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Question: Two meters scale, one of steel and the other of aluminum, agree at \(20^\circ C\) . Calculate the ra...

Two meters scale, one of steel and the other of aluminum, agree at 20∘C20^\circ C . Calculate the ratio of length of aluminum and steel at,
(a) 0∘C0^\circ C
(b) 40∘C40^\circ C and
(c) 100∘C100^\circ C
(α\alpha for steel =1.1×10−5∘C−11.1 \times {10^{ - 5}}^\circ {C^{ - 1}} and for aluminum = 2.3×10−5∘C−12.3 \times {10^{ - 5}}^\circ {C^{ - 1}})

Explanation

Solution

We can use the formula Lf=Li(1+αΔT){L_f} = {L_i}(1 + \alpha \Delta T), where Lf{L_f} and Li{L_i} is the final length and the initial length of the meter scale respectively, the formula for linear thermal expansion of a material.

Complete step by step solution:
Let the length of the meter scale of steel at 20∘C20^\circ C be Ls{L_s} and for that of aluminum be La{L_a}.
As given, the coefficient of linear expansion for steel and aluminum are αs=1.1×10−5∘C−1{\alpha _s} = 1.1 \times {10^{ - 5}}^\circ {C^{ - 1}} and αa=2.3×10−5∘C−1{\alpha _a} = 2.3 \times {10^{ - 5}}^\circ {C^{ - 1}} respectively.
(a) The ratio of aluminum-centimeter / steel-centimeter at 0∘C0^\circ C will be:
As given in the question:
Ls(20)=La(20){L_{s(20)}} = {L_{a(20)}}
Let the length of the meter scales of steel and aluminum be Ls(0){L_{s(0)}} and La(0){L_{a(0)}} respectively, the length of both the rods at 20∘C20^\circ C will be:
⇒Ls(20)=Ls(0)(1−αΔT)\Rightarrow {L_{s(20)}} = {L_{s(0)}}(1 - \alpha \Delta T)
⇒Ls(0)(1−α×20)\Rightarrow {L_{s(0)}}(1 - \alpha \times 20)
⇒Ls(0)(1−(1.1×10−5×20))\Rightarrow {L_{s(0)}}(1 - (1.1 \times {10^{ - 5}} \times 20))
⇒La(20)=La(0)(1−αΔT)\Rightarrow {L_{a(20)}} = {L_{a(0)}}(1 - \alpha \Delta T)
⇒La(0)(1−(α×20))\Rightarrow {L_{a(0)}}(1 - (\alpha \times 20))
⇒La(0)(1−(2.3×10−5×20))\Rightarrow {L_{a(0)}}(1 - (2.3 \times {10^{ - 5}} \times 20))
Since Ls(20)=La(20){L_{s(20)}} = {L_{a(20)}}
Therefore, La(0)(1−(2.3×10−5×20))=Ls(0)(1−(1.1×10−5×20)){L_{a(0)}}(1 - (2.3 \times {10^{ - 5}} \times 20)) = {L_{s(0)}}(1 - (1.1 \times {10^{ - 5}} \times 20))
⇒La(0)Ls(0)=1−(1.1×10−5×20)1−(2.3×10−5×20)\Rightarrow \dfrac{{{L_{a(0)}}}}{{{L_{s(0)}}}} = \dfrac{{1 - (1.1 \times {{10}^{ - 5}} \times 20)}}{{1 - (2.3 \times {{10}^{ - 5}} \times 20)}}
⇒La(0)Ls(0)=0.000240011\Rightarrow \dfrac{{{L_{a(0)}}}}{{{L_{s(0)}}}} = 0.000240011
Therefore, the ratio of aluminum-centimeter/steel-centimeter at 0∘C0^\circ C is 0.0002400110.000240011.

(b) The ratio of aluminum-centimeter / steel-centimeter at 40∘C40^\circ C will be:
Similarly, we can find out the ratio at 40∘C40^\circ C as we found out at 0∘C0^\circ C.
The length of the meter scales of steel and aluminum be Ls(40){L_{s(40)}} and La(40){L_{a(40)}} respectively, the length of both the rods at 20∘C20^\circ C will be:
⇒Ls(20)=Ls(40)(1+αΔT)\Rightarrow {L_{s(20)}} = {L_{s(40)}}(1 + \alpha \Delta T)
⇒Ls(40)(1+α×20)\Rightarrow {L_{s(40)}}(1 + \alpha \times 20)
⇒Ls(40)(1+(1.1×10−5×20))\Rightarrow {L_{s(40)}}(1 + (1.1 \times {10^{ - 5}} \times 20))
⇒La(20)=La(40)(1+αΔT)\Rightarrow {L_{a(20)}} = {L_{a(40)}}(1 + \alpha \Delta T)
⇒La(40)(1+(α×20))\Rightarrow {L_{a(40)}}(1 + (\alpha \times 20))
⇒La(40)(1+(2.3×10−5×20))\Rightarrow {L_{a(40)}}(1 + (2.3 \times {10^{ - 5}} \times 20))
Since Ls(20)=La(20){L_{s(20)}} = {L_{a(20)}}
Therefore, La(40)(1+(2.3×10−5×20))=Ls(40)(1+(1.1×10−5×20)){L_{a(40)}}(1 + (2.3 \times {10^{ - 5}} \times 20)) = {L_{s(40)}}(1 + (1.1 \times {10^{ - 5}} \times 20))
⇒La(40)Ls(40)=1+(1.1×10−5×20)1+(2.3×10−5×20)\Rightarrow \dfrac{{{L_{a(40)}}}}{{{L_{s(40)}}}} = \dfrac{{1 + (1.1 \times {{10}^{ - 5}} \times 20)}}{{1 + (2.3 \times {{10}^{ - 5}} \times 20)}}
⇒La(40)Ls(40)=1.00023995\Rightarrow \dfrac{{{L_{a(40)}}}}{{{L_{s(40)}}}} = 1.00023995
Therefore, the ratio of aluminum-centimeter/steel-centimeter at 40∘C40^\circ C is 1.000239951.00023995.

(c) The ratio of aluminum-centimeter / steel-centimeter at 100∘C100^\circ C will be:
The length of the meter scales of steel and aluminum be Ls(100){L_{s(100)}} and La(100){L_{a(100)}} respectively, the length of both the rods at 20∘C20^\circ C will be:
⇒Ls(20)=Ls(100)(1+αΔT)\Rightarrow {L_{s(20)}} = {L_{s(100)}}(1 + \alpha \Delta T)
⇒Ls(100)(1+α×80)\Rightarrow {L_{s(100)}}(1 + \alpha \times 80)
⇒Ls(100)(1+(1.1×10−5×80))\Rightarrow {L_{s(100)}}(1 + (1.1 \times {10^{ - 5}} \times 80))
⇒La(20)=La(100)(1+αΔT)\Rightarrow {L_{a(20)}} = {L_{a(100)}}(1 + \alpha \Delta T)
⇒La(100)(1+(α×80))\Rightarrow {L_{a(100)}}(1 + (\alpha \times 80))
⇒La(100)(1+(2.3×10−5×80))\Rightarrow {L_{a(100)}}(1 + (2.3 \times {10^{ - 5}} \times 80))
Since Ls(20)=La(20){L_{s(20)}} = {L_{a(20)}}
Therefore, La(100)(1+(2.3×10−5×80))=Ls(100)(1+(1.1×10−5×80)){L_{a(100)}}(1 + (2.3 \times {10^{ - 5}} \times 80)) = {L_{s(100)}}(1 + (1.1 \times {10^{ - 5}} \times 80))
⇒La(100)Ls(100)=1+(1.1×10−5×80)1+(2.3×10−5×80)\Rightarrow \dfrac{{{L_{a(100)}}}}{{{L_{s(100)}}}} = \dfrac{{1 + (1.1 \times {{10}^{ - 5}} \times 80)}}{{1 + (2.3 \times {{10}^{ - 5}} \times 80)}}
⇒La(100)Ls(100)=1.00095916\Rightarrow \dfrac{{{L_{a(100)}}}}{{{L_{s(100)}}}} = 1.00095916
Therefore, the ratio of aluminum-centimeter/steel-centimeter at 100∘C100^\circ C is 1.000959161.00095916.**

Note: The condition that both the aluminum meter scale and steel scale agree at 20∘C20^\circ C should be used in each case. ΔT\Delta T is the difference between the final and the initial temperature of the body and therefore affects the sign of the equation: the value would be negative if the body goes from higher temperature to lower one and positive for the other case.