At what temperature (in ^{circ} C) should a | vedclass

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(D) The change in length (diameter) required is $\Delta L = L_2 - L_1 = 6.241 \,cm - 6.230 \,cm = 0.011 \,cm$.The formula for linear thermal expansion

Vedclass · Accepted Answer

$152$

Vedclass · Answer

(D) The change in length (diameter) required is $\Delta L = L_2 - L_1 = 6.241 \,cm - 6.230 \,cm = 0.011 \,cm$.The formula for linear thermal expansion is $\Delta L = L_1 \alpha_L \Delta T$,where $\Delta T = T_f - T_i$.Substituting the given values: $0.011 = 6.230 	imes (1.4 	imes 10^{-5}) 	imes (T_f - 27)$.Solving for $\Delta T$: $\Delta T = \frac{0.011}{6.230 	imes 1.4 	imes 10^{-5}} = \frac{0.011 	imes 10^5}{8.722} \approx 126.11^{\circ} C$.Therefore,the final temperature $T_f = 27 + 126.11 = 153.11^{\circ} C$.Comparing with the given options,the closest value is $152.7^{\circ} C$.

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