### NEET Solved Paper 2017 Question 11

##### Question: A Carnot engine having an efficiency of $ \frac{1}{10} $ as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

#### Options:

A) 100 J

B) 1 J

C) 90 J

D) 99 J

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

- $ \beta =\frac{1-\eta }{\eta } $ $ =\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}} $ $ \beta =9 $ $ \beta =\frac{Q _2}{W} $ $ Q _2=9\times 10=90,J $