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Hint: Stress upon a material is given by the ratio of amount of force applied on area to the area. As the breaking stress is given, calculate the stress upon the wire taking the length of wire to be $ l $ and area of cross-section to be $ A $ . Then calculate force per unit area upon the wire which will give the value of breaking stress. Force upon the wire will only be its weight which will be vertically downward. Rearrange the formula to calculate the greatest length of the wire.
Formula used:
$ \begin{align}
& \text{Stress}=\dfrac{Force}{Area} \\
& \text{Density}=\dfrac{mass}{volume} \\
& \text{Volume}=length\times area \\
\end{align} $
Complete step-by-step answer:
If a wire of length $ l $ ,mass $ m $ and area of cross-section $ A $ hung from a fixed support then at equilibrium the only force upon the wire will be its weight which is vertically downward.
The stress on the wire is given by $ \text{Stress=}\dfrac{Force}{Area} $
Only force is its weight which is given by $ F=mg=\rho Vg $ where $ \rho =\text{density} $
So $ \text{Stress=}\dfrac{\rho Vg}{A} $
The rod has length $ l $ and area $ A $ so volume , $ V=Al $
Now, $ \text{Stress=}\dfrac{\rho A\lg }{A}=\rho \lg $
If $ \sigma $ is the breaking stress then
$ \begin{align}
& \sigma =\rho lg \\
& \Rightarrow l=\dfrac{\sigma }{\rho g} \\
\end{align} $
So the greatest length of wire that can be hung freely without breaking is $ \dfrac{\sigma }{\rho g} $
So, the correct answer is “Option A”.
Note: The force per unit area is defined as stress. The effect of stress on a body or material is called strain. The ratio of stress to strain gives the strength of a material. i.e. if the ratio of stress to strain is large then the material is a strong material and if the ratio of stress to strain is small then the material is called weak material. Tensile stress is the stress upon a material when it's under tension.
Formula used:
$ \begin{align}
& \text{Stress}=\dfrac{Force}{Area} \\
& \text{Density}=\dfrac{mass}{volume} \\
& \text{Volume}=length\times area \\
\end{align} $
Complete step-by-step answer:
If a wire of length $ l $ ,mass $ m $ and area of cross-section $ A $ hung from a fixed support then at equilibrium the only force upon the wire will be its weight which is vertically downward.
The stress on the wire is given by $ \text{Stress=}\dfrac{Force}{Area} $
Only force is its weight which is given by $ F=mg=\rho Vg $ where $ \rho =\text{density} $
So $ \text{Stress=}\dfrac{\rho Vg}{A} $
The rod has length $ l $ and area $ A $ so volume , $ V=Al $
Now, $ \text{Stress=}\dfrac{\rho A\lg }{A}=\rho \lg $
If $ \sigma $ is the breaking stress then
$ \begin{align}
& \sigma =\rho lg \\
& \Rightarrow l=\dfrac{\sigma }{\rho g} \\
\end{align} $
So the greatest length of wire that can be hung freely without breaking is $ \dfrac{\sigma }{\rho g} $
So, the correct answer is “Option A”.
Note: The force per unit area is defined as stress. The effect of stress on a body or material is called strain. The ratio of stress to strain gives the strength of a material. i.e. if the ratio of stress to strain is large then the material is a strong material and if the ratio of stress to strain is small then the material is called weak material. Tensile stress is the stress upon a material when it's under tension.
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