A metal rod of cross-sectional area 3 times 1 | vedclass

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(C) The thermal contraction that would occur due to cooling is $\Delta L = L \alpha \Delta T$.To prevent this contraction, the tensile force $F$ appli

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$30$

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(C) The thermal contraction that would occur due to cooling is $\Delta L = L \alpha \Delta T$.To prevent this contraction, the tensile force $F$ applied by the mass $m$ must produce an equal extension: $\Delta L = \frac{FL}{AY}$.Equating the two expressions: $L \alpha \Delta T = \frac{FL}{AY}$.Thus, the force required is $F = AY \alpha \Delta T$.Since $F = Mg$, we have $Mg = AY \alpha \Delta T$.Substituting the given values: $M = \frac{AY \alpha \Delta T}{g}$.$M = \frac{(3 	imes 10^{-6} \,m^{2}) 	imes (10^{11} \,N/m^{2}) 	imes (10^{-5} /K) 	imes (100 - 0) \,K}{10 \,m/s^{2}}$.$M = \frac{3 	imes 10^{0} 	imes 100}{10} = \frac{300}{10} = 30 \,kg$.

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