A) \[78{}^\circ \,W\]
B) \[117{}^\circ \,W\]
C) \[200{}^\circ \,W\]
D) \[139{}^\circ \,W\]
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) |
| \[\frac{-\text{(LFP}\,\text{for}\,\text{known}\,\text{scale)}}{{{\text{(UFP}-\text{LFP)}}_{\text{known}}}}\] |
| (Temperature on unknown scale) |
| \[\text{=}\frac{-\text{(LFP}\,\text{for}\,\,\text{unknown}\,\text{scale)}}{{{\text{(UFP}-\text{LFP)}}_{\text{known}}}}\] |
| 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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