# NEET Physics Thermodynamics Questions Solved

A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by 62°C, the efficiency of the engine is doubled. The temperatures of the source and sink are -

(1) 80°C, 37°C

(2) 95°C, 28°C

(3) 90°C, 37°C

(4) 99°C, 37°C

(4) Initially $\eta =\left(1-\frac{{T}_{2}}{{T}_{1}}\right)=\frac{W}{Q}=\frac{1}{6}$  ...(i)

Finally $\eta \text{'}=\left(1-\frac{{T}_{2}\text{'}}{{T}_{1}}\right)=\left(1-\frac{\left({T}_{2}-62\right)}{{T}_{1}}\right)=1-\frac{{T}_{2}}{{T}_{1}}+\frac{62}{{T}_{1}}$

$=\eta +\frac{62}{{T}_{1}}$ ....(ii)

It is given that $\eta \text{'}=2\eta .$ Hence solving equation (i) and (ii)

${T}_{1}=372\text{\hspace{0.17em}}K=99°C$ and ${T}_{2}=310K=37°C$

Difficulty Level:

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