Q137 / 160 · TS EAMCET 2013 · Physics
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$-96.3^{\circ} \mathrm{Y}$
- Correct answerC$-86.3^{\circ} \mathrm{Y}$
- D$-76.3^{\circ} Y$
Solution
In given condition
$$
\begin{aligned}
\frac{Y+160}{-50+160} & =\frac{340-273}{373-273} \\
\frac{Y+160}{110} & =\frac{67}{100} \\
Y+160 & =\frac{67 \times 110}{100} \\
Y & =73.7-160 \\
Y & =-86.3^{\circ} Y
\end{aligned}
$$
Hence, the correct answer is option: 'C'.