JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 11)
A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of
Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net
longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of
the material of the rod, is (nearly) equal to :
$${{3F} \over {\left( {\pi {r^2}YT} \right)}}$$
$${{6F} \over {\left( {\pi {r^2}YT} \right)}}$$
$${F \over {\left( {3\pi {r^2}YT} \right)}}$$
$${9F\left( {\pi {r^2}YT} \right)}$$
Explanation
Change in length due to temperature change,
$$\Delta $$$$l$$ = $$l$$$$\alpha $$$$\Delta $$T
$${{\Delta l} \over l}$$ = $$\alpha $$T [ Here $$\Delta $$T = T ]
Y = $${{{F \over {\pi {r^2}}}} \over {{{\Delta l} \over l}}}$$
= $${{{F \over {\pi {r^2}}}} \over {\alpha T}}$$
$$ \Rightarrow $$ Y = $${F \over {\pi {r^2}\alpha T}}$$
$$ \Rightarrow $$ $$\alpha $$ = $${F \over {\pi {r^2}YT}}$$
We know, The coefficient of volume expansion ($$\gamma $$) = 3$$\alpha $$
$$ \therefore $$ $$\gamma $$ = $${{3F} \over {\pi {r^2}YT}}$$
$$\Delta $$$$l$$ = $$l$$$$\alpha $$$$\Delta $$T
$${{\Delta l} \over l}$$ = $$\alpha $$T [ Here $$\Delta $$T = T ]
Y = $${{{F \over {\pi {r^2}}}} \over {{{\Delta l} \over l}}}$$
= $${{{F \over {\pi {r^2}}}} \over {\alpha T}}$$
$$ \Rightarrow $$ Y = $${F \over {\pi {r^2}\alpha T}}$$
$$ \Rightarrow $$ $$\alpha $$ = $${F \over {\pi {r^2}YT}}$$
We know, The coefficient of volume expansion ($$\gamma $$) = 3$$\alpha $$
$$ \therefore $$ $$\gamma $$ = $${{3F} \over {\pi {r^2}YT}}$$
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