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In an arbitrary scale of temperature, water freezes at $40 ° C$ and boils at $290 ° C$ . The boiling point of a liquid in its scale if it boils at $62 ° C$ is

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210 ° C

130 ° C

165 ° C

195 ° C

answer is C.

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Detailed Solution

Let t be the boiling point of the liquid on this scale. Then,

$\frac{t - 40}{290 - 40} = \frac{C - 0}{100 - 0}$

$\frac{t - 40}{290 - 40} = \frac{62}{100}$

$\Rightarrow C = \frac{62 \times 250}{100} + 40 = 195 ° C$

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