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(C) Given:Initial length $L = 100 \, m$Initial temperature $T_1 = 25^{\circ} C$Final temperature $T_2 = 40^{\circ} C$Change in temperature $\Delta T =

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$3.3$

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(C) Given:Initial length $L = 100 \, m$Initial temperature $T_1 = 25^{\circ} C$Final temperature $T_2 = 40^{\circ} C$Change in temperature $\Delta T = T_2 - T_1 = 40 - 25 = 15^{\circ} C$Coefficient of linear expansion $\alpha = 1.1 	imes 10^{-5} /^{\circ} C$Young's modulus $Y = 2 	imes 10^{11} \, Pa$Since the track is constrained from expanding,thermal stress is developed in the material.Thermal stress $\sigma$ is given by the formula:$\sigma = Y \cdot \alpha \cdot \Delta T$Substituting the values:$\sigma = (2 	imes 10^{11} \, Pa) 	imes (1.1 	imes 10^{-5} /^{\circ} C) 	imes (15^{\circ} C)$$\sigma = 2 	imes 1.1 	imes 15 	imes 10^{11-5} \, Pa$$\sigma = 33 	imes 10^6 \, Pa$$\sigma = 3.3 	imes 10^7 \, Pa$Thus,the stress on the track is $3.3 	imes 10^7 \, Pa$.

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