Correct Answer - Option 3 : 10 kW

__Concept:__

We know COP of a refrigerator is given by:

__\((COP)_{actual}=\frac{Refrigeration\ Effect}{Work\ Input}\)__

For reversible refrigerator, COP can be calculated by:

\((COP)_{ideal} = \frac{T_L}{T_H-T_L}\)

where TH = Absolute higher temperature, TL = Absolute lower temperature

__Calculation:__

__Given:__

TH = 24°C = 297 K, T = -9°C = 264 K, Refrigeration Effect = 20 kW, COPactual = \(\frac{1}{4}\) × COPideal

\(COP_{ideal} = \frac{264}{297-264}\)

\(COP_{ideal} = \frac{264}{33}\)

COPideal = 8

∴ COPactual = \(\frac{1}{4}\) × 8 = 2

Now, \((COP)_{actual}=\frac{Refrigeration\ Effect}{Work\ Input}\)

\(2=\frac{20}{Work\ Input}\)

Work Input = 10 kW

Hence t**he power required to drive the plant is** 10 kW.