Question

The coefficient of linear expansion for a certain metal varies with temperature as $\alpha( T )$. If $L _0$ is the initial length of the metal and the temperature of metal is changed from $T _0$ to $T$ $\left( T _0> T \right)$, then

• A. $L=L_0 \int \limits_{T_0}^T \alpha(T) d T$
• B. $L=L_0\left[1+\int\limits_{T_0}^T \alpha(T) d T\right]$
• C. $L=L_0\left[1-\int\limits_{T_0}^T \alpha(T) d T\right]$
• D. $L >L_0$

Answer 1

Answer: (B)

$As , \frac{ dL }{ L _0}=\alpha( T ) dT \int \limits_{ L _0}^{ L } dL = L _0 \int\limits_{ T _0}^{ T } \alpha( T ) dT$
$L - L _0= L _0 \int\limits_{ T _0}^{ T } \alpha( T ) dT L = L _0\left[1+\int\limits_{ T _0}^{ T } \alpha( T ) dT \right]$