The rate at which a spherical pill dissolves is described by the rate of chage of its volume with time dv/dt . Given that the rate of change of its volume directly proportional to its surface area , compute the time in which 80 % of pill of size 4 cm would dissolves. Given that it completely dissolves in 2 hrs

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2015-03-24T02:02:22+05:30

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Let R be the radius of the ball at any time.

   dV/dt = - K (4π R²),    K = a positive constant of proportionality.  As the volume is  decreasing we put a minus sign.

  V = 4/3 * π R³    at any time t.
  dV/dt = 4π/3 * 3 R² * dR/dt

     =>  4 π R² * dR/dt = - K 4 π R²
     =>  dR/dt = - K
     =>  R = - K t + R₀,      R₀ is an integration constant and is the radius at t= 0.

Given,   R₀ = 4 cm
given,   R = 0 in 2 hours  =>  0 = - K * 2 + 4  =>  K = 2 cm/hour

   80% of pill dissolves => remaining volume = 20% of initial volume
         R³ / R₀³ = 20/100 = 1/5
       R³ = 4³/5 = 12.8 cm³
       R = 2.34 cm

  Now,  R = - K t + R₀
     =>  t = (R₀ - R)/K = (4 - 2.34)/2  = 0.83 hours 
             = 49.82 minutes
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