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Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. Then the moment of inertia of the system about an axis along one of the sides of the square is

\[\begin {array} {1 1} (a)\;\frac{4}{5}Ma^2+2 Mb^2 & \quad (b)\;\frac{8}{5}Ma^2+2 Mb^2 \\ (c)\;\frac{8}{5}Ma^2+Mb^2 & \quad  (d)\;\frac{4}{5}Ma^2+4Mb^2 \end {array}\]

1 Answer

M.I of 1,4 spheres about $AB= 2 \bigg[\large\frac{2}{5} $$Ma^2\bigg]$
M.2 of 2,5 spheres about $BC= 2 \bigg[\large\frac{2}{5} $$Ma^2\bigg]$
2,3 spheres about $AB= \large\frac{4}{5} $$Ma^2+2 Mb^2$
$\therefore I= \large\frac{8}{5} $$Ma^2+2 Mb^2$
answered Nov 27, 2013 by meena.p

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