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(B) The angular momentum $(J)$ of a ring about its geometrical axis is given by $J = I \omega$. Since the moment of inertia of a ring is $I = M R^2$,w

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$5$

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(B) The angular momentum $(J)$ of a ring about its geometrical axis is given by $J = I \omega$. Since the moment of inertia of a ring is $I = M R^2$,we have $J = M R^2 \omega$. The relative error in $J$ is given by $\frac{\Delta J}{J} = \frac{\Delta M}{M} + 2 \frac{\Delta R}{R} + \frac{\Delta \omega}{\omega}$. To find the maximum percentage error,we multiply by $100$: $\frac{\Delta J}{J} 	imes 100 = \left( \frac{\Delta M}{M} 	imes 100 ight) + 2 \left( \frac{\Delta R}{R} 	imes 100 ight) + \left( \frac{\Delta \omega}{\omega} 	imes 100 ight)$. Given $\frac{\Delta M}{M} 	imes 100 = 2 \%$,$\frac{\Delta R}{R} 	imes 100 = 1 \%$,and $\frac{\Delta \omega}{\omega} 	imes 100 = 1 \%$. Substituting these values: $\frac{\Delta J}{J} 	imes 100 = 2 \% + 2(1 \%) + 1 \% = 2 \% + 2 \% + 1 \% = 5 \%$.

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 angular momentum $(J = I \omega)$ about its geometrical axis. (in $\%$)

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