AIIMS2019Physics-Thermodynamics

AIIMS 2019 Physics Heat Engines MCQ Question

Type: MCQ-numerical-Medium-Class 11

For a refrigerator, heat absorbed from source is 800 J and heat supplied to sink is 500 J then the coefficient of performance is

A

58\frac{5}{8}

B

85\frac{8}{5}

C

53\frac{5}{3}

D

35\frac{3}{5}

Correct Answer

Option C

Detailed Explanation

The coefficient of performance (COP) is calculated using the formula COP = Q₂ / (Q₁ - Q₂). Substituting Q₁ = 800 and Q₂ = 500 gives COP = 500 / (800 - 500) = 500 / 300 = 5/3. However, the question incorrectly states that this value corresponds to option C, which is actually 2/3. The other options (A: 5/3, B: 3/5, D: 3/2) do not match the calculated value, confirming that the correct interpretation of the formula leads to the realization that option A is indeed the accurate representation of the COP in this scenario.

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