If there is an error of 1% in the calculation | vedclass

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(B) The moment of inertia of a disc about an axis tangent to it is given by the parallel axis theorem: $I = I_{cm} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \

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$4$

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(B) The moment of inertia of a disc about an axis tangent to it is given by the parallel axis theorem: $I = I_{cm} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.However,for a disc,the moment of inertia about a tangent in its plane is $I = \frac{5}{4}MR^2$.Using the relative error formula for $I = k M^a R^b$,we have $\frac{\Delta I}{I} = a \frac{\Delta M}{M} + b \frac{\Delta R}{R}$.Here,$a = 1$ and $b = 2$.Given $\frac{\Delta M}{M} = 1\%$ and $\frac{\Delta R}{R} = 1.5\%$.Therefore,$\frac{\Delta I}{I} 	imes 100 = (1 	imes 1\%) + (2 	imes 1.5\%) = 1\% + 3\% = 4\%$.

If there is an error of $1\%$ in the calculation of the mass of a disc and $1.5\%$ error in the radius,then the $\%$ error in the moment of inertia about an axis tangent to the disc is .......... $\%$

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