Question

A cubical block of ice of mass m and edge length l is placed in a large tray of mass m if the ice melts how far does the centre of mass of the system ice plus tray comes down

Answer 1

Initially let the center of mass of the plate is at ZERO position

now the center of mass of the ice cube will be at the center of the cube

So its center of mass is at height "l/2"

now the center of mass of the system is given at the position

$y_{cm} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}$

$y_{cm} = \frac{m*0 + m*(l/2)}{m + m}$

$y_{cm} = \frac{l}{4}$

now when ice will melt then the center of mass of plate as well as the center of mass of water both are at ZERO position

now the center of mass of the system is also at ZERO

so the displacement of center of mass of the system will drop down by total distance l/4 downwards

Answer 2

hope it will help you

plz mark as brainest ☝️