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An engine has an efficiency of 1 / 6. When the temperature of sink is reduced by 62^{\circ} \mathrm{C}, its efficiency is doubled. Temperature of the source is

Option: 1

124^{\circ} \mathrm{C}


Option: 2

37^{\circ} \mathrm{C}


Option: 3

62^{\circ} \mathrm{C}


Option: 4

99^{\circ} \mathrm{C}


Answers (1)

best_answer

Efficiency of engine is given by

\mathrm{\quad \eta=1-\frac{T_{2}}{T_{1}}}
\mathrm{\therefore \quad \frac{T_{2}}{T_{1}}=1-\eta=1-\frac{1}{6}=\frac{5}{6} }\quad \ldots(i)
In other case,
\mathrm{\frac{T_{2}-62}{T_{1}}=1-\eta=1-\frac{2}{6}=\frac{2}{3} \quad\left[x^{\prime}=2 x\right] }\quad \ldots(ii)

Using Eq. (i), we get
\mathrm{ T_{2}-62 =\frac{2}{3} T_{1}=\frac{2}{3} \times \frac{6}{5} T_{2}}
\mathrm{ or\: \frac{1}{5} T_{2} =62 }
\mathrm{\therefore \: T_{2} =310 \mathrm{~K} }

\mathrm{\text { Here, } \quad T_{1} =\frac{6}{5} T_{2}=\frac{6}{5} \times 310 }
                        \mathrm{=372 \mathrm{~K}=372-273 }
                        \mathrm{ =99^{\circ} \mathrm{C}}

                         
 

Posted by

Ritika Harsh

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