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(A) The moment of inertia $I$ of a flywheel (disk) about an axis passing through its center is given by $I = \frac{1}{2}MR^2$,where $M$ is the mass an

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$2$

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(A) The moment of inertia $I$ of a flywheel (disk) about an axis passing through its center is given by $I = \frac{1}{2}MR^2$,where $M$ is the mass and $R$ is the radius.Since the diameter $D = 2R$,we have $R = D/2$. Substituting this into the formula,$I = \frac{1}{2}M(D/2)^2 = \frac{1}{8}MD^2$.This shows that $I \propto D^2$.Using the concept of percentage error,if $D$ increases by $1\%$,the percentage change in $I$ is given by $\frac{\Delta I}{I} 	imes 100 = 2 	imes (\frac{\Delta D}{D} 	imes 100)$.Substituting the given value,$\frac{\Delta I}{I} 	imes 100 = 2 	imes 1\% = 2\%$.Therefore,the percentage increase in the moment of inertia is $2\%$.

If the diameter of a flywheel is increased by $1\%$,then the percentage increase in its moment of inertia about an axis passing through its center will be ....... $\%$

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