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From a sphere of radius 1m, a sphere of radius 0.5 m is removed from the edge. The shift in the CM is

\[\begin {array} {1 1} (a)\;\frac{13}{16}\;m & \quad (b)\;\frac{16}{13}\;m \\ (c)\;\frac{14}{13}\;m & \quad  (d)\;\frac{1}{14}\;m \end {array}\]

Can you answer this question?
 
 

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$M_1=density \times Area$
$A=Area$
$A_1=A_2=A$
$Xcm$ from O:
$Xcm=\large\frac{d_1 A_1 \times L/A+ d_2 A_2 \times 3L/4}{(d_1+d_2)A}$
$Xcm=\large\frac{d_1 L + 3d_2 L}{4(d_1+d_2)A}$
answered Nov 21, 2013 by meena.p
 

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