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Two rods of the same cross-sectional area are attached end to end forming a total length of 1 m, at 25^(@)C. One rod in the combination is a 30 cm long copper rod. The composite rod, increases by 1.91 mm at 125^(@)C. If the composite rod is rigidly fixed between two walls so that no change in length may occur even with the rise in temperature, find the Young's modulus (Y) and the coefficient of linear expansion (alpha) for the second rod. alpha for copper =1.7 times 10^(-5@)C^(-1),Y for copper =1.3 times 10^(11)N*m^(-2). |
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Answer» Solution :Length of the 2nd rod at `25^(@)C` = 100 cm - length of the copper rod at `25^(@)C` = 100 - 30 = 70 cm. Increase in length of the copper rod for the rise in temperature. `""=30 times 1.7 times 10^(-5) times(125-25)cm` `"" =0.051 cm=0.51mm` Increase in length of the 2nd rod for the same rise in temperature = 1.91 - 0.51 = 1.4 mm = 0.14 cm If the coefficient of linear expansion of the MATERIAL of the second rod is `alpha,` `"" 0.14=70 times alpha times (125-15)` `therefore "" alpha=0.14/(70 times (125-25))=2.0 times 10^(-5@)C^(-1)` Again, thermal stress `=Yalpha(t_(2)-t_(1))`. For copper rod, thermal stress `=1.3 times 10^(11) times 1.7 times 10^(-5) times (t_(2)-t_(1))` and for the 2nd rod, the thermal stress `=Y times 2 times 10^(-5) times (t_(2)-t_(1)).` [Y = Young.s modulus for the second rod] Since the two RODS have the same area of cross-section, the thermal stresses should be equal. `therefore "" 1.3 times 10^(11) times 1.7 times 10^(-5) times (t_(2)-t_(1))=Y times 2 times 10^(-5) times (t_(2)-t_(1))` or, `"" Y=(1.3 times 10^(11) times 1.7 times 10^(-5))/(2 times 10^(-5))=1.105 times 10^(11) N*m^(-2).` |
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