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(A) The angular momentum $L$ of a rotating disc is given by $L = I\omega$,where $I = \frac{1}{2}MR^2$ is the moment of inertia.Since no external torqu

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Four times

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(A) The angular momentum $L$ of a rotating disc is given by $L = I\omega$,where $I = \frac{1}{2}MR^2$ is the moment of inertia.Since no external torque is applied,the angular momentum remains conserved,i.e.,$L = 	ext{constant}$.Therefore,$I_1\omega_1 = I_2\omega_2$.Substituting the expression for moment of inertia: $(\frac{1}{2}MR_1^2)\omega_1 = (\frac{1}{2}MR_2^2)\omega_2$.Given $R_2 = \frac{R_1}{2}$,we have $R_1^2\omega_1 = (\frac{R_1}{2})^2\omega_2$.$R_1^2\omega_1 = \frac{R_1^2}{4}\omega_2$.Solving for $\omega_2$,we get $\omega_2 = 4\omega_1$.Thus,the angular velocity becomes four times the original value.

If the radius of a rotating disc is suddenly halved while its mass remains the same,its angular velocity will become:

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