TS EAMCET · Physics · Thermal Properties of Matter
On a temperature scale \(Y\), water freezes at \(-160^{\circ} Y\) and boils at \(-50^{\circ} Y\). On this \(Y\) scale, a temperature of \(340 \mathrm{~K}\) is
- A \(-160.3^{\circ} \mathrm{Y}\)
- B \(-96.3^{\circ} \mathrm{Y}\)
- C \(-86.3^{\circ} \mathrm{Y}\)
- D \(-76.3^{\circ} Y\)
Answer & Solution
Correct Answer
(C) \(-86.3^{\circ} \mathrm{Y}\)
Step-by-step Solution
Detailed explanation
In given condition \[ \begin{aligned} \frac{Y+160}{-50+160} & =\frac{340-273}{373-273} \\ \frac{Y+160}{110} & =\frac{67}{100} \\ Y+160 & =\frac{67 \times 110}{100} \\ Y & =73.7-160 \\ Y & =-86.3^{\circ} Y \end{aligned} \]
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