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(A) The formula for the moment of inertia $(I)$ of a ring about its geometric axis is given by $I = M R^2$.To find the maximum percentage error,we use

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$4$

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(A) The formula for the moment of inertia $(I)$ of a ring about its geometric axis is given by $I = M R^2$.To find the maximum percentage error,we use the rule for propagation of errors in products and powers:$\frac{\Delta I}{I} 	imes 100 = \frac{\Delta M}{M} 	imes 100 + 2 \left( \frac{\Delta R}{R} 	imes 100 ight)$.Given:Percentage error in mass $\frac{\Delta M}{M} 	imes 100 = 2 \%$.Percentage error in radius $\frac{\Delta R}{R} 	imes 100 = 1 \%$.Substituting these values into the equation:$\frac{\Delta I}{I} 	imes 100 = 2 \% + 2(1 \%) = 2 \% + 2 \% = 4 \%$.Thus,the maximum percentage error in the measurement of the moment of inertia is $4 \%$.

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 moment of inertia $(I = M R^2)$ about its geometric axis. (in $\%$)

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