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JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                    On a linear temperature scale Y, water freezes at \[-{{160}^{o}}Y\]and boils at \[-{{50}^{o}}Y\]. On this Y scale, a temperature of 340 K would be read as : (water freezes at 273 K and boils at 373 K)                     JEE Main Online Paper (Held On 09 April 2013)

    A)                 \[-{{73.7}^{0}}Y\]                                            

    B)                 \[-{{233.7}^{0}}Y\]                          

    C)                 \[-{{86.3}^{0}}Y\]                            

    D)                        \[-{{103.3}^{0}}Y\]                

    Correct Answer: B

    Solution :

                    \[-{{160}^{0}}y=273\,K\]                 \[-{{50}^{0}}\,y=373\,K\]                 \[-{{110}^{0}}y=-100K\]                 ______________                 \[1K=1.\,{{1}^{0}}y\] \[\Rightarrow \]               \[\underset{\begin{smallmatrix}  \downarrow  \\   \end{smallmatrix}}{\mathop{373K}}\,=-{{50}^{0}}y\]                 \[340K={{h}^{0}}y\]                 __________                 33 K                 We know,                 1 K change \[\to \,\,{{1.1}^{0}}y\]change                 33 K change \[\to \,\,1.1\times 33\]                 \[={{33.6}^{0}}y\] change So,          \[h=-50-36.3\]                 \[=-{{86.3}^{0}}y\]                


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