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# Why are the answers not same?

Here is a question,

An isosceles triangle is formed with thin rod of length $${ l }_{ 1 }$$ and coefficient of linear expansion $${ \alpha }_{ 1 }$$ as the base and two thin rods each of length $${ l }_{ 2 }$$ and coefficient of linear expansion $${ \alpha }_{ 2 }$$ as the two sides. If the distance between the apex and the midpoint of the base remain unchanged as the temperature is varied, find a relation between $${ l }_{ 1 }$$, $${ l }_{ 2 }$$, $${ \alpha }_{ 1 }$$, $${ \alpha }_{ 2 }$$.

Here is how I figured a relation,

Case I

$${ l }^{ 2 }=\quad { { \left( { l }_{ 2 } \right) } }^{ 2 }-{ { \left( \frac { { l }_{ 1 } }{ 2 } \right) } }^{ 2 }$$

$$\Rightarrow { \Delta l }^{ 2 }=\quad { { \left( { \Delta l }_{ 2 } \right) } }^{ 2 }-{ { \left( \frac { { \Delta l }_{ 1 } }{ 2 } \right) } }^{ 2 }$$ -----$$(1)$$

$${ \Delta l }_{ 2 }={ l }_{ 2 }-{ l }_{ 2 }\left( 1+{ \alpha }_{ 2 }\Delta T \right) ={ l }_{ 2 }{ \alpha }_{ 2 }\Delta T$$ -----$$(2)$$

$${ \Delta l }_{ 1 }={ l }_{ 1 }-{ l }_{ 1 }\left( 1+{ \alpha }_{ 1 }\Delta T \right) ={ l }_{ 1 }{ \alpha }_{ 1 }\Delta T$$ ----$$(3)$$

$$(2)$$ & $$(3)$$ in $$(1)$$ and Since $$\Delta l =0$$,

$${ \left( { l }_{ 2 }{ \alpha }_{ 2 }\Delta T \right) }^{ 2 }={ \left( \frac { { l }_{ 1 }{ \alpha }_{ 1 }\Delta T }{ 2 } \right) }^{ 2 }$$

$$\Rightarrow { l }_{ 2 }{ \alpha }_{ 2 }\Delta T=\frac { { l }_{ 1 }{ \alpha }_{ 1 }\Delta T }{ 2 }$$

$$\boxed{\Rightarrow \frac { { l }_{ 1 } }{ { l }_{ 2 } } =2\frac { { \alpha }_{ 2 } }{ { \alpha }_{ 1 } }}$$

Case II

$$\left( l \right) ^{ 2 }=\quad { { \left( { l }_{ 2 } \right) } }^{ 2 }-{ { \left( \frac { { l }_{ 1 } }{ 2 } \right) } }^{ 2 }$$

Differentiating with respect to T,

$$0=2{ l }_{ 2 }\frac { d{ l }_{ 2 } }{ dT } -\frac { 1 }{ 4 } 2{ l }_{ 1 }\frac { d{ l }_{ 1 } }{ dT }$$

We know $$d{ l }_{ 2 }={ { l }_{ 2 }\alpha }_{ 2 }\Delta T$$ and $$d{ l }_{ 1 }={ { l }_{ 1 }\alpha }_{ 1 }\Delta T$$

$$\Rightarrow { l }_{ 1 }.\frac { { { l }_{ 1 }\alpha }_{ 1 } }{ 2 } .\Delta T=2{ { { l }_{ 2 }l }_{ 2 }\alpha }_{ 2 }\Delta T$$

$$\boxed{\Rightarrow \frac { { l }_{ 1 } }{ { l }_{ 2 } } =2\sqrt { \frac { { \alpha }_{ 2 } }{ { \alpha }_{ 1 } } } }$$

Note by Anandhu Raj
1 year, 6 months ago

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@Anandhu Raj The relation you wrote as equation $$1$$ is incorrect. See, when you move from $$l^{2} = {{ l }_{ 2 }}^{2} - \frac {{{ l }_{ 1 }}^{2}} { 4 }$$ to the next step, you cant directly compare their changes the same way.

Since $$l$$ always remains constant, therefore, after sometime, the new relation will be:

${ l }^{ 2 }={ ({ l }_{ 2 }+\Delta { l }_{ 2 }) }^{ 2 }-\frac { ({ { l }_{ 1 }+\Delta { l }_{ 1 } })^{ 2 } }{ 4 }$

Now, equating these two lengths, you get:

${ { l }_{ 2 } }^{ 2 }-\frac { { { l }_{ 1 } }^{ 2 } }{ 4 } ={ ({ l }_{ 2 }+\Delta { l }_{ 2 }) }^{ 2 }-\frac { ({ { l }_{ 1 }+\Delta { l }_{ 1 } })^{ 2 } }{ 4 } \\ { l }^{ 2 }-\frac { { { l }_{ 1 } }^{ 2 } }{ 4 } ={ { l }_{ 2 } }^{ 2 }+{ \Delta { l }_{ 2 } }^{ 2 }+2{ l }_{ 2 }\Delta { l }_{ 2 }-\frac { { l }^{ 1 }+{ \Delta { l }_{ 1 } }^{ 2 }+2{ l }_{ 1 }{ \Delta l }_{ 1 } }{ 4 }$

Now, since, we consider $$\Delta { l }_{ 1 }$$ and $$\Delta { l }_{ 2 }$$ to be negligible, their squares are neglected. Hence, we are left with the relation:

$4{ l }_{ 2 }\Delta { l }_{ 2 }={ l }_{ 1 }\Delta { l }_{ 1 }$

Now, using the relation, $$\Delta { l }_{ 1 }={ l }_{ 1 }{ \alpha }_{ 1 }\Delta T$$ and $$\Delta { l }_{ 2 }={ l }_{ 2 }{ \alpha }_{ 2 }\Delta T$$, we get:

$4{ l }^{ 2 }{ { \alpha }_{ 2 } }={ { l }_{ 1 } }^{ 2 }{ \alpha }_{ 1 }\\ \frac { { l }_{ 1 } }{ { l }_{ 2 } } =2\sqrt { \frac { { \alpha }_{ 2 } }{ { \alpha }_{ 1 } } }$

:) · 1 year, 6 months ago

Thanks! :) · 1 year, 6 months ago

@Abhineet Nayyar · 1 year, 6 months ago

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