A rod BC of negligible mass is fixed at end B | vedclass

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Question

(A) The thermal contraction of the rod is $\Delta L_{thermal} = L \alpha \Delta 	heta = 4 	imes (2.2 	imes 10^{-4}) 	imes 100 = 0.088 \ m = 8.8 \

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The thermal contraction of the rod is $\Delta L_{thermal} = L \alpha \Delta 	heta = 4 	imes (2.2 	imes 10^{-4}) 	imes 100 = 0.088 \ m = 8.8 \ cm$.Let $x$ be the actual decrease in length of the rod. The spring is stretched by $x$ due to the contraction of the rod.The force exerted by the spring is $F = Kx$.The stress in the rod is $\sigma = \frac{F}{A} = \frac{Kx}{A}$.The strain in the rod is $\epsilon = \frac{\Delta L_{thermal} - x}{L}$.Using Hooke's Law,$Y = \frac{\sigma}{\epsilon} = \frac{Kx/A}{(\Delta L_{thermal} - x)/L} = \frac{KxL}{A(\Delta L_{thermal} - x)}$.Rearranging for $x$: $Y A (\Delta L_{thermal} - x) = KxL \implies Y A \Delta L_{thermal} = x(YA + KL)$.$x = \frac{Y A \Delta L_{thermal}}{YA + KL} = \frac{(10^{11}) 	imes (4 	imes 10^{-4}) 	imes 0.088}{(10^{11} 	imes 4 	imes 10^{-4}) + (10^4 	imes 4)} = \frac{3520}{40000 + 40000} = \frac{3520}{44000} = 0.08 \ m = 8 \ cm$.Wait,re-evaluating the calculation: $YA = 4 	imes 10^7$,$KL = 4 	imes 10^4$. $x = \frac{4 	imes 10^7 	imes 0.088}{4 	imes 10^7 + 4 	imes 10^4} = \frac{3520000}{40040000} \approx 0.0879 \ m = 8.79 \ cm$.The closest integer is $9 \ cm$.

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