
If two rods of length $L$ and $2L$ having coefficients of linear expansion $\alpha $ and $2\alpha $ respectively are connected so that total length becomes $3L$ the average coefficient of linear expansion of the composite rod is $\dfrac{{x\alpha }}{3}$. Find $x$.
Answer
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Hint: We will use the concept linear expansion that refers to a fractional change in size of a material due to a change in temperature hence it is represented as $L = {L_0}\left( {1 + \alpha \Delta T} \right)$
Formula used:
$L = {L_0}\left( {1 + \alpha \Delta T} \right)$
Complete answer:
According to the question we have to find the value of $x$ when two rods of length $L$ and $2L$ having coefficients of linear expansion $\alpha $ and $2\alpha $ respectively are connected so that total length becomes $3L$ the average coefficient of linear expansion of the composite rod is $\dfrac{{x\alpha }}{3}$.
Hence,
$
\Delta L = \alpha L\Delta T \\
\Delta \left( {2L} \right) = 2\alpha \left( {2L} \right)\Delta T \\
\Delta \left( {3L} \right) = {\alpha _{composite}}\left( {3L} \right)\Delta T \\
\therefore \alpha L\Delta T + 2\alpha \left( {2L} \right)\Delta T = {\alpha _{composite}}\left( {3L} \right)\Delta T \\
$
$
\Rightarrow 5\alpha = 3{\alpha _{composite}} \\
\Rightarrow {\alpha _{composite}} = \dfrac{5}{3}\alpha \\
$
Comparing with $\dfrac{{5x}}{3}$, we get
$x = 5$
Note:
The two straight metallic stips each of thickness t and lengths L are riveted together. Their coefficients of linear expansions are ${\alpha _1}$ and ${\alpha _2}$. If they are heated through temperature $\Delta \theta $, the bimetallic strip will bend to form an arc of radius r, hence $r = \dfrac{t}{{\left( {{\alpha _1} - {\alpha _2}} \right)\Delta T}}$.
Formula used:
$L = {L_0}\left( {1 + \alpha \Delta T} \right)$
Complete answer:
According to the question we have to find the value of $x$ when two rods of length $L$ and $2L$ having coefficients of linear expansion $\alpha $ and $2\alpha $ respectively are connected so that total length becomes $3L$ the average coefficient of linear expansion of the composite rod is $\dfrac{{x\alpha }}{3}$.
Hence,
$
\Delta L = \alpha L\Delta T \\
\Delta \left( {2L} \right) = 2\alpha \left( {2L} \right)\Delta T \\
\Delta \left( {3L} \right) = {\alpha _{composite}}\left( {3L} \right)\Delta T \\
\therefore \alpha L\Delta T + 2\alpha \left( {2L} \right)\Delta T = {\alpha _{composite}}\left( {3L} \right)\Delta T \\
$
$
\Rightarrow 5\alpha = 3{\alpha _{composite}} \\
\Rightarrow {\alpha _{composite}} = \dfrac{5}{3}\alpha \\
$
Comparing with $\dfrac{{5x}}{3}$, we get
$x = 5$
Note:
The two straight metallic stips each of thickness t and lengths L are riveted together. Their coefficients of linear expansions are ${\alpha _1}$ and ${\alpha _2}$. If they are heated through temperature $\Delta \theta $, the bimetallic strip will bend to form an arc of radius r, hence $r = \dfrac{t}{{\left( {{\alpha _1} - {\alpha _2}} \right)\Delta T}}$.
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