The elongation in a metallic rod hinged at one end and rotating in a horizontal plane becomes four times of the initial value. The angular velocity of rotation becomes

$-{\int }_{\mathrm{T}}^0∆\mathrm{T} = {\int }_0^{\mathrm{l}}\frac{\mathrm{m}}{\mathrm{l}}{\mathrm{dx\omega }}^{2 }\mathrm{x} \Rightarrow \mathrm{T} = \frac{\mathrm{m}}{\mathrm{l}}{\mathrm{\omega }}^2\frac{{\mathrm{x}}^2}{2}$

$\Rightarrow \mathrm{Y} =\frac{\mathrm{Fl}}{\mathrm{A}∆\mathrm{l}}$

$\mathrm{\delta l} = \frac{\mathrm{Fl}}{\mathrm{AY}}$

$\mathrm{\delta l} = \frac{{\displaystyle \frac{\mathrm{m}}{\mathrm{l}}\frac{{\mathrm{\omega }}^2{\mathrm{x}}^2}{2}\mathrm{dx}}}{\mathrm{AY}}$

$∆\mathrm{l}=\int \mathrm{\delta l} = \frac{{\mathrm{m\omega }}^2{\mathrm{l}}^3}{ 6\mathrm{AYl}}$

$∆\mathrm{l} = \frac{{\mathrm{\rho \omega }}^2{\mathrm{l}}^3}{6\mathrm{Y}}$

$∆ \mathrm{l} \propto {\mathrm{\omega }}^2$

For elongation to become 4 times the final angular speed must be

${\mathrm{\omega }}_2 = 2{\mathrm{\omega }}_1$