To solve this problem, we need to set up an equation for the depth of water in the tank as a function of time, taking into account both the water flow in and the leakage out.

Given:

• The base dimensions of the tank are 10 cm by 10 cm.
• Water flows in at a rate of $25 \, \text{cm}^3$ per second.
• Water leaks out at a rate of $4h^2 \, \text{cm}^3$ per second, where $h$ is the depth of water in centimeters.
• We need to find the time taken for the water to reach a depth of 2 cm.

Step 1: Volume Flow Equations

• The volume of water in the tank at any time $t$ is given by the product of the base area and the depth $h(t)$: $V(t) = \text{base area} \times h(t) = 10 \times 10 \times h(t) = 100h(t)$

• The rate at which the volume changes, $\frac{dV}{dt}$, is the difference between the rate of water flowing in and the rate of water leaking out. This can be expressed as: $\frac{dV}{dt} = 25 - 4h^2$

Step 2: Relating Volume and Depth

We know that $V(t) = 100h(t)$, so $\frac{dV}{dt} = 100 \frac{dh}{dt}$. Substituting this into the volume flow equation, we get: $100 \frac{dh}{dt} = 25 - 4h^2$ Simplifying: $\frac{dh}{dt} = \frac{25 - 4h^2}{100}$

Step 3: Solving the Differential Equation

We need to solve the differential equation: $\frac{dh}{dt} = \frac{25 - 4h^2}{100}$ Separate the variables: $\frac{100}{25 - 4h^2} \, dh = dt$ Now integrate both sides. The left side is a bit tricky, so we use partial fractions or an appropriate substitution. Let's first simplify the integral: $\int \frac{100}{25 - 4h^2} \, dh = \int dt$

We can factor the denominator on the left side: $25 - 4h^2 = (5 - 2h)(5 + 2h)$ Using partial fraction decomposition, we can express $\frac{100}{25 - 4h^2}$ as a sum of simpler fractions. After integrating, we’ll have the solution for $h(t)$.

Step 4: Find the Time for Depth to Reach 2 cm

Finally, after solving for $h(t)$, we substitute $h = 2$ and find the corresponding time $t$.

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