AMU Medical AMU Solved Paper-2005

  • question_answer
    A beaker of hot water cools from \[75{}^\circ C\] to \[70{}^\circ C\] in ti min from \[70{}^\circ C\] to \[65{}^\circ C\] in \[{{t}_{1}}\] min and from \[65{}^\circ C\] to \[60{}^\circ C\] in (3 min. Then

    A) \[{{\text{r}}_{\text{1}}}={{\text{t}}_{\text{2}}}={{\text{t}}_{\text{3}}}\]                              

    B)  \[{{\text{r}}_{\text{1}}}\text{}{{\text{t}}_{\text{2}}}\text{}{{\text{t}}_{\text{3}}}\]

    C)  \[{{\text{r}}_{\text{1}}}\text{}{{\text{t}}_{\text{2}}}\text{}{{\text{t}}_{\text{3}}}\]                       

    D)  \[{{\text{r}}_{\text{1}}}\text{}{{\text{t}}_{\text{2}}}\text{}{{\text{t}}_{\text{3}}}\]

    Correct Answer: B

    Solution :

     From Newtons law of cooling, rate of cooling is given by                \[\frac{dQ}{dt}=K\,({{T}_{1}}-{{T}_{2}})\].            ... (i) where \[({{T}_{1}}-{{T}_{2}})\] is temperature difference. Since, temperature difference between \[{{75}^{o}}C\] and surrounding temperature is greater than the temperature difference between \[{{70}^{o}}C\]and surrounding temperature, hence \[{{t}_{1}}<{{t}_{2}}\]                                     ... (i) Similarly, the temperature difference between \[{{70}^{o}}C\] and surrounding temperature is greater than temperature difference between \[{{65}^{o}}C\] and surrounding temperature. Hence,                 \[{{t}_{2}}<{{t}_{3}}\]                                     ..? (ii) Hence,  \[{{t}_{1}}<{{t}_{2}}<{{t}_{3}}\]


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