Question

An aluminum rod (length ${{l}_{1}}$ and coefficient of linear expansion ${{\alpha }_{A}}$ and a steel rod (length ${{l}_{2}}$ and coefficient of linear expansion ${{\alpha }_{B}}$) are joined together. If the length of each rod increased by the same amount when their temperatures are raised by ${{t}^{o}}C$, then ${{l}_{1}}/\left( {{l}_{1}}+{{l}_{2}} \right)$ is:

Answer 1

**Hint : **Linear expansion denotes the expansion of length of the object under certain conditions that certain condition has many factors but the factors we are talking about here is temperature. Now as the question has the coefficient of linear expansions in it as the main component that ties both temperature and length it’s formula is: $\dfrac{\Delta L}{L}={{\alpha }_{L}}t$
where $\alpha$ is the coefficient of linear expansion, $\Delta L$ is the change in length and $t$ is the change in temperature, $L$ is the original length.

Complete step by step solution:
Now let us data for the two rods i.e. aluminum and steel rod.
For Aluminum the coefficient of linear expansion is ${{\alpha }_{A}}$.
The length of the rod of the aluminum is given as ${{L}_{1}}$.
The temperature when raised by ${{t}^{o}}C$ gives the linear expansion in terms of temperature as: $t=\dfrac{\Delta {{L}_{1}}}{{{\alpha }_{A}}{{L}_{1}}}$.
For Steel the coefficient of linear expansion is ${{\alpha }_{B}}$.
The length of the rod of the steel is given as ${{l}_{1}}$.
Similarly, when the temperature when raised by ${{t}^{o}}C$ gives the linear expansion in terms of temperature as: $t=\dfrac{\Delta {{L}_{2}}}{{{\alpha }_{B}}{{L}_{2}}}$.
Equating the temperature of both the aluminum and steel rod, we get the relationship between the coefficient of steel and aluminum rod is:
$t=\dfrac{\Delta {{L}_{1}}}{{{\alpha }_{A}}{{L}_{1}}}$ and $t=\dfrac{\Delta {{L}_{2}}}{{{\alpha }_{B}}{{L}_{2}}}$
$\Rightarrow \dfrac{\Delta {{L}_{1}}}{{{\alpha }_{A}}{{L}_{1}}}=\dfrac{\Delta {{L}_{2}}}{{{\alpha }_{B}}{{L}_{2}}}$
With $\Delta {{L}_{1}}$ and $\Delta {{L}_{2}}$ as negligible and valued equal to unity, the ratio of the original length with expansion coefficient as:
$\Rightarrow \dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{{{\alpha }_{B}}}{{{\alpha }_{A}}}$
The above ratio means ${{L}_{1}}={{\alpha }_{B}}$ and ${{L}_{2}}={{\alpha }_{A}}$. Hence, the value of ${{l}_{1}}/\left( {{l}_{1}}+{{l}_{2}} \right)$ after placing the value of ${{L}_{1}}$ and ${{L}_{2}}$ as ${{L}_{1}}={{\alpha }_{B}}$ and ${{L}_{2}}={{\alpha }_{A}}$, we get the value as:
$\Rightarrow \dfrac{{{l}_{1}}}{\left( {{l}_{1}}+{{l}_{2}} \right)}=\dfrac{{{\alpha }_{B}}}{\left( {{\alpha }_{B}}+{{\alpha }_{A}} \right)}$

Note: There are three types of expansion: Linear expansion where the substance or object’s length increases under influence of temperature, area expansion where the substance or object’s surface area increases under influence of temperature and lastly volumetric expansion where the substance or object’s volume increases under influence of temperature.