Question

By what temperature a rod should be heated so that it's length becomes $2l_0$ if initial length is $l_0$ . Assume that $\alpha$ (coefficient of linear expansion) in that range of temperature remains constant and melting point is very high.

• A. $\frac{2}{\alpha }$
• B. $\frac{1}{\alpha }$
• C. $\frac{ln2}{\alpha }$
• D. $\frac{ln1}{\alpha }$

Answer 1

The correct option is C $\frac{ln2}{\alpha }$

$\alpha =\frac{1}{L}\frac{dL}{dt}\stackrel{btw}{\to }{l}^{\prime }=l(1+\alpha \Delta T)$
Original form
Note:Here expansion is too large to use this approximation.
${}_{l_0}^{2l_0}\int \frac{dL}{L}{=}_{T_1}^{T_2}\int \alpha dT\Rightarrow lnL{|}_{l_0}^{2l_0}=\alpha (T_{\alpha }-T_1)\Rightarrow ln\left(2l_0\right)-ln{l}_0=\alpha \Delta T\Rightarrow \alpha \Delta T=ln2\Rightarrow \Delta T=\frac{ln2}{\alpha }$
A very large number btw