JEE Physics 2023 Question with Solution

Given: The faulty thermometer reads $5^\circ \text{C}$ at ice point and $95^\circ \text{C}$ at steam point.

Find: The correct temperature on the absolute scale when the faulty thermometer reads $41^\circ \text{C}$.

Using the linear calibration relation shown in the solution:

$$\frac{95^\circ \text{C} - 5^\circ \text{C}}{41^\circ \text{C} - 5^\circ \text{C}} = \frac{100^\circ \text{C} - 0^\circ \text{C}}{C - 0^\circ \text{C}}$$

Therefore,

$$C = \frac{90}{36} \times 100 = 40^\circ \text{C}$$

Now converting to absolute scale:

$$T = 273 + 40 = 313 \; \text{K}$$

Therefore, the correct temperature is $313 \; \text{K}$.

Given: A faulty thermometer has shifted fixed points, so its scale is not the true Celsius scale.

Find: The true temperature corresponding to a faulty reading of $41^\circ \text{C}$.

The interval from ice point to steam point on the faulty thermometer is:

$$95 - 5 = 90$$

The reading $41^\circ \text{C}$ is above the faulty ice point by:

$$41 - 5 = 36$$

So the true Celsius temperature is the same fraction of the true interval $0^\circ \text{C}$ to $100^\circ \text{C}$:

$$\frac{36}{90} = \frac{C}{100}$$

Hence,

$$C = \frac{36}{90} \times 100 = 40^\circ \text{C}$$

On the Kelvin scale,

$$T = 273 + 40 = 313 \; \text{K}$$

Therefore, the required answer is 313.