Question

A tank contains 100 litres of fresh water. A solution containing 1 gm/litre of salt runs into the tank at the rate of 1 lit/min. The homogenised mixture is pumped out of the tank at the rate of 3 lit/min. If T be the time when the amount of salt in the tank is maximum.

Answer 1

250/9 minutes

Answer 2

The problem describes a mixing process where salt solution enters a tank and the mixture is pumped out. We need to find the time when the amount of salt in the tank is maximum.

1. Define Variables and Rates: Let Q(t) be the amount of salt (in grams) in the tank at time t (in minutes). Let V(t) be the volume of the mixture (in litres) in the tank at time t.

• Initial Conditions:

  • At t = 0, V(0) = 100 litres (fresh water, so Q(0) = 0 gm).

• Inflow:

  • Concentration of incoming salt solution = 1 gm/litre.
  • Rate of inflow = 1 lit/min.
  • Rate of salt entering = (1 gm/litre) * (1 lit/min) = 1 gm/min.

• Outflow:

  • Rate of outflow = 3 lit/min.
  • Concentration of salt in the tank at time t = Q(t) / V(t) gm/litre.
  • Rate of salt leaving = (Q(t) / V(t)) * (3 lit/min) = 3Q(t) / V(t) gm/min.

2. Determine Volume V(t): The net rate of change of volume in the tank is (Inflow Rate - Outflow Rate): dV/dt = 1 - 3 = -2 lit/min. Since V(0) = 100, integrating dV/dt gives: V(t) = 100 - 2t

The process continues as long as V(t) > 0, so 100 - 2t > 0, which means t < 50 minutes.

3. Set up the Differential Equation for Q(t): The rate of change of salt in the tank is: dQ/dt = (Rate of salt entering) - (Rate of salt leaving) dQ/dt = 1 - (3Q / V(t)) Substitute V(t) = 100 - 2t: \frac{dQ}{dt} = 1 - \frac{3Q}{100 - 2t}

Rearrange into the standard form of a first-order linear differential equation, \frac{dQ}{dt} + P(t)Q = R(t): \frac{dQ}{dt} + \left(\frac{3}{100 - 2t}\right)Q = 1

4. Solve the Differential Equation:

• Integrating Factor (IF): IF = e^{\int P(t) dt} P(t) = \frac{3}{100 - 2t} \int P(t) dt = \int \frac{3}{100 - 2t} dt Let u = 100 - 2t, then du = -2 dt, so dt = -du/2. \int \frac{3}{u} \left(-\frac{du}{2}\right) = -\frac{3}{2} \int \frac{1}{u} du = -\frac{3}{2} \ln|100 - 2t| IF = e^{-\frac{3}{2} \ln(100 - 2t)} = e^{\ln((100 - 2t)^{-3/2})} = (100 - 2t)^{-3/2}

• General Solution: Q(t) \cdot IF = \int R(t) \cdot IF dt + C Q(t) (100 - 2t)^{-3/2} = \int 1 \cdot (100 - 2t)^{-3/2} dt + C To integrate \int (100 - 2t)^{-3/2} dt, again let u = 100 - 2t, du = -2 dt: \int u^{-3/2} \left(-\frac{du}{2}\right) = -\frac{1}{2} \left(\frac{u^{-1/2}}{-1/2}\right) = u^{-1/2} = \frac{1}{\sqrt{100 - 2t}} So, Q(t) (100 - 2t)^{-3/2} = \frac{1}{\sqrt{100 - 2t}} + C Multiply by (100 - 2t)^{3/2}: Q(t) = (100 - 2t) + C(100 - 2t)^{3/2}

• Apply Initial Condition: At t = 0, Q(0) = 0: 0 = (100 - 2 \cdot 0) + C(100 - 2 \cdot 0)^{3/2} 0 = 100 + C(100)^{3/2} 0 = 100 + C(10^2)^{3/2} 0 = 100 + C(10^3) 0 = 100 + 1000C 1000C = -100 \implies C = -\frac{1}{10}

• Specific Solution for Q(t): Q(t) = (100 - 2t) - \frac{1}{10}(100 - 2t)^{3/2}

5. Find Time T for Maximum Salt: To find the maximum amount of salt, set dQ/dt = 0: From the differential equation: \frac{dQ}{dt} = 1 - \frac{3Q}{100 - 2t} Setting dQ/dt = 0: 1 - \frac{3Q}{100 - 2t} = 0 1 = \frac{3Q}{100 - 2t} 3Q = 100 - 2t Q = \frac{100 - 2t}{3}

Now substitute this expression for Q back into the specific solution for Q(t): \frac{100 - 2t}{3} = (100 - 2t) - \frac{1}{10}(100 - 2t)^{3/2}

Let X = 100 - 2t. Since V(t) > 0, X > 0. \frac{X}{3} = X - \frac{1}{10}X^{3/2} Multiply by 30 to clear denominators: 10X = 30X - 3X^{3/2} 3X^{3/2} = 20X Since X > 0, we can divide by X: 3X^{1/2} = 20 3\sqrt{X} = 20 \sqrt{X} = \frac{20}{3} Square both sides: X = \left(\frac{20}{3}\right)^2 = \frac{400}{9}

Now substitute back X = 100 - 2t: 100 - 2t = \frac{400}{9} 2t = 100 - \frac{400}{9} 2t = \frac{900 - 400}{9} 2t = \frac{500}{9} t = \frac{250}{9} minutes.

This value of t (250/9 \approx 27.78 minutes) is less than 50 minutes, so it's a valid time within the tank's operational period. To confirm it's a maximum, we can check the second derivative d^2Q/dt^2. \frac{dQ}{dt} = 1 - \frac{3Q}{V} \frac{d^2Q}{dt^2} = -\frac{3}{V^2} \left(V \frac{dQ}{dt} - Q \frac{dV}{dt}\right) At dQ/dt = 0, this simplifies to: \frac{d^2Q}{dt^2} = -\frac{3}{V^2} \left(-Q \frac{dV}{dt}\right) = \frac{3Q}{V^2} \frac{dV}{dt} Since dV/dt = -2, \frac{d^2Q}{dt^2} = \frac{3Q}{V^2} (-2) = -\frac{6Q}{V^2}. As Q > 0 and V > 0, d^2Q/dt^2 is negative, confirming that t = 250/9 corresponds to a maximum amount of salt.

The time T when the amount of salt in the tank is maximum is 250/9 minutes.

The final answer is $\boxed{\text{250/9 minutes}}$