Question

If the rate of electricity is rupee one per unit, then, 20 h.p. electric motor, having approx. 100% efficiency will require how much cost (Rs.) on 100 hours running to irrigate the field ?
A 500
B 750
C 1500
D 2000

Answer 1

Hint: The total power-hour consumed is obtained by multiplying power of the machine with the number of hours. The cost for 1kWh power consumption is 1 Rupee. Hence the cost of required power-hour can be calculated. 1h.p. =0.746kW.

Complete step by step answer:
We know that 1 h.p. =0.746kW
Given,
power of the electric motor = 20 h.p.
   =20×0.746 kW
   =15kW
Now we need to calculate power-hour. The total power-hour consumed is obtained by multiplying power of the machine with the number of hours.
Given, number of hours =100
Therefore,
power-hour= power× number of hours
   =15×100
   =1500kWh
Now, the price for 1kWh is 1 Rupee.
So, cost of 1500kWh= 1×1500
   =1500 Rupees

So, the correct answer is “Option C”.

Additional Information:
Here the efficiency of the motor is said to be 100% which means it is considered as an ideal machine i.e. there is no dissipation of energy in any manner. For a practical machine efficiency is less than 100% due to friction and the parts are not weightless, elastic or perfectly smooth.

Note:
The rate of electricity is given in units of kWh power-hour. Sometimes instead of power consumed by the machine the total input power is given, then power consumed is the product of efficiency and power input.
Efficiency=Power consumed/ Power input.