A one metre steel wire of negligible mass and | vedclass

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(D) Given: Elongation of the wire, $\Delta l = 2 \,mm = 2 	imes 10^{-3} \,m$. Mass of the ball, $m = 1 \,kg$. Length of wire, $l = 1 \,m$. Area of cr

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$20 \,rad/s$

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(D) Given: Elongation of the wire, $\Delta l = 2 \,mm = 2 	imes 10^{-3} \,m$. Mass of the ball, $m = 1 \,kg$. Length of wire, $l = 1 \,m$. Area of cross-section of wire, $A = 0.01 \,cm^2 = 0.01 	imes 10^{-4} \,m^2 = 10^{-6} \,m^2$. Young's modulus of steel, $Y = 2 	imes 10^{11} \,N/m^2$.The tension force $T$ in the wire provides the necessary centripetal force for the ball: $T = m \omega^2 l$.From the definition of Young's modulus, $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{T/A}{\Delta l/l} = \frac{T l}{A \Delta l}$.Substituting $T = m \omega^2 l$ into the equation:$Y = \frac{(m \omega^2 l) l}{A \Delta l} = \frac{m \omega^2 l^2}{A \Delta l}$.Rearranging for $\omega$:$\omega^2 = \frac{Y A \Delta l}{m l^2} \implies \omega = \sqrt{\frac{Y A \Delta l}{m l^2}}$.Substituting the values:$\omega = \sqrt{\frac{(2 	imes 10^{11}) 	imes (10^{-6}) 	imes (2 	imes 10^{-3})}{1 	imes (1)^2}} = \sqrt{\frac{2 	imes 10^{11} 	imes 2 	imes 10^{-9}}{1}} = \sqrt{4 	imes 10^2} = \sqrt{400} = 20 \,rad/s$.

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