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(B) Given:Initial temperature $T_{1} = 27^{\circ}C$Initial length (diameter) of the iron ring $L_{T1} = 5.231\; m$Final length (diameter) required $L_

Vedclass · Accepted Answer

$218$

Vedclass · Answer

(B) Given:Initial temperature $T_{1} = 27^{\circ}C$Initial length (diameter) of the iron ring $L_{T1} = 5.231\; m$Final length (diameter) required $L_{T2} = 5.243\; m$Coefficient of linear expansion $\alpha = 1.20 	imes 10^{-5} K^{-1}$Using the formula for linear expansion:$L_{T2} = L_{T1} [1 + \alpha(T_{2} - T_{1})]$Substituting the values:$5.243 = 5.231 [1 + 1.20 	imes 10^{-5} (T_{2} - 27)]$Divide by $5.231$:$1.002294 = 1 + 1.20 	imes 10^{-5} (T_{2} - 27)$$0.002294 = 1.20 	imes 10^{-5} (T_{2} - 27)$$T_{2} - 27 = \frac{0.002294}{1.20 	imes 10^{-5}}$$T_{2} - 27 \approx 191.17$$T_{2} \approx 191.17 + 27 = 218.17^{\circ}C$Rounding to the nearest integer,the temperature is $218^{\circ}C$.

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