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(A) Since no external torque is applied,the angular momentum $L$ of the body remains conserved.$L = I_1 \omega_1 = I_2 \omega_2$We know that the momen

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$\sqrt{\omega_2} : \sqrt{\omega_1}$

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(A) Since no external torque is applied,the angular momentum $L$ of the body remains conserved.$L = I_1 \omega_1 = I_2 \omega_2$We know that the moment of inertia $I = MK^2$,where $M$ is the mass and $K$ is the radius of gyration.Substituting this into the conservation equation:$M K_1^2 \omega_1 = M K_2^2 \omega_2$$K_1^2 \omega_1 = K_2^2 \omega_2$$\frac{K_1^2}{K_2^2} = \frac{\omega_2}{\omega_1}$Taking the square root on both sides,we get the ratio of the radii of gyration:$\frac{K_1}{K_2} = \sqrt{\frac{\omega_2}{\omega_1}}$Thus,the ratio $K_1 : K_2 = \sqrt{\omega_2} : \sqrt{\omega_1}$.

The angular velocity of a body changes from $\omega_1$ to $\omega_2$ without the application of any external torque,but this change occurs due to a change in its moment of inertia. The ratio of the radii of gyration in the two cases will be:

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