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NEET PHYSICSEasy

Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? (g=10 m/s²)

A

12.3 kW

B

7.0 kW

C

10.2 kW

D

8.1 kW

Step-by-Step Solution

  1. Calculate Input Power (PinP_{in}): The input power is the rate at which gravitational potential energy is supplied by the falling water. The rate of mass flow is dmdt=15 kg/s\frac{dm}{dt} = 15\text{ kg/s}. Pin=d(mgh)dt=(dmdt)ghP_{in} = \frac{d(mgh)}{dt} = \left(\frac{dm}{dt}\right)gh Pin=(15 kg/s)(10 m/s2)(60 m)=9000 J/s=9000 W=9 kWP_{in} = (15\text{ kg/s})(10\text{ m/s}^2)(60\text{ m}) = 9000\text{ J/s} = 9000\text{ W} = 9\text{ kW}
  2. Calculate Output Power (PoutP_{out}): The problem states that 10% of the energy is lost to friction. Therefore, the turbine utilizes 90% of the input energy. Efficiency (η)=100%−10%=90%=0.9\text{Efficiency } (\eta) = 100\% - 10\% = 90\% = 0.9 Pout=η×Pin=0.9×9 kW=8.1 kWP_{out} = \eta \times P_{in} = 0.9 \times 9\text{ kW} = 8.1\text{ kW} [Class 11 Physics, Ch 6, Sec 6.10].
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