Why are the answers not same?

Here is a question,

An isosceles triangle is formed with thin rod of length l1{ l }_{ 1 } and coefficient of linear expansion α1{ \alpha }_{ 1 } as the base and two thin rods each of length l2{ l }_{ 2 } and coefficient of linear expansion α2{ \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 l1{ l }_{ 1 }, l2{ l }_{ 2 }, α1{ \alpha }_{ 1 }, α2{ \alpha }_{ 2 }.

Here is how I figured a relation,

Case I

l2=(l2)2(l12)2{ l }^{ 2 }=\quad { { \left( { l }_{ 2 } \right) } }^{ 2 }-{ { \left( \frac { { l }_{ 1 } }{ 2 } \right) } }^{ 2 }

Δl2=(Δl2)2(Δl12)2\Rightarrow { \Delta l }^{ 2 }=\quad { { \left( { \Delta l }_{ 2 } \right) } }^{ 2 }-{ { \left( \frac { { \Delta l }_{ 1 } }{ 2 } \right) } }^{ 2 } -----(1)(1)

Δl2=l2l2(1+α2ΔT)=l2α2ΔT{ \Delta l }_{ 2 }={ l }_{ 2 }-{ l }_{ 2 }\left( 1+{ \alpha }_{ 2 }\Delta T \right) ={ l }_{ 2 }{ \alpha }_{ 2 }\Delta T -----(2)(2)

Δl1=l1l1(1+α1ΔT)=l1α1ΔT{ \Delta l }_{ 1 }={ l }_{ 1 }-{ l }_{ 1 }\left( 1+{ \alpha }_{ 1 }\Delta T \right) ={ l }_{ 1 }{ \alpha }_{ 1 }\Delta T ----(3)(3)

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

(l2α2ΔT)2=(l1α1ΔT2)2{ \left( { l }_{ 2 }{ \alpha }_{ 2 }\Delta T \right) }^{ 2 }={ \left( \frac { { l }_{ 1 }{ \alpha }_{ 1 }\Delta T }{ 2 } \right) }^{ 2 }

l2α2ΔT=l1α1ΔT2\Rightarrow { l }_{ 2 }{ \alpha }_{ 2 }\Delta T=\frac { { l }_{ 1 }{ \alpha }_{ 1 }\Delta T }{ 2 }

l1l2=2α2α1\boxed{\Rightarrow \frac { { l }_{ 1 } }{ { l }_{ 2 } } =2\frac { { \alpha }_{ 2 } }{ { \alpha }_{ 1 } }}

Case II

(l)2=(l2)2(l12)2\left( l \right) ^{ 2 }=\quad { { \left( { l }_{ 2 } \right) } }^{ 2 }-{ { \left( \frac { { l }_{ 1 } }{ 2 } \right) } }^{ 2 }

Differentiating with respect to T,

0=2l2dl2dT142l1dl1dT0=2{ l }_{ 2 }\frac { d{ l }_{ 2 } }{ dT } -\frac { 1 }{ 4 } 2{ l }_{ 1 }\frac { d{ l }_{ 1 } }{ dT }

We know dl2=l2α2ΔTd{ l }_{ 2 }={ { l }_{ 2 }\alpha }_{ 2 }\Delta T and dl1=l1α1ΔTd{ l }_{ 1 }={ { l }_{ 1 }\alpha }_{ 1 }\Delta T

l1.l1α12.ΔT=2l2l2α2ΔT\Rightarrow { l }_{ 1 }.\frac { { { l }_{ 1 }\alpha }_{ 1 } }{ 2 } .\Delta T=2{ { { l }_{ 2 }l }_{ 2 }\alpha }_{ 2 }\Delta T

l1l2=2α2α1\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
3 years, 6 months ago

No vote yet
1 vote

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@Anandhu Raj The relation you wrote as equation 11 is incorrect. See, when you move from l2=l22l124l^{2} = {{ l }_{ 2 }}^{2} - \frac {{{ l }_{ 1 }}^{2}} { 4 } to the next step, you cant directly compare their changes the same way.

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

l2=(l2+Δl2)2(l1+Δl1)24{ l }^{ 2 }={ ({ l }_{ 2 }+\Delta { l }_{ 2 }) }^{ 2 }-\frac { ({ { l }_{ 1 }+\Delta { l }_{ 1 } })^{ 2 } }{ 4 }

Now, equating these two lengths, you get:

l22l124=(l2+Δl2)2(l1+Δl1)24l2l124=l22+Δl22+2l2Δl2l1+Δl12+2l1Δl14{ { 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 Δl1 \Delta { l }_{ 1 } and Δl2 \Delta { l }_{ 2 } to be negligible, their squares are neglected. Hence, we are left with the relation:

4l2Δl2=l1Δl14{ l }_{ 2 }\Delta { l }_{ 2 }={ l }_{ 1 }\Delta { l }_{ 1 }

Now, using the relation, Δl1=l1α1ΔT\Delta { l }_{ 1 }={ l }_{ 1 }{ \alpha }_{ 1 }\Delta T and Δl2=l2α2ΔT\Delta { l }_{ 2 }={ l }_{ 2 }{ \alpha }_{ 2 }\Delta T, we get:

4l2α2=l12α1l1l2=2α2α14{ l }^{ 2 }{ { \alpha }_{ 2 } }={ { l }_{ 1 } }^{ 2 }{ \alpha }_{ 1 }\\ \frac { { l }_{ 1 } }{ { l }_{ 2 } } =2\sqrt { \frac { { \alpha }_{ 2 } }{ { \alpha }_{ 1 } } }

:)

Abhineet Nayyar - 3 years, 6 months ago

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Thanks! :)

Anandhu Raj - 3 years, 6 months ago

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@Abhineet Nayyar

Rajdeep Dhingra - 3 years, 6 months ago

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