Question

A metal rod A of 25 cm length expands by 0.050 cm when its temperature is raised from $0^{\circ }C$ to ${100}^{\circ }C$. Another rod B of a different metal of length 40cm expands by 0.040 cm for the same rise in temperature. A third rod C of 50 cm length is made up of pieces of rods A and B placed end to end expands by 0.03 cm on heating from $0^{\circ }C$ to ${50}^{\circ }C$. Find the lengths of each portion of the composite rod.

Answer 1

Let ${\alpha }_1$ and ${\alpha }_2$ be the temperature coefficients of linear expansion of two rods of A and B respectively.

$\therefore △L_1=L_1{\alpha }_1\theta \Rightarrow {\alpha }_1=\frac{△L_1}{L_1\theta }△L_2=L_2{\alpha }_2\theta \Rightarrow {\alpha }_2=\frac{△L_2}{L_2\theta }$

Let the composite rod be made of $L_1^{\prime }$ length of A and $L_2^{\prime }$ length of B.

$\therefore L=L_1^{\prime }+L_2^{\prime }=50cm\left(1\right)$

When this rod is heated through $\theta$, then increase in length is,

$△L=({L^{\prime }}_1{\alpha }_1+{L^{\prime }}_2{\alpha }_2)\theta =({L^{\prime }}_1\frac{△L_1}{L_1\theta }+{L^{\prime }}_2\frac{△L_2}{L_2\theta })\theta ={L^{\prime }}_1\frac{△L_1}{L_1}+{L^{\prime }}_2\frac{△L_2}{L_2}$

Given,

$△L_1=0.05cm,△L_2=0.040cm△L=0.03\times 100/50=0.06{L^{\prime }}_1=25cm,{L^{\prime }}_2=40cm\therefore \frac{0.03}{50}\times 100={L^{\prime }}_1\frac{0.05}{25}+{L^{\prime }}_2\frac{0.04}{40}\Rightarrow 2{L^{\prime }}_1+{L^{\prime }}_2=60\left(2\right)$

From equation, $\left(1\right)$ and $\left(2\right)$

${L^{\prime }}_1=10cm{L^{\prime }}_2=40cm$

$L_1$ and $L_2$ represent the length of each portion of the composite rod.