}N$ and we have to calculate the breaking strength of rope having $1cm$ diameter made from the same material nylon.
Let’s say ${{F}_{1}}$ is the force in the rope of diameter ${{d}_{1}}=2cm$ and ${{F}_{2}}$ is the force in the rope of diameter ${{d}_{2}}=1cm$
Breaking force $F\propto {{r}^{2}}$
Where,
$r$ is the radius of the rope
Therefore,
$\dfrac{{{F}_{1}}}{{{F}_{2}}}=\dfrac{{{\left( {{r}_{1}} \right)}^{2}}}{{{\left( {{r}_{2}} \right)}^{2}}}$
Or,
${{F}_{2}}=\dfrac{{{\left( {{r}_{2}} \right)}^{2}}}{{{\left( {{r}_{1}} \right)}^{2}}}\times {{F}_{1}}$
Given that:
$\begin{align}
  & {{F}_{1}}=1.5\times {{10}^{5}}N \\
 & {{d}_{1}}=2cm\Rightarrow {{r}_{1}}=1cm \\
 & {{d}_{2}}=1cm\Rightarrow {{r}_{2}}=0.5cm \\
\end{align}$
Putting above values,
$\begin{align}
  & {{F}_{2}}=\dfrac{{{\left( 0.5 \right)}^{2}}}{{{\left( 1 \right)}^{2}}}\times 1.5\times {{10}^{5}} \\
 & {{F}_{2}}=0.25\times 1.5\times {{10}^{5}} \\
 & {{F}_{2}}=0.375\times {{10}^{5}}N \\
\end{align}$
The breaking strength of rope having $1cm$ diameter is $0.375\times {{10}^{5}}N$
Hence, the correct option is A.

Note: Breaking stress refers to the maximum force a rope can withstand before breaking apart. The ultimate maximum value of breaking stress just before the wire breaks is called breaking strength. Breaking strength of a rope is directly proportional to the cross-sectional area of the rope.