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(B) The formula for the rotational kinetic energy of a ring is $K = \frac{1}{2} I \omega^{2}$.Since the moment of inertia of a ring is $I = M R^{2}$,w

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$6$

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(B) The formula for the rotational kinetic energy of a ring is $K = \frac{1}{2} I \omega^{2}$.Since the moment of inertia of a ring is $I = M R^{2}$,we substitute this into the kinetic energy formula:$K = \frac{1}{2} (M R^{2}) \omega^{2} = \frac{1}{2} M R^{2} \omega^{2}$.The relative error formula is given by:$\frac{\Delta K}{K} = \frac{\Delta M}{M} + 2 \frac{\Delta R}{R} + 2 \frac{\Delta \omega}{\omega}$.To find the maximum percentage error,we multiply by $100$:$\frac{\Delta K}{K} 	imes 100 = \left( \frac{\Delta M}{M} 	imes 100 ight) + 2 \left( \frac{\Delta R}{R} 	imes 100 ight) + 2 \left( \frac{\Delta \omega}{\omega} 	imes 100 ight)$.Given values are $\frac{\Delta M}{M} 	imes 100 = 2 \%$,$\frac{\Delta R}{R} 	imes 100 = 1 \%$,and $\frac{\Delta \omega}{\omega} 	imes 100 = 1 \%$.Substituting these values:$\frac{\Delta K}{K} 	imes 100 = 2 \% + 2(1 \%) + 2(1 \%) = 2 \% + 2 \% + 2 \% = 6 \%$.

The maximum percentage errors in the measurement of mass $(M)$,radius $(R)$,and angular velocity $(\omega)$ of a ring are $2 \%$,$1 \%$,and $1 \%$ respectively. Find the maximum percentage error in the measurement of its rotational kinetic energy $(K = \frac{1}{2} I \omega^{2})$. (in $\%$)

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