First slide

Thermal expansion

Question

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

Moderate

A

$\sqrt{\frac{\ln 2}{a}}$
B

$\sqrt{\frac{\ln 4}{a}}$
C

$\frac{1}{a}$
D

$\frac{2}{a}$

Solution

expansion dl=l $α d t$
here $α = a t$
dl=l at dt
$\frac{d l}{l} = a t d t \int_{l}^{2 l} \frac{d l}{l} = \int_{0}^{t} a t d t_{}^{} {[\ln l]}_{l}^{2 l} = \frac{a t^{2}}{2} \ln 2 = \frac{a t^{2}}{2} \frac{\ln 4}{a} = t^{2} t = \sqrt{\frac{\ln 4}{a}}$

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