A cube of wood supporting 200gm mass just floats in water. When the mass is removed, the cube ruses by 2cm. What is the size of the cube?

Explanation

If, l=side of cube, h=height of cube above water and `rho`=density of wood.
Mass of the cube `P_(rho)`
Volume of cube in water=`I^(2)(l-h)`
Volume of the displaced water`=l^(2)(l-h)`
As the tube is floating
weight of cube + weight of wood=weight of liquid displaced
or `" " I^(3)+200=l^(2)(l-h)" "`....(2,10)
After the removal of 200 gm mass, the cube rises 2 cm
`=l^(2)xx{l-(h+2)}`
Volume of cube in water
or `" "l_(2)xx{l-(h-2)}=P rho" "`....(2.11)
Substituting the value of `l^(3)rho`from equation (2.11) in equation (2.10), we get
`Fxx{l-(h+2)}+200=l^(3)(l-h)`
or `" "l^(3)-l^(2)h--2F+200=l^(3)+l^(2)`
`2l^(2)=200 rArr l=10` cm
placed by stones `=w//rho cm^(3)`
As`rho gt1`, hence `w//rho lt w`, thus we have
Now `" "(W+w//rho) lt (W+w)`
This shows that the volume of water displaced in the second case is less than the volume of water displaced in the first case. Hence the level of water will come down.