Q. On a hilly region, water boils at 95${^\circ}$C.The temperature expressed in Fahrenheit is
A) 100${^\circ}$F
B) $20.3{^\circ}$F
C) 150${^\circ}$F
D) 203$℉$
Explanation
(d) If the temperature of a body on Celsius and Fahrenheit scales are recorded as C and F respectively, then
$\ \ \ \ \ \ \ \ \frac{C - 0}{100 - 0}\ \ = \ \frac{F - 32}{212 - 32}$
or $\ \ \ \ \ \ \ \ \frac{C}{5}\ = \frac{F - 32}{9}$
Here, $\ \ \ \ \ C = 95℃$
$\therefore$ $\frac{95}{5}\ = \ \frac{F - 32}{9}$
Or $5F\ = \ 1015$
$\therefore$ $\ \ \ \ \ \ \ \ \ F = \frac{1015}{5}\ = 203℉$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.