Half of the uniform rectangular plate of Length ‘L’ is made up of material of density $' d_{1}'$ and the other half with density $d_{2}$ . The perpendicular distance of center of mass from AB is

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$\frac{3 d_{1}}{d_{1} + d_{2}} \times \frac{L}{4}$

$\frac{2 d_{1} + 3 d_{2}}{d_{1} + d_{2}} \times \frac{L}{4}$

$\frac{d_{1} + 3 d_{2}}{d_{1} + d_{2}} \times \frac{L}{4}$

$\frac{3 d_{2}}{d_{1} + d_{2}} \times \frac{L}{4}$

answer is B.

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Detailed Solution

$m_{1} = \frac{L}{2} \times d_{1}$

$X_{1} = \frac{L}{4}$

$m_{2} = \frac{L}{2} x d_{2}$

$x_{2} = \frac{3 L}{4}$

$x_{c o m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$ $= \frac{\frac{L}{2} \times d_{1} \times \frac{L}{4} + \frac{L}{2} \times d_{2} \times \frac{3 L}{4}}{\frac{L}{2} d_{1} + \frac{L}{2} d_{2}}$

= $\frac{\frac{d_{1} L}{4} + \frac{3 d_{2} L}{4}}{\frac{d_{1} L}{2} + \frac{d_{2} L}{2}}$

$= \frac{d_{1} + 3 d_{2}}{d_{1} + d_{2}} \times \frac{L}{4}$

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