Question

A cube of ice of edge 4 cm is placed in an empty cylindrical glass of inner diameter 6 cm. Assume that the ice melts uniformly from each side so that it always retains its cubical shape. Remembering that ice is lighter than water, find the length of the edge of the ice cube at the instant it just leaves contact with the bottom of the glass.

Answer 1

Let the length of the edge of the ice block when it just leaves contact with the bottom of the glass be x and the height of water after melting be h.
Given:
Inner diameter of the cylindrical glass = 6 cm
∴ Inner radius, r = 3 cm
Edge of the ice cube = 4 cm
$$\begin{gathered} \mathrm{Weight}=\mathrm{Upward}\mathrm{thurst} \\ \Rightarrow (x{)}^3\times {\rho }_{\mathrm{ice}}\times g=(x{)}^2\times h\times {\mathrm{\rho }}_{\mathrm{w}}\times g \\ \Rightarrow h=(0.9)x \end{gathered}$$
Again, volume of the water left from the melting of the ice is given by
$$\begin{gathered} (4{)}^3-(x{)}^3=\mathrm{\pi }\times (r{)}^2\times h-x^{2}h[\mathrm{Amount}\mathrm{of}\mathrm{water}=\mathrm{\rho }(r_2-x_2)h] \\ \Rightarrow (4{)}^3-(\mathrm{x}{)}^3=\mathrm{\pi }\times (3{)}^2\times h-x^{2}h \\ \mathrm{Putting}h=0.9x,\mathrm{we}\mathrm{get}: \\ (4{)}^3-(\mathrm{x}{)}^3=\mathrm{\pi }\times (3{)}^2\times (0.9)\mathrm{x} \end{gathered}$$
On solving the above equation, we get:
x = 2.26 cm