A metal rod of length 20 , cm expands by 0.07 | vedclass

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(B) The formula for linear expansion is $\Delta L = L_0 \alpha \Delta 	heta$.For rod $A$: $0.075 = 20 	imes \alpha_A 	imes 100 \implies \alpha_A =

Vedclass · Accepted Answer

$10$

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(B) The formula for linear expansion is $\Delta L = L_0 \alpha \Delta 	heta$.For rod $A$: $0.075 = 20 	imes \alpha_A 	imes 100 \implies \alpha_A = \frac{0.075}{2000} = 3.75 	imes 10^{-5} \, ^\circ C^{-1}$.For rod $B$: $0.045 = 20 	imes \alpha_B 	imes 100 \implies \alpha_B = \frac{0.045}{2000} = 2.25 	imes 10^{-5} \, ^\circ C^{-1}$.Let the length of metal $A$ be $x$ and the length of metal $B$ be $(20 - x)$.The total expansion of the composite rod is the sum of the expansions of its parts:$0.060 = (x \alpha_A \Delta 	heta) + ((20 - x) \alpha_B \Delta 	heta)$$0.060 = [x(3.75 	imes 10^{-5}) + (20 - x)(2.25 	imes 10^{-5})] 	imes 100$$0.060 = [3.75x + 45 - 2.25x] 	imes 10^{-3}$$60 = 1.5x + 45$$1.5x = 15$$x = 10 \, cm$.

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