JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 5)
A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion $$\alpha $$/oC. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises by $$\Delta $$TK. Young's modulus, Y, for this metal is :
$${F \over {A\alpha \Delta T}}$$
$${F \over {A\alpha (\Delta T - 273)}}$$
$${F \over {2A\alpha \Delta T}}$$
$${{2F} \over {A\alpha \Delta T}}$$
Explanation
We know,
Young's Modulus, Y = $${{Stress} \over {Strain}}$$
Stress = $${F \over A}$$
Strain = $${{\Delta l} \over l}$$
$$ \therefore $$ Y = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$
We also know,
$$l$$f = $$l$$i (1 + $$\alpha $$$$\Delta $$T)
$$ \Rightarrow $$ $$l$$f = $$l$$i + $$l$$i$$\alpha $$$$\Delta $$T
$$ \Rightarrow $$ $${{{l_f} - {l_i}} \over {{l_i}}}$$ = $$\alpha $$$$\Delta T$$
$$ \Rightarrow $$ $${{\Delta l} \over {{l_i}}}$$ = $$\alpha $$$$\Delta $$T
$$ \therefore $$ y = $${{{F \over A}} \over {\alpha \Delta T}}$$
= $${F \over {A\alpha \Delta T}}$$
Young's Modulus, Y = $${{Stress} \over {Strain}}$$
Stress = $${F \over A}$$
Strain = $${{\Delta l} \over l}$$
$$ \therefore $$ Y = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$
We also know,
$$l$$f = $$l$$i (1 + $$\alpha $$$$\Delta $$T)
$$ \Rightarrow $$ $$l$$f = $$l$$i + $$l$$i$$\alpha $$$$\Delta $$T
$$ \Rightarrow $$ $${{{l_f} - {l_i}} \over {{l_i}}}$$ = $$\alpha $$$$\Delta T$$
$$ \Rightarrow $$ $${{\Delta l} \over {{l_i}}}$$ = $$\alpha $$$$\Delta $$T
$$ \therefore $$ y = $${{{F \over A}} \over {\alpha \Delta T}}$$
= $${F \over {A\alpha \Delta T}}$$
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