A large steel wheel is to be fitted onto a sh | vedclass

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(C) Initial temperature $T = 27^{\circ}C = 300\;K$.Initial diameter of the shaft $d_1 = 8.70\;cm$.Diameter of the hole in the wheel $d_2 = 8.69\;cm$.C

Vedclass · Accepted Answer

$-69$

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(C) Initial temperature $T = 27^{\circ}C = 300\;K$.Initial diameter of the shaft $d_1 = 8.70\;cm$.Diameter of the hole in the wheel $d_2 = 8.69\;cm$.Coefficient of linear expansion $\alpha = 1.20 	imes 10^{-5}\;K^{-1}$.The wheel will slip on the shaft when the diameter of the shaft becomes equal to the diameter of the hole,i.e.,$d_2 = 8.69\;cm$.The change in diameter is $\Delta d = d_2 - d_1 = 8.69 - 8.70 = -0.01\;cm$.Using the formula for linear expansion $\Delta d = d_1 \alpha \Delta T$,where $\Delta T = T_f - T_i$:$-0.01 = 8.70 	imes (1.20 	imes 10^{-5}) 	imes (T_f - 300)$.$T_f - 300 = \frac{-0.01}{8.70 	imes 1.20 	imes 10^{-5}} = \frac{-0.01}{0.0001044} \approx -95.78\;K$.$T_f = 300 - 95.78 = 204.22\;K$.Converting to Celsius: $T_f(^{\circ}C) = 204.22 - 273.15 \approx -68.93^{\circ}C$.Rounding to the nearest integer,the temperature is $-69^{\circ}C$.

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