A circular plate of diameter d is placed in contact with a square...

Let $\sigma$ be the mass per unit area of the plates.

Mass of the circular plate, $M_1 = \sigma \times \pi \left(\dfrac{d}{2}\right)^2 = \dfrac{\sigma \pi d^2}{4}$

Mass of the square plate, $M_2 = \sigma \times d^2$

Let the point of contact of the two plates be the origin $(0,0)$.

The centre of mass of the circular plate is at $x_1 = -\dfrac{d}{2}$.

The centre of mass of the square plate is at $x_2 = \dfrac{d}{2}$.

The $x$-coordinate of the centre of mass of the composite system is given by:

$X_{cm} = \dfrac{M_1 x_1 + M_2 x_2}{M_1 + M_2}$

Substituting the values, we get:

$X_{cm} = \dfrac{\left(\dfrac{\sigma \pi d^2}{4}\right) \left(-\dfrac{d}{2}\right) + (\sigma d^2) \left(\dfrac{d}{2}\right)}{\dfrac{\sigma \pi d^2}{4} + \sigma d^2}$

X_{cm} = \dfrac{\dfrac{d}{2} \left(1 - \dfrac{\pi}{4}\right)}{1 + \dfrac{\pi}{4}} =\dfrac{d}{2} \left(\dfrac{4 - \pi}{4 + \pi}\right)

Since $\pi \approx 3.14$, we have $4 - \pi > 0$, which implies $X_{cm} > 0$.

Because the point of contact is at $x = 0 and the square plate lies in the region$x > 0$, the positive value of$X_{cm}$indicates that the centre of mass of the system lies within the square plate.$