Question

A wire has a breaking stress of $6 \times {10^5}{\text{N}}{{\text{m}}^{ - 2}}$ and density of $3 \times {10^4}{\text{kg}}{{\text{m}}^{ - 3}}$. Find the length of the wire of the same material which will break under its own weight. ( $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ ).
A) 2000 m
B) 2500 m
C) 20 m
D) 2 m

Answer 1

The breaking stress of the wire refers to the force that has to be applied to the cross-sectional area of the given wire so that it breaks. Here it is mentioned that a wire of the same material has to break under its own weight. This suggests that the force to be applied is equal to the weight of the wire.

Formulas used:
The stress of an object is given by, $\sigma = \dfrac{F}{A}$ where $F$ is the applied force and $A$ is the area of the object.
The weight of a body is given by, $W = mg$ where $m$ is the mass of the body and $g$ is the acceleration due to gravity.

Complete step by step answer:
Step 1: Listing the parameters of the given wire.
The density of the wire is given to be $\rho = 3 \times {10^4}{\text{kg}}{{\text{m}}^{ - 3}}$ .
The breaking stress of the given wire is ${\sigma _{breaking}} = 6 \times {10^5}{\text{N}}{{\text{m}}^{ - 2}}$ .
It is also given that the acceleration due to gravity is $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ .
Then if the mass of the wire is $m$, the weight of the wire will be $W = mg$.
Let $l$ be the length of the wire.
Step 2: Expressing the relation for the breaking stress of the wire to determine its length.
We know that the breaking stress experienced by the wire can be expressed as
${\sigma _{breaking}} = \dfrac{F}{A}$ where $F$ is the force applied to break the wire and $A$ is the area of the wire.
As it is mentioned that it is a force equal to the weight of the wire that breaks the wire of the same material, we can express the breaking stress as ${\sigma _{breaking}} = \dfrac{W}{A} = \dfrac{{mg}}{A}$.
The above expression can also be written as ${\sigma _{breaking}} = \dfrac{{\rho A\lg }}{A}$ ------- (A)
$\Rightarrow l = \dfrac{{{\sigma _{breaking}}}}{{\rho g}}$ -------- (1)
Substituting for ${\sigma _{breaking}} = 6 \times {10^5}{\text{N}}{{\text{m}}^{ - 2}}$, $\rho = 3 \times {10^4}{\text{kg}}{{\text{m}}^{ - 3}}$and $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ in equation (1) we get,
$\Rightarrow l = \dfrac{{6 \times {{10}^5}}}{{3 \times {{10}^4} \times 10}} = 2{\text{m}}$

Thus the length of the wire will be $l = 2{\text{m}}$. So the correct option is D.

Note:
The mass of the wire in terms of its density and volume is expressed as $m = \rho V$; $V$ is the volume of the wire. Since the volume of any object can be expressed in terms of the length $l$ of the object and area $A$ of the object as $V = A \times l$, we substituted $m = \rho Al$ for the mass of the wire in equation (A).