JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2017
- Medium
- MCQ
A compressive force, \(F\) is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by \(\Delta T\). The net change in its length is zero. Let \(l\) be the length of the rod, \(A\) its area of cross- section, \(Y\) its Young's modulus, and \(\alpha \) its coefficient of linear expansion. Then, \(F\) is equal to
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(A\,Y\alpha \Delta T\)
Step-by-step Solution
Detailed explanation
Due to thermal exp., change in length \(\left( {\Delta l} \right)\)
\( = l\alpha \Delta T\) \(...(i)\)
\(Young's\,modulus (Y)\)
\( = \frac{{Normal\,stress}}{{Longitudinal\,strain}}\)
\(Y = \frac{{F/A}}{{\Delta l/l}} \Rightarrow \frac{{\Delta l}}{l} = \frac{F}{{AY}}\)
\(\Delta l = \frac{{Fl}}{{AY}}\)
\(From\,e{q^n}(i),\,\frac{{Fl}}{{AY}} = l\,\alpha \,\Delta T\)
\(F = AY\,\alpha \,\Delta T\)
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