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The coefficient of linear expansion of a rod changes from $\;\alpha_{1}\;$ to $\;\alpha_{2}\;$ from end to end for a length l of the rod . If the temperature changes from $\;0^{0}C\;$ to $\;t^{0}C\;$ , the change of length is

$(a)\;\large\frac{(\alpha_{2}-\alpha_{1})}{2}\;lt\qquad(b)\;\large\frac{(\alpha_{1}+\alpha_{2})}{2}\;lt\qquad(c)\;\sqrt{\alpha_{1} \alpha_{2}}\qquad(d)\;\large\frac{(\alpha_{1}-\alpha_{2})}{2}\;lt$

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Answer : $\;\large\frac{(\alpha_{1}+\alpha_{2})}{2}\;lt$
Explanation :
$\alpha\;$ at a distance x from one end then $\;\alpha=\large\frac{(\alpha_{2}-\alpha_{1})}{2} x + \alpha_{1}$
$\alpha = \alpha_{1} + \large\frac{(\alpha{2}-\alpha_{1})}{l}x$
Changes in length for element dx
$dl^{'}=[\alpha_{1}+ (\large\frac{\alpha_{2}-\alpha_{1}}{l}) x dx]\;\bigtriangleup t$
Therefore , total change in length
$\bigtriangleup l = \int_{0}^{l} [(\alpha_{1} + (\large\frac{\alpha_{2} - \alpha_{1}}{t}))]\;dx$
$=\alpha_{1} l + \large\frac{\alpha_{2}-\alpha_{1}}{l}\;\large\frac{l^2}{2}$
Note : -
Average $\;\alpha=(\large\frac{\alpha_{1}+\alpha_{2}}{2})$
$\bigtriangleup l=(\large\frac{\alpha_{1}+\alpha_{2}}{2}) l \bigtriangleup t$
answered Mar 24, 2014 by yamini.v

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