A tank leaks $0.1 \frac{\mathrm{m}^{3}}{\mathrm{hr}}$ of oil into a lake. The oil forms a semicircular disk with a thickness of $10^{-6}$ meters. How rapidly is the radius of the disk increasing when the radius is 150 meters?
The radius of the disk is increasing by $\mathrm{m} / \mathrm{hr}$ (round your answer to 3 decimal places)

Step 1 ：Given that the volume of oil leaking into the lake is \(0.1 m^3/hr\) and the thickness of the oil disk is \(10^{-6} m\), we can differentiate the volume equation with respect to time to find the rate at which the radius is increasing.

Step 2 ：The volume of a semicircular disk is given by the formula: \(V = \frac{1}{2} \pi r^2 h\)

Step 3 ：Differentiating the volume equation with respect to time, we get: \(\frac{dV}{dt} = \frac{1}{2} \pi h (2r \frac{dr}{dt})\)

Step 4 ：Substituting the given values into the equation, we get: \(0.1 = \frac{1}{2} \pi 10^{-6} (2 \times 150 \times \frac{dr}{dt})\)

Step 5 ：Solving for \(\frac{dr}{dt}\), we get: \(\frac{dr}{dt} = \frac{0.1}{\pi \times 10^{-6} \times 300}\)

Step 6 ：Calculating the above expression, we get: \(\frac{dr}{dt} = \frac{0.1}{3.14159 \times 0.0003}\)

Step 7 ：Finally, we find that the radius of the disk is increasing at a rate of \(\boxed{106.103}\) m/hr.