Question

A metal rod of $30\text{cm}$ length expands by $0.075\text{cm}$ when its temperature is raised from $0^{\circ }\text{C}$ to ${100}^{\circ }\text{C}$. Another rod of different metal of length $45\text{cm}$ expands by $0.045\text{cm}$ for the same rise in temperature. A composite rod $C$ made by joining $A$ and $B$ end to end expands by $0.040\text{cm}$ when its length is $45\text{cm}$ and it is heated from $0^{\circ }\text{C}$ to ${50}^{\circ }\text{C}$. Find the length of each portion of the composite rod.

• A. $l_A=23.3\text{cm},l_B=23.3\text{cm}$
• B. $l_A=23.3\text{cm},l_B=21.7\text{cm}$
• C. $l_A=21.7\text{cm},l_B=23.3\text{cm}$
• D. $l_A=25\text{cm},l_B=20\text{cm}$

Answer 1

The correct option is B $l_A=23.3\text{cm},l_B=21.7\text{cm}$

We know that,
$\Delta l=l\alpha \Delta T$ or
So, for metal $A$
$0.075=30\times {\alpha }_A\times 100$
or ${\alpha }_A=25\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$
and for metal $B$,
$0.045=45\times {\alpha }_B\times 100$
or ${\alpha }_B=10\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$
Let $l_A$ and $l_B$ be the lengths of the two sections of $C$.
Then $l_A+l_B=\mathrm{45....}\left(1\right)$
Now, expansion in the composite rod,
$0.04=l_A\times 25\times {10}^{-6}\times 50+l_B\times 10\times {10}^{-6}\times 50$
or $160=5l_A+2l_B....\left(2\right)$
On solving $\left(1\right)$ and $\left(2\right)$, we get
$l_A=23.3\text{cm}$
$l_B=21.7\text{cm}$