JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 16)
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 $$\ell $$ 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 :
$$\ell $$2 Y$$\alpha $$ $$\Delta $$T
$$\ell $$A Y$$\alpha $$ $$\Delta $$T
A Y$$\alpha $$ $$\Delta $$T
$${{AY} \over {\alpha \,\Delta T}}$$
Explanation
Because of thermal expansion, change in length
($$\Delta $$$$\ell $$) = $$\ell $$ $$\alpha $$ $$\Delta $$T . . . . .(1)
Because of compressive force, the compansion is $$\Delta $$$$\ell $$ ' ,
$$\therefore\,\,\,$$ Young's Modulus (y) = $${{{F \over A}} \over {{{\Delta \ell} \over \ell }}}$$
$$ \Rightarrow $$$$\,\,\,$$ F = YA $${{\Delta \ell '} \over \ell }$$
As net change in length is 0 . So,
$$\Delta \ell '$$ = $$\Delta \ell $$ = $$\ell \alpha \,\Delta T$$
$$\therefore\,\,\,$$ F = YA $$ \times $$ $${{\ell \alpha \,\Delta T} \over \ell }$$ = AY$$\alpha $$$$\Delta $$T
($$\Delta $$$$\ell $$) = $$\ell $$ $$\alpha $$ $$\Delta $$T . . . . .(1)
Because of compressive force, the compansion is $$\Delta $$$$\ell $$ ' ,
$$\therefore\,\,\,$$ Young's Modulus (y) = $${{{F \over A}} \over {{{\Delta \ell} \over \ell }}}$$
$$ \Rightarrow $$$$\,\,\,$$ F = YA $${{\Delta \ell '} \over \ell }$$
As net change in length is 0 . So,
$$\Delta \ell '$$ = $$\Delta \ell $$ = $$\ell \alpha \,\Delta T$$
$$\therefore\,\,\,$$ F = YA $$ \times $$ $${{\ell \alpha \,\Delta T} \over \ell }$$ = AY$$\alpha $$$$\Delta $$T
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