Question

The coefficient of performance of refrigerator, whose efficiency $25\%$ is
(A) 1
(B) 3
(C) 5
(D) 7

Answer 1

Hint The relation that relates the coefficient of performance and the efficiency is given by the formula, $C.O.P = \dfrac{{1 - \eta }}{\eta }$ . Now we can substitute the value of the efficiency in the formula, we get the value of the coefficient of performance.

Formula Used: In this solution we will be using the following formula,
$\Rightarrow C.O.P = \dfrac{{1 - \eta }}{\eta }$
where $C.O.P$ is the coefficient of performance and $\eta$ is the efficiency.

Complete step by step answer
The coefficient of friction of a refrigerator is the ratio of the cooling provided to the work required. And the efficiency of a refrigerator and the coefficient of performance are related by the formula, $C.O.P = \dfrac{{1 - \eta }}{\eta }$
In the question we are given that the efficiency of the refrigerator is $25\%$ . Therefore, we can write this as, $\eta = 25\%$
So we get,
$\Rightarrow \eta = \dfrac{{25}}{{100}} = 0.25$
Now we can substitute this value of the efficiency in the formula for the coefficient of performance. Hence we get the value of the coefficient of performance as,
$\Rightarrow C.O.P = \dfrac{{1 - 0.25}}{{0.25}}$
So on the numerator we get the value,
$\Rightarrow C.O.P = \dfrac{{0.75}}{{0.25}}$
Hence on doing the division we have,
$\Rightarrow C.O.P = 3$
Hence the coefficient of performance of the refrigerator is equal to 3.
Thus, the correct answer is option B.

Note
The relation between the coefficient of performance and the efficiency of the refrigerator is calculated by the following method. The efficiency of a refrigerator is given as, $\eta = \dfrac{{{T_H} - {T_C}}}{{{T_H}}}$ and the coefficient of performance of the refrigerator is given by the formula, $COP = \dfrac{{{T_C}}}{{{T_H} - {T_C}}}$ . Where ${T_C}$ and ${T_H}$ are the cold and hot temperature respectively. Now from the equation for COP we can write,
$\eta = \dfrac{1}{{{T_H}}} \times \dfrac{{{T_C}}}{{COP}}$ . Again from the equation of efficiency we can write, $1 - \eta = \dfrac{{{T_C}}}{{{T_H}}}$
On substituting the value we get,
$\eta = \dfrac{{1 - \eta }}{{COP}}$
Hence we get the COP as,
$COP = \dfrac{{1 - \eta }}{\eta }$