JEE Physics 2023 Question with Solution

Given: Two temperature scales $P$ and $Q$ are linearly related. Between lower and upper fixed points, scale $P$ has $150$ divisions and scale $Q$ has $100$ divisions.

Find: The conversion relation between $t_P$ and $t_Q$.

For a linear temperature scale,

$$\frac{\text{reading on scale} - \text{lower fixed point}}{\text{upper fixed point} - \text{lower fixed point}} = \text{constant}$$

So for the two scales,

$$\frac{t_P - 30}{180 - 30} = \frac{t_Q - 0}{100 - 0}$$

This gives,

$$\frac{t_P - 30}{150} = \frac{t_Q}{100}$$

Hence,

$$\frac{t_Q}{100} = \frac{t_P - 30}{150}$$

Therefore, the correct option is B.

The solution labels one approach as option D, but the derived equation matches option B in the listed options.

Given: Lower fixed point on scale $P$ is $30$, upper fixed point is $180$. On scale $Q$, the interval from lower to upper fixed point is $100$ divisions.

Find: The relation connecting $t_P$ and $t_Q$.

The temperature difference corresponding to the full fixed-point interval on scale $P$ is

$$180 - 30 = 150$$

The full fixed-point interval on scale $Q$ is

$$100 - 0 = 100$$

Therefore corresponding fractional readings must be equal:

$$\frac{t_P - 30}{150} = \frac{t_Q}{100}$$

Rewriting in the exact option form,

$$\frac{t_q}{100} = \frac{t_p - 30}{150}$$

So the matching option is B.