A tank consists of 50 litres of fresh water. Two litres of brine each litre containing 5 gms of dissolved salt are run into tank per minute, the mixture is kept uniform by stirring, and runs out at the rate of one litre per minute. If ‘m’ grams of salt are present in the tank after t minute, express ‘m’ in terms of t and find the amount of salt present after 10 minutes

Answer

To solve the problem, we need to set up a differential equation that describes the amount of salt in the tank over time. Let's break it down step by step.

### Step 1: Define the Variables
Let \( m(t) \) be the amount of salt in grams in the tank at time \( t \) (in minutes). Initially, the tank contains 50 liters of fresh water, which means \( m(0) = 0 \) grams of salt.

### Step 2: Determine the Input Rate of Salt
Brine is added to the tank at a rate of 2 liters per minute, with each liter containing 5 grams of salt. Therefore, the input rate of salt is:
\[ \text{Input Rate} = 2 \text{ liters/min} \times 5 \text{ grams/liter} = 10 \text{ grams/min} \]

### Step 3: Determine the Output Rate of Salt
The mixture in the tank is kept uniform, and it flows out at a rate of 1 liter per minute. The concentration of salt in the tank at time \( t \) is given by:
\[ \text{Concentration} = \frac{m(t)}{50 + t} \]
Thus, the output rate of salt is:
\[ \text{Output Rate} = \text{Concentration} \times \text{Flow Rate} = \frac{m(t)}{50 + t} \times 1 = \frac{m(t)}{50 + t} \text{ grams/min} \]

### Step 4: Set Up the Differential Equation
The rate of change of salt in the tank can be expressed as:
\[ \frac{dm}{dt} = \text{Input Rate} - \text{Output Rate} \]
Substituting the input and output rates:
\[ \frac{dm}{dt} = 10 - \frac{m(t)}{50 + t} \]

### Step 5: Rearranging the Equation
Rearranging gives us:
\[ \frac{dm}{dt} + \frac{m(t)}{50 + t} = 10 \]

### Step 6: Solve the Differential Equation
This is a linear first-order differential equation. We can use an integrating factor to solve it. The integrating factor \( \mu(t) \) is given by:
\[ \mu(t) = e^{\int \frac{1}{50 + t} dt} = e^{\ln(50 + t)} = 50 + t \]

Multiplying through by the integrating factor:
\[ (50 + t) \frac{dm}{dt} + m(t) = 10(50 + t) \]

This simplifies to:
\[ \frac{d}{dt} \left( m(t)(50 + t) \right) = 500 + 10t \]

### Step 7: Integrate Both Sides
Integrating both sides with respect to \( t \):
\[ m(t)(50 + t) = \int (500 + 10t) dt = 500t + 5t^2 + C \]

### Step 8: Solve for \( m(t) \)
Thus, we have:
\[ m(t)(50 + t) = 500t + 5t^2 + C \]
To find \( C \), we use the initial condition \( m(0) = 0 \):
\[ 0(50 + 0) = 500(0) + 5(0)^2 + C \implies C = 0 \]
So,
\[ m(t)(50 + t) = 500t + 5t^2 \]
Now, solving for \( m(t) \):
\[ m(t) = \frac{500t + 5t^2}{50 + t} \]

### Step 9: Find \( m(10) \)
To find the amount of salt present after 10 minutes:
\[ m(10) = \frac{500(10) + 5(10)^2}{50 + 10} = \frac{5000 + 500}{60} = \frac{5500}{60} = \frac{275}{3} \text{ grams} \]

### Final Answer
The amount of salt present in the tank after 10 minutes is \( \frac{275}{3} \) grams, which is approximately \( 91.67 \) grams.

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