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What per cent `T_(1)` is of `T_(2)` for a heat engine whose efficiency is `10%`A. `92%`B. `10%`C. `90%`D. `100%`

Answer» Correct Answer - C
`eta=1-(T_(1))/(T_(2))`
`(T_(1))/(T_(2))=1-eta=1-0.10=0.90=90%`
`T_(1)=90% T_(2)`


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