Question

A 100 gram ice block (p=1000kg/m3) is placed into a container that contains 400mL of a liquid. The liquid in the container has a density of 1100kg/m3. After the ice block has been added, and before any ice had melted, the new level of the liquid within the container is?

Answer 1

given
mass of ice cube, m_Ice = 100 g

= 0.100 kg

rho_liquid = 1100 kg/m^3

let V is the volume of the liquid displaced by the ice cube.

in the equilibrium,
buoyant force on the ice cube = weight of the ice cube
rho_liquid*V*g = m_Ice*g

rho_liquid*V = m_Ice

V = m_Ice/rho_liquid

= 0.1/1100

= 9.09*10^-5 m^3

= 9.09*10^-2 L

= 90.9 mL

if h is the initial height

increase in height of the level = h*90.9/400

= 0.227*h

new level = h + 0.227*h

= 1.227*h <<<<<<<<-------------Answer

Note : please comment for any clarification.