JEE Main & Advanced Physics Thermodynamical Processes Question Bank Topic Test - Thermodynamical Process (2021)

A carnot engine has the same efficiency between 800 K to 500 K and x K to 600 K. The value of x is

A) 1000 K

B) 960 K

C) 846 K

D) 754 K

Correct Answer: B

Solution :

[b] In first case, \[({{\eta }_{1}})=1-\frac{500}{800}=\frac{3}{8}\]

and in second case, \[({{\eta }_{2}})=1-\frac{600}{x}\]

Since \[{{\eta }_{1}}={{\eta }_{2}},\] therefore \[\frac{3}{8}=1-\frac{600}{x}\]

or \[\frac{600}{x}=1-\frac{3}{8}=\frac{5}{8}\]or\[x=\frac{600\times 8}{5}=960K\]