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Question: It has been given that the breaking stress of the aluminium is \(7.5\times {{10}^{7}}N{{m}^{-2}}\). ...

It has been given that the breaking stress of the aluminium is 7.5×107Nm27.5\times {{10}^{7}}N{{m}^{-2}}. The greatest length of the aluminium wire that can hang vertically without breaking will be given as,
(Density of the aluminium has been given as 2.7×103kgm32.7\times {{10}^{3}}kg{{m}^{-3}})
A.28.3×103m B.28.3×103m C.2.72×103m D.0.283×103m \begin{aligned} & A.28.3\times {{10}^{3}}m \\\ & B.28.3\times {{10}^{3}}m \\\ & C.2.72\times {{10}^{3}}m \\\ & D.0.283\times {{10}^{3}}m \\\ \end{aligned}

Explanation

Solution

The force with which the wire has been hung in the question can be found by taking the product of the density of the aluminium, area of the aluminium wire, length of the aluminium wire and the acceleration due to gravity. The breaking stress can be found by taking the ratio of the force to the area of the wire. These equations will help you in answering this question.

Complete answer:
The force with which the wire has been hung in the question can be found by taking the product of the density of the aluminium, area of the aluminium wire, length of the aluminium wire and the acceleration due to gravity. This can be written as an equation given as,
F=ρALgF=\rho ALg
The breaking stress can be found by taking the ratio of the force to the area of the wire. That is we can write that,
FA=σ\dfrac{F}{A}=\sigma
This can be obtained by rearranging the above obtained equation. That is we can write that,
σ=FA=ρLg\sigma =\dfrac{F}{A}=\rho Lg
Therefore the greatest length of the aluminium wire can be found as,
L=σρgL=\dfrac{\sigma }{\rho g}
It has been already mentioned in the question that,
The breaking stress of the wire is,
σ=7.5×107Nm2\sigma =7.5\times {{10}^{7}}N{{m}^{-2}}
Density of the aluminium wire can be shown as,
ρ=2.7×103kgm3\rho =2.7\times {{10}^{3}}kg{{m}^{-3}}
The acceleration due to gravity can be given the value,
g=9.8ms2g=9.8m{{s}^{-2}}
Substituting the values in the equation of the length can be written as,
L=7.5×1072.7×103×9.8=2.72×103mL=\dfrac{7.5\times {{10}^{7}}}{2.7\times {{10}^{3}}\times 9.8}=2.72\times {{10}^{3}}m

Hence the correct answer for the question has been obtained. It has been given as option C.

Note:
The stress given to a substance will be the force per unit area which is applied to the substance. The maximum stress a substance can overcome before it will break is referred to as the breaking stress or the ultimate tensile stress. Tensile which is meant by the substance will be under tension.