A 500 ,g ball is attached to one end of an al | vedclass

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(D) Let $L = 1.4 \,m$ be the unstretched length,$A = 0.5 \,mm^2 = 0.5 	imes 10^{-6} \,m^2$ be the area,$m = 0.5 \,kg$ be the mass,and $	heta = 30^{\

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$0.4$

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(D) Let $L = 1.4 \,m$ be the unstretched length,$A = 0.5 \,mm^2 = 0.5 	imes 10^{-6} \,m^2$ be the area,$m = 0.5 \,kg$ be the mass,and $	heta = 30^{\circ}$ be the angle with the horizontal.Let $r$ be the radius of the horizontal circle and $l$ be the stretched length of the wire.From geometry,$r = l \cos 	heta$ and $h = l \sin 	heta$.The forces acting on the ball are tension $T$ in the wire,gravity $mg$,and centripetal force $m \omega^2 r$.For vertical equilibrium: $T \sin 	heta = mg$.Thus,$T = \frac{mg}{\sin 30^{\circ}} = \frac{0.5 	imes 10}{0.5} = 10 \,N$.Using Young's modulus formula: $Y = \frac{T L}{A \Delta L}$,where $\Delta L$ is the extension.$\Delta L = \frac{T L}{A Y} = \frac{10 	imes 1.4}{0.5 	imes 10^{-6} 	imes 0.7 	imes 10^{11}}$.$\Delta L = \frac{14}{0.35 	imes 10^5} = \frac{14}{35000} = 0.0004 \,m$.Converting to $mm$: $\Delta L = 0.0004 	imes 1000 = 0.4 \,mm$.

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