Step 1: Setup
Volume: \(\pi r^2h = 125\pi\), so \(h = \frac{125}{r^2}\). The tank has no lid, so surface area \(S = \pi r^2 + 2\pi rh\).
Step 2: Single variable
\(S = \pi r^2 + \frac{250\pi}{r}\).
Step 3: Minimise
\(S' = 2\pi r - \frac{250\pi}{r^2} = 0\) gives \(r^3 = 125\), so \(r = 5\). \(S'' = 2\pi + \frac{500\pi}{r^3} > 0\), so this is a minimum.
Step 4: Value
\(S = 25\pi + 50\pi = 75\pi\). Option (A).
Final Answer:
The minimum surface area is 75 pi square cm.
\[ \boxed{\text{(A)}\ 75\pi} \]