In the shown planar frame made of thin uniform rods, the length of section A B and E F is l1 and its thermal linear coefficient of expansion is α1 . The length of section C D is l2 and its thermal linear coefficient of expansion is α2. C B and D E are of same length having thermal linear coefficient of expansion α2. Points A, B, E and F reside on same line, that is, sections A B and E F overlap. Then the ratio of (l1/l2), for which the distance between end A and end F remains the same at all temperatures, is:

Question

In the shown planar frame made of thin uniform rods, the length of section A B and E F is l1 and its thermal linear coefficient of expansion is α1 . The length of section C D is l2 and its thermal linear coefficient of expansion is α2. C B and D E are of same length having thermal linear coefficient of expansion α2. Points A, B, E and F reside on same line, that is, sections A B and E F overlap. Then the ratio of (l1/l2), for which the distance between end A and end F remains the same at all temperatures, is: (A) (α2/2 α1) (B) (2 α2/α1) (C) (2 α1/α2) (D) (α1/2 α2)

Tardigrade Admin · Accepted Answer

For distance between A and F to remain constant, extension in C D= extension in A B+ extension in E F ∴ Δ l2=2 Δ l1 ⇒ l2 α2 Δ θ2=2 l1 α1 Δ θ or (l1/l2)=(α2/2 α1)

$\frac{α _{2}}{2 α _{1}}$

$\frac{2 α _{2}}{α _{1}}$

$\frac{2 α _{1}}{α _{2}}$

$\frac{α _{1}}{2 α _{2}}$

For distance between $A$ and $F$ to remain constant, extension in $C D =$ extension in $A B +$ extension in $EF$
$∴ Δ l_{2} = 2Δ l_{1}$
$\Rightarrow l_{2} α_{2} Δ θ_{2} = 2 l_{1} α_{1} Δ θ$
or $\frac{l _{1}}{l _{2}} = \frac{α _{2}}{2 α _{1}}$

2α1​α2​​

α1​2α2​​

α2​2α1​​

2α2​α1​​

For distance between A and F to remain constant, extension in CD= extension in AB+ extension in EF ∴Δl2​=2Δl1​ ⇒l2​α2​Δθ2​=2l1​α1​Δθ or l2​l1​​=2α1​α2​​

$\frac{α _{2}}{2 α _{1}}$

$\frac{2 α _{2}}{α _{1}}$

$\frac{2 α _{1}}{α _{2}}$

$\frac{α _{1}}{2 α _{2}}$

For distance between $A$ and $F$ to remain constant, extension in $C D =$ extension in $A B +$ extension in $EF$
$∴ Δ l_{2} = 2Δ l_{1}$
$\Rightarrow l_{2} α_{2} Δ θ_{2} = 2 l_{1} α_{1} Δ θ$
or $\frac{l _{1}}{l _{2}} = \frac{α _{2}}{2 α _{1}}$