Question

The breaking stress of aluminium is \[7.5 \times {10^7}N{m^{ - 2}}\]. The greatest length of aluminium wire that can hang vertically without breaking is (Density of aluminium wire is\[2.7 \times {10^3}kg{m^{ - 3}}\]).
(A) \[283 \times {10^3}m\]
(B) \[28.3 \times {10^3}m\]
(C) \[2.72 \times {10^3}m\]
(D) \[0.283 \times {10^3}m\]

Answer 1

Hint Stress is the ratio of force applied to the area where Force can be written in terms of density and length using mass and density relations. Hence, we would get an expression of length and by substituting all known values, required value of wire length can be calculated.
Formula used: \[ \Rightarrow L = \dfrac{{stress}}{{\rho g}}\]
   Where, \[ \Rightarrow Stress = \dfrac{F}{A}\] and force =mg

Complete Step By Step Solution
Given that, the breaking stress of aluminium is \[7.5 \times {10^7}N{m^{ - 2}}\]. That means the maximum value of stress that the aluminium wire can bear is \[7.5 \times {10^7}N{m^{ - 2}}\] and beyond this if stress is applied on it, wire would break down. Also, we studied in Mechanical Properties of Solids that, Stress can be defined as the internal force of restitution per unit area of a deformed body to regain its original state.
Mathematically, \[ \Rightarrow Stress = \dfrac{{ForceApplied}}{{Area}}\]
\[ \Rightarrow Stress = \dfrac{F}{A}\]…………………………………………….eq.1
Now, an object must be hanging vertically on wire which can already assumed from the question, then force in such case can be given by,
\[ \Rightarrow Force = mg\]……………………………………………eq.2
Where, m = mass of the object hanging
g = acceleration due to gravity.

Since, the mass of an object is not mentioned in the question, then we must need to change the form of mass into another physical quantity. Carefully read the statement, we have density of wire then can use the relation between mass and density.
\[ \Rightarrow \rho (density) = \dfrac{{mass}}{{volume}}\]……………………………………….eq.3
Also, \[volume = area \times Length\]
\[volume = A \times L\]
By substituting value of volume in eq.3, we get
\[ \Rightarrow \rho (density) = \dfrac{{mass}}{{A \times L}}\]
\[ \Rightarrow \rho (density) \times A \times L = mass\]………………………………………..eq.4
Substitute value from eq.4 in eq.2, we get
\[ \Rightarrow Force = \rho ALg\]
Putting value of force from above equation in eq.1, then,
\[ \Rightarrow Stress = \dfrac{{\rho ALg}}{A}\]
\[ \Rightarrow Stress = \rho Lg\]
For finding value of L (length of wire),
\[ \Rightarrow L = \dfrac{{stress}}{{\rho g}}\]
Putting values, stress =\[7.5 \times {10^7}N{m^{ - 2}}\], g= 9.8m/\[{s^2}\], \[\rho = 2.7 \times {10^3}kg{m^{ - 3}}\]
\[ \Rightarrow L = \dfrac{{7.5 \times {{10}^7}}}{{2.7 \times {{10}^3} \times 9.8}}\]
\[ \Rightarrow L = \]\[2.8 \times {10^3}\]m
\[ \Rightarrow L = 2.8 \times {10^3} \approx 2.72 \times {10^3}\]m (nearest value matching the option)

Thus, the greatest length of aluminium wire that can hang vertically without breaking is \[2.72 \times {10^3}\]m.

Note Try to relate the unknown quantities (length) with known physical quantities (density, force, stress) step by step to avoid any error. Also, always check if all the given quantities are in the same unit system. Here, unit of stress is given as SI unit, similarly density is also given in kg\[{m^{ - 3}}\] , value of g is again substituted in m/\[{s^2}\]. A unit conversion needs to be done if the given quantity does not have the same unit system. Symbolic representation of stress is\[\sigma \].