Question

Coefficients of linear expansion of brass and steel rods are $α_1$ and $α_2$. Length of brass and steel rods are $l_1$ and $l_2$ respectively. If $(l_2−l_1)$ is maintained at all temperature, which one of the following relations holds good?
A. ${{\alpha }_{1}}{{l}_{2}}={{\alpha }_{2}}{{l}_{1}}$
B. ${{\alpha }_{1}}l_{2}^{2}={{\alpha }_{2}}l_{1}^{2}$
C. $\alpha _{1}^{2}{{l}_{2}}=\alpha _{2}^{2}{{l}_{1}}$
D. ${{\alpha }_{1}}{{l}_{1}}={{\alpha }_{1}}{{l}_{1}}$

Answer 1

Hint: Coefficient of linear expansion of a solid is defined as an increase in the length per unit original length at zero degrees per degree rise in temperature. It is found that an increase in the length of the rod is directly proportional to its original length and raise in temperature.

Complete step-by-step answer:
We know that when we heat the rod it gets expanded, therefore its length increases.
So new length is given by,
\[l'={{l}_{0}}(1+\alpha \vartriangle T)\]
Now the length of steel is given by
\[l_{steel}^{'}={{l}_{1}}(1+{{\alpha }_{1}}\vartriangle T)\]
And the length of brass is given by,
\[l_{brass}^{'}={{l}_{2}}(1+{{\alpha }_{2}}\vartriangle T)\]
It is given that different lengths of brass and length of steel must be $(l_2−l_1)$ is maintained at all temperatures.
\[l_{brass}^{'}-l_{steel}^{'}={{l}_{2}}-{{l}_{1}}\]
\[l_{brass}^{'}-l_{steel}^{'}={{l}_{2}}(1+{{\alpha }_{2}}\vartriangle T)-{{l}_{1}}(1+{{\alpha }_{1}}\vartriangle T)\]
\[l_{2}^{{}}-l_{1}^{{}}={{l}_{2}} + {{l}_{2}}{{\alpha }_{2}}\vartriangle T-{{l}_{1}}-{{l}_{1}}{{\alpha }_{1}}\vartriangle T\]

Cancelling $l_1$ and $l_2$, we get
\[{{l}_{2}}{{\alpha }_{2}}={{l}_{1}}{{\alpha }_{1}}\]
Answer is (D).

Note: When we heat any material like a rod, the molecule in the material gains some energy and it tries to move from their position, therefore, it gets vibrated. Because of vibration, it gets expanded. This expansion is known as superficial expansion only if length and breadth are increased. When a body is heated it gets a contract like plastic.