Question

The co-efficient of thermal expansion of a rod is temperature dependent and is given by the formula $=aT$, 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 $2l$ is:

• A. $\sqrt{\frac{ln2}{a}}$
• B. $\frac{2}{a}$
• C. $\frac{1}{a}$
• D. $\sqrt{\frac{ln4}{a}}$

Answer 1

The correct option is D $\sqrt{\frac{ln4}{a}}$

We know that:$\alpha =\frac{dl}{dT}$

Given: $\alpha =aT$

As $dl=\alpha ldT$

$\frac{dl}{l}=\alpha .dT$
$\frac{dl}{l}=aTdT$
${\int }_l^{2l}\frac{dl}{l}=a{\int }_0^{T}TdT$
$ln2=a\left(\frac{T^2}{2}\right)$
$T={\left[\frac{ln4}{a}\right]}^{\frac{1}{2}}$
Final answer (b)