A brass rod of length $$50$$ cm and diameter $$3.0$$ cm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $$250$$°C, if the original lengths are at $$40.0$$°C? (Coefficient$$ of linear expansion of brass $$=$$ $$2.0$$ $$x10-5/$$°C, steel $$=$$ $$1.2$$ x $$10-5/$$°$$C)

Question

A brass rod of length $$50$$ cm and diameter $$3.0$$ cm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $$250$$°C, if the original lengths are at $$40.0$$°C? (Coefficient$$ of linear expansion of brass $$=$$ $$2.0$$ $$x10-5/$$°C, steel $$=$$ $$1.2$$ x $$10-5/$$°$$C)

Infinity Learn · Accepted Answer

Change in temperature of each rod,∆T $$=$$ $$250$$ $$-$$ $$40$$ $$=$$ $$210$$°CClearly, $$change-in$$ length of the brass bar∆Lb $$=$$ α$$0L$$∆T $$=$$ (2.0$$ x $$10-5) x $$50$$ x $$210$$ $$=$$ $$0.21$$ cmand change in length of steel rod∆Ls $$=$$ αsL∆T $$=$$ (1.2$$×$$10-5)×$$50$$×$$210$$ $$=$$ $$0.126$$ cmSince the ends of the rod are free to expand, change in the length of the combined rod∆L $$=$$ ∆$$Lb+$$∆Ls $$=$$ $$0.21$$ $$+$$ $$0.13$$ $$=$$ $$0.34$$ cm

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A brass rod of length 50 cm and diameter 3.0 cm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? (Coefficient of linear expansion of brass = 2.0 x10-5/°C, steel = 1.2 x 10-5/°C)

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