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(B) Let the mass of the original square plate be $M$ and its area be $A_1 = a^2$. The center of mass is at the origin $(0,0)$.Let the mass of the remo

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$\frac{b^2}{4(a^2-b^2)} \sqrt{2}a$

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(B) Let the mass of the original square plate be $M$ and its area be $A_1 = a^2$. The center of mass is at the origin $(0,0)$.Let the mass of the removed square hole be $m$ and its area be $A_2 = b^2$. The center of mass of the hole is at $(a/4, a/4)$.Since the plate is uniform,the mass is proportional to the area. Thus,$M = \sigma a^2$ and $m = \sigma b^2$,where $\sigma$ is the surface mass density.The center of mass of the remaining part $(X_{cm}, Y_{cm})$ is given by:$X_{cm} = \frac{M(0) - m(a/4)}{M - m} = \frac{-\sigma b^2 (a/4)}{\sigma a^2 - \sigma b^2} = -\frac{b^2 a}{4(a^2 - b^2)}$Similarly,$Y_{cm} = -\frac{b^2 a}{4(a^2 - b^2)}$.The distance $R$ from the center $(0,0)$ is:$R = \sqrt{X_{cm}^2 + Y_{cm}^2} = \sqrt{2 \left( \frac{b^2 a}{4(a^2 - b^2)} \right)^2} = \frac{b^2 a}{4(a^2 - b^2)} \sqrt{2}$.

$A$ uniform square plate of side $a$ has a hole of side $b$ cut out of it as shown in the figure. The hole is centered at $(a/4, a/4)$ from the center of the square plate. The distance of the center of mass of the remaining part from the center of the square plate is:

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