Consider a Pelton wheel of 1 m diameter. The magnitude of relative velocity of water at the bucket inlet is same as the magnitude of relative velocity of water at the bucket exit. The absolute speed of water at the bucket inlet is 125.66 m s^-1. For maximum power output from the Pelton wheel, the rpm of the Pelton wheel is __________________ (rounded off to 1 decimal place).

The correct answer is 1200 (or 1200 rpm).

To find the rotational speed of the Pelton wheel for maximum power output, we analyze the relationship between the wheel's bucket speed and the absolute inlet velocity of the water jet.

Let:
- $D$ be the diameter of the Pelton wheel = 1 m
- $V_1$ be the absolute velocity of water at the inlet = 125.66 m s^-1
- $U$ be the peripheral velocity (bucket speed) of the wheel
- $N$ be the rotational speed of the wheel in rpm

For maximum power output (or maximum efficiency) of a Pelton wheel, the theoretical bucket speed $U$ should be half of the absolute inlet jet velocity $V_1$:

$U=\frac{V_1}{2}$

Substituting the given value of $V_1$:

$U=\frac{125.66}{2}=62.83\text{m s}{}^{-1}$

The relation between the bucket speed $U$, wheel diameter $D$, and rotational speed $N$ (in rpm) is given by:

$U=\frac{\pi DN}{60}$

Rearranging the equation to solve for $N$:

$N=\frac{60U}{\pi D}$

Substituting the values of $U$ and $D$ into the equation:

$N=\frac{60\times 62.83}{\pi \times 1}$

Using the value of $\pi \approx 3.14159$:

$N=\frac{3769.8}{3.14159}\approx 1200\text{rpm}$

Thus, the rotational speed of the Pelton wheel for maximum power output is 1200 rpm.