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(D) Let the radius of each disc be $R = \frac{a}{2}$.For the two discs lying on the axis $OO'$,the axis passes through their diameters. The moment of

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$3$

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(D) Let the radius of each disc be $R = \frac{a}{2}$.For the two discs lying on the axis $OO'$,the axis passes through their diameters. The moment of inertia of a disc about its diameter is $I_{diam} = \frac{1}{4} MR^2$.Thus,for the top and bottom discs,$I_1 = I_3 = \frac{1}{4} MR^2$.For the two discs on the sides,the axis $OO'$ is tangent to them. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $I_{cm} = \frac{1}{2} MR^2$ and $d = R$.So,$I_2 = I_4 = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.The total moment of inertia of the system is $I = I_1 + I_2 + I_3 + I_4 = 2 	imes (\frac{1}{4} MR^2) + 2 	imes (\frac{3}{2} MR^2) = \frac{1}{2} MR^2 + 3 MR^2 = \frac{7}{2} MR^2$.Substituting $R = \frac{a}{2}$,we get $I = \frac{7}{2} M(\frac{a}{2})^2 = \frac{7}{2} 	imes \frac{Ma^2}{4} = \frac{7}{8} Ma^2$.Wait,re-evaluating the geometry: The axis $OO'$ passes through the centers of the top and bottom discs (diameter) and is tangent to the side discs. The calculation $I = \frac{7}{2} MR^2$ is correct. With $R = a/2$,$I = \frac{7}{8} Ma^2 = \frac{3.5}{4} Ma^2$. Given the options and the standard form $\frac{x}{4}Ma^2$,if the side discs are at distance $R$ from the axis,$x=3.5$. Checking the provided solution logic: $I_2 = I_4 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$. This assumes the side discs rotate about an axis parallel to their diameter at distance $R$. Total $I = 2(\frac{1}{4}MR^2) + 2(\frac{5}{4}MR^2) = 3MR^2 = 3M(a/2)^2 = \frac{3}{4}Ma^2$. Thus $x=3$.

Four identical discs each of mass $M$ and diameter $a$ are arranged in a plane as shown in the figure. If the moment of inertia of the system about $OO'$ is $\frac{x}{4} Ma^2$,then the value of $x$ will be:

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