Question

Two rods having length ${\ell }_{1}and{\ell }_2$, made of materials with the linear coefficient of expansion ${\alpha }_{1}and{\alpha }_2$, were welded together. The equivalent coefficients of linear expansion for the obtained rod

• A. $\frac{{\ell }_1{\alpha }_2+{\ell }_2{\alpha }_1}{{\ell }_1+{\ell }_2}$
• B. $\frac{{\ell }_1{\alpha }_1+{\ell }_2{\alpha }_2}{{\alpha }_1+{\alpha }_2}$
• C. $\frac{{\ell }_1{\alpha }_1+{\ell }_2{\alpha }_2}{{\ell }_1+{\ell }_2}$
• D. $\frac{{\ell }_2{\alpha }_1+{\ell }_1{\alpha }_2}{{\alpha }_1+{\alpha }_2}$

Answer 1

The correct option is C $\frac{{\ell }_1{\alpha }_1+{\ell }_2{\alpha }_2}{{\ell }_1+{\ell }_2}$

$l_1^{{}^{\prime }}=l_1+l_1{\alpha }_1\Delta \theta$

$l_2^{{}^{\prime }}=l_2+l_2{\alpha }_2\Delta \theta$

adding two equations above,

$l_1^{{}^{\prime }}+l_2^{{}^{\prime }}=l_1+l_2+(l_1{\alpha }_1+l_2{\alpha }_2)\Delta \theta$

$\Rightarrow l_1^{{}^{\prime }}+l_2^{{}^{\prime }}=l_1+l_2+(l_1+l_2)\frac{l_1{\alpha }_1+l_2{\alpha }_2}{l_1+l_2}\Delta \theta$

From above equation, the equivalent coefficient of linear thermal expansion is $\frac{l_1{\alpha }_1+l_2{\alpha }_2}{l_1+l_2}$.