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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 Delta t^@ Cthen a) Find the force with which the rods acts on each other at higher temperature. b) Find the lengths of the rods at the higher temperature. Assume that there is no change in the cross-sectional area of the rods and the rods do not bend. There is no deformation of walls. |
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Answer» Solution :a) Due to heating the increases in LENGTH of the composite rod will be `(Delta l)_1 = l_1 alpha_1 Delta t + l_2 alpha_2 Delta t = (l_1 alpha_1 + l_2 alpha_2 ) Delta t `......(1) due to compressive force .F. from the walls, due to ELASTICITY, the decrease in length will be `(Delta l)_D = (Fl_1)/(AY_1) + (Fl_2)/(AY_2) = F/A (l_1/Y_1 + l_2Y_2)`.....(2) As the length of the composite rod remains unchanged the increase in length due to heating MUST be EQUAL to decrease in length due to compression. `(l_1 alpha_1 + l_2 alpha_2) Delta t = F/A (l_1/Y_1 + l_2Y_2)` ` rArr F = (A(l_1 alpha_1 + l_2 alpha_2)Delta t)/((l_1/Y_1 + l_2/Y_2))` b) As initially the length of one rod in `L_1`and due to heating its length increases by `(Delta l_1)_H = alpha_1 l_2 Delta t` ,while due to compression its length decreases by `(Delta l_1)_C = (Fl_1)/(AY_1)` so its final length` l_1^1 = l_1 + (Delta l_1)_H - (Delta l_1)_C` ` = l_1 + l_1 alpha_1 Delta t- (Fl_1)/(AY_1)` For the other rod `l_1^1 = l_2 + (Delta l_1)_H - (Delta l_2)_C` `l_2^1= l_2 + l_1 alpha_2 Delta t- (Fl_2)/(AY_2)` |
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