JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 12)
Two different wires having lengths L1 and L2,
and respective temperature coefficient of linear
expansion $$\alpha $$1 and $$\alpha $$2, are joined end-to-end.
Then the effective temperature coefficient of
linear expansion is :
$$2\sqrt {{\alpha _1}{\alpha _2}} $$
$$4{{{\alpha _1}{\alpha _2}} \over {{\alpha _1} + {\alpha _2}}}{{{L_2}{L_1}} \over {{{\left( {{L_2} + {L_1}} \right)}^2}}}$$
$${{{\alpha _1} + {\alpha _2}} \over 2}$$
$${{{\alpha _1}{L_1} + {\alpha _2}{L_2}} \over {{L_1} + {L_2}}}$$
Explanation
_5th_September_Evening_Slot_en_12_1.png)
L'1 = L1(1 + $$\alpha $$1$$\Delta $$T)
L'2 = L2(1 + $$\alpha $$2$$\Delta $$T)
L'eq = (L1 + L2) (1 + $$\alpha $$avg$$\Delta $$T)
$$ \therefore $$ (L1 + L2) (1 + $$\alpha $$avg$$\Delta $$T) = L1(1 + $$\alpha $$1$$\Delta $$T) + L2(1 + $$\alpha $$2$$\Delta $$T)
$$ \Rightarrow $$ (L1 + L2)$$\alpha $$avg = L1$$\alpha $$1 + L2$$\alpha $$2
$$ \Rightarrow $$ $$\alpha $$avg = $${{{\alpha _1}{L_1} + {\alpha _2}{L_2}} \over {{L_1} + {L_2}}}$$
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