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(A) Given: Mass of the wheel $M = 10 \ kg$,Radius $R = 10 \ cm = 0.1 \ m$,Force $F = 20 \ N$,Displacement $d = 1 \ m$.Since the wheel is supported by

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$20$

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(A) Given: Mass of the wheel $M = 10 \ kg$,Radius $R = 10 \ cm = 0.1 \ m$,Force $F = 20 \ N$,Displacement $d = 1 \ m$.Since the wheel is supported by spokes of negligible mass,its moment of inertia is $I = MR^2 = 10 	imes (0.1)^2 = 0.1 \ kg \ m^2$.The work done by the force $F$ is $W = F 	imes d = 20 	imes 1 = 20 \ J$.According to the work-energy theorem,the work done is equal to the change in rotational kinetic energy: $W = \Delta KE = \frac{1}{2} I \omega^2$.Substituting the values: $20 = \frac{1}{2} 	imes 0.1 	imes \omega^2$.$20 = 0.05 	imes \omega^2$.$\omega^2 = \frac{20}{0.05} = 400$.$\omega = \sqrt{400} = 20 \ rad/s$.

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