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The graph between two temperature scales `A` and `B` is shown in Fig. Between upper fixed point and lower fixed point there are `150` equal divisions on scales `A` and `100` on scale `B`. The relation between the temperature in two scales is given by_ A. `(t_(A)-180)/(100)=(t_(B))/(150)`B. `(t_(A)-30)/(150)=(t_(B))/(100)`C. `(t_(B)-180)/(150)=(t_(A))/(100)`D. `(t_(B)-40)/(100)=(t_A)/(1080)` |
Answer» Correct Answer - B From graph, we note that for scale A, the lowest fixed point than `0^(@)A` and the highest point is `180^(@)A`. For scale B, the lowest point is `0^(@)B` and the highest point is `100^(@)B`. Therefore, the relation `(t_(A)-30)/(150)= (t_(B)-0)/(100)= (t_(B))/(100)` is correct. |
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