
Two different wires having lengths \[{L_1}\] and \[{L_2}\], and respective temperature coefficient of linear expansion \[{\alpha _1}\] and \[{\alpha _2}\], are joined end-to-end. Then find the effective temperature coefficient of linear expansion.
A. \[\dfrac{{{\alpha _1}{L_1} + {\alpha _2}{L_2}}}{{{L_1} + {L_2}}}\]
B. \[2\sqrt {{\alpha _1}{\alpha _2}} \]
C. \[4\dfrac{{{\alpha _1}{\alpha _2}}}{{{\alpha _1} + {\alpha _2}}}\dfrac{{{L_2}{L_1}}}{{{{\left( {{L_2} + {L_1}} \right)}^2}}}\]
D. \[\dfrac{{{\alpha _1} + {\alpha _2}}}{2}\]
Answer
162.9k+ views
Hint:Before we proceed with the given problem. Let us understand about the temperature coefficient of linear expansion. When a material is heated and expands, the extent to which it expands is called the temperature coefficient of linear expansion.
Formula Used:
The formula to find the temperature coefficient of linear expansion is,
\[\Delta L = {\alpha _L}\Delta T\]
Where, \[\Delta L\] is change in length, \[\Delta T\] is change in temperature and \[{\alpha _L}\] is temperature coefficient of linear expansion.
Complete step by step solution:
Here we have two different wires of length \[{L_1}\] and \[{L_2}\], and respective coefficients linear expansions \[{\alpha _1}\] and \[{\alpha _2}\] are joined end to end, which means they have joined in series. Suppose the two wires are replaced by a single wire, then we need to find the effective temperature coefficient of linear expansion. Now let us see how this can be found.
In the case of the two wires, the total linear expansion is given by,
\[\Delta L = \Delta {L_1} + \Delta {L_2}\]
\[\Rightarrow \left( {{L_1} + {L_2}} \right){\alpha _{eff}}\Delta T = {L_1}{\alpha _1}\Delta T + {L_2}{\alpha _2}\Delta T\]
\[\therefore {\alpha _{eff}} = \dfrac{{{\alpha _1}{L_1} + {\alpha _2}{L_2}}}{{{L_1} + {L_2}}}\]
Therefore, the effective temperature coefficient of linear expansion is \[\dfrac{{{\alpha _1}{L_1} + {\alpha _2}{L_2}}}{{{L_1} + {L_2}}}\].
Hence, option A is the correct answer.
Note:The temperature coefficient of linear expansion depends on the original length of a material, change in temperature and the nature of the material. This coefficient of thermal expansion is used to predict the growth of materials in response to a known temperature change. If this coefficient of thermal expansion is larger for a material, the higher will be its expansion per degree temperature increase.
Formula Used:
The formula to find the temperature coefficient of linear expansion is,
\[\Delta L = {\alpha _L}\Delta T\]
Where, \[\Delta L\] is change in length, \[\Delta T\] is change in temperature and \[{\alpha _L}\] is temperature coefficient of linear expansion.
Complete step by step solution:
Here we have two different wires of length \[{L_1}\] and \[{L_2}\], and respective coefficients linear expansions \[{\alpha _1}\] and \[{\alpha _2}\] are joined end to end, which means they have joined in series. Suppose the two wires are replaced by a single wire, then we need to find the effective temperature coefficient of linear expansion. Now let us see how this can be found.
In the case of the two wires, the total linear expansion is given by,
\[\Delta L = \Delta {L_1} + \Delta {L_2}\]
\[\Rightarrow \left( {{L_1} + {L_2}} \right){\alpha _{eff}}\Delta T = {L_1}{\alpha _1}\Delta T + {L_2}{\alpha _2}\Delta T\]
\[\therefore {\alpha _{eff}} = \dfrac{{{\alpha _1}{L_1} + {\alpha _2}{L_2}}}{{{L_1} + {L_2}}}\]
Therefore, the effective temperature coefficient of linear expansion is \[\dfrac{{{\alpha _1}{L_1} + {\alpha _2}{L_2}}}{{{L_1} + {L_2}}}\].
Hence, option A is the correct answer.
Note:The temperature coefficient of linear expansion depends on the original length of a material, change in temperature and the nature of the material. This coefficient of thermal expansion is used to predict the growth of materials in response to a known temperature change. If this coefficient of thermal expansion is larger for a material, the higher will be its expansion per degree temperature increase.
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