Question

The coefficient of linear expansion ${}^{\prime }{\alpha }^{\prime }$ of a rod of length 2m varies with the distance x from the end of the rod as $\alpha ={\alpha }_0+{\alpha }_{1}x$ where ${\alpha }_0=1.76\times {10}^{-5}{}^{-1}and{\alpha }_1=1.2\times {10}^{-6}{m}^{-1}{}^{-1}$. The increase in the length of the rod, when heated through $100$ is :-

• A. $2cm$
• B. $3.76mm$
• C. $1.2mm$
• D. $Noneofthese$

Answer 1

The correct option is B $3.76mm$

$\alpha ={\alpha }_0+{\alpha }_{1}x$ where ${\alpha }_0=1.76\times {10}^{-5}{C}^{-1}$

and ${\alpha }_1=1.2\times {10}^{-6}{m}^{-1}{C}^{-1}$

and $\Delta T={100}^{\circ }C$

$\Delta l=\alpha .dx.\Delta T$

Net extension $l=\int \Delta l=\int \alpha .dx.\Delta T=\int ({\alpha }_0+{\alpha }_{1}x)dx.\Delta T$

$l=\Delta T({\int }_0^L{\alpha }_{0}dx+{\int }_0^L{\alpha }_{1}xdx)$

$=\Delta T({\alpha }_{0}L+\frac{{\alpha }_1{L}^2}{2})$

$=100(1.76\times {10}^{-5}\times 2+1.2\times {10}^{-6}\times \frac{4}{2})$

$=3.76\times {10}^{-3}$m

$=3.76$mm