A one metre steel wire of negligible mass and area of cross-section...

Given, elongation of the wire, $\Delta l=2 \mathrm{~mm}$ =2 $\times$10^{-3} $\mathrm{~m}$Mass of the ball, $m=1 \mathrm{~kg}$ Length of wire, $l=1 \mathrm{~m}$ Area of cross-sectional of wire, A=0.01 $\mathrm{~cm}^2=0.01$\times$10^{-4}$\mathrm{~m}$Young's modulus of steel,$Y=2 \times 10^{11} \mathrm{Nm}^{-2}\because$Tension force in wire,$T=m \omega^2 l\because$Stress$=\frac{\text { Tension }}{\text { Area }}=\frac{m \omega^2 l}{A}$Strain$=\frac{\Delta l}{l}=\frac{\text { stress }}{\text { Young's modulus }}$or$\quad \Delta l=\frac{m \omega^2 l^2}{Y A}$or$\quad \omega=\sqrt{\frac{Y A \Delta l}{m l^2}}$Putting the given values, we get$$\begin{aligned} \text{ and } =\sqrt{\frac{2 \times 10^{11} \times 0.01 \times 10^{-4} \times 2 \times 10^{-3}}{1 \times(1)^2}}\\ \omega \text{ and } =20 \mathrm{rad}/ \mathrm\sec^{-1} \end{aligned}$$