On a temperature scale $Y$, water freezes at $-160^{\circ} Y$ and boils at $-50^{\circ} Y$. On this $Y$…
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
- $-160.3^{\circ} \mathrm{Y}$
- $-96.3^{\circ} \mathrm{Y}$
- $-86.3^{\circ} \mathrm{Y}$
- $-76.3^{\circ} Y$
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}
$
Asked in: AP EAMCET 2013
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