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Half of the uniform rectangular plate of length 'L' is made of material of density $d_1$and the other half of density $d_2$. The horizontal distance of center of mass from AB is


\[\begin {array} {1 1} (a)\;\frac{2d_1+3d_2}{d_1+d_2} \times \frac{L}{4} & \quad (b)\;\frac{d_1+3d_2}{d_1+d_2} \times \frac{L}{4} \\ (c)\;\frac{3d_1}{d_1+d_2} \times \frac{L}{4} & \quad (d)\;\frac{3d_2}{d_1+d_2} \times \frac{L}{4} \end {array}\]

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$ (C)\;\frac{3d_1}{d_1+d_2} \times \frac{L}{4}$
Hence C is the correct answer.
answered Mar 24, 2014 by meena.p

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