Question

The co-efficient of thermal expansion of a rod is temperature dependent and is given by the formula $\alpha=a T$, where $a$ is a positive constant and $T$ in ${ }^{\circ} C$. If the length of the rod is $l$ at temperature $0^{\circ} C$, then the temperature at which the length will be $2 l$ is:

• A. $\sqrt{\frac{\ln 2}{\alpha}}$
• B. $\sqrt{\frac{\ln 4}{a}}$
• C. $\frac{1}{\alpha}$
• D. $\frac{2}{\alpha}$

Answer 1

Answer: (B)

As $d l =\alpha l d T$
$\therefore \int\limits_{l}^{2 l} \frac{d l}{l}=a \int\limits_{0}^{T} T d T$
$\ln 2=a \frac{T^{2}}{2}$
$\therefore T=\left[\frac{\ln 4}{a}\right]^{1 / 2}$