NEET-XII-Physics

14: Some Mechanical Properties of Matter

with Solutions - page 9
Qstn# iv-28 Prvs-QstnNext-Qstn
  • #28
    Consider an ice cube of edge 1.0 cm kept in a gravity-free hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.
    Ans : Given:
    Edge of the ice cube
    (a) = 1.0 cm
    The water that is formed due to the melting of ice acquires a spherical surface.
    In the absence of gravity, let the radius of the spherical surface be r.
    Volume of ice cube = volume of spherical surface of water
    `` \Rightarrow {a}^{3}=\frac{4}{3}\,\mathrm{\,\pi \,}{r}^{3}``
    `` \Rightarrow r={\left[\frac{3{\,\mathrm{\,a\,}}^{3}}{4\,\mathrm{\,\pi \,}}\right]}^{1/3}``
    Surface area of spherical water surface = 4Ï€r2
    `` =4\,\mathrm{\,\pi \,}{\left[\frac{3{a}^{3}}{4\,\mathrm{\,\pi \,}}\right]}^{2/3}``
    `` ={\left(36\,\mathrm{\,\pi \,}\right)}^{1/3}{\,\mathrm{\,cm\,}}^{2}``
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