From a water fall, water is falling at the rate of 100 kg/s on the blades of turbine. If the height of the fall is 100m then the power delivered to the turbine is approximately equal to

The correct option is 100 kW.

To find the power delivered to the turbine, we can calculate the rate at which gravitational potential energy is transferred to the turbine blades by the falling water.

First, let us identify the given values from the problem statement:
- Rate of flow of water, which is the mass of water falling per unit time: $\frac{dm}{dt}=100\text{kg/s}$
- Height of the waterfall: $h=100\text{m}$
- Acceleration due to gravity ($g$) is approximately $10{\text{m/s}}^2$

The potential energy ($E$) of a mass of water $m$ at a height $h$ is given by:
$E=mgh$

Power ($P$) is the rate at which energy is delivered or work is done over time ($t$). Thus, we differentiate energy with respect to time:
$P=\frac{dE}{dt}=\frac{d}{dt}\left(mgh\right)$

Since the height $h$ and gravity $g$ remain constant during the fall, we can rewrite the equation as:
$P=gh\left(\frac{dm}{dt}\right)$

Now, substituting the given values:
$P=10\text{m/s}{2}^{}\times 100\text{m}\times 100\text{kg/s}$
$P=\mathrm{100,000}\text{W}$

To express this power in kilowatts (kW), we divide the value in Watts by 1,000 (since 1 kW = 1,000 W):
$P=\frac{\mathrm{100,000}}{\mathrm{1,000}}\text{kW}=100\text{kW}$

Therefore, the power delivered to the turbine is approximately equal to 100 kW.