# NEET Physics Thermodynamics Questions Solved

An ideal heat engine working between temperature T1 and T2 has an efficiency η, the new efficiency if both the source and sink temperature are doubled, will be

(1) $\frac{\eta }{2}$

(2) η

(3) 2η

(4) 3η

(2) In first case ${\eta }_{1}=\frac{{T}_{1}-{T}_{2}}{{T}_{1}}$

In second case ${\eta }_{2}=\frac{2{T}_{1}-2{T}_{2}}{2{T}_{1}}$$=\frac{{T}_{1}-{T}_{2}}{{T}_{1}}=\eta$

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