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?

Options

A. `(mL)/((m + M))`
B. `(mL)/(2(m + M))`
C. `(mL)/(M)`
D. `(ML)/(m)`

Answer

B

Explanation

`(y_(c.m.))_(1) = (m (L)/(2) + M xx 0)/(m + M) = (m L)/(2( m + M))`
When ice becomes water , as tray is large , the height of water is negligible
`(y_(c.m.))_(2) = 0`
Shift `= (y_(c.m.))_(1) - (y_(c.m.))_(2) = (mL)/(2(m + M))`