A cannot engine with efficiency `eta` operates between two heat reservoirs with temperatures `T_(1)` and `T_(2)` , where `T_(1) gt T_(2)`. If only `T_(1)` is changed by `0.4%`, the change in efficiency is `Delta eta_(1)`, whereas if only `T_(2)` is changed by `0.2%`, the efficiency is changed by `Delta eta_(2)`. The ratio `(Delta eta_(1))/(Delta eta_(2))` is approximately.

Options

A. -2
B. -4
C. `+3`
D. `+4`

Answer

A

Explanation

Key Idea Efficiency of Carnot engine,
`eta=1-(T_(2))/(T_(1))`
where, `T_(1)=` source temperature and
`T_(2)=` sink temperature
If `T_(1)` is changed by 0.4% , then
`(Delta eta)/(eta)= (DeltaT_(1))/(T_(1))+(DeltaT_(2))/(T_(2))` ( From combination of the error)
`rArr (Delta eta_(1))/(eta)= (0.4)/(100)+0` ...(i)
Similarly, `T_(2)` is changed by 0.2%, then
`(Delta eta_(2))/(eta)=0+(0.2)/(100)` ...(ii)
So, from Eq. (i) and (ii), we get
`(Delta eta_(1))/(Delta eta_(2))=(0.4)/(0.2)=2`
Hence, option (a) is correct.