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(C) The rate of change of angular momentum is equal to the external torque applied,given by $	au = \frac{dL}{dt}$.Since $L = I\omega$,we have $\frac{

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$1$

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(C) The rate of change of angular momentum is equal to the external torque applied,given by $	au = \frac{dL}{dt}$.Since $L = I\omega$,we have $\frac{dL}{dt} = I \frac{d\omega}{dt} = I \alpha$.Given:Initial angular velocity $\omega_i = 6 \ rad \ s^{-1}$Final angular velocity $\omega_f = 21 \ rad \ s^{-1}$Time interval $\Delta t = 1.5 \ s$Moment of inertia $I = 100 \ g \ m^2 = 0.1 \ kg \ m^2$.First,calculate the angular acceleration $\alpha = \frac{\omega_f - \omega_i}{\Delta t} = \frac{21 - 6}{1.5} = \frac{15}{1.5} = 10 \ rad \ s^{-2}$.Now,calculate the rate of change of angular momentum: $\frac{dL}{dt} = I \alpha = 0.1 \ kg \ m^2 	imes 10 \ rad \ s^{-2} = 1 \ N \ m$ (or $kg \ m^2 \ s^{-2}$).Thus,the correct option is $C$.

The angular velocity of a body changes from $6 \ rad \ s^{-1}$ to $21 \ rad \ s^{-1}$ in a time of $1.5 \ s$. If the moment of inertia of the body is $100 \ g \ m^2$,then the rate of change of angular momentum of the body is (in $N \ m$)

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