Three uniform thin aluminium rods each of len | vedclass

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(D) Let the length of each rod be $L = 2 \,m$. The rods form an equilateral triangle. The centre of mass $(COM)$ of the system is at the centroid of t

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$0.2$

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(D) Let the length of each rod be $L = 2 \,m$. The rods form an equilateral triangle. The centre of mass $(COM)$ of the system is at the centroid of the triangle.For an equilateral triangle with side length $L$, the height is $h = \frac{\sqrt{3}}{2} L$. The $y$-coordinate of the $COM$ is $y_{COM} = \frac{1}{3} h = \frac{\sqrt{3}}{6} L$.Given $\gamma = 12 \sqrt{3} 	imes 10^{-6} \,K^{-1}$, the coefficient of linear expansion is $\alpha = \frac{\gamma}{3} = 4 \sqrt{3} 	imes 10^{-6} \,K^{-1}$.The change in length of each rod is $\Delta L = L \alpha \Delta T = 2 	imes (4 \sqrt{3} 	imes 10^{-6}) 	imes 50 = 400 \sqrt{3} 	imes 10^{-6} \,m = 4 \sqrt{3} 	imes 10^{-4} \,m$.The new length is $L' = L + \Delta L = L(1 + \alpha \Delta T)$.The new $y$-coordinate of the $COM$ is $y'_{COM} = \frac{\sqrt{3}}{6} L' = \frac{\sqrt{3}}{6} L(1 + \alpha \Delta T)$.The increase in $y$-coordinate is $\Delta y_{COM} = y'_{COM} - y_{COM} = \frac{\sqrt{3}}{6} L \alpha \Delta T$.Substituting the values: $\Delta y_{COM} = \frac{\sqrt{3}}{6} 	imes 2 	imes (4 \sqrt{3} 	imes 10^{-6}) 	imes 50 = \frac{\sqrt{3}}{3} 	imes 4 \sqrt{3} 	imes 50 	imes 10^{-6} = \frac{3}{3} 	imes 4 	imes 50 	imes 10^{-6} = 200 	imes 10^{-6} \,m = 0.2 	imes 10^{-3} \,m = 0.2 \,mm$.

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