Answer:
To solve this problem, we'll use the concept of exponential decay, where the rate of water draining is proportional to the amount of water present.
Let's denote the amount of water in the tank at time t as V(t) (in liters). We're given that the tank initially contains 100 liters of water, so V(0) = 100.
The rate of water draining is proportional to the amount of water present. Mathematically, we can express this as:
dV/dt = -kV
where dV/dt is the rate of change of water volume with respect to time, and k is the proportionality constant.
To solve this differential equation, we can separate variables and integrate:
∫(1/V) dV = -∫k dt
ln|V| = -kt + C
where C is the constant of integration.
Now, let's consider the given information: 30% of the water leaks out in the first 5 minutes. This means that after 5 minutes, the amount of water remaining is 70% of the initial volume, i.e., V(5) = 0.7 * 100 = 70 liters.
Substituting these values into the equation, we get:
ln|70| = -5k + C
Now, we need to find the value of the constant C. To do this, let's consider the condition at t = 0:
ln|100| = 0 + C
C = ln|100|
Substituting this value back into the equation, we have:
ln|70| = -5k + ln|100|
Simplifying further:
ln|70| - ln|100| = -5k
ln|70/100| = -5k
ln(0.7) = -5k
Now, we can solve for k:
k = -ln(0.7) / 5
Now that we have the value of k, we can find the amount of water left in the tank at t = 10 minutes. Substituting t = 10 into the equation, we have:
ln|V(10)| = -k * 10 + ln|100|
ln|V(10)| = -10 * (-ln(0.7) / 5) + ln|100|
ln|V(10)| = 2 * ln(0.7) + ln|100|
ln|V(10)| = ln(0.7^2) + ln|100|
ln|V(10)| = ln(0.49) + ln|100|
ln|V(10)| = ln(0.49 * 100)
ln|V(10)| = ln(49)
Now, we can solve for V(10):
V(10) = e^ln(49)
V(10) = 49 liters
Therefore, 10 minutes after the leak develops, there will be 49 liters of water left in the tank.