Question

Three rods of equal lengths are joined to form an equilateral triangle $ABC.D$ is the mid-point of $AB.$ The coefficient of linear expansion is ${\alpha }_1$ for material of $\mathrm{rod}AB$ and ${\alpha }_2$ for material of rods $AC$ and $BC.$ If the distance $DC$ 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)

$C{D}^2=(AC{)}^2-(AD{)}^2=l^2-{\left(\frac{l}{2}\right)}^2$
After small change in temperature
$C{D}^2=(AC{)}^2-(AD{)}^2$
$={\left[l\left(1+{\alpha }_{2}t\right)\right]}^2-{\left[\frac{l}{2}\left(1+{\alpha }_{1}t\right)\right]}^2$
$\therefore l^2-\frac{l^2}{4}=l^2\left[1+{\alpha }_2^2{t}^2+2{\alpha }_{2}t\right]$
$-\frac{l^2}{4}\left[1+{\alpha }_1^2{t}^2+2{\alpha }_{1}t\right]$
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 ${\alpha }_1=4{\alpha }_2$