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(C) The thermal stress developed in a wire fixed at both ends when cooled is given by the formula: $F = Y A \alpha \Delta T$.Given:Cross-sectional are

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$60$

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(C) The thermal stress developed in a wire fixed at both ends when cooled is given by the formula: $F = Y A \alpha \Delta T$.Given:Cross-sectional area $A = 3 \, mm^2 = 3 	imes 10^{-6} \, m^2$.Young's modulus $Y = 2 	imes 10^{11} \, N/m^2$.Coefficient of linear expansion $\alpha = 10^{-5} \, ^{\circ}C^{-1}$.Change in temperature $\Delta T = 20^{\circ}C - 10^{\circ}C = 10^{\circ}C$.Substituting these values into the formula:$F = (2 	imes 10^{11} \, N/m^2) 	imes (3 	imes 10^{-6} \, m^2) 	imes (10^{-5} \, ^{\circ}C^{-1}) 	imes (10^{\circ}C)$.$F = 2 	imes 3 	imes 10^{11 - 6 - 5 + 1} \, N$.$F = 6 	imes 10^1 \, N = 60 \, N$.Therefore,the tension in the wire is $60 \, N$.

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