Question

Breaking stress of a material is $2\times {{10}^{8}}N/{{m}^{2}}$. What maximum length of the wire of this material can be so that the wire does not break by its own weight? [Density of the material = $5\times {{10}^{3}}kg/{{m}^{3}}$]
A. 1 km
B. 2 km
C. 3 km
D. 4 km

Answer 1

Hint: Use the formula $Stress=\dfrac{Force}{Area}$, that is apply Hooke’s Law. Here force can be written as ‘mg’, where m is the mass. And mass can also be written as $m=\rho V$, and find the volume of wire by considering wire as cylindrical in shape.

Formula used:
$m=\rho V=\rho AL$
$Stress=\dfrac{Force}{Area}=\dfrac{mg}{A}=\dfrac{\rho ALg}{A}=\rho Lg$

Complete step by step answer:
Given,
Breaking stress of the material= $2\times {{10}^{8}}N/{{m}^{2}}$
density of the material ρ = $5\times {{10}^{3}}kg/{{m}^{3}}$
Let us consider the maximum length of the wire before breaking to be L.
Hence, its mass will be
$m=\rho V=\rho AL$
where V is volume of the wire, A its cross section area, and L is its length. Again,
$Stress=\dfrac{Force}{Area}=\dfrac{mg}{A}=\dfrac{\rho ALg}{A}=\rho Lg$
$L=\dfrac{Stress}{\rho g}=\dfrac{2\times {{10}^{8}}}{5\times {{10}^{3}}\times 9.8}\approx 4000m$
Hence, the correct option is D.

Additional Information:
Elasticity is a property of the material of a body due to which the body opposes any change in its shape or size when deforming forces are applied to it. However, they show this property up to a certain value of deforming force, beyond which they are unable to recover to its original state. Hence, the property of elasticity is completely lost, and the body is permanently deformed. The internal restoring force acting per unit area of the deformed body is called stress.

Note: In certain cases it is taken that the weight acts at the center of gravity. As such for a wire, the lower half pulls the upper half. Under this condition the length required to break as calculated above doubles.