The efficiency of a Carnot engine is mentioned to be as eighty percent. When the direction of the cycle be reversed, what will be the value of C.O.P. of reversed Carnot Cycle?
$\begin{align}
  & A.0.25 \\
 & B.0.5 \\
 & C.0.8 \\
 & D.1.25 \\
\end{align}$

Hint: The C.O.P of this condition will be equivalent to the ratio of the difference of one and the efficiency and the Carnot cycle to the efficiency of the Carnot cycle. Substitute the value of efficiency in this equation. This will help you in answering this question.

Complete answer:
It has been mentioned in the question that the efficiency of the Carnot cycle be,
$\eta =0.8$
If the direction of the cycle has been reversed, then the relation between the C.O.P and the efficiency can be found. The C.O.P of this condition will be equivalent to the ratio of the difference of one and the efficiency and the Carnot cycle to the efficiency of the Carnot cycle. This can be expressed in the form of an equation which can be written as,
$COP=\dfrac{1-\eta }{\eta }$
Substituting the values in this equation can be written as,
$COP=\dfrac{1-0.8}{0.8}=0.25$
Therefore the COP of the Carnot cycle has been calculated.

This answer has been mentioned as option A.

Note:
The Carnot engine is defined as a theoretically proven thermodynamic cycle. This cycle has been proposed by Leonard Carnot. It is providing the estimate of the maximum efficiency possible for a heat engine during the conversion process of heat into work and in the opposite manner also. The efficiency of this kind of engine will be independent of the nature of the working substance in the system and will be only dependent on the temperature of the hot and cold reservoirs. A Carnot cycle can be taken as an ideal reversible closed thermodynamic cycle.