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Two rods of different metals, having the same area of cross-section A, are placed end to end between two massive walls as shown in fig. If the temperature of both the rods are now raised by Deltat^(0)C then (a) Find the force with whichthe rods acts on each other at higher temperature. (b) Find the lengths of the rods at the higher temperature. Assume that there is so change in the corss-sectional area of the rodsdo not bend. There is no deformationof walls. |
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Answer» Solution :(a) Due to HEATING the increases in length of the composite rod will be `(Deltal)_(I)=l_(1) alpha_(1) Deltat +l_(2) alpha_(2)DELTA t=(l_(1)alpha_(1)+l_(2)alpha_(2)) Deltat …(1)` due to compressiveforce .F. from the walls, due to elasticity, the decreasein length will be `(Deltal)_(D) = (Fl_(1))/(AY_(1)) +(Fl_(2))/(AY_(2)) =(F)/(A) ((l_(1))/(Y_(1)) +(l_(2))/(Y_(2)))....(2)` As the lengthof the composite rod remainsunchangedthe increase in length due to heatingdue to heatingmustbe equalto decreasein lengthdue to compression. `(l_(1) alpha_(1) +l_(2)alpha_(2)) Deltat=(F)/(A) ((l_(1))/(Y_(1)) +(l_(2))/(Y_(2)))` `RARRF=(A(l_(1)alpha_(1) +l_(2) alpha_(2)) Deltat)/( ((l_(1))/(Y_(1)) +(l_(2))/(Y_(2))))` (b)As initiallythe lengthof one rod in `L_(1)` and due to heatingits length increases by `(Delta l_(1))_(H)=alpha_(1) l_(1)Deltat`, whiledue tocompression itslengthdecreases by `(Deltal_(1))_(C )=(Fl_(1))/(AY_(1))` so its final length`l_(1)^(1) =l_(1) +(Delta l_(1))_(H) -(Deltal_(1))_(C )` `=l_(1) +l_(1)alpha_(1)Deltat-(Fl_(1))/(AY_(1))` for the other rod`l_(2)^(1) =l_(2)+(Deltal_(2))_(H) -(Deltal_(2))_(C )` `l_(2)^(1) =l_(2) +l_(2) alpha_(2)Deltat-(Fl_(2))/(AY_(2))` |
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