(A) The power $P$ delivered to the turbine is given by the rate of change of potential energy.$P = \frac{d}{dt}(mgh) = \left( \frac{dm}{dt} \right) gh

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(A) The power $P$ delivered to the turbine is given by the rate of change of potential energy.$P = \frac{d}{dt}(mgh) = \left( \frac{dm}{dt} \right) gh$Given:Rate of mass flow $\frac{dm}{dt} = 100 \ kg/s$Height $h = 100 \ m$Acceleration due to gravity $g \approx 10 \ m/s^2$Substituting the values:$P = 100 \ kg/s \times 10 \ m/s^2 \times 100 \ m$$P = 100,000 \ W$Since $1 \ kW = 1000 \ W$,we have:$P = \frac{100,000}{1000} \ kW = 100 \ kW$.Thus,the correct option is $A$.

Class 11 Physics Work, Energy, Power and Collision Power

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

A
$100$
B
$10$
C
$1$
D
$1000$

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

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