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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 \(d{ l }_{ 1 }={ { l }_{ 1 }\alpha }_{ 1 }\Delta T\)

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

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

Why am I getting two answers?? Some one please help!

Note by Anandhu Raj
1 year, 8 months ago

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1 vote

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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:

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

:)

Abhineet Nayyar - 1 year, 8 months ago

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Thanks! :)

Anandhu Raj - 1 year, 8 months ago

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@Abhineet Nayyar

Rajdeep Dhingra - 1 year, 8 months ago

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