A cube with side length $100cm$ is filled with water and has a hole through which the water drains into a cylinder of radius $100cm.$ If the water level in the cube is falling at a rate of $1 \frac{cm}{s} ,$ how fast is the water level in the cylinder rising? 
1. **Determine the volume change rate in the cube:**
   - The side length of the cube is \(100 \, \text{cm}\).
   - The volume \(V_{\text{cube}}\) of the cube is given by \(V_{\text{cube}} = s^3\), where \(s\) is the side length.
   - The rate at which the water level in the cube is falling is \(1 \, \text{cm/s}\).
   - The area of the base of the cube is \(A_{\text{cube}} = s^2 = 100^2 = 10000 \, \text{cm}^2\).
   - The volume change rate in the cube is given by the product of the base area and the rate of change of the height:
     \[
     \frac{dV_{\text{cube}}}{dt} = A_{\text{cube}} \cdot \frac{dh_{\text{cube}}}{dt} = 10000 \, \text{cm}^2 \cdot (-1 \, \text{cm/s}) = -10000 \, \text{cm}^3/\text{s}
     \]

2. **Determine the volume change rate in the cylinder:**
   - The radius of the cylinder is \(100 \, \text{cm}\).
   - The volume \(V_{\text{cylinder}}\) of the cylinder is given by \(V_{\text{cylinder}} = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height.
   - The area of the base of the cylinder is \(A_{\text{cylinder}} = \pi r^2 = \pi \cdot 100^2 = 10000\pi \, \text{cm}^2\).
   - The volume change rate in the cylinder is equal to the volume change rate in the cube (since the water is draining from the cube into the cylinder):
     \[
     \frac{dV_{\text{cylinder}}}{dt} = -\frac{dV_{\text{cube}}}{dt} = 10000 \, \text{cm}^3/\text{s}
     \]

3. **Determine the rate at which the water level in the cylinder is rising:**
   - Let \(h_{\text{cylinder}}\) be the height of the water in the cylinder.
   - The volume change rate in the cylinder can also be expressed as:
     \[
     \frac{dV_{\text{cylinder}}}{dt} = A_{\text{cylinder}} \cdot \frac{dh_{\text{cylinder}}}{dt}
     \]
   - Substituting the known values:
     \[
     10000 \, \text{cm}^3/\text{s} = 10000\pi \, \text{cm}^2 \cdot \frac{dh_{\text{cylinder}}}{dt}
     \]
   - Solving for \(\frac{dh_{\text{cylinder}}}{dt}\):
     \[
     \frac{dh_{\text{cylinder}}}{dt} = \frac{10000 \, \text{cm}^3/\text{s}}{10000\pi \, \text{cm}^2} = \frac{1}{\pi} \, \text{cm/s}
     \]

The final answer is \(\boxed{\frac{1}{\pi}}\).