Gujarati
Hindi
10-1.Thermometry, Thermal Expansion and Calorimetry
normal

On a new scale of temperature (which is linear) and called the $W$ scale, the freezing and boiling points of water are $39\,^oW$ and $239\,^oW$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39\,^oC$ on the Celsius scale ? ............. $^\circ \mathrm{W}$

A

$200$

B

$139$

C

$78$

D

$117$

Solution

$\frac{39-0}{100-0}=\frac{W-39}{239-39} \Rightarrow W=117\,^{\circ} \mathrm{W}$

Standard 11
Physics

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