c
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}}$
$Y = \frac{{F/A}}{{\Delta l/l}} \Rightarrow \frac{{\Delta l}}{l} = \frac{F}{{AY}}$
$\Delta l = \frac{{Fl}}{{AY}}$
$From\,e{q^n}(i),\,\frac{{Fl}}{{AY}} = l\,\alpha \,\Delta T$
$F = AY\,\alpha \,\Delta T$