A cubical block of ice of maas m and edge L is placed in a large tray of maas M. If the ie melts, how far does the centre of maas of the system ''ice plus tray'' come down?

Explanation

In Fig. an ice block of mass `m` and edge `L` is placed in a large tray of mass `M.`
Point `O` is chosen as the origin of the system of ice block and the tray.
Let `x_(1) =` height of `CM` of tray above `O`,
`x_(2) =` height of `CM` of ice block above `O`.
Therefore, height of centre of mass of ice-tray system is
`x = (mx_(2) + Mx_(1))/(m + M)` ..(i)
When ice melts, mass `m` of water spreads on the surface of tray. As the tray is large, height of water spread is negligible. Therefore, centre of mass of water ia at a height `(x_(2) - L//2)` above `O`. Therefore, now height of centre of mass of the system is
`x' = (m(x_(2) - L//2) + M x_(1))/(m + M)`
The shift in the position of centre of mass of the system would be
`= x - x' = (mx_(2) + Mx_(1))/(m+M) - (m(x_(2) - L//2) + Mx_(1))/(m+ M)`
`x - x' = (mL)/(2(m + M))`