Question

A brass rod of length $50cm$ and diameter $3.0mm$ is joined to a steel rod of the same length and diameter. If the change in length of the combined rod at ${250}^{o}C$ is ($A\times {10}^{-2}cm$), given the original lengths are at ${20.0}^{o}C$, what is the sum of digits of $A$?
The ends of the rod are free to expand (Coefficient of linear expansion of brass $=2.0\times {10}^{-5}{K}^{-1}$, steel$=1.2\times {10}^{-5}{K}^{-1}$)

Answer 1

$△l_1=L\alpha △Q$ $△l_2=L\alpha △r$

$=\frac{1}{2}\times \frac{2}{{10}^5}\times 230$ $=\frac{1}{2}\times \frac{1.2}{{10}^5}\times 230$

$\Rightarrow △l-1+△l_2=\frac{1}{2}\left(\frac{2+1.2}{{10}^5}\right)\times 230$

$=0.368cm$

$=36.8\times {10}^{-2}cm$

$\Rightarrow A=36.8$

Sum of digits $=3+6+8$

$=17$

Hence, the answer is $17.$