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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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
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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