NEET AIPMT SOLVED PAPER SCREENING 2008

  • question_answer
    On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are \[{{39}^{\text{o}}}W\] and \[{{239}^{\text{o}}}W\] respectively. What will be the temperature on the new scale, corresponding to a temperature of \[{{39}^{\text{o}}}C\] on the Celsius scale?        

    A) \[{{78}^{\text{o}}}W\]

    B)                                        \[{{117}^{\text{o}}}W\]

    C) \[{{200}^{\text{o}}}W\]               

    D) \[{{139}^{\text{o}}}W\]

    Correct Answer: B

    Solution :

    In general, whenever we are to go from any known scale to any unknown scale, then we follow the equation     (Temperature on known scale) \[\frac{\text{-(LFP}\,\text{for}\,\text{known}\,\text{scale)}}{{{\text{(UFP-LFP)}}_{\text{known}}}}\] (Temperature on unknown scale) \[\text{=}\frac{\text{-(LFP}\,\text{for}\,\,\text{unknown}\,\text{scale)}}{{{\text{(UFP-LFP)}}_{\text{known}}}}\] or\[\frac{39-0}{100-0}=\frac{t-39}{239-39}\]or\[t={{117}^{o}}W\] Note: LFP \[\to \]Lower fixed point UFP\[\to \] Upper fixed point Alternative: \[\begin{matrix}    \text{10}{{\text{0}}^{\text{o}}}\text{C}  \\    {}  \\    {{\text{0}}^{\text{o}}}\text{C}  \\ \end{matrix}\underline{\overline{\underset{\downarrow }{\overset{\uparrow }{\mathop{100}}}\,\text{division}\underset{\downarrow }{\overset{\uparrow }{\mathop{200}}}\,}}\,\begin{matrix}    \begin{align}   & \text{Now}\,\text{scale} \\  & \text{23}{{\text{9}}^{\text{o}}}\text{W} \\ \end{align}  \\    \text{divisions}  \\    \text{3}{{\text{9}}^{\text{o}}}\text{W}  \\ \end{matrix}\] \[\therefore \]\[{{39}^{o}}C=39\times 2+39={{(78+39)}^{o}}W\] \[={{117}^{o}}W\]


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