Question

On a linear temperature scale Y, water freezes at $-{160}^{\circ }Y$ and boils at $-{50}^{\circ }Y$. On this Y scale, a temperature of 340 K would be read as: (water freezes at 273 k and boils at 373 K)

• A. $-{73.7}^{\circ }Y$
• B. $-{233.7}^{\circ }Y$
• C. $-{86.3}^{\circ }Y$
• D. $-{106.3}^{\circ }Y$

Answer 1

Answer: (C)

$-{86.3}^{\circ }Y$

Given,

Freezing point on $Y$ scale $=-{160}^{0}Y$

Boing point om $Y$ scale $=-{50}^{0}Y$

Freezing point on kelvin scale $=273K$

Boiling point on kelvin scale $=373K$

Here we have to find the temperature equivalent to $340K$ on $Y$ scale

Since the temperature scale is assumed to be linear, slope in two cases will be same. Hence,

$\frac{Y-(-160)}{-50-(-160)}=\frac{K-273}{373-273}$,

$\frac{Y+160}{110}=\frac{K-273}{100}$

$Y=\frac{11}{10}(K-273-160$

$Y=\frac{11}{10}(340-273)-160=-{86.3}^{0}Y$