JEE PYQ: Thermodynamics Question 43
Question 43 - 2019 (11 Jan Shift 2)
A thermometer graduated according to a linear scale reads a value $x_0$ when in contact with boiling water, and $x_0/3$ when in contact with ice. What is the temperature of an object in $°$C, if this thermometer in the contact with the object reads $x_0/2$?
(1) 25
(2) 60
(3) 40
(4) 35
Show Answer
Answer: (3)
Solution
Let required temperature $= T°$C
M.P. (0$°$C) corresponds to $\frac{x_0}{3}$, B.P. (100$°$C) corresponds to $x_0$
$T°C = \frac{x_0/2 - x_0/3}{x_0 - x_0/3} \times (100 - 0°C)$
$= \frac{x_0/6}{2x_0/3} \times 100 = \frac{x_0}{6} \times \frac{3}{2x_0} \times 100$
$= \frac{2x_0}{3 \times 6} \times 100 \Rightarrow x_0 = \frac{300}{2}$
$T°C = \frac{x_0}{6} = \frac{150}{6} = 25°$C…
Recalculating: $T = \frac{(x_0/2 - x_0/3)}{(x_0 - x_0/3)} \times 100 = \frac{x_0/6}{2x_0/3} \times 100 = \frac{1}{4} \times 100 = 25°$C
Actually the answer key says (3) which is 40. Let me recheck.
$T°C = \frac{\text{reading} - \text{LFP}}{\text{UFP} - \text{LFP}} \times 100 = \frac{x_0/2 - x_0/3}{x_0 - x_0/3} \times 100 = \frac{x_0/6}{2x_0/3} \times 100 = \frac{1}{4} \times 100 = 25°$C
Answer: (3) 40 per the answer key. The answer from the solution image gives 25$°$C.
BETA
