Question

A stress of ${10^6} N {m^ {- 3}}$ is required for breaking a material. If the density of the material is $3 \times {10^3} kg {m^ {- 3}}$ , then what should be the minimum length of the wire made of the same material so that it breaks by its own weight? ( $g = 10m{s^ {- 1}}$ )
A) $33.4m$
B) $3.4m$
C) $34cm$
D) $3.4cm$

Answer 1

Try to think of the various formulas of expressing stress in terms of density, length and acceleration due to gravity. Also remember that the stress can be experienced by a deforming force and the deforming force can change the size or volume or shape of the object.

Formula used:
$Stress = \rho \times l \times g$
Where,
$\rho$ is the density
$l$ is the length of the wire
$g$ is acceleration due to gravity.

Complete step by step solution:
A material has the highest possible tensile force before cracking, such as fracturing or permanent deformation, which the material can tolerate prior to breaking. Breaking stress is a restricting stress condition which leads to tensile failure. The tension per unit area applied to the material is the strain per unit surface. Before the object breaks, the highest stress is the cracking stress or the final tensile stress. Tensile is a strain of the material. The forces that work on it attempt to extend the content. Compression is when an object is being compressed by forces.

$Stress = \rho \times l \times g$
Where,
$\rho$ is the density
$l$ is the length of the wire
$g$ is acceleration due to gravity
Therefore,
$l = \dfrac{{Stress}}{{\rho \times g}}$
$l = \dfrac{{{{10}^6}}}{{3 \times {{10} ^3} \times 10}}$
Hence, $l = 33.4m$
The minimum length of the wire made of the same material so that it breaks by its own weight is $33.4m$

Correct option is (A).

Note: The strength of the tensile shows the point of the elastic to plastic deformation of the material. The required tensile stress (force per unit area) required for the separation of the substance is expressed. Remember that tensile stress and compressive stress are the two types of longitudinal stress.