Question

The coefficient of linear expansion for a certain metal varies with temperatures $\alpha \left( T \right)$. If ${L_o}$ is the initial length of metal and the temperature of metal changed from ${T_o}$ to $T({T_o} > T)$ then,
(A) $L = {L_o}\int_{{T_o}}^T {\alpha \left( T \right)} dT$
(B) $L = {L_o}\left[ {1 + \int_{{T_o}}^T {\alpha \left( T \right)} dT} \right]$
(C) $L = {L_o}\left[ {1 - \int_{{T_o}}^T {\alpha \left( T \right)} dT} \right]$
(D) $L > {L_o}$

Answer 1

Hint:
When any metal is heated or cooled then the change in length of the metal is directly proportional to the original length and the change in temperature of the metal, and is given by $dL = {L_o}\alpha \left( T \right)dT$ . So by integrating this over the limits of the initial and final temperature and length, we get the answer.

Complete step by step answer:
The change in length of a metal is directly proportional to the original length and the change in temperature of the metal. So if the original length of the metal is given by ${L_o}$ and the change in temperature is given by $dT$, then for a change in length $dL$ we get,
$\Rightarrow dL \propto {L_o}dT$
On removing the proportionality, we get a constant which is the coefficient of linear expansion and is given by, $\alpha \left( T \right)$. So the change in length is given by,
$\Rightarrow dL = {L_o}\alpha \left( T \right)dT$
Now let us consider the new length of the metal to be $L$ and the temperature changes from ${T_o}$ to $T$. Therefore we can integrate the above equation from the limits of length ${L_o}$ to $L$ and limits of temperature ${T_o}$ to $T$. Therefore, we get,
$\Rightarrow \int_{{L_o}}^L {dL} = \int_{{T_o}}^T {{L_o}\alpha \left( T \right)dT} $
In the R.H.S of the equation, we can take the ${L_o}$ out of the integration since it is constant. So we have,
$\Rightarrow \int_{{L_o}}^L {dL} = {L_o}\int_{{T_o}}^T {\alpha \left( T \right)dT} $
On integrating the L.H.S of the equation we get,
$\Rightarrow \left. L \right|_{{L_o}}^L = {L_o}\int_{{T_o}}^T {\alpha \left( T \right)dT} $
Therefore we get,
$\Rightarrow L - {L_o} = {L_o}\int_{{T_o}}^T {\alpha \left( T \right)dT} $
Taking the ${L_o}$ to the other side we get,
$\Rightarrow L = {L_o} + {L_o}\int_{{T_o}}^T {\alpha \left( T \right)dT} $
So taking the ${L_o}$ common on the R.H.S we get
$\Rightarrow L = {L_o}\left[ {1 + \int_{{T_o}}^T {\alpha \left( T \right)dT} } \right]$
Therefore, we get the final length of the metal as, $L = {L_o}\left[ {1 + \int_{{T_o}}^T {\alpha \left( T \right)dT} } \right]$
So the correct answer is option (B).

Note:
In the question, it is given that the initial temperature of the metal is less than the final temperature. So therefore the metal cools down and on cooling the metal contracts. So the final length, in that case, will be less than the initial length.