question_answer

A metal rod of cross-sectional area A and length L is fixed between two rigid walls. If the rod is heated by temperature \[\Delta \]T, then what shall be the reaction generated by walls given that Young's modulus of rod is Y and coefficient of thermal expansion is\[\alpha \]?

A)   \[\frac{\alpha AY(\Delta T)}{2}\]            

B)  \[\alpha AY(\Delta T)\]

C)  \[\frac{2\alpha (\Delta T)}{AY}\]             

D)  \[\frac{\alpha (\Delta T)}{2AY}\]

Correct Answer: B

Solution :

x Idea This question is based on the concept of thermal expansion and on Hooke's law. Combining these two concepts, students can easily solve this question. Due to heating, there shall be expansion in rod given by, \[\Delta {{L}_{1}}=\alpha (\Delta T)L\] But this expansion is resisted by rigid rod. If ff is the reaction produced then, stress in rod is\[\frac{R}{A}\] So,  \[\text{strain=}\frac{stress}{Y}\Rightarrow \frac{\Delta {{L}_{2}}}{L}=\frac{R/A}{Y}\] Y [\[\Delta {{L}_{2}}\], compression by stress due to wall] \[\Rightarrow \]\[\Delta {{L}_{2}}=\frac{R}{AY}.L\] Since walls are rigid so expansion due to heating should be equal to compression by stress due to walls \[\Rightarrow \Delta {{L}_{1}}=\Delta {{L}_{2}}\Rightarrow L\alpha .(\Delta T)=\frac{RL}{AY}\] \[\Rightarrow R=\alpha AY(\Delta T)\] TEST Edge This question is based on linear thermal expansion and on Hooke's law and could be asked separately.