Question

Three rods of equal lengths are joined to form an equilateral triangle $A B C . D$ is the mid-point of $A B .$ The coefficient of linear expansion is $\alpha_{1}$ for material of $\operatorname{rod} A B$ and $\alpha_{2}$ for material of rods $A C$ and $B C .$ If the distance $D C$ remains constant for small changes in temperature, then

• A. $\alpha_{1}=2 \alpha_{2}$
• B. $\alpha_{1}=4 \alpha_{2}$
• C. $\alpha_{1}=8 \alpha_{2}$
• D. $\alpha_{1}=\alpha_{2}$

Answer 1

Answer: (B)

$C D^{2}=(A C)^{2}-(A D)^{2}=l^{2}-\left(\frac{l}{2}\right)^{2}$
After small change in temperature
$C D^{2} =(A C)^{2}-(A D)^{2}$
$=\left[l\left(1+\alpha_{2} t\right)\right]^{2}-\left[\frac{l}{2}\left(1+\alpha_{1} t\right)\right]^{2}$
$\therefore l^{2}-\frac{l^{2}}{4}=l^{2}\left[1+\alpha_{2}^{2} t^{2}+2 \alpha_{2} t\right]$
$-\frac{l^{2}}{4}\left[1+\alpha_{1}^{2} t^{2}+2 \alpha_{1} t\right]$
Neglecting $\alpha_{2}^{2} t^{2}$ and $\alpha_{1}^{2} t^{2}$, (because very small quantity)
$0= l^{2}\left(2 \alpha_{2} t\right)-\frac{l^{2}}{4}\left(2 \alpha_{1} t\right)$
or $\alpha_{1}=4 \alpha_{2}$