2. Solved Papers
  3. Uttarakhand PMT

Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    A heat engine absorbs heat at\[327{}^\circ C\]and exhausts heat at\[127{}^\circ C\]. The efficiency of engine is \[\eta \] and the maximum amount of work performed by the engine per kilocalorie of heat input is W. Then,

    A)  \[\eta \]=0.38       

    B)  \[\eta \]=0.88

    C)  W=1596J     

    D)  W=1400J

    Correct Answer: D

    Solution :

     The efficiency of engine \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=1-\frac{127+273}{327+273}\] \[=\frac{3-2}{3}\] \[\eta =\frac{1}{3}\] Further, \[\eta =\frac{W}{Q}\] \[\Rightarrow \] \[W=Q\eta \] \[=\frac{1000}{3}cal\] \[\Rightarrow \] \[W=\frac{1000\times 4.2}{3}J\] \[\Rightarrow \] \[W=1400\,J\]


You need to login to perform this action.
You will be redirected in 3 sec spinner