Question

An iron rod of length 50 cm is joined to an aluminium rod of length 100cm. All measurements refer to \[20^\circ C\] . The coefficient of linear expansion of iron and aluminium are \[12 \times {10^{ - 6}}/^\circ C\] and \[24 \times {10^{ - 6}}/^\circ C\]respectively . The average coefficient of composite system at \[100^\circ C\] is
A. \[36 \times {10^{ - 6}}/\;^\circ C\]
B. \[12 \times {10^{ - 6}}/\;^\circ C\]
C. \[20 \times {10^{ - 6}}/\;^\circ C\]
D. \[48 \times {10^{ - 6}}/\;^\circ C\]

Answer 1

Hint: If the temperature of a body increases in general, its size also increases, if two rods of different materials are added together in end to end connection ( series combination ) then the combined increase in length is the sum of each individual increase in length.

Formulae Used:
 \[{L_{L = 50cmfinal}} = L(1 + \alpha \Delta T)\]
Where \[{L_{final}}\]= the length of rod after increasing the temperature \[\Delta T\].
\[\Delta T\]= Change in temperature.
\[\alpha \]= the coefficient of linear expansion .
\[L\]= initial length.

Complete step-by-step solution:
The length of the iron rod at \[100^\circ C\] is \[{L_{iro{n_{100}}}} = L(1 + \alpha \Delta T)\;cm\]; \[\Delta T = (100 - 20)\;^\circ \;C\]
The coefficient of linear expansion of iron is
\[\alpha = 12 \times {10^{ - 6}}\;^\circ {C^{ - 1}}\]
The length of iron rod at \[100^\circ C\]
\[{L_{iro{n_{100}}}} = 50(1 + 12 \times {10^{ - 6}}(100 - 20))\;cm\]
\[{L_{iro{n_{100}}}} = 50.048\;cm\]
The length of the iron rod at \[100^\circ C\] is
\[{L_{al{u_{100}}}} = L(1 + \alpha \Delta T)\;cm\]
Where \[L = 100cm\]; \[\Delta T = (100 - 20)\;^\circ \;C\]; \[\alpha = 24 \times {10^{ - 6}}\;^\circ {C^{ - 1}}\]
Putting the value in equation
\[{L_{al{u_{100}}}} = 100(1 + 24 \times {10^{ - 6}}(100 - 20))\;cm\]
\[{L_{al{u_{100}}}} = 100.192\;cm\]
When both iron rod and aluminium rod are combined in series
\[{L_{iron\; + \;alu}} = 50 + 100 = 150\;cm\]

Iron, L = 50 cm

Aluminium, L = 100 cm

Combined length of combination at \[100^\circ C\]is
\[{L_{combined}} = {L_{iron\; + \;alu}}(1 + {\alpha _{combined}}\Delta T)\;cm\]…………….(**)
\[{L_{iro{n_{100}}}} + {L_{al{u_{100}}}} = 50.048 + 100.192\;cm\]
\[{L_{iro{n_{100}}}} + {L_{al{u_{100}}}} = 150.24cm\]
Now equating all the equations
\[{L_{combined}} = {L_{iro{n_{100}}}} + {L_{al{u_{100}}}} = 150.24\;cm\]
\[{L_{combined}} = 150.24\;cm\]
Putting the value in equation (**)
\[150.24 = 150(1 + {\alpha _{combined}}(100 - 20))\]
Rearranging the equation
\[{\alpha _{combined}} = \dfrac{{0.24}}{{150(100 - 20)}}\;\;^\circ {C^{ - 1}}\]
\[{\alpha _{combined}} = 20 \times {10^{ - 6}}\;^\circ {C^{ - 1}}\]
Hence, the average coefficient of composite system at \[100^\circ C\] is \[{\alpha _{combined}} = 20 \times {10^{ - 6}}\;^\circ {C^{ - 1}}\]
Hence option ( C ) is the correct answer.

Note:- There are many applications of thermal expansion of metals in real life:
1. We provide a gap between two plates of railway tracks to give enough margin for expansion of those metal tracks.
2. Several mechanical fire alarm switches use bimetallic strips which work on the same principle of linear expansion of metals.
Expansion in liquids :- The anomalous expansion of water has a favourable effect for the animals living on water. Since the density of water is maximum at \[4^\circ C\], water at the bottom of the lakes remains at \[4^\circ C\] in winter even if that at the surface freezes. This allows marine animals to remain alive and move near the bottom.