# 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
5 years, 1 month 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:

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

:)

- 5 years, 1 month ago

Thanks! :)

- 5 years, 1 month ago