Answer
Δ h = 0.06 m
Explanation:
Given,
mass of ice cube = 0.11 Kg
inside radius of cylinder = 3.70 cm
density of water = 1000 Kg/m³
density of glycerin = 1260 Kg/m³
ice cube is floating on glycerin surface,
net force on the cube is equal to zero.
Buoyant force by glycerin is equal to the weight of cube
B = W
[tex]\rho_g V g = m g[/tex]
[tex]\rho_g (\pi r^2 h_1)= m[/tex]
[tex] h_1= \dfrac{m}{\rho_g\pi r^2}[/tex]
[tex] h_1= \dfrac{0.11}{1260\pi\times 0.037^2}[/tex]
h₁ = 0.0203 m
when ice melts it is completely turned into water
Volume of mass m of water
[tex]V = \dfrac{m}{\rho_w}[/tex]
[tex]\pi r^2 h_2 = \dfrac{m}{\rho_w}[/tex]
[tex]h_2 = \dfrac{m}{\rho_w\pi r^2}[/tex]
[tex]h_2 = \dfrac{0.11}{1000\pi\times 0.037^2}[/tex]
h₂ = 0.0803 m
distance by which height of liquid change
Δ h = h₂ - h₁
Δ h = 0.0803 - 0.0203
Δ h = 0.06 m
