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Q.

On a X temperature scale, water freezes at $- 125^{0} X$ and boils at $375.0 ° X$ . On a Y temperature scale water freezes at $- 70.0 °$ Y and boils at $- 30.0 °$ $Y$ . The value of temperature on X-scale equals to the temperature of $40.0 ° Y$ on Y-scale is -------------- $° X$

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Detailed Solution

From principle of calibration of thermometers

$\begin{array}{l} \frac{X - {(L E P)}_{1}}{{(U F P)}_{1} - {(L E P)}_{1}} = \frac{Y - {(L E P)}_{2}}{{(U F P)}_{2} - {(L E P)}_{2}} \\ \Rightarrow \frac{x + 125}{500} = \frac{40 + 70}{40} \end{array}$

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On a X temperature scale, water freezes at -1250X and boils at 375.0°X . On a Y temperature scale water freezes at −70.0° Y and boils at −30.0° Y. The value of temperature on X-scale equals to the temperature of 40.0° Y on Y-scale is -------------- °X