To calculate the rpm of the Pelton wheel for maximum power output, we need to consider the relationship between the velocity of water, the diameter of the Pelton wheel, and the rpm.
Given:
\[
D = 1 \ {m}, \quad V_1 = 125.66 \ {m/s}
\]
For maximum power output in a Pelton wheel, the bucket speed \( u \) is:
\[
u = \frac{V_1}{2} = \frac{125.66}{2} = 62.83 \ {m/s}
\]
The linear speed \( u \) is related to rotational speed \( N \) by:
\[
u = \frac{\pi D N}{60}
\Rightarrow N = \frac{60u}{\pi D}
\]
Substituting the known values:
\[
N = \frac{60 \cdot 62.83}{\pi \cdot 1} = \frac{3769.8}{\pi} \approx 1200.0 \ {rpm}
\]
% Correct Answer
Correct Answer:} \fbox{1200.0 rpm}