Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App