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Question: A steel rod of length 1 m is heated from \[{25^ \circ }C\] to \[{75^ \circ }C\]keeping its length co...

A steel rod of length 1 m is heated from 25∘C{25^ \circ }C to 75∘C{75^ \circ }Ckeeping its length constant. The longitudinal strain in the rod is: (given : coefficient of linear expansion of steel is 12×10−6C12 \times {10^{ - 6}}C)

Explanation

Solution

We will answer this question correctly by first understanding the meaning of strain. To solve this question we need to find the temperature difference as our first step and then substitute this into the formula of strain that relates the coefficient of linear expansion with temperature.
Formula used:
Strain=−ΔllStrain = \dfrac{{ - \Delta l}}{{l}}
Where l is the length of the rod.

Complete answer:
Let us first understand what strain is. Strain is a quantity that describes the relative deformation or change in shape and size of elastic, plastic, and fluid materials under applied forces in physical sciences and engineering.
We see that the coefficient of linear expansion has been given, let us know a little about that. When the temperature of a one-unit-long specimen is elevated by one degree, the coefficient of linear expansion is the change in length of the specimen.The expansion of various materials varies.
To give correct answer to this question,
Let's first see what all is given:
l=1ml = 1m,
T1=25∘C{T_1} = {25^ \circ }C
T2=75∘C{T_2} = {75^ \circ }C
α=12×10−6  C\alpha = 12 \times {10^{ - 6}}\;C (where α\alpha = coefficient of linear expansion of steel)
Also, Strain=−ΔllStrain = \dfrac{{ - \Delta l}}{{l}}
Or, Strain=−lαΔTlStrain = \dfrac{{ - l\alpha \Delta T}}{l}
Or, Strain=−αΔTStrain = - \alpha \Delta T
Lets find ΔT\Delta T first,
ΔT=T2−T1\Delta T = {T_2} - {T_1}
ΔT=75∘C−25∘C\Delta T = {75^ \circ }C - {25^ \circ }C
ΔT=50∘C\Delta T = {50^ \circ }C
Hence the strain will be given by:
Strain=−12×10−6×50Strain = - 12 \times {10^{ - 6}} \times 50
Or, Strain=−6×10−4Strain = - 6 \times {10^{ - 4}}.

Note:
Strain has no unit or it is an unit- less quantity as it is expressed in terms of ratio of change in dimension to original dimension.
Another way of solving can be : Δl=li(1+αΔT)\Delta l = {l_i}(1 + \alpha {{\Delta }}T)
Or, Δl=1(1+12×10−6×50)\Delta l = 1(1 + 12 \times {10^{ - 6}} \times 50)
Or, Δl=6×10−4\Delta l = 6 \times {10^{ - 4}}
As strain is : Strain=−ΔllStrain = \dfrac{{ - \Delta l}}{{l}}
And l=1ml = 1m
Hence Strain=−ΔlStrain = - \Delta l
Or, Strain=−6×10−4Strain = - 6 \times {10^{ - 4}}